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SC&S

Ship Construction & Stability

Hydrostatic stability, cross curves, GZ curves, drydocking, damage stability, ship structural members, welding, and load lines.

540 Qs 54 Papers 419 Repeated 96 Diagrams
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Q1 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 5x

(a) Explain in detail, how an underwater survey is carried out. (7)

(b) State the requirements to be fulfilled before an under -water survey is acceptable to the survey authority. (5)

(c) Construct a list of the items in order of importance that the underwater survey authority should include. (4)

Appeared In: Jul 2026 Oct 2024 Sep 2022 Jan 2019 Mar 2018
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Part (a)

An in-water survey, also known as Underwater Inspection in Lieu of Dry Docking (UWILD), involves a systematic and detailed examination of a vessel’s hull and underwater components while the ship remains afloat. The procedure includes the following steps:

  • The shipowner sends a request to the classification society surveyor, indicating the intention to perform an underwater survey.
  • A detailed plan of the ship's external hull features is submitted, showing the location of frames, bulkheads, welding lines, openings, etc.
  • The hull surface is cleaned before the survey to remove any marine growth or fouling that could obscure the inspection.
  • A diving company, approved by the classification society, is appointed to conduct the underwater inspection.

A self-propelled survey vehicle equipped with the following tools is used:

  • Long-Range Light TV Camera to aid in steering and checking hull deterioration.
  • High-resolution colour TV Camera to provide a close-up view of the hull coating and welded seams.
  • 35mm Still Camera to capture still images.
  • Ultrasonic Probe for measuring plate thickness.
  • Depth Meter and Speed Indicator to provide accurate data on the vehicle's depth and movement.
  • Umbilical Cable to connect the survey vehicle to the survey boat, transmitting power and information.

The survey boat is to be equipped with:

  • A control console with TV monitors.
  • Plate thickness printouts.
  • Audio and video cassette recorders.
  • Playback units.
  • Diver communication systems.
  • Vehicle control systems and associated instruments.

Operation:

  • The survey vehicle is taken underwater by a diver to the survey starting point.
  • Using TV monitors and shell expansion plans as guides, the vehicle is navigated over the hull, focusing on the bottom structure, sides, stern frame, rudder, propeller, bilge keel, and hull openings.
  • All images, data, and information are recorded and transmitted back to the survey boat.
  • Detailed pictures of the stern frame, rudder, propeller, bilge keel, and hull openings are captured.
  • Divers are deployed to measure stern tube bearing wear, pintle clearance, and inspect stern seals, anodes, and rudder stock palm coupling bolts.
  • All recorded video and audio, including conversations between the surveyor and drivers, thickness printouts, measurements, and pictures are analyzed to determine the detailed underwater condition of the vessel.
Part (b)

Before an in-water survey is accepted by the survey authority, the following conditions must be met:

The vessel's owner submits a request to the surveyor, including:

  • The proposed date and location for the survey.
  • General information about the diving company.
  • A declaration that the vessel has not suffered any damage due to grounding, collision, or other incidents.

The ship's master or owner’s representative must provide a declaration confirming:

  • Any suspected or actual damage to the hull since the last dry-docking.
  • The underwater portion of the hull is protected by a suitable paint scheme that is of adequate thickness and remains valid until the next dry-dock.
  • The survey site should be in a protected area with calm and clear water, ensuring good underwater visibility. Attention must be given to the effects of currents and tides.
  • The hull must be clean for the external survey. The surveyor must be satisfied with the method and quality of the pictorial presentation, ensuring that it provides a reliable assessment of the hull's condition.
  • The underwater examination should be conducted by an approved diving company using closed-circuit TV and two-way communication, which can be monitored by the surveyor.
  • The vessel should be in as light an operating condition as possible to facilitate the survey.
  • Means must be available for the surveyor to examine the outside shell plating above the waterline.
  • Any required repairs identified during the survey must be carried out to the satisfaction of the attending surveyor.
Part (c)

While the importance of each item may vary depending on the vessel and its specific requirements, below is a list of items that should be included in an underwater survey in order of importance:

  • Underwater Hull: General condition of the hull below the waterline.
  • Bottom and Shell Plating: Inspection for corrosion, damage, and fouling.
  • Shell Openings: Examination of openings such as sea chests, drain plugs, and overboard discharge points.
  • Stern Tube Oil Leaks: Check for leaks around the stern tube.
  • Propeller Blade: Inspection for damage, wear, and fouling.
  • Rudder: Inspection for damage, wear, and clearances.
  • Sea Chest Opening and Grating: Examination for blockages, damage, and fouling.
  • Anodes: Check the condition and effectiveness of cathodic protection anodes.
  • Bilge Keel: Inspection for damage and fouling.
  • Drain Plugs: Ensure all drain plugs are secure and in good condition.
  • Overboard Valve Openings: Check for proper operation and condition.
  • Forward Area: Inspection for any damage due to anchor and chain movement.
Q2 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 4x

If a ship is seriously damaged under water in way of a large fuel oil side bunker tank what is the immediate effect and what may ultimately happen? What features in the ship would enhance safety? (16)

Appeared In: Jul 2026 Aug 2022 Jun 2019 Feb 2018
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If a ship is seriously damaged underwater in the area of a large fuel oil side bunker tank, it will cause the following:

Bilging or Flooding of the Compartment:

  • Water will enter the damaged bunker tank, reaching the draught level.
  • The rate of flooding depends on the size of the breach and the water pressure at the depth of the damage.

Oil Leakage into the Sea:

  • If the tank contains fuel oil, oil will begin to leak out, causing pollution.
  • The extent of leakage depends on the tank’s contents (empty, half-full, or full) and the location of the damage.

List and Trim of the Vessel:

  • The ingress of water and loss of oil will create an imbalance, causing the vessel to list or trim.

If corrective actions are not taken, uncontrolled flooding and loss of stability could lead to capsizing or sinking of the vessel.

Features in the Ship to Enhance Safety:

  • Small Bunker Tank Sizes reduces the risk of extensive oil spillage and loss of stability.
  • Connectivity to transfer pumps allows the transfer of oil from the damaged tank to an empty tank, minimizing oil spillage and counteracting the loss of stability.
  • The tank’s size and location are designed to limit the effects of flooding, as per damage stability regulations.
  • Properly positioned transverse and longitudinal bulkheads enhance the subdivision factor, limiting water ingress to the damaged tank.
  • Ship’s Ballast System allows corrective ballasting to counteract the list or trim caused by the ingress of water.
  • Watertight Doors and Hatches prevent water from spreading to adjacent compartments.

Recommended Immediate Actions by Crew:

For Empty Tanks:

  • Quickly seal off the damaged tank by shutting all valves and isolating it from the transfer system to prevent water ingress into other parts of the vessel.

For Half-Empty or Full Tanks:

  • Initiate oil transfer to another empty tank to reduce oil leakage and stabilize the ship.
  • Monitor the water ingress and ensure the tank is filled to the draught level with seawater if necessary, using ballast to correct the list.
Q3 (16 Marks) Hull Construction πŸ”₯ Repeated 6x

(a) Sketch a transverse section through the hold space of a container ship hull. (8)

(b) Referring to the sketch in (a) describe how adequate structural strength is built into the hull. (8)

Appeared In: Jul 2026 Oct 2025 Jun 2025 Jul 2018 Mar 2018 Nov 2025
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Part (a)

Mid-ship half-section sketch of a container ship:

Part (b)

Structural Strength Features:

  • The deck plating is made thicker and uses higher tensile steel to withstand stresses caused by wide hatch openings and operational loads.
  • The deck, side shell, and longitudinal bulkheads are framed longitudinally. This arrangement combined with deep double bottom helps resist bending stresses due to hogging (upward bending) and sagging (downward bending) when the ship is under load.
  • The hatch coamings are made continuous to contribute to the overall longitudinal strength of the hull.
  • A torsion box is installed, running along the entire length of the ship from the machinery space bulkhead to the forward collision bulkhead. This structure provides the necessary torsional strength to counteract twisting forces acting on the hull during operation.
  • Deep web boxes are fitted at the ends of hatches, both at tank top and deck levels, to enhance transverse and torsional strength.
  • A deep double bottom is designed to withstand uplift forces caused by water pressure, especially when the ship is deeply loaded. It also provides additional strength to the hull structure.
  • Side girders are placed under container cells, with added transverse local stiffening. These elements distribute the concentrated loads from containers and increase overall stability.
Q4 (16 Marks) Ship Types & Design πŸ”₯ Repeated 3x

Discuss the importance of the following to be examined for meeting EEDI limitations: (16)

(a) Slimmer vessels with lower block coefficients

(b) Long-Stroke engines

(c) Low revolution large diameter propellers

Appeared In: Jul 2026 Feb 2024 Jan 2023
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The Energy Efficiency Design Index (EEDI) expresses the grams of CO2 emitted per tonne of cargo transported per nautical mile. Its level is directly proportional to the fuel consumed (and hence the propulsive power required) at the design speed, divided by the cargo capacity and speed. Anything that cuts the power needed to move a given deadweight at the design speed, or that raises propulsive efficiency, lowers the EEDI. The three methods below act on exactly those levers.

Part (a)

Slimmer vessels with lower block coefficients.

A fine hull with a low block coefficient (Cb) presents less displaced volume for a given length and thus a smaller wetted surface area, reducing the frictional resistance component at a given displacement. A fine (low Cb) forebody and afterbody also weaken the bow wave and the pressure (residual or wave-making) resistance at the design speed, because the water is pushed aside more gradually. Lower residual resistance means the installed propulsive power at the design speed is less, so less fuel is burnt per tonne-mile, lowering the EEDI. Being longer and narrower for the same displacement also raises waterline length, which reduces the length-related frictional and Froude-number-dependent resistance. The penalty is reduced cubic cargo capacity and somewhat less form stability, so the hull form is optimised rather than simply fined down. Lower block coefficient is therefore one of the strongest design levers for meeting EEDI limits.

Part (b)

Long-stroke engines.

A long-stroke (high stroke-to-bore ratio) slow-speed diesel extracts more work from each unit of fuel in the expansion stroke and achieves higher thermal (brake) efficiency, typically up to about 50 per cent, with a correspondingly lower specific fuel consumption per kWh. Because the stroke is long, the engine can turn slowly at the same piston speed, enabling direct coupling to a large slow-turning propeller with no reduction gearbox and no associated transmission losses. A more efficient engine burns less fuel for each kW it delivers, hence produces less CO2 per tonne-mile, which reduces the attained EEDI directly. Long-stroke engines also operate at low revolutions, which marries perfectly with the large-diameter, low-rev propeller of part (c).

Part (c)

Low-revolution large-diameter propellers.

Propeller open-water efficiency rises as the disc-area loading (thrust per unit swept area) falls. A large-diameter propeller turning slowly accelerates a large mass of water by a small amount, giving low disc loading and high efficiency for the same thrust and therefore less shaft power and less fuel per tonne of cargo. Large slow propellers also stay further away from cavitation, reducing blade erosion, vibration and noise. Because EEDI is fixed by the fuel (shaft power) needed at the design speed, optimising hull, engine and propeller together β€” a fine low-block hull at a low Froude number, a long-stroke low-speed engine, and a large-diameter low-rev propeller β€” minimises energy consumption per tonne of cargo and is the classic route to satisfying EEDI limitations.

Q5 (16 Marks) Hull Construction πŸ”₯ Repeated 8x

(a) Sketch the cross-section of a bulk carrier with either deep or shallow double bottom showing the type of framing used. (8)

(b) (i) Describe the corrosion problems experienced with ballast tanks.

(ii) State how such tanks are protected against extensive corrosion. (8)

Appeared In: Jul 2026 Mar 2026 Feb 2026 Dec 2025 Dec 2024 Oct 2024 Jan 2024 Jan 2023
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Part (a)

Mid-ship section of bulk carrier:

Part (b)

(i) Corrosion problems in ballast tanks

are significant and arise due to various factors:

  • Galvanic corrosion occurs due to the interaction between dissimilar metals, accelerated by the seawater environment and differential aeration.
  • Sulphate-reducing bacteria in river mud can cause localized pitting, leading to penetration of the bottom shell.
  • Dissolved oxygen in seawater reacts with metal surfaces to form rust, significantly contributing to structural deterioration.
  • The rate of corrosion peaks at a 3.5% salt concentration, typical of seawater.
  • Neglecting maintenance exacerbates the problem, allowing corrosion to progress unchecked.

(ii) Protection against extensive corrosion in ballast tanks involves the following measures:

  • Complete coating of the tank surfaces with properly selected marine-grade paint to create a protective barrier.
  • Installation of sacrificial anodes, designed to corrode preferentially and protect the tank structure.
  • Using large anodes with greater volume relative to surface area to ensure extended protection.
  • Regular gauging of plates to assess thickness and identify areas requiring reinforcement or repair.
  • Conducting regular inspections, cleaning the tanks, removing rust, and repainting to maintain structural integrity and prolong the lifespan of the ballast tanks.
Q6 (16 Marks) Ship Resistance & Propulsion

(a) Explain the term angle of loll and state what, if any dangers it poses to a vessel. (6)

(b) A propeller has a pitch ratio of 0.95. When turning at 120 rev/min the real slip is 30%, the wake fraction 0.28 and the ship speed 16 knots. The thrust is found to be 400 KN, the torque 270 KNm and the QPC 0.67.

Calculate: (10)

(i) Propeller diameter

(ii) Shaft Power

(iii) Propeller efficiency

(iv) Thrust Deduction Factor

Appeared In: Jul 2026
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Part (a)

Angle of Loll

The angle of loll is the angle of heel at which a ship with negative initial metacentric height (negative GM) comes to rest in still water, without any external heeling force such as wind, waves, or turning.

When a vessel has negative GM, it is unstable in the upright position because the centre of gravity (G) lies above the metacentre (M). As a result, any small disturbance causes the vessel to heel further instead of returning upright.

As the vessel heels, the centre of buoyancy (B) moves outboard towards the lower side. At a certain angle of heel, the line of action of buoyancy passes vertically through the centre of gravity (G), producing a small positive righting lever. The vessel then comes to rest at this angle, known as the angle of loll.

The ship may remain at this angle on either side and can suddenly loll or flop from one side to the other when disturbed by wind, waves, turning, or other external forces.

Dangers Posed by an Angle of Loll

  1. False sense of stability: Although the vessel appears to be stable because it remains at a constant angle of heel, it is actually in a dangerous condition with negative GM.
  2. Sudden lolling or flopping: The vessel may suddenly and violently swing from the angle of loll on one side to the same angle on the opposite side with little or no warning, especially when affected by beam seas, wind gusts, or turning.
  3. Reduced range of stability: The remaining range of positive righting lever is much smaller than normal. Consequently, even a relatively small additional heeling moment can cause the vessel to capsize.
  4. Risk due to free surface effect: Slack tanks further reduce the effective GM. Incorrect attempts to correct the condition, such as pumping ballast to one side or leaving tanks partially filled, increase the free surface effect and may worsen the angle of loll or even lead to capsize.
  5. Danger of capsize: If the vessel heels beyond the point where the righting arm again becomes zero (the point of vanishing stability), it will capsize.
  6. Oscillation about the angle of loll: Instead of rolling about the upright position, the vessel oscillates about the angle of loll, indicating that it is in an unstable equilibrium.

Corrective Actions

  • Do not attempt to correct the angle of loll by shifting ballast or cargo transversely, as this may cause the vessel to suddenly flop to the opposite side and worsen the condition.
  • The correct method is to lower the centre of gravity (G) and restore positive GM by:
    • Eliminating free surface effects by pressing tanks either completely full or completely empty (avoid slack tanks).
    • Pressing up double-bottom tanks to add weight low down and lower the centre of gravity.
  • Once positive GM is restored, the vessel will return to the upright position and regain true stability.
Q7 (16 Marks) Ship Stability

(a) What factors influence the frictional resistance of a ship and what formula is used to calculate this resistance. (6)

(b) A ship of 12000 tonne displacement has a rudder 15mΒ² in area, whose centre is 5m below the waterline. The metacentric height of the ship is 0.3m and the centre of buoyancy is 3.3m below the waterline. When travelling at 20 knots the rudder is turned through 30Β°. Find the initial angle of heel if the force Fn perpendicular to the plane of the rudder is given by: Fn = 577 AvΒ² sinΞ± N

Allow 20% for the race effect. (10)

Appeared In: Jul 2026
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Part (a)

Factors influencing frictional resistance, and the formula used.

Frictional (skin) resistance is caused by the shearing of water over the hull surface. The important factors are:

  • Speed (V): within normal service ranges frictional resistance rises approximately as V^1.8 to V^1.85, so speed strongly increases resistance.
  • Wetted surface area (S): the larger the wetted hull area the higher the resistance; it depends on length, breadth, draught, displacement and fullness of the ends.
  • Surface roughness and fouling: marine growth (barnacles, slime), corrosion pitting, weld beads, protruding plating and deteriorated antifouling all raise the friction coefficient sharply. Regular hull cleaning and good antifouling coatings reduce this.
  • Density of the water: resistance is proportional to density, so it is higher in sea water than in fresh water.
  • Viscosity and temperature: cooler, more viscous water gives a slightly higher frictional coefficient.
  • Hull length: the frictional coefficient varies with the length and Reynolds number.

The formula used is the frictional resistance line. In Admiralty-style form:

Rf = f and S and V^n

or the simplified relation R = 0.45 S V^1.83 (newtons)

Where f is the frictional coefficient, S the wetted surface area (m2) and V the ship speed. Modern practice uses the ITTC 1957 friction line, Rf = 0.5 Cf rho S V^2, where Cf = 0.075/(log10 Re - 2)^2 and Re is the Reynolds number; this permits accurate model-ship scaling.

Part (b)

Rudder heeling angle.

Data: displacement Delta=12000 t; rudder area A=15 m2; rudder centre 5 m below the waterline; GM=0.3 m; centre of buoyancy 3.3 m below the waterline; speed 20 knots; helm angle alpha=30 deg; apply 20% race allowance.

Speed converted: v=20 x 0.5144 = 10.29 m/s, so v2=105.84.

Normal rudder force Fn=577 A v2 sin(alpha) = 577 x 15 x 105.84 x 0.5 = 458,035 N.

Race effect (propeller race increases water velocity over the rudder) 20%: F=1.2 x 458,035 = 549,642 N.

The heeling couple is the force times its lever about the centre of lateral resistance, taken as the vertical distance of the rudder centre below the waterline, lever=5 m:

Heeling moment = 549,642 x 5 = 2,748,210 Nm.

Balancing against the ship's righting moment (displacement as a weight, W=12000 tonne-force):

tan(theta) = (F x lever)/(W x GM) = (549,642 x 5)/(12000 x 9810 x 0.3) = 2,748,210/35,316,000 = 0.07782.

Hence theta = atan(0.07782) = 4.45 deg.

Answer: the initial angle of heel is about 4.5 deg towards the rudder (outboard) side.

Q8 (16 Marks) Ship Resistance & Propulsion

(a) Why is it important in a tender ship to keep the double bottom tanks pressed up. (6)

(b) A ship of 6000 tonne displacement has a wetted surface area of 2500 mΒ² and a speed of 15 knots.

(i) Calculate the corresponding speed and wetted surface area of a similar ship of 2000 tonne displacement.

(ii) If the skin resistance is of the form R=0.45 SV^1.83 N ; find the resistance of the 6000 tonne ship (10)

Appeared In: Jul 2026
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Part (a)

Importance of pressing up double-bottom tanks in a tender (stiff, large GM) ship.

A tender ship has a small metacentric height and a long, slow roll period, giving uncomfortable motion, risk of rolling in resonance and reduced stability reserve. Pressing up (filling completely and venting air from) the double-bottom tanks achieves several things:

  • It lowers the centre of gravity of the ship. Water ballast occupies the very low double-bottom, so its addition lowers KG, which increases GM and makes the ship stiffer and more quickly self-righting.
  • It removes the free surface effect. A completely full tank (no free liquid surface) contributes no loss of GM, whereas a part-full tank produces a free-surface effect that reduces effective GM. In a tender ship this loss of margin can be dangerous, so tanks are pressed up to eliminate it.
  • It improves stability in that the weight is concentrated low and centrally, and it reduces the free-surface damage risk when other tanks are used.
  • Pressing up the double-bottom tanks also lowers the centre of buoyancy proportion and helps trim/heel correction and, because the tanks are deep (low) and of small individual breadth, even when ballasting the free-surface effect is small.

Hence for a tender ship ensuring double-bottom tanks are pressed up restores an adequate, safe GM and a more comfortable, lower-amplitude roll.

Part (b)

(i) Similar ship 6000 tonne and 2000 tonne.

For geometrically similar ships, the linear scale is the cube root of the displacement ratio:

Linear scale = (2000/6000)^(1/3) = (1/3)^(1/3) = 0.6934.

Speed scales as the square root of the linear dimension: V2 = V1 x sqrt(scale) = 15 x sqrt(0.6934) = 15 x 0.8326 = 12.49 knots. Answer: speed of the 2000 tonne ship is about 12.5 knots.

Wetted surface area scales as the square of the linear dimension:

S2 = S1 x scale^2 = 2500 x (0.6934)^2 = 2500 x 0.4808 = 1202 m2. Answer: wetted surface area about 1202 m2.

Part (b)

(ii) Resistance of the 6000 tonne ship.

Skin resistance R = 0.45 S V^1.83 N, taking V in knots and S in m2:

R = 0.45 x 2500 x 15^1.83.

15^1.83 = 141.9, so R = 0.45 x 2500 x 141.9 = 159,700 N.

Answer: resistance of the 6000 tonne ship is about 160 kN.

Q9 (16 Marks) Ship Stability

(a) Explain the effects on stability when a tank is partially filled with liquid. (6)

(b) A box barge 45 m long and 15 m wide floats at a level keel draught of 2 m in sea water, the load being uniformly distributed over the full length. Two masses, each of 30 tonne, are loaded at 10 m from each end and 50 tonne is evenly distributed between them. Sketch the shear force diagram and give the maximum shear force. (10)

Appeared In: Jul 2026
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Part (a)

Effect on stability when a tank is partially filled with liquid.

When a tank is part-full the free liquid surface is able to move as the ship heels, transferring liquid to the low side. This makes the centre of gravity of the tank's contents shift towards the low side, creating an additional heeling (virtually raising the centre of gravity) and thereby reducing the effective metacentric height. The loss of GM is:

reduction in GM = rho_liquid x i / Delta

where i is the second moment of area of the free liquid surface about its longitudinal centroidal axis (i = L B^3/12 for a rectangular surface) and Delta the ship displacement. The free-surface effect:

  • is independent of the quantity of liquid, depending only on the shape (area and breadth) of the free surface, so even a small amount can be damaging;
  • is proportional to the cube of the tank breadth, so wide tanks have the largest effect;
  • is eliminated when the tank is completely full (pressed up) or completely empty;
  • is a virtual loss of GM and does not affect the true KG, appearing as a reduction of effective GM and of righting levers (GZ).

It must therefore be taken into account in the metacentric and cross-curves, usually by reducing GM by the free-surface correction. In practice tanks are divided with longitudinal bulkheads (wash/centreline bulkheads) to reduce the breadth i, and press up or strip out tanks when the free surface is significant.

Part (b)

Shear-force diagram for the box barge.

Box barge 45 m long, 15 m wide, floats at a level keel draught of 2 m in sea water, load uniformly distributed over the full length. Two 30 tonne masses are loaded at 10 m from each end, and 50 tonne distributed evenly between them.

Buoyancy per metre, uniform over the full length: the box displaces LB.d.rho = 45x15x2 = 1350 m3 = 1350 x 1.025 = 1383.75 t. Buoyancy per metre b = 1383.75/45 = 30.75 t/m.

Weights:

Uniform (still floating hull and level-keel load) weight per metre w0 = 30.75 t/m (to give even keel before adding the point masses). The three added loads are then point loads.

The two 30 t masses at 10 m from ends and the 50 t at the middle (evenly distributed across the central 25 m length at 50/25 = 2 t/m between x=10m and x=35m). Shear force is the algebraic sum of the net load (weight - buoyancy) to one side.

For x from 0 to 10 m: only the uniform base load acts; since w0 balances b with no net load, the shear is essentially from the base alone - shear = 0 over the ends and the net jumps occur at the point loads.

Treated as a simply supported beam on uniform buoyancy with load diagram:

Net point loads cause shear jumps: at x=10 m a downward 30 t (jump in SF = -30 t); distributed 2 t/m over 10-35 m adds slope; at 35 m end of distributed load the accumulated shear returns and the symmetric load at 45-10=35m balances.

By symmetry, the maximum shear force occurs adjacent to one of the 30 t end masses. Consider the left side: total buoyancy of the left 10 m strip = 10 x 30.75 = 307.5 t; total weight of left 10 m (uniform base) = 10 x 30.75 = 307.5 t, so they exactly balance and the shear due to the base over the outer 10 m is zero. The net load diagram is therefore a series of point/partial loads sitting on an exactly balanced uniform basis.

Loaded mass summary: base uniform load = buoyancy (balanced). Added: 30 t end loads and a 50 t uniform load over the middle 25 m at 2 t/m.

For the left half (symmetry axis at 22.5 m): total added weight left of mid = 30 (at x=10) + 2 x 12.5 (50 t spread to mid) = 30+25 = 55 t. The reaction portion of buoyancy on the left half = 1383.75/2 = 691.9 t, of which the base weight matches. The point 30 t at 10 m therefore creates a shear to its right of SF = -30 t, and the 2 t/m distributed load adds -2 per metre, so just left of midspan the shear is -30 - 2x12.5 = -55 t.

By symmetry the maximum shear force occurs at the ends of the distributed 50 t section, and equals 30 + 50/2 = 55 t. Hence maximum shear force = 55 tonnes (55 t). Sketch: the shear-force diagram shows a step of -30 t at x=10 m, then a linearly falling line over x=10 to 35 m at -2 t/m reaching -55 t at midspan, and a mirror image rising on the after half -50-?with symmetric positive values, maximum 55 t.

Q10 (16 Marks) Ship Stability

The breadth of the upper edge of a deep tank bulkhead is 12 metres. The vertical heights of the bulkhead at equidistant intervals across it are 0, 3, 5, 6, 5, 3 and 0 metres respectively. Find the depth of the center of pressure below the waterline when the tank is filled to a head of 2 metres above the top of the tank. (16)

Appeared In: Jul 2026
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Centre of pressure on a deep tank bulkhead.

The bulkhead is a vertical plane whose top edge is 12 m wide and lies at a depth equal to the head, 2 m below the waterline. The vertical heights of the bulkhead at seven equidistant points across the 12 m breadth are 0, 3, 5, 6, 5, 3 and 0 m. With six equal intervals over 12 m the spacing is h=2 m.

For a submerged vertical plane surface, the centre of pressure lies at a depth below the waterline given by

depth of centre of pressure = (second moment of area about the waterline)/(first moment of area about the waterline), i.e. d = M2/M1.

For each station across the breadth the bulkhead bottom edge lies at depth (2 + height) below the waterline, while the top edge is at depth 2 m throughout.

First-moment integrand, element about the waterline (per unit breadth), integrating from depth 2 to (2+H):

M1_element = integral of z dz from 2 to (2+H) = [(z^2)/2] = 0.5[(2+H)^2 - 4] = 2H + 0.5H^2.

Second-moment integrand:

M2_element = integral of z^2 dz from 2 to (2+H) = [(z^3)/3] = (1/3)[(2+H)^3 - 8] = 4H + 2H^2 + H^3/3.

Values of H: 0,3,5,6,5,3,0:

For M1 (2H + 0.5 H2): 0,10.5,22.5,30,22.5,10.5,0.

For M2 (4H+2H2+H3/3): 0, 39, 111.67, 168, 111.67, 39, 0.

Using Simpson's rule with spacing 2 m and 7 ordinates:

M1 = (2/3)[ (0+0) + 4(10.5+30+10.5) + 2(22.5+22.5) ]

= (2/3)[0 + 4x51 + 2x45] = (2/3)[204+90] = (2/3)(294) = 196.

M2 = (2/3)[0 + 4(39+168+39) + 2(111.67+111.67)]

= (2/3)[4x246 + 2x223.33] = (2/3)[984+446.67] = (2/3)(1430.67) = 953.8.

Depth of centre of pressure below the waterline = M2/M1 = 953.8/196 = 4.87 m.

Answer: the centre of pressure is approximately 4.87 m below the waterline. (Since the top edge is 2 m below the waterline, the centre of pressure is about 2.87 m below the top edge of the bulkhead.)

Q1 (16 Marks) Ship Types & Design

(a) State the reasons for the freeboard requirement. (4)

(b) Explain the term condition of assignment. Explain how these are maintained for a ship. (4)

(c) Using a diagram indicate the freeboard of type A, type B, type B60, and type B100 vessels. (8)

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Part (a)

Reasons for the freeboard requirement.

Freeboard is the vertical distance from the waterline to the freeboard (upper) deck at side. It is required because it provides:

  • Reserve buoyancy: the volume of the hull above the waterline that keeps the ship afloat when waves or damage raise the external waterline. Adequate reserve buoyancy is essential for floating headroom and survivability.
  • Reduced shipping of water and deck wetness: a higher freeboard gives better watertight integrity against seas sweeping over the deck, protecting cargo, stability and the crew.
  • Improved stability and reduced angle of heel by keeping the deck edge and openings (doors, hatches, air pipes, vents) well above water, and by providing a larger intact range of stability (area under the GZ curve).
  • Comfort and safety in a seaway: better freeboard limits immersion in following/quartering seas and reduces green seas on deck, lowering risk of damage and of loss of stability.

The minimum freeboard is fixed by the International Load Line Convention, giving the load-line mark which must never be submerged.

Part (b)

Condition of assignment.

The condition of assignment is the series of requirements that a ship must satisfy to receive its load line certificate, ensuring the hull strength, watertight integrity and the protection of openings are commensurate with the assigned freeboard. It covers the scantlings and weathertightness of the freeboard deck, the strength of the superstructure and hatch covers, the provision and closing of watertight and weathertight openings (doors, scuttles, vents, air pipes, sounding pipes, sill heights, freeing ports), and the availability of bulwark freeing arrangements. These are maintained by:

  • keeping the hull structcuy sound and free from corrosion/defects;
  • maintaining all weathertight and watertight closures, gaskets, coamings and securing arrangements in good order;
  • ensuring proposed and deck fittings (gangways, cranes, lashings) do not open ill-hidden ports above the assigned deckline;
  • carrying out planned and classification surveys (annual/periodical) and maintaining the load line certificate, the certificate being renewed at the load line survey;
  • reporting any damage or alteration to the hull or openings so the surveyor can confirm the condition still satisfies the rules; unauthorised alterations that reduce the assigned freeboard are prohibited.
Part (c)

Freeboard types, A, B, B60 and B100.

(Provide a diagram: a ship's side at midship with the freeboard deck line, and four waterlines at different heights.)

  • Type A ships are designed to carry only liquid cargoes in bulk, with a closely guarded deck and high integrity (e.g. tankers). They get the lowest (smallest) freeboard because the spillover risk is small, giving maximum carrying capacity.
  • Type B ships are general cargo ships not covered by type A; they require a greater freeboard.
  • Type B60 and B100 are refined type B ships to which a reduced freeboard is permitted provided the ship satisfies additional strength and stability criteria - B60 allows 60% of the tabular difference between B and A freeboards to be taken off, B100 allows the full 100% reduction (i.e. the B freeboard may be reduced to the type A level) providing the prescribed additional conditions are met. Hence the freeboards order: Type A smallest, B100 next, B60 next, and plain Type B the largest freeboard.
Q2 (16 Marks) Machinery Space & Systems πŸ”₯ Repeated 5x

(a) Describe the double bottom and framing arrangement used in the machinery space to cope up with the concentrated loads and vibration, together with shaft and thrust block support. (10)

(b) Give reasons for the choice of thrust block position. (6)

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Part (a)

The construction of the double bottom in the machinery space regardless of the framing system has solid plate floors at every frame space under the main engine. Additional side girders are fitted outboard of the main engine seating, as required. The double-bottom height is usually increased to provide fuel oil, lubricating oil and fresh water tanks of suitable capacities. Shaft alignment also requires an increase in the double-bottom height or a raised seating, the former method usually being adopted.

Continuity of strength is ensured and maintained by gradually sloping the tank top height and internal structure to the required position. Additional support and stiffening is necessary for the main engines, boilers, etc., to provide a vibration-resistant solid platform capable of supporting the concentrated loads. On slow- speed diesel-engined ships, the tank top plating is increased to 40 mm thickness or thereabouts in way of the engine bedplate. This is achieved by using a special insert plate which is the length of the engine including the thrust block in size. Additional heavy girders are also fitted under this plate and in other positions under heavy machinery as required. Plating and girder material in the machinery spaces is of increased scantlings in the order of 10 per cent.

The method adopted.

A cellular void space within a ships structure is called a coffer dam. Like a bulkhead separates two spaces or divides a space into two, a coffer dam does the same with a larger degree of integrity since it incorporates a void which would contain any breach of either of the boundaries. This would contain the leakage and prevent it spreading into other areas. A cofferdam can be defined as an empty space separating compartments to prevent the contents of one compartment from entering another in case of leakage.

Part (b)

The thrust block is positioned close to the propulsion machinery (usually just aft of the main engine). Reasons are as follows:

  • The axial thrust generated by the propeller could cause deformation and misalignment of the shafting system if the thrust block were positioned far from the engine. Placing it close to the engine minimizes the length of shafting subject to this thrust, thus reducing potential for misalignment. The strong double bottom structure directly beneath provides necessary support to mitigate any deformation.
  • Differential expansion between the shaft and the hull due to temperature variations is a potential source of misalignment. Positioning the thrust block close to the engine helps to minimize the effect of this differential expansion.
  • The weight of the propeller and the dynamic forces it creates can lead to whirling of the tailshaft and misalignment if not properly managed. Positioning the thrust block near the engine helps to stabilize the shafting system and reduce the risk of these issues.
  • The substantial double-bottom structure under the main machinery provides an ideal, inherently strong foundation for the thrust block. This minimises the need for extensive additional reinforcement to support the thrust loads.
Q3 (16 Marks) Hull Construction πŸ”₯ Repeated 7x

(a) With reference to fatigue of engineering components explain the influence of stress level and cyclical frequency on expected operating life. (6)

(b) Explain the influence of material defects on the safe operating life of an engineering component. (4)

(c) State the factors which influence the possibility of fatigue cracking of a bed-plate transverse girder and explain how the risk of such cracking can be minimized. (6)

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Part (a)

Influence of Stress Level and Cyclic Frequency on Operating Life:

Fatigue is progressive and localised structural damage caused by cyclic loading, where the maximum stress is below the ultimate tensile strength. The relationship between stress level, cyclic frequency, and operating life depends on whether the fatigue is high-cycle/low-stress or low-cycle/high-stress.

High-cycle fatigue (low stress-high cycle):

  • This occurs at lower stress levels over a high number of cycles, resulting in elastic deformation. The component can withstand more cycles at these lower stress levels, and its life expectancy is determined by the S-N curve, which predicts the number of cycles before failure at a given stress level. For example, fatigue in turbocharger blowers often results from prolonged vibration over numerous cycles.

Low-cycle fatigue (high stress-low cycle):

  • This occurs at high-stress levels over fewer cycles, causing plastic deformation in the material. This type of fatigue is typically assessed by a strain curve. If the stress level increases, the component's operating life decreases, as higher stress accelerates the onset of failure. For example, air receivers filling automatically face high stress and experience fewer cycles before failure.

If stress levels or the number of cycles increase beyond the material’s capacity, failure will occur sooner. It is important to keep stress levels within allowable limits for extended component life.

Part (b)

Material defects can significantly reduce the safe operating life of engineering components because defects serve as stress concentrators that increase local stress around the defect. This leads to premature failure as the material cannot withstand the same level of cyclic stress as a defect-free component.

  • Surface roughness, porosity, inclusions, and abrupt section changes all create stress concentrations, lowering fatigue strength.
  • Coarse grain size, specific chemical compositions, and cold working introduce residual stresses that reduce fatigue resistance.
  • Corrosion, erosion, and decarbonisation weaken the material and accelerate fatigue crack initiation and propagation.
  • Faulty workmanship during assembly or processing introduces defects that may significantly shorten the component's life.
Part (c)

Factors Influencing Fatigue Cracking in Bedplate Transverse Girders:

  • Cylinder overload due to excess power puts excessive stress on the girders.
  • Incorrect crankshaft alignment induces uneven loading and stress concentrations.
  • Material defects, high residual stresses in welds, heat-affected zone hardening, and the presence of dissolved oxygen all reduce fatigue resistance.
  • Tank top deformation from pressurisation or overheating adds stress to the bedplate.

To minimise the risk of fatigue cracking:

(i) Constructional strength:

  • Bed plates are made up of M.S. plates with four steel casting, which are assembled and welded together so that the bed plate is strong longitudinally & transversely with good resistance to twisting along its length.
  • Longitudinal strength is obtained by fabricating each side of the bed plate in the form of a box girder.
  • The cast steel cross girder in which the main bearing is placed contributes to the bed plate's transverse strength and resistance against twisting along its length.
  • Resin cast chocks are used between the bedplate and the double bottom tank top to absorb the shocks & stress.

(ii) Maintenance:

  • Monthly checks on the bolt tension.
  • Monthly checks on engine load using power cards & measuring cylinder peak pressure.
  • Regular checking of tension for main bearing jack bolts as recommended by engine manufacturers.
  • Regular checks on crankshaft alignment by taking deflection & compare with recommended value.
  • By maintaining engine operations at specified load, temperature, pressure, speed, etc.
Q4 (16 Marks) Machinery Space & Systems πŸ”₯ Repeated 3x

With reference to a periodically unattended machinery space of a dry cargo vessel discuss the requirements for:

(a) Protection against flooding. (8)

(b) Control of propulsion machinery from the navigating bridge. (8)

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Part (a)

Essential requirements for any unattended machinery space (UMS) Ship to be able to sail at sea are enumerated in the SOLAS 1974 Chapter II-1, regulations 46 to regulation 53.

Requirements for Unattended Machinery Space (UMS) Ship:

1. Fire Precaution

  • Arrangements should be provided on the UMS ship to detect and give an alarm in case of fire.
  • In the boiler, air supply casing and uptake.
  • In scavenge space of propulsion machinery.
  • In engines of power, 2250 kW and above or cylinders having bore more than 300mm should be provided with an oil mist detector for the crankcase or bearing temperature monitor or either of two.

2. Centralized control & instruments are required in Machinery Space

  • UMS ships must have a centralised control room that is easily accessible and equipped with adequate instrumentation and equipment to monitor and operate all main and auxiliary machinery. A system must be provided to call the engineers to the machinery space in case of emergency

3. Protection against flooding:

  • UMS ships must have bilge wells that are located and designed to detect the accumulation of liquid at a normal angle of heel and trim and to accommodate the drainage of liquid during an unattended period. If the bilge pump starts automatically, an alarm must indicate that the flow of liquid pumped is more than the capacity of the pump.

4. Automatic Fire Detection

  • Alarms and detection should operate very rapidly and effectively. It should be placed at numerous well-sited places for quick response of the detectors.

5. Fire Extinguishing System

  • There should be arrangements for a fire extinguishing system other than the conventional hand extinguishers, which can be operated remotely from machinery space. The station must give control of emergency fire pumps, generators, valves, extinguishing media, etc.

6. Alarm System

  • A comprehensive alarm system must be provided for control & accommodation areas.

7. Automatic Start of Emergency Generator

  • Arrangements for the starting of an emergency generator and automatic connection to the bus bar must be provided in case of a blackout condition, apart from that, the following points are also to be noted.

8. Local hand control of essential machinery like steering, emergency generator starting, emergency start for main engine, etc. 8. Adequate settling tank storage capacity. 9. Regular testing & maintenance of machinery alarms & instruments.

Part (b)

(i) Protection against Flooding:

  • Bilge wells in UMS ships should be located and provided in such a manner that the accumulation of liquid is detected at a normal angle of heel and trim and should also have enough space to accommodate the drainage of liquid during unattended periods.
  • In the case of the automatic starting of the bilge pump, the alarm should be provided to indicate that the flow of liquid pumped is more than the capacity of the pump.

(ii) Control of Propulsion Machinery from Navigation Bridge:

  • The ship should be able to be controlled from the bridge under all sailing conditions. The bridge should be able to control the speed and direction of thrust and should be able to change the pitch in case of a controllable pitch propeller.
  • Emergency stops should be provided on navigating the bridge, independent of the bridge control system.
  • The remote operation of the propulsion should be possible from one location at a time; at such connection, interconnected control positions are permitted.
  • The number of consecutive automatic attempt which fails to start the propulsion machinery shall be limited to safeguard sufficient starting air pressure.
Q5 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 2x

(a) Describe, where on the hull plating would the following tests be carried out on the ships' hulls during drydock. (8)

(i) Hammer;

(ii) Hose

(b) Briefly identify which parts of the external plating of ships' hulls requires the closest attention. (8)

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Part (a)

Hull Plating Test Locations During Drydock:

(i) Hammer Test: The hammer test, a method of non-destructive testing, would be carried out across all areas of the hull plating. The surveyor would gently tap the plating to assess its condition, listening for sounds indicating potential damage like cracking or delamination. Particular attention would be given to areas showing signs of corrosion or previous repairs. Areas such as welds, bilge keels, and around penetrations (e.g., sea chests) would require extra scrutiny.

(ii) Hose Test: The hose test checks the watertight integrity of the hull. This would be performed on new welds on the shell plating and other areas where watertightness is required. This is done using a nozzle with a 12.5 mm diameter and 2 bar water pressure at a distance of 1.5 m. Specific locations would include welds, sea chests, and around valve penetrations. The exact location will depend on areas of concern identified during the visual inspection.

Part (b)

Parts of the External Plating Requiring the Closest Attention:

  • Welds: All weld joints, especially those in areas subject to stress or corrosion, must be carefully examined for cracks, porosity, and other defects.
  • Angles and other structural members: These need inspection for corrosion, cracking and distortion.
  • Bow: The bulbous bow, chain markings, and bow thruster area should be checked thoroughly.
  • Stern: The stern frame is subjected to slamming and must be examined closely for cracking and buckling.
  • Bilge Keel: This requires inspection for damage and proper attachment to the hull.
  • Openings in shell plating: Sea chests and ship side valves (overboard valves) must be inspected for leaks, corrosion, and proper operation.
  • Areas of previous damage or repair: These are especially susceptible to further corrosion or cracking and warrant close examination.
  • Areas exposed during previous dry-docking: Any areas that were exposed to the elements during the previous dry-dock require close examination for damage or corrosion.
  • Cathodic protection system: The effectiveness of the cathodic protection system and the condition of anodes need to be verified. Lack of anode consumption may indicate corrosion in other areas.
  • Rudder and Propeller: The condition of the rudder and propeller, including the welds, should be examined for any signs of damage or corrosion.
Q6 (16 Marks) Ship Stability πŸ”₯ Repeated 2x

(a) Explain what is meant by floodable length. (6)

(b) (i) Construct a graph from the following information:

Mean draft (m)

3.0

3.5

4.0

4.5

TPC (tonnes)

8.0

8.5

9.2

10.0

(ii) From this graph find the TPC's at draft of 3.2m, 3.7m, and 4.3m.

(iii) If the ship is floating at a mean draft of 4m, and then loads 50 tonnes of cargo, 10 tonnes of fresh water, and 25 tonnes of bunkers, whilst 45 tonnes of ballast are discharged, find the final mean draft. (10)

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Part (a)

Floodable length

is the amount, the ship can be flooded without the risk of sinking. i.e. the maximum length of the ship that can be flooded without submerging the margin line. The margin line is located 75mm below the bulkhead deck, providing a safety margin.

  • A floodable length curve can be drawn for different permeability.
  • It is used to test the after-bulkhead length so that the bulkhead length is sufficient to contain the flooding without sinking the vessel.
  • Factor of subdivision = Permissible length / Floodable length
Q7 (16 Marks) Hull Construction πŸ”₯ Repeated 2x

(a) Describe how bulkheads are tested. (6)

(b) A double bottom tank containing seawater is 6m long, 12m wide and 1m deep. The inlet pipe from the pump has its centre 75mm above the outer bottom. The pump has a pressure of 70kN/mΒ² and is left running indefinitely, Calculate the load on the tank top: (10)

(i) if there is no outlet

(ii) if the overflow pipe extends 5 m above the tank top.

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Part (a)

Testing for Water Tightness:

  • Hose Test: This involves applying a high-pressure water jet (2 bar from a distance of 1.5 meters, nozzle diameter at least 12mm) to the bulkhead's surface. This test directly checks for leaks. The test is typically done from the side where the stiffeners are attached.
  • Air Pressure Test: An air pressure test applies pressure (0.2 bar) for about an hour, detecting leaks that may not be apparent during the hose test. This usually happens prior to the application of protective coatings.
  • Structural Test: Visual inspection, especially of welding joints, is carried out. Non-Destructive Testing (NDT) is done where necessary. Tanks designed to hold liquids, which form subdivisions of the ship, are tested for tightness with a water head up to the deepest subdivision load line or to a head of 2/3 the depth from the top of the tank to the margin line, whichever is greater.
Q8 (16 Marks) Ship Types & Design πŸ”₯ Repeated 2x

(a) Define coefficient of fineness of waterplane area, block coefficient and midship coefficient. (6)

(b) A box shaped vessel has length, 100 m and breadth 12 m and floats at a range of drafts from 1m to 10m. Produce curves of KB, BM and KM. (10)

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Coefficient of Fineness of Waterplane Area (Cw)

This is the ratio of the ship's waterplane area to the area of a rectangle with the same length and breadth of the ship at the same waterline. It is given by:

$$C_{w}=\frac{WPA}{L\times B}$$

Where:

  • Cw​ = Coefficient of fineness of waterplane area
  • WPA = Waterplane area
  • L = Length of the ship
  • B = Breadth of the waterplane

Block Coefficient (Cb)

This is the ratio of the volume of displacement of a ship to that of a rectangular block having the same length, breadth, and draft as the ship. It is given by:

$$C_{b}=\frac{V}{L\times B\times D}$$

Where:

  • Cb = Block coefficient
  • V = Volume displaced by the ship
  • L = Length of the waterline
  • B = Breadth of the waterline
  • D = Draft

Midship Coefficient (Cm)

This is the ratio of the cross-sectional area at the midship section to the product of the ship's beam (breadth) and draft. It is given by:

$$C_{m}=\frac{A_{m}}{B\times D}$$

Where:

  • Cm​ = Midship coefficient
  • Am​ = Cross-sectional area at midship
  • B = Beam (breadth of the ship)
  • D = Draft
Q9 (16 Marks) Ship Stability πŸ”₯ Repeated 3x

(a) Explain the concept of dynamical stability. (6)

(b) A ship of 5000 tonne displacement has three rectangular double bottom tanks; A 12m long and 16m wide; B 14m long and 15m wide; C 14m long and 16m wide.

Calculate the free surface effect for any one tank and state in which order the tanks should be filled when making use of them for stability correction. (10)

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Part (a)

Dynamical Stability is defined as the amount of energy required to heel a ship from its upright equilibrium position to a specific angle of heel. It provides a measure of the vessel's stability by considering its behaviour in response to dynamic external forces, such as wind or waves.

  • The concept compares the heeling moment energy (from external forces) and the righting moment energy (from the ship's stability).
  • The ship will absorb the energy imparted by the heeling moment. If the righting energy is greater than the heeling energy, the ship will stabilize; otherwise, it may capsize.

Areas Under the Curve:

  • Area A: Represents the region where the heeling moment exceeds the righting moment (external energy > ship's stability).
  • Area B: Represents the region where the righting moment exceeds the heeling moment (ship's stability > external energy).
  • The balance of these areas determines whether the ship will right itself or continue to heel.

When exposed to heeling forces such as wind or waves, the vessel inclines and may roll over to a certain angle of heel. If the external force is applied instantaneously, the ship must have enough reserve dynamic stability to absorb the energy and return to an upright position. If the external force is constant, the ship will remain at an equilibrium angle where the righting moment equals the heeling moment.

This refers to the remaining righting energy available to counteract additional external forces. A higher reserve dynamic stability ensures the vessel can handle greater heeling forces without capsizing.

Q10 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 2x

A ship 120m long displaces 12000 tonne. The following data are available from trial results:

V (Knots)

10

11

12

13

14

15

sp (kW)

880

1155

1520

2010

2670

3600

(a) Draw the curve of Admiralty Coefficients on a base of speed

(b) Estimate the shaft power required for a similar ship 140m long at 14 knots. (16)

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Part (a)

Curve of Admiralty coefficients on a base of speed.

The Admiralty coefficient is defined as C = Delta^(2/3) V^3 / P, where Delta is displacement (tonnes), V ship speed, and P shaft power.

Chosen the parent ship: Delta=12000 t. Compute C at each trial speed.

Speed V=10: numerator = 12000^(2/3) x 10^3. 12000^(2/3)=524, 10^3=1000, so 524,000/880 = 595.

V=11: 524x1331=697,444; /1155 = 604.

V=12: 524x1728=905,472; /1520 = 596.

V=13: 524x2197=1,151,228; /2010 = 573.

V=14: 524x2744=1,437,856; /2670 = 538.

V=15: 524x3375=1,768,500; /3600 = 491.

Plotting C against V gives a curve that is roughly flat (595-604) at low speed and falls progressively at higher speed (to about 490 at 15 kn), because wave-making resistance and other power terms rise faster than V^3 at high Froude numbers. The curve demonstrates that the constant-Admiralty-coefficient assumption holds only near moderate speeds.

Part (b)

Shaft power for a similar ship 140 m long at 14 knots.

For geometrically similar ships the displacement scales as the cube of length:

Delta140 = 12000 x (140/120)^3 = 12000 x 1.5876 = 19051 t.

So Delta140^(2/3) = 19051^(2/3) = 714.

Corresponding speed: for similarity, V2/V1 = sqrt(L2/L1) = sqrt(1.1667) = 1.0801. For the 140 m ship running at 14 knots, the corresponding speed of the parent (120 m) ship is 14/1.0801 = 12.96 knots, which lies between the 13-knot trial point (C=573) - read C is about 570 from the curve.

Using the Admiralty coefficient at that point, C=570:

P140 = Delta140^(2/3) x V^3 / C = 714 x 14^3 / 570 = 714 x 2744 / 570 = 1,959,216 / 570 = 3437 kW.

Answer: the shaft power required is about 3440 kW (approximately 3.4 MW).

Q1 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 2x

(a) With reference to the underwater surface of a ship's hull

(i) Describe a hull plate roughness analyser system.

(ii) State the significance of the roughness profile and compare the typical roughness values for a new ship and a ship eight years old. (8)

(b) Which reference to the application of self-polishing paint in dry dock

(i) Describe the plate preparation necessary. (8)

(ii) State the defects that may occur in the if it is not correctly applied.

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Part (a)

With reference to the underwater surface of a hull:

(i) The Hull Roughness Analyzer (HRA) is a portable system designed to measure the roughness of a ship’s hull surface. It includes:

  • A portable microprocessor with a digital display and printout capability.
  • A handheld carriage equipped with a stylus measuring head. The stylus traces the hull surface, recording peaks and valleys to assess roughness.
  • The system measures 10 sample lengths of 50 mm during a single traverse over the hull. For each 50 mm sample, the highest peak to lowest valley is recorded as Rt(50).
  • Roughness surveys are conducted at approximately 100 locations on the hull, including the bow, stern, midship, and topping. Multiple traverses at each location provide numerous readings.
  • Mean Hull Roughness (MHR) is calculated as:
  • $$MHR\:=\:\frac{\sum Rt\left(50\right)}{n},\:where\:n\:is\:the\:number\:of\:readings\:at\:a\:station$$
  • Average Hull Roughness (AHR) is calculated as:
  • $$AHR\:\frac{\sum MHR}{m},\:where\:m\:is\:the\:total\:number\:of\:survey\:stations$$

(ii) Significance of Roughness Profile:

The roughness profile of the prepared hull surface impacts the performance of the applied coating and the overall operational efficiency of the vessel. A rough surface increases frictional resistance as the vessel moves through the water. This increased drag translates to higher power requirements for propulsion, leading to increased fuel consumption and operational costs. Furthermore, greater surface roughness contributes to increased carbon emissions, a concern under current MARPOL regulations. Therefore, a controlled and optimised roughness profile is essential for minimising frictional resistance, reducing fuel consumption and emissions, and maximising the longevity of the hull coating.

Typical Average Hull Roughness (AHR):

  • New Ship: 120-200 Β΅m (approx.).
  • Eight-Year-Old Ship: 300-400 Β΅m (approx.) due to annual deterioration of 20-40 Β΅m.
Part (b)

(i) Plate Preparation for Self-Polishing Paint Application:

Washing:

  • The hull surface must be thoroughly cleaned to remove all marine growth (algae, slime, etc.), accumulated salts, dirt, grease, and oil. High-pressure freshwater washing is the standard method for this initial cleaning. The goal is to present a clean substrate for subsequent stages.

Blasting:

  • Abrasive blasting is the preferred method for removing rust, defective paint, and any remaining contaminants. This process achieves a bare metal surface, essential for proper adhesion of the new coating. The extent of blasting (localized or full hull) depends on the condition of the existing surface. The intensity and type of abrasive used are carefully controlled to achieve the desired surface roughness profile.

Primer Application:

  • After blasting, the surface is again cleaned to remove any blasting debris. A primer coat is then applied to provide corrosion protection and to create an ideal surface for the subsequent topcoat adhesion. This primer acts as an intermediary layer, enhancing the bond between the substrate and the long-life coating system.

(ii) Defects in Self-Polishing Paint Application:

  • Blistering: Bubbles caused by trapped moisture.
  • Curtaining/Sagging: Paint sagging due to its own weight.
  • Dry Spray: Paint particles drying before reaching the surface.
  • Peeling/Flaking: Caused by moisture, dust, excessive film thickness, or incompatible paints.
  • Cissing/Crawling: Paint contracting immediately after application.
Q2 (16 Marks) Hull Construction πŸ”₯ Repeated 5x

With reference to membrane tanks for the carriage of liquefied gas at very low temperatures.

(a) Describe with the aid of a sketch, ONE method of building up the insulation: (6)

(b) State with reasons the alloy, which is used for the membrane. (4)

(c) Describe with the aid of a sketch, how the tanks are located and supported. (6)

(i) Longitudinally

(ii) Transversely

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Part (a)

Membrane tanks for liquefied gas carriage are built with a double hull throughout the cargo length. The insulation system includes the following:

  1. Primary Barrier: A thin membrane made of INVAR (36% Nickel, 64% Iron) with a thickness of 0.5 to 1.2 mm, forming the primary containment layer for the liquefied gas.
  2. Primary Insulation box: A 200 mm thick layer of granulated Perlite insulation packed in plywood boxes, surrounding the primary barrier. Perlite is siliconised to make it impervious to moisture.
  3. Secondary Barrier: A secondary membrane made of the same INVAR material is installed to prevent cargo leakage in case of primary barrier failure.
  4. Secondary Insulation box: Another 200 mm thick layer of granulated Perlite is placed above the secondary barrier to provide additional insulation and prevent thermal transfer.
Part (b)

Alloy used for the membrane:

Invar (36% Ni, 64% Fe) is used for both the primary and secondary barriers. The reason for this choice is its exceptionally low coefficient of thermal expansion. This eliminates the need for expansion joints or corrugations in the membrane design. In addition, Invar remains strong and does not become brittle at the very low temperatures experienced by the liquefied gas. The thin and lightweight nature of Invar maximizes the cargo-carrying capacity of the tank.

Part (c)

Membrane tanks are either independent or self-supporting

, meaning they don't form part of the ship's hull and don't contribute to the ship's structural strength. They can be spherical, cylindrical, or prismatic (box-shaped). Prismatic tanks usually have internal stiffeners like bulkheads, webs, girders, and stiffeners for added structural integrity.

(i) Longitudinally:

Tanks are positioned longitudinally within the ship's cargo hold. The tanks are supported longitudinally by anti-roll chocks and anti-lift chocks that resist forces caused by ship motions. These chocks ensure the tank remains stable even in rough sea conditions. They also act as thermal barriers, preventing the transfer of heat between the hull and the tank.

The chocks also serve as thermal barriers between the hull and cargo and are often constructed of wood or plastic materials.

(ii) Transversely: Transverse support is provided by anti-pitch chocks and support chocks, which stabilize the tank against lateral forces. These chocks are typically made of plywood or plastic and help prevent thermal stress between the tank and the ship’s structure.

Q3 (16 Marks) Hull Construction

State FOUR terms used to describe the conditions that relate to the distortion of a ship's hull undergoes in heavy seas, stating in EACH case the type of stresses involved and where the stresses occur. (16)

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Sagging: Sagging deformation are caused by the uneven distribution of weight and buoyancy in the length of the hull. If a ship is supported on either end by the crest of a wave, the forces exerted by the waves on the buoyancy of the vessel would tend to lift the ends, whilst the center of the vessel would suffer a loss of buoyancy and tend to sag. This deformation is known as sagging.

Type of Stress: Compressive stress on deck, tensile stress on keel.

Hogging: When the wave passes the reverse situation of hogging results, i.e. the middle of the vessel is supported on the crest whilst the two ends hang over the crest on either side. This action is known as hogging.

Type of Stress: Tensile stress on deck, compressive stress on keel

Racking: When a vessel is rolling in a seaway the transverse section will try to distort at the corners due to racking stresses.

Type of Stress: Shear stress on transverse bulkheads and hull structure

Torsional forces: If a vessel is subjected to pitching and rolling at the same time, i.e. taking the wave on either bow, the vessel tends to twist longitudinally. These are known as torsional stresses or forces.

Q4 (16 Marks) Ship Stability πŸ”₯ Repeated 3x

With respect to trim and stability, describe the following:

(a) Effects on centre of gravity of slack tanks. (4)

(b) Effect on stability of ice formation on superstructure. (4)

(c) Effects of wind and waves on ship's stability. (4)

(d) Effect of water absorption by deck cargo and retention of water on deck. (4)

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Part (a)

Slack tanks, or partially filled tanks, significantly impact a ship's stability due to the free surface effect

  • When tanks are partially filled (slack), the liquid inside moves freely as the ship heels.
  • This movement shifts the centre of gravity (G) laterally towards the heeling side.
  • The righting lever (GZ) decreases, resulting in reduced metacentric height (GM) and overall stability.
  • The virtual loss of GM is proportional to the breadth of the tank and the height of the free surface.

Consequences of Slack Tanks:

  • Increased angle of heel.
  • Reduced stability, making the ship tender and more prone to capsizing.

To Minimize the Effect:

  • Avoid slack tanks where possible; either fill tanks completely or empty them.
  • Design tanks with longitudinal divisions or swash bulkheads to limit liquid movement.
  • Use sluice valves in tanks to control liquid transfer.
  • Prioritize filling smaller tanks at the ship's bottom to lower G and improve stability.
Part (b)

Ice formation on superstructure:

Ice accumulating on the superstructure adds mass high up on the ship. This raises the centre of gravity (G), decreasing the metacentric height (GM). A lower GM reduces stability, making the ship more tender (more easily rolled) and increasing the period of roll. The ship becomes more susceptible to capsizing.

Part (c)

Effect of Wind and Waves

Wind:

  • High freeboard or tall superstructures increase windage, leading to a greater rolling effect.
  • Rolling caused by wind reduces stability, especially if the ship remains heeled for a prolonged period.

Waves:

  • Large waves, especially when the ship is on the crest, can cause a significant loss of stability due to reduced underwater buoyant volume.
  • The ship may develop excessive heeling or capsizing tendencies.
  • Long ships are more vulnerable to wave action due to greater surface exposure, further reducing stability.
Part (d)

Effect of Water Absorption by Deck Cargo and Retention of Water on Deck:

  • Water Absorption by Deck Cargo:
    • Certain types of deck cargo, such as timber or other absorbent materials, can take up water during a voyage.
    • This absorbed water increases the weight of the cargo, raising the centre of gravity (G) of the ship.
    • As G moves higher, GM reduces, leading to decreased stability.

    • Retention of Water on Deck:
      • Water that accumulates on the deck, such as from heavy rain or seawater, adds additional weight to the ship's upper structure.
      • This shifts G upward, reducing GM and stability.
      • Retained water may also cause a list if it collects asymmetrically, further impacting the ship's balance.

      • Consequences:
        • Increased risk of capsizing in rough seas.
        • Increased rolling and reduced righting ability.

        • Mitigation Measures:
          1. Ensure proper drainage systems are in place to quickly remove water from the deck.
          2. Regularly monitor and secure deck cargo to prevent excessive water absorption.
Q5 (16 Marks) Corrosion & Protection πŸ”₯ Repeated 4x

With reference to the prevention of hull corrosion discuss:

(a) Surface preparation and painting of new ship plates. (6)

(b) Design of the ships structure and its maintenance. (5)

(c) Cathodic protection by sacrificial anodes, of the internal and external areas of the ship. (5)

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Part (a)

Surface Preparation and Painting of New Ship Plates:

The process begins with the removal of mill scale, a thin layer of iron oxide that forms during the steel rolling process.

1. Removal of Mill Scale:

  • Weathering: Using wire brushes.
  • Pickling: Immersing plates in a weak solution of Hβ‚‚SOβ‚„ or HCl.
  • Shot Blasting: Abrasive blasting to remove rust and mill scale.
  • Flame Cleaning: Heating and cleaning the surface.

Freshwater Washing and Drying: Ensures the surface is clean and ready for priming.

2. A primer is then applied. This acts as a base layer for subsequent coatings and improves adhesion. Common types include epoxy primers with corrosion-inhibiting pigments like iron oxide, zinc, and calcium phosphates (though zinc content is reduced due to safety concerns).

3. Next comes the anticorrosive coating, the main layer offering corrosion protection. Common choices are two-component epoxies, coal tar epoxies, and epoxy or polyester coatings with glass flakes for added strength and water impermeability.

4. Finally, an antifouling coating prevents marine organism attachment. While tin-based paints were once common, environmental regulations have led to their replacement with copper-based, silicone-based, or non-TBT (Tributyltin) self-polishing coatings. These utilize seawater-soluble polymers. The number of antifouling layers (two or three) depends on the chosen system and desired lifespan. The dry film thickness (DFT) of the primer is closely monitored to prevent cracking and ensure effective protection.

Part (b)

Corrosion prevention begins at the design stage:

  • Avoid dissimilar metal joining to minimize galvanic corrosion.
  • Use high-quality steel plates based on the galvanic series.
  • Minimize areas where water can accumulate by ensuring proper drainage systems.
  • Ensure sufficient anodes are installed in optimal positions.
  • Consider installing Impressed Current Cathodic Protection (ICCP) and Marine Growth Prevention Systems (MGPS).

Maintenance:

  • Regular checks and maintenance of ICCP and MGPS systems.
  • Renewal of sacrificial anodes during dry-docking.
  • Cleaning, repairing, and repainting surfaces to maintain the protective coatings.
  • Ensuring proper drainage and inspecting for any areas of corrosion.
Part (c)

Cathodic Protection by Sacrificial Anodes:

Sacrificial anodes are less noble metals (more electro-negative) than the hull material. When immersed in seawater, they corrode preferentially, protecting the hull. Common materials include zinc, aluminium, and magnesium. They are welded to the hull for good electrical contact.

  • External surface: Large anodes provide complete protection until the next dry-docking.
  • Internal surface: Anodes are fitted inside tanks, filters, and coolers to prevent internal corrosion.
  • MGPS: These systems utilize sacrificial anodes to protect seawater lines and pumps, further preventing marine growth.
  • ICCP: Impressed current systems offer an alternative, using an external DC power source and permanent anodes (typically titanium or platinum) for continuous protection. These systems are self-regulating and adjust current output based on hull coating condition. Regular monitoring (e.g., monthly data analysis) is important for both sacrificial and impressed current systems.
Q6 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

(a) Describe how the distribution of mass within the ship affects the rolling period. (6)

(b) The righting moments of a ship at angles of heel of 0, 15Β°, 30Β°, 45Β°, and 60Β° are 0, 1690, 5430, 9360 and 9140 kN-m respectively. Calculate the dynamical stability at 60Β°. (10)

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Part (a)

The rolling period is influenced by the ship's metacentric height (GM) and radius of gyration (K), which are determined by the location of the masses onboard.

Masses at the bottom:

  • The centre of gravity (G) moves down.
  • Metacentric height (GM) increases, resulting in greater stability.
  • Rolling period decreases.

Masses at the top:

  • The centre of gravity (G) moves up.
  • Metacentric height (GM) decreases, reducing stability.
  • Rolling period increases.

Masses concentrated at the centre:

  • The radius of gyration (K) decreases.
  • Rolling period decreases.

Masses distributed away from the centre:

  • The radius of gyration (K) increases.
  • Rolling period increases.
Q7 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

(a) Describe briefly the significance of the factor of subdivision. (6)

(b) A ship 120m long has a light displacement of 4000 tonne and LCG in this condition 2.5m aft of midships. The following items are then added: (10)

Cargo 10000 tonne LCG 3.0m forward of midships

Fuel 1500 tonne LCG 2.0 m aft of midships

Water 400 tonne LCG 8.0m aft of midships

Stores 100 tonnes LCG 10.0m forward of midships

Using the following hydrostatic data, calculate the final draughts:

Draught (m)

Displacement (t)

MCT1cm (tm)

LCB from midships

LCF from midships

8.50

16650

183

1.94F

1.29A

8.00

15350

175

2.10F

0.60F

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Part (a)

Factor of Subdivision

The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

Permissible Length Formula:

$$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

Part (b)

Mass added

LCG from midship

Mass moments

F

A

10000

3.0 Fwd

30000

1500

2.0 Aft

3000

400

8.0 Aft

3200

100

10.0 Fwd

1000

4000

2.5 Aft

10000

16000

31000

16200

$$Excess\:moment=31000-16200=14800$$

$$LCG=\frac{\sum M}{\sum m}=\frac{14800}{16000}$$

$$LCG=0.925m\:fwd\:of\:midship$$

Draught

Displacement

MTC 1cm

LCB from midship

LCF from midship

8.5

16650

183

1.94 Fwd

1.20 Aft

8.25

16000

179

2.03 Fwd

0.57 Aft

8.0

15350

175

2.10 Fwd

0.06 Fwd

$$LCG=0.925m\:Fwd\:of\:midship$$

$$LCB=2.02m\:Fwd\:of\:midship$$

$$Trimming\:lever=LCB-LCG$$

$$2.02-0.925$$

$$=1.09m$$

$$Trimming\:moment=m\times d$$

$$=16000\times1.09$$

$$=17440tm$$

$$Change\:of\:trim=\frac{Trimming\:moment}{MCT_{1\operatorname{\mathrm{cm}}}}$$

$$=\frac{17440}{179}$$

$$97.43\operatorname{cm}$$

$$Draft\:fwd=8.25-\frac{97.43}{100\times120}\left(\frac{120}{2}+0.57\right)$$

$$=7.756m$$

$$Draft\:aft=8.25+\frac{97.43}{100\times120}\left(\frac{120}{2}-0.57\right)$$

$$=8.735m$$

Q8 (16 Marks) Ship Stability πŸ”₯ Repeated 2x

(a) Explain the effect of trim on tank soundings. (6)

(b) A ship of 6600 tonne displacement has KG 3.6m and KM 4.3m. A mass of 50 tonne is now lifted from the quay by one of the ship's derricks whose head is 18 m above the keel. The ship heels to a maximum of 9.5Β° while the mass is being transferred. Calculate the outreach of the derrick from the ship's centreline. (10)

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Part (a)

The effect of trim on tank soundings is significant because the reading obtained from the sounding pipe does not directly reflect the actual volume of liquid in the tank when the ship is trimmed. Typically, the sounding pipe is located at the aft end of the tank.

If the ship is trimmed by the stern, the liquid will settle more toward the aft end, resulting in a higher sounding than if the ship were on an even keel. Conversely, if the ship is trimmed by the bow, the liquid shifts forward, and the sounding obtained will be less, potentially underestimating the actual volume.

To ensure accurate volume measurement despite varying trim conditions, tank calibrations are carried out for different trims, and the data is compiled into a sounding book or trim correction table. When taking soundings, the trim is noted, and the corrected volume is obtained using these calibration tables, ensuring reliable tank content assessment regardless of trim.

Q9 (16 Marks) Ship Stability πŸ”₯ Repeated 2x

(a) Explain the effect on GM during the filling of a double–bottom tank (6)

(b) A ship of 8,000 tonnes displacement has KM 7.5 m, and KG 7.0 m. A double bottom tank is 12 meters long 15 meters wide and 1 meter deep. The tank is divided longitudinally at the centre line and both sides are full of salt water. Calculate the list if one side is pumped on until it is half empty. (10)

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Part (a)

During the filling of a double-bottom (DB) tank, which is located at the bottom of the ship, mass is added down in the vessel. As a result, the center of gravity (G) shifts downward, leading to an increase in GM (metacentric height) and, consequently, an improvement in the ship's stability.

However, if the ship is heeled or trimmed during filling, the liquid inside the tank may collect on the lower side, causing both the center of gravity (G) and the center of buoyancy (B) to shift to that side. This results in a virtual loss of GM and a reduction in stability.

Once the DB tanks are fully pressed up, the liquid is evenly distributed, the ship returns to an upright position, and the center of gravity remains at its lowest point, resulting in maximum GM and optimal stability.

Q10 (16 Marks) Ship Stability πŸ”₯ Repeated 2x

A ship of 9,900 tonnes displacement has KM = 7.3 m and KG = 6.4 m, she has yet to load two 50 tonne lifts with her own gear and the first lift is to be placed on deck on the inshore side (KG 9 m and center of gravity 6m out from the center line). When the derrick plumbs the quay its head is 15m above the keel an 12m out from the center line.

Calculate the maximum list during operation. (16)

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Maximum list during derrick operation.

Initial condition: displacement = 9900 t; KM = 7.3 m; KG = 6.4 m, so initial GM = KM - KG = 7.3 - 6.4 = 0.90 m.

The first 50 t lift is picked up off the quay (inshore side). While it is suspended at the derrick head, its position is above and outboard of the crane; as it swings, it acts as a weight at the head which is 15 m above the keel and 12 m out from the centreline. The maximum list occurs when the weight is suspended fully outboard (12 m out) before it is placed on deck at 6 m out, because the heeling moment is greatest then.

Effective heeling of the first lift. Treated the 50 t as suspended at (y=12 m outboard, z=15 m above keel):

  • Transverse shift of the centre of gravity, GG' = w x y / Delta = 50 x 12 / 9900 = 0.0606 m outboard (about 6.06 cm).
  • Vertical rise of the centre of gravity, since the load hangs at the derrick head above deck: compare to its final on-deck position at 6 m out, 9 m above keel. The movement from quay to head raises the effective load centre; the vertical rise of G = w x (rise)/(Delta). For the list the relevant GM is reduced by the raised KG.

A cleaner approach is to compute the moment. At maximum list the weight is fully outboard at 12 m; its KG contribution raises the centre of gravity by GGv = 50 x (15 - 9)/9900 = 50 x 6/9900 = 0.0303 m. New KG = 6.4 + 0.0303 = 6.4303 m; the effective GM falls to 7.3 - 6.4303 = 0.8697 m.

List, GGh = 50 x 12 / 9900 = 0.0606 m outboard.

tan(list) = GGh / GM = 0.0606 / 0.8697 = 0.0697.

So list = atan(0.0697) = 3.99 deg, about 4 deg.

Answer: the maximum list during the operation is about 4 degrees towards the outboard (derrick) side. (Note: after the first lift is placed on deck at 6 m out and the second lift is begun, the situation changes slightly; the critical maximum effect is when the suspended load is furthest outboard.)

Q1 (16 Marks) Hull Construction πŸ”₯ Repeated 8x

(a) Sketch the cross-section of a bulk carrier with either deep or shallow double bottom showing the type of framing used.

(b) (i) Describe the corrosion problems experienced with ballast tanks.

(ii) State how such tanks are protected against extensive corrosion.

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Part (a)

Mid-ship section of bulk carrier:

Part (b)

(i) Corrosion problems in ballast tanks

are significant and arise due to various factors:

  • Galvanic corrosion occurs due to the interaction between dissimilar metals, accelerated by the seawater environment and differential aeration.
  • Sulphate-reducing bacteria in river mud can cause localized pitting, leading to penetration of the bottom shell.
  • Dissolved oxygen in seawater reacts with metal surfaces to form rust, significantly contributing to structural deterioration.
  • The rate of corrosion peaks at a 3.5% salt concentration, typical of seawater.
  • Neglecting maintenance exacerbates the problem, allowing corrosion to progress unchecked.

(ii) Protection against extensive corrosion in ballast tanks involves the following measures:

  • Complete coating of the tank surfaces with properly selected marine-grade paint to create a protective barrier.
  • Installation of sacrificial anodes, designed to corrode preferentially and protect the tank structure.
  • Using large anodes with greater volume relative to surface area to ensure extended protection.
  • Regular gauging of plates to assess thickness and identify areas requiring reinforcement or repair.
  • Conducting regular inspections, cleaning the tanks, removing rust, and repainting to maintain structural integrity and prolong the lifespan of the ballast tanks.
Q2 (16 Marks) Surveys & Drydocking

Describe the process and preparation as per the classification society requirements for in-water survey of the underwater structure of a very large vessel.

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The purpose of the in-water survey is to obtain information on the condition of the hull and machinery of large vessels normally obtained from the Docking Survey. The survey may be carried out instead of any one of the two docking surveys, under the surveillance of a surveyor with the ship at a stable draught in sheltered waters, the in-water visibility is to be good and the underwater hull clean.

The survey requires a self-propelled, steerable survey vehicle fitted with a long-range TV camera to aid steering and check for hull distortion also a close-up high-resolution TV color camera to give a true picture of the state of coatings and for inspection of weld seams. In some cases a 35mm still camera is fitted. An ultrasonic probe is provided to measure plate thicknesses and other equipment includes a depth meter and speed indicator. Power is supplied and information is relayed using an umbilical from the vehicle to the survey boat

The survey boat usually houses a console containing TV monitors. plate thickness printout, audio cassette recorder, video recorder and playback unit, diver communication system, vehicle control system and associated instrumentation.

The survey vehicle is taken to the starting datum by a diver. With the aid of one of the TV monitors and using the shell expansion plan as a map, the vehicle may be guided, from the control console, over the bottom and sides of the hull by following weld runs and by reference to other features such as inlets and tank plugs. Pictures and navigational information are relayed back and video recorded along with plate thicknesses giving the surveyor an integrated visual record of all relevant information. In addition, a plate thickness print-out can be produced and/or an audio recording. The vehicle will also provide pictures of such items as the stern frame, rudder, propeller, bilge keels and hull openings although a diver may be used with a handheld camera for closer inspection of these items and also for inspection of plating on the turn of the bilge. Divers are used to measure stern bearing wear down and pintle clearances, and to inspect such things as stern seals, anodes, pintles and rudder stock couplings.

To facilitate underwater surveys plans must be submitted showing the external features of the hull below the sheer strake together with a key plan indicating the location of these features, also of reference points and the position of water-tight and oil-tight bulkheads. Notes are included on the proposed methods of marking and identifying plates. To assist divers color photographs should be provided of items such as shell openings, rudder closing plates and wear-down gauge plugs.

Provision should be made on the ship identifying bulkheads and frames above the waterline and also for establishing the identity and position of each propeller blade from inside the ship. The design of the ship must facilitate water inspection and repair, for example, sea inlets must be capable of being blanked off and drained to bilges, shell gratings hinged if practicable and anodes easily changed.

Q3 (16 Marks) General

Sketch and describe the construction of a bulbous bow. Why is such an arrangement fitted on a merchant ship?

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Construction and purpose of a bulbous bow.

Construction: The bulbous bow is a roughly pear-shaped or blunter solid forepeak projection from the stem, below the load waterline, formed by specially shaped shell plating. The bulb is built by a thick, heavily plated shell with internal framing made up of transverse webs (swash bulkheads/forepeak frames) and fore-and-aft stringers that connect it to the stem of the vessel and the side plating. It extends well below the waterline and often contains a forepeak collision-bulkhead behind it. The stanchion, stem bar and the bulb take the pounding and panting loads; the bulb shell plates are thicker at the leading edge and stiffened by the framing arrangement. The bulb usually joins the ship's side with a smooth radius, and the whole forepeak is subdivided and stiffened.

Why fitted: when a ship moves, water is pushed aside creating a bow wave. At moderate speeds (Froude number about 0.25 to 0.35 typical of merchant ships) the bulb creates its own system of waves, and by correct sizing and position the bulb's wave trough cancels part of the bow wave crest of the hull, reducing the wave-making (residual) resistance. The net result is a reduction in total resistance at the design speed, giving:

  • lower installed power and fuel consumption (better EEDI);
  • improved ship propulsion economics;
  • reduced bow wave and stern wave of the ship in general;
  • assistance in carrying forefoot buoyancy, helping the ship's longitudinal distribution of buoyancy and allowing finer, faster forms without a large bow wave penalty.

The bulb is essentially an interference device that cancels bow-wave energy, and is most beneficial at the design/ballast speeds of displacement vessels (bulk carriers, tankers, container ships, general cargo, ro-ro).

Q4 (16 Marks) Hull Construction πŸ”₯ Repeated 3x

With reference to the vessel hull panting and pounding

(a) Explain the causes and effects of panting and pounding, indicating the affected areas. (5)

(b) Describe the constructional details designed to resist panting and pounding. (5)

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Part (a)

Causes and Effects of Panting and Pounding, Including the Affected Areas

Panting

Panting refers to the in-and-out flexing or pulsation of the shell plating at the ends of a ship, mainly caused by variations in sea pressure during pitching motions in rough weather.

When a ship moves through heavy seas, especially while pitching, the water pressure acting on the bow and stern changes continuously. These fluctuating pressures cause the side shell plating to vibrate and flex. The effect is known as panting.

The forward end of the ship is affected more severely than the aft end because:

  • The bow directly encounters incoming waves while the ship is making headway.
  • Water pressure at the fore end changes rapidly as the bow rises and falls in the sea.

The affected areas are mainly:

  • Forward shell plating near the bow
  • Side shell plating at the fore peak region
  • To a lesser extent, the aft end of the ship

Effects of Panting

Panting can lead to:

  • Repeated flexing and vibration of shell plating
  • Fatigue stresses in plating and framing
  • Loosening of rivets or welded joints
  • Cracking or deformation of structural members
  • Reduction in structural strength if not properly reinforced

Therefore, special strengthening arrangements are provided at the forward and aft ends of the vessel to resist panting stresses.

Pounding

Pounding refers to the heavy impact experienced at the bottom forward part of the ship when the bow emerges from the water during pitching and then slams violently back onto the sea surface.

This condition usually occurs when:

  • The ship is pitching heavily in rough seas,
  • The fore part lifts clear of the water due to heaving and pitching motions,
  • The bow then falls heavily onto the wave surface.

The effect is most severe when the vessel is in the light ship condition, because the bow rises more easily out of the water.

The main area affected by pounding is:

  • The bottom shell plating in the forward region of the ship,
  • Especially near the forefoot and forward bottom structure.

The stern may also experience similar impacts from following seas, but usually to a lesser extent.

Effects of Pounding

Pounding produces:

  • Severe impact stresses on bottom plating
  • Buckling or deformation of bottom structure
  • Cracks in shell plating or framing
  • Structural fatigue and weakening
  • Damage to internal supporting members

To withstand these heavy impact loads, the forward bottom structure is specially strengthened.

Part (b)

Constructional Details Designed to Resist Panting and Pounding

Special structural arrangements are incorporated in ship construction to resist the stresses caused by panting and pounding.

Structural Arrangements to Resist Panting

The following strengthening arrangements are provided mainly at the bow and stern regions:

1. Panting Beams

Panting beams are horizontal beams fitted across the ship near the bow and stern.

Their purpose is to:

  • Support the side shell plating,
  • Reduce excessive flexing,
  • Increase structural rigidity against panting stresses.

2. Panting Stringers

Panting stringers are horizontal girders fitted along the ship side in the panting region.

They:

  • Connect frames together,
  • Provide additional stiffness to the shell plating,
  • Help distribute fluctuating sea pressure loads.

3. Close-Spaced Framing

Frames in the panting region are spaced closer together than in other parts of the ship.

This:

  • Provides better support to shell plating,
  • Reduces plate vibration and deformation.

Structural Arrangements to Resist Pounding

To resist pounding stresses at the forward bottom region, additional strengthening is provided.

1. Increased Bottom Plating Thickness

The shell plating near the keel and forward bottom region is made thicker.

Usually:

  • The first few strakes of bottom plating on either side of the keel are increased in thickness.

This enables the structure to withstand repeated impact loads.

2. Plate Floors

Solid plate floors are fitted at closer spacing in the forward bottom region.

These:

  • Strengthen the bottom structure,
  • Distribute pounding stresses more effectively,
  • Prevent deformation of bottom plating.

3. Additional Internal Reinforcement

Extra brackets, girders, and stiffeners may also be provided in the fore peak and bottom structure to improve strength and rigidity.

Q5 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 13x

Describe the relationship between frictional resistance and

(a) Ship speed

(b) The wetted surface area

(c) The surface roughness

(d) The length of the vessel.

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Frictional resistance arises due to the interaction between the ship's hull and the water as the vessel moves. This resistance is caused by eddying water adjacent to the hull that is drawn along with the ship. The frictional resistance is given by:

$$R_{f\:=\:}f\:.\:s\:.\:v^{n}$$

Where:

  • f = Coefficient of friction
  • s = Wetted surface area
  • v = Ship's speed in knots
  • n = Constant (1.82)
Part (a)

Ship Speed (v):

  • Frictional resistance increases with increasing ship speed. The relationship is not perfectly linear, but it's approximately described by the formula Rf = fsv^n (Rf ∝ v^n), where 'n' is a constant (approximately 1.82). This means that a small increase in speed results in a larger increase in frictional resistance.
Part (b)

Wetted Area (s):

  • Frictional resistance is directly proportional to the wetted surface area (Rf ∝ s). A larger wetted area (the part of the hull submerged in water) leads to greater frictional resistance. The wetted area increases with the ship's draught (depth in the water).
Part (c)

Surface Roughness:

  • Increased surface roughness increases frictional resistance. Roughness can be caused by fouling (marine growth), corrosion, or poor hull maintenance. Regular cleaning and maintenance of the hull surface help to minimise roughness and reduce resistance.
Part (d)

Length of the Vessel

  • Frictional resistance is influenced by the vessel's length. As the length increases, the wetted surface area grows, and the coefficient of friction (f) changes, leading to an increase in Rf. Longer vessels experience higher frictional resistance, but streamlined designs can mitigate the effect to some extent.
Q6 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

(a) Describe how the distribution of mass within the ship affects the rolling period. (6)

(b) The righting moments of a ship at angles of heel of 0, 150, 300, 450 and 600 are 0, 1690, 5430, 9360 and 9140 kNm respectively. Calculate the dynamical stability at 600. (10)

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Part (a)

The rolling period is influenced by the ship's metacentric height (GM) and radius of gyration (K), which are determined by the location of the masses onboard.

Masses at the bottom:

  • The centre of gravity (G) moves down.
  • Metacentric height (GM) increases, resulting in greater stability.
  • Rolling period decreases.

Masses at the top:

  • The centre of gravity (G) moves up.
  • Metacentric height (GM) decreases, reducing stability.
  • Rolling period increases.

Masses concentrated at the centre:

  • The radius of gyration (K) decreases.
  • Rolling period decreases.

Masses distributed away from the centre:

  • The radius of gyration (K) increases.
  • Rolling period increases.
Q7 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

(a) Describe briefly the significance of the factor of subdivision. (6)

(b) A ship 120m long has a light displacement of 4000 tonne and LCG in this condition 2.5m aft of midships. The following items are then added: (10)

Cargo 10000 tonne LCG 3.0m forward of midships

Fuel 1500 tonne LCG 2.0 m aft of midships

Water 400 tonne LCG 8.0m aft of midships

Stores 100 tonnes LCG 10.0m forward of midships

Using the following hydrostatic data, calculate the final draughts:

Draught (m)

Displacement (t)

MCT1cm (tm)

LCB from midships

LCF from midships

8.50

16650

183

1.94F

1.29A

8.00

15350

175

2.10F

0.60F

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Part (a)

Factor of Subdivision

The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

Permissible Length Formula:

$$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

Part (b)

Mass added

LCG from midship

Mass moments

F

A

10000

3.0 Fwd

30000

1500

2.0 Aft

3000

400

8.0 Aft

3200

100

10.0 Fwd

1000

4000

2.5 Aft

10000

16000

31000

16200

$$Excess\:moment=31000-16200=14800$$

$$LCG=\frac{\sum M}{\sum m}=\frac{14800}{16000}$$

$$LCG=0.925m\:fwd\:of\:midship$$

Draught

Displacement

MTC 1cm

LCB from midship

LCF from midship

8.5

16650

183

1.94 Fwd

1.20 Aft

8.25

16000

179

2.03 Fwd

0.57 Aft

8.0

15350

175

2.10 Fwd

0.06 Fwd

$$LCG=0.925m\:Fwd\:of\:midship$$

$$LCB=2.02m\:Fwd\:of\:midship$$

$$Trimming\:lever=LCB-LCG$$

$$2.02-0.925$$

$$=1.09m$$

$$Trimming\:moment=m\times d$$

$$=16000\times1.09$$

$$=17440tm$$

$$Change\:of\:trim=\frac{Trimming\:moment}{MCT_{1\operatorname{\mathrm{cm}}}}$$

$$=\frac{17440}{179}$$

$$97.43\operatorname{cm}$$

$$Draft\:fwd=8.25-\frac{97.43}{100\times120}\left(\frac{120}{2}+0.57\right)$$

$$=7.756m$$

$$Draft\:aft=8.25+\frac{97.43}{100\times120}\left(\frac{120}{2}-0.57\right)$$

$$=8.735m$$

Q8 (16 Marks) Ship Stability πŸ”₯ Repeated 3x

(a) Explain how the distribution of masses affects rolling and pitching. (6)

(b) A ship turns in a circle of radius 100 metres at a speed of 15 knots. The GM is 2/3 metres and BG is 1 metre. If g=981 cm/sec2 and 1 knot is equal to 1.8532 Km/hour, find the heel due to turning. (10)

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Part (a)

The rolling period is influenced by the ship's metacentric height (GM) and radius of gyration (K), which are determined by the location of the masses onboard.

Masses at the bottom:

  • The centre of gravity (G) moves down.
  • Metacentric height (GM) increases, resulting in greater stability.
  • Rolling period decreases.

Masses at the top:

  • The centre of gravity (G) moves up.
  • Metacentric height (GM) decreases, reducing stability.
  • Rolling period increases.

Masses concentrated at the centre:

  • The radius of gyration (K) decreases.
  • Rolling period decreases.

Masses distributed away from the centre:

  • The radius of gyration (K) increases.
  • Rolling period increases.
Part (b)

$$Radius\:\left(R\right)=100m$$

$$Speed\:\left(v\right)=15\:knots$$

$$GM=\frac23m$$

$$BG=1m$$

$$\tan\the\theta=\frac{v^2\times BG}{g\times r\times GM}$$

$$=\frac{\left(15\times0.514\right)^2\times1\times3}{9.81\times100\times2}$$

$$\theta=5.19\degree$$

Q9 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 4x

(a) Describe the effect of cavitations on the propeller blades. (6)

(b) A propeller 4.6m diameter has a pitch of 4.3m and boss diameter of 0.75m. The real slip is 28% at 95 rev/min. Calculate the speed of advance, thrust and thrust power. (10)

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Part (a)

Effect of cavitation on propeller blades:

Erosion:

  • Cavitation causes the formation and collapse of vapor bubbles on the propeller blade surface.
  • The collapse of these bubbles produces high-pressure shockwaves and microjets that erode the blade material, leading to surface pitting and progressive damage.

Vibration:

  • Uneven cavitation across the blades leads to imbalanced forces, causing vibrations in the propeller and the ship.
  • These vibrations can reduce the comfort of passengers and crew and stress the ship's structural components.

Noise:

  • The collapse of vapor bubbles generates loud noise, which can interfere with onboard communication and underwater sonar systems.
  • This noise is a significant concern for naval vessels and marine life.

Reduced Performance:

  • Cavitation reduces the efficiency of the propeller by causing loss of thrust and torque.
  • The presence of cavitation decreases the propeller’s ability to convert rotational energy into forward motion, lowering the ship's speed and increasing fuel consumption.

(b) Given:

$$D=4.6m$$

$$P=4.3m$$

$$d=0.75$$

$$S=28\%$$

$$n \space = \space 95 \space rev/ min$$

$$V_T \space = \space P \times N \times {{3600} \over 1852}$$

$$ = \space 4.3 \times {{95} \over 60} \times {{3600} \over 1852}$$

$$V_{T}=13.23knots$$

$$Real \space slip \space (S) \space = \space {{V_T - V_a} \over V_T}$$

$$0.28 \space = \space {{13.23 - V_a} \over 13.23}$$

$$V_{a}=9.52knots$$

$$Effective\:disc\:area\:\left(A\right)\:={{\pi}\over4}\left(D^2-d^2\right)$$

$$= {{\pi} \over 4} (4.6^2 - 0.75^2)$$

$$A=16.18m^2$$

$$Thrust\space=\space\rho AP^2n^2S$$

$$=1.025\times16.18\times4.3^2\times\left(\frac{95}{60}\right)^2\times0.28$$

$$T=215.25KN$$

$$Thrust \space power (T_p) \space = \space T \times V_a $$

$$215.25\times9.52\times\frac{1852}{3600}$$

$$T_{p}=1054.18KW$$

Q10 (16 Marks) Ship Stability πŸ”₯ Repeated 3x

(a) Describe the stability requirements of a ship for dry docking. (6)

(b) A ship 130m long displaces 14000 tonnes when floating at draughts of 7.5m forward and 8.10 m aft. GML 125m, TPC 18, LCF. 3m aft of midships. Calculate the final draughts when a mass of 180 tonne lying 40m aft of midships is removed from the ship. (10)

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Part (a)

For safe dry-docking, a ship must meet two key stability requirements:

  • Positive GM (Metacentric Height): The ship needs a positive GM. GM is the distance between the centre of gravity (G) and the metacentre (M). A positive GM indicates inherent stability; the ship will right itself if tilted. During dry-docking, the loss of buoyancy as the ship rests on the blocks reduces GM. Insufficient positive GM increases the risk of the ship heeling over or capsizing.
  • Trim by the stern: The vessel should be trimmed slightly by the stern (aft end lower than the bow) to ensure the aft end sits on the keel blocks first. This controlled settling minimises the risk of instability during the docking process. An even keel is generally preferred for the initial floating condition before the dry-docking procedure begins.
Q1 (16 Marks) Ship Stability πŸ”₯ Repeated 2x

(a) State the reasons for the freeboard requirement. (6)

(b) Explain the term condition of assignment and explain how these are maintained for a ship. (5)

(c) What is the difference between a Type "A" and a Type "B" ship. (5)

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Part (a)

Reasons for Freeboard Requirements:

Freeboard is the distance measured from the waterline to the upper edge of the deck plating at the side of the freeboard deck amidships.

  • Ensures the ship is seaworthy when fully loaded.
  • Provides reserve buoyancy, enabling the ship to rise as it passes through waves, keeping decks largely dry.
  • Enhances the ship's stability and increases its survivability in case of damage by allowing it to remain afloat longer, aiding crew escape or damage control.
Part (b)

Conditions of Assignment:

Conditions of Assignment are outlined in the Merchant Shipping Load Line Rules 1968 and must be satisfied before a ship is assigned freeboards and issued a load line certificate. These conditions address the practical need for openings (hatchways, doorways, vents, etc.) in the ship's hull and mandate appropriate protection and closure mechanisms for these openings.

Requirements Before Assigning Load Line:

  • The ship must have sufficient structural strength.
  • Adequate reserve buoyancy must be maintained.
  • Openings must be secured against water ingress.
  • Safety measures for the crew, such as guardrails and gangways, must be in place.

To ensure that the conditions of assignment are still current the following items can be checked and confirm to be without change or damage from when the ship was built.

  • Access openings in bulkheads, to ensure that they can be sealed and prevent flooding
  • Cargo and hatchways ensure they can be sealed to prevent flooding
  • Coamings of hatchways, sign of corrosion damage risk of failure would allow flooding
  • Protection of openings, can they all be sealed
  • Ventilator coamings not corroded as they would allow flooding to other arears
  • Air pipes can be shut in heavy weather and not corroded
  • Discharges, inlets and scuppers, all in good condition and operational
  • Side scuttles can be secured
  • Hull inspection, sea boxes and penetrations all checked for damage and corrosion free.
Part (c)

Difference between Type A and Type B ships:

TYPE β€œA” VESSELS:

  • A ship that is designed to carry only liquid cargoes in bulk and in which cargo tanks have only small access openings, closed by water-tight gasketed covers of steel or equivalent material.
  • The exposed deck must be one of high integrity.
  • It must have a high degree of safety against flooding, resulting from the low permeability of loaded cargo spaces and the degree of bulkhead subdivision usually provided.

TYPE β€œB” VESSELS: all ships that do not fall under type "A" vessels are type "B" ships. For these ships, it may be based on :

  • The vertical extent of damage is equal to the depth of the ship.
  • The penetration of damage is not more than 1/5 of the breadth moulded.
  • No main transverse bulkhead is damaged.
  • Ship’s Kg is assessed for homogenous loading of cargo holds and 50% of the designed capacity of consumable fluids and stores etc.
Q2 (16 Marks) Hull Construction πŸ”₯ Repeated 8x

(a) Sketch the cross-section of a bulk carrier with either deep or shallow double bottom showing the type of framing used. (8)

(b) (i) Describe the corrosion problems experienced with ballast tanks. (4)

(ii) State how such tanks are protected against extensive corrosion. (4)

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Part (a)

Mid-ship section of bulk carrier:

Part (b)

(i) Corrosion problems in ballast tanks

are significant and arise due to various factors:

  • Galvanic corrosion occurs due to the interaction between dissimilar metals, accelerated by the seawater environment and differential aeration.
  • Sulphate-reducing bacteria in river mud can cause localized pitting, leading to penetration of the bottom shell.
  • Dissolved oxygen in seawater reacts with metal surfaces to form rust, significantly contributing to structural deterioration.
  • The rate of corrosion peaks at a 3.5% salt concentration, typical of seawater.
  • Neglecting maintenance exacerbates the problem, allowing corrosion to progress unchecked.

(ii) Protection against extensive corrosion in ballast tanks involves the following measures:

  • Complete coating of the tank surfaces with properly selected marine-grade paint to create a protective barrier.
  • Installation of sacrificial anodes, designed to corrode preferentially and protect the tank structure.
  • Using large anodes with greater volume relative to surface area to ensure extended protection.
  • Regular gauging of plates to assess thickness and identify areas requiring reinforcement or repair.
  • Conducting regular inspections, cleaning the tanks, removing rust, and repainting to maintain structural integrity and prolong the lifespan of the ballast tanks.
Q3 (16 Marks) Hull Construction πŸ”₯ Repeated 2x

(a) With the aid of a sketch describe the method of attachment for a bilge keel and hence explain why provision is made to reduce the possibility of the hull being punctured in the event of damage to the keel. (6)

(b) State why the keel does not extend for the length of the ship. (5)

(c) Evaluate the effectiveness of bilge keels for large wall sided vessels. (5)

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Part (a)

Method of Bilge keel attachment to the hull:

Bilge keels are fitted port and starboard at the turn of the bilge. They do not extend outside the lines of the side and bottom shell. The bilge keels are attached directly in line with an internal stiffening member such as a girder or longitudinal. The bilge keel comprises a flat bar doubler welded directly to the shell, and an offset bulb plate (OBP) with 'scallops' cut in it is welded to the flat bar doubler. The ends of the bilge keels are tapered (minimum 3 in 1) and will end in line with transverse internal stiffening, such as a frame. Using a hull doubler protects the hull in case of damage to the keel, as the crack would not extend into the hull.

Part (b)

Bilge keels are not fitted for the full length of the vessel because:

  • The 'lever' to the ship's axis of rotation is reduced at the ends;
  • The ineffectiveness, the closer the bilge keel is to the rolling centre, increased resistance and more likelihood of damage.
  • The hydrodynamic effect would cause a large increase in resistance and fuel consumption;
  • At the aft end, the boundary layer is much thicker, and since the keel would not project through, it would have a much reduced effect.
Q4 (16 Marks) General πŸ”₯ Repeated 3x

(a) Draw a simple line diagram of the bow of a ship to show the position of the following components as parts of the ships mooring system: Hawse pipe, Cable stopper, Windlass and Cable lifter, Spurling pipe and Chain locker. (4)

(b) Describe the cable stopper and state its purpose (4)

(c) Show by means of a sketch how the anchor cable is attached to the ship. (4)

(d) Describe how the chain locker is drained of water, sand and mud. (4)

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Part (a)

At the bow of the ship, you'd typically find the following components:

  • Hawse pipes: These are openings in the hull through which the anchor chain passes.
  • Cable stopper: Positioned on the deck near the hawse pipes, the cable stopper is used to secure the anchor chain in place.
  • Windlass: Located on the deck near the hawse pipes, the windlass is a mechanical device used to raise and lower the anchor and its chain.
  • Cable lifts: These are devices used to guide the anchor chain from the windlass into the chain locker below deck.
  • Spurling pipe: A vertical pipe leading from the windlass to the chain locker, through which the anchor chain passes.
  • Chain locker: Below deck, typically located near the bow, the chain locker is a compartment where the anchor chain is stowed when not in use.
Part (b)

The cable stopper is a device used to secure the anchor chain in place once the anchor has been lowered or raised. Its purpose is to prevent the anchor chain from slipping or running out unintentionally, ensuring that the anchor remains securely in position.

Part (c)

The anchor cable is typically attached to the ship using a shackle or swivel at the end of the anchor chain. This attachment point allows the anchor chain to pivot freely as the anchor is raised or lowered, preventing twisting or tangling of the chain.

Part (d)

The chain locker is drained of water, sand, and mud through a drainage system that typically includes scupper holes or drain pipes located in the bottom of the locker. These drains allow any water or debris that accumulates in the chain locker to flow out of the compartment and overboard, ensuring that the anchor chain remains clean and free of obstructions. Additionally, regular maintenance and cleaning of the chain locker are important to prevent the buildup of sediment and ensure proper drainage.

Q5 (16 Marks) General πŸ”₯ Repeated 3x

With reference to membrane tanks for the carriage of liquefied gas at very low temperatures:

(a) Describe with a sketch one method of building up the insulation. (6)

(b) State which alloy is used for the membrane and the reason; (5)

(c) Explain why a secondary barrier is installed. (5)

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Part (a)

Membrane tanks for liquefied gas carriage are built with a double hull throughout the cargo length. The insulation system includes the following:

  1. Primary Barrier: A thin membrane made of INVAR (36% Nickel, 64% Iron) with a thickness of 0.5 to 1.2 mm, forming the primary containment layer for the liquefied gas.
  2. Primary Insulation box: A 200 mm thick layer of granulated Perlite insulation packed in plywood boxes, surrounding the primary barrier. Perlite is siliconised to make it impervious to moisture.
  3. Secondary Barrier: A secondary membrane made of the same INVAR material is installed to prevent cargo leakage in case of primary barrier failure.
  4. Secondary Insulation box: Another 200 mm thick layer of granulated Perlite is placed above the secondary barrier to provide additional insulation and prevent thermal transfer.
Part (b)

Alloy used for the membrane:

Invar (36% Ni, 64% Fe) is used for both the primary and secondary barriers. The reason for this choice is its exceptionally low coefficient of thermal expansion. This eliminates the need for expansion joints or corrugations in the membrane design. In addition, Invar remains strong and does not become brittle at the very low temperatures experienced by the liquefied gas. The thin and lightweight nature of Invar maximizes the cargo-carrying capacity of the tank.

Q6 (16 Marks) Ship Stability πŸ”₯ Repeated 9x

(a) Describe how the force on the ship's bottom and the GM vary when grounding takes place. (6)

(b) A ship of 8,000 tonnes displacement takes the ground on a sand bank on a falling tide at an even keel draft of 5.2 metres. KG 4.0 metres. The predicted depth of water over the sand bank at the following low water is 3.2 metres. Calculate the GM at this time assuming that the KM will then be 5.0 metres and that the mean TPC is 15 tonne (10)

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Part (a)

When a ship grounds, the forces on the bottom and the metacentric height (GM) change depending on the grounding's nature and the ship's position on the seabed.

If the ship grounds on a level bottom:

  • A ground reaction force acts vertically upwards from the seabed. This force counteracts part of the ship's weight and alters the distribution of buoyancy forces.
  • The ship’s centre of gravity (G) may appear to rise relative to the metacentre (M) because the upward ground reaction reduces the buoyancy force acting on the underwater volume.
  • This virtual rise in G reduces GM, potentially leading to a list.
  • If the list becomes excessive and the righting moment is insufficient, the ship may capsize.

If the ship grounds on a pinnacle:

  • The ship experiences two forces at the ship's bottom:
    • A downward force due to the ship’s weight.
    • An upward reaction force is concentrated on the pinnacle.
  • The resulting force between the grounding pressure and the ship’s centre of buoyancy shifts downward towards the pinnacle.
  • This situation is similar to when the ship's stern touches the keel block in a dry dock.
  • A virtual loss of GM occurs because the ship’s inclining moment may exceed the maximum righting lever.
  • If the inclining moment is too great, the ship may develop an excessive list or even capsize.

(b) Given:

$$Displacement,\:\Delta=8000\:tonnes$$

$$TPC=15\:tonnes$$

$$Initial\:Draught=5.2m$$

$$Final\:Draught=3.2m$$

$$Ship\:KG=4.0m$$

$$KM=5.0m$$

To find GM

$$Uptrust,\:P=TPC\times Fall\:in\:water\:level\:\left(cm\right)$$

$$=15\times\left(520-320\right)$$

$$=15\times200$$

$$P=3000\:tonnes$$

To Find Virtual loss of GM:

$$Virtual\:loss\:of\:GM_1=\frac{P\times KM}{\Delta}$$

$$=\frac{3000\times5}{8000}$$

$$=\frac{15000}{8000}$$

$$GM_1=1.88m$$

Actual KM = 5.0m (given)

$$Virutal\:KM=Actual\:KM-Virtual\:loss\:of\:GM_1$$

$$=5.0-1.88$$

$$=3.12$$

Similarly, Actual KG = 4.0m (given)

$$New\:GM=Virtual\:KM-\:Actual\:KG$$

$$=3.12-4.0$$

$$=-0.88$$

Q7 (16 Marks) Ship Stability πŸ”₯ Repeated 4x

(a) Describe stability requirement for dry-docking. (6)

(b) A box shaped vessel, 50 metres long x 10 metres wide, floats in salt water on an even keel at a draft of 4 metres. A center line longitudinal watertight bulkhead extends from end to end and for the full depth of the vessel. A compartment amidships on the starboard side is 15 metres long and contains cargo with permeability 30%. Calculate the list if this compartment is bilged. KG = 3 metres. (10)

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Part (a)

Stability requirements for dry-docking.

Before entering the dock the ship must have adequate and known stability. The requirements are:

  • The ship should be stable with an adequate GM, so that the small upward force (reaction) at the keel, building during docking, cannot produce a list. Dock support is provided on the keel blocks and any excessive list during docking would subject the structure and blocks to unequal loading.
  • As the water is pumped out, the reaction of the keel blocks reduces the effective displacement of the ship, lowering the KB and BM and hence GM; if all buoyancy were removed the ship would rest wholly on the blocks. The ship must retain sufficient GM throughout the docking operation (for example by keeping the GM such that the maximum inclination, where the righting moment at the critical point, does not become negative).
  • Ballast should be arranged to give even keel (or slight trim) and list-free condition; tanks should be pressed up or emptied to avoid free-surface effects, and draft read before docking.
  • Weight and trim must be such that the keel blocks contact evenly over the full keel length (avoid excessive trim that overloads the block crown or aft blocks).
  • Cargo/weights should not be changed and the dock must be level; adequate dock pumping should keep the vessel central over the keel line.

If a beam vessel develops instability while docking (transverse GM small or negative), it can heel and capsize on the blocks; hence a suitable stability margin (e.g. GM not less than a minimum) is insisted on and the docking weight, draft and trim are checked by stability data (e.g. from hydrostatics and the Docking Plan).

Part (b)

List when a starboard midship compartment is bilged.

Box-shaped vessel 50 m long, 10 m wide, floats on even keel in salt water at a draft of 4 m. A centreline longitudinal watertight bulkhead runs full depth/size. A midship compartment on the starboard side is 15 m long and has permeability 30 per cent. KG = 3 m. Calculate the list when bilged.

Waterplane area intact Aw = 50 x 10 = 500 m2; the flooded wing compartment has waterplane area = 15 x 5 = 75 m2 (only starboard half up to the centreline bulkhead).

Lost buoyancy volume of the compartment below the original waterline, with permeability 30%: Vlost = length x breadth x draft x permeability = 15 x 5 x 4 x 0.30 = 90 m3. Corresponding lost weight W = 90 x 1.025 = 92.25 t.

Sinkage: the region once flooded provides no increase of buoyancy; effective sinking waterplane = 500 - 75 = 425 m2. Mean sinkage = 90/425 = 0.212 m.

List: the lost buoyancy acts at the centroid of the lost volume, which is at 5/4 = 2.5 m out from the centreline (half-way between centreline bulkhead and side). Heeling moment = W x y = 92.25 x 2.5 = 230.6 t-m.

New GM. Using the lost-buoyancy method, displacement remains 2050 t. The second moment of area of the intact waterplane about the centreline loses the flooded starboard compartment: I(intact)=50x10^3/12 = 4166.7 m4; I(flooded)=15x5^3/3 = 625 m4; so I(new)=3541.7 m4. BM=new = I(new)/V = 3541.7/2000 = 1.771 m. KB = d/2 = 2 m (approximately, ignoring sinkage and list); GM = KB + BM - KG = 2 + 1.77 - 3 = 0.77 m.

List: tan(list) = heeling moment/(Delta x GM) = 230.6/(2050 x 0.77) = 230.6/1579 = 0.146. List = atan(0.146) = 8.3 deg.

Answer: the vessel lists about 8.3 deg to starboard.

Q8 (16 Marks) Ship Stability πŸ”₯ Repeated 11x

(a) Define longitudinal center of gravity (LCG) and longitudinal center of buoyancy (LCB). (6)

(b) A ship 120m long floats at draughts of 5.50m forward and 5.80m aft; MCT 1 cm 80 tonne m, TPC 13, LCF 2.5m forward of midships. Calculate the new draughts when a mass of 110 tonne is added 24m aft of midships. (10)

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Part (a)

Longitudinal Centre of Gravity (LCG):

  • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
  • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

Longitudinal Centre of Buoyancy (LCB):

  • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
  • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.

LCF in fwd and trim by stern

$$Bodily \space sinkage \space = \space {{mass \space added} \over TPC} \space = \space {{110} \over 13} \space$$

$$=\:8.5\operatorname{\mathrm{cm}}\:=0.085m$$

$$Trim \space = \space {{m \times d} \over MCT_{1cm}}$$

$$ = \space {{110 \times (24 + 2.5)} \over 80}$$

$$Trim=36.43\operatorname{cm}=0.364m\:$$

Change in fwd draught:

$$d_{F}=\frac{-t}{L}\left\lbrack\frac{L}{2}-LCF\right\rbrack$$

$$=\frac{-36.43}{120}\left\lbrack\frac{120}{2}-2.5\right\rbrack$$

$$=-17.45\operatorname{cm}=-0.1745m$$

Change in Aft draught:

$$d_{A}=\frac{+t}{L}\left\lbrack\frac{L}{2}+LCF\right\rbrack$$

$$=\frac{+36.43}{120}\left\lbrack\frac{120}{2}+2.5\right\rbrack$$

$$=+18.97\operatorname{cm}=0.189m$$

New draughts:

$$D_{F}=5.5+0.085-0.175=5.41m$$

$$D_{A}=5.8+0.085+0.18=6.065m$$

Q9 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 3x

(a) What are the main components of ship resistance that a vessel encounters while moving through water (6)

(b) The speed of a ship is increased to 18% above normal for 7.5 hours, then reduced to 9% below normal for 10 hours. The speed is then reduced for the remainder of the day so that the consumption in the day is the normal amount. Find the percentage difference between the distance travelled in that day and the normal distance travelled per day. (10)

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Main Components of Ship Resistance

When a vessel moves through water, it experiences total resistance, which is the combined effect of several opposing forces acting against its forward motion. These forces arise due to water viscosity, pressure distribution around the hull, wave formation, and external environmental conditions. For analysis, total resistance is divided into the following main components:

1. Frictional Resistance

Frictional resistance is caused by the viscosity of water acting along the ship’s wetted surface as the vessel moves forward. As water flows over the hull, a thin boundary layer forms, and shear stresses develop between the hull surface and adjacent water particles. This shear stress produces resistance.

Frictional resistance depends primarily on:

  • Wetted surface area – A larger underwater hull area increases the contact between hull and water, thereby increasing resistance.
  • Ship speed – Resistance increases rapidly with speed because higher velocity intensifies boundary layer shear forces.
  • Surface roughness of the hull – Fouling, corrosion, or rough coatings increase turbulence and significantly raise resistance.

At normal service speeds, frictional resistance forms the largest portion of total resistance, especially for slow and medium-speed vessels. It increases approximately with the square of the ship’s speed, making hull maintenance crucial for fuel efficiency.

2. Residual Resistance

Residual resistance is the portion of resistance remaining after subtracting frictional resistance from total resistance. It mainly arises from pressure effects and wave formation around the hull. Residual resistance consists of the following components:

(a) Wave-Making Resistance

Wave-making resistance is caused by the energy expended in generating surface waves at the bow and stern as the ship moves. When a vessel travels through water, it disturbs the free surface and creates a wave system that carries energy away from the ship.

This resistance becomes particularly significant at higher speeds because wave height and wave energy increase rapidly with speed.

Wave-making resistance depends on:

  • Hull form – Fuller hull shapes generally create larger waves.
  • Speed–length ratio (Froude number) – As the Froude number increases, wave-making resistance rises sharply.

At high speeds, wave-making resistance can become a dominant component of total resistance.

(b) Eddy-Making Resistance

Eddy-making resistance is caused by flow separation and the formation of vortices (eddies) around certain parts of the hull. When water flow cannot smoothly follow the hull contour, it separates and creates turbulent regions.

This commonly occurs around:

  • The stern region
  • Appendages
  • Areas with sudden changes in hull form

These turbulent eddies consume energy and increase resistance. Proper streamlining of the hull and stern design can significantly reduce eddy-making resistance.

3. Air Resistance

Air resistance is the force exerted by air on the portion of the ship above the waterline. As the ship moves, it must also overcome aerodynamic drag.

Air resistance depends on:

  • Wind speed and direction
  • Projected area above water
  • Shape of the superstructure

Although usually small compared to water resistance, it becomes significant for:

  • Container ships
  • Ro-Ro vessels
  • Ships operating in strong headwinds

For vessels with large exposed areas, air resistance can noticeably affect fuel consumption.

4. Appendage Resistance

Appendage resistance is caused by external fittings attached to the hull, such as:

  • Rudders
  • Bilge keels
  • Shaft brackets
  • Propeller bossings

These appendages increase the wetted surface area and disturb smooth water flow, thereby increasing both frictional and pressure resistance. In resistance calculations, appendage resistance is generally included as a separate correction added to frictional resistance.

5. Added (Special) Resistance

Added resistance refers to the additional resistance experienced in real sea conditions that is not present in calm-water trials.

It occurs due to:

  • Waves and swell, which cause pitching and heaving motions
  • Steering and yawing motions, which disturb steady flow around the hull

Although not considered in calm-water resistance analysis, added resistance is highly important in practical operations because it significantly affects power requirements and fuel consumption in rough weather.

Q10 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 8x

With reference to fixed pitch propellers:

(a) Explain Propeller Slip and Propeller Thrust. (6)

(b) The shaft power of a ship is 3000 kW, the ship's speed V is 13.2 knot. Propeller RPS is 1.27. propeller pitch is 5.5m and the speed of advance is 11 Knots. Find: (10)

(i) Real Slip

(ii) Wake fraction

(iii) Propeller thrust, when its efficiency, Ξ· = 70%

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Part (a)

Slip

is the difference between the theoretical distance the propeller should travel in one revolution and the actual distance the vessel travels.

$$Slip\:\left(\%\right)\:=\:\left(\frac{Engine\:distance\:-\:Ship^{\prime}s\:distance}{Engine\:distance}\right)\:\times100$$

Where Engine distance = no. of propeller revolutions * propeller pitch (usually calculated over a 24-hour period)

Propeller thrust: it is the force exerted by the propeller to move the vessel ahead and given by

$$Thrust\:\left(T\right)\:=\:\rho\times A\times P^2\times n^2\times S$$

Where,

ρ - Density

A - Area

P - Pitch

n - Revolution per second

S - Slip

Q1 (16 Marks) Hull Construction

With reference to the construction of refrigerated spaces:

(a) state suitable materials that can be used for insulating refrigerated spaces. (3)

(b) state the properties that an insulating material should possess. (3)

(c) sketch a section through a wall of a cold storage space explaining how the insulation is attached to the ship's structure. (10)

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Part (a)

Suitable materials used for insulating the spaces are

  • Armaflex (sheet or pipe insulation made from black flexible nitrile elastomer)
  • Calcium silicate
  • Polyurethane, pearlite, cork, glass fibre, rock wool, plastic foam
Part (b)

Properties insulating materials should posses are

  • Low thermal conductivity (high insulating capacity)
  • Thickness
  • Fire retardant
  • Moisture resistance
  • Vibration resistance
  • Vermin resistance (vermin - cockroaches and insects)
Part (c)

Internal lining required to retain and protect the insulation may be of galvanised iron, stainless steel or aluminium alloy. The linings are secured to timber grounds which are in turn connected to the steel structure. The linings are made air tight by coating the overlaps with a composition such as white lead and fitting sealing strips. This prevents heat transfer due to circulation of air and prevents moisture entering the insulation.

Q2 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 7x

With reference to dry docking, define the responsibilities of the Second Engineer and instructions to Junior Engineers:

(a) Prior to docking

(b) Whilst the vessel is in dry dock

(c) Prior to flooding and leaving the dock.

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(a) Prior to Docking:

Preliminary Preparation:

  • Review various plans, manuals, and previous drydock reports for reference.
  • Prepare a detailed repair list and ensure all required spares are accounted for.
  • Conduct an inventory of spares and requisition necessary items.
  • Gather required tools, including special tools like poker gauges, trammel gauges, gagging tools, etc.
  • Mark all overboard valves, NC (normally closed) and NO (normally open) valves clearly for easy identification.
  • Allocate jobs to team members and discuss the time schedule.
  • Conduct a safety meeting to highlight dry dock hazards and necessary precautions.

Before Entering Dry Dock:

  • Identify the power requirements and machinery to be operational during docking.
  • Check the shore connection box for proper functionality.
  • Record the soundings of FO (Fuel Oil), LO (Lubricating Oil), and DO (Diesel Oil) tanks.
  • Discharge contents from clean drain tanks and sewage tanks.
  • Carry out Economizer soot-blowing.
  • Change over Main Engine, Diesel Generators, and Boiler to Low Sulfur Gas Oil (LSGO).
  • Stop and clean purifiers.
  • Ensure the low sea chest is open and the high sea chest is shut.
  • Keep firefighting appliances (FFA) on standby.
  • Shut down all non-essential machinery, including MGPS (Marine Growth Prevention System) and FWG (Fresh Water Generator).

(b) Whilst the Vessel is in Dry Dock

Upon Arrival:

  • Connect shore power and supplies after ensuring safety checks are completed.
  • Start necessary equipment like cooling water, air compressors, air conditioning, and fridge compressors.
  • Check for jobs assigned by dry dock personnel and prepare accordingly.
  • Attend a safety meeting with dry dock personnel to understand local safety rules and procedures.

During Dry Docking::

  • Oversee and assist in:
    • Cleaning and inspecting the hull, rudder, sea chest, anodes, and propeller.
    • Measuring propeller drop, checking rudder clearances, and inspecting the stern tube bearing and seal.
    • Servicing underwater valves and overboard valves.
    • Inspecting anchor and cables conditions.
    • Overhauling deck machinery, cranes, elevators, and engine room equipment such as the Main Engine, Diesel Generators, Boiler, and Economizer.
    • Renewing pipes and valves as needed.
    • Performing electrical equipment maintenance and surveys.
    • Supervising service engineers for specific repair jobs.
  • Ensure tank cleaning, welding, and other repair works are completed according to the plan.
  • Run the standby diesel generator daily after starting the priming pump.

(c) Before Flooding and Leaving the Dry Dock

Final Checks:

  • Verify that all underwater fittings and drain plugs are securely in place.
  • Ensure all machinery has been boxed back and is ready for operation.
  • Check for any leakage in stern tube seals.
  • Take tank soundings to confirm proper levels.
  • Confirm the proper operation of all underwater valves, overboard valves, sea chests, and vents.
  • Inspect the stern tube tank for any irregularities.

System Restart:

  • Switch back to ship's power after confirming all systems are functional.
  • Test the proper operation of all machinery and systems to ensure the ship is fully operational before leaving the dock.
Q3 (16 Marks) Ship Stability πŸ”₯ Repeated 8x

(a) Explain what is meant by "permissible length" of compartments in passenger ships.

(b) Describe how the position of bulkheads is determined.

(c) Briefly describe the significance of the factor of subdivision

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Part (a)

Permissible Length

Permissible length refers to the maximum length of a ship's compartment that can be flooded while ensuring that the sinkage, heel, or trim remains within acceptable limits without exceeding the floodable length. It ensures the ship remains afloat and stable after flooding.

Permissible Length Formula:

$$Permissible\:lenght\:=\:Floodable\:length\:\times Factor\:of\:subdivision$$

The Factor of Subdivision depends on the ship's length and the nature of its service:

  • For passenger ships, the factor of subdivision is smaller compared to cargo ships.
  • Smaller compartments ensure enhanced safety in case of flooding.
Part (b)

Position of Bulkheads

The position of bulkheads is determined based on the need to comply with the "reasonable amount of damage" criterion in case of flooding:

  • Transverse Watertight Bulkheads should vertically extend up to the margin line.
  • At least one watertight longitudinal bulkhead must be located 20% of the ship's breadth inward from each side.
  • Bulkheads are positioned along the ship's length at intervals equal to the permissible length, but not exceeding the floodable length.
  • The maximum permissible compartment length is limited to 10.7 meters.
  • Collision Bulkhead must be located forward at a distance equal to the permissible length from the forward perpendicular.
  • The ER must be enclosed by two transverse watertight bulkheads, with an aft peak watertight bulkhead enclosing the forward part.
Part (c)

Factor of Subdivision

The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

Permissible Length Formula:

$$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

Q4 (16 Marks) Ship Stability πŸ”₯ Repeated 14x

Explain how the period of roll varies with:

(a) The amplitude of roll.

(b) The radius of gyration.

(c) The initial metacentric height.

(d) The location of masses in the ship

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The period of roll Tr of a ship is determined by the formula:

$$T_{r}=\frac{2\pi K}{\sqrt{g\times GM}}$$

where,

  • K is the radius of gyration (mass moment of inertia)
  • g is the acceleration due to gravity, and
  • GM is the metacentric height.
Part (a)

Amplitude of Roll:

  • The amplitude of roll, or the maximum angle of heel, does not affect the period of roll. While a larger amplitude implies a greater heeling moment and faster roll speed, the time taken to complete one full roll cycle remains constant for a given metacentric height and radius of gyration.
Part (b)

Radius of Gyration (K):

  • The period of roll is directly proportional to the radius of gyration. A larger radius of gyration (indicating a greater distribution of mass further from the ship's centre of rotation) leads to a longer period of roll. Conversely, a smaller radius of gyration (mass concentrated closer to the centre) results in a shorter period. The distribution of cargo significantly impacts K; cargo concentrated centrally minimises K and the roll period, while dispersed cargo maximises K and the roll period.
Part (c)

Initial Metacentric Height (GM):

  • The period of roll is inversely proportional to the square root of the metacentric height (GM). A larger GM (a stiffer ship) leads to a shorter roll period, as the ship quickly returns to its upright position. A smaller GM (a tender ship) results in a longer roll period, with slower return to the upright.
Part (d)

Location of Masses in the Ship:

The location of masses in the ship will effect the GM & K. So the period of roll will be affected.

  • If masses are at bottom, G moves down, GM ↑, period of roll ↓.
  • If masses are at top, G moves up, GM ↓, period of roll ↑.
  • If masses are concentrated at centre, K ↓, period of roll ↓.
  • If masses are away from centre, K ↑, period of roll ↑.
Q5 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 13x

Describe the relation between frictional resistance and

(a) Ship speed

(b) The wetted area

(c) The surface roughness

(d) The length of the vessel

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Frictional resistance arises due to the interaction between the ship's hull and the water as the vessel moves. This resistance is caused by eddying water adjacent to the hull that is drawn along with the ship. The frictional resistance is given by:

$$R_{f\:=\:}f\:.\:s\:.\:v^{n}$$

Where:

  • f = Coefficient of friction
  • s = Wetted surface area
  • v = Ship's speed in knots
  • n = Constant (1.82)
Part (a)

Ship Speed (v):

  • Frictional resistance increases with increasing ship speed. The relationship is not perfectly linear, but it's approximately described by the formula Rf = fsv^n (Rf ∝ v^n), where 'n' is a constant (approximately 1.82). This means that a small increase in speed results in a larger increase in frictional resistance.
Part (b)

Wetted Area (s):

  • Frictional resistance is directly proportional to the wetted surface area (Rf ∝ s). A larger wetted area (the part of the hull submerged in water) leads to greater frictional resistance. The wetted area increases with the ship's draught (depth in the water).
Part (c)

Surface Roughness:

  • Increased surface roughness increases frictional resistance. Roughness can be caused by fouling (marine growth), corrosion, or poor hull maintenance. Regular cleaning and maintenance of the hull surface help to minimise roughness and reduce resistance.
Part (d)

Length of the Vessel

  • Frictional resistance is influenced by the vessel's length. As the length increases, the wetted surface area grows, and the coefficient of friction (f) changes, leading to an increase in Rf. Longer vessels experience higher frictional resistance, but streamlined designs can mitigate the effect to some extent.
Q6 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

(a) How the distribution of mass within the ship affects the rolling period? (6)

(b) A ship of 14000 tonnes displacement is 125 m long and floats at draughts of 7.9 m forward and 8.5 m aft. The TPC is 19, GML 120 m, and LCF 3 m forward of midships. It is required to bring the vessel to an even keel draught of 8.5 m. Calculate the mass which should be added and the distance of the centre of the mass from midships. (10)

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Part (a)

The rolling period is influenced by the ship's metacentric height (GM) and radius of gyration (K), which are determined by the location of the masses onboard.

Masses at the bottom:

  • The centre of gravity (G) moves down.
  • Metacentric height (GM) increases, resulting in greater stability.
  • Rolling period decreases.

Masses at the top:

  • The centre of gravity (G) moves up.
  • Metacentric height (GM) decreases, reducing stability.
  • Rolling period increases.

Masses concentrated at the centre:

  • The radius of gyration (K) decreases.
  • Rolling period decreases.

Masses distributed away from the centre:

  • The radius of gyration (K) increases.
  • Rolling period increases.
Q7 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

(a) Explain how an increase of draught and of displacement influence rolling. (6)

(b) A pontoon has a constant cross-section as shown in Fig. Given below. The metacentric height is 2.5 m. Find the height of the centre of gravity above the keel. (10)

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Part (a)

In case of increase of draught and displacement, this is a case of loading being done.

Thus, when the ship is being loaded, its GM will start to decrease.

Now, the time period of roll is given by:

$$T_{r}=\frac{2\pi k}{\sqrt{GM.g}}$$

Where,

  • k = radius of gyration
  • GM = metacentric height
  • g = acceleration due to gravity

Now, since, GM has started to decrease, the Tr will start to increase.

Thus, the ship will now roll with greater time period. Thus, an increase in draught and displacement, influences rolling.

Q8 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

(a) Describe the fundamental principle of a propeller. (6)

(b) A propeller 6m in diameter has a pitch ratio of 0.9, BAR 0.48, and, when turning at 110 rev/min, has a real slip of 25% and wake fraction 0.30. If the propeller delivers a thrust of 300 kN and the propeller efficiency is 0.65, calculate: (10)

(a) Blade area

(b) Ship speed

(c) Thrust power

(d) Shaft power

(e) Torque

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Part (a)

A propeller is a type of fan, that transmits power by converting rotational motion into thrust. A pressure difference is produced between the forward and rear surface of the aerofoil shaped blade and the fluid is accelerated behind the blade. A marine propeller of this type is sometimes known as screw propeller or a screw.

Given:

$$D=6m$$

$$p=0.9$$

$$BAR=0.48$$

$$N=110rev\:per\min$$

$$real \space slip \space (s) \space = \space 25\%$$

$$W_{F}=0.3$$

$$Thrust \space = \space 300kN$$

$$Ξ·_{prop} \space = \space 0.65$$

(i) Blade area:

$$BAR \space = \space {{A_b} \over {{\pi} \over 4} D^2}$$

$$Blade \space area \space A_b \space = \space 0.48 \times {{\pi} \over 4} 6^2$$

$$Blade \space area \space = \space 13.57m^2 $$

$$p \space = \space {{P} \over D}$$

$$0.9 \space = \space {{P} \over 6}$$

$$Pitch \space p = \space 5.4m$$

$$V_{T}=P\times N\times\frac{3600}{1852}$$

$$V_{T}=5.4\times\frac{110}{60}\times\frac{3600}{1852}$$

$$V_{T}=19.24knots$$

$$Real \space slip \space S \space = \space {{V_T - V_a} \over V_T}$$

$$ 0.25 \space = \space {{19.24 - V_a} \over 19.24}$$

$$V_a \space = \space 14.42 knots$$

$$W_F \space = \space {{V - V_a} \over V}$$

$$0.30 \space = \space {{V - 14.42} \over V}$$

$$V=20.6knots$$

$$T_{p}\space=\space Thrust\times V_{a}\times\frac{1852}{3600}$$

$$T_p \space = \space Thrust \times 14.42 \times {{1852} \over 3600 }$$

$$T_p \space = \space 2225.48 $$

$$T_p \space = \space d_p \times Ξ·_{prop}$$

$$2225.48=d_{p}\times0.65$$

$$d_p \space = \space 3423.8kW$$

$$dp=2\pi NT$$

$$3423.8=2\times\pi\times\frac{110}{60}\times T$$

$$T=297.22KN$$

Q9 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

(a) Explain what is meant by: (6)

(i) Wave-making resistance

(ii) Frictional resistance

(iii) Eddy-making resistance

(b) When a ship is 800 nautical miles from port its speed is reduced by 20%, thereby reducing the daily fuel consumption by 42 tonnes and arriving in port with 50 tonnes on board. If the fuel consumption in t/h is given by the expression (0.136+0.001V^3) where V is the speed in knots, estimate: (10)

(i) The reduced consumption per day

(ii) The amount of fuel on board when the speed was reduced

(iii) The percentage decrease in consumption for the latter part of the voyage

(iv) The percentage increase in time for this latter period

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$$D=800nm$$

$$V_1=?knots$$

$$V_2=0.8V_1$$

$$DC_1=Cons\:per\:day\:at\:V_1$$

$$DC_2=Cons\:per\:day\:at\:V_2$$

$$DC_1-DC_2=42t$$

$$C=\left(0.136+0.001V^3\right)\:t\h$$

$$Therefore\:DC=24\left(0.136+0.001V^3\right)\:tonnes\:per\:day$$$$42=24\left\lbrack\left(0.136+0.001V_{1^{}}^3\right)-\left(0.136_{}+0.001\left(0.8V_1^3\right)\right)\right\rbrack$$

$$42=24\left(0.136+0.001V_1^3-0.136-0.512\times10^{-3}\times V_1^3\right)$$

$$42=24\left(0.001V_1^3-0.512\times10^{-3}\times V_1^3\right)$$

$$42=24\left(0.000488V_1^3\right)$$

$$V_1=\sqrt[3]{\frac{42}{24\times0.000488}}$$

$$V_1=15.31\:knots$$

$$V_2=0.8\times V_1$$

$$V_2=0.8\times15.31$$

$$V_2=12.245\:knots$$

$$\left(i\right)\:Reduced\:cons\:per\:day\:=\:\left(0.136+0.001V_2^3\right)\times24$$

$$=\left(0.136+0.001\times12.45^3\right)\times24$$

$$=49.57\:tonnes\:per\:day$$

$$Time\:taken\:for\:complete\:voyage\:of\:800nm$$

$$at\:V_2=\frac{800}{12.245\times24}=2.72\:days$$

$$Consumption\:=\:2.72\times49.57=134.93t\:\left(at\:reduced\:speed\right)$$

$$\left(ii\right)\:Fuel\:onboard=134.93+50$$

$$=184.93t\:\left(after\:speed\:reduction\right)$$

$$DC_1=DC_2+42$$

$$DC_1=49.57+42$$

$$DC_1=91.57t$$

$$Time\:taken\:for\:V_1=\frac{800}{15.31\times24}$$

$$=2.178days$$

$$Cons\:at\:V_1=91.57\times2.178$$

$$=199.38t$$

$$\left(iii\right)\:\%\:reduction\:in\:cons=\frac{199.38-134.93}{199.38}$$

$$=32.32\%$$

$$\left(iv\right)\:\%\:increase\:in\:time=\frac{2.72-2.178}{2.178}$$

$$=24.88\%$$

Q10 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

(a) Explain how to distinguish between list and loll and describe how to return the ship to the upright in each case. (6)

(b) A ship of 5000 tonnes displacement has a double bottom tank 12 m long. The 1/2 breadths of the top of the tank are 5, 4 m and 2 m respectively. The tank has a watertight centreline division. Calculate the free surface effect if the tank is partially full of freshwater on one side only. (10)

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Part (a)

Distinguishing between List and Loll:

1. List is caused by Uneven distribution of weight within the ship. List will have the below conditions:

  • GM is positive (GM > 0).
  • The centre of gravity (G) is off-centre.
  • The vessel is at equilibrium but inclines to one side due to uneven weight distribution, even without any external forces acting on it.
  • The vessel rolls around the angle of list.

Correction:

  • Redistribute the weight evenly to bring the centre of gravity (G) back in line with the centerline and metacentric height (M).

2. Loll is caused by High centre of gravity (G) leading to negative GM and is exacerbated by external forces, free surface effects, or poor distribution of weights. LOLL will have the below conditions:

  • GM is negative (GM < 0).
  • The centre of gravity (G) is on the centerline but too high, making the vessel inherently unstable.
  • The vessel flops or inclines to one side at an angle of loll, and it can incline equally to either side.
  • The vessel rolls unstably around the angle of loll.

Correction:

  • Reduce the centre of gravity by ballasting bottom tanks or removing weight from higher levels.
  • Minimize free surface effects by reducing the breadth of free surfaces in tanks.

Q1 (16 Marks) Hull Construction πŸ”₯ Repeated 8x

(a) Sketch the cross-section of a bulk carrier with either deep or shallow double bottom showing the type of framing used.

(b) (i) Describe the corrosion problems experienced with ballast tanks.

(ii) State how such tanks are protected against extensive corrosion.

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Part (a)

Mid-ship section of bulk carrier:

Part (b)

(i) Corrosion problems in ballast tanks

are significant and arise due to various factors:

  • Galvanic corrosion occurs due to the interaction between dissimilar metals, accelerated by the seawater environment and differential aeration.
  • Sulphate-reducing bacteria in river mud can cause localized pitting, leading to penetration of the bottom shell.
  • Dissolved oxygen in seawater reacts with metal surfaces to form rust, significantly contributing to structural deterioration.
  • The rate of corrosion peaks at a 3.5% salt concentration, typical of seawater.
  • Neglecting maintenance exacerbates the problem, allowing corrosion to progress unchecked.

(ii) Protection against extensive corrosion in ballast tanks involves the following measures:

  • Complete coating of the tank surfaces with properly selected marine-grade paint to create a protective barrier.
  • Installation of sacrificial anodes, designed to corrode preferentially and protect the tank structure.
  • Using large anodes with greater volume relative to surface area to ensure extended protection.
  • Regular gauging of plates to assess thickness and identify areas requiring reinforcement or repair.
  • Conducting regular inspections, cleaning the tanks, removing rust, and repainting to maintain structural integrity and prolong the lifespan of the ballast tanks.
Q2 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 7x

With reference to dry docking, define the responsibilities of the Second Engineer and instructions to Junior Engineers:

(a) Prior to docking.

(b) Whilst the vessel is in dry dock.

(c) Prior to flooding and leaving the dock.

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(a) Prior to Docking:

Preliminary Preparation:

  • Review various plans, manuals, and previous drydock reports for reference.
  • Prepare a detailed repair list and ensure all required spares are accounted for.
  • Conduct an inventory of spares and requisition necessary items.
  • Gather required tools, including special tools like poker gauges, trammel gauges, gagging tools, etc.
  • Mark all overboard valves, NC (normally closed) and NO (normally open) valves clearly for easy identification.
  • Allocate jobs to team members and discuss the time schedule.
  • Conduct a safety meeting to highlight dry dock hazards and necessary precautions.

Before Entering Dry Dock:

  • Identify the power requirements and machinery to be operational during docking.
  • Check the shore connection box for proper functionality.
  • Record the soundings of FO (Fuel Oil), LO (Lubricating Oil), and DO (Diesel Oil) tanks.
  • Discharge contents from clean drain tanks and sewage tanks.
  • Carry out Economizer soot-blowing.
  • Change over Main Engine, Diesel Generators, and Boiler to Low Sulfur Gas Oil (LSGO).
  • Stop and clean purifiers.
  • Ensure the low sea chest is open and the high sea chest is shut.
  • Keep firefighting appliances (FFA) on standby.
  • Shut down all non-essential machinery, including MGPS (Marine Growth Prevention System) and FWG (Fresh Water Generator).

(b) Whilst the Vessel is in Dry Dock

Upon Arrival:

  • Connect shore power and supplies after ensuring safety checks are completed.
  • Start necessary equipment like cooling water, air compressors, air conditioning, and fridge compressors.
  • Check for jobs assigned by dry dock personnel and prepare accordingly.
  • Attend a safety meeting with dry dock personnel to understand local safety rules and procedures.

During Dry Docking::

  • Oversee and assist in:
    • Cleaning and inspecting the hull, rudder, sea chest, anodes, and propeller.
    • Measuring propeller drop, checking rudder clearances, and inspecting the stern tube bearing and seal.
    • Servicing underwater valves and overboard valves.
    • Inspecting anchor and cables conditions.
    • Overhauling deck machinery, cranes, elevators, and engine room equipment such as the Main Engine, Diesel Generators, Boiler, and Economizer.
    • Renewing pipes and valves as needed.
    • Performing electrical equipment maintenance and surveys.
    • Supervising service engineers for specific repair jobs.
  • Ensure tank cleaning, welding, and other repair works are completed according to the plan.
  • Run the standby diesel generator daily after starting the priming pump.

(c) Before Flooding and Leaving the Dry Dock

Final Checks:

  • Verify that all underwater fittings and drain plugs are securely in place.
  • Ensure all machinery has been boxed back and is ready for operation.
  • Check for any leakage in stern tube seals.
  • Take tank soundings to confirm proper levels.
  • Confirm the proper operation of all underwater valves, overboard valves, sea chests, and vents.
  • Inspect the stern tube tank for any irregularities.

System Restart:

  • Switch back to ship's power after confirming all systems are functional.
  • Test the proper operation of all machinery and systems to ensure the ship is fully operational before leaving the dock.
Q3 (16 Marks) Ship Stability

With reference to Statutory Certification:

(a) State the reason for the freeboard requirements.

(b) (i) Explain the term conditions of assignments.

(ii) List the items that may be examined during a related survey.

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Part (a)

Reasons for Freeboard Requirements:

Freeboard is the distance measured from the waterline to the upper edge of the deck plating at the side of the freeboard deck amidships.

Purpose of Freeboard:

  • Ensures the ship is seaworthy when fully loaded.
  • Provides reserve buoyancy, enabling the ship to rise as it passes through waves, keeping decks largely dry.
  • Enhances the ship's stability and increases its survivability in case of damage by allowing it to remain afloat longer, aiding crew escape or damage control.
Part (b)

Conditions of Assignment:

Conditions of Assignment are outlined in the Merchant Shipping Load Line Rules 1968 and must be satisfied before a ship is assigned freeboards and issued a load line certificate. These conditions address the practical need for openings (hatchways, doorways, vents, etc.) in the ship's hull and mandate appropriate protection and closure mechanisms for these openings.

Requirements Before Assigning Load Line:

  • The ship must have sufficient structural strength.
  • Adequate reserve buoyancy must be maintained.
  • Openings must be secured against water ingress.
  • Safety measures for the crew, such as guardrails and gangways, must be in place.
Part (c)

Items Examined During a Load Line Survey.

  • Openings: This encompasses hatchways, machinery space openings, cargo ports, watertight doors, ventilators, air pipes, scuppers, freeing ports, side scuttles, and other openings in the freeboard and superstructure decks. The surveyor checks the condition of seals, gaskets, closing mechanisms, and overall watertightness.
  • Crew Protection: Safety features designed to protect the crew, such as guardrails, bulwarks, and gangways, are inspected for damage or deterioration and proper functionality.
  • Hull Structure: The hull itself is inspected for any damage, corrosion, or structural weaknesses that could compromise watertightness or strength. This often includes visual inspection for cracks, buckling, indentation, and paint adhesion. Bottom shell, bilge keel, stem and stern frames, rudder, sea chests, side ports, stern bearing, and propeller are all checked.
  • Other Systems: Depending on the scope of repairs, other systems may be examined, including but not limited to: pressure-vacuum valves, sounding pipes, air vents, access hatches, ventilation fans, dampers, weathertight doors and sealings, door securing arrangements, side scuttles, windows, and skylights. Testing, such as hose tests (to check watertight integrity), might also be conducted.
Q4 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 3x

State how and why the following machinery items are affected when the maximum service speed of a vessel is consistently maintained in heavy weather.

(a) Intermediate shafting,

(b) Propeller shafting,

(c) Shafting coupling bolts,

(d) Main thrust pads.

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Part (a)

Intermediate Shafting:

  • Torsional Stress: Resulting from the rapid changes in propeller speed (propeller racing) caused by variations in water resistance in heavy weather.
  • Compressive Stress: Due to the end thrust transmitted from the propeller.
  • Bending Stress: Caused by hull movements such as hogging (upward bending) and sagging (downward bending) in heavy seas.
Part (b)

Propeller Shafting:

  • Torque and Thrust: Caused by the propulsion forces transmitted through the shaft.
  • Torsional Stress: Due to the rapid fluctuations in engine speed when the propeller races.
  • Compressive Stress: Generated by the axial thrust from the propeller.
  • Bending Stress: Arises when the weight of the propeller acts on the shaft as the propeller emerges from the water during rough seas.
Part (c)

Shafting Coupling Bolts:

  • Bending Stress: From misalignments caused by hull deformations.
  • Shear Stress: Resulting from the whirling of the shaft and rapid engine speed changes during propeller racing.
  • Torsional Stress: Caused by the transmission of fluctuating torque.
  • Fatigue Failure: Due to the repeated application of fluctuating loads over time, particularly in heavy weather conditions.
Part (d)

Main Thrust Pads:

  • Stress from Hull Movements: Misalignment caused by hull hogging and sagging.
  • Load Fluctuations: Due to variations in propeller thrust during racing and rapid changes in sea conditions.
  • Axial and Torsional Vibration: Arising from inconsistent propulsion forces and shaft vibrations.
  • Surface Wear and Damage: Resulting from increased pressure and friction due to fluctuating thrust forces.
Q5 (16 Marks) Ship Stability πŸ”₯ Repeated 2x

(a) Describe the arrangement made in a main structural bulkhead for a watertight door aperture.

(b) Explain a procedure for ensuring that sliding watertight doors are operated safely.

(c) Differentiate between the categories of watertight door and state the regulation pertaining to each type.

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Part (a)

Arrangement made in a main structural bulkhead for a watertight door aperture.

To preserve the strength and watertight integrity of a main transverse bulkhead where a watertight door is fitted, the opening is bounded by a reinforced double-plate frame. Thicker compensating plates (or a vertical web frame) surround the aperture, with the door frame pieces doubling the plate thickness at the corners. The door opening edges are connected to strong horizontal and vertical stiffeners (intercostal framing) so that the bulkhead's ability to transmit shear and bending (it acts as a deep girder stiffening the hull) is not lost. The door itself is a steel watertight sliding or hinged door with a frame fitted with a resilient gasket/special joint; the edges are dressed and wired to the surrounding doubling plates, and the sill is raised above the deck to (in passenger ships) about a prescribed height. The framing around the opening is tied into the top and bottom stools/stringers, and where the aperture cuts the stiffening of the bulkhead, compensation is made by increased scantlings immediately adjacent.

Part (b)

Procedure for safely operating sliding watertight doors.

  • Only authorised personnel may operate them, and the closing mechanism (hydraulic or electric, power plus manual) must be tested before entering port/complex operations.
  • Before closing, check the doorway and its rails/wheels are clear of cargo, stores, obstructions and personnel; sound a warning (alarm) before closing.
  • The door must be fully open during cargo/engineering operations that require passage, and closed when the door is not in use, especially at sea when subdivision demands it.
  • On closing (or opening) by power, monitor the travel; never operate a sliding door manually against a heavy load or with persons in the way.
  • When a remote/central control is fitted, the interlock and indicator (open/closed/partly open) must be observed, and the action of the automatic watertight-closing (if fitted) verified.
  • After closing, check the door is fully shut and the securing (wedges/dogging) engaged; the indicator should confirm closed position.
  • Ensure the gasket is intact, the surrounding plates free of obstruction, and that the door is not used as a normal gangway when closed.
  • Record routine tests and keep the operating instruction posted.
Part (c)

Categories of watertight door and relevant regulations.

Watertight doors are categorised (per SOLAS Chapter II-1) by their type and location:

  • Class 1 hinged watertight doors, fitted in bulkheads in machinery spaces and working spaces, normally operated manually (may be required to be closed before the vessel proceeds to sea).
  • Class 2 sliding watertight doors, operated by power (remote) with local manual operation and capable of being closed from an accessible location above the bulkhead deck; these are fitted where frequent access is needed (e.g. main lobbies, machinery space trunks, cargo spaces in passenger ships).
  • Class 3 hinged door or sliding door of a certain design in the lower portion of the hull in specific locations.

Regulation (SOLAS II-1/13 and Reg.22 etc.): watertight doors in subdivision bulkheads below the bulkhead deck, and their closing arrangements, must comply with the regulations; all watertight doors below the bulkhead deck must be kept closed except when actually used for passage, and sliding doors fitted with power closing, voice alarm and remote indicator are required in certain fire/passenger contexts. The number of watertight doors below the bulkhead deck is minimised, they must have a status indicator at a control station, and hinges/operating gear must be such that they can be closed even with the vessel heeled. Non-wateredef: performance, and that they be opened/closed from both sides and remotely.

Q6 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

(a) Describe how the distribution of mass within the ship affects the rolling period. (6)

(b) The righting moments of a ship at angles of heel of 0, 15Β°, 30Β°, 45Β° and 60Β° are 0, 1690, 5430, 9360 and 9140 kNm respectively. Calculate the dynamical stability at 60Β°. (10)

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Part (a)

The rolling period is influenced by the ship's metacentric height (GM) and radius of gyration (K), which are determined by the location of the masses onboard.

Masses at the bottom:

  • The centre of gravity (G) moves down.
  • Metacentric height (GM) increases, resulting in greater stability.
  • Rolling period decreases.

Masses at the top:

  • The centre of gravity (G) moves up.
  • Metacentric height (GM) decreases, reducing stability.
  • Rolling period increases.

Masses concentrated at the centre:

  • The radius of gyration (K) decreases.
  • Rolling period decreases.

Masses distributed away from the centre:

  • The radius of gyration (K) increases.
  • Rolling period increases.
Q7 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

(a) Describe briefly the significance of the factor of subdivision. (6)

(b) A ship 120m long has a light displacement of 4000 tonne and LCG in this condition 2.5m aft of midships. The following items are then added: (10)

Cargo 10000 tonne LCG 3.0m forward of midships

Fuel 1500 tonne LCG 2.0 m aft of midships

Water 400 tonne LCG 8.0m aft of midships

Stores 100 tonnes LCG 10.0m forward of midships

Using the following hydrostatic data, calculate the final draughts:

Draught (m)

Displacement (t)

MCT1cm (tm)

LCB from midships

LCF from midships

8.50

16650

183

1.94F

1.29A

8.00

15350

175

2.10F

0.60F

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Part (a)

Factor of Subdivision

The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

Permissible Length Formula:

$$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

Part (b)

Mass added

LCG from midship

Mass moments

F

A

10000

3.0 Fwd

30000

1500

2.0 Aft

3000

400

8.0 Aft

3200

100

10.0 Fwd

1000

4000

2.5 Aft

10000

16000

31000

16200

$$Excess\:moment=31000-16200=14800$$

$$LCG=\frac{\sum M}{\sum m}=\frac{14800}{16000}$$

$$LCG=0.925m\:fwd\:of\:midship$$

Draught

Displacement

MTC 1cm

LCB from midship

LCF from midship

8.5

16650

183

1.94 Fwd

1.20 Aft

8.25

16000

179

2.03 Fwd

0.57 Aft

8.0

15350

175

2.10 Fwd

0.06 Fwd

$$LCG=0.925m\:Fwd\:of\:midship$$

$$LCB=2.02m\:Fwd\:of\:midship$$

$$Trimming\:lever=LCB-LCG$$

$$2.02-0.925$$

$$=1.09m$$

$$Trimming\:moment=m\times d$$

$$=16000\times1.09$$

$$=17440tm$$

$$Change\:of\:trim=\frac{Trimming\:moment}{MCT_{1\operatorname{\mathrm{cm}}}}$$

$$=\frac{17440}{179}$$

$$97.43\operatorname{cm}$$

$$Draft\:fwd=8.25-\frac{97.43}{100\times120}\left(\frac{120}{2}+0.57\right)$$

$$=7.756m$$

$$Draft\:aft=8.25+\frac{97.43}{100\times120}\left(\frac{120}{2}-0.57\right)$$

$$=8.735m$$

Q8 (16 Marks) Ship Stability πŸ”₯ Repeated 3x

(a) Explain how the distribution of masses affects rolling and pitching. (6)

(b) A ship turns in a circle of radius 100 metres at a speed of 15 knots. The GM is 2/3 metres and BG is 1 metre. If g = 981 cm/secΒ² and 1 knot is equal to 1.8532 Km/hour, find the heel due to turning. (10)

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Part (a)

The rolling period is influenced by the ship's metacentric height (GM) and radius of gyration (K), which are determined by the location of the masses onboard.

Masses at the bottom:

  • The centre of gravity (G) moves down.
  • Metacentric height (GM) increases, resulting in greater stability.
  • Rolling period decreases.

Masses at the top:

  • The centre of gravity (G) moves up.
  • Metacentric height (GM) decreases, reducing stability.
  • Rolling period increases.

Masses concentrated at the centre:

  • The radius of gyration (K) decreases.
  • Rolling period decreases.

Masses distributed away from the centre:

  • The radius of gyration (K) increases.
  • Rolling period increases.
Part (b)

$$Radius\:\left(R\right)=100m$$

$$Speed\:\left(v\right)=15\:knots$$

$$GM=\frac23m$$

$$BG=1m$$

$$\tan\the\theta=\frac{v^2\times BG}{g\times r\times GM}$$

$$=\frac{\left(15\times0.514\right)^2\times1\times3}{9.81\times100\times2}$$

$$\theta=5.19\degree$$

Q9 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 4x

(a) Describe the effect of cavitations on the propeller blades. (6)

(b) A propeller 4.6m diameter has a pitch of 4.3m and boss diameter of 0.75m. The real slip is 28% at 95 rev/min. Calculate the speed of advance, thrust and thrust power. (10)

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Part (a)

Effect of cavitation on propeller blades:

Erosion:

  • Cavitation causes the formation and collapse of vapor bubbles on the propeller blade surface.
  • The collapse of these bubbles produces high-pressure shockwaves and microjets that erode the blade material, leading to surface pitting and progressive damage.

Vibration:

  • Uneven cavitation across the blades leads to imbalanced forces, causing vibrations in the propeller and the ship.
  • These vibrations can reduce the comfort of passengers and crew and stress the ship's structural components.

Noise:

  • The collapse of vapor bubbles generates loud noise, which can interfere with onboard communication and underwater sonar systems.
  • This noise is a significant concern for naval vessels and marine life.

Reduced Performance:

  • Cavitation reduces the efficiency of the propeller by causing loss of thrust and torque.
  • The presence of cavitation decreases the propeller’s ability to convert rotational energy into forward motion, lowering the ship's speed and increasing fuel consumption.

(b) Given:

$$D=4.6m$$

$$P=4.3m$$

$$d=0.75$$

$$S=28\%$$

$$n \space = \space 95 \space rev/ min$$

$$V_T \space = \space P \times N \times {{3600} \over 1852}$$

$$ = \space 4.3 \times {{95} \over 60} \times {{3600} \over 1852}$$

$$V_{T}=13.23knots$$

$$Real \space slip \space (S) \space = \space {{V_T - V_a} \over V_T}$$

$$0.28 \space = \space {{13.23 - V_a} \over 13.23}$$

$$V_{a}=9.52knots$$

$$Effective\:disc\:area\:\left(A\right)\:={{\pi}\over4}\left(D^2-d^2\right)$$

$$= {{\pi} \over 4} (4.6^2 - 0.75^2)$$

$$A=16.18m^2$$

$$Thrust\space=\space\rho AP^2n^2S$$

$$=1.025\times16.18\times4.3^2\times\left(\frac{95}{60}\right)^2\times0.28$$

$$T=215.25KN$$

$$Thrust \space power (T_p) \space = \space T \times V_a $$

$$215.25\times9.52\times\frac{1852}{3600}$$

$$T_{p}=1054.18KW$$

Q10 (16 Marks) Ship Stability πŸ”₯ Repeated 3x

(a) Describe the stability requirements of a ship for dry-docking. (6)

(b) A ship 130m long displaces 14000 tonnes when floating at draughts of 7.5m forward and 8.10 m aft. GML 125m, TPC 18, LCF 3m aft of midships. Calculate the final draughts when a mass of 180 tonne lying 40m aft of midships is removed from the ship. (10)

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Part (a)

For safe dry-docking, a ship must meet two key stability requirements:

  • Positive GM (Metacentric Height): The ship needs a positive GM. GM is the distance between the centre of gravity (G) and the metacentre (M). A positive GM indicates inherent stability; the ship will right itself if tilted. During dry-docking, the loss of buoyancy as the ship rests on the blocks reduces GM. Insufficient positive GM increases the risk of the ship heeling over or capsizing.
  • Trim by the stern: The vessel should be trimmed slightly by the stern (aft end lower than the bow) to ensure the aft end sits on the keel blocks first. This controlled settling minimises the risk of instability during the docking process. An even keel is generally preferred for the initial floating condition before the dry-docking procedure begins.
Q1 (16 Marks) Hull Construction πŸ”₯ Repeated 10x

(a) Describe a method for the attachment of bilge keels. (5)

(b) State THREE reasons for not extending bilge keels to the entire length of the vessel. (6)

(c) Explain TWO principles of roll damping that bilge keels exploit. (5)

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Part (a)

Method for the attachment of bilge keels.

The bilge keel is a long fin attached externally along the turn of the bilge, running roughly in a fore-and-aft direction. Attachment method:

  • The bilge keel is fabricated as a light, flat bar or fabricated plate of length running over the middle portion of the hull. It is attached by one of two methods:

(i) riveted/bolted connection through doubler plates, in which the keel web is connected by a continuous fillet weld to a flat-bar or doubler section on the shell, or

(ii) welded directly to the shell by continuous or intermittent fillet welds.

  • Because the attachment is a potential source of high local stress concentration and fatigue cracking (the shell at the bilge is highly loaded in bending and docking), the weld is usually a full-strength continuous fillet or butt weld, and the keel is not welded to the shell with single small tack welds that crack under repeated relative movement (flexing).
  • The ends of the bilge keel are tapered and rolled to a fine radius and blended smoothly (scalloped/angled end) so as to reduce stress concentration at their termination; heavy strike plates (flat bar doubler) at the ends carry the connection.
  • The keel is set slightly at a small angle to the base plane so that its face lines up with the flow; it is normally fitted parallel to the load waterline and positioned at the turn of the bilge where it is not unduly loaded in docking (kept clear of the keel blocks).
  • Doubler plates and drainage are arranged, and the attachment is periodically inspected (especially the weld toes) because the bilge keel experiences severe fluctuating loads in a seaway.
Part (b)

THREE reasons for not extending bilge keels to the entire length.

  1. To avoid severe local stress concentrations and fatigue cracking at their ends/attachment; a full-length keel would place the highly-loaded ends in regions of the hull girder with large bending stresses, promoting cracking. Pounding/the connection would also be in heavily loaded way of the end bottoms.
  2. They would cause high drag/frictional resistance in the deep/fore and aft ends where the bulb and stem flow is disturbed; the ends are kept clear to allow the propeller/shaft area (aft) and bow (forward) flow and to reduce appendage resistance.
  3. The ends would foul the raised stake/Dfrenchman? no: the ends would interfere with docking (the keel blocks and cradle), with the propeller and with the sea-chests/ballast pipes, and would not be effective because at very great fineness near the ends the damping of roll is small. They also would add weight and cause the keel to hit the dock blocks or the ground.
Part (c)

TWO principles of roll damping that bilge keels exploit.

  1. Fluid drag / hydrodynamic damping: as the ship rolls, the bilge keel body moves transverse through the water, generating a resistance (both pressure and friction) opposing the roll velocity. Because the damping force is proportional to the velocity of the keel (highest at the bilge, which is far from the roll axis), the keel produces large moments that resist and dissipate the rolling energy, arresting and reducing the amplitude of roll. This is the principal roll-damping mechanism.
  2. Creation of eddies/vortices and flow separation: the sharp edges of the bilge keel shed eddies and cause wake and separated flow at the bilge, increasing the hydrodynamic damping and inducing a phase lag in the roll so that the passive system resists resonance and limits maximum roll angle. The energy of rolling is converted to eddying motion and hence heat, damping amplitude.

These mechanisms make the bilge keel an economical, passive roll damper that reduces roll amplitude, improves comfort and cargo security and reduces sloshing and parametric roll, without moving parts or power.

Q2 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 8x

Vessel has gone through very heavy weather. On arrival at safe anchorage, you are conducting your inspection to determine damages to hull.

(a) List the areas you will inspect. (3)

(b) List your findings of any significance. (6)

(c) Write a report to company suggesting repairs if any (7)

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Part (a)

Areas to Inspect After Heavy Weather

Upon arrival at safe anchorage, a thorough inspection of the vessel shall be carried out, both externally and internally, to assess any weather-related damage. Key areas include:

1. Hull and Main Deck

  • Shell plating along bow, midship, and stern sections (especially forepeak and flare areas).
  • Boot-top and bilge areas for dents or deformation.
  • Deck plating for buckling or cracked welds.
  • Bulwarks, rails, fairleads, chocks, and mooring fittings.

2. Forecastle and Forward Structure

  • Bosun store and chain locker for water ingress.
  • Windlass foundation, anchors, and hawse pipes for deformation or cracks.
  • Forepeak tank for leakage or pressure damage.

3. Cargo Holds / Tanks

  • Inspect for structural deformation, loose frames, or fractured stiffeners.
  • Check tank top plating and bilges for leakage.
  • Check watertight doors, gaskets, and vents.

4. Superstructure and Deck Fittings

  • Bridge wings, radar mast, funnel, vents, and deckhouses for cracks or loose fittings.
  • Lifeboat davits, securing arrangements, and deck cranes.

5. Underwater and Machinery Spaces

  • Rudder, propeller, and stern tube seals (via steering gear tests).
  • Sea chest gratings and overboard discharges.
  • Engine room bilges for any seawater ingress.

Part (b)

Typical Findings of Significance

  • Dents and indentations on shell plating at bow and forward port side due to heavy slamming.
  • Paint coating and corrosion protection partly peeled off near waterline and forepeak area.
  • Loose fairlead bolts and one cracked weld on starboard bulwark stanchion.
  • Deformed ventilator head on forecastle deck.
  • Minor leakage observed in forepeak tank during sounding check.
  • Bridge wing railing bent, likely from green sea impact.
  • Anchor chain links twisted and worn.
  • Lifeboat gripes loosened, requiring tightening and inspection.
  • No flooding reported; watertight integrity maintained overall.

Part (c)

Report to Company – Heavy Weather Damage Inspection

To: Superintendent / Technical Department

From: Name / Rank

Subject: Heavy Weather Damage Inspection Report

Date: [Insert date]

Vessel: [Insert vessel name]

Summary

The vessel experienced very heavy weather en route from [Port A] to [Port B], with significant pitching and rolling in seas up to [X] meters. On arrival at safe anchorage, a complete inspection of the hull and deck was carried out.

Findings

  • Bow and forecastle plating show minor dents, with no breach of watertight integrity.
  • Bulwark stanchion (starboard side) cracked at welded joint – requires repair.
  • Fairlead foundation bolts loosened – retightening and NDT inspection recommended.
  • Ventilator head on forecastle deformed – replacement advised.
  • Forepeak tank shows trace leakage at forward bulkhead – to be pressure tested.
  • Paint coating damage and corrosion exposure on bow area – recoating required.
  • Bridge wing railing bent – to be straightened or renewed.
  • All other structures and machinery appear satisfactory after testing.

Recommendations

  1. Conduct minor steel renewal and welding repairs at the next port with repair facilities.
  2. Carry out thickness measurements and NDT on affected hull areas.
  3. Recoat damaged paint areas to prevent corrosion.
  4. Replace deformed ventilator head and bent railing.
  5. Inspect anchor and chain for elongation; renew worn links.
  6. Pressure test forepeak tank after repairs.
  7. Submit class surveyor report if deemed necessary by the superintendent.

Conclusion

The vessel remains seaworthy but requires prompt attention to minor structural and fitting damages before the next voyage. Preventive maintenance and weather routing measures should be reviewed for future passages.

Signed:

Name / Rank

Signature

Q3 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 3x

With reference to ship's rudder state:

(a) Why a breached hollow rudder can add to fuel costs? (6)

(b) Why excessive pintle clearance should not be tolerated? (5)

(c) Why fitted bolts are used in connecting upper and lower stocks? (5)

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Ship’s Rudder – Effects of Defects and Design Features

(a) Why a Breached Hollow Rudder Can Add to Fuel Costs

  • A hollow rudder is designed to provide buoyancy at the stern of the ship.
  • When the rudder is breached, sea water enters the hollow space, resulting in loss of buoyancy.
  • The loss of buoyancy causes the stern to sit deeper in the water, increasing the aft draught.
  • Increased aft draught leads to the propeller operating deeper, which increases hydrodynamic resistance.
  • Water ingress into the rudder creates turbulence and disturbed water flow around the rudder and propeller.
  • To maintain the same ship speed, higher engine power is required, resulting in increased fuel consumption.

(b) Why Excessive Pintle Clearance Should Not Be Tolerated

  • Pintles support the rudder and fit into gudgeons with very small designed clearances.
  • Excessive clearance permits sideways and vertical movement of the rudder.
  • Such movement imposes additional bending stresses on the rudder stock.
  • In heavy seas, shock and impact loads increase wear and may cause misalignment.
  • This can lead to vibration, poor steering response, accelerated wear, and in extreme cases, structural failure of the rudder assembly.

(c) Why Fitted Bolts Are Used in Connecting Upper and Lower Rudder Stocks

  • The upper and lower rudder stocks are required to transmit high steering torque.
  • Fitted bolts provide a close, interference fit with no clearance in the bolt holes.
  • They ensure accurate alignment of the two stock sections.
  • Steering loads are transmitted in shear through the bolt shanks, rather than relying on friction alone.
  • This prevents relative movement, reduces stress concentration, and minimizes the risk of fatigue failure.
Q4 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 7x

With reference to International Load Line Statutory Certification,

(a) State the reasons for the freeboard requirements (6)

(b) (i) Explain the term "conditions of assignments" (5)

(ii) List the items that may be examined during a Load line survey after a vessel's major repairs in the dry dock. (5)

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Part (a)

Reasons for Freeboard Requirements:

Freeboard is the distance measured from the waterline to the upper edge of the deck plating at the side of the freeboard deck amidships.

Purpose of Freeboard:

  • Ensures the ship is seaworthy when fully loaded.
  • Provides reserve buoyancy, enabling the ship to rise as it passes through waves, keeping decks largely dry.
  • Enhances the ship's stability and increases its survivability in case of damage by allowing it to remain afloat longer, aiding crew escape or damage control.
Part (b)

(i) Conditions of Assignment:

Conditions of Assignment are outlined in the Merchant Shipping Load Line Rules 1968 and must be satisfied before a ship is assigned freeboards and issued a load line certificate. These conditions address the practical need for openings (hatchways, doorways, vents, etc.) in the ship's hull and mandate appropriate protection and closure mechanisms for these openings.

Requirements Before Assigning Load Line:

  • The ship must have sufficient structural strength.
  • Adequate reserve buoyancy must be maintained.
  • Openings must be secured against water ingress.
  • Safety measures for the crew, such as guardrails and gangways, must be in place.
Part (b)

(ii) Items Examined During a Load Line Survey After Major Repairs in Drydock:

  • Openings: This encompasses hatchways, machinery space openings, cargo ports, watertight doors, ventilators, air pipes, scuppers, freeing ports, side scuttles, and other openings in the freeboard and superstructure decks. The surveyor checks the condition of seals, gaskets, closing mechanisms, and overall watertightness.
  • Crew Protection: Safety features designed to protect the crew, such as guardrails, bulwarks, and gangways, are inspected for damage or deterioration and proper functionality.
  • Hull Structure: The hull itself is inspected for any damage, corrosion, or structural weaknesses that could compromise watertightness or strength. This often includes visual inspection for cracks, buckling, indentation, and paint adhesion. Bottom shell, bilge keel, stem and stern frames, rudder, sea chests, side ports, stern bearing, and propeller are all checked.
  • Other Systems: Depending on the scope of repairs, other systems may be examined, including but not limited to: pressure-vacuum valves, sounding pipes, air vents, access hatches, ventilation fans, dampers, weathertight doors and sealings, door securing arrangements, side scuttles, windows, and skylights. Testing, such as hose tests (to check watertight integrity), might also be conducted.
Q5 (16 Marks) Hull Construction πŸ”₯ Repeated 6x

(a) Sketch a transverse section through the hold space of a Bulk Carrier hull (8)

(b) Referring to the sketch in (a) describe how adequate structural strength is built into the hull. (8)

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Part (a)

Transverse Section Through the Hold Space of a Bulk Carrier

A transverse section through the cargo hold of a bulk carrier consists of the following principal structural members (as shown in the given sketch):

At the top, the strength deck plating forms the upper boundary of the hull girder. It is stiffened by deck longitudinals and connected to the hatch coaming, which increases deck strength around the hatch opening.

Below the deck on either side are the topside tanks (wing tanks). These are formed by:

  • Topside tank sloping plating
  • Topside tank longitudinal plating (vertical strake)
  • Topside tank sloping plate longitudinals
  • Topside tank transverse ring webs

The topside tanks taper inward toward the cargo hold and provide structural continuity between the deck and the side shell.

The side shell plating forms the vertical boundary of the hull and is supported by:

  • Side shell frames (transverse framing)
  • Side shell longitudinals

At the lower corners of the hold are the hopper tanks, consisting of:

  • Hopper tank sloping plating
  • Hopper tank sloping plate longitudinals
  • Hopper transverse ring webs

These sloping structures connect the side shell to the inner bottom and help direct cargo toward the centreline during discharge.

At the bottom of the hold is the inner bottom plating (tank top), supported by:

  • Inner bottom longitudinals
  • Double bottom floors
  • Double bottom girders

Below this lies the double bottom tank space, bounded externally by the bottom shell plating and centrally by the keel plate and duct keel.

The curved transition between bottom and side shell is formed by the bilge plating, which ensures smooth stress distribution between vertical and horizontal structures.

Part (b)

Adequate structural strength in a bulk carrier is achieved by combining longitudinal and transverse structural systems to resist global bending, shear forces, cargo pressure, and local stresses.

1. Hull Girder Strength (Longitudinal Strength)

  • The ship behaves like a beam subjected to wave-induced hogging and sagging. The strength deck plating forms the upper flange of the hull girder, while the bottom shell plating and keel structure form the lower flange.
  • The large vertical distance between deck and bottom increases the section modulus, enabling the hull to resist high longitudinal bending moments. Continuous deck, bottom, and side shell longitudinals further increase the moment of inertia and efficiently carry longitudinal stresses along the ship’s length.

2. Double Bottom Structure

  • The double bottom forms a rigid box girder at the base of the hull. The inner bottom plating supports cargo loads directly, while floors and girders distribute these loads to the side shell and keel.
  • This arrangement:
    • Resists vertical cargo pressure
    • Strengthens the lower flange of the hull girder
    • Provides protection against grounding damage
  • The double bottom tanks also contribute to structural stiffness by forming a closed cellular structure.

3. Side Shell and Transverse Framing

  • The side shell plating, supported by transverse frames and longitudinals, resists sea pressure externally and cargo pressure internally.
  • Transverse framing prevents local buckling of plating between stiffeners, while longitudinals ensure effective distribution of stresses along the ship’s length. This combined system balances local rigidity with overall flexibility.

4. Hopper and Topside Tanks (Box Structures)

  • The hopper tanks at the lower corners and topside tanks at the upper sides form strong triangular and trapezoidal box sections.
  • These structures:
    • Reduce unsupported plate spans
    • Transfer cargo loads smoothly to the side shell and double bottom
    • Minimize stress concentration at sharp corners
    • Increase torsional and transverse stiffness
  • By converting flat plate regions into closed box sections, they significantly enhance both local and global hull strength.

5. Transverse Ring Webs and Bulkhead Effect

  • The topside and hopper transverse ring webs act as deep transverse beams. They reinforce the sloping plates and maintain structural shape under heavy cargo loading.
  • In addition, transverse bulkheads (at hold boundaries) act as strong web frames, dividing the ship into rigid compartments and improving resistance to racking and shear deformation.

Q6 (16 Marks) Ship Stability πŸ”₯ Repeated 9x

(a) Describe how the force on the ship's bottom and the GM vary when grounding takes place. (6)

(b) A ship of 8,000 tonnes displacement takes the ground on a sand bank on a falling tide at an even keel draft of 5.2 metres. KG 4.0 metres. The predicted depth of water over the sand bank at the following low water is 3.2 metres. Calculate the GM at this time assuming that the KM will then be 5.0 metres and that the mean TPC is 15 tonne (10)

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Part (a)

When a ship grounds, the forces on the bottom and the metacentric height (GM) change depending on the grounding's nature and the ship's position on the seabed.

If the ship grounds on a level bottom:

  • A ground reaction force acts vertically upwards from the seabed. This force counteracts part of the ship's weight and alters the distribution of buoyancy forces.
  • The ship’s centre of gravity (G) may appear to rise relative to the metacentre (M) because the upward ground reaction reduces the buoyancy force acting on the underwater volume.
  • This virtual rise in G reduces GM, potentially leading to a list.
  • If the list becomes excessive and the righting moment is insufficient, the ship may capsize.

If the ship grounds on a pinnacle:

  • The ship experiences two forces at the ship's bottom:
    • A downward force due to the ship’s weight.
    • An upward reaction force is concentrated on the pinnacle.
  • The resulting force between the grounding pressure and the ship’s centre of buoyancy shifts downward towards the pinnacle.
  • This situation is similar to when the ship's stern touches the keel block in a dry dock.
  • A virtual loss of GM occurs because the ship’s inclining moment may exceed the maximum righting lever.
  • If the inclining moment is too great, the ship may develop an excessive list or even capsize.

(b) Given:

$$Displacement,\:\Delta=8000\:tonnes$$

$$TPC=15\:tonnes$$

$$Initial\:Draught=5.2m$$

$$Final\:Draught=3.2m$$

$$Ship\:KG=4.0m$$

$$KM=5.0m$$

To find GM

$$Uptrust,\:P=TPC\times Fall\:in\:water\:level\:\left(cm\right)$$

$$=15\times\left(520-320\right)$$

$$=15\times200$$

$$P=3000\:tonnes$$

To Find Virtual loss of GM:

$$Virtual\:loss\:of\:GM_1=\frac{P\times KM}{\Delta}$$

$$=\frac{3000\times5}{8000}$$

$$=\frac{15000}{8000}$$

$$GM_1=1.88m$$

Actual KM = 5.0m (given)

$$Virutal\:KM=Actual\:KM-Virtual\:loss\:of\:GM_1$$

$$=5.0-1.88$$

$$=3.12$$

Similarly, Actual KG = 4.0m (given)

$$New\:GM=Virtual\:KM-\:Actual\:KG$$

$$=3.12-4.0$$

$$=-0.88$$

Q7 (16 Marks) Ship Stability πŸ”₯ Repeated 4x

(a) Describe stability requirement for dry-docking. (6)

(b) A box shaped vessel, 50 metres long x 10 metres wide, floats in salt water on an even keel at a draft of 4 metres. A center line longitudinal watertight bulkhead extends from end to end and for the full depth of the vessel. A compartment amidships on the starboard side is 15 metres long and contains cargo with permeability 30%. Calculate the list if this compartment is bilged. KG = 3 metres. (10)

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Part (a)

Stability requirements for dry-docking.

Before entering the dock the ship must have adequate and known stability. The requirements are:

  • The ship should be stable with an adequate GM, so that the small upward force (reaction) at the keel, building during docking, cannot produce a list. Dock support is provided on the keel blocks and any excessive list during docking would subject the structure and blocks to unequal loading.
  • As the water is pumped out, the reaction of the keel blocks reduces the effective displacement of the ship, lowering the KB and BM and hence GM; if all buoyancy were removed the ship would rest wholly on the blocks. The ship must retain sufficient GM throughout the docking operation (for example by keeping the GM such that the maximum inclination, where the righting moment at the critical point, does not become negative).
  • Ballast should be arranged to give even keel (or slight trim) and list-free condition; tanks should be pressed up or emptied to avoid free-surface effects, and draft read before docking.
  • Weight and trim must be such that the keel blocks contact evenly over the full keel length (avoid excessive trim that overloads the block crown or aft blocks).
  • Cargo/weights should not be changed and the dock must be level; adequate dock pumping should keep the vessel central over the keel line.

If a beam vessel develops instability while docking (transverse GM small or negative), it can heel and capsize on the blocks; hence a suitable stability margin (e.g. GM not less than a minimum) is insisted on and the docking weight, draft and trim are checked by stability data (e.g. from hydrostatics and the Docking Plan).

Part (b)

List when a starboard midship compartment is bilged.

Box-shaped vessel 50 m long, 10 m wide, floats on even keel in salt water at a draft of 4 m. A centreline longitudinal watertight bulkhead runs full depth/size. A midship compartment on the starboard side is 15 m long and has permeability 30 per cent. KG = 3 m. Calculate the list when bilged.

Waterplane area intact Aw = 50 x 10 = 500 m2; the flooded wing compartment has waterplane area = 15 x 5 = 75 m2 (only starboard half up to the centreline bulkhead).

Lost buoyancy volume of the compartment below the original waterline, with permeability 30%: Vlost = length x breadth x draft x permeability = 15 x 5 x 4 x 0.30 = 90 m3. Corresponding lost weight W = 90 x 1.025 = 92.25 t.

Sinkage: the region once flooded provides no increase of buoyancy; effective sinking waterplane = 500 - 75 = 425 m2. Mean sinkage = 90/425 = 0.212 m.

List: the lost buoyancy acts at the centroid of the lost volume, which is at 5/4 = 2.5 m out from the centreline (half-way between centreline bulkhead and side). Heeling moment = W x y = 92.25 x 2.5 = 230.6 t-m.

New GM. Using the lost-buoyancy method, displacement remains 2050 t. The second moment of area of the intact waterplane about the centreline loses the flooded starboard compartment: I(intact)=50x10^3/12 = 4166.7 m4; I(flooded)=15x5^3/3 = 625 m4; so I(new)=3541.7 m4. BM=new = I(new)/V = 3541.7/2000 = 1.771 m. KB = d/2 = 2 m (approximately, ignoring sinkage and list); GM = KB + BM - KG = 2 + 1.77 - 3 = 0.77 m.

List: tan(list) = heeling moment/(Delta x GM) = 230.6/(2050 x 0.77) = 230.6/1579 = 0.146. List = atan(0.146) = 8.3 deg.

Answer: the vessel lists about 8.3 deg to starboard.

Q8 (16 Marks) Ship Stability πŸ”₯ Repeated 11x

(a) Define longitudinal center of gravity (LCG) and longitudinal center of buoyancy(LCB). (6)

(b) A ship 120m long floats at draughts of 5.50m forward and 5.80m aft; MCT1 cm 80 tonne m, TPC 13, LCF 2.5m forward of midships. Calculate the new draughts when a mass of 110 tonne is added 24m aft of midships. (10)

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Part (a)

Longitudinal Centre of Gravity (LCG):

  • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
  • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

Longitudinal Centre of Buoyancy (LCB):

  • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
  • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.

LCF in fwd and trim by stern

$$Bodily \space sinkage \space = \space {{mass \space added} \over TPC} \space = \space {{110} \over 13} \space$$

$$=\:8.5\operatorname{\mathrm{cm}}\:=0.085m$$

$$Trim \space = \space {{m \times d} \over MCT_{1cm}}$$

$$ = \space {{110 \times (24 + 2.5)} \over 80}$$

$$Trim=36.43\operatorname{cm}=0.364m\:$$

Change in fwd draught:

$$d_{F}=\frac{-t}{L}\left\lbrack\frac{L}{2}-LCF\right\rbrack$$

$$=\frac{-36.43}{120}\left\lbrack\frac{120}{2}-2.5\right\rbrack$$

$$=-17.45\operatorname{cm}=-0.1745m$$

Change in Aft draught:

$$d_{A}=\frac{+t}{L}\left\lbrack\frac{L}{2}+LCF\right\rbrack$$

$$=\frac{+36.43}{120}\left\lbrack\frac{120}{2}+2.5\right\rbrack$$

$$=+18.97\operatorname{cm}=0.189m$$

New draughts:

$$D_{F}=5.5+0.085-0.175=5.41m$$

$$D_{A}=5.8+0.085+0.18=6.065m$$

Q9 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 3x

(a) What are the main components of ship resistance that a vessel encounters while moving through water (6)

(b) The speed of a ship is increased to 18% above normal for 7.5 hours, then reduced to 9% below normal for 10 hours. The speed is then reduced for the remainder of the day so that the consumption for the day is the normal amount. Find the percentage difference between the distance travelled in that day and the normal distance travelled per day. (10)

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Main Components of Ship Resistance

When a vessel moves through water, it experiences total resistance, which is the combined effect of several opposing forces acting against its forward motion. These forces arise due to water viscosity, pressure distribution around the hull, wave formation, and external environmental conditions. For analysis, total resistance is divided into the following main components:

1. Frictional Resistance

Frictional resistance is caused by the viscosity of water acting along the ship’s wetted surface as the vessel moves forward. As water flows over the hull, a thin boundary layer forms, and shear stresses develop between the hull surface and adjacent water particles. This shear stress produces resistance.

Frictional resistance depends primarily on:

  • Wetted surface area – A larger underwater hull area increases the contact between hull and water, thereby increasing resistance.
  • Ship speed – Resistance increases rapidly with speed because higher velocity intensifies boundary layer shear forces.
  • Surface roughness of the hull – Fouling, corrosion, or rough coatings increase turbulence and significantly raise resistance.

At normal service speeds, frictional resistance forms the largest portion of total resistance, especially for slow and medium-speed vessels. It increases approximately with the square of the ship’s speed, making hull maintenance crucial for fuel efficiency.

2. Residual Resistance

Residual resistance is the portion of resistance remaining after subtracting frictional resistance from total resistance. It mainly arises from pressure effects and wave formation around the hull. Residual resistance consists of the following components:

(a) Wave-Making Resistance

Wave-making resistance is caused by the energy expended in generating surface waves at the bow and stern as the ship moves. When a vessel travels through water, it disturbs the free surface and creates a wave system that carries energy away from the ship.

This resistance becomes particularly significant at higher speeds because wave height and wave energy increase rapidly with speed.

Wave-making resistance depends on:

  • Hull form – Fuller hull shapes generally create larger waves.
  • Speed–length ratio (Froude number) – As the Froude number increases, wave-making resistance rises sharply.

At high speeds, wave-making resistance can become a dominant component of total resistance.

(b) Eddy-Making Resistance

Eddy-making resistance is caused by flow separation and the formation of vortices (eddies) around certain parts of the hull. When water flow cannot smoothly follow the hull contour, it separates and creates turbulent regions.

This commonly occurs around:

  • The stern region
  • Appendages
  • Areas with sudden changes in hull form

These turbulent eddies consume energy and increase resistance. Proper streamlining of the hull and stern design can significantly reduce eddy-making resistance.

3. Air Resistance

Air resistance is the force exerted by air on the portion of the ship above the waterline. As the ship moves, it must also overcome aerodynamic drag.

Air resistance depends on:

  • Wind speed and direction
  • Projected area above water
  • Shape of the superstructure

Although usually small compared to water resistance, it becomes significant for:

  • Container ships
  • Ro-Ro vessels
  • Ships operating in strong headwinds

For vessels with large exposed areas, air resistance can noticeably affect fuel consumption.

4. Appendage Resistance

Appendage resistance is caused by external fittings attached to the hull, such as:

  • Rudders
  • Bilge keels
  • Shaft brackets
  • Propeller bossings

These appendages increase the wetted surface area and disturb smooth water flow, thereby increasing both frictional and pressure resistance. In resistance calculations, appendage resistance is generally included as a separate correction added to frictional resistance.

5. Added (Special) Resistance

Added resistance refers to the additional resistance experienced in real sea conditions that is not present in calm-water trials.

It occurs due to:

  • Waves and swell, which cause pitching and heaving motions
  • Steering and yawing motions, which disturb steady flow around the hull

Although not considered in calm-water resistance analysis, added resistance is highly important in practical operations because it significantly affects power requirements and fuel consumption in rough weather.

Q10 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 8x

With reference to fixed pitch propellers:

(a) Explain Propeller Slip and Propeller Thrust. (6)

(b) The shaft power of a ship is 3000 KW, the ship's speed V is 13.2 knot. Propeller RPS is 1.27. propeller pitch is 5.5m and the speed of advance is 11 Knots. Find: (10)

(i) Real Slip

(ii) Wake fraction

(iii) Propeller thrust, when its efficiency, Ξ· = 70%

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Part (a)

Slip

is the difference between the theoretical distance the propeller should travel in one revolution and the actual distance the vessel travels.

$$Slip\:\left(\%\right)\:=\:\left(\frac{Engine\:distance\:-\:Ship^{\prime}s\:distance}{Engine\:distance}\right)\:\times100$$

Where Engine distance = no. of propeller revolutions * propeller pitch (usually calculated over a 24-hour period)

Propeller thrust: it is the force exerted by the propeller to move the vessel ahead and given by

$$Thrust\:\left(T\right)\:=\:\rho\times A\times P^2\times n^2\times S$$

Where,

ρ - Density

A - Area

P - Pitch

n - Revolution per second

S - Slip

Q1 (16 Marks) Hull Construction πŸ”₯ Repeated 10x

(a) Describe a method for the attachment of bilge keels. (5)

(b) State THREE reasons for not extending bilge keels the entire length of the vessel. (6)

(c) Explain TWO principles of roll damping that bilge keels exploit. (5)

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Part (a)

Method for the attachment of bilge keels.

The bilge keel is a long fin attached externally along the turn of the bilge, running roughly in a fore-and-aft direction. Attachment method:

  • The bilge keel is fabricated as a light, flat bar or fabricated plate of length running over the middle portion of the hull. It is attached by one of two methods:

(i) riveted/bolted connection through doubler plates, in which the keel web is connected by a continuous fillet weld to a flat-bar or doubler section on the shell, or

(ii) welded directly to the shell by continuous or intermittent fillet welds.

  • Because the attachment is a potential source of high local stress concentration and fatigue cracking (the shell at the bilge is highly loaded in bending and docking), the weld is usually a full-strength continuous fillet or butt weld, and the keel is not welded to the shell with single small tack welds that crack under repeated relative movement (flexing).
  • The ends of the bilge keel are tapered and rolled to a fine radius and blended smoothly (scalloped/angled end) so as to reduce stress concentration at their termination; heavy strike plates (flat bar doubler) at the ends carry the connection.
  • The keel is set slightly at a small angle to the base plane so that its face lines up with the flow; it is normally fitted parallel to the load waterline and positioned at the turn of the bilge where it is not unduly loaded in docking (kept clear of the keel blocks).
  • Doubler plates and drainage are arranged, and the attachment is periodically inspected (especially the weld toes) because the bilge keel experiences severe fluctuating loads in a seaway.
Part (b)

THREE reasons for not extending bilge keels to the entire length.

  1. To avoid severe local stress concentrations and fatigue cracking at their ends/attachment; a full-length keel would place the highly-loaded ends in regions of the hull girder with large bending stresses, promoting cracking. Pounding/the connection would also be in heavily loaded way of the end bottoms.
  2. They would cause high drag/frictional resistance in the deep/fore and aft ends where the bulb and stem flow is disturbed; the ends are kept clear to allow the propeller/shaft area (aft) and bow (forward) flow and to reduce appendage resistance.
  3. The ends would foul the raised stake/Dfrenchman? no: the ends would interfere with docking (the keel blocks and cradle), with the propeller and with the sea-chests/ballast pipes, and would not be effective because at very great fineness near the ends the damping of roll is small. They also would add weight and cause the keel to hit the dock blocks or the ground.
Part (c)

TWO principles of roll damping that bilge keels exploit.

  1. Fluid drag / hydrodynamic damping: as the ship rolls, the bilge keel body moves transverse through the water, generating a resistance (both pressure and friction) opposing the roll velocity. Because the damping force is proportional to the velocity of the keel (highest at the bilge, which is far from the roll axis), the keel produces large moments that resist and dissipate the rolling energy, arresting and reducing the amplitude of roll. This is the principal roll-damping mechanism.
  2. Creation of eddies/vortices and flow separation: the sharp edges of the bilge keel shed eddies and cause wake and separated flow at the bilge, increasing the hydrodynamic damping and inducing a phase lag in the roll so that the passive system resists resonance and limits maximum roll angle. The energy of rolling is converted to eddying motion and hence heat, damping amplitude.

These mechanisms make the bilge keel an economical, passive roll damper that reduces roll amplitude, improves comfort and cargo security and reduces sloshing and parametric roll, without moving parts or power.

Q2 (16 Marks) Hull Construction πŸ”₯ Repeated 10x

(a) Describe a method for the attachment of bilge keels. (5)

(b) State THREE reasons for not extending bilge keels the entire length of the vessel. (6)

(c) Explain TWO principles of roll damping that bilge keels exploit. (5)

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Part (a)

Method for the attachment of bilge keels.

The bilge keel is a long fin attached externally along the turn of the bilge, running roughly in a fore-and-aft direction. Attachment method:

  • The bilge keel is fabricated as a light, flat bar or fabricated plate of length running over the middle portion of the hull. It is attached by one of two methods:

(i) riveted/bolted connection through doubler plates, in which the keel web is connected by a continuous fillet weld to a flat-bar or doubler section on the shell, or

(ii) welded directly to the shell by continuous or intermittent fillet welds.

  • Because the attachment is a potential source of high local stress concentration and fatigue cracking (the shell at the bilge is highly loaded in bending and docking), the weld is usually a full-strength continuous fillet or butt weld, and the keel is not welded to the shell with single small tack welds that crack under repeated relative movement (flexing).
  • The ends of the bilge keel are tapered and rolled to a fine radius and blended smoothly (scalloped/angled end) so as to reduce stress concentration at their termination; heavy strike plates (flat bar doubler) at the ends carry the connection.
  • The keel is set slightly at a small angle to the base plane so that its face lines up with the flow; it is normally fitted parallel to the load waterline and positioned at the turn of the bilge where it is not unduly loaded in docking (kept clear of the keel blocks).
  • Doubler plates and drainage are arranged, and the attachment is periodically inspected (especially the weld toes) because the bilge keel experiences severe fluctuating loads in a seaway.
Part (b)

THREE reasons for not extending bilge keels to the entire length.

  1. To avoid severe local stress concentrations and fatigue cracking at their ends/attachment; a full-length keel would place the highly-loaded ends in regions of the hull girder with large bending stresses, promoting cracking. Pounding/the connection would also be in heavily loaded way of the end bottoms.
  2. They would cause high drag/frictional resistance in the deep/fore and aft ends where the bulb and stem flow is disturbed; the ends are kept clear to allow the propeller/shaft area (aft) and bow (forward) flow and to reduce appendage resistance.
  3. The ends would foul the raised stake/Dfrenchman? no: the ends would interfere with docking (the keel blocks and cradle), with the propeller and with the sea-chests/ballast pipes, and would not be effective because at very great fineness near the ends the damping of roll is small. They also would add weight and cause the keel to hit the dock blocks or the ground.
Part (c)

TWO principles of roll damping that bilge keels exploit.

  1. Fluid drag / hydrodynamic damping: as the ship rolls, the bilge keel body moves transverse through the water, generating a resistance (both pressure and friction) opposing the roll velocity. Because the damping force is proportional to the velocity of the keel (highest at the bilge, which is far from the roll axis), the keel produces large moments that resist and dissipate the rolling energy, arresting and reducing the amplitude of roll. This is the principal roll-damping mechanism.
  2. Creation of eddies/vortices and flow separation: the sharp edges of the bilge keel shed eddies and cause wake and separated flow at the bilge, increasing the hydrodynamic damping and inducing a phase lag in the roll so that the passive system resists resonance and limits maximum roll angle. The energy of rolling is converted to eddying motion and hence heat, damping amplitude.

These mechanisms make the bilge keel an economical, passive roll damper that reduces roll amplitude, improves comfort and cargo security and reduces sloshing and parametric roll, without moving parts or power.

Q3 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 3x

With reference to ship's rudder state:

(a) Why a breached hollow rudder can add to fuel costs? (6)

(b) Why excessive pintle clearance should not be tolerated? (5)

(c) Why fitted bolts are used in connecting upper and lower stocks? (5)

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Ship’s Rudder – Effects of Defects and Design Features

(a) Why a Breached Hollow Rudder Can Add to Fuel Costs

  • A hollow rudder is designed to provide buoyancy at the stern of the ship.
  • When the rudder is breached, sea water enters the hollow space, resulting in loss of buoyancy.
  • The loss of buoyancy causes the stern to sit deeper in the water, increasing the aft draught.
  • Increased aft draught leads to the propeller operating deeper, which increases hydrodynamic resistance.
  • Water ingress into the rudder creates turbulence and disturbed water flow around the rudder and propeller.
  • To maintain the same ship speed, higher engine power is required, resulting in increased fuel consumption.

(b) Why Excessive Pintle Clearance Should Not Be Tolerated

  • Pintles support the rudder and fit into gudgeons with very small designed clearances.
  • Excessive clearance permits sideways and vertical movement of the rudder.
  • Such movement imposes additional bending stresses on the rudder stock.
  • In heavy seas, shock and impact loads increase wear and may cause misalignment.
  • This can lead to vibration, poor steering response, accelerated wear, and in extreme cases, structural failure of the rudder assembly.

(c) Why Fitted Bolts Are Used in Connecting Upper and Lower Rudder Stocks

  • The upper and lower rudder stocks are required to transmit high steering torque.
  • Fitted bolts provide a close, interference fit with no clearance in the bolt holes.
  • They ensure accurate alignment of the two stock sections.
  • Steering loads are transmitted in shear through the bolt shanks, rather than relying on friction alone.
  • This prevents relative movement, reduces stress concentration, and minimizes the risk of fatigue failure.
Q4 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 7x

With reference to International Load Line Statutory Certification,

(a) State the reasons for the freeboard requirements. (6)

(b) (i) Explain the term "conditions of assignment". (5)

(ii) List the items that may be examined during a Load line survey after a vessel's major repairs in the dry dock. (5)

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Part (a)

Reasons for Freeboard Requirements:

Freeboard is the distance measured from the waterline to the upper edge of the deck plating at the side of the freeboard deck amidships.

Purpose of Freeboard:

  • Ensures the ship is seaworthy when fully loaded.
  • Provides reserve buoyancy, enabling the ship to rise as it passes through waves, keeping decks largely dry.
  • Enhances the ship's stability and increases its survivability in case of damage by allowing it to remain afloat longer, aiding crew escape or damage control.
Part (b)

(i) Conditions of Assignment:

Conditions of Assignment are outlined in the Merchant Shipping Load Line Rules 1968 and must be satisfied before a ship is assigned freeboards and issued a load line certificate. These conditions address the practical need for openings (hatchways, doorways, vents, etc.) in the ship's hull and mandate appropriate protection and closure mechanisms for these openings.

Requirements Before Assigning Load Line:

  • The ship must have sufficient structural strength.
  • Adequate reserve buoyancy must be maintained.
  • Openings must be secured against water ingress.
  • Safety measures for the crew, such as guardrails and gangways, must be in place.
Part (b)

(ii) Items Examined During a Load Line Survey After Major Repairs in Drydock:

  • Openings: This encompasses hatchways, machinery space openings, cargo ports, watertight doors, ventilators, air pipes, scuppers, freeing ports, side scuttles, and other openings in the freeboard and superstructure decks. The surveyor checks the condition of seals, gaskets, closing mechanisms, and overall watertightness.
  • Crew Protection: Safety features designed to protect the crew, such as guardrails, bulwarks, and gangways, are inspected for damage or deterioration and proper functionality.
  • Hull Structure: The hull itself is inspected for any damage, corrosion, or structural weaknesses that could compromise watertightness or strength. This often includes visual inspection for cracks, buckling, indentation, and paint adhesion. Bottom shell, bilge keel, stem and stern frames, rudder, sea chests, side ports, stern bearing, and propeller are all checked.
  • Other Systems: Depending on the scope of repairs, other systems may be examined, including but not limited to: pressure-vacuum valves, sounding pipes, air vents, access hatches, ventilation fans, dampers, weathertight doors and sealings, door securing arrangements, side scuttles, windows, and skylights. Testing, such as hose tests (to check watertight integrity), might also be conducted.
Q5 (16 Marks) Hull Construction πŸ”₯ Repeated 6x

(a) Sketch a transverse section through the hold space of a container ship hull (8)

(b) Referring to the sketch in (a) describe how adequate structural strength is built into the hull. (8)

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Part (a)

Mid-ship half-section sketch of a container ship:

Part (b)

Structural Strength Features:

  • The deck plating is made thicker and uses higher tensile steel to withstand stresses caused by wide hatch openings and operational loads.
  • The deck, side shell, and longitudinal bulkheads are framed longitudinally. This arrangement combined with deep double bottom helps resist bending stresses due to hogging (upward bending) and sagging (downward bending) when the ship is under load.
  • The hatch coamings are made continuous to contribute to the overall longitudinal strength of the hull.
  • A torsion box is installed, running along the entire length of the ship from the machinery space bulkhead to the forward collision bulkhead. This structure provides the necessary torsional strength to counteract twisting forces acting on the hull during operation.
  • Deep web boxes are fitted at the ends of hatches, both at tank top and deck levels, to enhance transverse and torsional strength.
  • A deep double bottom is designed to withstand uplift forces caused by water pressure, especially when the ship is deeply loaded. It also provides additional strength to the hull structure.
  • Side girders are placed under container cells, with added transverse local stiffening. These elements distribute the concentrated loads from containers and increase overall stability.
Q6 (16 Marks) Ship Stability πŸ”₯ Repeated 9x

(a) Describe how the force on the ship's bottom and the GM vary when grounding takes place. (6)

(b) A ship of 8,000 tonnes displacement takes the ground on a sand bank on a falling tide at an even keel draft of 5.2 metres. KG 4.0 metres. The predicted depth of water over the sandbank at the following low water is 3.2 metres. Calculate the GM at that time, assuming that the KM will then be 5.0 metres and that the mean TPC is 15 tonne. (10)

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Part (a)

When a ship grounds, the forces on the bottom and the metacentric height (GM) change depending on the grounding's nature and the ship's position on the seabed.

If the ship grounds on a level bottom:

  • A ground reaction force acts vertically upwards from the seabed. This force counteracts part of the ship's weight and alters the distribution of buoyancy forces.
  • The ship’s centre of gravity (G) may appear to rise relative to the metacentre (M) because the upward ground reaction reduces the buoyancy force acting on the underwater volume.
  • This virtual rise in G reduces GM, potentially leading to a list.
  • If the list becomes excessive and the righting moment is insufficient, the ship may capsize.

If the ship grounds on a pinnacle:

  • The ship experiences two forces at the ship's bottom:
    • A downward force due to the ship’s weight.
    • An upward reaction force is concentrated on the pinnacle.
  • The resulting force between the grounding pressure and the ship’s centre of buoyancy shifts downward towards the pinnacle.
  • This situation is similar to when the ship's stern touches the keel block in a dry dock.
  • A virtual loss of GM occurs because the ship’s inclining moment may exceed the maximum righting lever.
  • If the inclining moment is too great, the ship may develop an excessive list or even capsize.

(b) Given:

$$Displacement,\:\Delta=8000\:tonnes$$

$$TPC=15\:tonnes$$

$$Initial\:Draught=5.2m$$

$$Final\:Draught=3.2m$$

$$Ship\:KG=4.0m$$

$$KM=5.0m$$

To find GM

$$Uptrust,\:P=TPC\times Fall\:in\:water\:level\:\left(cm\right)$$

$$=15\times\left(520-320\right)$$

$$=15\times200$$

$$P=3000\:tonnes$$

To Find Virtual loss of GM:

$$Virtual\:loss\:of\:GM_1=\frac{P\times KM}{\Delta}$$

$$=\frac{3000\times5}{8000}$$

$$=\frac{15000}{8000}$$

$$GM_1=1.88m$$

Actual KM = 5.0m (given)

$$Virutal\:KM=Actual\:KM-Virtual\:loss\:of\:GM_1$$

$$=5.0-1.88$$

$$=3.12$$

Similarly, Actual KG = 4.0m (given)

$$New\:GM=Virtual\:KM-\:Actual\:KG$$

$$=3.12-4.0$$

$$=-0.88$$

Q7 (16 Marks) Ship Stability πŸ”₯ Repeated 4x

(a) Describe stability requirement for dry-docking. (6)

(b) A box shaped vessel, 50 metres long x 10 metres wide, floats in salt water on an even keel at a draft of 4 metres. A centre line longitudinal watertight bulkhead extends from end to end and for the full depth of the vessel. A compartment amidships on the starboard side 15 metres long and contains cargo with permeability 30%. Calculate the list if this compartment is bilged. KG = 3 metres. (10)

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Part (a)

Stability requirements for dry-docking.

Before entering the dock the ship must have adequate and known stability. The requirements are:

  • The ship should be stable with an adequate GM, so that the small upward force (reaction) at the keel, building during docking, cannot produce a list. Dock support is provided on the keel blocks and any excessive list during docking would subject the structure and blocks to unequal loading.
  • As the water is pumped out, the reaction of the keel blocks reduces the effective displacement of the ship, lowering the KB and BM and hence GM; if all buoyancy were removed the ship would rest wholly on the blocks. The ship must retain sufficient GM throughout the docking operation (for example by keeping the GM such that the maximum inclination, where the righting moment at the critical point, does not become negative).
  • Ballast should be arranged to give even keel (or slight trim) and list-free condition; tanks should be pressed up or emptied to avoid free-surface effects, and draft read before docking.
  • Weight and trim must be such that the keel blocks contact evenly over the full keel length (avoid excessive trim that overloads the block crown or aft blocks).
  • Cargo/weights should not be changed and the dock must be level; adequate dock pumping should keep the vessel central over the keel line.

If a beam vessel develops instability while docking (transverse GM small or negative), it can heel and capsize on the blocks; hence a suitable stability margin (e.g. GM not less than a minimum) is insisted on and the docking weight, draft and trim are checked by stability data (e.g. from hydrostatics and the Docking Plan).

Part (b)

List when a starboard midship compartment is bilged.

Box-shaped vessel 50 m long, 10 m wide, floats on even keel in salt water at a draft of 4 m. A centreline longitudinal watertight bulkhead runs full depth/size. A midship compartment on the starboard side is 15 m long and has permeability 30 per cent. KG = 3 m. Calculate the list when bilged.

Waterplane area intact Aw = 50 x 10 = 500 m2; the flooded wing compartment has waterplane area = 15 x 5 = 75 m2 (only starboard half up to the centreline bulkhead).

Lost buoyancy volume of the compartment below the original waterline, with permeability 30%: Vlost = length x breadth x draft x permeability = 15 x 5 x 4 x 0.30 = 90 m3. Corresponding lost weight W = 90 x 1.025 = 92.25 t.

Sinkage: the region once flooded provides no increase of buoyancy; effective sinking waterplane = 500 - 75 = 425 m2. Mean sinkage = 90/425 = 0.212 m.

List: the lost buoyancy acts at the centroid of the lost volume, which is at 5/4 = 2.5 m out from the centreline (half-way between centreline bulkhead and side). Heeling moment = W x y = 92.25 x 2.5 = 230.6 t-m.

New GM. Using the lost-buoyancy method, displacement remains 2050 t. The second moment of area of the intact waterplane about the centreline loses the flooded starboard compartment: I(intact)=50x10^3/12 = 4166.7 m4; I(flooded)=15x5^3/3 = 625 m4; so I(new)=3541.7 m4. BM=new = I(new)/V = 3541.7/2000 = 1.771 m. KB = d/2 = 2 m (approximately, ignoring sinkage and list); GM = KB + BM - KG = 2 + 1.77 - 3 = 0.77 m.

List: tan(list) = heeling moment/(Delta x GM) = 230.6/(2050 x 0.77) = 230.6/1579 = 0.146. List = atan(0.146) = 8.3 deg.

Answer: the vessel lists about 8.3 deg to starboard.

Q8 (16 Marks) Ship Stability πŸ”₯ Repeated 11x

(a) Define longitudinal center of gravity (LCG) and longitudinal center of buoyancy (LCB). (6)

(b) A ship 120m long floats at draughts of 5.50m forward and 5.80m aft. MCT Icm 80 tonne m, TPC 13, LCF 2.5m forward of midships. Calculate the new draughts when a mass of 110 tonne is added 24m aft of midships. (10)

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Part (a)

Longitudinal Centre of Gravity (LCG):

  • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
  • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

Longitudinal Centre of Buoyancy (LCB):

  • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
  • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.

LCF in fwd and trim by stern

$$Bodily \space sinkage \space = \space {{mass \space added} \over TPC} \space = \space {{110} \over 13} \space$$

$$=\:8.5\operatorname{\mathrm{cm}}\:=0.085m$$

$$Trim \space = \space {{m \times d} \over MCT_{1cm}}$$

$$ = \space {{110 \times (24 + 2.5)} \over 80}$$

$$Trim=36.43\operatorname{cm}=0.364m\:$$

Change in fwd draught:

$$d_{F}=\frac{-t}{L}\left\lbrack\frac{L}{2}-LCF\right\rbrack$$

$$=\frac{-36.43}{120}\left\lbrack\frac{120}{2}-2.5\right\rbrack$$

$$=-17.45\operatorname{cm}=-0.1745m$$

Change in Aft draught:

$$d_{A}=\frac{+t}{L}\left\lbrack\frac{L}{2}+LCF\right\rbrack$$

$$=\frac{+36.43}{120}\left\lbrack\frac{120}{2}+2.5\right\rbrack$$

$$=+18.97\operatorname{cm}=0.189m$$

New draughts:

$$D_{F}=5.5+0.085-0.175=5.41m$$

$$D_{A}=5.8+0.085+0.18=6.065m$$

Q9 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

(a) List the precautions necessary before an inclining experiment is carried out. (6)

(b) The speed of a ship is increased to 18% above normal for 7.5 hours, then reduced to 9% below normal for 10 hours. The speed is then reduced for the remainder of the day so that the consumption for the day is the normal amount. Find the percentage difference between the distance travelled in that day and the normal distance travelled per day. (10)

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Part (a)

Precautions before an inclining experiment.

  • The ship must be in a light, near-condition with all weights (cargo, bunkers, ballast, stores, fresh water, loose gear) accounted for and positioned; all such weights must be lined up on the centreline or paired, and their positions accurately recorded (or removed and weighed).
  • All tanks must be completely empty, pressed up (to prevent free surface) or their free-surface effect computed; no free liquid surfaces or sloshing should be present in the tanks; fuel/water/ballast tanks should be emptied or pressed up and kept gas-free.
  • Deck: no undischarged items, all cargo on board weighed and centred; the ship must be upright, at even keel or known initial trim, and floating freely (not touching piers, hangars, moorings, lines or holding fast - always moored with loose ropes and no gangway contact).
  • All personnel, apart from the inclining team and observers, should be docked? tensed, off the deck? person on board should not move about; boats and cranes set so as not to interfere.
  • Decide the draft marks (recording of forward, aft, midships mean) and water density (sample) at each set of readings, and assess the waterplane area and displacing conditions from hydrostatic data/docking information.
  • Prepare the inclining weights (movable ballast of known size) and their movement plan; ensure calm water, no currents or waves, and stable reference (plumb lines, clinometer/pendulum) freely available.
  • The ship should be at midships age at a suitable inclination, generally less than about 3 to 4 degrees, and the reference pendulum (plumb line or spirit level) checked for free movement before starting.
  • All weights should be moved smoothly amidships to starboard/port in a planned sequence (e.g. pendulum deflects), readings taken at each step.
Part (b)

Daily distance when speed is varied (consumption equal to normal).

Given speed increased 18% above normal (V1=1.18 Vn) for 7.5 h, then reduced to 9% below normal (V2=0.91 Vn) for 10 h, then a final speed V3 for the remainder of the day such that the total fuel for the 24 h day equals the normal day's consumption.

Fuel consumption is proportional to V^3 (per unit time). Normal day consumption = k x Vn^3 x 24.

Consumption in the three parts = k[V1^3 x 7.5 + V2^3 x 10 + V3^3 x 6.5], where 6.5 = 24 - 17.5 h.

Set equal to the normal consumption k x Vn^3 x 24:

1.18^3 x 7.5 + 0.91^3 x 10 + V3^3 x 6.5 = 1 x 24.

1.643 x 7.5 = 12.32; 0.7536 x 10 = 7.536; so 12.32 + 7.536 + 6.5 V3^3 = 24.

6.5 V3^3 = 24 - 19.86 = 4.14, so V3^3 = 0.6369, V3 = 0.8606 Vn.

Distances travelled in the day:

D = V1 x 7.5 + V2 x 10 + V3 x 6.5 = 1.18 x 7.5 + 0.91 x 10 + 0.8606 x 6.5

= 8.85 + 9.1 + 5.59 = 23.54 (as a multiple of distance per hour at Vn? Units: this is (Vn x hours) normalised by Vn, so in units of (Vn hours), normal daily distance = 24 x Vn).

Normal daily distance = 24 (in those units). Today's distance = 23.54.

Percentage difference = (23.54 - 24)/24 x 100 = -1.9%.

Answer: the distance travelled that day is about 1.9% less than the normal daily distance (approximately 0.46 Vn hours less). (Using the exact cubic relation the result is about -1.9%.)

Q10 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 8x

With reference to fixed pitch propellers:

(a) Explain Propeller Slip and Propeller thrust. (6)

(b) The shaft power of a ship is 3000 kW, the ship's speed V is 13.2 knot. Propeller rps is 1.27, propeller pitch is 5.5m and the speed of advance is 11 knots. Find: (10)

(i) Real slip

(ii) Wake fraction

(iii) Propeller thrust, when its efficiency, Ξ· = 70%

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Part (a)

Slip

is the difference between the theoretical distance the propeller should travel in one revolution and the actual distance the vessel travels.

$$Slip\:\left(\%\right)\:=\:\left(\frac{Engine\:distance\:-\:Ship^{\prime}s\:distance}{Engine\:distance}\right)\:\times100$$

Where Engine distance = no. of propeller revolutions * propeller pitch (usually calculated over a 24-hour period)

Propeller thrust: it is the force exerted by the propeller to move the vessel ahead and given by

$$Thrust\:\left(T\right)\:=\:\rho\times A\times P^2\times n^2\times S$$

Where,

ρ - Density

A - Area

P - Pitch

n - Revolution per second

S - Slip

Q1 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 8x

Vessel has gone through very heavy weather. On arrival at safe anchorage, you are conducting your inspections to determine damages to hull.

(a) List the areas you will inspect.

(b) List your findings of any significance.

(c) Write a report to company suggesting repairs if any

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Part (a)

Areas to Inspect After Heavy Weather

Upon arrival at safe anchorage, a thorough inspection of the vessel shall be carried out, both externally and internally, to assess any weather-related damage. Key areas include:

1. Hull and Main Deck

  • Shell plating along bow, midship, and stern sections (especially forepeak and flare areas).
  • Boot-top and bilge areas for dents or deformation.
  • Deck plating for buckling or cracked welds.
  • Bulwarks, rails, fairleads, chocks, and mooring fittings.

2. Forecastle and Forward Structure

  • Bosun store and chain locker for water ingress.
  • Windlass foundation, anchors, and hawse pipes for deformation or cracks.
  • Forepeak tank for leakage or pressure damage.

3. Cargo Holds / Tanks

  • Inspect for structural deformation, loose frames, or fractured stiffeners.
  • Check tank top plating and bilges for leakage.
  • Check watertight doors, gaskets, and vents.

4. Superstructure and Deck Fittings

  • Bridge wings, radar mast, funnel, vents, and deckhouses for cracks or loose fittings.
  • Lifeboat davits, securing arrangements, and deck cranes.

5. Underwater and Machinery Spaces

  • Rudder, propeller, and stern tube seals (via steering gear tests).
  • Sea chest gratings and overboard discharges.
  • Engine room bilges for any seawater ingress.

Part (b)

Typical Findings of Significance

  • Dents and indentations on shell plating at bow and forward port side due to heavy slamming.
  • Paint coating and corrosion protection partly peeled off near waterline and forepeak area.
  • Loose fairlead bolts and one cracked weld on starboard bulwark stanchion.
  • Deformed ventilator head on forecastle deck.
  • Minor leakage observed in forepeak tank during sounding check.
  • Bridge wing railing bent, likely from green sea impact.
  • Anchor chain links twisted and worn.
  • Lifeboat gripes loosened, requiring tightening and inspection.
  • No flooding reported; watertight integrity maintained overall.

Part (c)

Report to Company – Heavy Weather Damage Inspection

To: Superintendent / Technical Department

From: Name / Rank

Subject: Heavy Weather Damage Inspection Report

Date: [Insert date]

Vessel: [Insert vessel name]

Summary

The vessel experienced very heavy weather en route from [Port A] to [Port B], with significant pitching and rolling in seas up to [X] meters. On arrival at safe anchorage, a complete inspection of the hull and deck was carried out.

Findings

  • Bow and forecastle plating show minor dents, with no breach of watertight integrity.
  • Bulwark stanchion (starboard side) cracked at welded joint – requires repair.
  • Fairlead foundation bolts loosened – retightening and NDT inspection recommended.
  • Ventilator head on forecastle deformed – replacement advised.
  • Forepeak tank shows trace leakage at forward bulkhead – to be pressure tested.
  • Paint coating damage and corrosion exposure on bow area – recoating required.
  • Bridge wing railing bent – to be straightened or renewed.
  • All other structures and machinery appear satisfactory after testing.

Recommendations

  1. Conduct minor steel renewal and welding repairs at the next port with repair facilities.
  2. Carry out thickness measurements and NDT on affected hull areas.
  3. Recoat damaged paint areas to prevent corrosion.
  4. Replace deformed ventilator head and bent railing.
  5. Inspect anchor and chain for elongation; renew worn links.
  6. Pressure test forepeak tank after repairs.
  7. Submit class surveyor report if deemed necessary by the superintendent.

Conclusion

The vessel remains seaworthy but requires prompt attention to minor structural and fitting damages before the next voyage. Preventive maintenance and weather routing measures should be reviewed for future passages.

Signed:

Name / Rank

Signature

Q2 (16 Marks) Ship Stability πŸ”₯ Repeated 14x

Explain how the period of roll varies with:

(a) The amplitude of roll.

(b) The radius of gyration.

(c) The initial metacentric height.

(d) The location of masses in the ship

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The period of roll Tr of a ship is determined by the formula:

$$T_{r}=\frac{2\pi K}{\sqrt{g\times GM}}$$

where,

  • K is the radius of gyration (mass moment of inertia)
  • g is the acceleration due to gravity, and
  • GM is the metacentric height.
Part (a)

Amplitude of Roll:

  • The amplitude of roll, or the maximum angle of heel, does not affect the period of roll. While a larger amplitude implies a greater heeling moment and faster roll speed, the time taken to complete one full roll cycle remains constant for a given metacentric height and radius of gyration.
Part (b)

Radius of Gyration (K):

  • The period of roll is directly proportional to the radius of gyration. A larger radius of gyration (indicating a greater distribution of mass further from the ship's centre of rotation) leads to a longer period of roll. Conversely, a smaller radius of gyration (mass concentrated closer to the centre) results in a shorter period. The distribution of cargo significantly impacts K; cargo concentrated centrally minimises K and the roll period, while dispersed cargo maximises K and the roll period.
Part (c)

Initial Metacentric Height (GM):

  • The period of roll is inversely proportional to the square root of the metacentric height (GM). A larger GM (a stiffer ship) leads to a shorter roll period, as the ship quickly returns to its upright position. A smaller GM (a tender ship) results in a longer roll period, with slower return to the upright.
Part (d)

Location of Masses in the Ship:

The location of masses in the ship will effect the GM & K. So the period of roll will be affected.

  • If masses are at bottom, G moves down, GM ↑, period of roll ↓.
  • If masses are at top, G moves up, GM ↓, period of roll ↑.
  • If masses are concentrated at centre, K ↓, period of roll ↓.
  • If masses are away from centre, K ↑, period of roll ↑.
Q3 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 9x

With reference to Underwater Inspection in lieu of Dry docking (UWILD):

(a) Explain in detail, how an underwater survey is carried out.

(b) State the requirements to be fulfilled before an underwater survey is acceptable to the surveying authority.

(c) Construct a list of the items in order of importance that the underwater survey authority should include.

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(a) An underwater survey, also known as Underwater Inspection in Lieu of Dry Docking (UWILD), involves a systematic and detailed examination of a vessel’s hull and underwater components while the ship remains afloat. The procedure includes the following steps:
  • The shipowner sends a request to the classification society surveyor, indicating the intention to perform an underwater survey.
  • A detailed plan of the ship's external hull features is submitted, showing the location of frames, bulkheads, welding lines, openings, etc.
  • The hull surface is cleaned before the survey to remove any marine growth or fouling that could obscure the inspection.
  • A diving company, approved by the classification society, is appointed to conduct the underwater inspection.
  • A self-propelled survey vehicle equipped with the following tools is used:
    • Long-Range Light TV Camera to aid in steering and checking hull deterioration.
    • High-resolution colour TV Camera to provide a close-up view of the hull coating and welded seams.
    • 35mm Still Camera to capture still images.
    • Ultrasonic Probe for measuring plate thickness.
    • Depth Meter and Speed Indicator to provide accurate data on the vehicle's depth and movement.
    • Umbilical Cable to connect the survey vehicle to the survey boat, transmitting power and information.
  • The survey boat is to be equipped with:
    • A control console with TV monitors.
    • Plate thickness printouts.
    • Audio and video cassette recorders.
    • Playback units.
    • Diver communication systems.
    • Vehicle control systems and associated instruments.
  • Operation:
    • The survey vehicle is taken underwater by a diver to the survey starting point.
    • Using TV monitors and shell expansion plans as guides, the vehicle is navigated over the hull, focusing on the bottom structure, sides, stern frame, rudder, propeller, bilge keel, and hull openings.
    • All images, data, and information are recorded and transmitted back to the survey boat.
    • Detailed pictures of the stern frame, rudder, propeller, bilge keel, and hull openings are captured.
    • Divers are deployed to measure stern tube bearing wear, pintle clearance, and inspect stern seals, anodes, and rudder stock palm coupling bolts.
    • All recorded video and audio, including conversations between the surveyor and drivers, thickness printouts, measurements, and pictures are analyzed to determine the detailed underwater condition of the vessel.
    (b) Before an underwater survey is accepted by the surveying authority, the following conditions must be met:

    The vessel's owner submits a request to the surveyor, including:
    • The proposed date and location for the survey.
    • General information about the diving company.
    • A declaration that the vessel has not suffered any damage due to grounding, collision, or other incidents.
    The ship's master or owner’s representative must provide a declaration confirming:
    • Any suspected or actual damage to the hull since the last dry-docking.
    • The underwater portion of the hull is protected by a suitable paint scheme that is of adequate thickness and remains valid until the next dry-dock.

    • The survey site should be in a protected area with calm and clear water, ensuring good underwater visibility. Attention must be given to the effects of currents and tides.
    • The hull must be clean for the external survey. The surveyor must be satisfied with the method and quality of the pictorial presentation, ensuring that it provides a reliable assessment of the hull's condition.
    • The underwater examination should be conducted by an approved diving company using closed-circuit TV and two-way communication, which can be monitored by the surveyor.
    • The vessel should be in as light an operating condition as possible to facilitate the survey.
    • Means must be available for the surveyor to examine the outside shell plating above the waterline.
    • Any required repairs identified during the survey must be carried out to the satisfaction of the attending surveyor.
    (c) While the importance of each item may vary depending on the vessel and its specific requirements, below is a list of items that should be included in an underwater survey in order of importance:
    • Underwater Hull: General condition of the hull below the waterline.
    • Bottom and Shell Plating: Inspection for corrosion, damage, and fouling.
    • Shell Openings: Examination of openings such as sea chests, drain plugs, and overboard discharge points.
    • Stern Tube Oil Leaks: Check for leaks around the stern tube.
    • Propeller Blade: Inspection for damage, wear, and fouling.
    • Rudder: Inspection for damage, wear, and clearances.
    • Sea Chest Opening and Grating: Examination for blockages, damage, and fouling.
    • Anodes: Check the condition and effectiveness of cathodic protection anodes.
    • Bilge Keel: Inspection for damage and fouling.
    • Drain Plugs: Ensure all drain plugs are secure and in good condition.
    • Overboard Valve Openings: Check for proper operation and condition.
    • Forward Area: Inspection for any damage due to anchor and chain movement.
    Q4 (16 Marks) Hull Construction πŸ”₯ Repeated 10x

    Describe a method for the attachment of bilge keels. State THREE reasons for not extending bilge keels for the entire length of the vessel. Explain TWO principles of roll damping that bilge keels exploit.

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    Part (a)

    Method of Bilge keel attachment to the hull:

    Bilge keels are fitted port and starboard at the turn of the bilge. They do not extend outside the lines of the side and bottom shell. The bilge keels are attached directly in line with an internal stiffening member such as a girder or longitudinal. The bilge keel comprises a flat bar doubler welded directly to the shell, and an offset bulb plate (OBP) with 'scallops' cut in it is welded to the flat bar doubler. The ends of the bilge keels are tapered (minimum 3 in 1) and will end in line with transverse internal stiffening, such as a frame. Using a hull doubler protects the hull in case of damage to the keel, as the crack would not extend into the hull.

    Part (b)

    Bilge keels are not fitted for the full length of the vessel because:

    • The 'lever' to the ship's axis of rotation is reduced at the ends;
    • The ineffectiveness, the closer the bilge keel is to the rolling centre, increased resistance and more likelihood of damage.
    • The hydrodynamic effect would cause a large increase in resistance and fuel consumption;
    • At the aft end, the boundary layer is much thicker, and since the keel would not project through, it would have a much reduced effect.
    Part (c)

    Two principles of roll damping exploited by Bilge keels:

    Increased Roll Period:

    • Bilge keels increase the ship's roll period (the time it takes for the vessel to complete one roll). This is achieved by increasing the moment of inertia (K) of the vessel. A longer roll period means the vessel rolls more slowly, thus reducing the amplitude of the roll.

    $$T_{r}=2\pi\:\frac{k}{\sqrt{g\times GM}}$$

    Where:

    • g: acceleration due to gravity
    • GM: metacentric height
    • k: mass moment of inertia

    By increasing k, the ship's stability during roll improves.

    Hydrodynamic Resistance:

    • As the vessel rolls, the bilge keels move through the water, creating pressure differences. Water pressure on one side of the keel opposes the rolling motion, providing a damping effect. This is due to the interaction between the bilge keel and the water, creating hydrodynamic forces that counteract the rolling motion.
    Q5 (16 Marks) General πŸ”₯ Repeated 7x

    List SIX hazards that arise with the carriage of liquefied gas in bulk. Describe, with the aid of a sketch, the details of construction of a prismatic cargo tank within a gas carrier designed to carry liquefied gas (LPG)

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    Part (a)

    SIX hazards arising with the carriage of liquefied gas in bulk.

    1. Flammability / explosive hazard: the vapours are highly flammable and, mixed with air in the flammable range (between the lower and upper flammable limits), can be ignited by any ignition source causing fire or explosion.
    2. Extreme cold / cryogenic hazard: LPG is carried near -50 deg C boiling point; spillage causes severe cold burns (frostbite), embrittlement and fracture of steel structures and deck fittings, and condensation of moisture.
    3. Asphyxiation / toxicity: heavier-than-air gas displaces oxygen, causing oxygen deficiency and asphyxiation in enclosed and low-lying spaces; some gases are toxic or anaesthetic.
    4. Pressure hazard: liquefied gas develops high vapour pressure as temperature rises; over-pressure can rupture tanks, piping and relief valves; uncontrolled vapour release produces a BLEVE (boiling liquid expanding vapour explosion) risk.
    5. Heavy vapour cloud / dispersion risk: because the vapour is denser than air it flows along the deck and water in low patterns, travelling and finding ignition sources at a distance, and can accumulate in enclosed spaces, bilges and accommodation.
    6. Equipment/materials failure and corrosion: leakage from cargo piping, valves and flanges due to thermal contraction, corrosion and incompatible materials, plus the risk of liquid carry-over into pipelines and machinery spaces causing damage.

    (Also re-condensation, static electricity during loading). These require gas-tight equipment, gas detection, venting, cool-treatment and exclusion of ignition sources.

    Part (b)

    Construction of a prismatic cargo tank (free-standing prismatic tank) in a gas carrier (LPG).

    The tank is a prismatic (rectangular) self-supported tank built from cryogenic nickel-steel or aluminium plates, carried in the hull in a hold. Sketch: - a rectangular tank of flat or corrugated plate, supported on a mild-steel/chock support at the bottom (load-bearing), with vertical side hoppers and roofs, a dome at the top carrying cargo and pressure/vacuum valves, an access hatch and level gauging.

    Features:

    • It is designed as a self-supporting (type A or B) prismatic tank, adequately insulated externally (polyurethane foam panels) to keep the cargo cold and to reduce boil-off.
    • The tank bottom transfers loads through tank-support (wood/mild-steel) to seatings on the hull; the tank top is a dome through which pipes and vent lines pass. A drip tray / insulation and gas-tight secondary barrier is often fitted such that any leakage is contained.
    • The cargo is carried at atmospheric or slightly above pressure, and a relief valve (safety valve), pressure-vacuum valve, gas detection, and remote-operated valves are fitted.
    • The tank structure uses corrugated internal stiffening to withstand liquid sloshing and cryogenic temperature, with careful welded joints in cryogenic steel of notch-tough grade.
    • Venting systems, dry-pipe vent risers and flame arresters prevent accumulation of vapour; the tank is connected to the piping, with high-level and emergency shut-down (ESD) systems.

    The description should be accompanied by a sketch showing the tank seatings, insulation, dome, and secondary barrier. (Note: LPG carriers often use semi-pressurised or refrigerated prismatic or cylindrical tanks.)

    Q6 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 7x

    A ship of 15000 tonne displacement has an Admiralty Coefficient, based on shaft power, of 420. The mechanical efficiency of the machinery is 83%, shaft losses 6%, propeller efficiency 65% and QPC 0.71. At a particular speed the thrust power is 2550kW. Calculate: (16)

    (i) indicated power,

    (ii) effective power,

    (iii) ship speed.

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    Given:

    $$\Delta=15000t$$

    $$Shaft\:Power\:\left(SP\right)=420$$

    $$Transmission\:Efficiency=83\%$$

    $$Shaft\:losses=6\%$$

    $$Propeller\:Efficiency=65\%$$

    $$QPC=0.71$$

    $$Thrust\:Power=2550kW$$

    $$\left(\imaginaryI\right)\:Delivered\:Power\:\left(DP\right)=\frac{Thrust\:Power\:\left(TP\right)}{Propeller\:Efficiency\:\left(\eta P\right)}$$

    $$DP=\frac{2550}{0.65}$$

    $$DP=3923.07kW$$

    $$\left(ii\right)\:Shaft\:Power=\frac{Delivered\:Power\:\left(DP\right)}{Transmission\:Efficiency\:\left(\eta T\right)}\:$$

    $$SP=\frac{3923.07}{0.94}$$

    $$SP=4173.47kW$$

    $$\left(iii\right)\:Indicated\:Power=\frac{Shaft\:Power\:\left(SP\right)}{Mechanical\:Efficiency\:\left(\eta m\right)}$$

    $$IP=\frac{4173.47}{0.83}$$

    $$IP=5028.28kW$$

    $$\left(iv\right)\:Effective\:Power=DP\times QPC$$

    $$EP=3923.07\times0.71$$

    $$EP=2785.3797kW$$

    $$\left(v\right)\:Shaft\:Power=\frac{\Delta^{\frac23}\times V^3}{Admiralty\:Co-efficient}$$

    $$4173.47=\frac{15000^{\frac23}\times V^3}{420}$$

    $$V=14.23knots$$

    Q7 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Describe stability requirements for dry-docking. (6)

    (b) A ship of 8000t displacement floats upright in sea water, with KG= 7.6m, GM= 0.5m. A tank, whose kg is 0.6m above the keel and 3.5m from the center line, contains 100 t of water ballast. Neglecting the free surface effect, calculate the angle which the ship will heel, when the ballast water is pumped out. (10)

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    Part (a)

    For safe dry-docking, a ship must meet two key stability requirements:

    • Positive GM (Metacentric Height): The ship needs a positive GM. GM is the distance between the centre of gravity (G) and the metacentre (M). A positive GM indicates inherent stability; the ship will right itself if tilted. During dry-docking, the loss of buoyancy as the ship rests on the blocks reduces GM. Insufficient positive GM increases the risk of the ship heeling over or capsizing.
    • Trim by the stern: The vessel should be trimmed slightly by the stern (aft end lower than the bow) to ensure the aft end sits on the keel blocks first. This controlled settling minimises the risk of instability during the docking process. An even keel is generally preferred for the initial floating condition before the dry-docking procedure begins.
    Part (b)

    $$new \space KG \space = \space {{(8000 \times 7.6) - (100 \times 0.6)} \over 8000 - 100}$$

    $$New\:KG\:=\:7.689m$$

    $$New \space GM_1 \space = \space KM - KG$$

    $$= \space (7.6 + 0.5) - 7.689$$

    $$New\:GM\:=\:0.411m$$

    Angle of heel when 100t ballast is pumped out

    $$Tan\theta=\frac{m\times d}{\Delta GM}$$

    $$=\frac{100\times3.5}{7900\times0.411}$$

    $$Tan\theta=0.1077$$

    $$\theta=6^09^{^{\prime}}$$

    Q8 (16 Marks) Ship Stability πŸ”₯ Repeated 8x

    (a) Define longitudinal center of gravity (LCG) and longitudinal centre of buoyancy (LCB). (6)

    (b) The immersed cross-sectional areas of a ship 120 m long, commencing from aft are 2, 40, 79, 100, 103, 104, 104, 103, 97, 58 and 0 m2. Calculate: (10)

    (i) displacement

    (ii) longitudinal position of the centre of buoyancy.

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.
    Part (b)

    Given:

    $$Common \space interval \space (h) \space = \space {{L} \over h} \space = \space {{120} \over 10} \space = \space 12 $$

    Cross-sectional area

    SM

    Product of volume

    Lever

    Product of 1st moment

    2

    1

    2

    +5

    +10

    40

    4

    160

    +4

    +640

    79

    2

    158

    +3

    +474

    100

    4

    400

    +2

    +800

    103

    2

    206

    +1

    +206

    Ξ£MA = +2130

    104

    4

    416

    0

    0

    104

    2

    208

    -1

    -208

    103

    4

    412

    -2

    -824

    97

    2

    194

    -3

    -582

    58

    4

    232

    -4

    -928

    0

    1

    0

    -5

    0

    Ξ£βˆ‡ = 2388

    Ξ£MF = -2542

    $$Displacement \space = \space \rho \times {{h} \over 3} \times \sum βˆ‡ \space tonne $$

    $$=1.025\times{{12}\over3}\times2388$$

    $$Displacement \space = \space 9790.8 tonne$$

    Centre of buoyancy from midship (LCB)

    $$LCB\:=\:h\times({{\sum M_{A}+\sum M_{F}}\over\sum\nabla})$$

    $$=12\times({{2130-2542}\over2388})$$

    $$LCB \space = \space -2.07m fwd$$

    Q9 (16 Marks) Hull Construction πŸ”₯ Repeated 7x

    With respect to Buoyancy of a vessel:

    (a) What do you understand by reserve buoyancy what happen if the lost buoyancy is greater than the reserve buoyancy. (6)

    (b) A forward deep tank 12 m long extends from a longitudinal bulkhead to the ship's side. The widths of the tank surface measured from the longitudinal bulkhead at regular intervals are 10, 9, 7, 4 and 1 m. Calculate the second moment of area of the tank surface about a longitudinal axis passing through its centroid. (10)

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    Part (a)

    Reserve Buoyancy

    Reserve buoyancy is the watertight volume above the waterline of a vessel. It represents the potential buoyancy that can be utilised to keep the ship afloat when additional weight is added or if some buoyancy is lost due to damage (e.g., bilging).

    When a mass is added to the ship or buoyancy is lost (e.g., due to flooding of a compartment), the reserve buoyancy is converted into active buoyancy by an increase in draught.

    If the lost buoyancy (due to flooding or damage) is greater than the reserve buoyancy, the vessel will no longer have sufficient buoyant force to counteract its weight, causing it to sink.

    Part (b)
    Q10 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 6x

    A ship of length 140m, Breadth of 18.5m, draught of 8.1 m and a displacement of 17,025 tonnes in sea water, has a face pitch ratio of 0.673. The diameter of the propeller is 4.8m. The results of the speed trial show that true slip may be regarded as constant over a range of 9 to 13 knots and is 30%, w = 0.5Cb-0.05. If fuel used is 20t/day at 13 knots and fuel consumption/day varies as cube of speed of ship, determine the fuel consumption, when propeller runs at 110 pm. (16)

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    Given:

    $$Lenght,\:L=140m$$

    $$Breadth,\:B=18.5m$$

    $$Draught,\:d=8.1m$$

    $$Displacement,\:\Delta=17025tonnes$$

    $$Pitch\:ratio,\:p=0.673\operatorname{}$$

    $$Diameter\:of\:Propeller,\:D=4.8m$$

    $$\operatorname{Real\:Slip,\:R_{s}=30\%\:or\:0.3}$$

    $$Wake\:fraction,\:W=0.5C_{b}-0.05$$

    $$Consumption,\:C_2=\:20t\:per\:day\:$$

    $$Ship\:Speed,\:V_2=13\:knots$$

    $$\operatorname{Revolution,\:N}=110rpm$$

    $$cons\:per\:day\:\alpha\:V^3$$

    To find Fuel Consumption C1=?

    We know that,

    $$Displacement,\:\Delta=\nabla\times\rho$$

    $$17025=\nabla\times1.025$$

    $$\nabla=16609.76m^3$$

    $$Block\:Coefficient,\:C_{b}=\frac{\nabla}{L\times B\times d}$$

    $$C_{b}=\frac{16609.76}{140\times18.5\times8.1}$$

    $$C_{b}=0.792$$

    $$Wake\:Fraction,\:W=0.5C_{b}-0.05$$

    $$W=0.5\times0.792-0.05$$

    $$W=0.346$$

    $$Pitch\:ratio,\:p=\frac{P}{D}$$

    $$0.673=\frac{P}{4.8}$$

    $$P=4.8\times0.673$$

    $$P=3.23m$$

    $$Theoretical\:Speed,\:V_{t}=\frac{P\times N\times60}{1852}$$

    $$V_{t}=\frac{3.23\times110\times60}{1852}$$

    $$V_{t}=11.51knots$$

    Using, Real slip equation.

    $$\operatorname{\operatorname{Real\:Slip,\:R_{s}=\frac{V_{t}-V_{a}}{V_{t}}}}$$

    $$0.3=\frac{11.51-V_{a}}{11.51}$$

    $$V_{a}=11.51-11.51\times0.3$$

    $$V_{a}=11.51\left(1-0.3\right)$$

    $$V_{a}=11.51\times0.7$$

    $$V_{a}=8.057knots$$

    $$Wake\:fraction,\:W=\frac{V-V_{a}}{V}$$

    $$0.346=\frac{V-8.057}{V}$$

    $$0.346V=V-8.057$$

    $$V=\frac{8.057}{0.654}$$

    $$V=12.32knots$$

    $$cons\:per\:day\:\alpha\:V^3$$

    $$\frac{C_1}{C_2}=\left(\frac{V_1}{V_2}\right)^3$$

    $$\frac{C_1}{20}=\left(\frac{12.32}{13}\right)^3$$

    $$C_1=20\times\left(\frac{12.32}{13}\right)^3$$

    $$C_1=20\times0.851$$

    $$C_1=17.02t\:per\:day$$

    Q1 (16 Marks) Ship Stability πŸ”₯ Repeated 2x

    (a) Explain the six types of ship motions along different axes. Discuss the causes of these motions and their impact on the vessel’s stability and performance. (8)

    (b) Discuss the methods employed to reduce these motions and enhance the vessel’s stability and comfort. (8)

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    Part (a)

    The six ship motions along the three axes.

    A ship has six degrees of freedom, three translational and three rotational, about the three mutually perpendicular axes through the centre of gravity (longitudinal x, transverse y, vertical z):

    1. Surge - linear motion along the longitudinal (fore-and-aft) axis, caused by the variation of thrust and resistance, and by following/head seas pushing the hull. It affects speed and fuel consumption and can cause slamming of the bow in head seas.
    2. Sway - linear motion along the transverse axis, caused by beam seas, wind and rudder action; it produces lateral acceleration, cargo shift risk and can combine with roll.
    3. Heave - vertical linear motion along the vertical axis, caused by the passage of waves under the hull changing buoyancy; it produces vertical accelerations, slamming, and affects the effective GM and the comfort of crew and cargo.
    4. Roll - rotation about the longitudinal axis, caused by beam seas, wind, and the heeling moment of rudder or cargo; it is the most dangerous motion for stability because it can lead to large angles, cargo shift, and in resonance to parametric roll or capsize.
    5. Pitch - rotation about the transverse axis, caused by head and following seas; it produces bow/stern slamming, deck wetness, propeller emergence and affects speed and comfort.
    6. Yaw - rotation about the vertical axis, caused by asymmetric wave forces, wind and rudder; it affects course-keeping, increases resistance and can lead to broaching in following seas.

    Causes and impact: all motions are excited by wave forces, wind, and the ship's own propulsion; their amplitude depends on the encounter frequency relative to the ship's natural frequencies (resonance). They reduce stability margin (especially roll), increase structural loading (pitch/heave slamming), cause cargo damage and crew discomfort, and increase resistance and fuel consumption.

    Part (b)

    Methods to reduce these motions and enhance stability and comfort.

    • Increase GM (stiffen the ship) by lowering KG (ballast low, remove top weight) to reduce roll amplitude, but avoid excessive stiffness which gives short, uncomfortable periods.
    • Fit bilge keels, anti-roll tanks (passive/active U-tube tanks), fin stabilisers and gyro stabilisers to damp roll.
    • Use active/passive anti-pitching and anti-heave devices, and design the hull with fine ends and adequate freeboard to reduce slamming and deck wetness.
    • Operate at a speed and heading that avoids resonance (change course/speed to move the encounter frequency away from natural frequencies), and avoid beam seas for roll.
    • Use ballast and trim to improve stability and reduce motions; press up tanks to remove free surface.
    • Design with adequate GM, metacentric height and righting-lever (GZ) reserve, and use stabilisers, bilge keels and anti-rolling tanks; for comfort, reduce vertical accelerations by speed reduction and course changes.
    • Structural measures: adequate freeboard, flare, and bow form to reduce slamming; and operational measures such as weather routing.
    Q2 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 13x

    Describe the relationship between frictional resistance and

    (a) Ship speed. (4)

    (b) The wetted surface area. (4)

    (c) The surface roughness. (4)

    (d) The length of the vessel. (4)

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    Frictional resistance arises due to the interaction between the ship's hull and the water as the vessel moves. This resistance is caused by eddying water adjacent to the hull that is drawn along with the ship. The frictional resistance is given by:

    $$R_{f\:=\:}f\:.\:s\:.\:v^{n}$$

    Where:

    • f = Coefficient of friction
    • s = Wetted surface area
    • v = Ship's speed in knots
    • n = Constant (1.82)
    Part (a)

    Ship Speed (v):

    • Frictional resistance increases with increasing ship speed. The relationship is not perfectly linear, but it's approximately described by the formula Rf = fsv^n (Rf ∝ v^n), where 'n' is a constant (approximately 1.82). This means that a small increase in speed results in a larger increase in frictional resistance.
    Part (b)

    Wetted Area (s):

    • Frictional resistance is directly proportional to the wetted surface area (Rf ∝ s). A larger wetted area (the part of the hull submerged in water) leads to greater frictional resistance. The wetted area increases with the ship's draught (depth in the water).
    Part (c)

    Surface Roughness:

    • Increased surface roughness increases frictional resistance. Roughness can be caused by fouling (marine growth), corrosion, or poor hull maintenance. Regular cleaning and maintenance of the hull surface help to minimise roughness and reduce resistance.
    Part (d)

    Length of the Vessel

    • Frictional resistance is influenced by the vessel's length. As the length increases, the wetted surface area grows, and the coefficient of friction (f) changes, leading to an increase in Rf. Longer vessels experience higher frictional resistance, but streamlined designs can mitigate the effect to some extent.
    Q3 (16 Marks) Hull Construction

    During a voyage, sea water is found leaking into the engine room through a small hole in the ship side plating. What type of repairs will you carry out as an immediate action as a second engineer? What are the other actions and reporting procedures undertaken by the vessel to comply with class and flag requirements and procedures? How will you ensure that prompt and thorough repairs are undertaken as a permanent measure? (16)

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    Immediate, reporting and permanent repair actions for a small hole in the engine-room side plating.

    Part (a)

    Immediate action (as second engineer):

    • Raise the alarm and inform the bridge/chief engineer; stop or reduce the flooding by closing the sea inlet/outlet valves in the vicinity if the hole is near a sea chest, and isolate the affected area.
    • Attempt to stop or reduce the inflow: apply a temporary patch - a softwood plug, a wooden wedge, a collision mat, a steel plate patch with a rubber gasket held by a strongback and wedges, or a cement box (a wooden box filled with quick-setting cement) over the hole. Use a shore/strongback and wedges to hold the patch against the pressure.
    • Pump out the water using the bilge and ballast pumps, and use the emergency bilge suction; keep the water level down and monitor the bilge.
    • Reduce the pressure on the hole by trimming the ship (e.g. ballast to raise the damaged side or reduce the head) and by reducing speed to lower the dynamic pressure.
    • Isolate the compartment, close watertight doors, and keep the engine-room watch; monitor the level and the patch continuously.
    • If the hole is small and accessible, a temporary welded patch or a bolted patch with a gasket can be fitted once the area is dried and gas-free.
    Part (b)

    Other actions and reporting to comply with class and flag requirements:

    • Report the incident to the master, who reports to the company (DPA/technical superintendent) and, as required, to the flag state and the coastal state (if in territorial waters) and to the port state on arrival.
    • Make a full entry in the deck and engine log books, and record the damage, the temporary repair, and the water ingress.
    • Notify the classification society (class) of the damage; class will require a survey of the damage and the temporary repair, and will issue conditions of class or a recommendation for a permanent repair before the ship proceeds or within a specified time.
    • The ship should proceed to a port of refuge/repair as directed, and the temporary repair must be maintained and monitored during the voyage.
    • Comply with the ISM code (reporting, non-conformity, corrective action), and with the flag state's requirements for reporting marine casualties/incidents.
    Part (c)

    Ensuring prompt and thorough permanent repair:

    • Arrange for the permanent repair at a suitable repair yard or by an approved repairer, with the damaged plate renewed (renewal of the shell plate or a welded insert plate of the correct grade and thickness), the surrounding structure inspected for corrosion/fatigue, and the weld carried out to class-approved procedures by certified welders.
    • The repair must be surveyed and approved by the classification society surveyor, with NDT (ultrasonic, dye-penetrant) of the welds as required, and the plate thickness and material certified.
    • After repair, carry out a watertightness test (hose test or air test) and restore the protective coating (paint/antifouling) and any insulation.
    • Update the class records and the ship's drawings, and close out the condition of class; the permanent repair should be completed before the ship returns to unrestricted service, and the temporary patch removed.
    Q4 (16 Marks) General πŸ”₯ Repeated 3x

    (a) List SIX hazards associated with the carriage of liquefied gas in bulk. (8)

    (b) Sketch and describe the details of construction of a free-standing prismatic tank within a gas carrier designed to carry liquefied gas (LPG). (8)

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    Part (a)

    SIX hazards associated with the carriage of liquefied gas in bulk.

    1. Flammability/explosion: the vapour is highly flammable; in the flammable range with air it can be ignited by any ignition source, causing fire or explosion.
    2. Cryogenic (extreme cold) hazard: LPG is carried near -50 deg C; spillage causes severe cold burns, embrittlement and fracture of steel, and condensation of moisture.
    3. Asphyxiation/toxicity: heavier-than-air vapour displaces oxygen causing oxygen deficiency and asphyxiation in enclosed and low spaces; some gases are toxic.
    4. Pressure hazard: high vapour pressure with temperature rise can over-pressure tanks and piping; uncontrolled release can cause BLEVE (boiling liquid expanding vapour explosion).
    5. Heavy vapour cloud dispersion: the vapour is denser than air, flows along deck/water, travels to distant ignition sources and accumulates in enclosed spaces, bilges and accommodation.
    6. Equipment/materials failure and corrosion: leakage from piping, valves and flanges due to thermal contraction, corrosion and incompatible materials, plus liquid carry-over into machinery spaces causing damage.

    These require gas-tight equipment, gas detection, venting, exclusion of ignition sources and careful temperature/pressure control.

    Part (b)

    Construction of a free-standing prismatic tank within a gas carrier (LPG).

    A free-standing prismatic tank is a self-supporting rectangular tank built of cryogenic nickel-steel or aluminium, carried in a hold and supported on the hull. Sketch: a rectangular tank with flat or corrugated plates, a top dome carrying the cargo piping, safety and pressure/vacuum valves, access hatch and level gauging; the tank is supported on load-bearing seatings (wood/mild-steel chocks) at the bottom, with side hoppers and a roof, and is surrounded by insulation (polyurethane foam) to keep the cargo cold and reduce boil-off.

    Features:

    • Self-supporting (type A or B) prismatic tank, adequately insulated externally; a secondary barrier (drip tray/gas-tight) is fitted so any leakage is contained.
    • The tank bottom transfers loads through seatings to the hull; the tank top dome carries the cargo pipes, relief valve, pressure-vacuum valve, gas detection and remote-operated valves.
    • The tank is designed to withstand liquid sloshing and cryogenic temperature, with corrugated internal stiffening and notch-tough welded joints.
    • Venting systems, dry-pipe vent risers and flame arresters prevent vapour accumulation; high-level and emergency shut-down (ESD) systems are fitted.
    • The cargo is carried at atmospheric or slightly above pressure; a relief valve protects against over-pressure.

    The sketch should show the tank seatings, insulation, dome, and secondary barrier.

    Q5 (16 Marks) Hull Construction πŸ”₯ Repeated 7x

    (a) With reference to fatigue of hull structures explain the influence of stress level and cyclical frequency on expected operating life. (6)

    (b) Explain the influence of material defects on the safe operating life of forged components of stern fittings. (5)

    (c) State the factors which influence the possibility of fatigue cracking of a bed-plate transverse girder and explain how the risk of such cracking can be minimized. (5)

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    Q6 (16 Marks) Ship Stability πŸ”₯ Repeated 2x

    (a) List the variables which affect the force on a rudder. (6)

    (b) A triangular bulkhead is 7 m wide at the top and has a vertical depth of 8 m. Calculate the load on the bulkhead and the position of centre of pressure if the bulkhead is flooded with sea water on only side.

    (i) to the top edge

    (ii) with 4 m head to the top edge. (10)

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    when a rudder is turned from the centreline plane, to any angle, a force acts on the rudder, given by,

    $$F=kAV^2N$$

    Where,

    • k = Constant
    • A = Area of rudder in m2
    • V = Ship's speed in m/s

    The constant is dependent on size of rudder, rudder angle and density of water.

    Variables affecting force on rudder:

    • Rudder angle
    • Density of water
    • Ship speed

    These are the three variables, while all other are constant parameters i.e.

    • Area of rudder
    • Size of rudder.
    Q7 (16 Marks) Ship Stability

    (a) Explain the effect on GM during the filing of a double – bottom tank. (6)

    (b) The length of a ship is 18 times the draught. while the breadth is 2.1 times the draft. At the load water plane, the water plane area co-efficient is 0.83 and the difference between the TPC in sea water and the TPC in fresh water is 0.7. Determine the length of the ship and TPC in fresh water.

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    Part (a)

    Effect on GM during the filling of a double-bottom tank.

    As a double-bottom tank is filled, two effects act on GM:

    • The added weight of ballast is low in the ship (deep in the double bottom), so it lowers the centre of gravity, which increases GM (stiffening the ship). This is the dominant effect once the tank is full.
    • While the tank is partially full, a free surface exists and produces a free-surface effect that reduces the effective GM. The free-surface correction is rho x i / Delta, where i is the second moment of area of the free surface about its longitudinal axis (i = L B^3/12). For a wide double-bottom tank this can be significant.

    Hence during filling, the effective GM first falls (as the free surface develops) and then rises as the tank approaches full and the free surface disappears; when the tank is completely full (pressed up) the free-surface effect is zero and the GM is increased by the low weight. The net effect of a full double-bottom tank is an increase in GM (stiffer ship), but the transient free-surface loss during filling must be watched, especially in a tender ship.

    Part (b)

    Length of ship and TPC in fresh water.

    Given: length L = 18 x draught d; breadth B = 2.1 x d; waterplane area coefficient Cw = 0.83; difference between TPC in sea water and TPC in fresh water = 0.7.

    TPC = (waterplane area x density)/100. In sea water TPCsw = A x 1.025/100; in fresh water TPCfw = A x 1.000/100.

    Difference = A(1.025 - 1.000)/100 = A x 0.025/100 = 0.7.

    So A = 0.7 x 100/0.025 = 2800 m2.

    Cw = A/(L x B) = 0.83, so L x B = 2800/0.83 = 3373.5 m2.

    But L = 18d and B = 2.1d, so L x B = 18d x 2.1d = 37.8 d2 = 3373.5.

    d2 = 3373.5/37.8 = 89.25, so d = 9.45 m.

    L = 18 x 9.45 = 170.1 m.

    TPC in fresh water = A/100 = 2800/100 = 28.0 t/cm.

    Answer: length of ship about 170 m; TPC in fresh water = 28.0 t/cm.

    Q8 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 7x

    (a) What is meant by the Admiralty Coefficient and the Fuel Coefficient? (6)

    (b) A ship of 14900 tonne displacement has a shaft power of 4460 kW at 14.55 knots. The shaft power is reduced to 4120 kW and the fuel consumption at the same displacement is 541 kg/h. Calculate the fuel coefficient for the ship. (10)

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    The Admiralty Coefficient (C) is a method for estimating the propulsion power needed for a newly built ship. It's considered relatively constant for a given ship design. The formula is:

    $$C=\frac{\Delta^{2/3}\times V^3}{BP}$$

    Where:

    • C = Admiralty Coefficient
    • Ξ” = Displacement in tonnes (weight of the ship when fully loaded)
    • V = Speed in knots
    • BP = Brake power in kilowatts (power delivered by the ship's engine)

    A higher Admiralty Coefficient indicates a more efficient ship design, meaning it requires less power to achieve a given speed. Values typically range from 350 to 600.

    Fuel Coefficient:

    The Fuel Coefficient (F.C.) is used to calculate a ship's daily fuel oil consumption. The formula is:

    $$Daily\:fuel\:oil\:consumption\:=\:\frac{\Delta^{2/3}\times V^3}{FC}$$

    Where:

    • F.C. = Fuel Coefficient
    • Ξ” = Displacement in tonnes
    • V = Speed in knots

    The Fuel Coefficient can vary significantly, with typical values ranging from 40,000 to 120,000. A higher Fuel Coefficient implies greater fuel efficiency (lower daily fuel consumption) for a given speed and displacement.

    Part (b)

    $$admiraty\:coefficient\:\left(C\right)=\:\frac{\Delta^{2/3}V^3}{Shaft\:power}=\frac{\Delta^{\frac23}\times V^3}{SP}$$

    $$\frac{SP_1}{SP_2}=\frac{V_1^3}{V_2^3}$$

    $$\frac{4460}{4120}=\frac{14.55^3}{V_2^3}$$

    $$V_2=14.17kntos$$

    $$Fuel\:consumption\:per\:hour=541\operatorname{\mathrm{\:kg}}\:per\:hour$$

    $$Fuel\:consumption\:per\:day\:=\:541\times24=12.98t\:per\:day$$

    $$Fuel\:coefficient=\frac{\Delta^{\frac23}\times V_2^3}{Fuel\:consumption\:per\:day}$$

    $$=\:\frac{14900^{\frac23}\times14.17^3}{12.98}$$

    $$=132726.9$$

    Q9 (16 Marks) Ship Stability πŸ”₯ Repeated 2x

    (a) Describe the stability requirements of a ship for dry-docking. (6)

    (b) The Β½ ordinates of a waterplane at 15m intervals, commencing from aft, are 1, 7, 10.5, 11, 11, 10.5, 8, 4 and 0m. Calculate: (10)

    (i) TPC.

    (ii) Distance of the centre of flotation from midships.

    (iii) Second moment of area of the waterplane about a transverse axis through the centre of flotation.

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    Part (a)

    Stability requirements for dry-docking.

    Before docking, the ship must be stable with an adequate GM, and the docking operation must not cause a list or loss of stability. As the water is pumped out, the keel-block reaction reduces the effective buoyancy, lowering KB and BM and hence GM; the ship must retain sufficient GM throughout so that it does not heel over on the blocks. Requirements:

    • Adequate initial GM and even keel (or slight trim), with no list; ballast arranged to give even keel and to press up or empty tanks to avoid free surface.
    • The docking weight, draft and trim must be such that the keel blocks contact evenly over the full keel length; excessive trim overloads the aft blocks.
    • The ship must be central over the keel line and the dock level; the reaction builds gradually and the GM must remain positive at all stages.
    • No weights moved during docking; the stability data (hydrostatics, docking plan) used to check the condition.

    If the GM becomes too small or negative during docking, the vessel can heel and capsize on the blocks, so a stability margin is insisted on.

    Part (b)

    Waterplane calculations.

    The half-ordinates of a waterplane at 15 m intervals, commencing from aft, are 1, 7, 10.5, 11, 11, 10.5, 8, 4 and 0 m. There are 9 ordinates, so the waterplane length is 8 x 15 = 120 m.

    (i) TPC.

    Area A = 2 x (h/3)[y0 + y8 + 4(y1+y3+y5+y7) + 2(y2+y4+y6)]

    = 2 x (15/3)[1 + 0 + 4(7+11+10.5+4) + 2(10.5+11+8)]

    = 10[1 + 4x32.5 + 2x29.5] = 10[1 + 130 + 59] = 10 x 190 = 1900 m2.

    TPC (salt) = A x 1.025/100 = 1900 x 1.025/100 = 19.48 t/cm.

    (ii) Distance of the centre of flotation from midships.

    First moment of area about the after perpendicular:

    M = 2 x (h/3) x sum of (weighted y x distance). Using station distances 0,15,30,...,120 m:

    M = 2 x (15/3)[1x0 + 7x15 + 10.5x30 + 11x45 + 11x60 + 10.5x75 + 8x90 + 4x105 + 0x120] with Simpson weights (1,4,2,4,2,4,2,4,1):

    weighted sum = 1x0 + 4x(7x15) + 2x(10.5x30) + 4x(11x45) + 2x(11x60) + 4x(10.5x75) + 2x(8x90) + 4x(4x105) + 1x0

    = 0 + 4x105 + 2x315 + 4x495 + 2x660 + 4x787.5 + 2x720 + 4x420 + 0

    = 420 + 630 + 1980 + 1320 + 3150 + 1440 + 1680 = 10620.

    M = 2 x 5 x 10620 = 106,200 m3.

    Distance of centroid from AP = M/A = 106,200/1900 = 55.89 m. Midships is at 60 m from AP, so the centre of flotation is 60 - 55.89 = 4.11 m aft of midships.

    (iii) Second moment of area about a transverse axis through the centre of flotation.

    Second moment about AP: I_AP = 2 x (h/3) x sum of (weighted y x distance^2):

    weighted sum = 1x0 + 4x(7x225) + 2x(10.5x900) + 4x(11x2025) + 2x(11x3600) + 4x(10.5x5625) + 2x(8x8100) + 4x(4x11025) + 1x0

    = 0 + 4x1575 + 2x9450 + 4x22275 + 2x39600 + 4x59062.5 + 2x64800 + 4x44100 + 0

    = 6300 + 18900 + 89100 + 79200 + 236250 + 129600 + 176400 = 735,750.

    I_AP = 2 x 5 x 735,750 = 7,357,500 m4.

    Transfer to the centre of flotation: I_CF = I_AP - A x (distance from AP to CF)^2 = 7,357,500 - 1900 x (55.89)^2 = 7,357,500 - 1900 x 3123.7 = 7,357,500 - 5,935,000 = 1,422,500 m4.

    Answer: TPC = 19.48 t/cm; LCF = 4.11 m aft of midships; I about the transverse axis through CF = about 1.42 x 10^6 m4.

    Q10 (16 Marks) Ship Types & Design πŸ”₯ Repeated 2x

    (a) Describe the effect of cavitations on the propeller blades (6)

    (b) The following data are available from the hydrostatic curves of a vessel.

    Draught (m)

    KB (m)

    KM (m)

    I (m^4)

    4.9

    2.49

    10.73

    65250

    5.2

    2.61

    10.79

    68868

    Calculate the TPC at a draught of 5.05m. (10)

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    Part (a)

    Effect of cavitation on propeller blades:

    Erosion:

    • Cavitation causes the formation and collapse of vapor bubbles on the propeller blade surface.
    • The collapse of these bubbles produces high-pressure shockwaves and microjets that erode the blade material, leading to surface pitting and progressive damage.

    Vibration:

    • Uneven cavitation across the blades leads to imbalanced forces, causing vibrations in the propeller and the ship.
    • These vibrations can reduce the comfort of passengers and crew and stress the ship's structural components.

    Noise:

    • The collapse of vapor bubbles generates loud noise, which can interfere with onboard communication and underwater sonar systems.
    • This noise is a significant concern for naval vessels and marine life.

    Reduced Performance:

    • Cavitation reduces the efficiency of the propeller by causing loss of thrust and torque.
    • The presence of cavitation decreases the propeller’s ability to convert rotational energy into forward motion, lowering the ship's speed and increasing fuel consumption.
    Q1 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 3x

    State how and why the following machinery items are affected when the maximum service speed of a vessel is consistently maintained in heavy weather.

    (a) intermediate shafting.

    (b) propeller shafting.

    (c) shafting coupling bolts.

    (d) Main thrust pads.

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    Part (a)

    Intermediate Shafting:

    • Torsional Stress: Resulting from the rapid changes in propeller speed (propeller racing) caused by variations in water resistance in heavy weather.
    • Compressive Stress: Due to the end thrust transmitted from the propeller.
    • Bending Stress: Caused by hull movements such as hogging (upward bending) and sagging (downward bending) in heavy seas.
    Part (b)

    Propeller Shafting:

    • Torque and Thrust: Caused by the propulsion forces transmitted through the shaft.
    • Torsional Stress: Due to the rapid fluctuations in engine speed when the propeller races.
    • Compressive Stress: Generated by the axial thrust from the propeller.
    • Bending Stress: Arises when the weight of the propeller acts on the shaft as the propeller emerges from the water during rough seas.
    Part (c)

    Shafting Coupling Bolts:

    • Bending Stress: From misalignments caused by hull deformations.
    • Shear Stress: Resulting from the whirling of the shaft and rapid engine speed changes during propeller racing.
    • Torsional Stress: Caused by the transmission of fluctuating torque.
    • Fatigue Failure: Due to the repeated application of fluctuating loads over time, particularly in heavy weather conditions.
    Part (d)

    Main Thrust Pads:

    • Stress from Hull Movements: Misalignment caused by hull hogging and sagging.
    • Load Fluctuations: Due to variations in propeller thrust during racing and rapid changes in sea conditions.
    • Axial and Torsional Vibration: Arising from inconsistent propulsion forces and shaft vibrations.
    • Surface Wear and Damage: Resulting from increased pressure and friction due to fluctuating thrust forces.
    Q2 (16 Marks) Ship Stability πŸ”₯ Repeated 2x

    (a) Describe the arrangement made in a main structural bulkhead for a watertight door aperture.

    (b) Explain a procedure for ensuring that sliding watertight doors are operated safely.

    (c) Differentiate between the category of watertight door and state the regulation pertaining to each type.

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    Part (a)

    Arrangement made in a main structural bulkhead for a watertight door aperture.

    To preserve the strength and watertight integrity of a main transverse bulkhead where a watertight door is fitted, the opening is bounded by a reinforced double-plate frame. Thicker compensating plates (or a vertical web frame) surround the aperture, with the door frame pieces doubling the plate thickness at the corners. The door opening edges are connected to strong horizontal and vertical stiffeners (intercostal framing) so that the bulkhead's ability to transmit shear and bending (it acts as a deep girder stiffening the hull) is not lost. The door itself is a steel watertight sliding or hinged door with a frame fitted with a resilient gasket/special joint; the edges are dressed and wired to the surrounding doubling plates, and the sill is raised above the deck. The framing around the opening is tied into the top and bottom stools/stringers, and where the aperture cuts the stiffening of the bulkhead, compensation is made by increased scantlings immediately adjacent.

    Part (b)

    Procedure for safely operating sliding watertight doors.

    • Only authorised personnel may operate them, and the closing mechanism (hydraulic or electric, power plus manual) must be tested before entering port/complex operations.
    • Before closing, check the doorway and its rails/wheels are clear of cargo, stores, obstructions and personnel; sound a warning (alarm) before closing.
    • The door must be fully open during cargo/engineering operations that require passage, and closed when the door is not in use, especially at sea when subdivision demands it.
    • On closing (or opening) by power, monitor the travel; never operate a sliding door manually against a heavy load or with persons in the way.
    • When a remote/central control is fitted, the interlock and indicator (open/closed/partly open) must be observed, and the action of the automatic watertight-closing (if fitted) verified.
    • After closing, check the door is fully shut and the securing (wedges/dogging) engaged; the indicator should confirm closed position.
    • Ensure the gasket is intact, the surrounding plates free of obstruction, and that the door is not used as a normal gangway when closed.
    • Record routine tests and keep the operating instruction posted.
    Part (c)

    Categories of watertight door and relevant regulations.

    Watertight doors are categorised (per SOLAS Chapter II-1) by their type and location:

    • Class 1 hinged watertight doors, fitted in bulkheads in machinery spaces and working spaces, normally operated manually (may be required to be closed before the vessel proceeds to sea).
    • Class 2 sliding watertight doors, operated by power (remote) with local manual operation and capable of being closed from an accessible location above the bulkhead deck; these are fitted where frequent access is needed (e.g. main lobbies, machinery space trunks, cargo spaces in passenger ships).
    • Class 3 hinged or sliding door of a certain design in the lower portion of the hull in specific locations.

    Regulation (SOLAS II-1/13 and Reg.22 etc.): watertight doors in subdivision bulkheads below the bulkhead deck, and their closing arrangements, must comply with the regulations; all watertight doors below the bulkhead deck must be kept closed except when actually used for passage, and sliding doors fitted with power closing, voice alarm and remote indicator are required in certain fire/passenger contexts. The number of watertight doors below the bulkhead deck is minimised, they must have a status indicator at a control station, and hinges/operating gear must be such that they can be closed even with the vessel heeled. They must be capable of being opened and closed from both sides and remotely.

    Q3 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 4x

    (a) Describe the preparation necessary before the application (in dry dock) of sophisticated or approved long life coating to the underwater surface of the hull.

    (b) State the significance of the roughness profile.

    (c) List the different sophisticated coating which are available.

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    The preparation of a ship's underwater hull before applying a long-life coating in a dry dock involves a three-step process. This process addresses the removal of contaminants and the creation of a suitable surface profile.

    (i) Washing: The hull surface must be thoroughly cleaned to remove all marine growth (algae, slime, etc.), accumulated salts, dirt, grease, and oil. High-pressure freshwater washing is the standard method for this initial cleaning. The goal is to present a clean substrate for subsequent stages.

    (ii) Blasting: Abrasive blasting is the preferred method for removing rust, defective paint, and any remaining contaminants. This process achieves a bare metal surface, essential for proper adhesion of the new coating. The extent of blasting (localized or full hull) depends on the condition of the existing surface. The intensity and type of abrasive used are carefully controlled to achieve the desired surface roughness profile.

    (iii) Primer Application: After blasting, the surface is again cleaned to remove any blasting debris. A primer coat is then applied to provide corrosion protection and to create an ideal surface for the subsequent topcoat adhesion. This primer acts as an intermediary layer, enhancing the bond between the substrate and the long-life coating system.

    Part (a)

    Significance of Roughness Profile:

    The roughness profile of the prepared hull surface impacts the performance of the applied coating and the overall operational efficiency of the vessel. A rough surface increases frictional resistance as the vessel moves through the water. This increased drag translates to higher power requirements for propulsion, leading to increased fuel consumption and operational costs. Furthermore, greater surface roughness contributes to increased carbon emissions, a concern under current MARPOL regulations. Therefore, a controlled and optimized roughness profile is essential for minimizing frictional resistance, reducing fuel consumption and emissions, and maximizing the longevity of the hull coating.

    Part (b)

    Sophisticated hull coating systems comprise multiple layers designed to provide corrosion protection and antifouling properties.

    Wash Primer/Pretreatment Primer/Metal Conditioning Primer:

    • These primers act as a base layer, improving adhesion of subsequent layers. Common types include epoxy primers pigmented with iron oxide and corrosion inhibiting pigments (zinc and calcium phosphates, although zinc content is minimized due to safety concerns).

    Anticorrosive Coating:

    • This layer primarily provides corrosion protection to the underlying metal. Two-component epoxies, coal tar epoxies, and epoxy or polyester coatings incorporating glass flakes are frequently employed. Glass flakes enhance mechanical strength and water vapor impermeability.

    Antifouling Coating:

    • This layer prevents the attachment of marine organisms (fouling). Historically, tin-based paints were used, but due to environmental regulations, they have been largely replaced by copper-based, silicone-based, or non-TBT (Tributyltin) self-polishing antifouling coatings. These newer coatings typically use seawater-soluble polymers. The number of antifouling layers applied (two or three) depends on the specific system chosen and required longevity.
    Q4 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 9x

    With reference to Underwater Inspection in Lieu of Dry docking (UWILD)

    (a) Explain in detail, how an underwater survey is carried out.

    (b) State the requirements to be fulfilled before an underwater survey is acceptable to the surveying authority.

    (c) Construct a list of the items in order of importance that the underwater survey authority should include.

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    (a) An underwater survey, also known as Underwater Inspection in Lieu of Dry Docking (UWILD), involves a systematic and detailed examination of a vessel’s hull and underwater components while the ship remains afloat. The procedure includes the following steps:
    • The shipowner sends a request to the classification society surveyor, indicating the intention to perform an underwater survey.
    • A detailed plan of the ship's external hull features is submitted, showing the location of frames, bulkheads, welding lines, openings, etc.
    • The hull surface is cleaned before the survey to remove any marine growth or fouling that could obscure the inspection.
    • A diving company, approved by the classification society, is appointed to conduct the underwater inspection.
  • A self-propelled survey vehicle equipped with the following tools is used:
    • Long-Range Light TV Camera to aid in steering and checking hull deterioration.
    • High-resolution colour TV Camera to provide a close-up view of the hull coating and welded seams.
    • 35mm Still Camera to capture still images.
    • Ultrasonic Probe for measuring plate thickness.
    • Depth Meter and Speed Indicator to provide accurate data on the vehicle's depth and movement.
    • Umbilical Cable to connect the survey vehicle to the survey boat, transmitting power and information.
  • The survey boat is to be equipped with:
    • A control console with TV monitors.
    • Plate thickness printouts.
    • Audio and video cassette recorders.
    • Playback units.
    • Diver communication systems.
    • Vehicle control systems and associated instruments.
  • Operation:
    • The survey vehicle is taken underwater by a diver to the survey starting point.
    • Using TV monitors and shell expansion plans as guides, the vehicle is navigated over the hull, focusing on the bottom structure, sides, stern frame, rudder, propeller, bilge keel, and hull openings.
    • All images, data, and information are recorded and transmitted back to the survey boat.
    • Detailed pictures of the stern frame, rudder, propeller, bilge keel, and hull openings are captured.
    • Divers are deployed to measure stern tube bearing wear, pintle clearance, and inspect stern seals, anodes, and rudder stock palm coupling bolts.
    • All recorded video and audio, including conversations between the surveyor and drivers, thickness printouts, measurements, and pictures are analyzed to determine the detailed underwater condition of the vessel.
    (b) Before an underwater survey is accepted by the surveying authority, the following conditions must be met:

    The vessel's owner submits a request to the surveyor, including:
    • The proposed date and location for the survey.
    • General information about the diving company.
    • A declaration that the vessel has not suffered any damage due to grounding, collision, or other incidents.
    The ship's master or owner’s representative must provide a declaration confirming:
    • Any suspected or actual damage to the hull since the last dry-docking.
    • The underwater portion of the hull is protected by a suitable paint scheme that is of adequate thickness and remains valid until the next dry-dock.

    • The survey site should be in a protected area with calm and clear water, ensuring good underwater visibility. Attention must be given to the effects of currents and tides.
    • The hull must be clean for the external survey. The surveyor must be satisfied with the method and quality of the pictorial presentation, ensuring that it provides a reliable assessment of the hull's condition.
    • The underwater examination should be conducted by an approved diving company using closed-circuit TV and two-way communication, which can be monitored by the surveyor.
    • The vessel should be in as light an operating condition as possible to facilitate the survey.
    • Means must be available for the surveyor to examine the outside shell plating above the waterline.
    • Any required repairs identified during the survey must be carried out to the satisfaction of the attending surveyor.
    (c) While the importance of each item may vary depending on the vessel and its specific requirements, below is a list of items that should be included in an underwater survey in order of importance:
    • Underwater Hull: General condition of the hull below the waterline.
    • Bottom and Shell Plating: Inspection for corrosion, damage, and fouling.
    • Shell Openings: Examination of openings such as sea chests, drain plugs, and overboard discharge points.
    • Stern Tube Oil Leaks: Check for leaks around the stern tube.
    • Propeller Blade: Inspection for damage, wear, and fouling.
    • Rudder: Inspection for damage, wear, and clearances.
    • Sea Chest Opening and Grating: Examination for blockages, damage, and fouling.
    • Anodes: Check the condition and effectiveness of cathodic protection anodes.
    • Bilge Keel: Inspection for damage and fouling.
    • Drain Plugs: Ensure all drain plugs are secure and in good condition.
    • Overboard Valve Openings: Check for proper operation and condition.
    • Forward Area: Inspection for any damage due to anchor and chain movement.
    Q5 (16 Marks) General πŸ”₯ Repeated 2x

    With regard to the carriage of crude oil and its associated products:

    (a) (i) State the dangers involved.

    (ii) State what publications give guidance on safety.

    (b) Sketch and describe the operation of an explosimeter suitable for testing pump rooms or tank.

    (c) Define the terms lower and upper flammable limits illustrating your answer by means of a rough sketch of a hydrocarbon vapour oxygen graph.

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    Part (a)

    (i) Dangers involved in the carriage of crude oil and its products.

    • Fire and explosion: the vapours are highly flammable; in the flammable range with air they can be ignited by any ignition source (static, sparks, hot surfaces, smoking, electrical), causing fire or explosion.
    • Toxicity and health hazard: crude oil and its vapours are toxic; inhalation causes dizziness, nausea, asphyxiation and long-term health damage; skin contact causes irritation and dermatitis.
    • Asphyxiation: vapours displace oxygen in enclosed spaces (tanks, pump rooms), causing oxygen deficiency.
    • Environmental pollution: spillage causes marine pollution, requiring containment and clean-up.
    • Corrosion and material damage: crude oil and its sulphur content cause corrosion of tanks and piping.
    • Static electricity and sloshing: during loading/cleaning, static charge can build up and ignite vapour.
    Part (a)

    (ii) Publications giving guidance on safety.

    • The International Safety Guide for Oil Tankers and Terminals (ISGOTT).
    • The International Code for the Construction and Equipment of Ships Carrying Dangerous Chemicals in Bulk (IBC Code) and the MARPOL Convention.
    • The International Code of Safety for Ships Carrying Oil in Bulk (MARPOL Annex I).
    • The ship's own Safety Management System (SMS) and the Cargo Manual, plus the International Maritime Dangerous Goods (IMDG) Code for packaged products.
    • Company procedures and the tanker safety checklists (e.g. OCIMF Ship/Shore Safety Checklist).
    Part (b)

    Explosimeter (combustible gas indicator) for testing pump rooms or tanks.

    An explosimeter measures the concentration of flammable vapour in air as a percentage of the lower flammable limit (LFL). Sketch: a hand-held instrument with a sampling tube, a pump (aspirator bulb or motor), a meter calibrated 0-100% LFL, and a sensing element (a heated platinum filament or catalytic bead) in a chamber.

    Operation: the sample is drawn through the sampling tube into the instrument. The gas passes over a heated catalytic filament; if flammable vapour is present it burns on the filament, raising its temperature and changing its electrical resistance. This unbalances a Wheatstone bridge, producing a meter reading proportional to the vapour concentration, displayed as a percentage of the LFL. The instrument is calibrated against a known gas (e.g. pentane) and gives a reading; a reading above about 20-25% LFL indicates a dangerous condition and entry is prohibited. The instrument must be intrinsically safe (approved for use in flammable atmospheres) and used with a suitable sampling line lowered into the tank/pump room.

    Part (c)

    Lower and upper flammable limits, with a hydrocarbon vapour-oxygen graph.

    The lower flammable limit (LFL) is the minimum concentration of vapour in air below which the mixture is too lean to ignite; the upper flammable limit (UFL) is the maximum concentration above which the mixture is too rich to ignite. Between the LFL and UFL the mixture is flammable and will burn/explode if ignited. Sketch: a graph with the vertical axis as percentage of oxygen and the horizontal axis as percentage of hydrocarbon vapour (or a triangular diagram). The flammable range is the region bounded by the LFL and UFL lines and the oxygen line; the "too rich" region is above the UFL, the "too lean" region below the LFL, and the "oxygen deficient" region near the top where there is insufficient oxygen to support combustion. The graph shows that a mixture is only flammable within the shaded band between the LFL and UFL, and that inerting (reducing oxygen below about 8-11%) removes the flammable region entirely.

    Q6 (16 Marks) Ship Stability

    (a) Explain why the bilging of empty double-bottom or deep tanks below the waterline leads to an increase in GM. (6)

    (b) A ship of 10,000 tonnes displacement has GM = 0.5 metres. The period of roll in still water is 20 seconds. Find the new period of roll if a mass of 50 tonnes is discharged from a position 14 metres above the center of gravity. (10)

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    Part (a)

    Why bilging empty double-bottom or deep tanks below the waterline increases GM.

    When an empty tank below the waterline is bilged (flooded), the water that enters is at the same level as the sea and, being below the waterline, the added weight of water is exactly balanced by the added buoyancy of the extra volume displaced (the ship sinks slightly). The net effect is that the centre of gravity of the added water is low (in the tank, near the bottom of the ship), while the added buoyancy acts at the centre of buoyancy of the flooded volume, which is also low. Because the added weight is low, the overall centre of gravity KG is lowered, and because the added buoyancy is low, the centre of buoyancy KB is also lowered but by less than the weight effect. The result is that GM increases (the ship becomes stiffer). Additionally, once the tank is full there is no free surface, so there is no free-surface loss of GM. Hence bilging an empty low tank increases GM and improves stability (though it increases displacement and draft).

    Part (b)

    New period of roll after discharging a mass.

    Ship displacement 10,000 t, GM = 0.5 m, period of roll in still water T = 20 s. A mass of 50 t is discharged from a position 14 m above the centre of gravity.

    Radius of gyration: T = 2 pi k / sqrt(g GM), so k = T sqrt(g GM)/(2 pi) = 20 x sqrt(9.81 x 0.5)/(2 pi) = 20 x 2.2146/6.283 = 7.05 m.

    Discharging a weight above the CG lowers the CG: the CG falls by GG = w x h/(Delta - w) = 50 x 14/(10000 - 50) = 700/9950 = 0.0704 m. So the new GM = 0.5 + 0.0704 = 0.5704 m (KM unchanged).

    New period T' = 2 pi k / sqrt(g GM') = 6.283 x 7.05 / sqrt(9.81 x 0.5704) = 44.28 / sqrt(5.595) = 44.28/2.365 = 18.72 s.

    Answer: the new period of roll is about 18.7 s (the ship rolls faster because GM increased).

    Q7 (16 Marks) Hull Construction

    (a) Describe the ways in which an unstable ship can be made stable. (6)

    (b) When a mass of 25 tonnes is shifted 15m transversely across the deck of a ship of 8,000 tonnes displacement, it causes a deflection of 20cms in a plumb line 4m long. If the KM = 7.5m, calculate the KG. (10)

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    Part (a)

    Ways in which an unstable ship can be made stable.

    An unstable ship has a negative GM (the centre of gravity is above the metacentre), so it heels to a large angle. It can be made stable by:

    • Lowering the centre of gravity: add weight low (ballast in the double bottom), remove top weight (deck cargo, high tanks), or transfer weight from high to low positions.
    • Increasing the metacentric height: increase the waterplane area and beam (BM increases with the cube of beam), e.g. by increasing beam (rare) or by improving the waterplane; more practically, increase GM by lowering KG.
    • Removing free-surface effects: press up or empty tanks to eliminate the free-surface loss of GM.
    • Adding buoyancy high or increasing freeboard/reserve buoyancy to increase the range of stability and the righting levers.
    • Reducing the height of the centre of gravity by discharging high weights or by flooding low tanks (carefully, as this also increases displacement).
    • In an emergency, jettisoning top weight or transferring ballast to the double bottom to lower KG and restore positive GM.

    The most effective and common method is to lower KG (add low ballast, remove top weight) and to eliminate free surfaces.

    Part (b)

    Calculation of KG from a transverse shift.

    A mass of 25 t is shifted 15 m transversely across the deck of a ship of 8,000 t displacement, causing a deflection of 20 cm in a plumb line 4 m long. KM = 7.5 m.

    The plumb line deflection gives the angle of heel: tan(theta) = 0.20/4 = 0.05, so theta = 2.86 deg.

    The transverse shift of the centre of gravity: GG' = w x d / Delta = 25 x 15/8000 = 0.046875 m.

    At equilibrium the centre of gravity is over the centre of buoyancy, so tan(theta) = GG'/GM.

    GM = GG'/tan(theta) = 0.046875/0.05 = 0.9375 m.

    KG = KM - GM = 7.5 - 0.9375 = 6.5625 m.

    Answer: KG = 6.56 m.

    Q8 (16 Marks) Ship Resistance & Propulsion

    (a) Describe the process of correcting a negative GM. (6)

    (b) A ship 120m long displaces 10500 tonne and has a wetted surface area of 3000mΒ². At 15 knots the shaft power is 4100kW, propulsive coefficient 0.6 and 55% of the thrust is available to overcome frictional resistance. Calculate the shaft power required for a similar ship 140m long at the corresponding speed.

    f = 0.42 and n = 1.825. (10)

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    (a) Correcting a Negative GM

    A negative metacentric height (GM) means the metacentre (M) lies below the centre of gravity (G), making the ship unstable. If the ship is heeled by an external force, it will continue to heel further instead of returning to the upright position.

    To correct a negative GM:

    1. Lower the Centre of Gravity (G):
      • Transfer weights from higher positions to lower positions in the ship.
      • Load heavy ballast into double-bottom or low ballast tanks.
      • Remove or lower heavy weights carried on deck or in high spaces.
    2. Avoid Raising G Further:
      • Minimise loading of heavy cargo on deck.
      • Avoid lifting heavy loads with ship's cranes unless necessary.
    3. Reduce Free Surface Effect:
      • Press up slack tanks completely or empty them where possible.
      • This reduces the virtual rise of G caused by free surface.
    4. Check Stability Calculations:
      • Recalculate the ship's GM after ballast transfer or cargo redistribution.
      • Ensure a positive GM is achieved before sailing.

    Result:

    When the centre of gravity is lowered below the metacentre, GM becomes positive, restoring the ship's stability.

    (b) Shaft Power Required for a Similar 140 m Ship

    Given Data

    Parameter

    120 m Ship

    140 m Ship

    Length (L)

    120 m

    140 m

    Displacement (Ξ”)

    10,500 t

    Similar ship

    Wetted Surface Area (S)

    3000 mΒ²

    ?

    Shaft Power (SP₁)

    4100 kW

    ?

    Propulsive Coefficient (PC)

    0.6

    0.6

    Frictional Resistance

    55% of Total

    55% of Total

    Friction Coefficient (f)

    0.42

    0.42

    Speed Exponent (n)

    1.825

    1.825

    Step 1: Effective Power of 120 m Ship

    $$EP = PC \times SP$$

    $$EP = 0.6 \times 4100 = 2460\;kW$$

    Step 2: Total Resistance

    Since,

    $$EP = R_T \times V$$

    Convert 15 knots to m/s:

    $$V = 15 \times 0.514 = 7.71\;m/s$$

    $$R_T = \frac{2460}{7.71} = 319.066\;kN$$

    Step 3: Frictional and Residual Resistance

    Frictional resistance:

    $$R_F = 0.55R_T$$

    $$R_F = 0.55 \times 319.066 = 175.486\;kN$$

    Residual resistance:

    $$R_R = R_T - R_F$$

    $$R_R = 319.066 - 175.486 = 143.57\;kN$$

    Step 4: Scale Wetted Surface Area

    For similar ships,

    $$S \propto L^2$$

    $$S_2 = S_1\left(\frac{L_2}{L_1}\right)^2$$

    $$S_2 = 3000\left(\frac{140}{120}\right)^2$$

    $$S_2 = 4083.33\;m^2$$

    Step 5: Corresponding Speed

    For similar ships,

    $$V \propto \sqrt{L}$$

    $$V_2 = 15\sqrt{\frac{140}{120}}$$

    $$V_2 = 16.20\;knots$$

    Step 6: Frictional Resistance of 140 m Ship

    $$R_F = fSV^n$$

    $$R_F = 0.42 \times 4083.33 \times 16.20^{1.825}$$

    $$R_F = 227.984\;kN$$

    Step 7: Residual Resistance of 140 m Ship

    For similar ships,

    $$R_R \propto L^3$$

    $$R_{R2} = R_{R1}\left(\frac{L_2}{L_1}\right)^3$$

    $$R_{R2} = 143.57\left(\frac{140}{120}\right)^3$$

    $$R_{R2} = 227.984\;kN$$

    Step 8: Total Resistance

    $$R_T = R_F + R_R$$

    $$R_T = 276.457 + 227.984 = 504.441\;kN$$

    Step 9: Effective Power

    Convert 16.20 knots to m/s:

    $$V = 16.20 \times 0.514 = 8.33\;m/s$$

    $$EP = R_T \times V$$

    $$EP = 504.441 \times 8.33$$

    $$EP = 4200.346\;kW$$

    Step 10: Shaft Power

    $$SP = \frac{EP}{PC}$$

    $$SP = \frac{4200.346}{0.6}$$

    $$\boxed{SP \approx 7000.57\;kW}$$

    Answer

    The shaft power required for the similar 140 m ship at the corresponding speed is approximately

    7001 kW (β‰ˆ 7.0 MW).

    Q9 (16 Marks) Ship Stability

    (a) Explain why the rudder angle does not normally exceeds 35Β°. (6)

    (b) A ship of 12000 tonne displacement has a rudder 15mΒ² in area, whose centre is 5m below the waterline. The metacentric height of the ship is 0.3m and the centre of buoyancy is 3.3m below the waterline. When travelling at 20 knots the rudder is turned through 30Β°. Find the initial angle of heel if the force Fn is perpendicular to the plane of the rudder is given by: F = 577 AvΒ² sin Ξ±.

    Allow 20% for the race effect. (10)

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    Part (a)

    Why the rudder angle does not normally exceed 35 degrees.

    The rudder is most effective at moderate angles. As the helm angle increases, the lift (normal force) on the rudder rises up to about 30-35 degrees, beyond which the flow separates from the rudder surface, the lift falls off and the drag rises sharply (stall). The rudder force also produces a heeling moment and increases resistance. Beyond about 35 degrees the additional turning effect is negligible or negative, while the drag and heeling moment increase, so there is no benefit in a larger angle. Hence the maximum rudder angle is limited to about 35 degrees (SOLAS requires the rudder to be capable of being put hard over to 35 degrees in 28 seconds), giving the best turning performance with acceptable drag and heel.

    Part (b)

    Rudder heeling angle.

    Data: displacement 12,000 t; rudder area A = 15 m2; rudder centre 5 m below the waterline; GM = 0.3 m; centre of buoyancy 3.3 m below the waterline; speed 20 knots; helm angle 30 deg; force Fn = 577 A v2 sin(alpha); allow 20% race effect.

    Speed: v = 20 x 0.5144 = 10.29 m/s, v2 = 105.84.

    Fn = 577 x 15 x 105.84 x 0.5 = 458,035 N.

    Race effect 20%: F = 1.2 x 458,035 = 549,642 N.

    Heeling moment = F x lever = 549,642 x 5 = 2,748,210 Nm (lever = 5 m, the rudder centre below the waterline).

    tan(theta) = (F x lever)/(W x GM) = 2,748,210/(12000 x 9810 x 0.3) = 2,748,210/35,316,000 = 0.07782.

    theta = atan(0.07782) = 4.45 deg.

    Answer: initial angle of heel about 4.5 deg.

    Q10 (10 Marks) Ship Resistance & Propulsion

    (a) List the components of residual resistance.

    (b) The following data are available for a twin-screw vessel:

    V (knots)

    15

    16

    17

    18

    RP (kW)

    3000

    3750

    4700

    5650

    QPC

    0.73

    0.73

    0.72

    0.71

    Calculate the service speed if the brake power for engine is 3500kW. The transmission efficiency loss is 3% and the allowances for weather and appendages 30%. (10)

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    Part (a)

    Components of residual resistance.

    Residual resistance is the total resistance minus the frictional (skin) resistance. Its components are:

    • Wave-making resistance: the energy lost in creating the bow and stern wave systems, which depends on the Froude number and hull form.
    • Wave-breaking resistance: energy lost in the breaking of the bow wave.
    • Eddy-making resistance: energy lost in the formation of eddies at the stern, bilge keels, rudder and appendages.
    • Pressure (form) resistance: the resistance due to the pressure distribution over the hull, including the viscous pressure (form) drag.
    • Appendage resistance (in service): the additional resistance of the rudder, shaft brackets, bilge keels, etc.

    Residual resistance is largely independent of Reynolds number and is scaled from model tests by Froude's law.

    Part (b)

    Service speed of a twin-screw vessel.

    Data: V (knots) 15,16,17,18; RP (kW) 3000,3750,4700,5650; QPC 0.73,0.73,0.72,0.71. Brake power of engine = 3500 kW; transmission efficiency loss 3%; allowances for weather and appendages 30%.

    Delivered (shaft) power at the propeller: DHP = 3500 x (1 - 0.03) = 3395 kW.

    Effective power available at each speed = DHP x QPC. At 15 kn: 3395 x 0.73 = 2478 kW; at 16 kn: 2478 kW; at 17 kn: 3395 x 0.72 = 2444 kW; at 18 kn: 3395 x 0.71 = 2410 kW.

    The service allowance of 30% means the naked effective power required at the service speed is RP(V), and the available effective power must cover RP(V) x 1.30 (the ship must overcome 30% more resistance in service). Equating available EHP to 1.3 x RP(V):

    At 15 kn: available 2478, required 1.3 x 3000 = 3900 - not enough.

    The service speed is found where 1.3 x RP(V) = available EHP. Using the RP curve (approximately proportional to V^3, RP = 3000 x (V/15)^3) and available EHP about 2478 kW:

    1.3 x 3000 x (V/15)^3 = 2478, so (V/15)^3 = 2478/3900 = 0.6354, V/15 = 0.86, V = 12.9 knots.

    Answer: the service speed is about 13 knots (approximately 12.9-13.0 knots).

    Q1 (16 Marks) Hull Construction πŸ”₯ Repeated 10x

    (a) Describe a method for the attachment of bilge keels. (5)

    (b) State THREE reasons for not extending bilge keels the entire length of the vessel. (6)

    (c) Explain TWO principles of roll damping that bilge keels exploit. (5)

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    Part (a)

    Method of Bilge keel attachment to the hull:

    Bilge keels are fitted port and starboard at the turn of the bilge. They do not extend outside the lines of the side and bottom shell. The bilge keels are attached directly in line with an internal stiffening member such as a girder or longitudinal. The bilge keel comprises a flat bar doubler welded directly to the shell, and an offset bulb plate (OBP) with 'scallops' cut in it is welded to the flat bar doubler. The ends of the bilge keels are tapered (minimum 3 in 1) and will end in line with transverse internal stiffening, such as a frame. Using a hull doubler protects the hull in case of damage to the keel, as the crack would not extend into the hull.

    Part (b)

    Bilge keels are not fitted for the full length of the vessel because:

    • The 'lever' to the ship's axis of rotation is reduced at the ends;
    • The ineffectiveness, the closer the bilge keel is to the rolling centre, increased resistance and more likelihood of damage.
    • The hydrodynamic effect would cause a large increase in resistance and fuel consumption;
    • At the aft end, the boundary layer is much thicker, and since the keel would not project through, it would have a much reduced effect.
    Part (c)

    Two principles of roll damping exploited by Bilge keels:

    Increased Roll Period:

    • Bilge keels increase the ship's roll period (the time it takes for the vessel to complete one roll). This is achieved by increasing the moment of inertia (K) of the vessel. A longer roll period means the vessel rolls more slowly, thus reducing the amplitude of the roll.

    $$T_{r}=2\pi\:\frac{k}{\sqrt{g\times GM}}$$

    Where:

    • g: acceleration due to gravity
    • GM: metacentric height
    • k: mass moment of inertia

    By increasing k, the ship's stability during roll improves.

    Hydrodynamic Resistance:

    • As the vessel rolls, the bilge keels move through the water, creating pressure differences. Water pressure on one side of the keel opposes the rolling motion, providing a damping effect. This is due to the interaction between the bilge keel and the water, creating hydrodynamic forces that counteract the rolling motion.
    Q2 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 8x

    Vessel has gone through very heavy weather. On arrival at safe anchorage, you are conducting your inspection to determine damages to hull.

    (a) List the areas you will inspect. (3)

    (b) List your findings of any significance. (6)

    (c) Write a report to company suggesting repairs if any (7)

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    Part (a)

    Areas to Inspect After Heavy Weather

    Upon arrival at safe anchorage, a thorough inspection of the vessel shall be carried out, both externally and internally, to assess any weather-related damage. Key areas include:

    1. Hull and Main Deck

    • Shell plating along bow, midship, and stern sections (especially forepeak and flare areas).
    • Boot-top and bilge areas for dents or deformation.
    • Deck plating for buckling or cracked welds.
    • Bulwarks, rails, fairleads, chocks, and mooring fittings.

    2. Forecastle and Forward Structure

    • Bosun store and chain locker for water ingress.
    • Windlass foundation, anchors, and hawse pipes for deformation or cracks.
    • Forepeak tank for leakage or pressure damage.

    3. Cargo Holds / Tanks

    • Inspect for structural deformation, loose frames, or fractured stiffeners.
    • Check tank top plating and bilges for leakage.
    • Check watertight doors, gaskets, and vents.

    4. Superstructure and Deck Fittings

    • Bridge wings, radar mast, funnel, vents, and deckhouses for cracks or loose fittings.
    • Lifeboat davits, securing arrangements, and deck cranes.

    5. Underwater and Machinery Spaces

    • Rudder, propeller, and stern tube seals (via steering gear tests).
    • Sea chest gratings and overboard discharges.
    • Engine room bilges for any seawater ingress.

    Part (b)

    Typical Findings of Significance

    • Dents and indentations on shell plating at bow and forward port side due to heavy slamming.
    • Paint coating and corrosion protection partly peeled off near waterline and forepeak area.
    • Loose fairlead bolts and one cracked weld on starboard bulwark stanchion.
    • Deformed ventilator head on forecastle deck.
    • Minor leakage observed in forepeak tank during sounding check.
    • Bridge wing railing bent, likely from green sea impact.
    • Anchor chain links twisted and worn.
    • Lifeboat gripes loosened, requiring tightening and inspection.
    • No flooding reported; watertight integrity maintained overall.

    Part (c)

    Report to Company – Heavy Weather Damage Inspection

    To: Superintendent / Technical Department

    From: Name / Rank

    Subject: Heavy Weather Damage Inspection Report

    Date: [Insert date]

    Vessel: [Insert vessel name]

    Summary

    The vessel experienced very heavy weather en route from [Port A] to [Port B], with significant pitching and rolling in seas up to [X] meters. On arrival at safe anchorage, a complete inspection of the hull and deck was carried out.

    Findings

    • Bow and forecastle plating show minor dents, with no breach of watertight integrity.
    • Bulwark stanchion (starboard side) cracked at welded joint – requires repair.
    • Fairlead foundation bolts loosened – retightening and NDT inspection recommended.
    • Ventilator head on forecastle deformed – replacement advised.
    • Forepeak tank shows trace leakage at forward bulkhead – to be pressure tested.
    • Paint coating damage and corrosion exposure on bow area – recoating required.
    • Bridge wing railing bent – to be straightened or renewed.
    • All other structures and machinery appear satisfactory after testing.

    Recommendations

    1. Conduct minor steel renewal and welding repairs at the next port with repair facilities.
    2. Carry out thickness measurements and NDT on affected hull areas.
    3. Recoat damaged paint areas to prevent corrosion.
    4. Replace deformed ventilator head and bent railing.
    5. Inspect anchor and chain for elongation; renew worn links.
    6. Pressure test forepeak tank after repairs.
    7. Submit class surveyor report if deemed necessary by the superintendent.

    Conclusion

    The vessel remains seaworthy but requires prompt attention to minor structural and fitting damages before the next voyage. Preventive maintenance and weather routing measures should be reviewed for future passages.

    Signed:

    Name / Rank

    Signature

    Q3 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 3x

    With reference to ships rudder plate:

    (a) Why a breached hollow rudder can add to fuel costs? (6)

    (b) Why excessive pintle clearance should not be tolerated? (5)

    (c) Why fitted bolts are used in connecting upper and lower stocks? (5)

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    Ship’s Rudder – Effects of Defects and Design Features

    (a) Why a Breached Hollow Rudder Can Add to Fuel Costs

    • A hollow rudder is designed to provide buoyancy at the stern of the ship.
    • When the rudder is breached, sea water enters the hollow space, resulting in loss of buoyancy.
    • The loss of buoyancy causes the stern to sit deeper in the water, increasing the aft draught.
    • Increased aft draught leads to the propeller operating deeper, which increases hydrodynamic resistance.
    • Water ingress into the rudder creates turbulence and disturbed water flow around the rudder and propeller.
    • To maintain the same ship speed, higher engine power is required, resulting in increased fuel consumption.

    (b) Why Excessive Pintle Clearance Should Not Be Tolerated

    • Pintles support the rudder and fit into gudgeons with very small designed clearances.
    • Excessive clearance permits sideways and vertical movement of the rudder.
    • Such movement imposes additional bending stresses on the rudder stock.
    • In heavy seas, shock and impact loads increase wear and may cause misalignment.
    • This can lead to vibration, poor steering response, accelerated wear, and in extreme cases, structural failure of the rudder assembly.

    (c) Why Fitted Bolts Are Used in Connecting Upper and Lower Rudder Stocks

    • The upper and lower rudder stocks are required to transmit high steering torque.
    • Fitted bolts provide a close, interference fit with no clearance in the bolt holes.
    • They ensure accurate alignment of the two stock sections.
    • Steering loads are transmitted in shear through the bolt shanks, rather than relying on friction alone.
    • This prevents relative movement, reduces stress concentration, and minimizes the risk of fatigue failure.
    Q4 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 7x

    With reference to International Load Line Statutory Certification,

    (a) State the reasons for the freeboard requirements. (6)

    (b) (i) Explain the term "conditions of assignment". (5)

    (ii) List the items that may be examined during a Load line survey after a vessel's major repairs in the dry dock. (5)

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    Part (a)

    Reasons for Freeboard Requirements:

    Freeboard is the distance measured from the waterline to the upper edge of the deck plating at the side of the freeboard deck amidships.

    Purpose of Freeboard:

    • Ensures the ship is seaworthy when fully loaded.
    • Provides reserve buoyancy, enabling the ship to rise as it passes through waves, keeping decks largely dry.
    • Enhances the ship's stability and increases its survivability in case of damage by allowing it to remain afloat longer, aiding crew escape or damage control.
    Part (b)

    (i) Conditions of Assignment:

    Conditions of Assignment are outlined in the Merchant Shipping Load Line Rules 1968 and must be satisfied before a ship is assigned freeboards and issued a load line certificate. These conditions address the practical need for openings (hatchways, doorways, vents, etc.) in the ship's hull and mandate appropriate protection and closure mechanisms for these openings.

    Requirements Before Assigning Load Line:

    • The ship must have sufficient structural strength.
    • Adequate reserve buoyancy must be maintained.
    • Openings must be secured against water ingress.
    • Safety measures for the crew, such as guardrails and gangways, must be in place.
    Part (b)

    (ii) Items Examined During a Load Line Survey After Major Repairs in Drydock:

    • Openings: This encompasses hatchways, machinery space openings, cargo ports, watertight doors, ventilators, air pipes, scuppers, freeing ports, side scuttles, and other openings in the freeboard and superstructure decks. The surveyor checks the condition of seals, gaskets, closing mechanisms, and overall watertightness.
    • Crew Protection: Safety features designed to protect the crew, such as guardrails, bulwarks, and gangways, are inspected for damage or deterioration and proper functionality.
    • Hull Structure: The hull itself is inspected for any damage, corrosion, or structural weaknesses that could compromise watertightness or strength. This often includes visual inspection for cracks, buckling, indentation, and paint adhesion. Bottom shell, bilge keel, stem and stern frames, rudder, sea chests, side ports, stern bearing, and propeller are all checked.
    • Other Systems: Depending on the scope of repairs, other systems may be examined, including but not limited to: pressure-vacuum valves, sounding pipes, air vents, access hatches, ventilation fans, dampers, weathertight doors and sealings, door securing arrangements, side scuttles, windows, and skylights. Testing, such as hose tests (to check watertight integrity), might also be conducted.
    Q5 (16 Marks) Hull Construction πŸ”₯ Repeated 6x

    (a) Sketch a transverse section through the hold space of a container ship hull. (8)

    (b) Referring to the sketch in (a) describe how adequate structural strength is built into the hull. (8)

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    Part (a)

    Mid-ship half-section sketch of a container ship:

    Part (b)

    Structural Strength Features:

    • The deck plating is made thicker and uses higher tensile steel to withstand stresses caused by wide hatch openings and operational loads.
    • The deck, side shell, and longitudinal bulkheads are framed longitudinally. This arrangement combined with deep double bottom helps resist bending stresses due to hogging (upward bending) and sagging (downward bending) when the ship is under load.
    • The hatch coamings are made continuous to contribute to the overall longitudinal strength of the hull.
    • A torsion box is installed, running along the entire length of the ship from the machinery space bulkhead to the forward collision bulkhead. This structure provides the necessary torsional strength to counteract twisting forces acting on the hull during operation.
    • Deep web boxes are fitted at the ends of hatches, both at tank top and deck levels, to enhance transverse and torsional strength.
    • A deep double bottom is designed to withstand uplift forces caused by water pressure, especially when the ship is deeply loaded. It also provides additional strength to the hull structure.
    • Side girders are placed under container cells, with added transverse local stiffening. These elements distribute the concentrated loads from containers and increase overall stability.
    Q6 (16 Marks) Ship Stability πŸ”₯ Repeated 9x

    (a) Describe how the force on the ship's bottom and the GM vary when grounding takes place. (6)

    (b) A ship of 8,000 tonnes displacement takes the ground on a sand bank on a falling tide at an even keel draft of 5.2 metres. KG 4.0 metres. The predicted depth of water over the sand bank at the following low water is 3.2 metres. Calculate the GM at this time assuming that the KM will then be 5.0 metres and that the mean TPC is 15 tonne. (10)

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    Part (a)

    When a ship grounds, the forces on the bottom and the metacentric height (GM) change depending on the grounding's nature and the ship's position on the seabed.

    If the ship grounds on a level bottom:

    • A ground reaction force acts vertically upwards from the seabed. This force counteracts part of the ship's weight and alters the distribution of buoyancy forces.
    • The ship’s centre of gravity (G) may appear to rise relative to the metacentre (M) because the upward ground reaction reduces the buoyancy force acting on the underwater volume.
    • This virtual rise in G reduces GM, potentially leading to a list.
    • If the list becomes excessive and the righting moment is insufficient, the ship may capsize.

    If the ship grounds on a pinnacle:

    • The ship experiences two forces at the ship's bottom:
      • A downward force due to the ship’s weight.
      • An upward reaction force is concentrated on the pinnacle.
    • The resulting force between the grounding pressure and the ship’s centre of buoyancy shifts downward towards the pinnacle.
    • This situation is similar to when the ship's stern touches the keel block in a dry dock.
    • A virtual loss of GM occurs because the ship’s inclining moment may exceed the maximum righting lever.
    • If the inclining moment is too great, the ship may develop an excessive list or even capsize.

    (b) Given:

    $$Displacement,\:\Delta=8000\:tonnes$$

    $$TPC=15\:tonnes$$

    $$Initial\:Draught=5.2m$$

    $$Final\:Draught=3.2m$$

    $$Ship\:KG=4.0m$$

    $$KM=5.0m$$

    To find GM

    $$Uptrust,\:P=TPC\times Fall\:in\:water\:level\:\left(cm\right)$$

    $$=15\times\left(520-320\right)$$

    $$=15\times200$$

    $$P=3000\:tonnes$$

    To Find Virtual loss of GM:

    $$Virtual\:loss\:of\:GM_1=\frac{P\times KM}{\Delta}$$

    $$=\frac{3000\times5}{8000}$$

    $$=\frac{15000}{8000}$$

    $$GM_1=1.88m$$

    Actual KM = 5.0m (given)

    $$Virutal\:KM=Actual\:KM-Virtual\:loss\:of\:GM_1$$

    $$=5.0-1.88$$

    $$=3.12$$

    Similarly, Actual KG = 4.0m (given)

    $$New\:GM=Virtual\:KM-\:Actual\:KG$$

    $$=3.12-4.0$$

    $$=-0.88$$

    Q7 (16 Marks) Ship Stability πŸ”₯ Repeated 4x

    (a) Describe stability requirement for dry-docking. (6)

    (b) A box shaped vessel, 50 metres long x 10 metres wide, floats in salt water on an even keel at a draft of 4 metres. A center line longitudinal watertight bulkhead extends from end to end and for the full depth of the vessel. A compartment amidships on the starboard side is 15 metres long and contains cargo with permeability 30%. Calculate the list if this compartment is bilged. KG = 3 metres. (10)

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    Part (a)

    Stability requirements for dry-docking.

    Before entering the dock the ship must have adequate and known stability. The requirements are:

    • The ship should be stable with an adequate GM, so that the small upward force (reaction) at the keel, building during docking, cannot produce a list. Dock support is provided on the keel blocks and any excessive list during docking would subject the structure and blocks to unequal loading.
    • As the water is pumped out, the reaction of the keel blocks reduces the effective displacement of the ship, lowering the KB and BM and hence GM; if all buoyancy were removed the ship would rest wholly on the blocks. The ship must retain sufficient GM throughout the docking operation (for example by keeping the GM such that the maximum inclination, where the righting moment at the critical point, does not become negative).
    • Ballast should be arranged to give even keel (or slight trim) and list-free condition; tanks should be pressed up or emptied to avoid free-surface effects, and draft read before docking.
    • Weight and trim must be such that the keel blocks contact evenly over the full keel length (avoid excessive trim that overloads the block crown or aft blocks).
    • Cargo/weights should not be changed and the dock must be level; adequate dock pumping should keep the vessel central over the keel line.

    If a beam vessel develops instability while docking (transverse GM small or negative), it can heel and capsize on the blocks; hence a suitable stability margin (e.g. GM not less than a minimum) is insisted on and the docking weight, draft and trim are checked by stability data (e.g. from hydrostatics and the Docking Plan).

    Part (b)

    List when a starboard midship compartment is bilged.

    Box-shaped vessel 50 m long, 10 m wide, floats on even keel in salt water at a draft of 4 m. A centreline longitudinal watertight bulkhead runs full depth/size. A midship compartment on the starboard side is 15 m long and has permeability 30 per cent. KG = 3 m. Calculate the list when bilged.

    Waterplane area intact Aw = 50 x 10 = 500 m2; the flooded wing compartment has waterplane area = 15 x 5 = 75 m2 (only starboard half up to the centreline bulkhead).

    Lost buoyancy volume of the compartment below the original waterline, with permeability 30%: Vlost = length x breadth x draft x permeability = 15 x 5 x 4 x 0.30 = 90 m3. Corresponding lost weight W = 90 x 1.025 = 92.25 t.

    Sinkage: the region once flooded provides no increase of buoyancy; effective sinking waterplane = 500 - 75 = 425 m2. Mean sinkage = 90/425 = 0.212 m.

    List: the lost buoyancy acts at the centroid of the lost volume, which is at 5/4 = 2.5 m out from the centreline (half-way between centreline bulkhead and side). Heeling moment = W x y = 92.25 x 2.5 = 230.6 t-m.

    New GM. Using the lost-buoyancy method, displacement remains 2050 t. The second moment of area of the intact waterplane about the centreline loses the flooded starboard compartment: I(intact)=50x10^3/12 = 4166.7 m4; I(flooded)=15x5^3/3 = 625 m4; so I(new)=3541.7 m4. BM=new = I(new)/V = 3541.7/2000 = 1.771 m. KB = d/2 = 2 m (approximately, ignoring sinkage and list); GM = KB + BM - KG = 2 + 1.77 - 3 = 0.77 m.

    List: tan(list) = heeling moment/(Delta x GM) = 230.6/(2050 x 0.77) = 230.6/1579 = 0.146. List = atan(0.146) = 8.3 deg.

    Answer: the vessel lists about 8.3 deg to starboard.

    Q8 (16 Marks) Ship Stability πŸ”₯ Repeated 11x

    (a) Define longitudinal center of gravity (LCG) and longitudinal center of buoyancy(LCB). (6)

    (b) A ship 120m long floats at draughts of 5.50m forward and 5.80m aft; MCT1 cm 80 tonne m, TPC 13, LCF 2.5m forward of midships. Calculate the new draughts when a mass of 110 tonne is added 24m aft of midships. (10)

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.

    LCF in fwd and trim by stern

    $$Bodily \space sinkage \space = \space {{mass \space added} \over TPC} \space = \space {{110} \over 13} \space$$

    $$=\:8.5\operatorname{\mathrm{cm}}\:=0.085m$$

    $$Trim \space = \space {{m \times d} \over MCT_{1cm}}$$

    $$ = \space {{110 \times (24 + 2.5)} \over 80}$$

    $$Trim=36.43\operatorname{cm}=0.364m\:$$

    Change in fwd draught:

    $$d_{F}=\frac{-t}{L}\left\lbrack\frac{L}{2}-LCF\right\rbrack$$

    $$=\frac{-36.43}{120}\left\lbrack\frac{120}{2}-2.5\right\rbrack$$

    $$=-17.45\operatorname{cm}=-0.1745m$$

    Change in Aft draught:

    $$d_{A}=\frac{+t}{L}\left\lbrack\frac{L}{2}+LCF\right\rbrack$$

    $$=\frac{+36.43}{120}\left\lbrack\frac{120}{2}+2.5\right\rbrack$$

    $$=+18.97\operatorname{cm}=0.189m$$

    New draughts:

    $$D_{F}=5.5+0.085-0.175=5.41m$$

    $$D_{A}=5.8+0.085+0.18=6.065m$$

    Q9 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 3x

    (a) What are the main components of ship resistance that a vessel encounters while moving through water? (6)

    (b) The speed of a ship is increased to 18% above normal for 7.5 hours, then reduced to 9% below normal for 10 hours. The speed is then reduced for the remainder of the day so that the consumption for the day is the normal amount. Find the percentage difference between the distance travelled in that day and the normal distance travelled per day. (10)

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    Main Components of Ship Resistance

    When a vessel moves through water, it experiences total resistance, which is the combined effect of several opposing forces acting against its forward motion. These forces arise due to water viscosity, pressure distribution around the hull, wave formation, and external environmental conditions. For analysis, total resistance is divided into the following main components:

    1. Frictional Resistance

    Frictional resistance is caused by the viscosity of water acting along the ship’s wetted surface as the vessel moves forward. As water flows over the hull, a thin boundary layer forms, and shear stresses develop between the hull surface and adjacent water particles. This shear stress produces resistance.

    Frictional resistance depends primarily on:

    • Wetted surface area – A larger underwater hull area increases the contact between hull and water, thereby increasing resistance.
    • Ship speed – Resistance increases rapidly with speed because higher velocity intensifies boundary layer shear forces.
    • Surface roughness of the hull – Fouling, corrosion, or rough coatings increase turbulence and significantly raise resistance.

    At normal service speeds, frictional resistance forms the largest portion of total resistance, especially for slow and medium-speed vessels. It increases approximately with the square of the ship’s speed, making hull maintenance crucial for fuel efficiency.

    2. Residual Resistance

    Residual resistance is the portion of resistance remaining after subtracting frictional resistance from total resistance. It mainly arises from pressure effects and wave formation around the hull. Residual resistance consists of the following components:

    (a) Wave-Making Resistance

    Wave-making resistance is caused by the energy expended in generating surface waves at the bow and stern as the ship moves. When a vessel travels through water, it disturbs the free surface and creates a wave system that carries energy away from the ship.

    This resistance becomes particularly significant at higher speeds because wave height and wave energy increase rapidly with speed.

    Wave-making resistance depends on:

    • Hull form – Fuller hull shapes generally create larger waves.
    • Speed–length ratio (Froude number) – As the Froude number increases, wave-making resistance rises sharply.

    At high speeds, wave-making resistance can become a dominant component of total resistance.

    (b) Eddy-Making Resistance

    Eddy-making resistance is caused by flow separation and the formation of vortices (eddies) around certain parts of the hull. When water flow cannot smoothly follow the hull contour, it separates and creates turbulent regions.

    This commonly occurs around:

    • The stern region
    • Appendages
    • Areas with sudden changes in hull form

    These turbulent eddies consume energy and increase resistance. Proper streamlining of the hull and stern design can significantly reduce eddy-making resistance.

    3. Air Resistance

    Air resistance is the force exerted by air on the portion of the ship above the waterline. As the ship moves, it must also overcome aerodynamic drag.

    Air resistance depends on:

    • Wind speed and direction
    • Projected area above water
    • Shape of the superstructure

    Although usually small compared to water resistance, it becomes significant for:

    • Container ships
    • Ro-Ro vessels
    • Ships operating in strong headwinds

    For vessels with large exposed areas, air resistance can noticeably affect fuel consumption.

    4. Appendage Resistance

    Appendage resistance is caused by external fittings attached to the hull, such as:

    • Rudders
    • Bilge keels
    • Shaft brackets
    • Propeller bossings

    These appendages increase the wetted surface area and disturb smooth water flow, thereby increasing both frictional and pressure resistance. In resistance calculations, appendage resistance is generally included as a separate correction added to frictional resistance.

    5. Added (Special) Resistance

    Added resistance refers to the additional resistance experienced in real sea conditions that is not present in calm-water trials.

    It occurs due to:

    • Waves and swell, which cause pitching and heaving motions
    • Steering and yawing motions, which disturb steady flow around the hull

    Although not considered in calm-water resistance analysis, added resistance is highly important in practical operations because it significantly affects power requirements and fuel consumption in rough weather.

    Q10 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 8x

    With reference to fixed pitch propellers:

    (a) Explain propeller slip and propeller thrust. (6)

    (b) The shaft power of a ship is 3000 KW, the ship's speed V is 13.2 knot. Propeller RPS is 1.27. propeller pitch is 5.5m and the speed of advance is 11 Knots. Find: (10)

    (i) Real Slip

    (ii) Wake fraction

    (iii) Propeller thrust, when its efficiency, n = 70%

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    Part (a)

    Slip

    is the difference between the theoretical distance the propeller should travel in one revolution and the actual distance the vessel travels.

    $$Slip\:\left(\%\right)\:=\:\left(\frac{Engine\:distance\:-\:Ship^{\prime}s\:distance}{Engine\:distance}\right)\:\times100$$

    Where Engine distance = no. of propeller revolutions * propeller pitch (usually calculated over a 24-hour period)

    Propeller thrust: it is the force exerted by the propeller to move the vessel ahead and given by

    $$Thrust\:\left(T\right)\:=\:\rho\times A\times P^2\times n^2\times S$$

    Where,

    ρ - Density

    A - Area

    P - Pitch

    n - Revolution per second

    S - Slip

    Q1 (16 Marks) General

    (a) Define hogging and sagging with the help of neat sketches. Explain the causes of these structural conditions and when they are most likely to occur during a voyage. (8)

    (b) Discuss the effects of hogging and sagging on a ship’s structural integrity. What measures are taken in design, loading, and operation to minimize their impact? (8)

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    Part (a)

    Definition of hogging and sagging, causes and when they occur.

    Hogging: the ship bends so that the ends are lower than the middle - the keel is convex downwards (the deck is in tension, the bottom in compression). It occurs when the ends are heavier than the middle, i.e. when the buoyancy is concentrated amidships and the weight at the ends. This happens when the ship is light in the middle (e.g. empty midship holds) and heavy at the ends, or when the ship is in a wave crest amidships (a wave crest under the middle with troughs at the ends) - the "hogging" wave condition.

    Sagging: the ship bends so that the middle is lower than the ends - the keel is concave downwards (the deck is in compression, the bottom in tension). It occurs when the middle is heavier than the ends, i.e. when the weight is concentrated amidships and buoyancy at the ends. This happens when the ship is loaded heavily amidships (full midship holds, empty ends), or when the ship is in a wave trough amidships (a wave trough under the middle with crests at the ends) - the "sagging" wave condition.

    Causes: the difference between the weight distribution and the buoyancy distribution along the length. In still water, hogging/sagging arise from the loading pattern; in a seaway, the wave profile changes the buoyancy distribution, and the maximum hogging/sagging moments occur when the wave length is about equal to the ship length and the wave crest/trough is amidships. Hogging is most likely in ballast (light ship, empty midship) and sagging in the fully loaded condition with full midship holds.

    Part (b)

    Effects on structural integrity and measures to minimise.

    Effects: hogging and sagging produce longitudinal bending moments and shear forces in the hull girder. The deck and bottom are the extreme fibres and carry the highest bending stresses - in hogging the deck is in tension and the bottom in compression; in sagging the deck is in compression and the bottom in tension. Excessive bending can cause buckling of the deck or bottom plating, cracking, and fatigue, and can lead to structural failure. The shear forces are greatest near the quarter points. The hull girder must be designed to withstand the maximum still-water and wave bending moments.

    Measures:

    • Design: adequate section modulus of the hull girder (deep, strong deck and bottom, continuous longitudinal material), high-tensile steel where appropriate, and the hull girder designed for the maximum combined still-water and wave bending moment.
    • Loading: distribute cargo to avoid excessive hogging/sagging - avoid empty midship holds in ballast (use ballast in midship tanks), avoid overloading the ends or the middle, and follow the approved loading manual and the maximum still-water bending moment limits.
    • Operation: use ballast to reduce the still-water bending moment, avoid heavy weather that increases the wave bending moment, and reduce speed in heavy seas; monitor the hull stress (hull stress monitoring system) and the draft/trim.
    • Maintenance: inspect and maintain the deck and bottom structure, and repair corrosion and fatigue cracks promptly.
    Q2 (16 Marks) Ship Stability

    (a) Define the term angle of loll. Explain the conditions under which it occurs and support your answer with a diagram showing the stability curve and ship’s position at the angle of loll. (8)

    (b) Describe the actions to be taken to correct the angle of loll and restore positive stability. What precautions must be taken during the corrective process to avoid worsening the situation? (8)

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    (a) Angle of Loll

    The angle of loll is the angle at which a ship with an initial negative metacentric height (negative GM) comes to rest in still water. The ship is unstable in the upright position and therefore heels to either port or starboard until it reaches a position of neutral equilibrium.

    At the angle of loll:

    • GM = 0, or equivalently KG = KM.
    • The ship is in neutral stability and remains at rest at this angle in calm water.
    • If the ship is further heeled beyond the angle of loll by an external force, it may become unstable and capsize.

    Conditions Under Which Angle of Loll Occurs

    An angle of loll occurs when:

    • The ship has an initial negative GM, making it unstable in the upright condition.
    • This may result from:
      • Excessively high centre of gravity (KG).
      • Large free surface effect due to slack tanks.
      • Improper loading or shifting of weights.
    • Since the upright position is unstable, the ship heels to one side until sufficient form stability is developed, resulting in neutral stability at the angle of loll.

    Corrective Actions

    • First, determine whether the vessel is suffering from a list or an angle of loll.
    • For safety, always assume it is an angle of loll and take corrective action accordingly.
    • Calculate the contents of all tanks and identify any slack tanks, as these contribute to free surface effect.
    • In a listed condition, always try to lower the centre of gravity (CG) by discharging ballast from the high side first, where appropriate.
    • Begin ballasting the low-side tanks, preferably using small tanks to minimise free surface effect during filling.
    • Do not ballast the high-side tanks first, as the additional heeling moment may be sufficient to cause the ship to capsize.
    • Gradually fill the centre (midship) tanks, followed by the corresponding tanks on the opposite side.
    • The ship should gradually return to the upright position. If not, continue ballasting other tanks using the same method until adequate positive stability is restored.

    Important: During an angle of loll, never ballast the high-side tanks first, as the resulting heeling moment can rapidly shift the vessel to the opposite side and may lead to capsize.

    Q3 (16 Marks) Hull Construction πŸ”₯ Repeated 3x

    Discuss the need for adequate support of engine room gantry cranes, detailing the following

    (a) Sketch section through the engine room casing showing how the crane is supported by the ship structure. (6)

    (b) State what restricts the forward and aft limits of the crane and what is fitted to prevent the crane damaging the forward and aft bulkheads or casing. (5)

    (c) State the Second Engineer’s responsibilities for the engine room gantry crane. (5)

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    Part (a)

    Engine room gantry crane:

    Part (b)

    Forward and aft limits of crane:

    • Two limit switches are installed to automatically cut off the motor power when the crane reaches its forward or aft limits.
    • Mechanical Stoppers are fitted at the forward and aft ends of the crane rails. They act as a failsafe to prevent the crane from damaging the forward and aft bulkheads or casing if the limit switches fail. Rubber or spring buffers may also be added to absorb impact forces and reduce the risk of damage.
    Part (c)

    Second Engineer's responsibilities for the engine room gantry crane.

    • Inspect the crane, rails, mechanical stoppers, and support brackets for wear or damage.
    • Ensure limit switches are functional and properly aligned.
    • Check the condition of the gear case oil and renew it as required.
    • Inspect and maintain wire ropes and safety latches.
    • Ensure the Safe Working Load (SWL) is clearly marked and adhered to.
    • Verify the validity of the crane's test certificate.
    • Perform motor overhauls and measure insulation resistance.
    • Conduct brake tests and emergency stop function checks.

    Provide instructions to Junior Engineers:

    • Not exceeding the SWL of the crane.
    • Ensuring mechanical locks are removed before operating the crane.
    • Using proper Personal Protective Equipment (PPE).
    • Prohibiting personnel from standing beneath the crane during operation.
    Q4 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 9x

    With reference to Underwater Inspection in lieu of Dry docking (UWILD)

    (a) Explain in detail, how an underwater survey is carried out. (6)

    (b) State the requirements to be fulfilled before an underwater survey is acceptable to the survey authority. (5)

    (c) Construct a list of the items in order of importance that the underwater survey authority should include. (5)

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    (a) An underwater survey, also known as Underwater Inspection in Lieu of Dry Docking (UWILD), involves a systematic and detailed examination of a vessel’s hull and underwater components while the ship remains afloat. The procedure includes the following steps:
    • The shipowner sends a request to the classification society surveyor, indicating the intention to perform an underwater survey.
    • A detailed plan of the ship's external hull features is submitted, showing the location of frames, bulkheads, welding lines, openings, etc.
    • The hull surface is cleaned before the survey to remove any marine growth or fouling that could obscure the inspection.
    • A diving company, approved by the classification society, is appointed to conduct the underwater inspection.
  • A self-propelled survey vehicle equipped with the following tools is used:
    • Long-Range Light TV Camera to aid in steering and checking hull deterioration.
    • High-resolution colour TV Camera to provide a close-up view of the hull coating and welded seams.
    • 35mm Still Camera to capture still images.
    • Ultrasonic Probe for measuring plate thickness.
    • Depth Meter and Speed Indicator to provide accurate data on the vehicle's depth and movement.
    • Umbilical Cable to connect the survey vehicle to the survey boat, transmitting power and information.
  • The survey boat is to be equipped with:
    • A control console with TV monitors.
    • Plate thickness printouts.
    • Audio and video cassette recorders.
    • Playback units.
    • Diver communication systems.
    • Vehicle control systems and associated instruments.
  • Operation:
    • The survey vehicle is taken underwater by a diver to the survey starting point.
    • Using TV monitors and shell expansion plans as guides, the vehicle is navigated over the hull, focusing on the bottom structure, sides, stern frame, rudder, propeller, bilge keel, and hull openings.
    • All images, data, and information are recorded and transmitted back to the survey boat.
    • Detailed pictures of the stern frame, rudder, propeller, bilge keel, and hull openings are captured.
    • Divers are deployed to measure stern tube bearing wear, pintle clearance, and inspect stern seals, anodes, and rudder stock palm coupling bolts.
    • All recorded video and audio, including conversations between the surveyor and drivers, thickness printouts, measurements, and pictures are analyzed to determine the detailed underwater condition of the vessel.
    (b) Before an underwater survey is accepted by the surveying authority, the following conditions must be met:

    The vessel's owner submits a request to the surveyor, including:
    • The proposed date and location for the survey.
    • General information about the diving company.
    • A declaration that the vessel has not suffered any damage due to grounding, collision, or other incidents.
    The ship's master or owner’s representative must provide a declaration confirming:
    • Any suspected or actual damage to the hull since the last dry-docking.
    • The underwater portion of the hull is protected by a suitable paint scheme that is of adequate thickness and remains valid until the next dry-dock.

    • The survey site should be in a protected area with calm and clear water, ensuring good underwater visibility. Attention must be given to the effects of currents and tides.
    • The hull must be clean for the external survey. The surveyor must be satisfied with the method and quality of the pictorial presentation, ensuring that it provides a reliable assessment of the hull's condition.
    • The underwater examination should be conducted by an approved diving company using closed-circuit TV and two-way communication, which can be monitored by the surveyor.
    • The vessel should be in as light an operating condition as possible to facilitate the survey.
    • Means must be available for the surveyor to examine the outside shell plating above the waterline.
    • Any required repairs identified during the survey must be carried out to the satisfaction of the attending surveyor.
    (c) While the importance of each item may vary depending on the vessel and its specific requirements, below is a list of items that should be included in an underwater survey in order of importance:
    • Underwater Hull: General condition of the hull below the waterline.
    • Bottom and Shell Plating: Inspection for corrosion, damage, and fouling.
    • Shell Openings: Examination of openings such as sea chests, drain plugs, and overboard discharge points.
    • Stern Tube Oil Leaks: Check for leaks around the stern tube.
    • Propeller Blade: Inspection for damage, wear, and fouling.
    • Rudder: Inspection for damage, wear, and clearances.
    • Sea Chest Opening and Grating: Examination for blockages, damage, and fouling.
    • Anodes: Check the condition and effectiveness of cathodic protection anodes.
    • Bilge Keel: Inspection for damage and fouling.
    • Drain Plugs: Ensure all drain plugs are secure and in good condition.
    • Overboard Valve Openings: Check for proper operation and condition.
    • Forward Area: Inspection for any damage due to anchor and chain movement.
    Q5 (16 Marks) Ship Stability πŸ”₯ Repeated 3x

    Describe the effect of the following on the ship’s stability. (16)

    (a) Ice formation on superstructures

    (b) Effects of wind and waves

    (c) Changes that takes place during the ships voyage

    (d) Bilging of a compartment

    (e) While water is being pumped out from the dry dock.

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    Part (a)

    Ice formation on superstructure:

    Ice accumulating on the superstructure adds mass high up on the ship. This raises the centre of gravity (G), decreasing the metacentric height (GM). A lower GM reduces stability, making the ship more tender (more easily rolled) and increasing the period of roll. The ship becomes more susceptible to capsizing.

    Part (b)

    Effect of Wind and Waves

    Wind:

    • High freeboard or tall superstructures increase windage, leading to a greater rolling effect.
    • Rolling caused by wind reduces stability, especially if the ship remains heeled for a prolonged period.

    Waves:

    • Large waves, especially when the ship is on the crest, can cause a significant loss of stability due to reduced underwater buoyant volume.
    • The ship may develop excessive heeling or capsizing tendencies.
    • Long ships are more vulnerable to wave action due to greater surface exposure, further reducing stability.
    Part (c)

    Changes During a Voyage

    Consumption of fuel and water:

    • Stability depends on the location of consumed or emptied tanks. Loss from low-level tanks increases G and decreases stability, while consumption from high-level tanks lowers G, increasing stability.

    Ballast exchange or transfer:

    • Transferring or exchanging ballast affects GM based on tank locations. Removal of low ballast raises G, while adding ballast low down lowers G and improves stability.

    Sea conditions:

    • Rolling and pitching caused by waves and wind can disrupt stability and amplify heeling or capsizing risks.

    Ice formation:

    • Ice buildup on decks or superstructures raises G, reducing GM and stability.

    Shifting of cargo:

    • Movement of cargo can create a list or cause instability if the shift raises the centre of gravity or reduces symmetrical weight distribution.
    Part (d)

    Bilging of a compartment

    Side Compartments (Port/Starboard):

    • Bilging a side compartment creates a virtual loss of GM due to the asymmetric flooding. This results in excessive list and increases the danger of capsizing.

    Forward or Aft Compartments:

    • Bilging forward or aft of the midship causes trim by head or stern, respectively.
    • Reserve buoyancy is reduced or lost, significantly decreasing stability.
    • Complete loss of reserve buoyancy results in sinking.
    Part (e)

    Pumping water out of a dry dock:

    • During dry docking, as water is pumped out, the stern settles on the keel blocks first, creating an upthrust.
    • This upthrust causes a virtual reduction in GM, which can destabilize the ship.
    • If GM becomes negative, the vessel may heel to one side or slip off the keel blocks, potentially leading to capsizing.
    • It is critical to ensure a positive GM during the entire dry docking process to maintain stability.
    Q6 (16 Marks) Ship Stability πŸ”₯ Repeated 3x

    (a) Describe the stability requirements of a ship for dry-docking. (6)

    (b) A ship of 8000 tonne displacement, 110m long, floats in sea water of 1.024 t/m3 at draughts of 6m forward and 6.3 m aft. The TPC is 16, LCB 0.6 m aft of midships, LCF 3m aft of midships and MCT1cm 65 tonne m, the vessel now moves into fresh water of 1.000 t/m3. Calculate the distance a mass of 50 tonne must be moved to bring the vessel to an even keel and determine the final draught. (10)

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    Part (a)

    Stability requirements for dry-docking.

    Before docking, the ship must be stable with an adequate GM, and the docking operation must not cause a list or loss of stability. As the water is pumped out, the keel-block reaction reduces the effective buoyancy, lowering KB and BM and hence GM; the ship must retain sufficient GM throughout so that it does not heel over on the blocks. Requirements:

    • Adequate initial GM and even keel (or slight trim), with no list; ballast arranged to give even keel and to press up or empty tanks to avoid free surface.
    • The docking weight, draft and trim must be such that the keel blocks contact evenly over the full keel length; excessive trim overloads the aft blocks.
    • The ship must be central over the keel line and the dock level; the reaction builds gradually and the GM must remain positive at all stages.
    • No weights moved during docking; the stability data (hydrostatics, docking plan) used to check the condition.

    If the GM becomes too small or negative during docking, the vessel can heel and capsize on the blocks, so a stability margin is insisted on.

    Part (b)

    Moving a mass to bring the vessel to an even keel in fresh water.

    Ship 8,000 t displacement, 110 m long, floats in sea water (1.024 t/m3) at draughts 6.0 m forward and 6.3 m aft. TPC = 16, LCB 0.6 m aft of midships, LCF 3 m aft of midships, MCT1cm = 65 t-m. The vessel moves into fresh water (1.000 t/m3). Calculate the distance a 50 t mass must be moved to bring the vessel to an even keel, and the final draught.

    Step 1 - change of mean draught due to density.

    Waterplane area A = TPC x 100/rho = 16 x 100/1.024 = 1562.5 m2.

    Volume in sea water = 8000/1.024 = 7812.5 m3; volume in fresh water = 8000/1.000 = 8000 m3. Increase in volume = 187.5 m3.

    Increase in mean draught = 187.5/1562.5 = 0.12 m. New mean draught = 6.15 + 0.12 = 6.27 m.

    Step 2 - change of trim due to density.

    The added buoyancy (187.5 t) acts at the centre of flotation (3 m aft of midships), while the original centre of buoyancy is at LCB 0.6 m aft of midships. The moment about the centre of flotation = 187.5 x (3 - 0.6) = 187.5 x 2.4 = 450 t-m.

    Change of trim = moment/MCT1cm = 450/65 = 6.92 cm. Since the added buoyancy is aft of the original CB, it lifts the stern, reducing the stern-down trim. Original trim = 0.3 m (6.3 - 6.0) by the stern. New trim = 0.3 - 0.069 = 0.231 m by the stern.

    Step 3 - move the 50 t mass to bring the vessel to an even keel.

    To remove the 0.231 m (23.1 cm) stern trim, the required change of trim moment = 23.1 x 65 = 1502 t-m. For a 50 t mass, the distance moved = 1502/50 = 30.0 m. The mass must be moved forward (towards the bow) by 30 m.

    Final draught: since the displacement is unchanged, the even-keel draught equals the new mean draught = 6.27 m.

    Answer: move the 50 t mass 30 m forward; final even-keel draught = 6.27 m.

    Q7 (16 Marks) Ship Resistance & Propulsion

    (a) Describe the effect of cavitations on the propeller blades. (6)

    (b) A ship of 15000 tonne displacement has an Admiralty Coefficient, based on shaft power, of 420. The mechanical efficiency of the machinery is 83%, shaft losses 6%, propeller efficiency 65% and QPC 0.71. At a particular speed the thrust power is 2550 Kw. Calculate: (10)

    (i) Indicated power

    (ii) Effective power

    (iii) Ship speed.

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    Part (a)

    Effect of cavitation on the propeller blades.

    Cavitation is the formation and collapse of vapour bubbles on the propeller blade when the local pressure falls below the vapour pressure of water. Effects:

    • Erosion/pitting: the collapse of the bubbles on the blade surface produces high local pressures that erode the blade material, causing pitting, loss of material and reduced blade life.
    • Loss of thrust and efficiency: the vapour bubbles reduce the effective blade area and the density of the fluid, so the propeller produces less thrust for the same power and the efficiency falls.
    • Vibration and noise: the collapse of bubbles produces noise and vibration, which can be transmitted to the hull and cause discomfort and structural fatigue.
    • Reduced performance: cavitation limits the maximum thrust and speed, and can cause the propeller to "race" and lose grip.

    Cavitation is controlled by using a larger blade area (higher blade area ratio), a lower blade loading, a suitable pitch distribution, and by avoiding excessive speed and loading; the design should keep the cavitation number above the critical value.

    Part (b)

    Indicated power, effective power and ship speed.

    Ship 15,000 t displacement, Admiralty coefficient (based on shaft power) = 420. Mechanical efficiency 83%, shaft losses 6%, propeller efficiency 65%, QPC 0.71. At a particular speed the thrust power is 2550 kW.

    (i) Indicated power.

    Propeller efficiency = thrust power/delivered power, so delivered power = 2550/0.65 = 3923 kW.

    Shaft power = delivered power/(1 - shaft losses) = 3923/0.94 = 4173 kW.

    Indicated power = shaft power/mechanical efficiency = 4173/0.83 = 5028 kW.

    (ii) Effective power.

    QPC = effective power/delivered power, so effective power = 3923 x 0.71 = 2785 kW.

    (iii) Ship speed.

    Admiralty coefficient C = Delta^(2/3) V^3 / P_shaft. Delta^(2/3) = 15000^(2/3) = 608.

    V^3 = C x P_shaft/Delta^(2/3) = 420 x 4173/608 = 1,752,660/608 = 2882.7.

    V = 2882.7^(1/3) = 14.23 knots.

    Answer: indicated power about 5028 kW; effective power about 2785 kW; ship speed about 14.2 knots.

    Q8 (16 Marks) Hull Construction πŸ”₯ Repeated 7x

    (a) With respect to Buoyancy of a vessel:

    What do you understand by reserve buoyancy what happen if the lost buoyancy is greater than the reserve buoyancy? (6)

    (b) A forward deep tank 12 m long extends from a longitudinal bulkhead to the ship’s side. The widths of the tank surface measured from the longitudinal bulkhead at regular intervals are 10, 9, 7, 4 and 1 m. Calculate the second moment of area of the tank surface about a longitudinal axis passing through its centroid. (10)

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    Part (a)

    Reserve Buoyancy

    Reserve buoyancy is the watertight volume above the waterline of a vessel. It represents the potential buoyancy that can be utilised to keep the ship afloat when additional weight is added or if some buoyancy is lost due to damage (e.g., bilging).

    When a mass is added to the ship or buoyancy is lost (e.g., due to flooding of a compartment), the reserve buoyancy is converted into active buoyancy by an increase in draught.

    If the lost buoyancy (due to flooding or damage) is greater than the reserve buoyancy, the vessel will no longer have sufficient buoyant force to counteract its weight, causing it to sink.

    Part (b)
    Q9 (16 Marks) Ship Stability πŸ”₯ Repeated 8x

    (a) Define longitudinal centre of gravity (LCG) and longitudinal centre of buoyancy (LCB). (6)

    (b) The immersed cross-sectional areas of a ship 120m long, commencing from aft are 2, 40, 79, 100, 103, 104, 104, 103, 97, 58 and 0 m2. Calculate:

    (i) Displacement

    (ii) Longitudinal position of the centre of buoyancy. (10)

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.
    Part (b)

    Given:

    $$Common \space interval \space (h) \space = \space {{L} \over h} \space = \space {{120} \over 10} \space = \space 12 $$

    Cross-sectional area

    SM

    Product of volume

    Lever

    Product of 1st moment

    2

    1

    2

    +5

    +10

    40

    4

    160

    +4

    +640

    79

    2

    158

    +3

    +474

    100

    4

    400

    +2

    +800

    103

    2

    206

    +1

    +206

    Ξ£MA = +2130

    104

    4

    416

    0

    0

    104

    2

    208

    -1

    -208

    103

    4

    412

    -2

    -824

    97

    2

    194

    -3

    -582

    58

    4

    232

    -4

    -928

    0

    1

    0

    -5

    0

    Ξ£βˆ‡ = 2388

    Ξ£MF = -2542

    $$Displacement \space = \space \rho \times {{h} \over 3} \times \sum βˆ‡ \space tonne $$

    $$=1.025\times{{12}\over3}\times2388$$

    $$Displacement \space = \space 9790.8 tonne$$

    Centre of buoyancy from midship (LCB)

    $$LCB\:=\:h\times({{\sum M_{A}+\sum M_{F}}\over\sum\nabla})$$

    $$=12\times({{2130-2542}\over2388})$$

    $$LCB \space = \space -2.07m fwd$$

    Q10 (16 Marks) Hull Construction πŸ”₯ Repeated 2x

    (a) What is the effect on fuel consumption per unit time, if the ship’s speed is outside its operation range? (6)

    (b) An oil tanker 160m long and 22m beam floats at a draught of 9m in seawater. Cw is 0.865. The midships section is in the form of a rectangle with 1.2m radius at the bilges. A midships tank 10.5m long has twin longitudinal bulkheads and contains oil of 1.4 m3/t to a depth of 11.5m. The tank is holed to the sea for the whole of its transverse section. Find the new draught. (10)

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    Part (a)

    Effect on fuel consumption per unit time if the ship's speed is outside its operating range.

    Fuel consumption per unit time is proportional to the power developed, and power is approximately proportional to the cube of the speed (P proportional to V^3). If the ship is run above its design (operating) speed, the fuel consumption per hour rises very steeply (as V^3), so the specific fuel consumption per tonne-mile worsens and the voyage cost rises sharply. If the ship is run well below the operating range, the engine operates at a poor point on its specific fuel consumption curve (high SFC at low load), the propeller may be inefficient, and although the fuel per hour falls, the fuel per tonne-mile may not improve as much as expected; also the voyage time increases. Hence the most economical speed is near the design point where the SFC is minimum and the propulsive efficiency is good; running outside this range increases fuel consumption per unit of useful work (per tonne-mile) and per unit time at high speed.

    Part (b)

    New draught of the oil tanker when the midship tank is holed.

    Oil tanker 160 m long, 22 m beam, floats at a draught of 9 m in sea water. Cw = 0.865. The midship section is a rectangle with 1.2 m radius at the bilges. A midship tank 10.5 m long has twin longitudinal bulkheads and contains oil of 1.4 m3/t to a depth of 11.5 m. The tank is holed to the sea for the whole of its transverse section. Find the new draught.

    Waterplane area Aw = Cw x L x B = 0.865 x 160 x 22 = 3044.8 m2.

    Midship section area (rectangle with bilge radius r=1.2 m): Ams = B x d - (4 - pi) r^2 = 22 x 9 - 0.858 x 1.44 = 198 - 1.236 = 196.76 m2.

    The tank is 10.5 m long and spans the full beam (twin longitudinal bulkheads divide it, but it is holed for the whole transverse section). Volume of the tank below the original waterline = Ams x 10.5 = 196.76 x 10.5 = 2066 m3.

    The tank contains oil of density rho_o = 1/1.4 = 0.714 t/m3. When holed, sea water (1.025 t/m3) replaces the oil, so the net loss of buoyancy is the volume times the relative density difference:

    Vlost = 2066 x (1 - 0.714/1.025) = 2066 x (1 - 0.697) = 2066 x 0.303 = 626 m3.

    The flooded tank provides no increase of buoyancy, so the effective sinking waterplane = Aw - (10.5 x 22) = 3044.8 - 231 = 2813.8 m2.

    Sinkage = Vlost/effective waterplane = 626/2813.8 = 0.222 m.

    New draught = 9 + 0.22 = 9.22 m.

    Answer: the new draught is about 9.2 m.

    Q1 (16 Marks) Ship Stability πŸ”₯ Repeated 2x

    (a) Explain the six types of ship motions along different axes. Discuss the causes of these motions and their impact on the vessel's stability and performance. (8)

    (b) Discuss the methods employed to reduce these motions and enhance the vessel's stability and comfort. (8)

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    Part (a)

    The six ship motions along the three axes.

    A ship has six degrees of freedom, three translational and three rotational, about the three mutually perpendicular axes through the centre of gravity (longitudinal x, transverse y, vertical z):

    1. Surge - linear motion along the longitudinal (fore-and-aft) axis, caused by the variation of thrust and resistance, and by following/head seas pushing the hull. It affects speed and fuel consumption and can cause slamming of the bow in head seas.
    2. Sway - linear motion along the transverse axis, caused by beam seas, wind and rudder action; it produces lateral acceleration, cargo shift risk and can combine with roll.
    3. Heave - vertical linear motion along the vertical axis, caused by the passage of waves under the hull changing buoyancy; it produces vertical accelerations, slamming, and affects the effective GM and the comfort of crew and cargo.
    4. Roll - rotation about the longitudinal axis, caused by beam seas, wind, and the heeling moment of rudder or cargo; it is the most dangerous motion for stability because it can lead to large angles, cargo shift, and in resonance to parametric roll or capsize.
    5. Pitch - rotation about the transverse axis, caused by head and following seas; it produces bow/stern slamming, deck wetness, propeller emergence and affects speed and comfort.
    6. Yaw - rotation about the vertical axis, caused by asymmetric wave forces, wind and rudder; it affects course-keeping, increases resistance and can lead to broaching in following seas.

    Causes and impact: all motions are excited by wave forces, wind, and the ship's own propulsion; their amplitude depends on the encounter frequency relative to the ship's natural frequencies (resonance). They reduce stability margin (especially roll), increase structural loading (pitch/heave slamming), cause cargo damage and crew discomfort, and increase resistance and fuel consumption.

    Part (b)

    Methods to reduce these motions and enhance stability and comfort.

    • Increase GM (stiffen the ship) by lowering KG (ballast low, remove top weight) to reduce roll amplitude, but avoid excessive stiffness which gives short, uncomfortable periods.
    • Fit bilge keels, anti-roll tanks (passive/active U-tube tanks), fin stabilisers and gyro stabilisers to damp roll.
    • Use active/passive anti-pitching and anti-heave devices, and design the hull with fine ends and adequate freeboard to reduce slamming and deck wetness.
    • Operate at a speed and heading that avoids resonance (change course/speed to move the encounter frequency away from natural frequencies), and avoid beam seas for roll.
    • Use ballast and trim to improve stability and reduce motions; press up tanks to remove free surface.
    • Design with adequate GM, metacentric height and righting-lever (GZ) reserve, and use stabilisers, bilge keels and anti-rolling tanks; for comfort, reduce vertical accelerations by speed reduction and course changes.
    • Structural measures: adequate freeboard, flare, and bow form to reduce slamming; and operational measures such as weather routing.
    Q2 (16 Marks) Ship Stability

    Explain the GZ (Righting Arm) curve used in ship stability. Discuss its significance, the information that can be obtained from the curve, and its role in assessing a ship's stability under various conditions.

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    The GZ (righting arm) curve.

    The GZ curve is a graph of the righting lever GZ (the horizontal distance between the centre of gravity G and the centre of buoyancy B, i.e. the lever arm of the righting couple) plotted against the angle of heel. It is the fundamental curve of transverse stability.

    Significance: for any angle of heel, the righting moment = Delta x GZ, so the curve shows how the ship's ability to right itself varies with heel. It is obtained from the cross-curves of stability (GZ values for a fixed KG) corrected for the actual KG and free-surface effects.

    Information obtained from the curve:

    • The initial slope of the curve at the origin equals GM (the metacentric height), since GZ = GM sin(theta) for small angles.
    • The maximum righting lever (GZ max) and the angle at which it occurs (angle of maximum stability, typically 25-40 deg).
    • The range of stability: the angle from the upright to the angle of vanishing stability (where GZ returns to zero, typically 60-90 deg).
    • The area under the curve up to any angle, which is proportional to the dynamical stability (the work done in heeling the ship) and is used to assess stability in a seaway and against wind heeling.
    • The angle of loll (if the curve starts below the axis, indicating negative GM).
    • The effect of free surface, which reduces the curve by the free-surface correction.

    Role in assessing stability: the curve is compared with the statutory criteria (e.g. IMO intact stability criteria: area under the curve up to 30 deg not less than a value, area up to 40 deg, GZ max not less than 0.2 m at an angle not less than 30 deg, GM not less than 0.15 m, etc.). It is used to check the stability in all loading conditions, to determine the maximum permissible KG, and to assess the ship's behaviour in wind and waves. A ship with a large GM has a steep initial curve but a small range and a sharp, uncomfortable roll; a ship with a small GM has a shallow curve, a large range but a small righting moment. The curve therefore balances stiffness, range and dynamical stability.

    Q3 (16 Marks) Ship Stability

    Explain the different types of framing systems used in ship construction. Discuss their design, structural arrangement, and advantages in ensuring the vessel's strength and stability. (16)

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    Framing systems used in ship construction.

    The framing system is the arrangement of the transverse and longitudinal stiffeners that support the shell and deck plating. The three main systems are:

    1. Transverse framing system.

    In this system the principal stiffening members run transversely (athwartships), i.e. the frames are transverse rings spaced closely (about 0.6-0.9 m apart) running from the keel up the side to the deck, with the longitudinal members (stringers, keelsons) spaced widely. The shell plating is supported by the transverse frames. It is used for the ends of ships (fore and aft peaks) and for ships where the transverse strength is important and the length is moderate (e.g. general cargo ships, tugs, and the ends of most ships). Advantages: simple, cheap, good transverse strength, easy to build; the transverse frames resist the transverse pressure of the water and the racking loads. Disadvantage: the longitudinal strength is provided mainly by the shell plating and stringers, so it is less efficient for long, deep ships.

    1. Longitudinal framing system.

    In this system the principal stiffening members run fore-and-aft (longitudinally), closely spaced (about 0.6-0.9 m apart) along the deck, bottom and sides, with widely spaced transverse webs (transverse frames/ring frames) every few metres. The shell and deck plating is supported by the longitudinal stiffeners. It is used for the middle body of large ships (tankers, bulk carriers, container ships) where the longitudinal bending strength is critical. Advantages: high longitudinal strength and section modulus for a given weight, good resistance to buckling, and efficient use of material; the closely spaced longitudinals give a strong, stiff hull girder. Disadvantage: more complex and expensive to build, and the transverse strength must be provided by the widely spaced webs.

    1. Combined (mixed) framing system.

    This uses longitudinal framing in the middle body (where longitudinal bending is greatest) and transverse framing in the ends (where the hull is finer and transverse strength/racking is important). It combines the advantages of both: high longitudinal strength amidships and good transverse strength and economy at the ends. It is the most common arrangement for large merchant ships.

    Design and structural arrangement: the framing must support the plating against the hydrostatic and wave pressure, transmit the loads to the hull girder, and provide the required section modulus for longitudinal bending. The spacing, scantlings and the connection of the frames to the keel, stringers, deck and bulkheads are arranged to give continuity of strength and to avoid stress concentrations. The framing also provides the transverse rings that resist racking and the local loads of cargo and machinery.

    Advantages in ensuring strength and stability: the framing system determines the hull girder's section modulus (longitudinal strength), the local strength of the plating, and the distribution of material. A well-designed framing system gives adequate strength with minimum weight, resists buckling and fatigue, and maintains the watertight integrity and the stability of the hull.

    Q4 (16 Marks) General πŸ”₯ Repeated 3x

    (a) List SIX hazards associated with the carriage of liquefied gas in bulk. (8)

    (b) Sketch and describe the details of construction of a free-standing prismatic tank within a gas carrier designed to carry liquefied gas (LPG). (8)

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    Part (a)

    SIX hazards associated with the carriage of liquefied gas in bulk.

    1. Flammability/explosion: the vapour is highly flammable; in the flammable range with air it can be ignited by any ignition source, causing fire or explosion.
    2. Cryogenic (extreme cold) hazard: LPG is carried near -50 deg C; spillage causes severe cold burns, embrittlement and fracture of steel, and condensation of moisture.
    3. Asphyxiation/toxicity: heavier-than-air vapour displaces oxygen causing oxygen deficiency and asphyxiation in enclosed and low spaces; some gases are toxic.
    4. Pressure hazard: high vapour pressure with temperature rise can over-pressure tanks and piping; uncontrolled release can cause BLEVE (boiling liquid expanding vapour explosion).
    5. Heavy vapour cloud dispersion: the vapour is denser than air, flows along deck/water, travels to distant ignition sources and accumulates in enclosed spaces, bilges and accommodation.
    6. Equipment/materials failure and corrosion: leakage from piping, valves and flanges due to thermal contraction, corrosion and incompatible materials, plus liquid carry-over into machinery spaces causing damage.

    These require gas-tight equipment, gas detection, venting, exclusion of ignition sources and careful temperature/pressure control.

    Part (b)

    Construction of a free-standing prismatic tank within a gas carrier (LPG).

    A free-standing prismatic tank is a self-supporting rectangular tank built of cryogenic nickel-steel or aluminium, carried in a hold and supported on the hull. Sketch: a rectangular tank with flat or corrugated plates, a top dome carrying the cargo piping, safety and pressure/vacuum valves, access hatch and level gauging; the tank is supported on load-bearing seatings (wood/mild-steel chocks) at the bottom, with side hoppers and a roof, and is surrounded by insulation (polyurethane foam) to keep the cargo cold and reduce boil-off.

    Features:

    • Self-supporting (type A or B) prismatic tank, adequately insulated externally; a secondary barrier (drip tray/gas-tight) is fitted so any leakage is contained.
    • The tank bottom transfers loads through seatings to the hull; the tank top dome carries the cargo pipes, relief valve, pressure-vacuum valve, gas detection and remote-operated valves.
    • The tank is designed to withstand liquid sloshing and cryogenic temperature, with corrugated internal stiffening and notch-tough welded joints.
    • Venting systems, dry-pipe vent risers and flame arresters prevent vapour accumulation; high-level and emergency shut-down (ESD) systems are fitted.
    • The cargo is carried at atmospheric or slightly above pressure; a relief valve protects against over-pressure.

    The sketch should show the tank seatings, insulation, dome, and secondary barrier.

    Q5 (16 Marks) Hull Construction πŸ”₯ Repeated 7x

    (a) With reference to fatigue of hull structures explain the influence of stress level and cyclical frequency on expected operating life. (6)

    (b) Explain the influence of material defects on the safe operating life of forged components of stern fittings. (5)

    (c) State the factors which influence the possibility of fatigue cracking of a bed-plate transverse girder and explain how the risk of such cracking can be minimized. (5)

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    Q6 (16 Marks) Ship Stability πŸ”₯ Repeated 2x

    (a) List the variables which affect the force on a rudder. (6)

    (b) A triangular bulkhead is 7 m wide at the top and has a vertical depth of 8 m. Calculate the load on the bulkhead and the position of centre of pressure if the bulkhead is flooded with sea water on only side: (10)

    (i) To the top edge

    (ii) With 4 m head to the top edge.

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    when a rudder is turned from the centreline plane, to any angle, a force acts on the rudder, given by,

    $$F=kAV^2N$$

    Where,

    • k = Constant
    • A = Area of rudder in m2
    • V = Ship's speed in m/s

    The constant is dependent on size of rudder, rudder angle and density of water.

    Variables affecting force on rudder:

    • Rudder angle
    • Density of water
    • Ship speed

    These are the three variables, while all other are constant parameters i.e.

    • Area of rudder
    • Size of rudder.
    Q7 (16 Marks) Ship Stability πŸ”₯ Repeated 2x

    (a) Explain the effect on GM during the filing of a double - bottom tank (6)

    (b) An oil tanker 160m long and 22m beam floats at a draught of 9m in seawater. Cw is 0.865. The midship section is in the form of a rectangle with 1.2m radius at the bilges. A midship tank 10.5m long has twin longitudinal bulkheads and contains oil of 1.4m3/t to a depth of 11.5m. The tank is holed to the sea for the whole of its transverse section. Find the new draught. (10)

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    Part (a)

    Effect on GM during the filling of a double-bottom tank.

    As a double-bottom tank is filled, two effects act on GM:

    • The added weight of ballast is low in the ship (deep in the double bottom), so it lowers the centre of gravity, which increases GM (stiffening the ship). This is the dominant effect once the tank is full.
    • While the tank is partially full, a free surface exists and produces a free-surface effect that reduces the effective GM. The free-surface correction is rho x i / Delta, where i is the second moment of area of the free surface about its longitudinal axis (i = L B^3/12). For a wide double-bottom tank this can be significant.

    Hence during filling, the effective GM first falls (as the free surface develops) and then rises as the tank approaches full and the free surface disappears; when the tank is completely full (pressed up) the free-surface effect is zero and the GM is increased by the low weight. The net effect of a full double-bottom tank is an increase in GM (stiffer ship), but the transient free-surface loss during filling must be watched, especially in a tender ship.

    Part (b)

    New draught of the oil tanker when the midship tank is holed.

    Oil tanker 160 m long, 22 m beam, floats at a draught of 9 m in sea water. Cw = 0.865. The midship section is a rectangle with 1.2 m radius at the bilges. A midship tank 10.5 m long has twin longitudinal bulkheads and contains oil of 1.4 m3/t to a depth of 11.5 m. The tank is holed to the sea for the whole of its transverse section. Find the new draught.

    Waterplane area Aw = Cw x L x B = 0.865 x 160 x 22 = 3044.8 m2.

    Midship section area (rectangle with bilge radius r=1.2 m): Ams = B x d - (4 - pi) r^2 = 22 x 9 - 0.858 x 1.44 = 198 - 1.236 = 196.76 m2.

    Volume of the tank below the original waterline = Ams x 10.5 = 196.76 x 10.5 = 2066 m3.

    The tank contains oil of density rho_o = 1/1.4 = 0.714 t/m3. When holed, sea water (1.025 t/m3) replaces the oil, so the net loss of buoyancy is the volume times the relative density difference:

    Vlost = 2066 x (1 - 0.714/1.025) = 2066 x 0.303 = 626 m3.

    The flooded tank provides no increase of buoyancy, so the effective sinking waterplane = Aw - (10.5 x 22) = 3044.8 - 231 = 2813.8 m2.

    Sinkage = Vlost/effective waterplane = 626/2813.8 = 0.222 m.

    New draught = 9 + 0.22 = 9.22 m.

    Answer: the new draught is about 9.2 m.

    Q8 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 7x

    (a) What is meant by the Admiralty Coefficient and the Fuel Coefficient? (6)

    (b) A ship of 14900 tonnes displacement has a shaft power of 4460 kW at 14.55 knots. The shaft power is reduced to 4120 kW and the fuel consumption at the same displacement is 541 kg/h. Calculate the fuel coefficient for the ship. (10)

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    The Admiralty Coefficient (C) is a method for estimating the propulsion power needed for a newly built ship. It's considered relatively constant for a given ship design. The formula is:

    $$C=\frac{\Delta^{2/3}\times V^3}{BP}$$

    Where:

    • C = Admiralty Coefficient
    • Ξ” = Displacement in tonnes (weight of the ship when fully loaded)
    • V = Speed in knots
    • BP = Brake power in kilowatts (power delivered by the ship's engine)

    A higher Admiralty Coefficient indicates a more efficient ship design, meaning it requires less power to achieve a given speed. Values typically range from 350 to 600.

    Fuel Coefficient:

    The Fuel Coefficient (F.C.) is used to calculate a ship's daily fuel oil consumption. The formula is:

    $$Daily\:fuel\:oil\:consumption\:=\:\frac{\Delta^{2/3}\times V^3}{FC}$$

    Where:

    • F.C. = Fuel Coefficient
    • Ξ” = Displacement in tonnes
    • V = Speed in knots

    The Fuel Coefficient can vary significantly, with typical values ranging from 40,000 to 120,000. A higher Fuel Coefficient implies greater fuel efficiency (lower daily fuel consumption) for a given speed and displacement.

    Part (b)

    $$admiraty\:coefficient\:\left(C\right)=\:\frac{\Delta^{2/3}V^3}{Shaft\:power}=\frac{\Delta^{\frac23}\times V^3}{SP}$$

    $$\frac{SP_1}{SP_2}=\frac{V_1^3}{V_2^3}$$

    $$\frac{4460}{4120}=\frac{14.55^3}{V_2^3}$$

    $$V_2=14.17kntos$$

    $$Fuel\:consumption\:per\:hour=541\operatorname{\mathrm{\:kg}}\:per\:hour$$

    $$Fuel\:consumption\:per\:day\:=\:541\times24=12.98t\:per\:day$$

    $$Fuel\:coefficient=\frac{\Delta^{\frac23}\times V_2^3}{Fuel\:consumption\:per\:day}$$

    $$=\:\frac{14900^{\frac23}\times14.17^3}{12.98}$$

    $$=132726.9$$

    Q9 (16 Marks) Ship Stability πŸ”₯ Repeated 2x

    (a) Describe the stability requirements of a ship for dry-docking. (6)

    (b) The Β½ ordinates of a waterplane at 15m intervals, commencing from aft, are 1, 7, 10.5, 11, 11, 10.5, 8, 4 and 0m. Calculate: (10)

    (a) TPC

    (b) Distance of the centre of flotation from midships

    (c) Second moment of area of the waterplane about a transverse axis through the centre of flotation.

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    Part (a)

    Stability requirements for dry-docking.

    Before docking, the ship must be stable with an adequate GM, and the docking operation must not cause a list or loss of stability. As the water is pumped out, the keel-block reaction reduces the effective buoyancy, lowering KB and BM and hence GM; the ship must retain sufficient GM throughout so that it does not heel over on the blocks. Requirements:

    • Adequate initial GM and even keel (or slight trim), with no list; ballast arranged to give even keel and to press up or empty tanks to avoid free surface.
    • The docking weight, draft and trim must be such that the keel blocks contact evenly over the full keel length; excessive trim overloads the aft blocks.
    • The ship must be central over the keel line and the dock level; the reaction builds gradually and the GM must remain positive at all stages.
    • No weights moved during docking; the stability data (hydrostatics, docking plan) used to check the condition.

    If the GM becomes too small or negative during docking, the vessel can heel and capsize on the blocks, so a stability margin is insisted on.

    Part (b)

    Waterplane calculations.

    The half-ordinates of a waterplane at 15 m intervals, commencing from aft, are 1, 7, 10.5, 11, 11, 10.5, 8, 4 and 0 m. There are 9 ordinates, so the waterplane length is 8 x 15 = 120 m.

    (i) TPC.

    Area A = 2 x (h/3)[y0 + y8 + 4(y1+y3+y5+y7) + 2(y2+y4+y6)]

    = 2 x (15/3)[1 + 0 + 4(7+11+10.5+4) + 2(10.5+11+8)]

    = 10[1 + 4x32.5 + 2x29.5] = 10[1 + 130 + 59] = 10 x 190 = 1900 m2.

    TPC (salt) = A x 1.025/100 = 1900 x 1.025/100 = 19.48 t/cm.

    (ii) Distance of the centre of flotation from midships.

    First moment of area about the after perpendicular:

    M = 2 x (h/3) x sum of (weighted y x distance). Using station distances 0,15,30,...,120 m with Simpson weights (1,4,2,4,2,4,2,4,1):

    weighted sum = 1x0 + 4x(7x15) + 2x(10.5x30) + 4x(11x45) + 2x(11x60) + 4x(10.5x75) + 2x(8x90) + 4x(4x105) + 1x0

    = 0 + 420 + 630 + 1980 + 1320 + 3150 + 1440 + 1680 = 10620.

    M = 2 x 5 x 10620 = 106,200 m3.

    Distance of centroid from AP = M/A = 106,200/1900 = 55.89 m. Midships is at 60 m from AP, so the centre of flotation is 60 - 55.89 = 4.11 m aft of midships.

    (iii) Second moment of area about a transverse axis through the centre of flotation.

    Second moment about AP: I_AP = 2 x (h/3) x sum of (weighted y x distance^2):

    weighted sum = 1x0 + 4x(7x225) + 2x(10.5x900) + 4x(11x2025) + 2x(11x3600) + 4x(10.5x5625) + 2x(8x8100) + 4x(4x11025) + 1x0

    = 0 + 6300 + 18900 + 89100 + 79200 + 236250 + 129600 + 176400 = 735,750.

    I_AP = 2 x 5 x 735,750 = 7,357,500 m4.

    Transfer to the centre of flotation: I_CF = I_AP - A x (55.89)^2 = 7,357,500 - 1900 x 3123.7 = 7,357,500 - 5,935,000 = 1,422,500 m4.

    Answer: TPC = 19.48 t/cm; LCF = 4.11 m aft of midships; I about the transverse axis through CF = about 1.42 x 10^6 m4.

    Q10 (16 Marks) Ship Types & Design πŸ”₯ Repeated 2x

    (a) Describe the effect of cavitations on the propeller blades (6)

    (b) The following data are available from the hydrostatic curves of a vessel.

    Draught (m)

    KB (m)

    KM (m)

    I (m^4)

    4.9

    2.49

    10.73

    65250

    5.2

    2.61

    10.79

    68868

    Calculate the TPC at a draught of 5.05m. (10)

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    Part (a)

    Effect of cavitation on propeller blades:

    Erosion:

    • Cavitation causes the formation and collapse of vapor bubbles on the propeller blade surface.
    • The collapse of these bubbles produces high-pressure shockwaves and microjets that erode the blade material, leading to surface pitting and progressive damage.

    Vibration:

    • Uneven cavitation across the blades leads to imbalanced forces, causing vibrations in the propeller and the ship.
    • These vibrations can reduce the comfort of passengers and crew and stress the ship's structural components.

    Noise:

    • The collapse of vapor bubbles generates loud noise, which can interfere with onboard communication and underwater sonar systems.
    • This noise is a significant concern for naval vessels and marine life.

    Reduced Performance:

    • Cavitation reduces the efficiency of the propeller by causing loss of thrust and torque.
    • The presence of cavitation decreases the propeller’s ability to convert rotational energy into forward motion, lowering the ship's speed and increasing fuel consumption.
    Q1 (16 Marks) Hull Construction πŸ”₯ Repeated 5x

    With reference to membrane tanks for the carriage of liquefied gas at very low temperatures.

    (a) Describe with the aid of a sketch, ONE method of building up the insulation: (5)

    (b) State with reasons the alloy, which is used for the membrane. (5)

    (c) Describe with the aid of a sketch, how the tanks are located and supported. (6)

    (i) Longitudinally

    (ii) Transversely

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    Part (a)

    Membrane tanks for liquefied gas carriage are built with a double hull throughout the cargo length. The insulation system includes the following:

    1. Primary Barrier: A thin membrane made of INVAR (36% Nickel, 64% Iron) with a thickness of 0.7 to 1.2 mm, forming the primary containment layer for the liquefied gas.
    2. Primary Insulation box: A 230 mm thick layer of granulated Perlite insulation packed in plywood boxes, surrounding the primary barrier. Perlite is siliconised to make it impervious to moisture.
    3. Secondary Barrier: A secondary membrane made of the same INVAR material is installed to prevent cargo leakage in case of primary barrier failure.
    4. Secondary Insulation box: Another 300 mm thick layer of granulated Perlite is placed above the secondary barrier to provide additional insulation and prevent thermal transfer.
    Part (b)

    Alloy used for the membrane:

    Invar (36% Ni, 64% Fe) is used for both the primary and secondary barriers. The reason for this choice is its exceptionally low coefficient of thermal expansion. This eliminates the need for expansion joints or corrugations in the membrane design. In addition, Invar remains strong and does not become brittle at the very low temperatures experienced by the liquefied gas. The thin and lightweight nature of Invar maximizes the cargo-carrying capacity of the tank.

    Part (c)

    Membrane tanks are either independent or self-supporting

    , meaning they don't form part of the ship's hull and don't contribute to the ship's structural strength. They can be spherical, cylindrical, or prismatic (box-shaped). Prismatic tanks usually have internal stiffeners like bulkheads, webs, girders, and stiffeners for added structural integrity.

    (i) Longitudinally:

    Tanks are positioned longitudinally within the ship's cargo hold. The tanks are supported longitudinally by anti-roll chocks and anti-lift chocks that resist forces caused by ship motions. These chocks ensure the tank remains stable even in rough sea conditions. They also act as thermal barriers, preventing the transfer of heat between the hull and the tank.

    The chocks also serve as thermal barriers between the hull and cargo and are often constructed of wood or plastic materials.

    (ii) Transversely: Transverse support is provided by anti-pitch chocks and support chocks, which stabilize the tank against lateral forces. These chocks are typically made of plywood or plastic and help prevent thermal stress between the tank and the ship’s structure.

    Q2 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 7x

    With reference to dry docking, define the responsibilities of the Second Engineer and instructions to Junior Engineers: (16)

    (a) Prior to docking.

    (b) Whilst the vessel is in dry dock.

    (c) Prior to flooding and leaving the dock.

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    (a) Prior to Docking:

    Preliminary Preparation:

    • Review various plans, manuals, and previous drydock reports for reference.
    • Prepare a detailed repair list and ensure all required spares are accounted for.
    • Conduct an inventory of spares and requisition necessary items.
    • Gather required tools, including special tools like poker gauges, trammel gauges, gagging tools, etc.
    • Mark all overboard valves, NC (normally closed) and NO (normally open) valves clearly for easy identification.
    • Allocate jobs to team members and discuss the time schedule.
    • Conduct a safety meeting to highlight dry dock hazards and necessary precautions.

    Before Entering Dry Dock:

    • Identify the power requirements and machinery to be operational during docking.
    • Check the shore connection box for proper functionality.
    • Record the soundings of FO (Fuel Oil), LO (Lubricating Oil), and DO (Diesel Oil) tanks.
    • Discharge contents from clean drain tanks and sewage tanks.
    • Carry out Economizer soot-blowing.
    • Change over Main Engine, Diesel Generators, and Boiler to Low Sulfur Gas Oil (LSGO).
    • Stop and clean purifiers.
    • Ensure the low sea chest is open and the high sea chest is shut.
    • Keep firefighting appliances (FFA) on standby.
    • Shut down all non-essential machinery, including MGPS (Marine Growth Prevention System) and FWG (Fresh Water Generator).

    (b) Whilst the Vessel is in Dry Dock

    Upon Arrival:

    • Connect shore power and supplies after ensuring safety checks are completed.
    • Start necessary equipment like cooling water, air compressors, air conditioning, and fridge compressors.
    • Check for jobs assigned by dry dock personnel and prepare accordingly.
    • Attend a safety meeting with dry dock personnel to understand local safety rules and procedures.

    During Dry Docking::

    • Oversee and assist in:
      • Cleaning and inspecting the hull, rudder, sea chest, anodes, and propeller.
      • Measuring propeller drop, checking rudder clearances, and inspecting the stern tube bearing and seal.
      • Servicing underwater valves and overboard valves.
      • Inspecting anchor and cables conditions.
      • Overhauling deck machinery, cranes, elevators, and engine room equipment such as the Main Engine, Diesel Generators, Boiler, and Economizer.
      • Renewing pipes and valves as needed.
      • Performing electrical equipment maintenance and surveys.
      • Supervising service engineers for specific repair jobs.
    • Ensure tank cleaning, welding, and other repair works are completed according to the plan.
    • Run the standby diesel generator daily after starting the priming pump.

    (c) Before Flooding and Leaving the Dry Dock

    Final Checks:

    • Verify that all underwater fittings and drain plugs are securely in place.
    • Ensure all machinery has been boxed back and is ready for operation.
    • Check for any leakage in stern tube seals.
    • Take tank soundings to confirm proper levels.
    • Confirm the proper operation of all underwater valves, overboard valves, sea chests, and vents.
    • Inspect the stern tube tank for any irregularities.

    System Restart:

    • Switch back to ship's power after confirming all systems are functional.
    • Test the proper operation of all machinery and systems to ensure the ship is fully operational before leaving the dock.
    Q3 (16 Marks) Ship Stability πŸ”₯ Repeated 8x

    (a) Explain what is meant by β€œpermissible length” of compartments in passenger ships. (6)

    (b) Describe how the position of bulkheads is determined. (6)

    (c) Briefly describe the significance of the factor of subdivision. (4)

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    Part (a)

    Permissible Length

    Permissible length refers to the maximum length of a ship's compartment that can be flooded while ensuring that the sinkage, heel, or trim remains within acceptable limits without exceeding the floodable length. It ensures the ship remains afloat and stable after flooding.

    Permissible Length Formula:

    $$Permissible\:lenght\:=\:Floodable\:length\:\times Factor\:of\:subdivision$$

    The Factor of Subdivision depends on the ship's length and the nature of its service:

    • For passenger ships, the factor of subdivision is smaller compared to cargo ships.
    • Smaller compartments ensure enhanced safety in case of flooding.
    Part (b)

    Position of Bulkheads

    The position of bulkheads is determined based on the need to comply with the "reasonable amount of damage" criterion in case of flooding:

    • Transverse Watertight Bulkheads should vertically extend up to the margin line.
    • At least one watertight longitudinal bulkhead must be located 20% of the ship's breadth inward from each side.
    • Bulkheads are positioned along the ship's length at intervals equal to the permissible length, but not exceeding the floodable length.
    • The maximum permissible compartment length is limited to 10.7 meters.
    • Collision Bulkhead must be located forward at a distance equal to the permissible length from the forward perpendicular.
    • The ER must be enclosed by two transverse watertight bulkheads, with an aft peak watertight bulkhead enclosing the forward part.
    Part (c)

    Factor of Subdivision

    The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

    Permissible Length Formula:

    $$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

    A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

    Q4 (16 Marks) Ship Stability πŸ”₯ Repeated 14x

    Explain how the period of roll varies with

    (a) The amplitude of roll. (4)

    (b) The radius of gyration. (4)

    (c) The initial metacentric height. (4)

    (d) The location of masses in the ship. (4)

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    The period of roll Tr of a ship is determined by the formula:

    $$T_{r}=\frac{2\pi K}{\sqrt{g\times GM}}$$

    where,

    • K is the radius of gyration (mass moment of inertia)
    • g is the acceleration due to gravity, and
    • GM is the metacentric height.
    Part (a)

    Amplitude of Roll:

    • The amplitude of roll, or the maximum angle of heel, does not affect the period of roll. While a larger amplitude implies a greater heeling moment and faster roll speed, the time taken to complete one full roll cycle remains constant for a given metacentric height and radius of gyration.
    Part (b)

    Radius of Gyration (K):

    • The period of roll is directly proportional to the radius of gyration. A larger radius of gyration (indicating a greater distribution of mass further from the ship's centre of rotation) leads to a longer period of roll. Conversely, a smaller radius of gyration (mass concentrated closer to the centre) results in a shorter period. The distribution of cargo significantly impacts K; cargo concentrated centrally minimises K and the roll period, while dispersed cargo maximises K and the roll period.
    Part (c)

    Initial Metacentric Height (GM):

    • The period of roll is inversely proportional to the square root of the metacentric height (GM). A larger GM (a stiffer ship) leads to a shorter roll period, as the ship quickly returns to its upright position. A smaller GM (a tender ship) results in a longer roll period, with slower return to the upright.
    Part (d)

    Location of Masses in the Ship:

    The location of masses in the ship will effect the GM & K. So the period of roll will be affected.

    • If masses are at bottom, G moves down, GM ↑, period of roll ↓.
    • If masses are at top, G moves up, GM ↓, period of roll ↑.
    • If masses are concentrated at centre, K ↓, period of roll ↓.
    • If masses are away from centre, K ↑, period of roll ↑.
    Q5 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 13x

    Describe the relationship between frictional resistance and

    (a) Ship’s speed. (4)

    (b) the wetted area. (4)

    (c) surface roughness. (4)

    (d) The length of the vessel. (4)

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    Frictional resistance arises due to the interaction between the ship's hull and the water as the vessel moves. This resistance is caused by eddying water adjacent to the hull that is drawn along with the ship. The frictional resistance is given by:

    $$R_{f\:=\:}f\:.\:s\:.\:v^{n}$$

    Where:

    • f = Coefficient of friction
    • s = Wetted surface area
    • v = Ship's speed in knots
    • n = Constant (1.82)
    Part (a)

    Ship Speed (v):

    • Frictional resistance increases with increasing ship speed. The relationship is not perfectly linear, but it's approximately described by the formula Rf = fsv^n (Rf ∝ v^n), where 'n' is a constant (approximately 1.82). This means that a small increase in speed results in a larger increase in frictional resistance.
    Part (b)

    Wetted Area (s):

    • Frictional resistance is directly proportional to the wetted surface area (Rf ∝ s). A larger wetted area (the part of the hull submerged in water) leads to greater frictional resistance. The wetted area increases with the ship's draught (depth in the water).
    Part (c)

    Surface Roughness:

    • Increased surface roughness increases frictional resistance. Roughness can be caused by fouling (marine growth), corrosion, or poor hull maintenance. Regular cleaning and maintenance of the hull surface help to minimise roughness and reduce resistance.
    Part (d)

    Length of the Vessel

    • Frictional resistance is influenced by the vessel's length. As the length increases, the wetted surface area grows, and the coefficient of friction (f) changes, leading to an increase in Rf. Longer vessels experience higher frictional resistance, but streamlined designs can mitigate the effect to some extent.
    Q6 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Describe how the distribution of mass within the ship affects the rolling period? (6)

    (b) A ship of 14000 tonne displacement is 125 m long and floats at draughts of 7.9 m forward and 8.5 m aft. The TPC is 19, GML 120 m and LCF 3 m forward of midships. It is required to bring the vessel to an even keel draught of 8.5m. Calculate the mass which should be added and the distance of the centre of the mass from midships. (10)

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    Part (a)

    The rolling period is influenced by the ship's metacentric height (GM) and radius of gyration (K), which are determined by the location of the masses onboard.

    Masses at the bottom:

    • The centre of gravity (G) moves down.
    • Metacentric height (GM) increases, resulting in greater stability.
    • Rolling period decreases.

    Masses at the top:

    • The centre of gravity (G) moves up.
    • Metacentric height (GM) decreases, reducing stability.
    • Rolling period increases.

    Masses concentrated at the centre:

    • The radius of gyration (K) decreases.
    • Rolling period decreases.

    Masses distributed away from the centre:

    • The radius of gyration (K) increases.
    • Rolling period increases.
    Q7 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Explain how an increase of draught and of displacement influence rolling. (6)

    (b) A pontoon has a constant cross-section as shown in Fig. Given below. The metacentric height is 2.5 m. Find the height of the centre of gravity above the keel. (10)

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    Part (a)

    In case of increase of draught and displacement, this is a case of loading being done.

    Thus, when the ship is being loaded, its GM will start to decrease.

    Now, the time period of roll is given by:

    $$T_{r}=\frac{2\pi k}{\sqrt{GM.g}}$$

    Where,

    • k = radius of gyration
    • GM = metacentric height
    • g = acceleration due to gravity

    Now, since, GM has started to decrease, the Tr will start to increase.

    Thus, the ship will now roll with greater time period. Thus, an increase in draught and displacement, influences rolling.

    Q8 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    (a) Describe the fundamental principle of a propeller. (6)

    (b) A propeller 6m diameter has a pitch ratio of 0.9, BAR 0.48 and when turning at 110 rev/min, has a real slip of 25% and wake fraction 0.30. If the propeller delivers a thrust of 300 KN and the propeller efficiency is 0.65. Calculate: (10)

    (i) Blade area.

    (ii) Ship speed.

    (iii) Thrust power.

    (iv) Shaft power

    (v) Torque

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    Part (a)

    A propeller is a type of fan, that transmits power by converting rotational motion into thrust. A pressure difference is produced between the forward and rear surface of the aerofoil shaped blade and the fluid is accelerated behind the blade. A marine propeller of this type is sometimes known as screw propeller or a screw.

    Given:

    $$D=6m$$

    $$p=0.9$$

    $$BAR=0.48$$

    $$N=110rev\:per\min$$

    $$real \space slip \space (s) \space = \space 25\%$$

    $$W_{F}=0.3$$

    $$Thrust \space = \space 300kN$$

    $$Ξ·_{prop} \space = \space 0.65$$

    (i) Blade area:

    $$BAR \space = \space {{A_b} \over {{\pi} \over 4} D^2}$$

    $$Blade \space area \space A_b \space = \space 0.48 \times {{\pi} \over 4} 6^2$$

    $$Blade \space area \space = \space 13.57m^2 $$

    $$p \space = \space {{P} \over D}$$

    $$0.9 \space = \space {{P} \over 6}$$

    $$Pitch \space p = \space 5.4m$$

    $$V_{T}=P\times N\times\frac{3600}{1852}$$

    $$V_{T}=5.4\times\frac{110}{60}\times\frac{3600}{1852}$$

    $$V_{T}=19.24knots$$

    $$Real \space slip \space S \space = \space {{V_T - V_a} \over V_T}$$

    $$ 0.25 \space = \space {{19.24 - V_a} \over 19.24}$$

    $$V_a \space = \space 14.42 knots$$

    $$W_F \space = \space {{V - V_a} \over V}$$

    $$0.30 \space = \space {{V - 14.42} \over V}$$

    $$V=20.6knots$$

    $$T_{p}\space=\space Thrust\times V_{a}\times\frac{1852}{3600}$$

    $$T_p \space = \space Thrust \times 14.42 \times {{1852} \over 3600 }$$

    $$T_p \space = \space 2225.48 $$

    $$T_p \space = \space d_p \times Ξ·_{prop}$$

    $$2225.48=d_{p}\times0.65$$

    $$d_p \space = \space 3423.8kW$$

    $$dp=2\pi NT$$

    $$3423.8=2\times\pi\times\frac{110}{60}\times T$$

    $$T=297.22KN$$

    Q9 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    (a) Explain what is meant by: (6)

    (i) Wave-making resistance.

    (ii) Frictional resistance.

    (iii) Eddy-making resistance.

    (b) When a ship is 800 nautical miles from port its speed is reduced by 20%, thereby reducing the daily fuel consumption by 42 tonne and arriving in port with 50 tonne on board. If the fuel consumption in t/h is given by the expression (0.136+0.001 VΒ³) where V is the speed in knots, estimate: (10)

    (i) The reduced consumption per day.

    (ii) The amount of fuel on board when the speed was reduced

    (iii) The percentage decrease in consumption for the latter part of the voyage.

    (iv) The percentage increases in time for this latter period.

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    $$D=800nm$$

    $$V_1=?knots$$

    $$V_2=0.8V_1$$

    $$DC_1=Cons\:per\:day\:at\:V_1$$

    $$DC_2=Cons\:per\:day\:at\:V_2$$

    $$DC_1-DC_2=42t$$

    $$C=\left(0.136+0.001V^3\right)\:t\h$$

    $$Therefore\:DC=24\left(0.136+0.001V^3\right)\:tonnes\:per\:day$$$$42=24\left\lbrack\left(0.136+0.001V_{1^{}}^3\right)-\left(0.136_{}+0.001\left(0.8V_1^3\right)\right)\right\rbrack$$

    $$42=24\left(0.136+0.001V_1^3-0.136-0.512\times10^{-3}\times V_1^3\right)$$

    $$42=24\left(0.001V_1^3-0.512\times10^{-3}\times V_1^3\right)$$

    $$42=24\left(0.000488V_1^3\right)$$

    $$V_1=\sqrt[3]{\frac{42}{24\times0.000488}}$$

    $$V_1=15.31\:knots$$

    $$V_2=0.8\times V_1$$

    $$V_2=0.8\times15.31$$

    $$V_2=12.245\:knots$$

    $$\left(i\right)\:Reduced\:cons\:per\:day\:=\:\left(0.136+0.001V_2^3\right)\times24$$

    $$=\left(0.136+0.001\times12.45^3\right)\times24$$

    $$=49.57\:tonnes\:per\:day$$

    $$Time\:taken\:for\:complete\:voyage\:of\:800nm$$

    $$at\:V_2=\frac{800}{12.245\times24}=2.72\:days$$

    $$Consumption\:=\:2.72\times49.57=134.93t\:\left(at\:reduced\:speed\right)$$

    $$\left(ii\right)\:Fuel\:onboard=134.93+50$$

    $$=184.93t\:\left(after\:speed\:reduction\right)$$

    $$DC_1=DC_2+42$$

    $$DC_1=49.57+42$$

    $$DC_1=91.57t$$

    $$Time\:taken\:for\:V_1=\frac{800}{15.31\times24}$$

    $$=2.178days$$

    $$Cons\:at\:V_1=91.57\times2.178$$

    $$=199.38t$$

    $$\left(iii\right)\:\%\:reduction\:in\:cons=\frac{199.38-134.93}{199.38}$$

    $$=32.32\%$$

    $$\left(iv\right)\:\%\:increase\:in\:time=\frac{2.72-2.178}{2.178}$$

    $$=24.88\%$$

    Q10 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Explain how to distinguish between list and loll and describe how to return the ship to the upright in each case. (6)

    (b) A ship of 5000 tonne displacement has a double bottom tank 12m long. The Β½ breadths of the top of the tank are 5, 4 and 2m respectively. The tank has a watertight centreline division. Calculate the free surface effect if the tank is partially full of fresh water on one side only. (10)

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    Part (a)

    Distinguishing between List and Loll:

    1. List is caused by Uneven distribution of weight within the ship. List will have the below conditions:

    • GM is positive (GM > 0).
    • The centre of gravity (G) is off-centre.
    • The vessel is at equilibrium but inclines to one side due to uneven weight distribution, even without any external forces acting on it.
    • The vessel rolls around the angle of list.

    Correction:

    • Redistribute the weight evenly to bring the centre of gravity (G) back in line with the centerline and metacentric height (M).

    2. Loll is caused by High centre of gravity (G) leading to negative GM and is exacerbated by external forces, free surface effects, or poor distribution of weights. LOLL will have the below conditions:

    • GM is negative (GM < 0).
    • The centre of gravity (G) is on the centerline but too high, making the vessel inherently unstable.
    • The vessel flops or inclines to one side at an angle of loll, and it can incline equally to either side.
    • The vessel rolls unstably around the angle of loll.

    Correction:

    • Reduce the centre of gravity by ballasting bottom tanks or removing weight from higher levels.
    • Minimize free surface effects by reducing the breadth of free surfaces in tanks.

    Q1 (16 Marks) Ship Stability πŸ”₯ Repeated 2x

    (a) State the reasons for the freeboard requirement. (6)

    (b) Explain the term condition of assignment and explain how these are maintained for a ship. (5)

    (c) What is the difference between a Type "A" and a Type "B" ship. (5)

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    Part (a)

    Reasons for Freeboard Requirements:

    Freeboard is the distance measured from the waterline to the upper edge of the deck plating at the side of the freeboard deck amidships.

    • Ensures the ship is seaworthy when fully loaded.
    • Provides reserve buoyancy, enabling the ship to rise as it passes through waves, keeping decks largely dry.
    • Enhances the ship's stability and increases its survivability in case of damage by allowing it to remain afloat longer, aiding crew escape or damage control.
    Part (b)

    Conditions of Assignment:

    Conditions of Assignment are outlined in the Merchant Shipping Load Line Rules 1968 and must be satisfied before a ship is assigned freeboards and issued a load line certificate. These conditions address the practical need for openings (hatchways, doorways, vents, etc.) in the ship's hull and mandate appropriate protection and closure mechanisms for these openings.

    Requirements Before Assigning Load Line:

    • The ship must have sufficient structural strength.
    • Adequate reserve buoyancy must be maintained.
    • Openings must be secured against water ingress.
    • Safety measures for the crew, such as guardrails and gangways, must be in place.

    To ensure that the conditions of assignment are still current the following items can be checked and confirm to be without change or damage from when the ship was built.

    • Access openings in bulkheads, to ensure that they can be sealed and prevent flooding
    • Cargo and hatchways ensure they can be sealed to prevent flooding
    • Coamings of hatchways, sign of corrosion damage risk of failure would allow flooding
    • Protection of openings, can they all be sealed
    • Ventilator coamings not corroded as they would allow flooding to other arears
    • Air pipes can be shut in heavy weather and not corroded
    • Discharges, inlets and scuppers, all in good condition and operational
    • Side scuttles can be secured
    • Hull inspection, sea boxes and penetrations all checked for damage and corrosion free.
    Part (c)

    Difference between Type A and Type B ships:

    TYPE β€œA” VESSELS:

    • A ship that is designed to carry only liquid cargoes in bulk and in which cargo tanks have only small access openings, closed by water-tight gasketed covers of steel or equivalent material.
    • The exposed deck must be one of high integrity.
    • It must have a high degree of safety against flooding, resulting from the low permeability of loaded cargo spaces and the degree of bulkhead subdivision usually provided.

    TYPE β€œB” VESSELS: all ships that do not fall under type "A" vessels are type "B" ships. For these ships, it may be based on :

    • The vertical extent of damage is equal to the depth of the ship.
    • The penetration of damage is not more than 1/5 of the breadth moulded.
    • No main transverse bulkhead is damaged.
    • Ship’s Kg is assessed for homogenous loading of cargo holds and 50% of the designed capacity of consumable fluids and stores etc.
    Q2 (16 Marks) Hull Construction πŸ”₯ Repeated 8x

    (a) Sketch the cross-section of a bulk carrier with either deep or shallow double bottom showing the type of framing used. (8)

    (b) (i) Describe the corrosion problems experienced with ballast tanks. (4)

    (ii) State how such tanks are protected against extensive corrosion. (4)

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    Part (a)

    Mid-ship section of bulk carrier:

    Part (b)

    (i) Corrosion problems in ballast tanks

    are significant and arise due to various factors:

    • Galvanic corrosion occurs due to the interaction between dissimilar metals, accelerated by the seawater environment and differential aeration.
    • Sulphate-reducing bacteria in river mud can cause localized pitting, leading to penetration of the bottom shell.
    • Dissolved oxygen in seawater reacts with metal surfaces to form rust, significantly contributing to structural deterioration.
    • The rate of corrosion peaks at a 3.5% salt concentration, typical of seawater.
    • Neglecting maintenance exacerbates the problem, allowing corrosion to progress unchecked.

    (ii) Protection against extensive corrosion in ballast tanks involves the following measures:

    • Complete coating of the tank surfaces with properly selected marine-grade paint to create a protective barrier.
    • Installation of sacrificial anodes, designed to corrode preferentially and protect the tank structure.
    • Using large anodes with greater volume relative to surface area to ensure extended protection.
    • Regular gauging of plates to assess thickness and identify areas requiring reinforcement or repair.
    • Conducting regular inspections, cleaning the tanks, removing rust, and repainting to maintain structural integrity and prolong the lifespan of the ballast tanks.
    Q3 (16 Marks) Hull Construction πŸ”₯ Repeated 2x

    (a) With the aid of a sketch describe the method of attachment for a bilge keel and hence explain what protection is made to reduce the possibility of the shell being punctured in the event of damage to the keel. (6)

    (b) State why the keel does not extend for the length of the ship. (5)

    (c) Evaluate the effectiveness of bilge keels for large wall sided vessels. (5)

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    Part (a)

    Method of Bilge keel attachment to the hull:

    Bilge keels are fitted port and starboard at the turn of the bilge. They do not extend outside the lines of the side and bottom shell. The bilge keels are attached directly in line with an internal stiffening member such as a girder or longitudinal. The bilge keel comprises a flat bar doubler welded directly to the shell, and an offset bulb plate (OBP) with 'scallops' cut in it is welded to the flat bar doubler. The ends of the bilge keels are tapered (minimum 3 in 1) and will end in line with transverse internal stiffening, such as a frame. Using a hull doubler protects the hull in case of damage to the keel, as the crack would not extend into the hull.

    Part (b)

    Bilge keels are not fitted for the full length of the vessel because:

    • The 'lever' to the ship's axis of rotation is reduced at the ends;
    • The ineffectiveness, the closer the bilge keel is to the rolling centre, increased resistance and more likelihood of damage.
    • The hydrodynamic effect would cause a large increase in resistance and fuel consumption;
    • At the aft end, the boundary layer is much thicker, and since the keel would not project through, it would have a much reduced effect.
    Q4 (16 Marks) General πŸ”₯ Repeated 3x

    (a) Draw a simple line diagram of the bow of a ship to show the position of the following component parts of the ships anchoring system: Hawse pipe, Cable stopper, Windlass and Cable lifter, Spurling pipe and Chain locker. (4)

    (b) Describe the cable stopper and state its purpose. (4)

    (c) Show by means of a sketch how the anchor cable is attached to the ship. (4)

    (d) Describe how the chain locker is drained of water, sand and mud. (4)

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    Part (a)

    At the bow of the ship, you'd typically find the following components:

    • Hawse pipes: These are openings in the hull through which the anchor chain passes.
    • Cable stopper: Positioned on the deck near the hawse pipes, the cable stopper is used to secure the anchor chain in place.
    • Windlass: Located on the deck near the hawse pipes, the windlass is a mechanical device used to raise and lower the anchor and its chain.
    • Cable lifts: These are devices used to guide the anchor chain from the windlass into the chain locker below deck.
    • Spurling pipe: A vertical pipe leading from the windlass to the chain locker, through which the anchor chain passes.
    • Chain locker: Below deck, typically located near the bow, the chain locker is a compartment where the anchor chain is stowed when not in use.
    Part (b)

    The cable stopper is a device used to secure the anchor chain in place once the anchor has been lowered or raised. Its purpose is to prevent the anchor chain from slipping or running out unintentionally, ensuring that the anchor remains securely in position.

    Part (c)

    The anchor cable is typically attached to the ship using a shackle or swivel at the end of the anchor chain. This attachment point allows the anchor chain to pivot freely as the anchor is raised or lowered, preventing twisting or tangling of the chain.

    Part (d)

    The chain locker is drained of water, sand, and mud through a drainage system that typically includes scupper holes or drain pipes located in the bottom of the locker. These drains allow any water or debris that accumulates in the chain locker to flow out of the compartment and overboard, ensuring that the anchor chain remains clean and free of obstructions. Additionally, regular maintenance and cleaning of the chain locker are important to prevent the buildup of sediment and ensure proper drainage.

    Q5 (16 Marks) General πŸ”₯ Repeated 3x

    With reference to membrane tanks for the carriage of liquefied gas at very low temperatures:

    (a) Describe with a sketch one method of building up the insulation. (6)

    (b) State which alloy is used for the membrane and the reason; (5)

    (c) Explain why a secondary barrier is installed. (5)

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    Part (a)

    Membrane tanks for liquefied gas carriage are built with a double hull throughout the cargo length. The insulation system includes the following:

    1. Primary Barrier: A thin membrane made of INVAR (36% Nickel, 64% Iron) with a thickness of 0.5 to 1.2 mm, forming the primary containment layer for the liquefied gas.
    2. Primary Insulation box: A 200 mm thick layer of granulated Perlite insulation packed in plywood boxes, surrounding the primary barrier. Perlite is siliconised to make it impervious to moisture.
    3. Secondary Barrier: A secondary membrane made of the same INVAR material is installed to prevent cargo leakage in case of primary barrier failure.
    4. Secondary Insulation box: Another 200 mm thick layer of granulated Perlite is placed above the secondary barrier to provide additional insulation and prevent thermal transfer.
    Part (b)

    Alloy used for the membrane:

    Invar (36% Ni, 64% Fe) is used for both the primary and secondary barriers. The reason for this choice is its exceptionally low coefficient of thermal expansion. This eliminates the need for expansion joints or corrugations in the membrane design. In addition, Invar remains strong and does not become brittle at the very low temperatures experienced by the liquefied gas. The thin and lightweight nature of Invar maximizes the cargo-carrying capacity of the tank.

    Q6 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Describe how the distribution of mass within the ship affects the rolling period. (6)

    (b) The righting moments of a ship at angles of heel of 0, 15Β°, 30Β°, 45Β° and 60Β° are 0, 1690, 5430, 9360 and 9140 kNm respectively. Calculate the dynamical stability at 60Β°.

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    Part (a)

    The rolling period is influenced by the ship's metacentric height (GM) and radius of gyration (K), which are determined by the location of the masses onboard.

    Masses at the bottom:

    • The centre of gravity (G) moves down.
    • Metacentric height (GM) increases, resulting in greater stability.
    • Rolling period decreases.

    Masses at the top:

    • The centre of gravity (G) moves up.
    • Metacentric height (GM) decreases, reducing stability.
    • Rolling period increases.

    Masses concentrated at the centre:

    • The radius of gyration (K) decreases.
    • Rolling period decreases.

    Masses distributed away from the centre:

    • The radius of gyration (K) increases.
    • Rolling period increases.
    Q7 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Describe briefly the significance of the factor of subdivision.

    (b) A ship 120m long has a light displacement of 4000 tonne and LCG in this condition 2.5m aft of midships. The following items are then added:

    Cargo 10000 tonne LCG 3.0m forward of midships

    Fuel 1500 tonne LCG 2.0 m aft of midships

    Water 400 tonne LCG 8.0m aft of midships

    Stores 100 tonnes LCG 10.0m forward of midships

    Using the following hydrostatic data, calculate the final draughts:

    Draught (m)

    Displacement (t)

    MCT1cm (tm)

    LCB from midships

    LCF from midships

    8.50

    16650

    183

    1.94F

    1.29A

    8.00

    15350

    175

    2.10F

    0.60F

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    Part (a)

    Factor of Subdivision

    The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

    Permissible Length Formula:

    $$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

    A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

    Part (b)

    Mass added

    LCG from midship

    Mass moments

    F

    A

    10000

    3.0 Fwd

    30000

    1500

    2.0 Aft

    3000

    400

    8.0 Aft

    3200

    100

    10.0 Fwd

    1000

    4000

    2.5 Aft

    10000

    16000

    31000

    16200

    $$Excess\:moment=31000-16200=14800$$

    $$LCG=\frac{\sum M}{\sum m}=\frac{14800}{16000}$$

    $$LCG=0.925m\:fwd\:of\:midship$$

    Draught

    Displacement

    MTC 1cm

    LCB from midship

    LCF from midship

    8.5

    16650

    183

    1.94 Fwd

    1.20 Aft

    8.25

    16000

    179

    2.03 Fwd

    0.57 Aft

    8.0

    15350

    175

    2.10 Fwd

    0.06 Fwd

    $$LCG=0.925m\:Fwd\:of\:midship$$

    $$LCB=2.02m\:Fwd\:of\:midship$$

    $$Trimming\:lever=LCB-LCG$$

    $$2.02-0.925$$

    $$=1.09m$$

    $$Trimming\:moment=m\times d$$

    $$=16000\times1.09$$

    $$=17440tm$$

    $$Change\:of\:trim=\frac{Trimming\:moment}{MCT_{1\operatorname{\mathrm{cm}}}}$$

    $$=\frac{17440}{179}$$

    $$97.43\operatorname{cm}$$

    $$Draft\:fwd=8.25-\frac{97.43}{100\times120}\left(\frac{120}{2}+0.57\right)$$

    $$=7.756m$$

    $$Draft\:aft=8.25+\frac{97.43}{100\times120}\left(\frac{120}{2}-0.57\right)$$

    $$=8.735m$$

    Q8 (16 Marks) Ship Stability πŸ”₯ Repeated 3x

    (a) Explain how the distribution of masses affects rolling and pitching. (6)

    (b) A ship turns in a circle of radius 100 metres at a speed of 15 knots. The GM is 2/3 metres and BG is 1 metre. If g = 981 cm/sec2 and 1 knot is equal to 1.8532 Km/hour, find the heel due to turning. (10)

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    Part (a)

    The rolling period is influenced by the ship's metacentric height (GM) and radius of gyration (K), which are determined by the location of the masses onboard.

    Masses at the bottom:

    • The centre of gravity (G) moves down.
    • Metacentric height (GM) increases, resulting in greater stability.
    • Rolling period decreases.

    Masses at the top:

    • The centre of gravity (G) moves up.
    • Metacentric height (GM) decreases, reducing stability.
    • Rolling period increases.

    Masses concentrated at the centre:

    • The radius of gyration (K) decreases.
    • Rolling period decreases.

    Masses distributed away from the centre:

    • The radius of gyration (K) increases.
    • Rolling period increases.
    Part (b)

    $$Radius\:\left(R\right)=100m$$

    $$Speed\:\left(v\right)=15\:knots$$

    $$GM=\frac23m$$

    $$BG=1m$$

    $$\tan\the\theta=\frac{v^2\times BG}{g\times r\times GM}$$

    $$=\frac{\left(15\times0.514\right)^2\times1\times3}{9.81\times100\times2}$$

    $$\theta=5.19\degree$$

    Q9 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 4x

    (a) Describe the effect of cavitations on the propeller blades. (6)

    (b) A propeller 4.6m diameter has a pitch of 4.3m and boss diameter of 0.75m. The real slip is 28% at 95 rev/min. Calculate the speed of advance, thrust and thrust power. (10)

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    Part (a)

    Effect of cavitation on propeller blades:

    Erosion:

    • Cavitation causes the formation and collapse of vapor bubbles on the propeller blade surface.
    • The collapse of these bubbles produces high-pressure shockwaves and microjets that erode the blade material, leading to surface pitting and progressive damage.

    Vibration:

    • Uneven cavitation across the blades leads to imbalanced forces, causing vibrations in the propeller and the ship.
    • These vibrations can reduce the comfort of passengers and crew and stress the ship's structural components.

    Noise:

    • The collapse of vapor bubbles generates loud noise, which can interfere with onboard communication and underwater sonar systems.
    • This noise is a significant concern for naval vessels and marine life.

    Reduced Performance:

    • Cavitation reduces the efficiency of the propeller by causing loss of thrust and torque.
    • The presence of cavitation decreases the propeller’s ability to convert rotational energy into forward motion, lowering the ship's speed and increasing fuel consumption.

    (b) Given:

    $$D=4.6m$$

    $$P=4.3m$$

    $$d=0.75$$

    $$S=28\%$$

    $$n \space = \space 95 \space rev/ min$$

    $$V_T \space = \space P \times N \times {{3600} \over 1852}$$

    $$ = \space 4.3 \times {{95} \over 60} \times {{3600} \over 1852}$$

    $$V_{T}=13.23knots$$

    $$Real \space slip \space (S) \space = \space {{V_T - V_a} \over V_T}$$

    $$0.28 \space = \space {{13.23 - V_a} \over 13.23}$$

    $$V_{a}=9.52knots$$

    $$Effective\:disc\:area\:\left(A\right)\:={{\pi}\over4}\left(D^2-d^2\right)$$

    $$= {{\pi} \over 4} (4.6^2 - 0.75^2)$$

    $$A=16.18m^2$$

    $$Thrust\space=\space\rho AP^2n^2S$$

    $$=1.025\times16.18\times4.3^2\times\left(\frac{95}{60}\right)^2\times0.28$$

    $$T=215.25KN$$

    $$Thrust \space power (T_p) \space = \space T \times V_a $$

    $$215.25\times9.52\times\frac{1852}{3600}$$

    $$T_{p}=1054.18KW$$

    Q10 (16 Marks) Ship Stability πŸ”₯ Repeated 3x

    (a) Describe the stability requirements of a ship for dry-docking. (6)

    (b) A ship 130m long displaces 14000 tonnes when floating at draughts of 7.5m forward and 8.10 m aft. GM, 125m, TPC 18, LCF. 3m aft of midships. Calculate the final draughts when a mass of 180 tonne lying 40m aft of midships is removed from the ship. (10)

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    Part (a)

    For safe dry-docking, a ship must meet two key stability requirements:

    • Positive GM (Metacentric Height): The ship needs a positive GM. GM is the distance between the centre of gravity (G) and the metacentre (M). A positive GM indicates inherent stability; the ship will right itself if tilted. During dry-docking, the loss of buoyancy as the ship rests on the blocks reduces GM. Insufficient positive GM increases the risk of the ship heeling over or capsizing.
    • Trim by the stern: The vessel should be trimmed slightly by the stern (aft end lower than the bow) to ensure the aft end sits on the keel blocks first. This controlled settling minimises the risk of instability during the docking process. An even keel is generally preferred for the initial floating condition before the dry-docking procedure begins.
    Q1 (16 Marks) Hull Construction πŸ”₯ Repeated 5x

    With reference to membrane tanks for the carriage of liquefied gas at very low temperatures:

    (a) Describe with the aid of a sketch, ONE method of building up the insulation. (6)

    (b) State with reasons the alloy, which is used for the membrane. (4)

    (c) Describe with the aid of a sketch, how the tanks are located and supported: (6)

    (i) Longitudinally.

    (ii) Transversely.

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    Part (a)

    Membrane tanks for liquefied gas carriage are built with a double hull throughout the cargo length. The insulation system includes the following:

    1. Primary Barrier: A thin membrane made of INVAR (36% Nickel, 64% Iron) with a thickness of 0.7 to 1.2 mm, forming the primary containment layer for the liquefied gas.
    2. Primary Insulation box: A 230 mm thick layer of granulated Perlite insulation packed in plywood boxes, surrounding the primary barrier. Perlite is siliconised to make it impervious to moisture.
    3. Secondary Barrier: A secondary membrane made of the same INVAR material is installed to prevent cargo leakage in case of primary barrier failure.
    4. Secondary Insulation box: Another 300 mm thick layer of granulated Perlite is placed above the secondary barrier to provide additional insulation and prevent thermal transfer.
    Part (b)

    Alloy used for the membrane:

    Invar (36% Ni, 64% Fe) is used for both the primary and secondary barriers. The reason for this choice is its exceptionally low coefficient of thermal expansion. This eliminates the need for expansion joints or corrugations in the membrane design. In addition, Invar remains strong and does not become brittle at the very low temperatures experienced by the liquefied gas. The thin and lightweight nature of Invar maximizes the cargo-carrying capacity of the tank.

    Part (c)

    Membrane tanks are either independent or self-supporting

    , meaning they don't form part of the ship's hull and don't contribute to the ship's structural strength. They can be spherical, cylindrical, or prismatic (box-shaped). Prismatic tanks usually have internal stiffeners like bulkheads, webs, girders, and stiffeners for added structural integrity.

    (i) Longitudinally:

    Tanks are positioned longitudinally within the ship's cargo hold. The tanks are supported longitudinally by anti-roll chocks and anti-lift chocks that resist forces caused by ship motions. These chocks ensure the tank remains stable even in rough sea conditions. They also act as thermal barriers, preventing the transfer of heat between the hull and the tank.

    The chocks also serve as thermal barriers between the hull and cargo and are often constructed of wood or plastic materials.

    (ii) Transversely: Transverse support is provided by anti-pitch chocks and support chocks, which stabilize the tank against lateral forces. These chocks are typically made of plywood or plastic and help prevent thermal stress between the tank and the ship’s structure.

    Q2 (16 Marks) Surveys & Drydocking πŸ”₯ Repeated 7x

    With reference to dry docking, define the responsibilities of the Second Engineer and instructions to Junior Engineers: (16)

    (a) Prior to docking

    (b) Whilst the vessel is in dry dock

    (c) Prior to flooding and leaving the dock.

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    (a) Prior to Docking:

    Preliminary Preparation:

    • Review various plans, manuals, and previous drydock reports for reference.
    • Prepare a detailed repair list and ensure all required spares are accounted for.
    • Conduct an inventory of spares and requisition necessary items.
    • Gather required tools, including special tools like poker gauges, trammel gauges, gagging tools, etc.
    • Mark all overboard valves, NC (normally closed) and NO (normally open) valves clearly for easy identification.
    • Allocate jobs to team members and discuss the time schedule.
    • Conduct a safety meeting to highlight dry dock hazards and necessary precautions.

    Before Entering Dry Dock:

    • Identify the power requirements and machinery to be operational during docking.
    • Check the shore connection box for proper functionality.
    • Record the soundings of FO (Fuel Oil), LO (Lubricating Oil), and DO (Diesel Oil) tanks.
    • Discharge contents from clean drain tanks and sewage tanks.
    • Carry out Economizer soot-blowing.
    • Change over Main Engine, Diesel Generators, and Boiler to Low Sulfur Gas Oil (LSGO).
    • Stop and clean purifiers.
    • Ensure the low sea chest is open and the high sea chest is shut.
    • Keep firefighting appliances (FFA) on standby.
    • Shut down all non-essential machinery, including MGPS (Marine Growth Prevention System) and FWG (Fresh Water Generator).

    (b) Whilst the Vessel is in Dry Dock

    Upon Arrival:

    • Connect shore power and supplies after ensuring safety checks are completed.
    • Start necessary equipment like cooling water, air compressors, air conditioning, and fridge compressors.
    • Check for jobs assigned by dry dock personnel and prepare accordingly.
    • Attend a safety meeting with dry dock personnel to understand local safety rules and procedures.

    During Dry Docking::

    • Oversee and assist in:
      • Cleaning and inspecting the hull, rudder, sea chest, anodes, and propeller.
      • Measuring propeller drop, checking rudder clearances, and inspecting the stern tube bearing and seal.
      • Servicing underwater valves and overboard valves.
      • Inspecting anchor and cables conditions.
      • Overhauling deck machinery, cranes, elevators, and engine room equipment such as the Main Engine, Diesel Generators, Boiler, and Economizer.
      • Renewing pipes and valves as needed.
      • Performing electrical equipment maintenance and surveys.
      • Supervising service engineers for specific repair jobs.
    • Ensure tank cleaning, welding, and other repair works are completed according to the plan.
    • Run the standby diesel generator daily after starting the priming pump.

    (c) Before Flooding and Leaving the Dry Dock

    Final Checks:

    • Verify that all underwater fittings and drain plugs are securely in place.
    • Ensure all machinery has been boxed back and is ready for operation.
    • Check for any leakage in stern tube seals.
    • Take tank soundings to confirm proper levels.
    • Confirm the proper operation of all underwater valves, overboard valves, sea chests, and vents.
    • Inspect the stern tube tank for any irregularities.

    System Restart:

    • Switch back to ship's power after confirming all systems are functional.
    • Test the proper operation of all machinery and systems to ensure the ship is fully operational before leaving the dock.
    Q3 (16 Marks) Ship Stability πŸ”₯ Repeated 8x

    (a) Explain what is meant by "permissible length" of compartments in passenger ships. (6)

    (b) Describe how the position of bulkheads is determined. (5)

    (c) Briefly describe the significance of the factor of subdivision (5)

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    Part (a)

    Permissible Length

    Permissible length refers to the maximum length of a ship's compartment that can be flooded while ensuring that the sinkage, heel, or trim remains within acceptable limits without exceeding the floodable length. It ensures the ship remains afloat and stable after flooding.

    Permissible Length Formula:

    $$Permissible\:lenght\:=\:Floodable\:length\:\times Factor\:of\:subdivision$$

    The Factor of Subdivision depends on the ship's length and the nature of its service:

    • For passenger ships, the factor of subdivision is smaller compared to cargo ships.
    • Smaller compartments ensure enhanced safety in case of flooding.
    Part (b)

    Position of Bulkheads

    The position of bulkheads is determined based on the need to comply with the "reasonable amount of damage" criterion in case of flooding:

    • Transverse Watertight Bulkheads should vertically extend up to the margin line.
    • At least one watertight longitudinal bulkhead must be located 20% of the ship's breadth inward from each side.
    • Bulkheads are positioned along the ship's length at intervals equal to the permissible length, but not exceeding the floodable length.
    • The maximum permissible compartment length is limited to 10.7 meters.
    • Collision Bulkhead must be located forward at a distance equal to the permissible length from the forward perpendicular.
    • The ER must be enclosed by two transverse watertight bulkheads, with an aft peak watertight bulkhead enclosing the forward part.
    Part (c)

    Factor of Subdivision

    The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

    Permissible Length Formula:

    $$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

    A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

    Q4 (16 Marks) Ship Stability πŸ”₯ Repeated 14x

    Explain how the period of roll varies with

    (a) The amplitude of roll (4)

    (b) The radius of gyration (4)

    (c) The initial metacentric height (4)

    (d) The location of masses in the ship. (4)

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    The period of roll Tr of a ship is determined by the formula:

    $$T_{r}=\frac{2\pi K}{\sqrt{g\times GM}}$$

    where,

    • K is the radius of gyration (mass moment of inertia)
    • g is the acceleration due to gravity, and
    • GM is the metacentric height.
    Part (a)

    Amplitude of Roll:

    • The amplitude of roll, or the maximum angle of heel, does not affect the period of roll. While a larger amplitude implies a greater heeling moment and faster roll speed, the time taken to complete one full roll cycle remains constant for a given metacentric height and radius of gyration.
    Part (b)

    Radius of Gyration (K):

    • The period of roll is directly proportional to the radius of gyration. A larger radius of gyration (indicating a greater distribution of mass further from the ship's centre of rotation) leads to a longer period of roll. Conversely, a smaller radius of gyration (mass concentrated closer to the centre) results in a shorter period. The distribution of cargo significantly impacts K; cargo concentrated centrally minimises K and the roll period, while dispersed cargo maximises K and the roll period.
    Part (c)

    Initial Metacentric Height (GM):

    • The period of roll is inversely proportional to the square root of the metacentric height (GM). A larger GM (a stiffer ship) leads to a shorter roll period, as the ship quickly returns to its upright position. A smaller GM (a tender ship) results in a longer roll period, with slower return to the upright.
    Part (d)

    Location of Masses in the Ship:

    The location of masses in the ship will effect the GM & K. So the period of roll will be affected.

    • If masses are at bottom, G moves down, GM ↑, period of roll ↓.
    • If masses are at top, G moves up, GM ↓, period of roll ↑.
    • If masses are concentrated at centre, K ↓, period of roll ↓.
    • If masses are away from centre, K ↑, period of roll ↑.
    Q5 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 13x

    Describe the relationship between frictional resistance and

    (a) Ship speed (4)

    (b) The wetted area (4)

    (c) The surface roughness (4)

    (d) The length of the vessel (4)

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    Frictional resistance arises due to the interaction between the ship's hull and the water as the vessel moves. This resistance is caused by eddying water adjacent to the hull that is drawn along with the ship. The frictional resistance is given by:

    $$R_{f\:=\:}f\:.\:s\:.\:v^{n}$$

    Where:

    • f = Coefficient of friction
    • s = Wetted surface area
    • v = Ship's speed in knots
    • n = Constant (1.82)
    Part (a)

    Ship Speed (v):

    • Frictional resistance increases with increasing ship speed. The relationship is not perfectly linear, but it's approximately described by the formula Rf = fsv^n (Rf ∝ v^n), where 'n' is a constant (approximately 1.82). This means that a small increase in speed results in a larger increase in frictional resistance.
    Part (b)

    Wetted Area (s):

    • Frictional resistance is directly proportional to the wetted surface area (Rf ∝ s). A larger wetted area (the part of the hull submerged in water) leads to greater frictional resistance. The wetted area increases with the ship's draught (depth in the water).
    Part (c)

    Surface Roughness:

    • Increased surface roughness increases frictional resistance. Roughness can be caused by fouling (marine growth), corrosion, or poor hull maintenance. Regular cleaning and maintenance of the hull surface help to minimise roughness and reduce resistance.
    Part (d)

    Length of the Vessel

    • Frictional resistance is influenced by the vessel's length. As the length increases, the wetted surface area grows, and the coefficient of friction (f) changes, leading to an increase in Rf. Longer vessels experience higher frictional resistance, but streamlined designs can mitigate the effect to some extent.
    Q6 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) How the distribution of mass within the ship affects the rolling period? (6)

    (b) A ship of 14000 tonnes displacement is 125 m long and floats at draughts of 7.9 m forward and 8.5 m aft. The TPC is 19, GML 120 m, and LCF 3 m forward of midships. It is required to bring the vessel to an even keel draught of 8.5 m. Calculate the mass which should be added and the distance of the centre of the mass from midships. (10)

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    Part (a)

    The rolling period is influenced by the ship's metacentric height (GM) and radius of gyration (K), which are determined by the location of the masses onboard.

    Masses at the bottom:

    • The centre of gravity (G) moves down.
    • Metacentric height (GM) increases, resulting in greater stability.
    • Rolling period decreases.

    Masses at the top:

    • The centre of gravity (G) moves up.
    • Metacentric height (GM) decreases, reducing stability.
    • Rolling period increases.

    Masses concentrated at the centre:

    • The radius of gyration (K) decreases.
    • Rolling period decreases.

    Masses distributed away from the centre:

    • The radius of gyration (K) increases.
    • Rolling period increases.
    Q7 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Explain how an increase of draught and of displacement influence rolling. (6)

    (b) A pontoon has a constant cross-section as shown in Fig. Given below. The metacentric height is 2.5 m. Find the height of the centre of gravity above the keel. (10)

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    Part (a)

    In case of increase of draught and displacement, this is a case of loading being done.

    Thus, when the ship is being loaded, its GM will start to decrease.

    Now, the time period of roll is given by:

    $$T_{r}=\frac{2\pi k}{\sqrt{GM.g}}$$

    Where,

    • k = radius of gyration
    • GM = metacentric height
    • g = acceleration due to gravity

    Now, since, GM has started to decrease, the Tr will start to increase.

    Thus, the ship will now roll with greater time period. Thus, an increase in draught and displacement, influences rolling.

    Q8 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    (a) Describe the fundamental principle of a propeller. (6)

    (b) A propeller 6m in diameter has a pitch ratio of 0.9, BAR 0.48, and, when turning at 110 rev/min, has a real slip of 25% and wake fraction 0.30. If the propeller delivers a thrust of 300 kN and the propeller efficiency is 0.65, calculate: (10)

    (a) Blade area

    (b) Ship speed

    (c) Thrust power

    (d) Shaft power

    (e) Torque

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    Part (a)

    A propeller is a type of fan, that transmits power by converting rotational motion into thrust. A pressure difference is produced between the forward and rear surface of the aerofoil shaped blade and the fluid is accelerated behind the blade. A marine propeller of this type is sometimes known as screw propeller or a screw.

    Given:

    $$D=6m$$

    $$p=0.9$$

    $$BAR=0.48$$

    $$N=110rev\:per\min$$

    $$real \space slip \space (s) \space = \space 25\%$$

    $$W_{F}=0.3$$

    $$Thrust \space = \space 300kN$$

    $$Ξ·_{prop} \space = \space 0.65$$

    (i) Blade area:

    $$BAR \space = \space {{A_b} \over {{\pi} \over 4} D^2}$$

    $$Blade \space area \space A_b \space = \space 0.48 \times {{\pi} \over 4} 6^2$$

    $$Blade \space area \space = \space 13.57m^2 $$

    $$p \space = \space {{P} \over D}$$

    $$0.9 \space = \space {{P} \over 6}$$

    $$Pitch \space p = \space 5.4m$$

    $$V_{T}=P\times N\times\frac{3600}{1852}$$

    $$V_{T}=5.4\times\frac{110}{60}\times\frac{3600}{1852}$$

    $$V_{T}=19.24knots$$

    $$Real \space slip \space S \space = \space {{V_T - V_a} \over V_T}$$

    $$ 0.25 \space = \space {{19.24 - V_a} \over 19.24}$$

    $$V_a \space = \space 14.42 knots$$

    $$W_F \space = \space {{V - V_a} \over V}$$

    $$0.30 \space = \space {{V - 14.42} \over V}$$

    $$V=20.6knots$$

    $$T_{p}\space=\space Thrust\times V_{a}\times\frac{1852}{3600}$$

    $$T_p \space = \space Thrust \times 14.42 \times {{1852} \over 3600 }$$

    $$T_p \space = \space 2225.48 $$

    $$T_p \space = \space d_p \times Ξ·_{prop}$$

    $$2225.48=d_{p}\times0.65$$

    $$d_p \space = \space 3423.8kW$$

    $$dp=2\pi NT$$

    $$3423.8=2\times\pi\times\frac{110}{60}\times T$$

    $$T=297.22KN$$

    Q9 (16 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    (a) Explain what is meant by: (6)

    (i) Wave-making resistance

    (ii) Frictional resistance

    (iii) Eddy-making resistance

    (b) When a ship is 800 nautical miles from port its speed is reduced by 20%, thereby reducing the daily fuel consumption by 42 tonnes and arriving in port with 50 tonnes on board. If the fuel consumption in t/h is given by the expression (0.136+0.001V^3) where V is the speed in knots, estimate: (10)

    (i) The reduced consumption per day

    (ii) The amount of fuel on board when the speed was reduced

    (iii) The percentage decrease in consumption for the latter part of the voyage

    (iv) The percentage increase in time for this latter period

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    $$D=800nm$$

    $$V_1=?knots$$

    $$V_2=0.8V_1$$

    $$DC_1=Cons\:per\:day\:at\:V_1$$

    $$DC_2=Cons\:per\:day\:at\:V_2$$

    $$DC_1-DC_2=42t$$

    $$C=\left(0.136+0.001V^3\right)\:t\h$$

    $$Therefore\:DC=24\left(0.136+0.001V^3\right)\:tonnes\:per\:day$$$$42=24\left\lbrack\left(0.136+0.001V_{1^{}}^3\right)-\left(0.136_{}+0.001\left(0.8V_1^3\right)\right)\right\rbrack$$

    $$42=24\left(0.136+0.001V_1^3-0.136-0.512\times10^{-3}\times V_1^3\right)$$

    $$42=24\left(0.001V_1^3-0.512\times10^{-3}\times V_1^3\right)$$

    $$42=24\left(0.000488V_1^3\right)$$

    $$V_1=\sqrt[3]{\frac{42}{24\times0.000488}}$$

    $$V_1=15.31\:knots$$

    $$V_2=0.8\times V_1$$

    $$V_2=0.8\times15.31$$

    $$V_2=12.245\:knots$$

    $$\left(i\right)\:Reduced\:cons\:per\:day\:=\:\left(0.136+0.001V_2^3\right)\times24$$

    $$=\left(0.136+0.001\times12.45^3\right)\times24$$

    $$=49.57\:tonnes\:per\:day$$

    $$Time\:taken\:for\:complete\:voyage\:of\:800nm$$

    $$at\:V_2=\frac{800}{12.245\times24}=2.72\:days$$

    $$Consumption\:=\:2.72\times49.57=134.93t\:\left(at\:reduced\:speed\right)$$

    $$\left(ii\right)\:Fuel\:onboard=134.93+50$$

    $$=184.93t\:\left(after\:speed\:reduction\right)$$

    $$DC_1=DC_2+42$$

    $$DC_1=49.57+42$$

    $$DC_1=91.57t$$

    $$Time\:taken\:for\:V_1=\frac{800}{15.31\times24}$$

    $$=2.178days$$

    $$Cons\:at\:V_1=91.57\times2.178$$

    $$=199.38t$$

    $$\left(iii\right)\:\%\:reduction\:in\:cons=\frac{199.38-134.93}{199.38}$$

    $$=32.32\%$$

    $$\left(iv\right)\:\%\:increase\:in\:time=\frac{2.72-2.178}{2.178}$$

    $$=24.88\%$$

    Q10 (16 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Explain how to distinguish between list and loll and describe how to return the ship to the upright in each case. (6)

    (b) A ship of 5000 tonnes displacement has a double bottom tank 12 m long. The 1/2 breadths of the top of the tank are 5, 4 m and 2 m respectively. The tank has a watertight centreline division. Calculate the free surface effect if the tank is partially full of freshwater on one side only. (10)

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    Part (a)

    Distinguishing between List and Loll:

    1. List is caused by Uneven distribution of weight within the ship. List will have the below conditions:

    • GM is positive (GM > 0).
    • The centre of gravity (G) is off-centre.
    • The vessel is at equilibrium but inclines to one side due to uneven weight distribution, even without any external forces acting on it.
    • The vessel rolls around the angle of list.

    Correction:

    • Redistribute the weight evenly to bring the centre of gravity (G) back in line with the centerline and metacentric height (M).

    2. Loll is caused by High centre of gravity (G) leading to negative GM and is exacerbated by external forces, free surface effects, or poor distribution of weights. LOLL will have the below conditions:

    • GM is negative (GM < 0).
    • The centre of gravity (G) is on the centerline but too high, making the vessel inherently unstable.
    • The vessel flops or inclines to one side at an angle of loll, and it can incline equally to either side.
    • The vessel rolls unstably around the angle of loll.

    Correction:

    • Reduce the centre of gravity by ballasting bottom tanks or removing weight from higher levels.
    • Minimize free surface effects by reducing the breadth of free surfaces in tanks.

    Q1 (10 Marks) Hull Construction πŸ”₯ Repeated 8x

    (a) Describe a method for the attachment of bilge keels. (6)

    (b) State THREE reasons for not extending bilge keels to the entire length of the vessel. (5)

    (c) Explain TWO principles of roll damping that bilge keels exploit. (5)

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    Part (a)

    Method of Bilge keel attachment to the hull:

    Bilge keels are fitted port and starboard at the turn of the bilge. They do not extend outside the lines of the side and bottom shell. The bilge keels are attached directly in line with an internal stiffening member such as a girder or longitudinal. The bilge keel comprises a flat bar doubler welded directly to the shell, and an offset bulb plate (OBP) with 'scallops' cut in it is welded to the flat bar doubler. The ends of the bilge keels are tapered (minimum 3 in 1) and will end in line with transverse internal stiffening, such as a frame. Using a hull doubler protects the hull in case of damage to the keel, as the crack would not extend into the hull.

    Part (b)

    Bilge keels are not fitted for the full length of the vessel because:

    • The 'lever' to the ship's axis of rotation is reduced at the ends;
    • The ineffectiveness, the closer the bilge keel is to the rolling centre, increased resistance and more likelihood of damage.
    • The hydrodynamic effect would cause a large increase in resistance and fuel consumption;
    • At the aft end, the boundary layer is much thicker, and since the keel would not project through, it would have a much reduced effect.
    Part (c)

    Two principles of roll damping exploited by Bilge keels:

    Increased Roll Period:

    • Bilge keels increase the ship's roll period (the time it takes for the vessel to complete one roll). This is achieved by increasing the moment of inertia (K) of the vessel. A longer roll period means the vessel rolls more slowly, thus reducing the amplitude of the roll.

    $$T_{r}=2\pi\:\frac{k}{\sqrt{g\times GM}}$$

    Where:

    • g: acceleration due to gravity
    • GM: metacentric height
    • k: mass moment of inertia

    By increasing k, the ship's stability during roll improves.

    Hydrodynamic Resistance:

    • As the vessel rolls, the bilge keels move through the water, creating pressure differences. Water pressure on one side of the keel opposes the rolling motion, providing a damping effect. This is due to the interaction between the bilge keel and the water, creating hydrodynamic forces that counteract the rolling motion.
    Q2 (10 Marks) General πŸ”₯ Repeated 3x

    With reference to membrane tanks for the carriage of liquefied gas at very low temperatures:

    (a) Describe with a sketch one method of building up the insulation. (6)

    (b) State which alloy is used for the membrane and the reason. (5)

    (c) Explain why a secondary barrier is installed. (5)

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    Part (a)

    Membrane tanks for liquefied gas carriage are built with a double hull throughout the cargo length. The insulation system includes the following:

    1. Primary Barrier: A thin membrane made of INVAR (36% Nickel, 64% Iron) with a thickness of 0.5 to 1.2 mm, forming the primary containment layer for the liquefied gas.
    2. Primary Insulation box: A 200 mm thick layer of granulated Perlite insulation packed in plywood boxes, surrounding the primary barrier. Perlite is siliconised to make it impervious to moisture.
    3. Secondary Barrier: A secondary membrane made of the same INVAR material is installed to prevent cargo leakage in case of primary barrier failure.
    4. Secondary Insulation box: Another 200 mm thick layer of granulated Perlite is placed above the secondary barrier to provide additional insulation and prevent thermal transfer.
    Part (b)

    Alloy used for the membrane:

    Invar (36% Ni, 64% Fe) is used for both the primary and secondary barriers. The reason for this choice is its exceptionally low coefficient of thermal expansion. This eliminates the need for expansion joints or corrugations in the membrane design. In addition, Invar remains strong and does not become brittle at the very low temperatures experienced by the liquefied gas. The thin and lightweight nature of Invar maximizes the cargo-carrying capacity of the tank.

    Q3 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 5x

    (a) Explain in detail, how an underwater survey is carried out. (6)

    (b) State the requirements to be fulfilled before an under-water survey is acceptable to the survey authority. (5)

    (c) Construct a list of the items in order of importance that the underwater survey authority should include. (5)

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    Part (a)

    An in-water survey, also known as Underwater Inspection in Lieu of Dry Docking (UWILD), involves a systematic and detailed examination of a vessel’s hull and underwater components while the ship remains afloat. The procedure includes the following steps:

    • The shipowner sends a request to the classification society surveyor, indicating the intention to perform an underwater survey.
    • A detailed plan of the ship's external hull features is submitted, showing the location of frames, bulkheads, welding lines, openings, etc.
    • The hull surface is cleaned before the survey to remove any marine growth or fouling that could obscure the inspection.
    • A diving company, approved by the classification society, is appointed to conduct the underwater inspection.

    A self-propelled survey vehicle equipped with the following tools is used:

    • Long-Range Light TV Camera to aid in steering and checking hull deterioration.
    • High-resolution colour TV Camera to provide a close-up view of the hull coating and welded seams.
    • 35mm Still Camera to capture still images.
    • Ultrasonic Probe for measuring plate thickness.
    • Depth Meter and Speed Indicator to provide accurate data on the vehicle's depth and movement.
    • Umbilical Cable to connect the survey vehicle to the survey boat, transmitting power and information.

    The survey boat is to be equipped with:

    • A control console with TV monitors.
    • Plate thickness printouts.
    • Audio and video cassette recorders.
    • Playback units.
    • Diver communication systems.
    • Vehicle control systems and associated instruments.

    Operation:

    • The survey vehicle is taken underwater by a diver to the survey starting point.
    • Using TV monitors and shell expansion plans as guides, the vehicle is navigated over the hull, focusing on the bottom structure, sides, stern frame, rudder, propeller, bilge keel, and hull openings.
    • All images, data, and information are recorded and transmitted back to the survey boat.
    • Detailed pictures of the stern frame, rudder, propeller, bilge keel, and hull openings are captured.
    • Divers are deployed to measure stern tube bearing wear, pintle clearance, and inspect stern seals, anodes, and rudder stock palm coupling bolts.
    • All recorded video and audio, including conversations between the surveyor and drivers, thickness printouts, measurements, and pictures are analyzed to determine the detailed underwater condition of the vessel.
    Part (b)

    Before an in-water survey is accepted by the survey authority, the following conditions must be met:

    The vessel's owner submits a request to the surveyor, including:

    • The proposed date and location for the survey.
    • General information about the diving company.
    • A declaration that the vessel has not suffered any damage due to grounding, collision, or other incidents.

    The ship's master or owner’s representative must provide a declaration confirming:

    • Any suspected or actual damage to the hull since the last dry-docking.
    • The underwater portion of the hull is protected by a suitable paint scheme that is of adequate thickness and remains valid until the next dry-dock.
    • The survey site should be in a protected area with calm and clear water, ensuring good underwater visibility. Attention must be given to the effects of currents and tides.
    • The hull must be clean for the external survey. The surveyor must be satisfied with the method and quality of the pictorial presentation, ensuring that it provides a reliable assessment of the hull's condition.
    • The underwater examination should be conducted by an approved diving company using closed-circuit TV and two-way communication, which can be monitored by the surveyor.
    • The vessel should be in as light an operating condition as possible to facilitate the survey.
    • Means must be available for the surveyor to examine the outside shell plating above the waterline.
    • Any required repairs identified during the survey must be carried out to the satisfaction of the attending surveyor.
    Part (c)

    While the importance of each item may vary depending on the vessel and its specific requirements, below is a list of items that should be included in an underwater survey in order of importance:

    • Underwater Hull: General condition of the hull below the waterline.
    • Bottom and Shell Plating: Inspection for corrosion, damage, and fouling.
    • Shell Openings: Examination of openings such as sea chests, drain plugs, and overboard discharge points.
    • Stern Tube Oil Leaks: Check for leaks around the stern tube.
    • Propeller Blade: Inspection for damage, wear, and fouling.
    • Rudder: Inspection for damage, wear, and clearances.
    • Sea Chest Opening and Grating: Examination for blockages, damage, and fouling.
    • Anodes: Check the condition and effectiveness of cathodic protection anodes.
    • Bilge Keel: Inspection for damage and fouling.
    • Drain Plugs: Ensure all drain plugs are secure and in good condition.
    • Overboard Valve Openings: Check for proper operation and condition.
    • Forward Area: Inspection for any damage due to anchor and chain movement.
    Q4 (10 Marks) Hull Construction πŸ”₯ Repeated 8x

    (a) Sketch the cross-section of a bulk carrier with either deep or shallow double bottom showing the type of framing used. (8)

    (b) (i) Describe the corrosion problems experienced with ballast tanks. (4)

    (ii) State how such tanks are protected against extensive corrosion. (4)

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    Part (a)

    Mid-ship section of bulk carrier:

    Part (b)

    (i) Corrosion problems in ballast tanks

    are significant and arise due to various factors:

    • Galvanic corrosion occurs due to the interaction between dissimilar metals, accelerated by the seawater environment and differential aeration.
    • Sulphate-reducing bacteria in river mud can cause localized pitting, leading to penetration of the bottom shell.
    • Dissolved oxygen in seawater reacts with metal surfaces to form rust, significantly contributing to structural deterioration.
    • The rate of corrosion peaks at a 3.5% salt concentration, typical of seawater.
    • Neglecting maintenance exacerbates the problem, allowing corrosion to progress unchecked.

    (ii) Protection against extensive corrosion in ballast tanks involves the following measures:

    • Complete coating of the tank surfaces with properly selected marine-grade paint to create a protective barrier.
    • Installation of sacrificial anodes, designed to corrode preferentially and protect the tank structure.
    • Using large anodes with greater volume relative to surface area to ensure extended protection.
    • Regular gauging of plates to assess thickness and identify areas requiring reinforcement or repair.
    • Conducting regular inspections, cleaning the tanks, removing rust, and repainting to maintain structural integrity and prolong the lifespan of the ballast tanks.
    Q5 (10 Marks) Ship Stability

    With regard to ship construction details for transverse watertight bulkheads:

    (a) State the purpose of this type of bulkhead. (3)

    (b) State how the bulkheads are tested for water tightness. (3)

    (c) If it is necessary to penetrate the bulkhead, precaution must be taken to ensure that the watertight integrity and the strength of the bulkhead is maintained. With this in mind, describe, using simple sketches, how the following pass-through bulkheads. (10)

    (i) Main transmission shaft.

    (ii) Electrical cables.

    (iii) Fuel oil transfer pipes.

    (iv) Air and sounding pipes.

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    Part (a)

    The purpose of transverse watertight bulkheads includes:

    • Providing sufficient strength to contain flooding in the event of a compartment or one side of the bulkhead being bilged, preventing water from spreading.
    • Acting as hull-strengthening members by carrying a portion of the ship's vertical loading.
    • Separating incompatible cargos to prevent contamination or hazardous reactions.
    • Reducing transverse deformation of the ship, maintaining structural integrity.
    • Serving as effective barriers against the spread of fire in holds or machinery spaces.
    Part (b)

    Testing for Water Tightness:

    • Hose Test: This involves applying a high-pressure water jet (2 bar from a distance of 1.5 meters, nozzle diameter at least 12mm) to the bulkhead's surface. This test directly checks for leaks. The test is typically done from the side where the stiffeners are attached.
    • Air Pressure Test: An air pressure test applies pressure (0.2 bar) for about an hour, detecting leaks that may not be apparent during the hose test. This usually happens prior to the application of protective coatings.
    • Structural Test: Visual inspection, especially of welding joints, is carried out. Non-Destructive Testing (NDT) is done where necessary. Tanks designed to hold liquids, which form subdivisions of the ship, are tested for tightness with a water head up to the deepest subdivision load line or to a head of 2/3 the depth from the top of the tank to the margin line, whichever is greater.
    Part (c)

    Pass-Through Bulkheads:

    (i) Main Transmission Shaft:

    • The shaft is sealed using a stuffing box on both sides of the bulkhead. This consists of a gland that compresses packing material (e.g., braided flax, rubber) around the shaft, creating a watertight seal. The entire arrangement is secured with studs and wooden packing material between the bulkhead and the stuffing box to ensure a watertight fit.

    (ii) Electrical Cables:

    • The gland is typically composed of two halves (male and female) that fit together tightly. A sealing material (e.g., neoprene seal, felt washer) ensures a watertight seal around the cable. The gland is secured to the bulkhead to prevent leakage. The cable type and number needs consideration during the design of the gland to ensure adequate space and correct sealing material.

    (iii) Fuel Oil Transfer Pipes:

    • Fuel oil pipes are usually welded to the bulkhead. To accommodate expansion and contraction due to temperature changes, expansion bends (or omega loops) are incorporated into the piping system. Additionally, using smaller pipes connected to larger ones improves maintenance and repair accessibility.

    (iv) Air and Sounding Pipes:

    • One common method involves installing a doubler plate on both sides of the bulkhead around the pipe penetration area, improving the strength of the bulkhead. The pipe is then welded to the doubler plates. Alternative methods involve welding a flanged section of pipe directly to the bulkhead, ensuring a proper seal. A watertight seal around the penetration is required. Self-closing valves or blanking devices are often added as a safety measure to prevent flooding in case of a hull breach.
    Q6 (10 Marks) Ship Stability

    (a) Explain the purpose of non-watertight longitudinal subdivision of tanks. (6)

    (b) A ship 90 m long displaces 5200 tonne and floats at draughts of 4.95 m forward and 5.35 m aft when in sea water of 1023 Kg/m3. The waterplane area is 1100m2, GM, 95m, LCB 0.6m forward of midships and LCF 2.2m aft of midships. Calculate the new draughts when the vessel moves into fresh water of 1002 Kg/m3 (10)

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    Part (a)

    (b) Given:

    $$L\:=\:90m$$

    $$\Delta\:=\:5200\:tonnes$$

    $$d_{f}=\:4.95m$$

    $$d_{a}=\:5.35m$$

    $$\rho_{sw}=\:1023kg/m^3$$

    $$A_{w}=1100m^2$$

    $$G_{ML}=\:95cm$$

    $$LCB=\:0.6m\:fwd\:of\:midship\:$$

    $$LCF=\:2.2m\:aft\:of\:midship$$

    $$when\:vessel\:moves\:into\:fresh\:water\:of\:density\:1002kg/m^3$$

    $$new\:drafts\:=\:?$$

    $$Change\:in\:mean\:draft\:due\:to\:change\:in\:density\:$$

    $$=\:\frac{100\times\Delta}{A_{w}}\left\lbrack\frac{\rho_{s}\:-\:\rho_{r}}{\rho_{s}\times\rho_{r}}\right\rbrack$$

    $$=\:\frac{100\:\times5200}{1100}\left\lbrack\frac{1.023\:-\:1.002}{1.023\:\times1.002}\right\rbrack=\:9.68\times10^{-3}m$$

    $$Change\:in\:mean\:draft=\:9.7cm$$

    $$MCT_{1cm}=\frac{\Delta\:\times GM_{}_{L}}{100\:\times L}$$

    $$=\:\frac{5200\:\times95}{100\:\times90}$$

    $$MCT_{1cm}=\:54.88\:ton.\:m$$

    $$Change\:in\:trim\:when\:vessel\:moves\:from\:SW\:to\:FW$$

    $$=\:\frac{\Delta\times FB}{MCT_{1cm}}\left\lbrack\frac{\rho_{s}-\rho_{r}}{\rho_{s}}\right\rbrack$$

    $$FB\:=\:LCF\:+\:LCB$$

    $$FB\:=\:2.2\:+\:0.6\:=\:2.8m$$

    $$=\:\frac{5200\:\times2.8}{54.88}\left\lbrack\frac{1.023\:-\:1.002}{1.023}\right\rbrack\:=\:5.44\:\times10^{-3}$$

    $$Change\:in\:trim\:=\:5.45cm\:by\:head$$

    $$When\:trim\:by\:head,\:change\:in\:fwd\:draft$$

    $$d_{f}=\frac{+t}{L}\left\lbrack\frac{L}{2}+LCF\right\rbrack$$

    $$=\:\frac{5.45}{90}\left\lbrack\frac{90}{2}+2.2\right\rbrack$$

    $$d_{f}=\:2.858cm$$

    $$When\:trim\:by\:head,\:change\:in\:aft\:draft$$

    $$d_{a}=\:\frac{-t}{L}\left\lbrack\frac{L}{2}-LCF\right\rbrack$$

    $$=\:\frac{-5.45}{90}\left\lbrack\frac{90}{2}-2.2\right\rbrack$$

    $$d_{a}=\:-2.59cm$$

    New draught fwd = draft fwd + change in mean trim + change in fwd draft

    $$=\:4.95\:+\:0.097+0.02858\:$$

    $$D_{f}=\:5.076m$$

    $$New\:aft\:draft\:=\:5.35+0.097-0.0259\:$$

    $$D_{a}=5.421m$$

    $$New\:fwd\:draft\:D_{f}=5.076m$$

    $$New\:aft\:draft\:D_{a}=\:5.421m$$

    Q7 (10 Marks) Ship Resistance & Propulsion

    (a) Describe how thrust power is determined. (6)

    (b) The following information relates to a model propeller of 400mm pitch:

    Rev/min 400 450 500 550 600

    Thrust (N) 175 260 365 480 610

    Torque (Nm) 16.8 22.4 28.2 34.3 40.5

    (i) Plot curves of thrust and torque against rev/ min

    (ii) When the speed of advance of the model is 150 m/min and slip 0.20,

    Calculate the efficiency. (10)

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    Part (a)

    How thrust power is determined.

    Thrust power is the useful power developed by the propeller in producing thrust, given by:

    Thrust power = T x Va

    where T is the thrust of the propeller and Va is the speed of advance of the propeller through the water (the ship speed corrected for the wake). It is determined from the propeller thrust and the speed of advance. In model tests the thrust and torque are measured on the model propeller at various revolutions and speeds of advance, and the thrust power is computed as T x Va. The thrust power is less than the delivered power because of the propeller's own losses (the propeller efficiency = thrust power/delivered power). The thrust power is also related to the effective power by the hull efficiency (thrust deduction and wake).

    Part (b)

    Model propeller efficiency.

    Model propeller of 400 mm (0.4 m) pitch. Data: rev/min 400,450,500,550,600; thrust (N) 175,260,365,480,610; torque (Nm) 16.8,22.4,28.2,34.3,40.5.

    (i) Plot curves of thrust and torque against rev/min: both rise with rev/min, thrust from 175 N at 400 rpm to 610 N at 600 rpm, and torque from 16.8 to 40.5 Nm, both approximately linearly over the range.

    (ii) When the speed of advance of the model is 150 m/min and slip is 0.20, calculate the efficiency.

    Speed of advance Va = 150 m/min = 2.5 m/s.

    Slip = 0.20 means Va = pitch x n x (1 - slip), so pitch x n = Va/(1 - slip) = 2.5/0.8 = 3.125 m/s.

    n = 3.125/0.4 = 7.8125 rev/s = 468.75 rev/min.

    Interpolate thrust and torque at 468.75 rpm between 450 and 500 rpm (fraction = 0.375):

    T = 260 + 0.375 x (365 - 260) = 260 + 39.4 = 299.4 N.

    Q = 22.4 + 0.375 x (28.2 - 22.4) = 22.4 + 2.18 = 24.58 Nm.

    Thrust power = T x Va = 299.4 x 2.5 = 748.5 W.

    Delivered power = Q x omega = 24.58 x (2 pi x 7.8125) = 24.58 x 49.09 = 1206.6 W.

    Efficiency = thrust power/delivered power = 748.5/1206.6 = 0.620 = 62%.

    Answer: the propeller efficiency is about 62%.

    Q8 (10 Marks) Ship Stability

    (a) Explain why the amplitude of ship motion should be limited. (6)

    (b) A ship of 8100 tonne displacement floats upright in seawater, KG = 7.5m and GM = 0.45m. A tank, whose centre of gravity is 0.5m above the keel and 4m from the centreline, contains 100 tonne of water ballast. Neglecting free surface effect, calculate the angle of heel when the ballast is pumped out. (10)

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    Part (a)

    Why the amplitude of ship motion should be limited.

    Excessive amplitude of ship motion (especially roll, pitch and heave) is undesirable because:

    • It can lead to loss of stability and capsize, particularly in roll resonance or parametric roll.
    • It causes cargo shift, damage to cargo and lashings, and can endanger the crew.
    • It produces large accelerations that cause crew discomfort, seasickness and reduced efficiency, and can injure personnel.
    • It increases structural loading (slamming, racking) and can cause fatigue damage.
    • It increases resistance and fuel consumption, and reduces the ship's speed and course-keeping.
    • It can cause the propeller to emerge and the bow to slam, and can lead to broaching in following seas.

    Hence the amplitude is limited by adequate stability (GM), by damping devices (bilge keels, stabilisers, anti-roll tanks), and by operational measures (speed and course changes to avoid resonance).

    Part (b)

    Angle of heel when ballast is pumped out.

    Ship 8,100 t displacement, floats upright in sea water, KG = 7.5 m, GM = 0.45 m. A tank, whose centre of gravity is 0.5 m above the keel and 4 m from the centreline, contains 100 t of water ballast. Neglecting free-surface effect, calculate the angle of heel when the ballast is pumped out.

    KM = KG + GM = 7.5 + 0.45 = 7.95 m.

    Pumping out 100 t from (z = 0.5 m, y = 4 m):

    Transverse shift of G: GG' = w x y/(Delta - w) = 100 x 4/(8100 - 100) = 400/8000 = 0.05 m (towards the opposite side).

    Vertical shift of G: removing a low weight raises KG: GGv = w x (KG - z)/(Delta - w) = 100 x (7.5 - 0.5)/8000 = 100 x 7/8000 = 0.0875 m.

    New KG = 7.5 + 0.0875 = 7.5875 m. New GM = KM - KG = 7.95 - 7.5875 = 0.3625 m.

    Angle of heel: tan(theta) = GG'/GM = 0.05/0.3625 = 0.1379.

    theta = atan(0.1379) = 7.85 deg.

    Answer: the angle of heel is about 7.9 deg.

    Q9 (10 Marks) Ship Stability

    (a) Explain the effect of bilging a centerline compartment located away from amidships. (6)

    (b) A ship of 5000 tonne displacement has a double bottom tank 12m long. The Β½ breadths of the top of the tank are 5, 4 and 2m respectively. The tank has a watertight centreline division. Calculate the free surface effect if the tank is partially full of fresh water on one side only. (10)

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    Part (a)

    Effect of bilging a centreline compartment located away from amidships.

    Bilging a centreline compartment (one that spans the full width, with the centreline running through it) that is located away from amidships causes:

    • A loss of buoyancy in that compartment, so the ship sinks (increases draught) and, because the compartment is away from amidships, the loss of buoyancy is not symmetrically distributed about the centre of flotation, producing a change of trim (the ship trims towards the damaged end).
    • The loss of buoyancy also reduces the waterplane area and hence the BM and GM, so the stability is reduced.
    • Because the compartment is on the centreline, there is no list (the loss is symmetric about the centreline), but there is a change of trim and a reduction of GM.
    • The free-surface effect of the flooded water further reduces the effective GM.

    The ship must be able to survive the flooding with adequate residual stability and freeboard (damage stability criteria).

    Part (b)

    Free-surface effect of a partially full double-bottom tank.

    Ship of 5,000 t displacement has a double-bottom tank 12 m long. The half-breadths of the top of the tank are 5, 4 and 2 m respectively. The tank has a watertight centreline division. Calculate the free-surface effect if the tank is partially full of fresh water on one side only.

    Because of the centreline division, the free surface exists on one side only, and its breadth is the half-breadth (5, 4, 2 m) at the three stations over the 12 m length (spacing 6 m).

    Second moment of area of the free surface about its longitudinal axis:

    i = (1/12) x integral of (breadth^3) dx = (1/12) x (h/3)[b0^3 + b2^3 + 4 b1^3]

    = (1/12) x (6/3)[5^3 + 2^3 + 4 x 4^3] = (1/12) x 2[125 + 8 + 4 x 64] = (1/12) x 2[133 + 256] = (1/12) x 2 x 389 = (1/12) x 778 = 64.83 m4.

    Free-surface effect = rho x i/Delta = 1.000 x 64.83/5000 = 0.01297 m.

    Answer: the free-surface effect is about 0.013 m (13 mm) reduction in GM.

    Q10 (10 Marks) Ship Stability πŸ”₯ Repeated 3x

    (a) Describe stability requirements for dry-docking (6)

    (b) A ship of 8000 tonne displacement, 110 m long, floats in sea water of 1.04 t/m3 at draughts of 6 m forward and 6.3 m aft. The TPC is 16, LCB 0.6 m aft of midships, LCF 3 m aft of midships and MCT1 cm 65 tonne m. The vessel now moves into fresh water of 1.000 t/m3. Calculate the distance a mass of 50 tonne must be moved to bring the vessel to an even keel and determine the final draught.

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    Part (a)

    For safe dry-docking, a ship must meet two key stability requirements:

    • Positive GM (Metacentric Height): The ship needs a positive GM. GM is the distance between the centre of gravity (G) and the metacentre (M). A positive GM indicates inherent stability; the ship will right itself if tilted. During dry-docking, the loss of buoyancy as the ship rests on the blocks reduces GM. Insufficient positive GM increases the risk of the ship heeling over or capsizing.
    • Trim by the stern: The vessel should be trimmed slightly by the stern (aft end lower than the bow) to ensure the aft end sits on the keel blocks first. This controlled settling minimises the risk of instability during the docking process. An even keel is generally preferred for the initial floating condition before the dry-docking procedure begins.
    • Bilges: Ensure all bilges are dry and bilge tanks are empty, no unaccounted water onboard.
    • Soundings: Record all soundings of tanks including FW, Ballast, FO, LO, DO and make sure at the time undocking no change in these readings.
    • Upright: Vessel should always be upright during docking and undocking.
    Q1 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 2x

    (a) With reference to the underwater surface of a ship's hull: (16)

    (i) Describe a hull plate roughness analyser system

    (ii) State the significance of the roughness profile and compare the typical roughness values for a new ship and a ship eight years old;

    (b) Which reference to the application of self-polishing paint in dry dock:

    (i) Describe the plate preparation necessary

    (ii) State the defects that may occur in the paint coating if it is not correctly applied.

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    Part (a)

    With reference to the underwater surface of a hull:

    (i) The Hull Roughness Analyzer (HRA) is a portable system designed to measure the roughness of a ship’s hull surface. It includes:

    • A portable microprocessor with a digital display and printout capability.
    • A handheld carriage equipped with a stylus measuring head. The stylus traces the hull surface, recording peaks and valleys to assess roughness.
    • The system measures 10 sample lengths of 50 mm during a single traverse over the hull. For each 50 mm sample, the highest peak to lowest valley is recorded as Rt(50).
    • Roughness surveys are conducted at approximately 100 locations on the hull, including the bow, stern, midship, and topping. Multiple traverses at each location provide numerous readings.
    • Mean Hull Roughness (MHR) is calculated as:
    • $$MHR\:=\:\frac{\sum Rt\left(50\right)}{n},\:where\:n\:is\:the\:number\:of\:readings\:at\:a\:station$$
    • Average Hull Roughness (AHR) is calculated as:
    • $$AHR\:\frac{\sum MHR}{m},\:where\:m\:is\:the\:total\:number\:of\:survey\:stations$$

    (ii) Significance of Roughness Profile:

    The roughness profile of the prepared hull surface impacts the performance of the applied coating and the overall operational efficiency of the vessel. A rough surface increases frictional resistance as the vessel moves through the water. This increased drag translates to higher power requirements for propulsion, leading to increased fuel consumption and operational costs. Furthermore, greater surface roughness contributes to increased carbon emissions, a concern under current MARPOL regulations. Therefore, a controlled and optimised roughness profile is essential for minimising frictional resistance, reducing fuel consumption and emissions, and maximising the longevity of the hull coating.

    Typical Average Hull Roughness (AHR):

    • New Ship: 120-200 Β΅m (approx.).
    • Eight-Year-Old Ship: 300-400 Β΅m (approx.) due to annual deterioration of 20-40 Β΅m.
    Part (b)

    (i) Plate Preparation for Self-Polishing Paint Application:

    Washing:

    • The hull surface must be thoroughly cleaned to remove all marine growth (algae, slime, etc.), accumulated salts, dirt, grease, and oil. High-pressure freshwater washing is the standard method for this initial cleaning. The goal is to present a clean substrate for subsequent stages.

    Blasting:

    • Abrasive blasting is the preferred method for removing rust, defective paint, and any remaining contaminants. This process achieves a bare metal surface, essential for proper adhesion of the new coating. The extent of blasting (localized or full hull) depends on the condition of the existing surface. The intensity and type of abrasive used are carefully controlled to achieve the desired surface roughness profile.

    Primer Application:

    • After blasting, the surface is again cleaned to remove any blasting debris. A primer coat is then applied to provide corrosion protection and to create an ideal surface for the subsequent topcoat adhesion. This primer acts as an intermediary layer, enhancing the bond between the substrate and the long-life coating system.

    (ii) Defects in Self-Polishing Paint Application:

    • Blistering: Bubbles caused by trapped moisture.
    • Curtaining/Sagging: Paint sagging due to its own weight.
    • Dry Spray: Paint particles drying before reaching the surface.
    • Peeling/Flaking: Caused by moisture, dust, excessive film thickness, or incompatible paints.
    • Cissing/Crawling: Paint contracting immediately after application.
    Q2 (10 Marks) Hull Construction πŸ”₯ Repeated 2x

    With reference to membrane tanks for the carriage of liquefied gas at very low temperatures. (16)

    (a) Describe with a sketch one method of building up the insulation

    (b) State which alloy is used for the membrane and the reason

    (c) Explain why a secondary barrier is installed.

    (i) Longitudinally

    (ii) Transversely

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    Part (a)

    Membrane tanks

    for liquefied gas carriage are built with a double hull throughout the cargo length. The insulation system includes the following:

    1. Primary Barrier: A thin membrane made of INVAR (36% Nickel, 64% Iron) with a thickness of 0.5 to 1.2 mm, forming the primary containment layer for the liquefied gas.
    2. Primary Insulation box: A 200 mm thick layer of granulated Perlite insulation packed in plywood boxes, surrounding the primary barrier. Perlite is siliconised to make it impervious to moisture.
    3. Secondary Barrier: A secondary membrane made of the same INVAR material is installed to prevent cargo leakage in case of primary barrier failure.
    4. Secondary Insulation box: Another 200 mm thick layer of granulated Perlite is placed above the secondary barrier to provide additional insulation and prevent thermal transfer.
    Part (b)

    Alloy used for the membrane:

    Invar (36% Ni, 64% Fe) is used for both the primary and secondary barriers. The reason for this choice is its exceptionally low coefficient of thermal expansion. This eliminates the need for expansion joints or corrugations in the membrane design. In addition, Invar remains strong and does not become brittle at the very low temperatures experienced by the liquefied gas. The thin and lightweight nature of Invar maximizes the cargo-carrying capacity of the tank.

    Q3 (10 Marks) Hull Construction

    Describe a forced ventilation system for the machinery spaces and a natural ventilation system for a lower hold. Why hold ventilation is considered necessary? (16)

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    Forced ventilation for Machinery space:

    Cargo hatch Mechanical ventilation:

    Mechanical ventilation is achieved by using fans to circulate air through the hold. A mechanical ventilation system may have fans that are protected by covers.

    Cargo hatch Natural ventilation:

    Natural ventilation is achieved by opening the ventilator doors/ covers to allow the air to flow naturally. A natural ventilation system may have doors/ covers that are closed with gaskets.

    Why Hold Ventilation is Considered Necessary:

    The primary purpose of hold ventilation is to minimize damage to the cargo and ensure the safety of the crew and vessel.

    • Minimizing the Formation of Sweat (Dew Point Control):
    • Removing Hazardous Gases Emitted by Cargo:
    • Preventing Excessive Heating of Cargo:
    • Removing Taint:
    • Preventing the Hold from Becoming an Enclosed Space:
    • Preventing the Hold from Becoming a Vacuum Chamber:
    • Providing a Favorable Atmosphere for Crew Working in the Hold:
    Q4 (10 Marks) Ship Stability πŸ”₯ Repeated 3x

    With respect to trim and stability, describe the following: (16)

    (a) Effects on centre of gravity of slack tanks

    (b) Effect on stability of ice formation on superstructure

    (c) Effects of wind and waves on ship's stability

    (d) Effect of water absorption by deck cargo and retention of water on deck.

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    Part (a)

    Slack tanks, or partially filled tanks, significantly impact a ship's stability due to the free surface effect

    • When tanks are partially filled (slack), the liquid inside moves freely as the ship heels.
    • This movement shifts the centre of gravity (G) laterally towards the heeling side.
    • The righting lever (GZ) decreases, resulting in reduced metacentric height (GM) and overall stability.
    • The virtual loss of GM is proportional to the breadth of the tank and the height of the free surface.

    Consequences of Slack Tanks:

    • Increased angle of heel.
    • Reduced stability, making the ship tender and more prone to capsizing.

    To Minimize the Effect:

    • Avoid slack tanks where possible; either fill tanks completely or empty them.
    • Design tanks with longitudinal divisions or swash bulkheads to limit liquid movement.
    • Use sluice valves in tanks to control liquid transfer.
    • Prioritize filling smaller tanks at the ship's bottom to lower G and improve stability.
    Part (b)

    Ice formation on superstructure:

    Ice accumulating on the superstructure adds mass high up on the ship. This raises the centre of gravity (G), decreasing the metacentric height (GM). A lower GM reduces stability, making the ship more tender (more easily rolled) and increasing the period of roll. The ship becomes more susceptible to capsizing.

    Part (c)

    Effect of Wind and Waves

    Wind:

    • High freeboard or tall superstructures increase windage, leading to a greater rolling effect.
    • Rolling caused by wind reduces stability, especially if the ship remains heeled for a prolonged period.

    Waves:

    • Large waves, especially when the ship is on the crest, can cause a significant loss of stability due to reduced underwater buoyant volume.
    • The ship may develop excessive heeling or capsizing tendencies.
    • Long ships are more vulnerable to wave action due to greater surface exposure, further reducing stability.
    Part (d)

    Effect of Water Absorption by Deck Cargo and Retention of Water on Deck:

    • Water Absorption by Deck Cargo:
      • Certain types of deck cargo, such as timber or other absorbent materials, can take up water during a voyage.
      • This absorbed water increases the weight of the cargo, raising the centre of gravity (G) of the ship.
      • As G moves higher, GM reduces, leading to decreased stability.

      • Retention of Water on Deck:
        • Water that accumulates on the deck, such as from heavy rain or seawater, adds additional weight to the ship's upper structure.
        • This shifts G upward, reducing GM and stability.
        • Retained water may also cause a list if it collects asymmetrically, further impacting the ship's balance.

        • Consequences:
          • Increased risk of capsizing in rough seas.
          • Increased rolling and reduced righting ability.

          • Mitigation Measures:
            1. Ensure proper drainage systems are in place to quickly remove water from the deck.
            2. Regularly monitor and secure deck cargo to prevent excessive water absorption.
    Q5 (10 Marks) Corrosion & Protection πŸ”₯ Repeated 4x

    With reference to the prevention of hull corrosion discusses: (16)

    (a) Surface preparation and painting of new ship plates.

    (b) Design of the ships structure and its maintenance.

    (c) Cathodic protection by sacrificial anodes, of the internal and external areas of the ship.

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    Part (a)

    Surface Preparation and Painting of New Ship Plates:

    The process begins with the removal of mill scale, a thin layer of iron oxide that forms during the steel rolling process.

    1. Removal of Mill Scale:

    • Weathering: Using wire brushes.
    • Pickling: Immersing plates in a weak solution of Hβ‚‚SOβ‚„ or HCl.
    • Shot Blasting: Abrasive blasting to remove rust and mill scale.
    • Flame Cleaning: Heating and cleaning the surface.

    Freshwater Washing and Drying: Ensures the surface is clean and ready for priming.

    2. A primer is then applied. This acts as a base layer for subsequent coatings and improves adhesion. Common types include epoxy primers with corrosion-inhibiting pigments like iron oxide, zinc, and calcium phosphates (though zinc content is reduced due to safety concerns).

    3. Next comes the anticorrosive coating, the main layer offering corrosion protection. Common choices are two-component epoxies, coal tar epoxies, and epoxy or polyester coatings with glass flakes for added strength and water impermeability.

    4. Finally, an antifouling coating prevents marine organism attachment. While tin-based paints were once common, environmental regulations have led to their replacement with copper-based, silicone-based, or non-TBT (Tributyltin) self-polishing coatings. These utilize seawater-soluble polymers. The number of antifouling layers (two or three) depends on the chosen system and desired lifespan. The dry film thickness (DFT) of the primer is closely monitored to prevent cracking and ensure effective protection.

    Part (b)

    Corrosion prevention begins at the design stage:

    • Avoid dissimilar metal joining to minimize galvanic corrosion.
    • Use high-quality steel plates based on the galvanic series.
    • Minimize areas where water can accumulate by ensuring proper drainage systems.
    • Ensure sufficient anodes are installed in optimal positions.
    • Consider installing Impressed Current Cathodic Protection (ICCP) and Marine Growth Prevention Systems (MGPS).

    Maintenance:

    • Regular checks and maintenance of ICCP and MGPS systems.
    • Renewal of sacrificial anodes during dry-docking.
    • Cleaning, repairing, and repainting surfaces to maintain the protective coatings.
    • Ensuring proper drainage and inspecting for any areas of corrosion.
    Part (c)

    Cathodic Protection by Sacrificial Anodes:

    Sacrificial anodes are less noble metals (more electro-negative) than the hull material. When immersed in seawater, they corrode preferentially, protecting the hull. Common materials include zinc, aluminium, and magnesium. They are welded to the hull for good electrical contact.

    • External surface: Large anodes provide complete protection until the next dry-docking.
    • Internal surface: Anodes are fitted inside tanks, filters, and coolers to prevent internal corrosion.
    • MGPS: These systems utilize sacrificial anodes to protect seawater lines and pumps, further preventing marine growth.
    • ICCP: Impressed current systems offer an alternative, using an external DC power source and permanent anodes (typically titanium or platinum) for continuous protection. These systems are self-regulating and adjust current output based on hull coating condition. Regular monitoring (e.g., monthly data analysis) is important for both sacrificial and impressed current systems.
    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Describe how the distribution of mass within the ship affects the rolling period. (6)

    (b) The righting moments of a ship at angles of heel of 0, 15ΒΊ, 30Β°, 45ΒΊ, and 60Β° are 0, 1690, 5430, 9360 and 9140 k/Nm respectively. Calculate the dynamical stability at 60Β°. (10)

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    Part (a)

    The rolling period is influenced by the ship's metacentric height (GM) and radius of gyration (K), which are determined by the location of the masses onboard.

    Masses at the bottom:

    • The centre of gravity (G) moves down.
    • Metacentric height (GM) increases, resulting in greater stability.
    • Rolling period decreases.

    Masses at the top:

    • The centre of gravity (G) moves up.
    • Metacentric height (GM) decreases, reducing stability.
    • Rolling period increases.

    Masses concentrated at the centre:

    • The radius of gyration (K) decreases.
    • Rolling period decreases.

    Masses distributed away from the centre:

    • The radius of gyration (K) increases.
    • Rolling period increases.
    Q7 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Describe briefly the significance of the factor of subdivision. (6)

    (b) A ship 120m long has a light displacement of 4000 tonne and LCG in this condition 2.5m aft of midships. The following items are then added: (10)

    Cargo 10000 tonne LCG 3.0m forward of midships

    Fuel 1500 tonne LCG 2.0 m aft of midships

    Water 400 tonne LCG 8.0m aft of midships

    Stores 100 tonnes LCG 10.0m forward of midships

    Using the following hydrostatic data, calculate the final draughts:

    Draught (m)

    Displacement (t)

    MCT1cm (tm)

    LCB from midships

    LCF from midships

    8.50

    16650

    183

    1.94F

    1.29A

    8.00

    15350

    175

    2.10F

    0.60F

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    Part (a)

    Factor of Subdivision

    The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

    Permissible Length Formula:

    $$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

    A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

    Part (b)

    Mass added

    LCG from midship

    Mass moments

    F

    A

    10000

    3.0 Fwd

    30000

    1500

    2.0 Aft

    3000

    400

    8.0 Aft

    3200

    100

    10.0 Fwd

    1000

    4000

    2.5 Aft

    10000

    16000

    31000

    16200

    $$Excess\:moment=31000-16200=14800$$

    $$LCG=\frac{\sum M}{\sum m}=\frac{14800}{16000}$$

    $$LCG=0.925m\:fwd\:of\:midship$$

    Draught

    Displacement

    MTC 1cm

    LCB from midship

    LCF from midship

    8.5

    16650

    183

    1.94 Fwd

    1.20 Aft

    8.25

    16000

    179

    2.03 Fwd

    0.57 Aft

    8.0

    15350

    175

    2.10 Fwd

    0.06 Fwd

    $$LCG=0.925m\:Fwd\:of\:midship$$

    $$LCB=2.02m\:Fwd\:of\:midship$$

    $$Trimming\:lever=LCB-LCG$$

    $$2.02-0.925$$

    $$=1.09m$$

    $$Trimming\:moment=m\times d$$

    $$=16000\times1.09$$

    $$=17440tm$$

    $$Change\:of\:trim=\frac{Trimming\:moment}{MCT_{1\operatorname{\mathrm{cm}}}}$$

    $$=\frac{17440}{179}$$

    $$97.43\operatorname{cm}$$

    $$Draft\:fwd=8.25-\frac{97.43}{100\times120}\left(\frac{120}{2}+0.57\right)$$

    $$=7.756m$$

    $$Draft\:aft=8.25+\frac{97.43}{100\times120}\left(\frac{120}{2}-0.57\right)$$

    $$=8.735m$$

    Q8 (10 Marks) Ship Stability πŸ”₯ Repeated 2x

    (a) Explain the effect of trim on tank soundings. (6)

    (b) A ship of 6600 tonne displacement has KG 3.6m and KM 4.3m. A mass of 50 tonne is now lifted from the quay by one of the ship's derricks whose head is 18m above the keel. The ship heels to a maximum of 9.5Β° while the mass is being transferred. Calculate the outreach of the derrick from the ship's centreline. (10)

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    Part (a)

    The effect of trim on tank soundings is significant because the reading obtained from the sounding pipe does not directly reflect the actual volume of liquid in the tank when the ship is trimmed. Typically, the sounding pipe is located at the aft end of the tank.

    If the ship is trimmed by the stern, the liquid will settle more toward the aft end, resulting in a higher sounding than if the ship were on an even keel. Conversely, if the ship is trimmed by the bow, the liquid shifts forward, and the sounding obtained will be less, potentially underestimating the actual volume.

    To ensure accurate volume measurement despite varying trim conditions, tank calibrations are carried out for different trims, and the data is compiled into a sounding book or trim correction table. When taking soundings, the trim is noted, and the corrected volume is obtained using these calibration tables, ensuring reliable tank content assessment regardless of trim.

    Q9 (10 Marks) Ship Stability πŸ”₯ Repeated 2x

    (a) Explain the effect on GM during the filling of a double-bottom tank. (6)

    (b) A ship of 8,000 tonnes displacement has KM 7.5m., and KG 7.0 m. A double bottom tank is 12 metres long 15 metres wide and 1 metre deep. The tank is divided longitudinally at the center line and both sides are full of salt water. Calculate the list if one side is pumped out until it is half empty. (10)

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    Part (a)

    During the filling of a double-bottom (DB) tank, which is located at the bottom of the ship, mass is added down in the vessel. As a result, the center of gravity (G) shifts downward, leading to an increase in GM (metacentric height) and, consequently, an improvement in the ship's stability.

    However, if the ship is heeled or trimmed during filling, the liquid inside the tank may collect on the lower side, causing both the center of gravity (G) and the center of buoyancy (B) to shift to that side. This results in a virtual loss of GM and a reduction in stability.

    Once the DB tanks are fully pressed up, the liquid is evenly distributed, the ship returns to an upright position, and the center of gravity remains at its lowest point, resulting in maximum GM and optimal stability.

    Q10 (10 Marks) Ship Stability πŸ”₯ Repeated 2x

    A ship of 9,900 tonnes displacement has KM = 7.3 m and KG = 6.4 m, she has yet to load two 50 tonne lifts with her own gear and the first lift is to be placed on deck on the inshore side (KG 9 m and center of gravity 6m out from the center line). When the derrick plumbs the quay its head is 15m above the keel an 12m out from the center line.

    Calculate the maximum list during operation. (16)

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    Maximum list during derrick operation.

    Initial condition: displacement = 9900 t; KM = 7.3 m; KG = 6.4 m, so initial GM = KM - KG = 7.3 - 6.4 = 0.90 m.

    The first 50 t lift is picked up off the quay (inshore side). While it is suspended at the derrick head, its position is above and outboard of the crane; as it swings, it acts as a weight at the head which is 15 m above the keel and 12 m out from the centreline. The maximum list occurs when the weight is suspended fully outboard (12 m out) before it is placed on deck at 6 m out, because the heeling moment is greatest then.

    Effective heeling of the first lift. Treated the 50 t as suspended at (y=12 m outboard, z=15 m above keel):

    • Transverse shift of the centre of gravity, GG' = w x y / Delta = 50 x 12 / 9900 = 0.0606 m outboard (about 6.06 cm).
    • Vertical rise of the centre of gravity, since the load hangs at the derrick head above deck: compare to its final on-deck position at 6 m out, 9 m above keel. The movement from quay to head raises the effective load centre; the vertical rise of G = w x (rise)/(Delta). For the list the relevant GM is reduced by the raised KG.

    A cleaner approach is to compute the moment. At maximum list the weight is fully outboard at 12 m; its KG contribution raises the centre of gravity by GGv = 50 x (15 - 9)/9900 = 50 x 6/9900 = 0.0303 m. New KG = 6.4 + 0.0303 = 6.4303 m; the effective GM falls to 7.3 - 6.4303 = 0.8697 m.

    List, GGh = 50 x 12 / 9900 = 0.0606 m outboard.

    tan(list) = GGh / GM = 0.0606 / 0.8697 = 0.0697.

    So list = atan(0.0697) = 3.99 deg, about 4 deg.

    Answer: the maximum list during the operation is about 4 degrees towards the outboard (derrick) side. (Note: after the first lift is placed on deck at 6 m out and the second lift is begun, the situation changes slightly; the critical maximum effect is when the suspended load is furthest outboard.)

    Q1 (10 Marks) Ship Types & Design πŸ”₯ Repeated 3x

    Discuss the importance of the following to be examined for meeting EEDI limitations: (16)

    (a) Slimmer vessels with lower block coefficients

    (b) Long-Stroke engines

    (c) Low revolution large diameter propellers.

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    The Energy Efficiency Design Index (EEDI) expresses the grams of CO2 emitted per tonne of cargo transported per nautical mile. Its level is directly proportional to the fuel consumed (and hence the propulsive power required) at the design speed, divided by the cargo capacity and speed. Anything that cuts the power needed to move a given deadweight at the design speed, or that raises propulsive efficiency, lowers the EEDI. The three methods below act on exactly those levers.

    Part (a)

    Slimmer vessels with lower block coefficients.

    A fine hull with a low block coefficient (Cb) presents less displaced volume for a given length and thus a smaller wetted surface area, reducing the frictional resistance component at a given displacement. A fine (low Cb) forebody and afterbody also weaken the bow wave and the pressure (residual or wave-making) resistance at the design speed, because the water is pushed aside more gradually. Lower residual resistance means the installed propulsive power at the design speed is less, so less fuel is burnt per tonne-mile, lowering the EEDI. Being longer and narrower for the same displacement also raises waterline length, which reduces the length-related frictional and Froude-number-dependent resistance. The penalty is reduced cubic cargo capacity and somewhat less form stability, so the hull form is optimised rather than simply fined down. Lower block coefficient is therefore one of the strongest design levers for meeting EEDI limits.

    Part (b)

    Long-stroke engines.

    A long-stroke (high stroke-to-bore ratio) slow-speed diesel extracts more work from each unit of fuel in the expansion stroke and achieves higher thermal (brake) efficiency, typically up to about 50 per cent, with a correspondingly lower specific fuel consumption per kWh. Because the stroke is long, the engine can turn slowly at the same piston speed, enabling direct coupling to a large slow-turning propeller with no reduction gearbox and no associated transmission losses. A more efficient engine burns less fuel for each kW it delivers, hence produces less CO2 per tonne-mile, which reduces the attained EEDI directly. Long-stroke engines also operate at low revolutions, which marries perfectly with the large-diameter, low-rev propeller of part (c).

    Part (c)

    Low-revolution large-diameter propellers.

    Propeller open-water efficiency rises as the disc-area loading (thrust per unit swept area) falls. A large-diameter propeller turning slowly accelerates a large mass of water by a small amount, giving low disc loading and high efficiency for the same thrust and therefore less shaft power and less fuel per tonne of cargo. Large slow propellers also stay further away from cavitation, reducing blade erosion, vibration and noise. Because EEDI is fixed by the fuel (shaft power) needed at the design speed, optimising hull, engine and propeller together β€” a fine low-block hull at a low Froude number, a long-stroke low-speed engine, and a large-diameter low-rev propeller β€” minimises energy consumption per tonne of cargo and is the classic route to satisfying EEDI limitations.

    Q2 (10 Marks) General πŸ”₯ Repeated 3x

    (a) Draw a simple line diagram of the bow of a ship to show the position of the following component parts of the ships anchoring system. Hawse pipe, Cable stopper, Windlass and Cable lifter, Spurling pipe and Chain locker. (6)

    (b) Describe the cable stopper and state its purpose. (3)

    (c) Show by means of a sketch how the anchor cable is attached to the ship. (3)

    (d) Describe how the chain locker is drained of water, sand, and mud. (4)

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    Part (a)

    At the bow of the ship, you'd typically find the following components:

    • Hawse pipes: These are openings in the hull through which the anchor chain passes.
    • Cable stopper: Positioned on the deck near the hawse pipes, the cable stopper is used to secure the anchor chain in place.
    • Windlass: Located on the deck near the hawse pipes, the windlass is a mechanical device used to raise and lower the anchor and its chain.
    • Cable lifts: These are devices used to guide the anchor chain from the windlass into the chain locker below deck.
    • Spurling pipe: A vertical pipe leading from the windlass to the chain locker, through which the anchor chain passes.
    • Chain locker: Below deck, typically located near the bow, the chain locker is a compartment where the anchor chain is stowed when not in use.
    Part (b)

    The cable stopper is a device used to secure the anchor chain in place once the anchor has been lowered or raised. Its purpose is to prevent the anchor chain from slipping or running out unintentionally, ensuring that the anchor remains securely in position.

    Part (c)

    The anchor cable is typically attached to the ship using a shackle or swivel at the end of the anchor chain. This attachment point allows the anchor chain to pivot freely as the anchor is raised or lowered, preventing twisting or tangling of the chain.

    Part (d)

    The chain locker is drained of water, sand, and mud through a drainage system that typically includes scupper holes or drain pipes located in the bottom of the locker. These drains allow any water or debris that accumulates in the chain locker to flow out of the compartment and overboard, ensuring that the anchor chain remains clean and free of obstructions. Additionally, regular maintenance and cleaning of the chain locker are important to prevent the buildup of sediment and ensure proper drainage.

    Q3 (10 Marks) Ship Types & Design πŸ”₯ Repeated 3x

    (a) State the reasons for the freeboard requirement. (4)

    (b) Explain the term condition of assignment and explain how these are maintained for a ship. (4)

    (c) Using a diagram indicate the freeboard of type A, type B, type B60 and type B100 vessels giving an example of each type. (8)

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    Part (a)

    Reasons for Freeboard Requirements:

    Freeboard is the distance measured from the waterline to the upper edge of the deck plating at the side of the freeboard deck amidships.

    • Ensures the ship is seaworthy when fully loaded.
    • Provides reserve buoyancy, enabling the ship to rise as it passes through waves, keeping decks largely dry.
    • Enhances the ship's stability and increases its survivability in case of damage by allowing it to remain afloat longer, aiding crew escape or damage control.
    Part (b)

    Conditions of Assignment:

    Conditions of Assignment are outlined in the Merchant Shipping Load Line Rules 1968 and must be satisfied before a ship is assigned freeboards and issued a load line certificate. These conditions address the practical need for openings (hatchways, doorways, vents, etc.) in the ship's hull and mandate appropriate protection and closure mechanisms for these openings.

    Requirements Before Assigning Load Line:

    • The ship must have sufficient structural strength.
    • Adequate reserve buoyancy must be maintained.
    • Openings must be secured against water ingress.
    • Safety measures for the crew, such as guardrails and gangways, must be in place.

    To ensure that the conditions of assignment are still current the following items can be checked and confirm to be without change or damage from when the ship was built.

    • Access openings in bulkheads, to ensure that they can be sealed and prevent flooding
    • Cargo and hatchways ensure they can be sealed to prevent flooding
    • Coamings of hatchways, sign of corrosion damage risk of failure would allow flooding
    • Protection of openings, can they all be sealed
    • Ventilator coamings not corroded as they would allow flooding to other arears
    • Air pipes can be shut in heavy weather and not corroded
    • Discharges, inlets and scuppers, all in good condition and operational
    • Side scuttles can be secured
    • Hull inspection, sea boxes and penetrations all checked for damage and corrosion free.
    Part (c)

    Freeboard for Different Vessel Types:

    Type A Ships:

    • Designed for liquid cargo only (e.g., oil tankers).
    • Have the lowest freeboard among ship types due to their higher reserve buoyancy.

    Type B Ships:

    • Designed for general cargo (e.g., container ships).
    • Have a higher freeboard than Type A ships.

    Type B-60 Ships:

    • A variant of Type B ships, with freeboard reduced by up to 60% of the difference between Type A and B (e.g., OBO ships).

    Type B-100 Ships:

    • Type B ships with freeboard reduced by up to 100% of the difference between Type A and B (e.g., certain bulk carriers).
    Q4 (10 Marks) Machinery Space & Systems πŸ”₯ Repeated 5x

    (a) Describe the double bottom and framing arrangement used in the machinery space to cope up with the concentrated loads and vibration, together with shaft and thrust block support. (10)

    (b) Give reasons for the choice of thrust block position. (6)

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    Part (a)

    The construction of the double bottom in the machinery space regardless of the framing system has solid plate floors at every frame space under the main engine. Additional side girders are fitted outboard of the main engine seating, as required. The double-bottom height is usually increased to provide fuel oil, lubricating oil and fresh water tanks of suitable capacities. Shaft alignment also requires an increase in the double-bottom height or a raised seating, the former method usually being adopted.

    Continuity of strength is ensured and maintained by gradually sloping the tank top height and internal structure to the required position. Additional support and stiffening is necessary for the main engines, boilers, etc., to provide a vibration-resistant solid platform capable of supporting the concentrated loads. On slow- speed diesel-engined ships, the tank top plating is increased to 40 mm thickness or thereabouts in way of the engine bedplate. This is achieved by using a special insert plate which is the length of the engine including the thrust block in size. Additional heavy girders are also fitted under this plate and in other positions under heavy machinery as required. Plating and girder material in the machinery spaces is of increased scantlings in the order of 10 per cent.

    The method adopted.

    A cellular void space within a ships structure is called a coffer dam. Like a bulkhead separates two spaces or divides a space into two, a coffer dam does the same with a larger degree of integrity since it incorporates a void which would contain any breach of either of the boundaries. This would contain the leakage and prevent it spreading into other areas. A cofferdam can be defined as an empty space separating compartments to prevent the contents of one compartment from entering another in case of leakage.

    Part (b)

    The thrust block is positioned close to the propulsion machinery (usually just aft of the main engine). Reasons are as follows:

    • The axial thrust generated by the propeller could cause deformation and misalignment of the shafting system if the thrust block were positioned far from the engine. Placing it close to the engine minimizes the length of shafting subject to this thrust, thus reducing potential for misalignment. The strong double bottom structure directly beneath provides necessary support to mitigate any deformation.
    • Differential expansion between the shaft and the hull due to temperature variations is a potential source of misalignment. Positioning the thrust block close to the engine helps to minimize the effect of this differential expansion.
    • The weight of the propeller and the dynamic forces it creates can lead to whirling of the tailshaft and misalignment if not properly managed. Positioning the thrust block near the engine helps to stabilize the shafting system and reduce the risk of these issues.
    • The substantial double-bottom structure under the main machinery provides an ideal, inherently strong foundation for the thrust block. This minimises the need for extensive additional reinforcement to support the thrust loads.
    Q5 (10 Marks) Machinery Space & Systems πŸ”₯ Repeated 3x

    With reference to a periodically unattended machinery space of a dry cargo vessel discuss the requirements for:

    (a) Protection against flooding (8)

    (b) Control of propulsion machinery from the navigating bridge.

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    Part (a)

    Essential requirements for any unattended machinery space (UMS) Ship to be able to sail at sea are enumerated in the SOLAS 1974 Chapter II-1, regulations 46 to regulation 53.

    Requirements for Unattended Machinery Space (UMS) Ship:

    1. Fire Precaution

    • Arrangements should be provided on the UMS ship to detect and give an alarm in case of fire.
    • In the boiler, air supply casing and uptake.
    • In scavenge space of propulsion machinery.
    • In engines of power, 2250 kW and above or cylinders having bore more than 300mm should be provided with an oil mist detector for the crankcase or bearing temperature monitor or either of two.

    2. Centralized control & instruments are required in Machinery Space

    • UMS ships must have a centralised control room that is easily accessible and equipped with adequate instrumentation and equipment to monitor and operate all main and auxiliary machinery. A system must be provided to call the engineers to the machinery space in case of emergency

    3. Protection against flooding:

    • UMS ships must have bilge wells that are located and designed to detect the accumulation of liquid at a normal angle of heel and trim and to accommodate the drainage of liquid during an unattended period. If the bilge pump starts automatically, an alarm must indicate that the flow of liquid pumped is more than the capacity of the pump.

    4. Automatic Fire Detection

    • Alarms and detection should operate very rapidly and effectively. It should be placed at numerous well-sited places for quick response of the detectors.

    5. Fire Extinguishing System

    • There should be arrangements for a fire extinguishing system other than the conventional hand extinguishers, which can be operated remotely from machinery space. The station must give control of emergency fire pumps, generators, valves, extinguishing media, etc.

    6. Alarm System

    • A comprehensive alarm system must be provided for control & accommodation areas.

    7. Automatic Start of Emergency Generator

    • Arrangements for the starting of an emergency generator and automatic connection to the bus bar must be provided in case of a blackout condition, apart from that, the following points are also to be noted.

    8. Local hand control of essential machinery like steering, emergency generator starting, emergency start for main engine, etc. 8. Adequate settling tank storage capacity. 9. Regular testing & maintenance of machinery alarms & instruments.

    Part (b)

    (i) Protection against Flooding:

    • Bilge wells in UMS ships should be located and provided in such a manner that the accumulation of liquid is detected at a normal angle of heel and trim and should also have enough space to accommodate the drainage of liquid during unattended periods.
    • In the case of the automatic starting of the bilge pump, the alarm should be provided to indicate that the flow of liquid pumped is more than the capacity of the pump.

    (ii) Control of Propulsion Machinery from Navigation Bridge:

    • The ship should be able to be controlled from the bridge under all sailing conditions. The bridge should be able to control the speed and direction of thrust and should be able to change the pitch in case of a controllable pitch propeller.
    • Emergency stops should be provided on navigating the bridge, independent of the bridge control system.
    • The remote operation of the propulsion should be possible from one location at a time; at such connection, interconnected control positions are permitted.
    • The number of consecutive automatic attempt which fails to start the propulsion machinery shall be limited to safeguard sufficient starting air pressure.
    Q6 (10 Marks) Ship Stability

    (a) Explain the concept of dynamical stability. (6)

    (b) A vessel of 8000 tonne displacement has 75 tonne of cargo on the deck. It is lifted by a derrick whose head is 10.5m above the centre of gravity of the cargo, and placed in the lower hold 9m below the deck and 14m forward of its original position. Calculate the shift in the vessel's centre of gravity from its original position when the cargo is: (10)

    (i) Just clear of the deck,

    (ii) At the derrick head,

    (iii) In its final position

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    Part (a)

    Dynamical Stability is defined as the amount of energy required to heel a ship from its upright equilibrium position to a specific angle of heel. It provides a measure of the vessel's stability by considering its behaviour in response to dynamic external forces, such as wind or waves.

    • The concept compares the heeling moment energy (from external forces) and the righting moment energy (from the ship's stability).
    • The ship will absorb the energy imparted by the heeling moment. If the righting energy is greater than the heeling energy, the ship will stabilize; otherwise, it may capsize.

    Areas Under the Curve:

    • Area A: Represents the region where the heeling moment exceeds the righting moment (external energy > ship's stability).
    • Area B: Represents the region where the righting moment exceeds the heeling moment (ship's stability > external energy).
    • The balance of these areas determines whether the ship will right itself or continue to heel.

    When exposed to heeling forces such as wind or waves, the vessel inclines and may roll over to a certain angle of heel. If the external force is applied instantaneously, the ship must have enough reserve dynamic stability to absorb the energy and return to an upright position. If the external force is constant, the ship will remain at an equilibrium angle where the righting moment equals the heeling moment.

    This refers to the remaining righting energy available to counteract additional external forces. A higher reserve dynamic stability ensures the vessel can handle greater heeling forces without capsizing.

    Q7 (10 Marks) Ship Resistance & Propulsion

    (a) Describe how water tightness is maintained where bulkheads are pierced by longitudinal beams or pipes. (6)

    (b) A ship of 15000 tonne displacement has an Admiralty Coefficient, based on shaft power, of 420. The mechanical efficiency of the machinery is 83%, shaft losses 6%, propeller efficiency 65% and QPC 0.71. At a particular speed the thrust power is 2550kW. Calculate: (10)

    (i) Indicated power

    (ii) Effective power

    (iii) Ship speed

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    Part (a)

    Water tightness of

    bulkheads that are pierced by longitudinal beams or pipes:

    • Bulkheads must be made structurally watertight without the use of wood packing.
    • Pipes passing through bulkheads are either welded or fastened to the bulkhead using studs or hoses secured through tapped holes in the plating.
    • When a bulkhead is pierced by a longitudinal beam, the openings are kept as small as possible, and doubler plates are welded on each side to maintain watertight integrity.
    • Sealing materials and gaskets are used, particularly for pipes, cables, and similar penetrations.
    Q8 (10 Marks) Ship Stability

    (a) What is Prismatic Co-efficient (CP). Derive the formula CP = Cb/Cm, where Cb = Co-efficient of fineness and Cm = midship section area co-efficient. (6)

    (b) A watertight door is 1.2m high and 0.75m wide, with a 0.6m sill. The bulkhead is flooded with seawater to a depth of 3m on one side and 1.5m on the other side. Draw the load diagram and from it determine the resultant load and position of the centre of pressure on the door. (10)

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    Part (a)

    Prismatic coefficient is the ratio of the volume of displacement to the product of length and area of the immersed portion of the midship section

    Cb is the block co-efficient or co-efficient of fitness is the ratio of the volume of displacement to the product of length, breadth and draught

    $$\because\space C_{b}\space=\space{{\nabla}\over L\times B\times D}\:---\:1$$

    $$C_{p}\space=\space{{\nabla}\over L\times A_{m}}\:---\:2$$

    $$C_m \space = \space {{A_m} \over B \times D}$$

    $$A_{m}=\:C_{m}\times B\times D\:---\:3$$

    Substitute 3 in 2

    $$C_{p}\space=\space{{\nabla}\over C_{m}\times B\times D\ \times L}\:---\:4$$

    $$\nabla=C_{b}\times L\times B\times D\:---5\:\left(from\:equation\:1\right)$$

    Substitute 5 in 4

    $$C_{p}=\frac{C_{b}\times L\times B\times D}{C_{m}\times L\times B\times D}$$

    $$C_p \space = \space {{C_b} \over C_m}$$

    Q9 (10 Marks) Ship Resistance & Propulsion

    (a) Describe briefly the inclining experiment and explain how the results are used. (6)

    (b) A ship of 14900 tonne displacement has a shaft power of 4460 Kw at 14.55 knots. The shaft power is reduced to 4120 kW and the fuel consumption at the same displacement is 541 kg/h. Calculate the fuel coefficient for the ship (10)

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    Part (a)

    An inclining experiment

    is conducted to determine a ship's metacentric height (GM) and, consequently, the location of its centre of gravity (CG). Knowing the CG of an empty vessel allows for calculations of its position under various loading conditions.

    The inclining experiment is performed on:

    • Newly built ships.
    • After major alterations to the vessel.
    • As required by the classification society.

    Conducting the inclining experiment:

    • The ship should be in a sheltered location (like a dry dock) with mooring ropes slack, only essential personnel on board, all tanks either empty or full, and any loose weights removed or secured.
    • At least two pendulums (one forward, one aft) are used, ideally as long as possible and suspended from convenient points (e.g., under a hatch). A hood filled with water or oil is placed beneath each pendulum bob to dampen its swing for accuracy.
    • Four masses are positioned on the deck, two on each side of the midships, their centres as far from the centerline as possible. These masses are moved systematically: all four to one side, then all four to the other, and finally two on each side.
    • The pendulum deflection is recorded for each mass movement.
    • The average of these deflections is used to calculate the metacentric height (GM).
    Q10 (10 Marks) Ship Stability

    A ship of 14000 tonne displacement is 145m long and floats at draughts of 7.9 m forward and 8.5 m aft. The TPC is 19, GML 120 m and LCF 3 m forward of midships. It is required to bring the vessel to an even keel draught of 8.5m. Calculate the mass which should be added and the distance of the centre of the mass from midships. (16)

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    Mass to be added to bring the vessel to an even keel of 8.5 m.

    Ship 14,000 t displacement, 145 m long, floats at draughts 7.9 m forward and 8.5 m aft. TPC = 19, GML = 120 m, LCF 3 m forward of midships. It is required to bring the vessel to an even keel draught of 8.5 m. Calculate the mass to be added and the distance of its centre from midships.

    Current mean draught = (7.9 + 8.5)/2 = 8.2 m. Required even-keel draught = 8.5 m, so the mean draught must increase by 0.3 m.

    MCT1cm = Delta x GML/(100 x L) = 14000 x 120/(100 x 145) = 1,680,000/14,500 = 115.9 t-m per cm.

    Mass to add for the mean sinkage of 0.3 m (30 cm): w = 30 x TPC = 30 x 19 = 570 t.

    The current trim is 0.6 m by the stern (8.5 - 7.9). To bring the vessel to an even keel, the trim must be removed, i.e. a change of trim of 0.6 m (60 cm) is required. The change of trim moment = 60 x MCT1cm = 60 x 115.9 = 6954 t-m.

    The added mass must be placed forward of the centre of flotation so that it both sinks the vessel and removes the stern trim. Distance of the mass from the LCF: x = change of trim moment/w = 6954/570 = 12.2 m forward of the LCF.

    Since the LCF is 3 m forward of midships, the mass is at 12.2 + 3 = 15.2 m forward of midships.

    Answer: add 570 t with its centre 15.2 m forward of midships; the final even-keel draught is 8.5 m.

    Q1 (10 Marks) Hull Construction πŸ”₯ Repeated 8x

    (a) Sketch the cross-section of a bulk carrier with either deep or shallow double bottom showing the type of framing used (6)

    (b) Describe the corrosion problems experienced within the ballast tanks (5)

    (c) State how such tanks are protected against extensive corrosion. (5)

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    Part (a)

    Mid-ship section of bulk carrier:

    Part (b)

    (i) Corrosion problems in ballast tanks

    are significant and arise due to various factors:

    • Galvanic corrosion occurs due to the interaction between dissimilar metals, accelerated by the seawater environment and differential aeration.
    • Sulphate-reducing bacteria in river mud can cause localized pitting, leading to penetration of the bottom shell.
    • Dissolved oxygen in seawater reacts with metal surfaces to form rust, significantly contributing to structural deterioration.
    • The rate of corrosion peaks at a 3.5% salt concentration, typical of seawater.
    • Neglecting maintenance exacerbates the problem, allowing corrosion to progress unchecked.

    (ii) Protection against extensive corrosion in ballast tanks involves the following measures:

    • Complete coating of the tank surfaces with properly selected marine-grade paint to create a protective barrier.
    • Installation of sacrificial anodes, designed to corrode preferentially and protect the tank structure.
    • Using large anodes with greater volume relative to surface area to ensure extended protection.
    • Regular gauging of plates to assess thickness and identify areas requiring reinforcement or repair.
    • Conducting regular inspections, cleaning the tanks, removing rust, and repainting to maintain structural integrity and prolong the lifespan of the ballast tanks.
    Q2 (10 Marks) Hull Construction

    (a) Describe a method of the attachment of bilge keels. (5)

    (b) State three reasons for not extending bilge keels to the entire length of the vessel. (3)

    (c) Evaluate the effectiveness of bilge keels for large wall-sided vessels. (4)

    (d) Explain TWO principles of roll damping those bilge keels exploit. (4)

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    Part (a)

    Method of Bilge keel attachment to the hull:

    Bilge keels are fitted port and starboard at the turn of the bilge. They do not extend outside the lines of the side and bottom shell. The bilge keels are attached directly in line with an internal stiffening member such as a girder or longitudinal. The bilge keel comprises a flat bar doubler welded directly to the shell, and an offset bulb plate (OBP) with 'scallops' cut in it is welded to the flat bar doubler. The ends of the bilge keels are tapered (minimum 3 in 1) and will end in line with transverse internal stiffening, such as a frame. Using a hull doubler protects the hull in case of damage to the keel, as the crack would not extend into the hull.

    Part (b)

    Bilge keels are not fitted for the full length of the vessel because:

    • The 'lever' to the ship's axis of rotation is reduced at the ends;
    • The ineffectiveness, the closer the bilge keel is to the rolling centre, increased resistance and more likelihood of damage.
    • The hydrodynamic effect would cause a large increase in resistance and fuel consumption;
    • At the aft end, the boundary layer is much thicker, and since the keel would not project through, it would have a much reduced effect.
    Part (d)

    Two principles of roll damping exploited by Bilge keels:

    Increased Roll Period:

    • Bilge keels increase the ship's roll period (the time it takes for the vessel to complete one roll). This is achieved by increasing the moment of inertia (K) of the vessel. A longer roll period means the vessel rolls more slowly, thus reducing the amplitude of the roll.

    $$T_{r}=2\pi\:\frac{k}{\sqrt{g\times GM}}$$

    Where:

    • g: acceleration due to gravity
    • GM: metacentric height
    • k: mass moment of inertia

    By increasing k, the ship's stability during roll improves.

    Hydrodynamic Resistance:

    • As the vessel rolls, the bilge keels move through the water, creating pressure differences. Water pressure on one side of the keel opposes the rolling motion, providing a damping effect. This is due to the interaction between the bilge keel and the water, creating hydrodynamic forces that counteract the rolling motion.
    Q3 (10 Marks) Ship Stability πŸ”₯ Repeated 2x

    What is the significance of the area under the curve of statical stability or the GZ curve? Explain using a neat diagram, how this curve is used to assess the stability of the ship against a heeling arm (16)

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    Significance of area under GZ curve:

    The area under the statical stability or GZ curve up to any given angle ΞΈ, multiplied by the gravitational weight of the ship (displacement), represents the Dynamic Stability. This is the work done in heeling the ship to the angle ΞΈ, and it reflects the ship's ability to absorb the energy imparted by external forces such as waves and wind.

    The stability of a ship is directly related to the nature and value of its metacentric height (GM).

    • Since GM is closely associated with the Righting Lever (GZ) and the angle of heel, the statical stability curve is plotted between GZ and the angle of heel.

    Following information obtained from the curve can be used to assess the stability of the ship:

    Angle of Equilibrium

    • If the GZ curve intersects the horizontal axis (i.e., no righting lever) at the origin, the ship has inherent positive initial stability.

    Maximum GZ

    • This represents the ship's largest static heeling moment. The Maximum Righting Lever (GZ), when multiplied by the ship's displacement, gives the maximum heeling moment the ship can sustain without capsizing.

    Angle of deck Immersion

    • The angle at which the ship’s deck becomes submerged. This is often indicated as the point of inflection on the GZ curve, where the ship’s vulnerability to downflooding increases.

    Angle of Vanishing Stability

    • The angle at which the GZ curve intersects the horizontal axis, indicating the point where the righting lever becomes zero. Any heel beyond this angle results in negative stability, increasing the risk of capsizing.

    Range of Stability

    • The range is measured from the origin to the angle of vanishing stability, indicating the range of angles over which the ship has positive stability.

    Angle of Contraflexure

    • The angle of heel up to which the rate of increase of GZ with heel is rising. Beyond this point, while GZ may continue to increase, the rate of increase begins to diminish.

    Angle of Downflooding

    • The minimum angle of heel at which an external opening (without a watertight appliance) is submerged, allowing seawater ingress.
    • (Downflooding refers to seawater entering the hull or superstructure due to heel or immersion of the vessel.)

    Initial Metacentric Height (GM)

    • At 57.3 (1 radian), an ordinate is drawn at the origin of the GZ curve. A tangent to the curve is extended to intersect this ordinate, providing the value of the Initial Metacentric Height (GM).
    Q4 (10 Marks) Hull Construction

    (a) Explain the purposes of the collision bulkhead. Describe with the aid of sketches the construction of a collision bulkhead paying particular attention to the strength and attachment to the adjacent structure. (8)

    (b) Define "Pounding". Describe with the aid of sketches the arrangement provided to resist pounding. (8)

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    Part (a)

    Purposes of the collision bulkhead.

    The collision (forepeak) bulkhead is the first transverse watertight bulkhead, fitted near the bow. Its purposes are:

    • To limit the flooding in the event of a bow collision, so that the water is confined to the forepeak and the ship retains sufficient buoyancy and stability to survive.
    • To provide a watertight boundary at the forward end of the cargo/machinery spaces, protecting them from flooding.
    • To give the bow structural strength and to resist the impact loads of a collision and the pounding/panting loads.
    • To provide a strong transverse member that stiffens the fore end and supports the stem and the forepeak structure.
    Part (b)

    Construction of a collision bulkhead.

    The collision bulkhead is a strong transverse watertight bulkhead, built of plating with vertical stiffeners (or horizontal stringers) and a strong boundary connection. Sketch: a vertical transverse bulkhead at the forepeak, with the plating connected to the shell, the deck and the keel, and stiffened by vertical frames (or horizontal stringers) and a top and bottom stool. The bulkhead is made watertight and is connected to the stem and the side shell with continuous welds; the stiffeners are connected to the shell and deck to transmit the loads. The bulkhead is often corrugated or flat with vertical stiffeners, and is heavily constructed because it must resist the impact of a collision and the hydrostatic pressure of flooding. The attachment to the adjacent structure (shell, deck, keel) is made with continuous fillet or butt welds and the stiffeners are tied into the surrounding structure to give continuity of strength.

    Part (c)

    Regulations governing the position and construction.

    SOLAS requires a collision bulkhead to be fitted at the forward end, positioned at a distance from the forward perpendicular of not less than 5% and not more than 8% of the length (or 10 m, whichever is less) for ships, so that it is effective in limiting flooding. It must be watertight up to the freeboard deck, and no doors, manholes, ventilation ducts or other openings are permitted in it below the freeboard deck (except for a single watertight door in certain cases). The bulkhead must be of adequate strength and is subject to classification survey. The position and construction are governed by the SOLAS Convention and the classification society rules.

    Part (d)

    Pounding and the arrangement to resist it.

    Pounding (slamming) is the repeated impact of the bottom of the bow on the water as the ship pitches in a seaway, when the bow emerges and then slams down on the water. It produces severe local impact loads on the bottom and forefoot plating. To resist pounding, the fore end bottom plating is made thicker, the frames are more closely spaced, and additional stiffening (intercostal floors, stringers, and a strong forepeak structure) is provided in the region of the bow; the bottom is strengthened over the forward 10-15% of the length, and the stem and forefoot are heavily constructed. The arrangement (sketch) shows the thickened bottom shell, closely spaced frames and the strong forepeak framing that resist the pounding loads.

    Q5 (10 Marks) Hull Construction

    (a) State FOUR sources of excitation that may induce vibration into the main hull girder. (8)

    (b) Suggest methods for reducing the vibration levels induced by EACH of the exciting forces in (a). (8)

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    Part (a)

    FOUR sources of excitation that may induce vibration into the main hull girder.

    1. Propeller excitation: the propeller produces fluctuating forces (blade frequency and multiples) due to the varying pressure field and the wake, transmitted through the shaft and the water to the hull, causing vibration of the stern and the whole hull girder.
    2. Main engine excitation: the reciprocating and rotating parts of the engine produce unbalanced forces and moments (primary and secondary, and their harmonics) at the engine's firing frequency and multiples, transmitted through the engine seating to the hull.
    3. Wave excitation: the periodic wave forces on the hull (at the encounter frequency) excite the hull girder, particularly in heavy seas, causing springing and whipping.
    4. Auxiliary machinery and other rotating equipment: pumps, generators, compressors and other machinery produce unbalanced forces that are transmitted to the hull through their mountings.
    Part (b)

    Methods for reducing the vibration levels induced by each.

    1. Propeller: increase the clearance between the propeller and the hull (aperture), use a larger blade area and a suitable blade number to avoid resonance, fit a skew or a different blade design, and use a propeller boss cap fin; also avoid operating at resonant speeds.
    2. Main engine: balance the engine (fit balance weights, use a suitable firing order), fit a flexible coupling and a resilient engine mounting, tune the engine speed to avoid the hull's natural frequencies, and use a two-mass or tuned vibration damper on the crankshaft.
    3. Wave excitation: reduce speed and change course in heavy seas to avoid the resonant encounter frequency, and design the hull to have adequate stiffness and damping; use a hull stress/vibration monitoring system.
    4. Auxiliary machinery: fit resilient (anti-vibration) mountings, balance the rotating parts, isolate the machinery from the hull, and avoid operating at resonant speeds.
    Q6 (10 Marks) Ship Stability

    (a) With reference to dynamical stability, describe the effect of an increase in wind pressure when a vessel is at its maximum angle of roll to windward. (6)

    (b) An oil tanker 160 m long and 22 m beam floats at a draught of 9m in sea water. Cw is 0.865. The midship section is in the form of a rectangle with 1.2 m radius at the bilges. A midship tank 10.5 m long has twin longitudinal bulkheads and contains oil of 1.4 m3/t to a depth of 11.5 m. The tank is holed to the sea for the whole of its transverse section. Find the new draught. (10)

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    Part (a)

    Dynamical Stability is defined as the amount of energy required to heel a ship from its upright equilibrium position to a specific angle of heel. It provides a measure of the vessel's stability by considering its behaviour in response to dynamic external forces, such as wind or waves.

    • The concept compares the heeling moment energy (from external forces) and the righting moment energy (from the ship's stability).
    • The ship will absorb the energy imparted by the heeling moment. If the righting energy is greater than the heeling energy, the ship will stabilize; otherwise, it may capsize.

    Areas Under the Curve:

    • Area A: Represents the region where the heeling moment exceeds the righting moment (external energy > ship's stability).
    • Area B: Represents the region where the righting moment exceeds the heeling moment (ship's stability > external energy).
    • The balance of these areas determines whether the ship will right itself or continue to heel.

    When exposed to heeling forces such as wind or waves, the vessel inclines and may roll over to a certain angle of heel. If the external force is applied instantaneously, the ship must have enough reserve dynamic stability to absorb the energy and return to an upright position. If the external force is constant, the ship will remain at an equilibrium angle where the righting moment equals the heeling moment.

    This refers to the remaining righting energy available to counteract additional external forces. A higher reserve dynamic stability ensures the vessel can handle greater heeling forces without capsizing.

    Q7 (10 Marks) Ship Resistance & Propulsion

    (a) What factors influence the frictional resistance of a ship and what formula is used to calculate the resistance? (6)

    (b) A ship 120 m long displaces 10500 tonne and has a wetted surface area of 3000 m2. At 15 knots the shaft power is 4100 kW, propulsive coefficient 0.6 and 55% of the thrust is available to overcome frictional resistance. Calculate the shaft power required for a similar ship 140m long at the corresponding speed. f = 0.42, and n = 1.825. (10)

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    The frictional resistance of a ship depends on:

    • Speed of ship
    • Wetted surface area
    • Length of the ship
    • Roughness of hull
    • Density of water

    According to Froud's Formula

    $$R_{f}=fsv^{n}$$

    where,

    • f = Co-efficient which depends on
      • Length of ship
      • Roughness of hull
      • Density of water
    • S = Wetted surface area in m2
    • V = Speed of ship in knots
    • n = index (approx. 1.825)

    The value of f for mild steel is:

    $$f=0.417+\frac{0.773}{L+2.862}$$

    Q8 (10 Marks) Ship Stability

    (a) Explain the purpose of non-watertight longitudinal subdivision of tanks. (6)

    (b) A ship 160m long and 8700 tonne displacement floats at a waterline with half ordinates of 0, 2.4, 5.0, 7.3, 7.9, 8.0, 8.0, 7.7, 5.5, 2.8 and 0 m respectively. While floating at this waterline, the ship develops a list of 10Β° due to instability. Calculate the negative metacentric height when the vessel is upright in this condition. (10)

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    Part (a)

    Purpose of non-watertight longitudinal subdivision of tanks.

    Non-watertight longitudinal subdivision (longitudinal bulkheads or wash bulkheads that are not watertight) of tanks serves to:

    • Reduce the free-surface effect: by dividing a wide tank into narrower compartments, the second moment of area of the free surface (i = L B^3/12) is greatly reduced, so the free-surface loss of GM is reduced.
    • Reduce the sloshing of the liquid: the wash bulkheads damp the movement of the liquid in a partially full tank, reducing the dynamic loads on the tank structure and the effect on stability.
    • Provide structural support: the longitudinal bulkheads act as girders that stiffen the tank and the hull, and support the deck and bottom.
    • Reduce the free-surface effect during ballasting and cargo operations.

    The subdivision is non-watertight so that the liquid can flow between the compartments (for filling/emptying and for equalising), while still reducing the free-surface and sloshing effects.

    Part (b)

    Negative metacentric height from the angle of loll.

    Ship 160 m long, 8,700 t displacement, floats at a waterline with half-ordinates 0, 2.4, 5.0, 7.3, 7.9, 8.0, 8.0, 7.7, 5.5, 2.8 and 0 m (at 16 m intervals, 10 intervals = 160 m). The ship develops a list of 10 deg due to instability. Calculate the negative metacentric height when the vessel is upright.

    Waterplane area: A = 2 x (h/3)[y0 + y10 + 4(y1+y3+y5+y7+y9) + 2(y2+y4+y6+y8)]

    = 2 x (16/3)[0 + 0 + 4(2.4+7.3+8.0+7.7+2.8) + 2(5.0+7.9+8.0+5.5)]

    = 2 x 5.333[4 x 28.2 + 2 x 26.4] = 10.667[112.8 + 52.8] = 10.667 x 165.6 = 1766.4 m2.

    Second moment of area about the longitudinal axis (transverse stability): I = 2 x (h/3) x sum of (weighted y^3).

    y^3 values: 0, 13.824, 125, 389.017, 493.039, 512, 512, 456.533, 166.375, 21.952, 0.

    weighted sum = 1x0 + 4x13.824 + 2x125 + 4x389.017 + 2x493.039 + 4x512 + 2x512 + 4x456.533 + 2x166.375 + 4x21.952 + 1x0

    = 0 + 55.3 + 250 + 1556.1 + 986.1 + 2048 + 1024 + 1826.1 + 332.8 + 87.8 + 0 = 8166.2.

    I = 2 x (16/3) x 8166.2 = 10.667 x 8166.2 = 87,105 m4.

    Volume of displacement V = 8700/1.025 = 8487.8 m3.

    BM = I/V = 87,105/8487.8 = 10.26 m.

    At the angle of loll (10 deg) the righting lever is zero. For a wall-sided ship, tan(theta) = sqrt(-2 GM/BM):

    tan(10 deg) = 0.1763, tan^2 = 0.0311.

    -2 GM/BM = 0.0311, so GM = -0.0311 x 10.26/2 = -0.1596 m.

    Answer: the negative metacentric height is about -0.16 m.

    Q9 (10 Marks) Ship Stability

    (a) What is meant by the Admiralty Coefficient and the Fuel Coefficient? (6)

    (b) A rectangular watertight bulkhead 9 m high and 14.5 m wide has sea water on both sides, the height of water on one side being four times that on the other side. The resultant centre of pressure is 7m from the top of the bulkhead. Calculate: (10)

    (i) the depths of water

    (ii) the resultant load on the bulkhead

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    The Admiralty Coefficient (C) is a method for estimating the propulsion power needed for a newly built ship. It's considered relatively constant for a given ship design. The formula is:

    $$C=\frac{\Delta^{2/3}\times V^3}{BP}$$

    Where:

    • C = Admiralty Coefficient
    • Ξ” = Displacement in tonnes (weight of the ship when fully loaded)
    • V = Speed in knots
    • BP = Brake power in kilowatts (power delivered by the ship's engine)

    A higher Admiralty Coefficient indicates a more efficient ship design, meaning it requires less power to achieve a given speed. Values typically range from 350 to 600.

    Fuel Coefficient:

    The Fuel Coefficient (F.C.) is used to calculate a ship's daily fuel oil consumption. The formula is:

    $$Daily\:fuel\:oil\:consumption\:=\:\frac{\Delta^{2/3}\times V^3}{FC}$$

    Where:

    • F.C. = Fuel Coefficient
    • Ξ” = Displacement in tonnes
    • V = Speed in knots

    The Fuel Coefficient can vary significantly, with typical values ranging from 40,000 to 120,000. A higher Fuel Coefficient implies greater fuel efficiency (lower daily fuel consumption) for a given speed and displacement.

    Q10 (10 Marks) Ship Stability

    (a) Explain the term volumetric heeling moments. (6)

    (b) A ship 85 m long displaces 8100 tonne when floating in sea water at draughts of 5.25 m forward and 5.55 m aft. TPC 9.0, GMl 96 m, LCF 2 m aft of midships. It is decided to introduce water ballast to completely submerge the propeller and a draught aft of 5.85 m is required. A ballast tank 33 m aft of midships is available. Find the least amount of water required and the final draught forward. (10)

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    $$Let\:m=mass\:of\:ballast\:required$$

    $$MCT1_{\operatorname{\mathrm{cm}}}=\frac{\Delta\times GM_{L}}{100L}$$

    $$=\frac{8100\times96}{100\times85}$$

    $$=91.48ton.m$$

    $$Trimming\:moment=m\times d$$

    $$=m\left(33-2\right)$$

    $$=31m$$

    $$Change\:of\:trim=\frac{T\times M}{MCT_{1\operatorname{\mathrm{cm}}}}$$

    $$=\frac{31\times m}{91.48}cm\:by\:stern$$

    $$Bodily\:Sinkage=\frac{m}{TPC}$$

    $$=\frac{m}{9.0}$$

    $$Draft\:Aft\:d_{A^{}1}=d_{A}+\frac{m}{TPC}+\frac{t}{L}\left(\frac{L}{2}-x\right)$$$$5.85=5.55+\frac{m}{9\times100}+\frac{31m}{91.48\times100\times85}\left(\frac{85}{2}-2\right)$$

    $$5.85-5.55=2.7258\times10^{-3}m$$

    $$m=110\:tonnes$$

    $$Draft\:Fwd\:d_{F1^{}}=d_{F}+\frac{m}{TPC}-\frac{t}{L}\left(\frac{L}{2}+x\right)$$

    $$=5.25+\frac{110}{9\times100}-\frac{0.372}{85}\left(\frac{85}{2}+2\right)$$

    $$=5.25+0.1222-0.1951$$

    $$d_{F1^{}}=5.177m$$

    Q1 (10 Marks) Ship Types & Design πŸ”₯ Repeated 3x

    (a) State the reasons for the freeboard requirement. (4)

    (b) Explain the term condition of assignment and explain how these are maintained for a ship. (4)

    (c) Using a diagram indicate the freeboard of type A, type B, type B60 and type B100 vessels giving an example of each type. (8)

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    Part (a)

    Reasons for Freeboard Requirements:

    Freeboard is the distance measured from the waterline to the upper edge of the deck plating at the side of the freeboard deck amidships.

    • Ensures the ship is seaworthy when fully loaded.
    • Provides reserve buoyancy, enabling the ship to rise as it passes through waves, keeping decks largely dry.
    • Enhances the ship's stability and increases its survivability in case of damage by allowing it to remain afloat longer, aiding crew escape or damage control.
    Part (b)

    Conditions of Assignment:

    Conditions of Assignment are outlined in the Merchant Shipping Load Line Rules 1968 and must be satisfied before a ship is assigned freeboards and issued a load line certificate. These conditions address the practical need for openings (hatchways, doorways, vents, etc.) in the ship's hull and mandate appropriate protection and closure mechanisms for these openings.

    Requirements Before Assigning Load Line:

    • The ship must have sufficient structural strength.
    • Adequate reserve buoyancy must be maintained.
    • Openings must be secured against water ingress.
    • Safety measures for the crew, such as guardrails and gangways, must be in place.

    To ensure that the conditions of assignment are still current the following items can be checked and confirm to be without change or damage from when the ship was built.

    • Access openings in bulkheads, to ensure that they can be sealed and prevent flooding
    • Cargo and hatchways ensure they can be sealed to prevent flooding
    • Coamings of hatchways, sign of corrosion damage risk of failure would allow flooding
    • Protection of openings, can they all be sealed
    • Ventilator coamings not corroded as they would allow flooding to other arears
    • Air pipes can be shut in heavy weather and not corroded
    • Discharges, inlets and scuppers, all in good condition and operational
    • Side scuttles can be secured
    • Hull inspection, sea boxes and penetrations all checked for damage and corrosion free.
    Part (c)

    Freeboard for Different Vessel Types:

    Type A Ships:

    • Designed for liquid cargo only (e.g., oil tankers).
    • Have the lowest freeboard among ship types due to their higher reserve buoyancy.

    Type B Ships:

    • Designed for general cargo (e.g., container ships).
    • Have a higher freeboard than Type A ships.

    Type B-60 Ships:

    • A variant of Type B ships, with freeboard reduced by up to 60% of the difference between Type A and B (e.g., OBO ships).

    Type B-100 Ships:

    • Type B ships with freeboard reduced by up to 100% of the difference between Type A and B (e.g., certain bulk carriers).
    Q2 (10 Marks) Machinery Space & Systems πŸ”₯ Repeated 5x

    (a) Describe the double bottom and framing arrangement used in the machinery space to cope up with the concentrated loads and vibration, together with shaft and thrust block support. (10)

    (b) Give reasons for the choice of thrust block position. (6)

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    Part (a)

    The construction of the double bottom in the machinery space regardless of the framing system has solid plate floors at every frame space under the main engine. Additional side girders are fitted outboard of the main engine seating, as required. The double-bottom height is usually increased to provide fuel oil, lubricating oil and fresh water tanks of suitable capacities. Shaft alignment also requires an increase in the double-bottom height or a raised seating, the former method usually being adopted.

    Continuity of strength is ensured and maintained by gradually sloping the tank top height and internal structure to the required position. Additional support and stiffening is necessary for the main engines, boilers, etc., to provide a vibration-resistant solid platform capable of supporting the concentrated loads. On slow- speed diesel-engined ships, the tank top plating is increased to 40 mm thickness or thereabouts in way of the engine bedplate. This is achieved by using a special insert plate which is the length of the engine including the thrust block in size. Additional heavy girders are also fitted under this plate and in other positions under heavy machinery as required. Plating and girder material in the machinery spaces is of increased scantlings in the order of 10 per cent.

    The method adopted.

    A cellular void space within a ships structure is called a coffer dam. Like a bulkhead separates two spaces or divides a space into two, a coffer dam does the same with a larger degree of integrity since it incorporates a void which would contain any breach of either of the boundaries. This would contain the leakage and prevent it spreading into other areas. A cofferdam can be defined as an empty space separating compartments to prevent the contents of one compartment from entering another in case of leakage.

    Part (b)

    The thrust block is positioned close to the propulsion machinery (usually just aft of the main engine). Reasons are as follows:

    • The axial thrust generated by the propeller could cause deformation and misalignment of the shafting system if the thrust block were positioned far from the engine. Placing it close to the engine minimizes the length of shafting subject to this thrust, thus reducing potential for misalignment. The strong double bottom structure directly beneath provides necessary support to mitigate any deformation.
    • Differential expansion between the shaft and the hull due to temperature variations is a potential source of misalignment. Positioning the thrust block close to the engine helps to minimize the effect of this differential expansion.
    • The weight of the propeller and the dynamic forces it creates can lead to whirling of the tailshaft and misalignment if not properly managed. Positioning the thrust block near the engine helps to stabilize the shafting system and reduce the risk of these issues.
    • The substantial double-bottom structure under the main machinery provides an ideal, inherently strong foundation for the thrust block. This minimises the need for extensive additional reinforcement to support the thrust loads.
    Q3 (10 Marks) Hull Construction πŸ”₯ Repeated 5x

    (a) With reference to fatigue of engineering components explain the influence of stress level and cyclical frequency on expected operating life. (4)

    (b) Explain the influence of material defects on the safe operating life of an engineering component (6)

    (c) State the factors which influence the possibility of fatigue cracking of a bed-plate transverse girder and explain how the risk of such cracking can be minimized. (6)

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    Part (a)

    Influence of Stress Level and Cyclic Frequency on Operating Life:

    Fatigue is progressive and localised structural damage caused by cyclic loading, where the maximum stress is below the ultimate tensile strength. The relationship between stress level, cyclic frequency, and operating life depends on whether the fatigue is high-cycle/low-stress or low-cycle/high-stress.

    High-cycle fatigue (low stress-high cycle):

    • This occurs at lower stress levels over a high number of cycles, resulting in elastic deformation. The component can withstand more cycles at these lower stress levels, and its life expectancy is determined by the S-N curve, which predicts the number of cycles before failure at a given stress level. For example, fatigue in turbocharger blowers often results from prolonged vibration over numerous cycles.

    Low-cycle fatigue (high stress-low cycle):

    • This occurs at high-stress levels over fewer cycles, causing plastic deformation in the material. This type of fatigue is typically assessed by a strain curve. If the stress level increases, the component's operating life decreases, as higher stress accelerates the onset of failure. For example, air receivers filling automatically face high stress and experience fewer cycles before failure.

    If stress levels or the number of cycles increase beyond the material’s capacity, failure will occur sooner. It is important to keep stress levels within allowable limits for extended component life.

    Part (b)

    Material defects can significantly reduce the safe operating life of engineering components because defects serve as stress concentrators that increase local stress around the defect. This leads to premature failure as the material cannot withstand the same level of cyclic stress as a defect-free component.

    • Surface roughness, porosity, inclusions, and abrupt section changes all create stress concentrations, lowering fatigue strength.
    • Coarse grain size, specific chemical compositions, and cold working introduce residual stresses that reduce fatigue resistance.
    • Corrosion, erosion, and decarbonisation weaken the material and accelerate fatigue crack initiation and propagation.
    • Faulty workmanship during assembly or processing introduces defects that may significantly shorten the component's life.
    Part (c)

    Factors Influencing Fatigue Cracking in Bedplate Transverse Girders:

    • Cylinder overload due to excess power puts excessive stress on the girders.
    • Incorrect crankshaft alignment induces uneven loading and stress concentrations.
    • Material defects, high residual stresses in welds, heat-affected zone hardening, and the presence of dissolved oxygen all reduce fatigue resistance.
    • Tank top deformation from pressurisation or overheating adds stress to the bedplate.

    To minimise the risk of fatigue cracking:

    (i) Constructional strength:

    • Bed plates are made up of M.S. plates with four steel casting, which are assembled and welded together so that the bed plate is strong longitudinally & transversely with good resistance to twisting along its length.
    • Longitudinal strength is obtained by fabricating each side of the bed plate in the form of a box girder.
    • The cast steel cross girder in which the main bearing is placed contributes to the bed plate's transverse strength and resistance against twisting along its length.
    • Resin cast chocks are used between the bedplate and the double bottom tank top to absorb the shocks & stress.

    (ii) Maintenance:

    • Monthly checks on the bolt tension.
    • Monthly checks on engine load using power cards & measuring cylinder peak pressure.
    • Regular checking of tension for main bearing jack bolts as recommended by engine manufacturers.
    • Regular checks on crankshaft alignment by taking deflection & compare with recommended value.
    • By maintaining engine operations at specified load, temperature, pressure, speed, etc.
    Q4 (10 Marks) Machinery Space & Systems πŸ”₯ Repeated 3x

    With reference to a periodically unattended machinery space of a dry cargo vessel discusses the requirements for

    (a) Protection against flooding (8)

    (b) Control of propulsion machinery from the navigating bridge. (8)

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    Part (a)

    Essential requirements for any unattended machinery space (UMS) Ship to be able to sail at sea are enumerated in the SOLAS 1974 Chapter II-1, regulations 46 to regulation 53.

    Requirements for Unattended Machinery Space (UMS) Ship:

    1. Fire Precaution

    • Arrangements should be provided on the UMS ship to detect and give an alarm in case of fire.
    • In the boiler, air supply casing and uptake.
    • In scavenge space of propulsion machinery.
    • In engines of power, 2250 kW and above or cylinders having bore more than 300mm should be provided with an oil mist detector for the crankcase or bearing temperature monitor or either of two.

    2. Centralized control & instruments are required in Machinery Space

    • UMS ships must have a centralised control room that is easily accessible and equipped with adequate instrumentation and equipment to monitor and operate all main and auxiliary machinery. A system must be provided to call the engineers to the machinery space in case of emergency

    3. Protection against flooding:

    • UMS ships must have bilge wells that are located and designed to detect the accumulation of liquid at a normal angle of heel and trim and to accommodate the drainage of liquid during an unattended period. If the bilge pump starts automatically, an alarm must indicate that the flow of liquid pumped is more than the capacity of the pump.

    4. Automatic Fire Detection

    • Alarms and detection should operate very rapidly and effectively. It should be placed at numerous well-sited places for quick response of the detectors.

    5. Fire Extinguishing System

    • There should be arrangements for a fire extinguishing system other than the conventional hand extinguishers, which can be operated remotely from machinery space. The station must give control of emergency fire pumps, generators, valves, extinguishing media, etc.

    6. Alarm System

    • A comprehensive alarm system must be provided for control & accommodation areas.

    7. Automatic Start of Emergency Generator

    • Arrangements for the starting of an emergency generator and automatic connection to the bus bar must be provided in case of a blackout condition, apart from that, the following points are also to be noted.

    8. Local hand control of essential machinery like steering, emergency generator starting, emergency start for main engine, etc. 8. Adequate settling tank storage capacity. 9. Regular testing & maintenance of machinery alarms & instruments.

    Part (b)

    (i) Protection against Flooding:

    • Bilge wells in UMS ships should be located and provided in such a manner that the accumulation of liquid is detected at a normal angle of heel and trim and should also have enough space to accommodate the drainage of liquid during unattended periods.
    • In the case of the automatic starting of the bilge pump, the alarm should be provided to indicate that the flow of liquid pumped is more than the capacity of the pump.

    (ii) Control of Propulsion Machinery from Navigation Bridge:

    • The ship should be able to be controlled from the bridge under all sailing conditions. The bridge should be able to control the speed and direction of thrust and should be able to change the pitch in case of a controllable pitch propeller.
    • Emergency stops should be provided on navigating the bridge, independent of the bridge control system.
    • The remote operation of the propulsion should be possible from one location at a time; at such connection, interconnected control positions are permitted.
    • The number of consecutive automatic attempt which fails to start the propulsion machinery shall be limited to safeguard sufficient starting air pressure.
    Q5 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 2x

    (a) Describe, where on the hull plating would the following tests be carried out on the ships hulls during drydock.

    (i) Hammer

    (ii) Hose (8)

    (b) Briefly identify which parts of the external plating of ships hulls requires the closest attention. (8)

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    Part (a)

    Hull Plating Test Locations During Drydock:

    (i) Hammer Test: The hammer test, a method of non-destructive testing, would be carried out across all areas of the hull plating. The surveyor would gently tap the plating to assess its condition, listening for sounds indicating potential damage like cracking or delamination. Particular attention would be given to areas showing signs of corrosion or previous repairs. Areas such as welds, bilge keels, and around penetrations (e.g., sea chests) would require extra scrutiny.

    (ii) Hose Test: The hose test checks the watertight integrity of the hull. This would be performed on new welds on the shell plating and other areas where watertightness is required. This is done using a nozzle with a 12.5 mm diameter and 2 bar water pressure at a distance of 1.5 m. Specific locations would include welds, sea chests, and around valve penetrations. The exact location will depend on areas of concern identified during the visual inspection.

    Part (b)

    Parts of the External Plating Requiring the Closest Attention:

    • Welds: All weld joints, especially those in areas subject to stress or corrosion, must be carefully examined for cracks, porosity, and other defects.
    • Angles and other structural members: These need inspection for corrosion, cracking and distortion.
    • Bow: The bulbous bow, chain markings, and bow thruster area should be checked thoroughly.
    • Stern: The stern frame is subjected to slamming and must be examined closely for cracking and buckling.
    • Bilge Keel: This requires inspection for damage and proper attachment to the hull.
    • Openings in shell plating: Sea chests and ship side valves (overboard valves) must be inspected for leaks, corrosion, and proper operation.
    • Areas of previous damage or repair: These are especially susceptible to further corrosion or cracking and warrant close examination.
    • Areas exposed during previous dry-docking: Any areas that were exposed to the elements during the previous dry-dock require close examination for damage or corrosion.
    • Cathodic protection system: The effectiveness of the cathodic protection system and the condition of anodes need to be verified. Lack of anode consumption may indicate corrosion in other areas.
    • Rudder and Propeller: The condition of the rudder and propeller, including the welds, should be examined for any signs of damage or corrosion.
    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 2x

    (a) Explain what is meant by floodable length. (6)

    (b) (i) Construct a graph from the following information:

    Mean draft (m)

    3.0

    3.5

    4.0

    4.5

    TPC (tonnes)

    8.0

    8.5

    9.2

    10.0

    (ii) From this graph find the TPC's at draft of 3.2m, 3.7m, and 4.3m.

    (iii) If the ship is floating at a mean draft of 4m, and then loads 50 tonnes of cargo, 10 tonnes of fresh water, and 25 tonnes of bunkers, whilst 45 tonnes of ballast are discharged, find the final mean draft. (10)

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    Part (a)

    Floodable length

    is the amount, the ship can be flooded without the risk of sinking. i.e. the maximum length of the ship that can be flooded without submerging the margin line. The margin line is located 75mm below the bulkhead deck, providing a safety margin.

    • A floodable length curve can be drawn for different permeability.
    • It is used to test the after-bulkhead length so that the bulkhead length is sufficient to contain the flooding without sinking the vessel.
    • Factor of subdivision = Permissible length / Floodable length
    Q7 (10 Marks) Hull Construction πŸ”₯ Repeated 2x

    (a) Describe how bulkheads are tested. (6)

    (b) A double bottom tank containing seawater is 6m long, 12m wide and 1m deep. The inlet pipe from the pump has its center 75mm above the outer bottom. The pump has a pressure of 70 kN/m2 and is left running indefinitely. calculate the load on the tank top: (10)

    (i) If there is no outlet.

    (ii) If the overflow pipe extends 5m above the tank top.

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    Part (a)

    Testing for Water Tightness:

    • Hose Test: This involves applying a high-pressure water jet (2 bar from a distance of 1.5 meters, nozzle diameter at least 12mm) to the bulkhead's surface. This test directly checks for leaks. The test is typically done from the side where the stiffeners are attached.
    • Air Pressure Test: An air pressure test applies pressure (0.2 bar) for about an hour, detecting leaks that may not be apparent during the hose test. This usually happens prior to the application of protective coatings.
    • Structural Test: Visual inspection, especially of welding joints, is carried out. Non-Destructive Testing (NDT) is done where necessary. Tanks designed to hold liquids, which form subdivisions of the ship, are tested for tightness with a water head up to the deepest subdivision load line or to a head of 2/3 the depth from the top of the tank to the margin line, whichever is greater.
    Q8 (10 Marks) Ship Types & Design πŸ”₯ Repeated 2x

    (a) Define coefficient of fineness of waterplane area, block coefficient and midship. (6)

    (b) A box shaped vessel has length, 100m and breadth 12m and floats at a range of drafts from 1m to 10m. Produce curves of KB, BM and KM. (10)

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    Coefficient of Fineness of Waterplane Area (Cw)

    This is the ratio of the ship's waterplane area to the area of a rectangle with the same length and breadth of the ship at the same waterline. It is given by:

    $$C_{w}=\frac{WPA}{L\times B}$$

    Where:

    • Cw​ = Coefficient of fineness of waterplane area
    • WPA = Waterplane area
    • L = Length of the ship
    • B = Breadth of the waterplane

    Block Coefficient (Cb)

    This is the ratio of the volume of displacement of a ship to that of a rectangular block having the same length, breadth, and draft as the ship. It is given by:

    $$C_{b}=\frac{V}{L\times B\times D}$$

    Where:

    • Cb = Block coefficient
    • V = Volume displaced by the ship
    • L = Length of the waterline
    • B = Breadth of the waterline
    • D = Draft

    Midship Coefficient (Cm)

    This is the ratio of the cross-sectional area at the midship section to the product of the ship's beam (breadth) and draft. It is given by:

    $$C_{m}=\frac{A_{m}}{B\times D}$$

    Where:

    • Cm​ = Midship coefficient
    • Am​ = Cross-sectional area at midship
    • B = Beam (breadth of the ship)
    • D = Draft
    Q9 (10 Marks) Ship Stability πŸ”₯ Repeated 3x

    (a) Explain the concept of dynamical stability. (6)

    (b) A ship of 5000 tonne displacement has three rectangular double bottom tanks A: 12m long and 16m wide; Tank B: 14m long and 15m wide; C 14m long and 16m wide.

    calculate the free surface effect for any one tank and state in which order the tanks should be filled when making use of them for stability correction. (10)

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    Part (a)

    Dynamical Stability is defined as the amount of energy required to heel a ship from its upright equilibrium position to a specific angle of heel. It provides a measure of the vessel's stability by considering its behaviour in response to dynamic external forces, such as wind or waves.

    • The concept compares the heeling moment energy (from external forces) and the righting moment energy (from the ship's stability).
    • The ship will absorb the energy imparted by the heeling moment. If the righting energy is greater than the heeling energy, the ship will stabilize; otherwise, it may capsize.

    Areas Under the Curve:

    • Area A: Represents the region where the heeling moment exceeds the righting moment (external energy > ship's stability).
    • Area B: Represents the region where the righting moment exceeds the heeling moment (ship's stability > external energy).
    • The balance of these areas determines whether the ship will right itself or continue to heel.

    When exposed to heeling forces such as wind or waves, the vessel inclines and may roll over to a certain angle of heel. If the external force is applied instantaneously, the ship must have enough reserve dynamic stability to absorb the energy and return to an upright position. If the external force is constant, the ship will remain at an equilibrium angle where the righting moment equals the heeling moment.

    This refers to the remaining righting energy available to counteract additional external forces. A higher reserve dynamic stability ensures the vessel can handle greater heeling forces without capsizing.

    Q10 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 2x

    A ship 120m long displaces 12000 tonne. The following data are available from trial results:

    V (Knots)

    10

    11

    12

    13

    14

    15

    sp (kW)

    880

    1155

    1520

    2010

    2670

    3600

    (a) Draw the curve of Admiralty Coefficients on a base of speed

    (b) Estimate the shaft power required for a similar ship 140m long at 14 knots. (16)

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    Part (a)

    Curve of Admiralty coefficients on a base of speed.

    The Admiralty coefficient is defined as C = Delta^(2/3) V^3 / P, where Delta is displacement (tonnes), V ship speed, and P shaft power.

    Chosen the parent ship: Delta=12000 t. Compute C at each trial speed.

    Speed V=10: numerator = 12000^(2/3) x 10^3. 12000^(2/3)=524, 10^3=1000, so 524,000/880 = 595.

    V=11: 524x1331=697,444; /1155 = 604.

    V=12: 524x1728=905,472; /1520 = 596.

    V=13: 524x2197=1,151,228; /2010 = 573.

    V=14: 524x2744=1,437,856; /2670 = 538.

    V=15: 524x3375=1,768,500; /3600 = 491.

    Plotting C against V gives a curve that is roughly flat (595-604) at low speed and falls progressively at higher speed (to about 490 at 15 kn), because wave-making resistance and other power terms rise faster than V^3 at high Froude numbers. The curve demonstrates that the constant-Admiralty-coefficient assumption holds only near moderate speeds.

    Part (b)

    Shaft power for a similar ship 140 m long at 14 knots.

    For geometrically similar ships the displacement scales as the cube of length:

    Delta140 = 12000 x (140/120)^3 = 12000 x 1.5876 = 19051 t.

    So Delta140^(2/3) = 19051^(2/3) = 714.

    Corresponding speed: for similarity, V2/V1 = sqrt(L2/L1) = sqrt(1.1667) = 1.0801. For the 140 m ship running at 14 knots, the corresponding speed of the parent (120 m) ship is 14/1.0801 = 12.96 knots, which lies between the 13-knot trial point (C=573) - read C is about 570 from the curve.

    Using the Admiralty coefficient at that point, C=570:

    P140 = Delta140^(2/3) x V^3 / C = 714 x 14^3 / 570 = 714 x 2744 / 570 = 1,959,216 / 570 = 3437 kW.

    Answer: the shaft power required is about 3440 kW (approximately 3.4 MW).

    Q1 (10 Marks) Corrosion & Protection

    With reference to Crude Oil Carriers

    (a) Explain Each of the following.

    (i) Segregated Ballast Tanks. (3)

    (ii) Clean Ballast Tanks. (2)

    (iii) Protective Locations. (3)

    (b) (i) Explain the Crude oil Washing System for cargo tank cleaning. (4)

    (ii) State the advantages office Crude Oil Washing. (4)

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    Part (a)

    Crude oil carrier terms.

    (i) Segregated Ballast Tanks (SBT): dedicated ballast tanks that are completely separated from the cargo oil and the cargo oil piping system, used only for ballast. They are required by MARPOL to provide adequate ballast capacity without carrying ballast in cargo tanks, so that the cargo tanks are not contaminated and the risk of oil pollution from ballast discharge is eliminated. SBTs are located in protective positions (see iii) and are arranged so that the ship can operate safely in ballast.

    (ii) Clean Ballast Tanks (CBT): cargo tanks that have been cleaned and are used for ballast, but which are not dedicated; they are only used for ballast after the tanks have been cleaned to a standard such that the ballast water is clean (oil content below the discharge limit). CBT is an alternative to SBT permitted for existing tankers, but the tanks must be cleaned and the ballast water must meet the discharge criteria.

    (iii) Protective Locations (PL): the requirement that the segregated ballast tanks and other spaces be located so that, in the event of a collision or grounding, the cargo tanks are protected from damage and the outflow of oil is minimised. The SBTs are placed in the wing and bottom regions (protective locations) so that they absorb the impact of a collision or grounding, reducing the risk of cargo tank rupture and oil spillage.

    Part (b)

    (i) Crude Oil Washing (COW) system.

    COW is a method of cleaning cargo tanks using the crude oil itself. During discharge, high-pressure crude oil is sprayed from fixed washing machines (rotating nozzles) onto the tank walls and bottom, dissolving and washing the clinging oil and sludge back into the cargo being discharged. The washing is carried out in a sequence with the discharge, and the washings are collected in a slop tank. The system uses the ship's cargo pumps and a dedicated COW piping system with fixed washing machines, and is operated under a COW manual and safety procedures (inert gas, gas-free checks).

    (ii) Advantages of Crude Oil Washing.

    • Reduces the amount of oil left in the tank (better stripping), so more cargo is discharged and less sludge remains.
    • Reduces the need for water washing, so less oily water is produced and the risk of pollution is reduced.
    • Reduces corrosion of the tank structure (less water and sludge in contact with the steel).
    • Reduces the time and cost of tank cleaning.
    • Improves the safety of tank entry (less gas and sludge).
    • Reduces the amount of slop and the need for slop disposal.
    Q2 (10 Marks) Hull Construction πŸ”₯ Repeated 2x

    With respect to Induced Vibrations in a ships hull:

    (a) State FOUR sources of excitation that may induce vibration into the main hull girder. (8)

    (b) Suggest methods for reducing the vibration levels induced by EACH of the exciting forces in (a) (8)

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    Part (a)

    FOUR sources of excitation that may induce vibration into the main hull girder.

    1. Propeller excitation: the propeller produces fluctuating forces (blade frequency and multiples) due to the varying pressure field and the wake, transmitted through the shaft and the water to the hull, causing vibration of the stern and the whole hull girder.
    2. Main engine excitation: the reciprocating and rotating parts of the engine produce unbalanced forces and moments (primary and secondary, and their harmonics) at the engine's firing frequency and multiples, transmitted through the engine seating to the hull.
    3. Wave excitation: the periodic wave forces on the hull (at the encounter frequency) excite the hull girder, particularly in heavy seas, causing springing and whipping.
    4. Auxiliary machinery and other rotating equipment: pumps, generators, compressors and other machinery produce unbalanced forces that are transmitted to the hull through their mountings.
    Part (b)

    Methods for reducing the vibration levels induced by each.

    1. Propeller: increase the clearance between the propeller and the hull (aperture), use a larger blade area and a suitable blade number to avoid resonance, fit a skew or a different blade design, and use a propeller boss cap fin; also avoid operating at resonant speeds.
    2. Main engine: balance the engine (fit balance weights, use a suitable firing order), fit a flexible coupling and a resilient engine mounting, tune the engine speed to avoid the hull's natural frequencies, and use a tuned vibration damper on the crankshaft.
    3. Wave excitation: reduce speed and change course in heavy seas to avoid the resonant encounter frequency, and design the hull to have adequate stiffness and damping; use a hull stress/vibration monitoring system.
    4. Auxiliary machinery: fit resilient (anti-vibration) mountings, balance the rotating parts, isolate the machinery from the hull, and avoid operating at resonant speeds.
    Q3 (10 Marks) Ship Stability

    (a) Explain how to distinguish between list and loll and describe how to return the ship to the upright in each case. (8)

    (b) Explain the term Angle of loll and state the dangers it poses to a vessel. (8)

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    Part (a)

    Distinguishing between List and Loll:

    1. List is caused by Uneven distribution of weight within the ship. List will have the below conditions:

    • GM is positive (GM > 0).
    • The centre of gravity (G) is off-centre.
    • The vessel is at equilibrium but inclines to one side due to uneven weight distribution, even without any external forces acting on it.
    • The vessel rolls around the angle of list.

    Correction:

    • Redistribute the weight evenly to bring the centre of gravity (G) back in line with the centerline and metacentric height (M).

    2. Loll is caused by High centre of gravity (G) leading to negative GM and is exacerbated by external forces, free surface effects, or poor distribution of weights. LOLL will have the below conditions:

    • GM is negative (GM < 0).
    • The centre of gravity (G) is on the centerline but too high, making the vessel inherently unstable.
    • The vessel flops or inclines to one side at an angle of loll, and it can incline equally to either side.
    • The vessel rolls unstably around the angle of loll.

    Correction:

    • Reduce the centre of gravity by ballasting bottom tanks or removing weight from higher levels.
    • Minimize free surface effects by reducing the breadth of free surfaces in tanks.

    Q4 (10 Marks) General πŸ”₯ Repeated 7x

    List the hazards that arise with the carriage of liquefied gas in bulk. Describe, with the aid of a sketch the details of construction of a prismatic cargo tank within a gas carrier designed to carry liquefied gas. (16)

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    Hazards of Carriage of Liquefied Gas in Bulk and Construction of a Prismatic Cargo Tank

    Part (a)

    Six Hazards Associated with the Carriage of Liquefied Gas in Bulk

    1. Flammability and Explosion
      • Liquefied gases such as LPG vaporise rapidly when released.
      • The vapour can mix with air and form a highly flammable or explosive atmosphere, creating a serious risk of fire or explosion.
    2. Toxicity
      • Some liquefied gas cargoes, such as ammonia and vinyl chloride monomer, are highly toxic.
      • Exposure can cause serious poisoning through inhalation, ingestion or skin absorption.
    3. Asphyxiation
      • LPG vapours are generally heavier than air.
      • In the event of a leak, the vapour can collect in low-lying areas such as pump rooms and hold spaces, displacing oxygen and creating an asphyxiation hazard.
    4. Frostbite and Cold Burns
      • Liquefied gas cargoes are carried at very low temperatures, with fully refrigerated LPG being carried at approximately βˆ’50Β°C.
      • Contact with the liquid cargo or uninsulated pipes and equipment can cause severe frostbite and cold burns.
    5. Brittle Fracture
      • Ordinary ship hull steel, such as mild steel, can become brittle at very low temperatures.
      • If cold cargo leaks and comes into contact with unsuitable hull steel, it may cause cracking and catastrophic structural failure.
    6. Sloshing
      • Partially filled tanks have a free surface, allowing the liquid cargo to move violently with the ship's motion.
      • This can produce large dynamic impact loads on the tank walls and internal pump towers, potentially causing structural damage.
    Part (b)

    Construction of a Prismatic Cargo Tank for LPG

    Fully refrigerated LPG is typically carried in Independent Type A prismatic cargo tanks. These are self-supporting tanks that are independent of the ship's hull structure and do not contribute to the overall structural strength of the vessel.

    Sketch – Typical Prismatic Cargo Tank Arrangement

    Ship's Hull / Hold Space

    β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”

    β”‚ Secondary Barrier / Hull β”‚

    β”‚ β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚

    β”‚ β”‚ Thermal Insulation β”‚ β”‚

    β”‚ β”‚ β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚ β”‚

    β”‚ β”‚ β”‚ β”‚ β”‚ β”‚

    β”‚ β”‚ β”‚ LPG CARGO β”‚ β”‚ β”‚

    β”‚ β”‚ β”‚ PRISMATIC TANK β”‚ β”‚ β”‚

    β”‚ β”‚ β”‚ β”‚ β”‚ β”‚

    β”‚ β”‚ β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ β”‚ β”‚

    β”‚ β”‚ Primary Barrier β”‚ β”‚

    β”‚ β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ β”‚

    β”‚ Load-bearing supports β”‚

    β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜

    β”‚

    Double Bottom

    1. Shape and Structure

    • The tank is prismatic, meaning it is generally box-shaped with chamfered/angled top and bottom corners.
    • This shape allows the tank to closely follow the contours of the ship's inner hull and therefore maximises the available cargo capacity.
    • The tank is internally reinforced with frames and stiffeners to maintain structural integrity.
    • A centreline longitudinal bulkhead, together with transverse wash bulkheads, may be provided to strengthen the tank and reduce liquid movement and sloshing.

    2. Materials

    • LPG is carried at low temperatures, approximately βˆ’50Β°C for fully refrigerated LPG.
    • To prevent brittle fracture at these temperatures, the primary barrier/tank is constructed from suitable low-temperature-resistant materials, typically fine-grained carbon-manganese steel.

    3. Tank Supports and Chocks

    The cargo tank operates at a substantially different temperature from the ship's hull and therefore undergoes thermal expansion and contraction.

    • The tank is supported on the double bottom by load-bearing insulation blocks.
    • These may be made from specialised hardwood such as AzobΓ© or suitable synthetic materials.
    • The supports provide the necessary load-bearing capacity while reducing thermal transfer and structural stresses.
    • Anti-roll, anti-pitch and anti-flotation keys/chocks secure the tank against movement caused by the ship's motion.
    • At the same time, the arrangement allows the tank to expand and contract freely due to temperature changes.

    4. Insulation

    • The outside of the primary barrier is provided with high-efficiency thermal insulation.
    • Its purpose is to maintain the required low cargo temperature and minimise heat ingress and cargo boil-off.
    • Sprayed polyurethane foam (PUF) is commonly used as the insulation material.

    5. Secondary Barrier

    • Under the IGC Code, Type A tanks are required to have a complete secondary barrier capable of containing leaked cargo for up to 15 days, preventing the cold cargo from coming into contact with the outer hull.
    • In LPG carriers, the ship's inner hull is normally used as the secondary barrier.
    • The inner hull is also constructed from suitable fine-grained low-temperature steel so that it can withstand the low temperature if the primary cargo tank fails.
    • The space between the primary tank and secondary barrier, known as the hold space, is maintained with inert gas or dry air as applicable.
    Q5 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    With reference to Ship stability:

    (a) With the help of a neat sketch explain the relevant features of a G-Z curve. (8)

    (b) What are the effects of the below mentioned conditions on the G-Z curve: (8)

    (i) Increased freeboard,

    (ii) Increased beam, and

    (iii) Increased GM.

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    Part (a)

    Features of a GZ curve.

    The GZ curve is a graph of the righting lever GZ against the angle of heel. Its relevant features are:

    • The origin: at zero heel, GZ = 0.
    • The initial slope: the tangent to the curve at the origin equals GM (the metacentric height), since GZ = GM sin(theta) for small angles. A steeper initial slope means a larger GM.
    • The maximum righting lever (GZ max): the highest point of the curve, and the angle at which it occurs (the angle of maximum stability, typically 25-40 deg).
    • The range of stability: the angle from the upright to the angle of vanishing stability (where GZ returns to zero, typically 60-90 deg).
    • The area under the curve: proportional to the dynamical stability (the work done in heeling the ship), used to assess stability in a seaway and against wind heeling.
    • The angle of loll: if the curve starts below the axis (negative GZ at small angles), indicating a negative GM and an unstable ship that lolls to one side.
    • The effect of free surface: the curve is reduced by the free-surface correction.

    The curve is obtained from the cross-curves of stability corrected for the actual KG and free-surface effects, and is compared with the statutory criteria.

    Part (b)

    Effects of the following conditions on the GZ curve.

    (i) Increased freeboard: increasing the freeboard raises the deck edge and increases the reserve buoyancy, so the range of stability is increased (the angle of vanishing stability moves to a larger angle) and the area under the curve is increased. The initial slope (GM) is largely unchanged, but the curve is higher and extends further, giving greater dynamical stability and a larger range.

    (ii) Increased beam: increasing the beam increases the waterplane area and the BM (BM is proportional to the cube of the beam), so the initial slope (GM) increases and the curve is steeper at small angles. The maximum GZ is increased and occurs at a smaller angle, but the range of stability may be reduced (the angle of vanishing stability decreases) because the ship becomes stiffer and the deck edge immerses earlier. The area under the curve may be reduced at large angles.

    (iii) Increased GM: increasing the GM (e.g. by lowering KG) makes the initial slope steeper, so the curve rises more steeply at small angles and the maximum GZ is larger and occurs at a smaller angle. However, the range of stability is reduced (the angle of vanishing stability decreases) and the ship rolls more quickly and with a shorter period, which can be uncomfortable. The area under the curve at small angles increases but the overall range decreases.

    Q6 (10 Marks) Ship Stability

    (a) What is 'form stability' & 'weight stability'. (6)

    (b) The end bulkhead of the wing tank of an oil tanker has the following widths at 3m intervals. Commencing at the deck: 6.0, 6.0, 5.3, 3.6 and 0.6 m. Calculate the load on the bulkhead and the position of the centre of pressure if the tank is full of all rd 0.8. (10)

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    Part (a)

    Form stability and weight stability.

    Form stability is the stability that depends on the shape (form) of the hull, i.e. the position of the metacentre and the righting levers that arise from the geometry of the waterplane and the hull. It is measured by the metacentric height and the GZ curve, and depends on the beam, waterplane area, draught and the shape of the hull. A ship with a wide beam and a large waterplane area has high form stability (large BM and GM).

    Weight stability is the stability that depends on the position of the centre of gravity, i.e. the distribution of the weights in the ship. It is the effect of the KG on the stability: a low centre of gravity gives high weight stability (large GM), while a high centre of gravity reduces it. The total stability is the combination of the form stability (position of the metacentre) and the weight stability (position of the centre of gravity), since GM = KM - KG, where KM is the metacentric height above the keel (form) and KG is the height of the centre of gravity (weight).

    Part (b)

    Load on the end bulkhead of a wing tank.

    The end bulkhead of the wing tank of an oil tanker has widths at 3 m intervals, commencing at the deck: 6.0, 6.0, 5.3, 3.6 and 0.6 m. Calculate the load on the bulkhead and the position of the centre of pressure if the tank is full of oil of relative density 0.8.

    The bulkhead is a vertical plane of width varying with depth. The depth below the deck (and below the liquid surface, since the tank is full) at the five stations is 0, 3, 6, 9, 12 m (spacing 3 m).

    For a vertical plane surface, the load (total force) = rho g x (first moment of area about the liquid surface), and the centre of pressure is at depth = (second moment)/(first moment).

    First-moment integrand (per unit width) about the liquid surface: M1 = integral of z dz from 0 to H = H^2/2.

    Second-moment integrand: M2 = integral of z^2 dz from 0 to H = H^3/3.

    Widths b: 6.0, 6.0, 5.3, 3.6, 0.6. Depths H: 0, 3, 6, 9, 12.

    M1 values (b x H^2/2): 0, 6x4.5=27, 5.3x18=95.4, 3.6x40.5=145.8, 0.6x72=43.2.

    M2 values (b x H^3/3): 0, 6x9=54, 5.3x72=381.6, 3.6x243=874.8, 0.6x576=345.6.

    Using Simpson's rule with spacing 3 m and 5 ordinates:

    M1 = (3/3)[0 + 0.6 + 4(27 + 145.8) + 2(95.4)] = 1[0.6 + 4x172.8 + 190.8] = 0.6 + 691.2 + 190.8 = 882.6.

    M2 = (3/3)[0 + 345.6 + 4(54 + 874.8) + 2(381.6)] = 345.6 + 4x928.8 + 763.2 = 345.6 + 3715.2 + 763.2 = 4824.

    Load (force) = rho g x M1 = 0.8 x 1.025 x 9.81 x 882.6 (using sea-water density 1.025 t/m3 for the oil of RD 0.8, i.e. rho = 0.8 x 1.025 = 0.82 t/m3 = 820 kg/m3).

    Load = 820 x 9.81 x 882.6 = 7,099,000 N = 7099 kN, or in tonnes-force = 0.82 x 882.6 = 723.7 t.

    Centre of pressure depth below the liquid surface = M2/M1 = 4824/882.6 = 5.47 m.

    Answer: the load on the bulkhead is about 7100 kN (about 724 t), and the centre of pressure is about 5.5 m below the liquid surface (i.e. about 5.5 m below the deck).

    Q7 (10 Marks) Ship Stability

    (a) Explain how trim occurs and explain the effect of trim on tank soundings. (6)

    (b) A vessel of 8000 tonne displacement has 75 tonnes of cargo on the deck. It is lifted by a derrick whose head is 10.5m above the centre of gravity of the cargo and placed in the lower hold 9m below the deck and 14m forward of its original position. Calculate the shift in the vessel's centre of gravity from its original position when the cargo is (10)

    (i) Just clear of the deck

    (ii) At the derrick head

    (iii) In its final position.

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    Part (a)

    The effect of trim on tank soundings is significant because the reading obtained from the sounding pipe does not directly reflect the actual volume of liquid in the tank when the ship is trimmed. Typically, the sounding pipe is located at the aft end of the tank.

    If the ship is trimmed by the stern, the liquid will settle more toward the aft end, resulting in a higher sounding than if the ship were on an even keel. Conversely, if the ship is trimmed by the bow, the liquid shifts forward, and the sounding obtained will be less, potentially underestimating the actual volume.

    To ensure accurate volume measurement despite varying trim conditions, tank calibrations are carried out for different trims, and the data is compiled into a sounding book or trim correction table. When taking soundings, the trim is noted, and the corrected volume is obtained using these calibration tables, ensuring reliable tank content assessment regardless of trim.

    Q8 (10 Marks) Ship Stability

    (a) Explain why the bilging of empty double-bottom or deep tanks below the waterline leads to an increase in GM. (6)

    (b) A ship of 22000 tonne displacement is 160 m long and MCT1cm 280 tonne m, waterplane area 3060 m2 centre of buoyancy 1 m aft of midships and centre of flotation 4 m aft of midships. It floats in water of 1.007 t/ m3 at draughts of 8.15 m forward and 8.75 m aft. Calculate the new draughts if the vessel moves into sea water of 1.026 t/ m2. (10)

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    Part (a)

    Why bilging empty double-bottom or deep tanks below the waterline increases GM.

    When an empty tank below the waterline is bilged (flooded), the water that enters is at the same level as the sea and, being below the waterline, the added weight of water is exactly balanced by the added buoyancy of the extra volume displaced (the ship sinks slightly). The net effect is that the centre of gravity of the added water is low (in the tank, near the bottom of the ship), while the added buoyancy acts at the centre of buoyancy of the flooded volume, which is also low. Because the added weight is low, the overall centre of gravity KG is lowered, and because the added buoyancy is low, the centre of buoyancy KB is also lowered but by less than the weight effect. The result is that GM increases (the ship becomes stiffer). Additionally, once the tank is full there is no free surface, so there is no free-surface loss of GM. Hence bilging an empty low tank increases GM and improves stability (though it increases displacement and draft).

    Part (b)

    New draughts when the vessel moves into sea water.

    Ship 22,000 t displacement, 160 m long, MCT1cm = 280 t-m, waterplane area 3060 m2, centre of buoyancy 1 m aft of midships, centre of flotation 4 m aft of midships. It floats in water of 1.007 t/m3 at draughts of 8.15 m forward and 8.75 m aft. Calculate the new draughts if the vessel moves into sea water of 1.026 t/m3.

    Step 1 - change of mean draught.

    Volume in water of 1.007 = 22000/1.007 = 21847 m3; volume in sea water (1.026) = 22000/1.026 = 21442 m3. Decrease in volume = 405 m3.

    Decrease in mean draught = 405/3060 = 0.132 m. New mean draught = 8.45 - 0.132 = 8.318 m.

    Step 2 - change of trim.

    The decrease in buoyancy (405 t) acts at the centre of flotation (4 m aft of midships), while the centre of buoyancy is 1 m aft of midships. The moment about the centre of flotation = 405 x (4 - 1) = 405 x 3 = 1215 t-m.

    Change of trim = moment/MCT1cm = 1215/280 = 4.34 cm. Since the buoyancy is removed aft of the centre of buoyancy, the stern rises, reducing the stern trim. Original trim = 0.6 m (8.75 - 8.15) by the stern. New trim = 0.6 - 0.0434 = 0.5566 m by the stern.

    Step 3 - new draughts.

    The change of trim of 0.0434 m (4.34 cm) is distributed about the centre of flotation (4 m aft of midships). The length is 160 m, so the aft portion is 80 - 4 = 76 m and the forward portion is 80 + 4 = 84 m.

    Change in aft draught = 0.0434 x 76/160 = 0.0206 m (stern rises, so subtract).

    Change in forward draught = 0.0434 x 84/160 = 0.0228 m (bow rises, so subtract).

    New aft draught = 8.75 - 0.132 - 0.0206 = 8.597 m.

    New forward draught = 8.15 - 0.132 - 0.0228 = 7.995 m.

    Answer: new draughts are about 8.0 m forward and 8.6 m aft.

    Q9 (10 Marks) Ship Resistance & Propulsion

    (a) Define hull efficiency and propeller efficiency. (6)

    (b) When a propeller of 4.8m pitch turns at 110 rev/min, the apparent slip is found to be -S% and the real slip +1.5S%. If the wake speed is 25% of the ship speed, calculate the ship speed, the apparent slip and the real slip. (10)

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    Part (a)

    Hull efficiency and propeller efficiency.

    Hull efficiency is the ratio of the effective power (the power required to tow the hull) to the thrust power (the power developed by the propeller in producing thrust). It accounts for the wake and the thrust deduction:

    Hull efficiency = (1 - t)/(1 - w)

    where t is the thrust deduction factor and w is the wake fraction. It reflects how well the hull and propeller interact: the wake reduces the speed of advance of the propeller (reducing the power needed), while the thrust deduction increases the thrust required.

    Propeller efficiency (open-water efficiency) is the ratio of the thrust power to the delivered power:

    Propeller efficiency = T x Va / (2 pi n Q)

    where T is the thrust, Va the speed of advance, n the revolutions and Q the torque. It represents the efficiency of the propeller itself in converting the delivered power into thrust power, and depends on the propeller design (pitch, diameter, blade area) and the loading.

    Part (b)

    Ship speed, apparent slip and real slip.

    A propeller of 4.8 m pitch turns at 110 rev/min. The apparent slip is -S% and the real slip is +1.5S%. The wake speed is 25% of the ship speed. Calculate the ship speed, the apparent slip and the real slip.

    Pitch speed = pitch x rev/s = 4.8 x 110/60 = 8.8 m/s.

    Let the ship speed be V (m/s). The speed of advance Va = V x (1 - 0.25) = 0.75 V (wake = 25% of ship speed).

    Apparent slip = (pitch speed - V)/pitch speed = -S/100.

    Real slip = (pitch speed - Va)/pitch speed = 1.5 S/100.

    From the apparent slip: (8.8 - V)/8.8 = -S/100, so V = 8.8(1 + S/100).

    From the real slip: (8.8 - 0.75 V)/8.8 = 1.5 S/100, so 8.8 - 0.75 V = 8.8 x 1.5 S/100 = 0.132 S.

    Substitute V = 8.8(1 + S/100): 8.8 - 0.75 x 8.8(1 + S/100) = 0.132 S.

    8.8 - 6.6(1 + S/100) = 0.132 S.

    8.8 - 6.6 - 0.066 S = 0.132 S.

    2.2 = 0.198 S, so S = 11.11.

    Apparent slip = -11.11%; real slip = 1.5 x 11.11 = 16.67%.

    Ship speed V = 8.8(1 + 0.1111) = 8.8 x 1.1111 = 9.78 m/s = 9.78 x 1.944 = 19.0 knots.

    Answer: ship speed about 19.0 knots; apparent slip -11.1%; real slip +16.7%.

    Q10 (10 Marks) Ship Stability

    (a) Why is it important in a tender ship to keep the double bottom tanks pressed up? (6)

    (b) A double bottom tank is 23m long. The half breadths of the top of the tank are 5.5, 4.6, 4.3, 3.7 and 3.0m respectively. When the ship displaces 5350 tonnes, the loss in metacentric height due to free surface is 0.2m. Calculate the density of the liquid in the tank. (10)

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    Part (a)

    In a tender ship, which is characterized by a small metacentric height (GM), it is extremely important to keep the double bottom tanks pressed up. Tender ships have a high rolling period and tend to incline easily, taking longer to return to their upright position. If the double bottom tanks are only partially filled, the liquid inside them can move from side to side with the ship’s motion. This movement causes a shift in the center of gravity from G to a virtual position G1 due to the free surface effect, resulting in a virtual loss of GM.

    This reduction in GM diminishes the ship’s righting lever, changing it from (GZ sinΞΈ) to (G1Z1 sinΞΈ), which is smaller. As a result, the moment of statical stability is significantly reduced. In a tender ship, this is particularly dangerous, as it can lead to the ship reaching an angle of loll or even capsizing in severe conditions. Therefore, to minimize the free surface effect and maintain sufficient stability, it is crucial to keep the double bottom tanks either fully pressed up or completely empty, with pressed up being the preferred condition for safety in tender ships.

    Q1 (10 Marks) Hull Construction πŸ”₯ Repeated 5x

    With reference to membrane tanks for the carriage of liquefied gas at very low temperatures:

    (a) Describe with the aid of a sketch, ONE method of building up the insulation (6)

    (b) State with reasons the alloy, which is used for the membrane. (4)

    (c) Describe with the aid of a sketch, how the tanks are located and supported: (6)

    (i) Longitudinally

    (ii) Transversely

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    Part (a)

    Membrane tanks for liquefied gas carriage are built with a double hull throughout the cargo length. The insulation system includes the following:

    1. Primary Barrier: A thin membrane made of INVAR (36% Nickel, 64% Iron) with a thickness of 0.7 to 1.2 mm, forming the primary containment layer for the liquefied gas.
    2. Primary Insulation box: A 230 mm thick layer of granulated Perlite insulation packed in plywood boxes, surrounding the primary barrier. Perlite is siliconised to make it impervious to moisture.
    3. Secondary Barrier: A secondary membrane made of the same INVAR material is installed to prevent cargo leakage in case of primary barrier failure.
    4. Secondary Insulation box: Another 300 mm thick layer of granulated Perlite is placed above the secondary barrier to provide additional insulation and prevent thermal transfer.
    Part (b)

    Alloy used for the membrane:

    Invar (36% Ni, 64% Fe) is used for both the primary and secondary barriers. The reason for this choice is its exceptionally low coefficient of thermal expansion. This eliminates the need for expansion joints or corrugations in the membrane design. In addition, Invar remains strong and does not become brittle at the very low temperatures experienced by the liquefied gas. The thin and lightweight nature of Invar maximizes the cargo-carrying capacity of the tank.

    Part (c)

    Membrane tanks are either independent or self-supporting

    , meaning they don't form part of the ship's hull and don't contribute to the ship's structural strength. They can be spherical, cylindrical, or prismatic (box-shaped). Prismatic tanks usually have internal stiffeners like bulkheads, webs, girders, and stiffeners for added structural integrity.

    (i) Longitudinally:

    Tanks are positioned longitudinally within the ship's cargo hold. The tanks are supported longitudinally by anti-roll chocks and anti-lift chocks that resist forces caused by ship motions. These chocks ensure the tank remains stable even in rough sea conditions. They also act as thermal barriers, preventing the transfer of heat between the hull and the tank.

    The chocks also serve as thermal barriers between the hull and cargo and are often constructed of wood or plastic materials.

    (ii) Transversely: Transverse support is provided by anti-pitch chocks and support chocks, which stabilize the tank against lateral forces. These chocks are typically made of plywood or plastic and help prevent thermal stress between the tank and the ship’s structure.

    Q2 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 7x

    With reference to dry docking, define the responsibilities of the Second Engineer and instructions to Junior Engineers:

    (a) Prior to docking (6)

    (b) Whilst the vessel is in dry dock (5)

    (c) Prior to flooding and leaving the dock. (5)

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    (a) Prior to Docking:

    Preliminary Preparation:

    • Review various plans, manuals, and previous drydock reports for reference.
    • Prepare a detailed repair list and ensure all required spares are accounted for.
    • Conduct an inventory of spares and requisition necessary items.
    • Gather required tools, including special tools like poker gauges, trammel gauges, gagging tools, etc.
    • Mark all overboard valves, NC (normally closed) and NO (normally open) valves clearly for easy identification.
    • Allocate jobs to team members and discuss the time schedule.
    • Conduct a safety meeting to highlight dry dock hazards and necessary precautions.

    Before Entering Dry Dock:

    • Identify the power requirements and machinery to be operational during docking.
    • Check the shore connection box for proper functionality.
    • Record the soundings of FO (Fuel Oil), LO (Lubricating Oil), and DO (Diesel Oil) tanks.
    • Discharge contents from clean drain tanks and sewage tanks.
    • Carry out Economizer soot-blowing.
    • Change over Main Engine, Diesel Generators, and Boiler to Low Sulfur Gas Oil (LSGO).
    • Stop and clean purifiers.
    • Ensure the low sea chest is open and the high sea chest is shut.
    • Keep firefighting appliances (FFA) on standby.
    • Shut down all non-essential machinery, including MGPS (Marine Growth Prevention System) and FWG (Fresh Water Generator).

    (b) Whilst the Vessel is in Dry Dock

    Upon Arrival:

    • Connect shore power and supplies after ensuring safety checks are completed.
    • Start necessary equipment like cooling water, air compressors, air conditioning, and fridge compressors.
    • Check for jobs assigned by dry dock personnel and prepare accordingly.
    • Attend a safety meeting with dry dock personnel to understand local safety rules and procedures.

    During Dry Docking::

    • Oversee and assist in:
      • Cleaning and inspecting the hull, rudder, sea chest, anodes, and propeller.
      • Measuring propeller drop, checking rudder clearances, and inspecting the stern tube bearing and seal.
      • Servicing underwater valves and overboard valves.
      • Inspecting anchor and cables conditions.
      • Overhauling deck machinery, cranes, elevators, and engine room equipment such as the Main Engine, Diesel Generators, Boiler, and Economizer.
      • Renewing pipes and valves as needed.
      • Performing electrical equipment maintenance and surveys.
      • Supervising service engineers for specific repair jobs.
    • Ensure tank cleaning, welding, and other repair works are completed according to the plan.
    • Run the standby diesel generator daily after starting the priming pump.

    (c) Before Flooding and Leaving the Dry Dock

    Final Checks:

    • Verify that all underwater fittings and drain plugs are securely in place.
    • Ensure all machinery has been boxed back and is ready for operation.
    • Check for any leakage in stern tube seals.
    • Take tank soundings to confirm proper levels.
    • Confirm the proper operation of all underwater valves, overboard valves, sea chests, and vents.
    • Inspect the stern tube tank for any irregularities.

    System Restart:

    • Switch back to ship's power after confirming all systems are functional.
    • Test the proper operation of all machinery and systems to ensure the ship is fully operational before leaving the dock.
    Q3 (10 Marks) Ship Stability πŸ”₯ Repeated 8x

    (a) Explain what is meant by "permissible length" of compartments in passenger ships. (6)

    (b) Describe how the position of bulkheads is determined (5)

    (c) Describe briefly the significance of the factor of subdivision (5)

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    Part (a)

    Permissible Length

    Permissible length refers to the maximum length of a ship's compartment that can be flooded while ensuring that the sinkage, heel, or trim remains within acceptable limits without exceeding the floodable length. It ensures the ship remains afloat and stable after flooding.

    Permissible Length Formula:

    $$Permissible\:lenght\:=\:Floodable\:length\:\times Factor\:of\:subdivision$$

    The Factor of Subdivision depends on the ship's length and the nature of its service:

    • For passenger ships, the factor of subdivision is smaller compared to cargo ships.
    • Smaller compartments ensure enhanced safety in case of flooding.
    Part (b)

    Position of Bulkheads

    The position of bulkheads is determined based on the need to comply with the "reasonable amount of damage" criterion in case of flooding:

    • Transverse Watertight Bulkheads should vertically extend up to the margin line.
    • At least one watertight longitudinal bulkhead must be located 20% of the ship's breadth inward from each side.
    • Bulkheads are positioned along the ship's length at intervals equal to the permissible length, but not exceeding the floodable length.
    • The maximum permissible compartment length is limited to 10.7 meters.
    • Collision Bulkhead must be located forward at a distance equal to the permissible length from the forward perpendicular.
    • The ER must be enclosed by two transverse watertight bulkheads, with an aft peak watertight bulkhead enclosing the forward part.
    Part (c)

    Factor of Subdivision

    The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

    Permissible Length Formula:

    $$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

    A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

    Q4 (10 Marks) Ship Stability πŸ”₯ Repeated 14x

    Explain how the period of roll varies with

    (a) The amplitude of roll (4)

    (b) The radius of gyration (4)

    (c) The initial metacentric height (4)

    (d) The location of masses in the ship (4)

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    The period of roll Tr of a ship is determined by the formula:

    $$T_{r}=\frac{2\pi K}{\sqrt{g\times GM}}$$

    where,

    • K is the radius of gyration (mass moment of inertia)
    • g is the acceleration due to gravity, and
    • GM is the metacentric height.
    Part (a)

    Amplitude of Roll:

    • The amplitude of roll, or the maximum angle of heel, does not affect the period of roll. While a larger amplitude implies a greater heeling moment and faster roll speed, the time taken to complete one full roll cycle remains constant for a given metacentric height and radius of gyration.
    Part (b)

    Radius of Gyration (K):

    • The period of roll is directly proportional to the radius of gyration. A larger radius of gyration (indicating a greater distribution of mass further from the ship's centre of rotation) leads to a longer period of roll. Conversely, a smaller radius of gyration (mass concentrated closer to the centre) results in a shorter period. The distribution of cargo significantly impacts K; cargo concentrated centrally minimises K and the roll period, while dispersed cargo maximises K and the roll period.
    Part (c)

    Initial Metacentric Height (GM):

    • The period of roll is inversely proportional to the square root of the metacentric height (GM). A larger GM (a stiffer ship) leads to a shorter roll period, as the ship quickly returns to its upright position. A smaller GM (a tender ship) results in a longer roll period, with slower return to the upright.
    Part (d)

    Location of Masses in the Ship:

    The location of masses in the ship will effect the GM & K. So the period of roll will be affected.

    • If masses are at bottom, G moves down, GM ↑, period of roll ↓.
    • If masses are at top, G moves up, GM ↓, period of roll ↑.
    • If masses are concentrated at centre, K ↓, period of roll ↓.
    • If masses are away from centre, K ↑, period of roll ↑.
    Q5 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 13x

    Describe the relationship between frictional resistance and (16)

    (a) Ship speed

    (b) The wetted area

    (c) The surface roughness

    (d) The length of the vessel.

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    Frictional resistance arises due to the interaction between the ship's hull and the water as the vessel moves. This resistance is caused by eddying water adjacent to the hull that is drawn along with the ship. The frictional resistance is given by:

    $$R_{f\:=\:}f\:.\:s\:.\:v^{n}$$

    Where:

    • f = Coefficient of friction
    • s = Wetted surface area
    • v = Ship's speed in knots
    • n = Constant (1.82)
    Part (a)

    Ship Speed (v):

    • Frictional resistance increases with increasing ship speed. The relationship is not perfectly linear, but it's approximately described by the formula Rf = fsv^n (Rf ∝ v^n), where 'n' is a constant (approximately 1.82). This means that a small increase in speed results in a larger increase in frictional resistance.
    Part (b)

    Wetted Area (s):

    • Frictional resistance is directly proportional to the wetted surface area (Rf ∝ s). A larger wetted area (the part of the hull submerged in water) leads to greater frictional resistance. The wetted area increases with the ship's draught (depth in the water).
    Part (c)

    Surface Roughness:

    • Increased surface roughness increases frictional resistance. Roughness can be caused by fouling (marine growth), corrosion, or poor hull maintenance. Regular cleaning and maintenance of the hull surface help to minimise roughness and reduce resistance.
    Part (d)

    Length of the Vessel

    • Frictional resistance is influenced by the vessel's length. As the length increases, the wetted surface area grows, and the coefficient of friction (f) changes, leading to an increase in Rf. Longer vessels experience higher frictional resistance, but streamlined designs can mitigate the effect to some extent.
    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) How the distribution of mass within the ship affects the rolling period? (6)

    (b) A ship of 14000 tonne displacement is 125 m long and floats at draughts of 7.9 m forward and 8.5 m aft. The TPC is 19, GM. 120 m and LCF 3 m forward of midships. It is required to bring the vessel to an even keel draught of 8.5 m. Calculate the mass which should be added and the distance of the centre of the mass from midships. (10)

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    Part (a)

    The rolling period is influenced by the ship's metacentric height (GM) and radius of gyration (K), which are determined by the location of the masses onboard.

    Masses at the bottom:

    • The centre of gravity (G) moves down.
    • Metacentric height (GM) increases, resulting in greater stability.
    • Rolling period decreases.

    Masses at the top:

    • The centre of gravity (G) moves up.
    • Metacentric height (GM) decreases, reducing stability.
    • Rolling period increases.

    Masses concentrated at the centre:

    • The radius of gyration (K) decreases.
    • Rolling period decreases.

    Masses distributed away from the centre:

    • The radius of gyration (K) increases.
    • Rolling period increases.
    Q7 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Explain how increase of draught and of displacement influence rolling. (6)

    (b) A pontoon has a constant cross-section as shown in Fig. Given below The metacentric height is 2.5m. Find the height of the centre of gravity above the keel. (10)

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    Part (a)

    In case of increase of draught and displacement, this is a case of loading being done.

    Thus, when the ship is being loaded, its GM will start to decrease.

    Now, the time period of roll is given by:

    $$T_{r}=\frac{2\pi k}{\sqrt{GM.g}}$$

    Where,

    • k = radius of gyration
    • GM = metacentric height
    • g = acceleration due to gravity

    Now, since, GM has started to decrease, the Tr will start to increase.

    Thus, the ship will now roll with greater time period. Thus, an increase in draught and displacement, influences rolling.

    Q8 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    (a) Describe the fundamental principle of a propeller (6)

    (b) A propeller 6m diameter has a pitch ratio of 0.9, BAR 0.48 and, when turning at 110 rev/min, has a real slip of 25% and wake fraction 0.30. If the propeller delivers a thrust of 300 kN and propeller efficiency is 0.65, calculate: (10)

    (i) Blade area

    (ii) Ship speed

    (iii) Thrust power

    (iv) Shaft power

    (v) Torque.

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    Part (a)

    A propeller is a type of fan, that transmits power by converting rotational motion into thrust. A pressure difference is produced between the forward and rear surface of the aerofoil shaped blade and the fluid is accelerated behind the blade. A marine propeller of this type is sometimes known as screw propeller or a screw.

    Given:

    $$D=6m$$

    $$p=0.9$$

    $$BAR=0.48$$

    $$N=110rev\:per\min$$

    $$real \space slip \space (s) \space = \space 25\%$$

    $$W_{F}=0.3$$

    $$Thrust \space = \space 300kN$$

    $$Ξ·_{prop} \space = \space 0.65$$

    (i) Blade area:

    $$BAR \space = \space {{A_b} \over {{\pi} \over 4} D^2}$$

    $$Blade \space area \space A_b \space = \space 0.48 \times {{\pi} \over 4} 6^2$$

    $$Blade \space area \space = \space 13.57m^2 $$

    $$p \space = \space {{P} \over D}$$

    $$0.9 \space = \space {{P} \over 6}$$

    $$Pitch \space p = \space 5.4m$$

    $$V_{T}=P\times N\times\frac{3600}{1852}$$

    $$V_{T}=5.4\times\frac{110}{60}\times\frac{3600}{1852}$$

    $$V_{T}=19.24knots$$

    $$Real \space slip \space S \space = \space {{V_T - V_a} \over V_T}$$

    $$ 0.25 \space = \space {{19.24 - V_a} \over 19.24}$$

    $$V_a \space = \space 14.42 knots$$

    $$W_F \space = \space {{V - V_a} \over V}$$

    $$0.30 \space = \space {{V - 14.42} \over V}$$

    $$V=20.6knots$$

    $$T_{p}\space=\space Thrust\times V_{a}\times\frac{1852}{3600}$$

    $$T_p \space = \space Thrust \times 14.42 \times {{1852} \over 3600 }$$

    $$T_p \space = \space 2225.48 $$

    $$T_p \space = \space d_p \times Ξ·_{prop}$$

    $$2225.48=d_{p}\times0.65$$

    $$d_p \space = \space 3423.8kW$$

    $$dp=2\pi NT$$

    $$3423.8=2\times\pi\times\frac{110}{60}\times T$$

    $$T=297.22KN$$

    Q9 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    (a) Explain what is meant by: (6)

    (i) Wave-making resistance

    (ii) Frictional resistance

    (iii) Eddy-making resistance

    (b) When a ship is 800 nautical miles from port its speed is reduced by 20%, there by reducing the daily fuel consumption by 42 tonne and arriving in port with 50 tonne on board. If the fuel consumption in t/h is given by the expression (0.136+0.001 V^3) where V is the speed in knots, estimate: (10)

    (i) The reduced consumption per day.

    (ii) The amount of fuel on board when the speed was reduced.

    (iii) The percentage decrease in consumption for the latter part of the voyage.

    (iv) The percentage increases in time for this latter period.

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    $$D=800nm$$

    $$V_1=?knots$$

    $$V_2=0.8V_1$$

    $$DC_1=Cons\:per\:day\:at\:V_1$$

    $$DC_2=Cons\:per\:day\:at\:V_2$$

    $$DC_1-DC_2=42t$$

    $$C=\left(0.136+0.001V^3\right)\:t\h$$

    $$Therefore\:DC=24\left(0.136+0.001V^3\right)\:tonnes\:per\:day$$$$42=24\left\lbrack\left(0.136+0.001V_{1^{}}^3\right)-\left(0.136_{}+0.001\left(0.8V_1^3\right)\right)\right\rbrack$$

    $$42=24\left(0.136+0.001V_1^3-0.136-0.512\times10^{-3}\times V_1^3\right)$$

    $$42=24\left(0.001V_1^3-0.512\times10^{-3}\times V_1^3\right)$$

    $$42=24\left(0.000488V_1^3\right)$$

    $$V_1=\sqrt[3]{\frac{42}{24\times0.000488}}$$

    $$V_1=15.31\:knots$$

    $$V_2=0.8\times V_1$$

    $$V_2=0.8\times15.31$$

    $$V_2=12.245\:knots$$

    $$\left(i\right)\:Reduced\:cons\:per\:day\:=\:\left(0.136+0.001V_2^3\right)\times24$$

    $$=\left(0.136+0.001\times12.45^3\right)\times24$$

    $$=49.57\:tonnes\:per\:day$$

    $$Time\:taken\:for\:complete\:voyage\:of\:800nm$$

    $$at\:V_2=\frac{800}{12.245\times24}=2.72\:days$$

    $$Consumption\:=\:2.72\times49.57=134.93t\:\left(at\:reduced\:speed\right)$$

    $$\left(ii\right)\:Fuel\:onboard=134.93+50$$

    $$=184.93t\:\left(after\:speed\:reduction\right)$$

    $$DC_1=DC_2+42$$

    $$DC_1=49.57+42$$

    $$DC_1=91.57t$$

    $$Time\:taken\:for\:V_1=\frac{800}{15.31\times24}$$

    $$=2.178days$$

    $$Cons\:at\:V_1=91.57\times2.178$$

    $$=199.38t$$

    $$\left(iii\right)\:\%\:reduction\:in\:cons=\frac{199.38-134.93}{199.38}$$

    $$=32.32\%$$

    $$\left(iv\right)\:\%\:increase\:in\:time=\frac{2.72-2.178}{2.178}$$

    $$=24.88\%$$

    Q10 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Explain how to distinguish between list and loll and describe how to return the ship to the upright in each case. (6)

    (b) A ship of 5000 tonne displacement has a double bottom tank 12m long. The 1/2 breadths of the top of the tank are 5, 4 and 2m respectively. The tank has a watertight centreline division. Calculate the free surface effect if the tank is partially full of fresh water on one side only. (10)

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    Part (a)

    Distinguishing between List and Loll:

    1. List is caused by Uneven distribution of weight within the ship. List will have the below conditions:

    • GM is positive (GM > 0).
    • The centre of gravity (G) is off-centre.
    • The vessel is at equilibrium but inclines to one side due to uneven weight distribution, even without any external forces acting on it.
    • The vessel rolls around the angle of list.

    Correction:

    • Redistribute the weight evenly to bring the centre of gravity (G) back in line with the centerline and metacentric height (M).

    2. Loll is caused by High centre of gravity (G) leading to negative GM and is exacerbated by external forces, free surface effects, or poor distribution of weights. LOLL will have the below conditions:

    • GM is negative (GM < 0).
    • The centre of gravity (G) is on the centerline but too high, making the vessel inherently unstable.
    • The vessel flops or inclines to one side at an angle of loll, and it can incline equally to either side.
    • The vessel rolls unstably around the angle of loll.

    Correction:

    • Reduce the centre of gravity by ballasting bottom tanks or removing weight from higher levels.
    • Minimize free surface effects by reducing the breadth of free surfaces in tanks.

    Q1 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 4x

    Describe the preparation necessary before the application (in dry dock) of sophisticated or approved long life coating to the underwater surface of the hull.

    (a) State the significance of the roughness profile. (8)

    (b) List the different sophisticated coating which are available. (8)

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    The preparation of a ship's underwater hull before applying a long-life coating in a dry dock involves a three-step process. This process addresses the removal of contaminants and the creation of a suitable surface profile.

    (i) Washing: The hull surface must be thoroughly cleaned to remove all marine growth (algae, slime, etc.), accumulated salts, dirt, grease, and oil. High-pressure freshwater washing is the standard method for this initial cleaning. The goal is to present a clean substrate for subsequent stages.

    (ii) Blasting: Abrasive blasting is the preferred method for removing rust, defective paint, and any remaining contaminants. This process achieves a bare metal surface, essential for proper adhesion of the new coating. The extent of blasting (localized or full hull) depends on the condition of the existing surface. The intensity and type of abrasive used are carefully controlled to achieve the desired surface roughness profile.

    (iii) Primer Application: After blasting, the surface is again cleaned to remove any blasting debris. A primer coat is then applied to provide corrosion protection and to create an ideal surface for the subsequent topcoat adhesion. This primer acts as an intermediary layer, enhancing the bond between the substrate and the long-life coating system.

    Part (a)

    Significance of Roughness Profile:

    The roughness profile of the prepared hull surface impacts the performance of the applied coating and the overall operational efficiency of the vessel. A rough surface increases frictional resistance as the vessel moves through the water. This increased drag translates to higher power requirements for propulsion, leading to increased fuel consumption and operational costs. Furthermore, greater surface roughness contributes to increased carbon emissions, a concern under current MARPOL regulations. Therefore, a controlled and optimized roughness profile is essential for minimizing frictional resistance, reducing fuel consumption and emissions, and maximizing the longevity of the hull coating.

    Part (b)

    Sophisticated hull coating systems comprise multiple layers designed to provide corrosion protection and antifouling properties.

    Wash Primer/Pretreatment Primer/Metal Conditioning Primer:

    • These primers act as a base layer, improving adhesion of subsequent layers. Common types include epoxy primers pigmented with iron oxide and corrosion inhibiting pigments (zinc and calcium phosphates, although zinc content is minimized due to safety concerns).

    Anticorrosive Coating:

    • This layer primarily provides corrosion protection to the underlying metal. Two-component epoxies, coal tar epoxies, and epoxy or polyester coatings incorporating glass flakes are frequently employed. Glass flakes enhance mechanical strength and water vapor impermeability.

    Antifouling Coating:

    • This layer prevents the attachment of marine organisms (fouling). Historically, tin-based paints were used, but due to environmental regulations, they have been largely replaced by copper-based, silicone-based, or non-TBT (Tributyltin) self-polishing antifouling coatings. These newer coatings typically use seawater-soluble polymers. The number of antifouling layers applied (two or three) depends on the specific system chosen and required longevity.
    Q2 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    A rudder of a vessel requires extensive welding repairs and as Chief Engineer you are requested to supervise.

    (a) Suggest a suitable type of welding process. (6)

    (b) State, with reasons, FOUR common welding defects. (5)

    (c) State what tests may be carried out before returning the rudder to service. (5)

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    As Second Engineer, I would oversee the extensive welding repairs required for the vessel's rudder using the following plan:

    Part (a)

    Suitable Welding Process:

    Manual Metal Arc Welding (MMAW), also known as Shielded Metal Arc Welding (SMAW), is the most suitable process for this repair. The reasons are threefold:

    • MMAW is highly portable, allowing for on-site repair within the drydock. The process is adaptable to various welding positions (downhand, overhead, horizontal, vertical) – a necessity given the complex geometry of a rudder.
    • Assuming the rudder is constructed from standard steel, MMAW using readily available flux-coated electrodes provides good control, arc stability, and penetration. The flux coating protects the weld pool from atmospheric contamination during cooling.
    • MMAW requires relatively simple equipment and is less demanding in terms of operator skill compared to other processes like TIG or MIG. This translates to cost-effectiveness and allows for a wider pool of qualified welders.
    • If cast steel components are present, pre-heating will be necessary to minimize stress cracking, and specialized electrodes suited for the specific cast steel grade must be selected.

    During welding by the metal arc process, the following points must be observed:

    • Electrode Consumption Rate
    • Penetration
    • Slag Control
    • Arc Length and Sound
    Part (b)

    Four Common Welding Defects:

    1. Undercut: A groove formed along the edge of the weld bead, weakening the joint. Caused by excessive current, incorrect electrode angle, excessive travel speed, or improper electrode manipulation.

    2. Overlap: Molten weld metal flows over the parent metal without proper fusion. Caused by low current, slow travel speed, excessive arc length, or improper joint preparation.

    3. Slag Inclusion: Trapped slag within the weld metal, reducing its strength and potentially causing cracking. Caused by insufficient cleaning between passes, incorrect current, long arc length, slow travel speed, or too large an electrode diameter.

    4. Incomplete Penetration: The weld does not fully fuse the joint faces, resulting in a weak joint. Caused by insufficient current, incorrect joint preparation (too small a root gap or bevel angle), excessive travel speed, or too large an electrode diameter.

    Part (c)

    Tests Before Returning to Service:

    • A thorough visual examination of all welds to identify any surface defects like cracks, porosity, or lack of fusion.
    • NDT methods such as Magnetic Particle Inspection (MPI) or Dye Penetrant Inspection (DPI) will be employed to detect subsurface flaws that may not be visible during visual inspection. The specific NDT method chosen will depend on the type of steel and the accessibility of the weld areas.
    • The repaired rudder will undergo a hydrostatic pressure test. This involves filling the rudder with a water head of 2.46 meters and observing for any leaks. This confirms the watertight integrity of the welds and the overall rudder structure.
    Q3 (10 Marks) General πŸ”₯ Repeated 7x

    List SIX hazards that arise with the carriage of liquefied gas in bulk. Describe, with the aid of a sketch, the details of construction of a prismatic cargo tank within a gas carrier designed to carry liquefied gas (LPG). (16)

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    Hazards of Carriage of Liquefied Gas in Bulk and Construction of a Prismatic Cargo Tank

    Part (a)

    Six Hazards Associated with the Carriage of Liquefied Gas in Bulk

    1. Flammability and Explosion
      • Liquefied gases such as LPG vaporise rapidly when released.
      • The vapour can mix with air and form a highly flammable or explosive atmosphere, creating a serious risk of fire or explosion.
    2. Toxicity
      • Some liquefied gas cargoes, such as ammonia and vinyl chloride monomer, are highly toxic.
      • Exposure can cause serious poisoning through inhalation, ingestion or skin absorption.
    3. Asphyxiation
      • LPG vapours are generally heavier than air.
      • In the event of a leak, the vapour can collect in low-lying areas such as pump rooms and hold spaces, displacing oxygen and creating an asphyxiation hazard.
    4. Frostbite and Cold Burns
      • Liquefied gas cargoes are carried at very low temperatures, with fully refrigerated LPG being carried at approximately βˆ’50Β°C.
      • Contact with the liquid cargo or uninsulated pipes and equipment can cause severe frostbite and cold burns.
    5. Brittle Fracture
      • Ordinary ship hull steel, such as mild steel, can become brittle at very low temperatures.
      • If cold cargo leaks and comes into contact with unsuitable hull steel, it may cause cracking and catastrophic structural failure.
    6. Sloshing
      • Partially filled tanks have a free surface, allowing the liquid cargo to move violently with the ship's motion.
      • This can produce large dynamic impact loads on the tank walls and internal pump towers, potentially causing structural damage.
    Part (b)

    Construction of a Prismatic Cargo Tank for LPG

    Fully refrigerated LPG is typically carried in Independent Type A prismatic cargo tanks. These are self-supporting tanks that are independent of the ship's hull structure and do not contribute to the overall structural strength of the vessel.

    Sketch – Typical Prismatic Cargo Tank Arrangement

    Ship's Hull / Hold Space

    β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”

    β”‚ Secondary Barrier / Hull β”‚

    β”‚ β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚

    β”‚ β”‚ Thermal Insulation β”‚ β”‚

    β”‚ β”‚ β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚ β”‚

    β”‚ β”‚ β”‚ β”‚ β”‚ β”‚

    β”‚ β”‚ β”‚ LPG CARGO β”‚ β”‚ β”‚

    β”‚ β”‚ β”‚ PRISMATIC TANK β”‚ β”‚ β”‚

    β”‚ β”‚ β”‚ β”‚ β”‚ β”‚

    β”‚ β”‚ β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ β”‚ β”‚

    β”‚ β”‚ Primary Barrier β”‚ β”‚

    β”‚ β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ β”‚

    β”‚ Load-bearing supports β”‚

    β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜

    β”‚

    Double Bottom

    1. Shape and Structure

    • The tank is prismatic, meaning it is generally box-shaped with chamfered/angled top and bottom corners.
    • This shape allows the tank to closely follow the contours of the ship's inner hull and therefore maximises the available cargo capacity.
    • The tank is internally reinforced with frames and stiffeners to maintain structural integrity.
    • A centreline longitudinal bulkhead, together with transverse wash bulkheads, may be provided to strengthen the tank and reduce liquid movement and sloshing.

    2. Materials

    • LPG is carried at low temperatures, approximately βˆ’50Β°C for fully refrigerated LPG.
    • To prevent brittle fracture at these temperatures, the primary barrier/tank is constructed from suitable low-temperature-resistant materials, typically fine-grained carbon-manganese steel.

    3. Tank Supports and Chocks

    The cargo tank operates at a substantially different temperature from the ship's hull and therefore undergoes thermal expansion and contraction.

    • The tank is supported on the double bottom by load-bearing insulation blocks.
    • These may be made from specialised hardwood such as AzobΓ© or suitable synthetic materials.
    • The supports provide the necessary load-bearing capacity while reducing thermal transfer and structural stresses.
    • Anti-roll, anti-pitch and anti-flotation keys/chocks secure the tank against movement caused by the ship's motion.
    • At the same time, the arrangement allows the tank to expand and contract freely due to temperature changes.

    4. Insulation

    • The outside of the primary barrier is provided with high-efficiency thermal insulation.
    • Its purpose is to maintain the required low cargo temperature and minimise heat ingress and cargo boil-off.
    • Sprayed polyurethane foam (PUF) is commonly used as the insulation material.

    5. Secondary Barrier

    • Under the IGC Code, Type A tanks are required to have a complete secondary barrier capable of containing leaked cargo for up to 15 days, preventing the cold cargo from coming into contact with the outer hull.
    • In LPG carriers, the ship's inner hull is normally used as the secondary barrier.
    • The inner hull is also constructed from suitable fine-grained low-temperature steel so that it can withstand the low temperature if the primary cargo tank fails.
    • The space between the primary tank and secondary barrier, known as the hold space, is maintained with inert gas or dry air as applicable.
    Q4 (10 Marks) General

    With regard to the carriage of crude oil and its associated products:

    (a) (i) Sketch and describe the operation of an explosimeter suitable for testing pump rooms or tank

    (ii) state why false readings are likely to be given by the explosimeter.

    (b) State the publication that give guidance on safety.

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    Part (a)

    An Explosimeter is used to test for the presence of flammable gases or vapors in spaces like pump rooms or cargo tanks.

    • Consists of a Wheatstone bridge circuit powered by a battery.
    • Contains three fixed resistances and one resistance that is a hot filament within the combustion chamber.
    • A galvanometer measures the current flow in the circuit and is scaled to indicate gas concentration as a percentage of the Lower Flammable Limit (LFL) or in parts per million (ppm).
    • An aspirator bulb and flexible tube are used to draw a gas sample into the combustion chamber.

    Working:

    1. A gas sample is drawn into the combustion chamber using the aspirator bulb.
    2. If flammable gas is present, it burns on the hot filament, causing a rise in the filament’s temperature.
    3. The rise in temperature increases the filament’s resistance, unbalancing the Wheatstone bridge circuit.
    4. This unbalance generates a current, which is indicated on the galvanometer. The scale converts this reading into % LFL or ppm.

    (ii) Reasons for False Readings by an Explosimeter:

    1. If the oxygen content in the sampled space is insufficient or if inert gas is present, combustion on the filament may not occur, leading to inaccurate readings.
    2. The explosimeter is designed to detect gas up to the Lower Flammable Limit (LFL). If the gas concentration exceeds the instrument’s range, false or misleading readings may occur.
    3. If the instrument is not purged before use, residual gas from a previous reading may result in an inaccurate measurement.
    4. A drained or insufficiently charged battery can affect the accuracy of the readings.
    Part (b)

    Safety Guidance Publications:

    • IMO Resolution A.1050(27): Revised Recommendations for entering enclosed spaces aboard ships. This resolution provides detailed guidance on safe entry procedures into enclosed spaces.
    • SOLAS Chapter III, Regulation 19 (as of January 1st, 2015): This regulation addresses enclosed space drills, requiring them to be conducted every two months.
    • SOLAS Regulation XI-1/7 (as of July 1st, 2016): This regulation mandates that vessels carry portable instruments capable of measuring oxygen concentration, flammable gases, hydrogen sulfide (H2S), and carbon monoxide (CO).
    • Code of Safe Working Practices (COSWP), Chapter 17 is dedicated to "Enclosed Spaces", providing detailed guidance on the identification, risks, and safe procedures associated with entering and working in such environments.
    Q5 (10 Marks) Hull Construction πŸ”₯ Repeated 5x

    Give a reasoned opinion as to the validity of the following assertions concerning ship structure:

    (a) Crack propagation in propeller shaft 'A' brackets or spectacles frames is indicative of inadequate scantlings and strength (8)

    (b) The adequate provision of freeing ports is as critical to seaworthiness as watertight Integrity. (8)

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    Part (a)

    Crack propagation in propeller shaft 'A' brackets or spectacle frames - is it indicative of inadequate scantlings and strength?

    This assertion is only partly valid. Cracks in A-brackets or spectacle frames (the shaft brackets supporting the propeller shaft) are more commonly the result of fatigue due to fluctuating loads, stress concentrations at the bracket-to-hull connection, and the vibration and whipping of the shaft, rather than simply inadequate scantlings. The brackets are subject to severe alternating loads from the propeller and the shaft, and cracks typically initiate at stress raisers (sharp corners, weld toes, the bracket arm-to-hull connection) and propagate under fatigue. While inadequate scantlings or poor design (insufficient section, poor connection, sharp notches) can contribute, the primary cause is usually fatigue and stress concentration, aggravated by vibration, corrosion and the dynamic loads of the propeller. Hence the assertion is not fully valid: crack propagation is more indicative of fatigue and stress concentration than of inadequate strength alone, and the design should address the fatigue life, the connection detail and the avoidance of stress raisers, as well as the scantlings.

    Part (b)

    The adequate provision of freeing ports is as critical to seaworthiness as watertight integrity.

    This assertion is largely valid. Freeing ports (openings in the bulwark that allow water shipped on deck to drain overboard) are essential to seaworthiness because, if they are inadequate, water accumulating on the deck cannot drain, which:

    • increases the free-surface effect and the weight of water on deck, reducing stability and increasing the risk of capsize;
    • increases the deck load and the risk of structural damage;
    • reduces the reserve buoyancy and can lead to the ship becoming unstable.

    Watertight integrity (the ability of the hull and its openings to keep water out) is equally critical to seaworthiness, as it prevents flooding and loss of buoyancy. Both are essential: watertight integrity keeps water out, while freeing ports remove water that is shipped on deck. If either is inadequate, the ship's seaworthiness is compromised. Hence the assertion is valid - freeing ports are as critical to seaworthiness as watertight integrity, because they maintain the stability and buoyancy of the ship by removing deck water.

    Q6 (10 Marks) Ship Stability

    (a) With respect to Buoyancy of a vessel

    What do you understand by reserve buoyancy what happen if the lost buoyancy is greater than the reserve buoyancy?

    (b) A triangular bulkhead ty 7 m wide at the top and has a vertical depth of B m. Calculate the load on the bulkhead and the position of centre of pressure if the bulkhead is flooded with sea water on only side:(10)

    (i) To the top edge

    (ii) With 4 m head to the top edge.

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    Part (a)

    Reserve Buoyancy

    Reserve buoyancy is the watertight volume above the waterline of a vessel. It represents the potential buoyancy that can be utilised to keep the ship afloat when additional weight is added or if some buoyancy is lost due to damage (e.g., bilging).

    When a mass is added to the ship or buoyancy is lost (e.g., due to flooding of a compartment), the reserve buoyancy is converted into active buoyancy by an increase in draught.

    If the lost buoyancy (due to flooding or damage) is greater than the reserve buoyancy, the vessel will no longer have sufficient buoyant force to counteract its weight, causing it to sink.

    Q7 (10 Marks) Ship Stability πŸ”₯ Repeated 2x

    (a) Describe stability requirements for dry-docking. (6)

    (b) An oil tanker 160m long and 22m beam floats at a draught of 9m in seawater. Cw is 0.865. The midship section is in the form of a rectangle with 1.2m radius at the bilges. A midship tank 10.5m long has twin longitudinal bulkheads and contains oil of 1.4m3/t to a depth of 11.5m. The tank is holed to the sea for the whole of its transverse section. Find the new draught. (10)

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    Part (a)

    For safe dry-docking, a ship must meet two key stability requirements:

    • Positive GM (Metacentric Height): The ship needs a positive GM. GM is the distance between the centre of gravity (G) and the metacentre (M). A positive GM indicates inherent stability; the ship will right itself if tilted. During dry-docking, the loss of buoyancy as the ship rests on the blocks reduces GM. Insufficient positive GM increases the risk of the ship heeling over or capsizing.
    • Trim by the stern: The vessel should be trimmed slightly by the stern (aft end lower than the bow) to ensure the aft end sits on the keel blocks first. This controlled settling minimises the risk of instability during the docking process. An even keel is generally preferred for the initial floating condition before the dry-docking procedure begins.
    Q8 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 7x

    (a) What is meant by the Admiralty Coefficient and the Fuel Coefficient? (6)

    (b) A ship of 14900 tonne displacement has a shaft power of 4460 kW at 14.55 knots. The shaft power is reduced to 4120 kW and the fuel consumption at the same displacement is 541 kg/h. Calculate the fuel coefficient for the ship. (10)

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    The Admiralty Coefficient (C) is a method for estimating the propulsion power needed for a newly built ship. It's considered relatively constant for a given ship design. The formula is:

    $$C=\frac{\Delta^{2/3}\times V^3}{BP}$$

    Where:

    • C = Admiralty Coefficient
    • Ξ” = Displacement in tonnes (weight of the ship when fully loaded)
    • V = Speed in knots
    • BP = Brake power in kilowatts (power delivered by the ship's engine)

    A higher Admiralty Coefficient indicates a more efficient ship design, meaning it requires less power to achieve a given speed. Values typically range from 350 to 600.

    Fuel Coefficient:

    The Fuel Coefficient (F.C.) is used to calculate a ship's daily fuel oil consumption. The formula is:

    $$Daily\:fuel\:oil\:consumption\:=\:\frac{\Delta^{2/3}\times V^3}{FC}$$

    Where:

    • F.C. = Fuel Coefficient
    • Ξ” = Displacement in tonnes
    • V = Speed in knots

    The Fuel Coefficient can vary significantly, with typical values ranging from 40,000 to 120,000. A higher Fuel Coefficient implies greater fuel efficiency (lower daily fuel consumption) for a given speed and displacement.

    Part (b)

    $$admiraty\:coefficient\:\left(C\right)=\:\frac{\Delta^{2/3}V^3}{Shaft\:power}=\frac{\Delta^{\frac23}\times V^3}{SP}$$

    $$\frac{SP_1}{SP_2}=\frac{V_1^3}{V_2^3}$$

    $$\frac{4460}{4120}=\frac{14.55^3}{V_2^3}$$

    $$V_2=14.17kntos$$

    $$Fuel\:consumption\:per\:hour=541\operatorname{\mathrm{\:kg}}\:per\:hour$$

    $$Fuel\:consumption\:per\:day\:=\:541\times24=12.98t\:per\:day$$

    $$Fuel\:coefficient=\frac{\Delta^{\frac23}\times V_2^3}{Fuel\:consumption\:per\:day}$$

    $$=\:\frac{14900^{\frac23}\times14.17^3}{12.98}$$

    $$=132726.9$$

    Q9 (10 Marks) Hull Construction

    (a) What is Prismatic Co-efficient (CP). Derive the formula CP = Cb/cm where Cb = Coefficient of fineness and Cm = midship section area co-efficient. (6)

    (b) The 1/2 ordinates of a waterplane at 15m intervals, commencing from aft, are 1, 7, 10.5, 11, 11, 10.5, 8, 4 and 0m. Calculate:

    (a) TPC

    (b) Distance of the centre of flotation from midships

    (c) Second moment of area of the waterplane about a transverse axis through the centre of flotation.

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    Part (a)

    Prismatic coefficient (Cp) and its derivation.

    The prismatic coefficient Cp is the ratio of the volume of displacement to the volume of a prism having the same length as the ship and a cross-section equal to the midship section area:

    Cp = Volume of displacement/(L x Am)

    where L is the length and Am the midship section area.

    Derivation: The block coefficient Cb = Volume/(L x B x d), where B is the beam and d the draught. The midship section coefficient Cm = Am/(B x d). Therefore:

    Cb/Cm = [Volume/(L x B x d)] / [Am/(B x d)] = Volume/(L x Am) = Cp.

    Hence Cp = Cb/Cm. The prismatic coefficient indicates the fullness of the ends of the ship: a high Cp means full ends, a low Cp means fine ends. It is used in resistance calculations.

    Part (b)

    Waterplane calculations.

    The half-ordinates of a waterplane at 15 m intervals, commencing from aft, are 1, 7, 10.5, 11, 11, 10.5, 8, 4 and 0 m. There are 9 ordinates, so the waterplane length is 8 x 15 = 120 m.

    (i) TPC.

    Area A = 2 x (h/3)[y0 + y8 + 4(y1+y3+y5+y7) + 2(y2+y4+y6)]

    = 2 x (15/3)[1 + 0 + 4(7+11+10.5+4) + 2(10.5+11+8)]

    = 10[1 + 4x32.5 + 2x29.5] = 10[1 + 130 + 59] = 10 x 190 = 1900 m2.

    TPC (salt) = A x 1.025/100 = 1900 x 1.025/100 = 19.48 t/cm.

    (ii) Distance of the centre of flotation from midships.

    First moment of area about the after perpendicular:

    M = 2 x (h/3) x sum of (weighted y x distance). Using station distances 0,15,30,...,120 m with Simpson weights (1,4,2,4,2,4,2,4,1):

    weighted sum = 1x0 + 4x(7x15) + 2x(10.5x30) + 4x(11x45) + 2x(11x60) + 4x(10.5x75) + 2x(8x90) + 4x(4x105) + 1x0

    = 0 + 420 + 630 + 1980 + 1320 + 3150 + 1440 + 1680 = 10620.

    M = 2 x 5 x 10620 = 106,200 m3.

    Distance of centroid from AP = M/A = 106,200/1900 = 55.89 m. Midships is at 60 m from AP, so the centre of flotation is 60 - 55.89 = 4.11 m aft of midships.

    (iii) Second moment of area about a transverse axis through the centre of flotation.

    Second moment about AP: I_AP = 2 x (h/3) x sum of (weighted y x distance^2):

    weighted sum = 1x0 + 4x(7x225) + 2x(10.5x900) + 4x(11x2025) + 2x(11x3600) + 4x(10.5x5625) + 2x(8x8100) + 4x(4x11025) + 1x0

    = 0 + 6300 + 18900 + 89100 + 79200 + 236250 + 129600 + 176400 = 735,750.

    I_AP = 2 x 5 x 735,750 = 7,357,500 m4.

    Transfer to the centre of flotation: I_CF = I_AP - A x (55.89)^2 = 7,357,500 - 1900 x 3123.7 = 7,357,500 - 5,935,000 = 1,422,500 m4.

    Answer: TPC = 19.48 t/cm; LCF = 4.11 m aft of midships; I about the transverse axis through CF = about 1.42 x 10^6 m4.

    Q10 (10 Marks) Ship Stability πŸ”₯ Repeated 2x

    The following data are available from the hydrostatic curves of a vessel.

    Draught(m): 4.9 5.2

    KB(m): 2.49 2.61

    KM(m): 10.73 10.79

    I(m4): 65250 68860

    Calculate the TPC at a draught of 5.05 m. (16)

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    TPC at a draught of 5.05 m from hydrostatic data.

    Data: Draught 4.9 m: KB 2.49, KM 10.73, I 65250 m4. Draught 5.2 m: KB 2.61, KM 10.79, I 68860 m4.

    BM = KM - KB. At 4.9 m: BM = 10.73 - 2.49 = 8.24 m. At 5.2 m: BM = 10.79 - 2.61 = 8.18 m.

    Volume of displacement V = I/BM. At 4.9 m: V = 65250/8.24 = 7919 m3. At 5.2 m: V = 68860/8.18 = 8418 m3.

    Displacement = V x 1.025. At 4.9 m: 8117 t. At 5.2 m: 8629 t.

    The waterplane area between these draughts is the rate of change of volume with draught: A = (V2 - V1)/(draught change) = (8418 - 7919)/(5.2 - 4.9) = 499/0.3 = 1663 m2.

    TPC = A x 1.025/100 = 1663 x 1.025/100 = 17.05 t/cm.

    Answer: the TPC at a draught of 5.05 m is about 17.0 t/cm.

    Q1 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 3x

    With reference to dry docking, define the responsibilities of the Second Engineer: (16)

    (a) Prior to docking

    (b) Whilst the vessel is in dry dock

    (c) Prior to flooding and leaving the dock

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    (a) Prior to Docking:

    Preliminary Preparation:

    • Review various plans, manuals, and previous drydock reports for reference.
    • Prepare a detailed repair list and ensure all required spares are accounted for.
    • Conduct an inventory of spares and requisition necessary items.
    • Gather required tools, including special tools like poker gauges, trammel gauges, gagging tools, etc.
    • Mark all overboard valves, NC (normally closed) and NO (normally open) valves clearly for easy identification.
    • Allocate jobs to team members and discuss the time schedule.
    • Conduct a safety meeting to highlight dry dock hazards and necessary precautions.

    Before Entering Dry Dock:

    • Identify the power requirements and machinery to be operational during docking.
    • Check the shore connection box for proper functionality.
    • Record the soundings of FO (Fuel Oil), LO (Lubricating Oil), and DO (Diesel Oil) tanks.
    • Discharge contents from clean drain tanks and sewage tanks.
    • Carry out Economizer soot-blowing.
    • Change over Main Engine, Diesel Generators, and Boiler to Low Sulfur Gas Oil (LSGO).
    • Stop and clean purifiers.
    • Ensure the low sea chest is open and the high sea chest is shut.
    • Keep firefighting appliances (FFA) on standby.
    • Shut down all non-essential machinery, including MGPS (Marine Growth Prevention System) and FWG (Fresh Water Generator).

    (b) Whilst the Vessel is in Dry Dock

    Upon Arrival:

    • Connect shore power and supplies after ensuring safety checks are completed.
    • Start necessary equipment like cooling water, air compressors, air conditioning, and fridge compressors.
    • Check for jobs assigned by dry dock personnel and prepare accordingly.
    • Attend a safety meeting with dry dock personnel to understand local safety rules and procedures.

    During Dry Docking::

    • Oversee and assist in:
      • Cleaning and inspecting the hull, rudder, sea chest, anodes, and propeller.
      • Measuring propeller drop, checking rudder clearances, and inspecting the stern tube bearing and seal.
      • Servicing underwater valves and overboard valves.
      • Inspecting anchor and cables conditions.
      • Overhauling deck machinery, cranes, elevators, and engine room equipment such as the Main Engine, Diesel Generators, Boiler, and Economizer.
      • Renewing pipes and valves as needed.
      • Performing electrical equipment maintenance and surveys.
      • Supervising service engineers for specific repair jobs.
    • Ensure tank cleaning, welding, and other repair works are completed according to the plan.
    • Run the standby diesel generator daily after starting the priming pump.

    (c) Before Flooding and Leaving the Dry Dock

    Final Checks:

    • Verify that all underwater fittings and drain plugs are securely in place.
    • Ensure all machinery has been boxed back and is ready for operation.
    • Check for any leakage in stern tube seals.
    • Take tank soundings to confirm proper levels.
    • Confirm the proper operation of all underwater valves, overboard valves, sea chests, and vents.
    • Inspect the stern tube tank for any irregularities.

    System Restart:

    • Switch back to ship's power after confirming all systems are functional.
    • Test the proper operation of all machinery and systems to ensure the ship is fully operational before leaving the dock.
    Q2 (10 Marks) Ship Resistance & Propulsion

    With reference to rudder carrier bearings fitted to Merchant ships:

    (a) Sketch a bearing designed to transfer the full weight of the rudder to the ships structure. (6)

    (b) Describe the consequences if the rudder carrier bearing surfaces become heavily scored. (6)

    (c) Describe the consequences of and the action to be taken, if the carrier shatters. (4)

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    Part (a)

    Rudder carrier bearing:

    Part (b)

    Consequences if the rudder carrier bearing surfaces become heavily scored:

    • Increased vibration and operational instability.
    • Generation of noise during rudder movement.
    • Sluggish rudder movement due to impaired smooth operation of the bearing.
    • Overheating of the bearing, which may lead to further damage or failure.
    • In the worst case, sluggish to no movement at all, risking loss of rudder functionality.
    Part (c)

    Consequences if the carrier shatters:

    • Increased noise and vibration.
    • Rudder movement becomes sluggish or completely immobilised.
    • Risk of additional damage to associated components like rudder stock and steering gear.

    Actions:

    1. Supplement extra greasing to mitigate friction and reduce further damage.
    2. Use eye bolts and chain blocks to take the load off the rudder stock and minimize pressure on the shattered carrier.
    3. Avoid sudden or sharp rudder movements to prevent exacerbating the issue.
    4. Arrange for renewal or replacement of the rudder carrier bearing at the earliest opportunity.
    Q3 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 9x

    With reference to Underwater Inspection in lieu of Dry docking (UWILD):

    (a) Explain in detail, how an underwater survey is carried out. (6)

    (b) State the requirements to be fulfilled before an underwater survey is acceptable to the surveying authority. (6)

    (c) Construct a list of the items in order of importance that the underwater survey authority

    should include. (4)

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    (a) An underwater survey, also known as Underwater Inspection in Lieu of Dry Docking (UWILD), involves a systematic and detailed examination of a vessel’s hull and underwater components while the ship remains afloat. The procedure includes the following steps:
    • The shipowner sends a request to the classification society surveyor, indicating the intention to perform an underwater survey.
    • A detailed plan of the ship's external hull features is submitted, showing the location of frames, bulkheads, welding lines, openings, etc.
    • The hull surface is cleaned before the survey to remove any marine growth or fouling that could obscure the inspection.
    • A diving company, approved by the classification society, is appointed to conduct the underwater inspection.
  • A self-propelled survey vehicle equipped with the following tools is used:
    • Long-Range Light TV Camera to aid in steering and checking hull deterioration.
    • High-resolution colour TV Camera to provide a close-up view of the hull coating and welded seams.
    • 35mm Still Camera to capture still images.
    • Ultrasonic Probe for measuring plate thickness.
    • Depth Meter and Speed Indicator to provide accurate data on the vehicle's depth and movement.
    • Umbilical Cable to connect the survey vehicle to the survey boat, transmitting power and information.
  • The survey boat is to be equipped with:
    • A control console with TV monitors.
    • Plate thickness printouts.
    • Audio and video cassette recorders.
    • Playback units.
    • Diver communication systems.
    • Vehicle control systems and associated instruments.
  • Operation:
    • The survey vehicle is taken underwater by a diver to the survey starting point.
    • Using TV monitors and shell expansion plans as guides, the vehicle is navigated over the hull, focusing on the bottom structure, sides, stern frame, rudder, propeller, bilge keel, and hull openings.
    • All images, data, and information are recorded and transmitted back to the survey boat.
    • Detailed pictures of the stern frame, rudder, propeller, bilge keel, and hull openings are captured.
    • Divers are deployed to measure stern tube bearing wear, pintle clearance, and inspect stern seals, anodes, and rudder stock palm coupling bolts.
    • All recorded video and audio, including conversations between the surveyor and drivers, thickness printouts, measurements, and pictures are analyzed to determine the detailed underwater condition of the vessel.
    (b) Before an underwater survey is accepted by the surveying authority, the following conditions must be met:

    The vessel's owner submits a request to the surveyor, including:
    • The proposed date and location for the survey.
    • General information about the diving company.
    • A declaration that the vessel has not suffered any damage due to grounding, collision, or other incidents.
    The ship's master or owner’s representative must provide a declaration confirming:
    • Any suspected or actual damage to the hull since the last dry-docking.
    • The underwater portion of the hull is protected by a suitable paint scheme that is of adequate thickness and remains valid until the next dry-dock.

    • The survey site should be in a protected area with calm and clear water, ensuring good underwater visibility. Attention must be given to the effects of currents and tides.
    • The hull must be clean for the external survey. The surveyor must be satisfied with the method and quality of the pictorial presentation, ensuring that it provides a reliable assessment of the hull's condition.
    • The underwater examination should be conducted by an approved diving company using closed-circuit TV and two-way communication, which can be monitored by the surveyor.
    • The vessel should be in as light an operating condition as possible to facilitate the survey.
    • Means must be available for the surveyor to examine the outside shell plating above the waterline.
    • Any required repairs identified during the survey must be carried out to the satisfaction of the attending surveyor.
    (c) While the importance of each item may vary depending on the vessel and its specific requirements, below is a list of items that should be included in an underwater survey in order of importance:
    • Underwater Hull: General condition of the hull below the waterline.
    • Bottom and Shell Plating: Inspection for corrosion, damage, and fouling.
    • Shell Openings: Examination of openings such as sea chests, drain plugs, and overboard discharge points.
    • Stern Tube Oil Leaks: Check for leaks around the stern tube.
    • Propeller Blade: Inspection for damage, wear, and fouling.
    • Rudder: Inspection for damage, wear, and clearances.
    • Sea Chest Opening and Grating: Examination for blockages, damage, and fouling.
    • Anodes: Check the condition and effectiveness of cathodic protection anodes.
    • Bilge Keel: Inspection for damage and fouling.
    • Drain Plugs: Ensure all drain plugs are secure and in good condition.
    • Overboard Valve Openings: Check for proper operation and condition.
    • Forward Area: Inspection for any damage due to anchor and chain movement.
    Q4 (10 Marks) Hull Construction πŸ”₯ Repeated 2x

    With reference to membrane tanks for the carriage of liquefied gas at very low temperatures.

    (a) Describe with a sketch one method of building up the insulation. (6)

    (b) State which alloy is used for the membrane and the reason (6)

    (c) Explain why a secondary barrier is installed. (4)

    (i) Longitudinally

    (ii) Transversely

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    Part (a)

    Membrane tanks

    for liquefied gas carriage are built with a double hull throughout the cargo length. The insulation system includes the following:

    1. Primary Barrier: A thin membrane made of INVAR (36% Nickel, 64% Iron) with a thickness of 0.5 to 1.2 mm, forming the primary containment layer for the liquefied gas.
    2. Primary Insulation box: A 200 mm thick layer of granulated Perlite insulation packed in plywood boxes, surrounding the primary barrier. Perlite is siliconised to make it impervious to moisture.
    3. Secondary Barrier: A secondary membrane made of the same INVAR material is installed to prevent cargo leakage in case of primary barrier failure.
    4. Secondary Insulation box: Another 200 mm thick layer of granulated Perlite is placed above the secondary barrier to provide additional insulation and prevent thermal transfer.
    Part (b)

    Alloy used for the membrane:

    Invar (36% Ni, 64% Fe) is used for both the primary and secondary barriers. The reason for this choice is its exceptionally low coefficient of thermal expansion. This eliminates the need for expansion joints or corrugations in the membrane design. In addition, Invar remains strong and does not become brittle at the very low temperatures experienced by the liquefied gas. The thin and lightweight nature of Invar maximizes the cargo-carrying capacity of the tank.

    Q5 (10 Marks) Hull Construction πŸ”₯ Repeated 10x

    Describe a method for the attachment of bilge keels. State THREE reaons for not extending bilge keels for the entire length of the vessel. Explain TWO principles of roll damping that bilge keels exploit. (16)

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    Part (a)

    Method of Bilge keel attachment to the hull:

    Bilge keels are fitted port and starboard at the turn of the bilge. They do not extend outside the lines of the side and bottom shell. The bilge keels are attached directly in line with an internal stiffening member such as a girder or longitudinal. The bilge keel comprises a flat bar doubler welded directly to the shell, and an offset bulb plate (OBP) with 'scallops' cut in it is welded to the flat bar doubler. The ends of the bilge keels are tapered (minimum 3 in 1) and will end in line with transverse internal stiffening, such as a frame. Using a hull doubler protects the hull in case of damage to the keel, as the crack would not extend into the hull.

    Part (b)

    Bilge keels are not fitted for the full length of the vessel because:

    • The 'lever' to the ship's axis of rotation is reduced at the ends;
    • The ineffectiveness, the closer the bilge keel is to the rolling centre, increased resistance and more likelihood of damage.
    • The hydrodynamic effect would cause a large increase in resistance and fuel consumption;
    • At the aft end, the boundary layer is much thicker, and since the keel would not project through, it would have a much reduced effect.
    Part (c)

    Two principles of roll damping exploited by Bilge keels:

    Increased Roll Period:

    • Bilge keels increase the ship's roll period (the time it takes for the vessel to complete one roll). This is achieved by increasing the moment of inertia (K) of the vessel. A longer roll period means the vessel rolls more slowly, thus reducing the amplitude of the roll.

    $$T_{r}=2\pi\:\frac{k}{\sqrt{g\times GM}}$$

    Where:

    • g: acceleration due to gravity
    • GM: metacentric height
    • k: mass moment of inertia

    By increasing k, the ship's stability during roll improves.

    Hydrodynamic Resistance:

    • As the vessel rolls, the bilge keels move through the water, creating pressure differences. Water pressure on one side of the keel opposes the rolling motion, providing a damping effect. This is due to the interaction between the bilge keel and the water, creating hydrodynamic forces that counteract the rolling motion.
    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 3x

    (a) Describe stability requirements for dry-docking. (6)

    (b) A Ship of 8000 tonne displacement, 110 m long, floats in sea water of 1.024 t/m3 at draughts of 6 m forward and 6.3 m aft. The TPC is 16, LCB 0.6 m aft of midships, LCF 3 m aft of midships and MCT1 cm 65 tonne m. The vessel now moves into fresh water of 1.000 t/m3. Calculate the distance a mass of 50 tonne must be moved to bring the vessel to an even keel and determine the final draught. (10)

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    Part (a)

    For safe dry-docking, a ship must meet two key stability requirements:

    • Positive GM (Metacentric Height): The ship needs a positive GM. GM is the distance between the centre of gravity (G) and the metacentre (M). A positive GM indicates inherent stability; the ship will right itself if tilted. During dry-docking, the loss of buoyancy as the ship rests on the blocks reduces GM. Insufficient positive GM increases the risk of the ship heeling over or capsizing.
    • Trim by the stern: The vessel should be trimmed slightly by the stern (aft end lower than the bow) to ensure the aft end sits on the keel blocks first. This controlled settling minimises the risk of instability during the docking process. An even keel is generally preferred for the initial floating condition before the dry-docking procedure begins.
    • Bilges: Ensure all bilges are dry and bilge tanks are empty, no unaccounted water onboard.
    • Soundings: Record all soundings of tanks including FW, Ballast, FO, LO, DO and make sure at the time undocking no change in these readings.
    • Upright: Vessel should always be upright during docking and undocking.
    Q7 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 7x

    A ship of 15000 tonne displacement has an Admiralty Coefficient, based on shaft power, of 420. The mechanical efficiency of the machinery is 83%, shaft losses 6%, propeller efficiency 65% and QPC 0.71. At a particular speed the thrust power is 2550kW.

    Calculate: (16)

    (i) Indicated power

    (ii) Effective power

    (iii) Ship speed

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    Given:

    $$\Delta=15000t$$

    $$Shaft\:Power\:\left(SP\right)=420$$

    $$Transmission\:Efficiency=83\%$$

    $$Shaft\:losses=6\%$$

    $$Propeller\:Efficiency=65\%$$

    $$QPC=0.71$$

    $$Thrust\:Power=2550kW$$

    $$\left(\imaginaryI\right)\:Delivered\:Power\:\left(DP\right)=\frac{Thrust\:Power\:\left(TP\right)}{Propeller\:Efficiency\:\left(\eta P\right)}$$

    $$DP=\frac{2550}{0.65}$$

    $$DP=3923.07kW$$

    $$\left(ii\right)\:Shaft\:Power=\frac{Delivered\:Power\:\left(DP\right)}{Transmission\:Efficiency\:\left(\eta T\right)}\:$$

    $$SP=\frac{3923.07}{0.94}$$

    $$SP=4173.47kW$$

    $$\left(iii\right)\:Indicated\:Power=\frac{Shaft\:Power\:\left(SP\right)}{Mechanical\:Efficiency\:\left(\eta m\right)}$$

    $$IP=\frac{4173.47}{0.83}$$

    $$IP=5028.28kW$$

    $$\left(iv\right)\:Effective\:Power=DP\times QPC$$

    $$EP=3923.07\times0.71$$

    $$EP=2785.3797kW$$

    $$\left(v\right)\:Shaft\:Power=\frac{\Delta^{\frac23}\times V^3}{Admiralty\:Co-efficient}$$

    $$4173.47=\frac{15000^{\frac23}\times V^3}{420}$$

    $$V=14.23knots$$

    Q8 (10 Marks) Hull Construction πŸ”₯ Repeated 7x

    With respect to Buoyancy of a vessel:

    (a) What do you understand by reserve buoyancy what happen if the lost buoyancy is greater than the reserve buovancy. (6)

    (b) A forward deep tank 12 m long extends from a longitudinal bulkhead to the ship's side. The widths of the tank surface measured from the longitudinal bulkhead at regular intervals are 10, 9, 7, 4 and 1 m. Calculate the second moment of area of the tank surface about a longitudinal axis passing through its centroid (10)

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    Part (a)

    Reserve Buoyancy

    Reserve buoyancy is the watertight volume above the waterline of a vessel. It represents the potential buoyancy that can be utilised to keep the ship afloat when additional weight is added or if some buoyancy is lost due to damage (e.g., bilging).

    When a mass is added to the ship or buoyancy is lost (e.g., due to flooding of a compartment), the reserve buoyancy is converted into active buoyancy by an increase in draught.

    If the lost buoyancy (due to flooding or damage) is greater than the reserve buoyancy, the vessel will no longer have sufficient buoyant force to counteract its weight, causing it to sink.

    Part (b)
    Q9 (10 Marks) Ship Stability πŸ”₯ Repeated 8x

    (a) Define longitudinal center of gravity (LCG) and longitudinal centre of buoyancy (LCB).

    (b) The immersed cross sectional areas of a ship 120 m long, commencing from aft are 2, 40, 79, 100, 103, 104, 104, 103, 97, 58 and 0 m2. Calculate: (10)

    (i) Displacement

    (ii) Longitudinal position of the centre of buoyancy.

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.
    Part (b)

    Given:

    $$Common \space interval \space (h) \space = \space {{L} \over h} \space = \space {{120} \over 10} \space = \space 12 $$

    Cross-sectional area

    SM

    Product of volume

    Lever

    Product of 1st moment

    2

    1

    2

    +5

    +10

    40

    4

    160

    +4

    +640

    79

    2

    158

    +3

    +474

    100

    4

    400

    +2

    +800

    103

    2

    206

    +1

    +206

    Ξ£MA = +2130

    104

    4

    416

    0

    0

    104

    2

    208

    -1

    -208

    103

    4

    412

    -2

    -824

    97

    2

    194

    -3

    -582

    58

    4

    232

    -4

    -928

    0

    1

    0

    -5

    0

    Ξ£βˆ‡ = 2388

    Ξ£MF = -2542

    $$Displacement \space = \space \rho \times {{h} \over 3} \times \sum βˆ‡ \space tonne $$

    $$=1.025\times{{12}\over3}\times2388$$

    $$Displacement \space = \space 9790.8 tonne$$

    Centre of buoyancy from midship (LCB)

    $$LCB\:=\:h\times({{\sum M_{A}+\sum M_{F}}\over\sum\nabla})$$

    $$=12\times({{2130-2542}\over2388})$$

    $$LCB \space = \space -2.07m fwd$$

    Q10 (10 Marks) Hull Construction πŸ”₯ Repeated 2x

    (a) What is the effect on fuel consumption per unit time, if the ship's speed is outised its operating range?

    (b) An oil tanker 160m long and 22m beam floats at a draught of 9m in seawater, Cw is 0.865. The mid ship section is in the form of a rectangle with 1.2m radius at the bilges. A midship tank 10.5m long has twin longitudinal bulkheads and contains oil of 1.3m3/t to a depth of 11.5m. The tank is holed to the sea for the whole of its transverse section. Find the new draught.

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    The fuel consumption is dependent on power developed. The efficiency of the engine is measured in terms of Specific Fuel Oil Consumption (SFOC), which is consumption/ unit of power per hour. The SFOC of ship for different speeds of ship is:

    From the graph, it is seen that SFOC decreases upto a speed V2, which is most efficient for the ship. If the speed is increased beyond V2, then the power required is more and thus more fuel will be consumed to produce power to reach the increased fuel consumption.

    Generally, the range V1 to V2 will be given for most efficient operation, as the operating range.

    if this speed is exceeded, the fuel consumption per unit time will increase.

    Fuel consumption / unit time ∝ power developed ∝ shaft power.

    $$Shaft\:power\:\alpha\:\Delta^{\frac23}\times V^3^{};\:V=ship^{\prime}s\:speed$$

    $$Fuel\:consumption\:per\:unit\:time\:\alpha\:V^3$$

    Q1 (10 Marks) Hull Construction πŸ”₯ Repeated 8x

    (a) Sketch the cross-section of a bulk carrier with either deep or shallow double bottom showing the type of framing used

    (b) Describe the corrosion problems experienced with ballast tanks

    (c) State how such tanks are protected against extensive corrosion.

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    Part (a)

    Mid-ship section of bulk carrier:

    Part (b)

    (i) Corrosion problems in ballast tanks

    are significant and arise due to various factors:

    • Galvanic corrosion occurs due to the interaction between dissimilar metals, accelerated by the seawater environment and differential aeration.
    • Sulphate-reducing bacteria in river mud can cause localized pitting, leading to penetration of the bottom shell.
    • Dissolved oxygen in seawater reacts with metal surfaces to form rust, significantly contributing to structural deterioration.
    • The rate of corrosion peaks at a 3.5% salt concentration, typical of seawater.
    • Neglecting maintenance exacerbates the problem, allowing corrosion to progress unchecked.

    (ii) Protection against extensive corrosion in ballast tanks involves the following measures:

    • Complete coating of the tank surfaces with properly selected marine-grade paint to create a protective barrier.
    • Installation of sacrificial anodes, designed to corrode preferentially and protect the tank structure.
    • Using large anodes with greater volume relative to surface area to ensure extended protection.
    • Regular gauging of plates to assess thickness and identify areas requiring reinforcement or repair.
    • Conducting regular inspections, cleaning the tanks, removing rust, and repainting to maintain structural integrity and prolong the lifespan of the ballast tanks.
    Q2 (10 Marks) Ship Stability πŸ”₯ Repeated 3x

    With respect to trim and stability, describe the following

    (a) Effects on centre of gravity of slack tanks

    (b) Effect on stability of ice formation on superstructure

    (c) Effects of wind and waves on ship's stability

    (d) Effect of water absorption by deck cargo and retention of water on deck.

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    Part (a)

    Slack tanks, or partially filled tanks, significantly impact a ship's stability due to the free surface effect

    • When tanks are partially filled (slack), the liquid inside moves freely as the ship heels.
    • This movement shifts the centre of gravity (G) laterally towards the heeling side.
    • The righting lever (GZ) decreases, resulting in reduced metacentric height (GM) and overall stability.
    • The virtual loss of GM is proportional to the breadth of the tank and the height of the free surface.

    Consequences of Slack Tanks:

    • Increased angle of heel.
    • Reduced stability, making the ship tender and more prone to capsizing.

    To Minimize the Effect:

    • Avoid slack tanks where possible; either fill tanks completely or empty them.
    • Design tanks with longitudinal divisions or swash bulkheads to limit liquid movement.
    • Use sluice valves in tanks to control liquid transfer.
    • Prioritize filling smaller tanks at the ship's bottom to lower G and improve stability.
    Part (b)

    Ice formation on superstructure:

    Ice accumulating on the superstructure adds mass high up on the ship. This raises the centre of gravity (G), decreasing the metacentric height (GM). A lower GM reduces stability, making the ship more tender (more easily rolled) and increasing the period of roll. The ship becomes more susceptible to capsizing.

    Part (c)

    Effect of Wind and Waves

    Wind:

    • High freeboard or tall superstructures increase windage, leading to a greater rolling effect.
    • Rolling caused by wind reduces stability, especially if the ship remains heeled for a prolonged period.

    Waves:

    • Large waves, especially when the ship is on the crest, can cause a significant loss of stability due to reduced underwater buoyant volume.
    • The ship may develop excessive heeling or capsizing tendencies.
    • Long ships are more vulnerable to wave action due to greater surface exposure, further reducing stability.
    Part (d)

    Effect of Water Absorption by Deck Cargo and Retention of Water on Deck:

    • Water Absorption by Deck Cargo:
      • Certain types of deck cargo, such as timber or other absorbent materials, can take up water during a voyage.
      • This absorbed water increases the weight of the cargo, raising the centre of gravity (G) of the ship.
      • As G moves higher, GM reduces, leading to decreased stability.

      • Retention of Water on Deck:
        • Water that accumulates on the deck, such as from heavy rain or seawater, adds additional weight to the ship's upper structure.
        • This shifts G upward, reducing GM and stability.
        • Retained water may also cause a list if it collects asymmetrically, further impacting the ship's balance.

        • Consequences:
          • Increased risk of capsizing in rough seas.
          • Increased rolling and reduced righting ability.

          • Mitigation Measures:
            1. Ensure proper drainage systems are in place to quickly remove water from the deck.
            2. Regularly monitor and secure deck cargo to prevent excessive water absorption.
    Q3 (10 Marks) Ship Types & Design πŸ”₯ Repeated 3x

    (a) State the reasons for the freeboard requirement

    (b) Explain the term condition of assignment and explain how these are maintained for a ship.

    (c) Using a diagram indicate the freeboard of type A, type B, type B60 and type B100 vessels giving an example of each type.

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    Part (a)

    Reasons for Freeboard Requirements:

    Freeboard is the distance measured from the waterline to the upper edge of the deck plating at the side of the freeboard deck amidships.

    • Ensures the ship is seaworthy when fully loaded.
    • Provides reserve buoyancy, enabling the ship to rise as it passes through waves, keeping decks largely dry.
    • Enhances the ship's stability and increases its survivability in case of damage by allowing it to remain afloat longer, aiding crew escape or damage control.
    Part (b)

    Conditions of Assignment:

    Conditions of Assignment are outlined in the Merchant Shipping Load Line Rules 1968 and must be satisfied before a ship is assigned freeboards and issued a load line certificate. These conditions address the practical need for openings (hatchways, doorways, vents, etc.) in the ship's hull and mandate appropriate protection and closure mechanisms for these openings.

    Requirements Before Assigning Load Line:

    • The ship must have sufficient structural strength.
    • Adequate reserve buoyancy must be maintained.
    • Openings must be secured against water ingress.
    • Safety measures for the crew, such as guardrails and gangways, must be in place.

    To ensure that the conditions of assignment are still current the following items can be checked and confirm to be without change or damage from when the ship was built.

    • Access openings in bulkheads, to ensure that they can be sealed and prevent flooding
    • Cargo and hatchways ensure they can be sealed to prevent flooding
    • Coamings of hatchways, sign of corrosion damage risk of failure would allow flooding
    • Protection of openings, can they all be sealed
    • Ventilator coamings not corroded as they would allow flooding to other arears
    • Air pipes can be shut in heavy weather and not corroded
    • Discharges, inlets and scuppers, all in good condition and operational
    • Side scuttles can be secured
    • Hull inspection, sea boxes and penetrations all checked for damage and corrosion free.
    Part (c)

    Freeboard for Different Vessel Types:

    Type A Ships:

    • Designed for liquid cargo only (e.g., oil tankers).
    • Have the lowest freeboard among ship types due to their higher reserve buoyancy.

    Type B Ships:

    • Designed for general cargo (e.g., container ships).
    • Have a higher freeboard than Type A ships.

    Type B-60 Ships:

    • A variant of Type B ships, with freeboard reduced by up to 60% of the difference between Type A and B (e.g., OBO ships).

    Type B-100 Ships:

    • Type B ships with freeboard reduced by up to 100% of the difference between Type A and B (e.g., certain bulk carriers).
    Q4 (10 Marks) Ship Types & Design πŸ”₯ Repeated 3x

    Discuss the importance of the following to be examined for meeting EEDI limitations:

    (a) Slimmer vessels with lower block coefficients

    (b) Long-Stroke engines

    (c) Low revolution large diameter propellers.

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    The Energy Efficiency Design Index (EEDI) expresses the grams of CO2 emitted per tonne of cargo transported per nautical mile. Its level is directly proportional to the fuel consumed (and hence the propulsive power required) at the design speed, divided by the cargo capacity and speed. Anything that cuts the power needed to move a given deadweight at the design speed, or that raises propulsive efficiency, lowers the EEDI. The three methods below act on exactly those levers.

    Part (a)

    Slimmer vessels with lower block coefficients.

    A fine hull with a low block coefficient (Cb) presents less displaced volume for a given length and thus a smaller wetted surface area, reducing the frictional resistance component at a given displacement. A fine (low Cb) forebody and afterbody also weaken the bow wave and the pressure (residual or wave-making) resistance at the design speed, because the water is pushed aside more gradually. Lower residual resistance means the installed propulsive power at the design speed is less, so less fuel is burnt per tonne-mile, lowering the EEDI. Being longer and narrower for the same displacement also raises waterline length, which reduces the length-related frictional and Froude-number-dependent resistance. The penalty is reduced cubic cargo capacity and somewhat less form stability, so the hull form is optimised rather than simply fined down. Lower block coefficient is therefore one of the strongest design levers for meeting EEDI limits.

    Part (b)

    Long-stroke engines.

    A long-stroke (high stroke-to-bore ratio) slow-speed diesel extracts more work from each unit of fuel in the expansion stroke and achieves higher thermal (brake) efficiency, typically up to about 50 per cent, with a correspondingly lower specific fuel consumption per kWh. Because the stroke is long, the engine can turn slowly at the same piston speed, enabling direct coupling to a large slow-turning propeller with no reduction gearbox and no associated transmission losses. A more efficient engine burns less fuel for each kW it delivers, hence produces less CO2 per tonne-mile, which reduces the attained EEDI directly. Long-stroke engines also operate at low revolutions, which marries perfectly with the large-diameter, low-rev propeller of part (c).

    Part (c)

    Low-revolution large-diameter propellers.

    Propeller open-water efficiency rises as the disc-area loading (thrust per unit swept area) falls. A large-diameter propeller turning slowly accelerates a large mass of water by a small amount, giving low disc loading and high efficiency for the same thrust and therefore less shaft power and less fuel per tonne of cargo. Large slow propellers also stay further away from cavitation, reducing blade erosion, vibration and noise. Because EEDI is fixed by the fuel (shaft power) needed at the design speed, optimising hull, engine and propeller together β€” a fine low-block hull at a low Froude number, a long-stroke low-speed engine, and a large-diameter low-rev propeller β€” minimises energy consumption per tonne of cargo and is the classic route to satisfying EEDI limitations.

    Q5 (10 Marks) General πŸ”₯ Repeated 2x

    Regarding the carriage of crude oil and its associated products:

    (a) (i) State the dangers involved.

    (ii) State what publications give guidance on safety.

    (b) Sketch and describe the operation of an explosimeter suitable for testing pump rooms or tanks.

    (c) Define the terms lower and upper flammable limits illustrating your answer by means rough sketch of a hydrocarbon vapour oxygen graph.

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    Part (a)

    (i) Dangers involved in the carriage of crude oil and its products.

    • Fire and explosion: the vapours are highly flammable; in the flammable range with air they can be ignited by any ignition source (static, sparks, hot surfaces, smoking, electrical), causing fire or explosion.
    • Toxicity and health hazard: crude oil and its vapours are toxic; inhalation causes dizziness, nausea, asphyxiation and long-term health damage; skin contact causes irritation and dermatitis.
    • Asphyxiation: vapours displace oxygen in enclosed spaces (tanks, pump rooms), causing oxygen deficiency.
    • Environmental pollution: spillage causes marine pollution, requiring containment and clean-up.
    • Corrosion and material damage: crude oil and its sulphur content cause corrosion of tanks and piping.
    • Static electricity and sloshing: during loading/cleaning, static charge can build up and ignite vapour.
    Part (a)

    (ii) Publications giving guidance on safety.

    • The International Safety Guide for Oil Tankers and Terminals (ISGOTT).
    • The International Code for the Construction and Equipment of Ships Carrying Dangerous Chemicals in Bulk (IBC Code) and the MARPOL Convention.
    • The International Code of Safety for Ships Carrying Oil in Bulk (MARPOL Annex I).
    • The ship's own Safety Management System (SMS) and the Cargo Manual, plus the International Maritime Dangerous Goods (IMDG) Code for packaged products.
    • Company procedures and the tanker safety checklists (e.g. OCIMF Ship/Shore Safety Checklist).
    Part (b)

    Explosimeter (combustible gas indicator) for testing pump rooms or tanks.

    An explosimeter measures the concentration of flammable vapour in air as a percentage of the lower flammable limit (LFL). Sketch: a hand-held instrument with a sampling tube, a pump (aspirator bulb or motor), a meter calibrated 0-100% LFL, and a sensing element (a heated platinum filament or catalytic bead) in a chamber.

    Operation: the sample is drawn through the sampling tube into the instrument. The gas passes over a heated catalytic filament; if flammable vapour is present it burns on the filament, raising its temperature and changing its electrical resistance. This unbalances a Wheatstone bridge, producing a meter reading proportional to the vapour concentration, displayed as a percentage of the LFL. The instrument is calibrated against a known gas (e.g. pentane) and gives a reading; a reading above about 20-25% LFL indicates a dangerous condition and entry is prohibited. The instrument must be intrinsically safe (approved for use in flammable atmospheres) and used with a suitable sampling line lowered into the tank/pump room.

    Part (c)

    Lower and upper flammable limits, with a hydrocarbon vapour-oxygen graph.

    The lower flammable limit (LFL) is the minimum concentration of vapour in air below which the mixture is too lean to ignite; the upper flammable limit (UFL) is the maximum concentration above which the mixture is too rich to ignite. Between the LFL and UFL the mixture is flammable and will burn/explode if ignited. Sketch: a graph with the vertical axis as percentage of oxygen and the horizontal axis as percentage of hydrocarbon vapour (or a triangular diagram). The flammable range is the region bounded by the LFL and UFL lines and the oxygen line; the "too rich" region is above the UFL, the "too lean" region below the LFL, and the "oxygen deficient" region near the top where there is insufficient oxygen to support combustion. The graph shows that a mixture is only flammable within the shaded band between the LFL and UFL, and that inerting (reducing oxygen below about 8-11%) removes the flammable region entirely.

    Q6 (10 Marks) Ship Stability

    (a) Describe the effect of cavitations on the propeller blades (6)

    (b) A ship has a constant cross-section in the form of a triangle which floats apex down in sea water. The ship is 85 m long, 12 m wide at the deck and has a depth from keel to deck of 9 m. Draw the displacement curve using 1.25 m intervals of draught from the keel to the 7.5 m waterline. From this curve obtain the Displacement in fresh water at a draught of 6.50 m. (10)

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    Part (a)

    Effect of cavitation on propeller blades:

    Erosion:

    • Cavitation causes the formation and collapse of vapor bubbles on the propeller blade surface.
    • The collapse of these bubbles produces high-pressure shockwaves and microjets that erode the blade material, leading to surface pitting and progressive damage.

    Vibration:

    • Uneven cavitation across the blades leads to imbalanced forces, causing vibrations in the propeller and the ship.
    • These vibrations can reduce the comfort of passengers and crew and stress the ship's structural components.

    Noise:

    • The collapse of vapor bubbles generates loud noise, which can interfere with onboard communication and underwater sonar systems.
    • This noise is a significant concern for naval vessels and marine life.

    Reduced Performance:

    • Cavitation reduces the efficiency of the propeller by causing loss of thrust and torque.
    • The presence of cavitation decreases the propeller’s ability to convert rotational energy into forward motion, lowering the ship's speed and increasing fuel consumption.
    Q7 (10 Marks) Ship Resistance & Propulsion

    (a) Explain how wave profile affects the shear force and bending moment curves. (6)

    (b) The wetted surface area of a Container ship is 5946 sq. meter, when travelling at its service speed, the effective power required is 11250 KW with frictional resistance 74 % of the total resistance and specific fuel consumption of 0.22 Kg/kW h. To conserve fuel the ship speed is reduced by 10%, the daily fuel consumption is then found to be 83.0 tonne. Frictional coefficient in sea water is 1.432, Speed in m/s with index (n) 1.825. Propulsive coefficient may be assumed constant at 0.6.

    Determine (10)

    (i) The service speed of the ship.

    (ii) The percentage increase in specific fuel oil consumption when running at reduced speed.

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    Part (a)

    How the wave profile affects the shear force and bending moment curves.

    The wave profile changes the distribution of buoyancy along the ship. In still water the buoyancy is distributed according to the hull form, but in a seaway the waterline is no longer level: when a wave crest is amidships (hogging condition) the buoyancy is increased amidships and reduced at the ends, and when a wave trough is amidships (sagging condition) the buoyancy is reduced amidships and increased at the ends. This changes the net load (weight - buoyancy) distribution, which in turn changes the shear force and bending moment curves. The maximum hogging and sagging bending moments occur when the wave length is about equal to the ship length and the crest or trough is amidships. The wave-induced bending moment is added to the still-water bending moment to give the total bending moment, and the hull girder must be designed to withstand the maximum combined value. The shear force is also increased at the quarter points by the wave profile.

    Part (b)

    Service speed and increase in specific fuel consumption.

    Wetted surface area of a container ship = 5946 m2. At service speed the effective power required is 11,250 kW, with frictional resistance 74% of the total resistance and specific fuel consumption 0.22 kg/kWh. To conserve fuel the speed is reduced by 10%; the daily fuel consumption is then 83.0 t. Frictional coefficient in sea water = 1.432, speed in m/s with index n = 1.825. Propulsive coefficient constant at 0.6.

    (i) Service speed.

    Frictional resistance Rf = 1.432 x S x V^1.825 = 1.432 x 5946 x V^1.825.

    Total resistance = Rf/0.74.

    Effective power = R_total x V = 11,250,000 W.

    1.432 x 5946 x V^1.825/0.74 x V = 11,250,000.

    1.432 x 5946/0.74 = 11,506. So 11,506 x V^2.825 = 11,250,000.

    V^2.825 = 977.8, so V = 977.8^(1/2.825) = 11.44 m/s = 11.44 x 1.944 = 22.2 knots.

    Answer: the service speed is about 22.2 knots.

    (ii) Percentage increase in specific fuel consumption at reduced speed.

    Reduced speed = 0.9 x 11.44 = 10.30 m/s.

    Frictional resistance at reduced speed = 1.432 x 5946 x 10.30^1.825. 10.30^1.825 = 68.0, so Rf = 1.432 x 5946 x 68.0 = 579,000 N. Total resistance = 579,000/0.74 = 782,400 N.

    Effective power at reduced speed = 782,400 x 10.30 = 8,058,000 W = 8058 kW.

    Daily fuel at reduced speed = 83 t = 83,000 kg. Specific fuel consumption at reduced speed = 83,000/(8058 x 24) = 83,000/193,392 = 0.429 kg/kWh.

    At service speed the SFC = 0.22 kg/kWh. Percentage increase = (0.429 - 0.22)/0.22 x 100 = 95%.

    Answer: the specific fuel consumption increases by about 95% at the reduced speed.

    Q8 (10 Marks) Ship Stability

    (a) Explain the purpose of non-watertight longitudinal subdivision of tanks. (6)

    (b) A box-barge 30 m long and 9 m beam floats at a draught of 3 m. The centre of gravity lies on the centreline and KG is 3.50 m. A mass of 10 tonne, which is already on board, is now moved 6m across the ship.

    (i) Estimate the angle to which the vessel will heel, using the formula GZ = sinΞΈ (GM + 1/2BM tan2 ΞΈ)

    (ii) Compare the above result with the angle of heel obtained by the metacentric formula. (10)

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    Part (a)

    Purpose of non-watertight longitudinal subdivision of tanks.

    Non-watertight longitudinal subdivision (longitudinal bulkheads or wash bulkheads that are not watertight) of tanks serves to:

    • Reduce the free-surface effect: by dividing a wide tank into narrower compartments, the second moment of area of the free surface (i = L B^3/12) is greatly reduced, so the free-surface loss of GM is reduced.
    • Reduce the sloshing of the liquid: the wash bulkheads damp the movement of the liquid in a partially full tank, reducing the dynamic loads on the tank structure and the effect on stability.
    • Provide structural support: the longitudinal bulkheads act as girders that stiffen the tank and the hull, and support the deck and bottom.
    • Reduce the free-surface effect during ballasting and cargo operations.

    The subdivision is non-watertight so that the liquid can flow between the compartments (for filling/emptying and for equalising), while still reducing the free-surface and sloshing effects.

    Part (b)

    Angle of heel of a box-barge.

    A box-barge 30 m long and 9 m beam floats at a draught of 3 m. The centre of gravity lies on the centreline and KG is 3.50 m. A mass of 10 t, already on board, is moved 6 m across the ship.

    (i) Estimate the angle of heel using GZ = sin(theta)(GM + 1/2 BM tan^2 theta).

    Displacement = L x B x d x rho = 30 x 9 x 3 x 1.025 = 830.25 t.

    KB = d/2 = 1.5 m. BM = B^2/(12 d) = 81/(12 x 3) = 2.25 m. KM = 1.5 + 2.25 = 3.75 m. GM = KM - KG = 3.75 - 3.50 = 0.25 m.

    Transverse shift of G: GG' = w x d/Delta = 10 x 6/830.25 = 0.0723 m.

    At equilibrium, GG' = GZ = sin(theta)(GM + 0.5 BM tan^2 theta).

    0.0723 = sin(theta)(0.25 + 0.5 x 2.25 tan^2 theta) = sin(theta)(0.25 + 1.125 tan^2 theta).

    Solving iteratively: try theta = 13 deg. sin(13) = 0.225, tan(13) = 0.231, tan^2 = 0.0533. GZ = 0.225(0.25 + 1.125 x 0.0533) = 0.225(0.25 + 0.06) = 0.225 x 0.31 = 0.0698. Close to 0.0723.

    Try theta = 13.3 deg: sin = 0.230, tan = 0.236, tan^2 = 0.0557. GZ = 0.230(0.25 + 0.0627) = 0.230 x 0.3127 = 0.0719. Very close.

    So theta = 13.3 deg.

    (ii) Compare with the metacentric formula.

    Metacentric formula: tan(theta) = GG'/GM = 0.0723/0.25 = 0.2892, theta = atan(0.2892) = 16.1 deg.

    Answer: the wall-sided formula gives about 13.3 deg, while the metacentric (small-angle) formula gives about 16.1 deg. The difference is because the metacentric formula assumes small angles and a constant GM, while the wall-sided formula accounts for the change in BM with angle.

    Q9 (10 Marks) Ship Stability

    (a) Explain the concept of dynamical stability and describe the effect of an increase in wind pressure when a vessel is at its maximum angle of roll to windward (6)

    (b) A ship of 5000 tone displacement has three rectangular double bottom tanks; A 12 m long and 16 m wide; B 14 m long and 15 m wide; C 14 m long and 16 m wide. Calculate the free surface effect for any one tank and state in which order the tanks should be filled when making use of them for correction. (10)

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    Part (a)

    Dynamical Stability is defined as the amount of energy required to heel a ship from its upright equilibrium position to a specific angle of heel. It provides a measure of the vessel's stability by considering its behaviour in response to dynamic external forces, such as wind or waves.

    • The concept compares the heeling moment energy (from external forces) and the righting moment energy (from the ship's stability).
    • The ship will absorb the energy imparted by the heeling moment. If the righting energy is greater than the heeling energy, the ship will stabilize; otherwise, it may capsize.

    Areas Under the Curve:

    • Area A: Represents the region where the heeling moment exceeds the righting moment (external energy > ship's stability).
    • Area B: Represents the region where the righting moment exceeds the heeling moment (ship's stability > external energy).
    • The balance of these areas determines whether the ship will right itself or continue to heel.

    When exposed to heeling forces such as wind or waves, the vessel inclines and may roll over to a certain angle of heel. If the external force is applied instantaneously, the ship must have enough reserve dynamic stability to absorb the energy and return to an upright position. If the external force is constant, the ship will remain at an equilibrium angle where the righting moment equals the heeling moment.

    This refers to the remaining righting energy available to counteract additional external forces. A higher reserve dynamic stability ensures the vessel can handle greater heeling forces without capsizing.

    Part (b)

    $$Free\:surface\:effect=\frac{\rho_{_{L}_{}\times}i}{\rho_{S}\times\nabla}$$

    $$\rho_{L}=density\:of\:liquid\:in\:tank$$

    $$\rho_{S}=densituy\:of\:SW$$

    In this case, the density of the liquid to be filled is the same as the ship floats.

    $$therefore\:\rho_{L}=\rho_{S}$$

    $$i \space = \space {{LB^3} \over 12}$$

    Part (a)

    $$L=12m$$

    $$B=16m$$

    $$Free \space surface \space effect \space = \space {{{12 \times 16^3} \over 12} \over {{5000} \over 1.025}}$$

    $$=\frac{12\times16^3\times1.025}{5000\times12}$$

    $$= \space 0.84m$$

    Part (b)

    $$L=14m$$

    $$B=15m$$

    $$Free\:surface\:effect\:=\:\frac{\frac{14\times15^3}{12}}{\frac{5000}{1.025}}$$

    $$Free\space surface\space effect\space=\frac{14\times15^3\times1.025}{12\times5000}$$

    $$= \space 0.807m$$

    Part (c)

    $$L=14m$$

    $$B=16m$$

    $$Free \space surface \space effect \space = \space {{{14 \times 16^3} \over 12} \over {{5000} \over 1.025}}$$

    $$=\frac{14\times16^3\times1.025}{12\times5000}$$

    $$= \space 0.97m$$

    Tank with the lowest free surface effect should be filled first, so the tanks should be filled in order of B, A, then C

    Q10 (10 Marks) Ship Resistance & Propulsion

    (a) List the variables which affect the force on a rudder. (6)

    (b) A ship of 15000 tonne displacement has an Admiralty Coefficient, based on shaft power of 420. The mechanical efficiency of the machinery is 83%, shaft losses 6%, propeller Efficiency 65% and QPC 0.71. At a particular speed the thrust power is 2550kW.

    Calculate:

    (i) Indicated power

    (ii) Effective power

    (iii) Ship speed

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    Part (a)

    when a rudder is turned from the centreline plane, to any angle, a force acts on the rudder, given by,

    $$F=kAV^2N$$

    Where,

    • k = Constant
    • A = Area of rudder in m2
    • V = Ship's speed in m/s

    The constant is dependent on size of rudder, rudder angle and density of water.

    Variables affecting force on rudder:

    • Rudder angle
    • Density of water
    • Ship speed

    These are the three variables, while all other are constant parameters i.e.

    • Area of rudder
    • Size of rudder.

    (b) Given:

    $$\Delta=15000t$$

    $$Shaft\:Power\:\left(SP\right)=420$$

    $$Transmission\:Efficiency=83\%$$

    $$Shaft\:losses=6\%$$

    $$Propeller\:Efficiency=65\%$$

    $$QPC=0.71$$

    $$Thrust\:Power=2550kW$$

    $$\left(\imaginaryI\right)\:Delivered\:Power\:\left(DP\right)=\frac{Thrust\:Power\:\left(TP\right)}{Propeller\:Efficiency\:\left(\eta P\right)}$$

    $$DP=\frac{2550}{0.65}$$

    $$DP=3923.07kW$$

    $$\left(ii\right)\:Shaft\:Power=\frac{Delivered\:Power\:\left(DP\right)}{Transmission\:Efficiency\:\left(\eta T\right)}\:$$

    $$SP=\frac{3923.07}{0.83}$$

    $$SP=4726.59kW$$

    $$\left(iii\right)\:Indicated\:Power=\frac{Shaft\:Power\:\left(SP\right)}{Mechanical\:Efficiency\:\left(\eta m\right)}$$

    $$IP=\frac{4276.59}{0.83}$$

    $$IP=5694.68kW$$

    $$\left(iv\right)\:Effective\:Power=DP\times QPC$$

    $$EP=3923.07\times0.71$$

    $$EP=2785.3797kW$$

    $$\left(v\right)\:Shaft\:Power=\frac{\Delta^{\frac23}\times V^3}{Admiralty\:Co-efficient}$$

    $$4726.59=\frac{15000^{\frac23}\times V^3}{420}$$

    $$V=14.83\:knots$$

    Q1 (10 Marks) Hull Construction πŸ”₯ Repeated 5x

    With reference to membrane tank for the carriage of liquified gas at very low temperture

    (a) Describe with the aid of a sketch, ONE method of building up the insulation

    (b) State with reasons the alloy, which is used for the membrane

    (c) Describe with the aid of a sketch, how the tanks are located and supported

    (i) Longitudinally

    (ii) Transversly.

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    Part (a)

    Membrane tanks for liquefied gas carriage are built with a double hull throughout the cargo length. The insulation system includes the following:

    1. Primary Barrier: A thin membrane made of INVAR (36% Nickel, 64% Iron) with a thickness of 0.7 to 1.2 mm, forming the primary containment layer for the liquefied gas.
    2. Primary Insulation box: A 230 mm thick layer of granulated Perlite insulation packed in plywood boxes, surrounding the primary barrier. Perlite is siliconised to make it impervious to moisture.
    3. Secondary Barrier: A secondary membrane made of the same INVAR material is installed to prevent cargo leakage in case of primary barrier failure.
    4. Secondary Insulation box: Another 300 mm thick layer of granulated Perlite is placed above the secondary barrier to provide additional insulation and prevent thermal transfer.
    Part (b)

    Alloy used for the membrane:

    Invar (36% Ni, 64% Fe) is used for both the primary and secondary barriers. The reason for this choice is its exceptionally low coefficient of thermal expansion. This eliminates the need for expansion joints or corrugations in the membrane design. In addition, Invar remains strong and does not become brittle at the very low temperatures experienced by the liquefied gas. The thin and lightweight nature of Invar maximizes the cargo-carrying capacity of the tank.

    Part (c)

    Membrane tanks are either independent or self-supporting

    , meaning they don't form part of the ship's hull and don't contribute to the ship's structural strength. They can be spherical, cylindrical, or prismatic (box-shaped). Prismatic tanks usually have internal stiffeners like bulkheads, webs, girders, and stiffeners for added structural integrity.

    (i) Longitudinally:

    Tanks are positioned longitudinally within the ship's cargo hold. The tanks are supported longitudinally by anti-roll chocks and anti-lift chocks that resist forces caused by ship motions. These chocks ensure the tank remains stable even in rough sea conditions. They also act as thermal barriers, preventing the transfer of heat between the hull and the tank.

    The chocks also serve as thermal barriers between the hull and cargo and are often constructed of wood or plastic materials.

    (ii) Transversely: Transverse support is provided by anti-pitch chocks and support chocks, which stabilize the tank against lateral forces. These chocks are typically made of plywood or plastic and help prevent thermal stress between the tank and the ship’s structure.

    Q2 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 7x

    With reference to dry-docking, define the responsibilities of the Second Engineer and instructions to Junior Engineers:

    (a) Prior to docking

    (b) Whilst the vessel in dry dock

    (c) Prior to flooding and leaving the dock.

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    (a) Prior to Docking:

    Preliminary Preparation:

    • Review various plans, manuals, and previous drydock reports for reference.
    • Prepare a detailed repair list and ensure all required spares are accounted for.
    • Conduct an inventory of spares and requisition necessary items.
    • Gather required tools, including special tools like poker gauges, trammel gauges, gagging tools, etc.
    • Mark all overboard valves, NC (normally closed) and NO (normally open) valves clearly for easy identification.
    • Allocate jobs to team members and discuss the time schedule.
    • Conduct a safety meeting to highlight dry dock hazards and necessary precautions.

    Before Entering Dry Dock:

    • Identify the power requirements and machinery to be operational during docking.
    • Check the shore connection box for proper functionality.
    • Record the soundings of FO (Fuel Oil), LO (Lubricating Oil), and DO (Diesel Oil) tanks.
    • Discharge contents from clean drain tanks and sewage tanks.
    • Carry out Economizer soot-blowing.
    • Change over Main Engine, Diesel Generators, and Boiler to Low Sulfur Gas Oil (LSGO).
    • Stop and clean purifiers.
    • Ensure the low sea chest is open and the high sea chest is shut.
    • Keep firefighting appliances (FFA) on standby.
    • Shut down all non-essential machinery, including MGPS (Marine Growth Prevention System) and FWG (Fresh Water Generator).

    (b) Whilst the Vessel is in Dry Dock

    Upon Arrival:

    • Connect shore power and supplies after ensuring safety checks are completed.
    • Start necessary equipment like cooling water, air compressors, air conditioning, and fridge compressors.
    • Check for jobs assigned by dry dock personnel and prepare accordingly.
    • Attend a safety meeting with dry dock personnel to understand local safety rules and procedures.

    During Dry Docking::

    • Oversee and assist in:
      • Cleaning and inspecting the hull, rudder, sea chest, anodes, and propeller.
      • Measuring propeller drop, checking rudder clearances, and inspecting the stern tube bearing and seal.
      • Servicing underwater valves and overboard valves.
      • Inspecting anchor and cables conditions.
      • Overhauling deck machinery, cranes, elevators, and engine room equipment such as the Main Engine, Diesel Generators, Boiler, and Economizer.
      • Renewing pipes and valves as needed.
      • Performing electrical equipment maintenance and surveys.
      • Supervising service engineers for specific repair jobs.
    • Ensure tank cleaning, welding, and other repair works are completed according to the plan.
    • Run the standby diesel generator daily after starting the priming pump.

    (c) Before Flooding and Leaving the Dry Dock

    Final Checks:

    • Verify that all underwater fittings and drain plugs are securely in place.
    • Ensure all machinery has been boxed back and is ready for operation.
    • Check for any leakage in stern tube seals.
    • Take tank soundings to confirm proper levels.
    • Confirm the proper operation of all underwater valves, overboard valves, sea chests, and vents.
    • Inspect the stern tube tank for any irregularities.

    System Restart:

    • Switch back to ship's power after confirming all systems are functional.
    • Test the proper operation of all machinery and systems to ensure the ship is fully operational before leaving the dock.
    Q3 (10 Marks) Ship Stability πŸ”₯ Repeated 8x

    (a) Explain what is meant by "permissible length" of compartments in passanger ships

    (b) Describe how the position of bulkheads is determined

    (c) Briefly describe the significance of the factor of subdivision

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    Part (a)

    Permissible Length

    Permissible length refers to the maximum length of a ship's compartment that can be flooded while ensuring that the sinkage, heel, or trim remains within acceptable limits without exceeding the floodable length. It ensures the ship remains afloat and stable after flooding.

    Permissible Length Formula:

    $$Permissible\:lenght\:=\:Floodable\:length\:\times Factor\:of\:subdivision$$

    The Factor of Subdivision depends on the ship's length and the nature of its service:

    • For passenger ships, the factor of subdivision is smaller compared to cargo ships.
    • Smaller compartments ensure enhanced safety in case of flooding.
    Part (b)

    Position of Bulkheads

    The position of bulkheads is determined based on the need to comply with the "reasonable amount of damage" criterion in case of flooding:

    • Transverse Watertight Bulkheads should vertically extend up to the margin line.
    • At least one watertight longitudinal bulkhead must be located 20% of the ship's breadth inward from each side.
    • Bulkheads are positioned along the ship's length at intervals equal to the permissible length, but not exceeding the floodable length.
    • The maximum permissible compartment length is limited to 10.7 meters.
    • Collision Bulkhead must be located forward at a distance equal to the permissible length from the forward perpendicular.
    • The ER must be enclosed by two transverse watertight bulkheads, with an aft peak watertight bulkhead enclosing the forward part.
    Part (c)

    Factor of Subdivision

    The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

    Permissible Length Formula:

    $$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

    A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

    Q4 (10 Marks) Ship Stability πŸ”₯ Repeated 14x

    Explain how the period of roll varies with

    (a) The amplitude of roll

    (b) The radius of gyration

    (c) The initial metacentric height

    (d) The location of masses in the ship

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    The period of roll Tr of a ship is determined by the formula:

    $$T_{r}=\frac{2\pi K}{\sqrt{g\times GM}}$$

    where,

    • K is the radius of gyration (mass moment of inertia)
    • g is the acceleration due to gravity, and
    • GM is the metacentric height.
    Part (a)

    Amplitude of Roll:

    • The amplitude of roll, or the maximum angle of heel, does not affect the period of roll. While a larger amplitude implies a greater heeling moment and faster roll speed, the time taken to complete one full roll cycle remains constant for a given metacentric height and radius of gyration.
    Part (b)

    Radius of Gyration (K):

    • The period of roll is directly proportional to the radius of gyration. A larger radius of gyration (indicating a greater distribution of mass further from the ship's centre of rotation) leads to a longer period of roll. Conversely, a smaller radius of gyration (mass concentrated closer to the centre) results in a shorter period. The distribution of cargo significantly impacts K; cargo concentrated centrally minimises K and the roll period, while dispersed cargo maximises K and the roll period.
    Part (c)

    Initial Metacentric Height (GM):

    • The period of roll is inversely proportional to the square root of the metacentric height (GM). A larger GM (a stiffer ship) leads to a shorter roll period, as the ship quickly returns to its upright position. A smaller GM (a tender ship) results in a longer roll period, with slower return to the upright.
    Part (d)

    Location of Masses in the Ship:

    The location of masses in the ship will effect the GM & K. So the period of roll will be affected.

    • If masses are at bottom, G moves down, GM ↑, period of roll ↓.
    • If masses are at top, G moves up, GM ↓, period of roll ↑.
    • If masses are concentrated at centre, K ↓, period of roll ↓.
    • If masses are away from centre, K ↑, period of roll ↑.
    Q5 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 13x

    Describe the relationship between frictional resistance and

    (a) Ship speed

    (b) The wetted area

    (c) The surface roughness

    (d) The length of the vessel

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    Frictional resistance arises due to the interaction between the ship's hull and the water as the vessel moves. This resistance is caused by eddying water adjacent to the hull that is drawn along with the ship. The frictional resistance is given by:

    $$R_{f\:=\:}f\:.\:s\:.\:v^{n}$$

    Where:

    • f = Coefficient of friction
    • s = Wetted surface area
    • v = Ship's speed in knots
    • n = Constant (1.82)
    Part (a)

    Ship Speed (v):

    • Frictional resistance increases with increasing ship speed. The relationship is not perfectly linear, but it's approximately described by the formula Rf = fsv^n (Rf ∝ v^n), where 'n' is a constant (approximately 1.82). This means that a small increase in speed results in a larger increase in frictional resistance.
    Part (b)

    Wetted Area (s):

    • Frictional resistance is directly proportional to the wetted surface area (Rf ∝ s). A larger wetted area (the part of the hull submerged in water) leads to greater frictional resistance. The wetted area increases with the ship's draught (depth in the water).
    Part (c)

    Surface Roughness:

    • Increased surface roughness increases frictional resistance. Roughness can be caused by fouling (marine growth), corrosion, or poor hull maintenance. Regular cleaning and maintenance of the hull surface help to minimise roughness and reduce resistance.
    Part (d)

    Length of the Vessel

    • Frictional resistance is influenced by the vessel's length. As the length increases, the wetted surface area grows, and the coefficient of friction (f) changes, leading to an increase in Rf. Longer vessels experience higher frictional resistance, but streamlined designs can mitigate the effect to some extent.
    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) How the distribution of mass within the ship affects the rolling period (6)

    (b) A ship of 14000 tonne displacement is 125m long and floats at draught of 7.9m forward and 8.5m aft. The TPC is 19, GMl 120m and LCF 3m forward of midships. It is required to bring the vessel to an even keel draught of 8.5m. Calculate the mass which should be added and the distance of the centre of the mass from midships (10)

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    Part (a)

    The rolling period is influenced by the ship's metacentric height (GM) and radius of gyration (K), which are determined by the location of the masses onboard.

    Masses at the bottom:

    • The centre of gravity (G) moves down.
    • Metacentric height (GM) increases, resulting in greater stability.
    • Rolling period decreases.

    Masses at the top:

    • The centre of gravity (G) moves up.
    • Metacentric height (GM) decreases, reducing stability.
    • Rolling period increases.

    Masses concentrated at the centre:

    • The radius of gyration (K) decreases.
    • Rolling period decreases.

    Masses distributed away from the centre:

    • The radius of gyration (K) increases.
    • Rolling period increases.
    Q7 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Explain how increase of draught and of displacement influence rolling (6)

    (b) A pontoon has a constant cross-section as shown in Fig. given below. The metacentric height is 2.5m. Find the height of the centre of gravity above the keel (10)

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    Part (a)

    In case of increase of draught and displacement, this is a case of loading being done.

    Thus, when the ship is being loaded, its GM will start to decrease.

    Now, the time period of roll is given by:

    $$T_{r}=\frac{2\pi k}{\sqrt{GM.g}}$$

    Where,

    • k = radius of gyration
    • GM = metacentric height
    • g = acceleration due to gravity

    Now, since, GM has started to decrease, the Tr will start to increase.

    Thus, the ship will now roll with greater time period. Thus, an increase in draught and displacement, influences rolling.

    Q8 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    (a) Describe the fundamental principle of a propeller

    (b) A propeller 6m diameter has a pitch ratio of 0.9, BAR 0.48 and when turning at 110 rev/min, has a real slip of 25% and wake fraction 0.30. If the propeller delivers a thrust of 300kN and the propeller efficiency is 0.65, claculate (10)

    (i) Blade area

    (ii) Ship speed

    (iii) Thrust power

    (iv) Shaft power

    (v) Troque

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    Part (a)

    A propeller is a type of fan, that transmits power by converting rotational motion into thrust. A pressure difference is produced between the forward and rear surface of the aerofoil shaped blade and the fluid is accelerated behind the blade. A marine propeller of this type is sometimes known as screw propeller or a screw.

    Given:

    $$D=6m$$

    $$p=0.9$$

    $$BAR=0.48$$

    $$N=110rev\:per\min$$

    $$real \space slip \space (s) \space = \space 25\%$$

    $$W_{F}=0.3$$

    $$Thrust \space = \space 300kN$$

    $$Ξ·_{prop} \space = \space 0.65$$

    (i) Blade area:

    $$BAR \space = \space {{A_b} \over {{\pi} \over 4} D^2}$$

    $$Blade \space area \space A_b \space = \space 0.48 \times {{\pi} \over 4} 6^2$$

    $$Blade \space area \space = \space 13.57m^2 $$

    $$p \space = \space {{P} \over D}$$

    $$0.9 \space = \space {{P} \over 6}$$

    $$Pitch \space p = \space 5.4m$$

    $$V_{T}=P\times N\times\frac{3600}{1852}$$

    $$V_{T}=5.4\times\frac{110}{60}\times\frac{3600}{1852}$$

    $$V_{T}=19.24knots$$

    $$Real \space slip \space S \space = \space {{V_T - V_a} \over V_T}$$

    $$ 0.25 \space = \space {{19.24 - V_a} \over 19.24}$$

    $$V_a \space = \space 14.42 knots$$

    $$W_F \space = \space {{V - V_a} \over V}$$

    $$0.30 \space = \space {{V - 14.42} \over V}$$

    $$V=20.6knots$$

    $$T_{p}\space=\space Thrust\times V_{a}\times\frac{1852}{3600}$$

    $$T_p \space = \space Thrust \times 14.42 \times {{1852} \over 3600 }$$

    $$T_p \space = \space 2225.48 $$

    $$T_p \space = \space d_p \times Ξ·_{prop}$$

    $$2225.48=d_{p}\times0.65$$

    $$d_p \space = \space 3423.8kW$$

    $$dp=2\pi NT$$

    $$3423.8=2\times\pi\times\frac{110}{60}\times T$$

    $$T=297.22KN$$

    Q9 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    (a) Explain what is meant by

    (i) Wave-making resistance

    (ii) Frictional resistance

    (iii) Eddy-making resistance

    (b) When a ship is 800 nautical miles from port its speed is reduced by 20%, there by reducing the daily fuel consumption by 42 tonne and arriving in port with 50 tonne onboard. If the fuel consumption in t/h is given by the expression (0.136 + 0.001 v^3) where V is the speed in knots, estimate: (10)

    (i) The reduced consumption per day

    (ii) The amount of fuel onboard when the speed was reduced

    (iii) The percentage decrease in consumption for the latter part of the voyage

    (iv) The percentage increase in time for this latter period

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    $$D=800nm$$

    $$V_1=?knots$$

    $$V_2=0.8V_1$$

    $$DC_1=Cons\:per\:day\:at\:V_1$$

    $$DC_2=Cons\:per\:day\:at\:V_2$$

    $$DC_1-DC_2=42t$$

    $$C=\left(0.136+0.001V^3\right)\:t\h$$

    $$Therefore\:DC=24\left(0.136+0.001V^3\right)\:tonnes\:per\:day$$$$42=24\left\lbrack\left(0.136+0.001V_{1^{}}^3\right)-\left(0.136_{}+0.001\left(0.8V_1^3\right)\right)\right\rbrack$$

    $$42=24\left(0.136+0.001V_1^3-0.136-0.512\times10^{-3}\times V_1^3\right)$$

    $$42=24\left(0.001V_1^3-0.512\times10^{-3}\times V_1^3\right)$$

    $$42=24\left(0.000488V_1^3\right)$$

    $$V_1=\sqrt[3]{\frac{42}{24\times0.000488}}$$

    $$V_1=15.31\:knots$$

    $$V_2=0.8\times V_1$$

    $$V_2=0.8\times15.31$$

    $$V_2=12.245\:knots$$

    $$\left(i\right)\:Reduced\:cons\:per\:day\:=\:\left(0.136+0.001V_2^3\right)\times24$$

    $$=\left(0.136+0.001\times12.45^3\right)\times24$$

    $$=49.57\:tonnes\:per\:day$$

    $$Time\:taken\:for\:complete\:voyage\:of\:800nm$$

    $$at\:V_2=\frac{800}{12.245\times24}=2.72\:days$$

    $$Consumption\:=\:2.72\times49.57=134.93t\:\left(at\:reduced\:speed\right)$$

    $$\left(ii\right)\:Fuel\:onboard=134.93+50$$

    $$=184.93t\:\left(after\:speed\:reduction\right)$$

    $$DC_1=DC_2+42$$

    $$DC_1=49.57+42$$

    $$DC_1=91.57t$$

    $$Time\:taken\:for\:V_1=\frac{800}{15.31\times24}$$

    $$=2.178days$$

    $$Cons\:at\:V_1=91.57\times2.178$$

    $$=199.38t$$

    $$\left(iii\right)\:\%\:reduction\:in\:cons=\frac{199.38-134.93}{199.38}$$

    $$=32.32\%$$

    $$\left(iv\right)\:\%\:increase\:in\:time=\frac{2.72-2.178}{2.178}$$

    $$=24.88\%$$

    Q10 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Explain how to distinguish between list and loll and describe how to return the ship to the upright in each case (6)

    (b) A ship of 5000 tonne displacement has a double bottom tank 12m long. The 1/2 breaths of the top of the tank are 5, 4 and 2m respectively. The tank has a watertight centreline division. Claculate the free surface effect if the tank is partially full of fresh water on one side only (10)

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    Part (a)

    Distinguishing between List and Loll:

    1. List is caused by Uneven distribution of weight within the ship. List will have the below conditions:

    • GM is positive (GM > 0).
    • The centre of gravity (G) is off-centre.
    • The vessel is at equilibrium but inclines to one side due to uneven weight distribution, even without any external forces acting on it.
    • The vessel rolls around the angle of list.

    Correction:

    • Redistribute the weight evenly to bring the centre of gravity (G) back in line with the centerline and metacentric height (M).

    2. Loll is caused by High centre of gravity (G) leading to negative GM and is exacerbated by external forces, free surface effects, or poor distribution of weights. LOLL will have the below conditions:

    • GM is negative (GM < 0).
    • The centre of gravity (G) is on the centerline but too high, making the vessel inherently unstable.
    • The vessel flops or inclines to one side at an angle of loll, and it can incline equally to either side.
    • The vessel rolls unstably around the angle of loll.

    Correction:

    • Reduce the centre of gravity by ballasting bottom tanks or removing weight from higher levels.
    • Minimize free surface effects by reducing the breadth of free surfaces in tanks.

    Q1 (10 Marks) Hull Construction

    Container ships have very large cargo hatch openings.

    (a) Describe how this ship type is susceptible to torsion and how the structure is designed to combat torsional stress.

    (b) Describe the problem created by discontinuities in longitudinal structure.

    (c) State THREE points of discontinuity, in any ship type, describing how the problems are overcome

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    Part (a)

    Torsion in container ships and the design to combat it.

    Container ships have very large cargo hatch openings, which remove a large part of the deck and the upper side structure, so the hull girder is an open section that is weak in torsion. When the ship is loaded asymmetrically (containers of different weights on the two sides) or when it is in a seaway with the wave acting unevenly, the open section twists, producing torsional (warping) stresses in the structure. The structure is designed to combat torsion by:

    • Providing a strong closed section at the ends (the fore and aft peaks and the machinery space) which resist the torsion.
    • Using a strong double-bottom and a strong side structure, with the hatch coamings acting as longitudinal girders.
    • Fitting strong transverse (web) frames and a strong deck structure, and using the hatch coamings and the side shell to form a torsion box.
    • Using high-tensile steel and adequate scantlings in the deck and side to resist the warping stresses.
    • Ensuring the cargo is loaded symmetrically to minimise the torsional moment.
    Part (b)

    Problem created by discontinuities in longitudinal structure.

    Discontinuities in the longitudinal structure (e.g. at hatch corners, openings, changes in section, the ends of the superstructure, the break of the deck) create stress concentrations. At these points the longitudinal stress is concentrated, and the local stresses can be several times the average stress, leading to fatigue cracking and failure. The problem is that the longitudinal material is interrupted, so the load must be transferred around the opening, producing high local stresses at the corners and edges.

    Part (c)

    THREE points of discontinuity and how the problems are overcome.

    1. Hatch corners: the corners of large hatch openings are stress raisers. They are overcome by making the corners rounded (large radius), fitting doubling plates and soft patches, and providing adequate local reinforcement.
    2. The break of the deck / ends of the superstructure: the sudden change in section at the ends of the superstructure or the break of the deck creates stress concentrations. They are overcome by tapering the structure, fitting doubling plates, and providing adequate local reinforcement and smooth transitions.
    3. Openings in the shell or deck (e.g. doors, scuttles, air pipes, sea chests): these interrupt the longitudinal material. They are overcome by fitting compensating (doubling) plates around the openings, rounding the corners, and providing adequate local reinforcement.
    Q2 (10 Marks) General

    (a) Define critical temperature and boiling point and hence show how some liquefied gases may be transported fully pressurized, whilst others need to be carried fully refrigerated.

    (b) State the basic differences in construction of fully pressurized and fully refrigerated systems for the carriage of liquefied gas at sea.

    (c) Compare the membrane tank and independent tank systems of construction.

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    (a) Critical Temperature, Boiling Point, and Their Influence on Gas Carriage

    Critical Temperature (Tc):

    • The maximum temperature at which a gas can be liquefied by applying pressure alone.
    • Above this temperature, no amount of pressure will convert the gas into a liquid.

    Boiling Point (Tb):

    • The temperature at which a liquid changes to vapour at a given pressure.
    • At atmospheric pressure, this is called the normal boiling point.

    Relation to Liquefied Gas Transportation

    • If a gas has a critical temperature above ambient temperature, it can be liquefied by pressure alone and therefore carried in the fully pressurized condition.
      • Examples: Propane, Butane, Ammonia.
    • If a gas has a critical temperature below ambient temperature, pressure alone cannot liquefy it, so it must be cooled below its boiling point and carried as a fully refrigerated liquid.
      • Examples: LNG (methane), Ethylene.

      Hence:

      Fully Pressurized Carriage

      • Gases with relatively high critical temperature and higher boiling points
      • Can remain liquid at normal temperatures under moderate pressure (~17 bar)
      • Example: Butane
        • Tc = 152Β°C
        • Tb = –0.5Β°C

        Fully Refrigerated Carriage

        • Gases with very low critical temperature and very low boiling points
        • Must be kept at low temperature and near-atmospheric pressure
        • Example: Methane (LNG)
          • Tc = –82Β°C
          • Tb = –162Β°C

          (b) Basic Construction Differences

          Feature

          Fully Pressurized System

          Fully Refrigerated System

          Operating Pressure

          ~17 bar

          ~0.25 bar

          Operating Temperature

          Ambient

          –48Β°C to –163Β°C

          Tank Type

          Spherical or cylindrical independent tanks (Type C)

          Prismatic independent tanks (Type A or B)

          Construction Material

          High-strength carbon steel

          Low-temperature steels (nickel steel, aluminum alloy)

          Insulation

          Not required

          Heavy insulation to limit heat ingress

          Cargo Handling Equipment

          Simple piping, no refrigeration

          Reliquefaction or refrigeration plant required

          Typical Vessel Size

          Small (up to ~6,000 mΒ³)

          Large (up to 125,000 mΒ³ or more)

          (c) Membrane vs Independent Tank Systems

          Feature

          Membrane Tank System

          Independent Tank System

          Definition

          A thin membrane forms the cargo barrier, supported through insulation by the inner hull

          A self-supporting tank independent of the ship’s hull

          Construction Material

          Thin stainless steel or Invar (~1 mm) attached to insulation panels

          Type A (prismatic), Type B (spherical), Type C (cylindrical)

          Support

          Ship’s hull and insulation support the load

          Tank supports its own load; hull provides only positioning/protection

          Space Efficiency

          Very high β€” maximum utilization of hull volume

          Lower β€” space wasted between tank and hull

          Maintenance

          Repairs are complex; damage may require dry-docking

          Easier to inspect and repair

          Typical Use

          LNG carriers (GTT NO96, Mark III)

          LPG carriers, small LNG carriers

          Examples

          GTT membrane systems

          Moss spherical (Type B), Type C tanks

    Q3 (10 Marks) Ship Stability

    (a) Explain how a propeller blade may be eroded due to cavitation, describing the progressive nature of the damage

    (b) Outline the design features that may be considered to minimise cavitation.

    (c) State FOUR detrimental effects of propeller cavitation.

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    Part (a)

    Propeller blade erosion due to cavitation and its progressive nature

    Cavitation occurs when the local water pressure around the propeller blades falls below the vapour pressure of water, causing vapour bubbles to form. These bubbles, when carried into regions of higher pressure, collapse violently against the blade surface.

    Progressive nature of damage:

    1. Formation of vapour bubbles: Low-pressure zones, usually on the back (suction) side of the blade, form vapour cavities.
    2. Collapse of bubbles: As the bubbles move to higher-pressure regions, they implode.
    3. Shock waves and micro-jets: Each implosion generates high-energy shock waves and micro-jets that strike the blade.
    4. Surface pitting: Repeated bubble collapses remove tiny particles of metal, producing small pits.
    5. Material loss: Continuous pitting deepens the damage, causing roughness and erosion of the blade surface.
    6. Propeller imbalance: Progressive material loss changes blade shape and weight distribution, causing vibration and reduced efficiency.
    Part (b)

    Design features to minimise cavitation

    To minimise cavitation, propeller design focuses on maintaining sufficiently high blade surface pressures and ensuring smooth water flow. Key design considerations include:

    • Higher blade area ratio (BAR): Larger blade area reduces loading per unit area, preventing extreme pressure drop.
    • Optimised pitch and camber: Ensures smooth acceleration of water across the blade, avoiding sudden low-pressure regions.
    • Well-shaped and thinner blade sections: Streamlined profiles reduce turbulence and local suction peaks.
    • Improved blade tip design: Rounded or optimised tip geometry lowers tip vortices, which are common cavitation zones.
    • Reduced propeller RPM: Lower rotational speed reduces pressure fluctuations.
    • Adequate propeller submergence: Keeping the propeller deep enough prevents suction peaks near the surface.
    Part (c)

    Detrimental effects of propeller cavitation

    At least four major negative effects include:

    1. Erosion of blade surfaces β†’ leads to material loss and surface roughness.
    2. Reduced propeller efficiency β†’ disturbed flow causes reduction in thrust.
    3. Vibration and noise β†’ collapsing bubbles and imbalance generate noise and structural vibration.
    4. Damage to propeller and shafting β†’ prolonged cavitation may lead to blade cracks, imbalance, and stress on shaft bearings.
    Q4 (10 Marks) Ship Stability

    (a) Explain the procedure required to produce weight, buoyancy and load curves for a ship assumed to be floating in still water, stating any relevant features of the curves.

    (b) Describe how shear force and bending moment curves are produced from a load diagram, explaining how the features of EACH curve are connected

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    Part (a)

    Procedure to produce weight, buoyancy and load curves for a ship floating in still water.

    The ship is divided into a number of equal-length sections (stations) along its length. For each section:

    • Weight curve: the weight of the ship (hull, machinery, cargo, ballast, stores, etc.) in each section is determined and plotted as a stepped or smooth curve of weight per unit length against position along the ship. The total weight equals the displacement.
    • Buoyancy curve: the buoyancy (displacement) of each section is determined from the hull form (the volume of the section below the waterline, from the Bonjean curves or the hydrostatic data) and plotted as a curve of buoyancy per unit length against position. The total buoyancy equals the displacement.
    • Load curve: the load (net load) is the difference between the weight and the buoyancy in each section, i.e. load = weight - buoyancy. It is plotted as a curve of load per unit length against position. Where the weight exceeds the buoyancy the load is positive (downward), and where the buoyancy exceeds the weight the load is negative (upward).

    Relevant features: the weight and buoyancy curves are stepped (discontinuous) because the weights and buoyancy are lumped in sections; the load curve is the difference and shows where the ship is loaded or buoyant; the total area under the load curve is zero (the total weight equals the total buoyancy); the load curve is the basis for the shear force and bending moment curves.

    Part (b)

    How shear force and bending moment curves are produced from a load diagram.

    The shear force at any section is the algebraic sum of the loads (weight - buoyancy) to one side of the section, i.e. the integral of the load curve. Starting from one end (where the shear force is zero), the shear force is built up by integrating the load curve along the length; it is zero at the ends and reaches a maximum where the load curve changes sign (where the net load is zero). The shear force curve is the integral of the load curve.

    The bending moment at any section is the integral of the shear force curve, i.e. the algebraic sum of the moments of the loads to one side. The bending moment is zero at the ends and reaches a maximum where the shear force is zero (where the shear force changes sign). The bending moment curve is the integral of the shear force curve.

    Features connected: the load curve is the derivative of the shear force curve, and the shear force curve is the derivative of the bending moment curve. The maximum bending moment occurs where the shear force is zero, and the maximum shear force occurs where the load curve changes sign. The curves are used to check the hull girder strength.

    Q5 (10 Marks) Ship Stability

    (a) Explain how a force normal to the rudder is produced when the rudder is turned to a helm angle.

    (b) Define the term centre of effort as applied to a rudder.

    (c) Describe how the position of centre of effort changes as helm angle increases.

    (d) Explain the term balanced, describing the benefits of fitting a balanced rudder.

    (e) Describe, with the aid of a sketch, how an angle of heel is produced due to the force on the rudder.

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    Part (a)

    How a force normal to the rudder is produced when the rudder is turned to a helm angle.

    When the rudder is turned to a helm angle, the water flowing past the hull strikes the rudder at an angle of attack. The rudder acts as a hydrofoil: the flow over the two faces of the rudder is different, producing a pressure difference (higher pressure on the face towards the flow, lower pressure on the back face). This pressure difference produces a lift force normal to the plane of the rudder (the rudder normal force), which acts at the centre of effort. The normal force is approximately proportional to the rudder area, the square of the speed, and the sine of the helm angle: Fn = k A v^2 sin(alpha). This force, acting at the rudder, produces the turning moment on the ship and also a heeling moment.

    Part (b)

    Centre of effort as applied to a rudder.

    The centre of effort is the point on the rudder where the resultant of the normal (lift) force and the drag force acts. It is the point through which the total hydrodynamic force on the rudder is considered to act, and it is used to determine the turning moment and the torque on the rudder stock. The centre of effort is located at a certain distance from the rudder stock axis, and its position determines the balance of the rudder.

    Part (c)

    How the position of the centre of effort changes as the helm angle increases.

    As the helm angle increases, the normal force increases and the centre of effort moves. At small angles the centre of effort is near the leading edge; as the angle increases, the pressure distribution changes and the centre of effort moves towards the trailing edge (and slightly towards the leading edge for a balanced rudder). The movement of the centre of effort affects the torque on the rudder stock and the balance of the rudder.

    Part (d)

    Balanced rudder and its benefits.

    A balanced rudder is one in which part of the rudder area is forward of the rudder stock axis, so that the centre of effort is close to (or slightly ahead of) the stock axis. This reduces the torque required to turn the rudder, because the hydrodynamic force acts close to the stock, so the steering gear can be smaller and the rudder is easier to operate. The benefits are: reduced steering gear size and power, reduced torque on the stock, and easier, more responsive steering. The balance is such that the rudder is stable (tends to return to the centreline) and does not require excessive force to hold.

    Part (e)

    How an angle of heel is produced due to the force on the rudder.

    The rudder normal force acts at the rudder, which is below the waterline and at the stern. The force has a horizontal component that produces the turning moment, and because the force acts at a point below the centre of gravity (and at the stern), it produces a heeling moment about the longitudinal axis. The heeling moment = Fn x (vertical distance of the rudder centre below the centre of lateral resistance). This heeling moment heels the ship towards the side of the rudder (the ship heels outboard when turning). The angle of heel is given by tan(theta) = heeling moment/(Delta x GM). Sketch: the rudder force acting at the stern below the waterline, producing a heeling moment that heels the ship.

    Q6 (10 Marks) Ship Resistance & Propulsion

    (a) Describe the relationship between frictional resistance and (6)

    (i) Ship speed

    (ii) The wetted area

    (iii) The surface roughness

    (iv) The length of the vessel

    (b) A ship 150 m long and 8.5 m draught has a rudder whose area is one sixtieth of the middle-line plane and diameter of stock 320 mm. Calculate the maximum speed at which the vessel may travel if the maximum allowable stress is 70 MN/m2 the Centre of stock 0.9 m from the Centre of effort and the maximum rudder angle is 35 degrees.

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    Frictional resistance arises due to the interaction between the ship's hull and the water as the vessel moves. This resistance is caused by eddying water adjacent to the hull that is drawn along with the ship. The frictional resistance is given by:

    $$R_{f\:=\:}f\:.\:s\:.\:v^{n}$$

    Where:

    • f = Coefficient of friction
    • s = Wetted surface area
    • v = Ship's speed in knots
    • n = Constant (1.82)
    Part (a)

    Ship Speed (v):

    • Frictional resistance increases with increasing ship speed. The relationship is not perfectly linear, but it's approximately described by the formula Rf = fsv^n (Rf ∝ v^n), where 'n' is a constant (approximately 1.82). This means that a small increase in speed results in a larger increase in frictional resistance.
    Part (b)

    Wetted Area (s):

    • Frictional resistance is directly proportional to the wetted surface area (Rf ∝ s). A larger wetted area (the part of the hull submerged in water) leads to greater frictional resistance. The wetted area increases with the ship's draught (depth in the water).
    Part (c)

    Surface Roughness:

    • Increased surface roughness increases frictional resistance. Roughness can be caused by fouling (marine growth), corrosion, or poor hull maintenance. Regular cleaning and maintenance of the hull surface help to minimise roughness and reduce resistance.
    Part (d)

    Length of the Vessel

    • Frictional resistance is influenced by the vessel's length. As the length increases, the wetted surface area grows, and the coefficient of friction (f) changes, leading to an increase in Rf. Longer vessels experience higher frictional resistance, but streamlined designs can mitigate the effect to some extent.
    Q7 (10 Marks) Ship Resistance & Propulsion

    (a) Derive the Admiralty Coefficient formula and show how this may be modified to suit a fast Ship (6)

    (b) A 6m model of a ship has a wetted surface area of 7m2 and when towed in fresh water at 3knots, has a total resistance of 35N. Calculate the effective power of the ship, 120m long, at as corresponding speed.

    n= 1.825. f from formula SCF =1.15. (10)

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    Part (a)

    Derivation of the Admiralty coefficient and its modification for a fast ship.

    The Admiralty coefficient is derived from the relation that the power required to drive a ship is proportional to the displacement and the cube of the speed, and inversely proportional to the length:

    P = Delta^(2/3) V^3 / C

    where P is the shaft power, Delta the displacement, V the speed and C the Admiralty coefficient. The coefficient C is a measure of the efficiency of the hull and machinery; a higher C means a more efficient ship (less power for a given displacement and speed). The relation is derived from the assumption that the resistance is proportional to the wetted surface area (proportional to Delta^(2/3)) and to the square of the speed, and that the power is resistance x speed, giving P proportional to Delta^(2/3) V^3.

    For a fast ship, the resistance rises more steeply with speed (the wave-making resistance increases rapidly at high Froude numbers), so the simple V^3 relation underestimates the power. The Admiralty coefficient is therefore modified for a fast ship by using a higher power of the speed, e.g. P = Delta^(2/3) V^4 / C, or by using a speed-dependent coefficient. The modified form accounts for the increased wave-making resistance of fast ships.

    Part (b)

    Effective power of the ship from the model test.

    A 6 m model of a ship has a wetted surface area of 7 m2 and, when towed in fresh water at 3 knots, has a total resistance of 35 N. Calculate the effective power of the ship, 120 m long, at the corresponding speed. n = 1.825, f from the formula SCF = 1.15.

    Scale = 120/6 = 20.

    Corresponding speed: V_ship = V_model x sqrt(scale) = 3 x sqrt(20) = 3 x 4.472 = 13.42 knots.

    Ship resistance: R_ship = R_model x scale^3 x (rho_ship/rho_model) x SCF

    = 35 x 20^3 x (1.025/1.000) x 1.15 = 35 x 8000 x 1.025 x 1.15 = 35 x 8000 x 1.17875 = 330,050 N.

    Ship speed in m/s = 13.42 x 0.5144 = 6.90 m/s.

    Effective power = R x V = 330,050 x 6.90 = 2,277,000 W = 2277 kW.

    Answer: the effective power of the ship is about 2280 kW.

    Q8 (10 Marks) Ship Stability

    (a) Describe how the force on the ship's bottom and the GM vary when grounding takes place. (6)

    (b) The 1/2 ordinates of a waterplane at 15 m intervals, commencing from aft, are 1, 7, 10.5, 11, 11, 10.5, 8, 4 and 0 m. Calculate: (10)

    (i) Π’Π C

    (ii) Distance of the centre of flotation from midships

    (iii) Second moment of area of the waterplane about a transverse axis through the centre of flotation.

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    Part (a)

    When a ship grounds, the forces on the bottom and the metacentric height (GM) change depending on the grounding's nature and the ship's position on the seabed.

    If the ship grounds on a level bottom:

    • A ground reaction force acts vertically upwards from the seabed. This force counteracts part of the ship's weight and alters the distribution of buoyancy forces.
    • The ship’s centre of gravity (G) may appear to rise relative to the metacentre (M) because the upward ground reaction reduces the buoyancy force acting on the underwater volume.
    • This virtual rise in G reduces GM, potentially leading to a list.
    • If the list becomes excessive and the righting moment is insufficient, the ship may capsize.

    If the ship grounds on a pinnacle:

    • The ship experiences two forces at the ship's bottom:
      • A downward force due to the ship’s weight.
      • An upward reaction force is concentrated on the pinnacle.
    • The resulting force between the grounding pressure and the ship’s centre of buoyancy shifts downward towards the pinnacle.
    • This situation is similar to when the ship's stern touches the keel block in a dry dock.
    • A virtual loss of GM occurs because the ship’s inclining moment may exceed the maximum righting lever.
    • If the inclining moment is too great, the ship may develop an excessive list or even capsize.
    Q9 (10 Marks) Ship Stability

    (a) Define longitudinal center of gravity (LCG) and longitudinal centre of buoyancy (LCB) (6)

    (b) A vessel, when floating at a draught of 3.6 m has a displacement of 8172 tonne, KB 1.91 m and LCB 0.15 m aft of midships. From the following information, calculate the displacement, KB and position of the LCB for the vessel when floating at a draught of 1.2 m. (10)

    Draught: 1.2, 2.4, 3.6

    TPC: 23, 24.2, 25

    LCF from midships: 1.37F, 0.76A, 0.92A

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.
    Q10 (10 Marks) Ship Stability

    (a) Describe measures which may be taken to improve the stability or trim of a damaged ships (6)

    (b) A watertight bulkhead is 8 m high and is supported by vertical stiffeners 700 mm apart, connected at the tank top by brackets having 10 rivets 20 mm diameter. The bulkhead is flooded to its top edge with sea water. Determine:

    (a) Shearing force at top of stiffeners,

    (b) Shear stress in the rivets,

    (c) Position of zero shear.

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    Part (a)

    Measures to improve the Stability or Trim of a damaged ship:

    • Close all watertight doors, hatches, and openings to maintain the ship's watertight integrity.
    • Shut valves leading to the damaged compartment to isolate it.
    • Conduct a visual inspection to determine the extent of the damage.
    • Sound the flooded compartment to measure the water level and assess the severity of flooding.
    • Use bilge and ballast pumps to pump out water from the damaged compartment.
    • Transfer liquids in tanks to achieve a balanced condition and counteract the list caused by flooding.
    • Fill tanks on the opposite side of the damaged area to create a counteracting moment.
    • Fill slack tanks to reduce the free surface effect and improve stability.
    • Continuously monitor the ship's stability using the loadicator or other stability monitoring systems.
    • Inform the relevant authorities, including the shipowner and port officials, about the damage.
    • Request external help if necessary for towing, salvage, or additional pumping capacity.
    Q1 (10 Marks) General

    Describe with the aid of diagrammatic sketches the following systems used for transporting liquified gas in bulk:

    (a) Free-standing prismatic tanks

    (b) Membrane tanks

    (c) Free-standing spherical tanks

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    Systems for transporting liquefied gas in bulk.

    Part (a)

    Free-standing prismatic tanks.

    A free-standing prismatic tank is a self-supporting rectangular tank built of cryogenic nickel-steel or aluminium, carried in a hold and supported on the hull. Sketch: a rectangular tank with flat or corrugated plates, a top dome carrying the cargo piping, safety and pressure/vacuum valves, access hatch and level gauging; the tank is supported on load-bearing seatings (wood/mild-steel chocks) at the bottom, with side hoppers and a roof, and is surrounded by insulation (polyurethane foam) to keep the cargo cold and reduce boil-off. It is a self-supporting (type A or B) tank with a secondary barrier (drip tray/gas-tight) so any leakage is contained. The tank is designed to withstand liquid sloshing and cryogenic temperature, with corrugated internal stiffening and notch-tough welded joints. Venting systems, dry-pipe vent risers and flame arresters prevent vapour accumulation; high-level and emergency shut-down (ESD) systems are fitted.

    Part (b)

    Membrane tanks.

    A membrane tank is a thin, non-self-supporting membrane (a thin metallic or non-metallic lining) that is supported by the hull structure through insulation. Sketch: the hull forms the outer structure, and a thin membrane (e.g. stainless steel or Invar) is laid over a layer of insulation (polyurethane foam) on the inner surface of the hull. The membrane contains the liquid, and the insulation keeps the cargo cold and transfers the loads to the hull. The membrane is not self-supporting and relies entirely on the hull for its strength; it is used in large LNG carriers. The membrane is corrugated to accommodate thermal contraction, and a secondary barrier is provided. The system gives a high volumetric efficiency (the tank fills the hull) but requires very careful construction and inspection.

    Part (c)

    Free-standing spherical tanks.

    A free-standing spherical (Moss) tank is a self-supporting spherical tank built of aluminium or cryogenic steel, carried in the hull and supported on a cylindrical skirt (or equatorial ring) at the equator. Sketch: a sphere supported on a vertical cylindrical skirt attached to the hull, with the tank top carrying the cargo piping, relief valve, access hatch and level gauging; the tank is insulated externally. The spherical shape gives the minimum surface area for a given volume (low boil-off) and is structurally efficient (the sphere is strong in all directions), and it is self-supporting. The tank is supported at the equator, and the skirt transfers the loads to the hull. The spherical tank is used in LNG carriers and gives good structural efficiency but a lower volumetric efficiency than the membrane tank.

    Q2 (10 Marks) Ship Stability πŸ”₯ Repeated 8x

    (a) Explain what is meant by "Permissible length" of compartment in passenger ships

    (b) Describe how the position of bulkhead is determined

    (c) Describe briefly the significance of the factor of subdivision

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    Part (a)

    Permissible Length

    Permissible length refers to the maximum length of a ship's compartment that can be flooded while ensuring that the sinkage, heel, or trim remains within acceptable limits without exceeding the floodable length. It ensures the ship remains afloat and stable after flooding.

    Permissible Length Formula:

    $$Permissible\:lenght\:=\:Floodable\:length\:\times Factor\:of\:subdivision$$

    The Factor of Subdivision depends on the ship's length and the nature of its service:

    • For passenger ships, the factor of subdivision is smaller compared to cargo ships.
    • Smaller compartments ensure enhanced safety in case of flooding.
    Part (b)

    Position of Bulkheads

    The position of bulkheads is determined based on the need to comply with the "reasonable amount of damage" criterion in case of flooding:

    • Transverse Watertight Bulkheads should vertically extend up to the margin line.
    • At least one watertight longitudinal bulkhead must be located 20% of the ship's breadth inward from each side.
    • Bulkheads are positioned along the ship's length at intervals equal to the permissible length, but not exceeding the floodable length.
    • The maximum permissible compartment length is limited to 10.7 meters.
    • Collision Bulkhead must be located forward at a distance equal to the permissible length from the forward perpendicular.
    • The ER must be enclosed by two transverse watertight bulkheads, with an aft peak watertight bulkhead enclosing the forward part.
    Part (c)

    Factor of Subdivision

    The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

    Permissible Length Formula:

    $$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

    A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

    Q3 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 5x

    (a) Explain in detail, how an underwater survey is carried out

    (b) State the requirements to be fulfilled before an underwater survey is acceptible to the survey authority

    (c) Construct a list of the items in order of importance that the underwater survey authority should include

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    Part (a)

    An in-water survey, also known as Underwater Inspection in Lieu of Dry Docking (UWILD), involves a systematic and detailed examination of a vessel’s hull and underwater components while the ship remains afloat. The procedure includes the following steps:

    • The shipowner sends a request to the classification society surveyor, indicating the intention to perform an underwater survey.
    • A detailed plan of the ship's external hull features is submitted, showing the location of frames, bulkheads, welding lines, openings, etc.
    • The hull surface is cleaned before the survey to remove any marine growth or fouling that could obscure the inspection.
    • A diving company, approved by the classification society, is appointed to conduct the underwater inspection.

    A self-propelled survey vehicle equipped with the following tools is used:

    • Long-Range Light TV Camera to aid in steering and checking hull deterioration.
    • High-resolution colour TV Camera to provide a close-up view of the hull coating and welded seams.
    • 35mm Still Camera to capture still images.
    • Ultrasonic Probe for measuring plate thickness.
    • Depth Meter and Speed Indicator to provide accurate data on the vehicle's depth and movement.
    • Umbilical Cable to connect the survey vehicle to the survey boat, transmitting power and information.

    The survey boat is to be equipped with:

    • A control console with TV monitors.
    • Plate thickness printouts.
    • Audio and video cassette recorders.
    • Playback units.
    • Diver communication systems.
    • Vehicle control systems and associated instruments.

    Operation:

    • The survey vehicle is taken underwater by a diver to the survey starting point.
    • Using TV monitors and shell expansion plans as guides, the vehicle is navigated over the hull, focusing on the bottom structure, sides, stern frame, rudder, propeller, bilge keel, and hull openings.
    • All images, data, and information are recorded and transmitted back to the survey boat.
    • Detailed pictures of the stern frame, rudder, propeller, bilge keel, and hull openings are captured.
    • Divers are deployed to measure stern tube bearing wear, pintle clearance, and inspect stern seals, anodes, and rudder stock palm coupling bolts.
    • All recorded video and audio, including conversations between the surveyor and drivers, thickness printouts, measurements, and pictures are analyzed to determine the detailed underwater condition of the vessel.
    Part (b)

    Before an in-water survey is accepted by the survey authority, the following conditions must be met:

    The vessel's owner submits a request to the surveyor, including:

    • The proposed date and location for the survey.
    • General information about the diving company.
    • A declaration that the vessel has not suffered any damage due to grounding, collision, or other incidents.

    The ship's master or owner’s representative must provide a declaration confirming:

    • Any suspected or actual damage to the hull since the last dry-docking.
    • The underwater portion of the hull is protected by a suitable paint scheme that is of adequate thickness and remains valid until the next dry-dock.
    • The survey site should be in a protected area with calm and clear water, ensuring good underwater visibility. Attention must be given to the effects of currents and tides.
    • The hull must be clean for the external survey. The surveyor must be satisfied with the method and quality of the pictorial presentation, ensuring that it provides a reliable assessment of the hull's condition.
    • The underwater examination should be conducted by an approved diving company using closed-circuit TV and two-way communication, which can be monitored by the surveyor.
    • The vessel should be in as light an operating condition as possible to facilitate the survey.
    • Means must be available for the surveyor to examine the outside shell plating above the waterline.
    • Any required repairs identified during the survey must be carried out to the satisfaction of the attending surveyor.
    Part (c)

    While the importance of each item may vary depending on the vessel and its specific requirements, below is a list of items that should be included in an underwater survey in order of importance:

    • Underwater Hull: General condition of the hull below the waterline.
    • Bottom and Shell Plating: Inspection for corrosion, damage, and fouling.
    • Shell Openings: Examination of openings such as sea chests, drain plugs, and overboard discharge points.
    • Stern Tube Oil Leaks: Check for leaks around the stern tube.
    • Propeller Blade: Inspection for damage, wear, and fouling.
    • Rudder: Inspection for damage, wear, and clearances.
    • Sea Chest Opening and Grating: Examination for blockages, damage, and fouling.
    • Anodes: Check the condition and effectiveness of cathodic protection anodes.
    • Bilge Keel: Inspection for damage and fouling.
    • Drain Plugs: Ensure all drain plugs are secure and in good condition.
    • Overboard Valve Openings: Check for proper operation and condition.
    • Forward Area: Inspection for any damage due to anchor and chain movement.
    Q4 (10 Marks) Machinery Space & Systems πŸ”₯ Repeated 5x

    (a) Describe the double bottom and framing arrangement used in the machinery space to cope up with the concentrated loads and vibration, together with shaft and thrust block support

    (b) Give reason for the choice of thrust block position

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    Part (a)

    The construction of the double bottom in the machinery space regardless of the framing system has solid plate floors at every frame space under the main engine. Additional side girders are fitted outboard of the main engine seating, as required. The double-bottom height is usually increased to provide fuel oil, lubricating oil and fresh water tanks of suitable capacities. Shaft alignment also requires an increase in the double-bottom height or a raised seating, the former method usually being adopted.

    Continuity of strength is ensured and maintained by gradually sloping the tank top height and internal structure to the required position. Additional support and stiffening is necessary for the main engines, boilers, etc., to provide a vibration-resistant solid platform capable of supporting the concentrated loads. On slow- speed diesel-engined ships, the tank top plating is increased to 40 mm thickness or thereabouts in way of the engine bedplate. This is achieved by using a special insert plate which is the length of the engine including the thrust block in size. Additional heavy girders are also fitted under this plate and in other positions under heavy machinery as required. Plating and girder material in the machinery spaces is of increased scantlings in the order of 10 per cent.

    The method adopted.

    A cellular void space within a ships structure is called a coffer dam. Like a bulkhead separates two spaces or divides a space into two, a coffer dam does the same with a larger degree of integrity since it incorporates a void which would contain any breach of either of the boundaries. This would contain the leakage and prevent it spreading into other areas. A cofferdam can be defined as an empty space separating compartments to prevent the contents of one compartment from entering another in case of leakage.

    Part (b)

    The thrust block is positioned close to the propulsion machinery (usually just aft of the main engine). Reasons are as follows:

    • The axial thrust generated by the propeller could cause deformation and misalignment of the shafting system if the thrust block were positioned far from the engine. Placing it close to the engine minimizes the length of shafting subject to this thrust, thus reducing potential for misalignment. The strong double bottom structure directly beneath provides necessary support to mitigate any deformation.
    • Differential expansion between the shaft and the hull due to temperature variations is a potential source of misalignment. Positioning the thrust block close to the engine helps to minimize the effect of this differential expansion.
    • The weight of the propeller and the dynamic forces it creates can lead to whirling of the tailshaft and misalignment if not properly managed. Positioning the thrust block near the engine helps to stabilize the shafting system and reduce the risk of these issues.
    • The substantial double-bottom structure under the main machinery provides an ideal, inherently strong foundation for the thrust block. This minimises the need for extensive additional reinforcement to support the thrust loads.
    Q5 (10 Marks) Ship Stability πŸ”₯ Repeated 6x

    With reference to Roll-on, Roll-off ferries:

    (a) Describe the problem of free surface effect

    (b) Explain how it is intended that water should be cleared from car or cargo decks

    (c) Describe possible mehtods for improving the stability and survivability of these vessels

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    Part (a)

    The problem of free-surface effect in Ro-Ro ferries.

    Ro-Ro ferries have large, open car and cargo decks that extend over a large part of the ship. If water is shipped onto these decks (e.g. through the bow or stern doors, or in heavy weather), the water spreads over the large deck area, creating a very large free surface. The free-surface effect reduces the effective GM by rho x i/Delta, where i is the second moment of area of the free surface (i = L B^3/12 for a rectangular deck). Because the car deck is very wide and long, the free-surface effect is very large and can reduce the GM to a dangerously low value, causing the ship to lose stability and capsize. This is the principal stability problem of Ro-Ro ferries: the large open decks create a huge free-surface effect if flooded, and the ship can capsize rapidly.

    Part (b)

    How water should be cleared from car or cargo decks.

    Water on the car deck should be cleared by:

    • Providing adequate freeing arrangements (scuppers, freeing ports, drain valves) at the deck edge and at the ends, so that water can drain overboard.
    • Providing a camber (transverse slope) on the deck so that water runs to the sides and drains through the freeing ports.
    • Providing a longitudinal slope (sheer) so that water runs to the ends and drains.
    • Using bilge pumps and drainage systems to remove water that cannot drain overboard.
    • Ensuring the freeing ports are of adequate size and are not blocked by cargo or lashings.

    The freeing arrangements must be adequate to remove water quickly and prevent the build-up of a large free surface.

    Part (c)

    Methods for improving the stability and survivability of Ro-Ro ferries.

    • Lowering the centre of gravity by placing heavy weights low and ballast in the double bottom.
    • Increasing the GM by increasing the beam and the waterplane area, and by lowering KG.
    • Providing adequate freeboard and reserve buoyancy.
    • Subdividing the car deck with watertight bulkheads or providing a raised deck to limit the spread of water.
    • Providing adequate freeing ports and drainage to remove water quickly.
    • Fitting bilge keels and stabilisers to reduce roll.
    • Designing the ship to meet the damage stability criteria (survive flooding of a compartment).
    • Using a higher freeboard and a stronger, more watertight structure, and ensuring the bow and stern doors are watertight and properly secured.
    • Operating with adequate stability margins and following the loading manual.
    Q6 (10 Marks) Ship Stability

    (a) Define Centre of Buoyancy and show with the aid of sketches how a vessel which is stable will return to the upright after being heeled by an external force (6)

    (b) An oil tanker of 17000 tonne displacement has its centre of gravity 1m aft of midships and has 250 tonne of oil fuel in its forward deep tank 75m from midhships. This fuel is transferred to the after oil fuel bunker whose centre is 50m from midships. 200 tonne of fuel from the after bunker is now burned.

    Calculate the new position of the centre of gravity (10)

    (a) After the oil has been transferred

    (b) After the oil has been used

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    Part (a)

    Centre of buoyancy and how a stable vessel returns to the upright.

    The centre of buoyancy (B) is the centre of gravity of the displaced volume of water, i.e. the point through which the resultant of the buoyancy forces acts. For a floating ship it lies on the vertical through the centre of gravity when the ship is upright. When the ship is heeled by an external force, the shape of the displaced volume changes and the centre of buoyancy moves to the low side. The buoyancy force acts vertically upwards through the new centre of buoyancy, while the weight acts vertically downwards through the centre of gravity. If the centre of buoyancy moves such that the buoyancy force acts on the high side of the centre of gravity, a righting couple (Delta x GZ) is produced that returns the ship to the upright. Sketch: the ship heeled, with the centre of gravity G and the new centre of buoyancy B', the buoyancy force acting vertically through B' and the weight through G, producing a righting moment that returns the ship to the upright. The ship is stable if the metacentre M is above G (GM positive).

    Part (b)

    New position of the centre of gravity of an oil tanker.

    An oil tanker of 17,000 t displacement has its centre of gravity 1 m aft of midships and has 250 t of oil fuel in its forward deep tank 75 m from midships. This fuel is transferred to the after oil fuel bunker whose centre is 50 m from midships. 200 t of fuel from the after bunker is now burned. Calculate the new position of the centre of gravity (a) after the oil has been transferred, (b) after the oil has been used.

    Take aft as positive. Initial CG = +1 m (1 m aft of midships). Forward deep tank at -75 m (75 m forward), after bunker at +50 m.

    Part (a)

    After transfer of 250 t from -75 m to +50 m:

    The shift of the CG = w x (change in position)/Delta = 250 x (50 - (-75))/17000 = 250 x 125/17000 = 31,250/17,000 = 1.838 m aft.

    New CG = 1 + 1.838 = 2.838 m aft of midships.

    Part (b)

    After burning 200 t from the after bunker (+50 m):

    Removing 200 t from +50 m shifts the CG forward. The new displacement = 17,000 - 200 = 16,800 t.

    New CG = (17,000 x 2.838 - 200 x 50)/16,800 = (48,246 - 10,000)/16,800 = 38,246/16,800 = 2.277 m aft of midships.

    Answer: (a) CG is 2.84 m aft of midships after transfer; (b) CG is 2.28 m aft of midships after burning 200 t.

    Q7 (10 Marks) Ship Stability

    (a) Explain the term Angle of Loll and state the dangers it poses to a vessel. What action to be taken to correct angle of loll (6)

    (b) A ship of 22000 tonne displacement is 160m long and MCT 1cm 280 tonne m, waterplane area 3060 m2, centre of buoyancy 1m aft of midship and centre of flotation 4m aft of midships. It floats in water of 1.007 t/m3 at draughts of 8.15m forward and 8.75m aft. (10)

    (i) Calculate the new draughts if the vessel moves into sea water of 1.026 t/m3

    (ii) Calculate the metacentric height of the vessel

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    Angle of Loll refers to the angle of heel at which a ship with negative GM (i.e., unstable equilibrium) comes to rest due to the movement of the center of buoyancy beneath the center of gravity. When a ship with negative metacentric height heels to one sideβ€”usually due to an external force like wind or wave actionβ€”the center of buoyancy moves toward the lower side. As the heel increases, the center of buoyancy continues to shift outward, and if it moves directly under the center of gravity (G), the capsizing moment is momentarily nullified. The angle at which this occurs is called the angle of loll.

    Dangers it poses to a vessel:

    • The ship oscillates about the angle of loll rather than about the upright vertical, indicating unstable equilibrium.
    • If the center of buoyancy does not move sufficiently to position itself vertically beneath G, the vessel lacks restoring force and may capsize.
    • The condition is particularly hazardous because it mimics stability from appearance, but in reality, the ship is in a precarious balance.

    Actions to correct angle of loll:

    • Do not attempt to upright the ship by shifting weight transversely, as this can worsen the condition.
    • Press up double bottom tanks to add weight low down and lower the center of gravity, converting GM from negative to positive.
    • Once positive GM is restored, the vessel will return to upright and exhibit true stability.
    Q8 (10 Marks) Ship Stability

    (a) List the variables which affect the force on Rudder (6)

    (b) A ship of 12000 tonne displacement has a rudder 15m3 in area, whose centre is 5m below the waterline. The metacentric height of the ship is 0.3m and the centre of buoyancy is 3.3m below the waterline. When travelling at 20knots the rudder is turned through 30 degrees. Find the iniital angle of heel if the force Fr perpendicular to the plane of the rudder us given by Fr = 577 A v^2 Sin∝ N. Allow 20% for the race effect (10)

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    when a rudder is turned from the centreline plane, to any angle, a force acts on the rudder, given by,

    $$F=kAV^2N$$

    Where,

    • k = Constant
    • A = Area of rudder in m2
    • V = Ship's speed in m/s

    The constant is dependent on size of rudder, rudder angle and density of water.

    Variables affecting force on rudder:

    • Rudder angle
    • Density of water
    • Ship speed

    These are the three variables, while all other are constant parameters i.e.

    • Area of rudder
    • Size of rudder.
    Q9 (10 Marks) Ship Resistance & Propulsion

    (a) Explain the concept of Dynamical stability (6)

    (b) The wetted surface area of a container ship is 5946m2, when travelling at its service speed the effective power required is 11250 kW with frictional resistance 74% of the total resistance and specific fuel consumption of 0.22 kg/kW h. To conserve fuel, the ship speed is reduced by 10%, the daily fuel consumption is then found to be 83 tonne.

    Frictional coefficient in sea water is 1.432.

    Speed in m/s with index (n) 1.825

    Propulsive coefficient may be assumed constant at 0.6

    Determine (10)

    (i) The service speed of the ship

    (ii) The percentage increase in specific fuel consumption when running at reduced speed

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    Part (a)

    Dynamical Stability is defined as the amount of energy required to heel a ship from its upright equilibrium position to a specific angle of heel. It provides a measure of the vessel's stability by considering its behaviour in response to dynamic external forces, such as wind or waves.

    • The concept compares the heeling moment energy (from external forces) and the righting moment energy (from the ship's stability).
    • The ship will absorb the energy imparted by the heeling moment. If the righting energy is greater than the heeling energy, the ship will stabilize; otherwise, it may capsize.

    Areas Under the Curve:

    • Area A: Represents the region where the heeling moment exceeds the righting moment (external energy > ship's stability).
    • Area B: Represents the region where the righting moment exceeds the heeling moment (ship's stability > external energy).
    • The balance of these areas determines whether the ship will right itself or continue to heel.

    When exposed to heeling forces such as wind or waves, the vessel inclines and may roll over to a certain angle of heel. If the external force is applied instantaneously, the ship must have enough reserve dynamic stability to absorb the energy and return to an upright position. If the external force is constant, the ship will remain at an equilibrium angle where the righting moment equals the heeling moment.

    This refers to the remaining righting energy available to counteract additional external forces. A higher reserve dynamic stability ensures the vessel can handle greater heeling forces without capsizing.

    Q10 (10 Marks) Ship Stability

    A box shaped barge of uniform construction is 80m long, 12m beam and has a light displacement of 888 tonnes. The barge is loaded to a draught of 7m in sea water of density 1025 kg/m3 with cargo evenly distributed over two ennd compartments of equal length. The empty midship compartment extends to full width and depth of the barge is bilged and the draught increases to 10m.

    Determine (16)

    (a) The length of the midship compartment

    (b) The longitudinal still water bending moment at midships

    (c) In the loaded intact condition

    (d) In the new bilged condition

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    Box-shaped barge with a bilged midship compartment.

    A box-shaped barge of uniform construction is 80 m long, 12 m beam and has a light displacement of 888 t. The barge is loaded to a draught of 7 m in sea water (1025 kg/m3) with cargo evenly distributed over two end compartments of equal length. The empty midship compartment extends to full width and depth of the barge and is bilged, and the draught increases to 10 m. Determine (a) the length of the midship compartment, (b) the longitudinal still-water bending moment at midships (c) in the loaded intact condition, (d) in the new bilged condition.

    Part (a)

    Length of the midship compartment.

    Loaded displacement = 80 x 12 x 7 x 1.025 = 6888 t. Cargo = 6888 - 888 = 6000 t.

    When the midship compartment (length y) is bilged, the flooded compartment provides no buoyancy. The effective waterplane area = (80 - y) x 12. The lost buoyancy (the volume of the compartment up to the original draught) = y x 12 x 7. The sinkage to 10 m (increase of 3 m) must recover this lost buoyancy:

    Sinkage = lost volume/effective waterplane = (y x 12 x 7)/((80 - y) x 12) = 7y/(80 - y).

    Set equal to 3 m: 7y/(80 - y) = 3, so 7y = 240 - 3y, 10y = 240, y = 24 m.

    Answer: the midship compartment is 24 m long.

    Part (b)

    Longitudinal still-water bending moment at midships.

    The barge is a box of uniform construction (light weight uniformly distributed) with cargo in the two end compartments. The end compartments are each (80 - 24)/2 = 28 m long, carrying 3000 t each (6000/2), i.e. 3000/28 = 107.14 t/m. The light weight is 888/80 = 11.1 t/m uniformly. The buoyancy is uniform at 6888/80 = 86.1 t/m.

    Net load = weight - buoyancy. In the end compartments: weight = 11.1 + 107.14 = 118.24 t/m, buoyancy = 86.1 t/m, net load = 32.14 t/m (downward). In the midship compartment: weight = 11.1 t/m, buoyancy = 86.1 t/m, net load = -75 t/m (upward).

    The bending moment at midships is found by integrating the load. By symmetry, the maximum bending moment is at midships. Consider the left half (40 m): the net load is +32.14 t/m over the first 28 m and -75 t/m over the next 12 m (to midships).

    Shear force at midships = 0 (by symmetry). Bending moment at midships = integral of (shear) = integral of (net load x distance).

    Moment about midships of the left-half loads: the downward load of 32.14 t/m over 0-28 m and the upward load of 75 t/m over 28-40 m.

    Bending moment = 32.14 x 28 x (40 - 14) - 75 x 12 x (40 - 34) = 32.14 x 28 x 26 - 75 x 12 x 6 = 23,398 - 5,400 = 17,998 t-m.

    Answer: the still-water bending moment at midships in the loaded intact condition is about 18,000 t-m (hogging, since the ends are heavier).

    Part (c)

    In the new bilged condition.

    When the midship compartment is bilged, the flooded compartment is full of water, so its weight equals the buoyancy of the water it contains, and it provides no net load. The effective buoyancy is reduced in the flooded region. The bending moment is recomputed with the flooded compartment contributing no net load (the water in it balances the buoyancy). The result is a reduced bending moment at midships, because the heavy end loads are balanced by the buoyancy of the flooded compartment. The exact value depends on the distribution, but the bilged condition generally reduces the hogging moment at midships.

    Q1 (10 Marks) General

    What do you understand by margin line? With a neat diagram explain following terms:

    (a) Floodable length curve

    (b) Permissible length curve

    (c) Factor of subdivision

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    Part (a)

    Floodable length

    is the amount, the ship can be flooded without the risk of sinking. i.e. the maximum length of the ship that can be flooded without submerging the margin line. The margin line is located 75mm below the bulkhead deck, providing a safety margin.

    • A floodable length curve can be drawn for different permeability.
    • It is used to test the after-bulkhead length so that the bulkhead length is sufficient to contain the flooding without sinking the vessel.
    • Factor of subdivision = Permissible length / Floodable length
    Part (c)

    Factor of Subdivision

    The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

    Permissible Length Formula:

    $$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

    A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

    Q2 (10 Marks) Hull Construction

    With reference to cargo hatch covers on large container ships:

    (a) Describe how they are tested for water tightness

    (b) Explain how the weight of the hatch and containers is transferred to the ship's structure whilst allowing for deflections of the hull in a seaway

    (c) Describe, with the aid of a sketch, the type and location of damage that can occur due to wear of the hatch supporting arrangements.

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    Part (a)

    Cargo hatch cover can be tested for water tightness by Hose test. In this test, a water spray from a nozzle of 12mm diameter is sprayed over the joint of hold and cover from a distance of 1m to 1.5 m with a pressure of 0.5 m/sec water jet. Any leak of water inside the cargo hold will indicate a gap between the hold and the cover.

    Other methods include:

    • Chalk test - on top of bar that the hatch sealing rubber rests
    • Ultrasonic testers are available to check the seals
    • Inspect from inside of the closed hatch for any light penetrating
    Part (b)

    Weight-bearing pads are fitted on large hatch covers to relieve the load on the seals. Weight-bearing pads allow the hatch to flex as well. These pads allow the hatch to move, without compromising the watertight seal or structural integrity. This flexibility ensures that the loads are still transferred without stressing the hatch covers or the deck structure excessively

    Part (c)

    Wear of the support arrangements can put excessive loading on the seal, which will cause the failure of the sealing arrangement. The hatch coaming can be overloaded and could cause cracking in side plates and buckling of support stays

    Q3 (10 Marks) Ship Types & Design

    with reference to the classification of ships, explain each of the following:

    (a) Why ships are built to classification society rules.

    (b) The meaning of the notation 100A1.

    (c) How a ship remains in class throughout the life of the vessel.

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    Part (a)

    The cost of insurance of both ship and cargo depends largely upon the classification, the higher standard requiring smaller premiums. It is, therefore, to the advantage of the ship-owner to have a high-class ship. However, the classification societies are independent of the insurance companies. Each of the classification societies has its own rules, but there is a similarity between them, and they are used to determine the scantlings of the structural members.

    The scantlings of the structure are based on theory, but because a ship is a very complex structure, a `factor of experience' is introduced. The classification societies receive reports of all faults and failures in ships that carry their classification, and based on these reports, consistent faults in any particular type of ship may be analysed, and amendments made to the rules.

    The scantling plans are submitted to the classification societies for their approval before the detailed plans are drawn. The procedure should ensure the quality of build for the vessel, which, when built in accordance with the classification society rules, is assigned a class. This class applies as long as the ships are found under survey to be in a fit and efficient condition.

    Part (b)

    The notation ✠100 A1 means suitable for seagoing service on long international voyages built to the highest standard with a surveyor in attendance.

    • Class A is assigned to ships that are built in accordance with the rules or have equivalent strength.
    • Figure '1' is added (i.e. 100 A 1) when the equipment, consisting of anchors, cables, mooring ropes and towropes, is in good and efficient condition.
    • ✠ Is added to the `100 A 1' notation when a ship is fully built under Special Survey, i.e. when a surveyor is in attendance and examines the ship during all stages of the construction.
    Part (c)

    To ensure that the ship remains worthy of its classification, annual docking and special surveys are carried out by the surveyors. Docking surveys are approximately every 2Β½ years, whilst special surveys are carried out at intervals of 4 to 5 years also, any structural changes are to be agreed with the classification society before starting.

    Q4 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 4x

    If a ship is seriously damaged under water in way of a large fuel oil side bunker tank What is the immediate effect and what may ultimately happen? What features in the ship would enhance safety?

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    If a ship is seriously damaged underwater in the area of a large fuel oil side bunker tank, it will cause the following:

    Bilging or Flooding of the Compartment:

    • Water will enter the damaged bunker tank, reaching the draught level.
    • The rate of flooding depends on the size of the breach and the water pressure at the depth of the damage.

    Oil Leakage into the Sea:

    • If the tank contains fuel oil, oil will begin to leak out, causing pollution.
    • The extent of leakage depends on the tank’s contents (empty, half-full, or full) and the location of the damage.

    List and Trim of the Vessel:

    • The ingress of water and loss of oil will create an imbalance, causing the vessel to list or trim.

    If corrective actions are not taken, uncontrolled flooding and loss of stability could lead to capsizing or sinking of the vessel.

    Features in the Ship to Enhance Safety:

    • Small Bunker Tank Sizes reduces the risk of extensive oil spillage and loss of stability.
    • Connectivity to transfer pumps allows the transfer of oil from the damaged tank to an empty tank, minimizing oil spillage and counteracting the loss of stability.
    • The tank’s size and location are designed to limit the effects of flooding, as per damage stability regulations.
    • Properly positioned transverse and longitudinal bulkheads enhance the subdivision factor, limiting water ingress to the damaged tank.
    • Ship’s Ballast System allows corrective ballasting to counteract the list or trim caused by the ingress of water.
    • Watertight Doors and Hatches prevent water from spreading to adjacent compartments.

    Recommended Immediate Actions by Crew:

    For Empty Tanks:

    • Quickly seal off the damaged tank by shutting all valves and isolating it from the transfer system to prevent water ingress into other parts of the vessel.

    For Half-Empty or Full Tanks:

    • Initiate oil transfer to another empty tank to reduce oil leakage and stabilize the ship.
    • Monitor the water ingress and ensure the tank is filled to the draught level with seawater if necessary, using ballast to correct the list.
    Q5 (10 Marks) Corrosion & Protection πŸ”₯ Repeated 4x

    With reference to the prevention of hull corrosion, discuss:

    (a) Surface preparation and painting of new ship plates.

    (b) Design of the ships structure and its maintenance.

    (c) Cathodic protection by sacrificial anodes, of the internal and external areas of the ship

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    Part (a)

    Surface Preparation and Painting of New Ship Plates:

    The process begins with the removal of mill scale, a thin layer of iron oxide that forms during the steel rolling process.

    1. Removal of Mill Scale:

    • Weathering: Using wire brushes.
    • Pickling: Immersing plates in a weak solution of Hβ‚‚SOβ‚„ or HCl.
    • Shot Blasting: Abrasive blasting to remove rust and mill scale.
    • Flame Cleaning: Heating and cleaning the surface.

    Freshwater Washing and Drying: Ensures the surface is clean and ready for priming.

    2. A primer is then applied. This acts as a base layer for subsequent coatings and improves adhesion. Common types include epoxy primers with corrosion-inhibiting pigments like iron oxide, zinc, and calcium phosphates (though zinc content is reduced due to safety concerns).

    3. Next comes the anticorrosive coating, the main layer offering corrosion protection. Common choices are two-component epoxies, coal tar epoxies, and epoxy or polyester coatings with glass flakes for added strength and water impermeability.

    4. Finally, an antifouling coating prevents marine organism attachment. While tin-based paints were once common, environmental regulations have led to their replacement with copper-based, silicone-based, or non-TBT (Tributyltin) self-polishing coatings. These utilize seawater-soluble polymers. The number of antifouling layers (two or three) depends on the chosen system and desired lifespan. The dry film thickness (DFT) of the primer is closely monitored to prevent cracking and ensure effective protection.

    Part (b)

    Corrosion prevention begins at the design stage:

    • Avoid dissimilar metal joining to minimize galvanic corrosion.
    • Use high-quality steel plates based on the galvanic series.
    • Minimize areas where water can accumulate by ensuring proper drainage systems.
    • Ensure sufficient anodes are installed in optimal positions.
    • Consider installing Impressed Current Cathodic Protection (ICCP) and Marine Growth Prevention Systems (MGPS).

    Maintenance:

    • Regular checks and maintenance of ICCP and MGPS systems.
    • Renewal of sacrificial anodes during dry-docking.
    • Cleaning, repairing, and repainting surfaces to maintain the protective coatings.
    • Ensuring proper drainage and inspecting for any areas of corrosion.
    Part (c)

    Cathodic Protection by Sacrificial Anodes:

    Sacrificial anodes are less noble metals (more electro-negative) than the hull material. When immersed in seawater, they corrode preferentially, protecting the hull. Common materials include zinc, aluminium, and magnesium. They are welded to the hull for good electrical contact.

    • External surface: Large anodes provide complete protection until the next dry-docking.
    • Internal surface: Anodes are fitted inside tanks, filters, and coolers to prevent internal corrosion.
    • MGPS: These systems utilize sacrificial anodes to protect seawater lines and pumps, further preventing marine growth.
    • ICCP: Impressed current systems offer an alternative, using an external DC power source and permanent anodes (typically titanium or platinum) for continuous protection. These systems are self-regulating and adjust current output based on hull coating condition. Regular monitoring (e.g., monthly data analysis) is important for both sacrificial and impressed current systems.
    Q6 (10 Marks) Ship Resistance & Propulsion

    (a) Describe the effect of cavitation on (6)

    (i) The thrust and torque

    (ii) The propeller blades

    (b) A ship of 355190 tonne displacement is 325 m long, 56 m wide and floats in sea water of density 1025 kg/m3 at a draught of 22.4 m. The propeller has a diameter of 7.4 m, a pitch ratio of 0.85, and when rotating at 1.5 rev/s the real slip is 48.88% and the fuel consumption is 165 tonne per day. The Taylor wake fraction Wt, Wt=0.5Cb-0.05. Calculate: (10)

    (i) The ship speed in knots

    (ii) The reduced speed at which the ship should travel if the fuel consumption on a voyage is to be halved

    (iii) The length of the voyage if the extra time on passage is six days when travelling at the reduced speed

    (iv) The amount of fuel required on board before commencing on the voyage at the reduced speed.

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    Part (a)

    (ii) Effect of cavitation on propeller blades:

    Erosion:

    • Cavitation causes the formation and collapse of vapor bubbles on the propeller blade surface.
    • The collapse of these bubbles produces high-pressure shockwaves and microjets that erode the blade material, leading to surface pitting and progressive damage.

    Vibration:

    • Uneven cavitation across the blades leads to imbalanced forces, causing vibrations in the propeller and the ship.
    • These vibrations can reduce the comfort of passengers and crew and stress the ship's structural components.

    Noise:

    • The collapse of vapor bubbles generates loud noise, which can interfere with onboard communication and underwater sonar systems.
    • This noise is a significant concern for naval vessels and marine life.

    Reduced Performance:

    • Cavitation reduces the efficiency of the propeller by causing loss of thrust and torque.
    • The presence of cavitation decreases the propeller’s ability to convert rotational energy into forward motion, lowering the ship's speed and increasing fuel consumption.
    Q7 (10 Marks) Ship Stability

    (a) Why is an inclining experiment carried out? Write a short account of the method adopted. (6)

    (b) An inclining experiment was carried out on a ship of 8000 tonne displacement. The inclining ballast was moved transversely through 12 m and the deflections of a pendulum 5.5 m long, measured from the centreline, were as follows:

    3 tonne port to starboard 64 mm S

    3 tonne port to starboard 116 mm S

    Ballast restored 3 mm S

    3 tonne starboard to port 54 mm P

    3 tonne starboard to port 113 mm P

    Calculate the metacentric height of the vessel. (10)

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    Part (a)

    An inclining experiment

    is conducted to determine a ship's metacentric height (GM) and, consequently, the location of its centre of gravity (CG). Knowing the CG of an empty vessel allows for calculations of its position under various loading conditions.

    The inclining experiment is performed on:

    • Newly built ships.
    • After major alterations to the vessel.
    • As required by the classification society.

    Conducting the inclining experiment:

    • The ship should be in a sheltered location (like a dry dock) with mooring ropes slack, only essential personnel on board, all tanks either empty or full, and any loose weights removed or secured.
    • At least two pendulums (one forward, one aft) are used, ideally as long as possible and suspended from convenient points (e.g., under a hatch). A hood filled with water or oil is placed beneath each pendulum bob to dampen its swing for accuracy.
    • Four masses are positioned on the deck, two on each side of the midships, their centres as far from the centerline as possible. These masses are moved systematically: all four to one side, then all four to the other, and finally two on each side.
    • The pendulum deflection is recorded for each mass movement.
    • The average of these deflections is used to calculate the metacentric height (GM).
    Q8 (10 Marks) Hull Construction

    (a) Explain the use of KN curves. (6)

    (b) The 1/2 breadths of the load waterplane of a ship 150 m long commencing from aft, are 0.3, 3.8, 6.0, 7.7, 8.3, 9.0, 8.4, 7.8, 6.9, 4.7 and O m respectively. Calculate:

    (i) Area of waterplane

    (ii) Distance of centroid from amidships

    (iii) Second moment of area about a transverse axis through centroid.

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    Part (a)

    KN curves are stability curves used in naval architecture. They are created by plotting the righting lever (KN) against the angle of heel (ΞΈ) for a given displacement and an assumed KG (height of the centre of gravity above the keel) of 0. The righting lever is the distance between the centre of buoyancy and the centre of gravity, measured along a vertical line through the centre of gravity.

    The KN curve is then used to calculate the righting lever (GZ) for any other KG value using the following formula:

    GZ = KN - KG * sin ΞΈ

    Where:

    • GZ is the righting lever measured from the centre of gravity.
    • KN is the righting lever measured from the keel (obtained from the KN curve).
    • KG is the distance of the centre of gravity from the keel.
    • ΞΈ is the angle of heel.
    Q9 (10 Marks) Ship Stability

    (a) Describe how water tightness is maintained where bulkheads are pierced by longitudinal, beams or pipe. (6)

    (b) The length of a ship is 7.6 times the breadth, while the breadth is 2.85 times the draught. The block coefficient is 0.69, prismatic coefficient 0.735, waterplane area coefficient 0.81 and the wetted surface area 7000m2. The wetted surface area S is given by Denny's formula Calculate: (10)

    (i) Displacement in tonne

    (ii) Area of immersed midship section

    (iii) Waterplane areas

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    Part (a)

    Water tightness of

    bulkheads that are pierced by longitudinal beams or pipes:

    • Bulkheads must be made structurally watertight without the use of wood packing.
    • Pipes passing through bulkheads are either welded or fastened to the bulkhead using studs or hoses secured through tapped holes in the plating.
    • When a bulkhead is pierced by a longitudinal beam, the openings are kept as small as possible, and doubler plates are welded on each side to maintain watertight integrity.
    • Sealing materials and gaskets are used, particularly for pipes, cables, and similar penetrations.
    Q10 (10 Marks) Ship Stability

    A box shaped vessel is 20 m long and 10 m wide. The weight of the vessel is uniformly distributed throughout the length and the draught is 2.5 m. The vessel contains ten evenly spaced double bottom tanks. each having a depth of 1 m. Draw the shear force diagrams

    (a) With No. 1 and No. 10 tanks filled

    (b) With No. 3 and No. 8 tanks filled

    (c) With No. 5 and No. 6 tanks filled.

    Which ballast condition is to be preferred from the strength point of view. (16)

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    Shear-force diagrams for a box-shaped vessel with double-bottom tanks.

    A box-shaped vessel is 20 m long and 10 m wide. The weight of the vessel is uniformly distributed throughout the length and the draught is 2.5 m. The vessel contains ten evenly spaced double-bottom tanks, each having a depth of 1 m. Draw the shear-force diagrams for (a) No. 1 and No. 10 tanks filled, (b) No. 3 and No. 8 tanks filled, (c) No. 5 and No. 6 tanks filled. Which ballast condition is preferred from the strength point of view?

    The vessel is a box 20 m long, 10 m wide, draught 2.5 m. Displacement = 20 x 10 x 2.5 x 1.025 = 512.5 t. The weight is uniformly distributed (the light ship), and the ballast tanks add weight at their positions. Each tank is 2 m long (20/10), 10 m wide, 1 m deep, so each holds 2 x 10 x 1 x 1.025 = 20.5 t of water when full.

    The buoyancy is uniform at 512.5/20 = 25.625 t/m. The light weight is uniform at (512.5 - ballast)/20. When tanks are filled, the weight in those sections increases.

    Part (a)

    No. 1 and No. 10 tanks filled: the ballast is at the two ends (0-2 m and 18-20 m). The weight is increased at the ends, so the ends are heavier than the middle, producing a hogging condition (the ends sag, the middle is buoyant). The shear force diagram shows a positive shear at the ends building up, reaching a maximum near the quarter points, and the bending moment is hogging (positive) with a maximum at midships.

    Part (b)

    No. 3 and No. 8 tanks filled: the ballast is at 4-6 m and 14-16 m, i.e. near the quarter points. This produces a more even distribution, with the weight concentrated near the quarters. The shear force diagram shows the shear building up and the bending moment is smaller than in (a).

    Part (c)

    No. 5 and No. 6 tanks filled: the ballast is at 8-10 m and 10-12 m, i.e. amidships. The weight is concentrated amidships, producing a sagging condition (the middle is heavier, the ends are buoyant). The shear force diagram shows the shear building up and the bending moment is sagging (negative) with a maximum at midships.

    Preferred condition: from the strength point of view, the condition that produces the smallest bending moment is preferred. The condition with the ballast near the quarter points (No. 3 and No. 8) produces the smallest bending moment, because the weight is distributed to reduce the difference between the weight and buoyancy distributions. The end-filling (No. 1 and No. 10) produces a large hogging moment, and the midship-filling (No. 5 and No. 6) produces a large sagging moment. Hence the No. 3 and No. 8 condition is preferred.

    Q1 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 13x

    Describe the relationship between frictional resistance and

    (a) Ship's speed,

    (b) The wetted area

    (c) The surface roughness

    (d) The length of the vessel

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    Frictional resistance arises due to the interaction between the ship's hull and the water as the vessel moves. This resistance is caused by eddying water adjacent to the hull that is drawn along with the ship. The frictional resistance is given by:

    $$R_{f\:=\:}f\:.\:s\:.\:v^{n}$$

    Where:

    • f = Coefficient of friction
    • s = Wetted surface area
    • v = Ship's speed in knots
    • n = Constant (1.82)
    Part (a)

    Ship Speed (v):

    • Frictional resistance increases with increasing ship speed. The relationship is not perfectly linear, but it's approximately described by the formula Rf = fsv^n (Rf ∝ v^n), where 'n' is a constant (approximately 1.82). This means that a small increase in speed results in a larger increase in frictional resistance.
    Part (b)

    Wetted Area (s):

    • Frictional resistance is directly proportional to the wetted surface area (Rf ∝ s). A larger wetted area (the part of the hull submerged in water) leads to greater frictional resistance. The wetted area increases with the ship's draught (depth in the water).
    Part (c)

    Surface Roughness:

    • Increased surface roughness increases frictional resistance. Roughness can be caused by fouling (marine growth), corrosion, or poor hull maintenance. Regular cleaning and maintenance of the hull surface help to minimise roughness and reduce resistance.
    Part (d)

    Length of the Vessel

    • Frictional resistance is influenced by the vessel's length. As the length increases, the wetted surface area grows, and the coefficient of friction (f) changes, leading to an increase in Rf. Longer vessels experience higher frictional resistance, but streamlined designs can mitigate the effect to some extent.
    Q2 (10 Marks) Ship Stability πŸ”₯ Repeated 6x

    With reference to Roll-on, Roll-off ferries:

    (a) Describe the problem of free surface effect

    (b) Explain how it is intended that water should be cleared from car or cargo decks

    (c) Describe possible methods for improving the stability and survivability of these vessels.

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    Part (a)

    The problem of free-surface effect in Ro-Ro ferries.

    Ro-Ro ferries have large, open car and cargo decks that extend over a large part of the ship. If water is shipped onto these decks (e.g. through the bow or stern doors, or in heavy weather), the water spreads over the large deck area, creating a very large free surface. The free-surface effect reduces the effective GM by rho x i/Delta, where i is the second moment of area of the free surface (i = L B^3/12 for a rectangular deck). Because the car deck is very wide and long, the free-surface effect is very large and can reduce the GM to a dangerously low value, causing the ship to lose stability and capsize. This is the principal stability problem of Ro-Ro ferries: the large open decks create a huge free-surface effect if flooded, and the ship can capsize rapidly.

    Part (b)

    How water should be cleared from car or cargo decks.

    Water on the car deck should be cleared by:

    • Providing adequate freeing arrangements (scuppers, freeing ports, drain valves) at the deck edge and at the ends, so that water can drain overboard.
    • Providing a camber (transverse slope) on the deck so that water runs to the sides and drains through the freeing ports.
    • Providing a longitudinal slope (sheer) so that water runs to the ends and drains.
    • Using bilge pumps and drainage systems to remove water that cannot drain overboard.
    • Ensuring the freeing ports are of adequate size and are not blocked by cargo or lashings.

    The freeing arrangements must be adequate to remove water quickly and prevent the build-up of a large free surface.

    Part (c)

    Methods for improving the stability and survivability of Ro-Ro ferries.

    • Lowering the centre of gravity by placing heavy weights low and ballast in the double bottom.
    • Increasing the GM by increasing the beam and the waterplane area, and by lowering KG.
    • Providing adequate freeboard and reserve buoyancy.
    • Subdividing the car deck with watertight bulkheads or providing a raised deck to limit the spread of water.
    • Providing adequate freeing ports and drainage to remove water quickly.
    • Fitting bilge keels and stabilisers to reduce roll.
    • Designing the ship to meet the damage stability criteria (survive flooding of a compartment).
    • Using a higher freeboard and a stronger, more watertight structure, and ensuring the bow and stern doors are watertight and properly secured.
    • Operating with adequate stability margins and following the loading manual.
    Q3 (10 Marks) Ship Types & Design πŸ”₯ Repeated 6x

    (a) Considering the vessel as a compound beam define Bending moment shearing force. Which is the point of Maximum Bending Moment?

    (b) Sketch and Describe Hatch coming of a large bulk carrier.

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    Part (a)

    Bending moment and shearing force, and the point of maximum bending moment.

    Considering the vessel as a compound beam (the hull girder), the shearing force at any section is the algebraic sum of the vertical forces (loads) to one side of the section, i.e. the net load (weight - buoyancy) acting on that part. The bending moment at any section is the algebraic sum of the moments of the loads to one side, i.e. the integral of the shearing force. The shearing force is the rate of change of the bending moment, and the bending moment is the integral of the shearing force. The point of maximum bending moment occurs where the shearing force is zero (where the shearing force changes sign), which is usually at or near midships for a ship in still water, and at the point where the net load changes sign. The maximum bending moment is the largest hogging or sagging moment, and the hull girder must be designed to withstand it.

    Part (b)

    Hatch coaming of a large bulk carrier.

    The hatch coaming is the vertical structure around the hatch opening that raises the hatch above the deck to prevent water entering and to provide strength. Sketch: the hatch opening is bounded by a vertical coaming plate (about 600-900 mm high for a bulk carrier) welded to the deck, with a top flange (or a horizontal stiffener) and vertical stiffeners (brackets) connecting the coaming to the deck. The coaming is made of thick plate and is stiffened to resist the loads of the hatch cover and the cargo, and to provide the longitudinal strength of the deck (the coaming acts as a longitudinal girder). The hatch cover sits on the coaming with a gasket and is secured by cleats. The coaming corners are rounded and reinforced to avoid stress concentrations. The coaming provides the watertight seal for the hatch and contributes to the longitudinal strength of the hull girder.

    Q4 (10 Marks) Corrosion & Protection πŸ”₯ Repeated 4x

    With reference to the prevention of hull corrosion, discuss:

    (a) Surface preparation and painting of new ship plates.

    (b) Design of the ships structure and its maintenance.

    (c) Cathodic protection by sacrificial anodes, of the internal and external areas of the ship.

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    Part (a)

    Surface Preparation and Painting of New Ship Plates:

    The process begins with the removal of mill scale, a thin layer of iron oxide that forms during the steel rolling process.

    1. Removal of Mill Scale:

    • Weathering: Using wire brushes.
    • Pickling: Immersing plates in a weak solution of Hβ‚‚SOβ‚„ or HCl.
    • Shot Blasting: Abrasive blasting to remove rust and mill scale.
    • Flame Cleaning: Heating and cleaning the surface.

    Freshwater Washing and Drying: Ensures the surface is clean and ready for priming.

    2. A primer is then applied. This acts as a base layer for subsequent coatings and improves adhesion. Common types include epoxy primers with corrosion-inhibiting pigments like iron oxide, zinc, and calcium phosphates (though zinc content is reduced due to safety concerns).

    3. Next comes the anticorrosive coating, the main layer offering corrosion protection. Common choices are two-component epoxies, coal tar epoxies, and epoxy or polyester coatings with glass flakes for added strength and water impermeability.

    4. Finally, an antifouling coating prevents marine organism attachment. While tin-based paints were once common, environmental regulations have led to their replacement with copper-based, silicone-based, or non-TBT (Tributyltin) self-polishing coatings. These utilize seawater-soluble polymers. The number of antifouling layers (two or three) depends on the chosen system and desired lifespan. The dry film thickness (DFT) of the primer is closely monitored to prevent cracking and ensure effective protection.

    Part (b)

    Corrosion prevention begins at the design stage:

    • Avoid dissimilar metal joining to minimize galvanic corrosion.
    • Use high-quality steel plates based on the galvanic series.
    • Minimize areas where water can accumulate by ensuring proper drainage systems.
    • Ensure sufficient anodes are installed in optimal positions.
    • Consider installing Impressed Current Cathodic Protection (ICCP) and Marine Growth Prevention Systems (MGPS).

    Maintenance:

    • Regular checks and maintenance of ICCP and MGPS systems.
    • Renewal of sacrificial anodes during dry-docking.
    • Cleaning, repairing, and repainting surfaces to maintain the protective coatings.
    • Ensuring proper drainage and inspecting for any areas of corrosion.
    Part (c)

    Cathodic Protection by Sacrificial Anodes:

    Sacrificial anodes are less noble metals (more electro-negative) than the hull material. When immersed in seawater, they corrode preferentially, protecting the hull. Common materials include zinc, aluminium, and magnesium. They are welded to the hull for good electrical contact.

    • External surface: Large anodes provide complete protection until the next dry-docking.
    • Internal surface: Anodes are fitted inside tanks, filters, and coolers to prevent internal corrosion.
    • MGPS: These systems utilize sacrificial anodes to protect seawater lines and pumps, further preventing marine growth.
    • ICCP: Impressed current systems offer an alternative, using an external DC power source and permanent anodes (typically titanium or platinum) for continuous protection. These systems are self-regulating and adjust current output based on hull coating condition. Regular monitoring (e.g., monthly data analysis) is important for both sacrificial and impressed current systems.
    Q5 (10 Marks) Ship Types & Design

    With reference to Container Ships:

    (a) Describe the problem of Parametric Roll

    (b) Explain how this problem is suitably addressed in the design and operation of the vessel.

    (c) Describe possible methods for improving the stability of these vessels.

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    Part (a)

    The problem of parametric roll in container ships.

    Parametric roll is a phenomenon in which a ship in head or following seas develops a large, resonant roll even though the wave is not beam-on. It occurs because the ship's metacentric height (GM) and hence the roll natural frequency vary periodically as the wave passes along the hull: when a wave crest is amidships the waterplane area and GM are reduced, and when a trough is amidships the GM is increased. If the encounter frequency of the waves is about twice the natural roll frequency, the periodic variation of GM excites the roll in resonance, and the roll amplitude grows rapidly, even in moderate seas. Container ships are particularly susceptible because they have large, open hatch openings (reduced waterplane area and GM variation), a high centre of gravity (small GM), and a large bow flare and stern, which increase the variation of the waterplane area with the wave. Parametric roll can produce large roll angles (20-40 deg) that can cause cargo shift, container loss, and even capsize.

    Part (b)

    How the problem is addressed in design and operation.

    • Design: increase the GM (lower the centre of gravity, increase the beam), reduce the variation of the waterplane area with the wave (reduce the bow flare and the fullness of the ends), and provide adequate stability and damping (bilge keels, stabilisers). The hull form is designed to reduce the periodic variation of GM.
    • Operation: avoid the resonant conditions by changing speed and course so that the encounter frequency is not about twice the natural roll frequency; use the roll damping devices; monitor the roll and the sea state; and follow the operational guidance (e.g. the parametric roll guidance in the loading manual and the weather routing). Reducing speed in head seas and changing heading can avoid the resonance.
    Part (c)

    Methods for improving the stability of container ships.

    • Lowering the centre of gravity by placing heavy weights (machinery, ballast) low and by limiting the height of the container stacks.
    • Increasing the GM by increasing the beam and the waterplane area, and by lowering KG.
    • Providing adequate freeboard and reserve buoyancy.
    • Fitting bilge keels, anti-roll tanks and stabilisers to damp the roll.
    • Using ballast to improve the stability and to reduce the GM variation.
    • Designing the hull to reduce the susceptibility to parametric roll.
    • Following the loading manual and the stability criteria, and monitoring the stability in all conditions.
    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 3x

    (a) Describe how water tightness is maintained where bulkheads are pierced by longitudinal beams or pipes. (6)

    (b) A triangular bulkhead is 7m wide at the top and has a vertical depth of 8m. Calculate the load on the bulkhead and the position of centre of pressure if the bulkhead is flooded with sea water on only side:

    (i) To the top edge

    (ii) With 4m head to the top edge.

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    Part (a)

    Water tightness of

    bulkheads that are pierced by longitudinal beams or pipes:

    • Bulkheads must be made structurally watertight without the use of wood packing.
    • Pipes passing through bulkheads are either welded or fastened to the bulkhead using studs or hoses secured through tapped holes in the plating.
    • When a bulkhead is pierced by a longitudinal beam, the openings are kept as small as possible, and doubler plates are welded on each side to maintain watertight integrity.
    • Sealing materials and gaskets are used, particularly for pipes, cables, and similar penetrations.
    Part (b)

    $$\left(a\right)\:Load\:on\:bulkhead=\rho gAH$$

    $$=1025\times9.81\times\frac{7\times8}{2}\times\frac83$$

    $$=750.8\times10^3N$$

    $$=750.8KN$$

    $$Centre\:of\:pressure\:fron\:top=\frac12D=\frac12\times8$$

    $$=4m$$

    $$\left(b\right)\:Load\:on\:bulkhead=1025\times9.81\times\frac{7\times8}{2}\times\left(\frac83+4\right)\times$$

    $$=1.877\times10^6N$$

    $$=1.877MN$$

    $$For\:triangle\:I_{NA}=\frac{1}{36}BD^3$$

    $$Centre\:of\:pressure\:from\:surface\:of\:water\:=\frac{I_{NA}}{AH}+H$$

    $$=\frac{\frac{1}{36}\times7\times8^3}{\frac12\times7\times8\times\left(\frac83+4\right)}+\left(\frac83+4\right)$$

    $$=\frac{2\times7\times8^3}{36\times7\times8\times6.667}+6.667$$

    $$=7.20\:m$$

    $$Centre\:of\:pressure\:from\:top\:of\:bulkhead$$

    $$=7.20-4$$

    $$3.2m$$

    Q7 (10 Marks) Hull Construction πŸ”₯ Repeated 4x

    An oil tanker 160m long and 2m beam floats ta a draught of 9m in seawater. Cw is 0.865. The midship section is in the form of a rectangle with 1.2m radius at the bilges. A midship tank 10.5m long has twin longitudinal bulkheads and contains oil of 1.4m3/t to a depth of 11.5m. The tank is holed to the sea for the whole of its transverse section. Find the new draught. (16)

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    New draught of the oil tanker when the midship tank is holed.

    Oil tanker 160 m long, 22 m beam, floats at a draught of 9 m in sea water. Cw = 0.865. The midship section is a rectangle with 1.2 m radius at the bilges. A midship tank 10.5 m long has twin longitudinal bulkheads and contains oil of 1.4 m3/t to a depth of 11.5 m. The tank is holed to the sea for the whole of its transverse section. Find the new draught.

    Waterplane area Aw = Cw x L x B = 0.865 x 160 x 22 = 3044.8 m2.

    Midship section area (rectangle with bilge radius r=1.2 m): Ams = B x d - (4 - pi) r^2 = 22 x 9 - 0.858 x 1.44 = 198 - 1.236 = 196.76 m2.

    Volume of the tank below the original waterline = Ams x 10.5 = 196.76 x 10.5 = 2066 m3.

    The tank contains oil of density rho_o = 1/1.4 = 0.714 t/m3. When holed, sea water (1.025 t/m3) replaces the oil, so the net loss of buoyancy is the volume times the relative density difference:

    Vlost = 2066 x (1 - 0.714/1.025) = 2066 x 0.303 = 626 m3.

    The flooded tank provides no increase of buoyancy, so the effective sinking waterplane = Aw - (10.5 x 22) = 3044.8 - 231 = 2813.8 m2.

    Sinkage = Vlost/effective waterplane = 626/2813.8 = 0.222 m.

    New draught = 9 + 0.22 = 9.22 m.

    Answer: the new draught is about 9.2 m.

    Q8 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 7x

    What is meant by the Admiralty Coefficient and the Fuel Coefficient? (6)

    (a) A ship of 14900 tonne displacement has a shaft power of 4460 kW at 14.55 knots. The shaft power is reduced to 4120 kW and the fuel consumption at the same displacement is 541 kg/h. Calculate the fuel coefficient for the ship (10)

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    The Admiralty Coefficient (C) is a method for estimating the propulsion power needed for a newly built ship. It's considered relatively constant for a given ship design. The formula is:

    $$C=\frac{\Delta^{2/3}\times V^3}{BP}$$

    Where:

    • C = Admiralty Coefficient
    • Ξ” = Displacement in tonnes (weight of the ship when fully loaded)
    • V = Speed in knots
    • BP = Brake power in kilowatts (power delivered by the ship's engine)

    A higher Admiralty Coefficient indicates a more efficient ship design, meaning it requires less power to achieve a given speed. Values typically range from 350 to 600.

    Fuel Coefficient:

    The Fuel Coefficient (F.C.) is used to calculate a ship's daily fuel oil consumption. The formula is:

    $$Daily\:fuel\:oil\:consumption\:=\:\frac{\Delta^{2/3}\times V^3}{FC}$$

    Where:

    • F.C. = Fuel Coefficient
    • Ξ” = Displacement in tonnes
    • V = Speed in knots

    The Fuel Coefficient can vary significantly, with typical values ranging from 40,000 to 120,000. A higher Fuel Coefficient implies greater fuel efficiency (lower daily fuel consumption) for a given speed and displacement.

    Part (b)

    $$admiraty\:coefficient\:\left(C\right)=\:\frac{\Delta^{2/3}V^3}{Shaft\:power}=\frac{\Delta^{\frac23}\times V^3}{SP}$$

    $$\frac{SP_1}{SP_2}=\frac{V_1^3}{V_2^3}$$

    $$\frac{4460}{4120}=\frac{14.55^3}{V_2^3}$$

    $$V_2=14.17kntos$$

    $$Fuel\:consumption\:per\:hour=541\operatorname{\mathrm{\:kg}}\:per\:hour$$

    $$Fuel\:consumption\:per\:day\:=\:541\times24=12.98t\:per\:day$$

    $$Fuel\:coefficient=\frac{\Delta^{\frac23}\times V_2^3}{Fuel\:consumption\:per\:day}$$

    $$=\:\frac{14900^{\frac23}\times14.17^3}{12.98}$$

    $$=132726.9$$

    Q9 (10 Marks) Hull Construction πŸ”₯ Repeated 2x

    The 1⁄2 ordinates of a waterplane at 15m intervals, commencing from aft are 1, 7, 10.5, 11, 11, 10.5, 8.4 and Om. Calculate:

    (a) TPC (6)

    (b) Distance of the centre of flotation from midships (5)

    (c) Second moment of area of the waterplane about a transverse axis through the centre of flotation.(5)

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    1/2 Ord

    SM

    Product

    Lever

    Product

    Lever

    Product

    1

    1

    1

    _4

    4

    +4

    16

    7

    4

    28

    +3

    84

    +3

    252

    10.5

    2

    21

    +2

    42

    +2

    84

    11

    4

    44

    +1

    44

    +1

    44

    11

    2

    22

    0

    Ξ£MA = 174

    0

    0

    10.5

    4

    42

    -1

    -42

    -1

    42

    8

    2

    16

    -2

    -32

    -2

    64

    4

    4

    16

    -3

    -48

    -3

    144

    0

    1

    0

    -4

    0

    -4

    0

    Ξ£A = 190

    Ξ£MF = -122

    Ξ£I = +646

    $$Common\:interval\:\left(h\right)=15$$

    $$Waterplane\:area\:\left(A_{w}\right)=2\times\frac{h}{3}\sum A$$

    $$=2\times\frac{15}{3}\times190$$

    $$=1900m^2$$

    $$Longitudinal\:centre\:of\:flotation\:LCF=h\times\frac{\sum M_{A}+\sum M_{F}}{\sum A}$$

    $$=15\times\frac{174-122}{190}$$

    $$LCF=4.105m$$

    $$A_{w}=\frac{100\:\times\:TPC}{\rho}$$

    $$TPC=\frac{1.025\times1900}{100}$$

    $$TPC=19.475m^2$$

    $$Second\:moment\:about\:midships\:\left(I_{m}\right)=2\times\frac{h^3}{3}\sum I$$

    $$=2\times\frac{15^3}{3}\times646$$

    $$I_{m}=1453500m^4$$

    $$Second\:moment\:of\:area\:about\:centroid\:\left(I_{F}\right)=I_{m}-A_{w}\times TPC^2$$

    $$1453500-\left(1900\times41.05^2\right)$$

    $$I_{F}=1421483m^4$$

    Q10 (10 Marks) Ship Stability πŸ”₯ Repeated 2x

    The following data are available from the hydrostatic curves ofa vessel.

    Draught (m) 4.9 5.2

    KB (m) 2.49 2.61

    KM (m) 10.73 10.79

    I(m4) 65250 68860

    Calculate the TPC at a draught of 5.05m.

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    TPC at a draught of 5.05 m from hydrostatic data.

    Data: Draught 4.9 m: KB 2.49, KM 10.73, I 65250 m4. Draught 5.2 m: KB 2.61, KM 10.79, I 68860 m4.

    BM = KM - KB. At 4.9 m: BM = 10.73 - 2.49 = 8.24 m. At 5.2 m: BM = 10.79 - 2.61 = 8.18 m.

    Volume of displacement V = I/BM. At 4.9 m: V = 65250/8.24 = 7919 m3. At 5.2 m: V = 68860/8.18 = 8418 m3.

    Displacement = V x 1.025. At 4.9 m: 8117 t. At 5.2 m: 8629 t.

    The waterplane area between these draughts is the rate of change of volume with draught: A = (V2 - V1)/(draught change) = (8418 - 7919)/(5.2 - 4.9) = 499/0.3 = 1663 m2.

    TPC = A x 1.025/100 = 1663 x 1.025/100 = 17.05 t/cm.

    Answer: the TPC at a draught of 5.05 m is about 17.0 t/cm.

    Q1 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    With reference to Ship stability:

    (a) With the help of a neat sketch explain the relevant features of a G-Z curve.

    (b) What are the effects of the below mentioned conditions on the G-Z curve:

    (i) Increased freeboard

    (ii) Increased beam

    (iii) Increased GM

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    Part (a)

    Features of a GZ curve.

    The GZ curve is a graph of the righting lever GZ against the angle of heel. Its relevant features are:

    • The origin: at zero heel, GZ = 0.
    • The initial slope: the tangent to the curve at the origin equals GM (the metacentric height), since GZ = GM sin(theta) for small angles. A steeper initial slope means a larger GM.
    • The maximum righting lever (GZ max): the highest point of the curve, and the angle at which it occurs (the angle of maximum stability, typically 25-40 deg).
    • The range of stability: the angle from the upright to the angle of vanishing stability (where GZ returns to zero, typically 60-90 deg).
    • The area under the curve: proportional to the dynamical stability (the work done in heeling the ship), used to assess stability in a seaway and against wind heeling.
    • The angle of loll: if the curve starts below the axis (negative GZ at small angles), indicating a negative GM and an unstable ship that lolls to one side.
    • The effect of free surface: the curve is reduced by the free-surface correction.

    The curve is obtained from the cross-curves of stability corrected for the actual KG and free-surface effects, and is compared with the statutory criteria.

    Part (b)

    Effects of the following conditions on the GZ curve.

    (i) Increased freeboard: increasing the freeboard raises the deck edge and increases the reserve buoyancy, so the range of stability is increased (the angle of vanishing stability moves to a larger angle) and the area under the curve is increased. The initial slope (GM) is largely unchanged, but the curve is higher and extends further, giving greater dynamical stability and a larger range.

    (ii) Increased beam: increasing the beam increases the waterplane area and the BM (BM is proportional to the cube of the beam), so the initial slope (GM) increases and the curve is steeper at small angles. The maximum GZ is increased and occurs at a smaller angle, but the range of stability may be reduced (the angle of vanishing stability decreases) because the ship becomes stiffer and the deck edge immerses earlier. The area under the curve may be reduced at large angles.

    (iii) Increased GM: increasing the GM (e.g. by lowering KG) makes the initial slope steeper, so the curve rises more steeply at small angles and the maximum GZ is larger and occurs at a smaller angle. However, the range of stability is reduced (the angle of vanishing stability decreases) and the ship rolls more quickly and with a shorter period, which can be uncomfortable. The area under the curve at small angles increases but the overall range decreases.

    Q2 (10 Marks) Ship Stability πŸ”₯ Repeated 8x

    (a) Explain what is meant by permissible length of compartments in passenger ships.

    (b) Describe how the position of bulkheads is determined.

    (c) Describe briefly the significance of the factor of subdivision

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    Part (a)

    Permissible Length

    Permissible length refers to the maximum length of a ship's compartment that can be flooded while ensuring that the sinkage, heel, or trim remains within acceptable limits without exceeding the floodable length. It ensures the ship remains afloat and stable after flooding.

    Permissible Length Formula:

    $$Permissible\:lenght\:=\:Floodable\:length\:\times Factor\:of\:subdivision$$

    The Factor of Subdivision depends on the ship's length and the nature of its service:

    • For passenger ships, the factor of subdivision is smaller compared to cargo ships.
    • Smaller compartments ensure enhanced safety in case of flooding.
    Part (b)

    Position of Bulkheads

    The position of bulkheads is determined based on the need to comply with the "reasonable amount of damage" criterion in case of flooding:

    • Transverse Watertight Bulkheads should vertically extend up to the margin line.
    • At least one watertight longitudinal bulkhead must be located 20% of the ship's breadth inward from each side.
    • Bulkheads are positioned along the ship's length at intervals equal to the permissible length, but not exceeding the floodable length.
    • The maximum permissible compartment length is limited to 10.7 meters.
    • Collision Bulkhead must be located forward at a distance equal to the permissible length from the forward perpendicular.
    • The ER must be enclosed by two transverse watertight bulkheads, with an aft peak watertight bulkhead enclosing the forward part.
    Part (c)

    Factor of Subdivision

    The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

    Permissible Length Formula:

    $$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

    A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

    Q3 (10 Marks) Hull Construction πŸ”₯ Repeated 2x

    With the help of sketches explain the different types of strakes used in ship construction. What material is generally used for Hull plating and What are the tests carried out on Hull steel plating for certification as per class rules.

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    Different types of strakes used in ship construction, hull plating material, and tests on hull steel.

    Part (a)

    Strakes are the longitudinal rows of plates that make up the shell plating. The different types are:

    • Keel strake (garboard strake): the bottom strake adjacent to the keel, which is thicker because it is subject to grounding and docking loads.
    • Bottom strakes: the strakes between the keel strake and the bilge, forming the bottom of the hull.
    • Bilge strake: the strake at the turn of the bilge, which is thicker and subject to the bilge keel attachment and the docking loads.
    • Side strakes: the strakes on the side of the hull between the bilge and the sheer strake.
    • Sheer strake: the top strake at the deck edge, which is thicker because it is the extreme fibre of the hull girder and subject to the highest bending stresses.
    • Stringer strake: the strake at the deck edge (the deck stringer plate), which is thicker and connects the deck to the side.
    • Deck strakes: the strakes of the deck plating, with the deck stringer plate at the edge being the thickest.

    The strakes are arranged so that the thicker plates are placed where the stresses and loads are highest (keel, bilge, sheer strake, deck stringer).

    Part (b)

    Material generally used for hull plating: mild steel (ordinary shipbuilding steel, e.g. grade A, B, D, E) and, for larger or higher-strength ships, high-tensile steel (e.g. AH32, AH36, DH32, DH36, EH32, EH36). The steel is a carbon-manganese steel with good weldability and notch toughness.

    Part (c)

    Tests carried out on hull steel plating for certification as per class rules:

    • Chemical analysis: to verify the composition (carbon, manganese, sulphur, phosphorus, etc.).
    • Tensile test: to verify the yield strength, ultimate tensile strength and elongation.
    • Bend test: to verify the ductility and soundness of the plate.
    • Impact test (Charpy V-notch): to verify the notch toughness at the specified temperature (for grades D, E and high-tensile steels).
    • Ultrasonic or other non-destructive testing: to verify the internal soundness of the plate.
    • Dimensional and surface inspection: to verify the thickness, flatness and freedom from defects.

    The tests are carried out on samples from each cast/heat and are witnessed and certified by the classification society surveyor.

    Q4 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 13x

    Describe the relationship between frictional resistance and

    (a) Ship's speed

    (b) The wetted area

    (c) The surface roughness

    (d) The length of the vessel

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    Frictional resistance arises due to the interaction between the ship's hull and the water as the vessel moves. This resistance is caused by eddying water adjacent to the hull that is drawn along with the ship. The frictional resistance is given by:

    $$R_{f\:=\:}f\:.\:s\:.\:v^{n}$$

    Where:

    • f = Coefficient of friction
    • s = Wetted surface area
    • v = Ship's speed in knots
    • n = Constant (1.82)
    Part (a)

    Ship Speed (v):

    • Frictional resistance increases with increasing ship speed. The relationship is not perfectly linear, but it's approximately described by the formula Rf = fsv^n (Rf ∝ v^n), where 'n' is a constant (approximately 1.82). This means that a small increase in speed results in a larger increase in frictional resistance.
    Part (b)

    Wetted Area (s):

    • Frictional resistance is directly proportional to the wetted surface area (Rf ∝ s). A larger wetted area (the part of the hull submerged in water) leads to greater frictional resistance. The wetted area increases with the ship's draught (depth in the water).
    Part (c)

    Surface Roughness:

    • Increased surface roughness increases frictional resistance. Roughness can be caused by fouling (marine growth), corrosion, or poor hull maintenance. Regular cleaning and maintenance of the hull surface help to minimise roughness and reduce resistance.
    Part (d)

    Length of the Vessel

    • Frictional resistance is influenced by the vessel's length. As the length increases, the wetted surface area grows, and the coefficient of friction (f) changes, leading to an increase in Rf. Longer vessels experience higher frictional resistance, but streamlined designs can mitigate the effect to some extent.
    Q5 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    The palm of the rudder of a vessel requires extensive welding repairs and as Second Engineer you are requested to supervise.

    (a) Suggest a suitable type of welding process.

    (b) State, with reasons, FOUR common welding defects that can occur there.

    (c) State what tests may be carried out before returning the rudder to service.

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    As Second Engineer, I would oversee the extensive welding repairs required for the vessel's rudder using the following plan:

    Part (a)

    Suitable Welding Process:

    Manual Metal Arc Welding (MMAW), also known as Shielded Metal Arc Welding (SMAW), is the most suitable process for this repair. The reasons are threefold:

    • MMAW is highly portable, allowing for on-site repair within the drydock. The process is adaptable to various welding positions (downhand, overhead, horizontal, vertical) – a necessity given the complex geometry of a rudder.
    • Assuming the rudder is constructed from standard steel, MMAW using readily available flux-coated electrodes provides good control, arc stability, and penetration. The flux coating protects the weld pool from atmospheric contamination during cooling.
    • MMAW requires relatively simple equipment and is less demanding in terms of operator skill compared to other processes like TIG or MIG. This translates to cost-effectiveness and allows for a wider pool of qualified welders.
    • If cast steel components are present, pre-heating will be necessary to minimize stress cracking, and specialized electrodes suited for the specific cast steel grade must be selected.

    During welding by the metal arc process, the following points must be observed:

    • Electrode Consumption Rate
    • Penetration
    • Slag Control
    • Arc Length and Sound
    Part (b)

    Four Common Welding Defects:

    1. Undercut: A groove formed along the edge of the weld bead, weakening the joint. Caused by excessive current, incorrect electrode angle, excessive travel speed, or improper electrode manipulation.

    2. Overlap: Molten weld metal flows over the parent metal without proper fusion. Caused by low current, slow travel speed, excessive arc length, or improper joint preparation.

    3. Slag Inclusion: Trapped slag within the weld metal, reducing its strength and potentially causing cracking. Caused by insufficient cleaning between passes, incorrect current, long arc length, slow travel speed, or too large an electrode diameter.

    4. Incomplete Penetration: The weld does not fully fuse the joint faces, resulting in a weak joint. Caused by insufficient current, incorrect joint preparation (too small a root gap or bevel angle), excessive travel speed, or too large an electrode diameter.

    Part (c)

    Tests Before Returning to Service:

    • A thorough visual examination of all welds to identify any surface defects like cracks, porosity, or lack of fusion.
    • NDT methods such as Magnetic Particle Inspection (MPI) or Dye Penetrant Inspection (DPI) will be employed to detect subsurface flaws that may not be visible during visual inspection. The specific NDT method chosen will depend on the type of steel and the accessibility of the weld areas.
    • The repaired rudder will undergo a hydrostatic pressure test. This involves filling the rudder with a water head of 2.46 meters and observing for any leaks. This confirms the watertight integrity of the welds and the overall rudder structure.
    Q6 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 7x

    A ship of 15000 tonne displacement has an Admiraly Co-efficient, based on shaft power, of 420. The mechanical efficiency of the machinery is 83%, shaft losses 6%, propeller efficiency 65% and QPC 0.71. At a particular speed the thrust power is 2550kW.

    (a) Indicated power (6)

    (b) Effective power (6)

    (c) Ship speed (4)

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    Given:

    $$\Delta=15000t$$

    $$Shaft\:Power\:\left(SP\right)=420$$

    $$Transmission\:Efficiency=83\%$$

    $$Shaft\:losses=6\%$$

    $$Propeller\:Efficiency=65\%$$

    $$QPC=0.71$$

    $$Thrust\:Power=2550kW$$

    $$\left(\imaginaryI\right)\:Delivered\:Power\:\left(DP\right)=\frac{Thrust\:Power\:\left(TP\right)}{Propeller\:Efficiency\:\left(\eta P\right)}$$

    $$DP=\frac{2550}{0.65}$$

    $$DP=3923.07kW$$

    $$\left(ii\right)\:Shaft\:Power=\frac{Delivered\:Power\:\left(DP\right)}{Transmission\:Efficiency\:\left(\eta T\right)}\:$$

    $$SP=\frac{3923.07}{0.94}$$

    $$SP=4173.47kW$$

    $$\left(iii\right)\:Indicated\:Power=\frac{Shaft\:Power\:\left(SP\right)}{Mechanical\:Efficiency\:\left(\eta m\right)}$$

    $$IP=\frac{4173.47}{0.83}$$

    $$IP=5028.28kW$$

    $$\left(iv\right)\:Effective\:Power=DP\times QPC$$

    $$EP=3923.07\times0.71$$

    $$EP=2785.3797kW$$

    $$\left(v\right)\:Shaft\:Power=\frac{\Delta^{\frac23}\times V^3}{Admiralty\:Co-efficient}$$

    $$4173.47=\frac{15000^{\frac23}\times V^3}{420}$$

    $$V=14.23knots$$

    Q7 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) What is Prismatic Co-efficient (CP). (3)

    (b) Derive the formula Cp = Cb/Cm, where Cb = Co-efficient of fineness and Cm = midship section

    area co-efficient. (5)

    (c) The length of a ship is 18 times the draught, while the breadth is 2.1 times the draught. At the load water plane, the water plane area co-efficient is 0.83 and the difference between the TPC in sea water and the TPC in fresh water is 0.7. Determine the length of the ship and the TPC in fresh water. (8)

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    Part (a)

    Prismatic coefficient is the ratio of the volume of displacement to the product of length and area of the immersed portion of the midship section

    Cb is the block co-efficient or co-efficient of fitness is the ratio of the volume of displacement to the product of length, breadth and draught

    $$\because\space C_{b}\space=\space{{\nabla}\over L\times B\times D}\:---\:1$$

    $$C_{p}\space=\space{{\nabla}\over L\times A_{m}}\:---\:2$$

    $$C_m \space = \space {{A_m} \over B \times D}$$

    $$A_{m}=\:C_{m}\times B\times D\:---\:3$$

    Substitute 3 in 2

    $$C_{p}\space=\space{{\nabla}\over C_{m}\times B\times D\ \times L}\:---\:4$$

    $$\nabla=C_{b}\times L\times B\times D\:---5\:\left(from\:equation\:1\right)$$

    Substitute 5 in 4

    $$C_{p}=\frac{C_{b}\times L\times B\times D}{C_{m}\times L\times B\times D}$$

    $$C_p \space = \space {{C_b} \over C_m}$$

    Part (b)

    $$Length \space of \space ship \space = \space L$$

    $$Draught \space = \space {{L} \over 18 }$$

    $$breadth \space = \space 2.1 \times draught \space = \space 2.1 \times {{L} \over 18}$$

    $$TPC\:in\:SW\:=\:0.01025A_{w}$$

    $$TPC\:in\:FW\:=\:0.0100A_{w}$$

    $$0.01025A_{w}-0.0100A_{w}=0.7$$

    $$A_{w}=\frac{0.7}{2.5\times10^{-4}}=2800$$

    $$C_w \space = \space {{A_w} \over L \times B}$$$$0.83 \space = \space {{2800} \over L \times {{2.1L} \over 18}}$$

    $$L=170.04m$$

    $$TPC\:in\:FW=0.010\times A_{w}$$

    $$=\:0.0100\times2800$$

    $$TPC\:in\:FW=28$$

    Q8 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 6x

    A ship of length 140m, Breadth of 18.5m, draught of 8.1 m and a displacement of 17,025 tonnes in sea water, has a face pitch ratio of 0.673. The diameter of the propeller is 4.8m. The results of the speed trial show that true slip may be regarded as constant over a range of 9 to 13 knots and is 30%, w = 0.5Cb-0.05. If fuel used is 20t/day at 13 knots and fuel consumption/day varies as cube of speed of ship, determine the fuel consumption, when propeller runs at 110 pm. (16)

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    Given:

    $$Lenght,\:L=140m$$

    $$Breadth,\:B=18.5m$$

    $$Draught,\:d=8.1m$$

    $$Displacement,\:\Delta=17025tonnes$$

    $$Pitch\:ratio,\:p=0.673\operatorname{}$$

    $$Diameter\:of\:Propeller,\:D=4.8m$$

    $$\operatorname{Real\:Slip,\:R_{s}=30\%\:or\:0.3}$$

    $$Wake\:fraction,\:W=0.5C_{b}-0.05$$

    $$Consumption,\:C_2=\:20t\:per\:day\:$$

    $$Ship\:Speed,\:V_2=13\:knots$$

    $$\operatorname{Revolution,\:N}=110rpm$$

    $$cons\:per\:day\:\alpha\:V^3$$

    To find Fuel Consumption C1=?

    We know that,

    $$Displacement,\:\Delta=\nabla\times\rho$$

    $$17025=\nabla\times1.025$$

    $$\nabla=16609.76m^3$$

    $$Block\:Coefficient,\:C_{b}=\frac{\nabla}{L\times B\times d}$$

    $$C_{b}=\frac{16609.76}{140\times18.5\times8.1}$$

    $$C_{b}=0.792$$

    $$Wake\:Fraction,\:W=0.5C_{b}-0.05$$

    $$W=0.5\times0.792-0.05$$

    $$W=0.346$$

    $$Pitch\:ratio,\:p=\frac{P}{D}$$

    $$0.673=\frac{P}{4.8}$$

    $$P=4.8\times0.673$$

    $$P=3.23m$$

    $$Theoretical\:Speed,\:V_{t}=\frac{P\times N\times60}{1852}$$

    $$V_{t}=\frac{3.23\times110\times60}{1852}$$

    $$V_{t}=11.51knots$$

    Using, Real slip equation.

    $$\operatorname{\operatorname{Real\:Slip,\:R_{s}=\frac{V_{t}-V_{a}}{V_{t}}}}$$

    $$0.3=\frac{11.51-V_{a}}{11.51}$$

    $$V_{a}=11.51-11.51\times0.3$$

    $$V_{a}=11.51\left(1-0.3\right)$$

    $$V_{a}=11.51\times0.7$$

    $$V_{a}=8.057knots$$

    $$Wake\:fraction,\:W=\frac{V-V_{a}}{V}$$

    $$0.346=\frac{V-8.057}{V}$$

    $$0.346V=V-8.057$$

    $$V=\frac{8.057}{0.654}$$

    $$V=12.32knots$$

    $$cons\:per\:day\:\alpha\:V^3$$

    $$\frac{C_1}{C_2}=\left(\frac{V_1}{V_2}\right)^3$$

    $$\frac{C_1}{20}=\left(\frac{12.32}{13}\right)^3$$

    $$C_1=20\times\left(\frac{12.32}{13}\right)^3$$

    $$C_1=20\times0.851$$

    $$C_1=17.02t\:per\:day$$

    Q9 (10 Marks) Ship Stability πŸ”₯ Repeated 8x

    (a) Define longitudinal center of gravity (LCG) and longitudinal centre of buoyancy (LCB). (4)

    (b) The immersed cross-sectional areas of a ship 120 m long, commencing from aft are 2, 40, 79, 100, 103, 104, 104, 103, 97, 58 and 0 m2. Calculate:

    (i) Displacement (6)

    (ii) Longitudinal position of the centre of buoyancy. (6)

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.
    Q10 (10 Marks) Ship Stability πŸ”₯ Repeated 4x

    A ship 100 m long floats at a draft of 6m and in this condition the immersed cross sectional areas are as given in tables below. The equivalent base area (A) is required because of the fineness of the bottom shell

    Calculate each of the following:

    [Table will be here soon]

    (a) The equivalent base area value Ab (6)

    (b) The longitudinal position of the centre of buoyancy from midships (5)

    (c) The vertical position of the centre of buoyancy above the base (5)

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    Table for question:

    Section

    AP

    1

    2

    3

    4

    5

    FP

    Immersed cross sectional area

    12

    30

    65

    80

    70

    50

    0

    Draught (m)

    0

    0.6

    1.2

    2.4

    3.6

    4.8

    6.0

    Water plane area (m2)

    Ab

    560

    720

    880

    940

    1000

    1030

    Solution:

    Section

    CSA

    SM

    F volume

    Lever

    F moment

    AP

    12

    1

    12

    -3h

    -36h

    1

    30

    4

    120

    -2h

    -240h

    2

    65

    2

    130

    -1h

    -130h

    3

    80

    4

    320

    0

    0

    4

    70

    2

    140

    +1h

    140h

    5

    50

    4

    200

    +2h

    400h

    FP

    0

    1

    0

    +3h

    0

    Ξ£Fvol. = 992

    Ξ£Fmom. =134h

    $$Common\:interval\:\left(h\right)=\frac{L}{6}=\frac{100}{6}$$

    $$h=16.66m$$

    $$\sum F_{movement}=134\:\times16.66m^4$$

    $$\sum F_{moment}=2232.44m^4$$

    $$\nabla=\frac{h}{3}\times\sum F_{volume}$$

    $$\nabla=\frac{16.66}{3}\times992$$

    $$\nabla=5120.17m^3$$

    Q1 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 4x

    (a) Describe the preparation necessary before the application (in dry dock) of sophisticated or approved long life coating to the underwater surface of the hull (8)

    (b) State the significance of the roughness profile. (4)

    (c) List the different sophisticated coating which are available. (4)

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    The preparation of a ship's underwater hull before applying a long-life coating in a dry dock involves a three-step process. This process addresses the removal of contaminants and the creation of a suitable surface profile.

    (i) Washing: The hull surface must be thoroughly cleaned to remove all marine growth (algae, slime, etc.), accumulated salts, dirt, grease, and oil. High-pressure freshwater washing is the standard method for this initial cleaning. The goal is to present a clean substrate for subsequent stages.

    (ii) Blasting: Abrasive blasting is the preferred method for removing rust, defective paint, and any remaining contaminants. This process achieves a bare metal surface, essential for proper adhesion of the new coating. The extent of blasting (localized or full hull) depends on the condition of the existing surface. The intensity and type of abrasive used are carefully controlled to achieve the desired surface roughness profile.

    (iii) Primer Application: After blasting, the surface is again cleaned to remove any blasting debris. A primer coat is then applied to provide corrosion protection and to create an ideal surface for the subsequent topcoat adhesion. This primer acts as an intermediary layer, enhancing the bond between the substrate and the long-life coating system.

    Part (a)

    Significance of Roughness Profile:

    The roughness profile of the prepared hull surface impacts the performance of the applied coating and the overall operational efficiency of the vessel. A rough surface increases frictional resistance as the vessel moves through the water. This increased drag translates to higher power requirements for propulsion, leading to increased fuel consumption and operational costs. Furthermore, greater surface roughness contributes to increased carbon emissions, a concern under current MARPOL regulations. Therefore, a controlled and optimized roughness profile is essential for minimizing frictional resistance, reducing fuel consumption and emissions, and maximizing the longevity of the hull coating.

    Part (b)

    Sophisticated hull coating systems comprise multiple layers designed to provide corrosion protection and antifouling properties.

    Wash Primer/Pretreatment Primer/Metal Conditioning Primer:

    • These primers act as a base layer, improving adhesion of subsequent layers. Common types include epoxy primers pigmented with iron oxide and corrosion inhibiting pigments (zinc and calcium phosphates, although zinc content is minimized due to safety concerns).

    Anticorrosive Coating:

    • This layer primarily provides corrosion protection to the underlying metal. Two-component epoxies, coal tar epoxies, and epoxy or polyester coatings incorporating glass flakes are frequently employed. Glass flakes enhance mechanical strength and water vapor impermeability.

    Antifouling Coating:

    • This layer prevents the attachment of marine organisms (fouling). Historically, tin-based paints were used, but due to environmental regulations, they have been largely replaced by copper-based, silicone-based, or non-TBT (Tributyltin) self-polishing antifouling coatings. These newer coatings typically use seawater-soluble polymers. The number of antifouling layers applied (two or three) depends on the specific system chosen and required longevity.
    Q2 (10 Marks) Machinery Space & Systems πŸ”₯ Repeated 5x

    (a) Describe the double bottom and framing arrangement used in the machinery space to cope up with the concentrated loads and vibration, together with shaft and thrust block support. (10)

    (b) Give reasons for the choice of thrust block position. (6)

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    Part (a)

    The construction of the double bottom in the machinery space regardless of the framing system has solid plate floors at every frame space under the main engine. Additional side girders are fitted outboard of the main engine seating, as required. The double-bottom height is usually increased to provide fuel oil, lubricating oil and fresh water tanks of suitable capacities. Shaft alignment also requires an increase in the double-bottom height or a raised seating, the former method usually being adopted.

    Continuity of strength is ensured and maintained by gradually sloping the tank top height and internal structure to the required position. Additional support and stiffening is necessary for the main engines, boilers, etc., to provide a vibration-resistant solid platform capable of supporting the concentrated loads. On slow- speed diesel-engined ships, the tank top plating is increased to 40 mm thickness or thereabouts in way of the engine bedplate. This is achieved by using a special insert plate which is the length of the engine including the thrust block in size. Additional heavy girders are also fitted under this plate and in other positions under heavy machinery as required. Plating and girder material in the machinery spaces is of increased scantlings in the order of 10 per cent.

    The method adopted.

    A cellular void space within a ships structure is called a coffer dam. Like a bulkhead separates two spaces or divides a space into two, a coffer dam does the same with a larger degree of integrity since it incorporates a void which would contain any breach of either of the boundaries. This would contain the leakage and prevent it spreading into other areas. A cofferdam can be defined as an empty space separating compartments to prevent the contents of one compartment from entering another in case of leakage.

    Part (b)

    The thrust block is positioned close to the propulsion machinery (usually just aft of the main engine). Reasons are as follows:

    • The axial thrust generated by the propeller could cause deformation and misalignment of the shafting system if the thrust block were positioned far from the engine. Placing it close to the engine minimizes the length of shafting subject to this thrust, thus reducing potential for misalignment. The strong double bottom structure directly beneath provides necessary support to mitigate any deformation.
    • Differential expansion between the shaft and the hull due to temperature variations is a potential source of misalignment. Positioning the thrust block close to the engine helps to minimize the effect of this differential expansion.
    • The weight of the propeller and the dynamic forces it creates can lead to whirling of the tailshaft and misalignment if not properly managed. Positioning the thrust block near the engine helps to stabilize the shafting system and reduce the risk of these issues.
    • The substantial double-bottom structure under the main machinery provides an ideal, inherently strong foundation for the thrust block. This minimises the need for extensive additional reinforcement to support the thrust loads.
    Q3 (10 Marks) Ship Stability πŸ”₯ Repeated 14x

    Explain how the period of roll varies with:

    (a) The amplitude of roll (4)

    (b) The radius of gyration (4)

    (c) The initial metacentric height (4)

    (d) The location of masses in the ship (4)

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    The period of roll Tr of a ship is determined by the formula:

    $$T_{r}=\frac{2\pi K}{\sqrt{g\times GM}}$$

    where,

    • K is the radius of gyration (mass moment of inertia)
    • g is the acceleration due to gravity, and
    • GM is the metacentric height.
    Part (a)

    Amplitude of Roll:

    • The amplitude of roll, or the maximum angle of heel, does not affect the period of roll. While a larger amplitude implies a greater heeling moment and faster roll speed, the time taken to complete one full roll cycle remains constant for a given metacentric height and radius of gyration.
    Part (b)

    Radius of Gyration (K):

    • The period of roll is directly proportional to the radius of gyration. A larger radius of gyration (indicating a greater distribution of mass further from the ship's centre of rotation) leads to a longer period of roll. Conversely, a smaller radius of gyration (mass concentrated closer to the centre) results in a shorter period. The distribution of cargo significantly impacts K; cargo concentrated centrally minimises K and the roll period, while dispersed cargo maximises K and the roll period.
    Part (c)

    Initial Metacentric Height (GM):

    • The period of roll is inversely proportional to the square root of the metacentric height (GM). A larger GM (a stiffer ship) leads to a shorter roll period, as the ship quickly returns to its upright position. A smaller GM (a tender ship) results in a longer roll period, with slower return to the upright.
    Part (d)

    Location of Masses in the Ship:

    The location of masses in the ship will effect the GM & K. So the period of roll will be affected.

    • If masses are at bottom, G moves down, GM ↑, period of roll ↓.
    • If masses are at top, G moves up, GM ↓, period of roll ↑.
    • If masses are concentrated at centre, K ↓, period of roll ↓.
    • If masses are away from centre, K ↑, period of roll ↑.
    Q4 (10 Marks) Hull Construction πŸ”₯ Repeated 7x

    With reference to fatigue of engineering components,

    (a) Explain the influence of stress level and cyclical frequency on expected operating life. (5)

    (b) Explain the influence of material defects on the safe operating life of an engineering (5)

    (c) State the factors which influence the possibility of fatigue cracking of a bed-plate transverse girder and explain how the risk of such cracking can be minimized. (6)

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    Part (a)

    Influence of Stress Level and Cyclic Frequency on Operating Life:

    Fatigue is progressive and localised structural damage caused by cyclic loading, where the maximum stress is below the ultimate tensile strength. The relationship between stress level, cyclic frequency, and operating life depends on whether the fatigue is high-cycle/low-stress or low-cycle/high-stress.

    High-cycle fatigue (low stress-high cycle):

    • This occurs at lower stress levels over a high number of cycles, resulting in elastic deformation. The component can withstand more cycles at these lower stress levels, and its life expectancy is determined by the S-N curve, which predicts the number of cycles before failure at a given stress level. For example, fatigue in turbocharger blowers often results from prolonged vibration over numerous cycles.

    Low-cycle fatigue (high stress-low cycle):

    • This occurs at high-stress levels over fewer cycles, causing plastic deformation in the material. This type of fatigue is typically assessed by a strain curve. If the stress level increases, the component's operating life decreases, as higher stress accelerates the onset of failure. For example, air receivers filling automatically face high stress and experience fewer cycles before failure.

    If stress levels or the number of cycles increase beyond the material’s capacity, failure will occur sooner. It is important to keep stress levels within allowable limits for extended component life.

    Part (b)

    Material defects can significantly reduce the safe operating life of engineering components because defects serve as stress concentrators that increase local stress around the defect. This leads to premature failure as the material cannot withstand the same level of cyclic stress as a defect-free component.

    • Surface roughness, porosity, inclusions, and abrupt section changes all create stress concentrations, lowering fatigue strength.
    • Coarse grain size, specific chemical compositions, and cold working introduce residual stresses that reduce fatigue resistance.
    • Corrosion, erosion, and decarbonisation weaken the material and accelerate fatigue crack initiation and propagation.
    • Faulty workmanship during assembly or processing introduces defects that may significantly shorten the component's life.
    Part (c)

    Factors Influencing Fatigue Cracking in Bedplate Transverse Girders:

    • Cylinder overload due to excess power puts excessive stress on the girders.
    • Incorrect crankshaft alignment induces uneven loading and stress concentrations.
    • Material defects, high residual stresses in welds, heat-affected zone hardening, and the presence of dissolved oxygen all reduce fatigue resistance.
    • Tank top deformation from pressurisation or overheating adds stress to the bedplate.

    To minimise the risk of fatigue cracking:

    (i) Constructional strength:

    • Bed plates are made up of M.S. plates with four steel casting, which are assembled and welded together so that the bed plate is strong longitudinally & transversely with good resistance to twisting along its length.
    • Longitudinal strength is obtained by fabricating each side of the bed plate in the form of a box girder.
    • The cast steel cross girder in which the main bearing is placed contributes to the bed plate's transverse strength and resistance against twisting along its length.
    • Resin cast chocks are used between the bedplate and the double bottom tank top to absorb the shocks & stress.

    (ii) Maintenance:

    • Monthly checks on the bolt tension.
    • Monthly checks on engine load using power cards & measuring cylinder peak pressure.
    • Regular checking of tension for main bearing jack bolts as recommended by engine manufacturers.
    • Regular checks on crankshaft alignment by taking deflection & compare with recommended value.
    • By maintaining engine operations at specified load, temperature, pressure, speed, etc.
    Q5 (10 Marks) General πŸ”₯ Repeated 3x

    Draw and describe the construction of a forepeak tank. Explain how are the effects of panting and pounding taken care with the help of neat sketches?

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    Part (a)

    Causes and Effects of Panting and Pounding, Including the Affected Areas

    Panting

    Panting refers to the in-and-out flexing or pulsation of the shell plating at the ends of a ship, mainly caused by variations in sea pressure during pitching motions in rough weather.

    When a ship moves through heavy seas, especially while pitching, the water pressure acting on the bow and stern changes continuously. These fluctuating pressures cause the side shell plating to vibrate and flex. The effect is known as panting.

    The forward end of the ship is affected more severely than the aft end because:

    • The bow directly encounters incoming waves while the ship is making headway.
    • Water pressure at the fore end changes rapidly as the bow rises and falls in the sea.

    The affected areas are mainly:

    • Forward shell plating near the bow
    • Side shell plating at the fore peak region
    • To a lesser extent, the aft end of the ship

    Effects of Panting

    Panting can lead to:

    • Repeated flexing and vibration of shell plating
    • Fatigue stresses in plating and framing
    • Loosening of rivets or welded joints
    • Cracking or deformation of structural members
    • Reduction in structural strength if not properly reinforced

    Therefore, special strengthening arrangements are provided at the forward and aft ends of the vessel to resist panting stresses.

    Pounding

    Pounding refers to the heavy impact experienced at the bottom forward part of the ship when the bow emerges from the water during pitching and then slams violently back onto the sea surface.

    This condition usually occurs when:

    • The ship is pitching heavily in rough seas,
    • The fore part lifts clear of the water due to heaving and pitching motions,
    • The bow then falls heavily onto the wave surface.

    The effect is most severe when the vessel is in the light ship condition, because the bow rises more easily out of the water.

    The main area affected by pounding is:

    • The bottom shell plating in the forward region of the ship,
    • Especially near the forefoot and forward bottom structure.

    The stern may also experience similar impacts from following seas, but usually to a lesser extent.

    Effects of Pounding

    Pounding produces:

    • Severe impact stresses on bottom plating
    • Buckling or deformation of bottom structure
    • Cracks in shell plating or framing
    • Structural fatigue and weakening
    • Damage to internal supporting members

    To withstand these heavy impact loads, the forward bottom structure is specially strengthened.

    Part (b)

    Constructional Details Designed to Resist Panting and Pounding

    Special structural arrangements are incorporated in ship construction to resist the stresses caused by panting and pounding.

    Structural Arrangements to Resist Panting

    The following strengthening arrangements are provided mainly at the bow and stern regions:

    1. Panting Beams

    Panting beams are horizontal beams fitted across the ship near the bow and stern.

    Their purpose is to:

    • Support the side shell plating,
    • Reduce excessive flexing,
    • Increase structural rigidity against panting stresses.

    2. Panting Stringers

    Panting stringers are horizontal girders fitted along the ship side in the panting region.

    They:

    • Connect frames together,
    • Provide additional stiffness to the shell plating,
    • Help distribute fluctuating sea pressure loads.

    3. Close-Spaced Framing

    Frames in the panting region are spaced closer together than in other parts of the ship.

    This:

    • Provides better support to shell plating,
    • Reduces plate vibration and deformation.

    Structural Arrangements to Resist Pounding

    To resist pounding stresses at the forward bottom region, additional strengthening is provided.

    1. Increased Bottom Plating Thickness

    The shell plating near the keel and forward bottom region is made thicker.

    Usually:

    • The first few strakes of bottom plating on either side of the keel are increased in thickness.

    This enables the structure to withstand repeated impact loads.

    2. Plate Floors

    Solid plate floors are fitted at closer spacing in the forward bottom region.

    These:

    • Strengthen the bottom structure,
    • Distribute pounding stresses more effectively,
    • Prevent deformation of bottom plating.

    3. Additional Internal Reinforcement

    Extra brackets, girders, and stiffeners may also be provided in the fore peak and bottom structure to improve strength and rigidity.

    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) What is Prismatic Co-efficient (CP). Derive the formula Cp = Cb/Cm, where Cb = Co-efficient of

    fineness and Cm = midship section area co-efficient. (6)

    (b) The length of a ship is 18 times the draught, while the breadth is 2.1 times the draught. At the load water plane, the water plane area co-efficient is 0.83 and the difference between the TPC in sea water and the TPC in fresh water is 0.7. Determine the length of the ship and the TPC in fresh water (10)

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    Part (a)

    Prismatic coefficient is the ratio of the volume of displacement to the product of length and area of the immersed portion of the midship section

    Cb is the block co-efficient or co-efficient of fitness is the ratio of the volume of displacement to the product of length, breadth and draught

    $$\because\space C_{b}\space=\space{{\nabla}\over L\times B\times D}\:---\:1$$

    $$C_{p}\space=\space{{\nabla}\over L\times A_{m}}\:---\:2$$

    $$C_m \space = \space {{A_m} \over B \times D}$$

    $$A_{m}=\:C_{m}\times B\times D\:---\:3$$

    Substitute 3 in 2

    $$C_{p}\space=\space{{\nabla}\over C_{m}\times B\times D\ \times L}\:---\:4$$

    $$\nabla=C_{b}\times L\times B\times D\:---5\:\left(from\:equation\:1\right)$$

    Substitute 5 in 4

    $$C_{p}=\frac{C_{b}\times L\times B\times D}{C_{m}\times L\times B\times D}$$

    $$C_p \space = \space {{C_b} \over C_m}$$

    Part (b)

    $$Length \space of \space ship \space = \space L$$

    $$Draught \space = \space {{L} \over 18 }$$

    $$breadth \space = \space 2.1 \times draught \space = \space 2.1 \times {{L} \over 18}$$

    $$TPC\:in\:SW\:=\:0.01025A_{w}$$

    $$TPC\:in\:FW\:=\:0.0100A_{w}$$

    $$0.01025A_{w}-0.0100A_{w}=0.7$$

    $$A_{w}=\frac{0.7}{2.5\times10^{-4}}=2800$$

    $$C_w \space = \space {{A_w} \over L \times B}$$$$0.83 \space = \space {{2800} \over L \times {{2.1L} \over 18}}$$

    $$L=170.04m$$

    $$TPC\:in\:FW=0.010\times A_{w}$$

    $$=\:0.0100\times2800$$

    $$TPC\:in\:FW=28$$

    Q7 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 4x

    With respect to Ship Propulsion:

    (a) Explain the various efficiencies associated with propeller and shafting arrangement. (6)

    (b) When a propeller of 4.8 m pitch turns at 110 rpm, the apparent slip is found to be -S % and the real slip is 1.5 S %. If the wake speed is 25% of the ship speed, calculate the ship speed, apparent slip and the real slip.

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    Part (a)

    Efficiencies associated with the propeller and shafting arrangement.

    • Shaft (transmission) efficiency: the ratio of the delivered power at the propeller to the brake power of the engine, accounting for the losses in the shaft bearings, stern tube and any gearing. It is typically 0.97-0.99.
    • Propeller (open-water) efficiency: the ratio of the thrust power (T x Va) to the delivered power (2 pi n Q). It represents the efficiency of the propeller itself in converting the delivered power into thrust power.
    • Hull efficiency: the ratio of the effective power to the thrust power, = (1 - t)/(1 - w), where t is the thrust deduction factor and w the wake fraction. It accounts for the interaction between the hull and the propeller.
    • Quasi-propulsive coefficient (QPC): the ratio of the effective power to the delivered power, = hull efficiency x propeller efficiency. It is the overall efficiency of the propulsion system in converting the delivered power into effective (towing) power.
    • Overall (propulsive) efficiency: the ratio of the effective power to the brake power, = QPC x shaft efficiency. It is the overall efficiency of the engine-to-propeller-to-hull system.
    Part (b)

    Ship speed, apparent slip and real slip.

    A propeller of 4.8 m pitch turns at 110 rev/min. The apparent slip is -S% and the real slip is +1.5S%. The wake speed is 25% of the ship speed. Calculate the ship speed, the apparent slip and the real slip.

    Pitch speed = pitch x rev/s = 4.8 x 110/60 = 8.8 m/s.

    Let the ship speed be V (m/s). The speed of advance Va = V x (1 - 0.25) = 0.75 V.

    Apparent slip = (pitch speed - V)/pitch speed = -S/100.

    Real slip = (pitch speed - Va)/pitch speed = 1.5 S/100.

    From the apparent slip: (8.8 - V)/8.8 = -S/100, so V = 8.8(1 + S/100).

    From the real slip: (8.8 - 0.75 V)/8.8 = 1.5 S/100, so 8.8 - 0.75 V = 0.132 S.

    Substitute V = 8.8(1 + S/100): 8.8 - 6.6(1 + S/100) = 0.132 S.

    8.8 - 6.6 - 0.066 S = 0.132 S, so 2.2 = 0.198 S, S = 11.11.

    Apparent slip = -11.11%; real slip = 1.5 x 11.11 = 16.67%.

    Ship speed V = 8.8(1 + 0.1111) = 9.78 m/s = 9.78 x 1.944 = 19.0 knots.

    Answer: ship speed about 19.0 knots; apparent slip -11.1%; real slip +16.7%.

    Q8 (10 Marks) Ship Stability

    The following data are available from the hydrostatic curves of a vessel. (16)

    Draught (m) 4.9 5.2

    KB(m) 2.49 2.61

    KM (m) 10.73 10.79

    I(m4) 65250 68860

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    TPC at a draught of 5.05 m from hydrostatic data.

    Data: Draught 4.9 m: KB 2.49, KM 10.73, I 65250 m4. Draught 5.2 m: KB 2.61, KM 10.79, I 68860 m4.

    BM = KM - KB. At 4.9 m: BM = 10.73 - 2.49 = 8.24 m. At 5.2 m: BM = 10.79 - 2.61 = 8.18 m.

    Volume of displacement V = I/BM. At 4.9 m: V = 65250/8.24 = 7919 m3. At 5.2 m: V = 68860/8.18 = 8418 m3.

    Displacement = V x 1.025. At 4.9 m: 8117 t. At 5.2 m: 8629 t.

    The waterplane area between these draughts is the rate of change of volume with draught: A = (V2 - V1)/(draught change) = (8418 - 7919)/(5.2 - 4.9) = 499/0.3 = 1663 m2.

    TPC = A x 1.025/100 = 1663 x 1.025/100 = 17.05 t/cm.

    Answer: the TPC at a draught of 5.05 m is about 17.0 t/cm.

    Q9 (10 Marks) Ship Stability πŸ”₯ Repeated 3x

    (a) Explain the concept of dynamical stability. (6)

    (b) A ship of 5000 tonne displacement has three rectangular double bottom tanks; Tank A: 12m long and 16m wide; Tank B: 14m long and 15m wide; Tank C: 14m long and 16m wide.

    Calculate the free surface effect for any one tank and state in which order the tanks should be filled when making use of them for stability correction. (10)

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    Part (a)

    Dynamical Stability is defined as the amount of energy required to heel a ship from its upright equilibrium position to a specific angle of heel. It provides a measure of the vessel's stability by considering its behaviour in response to dynamic external forces, such as wind or waves.

    • The concept compares the heeling moment energy (from external forces) and the righting moment energy (from the ship's stability).
    • The ship will absorb the energy imparted by the heeling moment. If the righting energy is greater than the heeling energy, the ship will stabilize; otherwise, it may capsize.

    Areas Under the Curve:

    • Area A: Represents the region where the heeling moment exceeds the righting moment (external energy > ship's stability).
    • Area B: Represents the region where the righting moment exceeds the heeling moment (ship's stability > external energy).
    • The balance of these areas determines whether the ship will right itself or continue to heel.

    When exposed to heeling forces such as wind or waves, the vessel inclines and may roll over to a certain angle of heel. If the external force is applied instantaneously, the ship must have enough reserve dynamic stability to absorb the energy and return to an upright position. If the external force is constant, the ship will remain at an equilibrium angle where the righting moment equals the heeling moment.

    This refers to the remaining righting energy available to counteract additional external forces. A higher reserve dynamic stability ensures the vessel can handle greater heeling forces without capsizing.

    Q10 (10 Marks) Ship Stability πŸ”₯ Repeated 3x

    (a) Describe how water tightness is maintained where bulkheads are pierced by longitudinal beams or pipes. (6)

    (b) A triangular bulkhead is 7 m wide at the top and has a vertical depth of 8 m. Calculate the load on the bulkhead and the position of centre of pressure if the bulkhead is flooded with sea water on only side: (10)

    (i) To the top edge

    (ii) With 4 m head to the top edge.

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    Part (a)

    Water tightness of

    bulkheads that are pierced by longitudinal beams or pipes:

    • Bulkheads must be made structurally watertight without the use of wood packing.
    • Pipes passing through bulkheads are either welded or fastened to the bulkhead using studs or hoses secured through tapped holes in the plating.
    • When a bulkhead is pierced by a longitudinal beam, the openings are kept as small as possible, and doubler plates are welded on each side to maintain watertight integrity.
    • Sealing materials and gaskets are used, particularly for pipes, cables, and similar penetrations.
    Part (b)

    $$\left(a\right)\:Load\:on\:bulkhead=\rho gAH$$

    $$=1025\times9.81\times\frac{7\times8}{2}\times\frac83$$

    $$=750.8\times10^3N$$

    $$=750.8KN$$

    $$Centre\:of\:pressure\:fron\:top=\frac12D=\frac12\times8$$

    $$=4m$$

    $$\left(b\right)\:Load\:on\:bulkhead=1025\times9.81\times\frac{7\times8}{2}\times\left(\frac83+4\right)\times$$

    $$=1.877\times10^6N$$

    $$=1.877MN$$

    $$For\:triangle\:I_{NA}=\frac{1}{36}BD^3$$

    $$Centre\:of\:pressure\:from\:surface\:of\:water\:=\frac{I_{NA}}{AH}+H$$

    $$=\frac{\frac{1}{36}\times7\times8^3}{\frac12\times7\times8\times\left(\frac83+4\right)}+\left(\frac83+4\right)$$

    $$=\frac{2\times7\times8^3}{36\times7\times8\times6.667}+6.667$$

    $$=7.20\:m$$

    $$Centre\:of\:pressure\:from\:top\:of\:bulkhead$$

    $$=7.20-4$$

    $$3.2m$$

    Q1 (10 Marks) Ship Stability πŸ”₯ Repeated 6x

    With reference to Roll-on, Roll-off ferries:

    (a) Describe the problem of free surlace effect:

    (b) Explain how it is intended that water should be cleared from car or cargo decks

    (c) Describe possible methods for improving the stability and survivabilty of these vessels.

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    Part (a)

    The problem of free-surface effect in Ro-Ro ferries.

    Ro-Ro ferries have large, open car and cargo decks that extend over a large part of the ship. If water is shipped onto these decks (e.g. through the bow or stern doors, or in heavy weather), the water spreads over the large deck area, creating a very large free surface. The free-surface effect reduces the effective GM by rho x i/Delta, where i is the second moment of area of the free surface (i = L B^3/12 for a rectangular deck). Because the car deck is very wide and long, the free-surface effect is very large and can reduce the GM to a dangerously low value, causing the ship to lose stability and capsize. This is the principal stability problem of Ro-Ro ferries: the large open decks create a huge free-surface effect if flooded, and the ship can capsize rapidly.

    Part (b)

    How water should be cleared from car or cargo decks.

    Water on the car deck should be cleared by:

    • Providing adequate freeing arrangements (scuppers, freeing ports, drain valves) at the deck edge and at the ends, so that water can drain overboard.
    • Providing a camber (transverse slope) on the deck so that water runs to the sides and drains through the freeing ports.
    • Providing a longitudinal slope (sheer) so that water runs to the ends and drains.
    • Using bilge pumps and drainage systems to remove water that cannot drain overboard.
    • Ensuring the freeing ports are of adequate size and are not blocked by cargo or lashings.

    The freeing arrangements must be adequate to remove water quickly and prevent the build-up of a large free surface.

    Part (c)

    Methods for improving the stability and survivability of Ro-Ro ferries.

    • Lowering the centre of gravity by placing heavy weights low and ballast in the double bottom.
    • Increasing the GM by increasing the beam and the waterplane area, and by lowering KG.
    • Providing adequate freeboard and reserve buoyancy.
    • Subdividing the car deck with watertight bulkheads or providing a raised deck to limit the spread of water.
    • Providing adequate freeing ports and drainage to remove water quickly.
    • Fitting bilge keels and stabilisers to reduce roll.
    • Designing the ship to meet the damage stability criteria (survive flooding of a compartment).
    • Using a higher freeboard and a stronger, more watertight structure, and ensuring the bow and stern doors are watertight and properly secured.
    • Operating with adequate stability margins and following the loading manual.
    Q2 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    A rudder of a vessel requires extensive welding repairs and as Second Engineer you are requested to supervise the repairs:

    (a) Suggest a suitable type of welding process.

    (b) State, with reasons, FOUR common welding defects.

    (c) State what tests may be carried out before returning the rudder to service.

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    As Second Engineer, I would oversee the extensive welding repairs required for the vessel's rudder using the following plan:

    Part (a)

    Suitable Welding Process:

    Manual Metal Arc Welding (MMAW), also known as Shielded Metal Arc Welding (SMAW), is the most suitable process for this repair. The reasons are threefold:

    • MMAW is highly portable, allowing for on-site repair within the drydock. The process is adaptable to various welding positions (downhand, overhead, horizontal, vertical) – a necessity given the complex geometry of a rudder.
    • Assuming the rudder is constructed from standard steel, MMAW using readily available flux-coated electrodes provides good control, arc stability, and penetration. The flux coating protects the weld pool from atmospheric contamination during cooling.
    • MMAW requires relatively simple equipment and is less demanding in terms of operator skill compared to other processes like TIG or MIG. This translates to cost-effectiveness and allows for a wider pool of qualified welders.
    • If cast steel components are present, pre-heating will be necessary to minimize stress cracking, and specialized electrodes suited for the specific cast steel grade must be selected.

    During welding by the metal arc process, the following points must be observed:

    • Electrode Consumption Rate
    • Penetration
    • Slag Control
    • Arc Length and Sound
    Part (b)

    Four Common Welding Defects:

    1. Undercut: A groove formed along the edge of the weld bead, weakening the joint. Caused by excessive current, incorrect electrode angle, excessive travel speed, or improper electrode manipulation.

    2. Overlap: Molten weld metal flows over the parent metal without proper fusion. Caused by low current, slow travel speed, excessive arc length, or improper joint preparation.

    3. Slag Inclusion: Trapped slag within the weld metal, reducing its strength and potentially causing cracking. Caused by insufficient cleaning between passes, incorrect current, long arc length, slow travel speed, or too large an electrode diameter.

    4. Incomplete Penetration: The weld does not fully fuse the joint faces, resulting in a weak joint. Caused by insufficient current, incorrect joint preparation (too small a root gap or bevel angle), excessive travel speed, or too large an electrode diameter.

    Part (c)

    Tests Before Returning to Service:

    • A thorough visual examination of all welds to identify any surface defects like cracks, porosity, or lack of fusion.
    • NDT methods such as Magnetic Particle Inspection (MPI) or Dye Penetrant Inspection (DPI) will be employed to detect subsurface flaws that may not be visible during visual inspection. The specific NDT method chosen will depend on the type of steel and the accessibility of the weld areas.
    • The repaired rudder will undergo a hydrostatic pressure test. This involves filling the rudder with a water head of 2.46 meters and observing for any leaks. This confirms the watertight integrity of the welds and the overall rudder structure.
    Q3 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 7x

    With reference to International Load Line Statutory Certitication:

    (a) State the reason for freeboard requirements;

    (b) (i) Explain the term 'conditions of assignment'.

    (ii) List the items that may be examined during a Load line survey after a vessel's major repairs in the drydock.

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    Part (a)

    Reasons for Freeboard Requirements:

    Freeboard is the distance measured from the waterline to the upper edge of the deck plating at the side of the freeboard deck amidships.

    Purpose of Freeboard:

    • Ensures the ship is seaworthy when fully loaded.
    • Provides reserve buoyancy, enabling the ship to rise as it passes through waves, keeping decks largely dry.
    • Enhances the ship's stability and increases its survivability in case of damage by allowing it to remain afloat longer, aiding crew escape or damage control.
    Part (b)

    (i) Conditions of Assignment:

    Conditions of Assignment are outlined in the Merchant Shipping Load Line Rules 1968 and must be satisfied before a ship is assigned freeboards and issued a load line certificate. These conditions address the practical need for openings (hatchways, doorways, vents, etc.) in the ship's hull and mandate appropriate protection and closure mechanisms for these openings.

    Requirements Before Assigning Load Line:

    • The ship must have sufficient structural strength.
    • Adequate reserve buoyancy must be maintained.
    • Openings must be secured against water ingress.
    • Safety measures for the crew, such as guardrails and gangways, must be in place.
    Part (b)

    (ii) Items Examined During a Load Line Survey After Major Repairs in Drydock:

    • Openings: This encompasses hatchways, machinery space openings, cargo ports, watertight doors, ventilators, air pipes, scuppers, freeing ports, side scuttles, and other openings in the freeboard and superstructure decks. The surveyor checks the condition of seals, gaskets, closing mechanisms, and overall watertightness.
    • Crew Protection: Safety features designed to protect the crew, such as guardrails, bulwarks, and gangways, are inspected for damage or deterioration and proper functionality.
    • Hull Structure: The hull itself is inspected for any damage, corrosion, or structural weaknesses that could compromise watertightness or strength. This often includes visual inspection for cracks, buckling, indentation, and paint adhesion. Bottom shell, bilge keel, stem and stern frames, rudder, sea chests, side ports, stern bearing, and propeller are all checked.
    • Other Systems: Depending on the scope of repairs, other systems may be examined, including but not limited to: pressure-vacuum valves, sounding pipes, air vents, access hatches, ventilation fans, dampers, weathertight doors and sealings, door securing arrangements, side scuttles, windows, and skylights. Testing, such as hose tests (to check watertight integrity), might also be conducted.
    Q4 (10 Marks) Ship Types & Design πŸ”₯ Repeated 6x

    (a) Considering the vessel as a compound beam define Bending moment shearing force. Which is the point of Maximum Bending Moment?

    (b) Sketch and Describe Hatch coaming of a large bulk carrier.

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    Part (a)

    Bending moment and shearing force, and the point of maximum bending moment.

    Considering the vessel as a compound beam (the hull girder), the shearing force at any section is the algebraic sum of the vertical forces (loads) to one side of the section, i.e. the net load (weight - buoyancy) acting on that part. The bending moment at any section is the algebraic sum of the moments of the loads to one side, i.e. the integral of the shearing force. The shearing force is the rate of change of the bending moment, and the bending moment is the integral of the shearing force. The point of maximum bending moment occurs where the shearing force is zero (where the shearing force changes sign), which is usually at or near midships for a ship in still water, and at the point where the net load changes sign. The maximum bending moment is the largest hogging or sagging moment, and the hull girder must be designed to withstand it.

    Part (b)

    Hatch coaming of a large bulk carrier.

    The hatch coaming is the vertical structure around the hatch opening that raises the hatch above the deck to prevent water entering and to provide strength. Sketch: the hatch opening is bounded by a vertical coaming plate (about 600-900 mm high for a bulk carrier) welded to the deck, with a top flange (or a horizontal stiffener) and vertical stiffeners (brackets) connecting the coaming to the deck. The coaming is made of thick plate and is stiffened to resist the loads of the hatch cover and the cargo, and to provide the longitudinal strength of the deck (the coaming acts as a longitudinal girder). The hatch cover sits on the coaming with a gasket and is secured by cleats. The coaming corners are rounded and reinforced to avoid stress concentrations. The coaming provides the watertight seal for the hatch and contributes to the longitudinal strength of the hull girder.

    Q5 (10 Marks) Hull Construction πŸ”₯ Repeated 7x

    With reference to fatigue of engineering components:

    (a) Explain the influence of stress level and cyclical frequency on expected operating life

    (b) Explain the influence of material defects on the safe operating life of an engineering components

    (c) State the factors which influence the possibility of fatigue cracking of a bed-plate transverse girder and explain how the risk of such cracking can be minimised

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    Part (a)

    Influence of Stress Level and Cyclic Frequency on Operating Life:

    Fatigue is progressive and localised structural damage caused by cyclic loading, where the maximum stress is below the ultimate tensile strength. The relationship between stress level, cyclic frequency, and operating life depends on whether the fatigue is high-cycle/low-stress or low-cycle/high-stress.

    High-cycle fatigue (low stress-high cycle):

    • This occurs at lower stress levels over a high number of cycles, resulting in elastic deformation. The component can withstand more cycles at these lower stress levels, and its life expectancy is determined by the S-N curve, which predicts the number of cycles before failure at a given stress level. For example, fatigue in turbocharger blowers often results from prolonged vibration over numerous cycles.

    Low-cycle fatigue (high stress-low cycle):

    • This occurs at high-stress levels over fewer cycles, causing plastic deformation in the material. This type of fatigue is typically assessed by a strain curve. If the stress level increases, the component's operating life decreases, as higher stress accelerates the onset of failure. For example, air receivers filling automatically face high stress and experience fewer cycles before failure.

    If stress levels or the number of cycles increase beyond the material’s capacity, failure will occur sooner. It is important to keep stress levels within allowable limits for extended component life.

    Part (b)

    Material defects can significantly reduce the safe operating life of engineering components because defects serve as stress concentrators that increase local stress around the defect. This leads to premature failure as the material cannot withstand the same level of cyclic stress as a defect-free component.

    • Surface roughness, porosity, inclusions, and abrupt section changes all create stress concentrations, lowering fatigue strength.
    • Coarse grain size, specific chemical compositions, and cold working introduce residual stresses that reduce fatigue resistance.
    • Corrosion, erosion, and decarbonisation weaken the material and accelerate fatigue crack initiation and propagation.
    • Faulty workmanship during assembly or processing introduces defects that may significantly shorten the component's life.
    Part (c)

    Factors Influencing Fatigue Cracking in Bedplate Transverse Girders:

    • Cylinder overload due to excess power puts excessive stress on the girders.
    • Incorrect crankshaft alignment induces uneven loading and stress concentrations.
    • Material defects, high residual stresses in welds, heat-affected zone hardening, and the presence of dissolved oxygen all reduce fatigue resistance.
    • Tank top deformation from pressurisation or overheating adds stress to the bedplate.

    To minimise the risk of fatigue cracking:

    (i) Constructional strength:

    • Bed plates are made up of M.S. plates with four steel casting, which are assembled and welded together so that the bed plate is strong longitudinally & transversely with good resistance to twisting along its length.
    • Longitudinal strength is obtained by fabricating each side of the bed plate in the form of a box girder.
    • The cast steel cross girder in which the main bearing is placed contributes to the bed plate's transverse strength and resistance against twisting along its length.
    • Resin cast chocks are used between the bedplate and the double bottom tank top to absorb the shocks & stress.

    (ii) Maintenance:

    • Monthly checks on the bolt tension.
    • Monthly checks on engine load using power cards & measuring cylinder peak pressure.
    • Regular checking of tension for main bearing jack bolts as recommended by engine manufacturers.
    • Regular checks on crankshaft alignment by taking deflection & compare with recommended value.
    • By maintaining engine operations at specified load, temperature, pressure, speed, etc.
    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 7x

    (a) List the precautions necessary before an inclining experiment is carried out (6)

    (b) A box shaped vessel, 50 metres long x 10 metres wide, floats in salt water on an even keel at a draft of 4 metres. A center line longitudinal watertight bulkhead extends from end to end and for the full depth of the vessel. A compartment amidships on the starboard side is 15 metres long and contains cargo with permeability 30%. Calculate the list if this compartment is bilged. KG = 3 metres. (10)

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    Part (a)

    Precautions necessary before an Inclining experiment:

    • The ship should be in a sheltered position, such as a gravity dock, to minimize external influences from wind or waves.
    • Mooring ropes should be slack to allow the ship to move freely without restrictions.
    • Only workers essential to the experiment should be present onboard to avoid unnecessary weight and movement.
    • All tanks must either be completely empty or pressed up tight to eliminate the free surface effect, which can adversely affect stability calculations.
    • Any loose weights on the ship must either be removed or properly secured to prevent unintended movement during the experiment.
    • Ensure that the pendulums are long and properly suspended from stable points, such as underneath a hatch, to provide accurate deflection readings.
    • The test masses should be evenly distributed and placed as far from the centerline as possible to maximize measurable deflections.
    • The experiment should be carried out in calm weather to avoid the effects of wind, current, or waves on the vessel's stability.
    Q7 (10 Marks) Ship Stability πŸ”₯ Repeated 11x

    (a) Define longitudinal center of gravity (LCG) and longitudinal center of buoyancy (LCB). (6)

    (b) A ship 120m long floats at draughts of 5.50m forward and 5.80m aft; MCT 1cm 80 tonne m, TPC 13, LCF 2.5m forward of midships. Calculate the new draughts when a mass of 110 tonne is added 24m aft of midships.

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.

    LCF in fwd and trim by stern

    $$Bodily \space sinkage \space = \space {{mass \space added} \over TPC} \space = \space {{110} \over 13} \space$$

    $$=\:8.5\operatorname{\mathrm{cm}}\:=0.085m$$

    $$Trim \space = \space {{m \times d} \over MCT_{1cm}}$$

    $$ = \space {{110 \times (24 + 2.5)} \over 80}$$

    $$Trim=36.43\operatorname{cm}=0.364m\:$$

    Change in fwd draught:

    $$d_{F}=\frac{-t}{L}\left\lbrack\frac{L}{2}-LCF\right\rbrack$$

    $$=\frac{-36.43}{120}\left\lbrack\frac{120}{2}-2.5\right\rbrack$$

    $$=-17.45\operatorname{cm}=-0.1745m$$

    Change in Aft draught:

    $$d_{A}=\frac{+t}{L}\left\lbrack\frac{L}{2}+LCF\right\rbrack$$

    $$=\frac{+36.43}{120}\left\lbrack\frac{120}{2}+2.5\right\rbrack$$

    $$=+18.97\operatorname{cm}=0.189m$$

    New draughts:

    $$D_{F}=5.5+0.085-0.175=5.41m$$

    $$D_{A}=5.8+0.085+0.18=6.065m$$

    Q8 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    The speed of a ship is increased to 18% above normal for 7.5 hours, then reduced to 9% below normal for 10 hours. The speed is then reduced for the remainder of the day so that the consumption for the day is the normal amount. Find the percentage difference between the distance travelled in that day and the normal distance travelled per day.

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    Let normal speed be V and normal daily consumption be C. Consumption varies as the cube of speed, so C = k.V^3.

    Step 1 - Find the speed for the remainder of the day.

    For 7.5 hours the speed is 1.18V. Consumption in that period = k.(1.18V)^3 x 7.5/24 = k.V^3 x 1.643 x 0.3125 = 0.5134 k.V^3.

    For 10 hours the speed is 0.91V. Consumption = k.(0.91V)^3 x 10/24 = k.V^3 x 0.7536 x 0.4167 = 0.3140 k.V^3.

    Total consumption in first 17.5 hours = 0.5134 + 0.3140 = 0.8274 k.V^3.

    Remaining time in the day = 24 - 17.5 = 6.5 hours.

    For the day's total consumption to equal the normal amount k.V^3, the remaining consumption must be k.V^3 - 0.8274 k.V^3 = 0.1726 k.V^3.

    If the reduced speed is Vr, then k.Vr^3 x 6.5/24 = 0.1726 k.V^3.

    So Vr^3 = 0.1726 x 24/6.5 x V^3 = 0.6373 V^3.

    Vr = (0.6373)^(1/3) V = 0.8606 V.

    So the ship travels at 86.06% of normal speed for the last 6.5 hours.

    Step 2 - Find the distance travelled that day.

    Distance = speed x time.

    Normal distance per day = V x 24 = 24V.

    Actual distance = 1.18V x 7.5 + 0.91V x 10 + 0.8606V x 6.5

    = 8.85V + 9.10V + 5.594V = 23.544V.

    Step 3 - Percentage difference.

    Difference = 24V - 23.544V = 0.456V.

    Percentage difference = 0.456/24 x 100 = 1.9%.

    Answer: The distance travelled that day is 1.9% less than the normal distance per day.

    Q9 (10 Marks) Hull Construction πŸ”₯ Repeated 4x

    An oil tanker 160m long and 22m beam floats at a draught of 9m in seawater. Cw is 0.865. The midship section is in the form of a rectangle with 1.2m radius at the bilges. A midship tank 10.5m long has twin longitudinal bulkheads and contains oil of 1.4m3/t to a depth of 11.5m. The tank is holed to the sea for the whole of its transverse section. Find the new draught.

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    New draught of the oil tanker when the midship tank is holed.

    Oil tanker 160 m long, 22 m beam, floats at a draught of 9 m in sea water. Cw = 0.865. The midship section is a rectangle with 1.2 m radius at the bilges. A midship tank 10.5 m long has twin longitudinal bulkheads and contains oil of 1.4 m3/t to a depth of 11.5 m. The tank is holed to the sea for the whole of its transverse section. Find the new draught.

    Waterplane area Aw = Cw x L x B = 0.865 x 160 x 22 = 3044.8 m2.

    Midship section area (rectangle with bilge radius r=1.2 m): Ams = B x d - (4 - pi) r^2 = 22 x 9 - 0.858 x 1.44 = 198 - 1.236 = 196.76 m2.

    Volume of the tank below the original waterline = Ams x 10.5 = 196.76 x 10.5 = 2066 m3.

    The tank contains oil of density rho_o = 1/1.4 = 0.714 t/m3. When holed, sea water (1.025 t/m3) replaces the oil, so the net loss of buoyancy is the volume times the relative density difference:

    Vlost = 2066 x (1 - 0.714/1.025) = 2066 x 0.303 = 626 m3.

    The flooded tank provides no increase of buoyancy, so the effective sinking waterplane = Aw - (10.5 x 22) = 3044.8 - 231 = 2813.8 m2.

    Sinkage = Vlost/effective waterplane = 626/2813.8 = 0.222 m.

    New draught = 9 + 0.22 = 9.22 m.

    Answer: the new draught is about 9.2 m.

    Q10 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 7x

    (a) What is meant by the Admiralty Coefficient and the Fuel Coefficient?

    (b) A ship of 14900 tonne displacement has a shaft power of 4460 kW at 14.55 knots. The shaft power is reduced to 4120 kW and the fuel consumption at the same displacement is 541 kg/h. Calculate the fuel coefficient for the ship.

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    The Admiralty Coefficient (C) is a method for estimating the propulsion power needed for a newly built ship. It's considered relatively constant for a given ship design. The formula is:

    $$C=\frac{\Delta^{2/3}\times V^3}{BP}$$

    Where:

    • C = Admiralty Coefficient
    • Ξ” = Displacement in tonnes (weight of the ship when fully loaded)
    • V = Speed in knots
    • BP = Brake power in kilowatts (power delivered by the ship's engine)

    A higher Admiralty Coefficient indicates a more efficient ship design, meaning it requires less power to achieve a given speed. Values typically range from 350 to 600.

    Fuel Coefficient:

    The Fuel Coefficient (F.C.) is used to calculate a ship's daily fuel oil consumption. The formula is:

    $$Daily\:fuel\:oil\:consumption\:=\:\frac{\Delta^{2/3}\times V^3}{FC}$$

    Where:

    • F.C. = Fuel Coefficient
    • Ξ” = Displacement in tonnes
    • V = Speed in knots

    The Fuel Coefficient can vary significantly, with typical values ranging from 40,000 to 120,000. A higher Fuel Coefficient implies greater fuel efficiency (lower daily fuel consumption) for a given speed and displacement.

    Part (b)

    $$admiraty\:coefficient\:\left(C\right)=\:\frac{\Delta^{2/3}V^3}{Shaft\:power}=\frac{\Delta^{\frac23}\times V^3}{SP}$$

    $$\frac{SP_1}{SP_2}=\frac{V_1^3}{V_2^3}$$

    $$\frac{4460}{4120}=\frac{14.55^3}{V_2^3}$$

    $$V_2=14.17kntos$$

    $$Fuel\:consumption\:per\:hour=541\operatorname{\mathrm{\:kg}}\:per\:hour$$

    $$Fuel\:consumption\:per\:day\:=\:541\times24=12.98t\:per\:day$$

    $$Fuel\:coefficient=\frac{\Delta^{\frac23}\times V_2^3}{Fuel\:consumption\:per\:day}$$

    $$=\:\frac{14900^{\frac23}\times14.17^3}{12.98}$$

    $$=132726.9$$

    Q1 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 8x

    Vessel has gone through very heavy weather. On arrival at safe anchorage, you are conducting your inspection to determine damages to bull.

    (a) List the areas vou will inspect.

    (b) List your findings of any significance.

    (c) Write a report to company suggesting repairs if any

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    Part (a)

    Areas to Inspect After Heavy Weather

    Upon arrival at safe anchorage, a thorough inspection of the vessel shall be carried out, both externally and internally, to assess any weather-related damage. Key areas include:

    1. Hull and Main Deck

    • Shell plating along bow, midship, and stern sections (especially forepeak and flare areas).
    • Boot-top and bilge areas for dents or deformation.
    • Deck plating for buckling or cracked welds.
    • Bulwarks, rails, fairleads, chocks, and mooring fittings.

    2. Forecastle and Forward Structure

    • Bosun store and chain locker for water ingress.
    • Windlass foundation, anchors, and hawse pipes for deformation or cracks.
    • Forepeak tank for leakage or pressure damage.

    3. Cargo Holds / Tanks

    • Inspect for structural deformation, loose frames, or fractured stiffeners.
    • Check tank top plating and bilges for leakage.
    • Check watertight doors, gaskets, and vents.

    4. Superstructure and Deck Fittings

    • Bridge wings, radar mast, funnel, vents, and deckhouses for cracks or loose fittings.
    • Lifeboat davits, securing arrangements, and deck cranes.

    5. Underwater and Machinery Spaces

    • Rudder, propeller, and stern tube seals (via steering gear tests).
    • Sea chest gratings and overboard discharges.
    • Engine room bilges for any seawater ingress.

    Part (b)

    Typical Findings of Significance

    • Dents and indentations on shell plating at bow and forward port side due to heavy slamming.
    • Paint coating and corrosion protection partly peeled off near waterline and forepeak area.
    • Loose fairlead bolts and one cracked weld on starboard bulwark stanchion.
    • Deformed ventilator head on forecastle deck.
    • Minor leakage observed in forepeak tank during sounding check.
    • Bridge wing railing bent, likely from green sea impact.
    • Anchor chain links twisted and worn.
    • Lifeboat gripes loosened, requiring tightening and inspection.
    • No flooding reported; watertight integrity maintained overall.

    Part (c)

    Report to Company – Heavy Weather Damage Inspection

    To: Superintendent / Technical Department

    From: Name / Rank

    Subject: Heavy Weather Damage Inspection Report

    Date: [Insert date]

    Vessel: [Insert vessel name]

    Summary

    The vessel experienced very heavy weather en route from [Port A] to [Port B], with significant pitching and rolling in seas up to [X] meters. On arrival at safe anchorage, a complete inspection of the hull and deck was carried out.

    Findings

    • Bow and forecastle plating show minor dents, with no breach of watertight integrity.
    • Bulwark stanchion (starboard side) cracked at welded joint – requires repair.
    • Fairlead foundation bolts loosened – retightening and NDT inspection recommended.
    • Ventilator head on forecastle deformed – replacement advised.
    • Forepeak tank shows trace leakage at forward bulkhead – to be pressure tested.
    • Paint coating damage and corrosion exposure on bow area – recoating required.
    • Bridge wing railing bent – to be straightened or renewed.
    • All other structures and machinery appear satisfactory after testing.

    Recommendations

    1. Conduct minor steel renewal and welding repairs at the next port with repair facilities.
    2. Carry out thickness measurements and NDT on affected hull areas.
    3. Recoat damaged paint areas to prevent corrosion.
    4. Replace deformed ventilator head and bent railing.
    5. Inspect anchor and chain for elongation; renew worn links.
    6. Pressure test forepeak tank after repairs.
    7. Submit class surveyor report if deemed necessary by the superintendent.

    Conclusion

    The vessel remains seaworthy but requires prompt attention to minor structural and fitting damages before the next voyage. Preventive maintenance and weather routing measures should be reviewed for future passages.

    Signed:

    Name / Rank

    Signature

    Q2 (10 Marks) Ship Stability πŸ”₯ Repeated 14x

    Explain how the period of roll varies with

    (a) The amplitude of roll.

    (b) The radius of gyration.

    (c) The initial metacentric height.

    (d) The location of masses in the ship

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    The period of roll Tr of a ship is determined by the formula:

    $$T_{r}=\frac{2\pi K}{\sqrt{g\times GM}}$$

    where,

    • K is the radius of gyration (mass moment of inertia)
    • g is the acceleration due to gravity, and
    • GM is the metacentric height.
    Part (a)

    Amplitude of Roll:

    • The amplitude of roll, or the maximum angle of heel, does not affect the period of roll. While a larger amplitude implies a greater heeling moment and faster roll speed, the time taken to complete one full roll cycle remains constant for a given metacentric height and radius of gyration.
    Part (b)

    Radius of Gyration (K):

    • The period of roll is directly proportional to the radius of gyration. A larger radius of gyration (indicating a greater distribution of mass further from the ship's centre of rotation) leads to a longer period of roll. Conversely, a smaller radius of gyration (mass concentrated closer to the centre) results in a shorter period. The distribution of cargo significantly impacts K; cargo concentrated centrally minimises K and the roll period, while dispersed cargo maximises K and the roll period.
    Part (c)

    Initial Metacentric Height (GM):

    • The period of roll is inversely proportional to the square root of the metacentric height (GM). A larger GM (a stiffer ship) leads to a shorter roll period, as the ship quickly returns to its upright position. A smaller GM (a tender ship) results in a longer roll period, with slower return to the upright.
    Part (d)

    Location of Masses in the Ship:

    The location of masses in the ship will effect the GM & K. So the period of roll will be affected.

    • If masses are at bottom, G moves down, GM ↑, period of roll ↓.
    • If masses are at top, G moves up, GM ↓, period of roll ↑.
    • If masses are concentrated at centre, K ↓, period of roll ↓.
    • If masses are away from centre, K ↑, period of roll ↑.
    Q3 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 9x

    With reference to Underwater Inspection in lieu of Dry docking(UWILD):

    (a) Explain in detail, how an underwater survey is carried out.

    (b) State the requirements to be fulfilled before an underwater survey is acceptable to the surveying authority.

    (c) Construct a list of the items in order of importance that the underwater survey authority should include.

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    (a) An underwater survey, also known as Underwater Inspection in Lieu of Dry Docking (UWILD), involves a systematic and detailed examination of a vessel’s hull and underwater components while the ship remains afloat. The procedure includes the following steps:
    • The shipowner sends a request to the classification society surveyor, indicating the intention to perform an underwater survey.
    • A detailed plan of the ship's external hull features is submitted, showing the location of frames, bulkheads, welding lines, openings, etc.
    • The hull surface is cleaned before the survey to remove any marine growth or fouling that could obscure the inspection.
    • A diving company, approved by the classification society, is appointed to conduct the underwater inspection.
  • A self-propelled survey vehicle equipped with the following tools is used:
    • Long-Range Light TV Camera to aid in steering and checking hull deterioration.
    • High-resolution colour TV Camera to provide a close-up view of the hull coating and welded seams.
    • 35mm Still Camera to capture still images.
    • Ultrasonic Probe for measuring plate thickness.
    • Depth Meter and Speed Indicator to provide accurate data on the vehicle's depth and movement.
    • Umbilical Cable to connect the survey vehicle to the survey boat, transmitting power and information.
  • The survey boat is to be equipped with:
    • A control console with TV monitors.
    • Plate thickness printouts.
    • Audio and video cassette recorders.
    • Playback units.
    • Diver communication systems.
    • Vehicle control systems and associated instruments.
  • Operation:
    • The survey vehicle is taken underwater by a diver to the survey starting point.
    • Using TV monitors and shell expansion plans as guides, the vehicle is navigated over the hull, focusing on the bottom structure, sides, stern frame, rudder, propeller, bilge keel, and hull openings.
    • All images, data, and information are recorded and transmitted back to the survey boat.
    • Detailed pictures of the stern frame, rudder, propeller, bilge keel, and hull openings are captured.
    • Divers are deployed to measure stern tube bearing wear, pintle clearance, and inspect stern seals, anodes, and rudder stock palm coupling bolts.
    • All recorded video and audio, including conversations between the surveyor and drivers, thickness printouts, measurements, and pictures are analyzed to determine the detailed underwater condition of the vessel.
    (b) Before an underwater survey is accepted by the surveying authority, the following conditions must be met:

    The vessel's owner submits a request to the surveyor, including:
    • The proposed date and location for the survey.
    • General information about the diving company.
    • A declaration that the vessel has not suffered any damage due to grounding, collision, or other incidents.
    The ship's master or owner’s representative must provide a declaration confirming:
    • Any suspected or actual damage to the hull since the last dry-docking.
    • The underwater portion of the hull is protected by a suitable paint scheme that is of adequate thickness and remains valid until the next dry-dock.

    • The survey site should be in a protected area with calm and clear water, ensuring good underwater visibility. Attention must be given to the effects of currents and tides.
    • The hull must be clean for the external survey. The surveyor must be satisfied with the method and quality of the pictorial presentation, ensuring that it provides a reliable assessment of the hull's condition.
    • The underwater examination should be conducted by an approved diving company using closed-circuit TV and two-way communication, which can be monitored by the surveyor.
    • The vessel should be in as light an operating condition as possible to facilitate the survey.
    • Means must be available for the surveyor to examine the outside shell plating above the waterline.
    • Any required repairs identified during the survey must be carried out to the satisfaction of the attending surveyor.
    (c) While the importance of each item may vary depending on the vessel and its specific requirements, below is a list of items that should be included in an underwater survey in order of importance:
    • Underwater Hull: General condition of the hull below the waterline.
    • Bottom and Shell Plating: Inspection for corrosion, damage, and fouling.
    • Shell Openings: Examination of openings such as sea chests, drain plugs, and overboard discharge points.
    • Stern Tube Oil Leaks: Check for leaks around the stern tube.
    • Propeller Blade: Inspection for damage, wear, and fouling.
    • Rudder: Inspection for damage, wear, and clearances.
    • Sea Chest Opening and Grating: Examination for blockages, damage, and fouling.
    • Anodes: Check the condition and effectiveness of cathodic protection anodes.
    • Bilge Keel: Inspection for damage and fouling.
    • Drain Plugs: Ensure all drain plugs are secure and in good condition.
    • Overboard Valve Openings: Check for proper operation and condition.
    • Forward Area: Inspection for any damage due to anchor and chain movement.
    Q4 (10 Marks) Hull Construction

    Give a reasoned opinion as to the validity of the following assertions concering ship Structure

    (a) Crack propagation in propellers shaft 'A' bracket or spectacle frames is indicative of inadequate scantlings and strength.

    (b) The adequate provision of deck scuppers and freeing ports is as critical to the seaworthiness as watertight integrity.

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    Part (a)

    Crack propagation in propeller shaft 'A' bracket or spectacle frames - reasoned opinion.

    This assertion is only partly valid. Cracks in these members are not necessarily proof of inadequate scantlings or strength. In practice, cracks in A-brackets, spectacle frames and propeller shafts are far more commonly the result of:

    • Fatigue loading: these members are subject to continuous cyclic loading from the propeller, vibration and wave action. Fatigue cracks initiate at stress raisers such as sharp corners, weld toes, keyways, fillet radii and corrosion pits even when the general scantlings are perfectly adequate.
    • Stress concentration: poor detail design, sharp notches and abrupt changes of section concentrate stress locally and promote crack initiation.
    • Vibration: excessive propeller-induced vibration and whirling of the shaft can cause cracking.
    • Corrosion and erosion: pitting, cavitation damage and corrosion reduce the effective section and act as initiation points.
    • Material defects: inclusions, poor weld quality and inadequate heat treatment.
    • Misalignment and bearing wear.

    However, the assertion has some validity: if cracks are widespread, recurring, or accompanied by permanent set or buckling, they may indicate that the scantlings are inadequate for the duty, or that the design strength margin is insufficient. The correct approach is to investigate the cause (fatigue, vibration, corrosion, or genuine under-design) rather than assume inadequate scantlings. A single crack is usually a local fatigue or detail problem; repeated cracking in the same region may point to a genuine strength deficiency. So the assertion is not universally true - it is valid only when cracking is systematic and not attributable to fatigue, vibration or corrosion.

    Part (b)

    Deck scuppers and freeing ports as critical to seaworthiness as watertight integrity - reasoned opinion.

    This assertion is largely valid. Watertight integrity keeps water out of the hull, but the ability to shed water from the deck is equally important to seaworthiness. The reasons are:

    • Freeing ports allow water shipped on deck to drain rapidly. If water is retained on deck, it adds weight high up, raising the centre of gravity and reducing stability (free surface and added weight effect), and can lead to a dangerous reduction in metacentric height.
    • Retained deck water increases the effective displacement and can cause the vessel to become sluggish and unmanageable.
    • In heavy weather, water trapped on deck can shift and cause a sudden loss of stability or capsize, particularly in vessels with low freeboard.
    • Scuppers and freeing ports prevent accumulation of water that could enter the hull through openings, hatches and ventilators, thereby protecting watertight integrity itself.
    • Blocked or inadequate freeing ports are a recognised cause of loss of stability and capsizing of fishing vessels and small craft.

    Therefore, the provision of adequate, correctly sized and properly maintained freeing ports and scuppers is as critical to seaworthiness as watertight integrity. Both are essential; neither can compensate for the other. The assertion is valid.

    Q5 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    The palm of the riuder of a vessel requires extensive welding repairs and as Second Engineer you are requested to supervise.

    (a) Suggest a suitable type of welding process.

    (b) State, with reasons, FOUR common welding defects that can occur there

    (c) State what tests may be carried out before returning the rudder to service

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    As Second Engineer, I would oversee the extensive welding repairs required for the vessel's rudder using the following plan:

    Part (a)

    Suitable Welding Process:

    Manual Metal Arc Welding (MMAW), also known as Shielded Metal Arc Welding (SMAW), is the most suitable process for this repair. The reasons are threefold:

    • MMAW is highly portable, allowing for on-site repair within the drydock. The process is adaptable to various welding positions (downhand, overhead, horizontal, vertical) – a necessity given the complex geometry of a rudder.
    • Assuming the rudder is constructed from standard steel, MMAW using readily available flux-coated electrodes provides good control, arc stability, and penetration. The flux coating protects the weld pool from atmospheric contamination during cooling.
    • MMAW requires relatively simple equipment and is less demanding in terms of operator skill compared to other processes like TIG or MIG. This translates to cost-effectiveness and allows for a wider pool of qualified welders.
    • If cast steel components are present, pre-heating will be necessary to minimize stress cracking, and specialized electrodes suited for the specific cast steel grade must be selected.

    During welding by the metal arc process, the following points must be observed:

    • Electrode Consumption Rate
    • Penetration
    • Slag Control
    • Arc Length and Sound
    Part (b)

    Four Common Welding Defects:

    1. Undercut: A groove formed along the edge of the weld bead, weakening the joint. Caused by excessive current, incorrect electrode angle, excessive travel speed, or improper electrode manipulation.

    2. Overlap: Molten weld metal flows over the parent metal without proper fusion. Caused by low current, slow travel speed, excessive arc length, or improper joint preparation.

    3. Slag Inclusion: Trapped slag within the weld metal, reducing its strength and potentially causing cracking. Caused by insufficient cleaning between passes, incorrect current, long arc length, slow travel speed, or too large an electrode diameter.

    4. Incomplete Penetration: The weld does not fully fuse the joint faces, resulting in a weak joint. Caused by insufficient current, incorrect joint preparation (too small a root gap or bevel angle), excessive travel speed, or too large an electrode diameter.

    Part (c)

    Tests Before Returning to Service:

    • A thorough visual examination of all welds to identify any surface defects like cracks, porosity, or lack of fusion.
    • NDT methods such as Magnetic Particle Inspection (MPI) or Dye Penetrant Inspection (DPI) will be employed to detect subsurface flaws that may not be visible during visual inspection. The specific NDT method chosen will depend on the type of steel and the accessibility of the weld areas.
    • The repaired rudder will undergo a hydrostatic pressure test. This involves filling the rudder with a water head of 2.46 meters and observing for any leaks. This confirms the watertight integrity of the welds and the overall rudder structure.
    Q6 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 7x

    A ship of 15000 tonne displacement has an Admiralty Coefficient, based on shaft power, of 420. The mechanical efficiency of the machinery is 83%, shaft losses 6%, propeller efficiency 65% and QPC 0.71. At a particular speed the thrust power is 2550kW.

    Calculate: (16)

    (i) indicated power

    (ii) effective power

    (iii) ship speed

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    Given:

    $$\Delta=15000t$$

    $$Shaft\:Power\:\left(SP\right)=420$$

    $$Transmission\:Efficiency=83\%$$

    $$Shaft\:losses=6\%$$

    $$Propeller\:Efficiency=65\%$$

    $$QPC=0.71$$

    $$Thrust\:Power=2550kW$$

    $$\left(\imaginaryI\right)\:Delivered\:Power\:\left(DP\right)=\frac{Thrust\:Power\:\left(TP\right)}{Propeller\:Efficiency\:\left(\eta P\right)}$$

    $$DP=\frac{2550}{0.65}$$

    $$DP=3923.07kW$$

    $$\left(ii\right)\:Shaft\:Power=\frac{Delivered\:Power\:\left(DP\right)}{Transmission\:Efficiency\:\left(\eta T\right)}\:$$

    $$SP=\frac{3923.07}{0.94}$$

    $$SP=4173.47kW$$

    $$\left(iii\right)\:Indicated\:Power=\frac{Shaft\:Power\:\left(SP\right)}{Mechanical\:Efficiency\:\left(\eta m\right)}$$

    $$IP=\frac{4173.47}{0.83}$$

    $$IP=5028.28kW$$

    $$\left(iv\right)\:Effective\:Power=DP\times QPC$$

    $$EP=3923.07\times0.71$$

    $$EP=2785.3797kW$$

    $$\left(v\right)\:Shaft\:Power=\frac{\Delta^{\frac23}\times V^3}{Admiralty\:Co-efficient}$$

    $$4173.47=\frac{15000^{\frac23}\times V^3}{420}$$

    $$V=14.23knots$$

    Q7 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Describe stability requirements for dry-docking. (6)

    (b) A ship of 8000t displacement floats upright in sea water, with KG = 7.6m, GM = 0.5m. A tank, whose Kg is 0.6m above the keel and 3.5m from the center line contains 100 t of water ballast. Neglecting the free surface effect, calculate the angle which the ship will heel, when the ballast water is pumped out. (10)

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    Part (a)

    For safe dry-docking, a ship must meet two key stability requirements:

    • Positive GM (Metacentric Height): The ship needs a positive GM. GM is the distance between the centre of gravity (G) and the metacentre (M). A positive GM indicates inherent stability; the ship will right itself if tilted. During dry-docking, the loss of buoyancy as the ship rests on the blocks reduces GM. Insufficient positive GM increases the risk of the ship heeling over or capsizing.
    • Trim by the stern: The vessel should be trimmed slightly by the stern (aft end lower than the bow) to ensure the aft end sits on the keel blocks first. This controlled settling minimises the risk of instability during the docking process. An even keel is generally preferred for the initial floating condition before the dry-docking procedure begins.
    Part (b)

    $$new \space KG \space = \space {{(8000 \times 7.6) - (100 \times 0.6)} \over 8000 - 100}$$

    $$New\:KG\:=\:7.689m$$

    $$New \space GM_1 \space = \space KM - KG$$

    $$= \space (7.6 + 0.5) - 7.689$$

    $$New\:GM\:=\:0.411m$$

    Angle of heel when 100t ballast is pumped out

    $$Tan\theta=\frac{m\times d}{\Delta GM}$$

    $$=\frac{100\times3.5}{7900\times0.411}$$

    $$Tan\theta=0.1077$$

    $$\theta=6^09^{^{\prime}}$$

    Q8 (10 Marks) Ship Stability πŸ”₯ Repeated 8x

    (a) Define longitudinal center of gravity (LCG) and longitudinal centre of buoyancy (LCB). (6)

    (b) The immersed cross-sectional areas of a ship 120 m long, commencing from aft are 2, 40, 79, 100, 103, 104, 104, 103, 97, 58 and 0 m2.

    Calculate:

    (i) Displacement

    (ii) Longitudinal position of the centre of buoyancy.

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.
    Q9 (10 Marks) Hull Construction πŸ”₯ Repeated 7x

    With respect to Buoyancy of a vessel:

    (a) What do you understand by reserve buoyancy what happen if the lost buoyancy is greater than the reserve buoyancy. (6)

    (b) A forward deep tank 12 m long extends from a longitudinal bulkhead to the ship's side. The widths of the tank surface measured from the longitudinal bulkhead at regular intervals are 10, 9, 7, 4 and 1 m. Calculate the seeond moment of area of the tank surface about a longitudinal axis passing through its centroid. (10)

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    Part (a)

    Reserve Buoyancy

    Reserve buoyancy is the watertight volume above the waterline of a vessel. It represents the potential buoyancy that can be utilised to keep the ship afloat when additional weight is added or if some buoyancy is lost due to damage (e.g., bilging).

    When a mass is added to the ship or buoyancy is lost (e.g., due to flooding of a compartment), the reserve buoyancy is converted into active buoyancy by an increase in draught.

    If the lost buoyancy (due to flooding or damage) is greater than the reserve buoyancy, the vessel will no longer have sufficient buoyant force to counteract its weight, causing it to sink.

    Part (b)
    Q10 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 6x

    A ship of length 140m, Breadth of 18.5m, draught of 8.1 m and a displacement of 17,025 tonnes in sea water, has a face pitch ratio of 0.673. The diameter of the propeller is 4.8m. The results of the speed trial show that true slip may be regarded as constant over a range of 9 to 13 knots and is 30%. w = 0.5Cb-0.05. If fuel used is 20t/day at 13 knots and fuel consumption/day varies as cube of speed of ship. determine the fuel consumption, when propeller runs at 110 rpm.

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    Given:

    $$Lenght,\:L=140m$$

    $$Breadth,\:B=18.5m$$

    $$Draught,\:d=8.1m$$

    $$Displacement,\:\Delta=17025tonnes$$

    $$Pitch\:ratio,\:p=0.673\operatorname{}$$

    $$Diameter\:of\:Propeller,\:D=4.8m$$

    $$\operatorname{Real\:Slip,\:R_{s}=30\%\:or\:0.3}$$

    $$Wake\:fraction,\:W=0.5C_{b}-0.05$$

    $$Consumption,\:C_2=\:20t\:per\:day\:$$

    $$Ship\:Speed,\:V_2=13\:knots$$

    $$\operatorname{Revolution,\:N}=110rpm$$

    $$cons\:per\:day\:\alpha\:V^3$$

    To find Fuel Consumption C1=?

    We know that,

    $$Displacement,\:\Delta=\nabla\times\rho$$

    $$17025=\nabla\times1.025$$

    $$\nabla=16609.76m^3$$

    $$Block\:Coefficient,\:C_{b}=\frac{\nabla}{L\times B\times d}$$

    $$C_{b}=\frac{16609.76}{140\times18.5\times8.1}$$

    $$C_{b}=0.792$$

    $$Wake\:Fraction,\:W=0.5C_{b}-0.05$$

    $$W=0.5\times0.792-0.05$$

    $$W=0.346$$

    $$Pitch\:ratio,\:p=\frac{P}{D}$$

    $$0.673=\frac{P}{4.8}$$

    $$P=4.8\times0.673$$

    $$P=3.23m$$

    $$Theoretical\:Speed,\:V_{t}=\frac{P\times N\times60}{1852}$$

    $$V_{t}=\frac{3.23\times110\times60}{1852}$$

    $$V_{t}=11.51knots$$

    Using, Real slip equation.

    $$\operatorname{\operatorname{Real\:Slip,\:R_{s}=\frac{V_{t}-V_{a}}{V_{t}}}}$$

    $$0.3=\frac{11.51-V_{a}}{11.51}$$

    $$V_{a}=11.51-11.51\times0.3$$

    $$V_{a}=11.51\left(1-0.3\right)$$

    $$V_{a}=11.51\times0.7$$

    $$V_{a}=8.057knots$$

    $$Wake\:fraction,\:W=\frac{V-V_{a}}{V}$$

    $$0.346=\frac{V-8.057}{V}$$

    $$0.346V=V-8.057$$

    $$V=\frac{8.057}{0.654}$$

    $$V=12.32knots$$

    $$cons\:per\:day\:\alpha\:V^3$$

    $$\frac{C_1}{C_2}=\left(\frac{V_1}{V_2}\right)^3$$

    $$\frac{C_1}{20}=\left(\frac{12.32}{13}\right)^3$$

    $$C_1=20\times\left(\frac{12.32}{13}\right)^3$$

    $$C_1=20\times0.851$$

    $$C_1=17.02t\:per\:day$$

    Q1 (10 Marks) Ship Stability πŸ”₯ Repeated 6x

    With reference to Roll-on, Roll-off ferries:

    (a) Describe the problem of free surface effect

    (b) Explain how it is intended that water should be cleared from car or cargo decks

    (c) Describe possible methods for improving the stability and survivability of these vessels.

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    Part (a)

    The problem of free-surface effect in Ro-Ro ferries.

    Ro-Ro ferries have large, open car and cargo decks that extend over a large part of the ship. If water is shipped onto these decks (e.g. through the bow or stern doors, or in heavy weather), the water spreads over the large deck area, creating a very large free surface. The free-surface effect reduces the effective GM by rho x i/Delta, where i is the second moment of area of the free surface (i = L B^3/12 for a rectangular deck). Because the car deck is very wide and long, the free-surface effect is very large and can reduce the GM to a dangerously low value, causing the ship to lose stability and capsize. This is the principal stability problem of Ro-Ro ferries: the large open decks create a huge free-surface effect if flooded, and the ship can capsize rapidly.

    Part (b)

    How water should be cleared from car or cargo decks.

    Water on the car deck should be cleared by:

    • Providing adequate freeing arrangements (scuppers, freeing ports, drain valves) at the deck edge and at the ends, so that water can drain overboard.
    • Providing a camber (transverse slope) on the deck so that water runs to the sides and drains through the freeing ports.
    • Providing a longitudinal slope (sheer) so that water runs to the ends and drains.
    • Using bilge pumps and drainage systems to remove water that cannot drain overboard.
    • Ensuring the freeing ports are of adequate size and are not blocked by cargo or lashings.

    The freeing arrangements must be adequate to remove water quickly and prevent the build-up of a large free surface.

    Part (c)

    Methods for improving the stability and survivability of Ro-Ro ferries.

    • Lowering the centre of gravity by placing heavy weights low and ballast in the double bottom.
    • Increasing the GM by increasing the beam and the waterplane area, and by lowering KG.
    • Providing adequate freeboard and reserve buoyancy.
    • Subdividing the car deck with watertight bulkheads or providing a raised deck to limit the spread of water.
    • Providing adequate freeing ports and drainage to remove water quickly.
    • Fitting bilge keels and stabilisers to reduce roll.
    • Designing the ship to meet the damage stability criteria (survive flooding of a compartment).
    • Using a higher freeboard and a stronger, more watertight structure, and ensuring the bow and stern doors are watertight and properly secured.
    • Operating with adequate stability margins and following the loading manual.
    Q2 (10 Marks) Hull Construction πŸ”₯ Repeated 5x

    Give a reasoned opinion as to the validity of the following assertions concerning ship structure:

    (a) Crack propagation in propellers shaft 'A' bracket or spectacle frames is indicative of inadequate scantlings and strength.

    (b) The adequate provision of freeing ports is as critical to the seaworthiness as watertight integrity.

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    Part (a)

    Crack propagation in propeller shaft 'A' brackets or spectacle frames - is it indicative of inadequate scantlings and strength?

    This assertion is only partly valid. Cracks in A-brackets or spectacle frames (the shaft brackets supporting the propeller shaft) are more commonly the result of fatigue due to fluctuating loads, stress concentrations at the bracket-to-hull connection, and the vibration and whipping of the shaft, rather than simply inadequate scantlings. The brackets are subject to severe alternating loads from the propeller and the shaft, and cracks typically initiate at stress raisers (sharp corners, weld toes, the bracket arm-to-hull connection) and propagate under fatigue. While inadequate scantlings or poor design (insufficient section, poor connection, sharp notches) can contribute, the primary cause is usually fatigue and stress concentration, aggravated by vibration, corrosion and the dynamic loads of the propeller. Hence the assertion is not fully valid: crack propagation is more indicative of fatigue and stress concentration than of inadequate strength alone, and the design should address the fatigue life, the connection detail and the avoidance of stress raisers, as well as the scantlings.

    Part (b)

    The adequate provision of freeing ports is as critical to seaworthiness as watertight integrity.

    This assertion is largely valid. Freeing ports (openings in the bulwark that allow water shipped on deck to drain overboard) are essential to seaworthiness because, if they are inadequate, water accumulating on the deck cannot drain, which:

    • increases the free-surface effect and the weight of water on deck, reducing stability and increasing the risk of capsize;
    • increases the deck load and the risk of structural damage;
    • reduces the reserve buoyancy and can lead to the ship becoming unstable.

    Watertight integrity (the ability of the hull and its openings to keep water out) is equally critical to seaworthiness, as it prevents flooding and loss of buoyancy. Both are essential: watertight integrity keeps water out, while freeing ports remove water that is shipped on deck. If either is inadequate, the ship's seaworthiness is compromised. Hence the assertion is valid - freeing ports are as critical to seaworthiness as watertight integrity, because they maintain the stability and buoyancy of the ship by removing deck water.

    Q3 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 4x

    (a) What is free surface effect? How can be avoided or reduced.

    (b) Give the components of ships resistance while vessel is 'enroute'.

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    (a) Free Surface Effect:

    The free surface effect (FSE) is a reduction in the metacentric height (GM) of a vessel due to the movement of liquids within partially filled tanks when the ship heels. When a ship tilts, the liquid in a partially filled tank shifts to the lower side, causing the centre of gravity (CG) of the ship to move laterally. This lateral shift of the CG reduces the righting lever (GZ), effectively decreasing the ship's stability and increasing the angle of heel. This apparent loss of GM is the free surface effect.

    Minimizing Free Surface Effect:

    • Partially filled tanks should be avoided to minimize liquid movement.
    • Tanks should be designed with longitudinal divisions or swash bulkheads to limit the movement of liquid.
    • Install sluice valves to control the liquid movement between compartments in divided tanks.
    • Fill smaller tanks at the bottom of the ship first to lower the centre of gravity and improve stability.
    • Tanks should have reduced breadth to minimize the free surface's effect.
    Part (b)

    Components of ship’s resistance while vessel is en route

    When a ship moves through water, resistance opposes its motion. The ship must exert an equal force to maintain speed.

    Frictional resistance (Rf): This is caused by the friction between the hull and the water. The water immediately adjacent to the hull is dragged along, creating a boundary layer. This resistance depends on the water's viscosity, the ship's speed, and the wetted surface area of the hull. At lower speeds, frictional resistance can account for 70-90% of total resistance; however, at higher speeds, it can be less than 40%.

    Residuary resistance (Rr): This is the resistance that remains after subtracting the frictional resistance. These are:

    • Form drag: Resistance due to the shape of the hull and the flow of water around it, generating pressure differences.
    • Wave-making resistance: This is a major component at higher speeds. The ship creates waves, and energy is expended in this process.
    • Eddy resistance: Resistance caused by turbulent flow behind the ship, especially at sharp changes in the hull's shape. This is often minimized by optimizing the hull design.

    Air resistance (Ra): This resistance is generated by the ship moving through the air. It depends on the shape of the superstructure, the projected area above the waterline, and wind speed and direction. Air resistance is typically a smaller component, usually around 2% but can be up to 10% for large container ships with extensive superstructure.

    Total resistance (Rt): The total resistance experienced by the ship is the sum of frictional, residuary, and air resistance: Rt = Rf + Rr + Ra.

    Q4 (10 Marks) Ship Types & Design πŸ”₯ Repeated 4x

    (a) Draw and the mid ships section of an oil tanker with Double Hull & name each part.

    (b) What is Bow Flare? Why is it so important in Bulk Carriers?

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    Part (a)

    Mid-ship section of Oil tanker:

    Part (b)

    Bow flare

    is the outward curvature of a ship's side shell above the waterline at the forward end.

    Importance of Bow Flare in Bulk Carriers:

    • Bow flare enhances the reserve buoyancy at the forward end of the vessel, which improves seaworthiness by helping the ship ride over waves more effectively, especially when pitching in rough seas.
    • By dispersing water away from the ship, the bow flare reduces the amount of water shipped onto the deck during heavy weather.
    • The wider forecastle deck created by the bow flare allows for the installation of essential machinery such as windlasses, mooring equipment, and other fittings.
    • The bow flare shields the hull plating from damage caused by the anchor when it is raised or lowered.
    • A well-designed bow flare can reduce water resistance, leading to increased speed and better fuel efficiency.
    Q5 (10 Marks) Ship Types & Design πŸ”₯ Repeated 6x

    (a) Considering the vessel as a compound beam define Bending moment shearing force. Which is the point of Maximum Bending Moment?

    (b) Sketch and Describe Hatch coming of a large bulk carrier.

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    Part (a)

    Bending moment and shearing force, and the point of maximum bending moment.

    Considering the vessel as a compound beam (the hull girder), the shearing force at any section is the algebraic sum of the vertical forces (loads) to one side of the section, i.e. the net load (weight - buoyancy) acting on that part. The bending moment at any section is the algebraic sum of the moments of the loads to one side, i.e. the integral of the shearing force. The shearing force is the rate of change of the bending moment, and the bending moment is the integral of the shearing force. The point of maximum bending moment occurs where the shearing force is zero (where the shearing force changes sign), which is usually at or near midships for a ship in still water, and at the point where the net load changes sign. The maximum bending moment is the largest hogging or sagging moment, and the hull girder must be designed to withstand it.

    Part (b)

    Hatch coaming of a large bulk carrier.

    The hatch coaming is the vertical structure around the hatch opening that raises the hatch above the deck to prevent water entering and to provide strength. Sketch: the hatch opening is bounded by a vertical coaming plate (about 600-900 mm high for a bulk carrier) welded to the deck, with a top flange (or a horizontal stiffener) and vertical stiffeners (brackets) connecting the coaming to the deck. The coaming is made of thick plate and is stiffened to resist the loads of the hatch cover and the cargo, and to provide the longitudinal strength of the deck (the coaming acts as a longitudinal girder). The hatch cover sits on the coaming with a gasket and is secured by cleats. The coaming corners are rounded and reinforced to avoid stress concentrations. The coaming provides the watertight seal for the hatch and contributes to the longitudinal strength of the hull girder.

    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 9x

    (a) Describe how the force on the ship's bottom and the GM vary when grounding takes place. (6)

    (b) A ship of 8,000 tonnes displacement takes the ground on a sand bank on a falling tide at an even keel draft of 5.2 metres. KG 4.0 metres. The predicted depth of water over the sand bank at the following low water is 3.2 metres. Calculate the GM at this time assuming that the KM will then be 5.0 metres and that the mean TPC is 15 tonne (10)

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    Part (a)

    When a ship grounds, the forces on the bottom and the metacentric height (GM) change depending on the grounding's nature and the ship's position on the seabed.

    If the ship grounds on a level bottom:

    • A ground reaction force acts vertically upwards from the seabed. This force counteracts part of the ship's weight and alters the distribution of buoyancy forces.
    • The ship’s centre of gravity (G) may appear to rise relative to the metacentre (M) because the upward ground reaction reduces the buoyancy force acting on the underwater volume.
    • This virtual rise in G reduces GM, potentially leading to a list.
    • If the list becomes excessive and the righting moment is insufficient, the ship may capsize.

    If the ship grounds on a pinnacle:

    • The ship experiences two forces at the ship's bottom:
      • A downward force due to the ship’s weight.
      • An upward reaction force is concentrated on the pinnacle.
    • The resulting force between the grounding pressure and the ship’s centre of buoyancy shifts downward towards the pinnacle.
    • This situation is similar to when the ship's stern touches the keel block in a dry dock.
    • A virtual loss of GM occurs because the ship’s inclining moment may exceed the maximum righting lever.
    • If the inclining moment is too great, the ship may develop an excessive list or even capsize.

    (b) Given:

    $$Displacement,\:\Delta=8000\:tonnes$$

    $$TPC=15\:tonnes$$

    $$Initial\:Draught=5.2m$$

    $$Final\:Draught=3.2m$$

    $$Ship\:KG=4.0m$$

    $$KM=5.0m$$

    To find GM

    $$Uptrust,\:P=TPC\times Fall\:in\:water\:level\:\left(cm\right)$$

    $$=15\times\left(520-320\right)$$

    $$=15\times200$$

    $$P=3000\:tonnes$$

    To Find Virtual loss of GM:

    $$Virtual\:loss\:of\:GM_1=\frac{P\times KM}{\Delta}$$

    $$=\frac{3000\times5}{8000}$$

    $$=\frac{15000}{8000}$$

    $$GM_1=1.88m$$

    Actual KM = 5.0m (given)

    $$Virutal\:KM=Actual\:KM-Virtual\:loss\:of\:GM_1$$

    $$=5.0-1.88$$

    $$=3.12$$

    Similarly, Actual KG = 4.0m (given)

    $$New\:GM=Virtual\:KM-\:Actual\:KG$$

    $$=3.12-4.0$$

    $$=-0.88$$

    Q7 (10 Marks) Ship Stability πŸ”₯ Repeated 7x

    (a) List the precautions necessary before an inclining experiment is carried out.(6)

    (b) A box shaped vessel, 50 metres long x 10 metres wide, floats in salt water on an even keel at a draft of 4 metres. A center line longitudinal watertight bulkhead extends from end to end and for the full depth of the vessel. A compartment amidships on the starboard side is 15 metres long and contains cargo with permeability 30%. Calculate the list if this compartment is bilged. KG = 3 meters

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    Part (a)

    Precautions necessary before an Inclining experiment:

    • The ship should be in a sheltered position, such as a gravity dock, to minimize external influences from wind or waves.
    • Mooring ropes should be slack to allow the ship to move freely without restrictions.
    • Only workers essential to the experiment should be present onboard to avoid unnecessary weight and movement.
    • All tanks must either be completely empty or pressed up tight to eliminate the free surface effect, which can adversely affect stability calculations.
    • Any loose weights on the ship must either be removed or properly secured to prevent unintended movement during the experiment.
    • Ensure that the pendulums are long and properly suspended from stable points, such as underneath a hatch, to provide accurate deflection readings.
    • The test masses should be evenly distributed and placed as far from the centerline as possible to maximize measurable deflections.
    • The experiment should be carried out in calm weather to avoid the effects of wind, current, or waves on the vessel's stability.
    Q8 (10 Marks) Ship Stability πŸ”₯ Repeated 11x

    (a) Define longitudinal center of gravity (LCG) and longitudinal center of buoyancy (LCB). (6)

    (b) A ship 120m long floats at draughts of 5.50m forward and 5.80m aft; MCTI cm 80 tonne m, TPC 13, LCF 2.5m forward of midships. Calculate the new draughts when a mass of 110 tonne is added 24m aft of midships.

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.

    LCF in fwd and trim by stern

    $$Bodily \space sinkage \space = \space {{mass \space added} \over TPC} \space = \space {{110} \over 13} \space$$

    $$=\:8.5\operatorname{\mathrm{cm}}\:=0.085m$$

    $$Trim \space = \space {{m \times d} \over MCT_{1cm}}$$

    $$ = \space {{110 \times (24 + 2.5)} \over 80}$$

    $$Trim=36.43\operatorname{cm}=0.364m\:$$

    Change in fwd draught:

    $$d_{F}=\frac{-t}{L}\left\lbrack\frac{L}{2}-LCF\right\rbrack$$

    $$=\frac{-36.43}{120}\left\lbrack\frac{120}{2}-2.5\right\rbrack$$

    $$=-17.45\operatorname{cm}=-0.1745m$$

    Change in Aft draught:

    $$d_{A}=\frac{+t}{L}\left\lbrack\frac{L}{2}+LCF\right\rbrack$$

    $$=\frac{+36.43}{120}\left\lbrack\frac{120}{2}+2.5\right\rbrack$$

    $$=+18.97\operatorname{cm}=0.189m$$

    New draughts:

    $$D_{F}=5.5+0.085-0.175=5.41m$$

    $$D_{A}=5.8+0.085+0.18=6.065m$$

    Q9 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    (a) The speed of a ship is increased to 18% above normal for 7.5 hours, then reduced to 9% below normal for 10 hours. The speed is then reduced for the remainder of the day so that the consumption for the day is the normal amount. Find the percentage difference between the distance travelled in that day and the normal distance travelled per day.

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    Let normal speed be V and normal daily consumption be C. Consumption varies as the cube of speed, so C = k.V^3.

    Step 1 - Find the speed for the remainder of the day.

    For 7.5 hours the speed is 1.18V. Consumption in that period = k.(1.18V)^3 x 7.5/24 = k.V^3 x 1.643 x 0.3125 = 0.5134 k.V^3.

    For 10 hours the speed is 0.91V. Consumption = k.(0.91V)^3 x 10/24 = k.V^3 x 0.7536 x 0.4167 = 0.3140 k.V^3.

    Total consumption in first 17.5 hours = 0.5134 + 0.3140 = 0.8274 k.V^3.

    Remaining time in the day = 24 - 17.5 = 6.5 hours.

    For the day's total consumption to equal the normal amount k.V^3, the remaining consumption must be k.V^3 - 0.8274 k.V^3 = 0.1726 k.V^3.

    If the reduced speed is Vr, then k.Vr^3 x 6.5/24 = 0.1726 k.V^3.

    So Vr^3 = 0.1726 x 24/6.5 x V^3 = 0.6373 V^3.

    Vr = (0.6373)^(1/3) V = 0.8606 V.

    So the ship travels at 86.06% of normal speed for the last 6.5 hours.

    Step 2 - Find the distance travelled that day.

    Distance = speed x time.

    Normal distance per day = V x 24 = 24V.

    Actual distance = 1.18V x 7.5 + 0.91V x 10 + 0.8606V x 6.5

    = 8.85V + 9.10V + 5.594V = 23.544V.

    Step 3 - Percentage difference.

    Difference = 24V - 23.544V = 0.456V.

    Percentage difference = 0.456/24 x 100 = 1.9%.

    Answer: The distance travelled that day is 1.9% less than the normal distance per day.

    Q10 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 8x

    With reference to fixed pitch propellers:

    (a) Explain Propeller Slip and Propeller Thrust.

    (b) The shaft power of a ship is 3000 KW, the ship's speed V is 13.2 knot. Propeller rps is 1.27. Propeller pitch is 5.5m and the speed of advance is 11 Knots Find:

    (i) Real Slip

    (ii) Wake fraction

    (iii) Propeller thrust, when its efficiency. Ξ· = 70%

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    Part (a)

    Slip

    is the difference between the theoretical distance the propeller should travel in one revolution and the actual distance the vessel travels.

    $$Slip\:\left(\%\right)\:=\:\left(\frac{Engine\:distance\:-\:Ship^{\prime}s\:distance}{Engine\:distance}\right)\:\times100$$

    Where Engine distance = no. of propeller revolutions * propeller pitch (usually calculated over a 24-hour period)

    Propeller thrust: it is the force exerted by the propeller to move the vessel ahead and given by

    $$Thrust\:\left(T\right)\:=\:\rho\times A\times P^2\times n^2\times S$$

    Where,

    ρ - Density

    A - Area

    P - Pitch

    n - Revolution per second

    S - Slip

    Q1 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 2x

    If a ship is seriously damaged under water in way of a large fuel oil side bunker tank what is the immediate effect and what may ultimately happen? What features in the ship would enhance safety of the vessel and marine environment protection aspects in such a case?

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    If a ship is seriously damaged underwater in the area of a large fuel oil side bunker tank, it will cause the following:

    Bilging or Flooding of the Compartment:

    • Water will enter the damaged bunker tank, reaching the draught level.
    • The rate of flooding depends on the size of the breach and the water pressure at the depth of the damage.

    Oil Leakage into the Sea:

    • If the tank contains fuel oil, oil will begin to leak out, causing pollution.
    • The extent of leakage depends on the tank’s contents (empty, half-full, or full) and the location of the damage.

    List and Trim of the Vessel:

    • The ingress of water and loss of oil will create an imbalance, causing the vessel to list or trim.

    If corrective actions are not taken, uncontrolled flooding and loss of stability could lead to capsizing or sinking of the vessel.

    Features in the Ship to Enhance Safety:

    • Small Bunker Tank Sizes reduces the risk of extensive oil spillage and loss of stability.
    • Connectivity to transfer pumps allows the transfer of oil from the damaged tank to an empty tank, minimizing oil spillage and counteracting the loss of stability.
    • The tank’s size and location are designed to limit the effects of flooding, as per damage stability regulations.
    • Properly positioned transverse and longitudinal bulkheads enhance the subdivision factor, limiting water ingress to the damaged tank.
    • Ship’s Ballast System allows corrective ballasting to counteract the list or trim caused by the ingress of water.
    • Watertight Doors and Hatches prevent water from spreading to adjacent compartments.

    Recommended Immediate Actions by Crew:

    For Empty Tanks:

    • Quickly seal off the damaged tank by shutting all valves and isolating it from the transfer system to prevent water ingress into other parts of the vessel.

    For Half-Empty or Full Tanks:

    • Initiate oil transfer to another empty tank to reduce oil leakage and stabilize the ship.
    • Monitor the water ingress and ensure the tank is filled to the draught level with seawater if necessary, using ballast to correct the list.
    Q2 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 13x

    Describe the relationship between frictional resistance and

    (a) Ship's speed,

    (b) The wetted area

    (c) The surface roughness

    (d) The length of the vessel

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    Frictional resistance arises due to the interaction between the ship's hull and the water as the vessel moves. This resistance is caused by eddying water adjacent to the hull that is drawn along with the ship. The frictional resistance is given by:

    $$R_{f\:=\:}f\:.\:s\:.\:v^{n}$$

    Where:

    • f = Coefficient of friction
    • s = Wetted surface area
    • v = Ship's speed in knots
    • n = Constant (1.82)
    Part (a)

    Ship Speed (v):

    • Frictional resistance increases with increasing ship speed. The relationship is not perfectly linear, but it's approximately described by the formula Rf = fsv^n (Rf ∝ v^n), where 'n' is a constant (approximately 1.82). This means that a small increase in speed results in a larger increase in frictional resistance.
    Part (b)

    Wetted Area (s):

    • Frictional resistance is directly proportional to the wetted surface area (Rf ∝ s). A larger wetted area (the part of the hull submerged in water) leads to greater frictional resistance. The wetted area increases with the ship's draught (depth in the water).
    Part (c)

    Surface Roughness:

    • Increased surface roughness increases frictional resistance. Roughness can be caused by fouling (marine growth), corrosion, or poor hull maintenance. Regular cleaning and maintenance of the hull surface help to minimise roughness and reduce resistance.
    Part (d)

    Length of the Vessel

    • Frictional resistance is influenced by the vessel's length. As the length increases, the wetted surface area grows, and the coefficient of friction (f) changes, leading to an increase in Rf. Longer vessels experience higher frictional resistance, but streamlined designs can mitigate the effect to some extent.
    Q3 (10 Marks) Ship Stability

    With reference to collision bulkhead explain the following using sketches as required:

    (a) Purposes of collision bulkhead.

    (b) Construction of collision bulkhead.

    (c) Regulations governing the position and construction of such a bulkhead.

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    Part (a)

    Purposes of the collision bulkhead.

    • It is the foremost transverse watertight bulkhead, fitted near the bow.
    • Its primary purpose is to limit the flooding caused by a head-on collision or grounding damage to the forepeak, preventing water from flooding the main cargo or machinery spaces and thereby keeping the ship afloat.
    • It provides a watertight boundary so that damage to the bow does not compromise the rest of the hull.
    • It also acts as a structural stiffener for the bow region, resisting panting and pounding forces.
    • It protects the ship's buoyancy and stability in the event of forward damage.
    Part (b)

    Construction of the collision bulkhead.

    • It is a transverse watertight bulkhead of steel plating, stiffened by vertical stiffeners (usually angles or bulb plates) spaced about 600-750 mm apart.
    • The plating is generally thicker than other bulkheads because of the higher pressure head and the risk of damage.
    • Stiffeners are fitted on the after side (the dry side) so that the forward face is smooth, allowing water to run down and reducing the risk of damage to the stiffeners.
    • The bulkhead is connected to the shell plating, deck and bottom by efficient watertight connections, often with a larger radius at the bilge.
    • It extends from the keel (bottom shell) up to the freeboard deck, and in some designs to the superstructure deck.
    • No doors, manholes or openings are permitted below the freeboard deck in the collision bulkhead, except for a single pipe passing through with a valve operable from above the freeboard deck.
    • The stiffeners are connected to the deck and bottom by brackets or gussets to transmit the loads.
    Part (c)

    Regulations governing position and construction.

    • The collision bulkhead must be fitted at a distance from the forward perpendicular of not less than 5% and not more than 8% of the length of the ship (L), measured from the forward perpendicular. (For ships with a bulbous bow, the position is measured from the forward end of the length.)
    • It must be watertight up to the freeboard deck.
    • It must be fitted in all ships, and must extend to the freeboard deck.
    • No doors, manholes, ventilation ducts or other openings are permitted in the collision bulkhead below the freeboard deck.
    • A single pipe may pass through the collision bulkhead below the freeboard deck, provided it is fitted with a valve operable from above the freeboard deck, and the valve chest is secured to the bulkhead at the inboard end.
    • The bulkhead must be of adequate strength to withstand the pressure of water in the flooded forepeak, and is subject to classification society rules for plating thickness and stiffener spacing.
    • These requirements are set out in the International Convention on Load Lines and the SOLAS Convention, and are implemented through classification society rules.
    Q4 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    With reference to Ship stability:

    (a) With the help of a neat sketch explain the relevant features of a G-Z curve.

    (b) What are the effects of the below mentioned conditions on the G-Z curve:

    (i) Increased freeboard,

    (ii) Increased beam, and

    (iii) Increased GM.

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    Part (a)

    Features of a GZ curve.

    The GZ curve is a graph of the righting lever GZ against the angle of heel. Its relevant features are:

    • The origin: at zero heel, GZ = 0.
    • The initial slope: the tangent to the curve at the origin equals GM (the metacentric height), since GZ = GM sin(theta) for small angles. A steeper initial slope means a larger GM.
    • The maximum righting lever (GZ max): the highest point of the curve, and the angle at which it occurs (the angle of maximum stability, typically 25-40 deg).
    • The range of stability: the angle from the upright to the angle of vanishing stability (where GZ returns to zero, typically 60-90 deg).
    • The area under the curve: proportional to the dynamical stability (the work done in heeling the ship), used to assess stability in a seaway and against wind heeling.
    • The angle of loll: if the curve starts below the axis (negative GZ at small angles), indicating a negative GM and an unstable ship that lolls to one side.
    • The effect of free surface: the curve is reduced by the free-surface correction.

    The curve is obtained from the cross-curves of stability corrected for the actual KG and free-surface effects, and is compared with the statutory criteria.

    Part (b)

    Effects of the following conditions on the GZ curve.

    (i) Increased freeboard: increasing the freeboard raises the deck edge and increases the reserve buoyancy, so the range of stability is increased (the angle of vanishing stability moves to a larger angle) and the area under the curve is increased. The initial slope (GM) is largely unchanged, but the curve is higher and extends further, giving greater dynamical stability and a larger range.

    (ii) Increased beam: increasing the beam increases the waterplane area and the BM (BM is proportional to the cube of the beam), so the initial slope (GM) increases and the curve is steeper at small angles. The maximum GZ is increased and occurs at a smaller angle, but the range of stability may be reduced (the angle of vanishing stability decreases) because the ship becomes stiffer and the deck edge immerses earlier. The area under the curve may be reduced at large angles.

    (iii) Increased GM: increasing the GM (e.g. by lowering KG) makes the initial slope steeper, so the curve rises more steeply at small angles and the maximum GZ is larger and occurs at a smaller angle. However, the range of stability is reduced (the angle of vanishing stability decreases) and the ship rolls more quickly and with a shorter period, which can be uncomfortable. The area under the curve at small angles increases but the overall range decreases.

    Q5 (10 Marks) General

    Explain with sketches the terms hogging and sagging. Which structural members are affected due to these conditions? State the type of stresses these members are subjected to under such conditions.

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    Hogging and sagging are the two conditions of longitudinal bending of a ship's hull.

    Hogging:

    • The ship is supported by buoyancy at the ends (bow and stern) while the middle is relatively unsupported, or the weight is concentrated amidships.
    • The ends tend to rise and the middle tends to droop, so the keel is convex downwards (curves upward at the ends).
    • This occurs when the ship is light in the middle and heavy at the ends, or when the wave crest is at the ends and the trough is amidships (ship on a wave crest at bow and stern).
    • The deck is in tension (stretched) and the bottom is in compression.

    Sagging:

    • The ship is supported by buoyancy amidships while the ends are relatively unsupported, or the weight is concentrated at the ends.
    • The middle tends to sag and the ends tend to rise, so the keel is concave downwards (curves downward in the middle).
    • This occurs when the ship is heavy in the middle and light at the ends, or when the wave crest is amidships and the troughs are at the ends (ship on a wave crest amidships).
    • The deck is in compression and the bottom is in tension.

    Structural members affected and the stresses they are subjected to:

    • Deck plating and deck girders (stringers, longitudinals): in hogging the deck is in tension; in sagging the deck is in compression.
    • Bottom shell plating and bottom longitudinals, keel: in hogging the bottom is in compression; in sagging the bottom is in tension.
    • Side shell plating: subjected to shear stresses, with the maximum shear stress occurring near the neutral axis (approximately mid-depth of the hull).
    • Longitudinal bulkheads and girders: share the bending stresses, in tension or compression depending on their position relative to the neutral axis and the bending condition.
    • Transverse frames and bulkheads: resist the distortion of the cross-section (racking) and transmit the loads.
    • The neutral axis (approximately at mid-depth) is the level where the bending stress is zero; members above the neutral axis are in tension in hogging and compression in sagging, and vice versa for members below.

    The maximum bending moment occurs when the ship is on a wave of length approximately equal to the ship's length, with either a crest or trough amidships, and the resulting hogging or sagging stresses are the largest longitudinal stresses the hull experiences.

    Q6 (10 Marks) General πŸ”₯ Repeated 2x

    With the aid of sketches:

    (a) Explain various lines plan.

    (b) The half-breadths of waterplane of a ship of 120m length ad 15m breadth are given below

    Station: 0 1 2 3 4 5 6 7 8

    Half-breadth: 1.6 2.8 5.5 6.4 7.3 6.2 4.2 2.0 0

    Calculate:

    (i) Water plane area

    (ii) TPC in salt water

    (iii) Cw

    (iv) LCF from mid-ship

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    Part (a)

    Lines plan - explanation.

    The lines plan is a set of drawings that defines the shape of the hull. It consists of three principal views:

    • Sheer plan (profile): shows the hull as seen from the side. It contains the waterlines, the deck line, the sheer line and the profile of the stem and stern. The vertical positions of the waterlines and the longitudinal positions of the stations are shown.
    • Half-breadth plan (plan view): shows the hull as seen from above, drawn for one half of the ship (port or starboard) because the hull is symmetrical. It contains the waterlines (horizontal sections) and the deck line, showing the half-breadths at each station.
    • Body plan (end view): shows the transverse sections (stations) as seen from the bow and stern. The forward half of the stations is drawn on one side and the after half on the other. It contains the buttock lines and the diagonal lines.

    The three views are drawn in projection so that every point on the hull appears in all three views, allowing the shape to be fully defined and faired. The lines plan is used to calculate hydrostatic data, displacement, and to construct the ship.

    Part (b)

    Waterplane calculation.

    Length L = 120 m, breadth B = 15 m. Half-breadths at stations 0 to 8:

    Station: 0 1 2 3 4 5 6 7 8

    Half-breadth (m): 1.6 2.8 5.5 6.4 7.3 6.2 4.2 2.0 0

    Number of intervals n = 8, so the common interval h = L/n = 120/8 = 15 m.

    (i) Waterplane area.

    Using Simpson's First Rule with 8 intervals (9 ordinates), multipliers 1,4,2,4,2,4,2,4,1:

    Sum of products = 1x1.6 + 4x2.8 + 2x5.5 + 4x6.4 + 2x7.3 + 4x6.2 + 2x4.2 + 4x2.0 + 1x0

    = 1.6 + 11.2 + 11.0 + 25.6 + 14.6 + 24.8 + 8.4 + 8.0 + 0 = 105.2.

    Area of half waterplane = (h/3) x sum = (15/3) x 105.2 = 5 x 105.2 = 526 m2.

    Waterplane area (full) = 2 x 526 = 1052 m2.

    (ii) TPC in salt water.

    TPC = (waterplane area x density of sea water)/100 = (1052 x 1.025)/100 = 1078.3/100 = 10.78 tonne/cm.

    Answer: TPC = 10.78 tonne per cm.

    (iii) Cw (waterplane area coefficient).

    Cw = waterplane area / (L x B) = 1052 / (120 x 15) = 1052/1800 = 0.584.

    Answer: Cw = 0.584.

    (iv) LCF from midship.

    The centre of flotation is the centroid of the waterplane. Using the first moment about midship (station 4):

    Moment of products = sum of (multiplier x half-breadth x distance from midship in intervals).

    Distances from station 4 (in intervals): station 0 = -4, 1 = -3, 2 = -2, 3 = -1, 4 = 0, 5 = +1, 6 = +2, 7 = +3, 8 = +4.

    Moment = 1x1.6x(-4) + 4x2.8x(-3) + 2x5.5x(-2) + 4x6.4x(-1) + 2x7.3x0 + 4x6.2x(+1) + 2x4.2x(+2) + 4x2.0x(+3) + 1x0x(+4)

    = -6.4 - 33.6 - 22.0 - 25.6 + 0 + 24.8 + 16.8 + 24.0 + 0 = -22.0.

    Distance of centroid from midship = (moment/sum) x h = (-22.0/105.2) x 15 = -0.2092 x 15 = -3.14 m.

    The negative sign means the LCF is forward of midship.

    Answer: LCF is 3.14 m forward of midship.

    Q7 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    What is Prismatic Co-efficient (Cp).

    (a) Derive the formula Cp = Cb/Cm, where Cb = Co-efficient of fineness and Cm = midship section area co-efficient.

    (b) The length of a ship is 18 times the draught, while the breadth is 2.1 times the draught. At the load water plane, the water plane area co-efficient is 0.83 and the difference between the TPC in sea water and the TPC in fresh water is 0.7. Determine the length of the ship and the TPC in fresh water.

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    Part (a)

    Prismatic coefficient is the ratio of the volume of displacement to the product of length and area of the immersed portion of the midship section

    Cb is the block co-efficient or co-efficient of fitness is the ratio of the volume of displacement to the product of length, breadth and draught

    $$\because\space C_{b}\space=\space{{\nabla}\over L\times B\times D}\:---\:1$$

    $$C_{p}\space=\space{{\nabla}\over L\times A_{m}}\:---\:2$$

    $$C_m \space = \space {{A_m} \over B \times D}$$

    $$A_{m}=\:C_{m}\times B\times D\:---\:3$$

    Substitute 3 in 2

    $$C_{p}\space=\space{{\nabla}\over C_{m}\times B\times D\ \times L}\:---\:4$$

    $$\nabla=C_{b}\times L\times B\times D\:---5\:\left(from\:equation\:1\right)$$

    Substitute 5 in 4

    $$C_{p}=\frac{C_{b}\times L\times B\times D}{C_{m}\times L\times B\times D}$$

    $$C_p \space = \space {{C_b} \over C_m}$$

    Part (b)

    $$Length \space of \space ship \space = \space L$$

    $$Draught \space = \space {{L} \over 18 }$$

    $$breadth \space = \space 2.1 \times draught \space = \space 2.1 \times {{L} \over 18}$$

    $$TPC\:in\:SW\:=\:0.01025A_{w}$$

    $$TPC\:in\:FW\:=\:0.0100A_{w}$$

    $$0.01025A_{w}-0.0100A_{w}=0.7$$

    $$A_{w}=\frac{0.7}{2.5\times10^{-4}}=2800$$

    $$C_w \space = \space {{A_w} \over L \times B}$$$$0.83 \space = \space {{2800} \over L \times {{2.1L} \over 18}}$$

    $$L=170.04m$$

    $$TPC\:in\:FW=0.010\times A_{w}$$

    $$=\:0.0100\times2800$$

    $$TPC\:in\:FW=28$$

    Q8 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 4x

    With respeet to Ship Propulsion:

    (a) Explain the various efficiencies associated with propeller and shafting arrangement.

    (b) When a propeller of 4.8 m pitch turns at 110 pm, the apparent slip is found to be -S% and the real slip is 1.5 S%. If the wake speed is 25% of the ship speed, calculate the ship speed, apparent slip and the real slip

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    Part (a)

    Efficiencies associated with the propeller and shafting arrangement.

    • Shaft (transmission) efficiency: the ratio of the delivered power at the propeller to the brake power of the engine, accounting for the losses in the shaft bearings, stern tube and any gearing. It is typically 0.97-0.99.
    • Propeller (open-water) efficiency: the ratio of the thrust power (T x Va) to the delivered power (2 pi n Q). It represents the efficiency of the propeller itself in converting the delivered power into thrust power.
    • Hull efficiency: the ratio of the effective power to the thrust power, = (1 - t)/(1 - w), where t is the thrust deduction factor and w the wake fraction. It accounts for the interaction between the hull and the propeller.
    • Quasi-propulsive coefficient (QPC): the ratio of the effective power to the delivered power, = hull efficiency x propeller efficiency. It is the overall efficiency of the propulsion system in converting the delivered power into effective (towing) power.
    • Overall (propulsive) efficiency: the ratio of the effective power to the brake power, = QPC x shaft efficiency. It is the overall efficiency of the engine-to-propeller-to-hull system.
    Part (b)

    Ship speed, apparent slip and real slip.

    A propeller of 4.8 m pitch turns at 110 rev/min. The apparent slip is -S% and the real slip is +1.5S%. The wake speed is 25% of the ship speed. Calculate the ship speed, the apparent slip and the real slip.

    Pitch speed = pitch x rev/s = 4.8 x 110/60 = 8.8 m/s.

    Let the ship speed be V (m/s). The speed of advance Va = V x (1 - 0.25) = 0.75 V.

    Apparent slip = (pitch speed - V)/pitch speed = -S/100.

    Real slip = (pitch speed - Va)/pitch speed = 1.5 S/100.

    From the apparent slip: (8.8 - V)/8.8 = -S/100, so V = 8.8(1 + S/100).

    From the real slip: (8.8 - 0.75 V)/8.8 = 1.5 S/100, so 8.8 - 0.75 V = 0.132 S.

    Substitute V = 8.8(1 + S/100): 8.8 - 6.6(1 + S/100) = 0.132 S.

    8.8 - 6.6 - 0.066 S = 0.132 S, so 2.2 = 0.198 S, S = 11.11.

    Apparent slip = -11.11%; real slip = 1.5 x 11.11 = 16.67%.

    Ship speed V = 8.8(1 + 0.1111) = 9.78 m/s = 9.78 x 1.944 = 19.0 knots.

    Answer: ship speed about 19.0 knots; apparent slip -11.1%; real slip +16.7%.

    Q9 (10 Marks) Ship Stability πŸ”₯ Repeated 4x

    A ship 100 m long floats at a draft of 6m and in this condition the immersed cross sectional areas are as given in tables below. The equivalent base area (Ab) is required because of the fineness of the bottom shell

    Section: AP 1 2 3 4 5 6 FP

    Immeresed cross section area (m2): 12 30 65 80 70 50 0

    Draft(m): 0 0.6 1.2 2.4 3.6 4.8 6.0

    Waterplane area (m2): Ab 560 720 880 940 1000 1030

    Calculate each of the following:

    (a) The equivalent base are value Ab

    (b) The longitudinal position of the centre of buoyancy from midship

    (c) The vertical position of the centre of buoyancy above the base

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    Table for question:

    Section

    AP

    1

    2

    3

    4

    5

    FP

    Immersed cross sectional area

    12

    30

    65

    80

    70

    50

    0

    Draught (m)

    0

    0.6

    1.2

    2.4

    3.6

    4.8

    6.0

    Water plane area (m2)

    Ab

    560

    720

    880

    940

    1000

    1030

    Solution:

    Section

    CSA

    SM

    F volume

    Lever

    F moment

    AP

    12

    1

    12

    -3h

    -36h

    1

    30

    4

    120

    -2h

    -240h

    2

    65

    2

    130

    -1h

    -130h

    3

    80

    4

    320

    0

    0

    4

    70

    2

    140

    +1h

    140h

    5

    50

    4

    200

    +2h

    400h

    FP

    0

    1

    0

    +3h

    0

    Ξ£Fvol. = 992

    Ξ£Fmom. =134h

    $$Common\:interval\:\left(h\right)=\frac{L}{6}=\frac{100}{6}$$

    $$h=16.66m$$

    $$\sum F_{movement}=134\:\times16.66m^4$$

    $$\sum F_{moment}=2232.44m^4$$

    $$\nabla=\frac{h}{3}\times\sum F_{volume}$$

    $$\nabla=\frac{16.66}{3}\times992$$

    $$\nabla=5120.17m^3$$

    $$Longitudinal\:position\:of\:centre\:of\:buoyancy\:LCB\:=\:\frac{\sum F_{mom}}{\sum F_{vol}}$$

    $$=\frac{2232.44}{993}$$

    $$LCB=2.42m\:fwd$$

    Draught

    Aw

    SM

    F volume

    Lever

    F moment

    0

    Ab

    1/2

    1/2 Ab

    0

    0

    0.6

    560

    4/2

    1120

    0.6

    672

    1.2

    720

    1 2/4

    1080

    1.2

    1296

    2.4

    880

    4

    3520

    2.4

    8448

    3.6

    940

    2

    1880

    3.6

    6768

    4.8

    1000

    4

    4000

    4.8

    19200

    6.0

    1030

    1

    1030

    6.0

    6180

    Ξ£Fvol = 1/2Ab+12630

    Ξ£Fmom = 42564

    $$h=1.2$$

    $$\nabla=\frac{h}{3}\times\sum F_{vol}$$

    $$5120.17=\frac{1.2}{3}\times\left(\frac12A_{b}+12630\right)$$

    $$\frac{A_{b}}{2}=\frac{5120.17\times3}{1.2}-12630$$

    $$A_{b}=340.84m^2$$

    $$\sum F_{vol}=\left\lbrack\frac12\times340.85\right\rbrack+12630$$

    $$\sum F_{vol}=12800.425m^3$$

    $$Vertical\:position\:of\:centre\:of\:buoyancy=\frac{\sum F_{mom}}{\sum F_{vol}}$$

    $$=\frac{42564}{12800.425}$$

    $$KB=3.325m$$

    Q10 (10 Marks) Ship Resistance & Propulsion

    With reference to the wake of a vessel:

    (a) Explain Wake fraction and Quasi Propulsive co-efficient(OPC).

    (b) A ship travelling at 15.5 knots has a propeller of 5.5 m pitch turning at 95 rev/min. The thrust of the propeller is 380 KN and the delivered power 3540 KW. If the real slip is 20% and the thrust deduction factor 0.198, calculate the Quasi Propulsive Coefficient (QPC) and the wake fraction.

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    Part (a)

    Wake fraction and Quasi Propulsive Coefficient (QPC).

    Wake fraction (w):

    • When a ship moves through the water, the water in the wake region behind the hull is carried forward with the ship. This is the wake.
    • The wake fraction is the ratio of the speed of the wake (the water carried forward) to the ship's speed. It is defined as w = (V - Va)/V, where V is the ship speed and Va is the speed of advance of the propeller relative to the water.
    • The wake is made up of frictional wake (water dragged along by the hull), potential wake (due to the pressure field around the hull) and wave wake.
    • The wake fraction is beneficial because it reduces the effective speed of the water entering the propeller, reducing the power required.

    Quasi Propulsive Coefficient (QPC):

    • The QPC is the ratio of the effective power (the power required to tow the hull, i.e. the useful work done in propelling the ship) to the delivered power (the power delivered to the propeller).
    • QPC = Effective Power / Delivered Power.
    • It is a measure of the overall efficiency of the propulsion system, combining the hull efficiency, the propeller (open water) efficiency and the relative rotative efficiency.
    • QPC = hull efficiency x propeller efficiency x relative rotative efficiency.
    • The hull efficiency = (1 - t)/(1 - w), where t is the thrust deduction factor and w is the wake fraction.
    Part (b)

    Calculation.

    Ship speed V = 15.5 knots = 15.5 x 1852/3600 = 7.968 m/s.

    Propeller pitch P = 5.5 m, rev/min = 95, so rev/s = 95/60 = 1.5833 rev/s.

    Pitch speed (theoretical speed of advance) = P x rev/s = 5.5 x 1.5833 = 8.708 m/s.

    Real slip = 20% = 0.20.

    Real slip = (Pitch speed - Va)/Pitch speed, so Va = Pitch speed x (1 - real slip) = 8.708 x 0.80 = 6.966 m/s.

    Wake fraction w = (V - Va)/V = (7.968 - 6.966)/7.968 = 1.002/7.968 = 0.1258.

    Answer: Wake fraction = 0.126 (12.6%).

    Thrust T = 380 kN = 380000 N.

    Thrust power = T x Va = 380000 x 6.966 = 2647080 W = 2647 kW.

    Delivered power PD = 3540 kW.

    Propeller efficiency (open water) = Thrust power/Delivered power = 2647/3540 = 0.7477.

    Thrust deduction factor t = 0.198.

    Hull efficiency = (1 - t)/(1 - w) = (1 - 0.198)/(1 - 0.1258) = 0.802/0.8742 = 0.9174.

    QPC = hull efficiency x propeller efficiency = 0.9174 x 0.7477 = 0.686.

    Answer: QPC = 0.686.

    (Note: QPC can also be found as QPC = (T x V)/(PD) x (1-t) ... but the standard method above gives QPC = 0.686.)

    Q1 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 9x

    With reference to Underwater Inspection in lieu of Dry docking (UWILD):

    (a) Explain in detail, how an underwater survey is carried out.

    (b) State the requirements to be fulfilled before an underwater survey is acceptable to the surveying authority

    (c) Construct a list of the items in order of importance that the underwater survey authority should include.

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    (a) An underwater survey, also known as Underwater Inspection in Lieu of Dry Docking (UWILD), involves a systematic and detailed examination of a vessel’s hull and underwater components while the ship remains afloat. The procedure includes the following steps:
    • The shipowner sends a request to the classification society surveyor, indicating the intention to perform an underwater survey.
    • A detailed plan of the ship's external hull features is submitted, showing the location of frames, bulkheads, welding lines, openings, etc.
    • The hull surface is cleaned before the survey to remove any marine growth or fouling that could obscure the inspection.
    • A diving company, approved by the classification society, is appointed to conduct the underwater inspection.
  • A self-propelled survey vehicle equipped with the following tools is used:
    • Long-Range Light TV Camera to aid in steering and checking hull deterioration.
    • High-resolution colour TV Camera to provide a close-up view of the hull coating and welded seams.
    • 35mm Still Camera to capture still images.
    • Ultrasonic Probe for measuring plate thickness.
    • Depth Meter and Speed Indicator to provide accurate data on the vehicle's depth and movement.
    • Umbilical Cable to connect the survey vehicle to the survey boat, transmitting power and information.
  • The survey boat is to be equipped with:
    • A control console with TV monitors.
    • Plate thickness printouts.
    • Audio and video cassette recorders.
    • Playback units.
    • Diver communication systems.
    • Vehicle control systems and associated instruments.
  • Operation:
    • The survey vehicle is taken underwater by a diver to the survey starting point.
    • Using TV monitors and shell expansion plans as guides, the vehicle is navigated over the hull, focusing on the bottom structure, sides, stern frame, rudder, propeller, bilge keel, and hull openings.
    • All images, data, and information are recorded and transmitted back to the survey boat.
    • Detailed pictures of the stern frame, rudder, propeller, bilge keel, and hull openings are captured.
    • Divers are deployed to measure stern tube bearing wear, pintle clearance, and inspect stern seals, anodes, and rudder stock palm coupling bolts.
    • All recorded video and audio, including conversations between the surveyor and drivers, thickness printouts, measurements, and pictures are analyzed to determine the detailed underwater condition of the vessel.
    (b) Before an underwater survey is accepted by the surveying authority, the following conditions must be met:

    The vessel's owner submits a request to the surveyor, including:
    • The proposed date and location for the survey.
    • General information about the diving company.
    • A declaration that the vessel has not suffered any damage due to grounding, collision, or other incidents.
    The ship's master or owner’s representative must provide a declaration confirming:
    • Any suspected or actual damage to the hull since the last dry-docking.
    • The underwater portion of the hull is protected by a suitable paint scheme that is of adequate thickness and remains valid until the next dry-dock.

    • The survey site should be in a protected area with calm and clear water, ensuring good underwater visibility. Attention must be given to the effects of currents and tides.
    • The hull must be clean for the external survey. The surveyor must be satisfied with the method and quality of the pictorial presentation, ensuring that it provides a reliable assessment of the hull's condition.
    • The underwater examination should be conducted by an approved diving company using closed-circuit TV and two-way communication, which can be monitored by the surveyor.
    • The vessel should be in as light an operating condition as possible to facilitate the survey.
    • Means must be available for the surveyor to examine the outside shell plating above the waterline.
    • Any required repairs identified during the survey must be carried out to the satisfaction of the attending surveyor.
    (c) While the importance of each item may vary depending on the vessel and its specific requirements, below is a list of items that should be included in an underwater survey in order of importance:
    • Underwater Hull: General condition of the hull below the waterline.
    • Bottom and Shell Plating: Inspection for corrosion, damage, and fouling.
    • Shell Openings: Examination of openings such as sea chests, drain plugs, and overboard discharge points.
    • Stern Tube Oil Leaks: Check for leaks around the stern tube.
    • Propeller Blade: Inspection for damage, wear, and fouling.
    • Rudder: Inspection for damage, wear, and clearances.
    • Sea Chest Opening and Grating: Examination for blockages, damage, and fouling.
    • Anodes: Check the condition and effectiveness of cathodic protection anodes.
    • Bilge Keel: Inspection for damage and fouling.
    • Drain Plugs: Ensure all drain plugs are secure and in good condition.
    • Overboard Valve Openings: Check for proper operation and condition.
    • Forward Area: Inspection for any damage due to anchor and chain movement.
    Q2 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 8x

    Vessel has gone through very heavy weather. On arrival at safe anchorage, you are conducting your inspection to determine damages to hull

    (a) List the areas you will inspect.

    (b) List your findings of any significance.

    (c) Write a report to company suggesting repairs if any

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    Part (a)

    Areas to Inspect After Heavy Weather

    Upon arrival at safe anchorage, a thorough inspection of the vessel shall be carried out, both externally and internally, to assess any weather-related damage. Key areas include:

    1. Hull and Main Deck

    • Shell plating along bow, midship, and stern sections (especially forepeak and flare areas).
    • Boot-top and bilge areas for dents or deformation.
    • Deck plating for buckling or cracked welds.
    • Bulwarks, rails, fairleads, chocks, and mooring fittings.

    2. Forecastle and Forward Structure

    • Bosun store and chain locker for water ingress.
    • Windlass foundation, anchors, and hawse pipes for deformation or cracks.
    • Forepeak tank for leakage or pressure damage.

    3. Cargo Holds / Tanks

    • Inspect for structural deformation, loose frames, or fractured stiffeners.
    • Check tank top plating and bilges for leakage.
    • Check watertight doors, gaskets, and vents.

    4. Superstructure and Deck Fittings

    • Bridge wings, radar mast, funnel, vents, and deckhouses for cracks or loose fittings.
    • Lifeboat davits, securing arrangements, and deck cranes.

    5. Underwater and Machinery Spaces

    • Rudder, propeller, and stern tube seals (via steering gear tests).
    • Sea chest gratings and overboard discharges.
    • Engine room bilges for any seawater ingress.

    Part (b)

    Typical Findings of Significance

    • Dents and indentations on shell plating at bow and forward port side due to heavy slamming.
    • Paint coating and corrosion protection partly peeled off near waterline and forepeak area.
    • Loose fairlead bolts and one cracked weld on starboard bulwark stanchion.
    • Deformed ventilator head on forecastle deck.
    • Minor leakage observed in forepeak tank during sounding check.
    • Bridge wing railing bent, likely from green sea impact.
    • Anchor chain links twisted and worn.
    • Lifeboat gripes loosened, requiring tightening and inspection.
    • No flooding reported; watertight integrity maintained overall.

    Part (c)

    Report to Company – Heavy Weather Damage Inspection

    To: Superintendent / Technical Department

    From: Name / Rank

    Subject: Heavy Weather Damage Inspection Report

    Date: [Insert date]

    Vessel: [Insert vessel name]

    Summary

    The vessel experienced very heavy weather en route from [Port A] to [Port B], with significant pitching and rolling in seas up to [X] meters. On arrival at safe anchorage, a complete inspection of the hull and deck was carried out.

    Findings

    • Bow and forecastle plating show minor dents, with no breach of watertight integrity.
    • Bulwark stanchion (starboard side) cracked at welded joint – requires repair.
    • Fairlead foundation bolts loosened – retightening and NDT inspection recommended.
    • Ventilator head on forecastle deformed – replacement advised.
    • Forepeak tank shows trace leakage at forward bulkhead – to be pressure tested.
    • Paint coating damage and corrosion exposure on bow area – recoating required.
    • Bridge wing railing bent – to be straightened or renewed.
    • All other structures and machinery appear satisfactory after testing.

    Recommendations

    1. Conduct minor steel renewal and welding repairs at the next port with repair facilities.
    2. Carry out thickness measurements and NDT on affected hull areas.
    3. Recoat damaged paint areas to prevent corrosion.
    4. Replace deformed ventilator head and bent railing.
    5. Inspect anchor and chain for elongation; renew worn links.
    6. Pressure test forepeak tank after repairs.
    7. Submit class surveyor report if deemed necessary by the superintendent.

    Conclusion

    The vessel remains seaworthy but requires prompt attention to minor structural and fitting damages before the next voyage. Preventive maintenance and weather routing measures should be reviewed for future passages.

    Signed:

    Name / Rank

    Signature

    Q3 (10 Marks) Ship Stability πŸ”₯ Repeated 14x

    Explain how the period of roll varies with:

    (a) The amplitude of roll.

    (b) The radius of gyration.

    (c) The initial metacentric height.

    (d) The location of masses in the ship.

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    The period of roll Tr of a ship is determined by the formula:

    $$T_{r}=\frac{2\pi K}{\sqrt{g\times GM}}$$

    where,

    • K is the radius of gyration (mass moment of inertia)
    • g is the acceleration due to gravity, and
    • GM is the metacentric height.
    Part (a)

    Amplitude of Roll:

    • The amplitude of roll, or the maximum angle of heel, does not affect the period of roll. While a larger amplitude implies a greater heeling moment and faster roll speed, the time taken to complete one full roll cycle remains constant for a given metacentric height and radius of gyration.
    Part (b)

    Radius of Gyration (K):

    • The period of roll is directly proportional to the radius of gyration. A larger radius of gyration (indicating a greater distribution of mass further from the ship's centre of rotation) leads to a longer period of roll. Conversely, a smaller radius of gyration (mass concentrated closer to the centre) results in a shorter period. The distribution of cargo significantly impacts K; cargo concentrated centrally minimises K and the roll period, while dispersed cargo maximises K and the roll period.
    Part (c)

    Initial Metacentric Height (GM):

    • The period of roll is inversely proportional to the square root of the metacentric height (GM). A larger GM (a stiffer ship) leads to a shorter roll period, as the ship quickly returns to its upright position. A smaller GM (a tender ship) results in a longer roll period, with slower return to the upright.
    Part (d)

    Location of Masses in the Ship:

    The location of masses in the ship will effect the GM & K. So the period of roll will be affected.

    • If masses are at bottom, G moves down, GM ↑, period of roll ↓.
    • If masses are at top, G moves up, GM ↓, period of roll ↑.
    • If masses are concentrated at centre, K ↓, period of roll ↓.
    • If masses are away from centre, K ↑, period of roll ↑.
    Q4 (10 Marks) General πŸ”₯ Repeated 7x

    List SIX hazards that arise with the carriage of liquefied gas in bulk. Describe, with the aid of a sketch, the details of construction of a prismatic cargo tank within a gas carrier designed to carry liquefied gas (LPG)

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    Hazards of Carriage of Liquefied Gas in Bulk and Construction of a Prismatic Cargo Tank

    Part (a)

    Six Hazards Associated with the Carriage of Liquefied Gas in Bulk

    1. Flammability and Explosion
      • Liquefied gases such as LPG vaporise rapidly when released.
      • The vapour can mix with air and form a highly flammable or explosive atmosphere, creating a serious risk of fire or explosion.
    2. Toxicity
      • Some liquefied gas cargoes, such as ammonia and vinyl chloride monomer, are highly toxic.
      • Exposure can cause serious poisoning through inhalation, ingestion or skin absorption.
    3. Asphyxiation
      • LPG vapours are generally heavier than air.
      • In the event of a leak, the vapour can collect in low-lying areas such as pump rooms and hold spaces, displacing oxygen and creating an asphyxiation hazard.
    4. Frostbite and Cold Burns
      • Liquefied gas cargoes are carried at very low temperatures, with fully refrigerated LPG being carried at approximately βˆ’50Β°C.
      • Contact with the liquid cargo or uninsulated pipes and equipment can cause severe frostbite and cold burns.
    5. Brittle Fracture
      • Ordinary ship hull steel, such as mild steel, can become brittle at very low temperatures.
      • If cold cargo leaks and comes into contact with unsuitable hull steel, it may cause cracking and catastrophic structural failure.
    6. Sloshing
      • Partially filled tanks have a free surface, allowing the liquid cargo to move violently with the ship's motion.
      • This can produce large dynamic impact loads on the tank walls and internal pump towers, potentially causing structural damage.
    Part (b)

    Construction of a Prismatic Cargo Tank for LPG

    Fully refrigerated LPG is typically carried in Independent Type A prismatic cargo tanks. These are self-supporting tanks that are independent of the ship's hull structure and do not contribute to the overall structural strength of the vessel.

    Sketch – Typical Prismatic Cargo Tank Arrangement

    Ship's Hull / Hold Space

    β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”

    β”‚ Secondary Barrier / Hull β”‚

    β”‚ β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚

    β”‚ β”‚ Thermal Insulation β”‚ β”‚

    β”‚ β”‚ β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚ β”‚

    β”‚ β”‚ β”‚ β”‚ β”‚ β”‚

    β”‚ β”‚ β”‚ LPG CARGO β”‚ β”‚ β”‚

    β”‚ β”‚ β”‚ PRISMATIC TANK β”‚ β”‚ β”‚

    β”‚ β”‚ β”‚ β”‚ β”‚ β”‚

    β”‚ β”‚ β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ β”‚ β”‚

    β”‚ β”‚ Primary Barrier β”‚ β”‚

    β”‚ β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ β”‚

    β”‚ Load-bearing supports β”‚

    β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜

    β”‚

    Double Bottom

    1. Shape and Structure

    • The tank is prismatic, meaning it is generally box-shaped with chamfered/angled top and bottom corners.
    • This shape allows the tank to closely follow the contours of the ship's inner hull and therefore maximises the available cargo capacity.
    • The tank is internally reinforced with frames and stiffeners to maintain structural integrity.
    • A centreline longitudinal bulkhead, together with transverse wash bulkheads, may be provided to strengthen the tank and reduce liquid movement and sloshing.

    2. Materials

    • LPG is carried at low temperatures, approximately βˆ’50Β°C for fully refrigerated LPG.
    • To prevent brittle fracture at these temperatures, the primary barrier/tank is constructed from suitable low-temperature-resistant materials, typically fine-grained carbon-manganese steel.

    3. Tank Supports and Chocks

    The cargo tank operates at a substantially different temperature from the ship's hull and therefore undergoes thermal expansion and contraction.

    • The tank is supported on the double bottom by load-bearing insulation blocks.
    • These may be made from specialised hardwood such as AzobΓ© or suitable synthetic materials.
    • The supports provide the necessary load-bearing capacity while reducing thermal transfer and structural stresses.
    • Anti-roll, anti-pitch and anti-flotation keys/chocks secure the tank against movement caused by the ship's motion.
    • At the same time, the arrangement allows the tank to expand and contract freely due to temperature changes.

    4. Insulation

    • The outside of the primary barrier is provided with high-efficiency thermal insulation.
    • Its purpose is to maintain the required low cargo temperature and minimise heat ingress and cargo boil-off.
    • Sprayed polyurethane foam (PUF) is commonly used as the insulation material.

    5. Secondary Barrier

    • Under the IGC Code, Type A tanks are required to have a complete secondary barrier capable of containing leaked cargo for up to 15 days, preventing the cold cargo from coming into contact with the outer hull.
    • In LPG carriers, the ship's inner hull is normally used as the secondary barrier.
    • The inner hull is also constructed from suitable fine-grained low-temperature steel so that it can withstand the low temperature if the primary cargo tank fails.
    • The space between the primary tank and secondary barrier, known as the hold space, is maintained with inert gas or dry air as applicable.
    Q5 (10 Marks) Hull Construction πŸ”₯ Repeated 10x

    Describe a method for the attachment of bilge keels. State THREE reasons for not extending bilge keels for the entire length of the vessel. Explain TWO principles of roll damping that bilge keels exploit.

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    Part (a)

    Method of Bilge keel attachment to the hull:

    Bilge keels are fitted port and starboard at the turn of the bilge. They do not extend outside the lines of the side and bottom shell. The bilge keels are attached directly in line with an internal stiffening member such as a girder or longitudinal. The bilge keel comprises a flat bar doubler welded directly to the shell, and an offset bulb plate (OBP) with 'scallops' cut in it is welded to the flat bar doubler. The ends of the bilge keels are tapered (minimum 3 in 1) and will end in line with transverse internal stiffening, such as a frame. Using a hull doubler protects the hull in case of damage to the keel, as the crack would not extend into the hull.

    Part (b)

    Bilge keels are not fitted for the full length of the vessel because:

    • The 'lever' to the ship's axis of rotation is reduced at the ends;
    • The ineffectiveness, the closer the bilge keel is to the rolling centre, increased resistance and more likelihood of damage.
    • The hydrodynamic effect would cause a large increase in resistance and fuel consumption;
    • At the aft end, the boundary layer is much thicker, and since the keel would not project through, it would have a much reduced effect.
    Part (c)

    Two principles of roll damping exploited by Bilge keels:

    Increased Roll Period:

    • Bilge keels increase the ship's roll period (the time it takes for the vessel to complete one roll). This is achieved by increasing the moment of inertia (K) of the vessel. A longer roll period means the vessel rolls more slowly, thus reducing the amplitude of the roll.

    $$T_{r}=2\pi\:\frac{k}{\sqrt{g\times GM}}$$

    Where:

    • g: acceleration due to gravity
    • GM: metacentric height
    • k: mass moment of inertia

    By increasing k, the ship's stability during roll improves.

    Hydrodynamic Resistance:

    • As the vessel rolls, the bilge keels move through the water, creating pressure differences. Water pressure on one side of the keel opposes the rolling motion, providing a damping effect. This is due to the interaction between the bilge keel and the water, creating hydrodynamic forces that counteract the rolling motion.
    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 8x

    (a) Define longitudinal center of gravity (LCG) and longitudinal centre of buoyancy (LCB). (6)

    (b) The immersed cross-sectional areas of a ship 120 m long. commencing from aft are 2, 40, 79, 100, 103, 104, 104, 103, 97, 58 and 0 m2. Calculate: (10)

    (i) Displacement

    (ii) Longitudinal position of the centre of buoyancy.

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.
    Part (b)

    Given:

    $$Common \space interval \space (h) \space = \space {{L} \over h} \space = \space {{120} \over 10} \space = \space 12 $$

    Cross-sectional area

    SM

    Product of volume

    Lever

    Product of 1st moment

    2

    1

    2

    +5

    +10

    40

    4

    160

    +4

    +640

    79

    2

    158

    +3

    +474

    100

    4

    400

    +2

    +800

    103

    2

    206

    +1

    +206

    Ξ£MA = +2130

    104

    4

    416

    0

    0

    104

    2

    208

    -1

    -208

    103

    4

    412

    -2

    -824

    97

    2

    194

    -3

    -582

    58

    4

    232

    -4

    -928

    0

    1

    0

    -5

    0

    Ξ£βˆ‡ = 2388

    Ξ£MF = -2542

    $$Displacement \space = \space \rho \times {{h} \over 3} \times \sum βˆ‡ \space tonne $$

    $$=1.025\times{{12}\over3}\times2388$$

    $$Displacement \space = \space 9790.8 tonne$$

    Centre of buoyancy from midship (LCB)

    $$LCB\:=\:h\times({{\sum M_{A}+\sum M_{F}}\over\sum\nabla})$$

    $$=12\times({{2130-2542}\over2388})$$

    $$LCB \space = \space -2.07m fwd$$

    Q7 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 7x

    A ship of 15000 tonne displacement has an Admiralty Coefficient, based on shaft power of 420. The mechanical efficiency of the machinery is 83%, shaft losses 6%, propeller efficiency 65% and QPC 0.71. At a particular speed the thrust power is 2550 kW. Calculate: (16)

    (i) Indicated power.

    (ii) Effective power.

    (iii) Ship speed.

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    Given:

    $$\Delta=15000t$$

    $$Shaft\:Power\:\left(SP\right)=420$$

    $$Transmission\:Efficiency=83\%$$

    $$Shaft\:losses=6\%$$

    $$Propeller\:Efficiency=65\%$$

    $$QPC=0.71$$

    $$Thrust\:Power=2550kW$$

    $$\left(\imaginaryI\right)\:Delivered\:Power\:\left(DP\right)=\frac{Thrust\:Power\:\left(TP\right)}{Propeller\:Efficiency\:\left(\eta P\right)}$$

    $$DP=\frac{2550}{0.65}$$

    $$DP=3923.07kW$$

    $$\left(ii\right)\:Shaft\:Power=\frac{Delivered\:Power\:\left(DP\right)}{Transmission\:Efficiency\:\left(\eta T\right)}\:$$

    $$SP=\frac{3923.07}{0.94}$$

    $$SP=4173.47kW$$

    $$\left(iii\right)\:Indicated\:Power=\frac{Shaft\:Power\:\left(SP\right)}{Mechanical\:Efficiency\:\left(\eta m\right)}$$

    $$IP=\frac{4173.47}{0.83}$$

    $$IP=5028.28kW$$

    $$\left(iv\right)\:Effective\:Power=DP\times QPC$$

    $$EP=3923.07\times0.71$$

    $$EP=2785.3797kW$$

    $$\left(v\right)\:Shaft\:Power=\frac{\Delta^{\frac23}\times V^3}{Admiralty\:Co-efficient}$$

    $$4173.47=\frac{15000^{\frac23}\times V^3}{420}$$

    $$V=14.23knots$$

    Q8 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Describe stability requirements for dry-docking. (6)

    (b) A ship of 8000t displacement floats upright in sea water, with KG = 7.6m. GM = 0.5m. A tank, whose Kg is 0.6m above the keel and 3.5m from the center line contains 100 t of water ballast. Neglecting the free surface effect. Calculate the angle which the ship will heel, when the ballast water is pumped out. (10)

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    Part (a)

    For safe dry-docking, a ship must meet two key stability requirements:

    • Positive GM (Metacentric Height): The ship needs a positive GM. GM is the distance between the centre of gravity (G) and the metacentre (M). A positive GM indicates inherent stability; the ship will right itself if tilted. During dry-docking, the loss of buoyancy as the ship rests on the blocks reduces GM. Insufficient positive GM increases the risk of the ship heeling over or capsizing.
    • Trim by the stern: The vessel should be trimmed slightly by the stern (aft end lower than the bow) to ensure the aft end sits on the keel blocks first. This controlled settling minimises the risk of instability during the docking process. An even keel is generally preferred for the initial floating condition before the dry-docking procedure begins.
    Part (b)

    $$new \space KG \space = \space {{(8000 \times 7.6) - (100 \times 0.6)} \over 8000 - 100}$$

    $$New\:KG\:=\:7.689m$$

    $$New \space GM_1 \space = \space KM - KG$$

    $$= \space (7.6 + 0.5) - 7.689$$

    $$New\:GM\:=\:0.411m$$

    Angle of heel when 100t ballast is pumped out

    $$Tan\theta=\frac{m\times d}{\Delta GM}$$

    $$=\frac{100\times3.5}{7900\times0.411}$$

    $$Tan\theta=0.1077$$

    $$\theta=6^09^{^{\prime}}$$

    Q9 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 6x

    A ship of length 140m, Breadth of 18.5m, draught of 8.1 m and a displacement of 17,025 tonnes in sea water, has a face pitch ratio of 0.673. The diameter of the propeller is 4.8m. The results of the speed trial show that true slip may be regarded as constant over a range of 9 to 13 knots and is 30%. w = 0.5Cb-0.05. If fuel used is 20t/day at 13 knots and fuel consumption/day varies as cube of speed of ship. determine the fuel consumption, when propeller runs at 110 rpm.

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    Given:

    $$Lenght,\:L=140m$$

    $$Breadth,\:B=18.5m$$

    $$Draught,\:d=8.1m$$

    $$Displacement,\:\Delta=17025tonnes$$

    $$Pitch\:ratio,\:p=0.673\operatorname{}$$

    $$Diameter\:of\:Propeller,\:D=4.8m$$

    $$\operatorname{Real\:Slip,\:R_{s}=30\%\:or\:0.3}$$

    $$Wake\:fraction,\:W=0.5C_{b}-0.05$$

    $$Consumption,\:C_2=\:20t\:per\:day\:$$

    $$Ship\:Speed,\:V_2=13\:knots$$

    $$\operatorname{Revolution,\:N}=110rpm$$

    $$cons\:per\:day\:\alpha\:V^3$$

    To find Fuel Consumption C1=?

    We know that,

    $$Displacement,\:\Delta=\nabla\times\rho$$

    $$17025=\nabla\times1.025$$

    $$\nabla=16609.76m^3$$

    $$Block\:Coefficient,\:C_{b}=\frac{\nabla}{L\times B\times d}$$

    $$C_{b}=\frac{16609.76}{140\times18.5\times8.1}$$

    $$C_{b}=0.792$$

    $$Wake\:Fraction,\:W=0.5C_{b}-0.05$$

    $$W=0.5\times0.792-0.05$$

    $$W=0.346$$

    $$Pitch\:ratio,\:p=\frac{P}{D}$$

    $$0.673=\frac{P}{4.8}$$

    $$P=4.8\times0.673$$

    $$P=3.23m$$

    $$Theoretical\:Speed,\:V_{t}=\frac{P\times N\times60}{1852}$$

    $$V_{t}=\frac{3.23\times110\times60}{1852}$$

    $$V_{t}=11.51knots$$

    Using, Real slip equation.

    $$\operatorname{\operatorname{Real\:Slip,\:R_{s}=\frac{V_{t}-V_{a}}{V_{t}}}}$$

    $$0.3=\frac{11.51-V_{a}}{11.51}$$

    $$V_{a}=11.51-11.51\times0.3$$

    $$V_{a}=11.51\left(1-0.3\right)$$

    $$V_{a}=11.51\times0.7$$

    $$V_{a}=8.057knots$$

    $$Wake\:fraction,\:W=\frac{V-V_{a}}{V}$$

    $$0.346=\frac{V-8.057}{V}$$

    $$0.346V=V-8.057$$

    $$V=\frac{8.057}{0.654}$$

    $$V=12.32knots$$

    $$cons\:per\:day\:\alpha\:V^3$$

    $$\frac{C_1}{C_2}=\left(\frac{V_1}{V_2}\right)^3$$

    $$\frac{C_1}{20}=\left(\frac{12.32}{13}\right)^3$$

    $$C_1=20\times\left(\frac{12.32}{13}\right)^3$$

    $$C_1=20\times0.851$$

    $$C_1=17.02t\:per\:day$$

    Q10 (10 Marks) Hull Construction πŸ”₯ Repeated 7x

    With respect to Buoyancy of a vessel:

    (a) What do you understand by reserve buoyancy what happen if the lost buoyancy is greater than the reserve buoyancy. (6)

    (b) A forward deep tank 12 m long extends from a longitudinal bulkhead to the ship's side. The widths of the tank surface measured from the longitudinal bulkhead at regular intervals are 10, 9, 7, 4 and 1 m. Calculate the second moment of area of the tank surface about a longitudinal axis passing through its centroid. (10)

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    Part (a)

    Reserve Buoyancy

    Reserve buoyancy is the watertight volume above the waterline of a vessel. It represents the potential buoyancy that can be utilised to keep the ship afloat when additional weight is added or if some buoyancy is lost due to damage (e.g., bilging).

    When a mass is added to the ship or buoyancy is lost (e.g., due to flooding of a compartment), the reserve buoyancy is converted into active buoyancy by an increase in draught.

    If the lost buoyancy (due to flooding or damage) is greater than the reserve buoyancy, the vessel will no longer have sufficient buoyant force to counteract its weight, causing it to sink.

    Part (b)
    Q1 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 13x

    Describe the relationship between frictional resistance and

    (a) Ship's speed

    (b) The wetted area

    (c) The surface roughness

    (d) The length of the vessel

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    Frictional resistance arises due to the interaction between the ship's hull and the water as the vessel moves. This resistance is caused by eddying water adjacent to the hull that is drawn along with the ship. The frictional resistance is given by:

    $$R_{f\:=\:}f\:.\:s\:.\:v^{n}$$

    Where:

    • f = Coefficient of friction
    • s = Wetted surface area
    • v = Ship's speed in knots
    • n = Constant (1.82)
    Part (a)

    Ship Speed (v):

    • Frictional resistance increases with increasing ship speed. The relationship is not perfectly linear, but it's approximately described by the formula Rf = fsv^n (Rf ∝ v^n), where 'n' is a constant (approximately 1.82). This means that a small increase in speed results in a larger increase in frictional resistance.
    Part (b)

    Wetted Area (s):

    • Frictional resistance is directly proportional to the wetted surface area (Rf ∝ s). A larger wetted area (the part of the hull submerged in water) leads to greater frictional resistance. The wetted area increases with the ship's draught (depth in the water).
    Part (c)

    Surface Roughness:

    • Increased surface roughness increases frictional resistance. Roughness can be caused by fouling (marine growth), corrosion, or poor hull maintenance. Regular cleaning and maintenance of the hull surface help to minimise roughness and reduce resistance.
    Part (d)

    Length of the Vessel

    • Frictional resistance is influenced by the vessel's length. As the length increases, the wetted surface area grows, and the coefficient of friction (f) changes, leading to an increase in Rf. Longer vessels experience higher frictional resistance, but streamlined designs can mitigate the effect to some extent.
    Q2 (10 Marks) General πŸ”₯ Repeated 3x

    Draw and describe the construction of a forepeak tank. Explain how are the effects of panting and pounding taken care with the help neat sketches?

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    Part (a)

    Causes and Effects of Panting and Pounding, Including the Affected Areas

    Panting

    Panting refers to the in-and-out flexing or pulsation of the shell plating at the ends of a ship, mainly caused by variations in sea pressure during pitching motions in rough weather.

    When a ship moves through heavy seas, especially while pitching, the water pressure acting on the bow and stern changes continuously. These fluctuating pressures cause the side shell plating to vibrate and flex. The effect is known as panting.

    The forward end of the ship is affected more severely than the aft end because:

    • The bow directly encounters incoming waves while the ship is making headway.
    • Water pressure at the fore end changes rapidly as the bow rises and falls in the sea.

    The affected areas are mainly:

    • Forward shell plating near the bow
    • Side shell plating at the fore peak region
    • To a lesser extent, the aft end of the ship

    Effects of Panting

    Panting can lead to:

    • Repeated flexing and vibration of shell plating
    • Fatigue stresses in plating and framing
    • Loosening of rivets or welded joints
    • Cracking or deformation of structural members
    • Reduction in structural strength if not properly reinforced

    Therefore, special strengthening arrangements are provided at the forward and aft ends of the vessel to resist panting stresses.

    Pounding

    Pounding refers to the heavy impact experienced at the bottom forward part of the ship when the bow emerges from the water during pitching and then slams violently back onto the sea surface.

    This condition usually occurs when:

    • The ship is pitching heavily in rough seas,
    • The fore part lifts clear of the water due to heaving and pitching motions,
    • The bow then falls heavily onto the wave surface.

    The effect is most severe when the vessel is in the light ship condition, because the bow rises more easily out of the water.

    The main area affected by pounding is:

    • The bottom shell plating in the forward region of the ship,
    • Especially near the forefoot and forward bottom structure.

    The stern may also experience similar impacts from following seas, but usually to a lesser extent.

    Effects of Pounding

    Pounding produces:

    • Severe impact stresses on bottom plating
    • Buckling or deformation of bottom structure
    • Cracks in shell plating or framing
    • Structural fatigue and weakening
    • Damage to internal supporting members

    To withstand these heavy impact loads, the forward bottom structure is specially strengthened.

    Part (b)

    Constructional Details Designed to Resist Panting and Pounding

    Special structural arrangements are incorporated in ship construction to resist the stresses caused by panting and pounding.

    Structural Arrangements to Resist Panting

    The following strengthening arrangements are provided mainly at the bow and stern regions:

    1. Panting Beams

    Panting beams are horizontal beams fitted across the ship near the bow and stern.

    Their purpose is to:

    • Support the side shell plating,
    • Reduce excessive flexing,
    • Increase structural rigidity against panting stresses.

    2. Panting Stringers

    Panting stringers are horizontal girders fitted along the ship side in the panting region.

    They:

    • Connect frames together,
    • Provide additional stiffness to the shell plating,
    • Help distribute fluctuating sea pressure loads.

    3. Close-Spaced Framing

    Frames in the panting region are spaced closer together than in other parts of the ship.

    This:

    • Provides better support to shell plating,
    • Reduces plate vibration and deformation.

    Structural Arrangements to Resist Pounding

    To resist pounding stresses at the forward bottom region, additional strengthening is provided.

    1. Increased Bottom Plating Thickness

    The shell plating near the keel and forward bottom region is made thicker.

    Usually:

    • The first few strakes of bottom plating on either side of the keel are increased in thickness.

    This enables the structure to withstand repeated impact loads.

    2. Plate Floors

    Solid plate floors are fitted at closer spacing in the forward bottom region.

    These:

    • Strengthen the bottom structure,
    • Distribute pounding stresses more effectively,
    • Prevent deformation of bottom plating.

    3. Additional Internal Reinforcement

    Extra brackets, girders, and stiffeners may also be provided in the fore peak and bottom structure to improve strength and rigidity.

    Q3 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 2x

    If a ship is seriously damaged under water in way of a large fuel oil side bunker tank what is the immediate effect and what may ultimately happen? What features in the ship would enhance safety of the vessel and marine environmental protection aspects in such a case?

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    If a ship is seriously damaged underwater in the area of a large fuel oil side bunker tank, it will cause the following:

    Bilging or Flooding of the Compartment:

    • Water will enter the damaged bunker tank, reaching the draught level.
    • The rate of flooding depends on the size of the breach and the water pressure at the depth of the damage.

    Oil Leakage into the Sea:

    • If the tank contains fuel oil, oil will begin to leak out, causing pollution.
    • The extent of leakage depends on the tank’s contents (empty, half-full, or full) and the location of the damage.

    List and Trim of the Vessel:

    • The ingress of water and loss of oil will create an imbalance, causing the vessel to list or trim.

    If corrective actions are not taken, uncontrolled flooding and loss of stability could lead to capsizing or sinking of the vessel.

    Features in the Ship to Enhance Safety:

    • Small Bunker Tank Sizes reduces the risk of extensive oil spillage and loss of stability.
    • Connectivity to transfer pumps allows the transfer of oil from the damaged tank to an empty tank, minimizing oil spillage and counteracting the loss of stability.
    • The tank’s size and location are designed to limit the effects of flooding, as per damage stability regulations.
    • Properly positioned transverse and longitudinal bulkheads enhance the subdivision factor, limiting water ingress to the damaged tank.
    • Ship’s Ballast System allows corrective ballasting to counteract the list or trim caused by the ingress of water.
    • Watertight Doors and Hatches prevent water from spreading to adjacent compartments.

    Recommended Immediate Actions by Crew:

    For Empty Tanks:

    • Quickly seal off the damaged tank by shutting all valves and isolating it from the transfer system to prevent water ingress into other parts of the vessel.

    For Half-Empty or Full Tanks:

    • Initiate oil transfer to another empty tank to reduce oil leakage and stabilize the ship.
    • Monitor the water ingress and ensure the tank is filled to the draught level with seawater if necessary, using ballast to correct the list.
    Q4 (10 Marks) Hull Construction πŸ”₯ Repeated 3x

    Discuss the need for adequate support of engine room gantry cranes, detailing the following.

    (a) Sketch section through the engine room casing showing how the crane is supported by the ship structure.

    (b) State what restricts the forward and aft limits of the crane and what is fitted to prevent the crane damaging the forward and aft bulkheads or casing.

    (c) State the Second Engineer's responsibilittes for the engine room gantry crane.

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    Part (a)

    Engine room gantry crane:

    Part (b)

    Forward and aft limits of crane:

    • Two limit switches are installed to automatically cut off the motor power when the crane reaches its forward or aft limits.
    • Mechanical Stoppers are fitted at the forward and aft ends of the crane rails. They act as a failsafe to prevent the crane from damaging the forward and aft bulkheads or casing if the limit switches fail. Rubber or spring buffers may also be added to absorb impact forces and reduce the risk of damage.
    Part (c)

    Second Engineer's responsibilities for the engine room gantry crane.

    • Inspect the crane, rails, mechanical stoppers, and support brackets for wear or damage.
    • Ensure limit switches are functional and properly aligned.
    • Check the condition of the gear case oil and renew it as required.
    • Inspect and maintain wire ropes and safety latches.
    • Ensure the Safe Working Load (SWL) is clearly marked and adhered to.
    • Verify the validity of the crane's test certificate.
    • Perform motor overhauls and measure insulation resistance.
    • Conduct brake tests and emergency stop function checks.

    Provide instructions to Junior Engineers:

    • Not exceeding the SWL of the crane.
    • Ensuring mechanical locks are removed before operating the crane.
    • Using proper Personal Protective Equipment (PPE).
    • Prohibiting personnel from standing beneath the crane during operation.
    Q5 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    With reference to Ship stability:

    (a) With the help of a neat sketch explain the relevant features of a G-Z. curve.

    (b) What are the effects of the below mentioned conditions on the G-Z curve:

    (i) Increased freeboard

    (ii) Increased beam, and

    (iii) Increased GM.

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    Part (a)

    Features of a GZ curve.

    The GZ curve is a graph of the righting lever GZ against the angle of heel. Its relevant features are:

    • The origin: at zero heel, GZ = 0.
    • The initial slope: the tangent to the curve at the origin equals GM (the metacentric height), since GZ = GM sin(theta) for small angles. A steeper initial slope means a larger GM.
    • The maximum righting lever (GZ max): the highest point of the curve, and the angle at which it occurs (the angle of maximum stability, typically 25-40 deg).
    • The range of stability: the angle from the upright to the angle of vanishing stability (where GZ returns to zero, typically 60-90 deg).
    • The area under the curve: proportional to the dynamical stability (the work done in heeling the ship), used to assess stability in a seaway and against wind heeling.
    • The angle of loll: if the curve starts below the axis (negative GZ at small angles), indicating a negative GM and an unstable ship that lolls to one side.
    • The effect of free surface: the curve is reduced by the free-surface correction.

    The curve is obtained from the cross-curves of stability corrected for the actual KG and free-surface effects, and is compared with the statutory criteria.

    Part (b)

    Effects of the following conditions on the GZ curve.

    (i) Increased freeboard: increasing the freeboard raises the deck edge and increases the reserve buoyancy, so the range of stability is increased (the angle of vanishing stability moves to a larger angle) and the area under the curve is increased. The initial slope (GM) is largely unchanged, but the curve is higher and extends further, giving greater dynamical stability and a larger range.

    (ii) Increased beam: increasing the beam increases the waterplane area and the BM (BM is proportional to the cube of the beam), so the initial slope (GM) increases and the curve is steeper at small angles. The maximum GZ is increased and occurs at a smaller angle, but the range of stability may be reduced (the angle of vanishing stability decreases) because the ship becomes stiffer and the deck edge immerses earlier. The area under the curve may be reduced at large angles.

    (iii) Increased GM: increasing the GM (e.g. by lowering KG) makes the initial slope steeper, so the curve rises more steeply at small angles and the maximum GZ is larger and occurs at a smaller angle. However, the range of stability is reduced (the angle of vanishing stability decreases) and the ship rolls more quickly and with a shorter period, which can be uncomfortable. The area under the curve at small angles increases but the overall range decreases.

    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    What is Prismatic Co-efficient (Cp).

    (a) Derive the formula Cp = Cb/Cm, where Cb = Co-efficient of fineness and Cm = midship section area co-efficient.

    (b) The length of a ship is 18 times the draught, while the breadth is 2.1 times the draught. At the load water plane, the water plane area co-efficient is 0.83 and the difference between the TPC in sea water and the TPC in fresh water is 0.7. Determine the length of the ship and the TPC in fresh water.

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    Part (a)

    Prismatic coefficient is the ratio of the volume of displacement to the product of length and area of the immersed portion of the midship section

    Cb is the block co-efficient or co-efficient of fitness is the ratio of the volume of displacement to the product of length, breadth and draught

    $$\because\space C_{b}\space=\space{{\nabla}\over L\times B\times D}\:---\:1$$

    $$C_{p}\space=\space{{\nabla}\over L\times A_{m}}\:---\:2$$

    $$C_m \space = \space {{A_m} \over B \times D}$$

    $$A_{m}=\:C_{m}\times B\times D\:---\:3$$

    Substitute 3 in 2

    $$C_{p}\space=\space{{\nabla}\over C_{m}\times B\times D\ \times L}\:---\:4$$

    $$\nabla=C_{b}\times L\times B\times D\:---5\:\left(from\:equation\:1\right)$$

    Substitute 5 in 4

    $$C_{p}=\frac{C_{b}\times L\times B\times D}{C_{m}\times L\times B\times D}$$

    $$C_p \space = \space {{C_b} \over C_m}$$

    Part (b)

    $$Length \space of \space ship \space = \space L$$

    $$Draught \space = \space {{L} \over 18 }$$

    $$breadth \space = \space 2.1 \times draught \space = \space 2.1 \times {{L} \over 18}$$

    $$TPC\:in\:SW\:=\:0.01025A_{w}$$

    $$TPC\:in\:FW\:=\:0.0100A_{w}$$

    $$0.01025A_{w}-0.0100A_{w}=0.7$$

    $$A_{w}=\frac{0.7}{2.5\times10^{-4}}=2800$$

    $$C_w \space = \space {{A_w} \over L \times B}$$$$0.83 \space = \space {{2800} \over L \times {{2.1L} \over 18}}$$

    $$L=170.04m$$

    $$TPC\:in\:FW=0.010\times A_{w}$$

    $$=\:0.0100\times2800$$

    $$TPC\:in\:FW=28$$

    Q7 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 4x

    With respect to ship propulsion:

    (a) Explain the various effeciencies associated with propeller and shafting arrangement

    (b) When a propeller of 4.8m pitch turns at 110 rpm, the apparent ship is found to be -S% and the real slip is 1.5 S%. If the wake speed is 25% of the ship speed, calculate the ship speed, apparent slip and real slip.

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    Part (a)

    Efficiencies associated with the propeller and shafting arrangement.

    • Shaft (transmission) efficiency: the ratio of the delivered power at the propeller to the brake power of the engine, accounting for the losses in the shaft bearings, stern tube and any gearing. It is typically 0.97-0.99.
    • Propeller (open-water) efficiency: the ratio of the thrust power (T x Va) to the delivered power (2 pi n Q). It represents the efficiency of the propeller itself in converting the delivered power into thrust power.
    • Hull efficiency: the ratio of the effective power to the thrust power, = (1 - t)/(1 - w), where t is the thrust deduction factor and w the wake fraction. It accounts for the interaction between the hull and the propeller.
    • Quasi-propulsive coefficient (QPC): the ratio of the effective power to the delivered power, = hull efficiency x propeller efficiency. It is the overall efficiency of the propulsion system in converting the delivered power into effective (towing) power.
    • Overall (propulsive) efficiency: the ratio of the effective power to the brake power, = QPC x shaft efficiency. It is the overall efficiency of the engine-to-propeller-to-hull system.
    Part (b)

    Ship speed, apparent slip and real slip.

    A propeller of 4.8 m pitch turns at 110 rev/min. The apparent slip is -S% and the real slip is +1.5S%. The wake speed is 25% of the ship speed. Calculate the ship speed, the apparent slip and the real slip.

    Pitch speed = pitch x rev/s = 4.8 x 110/60 = 8.8 m/s.

    Let the ship speed be V (m/s). The speed of advance Va = V x (1 - 0.25) = 0.75 V.

    Apparent slip = (pitch speed - V)/pitch speed = -S/100.

    Real slip = (pitch speed - Va)/pitch speed = 1.5 S/100.

    From the apparent slip: (8.8 - V)/8.8 = -S/100, so V = 8.8(1 + S/100).

    From the real slip: (8.8 - 0.75 V)/8.8 = 1.5 S/100, so 8.8 - 0.75 V = 0.132 S.

    Substitute V = 8.8(1 + S/100): 8.8 - 6.6(1 + S/100) = 0.132 S.

    8.8 - 6.6 - 0.066 S = 0.132 S, so 2.2 = 0.198 S, S = 11.11.

    Apparent slip = -11.11%; real slip = 1.5 x 11.11 = 16.67%.

    Ship speed V = 8.8(1 + 0.1111) = 9.78 m/s = 9.78 x 1.944 = 19.0 knots.

    Answer: ship speed about 19.0 knots; apparent slip -11.1%; real slip +16.7%.

    Q8 (10 Marks) General πŸ”₯ Repeated 2x

    With the aid of sketches:

    (a) Explain various lines plan

    (b) The half-breadths of waterplane of a ship of 120m length and 15m breadth are given below:

    Section: 0 1 2 3 4 5 6 7 8

    Half-breadth: 1.6 2.8 5.5 6.4 7.3 6.2 4.2 2.0 0

    Calculate:

    (i) Water plane area

    (ii) TPC in salt water

    (iii) Cw

    (iv) LCF from mid-ship

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    Part (a)

    Lines plan - explanation.

    The lines plan is a set of drawings that defines the shape of the hull. It consists of three principal views:

    • Sheer plan (profile): shows the hull as seen from the side. It contains the waterlines, the deck line, the sheer line and the profile of the stem and stern. The vertical positions of the waterlines and the longitudinal positions of the stations are shown.
    • Half-breadth plan (plan view): shows the hull as seen from above, drawn for one half of the ship (port or starboard) because the hull is symmetrical. It contains the waterlines (horizontal sections) and the deck line, showing the half-breadths at each station.
    • Body plan (end view): shows the transverse sections (stations) as seen from the bow and stern. The forward half of the stations is drawn on one side and the after half on the other. It contains the buttock lines and the diagonal lines.

    The three views are drawn in projection so that every point on the hull appears in all three views, allowing the shape to be fully defined and faired. The lines plan is used to calculate hydrostatic data, displacement, and to construct the ship.

    Part (b)

    Waterplane calculation.

    Length L = 120 m, breadth B = 15 m. Half-breadths at stations 0 to 8:

    Station: 0 1 2 3 4 5 6 7 8

    Half-breadth (m): 1.6 2.8 5.5 6.4 7.3 6.2 4.2 2.0 0

    Number of intervals n = 8, so the common interval h = L/n = 120/8 = 15 m.

    (i) Waterplane area.

    Using Simpson's First Rule with 8 intervals (9 ordinates), multipliers 1,4,2,4,2,4,2,4,1:

    Sum of products = 1x1.6 + 4x2.8 + 2x5.5 + 4x6.4 + 2x7.3 + 4x6.2 + 2x4.2 + 4x2.0 + 1x0

    = 1.6 + 11.2 + 11.0 + 25.6 + 14.6 + 24.8 + 8.4 + 8.0 + 0 = 105.2.

    Area of half waterplane = (h/3) x sum = (15/3) x 105.2 = 5 x 105.2 = 526 m2.

    Waterplane area (full) = 2 x 526 = 1052 m2.

    (ii) TPC in salt water.

    TPC = (waterplane area x density of sea water)/100 = (1052 x 1.025)/100 = 1078.3/100 = 10.78 tonne/cm.

    Answer: TPC = 10.78 tonne per cm.

    (iii) Cw (waterplane area coefficient).

    Cw = waterplane area / (L x B) = 1052 / (120 x 15) = 1052/1800 = 0.584.

    Answer: Cw = 0.584.

    (iv) LCF from midship.

    The centre of flotation is the centroid of the waterplane. Using the first moment about midship (station 4):

    Moment of products = sum of (multiplier x half-breadth x distance from midship in intervals).

    Distances from station 4 (in intervals): station 0 = -4, 1 = -3, 2 = -2, 3 = -1, 4 = 0, 5 = +1, 6 = +2, 7 = +3, 8 = +4.

    Moment = 1x1.6x(-4) + 4x2.8x(-3) + 2x5.5x(-2) + 4x6.4x(-1) + 2x7.3x0 + 4x6.2x(+1) + 2x4.2x(+2) + 4x2.0x(+3) + 1x0x(+4)

    = -6.4 - 33.6 - 22.0 - 25.6 + 0 + 24.8 + 16.8 + 24.0 + 0 = -22.0.

    Distance of centroid from midship = (moment/sum) x h = (-22.0/105.2) x 15 = -0.2092 x 15 = -3.14 m.

    The negative sign means the LCF is forward of midship.

    Answer: LCF is 3.14 m forward of midship.

    Q9 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 2x

    A ship 150 m in length, 24 m breadth, displaces 25000 tonne when floating at a draft of 9 m in sea water of density 1025 kg/m?. The ship's propeller has a diameter of 5.8 m, a pitch ratio of 0.9 and a blade area ratio of 0.45. With the propeller operating at 2 revs/sec, the following results were recorded:

    Apparent slip = 0.06

    Thrust power = 3800 Kw

    Propeller efficiency = 64%

    The taylor wake fraction Wt = 0.5Cb-0.05

    Calculate each of the following for the above condition:

    (a) The ship's speed

    (b) The real slip ratio

    (c) The thrust per unit area of blade surface

    (d) The torque delivered to the propeller

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    Given:

    $$L \space = \space 150m$$

    $$B \space = \space 24m$$

    $$D \space = \space 9m$$

    $$\Delta=25000t$$

    $$d \space = \space 5.8m$$

    $$p \space = \space 0.9m$$

    $$BAR \space = \space 0.45m$$

    $$n \space = \space 2 \space rev/sec$$

    $$Apparent \space slip \space = \space 0.06m$$

    $$T_p \space = \space 3800kw$$

    $$Ξ·_{prop} \space = \space 64\%$$

    $$W_f \space = \space 0.5C_b - 0.05$$

    $$p\space=\space{{P}\over d}\space$$

    $$0.9 \space = \space {{P} \over 5.8}$$

    $$Pitch \space = \space 5.22m$$

    $$V_T \space = \space {{P \times N \times 3600} \over 1852}$$

    $$V_T P \space = \space 20.29 \space knots$$

    $$App. \space slip \space = \space {{V_T - V} \over V_T}$$

    $$0.06\space=\space{{20.29-V}\over20.29}$$

    $$Ship's \space speed \space (V) \space = \space 19.07 \space knots $$

    $$C_{b}=\frac{\Delta}{L\times B\times D\times\rho}$$

    $$=\frac{25000}{150\times24\times9\times1.025}$$

    $$C_{b}=0.752$$

    $$given, \space W_f \space = \space 0.5C_b - 0.05$$

    $$W_{f}=0.5\times0.752-0.05$$

    $$W_{f}=0.326$$

    $$W_F \space = \space {{V - V_a} \over V}$$

    $$0.326 \space = \space {{19.07 - V_a} \over 19.07}$$

    $$V_a \space = \space 12.85Knots$$

    $$Real\space slip\space=\space{{V_T - V_a} \over V_T}\space\space{}$$

    $$=\frac{20.29-12.85}{20.29}$$

    $$Real \space slip \space = \space 0.367 $$

    $$Thrust \space power \space (T_p) \space = \space d_p \times Ξ·_{prop}$$

    $$3800=d_{p}\times0.64$$

    $$d_{p}=5937.5kw$$

    $$Thrust \space power \space T_p \space = \space Thrust \times V_a $$

    $$3800 \space = \space Thrust \times 12.85 \times {{1852} \over 3600}$$

    $$Thrust=574.83kw$$

    $$Blade \space area \space = \space {{\pi} \over 4}d^2 \times BAR $$

    $$=\space{{\pi}\over4}\times5.8^2\times0.45$$

    $$Blade \space area \space = \space 11.88m^2$$

    $$Thrust \space per \space area \space = \space {{574.82} \over 11.88} \space = \space 48.34 \space KN/m^2$$

    Q10 (10 Marks) Ship Stability πŸ”₯ Repeated 4x

    A ship 100 m long floats at a draft of 6m and in this condition the immersed cross sectional areas are as given in tables below. The equivalent base area (Ab) is required because of the fineness of the bottom shell

    Section: AP 1 2 3 4 5 6 FP

    Immeresed cross section area (m2): 12 30 65 80 70 50 0

    Draft(m): 0 0.6 1.2 2.4 3.6 4.8 6.0

    Waterplane area (m2): Ab 560 720 880 940 1000 1030

    Calculate each of the following:

    (a) The equivalent base are value Ab

    (b) The longitudinal position of the centre of buoyancy from midship

    (c) The vertical position of the centre of buoyancy above the base

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    Table for question:

    Section

    AP

    1

    2

    3

    4

    5

    FP

    Immersed cross sectional area

    12

    30

    65

    80

    70

    50

    0

    Draught (m)

    0

    0.6

    1.2

    2.4

    3.6

    4.8

    6.0

    Water plane area (m2)

    Ab

    560

    720

    880

    940

    1000

    1030

    Solution:

    Section

    CSA

    SM

    F volume

    Lever

    F moment

    AP

    12

    1

    12

    -3h

    -36h

    1

    30

    4

    120

    -2h

    -240h

    2

    65

    2

    130

    -1h

    -130h

    3

    80

    4

    320

    0

    0

    4

    70

    2

    140

    +1h

    140h

    5

    50

    4

    200

    +2h

    400h

    FP

    0

    1

    0

    +3h

    0

    Ξ£Fvol. = 992

    Ξ£Fmom. =134h

    $$Common\:interval\:\left(h\right)=\frac{L}{6}=\frac{100}{6}$$

    $$h=16.66m$$

    $$\sum F_{movement}=134\:\times16.66m^4$$

    $$\sum F_{moment}=2232.44m^4$$

    $$\nabla=\frac{h}{3}\times\sum F_{volume}$$

    $$\nabla=\frac{16.66}{3}\times992$$

    $$\nabla=5120.17m^3$$

    $$Longitudinal\:position\:of\:centre\:of\:buoyancy\:LCB\:=\:\frac{\sum F_{mom}}{\sum F_{vol}}$$

    $$=\frac{2232.44}{993}$$

    $$LCB=2.42m\:fwd$$

    Draught

    Aw

    SM

    F volume

    Lever

    F moment

    0

    Ab

    1/2

    1/2 Ab

    0

    0

    0.6

    560

    4/2

    1120

    0.6

    672

    1.2

    720

    1 2/4

    1080

    1.2

    1296

    2.4

    880

    4

    3520

    2.4

    8448

    3.6

    940

    2

    1880

    3.6

    6768

    4.8

    1000

    4

    4000

    4.8

    19200

    6.0

    1030

    1

    1030

    6.0

    6180

    Ξ£Fvol = 1/2Ab+12630

    Ξ£Fmom = 42564

    $$h=1.2$$

    $$\nabla=\frac{h}{3}\times\sum F_{vol}$$

    $$5120.17=\frac{1.2}{3}\times\left(\frac12A_{b}+12630\right)$$

    $$\frac{A_{b}}{2}=\frac{5120.17\times3}{1.2}-12630$$

    $$A_{b}=340.84m^2$$

    $$\sum F_{vol}=\left\lbrack\frac12\times340.85\right\rbrack+12630$$

    $$\sum F_{vol}=12800.425m^3$$

    $$Vertical\:position\:of\:centre\:of\:buoyancy=\frac{\sum F_{mom}}{\sum F_{vol}}$$

    $$=\frac{42564}{12800.425}$$

    $$KB=3.325m$$

    Q1 (10 Marks) Ship Stability πŸ”₯ Repeated 6x

    With reference to Roll-on, Roll-off ferries:

    (a) Describe the problem of free surlace effect:

    (b) Explain how it is intended that water should be cleared from car or cargo decks

    (c) Describe possible methods for improving the stability and survivabilty of these vessels.

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    Part (a)

    The problem of free-surface effect in Ro-Ro ferries.

    Ro-Ro ferries have large, open car and cargo decks that extend over a large part of the ship. If water is shipped onto these decks (e.g. through the bow or stern doors, or in heavy weather), the water spreads over the large deck area, creating a very large free surface. The free-surface effect reduces the effective GM by rho x i/Delta, where i is the second moment of area of the free surface (i = L B^3/12 for a rectangular deck). Because the car deck is very wide and long, the free-surface effect is very large and can reduce the GM to a dangerously low value, causing the ship to lose stability and capsize. This is the principal stability problem of Ro-Ro ferries: the large open decks create a huge free-surface effect if flooded, and the ship can capsize rapidly.

    Part (b)

    How water should be cleared from car or cargo decks.

    Water on the car deck should be cleared by:

    • Providing adequate freeing arrangements (scuppers, freeing ports, drain valves) at the deck edge and at the ends, so that water can drain overboard.
    • Providing a camber (transverse slope) on the deck so that water runs to the sides and drains through the freeing ports.
    • Providing a longitudinal slope (sheer) so that water runs to the ends and drains.
    • Using bilge pumps and drainage systems to remove water that cannot drain overboard.
    • Ensuring the freeing ports are of adequate size and are not blocked by cargo or lashings.

    The freeing arrangements must be adequate to remove water quickly and prevent the build-up of a large free surface.

    Part (c)

    Methods for improving the stability and survivability of Ro-Ro ferries.

    • Lowering the centre of gravity by placing heavy weights low and ballast in the double bottom.
    • Increasing the GM by increasing the beam and the waterplane area, and by lowering KG.
    • Providing adequate freeboard and reserve buoyancy.
    • Subdividing the car deck with watertight bulkheads or providing a raised deck to limit the spread of water.
    • Providing adequate freeing ports and drainage to remove water quickly.
    • Fitting bilge keels and stabilisers to reduce roll.
    • Designing the ship to meet the damage stability criteria (survive flooding of a compartment).
    • Using a higher freeboard and a stronger, more watertight structure, and ensuring the bow and stern doors are watertight and properly secured.
    • Operating with adequate stability margins and following the loading manual.
    Q2 (10 Marks) Hull Construction πŸ”₯ Repeated 5x

    Give a reasoned opinion as to the validity of the following assertions concering ship Structure

    (a) Crack propagation in propellers shaft 'A' bracket or spectacle frames is indicative of inadequate scantlings and strength.

    (b) The adequate provision of freeing ports is as critical to the seaworthiness as watertight integrity.

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    Part (a)

    Crack propagation in propeller shaft 'A' brackets or spectacle frames - is it indicative of inadequate scantlings and strength?

    This assertion is only partly valid. Cracks in A-brackets or spectacle frames (the shaft brackets supporting the propeller shaft) are more commonly the result of fatigue due to fluctuating loads, stress concentrations at the bracket-to-hull connection, and the vibration and whipping of the shaft, rather than simply inadequate scantlings. The brackets are subject to severe alternating loads from the propeller and the shaft, and cracks typically initiate at stress raisers (sharp corners, weld toes, the bracket arm-to-hull connection) and propagate under fatigue. While inadequate scantlings or poor design (insufficient section, poor connection, sharp notches) can contribute, the primary cause is usually fatigue and stress concentration, aggravated by vibration, corrosion and the dynamic loads of the propeller. Hence the assertion is not fully valid: crack propagation is more indicative of fatigue and stress concentration than of inadequate strength alone, and the design should address the fatigue life, the connection detail and the avoidance of stress raisers, as well as the scantlings.

    Part (b)

    The adequate provision of freeing ports is as critical to seaworthiness as watertight integrity.

    This assertion is largely valid. Freeing ports (openings in the bulwark that allow water shipped on deck to drain overboard) are essential to seaworthiness because, if they are inadequate, water accumulating on the deck cannot drain, which:

    • increases the free-surface effect and the weight of water on deck, reducing stability and increasing the risk of capsize;
    • increases the deck load and the risk of structural damage;
    • reduces the reserve buoyancy and can lead to the ship becoming unstable.

    Watertight integrity (the ability of the hull and its openings to keep water out) is equally critical to seaworthiness, as it prevents flooding and loss of buoyancy. Both are essential: watertight integrity keeps water out, while freeing ports remove water that is shipped on deck. If either is inadequate, the ship's seaworthiness is compromised. Hence the assertion is valid - freeing ports are as critical to seaworthiness as watertight integrity, because they maintain the stability and buoyancy of the ship by removing deck water.

    Q3 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 4x

    (a) What is free surface effect? How can be avoided or reduced.

    (b) Give the components of ships resistance while vessel is 'enroute'.

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    (a) Free Surface Effect:

    The free surface effect (FSE) is a reduction in the metacentric height (GM) of a vessel due to the movement of liquids within partially filled tanks when the ship heels. When a ship tilts, the liquid in a partially filled tank shifts to the lower side, causing the centre of gravity (CG) of the ship to move laterally. This lateral shift of the CG reduces the righting lever (GZ), effectively decreasing the ship's stability and increasing the angle of heel. This apparent loss of GM is the free surface effect.

    Minimizing Free Surface Effect:

    • Partially filled tanks should be avoided to minimize liquid movement.
    • Tanks should be designed with longitudinal divisions or swash bulkheads to limit the movement of liquid.
    • Install sluice valves to control the liquid movement between compartments in divided tanks.
    • Fill smaller tanks at the bottom of the ship first to lower the centre of gravity and improve stability.
    • Tanks should have reduced breadth to minimize the free surface's effect.
    Part (b)

    Components of ship’s resistance while vessel is en route

    When a ship moves through water, resistance opposes its motion. The ship must exert an equal force to maintain speed.

    Frictional resistance (Rf): This is caused by the friction between the hull and the water. The water immediately adjacent to the hull is dragged along, creating a boundary layer. This resistance depends on the water's viscosity, the ship's speed, and the wetted surface area of the hull. At lower speeds, frictional resistance can account for 70-90% of total resistance; however, at higher speeds, it can be less than 40%.

    Residuary resistance (Rr): This is the resistance that remains after subtracting the frictional resistance. These are:

    • Form drag: Resistance due to the shape of the hull and the flow of water around it, generating pressure differences.
    • Wave-making resistance: This is a major component at higher speeds. The ship creates waves, and energy is expended in this process.
    • Eddy resistance: Resistance caused by turbulent flow behind the ship, especially at sharp changes in the hull's shape. This is often minimized by optimizing the hull design.

    Air resistance (Ra): This resistance is generated by the ship moving through the air. It depends on the shape of the superstructure, the projected area above the waterline, and wind speed and direction. Air resistance is typically a smaller component, usually around 2% but can be up to 10% for large container ships with extensive superstructure.

    Total resistance (Rt): The total resistance experienced by the ship is the sum of frictional, residuary, and air resistance: Rt = Rf + Rr + Ra.

    Q4 (10 Marks) Ship Types & Design πŸ”₯ Repeated 4x

    (a) Draw and the mid ships section of an oil tanker with Double Hull & name cach part.

    (b) What is Bow Flare? Why is it so important in Bulk Carriers?

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    Part (a)

    Mid-ship section of Oil tanker:

    Part (b)

    Bow flare

    is the outward curvature of a ship's side shell above the waterline at the forward end.

    Importance of Bow Flare in Bulk Carriers:

    • Bow flare enhances the reserve buoyancy at the forward end of the vessel, which improves seaworthiness by helping the ship ride over waves more effectively, especially when pitching in rough seas.
    • By dispersing water away from the ship, the bow flare reduces the amount of water shipped onto the deck during heavy weather.
    • The wider forecastle deck created by the bow flare allows for the installation of essential machinery such as windlasses, mooring equipment, and other fittings.
    • The bow flare shields the hull plating from damage caused by the anchor when it is raised or lowered.
    • A well-designed bow flare can reduce water resistance, leading to increased speed and better fuel efficiency.
    Q5 (10 Marks) Ship Types & Design πŸ”₯ Repeated 6x

    (a) Considering the vessel as a compound beam define Bending moment shearing force. Which is the point of Maximum Bending Moment?

    (b) Sketch and Describe Hatch coaming of a large bulk carrier.

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    Part (a)

    Bending moment and shearing force, and the point of maximum bending moment.

    Considering the vessel as a compound beam (the hull girder), the shearing force at any section is the algebraic sum of the vertical forces (loads) to one side of the section, i.e. the net load (weight - buoyancy) acting on that part. The bending moment at any section is the algebraic sum of the moments of the loads to one side, i.e. the integral of the shearing force. The shearing force is the rate of change of the bending moment, and the bending moment is the integral of the shearing force. The point of maximum bending moment occurs where the shearing force is zero (where the shearing force changes sign), which is usually at or near midships for a ship in still water, and at the point where the net load changes sign. The maximum bending moment is the largest hogging or sagging moment, and the hull girder must be designed to withstand it.

    Part (b)

    Hatch coaming of a large bulk carrier.

    The hatch coaming is the vertical structure around the hatch opening that raises the hatch above the deck to prevent water entering and to provide strength. Sketch: the hatch opening is bounded by a vertical coaming plate (about 600-900 mm high for a bulk carrier) welded to the deck, with a top flange (or a horizontal stiffener) and vertical stiffeners (brackets) connecting the coaming to the deck. The coaming is made of thick plate and is stiffened to resist the loads of the hatch cover and the cargo, and to provide the longitudinal strength of the deck (the coaming acts as a longitudinal girder). The hatch cover sits on the coaming with a gasket and is secured by cleats. The coaming corners are rounded and reinforced to avoid stress concentrations. The coaming provides the watertight seal for the hatch and contributes to the longitudinal strength of the hull girder.

    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 9x

    (a) Describe how the force on the ship's bottom and the GM vary when grounding (6)

    (b) A ship of 8,000 tonnes displacement takes the ground on a sand bank on a falling tide at an even keel draft of 5.2 metres. KG 4.0 metres. The predicted depth of water over the sand bank at the following low water is 3.2 metres. Calculate the GM at this time assuming that the KM will then be 5.0 metres and that the mean TPC is 15 tonne (10)

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    Part (a)

    When a ship grounds, the forces on the bottom and the metacentric height (GM) change depending on the grounding's nature and the ship's position on the seabed.

    If the ship grounds on a level bottom:

    • A ground reaction force acts vertically upwards from the seabed. This force counteracts part of the ship's weight and alters the distribution of buoyancy forces.
    • The ship’s centre of gravity (G) may appear to rise relative to the metacentre (M) because the upward ground reaction reduces the buoyancy force acting on the underwater volume.
    • This virtual rise in G reduces GM, potentially leading to a list.
    • If the list becomes excessive and the righting moment is insufficient, the ship may capsize.

    If the ship grounds on a pinnacle:

    • The ship experiences two forces at the ship's bottom:
      • A downward force due to the ship’s weight.
      • An upward reaction force is concentrated on the pinnacle.
    • The resulting force between the grounding pressure and the ship’s centre of buoyancy shifts downward towards the pinnacle.
    • This situation is similar to when the ship's stern touches the keel block in a dry dock.
    • A virtual loss of GM occurs because the ship’s inclining moment may exceed the maximum righting lever.
    • If the inclining moment is too great, the ship may develop an excessive list or even capsize.

    (b) Given:

    $$Displacement,\:\Delta=8000\:tonnes$$

    $$TPC=15\:tonnes$$

    $$Initial\:Draught=5.2m$$

    $$Final\:Draught=3.2m$$

    $$Ship\:KG=4.0m$$

    $$KM=5.0m$$

    To find GM

    $$Uptrust,\:P=TPC\times Fall\:in\:water\:level\:\left(cm\right)$$

    $$=15\times\left(520-320\right)$$

    $$=15\times200$$

    $$P=3000\:tonnes$$

    To Find Virtual loss of GM:

    $$Virtual\:loss\:of\:GM_1=\frac{P\times KM}{\Delta}$$

    $$=\frac{3000\times5}{8000}$$

    $$=\frac{15000}{8000}$$

    $$GM_1=1.88m$$

    Actual KM = 5.0m (given)

    $$Virutal\:KM=Actual\:KM-Virtual\:loss\:of\:GM_1$$

    $$=5.0-1.88$$

    $$=3.12$$

    Similarly, Actual KG = 4.0m (given)

    $$New\:GM=Virtual\:KM-\:Actual\:KG$$

    $$=3.12-4.0$$

    $$=-0.88$$

    Q7 (10 Marks) Ship Stability πŸ”₯ Repeated 7x

    (a) List the precautions necessary before an inclining experiment is carried out. (6)

    (b) A box shaped vessel, 50 metres long x 10 metres wide, floats in salt water on an even keel at a draft of 4 metres. A center line longitudinal watertight bulkhead extends from end to end and for the full depth of the vessel. A compartment amidships on the starboard side is 15 metres long and contains cargo with permeability 30%. Calculate the list if this compartment is bilged. KG = 3 meters (10)

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    Part (a)

    Precautions necessary before an Inclining experiment:

    • The ship should be in a sheltered position, such as a gravity dock, to minimize external influences from wind or waves.
    • Mooring ropes should be slack to allow the ship to move freely without restrictions.
    • Only workers essential to the experiment should be present onboard to avoid unnecessary weight and movement.
    • All tanks must either be completely empty or pressed up tight to eliminate the free surface effect, which can adversely affect stability calculations.
    • Any loose weights on the ship must either be removed or properly secured to prevent unintended movement during the experiment.
    • Ensure that the pendulums are long and properly suspended from stable points, such as underneath a hatch, to provide accurate deflection readings.
    • The test masses should be evenly distributed and placed as far from the centerline as possible to maximize measurable deflections.
    • The experiment should be carried out in calm weather to avoid the effects of wind, current, or waves on the vessel's stability.
    Q8 (10 Marks) Ship Stability πŸ”₯ Repeated 11x

    (a) Define longitudinal center of gravity (LCG) and longitudinal center of buoyancy (LCB). (6)

    (b) A ship 120m long floats at draughts of 5.50m forward and 5.80m aft; MCT 1cm 80 tonne m, TPC 13, LCF 2.5m forward of midships. Calculate the new draughts when a mass of 110 tonne is added 24m aft of midships. (10)

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.

    LCF in fwd and trim by stern

    $$Bodily \space sinkage \space = \space {{mass \space added} \over TPC} \space = \space {{110} \over 13} \space$$

    $$=\:8.5\operatorname{\mathrm{cm}}\:=0.085m$$

    $$Trim \space = \space {{m \times d} \over MCT_{1cm}}$$

    $$ = \space {{110 \times (24 + 2.5)} \over 80}$$

    $$Trim=36.43\operatorname{cm}=0.364m\:$$

    Change in fwd draught:

    $$d_{F}=\frac{-t}{L}\left\lbrack\frac{L}{2}-LCF\right\rbrack$$

    $$=\frac{-36.43}{120}\left\lbrack\frac{120}{2}-2.5\right\rbrack$$

    $$=-17.45\operatorname{cm}=-0.1745m$$

    Change in Aft draught:

    $$d_{A}=\frac{+t}{L}\left\lbrack\frac{L}{2}+LCF\right\rbrack$$

    $$=\frac{+36.43}{120}\left\lbrack\frac{120}{2}+2.5\right\rbrack$$

    $$=+18.97\operatorname{cm}=0.189m$$

    New draughts:

    $$D_{F}=5.5+0.085-0.175=5.41m$$

    $$D_{A}=5.8+0.085+0.18=6.065m$$

    Q9 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    The speed of a ship is increased to 18% above normal for 7.5 hours, then reduced to 9% below normal for 10 hours. The speed is then reduced for the remainder of the day so that the consumption for the day is the normal amount. Find the percentage difference between the distance travelled in that day and the normal distance travelled per day.

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    Let normal speed be V and normal daily consumption be C. Consumption varies as the cube of speed, so C = k.V^3.

    Step 1 - Find the speed for the remainder of the day.

    For 7.5 hours the speed is 1.18V. Consumption in that period = k.(1.18V)^3 x 7.5/24 = k.V^3 x 1.643 x 0.3125 = 0.5134 k.V^3.

    For 10 hours the speed is 0.91V. Consumption = k.(0.91V)^3 x 10/24 = k.V^3 x 0.7536 x 0.4167 = 0.3140 k.V^3.

    Total consumption in first 17.5 hours = 0.5134 + 0.3140 = 0.8274 k.V^3.

    Remaining time in the day = 24 - 17.5 = 6.5 hours.

    For the day's total consumption to equal the normal amount k.V^3, the remaining consumption must be k.V^3 - 0.8274 k.V^3 = 0.1726 k.V^3.

    If the reduced speed is Vr, then k.Vr^3 x 6.5/24 = 0.1726 k.V^3.

    So Vr^3 = 0.1726 x 24/6.5 x V^3 = 0.6373 V^3.

    Vr = (0.6373)^(1/3) V = 0.8606 V.

    So the ship travels at 86.06% of normal speed for the last 6.5 hours.

    Step 2 - Find the distance travelled that day.

    Distance = speed x time.

    Normal distance per day = V x 24 = 24V.

    Actual distance = 1.18V x 7.5 + 0.91V x 10 + 0.8606V x 6.5

    = 8.85V + 9.10V + 5.594V = 23.544V.

    Step 3 - Percentage difference.

    Difference = 24V - 23.544V = 0.456V.

    Percentage difference = 0.456/24 x 100 = 1.9%.

    Answer: The distance travelled that day is 1.9% less than the normal distance per day.

    Q10 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 8x

    With reference to fixed pitch propellers:

    (a) Explain Propeller Slip and Propeller Thrust.

    (b) The shaft power of a ship is 3000 KW, the ship's speed V is 13.2 knot. Propeller rps is 1.27. propeller pitch is 5.5m and the speed of advance is 11 Knots. Find:

    (i) Real Slip

    (ii) Wake fraction

    (iii) Propeller thrust, when its efficiency, Ξ· = 70%

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    Part (a)

    Slip

    is the difference between the theoretical distance the propeller should travel in one revolution and the actual distance the vessel travels.

    $$Slip\:\left(\%\right)\:=\:\left(\frac{Engine\:distance\:-\:Ship^{\prime}s\:distance}{Engine\:distance}\right)\:\times100$$

    Where Engine distance = no. of propeller revolutions * propeller pitch (usually calculated over a 24-hour period)

    Propeller thrust: it is the force exerted by the propeller to move the vessel ahead and given by

    $$Thrust\:\left(T\right)\:=\:\rho\times A\times P^2\times n^2\times S$$

    Where,

    ρ - Density

    A - Area

    P - Pitch

    n - Revolution per second

    S - Slip

    Q1 (10 Marks) Hull Construction πŸ”₯ Repeated 3x

    Discuss the need for adequate support of engine room gantry cranes, detailing the following.

    (a) Sketch section through the engine room casing showing how the crane is supported by the ship structure.

    (b) State what restricts the forward and aft limits of the crane and what is fitted to prevent the crane damaging the forward and aft bulkheads or casing.

    (c) State the Second Engineer's responsibilities for the engine room gantry crane.

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    Part (a)

    Engine room gantry crane:

    Part (b)

    Forward and aft limits of crane:

    • Two limit switches are installed to automatically cut off the motor power when the crane reaches its forward or aft limits.
    • Mechanical Stoppers are fitted at the forward and aft ends of the crane rails. They act as a failsafe to prevent the crane from damaging the forward and aft bulkheads or casing if the limit switches fail. Rubber or spring buffers may also be added to absorb impact forces and reduce the risk of damage.
    Part (c)

    Second Engineer's responsibilities for the engine room gantry crane.

    • Inspect the crane, rails, mechanical stoppers, and support brackets for wear or damage.
    • Ensure limit switches are functional and properly aligned.
    • Check the condition of the gear case oil and renew it as required.
    • Inspect and maintain wire ropes and safety latches.
    • Ensure the Safe Working Load (SWL) is clearly marked and adhered to.
    • Verify the validity of the crane's test certificate.
    • Perform motor overhauls and measure insulation resistance.
    • Conduct brake tests and emergency stop function checks.

    Provide instructions to Junior Engineers:

    • Not exceeding the SWL of the crane.
    • Ensuring mechanical locks are removed before operating the crane.
    • Using proper Personal Protective Equipment (PPE).
    • Prohibiting personnel from standing beneath the crane during operation.
    Q2 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 8x

    Vessel has gone through very heavy weather. On arrival at safe anchorage, you are conducting your inspection to determine damages to hull.

    (a) List the areas you will inspect.

    (b) List your findings of any significance.

    (c) Write a report to company suggesting repairs if any

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    Part (a)

    Areas to Inspect After Heavy Weather

    Upon arrival at safe anchorage, a thorough inspection of the vessel shall be carried out, both externally and internally, to assess any weather-related damage. Key areas include:

    1. Hull and Main Deck

    • Shell plating along bow, midship, and stern sections (especially forepeak and flare areas).
    • Boot-top and bilge areas for dents or deformation.
    • Deck plating for buckling or cracked welds.
    • Bulwarks, rails, fairleads, chocks, and mooring fittings.

    2. Forecastle and Forward Structure

    • Bosun store and chain locker for water ingress.
    • Windlass foundation, anchors, and hawse pipes for deformation or cracks.
    • Forepeak tank for leakage or pressure damage.

    3. Cargo Holds / Tanks

    • Inspect for structural deformation, loose frames, or fractured stiffeners.
    • Check tank top plating and bilges for leakage.
    • Check watertight doors, gaskets, and vents.

    4. Superstructure and Deck Fittings

    • Bridge wings, radar mast, funnel, vents, and deckhouses for cracks or loose fittings.
    • Lifeboat davits, securing arrangements, and deck cranes.

    5. Underwater and Machinery Spaces

    • Rudder, propeller, and stern tube seals (via steering gear tests).
    • Sea chest gratings and overboard discharges.
    • Engine room bilges for any seawater ingress.

    Part (b)

    Typical Findings of Significance

    • Dents and indentations on shell plating at bow and forward port side due to heavy slamming.
    • Paint coating and corrosion protection partly peeled off near waterline and forepeak area.
    • Loose fairlead bolts and one cracked weld on starboard bulwark stanchion.
    • Deformed ventilator head on forecastle deck.
    • Minor leakage observed in forepeak tank during sounding check.
    • Bridge wing railing bent, likely from green sea impact.
    • Anchor chain links twisted and worn.
    • Lifeboat gripes loosened, requiring tightening and inspection.
    • No flooding reported; watertight integrity maintained overall.

    Part (c)

    Report to Company – Heavy Weather Damage Inspection

    To: Superintendent / Technical Department

    From: Name / Rank

    Subject: Heavy Weather Damage Inspection Report

    Date: [Insert date]

    Vessel: [Insert vessel name]

    Summary

    The vessel experienced very heavy weather en route from [Port A] to [Port B], with significant pitching and rolling in seas up to [X] meters. On arrival at safe anchorage, a complete inspection of the hull and deck was carried out.

    Findings

    • Bow and forecastle plating show minor dents, with no breach of watertight integrity.
    • Bulwark stanchion (starboard side) cracked at welded joint – requires repair.
    • Fairlead foundation bolts loosened – retightening and NDT inspection recommended.
    • Ventilator head on forecastle deformed – replacement advised.
    • Forepeak tank shows trace leakage at forward bulkhead – to be pressure tested.
    • Paint coating damage and corrosion exposure on bow area – recoating required.
    • Bridge wing railing bent – to be straightened or renewed.
    • All other structures and machinery appear satisfactory after testing.

    Recommendations

    1. Conduct minor steel renewal and welding repairs at the next port with repair facilities.
    2. Carry out thickness measurements and NDT on affected hull areas.
    3. Recoat damaged paint areas to prevent corrosion.
    4. Replace deformed ventilator head and bent railing.
    5. Inspect anchor and chain for elongation; renew worn links.
    6. Pressure test forepeak tank after repairs.
    7. Submit class surveyor report if deemed necessary by the superintendent.

    Conclusion

    The vessel remains seaworthy but requires prompt attention to minor structural and fitting damages before the next voyage. Preventive maintenance and weather routing measures should be reviewed for future passages.

    Signed:

    Name / Rank

    Signature

    Q3 (10 Marks) General πŸ”₯ Repeated 14x

    Explain how the period of roll varies with

    (a) The amplitude of roll.

    (b) the radius of gyration.

    (c) The initial meracentric height.

    (d) The location of masses in the ship

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    The period of roll Tr of a ship is determined by the formula:

    $$T_{r}=\frac{2\pi K}{\sqrt{g\times GM}}$$

    where,

    • K is the radius of gyration (mass moment of inertia)
    • g is the acceleration due to gravity, and
    • GM is the metacentric height.
    Part (a)

    Amplitude of Roll:

    • The amplitude of roll, or the maximum angle of heel, does not affect the period of roll. While a larger amplitude implies a greater heeling moment and faster roll speed, the time taken to complete one full roll cycle remains constant for a given metacentric height and radius of gyration.
    Part (b)

    Radius of Gyration (K):

    • The period of roll is directly proportional to the radius of gyration. A larger radius of gyration (indicating a greater distribution of mass further from the ship's centre of rotation) leads to a longer period of roll. Conversely, a smaller radius of gyration (mass concentrated closer to the centre) results in a shorter period. The distribution of cargo significantly impacts K; cargo concentrated centrally minimises K and the roll period, while dispersed cargo maximises K and the roll period.
    Part (c)

    Initial Metacentric Height (GM):

    • The period of roll is inversely proportional to the square root of the metacentric height (GM). A larger GM (a stiffer ship) leads to a shorter roll period, as the ship quickly returns to its upright position. A smaller GM (a tender ship) results in a longer roll period, with slower return to the upright.
    Part (d)

    Location of Masses in the Ship:

    The location of masses in the ship will effect the GM & K. So the period of roll will be affected.

    • If masses are at bottom, G moves down, GM ↑, period of roll ↓.
    • If masses are at top, G moves up, GM ↓, period of roll ↑.
    • If masses are concentrated at centre, K ↓, period of roll ↓.
    • If masses are away from centre, K ↑, period of roll ↑.
    Q4 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 13x

    Describe the relationship between frictional resistance and

    (a) Ship's speed

    (b) The wetted area

    (c) The surface roughness

    (d) The length of the vessel

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    Frictional resistance arises due to the interaction between the ship's hull and the water as the vessel moves. This resistance is caused by eddying water adjacent to the hull that is drawn along with the ship. The frictional resistance is given by:

    $$R_{f\:=\:}f\:.\:s\:.\:v^{n}$$

    Where:

    • f = Coefficient of friction
    • s = Wetted surface area
    • v = Ship's speed in knots
    • n = Constant (1.82)
    Part (a)

    Ship Speed (v):

    • Frictional resistance increases with increasing ship speed. The relationship is not perfectly linear, but it's approximately described by the formula Rf = fsv^n (Rf ∝ v^n), where 'n' is a constant (approximately 1.82). This means that a small increase in speed results in a larger increase in frictional resistance.
    Part (b)

    Wetted Area (s):

    • Frictional resistance is directly proportional to the wetted surface area (Rf ∝ s). A larger wetted area (the part of the hull submerged in water) leads to greater frictional resistance. The wetted area increases with the ship's draught (depth in the water).
    Part (c)

    Surface Roughness:

    • Increased surface roughness increases frictional resistance. Roughness can be caused by fouling (marine growth), corrosion, or poor hull maintenance. Regular cleaning and maintenance of the hull surface help to minimise roughness and reduce resistance.
    Part (d)

    Length of the Vessel

    • Frictional resistance is influenced by the vessel's length. As the length increases, the wetted surface area grows, and the coefficient of friction (f) changes, leading to an increase in Rf. Longer vessels experience higher frictional resistance, but streamlined designs can mitigate the effect to some extent.
    Q5 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 4x

    If a ship is seriously damaged under water in way of a large fuel oil side bunker tank what is the immediate effect and what may ultimately happen? What features in the ship would enhance safety?

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    If a ship is seriously damaged underwater in the area of a large fuel oil side bunker tank, it will cause the following:

    Bilging or Flooding of the Compartment:

    • Water will enter the damaged bunker tank, reaching the draught level.
    • The rate of flooding depends on the size of the breach and the water pressure at the depth of the damage.

    Oil Leakage into the Sea:

    • If the tank contains fuel oil, oil will begin to leak out, causing pollution.
    • The extent of leakage depends on the tank’s contents (empty, half-full, or full) and the location of the damage.

    List and Trim of the Vessel:

    • The ingress of water and loss of oil will create an imbalance, causing the vessel to list or trim.

    If corrective actions are not taken, uncontrolled flooding and loss of stability could lead to capsizing or sinking of the vessel.

    Features in the Ship to Enhance Safety:

    • Small Bunker Tank Sizes reduces the risk of extensive oil spillage and loss of stability.
    • Connectivity to transfer pumps allows the transfer of oil from the damaged tank to an empty tank, minimizing oil spillage and counteracting the loss of stability.
    • The tank’s size and location are designed to limit the effects of flooding, as per damage stability regulations.
    • Properly positioned transverse and longitudinal bulkheads enhance the subdivision factor, limiting water ingress to the damaged tank.
    • Ship’s Ballast System allows corrective ballasting to counteract the list or trim caused by the ingress of water.
    • Watertight Doors and Hatches prevent water from spreading to adjacent compartments.

    Recommended Immediate Actions by Crew:

    For Empty Tanks:

    • Quickly seal off the damaged tank by shutting all valves and isolating it from the transfer system to prevent water ingress into other parts of the vessel.

    For Half-Empty or Full Tanks:

    • Initiate oil transfer to another empty tank to reduce oil leakage and stabilize the ship.
    • Monitor the water ingress and ensure the tank is filled to the draught level with seawater if necessary, using ballast to correct the list.
    Q6 (10 Marks) Ship Resistance & Propulsion

    The following data applies to a ship operating on a particular voyage with a propeller of 6m diameter having a pitch ratio of 0.95.

    Propeller speed = 1.8rev/sec

    Real slip = 34%

    Apparent slip = 7%

    shaft power = 10000 kw

    Specitic fuel consumption = 0.22 kg/kw hr

    Caculate each of the tollowin:

    (a) The ship speed in knots (3)

    (b) The Taylor wake fraction (4)

    (c) The reduced speed at which the ship should travel in order to halve the voyage consumption (2)

    (d) The voyage consumption if the vovage takes 3 days longer at the reduced speed (4)

    (e) The amount of fuel required for the voyage at the reduced speed (3)

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    Given: propeller diameter D = 6 m, pitch ratio = 0.95, so pitch P = 0.95 x 6 = 5.7 m.

    Propeller speed = 1.8 rev/s.

    Real slip = 34% = 0.34, Apparent slip = 7% = 0.07.

    Shaft power = 10000 kW, SFC = 0.22 kg/kW-hr.

    Part (a)

    Ship speed in knots.

    Pitch speed (theoretical speed of advance) = P x rev/s = 5.7 x 1.8 = 10.26 m/s.

    Apparent slip = (Pitch speed - V)/Pitch speed, where V is the ship speed.

    So V = Pitch speed x (1 - apparent slip) = 10.26 x (1 - 0.07) = 10.26 x 0.93 = 9.542 m/s.

    Ship speed in knots = 9.542 x 3600/1852 = 34351/1852 = 18.55 knots.

    Answer: Ship speed = 18.55 knots.

    Part (b)

    Taylor wake fraction.

    Real slip = (Pitch speed - Va)/Pitch speed, where Va is the speed of advance.

    Va = Pitch speed x (1 - real slip) = 10.26 x (1 - 0.34) = 10.26 x 0.66 = 6.772 m/s.

    Taylor wake fraction w = (V - Va)/V = (9.542 - 6.772)/9.542 = 2.770/9.542 = 0.2903.

    Answer: Taylor wake fraction = 0.290 (29%).

    Part (c)

    Reduced speed to halve the voyage consumption.

    Consumption varies as the cube of speed. To halve the consumption, the speed must be reduced by the cube root of 0.5.

    Reduced speed = V x (0.5)^(1/3) = 18.55 x 0.7937 = 14.72 knots.

    Answer: Reduced speed = 14.72 knots.

    Part (d)

    Voyage consumption if the voyage takes 3 days longer at reduced speed.

    At normal speed 18.55 knots, the daily consumption = shaft power x 24 x SFC = 10000 x 24 x 0.22 = 52800 kg/day = 52.8 tonne/day.

    At reduced speed, consumption per day = 52.8 x (0.5) = 26.4 tonne/day (since speed cubed halved).

    Let the normal voyage time be T days. Distance = 18.55 x 24 x T (nautical miles).

    At reduced speed 14.72 knots, the time taken = Distance/(14.72 x 24) = (18.55 x 24 x T)/(14.72 x 24) = 18.55/14.72 x T = 1.2602 T days.

    The voyage takes 3 days longer, so 1.2602 T = T + 3, giving 0.2602 T = 3, T = 11.53 days.

    Voyage consumption at reduced speed = 26.4 x 1.2602 x 11.53 = 26.4 x 14.53 = 383.6 tonne.

    Answer: Voyage consumption at reduced speed = 383.6 tonne.

    Part (e)

    Amount of fuel required for the voyage at reduced speed.

    This is the same as (d), the total fuel for the voyage at reduced speed.

    Answer: 383.6 tonne of fuel required for the voyage at reduced speed.

    Q7 (10 Marks) Ship Stability

    A ship of length 120m displaces 11750 tonne when floating in sea water of density 1025 kg/m3. The centre of gravity is 2m above the centre of buovanoy and the waterplane is defined by the following equidistant half breadths given in table below:

    Section: AP 1 2 3 4 5 6 7 FP

    Half-breadth (m) 3.3 6.8 7.6 8.1 8.1 8.0 6.6 2.8 0

    Calculate EACH of the following:

    (a) The area of the waterplane (3)

    (b) The position of the centroid of the waterplane from midships (3)

    (c) The second moment of area of the waterplane about a transverse axis through the centroid (5)

    (d) The moment to change trim one centimetre (MCT1cm) (5)

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    Ship length L = 120 m, displacement = 11750 tonne, sea water density 1025 kg/m3.

    KG - KB = 2 m (centre of gravity 2 m above centre of buoyancy).

    Half-breadths at sections AP(0),1,2,3,4,5,6,7,FP(8):

    Section: 0 1 2 3 4 5 6 7 8

    Half-breadth (m): 3.3 6.8 7.6 8.1 8.1 8.0 6.6 2.8 0

    Number of intervals n = 8, common interval h = 120/8 = 15 m.

    Part (a)

    Area of the waterplane.

    Simpson's First Rule, multipliers 1,4,2,4,2,4,2,4,1:

    Sum = 1x3.3 + 4x6.8 + 2x7.6 + 4x8.1 + 2x8.1 + 4x8.0 + 2x6.6 + 4x2.8 + 1x0

    = 3.3 + 27.2 + 15.2 + 32.4 + 16.2 + 32.0 + 13.2 + 11.2 + 0 = 150.7.

    Half waterplane area = (h/3) x sum = (15/3) x 150.7 = 5 x 150.7 = 753.5 m2.

    Full waterplane area = 2 x 753.5 = 1507 m2.

    Answer: Waterplane area = 1507 m2.

    Part (b)

    Position of the centroid of the waterplane from midships.

    Midship is at station 4. First moment about station 4:

    Distances in intervals: station 0 = -4, 1 = -3, 2 = -2, 3 = -1, 4 = 0, 5 = +1, 6 = +2, 7 = +3, 8 = +4.

    Moment = 1x3.3x(-4) + 4x6.8x(-3) + 2x7.6x(-2) + 4x8.1x(-1) + 2x8.1x0 + 4x8.0x(+1) + 2x6.6x(+2) + 4x2.8x(+3) + 1x0x(+4)

    = -13.2 - 81.6 - 30.4 - 32.4 + 0 + 32.0 + 26.4 + 33.6 + 0 = -65.6.

    Distance of centroid from midship = (moment/sum) x h = (-65.6/150.7) x 15 = -0.4353 x 15 = -6.53 m.

    Negative means forward of midship.

    Answer: Centroid (LCF) is 6.53 m forward of midship.

    Part (c)

    Second moment of area of the waterplane about a transverse axis through the centroid.

    Second moment about midship (station 4) using the product of (multiplier x half-breadth x distance^2):

    Moment2 = 1x3.3x16 + 4x6.8x9 + 2x7.6x4 + 4x8.1x1 + 2x8.1x0 + 4x8.0x1 + 2x6.6x4 + 4x2.8x9 + 1x0x16

    = 52.8 + 244.8 + 60.8 + 32.4 + 0 + 32.0 + 52.8 + 100.8 + 0 = 576.4.

    Second moment of half waterplane about midship = (1/3) x h^3 x sum = (1/3) x 15^3 x 576.4 = (1/3) x 3375 x 576.4 = 1125 x 576.4 = 648450 m4.

    Full waterplane second moment about midship = 2 x 648450 = 1296900 m4.

    Using the parallel axis theorem to transfer to the centroid:

    I about centroid = I about midship - A x (distance)^2

    = 1296900 - 1507 x (6.53)^2 = 1296900 - 1507 x 42.64 = 1296900 - 64258 = 1232642 m4.

    Answer: Second moment of area about the transverse axis through the centroid = 1.233 x 10^6 m4.

    Part (d)

    MCT 1 cm.

    MCT 1 cm = (displacement x GML)/(100 x L), where GML is the longitudinal metacentric height.

    GML = KML - KG. KML = KB + BML.

    BML = I / Volume of displacement.

    Volume of displacement = 11750/1.025 = 11463.4 m3.

    BML = 1232642/11463.4 = 107.53 m.

    KB is not given directly, but KG - KB = 2 m. We need KB.

    Using the waterplane and displacement, KB can be estimated, but a standard approach:

    KML = KB + BML. GML = KML - KG = (KB + BML) - (KB + 2) = BML - 2 = 107.53 - 2 = 105.53 m.

    MCT 1 cm = (11750 x 105.53)/(100 x 120) = 1240000/12000 = 103.3 tonne-m per cm.

    Answer: MCT 1 cm = 103.3 tonne-m per cm.

    Q8 (10 Marks) Ship Stability

    A box shaped vessel is 80m long, 12m wide and floats at a draft of 4m. A full width midship compartment 15m long is bilged and this results in the draft increasing to 4.5m. Calculate each of the tollowing

    (a) The permeability of the compartment (4)

    (b) The change in metacentric height due to bilging (12)

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    Box-shaped vessel: length L = 80 m, breadth B = 12 m, draft d = 4 m.

    A full-width midship compartment 15 m long is bilged, and the draft increases to 4.5 m.

    Part (a)

    Permeability of the compartment.

    When a compartment is bilged, the increase in draft is due to the loss of buoyancy of the compartment.

    The volume of lost buoyancy = the volume of water entering the compartment = (permeability x compartment volume).

    The increase in draft causes an increase in displacement equal to the weight of water entering.

    Increase in draft = 4.5 - 4.0 = 0.5 m.

    The increase in displacement = L x B x increase in draft x density = 80 x 12 x 0.5 x 1.025 = 492 tonne.

    This equals the weight of water entering the compartment.

    Weight of water entering = permeability x (compartment volume) x density = permeability x (15 x 12 x 4) x 1.025.

    So 492 = permeability x 720 x 1.025 = permeability x 738.

    Permeability = 492/738 = 0.667.

    Answer: Permeability = 0.667 (66.7%).

    Part (b)

    Change in metacentric height due to bilging.

    The bilged compartment is full width, so the waterplane area is reduced by the area of the compartment at the waterline.

    Original waterplane area = L x B = 80 x 12 = 960 m2.

    The compartment is full width (12 m) and 15 m long, so the lost waterplane area = 15 x 12 = 180 m2.

    Effective waterplane area after bilging = 960 - 180 = 780 m2.

    Original displacement = L x B x d x density = 80 x 12 x 4 x 1.025 = 3936 tonne.

    Original KB = d/2 = 4/2 = 2 m.

    Original BM = I/V = (L x B^3/12)/(L x B x d) = B^2/(12 x d) = 144/(12 x 4) = 3 m.

    Original KM = KB + BM = 2 + 3 = 5 m.

    After bilging, the draft is 4.5 m.

    New KB = 4.5/2 = 2.25 m.

    New BM = I/V. The second moment of area of the effective waterplane = (L x B^3/12) - (15 x 12^3/12) = (80 x 1728/12) - (15 x 1728/12) = (80 x 144) - (15 x 144) = 11520 - 2160 = 9360 m4.

    New volume of displacement = 3936/1.025 = 3840 m3 (unchanged, since the ship's weight is unchanged).

    New BM = 9360/3840 = 2.4375 m.

    New KM = new KB + new BM = 2.25 + 2.4375 = 4.6875 m.

    The change in metacentric height = new KM - original KM = 4.6875 - 5.0 = -0.3125 m.

    Answer: The metacentric height is reduced by 0.3125 m (GM decreases by 0.3125 m).

    Q9 (10 Marks) Ship Stability πŸ”₯ Repeated 4x

    A ship 100 m long floats at a draft of 6m and in this condition the immersed cross-sectional areas are as given in tables below. The equivalent base area (Ab) Is required because of the fineness of the bolton shell

    Calculate each of the following.

    (a) The equivalent base area value Ab

    (b) The longitudinal position of the centre of buoyancy from midships

    (c) The vertical position of the centre of buoyancy above the base

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    Table for question:

    Section

    AP

    1

    2

    3

    4

    5

    FP

    Immersed cross sectional area

    12

    30

    65

    80

    70

    50

    0

    Draught (m)

    0

    0.6

    1.2

    2.4

    3.6

    4.8

    6.0

    Water plane area (m2)

    Ab

    560

    720

    880

    940

    1000

    1030

    Solution:

    Section

    CSA

    SM

    F volume

    Lever

    F moment

    AP

    12

    1

    12

    -3h

    -36h

    1

    30

    4

    120

    -2h

    -240h

    2

    65

    2

    130

    -1h

    -130h

    3

    80

    4

    320

    0

    0

    4

    70

    2

    140

    +1h

    140h

    5

    50

    4

    200

    +2h

    400h

    FP

    0

    1

    0

    +3h

    0

    Ξ£Fvol. = 992

    Ξ£Fmom. =134h

    $$Common\:interval\:\left(h\right)=\frac{L}{6}=\frac{100}{6}$$

    $$h=16.66m$$

    $$\sum F_{movement}=134\:\times16.66m^4$$

    $$\sum F_{moment}=2232.44m^4$$

    $$\nabla=\frac{h}{3}\times\sum F_{volume}$$

    $$\nabla=\frac{16.66}{3}\times992$$

    $$\nabla=5120.17m^3$$

    Q10 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 2x

    A ship 150 m in length, 24 m breadth, displaces 25000 tonne when floating at a draft of 9 m in, sea water of density 1025 kg/m3. The ship's propeller has a diameter of 5.8 m, a pitch ratio of 0.9 and a blade area ratio of 0.45. With the propeller operating at 2 revs/sec, the following results were

    recorded:

    Apparent slip = 0.06

    Thrust power = 3800 Kw

    Propeller efficiency = 64%

    The taylor wake fraction Wt = 0.5Cb-0.05

    Calculate each of the following for the above condition:

    (a) The ship's speed

    (b) The real slip ratio

    (c) The thrust per unit are of blade surface

    (d) The torque delivered to the propeller

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    Given:

    $$L \space = \space 150m$$

    $$B \space = \space 24m$$

    $$D \space = \space 9m$$

    $$\Delta=25000t$$

    $$d \space = \space 5.8m$$

    $$p \space = \space 0.9m$$

    $$BAR \space = \space 0.45m$$

    $$n \space = \space 2 \space rev/sec$$

    $$Apparent \space slip \space = \space 0.06m$$

    $$T_p \space = \space 3800kw$$

    $$Ξ·_{prop} \space = \space 64\%$$

    $$W_f \space = \space 0.5C_b - 0.05$$

    $$p\space=\space{{P}\over d}\space$$

    $$0.9 \space = \space {{P} \over 5.8}$$

    $$Pitch \space = \space 5.22m$$

    $$V_T \space = \space {{P \times N \times 3600} \over 1852}$$

    $$V_T P \space = \space 20.29 \space knots$$

    $$App. \space slip \space = \space {{V_T - V} \over V_T}$$

    $$0.06\space=\space{{20.29-V}\over20.29}$$

    $$Ship's \space speed \space (V) \space = \space 19.07 \space knots $$

    $$C_{b}=\frac{\Delta}{L\times B\times D\times\rho}$$

    $$=\frac{25000}{150\times24\times9\times1.025}$$

    $$C_{b}=0.752$$

    $$given, \space W_f \space = \space 0.5C_b - 0.05$$

    $$W_{f}=0.5\times0.752-0.05$$

    $$W_{f}=0.326$$

    $$W_F \space = \space {{V - V_a} \over V}$$

    $$0.326 \space = \space {{19.07 - V_a} \over 19.07}$$

    $$V_a \space = \space 12.85Knots$$

    $$Real\space slip\space=\space{{V_T - V_a} \over V_T}\space\space{}$$

    $$=\frac{20.29-12.85}{20.29}$$

    $$Real \space slip \space = \space 0.367 $$

    $$Thrust \space power \space (T_p) \space = \space d_p \times Ξ·_{prop}$$

    $$3800=d_{p}\times0.64$$

    $$d_{p}=5937.5kw$$

    $$Thrust \space power \space T_p \space = \space Thrust \times V_a $$

    $$3800 \space = \space Thrust \times 12.85 \times {{1852} \over 3600}$$

    $$Thrust=574.83kw$$

    $$Blade \space area \space = \space {{\pi} \over 4}d^2 \times BAR $$

    $$=\space{{\pi}\over4}\times5.8^2\times0.45$$

    $$Blade \space area \space = \space 11.88m^2$$

    $$Thrust \space per \space area \space = \space {{574.82} \over 11.88} \space = \space 48.34 \space KN/m^2$$

    Q1 (10 Marks) Ship Stability πŸ”₯ Repeated 14x

    Explain how the period of roll varies with

    (a) The amplitude of roll.

    (b) The radius of gyration.

    (c) The initial metacentric height.

    (d) The location of masses in the ship

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    The period of roll Tr of a ship is determined by the formula:

    $$T_{r}=\frac{2\pi K}{\sqrt{g\times GM}}$$

    where,

    • K is the radius of gyration (mass moment of inertia)
    • g is the acceleration due to gravity, and
    • GM is the metacentric height.
    Part (a)

    Amplitude of Roll:

    • The amplitude of roll, or the maximum angle of heel, does not affect the period of roll. While a larger amplitude implies a greater heeling moment and faster roll speed, the time taken to complete one full roll cycle remains constant for a given metacentric height and radius of gyration.
    Part (b)

    Radius of Gyration (K):

    • The period of roll is directly proportional to the radius of gyration. A larger radius of gyration (indicating a greater distribution of mass further from the ship's centre of rotation) leads to a longer period of roll. Conversely, a smaller radius of gyration (mass concentrated closer to the centre) results in a shorter period. The distribution of cargo significantly impacts K; cargo concentrated centrally minimises K and the roll period, while dispersed cargo maximises K and the roll period.
    Part (c)

    Initial Metacentric Height (GM):

    • The period of roll is inversely proportional to the square root of the metacentric height (GM). A larger GM (a stiffer ship) leads to a shorter roll period, as the ship quickly returns to its upright position. A smaller GM (a tender ship) results in a longer roll period, with slower return to the upright.
    Part (d)

    Location of Masses in the Ship:

    The location of masses in the ship will effect the GM & K. So the period of roll will be affected.

    • If masses are at bottom, G moves down, GM ↑, period of roll ↓.
    • If masses are at top, G moves up, GM ↓, period of roll ↑.
    • If masses are concentrated at centre, K ↓, period of roll ↓.
    • If masses are away from centre, K ↑, period of roll ↑.
    Q2 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 8x

    Vessel has gone through very heavy weather on arrival at safe anchorage, you are conducting your inspection to determine damage of hull.

    (a) List the areas you will inspect.

    (b) List your findings of any significance.

    (c) Write a report to company suggesting repairs if any

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    Part (a)

    Areas to Inspect After Heavy Weather

    Upon arrival at safe anchorage, a thorough inspection of the vessel shall be carried out, both externally and internally, to assess any weather-related damage. Key areas include:

    1. Hull and Main Deck

    • Shell plating along bow, midship, and stern sections (especially forepeak and flare areas).
    • Boot-top and bilge areas for dents or deformation.
    • Deck plating for buckling or cracked welds.
    • Bulwarks, rails, fairleads, chocks, and mooring fittings.

    2. Forecastle and Forward Structure

    • Bosun store and chain locker for water ingress.
    • Windlass foundation, anchors, and hawse pipes for deformation or cracks.
    • Forepeak tank for leakage or pressure damage.

    3. Cargo Holds / Tanks

    • Inspect for structural deformation, loose frames, or fractured stiffeners.
    • Check tank top plating and bilges for leakage.
    • Check watertight doors, gaskets, and vents.

    4. Superstructure and Deck Fittings

    • Bridge wings, radar mast, funnel, vents, and deckhouses for cracks or loose fittings.
    • Lifeboat davits, securing arrangements, and deck cranes.

    5. Underwater and Machinery Spaces

    • Rudder, propeller, and stern tube seals (via steering gear tests).
    • Sea chest gratings and overboard discharges.
    • Engine room bilges for any seawater ingress.

    Part (b)

    Typical Findings of Significance

    • Dents and indentations on shell plating at bow and forward port side due to heavy slamming.
    • Paint coating and corrosion protection partly peeled off near waterline and forepeak area.
    • Loose fairlead bolts and one cracked weld on starboard bulwark stanchion.
    • Deformed ventilator head on forecastle deck.
    • Minor leakage observed in forepeak tank during sounding check.
    • Bridge wing railing bent, likely from green sea impact.
    • Anchor chain links twisted and worn.
    • Lifeboat gripes loosened, requiring tightening and inspection.
    • No flooding reported; watertight integrity maintained overall.

    Part (c)

    Report to Company – Heavy Weather Damage Inspection

    To: Superintendent / Technical Department

    From: Name / Rank

    Subject: Heavy Weather Damage Inspection Report

    Date: [Insert date]

    Vessel: [Insert vessel name]

    Summary

    The vessel experienced very heavy weather en route from [Port A] to [Port B], with significant pitching and rolling in seas up to [X] meters. On arrival at safe anchorage, a complete inspection of the hull and deck was carried out.

    Findings

    • Bow and forecastle plating show minor dents, with no breach of watertight integrity.
    • Bulwark stanchion (starboard side) cracked at welded joint – requires repair.
    • Fairlead foundation bolts loosened – retightening and NDT inspection recommended.
    • Ventilator head on forecastle deformed – replacement advised.
    • Forepeak tank shows trace leakage at forward bulkhead – to be pressure tested.
    • Paint coating damage and corrosion exposure on bow area – recoating required.
    • Bridge wing railing bent – to be straightened or renewed.
    • All other structures and machinery appear satisfactory after testing.

    Recommendations

    1. Conduct minor steel renewal and welding repairs at the next port with repair facilities.
    2. Carry out thickness measurements and NDT on affected hull areas.
    3. Recoat damaged paint areas to prevent corrosion.
    4. Replace deformed ventilator head and bent railing.
    5. Inspect anchor and chain for elongation; renew worn links.
    6. Pressure test forepeak tank after repairs.
    7. Submit class surveyor report if deemed necessary by the superintendent.

    Conclusion

    The vessel remains seaworthy but requires prompt attention to minor structural and fitting damages before the next voyage. Preventive maintenance and weather routing measures should be reviewed for future passages.

    Signed:

    Name / Rank

    Signature

    Q3 (10 Marks) Hull Construction πŸ”₯ Repeated 10x

    Describe a method for the attachment of bilge keels. State THREE reasons for not extending bilge keels for the entire length of the vessel. Explain TWO principles of roll damping that bilge keels exploit.

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    Part (a)

    Method of Bilge keel attachment to the hull:

    Bilge keels are fitted port and starboard at the turn of the bilge. They do not extend outside the lines of the side and bottom shell. The bilge keels are attached directly in line with an internal stiffening member such as a girder or longitudinal. The bilge keel comprises a flat bar doubler welded directly to the shell, and an offset bulb plate (OBP) with 'scallops' cut in it is welded to the flat bar doubler. The ends of the bilge keels are tapered (minimum 3 in 1) and will end in line with transverse internal stiffening, such as a frame. Using a hull doubler protects the hull in case of damage to the keel, as the crack would not extend into the hull.

    Part (b)

    Bilge keels are not fitted for the full length of the vessel because:

    • The 'lever' to the ship's axis of rotation is reduced at the ends;
    • The ineffectiveness, the closer the bilge keel is to the rolling centre, increased resistance and more likelihood of damage.
    • The hydrodynamic effect would cause a large increase in resistance and fuel consumption;
    • At the aft end, the boundary layer is much thicker, and since the keel would not project through, it would have a much reduced effect.
    Part (c)

    Two principles of roll damping exploited by Bilge keels:

    Increased Roll Period:

    • Bilge keels increase the ship's roll period (the time it takes for the vessel to complete one roll). This is achieved by increasing the moment of inertia (K) of the vessel. A longer roll period means the vessel rolls more slowly, thus reducing the amplitude of the roll.

    $$T_{r}=2\pi\:\frac{k}{\sqrt{g\times GM}}$$

    Where:

    • g: acceleration due to gravity
    • GM: metacentric height
    • k: mass moment of inertia

    By increasing k, the ship's stability during roll improves.

    Hydrodynamic Resistance:

    • As the vessel rolls, the bilge keels move through the water, creating pressure differences. Water pressure on one side of the keel opposes the rolling motion, providing a damping effect. This is due to the interaction between the bilge keel and the water, creating hydrodynamic forces that counteract the rolling motion.
    Q4 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 9x

    With reference to Underwater Inspection in lieu of Dry docking(UWILD):

    (a) Explain in detail, how an underwater survey is carried out.

    (b) State the requirements to be fulfilled before an underwater survey is acceptable to the surveying authority.

    (c) Construct a list of the items in order of importance that the underwater survey authority should include.

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    (a) An underwater survey, also known as Underwater Inspection in Lieu of Dry Docking (UWILD), involves a systematic and detailed examination of a vessel’s hull and underwater components while the ship remains afloat. The procedure includes the following steps:
    • The shipowner sends a request to the classification society surveyor, indicating the intention to perform an underwater survey.
    • A detailed plan of the ship's external hull features is submitted, showing the location of frames, bulkheads, welding lines, openings, etc.
    • The hull surface is cleaned before the survey to remove any marine growth or fouling that could obscure the inspection.
    • A diving company, approved by the classification society, is appointed to conduct the underwater inspection.
  • A self-propelled survey vehicle equipped with the following tools is used:
    • Long-Range Light TV Camera to aid in steering and checking hull deterioration.
    • High-resolution colour TV Camera to provide a close-up view of the hull coating and welded seams.
    • 35mm Still Camera to capture still images.
    • Ultrasonic Probe for measuring plate thickness.
    • Depth Meter and Speed Indicator to provide accurate data on the vehicle's depth and movement.
    • Umbilical Cable to connect the survey vehicle to the survey boat, transmitting power and information.
  • The survey boat is to be equipped with:
    • A control console with TV monitors.
    • Plate thickness printouts.
    • Audio and video cassette recorders.
    • Playback units.
    • Diver communication systems.
    • Vehicle control systems and associated instruments.
  • Operation:
    • The survey vehicle is taken underwater by a diver to the survey starting point.
    • Using TV monitors and shell expansion plans as guides, the vehicle is navigated over the hull, focusing on the bottom structure, sides, stern frame, rudder, propeller, bilge keel, and hull openings.
    • All images, data, and information are recorded and transmitted back to the survey boat.
    • Detailed pictures of the stern frame, rudder, propeller, bilge keel, and hull openings are captured.
    • Divers are deployed to measure stern tube bearing wear, pintle clearance, and inspect stern seals, anodes, and rudder stock palm coupling bolts.
    • All recorded video and audio, including conversations between the surveyor and drivers, thickness printouts, measurements, and pictures are analyzed to determine the detailed underwater condition of the vessel.
    (b) Before an underwater survey is accepted by the surveying authority, the following conditions must be met:

    The vessel's owner submits a request to the surveyor, including:
    • The proposed date and location for the survey.
    • General information about the diving company.
    • A declaration that the vessel has not suffered any damage due to grounding, collision, or other incidents.
    The ship's master or owner’s representative must provide a declaration confirming:
    • Any suspected or actual damage to the hull since the last dry-docking.
    • The underwater portion of the hull is protected by a suitable paint scheme that is of adequate thickness and remains valid until the next dry-dock.

    • The survey site should be in a protected area with calm and clear water, ensuring good underwater visibility. Attention must be given to the effects of currents and tides.
    • The hull must be clean for the external survey. The surveyor must be satisfied with the method and quality of the pictorial presentation, ensuring that it provides a reliable assessment of the hull's condition.
    • The underwater examination should be conducted by an approved diving company using closed-circuit TV and two-way communication, which can be monitored by the surveyor.
    • The vessel should be in as light an operating condition as possible to facilitate the survey.
    • Means must be available for the surveyor to examine the outside shell plating above the waterline.
    • Any required repairs identified during the survey must be carried out to the satisfaction of the attending surveyor.
    (c) While the importance of each item may vary depending on the vessel and its specific requirements, below is a list of items that should be included in an underwater survey in order of importance:
    • Underwater Hull: General condition of the hull below the waterline.
    • Bottom and Shell Plating: Inspection for corrosion, damage, and fouling.
    • Shell Openings: Examination of openings such as sea chests, drain plugs, and overboard discharge points.
    • Stern Tube Oil Leaks: Check for leaks around the stern tube.
    • Propeller Blade: Inspection for damage, wear, and fouling.
    • Rudder: Inspection for damage, wear, and clearances.
    • Sea Chest Opening and Grating: Examination for blockages, damage, and fouling.
    • Anodes: Check the condition and effectiveness of cathodic protection anodes.
    • Bilge Keel: Inspection for damage and fouling.
    • Drain Plugs: Ensure all drain plugs are secure and in good condition.
    • Overboard Valve Openings: Check for proper operation and condition.
    • Forward Area: Inspection for any damage due to anchor and chain movement.
    Q5 (10 Marks) General πŸ”₯ Repeated 7x

    List SIX hazards that arise with the carriage of liquefied gas in bulk. Describe, with the aid of a sketch. the details of construction of a prismatic cargo tank within a gas carrier designed to carry liquefied gas (LPG)

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    Hazards of Carriage of Liquefied Gas in Bulk and Construction of a Prismatic Cargo Tank

    Part (a)

    Six Hazards Associated with the Carriage of Liquefied Gas in Bulk

    1. Flammability and Explosion
      • Liquefied gases such as LPG vaporise rapidly when released.
      • The vapour can mix with air and form a highly flammable or explosive atmosphere, creating a serious risk of fire or explosion.
    2. Toxicity
      • Some liquefied gas cargoes, such as ammonia and vinyl chloride monomer, are highly toxic.
      • Exposure can cause serious poisoning through inhalation, ingestion or skin absorption.
    3. Asphyxiation
      • LPG vapours are generally heavier than air.
      • In the event of a leak, the vapour can collect in low-lying areas such as pump rooms and hold spaces, displacing oxygen and creating an asphyxiation hazard.
    4. Frostbite and Cold Burns
      • Liquefied gas cargoes are carried at very low temperatures, with fully refrigerated LPG being carried at approximately βˆ’50Β°C.
      • Contact with the liquid cargo or uninsulated pipes and equipment can cause severe frostbite and cold burns.
    5. Brittle Fracture
      • Ordinary ship hull steel, such as mild steel, can become brittle at very low temperatures.
      • If cold cargo leaks and comes into contact with unsuitable hull steel, it may cause cracking and catastrophic structural failure.
    6. Sloshing
      • Partially filled tanks have a free surface, allowing the liquid cargo to move violently with the ship's motion.
      • This can produce large dynamic impact loads on the tank walls and internal pump towers, potentially causing structural damage.
    Part (b)

    Construction of a Prismatic Cargo Tank for LPG

    Fully refrigerated LPG is typically carried in Independent Type A prismatic cargo tanks. These are self-supporting tanks that are independent of the ship's hull structure and do not contribute to the overall structural strength of the vessel.

    Sketch – Typical Prismatic Cargo Tank Arrangement

    Ship's Hull / Hold Space

    β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”

    β”‚ Secondary Barrier / Hull β”‚

    β”‚ β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚

    β”‚ β”‚ Thermal Insulation β”‚ β”‚

    β”‚ β”‚ β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚ β”‚

    β”‚ β”‚ β”‚ β”‚ β”‚ β”‚

    β”‚ β”‚ β”‚ LPG CARGO β”‚ β”‚ β”‚

    β”‚ β”‚ β”‚ PRISMATIC TANK β”‚ β”‚ β”‚

    β”‚ β”‚ β”‚ β”‚ β”‚ β”‚

    β”‚ β”‚ β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ β”‚ β”‚

    β”‚ β”‚ Primary Barrier β”‚ β”‚

    β”‚ β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ β”‚

    β”‚ Load-bearing supports β”‚

    β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜

    β”‚

    Double Bottom

    1. Shape and Structure

    • The tank is prismatic, meaning it is generally box-shaped with chamfered/angled top and bottom corners.
    • This shape allows the tank to closely follow the contours of the ship's inner hull and therefore maximises the available cargo capacity.
    • The tank is internally reinforced with frames and stiffeners to maintain structural integrity.
    • A centreline longitudinal bulkhead, together with transverse wash bulkheads, may be provided to strengthen the tank and reduce liquid movement and sloshing.

    2. Materials

    • LPG is carried at low temperatures, approximately βˆ’50Β°C for fully refrigerated LPG.
    • To prevent brittle fracture at these temperatures, the primary barrier/tank is constructed from suitable low-temperature-resistant materials, typically fine-grained carbon-manganese steel.

    3. Tank Supports and Chocks

    The cargo tank operates at a substantially different temperature from the ship's hull and therefore undergoes thermal expansion and contraction.

    • The tank is supported on the double bottom by load-bearing insulation blocks.
    • These may be made from specialised hardwood such as AzobΓ© or suitable synthetic materials.
    • The supports provide the necessary load-bearing capacity while reducing thermal transfer and structural stresses.
    • Anti-roll, anti-pitch and anti-flotation keys/chocks secure the tank against movement caused by the ship's motion.
    • At the same time, the arrangement allows the tank to expand and contract freely due to temperature changes.

    4. Insulation

    • The outside of the primary barrier is provided with high-efficiency thermal insulation.
    • Its purpose is to maintain the required low cargo temperature and minimise heat ingress and cargo boil-off.
    • Sprayed polyurethane foam (PUF) is commonly used as the insulation material.

    5. Secondary Barrier

    • Under the IGC Code, Type A tanks are required to have a complete secondary barrier capable of containing leaked cargo for up to 15 days, preventing the cold cargo from coming into contact with the outer hull.
    • In LPG carriers, the ship's inner hull is normally used as the secondary barrier.
    • The inner hull is also constructed from suitable fine-grained low-temperature steel so that it can withstand the low temperature if the primary cargo tank fails.
    • The space between the primary tank and secondary barrier, known as the hold space, is maintained with inert gas or dry air as applicable.
    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Describe stability requirements for dry-docking. (6)

    (b) A ship of 8000t displacement floats upright in sea water, with KG = 7.6m, GM = 0.5m. A tank, whose Kg is 0.6m above the keel and 3.5m from the center line contains 100 t of water ballast. Neglecting the free surface effect, calculate the angle which the ship will heel, when the ballast water is pumped out. (10)

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    Part (a)

    For safe dry-docking, a ship must meet two key stability requirements:

    • Positive GM (Metacentric Height): The ship needs a positive GM. GM is the distance between the centre of gravity (G) and the metacentre (M). A positive GM indicates inherent stability; the ship will right itself if tilted. During dry-docking, the loss of buoyancy as the ship rests on the blocks reduces GM. Insufficient positive GM increases the risk of the ship heeling over or capsizing.
    • Trim by the stern: The vessel should be trimmed slightly by the stern (aft end lower than the bow) to ensure the aft end sits on the keel blocks first. This controlled settling minimises the risk of instability during the docking process. An even keel is generally preferred for the initial floating condition before the dry-docking procedure begins.
    Part (b)

    $$new \space KG \space = \space {{(8000 \times 7.6) - (100 \times 0.6)} \over 8000 - 100}$$

    $$New\:KG\:=\:7.689m$$

    $$New \space GM_1 \space = \space KM - KG$$

    $$= \space (7.6 + 0.5) - 7.689$$

    $$New\:GM\:=\:0.411m$$

    Angle of heel when 100t ballast is pumped out

    $$Tan\theta=\frac{m\times d}{\Delta GM}$$

    $$=\frac{100\times3.5}{7900\times0.411}$$

    $$Tan\theta=0.1077$$

    $$\theta=6^09^{^{\prime}}$$

    Q7 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 7x

    A ship of 15000 tonne displacement has an Admiralty Coefficient, based on shaft power, of 420. The mechanical efficiency of the machinery is 83%, shaft losses 6%, propeller efficiency 65% and QPC 0.71. At a particular speed the thrust power is 2550kW. Calculate: (16)

    (i) indicated power

    (ii) effective power

    (iii) ship speed.

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    Given:

    $$\Delta=15000t$$

    $$Shaft\:Power\:\left(SP\right)=420$$

    $$Transmission\:Efficiency=83\%$$

    $$Shaft\:losses=6\%$$

    $$Propeller\:Efficiency=65\%$$

    $$QPC=0.71$$

    $$Thrust\:Power=2550kW$$

    $$\left(\imaginaryI\right)\:Delivered\:Power\:\left(DP\right)=\frac{Thrust\:Power\:\left(TP\right)}{Propeller\:Efficiency\:\left(\eta P\right)}$$

    $$DP=\frac{2550}{0.65}$$

    $$DP=3923.07kW$$

    $$\left(ii\right)\:Shaft\:Power=\frac{Delivered\:Power\:\left(DP\right)}{Transmission\:Efficiency\:\left(\eta T\right)}\:$$

    $$SP=\frac{3923.07}{0.94}$$

    $$SP=4173.47kW$$

    $$\left(iii\right)\:Indicated\:Power=\frac{Shaft\:Power\:\left(SP\right)}{Mechanical\:Efficiency\:\left(\eta m\right)}$$

    $$IP=\frac{4173.47}{0.83}$$

    $$IP=5028.28kW$$

    $$\left(iv\right)\:Effective\:Power=DP\times QPC$$

    $$EP=3923.07\times0.71$$

    $$EP=2785.3797kW$$

    $$\left(v\right)\:Shaft\:Power=\frac{\Delta^{\frac23}\times V^3}{Admiralty\:Co-efficient}$$

    $$4173.47=\frac{15000^{\frac23}\times V^3}{420}$$

    $$V=14.23knots$$

    Q8 (10 Marks) Hull Construction πŸ”₯ Repeated 7x

    With respect to Buoyancy of a vessel:

    (a) What do you understand by reserve buoyancy what happen if the lost buoyancy is greater than the reserve buoyancy. (6)

    (b) A forward deep tank 12 m long extends from a longitudinal bulkhead to the ship's side. The widths of the tank surface measured from the longitudinal bulkhead at regular intervals are 10, 9, 7, 4 and 1 m. Calculate the second moment of area of the tank surface about a longitudinal axis passing through its centroid. (10)

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    Part (a)

    Reserve Buoyancy

    Reserve buoyancy is the watertight volume above the waterline of a vessel. It represents the potential buoyancy that can be utilised to keep the ship afloat when additional weight is added or if some buoyancy is lost due to damage (e.g., bilging).

    When a mass is added to the ship or buoyancy is lost (e.g., due to flooding of a compartment), the reserve buoyancy is converted into active buoyancy by an increase in draught.

    If the lost buoyancy (due to flooding or damage) is greater than the reserve buoyancy, the vessel will no longer have sufficient buoyant force to counteract its weight, causing it to sink.

    Part (b)
    Q9 (10 Marks) Ship Stability πŸ”₯ Repeated 8x

    (a) Define longitudinal center of gravity (LCG) and longitudinal centre of buoyancy (LCB). (6)

    (b) The immersed cross-sectional areas of a ship 120 m long, commencing from aft are 2, 40, 79, 100, 103, 104, 104, 103, 97, 58 and 0 m2. Calculate: (10)

    (i) Displacement

    (ii) Longitudinal position of the centre of buoyancy.

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.
    Q10 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 6x

    A ship of length 140m, Breadth of 18.5m, draught of 8.1m and a displacement of 17,025 tonnes in sea water, has a face pitch ratio of 0.673. The diameter of the propeller is 4.8m. The results of the speed trial show that true slip may be regarded as constant over a range of 9 to 13 knots and is 30%, w = 0.5Cb-0.05. If fuel used is 20t/day at 13 knots and fuel consumption/day varies as cube of speed of ship, determine the fuel consumption, when propeller runs at 110 rpm.

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    Given:

    $$Lenght,\:L=140m$$

    $$Breadth,\:B=18.5m$$

    $$Draught,\:d=8.1m$$

    $$Displacement,\:\Delta=17025tonnes$$

    $$Pitch\:ratio,\:p=0.673\operatorname{}$$

    $$Diameter\:of\:Propeller,\:D=4.8m$$

    $$\operatorname{Real\:Slip,\:R_{s}=30\%\:or\:0.3}$$

    $$Wake\:fraction,\:W=0.5C_{b}-0.05$$

    $$Consumption,\:C_2=\:20t\:per\:day\:$$

    $$Ship\:Speed,\:V_2=13\:knots$$

    $$\operatorname{Revolution,\:N}=110rpm$$

    $$cons\:per\:day\:\alpha\:V^3$$

    To find Fuel Consumption C1=?

    We know that,

    $$Displacement,\:\Delta=\nabla\times\rho$$

    $$17025=\nabla\times1.025$$

    $$\nabla=16609.76m^3$$

    $$Block\:Coefficient,\:C_{b}=\frac{\nabla}{L\times B\times d}$$

    $$C_{b}=\frac{16609.76}{140\times18.5\times8.1}$$

    $$C_{b}=0.792$$

    $$Wake\:Fraction,\:W=0.5C_{b}-0.05$$

    $$W=0.5\times0.792-0.05$$

    $$W=0.346$$

    $$Pitch\:ratio,\:p=\frac{P}{D}$$

    $$0.673=\frac{P}{4.8}$$

    $$P=4.8\times0.673$$

    $$P=3.23m$$

    $$Theoretical\:Speed,\:V_{t}=\frac{P\times N\times60}{1852}$$

    $$V_{t}=\frac{3.23\times110\times60}{1852}$$

    $$V_{t}=11.51knots$$

    Using, Real slip equation.

    $$\operatorname{\operatorname{Real\:Slip,\:R_{s}=\frac{V_{t}-V_{a}}{V_{t}}}}$$

    $$0.3=\frac{11.51-V_{a}}{11.51}$$

    $$V_{a}=11.51-11.51\times0.3$$

    $$V_{a}=11.51\left(1-0.3\right)$$

    $$V_{a}=11.51\times0.7$$

    $$V_{a}=8.057knots$$

    $$Wake\:fraction,\:W=\frac{V-V_{a}}{V}$$

    $$0.346=\frac{V-8.057}{V}$$

    $$0.346V=V-8.057$$

    $$V=\frac{8.057}{0.654}$$

    $$V=12.32knots$$

    $$cons\:per\:day\:\alpha\:V^3$$

    $$\frac{C_1}{C_2}=\left(\frac{V_1}{V_2}\right)^3$$

    $$\frac{C_1}{20}=\left(\frac{12.32}{13}\right)^3$$

    $$C_1=20\times\left(\frac{12.32}{13}\right)^3$$

    $$C_1=20\times0.851$$

    $$C_1=17.02t\:per\:day$$

    Q1 (10 Marks) Hull Construction πŸ”₯ Repeated 5x

    Give a reasoned opinion as to the validity of the following assertions concerning ship structure:

    (a) Crack propagation in propellers shaft 'A' bracket or spectacle frames is indicative of inadequate scantlings and strength.

    (b) The adequate provision of freeing ports is as critical to the seaworthiness as watertight integrity.

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    Part (a)

    Crack propagation in propeller shaft 'A' brackets or spectacle frames - is it indicative of inadequate scantlings and strength?

    This assertion is only partly valid. Cracks in A-brackets or spectacle frames (the shaft brackets supporting the propeller shaft) are more commonly the result of fatigue due to fluctuating loads, stress concentrations at the bracket-to-hull connection, and the vibration and whipping of the shaft, rather than simply inadequate scantlings. The brackets are subject to severe alternating loads from the propeller and the shaft, and cracks typically initiate at stress raisers (sharp corners, weld toes, the bracket arm-to-hull connection) and propagate under fatigue. While inadequate scantlings or poor design (insufficient section, poor connection, sharp notches) can contribute, the primary cause is usually fatigue and stress concentration, aggravated by vibration, corrosion and the dynamic loads of the propeller. Hence the assertion is not fully valid: crack propagation is more indicative of fatigue and stress concentration than of inadequate strength alone, and the design should address the fatigue life, the connection detail and the avoidance of stress raisers, as well as the scantlings.

    Part (b)

    The adequate provision of freeing ports is as critical to seaworthiness as watertight integrity.

    This assertion is largely valid. Freeing ports (openings in the bulwark that allow water shipped on deck to drain overboard) are essential to seaworthiness because, if they are inadequate, water accumulating on the deck cannot drain, which:

    • increases the free-surface effect and the weight of water on deck, reducing stability and increasing the risk of capsize;
    • increases the deck load and the risk of structural damage;
    • reduces the reserve buoyancy and can lead to the ship becoming unstable.

    Watertight integrity (the ability of the hull and its openings to keep water out) is equally critical to seaworthiness, as it prevents flooding and loss of buoyancy. Both are essential: watertight integrity keeps water out, while freeing ports remove water that is shipped on deck. If either is inadequate, the ship's seaworthiness is compromised. Hence the assertion is valid - freeing ports are as critical to seaworthiness as watertight integrity, because they maintain the stability and buoyancy of the ship by removing deck water.

    Q2 (10 Marks) Ship Stability πŸ”₯ Repeated 6x

    With reference to Roll-on. Roll-off ferries:

    (a) Describe the problem of free surface effect:

    (b) Explain how it is intended that water should be cleared from car or cargo decks:

    (c) Describe possible methods for improving the stability and survivability of these vessels.

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    Part (a)

    The problem of free-surface effect in Ro-Ro ferries.

    Ro-Ro ferries have large, open car and cargo decks that extend over a large part of the ship. If water is shipped onto these decks (e.g. through the bow or stern doors, or in heavy weather), the water spreads over the large deck area, creating a very large free surface. The free-surface effect reduces the effective GM by rho x i/Delta, where i is the second moment of area of the free surface (i = L B^3/12 for a rectangular deck). Because the car deck is very wide and long, the free-surface effect is very large and can reduce the GM to a dangerously low value, causing the ship to lose stability and capsize. This is the principal stability problem of Ro-Ro ferries: the large open decks create a huge free-surface effect if flooded, and the ship can capsize rapidly.

    Part (b)

    How water should be cleared from car or cargo decks.

    Water on the car deck should be cleared by:

    • Providing adequate freeing arrangements (scuppers, freeing ports, drain valves) at the deck edge and at the ends, so that water can drain overboard.
    • Providing a camber (transverse slope) on the deck so that water runs to the sides and drains through the freeing ports.
    • Providing a longitudinal slope (sheer) so that water runs to the ends and drains.
    • Using bilge pumps and drainage systems to remove water that cannot drain overboard.
    • Ensuring the freeing ports are of adequate size and are not blocked by cargo or lashings.

    The freeing arrangements must be adequate to remove water quickly and prevent the build-up of a large free surface.

    Part (c)

    Methods for improving the stability and survivability of Ro-Ro ferries.

    • Lowering the centre of gravity by placing heavy weights low and ballast in the double bottom.
    • Increasing the GM by increasing the beam and the waterplane area, and by lowering KG.
    • Providing adequate freeboard and reserve buoyancy.
    • Subdividing the car deck with watertight bulkheads or providing a raised deck to limit the spread of water.
    • Providing adequate freeing ports and drainage to remove water quickly.
    • Fitting bilge keels and stabilisers to reduce roll.
    • Designing the ship to meet the damage stability criteria (survive flooding of a compartment).
    • Using a higher freeboard and a stronger, more watertight structure, and ensuring the bow and stern doors are watertight and properly secured.
    • Operating with adequate stability margins and following the loading manual.
    Q3 (10 Marks) Ship Types & Design πŸ”₯ Repeated 4x

    (a) Draw and the mid ships section of an oil tanker with Double Hull & name each part.

    (b) What is Bow Flare? Why is it so important in Bulk Carriers?

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    Part (a)

    Mid-ship section of Oil tanker:

    Part (b)

    Bow flare

    is the outward curvature of a ship's side shell above the waterline at the forward end.

    Importance of Bow Flare in Bulk Carriers:

    • Bow flare enhances the reserve buoyancy at the forward end of the vessel, which improves seaworthiness by helping the ship ride over waves more effectively, especially when pitching in rough seas.
    • By dispersing water away from the ship, the bow flare reduces the amount of water shipped onto the deck during heavy weather.
    • The wider forecastle deck created by the bow flare allows for the installation of essential machinery such as windlasses, mooring equipment, and other fittings.
    • The bow flare shields the hull plating from damage caused by the anchor when it is raised or lowered.
    • A well-designed bow flare can reduce water resistance, leading to increased speed and better fuel efficiency.
    Q4 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 4x

    (a) What is free surface effect? How can be avoided or reduced.

    (b) Give the components of ships resistance while vessel is 'enroute'.

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    (a) Free Surface Effect:

    The free surface effect (FSE) is a reduction in the metacentric height (GM) of a vessel due to the movement of liquids within partially filled tanks when the ship heels. When a ship tilts, the liquid in a partially filled tank shifts to the lower side, causing the centre of gravity (CG) of the ship to move laterally. This lateral shift of the CG reduces the righting lever (GZ), effectively decreasing the ship's stability and increasing the angle of heel. This apparent loss of GM is the free surface effect.

    Minimizing Free Surface Effect:

    • Partially filled tanks should be avoided to minimize liquid movement.
    • Tanks should be designed with longitudinal divisions or swash bulkheads to limit the movement of liquid.
    • Install sluice valves to control the liquid movement between compartments in divided tanks.
    • Fill smaller tanks at the bottom of the ship first to lower the centre of gravity and improve stability.
    • Tanks should have reduced breadth to minimize the free surface's effect.
    Part (b)

    Components of ship’s resistance while vessel is en route

    When a ship moves through water, resistance opposes its motion. The ship must exert an equal force to maintain speed.

    Frictional resistance (Rf): This is caused by the friction between the hull and the water. The water immediately adjacent to the hull is dragged along, creating a boundary layer. This resistance depends on the water's viscosity, the ship's speed, and the wetted surface area of the hull. At lower speeds, frictional resistance can account for 70-90% of total resistance; however, at higher speeds, it can be less than 40%.

    Residuary resistance (Rr): This is the resistance that remains after subtracting the frictional resistance. These are:

    • Form drag: Resistance due to the shape of the hull and the flow of water around it, generating pressure differences.
    • Wave-making resistance: This is a major component at higher speeds. The ship creates waves, and energy is expended in this process.
    • Eddy resistance: Resistance caused by turbulent flow behind the ship, especially at sharp changes in the hull's shape. This is often minimized by optimizing the hull design.

    Air resistance (Ra): This resistance is generated by the ship moving through the air. It depends on the shape of the superstructure, the projected area above the waterline, and wind speed and direction. Air resistance is typically a smaller component, usually around 2% but can be up to 10% for large container ships with extensive superstructure.

    Total resistance (Rt): The total resistance experienced by the ship is the sum of frictional, residuary, and air resistance: Rt = Rf + Rr + Ra.

    Q5 (10 Marks) Ship Types & Design πŸ”₯ Repeated 6x

    (a) Considering the vessel as a compound beam define Bending moment shearing force. Which is the point of Maximum Bending Moment?

    (b) Sketch and Describe Hatch coaming of a large bulk carrier.

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    Part (a)

    Bending moment and shearing force, and the point of maximum bending moment.

    Considering the vessel as a compound beam (the hull girder), the shearing force at any section is the algebraic sum of the vertical forces (loads) to one side of the section, i.e. the net load (weight - buoyancy) acting on that part. The bending moment at any section is the algebraic sum of the moments of the loads to one side, i.e. the integral of the shearing force. The shearing force is the rate of change of the bending moment, and the bending moment is the integral of the shearing force. The point of maximum bending moment occurs where the shearing force is zero (where the shearing force changes sign), which is usually at or near midships for a ship in still water, and at the point where the net load changes sign. The maximum bending moment is the largest hogging or sagging moment, and the hull girder must be designed to withstand it.

    Part (b)

    Hatch coaming of a large bulk carrier.

    The hatch coaming is the vertical structure around the hatch opening that raises the hatch above the deck to prevent water entering and to provide strength. Sketch: the hatch opening is bounded by a vertical coaming plate (about 600-900 mm high for a bulk carrier) welded to the deck, with a top flange (or a horizontal stiffener) and vertical stiffeners (brackets) connecting the coaming to the deck. The coaming is made of thick plate and is stiffened to resist the loads of the hatch cover and the cargo, and to provide the longitudinal strength of the deck (the coaming acts as a longitudinal girder). The hatch cover sits on the coaming with a gasket and is secured by cleats. The coaming corners are rounded and reinforced to avoid stress concentrations. The coaming provides the watertight seal for the hatch and contributes to the longitudinal strength of the hull girder.

    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 7x

    (a) List the precautions necessary before an inclining experiment is carried out. (6)

    (b) A box shaped vessel 50 metres long x 10 metres wide. floats in salt water on an even keel at a draft of 4 metres. A center line longitudinal watertight bulkhead extends from end to end and for the full depth of the vessel. A compartment amidships on the starboard side is 15 metres long and contains cargo with permeability 30%. Calculate the list if this compartment is bilged. KG = 3 metres. (10)

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    Part (a)

    Precautions necessary before an Inclining experiment:

    • The ship should be in a sheltered position, such as a gravity dock, to minimize external influences from wind or waves.
    • Mooring ropes should be slack to allow the ship to move freely without restrictions.
    • Only workers essential to the experiment should be present onboard to avoid unnecessary weight and movement.
    • All tanks must either be completely empty or pressed up tight to eliminate the free surface effect, which can adversely affect stability calculations.
    • Any loose weights on the ship must either be removed or properly secured to prevent unintended movement during the experiment.
    • Ensure that the pendulums are long and properly suspended from stable points, such as underneath a hatch, to provide accurate deflection readings.
    • The test masses should be evenly distributed and placed as far from the centerline as possible to maximize measurable deflections.
    • The experiment should be carried out in calm weather to avoid the effects of wind, current, or waves on the vessel's stability.
    Q7 (10 Marks) Ship Stability πŸ”₯ Repeated 9x

    (a) Describe how the force on the ship's bottom and the GM vary when grounding takes place. (6)

    (b) A ship of 8,000 tonnes displacement takes the ground on a sand bank on a falling tide at an even keel draft of 5.2 metres. KG 4.0 metres. The predicted depth of water over the sand bank at the following low water is 3.2 metres. Calculate the GM at this time assuming that the KM will then be 5.0 metres and that the mean TPC is 15 tonne (10)

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    Part (a)

    When a ship grounds, the forces on the bottom and the metacentric height (GM) change depending on the grounding's nature and the ship's position on the seabed.

    If the ship grounds on a level bottom:

    • A ground reaction force acts vertically upwards from the seabed. This force counteracts part of the ship's weight and alters the distribution of buoyancy forces.
    • The ship’s centre of gravity (G) may appear to rise relative to the metacentre (M) because the upward ground reaction reduces the buoyancy force acting on the underwater volume.
    • This virtual rise in G reduces GM, potentially leading to a list.
    • If the list becomes excessive and the righting moment is insufficient, the ship may capsize.

    If the ship grounds on a pinnacle:

    • The ship experiences two forces at the ship's bottom:
      • A downward force due to the ship’s weight.
      • An upward reaction force is concentrated on the pinnacle.
    • The resulting force between the grounding pressure and the ship’s centre of buoyancy shifts downward towards the pinnacle.
    • This situation is similar to when the ship's stern touches the keel block in a dry dock.
    • A virtual loss of GM occurs because the ship’s inclining moment may exceed the maximum righting lever.
    • If the inclining moment is too great, the ship may develop an excessive list or even capsize.

    (b) Given:

    $$Displacement,\:\Delta=8000\:tonnes$$

    $$TPC=15\:tonnes$$

    $$Initial\:Draught=5.2m$$

    $$Final\:Draught=3.2m$$

    $$Ship\:KG=4.0m$$

    $$KM=5.0m$$

    To find GM

    $$Uptrust,\:P=TPC\times Fall\:in\:water\:level\:\left(cm\right)$$

    $$=15\times\left(520-320\right)$$

    $$=15\times200$$

    $$P=3000\:tonnes$$

    To Find Virtual loss of GM:

    $$Virtual\:loss\:of\:GM_1=\frac{P\times KM}{\Delta}$$

    $$=\frac{3000\times5}{8000}$$

    $$=\frac{15000}{8000}$$

    $$GM_1=1.88m$$

    Actual KM = 5.0m (given)

    $$Virutal\:KM=Actual\:KM-Virtual\:loss\:of\:GM_1$$

    $$=5.0-1.88$$

    $$=3.12$$

    Similarly, Actual KG = 4.0m (given)

    $$New\:GM=Virtual\:KM-\:Actual\:KG$$

    $$=3.12-4.0$$

    $$=-0.88$$

    Q8 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    The speed of a ship is increased to 18% above normal for 7-5 hours, then reduced to 9% below normal for 10 hours. The speed is then reduced for the remainder of the day so that the consumption for the day is the normal amount. Find the percentage difference between the distance travelled in that day and the normal distance travelled per day.

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    Let normal speed be V and normal daily consumption be C. Consumption varies as the cube of speed, so C = k.V^3.

    Step 1 - Find the speed for the remainder of the day.

    For 7.5 hours the speed is 1.18V. Consumption in that period = k.(1.18V)^3 x 7.5/24 = k.V^3 x 1.643 x 0.3125 = 0.5134 k.V^3.

    For 10 hours the speed is 0.91V. Consumption = k.(0.91V)^3 x 10/24 = k.V^3 x 0.7536 x 0.4167 = 0.3140 k.V^3.

    Total consumption in first 17.5 hours = 0.5134 + 0.3140 = 0.8274 k.V^3.

    Remaining time in the day = 24 - 17.5 = 6.5 hours.

    For the day's total consumption to equal the normal amount k.V^3, the remaining consumption must be k.V^3 - 0.8274 k.V^3 = 0.1726 k.V^3.

    If the reduced speed is Vr, then k.Vr^3 x 6.5/24 = 0.1726 k.V^3.

    So Vr^3 = 0.1726 x 24/6.5 x V^3 = 0.6373 V^3.

    Vr = (0.6373)^(1/3) V = 0.8606 V.

    So the ship travels at 86.06% of normal speed for the last 6.5 hours.

    Step 2 - Find the distance travelled that day.

    Distance = speed x time.

    Normal distance per day = V x 24 = 24V.

    Actual distance = 1.18V x 7.5 + 0.91V x 10 + 0.8606V x 6.5

    = 8.85V + 9.10V + 5.594V = 23.544V.

    Step 3 - Percentage difference.

    Difference = 24V - 23.544V = 0.456V.

    Percentage difference = 0.456/24 x 100 = 1.9%.

    Answer: The distance travelled that day is 1.9% less than the normal distance per day.

    Q9 (10 Marks) Ship Stability πŸ”₯ Repeated 11x

    (a) Define longitudinal center of gravity (LCG) and longitudinal center of buoyancy (LCB). (6)

    (b) A ship 120m long floats at draughts of 5.50m forward and 5.80m aft: MCT 1cm 80 tonne m. TPC 13. LCF 2.5m forward of midships. Calculate the new draughts when a mass of 110 tonne is added 24m aft of midships.

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.

    LCF in fwd and trim by stern

    $$Bodily \space sinkage \space = \space {{mass \space added} \over TPC} \space = \space {{110} \over 13} \space$$

    $$=\:8.5\operatorname{\mathrm{cm}}\:=0.085m$$

    $$Trim \space = \space {{m \times d} \over MCT_{1cm}}$$

    $$ = \space {{110 \times (24 + 2.5)} \over 80}$$

    $$Trim=36.43\operatorname{cm}=0.364m\:$$

    Change in fwd draught:

    $$d_{F}=\frac{-t}{L}\left\lbrack\frac{L}{2}-LCF\right\rbrack$$

    $$=\frac{-36.43}{120}\left\lbrack\frac{120}{2}-2.5\right\rbrack$$

    $$=-17.45\operatorname{cm}=-0.1745m$$

    Change in Aft draught:

    $$d_{A}=\frac{+t}{L}\left\lbrack\frac{L}{2}+LCF\right\rbrack$$

    $$=\frac{+36.43}{120}\left\lbrack\frac{120}{2}+2.5\right\rbrack$$

    $$=+18.97\operatorname{cm}=0.189m$$

    New draughts:

    $$D_{F}=5.5+0.085-0.175=5.41m$$

    $$D_{A}=5.8+0.085+0.18=6.065m$$

    Q10 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 8x

    With reference to fixed pitch propellers:

    (a) Explain Propeller Slip and Propeller Thrust.

    (b) The shaft power of a ship is 3000 KW, the ship's speed V is 13.2 knot. Propeller rps is 1.27. Propeller pitch is 5.5m and the speed of advance is 11 Knots Find:

    (i) Real Slip

    (ii) Wake fraction

    (iii) Propeller thrust, when its efficiency. Ξ· = 70%

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    Part (a)

    Slip

    is the difference between the theoretical distance the propeller should travel in one revolution and the actual distance the vessel travels.

    $$Slip\:\left(\%\right)\:=\:\left(\frac{Engine\:distance\:-\:Ship^{\prime}s\:distance}{Engine\:distance}\right)\:\times100$$

    Where Engine distance = no. of propeller revolutions * propeller pitch (usually calculated over a 24-hour period)

    Propeller thrust: it is the force exerted by the propeller to move the vessel ahead and given by

    $$Thrust\:\left(T\right)\:=\:\rho\times A\times P^2\times n^2\times S$$

    Where,

    ρ - Density

    A - Area

    P - Pitch

    n - Revolution per second

    S - Slip

    Q1 (10 Marks) Ship Stability πŸ”₯ Repeated 3x

    Describe the effect of the following on the ship's stability;

    (a) Ice formation on superstructure

    (b) Effect of wind and waves

    (c) Changes that takes place during the ships voyage

    (d) Bilging of a compartment

    (d) While water is being pumped out from the dry dock

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    Part (a)

    Ice formation on superstructure:

    Ice accumulating on the superstructure adds mass high up on the ship. This raises the centre of gravity (G), decreasing the metacentric height (GM). A lower GM reduces stability, making the ship more tender (more easily rolled) and increasing the period of roll. The ship becomes more susceptible to capsizing.

    Part (b)

    Effect of Wind and Waves

    Wind:

    • High freeboard or tall superstructures increase windage, leading to a greater rolling effect.
    • Rolling caused by wind reduces stability, especially if the ship remains heeled for a prolonged period.

    Waves:

    • Large waves, especially when the ship is on the crest, can cause a significant loss of stability due to reduced underwater buoyant volume.
    • The ship may develop excessive heeling or capsizing tendencies.
    • Long ships are more vulnerable to wave action due to greater surface exposure, further reducing stability.
    Part (c)

    Changes During a Voyage

    Consumption of fuel and water:

    • Stability depends on the location of consumed or emptied tanks. Loss from low-level tanks increases G and decreases stability, while consumption from high-level tanks lowers G, increasing stability.

    Ballast exchange or transfer:

    • Transferring or exchanging ballast affects GM based on tank locations. Removal of low ballast raises G, while adding ballast low down lowers G and improves stability.

    Sea conditions:

    • Rolling and pitching caused by waves and wind can disrupt stability and amplify heeling or capsizing risks.

    Ice formation:

    • Ice buildup on decks or superstructures raises G, reducing GM and stability.

    Shifting of cargo:

    • Movement of cargo can create a list or cause instability if the shift raises the centre of gravity or reduces symmetrical weight distribution.
    Part (d)

    Bilging of a compartment

    Side Compartments (Port/Starboard):

    • Bilging a side compartment creates a virtual loss of GM due to the asymmetric flooding. This results in excessive list and increases the danger of capsizing.

    Forward or Aft Compartments:

    • Bilging forward or aft of the midship causes trim by head or stern, respectively.
    • Reserve buoyancy is reduced or lost, significantly decreasing stability.
    • Complete loss of reserve buoyancy results in sinking.
    Part (e)

    Pumping water out of a dry dock:

    • During dry docking, as water is pumped out, the stern settles on the keel blocks first, creating an upthrust.
    • This upthrust causes a virtual reduction in GM, which can destabilize the ship.
    • If GM becomes negative, the vessel may heel to one side or slip off the keel blocks, potentially leading to capsizing.
    • It is critical to ensure a positive GM during the entire dry docking process to maintain stability.
    Q2 (10 Marks) General

    Describe the construction of a forepeak tank. How are the eftects of panting and pounding taken care?

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    Construction of a forepeak tank.

    • The forepeak tank is the tank formed between the collision bulkhead and the bow (stem) of the ship, extending from the keel up to the freeboard deck or the tank top.
    • It is bounded aft by the collision bulkhead, forward by the stem and bow plating, and at the sides by the shell plating.
    • It is used for fresh water, ballast or occasionally for oil, and is fitted with a deep tank arrangement.
    • The tank is fitted with a suction pipe, an air pipe and a sounding pipe. The suction is arranged so that the tank can be completely emptied.
    • A manhole is provided for access, and the tank is fitted with a vent.
    • The plating is stiffened by vertical and horizontal stiffeners. Because the tank is at the bow, the stiffening must resist panting and pounding.
    • The collision bulkhead forms the after boundary and is a watertight bulkhead with no openings below the freeboard deck.
    • The tank top (if a double bottom is present) or the bottom shell forms the lower boundary.

    How the effects of panting and pounding are taken care of:

    Panting:

    • Panting is the in-and-out flexing (breathing) of the bow shell plating caused by the variation of water pressure as the ship pitches and the bow alternately rises and falls in waves.
    • To resist panting, the bow region is stiffened with:
    • Panting stringers: horizontal girders fitted on the inside of the shell plating at intervals, running longitudinally.
    • Panting beams: transverse beams fitted at intervals.
    • Additional vertical stiffeners and increased plate thickness in the bow region.
    • The collision bulkhead itself acts as a major stiffener.
    • Deep floors and increased frame spacing reduction in the forepeak.

    Pounding (slamming):

    • Pounding is the impact of the bow on the water surface when the ship pitches and the bow emerges and then slams down onto the water, causing severe impact loads on the bottom and bow structure.
    • To resist pounding, the following are provided:
    • Increased plate thickness in the bottom and bow region.
    • Closer spacing of frames and floors in the forepeak.
    • Deep floors (solid floors) fitted at every frame in the forepeak region.
    • Additional longitudinal stiffening and stringers.
    • The collision bulkhead and the forepeak structure are made stronger to absorb the impact loads.
    • In some ships, the forepeak is kept empty or partially filled to reduce the impact, and the structure is designed to withstand the slamming pressures.

    The combination of panting stringers, deep floors, closer frame spacing and increased plate thickness in the forepeak region effectively resists both panting and pounding.

    Q3 (10 Marks) General

    Describe the phenomenon of parametric rolling. what are the measures taken to prevent parametric rolling of ships.

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    Parametric rolling - phenomenon.

    • Parametric rolling is a phenomenon in which a ship experiences large-amplitude rolling motion due to the periodic variation of its stability (metacentric height, GM) as it moves through waves, even when there is no direct beam-sea excitation.
    • It occurs when the ship is in following or head seas with a wave length comparable to the ship's length, and the ship's natural roll period is approximately twice the wave encounter period (i.e. the ship rolls once for every two wave encounters).
    • As the ship moves through the wave, the waterplane area and hence the metacentric height (GM) change periodically. When the wave crest is amidships, the waterplane area is reduced (for a ship with a fine bow and stern), reducing GM; when the trough is amidships, the waterplane area is increased, increasing GM.
    • This periodic variation of GM causes the restoring moment to vary, and when the encounter frequency is about twice the natural roll frequency, the roll motion is parametrically excited and grows rapidly.
    • The roll amplitude can build up very quickly to large angles (often 30-40 degrees or more) even in moderate seas, because the damping is insufficient to limit the growth.
    • Parametric rolling is particularly dangerous for container ships and other ships with large bow flare and fine ends, and for ships with low GM (high stability is not necessarily protective). It can cause cargo shifting, container loss, structural damage and endanger the crew.

    Measures to prevent parametric rolling:

    • Avoid the critical speed and heading: alter course and/or speed so that the wave encounter period is not approximately twice the natural roll period. This is the most effective operational measure.
    • Change the natural roll period: adjust the GM (e.g. by ballasting or changing the loading condition) so that the natural roll period is not in the critical range relative to the wave encounter period.
    • Reduce speed in following/quartering seas to change the encounter frequency.
    • Use active or passive roll stabilisation systems: anti-rolling tanks, fin stabilisers, or bilge keels (bilge keels increase roll damping and reduce the amplitude).
    • Increase roll damping: bilge keels are effective in limiting the build-up of parametric rolling.
    • Route planning: use weather routing to avoid sea states and headings that are prone to parametric rolling.
    • For container ships, careful stowage to maintain adequate GM and avoid excessive metacentric height that could make the ship more susceptible.
    • Operational guidance: follow the ship-specific parametric rolling guidance and the IMO guidance on avoiding dangerous situations in following and quartering seas.
    • Some ships are fitted with parametric rolling detection and warning systems that alert the crew to take corrective action.
    Q4 (10 Marks) Hull Construction πŸ”₯ Repeated 2x

    With the help of sketches explain the different types of strakes used in ship construction. What material is generally used for Hull plating and What are the tests carried out on Hull steel plating for certification as per class rules.

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    Different types of strakes used in ship construction, hull plating material, and tests on hull steel.

    Part (a)

    Strakes are the longitudinal rows of plates that make up the shell plating. The different types are:

    • Keel strake (garboard strake): the bottom strake adjacent to the keel, which is thicker because it is subject to grounding and docking loads.
    • Bottom strakes: the strakes between the keel strake and the bilge, forming the bottom of the hull.
    • Bilge strake: the strake at the turn of the bilge, which is thicker and subject to the bilge keel attachment and the docking loads.
    • Side strakes: the strakes on the side of the hull between the bilge and the sheer strake.
    • Sheer strake: the top strake at the deck edge, which is thicker because it is the extreme fibre of the hull girder and subject to the highest bending stresses.
    • Stringer strake: the strake at the deck edge (the deck stringer plate), which is thicker and connects the deck to the side.
    • Deck strakes: the strakes of the deck plating, with the deck stringer plate at the edge being the thickest.

    The strakes are arranged so that the thicker plates are placed where the stresses and loads are highest (keel, bilge, sheer strake, deck stringer).

    Part (b)

    Material generally used for hull plating: mild steel (ordinary shipbuilding steel, e.g. grade A, B, D, E) and, for larger or higher-strength ships, high-tensile steel (e.g. AH32, AH36, DH32, DH36, EH32, EH36). The steel is a carbon-manganese steel with good weldability and notch toughness.

    Part (c)

    Tests carried out on hull steel plating for certification as per class rules:

    • Chemical analysis: to verify the composition (carbon, manganese, sulphur, phosphorus, etc.).
    • Tensile test: to verify the yield strength, ultimate tensile strength and elongation.
    • Bend test: to verify the ductility and soundness of the plate.
    • Impact test (Charpy V-notch): to verify the notch toughness at the specified temperature (for grades D, E and high-tensile steels).
    • Ultrasonic or other non-destructive testing: to verify the internal soundness of the plate.
    • Dimensional and surface inspection: to verify the thickness, flatness and freedom from defects.

    The tests are carried out on samples from each cast/heat and are witnessed and certified by the classification society surveyor.

    Q5 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    With reference to Ship stability:

    (a) With the help of a neat sketch explain the relevant features of a G-Z curve.

    (b) What are the effects of the below mentioned conditions on the G-Z curve:

    (i) Increased freeboard

    (ii) Increased beam, and

    (iii) Increased GM.

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    Part (a)

    Features of a GZ curve.

    The GZ curve is a graph of the righting lever GZ against the angle of heel. Its relevant features are:

    • The origin: at zero heel, GZ = 0.
    • The initial slope: the tangent to the curve at the origin equals GM (the metacentric height), since GZ = GM sin(theta) for small angles. A steeper initial slope means a larger GM.
    • The maximum righting lever (GZ max): the highest point of the curve, and the angle at which it occurs (the angle of maximum stability, typically 25-40 deg).
    • The range of stability: the angle from the upright to the angle of vanishing stability (where GZ returns to zero, typically 60-90 deg).
    • The area under the curve: proportional to the dynamical stability (the work done in heeling the ship), used to assess stability in a seaway and against wind heeling.
    • The angle of loll: if the curve starts below the axis (negative GZ at small angles), indicating a negative GM and an unstable ship that lolls to one side.
    • The effect of free surface: the curve is reduced by the free-surface correction.

    The curve is obtained from the cross-curves of stability corrected for the actual KG and free-surface effects, and is compared with the statutory criteria.

    Part (b)

    Effects of the following conditions on the GZ curve.

    (i) Increased freeboard: increasing the freeboard raises the deck edge and increases the reserve buoyancy, so the range of stability is increased (the angle of vanishing stability moves to a larger angle) and the area under the curve is increased. The initial slope (GM) is largely unchanged, but the curve is higher and extends further, giving greater dynamical stability and a larger range.

    (ii) Increased beam: increasing the beam increases the waterplane area and the BM (BM is proportional to the cube of the beam), so the initial slope (GM) increases and the curve is steeper at small angles. The maximum GZ is increased and occurs at a smaller angle, but the range of stability may be reduced (the angle of vanishing stability decreases) because the ship becomes stiffer and the deck edge immerses earlier. The area under the curve may be reduced at large angles.

    (iii) Increased GM: increasing the GM (e.g. by lowering KG) makes the initial slope steeper, so the curve rises more steeply at small angles and the maximum GZ is larger and occurs at a smaller angle. However, the range of stability is reduced (the angle of vanishing stability decreases) and the ship rolls more quickly and with a shorter period, which can be uncomfortable. The area under the curve at small angles increases but the overall range decreases.

    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    What is Prismatic Co-efficient (CP).

    (a) Derive the formula Cp= Cb/Cm, where Cb = Co-efficient of fineness and Cm = midship section area co-efficient.

    (b) The length of a ship is 18 times the draught, while the breadth is 2.1 times the draught. At the load water plane, the water plane area co-efficient is 0.83 and the difference between the TPC in sea water and the TPC in fresh water is 0.7. Determine the length of the ship and the TPC in fresh water.

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    Part (a)

    Prismatic coefficient is the ratio of the volume of displacement to the product of length and area of the immersed portion of the midship section

    Cb is the block co-efficient or co-efficient of fitness is the ratio of the volume of displacement to the product of length, breadth and draught

    $$\because\space C_{b}\space=\space{{\nabla}\over L\times B\times D}\:---\:1$$

    $$C_{p}\space=\space{{\nabla}\over L\times A_{m}}\:---\:2$$

    $$C_m \space = \space {{A_m} \over B \times D}$$

    $$A_{m}=\:C_{m}\times B\times D\:---\:3$$

    Substitute 3 in 2

    $$C_{p}\space=\space{{\nabla}\over C_{m}\times B\times D\ \times L}\:---\:4$$

    $$\nabla=C_{b}\times L\times B\times D\:---5\:\left(from\:equation\:1\right)$$

    Substitute 5 in 4

    $$C_{p}=\frac{C_{b}\times L\times B\times D}{C_{m}\times L\times B\times D}$$

    $$C_p \space = \space {{C_b} \over C_m}$$

    Part (b)

    $$Length \space of \space ship \space = \space L$$

    $$Draught \space = \space {{L} \over 18 }$$

    $$breadth \space = \space 2.1 \times draught \space = \space 2.1 \times {{L} \over 18}$$

    $$TPC\:in\:SW\:=\:0.01025A_{w}$$

    $$TPC\:in\:FW\:=\:0.0100A_{w}$$

    $$0.01025A_{w}-0.0100A_{w}=0.7$$

    $$A_{w}=\frac{0.7}{2.5\times10^{-4}}=2800$$

    $$C_w \space = \space {{A_w} \over L \times B}$$$$0.83 \space = \space {{2800} \over L \times {{2.1L} \over 18}}$$

    $$L=170.04m$$

    $$TPC\:in\:FW=0.010\times A_{w}$$

    $$=\:0.0100\times2800$$

    $$TPC\:in\:FW=28$$

    Q7 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 4x

    With respeet to Ship Propulsion:

    (a) Explain the various efficiencies associated with propeller and shafting arrangement.

    (b) When a propeller of 4.8 m pitch turns at 110 pm, the apparent slip is found to be -S% and the real slip is 1.5 S%. If the wake speed is 25% of the ship speed, calculate the ship speed, apparent slip and the real slip

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    Part (a)

    Efficiencies associated with the propeller and shafting arrangement.

    • Shaft (transmission) efficiency: the ratio of the delivered power at the propeller to the brake power of the engine, accounting for the losses in the shaft bearings, stern tube and any gearing. It is typically 0.97-0.99.
    • Propeller (open-water) efficiency: the ratio of the thrust power (T x Va) to the delivered power (2 pi n Q). It represents the efficiency of the propeller itself in converting the delivered power into thrust power.
    • Hull efficiency: the ratio of the effective power to the thrust power, = (1 - t)/(1 - w), where t is the thrust deduction factor and w the wake fraction. It accounts for the interaction between the hull and the propeller.
    • Quasi-propulsive coefficient (QPC): the ratio of the effective power to the delivered power, = hull efficiency x propeller efficiency. It is the overall efficiency of the propulsion system in converting the delivered power into effective (towing) power.
    • Overall (propulsive) efficiency: the ratio of the effective power to the brake power, = QPC x shaft efficiency. It is the overall efficiency of the engine-to-propeller-to-hull system.
    Part (b)

    Ship speed, apparent slip and real slip.

    A propeller of 4.8 m pitch turns at 110 rev/min. The apparent slip is -S% and the real slip is +1.5S%. The wake speed is 25% of the ship speed. Calculate the ship speed, the apparent slip and the real slip.

    Pitch speed = pitch x rev/s = 4.8 x 110/60 = 8.8 m/s.

    Let the ship speed be V (m/s). The speed of advance Va = V x (1 - 0.25) = 0.75 V.

    Apparent slip = (pitch speed - V)/pitch speed = -S/100.

    Real slip = (pitch speed - Va)/pitch speed = 1.5 S/100.

    From the apparent slip: (8.8 - V)/8.8 = -S/100, so V = 8.8(1 + S/100).

    From the real slip: (8.8 - 0.75 V)/8.8 = 1.5 S/100, so 8.8 - 0.75 V = 0.132 S.

    Substitute V = 8.8(1 + S/100): 8.8 - 6.6(1 + S/100) = 0.132 S.

    8.8 - 6.6 - 0.066 S = 0.132 S, so 2.2 = 0.198 S, S = 11.11.

    Apparent slip = -11.11%; real slip = 1.5 x 11.11 = 16.67%.

    Ship speed V = 8.8(1 + 0.1111) = 9.78 m/s = 9.78 x 1.944 = 19.0 knots.

    Answer: ship speed about 19.0 knots; apparent slip -11.1%; real slip +16.7%.

    Q8 (10 Marks) General

    With the aid of sketches:

    (a) Explain various lines plan.

    (b) The half-breadths of waterplane of a ship of 120m length ad 15m breadth are given below:

    (table will be here soon)

    Calculate

    (i) Water plane area

    (ii) TPC in salt water

    (iii) Cw

    (iv) LCF from Mid-ship

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    Part (a)

    Lines plan - explanation.

    The lines plan is a set of drawings that defines the shape of the hull. It consists of three principal views:

    • Sheer plan (profile): shows the hull as seen from the side. It contains the waterlines, the deck line, the sheer line and the profile of the stem and stern. The vertical positions of the waterlines and the longitudinal positions of the stations are shown.
    • Half-breadth plan (plan view): shows the hull as seen from above, drawn for one half of the ship (port or starboard) because the hull is symmetrical. It contains the waterlines (horizontal sections) and the deck line, showing the half-breadths at each station.
    • Body plan (end view): shows the transverse sections (stations) as seen from the bow and stern. The forward half of the stations is drawn on one side and the after half on the other. It contains the buttock lines and the diagonal lines.

    The three views are drawn in projection so that every point on the hull appears in all three views, allowing the shape to be fully defined and faired. The lines plan is used to calculate hydrostatic data, displacement, and to construct the ship.

    Part (b)

    Waterplane calculation.

    Length L = 120 m, breadth B = 15 m. Half-breadths at stations 0 to 8:

    Station: 0 1 2 3 4 5 6 7 8

    Half-breadth (m): 1.6 2.8 5.5 6.4 7.3 6.2 4.2 2.0 0

    Number of intervals n = 8, so the common interval h = L/n = 120/8 = 15 m.

    (i) Waterplane area.

    Using Simpson's First Rule with 8 intervals (9 ordinates), multipliers 1,4,2,4,2,4,2,4,1:

    Sum of products = 1x1.6 + 4x2.8 + 2x5.5 + 4x6.4 + 2x7.3 + 4x6.2 + 2x4.2 + 4x2.0 + 1x0

    = 1.6 + 11.2 + 11.0 + 25.6 + 14.6 + 24.8 + 8.4 + 8.0 + 0 = 105.2.

    Area of half waterplane = (h/3) x sum = (15/3) x 105.2 = 5 x 105.2 = 526 m2.

    Waterplane area (full) = 2 x 526 = 1052 m2.

    (ii) TPC in salt water.

    TPC = (waterplane area x density of sea water)/100 = (1052 x 1.025)/100 = 1078.3/100 = 10.78 tonne/cm.

    Answer: TPC = 10.78 tonne per cm.

    (iii) Cw (waterplane area coefficient).

    Cw = waterplane area / (L x B) = 1052 / (120 x 15) = 1052/1800 = 0.584.

    Answer: Cw = 0.584.

    (iv) LCF from midship.

    The centre of flotation is the centroid of the waterplane. Using the first moment about midship (station 4):

    Distances from station 4 (in intervals): station 0 = -4, 1 = -3, 2 = -2, 3 = -1, 4 = 0, 5 = +1, 6 = +2, 7 = +3, 8 = +4.

    Moment = 1x1.6x(-4) + 4x2.8x(-3) + 2x5.5x(-2) + 4x6.4x(-1) + 2x7.3x0 + 4x6.2x(+1) + 2x4.2x(+2) + 4x2.0x(+3) + 1x0x(+4)

    = -6.4 - 33.6 - 22.0 - 25.6 + 0 + 24.8 + 16.8 + 24.0 + 0 = -22.0.

    Distance of centroid from midship = (moment/sum) x h = (-22.0/105.2) x 15 = -0.2092 x 15 = -3.14 m.

    The negative sign means the LCF is forward of midship.

    Answer: LCF is 3.14 m forward of midship.

    Q9 (10 Marks) Ship Stability

    With reference to inclining experiments:

    (a) List the precautions necessary before an inclining experiment is carried out. (6)

    (b) A box shaped vessel, 50 metres long x 10 metres wide, floats in salt water on an even keel at a draft of 4 metres. A center line longitudinal watertight bulkhead extends from end to end and for the full depth of the vessel. A compartment amidships on the starboard side is 15 metres long and contains cargo with permeability 30%. Calculate the list if this compartment is bilged. KG = 3 (10)

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    Part (a)

    Precautions necessary before an Inclining experiment:

    • The ship should be in a sheltered position, such as a gravity dock, to minimize external influences from wind or waves.
    • Mooring ropes should be slack to allow the ship to move freely without restrictions.
    • Only workers essential to the experiment should be present onboard to avoid unnecessary weight and movement.
    • All tanks must either be completely empty or pressed up tight to eliminate the free surface effect, which can adversely affect stability calculations.
    • Any loose weights on the ship must either be removed or properly secured to prevent unintended movement during the experiment.
    • Ensure that the pendulums are long and properly suspended from stable points, such as underneath a hatch, to provide accurate deflection readings.
    • The test masses should be evenly distributed and placed as far from the centerline as possible to maximize measurable deflections.
    • The experiment should be carried out in calm weather to avoid the effects of wind, current, or waves on the vessel's stability.
    Q10 (10 Marks) Ship Stability

    With reference to subdivisional stability:

    (a) Describe briefly the significance of the factor of subdivision. (6)

    (b) A ship of 8000t displacement floats upright in sea water, with KG = 7.6m, GM = 0.5m. A tank, whose Kg is 0.6m above the keel and 3.5m from the center line contains 100 t of water ballast. Neglecting the free surface effect, calculate the angle which the ship will heel, when the ballast water is pumped out. (10)

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    Part (a)

    Factor of Subdivision

    The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

    Permissible Length Formula:

    $$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

    A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

    Q1 (10 Marks) Ship Stability πŸ”₯ Repeated 3x

    Describe the effect of the following on the ship's stability;

    (a) Ice formation on superstructure

    (b) Effect of wind and waves

    (c) Changes that takes place during the ships voyage

    (d) Bilging of a compartment

    (e) While water is being pumped out from the dry dock

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    Part (a)

    Ice formation on superstructure:

    Ice accumulating on the superstructure adds mass high up on the ship. This raises the centre of gravity (G), decreasing the metacentric height (GM). A lower GM reduces stability, making the ship more tender (more easily rolled) and increasing the period of roll. The ship becomes more susceptible to capsizing.

    Part (b)

    Effect of Wind and Waves

    Wind:

    • High freeboard or tall superstructures increase windage, leading to a greater rolling effect.
    • Rolling caused by wind reduces stability, especially if the ship remains heeled for a prolonged period.

    Waves:

    • Large waves, especially when the ship is on the crest, can cause a significant loss of stability due to reduced underwater buoyant volume.
    • The ship may develop excessive heeling or capsizing tendencies.
    • Long ships are more vulnerable to wave action due to greater surface exposure, further reducing stability.
    Part (c)

    Changes During a Voyage

    Consumption of fuel and water:

    • Stability depends on the location of consumed or emptied tanks. Loss from low-level tanks increases G and decreases stability, while consumption from high-level tanks lowers G, increasing stability.

    Ballast exchange or transfer:

    • Transferring or exchanging ballast affects GM based on tank locations. Removal of low ballast raises G, while adding ballast low down lowers G and improves stability.

    Sea conditions:

    • Rolling and pitching caused by waves and wind can disrupt stability and amplify heeling or capsizing risks.

    Ice formation:

    • Ice buildup on decks or superstructures raises G, reducing GM and stability.

    Shifting of cargo:

    • Movement of cargo can create a list or cause instability if the shift raises the centre of gravity or reduces symmetrical weight distribution.
    Part (d)

    Bilging of a compartment

    Side Compartments (Port/Starboard):

    • Bilging a side compartment creates a virtual loss of GM due to the asymmetric flooding. This results in excessive list and increases the danger of capsizing.

    Forward or Aft Compartments:

    • Bilging forward or aft of the midship causes trim by head or stern, respectively.
    • Reserve buoyancy is reduced or lost, significantly decreasing stability.
    • Complete loss of reserve buoyancy results in sinking.
    Part (e)

    Pumping water out of a dry dock:

    • During dry docking, as water is pumped out, the stern settles on the keel blocks first, creating an upthrust.
    • This upthrust causes a virtual reduction in GM, which can destabilize the ship.
    • If GM becomes negative, the vessel may heel to one side or slip off the keel blocks, potentially leading to capsizing.
    • It is critical to ensure a positive GM during the entire dry docking process to maintain stability.
    Q2 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 5x

    (a) Explain in detail, how an underwater survey is carried out.

    (b) State the requirements to be fulfilled before an underwater survey is acceptable to the survey authority.

    (c) Construct a list of the items in order of importance that the underwater survey authority should include.

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    Part (a)

    An in-water survey, also known as Underwater Inspection in Lieu of Dry Docking (UWILD), involves a systematic and detailed examination of a vessel’s hull and underwater components while the ship remains afloat. The procedure includes the following steps:

    • The shipowner sends a request to the classification society surveyor, indicating the intention to perform an underwater survey.
    • A detailed plan of the ship's external hull features is submitted, showing the location of frames, bulkheads, welding lines, openings, etc.
    • The hull surface is cleaned before the survey to remove any marine growth or fouling that could obscure the inspection.
    • A diving company, approved by the classification society, is appointed to conduct the underwater inspection.

    A self-propelled survey vehicle equipped with the following tools is used:

    • Long-Range Light TV Camera to aid in steering and checking hull deterioration.
    • High-resolution colour TV Camera to provide a close-up view of the hull coating and welded seams.
    • 35mm Still Camera to capture still images.
    • Ultrasonic Probe for measuring plate thickness.
    • Depth Meter and Speed Indicator to provide accurate data on the vehicle's depth and movement.
    • Umbilical Cable to connect the survey vehicle to the survey boat, transmitting power and information.

    The survey boat is to be equipped with:

    • A control console with TV monitors.
    • Plate thickness printouts.
    • Audio and video cassette recorders.
    • Playback units.
    • Diver communication systems.
    • Vehicle control systems and associated instruments.

    Operation:

    • The survey vehicle is taken underwater by a diver to the survey starting point.
    • Using TV monitors and shell expansion plans as guides, the vehicle is navigated over the hull, focusing on the bottom structure, sides, stern frame, rudder, propeller, bilge keel, and hull openings.
    • All images, data, and information are recorded and transmitted back to the survey boat.
    • Detailed pictures of the stern frame, rudder, propeller, bilge keel, and hull openings are captured.
    • Divers are deployed to measure stern tube bearing wear, pintle clearance, and inspect stern seals, anodes, and rudder stock palm coupling bolts.
    • All recorded video and audio, including conversations between the surveyor and drivers, thickness printouts, measurements, and pictures are analyzed to determine the detailed underwater condition of the vessel.
    Part (b)

    Before an in-water survey is accepted by the survey authority, the following conditions must be met:

    The vessel's owner submits a request to the surveyor, including:

    • The proposed date and location for the survey.
    • General information about the diving company.
    • A declaration that the vessel has not suffered any damage due to grounding, collision, or other incidents.

    The ship's master or owner’s representative must provide a declaration confirming:

    • Any suspected or actual damage to the hull since the last dry-docking.
    • The underwater portion of the hull is protected by a suitable paint scheme that is of adequate thickness and remains valid until the next dry-dock.
    • The survey site should be in a protected area with calm and clear water, ensuring good underwater visibility. Attention must be given to the effects of currents and tides.
    • The hull must be clean for the external survey. The surveyor must be satisfied with the method and quality of the pictorial presentation, ensuring that it provides a reliable assessment of the hull's condition.
    • The underwater examination should be conducted by an approved diving company using closed-circuit TV and two-way communication, which can be monitored by the surveyor.
    • The vessel should be in as light an operating condition as possible to facilitate the survey.
    • Means must be available for the surveyor to examine the outside shell plating above the waterline.
    • Any required repairs identified during the survey must be carried out to the satisfaction of the attending surveyor.
    Part (c)

    While the importance of each item may vary depending on the vessel and its specific requirements, below is a list of items that should be included in an underwater survey in order of importance:

    • Underwater Hull: General condition of the hull below the waterline.
    • Bottom and Shell Plating: Inspection for corrosion, damage, and fouling.
    • Shell Openings: Examination of openings such as sea chests, drain plugs, and overboard discharge points.
    • Stern Tube Oil Leaks: Check for leaks around the stern tube.
    • Propeller Blade: Inspection for damage, wear, and fouling.
    • Rudder: Inspection for damage, wear, and clearances.
    • Sea Chest Opening and Grating: Examination for blockages, damage, and fouling.
    • Anodes: Check the condition and effectiveness of cathodic protection anodes.
    • Bilge Keel: Inspection for damage and fouling.
    • Drain Plugs: Ensure all drain plugs are secure and in good condition.
    • Overboard Valve Openings: Check for proper operation and condition.
    • Forward Area: Inspection for any damage due to anchor and chain movement.
    Q3 (10 Marks) Ship Stability πŸ”₯ Repeated 7x

    (a) List the precautions necessary before an inclining experiment is carried out. (6)

    (b) A box shaped vessel, 50 metres long x 10 metres wide, floats in salt water on an even keel at a draft of 4 metres. A center line longitudinal watertight bulkhead extends from end to end and for the full depth of the vessel. A compartment amidships on the starboard side is 15 metres long and contains cargo with permeability 30%. Calculate the list if this compartment is bilged. KG = 3 metres. (10)

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    Part (a)

    Precautions necessary before an Inclining experiment:

    • The ship should be in a sheltered position, such as a gravity dock, to minimize external influences from wind or waves.
    • Mooring ropes should be slack to allow the ship to move freely without restrictions.
    • Only workers essential to the experiment should be present onboard to avoid unnecessary weight and movement.
    • All tanks must either be completely empty or pressed up tight to eliminate the free surface effect, which can adversely affect stability calculations.
    • Any loose weights on the ship must either be removed or properly secured to prevent unintended movement during the experiment.
    • Ensure that the pendulums are long and properly suspended from stable points, such as underneath a hatch, to provide accurate deflection readings.
    • The test masses should be evenly distributed and placed as far from the centerline as possible to maximize measurable deflections.
    • The experiment should be carried out in calm weather to avoid the effects of wind, current, or waves on the vessel's stability.
    Q4 (10 Marks) Ship Stability

    (a) Describe briefly the significance of the factor of subdivision. (6)

    (b) A ship of 8000t displacement floats upright in sea water, with KG = 7.6m, GM = 0.5m. A tank, whose Kg is 0.6m above the keel and 3.5m from the center line contains 100 t of water ballast. Neglecting the free surface effect, calculate the angle which the ship will heel, when the ballast water is pumped out. (10)

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    Part (a)

    Factor of Subdivision

    The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

    Permissible Length Formula:

    $$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

    A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

    Q5 (10 Marks) Ship Stability

    (a) Explain why the GM must remain positive until the critical instant at which the ship takes the blocks overall. (6)

    (b) A ship of displacement 10,010 tones has a container of 10t at Kg = 7.5m. The container is shifted transversely. A pendulum of length 7.5m deflects through 13.5cm. GM of the ship = 0.76m, KM = 6.7m. Find the distance through which the container is shifted. Also find the new KG if the

    container is removed. (10)

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    Part (a)

    For safe dry-docking, a ship must meet two key stability requirements:

    • Positive GM (Metacentric Height): The ship needs a positive GM. GM is the distance between the centre of gravity (G) and the metacentre (M). A positive GM indicates inherent stability; the ship will right itself if tilted. During dry-docking, the loss of buoyancy as the ship rests on the blocks reduces GM. Insufficient positive GM increases the risk of the ship heeling over or capsizing.
    • Trim by the stern: The vessel should be trimmed slightly by the stern (aft end lower than the bow) to ensure the aft end sits on the keel blocks first. This controlled settling minimises the risk of instability during the docking process. An even keel is generally preferred for the initial floating condition before the dry-docking procedure begins.

    At the critical moment, the centre of gravity G rises faster than the metacentre M, so G catches up with M, and the vessel stability is compromised. So, GM needs to be positive until the critical moment.

    Q6 (10 Marks) Ship Stability

    (a) With reference to dynamical stability, describe the effect of an increase in wind pressure when a vessel is at its maximum angle of roll to windward (6)

    (b) A ship of 15000 tonne displacement has righting levers of 0, 0.38, 1.0, 1.41 and 1.2 m at angles of heel of 0Β°, 15Β°, 30Β°, 45Β° and 60Β° respectively and an assumed KG of 7.0 m. The vessel is loaded to this displacement but the KG is found to be 6.80 m and GM 1.5 m. (10)

    (i) Draw the amended stability curve

    (ii) Estimate the dynamic stability at 60Β°

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    Part (a)

    Dynamical Stability is defined as the amount of energy required to heel a ship from its upright equilibrium position to a specific angle of heel. It provides a measure of the vessel's stability by considering its behaviour in response to dynamic external forces, such as wind or waves.

    • The concept compares the heeling moment energy (from external forces) and the righting moment energy (from the ship's stability).
    • The ship will absorb the energy imparted by the heeling moment. If the righting energy is greater than the heeling energy, the ship will stabilize; otherwise, it may capsize.

    Areas Under the Curve:

    • Area A: Represents the region where the heeling moment exceeds the righting moment (external energy > ship's stability).
    • Area B: Represents the region where the righting moment exceeds the heeling moment (ship's stability > external energy).
    • The balance of these areas determines whether the ship will right itself or continue to heel.

    When exposed to heeling forces such as wind or waves, the vessel inclines and may roll over to a certain angle of heel. If the external force is applied instantaneously, the ship must have enough reserve dynamic stability to absorb the energy and return to an upright position. If the external force is constant, the ship will remain at an equilibrium angle where the righting moment equals the heeling moment.

    This refers to the remaining righting energy available to counteract additional external forces. A higher reserve dynamic stability ensures the vessel can handle greater heeling forces without capsizing.

    Q7 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 7x

    What is meant by the Admiralty Coefficient and the Fuel Coefficient?

    A ship of 14900 tonne displacement has a shaft power of 4460 kW at 14.55 knots. The shaft power is reduced to 4120 kW and the fuel consumption at the same displacement is 541 kg/h. Calculate the fuel coefficient for the ship.

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    The Admiralty Coefficient (C) is a method for estimating the propulsion power needed for a newly built ship. It's considered relatively constant for a given ship design. The formula is:

    $$C=\frac{\Delta^{2/3}\times V^3}{BP}$$

    Where:

    • C = Admiralty Coefficient
    • Ξ” = Displacement in tonnes (weight of the ship when fully loaded)
    • V = Speed in knots
    • BP = Brake power in kilowatts (power delivered by the ship's engine)

    A higher Admiralty Coefficient indicates a more efficient ship design, meaning it requires less power to achieve a given speed. Values typically range from 350 to 600.

    Fuel Coefficient:

    The Fuel Coefficient (F.C.) is used to calculate a ship's daily fuel oil consumption. The formula is:

    $$Daily\:fuel\:oil\:consumption\:=\:\frac{\Delta^{2/3}\times V^3}{FC}$$

    Where:

    • F.C. = Fuel Coefficient
    • Ξ” = Displacement in tonnes
    • V = Speed in knots

    The Fuel Coefficient can vary significantly, with typical values ranging from 40,000 to 120,000. A higher Fuel Coefficient implies greater fuel efficiency (lower daily fuel consumption) for a given speed and displacement.

    Part (b)

    $$admiraty\:coefficient\:\left(C\right)=\:\frac{\Delta^{2/3}V^3}{Shaft\:power}=\frac{\Delta^{\frac23}\times V^3}{SP}$$

    $$\frac{SP_1}{SP_2}=\frac{V_1^3}{V_2^3}$$

    $$\frac{4460}{4120}=\frac{14.55^3}{V_2^3}$$

    $$V_2=14.17kntos$$

    $$Fuel\:consumption\:per\:hour=541\operatorname{\mathrm{\:kg}}\:per\:hour$$

    $$Fuel\:consumption\:per\:day\:=\:541\times24=12.98t\:per\:day$$

    $$Fuel\:coefficient=\frac{\Delta^{\frac23}\times V_2^3}{Fuel\:consumption\:per\:day}$$

    $$=\:\frac{14900^{\frac23}\times14.17^3}{12.98}$$

    $$=132726.9$$

    Q8 (10 Marks) Ship Types & Design

    (a) Define the purpose of cofferdams.

    (b) Sate where cofferdams are most likely to be found on:

    (i) Dry cargo ships

    (ii) Oil tankers

    (c) (i) State what information is available about danger of entering void spaces.

    (ii) Identify, with reasons, the precaution to be observed before and during entry to cofferdams

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    Part (a)

    Purpose of Cofferdams:

    Cofferdams act as barriers, preventing leakage between tanks containing different liquids (e.g., fuel oil, diesel oil, lubricating oil, freshwater, and ballast water). This prevents contamination. They also facilitate tank inspection. Regular sounding of cofferdams is necessary for detecting leaks. Additionally, cofferdams contribute to the vessel's reserve buoyancy and can provide access to the main engine's holding-down bolts.

    Part (b)

    Location of Cofferdams:

    (i) Dry Cargo Ships: Cofferdams on dry cargo ships are typically located between fuel and lubricating oil tanks, between fuel oil, fresh water and ballast tanks, and around the main engine's lubricating oil drain tank.

    (ii) Oil Tankers: Oil tankers have cofferdams in locations similar to dry cargo ships (between fuel tanks and other liquid tanks) but also importantly between cargo spaces and the machinery space (or pump room). Again, a cofferdam is usually present around the main engine's lubricating oil drain tank.

    Part (c)

    Dangers of entering void space:

    (i) Information About Void Space Hazards:

    Void spaces are classified as enclosed spaces with potential risks due to:

    • Limited openings for entry and exit.
    • Poor natural ventilation, which may cause a buildup of hazardous gases.
    • Lack of continuous worker occupancy, making them prone to:
      • Toxic or flammable gases (e.g., hydrocarbons, Hβ‚‚S).
      • Oxygen deficiency.
      • Cargo vapours that may pose health and safety risks.
    • According to SOLAS Chapter 3, Regulation 19, entry into such spaces requires regular enclosed space entry drills and rescue training every two months.

    (ii) Precautions Before and During Entry

    Before Entry:

    • An enclosed space entry permit must be filled and signed by a responsible officer.
    • Use portable instruments to measure:
      • Oxygen concentration (should be above 20%).
      • Hydrocarbons (should be below 1% of the lower explosive limit (LEL)).
      • Toxic gases (should be nil).
      1. Ensure the space is properly ventilated before entry and continuously during work.
      2. Provide adequate lighting inside the space.
      3. Conduct and document a risk assessment.
      4. Prepare rescue equipment, including breathing apparatus (BA sets) and a first aid kit, at the entrance.
      5. Establish clear communication protocols and intervals for contact.
      6. Assign a responsible person at the entrance and inform the duty officer.

      During Entry:

      • Carry a portable atmosphere testing instrument and check the space’s atmosphere at regular intervals.
      • Ensure continuous ventilation throughout the duration of the work or inspection.
      • Maintain regular communication with the supervising personnel.
      • Observe for any abnormalities, such as dizziness or unusual smells, and exit immediately if conditions become unsafe.
    Q9 (10 Marks) Hull Construction

    Describe the arrangement of tank top and double bottom in the machinery space making particular reference to the structure and scantlings below the main engine. Show the method adopted in the arrangement of D.B. tanks to avoid contamination of fresh water, fuel oil and lube oil stored in D.B. tanks.

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    The construction of the double bottom in the machinery space regardless of the framing system has solid plate floors at every frame space under the main engine. Additional side girders are fitted outboard of the main engine seating, as required. The double-bottom height is usually increased to provide fuel oil, lubricating oil and fresh water tanks of suitable capacities. Shaft alignment also requires an increase in the double-bottom height or a raised seating, the former method usually being adopted.

    Continuity of strength is ensured and maintained by gradually sloping the tank top height and internal structure to the required position. Additional support and stiffening is necessary for the main engines, boilers, etc., to provide a vibration-resistant solid platform capable of supporting the concentrated loads. On slow- speed diesel-engined ships, the tank top plating is increased to 40 mm thickness or thereabouts in way of the engine bedplate. This is achieved by using a special insert plate which is the length of the engine including the thrust block in size. Additional heavy girders are also fitted under this plate and in other positions under heavy machinery as required. Plating and girder material in the machinery spaces is of increased scantlings in the order of 10 per cent.

    The method adopted.

    A cellular void space within a ships structure is called a coffer dam. Like a bulkhead separates two spaces or divides a space into two, a coffer dam does the same with a larger degree of integrity since it incorporates a void which would contain any breach of either of the boundaries. This would contain the leakage and prevent it spreading into other areas. A cofferdam can be defined as an empty space separating compartments to prevent the contents of one compartment from entering another in case of leakage.

    Q10 (10 Marks) Ship Stability

    (a) Explain the effects on stability when a tank is partially filled with liquid. (6)

    (b) A ship of 11200 tonne displacement has a double bottom tank containing oil, whose centre of gravity is 16.5m forward and 6.6m below the centre of gravity of the ship. When the oil is used the ship's centre of gravity moves 380mm. Calculate: (10)

    (i) The mass of oil used

    (ii) The angles which the centre of gravity moves relative to the horizontal.

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    Part (a)

    Slack tanks, or partially filled tanks, significantly impact a ship's stability due to the free surface effect

    • When tanks are partially filled (slack), the liquid inside moves freely as the ship heels.
    • This movement shifts the centre of gravity (G) laterally towards the heeling side.
    • The righting lever (GZ) decreases, resulting in reduced metacentric height (GM) and overall stability.
    • The virtual loss of GM is proportional to the breadth of the tank and the height of the free surface.

    Consequences of Slack Tanks:

    • Increased angle of heel.
    • Reduced stability, making the ship tender and more prone to capsizing.

    To Minimize the Effect:

    • Avoid slack tanks where possible; either fill tanks completely or empty them.
    • Design tanks with longitudinal divisions or swash bulkheads to limit liquid movement.
    • Use sluice valves in tanks to control liquid transfer.
    • Prioritize filling smaller tanks at the ship's bottom to lower G and improve stability.
    Q1 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 3x

    With reference to dry docking, define the responsibilities of the Second Engineer:

    (a) Prior to docking

    (b) Whilst the vessel is in dry dock

    (c) Prior to flooding and leaving the dock

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    (a) Prior to Docking:

    Preliminary Preparation:

    • Review various plans, manuals, and previous drydock reports for reference.
    • Prepare a detailed repair list and ensure all required spares are accounted for.
    • Conduct an inventory of spares and requisition necessary items.
    • Gather required tools, including special tools like poker gauges, trammel gauges, gagging tools, etc.
    • Mark all overboard valves, NC (normally closed) and NO (normally open) valves clearly for easy identification.
    • Allocate jobs to team members and discuss the time schedule.
    • Conduct a safety meeting to highlight dry dock hazards and necessary precautions.

    Before Entering Dry Dock:

    • Identify the power requirements and machinery to be operational during docking.
    • Check the shore connection box for proper functionality.
    • Record the soundings of FO (Fuel Oil), LO (Lubricating Oil), and DO (Diesel Oil) tanks.
    • Discharge contents from clean drain tanks and sewage tanks.
    • Carry out Economizer soot-blowing.
    • Change over Main Engine, Diesel Generators, and Boiler to Low Sulfur Gas Oil (LSGO).
    • Stop and clean purifiers.
    • Ensure the low sea chest is open and the high sea chest is shut.
    • Keep firefighting appliances (FFA) on standby.
    • Shut down all non-essential machinery, including MGPS (Marine Growth Prevention System) and FWG (Fresh Water Generator).

    (b) Whilst the Vessel is in Dry Dock

    Upon Arrival:

    • Connect shore power and supplies after ensuring safety checks are completed.
    • Start necessary equipment like cooling water, air compressors, air conditioning, and fridge compressors.
    • Check for jobs assigned by dry dock personnel and prepare accordingly.
    • Attend a safety meeting with dry dock personnel to understand local safety rules and procedures.

    During Dry Docking::

    • Oversee and assist in:
      • Cleaning and inspecting the hull, rudder, sea chest, anodes, and propeller.
      • Measuring propeller drop, checking rudder clearances, and inspecting the stern tube bearing and seal.
      • Servicing underwater valves and overboard valves.
      • Inspecting anchor and cables conditions.
      • Overhauling deck machinery, cranes, elevators, and engine room equipment such as the Main Engine, Diesel Generators, Boiler, and Economizer.
      • Renewing pipes and valves as needed.
      • Performing electrical equipment maintenance and surveys.
      • Supervising service engineers for specific repair jobs.
    • Ensure tank cleaning, welding, and other repair works are completed according to the plan.
    • Run the standby diesel generator daily after starting the priming pump.

    (c) Before Flooding and Leaving the Dry Dock

    Final Checks:

    • Verify that all underwater fittings and drain plugs are securely in place.
    • Ensure all machinery has been boxed back and is ready for operation.
    • Check for any leakage in stern tube seals.
    • Take tank soundings to confirm proper levels.
    • Confirm the proper operation of all underwater valves, overboard valves, sea chests, and vents.
    • Inspect the stern tube tank for any irregularities.

    System Restart:

    • Switch back to ship's power after confirming all systems are functional.
    • Test the proper operation of all machinery and systems to ensure the ship is fully operational before leaving the dock.
    Q2 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 3x

    State how and why the following machinery items are affected when the maximum service speed of a vessel is consistently maintained in heavy weather:

    (a) Intermediate shafting.

    (b) Propeller shafting

    (c) Shafting coupling bolts.

    (d) Main thrust pads.

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    Part (a)

    Intermediate Shafting:

    • Torsional Stress: Resulting from the rapid changes in propeller speed (propeller racing) caused by variations in water resistance in heavy weather.
    • Compressive Stress: Due to the end thrust transmitted from the propeller.
    • Bending Stress: Caused by hull movements such as hogging (upward bending) and sagging (downward bending) in heavy seas.
    Part (b)

    Propeller Shafting:

    • Torque and Thrust: Caused by the propulsion forces transmitted through the shaft.
    • Torsional Stress: Due to the rapid fluctuations in engine speed when the propeller races.
    • Compressive Stress: Generated by the axial thrust from the propeller.
    • Bending Stress: Arises when the weight of the propeller acts on the shaft as the propeller emerges from the water during rough seas.
    Part (c)

    Shafting Coupling Bolts:

    • Bending Stress: From misalignments caused by hull deformations.
    • Shear Stress: Resulting from the whirling of the shaft and rapid engine speed changes during propeller racing.
    • Torsional Stress: Caused by the transmission of fluctuating torque.
    • Fatigue Failure: Due to the repeated application of fluctuating loads over time, particularly in heavy weather conditions.
    Part (d)

    Main Thrust Pads:

    • Stress from Hull Movements: Misalignment caused by hull hogging and sagging.
    • Load Fluctuations: Due to variations in propeller thrust during racing and rapid changes in sea conditions.
    • Axial and Torsional Vibration: Arising from inconsistent propulsion forces and shaft vibrations.
    • Surface Wear and Damage: Resulting from increased pressure and friction due to fluctuating thrust forces.
    Q3 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 9x

    With reference to Underwater Inspection in lieu of Dry docking(UWILD):

    (a) Explain in detail, how an underwater survey is carried out.

    (b) State the requirements to be fulfilled before an underwater survey is acceptable to the surveying authority.

    (c) Construct a list of the items in order of importance that the underwater survey authority should include.

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    (a) An underwater survey, also known as Underwater Inspection in Lieu of Dry Docking (UWILD), involves a systematic and detailed examination of a vessel’s hull and underwater components while the ship remains afloat. The procedure includes the following steps:
    • The shipowner sends a request to the classification society surveyor, indicating the intention to perform an underwater survey.
    • A detailed plan of the ship's external hull features is submitted, showing the location of frames, bulkheads, welding lines, openings, etc.
    • The hull surface is cleaned before the survey to remove any marine growth or fouling that could obscure the inspection.
    • A diving company, approved by the classification society, is appointed to conduct the underwater inspection.
  • A self-propelled survey vehicle equipped with the following tools is used:
    • Long-Range Light TV Camera to aid in steering and checking hull deterioration.
    • High-resolution colour TV Camera to provide a close-up view of the hull coating and welded seams.
    • 35mm Still Camera to capture still images.
    • Ultrasonic Probe for measuring plate thickness.
    • Depth Meter and Speed Indicator to provide accurate data on the vehicle's depth and movement.
    • Umbilical Cable to connect the survey vehicle to the survey boat, transmitting power and information.
  • The survey boat is to be equipped with:
    • A control console with TV monitors.
    • Plate thickness printouts.
    • Audio and video cassette recorders.
    • Playback units.
    • Diver communication systems.
    • Vehicle control systems and associated instruments.
  • Operation:
    • The survey vehicle is taken underwater by a diver to the survey starting point.
    • Using TV monitors and shell expansion plans as guides, the vehicle is navigated over the hull, focusing on the bottom structure, sides, stern frame, rudder, propeller, bilge keel, and hull openings.
    • All images, data, and information are recorded and transmitted back to the survey boat.
    • Detailed pictures of the stern frame, rudder, propeller, bilge keel, and hull openings are captured.
    • Divers are deployed to measure stern tube bearing wear, pintle clearance, and inspect stern seals, anodes, and rudder stock palm coupling bolts.
    • All recorded video and audio, including conversations between the surveyor and drivers, thickness printouts, measurements, and pictures are analyzed to determine the detailed underwater condition of the vessel.
    (b) Before an underwater survey is accepted by the surveying authority, the following conditions must be met:

    The vessel's owner submits a request to the surveyor, including:
    • The proposed date and location for the survey.
    • General information about the diving company.
    • A declaration that the vessel has not suffered any damage due to grounding, collision, or other incidents.
    The ship's master or owner’s representative must provide a declaration confirming:
    • Any suspected or actual damage to the hull since the last dry-docking.
    • The underwater portion of the hull is protected by a suitable paint scheme that is of adequate thickness and remains valid until the next dry-dock.

    • The survey site should be in a protected area with calm and clear water, ensuring good underwater visibility. Attention must be given to the effects of currents and tides.
    • The hull must be clean for the external survey. The surveyor must be satisfied with the method and quality of the pictorial presentation, ensuring that it provides a reliable assessment of the hull's condition.
    • The underwater examination should be conducted by an approved diving company using closed-circuit TV and two-way communication, which can be monitored by the surveyor.
    • The vessel should be in as light an operating condition as possible to facilitate the survey.
    • Means must be available for the surveyor to examine the outside shell plating above the waterline.
    • Any required repairs identified during the survey must be carried out to the satisfaction of the attending surveyor.
    (c) While the importance of each item may vary depending on the vessel and its specific requirements, below is a list of items that should be included in an underwater survey in order of importance:
    • Underwater Hull: General condition of the hull below the waterline.
    • Bottom and Shell Plating: Inspection for corrosion, damage, and fouling.
    • Shell Openings: Examination of openings such as sea chests, drain plugs, and overboard discharge points.
    • Stern Tube Oil Leaks: Check for leaks around the stern tube.
    • Propeller Blade: Inspection for damage, wear, and fouling.
    • Rudder: Inspection for damage, wear, and clearances.
    • Sea Chest Opening and Grating: Examination for blockages, damage, and fouling.
    • Anodes: Check the condition and effectiveness of cathodic protection anodes.
    • Bilge Keel: Inspection for damage and fouling.
    • Drain Plugs: Ensure all drain plugs are secure and in good condition.
    • Overboard Valve Openings: Check for proper operation and condition.
    • Forward Area: Inspection for any damage due to anchor and chain movement.
    Q4 (10 Marks) Ship Stability πŸ”₯ Repeated 14x

    Explain how the period of roll varies with:

    (a) The amplitude of roll.

    (b) The radius of gyration.

    (c) The initial metacentric height.

    (d) The location of masses in the ship

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    The period of roll Tr of a ship is determined by the formula:

    $$T_{r}=\frac{2\pi K}{\sqrt{g\times GM}}$$

    where,

    • K is the radius of gyration (mass moment of inertia)
    • g is the acceleration due to gravity, and
    • GM is the metacentric height.
    Part (a)

    Amplitude of Roll:

    • The amplitude of roll, or the maximum angle of heel, does not affect the period of roll. While a larger amplitude implies a greater heeling moment and faster roll speed, the time taken to complete one full roll cycle remains constant for a given metacentric height and radius of gyration.
    Part (b)

    Radius of Gyration (K):

    • The period of roll is directly proportional to the radius of gyration. A larger radius of gyration (indicating a greater distribution of mass further from the ship's centre of rotation) leads to a longer period of roll. Conversely, a smaller radius of gyration (mass concentrated closer to the centre) results in a shorter period. The distribution of cargo significantly impacts K; cargo concentrated centrally minimises K and the roll period, while dispersed cargo maximises K and the roll period.
    Part (c)

    Initial Metacentric Height (GM):

    • The period of roll is inversely proportional to the square root of the metacentric height (GM). A larger GM (a stiffer ship) leads to a shorter roll period, as the ship quickly returns to its upright position. A smaller GM (a tender ship) results in a longer roll period, with slower return to the upright.
    Part (d)

    Location of Masses in the Ship:

    The location of masses in the ship will effect the GM & K. So the period of roll will be affected.

    • If masses are at bottom, G moves down, GM ↑, period of roll ↓.
    • If masses are at top, G moves up, GM ↓, period of roll ↑.
    • If masses are concentrated at centre, K ↓, period of roll ↓.
    • If masses are away from centre, K ↑, period of roll ↑.
    Q5 (10 Marks) Hull Construction πŸ”₯ Repeated 2x

    With respect to Induced Vibrations in a ships hull:

    (a) State FOUR sources of excitation that may induce vibration into the main hull girder.

    (b) Suggest methods for reducing the vibration levels induced by EACH of the exciting forces in (a).

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    Part (a)

    FOUR sources of excitation that may induce vibration into the main hull girder.

    1. Propeller excitation: the propeller produces fluctuating forces (blade frequency and multiples) due to the varying pressure field and the wake, transmitted through the shaft and the water to the hull, causing vibration of the stern and the whole hull girder.
    2. Main engine excitation: the reciprocating and rotating parts of the engine produce unbalanced forces and moments (primary and secondary, and their harmonics) at the engine's firing frequency and multiples, transmitted through the engine seating to the hull.
    3. Wave excitation: the periodic wave forces on the hull (at the encounter frequency) excite the hull girder, particularly in heavy seas, causing springing and whipping.
    4. Auxiliary machinery and other rotating equipment: pumps, generators, compressors and other machinery produce unbalanced forces that are transmitted to the hull through their mountings.
    Part (b)

    Methods for reducing the vibration levels induced by each.

    1. Propeller: increase the clearance between the propeller and the hull (aperture), use a larger blade area and a suitable blade number to avoid resonance, fit a skew or a different blade design, and use a propeller boss cap fin; also avoid operating at resonant speeds.
    2. Main engine: balance the engine (fit balance weights, use a suitable firing order), fit a flexible coupling and a resilient engine mounting, tune the engine speed to avoid the hull's natural frequencies, and use a tuned vibration damper on the crankshaft.
    3. Wave excitation: reduce speed and change course in heavy seas to avoid the resonant encounter frequency, and design the hull to have adequate stiffness and damping; use a hull stress/vibration monitoring system.
    4. Auxiliary machinery: fit resilient (anti-vibration) mountings, balance the rotating parts, isolate the machinery from the hull, and avoid operating at resonant speeds.
    Q6 (10 Marks) Ship Resistance & Propulsion

    The daily fuel consumption of a ship at 17 knots is 42 tonne. Calculate the speed of the ship if the consumption is reduced to 28 tonne per day, and the specific consumption at the reduced speed is 18% more than at 17 knots.

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    Daily fuel consumption at 17 knots = 42 tonne.

    Consumption varies as the cube of speed, but here the specific consumption (consumption per unit power) also changes.

    Let the normal specific consumption be s. At the reduced speed the specific consumption is 1.18s.

    Consumption per day = k x V^3 x (specific consumption), where k is a constant.

    At 17 knots: 42 = k x 17^3 x s.

    At the reduced speed V: 28 = k x V^3 x 1.18s.

    Dividing the second equation by the first:

    28/42 = (V^3/17^3) x 1.18.

    0.6667 = (V^3/4913) x 1.18.

    V^3/4913 = 0.6667/1.18 = 0.5650.

    V^3 = 0.5650 x 4913 = 2775.8.

    V = (2775.8)^(1/3) = 14.05 knots.

    Answer: The ship speed at the reduced consumption = 14.05 knots.

    Q7 (10 Marks) Ship Stability

    A ship 90 m long displaces 5200 tone and floats at draughts of 4.95 m forward and 5.35 m aft when in sea water of 1023 Kg/m2. The waterplane area is 1100m2, GMl 95m, LCB 0.6m forward of midships and LCF 2.2m aft of midships. Calculate the new draughts when the vessel moves into fresh water of 1002 Kg/m3

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    Given:

    $$L\:=\:90m$$

    $$\Delta\:=\:5200\:tonnes$$

    $$d_{f}=\:4.95m$$

    $$d_{a}=\:5.35m$$

    $$\rho_{sw}=\:1023kg/m^3$$

    $$A_{w}=1100m^2$$

    $$G_{ML}=\:95cm$$

    $$LCB=\:0.6m\:fwd\:of\:midship\:$$

    $$LCF=\:2.2m\:aft\:of\:midship$$

    $$when\:vessel\:moves\:into\:fresh\:water\:of\:density\:1002kg/m^3$$

    $$new\:drafts\:=\:?$$

    $$\mathbf{Change\:in\:mean\:draft\:due\:to\:change\:in\:density\:}$$

    $$=\:\frac{100\times\Delta}{A_{w}}\left\lbrack\frac{\rho_{s}\:-\:\rho_{r}}{\rho_{s}\times\rho_{r}}\right\rbrack$$

    $$=\:\frac{100\:\times5200}{1100}\left\lbrack\frac{1.023\:-\:1.002}{1.023\:\times1.002}\right\rbrack=\:9.68\times10^{-3}m$$

    $$Change\:in\:mean\:draft=\:9.7cm$$

    $$MCT_{1cm}=\frac{\Delta\:\times GM_{}_{L}}{100\:\times L}$$

    $$=\:\frac{5200\:\times95}{100\:\times90}$$

    $$MCT_{1cm}=\:54.88\:ton.\:m$$

    $$\mathbf{Change\:in\:trim\:when\:vessel\:moves\:from\:SW\:to\:FW}$$

    $$=\:\frac{\Delta\times FB}{MCT_{1cm}}\left\lbrack\frac{\rho_{s}-\rho_{r}}{\rho_{s}}\right\rbrack$$

    $$FB\:=\:LCF\:+\:LCB$$

    $$FB\:=\:2.2\:+\:0.6\:=\:2.8m$$

    $$=\:\frac{5200\:\times2.8}{54.88}\left\lbrack\frac{1.023\:-\:1.002}{1.023}\right\rbrack\:=\:5.44\:\times10^{-3}$$

    $$Change\:in\:trim\:=\:5.45cm\:by\:head$$

    $$\bm{When\:trim\:by\:head,\:change\:in\:fwd\:draft}$$

    $$d_{f}=\frac{+t}{L}\left\lbrack\frac{L}{2}+LCF\right\rbrack$$

    $$=\:\frac{5.45}{90}\left\lbrack\frac{90}{2}+2.2\right\rbrack$$

    $$d_{f}=\:2.858cm$$

    $$\bm{When\:trim\:by\:head,\:change\:in\:aft\:draft}$$

    $$d_{a}=\:\frac{-t}{L}\left\lbrack\frac{L}{2}-LCF\right\rbrack$$

    $$=\:\frac{-5.45}{90}\left\lbrack\frac{90}{2}-2.2\right\rbrack$$

    $$d_{a}=\:-2.59cm$$

    New draught fwd = draft fwd + change in mean trim + change in fwd draft

    $$=\:4.95\:+\:0.097+0.02858\:$$

    $$D_{f}=\:5.076m$$

    $$New\:aft\:draft\:=\:5.35+0.097-0.0259\:$$

    $$D_{a}=5.421m$$

    $$\bm{New\:fwd\:draft\:D_{f}=5.076m}$$

    $$\bm{New\:aft\:draft\:D_{a}=\:5.421m}$$

    Q8 (10 Marks) Hull Construction

    With respect to Inclining Experiments onboard vessels:

    (a) Sketch and Describe briefly the inclining experiment and explain how the results are used. (6)

    (b) A forward deep tank 12 m long extends from a longitudinal bulkhead to the ship's side. The widths of the tank surface measured from the longitudinal bulkhead at regular intervals are 10, 9, 7, 4 and 1 m. Calculate the second moment of area of the tank surface about a longitudinal axis passing through its centroid (10)

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    Part (a)

    An inclining experiment

    is conducted to determine a ship's metacentric height (GM) and, consequently, the location of its centre of gravity (CG). Knowing the CG of an empty vessel allows for calculations of its position under various loading conditions.

    The inclining experiment is performed on:

    • Newly built ships.
    • After major alterations to the vessel.
    • As required by the classification society.

    Conducting the inclining experiment:

    • The ship should be in a sheltered location (like a dry dock) with mooring ropes slack, only essential personnel on board, all tanks either empty or full, and any loose weights removed or secured.
    • At least two pendulums (one forward, one aft) are used, ideally as long as possible and suspended from convenient points (e.g., under a hatch). A hood filled with water or oil is placed beneath each pendulum bob to dampen its swing for accuracy.
    • Four masses are positioned on the deck, two on each side of the midships, their centres as far from the centerline as possible. These masses are moved systematically: all four to one side, then all four to the other, and finally two on each side.
    • The pendulum deflection is recorded for each mass movement.
    • The average of these deflections is used to calculate the metacentric height (GM).
    Q9 (10 Marks) Ship Stability

    A ship 160m long and 8700 tonne displacement floats at a waterline with Station

    AP 1/2 1 2 3 4 5 6 7 71/2 FP

    1/2 ordinate 0 2.4 5.0 7.3 7.9 8.0 8.0 7.7 5.5 2.8 0m

    While floating at this waterline, the ship develops a list of 10Β° due to instability.

    Calculate the negative metacentric height when the vessel is upright in this condition.

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    1/2 ord

    1/2 ord^3

    SM

    Product for second moment

    0

    0

    1

    0

    2.4

    13.82

    4

    55.28

    5.0

    125

    2

    250

    7.3

    389.02

    4

    1556.08

    7.9

    493.04

    2

    986.08

    8.0

    512

    4

    2048

    8.0

    512

    2

    1024

    7.7

    456.53

    4

    1826.12

    5.5

    166.38

    2

    332.76

    2.8

    21.95

    4

    87.8

    0

    0

    1

    0

    $$Common \space interval \space (h) \space = \space {{L} \over h} \space = \space {{160} \over 10} \space = \space 16 $$

    Second moment of area of waterplane about centreline

    $$=\:2\times{{h} \over9}\times\sum I_{CL}$$

    $$=\:2\times{{16} \over9}\times8166.12$$

    $$=\:29035.1m^4$$

    $$BM=\rho\:\times\frac{\sum I_{CL}}{\Delta}$$

    $$=\:\frac{1.025\times29035.1}{8700}$$

    $$=\:3.421m$$

    $$GZ\:=\:\sin\theta\:\left\lbrack GM\:+\:\frac12BM\tan^2\theta\right\rbrack$$

    $$at\:angle\:of\:LOLL,\:GZ=0$$

    $$0=\sin\theta\left\lbrack GM\:+\:\frac12BM\tan^2\theta\right\rbrack$$

    $$GM + {{1} \over 2} BM \times tan^2\theta \space = 0$$

    $$GM=\:-\frac12BM\:\times\tan^2\theta$$

    $$GM=\:-\frac12\times3.421\times\tan^210$$

    $$GM=-0.0531m$$

    $$Negative\:metacentric\:height\:GM\:=\:-\:0.0531m$$

    Q10 (10 Marks) Hull Construction πŸ”₯ Repeated 4x

    An oil tanker 160m long and 22m beam floats at a draught of 9m in seawater. Cw is 0.865. The midship section is in the form of a rectangle with 1.2m radius at the bilges. A midship tank 10.5m long has twin longitudinal bulkheads and contains oil of 1.4m3/t to a depth of 11.5m. The tank is holed to the sea for the whole of its transverse section. Find the new draught.

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    New draught of the oil tanker when the midship tank is holed.

    Oil tanker 160 m long, 22 m beam, floats at a draught of 9 m in sea water. Cw = 0.865. The midship section is a rectangle with 1.2 m radius at the bilges. A midship tank 10.5 m long has twin longitudinal bulkheads and contains oil of 1.4 m3/t to a depth of 11.5 m. The tank is holed to the sea for the whole of its transverse section. Find the new draught.

    Waterplane area Aw = Cw x L x B = 0.865 x 160 x 22 = 3044.8 m2.

    Midship section area (rectangle with bilge radius r=1.2 m): Ams = B x d - (4 - pi) r^2 = 22 x 9 - 0.858 x 1.44 = 198 - 1.236 = 196.76 m2.

    Volume of the tank below the original waterline = Ams x 10.5 = 196.76 x 10.5 = 2066 m3.

    The tank contains oil of density rho_o = 1/1.4 = 0.714 t/m3. When holed, sea water (1.025 t/m3) replaces the oil, so the net loss of buoyancy is the volume times the relative density difference:

    Vlost = 2066 x (1 - 0.714/1.025) = 2066 x 0.303 = 626 m3.

    The flooded tank provides no increase of buoyancy, so the effective sinking waterplane = Aw - (10.5 x 22) = 3044.8 - 231 = 2813.8 m2.

    Sinkage = Vlost/effective waterplane = 626/2813.8 = 0.222 m.

    New draught = 9 + 0.22 = 9.22 m.

    Answer: the new draught is about 9.2 m.

    Q1 (10 Marks) Hull Construction πŸ”₯ Repeated 5x

    Give a reasoned opinion as to the validity of the following assertions concerning ship structure:

    (a) Crack propagation in propellers shaft 'A' bracket or spectacle frames is indicative of inadequate scantlings and strength.

    (b) The adequate provision of freeing ports is as critical to the seaworthiness as watertight integrity.

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    Part (a)

    Crack propagation in propeller shaft 'A' brackets or spectacle frames - is it indicative of inadequate scantlings and strength?

    This assertion is only partly valid. Cracks in A-brackets or spectacle frames (the shaft brackets supporting the propeller shaft) are more commonly the result of fatigue due to fluctuating loads, stress concentrations at the bracket-to-hull connection, and the vibration and whipping of the shaft, rather than simply inadequate scantlings. The brackets are subject to severe alternating loads from the propeller and the shaft, and cracks typically initiate at stress raisers (sharp corners, weld toes, the bracket arm-to-hull connection) and propagate under fatigue. While inadequate scantlings or poor design (insufficient section, poor connection, sharp notches) can contribute, the primary cause is usually fatigue and stress concentration, aggravated by vibration, corrosion and the dynamic loads of the propeller. Hence the assertion is not fully valid: crack propagation is more indicative of fatigue and stress concentration than of inadequate strength alone, and the design should address the fatigue life, the connection detail and the avoidance of stress raisers, as well as the scantlings.

    Part (b)

    The adequate provision of freeing ports is as critical to seaworthiness as watertight integrity.

    This assertion is largely valid. Freeing ports (openings in the bulwark that allow water shipped on deck to drain overboard) are essential to seaworthiness because, if they are inadequate, water accumulating on the deck cannot drain, which:

    • increases the free-surface effect and the weight of water on deck, reducing stability and increasing the risk of capsize;
    • increases the deck load and the risk of structural damage;
    • reduces the reserve buoyancy and can lead to the ship becoming unstable.

    Watertight integrity (the ability of the hull and its openings to keep water out) is equally critical to seaworthiness, as it prevents flooding and loss of buoyancy. Both are essential: watertight integrity keeps water out, while freeing ports remove water that is shipped on deck. If either is inadequate, the ship's seaworthiness is compromised. Hence the assertion is valid - freeing ports are as critical to seaworthiness as watertight integrity, because they maintain the stability and buoyancy of the ship by removing deck water.

    Q2 (10 Marks) Hull Construction

    With respect to the sacrificial anodes fitted to a ship's hull.

    (a) State the purpose of fitting anodes to a hull structure.

    (b) Describe with the aid of sketch, how anodes may be attached to the hull.

    (c) Upon inspection in dry-dock it is found that the anodes have not wasted and areas of hull structure have experienced severe corrosion. Explain possible reasons for this situation

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    Part (a)

    Purpose of Sacrificial Anodes:

    Sacrificial anodes are fitted to a ship's hull to prevent corrosion. Seawater is highly corrosive due to its salt content, which acts as an electrolyte, facilitating the oxidation of iron (mild steel) and the formation of rust. Zinc anodes, being more reactive than steel, act as a sacrificial anode. They corrode preferentially, protecting the steel hull from corrosion. This is a form of cathodic protection.

    Part (b)

    Attachment of anode to the hull:

    Part (c)

    Reasons for Unwasted Anodes and Hull Corrosion:

    • The anodes may have been manufactured from substandard material that does not possess the required electrochemical properties to act as a sacrificial metal.
    • If the wrong type of material (e.g., aluminium or magnesium instead of zinc, or improperly alloyed zinc) was supplied or installed, the anodes may not function as intended.
    • If the anodes are not electrically connected to the hull (e.g., due to improper installation or poor contact), they cannot provide cathodic protection.
    • Incorrect placement of the anodes may leave sections of the hull unprotected, leading to corrosion in those areas.
    • Poor-quality or damaged coatings on the hull can expose bare steel, which may corrode faster than the anodes can protect it.
    • Stray currents from other sources (e.g., from shore connections, dock equipment, or adjacent vessels) may bypass the anodes and accelerate hull corrosion.
    • The anodes require continuous exposure to seawater to create the necessary electrochemical reaction. If seawater flow is obstructed, the anodes may remain inactive.
    Q3 (10 Marks) Ship Types & Design πŸ”₯ Repeated 4x

    (a) Draw and the mid ships section of an oil tanker with Double Hull & name each part.

    (b) What is Bow Flare? Why is it so important in Bulk Carriers?

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    Part (a)

    Mid-ship section of Oil tanker:

    Part (b)

    Bow flare

    is the outward curvature of a ship's side shell above the waterline at the forward end.

    Importance of Bow Flare in Bulk Carriers:

    • Bow flare enhances the reserve buoyancy at the forward end of the vessel, which improves seaworthiness by helping the ship ride over waves more effectively, especially when pitching in rough seas.
    • By dispersing water away from the ship, the bow flare reduces the amount of water shipped onto the deck during heavy weather.
    • The wider forecastle deck created by the bow flare allows for the installation of essential machinery such as windlasses, mooring equipment, and other fittings.
    • The bow flare shields the hull plating from damage caused by the anchor when it is raised or lowered.
    • A well-designed bow flare can reduce water resistance, leading to increased speed and better fuel efficiency.
    Q4 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 4x

    (a) What is free surface effect? How can be avoided or reduced.

    (b) Give the components of ships resistance while vessel is 'enroute'.

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    (a) Free Surface Effect:

    The free surface effect (FSE) is a reduction in the metacentric height (GM) of a vessel due to the movement of liquids within partially filled tanks when the ship heels. When a ship tilts, the liquid in a partially filled tank shifts to the lower side, causing the centre of gravity (CG) of the ship to move laterally. This lateral shift of the CG reduces the righting lever (GZ), effectively decreasing the ship's stability and increasing the angle of heel. This apparent loss of GM is the free surface effect.

    Minimizing Free Surface Effect:

    • Partially filled tanks should be avoided to minimize liquid movement.
    • Tanks should be designed with longitudinal divisions or swash bulkheads to limit the movement of liquid.
    • Install sluice valves to control the liquid movement between compartments in divided tanks.
    • Fill smaller tanks at the bottom of the ship first to lower the centre of gravity and improve stability.
    • Tanks should have reduced breadth to minimize the free surface's effect.
    Part (b)

    Components of ship’s resistance while vessel is en route

    When a ship moves through water, resistance opposes its motion. The ship must exert an equal force to maintain speed.

    Frictional resistance (Rf): This is caused by the friction between the hull and the water. The water immediately adjacent to the hull is dragged along, creating a boundary layer. This resistance depends on the water's viscosity, the ship's speed, and the wetted surface area of the hull. At lower speeds, frictional resistance can account for 70-90% of total resistance; however, at higher speeds, it can be less than 40%.

    Residuary resistance (Rr): This is the resistance that remains after subtracting the frictional resistance. These are:

    • Form drag: Resistance due to the shape of the hull and the flow of water around it, generating pressure differences.
    • Wave-making resistance: This is a major component at higher speeds. The ship creates waves, and energy is expended in this process.
    • Eddy resistance: Resistance caused by turbulent flow behind the ship, especially at sharp changes in the hull's shape. This is often minimized by optimizing the hull design.

    Air resistance (Ra): This resistance is generated by the ship moving through the air. It depends on the shape of the superstructure, the projected area above the waterline, and wind speed and direction. Air resistance is typically a smaller component, usually around 2% but can be up to 10% for large container ships with extensive superstructure.

    Total resistance (Rt): The total resistance experienced by the ship is the sum of frictional, residuary, and air resistance: Rt = Rf + Rr + Ra.

    Q5 (10 Marks) Ship Types & Design πŸ”₯ Repeated 6x

    (a) Considering the vessel as a compound beam define Bending moment shearing force. Which is the point of Maximum Bending Moment?

    (b) Sketch and Describe Hatch coaming of a large bulk carrier.

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    Part (a)

    Bending moment and shearing force, and the point of maximum bending moment.

    Considering the vessel as a compound beam (the hull girder), the shearing force at any section is the algebraic sum of the vertical forces (loads) to one side of the section, i.e. the net load (weight - buoyancy) acting on that part. The bending moment at any section is the algebraic sum of the moments of the loads to one side, i.e. the integral of the shearing force. The shearing force is the rate of change of the bending moment, and the bending moment is the integral of the shearing force. The point of maximum bending moment occurs where the shearing force is zero (where the shearing force changes sign), which is usually at or near midships for a ship in still water, and at the point where the net load changes sign. The maximum bending moment is the largest hogging or sagging moment, and the hull girder must be designed to withstand it.

    Part (b)

    Hatch coaming of a large bulk carrier.

    The hatch coaming is the vertical structure around the hatch opening that raises the hatch above the deck to prevent water entering and to provide strength. Sketch: the hatch opening is bounded by a vertical coaming plate (about 600-900 mm high for a bulk carrier) welded to the deck, with a top flange (or a horizontal stiffener) and vertical stiffeners (brackets) connecting the coaming to the deck. The coaming is made of thick plate and is stiffened to resist the loads of the hatch cover and the cargo, and to provide the longitudinal strength of the deck (the coaming acts as a longitudinal girder). The hatch cover sits on the coaming with a gasket and is secured by cleats. The coaming corners are rounded and reinforced to avoid stress concentrations. The coaming provides the watertight seal for the hatch and contributes to the longitudinal strength of the hull girder.

    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 7x

    (a) List the precautions necessary before an inclining experiment is carried out. (6)

    (b) A box shaped vessel, 50 metres long x 10 metres wide, floats in salt water on an even keel at a draft of 4 metres. A center line longitudinal watertight bulkhead extends from end to end and for the full depth of the vessel. A compartment amidships on the starboard side is 15 metres long and contains cargo with permeability 30%. Calculate the list if this compartment is bilged. KG = 3 metres. (10)

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    Part (a)

    Precautions necessary before an Inclining experiment:

    • The ship should be in a sheltered position, such as a gravity dock, to minimize external influences from wind or waves.
    • Mooring ropes should be slack to allow the ship to move freely without restrictions.
    • Only workers essential to the experiment should be present onboard to avoid unnecessary weight and movement.
    • All tanks must either be completely empty or pressed up tight to eliminate the free surface effect, which can adversely affect stability calculations.
    • Any loose weights on the ship must either be removed or properly secured to prevent unintended movement during the experiment.
    • Ensure that the pendulums are long and properly suspended from stable points, such as underneath a hatch, to provide accurate deflection readings.
    • The test masses should be evenly distributed and placed as far from the centerline as possible to maximize measurable deflections.
    • The experiment should be carried out in calm weather to avoid the effects of wind, current, or waves on the vessel's stability.
    Q7 (10 Marks) Ship Stability πŸ”₯ Repeated 9x

    (a) Describe how the force on the ship's bottom and the GM vary when grounding takes place (6)

    (b) A ship of 8,000 tones displacement takes the ground on a sand bank on a falling tide at an even keel draft of 5.2 metres. KG 4.0 metres. The predicted depth of water over the sand bank at the following low water is 3.2 metres. Calculate the GM at this time assuming that the KM will then be 5.0 metres and that the mean TPC is 15 tonne (10)

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    Part (a)

    When a ship grounds, the forces on the bottom and the metacentric height (GM) change depending on the grounding's nature and the ship's position on the seabed.

    If the ship grounds on a level bottom:

    • A ground reaction force acts vertically upwards from the seabed. This force counteracts part of the ship's weight and alters the distribution of buoyancy forces.
    • The ship’s centre of gravity (G) may appear to rise relative to the metacentre (M) because the upward ground reaction reduces the buoyancy force acting on the underwater volume.
    • This virtual rise in G reduces GM, potentially leading to a list.
    • If the list becomes excessive and the righting moment is insufficient, the ship may capsize.

    If the ship grounds on a pinnacle:

    • The ship experiences two forces at the ship's bottom:
      • A downward force due to the ship’s weight.
      • An upward reaction force is concentrated on the pinnacle.
    • The resulting force between the grounding pressure and the ship’s centre of buoyancy shifts downward towards the pinnacle.
    • This situation is similar to when the ship's stern touches the keel block in a dry dock.
    • A virtual loss of GM occurs because the ship’s inclining moment may exceed the maximum righting lever.
    • If the inclining moment is too great, the ship may develop an excessive list or even capsize.

    (b) Given:

    $$Displacement,\:\Delta=8000\:tonnes$$

    $$TPC=15\:tonnes$$

    $$Initial\:Draught=5.2m$$

    $$Final\:Draught=3.2m$$

    $$Ship\:KG=4.0m$$

    $$KM=5.0m$$

    To find GM

    $$Uptrust,\:P=TPC\times Fall\:in\:water\:level\:\left(cm\right)$$

    $$=15\times\left(520-320\right)$$

    $$=15\times200$$

    $$P=3000\:tonnes$$

    To Find Virtual loss of GM:

    $$Virtual\:loss\:of\:GM_1=\frac{P\times KM}{\Delta}$$

    $$=\frac{3000\times5}{8000}$$

    $$=\frac{15000}{8000}$$

    $$GM_1=1.88m$$

    Actual KM = 5.0m (given)

    $$Virutal\:KM=Actual\:KM-Virtual\:loss\:of\:GM_1$$

    $$=5.0-1.88$$

    $$=3.12$$

    Similarly, Actual KG = 4.0m (given)

    $$New\:GM=Virtual\:KM-\:Actual\:KG$$

    $$=3.12-4.0$$

    $$=-0.88$$

    Q8 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 7x

    With reference to International Load Line Statutory Certification:

    (a) State the reason for the freeboard requirements.

    (b) Explain the term conditions of assignments.

    (c) List the items that may be examined during a Load line survey after major repairs in the drydock.

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    Part (a)

    Reasons for Freeboard Requirements:

    Freeboard is the distance measured from the waterline to the upper edge of the deck plating at the side of the freeboard deck amidships.

    Purpose of Freeboard:

    • Ensures the ship is seaworthy when fully loaded.
    • Provides reserve buoyancy, enabling the ship to rise as it passes through waves, keeping decks largely dry.
    • Enhances the ship's stability and increases its survivability in case of damage by allowing it to remain afloat longer, aiding crew escape or damage control.
    Part (b)

    (i) Conditions of Assignment:

    Conditions of Assignment are outlined in the Merchant Shipping Load Line Rules 1968 and must be satisfied before a ship is assigned freeboards and issued a load line certificate. These conditions address the practical need for openings (hatchways, doorways, vents, etc.) in the ship's hull and mandate appropriate protection and closure mechanisms for these openings.

    Requirements Before Assigning Load Line:

    • The ship must have sufficient structural strength.
    • Adequate reserve buoyancy must be maintained.
    • Openings must be secured against water ingress.
    • Safety measures for the crew, such as guardrails and gangways, must be in place.
    Part (b)

    (ii) Items Examined During a Load Line Survey After Major Repairs in Drydock:

    • Openings: This encompasses hatchways, machinery space openings, cargo ports, watertight doors, ventilators, air pipes, scuppers, freeing ports, side scuttles, and other openings in the freeboard and superstructure decks. The surveyor checks the condition of seals, gaskets, closing mechanisms, and overall watertightness.
    • Crew Protection: Safety features designed to protect the crew, such as guardrails, bulwarks, and gangways, are inspected for damage or deterioration and proper functionality.
    • Hull Structure: The hull itself is inspected for any damage, corrosion, or structural weaknesses that could compromise watertightness or strength. This often includes visual inspection for cracks, buckling, indentation, and paint adhesion. Bottom shell, bilge keel, stem and stern frames, rudder, sea chests, side ports, stern bearing, and propeller are all checked.
    • Other Systems: Depending on the scope of repairs, other systems may be examined, including but not limited to: pressure-vacuum valves, sounding pipes, air vents, access hatches, ventilation fans, dampers, weathertight doors and sealings, door securing arrangements, side scuttles, windows, and skylights. Testing, such as hose tests (to check watertight integrity), might also be conducted.
    Q9 (10 Marks) Ship Stability πŸ”₯ Repeated 11x

    (a) Define longitudinal center of gravity (LCG) and longitudinal center of buoyancy (LCB). (6)

    (b) A ship 120m long floats at draughts of 5.50m forward and 5.80m aft; MCTI cm 80 tonne m, TPC 13, LCF 2.5m forward of midships. Calculate the new draughts when a mass of 110 tonne is added 24m aft of midships.

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.

    LCF in fwd and trim by stern

    $$Bodily \space sinkage \space = \space {{mass \space added} \over TPC} \space = \space {{110} \over 13} \space$$

    $$=\:8.5\operatorname{\mathrm{cm}}\:=0.085m$$

    $$Trim \space = \space {{m \times d} \over MCT_{1cm}}$$

    $$ = \space {{110 \times (24 + 2.5)} \over 80}$$

    $$Trim=36.43\operatorname{cm}=0.364m\:$$

    Change in fwd draught:

    $$d_{F}=\frac{-t}{L}\left\lbrack\frac{L}{2}-LCF\right\rbrack$$

    $$=\frac{-36.43}{120}\left\lbrack\frac{120}{2}-2.5\right\rbrack$$

    $$=-17.45\operatorname{cm}=-0.1745m$$

    Change in Aft draught:

    $$d_{A}=\frac{+t}{L}\left\lbrack\frac{L}{2}+LCF\right\rbrack$$

    $$=\frac{+36.43}{120}\left\lbrack\frac{120}{2}+2.5\right\rbrack$$

    $$=+18.97\operatorname{cm}=0.189m$$

    New draughts:

    $$D_{F}=5.5+0.085-0.175=5.41m$$

    $$D_{A}=5.8+0.085+0.18=6.065m$$

    Q10 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 8x

    With reference to fixed pitch propellers:

    (a) Explain Propeller Slip and Propeller Thrust.

    (b) The shaft power of a ship is 3000 KW, the ship's speed V is 13.2 knot. Propeller rps is 1.27. propeller pitch is 5.5m and the speed of advance is 11 Knots. Find:

    (i) Real Slip

    (ii) Wake fraction

    (iii) Propeller thrust, when its efficiency, n = 70%

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    Part (a)

    Slip

    is the difference between the theoretical distance the propeller should travel in one revolution and the actual distance the vessel travels.

    $$Slip\:\left(\%\right)\:=\:\left(\frac{Engine\:distance\:-\:Ship^{\prime}s\:distance}{Engine\:distance}\right)\:\times100$$

    Where Engine distance = no. of propeller revolutions * propeller pitch (usually calculated over a 24-hour period)

    Propeller thrust: it is the force exerted by the propeller to move the vessel ahead and given by

    $$Thrust\:\left(T\right)\:=\:\rho\times A\times P^2\times n^2\times S$$

    Where,

    ρ - Density

    A - Area

    P - Pitch

    n - Revolution per second

    S - Slip

    Q1 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 9x

    With reference to Underwater Inspection in lieu of Dry docking (UWILD):

    (a) Explain in detail, how an underwater survey is carried out.

    (b) State the requirements to be fulfilled before an underwater survey is acceptable to the surveying authority.

    (c) Construct a list of the items in order of importance that the underwater survey authority should include

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    (a) An underwater survey, also known as Underwater Inspection in Lieu of Dry Docking (UWILD), involves a systematic and detailed examination of a vessel’s hull and underwater components while the ship remains afloat. The procedure includes the following steps:
    • The shipowner sends a request to the classification society surveyor, indicating the intention to perform an underwater survey.
    • A detailed plan of the ship's external hull features is submitted, showing the location of frames, bulkheads, welding lines, openings, etc.
    • The hull surface is cleaned before the survey to remove any marine growth or fouling that could obscure the inspection.
    • A diving company, approved by the classification society, is appointed to conduct the underwater inspection.
  • A self-propelled survey vehicle equipped with the following tools is used:
    • Long-Range Light TV Camera to aid in steering and checking hull deterioration.
    • High-resolution colour TV Camera to provide a close-up view of the hull coating and welded seams.
    • 35mm Still Camera to capture still images.
    • Ultrasonic Probe for measuring plate thickness.
    • Depth Meter and Speed Indicator to provide accurate data on the vehicle's depth and movement.
    • Umbilical Cable to connect the survey vehicle to the survey boat, transmitting power and information.
  • The survey boat is to be equipped with:
    • A control console with TV monitors.
    • Plate thickness printouts.
    • Audio and video cassette recorders.
    • Playback units.
    • Diver communication systems.
    • Vehicle control systems and associated instruments.
  • Operation:
    • The survey vehicle is taken underwater by a diver to the survey starting point.
    • Using TV monitors and shell expansion plans as guides, the vehicle is navigated over the hull, focusing on the bottom structure, sides, stern frame, rudder, propeller, bilge keel, and hull openings.
    • All images, data, and information are recorded and transmitted back to the survey boat.
    • Detailed pictures of the stern frame, rudder, propeller, bilge keel, and hull openings are captured.
    • Divers are deployed to measure stern tube bearing wear, pintle clearance, and inspect stern seals, anodes, and rudder stock palm coupling bolts.
    • All recorded video and audio, including conversations between the surveyor and drivers, thickness printouts, measurements, and pictures are analyzed to determine the detailed underwater condition of the vessel.
    (b) Before an underwater survey is accepted by the surveying authority, the following conditions must be met:

    The vessel's owner submits a request to the surveyor, including:
    • The proposed date and location for the survey.
    • General information about the diving company.
    • A declaration that the vessel has not suffered any damage due to grounding, collision, or other incidents.
    The ship's master or owner’s representative must provide a declaration confirming:
    • Any suspected or actual damage to the hull since the last dry-docking.
    • The underwater portion of the hull is protected by a suitable paint scheme that is of adequate thickness and remains valid until the next dry-dock.

    • The survey site should be in a protected area with calm and clear water, ensuring good underwater visibility. Attention must be given to the effects of currents and tides.
    • The hull must be clean for the external survey. The surveyor must be satisfied with the method and quality of the pictorial presentation, ensuring that it provides a reliable assessment of the hull's condition.
    • The underwater examination should be conducted by an approved diving company using closed-circuit TV and two-way communication, which can be monitored by the surveyor.
    • The vessel should be in as light an operating condition as possible to facilitate the survey.
    • Means must be available for the surveyor to examine the outside shell plating above the waterline.
    • Any required repairs identified during the survey must be carried out to the satisfaction of the attending surveyor.
    (c) While the importance of each item may vary depending on the vessel and its specific requirements, below is a list of items that should be included in an underwater survey in order of importance:
    • Underwater Hull: General condition of the hull below the waterline.
    • Bottom and Shell Plating: Inspection for corrosion, damage, and fouling.
    • Shell Openings: Examination of openings such as sea chests, drain plugs, and overboard discharge points.
    • Stern Tube Oil Leaks: Check for leaks around the stern tube.
    • Propeller Blade: Inspection for damage, wear, and fouling.
    • Rudder: Inspection for damage, wear, and clearances.
    • Sea Chest Opening and Grating: Examination for blockages, damage, and fouling.
    • Anodes: Check the condition and effectiveness of cathodic protection anodes.
    • Bilge Keel: Inspection for damage and fouling.
    • Drain Plugs: Ensure all drain plugs are secure and in good condition.
    • Overboard Valve Openings: Check for proper operation and condition.
    • Forward Area: Inspection for any damage due to anchor and chain movement.
    Q2 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 8x

    Vessel has gone through very heavy weather. On arrival at safe anchorage, you are conducting your inspection to determine damages to hull.

    (a) List the areas you will inspect.

    (b) List your findings of any significance.

    Write a report to company suggesting repairs if any

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    Part (a)

    Areas to Inspect After Heavy Weather

    Upon arrival at safe anchorage, a thorough inspection of the vessel shall be carried out, both externally and internally, to assess any weather-related damage. Key areas include:

    1. Hull and Main Deck

    • Shell plating along bow, midship, and stern sections (especially forepeak and flare areas).
    • Boot-top and bilge areas for dents or deformation.
    • Deck plating for buckling or cracked welds.
    • Bulwarks, rails, fairleads, chocks, and mooring fittings.

    2. Forecastle and Forward Structure

    • Bosun store and chain locker for water ingress.
    • Windlass foundation, anchors, and hawse pipes for deformation or cracks.
    • Forepeak tank for leakage or pressure damage.

    3. Cargo Holds / Tanks

    • Inspect for structural deformation, loose frames, or fractured stiffeners.
    • Check tank top plating and bilges for leakage.
    • Check watertight doors, gaskets, and vents.

    4. Superstructure and Deck Fittings

    • Bridge wings, radar mast, funnel, vents, and deckhouses for cracks or loose fittings.
    • Lifeboat davits, securing arrangements, and deck cranes.

    5. Underwater and Machinery Spaces

    • Rudder, propeller, and stern tube seals (via steering gear tests).
    • Sea chest gratings and overboard discharges.
    • Engine room bilges for any seawater ingress.

    Part (b)

    Typical Findings of Significance

    • Dents and indentations on shell plating at bow and forward port side due to heavy slamming.
    • Paint coating and corrosion protection partly peeled off near waterline and forepeak area.
    • Loose fairlead bolts and one cracked weld on starboard bulwark stanchion.
    • Deformed ventilator head on forecastle deck.
    • Minor leakage observed in forepeak tank during sounding check.
    • Bridge wing railing bent, likely from green sea impact.
    • Anchor chain links twisted and worn.
    • Lifeboat gripes loosened, requiring tightening and inspection.
    • No flooding reported; watertight integrity maintained overall.

    Part (c)

    Report to Company – Heavy Weather Damage Inspection

    To: Superintendent / Technical Department

    From: Name / Rank

    Subject: Heavy Weather Damage Inspection Report

    Date: [Insert date]

    Vessel: [Insert vessel name]

    Summary

    The vessel experienced very heavy weather en route from [Port A] to [Port B], with significant pitching and rolling in seas up to [X] meters. On arrival at safe anchorage, a complete inspection of the hull and deck was carried out.

    Findings

    • Bow and forecastle plating show minor dents, with no breach of watertight integrity.
    • Bulwark stanchion (starboard side) cracked at welded joint – requires repair.
    • Fairlead foundation bolts loosened – retightening and NDT inspection recommended.
    • Ventilator head on forecastle deformed – replacement advised.
    • Forepeak tank shows trace leakage at forward bulkhead – to be pressure tested.
    • Paint coating damage and corrosion exposure on bow area – recoating required.
    • Bridge wing railing bent – to be straightened or renewed.
    • All other structures and machinery appear satisfactory after testing.

    Recommendations

    1. Conduct minor steel renewal and welding repairs at the next port with repair facilities.
    2. Carry out thickness measurements and NDT on affected hull areas.
    3. Recoat damaged paint areas to prevent corrosion.
    4. Replace deformed ventilator head and bent railing.
    5. Inspect anchor and chain for elongation; renew worn links.
    6. Pressure test forepeak tank after repairs.
    7. Submit class surveyor report if deemed necessary by the superintendent.

    Conclusion

    The vessel remains seaworthy but requires prompt attention to minor structural and fitting damages before the next voyage. Preventive maintenance and weather routing measures should be reviewed for future passages.

    Signed:

    Name / Rank

    Signature

    Q3 (10 Marks) Ship Stability πŸ”₯ Repeated 14x

    Explain how the period of roll varies with:

    (a) The amplitude of roll.

    (b) The radius of gyration.

    (c) The initial metacentric height.

    (d) The location of masses in the ship

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    The period of roll Tr of a ship is determined by the formula:

    $$T_{r}=\frac{2\pi K}{\sqrt{g\times GM}}$$

    where,

    • K is the radius of gyration (mass moment of inertia)
    • g is the acceleration due to gravity, and
    • GM is the metacentric height.
    Part (a)

    Amplitude of Roll:

    • The amplitude of roll, or the maximum angle of heel, does not affect the period of roll. While a larger amplitude implies a greater heeling moment and faster roll speed, the time taken to complete one full roll cycle remains constant for a given metacentric height and radius of gyration.
    Part (b)

    Radius of Gyration (K):

    • The period of roll is directly proportional to the radius of gyration. A larger radius of gyration (indicating a greater distribution of mass further from the ship's centre of rotation) leads to a longer period of roll. Conversely, a smaller radius of gyration (mass concentrated closer to the centre) results in a shorter period. The distribution of cargo significantly impacts K; cargo concentrated centrally minimises K and the roll period, while dispersed cargo maximises K and the roll period.
    Part (c)

    Initial Metacentric Height (GM):

    • The period of roll is inversely proportional to the square root of the metacentric height (GM). A larger GM (a stiffer ship) leads to a shorter roll period, as the ship quickly returns to its upright position. A smaller GM (a tender ship) results in a longer roll period, with slower return to the upright.
    Part (d)

    Location of Masses in the Ship:

    The location of masses in the ship will effect the GM & K. So the period of roll will be affected.

    • If masses are at bottom, G moves down, GM ↑, period of roll ↓.
    • If masses are at top, G moves up, GM ↓, period of roll ↑.
    • If masses are concentrated at centre, K ↓, period of roll ↓.
    • If masses are away from centre, K ↑, period of roll ↑.
    Q4 (10 Marks) General πŸ”₯ Repeated 7x

    List SIX hazards that arise with the carriage of liquefied gas in bulk. Describe, with the aid of a sketch. the details of construction of a prismatic cargo tank within a gas carrier designed to carry liquefied gas (LPG)

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    Hazards of Carriage of Liquefied Gas in Bulk and Construction of a Prismatic Cargo Tank

    Part (a)

    Six Hazards Associated with the Carriage of Liquefied Gas in Bulk

    1. Flammability and Explosion
      • Liquefied gases such as LPG vaporise rapidly when released.
      • The vapour can mix with air and form a highly flammable or explosive atmosphere, creating a serious risk of fire or explosion.
    2. Toxicity
      • Some liquefied gas cargoes, such as ammonia and vinyl chloride monomer, are highly toxic.
      • Exposure can cause serious poisoning through inhalation, ingestion or skin absorption.
    3. Asphyxiation
      • LPG vapours are generally heavier than air.
      • In the event of a leak, the vapour can collect in low-lying areas such as pump rooms and hold spaces, displacing oxygen and creating an asphyxiation hazard.
    4. Frostbite and Cold Burns
      • Liquefied gas cargoes are carried at very low temperatures, with fully refrigerated LPG being carried at approximately βˆ’50Β°C.
      • Contact with the liquid cargo or uninsulated pipes and equipment can cause severe frostbite and cold burns.
    5. Brittle Fracture
      • Ordinary ship hull steel, such as mild steel, can become brittle at very low temperatures.
      • If cold cargo leaks and comes into contact with unsuitable hull steel, it may cause cracking and catastrophic structural failure.
    6. Sloshing
      • Partially filled tanks have a free surface, allowing the liquid cargo to move violently with the ship's motion.
      • This can produce large dynamic impact loads on the tank walls and internal pump towers, potentially causing structural damage.
    Part (b)

    Construction of a Prismatic Cargo Tank for LPG

    Fully refrigerated LPG is typically carried in Independent Type A prismatic cargo tanks. These are self-supporting tanks that are independent of the ship's hull structure and do not contribute to the overall structural strength of the vessel.

    Sketch – Typical Prismatic Cargo Tank Arrangement

    Ship's Hull / Hold Space

    β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”

    β”‚ Secondary Barrier / Hull β”‚

    β”‚ β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚

    β”‚ β”‚ Thermal Insulation β”‚ β”‚

    β”‚ β”‚ β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚ β”‚

    β”‚ β”‚ β”‚ β”‚ β”‚ β”‚

    β”‚ β”‚ β”‚ LPG CARGO β”‚ β”‚ β”‚

    β”‚ β”‚ β”‚ PRISMATIC TANK β”‚ β”‚ β”‚

    β”‚ β”‚ β”‚ β”‚ β”‚ β”‚

    β”‚ β”‚ β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ β”‚ β”‚

    β”‚ β”‚ Primary Barrier β”‚ β”‚

    β”‚ β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ β”‚

    β”‚ Load-bearing supports β”‚

    β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜

    β”‚

    Double Bottom

    1. Shape and Structure

    • The tank is prismatic, meaning it is generally box-shaped with chamfered/angled top and bottom corners.
    • This shape allows the tank to closely follow the contours of the ship's inner hull and therefore maximises the available cargo capacity.
    • The tank is internally reinforced with frames and stiffeners to maintain structural integrity.
    • A centreline longitudinal bulkhead, together with transverse wash bulkheads, may be provided to strengthen the tank and reduce liquid movement and sloshing.

    2. Materials

    • LPG is carried at low temperatures, approximately βˆ’50Β°C for fully refrigerated LPG.
    • To prevent brittle fracture at these temperatures, the primary barrier/tank is constructed from suitable low-temperature-resistant materials, typically fine-grained carbon-manganese steel.

    3. Tank Supports and Chocks

    The cargo tank operates at a substantially different temperature from the ship's hull and therefore undergoes thermal expansion and contraction.

    • The tank is supported on the double bottom by load-bearing insulation blocks.
    • These may be made from specialised hardwood such as AzobΓ© or suitable synthetic materials.
    • The supports provide the necessary load-bearing capacity while reducing thermal transfer and structural stresses.
    • Anti-roll, anti-pitch and anti-flotation keys/chocks secure the tank against movement caused by the ship's motion.
    • At the same time, the arrangement allows the tank to expand and contract freely due to temperature changes.

    4. Insulation

    • The outside of the primary barrier is provided with high-efficiency thermal insulation.
    • Its purpose is to maintain the required low cargo temperature and minimise heat ingress and cargo boil-off.
    • Sprayed polyurethane foam (PUF) is commonly used as the insulation material.

    5. Secondary Barrier

    • Under the IGC Code, Type A tanks are required to have a complete secondary barrier capable of containing leaked cargo for up to 15 days, preventing the cold cargo from coming into contact with the outer hull.
    • In LPG carriers, the ship's inner hull is normally used as the secondary barrier.
    • The inner hull is also constructed from suitable fine-grained low-temperature steel so that it can withstand the low temperature if the primary cargo tank fails.
    • The space between the primary tank and secondary barrier, known as the hold space, is maintained with inert gas or dry air as applicable.
    Q5 (10 Marks) Hull Construction πŸ”₯ Repeated 10x

    Describe a method for the attachment of bilge keels. State THREE reasons for not extending bilge keels for the entire length of the vessel. Explain TWO principles of roll damping that bilge keels exploit.

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    Part (a)

    Method of Bilge keel attachment to the hull:

    Bilge keels are fitted port and starboard at the turn of the bilge. They do not extend outside the lines of the side and bottom shell. The bilge keels are attached directly in line with an internal stiffening member such as a girder or longitudinal. The bilge keel comprises a flat bar doubler welded directly to the shell, and an offset bulb plate (OBP) with 'scallops' cut in it is welded to the flat bar doubler. The ends of the bilge keels are tapered (minimum 3 in 1) and will end in line with transverse internal stiffening, such as a frame. Using a hull doubler protects the hull in case of damage to the keel, as the crack would not extend into the hull.

    Part (b)

    Bilge keels are not fitted for the full length of the vessel because:

    • The 'lever' to the ship's axis of rotation is reduced at the ends;
    • The ineffectiveness, the closer the bilge keel is to the rolling centre, increased resistance and more likelihood of damage.
    • The hydrodynamic effect would cause a large increase in resistance and fuel consumption;
    • At the aft end, the boundary layer is much thicker, and since the keel would not project through, it would have a much reduced effect.
    Part (c)

    Two principles of roll damping exploited by Bilge keels:

    Increased Roll Period:

    • Bilge keels increase the ship's roll period (the time it takes for the vessel to complete one roll). This is achieved by increasing the moment of inertia (K) of the vessel. A longer roll period means the vessel rolls more slowly, thus reducing the amplitude of the roll.

    $$T_{r}=2\pi\:\frac{k}{\sqrt{g\times GM}}$$

    Where:

    • g: acceleration due to gravity
    • GM: metacentric height
    • k: mass moment of inertia

    By increasing k, the ship's stability during roll improves.

    Hydrodynamic Resistance:

    • As the vessel rolls, the bilge keels move through the water, creating pressure differences. Water pressure on one side of the keel opposes the rolling motion, providing a damping effect. This is due to the interaction between the bilge keel and the water, creating hydrodynamic forces that counteract the rolling motion.
    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 8x

    (a) Define longitudinal center of gravity (LCG) and longitudinal centre of buoyancy (LCB). (6)

    (b) The immersed cross-sectional areas of a ship 120 m long, commencing from aft are 2, 40, 79, 100, 103, 104, 104, 103, 97, 58 and 0 m2.

    Calculate:

    (i) Displacement

    (ii) Longitudinal position of the centre of buoyancy.

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.
    Q7 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 7x

    A ship of 15000 tonne displacement has an Admiralty Coefficient, based on shaft power, of 420. The mechanical efficiency of the machinery is 83%, shaft losses 6%, propeller efficiency 65% and QPC 0.71. At a particular speed the thrust power is 2550kW.

    Calculate: (16)

    (i) indicated power

    (ii) effective power

    (iii) ship speed

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    Given:

    $$\Delta=15000t$$

    $$Shaft\:Power\:\left(SP\right)=420$$

    $$Transmission\:Efficiency=83\%$$

    $$Shaft\:losses=6\%$$

    $$Propeller\:Efficiency=65\%$$

    $$QPC=0.71$$

    $$Thrust\:Power=2550kW$$

    $$\left(\imaginaryI\right)\:Delivered\:Power\:\left(DP\right)=\frac{Thrust\:Power\:\left(TP\right)}{Propeller\:Efficiency\:\left(\eta P\right)}$$

    $$DP=\frac{2550}{0.65}$$

    $$DP=3923.07kW$$

    $$\left(ii\right)\:Shaft\:Power=\frac{Delivered\:Power\:\left(DP\right)}{Transmission\:Efficiency\:\left(\eta T\right)}\:$$

    $$SP=\frac{3923.07}{0.94}$$

    $$SP=4173.47kW$$

    $$\left(iii\right)\:Indicated\:Power=\frac{Shaft\:Power\:\left(SP\right)}{Mechanical\:Efficiency\:\left(\eta m\right)}$$

    $$IP=\frac{4173.47}{0.83}$$

    $$IP=5028.28kW$$

    $$\left(iv\right)\:Effective\:Power=DP\times QPC$$

    $$EP=3923.07\times0.71$$

    $$EP=2785.3797kW$$

    $$\left(v\right)\:Shaft\:Power=\frac{\Delta^{\frac23}\times V^3}{Admiralty\:Co-efficient}$$

    $$4173.47=\frac{15000^{\frac23}\times V^3}{420}$$

    $$V=14.23knots$$

    Q8 (10 Marks) Ship Stability πŸ”₯ Repeated 5x

    (a) Describe stability requirements for dry-docking. (6)

    (b) A ship of 8000t displacement floats upright in sea water, with KG = 7.6m, GM = 0.5m. A tank, whose Kg is 0.6m above the keel and 3.5m from the center line contains 100 t of water ballast. Neglecting the free surface effect, calculate the angle which the ship will heel, when the ballast water is pumped out. (10)

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    Part (a)

    For safe dry-docking, a ship must meet two key stability requirements:

    • Positive GM (Metacentric Height): The ship needs a positive GM. GM is the distance between the centre of gravity (G) and the metacentre (M). A positive GM indicates inherent stability; the ship will right itself if tilted. During dry-docking, the loss of buoyancy as the ship rests on the blocks reduces GM. Insufficient positive GM increases the risk of the ship heeling over or capsizing.
    • Trim by the stern: The vessel should be trimmed slightly by the stern (aft end lower than the bow) to ensure the aft end sits on the keel blocks first. This controlled settling minimises the risk of instability during the docking process. An even keel is generally preferred for the initial floating condition before the dry-docking procedure begins.
    Part (b)

    $$new \space KG \space = \space {{(8000 \times 7.6) - (100 \times 0.6)} \over 8000 - 100}$$

    $$New\:KG\:=\:7.689m$$

    $$New \space GM_1 \space = \space KM - KG$$

    $$= \space (7.6 + 0.5) - 7.689$$

    $$New\:GM\:=\:0.411m$$

    Angle of heel when 100t ballast is pumped out

    $$Tan\theta=\frac{m\times d}{\Delta GM}$$

    $$=\frac{100\times3.5}{7900\times0.411}$$

    $$Tan\theta=0.1077$$

    $$\theta=6^09^{^{\prime}}$$

    Q9 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 6x

    A ship of length 140m, Breadth of 18.5m, draught of 8.1m and a displacement of 17,025 tonnes in sea water, has a face pitch ratio of 0.673. The diameter of the propeller is 4.8m. The results of the speed trial show that true slip may be regarded as constant over a range of 9 to 13 knots and is 30%, w = 0.5Cb - 0.05. If fuel used is 20t/day at 13 knots and fuel consumption/day varies as cube of speed of ship, determine the fuel consumption, when propeller runs at 110 pm.

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    Given:

    $$Lenght,\:L=140m$$

    $$Breadth,\:B=18.5m$$

    $$Draught,\:d=8.1m$$

    $$Displacement,\:\Delta=17025tonnes$$

    $$Pitch\:ratio,\:p=0.673\operatorname{}$$

    $$Diameter\:of\:Propeller,\:D=4.8m$$

    $$\operatorname{Real\:Slip,\:R_{s}=30\%\:or\:0.3}$$

    $$Wake\:fraction,\:W=0.5C_{b}-0.05$$

    $$Consumption,\:C_2=\:20t\:per\:day\:$$

    $$Ship\:Speed,\:V_2=13\:knots$$

    $$\operatorname{Revolution,\:N}=110rpm$$

    $$cons\:per\:day\:\alpha\:V^3$$

    To find Fuel Consumption C1=?

    We know that,

    $$Displacement,\:\Delta=\nabla\times\rho$$

    $$17025=\nabla\times1.025$$

    $$\nabla=16609.76m^3$$

    $$Block\:Coefficient,\:C_{b}=\frac{\nabla}{L\times B\times d}$$

    $$C_{b}=\frac{16609.76}{140\times18.5\times8.1}$$

    $$C_{b}=0.792$$

    $$Wake\:Fraction,\:W=0.5C_{b}-0.05$$

    $$W=0.5\times0.792-0.05$$

    $$W=0.346$$

    $$Pitch\:ratio,\:p=\frac{P}{D}$$

    $$0.673=\frac{P}{4.8}$$

    $$P=4.8\times0.673$$

    $$P=3.23m$$

    $$Theoretical\:Speed,\:V_{t}=\frac{P\times N\times60}{1852}$$

    $$V_{t}=\frac{3.23\times110\times60}{1852}$$

    $$V_{t}=11.51knots$$

    Using, Real slip equation.

    $$\operatorname{\operatorname{Real\:Slip,\:R_{s}=\frac{V_{t}-V_{a}}{V_{t}}}}$$

    $$0.3=\frac{11.51-V_{a}}{11.51}$$

    $$V_{a}=11.51-11.51\times0.3$$

    $$V_{a}=11.51\left(1-0.3\right)$$

    $$V_{a}=11.51\times0.7$$

    $$V_{a}=8.057knots$$

    $$Wake\:fraction,\:W=\frac{V-V_{a}}{V}$$

    $$0.346=\frac{V-8.057}{V}$$

    $$0.346V=V-8.057$$

    $$V=\frac{8.057}{0.654}$$

    $$V=12.32knots$$

    $$cons\:per\:day\:\alpha\:V^3$$

    $$\frac{C_1}{C_2}=\left(\frac{V_1}{V_2}\right)^3$$

    $$\frac{C_1}{20}=\left(\frac{12.32}{13}\right)^3$$

    $$C_1=20\times\left(\frac{12.32}{13}\right)^3$$

    $$C_1=20\times0.851$$

    $$C_1=17.02t\:per\:day$$

    Q10 (10 Marks) Hull Construction πŸ”₯ Repeated 7x

    With respect to Buoyancy of a vessel:

    (a) What do you understand by reserve buoyancy what happen if the lost buoyancy is greater than the reserve buoyancy. (6)

    (b) A forward deep tank 12 m long extends from a longitudinal bulkhead to the ship's side. The widths of the tank surface measured from the longitudinal bulkhead at regular intervals are 10, 9, 7, 4 and 1 m. Calculate the seeond moment of area of the tank surface about a longitudinal axis passing through its centroid. (10)

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    Part (a)

    Reserve Buoyancy

    Reserve buoyancy is the watertight volume above the waterline of a vessel. It represents the potential buoyancy that can be utilised to keep the ship afloat when additional weight is added or if some buoyancy is lost due to damage (e.g., bilging).

    When a mass is added to the ship or buoyancy is lost (e.g., due to flooding of a compartment), the reserve buoyancy is converted into active buoyancy by an increase in draught.

    If the lost buoyancy (due to flooding or damage) is greater than the reserve buoyancy, the vessel will no longer have sufficient buoyant force to counteract its weight, causing it to sink.

    Part (b)
    Q1 (10 Marks) Ship Resistance & Propulsion

    (a) Explain the purpose of the rudder carrier and pintles. (8)

    (b) The speed of a ship is increased to 18% above normal for 7.5 hours, then reduced to 9% below normal for 10 hours. The speed is then reduced for the remainder of the day so that the consumption for the day is the normal amount. Find the percentage difference between the distance travelled in that day and the normal distance travelled per day. (8)

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    Part (a)

    Rudder Carrier:

    The rudder carrier is a structure or assembly that supports the weight and movement of the ship's rudder. It serves as a mounting point for the rudder and ensures its stability and proper functioning. The carrier is typically located at the stern of the ship, where the rudder is attached.

    Pintle: The pintle is a vertical pin or pivot that connects the rudder to the rudder carrier. It acts as the primary hinge point, allowing the rudder to rotate horizontally for steering purposes. The pintle provides a secure and stable connection between the rudder and the carrier, allowing controlled movement of the rudder.

    Q2 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 7x

    With reference to dry docking, define the responsibilities of the Second Engineer and the instructions he needs to give to Junior Engineers:

    (a) Prior to docking

    (b) Whilst the vessel is in dry dock

    (c) Prior to flooding and leaving the dock

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    (a) Prior to Docking:

    Preliminary Preparation:

    • Review various plans, manuals, and previous drydock reports for reference.
    • Prepare a detailed repair list and ensure all required spares are accounted for.
    • Conduct an inventory of spares and requisition necessary items.
    • Gather required tools, including special tools like poker gauges, trammel gauges, gagging tools, etc.
    • Mark all overboard valves, NC (normally closed) and NO (normally open) valves clearly for easy identification.
    • Allocate jobs to team members and discuss the time schedule.
    • Conduct a safety meeting to highlight dry dock hazards and necessary precautions.

    Before Entering Dry Dock:

    • Identify the power requirements and machinery to be operational during docking.
    • Check the shore connection box for proper functionality.
    • Record the soundings of FO (Fuel Oil), LO (Lubricating Oil), and DO (Diesel Oil) tanks.
    • Discharge contents from clean drain tanks and sewage tanks.
    • Carry out Economizer soot-blowing.
    • Change over Main Engine, Diesel Generators, and Boiler to Low Sulfur Gas Oil (LSGO).
    • Stop and clean purifiers.
    • Ensure the low sea chest is open and the high sea chest is shut.
    • Keep firefighting appliances (FFA) on standby.
    • Shut down all non-essential machinery, including MGPS (Marine Growth Prevention System) and FWG (Fresh Water Generator).

    (b) Whilst the Vessel is in Dry Dock

    Upon Arrival:

    • Connect shore power and supplies after ensuring safety checks are completed.
    • Start necessary equipment like cooling water, air compressors, air conditioning, and fridge compressors.
    • Check for jobs assigned by dry dock personnel and prepare accordingly.
    • Attend a safety meeting with dry dock personnel to understand local safety rules and procedures.

    During Dry Docking::

    • Oversee and assist in:
      • Cleaning and inspecting the hull, rudder, sea chest, anodes, and propeller.
      • Measuring propeller drop, checking rudder clearances, and inspecting the stern tube bearing and seal.
      • Servicing underwater valves and overboard valves.
      • Inspecting anchor and cables conditions.
      • Overhauling deck machinery, cranes, elevators, and engine room equipment such as the Main Engine, Diesel Generators, Boiler, and Economizer.
      • Renewing pipes and valves as needed.
      • Performing electrical equipment maintenance and surveys.
      • Supervising service engineers for specific repair jobs.
    • Ensure tank cleaning, welding, and other repair works are completed according to the plan.
    • Run the standby diesel generator daily after starting the priming pump.

    (c) Before Flooding and Leaving the Dry Dock

    Final Checks:

    • Verify that all underwater fittings and drain plugs are securely in place.
    • Ensure all machinery has been boxed back and is ready for operation.
    • Check for any leakage in stern tube seals.
    • Take tank soundings to confirm proper levels.
    • Confirm the proper operation of all underwater valves, overboard valves, sea chests, and vents.
    • Inspect the stern tube tank for any irregularities.

    System Restart:

    • Switch back to ship's power after confirming all systems are functional.
    • Test the proper operation of all machinery and systems to ensure the ship is fully operational before leaving the dock.
    Q3 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 4x

    (a) Describe the effect of cavitation on the propeller blades. (8)

    (b) A propeller 4.6m diameter has a pitch of 4.3m and boss diameter of 0.75m. The real slip is 28% at 95 rev/min. Calculate the speed of advance, thrust and thrust power.

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    Part (a)

    Effect of cavitation on propeller blades:

    Erosion:

    • Cavitation causes the formation and collapse of vapor bubbles on the propeller blade surface.
    • The collapse of these bubbles produces high-pressure shockwaves and microjets that erode the blade material, leading to surface pitting and progressive damage.

    Vibration:

    • Uneven cavitation across the blades leads to imbalanced forces, causing vibrations in the propeller and the ship.
    • These vibrations can reduce the comfort of passengers and crew and stress the ship's structural components.

    Noise:

    • The collapse of vapor bubbles generates loud noise, which can interfere with onboard communication and underwater sonar systems.
    • This noise is a significant concern for naval vessels and marine life.

    Reduced Performance:

    • Cavitation reduces the efficiency of the propeller by causing loss of thrust and torque.
    • The presence of cavitation decreases the propeller’s ability to convert rotational energy into forward motion, lowering the ship's speed and increasing fuel consumption.

    (b) Given:

    $$D=4.6m$$

    $$P=4.3m$$

    $$d=0.75$$

    $$S=28\%$$

    $$n \space = \space 95 \space rev/ min$$

    $$V_T \space = \space P \times N \times {{3600} \over 1852}$$

    $$ = \space 4.3 \times {{95} \over 60} \times {{3600} \over 1852}$$

    $$V_{T}=13.23knots$$

    $$Real \space slip \space (S) \space = \space {{V_T - V_a} \over V_T}$$

    $$0.28 \space = \space {{13.23 - V_a} \over 13.23}$$

    $$V_{a}=9.52knots$$

    $$Effective\:disc\:area\:\left(A\right)\:={{\pi}\over4}\left(D^2-d^2\right)$$

    $$= {{\pi} \over 4} (4.6^2 - 0.75^2)$$

    $$A=16.18m^2$$

    $$Thrust\space=\space\rho AP^2n^2S$$

    $$=1.025\times16.18\times4.3^2\times\left(\frac{95}{60}\right)^2\times0.28$$

    $$T=215.25KN$$

    $$Thrust \space power (T_p) \space = \space T \times V_a $$

    $$215.25\times9.52\times\frac{1852}{3600}$$

    $$T_{p}=1054.18KW$$

    Q4 (10 Marks) Ship Stability

    When a ship moves into water of different density, there will a change in its draught and trim. List out the minimum information required to calculate the final draughts. Also briefly outline the procedure involved. (16)

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    If the specific gravity of the water in which the ship is floating changes without any changes to the ship's displacement, the ship's draft will change. The draft will change because the ship must displace the same mass of water, which no longer has the same density.

    $$\Delta=V\times\rho$$

    If the displacement ( Ξ” ) remains constant and the density ( ρ ) changes, the underwater volume must change; therefore, the ship's draft will change automatically.

    If a ship goes from fresh water to salt water, buoyancy will increase and the draft will decrease. Inversely, a ship that goes from salt water to fresh water will see its draft increase.

    A ship that loads to its marks (summer load lines S ) in salt water, will see its draft increase as it goes into fresh water and will exceed its marks. This situation is accepted because the absence of heavy weather on bodies of fresh water compensates for the ship's increased draft. This situation has even been made official by adding an additional mark ( F ) to the load lines.

    Minimum Information Required to Calculate Final Draughts:

    1. Displacement (Ξ”): The mass of water displaced by the ship in tons.
    2. Densities (ρ₁, ρ₂): Densities of the initial and final liquids (e.g., seawater = 1.025 t/mΒ³, freshwater = 1.0 t/mΒ³).
    3. TPC (Tonne Per Centimeter Immersion): The mass required to change the mean draught by 1 cm.
    4. MCTC (Moment to Change Trim by 1 cm): The moment required to cause a 1 cm change in trim.
    5. LCB and LCF (Longitudinal Center of Buoyancy and Floatation): To determine the trimming effect.
    6. Length of Vessel (L): Used in trim calculations.

    Procedure to Calculate Final Draughts:

    1. Determine Change in Mean Draught:

    $$Change\:in\:draught\:=\:\frac{\Delta}{TPC}\:\times\:\frac{\rho_1-\rho_2}{\rho_2}\:\left(in\:cm\right)$$

    2. Determine Change in Trim:

    $$Change\:in\:trim\:=\:\frac{\Delta\:\times\:FB}{MCTC}\times\frac{\rho_1-\rho_2}{\rho_2}$$

    where:

    $$FB\:=\:LCB\:+\:LCF$$

    3. Determine the change in forward/aft draught:

    $$Change\:in\:Fwd/Aft\:=\:\frac{Change\:in\:trim}{Length\:of\:vessel}\:\times Distance\:from\:CF$$

    4. Calculate the new forward and aft draughts

    $$New\:draught\:Fwd/Aft\:=\:Even\:keel\:draught\:-\:change\:in\:mean\:draught\:+\:change\:in\:Fwd/aft$$

    Q5 (10 Marks) Hull Construction πŸ”₯ Repeated 10x

    Describe a method for the attachment of bilge keels. State THREE reasons for not extending bilge keels for the entire length of the vessel. Explain TWO principles of roll damping that bilge keels exploit.

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    Part (a)

    Method of Bilge keel attachment to the hull:

    Bilge keels are fitted port and starboard at the turn of the bilge. They do not extend outside the lines of the side and bottom shell. The bilge keels are attached directly in line with an internal stiffening member such as a girder or longitudinal. The bilge keel comprises a flat bar doubler welded directly to the shell, and an offset bulb plate (OBP) with 'scallops' cut in it is welded to the flat bar doubler. The ends of the bilge keels are tapered (minimum 3 in 1) and will end in line with transverse internal stiffening, such as a frame. Using a hull doubler protects the hull in case of damage to the keel, as the crack would not extend into the hull.

    Part (b)

    Bilge keels are not fitted for the full length of the vessel because:

    • The 'lever' to the ship's axis of rotation is reduced at the ends;
    • The ineffectiveness, the closer the bilge keel is to the rolling centre, increased resistance and more likelihood of damage.
    • The hydrodynamic effect would cause a large increase in resistance and fuel consumption;
    • At the aft end, the boundary layer is much thicker, and since the keel would not project through, it would have a much reduced effect.
    Part (c)

    Two principles of roll damping exploited by Bilge keels:

    Increased Roll Period:

    • Bilge keels increase the ship's roll period (the time it takes for the vessel to complete one roll). This is achieved by increasing the moment of inertia (K) of the vessel. A longer roll period means the vessel rolls more slowly, thus reducing the amplitude of the roll.

    $$T_{r}=2\pi\:\frac{k}{\sqrt{g\times GM}}$$

    Where:

    • g: acceleration due to gravity
    • GM: metacentric height
    • k: mass moment of inertia

    By increasing k, the ship's stability during roll improves.

    Hydrodynamic Resistance:

    • As the vessel rolls, the bilge keels move through the water, creating pressure differences. Water pressure on one side of the keel opposes the rolling motion, providing a damping effect. This is due to the interaction between the bilge keel and the water, creating hydrodynamic forces that counteract the rolling motion.
    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 11x

    (a) Define longitudinal center of gravity (LCG) and longitudinal center of buoyancy (LCB). (6)

    (b) A ship 120m long floats at draughts of 5.50m forward and 5.80m aft; MCT 1cm 80 tonne m, TPC 13, LCF 2.5m forward of midships. Calculate the new draughts when a mass of 110 tonne is added 24m aft of midships.

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.

    LCF in fwd and trim by stern

    $$Bodily \space sinkage \space = \space {{mass \space added} \over TPC} \space = \space {{110} \over 13} \space$$

    $$=\:8.5\operatorname{\mathrm{cm}}\:=0.085m$$

    $$Trim \space = \space {{m \times d} \over MCT_{1cm}}$$

    $$ = \space {{110 \times (24 + 2.5)} \over 80}$$

    $$Trim=36.43\operatorname{cm}=0.364m\:$$

    Change in fwd draught:

    $$d_{F}=\frac{-t}{L}\left\lbrack\frac{L}{2}-LCF\right\rbrack$$

    $$=\frac{-36.43}{120}\left\lbrack\frac{120}{2}-2.5\right\rbrack$$

    $$=-17.45\operatorname{cm}=-0.1745m$$

    Change in Aft draught:

    $$d_{A}=\frac{+t}{L}\left\lbrack\frac{L}{2}+LCF\right\rbrack$$

    $$=\frac{+36.43}{120}\left\lbrack\frac{120}{2}+2.5\right\rbrack$$

    $$=+18.97\operatorname{cm}=0.189m$$

    New draughts:

    $$D_{F}=5.5+0.085-0.175=5.41m$$

    $$D_{A}=5.8+0.085+0.18=6.065m$$

    Q7 (10 Marks) Ship Stability πŸ”₯ Repeated 2x

    What is the significance of the area under the curve of statical stability or the GZ curve? Explain using a neat diagram, how this curve is used to assess the stability of the ship against a heeling arm.

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    Significance of area under GZ curve:

    The area under the statical stability or GZ curve up to any given angle ΞΈ, multiplied by the gravitational weight of the ship (displacement), represents the Dynamic Stability. This is the work done in heeling the ship to the angle ΞΈ, and it reflects the ship's ability to absorb the energy imparted by external forces such as waves and wind.

    The stability of a ship is directly related to the nature and value of its metacentric height (GM).

    • Since GM is closely associated with the Righting Lever (GZ) and the angle of heel, the statical stability curve is plotted between GZ and the angle of heel.

    Following information obtained from the curve can be used to assess the stability of the ship:

    Angle of Equilibrium

    • If the GZ curve intersects the horizontal axis (i.e., no righting lever) at the origin, the ship has inherent positive initial stability.

    Maximum GZ

    • This represents the ship's largest static heeling moment. The Maximum Righting Lever (GZ), when multiplied by the ship's displacement, gives the maximum heeling moment the ship can sustain without capsizing.

    Angle of deck Immersion

    • The angle at which the ship’s deck becomes submerged. This is often indicated as the point of inflection on the GZ curve, where the ship’s vulnerability to downflooding increases.

    Angle of Vanishing Stability

    • The angle at which the GZ curve intersects the horizontal axis, indicating the point where the righting lever becomes zero. Any heel beyond this angle results in negative stability, increasing the risk of capsizing.

    Range of Stability

    • The range is measured from the origin to the angle of vanishing stability, indicating the range of angles over which the ship has positive stability.

    Angle of Contraflexure

    • The angle of heel up to which the rate of increase of GZ with heel is rising. Beyond this point, while GZ may continue to increase, the rate of increase begins to diminish.

    Angle of Downflooding

    • The minimum angle of heel at which an external opening (without a watertight appliance) is submerged, allowing seawater ingress.
    • (Downflooding refers to seawater entering the hull or superstructure due to heel or immersion of the vessel.)

    Initial Metacentric Height (GM)

    • At 57.3 (1 radian), an ordinate is drawn at the origin of the GZ curve. A tangent to the curve is extended to intersect this ordinate, providing the value of the Initial Metacentric Height (GM).
    Q8 (10 Marks) Ship Stability

    (a) Explain the use of KN curves. (8)

    (b) A ship of 12000 tonne displacement has a rudder 15m2 in area, whose centre is 5m below the waterline. The metacentric height of the ship is 0.3m and the centre of buovancy is 3.3m below the waterline. When travelling at 20 knots the rudder is turned through 30°. Find the initial angle of heel if the force Fn perpendicular to the plane of the rudder is given by: Fn = 577 Av2 sin∝ N

    Allow 20% for the race effect. (8)

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    Part (a)

    KN curves are stability curves used in naval architecture. They are created by plotting the righting lever (KN) against the angle of heel (ΞΈ) for a given displacement and an assumed KG (height of the centre of gravity above the keel) of 0. The righting lever is the distance between the centre of buoyancy and the centre of gravity, measured along a vertical line through the centre of gravity.

    The KN curve is then used to calculate the righting lever (GZ) for any other KG value using the following formula:

    GZ = KN - KG * sin ΞΈ

    Where:

    • GZ is the righting lever measured from the centre of gravity.
    • KN is the righting lever measured from the keel (obtained from the KN curve).
    • KG is the distance of the centre of gravity from the keel.
    • ΞΈ is the angle of heel.
    Q9 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 7x

    With reference to International Load Line Statutory Certification:

    (a) State the reason for the freeboard requirements

    (b) Explain the term conditions of assignments.

    (c) List the items that may be examined during a Load line survey after major repairs in the drydock.

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    Part (a)

    Reasons for Freeboard Requirements:

    Freeboard is the distance measured from the waterline to the upper edge of the deck plating at the side of the freeboard deck amidships.

    Purpose of Freeboard:

    • Ensures the ship is seaworthy when fully loaded.
    • Provides reserve buoyancy, enabling the ship to rise as it passes through waves, keeping decks largely dry.
    • Enhances the ship's stability and increases its survivability in case of damage by allowing it to remain afloat longer, aiding crew escape or damage control.
    Part (b)

    (i) Conditions of Assignment:

    Conditions of Assignment are outlined in the Merchant Shipping Load Line Rules 1968 and must be satisfied before a ship is assigned freeboards and issued a load line certificate. These conditions address the practical need for openings (hatchways, doorways, vents, etc.) in the ship's hull and mandate appropriate protection and closure mechanisms for these openings.

    Requirements Before Assigning Load Line:

    • The ship must have sufficient structural strength.
    • Adequate reserve buoyancy must be maintained.
    • Openings must be secured against water ingress.
    • Safety measures for the crew, such as guardrails and gangways, must be in place.
    Part (b)

    (ii) Items Examined During a Load Line Survey After Major Repairs in Drydock:

    • Openings: This encompasses hatchways, machinery space openings, cargo ports, watertight doors, ventilators, air pipes, scuppers, freeing ports, side scuttles, and other openings in the freeboard and superstructure decks. The surveyor checks the condition of seals, gaskets, closing mechanisms, and overall watertightness.
    • Crew Protection: Safety features designed to protect the crew, such as guardrails, bulwarks, and gangways, are inspected for damage or deterioration and proper functionality.
    • Hull Structure: The hull itself is inspected for any damage, corrosion, or structural weaknesses that could compromise watertightness or strength. This often includes visual inspection for cracks, buckling, indentation, and paint adhesion. Bottom shell, bilge keel, stem and stern frames, rudder, sea chests, side ports, stern bearing, and propeller are all checked.
    • Other Systems: Depending on the scope of repairs, other systems may be examined, including but not limited to: pressure-vacuum valves, sounding pipes, air vents, access hatches, ventilation fans, dampers, weathertight doors and sealings, door securing arrangements, side scuttles, windows, and skylights. Testing, such as hose tests (to check watertight integrity), might also be conducted.
    Q10 (10 Marks) Hull Construction πŸ”₯ Repeated 6x

    (a) Sketch a transverse section through the hold space of a container ship hull.

    (b) Referring to the sketch in (a) describe how adequate structural strength is built into the hull.

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    Part (a)

    Mid-ship half-section sketch of a container ship:

    Part (b)

    Structural Strength Features:

    • The deck plating is made thicker and uses higher tensile steel to withstand stresses caused by wide hatch openings and operational loads.
    • The deck, side shell, and longitudinal bulkheads are framed longitudinally. This arrangement combined with deep double bottom helps resist bending stresses due to hogging (upward bending) and sagging (downward bending) when the ship is under load.
    • The hatch coamings are made continuous to contribute to the overall longitudinal strength of the hull.
    • A torsion box is installed, running along the entire length of the ship from the machinery space bulkhead to the forward collision bulkhead. This structure provides the necessary torsional strength to counteract twisting forces acting on the hull during operation.
    • Deep web boxes are fitted at the ends of hatches, both at tank top and deck levels, to enhance transverse and torsional strength.
    • A deep double bottom is designed to withstand uplift forces caused by water pressure, especially when the ship is deeply loaded. It also provides additional strength to the hull structure.
    • Side girders are placed under container cells, with added transverse local stiffening. These elements distribute the concentrated loads from containers and increase overall stability.
    Q1 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 4x

    Describe the preparation necessary before the application (in dry dock) of sophisticated or approved long life coating to the underwater surface of the hull.

    (a) State the significance of the roughness profile.

    (b) List the different sophisticated coating which are available.

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    The preparation of a ship's underwater hull before applying a long-life coating in a dry dock involves a three-step process. This process addresses the removal of contaminants and the creation of a suitable surface profile.

    (i) Washing: The hull surface must be thoroughly cleaned to remove all marine growth (algae, slime, etc.), accumulated salts, dirt, grease, and oil. High-pressure freshwater washing is the standard method for this initial cleaning. The goal is to present a clean substrate for subsequent stages.

    (ii) Blasting: Abrasive blasting is the preferred method for removing rust, defective paint, and any remaining contaminants. This process achieves a bare metal surface, essential for proper adhesion of the new coating. The extent of blasting (localized or full hull) depends on the condition of the existing surface. The intensity and type of abrasive used are carefully controlled to achieve the desired surface roughness profile.

    (iii) Primer Application: After blasting, the surface is again cleaned to remove any blasting debris. A primer coat is then applied to provide corrosion protection and to create an ideal surface for the subsequent topcoat adhesion. This primer acts as an intermediary layer, enhancing the bond between the substrate and the long-life coating system.

    Part (a)

    Significance of Roughness Profile:

    The roughness profile of the prepared hull surface impacts the performance of the applied coating and the overall operational efficiency of the vessel. A rough surface increases frictional resistance as the vessel moves through the water. This increased drag translates to higher power requirements for propulsion, leading to increased fuel consumption and operational costs. Furthermore, greater surface roughness contributes to increased carbon emissions, a concern under current MARPOL regulations. Therefore, a controlled and optimized roughness profile is essential for minimizing frictional resistance, reducing fuel consumption and emissions, and maximizing the longevity of the hull coating.

    Part (b)

    Sophisticated hull coating systems comprise multiple layers designed to provide corrosion protection and antifouling properties.

    Wash Primer/Pretreatment Primer/Metal Conditioning Primer:

    • These primers act as a base layer, improving adhesion of subsequent layers. Common types include epoxy primers pigmented with iron oxide and corrosion inhibiting pigments (zinc and calcium phosphates, although zinc content is minimized due to safety concerns).

    Anticorrosive Coating:

    • This layer primarily provides corrosion protection to the underlying metal. Two-component epoxies, coal tar epoxies, and epoxy or polyester coatings incorporating glass flakes are frequently employed. Glass flakes enhance mechanical strength and water vapor impermeability.

    Antifouling Coating:

    • This layer prevents the attachment of marine organisms (fouling). Historically, tin-based paints were used, but due to environmental regulations, they have been largely replaced by copper-based, silicone-based, or non-TBT (Tributyltin) self-polishing antifouling coatings. These newer coatings typically use seawater-soluble polymers. The number of antifouling layers applied (two or three) depends on the specific system chosen and required longevity.
    Q2 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 5x

    A rudder of a vessel requires extensive welding repairs and as Chief Engineer you are requested to supervise

    (a) Suggest a suitable type of welding process.

    (b) State, with reasons, FOUR conmon welding delects

    (c) State what tests may be carried out before returning the rudder to service.

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    As Second Engineer, I would oversee the extensive welding repairs required for the vessel's rudder using the following plan:

    Part (a)

    Suitable Welding Process:

    Manual Metal Arc Welding (MMAW), also known as Shielded Metal Arc Welding (SMAW), is the most suitable process for this repair. The reasons are threefold:

    • MMAW is highly portable, allowing for on-site repair within the drydock. The process is adaptable to various welding positions (downhand, overhead, horizontal, vertical) – a necessity given the complex geometry of a rudder.
    • Assuming the rudder is constructed from standard steel, MMAW using readily available flux-coated electrodes provides good control, arc stability, and penetration. The flux coating protects the weld pool from atmospheric contamination during cooling.
    • MMAW requires relatively simple equipment and is less demanding in terms of operator skill compared to other processes like TIG or MIG. This translates to cost-effectiveness and allows for a wider pool of qualified welders.
    • If cast steel components are present, pre-heating will be necessary to minimize stress cracking, and specialized electrodes suited for the specific cast steel grade must be selected.

    During welding by the metal arc process, the following points must be observed:

    • Electrode Consumption Rate
    • Penetration
    • Slag Control
    • Arc Length and Sound
    Part (b)

    Four Common Welding Defects:

    1. Undercut: A groove formed along the edge of the weld bead, weakening the joint. Caused by excessive current, incorrect electrode angle, excessive travel speed, or improper electrode manipulation.

    2. Overlap: Molten weld metal flows over the parent metal without proper fusion. Caused by low current, slow travel speed, excessive arc length, or improper joint preparation.

    3. Slag Inclusion: Trapped slag within the weld metal, reducing its strength and potentially causing cracking. Caused by insufficient cleaning between passes, incorrect current, long arc length, slow travel speed, or too large an electrode diameter.

    4. Incomplete Penetration: The weld does not fully fuse the joint faces, resulting in a weak joint. Caused by insufficient current, incorrect joint preparation (too small a root gap or bevel angle), excessive travel speed, or too large an electrode diameter.

    Part (c)

    Tests Before Returning to Service:

    • A thorough visual examination of all welds to identify any surface defects like cracks, porosity, or lack of fusion.
    • NDT methods such as Magnetic Particle Inspection (MPI) or Dye Penetrant Inspection (DPI) will be employed to detect subsurface flaws that may not be visible during visual inspection. The specific NDT method chosen will depend on the type of steel and the accessibility of the weld areas.
    • The repaired rudder will undergo a hydrostatic pressure test. This involves filling the rudder with a water head of 2.46 meters and observing for any leaks. This confirms the watertight integrity of the welds and the overall rudder structure.
    Q3 (10 Marks) Hull Construction πŸ”₯ Repeated 5x

    With reference to fatigue of engineering components explain the influence of stress level and cyclical frequency on expected operating life.

    (a) Explain the influence of material defects on the safe operating life of an engineering component.

    (b) State the factors which influence the possibility of fatigue cracking of a bed-plate transverse girder and explain how the risk of such cracking can be minimized.

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    Part (a)

    Influence of Stress Level and Cyclic Frequency on Operating Life:

    Fatigue is progressive and localised structural damage caused by cyclic loading, where the maximum stress is below the ultimate tensile strength. The relationship between stress level, cyclic frequency, and operating life depends on whether the fatigue is high-cycle/low-stress or low-cycle/high-stress.

    High-cycle fatigue (low stress-high cycle):

    • This occurs at lower stress levels over a high number of cycles, resulting in elastic deformation. The component can withstand more cycles at these lower stress levels, and its life expectancy is determined by the S-N curve, which predicts the number of cycles before failure at a given stress level. For example, fatigue in turbocharger blowers often results from prolonged vibration over numerous cycles.

    Low-cycle fatigue (high stress-low cycle):

    • This occurs at high-stress levels over fewer cycles, causing plastic deformation in the material. This type of fatigue is typically assessed by a strain curve. If the stress level increases, the component's operating life decreases, as higher stress accelerates the onset of failure. For example, air receivers filling automatically face high stress and experience fewer cycles before failure.

    If stress levels or the number of cycles increase beyond the material’s capacity, failure will occur sooner. It is important to keep stress levels within allowable limits for extended component life.

    Part (b)

    Material defects can significantly reduce the safe operating life of engineering components because defects serve as stress concentrators that increase local stress around the defect. This leads to premature failure as the material cannot withstand the same level of cyclic stress as a defect-free component.

    • Surface roughness, porosity, inclusions, and abrupt section changes all create stress concentrations, lowering fatigue strength.
    • Coarse grain size, specific chemical compositions, and cold working introduce residual stresses that reduce fatigue resistance.
    • Corrosion, erosion, and decarbonisation weaken the material and accelerate fatigue crack initiation and propagation.
    • Faulty workmanship during assembly or processing introduces defects that may significantly shorten the component's life.
    Part (c)

    Factors Influencing Fatigue Cracking in Bedplate Transverse Girders:

    • Cylinder overload due to excess power puts excessive stress on the girders.
    • Incorrect crankshaft alignment induces uneven loading and stress concentrations.
    • Material defects, high residual stresses in welds, heat-affected zone hardening, and the presence of dissolved oxygen all reduce fatigue resistance.
    • Tank top deformation from pressurisation or overheating adds stress to the bedplate.

    To minimise the risk of fatigue cracking:

    (i) Constructional strength:

    • Bed plates are made up of M.S. plates with four steel casting, which are assembled and welded together so that the bed plate is strong longitudinally & transversely with good resistance to twisting along its length.
    • Longitudinal strength is obtained by fabricating each side of the bed plate in the form of a box girder.
    • The cast steel cross girder in which the main bearing is placed contributes to the bed plate's transverse strength and resistance against twisting along its length.
    • Resin cast chocks are used between the bedplate and the double bottom tank top to absorb the shocks & stress.

    (ii) Maintenance:

    • Monthly checks on the bolt tension.
    • Monthly checks on engine load using power cards & measuring cylinder peak pressure.
    • Regular checking of tension for main bearing jack bolts as recommended by engine manufacturers.
    • Regular checks on crankshaft alignment by taking deflection & compare with recommended value.
    • By maintaining engine operations at specified load, temperature, pressure, speed, etc.
    Q4 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 7x

    With reference to International Load Line Statutory Certitication:

    (a) State the reason for freeboard requirements;

    (b) (i) Explain the term 'conditions of assignment'.

    (ii) List the items that may be examined during a Load line survey after a vessel's major repairs in the drydock.

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    Part (a)

    Reasons for Freeboard Requirements:

    Freeboard is the distance measured from the waterline to the upper edge of the deck plating at the side of the freeboard deck amidships.

    Purpose of Freeboard:

    • Ensures the ship is seaworthy when fully loaded.
    • Provides reserve buoyancy, enabling the ship to rise as it passes through waves, keeping decks largely dry.
    • Enhances the ship's stability and increases its survivability in case of damage by allowing it to remain afloat longer, aiding crew escape or damage control.
    Part (b)

    (i) Conditions of Assignment:

    Conditions of Assignment are outlined in the Merchant Shipping Load Line Rules 1968 and must be satisfied before a ship is assigned freeboards and issued a load line certificate. These conditions address the practical need for openings (hatchways, doorways, vents, etc.) in the ship's hull and mandate appropriate protection and closure mechanisms for these openings.

    Requirements Before Assigning Load Line:

    • The ship must have sufficient structural strength.
    • Adequate reserve buoyancy must be maintained.
    • Openings must be secured against water ingress.
    • Safety measures for the crew, such as guardrails and gangways, must be in place.
    Part (b)

    (ii) Items Examined During a Load Line Survey After Major Repairs in Drydock:

    • Openings: This encompasses hatchways, machinery space openings, cargo ports, watertight doors, ventilators, air pipes, scuppers, freeing ports, side scuttles, and other openings in the freeboard and superstructure decks. The surveyor checks the condition of seals, gaskets, closing mechanisms, and overall watertightness.
    • Crew Protection: Safety features designed to protect the crew, such as guardrails, bulwarks, and gangways, are inspected for damage or deterioration and proper functionality.
    • Hull Structure: The hull itself is inspected for any damage, corrosion, or structural weaknesses that could compromise watertightness or strength. This often includes visual inspection for cracks, buckling, indentation, and paint adhesion. Bottom shell, bilge keel, stem and stern frames, rudder, sea chests, side ports, stern bearing, and propeller are all checked.
    • Other Systems: Depending on the scope of repairs, other systems may be examined, including but not limited to: pressure-vacuum valves, sounding pipes, air vents, access hatches, ventilation fans, dampers, weathertight doors and sealings, door securing arrangements, side scuttles, windows, and skylights. Testing, such as hose tests (to check watertight integrity), might also be conducted.
    Q5 (10 Marks) Hull Construction πŸ”₯ Repeated 7x

    (a) With reference to fatigue failure of engineering components explain the influence of stress level and cyclical frequency on expected operating life.

    (b) Explain the influence of material defects on the safe operating life of an engineering component.

    (c) State the factors which influence the possibility of fatigue cracking of a bed-plate transverse girder and explain how the risk of such cracking can be minimized.

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    Part (a)

    Influence of Stress Level and Cyclic Frequency on Operating Life:

    Fatigue is progressive and localised structural damage caused by cyclic loading, where the maximum stress is below the ultimate tensile strength. The relationship between stress level, cyclic frequency, and operating life depends on whether the fatigue is high-cycle/low-stress or low-cycle/high-stress.

    High-cycle fatigue (low stress-high cycle):

    • This occurs at lower stress levels over a high number of cycles, resulting in elastic deformation. The component can withstand more cycles at these lower stress levels, and its life expectancy is determined by the S-N curve, which predicts the number of cycles before failure at a given stress level. For example, fatigue in turbocharger blowers often results from prolonged vibration over numerous cycles.

    Low-cycle fatigue (high stress-low cycle):

    • This occurs at high-stress levels over fewer cycles, causing plastic deformation in the material. This type of fatigue is typically assessed by a strain curve. If the stress level increases, the component's operating life decreases, as higher stress accelerates the onset of failure. For example, air receivers filling automatically face high stress and experience fewer cycles before failure.

    If stress levels or the number of cycles increase beyond the material’s capacity, failure will occur sooner. It is important to keep stress levels within allowable limits for extended component life.

    Part (b)

    Material defects can significantly reduce the safe operating life of engineering components because defects serve as stress concentrators that increase local stress around the defect. This leads to premature failure as the material cannot withstand the same level of cyclic stress as a defect-free component.

    • Surface roughness, porosity, inclusions, and abrupt section changes all create stress concentrations, lowering fatigue strength.
    • Coarse grain size, specific chemical compositions, and cold working introduce residual stresses that reduce fatigue resistance.
    • Corrosion, erosion, and decarbonisation weaken the material and accelerate fatigue crack initiation and propagation.
    • Faulty workmanship during assembly or processing introduces defects that may significantly shorten the component's life.
    Part (c)

    Factors Influencing Fatigue Cracking in Bedplate Transverse Girders:

    • Cylinder overload due to excess power puts excessive stress on the girders.
    • Incorrect crankshaft alignment induces uneven loading and stress concentrations.
    • Material defects, high residual stresses in welds, heat-affected zone hardening, and the presence of dissolved oxygen all reduce fatigue resistance.
    • Tank top deformation from pressurisation or overheating adds stress to the bedplate.

    To minimise the risk of fatigue cracking:

    (i) Constructional strength:

    • Bed plates are made up of M.S. plates with four steel casting, which are assembled and welded together so that the bed plate is strong longitudinally & transversely with good resistance to twisting along its length.
    • Longitudinal strength is obtained by fabricating each side of the bed plate in the form of a box girder.
    • The cast steel cross girder in which the main bearing is placed contributes to the bed plate's transverse strength and resistance against twisting along its length.
    • Resin cast chocks are used between the bedplate and the double bottom tank top to absorb the shocks & stress.

    (ii) Maintenance:

    • Monthly checks on the bolt tension.
    • Monthly checks on engine load using power cards & measuring cylinder peak pressure.
    • Regular checking of tension for main bearing jack bolts as recommended by engine manufacturers.
    • Regular checks on crankshaft alignment by taking deflection & compare with recommended value.
    • By maintaining engine operations at specified load, temperature, pressure, speed, etc.
    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 3x

    (a) Describe how water tightness is maintained where bulkheads are pierced by longitudinal beams or pipes.

    (b) A triangular bulkhead is 7 m wide at the top and has a vertical depth of & m. Calculate the load on the bulkhead and the position of centre of pressure if the bulkhead is flooded with sea water on only side: (10)

    (i) To the top edge

    (ii) With 4 m head to the top edge.

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    Part (a)

    Water tightness of

    bulkheads that are pierced by longitudinal beams or pipes:

    • Bulkheads must be made structurally watertight without the use of wood packing.
    • Pipes passing through bulkheads are either welded or fastened to the bulkhead using studs or hoses secured through tapped holes in the plating.
    • When a bulkhead is pierced by a longitudinal beam, the openings are kept as small as possible, and doubler plates are welded on each side to maintain watertight integrity.
    • Sealing materials and gaskets are used, particularly for pipes, cables, and similar penetrations.
    Part (b)

    $$\left(a\right)\:Load\:on\:bulkhead=\rho gAH$$

    $$=1025\times9.81\times\frac{7\times8}{2}\times\frac83$$

    $$=750.8\times10^3N$$

    $$=750.8KN$$

    $$Centre\:of\:pressure\:fron\:top=\frac12D=\frac12\times8$$

    $$=4m$$

    $$\left(b\right)\:Load\:on\:bulkhead=1025\times9.81\times\frac{7\times8}{2}\times\left(\frac83+4\right)\times$$

    $$=1.877\times10^6N$$

    $$=1.877MN$$

    $$For\:triangle\:I_{NA}=\frac{1}{36}BD^3$$

    $$Centre\:of\:pressure\:from\:surface\:of\:water\:=\frac{I_{NA}}{AH}+H$$

    $$=\frac{\frac{1}{36}\times7\times8^3}{\frac12\times7\times8\times\left(\frac83+4\right)}+\left(\frac83+4\right)$$

    $$=\frac{2\times7\times8^3}{36\times7\times8\times6.667}+6.667$$

    $$=7.20\:m$$

    $$Centre\:of\:pressure\:from\:top\:of\:bulkhead$$

    $$=7.20-4$$

    $$3.2m$$

    Q7 (10 Marks) Hull Construction πŸ”₯ Repeated 4x

    An oil tanker 160m long and 22m beam floats at a draught of 9m in seawater. Cw is 0.865. The midship section is in the form of a rectangle with 1.2m radius at the bilges. A midship tank 10.5m long has twin longitudinal bulkheads and contains oil of 1.4m3/t to a depth of 11.5m. The tank is holed to the sea for the whole of its transverse section. Find the new draught.

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    New draught of the oil tanker when the midship tank is holed.

    Oil tanker 160 m long, 22 m beam, floats at a draught of 9 m in sea water. Cw = 0.865. The midship section is a rectangle with 1.2 m radius at the bilges. A midship tank 10.5 m long has twin longitudinal bulkheads and contains oil of 1.4 m3/t to a depth of 11.5 m. The tank is holed to the sea for the whole of its transverse section. Find the new draught.

    Waterplane area Aw = Cw x L x B = 0.865 x 160 x 22 = 3044.8 m2.

    Midship section area (rectangle with bilge radius r=1.2 m): Ams = B x d - (4 - pi) r^2 = 22 x 9 - 0.858 x 1.44 = 198 - 1.236 = 196.76 m2.

    Volume of the tank below the original waterline = Ams x 10.5 = 196.76 x 10.5 = 2066 m3.

    The tank contains oil of density rho_o = 1/1.4 = 0.714 t/m3. When holed, sea water (1.025 t/m3) replaces the oil, so the net loss of buoyancy is the volume times the relative density difference:

    Vlost = 2066 x (1 - 0.714/1.025) = 2066 x 0.303 = 626 m3.

    The flooded tank provides no increase of buoyancy, so the effective sinking waterplane = Aw - (10.5 x 22) = 3044.8 - 231 = 2813.8 m2.

    Sinkage = Vlost/effective waterplane = 626/2813.8 = 0.222 m.

    New draught = 9 + 0.22 = 9.22 m.

    Answer: the new draught is about 9.2 m.

    Q8 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 7x

    (a) What is meant by the Admiralty Coefficient and the Fuel Coefficient?

    (b) A ship of 14900 tonne displacement has a shaft power of 4460 kW at 14.55 knots. The shaft power is reduced to 4120 kW and the fuel consumption at the same displacement is 541 kg/h. Calculate the fuel coefficient for the ship.

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    The Admiralty Coefficient (C) is a method for estimating the propulsion power needed for a newly built ship. It's considered relatively constant for a given ship design. The formula is:

    $$C=\frac{\Delta^{2/3}\times V^3}{BP}$$

    Where:

    • C = Admiralty Coefficient
    • Ξ” = Displacement in tonnes (weight of the ship when fully loaded)
    • V = Speed in knots
    • BP = Brake power in kilowatts (power delivered by the ship's engine)

    A higher Admiralty Coefficient indicates a more efficient ship design, meaning it requires less power to achieve a given speed. Values typically range from 350 to 600.

    Fuel Coefficient:

    The Fuel Coefficient (F.C.) is used to calculate a ship's daily fuel oil consumption. The formula is:

    $$Daily\:fuel\:oil\:consumption\:=\:\frac{\Delta^{2/3}\times V^3}{FC}$$

    Where:

    • F.C. = Fuel Coefficient
    • Ξ” = Displacement in tonnes
    • V = Speed in knots

    The Fuel Coefficient can vary significantly, with typical values ranging from 40,000 to 120,000. A higher Fuel Coefficient implies greater fuel efficiency (lower daily fuel consumption) for a given speed and displacement.

    Part (b)

    $$admiraty\:coefficient\:\left(C\right)=\:\frac{\Delta^{2/3}V^3}{Shaft\:power}=\frac{\Delta^{\frac23}\times V^3}{SP}$$

    $$\frac{SP_1}{SP_2}=\frac{V_1^3}{V_2^3}$$

    $$\frac{4460}{4120}=\frac{14.55^3}{V_2^3}$$

    $$V_2=14.17kntos$$

    $$Fuel\:consumption\:per\:hour=541\operatorname{\mathrm{\:kg}}\:per\:hour$$

    $$Fuel\:consumption\:per\:day\:=\:541\times24=12.98t\:per\:day$$

    $$Fuel\:coefficient=\frac{\Delta^{\frac23}\times V_2^3}{Fuel\:consumption\:per\:day}$$

    $$=\:\frac{14900^{\frac23}\times14.17^3}{12.98}$$

    $$=132726.9$$

    Q9 (10 Marks) Hull Construction πŸ”₯ Repeated 2x

    The 1/2 ordinates of a waterplane at 15m intervals, commencing from aft, are 1, 7, 10.5, 11, 11, 10.5, 8, 4 and Om.

    Calculate:

    (a) TPC

    (b) Distance of the centre of flotation from midships

    (c) Second moment of area of the waterplane about a transverse axis through the centre of flotation.

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    1/2 Ord

    SM

    Product

    Lever

    Product

    Lever

    Product

    1

    1

    1

    _4

    4

    +4

    16

    7

    4

    28

    +3

    84

    +3

    252

    10.5

    2

    21

    +2

    42

    +2

    84

    11

    4

    44

    +1

    44

    +1

    44

    11

    2

    22

    0

    Ξ£MA = 174

    0

    0

    10.5

    4

    42

    -1

    -42

    -1

    42

    8

    2

    16

    -2

    -32

    -2

    64

    4

    4

    16

    -3

    -48

    -3

    144

    0

    1

    0

    -4

    0

    -4

    0

    Ξ£A = 190

    Ξ£MF = -122

    Ξ£I = +646

    $$Common\:interval\:\left(h\right)=15$$

    $$Waterplane\:area\:\left(A_{w}\right)=2\times\frac{h}{3}\sum A$$

    $$=2\times\frac{15}{3}\times190$$

    $$=1900m^2$$

    $$Longitudinal\:centre\:of\:flotation\:LCF=h\times\frac{\sum M_{A}+\sum M_{F}}{\sum A}$$

    $$=15\times\frac{174-122}{190}$$

    $$LCF=4.105m$$

    $$A_{w}=\frac{100\:\times\:TPC}{\rho}$$

    $$TPC=\frac{1.025\times1900}{100}$$

    $$TPC=19.475m^2$$

    $$Second\:moment\:about\:midships\:\left(I_{m}\right)=2\times\frac{h^3}{3}\sum I$$

    $$=2\times\frac{15^3}{3}\times646$$

    $$I_{m}=1453500m^4$$

    $$Second\:moment\:of\:area\:about\:centroid\:\left(I_{F}\right)=I_{m}-A_{w}\times TPC^2$$

    $$1453500-\left(1900\times41.05^2\right)$$

    $$I_{F}=1421483m^4$$

    Q10 (10 Marks) Ship Stability

    0. The following data are available from the hydrostatic curves of a vessel.

    (Table will be here soon)

    Calculate the TPC at a draught of 5.05 m

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    The question refers to a table of hydrostatic data from which the TPC at a draught of 5.05 m is to be calculated. The table is not reproduced here, but the standard method is as follows.

    Method:

    • The TPC (tonnes per centimetre immersion) is the change in displacement (in tonnes) for a 1 cm change in draught.
    • TPC = (waterplane area x density of sea water)/100, where the waterplane area is in m2 and the density is 1.025 tonne/m3 for sea water.
    • Alternatively, TPC = (change in displacement)/(change in draught in cm) between two adjacent draughts in the hydrostatic table.
    • To find the TPC at 5.05 m, take the waterplane area at the draught of 5.05 m from the hydrostatic curves (or interpolate between the waterplane areas at the draughts immediately above and below 5.05 m), and apply the formula.

    Working example (assuming the table gives waterplane area at draughts):

    • If the waterplane area at 5.05 m is A m2, then TPC = (A x 1.025)/100 tonne/cm.
    • If the table gives displacement at successive draughts, then TPC at 5.05 m = (displacement at 5.10 m - displacement at 5.00 m)/10, i.e. the change in displacement for a 10 cm change divided by 10.

    Since the specific table values are not available in this question paper, the candidate should read the waterplane area (or the displacement change) at 5.05 m from the given hydrostatic data and apply the formula above to obtain the TPC. The answer will be the TPC in tonne/cm at a draught of 5.05 m.

    Q1 (10 Marks) Hull Construction πŸ”₯ Repeated 6x

    (a) Sketch a transverse section through the hold space of a container ship hull

    (b) Referring to the sketch in (a) describe how adequate structural strength is built into the hull

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    Part (a)

    Mid-ship half-section sketch of a container ship:

    Part (b)

    Structural Strength Features:

    • The deck plating is made thicker and uses higher tensile steel to withstand stresses caused by wide hatch openings and operational loads.
    • The deck, side shell, and longitudinal bulkheads are framed longitudinally. This arrangement combined with deep double bottom helps resist bending stresses due to hogging (upward bending) and sagging (downward bending) when the ship is under load.
    • The hatch coamings are made continuous to contribute to the overall longitudinal strength of the hull.
    • A torsion box is installed, running along the entire length of the ship from the machinery space bulkhead to the forward collision bulkhead. This structure provides the necessary torsional strength to counteract twisting forces acting on the hull during operation.
    • Deep web boxes are fitted at the ends of hatches, both at tank top and deck levels, to enhance transverse and torsional strength.
    • A deep double bottom is designed to withstand uplift forces caused by water pressure, especially when the ship is deeply loaded. It also provides additional strength to the hull structure.
    • Side girders are placed under container cells, with added transverse local stiffening. These elements distribute the concentrated loads from containers and increase overall stability.
    Q2 (10 Marks) Ship Resistance & Propulsion πŸ”₯ Repeated 13x

    Describe the relation ship between frictional resistance and

    (a) Ship speed

    (b) The wetted area

    (c) The surface roughness

    (d) The length of the vessel

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    Frictional resistance arises due to the interaction between the ship's hull and the water as the vessel moves. This resistance is caused by eddying water adjacent to the hull that is drawn along with the ship. The frictional resistance is given by:

    $$R_{f\:=\:}f\:.\:s\:.\:v^{n}$$

    Where:

    • f = Coefficient of friction
    • s = Wetted surface area
    • v = Ship's speed in knots
    • n = Constant (1.82)
    Part (a)

    Ship Speed (v):

    • Frictional resistance increases with increasing ship speed. The relationship is not perfectly linear, but it's approximately described by the formula Rf = fsv^n (Rf ∝ v^n), where 'n' is a constant (approximately 1.82). This means that a small increase in speed results in a larger increase in frictional resistance.
    Part (b)

    Wetted Area (s):

    • Frictional resistance is directly proportional to the wetted surface area (Rf ∝ s). A larger wetted area (the part of the hull submerged in water) leads to greater frictional resistance. The wetted area increases with the ship's draught (depth in the water).
    Part (c)

    Surface Roughness:

    • Increased surface roughness increases frictional resistance. Roughness can be caused by fouling (marine growth), corrosion, or poor hull maintenance. Regular cleaning and maintenance of the hull surface help to minimise roughness and reduce resistance.
    Part (d)

    Length of the Vessel

    • Frictional resistance is influenced by the vessel's length. As the length increases, the wetted surface area grows, and the coefficient of friction (f) changes, leading to an increase in Rf. Longer vessels experience higher frictional resistance, but streamlined designs can mitigate the effect to some extent.
    Q3 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 3x

    With reference to dry docking, define the responsibilities of the Second Engineer:

    (a) Priot to docking

    (b) Whilst the vessel is in dry dock

    (c) Prior to flooding and leaving the dock.

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    (a) Prior to Docking:

    Preliminary Preparation:

    • Review various plans, manuals, and previous drydock reports for reference.
    • Prepare a detailed repair list and ensure all required spares are accounted for.
    • Conduct an inventory of spares and requisition necessary items.
    • Gather required tools, including special tools like poker gauges, trammel gauges, gagging tools, etc.
    • Mark all overboard valves, NC (normally closed) and NO (normally open) valves clearly for easy identification.
    • Allocate jobs to team members and discuss the time schedule.
    • Conduct a safety meeting to highlight dry dock hazards and necessary precautions.

    Before Entering Dry Dock:

    • Identify the power requirements and machinery to be operational during docking.
    • Check the shore connection box for proper functionality.
    • Record the soundings of FO (Fuel Oil), LO (Lubricating Oil), and DO (Diesel Oil) tanks.
    • Discharge contents from clean drain tanks and sewage tanks.
    • Carry out Economizer soot-blowing.
    • Change over Main Engine, Diesel Generators, and Boiler to Low Sulfur Gas Oil (LSGO).
    • Stop and clean purifiers.
    • Ensure the low sea chest is open and the high sea chest is shut.
    • Keep firefighting appliances (FFA) on standby.
    • Shut down all non-essential machinery, including MGPS (Marine Growth Prevention System) and FWG (Fresh Water Generator).

    (b) Whilst the Vessel is in Dry Dock

    Upon Arrival:

    • Connect shore power and supplies after ensuring safety checks are completed.
    • Start necessary equipment like cooling water, air compressors, air conditioning, and fridge compressors.
    • Check for jobs assigned by dry dock personnel and prepare accordingly.
    • Attend a safety meeting with dry dock personnel to understand local safety rules and procedures.

    During Dry Docking::

    • Oversee and assist in:
      • Cleaning and inspecting the hull, rudder, sea chest, anodes, and propeller.
      • Measuring propeller drop, checking rudder clearances, and inspecting the stern tube bearing and seal.
      • Servicing underwater valves and overboard valves.
      • Inspecting anchor and cables conditions.
      • Overhauling deck machinery, cranes, elevators, and engine room equipment such as the Main Engine, Diesel Generators, Boiler, and Economizer.
      • Renewing pipes and valves as needed.
      • Performing electrical equipment maintenance and surveys.
      • Supervising service engineers for specific repair jobs.
    • Ensure tank cleaning, welding, and other repair works are completed according to the plan.
    • Run the standby diesel generator daily after starting the priming pump.

    (c) Before Flooding and Leaving the Dry Dock

    Final Checks:

    • Verify that all underwater fittings and drain plugs are securely in place.
    • Ensure all machinery has been boxed back and is ready for operation.
    • Check for any leakage in stern tube seals.
    • Take tank soundings to confirm proper levels.
    • Confirm the proper operation of all underwater valves, overboard valves, sea chests, and vents.
    • Inspect the stern tube tank for any irregularities.

    System Restart:

    • Switch back to ship's power after confirming all systems are functional.
    • Test the proper operation of all machinery and systems to ensure the ship is fully operational before leaving the dock.
    Q4 (10 Marks) Ship Stability πŸ”₯ Repeated 14x

    Explain how the period of roll varies with

    (a) The amplitude of roll

    (b) The radius of gyration

    (c) The initial metacentric height

    (d) The location of masses in the ship

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    The period of roll Tr of a ship is determined by the formula:

    $$T_{r}=\frac{2\pi K}{\sqrt{g\times GM}}$$

    where,

    • K is the radius of gyration (mass moment of inertia)
    • g is the acceleration due to gravity, and
    • GM is the metacentric height.
    Part (a)

    Amplitude of Roll:

    • The amplitude of roll, or the maximum angle of heel, does not affect the period of roll. While a larger amplitude implies a greater heeling moment and faster roll speed, the time taken to complete one full roll cycle remains constant for a given metacentric height and radius of gyration.
    Part (b)

    Radius of Gyration (K):

    • The period of roll is directly proportional to the radius of gyration. A larger radius of gyration (indicating a greater distribution of mass further from the ship's centre of rotation) leads to a longer period of roll. Conversely, a smaller radius of gyration (mass concentrated closer to the centre) results in a shorter period. The distribution of cargo significantly impacts K; cargo concentrated centrally minimises K and the roll period, while dispersed cargo maximises K and the roll period.
    Part (c)

    Initial Metacentric Height (GM):

    • The period of roll is inversely proportional to the square root of the metacentric height (GM). A larger GM (a stiffer ship) leads to a shorter roll period, as the ship quickly returns to its upright position. A smaller GM (a tender ship) results in a longer roll period, with slower return to the upright.
    Part (d)

    Location of Masses in the Ship:

    The location of masses in the ship will effect the GM & K. So the period of roll will be affected.

    • If masses are at bottom, G moves down, GM ↑, period of roll ↓.
    • If masses are at top, G moves up, GM ↓, period of roll ↑.
    • If masses are concentrated at centre, K ↓, period of roll ↓.
    • If masses are away from centre, K ↑, period of roll ↑.
    Q5 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 5x

    (a) Explain in detail, how an under water survey is carried out.

    (b) State the requirements to be fulfilled before an underwater survey is acceptable to the surveying authority.

    (c) Construct a list of the items in order of importance that the underwater survey authority should include

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    Part (a)

    An in-water survey, also known as Underwater Inspection in Lieu of Dry Docking (UWILD), involves a systematic and detailed examination of a vessel’s hull and underwater components while the ship remains afloat. The procedure includes the following steps:

    • The shipowner sends a request to the classification society surveyor, indicating the intention to perform an underwater survey.
    • A detailed plan of the ship's external hull features is submitted, showing the location of frames, bulkheads, welding lines, openings, etc.
    • The hull surface is cleaned before the survey to remove any marine growth or fouling that could obscure the inspection.
    • A diving company, approved by the classification society, is appointed to conduct the underwater inspection.

    A self-propelled survey vehicle equipped with the following tools is used:

    • Long-Range Light TV Camera to aid in steering and checking hull deterioration.
    • High-resolution colour TV Camera to provide a close-up view of the hull coating and welded seams.
    • 35mm Still Camera to capture still images.
    • Ultrasonic Probe for measuring plate thickness.
    • Depth Meter and Speed Indicator to provide accurate data on the vehicle's depth and movement.
    • Umbilical Cable to connect the survey vehicle to the survey boat, transmitting power and information.

    The survey boat is to be equipped with:

    • A control console with TV monitors.
    • Plate thickness printouts.
    • Audio and video cassette recorders.
    • Playback units.
    • Diver communication systems.
    • Vehicle control systems and associated instruments.

    Operation:

    • The survey vehicle is taken underwater by a diver to the survey starting point.
    • Using TV monitors and shell expansion plans as guides, the vehicle is navigated over the hull, focusing on the bottom structure, sides, stern frame, rudder, propeller, bilge keel, and hull openings.
    • All images, data, and information are recorded and transmitted back to the survey boat.
    • Detailed pictures of the stern frame, rudder, propeller, bilge keel, and hull openings are captured.
    • Divers are deployed to measure stern tube bearing wear, pintle clearance, and inspect stern seals, anodes, and rudder stock palm coupling bolts.
    • All recorded video and audio, including conversations between the surveyor and drivers, thickness printouts, measurements, and pictures are analyzed to determine the detailed underwater condition of the vessel.
    Part (b)

    Before an in-water survey is accepted by the survey authority, the following conditions must be met:

    The vessel's owner submits a request to the surveyor, including:

    • The proposed date and location for the survey.
    • General information about the diving company.
    • A declaration that the vessel has not suffered any damage due to grounding, collision, or other incidents.

    The ship's master or owner’s representative must provide a declaration confirming:

    • Any suspected or actual damage to the hull since the last dry-docking.
    • The underwater portion of the hull is protected by a suitable paint scheme that is of adequate thickness and remains valid until the next dry-dock.
    • The survey site should be in a protected area with calm and clear water, ensuring good underwater visibility. Attention must be given to the effects of currents and tides.
    • The hull must be clean for the external survey. The surveyor must be satisfied with the method and quality of the pictorial presentation, ensuring that it provides a reliable assessment of the hull's condition.
    • The underwater examination should be conducted by an approved diving company using closed-circuit TV and two-way communication, which can be monitored by the surveyor.
    • The vessel should be in as light an operating condition as possible to facilitate the survey.
    • Means must be available for the surveyor to examine the outside shell plating above the waterline.
    • Any required repairs identified during the survey must be carried out to the satisfaction of the attending surveyor.
    Part (c)

    While the importance of each item may vary depending on the vessel and its specific requirements, below is a list of items that should be included in an underwater survey in order of importance:

    • Underwater Hull: General condition of the hull below the waterline.
    • Bottom and Shell Plating: Inspection for corrosion, damage, and fouling.
    • Shell Openings: Examination of openings such as sea chests, drain plugs, and overboard discharge points.
    • Stern Tube Oil Leaks: Check for leaks around the stern tube.
    • Propeller Blade: Inspection for damage, wear, and fouling.
    • Rudder: Inspection for damage, wear, and clearances.
    • Sea Chest Opening and Grating: Examination for blockages, damage, and fouling.
    • Anodes: Check the condition and effectiveness of cathodic protection anodes.
    • Bilge Keel: Inspection for damage and fouling.
    • Drain Plugs: Ensure all drain plugs are secure and in good condition.
    • Overboard Valve Openings: Check for proper operation and condition.
    • Forward Area: Inspection for any damage due to anchor and chain movement.
    Q6 (10 Marks) Ship Stability πŸ”₯ Repeated 9x

    (a) Describe how the force on the ship's bottom and the GM vary when grounding takes place. (6)

    (b) A ship of 8,000 tonnes displacement takes the ground on a sand bank on a falling tide at an even keel draft of 5.2 metres. KG 4.0 metres. The predicted depth of water over the sand bank at the following low water is 3.2 metres. Calculate the GM at this time assuming that the KM will then be 5.0 metres and that the mean TPC is 15 tonne (10)

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    Part (a)

    When a ship grounds, the forces on the bottom and the metacentric height (GM) change depending on the grounding's nature and the ship's position on the seabed.

    If the ship grounds on a level bottom:

    • A ground reaction force acts vertically upwards from the seabed. This force counteracts part of the ship's weight and alters the distribution of buoyancy forces.
    • The ship’s centre of gravity (G) may appear to rise relative to the metacentre (M) because the upward ground reaction reduces the buoyancy force acting on the underwater volume.
    • This virtual rise in G reduces GM, potentially leading to a list.
    • If the list becomes excessive and the righting moment is insufficient, the ship may capsize.

    If the ship grounds on a pinnacle:

    • The ship experiences two forces at the ship's bottom:
      • A downward force due to the ship’s weight.
      • An upward reaction force is concentrated on the pinnacle.
    • The resulting force between the grounding pressure and the ship’s centre of buoyancy shifts downward towards the pinnacle.
    • This situation is similar to when the ship's stern touches the keel block in a dry dock.
    • A virtual loss of GM occurs because the ship’s inclining moment may exceed the maximum righting lever.
    • If the inclining moment is too great, the ship may develop an excessive list or even capsize.

    (b) Given:

    $$Displacement,\:\Delta=8000\:tonnes$$

    $$TPC=15\:tonnes$$

    $$Initial\:Draught=5.2m$$

    $$Final\:Draught=3.2m$$

    $$Ship\:KG=4.0m$$

    $$KM=5.0m$$

    To find GM

    $$Uptrust,\:P=TPC\times Fall\:in\:water\:level\:\left(cm\right)$$

    $$=15\times\left(520-320\right)$$

    $$=15\times200$$

    $$P=3000\:tonnes$$

    To Find Virtual loss of GM:

    $$Virtual\:loss\:of\:GM_1=\frac{P\times KM}{\Delta}$$

    $$=\frac{3000\times5}{8000}$$

    $$=\frac{15000}{8000}$$

    $$GM_1=1.88m$$

    Actual KM = 5.0m (given)

    $$Virutal\:KM=Actual\:KM-Virtual\:loss\:of\:GM_1$$

    $$=5.0-1.88$$

    $$=3.12$$

    Similarly, Actual KG = 4.0m (given)

    $$New\:GM=Virtual\:KM-\:Actual\:KG$$

    $$=3.12-4.0$$

    $$=-0.88$$

    Q7 (10 Marks) Ship Stability πŸ”₯ Repeated 7x

    (a) List the precautions necessary before an inclining experiment is carried out (6)

    (b) A box shaped vessel, 50 metres long x 10 metres wide, floats in salt water on an even keel at a draft of 4 metres. A center line longitudinal watertight bulkhead extends from end to end and for the full depth of the vessel. A compartment amidships on the starboard side is 15 metres long and contains cargo with permeability 30%. Calculate the list if this compartment is bilged. KG = 3 metres. (10)

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    Part (a)

    Precautions necessary before an Inclining experiment:

    • The ship should be in a sheltered position, such as a gravity dock, to minimize external influences from wind or waves.
    • Mooring ropes should be slack to allow the ship to move freely without restrictions.
    • Only workers essential to the experiment should be present onboard to avoid unnecessary weight and movement.
    • All tanks must either be completely empty or pressed up tight to eliminate the free surface effect, which can adversely affect stability calculations.
    • Any loose weights on the ship must either be removed or properly secured to prevent unintended movement during the experiment.
    • Ensure that the pendulums are long and properly suspended from stable points, such as underneath a hatch, to provide accurate deflection readings.
    • The test masses should be evenly distributed and placed as far from the centerline as possible to maximize measurable deflections.
    • The experiment should be carried out in calm weather to avoid the effects of wind, current, or waves on the vessel's stability.
    Q8 (10 Marks) Ship Stability πŸ”₯ Repeated 11x

    (a) Define longitudinal center of gravity (LCG) and longitudinal centre of buoyancy (LCB). (6)

    (b) A ship 120m long floats at draughts of 5.50m forward and 5.80m aft; MCT 1cm 80 tonne m, TPC 13. LCF 2.5m forward of midships. Calculate the new draughts when a mass of 110 tonne is added 24m alt of midships.

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    Part (a)

    Longitudinal Centre of Gravity (LCG):

    • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
    • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

    Longitudinal Centre of Buoyancy (LCB):

    • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
    • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.

    LCF in fwd and trim by stern

    $$Bodily \space sinkage \space = \space {{mass \space added} \over TPC} \space = \space {{110} \over 13} \space$$

    $$=\:8.5\operatorname{\mathrm{cm}}\:=0.085m$$

    $$Trim \space = \space {{m \times d} \over MCT_{1cm}}$$

    $$ = \space {{110 \times (24 + 2.5)} \over 80}$$

    $$Trim=36.43\operatorname{cm}=0.364m\:$$

    Change in fwd draught:

    $$d_{F}=\frac{-t}{L}\left\lbrack\frac{L}{2}-LCF\right\rbrack$$

    $$=\frac{-36.43}{120}\left\lbrack\frac{120}{2}-2.5\right\rbrack$$

    $$=-17.45\operatorname{cm}=-0.1745m$$

    Change in Aft draught:

    $$d_{A}=\frac{+t}{L}\left\lbrack\frac{L}{2}+LCF\right\rbrack$$

    $$=\frac{+36.43}{120}\left\lbrack\frac{120}{2}+2.5\right\rbrack$$

    $$=+18.97\operatorname{cm}=0.189m$$

    New draughts:

    $$D_{F}=5.5+0.085-0.175=5.41m$$

    $$D_{A}=5.8+0.085+0.18=6.065m$$

    Q9 (10 Marks) Ship Stability

    (a) Explain the concept of dynamical stability (6)

    (b) A ship of 12000 tonne displacement has a rudder 15m3 in area, whose centre is 5m below the waterline. The metacentric height of the ship is 0.3m and the centre of buoyancy is 3 3m below the waterline. When travelling at 20 knots the rudder is turned through 30°. Find the initial angle of heel if the force Fn perpendicular to the plane of the rudder is given by Fn = 577 AV2 sin∝ N

    Allow 20% for the race effcet. (10)

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    Part (a)

    Dynamical Stability is defined as the amount of energy required to heel a ship from its upright equilibrium position to a specific angle of heel. It provides a measure of the vessel's stability by considering its behaviour in response to dynamic external forces, such as wind or waves.

    • The concept compares the heeling moment energy (from external forces) and the righting moment energy (from the ship's stability).
    • The ship will absorb the energy imparted by the heeling moment. If the righting energy is greater than the heeling energy, the ship will stabilize; otherwise, it may capsize.

    Areas Under the Curve:

    • Area A: Represents the region where the heeling moment exceeds the righting moment (external energy > ship's stability).
    • Area B: Represents the region where the righting moment exceeds the heeling moment (ship's stability > external energy).
    • The balance of these areas determines whether the ship will right itself or continue to heel.

    When exposed to heeling forces such as wind or waves, the vessel inclines and may roll over to a certain angle of heel. If the external force is applied instantaneously, the ship must have enough reserve dynamic stability to absorb the energy and return to an upright position. If the external force is constant, the ship will remain at an equilibrium angle where the righting moment equals the heeling moment.

    This refers to the remaining righting energy available to counteract additional external forces. A higher reserve dynamic stability ensures the vessel can handle greater heeling forces without capsizing.

    Q10 (10 Marks) Ship Resistance & Propulsion

    (a) Describe the relationship between frictional resistanee and:

    (i) Ship speed

    (ii) The wetted area

    (iii) The surface roughness

    (iv) The length of the vessel

    (b) A ship travels at 15 kmots and has a QPC or 0.865 with a delivered power of 2600 kW. The apparent slip and 5% and the real slip is 28%. Calculate the total resistance and the wake fraction

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    Frictional resistance arises due to the interaction between the ship's hull and the water as the vessel moves. This resistance is caused by eddying water adjacent to the hull that is drawn along with the ship. The frictional resistance is given by:

    $$R_{f\:=\:}f\:.\:s\:.\:v^{n}$$

    Where:

    • f = Coefficient of friction
    • s = Wetted surface area
    • v = Ship's speed in knots
    • n = Constant (1.82)
    Part (a)

    Ship Speed (v):

    • Frictional resistance increases with increasing ship speed. The relationship is not perfectly linear, but it's approximately described by the formula Rf = fsv^n (Rf ∝ v^n), where 'n' is a constant (approximately 1.82). This means that a small increase in speed results in a larger increase in frictional resistance.
    Part (b)

    Wetted Area (s):

    • Frictional resistance is directly proportional to the wetted surface area (Rf ∝ s). A larger wetted area (the part of the hull submerged in water) leads to greater frictional resistance. The wetted area increases with the ship's draught (depth in the water).
    Part (c)

    Surface Roughness:

    • Increased surface roughness increases frictional resistance. Roughness can be caused by fouling (marine growth), corrosion, or poor hull maintenance. Regular cleaning and maintenance of the hull surface help to minimise roughness and reduce resistance.
    Part (d)

    Length of the Vessel

    • Frictional resistance is influenced by the vessel's length. As the length increases, the wetted surface area grows, and the coefficient of friction (f) changes, leading to an increase in Rf. Longer vessels experience higher frictional resistance, but streamlined designs can mitigate the effect to some extent.
    Q1 (10 Marks) Ship Resistance & Propulsion

    (a) Describe how a ship's propeller produces thrust.

    (b) Explain how the action of producing thrust may lead to propeller cavitation.

    (c) State the origin of vortex cavitation from the propeller cone.

    (d) Explain how a propeller blade may be eroded due to cavitation, describing the progressive nature of damage.

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    Part (a)

    How a propeller produces thrust:

    A ship's propeller generates thrust by rotating its blades, which displace water. This displacement creates a void, causing water to rush in and fill the space. This movement results in a pressure difference. Water accelerates from the front of the propeller blades to the back, forming a column of water slightly larger than the propeller itself. The increased velocity creates a water jet behind the propeller. This jet adds momentum and thrust to the water, propelling the vessel forward.

    Part (b)

    How producing thrust may lead to propeller cavitation:

    Cavitation occurs when the local pressure in the liquid drops below the saturated vapor pressure of water, leading to the formation of vapor bubbles. When producing thrust:

    1. The propeller creates areas of high and low pressure due to the pressure differential on the blades.
    2. If the pressure in the low-pressure areas (typically on the suction side of the blades) falls below the vapor pressure, vapor bubbles form.
    3. These bubbles collapse violently when they move into high-pressure regions, producing shockwaves and leading to cavitation. This phenomenon is exacerbated by sudden variations in local pressure caused by blade movement.
    Part (c)

    Origin of vortex cavitation from the propeller cone:

    Vortex cavitation originates from the vortices shed at the propeller blade tips and edges due to low-pressure regions.

    • It occurs when there is a high angle of incidence between the water flow direction and the blade's leading edge.
    • The low pressure within these vortices causes vapor bubble formation.
    • This type of cavitation is heavily influenced by the vessel's wake field and the propeller’s design, especially the blade tip geometry.
    Part (d)

    Propeller blade erosion due to cavitation:

    Cavitation causes erosion of propeller blades through a progressive damage process. The high-pressure shockwaves created by the collapsing bubbles erode the blade surface material. This process is a type of mechanical corrosion. Initially, the damage may be minor pitting. However, as the process continues, the pits grow larger and deeper, leading to significant material loss and potentially structural weakening of the blade. The erosion is progressive, with small defects growing over time. This necessitates regular inspections and potential repairs or replacements to prevent failure.

    Q2 (10 Marks) Ship Resistance & Propulsion

    (a) Describe TWO functions that trial data fullfils on a newly built ship, other than for satisfying owners of ship performance at sea.

    (b) State the TWO types of speed trial carried out.

    (c) State the requirements of a measured mile trials course.

    (d) List the conditions to be satisfied on a speed trials run.

    (e) Explain why trial runs are carried out in double runs

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    Part (a)

    Two Functions of Trial Data (Excluding Owner Satisfaction)

    • Establishing Baseline Performance Benchmarks: Trial data creates the definitive "ideal" baseline for the ship's performance. Throughout the vessel's operational life, engineers use this baseline to monitor hull fouling, propeller surface degradation, and main engine wear by comparing current fuel consumption and speed against the original trial data.
    • Compiling Statutory Maneuvering Data: The data extracted during trials is legally required to produce the ship's Maneuvering Booklet and the Wheelhouse Poster (as mandated by IMO/SOLAS). This provides the Master and Pilots with critical, factual data regarding the ship’s turning circles, crash-stop distances, and overall handling characteristics.
    Part (b)

    Two Types of Speed Trials

    1. Progressive Speed Trials: Conducted sequentially at several different engine power levels (e.g., 50%, 75%, 85%, and 100% Maximum Continuous Rating). This data is used to plot the ship's fundamental Speed-Power and Speed-RPM curves.
    2. Maximum / Endurance Speed Trials: Conducted at full power (MCR) for a sustained period to verify the vessel's top contractual speed and ensure the propulsion machinery can operate reliably at maximum load without overheating or failure.
    Part (c)

    Requirements of a Measured Mile Trials Course

    • Adequate Water Depth: The water must be deep enough to completely avoid "shallow water effects," which alter the ship's wave-making characteristics, increase hydrodynamic resistance, and artificially reduce speed.
    • Sheltered Environment: The area should be naturally protected from heavy swells, large waves, and strong prevailing winds to ensure weather does not skew the performance data.
    • Distinct Shore Marks: If using traditional methods, the course must have highly visible, precisely surveyed transit posts on the shore indicating the exact beginning and end of the nautical mile.
    • Predictable/Minimal Currents: The location should have negligible tidal streams or currents that run strictly parallel to the course.
    • Low Traffic Density: The area must be clear of dense commercial shipping to allow the vessel to make long, uninterrupted, straight-line approach runs.
    Part (d)

    Conditions to be Satisfied on a Speed Trials Run

    • Stabilized Approach: The vessel must be on a steady course with engine RPM and temperatures fully stabilized well before crossing the starting line of the measured distance.
    • Constant Engine Settings: The fuel rack position/index and engine RPM must remain strictly constant for the entire duration of the run.
    • Minimal Helm Movement: The rudder must be kept as close to amidships as possible. Excessive steering or large rudder angles induce drag and will artificially lower the recorded speed.
    • Specified Draft and Trim: The ship must be accurately ballasted or loaded to the precise draft and trim stipulated in the shipbuilding contract.
    • Permissible Weather: Wind force and sea states must be below agreed-upon maximum limits (typically not exceeding Beaufort force 4).
    Part (e)

    Why Trial Runs are Carried Out in Double Runs

    Trial runs are conducted in double runs (consecutive runs back and forth along the exact same reciprocal course) to eliminate the effects of wind and tidal currents. A single run only measures the Speed Over Ground (SOG). If a ship has a current pushing it from behind, its SOG will be artificially high. By turning around and running the exact same course in the opposite direction at the exact same engine settings, that same current will now be pushing against the ship. By taking the mathematical mean of these reciprocal runs (often using the "mean of means" method over multiple double runs), engineers can average out the environmental variables and calculate the ship's true speed through the water.

    Q3 (10 Marks) Hull Construction

    With reference to the carriage by sea of hazardous chemicals in bulk:

    (a) Explain how the protection of the internal structure is achieved;

    (b) outline the salety precautions to be observed by crew members to ensure personal safety

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    Part (a)

    Protection of the internal structure when carrying hazardous chemicals in bulk.

    The internal structure of the ship is protected from the corrosive and damaging effects of the chemicals by the following means:

    • Coating: the internal surfaces of the cargo tanks (tank plating, bulkheads, frames, stiffeners and piping) are coated with a suitable protective coating, such as epoxy, zinc silicate or other chemical-resistant paint, selected to be compatible with the cargoes to be carried. The coating is applied to a specified film thickness and is inspected and maintained.
    • Corrosion-resistant material: the tank structure may be made of, or lined with, corrosion-resistant material such as stainless steel, or fitted with a corrosion-resistant lining.
    • Cathodic protection: sacrificial anodes or impressed current systems may be fitted to protect the steel from corrosion.
    • Inerting and drying: the tanks are kept inerted (with inert gas) and dry to prevent corrosion and to avoid reactions between the cargo and moisture or oxygen.
    • Compatibility: only compatible cargoes are carried in a tank, and the tank is thoroughly cleaned, gas-freed and dried between different cargoes to prevent chemical attack and reactions.
    • Temperature control: cargoes that are corrosive at high temperature are carried with temperature control to limit the rate of corrosion.
    • Venting and vapour control: proper venting prevents the build-up of corrosive vapours.
    • The IBC Code (International Code for the Construction and Equipment of Ships Carrying Dangerous Chemicals in Bulk) specifies the materials, coatings and construction requirements for the tanks based on the properties of the cargo.
    Part (b)

    Safety precautions for crew members to ensure personal safety.

    • Wear appropriate personal protective equipment (PPE): chemical-resistant gloves, boots, goggles or face shield, and chemical-resistant protective clothing (coveralls, apron) appropriate to the cargo.
    • Use respiratory protection: self-contained breathing apparatus (SCBA) or suitable respirators when there is a risk of toxic or corrosive vapours, and when entering tanks or enclosed spaces.
    • Follow the entry procedures for enclosed spaces: gas testing, ventilation, use of the entry permit system, and having a standby person at the entrance.
    • Avoid skin and eye contact with the cargo; wash immediately if contact occurs, using the emergency shower and eyewash.
    • Do not eat, drink or smoke in the cargo handling areas.
    • Follow the cargo handling procedures and the ship's safety management system (SMS).
    • Be aware of the cargo's properties from the Material Safety Data Sheet (MSDS) and the cargo information provided.
    • Use proper tools and avoid sparks; follow the hot work permit system.
    • Report any leaks, spills or unusual conditions immediately.
    • Ensure adequate ventilation of the cargo spaces and accommodation.
    • Follow the emergency procedures and know the location of safety equipment, emergency escape routes and the emergency plan for the specific cargo.
    Q4 (10 Marks) General πŸ”₯ Repeated 8x

    (a) Explain what is meant by permissible length of compartments in passenger ships

    (b) Describe how the position of hulkhoads is determined.

    (c) Deseribe briefly the significance of the factor of subdivision

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    Part (a)

    Permissible Length

    Permissible length refers to the maximum length of a ship's compartment that can be flooded while ensuring that the sinkage, heel, or trim remains within acceptable limits without exceeding the floodable length. It ensures the ship remains afloat and stable after flooding.

    Permissible Length Formula:

    $$Permissible\:lenght\:=\:Floodable\:length\:\times Factor\:of\:subdivision$$

    The Factor of Subdivision depends on the ship's length and the nature of its service:

    • For passenger ships, the factor of subdivision is smaller compared to cargo ships.
    • Smaller compartments ensure enhanced safety in case of flooding.
    Part (b)

    Position of Bulkheads

    The position of bulkheads is determined based on the need to comply with the "reasonable amount of damage" criterion in case of flooding:

    • Transverse Watertight Bulkheads should vertically extend up to the margin line.
    • At least one watertight longitudinal bulkhead must be located 20% of the ship's breadth inward from each side.
    • Bulkheads are positioned along the ship's length at intervals equal to the permissible length, but not exceeding the floodable length.
    • The maximum permissible compartment length is limited to 10.7 meters.
    • Collision Bulkhead must be located forward at a distance equal to the permissible length from the forward perpendicular.
    • The ER must be enclosed by two transverse watertight bulkheads, with an aft peak watertight bulkhead enclosing the forward part.
    Part (c)

    Factor of Subdivision

    The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

    Permissible Length Formula:

    $$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

    A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

    Q5 (10 Marks) Surveys & Drydocking πŸ”₯ Repeated 4x

    If a ship is seriously damaged under water in way of a large fuel oil side bunker tank, what is the immediate effect and what may ultimately happen? What features in the ship would enhance safety?

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    If a ship is seriously damaged underwater in the area of a large fuel oil side bunker tank, it will cause the following:

    Bilging or Flooding of the Compartment:

    • Water will enter the damaged bunker tank, reaching the draught level.
    • The rate of flooding depends on the size of the breach and the water pressure at the depth of the damage.

    Oil Leakage into the Sea:

    • If the tank contains fuel oil, oil will begin to leak out, causing pollution.
    • The extent of leakage depends on the tank’s contents (empty, half-full, or full) and the location of the damage.

    List and Trim of the Vessel:

    • The ingress of water and loss of oil will create an imbalance, causing the vessel to list or trim.

    If corrective actions are not taken, uncontrolled flooding and loss of stability could lead to capsizing or sinking of the vessel.

    Features in the Ship to Enhance Safety:

    • Small Bunker Tank Sizes reduces the risk of extensive oil spillage and loss of stability.
    • Connectivity to transfer pumps allows the transfer of oil from the damaged tank to an empty tank, minimizing oil spillage and counteracting the loss of stability.
    • The tank’s size and location are designed to limit the effects of flooding, as per damage stability regulations.
    • Properly positioned transverse and longitudinal bulkheads enhance the subdivision factor, limiting water ingress to the damaged tank.
    • Ship’s Ballast System allows corrective ballasting to counteract the list or trim caused by the ingress of water.
    • Watertight Doors and Hatches prevent water from spreading to adjacent compartments.

    Recommended Immediate Actions by Crew:

    For Empty Tanks:

    • Quickly seal off the damaged tank by shutting all valves and isolating it from the transfer system to prevent water ingress into other parts of the vessel.

    For Half-Empty or Full Tanks:

    • Initiate oil transfer to another empty tank to reduce oil leakage and stabilize the ship.
    • Monitor the water ingress and ensure the tank is filled to the draught level with seawater if necessary, using ballast to correct the list.
    Q6 (10 Marks) Ship Resistance & Propulsion

    The force acting normal to the centerine plane of a roulder is given by the expression: Fn = 15.5 Av2∝ newtons.

    (drawing will be here soon)

    Calculate EACH of the following:

    (a) The diameter of the radder stock required for a maximum allowable stress of 77 MN/m2

    (b) The drag component of the rudder force when the rudder is put hard over at full speed.

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    The force acting normal to the centreline plane of the rudder is given by:

    Fn = 15.5 x A x V^2 x sin(alpha) newtons,

    where A is the rudder area (m2), V is the speed (m/s) and alpha is the rudder angle.

    The question refers to a drawing that is not reproduced here, but the standard method is as follows.

    Part (a)

    Diameter of the rudder stock for a maximum allowable stress of 77 MN/m2.

    • The rudder stock is subjected to a bending moment and a torque due to the rudder force.
    • The normal force Fn acts at the centre of pressure of the rudder, at a distance from the stock, producing a bending moment M = Fn x (distance from stock to centre of pressure).
    • The torque on the stock is produced by the component of the force and the tiller arrangement; the maximum torque occurs when the rudder is hard over.
    • The equivalent bending moment (or the combined stress) is calculated, and the stock diameter is found from the bending/torsion formula.
    • For a solid circular shaft, the section modulus Z = (pi x d^3)/32, and the bending stress = M/Z.
    • Combining bending and torsion, the equivalent bending moment Me = (M + sqrt(M^2 + T^2))/2, and the diameter is found from:

    d^3 = (32 x Me)/(pi x sigma_allowable), where sigma_allowable = 77 MN/m2 = 77 x 10^6 N/m2.

    • The diameter is then d = (32 x Me/(pi x 77 x 10^6))^(1/3) metres.
    • The calculated diameter is rounded up to the next standard size and checked against classification society rules.
    Part (b)

    Drag component of the rudder force when the rudder is put hard over at full speed.

    • The drag component is the component of the rudder force acting in the direction of the ship's motion (i.e. along the centreline, opposing the motion).
    • The normal force Fn acts normal to the rudder plane. When the rudder is at an angle alpha to the centreline, the drag component (the component along the ship's centreline) is:

    Drag = Fn x sin(alpha).

    • At hard over, alpha is typically 35 degrees, so Drag = Fn x sin(35) = Fn x 0.574.
    • The drag component increases the resistance of the ship and is a maximum at hard over and full speed.
    • The value is obtained by substituting the rudder area, the full speed and the hard-over angle into the expression.

    (Note: the specific numerical values of rudder area, speed and angle are given in the drawing referred to in the question; the candidate should substitute them into the expressions above to obtain the numerical answers.)

    Q7 (10 Marks) Ship Resistance & Propulsion

    (a) The residuary resistance of a 1/25 scale model of a ship is 7.68N when tested at 1.646 m/s in fresh water of density 1000 Kg/m3. The frictional resistance of the ship at 12 knots in sea water density 1025 kg/m3 is 148KN. Frictional resistance can be assumed to vary wild speed to the power 1.825

    Calculate the effectie power (naked) for the ship at the speed currespon ding to the model test

    (b) The following additional data apply to the ship operating in service aat the curresponding speed calculated in (a) with a propeller having a pitch of 4.8m.

    Appendage and weather alowance = 24%

    Quasi-propulsive coefficient (QPC) = 0.71

    Propeller speed = 1.85 rev/sec

    Taylor wake fraction = 0.3

    Propeller thrust = 650kN

    Calculate EACH of the following

    (i) The torque delivered to the propeller

    (ii) The propeller efficiency

    (iii) The real slip ratio

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    Part (a)

    Effective power (naked) for the ship at the speed corresponding to the model test.

    Model scale = 1/25, so the linear scale ratio lambda = 25.

    Model residuary resistance Rr(m) = 7.68 N at model speed 1.646 m/s in fresh water (density 1000 kg/m3).

    Froude's law of comparison: the residuary resistance of the ship is related to that of the model by:

    Rr(ship) = Rr(model) x (density_ship/density_model) x lambda^3.

    Rr(ship) = 7.68 x (1025/1000) x 25^3 = 7.68 x 1.025 x 15625 = 7.68 x 16015.6 = 123000 N = 123 kN.

    Ship speed corresponding to model speed (Froude scaling): Vs = Vm x sqrt(lambda) = 1.646 x sqrt(25) = 1.646 x 5 = 8.23 m/s.

    Ship speed in knots = 8.23 x 3600/1852 = 29628/1852 = 16.0 knots.

    Frictional resistance of the ship at 12 knots = 148 kN. Frictional resistance varies as speed^1.825.

    Frictional resistance at 16 knots = 148 x (16/12)^1.825 = 148 x (1.3333)^1.825.

    (1.3333)^1.825 = e^(1.825 x ln 1.3333) = e^(1.825 x 0.2877) = e^0.5251 = 1.6905.

    Frictional resistance at 16 knots = 148 x 1.6905 = 250.2 kN.

    Total naked resistance at 16 knots = residuary + frictional = 123 + 250.2 = 373.2 kN.

    Effective power (naked) = resistance x speed = 373200 x 8.23 = 3071436 W = 3071 kW.

    Answer: Effective power (naked) = 3071 kW.

    Part (b)

    Service data at the corresponding speed (16 knots = 8.23 m/s).

    Appendage and weather allowance = 24%, QPC = 0.71, propeller pitch = 4.8 m, propeller speed = 1.85 rev/s, Taylor wake fraction = 0.3, propeller thrust = 650 kN.

    (i) Torque delivered to the propeller.

    Total resistance in service = naked resistance x (1 + allowance) = 373.2 x 1.24 = 462.8 kN.

    Thrust required = total resistance/(1 - t), where t is the thrust deduction factor. However, t is not given; the propeller thrust is given as 650 kN, so we use this.

    Delivered power PD = Effective power (service)/QPC = (462.8 x 8.23)/0.71 = 3808.8/0.71 = 5364 kW.

    Torque Q = PD/(2 x pi x rev/s) = 5364000/(2 x pi x 1.85) = 5364000/11.6239 = 461464 N-m = 461.5 kN-m.

    Answer: Torque delivered to the propeller = 461.5 kN-m.

    (ii) Propeller efficiency.

    Speed of advance Va = V x (1 - w) = 8.23 x (1 - 0.3) = 8.23 x 0.7 = 5.761 m/s.

    Thrust power = T x Va = 650000 x 5.761 = 3744650 W = 3744.7 kW.

    Propeller efficiency (open water) = Thrust power/Delivered power = 3744.7/5364 = 0.698.

    Answer: Propeller efficiency = 0.698 (69.8%).

    (iii) Real slip ratio.

    Pitch speed = P x rev/s = 4.8 x 1.85 = 8.88 m/s.

    Real slip = (Pitch speed - Va)/Pitch speed = (8.88 - 5.761)/8.88 = 3.119/8.88 = 0.3512.

    Answer: Real slip ratio = 0.351 (35.1%).

    Q8 (10 Marks) Ship Resistance & Propulsion

    A ship consumes an average of 70 tonnes of fuel per day on main engines at speed of 17 knots. The fuel consumption for auxiliary purposes is 8 tonnes per day. When 800 nautical miles from port it is found that only 140 tonnes of fuel renmains on boarrd and this will be insufficient to reach port at the normal speed. Determine the speed at which the ship should travel to complete the voyage with 20 tonne fuel remaining.

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    Main engine consumption at 17 knots = 70 tonne/day.

    Auxiliary consumption = 8 tonne/day.

    Total consumption at 17 knots = 70 + 8 = 78 tonne/day.

    Distance to port = 800 nautical miles.

    Fuel remaining = 140 tonne. We must complete the voyage with 20 tonne remaining, so fuel available for the voyage = 140 - 20 = 120 tonne.

    Main engine consumption varies as the cube of speed. Auxiliary consumption is constant (8 tonne/day).

    Let the reduced speed be V knots.

    Main engine consumption at speed V = 70 x (V/17)^3 tonne/day.

    Total consumption at speed V = 70 x (V/17)^3 + 8 tonne/day.

    Time to cover 800 nm at speed V = 800/V days.

    Total fuel used for the voyage = (70 x (V/17)^3 + 8) x (800/V) = 120.

    So 800 x 70 x (V/17)^3 / V + 800 x 8/V = 120.

    800 x 70 x V^2/17^3 + 6400/V = 120.

    800 x 70 x V^2/4913 + 6400/V = 120.

    56000 x V^2/4913 + 6400/V = 120.

    11.398 V^2 + 6400/V = 120.

    Multiply through by V: 11.398 V^3 + 6400 = 120 V.

    11.398 V^3 - 120 V + 6400 = 0.

    This is a cubic. Solve by trial:

    Try V = 14: 11.398 x 2744 - 120 x 14 + 6400 = 31276 - 1680 + 6400 = 35996 (too high, positive).

    Try V = 8: 11.398 x 512 - 960 + 6400 = 5836 - 960 + 6400 = 11276 (positive).

    Try V = 6: 11.398 x 216 - 720 + 6400 = 2462 - 720 + 6400 = 8142 (positive).

    Try V = 4: 11.398 x 64 - 480 + 6400 = 729 - 480 + 6400 = 6649 (positive).

    Try V = 2: 11.398 x 8 - 240 + 6400 = 91 - 240 + 6400 = 6251 (positive).

    The equation 11.398 V^3 - 120 V + 6400 = 0 has no positive root near these values because the constant term is large. This indicates the fuel is insufficient at any reasonable speed, OR the interpretation should be reconsidered.

    Reconsidering: The main engine consumption varies as the cube of speed, but the auxiliary consumption is constant per day. Let us check whether 120 tonne is sufficient.

    At the lowest practical speed, say 5 knots: time = 800/5 = 160 days. Main engine consumption = 70 x (5/17)^3 = 70 x 0.0254 = 1.78 tonne/day. Auxiliary = 8 tonne/day. Total = 9.78 tonne/day. Fuel for 160 days = 1565 tonne - far more than 120 tonne. So the fuel is grossly insufficient.

    This suggests the intended interpretation is that the auxiliary consumption is included in the 70 tonne, or that the problem expects the main engine consumption to be the only variable and the auxiliary is small. Given the numbers, the intended answer is obtained by assuming total consumption varies as the cube of speed (i.e. treating the 78 tonne/day as varying with speed cubed):

    78 x (V/17)^3 x (800/V) = 120.

    78 x 800 x V^2/4913 = 120.

    62400 x V^2/4913 = 120.

    V^2 = 120 x 4913/62400 = 589560/62400 = 9.448.

    V = 3.07 knots.

    This is unrealistically low, indicating the data is not fully consistent. However, the standard textbook method for this type of problem treats the total daily consumption as varying with the cube of speed:

    Total consumption at 17 knots = 78 tonne/day.

    Consumption at speed V = 78 x (V/17)^3 tonne/day.

    Time = 800/V days.

    Fuel used = 78 x (V/17)^3 x 800/V = 120.

    78 x 800 x V^2/17^3 = 120.

    62400 x V^2/4913 = 120.

    V^2 = 120 x 4913/62400 = 9.448.

    V = 3.07 knots.

    Given the inconsistency, the most defensible answer using the standard assumption (total consumption varies as cube of speed) is:

    Answer: Speed = 3.07 knots.

    Note: This result is very low and indicates that with only 120 tonne available for 800 nm, the ship cannot realistically reach port at a normal speed; the calculation shows the severe fuel shortage. In practice the ship would need to reduce speed drastically or obtain additional fuel.

    Q9 (10 Marks) Ship Stability

    At a draught of 1.0 m in sea water density 1025 kg/m3 the dispalcement of a ship is 900 tonne and the height of the centre of buoyancy above the keel (KB) is 0.6 m. Values of tonne per centimeter immersion (TPC) in sea water for a range of draught are given in the table.

    (Table will be here soon)

    (a) Calcualte EACH of the following for a draught of 6.0m in sea water

    (i) The displacement

    (ii) The height of the centre of buoyancy above the keel

    (b) At a draught of 6.0m, the height of the longitudinal metacentre above the keel (KML) is 128m and the second moment of are of the waterplane above a transverse axis through midship is 996728m4. The centre of flotation is aft of midships. Calculate the distance of the centre of flotation (LCF) from midship.

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    At draught 1.0 m in sea water (density 1025 kg/m3): displacement = 900 tonne, KB = 0.6 m.

    TPC values in sea water for a range of draughts are given in the table (not reproduced here). The standard method is as follows.

    Part (a)

    For a draught of 6.0 m in sea water:

    (i) Displacement.

    • The displacement at 6.0 m is found by integrating the TPC curve (or the waterplane area) from 1.0 m to 6.0 m and adding the displacement at 1.0 m.
    • Displacement at 6.0 m = displacement at 1.0 m + integral of (TPC x 100) over the draught from 1.0 to 6.0 m.
    • Using Simpson's rule on the TPC values at the draughts given in the table, the added displacement = (h/3) x sum of (TPC x multiplier) x 100, where h is the draught interval in metres.
    • Displacement at 6.0 m = 900 + added displacement (tonne).

    (ii) Height of the centre of buoyancy above the keel (KB) at 6.0 m.

    • KB is found from the moment of the displacement about the keel.
    • KB at 6.0 m = (moment of displacement about keel)/(total displacement).
    • The moment of the added displacement about the keel is found by integrating the TPC curve with the lever arm (draught) using Simpson's rule, and adding the moment of the initial 900 tonne at KB 0.6 m.
    • KB = (900 x 0.6 + moment of added displacement)/(total displacement at 6.0 m).
    Part (b)

    Distance of the centre of flotation (LCF) from midship at draught 6.0 m.

    Given: KML = 128 m, second moment of area of the waterplane about a transverse axis through midship I = 996728 m4, centre of flotation is aft of midships.

    The longitudinal metacentre above the keel: KML = KB + BML.

    BML = I/V, where V is the volume of displacement = displacement/density = (displacement at 6.0 m)/1.025 m3.

    So BML = 996728/V.

    KML = KB + BML = 128 m, so BML = 128 - KB.

    The distance of the LCF from midship is found from the relationship between the second moment of area about midship and about the centroid (LCF):

    I about centroid = I about midship - A x (LCF distance)^2.

    The second moment of area about the centroid (through the LCF) is related to BML by BML = I_about_LCF/V.

    So I_about_LCF = BML x V.

    Then (LCF distance)^2 = (I about midship - I about LCF)/A, where A is the waterplane area at 6.0 m (found from the TPC: A = TPC x 100/1.025).

    LCF distance = sqrt((996728 - I_about_LCF)/A) metres, aft of midships.

    The numerical values are obtained by substituting the displacement, KB, waterplane area and BML found in part (a) and from the given data. The LCF is aft of midships as stated.

    Q10 (10 Marks) Ship Stability

    A ship of 10000 tonne displacement floats in sea water of density 1025 kg/m3 at a draught of 6m. A rectangular tank 1Om long and 8 m wide is partially full of oil fuel of density 900 kg/m3. In this condition, the KG of the ship is 6.25m.

    Other hydrostatic data for the above condition are:

    Centre of buoyancy above the keel (KB) = 3.325m

    Transverse metacentre above the centre of buovancy (BM) = 4.865m

    Tonnes per centimeter immersion (TPC) = 20.5

    Calculate the change in effective metacentric height when a rectangular tank 12m long. 10m wide and 6m deep, with its base 1m above the keel, is filled to a depth of 5m with, sea water ballast.

    Note: Assume the ship to be wall-sided over the affected range of draught.

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    Ship displacement = 10000 tonne, sea water density 1025 kg/m3, draught 6 m.

    KG = 6.25 m, KB = 3.325 m, BM = 4.865 m, TPC = 20.5.

    A rectangular tank 12 m long, 10 m wide, 6 m deep, with its base 1 m above the keel, is filled to a depth of 5 m with sea water ballast.

    Step 1 - Weight of ballast added.

    Volume of ballast = 12 x 10 x 5 = 600 m3.

    Weight of ballast = 600 x 1.025 = 615 tonne.

    New displacement = 10000 + 615 = 10615 tonne.

    Step 2 - Change in draught.

    Increase in draught = weight added/TPC = 615/20.5 = 30 cm = 0.30 m.

    New draught = 6 + 0.30 = 6.30 m.

    Step 3 - New KB.

    The ship is wall-sided, so the increase in KB = increase in draught/2 = 0.30/2 = 0.15 m.

    New KB = 3.325 + 0.15 = 3.475 m.

    Step 4 - New BM.

    BM = I/V. For a wall-sided ship, I (second moment of area of waterplane) is constant, so BM varies inversely with volume.

    Original volume V1 = 10000/1.025 = 9756.1 m3.

    Original BM = 4.865 m, so I = BM x V1 = 4.865 x 9756.1 = 47463 m4.

    New volume V2 = 10615/1.025 = 10356.1 m3.

    New BM = I/V2 = 47463/10356.1 = 4.583 m.

    Step 5 - New KG.

    The ballast is added at a height above the keel. The centre of gravity of the ballast is at the centre of the 5 m depth, i.e. at 1 + 5/2 = 3.5 m above the keel.

    New KG = (10000 x 6.25 + 615 x 3.5)/10615 = (62500 + 2152.5)/10615 = 64652.5/10615 = 6.091 m.

    Step 6 - New KM and GM.

    New KM = new KB + new BM = 3.475 + 4.583 = 8.058 m.

    New GM = new KM - new KG = 8.058 - 6.091 = 1.967 m.

    Step 7 - Original GM.

    Original KM = KB + BM = 3.325 + 4.865 = 8.190 m.

    Original GM = KM - KG = 8.190 - 6.25 = 1.940 m.

    Step 8 - Change in effective metacentric height.

    Change in GM = new GM - original GM = 1.967 - 1.940 = +0.027 m.

    Answer: The effective metacentric height increases by 0.027 m (GM increases from 1.940 m to 1.967 m).