Q10 (10 Marks) Ship Resistance & Propulsion
SC&S • Written Exam

(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)

Appeared In: Jul 2025

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

Exam Ready
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).

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