Q7 (16 Marks) Ship Resistance & Propulsion
SC&S • Written Exam

(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.

Appeared In: Apr 2025

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

Exam Ready
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.

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