Q3 (10 Marks) Ship Resistance & Propulsion 🔥 Repeated 4x in exams
SC&S • Written Exam

(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.

Appeared In: Mar 2026Dec 2025Dec 2024Jul 2018

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

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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:

$$D=4.6m$$

$$P=4.3m$$

$$d=0.75$$

$$S=28\%$$

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

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

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

$$V_{T}=13.23knots$$

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

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

$$V_{a}=9.52knots$$

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

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

$$A=16.18m^2$$

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

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

$$T=215.25KN$$

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

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

$$T_{p}=1054.18KW$$

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