Q8 (10 Marks) Ship Resistance & Propulsion 🔥 Repeated 5x in exams
SC&S • Written Exam

(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

Appeared In: Jan 2026Jan 2025Nov 2024Oct 2023Dec 2022

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

Exam Ready
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:

$$D=6m$$

$$p=0.9$$

$$BAR=0.48$$

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

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

$$W_{F}=0.3$$

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

$$η_{prop} \space = \space 0.65$$

(i) Blade area:

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

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

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

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

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

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

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

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

$$V_{T}=19.24knots$$

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

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

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

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

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

$$V=20.6knots$$

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

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

$$T_p \space = \space 2225.48 $$

$$T_p \space = \space d_p \times η_{prop}$$

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

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

$$dp=2\pi NT$$

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

$$T=297.22KN$$

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