Q7 (10 Marks) Ship Resistance & Propulsion
SC&S • Written Exam

(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)

Appeared In: Oct 2024

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

Exam Ready
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%.

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