Q7 (10 Marks) Ship Resistance & Propulsion 🔥 Repeated 7x in exams
SC&S • Written Exam

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 2550kW. Calculate: (16)

(i) indicated power

(ii) effective power

(iii) ship speed.

Appeared In: Sep 2025Apr 2023Feb 2021Dec 2019Sep 2019Apr 2019Aug 2018

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

Exam Ready

Given:

$$\Delta=15000t$$

$$Shaft\:Power\:\left(SP\right)=420$$

$$Transmission\:Efficiency=83\%$$

$$Shaft\:losses=6\%$$

$$Propeller\:Efficiency=65\%$$

$$QPC=0.71$$

$$Thrust\:Power=2550kW$$

$$\left(\imaginaryI\right)\:Delivered\:Power\:\left(DP\right)=\frac{Thrust\:Power\:\left(TP\right)}{Propeller\:Efficiency\:\left(\eta P\right)}$$

$$DP=\frac{2550}{0.65}$$

$$DP=3923.07kW$$

$$\left(ii\right)\:Shaft\:Power=\frac{Delivered\:Power\:\left(DP\right)}{Transmission\:Efficiency\:\left(\eta T\right)}\:$$

$$SP=\frac{3923.07}{0.94}$$

$$SP=4173.47kW$$

$$\left(iii\right)\:Indicated\:Power=\frac{Shaft\:Power\:\left(SP\right)}{Mechanical\:Efficiency\:\left(\eta m\right)}$$

$$IP=\frac{4173.47}{0.83}$$

$$IP=5028.28kW$$

$$\left(iv\right)\:Effective\:Power=DP\times QPC$$

$$EP=3923.07\times0.71$$

$$EP=2785.3797kW$$

$$\left(v\right)\:Shaft\:Power=\frac{\Delta^{\frac23}\times V^3}{Admiralty\:Co-efficient}$$

$$4173.47=\frac{15000^{\frac23}\times V^3}{420}$$

$$V=14.23knots$$

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