Q8 (10 Marks) Ship Resistance & Propulsion 🔥 Repeated 7x in exams
SC&S • Written Exam

(a) What is meant by the Admiralty Coefficient and the Fuel Coefficient? (6)

(b) A ship of 14900 tonne displacement has a shaft power of 4460 kW at 14.55 knots. The shaft power is reduced to 4120 kW and the fuel consumption at the same displacement is 541 kg/h. Calculate the fuel coefficient for the ship. (10)

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Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

Exam Ready

The Admiralty Coefficient (C) is a method for estimating the propulsion power needed for a newly built ship. It's considered relatively constant for a given ship design. The formula is:

$$C=\frac{\Delta^{2/3}\times V^3}{BP}$$

Where:

  • C = Admiralty Coefficient
  • Δ = Displacement in tonnes (weight of the ship when fully loaded)
  • V = Speed in knots
  • BP = Brake power in kilowatts (power delivered by the ship's engine)

A higher Admiralty Coefficient indicates a more efficient ship design, meaning it requires less power to achieve a given speed. Values typically range from 350 to 600.

Fuel Coefficient:

The Fuel Coefficient (F.C.) is used to calculate a ship's daily fuel oil consumption. The formula is:

$$Daily\:fuel\:oil\:consumption\:=\:\frac{\Delta^{2/3}\times V^3}{FC}$$

Where:

  • F.C. = Fuel Coefficient
  • Δ = Displacement in tonnes
  • V = Speed in knots

The Fuel Coefficient can vary significantly, with typical values ranging from 40,000 to 120,000. A higher Fuel Coefficient implies greater fuel efficiency (lower daily fuel consumption) for a given speed and displacement.

Part (b)

$$admiraty\:coefficient\:\left(C\right)=\:\frac{\Delta^{2/3}V^3}{Shaft\:power}=\frac{\Delta^{\frac23}\times V^3}{SP}$$

$$\frac{SP_1}{SP_2}=\frac{V_1^3}{V_2^3}$$

$$\frac{4460}{4120}=\frac{14.55^3}{V_2^3}$$

$$V_2=14.17kntos$$

$$Fuel\:consumption\:per\:hour=541\operatorname{\mathrm{\:kg}}\:per\:hour$$

$$Fuel\:consumption\:per\:day\:=\:541\times24=12.98t\:per\:day$$

$$Fuel\:coefficient=\frac{\Delta^{\frac23}\times V_2^3}{Fuel\:consumption\:per\:day}$$

$$=\:\frac{14900^{\frac23}\times14.17^3}{12.98}$$

$$=132726.9$$

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