The daily fuel consumption of a ship at 17 knots is 42 tonne. Calculate the speed of the ship if the consumption is reduced to 28 tonne per day, and the specific consumption at the reduced speed is 18% more than at 17 knots.
Structured for DG Shipping MEO Class II examination scoring criteria.
Daily fuel consumption at 17 knots = 42 tonne.
Consumption varies as the cube of speed, but here the specific consumption (consumption per unit power) also changes.
Let the normal specific consumption be s. At the reduced speed the specific consumption is 1.18s.
Consumption per day = k x V^3 x (specific consumption), where k is a constant.
At 17 knots: 42 = k x 17^3 x s.
At the reduced speed V: 28 = k x V^3 x 1.18s.
28/42 = (V^3/17^3) x 1.18.
0.6667 = (V^3/4913) x 1.18.
V^3/4913 = 0.6667/1.18 = 0.5650.
V^3 = 0.5650 x 4913 = 2775.8.
V = (2775.8)^(1/3) = 14.05 knots.
Answer: The ship speed at the reduced consumption = 14.05 knots.