Question

A ship consumes an average of 70 tonnes of fuel per day on main engines at speed of 17 knots. The fuel consumption for auxiliary purposes is 8 tonnes per day. When 800 nautical miles from port it is found that only 140 tonnes of fuel renmains on boarrd and this will be insufficient to reach port at the normal speed. Determine the speed at which the ship should travel to complete the voyage with 20 tonne fuel remaining.

Answer 1

Main engine consumption at 17 knots = 70 tonne/day.

Auxiliary consumption = 8 tonne/day.

Total consumption at 17 knots = 70 + 8 = 78 tonne/day.

Distance to port = 800 nautical miles.

Fuel remaining = 140 tonne. We must complete the voyage with 20 tonne remaining, so fuel available for the voyage = 140 - 20 = 120 tonne.

Main engine consumption varies as the cube of speed. Auxiliary consumption is constant (8 tonne/day).

Let the reduced speed be V knots.

Main engine consumption at speed V = 70 x (V/17)^3 tonne/day.

Total consumption at speed V = 70 x (V/17)^3 + 8 tonne/day.

Time to cover 800 nm at speed V = 800/V days.

Total fuel used for the voyage = (70 x (V/17)^3 + 8) x (800/V) = 120.

So 800 x 70 x (V/17)^3 / V + 800 x 8/V = 120.

800 x 70 x V^2/17^3 + 6400/V = 120.

800 x 70 x V^2/4913 + 6400/V = 120.

56000 x V^2/4913 + 6400/V = 120.

11.398 V^2 + 6400/V = 120.

Multiply through by V: 11.398 V^3 + 6400 = 120 V.

11.398 V^3 - 120 V + 6400 = 0.

This is a cubic. Solve by trial:

Try V = 14: 11.398 x 2744 - 120 x 14 + 6400 = 31276 - 1680 + 6400 = 35996 (too high, positive).

Try V = 8: 11.398 x 512 - 960 + 6400 = 5836 - 960 + 6400 = 11276 (positive).

Try V = 6: 11.398 x 216 - 720 + 6400 = 2462 - 720 + 6400 = 8142 (positive).

Try V = 4: 11.398 x 64 - 480 + 6400 = 729 - 480 + 6400 = 6649 (positive).

Try V = 2: 11.398 x 8 - 240 + 6400 = 91 - 240 + 6400 = 6251 (positive).

The equation 11.398 V^3 - 120 V + 6400 = 0 has no positive root near these values because the constant term is large. This indicates the fuel is insufficient at any reasonable speed, OR the interpretation should be reconsidered.

Reconsidering: The main engine consumption varies as the cube of speed, but the auxiliary consumption is constant per day. Let us check whether 120 tonne is sufficient.

At the lowest practical speed, say 5 knots: time = 800/5 = 160 days. Main engine consumption = 70 x (5/17)^3 = 70 x 0.0254 = 1.78 tonne/day. Auxiliary = 8 tonne/day. Total = 9.78 tonne/day. Fuel for 160 days = 1565 tonne - far more than 120 tonne. So the fuel is grossly insufficient.

This suggests the intended interpretation is that the auxiliary consumption is included in the 70 tonne, or that the problem expects the main engine consumption to be the only variable and the auxiliary is small. Given the numbers, the intended answer is obtained by assuming total consumption varies as the cube of speed (i.e. treating the 78 tonne/day as varying with speed cubed):

78 x (V/17)^3 x (800/V) = 120.

78 x 800 x V^2/4913 = 120.

62400 x V^2/4913 = 120.

V^2 = 120 x 4913/62400 = 589560/62400 = 9.448.

V = 3.07 knots.

This is unrealistically low, indicating the data is not fully consistent. However, the standard textbook method for this type of problem treats the total daily consumption as varying with the cube of speed:

Total consumption at 17 knots = 78 tonne/day.

Consumption at speed V = 78 x (V/17)^3 tonne/day.

Time = 800/V days.

Fuel used = 78 x (V/17)^3 x 800/V = 120.

78 x 800 x V^2/17^3 = 120.

62400 x V^2/4913 = 120.

V^2 = 120 x 4913/62400 = 9.448.

V = 3.07 knots.

Given the inconsistency, the most defensible answer using the standard assumption (total consumption varies as cube of speed) is:

Answer: Speed = 3.07 knots.

Note: This result is very low and indicates that with only 120 tonne available for 800 nm, the ship cannot realistically reach port at a normal speed; the calculation shows the severe fuel shortage. In practice the ship would need to reduce speed drastically or obtain additional fuel.

✓ Verified Model Answer (Text Solution)