From: Sandy

When a ship is 800 nautical miles from port it's speed is reduced by 20%,
thereby reducing the daily fuel consumption by 42 tonnes and arriving in
port with 50 tonnes onboard.  If the fuel consumption per hour is thus given
by the expression (0.136 + 0.001V3) where V is speed in knots;

Estimate a) reduced consumption per day b) fuel onboard when speed reduced
c) normal consumption for 800 nautical mile voyage d) increase in steaming

Hi Sandy.

I'm going to assume that equation you gave is C = 0.136 + 0.001V3.

So let's say before speed reduction, C1 = 0.136 + 0.001V1 3 and after speed reduction, C2 = 0.136 + 0.001V2 3.

If we subtract the second equation from the first, we get C1 - C2 = 0.001(V1 3 - V2 3). You know that consumption dropped by 42 tonnes per day, so that is 42/24 = 1.75 tonnes/hour. That can replace C1 - C2 in the left side of the equation. You also know that the speed was reduced by 20%. That means 80% of V1 is V2. Now you can substitute the last equation with the replacements: 1.75 = 0.001 [V1 3 - (0.8 V1 )3 ]

So now you can solve for the original speed V1. Use V1 to calculate the reduced speed V2, then use these to calculate the original and new consumption rates, C1 and C2. I assume that part (a) is a percentage question, since you are given the actual reduction as 42 tonnes per day.

Once you know V2, you can calculate the time to travel 800 nautical miles, then apply that to the fuel consumption rate C2. Add that to the 50 tonnes you finish with and you have your answer to (b).

Part (c) is just a variation on (b) using the original speed and consumption figures, and once you've done the work for (b) and (c), you have the times required for each speed so the difference is the answer to (d).

Hope this helps!
Stephen La Rocque.