21.NA December 2018 Q.6

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.

If you are noticing some error in problems kindly comment below.Thanks

Given

Speed 2 , V2 = 17 knots
Consumption 2, C2 = 42 t
Consumption 1 , C1 = 28 t

To find

1.Speed 1 , V1 

Solution

We know that Fuel consumption / Unit time ∝ Speed³

C1 / C2 = (V1 / V2)³   -------------------------------------- (1)

But it is also given that specific consumption at the reduced speed is 18% more than at 17 knots. That means 118 %.

So the speed ratio will also be more than 18 % of their initial value.

Considering above fact and applying in (1)

C1 / C2 = 1.18 x (V1 / V2)³

 28 / 42 = 1.18 x (V1 / 17)³

   0.667 = 1.18 x (V1 / 17)³

   
   0.565 = (V1 / 17)³

Taking Cube root on both sides

   з√ 0.565 = V1 / 17

      0.8267 = V1 / 17

 Speed V1 = 14.05 Knots