# Fuel consumption formula for license exam

Can anyone show me how to work this fuel problem,
Your vessel consumes 274 bbls of fuel per day at a speed of 17.5 kts. What will be the fuel consumption of your vessel at 13.5 kts.

I think you’re missing some information to solve this problem accurately.

cons1/cons2=(v1/v2)^3.

cons1 = 274 bbl/day
v1 = 17.5 kts
v2 = 13.5 kts

cons2 = 125.8 bbls /day

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Thats correct but not understanding how u figuring this problem. How would you do it on a calculator?

Using algebra.

cons1/cons2=(v1/v2)^3.

first flip it

con2/con1 = (v2/v1)^3

divide v2 by v1
v2/v1 = 13.5/17.5 = 0.7714

get the cube x^3 = xxx
0.7714 * 0.7714 * 0.7714 = 0.4591

Now you have:
con2/con1 = 0.4591

multiply both sides by con1
con2 = (0.4591 * 274)

con2 = 125.8

Thanks so much

39 posts were split to a new topic: Fuel Consumption Not for License Exam

Would u use same formula if question asked u if u have enough fuel to travel at 17kts for 775 miles. What speed would u have to go to travel 977 miles. With same amount of fuel.

Shipgr
Were u answering my question or dutchie?

No, the formulas are in Merchant Marine Officer’s Handbook, the chapter Engineering for Deck officer. Also maybe in Richard Plant’s Formula for the Mariner.

This is the formula you need:

Con1/Con2 = (S1^2 *D1) / (S2^2 * D2)

Thank you sir. Yall been very helpful

Maybe I’m getting myself lost in the numbers, but I don’t see how that formula works for this problem since it doesn’t account for the cube-root curve for the Consumption. The question is asking what speed2 is, and you need speed2 to calculate consumption 2. So you have two missing variables, no?

I just used your original con2/con1 = (v2/v1)^3 to solve for consumption@17kt which is 251 bbl/d. At 17kt it would take you 1.89days to go 775nm, multiply by consumption and you’d burn 477bbl.

The question then becomes what speed must you go to burn only 477bbl and still make it 977nm. I would plot the consumption for various speeds to interpolate the one that gets you that distance burning less than 477bbl and I get about 15.1 kts. So any speed less than that and you won’t run out of fuel.

Did I make that way more difficult than it had to be?

The missing “s” (squared accounts for the first two) is hiding in the d because s*t = d

consumption2 is given as the same as con1 so Con1/con2 = 1

My copy of the Merchant Maine Officer’s Handbook is a 1980 reprint of the 4th ed, 1965.

This is what it says about the formula: page 19-12

Within 3 knots on either side of the normal speed, in vessels capable of 15 knots, the consumption of fuel varies approximately as the cube of the speed.

On the question i have it didnt give any bbls heres the exact question. Your vessel arrives in port with sufficient fuel to steam 775 miles at 17kts. If you are unable to take on bunkers, at what speed must you proceed to reach your next port, 977 miles distant?.
The correct answer is 15.1 kts.

KC’s formula is the one you need and the understanding that con1/con2=1

So

1 = (S1^2 *D1) / (S2^2 * D2)

(Multiply denominator on both sides of equation)

(S2^2 * D2)= (S1^2 *D1)

divide both sides by D2 and take square root

S2 =Sqrt [(S1^2 *D1)/D2]

Dont follow how u have a second speed, when thats what im looking for, but please correct me if im wrong. Problem only has 17kts to work with.

S1=17 knots, you have D1, you have D2 and want to find S2 which you say is 15.1 knots. To solve for a quantity in algebra, you rearrange the formulas to isolate the unknown on one side of the equation and solve using the knowns.

This is simpler if it can be seen that Con1/Con1 = 1 but if not then a little longer way round: there are two unknowns and two equations.

1. This is from the book: Con1/Con2 = (S1^2 *D1) / (S2^2 * D2)
2. This is given: Con1 = Con2

Given Con1 = Con2 then #2 equation can substitute: Con1/Con1 = (S1^2 *D1) / (S2^2 * D2)

For Con1 ≠ 0 :then Con1/Con1 = 1

1 = (S1^2 *D1) / (S2^2 * D2)

Then rearrange to solve for S2 which has already been shown.

D1 = 775 miles
D2 = 977 miles
S1 = 17 kts

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This guy did a great explanation of fuel conservation problems including a video. There are also some additional tips for the exam.