fuel consumption VS speed change

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QAMER
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fuel consumption VS speed change

Post by QAMER »

Hi anyone know how to solve this type of question formula/calculation . THANKS



question: If your ship burns 8 tons of fuel per hour at 15 knots, how many tons per hour will it burn at 21 knots?
11.2 tons
22.0 tons
15.7 tons
12.9 tons
Answer Was: B
Your Answer A
Wrong
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The Dieselduck
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Re: fuel consumption VS speed change

Post by The Dieselduck »

My head is not much into this kind of puzzle these days, but it seems to me the question is missing some information to be able to answer correctly. You would have to know more info to calculate, like coefficent of hull friction and such... I know, from experience that fuel consumption goes up dramatically but the exact formula I forget right this minute.
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JK
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Re: fuel consumption VS speed change

Post by JK »

I played around with it while I was suppose to be working on something else and like Martin says, there must be something missing. Then again, in my job we don't buy the fuel so I don't worry too much about it.
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Sébastien
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Re: fuel consumption VS speed change

Post by Sébastien »

Reed's volume 4 chapter 7
It can be assumed that a major portion of the designed operational speed range is proportional, and that for any two points within this range the specific fuel consumption is constant. On the same trip, it can be assumed that the ships' displacement is also constant. You are left with cons1/cons2=(v1/v2)3. In your example cons2= 8* (21/15)3 = 8*2.744 = 21.95 tons
QAMER
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fuel consumption VS speed change

Post by QAMER »

Thanks to all of you specially SEBASTIEN.
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The Dieselduck
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Re: fuel consumption VS speed change

Post by The Dieselduck »

very sharp indeed sebastien, thank you. Its so simple I feel stupid.. ehehehe
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Re: fuel consumption VS speed change

Post by KWOBGUS »

Hey I was just curious where you got 3 as a multiplication factor. Thanks!
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JollyJack
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Re: fuel consumption VS speed change

Post by JollyJack »

3 is not a multiplication factor, it's cons1/cons2=(v1/v2) cubed
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