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Thread: Power of water pump engine

  1. #1
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    Question Power of water pump engine

    The following question is from an old UK A-Level Applied Mathematics textbook. I have tried to solve this, but the answer I get is different from that given in the book. Could anyone please see what I'm doing wrong, or is the book answer wrong?

    An engine is pumping water from a large tank and delivering it through a pipe of diameter 0.04 m at a rate of 100 litre s-1. Find the work done by the engine in one second.

    This is my answer:

    100 litres = 0.1 m3, so the length of pipe l filled in one second is given by

    Power of water pump engine-eqn1.jpg

    The work done in one second is therefore given by

    Power of water pump engine-eqn2.jpg


    The book gives the answer as 317 kJ. What am I doing wrong?

    Many thanks
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  2. #2
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    Re: Power of water pump engine

    1 liter of water weighs 1kg. 100 liters of water weighs 100kg. So, you are saying that the force needed to move the water is 100g, and it is moving it for a distance of 250/pi meters. I do not think that is correct. I believe you should be figuring out how much force is needed to get 100kg of water from rest to a distance of 250/pi m away in 1 second.

    F = ma

    s = at^2

    At $t=0$, you have $s=0$. At $t=1$, you have $s = \dfrac{250}{\pi}$. So, it seems like your acceleration is $\dfrac{250}{\pi}\, \dfrac{\text{m}}{\text{s}^2}$. So, your force is:

    F= 100 \cdot \dfrac{250}{\pi}

    Then, work is:

    100 \left( \dfrac{250}{\pi} \right)^2 \approx 633.3 \text{ kJ}

    I get roughly double what the book gets. Not sure why. Perhaps I needed to divide something by 2.
    Last edited by SlipEternal; Nov 14th 2017 at 06:09 AM.
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  3. #3
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    [SOLVED] Power of water pump engine

    Thanks, that makes sense. I also think that the force is not constant throughout the whole distance, as not all of the 100 litres needs to be moved through the entire distance.

    So, the initial force is zero and the final force is F = 100 * 250/pi. Therefore the average force throughout the whole of the distance is F.

    So then, work is (100)(250/pi)^2 = 317 kJ
    Last edited by enceladus; Nov 14th 2017 at 07:58 AM.
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  4. #4
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    Re: Power of water pump engine

    The question, as stated, makes no sense to me. Just moving water through a distance does not necessarily require any work at all! In fact if the water is moving down hill, you can get work out of it. Are we to assume that the water is moving directly upward?
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  5. #5
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    Re: Power of water pump engine

    Yes, I see your point. I think we must assume the water is moving directly upwards, otherwise there's no point to the question!
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