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Work done pumping a reservoir

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data
    a reservoir shaped like a right circular cone, point down, 20ft across the top and 8 ft deep, is full of water. how much work does it take to pump the water to a level 6ft above the top?


    2. Relevant equations



    3. The attempt at a solution
    so i placed the cone with the point at the origin. then the ΔV= πr^2h
    since the cone is 8 ft deep and has a radius of 10 ft, then the slope of the line of the side of the cone is (8/10)x or (4/5)x, so (5/4)y=x, therefore the radius will be (5/4)y
    so then ΔV=π(25/16)y^2dy
    the force required is then equal to the weight per unit volume*volume and my books says that water is 62.4lb/ft^3 so the force is 62.4(25/16)πy^2dy
    then the ΔW=62.4(25/16)πy^2(6-y)dy since were pumping out 6ft of water?
    so to calculate the total work i have to calculate 62.4(25/16)π[itex]\int_{0}^{8}(6y^2-y^3)dy[/itex], but this gives me 0 work done, and i dont see where i went wrong in setting up the integral, but i think it might be the limits of integration or the distance through which the force must act.
     
  2. jcsd
  3. Dec 8, 2011 #2

    Simon Bridge

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    work against gravity to lift mass element dm a height z is g.z.dm where dm = \rho dV

    I read the question not that you are pumping out 6' of water but that you are pumping all the water "to a level 6' above the top". Thus the water in the top-most layer is raised 6', and the bit in the point has to be raised 8+6=14'.
     
  4. Dec 9, 2011 #3
    yeah i realized it was 14 when i was working on it at the school library, but thanks for responding anyways
     
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