1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    User Avatar
    Science Advisor
    Homework Helper

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook