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Work required to pump - Very tricky

  1. Dec 6, 2009 #1
    A right cylindrical tank is filled with sea water. The tank has a radius of 2 feet and a height of 8 feet. If the water level is now 2 feet below the top of the tank, how much work will be required to pump the sea water to the top of the tank? (The weight-density of seawater is 64 Ib/ft3.)

    The answer is 24,127 ft-Ibs, can anyone figure out why?
     
  2. jcsd
  3. Dec 6, 2009 #2

    rock.freak667

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    How do you think you need to approach the problem?
     
  4. Dec 6, 2009 #3
    deltaV = pi(r)^2(thickness) --this is of the cylindrical pieces you cut out of the whole
    so r = 2

    deltaV = 4pi(deltaY)

    F(y) = weight = 64 Ib/ft3 (4pi(deltaY))

    ..and that's pretty much all i can think of?
    It's very confusing, I'm sorry.
     
  5. Dec 6, 2009 #4

    rock.freak667

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    The water fills the tank to 6ft. You can find the volume of water using the volume of a cylinder [itex]V=\pi r^2 h[/itex]. Give volume and density you can get the mass of water. You just need to get the energy needed to raise the water a height of 2 ft.
     
  6. Dec 6, 2009 #5
    so using the volume equation i get V = 4pi(6) = 24pi

    Then to get the force i use the work equation?

    so,

    W = fnint ((64 Ib/ft3)(24pi),x,0,2)

    Is that it?
     
  7. Dec 7, 2009 #6

    HallsofIvy

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    No, that's wrong. The water at the top of the tank only has to be raised 2 feet but the water at the bottom of the tank has to be raised 8 feet. That's why IntegrateMe is talking about "cylindrical pieces". Each "piece" has area [/itex]4\pi[/itex] square feet and, if we take the "thickness" of each "piece" to be dx, volume [itex]4\pi dx[/itex] cubic feet and so mass [itex](64)(4\pi dx)= 256\pi dx[/itex] pounds. If that pieces is at height "x", it has to be raised 8- x feet. The work done in raising that piece is [itex]256\pi (8- x)dx[/itex] and so the total work done is
    [tex]256\pi\int_0^6 (8- x)dx[/itex] "foot-pounds".

    Because the density is constant, you could look at the average distance the water must be lifted. The cylinder also has constant cross section so the "average" height of the water is 1/2 way up, at (0+ 6)/2= 3 feet. The water has to be lifted and "average" distance of 8- 3= 5 feet. Find the weight of the entire cylinder of water and multiply by 5 feet, not 2. That will give the same answer as the integral.
     
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