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Water pumped from a cylinder

  1. Feb 20, 2007 #1
    1. The problem statement, all variables and given/known data
    Water in a cylinder of heigth 10 ft and radius 4 ft is to be pumped out.
    Find the work required if the tank is full of water and the water must be
    pumped to a height 5 ft above the top of the tank


    2. Relevant equations



    3. The attempt at a solution
    I tried to slice this cylinder horizontally, so I get:

    Force of gravity on slice = 62.4 * 16 * phi * delta h
    Work done of slice = Force of gravity * distance

    I am confused on what is the distance here, is it 5?
     
  2. jcsd
  3. Feb 20, 2007 #2
    for that slice where did you get 62.4 and 16 from an who is phi??

    think about this - for some slice at some height x within the cylinder, this volume must be displaced what distance??

    Also since you are dealing with a continuous volume of water you will need to use this formula

    [tex] W = \int F\cdot dl [/tex]
     
  4. Feb 20, 2007 #3
    62.4 is the density of water and 16 phi is the area of each slice which is actually a circle. the formula of a circle is phi*r^2 and the radius here is 4
     
  5. Feb 21, 2007 #4
    i think you mean pi not phi
     
  6. Feb 21, 2007 #5
    sorry I mean pi, that's right! sorry for the mistake
     
  7. Feb 21, 2007 #6

    HallsofIvy

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    The whole point of "slicing horizontally" is that different "slices of water", a different depths have to be raised different distances! Since you are using "delta h" as the infinitesmal thickness, you are taking h to be either the depth (distance below the top of the tank) or the height (distance above the bottom of the tank. If it is the former, then the distance you need to lift that "slice of water" (Have you ever tried to slice water?!) is to the top of the tank and then 5 feet: h+ 5. Integrate with h going from 0 to 10. If it is the latter, then the distance to the top of the tank is 10-h and the total distance to 5 feet above the tank is 10-h+ 5=15- h. Again integrate with h going from 0 to 10. In fact, try it both ways and see that you get the same answer!
     
  8. Feb 21, 2007 #7
    so what you mean here that I can choose whether distance and it will give the same result?? My equation of force of gravity is correct right?? I just then have to multiply it by the distance, either h+5 or 15-h and take the integral?
     
  9. Feb 21, 2007 #8

    HallsofIvy

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    Yes, that was pretty much what I said. Since 62.4*16= 998.4, your formula for the force of gravity on a "slice of water" will be [itex]998.4\pi dh[/itex].

    If h is the distance below the top of the tank, then h+ 5 is the distance that water must be pumped so the work done on that water is [itex]998.4\pi (h+ 5)dh[/itex] and so the total work done is
    [tex]998.4\pi\int_0^{10} (h+5) dh[/tex].

    If, on the other hand, you choose h to represent the distance above the bottom of the tank, the distance to the top of the tank is 10- h and so the distance that water must be raised is 10-h+ 5= 15-h. Now the work done on that "slice of water" is [itex]998.4\pi (15- h)dh[/itex] and the total work done is
    [tex]998.4\pi\int_0^{10} (15- h) dh[/tex].

    You should find that those two calculations give exactly the same answer.
     
  10. Feb 21, 2007 #9
    I think I made a little mistake here, should the Force on gravity on slice also be multiplied by g (gravity)??
     
  11. Feb 21, 2007 #10
    No it shouldn't be multiplied by by g because 62.4 is the weight density and weight is a unit of force.
     
  12. Feb 21, 2007 #11
    are you sure about this? because on the book there are some examples that involved counting the work and they multiply this by g??
     
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