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Pumping a cylindrical tank

  1. Dec 5, 2007 #1
    A storage tank is a right circular cylinder 20ft long and 8 ft in diameter with its axis horizontal. If the tank is half full of olive oil weighing 57 lb/ft^3, find the work done in emptying it through a pipe that runs from the bottom of the tank to an outlet that is 6 ft above the top of the tank.

    My work
    I set the bottom of the tank at the point (0,0) & then drew a circle with a radius of 4.
    then solved the the circle's equation for x, x=[tex]\sqrt{(16-y^2)}[/tex]
    10-y should be the distance the work must do
    W=2280 [tex]\int[/tex](10-y)[tex]\sqrt{(16-y^2)}[/tex]
    Then I distributed the (10-y)
    W=22800[tex]\int[/tex][tex]\sqrt{(16-y^2)}[/tex]- 2280[tex]\int[/tex]y[tex]\sqrt{(16-y^2)}[/tex]

    For part two, I set u=16-y^2 & got
    W=22800[tex]\int[/tex][tex]\sqrt{(16-y^2)}[/tex]+ 1140[tex]\int[/tex]y[tex]\sqrt{(16-y^2)}[/tex]

    This is the part where I get lost, I did it a bit differently from the solutions manual, but at this point the manual shows inserting 4[tex]\pi[/tex] as follows
    22800(4[tex]\pi[/tex])+ 1140[tex]\int[/tex]y[tex]\sqrt{(16-y^2)}[/tex] (evaluated from 0 to -4), I tried from (0 to 4) in my solution.

    I don't understand why the solutions manual is doing this particular step, it says its the area of a semicircle, but can anyone explain why I use it in this problem & where the integral disappears to?
  2. jcsd
  3. Dec 5, 2007 #2

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    The problem can be solved easily in a different way if you calculate position of the centre of mass of the oil by integration. After that, work done = Mgh, where M is the total mass and h the height the CM has to rise. This would be the Physicist's approach.
  4. Dec 5, 2007 #3
    Wouldn't you have to calculate it for a 3d object though? I don't have those skills yet. So far all I've dealt with was thin plats, 2d objects.
  5. Dec 6, 2007 #4

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    Yes, for 3d objects, but you have to do that anyway. And because of symmetry, the integration would be only for 2d.

    I don't notice any value of g in your calcs? Also, I'm not very sure what you are trying to do.

    Let me know if you need more help, but after explaining what is the method you are following.
  6. Dec 6, 2007 #5
    The value of g should already be in the 57 lbs, as weight. Guess I'm kind of lost myself, but up unto there, I everything the solution manual has.
  7. Jul 9, 2008 #6
    Hi everyone.

    I have a very similar problem to the one kuahji posted, but my problem is a storage tank completely full of oil. Would I then calculate using (6+8 -y) or (14-y) or would I use the same (10-y)? Also, wouldn't my integral be from -4 to 4 ( or 2* [0 to 4])? Thanks for any help!
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