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Work problem, easier inegral?

  1. Sep 26, 2004 #1
    I am working on a problem involving a semicircular tank underground, and how much work is required to pump water out of the top. i set everything as i see correctly, but cannot computer the definite integral. i only know basic substitution and to some degree integration by parts, but that is apparently not required for this problem. the integral i am having problems with is (def integral, from zero to 2, of) [(3-y)(sqrt(4y - y^2))] dy.
    i narrowed it down to the fact i can't integrate sqrt (4y - y^2), which i found out is incredibly difficult and complicated. so i'm thinking there is either an easier way to inegrate this that i'm unaware of, or a way to set the problem up so that i'm not dealing with square roots. i have been working on this for over 5 hours now, in the past two days, with no good results.

    the problem is something like this. semicircular tank underground, lenght of 8, radius of 2. the tank is laying horizontal to the horizon, with the flat part of the semicircle at the top. there is a nozzle 1 ft high that the water has to go through.

    i set it up like this:
    work = force x distance = volume x density x acceleration due to gravity x distance.

    so i computed the dimensions of the slab, 8 * dy * 2(sqrt[4y - y^2])

    multiplied this by 62.5 (density of water), 32 (accel. of gravity), and (3-y) (distance the slab of water travels, (2-y)+1, since there is a 1 ft nozzle).

    this left me with 32000 times the inegral i listed above. sorry this is so long, but i'm really stuck with this one. thanks for any advice
     
  2. jcsd
  3. Sep 26, 2004 #2

    Tide

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    You didn't mention how fast the water flows out of the nozzle.

    Wouldn't it be easier to imagine all the mass of the water concentrated at its center of gravity and calculate the work required to lift it to the desired location?
     
  4. Sep 26, 2004 #3
    doesn't matter how fast it flows out, same amount of work is required. also, i can't do the center of gravity approach for two reasons, 1) i was never exposed to anything having to do with center of gravity and most importantly 2) the layers of water at the top of the tank have less distance to travel than the layers towards the bottom, so each layer requires a different amount of work to pump out of the tank.
     
  5. Sep 26, 2004 #4

    Tide

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    Actually, the flow could matter since it requires work/energy to accelerate the water that is initally at rest. However, if the flow rate is sufficiently small then the kinetic energy component can be neglected.

    Basically, your problem is the integration which you can do: I get, for the indefinite integral,
    [tex]\sin^{-1} \frac {y-2}{2} + \sqrt {y(4-y)}[/tex]

    I think you would have gotten a much simpler integral if you had taken the y = 0 level to be at the top of tank.
     
  6. Sep 26, 2004 #5

    pig

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    Could you draw a picture? I got a different formula, but I'm not sure I understand the problem...

    The way I understand it, you have a half-cylinder tank, and the large flat part is facing up? Wouldn't then dV be 8*2sqrt(4-y^2)dy? Where does the y in 4y come from?

    And since distance is y+1, you get:

    dW = density*acceleration*16*sqrt(4-y^2)*(y+1)*dy
    W = density*acceleration*16*[integral from 0 to 2]((y+1)sqrt(4-y^2))

    It's almost 5 AM here so I have no time to integrate it by hand, but wolfram integrator says the indefinite integral is -1/3*(4-y^2)^(3/2). The definite one then gives 8/3.. Multiply with 16, density and acceleration and that should be it...
     
    Last edited: Sep 26, 2004
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