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Time it takes for a water tank to empty

  1. Oct 27, 2012 #1
    1. The problem statement, all variables and given/known data
    A cylindrical tank of diameter 2R contains water to a depth d. A small hole of diameter 2r is opened in the bottom of the tank. r<<R, so the tank drains slowly. Find an expression for the time it takes to drain the tank completely.


    2. Relevant equations
    [tex]p_1+\frac{1}{2}ρv_1^2+ρgh_1=p_2+\frac{1}{2}ρv_2^2+ρgh_2\\
    Q=vA=\frac{\delta V}{\delta t}[/tex]


    3. The attempt at a solution
    I believe, since ##p_1## and ##p_2## are the same, that Bernoulli's equation becomes ##2\rho gd=v_2^2##. I am assuming I need to use the equation of volume rate of flow for time, but then I would need the velocity ##v_2##. But how do I solve for time from that? How am I to find Q?
     
    Last edited: Oct 27, 2012
  2. jcsd
  3. Oct 27, 2012 #2

    TSny

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    Hello. Hmm, you have a factor of 2 in the third term on the left of Bernoulli's equation. Check to make sure that's right.

    It would help if you told us where you're picking points 1 and 2.

    What is the justification for cancelling the terms that involve h1 and h2?
     
  4. Oct 27, 2012 #3
    Whoooops I should have reread my post. Typos everywhere. I've edited now; thanks for pointing them out.
     
  5. Oct 27, 2012 #4

    TSny

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    OK, so I presume point 2 is at the little hole at the bottom and that you made the approximation ##v_1≈0##. Still looks like a little error (or typo) in your expression for ##v_2^2##. Can you find it?
     
  6. Oct 27, 2012 #5
    Ah the ##\rho ## should have canceled out there as well. So now that I know what the velocity is, can I then use ##vA=\frac {\Delta V}{\Delta t}?
     
  7. Oct 27, 2012 #6

    haruspex

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    Not quite. Check that again.
    If ##v_2## is the linear flow rate out of the hole, what is the volume flow rate? What does that then tell you about how fast the depth in the tank changes?
     
  8. Oct 27, 2012 #7

    TSny

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    Yes, Can you express the volume in terms of the depth ##d##? Instead of using finite differences Δ, you will want to use instantaneous rates of change (you're heading towards a differential equation). It might be better to let the depth of the water be denoted by ##y## or ##h## instead of ##d## in case you need to express the rate of change of the depth as a derivative.
     
    Last edited: Oct 27, 2012
  9. Oct 27, 2012 #8
    Alright, so if I have ##vA=\frac{dV}{dt}=A\frac{dy}{dt}##, then plug in y from ##v=\sqrt{2gy}## it looks like ##\frac{dV}{dt}=A\frac{d}{dt}(\frac{v^2}{2g})## which I believe gives me...
    $$Q=\frac{dV}{dt}=\frac{Av}{g}$$

    Can I say that Volume(final) = Volume(initial) + QΔt and solve for Δt? Or is it more complicated than this...
     
  10. Oct 27, 2012 #9

    TSny

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    Stick with ##y## as the dependent variable and ##t## as the independent variable. Can you find an expression for the rate of change of ##y##?
     
  11. Oct 27, 2012 #10

    TSny

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    Careful here. Are the two areas ##A## the same in this equation?
     
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