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Draining a Tank

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose water is leaking from a tank through a circular hole of area Ah at its bottom. When water leaks through a hole, friction and contraction of the stream near the hole reduce the volume of water leaving the tank per second to [tex]cA_{h}\sqrt{2gh}[/tex], where c (0 < c < 1) is a empirical constant. Determine a differential equation for the height h of water at time t for the cubical tank shown. The radius of the hole is 2 in., and g = 32ft/s2

    9zwqhw.jpg

    2. Relevant equations

    [tex]cA_{h}\sqrt{2gh}[/tex]

    [tex]\frac{dV}{dt} = -cA_{h}\sqrt{2gh}[/tex]

    [tex]A_{h} = (\pi)r^{2} = (\pi)2^{2} = 4\pi[/tex]

    3. The attempt at a solution

    I think have to find the volume and take the derivative of it with respect to time.

    [tex]V = h^{3}[/tex] not sure if this is the right volume

    [tex]\frac{dV}{dt} = 3h^{2}\frac{dh}{dt}[/tex]

    [tex]3h^{2}\frac{dh}{dt} = -cA_{h}\sqrt{2gh}[/tex]

    [tex]\frac{dh}{dt} = \frac{-cA_{h}\sqrt{(2)(32)h}}{3h^{2}}[/tex]

    [tex]\frac{dh}{dt} = \frac{-cA_{h}8\sqrt{h}}{3h^{2}}[/tex]

    [tex]\frac{dh}{dt} = \frac{-c32\pi\sqrt{h}}{3h^{2}}[/tex]
     
  2. jcsd
  3. Sep 23, 2009 #2

    tiny-tim

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    Hi KillerZ! :smile:
    No, V is proportional to h, isn't it?
     
  4. Sep 23, 2009 #3
    Is it something like V = (a)(b)(h) then? So a and b both are 10 but h is changing.
     
  5. Sep 23, 2009 #4

    tiny-tim

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    That's right! :smile:
     
  6. Sep 23, 2009 #5
    so like this:

    [tex]A_{h} = (\pi)r^{2} = (\pi)(\frac{2}{12})^{2} = \frac{4}{144}\pi[/tex] I missed converting in to ft in the first post.

    [tex](a) = (b) = 10[/tex]

    [tex]V = (a)(b)(h) = 100h[/tex]

    [tex]\frac{dV}{dt} = 100\frac{dh}{dt}[/tex]

    [tex]100\frac{dh}{dt} = -cA_{h}\sqrt{2gh}[/tex]

    [tex]\frac{dh}{dt} = \frac{-cA_{h}\sqrt{(2)(32)h}}{100}[/tex]

    [tex]\frac{dh}{dt} = \frac{-cA_{h}8\sqrt{h}}{100}[/tex]

    [tex]\frac{dh}{dt} = \frac{-c\frac{32}{144}\pi\sqrt{h}}{100}[/tex]

    [tex]\frac{dh}{dt} = \frac{-c\pi\sqrt{h}}{450}[/tex]
     
    Last edited: Sep 23, 2009
  7. Sep 23, 2009 #6

    tiny-tim

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    Looks good! :smile:
     
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