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Fluid flow

  1. Feb 5, 2004 #1
    A large storage tank is filled with water. Neglecting viscosity, show that the speed of water emerging through a hole in the side a distance h below the surface is [itex] v = \sqrt{2gh} [/tex].
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  3. Feb 6, 2004 #2


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    Please show us what you have to work with and what you have tried on this problem.
  4. Feb 6, 2004 #3


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    Square the whole thing and it starts to look familiar.

    [tex]V^2 = 2gh[/tex]

    Remember that gravity applies a downward force on the fluid and that a force on any side of a fluid is applied to ALL sides of the fluid.
  5. Feb 6, 2004 #4


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    Good! A good solid hint without completely solving the problem!
  6. Feb 6, 2004 #5
    Write the Bernoulli's Theorem at the Point just inside the tank and just outside the tank.
  7. Feb 9, 2004 #6
    If I use Bernoulli's Equation, one side of it says
    [tex] P+ \rho gh + \frac{1}{2}\rho v^2 [/tex].
    What does the other side of the equation look like?

    If the other side is 0,
    [tex] P = \rho gh [/tex]
    [tex]2\rho gh = -\frac{1}{2}\rho v^2 [/tex]
    But since h is going below the surface the negative is expected and can be ignored. Then-
    [tex] 2\rho gh = \frac{1}{2}\rho v^2 [/tex]
    The density of water is 1 g/cm^3.
    [tex] 2gh = \frac{1}{2} v^2 [/tex]
    [tex]4gh = v^2 [/tex]
    But then I get
    [tex] v = 2\sqrt{gh} [/tex], not [tex] v = \sqrt{2gh} [/tex]
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