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Speed of Efflux

  1. Sep 28, 2010 #1
    1. The problem statement, all variables and given/known data
    The container of uniform cross section area A holds two immiscible non-viscous incompressible liquids of densities d and 2d each of height H/2. A tiny hole of area S<<A is punched on the vertical side of the container at height h(<H/2)

    Find the speed of efflux.

    2. Relevant equations

    3. The attempt at a solution

    I know how to find it when there is only one liquid present.
    What is the approach in this case?

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  3. Sep 28, 2010 #2


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    The pressure difference at the hole is
    P - Po = g[H/2*d + (H/2-h)*2d ]

    The density of the liquid flowing out is 2d. Hence the velocity of the liquid is

    v = sqrt[2(P - Po)/2d]

    Substitute the values and find v.
    Last edited: Sep 29, 2010
  4. Sep 29, 2010 #3
    how did you get H/4*2d? the height 'h' is a variable.

    After some time density of liquid flowing out will be d. But we are not taking it into account. Why is it so?
  5. Sep 29, 2010 #4


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    It should be (H/2 - h)*2d

    From the expression we get only the initial velocity. As the level of the liquid decreases, the velocity will also change. When the liquid with density d flows out, again velocity will change.
  6. Sep 29, 2010 #5
    Thanks alot, Sir.
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