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Bernoulli's equation and velocity

  1. Dec 10, 2014 #1
    What happens when you have more than one hole in a vessel containing liquid?
    Can you help me finding the velocities at different holes if they are at some height from level of liquid at a particular instant?
    I know what happens when you put a small hole in a tank containing liquid at a height h from the top level of liquid(which decreases).
    P0+ 0.5pv^2 + pgh = P0+ 0.5pV^2
    Assuming v (small v) the velocity of liquid on top is very small,
    ##V=root(2gh)##
    Should I take Bernoulli equation separately?
     
  2. jcsd
  3. Dec 10, 2014 #2

    Nathanael

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    Yes you can apply the Bernoulli equation separately. (Bernoulli equation essentially says that the energy-per-volume is the same throughout the liquid, so the energy per volume at the top of the liquid will be the same as at each hole.)

    If the speed of the liquid on top is not negligible, then you would have to consider that adding an extra hole would increase the speed of the top of the liquid.
     
  4. Dec 10, 2014 #3
    Thank you for the reply. When I consider that velocity of liquid on top is not negligible should I first find the relation between velocity of liquid in each hole and the velocity of top liquid and then apply BE? Like in a time dt if the level of liquid decrease by dy,
    then the volume that has decreased is Ady. Then from each hole a total volume of dV=Ady has to expelled in dt time. If area of cross section is given then can I apply principle of continuity equation for liquid at to and each hole like this:
    ##Ady/dt = a1dv1/dt##
     
  5. Dec 10, 2014 #4

    boneh3ad

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    If the level of liquid in your container is changing appreciably, then the flow is no longer steady and Bernoulli's equation no longer strictly applies and you would need to use the unsteady form of the Bernoulli equation, which is not as easy to work with:
    [tex]\rho\dfrac{\partial \phi}{\partial t} + \dfrac{1}{2}\rho\left( \vec{\nabla \phi}\cdot\vec{\nabla\phi} \right)^2 + p + \rho g z = f(t).[/tex]
     
  6. Dec 11, 2014 #5
    Why cant it be steady?
     
  7. Dec 11, 2014 #6

    boneh3ad

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    If your liquid level is dropping then your pressure is time-varying.
     
  8. Dec 11, 2014 #7
    For a particular time I can write v=dy/dt. I will then put this in the simple BE and I will replace pgh with pgy(t). Why cant I do that?
     
  9. Dec 11, 2014 #8

    boneh3ad

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    Bernoulli's equation is derived under the assumption that the flow is steady. It comes from the fact that at some point you reach the point where you have
    [tex]\dfrac{d}{dt}\left( \dfrac{1}{2}\rho v^2 + p + \rho g z \right) = 0[/tex]
    so that when you integrate this, the result is
    [tex]\dfrac{1}{2}\rho v^2 + p + \rho g z = \textrm{constant}.[/tex]
    If the flow is not steady, then you can't set the time derivative to zero and you now have to deal with the fact that
    [tex]\dfrac{1}{2}\rho v^2 + p + \rho g z \neq \textrm{constant},[/tex]
    meaning that even if you add in variables that are a function of time, Bernoulli's equation as it is typically written does not apply.
     
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