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Applying Bernoulli's to this geometry.

  1. Feb 23, 2014 #1
    I have attached the geometry of interest with some parts of the solution. The geometry is a vessel that is half of a sphere with an orifice at the bottom.

    The first expression that they have written, the "A*(2*g*z)^0.5=....." is from conservation of flow rate. 2*g*z is the velocity at the inlet of the orifice.

    I don't understand how they got that velocity since there should be a pressure drop from z to the orifice opening.
    Basically it just seems like they did Potential energy (@ z) = Kinetic Energy at orifice opening.
    So that gives g*z=0.5*v^2=>v=(2*g*z)^0.5

    but isn't there a pressure gain as well of density*gravity*z?
    The only way they could have gotten that velocity is if the pressure at both locations are the same.
     

    Attached Files:

  2. jcsd
  3. Feb 23, 2014 #2

    SteamKing

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    This problem is an application of Torricelli's Law, which is a particular form of the more general Bernoulli equation:

    http://en.wikipedia.org/wiki/Torricelli's_law

    The pressure at the open top of the sphere and at the bottom where the outlet is are approximately equal to each other and to the ambient atmospheric pressure.
     
  4. Feb 23, 2014 #3
    Thanks for the reply. What you stated above I agree with, but isn't that different from what the drawing shows? it looks like at height z is at a location that is not exposed to the ambient pressure and same with at z=0 (orifice opening)? That's what has me confused.

    If Z is indeed at the top, wouldn't Z=R?

    Thus the expression should be 2*g*R, instead of 2*g*z.

    It looks like z is some arbitary height, at z=R (P=Patm), but at other points below R, it should be Patm+(density*gravity*(R-z)) from the hydrostatic pressure relation.
     
  5. Feb 23, 2014 #4

    rude man

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    I'm wondering why the velocity on the surface v1 = 0. Seems to me that v1 is finite (= -dz/dt). Maybe it's deliberately ignored. But if not, (v1^2)/2 = (v2^2)/2 - gz > 0.
     
  6. Feb 23, 2014 #5

    rude man

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    I think z is the instantaneous level of the fluid.
     
  7. Feb 23, 2014 #6
    In all the classes I've taken, we assume the velocity at the free space is Zero.
     
  8. Feb 23, 2014 #7
    Sorry, what does that mean?
     
  9. Feb 23, 2014 #8
    Hi rude man,

    The cross sectional area at the top of a huge tank like this is usually assumed to be much larger than the cross sectional area of the oriface, so the kinetic energy at the top is usually neglected in Bernoulli.

    chet
     
  10. Feb 23, 2014 #9

    rude man

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    It means the fluid surface is at z.
     
  11. Feb 23, 2014 #10

    rude man

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    Hi Chet,
    I defer to the majority ...

    PS congrats for making Sci Advisor! Richly deserved!
     
    Last edited: Feb 23, 2014
  12. Feb 23, 2014 #11
    Hmmmm I see, that would explain velocity=0 there and P=Patm. What about at z=0. Would you make an assumption that's the bottom surface (lack of a better term)?
     
  13. Feb 24, 2014 #12

    rude man

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    Yes.
    Look at the given equation r^2 + (R - z)^2 = R^2
    so if z=0, r=0 as the figure corroborates.
    It also states dz/dt < 0 which again implies z=0 at the bottom.

    The figure should have indicated z as going + from the bottom to the surface instead of an ambiguous two-way arrow.
     
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