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Air Flow

  1. Jul 31, 2006 #1
    Hello,

    I am trying to find a formula for calculating air flow through a hole. If I have a container with a known volume and a pump that can maintain a specific vacuum in the container, is there a way to calculate the amount air that would flow through a leak of a known size?

    Thanks
    Solar1
     
  2. jcsd
  3. Aug 1, 2006 #2

    Andrew Mason

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    Yes.

    Pressure is a measure of the energy density of the gas. The energy density of the gas on the outside is the outside pressure. Since energy is conserved in the process of the gas entering the vaccum (lower pressure) through the hole, it must gain kinetic energy ie. speed up.

    The energy per unit volume of the gas flowing through the hole is [itex]P_{outside} = \frac{1}{2}\rho v^2 + P_{inside}[/itex].

    where [itex]\rho[/itex] is the mass/unit volume of the air as it is passing through the hole, which is the same as the density of the air on the outside.

    AM
     
    Last edited: Aug 1, 2006
  4. Aug 1, 2006 #3

    Q_Goest

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  5. Aug 1, 2006 #4

    Andrew Mason

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  6. Aug 1, 2006 #5

    Q_Goest

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    Hi Andrew. I respectfully disagree with you. Bernoulli's is incorrect here. It does not provide any way of calculating the flow rate given the geometry of an orifice (ie: discharge coefficient). It also does not take into account choked versus unchoked flow and sonic conditions being present. In the case of a vacuum vessel leaking from atmosphere, you will have choked flow and a sonic shock at the orifice vena contracta. Bernoulli's is only an energy conservation equation.
     
  7. Aug 1, 2006 #6

    Andrew Mason

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    Well you could use Bernoulli's equation for compressible flow, but I don't see why there would be compression of air before the air passes through the hole. Expansion of the air after it enters the vacuum is irrelevant to the flow through the hole. The pump keeps the vacuum pressure constant.

    All you need to do is analyse a slice of air of area A and thickness ds that is pushed through the opening. The work done on that slice by the pressure difference is [itex]W = \Delta PdV = (P_{outside}-0)Ads[/itex]. Since this energy must increase the kinetic energy of the air slice, [itex]PAds = \frac{1}{2}dmv^2 = \frac{1}{2}\rho Adsv^2[/itex]. This simplifies to [itex]P = \frac{1}{2}\rho v^2[/itex]

    AM
     
  8. Aug 2, 2006 #7

    FredGarvin

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    One of the three stipulations to the derivation of Bernoulli is incompressible flow. Q is absolutely correct in that in this case it is very easy to have choked flow. Also, by that time compressibility effects are there as well. You could have a pressure factor of 2 across that orifice and not quite be choked flow, but the density is definitely going to be effected, which effects your mass flow rate (it should be noted that since it appears that a vacuum is on the outlet of the orifice, mass flow will choke as well). However, since it is so easy, I'd take a stab at it using Bernoulli as a first pass just to see if it's in the ball park. It wouldn't hurt. Perhaps the OP is keeping the delta P very small across the orifice and it may be applicable. We don't really know right now.

    As much as I hate to advertise for M.B., his url referenced by Q is probably the best way to go. It covers all the bases. It does have the one sticking point which is estimating a discharge coefficient for the orifice.
     
    Last edited: Aug 2, 2006
  9. Aug 2, 2006 #8

    Q_Goest

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    Thanks Fred.
     
  10. Aug 2, 2006 #9

    FredGarvin

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    No thanks needed, but you're welcome. You jarred my memory on this type of set up. I haven't had to deal with this in quite a while (getting rusty).
     
  11. Aug 4, 2006 #10

    Clausius2

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    Sure, you will find that the ratio of pressures is proportional somehow to the local mach number, assuming that the flow is isentropic:


    [tex]\frac{P_o}{P}=\left(1+\frac{\gamma-1}{2}M^2\right)^{\gamma/(\gamma-1)}[/tex]

    If [tex]M=\frac{U}{a}=\sqrt{\frac{1}{\gamma}\frac{\rho U^2}{P}}<<1[/tex] then the flow can be regarded as approximately incompressible at first order in [tex]M^2[/tex]:

    [tex]\frac{P_o}{P}=1+\frac{1}{2}\gamma M^2+\frac{1}{8}\gamma M^4+O(M^6)=1+\frac{1}{2}\frac{\rho U^2}{P}+\frac{1}{8}\frac{\rho^2 U^4}{\gamma P^2}+O(M^6)[/tex]

    Notice that the first two terms in this Mach number asymptotic expansion are just the classical Bernoulli equation for incompressible flow. Now, [tex]M^2\sim \rho U^2/P\sim \Delta P/P[/tex]. Thus the classical Bernoulli equation is only valid if [tex]\Delta P/P<<1[/tex].

    Therefore Qgoest and Fred Garvin are right. In this example, particularly dealing with vacuum, one may expect a great variation of pressure across the hole.
     
  12. Aug 4, 2006 #11

    FredGarvin

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    Hey Clausius! Long time no see. Hope things are going well out west.

    Out of curiosity, what is [tex]O[/tex] in the above equation? It's not ringing any bells.
     
  13. Aug 4, 2006 #12

    Clausius2

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    Hey man,

    It's an [tex]O[/tex] of [tex]O[/tex]key Cool. :rofl:

    Now seriously, that symbol is called a big "O of Landau". Indicates that the next term iin the expansion is as large as x (of order x): [tex]O(x)[/tex].
     
  14. Aug 4, 2006 #13

    FredGarvin

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    Ahhhh. I see. Thanks.
     
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