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Speed of an exiting gas from a pressured deposit

  1. Sep 5, 2008 #1

    Please, what is the formula to calculate the speed (velocity) of an exiting flow of gas from a pressured deposit?

    For example, it is supposed a deposit, that contains gas CO2, at 10 atmospheres of pressure (980.665 KPa), with a density of 18.7 Kg/m^3, at 288.15 K (59 F) of temperature. The molar mass of the CO2 is 0.044 Kg/mol.

    How much would be the speed (velocity) during the first second of an exiting flow of gas from the deposit to the atmosphere, through a 10 cm diameter hole made in the deposit, being the exterior atmosphere at 1 atmosphere of pressure and 288.15 K of temperature as well?

    Please, image a hypothetical huge deposit, so the loss of mass of the first second exiting gas doesn’t make any perceptible difference in the density of the total mass of the deposit (so neither the pressure changes), and doesn’t matter any other variable as humidity, external wind, resistance or viscosity.

    I have been several days looking for this formula, and I couldn’t find it. Any link will help.

    Thanks in advance.

    PS - [[ Maybe this is ridicules, but I supposed that the difference of 9 atmospheres of pressure, it is similar to the energy that a mass of 1 cubic meter of water will have when falls vertically 90 meters. (1 atmosphere of pressure it is similar than the hydrostatic pressure at 10 meters of depth in water). So I supposed that the speed of the exiting gas will be similar to the speed you get in a free fall of 90 meters. I estimated 42.02 m/s. But I am afraid this is completely wrong. ]].
  2. jcsd
  3. Sep 6, 2008 #2
    Well, you just calculate the force:

    [tex]F = P S[/tex]

    Then, according to a modified version of Newton's 2nd law,

    [tex] F = v\frac{\Delta m}{\Delta t} = \rho S v \frac{\Delta x}{\Delta t} = \rho S v^2[/tex]

    And so, the velocity is

    [tex]v = \sqrt{\frac{P}{\rho}}[/tex]
  4. Sep 7, 2008 #3
    Hi Irid,

    Thanks for your answer. Before you answer me I posted also in other forum in order to compare results. This problem seems to be a little bite tricky.

    I have tried the formula you suggested and I have got this:

    [tex]P[/tex]: 980665 Pa
    [tex]\rho[/tex]: 18.7 Kg/m[tex]\overline{}3[/tex]

    [tex]\nu[/tex] = 229 m/s

    I have also get this other answers: http://es.answers.yahoo.com/question/index?qid=20080906011906AA9OJzs&r=w and from this answer these other links:


    Being all of them relatives to "Chocked Flow", that I think is what the problem is going about.

    I have tried two of them, getting a similar result that "Krchwey" answered in the other forum: approximately 16 Kg/s that means a volume of 0.8556 m^3, that will results in a speed flow of 109 m/s approximately.

    This is the formula I have used:

    Q = C A [K d P]^(1/2) [2/(K+1)]^(K+1)/(2K-2)

    C= 0.72
    A= 0.007853981 m^2
    K= Cp/Cv = 815 / 626 = 1.3 J/Kg
    d= 18.7 Kg/m^3

    Q = 0.005654866 * 4882.608 * 0.585227 = 16.158 Kg/s

    My result is a little bite different, but is close.

    Still I am confused, and I am not sure if this is the correct answer. I have to check everything again, and I will let you know.

    Best and thanks for your answer.
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