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Simplified depressurisation time

  1. Mar 19, 2014 #1
    I have a gas cylinder pressurised to 16,600kPa and it holds 9.2m^3 of Nitrogen.

    There's a regulator that steps the gas pressure down to 2.8kPa above ambient and exhausts through a 1/4 NPT fitting (A=32mm²?).

    Assuming the gas release is slow enough to neglect temperature changes, is there a simplified equation to calculate how long until the cylinder is empty?

    A search of similar threads here all have high flow rate and temperature complications.
  2. jcsd
  3. Mar 21, 2014 #2


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    I'm not sure you are using the correct area for the 1/4" NPT fitting. The ID is usually 0.54 inches or 13.716 mm, which would give an area of about 147 mm^2.

    In any event, the flow out of this fitting is going to be choked, given the ratio of the internal pressure of the cylinder to ambient pressure. When the flow is choked, the speed of the gas has reached Mach 1 (the speed of sound for that gas at ambient temperature) and the flow rate has reached a maximum. This flow rate will stay constant until the pressure in the cylinder declines below a certain amount, and then the rate will drop off until the pressure inside and outside the cylinder has equalized.

    Does the nitrogen flow just thru the regulator and the fitting, or is there any additional piping or tubing involved?

    And just to confirm, the pressure in the cylinder is 16,600,000 Pa (2400 psi give or take)?
  4. Mar 21, 2014 #3
    The ID of a 1/4" NPT fitting is, for standard fittings, 0.28" or 7.1 mm.
    So ~39.6 mm^2

    The rest of the post is super informative for the OP, just wanted to chime in for that point.
  5. Mar 21, 2014 #4


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    There exists a nice, clean, analytical solution for the mass flow given a total pressure, temperature, and cross-section. If you assume constant temperature (and area, obviously), you can set that equal to the mass flow rate, [itex]\frac{dm}{dt}[/itex], and you are left with a simple first order ODE with a constant coefficient of the form
    [tex]\dfrac{dm}{dt} = -Cm[/tex]
    where [itex]C[/itex] is a constant (from the equation I mentioned earlier). That should be super easy to solve and can give you the time until you reach the point the flow unchokes, at which point you will have to use the changing pressure differential to get the answer. This all neglects viscosity and any pressure loss through your regulator, but it should give you a pretty reasonable answer, especially at pressures well above where the flow unchokes and you don't have to worry so much about losses through the regulator. It's probably best if you recast the variables in terms of pressure, though, for your purposes.
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