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Estimating Gas Flow Rate Through a Line Puncture or Blowdown Orifice

  1. Jul 6, 2011 #1
    The Pipeline Rules of Thumb Handbook uses the equation Q = D^2*P to estimate the flow rate of escaping gas from a pressured line (see link below for excerpt) under standard conditions. Where Q is flow rate in Mcf/hr, D is diameter of the nipple or orifice in inches, and P is the absolute pressure in psi at a near point upstream from the opening.

    Looking at the equation with respect to units, I see that the right side has an area multiplied by a pressure, which results in a force.

    in^2 * lbs/in^2 = lbs ..... = Mcf/hr ????

    But according to the book, we end up with Mcf/hr...not lbs.

    Obviously PiplineROT did some sort of approximation and cancellation here that they did not explain, but what is it? How does this make sense?

    http://books.google.com/books?id=2D...as blown off through a line puncture&f=false"
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Jul 6, 2011 #2


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    Hi engineer, welcome to the board. You're right, the units don't come out. But I checked their example and tried it with a few different pressures and diameters and found the equation they give is reasonably accurate.

    One caveat, the equation assumes ambient temperature.
  4. Jul 7, 2011 #3
    Thanks for your help. I figured it had to be right, I was just curious/confused about their logic.

    What equation did you use to check it? I was using Bernoulli's and getting flows significantly lower, but then I realized that I was analyzing a compressible substance (Natural Gas) and Bernoulli doesn't apply. So did you use Euler's then? Or do you have a flow calculator?

    Thanks again.
  5. Jul 7, 2011 #4


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  6. Jul 7, 2011 #5
    Thanks. Although I'm still having some trouble creating the same asnwers as this approximation. Maybe you can point out where I'm going wrong?

    For my case, the back pressure is 414.7 psia, 4" orifice diameter, with natural gas (SG = .6, density = .0014 slugs/cf, k = 1.27).

    A = 3.14*(2 in)2 = 12.56 in2 = .0872 sqft
    P = 414.7 psi = 59717 lbs/sqft

    Using the approximation,

    Q = D2*P = 42*414.7 = 6635 Mcf/hr

    Using the equation from your link, and C = 1

    mdot = .0872*sqrt(1.27*0.0014*59717*(2/2.27)2.27/.27) [sqft*sqrt(slugs2/(s2*ft2)]
    mdot = .5278 [slugs/s]

    Q = mdot/density = .5278/.0014 [slugs*cf/(s*slugs)] = 377 cf/s = 1357 Mcf/hr

    Using metric units and then converting at the end I still get 1355 Mcf/hr. Where is the mistake? Or is my pressure too low for the approximation? Idealizing C = 1 may not necessarily be correct, but any other value would just make my answer be even smaller; so the problem must be somewhere else...
    Last edited: Jul 7, 2011
  7. Jul 7, 2011 #6


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    C is discharge coefficient. Suggest using 0.6 to 0.8 as a ballpark.
    Check your density. Seems way low. (rho is real gas density at P and T)
    SG isn't needed.

    The value from D2P is going to be a bit high. I get a lower value depending on what C is used, but for a ballpark guestimate using a very simple equation such as D2P, I would expect +/- 20% or so.

    Note also this is for choked flow. If downstream pressure is greater than ~ 220 psig, it won't be choked and a different equation must be used.
  8. Jul 7, 2011 #7
    I've double checked the density, by both using SG and looking up it's listed rho, and it is correct.

    However, you're right that the flow wont be choked in my case. So I'll have to go to Euler's equations from here for a reality check. Thanks for your help.
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