Gas leak calculations

[Pipeliner1214]

OK so I have 300 feet of 16 in ID pipe blocked off and bled down. The bottom side 16inch valve is leaking through. I'm trying to calculate how much cfm is leaking or anything relevant. If the 300 ft section is blocked off and bled down then 30 minutes later I have 100 psi on the line. What calculations could I do to get information on what rate is the leak? Thanks

 

[zdas04]

Take a look at faq378-1864

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[Latexman]

PV=nRT, some pipe volume calcs., some ASSuMEing, and knowing it took 30 minutes to get to 100 psi will get you there.

Good luck,
Latexman

Technically, the glass is always full - 1/2 air and 1/2 water.

 

[LittleInch]

Also depends on what the pressure is on the other side of the valve. So long as this is above 200 psig, then your flow should be choked and not affected by the ds pressure rise. Otherwise time the pressure rise to a lower figure.

That sounds like a pretty substantial leak

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[hamid38]

well it depends on the pressure in your pipeline and maybe fluid density, for natural gas:

V_((mcf))=(D_((in))^2 * P_((psig))* L_((feet))*0.372) / 10^6

 

[Pipeliner1214]

Well its 151 feet of pipe and there is 1100 psi on the other side of the valve. I'm no engineer I'm just going above and beyond getting extra information since our engineers aren't working right now. Thanks

 

[zdas04]

hamid38,
You are joking, right? Thought you'd throw in some nonsense to get a laugh?

Your equation works out (assuming that I can pretend that psig can be called lbf/in^2) lbf*ft which is a torque, not volume. I can't even guess what the 0.372 is, but you seem to be missing a pi/4.

"mcf" only means (if I ignore your capitalization) "thousand cubic feet", but you divide by a million?

Let's pretend that you are talking about a quantity relative to local atmospheric pressure for some reason, then P(psig)=P(psia)-Atmospheric Pressure (psia).

At best it is some sort of indication of what is there, but he's looking for a rate.

The beauty of this site is the immediate peer review, I hope that anyone who stumbles across your gibberish reads down to this review of it.



David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist