Pipeline Storage
Pipeline Storage
(OP)
We have a pipeline that we had shut-in for a few days and then began flow. It had built up to a pressure of 800 psig before it was opened back up. Normal line pressure is 60 psig. I am curious to know how to calculate how much gas storage we had built up in the line. I just want to have an idea of how much of that initial delivery after the shut-in was due to pipeline storage.
gas gravity of .59 and line temperature of 60 deg F. The pipeline diameter is 8" and length is 20 miles.
gas gravity of .59 and line temperature of 60 deg F. The pipeline diameter is 8" and length is 20 miles.





RE: Pipeline Storage
Vol(normal)=(pi/4)ID^2*length*(60+atm)/14.7
(which assumes that normal flowing temperature is 60F and the change in compressibility is negligible). Then at the elevated pressure you replace the "60" with "800" and decide if the compressibility change is material (it probably is). If I assume sea level and that compressibility is not material then I get 186 MSCF at 60 psig and 2,033 MSCF at 800 psig.
David Simpson, PE
MuleShoe Engineering
www.muleshoe-eng.com
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RE: Pipeline Storage
PV / zRT (at standard conditions) = PV / zRT (at 60 & 800 psig)
14.7 * (pi*r^2*L) /(z*R*60F) = (800 psig +14.7 psia) * V2 / (z*R*60F) <-- Solve for V2
L in feet
r in feet
Thank you for your inital fast response. I really appreciate it.
RE: Pipeline Storage
P1V1/(R*Z1*T1)=P2V2/(R*Z2*T2)
So if Z and T are constant, you can Solve for V2 (take the second point to be standard conditions) Note that pi*r^2=pi*(D/2)^2=(pi/4)D^2, real world problems rarely have radius:
V2=V1*P1/P2=(pi/4)ID^2*Len*(P(flowing)/P(std))
If the change in compressibility is material then it becomes:
V2=V1*(P1*Z2)/(P2*Z1)=(pi/4)ID^2*Len*(P(flowing)/P(std))*Z(std)/Z(flowing)
That gives you V2 in standard units. Solve this for 60 psig and 800 psig and if you do all the conversions right you should get pretty close to the numbers I posted above for the immaterial compressibility change case (I let MathCAD do the unit conversions so I'm pretty confident in them).
David
RE: Pipeline Storage
Thanks again!
RE: Pipeline Storage
The equation is really:
PV=ZnRT or more usefully P=rho*R*T*Z so P*(V/m)=ZRT
Now, to convert this to standard conditions, you are creating an imaginary "volume" that would contain the same mass at STP as your pipe holds at elevated pressure--but the mass is the same in both cases so it cancels and you end up with
V(std or Pretend) = V(physical)*(P(flowing)/P(std))*(Z(std)/Z(flowing))
The notation that I chose to use above the "2" subscript is "standard conditions", that is why you solve for "V2" or the imaginary volume.
David
RE: Pipeline Storage
RE: Pipeline Storage