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Pipeline Storage
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Source：Internet Author：Unknow Pubdate：2010-02-21
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beauu (Petroleum)
18 Jan 10 15:42
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.
zdas04 (Mechanical)
18 Jan 10 16:10
You need to do the calculation in either mass units or in "standard" volume units. For a "I just want to know" project you can use any standard you want (say standard pressure is 14.7 psia and standard temp is 60F). Then normal storage is: 字串8
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
Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips Fora.
"It is always a poor idea to ask your Bridge Club for medical advice or a collection of geek engineers for legal advice"
beauu (Petroleum)
18 Jan 10 16:36
Thank you for your help, although I am not quite coming up with the values you have posted. If I am trying to solve the problem not assuming z-factor is negligible could I use the ideal gas law. 字串7
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.
zdas04 (Mechanical)
18 Jan 10 17:16
My approach was using the Real Gas modification of the Ideal Gas Law:
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). 字串1
David
beauu (Petroleum)
18 Jan 10 17:47
Thank you David. One last question though.. If we are to take the second point as standard conditions (P2 being pressure at 14.7 psia), why is it that V2 is the volume at the varied pressures? I would assume that P(std)* V (std) = P2 (800 psig or 60 psig) * V2.
Thanks again!
zdas04 (Mechanical)
18 Jan 10 18:22
I've been doing this so long that I bastardize the equations without even thinking about it.
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 字串5
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
beauu (Petroleum)
18 Jan 10 18:25
Hey- you're great man! Thank you!
dcasto (Chemical)
18 Jan 10 23:57
thanks, kids today forget the positive feedbacks like stars....
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