Gas flowline volume calculation

[jwag]

I can't find a gas flowline volume calculation anywhere. Does anyone have a quick and dirty method. I have flow rate 10mmcfd, pressure 1000 psi, standard temp, pipe ID 4", and length (9088 ft).
I want to know how much gas is in this flowline while flowing. How much when I shut it in (on both ends)?

I will use this volume to estimate how much petroleum condensate will drop out upon shut-in of the system.

The well currently produces 80 bpd of condensate(60 api) when flowing at a 10MMCF/D rate.

Anyone?

 

[mbeychok]

.
Jwag:

You haven't explained your question very well. So I will assume that the given data are:

(1) The pipe has an inside diameter (D) of 4 inches
(2) The gas pressure (P) within the pipe is 1000 psia (you didn't say whether it was psia or psig)
(3) The gas temperature (T) within the pipe is 60 [°]F (you said it was at standard temperature but you failed to define your standard)
(4) The closed-in length of the gas pipe will be 9088 ft

Using the above information, and assuming that the volume of entrained liquid condensate is negligible relative to the volume of gas, the gas volume enclosed in the pipe will be:

(Pi)(D²/4)(pipe length in ft)/(144 in²/ft²)
= (3.1416)(4²/4)(9088)/(144) = 793.1 ft³ at 60 [°]F and 1000 psia

At atmospheric pressure (14.696 psia) and 60 [°]F, that is equivalent to:

(793.1)(1000/14.696) = 53,967 ft³

At atmospheric pressure and 60 [°]F, 1 lb-mol of any gas equals 379.48 ft³ of that gas. Thus, the lb-mols of gas enclosed in the pipe will be:

53,967/379.48 = 142.2 lb-mols

You haven't given us the gas molecular weight, but assuming that it is natural gas with a molecular weight of 18, the pounds (lbs) of gas enclosed in the pipe will be:

(142.2 lb-mols)(18 lbs/lb-mol) = 2559.6 lbs

I will leave it to you to estimate how much of the associated condensate drops out in the enclosed pipe.



Milton Beychok
(Contact me at www.air-dispersion.com)

 

[jwag]

mbeychok,

Thanks, this has gotten me started. Your assumptions about standard temp is correct.

One question: What constant or what value did you use to take into effect the gas being under 1000 psia of pressure to get your volume of 793 cu ft (at 60 deg F and 1000 psia)?
You used the formula: (Pi)(D2/4)(pipe length in ft)/(144 in2/ft2)

Thanks,

Jwag

 

[quark]

No constant is required to calculate the volume of gas at 1000psi. It is just the pipe volume (in layman's terms it is pipe water volume). When you convert it to atmospheric pressure then, formula given by mbeychok is P1V1 = P2V2. If initial pressure is 1000 psig then you should get 793.1 x (1000+14.696)/14.696

 

[mbeychok]

Jwag:

As Quark has said, there is no constant involved in calculating the internal volume of a pipe. The equation I used is simply the (internal cross-sectional area of the pipe) multiplied by (the length of the pipe).

(Pi)(D²/4) is the internal cross-section of the pipe in square inches (where D is in inches). That is divided by 144 to convert it to square feet, which is then multiplied by the pipe length in feet to obtain the cubic feet of volume.

As for the 1000 psia, you said that was the gas pressure within the pipe and I assumed that would be the gas pressure in the closed-in section of pipe.



Milton Beychok
(Contact me at www.air-dispersion.com)

 

[sailoday28]

With a flowing gas, there is a pressure drop. If the process is isothermal (or adiabatic), then the density can be determined as a function of distance from the first block valve. The mass of gas in the pipe is then determined from the integral of rho*A*dx, where rho is the denity, A, the pipe flow area and dx, the incremental distance. The limits of the integral are 0 to 9088 ft.
Note this is the mass while flow occurs.
How fast/slow and order of closure of the blocking valves will affect the final pressure and mass within the blocked area. For example closure of the downstream valve in long piping with low friction losses will give a hammer pressure at that valve similar to instant valve closure.