How do you calculate the volume of natural gas in a pipeline?

Answer 1

You could use the following Rule of Thumb: Multiply the square of the inside diameter, in inches, by the gauge pressure, in psi; multiply this by 0.372; the answer is the approximate number of cubic ft of gas (standard conditions) in 1,000 ft of line, e.g 7 inch ID pipeline, 8km long, operating at 65 barg:

8 km = 26246 ft = 26.246 kft

65 Bar = 942.7 psi

so, 7 x 7 x 942.7 x 0.372 = 17183.5356 scf / 1000 ft of line, so total gas in 26246 ft of line = 17183.5356x26.246=450999.0753576 = 0.45 MMSCF of gas.

Or, you could use, Pv=znRT, assume a z of, say 0.9 , lets say the pipeline is at 6 deg (normal temperature for a shutin subsea pipeline in the North Sea)

65x (Pi x 0.1778 x 0.1778 x 8000/4) = 0.9 x n x 8.314x10-5)x (273+6)

n = 618523 moles

1 kmol of a gas occupies 22.441 Nm3 at standard conditions

t.f 618.523 kmol should ocupy 618.523 x 22.441 = 13880.274643 Nm3.

1 Nm3 = 37.326 SCF, therefore, 13880.274 Nm3 = 518095.131324618 SCF = 0.52 MMSCF.(Nm3 to SCF conversion seems to have different factors, I've seen it range from 34.89 to 38.9!!!).

Not very far from the 0.45 the rule of thumb calculated!!!. It should be noted that the standard volume is independent of the particular gas in the pipeline, so we don't need to knwo the MW or density. Any gas at a given P&T will have the same number of moles (and hence standard cubic feet), the actual mass in kg will ofcourse depend on the molecular weight. Riz

If you mean standard or normal volume, in case the pressure is considerably higher than the atmospheric value, you need to use an expression for the compressibility factor or take it from a table, depending on the values of pressure and temperature in the pipeline. You can then use a state equation for the gas (knowing its molecular mass), from which you'll be able to calculate its density at working conditions. By multiplying the density by the physical volume of the pipeline pi*D^2/4*L (L=length, D=diameter) you obtain the mass of gas, which divided by the standard or normal density gives you the desired volume.