Question

A sweet, water-saturated natural gas enters a compressor at 50 psia and 100F and leaves at 200 psia and 300F.

a. Is there any chance of water condensing in the compressor? If so, how much? Answer in lb H2O/MMscf gas compressed.

b. Estimate the mole fraction of water vapor in the gas stream leaving the compressor.

c. The natural gas is then cooled to 190 psia and 100F before entering the second stge of compression. Should a vapor-liquid scrubber be installed between the intercooler and the second stage of compression? If so, how much water (lb H2O/MMscf) will be removed? Will any hydrocarbon liquids condense?

Answer 1

a) Using Bukacek correlation, For sweet natural gas

Where Pv is the Vapor pressure of Water in psia at T

At 100 F, Pv = 0.95 psia

lb water/(MMSCF Natural Gas) = 904

At 100 F, Pv = 0.95 psia

@300F & 50 psia, lb water/(MMSCFT Natural Gas) = 904

At 50 F, Pv = 66.98 psia

@300F & 200 psia, lb water/(MMSCF Natural Gas) = 15803

To calculate no. of moles in 1 mmscf @100 F and 50 psia, Applying Ideal Gas Law

Therefore, PV = nRT

P=50 psia = 3.4 atm

1 MMSCF = 28252*10^3 litres

T=100 F = 310 K

R=0.082 litre atm/mol/K

Therefore, n = 3.4*28252*10^2/.082/310 = 377878 moles

@ 300 F and 200 psia, using same n as above, finding volume using Ideal gas law

P= 200 psia = 13.6 atm

T=300F = 422 K

Therefore, V = 377878*.082*422/13.6 = 961477 Litres =.033 MMSCF

Therefore, Total water content in outlet stream = (lb water/(MMSCF Natural Gas))*.033 = 15803*.033 = 521.49

Since Inlet 1 MMSCF Sweet gas contains water = 904 lb

Outlet compressed sweet gas contains water = 521.59 lb

Therefore, Compressed water = (904-521.59) = 382.41 lb/MMSCF sweet Natural gas entered

b) As calculated above, lb of Water vapor in the Outlet stream = 521.49

1 lb = 453 gm

Molecular weight of water = 18 g/mol

Therefore, moles of water vapor in the outlet stream = 521.49*453/18 = 13124

Moles of Sweet Gas as calculated above = 377878

Therefore, Mole fraction of water vapor in Outlet stream = (13124/(13124+377878)) = .03

c) @ 100 F and 190 psia, Using the above formula, lb water/(MMSCF Natural Gas) = 240.38

190 psia = 12.92 atm

Using Ideal law, V = 377878*.082*310/12.92 = 743472 litres = .026 MMSCF

Therefore, Water vapor content in Sweet gas @100 F and 190 psia = 240.38*.026 =6.24 lb

As calculated above, Water vapor content in the Outlet stream of 1st compressor = 521.49 lb

Therefore, Water removed in Condensor = (521.49 - 6.24) = 515.25 lb /(MMSCF Sweet Gas in 1st compressor inlet)

At the inlet of intercooler, Natural gas was @300 F, 200 psia,

At the outlet of Intercooler, Naturas gas was @ 100 F, 190 psia

As can be seen from above graph, Inlet lies in yellow region, i..e Gas phase. Outlet lies in Green region, i.e. Liquid+Gas phase.

Hence Hydrocarbon (i.e. Sweet Natural Gas) will also get condensed in intercooler