Question

It is desired to produce liquefied natural gas (LNG), which we consider to be pure methane, from that gas at 1 bar and 280 K (conditions “1” in the Linde process of the diagram). Leaving the cooler (“enfriador”) methane is at 100 bar and 210 K. The flash drum (“separador”) is adiabatic and operates at 1 bar. The recycled methane leaving the heat exchanger is at 1 bar and 200 K. The compressor can be assumed to operate reversibly and adiabatically. However, because of the large pressure change, a three-stage compressor with intercooling is used. The first stage compresses the gas from 1 bar to 5 bar, the second stage from 5 bar to 25 bar, and the third stage from 25 bar to 100 bar. Between stages the gas is isobarically cooled to 280 K.

(a)For each point in the Linde diagram write the conditions you know, and find the corresponding points in the H-P diagram for methane.

(b)Calculate the fractions of vapor and liquid leaving the flash drum.

(c)Calculate the amount of compressor work required for each kg of LNG produced.

Answer 1

Assume the Process are steady state and Reversible adiabatic

Mass Balance:

M_in + M_out =0 (Steady Flow) à M_out = - M_in = - M

Energy Balance:

M_inH_in + M_outH_out +W =0 à W = M (H_out –H_in )

Entropy Balance:

M_inS_in + M_outS_out =0 à S_out = S_in

From tables

At P=1 bar and T =280K Enthalpy H_in = 940 KJ/kg and Entropy S_in = 7.2 KJ/kgK

At Pressure P = 5bar and for constant Entropy process

Enthalpy H_out = 1195 KJ/kg and T_out = 388K

First stage work

W(First stage) = 1195 – 940 = 225 KJ/kg

After the intercooling the Temperature of methane stream is 280K So for Second compression stage

At P=5 bar and T =280K Enthalpy H_in = 938 KJ/kg and Entropy S_in = 6.35 KJ/kgK

At Pressure P = 25bar and for constant Entropy process

Enthalpy H_out = 1180 KJ/kg and T_out = 386K

Second stage work

W(Second stage) = 1180 -938 =242 KJ/kg

Similarly after the intercooling the Temperature of methane for third stage compression is found as

At P=25 bar and T =280K Enthalpy H_in = 915 KJ/kg and Entropy S_in = 5.5 KJ/kgK

At Pressure P = 100bar and for constant Entropy process

Enthalpy H_out = 1140 KJ/kg and T_out = 383K

Third stage work

W(Third stage) = 1140 -915 =225 KJ/kg

1. Total Compressor work = 255 +242+225 = 722 KJ/kg

Partb)

The Liquefaction process is a Joul Thomson Expansion and therefore occurs at constant enthalpy. The Enthalpy of methane leaving the cooler at bar and 210K is 493Kj/kg

And At 1 bar the Enthalpy of saturated vapor is 582 KJ/kg and that of liquid is 71Kj/kg

From Energy balance

H_in = H_out

493 = (1-x) *71 + x*582

So the dryness fraction x = 0.826 it is the fraction of vapor leaving the Drum

1-x = 1-0.826 =0.174 is the fraction of methane that has been Liquefied