Question

It is desired to compress ethylene gas [MW = 28, k = 1.3, cp = 0.357 Btu/(lbm-°F)] from 1 atm and 80°F to 10,000 psia. Assuming ideal gas behavior, calculate the compression work required per pound of ethylene under the following conditions: a. A single stage isothermal compressor. b. A four stage adiabatic compressor with interstage cooling to 80°F and optimum interstage pressures c. A four stage adiabatic compressor with no intercooling, assuming the same interstage pressures as (b) and 100% efficiency.

Answer 1

800F= (80-32)/1.8 deg.c =26.67 deg.c 14.7 Psia= 1atm

1. For isothermal compression work done (W) per mole= RTln (P2/P1)   where P2= 10000/14.7 atm and P1= 1 atm an T= 26.67+273.15=299.82K   W = RTln (P2/P1)= 8.314*(26.67+273.15) *ln (10000/14.7) =16258.64 Joules

2. for interstage cooling Optimum pressure ratio for each stage = (10000/14.7) ^1/4 =5.1

for interstage cooling T2/T1= (P2/P1) ^R/CP = (P2/P1) ^K-1 Where K= 1.3, T2/T1= (5.1)^0.3 =1.63

T2= (26.67+273.15)*1.63=488.71 K

Work dones per stage= R*(T2-T1)/ (K-1)= 8.314*(488.71-299.82)/0.3=5235 Joules

since there are four stages work done = 4*5235= 20940joules/mole

3. Let T2, T3, T4 and T5 are temperatures without interstages.

without intercooling T2= 488.71 work done for 1st stage= 5235 joules

for second stage T3/T2= (5.1)^0.3, T3= 488.71*1.63=797 K work done = 8.314*(797-488.71)/0.3 =8544 Joules

for third stage T4/T3= (5.1)^0.3 , T4= 1.63*797=1299K Work done = 8.314*(1299-797)/0.3 =13912 joules

for 4th stage T5/T4= 1.63 and T5= 1.63*1299=2111.37 and work done = 8.314*(2111.37-1299)/0.3= 22677 Joules

Total work done= 5235+8544+13912+22677 =50368 Joules/mole.