Question

Carbon dioxide (CO₂) at 1 bar, 300 K enters a compressor operating at steady state and is compressed adiabatically to an exit state of 10 bar, 540 K. The CO₂ is modeled as an ideal gas, and kinetic and potential energy effects are negligible.


For the compressor, determine:

(a) the work input, in kJ per kg of CO₂ flowing,

(b) the rate of entropy production, in kJ/K per kg of CO₂ flowing, and

(c) the percent isentropic compressor efficiency.

Answer 1

Compressor inlet condition :

Temperature T1 = 300K

Pressure P1 = 1 bar

Compressor outlet condition :

T2 = 540 K

P2 = 10 bar

Assumptions : 1. Steady state process.

2. Negligible kinetic and potential energy changes.

3. Adiabatic process.

a) For compressor work input :

W = nYR(T1 - T2) /(Y-1)

Y - specific heat ratio for co2 = 1.28

Work input required per 1kg.

Moles n = 1000 gm/(44 gm/mol) = 22.72 mol

W = 22.72 mol*1.28*8.314J/mol.k * (300 - 540)/(1.28-1)

W = - 207243.79 J

W = -207.243 kJ

And this work is per kg of CO2 flowing, W = -207.243 kJ/kg

(-) sign shows work required for compression.

Ans: W = -207.243 kJ/kg

b) rate of production of entropy :

For compressor entropy change formula

(ΔS)system = Cpln(T2/T1) - Rln(P2/P1)

= 0.946 kj/kg.k * ln(540/300) - (8.314kJ/kmol.k)/44kg/kmol)*ln(10/1)

(ΔS)system = 0.556 - 0.435= 0.121 kJ/kg.k

Change in entropy (ΔS)system = 0.121 kJ/kg.k

compressor is insulated then no heat transfer Q = 0

Entropy  change for surrounding( ΔS)surr = Q/Tsurr = 0

Total entropy generation,

(ΔS)gen = ΔS)system + ΔS)surr = 0.121 + 0 = 0.121 kJ/k.kg

Ans : 0.121 kJ/kg.k

c) isentropic compressor efficiency :

=( isentropic work input/real work input)*100

=[ (T1 - T2isent) /(T1 - T2)]*100

By isentropic process,

T2isent = T1*(P2/P1)^(Y-1)/Y

= 300*(10/1)^(1.28-1)/1.28

= 496.44 K

T2isent = 496.44 K

Wisent = YnR(T1 - T2isent) /(Y-1)

= 1.28*22.72*8.314*(300 - 496.44)/(1.28-1)

= 169633.47 J

= 169.633 kJ

isentropic efficiency = (169.633)/(207.243)*100

= 81.85%

or

isentropic efficiency =[ (300- 496.44)/(300-540)]*100= 81.85%

Ans : 81.85%