Question

One mole of an ideal gas (C_P/R=7/2), is compressed in a steady-flow compressor from 2.5 bar and 25°C to 6.5 bar and 120°C. The compressor rejects 0.5 kJ as heat to the surrounding at 293K.

Calculate:

1.     The enthalpy change of the gas (in kJ)

2.     The entropy change of the gas (in J.mol^-1)

3.     The work required for the compression (in kJ)

4.     The ideal work of the process (in kJ)

5.     The thermodynamic efficiency

The lost work (in kJ)

Answer 1

One mol of ideal gas Cp = 7/2 R is compressed in a steady flow compressor from 2.5 bar and 25 C to 6.5 bar and 120C.

Compressor rejects heat to the surrounding at 293 K, Q = 0.5 kJ

T1 = 25 C = 273 +25 = 298 k and T2 = 120 C = 120 + 273 = 393 K

P1 = 2.5 bar and P2 = 6.5 bar

Moles of gas n = 1 mol

1. Enthalpy change : we know formula for change in enthalpy,

ΔH = nCp(T2 - T1) = 1mol*7/2 * R *(120 - 25) = 1mol*3.5 * 8.314 J/mol*k * 95 K = 2764.405 J

change in enthalpy ΔH = 2764.405 J = 2.764 kJ

ans: 2.764 kJ

2) entropy change for gas ,

Gas treated as system and for compressor change in entropy formula,

(ΔS)gas = nCp*ln(T2/T1) - nR*ln(P2/P1)

= 1mol*3.5*8.314J/mol.k * ln(393/298) - 1mol*8.314 J/mol.k * ln(6.5/2.5)

= 8.052 - 7.944

= 0.1078 J/K

(ΔS)gas = 0.1078 J/K

entropy change for gas ΔS = 0.1078 J

Entropy change for surrounding: surrounding receive heat Q = 500 J.

(ΔS)surr = Q/Tsurr = 500 J/293K = 1.706 J/K

Total entropy change :

ΔSt =( ΔS)system +( ΔS)surr = 0.1078 + 1.706 = 1.8138 J/K

Total Entropy change for irreversible process is nonzero. It means compressor work is irreversible. If it is reversible then total entropy change is  zero.

3) here we can see that Y = Cp/Cv = 3.5R/(Cp - R) = 3.5/2.5 = 1.4

Specific heat ratio 1.4 for air.

So ideal gas is air.

Applying energy balance for compressor,

ΔK. E + ΔP. E +ΔH = Q - Ws

Change in kinetic energy and potential energy is negligible.

ΔH = Q - Ws

Ws = Q - ΔH = 0.5 - 2.764 = - 2.264 kJ

work required for compression Wactual  = - 2.264 kJ

(-) sign shows work given to the compressor.

4) ideal work required for the compression :

W = Y*n*R(T1 - T2)/(Y -1)

Y - specific heat ratio = Cp/Cv = 7/2R/(Cp-R) = 3.5R/2.5R = 1.4

W = 1.4*1mol*8.314 J/mol.k *(25 -120 )/(1.4-1)

W = - 2764.405 J

Ideal Work done for compressor W =- 2.764 kJ

Wideal = - 2.764 kJ

5) thermodynamic efficiency =( actual Work required /ideal work required) *100

=( 2.264/2.764 )*100

= 81.91%

thermodynamic efficiency = 81.91 %

lost work =Wideal - Wactual = heat rejects = 0.5 kJ