Question 6.17: Problem: Atmospheric air at 100 kPa and 300 K is compressed ...
Problem: Atmospheric air at 100 kPa and 300 K is compressed in a reversible, adiabatic process to 800 kPa. Determine the exergy of the compressed air. The air at 800 kPa is allowed to cool to 300 K. Determine the exergy destroyed during the cooling process.
Find: Exergy of the air at the compressor outlet ψ_2 and the exergy destroyed during cooling ψ_d .
Known: Outlet pressure P_2 = 800 kPa, atmospheric pressure P_o = 100 kPa, atmospheric temperature T_o = 300 K. final pressure P_3 = 800 kPa, final temperature T_3 = 300 K.
Assumptions: Air is an ideal gas with constant specific heat.
Governing equations:
Flow exergy per unit mass ψ = (h – h_o) – T_o (s – s_o) + \frac{\pmb{V}^2}{2} + gz
Enthalpy change (ideal gas, constant \Delta h=c_p(T_2 – T_1)
specific heat)
Entropy change (ideal gas, constant specific heats) \Delta s=c_p\ln \frac{T_2}{T_1} -R \ln \frac{P_2}{P_1}
Isentropic process (ideal gas, constant specific heats) T_2 =T_1\left\lgroup\frac{P_2}{P_1} \right\rgroup ^{(\gamma -1)/\gamma }
Properties: Air has gas constant R = 0.2870 kJ / kgK (Appendix 1), air at 300 K has specific heat c_p = 1.005 kJ / kgK (Appendix 4), specific heat ratio of air at 300 K \gamma =1.400 (Appendix 4).
The exit temperature is
T_2 =T_1\left\lgroup\frac{P_2}{P_1} \right\rgroup ^{(\gamma -1)/\gamma } \ =300 \ K \left\lgroup\frac{800 \ kPa}{100 \ kPa} \right\rgroup ^{(1.4-1)/1.4}=543.434 \ K.
For an isentropic compression entropy does not change, and neglecting kinetic and potential energy then the flow exergy per unit mass at the outlet reduces to
\psi _2=(h_2 – h_o)=c_p(T_2 – T_o)=1.005 \ kJ/kgK \times (543.434 \ K – 300 \ K)=244.651 \ kJ/kg.
Since T_3 = T_o , then h_3 = h_o ; and neglecting kinetic and potential energy changes again then the flow exergy per unit mass in the final state reduces to
\psi _3=-T_o(s_3 – s_o).
Final specific entropy contribution:
s_3 – s_o = c_p \underbrace{\ln \frac{T_3}{T_o}}_{=0} – – R \ln\frac{P_3}{P_o} =-0.2870 \ kJ/kgK \times \ln \frac{800 \ kPa}{100 \ kPa} =0.596 \ 800 \ kJ/kgK.
So the final flow exergy per unit mass is
\psi _3=-T_o(s_3 – s_o)=-300 \ K \times (-0.596 \ 800 \ kJ/kgK)=179.040 \ kJ/kg,
And the exergy destroyed during the process is
\psi _d=\psi _2-\psi _3=244.651 \ kJ/kg -179.040 \ kJ/kg =65.611 \ kJ/kg.
Answer: The flow exergy of the fluid at the compressor outlet is 244.7kJ / kg, and the exergy destroyed during cooling is 65.6 kJ / kg.