Question 6.16: Problem: Air enters a gas turbine at 1100 K and 800 kPa with...
Problem: Air enters a gas turbine at 1100 K and 800 kPa with a velocity of 80 m / s. Determine the exergy of the air at the inlet to the turbine. Assume the atmosphere is at 300 K and 100 kPa.
Find: Flow exergy ψ of the air at the turbine inlet.
Known: Air pressure P = 800 kPa, air temperature T = 1100 K, air velocity V = 80 m / s, atmospheric pressure P_o = 100 kPa, atmospheric temperature T_o = 300 K.
Assumptions: Air is an ideal gas, negligible contribution from potential energy.
Properties: Air has gas constant R = 0.2870 kJ / kgK (Appendix 1), specific enthalpy at 1100 K h(1100 K) = 1161.07 kJ / kg (Appendix 7), specific enthalpy h(300 K) = 300.19 kJ / kg (Appendix 7), specific entropy sº (1100 K) = 3.07732 kJ / kgK (Appendix 7), specific entropy sº (300 K) = 1.70203 kJ / kgK (Appendix 7).
Governing equations:
Flow exergy per unit mass \psi =(h-h_o) – T_o (s-s_o) + \frac{\pmb{V}^2}{2} +\underbrace{gz}_{=0}
Entropy change (ideal gas) \Delta s =s^\circ (T_2)-s^\circ (T_1)-R \ln \frac{P_s}{P_1}
Specific enthalpy contribution:
h-h_o =h(1100 \ K) – h(300 \ K)=1161.07 \ kJ/kg – 300.19 \ kJ/kg =860.880 \ kJ/kg.
Specific entropy contribution: s – s_o =s^\circ (1100 \ K)-s^\circ (300 \ K)-R \ln \frac{P}{P_o}
=3.07732 \ kJ/kgK – 1.70203 \ kJ/kgK -0.2870 \ kJ/kgK \times \ln \frac{800 \ kPa}{100 \ kPa} =0.778490 \ kJ/kgK
Total flow exergy:
\psi = 860.880 \ kJ/kg – 300 \ K \times 0.778490 \ kJ/kgK + \frac{(80 \ m/s)^2}{2 \times 1000 \ J/kJ} =630.533 \ kJ/kg.
Answer: The flow exergy of the fluid at the turbine inlet is 630.5 kJ / kg.