Question 10.3: Problem: An air standard Brayton cycle with a compressor pre...
Problem: An air standard Brayton cycle with a compressor pressure ratio of 10 takes in air at 100 kPa and 300 K and a mass flow rate of 5 kg / s. The air leaves the combustor at 1260 K. Find the efficiency of the cycle and the net power output. Use air tables to find the properties of air.
Find: Efficiency η_{th,Brayton} of the cycle, net power output \dot{W} _{net} of the cycle.
Known: Air standard Brayton cycle, compressor pressure ratio r_p = 10, initial pressure P_1 = 100 kPa, initial temperature T_1 = 300 K, mass flow rate of air \dot{m} = 5 kg s/ , air temperature at combustor exit T_3 = 1260 K.
Diagram: Figure E10.3.
Assumptions: Air is an ideal gas with temperature dependent properties.
Governing Equation:
Efficiency of Brayton cycle \eta _{th,Brayton}=\frac{w_{net}}{q_H} =1-\frac{q_C}{q_H} =1-\frac{h_4-h_1}{h_3-h_2}
From the ideal gas tables (Appendix 7) at T_1 = 300 K, specific enthalpy h_1 = 300.19 kJ / kg and relative pressure P_{r1} = 1.3860.
For the isentropic process 1→2, using the given pressure ratio,
P_{r2}=P_{r1}\frac{P_2}{P_1} =1.3860 \times 10 =13.860.
Interpolating in the ideal gas tables, specific enthalpy h_2 = 579.8650 kJ / kg.
From the ideal gas tables (Appendix 7) at T_3 = 1260 K, specific enthalpy h_3 = 1348.55 kJ / kg, and relative pressure P_{r3} = 290.8.
For the isentropic process 3→4, using the inverse of the pressure ratio,
P_{r4}=P_{r3}\frac{P_4}{P_3} =\frac{290.8}{10} =29.080.
Interpolating in the ideal gas tables, specific enthalpy h_4 = 715.1786 kJ / kg. The Brayton cycle efficiency is then
\eta _{th,Brayton}=1-\frac{h_4-h_1}{h_3-h_2} =1-\frac{715.1786 \ kJ/kg – 300.19 \ kJ/kg}{1348.55 \ kJ/kg -579.8650 \ kJ/kg} =0.460 \ 131 \ 8.
The net work output per kilogram of air is
w_{net}=q_H – q_C=(h_3-h_2)-(h_4-h_1), \\ w_{net}=(1348.55 \ kJ/kg – 579.8650 \ kJ/kg) – (715.1786 \ kJ/kg – 300.19 \ kJ/kg) \\ = 353.6964 \ kJ/kg.
Finally, the net power output is
\dot{W} _{net}=\dot{m} w_{net}=5 \ kg/s \times 353.6964 \ kJ/kg = 1768.482 \ kW.
Answer: The cycle efficiency is 46.0% and the net power output is 1768.5 kW.