Question 10.5: Problem: A reheat stage is added to the Brayton cycle of Exa...
Problem: A reheat stage is added to the Brayton cycle of Example 10.3. The air heated in the combustor to 1260 K is expanded in the first turbine to a pressure of 300 kPa, reheated to 1260 K, and then expanded to 100 kPa. Find the power output and efficiency of this cycle.
Find: Power output \dot{W} _{net} of new cycle, efficiency η_{th,Brayton} of the cycle from Example 10.3 with reheat.
Known: Example 10.3 Brayton cycle with reheat, air temperature at combustor exit T_3 = T_5 = 1260 K, reheat pressure P_4 = P_5 = 0.5 MPa.
Diagram: Figure E10.5.
Assumptions: Air is an ideal gas with temperature dependent properties.
Governing Equation:
Efficiency of Brayton cycle \eta _{th,Brayton}=\frac{w_{net}}{q_H} =\frac{(h_3-h_4) + (h_5 – h_6) – (h_2 – h_1)}{(h_3-h_2)+(h_5-h_4)}
From the solution to Example 10.3, specific enthalpies h_1 = 300.19 kJ / kg, h_2 = 579.8650 kJ / kg and h_3 = 1348.55 kJ / kg.
From the ideal gas tables (Appendix 7) at T_3 = 1260 K, relative pressure P_{r3} = 290.8. For the isentropic process 3→4,
P_{r4}=P_{r3}\frac{P_4}{P_3} =290.8 \times \left\lgroup\frac{0.3 \ MPa}{1.0 MPa} \right\rgroup =87.24.
Interpolating in the ideal gas tables, specific enthalpy h_4 = 971.5602 kJ / kg.
From the ideal gas tables (Appendix 7) at T_5 = 1260 K, specific enthalpy h_5 = 1348.55 kJ / kg, relative pressure P_{r5} = 290.8.
For the isentropic process 5→6,
P_{r6}=P_{r5}\frac{P_6}{P_5} =290.8 \times \left\lgroup\frac{0.1 \ MPa}{0.3 MPa} \right\rgroup =96.93.
Interpolating in the ideal gas tables, specific enthalpy h_6 = 1000.3550 kJ / kg.
The net work output per kilogram of air is
w_{net}=(h_3-h_4)+(h_5-h_6) – (h_2-h_1), \\ w_{net}=(1348.55 \ kJ/kg – 971.5602 \ kJ/kg) + (1348.55 \ kJ/kg -1000.3550 \ kJ/kg) \\ – (579.8650 \ kJ/kg – 300.19 \ kJ/kg) = 445.5098 \ kJ/kg.
Then the net power output is
\dot{W} _{net}=\dot{m} w_{net}=5 \ kg/s \times 445.5098 \ kJ/kg=2227.549 \ kW.
The heat added per kilogram of air is
q_H=(h_3-h_2)+(h_5-h_4),
q_H=(1348.55 \ kJ/kg-579.8650 \ kJ/kg)+ (1348.55 \ kJ/kg – 1117.917 \ kJ/kg)=999.3180 \ kJ/kg.
The new Brayton cycle efficiency is
\eta _{th,Brayton}=\frac{w_{net}}{q_H} =\frac{445.5098 \ kJ/kg}{999.3180 \ kJ/kg} =0.4458.
Answer: The cycle efficiency is 44.6% and the net power output is 2227.5 kW.