Question 3 of 9: Regenerative Brayton Cycle — Exergy Destruction by Process

Question 3: Regenerative Brayton Cycle — Exergy Destruction by Process (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Compressor inlet (state 1): $T_1=310$ K, $P_1=100$ kPa. Pressure ratio $=7$ ($P_2=700$ kPa). Turbine inlet (state 3): $T_3=1150$ K. $\eta_c=0.75$, $\eta_t=0.82$, regenerator effectiveness $\varepsilon=0.65$. Variable specific heats (air evaluated via high-accuracy ideal-gas-limit properties, root-solved for isentropic temperatures rather than a printed $s^\circ(T)$ table). Combustor heat source $T_H=1800$ K, sink $T_L=T_0=310$ K.

Find. (a) $T_4$; (b) $w_{net}$ [kJ/kg]; (c) $\eta_{th}$; (d) $\dot X_{dest}$ for each of compressor, combustor, turbine, regenerator, and heat rejection.

Approach

Use variable-specific-heat air properties with root-finding on temperature to satisfy each isentropic pressure ratio, then apply the given component efficiencies to get the actual compressor- and turbine-exit states. Size the regenerator from its effectiveness definition to fix the combustor-inlet state, then evaluate net work, heat input, and thermal efficiency. Exergy destruction in each process uses the full ideal-gas entropy $s(T,P)=s^\circ(T)-R\ln(P/P_{ref})$ (the pressure term matters whenever a process crosses two different pressure levels, e.g. the compressor and turbine).

1. Compressor (1→2). Isentropic exit at 700 kPa: $T_{2s}=263.57\ ^\circ$C, giving $w_{c,s}=231.0$ kJ/kg. Actual: $$w_{c,a}=\frac{w_{c,s}}{\eta_c}=\frac{231.0}{0.75}=\boxed{307.98\ \text{kJ/kg}},\qquad T_2=337.23\ ^\circ\text{C},\ \ h_2=744.34\ \text{kJ/kg}.$$
2. Turbine (3→4), 700→100 kPa. Isentropic exit $T_{4s}=425.41\ ^\circ$C, giving $w_{t,s}=507.5$ kJ/kg. Actual: $$w_{t,a}=\eta_t\,w_{t,s}=0.82\times507.5=\boxed{416.15\ \text{kJ/kg}},\qquad T_4=\boxed{509.63\ ^\circ\text{C}}\ \text{(part a)},\ \ h_4=929.52\ \text{kJ/kg}.$$
3. Regenerator: cold-side exit (combustor inlet). Effectiveness relates the actual cold-side gain to the maximum possible (heating the compressor-exit air all the way to the turbine-exit temperature $T_4$): $$h_x=h_2+\varepsilon(h_4-h_2)=744.34+0.65\times(929.52-744.34)=\boxed{864.71\ \text{kJ/kg}}\ (T_x=450.04\ ^\circ\text{C}).$$ Regenerator energy balance (equal mass flow both sides) fixes the hot-side exit: $$h_5=h_4-(h_x-h_2)=929.52-120.37=809.16\ \text{kJ/kg}\ (T_5=398.33\ ^\circ\text{C}).$$
4. Net specific work (part b). $$w_{net}=w_{t,a}-w_{c,a}=416.15-307.98=\boxed{108.17\ \text{kJ/kg}}.$$
5. Combustor heat input and thermal efficiency (part c). Heat is added only from $T_x$ to $T_3$, since the regenerator has already preheated the compressor-exit air: $$q_{in}=h_3-h_x=1345.67-864.71=\boxed{480.96\ \text{kJ/kg}}.$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{108.17}{480.96}=\boxed{0.2249\ (22.5\%)}.$$
6. Exergy destruction by process (part d). Using full entropy $s(T,P)=s^\circ(T)-R\ln(P/P_{ref})$ at each state's own pressure (100 or 700 kPa): $$\dot X_{c}/\dot m=T_0(s_2-s_1)=310\times(4.0579-3.9235)=\boxed{41.67\ \text{kJ/kg}}\quad\text{(compressor)}$$ $$\dot X_{comb}/\dot m=T_0\!\left[(s_3-s_x)-\frac{q_{in}}{T_H}\right]=310\times\left[(4.7599-4.2388)-\frac{480.96}{1800}\right]=\boxed{78.72\ \text{kJ/kg}}\quad\text{(combustor)}$$ $$\dot X_{t}/\dot m=T_0(s_4-s_3)=310\times(4.8834-4.7599)=\boxed{38.27\ \text{kJ/kg}}\quad\text{(turbine)}$$ $$\dot X_{regen}/\dot m=T_0[(s_x-s_2)+(s_5-s_4)]=310\times[0.1809-0.1658]=\boxed{4.67\ \text{kJ/kg}}\quad\text{(regenerator)}$$ $$\dot X_{reject}/\dot m=T_0\!\left[(s_1-s_5)+\frac{q_{out}}{T_0}\right]=\boxed{126.63\ \text{kJ/kg}}\quad\text{(heat rejection, }q_{out}=h_5-h_1\text{)}$$ All five terms are positive, as required by the second law for an irreversible cycle.