Regenerative Brayton Cycle — Variable Specific Heats and Exergy: 04-BS-10, May 2013, Q2

Question 2: Regenerative Brayton Cycle — Variable Specific Heats and Exergy (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. Air-standard regenerative Brayton cycle. Pressure ratio $r_p=10$; $T_{min}=T_1=300$ K, $T_{max}=T_4=1200$ K (turbine inlet); compressor isentropic efficiency $\eta_C=75\%$; turbine isentropic efficiency $\eta_T=80\%$; regenerator effectiveness $\varepsilon=70\%$; heat-source temperature $T_H=1200$ K; heat-sink/dead-state temperature $T_0=300$ K. Variable specific heats (real air property data, not the cold-air-standard constant-$c_p$ assumption) are used throughout, exactly as the printed ideal-gas air table would be used in a closed-book exam.

Quantity	Value
$r_p=P_2/P_1$	10
$T_1$	300 K
$T_4$	1200 K
$\eta_C,\ \eta_T$	0.75, 0.80
$\varepsilon_{regen}$	0.70

Find. (a) $q_{in}$; (b) $w_{net}$; (c) $\eta_{th}$; (d) exergy destruction of each process and the total; (e) $\eta_{II}$.

Fig. Q2 — T–s path of the regenerative Brayton cycle (1→2 real compression, 2→3 regenerator cold side, 3→4 combustor heat addition, 4→5 real expansion, 5→6 regenerator hot side, 6→1 heat rejection to the sink), plotted against entropy measured relative to state 1.

Approach

Work state-by-state around the cycle using the actual (real, variable-$c_p$) air enthalpy $h(T)$ and standard-state entropy function $s^\circ(T)$: fix the compressor and turbine exits from their isentropic targets and stated efficiencies, close the regenerator with its effectiveness definition, then take heat-addition, net-work and efficiency directly from the resulting enthalpies, and exergy destruction from $T_0$ times the entropy generated in each of the five processes (compressor, regenerator, combustor, turbine, and the closing heat rejection to the sink).

Compressor (1→2), actual. The isentropic exit temperature from $s^\circ(T_{2s})-s^\circ(T_1)=R\ln(P_2/P_1)$ is $T_{2s}=573.7$ K, giving $h_{2s}=705.85$ kJ/kg and ideal work $w_{s}=279.55$ kJ/kg. With $\eta_C=0.75$: $$w_{C}=\frac{h_{2s}-h_1}{\eta_C}=\frac{279.55}{0.75}=\boxed{372.73\ \text{kJ/kg}}\quad\Rightarrow\quad h_2=h_1+w_C=799.03\ \text{kJ/kg}\ (T_2=662.0\ \text{K}).$$
Turbine (4→5), actual. Similarly, $T_{5s}=665.1$ K from $s^\circ(T_{5s})-s^\circ(T_4)=R\ln(P_1/P_2)$, $h_{5s}=802.35$ kJ/kg, ideal work $w_s=601.86$ kJ/kg. With $\eta_T=0.80$: $$w_T=\eta_T(h_4-h_{5s})=0.80\times601.86=\boxed{481.49\ \text{kJ/kg}}\quad\Rightarrow\quad h_5=h_4-w_T=922.72\ \text{kJ/kg}\ (T_5=776.6\ \text{K}).$$
Regenerator, effectiveness definition. $\varepsilon=(h_3-h_2)/(h_5-h_2)$, so the cold-side (combustor-inlet) enthalpy is $$h_3=h_2+\varepsilon(h_5-h_2)=799.03+0.70(922.72-799.03)=885.61\ \text{kJ/kg}\ (T_3=742.5\ \text{K}).$$ An energy balance on the (adiabatic) regenerator gives the hot-side exit: $h_6=h_5-(h_3-h_2)=836.14$ kJ/kg ($T_6=696.7$ K).
Heat addition rate (part a). $$q_{in}=h_4-h_3=1404.21-885.61=\boxed{518.60\ \text{kJ/kg}}.$$
Net specific work and thermal efficiency (parts b, c). $$w_{net}=w_T-w_C=481.49-372.73=\boxed{108.76\ \text{kJ/kg}},\qquad \eta_{th}=\frac{w_{net}}{q_{in}}=\boxed{20.97\%}.$$
Exergy destruction, process by process (part d). Using $T_0\Delta s$ for each irreversible/finite-$\Delta T$ step (actual $\Delta s$ includes the pressure term $-R\ln(P_{out}/P_{in})$ for the compressor/turbine, and the combustor is heat addition from a source at $T_H$): $$X_{dest,C}=T_0[(s_2-s_1)]=45.31\ \text{kJ/kg},\qquad X_{dest,T}=T_0[(s_5-s_4)]=50.18\ \text{kJ/kg},$$ $$X_{dest,combustor}=T_0\!\left[(s_4-s_3)-\frac{q_{in}}{T_H}\right]=33.08\ \text{kJ/kg},\qquad X_{dest,regen}=T_0[(s_3-s_2)+(s_6-s_5)]=1.73\ \text{kJ/kg}.$$ The cycle closes with heat rejection 6→1 to the sink at $T_{sink}=T_0=300$ K, $q_{out}=h_6-h_1=409.84$ kJ/kg — a genuine cycle process and by far the largest destroyer of exergy here, because the exhaust leaves the regenerator still at 696.7 K and is cooled all the way to 300 K against a 300 K sink: $$X_{dest,rejection}=T_0\!\left[(s_1-s_6)+\frac{q_{out}}{T_{sink}}\right]=149.88\ \text{kJ/kg}.$$ $$X_{dest,total}=45.31+33.08+50.18+1.73+149.88=\boxed{280.19\ \text{kJ/kg}}.$$
Second-law efficiency (part e). The exergy supplied by the heat source is $X_{in}=q_{in}(1-T_0/T_H)=388.95$ kJ/kg, so $$\eta_{II}=\frac{w_{net}}{X_{in}}=\frac{108.76}{388.95}=\boxed{27.96\%}.$$ This also checks the part-(d) tally: the exergy supplied must be either converted to work or destroyed, $X_{in}-w_{net}=388.95-108.76=280.19$ kJ/kg, which matches the five-process total exactly, and equivalently $\eta_{II}=1-X_{dest,total}/X_{in}=1-280.19/388.95=27.96\%$. An exergy list that leaves out the heat-rejection process would fail this closure.

Quantity	Result
(a) $q_{in}$	518.60 kJ/kg
(b) $w_{net}$	108.76 kJ/kg
(c) $\eta_{th}$	20.97%
(d) $X_{dest}$: C / comb. / T / regen. / rejection / total	45.31 / 33.08 / 50.18 / 1.73 / 149.88 / 280.19 kJ/kg
(e) $\eta_{II}$	27.96%

Concept & Theory

Topic. Air-standard regenerative Brayton cycle with non-ideal turbomachinery, evaluated with temperature-dependent specific heats and second-law (exergy) accounting.

Key relations. Ideal-gas entropy change $s_2-s_1=s^\circ(T_2)-s^\circ(T_1)-R\ln(P_2/P_1)$; regenerator effectiveness $\varepsilon=(h_3-h_2)/(h_5-h_2)$; exergy destruction of a process exchanging heat with a source/sink at fixed $T$: $X_{dest}=T_0[\Delta s_{system}-q/T_{reservoir}]$ (source term subtracted, sink term added since heat leaves the system).

Why this works. Regeneration recovers turbine-exhaust exergy that would otherwise be dumped to the sink, preheating the compressor discharge and cutting the fuel-equivalent heat addition $q_{in}$ — but it can never make the cycle reversible, because effectiveness $<100\%$ leaves a finite temperature difference (and hence entropy generation) between the two regenerator streams, alongside the unavoidable turbomachinery irreversibility already built into $\eta_C,\eta_T<1$.

Common pitfall. Listing only the four hardware components (compressor, combustor, turbine, regenerator) and calling their sum the total exergy destruction — the closing process 6→1, in which the still-hot exhaust is cooled to the dead state against a 300 K sink, is a process of the cycle too and here destroys 149.88 kJ/kg, more than the other four combined. The free test is the closure $\sum X_{dest}=X_{in}-w_{net}$: omit heat rejection and the tally comes to 130.31 instead of 280.19. (Second pitfall: using $c_p$ at the inlet or a single average temperature instead of the actual variable-$c_p$ property data across the full 300–1200 K span, and dropping the $R\ln(P_2/P_1)$ pressure term from the entropy change.)

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 9 (Gas Power Cycles) & Ch. 8 (Exergy Analysis of Closed/Steady-Flow Systems).

Question 2 of 9: Regenerative Brayton Cycle — Variable Specific Heats and Exergy

Question 2: Regenerative Brayton Cycle — Variable Specific Heats and Exergy (20 marks)

Approach