Regenerative Brayton Cycle — Constant Specific Heats: 04-BS-10, December 2013, Q3

Question 3: Regenerative Brayton Cycle — Constant Specific Heats (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, cold-air-standard properties ($c_p=1.005$ kJ/kg·K, $k=1.4$). Pressure ratio $r_p=10$; $T_1=300$ K (compressor inlet, cycle minimum); $T_4=1200$ K (turbine inlet, cycle maximum). Compressor isentropic efficiency $\eta_C=75\%$; turbine isentropic efficiency $\eta_T=80\%$; regenerator effectiveness $\varepsilon=70\%$. Source $T_H=1200$ K, sink/dead-state $T_0=300$ K.

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

Find. (a) $T_5$ (turbine exit); (b) $w_{net}$; (c) $\eta_{th}$; (d) $\eta_{II}$.

Fig. Q3 — 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), plotted against entropy measured relative to state 1.

Approach

Work state-by-state with constant specific heats: find the isentropic compressor/turbine exit temperatures from $T_2=T_1r_p^{(k-1)/k}$ and $T_5=T_4r_p^{-(k-1)/k}$, apply the stated isentropic efficiencies to get the actual exits, close the regenerator with its effectiveness definition, then take heat addition, net work, thermal efficiency and second-law efficiency directly.

Compressor (1→2), actual. Isentropic exit: $T_{2s}=T_1r_p^{(k-1)/k}=300\times10^{0.2857}=579.21$ K, so $w_{s,C}=c_p(T_{2s}-T_1)=280.61$ kJ/kg. With $\eta_C=0.75$: $$w_C=\frac{w_{s,C}}{\eta_C}=\frac{280.61}{0.75}=374.14\ \text{kJ/kg}\quad\Rightarrow\quad T_2=T_1+\frac{w_C}{c_p}=300+372.28=672.28\ \text{K}.$$
Turbine (4→5), actual. Isentropic exit: $T_{5s}=T_4r_p^{-(k-1)/k}=1200/10^{0.2857}=621.54$ K, so $w_{s,T}=c_p(T_4-T_{5s})=581.36$ kJ/kg. With $\eta_T=0.80$: $$w_T=\eta_T\,w_{s,T}=0.80\times581.36=465.08\ \text{kJ/kg}\quad\Rightarrow\quad T_5=T_4-\frac{w_T}{c_p}=1200-462.77=\boxed{737.23\ \text{K}}\ \ (\text{part a}).$$
Regenerator, effectiveness definition. $\varepsilon=(T_3-T_2)/(T_5-T_2)$, so the combustor-inlet temperature is $$T_3=T_2+\varepsilon(T_5-T_2)=672.28+0.70\times(737.23-672.28)=\boxed{717.74\ \text{K}}.$$
Net specific work (part b). $$w_{net}=w_T-w_C=465.08-374.14=\boxed{90.94\ \text{kJ/kg}}.$$
Heat addition and thermal efficiency (part c). $$q_{in}=c_p(T_4-T_3)=1.005\times(1200-717.74)=484.67\ \text{kJ/kg}.$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{90.94}{484.67}=\boxed{18.76\%}.$$
Second-law efficiency (part d). The exergy supplied by the source at $T_H$ for the actual heat added $q_{in}$ is $X_{in}=q_{in}(1-T_0/T_H)$: $$X_{in}=484.67\times\left(1-\frac{300}{1200}\right)=363.50\ \text{kJ/kg},\qquad \eta_{II}=\frac{w_{net}}{X_{in}}=\frac{90.94}{363.50}=\boxed{25.02\%}.$$

Quantity	Result
(a) $T_5$ (turbine exit)	737.23 K
(b) $w_{net}$	90.94 kJ/kg
(c) $\eta_{th}$	18.76%
(d) $\eta_{II}$	25.02%

Concept & Theory

Topic. Air-standard regenerative Brayton cycle, cold-air-standard (constant specific heat) analysis with non-ideal turbomachinery and second-law accounting.

Key relations. Cold-air-standard isentropic relation $T_2/T_1=(P_2/P_1)^{(k-1)/k}$; isentropic efficiencies $\eta_C=w_s/w_a$ (compressor), $\eta_T=w_a/w_s$ (turbine); regenerator effectiveness $\varepsilon=(T_3-T_2)/(T_5-T_2)$; exergy supplied by a finite-temperature source, $X_{in}=q_{in}(1-T_0/T_H)$.

Why this works. Regeneration preheats the compressor discharge with turbine exhaust that would otherwise be wasted to the sink, cutting the external heat $q_{in}$ needed to reach $T_4$ without changing $w_{net}$ — but because $\varepsilon<100\%$ and both turbomachines have efficiencies below unity, the cycle is far from reversible, which is why $\eta_{II}$ (25%) is much lower than $\eta_{th}$ relative to the Carnot ceiling.

Common pitfall. This variant explicitly specifies constant (room-temperature) specific heats — using the variable-property air table instead (appropriate for other problems, e.g. Q4 below) would shift every temperature by tens of kelvin. Also, part (a) asks for the actual turbine-exit temperature $T_5$, not the isentropic $T_{5s}$.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 9 (Gas Power Cycles) & Ch. 8 (Exergy Analysis).

Question 3 of 9: Regenerative Brayton Cycle — Constant Specific Heats

Question 3: Regenerative Brayton Cycle — Constant Specific Heats (20 marks)

Approach