Cold-Air-Standard Regenerative Brayton Cycle: 04-BS-10, December 2018, Q2

Question 2: Cold-Air-Standard Regenerative Brayton Cycle (20 marks)

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Given. $T_1=300$ K, $P_1=100$ kPa, $\dot m=6$ kg/s, $r_p=10$, $T_3=1400$ K, $\eta_c=\eta_t=0.80$, regenerator effectiveness $\varepsilon=0.80$, cold-air-standard ($c_p=1.005$ kJ/kg·K, $k=1.4$, constant properties), $T_0=300$ K.

Find. (a) $\eta_{th}$; (b) back work ratio; (c) $\dot W_{net}$ [kW]; (d) $\dot X_{dest}$ in the compressor, turbine and regenerator [kW].

Fig. Q2 — T–s cycle (ideal gas, no saturation dome): 1→2 actual compression, 2→x regenerator cold-side preheat, x→3 combustor heat addition, 3→4 actual expansion, 4→5 regenerator hot-side cool-down, 5→1 heat rejection.

Approach

Work each turbomachine from its isentropic-efficiency definition using the closed-form cold-air power law $T_{2s}/T_1=r_p^{(k-1)/k}$, fix the regenerator's cold-side exit temperature from its effectiveness definition, then take energy balances for $\eta_{th}$, back work ratio and net power. Part (d) uses $T_0\Delta s$ for each component, with $\Delta s$ built from $c_p\ln(T_2/T_1)-R\ln(P_2/P_1)$ so the pressure-drop term is correctly zero across the (constant-pressure) regenerator legs.

Compressor (1→2), actual. $$T_{2s}=T_1\,r_p^{(k-1)/k}=300\times10^{0.2857}=579.2\ \text{K},\qquad w_{c,a}=\frac{c_p(T_{2s}-T_1)}{\eta_c}=\frac{1.005(579.2-300)}{0.80}=350.7\ \text{kJ/kg}$$ $$T_2=T_1+\frac{w_{c,a}}{c_p}=300+\frac{350.7}{1.005}=\boxed{649.0\ \text{K}}.$$
Turbine (3→4), actual. $$T_{4s}=\frac{T_3}{r_p^{(k-1)/k}}=\frac{1400}{1.9307}=725.1\ \text{K},\qquad w_{t,a}=\eta_t\,c_p(T_3-T_{4s})=0.80\times1.005(1400-725.1)=542.5\ \text{kJ/kg}$$ $$T_4=T_3-\frac{w_{t,a}}{c_p}=1400-\frac{542.5}{1.005}=\boxed{860.1\ \text{K}}.$$
Regenerator. Effectiveness $\varepsilon=(T_x-T_2)/(T_4-T_2)$ gives the cold-side exit; an energy balance (equal $\dot m c_p$ both sides) gives the hot-side exit: $$T_x=T_2+\varepsilon(T_4-T_2)=649.0+0.80(860.1-649.0)=\boxed{817.9\ \text{K}}$$ $$T_5=T_4-(T_x-T_2)=860.1-(817.9-649.0)=\boxed{691.2\ \text{K}}.$$
Thermal efficiency (part a) and back work ratio (part b). $$q_{in}=c_p(T_3-T_x)=1.005(1400-817.9)=585.0\ \text{kJ/kg},\qquad w_{net}=w_{t,a}-w_{c,a}=542.5-350.7=191.8\ \text{kJ/kg}$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{191.8}{585.0}=\boxed{0.3279\ (32.8\%)},\qquad \text{bwr}=\frac{w_{c,a}}{w_{t,a}}=\frac{350.7}{542.5}=\boxed{0.6464}.$$
Net power (part c). $$\dot W_{net}=\dot m\,w_{net}=6\times191.8=\boxed{1151.0\ \text{kW}}.$$
Exergy destruction (part d). Each component's entropy change combines a temperature term and (where the pressure changes) a pressure term: $$\Delta s_c=c_p\ln\frac{T_2}{T_1}-R\ln\frac{P_2}{P_1}=1.005\ln\frac{649.0}{300}-0.287\ln(10)=0.7752-0.6608=0.1147\ \text{kJ/kg}\cdot\text{K}$$ $$\Delta s_t=c_p\ln\frac{T_4}{T_3}-R\ln\frac{P_1}{P_2}=1.005\ln\frac{860.1}{1400}-0.287\ln(0.1)=-0.4899+0.6608=0.1712\ \text{kJ/kg}\cdot\text{K}$$ $$\Delta s_{regen}=c_p\ln\frac{T_x}{T_2}+c_p\ln\frac{T_5}{T_4}=0.2326-0.2198=0.0128\ \text{kJ/kg}\cdot\text{K}$$ $$\dot X_c=\dot m\,T_0\,\Delta s_c=6\times300\times0.1147=\boxed{206.4\ \text{kW}}$$ $$\dot X_t=\dot m\,T_0\,\Delta s_t=6\times300\times0.1712=\boxed{308.2\ \text{kW}}$$ $$\dot X_{regen}=\dot m\,T_0\,\Delta s_{regen}=6\times300\times0.0128=\boxed{23.0\ \text{kW}}.$$

Quantity	Result
(a) $\eta_{th}$	0.3279 (32.8%)
(b) Back work ratio	0.6464
(c) $\dot W_{net}$	1151.0 kW
(d) $\dot X_{c}$, $\dot X_{t}$, $\dot X_{regen}$	206.4, 308.2, 23.0 kW

Concept & Theory

Topic. Non-ideal, cold-air-standard regenerative Brayton cycle with a component-by-component (rather than whole-cycle) exergy destruction breakdown.

Key relations. Cold-air isentropic power law $T_{2s}/T_1=r_p^{(k-1)/k}$; actual work from isentropic efficiency $w_{c,a}=w_{c,s}/\eta_c$, $w_{t,a}=\eta_t\,w_{t,s}$; regenerator effectiveness $\varepsilon=(T_x-T_2)/(T_4-T_2)$; component exergy destruction $\dot X_{dest}=\dot m\,T_0\,\Delta s$ with $\Delta s=c_p\ln(T_{out}/T_{in})-R\ln(P_{out}/P_{in})$.

Why this works. Because the compressor and turbine are each adiabatic, their entropy generation equals their entropy change outright ($\dot S_{gen}=\dot m\Delta s$, no $Q/T$ term); the regenerator is also adiabatic to the surroundings (heat only crosses internally between its two streams), so its total destruction is simply the sum of the cold-side gain and the hot-side loss in entropy, each evaluated at its own constant pressure.

Common pitfall. Dropping the $-R\ln(P_2/P_1)$ term from the compressor/turbine entropy change (it is easy to reason only in terms of the $T$ ratio) — doing so would make the compressor's destruction come out far too large and could even flip its sign, since the pressure rise by itself lowers entropy for a fixed $T$ ratio.

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

Question 2 of 9: Cold-Air-Standard Regenerative Brayton Cycle

Question 2: Cold-Air-Standard Regenerative Brayton Cycle (20 marks)

Approach