Regenerative Gas-Turbine (Brayton) Cycle: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, December 2013, Q3

Question 3: Regenerative Gas-Turbine (Brayton) Cycle

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. Open air-standard Brayton cycle with a regenerator: compressor inlet 100 kPa/27°C, pressure ratio 5, turbine inlet 984°C, compressor and turbine each 80% efficient, regenerator effectiveness 75%, fuel mass neglected (air-standard throughout).

Given data
Quantity	Symbol	Value
Compressor inlet pressure	$P_1$	100 kPa
Compressor inlet temperature	$T_1$	$27\,{}^{\circ}\text{C}$ (300.15 K)
Pressure ratio	$r_p$	5
Turbine inlet temperature	$T_3$	$984\,{}^{\circ}\text{C}$ (1257.15 K)
Compressor efficiency	$\eta_c$	80%
Turbine efficiency	$\eta_t$	80%
Regenerator effectiveness	$\varepsilon$	75%

Find. The cycle thermal efficiency $\eta_{th}$, with the process sketched on a $T$–$s$ diagram.

T–s sketch of the regenerative Brayton cycle (entropy shown relative to state 1 via $\Delta s = C_p\ln(T/T_1)-R\ln(P/P_1)$, cold-air-standard).

Approach. Use cold-air-standard properties ($C_p=1.005$ kJ/kg·K, $\gamma=1.4$) throughout. Find the isentropic compressor and turbine exit temperatures from the pressure ratio, apply the component efficiencies to get the actual exit temperatures, use the regenerator effectiveness to fix the actual combustor-inlet temperature $T_{2'}$, then form $\eta_{th}=w_{net}/q_{in}$ from the four temperatures.

Compressor (1→2). Isentropic exit temperature from the pressure ratio: $$T_{2s} = T_1\,r_p^{(\gamma-1)/\gamma} = 300.15\times5^{0.2857} = 475.38\text{ K}$$ Actual exit temperature from the compressor efficiency $\eta_c=(T_{2s}-T_1)/(T_2-T_1)$: $$T_2 = T_1 + \frac{T_{2s}-T_1}{\eta_c} = 300.15+\frac{475.38-300.15}{0.80} = 519.19\text{ K}$$
Turbine (3→4). Isentropic exit temperature: $$T_{4s} = T_3\left(\frac{1}{r_p}\right)^{(\gamma-1)/\gamma} = 1257.15\times5^{-0.2857} = 793.75\text{ K}$$ Actual exit temperature from the turbine efficiency $\eta_t=(T_3-T_4)/(T_3-T_{4s})$: $$T_4 = T_3 - \eta_t(T_3-T_{4s}) = 1257.15-0.80(1257.15-793.75) = 886.43\text{ K}$$
Regenerator — combustor-inlet temperature. Effectiveness measures how close the cold-side (compressor-discharge) air gets to the hot-side (turbine-exhaust) inlet temperature, $\varepsilon=(T_{2'}-T_2)/(T_4-T_2)$: $$T_{2'} = T_2+\varepsilon(T_4-T_2) = 519.19+0.75(886.43-519.19) = 794.62\text{ K}$$
Work and heat terms, and thermal efficiency. With combustion heat added only from $T_{2'}$ to $T_3$ (the regenerator supplies the rest for free): $$q_{in} = C_p(T_3-T_{2'}) = 1.005\times(1257.15-794.62) = 464.84\text{ kJ/kg}$$ $$w_{comp} = C_p(T_2-T_1) = 1.005\times219.04 = 220.14\text{ kJ/kg}, \qquad w_{turb} = C_p(T_3-T_4) = 1.005\times370.72 = 372.58\text{ kJ/kg}$$ $$w_{net} = w_{turb}-w_{comp} = 372.58-220.14 = 152.44\text{ kJ/kg}$$ $$\boxed{\eta_{th} = \frac{w_{net}}{q_{in}} = \frac{152.44}{464.84} = 32.79\%}$$

Check: Cold-air-standard properties ($C_p$, $\gamma$ constant at their room-temperature values) are used throughout, per this exam's own instruction to treat "all processes as ideal" — a variable-specific-heat (gas-table) analysis would shift the answer by a few percentage points but is not what the given data supports without an air table.

Question 3 — results
Quantity	Value
$T_2$ (compressor exit, actual)	519.19 K ($246.04\,{}^{\circ}\text{C}$)
$T_4$ (turbine exit, actual)	886.43 K ($613.28\,{}^{\circ}\text{C}$)
$T_{2'}$ (regenerator cold-side exit)	794.62 K ($521.47\,{}^{\circ}\text{C}$)
Heat added $q_{in}$	464.84 kJ/kg
Net specific work $w_{net}$	152.44 kJ/kg
Cycle thermal efficiency $\eta_{th}$	32.79%

Topic. Air-standard regenerative Brayton cycle with non-ideal compressor and turbine.

Key relations. Isentropic $T$-$P$ relation $T_2/T_1=(P_2/P_1)^{(\gamma-1)/\gamma}$; compressor efficiency $\eta_c=(T_{2s}-T_1)/(T_2-T_1)$; turbine efficiency $\eta_t=(T_3-T_4)/(T_3-T_{4s})$; regenerator effectiveness $\varepsilon=(T_{2'}-T_2)/(T_4-T_2)$; $\eta_{th}=w_{net}/q_{in}$.

Why this works. A regenerator recovers exhaust heat that would otherwise be rejected to raise the compressor-discharge air toward the turbine-exhaust temperature before combustion, cutting the fuel heat $q_{in}$ needed for the same net work — this is why $\eta_{th}$ here (32.8%) beats a simple, non-regenerative cycle at the same pressure ratio and component efficiencies (which would sit closer to 20–24% at $r_p=5$). Regeneration only helps when $T_4>T_2$ (true here, 886 K vs 519 K); at high pressure ratios where $T_2$ exceeds $T_4$, a regenerator would need to run backwards and effectiveness becomes meaningless.

Common pitfall. Applying the regenerator effectiveness formula with $T_1$ or $T_3$ in place of the actual (not isentropic) compressor/turbine exit temperatures $T_2$, $T_4$; and computing $q_{in}$ from $T_3-T_2$ instead of $T_3-T_{2'}$, which ignores the regenerator's whole purpose and overstates the fuel heat required.

Source. Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 9 (gas-power cycles, Brayton cycle with regeneration).

Question 3 of 8: Regenerative Gas-Turbine (Brayton) Cycle

Question 3: Regenerative Gas-Turbine (Brayton) Cycle