Regenerative Gas-Turbine (Brayton) Cycle: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2013, Q3

Question 3: Regenerative Gas-Turbine (Brayton) Cycle (equal value)

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 Brayton cycle with regeneration: $T_1=27\ ^\circ\text{C}=300.15$ K, $P_1=100$ kPa, pressure ratio $r_p=5$, turbine inlet $T_3=984\ ^\circ\text{C}=1257.15$ K. Component efficiencies $\eta_c=\eta_t=0.80$, regenerator effectiveness $\varepsilon=0.75$. Air: $c_p=1.005\ \text{kJ/kg}\cdot\text{K}$, $\gamma=1.4$, so $(\gamma-1)/\gamma=0.2857$.

Find. The cycle thermal efficiency $\eta_\text{th}$, and a $T$–$s$ sketch of the process.

Figure 3 — T–s sketch: 1→2 actual compression, 2→2′ regenerative preheat (blue), 2′→3 combustion, 3→4 actual expansion, 4→4′ regenerator hot side (blue), 4′→1 exhaust. The two blue legs are the same heat transfer (dashed link), so with equal $\dot m c_p$ the exhaust drops by exactly the rise on the air side: $T_{4'}=T_4-(T_{2'}-T_2)=886-275=611$ K.

Approach. Get the isentropic compressor and turbine end temperatures from the pressure ratio, apply the 80 % efficiencies for the real states, use the 75 % regenerator effectiveness to fix the air temperature entering the combustor, then form $\eta_\text{th}=w_\text{net}/q_\text{in}$.

Compressor exit. Isentropic $T_{2s}=T_1 r_p^{(\gamma-1)/\gamma}=300.15(5)^{0.2857}=475.4$ K; with $\eta_c=0.80$,$$T_2=T_1+\frac{T_{2s}-T_1}{\eta_c}=300.15+\frac{175.2}{0.80}=519.2\ \text{K}.$$
Turbine exit. Isentropic $T_{4s}=T_3/r_p^{(\gamma-1)/\gamma}=1257.15/1.5838=793.8$ K; with $\eta_t=0.80$,$$T_4=T_3-\eta_t(T_3-T_{4s})=1257.15-0.80(463.4)=886.4\ \text{K}.$$
Regenerator (combustor-inlet temperature). The effectiveness raises the compressed air from $T_2$ toward the turbine-exhaust temperature $T_4$: $T_{2'}=T_2+\varepsilon(T_4-T_2)=519.2+0.75(367.2)=\boxed{794.6\ \text{K}}.$
Specific works and heat input. $w_c=c_p(T_2-T_1)=1.005(219.0)=220.1$; $w_t=c_p(T_3-T_4)=1.005(370.7)=372.6$; net $w_\text{net}=152.5$ kJ/kg. Heat is added only from $2'$ to $3$: $q_\text{in}=c_p(T_3-T_{2'})=1.005(462.5)=464.8$ kJ/kg.
Thermal efficiency.$$\eta_\text{th}=\frac{w_\text{net}}{q_\text{in}}=\frac{152.5}{464.8}=\boxed{0.328\;(32.8\%)}.$$

Quantity	Result
Compressor / turbine exit T	519.2 K / 886.4 K
Combustor-inlet T (after regen)	794.6 K
Net work / heat input	152.5 / 464.8 kJ/kg
Cycle thermal efficiency	≈ 32.8 %

Concept & Theory

Topic. Air-standard regenerative Brayton cycle with component efficiencies.

Key relations. Isentropic $T_2/T_1=r_p^{(\gamma-1)/\gamma}$; $\eta_c=(T_{2s}-T_1)/(T_2-T_1)$, $\eta_t=(T_3-T_4)/(T_3-T_{4s})$; regenerator $\varepsilon=(T_{2'}-T_2)/(T_4-T_2)$; $\eta_\text{th}=w_\text{net}/q_\text{in}$.

Why this works. The regenerator transfers heat from the hot turbine exhaust to the cooler compressed air, so combustion only has to supply the enthalpy rise from $T_{2'}$ to $T_3$ — less fuel for the same net work, raising efficiency. Neglecting the fuel mass lets a single air stream with constant $c_p$ carry every energy term.

Common pitfall. Adding heat from $T_2$ (not $T_{2'}$) in $q_\text{in}$, which ignores the regenerator entirely; or using isentropic temperatures in the work terms instead of the real (efficiency-corrected) $T_2$, $T_4$.

Source. Çengel & Boles, Thermodynamics §9 (Brayton cycle with regeneration; compressor and turbine isentropic efficiencies).

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

Question 3: Regenerative Gas-Turbine (Brayton) Cycle (equal value)