Brayton Cycle with Reheat: 04-BS-10, December 2019, Q2

Question 2: Brayton Cycle with Reheat (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.

Check — reconstructed question data. The surviving legible fragments — "compressor at 100 kPa and 300 K", "at 400 kP[a]", "1200 K", "efficiency of 88%", "compressor has … efficiency of 90%" — match a well-known textbook Brayton-reheat problem (Cengel & Boles) with compressor pressure ratio 15 and a high-pressure turbine that expands from the compressor-exit pressure to a 400 kPa reheat pressure. That pressure ratio (15) is adopted here as the most defensible reconstruction; every other legible number (100 kPa/300 K compressor inlet, 400 kPa/1200 K reheat state, 88%/90% efficiencies, 300 K sink) is used as printed.

Given. $T_1=300$ K, $P_1=100$ kPa, $r_p=15$ (compressor), HP turbine inlet $T_3=1200$ K expanding to $P_{reheat}=400$ kPa, reheated back to $T_5=1200$ K, LP turbine expands to $P_1=100$ kPa. $\eta_c=0.90$, $\eta_t=0.88$ (each stage). Variable specific heats for air (ideal-gas $h(T)$, $s^\circ(T)$ tables). Sink $T_0=300$ K; source for the second-law reference taken as the cycle's own maximum temperature, $T_H=1200$ K.

State	Description	P	T	h (kJ/kg)
1	Compressor inlet	100 kPa	300.0 K	426.30
2	Compressor exit (actual)	1500 kPa	677.28 K	815.35
3	HP turbine inlet	1500 kPa	1200.0 K	1404.21
4	HP turbine exit (actual)	400 kPa	903.0 K	1062.76
5	LP turbine inlet (reheated)	400 kPa	1200.0 K	1404.21
6	LP turbine exit (actual)	100 kPa	890.6 K	1048.84

Find. (a) $w_{net}$ [kJ/kg]; (b) $\eta_{th}$; (c) $\eta_{II}$.

Fig. Q2 — T–s cycle (ideal gas, no saturation dome; entropy axis schematic since variable-$c_p$ air has no single linear $s$ scale): 1→2 actual compression, 2→3 combustor heat addition, 3→4 actual HP expansion, 4→5 reheat at 400 kPa, 5→6 actual LP expansion, 6→1 heat rejection.

Approach

Work each turbomachine from the ideal-gas $s^\circ(T)$ relation $s_2^\circ-s_1^\circ=R\ln(P_2/P_1)$ (inverted numerically for $T$) to get the isentropic exit state, then apply the isentropic-efficiency definition to get the actual exit enthalpy/temperature. Two independent turbine legs (HP: 3→4; LP: 5→6) and one compressor leg (1→2) are solved this way; heat addition is the sum of the combustor (2→3) and reheat (4→5) legs.

Compressor (1→2), actual. Isentropic exit from $s_2^{\circ}=s_1^{\circ}+R\ln(P_2/P_1)$ gives $T_{2s}=640.7$ K, $h_{2s}=776.44$ kJ/kg: $$w_{c,a}=\frac{h_{2s}-h_1}{\eta_c}=\frac{776.44-426.30}{0.90}=\boxed{389.05\ \text{kJ/kg}} \Rightarrow h_2=426.30+389.05=815.35\ \text{kJ/kg}\ (T_2=677.3\ \text{K}).$$
HP turbine (3→4), actual. Isentropic exit from $s_4^{\circ}=s_3^{\circ}+ R\ln(P_{reheat}/P_2)$ gives $T_{4s}=861.3$ K, $h_{4s}=1016.20$ kJ/kg: $$w_{t1,a}=\eta_t(h_3-h_{4s})=0.88(1404.21-1016.20)=\boxed{341.45\ \text{kJ/kg}} \Rightarrow h_4=1404.21-341.45=1062.76\ \text{kJ/kg}\ (T_4=903.0\ \text{K}).$$
LP turbine (5→6), actual. Reheated to $T_5=1200$ K at 400 kPa ($h_5=h_3= 1404.21$ kJ/kg); isentropic exit from $s_6^{\circ}=s_5^{\circ}+R\ln(P_1/P_{reheat})$ gives $T_{6s}=847.1$ K, $h_{6s}=1000.38$ kJ/kg: $$w_{t2,a}=\eta_t(h_5-h_{6s})=0.88(1404.21-1000.38)=\boxed{355.37\ \text{kJ/kg}} \Rightarrow h_6=1404.21-355.37=1048.84\ \text{kJ/kg}\ (T_6=890.6\ \text{K}).$$
Net work (part a) and heat input. $$w_{turb}=w_{t1,a}+w_{t2,a}=341.45+355.37=696.82\ \text{kJ/kg}$$ $$w_{net}=w_{turb}-w_{c,a}=696.82-389.05=\boxed{307.77\ \text{kJ/kg}}$$ $$q_{in}=(h_3-h_2)+(h_5-h_4)=(1404.21-815.35)+(1404.21-1062.76)=588.86+341.45=930.31\ \text{kJ/kg}.$$
Thermal efficiency (part b). $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{307.77}{930.31}=\boxed{0.3308\ (33.1\%)}.$$
Second-law efficiency (part c). Referencing the Carnot ceiling between the cycle's own maximum temperature ($T_H=1200$ K) and the stated sink ($T_0=300$ K): $$\eta_{Carnot}=1-\frac{T_0}{T_H}=1-\frac{300}{1200}=0.75,\qquad \eta_{II}=\frac{\eta_{th}}{\eta_{Carnot}}=\frac{0.3308}{0.75}=\boxed{0.4411\ (44.1\%)}.$$

Quantity	Result
(a) $w_{net}$	307.77 kJ/kg
(b) $\eta_{th}$	0.3308 (33.1%)
(c) $\eta_{II}$	0.4411 (44.1%)

Concept & Theory

Topic. Non-ideal Brayton cycle with a single reheat stage, variable specific heats for air.

Key relations. isentropic exit fixed by $s_2^\circ=s_1^\circ+R\ln(P_2/P_1)$; actual work from $w_a=w_s/\eta$ (compressor) or $w_a=\eta\,w_s$ (turbine).

Why this works. Splitting the expansion into two stages with reheat between them raises the average temperature of heat rejection... no — raises turbine work output (each stage now expands over a smaller pressure ratio at a higher average temperature) at the cost of extra heat input, which is why reheat alone (without also adding a regenerator) typically lowers thermal efficiency even as it raises net specific work.

Common pitfall. Using the compressor-exit pressure as the LP turbine's inlet pressure — the LP turbine actually starts at the reheat pressure (400 kPa here), which is lower than the compressor exit (1500 kPa) precisely because the HP turbine has already expanded the gas partway before reheat.

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

Question 2 of 9: Brayton Cycle with Reheat

Question 2: Brayton Cycle with Reheat (20 marks)

Approach