Question 3 of 9: Brayton Cycle with Reheat, Variable Specific Heats

Question 3: Brayton Cycle with Reheat, Variable 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. Compressor: 100 kPa, 300 K $\to$ 400 kPa, $\eta_c=0.90$. Turbine stage 1: 400 kPa, 1200 K $\to$ 200 kPa, $\eta_t=0.88$. Reheat at 200 kPa back to 1200 K. Turbine stage 2: 200 kPa, 1200 K $\to$ 100 kPa, $\eta_t=0.88$. $T_0=300$ K (sink), $T_H=1200$ K (source). Variable specific heats.

Find. (a) $w_{net}$ [kJ/kg]; (b) $\eta_{th}$; (c) % increase in $w_{net}$ vs. single-stage expansion, no reheat; (d) $\eta_{II}$.

Approach

Solve the compressor and each turbine stage independently via the variable-specific-heat isentropic relation $s^\circ(T_2)-s^\circ(T_1)=R\ln(P_2/P_1)$ (root-solved), apply the stated efficiency to get each actual work term, then sum the two heat-addition legs (main combustor plus reheater) for $q_{in}$. Part (c) repeats the turbine calculation as a single 4:1 expansion with no reheat for comparison; part (d) uses the source/sink temperatures to convert $q_{in}$ to an exergy input.

1. Compressor (1→2), 100→400 kPa. $s^\circ_1$ at 300 K, isentropic exit $T_{2s}=444.59$ K gives $w_{c,s}=146.31$ kJ/kg; actual: $$w_{c,a}=\frac{w_{c,s}}{\eta_c}=\frac{146.31}{0.90}=\boxed{162.56\ \text{kJ/kg}},\qquad T_2=460.50\ \text{K (from } h_2=h_1+w_{c,a}\text{)}.$$
2. Turbine stage 1 (3→4), 400→200 kPa. At $T_3=1200$ K, isentropic exit $T_{4s}=1010.73$ K gives $w_{t1,s}=219.44$ kJ/kg; actual: $$w_{t1,a}=\eta_t\,w_{t1,s}=0.88\times219.44=\boxed{193.11\ \text{kJ/kg}},\qquad T_4=1033.72\ \text{K}.$$
3. Reheat and turbine stage 2 (5→6), 200→100 kPa. Reheat restores $T_5=1200$ K at 200 kPa. Because this stage's pressure ratio (200/100=2) exactly equals stage 1's (400/200=2) and the inlet temperature is again 1200 K, the two stages are identical: $$w_{t2,a}=\eta_t\,w_{t2,s}=0.88\times219.44=\boxed{193.11\ \text{kJ/kg}},\qquad T_6=1033.72\ \text{K}.$$
4. Net work and thermal efficiency (parts a, b). $$w_{net}=w_{t1,a}+w_{t2,a}-w_{c,a}=193.11+193.11-162.56=\boxed{223.65\ \text{kJ/kg}}\quad\textbf{(a)}.$$ Heat input has two legs, the combustor ($2\to3$) and the reheater ($4\to5$): $$q_{in}=(h_3-h_2)+(h_5-h_4)=815.35+193.11=1008.46\ \text{kJ/kg}.$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{223.65}{1008.46}=\boxed{0.2218\ (22.2\%)}\quad\textbf{(b)}.$$
5. Single-stage comparison, no reheat (part c). Expanding directly from 400 kPa, 1200 K to 100 kPa in one stage (pressure ratio 4) with the same $\eta_t=0.88$: isentropic exit $T_s=847.07$ K gives $w_{t,single,a}=355.37$ kJ/kg, so $$w_{net,single}=355.37-162.56=192.81\ \text{kJ/kg}.$$ $$\%\ \text{increase}=\frac{223.65-192.81}{192.81}\times100=\boxed{16.00\%}\quad\textbf{(c)}.$$ Reheat increases the net work (splitting the expansion and reheating keeps the average expansion temperature higher) but, as seen in part (b), it also raises $q_{in}$ enough that $\eta_{th}$ here is actually below the ideal single-stage value of $1-(1/4)^{0.4/1.4}\approx32.6\%$ — reheat alone (without a regenerator) trades some thermal efficiency for more net work per unit mass flow.
6. Second-law efficiency (part d). Exergy input using the source/sink temperatures: $$\dot X_{in}=q_{in}\left(1-\frac{T_0}{T_H}\right)=1008.46\times\left(1-\frac{300}{1200}\right)=756.34\ \text{kJ/kg}.$$ $$\eta_{II}=\frac{w_{net}}{X_{in}}=\frac{223.65}{756.34}=\boxed{0.2957\ (29.6\%)}\quad\textbf{(d)}.$$