Question 2 of 9: Two-Stage Intercooled/Reheated Regenerative Brayton Cycle

Question 2: Two-Stage Intercooled/Reheated Regenerative Brayton Cycle (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. Ideal cycle (isentropic compressor/turbine stages), constant/cold-air-standard properties ($c_p=1.005$ kJ/kg·K, $k=1.4$). Pressure ratio per stage $r_p=3$ (overall compression and expansion ratio $=9$). Each compressor stage inlet $T=300$ K; each turbine stage inlet $T=1200$ K (i.e. the intercooler returns the air to 300 K before the second compressor stage, and the reheater returns it to 1200 K before the second turbine stage). Regenerator effectiveness $\varepsilon=0.75$. Source $T_H=1200$ K, sink $T_L=300$ K (taken as $T_0$).

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

Approach

Because both compressor stages share the same inlet temperature and pressure ratio, and both turbine stages share the same inlet temperature and pressure ratio, each pair of stages does identical work — compute one compressor stage and one turbine stage, then double. Find the regenerator's cold-side exit temperature from its effectiveness definition, add up the two heat-input legs (primary combustor + reheater), and form the four requested ratios.

1. Compressor stage (isentropic, $r_p=3$). $$T_{2s}=T_1\,r_p^{(k-1)/k}=300\times3^{0.2857}=410.62\ \text{K}\ (137.47\ ^\circ\text{C}).$$ $$w_{c,stage}=c_p(T_{2s}-T_1)=1.005\times110.62=111.17\ \text{kJ/kg},\qquad w_{c,total}=2\times111.17=222.35\ \text{kJ/kg}.$$
2. Turbine stage (isentropic, $r_p=3$). $$T_{exit,s}=\frac{T_{in}}{r_p^{(k-1)/k}}=\frac{1200}{3^{0.2857}}=876.72\ \text{K}\ (603.57\ ^\circ\text{C}).$$ $$w_{t,stage}=c_p(T_{in}-T_{exit,s})=1.005\times323.28=324.90\ \text{kJ/kg},\qquad w_{t,total}=2\times324.90=649.79\ \text{kJ/kg}.$$
3. Back work ratio and net work (parts a, b). $$\text{BWR}=\frac{w_{c,total}}{w_{t,total}}=\frac{222.35}{649.79}=\boxed{0.3422}.$$ $$w_{net}=w_{t,total}-w_{c,total}=649.79-222.35=\boxed{427.44\ \text{kJ/kg}}.$$
4. Regenerator cold-side exit temperature. Compressor-final exit $T_4=T_{2s}=410.62$ K; turbine-final exit $T_9=T_{exit,s}=876.72$ K. $$T_5=T_4+\varepsilon(T_9-T_4)=410.62+0.75\times(876.72-410.62)=760.20\ \text{K}\ (487.05\ ^\circ\text{C}).$$
5. Total heat input and thermal efficiency (part c). Heat is added in the primary combustor (5→6) and again in the reheater (7→8, which by symmetry needs the same duty as Turbine-I's own work since it restores $T_7=876.72$ K back to $1200$ K): $$q_{in}=c_p(T_6-T_5)+c_p(T_8-T_7)=1.005\times(1200-760.20)+1.005\times(1200-876.72)=442.00+324.90=766.90\ \text{kJ/kg}.$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{427.44}{766.90}=\boxed{0.5574\ (55.7\%)}.$$
6. Second-law efficiency (part d). With $T_0=T_L=300$ K and source $T_H=1200$ K: $$x_{in}=q_{in}\left(1-\frac{T_0}{T_H}\right)=766.90\times\left(1-\frac{300}{1200}\right)=575.18\ \text{kJ/kg}.$$ $$\eta_{II}=\frac{w_{net}}{x_{in}}=\frac{427.44}{575.18}=\boxed{0.7432\ (74.3\%)}.$$