Question 2 of 9: Regenerative Gas Turbine, Two-Stage Intercooled Compressor

Question 2: Regenerative Gas Turbine, Two-Stage Intercooled Compressor (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. Volumetric flow $\dot V_1=10$ m$^3$/min at $P_1=100$ kPa, $T_1=300$ K. Stage-1 exit / intercooler $P=300$ kPa; intercooler returns air to $T=300$ K. Overall compressor exit $P_3=1$ MPa. $\eta_c=0.85$ (each stage). Turbine inlet $T=1300$ K at 1 MPa, $\eta_t=0.87$, expanding to 100 kPa. Regenerator effectiveness $\varepsilon=0.80$. Variable specific heats (air evaluated via high-accuracy ideal-gas-limit properties, root-solved for isentropic temperatures). $T_0=300$ K for part (e); source $T_H=1200$ K, sink $T_L=300$ K for part (f).

Find. (a) $\dot W_c$ [kW]; (b) $\eta_{th}$; (c) $\dot Q_{in}$ [kW]; (d) $\dot W_{net}$ [kW]; (e) $\dot X_{dest}$ for each compressor stage and the turbine [kJ/s]; (f) $\eta_{II}$.

Approach

Convert the given volumetric flow to a mass flow rate via the ideal-gas law at the inlet state. Solve each compressor stage independently (both use the same variable-cp root-finding method: match the isentropic condition $s^\circ(T_{2s})=s^\circ(T_1)+R\ln(P_2/P_1)$, then divide the isentropic work by the stage efficiency), since the intercooler resets the second stage's inlet temperature back to 300 K. Solve the turbine the same way in reverse (multiplying by efficiency). Size the regenerator from its effectiveness definition to fix the combustor-only heat input, then compute thermal efficiency, exergy destruction (compressor/turbine only, per part e), and second-law efficiency.

1. Mass flow rate. $$\dot m=\frac{P_1\dot V_1}{RT_1}=\frac{100\times(10/60)}{0.287\times300}=\boxed{0.1936\ \text{kg/s}}.$$
2. Compressor stage 1 (1→2), 100→300 kPa. Isentropic exit $T_{2s}=409.95$ K gives $w_{c1,s}=111.05$ kJ/kg. Actual: $$w_{c1,a}=\frac{111.05}{0.85}=130.65\ \text{kJ/kg},\qquad T_2=429.22\ \text{K}\ (156.07\ ^\circ\text{C}).$$
3. Intercooler and compressor stage 2 (2i→3), 300→1000 kPa. Intercooled back to $T_{2i}=300$ K. Isentropic exit $T_{3s}=422.33$ K gives $w_{c2,s}=123.63$ kJ/kg. Actual: $$w_{c2,a}=\frac{123.63}{0.85}=145.45\ \text{kJ/kg},\qquad T_3=443.75\ \text{K}\ (170.60\ ^\circ\text{C}).$$
4. Total compressor power (part a). $$w_{c,total}=w_{c1,a}+w_{c2,a}=130.65+145.45=276.09\ \text{kJ/kg}.$$ $$\dot W_c=\dot m\,w_{c,total}=0.1936\times276.09=\boxed{53.44\ \text{kW}}.$$
5. Turbine (4→5), 1000→100 kPa. Isentropic exit $T_{5s}=726.33$ K gives $w_{t,s}=654.26$ kJ/kg. Actual: $$w_{t,a}=0.87\times654.26=569.20\ \text{kJ/kg},\qquad T_5=804.32\ \text{K}\ (531.17\ ^\circ\text{C}).$$
6. Net power (part d). $$w_{net}=w_{t,a}-w_{c,total}=569.20-276.09=293.11\ \text{kJ/kg}.$$ $$\dot W_{net}=\dot m\,w_{net}=0.1936\times293.11=\boxed{56.74\ \text{kW}}.$$
7. Regenerator and heat addition (part c) / thermal efficiency (part b). Effectiveness fixes the combustor-inlet enthalpy from the cold-side (compressor-exit) and hot-side (turbine-exit) enthalpies: $$h_x=h_3+\varepsilon(h_5-h_3)=571.75+0.80\times(953.16-571.75)=876.88\ \text{kJ/kg}\ (T_x=461.29\ ^\circ\text{C}).$$ $$q_{in}=h_4-h_x=1300\text{K enthalpy}-876.88=645.49\ \text{kJ/kg}.$$ $$\dot Q_{in}=\dot m\,q_{in}=0.1936\times645.49=\boxed{124.95\ \text{kW}}.$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{293.11}{645.49}=\boxed{0.4541\ (45.4\%)}.$$
8. Exergy destruction, compressor stages and turbine only (part e). Using full ideal-gas entropy $s(T,P)=s^\circ(T)-R\ln(P/P_{ref})$ at each state's own pressure: $$\dot X_{c1}=\dot m\,T_0(s_2-s_1)=0.1936\times300\times(3.9372-3.8905)=\boxed{2.713\ \text{kJ/s}}$$ $$\dot X_{c2}=\dot m\,T_0(s_3-s_{2i})=0.1936\times300\times(3.6256-3.5752)=\boxed{2.926\ \text{kJ/s}}$$ $$\dot X_{t}=\dot m\,T_0(s_5-s_4)=0.1936\times300\times(4.9132-4.8019)=\boxed{6.458\ \text{kJ/s}}$$ All three are positive, as required for real (irreversible) adiabatic processes.
9. Second-law efficiency (part f). $\eta_{rev}=1-T_L/T_H=1-300/1200=0.75$. $$\eta_{II}=\frac{\eta_{th}}{\eta_{rev}}=\frac{0.4541}{0.75}=\boxed{0.6055\ (60.5\%)}.$$