Regenerative Gas Turbine, Two-Stage Intercooled Compressor: 04-BS-10, May 2016, Q2

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).

State	Description	T	P	s (kJ/kg·K)
1	Compressor stage-1 inlet	26.85°C (300 K)	100 kPa	3.891
2	Stage-1 exit (actual)	156.07°C	300 kPa	3.937
2i	Intercooler exit	26.85°C (300 K)	300 kPa	3.574
3	Stage-2 exit (actual, combustor inlet before regen.)	170.60°C	1 MPa	3.623
4	Turbine inlet	1026.85°C (1300 K)	1 MPa	4.802
5	Turbine exit (actual)	531.17°C	100 kPa	4.913

Find. (a) $\dot W_c$ [kW]; (b) $\eta_{th}$; (c) $\dot Q_{in}$ [kW]; (d) $\dot W_{net}$ [kW].

Fig. Q2 — T–s state points for the 2-stage intercooled, regenerated Brayton cycle (1→2 stage-1 compressor, 2→2i intercooler, 2i→3 stage-2 compressor, 3→4 regenerator+combustor, 4→5 turbine). No saturation dome for an ideal-gas working fluid.

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 and net power.

Mass flow rate. $$\dot m=\frac{P_1\dot V_1}{RT_1}=\frac{100\times(10/60)}{0.287\times300}=\boxed{0.1936\ \text{kg/s}}.$$
Compressor stage 1 (1→2), 100→300 kPa. Isentropic exit $T_{2s}=409.95$ K gives $w_{c1,s}=111.05$ kJ/kg. Actual: $$w_{c1,a}=\frac{111.05}{0.85}=130.65\ \text{kJ/kg},\qquad T_2=429.22\ \text{K}\ (156.07\ ^\circ\text{C}).$$
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: $$w_{c2,a}=\frac{123.63}{0.85}=145.45\ \text{kJ/kg},\qquad T_3=443.75\ \text{K}\ (170.60\ ^\circ\text{C}).$$
Total compressor power (part a). $$w_{c,total}=w_{c1,a}+w_{c2,a}=130.65+145.45=276.09\ \text{kJ/kg}.$$ $$\dot W_c=\dot m\,w_{c,total}=0.1936\times276.09=\boxed{53.44\ \text{kW}}.$$
Turbine (4→5), 1000→100 kPa. Isentropic exit $T_{5s}=726.33$ K gives $w_{t,s}=654.26$ kJ/kg. Actual: $$w_{t,a}=0.87\times654.26=569.20\ \text{kJ/kg},\qquad T_5=804.32\ \text{K}\ (531.17\ ^\circ\text{C}).$$
Net power (part d). $$w_{net}=w_{t,a}-w_{c,total}=569.20-276.09=293.11\ \text{kJ/kg}.$$ $$\dot W_{net}=\dot m\,w_{net}=0.1936\times293.11=\boxed{56.74\ \text{kW}}.$$
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: $$h_x=h_3+\varepsilon(h_5-h_3)=571.75+0.80\times(953.16-571.75)=876.88\ \text{kJ/kg}\ (T_x=461.29\ ^\circ\text{C}).$$ $$q_{in}=h_4-h_x=1300\text{K enthalpy}-876.88=645.49\ \text{kJ/kg}.$$ $$\dot Q_{in}=\dot m\,q_{in}=0.1936\times645.49=\boxed{124.95\ \text{kW}}.$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{293.11}{645.49}=\boxed{0.4541\ (45.4\%)}.$$

Quantity	Result
(a) $\dot W_c$	53.44 kW
(b) $\eta_{th}$	0.4541 (45.4%)
(c) $\dot Q_{in}$	124.95 kW
(d) $\dot W_{net}$	56.74 kW

Concept & Theory

Topic. Regenerative Brayton cycle with a two-stage intercooled compressor and a single (non-reheat) turbine stage, variable specific heats.

Key relations. Variable-cp isentropic condition: $s^\circ(T_{2s})-s^\circ(T_1)=R\ln(P_2/P_1)$, solved by root-finding rather than a constant-$k$ formula. Regenerator effectiveness: $\varepsilon=(h_x-h_3)/(h_5-h_3)$.

Why this works. The intercooler is idealized as returning the air to exactly the inlet temperature (300 K) before the second stage, so each compressor stage can be solved independently with the same inlet temperature — only the pressure ratio differs (3 for stage 1, 10/3 for stage 2). Intercooling lowers total compressor work below a single-stage compression to the same final pressure, and regeneration recovers turbine-exhaust heat to cut the fresh heat input needed from the combustor, both of which raise thermal efficiency above a simple (non-regenerative, non-intercooled) Brayton cycle at the same pressure ratio.

Common pitfall. Using the overall pressure ratio (10) for each compressor stage instead of the actual stage pressure ratios (3 for stage 1, 10/3 for stage 2, since the intercooler sits at 300 kPa, not at the geometric mean of 100 and 1000 kPa).

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

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

Question 2: Regenerative Gas Turbine, Two-Stage Intercooled Compressor (20 marks)

Approach