Reheat Rankine Cycle with Second-Law Efficiency: 04-BS-10, May 2018, Q1

Question 1: Reheat Rankine Cycle with Second-Law Efficiency (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. Turbine-1 inlet (state C): $P=8$ MPa, $T=475\ ^\circ$C. Reheat to 475°C at 0.8 MPa (state E) before turbine-2, which expands to the condenser pressure of 7.5 kPa (states A/F). $\dot m=20$ kg/s. $\eta_{t1}=\eta_{t2}=0.88$, $\eta_p=0.80$. Source $T_H=1500$ K, sink $T_0=300$ K.

State	Description	P	h (kJ/kg)	s (kJ/kg·K)
A	Condenser exit, sat. liquid	7.5 kPa	168.75	0.5763
B	Pump exit (actual)	8 MPa	178.80	0.5827
C	Turbine-1 inlet	8 MPa, 475°C	3337.07	6.6445
D	Turbine-1 exit (actual)	0.8 MPa	2829.88	6.7967
E	Reheat exit / turbine-2 inlet	0.8 MPa, 475°C	3427.41	7.7984
F	Turbine-2 exit (actual)	7.5 kPa	2551.85	8.1793

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

Fig. Q1 — T–s state points for the reheat Rankine cycle (A→B actual pump, B→C boiler, C→D actual turbine 1, D→E reheater, E→F actual turbine 2, F→A condenser). State F lands just inside the saturation dome ($x_F=0.9908$) despite the reheat pushing state E well into the superheated region.

Approach

Fix the condenser-exit and both turbine-inlet states directly from the given pressures and temperatures, then run the pump and each turbine stage isentropically to get the ideal work, divide/multiply by the stated efficiency to get the actual work, and back out each actual exit enthalpy. Summing the two boiler-side heat additions and weighting by the given mass flow rate gives (a)–(c); part (d) treats the given source/sink temperatures as the exergy reference for a second-law efficiency.

Condenser exit and actual pump. Saturated liquid at 7.5 kPa: $h_A=168.75$ kJ/kg, $s_A=0.5763$ kJ/kg·K ($T_{sat}=40.29\ ^\circ$C). Isentropic pump exit at 8 MPa gives $w_{p,s}=8.043$ kJ/kg; actual: $$w_{p,a}=\frac{w_{p,s}}{\eta_p}=\frac{8.043}{0.80}=\boxed{10.05\ \text{kJ/kg}},\qquad h_B=178.80\ \text{kJ/kg}.$$
Turbine 1 (C→D), 8→0.8 MPa. At 8 MPa, 475°C: $h_C=3337.07$ kJ/kg, $s_C=6.6445$ kJ/kg·K. Isentropic exit at 0.8 MPa gives $w_{t1,s}=576.35$ kJ/kg; actual: $$w_{t1,a}=\eta_{t1}\,w_{t1,s}=0.88\times576.35=\boxed{507.19\ \text{kJ/kg}},\qquad h_D=2829.88\ \text{kJ/kg}\ (195.7\ ^\circ\text{C, superheated}).$$
Reheat and turbine 2 (E→F), 0.8→0.0075 MPa. At 0.8 MPa, 475°C: $h_E=3427.41$ kJ/kg, $s_E=7.7984$ kJ/kg·K. Isentropic exit at 7.5 kPa is wet, giving $w_{t2,s}=994.95$ kJ/kg; actual: $$w_{t2,a}=\eta_{t2}\,w_{t2,s}=0.88\times994.95=\boxed{875.56\ \text{kJ/kg}},\qquad h_F=2551.85\ \text{kJ/kg}\ (x_F=0.9908).$$
Heat input rate (part b). Two heat-addition legs: the boiler ($B\to C$) and the reheater ($D\to E$). $$q_{in}=(h_C-h_B)+(h_E-h_D)=(3337.07-178.80)+(3427.41-2829.88)=3158.27+597.53=3755.80\ \text{kJ/kg}.$$ $$\dot Q_{in}=\dot m\,q_{in}=20\times3755.80=\boxed{75{,}116\ \text{kJ/s}}\ (75.1\ \text{MW}).$$
Net power output (part a). $$w_{net}=w_{t1,a}+w_{t2,a}-w_{p,a}=507.19+875.56-10.05=1372.69\ \text{kJ/kg}.$$ $$\dot W_{net}=\dot m\,w_{net}=20\times1372.69=\boxed{27{,}454\ \text{kW}}\ (27.5\ \text{MW}).$$ Cross-check via condenser leg: $q_{out}=h_F-h_A=2551.85-168.75=2383.11$ kJ/kg, $\dot Q_{out}=20\times2383.11=47{,}662$ kW, and $\dot Q_{in}-\dot Q_{out}=75{,}116-47{,}662=27{,}454$ kW $=\dot W_{net}$ ✓.
Thermal efficiency (part c). $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{1372.69}{3755.80}=\boxed{0.3655\ (36.6\%)}.$$
Second-law efficiency (part d). Treat the source temperature as the reservoir supplying $\dot Q_{in}$ and the sink temperature as the dead-state reference $T_0$: the maximum (exergy) work obtainable from that heat rate is $\dot X_{in}=\dot Q_{in}(1-T_0/T_H)$. $$\dot X_{in}=75{,}116\times\left(1-\frac{300}{1500}\right)=75{,}116\times0.80=\boxed{60{,}093\ \text{kW}}.$$ $$\eta_{II}=\frac{\dot W_{net}}{\dot X_{in}}=\frac{27{,}454}{60{,}093}=\boxed{0.4569\ (45.7\%)}.$$

Quantity	Result
(a) $\dot W_{net}$	27,454 kW (27.5 MW)
(b) $\dot Q_{in}$	75,116 kJ/s (75.1 MW)
(c) $\eta_{th}$	0.3655 (36.6%)
(d) $\eta_{II}$	0.4569 (45.7%)

Concept & Theory

Topic. Non-ideal reheat Rankine cycle sized from a directly-given mass flow rate, with a whole-cycle second-law (exergetic) efficiency evaluated against stated source/sink temperatures.

Key relations. Actual component work from isentropic efficiency: $w_{t,a}=\eta_t w_{t,s}$, $w_{p,a}=w_{p,s}/\eta_p$. Exergy supplied by a heat-input rate drawn from a reservoir at $T_H$, referenced to a sink/dead-state at $T_0$: $\dot X_{in}=\dot Q_{in}(1-T_0/T_H)$; second-law efficiency $\eta_{II}=\dot W_{net}/\dot X_{in}$.

Why this works. $\eta_{II}$ measures how close the actual net power comes to the theoretical maximum obtainable from the same heat-input rate if it were all supplied reversibly at the stated source temperature and rejected at the stated sink temperature — it is bounded above by the Carnot ceiling between $T_H$ and $T_0$ ($1-300/1500=80\%$ here), and the gap between $\eta_{th}$ (36.6%) and that ceiling is why $\eta_{II}$ (45.7%) is materially higher than $\eta_{th}$ itself.

Common pitfall. Confusing the source/sink temperatures given for the second-law-efficiency calculation with the cycle's own boiler/condenser saturation temperatures — $T_H=1500$ K and $T_0=300$ K are external reservoir temperatures for the exergy accounting, not state-point temperatures on the T–s diagram.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 10 (Vapor Power Cycles) & Ch. 8 (Exergy).

Question 1 of 9: Reheat Rankine Cycle with Second-Law Efficiency

Question 1: Reheat Rankine Cycle with Second-Law Efficiency (20 marks)

Approach