Rankine Cycle with Non-Isentropic Turbine and Pump: 04-BS-10, December 2015, Q1

Question 1: Rankine Cycle with Non-Isentropic Turbine and Pump (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 inlet (state 3): $P=8$ MPa, $T=500\ ^\circ$C. Condenser pressure (states 1/4) $=10$ kPa. $\eta_t=0.85$, $\eta_p=0.70$. Net power $\dot W_{net}=105$ kW. Dead state $T_0=288$ K, $p_0=100$ kPa. Heat source $T_H=1000$ K, sink $T_L=288$ K. Cooling water enters at 15°C, exits at 35°C.

State	Description	P	h (kJ/kg)	s (kJ/kg·K)
1	Condenser exit, sat. liquid	10 kPa	191.81	0.6492
2	Pump exit (actual)	8 MPa	203.32	0.6600
3	Turbine inlet	8 MPa, 500°C	3399.49	6.7266
4	Turbine exit (actual)	10 kPa	2320.62	7.3235

Find. (a) $\dot m$ [kg/s]; (b) $\dot Q_{boiler}$ [kW]; (c) $\eta_{th}$; (d) $\dot m_{cw}$ [kg/h]; (e) $\eta_{II}$; (f) $\dot X_{dest}$ for pump, boiler, turbine, condenser [kJ/s].

Fig. Q1 — T–s state points for the simple Rankine cycle (1→2 actual pump, 2→3 boiler, 3→4 actual turbine, 4→1 condenser). States 1 and 4 sit on the 10 kPa isobar ($T=45.8\ ^\circ$C); state 4 is wet ($x_4=0.890$).

Approach

Fix the condenser-exit and turbine-inlet states directly from the given pressures/temperature, then run each of the pump and turbine isentropically to get the ideal work, divide/multiply by the stated efficiency to get the actual work, and back out the actual exit enthalpy. Net specific work then fixes the mass flow rate from the given total net power; boiler duty, thermal efficiency, cooling-water flow, and the exergy/second-law figures all follow from the four state points.

Condenser exit and actual pump. Saturated liquid at 10 kPa: $h_1=191.81$ kJ/kg, $s_1=0.6492$ kJ/kg·K ($T_{sat}=45.8\ ^\circ$C). Isentropic pump exit at 8 MPa gives $w_{p,s}=8.058$ kJ/kg; actual: $$w_{p,a}=\frac{w_{p,s}}{\eta_p}=\frac{8.058}{0.70}=\boxed{11.512\ \text{kJ/kg}},\qquad h_2=203.32\ \text{kJ/kg},\ \ s_2=0.6600\ \text{kJ/kg}\cdot\text{K}.$$
Turbine inlet and actual turbine. At 8 MPa, 500°C: $h_3=3399.49$ kJ/kg, $s_3=6.7266$ kJ/kg·K. Isentropic exit at 10 kPa gives $w_{t,s}=1269.26$ kJ/kg; actual: $$w_{t,a}=\eta_t\,w_{t,s}=0.85\times1269.26=\boxed{1078.87\ \text{kJ/kg}},\qquad h_4=2320.62\ \text{kJ/kg}\ (x_4=0.890),\ \ s_4=7.3235\ \text{kJ/kg}\cdot\text{K}.$$
Net specific work and mass flow rate (part a). $$w_{net}=w_{t,a}-w_{p,a}=1078.87-11.512=1067.36\ \text{kJ/kg}.$$ $$\dot m=\frac{\dot W_{net}}{w_{net}}=\frac{105}{1067.36}=\boxed{0.09837\ \text{kg/s}}.$$
Boiler heat-transfer rate (part b) and thermal efficiency (part c). $$q_{in}=h_3-h_2=3399.49-203.32=3196.17\ \text{kJ/kg}.$$ $$\dot Q_{boiler}=\dot m\,q_{in}=0.09837\times3196.17=\boxed{314.42\ \text{kW}}.$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{1067.36}{3196.17}=\boxed{0.3339\ (33.4\%)}.$$
Cooling-water flow rate (part d). Condenser duty per kg steam: $q_{out}=h_4-h_1=2320.62-191.81=2128.81$ kJ/kg, so $\dot Q_{out}=\dot m\,q_{out}=209.42$ kW. With $c_{p,w}=4.181$ kJ/kg·K (liquid water near 25°C): $$\dot m_{cw}=\frac{\dot Q_{out}}{c_{p,w}(35-15)}=\frac{209.42}{4.181\times20}=2.504\ \text{kg/s} =\boxed{9015\ \text{kg/h}}.$$
Second-law efficiency (part e). Carnot efficiency between the stated source and sink: $\eta_{rev}=1-T_L/T_H=1-288/1000=0.712$. $$\eta_{II}=\frac{\eta_{th}}{\eta_{rev}}=\frac{0.3339}{0.712}=\boxed{0.4690\ (46.9\%)}.$$
Exergy destruction by process (part f). Pump and turbine are adiabatic, so all their irreversibility appears as $\Delta s$; the boiler draws heat from the source at $T_H$ and the condenser rejects heat to the sink at $T_L$: $$\dot X_{pump}=\dot m\,T_0(s_2-s_1)=0.09837\times288\times(0.6600-0.6492)=\boxed{0.306\ \text{kJ/s}}$$ $$\dot X_{boiler}=\dot m\,T_0\!\left[(s_3-s_2)-\frac{q_{in}}{T_H}\right] =0.09837\times288\times\left[(6.7266-0.6600)-\frac{3196.17}{1000}\right]=\boxed{81.32\ \text{kJ/s}}$$ $$\dot X_{turb}=\dot m\,T_0(s_4-s_3)=0.09837\times288\times(7.3235-6.7266)=\boxed{16.91\ \text{kJ/s}}$$ $$\dot X_{cond}=\dot m\,T_0\!\left[(s_1-s_4)+\frac{q_{out}}{T_L}\right] =0.09837\times288\times\left[(0.6492-7.3235)+\frac{2128.81}{288}\right]=\boxed{20.33\ \text{kJ/s}}$$ All four terms are positive; their sum (118.87 kJ/s) matches the whole-cycle exergy balance $\dot m\,q_{in}(1-T_0/T_H)-\dot W_{net}$ to within rounding, a useful cross-check.

Quantity	Result
(a) $\dot m$	0.09837 kg/s
(b) $\dot Q_{boiler}$	314.42 kW
(c) $\eta_{th}$	0.3339 (33.4%)
(d) $\dot m_{cw}$	9015 kg/h
(e) $\eta_{II}$	0.4690 (46.9%)
(f) $\dot X_{dest}$: pump	0.306 kJ/s
(f) $\dot X_{dest}$: boiler	81.32 kJ/s
(f) $\dot X_{dest}$: turbine	16.91 kJ/s
(f) $\dot X_{dest}$: condenser	20.33 kJ/s

Concept & Theory

Topic. Simple (non-regenerative, non-reheat) Rankine cycle with a non-isentropic pump AND turbine, sized from a given total net power, with a full process-by-process exergy breakdown against separately stated source/sink temperatures.

Key relations. Actual component work from isentropic work and efficiency: $w_{t,a}=\eta_t w_{t,s}$, $w_{p,a}=w_{p,s}/\eta_p$. Exergy destruction of a heat-transfer process crossing a finite-temperature boundary: $\dot X_{dest}=\dot m\,T_0\!\left[\Delta s\mp q/T_{boundary}\right]$ (minus for heat entering, plus for heat leaving).

Why this works. Because the mass flow rate is not given directly, it must be back-solved from the given total net power divided by the specific net work — every other extensive quantity (boiler duty, cooling-water flow, exergy rates) is then just that mass flow rate times the appropriate specific quantity. Second-law efficiency compares the actual thermal efficiency to the Carnot efficiency of a hypothetical reversible engine operating between the stated source and sink, not between the cycle's own boiler/condenser temperatures.

Common pitfall. Assuming an ideal (isentropic) cycle has zero exergy destruction in its turbine and pump — that shortcut only holds for a fully ideal cycle; here both components are explicitly non-isentropic (85%/70%), so they each generate entropy and destroy exergy just like the heat-transfer processes.

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

Question 1 of 9: Rankine Cycle with Non-Isentropic Turbine and Pump

Question 1: Rankine Cycle with Non-Isentropic Turbine and Pump (20 marks)

Approach