Question 1 of 9: Reheat Rankine Cycle

Question 1: Reheat Rankine 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. Turbine-1 inlet (state C): $P=8$ MPa, $T=480\ ^\circ$C. Reheat to 480 °C at 0.7 MPa (state E) before turbine-2, which expands to the condenser pressure of 8 kPa (states A/F). $\dot m=2.63\times10^5$ kg/h. $\eta_{t1}=\eta_{t2}=0.88$, $\eta_p=0.80$.

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

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 (main boiler and reheater) and the single condenser heat rejection, weighted by the given mass flow rate, gives the four required rates.

1. Condenser exit and actual pump. Saturated liquid at 8 kPa: $h_A=173.84$ kJ/kg, $s_A=0.5925$ kJ/kg·K ($T_{sat}=41.51\ ^\circ$C). Isentropic pump exit at 8 MPa gives $w_{p,s}=8.046$ kJ/kg; actual: $$w_{p,a}=\frac{w_{p,s}}{\eta_p}=\frac{8.046}{0.80}=\boxed{10.058\ \text{kJ/kg}},\qquad h_B=183.90\ \text{kJ/kg}.$$
2. Turbine 1 (C→D), 8→0.7 MPa. At 8 MPa, 480°C: $h_C=3349.65$ kJ/kg, $s_C=6.6613$ kJ/kg·K. Isentropic exit at 0.7 MPa gives $w_{t1,s}=607.0$ kJ/kg; actual: $$w_{t1,a}=\eta_{t1}\,w_{t1,s}=0.88\times607.0=\boxed{534.12\ \text{kJ/kg}},\qquad h_D=2815.52\ \text{kJ/kg}\ (T_D=186.93\ ^\circ\text{C, superheated}).$$
3. Reheat and turbine 2 (E→F), 0.7→0.008 MPa. At 0.7 MPa, 480°C: $h_E=3439.31$ kJ/kg, $s_E=7.8755$ kJ/kg·K. Isentropic exit at 8 kPa is wet ($h_{Fs}=2465.5$ kJ/kg, well inside the dome), giving $w_{t2,s}=973.79$ kJ/kg; actual: $$w_{t2,a}=\eta_{t2}\,w_{t2,s}=0.88\times973.79=\boxed{856.95\ \text{kJ/kg}},\qquad h_F=2582.36\ \text{kJ/kg}.$$ Checking the actual exit state: at 8 kPa the saturated-vapor enthalpy is $h_g=2576.21$ kJ/kg, and $h_F=2582.36>h_g$, so the 88%-efficient turbine, despite an isentropic exit deep in the wet region ($x_{Fs}\approx0.93$), lands the actual exit state just outside the dome — superheated at $T_F=44.70\ ^\circ$C, only $3.2\ ^\circ$C above $T_{sat}(8\ \text{kPa})=41.51\ ^\circ$C. This is a useful reminder that turbine inefficiency always moves the actual exit state toward higher enthalpy (up and to the right on the dome), which can flip a nominally-wet isentropic exit into an actually-dry one.
4. Heat input rate (part a). Two heat-addition legs: the boiler ($B\to C$) and the reheater ($D\to E$). $$q_{in}=(h_C-h_B)+(h_E-h_D)=(3349.65-183.90)+(3439.31-2815.52)=3165.75+623.79=3789.54\ \text{kJ/kg}.$$ $$\dot Q_{in}=\dot m\,q_{in}=73.056\times3789.54=\boxed{276{,}846\ \text{kW}}\ (276.8\ \text{MW}).$$
5. Heat rejected rate (part b). Single condenser leg, $F\to A$: $$q_{out}=h_F-h_A=2582.36-173.84=2408.52\ \text{kJ/kg}.$$ $$\dot Q_{out}=\dot m\,q_{out}=73.056\times2408.52=\boxed{175{,}956\ \text{kW}}\ (176.0\ \text{MW}).$$
6. Net power output (part c). $$w_{net}=w_{t1,a}+w_{t2,a}-w_{p,a}=534.12+856.95-10.06=1381.01\ \text{kJ/kg}.$$ $$\dot W_{net}=\dot m\,w_{net}=73.056\times1381.01=\boxed{100{,}890\ \text{kW}}\ (100.9\ \text{MW}).$$ Cross-check: $\dot Q_{in}-\dot Q_{out}=276{,}846-175{,}956=100{,}890$ kW $=\dot W_{net}$, confirming the overall energy balance.
7. Thermal efficiency (part d). $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{1381.01}{3789.54}=\boxed{0.3644\ (36.4\%)}.$$