Question 1 of 9: Ideal Reheat Rankine Cycle with Full Exergy Breakdown

Question 1: Ideal Reheat Rankine Cycle with Full Exergy Breakdown (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. Ideal (isentropic turbines and pump) reheat Rankine cycle, water. $P_1=8$ MPa, $T_1=500\ ^\circ$C (HP turbine inlet); expands to $P_2=2$ MPa; reheated to $T_3=500\ ^\circ$C at 2 MPa; LP turbine expands to $P_{cond}=7.5$ kPa. $\dot m=2630$ kg/h$=0.7306$ kg/s. $T_0=300$ K, source $T_H=1500$ K, sink $T_{sink}=300$ K.

Find. (a) $\dot Q_{in}$; (b) $\dot Q_{out}$; (c) $\dot W_{net}$; (d) $\eta_{th}$; (e) $\dot X_{dest}$ by process; (f) $\eta_{II}$.

Approach

Fix all six states from the given pressures/temperatures and the isentropic-turbine/pump condition, then convert specific quantities to rates using $\dot m=0.7306$ kg/s. Because the cycle is ideal, the two turbines and the pump are individually isentropic and therefore destroy zero exergy; every process irreversibility is concentrated in the three finite-temperature-difference heat-transfer processes (boiler, reheater, condenser).

1. Heat input rate (part a). $q_{boiler}=h_1-h_6=3399.49-176.79=3222.70$ kJ/kg; $q_{reheat}=h_3-h_2=3468.24-3000.42=467.82$ kJ/kg; $q_{in}=3690.52$ kJ/kg. $$\dot Q_{in}=\dot m\,q_{in}=0.7306\times3690.52=\boxed{2696.1\ \text{kW}}.$$
2. Heat rejected rate (part b). $q_{out}=h_4-h_5=2318.15-168.75=2149.40$ kJ/kg. $$\dot Q_{out}=\dot m\,q_{out}=0.7306\times2149.40=\boxed{1570.3\ \text{kW}}.$$
3. Net power output (part c). $w_{turb}=(h_1-h_2)+(h_3-h_4)=399.07+1150.09=1549.16$ kJ/kg; $w_{pump}=h_6-h_5=8.04$ kJ/kg; $w_{net}=1541.12$ kJ/kg. $$\dot W_{net}=\dot m\,w_{net}=0.7306\times1541.12=\boxed{1125.9\ \text{kW}}\quad(=\dot Q_{in}-\dot Q_{out}\ \checkmark).$$
4. Thermal efficiency (part d). $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{1541.12}{3690.52}=\boxed{0.4176\ (41.8\%)}.$$
5. Exergy destruction by process (part e). Turbines and pump are isentropic $\Rightarrow\dot X_{dest,turb}=\dot X_{dest,pump}=0$. For the heat-transfer processes, $\dot X_{dest}=T_0\dot m\big[\Delta s-q/T_{reservoir}\big]$ (boiler/reheat draw from the source at $T_H=1500$ K; the condenser rejects to the sink at $T_{sink}=300$ K): $$\dot X_{dest,boiler}=300\times0.7306\times\left[(s_1-s_6)-\frac{q_{boiler}}{1500}\right] =300\times0.7306\times[6.1503-2.1485]=\boxed{877.1\ \text{kW}}.$$ $$\dot X_{dest,reheat}=300\times0.7306\times\left[(s_3-s_2)-\frac{q_{reheat}}{1500}\right] =300\times0.7306\times[0.7071-0.3119]=\boxed{86.6\ \text{kW}}.$$ $$\dot X_{dest,cond}=300\times0.7306\times\left[(s_5-s_4)+\frac{q_{out}}{300}\right] =300\times0.7306\times[-6.8574+7.1647]=\boxed{67.3\ \text{kW}}.$$ Total: $877.1+86.6+67.3+0+0=\boxed{1031.0\ \text{kW}}$.
6. Second-law efficiency (part f). Exergy supplied by the heat source: $$\dot X_{in}=\dot Q_{in}\left(1-\frac{T_0}{T_H}\right)=2696.1\times\left(1-\frac{300}{1500}\right)=2156.9\ \text{kW}.$$ $$\eta_{II}=\frac{\dot W_{net}}{\dot X_{in}}=\frac{1125.9}{2156.9}=\boxed{0.5220\ (52.2\%)} \quad(\text{check: }\dot W_{net}+\dot X_{dest,total}=1125.9+1031.0=2156.9=\dot X_{in}\ \checkmark).$$