Ideal Reheat Rankine Cycle, Exergy Destruction by Process: 04-BS-10, Undated paper, Q1

Question 1: Ideal Reheat Rankine Cycle, Exergy Destruction by Process (Part A – 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. Turbine-1 inlet (state 1): $P_1=8$ MPa, $T_1=500\ ^\circ\text{C}$. Reheat pressure $P_2=2$ MPa. Turbine-2 inlet (state 3): $P_3=2$ MPa, $T_3=500\ ^\circ\text{C}$. Condenser $P_4=7.5$ kPa. $\dot m=2630$ kg/h$=0.7306$ kg/s. Source $T_H=1500$ K, sink/dead-state $T_0=300$ K.

State	Description	P	T	h (kJ/kg)	s (kJ/kg·K)
1	Turbine-1 inlet	8 MPa	500.0°C	3399.49	6.7266
2	Turbine-1 exit / reheat inlet	2 MPa	289.83°C	3000.42	6.7266
3	Turbine-2 inlet	2 MPa	500.0°C	3468.24	7.4337
4	Turbine-2 exit ($x=0.8936$)	7.5 kPa	40.29°C	2318.15	7.4337
5	Condenser exit, sat. liquid	7.5 kPa	40.29°C	168.75	0.5763
6	Pump exit	8 MPa	—	176.79	0.5763

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

Fig. Q1 — T–s state points for the reheat Rankine cycle (schematic dome). $6\to1$ boiler, $1\to2$ turbine 1, $2\to3$ reheat, $3\to4$ turbine 2, $4\to5$ condenser, $5\to6$ pump.

Approach

Fix all six states (IAPWS water), then apply first-law energy balances around the boiler/reheater (heat in), condenser (heat out), and the two turbines less the pump (net work); since every device is ideal (isentropic), only the three heat-transfer processes destroy exergy.

Heat input rate (part a). Heat is added in two steps: the main boiler $6\to1$ and the reheater $2\to3$. $$q_{in}=(h_1-h_6)+(h_3-h_2)=(3399.49-176.79)+(3468.24-3000.42)=3690.52\ \text{kJ/kg}.$$ $$\dot Q_{in}=\dot m\,q_{in}=0.7306\times3690.52=\boxed{2696.1\ \text{kW}}.$$
Heat rejected rate (part b). $q_{out}=h_4-h_5=2318.15-168.75=2149.40$ kJ/kg. $$\dot Q_{out}=\dot m\,q_{out}=0.7306\times2149.40=\boxed{1570.3\ \text{kW}}.$$
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. $$\dot W_{net}=\dot m\,w_{net}=0.7306\times1541.12=\boxed{1125.9\ \text{kW}}\quad(=\dot Q_{in}-\dot Q_{out}\ \checkmark).$$
Thermal efficiency (part d). $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{1541.12}{3690.52}=\boxed{0.4176\ (41.8\%)}.$$
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, $\dot X_{dest}=\dot m\,T_0\,\Delta s_{gen}$: $$\dot X_{dest,boiler}=T_0\dot m\left[(s_1-s_6)-\frac{q_{boiler}}{T_H}\right] =300\times0.7306\times\left[(6.7266-0.5763)-\frac{3222.70}{1500}\right]=\boxed{877.1\ \text{kW}}.$$ $$\dot X_{dest,reheat}=T_0\dot m\left[(s_3-s_1)-\frac{q_{reheat}}{T_H}\right] =300\times0.7306\times\left[(7.4337-6.7266)-\frac{467.82}{1500}\right]=\boxed{86.6\ \text{kW}}.$$ $$\dot X_{dest,cond}=T_0\dot m\left[(s_5-s_3)+\frac{q_{out}}{T_0}\right] =300\times0.7306\times\left[(0.5763-7.4337)+\frac{2149.40}{300}\right]=\boxed{67.3\ \text{kW}}.$$ Total: $877.1+86.6+67.3+0+0=\boxed{1031.0\ \text{kW}}$.
Second-law efficiency (part f). Exergy supplied by the heat source: $$\dot X_{in}=\dot Q_{in}\left(1-\frac{T_0}{T_H}\right)=2696.1\times\left(1-\frac{300}{1500}\right)=2156.9\ \text{kW}.$$ $$\eta_{II}=\frac{\dot W_{net}}{\dot X_{in}}=\frac{1125.9}{2156.9}=\boxed{0.5220\ (52.2\%)} \quad(\text{check: }\dot W_{net}+\dot X_{dest,total}=1125.9+1031.0=2156.9=\dot X_{in}\ \checkmark).$$

Quantity	Value
(a) $\dot Q_{in}$	2696.1 kW
(b) $\dot Q_{out}$	1570.3 kW
(c) $\dot W_{net}$	1125.9 kW
(d) $\eta_{th}$	41.76%
(e) $\dot X_{dest}$: boiler / reheat / condenser / turbines / pump	877.1 / 86.6 / 67.3 / 0 / 0 kW (total 1031.0 kW)
(f) $\eta_{II}$	52.20%

Topic. Ideal reheat Rankine cycle — energy and exergy (2nd-law) analysis.

Key relations. $\eta_{th}=w_{net}/q_{in}$; $\dot X_{dest}=T_0\dot S_{gen}$ with $\dot S_{gen}=\dot m\Delta s-\dot Q/T_{reservoir}$ for each control volume; $\eta_{II}=w_{net}/x_{in}$ with $x_{in}=q_{in}(1-T_0/T_H)$.

Why this works. Reheat raises the average temperature of heat addition and keeps the turbine-2 exit quality high (here $x_4=0.894$), both of which raise $\eta_{th}$ relative to a simple Rankine cycle at the same boiler/condenser pressures. Because every device here is modelled as internally reversible, ALL of the cycle's irreversibility is concentrated in the three finite- temperature-difference heat-transfer steps — this is exactly what the exergy balance in part (e) isolates.

Common pitfall. Forgetting that the reheat step is a SECOND, separate heat-input process with its own (smaller) exergy destruction — lumping it into the boiler term understates the process count and mis-splits the 1031.0 kW total.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., §10-5 (reheat Rankine cycle) and §8-7 (exergy analysis of a cycle).

Question 1 of 9: Ideal Reheat Rankine Cycle, Exergy Destruction by Process

Question 1: Ideal Reheat Rankine Cycle, Exergy Destruction by Process (Part A – 20 marks)

Approach