Regenerative Rankine Cycle with One Closed Feedwater Heater: 04-BS-10, May 2015, Q1

Question 1: Regenerative Rankine Cycle with One Closed Feedwater Heater (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 1): $P=120$ bar, $T=520\ ^\circ$C. Extraction pressure (state 2) $=10$ bar. Condenser pressure (states 3/4) $=0.06$ bar. Feedwater heater (FWH) feedwater exit (state 6): $120$ bar, $170\ ^\circ$C. FWH condensate exit (state 7): saturated liquid at $10$ bar, throttled through a trap to the condenser. Isentropic turbine stages and pump. Mass flow into the first-stage turbine $\dot m_1=10^6$ kg/h $=277.78$ kg/s. Source $T_H=1500$ K, sink $T_L=300$ K, $T_0=300$ K.

State	Description	P	T	h (kJ/kg)	s (kJ/kg·K)
1	Turbine 1 inlet	120 bar	520.0°C	3403.39	6.5585
2	Extraction (isentropic)	10 bar	179.9°C	2765.09	6.5585
3	Turbine 2 exit (isentropic)	0.06 bar	36.2°C	2018.98	6.5585
4	Condenser exit, sat. liquid	0.06 bar	36.2°C	151.48	0.5208
5	Pump exit (isentropic)	120 bar	—	163.52	0.5208
6	FWH feedwater exit	120 bar	170.0°C	725.31	2.0276
7	FWH condensate exit, sat. liquid	10 bar	179.9°C	762.52	2.1381

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

Fig. Q1 — T–s state points for the regenerative Rankine cycle (1→2→3 single isentropic turbine expansion with extraction at 2, 4 condenser exit, 5 pump exit, 6 FWH feedwater exit).

Approach

Fix the turbine-inlet state and follow the single isentrope down through the extraction pressure to the condenser pressure (states 1→2→3), since extraction only splits the flow without changing its entropy. Fix the condenser-exit and (isentropic) pump-exit states, then close the closed-FWH energy balance to solve for the extraction fraction $y$ — the entire feedwater flow passes through the heater's tubes (state 5→6), while only the fraction $y$ of extracted steam condenses around them (state 2→7) before being trapped into the condenser. Net work, heat input, and the requested exergy/second-law quantities follow directly.

Turbine-inlet and extraction states. At 120 bar, 520°C: $h_1=3403.39$ kJ/kg, $s_1=6.5585$ kJ/kg·K. Following $s=s_1$ down to 10 bar: $h_2=2765.09$ kJ/kg ($T_2=179.9\ ^\circ$C). Continuing the same isentrope to the condenser pressure (0.06 bar): $h_3=2018.98$ kJ/kg.
Condenser exit and (isentropic) pump. Saturated liquid at 0.06 bar: $h_4=151.48$ kJ/kg, $s_4=0.5208$ kJ/kg·K. Isentropic pump exit at 120 bar: $$w_p=h_5-h_4=163.52-151.48=\boxed{12.04\ \text{kJ/kg}}.$$
Closed FWH: solve for the extraction fraction $y$. All feedwater ($\dot m=1$ basis) passes through the heater tubes (5→6); the extracted fraction $y$ condenses around them and exits as saturated liquid at 10 bar (state 7). Energy balance $y(h_2-h_7)=1\cdot(h_6-h_5)$: $$y=\frac{h_6-h_5}{h_2-h_7}=\frac{725.31-163.52}{2765.09-762.52}=\boxed{0.28054}.$$
Turbine work and net specific work. The full flow expands 1→2, then only the un-extracted fraction $(1-y)$ continues 2→3: $$w_t=(h_1-h_2)+(1-y)(h_2-h_3)=638.30+(1-0.28054)(746.11)=1175.10\ \text{kJ/kg}.$$ $$w_{net}=w_t-w_p=1175.10-12.04=\boxed{1163.06\ \text{kJ/kg}}.$$
Boiler heat input and thermal efficiency (part a). The boiler heats the full feedwater flow from the FWH exit (state 6) to the turbine inlet (state 1): $$q_{in}=h_1-h_6=3403.39-725.31=\boxed{2678.07\ \text{kJ/kg}}.$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{1163.06}{2678.07}=\boxed{0.4343\ (43.4\%)}.$$
Net power output (part b). With $\dot m_1=277.78$ kg/s into the first-stage turbine: $$\dot W_{net}=\dot m_1\,w_{net}=277.78\times1163.06=\boxed{323{,}071\ \text{kW}\ (323.1\ \text{MW})}.$$
Exergy destruction in the boiler (part c). Modeling the boiler/superheat heat addition (state 6→1) as supplied from a source at $T_H=1500$ K, with the full feedwater flow $\dot m_1$ passing through the boiler: $$\dot X_{dest,boiler}=\dot m_1\,T_0\!\left[(s_1-s_6)-\frac{q_{in}}{T_H}\right] =277.78\times300\times\left[(6.5585-2.0276)-\frac{2678.07}{1500}\right]=\boxed{228{,}788\ \text{kJ/s}}.$$
Second-law efficiency of the cycle (part d). The exergy supplied to the working fluid by the boiler heat input, referenced to $T_0=300$ K: $$x_{in}=q_{in}\left(1-\frac{T_0}{T_H}\right)=2678.07\times\left(1-\frac{300}{1500}\right)=2142.46\ \text{kJ/kg}.$$ $$\eta_{II}=\frac{w_{net}}{x_{in}}=\frac{1163.06}{2142.46}=\boxed{0.5429\ (54.3\%)}.$$

Quantity	Result
(a) $\eta_{th}$	0.4343 (43.4%)
(b) $\dot W_{net}$	323,071 kW (323.1 MW)
(c) $\dot X_{dest,boiler}$	228,788 kJ/s
(d) $\eta_{II}$	0.5429 (54.3%)

Concept & Theory

Topic. Regenerative Rankine cycle with a single closed feedwater heater (FWH) and a trap-drained condensate line, plus boiler-side exergy destruction and overall second-law efficiency.

Key relations. Closed-FWH energy balance: $y(h_{extract}-h_{drain})=1\cdot (h_{fw,out}-h_{fw,in})$, since the feedwater and the extraction steam never mix (separated by tube walls). Exergy destruction of a heat-transfer process from a finite-temperature source: $\dot X_{dest}=\dot m\,T_0[(s_{out}-s_{in})-q/T_H]$. Second-law efficiency: $\eta_{II}=w_{net}/[q_{in}(1-T_0/T_H)]$.

Why this works. Because the closed FWH keeps the two streams separate, the feedwater pump only needs to raise the condensate to boiler pressure once (a single pump, unlike an open FWH's two-pump arrangement); the extraction steam's own condensate is simply trapped (throttled) down to condenser pressure rather than pumped, since it never needs to reach boiler pressure.

Common pitfall. Assuming the full flow reaches the boiler at the FWH's exit enthalpy directly without checking that the FWH-side energy balance actually determines a physical extraction fraction $y\in(0,1)$ first — here $y=0.281$, a plausible extraction fraction; a value outside $(0,1)$ would signal a modeling error (e.g., swapped which stream is basis 1 vs. $y$).

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

Question 1 of 9: Regenerative Rankine Cycle with One Closed Feedwater Heater

Question 1: Regenerative Rankine Cycle with One Closed Feedwater Heater (20 marks)

Approach