Ideal Regenerative Rankine Cycle — Closed + Open Feedwater Heaters, with Second-Law Efficiency: 04-BS-10, May 2014, Q1

Question 1: Ideal Regenerative Rankine Cycle — Closed + Open Feedwater Heaters, with Second-Law Efficiency (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 8) $P=12.5$ MPa, $T=550\ ^\circ$C. First extraction (state 9) at $P_{e1}=0.8$ MPa to the closed FWH; second extraction (state 10) at $P_{e2}=0.3$ MPa to the open FWH; condenser (state 11 in, state 1 out) $P=10$ kPa. Closed-FWH feedwater exit (state 5): $170\ ^\circ$C, 12.5 MPa. Closed-FWH drain (state 6) leaves as saturated liquid at 0.8 MPa and is trapped (throttled) into the open FWH at state 7. Open-FWH exit (state 3): saturated liquid at 0.3 MPa. Ideal cycle: isentropic turbine stages and pumps. Net power $\dot W_{net}=150$ MW. Source $T_H=1500$ K, sink $T_L=300$ K, dead-state $T_0=300$ K.

State	Description	P	T	h (kJ/kg)	s (kJ/kg·K)
8	Turbine inlet	12.5 MPa	550.0°C	3476.51	6.6317
9	1st extraction (closed FWH)	0.8 MPa	170.4°C	2755.03	6.6317
10	2nd extraction (open FWH)	0.3 MPa	133.5°C	2578.51	6.6317
11	Turbine exit (to condenser)	10 kPa	45.8°C	2099.96	6.6317
1	Condenser exit, sat. liquid	10 kPa	45.8°C	191.81	0.6492
2	Pump I exit	0.3 MPa	45.8°C	192.10	0.6492
3	Open FWH exit, sat. liquid	0.3 MPa	133.5°C	561.43	1.6717
4	Pump II exit	12.5 MPa	134.7°C	574.48	1.6717
5	Closed FWH feedwater exit (given)	12.5 MPa	170.0°C	725.60	2.0270
6	Closed FWH drain, sat. liquid	0.8 MPa	170.4°C	720.86	2.0457
7	After trap, into open FWH	0.3 MPa	170.4°C	720.86	2.0457

Find. (a) $w_{net}$ [kJ/kg]; (b) $\dot m$ entering the first-stage turbine [kg/h]; (c) $\eta_{th}$; (d) $\eta_{II}$.

Fig. Q1 — T–s state points for the regenerative Rankine cycle with closed + open feedwater heaters (2→3 and 6→7 are mixing/throttling steps drawn as straight connectors for continuity rather than single-stream process lines).

Approach

Fix the turbine-inlet state and follow the isentrope through both extraction pressures and the condenser pressure to get states 9, 10 and 11. Fix the two feedwater-heater exit/drain states from the given temperatures and saturation conditions, pump each stream to its next pressure, then close the closed-FWH energy balance for the extraction fraction $y$ and the open-FWH mixing balance for $z$. Net specific work, mass flow rate, thermal efficiency and second-law efficiency follow directly.

Turbine-inlet state and isentropic expansion legs. State 8: $h_8=3476.51$ kJ/kg, $s_8=6.6317$ kJ/kg·K. Following $s=s_8$ down to each pressure: $h_9=2755.03$ kJ/kg (0.8 MPa), $h_{10}=2578.51$ kJ/kg (0.3 MPa), $h_{11}=2099.96$ kJ/kg (10 kPa).
Condenser exit and Pump I (1→2). Saturated liquid at 10 kPa: $h_1=191.81$ kJ/kg, $s_1=0.6492$ kJ/kg·K. Pumped isentropically to 0.3 MPa: $h_2=192.10$ kJ/kg.
Open FWH exit and Pump II (3→4). Saturated liquid at 0.3 MPa: $h_3=561.43$ kJ/kg, $s_3=1.6717$ kJ/kg·K. Pumped isentropically to 12.5 MPa: $h_4=574.48$ kJ/kg.
Closed-FWH states. Feedwater exit (state 5, given): $170\ ^\circ$C, 12.5 MPa ⇒ $h_5=725.60$ kJ/kg. Drain (state 6), saturated liquid at 0.8 MPa: $h_6=720.86$ kJ/kg; trapped (isenthalpic) into the open FWH, $h_7=h_6=720.86$ kJ/kg.
Closed-FWH energy balance ⇒ extraction fraction $y$. Tube-side feedwater ($\dot m=1$, normalized) rises from $h_4$ to $h_5$; the extraction-steam stream $y$ condenses from $h_9$ to $h_6$: $$y(h_9-h_6)=(h_5-h_4)\ \Rightarrow\ y=\frac{725.60-574.48}{2755.03-720.86}=\boxed{0.07429}.$$
Open-FWH mixing balance ⇒ extraction fraction $z$. The open FWH mixes extraction $z$ (at $h_{10}$), the trapped closed-FWH drain $y$ (at $h_7$), and the remaining $(1-y-z)$ from Pump I (at $h_2$), to leave as saturated liquid at $h_3$: $$z\,h_{10}+y\,h_7+(1-y-z)h_2=h_3$$ $$z=\frac{h_3-h_2+y(h_2-h_7)}{h_{10}-h_2}=\frac{561.43-192.10+0.07429\times(192.10-720.86)}{2578.51-192.10}=\boxed{0.13830}.$$
Turbine and pump work, net specific work (part a). $$w_{turb}=(h_8-h_9)+(1-y)(h_9-h_{10})+(1-y-z)(h_{10}-h_{11})=1261.71\ \text{kJ/kg}.$$ $$w_{pI}=(1-y-z)(h_2-h_1)=0.231\ \text{kJ/kg},\qquad w_{pII}=h_4-h_3=13.055\ \text{kJ/kg}.$$ $$w_{net}=w_{turb}-w_{pI}-w_{pII}=1261.71-0.23-13.06=\boxed{1248.42\ \text{kJ/kg}}.$$
Mass flow rate entering the first-stage turbine (part b). $$\dot m=\frac{\dot W_{net}}{w_{net}}=\frac{150{,}000\ \text{kW}}{1248.42\ \text{kJ/kg}}\times3600\ \text{s/h}=\boxed{432{,}547\ \text{kg/h}}.$$
Thermal efficiency (part c). Heat is added only in the boiler, $q_{in}=h_8-h_5$: $$q_{in}=3476.51-725.60=2750.92\ \text{kJ/kg},\qquad \eta_{th}=\frac{w_{net}}{q_{in}}=\frac{1248.42}{2750.92}=\boxed{0.4538\ (45.4\%)}.$$
Second-law (exergy) efficiency (part d). Modeling the heat addition as supplied from a source at $T_H=1500$ K, the exergy content of $q_{in}$ is $$x_{in}=q_{in}\left(1-\frac{T_0}{T_H}\right)=2750.92\times\left(1-\frac{300}{1500}\right)=2200.73\ \text{kJ/kg}.$$ $$\eta_{II}=\frac{w_{net}}{x_{in}}=\frac{1248.42}{2200.73}=\boxed{0.5673\ (56.7\%)}.$$

Quantity	Result
(a) $w_{net}$	1248.42 kJ/kg
(b) $\dot m$ (turbine inlet)	432,547 kg/h
(c) $\eta_{th}$	0.4538 (45.4%)
(d) $\eta_{II}$	0.5673 (56.7%)

Concept & Theory

Topic. Ideal regenerative Rankine cycle with one closed and one open feedwater heater in series, plus second-law (exergy) efficiency of a power cycle.

Key relations. Closed FWH energy balance (two separate streams, no mixing): $y(h_{extract}-h_{drain})=(h_{fw,out}-h_{fw,in})$. Open FWH mixing balance (single mixed stream): $\sum \dot m_i h_i = \dot m_{out}h_{out}$. Second-law efficiency of a heat engine: $\eta_{II}=w_{net}/[q_{in}(1-T_0/T_H)]=\eta_{th}/\eta_{Carnot}$.

Why this works. Extracting steam to preheat the feedwater raises the average temperature of heat addition in the boiler without raising peak pressure or temperature, which is exactly what regeneration is for; solving the closed heater's two-stream balance first (it has only one unknown, $y$) before the open heater's three-stream balance (which needs $y$ as an input) is the only balance order that avoids two simultaneous equations in two unknowns.

Common pitfall. Assuming the second-law efficiency equals $\eta_{th}$ divided by the sink-referenced Carnot efficiency $(1-T_L/T_H)$. The correct denominator uses the actual dead state $T_0$ (given here as 300 K, coincidentally equal to $T_L$), not the source/sink pair directly; in a problem where $T_0\ne T_L$ the two would differ.

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

Question 1 of 9: Ideal Regenerative Rankine Cycle — Closed + Open Feedwater Heaters, with Second-Law Efficiency

Question 1: Ideal Regenerative Rankine Cycle — Closed + Open Feedwater Heaters, with Second-Law Efficiency (20 marks)

Approach