Ideal Regenerative Rankine Cycle — Closed + Open Feedwater Heaters: 04-BS-10, December 2013, Q2

Question 2: Ideal Regenerative Rankine Cycle — Closed + Open Feedwater Heaters (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 6/A) $P=12.5$ MPa, $T=550\ ^\circ$C. First extraction at $P_{e1}=0.8$ MPa (to closed FWH); second extraction at $P_{e2}=0.3$ MPa (to open FWH); condenser $P=10$ kPa. Closed-FWH feedwater exit: $170\ ^\circ$C, 12.5 MPa. Closed-FWH drain leaves as saturated liquid at 0.8 MPa and is trapped (throttled) into the open FWH. Open-FWH exit: saturated liquid at 0.3 MPa. Ideal cycle: isentropic turbine stages and pumps. Net power $\dot W_{net}=150$ MW. (Normalizing the total flow entering the turbine to 1 kg, $y$ = fraction extracted at 0.8 MPa, $z$ = fraction extracted at 0.3 MPa; the remaining $1-y-z$ reaches the condenser, and all fractions reunite at the open FWH before Pump II sends the full flow through the closed FWH to the boiler.)

State	Description	P, T / x
A (=6)	Turbine inlet	12.5 MPa, 550°C
1	Condenser exit, sat. liquid	10 kPa
2	Pump I exit	0.3 MPa
3	Open FWH exit, sat. liquid	0.3 MPa
4	Pump II exit	12.5 MPa
5	Closed FWH exit / boiler inlet	12.5 MPa, 170°C
7	First extraction (isentropic from A)	0.8 MPa
8	Second extraction (isentropic from A)	0.3 MPa
9	Turbine exit / condenser inlet	10 kPa

Find. (a) $w_{net}$ (kJ/kg); (b) $\dot m$ (kg/h); (c) $\eta_{th}$.

Fig. Q2 — T–s path of the main feedwater loop (1→2 Pump I · 2→3 open FWH mixing · 3→4 Pump II · 4→5 closed FWH · 5→6 boiler · 6→7→8→9 isentropic turbine expansion, constant $s$, with extractions at 7 and 8 · 9→1 condenser).

Approach

Fix the turbine-inlet state and take all three turbine exit/extraction states off the same isentrope (ideal turbine). Compute the two pump exits, then solve the closed FWH's energy balance for extraction fraction $y$ (the only unknown in a single heat exchanger with known inlet/outlet temperatures), then the open FWH's mixing energy balance for $z$. Net turbine work, pump work, heat input and mass flow rate follow directly.

Turbine inlet and isentropic expansion line. State A: $h_A=3476.51$ kJ/kg, $s_A=6.6317$ kJ/kg·K. Holding $s=s_A$: at 0.8 MPa, $h_7=2755.03$ kJ/kg ($T_7=170.4\ ^\circ$C); at 0.3 MPa, $h_8=2578.51$ kJ/kg ($T_8=133.5\ ^\circ$C); at 10 kPa, $h_9=2099.96$ kJ/kg (quality $x_9=0.798$).
Condenser exit and Pump I (1→2). $h_1=191.81$ kJ/kg, $s_1=0.6493$ kJ/kg·K (sat. liquid, 10 kPa). Isentropic pump to 0.3 MPa: $h_2=192.10$ kJ/kg (negligible pump-I work, $w_{p,I}=0.29$ kJ/kg).
Open FWH exit and Pump II (3→4). $h_3=561.43$ kJ/kg, $s_3=1.6717$ kJ/kg·K (sat. liquid, 0.3 MPa). Isentropic pump to 12.5 MPa: $h_4=574.48$ kJ/kg ($w_{p,II}=13.06$ kJ/kg).
Closed FWH energy balance ⇒ extraction fraction $y$. The full feedwater flow (fraction 1) is heated from $h_4$ to the stated exit $h_5=725.60$ kJ/kg (170°C, 12.5 MPa) by extraction steam $y$ condensing from $h_7$ to saturated liquid at 0.8 MPa, $h_{f,0.8}=720.86$ kJ/kg: $$y=\frac{h_5-h_4}{h_7-h_{f,0.8}}=\frac{725.60-574.48}{2755.03-720.86}=\boxed{0.0743}.$$
Open FWH mixing balance ⇒ extraction fraction $z$. Three streams enter the open FWH at 0.3 MPa — the main condensate $(1-y-z)$ at $h_2$, the second extraction $z$ at $h_8$, and the closed-FWH drain $y$ throttled (isenthalpically) from $h_{f,0.8}$ — and the combined flow (fraction 1) leaves as saturated liquid $h_3$: $$(1-y-z)h_2+z\,h_8+y\,h_{f,0.8}=h_3$$ $$z=\frac{h_3-h_2+y(h_2-h_{f,0.8})}{h_8-h_2}=\boxed{0.1383}.$$
Turbine work, pump work, net work (part a). $$w_{turb}=(h_A-h_7)+(1-y)(h_7-h_8)+(1-y-z)(h_8-h_9)=1261.71\ \text{kJ/kg}.$$ $$w_{pumps}=(1-y-z)(h_2-h_1)+(h_4-h_3)=0.23+13.06=13.29\ \text{kJ/kg}.$$ $$w_{net}=w_{turb}-w_{pumps}=1261.71-13.29=\boxed{1248.42\ \text{kJ/kg}}.$$
Mass flow rate (part b). $$\dot m=\frac{\dot W_{net}}{w_{net}}=\frac{150{,}000\ \text{kW}}{1248.42\ \text{kJ/kg}}=120.15\ \text{kg/s}=\boxed{432{,}547\ \text{kg/h}}.$$
Heat input and thermal efficiency (part c). The boiler receives feedwater at $h_5$ (post closed-FWH) and heats it to $h_A$: $$q_{in}=h_A-h_5=3476.51-725.60=2750.92\ \text{kJ/kg}.$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{1248.42}{2750.92}=\boxed{45.38\%}.$$

Quantity	Result
(a) $w_{net}$	1248.42 kJ/kg
(b) $\dot m$	432,547 kg/h (120.15 kg/s)
(c) $\eta_{th}$	45.38%

Concept & Theory

Topic. Ideal regenerative Rankine cycle with one closed and one open feedwater heater in series.

Key relations. Closed-FWH energy balance (no mixing, two separate streams): $y=(h_{fw,out}-h_{fw,in})/(h_{extr}-h_{f,drain})$. Open-FWH energy balance (direct-contact mixing, all streams at the heater pressure): $\sum(\text{fraction}\times h_{in})=1\times h_{out}$. A trap throttles the closed-FWH condensate into the open FWH isenthalpically (no temperature/enthalpy change, only pressure drops).

Why this works. Extracting steam at two pressures and using it to preheat the feedwater raises the average temperature at which heat is added in the boiler without adding any heat-rejection penalty, which is exactly what regeneration is for; the closed heater lets the feedwater reach a higher temperature (170°C, above the 133.5°C saturation temperature at the open heater's 0.3 MPa) without mixing steam directly into the high-pressure feed stream.

Common pitfall. Solving for $z$ before $y$: the open-FWH balance already contains the closed-FWH drain term $y\,h_{f,0.8}$, so $y$ must be found first from the closed heater alone. Also, "net power output, in kJ/kg" in part (a) is specific net work, not power — the actual power (150 MW) is used only to back out the mass flow rate in part (b).

Source. Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 8th ed., Ch. 8 (Vapor Power Systems, regenerative cycles with FWHs).

Question 2 of 9: Ideal Regenerative Rankine Cycle — Closed + Open Feedwater Heaters

Question 2: Ideal Regenerative Rankine Cycle — Closed + Open Feedwater Heaters (20 marks)

Approach