Question 1 of 9: Regenerative Rankine Cycle, Closed + Open Feedwater Heaters

Question 1: 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-1 inlet (state A): $P=12$ MPa, $T=480\ ^\circ$C. First extraction at 2 MPa to the closed heater (state e1), second extraction at 0.3 MPa to the open heater (state e2), condenser at 6 kPa. Feedwater leaves the closed heater at $210\ ^\circ$C, 12 MPa; its drain leaves as saturated liquid at 2 MPa and is trapped (throttled) into the open heater. Saturated liquid at 0.3 MPa leaves the open heater. All pumps and turbine stages isentropic. Source $T_H=1000$ K, sink $T_0=288$ K.

Find. (a) $q_{in}$ [kJ/kg]; (b) $w_{net}$ [kJ/kg]; (c) $\eta_{th}$; (d) $\eta_{II}$.

Approach

Fix state A and both extraction states from the constant entropy of the turbine line, get the condenser-exit and heater-exit states from the given pressures/temperatures, then close two energy balances — the closed heater (feedwater vs. condensing extraction steam) and the open heater (mixing chamber) — to solve for the two extraction fractions $y$ (closed) and $z$ (open). Turbine work, pump work, and the boiler heat input then follow directly per kg of steam entering the first stage.

1. Closed feedwater heater energy balance (solve for $y$). The full feedwater flow is heated from $h_{pII,out}=573.95$ to $h_{fw,out}=901.29$ kJ/kg by the extracted steam ($y$ kg) condensing from $h_{e1}=2837.44$ to its saturated-liquid drain $h_{f,e1}=908.50$ kJ/kg: $$y\,(h_{e1}-h_{f,e1})=h_{fw,out}-h_{pII,out}\ \Rightarrow\ y=\frac{901.29-573.95}{2837.44-908.50}=\boxed{0.1697}.$$
2. Open feedwater heater (mixing chamber) energy balance (solve for $z$). The drain from the closed heater is throttled to 0.3 MPa (isenthalpic, still $h_{f,e1}=908.50$) and joins the pumped condensate ($1-y-z$ kg at $h_{pI,out}=151.77$) and the second extraction ($z$ kg at $h_{e2}=2491.85$) to leave as 1 kg of saturated liquid at $h_{open,out}=561.43$ kJ/kg: $$(1-y-z)h_{pI,out}+z\,h_{e2}+y\,h_{f,e1}=h_{open,out}$$ $$z=\frac{h_{open,out}-h_{pI,out}-y\,(h_{f,e1}-h_{pI,out})}{h_{e2}-h_{pI,out}} =\frac{561.43-151.77-0.1697(908.50-151.77)}{2491.85-151.77}=\boxed{0.1202}.$$
3. Turbine work per kg entering the first stage. Full flow expands A→e1, $(1-y)$ expands e1→e2, and $(1-y-z)$ continues e2→condenser: $$w_{turb}=(h_A-h_{e1})+(1-y)(h_{e1}-h_{e2})+(1-y-z)(h_{e2}-h_{turb,exit})$$ $$=(3295.29-2837.44)+0.8303(2837.44-2491.85)+0.7101(2491.85-1975.71) =457.85+287.20+366.36=\boxed{1111.31\ \text{kJ/kg}}.$$
4. Pump work per kg entering the first stage. Pump I moves only the condensate fraction $(1-y-z)$ from the condenser to 0.3 MPa; Pump II moves the full stream from 0.3 to 12 MPa: $$w_{pI}=(1-y-z)(h_{pI,out}-h_{cond,out})=0.7101(151.77-151.48)=0.21\ \text{kJ/kg}$$ $$w_{pII}=h_{pII,out}-h_{open,out}=573.95-561.43=12.52\ \text{kJ/kg}.$$
5. Net power output (part b) and boiler heat input (part a). $$w_{net}=w_{turb}-w_{pI}-w_{pII}=1111.31-0.21-12.52=\boxed{1098.57\ \text{kJ/kg}}$$ $$q_{in}=h_A-h_{fw,out}=3295.29-901.29=\boxed{2393.99\ \text{kJ/kg}}.$$
6. Thermal efficiency (part c). $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{1098.57}{2393.99}=\boxed{0.4589\ (45.9\%)}.$$
7. Second-law efficiency (part d). Treat $T_H=1000$ K as the source supplying $q_{in}$ and $T_0=288$ K as the sink; the Carnot ceiling between them bounds $\eta_{th}$: $$\eta_{Carnot}=1-\frac{T_0}{T_H}=1-\frac{288}{1000}=0.7120,\qquad \eta_{II}=\frac{\eta_{th}}{\eta_{Carnot}}=\frac{0.4589}{0.7120}=\boxed{0.6445\ (64.5\%)}.$$