Question 2 of 9: Regenerative Rankine Cycle, One Open Feedwater Heater

Question 2: Regenerative Rankine Cycle, One Open 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-1 inlet: $P=10$ MPa, $T=500\ ^\circ$C. Extraction/FWH pressure $P_e=1$ MPa; condenser $P_{cond}=7.5$ kPa. Ideal cycle (isentropic turbine and pumps, no stated component efficiencies). Saturated liquid leaves the open FWH. $\dot m=95$ kg/s (entering turbine 1). Source $T_H=1500$ K, sink $T_0=300$ K.

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

Approach

Because the cycle is ideal, every turbine/pump leg is isentropic, so all states share $s_A$ except the two pump legs (which each carry their own inlet entropy forward unchanged). The open FWH's energy balance — extracted steam plus pumped condensate mixing to saturated liquid at the extraction pressure — fixes the extraction fraction $y$; everything else follows from mass- and energy-weighted sums over the split streams.

1. Turbine-1 inlet and isentropic expansion. At 10 MPa, 500°C: $h_A=3375.13$ kJ/kg, $s_A=6.5995$ kJ/kg·K. Following the isentrope to the extraction pressure (1 MPa, $T_{sat}=179.88\ ^\circ$C, still superheated) gives $h_{e1}=2783.67$ kJ/kg; continuing to the condenser pressure (7.5 kPa) gives $h_{e2}=2056.66$ kJ/kg at quality $x_{e2}=\boxed{0.7849}$.
2. Condenser and pump I (cond.out→pI.out), 7.5 kPa→1 MPa. Saturated liquid at 7.5 kPa: $h=168.75$ kJ/kg, $s=0.5763$ kJ/kg·K. Isentropic pump I to 1 MPa: $$w_{pI}=h_{pI,out}-h_{cond,out}=169.75-168.75=\boxed{1.00\ \text{kJ/kg}}.$$
3. Open feedwater heater energy balance. A fraction $y$ of the total flow arrives as extracted steam at $h_{e1}=2783.67$ kJ/kg; the remaining $(1-y)$ arrives as pumped condensate at $h_{pI,out}=169.75$ kJ/kg. The mixture leaves as saturated liquid at 1 MPa, $h_f=762.52$ kJ/kg: $$y\,h_{e1}+(1-y)\,h_{pI,out}=h_f \;\Rightarrow\; y=\frac{h_f-h_{pI,out}}{h_{e1}-h_{pI,out}}=\frac{762.52-169.75}{2783.67-169.75}=\boxed{0.2268}.$$
4. Pump II (FWH.out→pII.out), 1→10 MPa, full flow. Isentropic compression of the saturated liquid ($s_f=2.1381$ kJ/kg·K) to the boiler pressure: $$w_{pII}=h_{pII,out}-h_f=772.63-762.52=\boxed{10.12\ \text{kJ/kg}}.$$
5. Heat input rate (part a). The boiler heats the full flow from the pump-II exit to the turbine-1 inlet state: $$q_{in}=h_A-h_{pII,out}=3375.13-772.63=2602.49\ \text{kJ/kg}.$$ $$\dot Q_{in}=\dot m\,q_{in}=95\times2602.49=\boxed{247{,}237\ \text{kJ/s}}\ (247.2\ \text{MW}).$$
6. Net power output (part b). Turbine work is full flow through the first leg, then $(1-y)$ through the second; pump work is $(1-y)$ through pump I (only the condensate fraction) plus the full flow through pump II: $$w_{turb}=(h_A-h_{e1})+(1-y)(h_{e1}-h_{e2})=(3375.13-2783.67)+0.7732\times(2783.67-2056.66) =591.46+562.14=1153.60\ \text{kJ/kg}.$$ $$w_{pump}=(1-y)\,w_{pI}+w_{pII}=0.7732\times1.00+10.12=0.77+10.12=10.89\ \text{kJ/kg}.$$ $$w_{net}=w_{turb}-w_{pump}=1153.60-10.89=1142.71\ \text{kJ/kg}.$$ $$\dot W_{net}=\dot m\,w_{net}=95\times1142.71=\boxed{108{,}557\ \text{kW}}\ (108.6\ \text{MW}).$$
7. Thermal efficiency (part c). $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{1142.71}{2602.49}=\boxed{0.4391\ (43.9\%)}.$$
8. Second-law efficiency (part d). Same exergy-input construction as Q1, with the stated $T_H=1500$ K, $T_0=300$ K: $$\dot X_{in}=\dot Q_{in}\left(1-\frac{T_0}{T_H}\right)=247{,}237\times0.80=\boxed{197{,}790\ \text{kW}}.$$ $$\eta_{II}=\frac{\dot W_{net}}{\dot X_{in}}=\frac{108{,}557}{197{,}790}=\boxed{0.5489\ (54.9\%)}.$$