Throttled reheat cycle with 50 bar steam

Data: boiler 50 bar, 450 C; throttled to 30 bar; HP stage 30 bar to 3 bar, exit dry saturated; reheat at 3 bar to 300 C; LP stage 3 bar to 0.06 bar, exit dry saturated. Values from steam tables (IAPWS-IF97).

At 3 bar: $h_f = 561.46$, $h_{fg} = 2163.44$, $s_f = 1.6718$, $s_{fg} = 5.3198$. At 0.06 bar: $h_{fg} = 2415.17$, $s_{fg} = 7.8083$, $v_f = 0.001006$.

 1 (50 bar) --throttle--> 1' (30 bar) --HP stage-->
 2 (3 bar) --reheat--> 3 (3 bar) --LP stage-->
 4 (0.06 bar) --condenser--> 5 --pump--> boiler

HP stage efficiency

Throttling is the governing device of the HP stage, so the isentropic reference is taken from the stop-valve state 1 (50 bar, 450 C) to 3 bar:

$$\begin{aligned} x_{2s} &= \frac{6.8208 - 1.6718}{5.3198} = 0.9679 \\ h_{2s} &= 561.46 + 0.9679 \times 2163.44 = 2655.4\ \text{kJ/kg} \\ \eta_{HP} &= \frac{h_1 - h_2}{h_1 - h_{2s}} = \frac{3317.0 - 2724.9}{3317.0 - 2655.4} = \frac{592.1}{661.6} = 0.895 \end{aligned}$$

Note: if the reference is taken from the throttled state 1' ($s = 7.0468 > s_g$ at 3 bar), the isentropic end point is superheated with $h = 2747.6$ kJ/kg, which is higher than the given actual exit (2724.9). That would give an efficiency above 100%, so with these data the stage efficiency must be based on the stop-valve state (it then includes the throttling loss).

LP stage efficiency

$$\begin{aligned} x_{4s} &= \frac{7.7037 - 0.5209}{7.8083} = 0.9199 \\ h_{4s} &= 151.49 + 0.9199 \times 2415.17 = 2373.2\ \text{kJ/kg} \\ \eta_{LP} &= \frac{h_3 - h_4}{h_3 - h_{4s}} = \frac{3069.6 - 2566.7}{3069.6 - 2373.2} = \frac{502.9}{696.4} = 0.722 \end{aligned}$$

Overall cycle efficiency

$$\begin{aligned} W_{HP} &= 3317.0 - 2724.9 = 592.1\ \text{kJ/kg} \\ W_{LP} &= 3069.6 - 2566.7 = 502.9\ \text{kJ/kg} \\ W_p &= 0.001006 (5000 - 6) = 5.03\ \text{kJ/kg} \\ Q_s &= (h_1 - h_5 - W_p) + (h_3 - h_2) \\ &= (3317.0 - 151.49 - 5.03) + (3069.6 - 2724.9) \\ &= 3160.5 + 344.7 = 3505.2\ \text{kJ/kg} \\ \eta &= \frac{592.1 + 502.9 - 5.03}{3505.2} = \frac{1090.0}{3505.2} = 0.311 \end{aligned}$$

(Neglecting pump work: $\eta = 1095.0 / 3510.3 = 0.312$.)

Work ratio HP : LP

$$\frac{W_{HP}}{W_{LP}} = \frac{592.1}{502.9} = 1.177$$

Answer: $\eta_{HP} \approx 89.5\%$, $\eta_{LP} \approx 72.2\%$, overall $\eta \approx 31.1\%$, $W_{HP}/W_{LP} \approx 1.18$.