Question 12.25: A steam power plant equipped with regenerative as well as re...
A steam power plant equipped with regenerative as well as reheat arrangement is supplied with steam to the H.P. turbine at 80 bar 470°C. For feed heating, a part of steam is extracted at 7 bar and remainder of the steam is reheated to 350°C in a reheater and then expanded in L.P. turbine down to 0.035 bar. Determine :
(i) Amount of steam bled-off for feed heating,
(ii) Amount of steam supplied to L.P. turbine,
(iii) Heat supplied in the boiler and reheater
(iv) Cycle efficiency, and
(v) Power developed by the system.
The steam supplied by the boiler is 50 kg/s.
The schematic arrangement is the steam power plant of shown in Fig. 12.40 (a) and the processes are represented on h-s diagram as shown in Fig. 12.40 (b).
From h-s chart and steam tables, we have enthalpies at different points as follows :
\left.\begin{array}{ll}h_1=3315 \mathrm{~kJ} / \mathrm{kg} ; & h_2=2716 \mathrm{~kJ} / \mathrm{kg} \\h_3=3165 \mathrm{~kJ} / \mathrm{kg} ; & h_4=2236 \mathrm{~kJ} / \mathrm{kg}\end{array}\right\} From h-s chart
h_{f 6}=h_{f 2}=697.1 kJ/kg ; h_{f 5}=h_{f 4}=101.9 101.9 kJ/kg } From steam table.
(i) Amount of steam bled off for feed heating :
Considering energy balance at regenerator, we have :
Heat lost by steam = Heat gained by water
m\left(h_2-h_{f 6}\right)=(1-m)\left(h_{f 6}-h_{f 5}\right)or m\left(h_2-h_{f 2}\right)=(1-m)\left(h_{f 2}-h_{f 4}\right) \left[\because \quad h_{f 6}=h_{f 2} ; h_{f 5}=h_{f 4}\right]
or m(2716 – 697.1) = (1 – m) (697.1 – 111.9)
or 2018.9 m = 585.2 (1 – m)
∴ m = 0.225 g of steam supplied
Hence amount of steam bled off is 22.5% of steam generated by the boiler.
(ii) Amount of steam supplied to L.P. turbine :
Amount of steam supplied to L.P. turbine
= 100 – 22.5
= 77.5% of the steam generated by the boiler.
(iii) Heat supplied in the boiler and reheater
Heat supplied in the boiler per kg of steam generated
=h_1-h_{f 6}=3315-697.1=2617.9 kJ/kg. \left(\because h_{f 6}=h_{f 2}\right)
Heat supplied in the reheater per kg of steam generated
=(1-m)\left(h_3-h_2\right)= (1 – 0.225) (3165 – 2716) = 347.97 kJ/kg.
Total amount of heat supplied by the boiler and reheater per kg of steam generated,
Q_s=2617.9+347.97=2965.87 kJ/kg
(iv) Cycle efficiency, \eta_{\text {cycle }} :
Amount of work done by per kg of steam generated by the boiler,
W=1\left(h_1-h_2\right)+(1-m)\left(h_3-h_4\right), Neglecting pump work
= (3315 – 2716) + (1 – 0.225) (3165 – 2236) \simeq 1319 kJ/kg
∴ \eta_{\text {cycle }}=\frac{W}{Q_s}=\frac{1319}{2965.87}=0.4447 or 44.47%
(v) Power developed by the system :
Power developed by the system
=m_s \times W=50 \times 1319 \mathrm{~kJ} / \mathrm{s}=\frac{50 \times 1319}{1000}= 65.95 MW
