Question 12.24: A steam power plant of 110 MW capacity is equipped with rege...

The schematic arrangement of the plant is shown in Fig. 12.39 (a) and the processes are represented on h-s chart in Fig. 12.39 (b).

Given : Capacity of plant = 110 MW ;

• Locate point 1 corresponding to the condition p_1=80 bar and t_1=350^{\circ} C, on the h-s chart.

• Locate point 2 by drawing vertical line through point 1 till it cuts the 7 bar pressure line.

• Locate point 3 as the cross point of 7 bar and 350°C temperature line.

• Locate point 4 by drawing vertical line through the point 3 till it cuts the 0.4 bar pressure line.

From h-s chart, we find :

h_1=2985 kJ/kg ; h_2=2520 kJ/kg ;

h_3=3170 kJ/kg ; h_4=2555 kJ/kg.

Also, from steam tables, we have :

h_{f 2} (at 7 bar) = 697.1 kJ/kg ;  h_{f 4} (at 0.4 bar) = 317.7 kJ/kg.

(i) The ratio of steam bled to steam generated :

Consider energy/heat balance of feed heater :

Heat lost by m kg of steam = Heat gained by (1 – m) kg of condensed steam

m(2520 – 697.1) = (1 – m) (697.1 – 317.7)

1822.9 m = (1 – m) × 379.4

∴ m = 0.172 kg

i.e. Amount of steam bled per kg of steam supplied to the turbine =

0.172 kg

∴ \frac{\text { Steam generated }}{\text { Steam bled }}=\frac{1}{0.172}=5.814.

(ii) The boiler generating capacity :

If m_s is the mass of steam supplied to the power plant per second, then the work developed is given by :

or, m_s(2985-2520)+m_s(1-0.172)(3170-2555)=110 \times 10^3

or, m_s(465+509.22)=110 \times 10^3

∴ m_s=112.91 kg/s or 406.48 tonnes/hour

(iii) Thermal efficiency of the cycle, \eta_{\text {thermal }} :

= \frac{(2985-2520)+(1-0.172)(3170-2555)}{(2985-697.1)+(1-0.172)(3170-2520)}

= \frac{974.22}{2826.1}=0.3447 or 34.47%.