Question 12.23: The following data relate to a regenerative steam power plan...
The following data relate to a regenerative steam power plant generating 22500 kW energy, the alternator directly coupled to steam turbine :
Condition of steam supplied to the steam turbine … 60 bar, 450°C
Condenser vacuum … 707.5 mm
Pressure at which steam is bled from the steam turbine … 3 bar
Turbine efficiency of each portion of expansion … 87 per cent
Boiler efficiency … 86 per cent
Alternator efficiency … 94 per cent
Mechanical efficiency from turbine to generator … 97 per cent
Neglecting the pump work in calculating the input to the boiler, determine :
(i) The steam bled per kg of steam supplied to the turbine.
(ii) The steam generated per hour if the 9 percent of the generator output is used to run the pumps.
(iii) The overall efficiency of the plant.
The schematic arrangement of the steam power plant is shown in Fig. 12.38 (a), while the conditions of the fluid passing through the components are represented on T-s and h-s diagrams as shown in Figs. 12.38 (b) and (c). The conditions of the fluid entering and leaving the pump are shown by the same point as the rise in temperature due to pump work is neglected.
Given : Power generated = 22500 kW ;
p_1=60 \text { bar } ; t_1=450^{\circ} C ; p_2\left(=p_2^{\prime}\right)=3 bar ;
p_3\left(=p_3{ }^{\prime}\right)=\frac{760-707.5}{760} \times 1.013=0.07 \text { bar } ; \eta_{\text {turbine (each portion) }}=87 \% ;
\eta_{\text {boiler }}=86 \% ; \eta_{\text {alt. }}=94 \%, \eta_{\text {mech. }}=97 \%• Locate point 1 corresponding to the conditions : p_1=60 \text { bar } ; t_1=450^{\circ} C on the h-s chart (Mollier chart).
From h-s chart ; we find : h_1=3300 kJ/kg.
• Draw vertical line through point 1 till it cuts the 3 bar pressure line, then locate point 2.
∴ h_2=2607 kJ/kg
Now, \eta_{\text {turbine }}=0.87=\frac{h_1-h_2^{\prime}}{h_1-h_2} \quad \text { or } \quad 0.87=\frac{3300-h_2^{\prime}}{3300-2607}
∴ h_2^{\prime}=2697 kJ/kg
• Locate the point 2 on the h-s chart as enthalpy and pressure are known and then draw a vertical line through the point 2 till it cuts the 0.07 bar pressure line and then locate the point 3.
∴ h_3=2165 kJ/kg
Again, \eta_{\text {turbine }}=0.87=\frac{h_2^{\prime}-h_3^{\prime}}{h_2{ }^{\prime}-h_3} \quad \text { or } \quad 0.87=\frac{2697-h_3{ }^{\prime}}{2697-2165}
∴ h_3^{\prime}=2234 kJ/kg
From steam tables, corresponding to pressures 3 bar and 0.02 bar, the saturated liquid heats at points 4 and 5 are :
h_{f 4}=163.4 kJ/kg ; h_{f 5}=561.4 kJ/kg.
(i) The steam bled per kg of steam supplied to the turbine, m :
Considering the energy balance for feed heater we have ;
m\left(h_2{ }^{\prime}-h_{f 5}\right)=(1-m)\left(h_{f 5}-h_{f 4}\right)or m(2697-561.4)=(1-m)(561.4-163.4)
or 2135.6 m=398(1-m)
∴ m = 0.157 kJ/kg of steam generated.
(ii) Steam generated per hour :
Work developed per kg of steam in the turbine
=1\left(h_1-h_2{ }^{\prime}\right)+(1-m)\left(h_2{ }^{\prime}-h_3{ }^{\prime}\right)= (3300 – 2697) + (1 – 0.157) (2697 – 2234) = 993.3 kJ/kg
Actual work developed by the turbine
=\frac{22500}{\eta_{\text {alt. }} \times \eta_{\text {mech. }}}=\frac{22500}{0.94 \times 0.97}=24676.5 kW
∴ Steam generated per hour = \frac{24676.5}{993.3} \times \frac{3600}{1000} tonnes/h = 89.43 tonnes/h.
(iii) The overall efficiency of the plant, \eta_{\text {overall }} :
Net power available deducting pump power
= 22500 (1 – 0.09) = 20475 kW
Heat supplied in the boiler =\frac{89.43 \times 1000\left(h_1-h_{f 5}\right)}{0.86} kJ/h
= \frac{89.43 \times 1000(3300-561.4)}{0.86 \times 3600} kW = 79106.3 kW
∴ \eta_{\text {overall }}=\frac{\text { Net power available }}{\text { Heat supplied by the boiler }}
= \frac{20475}{79106.3}=0.2588 \text { or } 2 5 . 8 8 \%.
