A steam power plant of 110 MW capacity is equipped with regenerative as well as reheat arrangement. The steam is supplied at 80 bar and 55°C of superheat. The steam is extracted at 7 bar for feed heating and remaining steam is reheated to 350°C, and then expanded to 0.4 bar in the L.P. stage.

Question

A steam power plant of 110 MW capacity is equipped with regenerative as well as reheat arrangement. The steam is supplied at 80 bar and 55°C of superheat. The steam is extracted at 7 bar for feed heating and remaining steam is reheated to 350°C, and then expanded to 0.4 bar in the L.P. stage.Assume indirect type of feed heaters. Determine :

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

(ii) The boiler generating capacity in tonnes of steam/hour, and

(iii) Thermal efficiency of the cycle.

Assume no losses and ideal processes of expansion.

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

(ii) The boiler generating capacity in tonnes of steam/hour, and

(iii) Thermal efficiency of the cycle.

Assume no losses and ideal processes of expansion.

Accepted Answer

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 ;

[latex]\left.t_1=350^{\circ} C ext { i.e., } t_s ext { at } 80 ext { bar } \simeq 295^{\circ} C +55^{\circ} C =350^{\circ} C ight)[/latex]

[latex]p_2=p_3=7 ext { bar } ; t_3=350^{\circ} C ; p_4=0.4 ext { bar }[/latex]

• Locate point 1 corresponding to the condition [latex]p_1=80[/latex] bar and [latex]t_1=350^{\circ} C[/latex], 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 :

[latex]h_1=2985[/latex] kJ/kg ; [latex]h_2=2520[/latex] kJ/kg ;

[latex]h_3=3170[/latex] kJ/kg ; [latex]h_4=2555[/latex] kJ/kg.

Also, from steam tables, we have :

[latex]h_{f 2}[/latex] (at 7 bar) = 697.1 kJ/kg ; [latex]h_{f 4}[/latex] (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

[latex]m\left(h_2-h_{f 2} ight)=(1-m)\left(h_{f 2}-h_{f 4} ight)[/latex]

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

∴ [latex]\frac{ ext { Steam generated }}{ ext { Steam bled }}=\frac{1}{0.172}=5.814[/latex].

(ii) The boiler generating capacity :

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

[latex]m_s\left(h_1-h_2 ight)+m_s(1-m)\left(h_3-h_4 ight)=110 imes 10^3[/latex]

or, [latex]m_s(2985-2520)+m_s(1-0.172)(3170-2555)=110 imes 10^3[/latex]

or, [latex]m_s(465+509.22)=110 imes 10^3[/latex]

∴ [latex]m_s=112.91[/latex] kg/s or 406.48 tonnes/hour

(iii) Thermal efficiency of the cycle, [latex]\eta_{ ext {thermal }}[/latex] :

[latex]\eta_{ ext {thermal }}=\frac{ ext { Output } / kg ext { of steam }}{ ext { Input } / kg ext { of steam }}=\frac{\left(h_1-h_2 ight)+(1-m)\left(h_3-h_4 ight)}{\left(h_1-h_{f_2} ight)+(1-m)\left(h_3-h_2 ight)}[/latex]

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

= [latex]\frac{974.22}{2826.1}=0.3447[/latex] or 34.47%.

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

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