A steam turbine plant developing 120 MW of electrical output is equipped with reheating and regenerative feed heating arrangement consisting of two feed heaters—one surface type on H.P. side and other direct contact type on L.P. side. The steam conditions before the steam stop valve are 100 bar and

Question

A steam turbine plant developing 120 MW of electrical output is equipped with reheating and regenerative feed heating arrangement consisting of two feed heaters—one surface type on H.P. side and other direct contact type on L.P. side. The steam conditions before the steam stop valve are 100 bar and 530°C. A pressure drop of 5 bar takes place due to throttling in valves.

Steam exhausts from the H.P. turbine at 25 bar. A small quantity of steam is bled off at 25 bar for H.P. surface heater for feed heating and the remaining is reheated in a reheater to 550°C and the steam enters at 22 bar in L.P. turbine for further expansion. Another small quantity of steam is bled off at pressure 6 bar for the L.P. heater and the rest of steam expands up to the back pressure of 0.05 bar. The drain from the H.P. heater is led to the L.P. heater and the combined feed from the L.P. heater is pumped to the high-pressure feed heater and finally to the boiler with the help of boiler feed pump.

The component efficiencies are : Turbine efficiency 85%, pump efficiency 90%, generator efficiency 96%, boiler efficiency 90% and mechanical efficiency 95%. It may be assumed that the feed-water is heated up to the saturation temperature at the prevailing pressure in feed heater. Work out the following :

(i) Sketch the feed heating system and show the process on T-s and h-s diagrams.

(ii) Amounts of steam bled off.

(iii) Overall thermal efficiency of turbo-alternator considering pump work.

(iv) Specific steam consumption in kg/kWh.

Accepted Answer

(i) The schematic arrangement including feed heating system, and T-s and h-s diagrams of the process are shown in Figs. 12.21 and 12.22 respectively.

(ii) Amounts of bled off. The enthalpies at various state points as read from h-s diagram/steam tables, in kJ/kg, are :

[latex]h_1=h_2=3460[/latex]

[latex]h_3{ }^{\prime}=3050, ext { and } \quad herefore \quad h_3=3460-0.85(3460-3050)=3111.5[/latex]

[latex]h_4=3585[/latex]

[latex]h_5^{\prime}=3140, ext { and } \quad herefore \quad h_5=3585-0.85(3585-3140)=3207[/latex]

[latex]h_6^{\prime}=2335, ext { and } \quad herefore \quad h_6=3207-0.85(3207-2335)=2466[/latex]

[latex]h_7=137.8[/latex] kJ/kg ([latex]h_f[/latex] at 0.05 bar)

[latex]h_8=h_{10}=962[/latex] 962 kJ/kg ([latex]h_f[/latex] at 25 bar)

and [latex]h_9=670.4[/latex] ([latex]h_f[/latex] at 6 bar).

Enthalpy balance for surface heater :

[latex]m_1 h_3+h_9=m_1 h_8+h_{10}[/latex], neglecting pump work

or [latex]m_1=\frac{h_{10}-h_9}{h_3-h_8}=\frac{962-670.4}{3111.5-962}=0.13566[/latex] kg

Enthalpy balance for contact heater :

[latex]m_2 h_5+\left(1-m_1-m_2 ight) h_7+m_1 h_8=h_9[/latex], neglecting pump work

or [latex]m_2 imes 3207+\left(1-0.13566-m_2 ight) imes 137.8+0.13566 imes 962=670.4[/latex]

or [latex]m_2=0.1371[/latex] kg.

Pump Work. Take specific volume of water as 0.001 m³/kg.

[latex]\left(W_{ ext {pump }} ight)_{ ext {L.P. }}=\left(1-m_1-m_2 ight)(6-0.05) imes 0.001 imes 10^2[/latex]

[latex]=(1-0.13566-0.1371) imes 5.95 imes 0.1=0.4327[/latex] kJ/kg.

[latex]\left(W_{ ext {pump }} ight)_{ ext {H.P. }}=1 imes(100-6) imes 0.001 imes 10^2=9.4[/latex] kJ/kg

Total pump work (actual) [latex]=\frac{0.4327+9.4}{0.9}=10.925[/latex] kJ/kg

Turbine output (indicated) [latex]=\left(h_2-h_3 ight)+\left(1-m_1 ight)\left(h_4-h_5 ight)+\left(1-m_1-m_2 ight)\left(h_5-h_6 ight)[/latex]

= (3460 – 3111.5) + (1 – 0.13566)(3585 – 3207)

+ (1 – 0.13566 – 0.1371)(3207 – 2466)

= 1214.105 kJ/kg

Net electrical output = (Indicated work – Pump work) [latex] imes \eta_{ ext {mech. }} imes \eta_{g e n}[/latex].

= (1214.105 – 10.925) × 0.9 × 0.96 = 1039.55 kJ/kg

[Note. All the above calculations are for 1 kg of main (boiler) flow.]

∴ Main steam flow rate [latex]=\frac{120 imes 10^3 imes 3600}{1039.55}=4.155 imes 10^5[/latex] kJ/h.

Amounts of bled off are :

(a) Surface (high pressure) heater,

= 0.13566 kg/kg of boiler flow

or [latex]=0.13566 imes 4.155 imes 10^5[/latex]

i.e., [latex]=5.6367 imes 10^4[/latex] kg/h.

(b) Direct contact (low pressure) heater

= 0.1371 kg/kg of boiler flow

or [latex]=0.1371 imes 4.155 imes 10^5[/latex]

i.e., [latex]=5.697 imes 10^4[/latex] kg/h.

(iii) Overall thermal efficiency, [latex]\eta_{ ext {overall }}[/latex] :

Heat input in boiler [latex]=\frac{h_1-h_{10}}{\eta_{ ext {boiler }}}=\frac{3460-962}{0.9}[/latex]

= 2775.6 kJ/kg of boiler flow.

Heat input in reheater [latex]=\frac{h_4-h_3}{\eta_{ ext {boiler }}}=\frac{3585-3111.5}{0.9}=526.1[/latex] kJ/kg of boiler flow

∴ [latex]\eta_{ ext {overall }}=\frac{1039.55}{2775.6+526.1} imes 100= 3 1 . 4 8 \%[/latex].

(iv) Specific steam consumption :

Specific steam consumption [latex]=\frac{4.155 imes 10^5}{120 imes 10^3}= 3 . 4 6 2 5[/latex] kg/kWh.

Question 11.27: A steam turbine plant developing 120 MW of electrical output...

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