Steam at a pressure of 20 bar and 250°C enters a turbine and leaves it finally at a pressure of 0.05 bar. Steam is bled off at pressures of 5.0, 1.5 and 0.3 bar. Assuming (i) that the condensate is heated in each heater upto the saturation temperature of the steam in that heater, (ii) that the

Question

Steam at a pressure of 20 bar and 250°C enters a turbine and leaves it finally at a pressure of 0.05 bar. Steam is bled off at pressures of 5.0, 1.5 and 0.3 bar. Assuming (i) that the condensate is heated in each heater upto the saturation temperature of the steam in that heater, (ii) that the drain water from each heater is cascaded through a trap into the next heater on the low pressure side of it, (iii) that the combined drains from the heater operating at 0.3 bar are cooled in a drain cooler to condenser temperature, calculate the following :

(i) Mass of bled steam for each heater per kg of steam entering the turbine

(ii) Thermal efficiency of the cycle,

(iii) Thermal efficiency of the Rankine cycle

(iv) Theoretical gain due to regenerative feed heating,

(v) Steam consumption in kg/kWh with or without regenerative feed heating, and

(vi) Quantity of steam passing through the last stage nozzle of a 50000 kW turbine with and without regenerative feed heating.

Accepted Answer

Refer Fig. 12.20 (a), (b).

From Mallier Chart : [latex]h_0=2905[/latex] kJ/kg, [latex]h_1=2600[/latex] kJ/kg, [latex]h_2=2430[/latex] kJ/kg

[latex]h_3=2210[/latex] kJ/kg, [latex]h_4=2000[/latex] kJ/kg

From steam tables :

At 5 bar : [latex]h_{f_1}=640.1[/latex] kJ/kg

At 1.5 bar : [latex]h_{f_2}=467.1[/latex] kJ/kg

At 0.3 bar : [latex]h_{f_3}=289.3[/latex] kJ/kg

At 0.05 bar : [latex]h_{f_4}=137.8[/latex] kJ/kg.

(i) Mass of bled steam for each heater per kg of steam :

Using heat balance equation :

At heater No. 1 :

[latex]m_1 h_1+h_{f_2}=m_1 h_{f_1}+h_{f_1}[/latex]

∴ [latex]m _1=\frac{h_{f_1}-h_{f_2}}{h_1-h_{f_1}}=\frac{640.1-467.1}{2600-640.1}[/latex]

= 0.088 kJ/kg of entering steam.

At heater No. 2 :

[latex]m_2 h_2+h_{f_3}+m_1 h_{f_1}=h_{f_2}+\left(m_1+m_2 ight) h_{f_2}[/latex]

[latex]m _2=\frac{\left(h_{f_2}+h_{f_3} ight)-m_1\left(h_{f_1}-h_{f_2} ight)}{\left(h_2-h_{f_2} ight)}[/latex]

[latex]=\frac{(467.1-289.3)-0.088(640.1-467.1)}{(2430-467.1)}=\frac{162.57}{1962.9}[/latex]

= 0.0828 kJ/kg of entering steam.

At heater No. 3 :

[latex]m_3 h_3+h_{f_5}+\left(m_1+m_2 ight) h_{f_2}=h_{f_3}+\left(m_1+m_2+m_3 ight) h_{f_3}[/latex] ...(i)

At drain cooler :

[latex]\left(m_1+m_2+m_3 ight) h_{f_3}+h_{f_4}=h_{f_5}+\left(m_1+m_2+m_3 ight) h_{f_4}[/latex]

∴ [latex]h_{f_5}=\left(m_1+m_2+m_3 ight)\left(h_{f_3}-h_{f_4} ight)+h_{f_4}[/latex] ...(ii)

Inserting the value of [latex]h_{f_5}[/latex] in eqn. (i), we get

[latex]m_3 h_3+\left(m_1+m_2+m_3 ight)\left(h_{f_3}-h_{f_4} ight)+h_{f_4}+\left(m_1+m_2 ight) h_{f_2}=h_{f_3}+\left(m_1+m_2+m_3 ight) h_{f_3}[/latex]

∴ [latex]m _3=\frac{\left(h_{f_3}-h_{f_4} ight)-\left(m_1+m_2 ight)\left(h_{f_2}-h_{f_4} ight)}{h_3-h_{f_4}}[/latex]

[latex]=\frac{(289.3-137.8)-(0.088+0.0828)(467.1-137.8)}{(2210-137.8)}[/latex]

[latex]=\frac{151.5-56.24}{2072.2}= 0 . 0 4 6[/latex] kJ/kg of entering steam.

Work done/kg (neglecting pump work)

[latex]=\left(h_0-h_1 ight)+\left(1-m_1 ight)\left(h_1-h_2 ight)+\left(1-m_1-m_2 ight)\left(h_2-h_3 ight)+\left(1-m_1-m_2-m_3 ight)\left(h_3-h_4 ight)[/latex]

= (2905 – 2600) + (1 – 0.088) (2600 – 2430) + (1 – 0.088 – 0.0828) (2430 – 2210)

+ (1 – 0.088 – 0.0828 – 0.046) (2210 – 2000)

= 305 + 155.04 + 182.42 + 164.47 = 806.93 kJ/kg

Heat supplied/kg [latex]=h_0-h_{f_1}=2905-640.1=2264.9[/latex] kJ/kg.

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

[latex]\eta_{ ext {Thermal }}=\frac{ ext { Work done }}{ ext { Heat supplied }}=\frac{806.93}{2264.9}= 0 . 3 5 6 3 ext { or } 3 5 . 6 3 \%[/latex].

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

[latex]\eta_{ ext {Rankine }}=\frac{h_0-h_4}{h_0-h_{f_4}}=\frac{2905-2000}{2905-137.8}=0.327 ext { or } 32.7 \%[/latex].

(iv) Theoretical gain due to regenerative feed heating

[latex]=\frac{35.63-32.7}{35.63}=0.0822 ext { or } 8.22 \%[/latex].

(v) Steam consumption with regenerative feed heating

[latex]=\frac{1 imes 3600}{ ext { Work done } / kg }=\frac{1 imes 3600}{806.93}= 4 . 4 6[/latex] kg/kWh.

Steam consumption without regenerative feed heating

[latex]=\frac{1 imes 3600}{ ext { Work done } / kg ext { without regeneration }}=\frac{1 imes 3600}{h_0-h_4}[/latex]

= [latex]\frac{1 imes 3600}{2905-2000}= 3 . 9 7[/latex] kg/kWh.

(vi) Quantity of steam passing through the last stage of a 50000 kW turbine with regenerative feed-heating

[latex]=4.46\left(1-m_1-m_2-m_3 ight) imes 50000[/latex]

= 4.46 (1 – 0.088 – 0.0828 – 0.046) × 50000 = 174653.6 kg/h.

Same without regenerative arrangement

= 3.97 × 50000 = 198500 kg/h.

Question 12.16: Steam at a pressure of 20 bar and 250°C enters a turbine and...

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