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

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

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

h_3=2210 kJ/kg, h_4=2000 kJ/kg

From steam tables :

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

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

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

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

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

Using heat balance equation :

At heater No. 1 :

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

= 0.088 kJ/kg of entering steam.

At heater No. 2 :

m _2=\frac{\left(h_{f_2}+h_{f_3}\right)-m_1\left(h_{f_1}-h_{f_2}\right)}{\left(h_2-h_{f_2}\right)}

=\frac{(467.1-289.3)-0.088(640.1-467.1)}{(2430-467.1)}=\frac{162.57}{1962.9}

= 0.0828 kJ/kg of entering steam.

At heater No. 3 :

m_3 h_3+h_{f_5}+\left(m_1+m_2\right) h_{f_2}=h_{f_3}+\left(m_1+m_2+m_3\right) h_{f_3} …(i)

At drain cooler :

∴ h_{f_5}=\left(m_1+m_2+m_3\right)\left(h_{f_3}-h_{f_4}\right)+h_{f_4} …(ii)

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

∴ m _3=\frac{\left(h_{f_3}-h_{f_4}\right)-\left(m_1+m_2\right)\left(h_{f_2}-h_{f_4}\right)}{h_3-h_{f_4}}

=\frac{(289.3-137.8)-(0.088+0.0828)(467.1-137.8)}{(2210-137.8)}

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

Work done/kg (neglecting pump work)

= (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 =h_0-h_{f_1}=2905-640.1=2264.9 kJ/kg.

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

\eta_{\text {Thermal }}=\frac{\text { Work done }}{\text { Heat supplied }}=\frac{806.93}{2264.9}= 0 . 3 5 6 3 \text { or } 3 5 . 6 3 \%.

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

\eta_{\text {Rankine }}=\frac{h_0-h_4}{h_0-h_{f_4}}=\frac{2905-2000}{2905-137.8}=0.327 \text { or } 32.7 \%.

(iv) Theoretical gain due to regenerative feed heating

=\frac{35.63-32.7}{35.63}=0.0822 \text { or } 8.22 \%.

(v) Steam consumption with regenerative feed heating

=\frac{1 \times 3600}{\text { Work done } / kg }=\frac{1 \times 3600}{806.93}= 4 . 4 6 kg/kWh.

Steam consumption without regenerative feed heating

= \frac{1 \times 3600}{2905-2000}= 3 . 9 7 kg/kWh.

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

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

Same without regenerative arrangement

= 3.97 × 50000 = 198500 kg/h.