Question 10.9: A steam power plant operates on ideal Rankine cycle using re......
A steam power plant operates on ideal Rankine cycle using reheater and regenerative feed water heaters. It has one open feed heater. Steam is supplied at 150 bar and 600°C. The condenser pressure is 0.1 bar. Some steam is extracted from the turbine at 40 bar for closed feed water heater and remaining steam is reduced at 40 bar to 600°C. Extracted steam is completely condensed in this closed feed water heater and is pumped to 150 bar before mixing with the feed water heater. Steam for the open feed water heater is bled from L.P. turbine at 5 bar. Determine:
(a) Fraction of steam extracted from the turbines at each bled heater, and
(b) Thermal efficiency of the system. (AMIE Summer, 1999)
The arrangement of the components and the processes are represented on H-S diagram as shown in Figures 10.23 a and b.
From H-S chart and steam tables we have enthalpies at different points as follows:
H_1 = 3,578 kJ/kg; H_2 = 3,140 kJ/kg;
H_3 = 3,678 kJ/kg; H_4 = 3,000 kJ/kg;
H_5 = 2,330 kJ/kg
H_{f1} (at 150 bar) = 1,611 kJ/kg
H_{f2} (at 40 bar) = 1,087.4 kJ/kg; H_{f4} (at 5 bar) = 640.1 kJ/kg
H_{f5} = H_{f6} (at 0.1 bar) = 191.8 kJ/kg; Steam tables
(a) Fraction of steam extracted from the turbines at each bled heater m_1, m_2:
Considering energy balance for closed feed heater, we have:
m_1(H_2 – H_{f2}) = (1 – m_1) (H_{f2} – H_{f4})
m_1 (3,140 – 1,087.4) = (1 – m_1)(1,087.4 – 640.1)
or 2,052.6 m_1 = (1 – m_1) × 447.3
m_1 = 0.179 kg/kg of steam supplied by the boiler.
Considering energy balance for open feed heater, we have:
m_2(H_4 – H_{f4}) = (1 – m_1 – m_2)(H_{f4} – H_{f6})
or m_2(H_4 – H_{f4}) = (1 – m_1 – m_2)(H_{f4} – H_{f5}) (H_{f6} = H_{f5})
or m_2(3000 – 640.1) = (1 – 0.179 – m_2) (640.1 – 191.8)
or 2359.9 m_2 = (0.821 – m_2) × 448.3 = 368.05 – 448.3 m_2
m_2 = 0.131 kg/kg of steam supplied by boiler.
(b) Thermal efficiency of the system, \eta _{therma}:
Total work done per kg of steam supplied by the boiler
= 1 × (H_1 – H_2) + (1 – m_1) (H_3 – H_4) + (1 – m_1 – m_2) (H_4 – H_5)
= (3578 – 3140) + (1 – 0.179) (3678 – 3000) + (1 – 0.179 – 0.131)(3000 – 2330)
= 438 + 556.64 + 462.3 = 1456.94 kJ/kg
Work done by the pump P_1
W_{P1} = V_{W1} (1 – m_1 – m_2)(5 – 0.1) × 10^5 × 10^{–3} kJ/kg
= 0.001 × (1 – 0.179 – 0.131)(5 – 0.1) × 10^5 × 10^{–3} = 0.338 kJ/kg
Work done by the pump P2,
W_{P2} = V_{W2} (1 – m_1)(150 – 5) × 10^5 × 10^{–3} kJ/kg = 11.9 kJ/kg
Work done by pump P_3,
W_{P3} = V_{W3} × m_1 × (150 – 40) × 10^5 × 10^{–3} = 1.97 kJ/kg
Total pump work = W_{P1} + W_{P2} + W_{P3} = 0.338 + 11.9 + 1.97 = 14.21 kJ/kg of steam supplied by boiler
Net work done by the turbine per kg of steam supplied by the boiler,
W_{net} = 1,456.94 – 14.21 = 1,442.73 kJ/kg
Heat of feed water entering the boiler = (1 – m_1) × 1,611 + m_1 × 1,611 = 1,611 kJ/kg
Heat supplied by the boiler per kg of steam,
QS_1 = H_1 – 1,610 = 3,578 – 1,610 = 1,968 kJ/kg
QS_2 = Heat supplied in the reheater = (1 – m_1) (H_3 – H_2) = (1 – 0.179)(3,678 – 3,140)
= 441.7 kJ/kg of steam supplied by the boiler
QS_T (Total heat supplied) = QS_1 + QS_2 = 1,968 + 441.7 = 2,409.7 kJ/kg
\eta _{thermal} = \frac{W_{net}}{QS_t} = 0.5987 or 59.87%.
