Vapour Power Cycles
By: S K Mondal
Chapter 12
∴ Wnet = 931 kJ/kg Heat supplied (Q) = m(h1– h8) + (1 – m) (h1 – h6) = 2609.5 kJ/kg
931 × 100% = 35.68% with turbine exhaust quality 0.8186 2609.5 If No separation is taking place, Then is quality of exhaust is x ⇒ x = 0.715 Then 6.1837 = 0.423 + x × 8.052 ∴η=
∴
h4 = hf + x × hfg = 1862 kJ/kg
∴
WT = h1-h4 = 941.28 kJ/kg WP = WP5 − 6 = 3 kJ/kg
∴ Wnet = 938.28 kJ/kg ∴ Heat input, Q = h1 – h6 = 2677.8 kJ/kg 938.28 ∴ η= × 100% = 35% 2677.8
Q.12.17
Solution:
The net power output of an ideal regenerative-reheat steam cycle is 80MW. Steam enters the h.p. turbine at 80 bar, 500°C and expands till it becomes saturated vapour. Some of the steam then goes to an open feedwater heater and the balance is reheated to 400°C, after which it expands in the I.p. turbine to 0.07 bar. Compute (a) the reheat pressure, (b) the steam flow rate to the h.p. turbine, and (c) the cycle efficiency. Neglect pump work. (Ans. 6.5 bar, 58.4 kg/s, 43.7%) From S.T of 80 bar 500°C h1 = 3398.3 kJ/kg s1 = 6.724 kJ/kg – K s2 = 6.725 at 6.6 bar so Reheat pr. 6.6 bar
1 80 bar 500°C 3 400°C 8
T
7
6
m kg
(1 – m) kg
2
0.07 bar 5
(1 – m) kg
4
S ∴ h2 = 2759.5 kJ/kg h3 = 3270.3 + 0.6(3268.7 – 3270.3) = 3269.3 kJ/kg s3 = 7.708 + 0.6 (7.635 – 7.708)
= 7.6643 kJ/kg – K At 0.07 bar hf = 163.4,
hfg = 2409.1
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