Properties of Pure Substances
By: S K Mondal • m 1 = 3 kg/min 1 10 bar 250°C
Chapter 9
3
4
1
5 5 bar
1 10 bar x2 = 0.75 • = 5 kg/min 1 m
3
4
p3 = 10 bar • = 8 kg/min m 3 t3 = 180°c
2
h3 = 762.6 + x 3 × 2013.6 or x 3 = 0.87474
(b)
h 4 = 2524 kJ/kg = 640.1 + x 4 × 2107.4 x 4 = 0.89395 s4 = 1.8604 + x 4 × 4.9588 = 6.2933 kJ/kg – K s5 = s4
∴
5
t4 = 151.8°c
(a)
∴
2 bar
at 2 bar quality of steam 6.2933 = 1.5301 + x 5 × 5.5967 x 5 = 0.851
∴
(c)
h5 = 504.7 + 0.851 × 2201.6 = 2378.4 kJ/kg s3 = sf + x 3 s fg = 2.1382 + 0.89395 × 4.4446 = 6.111451 kJ/kg Δs = s4 – s3 = 6.2933 – 6.11145 = 0.18185 kJ/kg – K
(d)
V=
(e)
A= ⇒
Q9.22
Solution:
2000(2524 − 2378.4) = 540 m/s
mv V m × x 5 . 0.885 8 0.885 2 = × 0.851 × m = 1.86 cm2 V 60 540
Steam of 65 bar, 400°C leaves the boiler to enter a steam turbine fitted with a throttle governor. At a reduced load, as the governor takes action, the pressure of steam is reduced to 59 bar by throttling before it is admitted to the turbine. Evaluate the availabilities of steam before and after the throttling process and the irreversibility due to it. (Ans. I = 21 kJ/kg) From Steam Table h1 = 3167.65 kJ/kg h 2 = 3167.65 kJ/kg s1 = 6.4945 kJ/kg-K t 2 = 396.6º C
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