Chemical Engineering - Chemical Engineering Thermodynamics - Discussion
Discussion Forum : Chemical Engineering Thermodynamics - Section 9 (Q.No. 41)
41.
Steam undergoes isentropic expansion in a turbine from 5000 kPa and 400°C (entropy = 6.65 kJ/kg K) to 150 kPa) (entropy of saturated liquid = 1.4336 kJ/kg . K, entropy of saturated vapour = 7.2234 kJ/kg. K) The exit condition of steam is
Discussion:
3 comments Page 1 of 1.
Taisir said:
1 month ago
sf<s2<sg⇒Steam is in the **wet (two-phase)** region at exit.
s2 = sf + x(sg−sf).
Substitute known values:
6.65 = 1.4336 + x(7.2234 − 1.4336)
6.65 = 1.4336 + x(5.7898).
x = 5.78986.65 − 1.4336
= 5.78985.2164≈0.901.
So, the quality is approximately 0.9 or 90% vapor, 10% liquid.
s2 = sf + x(sg−sf).
Substitute known values:
6.65 = 1.4336 + x(7.2234 − 1.4336)
6.65 = 1.4336 + x(5.7898).
x = 5.78986.65 − 1.4336
= 5.78985.2164≈0.901.
So, the quality is approximately 0.9 or 90% vapor, 10% liquid.
Ayush said:
5 years ago
Yes, I think Bharath is right.
Bharath said:
6 years ago
The correct answer is wet steam with quality 0.9 only then will the entropy be the same.
(3)
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