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

A. Superheated vapour

B. Partially condensed vapour with quality of 0.9

C. Saturated vapour

D. Partially condensed vapour with quality of 0.1

Answer: Option A

Solution(By Examveda Team)

For an isentropic process $$s{\text{ }}initial = s{\text{ }}final{\text{ }}\left( {exit} \right)$$
$$\eqalign{ & \Rightarrow 6.65 = {S_{liq}} + x\left( {{S_{liq}} - {\text{ }}{S_v}} \right) \cr & \Rightarrow 6.63 = 1.4336 + x\left( {7.2234 - 1.4336} \right) \cr & \Rightarrow x = 0.9 \cr} $$
So, we are getting partially condensed vapour with quality = 0.9.