Question

Steam initially at 20 MPa, T = 525°C is expanded adiabatically and reversibly to 0.2 MPa. Compute final quality.

Answer 1

Adiabatic reversible process is isentropic process

Initial entropy of steam = final entropy of steam

At T = 525°C and P = 20 MPa, steam is superheated

Specific entropy of inlet steam

S1 = 6.245 kJ/kg

At P = 0.2 MPa, steam is saturated

Specific entropy of outlet saturated water

Sf = 1.53 kJ/kg

Specific entropy of outlet saturated steam

Sg = 7.126 kJ/kg

S2 = xSg + (1-x)Sf

= 7.126x + (1-x)1.53

For isentropic process

6.245 = 7.126x + (1-x)1.53

6.245 = 7.126x + 1.53 - 1.53x

4.715 = 5.596x

Steam quality x = 0.8425 = 84.25%