Question

Saturated water vapor at 60°C has its pressure decreased to increase the volume by 10% keeping the temperature constant. To what pressure should it be expanded? 

Answer 1

Initial state: v = 7.6707 m³/kg

Final state: v = 1.10 × v_g = 1.1 × 7.6707 = 8.4378 m³/kg 

Interpolate at 60°C between saturated (P = 19.94 kPa) and superheated vapor

P = 10 kPa

P ≅ 19.941 + (10 − 19.941) 8.4378 − 7.6707/15.3345 − 7.6707

= 18.9 kPa  

Comment: T,v ⇒ P = 18 kPa (software) v is not  linear in P, more like 1/P, so the linear interpolation in P is not very accurate.