0.5 kg of water at 120°C occupies a volume of 0.12 m3. The specific volume of saturated liquid water, vf, at 120oC is 0.0010603 m3/kg, and the specific volume of saturated water vapor, vg, at 120°C is 0.8919 m3/kg. Find the quality of the saturated mixture.

Question

Given: m = 0.5 kg; V = 0.12 m3; vf = 0.0010603 m3/kg; vg = 0.8919 m3/kg

What will be an ideal response?

Answer 1

v = V/m = 0.24 m3/kg
x = (v – vf) / (vg – vf) = 0.268