Answer:
a) 0.478
4
b) 1.45 × 10⁻⁶
Explanation:
A two phase liquid-vapor mixture
a) Given:
Temperature of H.O = 100°C
Specific volume of mixture, v = 0.8
now, we know
The specific volume of water at 100°C, vf = 0.001043 m³/kg
and,specific volume of steam At 100°C , vg = 1.671 m³/kg
also,
Specific volume, v = [ vf + x × (vg - vf ) ]
here, x is the quality
therefore, we get
0.8 = 0.001043 + [ x × (1.671 - 0.001043) ]
or
0.798957 = x × 1.669957
or
x = 0.478
4
b) Temperature of refrigerant = 0°C
Specific volume of refrigerant, v = 0.7721 cm³/g = 0.7721 × 10⁻³ m³/kg
Specific volume of fluid, vf = 7.72 × 10⁻⁴ m³/kg
Specific volume of gas, vg = 0.06930 m³/kg
Now,
Specific volume of two-phase mixture, v = [ vf + x × (vg - vf ) ]
on substituting the values in the above equation, we get
0.7721 × 10⁻³ = 7.72 × 10⁻⁴ + [ x × (0.06930 - 7.72 × 10⁻⁴ ) ]
or
x = 1.45 × 10⁻⁶
