Determine the quality of a two-phase liquid-vapor mixture of (a) h. o at 100°c with a specific volume of 0.8 m /kg. (b) refrigerant 134a at 0°c with a specific volume of 0.7721 cm/g.

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

the answer for sodium is 0.186

the answer for rubidium is 0.251