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

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.

the answer is option a.