Question

What is the quality of a two-phase liquid-vapor mixture of H20 at 20 °C with a specific volume of 10 m^3/kg?

Answer 1

Answer:

Quality of vapor is equal to 17.3%.

Explanation:

We know that if we know only one property in side the wet region then we will find the other property by using steam property table.

So pressure at saturation temperature 20°C

$v_f= 0.001\frac{m^3}{Kg} ,v_g= 57.76\frac{m^3}{Kg}$

So specific volume v

$v=v_f+x(v_g-v_f)\frac{m^3}{Kg}$  

Where x is quality of mixture

Now putting the values ,given that $v=10m^3/kg$

$10=0.001+x(57.76-0.001)\frac{m^3}{Kg}$  

x=0.173

So quality of vapor is equal to 17.3%.