Answer:
initial quality = 0.3690
heat transfer = 979.63 kJ/kg
Explanation:
Given data:
volume of tank 0.45^3
weight of water 12 kg
Initial pressure 20 bar
final pressure 4 bar
Specific volume [tex]v = \frac {0.45}{12} = 0.0375 m^3/kg[/tex]
At Pressure = 20 bar, from saturated water table
[tex]v_f = 0.01177 m^/kg[/tex]
[tex]v_g = 0.099587 m^3/kg[/tex]
[tex]x = \frac{v -v_f}{v_g -v_f} = \frac{0.0375 - 0.001177}{0.099587 - 0.001177}[/tex]
inital quality is x =0.3690
Heat transfer is calculated as
[tex]u_1 = h_f + x(h_g - h_f) = v_f + x( h_{fg})[/tex]
from saturated water table, for pressure 20 bar ,
[tex]h_f = 908.79 kJ/kg, h_{fg} = 1890.7 kJ/kg[/tex]
=908.79 + 0.0357(1890.7)
= 979.63 kJ/kg
