Solution :
Given :
Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa
The phase diagram is provided below.
a). The phase is a standard mixture.
b). At pressure, p = 200 kPa, T = [tex]$T_{saturated}$[/tex]
Temperature = 120.21°C
c). Specific volume
[tex]$v_{f}= 0.001061, \ \ v_g=0.88578 \ m^3/kg$[/tex]
[tex]$v_x=v_f+x(v_g-v_f)$[/tex]
[tex]$=0.001061+0.7(0.88578-0.001061)$[/tex]
[tex]$=0.62036 \ m^3/kg$[/tex]
d). Specific energy ([tex]u_x[/tex])
[tex]$u_f=504.5 \ kJ/kg, \ \ u_{fg}=2024.6 \ kJ/kg$[/tex]
[tex]$u_x=504.5 + 0.7(2024.6)$[/tex]
[tex]$=1921.72 \ kJ/kg$[/tex]
e). Specific enthalpy [tex]$(h_x)$[/tex]
At [tex]$h_f = 504.71, \ \ h_{fg} = 2201.6$[/tex]
[tex]h_x=504.71+(0.7\times 2201.6)[/tex]
[tex]$= 2045.83 \ kJ/kg$[/tex]
f). Enthalpy at m = 0.5 kg
[tex]$H=mh_x$[/tex]
[tex]$= 0.5 \times 2045.83$[/tex]
= 1022.91 kJ
