Answer
$T_3=39.37°C$
$x_3=0.307$
Work Step by Step
From tables A-11 to A-13:
Inlet 1 ($P_1=1\ MPa, T_1=20°C$): $h_1=79.32\ kJ/kg$
Inlet 2 ($P_2=1\ MPa, T_2=80°C$): $h_2=314.27\ kJ/kg$
From the energy balance:
$\dot{m}_1h_1+\dot{m}_2h_2=(\dot{m}_1+\dot{m}_2)h_3$
Since $\dot{m}_1=2\dot{m}_2$
$2h_1+h_2=3h_3$
So $h_3=157.64\ kJ/kg$
At 1 MPa, saturation: $T_3=39.37°C,\ h_{3,G}=271.04\ kJ/kg,\ h_{3,L}=107.34\ kJ/kg$
$h_3=h_{3,L}+x_3(h_{3,G}-h_{3,L})$
$x_3=0.307$
