Chapter 5 - Mass and Energy Analysis of Control Volumes - Problems - Page 257: 5-64

$P_2=132.82\ kPa$ $u_2=80.7\ kJ/kg$

From tables A-11 to A-13: Inlet ($P_1=0.8\ MPa, T_1=25°C$): $h_1=86.41\ kJ/kg$ Outlet ($T_2=-20°C, h_2=h_1$): $P_2=132.82\ kPa,\ h_{2,L}=25.47\ kJ/kg,\ h_{2,G}=238.43\ kJ/kg,$ $\ \ \ \ \ \ u_{2,L}=25.37\ kJ/kg,\ u_{2,G}=218.86\ kJ/kg$ $h_2=h_{2,L}+x_2(h_{2,G}-h_{2,L})$ So $x_2=0.2862$ $u_2=u_{2,L}+x_2(u_{2,G}-u_{2,L})$ Hence $u_2=80.7\ kJ/kg$