Chapter 5 - Mass and Energy Analysis of Control Volumes - Problems - Page 257: 5-72

$r=3.73$

From tables A-4 to A-6: Inlet 1 ($P_1=1\ MPa, T_1=50°C$): $h_1=209.34\ kJ/kg$ Inlet 2 ($P_2=1\ MPa, T_1=200°C$): $h_2=2828.3\ kJ/kg$ Outlet ($P_3=1\ MPa$, sat. liquid): $h_3=762.51\ kJ/kg$ From the energy balance: $\dot{m}_1h_1+\dot{m}_2h_2=(\dot{m}_1+\dot{m}_2)h_3$ With $r=\dot{m}_1/\dot{m}_2$ $rh_1+h_2=(r+1)h_3$ $r =\frac{h_2-h_3}{h_3-h_1}$ Hence $r=3.73$