Answer
$\eta=60.3\%$
Work Step by Step
From table A-4 to A-6:
Inlet ($P_1=4\ MPa, T_1=350°C$): $s_1=6.5843\ kJ/kg.K,\ h_1=3093.3\ kJ/kg$
Actual outlet ($P_2=120\ kPa, x_2=1$): $h_2=2683.1\ kJ/kg$
Isentropic outlet ($P_2=120\ kPa, s_{2,s}=s_1$): $x_{2,s}=0.8798,\ h_{2,s}=2413.4\ kJ/kg$
Given
$\dot{W}_a=\dot{m}(h_1-h_2),\quad \dot{W}_s=\dot{m}(h_1-h_{2,s}),\quad\eta=\dot{W}_a/\dot{W}_s$
$\eta=\frac{h_1-h_2}{h_1-h_{2,s}}=0.603=60.3\%$
