Answer
a) $\Delta KE=-23.4\ kJ$
b) $\dot{W}_s=12123\ kW=12.1MW$
c) $A_1=0.0130m²$
Work Step by Step
From tables A-4 to A-6:
Inlet ($P_1=4\ MPa, T_1=500°C$): $v_1=0.086442\ m³/kg,\ h_1=3446.0\ kJ/kg$
Outlet ($P_2=30\ kPa, x_2=0.92$): $h_2=2437.7\ kJ/kg$
The change in kinetic energy is given by:
$\Delta KE=\dot{m}\frac{\mathcal{V}_2^2-\mathcal{V}_1^2}{2}$
Given $\dot{m}=12\ kg/s,\ \mathcal{V}_2=50\ m/s,\ \mathcal{V}_1=80\ m/s$:
$\Delta KE=-23.4\ kJ$
From the energy balance:
$\dot{m}(h_1+\mathcal{V}_1^2/2)=\dot{m}(h_2+\mathcal{V}_2^2/2)+\dot{W}_s$
$\dot{W}_s=12123\ kW=12.1MW$
From the material balance:
$\dot{m}=\frac{A_1\mathcal{V}_1}{v_1}$
$A_1=0.0130m²$
