Answer
a) $\dot{m}=4.085\ kg/s$
b) $\mathcal{V}_2=589.5\ m/s$
c) $A_2=0.00087 m²=8.7\ cm²$
Work Step by Step
From table A-6:
Inlet ($P_1=4\ MPa, T_1=400°C$): $v_1=0.07343\ m³/kg,\ h_1=3214.5\ kJ/kg$
Outlet ($P_2=2\ MPa, T_2=300°C$): $v_2=0.12551\ m³/kg,\ h_2=3024.2\ kJ/kg$
From the material balance:
$\dot{m}=\frac{A_1\mathcal{V}_1}{v_1}$
Given $A_1=0.005\ m²,\ \mathcal{V}_1=60\ m/s$
$\dot{m}=4.085\ kg/s$
From the energy balance:
$\dot{m}(h_1+\mathcal{V}_1^2/2)=\dot{m}(h_2+\mathcal{V}_2^2/2)+\dot{Q}$
Given $\dot{Q}=75\ kW$
and solving for the outlet velocity: $\mathcal{V}_2=589.5\ m/s$
From the material balance:
$\dot{m}=\frac{A_2\mathcal{V}_2}{v_2}$
solving for the area: $A_2=0.00087 m²=8.7\ cm²$
