Answer
a) $\mathcal{V}_2=82.06\ m/s$
b) $\dot{m}=0.2984\ kg/s$
Work Step by Step
From tables A-11 to A-13:
Inlet (600 kPa, sat. vapor): $v_1=0.034355\ m³/kg,\ h_1=262.46\ kJ/kg$
Outlet (700 kPa, sat. vapor): $v_2=0.031696\ m³/kg,\ h_2=278.59\ kJ/kg$
From the material balance:
$\frac{A_1\mathcal{V}_1}{v_1}=\frac{A_2\mathcal{V}_2}{v_2}$
Given $A_2/A_1=1.8,\ \mathcal{V}_1=160\ m/s$
$\mathcal{V}_2=82.06\ m/s$
From the energy balance:
$\dot{Q}+\dot{m}(h_1+\mathcal{V}_1^2/2)=\dot{m}(h_2+\mathcal{V}_2^2/2)$
Given $\dot{Q}=2kW$,
we can solve for the mass flowrate: $\dot{m}=0.2984\ kg/s$
