Answer
$\dot{m}=0.181\ kg/s$
Work Step by Step
From table A-6:
Inlet ($P_1=0.2\ MPa, T_1=150°C$): $h_1=2769.1\ kJ/kg$
Outlet ($P_2=75\ kPa, x_2=1$): $v_2=2.2172\ m³/kg,\ h_2=2662.4\ kJ/kg$
From the energy balance:
$\dot{m}(h_1+\mathcal{V}_1^2/2)=\dot{m}(h_2+\mathcal{V}_2^2/2)+\dot{Q}$
$h_1+\mathcal{V}_1^2/2=h_2+\mathcal{V}_2^2/2+q$
Given $\mathcal{V}_1\approx0,\ q=26\ kJ/kg$:
$\mathcal{V}_2=401.7\ m/s$
The mass flowrate is given by:
$\dot{m}=\frac{A_2\mathcal{V}_2}{v_2}$
Since $A_2=0.001\ m²$:
$\dot{m}=0.181\ kg/s$
