Answer
a) $\dot{m}=46.88\ kg/s$
b) $\dot{W}_s=18300\ kW=18.3\ MW$
Work Step by Step
From the energy balance:
$\dot{m}(h_1+\mathcal{V}_1^2/2)=\dot{m}(h_2+\mathcal{V}_2^2/2)+\dot{W}_s$
With $h_2=h_1+c_p(T_2-T_1)$
$\dot{W}_s=-\dot{m}(c_p(T_2-T_1)+(\mathcal{V}_2^2-\mathcal{V}_1^2)/2)$
From the material balance:
$\dot{m}=\frac{P_1A_1\mathcal{V}_1}{RT_1}$
Given $P_1=1300\ kPa,\ A_1=0.2\ m²,\ \mathcal{V}_1=40\ m/s,\ R=0.287\ kJ/kg.K,\ T_1=773\ K$:
$\dot{m}=46.88\ kg/s$
For the outlet $P_2=100\ kPa,\ A_2=1\ m²,\ T_2=400\ K$:
$\mathcal{V}_2=53.82\ m/s$
Back to the energy balance, with $c_p=1.048\ kJ/kg.K$:
$\dot{W}_s=18300\ kW=18.3\ MW$
