Answer
a) $\dot{m}=10.42\ lbm/s$
b) $\dot{W}_s=1486.5\ Btu/s$
Work Step by Step
From the material balance, for ideal gases:
$\dot{m}=\frac{P_1A_1\mathcal{V}_1}{RT_1}$
Given $P_1=150\ psia,\ A_1=0.1\ ft²,\ \mathcal{V}_1=350\ ft/s,\ R=0.3704\ psia.ft³/lbm.°R,\ T_1=1360°R$:
$\dot{m}=10.42\ lbm/s$
From the energy balance:
$\dot{m}(h_1+\mathcal{V}_1^2/2)=\dot{m}(h_2+\mathcal{V}_2^2/2)+\dot{W}_s$
Given $h_2=h_1+c_p(T_2-T_1),\ c_p=0.25\ Btu/lbm°R,\ T_2=760°F,\ \mathcal{V}_2=700\ ft/s$
and solving for the work $\dot{W}_s=1486.5\ Btu/s$
