Answer
a) $T_2=507°R$
b) $A_2=0.048\ ft²$
Work Step by Step
The energy balance for the system:
$\dot{e}_{in}=\dot{e}_{out}$
$h_1+\nu_1^2/2=\dot{q}+h_2+\nu_2^2/2$
From Table A-17E (50psia, 140°F): $h_1=143.47\ Btu/lbm$
$\nu_1=150\ ft/s,\ \nu_2=900\ ft/s,\ \dot{q}=6.5\ Btu/lbm$
Hence $h_2=121.2\ Btu/lbm$
Back to Table A-17E: $T_2=507°R$
From the material balance:
$\frac{P_1}{RT_1}A_1\nu_1=\frac{P_2}{RT_2}A_2\nu_2$
With $P_2=14.7\ psia, A_1=0.1\ ft²$
$A_2=0.048\ ft²$
