Answer
$\dot{V}_{1,in}=300\ ft³/s$
Work Step by Step
From tables A-4E to A-6E, for steam:
Inlet ($P=30\ psia, T=400°F$): $h=1237.9\ Btu/lbm$
Outlet ($P=25\ psia, T=212°F$): $h=180.21\ Btu/lbm$
The energy balance for the steam ($\dot{m}_2=15\ lbm/min$):
$\dot{Q}=-\dot{m}_2(h_{2,out}-h_{2,in})$
$\dot{Q}=264.42\ Btu/s$
The energy balance for the air($c_{p_1}=0.240\ Btu/lbm.°R,\ T_{1,out}=130°F,\ T_{1,in}=80°F$):
$\dot{Q}=\dot{m}_1c_{p_1}(T_{1,out}-T_{1,in})$
$\dot{m}_1=22.04\ lbm/s$
For ideal gases:
$\dot{m}=\frac{P\dot{V}}{RT}$
With ($P=14.7\ psia,\ T=540°R,\ R=0.3704\ psia.ft³/lbm.°R$)
$\dot{V}_{1,in}=300\ ft³/s$
