Answer
a) $\dot{m}=6.36\ lbm/s$
b) $T_2=801°R= 341°F$
Work Step by Step
From table A-17E:
Inlet ($T_1=60°F$): $h_1=124.27\ Btu/lbm$
From the material balance:
$\dot{m}=\frac{P_1\dot{V}_1}{RT_1}$
Given the values of $P_1=14.7\ psia,\ R=0.3704\ psia.ft³/lbm.°R,\ \dot{V}_1=5000ft³/min $
$\dot{m}=6.36\ lbm/s$
From the energy balance:
$\dot{W}_s+\dot{m}h_1=\dot{m}h_2+\dot{Q}$
$\dot{W}_s=\dot{m}(h_2-h_1+q)$
With $\dot{W}_s=700\ hp=494.8\ Btu/s,\ q=10\ Btu/lbm$:
$h_2=192.06\ Btu/lbm$
Hence, from table A-17E: $T_2=801°R= 341°F$
