Answer
a) $\dot{m}=1.765\times10^{-3}\ lbm/s$, $\nu=34.1\ ft/s$
b) $e_{flow}=74.4\ Btu/lbm$, $\theta=1156.2\ Btu/lbm$
c)$\dot{E}=2.04\ Btu/s$
Work Step by Step
The properties of the saturated liquid and vapor at 20psia:
$v_L=0.01683\ ft³/lbm,\ v_G=20.093\ ft³/lbm,\ u_G=1081.8\ Btu/lbm,\ h_G=1156.2\ Btu/lbm$
For a decrease in 0.6 gal in liquid volume in 45 min:
$m=\frac{\Delta V_L}{v_L}=4.766\ lbm$
$\dot{m}=m/\Delta t=1.765\times10^{-3}\ lbm/s$
With an area of 0.15 in²:
$\dot{m}v_G=A_c\nu$
$\nu=34.1\ ft/s$
Since $h=u+Pv\rightarrow e_{flow}=Pv=h-u$
For the steam $e_{flow}=74.4\ Btu/lbm$
$\theta=h+ke+pe\approx h=1156.2\ Btu/lbm$
$\dot{E}=\dot{m}\theta\approx \dot{m}h=2.04\ Btu/s$
