Answer
a) $\dot{Q}=1958\ Btu/s$
b) $\mathcal{V}=5.02\ ft/s$
Work Step by Step
From the energy balance in the steam:
$\dot{m}h_1=\dot{m}h_2+\dot{Q}$
$\dot{Q}=\dot{m}\Delta h_v$
Given $\dot{m}=6800\ lbm/h,\ \Delta h_v=1036.7\ Btu/lbm$
$\dot{Q}=1958\ Btu/s$
From the energy balance in the water:
$\dot{Q}+\dot{m}h_1=\dot{m}h_2$
$\dot{Q}=\dot{m}c_p\Delta T$
$\dot{Q}=\rho \frac{\pi D^2}4 \mathcal{V}c_p\Delta T$
Given $\rho=62.1\ lbm/ft³,\ D=1\ in,\ c_p=1.00\ Btu/lbm.°F,\ \Delta T=8°F$
$\mathcal{V}=5.02\ ft/s$
