Chapter 5 - Mass and Energy Analysis of Control Volumes - Problems - Page 258: 5-77E

$\dot{m}=0.642\ lbm/s$

The energy balance for the cold side: $\dot{Q}+\dot{m}h_{1,in}=\dot{m}h_{1,out}$ $\dot{Q}=\dot{m}c_{p_1}(T_{1,out}-T_{1,in})$ Given $\dot{m}=45\ lbm/s,\ c_{p_1}=1.0\ Btu/lbm.°F,\ T_{1,out}=65°F,\ T_{1,in}=50°F$: $\dot{Q}=675\ Btu/s$ For the hot side ($\Delta h_{v,2}=1050.9\ Btu/lbm$): $\dot{Q}=\dot{m}\Delta h_{v,2}$ $\dot{m}=0.642\ lbm/s$