Chapter 5 - Mass and Energy Analysis of Control Volumes - Problems - Page 260: 5-91

$\dot{W}_s=0.07\ Btu/s=0.074\ kW$

From the energy balance: $\dot{W}_s+\dot{m}h_1=\dot{m}h_2$ $\dot{W}_s=\dot{m}c_p(T_2-T_1)$ $\dot{m}=\frac{P_1\dot{V}_1}{RT_1}$ Given $P_1=14.7\ psia,\ \dot{V}_1=0.3\ ft³/s,\ R=0.3704\ psia.ft³/lbm.°R,\ T_1=70°F=530°R$ $\dot{m}=0.02246\ lbm/s$ And since $c_p=0.240\ Btu/lbm.°R,\ T_2=83°F$: $\dot{W}_s=0.07\ Btu/s=0.074\ kW$