Chapter 2 - Energy, Energy Transfer, and General Energy Analysis - Problems - Page 100: 2-48E

$E_{e}=151.43W$

Since the fan transform electrical energy in kinetic energy: $E_{e}=E_{k}=\frac{1}{2}m_{air} v^2$ $m_{air}=\rho_{air}vA=(0.075\frac{lbm}{ft^3})*(22\frac{ft}{s})*(3ft*3ft)=14.85\frac{lbm}{s}$ $E_{e}=(\frac{1}{2}*14.85\frac{lbm}{s}*(22\frac{ft}{s})^2)*(\frac{\frac{1Btu}{lbm}}{25037\frac{ft^2}{s^2}})*(\frac{1055J}{1Btu})=151.43W$