Chapter 5 - Mass and Energy Analysis of Control Volumes - Problems - Page 270: 5-182

a) $\dot{W}_e=46.8\ kW$ b) $\mathcal{V}_1=5.43\ m/min$

$\dot{W}_e+\dot{m}h_1=\dot{m}h_2$ $\dot{W}_e=\dot{m}c_p(T_2-T_1)$ $\dot{m}=\rho.\dot{V}_1,\quad \rho=1000\ kg/m^3,\ \dot{V}_1=0.024\ m^3/min$ $\dot{m}=24\ kg/min,\ c_p=4.18\ kJ/kg.°C,\ T_2=48°C,\ T_1=20°C$ $\dot{W}_e=46.8\ kW$ $\dot{V}_1=A_1\mathcal{V}_1,\ A_1=\pi D^2/4,\ D=7.5\ cm$ $\mathcal{V}_1=5.43\ m/min$