Chapter 5 - Mass and Energy Analysis of Control Volumes - Problems - Page 260: 5-90

$I=49.7\ A$

From the energy balance: $\dot{W}_e+\dot{m}h_1=\dot{m}h_2$ $V.I=\dot{m}c_p(T_2-T_1)$ $\dot{m}=\frac{P_1\dot{V}_1}{RT_1}$ Given $P_1=100\ kPa,\ \dot{V}_1=0.3\ m³/s,\ R=0.287\ kJ/kg.K,\ T_1=15°C=288\ K$: $\dot{m}=0.3629\ kg/s$ Since $c_p=1.005\ kJ/kg.°C,\ T_2=30°C,\ V=110\ V$ $I=49.7\ A$