• # question_answer The atmosphere air at dry bulb temperature of $15{}^\circ C$ enters a heating coil maintained at $40{}^\circ C.$ The air leaves the heating coil at $25{}^\circ C.$ The bypass factor of the heating coil is: A) 0.376               B) 0.476C) 0.6                               D) 0.67

BPF $=\frac{40-25}{40-15}=\frac{15}{25}=0.6$