An electric motor under steady load draws 9.7 amperes at 110 volts, delivering 1.25(hp) of mechanical energy. What is the rate of heat transfer from the motor, in kW?

Question

What will be an ideal response?

Answer 1

At steady state, the electrical power input to the motor has to be balanced by the work and heat output of the motor. At steady-state, for a closed system, the internal energy is constant:



The work is equal to the difference between the electrical work done on the system (positive) and the mechanical work done by the system (negative). Thinking way, way, back to first-year physics, we remember that electrical power (work per time) is the product of current and voltage (P = IV), and that the product of volts times amps is equal to watts. Thus the energy input to the motor (electrical work) is P = 9.7 A·110 V = 1067 V–A = 1067 W. The work output is 1.25 hp, and from table A1 we see that 1 kW = 1000 W = 1.341 hp. So, the work output is 1.25 hp/ (1.341 kW/hp) = 0.932 kW = 932 W. The heat transfer rate (the energy removed from the motor as heat) is equal to the difference between the electrical energy input and the mechanical work output:

QQ? = – ? = ?(1067 ? 932) = ?135 W

135 W is removed from the motor as heat.