What is the rate of heat transfer from an electric motor that draws 9.7 amperes at 110 volts and delivers 1.25 horsepower of mechanical energy?
The rate of heat transfer from the electric motor can be calculated by finding the difference between the total power consumed by the motor and the mechanical power output. In this case, the motor consumes approximately 1.067 kilowatts of electrical power and delivers 0.932 kilowatts of mechanical power. Hence, the rate of heat transfer, which is the heat produced and dissipated by the motor, is about 0.135 kilowatts.
Calculation Method:
Power Consumed by the Motor:
To find the total power consumed by the motor, we use the formula P = IV (where I is the current in amperes and V is the voltage). Given that the motor draws 9.7 amperes at 110 volts, we have:
P = 9.7 A * 110 V = 1067 watts, or approximately 1.067 kilowatts.
Mechanical Power Output of the Motor:
The mechanical power output of the motor is given as 1.25 horsepower, which is equivalent to about 0.932 kilowatts.
Rate of Heat Transfer:
The rate of heat transfer from the motor is the difference between the power consumed by the motor and the mechanical power output. Therefore, the heat generated and dissipated by the motor is:
1.067 kW - 0.932 kW = 0.135 kW (where 1 kilowatt is equal to 1 kilojoule per second).
By calculating the rate of heat transfer from the motor, we can understand the efficiency and thermal characteristics of the electric motor under steady load conditions.