Question

A three-phase induction motor of 480 V, 60 Hz and 50 hp is supplied with 60 A with a power factor of 0.85 in delay. The losses of copper from the stator are 2 kW and the losses in the copper from the rotor are 700 W. The losses due to friction and friction with air are 600 W, the losses in the core are 1 800 W and the miscellaneous ones are negligible. Find the following amounts:

a) Input power
b) Power gap
c) Developed power
d) Output power
(e) Efficiency

Answer 1

The input power of the motor is 24,624W. The power gap is 19,224W, the developed power is 18,524W, and the output power is 16,524W. The efficiency of the motor is 67.10%.

Step-by-step explanation:

To find the input power of the motor, we need to multiply the line voltage (480V) by the line current (60A) and the power factor (0.85). Therefore:

Input power = 480V x 60A x 0.85 = 24,624W

The power gap can be calculated by subtracting the losses from the input power:

Power gap = Input power - Copper losses - Friction losses - Core losses = 24,624W - 2kW - 600W - 1,800W = 19,224W

The developed power is the power gap minus the rotor copper losses:

Developed power = Power gap - Rotor copper losses = 19,224W - 700W = 18,524W

The output power is the developed power minus the stator copper losses:

Output power = Developed power - Stator copper losses = 18,524W - 2kW = 16,524W

Finally, the efficiency can be calculated by dividing the output power by the input power:

Efficiency = (Output power / Input power) x 100% = (16,524W / 24,624W) x 100% = 67.10%