This document discusses the calculation of power input, efficiency, and an example problem for an induction motor. It defines the power input formula as Pin = √3×VL×IL×cosθ, where Pin is power in watts, V is supply voltage, I is stator current, and θ is the power factor. Efficiency is defined as the output power divided by the input power expressed as a percentage. The example calculates the efficiency of a 40 kW, 415V, 70A motor with 0.88 power factor to be 90.3%.

8. Power in = √3×VL×IL×cosø

9. Where:
Power in (Pin) = Watts or kilowatts
V = Supply voltage volts
I = Stator Current in amperes
ø = Power factor

10. Efficiency is a comparison
between input and output power
expressed as a percentage.

12. Example Calculation for Efficiency:
A 40 kW 415 volt three phase induction
motor requires a full load current of 70
amps per terminal at a power factor of
0.88. What is the motorʼs full load
efficiency?

13. Answer:
The motor has a power output of 40 kW
the motor has an input of power
determined by Pin = √3 x VL x IL x Cos ø

14. Pin = √3×415×70×.88 = 44.278 kW

15. Efficiency is Power out x 100 %
Power in

16. Therefore efficiency is
40, 000 x 100 = 90.3%
44,278