Induction motor efficiency

Only the motor manufacturer can answer that and it will vary by motor design. But generally, the efficiency only drops a few percent even in the worst cases.

Normally you can ask (or buy) a motor type test report. In this report there are different graphs.

1) tests at no-load (testing while the supply voltage increases)

• cos fi against motor voltage
• input electrical power against motor voltage
• motor current against motor voltage

2) locked rotor tests (testing while the supply voltage increases)

• cos fi against motor voltage
• rotor torque against motor voltage
• input electrical power against motor voltage
• motor current against motor voltage

3) load characteristic tests (supply voltage equals nominal motor voltage and mechanical output power increases from 0 to 200% of the nominal motor power)

• cos fi against motor output power
• efficiency against motor output power
• slip (shaft speed decreasing ?) against motor output power
• motor current against motor output power
• input electrical power against motor output power

4) control characteristic tests (shaft torque equals to nominal motor torque and the supply voltage increases from 80% tot 120% of the nominal motor voltage)

• cos fi against motor output power
• efficiency against motor output power
• slip (shaft speed decreasing ?) against motor output power
• motor current against motor output power
• input electrical power against motor output power

5) cooling down or warmin up test ( at nominal motor torque and nominal motor supply voltage)

• temperature of the windings against time

6) a motor torque against motor speed graph

7) a spectrum of the motor sound generating

8) a spectrum of the motor vibration generating

As you can see, there is a lot of documentation about a motor. These tests are made up for every model of motor (output power, number of poles, frequency).

Good luck

(without any form of warrenty)

In a squirrel cage IM if you decrease the load the rotation speed will about the same. That means that the mechanical loss also will be the same so it will consume higher part of energy consumed by the motor. At the same time the motor current will be lower so both the iron loss and the copper loss will be lower. If you measure the power consumption of a free-running motor, that will be the motor's own loss ("zero efficiency"). Now you can draw a diagram and can interpolate between the free-running and the full load.