Induction motor magnetising current and power factor at duty point

[geefer]

We have several Squirel cage induction motors running on our plant at well below their power rating e.g. 345kW 3.3kV four pole 50Hz motor full load current 75 amps. with a power factor of 0.8 at full rated power. I understand that if you run a motor at well below its rated capacity
(in this case absorbed power 146 kW ) The power factor drops markedly . can anyone help with the formulae and calculations to work out the power factor at duty point and the magnetising current.

 

[ijl]

Use the equivalent circuit of the motor. You'll find the circuit in books on power system analysis. The circuit and a program for the parameters can also be found at

http://pp.kpnet.fi/ijl/prog.htm

Alternatively, use a computer program for power system analysis to calculate the current etc, for example at

http://pp.kpnet.fi/ijl

 

[Skogsgurra]

Simply put, the magnetizing current is constant when voltage is constant.

It can be calculated as In*sqrt((1-cos(phi)^2). In your example the magnetizing current is 75*sqrt(1-.8^2) = 75*.6 = 45 A while the torque producing current is 75*.8 = 60 A.

This current stays roughly constant when shaft load changes so, at 146 kW, your torque producing current will be 60*146/345 = 25.4 A. Your total current will then be around 65 A and the PF is 25.4/65 or around .4 not so good.

This is a rough calculation, but the result should not be too far off the actual value.



Gunnar Englund

http://www.gke.org

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