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# Induction motor analysis

QuickField simulation example

This is a simulation example from Compumag TEAM Workshop Problem 30*.

Problem Type
Plane-parallel problem of AC magnetics.

Geometry

Given
Current density in a coil j = 310 A/cm² (r.m.s)
Frequency f = 60 Hz.
Aluminum conductivity 37.2 MS/m.
Relative permeability of steel μr=30.
Rotor steel conductivity 1.6 MS/m.
Stator steel is laminated and has a zero conductivity.

Calculate the torque, rotor steel and aluminum Joule heat losses and compare the results with a reference value.

Solution
To simulate the motion we will attenuate the AC problem frequency: fslip = slip * f,
where slip = 1 - (rotor angular frequency / synchronous angular frequency).
Synchronous angular frequency is 2πf = 2*3.142*60 = 377 rad/s.

Rotor angular frequency to frequency fslip conversion table
Rotor
angular
frequency,
Slip Frequency fslip, Hz
0160
2000.46928.2
400-0.061-3.66
600-0.592-35.5
800-1.122-67.3
1000-1.653-99.2
1200-2.183-131

It is not possible to specify the negative frequency in the problem. Instead we specify |fslip|, but consider the calculated torque value to be negative.

Results

Eddy current distribution at 28.2 Hz (equivalent to rotor angular frequency of 200 rad/s). Torque is 6 N*m, rotor steel losses are 15.7 W, rotor aluminum losses are 1061 W.

Reference
*Compumag TEAM Workshop