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.
Task
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, rad/s | Slip | Frequency fslip, Hz |
|---|---|---|
| 0 | 1 | 60 |
| 200 | 0.469 | 28.2 |
| 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
- Download simulation files (files may be viewed using any QuickField Edition).