Equivalent airgap
QuickField simulation example
Engineering question
How to find equivalent airgap Carter coefficient in electrical machines?
Engineering answer
Set up a plane-parallel QuickField DC Magnetics problem for a slotted electrical machine model and evaluate equivalent airgap Carter coefficient from computed field results.
Typical applications
slotted electrical machine airgaps, rotating machine airgaps, stator-rotor magnetic gaps
- Download simulation files (files may be viewed using any QuickField Edition).
Problem Type
Plane-parallel problem of DC magnetics.
Geometry
Given
Stator slots number Z1 = 48
Rotor slots number Z2 = 36
Number of phases m=3
Number of poles 2p=4
The B-H curve for the stator and the rotor:
Task
Calculate Carter coefficient.
Solution
The Carter coefficient is a ratio of the fundamental airgap flux density calculated for double slotted and smooth air gap.*
We are going to fill slots with steel to get a smooth air gap. In this case there will be no place for the winding. We replace real winding with the current layer placed in the air gap. The same current layer is used for the slotted configuration:
Ht(phi) = 30000 * cos(p * phi) [A/m], where p=2 is a pole's pairs.
We are not interested in induced voltages and eddy currents so the analysis is carried on using the DC Magnetic formulation.
QuickField built-in Harmonic browser add-in is used to calculate the flux density fundamental harmonic amplitude.
Results
Airgap flux density fundamental harmonic is 1.18T for smooth stator and 0.80T for slotted stator.
Carter coefficient = 1.18 / 0.80 = 1.47
Field picture of the flux density distribution and the fundamental harmonic magnitude.
Reference
* Ion Boldea, Syed A. Nasar. The Induction Machines Design Handbook, p.88.