Electrical generator winding loss

QuickField simulation example

We model a synchronous machine with an outer rotor. There are permanent magnets glued to the rotor. There is a 3-phase concentrated winding on the stator poles. The task is to calculate eddy current losses in the conductors.The task is divided in two parts:
1. First we simulate a set of DC magnetic problems with various rotor positions. The stator winding is disconnected and carries no current. The result of the analysis is the flux linkage with conductors vs rotor position dependency, which is transformed to EMF vs time dependency.
2. Then we replace the permanent magnet with the current layer and run an AC magnetic problem. The actual magnetic flux density distribution calculated in the DC magnetic problem is replaced with a sinusoidal flux density distribution. So we take into account only the unity harmonic. The current layer produces the running wave of the magnetic field in the air gap. The layer parameters are adjusted so that the EMF induced matches the unity harmonic that was calculated in the DC magnetic problem.

Problem Type
Plane-parallel problem of DC magnetics and AC magnetics.

Geometry:
Motor z-length is 65 mm.

Given
Rotational speed is 1840 rpm (192.68 rad/s).
Number of turns per pole is 13.
Permanent magnet coercive force is 750 kA/m, residual flux density is 1.04 T.
Steel magnetic permeability is 1000.

Task
Calculate the eddy current losses in the winding

Solution
A set of DC magnetic problems with different rotor positions is automatically generated and solved using LabelMover.
The flux linkage vs. rotor position dependency is converted to EMF vs. time dependency: $$\mathrm{EMF}=\; -\frac{\mathrm{d\Psi }}{\mathrm{dt}}=\; -\frac{\mathrm{d\Psi }}{\mathrm{d(angle)}}*\frac{\mathrm{d(angle)}}{\mathrm{d(t)}}=\; -\frac{\mathrm{\Delta \; \Psi }}{\mathrm{\Delta \; angle}}*\mathrm{rotational\; speed}$$

In AC problem we specify the frequency of the generated voltage: f = 1840/60 * (permanent magnets number /2) = 306.66 Hz.
Equation H_t * cos (ωt + phase) defines the running wave, where H_t=290 kA/m is the magnitude of the distributed current and phase depends on the coordinates.
In QuickField AC magnetic problems all terms are functions of ωt, so in input parameters we need to specify only the H_t and the phase = 10*phi, where phi is a built-in angular coordinate, (permanent magnets number /2) = 10

Results

EMF unity harmonic is 29.35 V per pole. There are 8 pole windings connected in series, so the total voltage is 235 V (magnitude) per phase.
Next we replace the permanent magnets with the current layer that induces in the stator armature the same magnitude of the EMF unity harmonic.

In the AC magnetic problem the EMF is 234 V, which is close to the value calculated in the DC magnetic problem.
Eddy current losses in the conductor close to the air gap is 0.24 W per 65 mm on the slot length.

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