Magnetic Potential inside an Electric Motor

Find a fit for measured BH-curve data with a Gaussian model to solve a magnetostatic problem

To compute the torque of an electric motor, the magnetostatic field lines are computed. Consider a nonlinear Poisson-type equation

-∇·(-1/μ(X,

∇

u)∇u)=

as a model, where permeability

μ(X,

∇

is nonlinear and its values also differ in the different parts of the device. Here,

is the magnetic potential that is related to the magnetic field strength

and

is the current density in the

direction. The magnetic field strengths

and

are related through

and the permeability

is a function of the magnetic field

itself.

Get

B/H

curve data from measurements:

In[1]:=

HField={10,25,50,75,108,137,160,180,200,235,285,320,365,480,560,660,800,1000,1150,1550,1840,2250,2800,3450,4350,5500,7707,11863,15839,21076,27281,35289,45642,59046,76413,98907,128000,134764,141880,149364,157235};BField={0.0025,0.0066,0.0151,0.0264,0.05,0.1,0.15,0.2,0.25,0.35,0.5,0.6,0.7,0.9,1.,1.1,1.2,1.3,1.354,1.45,1.5,1.55,1.6,1.65,1.7,1.75,1.81,1.89,1.945,2.,2.05,2.1,2.15,2.2,2.25,2.3,2.35,2.36,2.37,2.38,2.39};BHData=Transpose[{BField,BField/HField}];

Use two exponentials to fit the data:

In[2]:=

fit=Module[{x,a1,a2,b1,b2,c1,c2,model,fitData},model=a1Exp[-(x-b1)^2/(c1^2)]+a2/(1+Exp[(x-b2)^2/(c2^2)])^2;fitData=FindFit[BHData,model,{a1,b1,c1,a2,b2,c2},x];Function@@{x,model/.fitData}];

Visualize the fit and the data:

In[3]:=

ShowListPlot[BHData],Plot[fit[x],{x,0,3}],



Out[3]=

Import and visualize the geometry of the motor. In the following graphic, the stator is shown in gray and the rotor in red. The orange coil will carry a positive current, while the yellow coil will carry a negative current. The light orange coils will not carry a current:

In[4]:=

Needs["NDSolve`FEM`"]mesh=Import

CloudObject[

]

;mesh"Wireframe""MeshElementStyle"

styles

...



Out[4]=

The permeability

μ(X,

∇

depends on the intensity of the magnetic field.

in the

and

directions is assumed to be zero. Thus the strength of the field is equal to the norm of the gradient of

. Set up the gradient of the magnetic field and avoid it becoming zero by adding a small positive number:

In[5]:=

B2Norm=

(0,1)

[x,y]

(1,0)

[x,y]

+$MachineEpsilon

Out[5]=

2.22045×

-16

(0,1)

[x,y]

(1,0)

[x,y]

Specify a different permeability in different parts of the device, so that it follows the fit inside the part and represents the permeability of free space outside:

In[6]:=

ν=Piecewise{{-1/fit[B2Norm],ElementMarker2||ElementMarker3}},-1QuantityMagnitude@UnitConvert[

]*IdentityMatrix[2];

Set up the current in the coils:

In[7]:=

jz=Piecewise[{{10,ElementMarker4},{-10,ElementMarker6}},0];

Solve the equation:

In[8]:=

usol=NDSolveValue

∇

{x,y}

ν.

∇

{x,y}

u[x,y]jz,DirichletCondition[u[x,y]0,x^2+y^2≥0.95],u,{x,y}∈mesh;

Visualize the magnetic potential and its gradient:

In[9]:=

{minsol,maxsol}=MinMax[usol["ValuesOnGrid"]];

In[10]:=

ShowContourPlotusol[x,y],{x,y}∈mesh,

,ToBoundaryMesh[mesh]["Wireframe"],VectorPlotEvaluate[{{0,1},{-1,0}}.Grad[usol[x,y],{x,y}]],{x,y}∈mesh,

,



Out[10]=

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Contributed by: Wolfram Staff

Magnetic Potential inside an Electric Motor

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