FEA of Electromagnetics

1
FEA of Electromagnetics

• Contents
• Governing Equation for Magnetostatic field
• Boundary Condition
• The Galerkin Method
• Shape Function
• Finite Element Method for Magnetostatic
• Discretization
• Example 1 Magnetostatic field
• Example 2 Convergence Test
• Example 3 C-core Actuator

Reference   1 S. J. Salon, Finite Element
Analysis of Electrical Machines, Kluwer Academic
Publishers, Second Printing 1998 2 J. K.
Sykulski, Computational Magnetics, Chaman
Hall, 1995
2
Governing Equation for Magnetostatic
Maxwell Equations for Magnetostatic Field
Introduction of Vector Potential
(2)
where H, J, B, µ, ? are magnetic flux intensity
(A/m), current density (A/m2), magnetic flux
density (Tesla), permeability, and reluctivity,
respectively.
(5)
Single Governing Equation
Permanent Magnet
(3)
where M is the magnetization vector (A/m)
(4)
where Br is called the residual or remanent flux
density and Hc is called the coercive force.
3
Boundary Condition
Neumann condition
Homogeneous dirichlet condition
4
The Galerkin Method
The method of weighted residuals (MWR) can be
applied as follows. We begin with an operator
equation
(6)
(7)
The MWR now requires that the integral of the
projection of the residual on a specified
weighting function (W) is zero over the domain of
interest. We will choose the weighting function
to have the same form as the finite element shape
function (N).
(8)
This is known as the Galerkin method and will
yield the same finite element equations as the
variational method.
5
Shape Function Triangular element (1)
We may express the vector potential at any point
in the triangle as
(9)
2-D
where C1, C2, and C3 are constants to be
determined.
Using Kramers rule
(10)
(11)
where ? is the area of the triangle.
6
Shape Function Triangular element (2)
Using these results we may now express  as
(12)
where
The coefficients of the nodal potentials in (12)
are called shape functions. The potential can be
expressed as the sum of the shape functions times
the nodal potential.
(13)
(14)
7
Shape Function Triangular element (3)
Taking derivatives with respect to x and y
(15)
and
(16)
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Finite Element Method for magnetostatic (1)
Using (1) and Galerkins method
(17)
As before, we choose the weighting function W to
be the same as the element shape function, N.
Using the vector identity
(18)
We can write the first term of (17) as
(19)
The last term of (19) can be written as a line
integral using the divergence theorem
(20)
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Finite Element Method for magnetostatic (2)
Using identities
(21)
The line integral becomes
(22)
If we choose to ignore this line integral
(because of the flux normal and parallel boundary
condition), the integral must be zero for all
choices of N so that quantity in brackets must be
zero. Since this quantity is the tangential
component of H, we have imposed a homogeneous
Neumann boundary condition. We are left with
(23)
Substituting the curl of the weighting function
in the first term on the right hand side of (23),
we obtain
(24)
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Discretization (1)
The left side of (24) becomes
(25)
We recognize that
(26)
Substituting
Stiffness Matrix
(27)
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Discretization (2)
The last term on the right side of (24), forcing
function, becomes
(28)
(29)
Substituting the values that we found for ai, bi,
and ci (equation 13), this becomes
(30)
Assuming that J is constant over the triangle, we
get
(31)
12
Discretization (3)
The term representing the permanent magnet is
(32)
Using (16) and (32)
(33)
From vector analysis, the curl of vector
potential in cartesian coordinate is
(34)
13
Example 1 Magnetostatic (1)
(0,2)
(2,2)
(3,2)
Element 1 is magnetic steel with relative
permeability Element 2 is air (
) Element 3 is a conductor ( )with a
uniform current source (
) All lengths are in meters We will set
homogeneous Dirichlet boundary conditions, A0,
at nodes 1 and 2
4
2
3
1
5
(3,0)
(0,0)

• Problem
• Make the stiffness matrix
• Calculate the vector potentials
• Calculate the y component of flux density of the
  element 1

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Example 1 Magnetostatic (2) - Solution
The stiffness matrix for element 1 (ijk132)
becomes
(e1-1)
Similarly the stiffness matrix for element 2
(ijk153) and 3 (ijk354) becomes
(e1-2)
(e1-3)
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Example 1 Magnetostatic (3) - Solution
We now assemble the three element matrices into a
system matrix (12345) which will be (5?5)
(e1-4)
This matrix is singular as the boundary
conditions have not yet been applied. The right
side of (24) is a vector composed of the input
current. We distribute the total element current
equally among the three nodes (354)
(e1-5)
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Example 1 Magnetostatic (4) - Solution
We now apply the boundary conditions. A simple
method is to zero out the rows and columns
corresponding to the nodes which are set to zero
and put a 1 on the diagonal for (e1-4). The final
global equation is
(e1-6)
The vector potential solution is
(e1-7)
17
Example 1 Magnetostatic (5) - Solution
The vector potential, in terms of the shape
functions, is
(e1-8)
or
(e1-9)
For example, for element 1, the y component of
flux density is
(e1-10)
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Example 2 Convergence Test (1)
UNIT m
19
Example 2 Convergence Test (2)
Problem
1. Convergence Test by changing the Element edge
length -      At 0.015, 0.01, 0.005, 0.003 m 2.
FEA -      Plot of EquiFlux Lines -      Plot of
Flux Density (B) -      Flux density (B) in the
Midpoint of Airgap (Pick the node) 3. Analytical
Solution by using the Magnetic Circuit Analysis
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Example 3 C-core Actuator (1)

• CASE 1 Air-gap of 5mm No current in the coil
• CASE 2 Air-gap of 5mm Current in the coil
• - CASE 3 Air-gap of 2mm No current in the coil

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Example 3 C-core Actuator (2)
UNIT mm
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Example 3 C-core Actuator (3)
Material properties
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Example 3 C-core Actuator (4)
Problem

• FE Modeling by using FLUX2D and ANSYS
• 2. CASE
• - CASE 1 Air-gap of 5mm No current in
  the coil
• - CASE 2 Air-gap of 5mm Current in
  the coil
• - CASE 3 Air-gap of 2mm No current in
  the coil
• 3. Post-Processing
• - Displaying the equiflux lines
• - Displaying a color shading of the flux
  density
• - Displaying the magnetic force vectors
• - Computation of the force on the blade