Homogenization of the

Eddy Current Problem in 2D

Karl Hollaus and Joachim Schoberl

Vienna University of Technology

Institute for Analysis and Scientific Computing

September 20, 2010

Laplace Problem

Motivation:

Fig.: FE-Model with 100laminations.

Fig.: FE-Model with 100 laminations(Detail, lower right corner).

Finite element models of laminated media lead to large equation systems.

Homogenization overcomes this problem!

Contents

Motivation

Electrostatic problemI Boundary value problem

I Weak form

I Multiscal ansatz

I Homogenization

I Numerical example

Eddy current problemI Boundary value problem

I Weak form

I Multiscal ansatz

I Homogenization

I Numerical example

Contents

Motivation

Electrostatic problemI Boundary value problem

I Weak form

I Multiscal ansatz

I Homogenization

I Numerical example

Eddy current problemI Boundary value problem

I Weak form

I Multiscal ansatz

I Homogenization

I Numerical example

Contents

Motivation

Electrostatic problemI Boundary value problem

I Weak form

I Multiscal ansatz

I Homogenization

I Numerical example

Eddy current problemI Boundary value problem

I Weak form

I Multiscal ansatz

I Homogenization

I Numerical example

Contents

Motivation

Electrostatic problemI Boundary value problem

I Weak form

I Multiscal ansatz

I Homogenization

I Numerical example

Eddy current problemI Boundary value problem

I Weak form

I Multiscal ansatz

I Homogenization

I Numerical example

Electrostatic ProblemLaminated medium:

Fig.: Problem model.

λi ... Material parametern ... No. laminationsd = d1 + d2

ff ... Fill factorff = d1

d1+d2

For example:Electrostatics

λ=ε ... Electric permittivityu=V ... Electric scalar potential

Electrostatic Problem

Boundary value problem:

Fig.: Boundary value problem.

Ω ... Domain

Ω = Ω0 ∪ Ωm

Γ ... Boundary

Γ = ΓD ∪ ΓN

−∇(λ(x , y)∇u(x , y)) = 0 in Ω (1)

u = uD on ΓD (2)

λ∂u

∂n· n = α on ΓN (3)

Electrostatic Problem

Weak Form:

Multiplication of (1) by a test function v and integration over Ω yields

−∫

Ω

∇(λ∇(u))v dΩ = 0

and integration by parts leads to∫Ω

λ∇u · ∇v dΩ−∫S

λ∂u

∂nv · ds = 0.

Considering boundary conditions (2, 3) and v = 0 on ΓD yields the weakform ∫

Ω

λ∇u · ∇v dΩ = 0. (4)

Electrostatic Problem

Reference solution:

Fig.: Reference solution u of one halfe of the domain.

Reference solution: Laminations are modeled individually!

Electrostatic Problem

Reference solution:

Fig.: Reference solution.

Fig.: Solution u, mean value u0 andenvelope u1 along x at y = 0.

Fig.: Periodic micro shapefunction φ(x), one periode isshown.

Electrostatic Problem

Two scale ansatz:

Thus, the following ansatz can be made:

u(x , y) = u0(x , y) + φ

(x

p

)u1(x , y) (5)

u0 ... mean valueu1 ... envelope of the staggered partφ ... periodic micro shape function

Inserting (5) into the bilinear form (4) and caring out simplemanipulations leads to∫

Ω

λ(∇u0 +∇φu1 + φ∇u1) · (∇v0 +∇φv1 + φ∇v1) dΩ = 0. (6)

Electrostatic Problem

Homogenization of the weak form:

The finite element matrix would be calculated by

∫ΩFE

∂u0

∂x∂u0

∂y

u1∂u1

∂x∂u1

∂y

T

λ 0 λφx λφ 00 λ 0 0 λφλφx 0 λφ2

x λφφx 0λφ 0 λφφx λφ2 00 λφ 0 0 λφ2

∂v0

∂x∂v0

∂y

v1∂v1

∂x∂v1

∂y

dΩ.

With φx = ∇φ.

The micro shape function φ is a highly oscillating function.

To homogenize the weak form, the coefficients λ, λ∇φ, λφ, etc. areaveraged over the periode d :

λ =1

d

∫ d

0

λ

(x

p

)dx =

λ1d1 + λ2d2

d

Electrostatic Problem

Homogenization of the weak form:

λφx =1

d

∫ d

0

λ

(x

p

)φx

(x

p

)dx = 2

λ1 − λ2

d

λφ =1

d

∫ d

0

λ

(x

p

)φ

(x

p

)dx = 0

λφ2x =

1

d

∫ d

0

λ

(x

p

)φx

(x

p

)φx

(x

p

)dx =

4

d(λ1

d1+λ2

d2)

λφxφ =1

d

∫ d

0

λ

(x

p

)φx

(x

p

)φ

(x

p

)dx = 0

λφ2 =1

d

∫ d

0

λ

(x

p

)φ

(x

p

)φ

(x

p

)dx =

λ1d1 + λ2d2

3d

Electrostatic Problem

Homogenization of the weak form:

Homogenized finite element matrix:

∫ΩFE

∂u0

∂x∂u0

∂y

u1∂u1

∂x∂u1

∂y

T

λ 0 λφx 0 00 λ 0 0 0

λφx 0 λφ2x 0 0

0 0 0 λφ2 0

0 0 0 0 λφ2

∂v0

∂x∂v0

∂y

v1∂v1

∂x∂v1

∂y

dΩ.

The homogenized medium can be modeled like a bulk by a coarse finiteelement mesh!

Choice for the ansatz: u0 ∈ H1(Ω), u1 ∈ L2(Ωm)

Electrostatic Problem in 2D

Numerical Example:

Fig.: Numerical model.

ff ... Fill factorff = d1

d1+d2

ff = 0.9

λ0 = 1

λ1 = 1000

λ2 = 1

ub − ua = 2

No.laminations = 10

Electrostatic Problem

Comparison of the results:

Fig.: Reference solution. Fig.: Homogenized solution.

Electrostatic Problem

Comparison of the results: flux density, x-component

Fig.: Reference solution. Fig.: Homogenized solution.

Electrostatic Problem

Comparison of the results: flux density, y-component

Fig.: Reference solution. Fig.: Homogenized solution.

”Energy” stored in the media:

Wlam = 0.586 and Whom = 0.608

Eddy Current Problem in 2D

Boundary value problem:

Fig.: Boundary value problem.

Ω ... DomainΩ = Ω0 ∪ Ωm

Γ ... Boundary

Γ = ΓH ∪ ΓB

µ ... Magnetic permeability

σ ... Electric conductivity

J 0 ... Impressed current density in a coil

B ... Magnetic flux density

Assumptions:- Linear material properties

- Time harmonic case

- Steady state

Eddy Current Problem

Maxwell’s equations for the eddy current problem:

curlH = J in Ωm (7)

curlE = −jωB (8)

divB = 0 (9)

J = σE (10)

B = µH (11)

curlH = J 0 in Ω0 (12)

divB = 0 (13)

B = µH (14)

H × n = K on ΓH (15)

B · n = b on ΓB (16)

Eddy Current Problem

Boundary value problem:

Magnetic vector potential:

B = curlA (17)

Faraday’s law:

E = −jωA. (18)

Ampere’s law:

curl(µ−1curlA)− jωσA = J 0 in Ω (19)

Boundary conditions:

µ−1curlA× n = K on ΓH (20)

A× n = α on ΓB . (21)

Eddy Current Problem

Weak form:

∫Ω

(curl(µ−1curlA) + jωσA)v dΩ =

∫Ω0

J 0v dΩ. (22)

(23)

Integration by parts yields

∫Ω

µ−1curlAcurlv dΩ +

∫Γ

µ−1curlA× v dΓ +

∫Ω

jωσAv dΩ =

∫Ω0

J 0v dΩ

Weak form with homogeneous boundary condition on ΓH :∫Ω

µ−1curlAcurlv dΩ + jω

∫Ω

σAv dΩ =

∫Ω0

J 0v dΩ. (24)

Eddy Current Problem

Multiscale ansatz:

Fig.: Eddy currents in laminations, 2D problem.

A = A0 + φ

(0

A1

)+

∇(φw)φ∇w

(25)

A0 represents the mean value of the solution

A1 considers J y and

w models J x at the end of the laminates

Eddy Current Problem

Modified weak form:

Inserting the approaches (27) into the weak form (26) yields:∫Ω

µ−1curl(A0 + φ(0,A1)T +∇(φw)

)curl

(v0 + φ(0, v1)T +∇(φq)

)dΩ

+jω

∫Ω

σ(A0 + φ(0,A1)T +∇(φw)

) (v0 + φ(0, v1)T +∇(φq)

)dΩ =

∫Ω0

J 0v dΩ

(26)∫Ω

µ−1curl(A0 + φ(0,A1)T + φ∇w

)curl

(v0 + φ(0, v1)T + φ∇q

)dΩ

+jω

∫Ω

σ(A0 + φ(0,A1)T + φ∇w

) (v0 + φ(0, v1)T + φ∇q

)dΩ =

∫Ω0

J 0v dΩ

(27)

Eddy Current Problem

Finite element matrix: ”Stiffness”

Multiscale ansatz:

A = A0 + φ(0,A1)T +∇(φw)

Bilinearform: ∫Ω

µ−1curlAcurlv dΩ

Homogenized finite element matrix:∫ΩFE

(curlA0

A1

)T (ν νφxνφx νφ2

x

)(curlv0

v1

)dΩ,

with ν = µ−1.

Eddy Current Problem

Finite element matrix: ”Mass”

Multiscale ansatz:

A = A0 + φ(0,A1)T +∇(φw)

Bilinearform:

jω

∫Ω

σAv dΩ

Homogenized finite element matrix:

jω

∫ΩFE

(A0)x(A0)y

A1

w∂w∂x∂w∂y

T

σ 0 0 σφx σφ 00 σ σφ 0 0 σφ

0 σφ σφ2 0 0 σφ2

σφx 0 0 σφ2x σφxφ 0

σφ 0 0 σφxφ σφ2 0

0 σφ σφ2 0 0 σφ2

(v0)x(v0)y

v1

q∂q∂x∂q∂y

dΩ

Eddy Current Problem

Finite element matrix: ”Stiffness”

Multiscale ansatz:

A = A0 + φ(0,A1)T + φ∇w

Bilinearform: ∫Ω

µ−1curlAcurlv dΩ

Homogenized finite element matrix:

∫ΩFE

curlA0

A1∂w∂y

T ν νφx νφxνφx νφ2

x νφ2x

νφx νφ2x νφ2

x

curlv0

v1∂q∂y

dΩ,

with ν = µ−1.

Eddy Current Problem

Finite element matrix: ”Mass”

Multiscale ansatz:

A = A0 + φ(0,A1)T + φ∇w

Bilinearform:

jω

∫Ω

σAv dΩ

Homogenized finite element matrix:

jω

∫ΩFE

(A0)x(A0)y

A1∂w∂x∂w∂y

T

σ 0 0 σφ 00 σ σφ 0 σφ

0 σφ σφ2 0 σφ2

σφ 0 0 σφ2 0

0 σφ σφ2 0 σφ2

(v0)x(v0)y

v1∂q∂x∂q∂y

dΩ

Eddy Current Problem

Multiscale ansatz:

A = A0 + φ

(0

A1

)+

∇(φw)φ∇w

We propose the following choice:

A0, v0 ∈ H(curl ,Ω)

A1, v1 ∈ L2(Ωm)

w , q ∈ H1(Ωm)

φ ∈ Hper (Ωm)

Again, a coarse finite element mesh suffices for the homogenized medium!

Eddy Current Problem in 2D

Numerical example:

Fig.: Numerical example,dimensions in mm.

α = 1.0Vs/mff = 0.9

µ0 = 4π10−7Vs/Am

µ = µrµ0

µr1 = 1000

σ1 = 2 · 106S/m

f = 50Hz

δ ... Penetration depth

δ = 1.6mm

Eddy Current Problem

Comparison of the results: Eddy current density J

Fig.: Reference solution. Fig.: Homogenized solution.

... solution in the upper left corner.

d1 = 1.8mm

10 laminations

Eddy Current Problem

Comparison of the results: Losses and computational costs

Table: Eddy current losses

Losses in W/md in mm Laminations Ansatz ∇(φw) Ansatz φ∇(w)

0.5 11.59 11.58 11.481.0 45.70 45.46 44.742.0 177.2 174.16 169.4

Table: Computational costs for d=0.5mm and 40 laminations

Model FE NDOFLaminated 104 452 783 555

Homogenized 1 286 11 117

FE ... No. finite elementsNDOF ... No. degrees of freedom

Thank you for your attention!