Convergence tests

Final algorithm.

Finally, we have three unknown elds (one vector-eldH^c, two scalar eldsφ, p^c) instead of two (H^c, φ).

We introduce a new nite element space to approximate the new scalar unknown p^c: X_h^p,2D :=

n

p_h ∈L²(Ωc)/ p_h∈ C⁰(Ωc), p_h(TK)∈P`p, ∀K ∈ F_h^c, p_h = 0 on∂Ωc

o , X_h^p:=

( p=

M

X

m=−M

p^m_h(r, z)e^imθ /∀m= 1. . . , M, p^m ∈X_h^p,2D and p^m_h =p^−m_h )

Here `_p is an integer in {1,2}.

The nal form of the algorithm is the following : after proper initialization, we solve for H^c,n+1 ∈ X^H_h, φⁿ⁺¹ ∈X_h^φ and p^c,n+1 ∈X_h^p so that the following holds for all b ∈X^H_h, ψ ∈ X_h^φ, q∈X_h^p

Z

Ωc

µ^cDH^c,n+1

∆t ·b+ Z

Ωv

µ^v∇Dφⁿ⁺¹

∆t ·∇ϕ+L (H^c,n+1, φⁿ⁺¹),(b, ϕ) +P(φⁿ⁺¹, ϕ) +D (H^c,n+1, p^c,n+1),(b, q)

=R_n(b, ϕ) (D.3.11)

whereP denotes the stabilizing bilinear form dened by P(φ, ψ) =

Z

Ωv

µ^v∇φ·∇ψ− Z

∂Ωv

µ^vψn·∇φ, andD is dened by

D((H, p),(b, q)) :=β0



 Z

Ωc

µ^c∇p·b− Z

Ωc

µ^cH·∇q+ X

K∈F_h^c

Z

K^3D

h^2(1−α)_K ∇p·∇q+s(H^c,n+1,b)



, where the last bilinear form sis dened by

s(H,b) := X

K∈F_h^c

Z

K^3D

h^2α_K∇·(µ^cH)∇·(µ^cb).

P accounts for the addition ofp^v andD is a discrete approximation for the weak formulation of (D.3.9). Finally, s is a stabilization term that makes the discrete formulation well-posed irrespective of the polynomial degree of the approximation forp^c. The coecient β0 is scaled as follows:

β0 =γ0/(R_mmin

x∈Ωc

(σ(x))),

with γ0 = 1. This scaling can be justied by arguments from the Interior Penalty theory [7, 65, 66].

D.4.1 The L-shape domain

We rst illustrate the positive eect of the magnetic pressure in the case of steady-state regime in a non-smooth and non-convex domain. The setting is the following: we consider the conducting L-shape domain (two dimensional case)

(D.4.1) Ω = Ω_c= (−1,+1)²\([0,+1]×[−1,0]).

with no insulating region,Ω_v =∅, (cf. Fig. D.2). We takeµ= 1,u˜ = 0, and σ= 1.

Ω

-6

−1 0 1

−1 0 1

Figure D.2: Two-dimensional L-shape domain with constant µ^c

Boundary value problem

Consider the following boundary value problem in the above denedL-shape domain: ndH such that

(D.4.2) ∇×∇×H= 0, ∇·H= 0, H×n|_Γ=G×n,

where the Cartesian components of the boundary data Gare given by

(D.4.3) G(r, θ) = 2

3r⁻¹³

−sin(^θ₃) cos(^θ₃)

,

and (r, θ) are the polar coordinates centered at the re-entrant corner of the domain. The solution to the above problem isH=∇ψ, whereψ(r, θ) =r²³ sin(²₃θ).

Five quasi-uniform (non-nested) Delaunay meshes are considered of mesh-sizesh= 1/10, 1/20, 1/40, 1/80, 1/160, respectively. The meshes are composed of triangles. Two types of approximation are tested; we use P1 elements in the rst case and P2 elements in the second case. The magnetic eld and the magnetic pressure are approximated using equal order polynomials in each case.

Denoting byHh the approximate magnetic eld, we report in Table D.1 the relative errors kH_h −Hk_L2/kHk_L2 for α = 0.75 and α = 1. Table D.1 also shows the computed order of convergence (COC). Convergence is observed for the P1 and P2 approximations. The best possible convergence rate is ²₃ and this rate is achieved numerically when usingP2 elements.

h P1 P2

α= 0.75 α = 1 α= 0.75 α= 1

Rel. Error COC Rel. Error COC Rel. Error COC Rel. Error COC 0.1 2.39010⁻¹ N/A 2.30310⁻¹ N/A 1.29010⁻¹ N/A 1.11010⁻¹ N/A 0.05 1.84310⁻¹ 0.38 1.82610⁻¹ 0.34 8.17810⁻² 0.66 7.01610⁻² 0.66 0.025 1.40510⁻¹ 0.39 1.36710⁻¹ 0.42 5.97810⁻² 0.45 5.01710⁻² 0.48 0.0125 1.03110⁻¹ 0.45 1.01010⁻¹ 0.44 3.75910⁻² 0.67 3.19110⁻² 0.65 0.00625 7.54410⁻² 0.45 7.65610⁻² 0.4 2.23210⁻² 0.75 1.93810⁻² 0.72 Table D.1: L²(Ω) relative errors and computed order of convergence for the boundary value

problem (D.4.2)-(D.4.3) usingP1elements (2nd and 3rd columns) andP2 elements (4th and 5th columns) with α = 0.75 and α= 1; h is the typical diameter of the Delaunay meshes.

Eigenvalue problem

We now study Ohmic decay in the conducting L-shape domain. Assuming that the magnetic eld has the following behavior H(x, t) =H(x)e^−λt, where λ >0, we are lead to consider the following eigenvalue problem: nd (λ,H) such that

(D.4.4) ∇×∇×H=λH, ∇·H= 0, H×n|_Γ= 0,

Approximations of the rst ve eigenvalues with10⁻¹¹ tolerance are provided in [40]: λ₁ ≈ 1.47562182408,λ2 ≈3.53403136678,λ3 =λ4 =π² ≈9.86960440109, andλ5≈11.3894793979. We solve (D.4.4) using ARPACK [91] with a relative tolerance of 10⁻⁸.

Table D.2 shows the rst eigenvalue computed with α = 0.9 on ve quasi-uniform (non-nested) Delaunay meshes of mesh-sizes 1/10, 1/20, 1/40, 1/80, 1/160, respectively. As ex-plained in [17], taking αclose to 1improves the convergence rate on the rst eigenvalue. The method is clearly convergent although the eigenvector has a strong unbounded singularity.

h P1 P2

λ₁ Rel. Error COC λ₁ Rel. Error COC 0.1 1.555 5.25610⁻² N/A 1.508 2.19210⁻² N/A 0.05 1.541 4.35310⁻² 0.27 1.493 1.16710⁻² 0.9 0.025 1.522 3.09410⁻² 0.49 1.487 7.37110⁻³ 0.66 0.0125 1.507 2.12610⁻² 0.54 1.481 3.72610⁻³ 0.98

0.00625 1.497 1.46510⁻² 0.54 - - N/A

Table D.2: Relative errors and COC forλ1 using P1 elements and P2 elements with α= 0.9. The symbol - indicates that the pair (Linear Solver + ARPACK) did not converge with the assigned tolerances.

Table D.3 shows the rst ve eigenvalues computed with α = 0.7 on ve quasi-uniform (non-nested) Delaunay meshes of mesh-sizes1/10,1/20,1/40,1/80,1/160, respectively. Here again we observe convergence and there is no spurious eigenvalue. As expected the worst rate of convergence is observed for the rst eigenvalue which corresponds to the most singular eigenvector. The second eigenvector is inH¹(Ω), the third and fourth eigenvectors are analytic, the fth one has a strong unbounded singularity. The theory developed in [17] shows that

the accuracy of the method improves whenα→1, but the absence of spurious eigenvalues is assured only forα <1. This phenomenon can be observed on the rst eigenvalue by comparing Table D.2 and Table D.3. The COC stalls for the eigenvalues λ₃ and λ₄ using P2 since the accuracy of the computed eigenvalues is limited by the tolerance in ARPACK (10⁻⁸).

h P1 P2

λ1 Rel. Error COC λ1 Rel. Error COC 0.1 1.930 2.66810⁻¹ N/A 1.707 1.45210⁻¹ N/A 0.05 1.845 2.22410⁻¹ 0.26 1.623 9.52210⁻² 0.61 0.025 1.765 1.78810⁻¹ 0.32 1.586 7.24010⁻² 0.4 0.0125 1.696 1.38910⁻¹ 0.36 1.545 4.61410⁻² 0.65

0.006256 1.644 1.08010⁻¹ 0.36 - - N/A

h P1 P2

λ₂ Rel. Error COC λ₂ Rel. Error COC 0.1 3.573 1.10110⁻² N/A 3.537 8.26610⁻⁴ N/A 0.05 3.551 4.71610⁻³ 1.22 3.535 2.38010⁻⁴ 1.8 0.025 3.540 1.57810⁻³ 1.58 3.534 6.64010⁻⁵ 1.8 0.0125 3.536 6.24510⁻⁴ 1.33 3.534 1.72610⁻⁵ 1.9

0.006256 3.535 2.76810⁻⁴ 1.17 - - N/A

h P1 P2

λ3 Rel. Error COC λ3 Rel. Error COC 0.1 5.450 5.77010⁻¹ N/A 7.828 2.30710⁻¹ N/A 0.05 7.852 2.27710⁻¹ 1.34 9.870 3.79910⁻⁷ 19.21 0.025 9.873 3.07510⁻⁴ 2.89 9.870 3.85610⁻⁸ 3.3 0.0125 9.870 7.71410⁻⁵ 2.0 9.870 3.44410⁻⁸ 0.16

0.006256 9.870 1.93410⁻⁵ 2.0 - - N/A

h P1 P2

λ4 Rel. Error COC λ4 Rel. Error COC 0.1 5.455 5.76110⁻¹ N/A 7.841 2.29110⁻¹ N/A 0.05 7.858 2.27010⁻¹ 1.34 9.870 4.71210⁻⁷ 18.9 0.025 9.873 3.10010⁻⁴ 9.52 9.870 3.85610⁻⁸ 3.61 0.0125 9.870 7.76810⁻⁵ 2.0 9.870 1.99010⁻⁸ 0.95

0.006256 9.870 1.93510⁻⁵ 2.0 - - N/A

h P1 P2

λ₅ Rel. Error COC λ₅ Rel. Error COC 0.1 5.506 6.96410⁻¹ N/A 7.903 3.61410⁻¹ N/A 0.05 7.877 3.64610⁻¹ 0.93 11.39 2.37410⁻⁵ 13.89 0.025 11.39 4.32610⁻⁴ 9.72 11.39 7.78610⁻⁶ 1.61 0.0125 11.39 1.45710⁻⁴ 1.57 11.39 2.16810⁻⁶ 1.85

0.006256 11.39 5.30310⁻⁵ 1.46 - - N/A

Table D.3: First ve eigenvalues usingP1 elements andP2elements withα= 0.7. The symbol - indicates that the pair (Linear Solver + ARPACK) did not converge with the assigned tolerances.

D.4.2 Induction in a composite sphere

We now turn our attention to three-dimensional induction problems with discontinuous per-meability elds.

Description of the problem

The domain isΩ :=R³ and the conductor is composed of two concentric spheres centered at 0. The radius of the inner sphere, say Ω1, is R1 and its magnetic permeability is µ1. The radius of the outer conducting sphere, sayΩ₂, isR₂ and its magnetic permeability isµ₂. This composite sphere is surrounded by vacuum of magnetic permeabilityµ0. The magnetic eld at innity is the vertical uniform eldH0:=H0ez. The magnetic eld solves

(D.4.5) ∇×H= 0, ∇·(µH) = 0, lim

kxk→+∞H₀(x) =H₀e_z.

This problem has an analytical steady state solution which is derived in [45] and which we briey recall for the sake of completeness.

There is a scalar potentialψso thatH=∇ψinR³, andψsolves∇·(µ∇ψ) = 0inR³ with

∇ψ → H₀e_z at innity. Using the spherical coordinates (%, ϑ, θ), where % is the distance to the origin,ϑ∈[0, π]is the colatitude and θ∈[0,2π) is the azimuth, the potential is given by

(D.4.6) ψ(%, ϑ, θ) =











−A%cosϑ, for %≤R₁

−

B%+C^R_%2³¹

cosϑ for R₁≤%≤R₂

−

D^R_%2³¹ −H0%

cosϑ for R2≤%,

where A, B, C and D are constants. The constants can be computed by enforcing ψ and µ∂%ψ to be continuous across Σµ and Σ, (the continuity ofψ guarantees that the tangential components of the magnetic eld H are continuous and the continuity of µ∂%ψ guarantees that the normal component of the magnetic induction µH is continuous). To simplify the expressions ofA,B,C and Dwe assume that µ1 =µ0 and we abuse the notation by setting µ:=µ2/µ0. Then,

A=− 9µH₀

(2µ+ 1)(µ+ 2)−2(µ−1)² R1

R2

3

D=

(2µ+ 1)(µ−1)

R2

R1

3

−1

H0

(2µ+ 1)(µ+ 2)−2(µ−1)² R1

R2

3

B = 1 3

2 + 1

µ

A, C = 1 3

1− 1

µ

A.

The magnetic eld in Ω₁ is H|_Ω1 =−Ae_z. Whether the spheres are composed of conducting material or not does not matter since the conductivity coecient does not appear in any formula. As a result, the inner sphere can be viewed from two dierent perspectives: we can either consider Ω₁ to be part of the conducting medium (with µ₁ = µ₀), in which case Ωc = Ω1 ∪Ω2, or we can consider Ω1 to be part of the non-conducting medium, in which

µ h H, L² COC ∇×H, L² COC ∇·(µ^cH^c), L² COC φ, H¹ COC

2

0.16 1.68810⁻³ - 7.32810⁻³ - 2.66510⁻² - 9.53610⁻⁵ -0.08 2.69110⁻⁴ 2.65 2.09410⁻³ 1.81 1.06810⁻² 1.32 2.01810⁻⁵ 2.24 0.04 3.89810⁻⁵ 2.79 4.88910⁻⁴ 2.10 3.83110⁻³ 1.48 3.43110⁻⁶ 2.56 0.02 7.08810⁻⁶ 2.46 1.23910⁻⁴ 1.98 1.48010⁻³ 1.37 5.94510⁻⁷ 2.53 0.01 1.36310⁻⁶ 2.38 3.11410⁻⁵ 1.99 5.98010⁻⁴ 1.31 1.03210⁻⁷ 2.53

20

0.16 8.04410⁻³ - 3.72910⁻² - 1.31410⁻² - 3.21810⁻⁴ -0.08 1.00410⁻³ 3.00 6.18010⁻³ 2.59 6.69910⁻³ 0.97 7.06510⁻⁵ 2.19 0.04 1.08910⁻⁴ 3.21 4.27310⁻⁴ 3.85 1.84510⁻³ 1.86 1.25310⁻⁵ 2.50 0.02 2.04810⁻⁵ 2.41 4.57010⁻⁵ 3.22 4.85610⁻⁴ 1.93 2.22010⁻⁶ 2.50 0.01 3.83210⁻⁶ 2.42 1.06910⁻⁵ 2.10 1.31010⁻⁴ 1.89 3.88510⁻⁷ 2.51

200

0.16 1.06710⁻¹ - 3.72810⁻¹ - 3.87610⁻³ - 3.98410⁻⁴ -0.08 2.43910⁻² 2.13 9.23910⁻² 2.01 2.62010⁻³ 0.57 8.33110⁻⁵ 2.26 0.04 4.32110⁻³ 2.50 1.57110⁻² 2.56 1.07610⁻³ 1.28 1.44410⁻⁵ 2.53 0.02 6.54710⁻⁴ 2.72 2.23310⁻³ 2.81 4.11410⁻⁴ 1.39 2.57710⁻⁶ 2.49 0.01 9.00810⁻⁵ 2.86 2.95610⁻⁴ 2.92 1.22310⁻⁴ 1.75 4.53610⁻⁷ 2.51

Table D.4: Case 1, P2/P2; one iteration (∆t= 10⁹); α= 0.75

case Ω_c = Ω₂. Both cases are described by the same steady solution but the numerical approximations computed by our method are computed dierently.

Note that A → 0, B → 0, C → 0, D → (R₂/R₁)³H₀, and µB → 3H₀/(1−(R₁/R₂)³), µC → 6H0/(1−(R1/R2)³) when µ → ∞; as a result, the magnetic eld tends to zero in Ω₁∪Ω₂ but the magnetic induction converges to a non-zero limit in Ω₂ when µ→ ∞. The magnetic eld penetrates more or less in the spheres depending on the value of µ, and it is completely expelled from the spheres in the limitµ→ ∞.

Case 1: Inner sphere is a conductor

We assume that Ωc = Ω1 ∪Ω2, i.e., the conducting medium is composed of the inner and the outer spheres. We takeL := R₂ as reference length scale and we set R₁ = ¹₂R₂. We set H := H0 to non-dimensionalize the magnetic eld since there is no velocity to construct a reference magnetic eld.

The innite vacuum region is truncated at % = 10R2. We enforce the time-independent Dirichlet condition φ = H₀z := H₀%cosϑ at the outer boundary of the vacuum region, Γ_v. The steady solution is computed in one time step using∆t= 10⁹. (Recall that the steady-state problem is now well-posed thanks to our introducing the magnetic pressure.)

The above problem is solved using various uniformly rened meshes and various values of µ. The stabilizing exponentα is equal to0.75. The magnetic pressure is approximated using P1 elements, the magnetic eld is approximated usingP2 elements, and the scalar potential is approximated usingP2 elements. For each computation we measure the relative error on H^c,

∇×H^c, ∇·(µ^cH^c) in the L²(Ωc)-norm, and the error on φin the H¹(Ωv)-norm. The results are reported in Table D.4. The method converges well in the rangeµ∈[2,200].

Figure D.3 shows the computed solution for µ = 200. We observe that the radial com-ponent H_r^c (panel (a)) is continuous at (% = R₁ and ϑ = 0, ϑ = π) and that the vertical component H_z^c (panel (b)) is continuous at (% = R1, ϑ = π/2). This shows that the IP

method enforces well the continuity of the tangential component of the magnetic eld. The panel (c) shows the magnetic eld lines of H^c (0 ≤ % ≤ R₂) and those of ∇φ (R₂ ≤ %).

The magnetic lines in the vacuum region arrive nearly perpendicularly at the ferromagnetic interface. This phenomenon is a feature ofµ→ ∞.

(a) (b) (c)

Figure D.3: Steady solution for a composite sphere embedded in a vertical uniform magnetic eld: (a-b) Hr and Hz for conducting inner and outer spheres with relative per-meabilityµ= 200; (c) magnetic eld lines.

Case 2: Hollow sphere

We use the same geometric setting as in case 1, but we now assume that the inner sphere is an insulator, i.e.,Ω_c= Ω₂. The non-conducting medium,Ω_v, is composed of the inner sphere Ω₁ plus the spherical annulus%∈(R₂,10R₂). The exact solution to this problem is the same as in case 1.

We repeat the same convergence tests as in case 1. The results are reported in Table D.5.

We observe that the method converges well in the rangeµ∈[2,200]and that the convergence rates are almost identical to those shown in Table D.4.

D.4.3 Induction in rotating devices

We test in this section the proposed method on rotating conductors embedded in a uniform external magnetic eld. We make two numerical tests: the rst one assesses the robustness of the method with respect to geometrical singularities and the second one assesses the robustness of the method with respect to high permeability contrasts. These tests have been preformed withα= 0.7.

Induction in a nite rotating solid cylinder

LetΩc be a conducting cylinder of non-dimensional radiusR = 1and height L^c_z = 1.6. This cylinder is embedded in vacuum inR³ and rotates about the z-axis with angular speed$= 1.(The reference velocityU is equal to the product of the radius of the cylinder and the angular

µ h H, L² COC ∇×H, L² COC ∇·(µ^cH^c), L² COC φ, H¹ COC

2

0.16 1.59010⁻³ - 7.31410⁻³ - 2.42410⁻² - 1.36210⁻⁴ -0.08 2.91310⁻⁴ 2.45 2.00210⁻³ 1.87 9.42310⁻³ 1.36 2.67910⁻⁵ 2.35 0.04 2.89810⁻⁵ 3.33 4.52510⁻⁴ 2.15 3.28510⁻³ 1.52 3.92410⁻⁶ 2.77 0.02 4.91010⁻⁶ 2.56 1.08810⁻⁴ 2.06 1.18910⁻³ 1.47 6.69410⁻⁷ 2.55 0.01 1.10910⁻⁶ 2.15 2.66510⁻⁵ 2.03 4.63710⁻⁴ 1.36 1.16210⁻⁷ 2.53

20

0.16 9.41810⁻³ - 3.92410⁻² - 1.28210⁻² - 3.42310⁻⁴ -0.08 1.49410⁻³ 2.66 6.74910⁻³ 2.54 6.62710⁻³ 0.95 7.26110⁻⁵ 2.24 0.04 1.95210⁻⁴ 2.94 5.85910⁻⁴ 3.53 1.83210⁻³ 1.85 1.24510⁻⁵ 2.54 0.02 2.40910⁻⁵ 3.02 7.07510⁻⁵ 3.05 4.81910⁻⁴ 1.93 2.20310⁻⁶ 2.50 0.01 2.88910⁻⁶ 3.06 1.25510⁻⁵ 2.49 1.29110⁻⁴ 1.90 3.86210⁻⁷ 2.51

200

0.16 1.09810⁻¹ - 3.93410⁻¹ - 3.86110⁻³ - 4.01310⁻⁴ -0.08 2.47410⁻² 2.15 9.84710⁻² 2.00 2.59610⁻³ 0.57 8.38010⁻⁵ 2.26 0.04 4.41510⁻³ 2.49 1.74010⁻² 2.50 1.06710⁻³ 1.28 1.47210⁻⁵ 2.51 0.02 7.45110⁻⁴ 2.57 2.65810⁻³ 2.71 4.09110⁻⁴ 1.38 2.64210⁻⁶ 2.48 0.01 1.21110⁻⁴ 2.62 3.99910⁻⁴ 2.73 1.21710⁻⁴ 1.75 4.66810⁻⁷ 2.50

Table D.5: Case 2, P2/P2; one iteration (∆t= 10⁹); α= 0.75

velocity.) The non-dimensional conductivity is σ = 1 and the magnetic Reynolds number is Rm= 100. The non-dimensional magnetic permeability in the entire electromagnetic domain is constant and equal to one, i.e.,µ^c=µ^v= 1. The imposed magnetic eld at innity isH₀e_x. This is a benchmark test case thoroughly investigated in [105].

The time-dependent problem is solved with initial data H0 =H0ex on a Delaunay mesh which is quasi-uniform in the conducting region and of mesh-size h = 1/100. We use P2

elements for both the magnetic eld and the magnetic potential. The magnetic pressure is approximated using P1 elements. The time step is ∆t = 5 10⁻². The truncated numerical domain is Ω = {r ∈ (0,1.6), θ ∈ [0,2π], z ∈ (−4,4)} and the non-conducting domain is Ω_v = Ω\Ω_c. The imposed boundary condition on Γ_v is φ|_Γv = H₀rcosθ. The only active Fourier mode ism= 1.

The time evolution of the magnetic energy is shown in Figure D.4(a). The graph shows oscillations that correspond to reconnections of the magnetic lines. Figure D.4(b) shows the radial prole of Hz at z = 0.8 in the meridian plane θ = 0 at steady state. Note that the point r = 1, θ = 0, z = 0.8 is located on the upper sharp edge of the cylinder. The prole is compared with that obtained in [105]. The agreement is excellent considering that the gradient of the solution is discontinuous at the edges of the cylinder.

Figures D.5(a), D.5(b) show the contour lines of them= 1 azimuthal Fourier mode ofH_θ at t= 100. Observe thatH_θ is symmetric with respect to the equatorial plane. Plotting the contour lines ofHθ emphasizes the skin eect. The lines shown in Figures D.5(a), D.5(b) are very close to those reported in Figure 5 from [105] even at the corners. Figure D.5(c) shows the streamlines of the Fourier modem= 1 of the electric current in the cylinder. The current is mainly contained in a thin layer (of the order of the skin depth). It varies smoothly in the azimuthal direction and bends sharply at the corners. This behavior is a direct consequence of the presence of the cylindrical interface with vacuum. The current creates thez-component of the magnetic eld and is responsible for the strong extremum ofHz at the sharp edges of

0.05 0.1 0.15 0.2 0.25 0.3 0.35

0 20 40 60 80 100

Emag

(a) Time evolution of the magnetic energyTime -0.1

0 0.1 0.2 0.3 0.4

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

-Hzc

r

LMW

(b) Radial prole ofHz(r, z= 0.8)

Figure D.4: Induction in a nite rotating solid cylinder at R_m = 100. 'LMW' is the result from [105], 'FEM' is our result withP2nite elements forHandP2 nite elements forφ withh= 1/100.

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0.0497 0.0497

0.0497 0.0497

(b)θ=π/2. (c) Electric current

Figure D.5: Rotating cylinder at Rm = 100 at steady state. Contours of Hθ of the m = 1 mode in azimuthal planes and streamlines of the electric current of the m = 1 mode colored by the norm of the current.

the cylinder (r= 1, θ∈[0,2π], z=±0.8)

Induction in counter-rotating disks made of soft iron

In order to measure the impact of soft iron disks on induction elds, we now consider two counter-rotating disks embedded in a cylindrical conductor which is itself embedded in vacuum.

This test case is a qualitative illustration of the Cadarache VKS2 uid dynamo studied in more details in Section D.5.

The conducting domain is a cylinder of non-dimensional radius R = 1 and of rectan-gular cross section of non-dimensional height L = 2.55: Ωc = {(r, θ, z);r ∈ [0,1), z ∈

(−1.275,1.275), θ ∈ [0,2π)}. Two counter-rotating disks, Ω^topc , Ω^bot_c , are embedded in Ωc. The upper rotating conducting disk is a cylinder whose cross section is dened as follows:

(0.775≤z≤0.975 if r ≤0.65, (r−0.65)²+ (z−0.875)² ≤(0.1)² if r ≥0.65.

The lower rotating conducting disk is the image by reection about the equatorial plane z = 0 of the upper disk. There is no analytical solution to this problem, but asymptotic solutions are given in [69] assuming that the disks are of rectangular cross section. The upper and lower disks rotate with non-dimensional angular speed $_top =−1 and $_bot = 1, respectively. The non-dimensional magnetic permeability and conductivity of the non-rotating solid container,Ωc\(Ω^top_c ∪Ω^bot_c ), are µ0 = 1 and σ0 = 1, respectively. The non-dimensional magnetic permeability and conductivity of the two counter-rotating disks,Ω^topc ∪Ω^bot_c , are µ_d and σd, respectively. The non-dimensional magnetic permeability of the vacuum is µ0 = 1.

The imposed velocity eld inΩc is

˜

uⁿ⁺¹(x) =







0 inΩ_c\(Ω^top_c ∪Ω^bot_c )

$_topez×x inΩ^topc

$_bote_z×x inΩ^bot_c

The device is placed in a transverse uniform magnetic eldH₀:=H₀e_x=H₀(cosθe_r−sinθe_θ) and we look for the steady state solution in two cases: (a) µ_d = 200µ₀, σ_d = 1; (b)µ_d = 1, σd= 200σ0. In both cases the eective magnetic Reynolds number is the same for the disks R^disks_m =µ_dσ_d$_botR² = 200µ₀σ₀.

For computational purposes the vacuum region is truncated and restricted to the sphere of non-dimensional radius Rv = 10. The time-independent Dirichlet condition φ= H0x :=

H₀rcosθ is enforced at the outer boundary of the vacuum region,Γ_v. The steady solution is computed by advancing (D.3.11) in time until convergence to steady state is reached. We use theP2/P2 nite element pair forH andφand P1 elements for the magnetic pressure.

Some three-dimensional representations of the computed solutions are shown in Figure D.6.

Panels (a) and (d) show some magnetic eld streamlines near the top disk seen from the side of the cylinder. Panels (b) and (e) show the same magnetic eld streamlines seen from the top of the cylinder. Panels (c) and (f) show the contour of the magnetic energy corresponding to10%of the maximum energy. The top panels correspond to the solution with µ_d=µ₀ and σd = 200σ0 and the bottom panels (c,d,e) correspond to the solution with µd = 200µ0 and σ_d = σ0. The two steady solutions are very dierent although the two congurations have the same eective magnetic Reynolds number. When the disks are non-ferromagnetic, the magnetic eld lines are distorted horizontally due to the eddy current in each disk. When the disks are ferromagnetic, the eld lines are distorted inside the disks but also outside as they connect nearly perpendicularly to the disks.

Dans le document Analyse théorique et numérique des équations de la magnétohydrodynamique : (Page 164-173)