Coupling of eddy-current and circuit problems

Transcription

1 Coupling of eddy-current and circuit problems Ana Alonso Rodríguez*, ALBERTO VALLI*, Rafael Vázquez Hernández** * Department of Mathematics, University of Trento ** Department of Applied Mathematics, University of Santiago de Compostela Coupling of eddy-currentand circuit problems p.1/28

2 Time-harmonic eddy-current equations Starting from Maxwell equations, assuming a sinusoidal dependence on time and disregarding displacement currents one obtains the so-called time-harmonic eddy-current problem { curlh σe = 0 in Ω curle + iωµh = 0 in Ω. (1) Coupling of eddy-currentand circuit problems p.2/28

3 Time-harmonic eddy-current equations Starting from Maxwell equations, assuming a sinusoidal dependence on time and disregarding displacement currents one obtains the so-called time-harmonic eddy-current problem { curlh σe = 0 in Ω curle + iωµh = 0 in Ω. (1) Here H and E are the magnetic and electric fields, respectively σ and µ are the electric conductivity and the magnetic permeability, respectively ω 0 is the frequency. Coupling of eddy-currentand circuit problems p.2/28

4 Time-harmonic eddy-current equations (cont d) [As shown in the previous talk, in an insulator one has σ = 0, therefore E is not uniquely determined in that region (E + ψ is still a solution). Some additional conditions ("gauge" conditions) are thus necessary: as in the insulator Ω I we have no charges, we impose div(ǫe) = 0 in Ω I, (2) where ǫ is the electric permittivity.] Coupling of eddy-currentand circuit problems p.3/28

5 Geometry The physical domain Ω R 3 is a box", and the conductor Ω C is simply-connected with Ω C Ω = Γ E Γ J, where Γ E and Γ J are connected and disjoint surfaces on Ω ( electric ports"). Notation: Γ = Ω C Ω I, Ω = Γ E Γ J Γ D, Ω C = Γ E Γ J Γ, Ω I = Γ D Γ. Γ D Γ J Γ Ξ Γ E Coupling of eddy-currentand circuit problems p.4/28

6 Boundary conditions We will distinguish among three types of boundary conditions. Coupling of eddy-currentand circuit problems p.5/28

7 Boundary conditions We will distinguish among three types of boundary conditions. Electric (Case A). One imposes E n = 0 on Ω. Coupling of eddy-currentand circuit problems p.5/28

8 Boundary conditions We will distinguish among three types of boundary conditions. Electric (Case A). One imposes E n = 0 on Ω. Magnetic (Case B). One imposes E n = 0 on Γ E Γ J, H n = 0 and ǫe n = 0 on Γ D. Coupling of eddy-currentand circuit problems p.5/28

9 Boundary conditions We will distinguish among three types of boundary conditions. Electric (Case A). One imposes E n = 0 on Ω. Magnetic (Case B). One imposes E n = 0 on Γ E Γ J, H n = 0 and ǫe n = 0 on Γ D. No-flux (Case C) [Bossavit, 2000]. One imposes E n = 0 on Γ E Γ J, µh n = 0 and ǫe n = 0 on Γ D. Coupling of eddy-currentand circuit problems p.5/28

10 Voltage and current intensity When one wants to couple the eddy-current problem with a circuit problem, one has to consider, as the only external datum that determines the solution, a voltage V or a current intensity I 0. Coupling of eddy-currentand circuit problems p.6/28

11 Voltage and current intensity When one wants to couple the eddy-current problem with a circuit problem, one has to consider, as the only external datum that determines the solution, a voltage V or a current intensity I 0. Question: Coupling of eddy-currentand circuit problems p.6/28

12 Voltage and current intensity When one wants to couple the eddy-current problem with a circuit problem, one has to consider, as the only external datum that determines the solution, a voltage V or a current intensity I 0. Question: how can we formulate the eddy-current problems when the excitation is given by a voltage or by a current intensity? Coupling of eddy-currentand circuit problems p.6/28

13 Voltage and current intensity When one wants to couple the eddy-current problem with a circuit problem, one has to consider, as the only external datum that determines the solution, a voltage V or a current intensity I 0. Question: how can we formulate the eddy-current problems when the excitation is given by a voltage or by a current intensity? This is a delicate point, as eddy-current problems, for the two cases A and B, have a unique solution already before a voltage or a current intensity is assigned! Coupling of eddy-currentand circuit problems p.6/28

14 Poynting Theorem (energy balance) In fact one has: Uniqueness theorem. In the cases A and B for the solution of the eddy-current problem (1) the magnetic field H in Ω and the electric field E C in Ω C are uniquely determined. [Adding the "gauge" conditions, also the electric field E I in Ω I is uniquely determined.] Coupling of eddy-currentand circuit problems p.7/28

15 Poynting Theorem (energy balance) In fact one has: Uniqueness theorem. In the cases A and B for the solution of the eddy-current problem (1) the magnetic field H in Ω and the electric field E C in Ω C are uniquely determined. [Adding the "gauge" conditions, also the electric field E I in Ω I is uniquely determined.] Proof. Multiply the Faraday equation by H, integrate in Ω and integrate by parts: it holds 0 = Ω curle H + Ω iωµh H = Ω E curlh + Ω iωµh H + Ω n E H. Coupling of eddy-currentand circuit problems p.7/28

16 Poynting Theorem (energy balance) In fact one has: Uniqueness theorem. In the cases A and B for the solution of the eddy-current problem (1) the magnetic field H in Ω and the electric field E C in Ω C are uniquely determined. [Adding the "gauge" conditions, also the electric field E I in Ω I is uniquely determined.] Proof. Multiply the Faraday equation by H, integrate in Ω and integrate by parts: it holds 0 = Ω curle H + Ω iωµh H = Ω E curlh + Ω iωµh H + Ω n E H. Remembering that curlh I = 0 in Ω I and replacing curlh C with σe C, one has the Poynting Theorem (energy balance) Coupling of eddy-currentand circuit problems p.7/28

17 Poynting Theorem (energy balance) (cont d) Ω C σe C E C + Ω iωµh H = Ω n E H. Coupling of eddy-currentand circuit problems p.8/28

18 Poynting Theorem (energy balance) (cont d) Ω C σe C E C + Ω iωµh H = Ω n E H. The term on Ω is clearly vanishing in the cases A and B. Coupling of eddy-currentand circuit problems p.8/28

19 Poynting Theorem for the case C In the case C, instead, since div τ (E n) = iωµh n = 0 on Ω, one has and therefore E n = grad W n on Ω, Ω n E H = Ω H n gradw = Ω div(h n)w = Ω curlh nw = W Γ J Γ J curlh C n, as curlh I = 0 in Ω I, and we have denoted by W ΓJ the (constant) value of the potential W on the electric port Γ J (whereas W ΓE = 0). Coupling of eddy-currentand circuit problems p.9/28

20 Poynting Theorem for the case C (cont d) In this case a degree of freedom is indeed still free (either the voltage W ΓJ, that will be denoted by V, or else the current intensity Γ J curlh C n in Ω C, that will be denoted by I 0 ). Coupling of eddy-currentand circuit problems p.10/28

21 Poynting Theorem for the case C (cont d) In this case a degree of freedom is indeed still free (either the voltage W ΓJ, that will be denoted by V, or else the current intensity Γ J curlh C n in Ω C, that will be denoted by I 0 ). We can thus conclude that the only meaningful boundary value problem is the one with assigned no-flux boundary conditions: the case C. Coupling of eddy-currentand circuit problems p.10/28

22 The case C: variational formulation How can we formulate the problem when the voltage or the current intensity are assigned? [Alonso Rodríguez, Valli and Vázquez Hernández, 2009] [Other approaches: Bíró, Preis, Buchgraber and Tičar, 2004; Bermúdez, Rodríguez and Salgado, 2005] Coupling of eddy-currentand circuit problems p.11/28

23 The case C: variational formulation How can we formulate the problem when the voltage or the current intensity are assigned? [Alonso Rodríguez, Valli and Vázquez Hernández, 2009] [Other approaches: Bíró, Preis, Buchgraber and Tičar, 2004; Bermúdez, Rodríguez and Salgado, 2005] This orthogonal decomposition result turns out to be useful: each vector function v I can be decomposed as v I = µ 1 I curlq I + gradψ I + αρ I, where ρ I is a harmonic field, namely, it belongs to the space H µi (Ω I ) := {v I (L 2 (Ω I )) 3 curlv I = 0, div(µ I v I ) = 0, µ I v I n = 0 on Ω I }. Coupling of eddy-currentand circuit problems p.11/28

24 The case C: variational formulation (cont d) The harmonic field ρ I is known from the data of the problem, and satisfies Γ J ρ I dτ = 1; moreover, if the vector field v I satisfies curlv I = 0, it follows q I = 0 and therefore α = Γ J v I dτ. Coupling of eddy-currentand circuit problems p.12/28

25 The case C: variational formulation (cont d) The harmonic field ρ I is known from the data of the problem, and satisfies Γ J ρ I dτ = 1; moreover, if the vector field v I satisfies curlv I = 0, it follows q I = 0 and therefore α = Γ J v I dτ. In particular, setting H I = grad ψ I + α I ρ I, from the Stokes Theorem one has I 0 = curlh C n C = H C dτ = H I dτ = α I, Γ J Γ J Γ J hence H I = gradψ I + I 0 ρ I. (3) Coupling of eddy-currentand circuit problems p.12/28

26 The case C: variational formulation (cont d) The harmonic field ρ I is known from the data of the problem, and satisfies Γ J ρ I dτ = 1; moreover, if the vector field v I satisfies curlv I = 0, it follows q I = 0 and therefore α = Γ J v I dτ. In particular, setting H I = grad ψ I + α I ρ I, from the Stokes Theorem one has I 0 = curlh C n C = H C dτ = H I dτ = α I, Γ J Γ J Γ J hence H I = gradψ I + I 0 ρ I. (3) We want to provide a "coupled" variational formulation, in terms of E C in Ω C and of H I in Ω I. Coupling of eddy-currentand circuit problems p.12/28

27 The case C: variational formulation (cont d) Inserting the Faraday equation into the Ampère equation in Ω C we find Ω C µ 1 C curle C curlw C + iω Ω C σe C w C iω Γ w C n C H I = 0. (4) Coupling of eddy-currentand circuit problems p.13/28

28 The case C: variational formulation (cont d) Inserting the Faraday equation into the Ampère equation in Ω C we find Ω C µ 1 C curle C curlw C + iω Ω C σe C w C iω Γ w C n C H I = 0. (4) Instead, the Faraday equation in Ω I gives iω µ I H I grad ϕ I + E C n C gradϕ I = 0 (5) Ω I Γ Coupling of eddy-currentand circuit problems p.13/28

29 The case C: variational formulation (cont d) Inserting the Faraday equation into the Ampère equation in Ω C we find Ω C µ 1 C curle C curlw C + iω Ω C σe C w C iω Γ w C n C H I = 0. (4) Instead, the Faraday equation in Ω I gives iω µ I H I grad ϕ I + E C n C gradϕ I = 0 (5) Ω I Γ and iω Ω I µ I H I ρ I + Γ E C n C ρ I = V. (6) Coupling of eddy-currentand circuit problems p.13/28

30 The case C: variational formulation (cont d) Here we have to note that Γ D E I n I ρ I = Γ D gradw n I ρ I = Γ D div τ (ρ I n I )W + V Γ J ρ I dτ = V. Coupling of eddy-currentand circuit problems p.14/28

31 The case C: variational formulation (cont d) Here we have to note that Γ D E I n I ρ I = Γ D gradw n I ρ I = Γ D div τ (ρ I n I )W + V Γ J ρ I dτ = V. Using (3) in (4), (5) and (6) one has Ω C µ 1 C curle C curlw C + iω Ω C σe C w C iω Γ w C n C grad ψ I iωi 0 Γ w C n C ρ I = 0 (7) iω E C n C grad ϕ I +ω Ω 2 µ I grad ψ I grad ϕ I = 0 (8) I Γ iωq Γ E C n C ρ I + ω 2 I 0 Q µ I ρ I ρ I = iωv Q. (9) Ω I Coupling of eddy-currentand circuit problems p.14/28

32 The case C: existence and uniqueness If V is given, one solves (7), (8), (9) and determines E C, ψ I and I 0 (hence H C and H I ). Coupling of eddy-currentand circuit problems p.15/28

33 The case C: existence and uniqueness If V is given, one solves (7), (8), (9) and determines E C, ψ I and I 0 (hence H C and H I ). If I 0 is given, one solves (7), (8) and determines E C and ψ I (hence H C and H I ); then from (9) one can also compute V. Coupling of eddy-currentand circuit problems p.15/28

34 The case C: existence and uniqueness If V is given, one solves (7), (8), (9) and determines E C, ψ I and I 0 (hence H C and H I ). If I 0 is given, one solves (7), (8) and determines E C and ψ I (hence H C and H I ); then from (9) one can also compute V. Both problems are well-posed, namely, they have a unique solution, since the associated sesquilinear form is coercive (thus one can apply the Lax Milgram Lemma). Coupling of eddy-currentand circuit problems p.15/28

35 The case C: existence and uniqueness If V is given, one solves (7), (8), (9) and determines E C, ψ I and I 0 (hence H C and H I ). If I 0 is given, one solves (7), (8) and determines E C and ψ I (hence H C and H I ); then from (9) one can also compute V. Both problems are well-posed, namely, they have a unique solution, since the associated sesquilinear form is coercive (thus one can apply the Lax Milgram Lemma). Moreover, it is simple to propose an approximation method based on finite elements, of "edge" type for E C in Ω C and of (scalar) nodal type for ψ I in Ω I. Convergence is assured by the Céa Lemma. Coupling of eddy-currentand circuit problems p.15/28

36 The case C: existence and uniqueness If V is given, one solves (7), (8), (9) and determines E C, ψ I and I 0 (hence H C and H I ). If I 0 is given, one solves (7), (8) and determines E C and ψ I (hence H C and H I ); then from (9) one can also compute V. Both problems are well-posed, namely, they have a unique solution, since the associated sesquilinear form is coercive (thus one can apply the Lax Milgram Lemma). Moreover, it is simple to propose an approximation method based on finite elements, of "edge" type for E C in Ω C and of (scalar) nodal type for ψ I in Ω I. Convergence is assured by the Céa Lemma. [However, an efficient implementation demands to replace the harmonic field ρ I with an easily computable function.] Coupling of eddy-currentand circuit problems p.15/28

37 Physical interpretation Note: the physical interpretation of equation (9) is that E C dr + iω µ I H I n Ξ = V, γ where γ = Ξ Γ is oriented from Γ J to Γ E, and n Ξ is directed in such a way that γ is clockwise oriented with respect to it. In other words, if it is possible to determine the electric field E I in Ω I satisfying the Faraday equation, it follows that γ E I dr = V, where γ = Ξ Γ D is oriented from Γ E to Γ J : hence (9) is indeed determining the voltage drop between the electric ports. Ξ Coupling of eddy-currentand circuit problems p.16/28

38 Physical interpretation (cont d) This explains from another point of view why, when the source is a voltage drop or a current intensity, it is not possible to assume the electric boundary conditions E n = 0 on Ω. Coupling of eddy-currentand circuit problems p.17/28

39 Physical interpretation (cont d) This explains from another point of view why, when the source is a voltage drop or a current intensity, it is not possible to assume the electric boundary conditions E n = 0 on Ω. In fact, in that case one would have γ E I dr = 0, hence from (9) iω Ξ µ IH I n Ξ = V + γ E C dr = V + γ γ E dr = V + Ξ E dr, with Ξ clockwise oriented with respect n Ξ : due to the term V the Faraday equation would be violated on Ξ! Coupling of eddy-currentand circuit problems p.17/28

40 Numerical results for the Case C Coming back to the case C and to its variational formulation (7), (8), (9), we use edge finite elements of the lowest degree (a + b x in each element) for approximating E C, and scalar piecewise-linear elements for approximating ψ I. Coupling of eddy-currentand circuit problems p.18/28

41 Numerical results for the Case C Coming back to the case C and to its variational formulation (7), (8), (9), we use edge finite elements of the lowest degree (a + b x in each element) for approximating E C, and scalar piecewise-linear elements for approximating ψ I. The problem description is the following: the conductor Ω C and the whole domain Ω are two coaxial cylinders of radius R C and R D, respectively, and height L. Assuming that σ and µ are scalar constants, the exact solution for an assigned current intensity I 0 is known (through suitable Bessel functions), and also the basis function ρ I is known, thus from (9) one easily computes the voltage V, too. Coupling of eddy-currentand circuit problems p.18/28

42 Numerical results for the Case C (cont d) We have the following data: R C R D = 0.25 m = 0.5 m L = 0.25 m σ = S/m µ = 4π 10 7 H/m ω = 2π 50 rad/s and I 0 = 10 4 A or V = i [the voltage corresponds to the current intensity I 0 = 10 4 A]. Coupling of eddy-currentand circuit problems p.19/28

43 Numerical results for the Case C (cont d) The relative errors (for E C in H(curl; Ω C ) and for H I in L 2 (Ω I )) with respect to the number of degrees of freedom are given by: Coupling of eddy-currentand circuit problems p.20/28

44 Numerical results for the Case C (cont d) The relative errors (for E C in H(curl; Ω C ) and for H I in L 2 (Ω I )) with respect to the number of degrees of freedom are given by: Elements DoF e E e H e V Coupling of eddy-currentand circuit problems p.20/28

45 Numerical results for the Case C (cont d) The relative errors (for E C in H(curl; Ω C ) and for H I in L 2 (Ω I )) with respect to the number of degrees of freedom are given by: Elements DoF e E e H e V Elements DoF e E e H e I Coupling of eddy-currentand circuit problems p.20/28

46 Numerical results for the Case C (cont d) On a graph: for assigned current intensity Relative errors Rel. error E C Rel. error H D Rel. error V y = Ch y=ch Number of d.o.f. Coupling of eddy-currentand circuit problems p.21/28

47 Numerical results for the Case C (cont d) for assigned voltage Relative errors Rel. error E C Rel. error H D Rel. error I y = Ch y=ch Number of d.o.f. Coupling of eddy-currentand circuit problems p.22/28

48 Numerical results for the Case C (cont d) A more realistic problem, considered by Bermúdez, Rodríguez and Salgado, 2005, is that of a cylindrical electric furnace with three electrodes ELSA [dimensions: furnace height 2 m; furnace diameter 8.88 m; electrode height 1.25 m; electrode diameter 1 m; distance of the center of the electrode from the wall 3 m]. Coupling of eddy-currentand circuit problems p.23/28

49 Numerical results for the Case C (cont d) A more realistic problem, considered by Bermúdez, Rodríguez and Salgado, 2005, is that of a cylindrical electric furnace with three electrodes ELSA [dimensions: furnace height 2 m; furnace diameter 8.88 m; electrode height 1.25 m; electrode diameter 1 m; distance of the center of the electrode from the wall 3 m]. The three electrodes ELSA are constituted by a graphite core of 0.4 m of diameter, and by an outer part of Söderberg paste. The electric current enters the electrodes through horizontal copper bars of rectangular section (0.07 m 0.25 m), connecting the top of the electrode with the external boundary. Coupling of eddy-currentand circuit problems p.23/28

50 Numerical results for the Case C (cont d) A more realistic problem, considered by Bermúdez, Rodríguez and Salgado, 2005, is that of a cylindrical electric furnace with three electrodes ELSA [dimensions: furnace height 2 m; furnace diameter 8.88 m; electrode height 1.25 m; electrode diameter 1 m; distance of the center of the electrode from the wall 3 m]. The three electrodes ELSA are constituted by a graphite core of 0.4 m of diameter, and by an outer part of Söderberg paste. The electric current enters the electrodes through horizontal copper bars of rectangular section (0.07 m 0.25 m), connecting the top of the electrode with the external boundary. Data: σ = 10 6 S/m for graphite, σ = 10 4 S/m for Söderberg paste, σ = S/m for copper, µ = 4π 10 7 H/m, ω = 2π 50 rad/s, I 0 = A for each electrode. Coupling of eddy-currentand circuit problems p.23/28

51 Numerical results for the Case C (cont d) The value of the magnetic "potential" in the insulator: the magnetic field is the gradient of the represented function (not taking into account the jump surfaces). Coupling of eddy-currentand circuit problems p.24/28

52 Numerical results for the Case C (cont d) The magnitude of the current density σe C on a horizontal section of one electrode. Coupling of eddy-currentand circuit problems p.25/28

53 Numerical results for the Case C (cont d) The magnitude of the current density σe C on a vertical section of one electrode. Coupling of eddy-currentand circuit problems p.26/28

54 References A. Alonso Rodríguez, A. Valli and R. Vázquez Hernández: A formulation of the eddy-current problem in the presence of electric ports. Numer. Math., 113 (2009), A. Bermúdez, R. Rodríguez and P. Salgado: Numerical solution of eddy-current problems in bounded domains using realistic boundary conditions. Comput. Methods Appl. Mech. Engrg., 194 (2005), O. Bíró, K. Preis, G. Buchgraber and I. Tičar: Voltage-driven coils in finite-element formulations using a current vector and a magnetic scalar potential, IEEE Trans. Magn., 40 (2004), A. Bossavit: Most general non-local boundary conditions for the Maxwell equations in a bounded region. COMPEL, 19 (2000), Coupling of eddy-currentand circuit problems p.27/28

55 Additional references A. Alonso Rodríguez and A. Valli: Voltage and current excitation for time-harmonic eddy-current problem. SIAM J. Appl. Math., 68 (2008), P. Dular, C. Geuzaine and W. Legros: A natural method for coupling magnetodynamic H-formulations and circuits equations. IEEE Trans. Magn., 35 (1999), R. Hiptmair and O. Sterz: Current and voltage excitations for the eddy current model. Int. J. Numer. Modelling, 18 (2005), J. Rappaz, M. Swierkosz and C.Trophime: Un modèle mathématique et numérique pour un logiciel de simulation tridimensionnelle d induction électromagnétique. Report 05.99, Département de Mathématiques, École Polytechnique Fédérale de Lausanne, Coupling of eddy-currentand circuit problems p.28/28