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Advances in Computational Mathematics (2007) 26: 39–62

 Springer 2006

Transient numerical simulation of a thermoelectrical problem in cylindrical induction heating furnaces A. Bermúdez, D. Gómez, M.C. Muñiz and P. Salgado Departamento de Matemática Aplicada, Universidad de Santiago de Compostela, Spain E-mail: {mabermud;malola;mcarmen;mpilar}@usc.es

Received 28 April 2004; accepted 26 December 2004 Communicated by J. Carnicer and J.M. Peña

Dedicated to Mariano Gasca on his 60th birthday

This paper concerns the mathematical modelling and numerical solution of thermoelectrical phenomena taking place in an axisymmetric induction heating furnace. We formulate the problem in a two-dimensional domain and propose a finite element method and an iterative algorithm for its numerical solution. We also provide a family of one-dimensional analytical solutions which are used to test the two-dimensional code and to predict the behaviour of the furnace under special conditions. Some numerical results for an industrial furnace used in silicon purification are shown. Keywords: numerical methods, simulation, induction heating, eddy currents, electromagnetics, finite elements, phase change, analytical solution Mathematics subject classifications (2000): 35J25, 35K20, 65K05, 65N30

1.

Introduction

Induction heating is a method by which electrically conducting materials (generally metals) are heated by a non-contact method in an alternating electromagnetic field. An induction heating system consists basically of one or several inductors and metallic workpieces to be heated. The inductors are supplied with alternating current which induces eddy currents inside the component being heated due to Faraday’s law. This technique is widely used in the metallurgical industry in an important number of applications such as metal smelting, preheating for operations of welding, purification systems and, in general, in those processes needing a high speed of heating in located zones of a piece of a conductive material. The overall process is very complex and involves different physical phenomena: electromagnetic, thermal with phase change and hydrodynamic in the liquid metal.

40

Bermúdez et al. / Numerical simulation of induction heating furnaces

In the last years, numerical simulation has been revealed as an important tool to design the induction heating system and to know its behaviour, thus allowing to reduce considerably the number of trial–error procedures in plant which usually are long and costly. From the mathematical point of view, the full problem consists of a coupled nonlinear system of partial differential equations which arises from a thermal-magnetohydrodynamic problem. More precisely, the electromagnetic model is an eddy current problem, the thermal one is obtained from the heat transfer equation in transient state with phase change (Stefan problem) and the hydrodynamic model is obtained from the incompressible Navier–Stokes equations. All of the models are coupled because the physical parameters depend on temperature, the heat source in the thermal problem is the Joule effect, the liquid domain of the hydrodynamic model depends on temperature and the velocity of the liquid metal determines convective heat transfer and it also appears in the Ohm’s law. In the literature we can find several publications devoted to solve numerically some of these problems (see, for instance, [4,5,8,12–14]). At our knowledge, none of these works carry out full coupling of electromagnetism, heat transfer and hydrodynamics, but they only focus on the eletromagnetic/heat transfer coupling [4,5,12] or on the electromagnetic/hydrodynamic coupling [13]. As a first step, in this paper we introduce and numerically solve the thermoelectrical problem arising in a cylindrical induction heating furnace. The coupling with the hydrodynamic problem will be the object of a forthcoming paper. The main contribution with respect to previous works like [4,5,12] is that we consider change of state in the workpiece which leads to a multi-valued nonlinearity in the heat equation. We also include the analytical solution of a simplified problem which can be considered as first approximation of a real situation. It has been used in the numerical tests to validate the computer codes. The outline of the paper is the following. In section 2, by assuming cylindrical symmetry, we describe the problem in a two-dimensional domain and propose a finite element method for its numerical solution. The basic equations of the electromagnetic model are expressed in terms of the magnetic vector potential. Moreover, in order to take into account the phase change, we write the heat transfer equation in terms of the enthalpy. In section 3, we propose an iterative algorithm for the numerical solution of the resulting nonlinear problem. On the other hand, to handle the coupling between the thermal and the electromagnetic problems we propose a fixed point algorithm. Finally, in section 4, we present numerical results obtained for an industrial furnace used in silicon purification with our two-dimensional code. We complete the work by giving, in the appendix, the analytical solution of the thermoelectrical problem under simplified assumptions on geometry and data. 2.

Statement of the problem. Mathematical modelling

We consider an induction furnace consisting of an induction coil surrounding a workpiece as the one sketched in figure 1. The workpiece consists of a crucible containing the metal to be heated. The current flowing through the coil produces an electromag-


41

Figure 1. Sketch of the induction furnace and radial section.

netic field. This, in turn, induces the so-called eddy currents in the workpiece which, due to the Joule effect, produce heat that melts the metal. The goal is to compute the distribution of heat in the workpiece caused by the eddy currents. We shall denote by 0 the radial section of the workpiece to be heated and by 1 , 2 , . . . , m the radial sections of the turns of the coil. In fact, to be able to consider the problem in an axisymmetric setting, the induction coil is replaced by m cylindrical rings. Moreover, a will be the air around the conductors in such a way that = a ∪ 0 ∪ 1 ∪ · · · ∪ m will denote the two-dimensional domain of the model (see figure 2). In principle, is a half-plane and we should impose “boundary conditions” at infinite. Although this problem can be handled by using boundary elements, for the sake of simplicity, in the present paper we cut the domain and introduce suitable boundary conditions on the artificial boundary (see section 2.1.1 below). 2.1. The electromagnetic submodel Since alternating currents are considered, all of the fields have the following form: (1) F(x, t) = Re eiωt F(x) where t is time, x ∈ R3 is the space position, ω is the angular frequency, i the imaginary unit number and F(x) is the complex amplitude of the field. Moreover, in the low-frequency harmonic regime, the term in Ampere’s law including the electric displacement can be neglected. This quasi-static assumption is reasonable if the current frequency ω is low enough as it happens in industrial furnaces (for a discussion of parameter ranges in which this model is valid one can see, for instance, the book by Bossavit [3]).

42


Figure 2. Computational domain for the electromagnetic problem.

Under these assumptions, the Maxwell’s equations reduce to the so-called eddy current model curl H = J, i ωB + curl E = 0, div B = 0, div D = ,

(2) (3) (4) (5)

to which we have to add the data of the intensity current, I, flowing in the coil. In (2)–(5), H, J, B, E and D are the complex amplitudes associated with the magnetic field, the current density, the magnetic induction, the electric field and the electric displacement, respectively; is the charge density. The system (2)–(5) above needs to be completed by the constitutive relations B = µH, D = εE,

(6) (7)

which take into account material properties: µ is the magnetic permeability and ε is the electric permittivity. We also need the Ohm’s law σ E inside conductors, J= (8) 0 in air, where σ is the electric conductivity. Remark 1. We notice that, since ω = 0, equation (4) follows from equation (3). Moreover charge density is not a data in conductors but a result. Indeed, as we will see later, equations (2) and (3) can be solved independently of (5) leading to H in the whole


43

domain and J in conductors. Then E can be uniquely determined in conductors by E = J/σ . Furthermore, E and equation (5) determine the charge density in conductors. However, in order to determine the electric field in the dielectrics we need the charge density and then E is determined by solving the equations div(εE) = , curl E = −i ωB, with suitable boundary conditions. Anyway, since J is the relevant field in induction heating problems, solving for E in the dielectrics is not required. As we already said, we are interested in solving the problem using a cylindrical coordinate system (r, θ, z) with the z-axis coinciding with the axis of the cylinder. We denote er , eθ and ez the local unit vectors in the corresponding coordinate directions. Now we assume cylindrical symmetry, i.e., we suppose that no field depends on the angular variable θ . We further assume that the current density field has non-zero component only in the tangential direction eθ , i.e., J(r, θ, z) = Jθ (r, z) eθ . Because of the assumed conditions on J, only the θ -component of the electric field E = J/σ does not vanish in the conductor. Given a vector field F = Fr (r, θ, z)er + Fθ (r, θ, z)eθ + Fz (r, θ, z)ez , we recall that ∂Fz ∂Fr 1 ∂(rFθ ) 1 ∂Fr 1 ∂Fz ∂Fθ − − − (9) curl F = er + eθ + ez . r ∂θ ∂z ∂z ∂r r ∂r r ∂θ Hence equations (2), (3) and (8) imply H(r, θ, z) = Hr (r, z)er + Hz (r, z)ez , B(r, θ, z) = Br (r, z)er + Bz (r, z)ez , E(r, θ, z) = Eθ (r, z)eθ ,

(10) (11) (12)

that is, the electric field E is directed in the θ -direction, whereas the magnetic field H is on the (r, z)-plane. A well-known result allows us to conclude from (4) that B = curl A,

(13)

where A is the magnetic vector potential. For the sake of uniqueness we take A to be divergence-free (Coulomb gauge). From (11), (13) and (9) we deduce A(r, θ, z) = Aθ (r, z)eθ ,

(14)

and Br (r, z) = −

∂Aθ , ∂z

(15)

44


Bz (r, z) =

1 ∂(rAθ ) . r ∂r

(16)

Taking into account (13), we deduce from (3)

or, equivalently,

curl(i ωA + E) = 0,

(17)

curl i ωAθ + σ −1 Jθ eθ = 0

(18)

from which it follows that ∂ i ωAθ + σ −1 Jθ = 0, ∂z ∂ r i ωAθ + σ −1 Jθ = 0. ∂r Hence we deduce there exist constants Ck ∈ C, k = 0, . . . , m, such that i ωAθ + σ −1 Jθ =

Ck r

in k ,

(19)

recalling that k denotes every connected component of the conductor. On the other hand, (2), (6), (14) and (19) combined together yield 1 ∂(rAθ ) Ck 1 ∂Aθ er + ez = σ −i ωAθ + eθ curl − µ ∂z µr ∂r r and, taking again into account the expression (9) for the curl operator, we get σ 1 ∂(rAθ ) ∂ 1 ∂Aθ ∂ + + i ωσ Aθ = Ck , − ∂r µr ∂r ∂z µ ∂z r in any connected component of the conducting domain and 1 ∂(rAθ ) ∂ 1 ∂Aθ ∂ + =0 − ∂r µr ∂r ∂z µ ∂z

(20)

(21)

in the air. In order to solve equations (20) and (21) we assume that the intensities going through each cylindrical ring are given data. Thus we add to the model the following equations Jθ dr dz = Ik , k = 1, . . . , m, k

Ik being the intensity traversing k . Constants Ck above can be considered as Lagrange multipliers associated with these equations. Physically, they represent electric potentials divided by 2π (see [16], for instance). Of course, all intensities Ik are the same in the case of a coil.


45

2.1.1. Electromagnetic boundary conditions Notice that the computational domain is an unbounded domain. In order to apply a standard finite element method, a common way of solving this problem is to approach the infinite domain by a “sufficiently large” bounded one with suitable conditions on the boundary. Thus, we shall consider a rectangular box in the (r, z)-plane surrounding the induction heating system, and large enough for the magnetic field to be small at the boundaries of the box. We shall denote by A the boundary of this rectangle and we set A = DA ∪ NA ∪ RA (see figure 2). Following [4] (see also remark 2 below), on the boundary RA of the box which is parallel to the symmetry axis, we impose the Robin condition ∂(rAθ ) + Aθ = 0. ∂r

(22)

For those boundaries NA of the box which are perpendicular to the symmetry axis, we set the condition ∂(rAθ ) = 0, (23) ∂z which stems from the fact that the radial component of the magnetic field is close to zero on these boundaries (see [4]). Finally, the natural symmetry condition along the revolution axis DA is Aθ = 0.

(24)

We could also use exact boundary conditions by using integral equations methods (see [14,15]), but the previous approximations are sufficiently accurate and simpler. Remark 2. From the Biot–Savart law, it is straightforward to deduce that field B behaves like 1/(r 3 + z3 ) far from the conductors (see [10], for instance). Hence, for r large enough, the behaviour of Aθ can be considered to be similar to 1/r 2 . This fact leads to impose the Robin condition (22) on the artificial boundary RA . 2.1.2. Computation of constants Ck of the electromagnetic problem This section deals with the procedure that we have adopted to compute constants Ck , k = 0, . . . , m, which appear in the electromagnetic model by assuming that current intensities are given. This is the same method as the one described in [11]. Following [16], any simply-connected axisymmetric region must have a non-empty intersection with the Oz axis, where r is zero. Since the current density cannot be infinite, equation (19) implies that Ck is zero in those of the regions k that are simplyconnected. In particular, C0 = 0 in the workpiece. On the other hand, it is clear that, since constants Ck are the only parameters that make it possible to prescribe the injected currents, one should choose them in such a way that the computed total current intensities coincide with the prescribed ones. More precisely, the algorithm for the computation of the constants is as follows:

46


0 = 0 in the workpiece (induced material). 1. Set C 2. Loop in the number of inductors j = 1, . . . , m. Step j : k = δkj , k = 1, . . . , m. (a) Set C j (b) Solve equation (20) to obtain A˜ θ . Notice that, for each iteration j , the matrix of the linear discrete problem arising from a fixed spatial discretization of equation (20) is the same. Hence, the numerical cost is not very important.

(c) Compute ˜Jj · n dS = ˜Ijk = k

j J˜θ dr dz =

k

Ck j ˜ σ − iωAθ dr dz, r k

k = 1, . . . , m.

3. Compute constants Ck as the solution of the linear system m

˜Ijk Cj = Ik ,

k = 1, . . . , m,

j =1

where Ik is the given intensity which goes through the connected component k , k = 1, . . . , m, of the inductor. 4. Finally, compute the solution with the new constants. 2.2. The thermal submodel The above model must be coupled with the heat equation in order to study the thermal effects of the electromagnetic phenomena in the workpiece. Neglecting convective heat transfer, a mathematical model is given by equation ρ(x, T ) c(x, T )

|J|2 ∂T − div k(x, T ) grad T = ∂t 2σ

in 0 ,

(25)

where T is the temperature, ρ the density, c the specific heat and k the thermal conductivity, which also depend on temperature. We also assume that other material properties as the magnetic permeability µ and the electric conductivity σ may depend on temperature. The source term on the right-hand side of (25) represents the heat released by the electric current due to the Joule effect which is obtained from (19). It makes a coupling between both the thermal and the electromagnetic submodels. Remark 3. Notice that the time scale for the variation of the electromagnetic field is much smaller than the one for variation of temperature. Indeed, the physical parameters used in a typical industrial situation give a time scale for temperature of the order of 60 seconds while for the electromagnetic problem it is of the order of 1.e−5 seconds. Thus, we may consider the eddy current model to compute the electromagnetic field


47

in the frequency domain and then the heat source due to Joule effect is determined by taking the mean value on a cycle, namely, 2π/ω ω J (x, t) · E(x, t) dt, (26) 2π 0 J and E being the current density and the electric field, respectively. By using (1) it is not difficult to prove that it coincides with |J|2 /(2σ ). In general, the metal in the crucible can undergo a phase change during the heating process. As a consequence, equation (25) must be corrected to take into account the latent heat involved in this phenomenon. This can be done by introducing an enthalpy function similar to that used for Stefan problems (see [9]). More precisely, the heat transfer equation is rewritten in the form |J|2 ∂e − div(k grad T ) = , ∂t 2σ

(27)

where e denotes the enthalpy density which is expressed as a function of temperature by e(x, t) ∈ H x, T (x, t) , (28) and H(x, T ) is the multi-valued function given by  T    ρ(x, s)c(x, s) ds, T < TS (x),   0   T T    ρ(x, s)c(x, s) ds, ρ(x, s)c(x, s) ds + ρ(x, TS )L(x) , H(x, T ) = 0 0    T = TS (x),   T      ρ(x, s)c(x, s) ds + ρ(x, TS )L(x), T > TS (x),

(29)

0

L being the latent heat, i.e., the heat per unit mass necessary to achieve the change of state at temperature TS . Remark 4. We notice that H is a multivalued function rather than discontinuous. Indeed, an element of volume at the solidification temperature may have any enthalpy in the T T interval [ 0 S ρc ds, 0 S ρc ds + ρL]. Assuming cylindrical symmetry and the fact that T only depends on spatial coordinates r and z, the div and grad operators are given by ∂T ∂T er + eθ , ∂r ∂z 1 ∂(rur ) ∂uz + , div u(r, z) = r ∂r ∂z

grad T (r, z) =

48


and hence the heat equation (27) becomes ∂T ∂ ∂T |Jθ |2 ∂e 1 ∂ − rk(r, z, T ) − k(r, z, T ) = . ∂t r ∂r ∂r ∂z ∂z 2σ

(30)

Notice that, from equation (19), we deduce Jθ = −i ωσ Aθ

in 0

(31)

because C0 = 0 in 0 . 2.2.1. Thermal boundary conditions Let us denote by T the boundary of the radial section of the workpiece, which is the computational domain for the thermal problem. By RT we denote the part of the boundary T which is not on the symmetry axis. Equation (30) above is completed by the following radiation–convection condition on the boundary RT (see figure 3): k(x, T )

∂T = η(Tc − T ) + γ Tr4 − T 4 , ∂n

(32)

where η is the coefficient of convective heat transfer, Tc and Tr are the external convection and radiation temperatures, respectively, the coefficient γ is the product of emissivity by Stefan–Boltzmann constant and n is the outward unit normal vector to the boundary. Besides, on the axis T \ RT we set k(x, T )

∂T = 0. ∂n

Figure 3. Computational domain for the thermal problem.


3.

49

Weak formulation and discretization

In this section we introduce a weak formulation and a time–space discretization of the thermoelectrical problem. For the sake of simplicity, in what follows we shall drop index θ in notation for Aθ and Jθ . 3.1. Time discretization We consider a time interval [0, tf ] for numerical simulation and a uniform mesh = {t 0 , . . . , t M }, 0 = t 0 < t 1 < · · · < t M = t f . To integrate the equation in time, we use a one-step implicit scheme. We approximate the value of the partial derivative of enthalpy ∂e/∂t ((r, z), t) at (r, z) ∈ 0 and t = t n+1 by the two-point finite difference formula en+1 (r, z) − en (r, z) ∂e (r, z), t n+1 ≈ , ∂t t where t = t n+1 − t n . Multiplying equation (30) discretized in time by a test function and integrating in the section 0 , we obtain, after using Green’s formula, the following weak formulation of the semi-discretized thermal problem: (WTP) For each n = 0, 1, . . . , M − 1, find a function T n+1 such that 1 n+1 e W r dr dz + k r, z, T n+1 grad T n+1 · grad W r dr dz 0 t 0 4 n+1 4 1 n n+1 e W r dr dz η Tc − T + γ Tr − T W r d + = RT 0 t n+1 2 1 J W r dr dz, + n+1 ) 0 2σ (r, z, T for all test function W . Similarly, an approximation of field A at time t n+1 is determined as a solution of the following weak formulation of the electromagnetic problem, which has been obtained from (20) and boundary conditions (22)–(24) using that σ = 0 in the air: (WEP) Find a complex function An+1 such that An+1 = 0 on DA and furthermore ∂(rAn+1 ) 1 ∂(rG) ∂An+1 ∂G 1 1 + r dr dz µ(r, z, T n+1 )r ∂r r ∂r µ(r, z, T n+1 ) ∂z ∂z 1 n+1 n+1 An+1 G d i ωσ (r, z, T )A Gr dr dz + + n+1 ) A R µ(r, z, T m

= σ (r, z, T n+1 )Ck G dr dz, k=1

k

for all function test G null on DA . In the above equality, G denotes the complex conjugate of G.

50


3.2. Space discretization Problems (WTP) and (WEP) have been spatially discretized by a finite element method. More precisely, we consider continuous piecewise linear finite element spaces associated with triangular meshes of the domain for both fields A and T . The electromagnetic problem is discretized and solved in the workpiece, air and inductors while the heat transfer equation is only solved in the workpiece. We notice that, at each time step, a coupled nonlinear system must be solved. Indeed, the heat source depends on the solution of the electromagnetic problem. On the other hand, parameters k, σ and µ depend on temperature and so does enthalpy. Furthermore, the radiation–convection boundary condition depends on T 4 . To handle the coupling between the thermal and the electromagnetic problem we propose a fixed point algorithm which will be described in the next section. 3.3. An iterative algorithm to solve the discretized problem To deal with the multi-valued nonlinear dependence of enthalpy with respect to temperature, we have used an iterative algorithm based on a fixed point argument, similar to that used in [2]. It takes into account the following result (see [1]): Lemma 3.1. Let H be a (possibly multi-valued) maximal monotone operator. Then the following statements are equivalent: • p ∈ Hα (s), • p = Hλα (s + λp), where α and λ are selected parameters with λα 12 and Hλα is the Yosida approximation of the operator Hα = H − αI , which is defined by Hλα (s) =

[I − ((1 − λα)I + λH)−1 ](s) , λ

s ∈ R.

We recall that Hλα is a single-valued Lipschitz-continuous function. At time step n + 1 we introduce the new function pn+1 = en+1 − αT n+1 . According to (28),

pn+1 (r, z) ∈ H (r, z), T n+1 (r, z) − αT n+1 (r, z)

and hence lemma 3.1 stablishes that 1 . pn+1 (r, z) = Hλα (r, z), T n+1 (r, z) + λpn+1 (r, z) for 0 < λ 2α

(33)

Now, in order to solve (WTP), the idea is to replace en+1 by pn+1 + αT n+1 . Then, to determine both T n+1 and pn+1 we introduce an iterative process: we start with an initial guess of pn+1 , for instance pn , and update it by using formula (33) at each iteration.


51

On the other hand, and in order to solve the nonlinearity due to term T 4 in thermal boundary condition (32), we consider the maximal monotone operator G(T ) = |T |T 3 , which coincides with T 4 for T > 0. We introduce q = |T |T 3 − δT .

(34)

Notice that q = G δ (T ) and therefore, according to lemma 3.1, we have q = Gβδ (T + βq) where G, δ and β play the role of H, α and λ, respectively, in lemma 3.1. Hence, for T > 0, we can transform the expression η(Tc − T ) + γ Tr4 − T 4 ,

(35)

(36)

appearing in the right-hand side of (32), as follows: η(Tc − T ) + γ Tr4 − T 4 = ηTc − ηT + γ Tr4 − γ |T |T 3 + γ δT − γ δT = −(η + γ δ)T + ηTc + γ Tr4 − γ |T |T 3 − δT = −(η + γ δ)T + ηTc + γ Tr4 − γ q. In order to introduce an iterative algorithm and solve a linear system in each iteration, we replace (36) in the weak formulation (WTP) by the previous expression. The term including function q will be on the right-hand side of the heat transfer equation and it will be updated by using equality (35). In what follows, we describe the whole algorithm which, in fact, consists of three nested loops (see sketch in figure 4):

Figure 4. Scheme of the algorithm.

52


1. Initial time step. At the beginning, T 0 is given. Then A0 is computed as the solution of the linear equation 1 ∂A0 ∂G 1 ∂(rA0 ) 1 ∂(rG) + r dr dz 0 ∂r r ∂r µ(T 0 ) ∂z ∂z µ(T )r m

1 0 A i ωσ (T 0 )A0 Gr dr dz + G d = σ (T 0 ) Ck G dr dz, + 0) A µ(T R k=1 k ∀G test function such that G = 0 on DA . 2. (n + 1)th time step. Let us suppose An and T n are known. Then, at time t n+1 , func, Tsn+1 constructed tions An+1 and T n+1 are obtained as the limit of sequences An+1 s with the following iterative algorithm: = An , T0n+1 = T n , p0n+1 = pn and q0n+1 = q n . (a) Initialization: An+1 0 n+1 n+1 and Tsn+1 as (b) Iteration s: An+1 s−1 , Ts−1 are known. We successively determine As follows:

is the solution of • An+1 s ∂(rAn+1 ∂An+1 1 ) 1 ∂(rG) 1 ∂G s s r dr dz + n+1 n+1 ∂r r ∂r µ(TS−1 ) ∂z ∂z µ(TS−1 )r 1 n+1 n+1 An+1 + i ωσ (TS−1 )As Gr dr dz + G d s n+1 A R µ(TS−1 ) m

n+1 σ (TS−1 )Ck G dr dz, = k=1

k

∀G test function such that G = 0 on DA . n+1 n+1 n+1 • Tsn+1 , psn+1 and qsn+1 are the limits of sequences Ts,k , ps,k and qs,k computed by n+1 n+1 n+1 n+1 n+1 n+1 = Ts−1 , ps,0 = ps−1 and qs,0 = qs−1 . (a) Initial step: Ts,0 n+1 n+1 n+1 n+1 n+1 , ps,k−1 and qs,k−1 are known. To calculate Ts,k , ps,k and (b) k th step: Ts,k−1 n+1 qs,k we proceed as follows: n+1 is the solution of the linear problem (b.1) Firstly, Ts,k n+1 n+1 α n+1 Ts,k W r dr dz + k Ts−1 grad Ts,k · grad W r dr dz t 0 0 n 1 n+1 n+1 (η + γ δ)Ts,k W r d = e − ps,k−1 W r dr dz + t 0 RT


+ +

RT

ηTc + γ Tr4 W rd − 1

n+1 0 2σ (Ts−1 )

RT

n+1 γ qs,k−1 W r d

n+1 2 J W r dr dz, s

53

for all test functions W. (37)

n+1 n+1 and qs,k are calculated, at vertices (ri , zi ) of the mesh, (b.2) Then ps,k by formulas

n+1 n+1 n+1 (ri , zi ) = Hλα (ri , zi ), Ts,k (ri , zi ) + λps,k−1 (ri , zi ) , ps,k n+1 n+1 n+1 (ri , zi ) = Gβδ Ts,k (ri , zi ) + β qs,k−1 (ri , zi ) . qs,k

4.

Numerical results

This section is devoted to present some numerical results concerning the above algorithm which has been implemented in a computer by using a Fortran program. Two different cases are considered: the first one is a comparison of the results of the twodimensional code with the analytical solution obtained in the appendix. The second one concerns some numerical experiments simulating the behaviour of an industrial furnace. 4.1. Analytical test We consider a steady state problem corresponding to an infinite cylinder made of graphite surrounded by an infinite and extremely thin coil, that is to say, the same geometry described in the appendix with only one material in the workpiece. It is important to notice that, in this case, we know the analytical solution of the electromagnetic and thermal problems and, in particular, the magnetic vector potential in the whole space (see appendix). Thus, in order to solve the two-dimensional model we can impose the exact boundary condition A = βext /Rext , where Rext denotes the width of the rectangular box enclosing the domain for the finite element computations and βext is computed by solving the system (A.13)–(A.20) for the one-dimensional electromagnetic model (see appendix). The geometrical data and physical parameters used for this problem are displayed in table 1. Notice that these data do not take into account the temperature dependence. The numerical method has been used on several successively refined twodimensional meshes, and we have compared the obtained numerical solutions with the analytical one. Figure 5 shows the log-log plot of the errors for the computed field A and the temperature T , respectively, versus the number of degrees of freedom for the same meshes. We have computed the error in L2 -norm and we can observe a quadratic dependence on the meshsize h for the two fields.

54

Bermúdez et al. / Numerical simulation of induction heating furnaces Table 1 Analytical test. Geometrical data and physical parameters. Radius of crucible: Radius of the induction coil: Rext : Frequency: RMS intensity/unit of length: Electrical conductivity of graphite: Magnetic permeability of vacuum:

0.07 m 0.09 m 0.2 m 3700 Hz 30460 Am−1 240000 (Ohm m)−1 4π · 10−7 Hm−1

Figure 5. Error versus number of d.o.f. (log-log scale). Fields A (left) and T (right). Table 2 Industrial furnace. Geometrical data and physical parameters. Number of coil turns: Inner radius of crucible: Outer radius of crucible: Radius of the induction coil: Crucible height: Height of silicon inside the crucible: RMS coil current (in each turn): Frequency: Ambient temperature:

8 0.05 m 0.07 m 0.09 m 0.26 m 0.057 m 3000 A 3700 Hz 30◦ C

4.2. Industrial furnace In this test, we consider two materials because the piece to be heated is silicon powder contained in a graphite crucible, both of them initially at 30◦ C. The geometry and the mesh used for this test are shown in figures 6 and 7. The geometrical data and process parameters are summarized in table 2. The physical properties of silicon and graphite depend on temperature and have been obtained from literature.


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Figure 6. Geometry of the induction furnace.

Figure 7. Detail of the mesh for the domain of the electromagnetic problem.

Figures 8–10 show the modulus of the current density and the temperature in the workpiece (silicon and graphite) for times 20, 120 and 360 seconds, respectively. During the first two minutes, the temperature of the workpiece increases fast and after 6 minutes the silicon undergoes a phase change (at 1412◦ C) and becomes liquid. Notice that, at ambient temperature, silicon is not an electric conductor and then the induced current density concentrates on graphite. As silicon temperature increases, so does its electrical

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Figure 8. Modulus of current density (left) and temperature (right) in the workpiece (t = 20 s).


conductivity and the induced current density on √ its surface as figure 10 shows. Notice that the skin depth, which is given by formula 2/(ωµσ ), decreases when the electrical conductivity increases. In figures 11 and 12 we have represented the evolution in time of the temperature and current density for two different points in the silicon: a point P located at the axis and a point Q close to the crucible (see figure 6). We emphasize the fact that the modulus of the current density at points P and Q is close to zero around the first 100 seconds of the process and then it grows up to remains almost constant.


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Figure 11. Evolution of temperature and current density at point P.

Appendix. Analytical solution of a simplified problem We present a simplified problem having an analytical solution, which has been used for validation purposes. For the thermal problem we will not consider the transient model but the corresponding steady state one. Let us consider an infinite cylinder consisting of a core material surrounded by a crucible and an extremely thin coil. The crucible itself may consist of several concentric layers made with different materials. The multi-turn coil is modelled as a continuous single coil with a uniform initial current intensity. Let R1 , . . . , RN be the respective radii of the core and crucible materials and RN +1 the radius of the coil (see figure 13 for N = 2). We also define R0 = 0.

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Figure 12. Evolution of temperature and current density at point Q.

Figure 13. Sketch of the domain for the one-dimensional problem (N = 2).

In order to compute the analytical solution, we assume that thermal conductivity, electrical conductivity and magnetic permeability are constant in each material, i.e., k = kj ,

σ = σj ,

µ = µj

if Rj −1 < r < Rj ,

j = 1, . . . , N.

We shall denote µ0 the magnetic permeability of the vacuum. Solution of the electromagnetic problem Let us denote Aj = A|(Rj −1 ,Rj ) , j = 1, . . . , N + 1, Aext = A|(RN+1 ,+∞) .


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We will assume that electric current flows in the coil in the eθ direction and that it is uniformly distributed in the ez direction. Then, taking into account that we only have one connected component, equations (20), (21) become 1 d 1 d(rAj ) − + i ωσj Aj = 0 if Rj −1 < r < Rj , j = 1, . . . , N, (A.1) µj dr r dr 1 d 1 d(rAN +1 ) = 0 if RN < r < RN +1 , (A.2) − µ0 dr r dr 1 d 1 d(rAext ) (A.3) = 0 if r > RN +1 , − µ0 dr r dr with boundary conditions A1 (r) is bounded as r → 0, 1 Aext (r) = O as r → ∞. r

(A.4) (A.5)

Equation (A.5) follows from the Biot–Savart law in the case of an infinitely long conductor. Now we must add interface conditions at the boundaries of the conductors and at the inductor position. On the one hand, equations (13) and (14) imply that A must be continuous. Then we set Aj (Rj ) = Aj +1 (Rj ), j = 1, . . . , N, AN +1 (RN +1 ) = Aext (RN +1 ).

(A.6) (A.7)

Moreover, we have 1 1 d(rAj +1 ) 1 1 d(rAj ) (Rj ) = (Rj ), µj r dr µj +1 r dr

j = 1, . . . , N.

(A.8)

Finally, if we denote by I the intensity per unit length in the induction coil, we have 1 1 d(rAN +1 ) d(rAext ) (RN +1 ) = I. (A.9) − µ0 r dr dr Consequently, the model consists of equations (A.1)–(A.3) with boundary conditions (A.4), (A.5) and interface conditions (A.6)–(A.9). In order to solve equation (A.1) forj = 1, . . . , N we perform the change of variable x = rγj in (Rj −1 , Rj ), where γj = i ωµj σj . Then we get x2

dA˜ j 2 d2 A˜ j − x + x + 1 A˜ j = 0 in (Rj −1 , Rj ), dx 2 dx

(A.10)

where A˜ j (x) = Aj (x/γj ). Equation (A.10) is a Bessel equation, the general solution of which is given by A˜ j (x) = αj I1 (x) + βj K1 (x),

60


where I1 and K1 are the modified Bessel function of the first and second kind, respectively. Simple integration of equations (A.2) and (A.3) yields βN +1 r , AN +1 = µ0 αN +1 + 2 r βext r . Aext = µ0 αext + 2 r

(A.11) (A.12)

By using boundary conditions (A.4), (A.5) we first deduce that β1 = 0 for A1 to be bounded as r → 0 and αext = 0. Besides, from (A.9) we obtain that αN +1 = I.

(A.13)

On the other hand, interface conditions (A.7) imply α1 I1 (γ1 R1 ) − α2 I1 (γ2 R1 ) − β2 K1 (γ2 R1 ) = 0, αj I1 (γj Rj ) + βj K1 (γj Rj ) − αj +1 I1 (γj +1 Rj ) − βj +1 K1 (γj +1 Rj ) = 0, for j = 2, . . . , N − 1, µ0 βN +1 αN I1 (γN RN ) + βN K1 (γN RN ) − = 0, RN αN +1 − 2 RN βN +1 βext µ0 RN +1 αN +1 + − = 0. 2 RN +1 RN +1

(A.14) (A.15) (A.16) (A.17)

Finally, from (A.8) we deduce, for j = 1, α1 I1 (γ1 R1 ) + R1 γ1 I1 (γ1 R1 ) = α2 I1 (γ2 R1 ) + R1 γ2 I1 (γ2 R1 ) + β2 K1 (γ2 R1 ) + R1 γ2 K1 (γ2 R1 ) , (A.18) for j = 2, . . . , N − 1, αj I1 (γj Rj ) + Rj γj I1 (γj Rj ) + βj K1 (γj Rj ) + Rj γj K1 (γj Rj ) = αj +1 I1 (γj +1 Rj ) + Rj γj +1 I1 (γj +1 Rj ) + βj +1 K1 (γj +1 Rj ) + Rj γj +1 K1 (γj +1 Rj ) ,

(A.19)

and, for j = N , αN I1 (γN RN ) + RN γN I1 (γN RN ) + βN K1 (γN RN ) + RN γN K1 (γN RN ) = µ0 RN αN +1 . (A.20) Thus we obtain a linear system of order 2N + 2 for unknowns α1 , βext and αj , βj , j = 2, . . . , N + 1.


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Solution of the thermal problem Once Aj has been computed, and taking equation (31) into account, the Joule effect fj (r) is given by fj (r) =

ω2 σj |Aj |2 , 2

Rj −1 < r < Rj , j = 1, . . . , N.

By using the expression of the divergence operator in cylindrical coordinates, the steady state heat equation is k d rdT = fj (r), − r dr dr and therefore,

d r dT r = − fj (r) in (0, RN ). dr dr k

(A.21)

Let us denote Tj = T |(Rj −1 ,Rj ) , j = 1, . . . , N. Equation (A.21) is completed with the transmission conditions Tj (Rj ) = Tj +1 (Rj ), dTj dTj +1 (Rj ) = kj +1 (Rj ), kj dr dr

j = 1, . . . , N,

(A.22)

j = 2, . . . , N.

(A.23)

For j = 1, . . . , N, integration of equation (A.21) between Rj −1 and r, with Rj −1 r Rj yields dTj 1 r dTj (r) − Rj −1 (Rj −1 ) = − r sfj (s) ds. (A.24) dr dr kj Rj −1 Taking into account continuity conditions (A.23), we can write R1 R2 r dTj 1 1 sf1 (s) ds + sf2 (s) ds + · · · + sfN (s) ds . (r) = − dr kj r 0 R1 Rj −1

(A.25)

Then, for j = 1, . . . , N, we integrate (A.25) between Rj and r, with Rj −1 r Rj , and we obtain j −1 r s 1 r 1 Ri 1 ufi (u) du + ufj (u) du . Tj (r) = Tj (Rj ) − kj i=1 Rj s Ri−1 Rj s Rj −1 We compute T (RN ) from the radiation boundary condition at r = RN kN

dT (RN ) = η Tc − T (RN ) + γ Tr4 − T (RN )4 . dr

(A.26)

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More precisely, T (RN ) is computed as the numerical solution of the equation N 1 Rj s fj (s) ds = η Tc − T (RN ) + γ Tr4 − T (RN )4 . − RN j =1 Rj −1 Once T (RN ) is known, we can use expressions (A.26) taking this value into account and recursive application of continuity conditions (A.22). Acknowledgements This work has been partly supported by MCYT-FEDER (Spain) through research project DPI2003-01316 and Ferroatlántica I+D company under contract. The authors also express their gratitude to the referees for their useful remarks. References [1] A. Bermúdez and C. Moreno, Duality methods for solving variational inequalities, Comput. Math. Appl. 7 (1994) 43–58. [2] A. Bermúdez, J. Bullón, F. Pena and P. Salgado, A numerical method for transient simulation of metallurgical compound electrodes, Finite Elem. Anal. Des. 39 (2003) 283–299. [3] A. Bossavit, Computational Electromagnetism (Academic Press, San Diego, CA, 1998). [4] C. Chaboudez, S. Clain, R. Glardon, D. Mari, J. Rappaz and M. Swierkosz, Numerical modeling in induction heating for axisymmetric geometries, IEEE Trans. Magn. 33 (1997) 739–745. [5] S. Clain, J. Rappaz, M. Swierkosz and R. Touzani, Numerical modelling of induction heating for two-dimensional geometries, Math. Models Methods Appl. Sci. 3 (1993) 805–822. [6] R. Dautray and J.-L. Lions, Analyse Mathématique et Calcul Numérique pour les Sciences et les Techniques (Masson, Paris, 1988). [7] Y. Delannoy, C. Alemany, K.-I. Li, P. Proulx and C. Trassy, Plasma-refining process to provide solargrade silicon, Solar Energy Materials Solar Cells 72 (2002) 69–75. [8] L.R. Egan and E.P. Furlani, A computer simulation of an induction heating system, IEEE Trans. Magn. 27(5) (1991) 4343–4354. [9] C. Elliot and J.R. Ockendon, Weak and Variational Methods for Free Boundary Problems (Pitman, London, 1985). [10] C.T.A. Johnk, Engineering Electromagnetic Fields and Waves (Wiley, New York, 1988). [11] O. Klein and P. Philip, modeling of induction heating with prescribed current, voltage or power, IEEE Trans. Magn. 38(3) (2002) 1519–1523. [12] P. Massé, B. Morel and T. Breville, A finite element prediction correction scheme for magneto-thermal coupled problem during curie transition, IEEE Trans. Magn. 21(5) (1985) 1871–1873. [13] T.T. Natarajan and N. El-Kaddah, electromagnetically driven flow in induction stirring systems, IEEE Trans. Magn. 35(3) (1999) 1773–1776. [14] L. Pichon and A. Razek, Hybrid finite-element method and boundary-element method using timestepping for eddy-current calculation in axisymmetric problems, IEE Proceedings 136(4) (1989) 217– 222. [15] L. Pichon and A. Razek, Force calculation in axisymmetric induction devices using a hybrid FEMBEM technique, IEEE Trans. Magn. 26(2) (1990) 1050–1053. [16] R. Rappaz and M. Swierkosz, Mathematical modelling and numerical simulation of induction heating processes, Appl. Math. Comput. Sci. 6(2) (1996) 207–221.