The mathematical equations of the HiFiMagnet

First of all, we start with the standart heat equation:

\[ \rho C_{p}\frac{\partial T}{\partial t} - \nabla.(k \nabla T)=P \]

The only source of heat considered is the Joule effect, expressed by :

\[ P=\textbf{j.E} \]

Where \(\textbf{j}=\sigma\textbf{E}\) (from the Ohm’s law). But \(\exists V | \textbf{E}=-\nabla V\)

We also consider that T is time independent, so, finally, we have :

\[ - \nabla.(k \nabla T)=\sigma\nabla V . \nabla V \]

Coefficients \(\sigma\) and k are temperature-dependent as shown in this equations :

The current is due to a difference of potential. We model this with the Dirichlet conditions :

We consider the air and the cooling water as they are electrically insulating. On other surfaces we impose homogeneous Neumann conditions as no current "flows" out these surfaces (j.n=0).

\(-\sigma(T)\nabla V.n=0\) for all surfaces except V_in and V_out

On the cooled surfaces, we set the heat transfer coefficient h, determined as follow :

\(h=\frac{k(T)N_{u}}{D_{h}}\) with N_u the Nusselt number and D_h the hydraulic diameter.

First of all, we start with two of the four Maxwell’s equations :

\[ \left\{ \begin{array}{cc} \nabla\times\textbf{H}=\textbf{j}\\ \nabla.\textbf{B}=0 \end{array} \right. \]

We also can link the magnetic induction (B) and the magnetic field (H) using the permeability (\(\mu\)) as :

\[ \textbf{B}=\mu\textbf{H} \]

with \(\mu=\mu_{r}\mu_{0}\)

From the differential operators, we know that \(\nabla.\nabla\times\textbf{A}=0\) \(\forall\) A.

Thus, since \(\nabla.\textbf{B}=0\) there exists A such as \(\textbf{B}=\nabla\times \textbf{A}\)

Finally, we can write :

\[ \nabla\times\left(\frac{1}{\mu}\nabla\times\textbf{A}\right)=\textbf{j} \]

The Finite Element Method is detailed in the chapter The Mathematics of Feel++

In our case, the variational formulation consists in resolve :

\[ \int_{\Omega\cup\Omega_{c}}\frac{1}{\mu}\textbf{A}.\nabla\times\varphi+\int_{\partial\Omega_{c}}\frac{1}{\mu}\left(\nabla\times\textbf{A}\times\textbf{n}\right).\varphi=\int_{\Omega}\textbf{j}.\varphi \]

With \(\Omega\) being the volume of our geometry and \(\Omega_{c}\) the volume inside (for example the air). \(\textbf{A}\in\{\textbf{v}\in L^{2}(\Omega\cup\Omega_{c}); \nabla\times\textbf{v}\in L^{2}(\Omega\cup\Omega_{c})\}\) corresponding of H_curl .

The current density is located in some finite region in space, actually \(\Omega\), this involves considering B as zero at infinity. But, with the Finite Element Method,we discretize the domain so we impose the domain to be of finite dimension. In fact, this domain is composed by \(\Omega\) (the conductor) and a box inside, \(\Omega_{c}\), whose boundaries model the infinity. The size of this box can be set manually or be calculated for more precision and validity. The boundary condition is expressed with the magnetic potential (A) like this :

\[ \textbf{A}\times\textbf{n}=0\, on\, \partial\Omega_{c} \]

Our materials allow us to only consider the vacuum permeability, so \(\mu=\mu_{0}=4\pi.10^{-7} kg.m.A^{-2} .s^{-2}\).

We need some preconditions :

First, we start with the equation of motion, becoming the equilibrium equation :

\[ div(\bar{\bar{\sigma}})+\textbf{f}=0 \]

With \(\bar{\bar{\sigma}}\) the stress tensor and f the volume forces applied on the conductor. As we search the displacement vector u, we introduce the tensor of small deformation \(\bar{\bar{\epsilon}}\)

\[ \bar{\bar{\epsilon}}=\frac{1}{2}(\nabla\textbf{u}+\nabla\textbf{u}^{T}) \]

We divide the stress tensor in two terms : \(\bar{\bar{\sigma}}^{E}\) given by the Hooke’s law for the small deformation and \(\bar{\bar{\sigma}}^{T}\) bring by the Joule eating.

\[ \bar{\bar{\sigma}}(\bar{\bar{\epsilon}})=\bar{\bar{\sigma}}^{E} (\bar{\bar{\epsilon}})+\bar{\bar{\sigma}}^{T}(\bar{\bar{\epsilon}}) \]

This terms are define so :

With :

The Finite Element Methid is detailed in the chapter The Mathematics of Fell++

In our case, the variational formulation consists in resolve :

\[ \frac{E}{1+\nu}\int_{\Omega}Tr\lgroup\frac{1}{2}(\nabla\textbf{u}+\nabla\textbf{u}^{T})\rgroup+\frac{E\nu}{(1+\nu)(1-2\nu)}\int_{\Omega}(\nabla.\textbf{u})(\nabla.\varphi)=\int_{\Omega}\textbf{f}.\varphi+\int_{\partial\Omega_{P}}\textbf{g}.\varphi+\int_{\Omega}\frac{E\alpha_{T}}{1-2\nu}(T-T_{0})(\nabla.\varphi) \] for all \(\varphi\in H_{1,\varphi}^{d}(\Omega)\) With \(\circ\) the element-wise product.