Validations

1. Introduction

Two simple test cases are used to validate the formulation:

  • a single magnet,

  • a set of 2 magnets.

We will consider simple geometries for the magnets. They are modeled as rectangular cross section torus. This allows us to derive analytical expressions for the resistance \(R\) and inductances \(L\) of the magnets. For each case, we will model the powering and a powerfailure. The computed total current flowing into the magnets will be compared to the results obtained by a simple equivalent \(RL\) circuit. The circuit may be either modeled by a simple ODE solver or using analytical methods (aka exponential Matrices). For other quantities, we will compared our simulation results with other Finite Element Solver. We have chosen getdp with a MQS formulation for axisymetric geometries.

NB: getdp model and test geometry are accessible here. To reproduce the results follow the procedure in this appendix.

2. A solenoidal magnet

2.1. Setup

The magnet is centered on the origin of the cylindrical frame as shown bellow.

geometry

The geometry is defined by the following parameters:

Name Description Value Unit

\(r_i\)

internal radius

75.

\(mm\)

\(r_e\)

external radius

100.2

\(mm\)

\(h\)

heigth

50.

\(mm\)

\(r_\text{infty}\)

internal radius

500.

\(mm\)

The magnet is supposed to be made of some conducting amagnetic material:

Name Description Marker Value Unit

\(\sigma\)

electric conductivity

Ind

\(58.e3\)

\(S.mm^{-1}\)

\(\mu_r\)

relative magnetic permeability

Ind

\(1\)

The applied source electrical potential has the following form:

\(V_D\) electrical potential \(1*t/(0.1*\tau)\) \(V\) \(0 <= t < 0.1\,\tau\)

\(V_D\)

electrical potential

1

\(V\)

\( 0.1 \, \tau <= t <= 0.5\,\tau\)

\(V_D\)

electrical potential

0

\(V\)

\(t>\tau\)

The boundary conditions for the electromagnetic problem are:

Marker Type Value

\(Oz\) Axis

Dirichlet

\(\mathbf{0}\)

Infty

Dirichlet

\(\mathbf{0}\)

2.2. Resistance \(R\) and Self Inductance \(L\)

The resistance is defined as the ration of the applied electrical potential difference over the total current. In case of a rectangular cross section \( (r_1,r_2) \times (z_1,z_2) \) torus, we can show that:

\[R = \frac{2 \pi r_1 \rho}{r_1 ln(r_2/r_1)\,(z_2-z_1)}\]

with \(\rho=1/\sigma\) the resistivity of the material composing the torus. For details on this expression, see feelpp electric toolbox test case.

As for the self-inductance, we recall the defintion of the stored magnetic energy:

\[E = \frac{1}{2} L I^2 = \frac{1}{2} \int_\mathbf{R^3} \mathbf{B} \times \mathbf{H}\, d\Omega\]

to continue…​

2.3. Equivalent circuit model

From a macroscopic point of view, the studied system is simply equivalent to a \(RL\) circuit modeled by:

\[U(t) = R I(t) + L \, \frac{dI}{dt}\]
Value Unit

\(R\)

\(7.5313 10^{-6}\)

Ohm

\(L\)

\(1.9204 10^{-7}\)

Henry

2.4. Results

The normalized computed electric potential, current and magnetic field at the Origin are plotted bellow:

results

We use the expected values of the applied electric potential, current and magnetic field for the transient regime (aka t):

Value Unit

\(V\)

1

V

\(I\)

135069

A

\(B_z(\mathbf{O})\)

0.944

T

3. 2 solenoidal magnets

For sake of simplicity, we consider 2 solenoid magnets similar to the one described in previous section stacked as shown bellow.

insert a figure geometry

3.1. Mutual Inductance \(M\)

Obviously, the 2 magnets have the same resistance and self-inductance as they have the same geometry and are made of the same material. We only need then to compute the so-called mutual inductance \(M\).

As before, we start wih the stored magnetic energy \(E\) of the system:

\[E = \frac{1}{2} \sum_k L_k I_k^2 + \sum_k \sum_{l \neq k} M_{k,l} I_k I_l\]

to continue…​

3.2. Equivalent circuit model

The equivalent circuit is this time similar to a transformer circuit. Thus, it may be modeled as:

\[\begin{align} U_1(t) &=& R_1 I_1(t) + L_1 \, \frac{dI_1}{dt}+ M\,\frac{dI_2}{dt}, \\ U_2(t) &=& R_2 I_2(t) + M\, \frac{dI_1}{dt}+ L_2\,\frac{dI_2}{dt} \end{align}\]

In our case, we have \(R_1=R_2\) and \(L_1=L_2\).

Value Unit

\(R\)

\(7.5313 10^{-6}\)

Ohm

\(L\)

\(1.9204 10^{-7}\)

Henry

\(M\)

\(2.6487 10^{-8}\)

Henry

3.3. Results

The normalized computed electric potential, current and magnetic field at the Origin are plotted bellow:

results

We use the expected values of the applied electric potential, current and magnetic field for the transient regime (aka t):

Value Unit

\(V\)

1

V

\(I\)

135069

A

\(B_z(\mathbf{O})\)

0.8507

T