QuarterTurn

In this example, we will model a stranded conductor. An electrical potential \(V_0\) is applied to the entry/exit of the conductor which is also cooled by a force flow.

We will compute the temperature field and the magnetic field generated. The geometry of the conductor is chosen such that we can derive the analytical expression for both the temperature and the magnetic fields.

1. Geometry

The conductor consists in a 1/4th of a rectangular cross section torus.

2. Input parameters

Name Description Value Unit

\(r_i\)

internal radius

1

\(m\)

\(r_e\)

external radius

2

\(m\)

\(\delta\)

angle

\(\pi/2\)

\(rad\)

\(V_D\)

electrical potential

9

\(V\)

\(h_i\)

internal transfer coefficient

\(60e3\)

\(W\cdot m^{-2}\cdot K^{-1}\)

\(T_{wi}\)

internal water temperature

303

\(K\)

\(h_e\)

external transfer coefficient

\(58e3\)

\(W\cdot m^{-2}\cdot K^{-1}\)

\(T_{we}\)

external water temperature

293

\(K\)

2.1. Model & Toolbox

This problem is fully described by a fully coupled model involving:

  • thermoelectric model,

  • magnetostatic model.

2.2. Materials

Name Description Marker Value Unit

\(\sigma\)

electric conductivity

omega

\(4.8e7\)

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

2.3. Boundary conditions

The boundary conditions for the electrical probleme are introduced as simple Dirichlet boundary conditions for the electric potential on the entry/exit of the conductor. For the remaining faces, as no current is flowing througth these faces, we add Homogeneous Neumann conditions.

Marker Type Value

V0

Dirichlet

0

V1

Dirichlet

\(V_D\)

Rint, Rext, top*, bottom*

Neumann

0

As for the heat equation, the forced water cooling is modeled by robin boundary condition with \(Tw\) the temperature of the coolant and \(h\) an heat exchange coefficient.

Marker Type Value

Rint

Robin

\(h_i(T-T_{wi})\)

Rext

Robin

\(h_e(T-T_{we})\)

V0, V1, top*, bottom*

Neumann

0

For the Magnetostatic model, the boundary conditions are given below: - \(\mathbf{A} = \mathbf{0}\) at the infinity - \(\mathbf{A} \times \mathbf{n} = 0 \) on symetry plane

3. Outputs

4. Verification Benchmark

The analytical solutions for the termoelectric problem are given by:

\[\begin{align*} V&=\frac{V_D}{\delta}\theta=\frac{V_D}{\delta}\operatorname{atan2}(y,x)\\ \mathbf{E}&=\left( -\frac{V_D}{\delta}\frac{y}{x^2+y^2}, \frac{V_D}{\delta}\frac{x}{x^2+y^2}\right)\\ T&=A\log(r)^2+B\log(r)+C=A\log\left(\sqrt{x^2+y^2}\right)^2+B\log\left(\sqrt{x^2+y^2}\right)+C\\ A&=-\frac{\sigma}{2k}\left(\frac{V_D}{\delta}\right)^2\\ B&=\frac{B_e-B_i}{D}\\ C&=\frac{C_e-C_i}{D}\\ B_e&=2T_{we}\delta^2h_eh_ikr_er_i + V_D^2h_eh_ir_er_i\sigma\log(r_e)^2 + V_D^2h_ikr_i\sigma\log(r_e^2)\\ B_i&=2T_{wi}\delta^2h_eh_ikr_er_i + V_D^2h_eh_ir_er_i\sigma\log(r_i)^2 - V_D^2h_ekr_e\sigma\log(r_i^2)\\ C_e&=(h_er_e\log(r_e) + k)(2T_{wi}\delta^2h_ikr_i + V_D^2h_ir_i\sigma\log(r_i)^2 - V_D^2k\sigma\log(r_i^2))\\ C_i&=(h_ir_i\log(r_i) - k)(2T_{we}\delta^2h_ekr_e + V_D^2h_er_e\sigma\log(r_e)^2 + V_D^2k\sigma\log(r_e^2))\\ D&=2\delta^2k(h_eh_ir_er_i\log(r_e) - h_eh_ir_er_i\log(r_i) + h_ekr_e + h_ikr_i) \end{align*}\]

For simple conductor geometry, analytical expressions of the magnetic field along the Z-Axis may be found in physics textbooks. The expression of the magnetic field in \(\mathbb{R}^3\) is more difficult to derive but may be found in several papers, like \cite{Jackson99}.
As a classical result, we consider only the magnetic field along the Z-Axis, which analytical expression is given bellow:

\[\begin{equation*} B_z(z)=\frac{1}{2} \mu_0 J(r_i) \left[\left[log(r^2+t^2)\right]_{r=r_1}^{r=r_2}\right]_{t=z-H}^{t=z+H} \end{equation*}\]

with:

\[J(r_i) = -\frac{\sigma} \frac{4*V_D}/{2\,\pi\,r_i}.\]

The factor \(4\) stems from the geometry as we are considering only 1/4th of the torus.