Thermal analysis of eddy-current magnetic couplings has been performed by very few authors [3, 4], although this phenomenon has a great impact on the coupling performances [14]. The thermal analysis is usually based on thermal network models, which give accurate results with less computational time than a purely numerical method i.e. finite element analysis. This is an important issue if the model has to be used in a design optimization procedure. Analytical models that can be found in the literature usually consider both heat conduction and heat convection in order to obtain the temperature distribution in each part of the magnetic coupling. Nevertheless, it was shown in [14], and confirmed by the results of Figure 4, that the temperature distribution in the copper and in the back-iron is almost homogeneous. This is particularly true when the magnetic coupling is used in the normal working range area, which corresponds to low slip values.

Then, a simple but realistic thermal model is considered in this paper. The following assumptions are adopted:

–

Steady state conditions.

–

The copper disc and back-iron thermal conductivities are high enough to neglect their thermal resistances.

–

Only convection heat transfer through the external surfaces, whose convective heat transfer coefficients are noted *h*_{1} and *h*_{2}, is then considered.

–

No heat exchange through the air-gap owing to its low film coefficient and air thermal conductivity [16].

In addition, this last assumption will give an excess value for the temperature in the copper, which goes in the right direction for the design.

The temperature of the copper *Θ* is computed by:

$$\begin{array}{}\mathit{\Theta}={\mathit{\Theta}}_{a}+\frac{{P}_{J}}{{h}_{1}\phantom{\rule{thinmathspace}{0ex}}{S}_{1}+{h}_{2}\phantom{\rule{thinmathspace}{0ex}}{S}_{2}}\end{array}$$(7)

where *P*_{J} = *T*_{e} · *s* *Ω*_{1} is the eddy current Joule losses, *Θ*_{a} is the ambient temperature, which is fixed to 25°C, *S*_{1} is the area of the cylinder surface normal to the radial direction, *S*_{2} is the area of the cylinder surface normal to the axial direction:

$$\begin{array}{}{S}_{1}=2\phantom{\rule{thinmathspace}{0ex}}\pi \phantom{\rule{thinmathspace}{0ex}}{R}_{3}\phantom{\rule{thinmathspace}{0ex}}(d+e),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{2}=\pi \phantom{\rule{thinmathspace}{0ex}}{R}_{3}^{2}\end{array}$$(8)

In order to obtain a good prediction for the copper temperature, it is important to have a good estimation of the convective heat transfer coefficients *h*_{1} and *h*_{2}. The *h*_{1} coefficient is usually determined by dimensionless analysis where the Nusselt number *N*_{u} has to be computed for the considered convection surface:

$$\begin{array}{}{h}_{1}=\frac{k\phantom{\rule{thinmathspace}{0ex}}{N}_{u}}{2\phantom{\rule{thinmathspace}{0ex}}{R}_{3}}\end{array}$$(9)

where *k* is the thermal conductivity of the fluid (air) at ambient temperature, and *R*_{3} is the outer radius of the copper. For forced convection systems, the Nusselt number is usually given as a power law of the Reynolds number *R*_{e} and the Prandtl number *P*_{r}:

$$\begin{array}{}{N}_{u}={a}_{1}{R}_{e}^{{a}_{2}}{P}_{r}^{{a}_{3}}\end{array}$$(10)

The values of the coefficients *a*_{1}, *a*_{2}, and *a*_{3} depend on the fluid flow conditions. In our study, we have always set *R*_{e} < 5·10^{5} and then *a*_{1} = 0.084, *a*_{2} = 0.70, and *a*_{3} = 0.35 (see [15] for details). The Reynolds and the Prandtl numbers that are used for calculating the Nusselt number are given by:

$$\begin{array}{}{R}_{e}=2\frac{\rho}{\mu}{\mathit{\Omega}}_{2}\phantom{\rule{thinmathspace}{0ex}}{R}_{3}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{P}_{r}=\frac{\mu \phantom{\rule{thinmathspace}{0ex}}{c}_{p}}{k}\end{array}$$(11)

The thermal properties of air are given in .

Table 2 Thermal properties for air at ambient temperature (25°C)

The coefficient *h*_{2} can be determined by following the same method, but for a disk-type surface. In [15], a mean value for *h*_{2} is given as:

$$\begin{array}{}{h}_{2}=\frac{2\phantom{\rule{thinmathspace}{0ex}}{b}_{1}}{2\phantom{\rule{thinmathspace}{0ex}}{b}_{2}+1}\cdot \frac{k}{{R}_{3}}{\left(\frac{\rho}{\mu}{\mathit{\Omega}}_{2}{R}_{3}^{2}\right)}^{{b}_{2}}{P}_{r}^{{b}_{3}}\end{array}$$(12)

where *b*_{1} = 0.024, *b*_{2} = 0.8 and *b*_{3} = 0.6.

From (9) to (12), we can note that the convective heat coefficients *h*_{1} and *h*_{2} depend on the coupling geometry and the rotating speed.

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