Internal (hot spot) temperature of core

In an iron core (or any part where losses are produced inside the material), heat must flow through the material to reach the cooling surface. The inside is therefore hotter than the surface. The highest temperature, at the centre, is the hot spot, and it decides the safety of nearby insulation.

Step 1: heat flow in one direction

Assume:

• $p$ = heat (loss) produced per unit volume, W/m³, uniform
• $\rho$ = thermal resistivity of the material, °C·m/W ($\rho = 1/k$)
• heat flows only in the $x$ direction (the other dimensions are large), and both faces are at the same surface temperature $\theta_s$.

By symmetry the hottest point is at the centre plane, $x = 0$, where no heat crosses. Consider a plane at distance $x$ from the centre. All heat produced between the centre and this plane must cross it. Per unit area:

$$Q_x = p\,x \quad \text{W/m}^2$$

By Fourier's law, the temperature gradient is

$$\frac{d\theta}{dx} = -\rho\, Q_x = -\rho\, p\, x$$

Integrating from the centre (temperature $\theta_h$):

$$\begin{aligned} \theta(x) &= \theta_h - \frac{\rho\, p\, x^2}{2} \end{aligned}$$

So the temperature is parabolic across the thickness. At the surface, $x = t/2$, $\theta = \theta_s$:

$$\theta_h - \theta_s = \frac{\rho\, p}{2}\left(\frac{t}{2}\right)^2 = \frac{p\,\rho\, t^2}{8}$$

 temperature
     ^        theta_h (hot spot)
     |         _..--.._
     |      .-'        '-.
     |    .'              '.
     |   /                  \
     |  theta_s        theta_s
     +--|---------+---------|--> x
      -t/2        0        +t/2
        <------ thickness t ------>

Step 2: core with heat flow in two directions

A laminated core block can lose heat in two directions:

• along the laminations (direction $x$, length $l_x$ between cooling surfaces or ducts), where thermal resistivity $\rho_x$ is low (through iron);
• across the laminations (direction $y$, length $l_y$), where thermal resistivity $\rho_y$ is much higher, because heat must cross the insulation (varnish, oxide) between sheets.

        cooling surface
   +---------------------+  ^
   | ==================  |  |
   | ==================  |  | l_y (across lam.)
   | ==================  |  |
   +---------------------+  v
   <-------- l_x -------->   (along laminations)

Let $p_x$ and $p_y$ be the parts of $p$ carried away in each direction, with $p_x + p_y = p$. Both paths start at the same hot spot and end at the same surface temperature, so by Step 1:

$$\theta_h - \theta_s = \frac{p_x \rho_x l_x^2}{8} = \frac{p_y \rho_y l_y^2}{8}$$

Hence $p_x = \dfrac{8(\theta_h-\theta_s)}{\rho_x l_x^2}$ and $p_y = \dfrac{8(\theta_h-\theta_s)}{\rho_y l_y^2}$. Adding:

$$p = 8(\theta_h-\theta_s)\left[\frac{1}{\rho_x l_x^2} + \frac{1}{\rho_y l_y^2}\right]$$ $$\theta_h - \theta_s = \frac{p}{8\left[\dfrac{1}{\rho_x l_x^2} + \dfrac{1}{\rho_y l_y^2}\right]}$$

This is the expression for the internal (hot-spot) temperature rise of the core above its surface. The hot-spot temperature itself is $\theta_h = \theta_s + (\theta_h - \theta_s)$, where $\theta_s$ is found from the surface dissipation, $\theta_s - \theta_a = \dfrac{\text{loss}}{S\lambda}$.

Conclusions

• The temperature difference is proportional to the square of the length of the heat path. Splitting the core into packets with ventilating ducts reduces it sharply.
• Since $\rho_y \gg \rho_x$, most heat flows along the laminations. Ducts placed across the laminations, which cut $l_x$, are therefore the most effective.