Internal temperature of a homogeneous material

When losses are produced inside a body, heat must flow through the material to reach its cooling surfaces. The interior is therefore hotter than the surface. The highest (internal or hot-spot) temperature is found as follows.

Assumptions

• A block of homogeneous material of thickness $t$, length $l$ and width $w$, with $l$ and $w$ much larger than $t$. Heat therefore flows only across the thickness ($x$ direction) and leaves equally from the two large faces.
• Uniform loss $p$ W/m³; thermal resistivity $\rho$ °C·m/W (constant).
• Surface heat dissipation coefficient $\lambda$ W/m²°C; ambient temperature $\theta_a$.

          face 1 (area l x w)
   +--------------------------------+   ^
   |   heat <---  centre  ---> heat |   | t
   +--------------------------------+   v
          face 2 (area l x w)

Step 1: surface temperature

Total heat produced $= p\,l\,w\,t$. It leaves from two faces of total area $2lw$:

$$\begin{aligned} p\,l\,w\,t &= 2\,l\,w\,\lambda\,(\theta_s - \theta_a) \\ \theta_s &= \theta_a + \frac{p\,t}{2\lambda} \end{aligned}$$

Step 2: temperature rise inside

Take $x$ from the mid-plane. Heat produced between the mid-plane and a plane at $x$ crosses that plane. Its flow per unit area is $p\,x$. By Fourier's law:

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

At $x = t/2$, $\theta = \theta_s$:

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

Step 3: internal (hot-spot) temperature

$$\theta_h = \theta_a + \frac{p\,t}{2\lambda} + \frac{p\,\rho\,t^2}{8}$$

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

Remarks

• The first term is the rise of the surface above ambient, set by the cooling. The second is the internal rise, set by conduction.
• The internal rise varies as $t^2$. Dividing the block with a cooling duct at mid-thickness reduces it to one-quarter, which is why cores are split into packets.
• If heat flows in two directions (thickness $t$ and width $w$, with resistivities $\rho_t$, $\rho_w$), the internal rise becomes

$$\theta_h - \theta_s = \frac{p}{8\left[\dfrac{1}{\rho_t t^2} + \dfrac{1}{\rho_w w^2}\right]}$$