Hot spot temperature of a plate

A hot spot is the point of highest temperature inside a body in which heat is produced, such as a core, a coil or a conductor. Heat produced inside must flow outwards to the cooling surfaces, so the interior is hotter than the surface. Insulation near the hot spot ages fastest, so the hot-spot temperature, not the average, must be kept within the limit of the insulation class.

Derivation for a plate of thickness $t$

Consider a plate whose length and width are large compared with its thickness $t$, cooled equally on both faces. Then heat flows only across the thickness. 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 ------>

Result

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

where $\theta_h$ is the hottest-spot temperature (at the mid-plane) and $\theta_s$ the surface temperature.

Points to note

• The temperature profile across the plate is a parabola with its peak at the centre.
• The hot-spot rise above the surface is proportional to $t^2$. Halving the thickness, for example by a cooling duct in the middle, reduces it to one-quarter.
• It is also proportional to the loss density $p$ and to the thermal resistivity $\rho$. Materials with poor thermal conductivity (insulation, or laminated iron across the sheets) produce large internal gradients.
• If only one face is cooled, the hot spot is at the insulated face, and $\theta_h - \theta_s = p\rho t^2/2$ (the same formula with thickness $2t$).