Temperature gradients in slot conductors

A conductor in a slot produces copper loss $p = J^2\rho_e$ per unit volume. This heat can leave by two paths: across the slot insulation into the iron (then to the cooling air), or along the conductor to the overhang, where it is dissipated to the air. Which path dominates depends on the thickness of the slot insulation and the length of the overhang. This gives two limiting cases.

Case 1: slot insulation very thick (heat flows along the conductor)

Assumptions

• The conductor carries current density $J$. Loss per unit volume is $p = J^2\rho_e$, where $\rho_e$ is the electrical resistivity.
• The slot insulation is very thick (a poor conductor of heat). No heat flows sideways into the iron; all heat produced in the slot portion flows along the conductor to the overhang (end connections), where it is dissipated.
• $\rho_t$ = thermal resistivity of the conductor (copper), $A$ = conductor cross-section, $L$ = embedded (slot) length, $\theta_o$ = temperature of the conductor at the slot ends (overhang).

   overhang      slot portion (length L)      overhang
  ~~~~~~~~~ |============================| ~~~~~~~~~
            |<--- heat    hottest  heat --->|
           x=-L/2         x = 0          x=+L/2
               temperature: parabola, max at centre

Derivation By symmetry the hottest point is at the centre of the slot length ($x = 0$). Heat produced between the centre and a section at distance $x$ flows through that section:

$$Q_x = p\,A\,x$$

Fourier's law along the conductor gives

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

Integrating from the centre ($\theta = \theta_m$ at $x = 0$):

$$\theta = \theta_m - \frac{\rho_t\,p\,x^2}{2}$$

At the slot ends, $x = L/2$, $\theta = \theta_o$:

$$\theta_m - \theta_o = \frac{p\,\rho_t\,L^2}{8} = \frac{J^2\rho_e\,\rho_t\,L^2}{8}$$

So the temperature varies parabolically along the conductor. The gradient at any point is $\rho_t p x$, which is zero at the centre and largest at the slot ends.

Case 2: slot insulation thin (heat flows across the insulation)

If the insulation is thin, almost all the heat of the embedded part flows radially through it to the teeth and core. The conductor temperature is then nearly uniform along its length, and the temperature drop occurs across the insulation.

  iron | insulation | copper (theta_c)
       |  t_i       |
  theta_i <-- heat --

Heat produced in length $L$: $Q = p A L$. It flows through insulation of thickness $t_i$, thermal resistivity $\rho_i$ and area $S_p L$, where $S_p$ is the perimeter of the slot insulation:

$$\theta_c - \theta_{iron} = Q \times \frac{\rho_i\,t_i}{S_p L} = \frac{p\,A\,\rho_i\,t_i}{S_p}$$

The gradient is now linear across the insulation: $\dfrac{d\theta}{dy} = \dfrac{p A \rho_i}{S_p}$.

Summary

In practice both paths act together. Thin, high-conductivity insulation (mica-epoxy) and ventilating ducts keep the conductor hot spot low.