Temperature gradient in slot conductors (thick insulation)

When the slot insulation is very thick, heat from the conductor cannot pass through it into the core. Heat therefore flows axially along the copper to the overhang, and a temperature gradient appears along the length of 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.

Conclusions

• The temperature rise of the centre over the ends is proportional to $J^2$ and to $L^2$. Long cores with high current density (for example turbo-alternators) have large axial gradients.
• Copper has a very low thermal resistivity, so in normal machines this rise is small. Its importance is that it shows heat cannot pass through thick insulation easily, so large machines need ventilating ducts and direct conductor cooling.