Temperature gradient in slot: two cases

A slot conductor produces loss $p = J^2\rho_e$ per unit volume ($J$ = current density, $\rho_e$ = electrical resistivity). The heat leaves along the conductor to the overhang, or across the slot insulation to the iron. Part (ii) of the question is cut off in the source; it is read here as the case where the overhang is not considerable (short, with little cooling), so that the heat must leave through the slot insulation.

(i) Slot insulation thick

No heat passes through the insulation, so heat flows axially to the overhang. With slot length $L$, copper thermal resistivity $\rho_t$ and the centre at $x = 0$:

$$\begin{aligned} \frac{d\theta}{dx} &= -\rho_t\,p\,x \\ \theta_{max} - \theta_o &= \frac{p\,\rho_t\,L^2}{8} \end{aligned}$$

The gradient is zero at the slot centre and largest at the slot ends. The temperature varies parabolically along the conductor, with its maximum at the middle.

(ii) Overhang not considerable (heat through slot insulation)

Little heat can flow to the short overhang, so the heat of the slot portion, $Q = p A L$, crosses the insulation (thickness $t_i$, thermal resistivity $\rho_i$, perimeter $S_p$, area $S_p L$):

$$\begin{aligned} \theta_c - \theta_{iron} &= Q\,\frac{\rho_i\,t_i}{S_p L} = \frac{p\,A\,\rho_i\,t_i}{S_p} \\ \frac{d\theta}{dy} &= \frac{p\,A\,\rho_i}{S_p}\ \ (\text{constant}) \end{aligned}$$

The conductor is at nearly uniform temperature along its length. The gradient is linear across the insulation thickness.