Heat conducted across field coil former

Heat flows by conduction from the winding, through the former (2.5 mm) and then through the air space (1.5 mm), into the pole core. The two layers are in series, so their thermal resistances add.

Area of heat flow

The heat passes across the inner surface of the former:

$$\begin{aligned} \text{Perimeter} &= 2(150 + 250) = 800\ \text{mm} = 0.8\ \text{m}\\ S &= \text{perimeter} \times \text{winding height} = 0.8 \times 0.2 = 0.16\ \text{m}^2 \end{aligned}$$

Thermal resistances

Thermal resistance of a layer: $R = \dfrac{t}{kS}$ (°C/W).

$$\begin{aligned} R_{former} &= \frac{0.0025}{0.166 \times 0.16} = 0.0941\ ^\circ\text{C/W}\\ R_{air} &= \frac{0.0015}{0.05 \times 0.16} = 0.1875\ ^\circ\text{C/W}\\ R_{total} &= 0.0941 + 0.1875 = 0.2816\ ^\circ\text{C/W} \end{aligned}$$

Heat conducted

$$Q = \frac{\theta}{R_{total}} = \frac{40}{0.2816} = 142.0\ \text{W}$$

 winding | former | air |  pole core
  (hot)  | 2.5 mm |1.5mm|
   ----> heat flow ---->

The air space has about twice the resistance of the former, so a tight fit between former and core improves cooling.

Answer: heat conducted from winding to core ≈ 142 W.