---
title: "Heat conducted across field coil former"
subject: "Electric Machine Design"
chapter: 2
key: "semester_5/electric_machine_design:field-coil-former-heat"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/field-coil-former-heat
site: BCT Engineering
---

# Heat conducted across field coil former

Electric Machine Design · Chapter 2: Heating and cooling of electric machine

- Asked in: 2073 Chaitra, 6 marks
- Video: [8. Heating -  Problem 1](https://www.youtube.com/watch?v=dTD8RFG0k5I) — Nisha Prakash
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/field-coil-former-heat
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2

## Question

The internal dimensions of the former of field coil of d.c. generator are 150×250 mm². The former is 2.5 mm thick. Calculate the heat conducted across the former from winding to core if there is an air space 1.5 mm wide between the former and the pole core. The thermal conductivity of former and air is 0.166 and 0.05 W/m-°C respectively. The winding height is 200 mm and the temperature rise is 40°C.

## Short answer

Heat flow area = 0.16 m². Series thermal resistances of former and air are 0.0941 and 0.1875 °C/W, total 0.2816 °C/W. For 40 °C rise, conducted heat = 40/0.2816 ≈ 142 W.

## Answer

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

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

```text
 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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The former and the air space are in series, so:
   A. Their conductivities are added
   B. Their thermal resistances are added
   C. Their thermal resistances are averaged
   D. Only the air layer counts
   Answer: B. Heat passes through both layers in turn, so R_total = R_former + R_air. (Key point: Series resistances)

2. The heat flow area S is:
   A. 150 x 250 mm²
   B. Perimeter 0.8 m x winding height 0.2 m = 0.16 m²
   C. Perimeter x thickness
   D. Winding height squared
   Answer: B. Heat crosses the sides of the former: 2(150 + 250) mm x 200 mm. (Key point: Heat flow area)

3. The thermal resistance of a layer is:
   A. k S/t
   B. k t/S
   C. t/(k S)
   D. S/(k t)
   Answer: C. R = t/(kS), giving 0.0941 for the former and 0.1875 for the air. (Key point: Resistance formula)

4. Heat conducted Q = θ/R_total is about:
   A. 142 W
   B. 40 W
   C. 0.28 W
   D. 14 W
   Answer: A. Q = 40/0.2816 = 142 W; the air layer has about twice the resistance of the former. (Key point: Result)

## Related in Electric Machine Design

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