---
title: "Maximum temperature of slot conductor"
subject: "Electric Machine Design"
chapter: 2
key: "semester_5/electric_machine_design:max-conductor-temperature-slot"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/max-conductor-temperature-slot
site: BCT Engineering
---

# Maximum temperature of slot conductor

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

- Asked in: 2078 Kartik, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/max-conductor-temperature-slot
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2

## Question

If the ambient temperature is θa, determine the maximum temperature of conductor placed in slot if the slot insulation is comparatively very thick.

## Short answer

Overhang temperature: theta_o = theta_a + p A(L+2l_o)/(2 l_o S_p lambda). Slot centre adds rise p rho_t L^2/8. Thus theta_max = theta_a + J^2 rho_e A(L+2l_o)/(2 l_o S_p lambda) + J^2 rho_e rho_t L^2/8.

## Answer

When the slot insulation is comparatively very thick, no heat flows from the conductor into the iron. All copper loss of the slot portion flows **along the conductor** to the overhang, and the overhang dissipates the whole loss of the conductor to the air. The maximum temperature is at the middle of the slot length.

#### Notation
- $J$ = current density, $\rho_e$ = electrical resistivity, so loss per unit volume $p = J^2\rho_e$
- $A$ = conductor cross-section, $S_p$ = perimeter of the overhang surface (W dissipated per unit length per °C is $S_p\lambda$)
- $L$ = slot (embedded) length, $l_o$ = overhang length at each end, $\rho_t$ = thermal resistivity of copper
- $\lambda$ = specific heat dissipation of the overhang surface (W/m²°C), $\theta_a$ = ambient temperature

```text
  l_o            L (slot)            l_o
 ~~~~~ |=====================| ~~~~~
 theta_o      theta_max        theta_o
```

#### Step 1: overhang temperature
The total loss of one conductor, $p A (L + 2l_o)$, is dissipated from the two overhang surfaces, of area $2 l_o S_p$ (overhang temperature taken as uniform):

$$
\begin{aligned}
p A (L + 2 l_o) &= 2 l_o S_p \lambda (\theta_o - \theta_a) \\
\theta_o &= \theta_a + \frac{p A (L + 2 l_o)}{2 l_o S_p \lambda}
\end{aligned}
$$

#### Step 2: rise of slot portion above the overhang
At distance $x$ from the slot centre, the heat crossing the conductor section is $p A x$. So

$$
\frac{d\theta}{dx} = -\rho_t\,p\,x \;\Rightarrow\; \theta = \theta_{max} - \frac{\rho_t p x^2}{2}
$$

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

$$
\theta_{max} - \theta_o = \frac{p\,\rho_t\,L^2}{8}
$$

#### Step 3: maximum conductor temperature
$$
\theta_{max} = \theta_a + \frac{J^2\rho_e A (L + 2 l_o)}{2 l_o S_p \lambda} + \frac{J^2\rho_e\,\rho_t\,L^2}{8}
$$

The maximum temperature is the ambient, plus the rise of the overhang (set by its cooling), plus the parabolic rise along the slot. A long core or a high current density increases it sharply. This is why large machines use ventilating ducts, thin high-conductivity insulation or direct conductor cooling.

## Check yourself

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

1. Where is the maximum conductor temperature with thick slot insulation?
   A. At the iron surface
   B. At the overhang
   C. At the middle of the slot length
   D. At the slot ends
   Answer: C. Heat flows outward from the centre to the overhangs, so the centre is the hottest. (Key point: Where maximum occurs)

2. How is the overhang temperature found?
   A. From the core loss
   B. Total conductor loss p A (L + 2 l_o) equals heat from overhang surface 2 l_o S_p lambda (theta_o - theta_a)
   C. From the slot insulation thickness
   D. Assumed equal to ambient
   Answer: B. The overhang dissipates the whole loss of the conductor to the air. (Key point: Overhang temperature)

3. What is the rise of the slot centre above the overhang?
   A. p rho_t L^2 / 8
   B. p rho_t L / 2
   C. p L^2/(8 rho_t)
   D. p A rho_t
   Answer: A. Integrating d theta/dx = -rho_t p x gives p rho_t L^2/8. (Key point: Rise in slot)

## Related in Electric Machine Design

- [Temperature gradient in slot conductors (thick insulation)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Atemperature-gradient-slot-thick) — Electric Machine Design, chapter 2, asked 2×
- [Temperature gradient in slot: two cases](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Atemperature-gradient-slot-overhang) — Electric Machine Design, chapter 2
- [Temperature difference in slot copper strips](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aslot-strips-temperature-difference) — Electric Machine Design, chapter 2
- [Temperature gradients in slot conductors](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Atemperature-gradient-slot-explain) — Electric Machine Design, chapter 2, asked 2×

## Related in other subjects

- [Ampacity at new ambient and max temperature](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aampacity-change-new-temperatures) — Transmission and Distribution System Design, chapter 4
- [T/F: ampacity vs ambient temperature](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atf-ampacity-ambient-temperature) — Transmission and Distribution System Design, chapter 4, asked 2×