---
title: "Temperature gradient in slot conductors (thick insulation)"
subject: "Electric Machine Design"
chapter: 2
key: "semester_5/electric_machine_design:temperature-gradient-slot-thick"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/temperature-gradient-slot-thick
site: BCT Engineering
---

# Temperature gradient in slot conductors (thick insulation)

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

- Asked in: 2082 Baishakh, 4 marks; 2071 Shrawan, 6 marks (asked 2 times)
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/temperature-gradient-slot-thick
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2

## Question

Derive the expression of temperature gradient in conductors placed in slot if slot insulation is very thick.

## Short answer

With very thick slot insulation, heat flows axially along the conductor. If current density is J, the centre temperature rise above slot ends is θ_m − θ_o = J²ρ_e ρ_t L²/8, where ρ_e and ρ_t are electrical and thermal resistivities of copper, and L is the embedded slot length. The profile is parabolic.

## Answer

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).

```text
   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.

## Check yourself

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

1. With very thick slot insulation, in which direction does the heat of the slot conductor flow?
   A. Axially along the copper to the overhang
   B. Into the air gap
   C. It does not flow
   D. Across the insulation to the iron
   Answer: A. Thick insulation blocks sideways flow, so all heat goes along the conductor to the overhang. (Key point: Heat path)

2. What is the rise of the conductor centre over the slot ends?
   A. J^2 rho_e rho_t L / 2
   B. J rho_e rho_t L^2 / 8
   C. J^2 rho_e L / rho_t
   D. J^2 rho_e rho_t L^2 / 8
   Answer: D. theta_m - theta_o = p rho_t L^2/8 with p = J^2 rho_e. (Key point: Result)

3. Where is the temperature gradient along the conductor zero and where largest?
   A. Zero at the overhang
   B. Zero at slot ends, largest at centre
   C. Zero at centre, largest at slot ends
   D. Uniform everywhere
   Answer: C. d theta/dx = rho_t p x, so it is zero at x = 0 and largest at x = L/2. (Key point: Gradient)

## Related in Electric Machine Design

- [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 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×
- [Maximum temperature of slot conductor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Amax-conductor-temperature-slot) — 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 gradient in an iron core](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Atemperature-gradient-iron-core) — Electric Machine Design, chapter 2

## Related in other subjects

- [Cantilever tip deflection, temperature gradient](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Abeam-temperature-gradient-cantilever) — Theory of Structures I, chapter 3
- [Derive expression for tile drain spacing](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Atile-drain-spacing-derivation) — Irrigation and Drainage Engineering, chapter 9, asked 8×
- [Derive development length expression](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aderive-development-length-expression) — Design of RCC Structure, chapter 6, asked 6×
- [Derive PET expression by Penman's method](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aderive-penman-pet) — Engineering Hydrology, chapter 3