---
title: "Temperature difference in slot copper strips"
subject: "Electric Machine Design"
chapter: 2
key: "semester_5/electric_machine_design:slot-strips-temperature-difference"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/slot-strips-temperature-difference
site: BCT Engineering
---

# Temperature difference in slot copper strips

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

- Asked in: 2072 Chaitra, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/slot-strips-temperature-difference
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2

## Question

A generator has open slots each containing 3 strips of copper in the arrangement of three strips along the depth of the slot. The length of the slot portion of the conductor is 0.5 m, each strip has a cross section of 8×10 mm² and current density in the strip is 4 A/mm². The electrical and thermal resistivities of the copper are 0.021×10⁻⁶ Ω-m and 0.0025 Ω-m respectively. The insulation between the strips and the slot walls is taken as 4 mm thick and has a thermal resistivity of 3 Ω-m. Calculate the temperature difference for the following cases: i) Between the centre of the embedded portion of the strip and the overhang ii) Between the conductor and the slot walls

## Short answer

Heat generation density = 3.36×10⁵ W/m³. (i) Centre‑to‑overhang: θ = (q ρ_th L²)/8 = 26.25 °C. (ii) Total slot loss 40.32 W, heat flow area 0.034 m², insulation resistance 0.353 °C/W, giving temperature difference ≈ 14.2 °C.

## Answer

Assumption: each strip is 8 mm wide (slot width) and 10 mm deep, so the three strips fill a depth of 30 mm. Heat in the slot portion flows (i) along the copper to the cooler overhang, and (ii) across the slot insulation to the iron.

Heat produced per unit volume of copper:

$$
q = J^2\rho = (4\times10^6)^2 \times 0.021\times10^{-6} = 3.36\times10^5\ \text{W/m}^3
$$

#### i) Centre of embedded portion to overhang
Take $x$ from the centre of the slot length $L$. Heat crossing a section at $x$ is $qAx$, and over $dx$ the temperature drop is $d\theta = qAx \cdot \rho_t\,dx/A = q\rho_t x\,dx$. Integrating from 0 to $L/2$:

$$
\begin{aligned}
\theta_1 &= \frac{q\rho_t}{2}\left(\frac{L}{2}\right)^2 = \frac{q\rho_t L^2}{8}\\
&= \frac{3.36\times10^5 \times 0.0025 \times 0.5^2}{8} = 26.25\ ^\circ\text{C}
\end{aligned}
$$

#### ii) Conductor to slot walls
Loss in the three strips (slot portion):

$$
\begin{aligned}
V &= 3 \times (8\times10\times10^{-6}) \times 0.5 = 1.2\times10^{-4}\ \text{m}^3\\
P &= qV = 3.36\times10^5 \times 1.2\times10^{-4} = 40.32\ \text{W}
\end{aligned}
$$

Heat leaves through the two sides and the bottom of the slot (the open top is assumed not to dissipate):

$$
S = (2 \times 30 + 8)\times10^{-3} \times 0.5 = 0.034\ \text{m}^2
$$

Thermal resistance of insulation:

$$
R = \frac{\rho_t\,t}{S} = \frac{3 \times 0.004}{0.034} = 0.353\ ^\circ\text{C/W}
$$

$$
\theta_2 = PR = 40.32 \times 0.353 = 14.23\ ^\circ\text{C}
$$

```text
   open top (wedge)
  |  [ strip 1 ]  |
  |  [ strip 2 ]  |  <- 4 mm insulation
  |  [ strip 3 ]  |     on sides and bottom
  |_______________|
     slot wall
```

**Answer: (i) ≈ 26.25 °C between the centre of the embedded part and the overhang; (ii) ≈ 14.2 °C between conductors and slot walls.**

## Check yourself

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

1. The heat produced per unit volume of the copper is:
   A. J ρ
   B. J^2 ρ
   C. ρ/J^2
   D. J/ρ
   Answer: B. q = J²ρ = (4x10^6)² x 0.021x10^-6 = 3.36x10^5 W/m³. (Key point: Heat per volume)

2. The temperature difference between the slot centre and the overhang is:
   A. q ρt L^2/2
   B. q ρt L^2/8
   C. q ρt L
   D. q ρt L^2
   Answer: B. Integrating q ρt x dx from 0 to L/2 gives q ρt L²/8 = 26.25 °C. (Key point: Axial drop)

3. For the conductor-to-wall difference, the surface through which heat leaves is:
   A. All four sides of the strips
   B. Two sides plus the bottom, with the open top ignored
   C. Only the bottom
   D. Only the two sides
   Answer: B. S = (2 x 30 + 8) mm x 0.5 m = 0.034 m². (Key point: Cooling surface)

4. Which relation gives the drop across the slot insulation?
   A. θ = P x ρt t/S
   B. θ = P/(ρt t S)
   C. θ = P ρt S/t
   D. θ = ρt t S
   Answer: A. R = ρt t/S = 0.353 °C/W and θ = PR = 40.32 x 0.353 = 14.2 °C. (Key point: Insulation drop)

## Related in Electric Machine Design

- [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
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- [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×
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- [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

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