---
title: "Internal (hot spot) temperature of core"
subject: "Electric Machine Design"
chapter: 2
key: "semester_5/electric_machine_design:hot-spot-core-temperature"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/hot-spot-core-temperature
site: BCT Engineering
---

# Internal (hot spot) temperature of core

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

- Asked in: 2080 Bhadra, 6 marks; 2075 Chaitra, 6 marks; 2075 Asoj, 8 marks; 2073 Chaitra, 6 marks; 2070 Chaitra, 6 marks; 2069 Asar, 4 marks (asked 6 times)
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/hot-spot-core-temperature
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2

## Question

Derive the expression for internal temperature (hot spot) of the core.

## Short answer

For a core with two-directional heat flow, along laminations (x) and across (y), the hot-spot temperature rise above surface is θ_h − θ_s = p / [8(1/(ρ_x lx^2) + 1/(ρ_y ly^2))], where p is loss per unit volume, ρ_x, ρ_y are thermal resistivities, and lx, ly are distances between cooling surfaces or ducts in each direction.

## Answer

In an iron core (or any part where losses are produced inside the material), heat must flow through the material to reach the cooling surface. The inside is therefore hotter than the surface. The highest temperature, at the centre, is the **hot spot**, and it decides the safety of nearby insulation.

#### Step 1: heat flow in one direction
Assume:
- $p$ = heat (loss) produced per unit volume, W/m³, uniform
- $\rho$ = thermal resistivity of the material, °C·m/W ($\rho = 1/k$)
- heat flows only in the $x$ direction (the other dimensions are large), and both faces are at the same surface temperature $\theta_s$.

By symmetry the hottest point is at the centre plane, $x = 0$, where no heat crosses. Consider a plane at distance $x$ from the centre. All heat produced between the centre and this plane must cross it. Per unit area:

$$
Q_x = p\,x \quad \text{W/m}^2
$$

By Fourier's law, the temperature gradient is

$$
\frac{d\theta}{dx} = -\rho\, Q_x = -\rho\, p\, x
$$

Integrating from the centre (temperature $\theta_h$):

$$
\begin{aligned}
\theta(x) &= \theta_h - \frac{\rho\, p\, x^2}{2}
\end{aligned}
$$

So the temperature is **parabolic** across the thickness. At the surface, $x = t/2$, $\theta = \theta_s$:

$$
\theta_h - \theta_s = \frac{\rho\, p}{2}\left(\frac{t}{2}\right)^2 = \frac{p\,\rho\, t^2}{8}
$$

```text
 temperature
     ^        theta_h (hot spot)
     |         _..--.._
     |      .-'        '-.
     |    .'              '.
     |   /                  \
     |  theta_s        theta_s
     +--|---------+---------|--> x
      -t/2        0        +t/2
        <------ thickness t ------>
```

#### Step 2: core with heat flow in two directions
A laminated core block can lose heat in two directions:
- **along the laminations** (direction $x$, length $l_x$ between cooling surfaces or ducts), where thermal resistivity $\rho_x$ is low (through iron);
- **across the laminations** (direction $y$, length $l_y$), where thermal resistivity $\rho_y$ is much higher, because heat must cross the insulation (varnish, oxide) between sheets.

```text
        cooling surface
   +---------------------+  ^
   | ==================  |  |
   | ==================  |  | l_y (across lam.)
   | ==================  |  |
   +---------------------+  v
   <-------- l_x -------->   (along laminations)
```

Let $p_x$ and $p_y$ be the parts of $p$ carried away in each direction, with $p_x + p_y = p$. Both paths start at the same hot spot and end at the same surface temperature, so by Step 1:

$$
\theta_h - \theta_s = \frac{p_x \rho_x l_x^2}{8} = \frac{p_y \rho_y l_y^2}{8}
$$

Hence $p_x = \dfrac{8(\theta_h-\theta_s)}{\rho_x l_x^2}$ and $p_y = \dfrac{8(\theta_h-\theta_s)}{\rho_y l_y^2}$. Adding:

$$
p = 8(\theta_h-\theta_s)\left[\frac{1}{\rho_x l_x^2} + \frac{1}{\rho_y l_y^2}\right]
$$

$$
\theta_h - \theta_s = \frac{p}{8\left[\dfrac{1}{\rho_x l_x^2} + \dfrac{1}{\rho_y l_y^2}\right]}
$$

This is the expression for the **internal (hot-spot) temperature rise** of the core above its surface. The hot-spot temperature itself is $\theta_h = \theta_s + (\theta_h - \theta_s)$, where $\theta_s$ is found from the surface dissipation, $\theta_s - \theta_a = \dfrac{\text{loss}}{S\lambda}$.

#### Conclusions
- The temperature difference is proportional to the **square of the length** of the heat path. Splitting the core into packets with **ventilating ducts** reduces it sharply.
- Since $\rho_y \gg \rho_x$, most heat flows along the laminations. Ducts placed across the laminations, which cut $l_x$, are therefore the most effective.

## Check yourself

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

1. Where is the hot spot of an iron core and why?
   A. At the centre, because heat made inside must flow through the iron to the cooling surface
   B. At the ducts
   C. At the bolts
   D. At the surface, because of cooling
   Answer: A. The inside is hotter than the surface, and by symmetry the centre plane has no heat crossing it. (Key point: Hot spot location)

2. For heat flow in one direction through thickness t, what is the hot-spot rise over the surface?
   A. p rho t / 2
   B. p rho t^2 / 2
   C. p t / (8 rho)
   D. p rho t^2 / 8
   Answer: D. Integrating d theta/dx = -rho p x from the centre to x = t/2 gives p rho t^2/8. (Key point: Single direction result)

3. Which expression gives the hot-spot rise for heat flow along and across laminations?
   A. p (rho_x + rho_y)/8
   B. p / (rho_x l_x + rho_y l_y)
   C. p / (8[1/(rho_x l_x^2) + 1/(rho_y l_y^2)])
   D. 8p [rho_x l_x^2 + rho_y l_y^2]
   Answer: C. Equating the rise for both paths and adding p_x + p_y = p gives this formula. (Key point: Two directions)

4. Why are ducts across the laminations most effective in reducing the hot-spot rise?
   A. Because rho_y is small
   B. Because most heat flows along the laminations (rho_y >> rho_x), and ducts shorten that path l_x; the rise varies as length squared
   C. Because ducts raise the loss
   D. Because heat flows only across the laminations
   Answer: B. The rise is proportional to the square of path length, and ducts cut the main path. (Key point: Cooling ducts)

## Related in Electric Machine Design

- [Hot spot temperature of a plate](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Ahot-spot-plate) — Electric Machine Design, chapter 2, asked 2×
- [Internal temperature of a homogeneous material](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Ainternal-temperature-homogeneous-material) — 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

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