---
title: "Temperature gradient in an iron core"
subject: "Electric Machine Design"
chapter: 2
key: "semester_5/electric_machine_design:temperature-gradient-iron-core"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/temperature-gradient-iron-core
site: BCT Engineering
---

# Temperature gradient in an iron core

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

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

## Question

Explain the temperature gradient in an iron core with necessary figure and expression.

## Short answer

In an iron core, heat generated flows to cooling surfaces. Due to lamination, thermal resistivity across sheets is high, so heat prefers to flow along the laminations. The temperature rise from hot spot to surface follows p / [8(1/(ρ_x l_x^2) + 1/(ρ_y l_y^2))], where l_x, l_y are core dimensions and ρ_x, ρ_y are resistivities.

## Answer

Iron losses (hysteresis and eddy) are produced throughout the volume of the core. This heat must flow through the iron to the cooling surfaces and ducts, so the centre of each core packet is hotter than its surface. The resulting variation of temperature inside the core is the **temperature gradient**.

#### Anisotropy of a laminated core
- **Along the laminations**, heat flows through continuous iron, so the thermal resistivity $\rho_x$ is low.
- **Across the laminations**, heat must cross the insulation (varnish or oxide) and the air films between sheets, so the thermal resistivity $\rho_y$ is many times higher (roughly 20–50 times).

```text
       cooling surface / duct
   +----------------------------+  ^
   | ========================== |  |
   | ======== hot spot ======== |  | l_y (across)
   | ========================== |  |
   +----------------------------+  v
   <----------- l_x ------------>  (along)
```

#### One-direction flow
For loss $p$ per unit volume flowing across a length $l$ between two cooled faces, consider a plane at $x$ from the centre. Heat crossing it per unit area is $p x$, so $\dfrac{d\theta}{dx} = -\rho p x$ and

$$
\theta_h - \theta_s = \frac{p\,\rho\,l^2}{8}
$$

The temperature is parabolic, with the hot spot at the centre.

#### Two-direction flow (actual core)
Let the loss divide into $p_x$ (along) and $p_y$ (across). Both paths share the same hot spot and surface temperature:

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

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

#### Conclusions
- Most heat flows **along** the laminations because $\rho_x$ is small.
- The hot-spot rise varies as the square of the path length. Ventilating ducts that divide the core into packets of 40–80 mm therefore reduce the gradient sharply.
- Iron loss density $p$ should be kept moderate (a suitable $B_m$) in large cores.

## Check yourself

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

1. Why is the centre of an iron core packet hotter than its surface?
   A. Iron loss is produced throughout the volume and must flow through the iron to the cooling surface
   B. Only surface losses exist
   C. Heat is absorbed by the centre
   D. The centre has less iron
   Answer: A. Hysteresis and eddy losses are internal, so there is a gradient. (Key point: Cause)

2. Why is thermal resistivity across the laminations much higher than along them?
   A. The iron is thicker across
   B. Laminations are thinner across
   C. The iron is magnetic
   D. Heat must cross the insulation and air films between sheets
   Answer: D. Across the laminations the heat crosses the varnish/oxide layers, so rho_y is 20-50 times rho_x. (Key point: Anisotropy)

3. How does the hot-spot rise depend on the heat path length?
   A. Inversely
   B. Directly
   C. As its square, so ventilating ducts dividing the core into packets help greatly
   D. It is independent
   Answer: C. theta_h - theta_s = p rho l^2/8, so shortening l reduces it sharply. (Key point: Effect of length)

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