---
title: "Ventilating ducts across laminations"
subject: "Electric Machine Design"
chapter: 2
key: "semester_5/electric_machine_design:ventilating-ducts-laminations"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/ventilating-ducts-laminations
site: BCT Engineering
---

# Ventilating ducts across laminations

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

- Asked in: 2072 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/ventilating-ducts-laminations
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2

## Question

"Ventilating ducts are kept across the lamination to reduce the temperature rise within the material in electrical equipment." Justify the statement.

## Short answer

Ventilating ducts divide the core into thinner packets, shortening the heat path length l. Because the hot-spot rise varies as l^2, one duct reduces it to a quarter, n ducts to 1/(n+1)^2. Ducts also provide extra cooling surface and allow heat to flow along the laminations, bypassing the high thermal resistance across the insulation.

## Answer

The statement is true. Iron loss is produced throughout the volume of a core, and the heat must flow through the iron to a cooling surface. The internal (hot-spot) temperature rise depends on the **square of the heat path length**. Ventilating ducts shorten this path and add cooling surface, so they reduce the temperature rise inside the material sharply.

#### Justification
For a core packet of length $l$ between two cooled faces, with loss $p$ per unit volume and thermal resistivity $\rho$, the hot spot is at the middle:

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

If the core is divided by $n$ radial ducts into $(n+1)$ packets, each packet has length $\dfrac{l}{n+1}$:

$$
\theta_h' - \theta_s = \frac{p\,\rho}{8}\left(\frac{l}{n+1}\right)^2 = \frac{\theta_h - \theta_s}{(n+1)^2}
$$

For example, one central duct ($n = 1$) reduces the internal rise to **one-quarter**. Three ducts reduce it to one-sixteenth.

```text
 without duct            with one duct
 +-----------+          +----+  +----+
 |    hot    |          |    |  |    |
 +-----------+          +----+  +----+
   rise ~ l^2           rise ~ (l/2)^2 = 1/4
                               ^ air
```

#### Why the ducts cut across the laminations
- Heat flows easily **along** the laminations (through iron), but with difficulty **across** them, through the insulation between sheets, where the thermal resistivity is much higher.
- A duct "across the lamination" (between packets, at right angles to the core length) gives every sheet its own short path along its plane to cooling air, so heat does not have to cross many insulation layers.

#### Further benefits
1. Extra cooling surface: the duct faces dissipate heat, so $\theta_s$ (surface temperature) also falls.
2. Air is blown through the ducts by the rotor (radial ventilation), raising the heat transfer coefficient.
3. Lower core temperature protects the winding insulation and permits higher flux density or current density, giving more output from the same frame.

Typical practice: packets of 40–80 mm with ducts about 10 mm wide in the stator and rotor cores of large machines, and oil ducts in transformer cores.

## Check yourself

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

1. The internal hot-spot temperature rise of a core packet varies with the heat path length l as:
   A. l
   B. l^2
   C. 1/l
   D. sqrt(l)
   Answer: B. θh - θs = pρl²/8, so it depends on the square of the path length. (Key point: Hot spot dependence)

2. With n ducts dividing the core into (n+1) packets, the internal rise becomes:
   A. original/(n+1)
   B. original/(n+1)^2
   C. original/n
   D. original x (n+1)
   Answer: B. Each packet has length l/(n+1), so the rise falls by (n+1)². One duct gives one-quarter. (Key point: Effect of ducts)

3. Why are the ducts placed across the laminations?
   A. Heat flows easily along the sheets but poorly across the insulation, so each sheet gets a short path to cooling air
   B. Heat flows easily across the insulation between sheets
   C. To reduce the weight of the core only
   D. To increase eddy currents
   Answer: A. Thermal resistivity across the insulation is high, so ducts let heat go along the plane of each sheet. (Key point: Direction of heat flow)

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