---
title: "Stacking factor and window space factor"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:stacking-window-space-factor"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/stacking-window-space-factor
site: BCT Engineering
---

# Stacking factor and window space factor

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2069 Asar, 6 marks
- Video: [Transformer degine| window space factor | window area #electricalengineering #electronic](https://www.youtube.com/watch?v=wGsB_RQ9rX0) — BECURRENT
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/stacking-window-space-factor
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

What is "stacking factor" and "window space factor" in a transformer? How will you select appropriate value of window space factor for transformers of different capacities? Why higher voltage machines have lower window space factor?

## Short answer

Stacking factor is net/gross core area ≈0.9. Window space factor Kw is copper area/window area, chosen empirically, e.g., Kw=8/(30+kV) for 50–200 kVA. High voltage reduces Kw because thicker conductor insulation and larger clearances occupy more window.

## Answer

#### Stacking factor
The core is built of thin laminations insulated from each other by varnish or oxide. Some of the gross cross-section is therefore insulation, not iron.

$$
\text{Stacking factor } K_i = \frac{\text{net iron area } A_i}{\text{gross core area } A_{gi}}
$$

Typical value is about 0.9 (0.88–0.92), depending on lamination thickness and insulation coating. Thinner laminations give a lower stacking factor.

#### Window space factor
The window holds conductors plus their insulation, the insulation between layers and windings, ducts and clearances.

$$
\text{Window space factor } K_w = \frac{\text{copper area in window}}{\text{total window area } A_w}
$$

#### Selecting $K_w$ for different ratings
$K_w$ depends mainly on voltage rating and output. Empirical formulas (Sawhney) for the HV voltage in kV:

| Rating | $K_w$ |
|---|---|
| About 50–200 kVA | $\dfrac{8}{30 + kV}$ |
| About 1000 kVA | $\dfrac{10}{30 + kV}$ |
| Larger ratings | $\dfrac{12}{30 + kV}$ |

Example: an 11 kV, 100 kVA transformer has $K_w = 8/(30 + 11) = 0.195$. $K_w$ is roughly 0.1–0.4 for most transformers.

Larger units use bigger conductors (strips), so the insulation is a smaller fraction of the area and $K_w$ rises with kVA for the same voltage.

#### Why higher voltage transformers have lower $K_w$
- Insulation thickness on conductors, between layers, between LV and HV, and to the yoke must increase with voltage.
- Larger clearances and oil ducts are needed to withstand impulse and test voltages.
- HV windings carry small currents, so conductors are thin; the insulation covering forms a large fraction of each conductor's area.

So a larger part of the window is occupied by insulation and space, and the copper fraction ($K_w$) falls, as the formula $K_w \propto 1/(30 + kV)$ shows.

## Check yourself

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

1. Stacking factor is:
   A. Gross / net
   B. Copper / window
   C. Iron / copper
   D. Net iron area / gross core area
   Answer: D. Typical 0.9. (Key point: Stacking factor)

2. Window space factor is:
   A. Copper area in window / total window area
   B. Window area / core area
   C. Iron area / window
   D. Insulation / copper
   Answer: A. The rest is insulation, ducts and clearances. (Key point: Window space factor)

3. Kw typically:
   A. Decreases with kVA
   B. Increases with kVA for the same voltage
   C. Is independent
   D. Is always 1
   Answer: B. Larger units use bigger conductors so insulation share falls. (Key point: Selection by rating)

4. Why do higher voltage machines have lower Kw?
   A. Less current
   B. More copper
   C. More insulation, clearances and ducts are required
   D. Smaller core
   Answer: C. Kw ~ 1/(30+kV). (Key point: High voltage lowers Kw)

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