---
title: "Design: 100 kVA single phase, square core"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:design-100kva-single-phase-square"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-100kva-single-phase-square
site: BCT Engineering
---

# Design: 100 kVA single phase, square core

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2073 Chaitra, 14 marks
- Video: [Transformer Design solved example single phase|core|window|area|winding|specific magnetic loading](https://www.youtube.com/watch?v=JpbLOnyhKKM) — ShreeDevi Classes
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-100kva-single-phase-square
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

Determine the (i) overall dimension of the core, (ii) overall dimension of frame and (iii) number of turns and the cross sectional area of conductors in the primary and secondary windings of a 100 kVA, 2200/480 V single phase core type transformer to operate at frequency of 50 Hz, assuming the following data: Constant for voltage per turn = 0.75; Maximum flux density = 1.2 Wb/m²; Core type = Square; Stacking factor = 0.9; Ratio of height to width of window = 2; Window space factor = 0.28; Current density = 2.5 A/mm²

## Short answer

Square limb side a=177.6 mm (d=250.1 mm), window 138×276 mm. Frame: height 699 mm, width 566 mm, depth 178 mm. HV winding: 293 turns, 18.18 mm²; LV winding: 64 turns, 83.33 mm².

## Answer

Standard method (A.K. Sawhney, *A Course in Electrical Machine Design*): $E_t = K\sqrt{Q}$, net core area ratio for the core type, output equation for the window, yoke depth = width of largest stamping. Assumed: yoke area 1.2 × core area (hot-rolled steel).

#### (i) Overall dimensions of core
Voltage per turn, $E_t = K\sqrt{Q}$ ($Q$ in kVA):

$$
\begin{aligned}
E_t &= 0.75\sqrt{100} = 7.500\ \text{V}\\
\phi_m &= \frac{E_t}{4.44f} = \frac{7.500}{4.44\times50} = 0.03378\ \text{Wb}\\
A_i &= \frac{\phi_m}{B_m} = \frac{0.03378}{1.2} = 0.02815\ \text{m}^2 = 281.5\ \text{cm}^2
\end{aligned}
$$

For a square core with stacking factor 0.9, $A_i = 0.45d^2$ (standard ratio, Sawhney):

$$
\begin{aligned}
d &= \sqrt{\frac{A_i}{0.45}} = \sqrt{\frac{0.02815}{0.45}} = 0.2501\ \text{m} = 250.1\ \text{mm}\\
a &= 0.71d = 177.6\ \text{mm}
\end{aligned}
$$

Gross core area $A_{gi} = A_i/0.9 = 0.03128$ m².

Output equation (single phase): $Q = 2.22fB_m\delta K_wA_wA_i\times10^{-3}$

$$
\begin{aligned}
A_w &= \frac{Q}{2.22fB_m\delta K_wA_i\times10^{-3}}\\
&= \frac{100}{2.22\times50\times1.2\times2.5\times10^6\times0.28\times0.02815\times10^{-3}}\\
&= 0.03810\ \text{m}^2\\
W_w &= \sqrt{\frac{A_w}{2}} = 138.0\ \text{mm},\quad H_w = 2W_w = 276.0\ \text{mm}\\
D &= d + W_w = 388.1\ \text{mm (distance between adjacent limb centres)}
\end{aligned}
$$

Yoke:

Yoke area 20% more than core area (hot-rolled steel, to reduce yoke loss and magnetising current). Yoke depth = width of largest stamping:

$$
\begin{aligned}
A_y &= 1.2A_i = 0.03378\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.03754\ \text{m}^2\\
D_y &= a = 177.6\ \text{mm}\\
H_y &= \frac{A_{gy}}{D_y} = \frac{0.03754}{0.1776} = 211.4\ \text{mm}
\end{aligned}
$$

#### (ii) Overall dimensions of frame
$$
\begin{aligned}
H &= H_w + 2H_y = 276.0 + 2(211.4) = 698.8\ \text{mm}\\
W &= D + a = 388.1 + 177.6 = 565.7\ \text{mm}\\
\text{Depth} &= D_y = a = 177.6\ \text{mm}
\end{aligned}
$$

```text
 +-----------------------+  ---
 |       top yoke        |   ^
 +-----+-----------+-----+   |
 |limb |  window   |limb |   H
 |     |  Ww x Hw  |     |   |
 +-----+-----------+-----+   |
 |      bottom yoke      |   v
 +-----------------------+  ---
 |<--------- W --------->|
    |<----- D ----->|
```

#### (iii) Turns and conductor areas
Currents: $I_{LV} = \frac{Q}{V_{LV}} = 208.33$ A; $I_{HV} = 45.455$ A.

$$
\begin{aligned}
T_{LV} &= \frac{V_{LV}}{E_t} = \frac{480.0}{7.500} = 64.00 \approx 64\\
E_t\ (\text{revised}) &= \frac{480.0}{64} = 7.500\ \text{V}\\
T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 293.3 \approx 293\\
a_{LV} &= \frac{I_{LV}}{\delta} = 83.33\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 18.182\ \text{mm}^2
\end{aligned}
$$

| Winding | Voltage (V) | Current (A) | Turns | Conductor area (mm²) |
|---|---|---|---|---|
| Primary (HV) | 2200 | 45.45 | 293 | 18.18 |
| Secondary (LV) | 480 | 208.33 | 64 | 83.33 |

The LV conductor (83.3 mm²) would be a rectangular strip (for example several strips in parallel); the HV conductor is a round or small rectangular wire.

**Answer: $d$ = 250.1 mm ($a$ = 177.6 mm), window 138 × 276 mm; frame 699 × 566 × 178 mm; HV 293 turns of 18.18 mm², LV 64 turns of 83.33 mm².**

## Check yourself

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

1. For a square core, how are E_t and d obtained here (K = 0.75, Q = 100 kVA)?
   A. E_t = K sqrt(Q) = 7.5 V; A_i = 0.45 d^2 gives d = 250 mm
   B. E_t = K x Q = 75 V; A_i = 0.56 d^2 gives d = 224 mm
   C. E_t = K sqrt(Q) = 7.5 V; A_i = 0.56 d^2 gives d = 224 mm
   D. E_t = K/sqrt(Q) = 0.075 V; A_i = 0.45 d^2 gives d = 250 mm
   Answer: A. E_t = 0.75 x sqrt(100) = 7.5 V, giving A_i = 0.02815 m^2. For a square core A_i = 0.45 d^2, so d = 250.1 mm. (Key point: Voltage per turn and square core diameter)

2. How are the window dimensions found?
   A. From the core diameter only, W_w = d
   B. A_w from Q = 2.22 f B_m delta K_w A_w A_i x 10^-3; then W_w = sqrt(A_w/2) = 138 mm and H_w = 2 W_w = 276 mm
   C. A_w from the three-phase equation with 3.33; W_w = A_w/2
   D. A_w = A_i/K_w then H_w = W_w
   Answer: B. A single phase transformer uses the 2.22 output equation to give A_w = 0.0381 m^2. With H_w/W_w = 2, W_w = sqrt(A_w/2) = 138 mm and H_w = 276 mm. (Key point: Window area from output equation)

3. How is the yoke height and frame width obtained?
   A. Yoke area = core area, so H_y = 138 mm; W = D
   B. Yoke area is 1.2 A_i, depth D_y = a (largest stamping width); H_y = A_gy/D_y = 211 mm; W = D + a
   C. Yoke area = 0.8 A_i; H_y = A_y/d; W = 2D + a
   D. Yoke depth = window width, H_y = A_w/D_y; W = d + 2a
   Answer: B. The hot-rolled yoke has 20% more area than the core, gross A_gy = 0.03754 m^2. Dividing by D_y = a = 177.6 mm gives H_y = 211.4 mm, and W = D + a = 565.7 mm for single phase. (Key point: Yoke dimensions and frame size)

4. What are the LV turns and the conductor areas?
   A. T_LV = 64; a_LV = 83.33 mm^2; T_HV = 293; a_HV = 18.18 mm^2
   B. T_LV = 293; a_LV = 18.18 mm^2; T_HV = 64; a_HV = 83.33 mm^2
   C. T_LV = 64; a_LV = 18.18 mm^2; T_HV = 293; a_HV = 83.33 mm^2
   D. T_LV = 32; a_LV = 83.33 mm^2; T_HV = 147; a_HV = 18.18 mm^2
   Answer: A. T_LV = 480/7.5 = 64 and T_HV = 64 x 2200/480 = 293. Conductor area = I/delta, so LV 208.33/2.5 = 83.33 mm^2 and HV 45.45/2.5 = 18.18 mm^2. (Key point: Turns and conductor areas)

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