---
title: "Design: 125 kVA single phase transformer"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:design-125kva-single-phase-windings"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-125kva-single-phase-windings
site: BCT Engineering
---

# Design: 125 kVA single phase transformer

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2072 Chaitra, 18 marks
- Video: [Three Phase Transformer complete design core, window, winding, space factor, specific loading, turns](https://www.youtube.com/watch?v=O7FdTj5wCV0) — 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-125kva-single-phase-windings
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

Design a 125 kVA, 50 Hz, 6600/400 V, 1-phase, core type oil immersed natural cooled distribution transformer. Given that: Maximum flux density in the core = 1.35 Wb/m²; Current density = 2.75 A/mm²; Core type = cruciform two stepped; Window space factor = 0.30; Stacking factor = 0.9; Ratio of window height to width 2.5; Take hot rolled steel sheet and area of yoke is 20% greater than area of core; Axial depth of L.V. winding = 268 mm; Axial depth of H.V. winding = 276 mm; Inside diameter of LV winding = 255 mm; Radial depth of LV winding = 14 mm; Radial depth of HV winding = 18 mm; Width of insulation between LV and HV = 11 mm; Outside diameter of LV winding = 293 mm; Inside diameter of HV winding = 314 mm; Outside diameter of HV winding = 351 mm. Calculate: i) Dimension of the core, window and yoke ii) Overall dimension of the frame iii) Leakage reactance of the transformer. Appropriate values for additional data required may be assumed if necessary.

## Short answer

Core diameter 223.5 mm, limb 190.0×118.5 mm; window 137.5×300.6 mm (adjusted for coil clearance); yoke depth 190.0 mm, height 196.3 mm. Overall frame 693.3×551.0×190.0 mm. Leakage reactance calculation is not fully provided.

## 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. In a single-phase core type transformer half of each winding is on each limb, so the ampere-turns per limb are $I_{HV}T_{HV}/2$. $K$ = 0.75 assumed (single-phase core type), $\rho = 0.021\ \Omega$-mm²/m. Winding data are given: LV 255/293 mm (inside/outside diameter), HV 314/351 mm, radial depths $b_1$ = 14 mm, $b_2$ = 18 mm, gap $a$ = 11 mm. Axial length taken as the mean of 268 and 276 mm, $L_c$ = 272 mm.

#### i) Dimensions of core, window and yoke
Voltage per turn, $E_t = K\sqrt{Q}$ ($Q$ in kVA):

$$
\begin{aligned}
E_t &= 0.75\sqrt{125} = 8.385\ \text{V}\\
\phi_m &= \frac{E_t}{4.44f} = \frac{8.385}{4.44\times50} = 0.03777\ \text{Wb}\\
A_i &= \frac{\phi_m}{B_m} = \frac{0.03777}{1.35} = 0.02798\ \text{m}^2 = 279.8\ \text{cm}^2
\end{aligned}
$$

For a cruciform (two-stepped) core with stacking factor 0.9, $A_i = 0.56d^2$ (standard ratio, Sawhney):

$$
\begin{aligned}
d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.02798}{0.56}} = 0.2235\ \text{m} = 223.5\ \text{mm}\\
a &= 0.85d = 190.0\ \text{mm}\\
b &= 0.53d = 118.5\ \text{mm}
\end{aligned}
$$

Gross core area $A_{gi} = A_i/0.9 = 0.03109$ m².

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

$$
\begin{aligned}
A_w &= \frac{Q}{2.22fB_m\delta K_wA_i\times10^{-3}}\\
&= \frac{125}{2.22\times50\times1.35\times2.75\times10^6\times0.3\times0.02798\times10^{-3}}\\
&= 0.03614\ \text{m}^2\\
W_w &= \sqrt{\frac{A_w}{2.5}} = 120.2\ \text{mm},\quad H_w = 2.5W_w = 300.6\ \text{mm}\\
D &= d + W_w = 343.8\ \text{mm (distance between adjacent limb centres)}
\end{aligned}
$$

Check: $d$ = 224 mm fits inside the LV (255 mm) and coil height 276 mm < $H_w$. But HV outside diameter 351 mm > $D$ = 344 mm, so the window is widened (10 mm between the HV coils on the two limbs):

$$
\begin{aligned}
D &= 351 + 10 = 361\ \text{mm},\quad W_w = D - d = 137.5\ \text{mm},\quad H_w = 300.6\ \text{mm}
\end{aligned}
$$

Yoke:

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

$$
\begin{aligned}
A_y &= 1.2A_i = 0.03357\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.03731\ \text{m}^2\\
D_y &= a = 190.0\ \text{mm}\\
H_y &= \frac{A_{gy}}{D_y} = \frac{0.03731}{0.1900} = 196.3\ \text{mm}
\end{aligned}
$$

#### ii) Overall dimensions of frame
$$
\begin{aligned}
H &= H_w + 2H_y = 300.6 + 2(196.3) = 693.3\ \text{mm}\\
W &= D + a = 361.0 + 190.0 = 551.0\ \text{mm}\\
\text{Depth} &= D_y = a = 190.0\ \text{mm}
\end{aligned}
$$

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

#### iii) Leakage reactance
Currents: $I_{LV} = \frac{Q}{V_{LV}} = 312.50$ A; $I_{HV} = 18.939$ A.

$$
\begin{aligned}
T_{LV} &= \frac{V_{LV}}{E_t} = \frac{400.0}{8.385} = 47.70 \approx 48\\
E_t\ (\text{revised}) &= \frac{400.0}{48} = 8.333\ \text{V}\\
T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 792.0 \approx 792\\
a_{LV} &= \frac{I_{LV}}{\delta} = 113.64\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 6.887\ \text{mm}^2
\end{aligned}
$$

Radial depths: $b_1 = 14$ mm (LV), $b_2 = 18$ mm (HV), duct $a = 11$ mm. 

Mean lengths of turn:

$$
\begin{aligned}
L_{mt,LV} &= \pi\frac{255.0 + 293.0}{2} = 0.8608\ \text{m}\\
L_{mt,HV} &= \pi\frac{314.0 + 351.0}{2} = 1.0446\ \text{m}\\
L_{mt} &= \pi\frac{255.0 + 351.0}{2} = 0.9519\ \text{m (both windings, for reactance)}
\end{aligned}
$$

Leakage reactance (per unit), with $AT$ per limb $= 7500$ and axial winding length $L_c = 272.0$ mm:

$$
\begin{aligned}
\varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\
&= 2\pi(50)(4\pi\times10^{-7})\frac{7500}{0.2720}\cdot\frac{0.9519}{8.333}\left(11 + \frac{14+18}{3}\right)\times10^{-3}\\
&= 0.0269\ \text{p.u.}\\
X_{HV} &= \varepsilon_x\frac{V_{HV}}{I_{HV}} = 0.0269\times\frac{6600.0}{18.94} = 9.388\ \Omega\ \text{(referred to HV)}\\
X_{LV} &= X_{HV}\left(\frac{T_{LV}}{T_{HV}}\right)^2 = 0.03448\ \Omega
\end{aligned}
$$

**Answer: $d$ = 224 mm; window 137 × 301 mm; yoke 196 × 190 mm; frame 693 × 551 × 190 mm; leakage reactance ≈ 9.39 Ω referred to HV (2.69%).**

## Check yourself

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

1. Which output equation is used to find the window area of this single-phase transformer?
   A. Q = 3.33 f Bm delta Kw Aw Ai x 10^-3
   B. Q = 11 Bav ac Kw D^2 L ns x 10^-3
   C. Q = 2.22 f Bm delta Kw Aw Ai x 10^-3
   D. Q = 4.44 f Bm Ai x 10^-3
   Answer: C. For a single-phase core-type transformer the constant is 2.22; 3.33 is for the three-phase case. (Key point: Single-phase output equation)

2. The calculated D = 343.8 mm but the HV outside diameter is 351 mm. What is done?
   A. D is increased to 351 + 10 = 361 mm so the HV coils on the two limbs clear each other
   B. The HV winding is reduced to 340 mm diameter
   C. The core diameter d is increased until it matches 351 mm
   D. The window height is reduced to 280 mm
   Answer: A. The HV coils on adjacent limbs would overlap, so the limb-centre distance is widened to D = 351 + 10 = 361 mm, making Ww = D - d = 137.5 mm. (Key point: Widening the window)

3. How are the overall width and depth of the single-phase frame found?
   A. W = D + a, depth = a (width of largest stamping)
   B. W = 2D + a, depth = d
   C. W = D, depth = Hy
   D. W = Ww + 2a, depth = Hw
   Answer: A. Width is limb-centre distance plus the largest stamping width, W = 361 + 190 = 551 mm, and the depth equals a = 190 mm. (Key point: Frame dimensions)

4. With Et = 8.385 V, TLV = 47.7 is rounded to 48. What are the revised Et and HV turns?
   A. Et = 8.333 V, THV = 48*6600/400 = 792
   B. Et = 8.385 V, THV = 48*400/6600 = 2.9
   C. Et = 8.333 V, THV = 48*6600/400 = 79
   D. Et = 9.1 V, THV = 726
   Answer: A. Revised Et = 400/48 = 8.333 V and THV = TLV*VHV/VLV = 48*16.5 = 792. (Key point: Turns on LV and HV)

## Close variants set in other exams

- [Design: 150 kVA single phase transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-150kva-single-phase-windings) — Electric Machine Design, chapter 3, asked 4×
- [Design: 125 kVA 6600/400 V three phase transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-125kva-three-phase-windings) — Electric Machine Design, chapter 3, asked 4×

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