---
title: "Design: 150 kVA single phase transformer"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:design-150kva-single-phase-windings"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-150kva-single-phase-windings
site: BCT Engineering
---

# Design: 150 kVA single phase transformer

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2072 Kartik, 20 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-150kva-single-phase-windings
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

Design a 150 kVA, 50 Hz, 6600/400 V, 1-phase, core type oil immersed natural cooled distribution transformer. Given that: Maximum flux density in core = 1.35 Wb/m²; Current density = 2.75 A/mm²; Core type = cruciform two stepped; Window space factor = 0.27; 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 LV winding = 268 mm; Axial depth of HV 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) Dimensions of the core, window and yoke ii) Overall dimensions of the frame iii) Leakage reactance of the transformer iv) Draw overall dimension of the transformer

## Short answer

Core diameter 233.9 mm, limb 198.9×124.0 mm; window 132.6×331.6 mm; yoke depth 198.9 mm, height 205.5 mm. Overall frame 742.6×565.4×198.9 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 $I_{HV}T_{HV}/2$. $K$ is not given; $K$ = 0.75 is taken (single-phase core type). 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):

$$
\begin{aligned}
E_t &= 0.75\sqrt{150} = 9.186\ \text{V}\\
\phi_m &= \frac{E_t}{4.44f} = \frac{9.186}{4.44\times50} = 0.04138\ \text{Wb}\\
A_i &= \frac{\phi_m}{B_m} = \frac{0.04138}{1.35} = 0.03065\ \text{m}^2 = 306.5\ \text{cm}^2
\end{aligned}
$$

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

$$
\begin{aligned}
d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.03065}{0.56}} = 0.2339\ \text{m} = 233.9\ \text{mm}\\
a &= 0.85d = 198.9\ \text{mm}\\
b &= 0.53d = 124.0\ \text{mm}
\end{aligned}
$$

Gross core area $A_{gi} = A_i/0.9 = 0.03405$ m².

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

$$
\begin{aligned}
A_w &= \frac{Q}{2.22fB_m\delta K_wA_i\times10^{-3}}\\
&= \frac{150}{2.22\times50\times1.35\times2.75\times10^6\times0.27\times0.03065\times10^{-3}}\\
&= 0.04399\ \text{m}^2\\
W_w &= \sqrt{\frac{A_w}{2.5}} = 132.6\ \text{mm},\quad H_w = 2.5W_w = 331.6\ \text{mm}\\
D &= d + W_w = 366.6\ \text{mm (distance between adjacent limb centres)}
\end{aligned}
$$

Check: $d$ = 234 mm < LV inside diameter 255 mm; HV outside diameter 351 mm < $D$ = 367 mm; coil height 276 mm < $H_w$ = 332 mm. The given windings fit.

Yoke:

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

$$
\begin{aligned}
A_y &= 1.2A_i = 0.03678\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.04087\ \text{m}^2\\
D_y &= a = 198.9\ \text{mm}\\
H_y &= \frac{A_{gy}}{D_y} = \frac{0.04087}{0.1989} = 205.5\ \text{mm}
\end{aligned}
$$

#### ii) Overall dimensions of frame
$$
\begin{aligned}
H &= H_w + 2H_y = 331.6 + 2(205.5) = 742.6\ \text{mm}\\
W &= D + a = 366.6 + 198.9 = 565.4\ \text{mm}\\
\text{Depth} &= D_y = a = 198.9\ \text{mm}
\end{aligned}
$$

#### iii) Leakage reactance
Currents: $I_{LV} = \frac{Q}{V_{LV}} = 375.00$ A; $I_{HV} = 22.727$ A.

$$
\begin{aligned}
T_{LV} &= \frac{V_{LV}}{E_t} = \frac{400.0}{9.186} = 43.55 \approx 44\\
E_t\ (\text{revised}) &= \frac{400.0}{44} = 9.091\ \text{V}\\
T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 726.0 \approx 726\\
a_{LV} &= \frac{I_{LV}}{\delta} = 136.36\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 8.264\ \text{mm}^2
\end{aligned}
$$

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 $= 8250$ and axial winding length $L_c = 272.0$ mm:

$$
\begin{aligned}
\varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\
&= 2\pi(50)(4\pi\times10^{-7})\frac{8250}{0.2720}\cdot\frac{0.9519}{9.091}\left(11 + \frac{14+18}{3}\right)\times10^{-3}\\
&= 0.0272\ \text{p.u.}\\
X_{HV} &= \varepsilon_x\frac{V_{HV}}{I_{HV}} = 0.0272\times\frac{6600.0}{22.73} = 7.889\ \Omega\ \text{(referred to HV)}\\
X_{LV} &= X_{HV}\left(\frac{T_{LV}}{T_{HV}}\right)^2 = 0.02898\ \Omega
\end{aligned}
$$

#### iv) Overall dimensions (sketch)
```text
 +-----------------------+  ---
 |  yoke, Hy = 206 mm    |   ^
 +-----+-----------+-----+   |
 |limb |  window   |limb |   H = 743 mm
 |d=234| 133 x 332 |     |   |
 +-----+-----------+-----+   |
 |        yoke           |   v
 +-----------------------+  ---
 |<--------- W --------->|
 W = 565 mm, D = 367 mm (limb centres)
 Depth = 199 mm
```

**Answer: $d$ = 234 mm; window 133 × 332 mm; yoke 206 × 199 mm; frame 743 × 565 × 199 mm; leakage reactance ≈ 7.89 Ω referred to HV (2.72%).**

## Check yourself

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

1. For the 150 kVA single-phase core-type transformer, how is the voltage per turn found and what is its value (K = 0.75)?
   A. Et = K*sqrt(Q) = 0.75*sqrt(150) = 9.19 V
   B. Et = K*Q = 0.75*150 = 112.5 V
   C. Et = K*sqrt(Q) = 0.75*sqrt(150)/2 = 4.59 V
   D. Et = 4.44*f*Bm = 4.44*50*1.35 = 300 V
   Answer: A. Voltage per turn is Et = K*sqrt(Q) with Q in kVA, giving 0.75*12.25 = 9.186 V. (Key point: Voltage per turn)

2. For a cruciform (two-stepped) core with stacking factor 0.9, how is the circumscribing circle diameter d obtained from the net iron area Ai?
   A. Ai = 0.785 d^2
   B. Ai = 0.45 d^2
   C. Ai = 0.9 d^2
   D. Ai = 0.56 d^2, so d = sqrt(Ai/0.56) = 234 mm
   Answer: D. The standard ratio for a two-stepped core with stacking factor 0.9 is Ai = 0.56 d^2, so d = sqrt(0.03065/0.56) = 234 mm. (Key point: Core area and diameter)

3. Why is the yoke made 20% larger in area than the core, and how is its depth chosen?
   A. To carry more winding current; the depth equals the window height
   B. To lower the yoke flux density and so reduce yoke loss and magnetising current; depth equals the width of the largest stamping
   C. To make the frame heavier for stability; depth equals the core diameter d
   D. To compensate for the stacking factor; depth equals the window width
   Answer: B. A larger yoke area gives a lower flux density, reducing iron loss and magnetising current. The yoke depth is the width of the largest stamping (a = 0.85d). (Key point: Yoke area and depth)

4. In the leakage reactance formula the ampere-turns per limb are taken as I_HV*T_HV/2. Why?
   A. Because the current divides equally between two parallel HV paths
   B. Because only half the HV winding carries current at any time
   C. Because half of each winding is on each of the two limbs of a single-phase core-type transformer
   D. Because the flux is halved in the yoke
   Answer: C. Each winding is split equally over the two limbs, so each limb carries half the turns and hence half the total ampere-turns. (Key point: Ampere-turns per limb in leakage reactance)

## Close variants set in other exams

- [Design: 125 kVA single phase transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-125kva-single-phase-windings) — Electric Machine Design, chapter 3, asked 3×
- [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×
- [Design: 100 kVA 11/132 kV transformer with winding data](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-100kva-11-132kv) — Electric Machine Design, chapter 3, asked 3×

## Related in Electric Machine Design

- [Design: 150 kVA 6600/400 V distribution transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-150kva-6600-400-regulation) — Electric Machine Design, chapter 3, asked 5×
- [Design: 500 kVA single phase transformer with tubes](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-500kva-single-phase-tubes) — Electric Machine Design, chapter 3, asked 2×
- [Design: 150 kVA 11000/420 V distribution transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-150kva-11000-420) — Electric Machine Design, chapter 3, asked 3×
- [Design: 100 kVA 2200/480 V single phase transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-100kva-single-phase-ducts) — Electric Machine Design, chapter 3, asked 3×
- [Design: 150 kVA 11000/440 V transformer with tubes](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-150kva-11000-440-tubes) — Electric Machine Design, chapter 3
- [Design: 100 kVA 4000/433 V distribution transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-100kva-4000-433) — Electric Machine Design, chapter 3, asked 4×