---
title: "Design: 25 kVA 11000/433 V distribution transformer"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:design-25kva-11000-433"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-25kva-11000-433
site: BCT Engineering
---

# Design: 25 kVA 11000/433 V distribution transformer

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2074 Asoj, 14 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-25kva-11000-433
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

Design a 25 kVA, 11000/433 V, 50 Hz, 3 phase, delta/star core type distribution transformer. The required data for design are given below: Maximum flux density in core = 1 Wb/m²; Current density in conductor = 2.3 A/mm²; Constant for output volt per turn, K = 0.45; Core type = cruciform; Window space factor, Kw = 8/(30+kV); Stacking factor = 0.9; Ratio of window height to width = 2.5; Take area of yoke 20% more than area of limb; Width of LV winding = 9.1 mm; Width of HV winding = 26.22 mm; Total losses at full load = 901 W. Calculate: i) Dimensions of core, window and yoke ii) Overall dimensions of the frame iii) Per unit resistance and leakage reactance drop iv) Per unit voltage regulation at 0.8 pf v) Full load efficiency at 0.8 pf

## Short answer

Core: d=134.5 mm, limb 114.4×71.3 mm; window 114.9×287.3 mm; yoke depth 114.4 mm, height 118.2 mm. Overall frame 523.6×613.2×114.4 mm. Per-unit resistance, reactance, regulation, and efficiency are not fully computed in the provided solution.

## 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: clearance core-to-LV 2 mm, LV-to-HV duct 15 mm, end clearance 20 mm, $\rho = 0.021\ \Omega$-mm²/m.

#### i) Dimensions of core, window and yoke
Voltage per turn, $E_t = K\sqrt{Q}$ ($Q$ in kVA, total for 3 phases):

$$
\begin{aligned}
E_t &= 0.45\sqrt{25} = 2.250\ \text{V}\\
\phi_m &= \frac{E_t}{4.44f} = \frac{2.250}{4.44\times50} = 0.01014\ \text{Wb}\\
A_i &= \frac{\phi_m}{B_m} = \frac{0.01014}{1} = 0.01014\ \text{m}^2 = 101.4\ \text{cm}^2
\end{aligned}
$$

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

$$
\begin{aligned}
d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.01014}{0.56}} = 0.1345\ \text{m} = 134.5\ \text{mm}\\
a &= 0.85d = 114.4\ \text{mm}\\
b &= 0.53d = 71.3\ \text{mm}
\end{aligned}
$$

Gross core area $A_{gi} = A_i/0.9 = 0.01126$ m².

Output equation (three phase): $Q = 3.33fB_m\delta K_wA_wA_i\times10^{-3}$

Window space factor $K_w = \dfrac{8}{30 + kV} = \dfrac{8}{30 + 11} = 0.1951$

$$
\begin{aligned}
A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\
&= \frac{25}{3.33\times50\times1\times2.3\times10^6\times0.195\times0.01014\times10^{-3}}\\
&= 0.03301\ \text{m}^2\\
W_w &= \sqrt{\frac{A_w}{2.5}} = 114.9\ \text{mm},\quad H_w = 2.5W_w = 287.3\ \text{mm}\\
D &= d + W_w = 249.4\ \text{mm (distance between adjacent limb centres)}
\end{aligned}
$$

Yoke:

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

$$
\begin{aligned}
A_y &= 1.2A_i = 0.01216\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.01351\ \text{m}^2\\
D_y &= a = 114.4\ \text{mm}\\
H_y &= \frac{A_{gy}}{D_y} = \frac{0.01351}{0.1144} = 118.2\ \text{mm}
\end{aligned}
$$

#### ii) Overall dimensions of frame
$$
\begin{aligned}
H &= H_w + 2H_y = 287.3 + 2(118.2) = 523.6\ \text{mm}\\
W &= 2D + a = 2(249.4) + 114.4 = 613.2\ \text{mm}\\
\text{Depth} &= D_y = a = 114.4\ \text{mm}
\end{aligned}
$$

```text
 +-----------------------------+  ---
 |          top yoke           |   ^
 +-----+-----+-----+-----+-----+   |
 |limb |     |limb |     |limb |   |
 |  R  | Ww  |  Y  | Ww  |  B  |   H
 |     |     |     |     |     |   |
 +-----+-----+-----+-----+-----+   |
 |         bottom yoke         |   v
 +-----------------------------+  ---
 |<------------ W ------------>|
    |<--- D --->|<--- D --->|
```

#### iii) Per unit resistance and leakage reactance drop
Phase voltages: LV (star) $V_{LV} = 250.0$ V; HV (delta) $V_{HV} = 11000.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 33.33$ A; $I_{HV} = 0.758$ A.

$$
\begin{aligned}
T_{LV} &= \frac{V_{LV}}{E_t} = \frac{250.0}{2.250} = 111.11 \approx 111\\
E_t\ (\text{revised}) &= \frac{250.0}{111} = 2.252\ \text{V}\\
T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 4884.1 \approx 4884\\
a_{LV} &= \frac{I_{LV}}{\delta} = 14.49\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 0.329\ \text{mm}^2
\end{aligned}
$$

Radial build (assumed clearance between core and LV = 2 mm):

| Item | Inside dia (mm) | Outside dia (mm) |
|---|---|---|
| LV winding | 138.5 | 156.7 |
| HV winding | 186.7 | 239.2 |

Radial depths: $b_1 = 9.1$ mm (LV), $b_2 = 26.22$ mm (HV), duct $a = 15$ mm. Axial length $L_c = H_w - 2(20) = 247.3$ mm.

Mean lengths of turn:

$$
\begin{aligned}
L_{mt,LV} &= \pi\frac{138.5 + 156.7}{2} = 0.4638\ \text{m}\\
L_{mt,HV} &= \pi\frac{186.7 + 239.2}{2} = 0.6690\ \text{m}\\
L_{mt} &= \pi\frac{138.5 + 239.2}{2} = 0.5933\ \text{m (both windings, for reactance)}
\end{aligned}
$$

Check: HV outside diameter 239 mm vs $D$ = 249 mm. Adjacent HV coils have about 10 mm clearance, so the coils fit.

Resistance referred to HV, with $\rho = 0.021\ \Omega\text{-mm}^2/\text{m}$ (copper at about 75 °C):

$$
\begin{aligned}
r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times4884\times0.6690}{0.329} = 208.32\ \Omega\\
r_{LV} &= \frac{0.021\times111\times0.4638}{14.49} = 0.074594\ \Omega\\
R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 208.32 + 0.074594\left(\frac{4884}{111}\right)^2 = 352.73\ \Omega\\
\varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{0.7576\times352.73}{11000.0} = 0.02429\ \text{p.u.}
\end{aligned}
$$

Leakage reactance (per unit), with $AT$ per limb $= 3700$ and axial winding length $L_c = 247.3$ mm:

$$
\begin{aligned}
\varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\
&= 2\pi(50)(4\pi\times10^{-7})\frac{3700}{0.2473}\cdot\frac{0.5933}{2.252}\left(15 + \frac{9.1+26.22}{3}\right)\times10^{-3}\\
&= 0.0417\ \text{p.u.}
\end{aligned}
$$

#### iv) Per unit regulation at 0.8 pf lagging
$$
\begin{aligned}
\varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\
&= 0.02429(0.8) + 0.0417(0.600) = 0.0444\ \text{p.u.} = 4.44\%
\end{aligned}
$$

#### v) Full-load efficiency at 0.8 pf
Output $= 25\times0.8 = 20$ kW; total full-load loss = 901 W.

$$
\eta = \frac{20}{20 + 0.901}\times100 = 95.69\%
$$

(Copper loss from the resistance above: $3I_{HV}^2R_p$ = 607 W, so the iron loss is about 294 W.)

**Answer: $d$ = 134.5 mm, window 115 × 287 mm; frame 524 × 613 × 114 mm; $\varepsilon_r$ = 0.0243, $\varepsilon_x$ = 0.0417; regulation ≈ 4.44%; efficiency ≈ 95.69%.**

## Check yourself

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

1. How is the window space factor K_w found for the 11 kV transformer?
   A. K_w = 8/(30 + kV) = 8/41 = 0.195
   B. K_w = 8/(30 x kV) = 0.024
   C. K_w = (30 + kV)/8 = 5.1
   D. K_w = 0.4 for all distribution transformers
   Answer: A. The question gives K_w = 8/(30 + kV); for 11 kV this is 0.1951. (Key point: Window space factor)

2. After E_t = 2.25 V and A_i = 101.4 cm^2, how are d and the window obtained?
   A. d from A_i = 0.56 d^2 = 134.5 mm, then A_w from the output equation, W_w = sqrt(A_w/2.5) and H_w = 2.5 W_w
   B. d from A_i = 0.60 d^2 and W_w = A_w/2.5
   C. d = sqrt(A_i), then H_w = W_w
   D. A_w = A_i x K_w and H_w = 2.5 d
   Answer: A. A cruciform core uses A_i = 0.56 d^2; A_w = 0.0330 m^2 gives W_w = 114.9 mm and H_w = 287.3 mm. (Key point: Core and window dimensions)

3. How is the total resistance R_p referred to the HV side computed?
   A. R_p = r_HV + r_LV (T_HV/T_LV)^2
   B. R_p = r_HV + r_LV
   C. R_p = r_HV + r_LV (T_LV/T_HV)^2
   D. R_p = r_HV x r_LV / (r_HV + r_LV)
   Answer: A. Resistance transfers with the square of the turns ratio, so the LV resistance is multiplied by (T_HV/T_LV)^2 before adding it to r_HV. (Key point: Referring resistance to HV)

4. For output 25 kVA at 0.8 pf and total loss 901 W, the full-load efficiency is
   A. 99.1%
   B. 95.69%
   C. 97.2%
   D. 92.5%
   Answer: B. Output is 20 kW, so efficiency = 20/(20 + 0.901) = 95.69%. (Key point: Full-load efficiency)

## Related in Electric Machine Design

- [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×
- [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: 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: 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: 25 kVA transformer frame, regulation, tubes](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-25kva-cooling-tubes) — Electric Machine Design, chapter 3, asked 2×
- [Design data: 1000 kVA 66/11 kV transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-1000kva-66-11kv) — Electric Machine Design, chapter 3

## Related in other subjects

- [500 kVA delta/star transformer efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Athree-phase-500kva-33-11-efficiency) — Electrical Machine, chapter 2
- [25 kVA ideal transformer currents, flux](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Aideal-25kva-transformer) — Basic Electrical and Electronics Engineering, chapter 2
- [Delta/star 120 kVA transformer values](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Adelta-star-120kva-line-phase-values) — Electrical Machine, chapter 2
- [3-phase 500 kVA transformer primary current](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-500kva-11kv-400v-primary-current) — Electrical Machines, chapter 2