---
title: "Design: 100 kVA 4000/433 V distribution transformer"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:design-100kva-4000-433"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-100kva-4000-433
site: BCT Engineering
---

# Design: 100 kVA 4000/433 V distribution transformer

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2078 Bhadra, 20 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-100kva-4000-433
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

The design parameters for a 100 kVA, 4000/433 V, 50 Hz, 3-phase, Δ/Y core type oil immersed natural cooled distribution transformer are given below. Max. flux density in core = 1.3 Wb/m²; Current density in conductor = 2.5 A/mm²; Constant for output voltage per turn = 0.45; Core type = cruciform; Window space factor = 0.25; Stacking factor = 0.9; Ratio of window height to width = 2.5; Width of LV winding = 20 mm; Width of HV winding = 25 mm; Width of duct between HV and LV = 15 mm. Take hot rolled steel and area of yoke is 20% greater than area of core. Calculate: (i) Dimension of the core, window and yoke (ii) Overall dimension of the frame (iii) Per unit voltage regulation at 0.8 pf

## Short answer

Cruciform core d=167 mm, stampings 142×88 mm; window 344 mm high × 138 mm wide; yoke depth 142 mm, height 147 mm; overall frame 637×751×142 mm. Per unit voltage regulation at 0.8 pf lagging is 0.0423 (4.23%).

## 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: core-to-LV clearance 5 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):

$$
\begin{aligned}
E_t &= 0.45\sqrt{100} = 4.500\ \text{V}\\
\phi_m &= \frac{E_t}{4.44f} = \frac{4.500}{4.44\times50} = 0.02027\ \text{Wb}\\
A_i &= \frac{\phi_m}{B_m} = \frac{0.02027}{1.3} = 0.01559\ \text{m}^2 = 155.9\ \text{cm}^2
\end{aligned}
$$

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

$$
\begin{aligned}
d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.01559}{0.56}} = 0.1669\ \text{m} = 166.9\ \text{mm}\\
a &= 0.85d = 141.8\ \text{mm}\\
b &= 0.53d = 88.4\ \text{mm}
\end{aligned}
$$

Gross core area $A_{gi} = A_i/0.9 = 0.01733$ m².

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

$$
\begin{aligned}
A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\
&= \frac{100}{3.33\times50\times1.3\times2.5\times10^6\times0.25\times0.01559\times10^{-3}}\\
&= 0.04741\ \text{m}^2\\
W_w &= \sqrt{\frac{A_w}{2.5}} = 137.7\ \text{mm},\quad H_w = 2.5W_w = 344.3\ \text{mm}\\
D &= d + W_w = 304.6\ \text{mm (distance between adjacent limb centres)}
\end{aligned}
$$

Yoke:

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

$$
\begin{aligned}
A_y &= 1.2A_i = 0.01871\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.02079\ \text{m}^2\\
D_y &= a = 141.8\ \text{mm}\\
H_y &= \frac{A_{gy}}{D_y} = \frac{0.02079}{0.1418} = 146.6\ \text{mm}
\end{aligned}
$$

#### (ii) Overall dimensions of frame
$$
\begin{aligned}
H &= H_w + 2H_y = 344.3 + 2(146.6) = 637.4\ \text{mm}\\
W &= 2D + a = 2(304.6) + 141.8 = 751.0\ \text{mm}\\
\text{Depth} &= D_y = a = 141.8\ \text{mm}
\end{aligned}
$$

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

#### (iii) Per unit voltage regulation at 0.8 pf lagging
**Turns and conductors**

Phase voltages: LV (star) $V_{LV} = 250.0$ V; HV (delta) $V_{HV} = 4000.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 133.34$ A; $I_{HV} = 8.333$ A.

$$
\begin{aligned}
T_{LV} &= \frac{V_{LV}}{E_t} = \frac{250.0}{4.500} = 55.55 \approx 56\\
E_t\ (\text{revised}) &= \frac{250.0}{56} = 4.464\ \text{V}\\
T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 896.0 \approx 896\\
a_{LV} &= \frac{I_{LV}}{\delta} = 53.33\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 3.333\ \text{mm}^2
\end{aligned}
$$

**Winding build**

Radial build (assumed clearance between core and LV = 5 mm):

| Item | Inside dia (mm) | Outside dia (mm) |
|---|---|---|
| LV winding | 176.9 | 216.9 |
| HV winding | 246.9 | 296.9 |

Radial depths: $b_1 = 20$ mm (LV), $b_2 = 25$ mm (HV), duct $a = 15$ mm. Axial length $L_c = H_w - 2(20) = 304.3$ mm.

Mean lengths of turn:

$$
\begin{aligned}
L_{mt,LV} &= \pi\frac{176.9 + 216.9}{2} = 0.6185\ \text{m}\\
L_{mt,HV} &= \pi\frac{246.9 + 296.9}{2} = 0.8541\ \text{m}\\
L_{mt} &= \pi\frac{176.9 + 296.9}{2} = 0.7441\ \text{m (both windings, for reactance)}
\end{aligned}
$$

Check: HV outside diameter 296.9 mm < $D$ = 304.6 mm (about 8 mm between adjacent coils; just adequate at 4 kV).

**Resistance**

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

$$
\begin{aligned}
r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times896\times0.8541}{3.333} = 4.8212\ \Omega\\
r_{LV} &= \frac{0.021\times56\times0.6185}{53.33} = 0.013637\ \Omega\\
R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 4.8212 + 0.013637\left(\frac{896}{56}\right)^2 = 8.3122\ \Omega\\
\varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{8.333\times8.3122}{4000.0} = 0.01732\ \text{p.u.}
\end{aligned}
$$

**Reactance**

Leakage reactance (per unit), with $AT$ per limb $= 7467$ and axial winding length $L_c = 304.3$ mm:

$$
\begin{aligned}
\varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\
&= 2\pi(50)(4\pi\times10^{-7})\frac{7467}{0.3043}\cdot\frac{0.7441}{4.464}\left(15 + \frac{20+25}{3}\right)\times10^{-3}\\
&= 0.0484\ \text{p.u.}
\end{aligned}
$$

**Regulation**

$$
\begin{aligned}
\varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\
&= 0.01732(0.8) + 0.0484(0.600) = 0.0429\ \text{p.u.} = 4.29\%
\end{aligned}
$$

**Answer: $d$ = 167 mm, window 138 × 344 mm, yoke 147 × 142 mm; frame 637 × 751 × 142 mm; regulation ≈ 4.29%.**

## Check yourself

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

1. How is the net core area found for this 100 kVA transformer (K = 0.45, B_m = 1.3 T)?
   A. E_t = 0.45 x 100 = 45 V; A_i = 45/(1.3 x 50)
   B. E_t = 0.45 sqrt(100) = 4.5 V; phi_m = 4.5/(4.44 x 50) = 0.02027 Wb; A_i = phi_m/B_m = 0.01559 m^2
   C. E_t = 4.5 V; A_i = E_t/(4.44 f B_m) x 3
   D. E_t = 0.45 sqrt(100) = 4.5 V; A_i = phi_m x B_m
   Answer: B. E_t = K sqrt(Q) = 4.5 V and phi_m = E_t/(4.44 f) = 0.02027 Wb. Net core area is A_i = phi_m/B_m = 155.9 cm^2. (Key point: Voltage per turn and core area)

2. After A_w = 0.04741 m^2 is found, how are the window dimensions obtained for ratio 2.5?
   A. W_w = A_w/2.5 = 190 mm
   B. W_w = sqrt(A_w/2.5) = 137.7 mm and H_w = 2.5 W_w = 344 mm
   C. H_w = sqrt(A_w/2.5) = 137.7 mm and W_w = 344 mm
   D. W_w = sqrt(2.5 A_w) = 344 mm and H_w = 137.7 mm
   Answer: B. A_w = H_w x W_w = 2.5 W_w^2, so W_w = sqrt(0.04741/2.5) = 137.7 mm and H_w = 344.3 mm. (Key point: Window height and width)

3. Which set gives the overall frame size of this three-phase core?
   A. H = H_w + 2 H_y = 637 mm; W = 2D + a = 751 mm; depth = a = 142 mm
   B. H = H_w + H_y = 491 mm; W = D + a; depth = d
   C. H = H_w + 2 H_y = 637 mm; W = D + a = 446 mm; depth = d
   D. H = 2 H_w + H_y; W = 2D + a; depth = D_y = H_y
   Answer: A. Two yokes give H = 344.3 + 2 x 146.6 = 637.4 mm. Three limbs with two spacings of D = 304.6 mm give W = 751 mm, and the depth equals the largest stamping width a = 141.8 mm. (Key point: Frame dimensions)

4. Which statement matches the method used to reach 4.29% regulation?
   A. T_LV = 433/4.5 = 96 because LV uses line voltage
   B. T_LV = 250/4.5 = 56 (phase voltage of star LV); regulation = er cos(phi) + ex sin(phi) = 0.01732(0.8) + 0.0484(0.6)
   C. Regulation = er + ex = 6.6%
   D. Regulation = er sin(phi) + ex cos(phi) = 0.01732(0.6) + 0.0484(0.8)
   Answer: B. The LV star winding has phase voltage 433/sqrt(3) = 250 V, giving 56 turns. Regulation is er cos(phi) + ex sin(phi) = 0.01386 + 0.02904 = 0.0429, i.e. 4.29%. (Key point: Regulation at 0.8 pf and star winding turns)

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