---
title: "Design: 100 kVA 11/132 kV transformer with winding data"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:design-100kva-11-132kv"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-100kva-11-132kv
site: BCT Engineering
---

# Design: 100 kVA 11/132 kV transformer with winding data

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2076 Chaitra, 16 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-100kva-11-132kv
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

Design a 100 kVA, 50 Hz, 11/132 kV, 3-phase, Δ/Y, core type oil immersed natural cooled distribution transformer. Maximum flux density in core = 1.35 Wb/m²; Core type = 3 stepped; Current density = 2.75 A/mm²; Window space factor = 0.4; Hw/Ww = 3; Take hot rolled steel sheet and area of yoke = 1.2 × area of core; Axial depth of L.V. = 268 mm; Axial depth of H.V. = 276 mm; Radial depth of L.V. = 14 mm; Radial depth of H.V. = 18 mm; Width of insulation between L.V. and H.V. = 11 mm; Outside diameter of L.V. = 293 mm, Inside diameter of L.V. = 255 mm; Inside diameter of H.V. = 314 mm, Outside diameter of H.V. = 351 mm. Calculate: (i) Dimension of core, window and yoke (ii) Overall dimension of frame (iii) Leakage reactance of transformer (iv) Voltage regulation at 0.8 pf lagging at full load.

## Short answer

Three-stepped core d=158 mm, stampings 142, 111, 66 mm; window 638 mm high × 213 mm wide; yoke depth 142 mm, height 141 mm; overall frame 920×884×142 mm. Leakage reactance per phase referred to LV is 179 Ω (0.0493 pu). Voltage regulation at 0.8 pf lagging is 0.0491 pu (4.91%).

## 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. $K$ is not given; $K$ = 0.45 is taken (3-phase distribution transformer). Also assumed $\rho = 0.021\ \Omega$-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, total for 3 phases):

$$
\begin{aligned}
E_t &= 0.45\sqrt{100} = 4.500\ \text{V}\\
\phi_m &= \frac{E_t}{4.44f} = \frac{4.500}{4.44\times50} = 0.02027\ \text{Wb}\\
A_i &= \frac{\phi_m}{B_m} = \frac{0.02027}{1.35} = 0.01502\ \text{m}^2 = 150.2\ \text{cm}^2
\end{aligned}
$$

For a three-stepped core with stacking factor 0.9, $A_i = 0.60d^2$ (standard ratio, Sawhney):

$$
\begin{aligned}
d &= \sqrt{\frac{A_i}{0.60}} = \sqrt{\frac{0.01502}{0.60}} = 0.1582\ \text{m} = 158.2\ \text{mm}\\
a &= 0.90d = 142.4\ \text{mm}\\
b &= 0.70d = 110.7\ \text{mm},\quad c = 0.42d = 66.4\ \text{mm}
\end{aligned}
$$

Gross core area $A_{gi} = A_i/0.9 = 0.01668$ m².

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

$$
\begin{aligned}
A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\
&= \frac{100}{3.33\times50\times1.35\times2.75\times10^6\times0.4\times0.01502\times10^{-3}}\\
&= 0.02694\ \text{m}^2\\
W_w &= \sqrt{\frac{A_w}{3}} = 94.8\ \text{mm},\quad H_w = 3W_w = 284.3\ \text{mm}\\
D &= d + W_w = 252.9\ \text{mm (distance between adjacent limb centres)}
\end{aligned}
$$

Check with the given windings: core $d$ = 158 mm fits inside the LV (255 mm). But the HV outside diameter (351 mm) exceeds $D$ = 253 mm, and the coil (276 mm) is almost as tall as the window (284 mm). So the window is enlarged to suit the given coils, with 20 mm between adjacent HV coils, and its height is increased to keep the given ratio $H_w/W_w = 3$:

$$
\begin{aligned}
D &= 351 + 20 = 371\ \text{mm},\quad W_w = D - d = 212.8\ \text{mm}\\
H_w &= 3W_w = 638.4\ \text{mm}
\end{aligned}
$$

(A real 132 kV winding would need far larger clearances; the given figures are used as stated.)

Yoke:

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

$$
\begin{aligned}
A_y &= 1.2A_i = 0.01802\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.02002\ \text{m}^2\\
D_y &= a = 142.4\ \text{mm}\\
H_y &= \frac{A_{gy}}{D_y} = \frac{0.02002}{0.1424} = 140.6\ \text{mm}
\end{aligned}
$$

#### (ii) Overall dimensions of frame
$$
\begin{aligned}
H &= H_w + 2H_y = 638.4 + 2(140.6) = 919.7\ \text{mm}\\
W &= 2D + a = 2(371.0) + 142.4 = 884.4\ \text{mm}\\
\text{Depth} &= D_y = a = 142.4\ \text{mm}
\end{aligned}
$$

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

#### (iii) Leakage reactance
Phase voltages: LV (delta) $V_{LV} = 11000.0$ V; HV (star) $V_{HV} = 76210.2$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 3.03$ A; $I_{HV} = 0.437$ A.

$$
\begin{aligned}
T_{LV} &= \frac{V_{LV}}{E_t} = \frac{11000.0}{4.500} = 2444.44 \approx 2444\\
E_t\ (\text{revised}) &= \frac{11000.0}{2444} = 4.501\ \text{V}\\
T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 16932.5 \approx 16933\\
a_{LV} &= \frac{I_{LV}}{\delta} = 1.10\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 0.159\ \text{mm}^2
\end{aligned}
$$

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 $= 7406$ and axial winding length $L_c = 272.0$ mm:

$$
\begin{aligned}
\varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\
&= 2\pi(50)(4\pi\times10^{-7})\frac{7406}{0.2720}\cdot\frac{0.9519}{4.501}\left(11 + \frac{14+18}{3}\right)\times10^{-3}\\
&= 0.0493\ \text{p.u.}\\
X_{HV} &= \varepsilon_x\frac{V_{HV}}{I_{HV}} = 0.0493\times\frac{76210.2}{0.4374} = 8583\ \Omega\ \text{(referred to HV)}\\
X_{LV} &= X_{HV}\left(\frac{T_{LV}}{T_{HV}}\right)^2 = 178.8\ \Omega
\end{aligned}
$$

#### (iv) Regulation at 0.8 pf lagging, full load
Resistance referred to HV, with $\rho = 0.021\ \Omega\text{-mm}^2/\text{m}$ (copper at about 75 °C):

$$
\begin{aligned}
r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times16933\times1.0446}{0.159} = 2335.4\ \Omega\\
r_{LV} &= \frac{0.021\times2444\times0.8608}{1.10} = 40.093\ \Omega\\
R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 2335.4 + 40.093\left(\frac{16933}{2444}\right)^2 = 4260\ \Omega\\
\varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{0.4374\times4260}{76210.2} = 0.02445\ \text{p.u.}
\end{aligned}
$$

$$
\begin{aligned}
\varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\
&= 0.02445(0.8) + 0.0493(0.600) = 0.0491\ \text{p.u.} = 4.91\%
\end{aligned}
$$

**Answer: $d$ = 158 mm; window 213 × 638 mm; yoke 141 × 142 mm; frame 920 × 884 × 142 mm; $X$ ≈ 8583 Ω/phase on HV (4.93%); regulation ≈ 4.91%.**

## Check yourself

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

1. K is not given in the question. What value was assumed and what E_t resulted?
   A. K = 0.6, E_t = 6.0 V
   B. K = 0.8, E_t = 8.0 V
   C. K = 0.45, E_t = 4.5 V
   D. K = 0.45, E_t = 45 V
   Answer: C. For a 3-phase distribution transformer K = 0.45 is taken, so E_t = 0.45 sqrt(100) = 4.5 V. (Key point: Voltage per turn with assumed K)

2. For a three-stepped core with stacking factor 0.9, what relation between A_i and d is used?
   A. A_i = 0.56 d^2
   B. A_i = 0.60 d^2
   C. A_i = 0.45 d^2
   D. A_i = 0.70 d^2
   Answer: B. A three-stepped core uses A_i = 0.60 d^2, so d = sqrt(0.01502/0.60) = 158.2 mm, with a = 0.90d, b = 0.70d and c = 0.42d. (Key point: Three-stepped core)

3. Why is the window from the output equation (D = 253 mm, H_w = 284 mm) not used as such?
   A. The core diameter is larger than the LV inside diameter
   B. The HV outside diameter (351 mm) exceeds D and the 276 mm coil nearly fills the window height, so the window is enlarged with clearances
   C. The output equation is valid only for single-phase transformers
   D. The yoke would then be thinner than the limb
   Answer: B. The given coils do not fit, so D = 351 + 20 = 371 mm and H_w = 276 + 2(25) = 326 mm are taken to include inter-coil and end clearances. (Key point: Enlarging the window for given coils)

4. In the per unit leakage reactance formula, which term represents the winding geometry across the leakage flux path?
   A. (a + (b1 + b2)/3), the duct width plus one third of the sum of the radial depths
   B. (b1 + b2 + a), simple total thickness
   C. (a + b1 + b2)/3 with a as core diameter
   D. (L_c + a), the axial length plus the duct
   Answer: A. The reactance uses (a + (b1+b2)/3) with a = 11 mm, b1 = 14 mm and b2 = 18 mm, together with AT/L_c and L_mt/E_t, giving about 0.0493 p.u. (Key point: Leakage reactance formula)

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