---
title: "Design: 4000 kVA 66/11 kV power transformer"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:design-4000kva-66-11kv"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-4000kva-66-11kv
site: BCT Engineering
---

# Design: 4000 kVA 66/11 kV power transformer

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2074 Chaitra, 18 marks; 2070 Asar, 16 marks (asked 2 times)
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-4000kva-66-11kv
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

For a 4000 kVA, 3 phase, 50 Hz, 66 kV/11 kV, delta/delta, core type, oil immersed natural cooled power transformer the design data are: Max flux density in core = 1.6 Wb/m²; Constant for output voltage per turn = 0.6; Resistivity of copper = 0.021 Ω-mm²/m; Core type = Cruciform; Current density in conductors = 2.5 A/mm²; Window space factor = 0.22; Stacking factor = 0.9; Ratio of window height to width = 2.75; Take hot rolled steel and area of yoke is 20% greater than area of core; Width of duct between LV and core = 10 mm; Width of LV winding = 50 mm; Width of HV winding = 50 mm; Width of duct between LV and HV = 20 mm. Assuming all the other required parameters, calculate: (i) Overall core dimension (ii) Overall dimension of frame (iii) Per unit resistance and leakage reactance drop (iv) Per unit voltage regulation at 0.8 pf

## Short answer

Core: d = 436.8 mm, a = 371.3 mm, b = 231.5 mm. Frame: H = 1605.6 mm, W = 1854.5 mm, depth = 371.3 mm. Per-unit resistance drop ≈ 0.005, leakage reactance drop ≈ 0.048. Per-unit voltage regulation at 0.8 pf lag ≈ 0.033 (3.3%).

## Answer

Standard method (A.K. Sawhney). Assumptions: hot-rolled steel, yoke area = 1.2 × core area, yoke depth = largest stamping width, $\rho = 0.021\ \Omega$-mm²/m, end clearance 50 mm at each end of the 66 kV winding.

#### (i) Overall core dimensions
Voltage per turn, $E_t = K\sqrt{Q}$ ($Q$ in kVA, total for 3 phases):

$$
\begin{aligned}
E_t &= 0.6\sqrt{4000} = 37.947\ \text{V}\\
\phi_m &= \frac{E_t}{4.44f} = \frac{37.947}{4.44\times50} = 0.17093\ \text{Wb}\\
A_i &= \frac{\phi_m}{B_m} = \frac{0.17093}{1.6} = 0.10683\ \text{m}^2 = 1068.3\ \text{cm}^2
\end{aligned}
$$

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

$$
\begin{aligned}
d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.10683}{0.56}} = 0.4368\ \text{m} = 436.8\ \text{mm}\\
a &= 0.85d = 371.3\ \text{mm}\\
b &= 0.53d = 231.5\ \text{mm}
\end{aligned}
$$

Gross core area $A_{gi} = A_i/0.9 = 0.11870$ m².

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

$$
\begin{aligned}
A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\
&= \frac{4000}{3.33\times50\times1.6\times2.5\times10^6\times0.22\times0.10683\times10^{-3}}\\
&= 0.25554\ \text{m}^2\\
W_w &= \sqrt{\frac{A_w}{2.75}} = 304.8\ \text{mm},\quad H_w = 2.75W_w = 838.3\ \text{mm}\\
D &= d + W_w = 741.6\ \text{mm (distance between adjacent limb centres)}
\end{aligned}
$$

**Yoke**

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

$$
\begin{aligned}
A_y &= 1.2A_i = 0.12820\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.14244\ \text{m}^2\\
D_y &= a = 371.3\ \text{mm}\\
H_y &= \frac{A_{gy}}{D_y} = \frac{0.14244}{0.3713} = 383.7\ \text{mm}
\end{aligned}
$$

#### (ii) Overall dimensions of frame
$$
\begin{aligned}
H &= H_w + 2H_y = 838.3 + 2(383.7) = 1605.6\ \text{mm}\\
W &= 2D + a = 2(741.6) + 371.3 = 1854.5\ \text{mm}\\
\text{Depth} &= D_y = a = 371.3\ \text{mm}
\end{aligned}
$$

```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
**Turns and conductors** (both windings delta):

Phase voltages: LV (delta) $V_{LV} = 11000.0$ V; HV (delta) $V_{HV} = 66000.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 121.21$ A; $I_{HV} = 20.202$ A.

$$
\begin{aligned}
T_{LV} &= \frac{V_{LV}}{E_t} = \frac{11000.0}{37.947} = 289.88 \approx 290\\
E_t\ (\text{revised}) &= \frac{11000.0}{290} = 37.931\ \text{V}\\
T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 1740.0 \approx 1740\\
a_{LV} &= \frac{I_{LV}}{\delta} = 48.48\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 8.081\ \text{mm}^2
\end{aligned}
$$

**Winding dimensions**

Radial build (given clearance between core and LV = 10 mm):

| Item | Inside dia (mm) | Outside dia (mm) |
|---|---|---|
| LV winding | 456.8 | 556.8 |
| HV winding | 596.8 | 696.8 |

Radial depths: $b_1 = 50$ mm (LV), $b_2 = 50$ mm (HV), duct $a = 20$ mm. Axial length $L_c = H_w - 2(50) = 738.3$ mm.

Mean lengths of turn:

$$
\begin{aligned}
L_{mt,LV} &= \pi\frac{456.8 + 556.8}{2} = 1.5921\ \text{m}\\
L_{mt,HV} &= \pi\frac{596.8 + 696.8}{2} = 2.0319\ \text{m}\\
L_{mt} &= \pi\frac{456.8 + 696.8}{2} = 1.8120\ \text{m (both windings, for reactance)}
\end{aligned}
$$

Check: HV outside diameter 697 mm < $D$ = 742 mm, leaving about 45 mm between adjacent HV coils.

**Per unit resistance**

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

$$
\begin{aligned}
r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times1740\times2.0319}{8.081} = 9.1879\ \Omega\\
r_{LV} &= \frac{0.021\times290\times1.5921}{48.48} = 0.19998\ \Omega\\
R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 9.1879 + 0.19998\left(\frac{1740}{290}\right)^2 = 16.387\ \Omega\\
\varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{20.2\times16.387}{66000.0} = 0.00502\ \text{p.u.}
\end{aligned}
$$

**Per unit leakage reactance**

Leakage reactance (per unit), with $AT$ per limb $= 35152$ and axial winding length $L_c = 738.3$ mm:

$$
\begin{aligned}
\varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\
&= 2\pi(50)(4\pi\times10^{-7})\frac{35152}{0.7383}\cdot\frac{1.8120}{37.931}\left(20 + \frac{50+50}{3}\right)\times10^{-3}\\
&= 0.0479\ \text{p.u.}
\end{aligned}
$$

#### (iv) Per unit regulation at 0.8 pf lagging
$$
\begin{aligned}
\varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\
&= 0.00502(0.8) + 0.0479(0.600) = 0.0327\ \text{p.u.} = 3.27\%
\end{aligned}
$$

**Answer: core $d$ = 437 mm; window 305 × 838 mm; frame 1606 × 1854 × 371 mm; $\varepsilon_r$ = 0.0050, $\varepsilon_x$ = 0.0479 p.u.; regulation ≈ 3.27%.**

## Check yourself

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

1. For 4000 kVA with K = 0.6, voltage per turn Et is about:
   A. 24.0 V
   B. 37.9 V
   C. 60.0 V
   D. 12.6 V
   Answer: B. 0.6 x sqrt(4000) = 0.6 x 63.25 = 37.95 V. (Key point: Voltage per turn)

2. For a cruciform core, the net area and largest stamping width are:
   A. Ai = 0.45d^2, a = 0.71d
   B. Ai = 0.60d^2, a = 0.90d
   C. Ai = 0.56d^2, a = 0.85d
   D. Ai = 0.62d^2, a = 0.93d
   Answer: C. A cruciform core has Ai = 0.56d^2 and a = 0.85d, b = 0.53d. (Key point: Cruciform core dimensions)

3. Using hot-rolled steel, the yoke area is made:
   A. Equal to limb area
   B. 20% greater than limb area
   C. 50% less than limb area
   D. Double the limb area
   Answer: B. A larger yoke lowers yoke flux density, loss and magnetising current. (Key point: Yoke area)

4. Given er = 0.00502 and ex = 0.0479 at 0.8 pf lag, regulation is about:
   A. 4.81%
   B. 3.27%
   C. 0.50%
   D. 5.28%
   Answer: B. 0.00502(0.8) + 0.0479(0.6) = 0.0327 p.u. (Key point: Regulation result)

## Close variants set in other exams

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