---
title: "Design data: 1000 kVA 66/11 kV transformer"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:design-1000kva-66-11kv"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-1000kva-66-11kv
site: BCT Engineering
---

# Design data: 1000 kVA 66/11 kV transformer

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2080 Bhadra, 18 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-1000kva-66-11kv
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

The design parameters for a 1000 kVA, 50 Hz, 66/11 kV, 3-phase, delta/delta, core type, oil immersed natural cooled power transformer are given below. Constant for output voltage per turn = 0.6; Maximum flux density in the core = 1.45 Wb/m²; Current density in conductor = 2.75 A/mm²; Core type = cruciform two stepped; Window space factor = 0.13; Stacking factor = 0.9; Ratio of window height to width = 2.75; Take hot rolled steel and area of yoke is 20% greater than gross area of core; Width of LV winding = 40 mm; Width of HV winding = 50 mm; Width of duct between HV and LV winding = 20 mm; Resistivity of copper = 0.021 Ω-mm²/m. [The list of quantities to calculate is not shown in the scanned paper.]

## Short answer

Core d=324.4 mm, cruciform (a=275.8, b=171.9 mm); window 267.4×735.2 mm; yoke depth 275.8 mm height 285.0 mm; frame H=1305.2 mm, W=1459.3 mm, depth 275.8 mm.

## Answer

The list of quantities is missing from the paper, so the usual set is worked out: core, window and yoke; overall frame; per unit resistance and reactance; regulation at 0.8 pf lagging. 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 10 mm, end clearance 50 mm (66 kV), $\rho = 0.021\ \Omega$-mm²/m.

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

$$
\begin{aligned}
E_t &= 0.6\sqrt{1000} = 18.974\ \text{V}\\
\phi_m &= \frac{E_t}{4.44f} = \frac{18.974}{4.44\times50} = 0.08547\ \text{Wb}\\
A_i &= \frac{\phi_m}{B_m} = \frac{0.08547}{1.45} = 0.05894\ \text{m}^2 = 589.4\ \text{cm}^2
\end{aligned}
$$

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

$$
\begin{aligned}
d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.05894}{0.56}} = 0.3244\ \text{m} = 324.4\ \text{mm}\\
a &= 0.85d = 275.8\ \text{mm}\\
b &= 0.53d = 171.9\ \text{mm}
\end{aligned}
$$

Gross core area $A_{gi} = A_i/0.9 = 0.06549$ m².

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

$$
\begin{aligned}
A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\
&= \frac{1000}{3.33\times50\times1.45\times2.75\times10^6\times0.13\times0.05894\times10^{-3}}\\
&= 0.19657\ \text{m}^2\\
W_w &= \sqrt{\frac{A_w}{2.75}} = 267.4\ \text{mm},\quad H_w = 2.75W_w = 735.2\ \text{mm}\\
D &= d + W_w = 591.8\ \text{mm (distance between adjacent limb centres)}
\end{aligned}
$$

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

$$
\begin{aligned}
A_y &= 1.2A_i = 0.07073\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.07859\ \text{m}^2\\
D_y &= a = 275.8\ \text{mm}\\
H_y &= \frac{A_{gy}}{D_y} = \frac{0.07859}{0.2758} = 285.0\ \text{mm}
\end{aligned}
$$

#### Overall dimensions of frame
$$
\begin{aligned}
H &= H_w + 2H_y = 735.2 + 2(285.0) = 1305.2\ \text{mm}\\
W &= 2D + a = 2(591.8) + 275.8 = 1459.3\ \text{mm}\\
\text{Depth} &= D_y = a = 275.8\ \text{mm}
\end{aligned}
$$

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

#### Turns and conductors (delta/delta)
Phase voltages: LV (delta) $V_{LV} = 11000.0$ V; HV (delta) $V_{HV} = 66000.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 30.30$ A; $I_{HV} = 5.051$ A.

$$
\begin{aligned}
T_{LV} &= \frac{V_{LV}}{E_t} = \frac{11000.0}{18.974} = 579.75 \approx 580\\
E_t\ (\text{revised}) &= \frac{11000.0}{580} = 18.966\ \text{V}\\
T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 3480.0 \approx 3480\\
a_{LV} &= \frac{I_{LV}}{\delta} = 11.02\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 1.837\ \text{mm}^2
\end{aligned}
$$

#### Winding dimensions
Radial build (assumed clearance between core and LV = 10 mm):

| Item | Inside dia (mm) | Outside dia (mm) |
|---|---|---|
| LV winding | 344.4 | 424.4 |
| HV winding | 464.4 | 564.4 |

Radial depths: $b_1 = 40$ mm (LV), $b_2 = 50$ mm (HV), duct $a = 20$ mm. Axial length $L_c = H_w - 2(50) = 635.2$ mm.

Mean lengths of turn:

$$
\begin{aligned}
L_{mt,LV} &= \pi\frac{344.4 + 424.4}{2} = 1.2077\ \text{m}\\
L_{mt,HV} &= \pi\frac{464.4 + 564.4}{2} = 1.6161\ \text{m}\\
L_{mt} &= \pi\frac{344.4 + 564.4}{2} = 1.4276\ \text{m (both windings, for reactance)}
\end{aligned}
$$

Check: HV outside diameter 564 mm vs $D$ = 592 mm. Adjacent HV coils have about 27 mm clearance, so the coils fit.

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

$$
\begin{aligned}
r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times3480\times1.6161}{1.837} = 64.309\ \Omega\\
r_{LV} &= \frac{0.021\times580\times1.2077}{11.02} = 1.3349\ \Omega\\
R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 64.309 + 1.3349\left(\frac{3480}{580}\right)^2 = 112.37\ \Omega\\
\varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{5.051\times112.37}{66000.0} = 0.00860\ \text{p.u.}
\end{aligned}
$$

#### Per unit leakage reactance
Leakage reactance (per unit), with $AT$ per limb $= 17576$ and axial winding length $L_c = 635.2$ mm:

$$
\begin{aligned}
\varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\
&= 2\pi(50)(4\pi\times10^{-7})\frac{17576}{0.6352}\cdot\frac{1.4276}{18.966}\left(20 + \frac{40+50}{3}\right)\times10^{-3}\\
&= 0.0411\ \text{p.u.}
\end{aligned}
$$

#### Regulation at 0.8 pf lagging
$$
\begin{aligned}
\varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\
&= 0.00860(0.8) + 0.0411(0.600) = 0.0315\ \text{p.u.} = 3.15\%
\end{aligned}
$$

**Answer: $d$ = 324 mm; window 267 × 735 mm; frame 1305 × 1459 × 276 mm; $\varepsilon_r$ = 0.0086, $\varepsilon_x$ = 0.0411; regulation ≈ 3.15%.**

## Check yourself

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

1. For E_t = 18.97 V, B_m = 1.45 T, how is the core diameter found?
   A. phi_m = 18.97/(4.44 x 50) = 0.0855 Wb; A_i = 0.0589 m^2; d = sqrt(A_i/0.56) = 324 mm
   B. phi_m = 18.97 x 4.44 x 50; d = sqrt(A_i/0.45)
   C. A_i = 0.0589 m^2; d = A_i/0.56 = 105 mm
   D. phi_m = 18.97/(4.44 x 50); A_i = phi_m x B_m; d = sqrt(A_i/0.56)
   Answer: A. phi_m = E_t/(4.44 f) = 0.08547 Wb, A_i = phi_m/B_m = 0.05894 m^2, and the two-stepped core has A_i = 0.56 d^2, so d = 324.4 mm. (Key point: Core diameter for cruciform core)

2. With window height to width ratio 2.75, how is the window sized?
   A. W_w = sqrt(A_w/2.75) = 267 mm and H_w = 2.75 W_w = 735 mm, where A_w comes from the three-phase output equation
   B. W_w = A_w/2.75 and H_w = W_w
   C. W_w = 2.75 sqrt(A_w) and H_w = sqrt(A_w)
   D. W_w = sqrt(2.75 A_w) = 735 mm and H_w = 267 mm
   Answer: A. A_w = 0.19657 m^2 from Q = 3.33 f B_m delta K_w A_w A_i x 10^-3. Then H_w = 2.75 W_w gives W_w = sqrt(A_w/2.75) = 267.4 mm and H_w = 735.2 mm. (Key point: Window dimensions)

3. How is the overall frame size of this three-limb core found?
   A. H = H_w + H_y and W = D + a
   B. H = H_w + 2 H_y, W = 2D + a, depth = D_y = a
   C. H = 2 H_w + H_y, W = 3D, depth = d
   D. H = H_w + 2 H_y, W = 3d, depth = H_y
   Answer: B. There is a top and bottom yoke, so H = H_w + 2 H_y = 1305 mm. Three limbs with two window gaps give W = 2D + a = 1459 mm, and the depth equals the largest stamping width a. (Key point: Frame dimensions of three-limb core)

4. How are the turns calculated for this 66/11 kV delta/delta transformer?
   A. T_LV = 11000/18.97 = 580 and T_HV = 580 x 66000/11000 = 3480, using full line voltages since both are delta
   B. T_LV = 11000/(sqrt(3) x 18.97) and T_HV = T_LV/6
   C. T_HV = 580 and T_LV = 3480
   D. T_LV = 66000/18.97 = 3480 and T_HV = 580
   Answer: A. In delta the phase voltage equals the line voltage, so T_LV = 11000/18.97 = 580. The HV turns follow by the voltage ratio: 580 x 6 = 3480. (Key point: Turns in a delta/delta transformer)

## Related in Electric Machine Design

- [Design: 500 kVA 66/11 kV power transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-500kva-66-11kv) — Electric Machine Design, chapter 3, asked 6×
- [Design: 4000 kVA 66/11 kV power transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-4000kva-66-11kv) — Electric Machine Design, chapter 3, asked 5×
- [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×
- [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: 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 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×

## Related in other subjects

- [Design: 250 m head plant, 66 kV double circuit](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2071-250m-0-92mh) — Power Plant Design, chapter 4, asked 13×
- [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
- [Design: 280 m head plant, 66 kV](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2075-280m-66kv) — Power Plant Design, chapter 4, asked 12×
- [Delta capacitor per phase for 50 kW motor](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Acapacitor-delta-415v-50kw-0-72) — Basic Electrical Engineering, chapter 8