---
title: "Design: 150 kVA 6600/400 V distribution transformer"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:design-150kva-6600-400-regulation"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-150kva-6600-400-regulation
site: BCT Engineering
---

# Design: 150 kVA 6600/400 V distribution transformer

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2079 Bhadra, 18 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-150kva-6600-400-regulation
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

The design parameters for a 150 kVA, 50 Hz, 6600/400 V, 3-phase, delta/star, core type, oil immersed natural cooled distribution transformer are given below. Max. flux density in core = 1.35 Wb/m²; Current density in conductor = 2.75 A/mm²; Constant for output voltage per turn = 0.45; Core type = cruciform two stepped; Window space factor = 0.27; Stacking factor = 0.9; Ratio of window height to width = 2.5; Take hot rolled steel and area of yoke is 20% greater than gross area of core; Width of LV, HV winding and duct between them are 20 mm, 25 mm and 15 mm respectively. Assuming all other required parameters, calculate: (i) Dimension of the core, window and yoke (ii) Overall dimension of the frame (iii) Per unit resistance and leakage reactance drop (iv) Per unit voltage regulation at 0.85 pf

## Short answer

Core d=181.2 mm, cruciform (a=154.0, b=96.0 mm); window 139.8×349.5 mm; yoke depth 154.0 mm height 159.2 mm; frame H=667.9 mm, W=796.1 mm, depth 154.0 mm. Per-unit R and X drops partially computed; regulation not finalized.

## 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):

$$
\begin{aligned}
E_t &= 0.45\sqrt{150} = 5.511\ \text{V}\\
\phi_m &= \frac{E_t}{4.44f} = \frac{5.511}{4.44\times50} = 0.02483\ \text{Wb}\\
A_i &= \frac{\phi_m}{B_m} = \frac{0.02483}{1.35} = 0.01839\ \text{m}^2 = 183.9\ \text{cm}^2
\end{aligned}
$$

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

$$
\begin{aligned}
d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.01839}{0.56}} = 0.1812\ \text{m} = 181.2\ \text{mm}\\
a &= 0.85d = 154.0\ \text{mm}\\
b &= 0.53d = 96.0\ \text{mm}
\end{aligned}
$$

Gross core area $A_{gi} = A_i/0.9 = 0.02043$ m².

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

$$
\begin{aligned}
A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\
&= \frac{150}{3.33\times50\times1.35\times2.75\times10^6\times0.27\times0.01839\times10^{-3}}\\
&= 0.04887\ \text{m}^2\\
W_w &= \sqrt{\frac{A_w}{2.5}} = 139.8\ \text{mm},\quad H_w = 2.5W_w = 349.5\ \text{mm}\\
D &= d + W_w = 321.0\ \text{mm (distance between adjacent limb centres)}
\end{aligned}
$$

Yoke:

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

$$
\begin{aligned}
A_y &= 1.2A_i = 0.02207\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.02452\ \text{m}^2\\
D_y &= a = 154.0\ \text{mm}\\
H_y &= \frac{A_{gy}}{D_y} = \frac{0.02452}{0.1540} = 159.2\ \text{mm}
\end{aligned}
$$

#### (ii) Overall dimensions of frame
$$
\begin{aligned}
H &= H_w + 2H_y = 349.5 + 2(159.2) = 667.9\ \text{mm}\\
W &= 2D + a = 2(321.0) + 154.0 = 796.1\ \text{mm}\\
\text{Depth} &= D_y = a = 154.0\ \text{mm}
\end{aligned}
$$

```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**

Phase voltages: LV (star) $V_{LV} = 230.9$ V; HV (delta) $V_{HV} = 6600.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 216.51$ A; $I_{HV} = 7.576$ A.

$$
\begin{aligned}
T_{LV} &= \frac{V_{LV}}{E_t} = \frac{230.9}{5.511} = 41.90 \approx 42\\
E_t\ (\text{revised}) &= \frac{230.9}{42} = 5.499\ \text{V}\\
T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 1200.3 \approx 1200\\
a_{LV} &= \frac{I_{LV}}{\delta} = 78.73\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 2.755\ \text{mm}^2
\end{aligned}
$$

**Winding build**

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

| Item | Inside dia (mm) | Outside dia (mm) |
|---|---|---|
| LV winding | 191.2 | 231.2 |
| HV winding | 261.2 | 311.2 |

Radial depths: $b_1 = 20$ mm (LV), $b_2 = 25$ mm (HV), duct $a = 15$ mm. Axial length $L_c = H_w - 2(20) = 309.5$ mm.

Mean lengths of turn:

$$
\begin{aligned}
L_{mt,LV} &= \pi\frac{191.2 + 231.2}{2} = 0.6635\ \text{m}\\
L_{mt,HV} &= \pi\frac{261.2 + 311.2}{2} = 0.8992\ \text{m}\\
L_{mt} &= \pi\frac{191.2 + 311.2}{2} = 0.7892\ \text{m (both windings, for reactance)}
\end{aligned}
$$

Check: HV outside diameter 311 mm vs $D$ = 321 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):

$$
\begin{aligned}
r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times1200\times0.8992}{2.755} = 8.2252\ \Omega\\
r_{LV} &= \frac{0.021\times42\times0.6635}{78.73} = 0.0074337\ \Omega\\
R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 8.2252 + 0.0074337\left(\frac{1200}{42}\right)^2 = 14.294\ \Omega\\
\varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{7.576\times14.294}{6600.0} = 0.01641\ \text{p.u.}
\end{aligned}
$$

Leakage reactance (per unit), with $AT$ per limb $= 9091$ and axial winding length $L_c = 309.5$ mm:

$$
\begin{aligned}
\varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\
&= 2\pi(50)(4\pi\times10^{-7})\frac{9091}{0.3095}\cdot\frac{0.7892}{5.499}\left(15 + \frac{20+25}{3}\right)\times10^{-3}\\
&= 0.0499\ \text{p.u.}
\end{aligned}
$$

#### (iv) Per unit regulation at 0.85 pf lagging
$$
\begin{aligned}
\varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\
&= 0.01641(0.85) + 0.0499(0.527) = 0.0402\ \text{p.u.} = 4.02\%
\end{aligned}
$$

**Answer: $d$ = 181 mm, window 140 × 350 mm, yoke 159 × 154 mm; frame 668 × 796 × 154 mm; $\varepsilon_r$ = 0.0164, $\varepsilon_x$ = 0.0499; regulation ≈ 4.02%.**

## Check yourself

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

1. For this cruciform two-stepped core, what is d and how is it found?
   A. d = sqrt(0.01839/0.45) = 202 mm
   B. d = sqrt(0.01839/0.56) = 181 mm, with a = 0.85 d and b = 0.53 d
   C. d = 0.01839/0.56 = 33 mm
   D. d = sqrt(0.01839/0.56) = 181 mm, with a = 0.53 d and b = 0.85 d
   Answer: B. E_t = 5.511 V gives A_i = 0.01839 m^2. A_i = 0.56 d^2 gives d = 181.2 mm, with the largest stamping a = 0.85 d = 154 mm and b = 0.53 d = 96 mm. (Key point: Core diameter and steps)

2. How is the yoke height H_y found?
   A. Yoke area = 1.2 A_i; gross area = A_y/0.9 = 0.02452 m^2; H_y = A_gy/D_y with D_y = a = 154 mm, giving 159 mm
   B. H_y = window width = 140 mm
   C. Yoke area = A_i; H_y = A_i/d
   D. H_y = A_gy/d with d = 181 mm, giving 135 mm
   Answer: A. Yoke area is 20% more than core area and divided by stacking factor 0.9 to get gross area. The yoke depth equals the widest stamping a, so H_y = 0.02452/0.1540 = 159.2 mm. (Key point: Yoke sizing)

3. Which winding phase voltages are used for turns and currents?
   A. LV 400 V and HV 6600/sqrt(3)
   B. LV 400/sqrt(3) = 230.9 V (star) and HV 6600 V (delta)
   C. Both use line voltage, 400 V and 6600 V
   D. LV 400 V (star) and HV 6600/sqrt(3) (delta)
   Answer: B. A star winding has phase voltage = line/sqrt(3) = 230.9 V, while the delta winding has phase voltage equal to the line voltage 6600 V. So T_LV = 230.9/5.511 = 42. (Key point: Phase voltages in delta/star)

4. How does the 4.02% regulation at 0.85 pf arise?
   A. er sin(phi) + ex cos(phi) = 0.0430
   B. er cos(phi) + ex sin(phi) = 0.01641(0.85) + 0.0499(0.527) = 0.0402
   C. er + ex = 0.0663
   D. ex cos(phi) = 0.0424
   Answer: B. Regulation = er cos(phi) + ex sin(phi) using cos = 0.85 and sin = 0.527. This gives 0.01395 + 0.0263 = 0.0402, i.e. 4.02%. (Key point: Total regulation)

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