---
title: "Design: 150 kVA 11000/420 V distribution transformer"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:design-150kva-11000-420"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-150kva-11000-420
site: BCT Engineering
---

# Design: 150 kVA 11000/420 V distribution transformer

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2071 Shrawan, 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-150kva-11000-420
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

Calculate (i) Overall dimension of core (ii) Overall dimension of frame (iii) Per unit resistance and leakage reactance drop and (iv) Per unit voltage regulation at 0.85 power factor, for a 150 kVA, 11000/420 V, 50 Hz, 3-phase, Δ/Y core type, oil immersed natural cooled distribution transformer from the following data: Maximum flux density = 1.35 Wb/m², constant for output voltage per turn = 0.45, current density in conductors = 2.5 A/mm², core type = cruciform, window space factor = 0.25, stacking factor = 0.9, ratio of height of window to width = 2.5, width of LV winding, HV winding and duct in between = 20 mm, 25 mm, 15 mm respectively.

## Short answer

Core: cruciform, d=181.2 mm, window 152.4×381 mm, limb centres 333.6 mm, yoke depth 154.0 mm, height 159.2 mm. Frame overall: 821.3 mm wide, 699.4 mm high, 154 mm deep. Per-unit resistance, leakage reactance drop, and regulation were not completed in the provided solution.

## 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: yoke area 1.2 × core area (hot-rolled steel), core-to-LV clearance 5 mm, end clearance 20 mm, $\rho = 0.021\ \Omega$-mm²/m.

#### (i) Overall dimensions of core
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): $Q = 3.33fB_m\delta K_wA_wA_i\times10^{-3}$

$$
\begin{aligned}
A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\
&= \frac{150}{3.33\times50\times1.35\times2.5\times10^6\times0.25\times0.01839\times10^{-3}}\\
&= 0.05806\ \text{m}^2\\
W_w &= \sqrt{\frac{A_w}{2.5}} = 152.4\ \text{mm},\quad H_w = 2.5W_w = 381.0\ \text{mm}\\
D &= d + W_w = 333.6\ \text{mm (distance between adjacent limb centres)}
\end{aligned}
$$

Yoke:

Yoke area is 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
$$
\begin{aligned}
H &= H_w + 2H_y = 381.0 + 2(159.2) = 699.4\ \text{mm}\\
W &= 2D + a = 2(333.6) + 154.0 = 821.3\ \text{mm}\\
\text{Depth} &= D_y = a = 154.0\ \text{mm}
\end{aligned}
$$

```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
Phase voltages: LV (star) $V_{LV} = 242.5$ V; HV (delta) $V_{HV} = 11000.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 206.20$ A; $I_{HV} = 4.545$ A.

$$
\begin{aligned}
T_{LV} &= \frac{V_{LV}}{E_t} = \frac{242.5}{5.511} = 44.00 \approx 44\\
E_t\ (\text{revised}) &= \frac{242.5}{44} = 5.511\ \text{V}\\
T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 1996.0 \approx 1996\\
a_{LV} &= \frac{I_{LV}}{\delta} = 82.48\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 1.818\ \text{mm}^2
\end{aligned}
$$

Radial build (assumed clearance between core and LV = 5 mm). Each radial gap or depth appears on both sides of the diameter, so it is added twice:

- LV inside diameter $= 181.2 + 2(5) = 191.2$ mm
- LV outside diameter $= 191.2 + 2(20) = 231.2$ mm
- HV inside diameter $= 231.2 + 2(15) = 261.2$ mm
- HV outside diameter $= 261.2 + 2(25) = 311.2$ 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 $L_c = H_w - 2(20) = 341.0$ 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 $D$ = 334 mm. Adjacent HV coils have about 22 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):

$$
\begin{aligned}
r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times1996\times0.8992}{1.818} = 20.729\ \Omega\\
r_{LV} &= \frac{0.021\times44\times0.6635}{82.48} = 0.0074337\ \Omega\\
R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 20.729 + 0.0074337\left(\frac{1996}{44}\right)^2 = 36.027\ \Omega\\
\varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{4.545\times36.027}{11000.0} = 0.01489\ \text{p.u.}
\end{aligned}
$$

Leakage reactance (per unit), with $AT$ per limb $= 9073$ and axial winding length $L_c = 341.0$ mm:

$$
\begin{aligned}
\varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\
&= 2\pi(50)(4\pi\times10^{-7})\frac{9073}{0.3410}\cdot\frac{0.7892}{5.511}\left(15 + \frac{20+25}{3}\right)\times10^{-3}\\
&= 0.0451\ \text{p.u.}
\end{aligned}
$$

#### (iv) Per unit regulation at 0.85 pf lagging
$$
\begin{aligned}
\varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\
&= 0.01489(0.85) + 0.0451(0.527) = 0.0364\ \text{p.u.} = 3.64\%
\end{aligned}
$$

**Answer: $d$ = 181 mm, window 152 × 381 mm; frame 699 × 821 × 154 mm; $\varepsilon_r$ = 0.0149, $\varepsilon_x$ = 0.0451; regulation ≈ 3.64%.**

## Check yourself

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

1. For this 3-phase transformer with K = 0.45, which is correct?
   A. Et = 0.45*sqrt(50) with Q per phase and the 2.22 constant
   B. Et = 0.45*sqrt(150) = 5.51 V with total Q, and the 3.33 constant in the output equation
   C. Et = 0.45*150 = 67.5 V with the 3.33 constant
   D. Et = 0.45*sqrt(150) = 5.51 V with the 2.22 constant
   Answer: B. Q is the total three-phase kVA; the three-phase output equation uses 3.33 f Bm delta Kw Aw Ai. (Key point: Three-phase voltage per turn and output equation)

2. For a three-limb, 3-phase core, what is the overall frame width?
   A. W = D + a
   B. W = 3D
   C. W = D + 2a
   D. W = 2D + a
   Answer: D. Three limbs are separated by two limb-centre distances D, plus the width a of the outer limb, so W = 2D + a = 821 mm. (Key point: Three-limb frame width)

3. With a delta HV of 11000 V and star LV of 420 V, TLV = 44. What is THV?
   A. 44*11000/420 = 1152
   B. 44*11000/242.5 = 1996, using the star phase voltage 242.5 V
   C. 44*11000/(420*1.732) = 665
   D. 44*242.5/11000 = 1
   Answer: B. The turns ratio uses phase voltages: LV star phase voltage is 420/sqrt3 = 242.5 V and HV delta phase voltage is 11000 V, giving 1996 turns. (Key point: Phase voltages and turns)

4. How is the per-unit regulation at 0.85 lagging pf obtained?
   A. eps = er*sin(phi) + ex*cos(phi)
   B. eps = er + ex
   C. eps = er*cos(phi) + ex*sin(phi) = 0.0149*0.85 + 0.0451*0.527 = 3.64%
   D. eps = ex/er = 3.03
   Answer: C. Regulation is the resistive drop times cos(phi) plus the reactive drop times sin(phi), giving about 3.64%. (Key point: Voltage regulation)

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