---
title: "Design: 100 kVA 2200/480 V single phase transformer"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:design-100kva-single-phase-ducts"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-100kva-single-phase-ducts
site: BCT Engineering
---

# Design: 100 kVA 2200/480 V single phase transformer

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2082 Baishakh, 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-single-phase-ducts
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

Design a 100 kVA, 2,200/480 V, 50 Hz, single phase, core type oil immersed natural cooled transformer. The required data for design are given below: Voltage per turn = 7.5; Maximum flux density in the core = 1.2 Wb/m²; Core type = Cruciform; Current density = 2.5 A/mm²; Window space factor = 0.28; Stacking factor = 0.9; Ratio of window height to width = 2. And, width of duct between LV and core, LV winding, HV winding and duct between HV and LV are 5 mm, 25 mm, 30 mm, 10 mm respectively. Assuming all other required parameters, calculate: (i) Overall dimension of core. (ii) Overall dimension of frame. (iii) Per unit resistance and leakage reactance drop. (iv) Per unit regulation at 0.8 power factor.

## Short answer

Limb diameter 224.2 mm, largest stamping 190.6 mm, window 150×300 mm. Frame dimensions: height 694 mm, width 565 mm, depth 191 mm.

## 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. In a single-phase core type transformer half of each winding is on each limb, so the ampere-turns per limb are $I_{HV}T_{HV}/2$. Assumed: yoke area 1.2 × core area (hot-rolled steel), $\rho = 0.021\ \Omega$-mm²/m, end clearance 20 mm at each end of the windings.

#### (i) Overall dimensions of core
Voltage per turn is given: $E_t = 7.50$ V.

$$
\begin{aligned}
\phi_m &= \frac{E_t}{4.44f} = \frac{7.500}{4.44\times50} = 0.03378\ \text{Wb}\\
A_i &= \frac{\phi_m}{B_m} = \frac{0.03378}{1.2} = 0.02815\ \text{m}^2 = 281.5\ \text{cm}^2
\end{aligned}
$$

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

$$
\begin{aligned}
d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.02815}{0.56}} = 0.2242\ \text{m} = 224.2\ \text{mm}\\
a &= 0.85d = 190.6\ \text{mm}\\
b &= 0.53d = 118.8\ \text{mm}
\end{aligned}
$$

Gross core area $A_{gi} = A_i/0.9 = 0.03128$ m².

Output equation (single phase): $Q = 2.22fB_m\delta K_wA_wA_i\times10^{-3}$

$$
\begin{aligned}
A_w &= \frac{Q}{2.22fB_m\delta K_wA_i\times10^{-3}}\\
&= \frac{100}{2.22\times50\times1.2\times2.5\times10^6\times0.28\times0.02815\times10^{-3}}\\
&= 0.03810\ \text{m}^2\\
W_w &= \sqrt{\frac{A_w}{2}} = 138.0\ \text{mm},\quad H_w = 2W_w = 276.0\ \text{mm}\\
D &= d + W_w = 362.2\ \text{mm (distance between adjacent limb centres)}
\end{aligned}
$$

**Winding build** (LV next to core):

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

| Item | Inside dia (mm) | Outside dia (mm) |
|---|---|---|
| LV winding | 234.2 | 284.2 |
| HV winding | 304.2 | 364.2 |

Radial depths: $b_1 = 25$ mm (LV), $b_2 = 30$ mm (HV), duct $a = 10$ mm. Axial length of coil, from the output-equation window height: $L_c = H_w - 2(20) = 236.0$ mm.

Mean lengths of turn:

$$
\begin{aligned}
L_{mt,LV} &= \pi\frac{234.2 + 284.2}{2} = 0.8144\ \text{m}\\
L_{mt,HV} &= \pi\frac{304.2 + 364.2}{2} = 1.0500\ \text{m}\\
L_{mt} &= \pi\frac{234.2 + 364.2}{2} = 0.9400\ \text{m (both windings, for reactance)}
\end{aligned}
$$

Check: the HV outside diameter is 364.2 mm but $D$ from the output equation is only 362.2 mm, so the coils of adjacent limbs would overlap. The window is widened so that there is 10 mm clearance between adjacent HV coils, and its height is increased to keep the given ratio $H_w/W_w = 2$. The coil height stays at $L_c$ = 236.0 mm.

$$
\begin{aligned}
D &= d_{o,HV} + 10 = 374.2\ \text{mm}\\
W_w &= D - d = 150.0\ \text{mm},\quad H_w = 2W_w = 300.0\ \text{mm}
\end{aligned}
$$

Yoke:

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

$$
\begin{aligned}
A_y &= 1.2A_i = 0.03378\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.03754\ \text{m}^2\\
D_y &= a = 190.6\ \text{mm}\\
H_y &= \frac{A_{gy}}{D_y} = \frac{0.03754}{0.1906} = 197.0\ \text{mm}
\end{aligned}
$$

#### (ii) Overall dimensions of frame
$$
\begin{aligned}
H &= H_w + 2H_y = 300.0 + 2(197.0) = 693.9\ \text{mm}\\
W &= D + a = 374.2 + 190.6 = 564.8\ \text{mm}\\
\text{Depth} &= D_y = a = 190.6\ \text{mm}
\end{aligned}
$$

```text
 +-----------------------+  ---
 |       top yoke        |   ^
 +-----+-----------+-----+   |
 |limb |  window   |limb |   H
 |     |  Ww x Hw  |     |   |
 +-----+-----------+-----+   |
 |      bottom yoke      |   v
 +-----------------------+  ---
 |<--------- W --------->|
    |<----- D ----->|
```

#### (iii) Per unit resistance and leakage reactance drop
Currents: $I_{LV} = \frac{Q}{V_{LV}} = 208.33$ A; $I_{HV} = 45.455$ A.

$$
\begin{aligned}
T_{LV} &= \frac{V_{LV}}{E_t} = \frac{480.0}{7.500} = 64.00 \approx 64\\
E_t\ (\text{revised}) &= \frac{480.0}{64} = 7.500\ \text{V}\\
T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 293.3 \approx 293\\
a_{LV} &= \frac{I_{LV}}{\delta} = 83.33\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 18.182\ \text{mm}^2
\end{aligned}
$$

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

$$
\begin{aligned}
r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times293\times1.0500}{18.182} = 0.35533\ \Omega\\
r_{LV} &= \frac{0.021\times64\times0.8144}{83.33} = 0.013134\ \Omega\\
R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 0.35533 + 0.013134\left(\frac{293}{64}\right)^2 = 0.6306\ \Omega\\
\varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{45.45\times0.6306}{2200.0} = 0.01303\ \text{p.u.}
\end{aligned}
$$

Leakage reactance (per unit), with $AT$ per limb $= 6659$ and axial winding length $L_c = 236.0$ mm:

$$
\begin{aligned}
\varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\
&= 2\pi(50)(4\pi\times10^{-7})\frac{6659}{0.2360}\cdot\frac{0.9400}{7.500}\left(10 + \frac{25+30}{3}\right)\times10^{-3}\\
&= 0.0396\ \text{p.u.}
\end{aligned}
$$

#### (iv) Per unit regulation at 0.8 pf lagging
$$
\begin{aligned}
\varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\
&= 0.01303(0.8) + 0.0396(0.600) = 0.0342\ \text{p.u.} = 3.42\%
\end{aligned}
$$

**Answer: $d$ = 224 mm, $A_i$ = 281.5 cm²; window 150 × 300 mm; frame 694 × 565 × 191 mm; $\varepsilon_r$ = 0.0130, $\varepsilon_x$ = 0.0396; regulation ≈ 3.42%.**

## Check yourself

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

1. How is the net core area A_i found from the given voltage per turn of 7.5 V at 50 Hz and B_m = 1.2 T?
   A. phi_m = Et/(4.44 f) = 0.03378 Wb, then A_i = phi_m/B_m = 0.02815 m^2
   B. A_i = Et x B_m = 9 m^2
   C. phi_m = Et/(2 pi f) then A_i = phi_m x B_m
   D. A_i = Q/(B_m x f) directly from the kVA rating
   Answer: A. EMF per turn Et = 4.44 f phi_m, so phi_m = 7.5/(4.44 x 50) = 0.03378 Wb. Dividing by B_m gives A_i = 0.02815 m^2. (Key point: Net core area from voltage per turn)

2. For a cruciform (two-stepped) core, how is the circumscribing circle diameter d obtained, and what is it here?
   A. d = sqrt(A_i/0.45) = 250 mm
   B. d = A_i/0.9 = 31 mm
   C. d = sqrt(A_i/0.56) = 224 mm, using A_i = 0.56 d^2
   D. d = sqrt(A_i/0.56) = 224 m
   Answer: C. A two-stepped core with stacking factor 0.9 has A_i = 0.56 d^2, so d = sqrt(0.02815/0.56) = 224.2 mm. The 0.45 ratio belongs to a square core. (Key point: Cruciform core diameter)

3. Why is the window width increased from 138 mm to 150 mm after the winding build is worked out?
   A. The yoke area has to be raised by 20%
   B. The HV outside diameter (364.2 mm) exceeds the limb spacing D (362.2 mm), so adjacent HV coils would overlap; D is set to 364.2 + 10 = 374.2 mm
   C. The window space factor of 0.28 is too low and must be raised
   D. The LV winding is wider than the window height
   Answer: B. The first D = d + Ww = 362.2 mm is smaller than the HV coil outer diameter, so the coils would clash. Allowing 10 mm clearance gives D = 374.2 mm and Ww = 150 mm. (Key point: Window widened for HV coil clearance)

4. With per unit resistance 0.01303 and reactance 0.0396, how is regulation at 0.8 pf lagging found?
   A. 0.01303 x 0.6 + 0.0396 x 0.8 = 3.95%
   B. 0.01303 + 0.0396 = 5.26%
   C. 0.01303 x 0.8 + 0.0396 x 0.6 = 3.42%
   D. 0.0396 x 0.8 - 0.01303 x 0.6 = 2.4%
   Answer: C. Regulation = er cos(phi) + ex sin(phi) for lagging load, with cos = 0.8 and sin = 0.6. This gives 0.0104 + 0.0238 = 0.0342, i.e. 3.42%. (Key point: Regulation from resistance and reactance)

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