---
title: "Design: 25 kVA transformer frame, regulation, tubes"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:design-25kva-cooling-tubes"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-25kva-cooling-tubes
site: BCT Engineering
---

# Design: 25 kVA transformer frame, regulation, tubes

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2076 Asoj, 6+6+4 marks; 2070 Chaitra, 6+6+4 marks (asked 2 times)
- 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-25kva-cooling-tubes
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

For the design of a 25 kVA, 50 Hz, 11/0.433 kV, delta/star, 3-phase, core type, oil immersed, naturally cooled distribution transformer, the mean temperature of oil is not to exceed 35°C. The following parameters are chosen: Bm = 1.0 Wb/m², δ = 2.3 A/mm², constant for volt per turn = 0.45, type of core - cruciform, Kw = 0.18, Hw/Ww = 2.5, total loss at full load = 1.2 kW.
Winding dimensions:

| | Inside diameter (mm) | Outside diameter (mm) | Conductor area (mm²) |
|---|---|---|---|
| LV | 138 | 156.2 | 14.9 |
| HV | 186.2 | 239 | 0.312 |

Lc = 253 mm, ρ = 0.021 Ω-mm²/m. Take dimension: Ht = 950 mm, Wt = 840 mm, Lt = 350 mm.
i) Calculate overall dimensions of frame. ii) Calculate per unit regulation at full load and 0.8 pf (lag). iii) Calculate the minimum number of tubes of diameter 50 mm with average length of 1.35 m required for maintaining the mean temperature within the permissible limit. The rate of heat dissipation from plain wall is 6.5 and 6 W/m²-°C for convection and radiation respectively. The provision of tube improves the rate of heat dissipation by 35%.

## Short answer

Frame overall dimensions: H = 535.5 mm, W = 622.7 mm, depth = 114.4 mm. Per-unit voltage regulation at 0.8 pf lag is 0.0442 (4.42%). Minimum number of 50 mm diameter, 1.35 m long tubes required to maintain oil temperature within limit is 2.

## Answer

Design follows the standard method (A.K. Sawhney): $E_t = K\sqrt{Q}$, output equation for $A_w$, yoke 20% larger than core (hot-rolled steel assumed), $\rho = 0.021\ \Omega$-mm²/m.

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

$$
\begin{aligned}
E_t &= 0.45\sqrt{25} = 2.250\ \text{V}\\
\phi_m &= \frac{E_t}{4.44f} = \frac{2.250}{4.44\times50} = 0.01014\ \text{Wb}\\
A_i &= \frac{\phi_m}{B_m} = \frac{0.01014}{1} = 0.01014\ \text{m}^2 = 101.4\ \text{cm}^2
\end{aligned}
$$

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

$$
\begin{aligned}
d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.01014}{0.56}} = 0.1345\ \text{m} = 134.5\ \text{mm}\\
a &= 0.85d = 114.4\ \text{mm}\\
b &= 0.53d = 71.3\ \text{mm}
\end{aligned}
$$

Gross core area $A_{gi} = A_i/0.9 = 0.01126$ m².

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

$$
\begin{aligned}
A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\
&= \frac{25}{3.33\times50\times1\times2.3\times10^6\times0.18\times0.01014\times10^{-3}}\\
&= 0.03578\ \text{m}^2\\
W_w &= \sqrt{\frac{A_w}{2.5}} = 119.6\ \text{mm},\quad H_w = 2.5W_w = 299.1\ \text{mm}\\
D &= d + W_w = 254.2\ \text{mm (distance between adjacent limb centres)}
\end{aligned}
$$

Check: HV outside diameter (239 mm) < $D$ (254.2 mm), so adjacent HV coils have about 15 mm clearance. LV inside diameter 138 mm > $d$, so the windings fit.

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

$$
\begin{aligned}
A_y &= 1.2A_i = 0.01216\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.01351\ \text{m}^2\\
D_y &= a = 114.4\ \text{mm}\\
H_y &= \frac{A_{gy}}{D_y} = \frac{0.01351}{0.1144} = 118.2\ \text{mm}
\end{aligned}
$$

#### i) Overall dimensions of frame
$$
\begin{aligned}
H &= H_w + 2H_y = 299.1 + 2(118.2) = 535.5\ \text{mm}\\
W &= 2D + a = 2(254.2) + 114.4 = 622.7\ \text{mm}\\
\text{Depth} &= D_y = a = 114.4\ \text{mm}
\end{aligned}
$$

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

#### ii) Per unit regulation at full load, 0.8 pf lagging
**Turns and currents**

Phase voltages: LV (star) $V_{LV} = 250.0$ V; HV (delta) $V_{HV} = 11000.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 33.33$ A; $I_{HV} = 0.758$ A.

$$
\begin{aligned}
T_{LV} &= \frac{V_{LV}}{E_t} = \frac{250.0}{2.250} = 111.11 \approx 111\\
E_t\ (\text{revised}) &= \frac{250.0}{111} = 2.252\ \text{V}\\
T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 4884.1 \approx 4884\\
a_{LV} &= \frac{I_{LV}}{\delta} = 14.49\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 0.329\ \text{mm}^2
\end{aligned}
$$

(The given conductor areas, 14.9 mm² and 0.312 mm², match these values.)

**Resistance**

Radial depths: $b_1 = 9.1$ mm (LV), $b_2 = 26.4$ mm (HV), duct $a = 15$ mm. Axial length $L_c = 253$ mm (given).

Mean lengths of turn:

$$
\begin{aligned}
L_{mt,LV} &= \pi\frac{138.0 + 156.2}{2} = 0.4621\ \text{m}\\
L_{mt,HV} &= \pi\frac{186.2 + 239.0}{2} = 0.6679\ \text{m}\\
L_{mt} &= \pi\frac{138.0 + 239.0}{2} = 0.5922\ \text{m (both windings, for reactance)}
\end{aligned}
$$

Using the given conductor areas:

$$
\begin{aligned}
r_{HV} &= \frac{0.021\times4884\times0.6679}{0.312} = 219.6\ \Omega\\
r_{LV} &= \frac{0.021\times111\times0.4621}{14.9} = 0.0723\ \Omega\\
R_p &= 219.6 + 0.0723\left(\frac{4884}{111}\right)^2 = 359.5\ \Omega\\
\varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{0.7576\times359.5}{11000} = 0.0248
\end{aligned}
$$

**Reactance**

Leakage reactance (per unit), with $AT$ per limb $= 3700$ and axial winding length $L_c = 253.0$ mm:

$$
\begin{aligned}
\varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\
&= 2\pi(50)(4\pi\times10^{-7})\frac{3700}{0.2530}\cdot\frac{0.5922}{2.252}\left(15 + \frac{9.1+26.4}{3}\right)\times10^{-3}\\
&= 0.0407\ \text{p.u.}
\end{aligned}
$$

**Regulation**

$$
\begin{aligned}
\varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi = 0.0248(0.8) + 0.0407(0.6)\\
&= 0.0442\ \text{p.u.} = 4.42\%
\end{aligned}
$$

#### iii) Number of cooling tubes
Total loss $P = 1200$ W, allowed mean oil rise $\theta = 35\ ^\circ$C, tube diameter 50 mm, length 1.35 m.

Tank walls (top and bottom neglected), dissipation 12.5 W/m²°C (6 radiation + 6.5 convection):

$$
\begin{aligned}
S_t &= 2(W_t + L_t)H_t = 2.2610\ \text{m}^2\\
\theta_{plain} &= \frac{1200}{12.5S_t} = 42.46\ ^\circ\text{C}
\end{aligned}
$$

This exceeds 35 °C, so tubes are needed. Tubes improve convection by 35%: $6.5\times1.35 \approx 8.8$ W/m²°C.

$$
\begin{aligned}
A_{tubes} &= \frac{1}{8.8}\left(\frac{P}{\theta} - 12.5S_t\right) = \frac{1}{8.8}\left(\frac{1200}{35} - 12.5\times2.2610\right) = 0.684\ \text{m}^2\\
\text{Area of one tube} &= \pi d_tl_t = \pi\times0.05\times1.35 = 0.2121\ \text{m}^2\\
n_t &= \frac{0.684}{0.2121} = 3.23 \approx 4\ \text{tubes}
\end{aligned}
$$

**Answer: frame 535 mm high × 623 mm wide × 114 mm deep; regulation ≈ 4.42%; minimum 4 tubes (50 mm dia, 1.35 m).**

## Check yourself

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

1. For the 25 kVA transformer, Et = K sqrt(Q) with K = 0.45 gives:
   A. 1.125 V
   B. 2.25 V
   C. 4.5 V
   D. 11.25 V
   Answer: B. 0.45 x sqrt(25) = 0.45 x 5 = 2.25 V. (Key point: Voltage per turn)

2. The overall frame height is given by:
   A. Hw + Hy
   B. Hw + 2Hy
   C. 2Hw + Hy
   D. Hw + 2D
   Answer: B. Two yokes, top and bottom, add to window height: 299.1 + 2(118.2) = 535.5 mm. (Key point: Frame height)

3. Per unit regulation at 0.8 pf lagging is calculated as:
   A. er cos(phi) - ex sin(phi)
   B. er sin(phi) + ex cos(phi)
   C. er cos(phi) + ex sin(phi)
   D. er + ex
   Answer: C. 0.0248(0.8) + 0.0407(0.6) = 4.42%. (Key point: Regulation formula)

4. Why are tubes needed here, and how many?
   A. Plain tank rise is 42.46 C which exceeds 35 C; 4 tubes
   B. Plain tank rise is 20 C; 2 tubes
   C. Plain tank rise is 35 C; none
   D. Plain tank rise is 42.46 C; 10 tubes
   Answer: A. The plain-wall rise exceeds the limit; tube area 0.684 m2 / 0.2121 m2 = 3.23, rounded up to 4. (Key point: Tube requirement)

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