---
title: "Design: 500 kVA single phase transformer with tubes"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:design-500kva-single-phase-tubes"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-500kva-single-phase-tubes
site: BCT Engineering
---

# Design: 500 kVA single phase transformer with tubes

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2075 Asoj, 20 marks
- Video: [Transformer Design solved example single phase|core|window|area|winding|specific magnetic loading](https://www.youtube.com/watch?v=JpbLOnyhKKM) — 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-500kva-single-phase-tubes
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

For a 500 kVA, 50 Hz, 6600/400 V, single phase core type, oil immersed, natural cooled power transformer, the design parameters are: Constant for output voltage per turn = 0.8; Resistivity of copper = 0.021 Ω-mm²/m; Maximum flux density in the core = 1.5 Wb/m²; Current density = 2.75 A/mm²; Core type = Cruciform; Window space factor = 0.27; Stacking factor = 0.9; Ratio of window height to width = 2.5; Ratio of yoke height to width = 1; Axial depth of LV winding = 402 mm; Axial depth of HV winding = 377.5 mm; Inside diameter of LV winding = 310 mm; Outer diameter of LV winding = 348 mm; Inside diameter of HV winding = 360 mm; Outside diameter of HV winding = 418 mm. Calculate: i) Dimension of the core, window and yoke ii) Overall dimension of the frame iii) Per unit regulation at 0.8 pf lagging iv) Taking iron loss = 1460 W, copper loss = 3865 W at full load, height of tank = 1.6 m, length of tank = 1.05 m, width of tank = 0.62 m, find the temperature rise. If the mean temperature rise of oil is not to rise 35°C, find the necessary number of tubes and also show its arrangement.

## Short answer

Core diameter 309.7 mm, window 173.5 × 433.8 mm, yoke 263.3 mm square. Overall frame 960.4 × 746.5 × 263.3 mm. Plain tank temperature rise 79.7 °C; 42 tubes of 50 mm diameter, 1.5 m long keep mean oil rise within 35 °C (14 per long side, 7 per short side).

## Answer

Standard method (A.K. Sawhney). Half of each winding is on each limb, so ampere-turns per limb are $I_{HV}T_{HV}/2$. Yoke: $H_y/D_y = 1$ with depth equal to the largest stamping width. Axial length for reactance = mean of 402 and 377.5 mm = 389.75 mm.

#### i) Dimensions of core, window and yoke
Voltage per turn, $E_t = K\sqrt{Q}$ ($Q$ in kVA):

$$
\begin{aligned}
E_t &= 0.8\sqrt{500} = 17.889\ \text{V}\\
\phi_m &= \frac{E_t}{4.44f} = \frac{17.889}{4.44\times50} = 0.08058\ \text{Wb}\\
A_i &= \frac{\phi_m}{B_m} = \frac{0.08058}{1.5} = 0.05372\ \text{m}^2 = 537.2\ \text{cm}^2
\end{aligned}
$$

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

$$
\begin{aligned}
d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.05372}{0.56}} = 0.3097\ \text{m} = 309.7\ \text{mm}\\
a &= 0.85d = 263.3\ \text{mm}\\
b &= 0.53d = 164.2\ \text{mm}
\end{aligned}
$$

Gross core area $A_{gi} = A_i/0.9 = 0.05969$ m².

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

$$
\begin{aligned}
A_w &= \frac{Q}{2.22fB_m\delta K_wA_i\times10^{-3}}\\
&= \frac{500}{2.22\times50\times1.5\times2.75\times10^6\times0.27\times0.05372\times10^{-3}}\\
&= 0.07529\ \text{m}^2\\
W_w &= \sqrt{\frac{A_w}{2.5}} = 173.5\ \text{mm},\quad H_w = 2.5W_w = 433.8\ \text{mm}\\
D &= d + W_w = 483.3\ \text{mm (distance between adjacent limb centres)}
\end{aligned}
$$

Check: HV outside diameter 418 mm < $D$ = 483 mm and coil height 402 mm < $H_w$ = 434 mm, so the windings fit. The LV inside diameter (310 mm) only just clears the core circle (309.7 mm), so the LV is wound directly on a thin insulating cylinder over the core.

**Yoke**

Yoke height equal to its depth ($H_y/D_y = 1$), with depth equal to the largest stamping width:

$$
\begin{aligned}
D_y &= H_y = a = 263.3\ \text{mm}\\
A_{gy} &= a^2 = 0.06931\ \text{m}^2,\quad A_y = 0.9A_{gy} = 0.06238\ \text{m}^2
\end{aligned}
$$

#### ii) Overall dimensions of frame
$$
\begin{aligned}
H &= H_w + 2H_y = 433.8 + 2(263.3) = 960.4\ \text{mm}\\
W &= D + a = 483.3 + 263.3 = 746.5\ \text{mm}\\
\text{Depth} &= D_y = a = 263.3\ \text{mm}
\end{aligned}
$$

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

#### iii) Per unit regulation at 0.8 pf lagging
Currents: $I_{LV} = \frac{Q}{V_{LV}} = 1250.00$ A; $I_{HV} = 75.758$ A.

$$
\begin{aligned}
T_{LV} &= \frac{V_{LV}}{E_t} = \frac{400.0}{17.889} = 22.36 \approx 22\\
E_t\ (\text{revised}) &= \frac{400.0}{22} = 18.182\ \text{V}\\
T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 363.0 \approx 363\\
a_{LV} &= \frac{I_{LV}}{\delta} = 454.55\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 27.548\ \text{mm}^2
\end{aligned}
$$

Radial depths: $b_1 = 19$ mm (LV), $b_2 = 29$ mm (HV), duct $a = 6$ mm. $L_c$ = 389.75 mm.

Mean lengths of turn:

$$
\begin{aligned}
L_{mt,LV} &= \pi\frac{310.0 + 348.0}{2} = 1.0336\ \text{m}\\
L_{mt,HV} &= \pi\frac{360.0 + 418.0}{2} = 1.2221\ \text{m}\\
L_{mt} &= \pi\frac{310.0 + 418.0}{2} = 1.1435\ \text{m (both windings, for reactance)}
\end{aligned}
$$

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

$$
\begin{aligned}
r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times363\times1.2221}{27.548} = 0.33817\ \Omega\\
r_{LV} &= \frac{0.021\times22\times1.0336}{454.55} = 0.0010505\ \Omega\\
R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 0.33817 + 0.0010505\left(\frac{363}{22}\right)^2 = 0.62418\ \Omega\\
\varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{75.76\times0.62418}{6600.0} = 0.00716\ \text{p.u.}
\end{aligned}
$$

Leakage reactance (per unit), with $AT$ per limb $= 13750$ and axial winding length $L_c = 389.8$ mm:

$$
\begin{aligned}
\varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\
&= 2\pi(50)(4\pi\times10^{-7})\frac{13750}{0.3897}\cdot\frac{1.1435}{18.182}\left(6 + \frac{19+29}{3}\right)\times10^{-3}\\
&= 0.0193\ \text{p.u.}
\end{aligned}
$$

$$
\begin{aligned}
\varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\
&= 0.00716(0.8) + 0.0193(0.600) = 0.0173\ \text{p.u.} = 1.73\%
\end{aligned}
$$

#### iv) Temperature rise and cooling tubes
Total loss $P = 1460 + 3865 = 5325$ W. Tubes assumed 50 mm in diameter and 1.5 m long (tank height 1.6 m).

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

$$
\begin{aligned}
S_t &= 2(W_t + L_t)H_t = 5.3440\ \text{m}^2\\
\theta_{plain} &= \frac{5325}{12.5S_t} = 79.72\ ^\circ\text{C}
\end{aligned}
$$

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

$$
\begin{aligned}
A_{tubes} &= \frac{1}{8.8}\left(\frac{P}{\theta} - 12.5S_t\right) = \frac{1}{8.8}\left(\frac{5325}{35} - 12.5\times5.3440\right) = 9.698\ \text{m}^2\\
\text{Area of one tube} &= \pi d_tl_t = \pi\times0.05\times1.5 = 0.2356\ \text{m}^2\\
n_t &= \frac{9.698}{0.2356} = 41.16 \approx 42\ \text{tubes}
\end{aligned}
$$

**Arrangement of tubes:** with 50 mm tubes at 75 mm centre-to-centre spacing, one row round the tank can hold about $1050/75 = 14$ tubes on each long side and $620/75 \approx 8$ on each short side. So 42 tubes are placed in a single row: 14 on each long side and 7 on each short side ($2\times14 + 2\times7 = 42$).

```text
        o o o o o o o o o o o o o o    (14)
      +-------------------------------+
   o  |                               |  o
   o  |                               |  o
   o  |        TANK (top view)        |  o
(7)o  |      1.05 m  x  0.62 m        |  o (7)
   o  |                               |  o
   o  |                               |  o
   o  |                               |  o
      +-------------------------------+
        o o o o o o o o o o o o o o    (14)
   o = cooling tube, 50 mm dia, 75 mm pitch
```

**Answer: $d$ = 310 mm; window 174 × 434 mm; yoke 263 × 263 mm; frame 960 × 747 × 263 mm; regulation ≈ 1.73%; plain tank rise ≈ 79.7 °C; 42 tubes needed.**

## Check yourself

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

1. Which output equation is used for this single-phase transformer?
   A. Q = 3.33 f B_m delta K_w A_w A_i x 10^-3
   B. Q = 2.22 f B_m delta K_w A_w A_i x 10^-3
   C. Q = 4.44 f B_m A_i x 10^-3
   D. Q = 1.11 f B_m delta A_w A_i x 10^-3
   Answer: B. A single-phase core-type transformer uses the constant 2.22 (the 3.33 applies to three-phase), giving A_w = 0.0753 m^2. (Key point: Single-phase output equation)

2. With H_y/D_y = 1, D = 483.3 mm and a = 263.3 mm, how is the yoke and frame found?
   A. D_y = H_y = a = 263.3 mm, H = H_w + 2H_y, W = D + a
   B. D_y = H_y = d, W = 2D + a
   C. H_y = 0.5 a, W = D + a
   D. D_y = b, W = 2D
   Answer: A. Yoke depth is the largest stamping width a and H_y = D_y. A single-phase core has two limbs, so W = D + a = 746.5 mm and H = 433.8 + 526.6 = 960.4 mm. (Key point: Yoke and frame for single phase)

3. Regulation is found from which combination at 0.8 pf lagging?
   A. epsilon = epsilon_r sin(phi) + epsilon_x cos(phi)
   B. epsilon = epsilon_r cos(phi) + epsilon_x sin(phi) = 1.73%
   C. epsilon = epsilon_r + epsilon_x = 2.6%
   D. epsilon = epsilon_x cos(phi) - epsilon_r sin(phi)
   Answer: B. epsilon = 0.00716 x 0.8 + 0.0193 x 0.6 = 0.0173 p.u., i.e. 1.73%. (Key point: Ampere-turns and regulation)

4. With P = 5325 W, theta = 35 C, tank wall area S_t = 5.344 m^2, how is the tube area found?
   A. A_t = (1/8.8)(P/theta - 12.5 S_t) = 9.7 m^2, so about 42 tubes
   B. A_t = P/(theta x 12.5) = 12.2 m^2, so 52 tubes
   C. A_t = (1/12.5)(P/theta - 8.8 S_t) = 6.9 m^2, so 30 tubes
   D. A_t = P/(8.8 S_t) = 113 m^2
   Answer: A. The tank walls dissipate 12.5 W/m2 C and tubes 8.8 W/m2 C. The remaining loss 152.1 - 66.8 W/C is divided by 8.8, giving 9.698 m^2, i.e. 41.16, rounded up to 42 tubes of 0.2356 m^2. (Key point: Tube area and number)

## Close variants set in other exams

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