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

# Temperature rise and tubes for 500 kVA transformer

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2082 Chaitra (new course), 16 marks
- Video: [DEA | Design of Cooling Tubes in Transormer](https://www.youtube.com/watch?v=AUybfuY4-TM) — FOR STUDENTS
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/tank-tubes-500kva-single-phase
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

For a 500 kVA, 50 Hz, 6600/400, single phase core type, oil immersed, natural cooled power transformer, the design parameters are: Constant for output voltage per turn (K) = 0.8; Resistivity of copper = 0.021 Ω-mm²/m; Maximum flux density in core (Bm) = 1.5 Wb/m²; Current density = 2.75 A/mm²; Core type = Cruciform; Window space factor (Kw) = 0.27; Stacking factor (Ki) = 0.9; Ratio of height to width of windows (Hw/Ww) = 2.5; Ratio of yoke height to width (Hy/Dy) = 1; Axial depth of LV winding = 402 mm; Axial depth of HV winding = 377.5 mm; Inside diameter of LV winding = 310 mm; Outside diameter of LV winding = 348 mm; Inside diameter of HV winding = 360 mm; Outside diameter of HV winding = 418 mm; 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 rise of oil is not to rise 35°C, find the necessary number of tubes and show its arrangement.

## Short answer

Plain tank rise 79.7 °C. Required tube area 9.698 m². With 50 mm diameter, 1.5 m long tubes, 42 needed. Arrangement: 14 tubes on each long side, 7 on each short side in a single row.

## Answer

Only the losses and tank size are needed for this part; the core and winding data are used in the full design of the same transformer.

#### Basis
- A plain tank wall loses heat by radiation (≈ 6 W/m²°C) and natural convection (≈ 6.5 W/m²°C): total 12.5 W/m²°C.
- Tubes screen each other, so they add almost nothing by radiation, but they improve convection by about 35%: $6.5\times1.35 \approx 8.8$ W/m²°C.
- Top and bottom of the tank are neglected.

#### Temperature rise with a plain tank
Total loss $P = P_i + P_c = 1460 + 3865 = 5325$ W.

$$
\begin{aligned}
S_t &= 2(L_t + W_t)H_t = 2(1.05 + 0.62)\times1.6 = 5.344\ \text{m}^2\\
\theta &= \frac{P}{12.5S_t} = \frac{5325}{12.5\times5.344} = 79.7\ ^\circ\text{C}
\end{aligned}
$$

This is far above the permitted mean oil rise of 35 °C, so cooling tubes are needed.

#### Number of tubes
Let the total tube area be $A_t$:

$$
\begin{aligned}
P &= \theta\,(12.5S_t + 8.8A_t)\\
A_t &= \frac{1}{8.8}\left(\frac{5325}{35} - 12.5\times5.344\right) = \frac{1}{8.8}(152.14 - 66.80) = 9.698\ \text{m}^2
\end{aligned}
$$

Assume tubes of 50 mm diameter and 1.5 m mean length (tank height 1.6 m):

$$
\begin{aligned}
\text{Area of one tube} &= \pi\times0.05\times1.5 = 0.2356\ \text{m}^2\\
n_t &= \frac{9.698}{0.2356} = 41.16 \approx 42\ \text{tubes}
\end{aligned}
$$

Check: $35(12.5\times5.344 + 8.8\times42\times0.2356) = 5386$ W ≥ 5325 W.

**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: temperature rise with plain tank ≈ 79.7 °C; 42 tubes of 50 mm diameter, 1.5 m long, keep the mean oil rise within 35 °C.**

## Check yourself

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

1. What heat dissipation per m^2 per degree C is taken for a plain tank wall?
   A. 6.5 W/m2 C from convection only
   B. 12.5 W/m2 C (about 6 radiation + 6.5 convection)
   C. 8.8 W/m2 C
   D. 35 W/m2 C
   Answer: B. A plain wall loses heat both by radiation (about 6) and natural convection (about 6.5), total 12.5 W/m2 C. (Key point: Plain tank dissipation coefficient)

2. What is the temperature rise with a plain tank for total loss 5325 W and S_t = 5.344 m^2 (top and bottom neglected)?
   A. About 35 C
   B. About 79.7 C
   C. About 40 C
   D. About 159 C
   Answer: B. theta = P/(12.5 S_t) = 5325/(12.5 x 5.344) = 79.7 C, which exceeds the 35 C limit, so tubes are needed. (Key point: Plain tank temperature rise)

3. How do the cooling tubes affect heat dissipation?
   A. They add radiation but no convection
   B. They improve convection by about 35% (6.5 x 1.35 = 8.8) and add almost no radiation since tubes screen each other
   C. They double both radiation and convection
   D. They only increase the tank wall area, with the same 12.5 coefficient
   Answer: B. Tubes screen each other so radiation is not increased, but they improve convection by about 35%, hence 8.8 W/m2 C. (Key point: Effect of tubes)

## Close variants set in other exams

- [Design: 500 kVA single phase transformer with tubes](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-500kva-single-phase-tubes) — Electric Machine Design, chapter 3, asked 2×

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