---
title: "Transformer heating and cooling time constants (25°C/40°C)"
subject: "Electric Machine Design"
chapter: 2
key: "semester_5/electric_machine_design:transformer-cooling-constant-45c"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/transformer-cooling-constant-45c
site: BCT Engineering
---

# Transformer heating and cooling time constants (25°C/40°C)

Electric Machine Design · Chapter 2: Heating and cooling of electric machine

- Asked in: 2071 Chaitra, 6 marks
- Video: [Module_1_Part_5_ Heating of Electrical Machines(Problems)](https://www.youtube.com/watch?v=6U22xE42xpE) — Jyothi G K
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/transformer-cooling-constant-45c
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2

## Question

The rise in temperature of a transformer after one hour and two hours of starting from cold conditions are 25°C and 40°C respectively. Determine its final steady temperature rise and the heating time constant. If its temperature falls from the final steady value to 45°C in 90 minutes when disconnected from the operation, determine its cooling time constant. The ambient temperature is 30°C.

## Short answer

Solve heating equation: a = e^(-1/τ) from 25=θm(1-a), 40=θm(1-a²) gives a=0.6, θm=62.5°C, τh=1.958h (117.5min). Cooling from rise 62.5°C to 15°C in 1.5h yields τc=1.051h (63.1min). Final steady temperature rise 62.5°C, heating time constant 1.958h, cooling time constant 1.051h.

## Answer

#### Given
- From cold: $\theta_1 = 25$ °C after 1 h, $\theta_2 = 40$ °C after 2 h
- On disconnection, temperature falls from its final steady value to **45 °C** in 90 min (1.5 h); ambient = 30 °C

#### Heating: final rise and heating time constant
$\theta = \theta_m(1 - e^{-t/\tau_h})$. Let $a = e^{-1/\tau_h}$:

$$
\begin{aligned}
25 &= \theta_m(1 - a), \qquad 40 = \theta_m(1 - a^2) \\
1 + a &= \frac{40}{25} = 1.6 \;\Rightarrow\; a = 0.6 \\
\theta_m &= \frac{25}{1 - 0.6} = 62.5\ ^\circ\text{C} \\
\tau_h &= \frac{1}{\ln(1/0.6)} = \frac{1}{0.5108} = 1.958\ \text{h} \approx 117.5\ \text{min}
\end{aligned}
$$

#### Cooling time constant
Final temperature = 30 + 62.5 = 92.5 °C (rise 62.5 °C). After 1.5 h the temperature is 45 °C, a **rise of 45 − 30 = 15 °C**.

$$
\begin{aligned}
15 &= 62.5\,e^{-1.5/\tau_c} \\
\tau_c &= \frac{1.5}{\ln(62.5/15)} = \frac{1.5}{\ln 4.1667} = \frac{1.5}{1.4271} = 1.051\ \text{h} \approx 63.1\ \text{min}
\end{aligned}
$$

**Answer: final steady temperature rise = 62.5 °C; heating time constant = 1.958 h (117.5 min); cooling time constant = 1.051 h (63.1 min).**

## Check yourself

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

1. With rises 25 °C at 1 h and 40 °C at 2 h, which step gives a = e^(-1/τ)?
   A. 1 + a = 40/25 = 1.6
   B. 1 - a = 1.6
   C. a = 25/40
   D. a^2 = 1.6
   Answer: A. Dividing θ2 = θm(1 - a²) by θ1 = θm(1 - a) gives 1 + a = 1.6, so a = 0.6. (Key point: Heating method)

2. What is the final steady temperature rise?
   A. 40 °C
   B. 92.5 °C
   C. 62.5 °C
   D. 15 °C
   Answer: C. θm = 25/(1 - 0.6) = 62.5 °C. (Key point: Final rise)

3. When the temperature falls to 45 °C with ambient 30 °C, the rise used in the cooling equation is:
   A. 45 °C
   B. 75 °C
   C. 30 °C
   D. 15 °C
   Answer: D. Rise is measured above ambient: 45 - 30 = 15 °C, with 15 = 62.5 e^(-1.5/τc). (Key point: Cooling rise above ambient)

## Close variants set in other exams

- [Transformer heating and cooling time constants (25°C/37.5°C)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Atransformer-heating-cooling-constants) — Electric Machine Design, chapter 2, asked 3×

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