--- title: "Transformer heating and cooling time constants (25°C/37.5°C)" subject: "Electric Machine Design" chapter: 2 key: "semester_5/electric_machine_design:transformer-heating-cooling-constants" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/transformer-heating-cooling-constants site: BCT Engineering --- # Transformer heating and cooling time constants (25°C/37.5°C) Electric Machine Design · Chapter 2: Heating and cooling of electric machine - Asked in: 2081 Bhadra, 6 marks; 2080 Bhadra, 8 marks (asked 2 times) - 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-heating-cooling-constants - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2 ## Question The temperature rise of a transformer is 25°C after one hour and 37.5°C after two hours of starting from cold conditions. Calculate the final steady state temperature rise and the heating time constant. If its temperature falls from the final steady value to 40°C in 2.5 hours when disconnected, calculate its cooling time constant. The ambient temperature is 30°C. ## Short answer From rises: 25°C at 1 h, 37.5°C at 2 h, solve 1 + e^(-1/tau_h) = 1.5 → e^(-1/tau_h)=0.5 → theta_m=50°C, heating time constant 1.443 h. Cooling: from 80°C to 40°C in 2.5 h with ambient 30°C gives 10 = 50 e^(-2.5/tau_c) → cooling time constant 1.553 h. ## Answer #### Given - Temperature rise from cold: $\theta_1 = 25$ °C at $t = 1$ h, $\theta_2 = 37.5$ °C at $t = 2$ h - On disconnection, the temperature falls from its final steady value to **40 °C** in 2.5 h; ambient = 30 °C. #### Heating: final rise and heating time constant From cold, $\theta = \theta_m(1 - e^{-t/\tau_h})$. Let $a = e^{-1/\tau_h}$: $$ \begin{aligned} 25 &= \theta_m(1 - a) \\ 37.5 &= \theta_m(1 - a^2) = \theta_m(1-a)(1+a) \end{aligned} $$ Dividing: $$ \begin{aligned} 1 + a &= \frac{37.5}{25} = 1.5 \;\Rightarrow\; a = 0.5 \\ \theta_m &= \frac{25}{1 - 0.5} = 50\ ^\circ\text{C} \\ \tau_h &= \frac{1}{\ln(1/a)} = \frac{1}{\ln 2} = 1.443\ \text{h} \approx 86.6\ \text{min} \end{aligned} $$ #### Cooling time constant Final steady temperature = 30 + 50 = 80 °C (rise 50 °C). After 2.5 h the temperature is 40 °C, i.e. a **rise of 40 − 30 = 10 °C** above ambient. $$ \begin{aligned} \theta &= \theta_i\, e^{-t/\tau_c} \\ 10 &= 50\, e^{-2.5/\tau_c} \\ \tau_c &= \frac{2.5}{\ln(50/10)} = \frac{2.5}{\ln 5} = \frac{2.5}{1.6094} = 1.553\ \text{h} \approx 93.2\ \text{min} \end{aligned} $$ **Answer: final steady temperature rise = 50 °C; heating time constant = 1.443 h (86.6 min); cooling time constant = 1.553 h (93.2 min).** The cooling time constant is larger than the heating one, as expected, because heat dissipation is poorer when the transformer is not loaded (weaker oil circulation). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are theta_m and the heating time constant found from rises at 1 h and 2 h? A. Use the cooling curve B. Divide the two equations to get 1 + a = 1.5 where a = e^(-1/tau) C. Add the two equations D. Take their difference only Answer: B. theta1 = theta_m(1 - a) and theta2 = theta_m(1 - a^2), so the ratio gives 1 + a. (Key point: Method) 2. What is the final steady temperature rise? A. 50 °C B. 25 °C C. 75 °C D. 37.5 °C Answer: A. a = 0.5, so theta_m = 25/(1 - 0.5) = 50 °C. (Key point: Final rise) 3. What is the heating time constant? A. 0.5 h B. 2 h C. 1 h D. 1.443 h (86.6 min) Answer: D. a = e^(-1/tau) = 0.5 gives tau = 1/ln2 = 1.443 h. (Key point: Heating constant) 4. For the cooling calculation, what rise is used when the temperature falls to 40 °C with ambient 30 °C? A. 50 °C B. 40 °C C. 10 °C D. 70 °C Answer: C. The cooling follows the rise above ambient: 10 = 50 e^(-2.5/tau_c), giving tau_c = 1.553 h. (Key point: Cooling initial rise) ## Close variants set in other exams - [Transformer heating and cooling time constants (25°C/40°C)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Atransformer-cooling-constant-45c) — Electric Machine Design, chapter 2, asked 3× ## Related in Electric Machine Design - [Machine temperature after one hour of heating](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Amachine-temperature-after-hour) — Electric Machine Design, chapter 2 - [Heating constant and 1-hour rating (400 kVA)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aoil-temperature-400kva-rating) — Electric Machine Design, chapter 2, asked 2× - [Heating constant and 1-hour rating (40 kVA)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aoil-temperature-40kva-rating) — Electric Machine Design, chapter 2, asked 2× - [Heating and cooling time constants](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aheating-cooling-time-constants) — Electric Machine Design, chapter 2, asked 3× - [New kVA rating with fan cooling (150 kVA)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Afan-cooling-150kva-rating) — Electric Machine Design, chapter 2 - [Temperature rise-time curve under heating](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aheating-curve-time-constant) — Electric Machine Design, chapter 2