--- title: "Heating constant and 1-hour rating (400 kVA)" subject: "Electric Machine Design" chapter: 2 key: "semester_5/electric_machine_design:oil-temperature-400kva-rating" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/oil-temperature-400kva-rating site: BCT Engineering --- # Heating constant and 1-hour rating (400 kVA) Electric Machine Design · Chapter 2: Heating and cooling of electric machine - Asked in: 2078 Kartik, 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/oil-temperature-400kva-rating - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2 ## Question A 400 kVA 1100/400V three phase transformer is working in an ambient temperature of 35°C on full load, its oil temperature is recorded as follows: 59°C after 1.5 hour and 71°C after 3 hours. Its full load copper losses is 2 times the iron losses. Calculate its heating time constant, final steady temperature rise and 1 hour rating. ## Short answer Oil rises: 24°C at 1.5h, 36°C at 3h (ambient 35°C). x=e^(-1.5/τ)=0.5 → τ=2.164h, θm=48°C. 1-hour rating: 48=θm'(1-e^(-1/2.164)) → θm'=129.72°C. Pc=2Pi, full-load losses 3Pi; losses ratio 129.72/48 gives k²=3.554, k=1.885, rating 754kVA. τ=2.164h, final rise 48°C, 1-hour rating 754kVA. ## Answer Temperature rise above ambient (35 °C): - after 1.5 h: $\theta_1 = 59 - 35 = 24\ ^\circ$C - after 3 h: $\theta_2 = 71 - 35 = 36\ ^\circ$C Heating from cold: $\theta = \theta_m(1-e^{-t/\tau})$. #### i) Heating time constant Let $x = e^{-1.5/\tau}$. Since 3 h = 2 × 1.5 h: $$ \begin{aligned} \frac{\theta_2}{\theta_1} &= \frac{1-x^2}{1-x} = 1 + x = \frac{36}{24} = 1.5 \;\Rightarrow\; x = 0.5\\ \tau &= \frac{1.5}{\ln 2} = 2.164\ \text{h} \end{aligned} $$ #### ii) Final steady temperature rise $$ \theta_m = \frac{24}{1-0.5} = 48\ ^\circ\text{C} $$ (Final oil temperature = 35 + 48 = 83 °C.) #### iii) One-hour rating The 1-hour rating is the load that, starting cold, brings the rise to 48 °C in exactly 1 hour. Let its final rise be $\theta_m'$: $$ \begin{aligned} 48 &= \theta_m'\left(1-e^{-1/2.164}\right) = \theta_m'(1-0.630) = 0.370\,\theta_m'\\ \theta_m' &= 129.72\ ^\circ\text{C} \end{aligned} $$ Final rise ∝ losses. With $P_c = 2P_i$ at full load, losses at load fraction $k$ are $P_i(1+2k^2)$, and at full load $3P_i$: $$ \begin{aligned} \frac{1+2k^2}{3} &= \frac{129.72}{48} = 2.702\\ k^2 &= \frac{3(2.702)-1}{2} = 3.554 \;\Rightarrow\; k = 1.885 \end{aligned} $$ One-hour rating $= 1.885 \times 400 = 754$ kVA. **Answer: $\tau$ = 2.164 h, final steady rise = 48 °C, 1-hour rating ≈ 754 kVA (1.885 × full load).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With ambient 35 °C, the rises used for the oil readings 59 °C and 71 °C are: A. 59 and 71 B. 24 and 36 C. 94 and 106 D. 35 and 35 Answer: B. Subtract ambient: 59 - 35 = 24 °C and 71 - 35 = 36 °C. (Key point: Rise above ambient) 2. Since 3 h is twice 1.5 h, the ratio 36/24 equals: A. x^2 B. 1 - x C. 1 + x, with x = e^(-1.5/τ) = 0.5 D. x Answer: C. (1 - x²)/(1 - x) = 1 + x = 1.5, so x = 0.5, τ = 2.164 h, θm = 48 °C. (Key point: Time constant) 3. The 1-hour rating is the load that: A. Reaches 48 °C steady state B. Is carried for 1 h with no temperature rise C. Equals the full load D. From cold gives a rise of 48 °C after exactly 1 hour Answer: D. Its final rise is 48/0.370 = 129.7 °C, and since losses are proportional to rise this gives k = 1.885, about 754 kVA. (Key point: One-hour rating) ## Close variants set in other exams - [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× ## 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 - [Design: 125 kVA 6600/400 V three phase transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-125kva-three-phase-windings) — Electric Machine Design, chapter 3, asked 4× - [New kVA rating with fan cooling (400 kVA)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Afan-cooling-400kva-rating) — Electric Machine Design, chapter 2, asked 2× - [Temperature rise and tubes for 500 kVA transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Atank-tubes-500kva-single-phase) — Electric Machine Design, chapter 3, asked 2× - [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× - [One hour rating of a 20 hp induction motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Ainduction-motor-one-hour-rating) — Electric Machine Design, chapter 2, asked 2× ## Related in other subjects - [500 kVA transformer efficiency full/half load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Atransformer-500kva-efficiency-full-half) — Electrical Machine, chapter 2 - [500 kVA transformer efficiency, full and half load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-500kva-6600-400-efficiency) — Electrical Machines, chapter 2 - [Full-load efficiency of 500 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A500kva-90pc-half-load-full-load) — Electrical Machines I, chapter 2