--- title: "New kVA rating with fan cooling (400 kVA)" subject: "Electric Machine Design" chapter: 2 key: "semester_5/electric_machine_design:fan-cooling-400kva-rating" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/fan-cooling-400kva-rating site: BCT Engineering --- # New kVA rating with fan cooling (400 kVA) Electric Machine Design · Chapter 2: Heating and cooling of electric machine - Asked in: 2074 Chaitra, 8 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/fan-cooling-400kva-rating - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2 ## Question A 400 kVA transformer has its maximum efficiency at 80% of full load. During a short full load heat run, the temperature rise after one hour and two hours is observed to be 24°C and 34°C respectively. Find the thermal time constant and final steady temperature rise of the transformer. If, by use of a fan, the cooling is improved so that the rate of heat dissipation per unit area per degree rise in temperature is increased by 15%, find the new kVA rating possible (i) For the same final temperature rise as before (ii) If allowable temperature rise is taken as 50°C ## Short answer From heat run 24°C (1h) and 34°C (2h): τ=1.14h (68.5min), θm=41.14°C. Max eff at 80% → Pi=0.64Pc, losses 1.64Pc. Improved cooling by 15%: same rise → losses ratio 1.15×1.64 → x=1.1162, rating 446.5kVA; 50°C rise → 1.15×50/41.14×1.64 → x=1.2853, rating 514.1kVA. Final rise 41.1°C; ratings 446kVA and 514kVA. ## Answer #### Given - Rating $S$ = 400 kVA; maximum efficiency at 80 % of full load - Heat run at full load: $\theta_1 = 24$ °C after 1 h, $\theta_2 = 34$ °C after 2 h - Fan cooling raises heat dissipation per unit area per °C ($\lambda$) by 15 % #### Step 1: time constant and final temperature rise With $a = e^{-1/\tau}$, $\theta_1 = \theta_m(1-a)$ and $\theta_2 = \theta_m(1-a^2)$: $$ \begin{aligned} 1 + a &= \frac{34}{24} = 1.4167 \;\Rightarrow\; a = 0.4167 \\ \tau &= \frac{1}{\ln(1/0.4167)} = \frac{1}{\ln 2.4} = 1.142\ \text{h} \approx 68.5\ \text{min} \\ \theta_m &= \frac{24}{1 - 0.4167} = 41.14\ ^\circ\text{C} \end{aligned} $$ #### Step 2: losses in terms of full-load copper loss $P_c$ At maximum efficiency, copper loss = iron loss: $$ \begin{aligned} P_i &= (0.80)^2 P_c = 0.6400\,P_c \\ \text{Full-load losses} &= P_i + P_c = 1.6400\,P_c \end{aligned} $$ The final rise is $\theta_m = \dfrac{\text{losses}}{S\lambda}$, so the losses allowed $\propto \lambda\,\theta_m$. At a new load of $x$ times full load, the losses are $P_i + x^2 P_c$ (iron loss is constant). #### (i) Same final temperature rise ($41.14$ °C) Losses allowed rise by 15 %: $$ \begin{aligned} 0.6400 + x^2 &= 1.15 \times 1.6400 = 1.8860 \\ x^2 &= 1.2460 \;\Rightarrow\; x = 1.1162 \\ \text{New rating} &= 1.1162 \times 400 = 446.5\ \text{kVA} \end{aligned} $$ #### (ii) Allowable temperature rise 50 °C Losses allowed are now in the ratio $1.15 \times \dfrac{50}{41.14}$: $$ \begin{aligned} 0.6400 + x^2 &= 1.15 \times \frac{50}{41.14} \times 1.6400 = 2.2920 \\ x^2 &= 1.6520 \;\Rightarrow\; x = 1.2853 \\ \text{New rating} &= 1.2853 \times 400 = 514.1\ \text{kVA} \end{aligned} $$ **Answer: time constant ≈ 1.14 h (68.5 min); final rise ≈ 41.1 °C; new rating (i) ≈ 446 kVA, (ii) ≈ 514 kVA.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. From 24 °C at 1 h and 34 °C at 2 h, the final steady rise is about: A. 34 °C B. 50 °C C. 24 °C D. 41.1 °C Answer: D. θm = 24/(1 - 0.4167) = 41.14 °C. (Key point: Final rise) 2. With maximum efficiency at 80 % of full load, iron loss in terms of full-load copper loss Pc is: A. 0.64 Pc B. 0.8 Pc C. 1.64 Pc D. 0.2 Pc Answer: A. Iron loss = 0.8² Pc = 0.64 Pc, so full-load losses are 1.64 Pc. (Key point: Iron loss) 3. For case (ii) with a 50 °C allowed rise, the allowed losses are scaled by: A. 1.15 only B. 50/41.14 only C. 1.15 x 50/41.14 D. 41.14/50 Answer: C. Losses are proportional to λθm, so both the 15 % gain and the higher rise count. (Key point: Higher allowed rise) 4. Case (ii) gives a rating of about: A. 446 kVA B. 400 kVA C. 600 kVA D. 514 kVA Answer: D. 0.64 + x² = 2.292 gives x = 1.2853, so 514 kVA. (Key point: Result) ## Close variants set in other exams - [New kVA rating with fan cooling (350 kVA)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Afan-cooling-350kva-rating) — Electric Machine Design, chapter 2, asked 2× ## Related in Electric Machine Design - [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 - [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× - [Full-load heat run and 50% overload time](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Afull-load-overload-20-32) — Electric Machine Design, chapter 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× - [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 - [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× ## Related in other subjects - [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 - [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 - [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 - [Full-load and half-load efficiency of 100 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A100kva-1200w-960w-efficiency) — Electrical Machines I, chapter 2