--- title: "New kVA rating with fan cooling (350 kVA)" subject: "Electric Machine Design" chapter: 2 key: "semester_5/electric_machine_design:fan-cooling-350kva-rating" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/fan-cooling-350kva-rating site: BCT Engineering --- # New kVA rating with fan cooling (350 kVA) Electric Machine Design · Chapter 2: Heating and cooling of electric machine - Asked in: 2079 Bhadra, 10 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-350kva-rating - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2 ## Question A 350 kVA transformer has its maximum efficiency at 85% 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 thermal time constant and final steady temperature rise of the transformer. If by use of fan, cooling is improved so that heat dissipation per unit area per degree rise in temperature is increased by 15%, find the new kVA rating possible (i) for same final temperature rise as before (ii) if the allowable temperature rise is taken as 50°C. ## Short answer Using 24°C at 1h and 34°C at 2h: a=0.4167, τ=1.14h (68.5min), θm=41.14°C. Pi=0.7225Pc, full-load losses 1.7225Pc. With 15% better cooling: same rise allows 1.15×losses → rating 392.6kVA; 50°C rise gives 1.15×50/41.14×losses → 454.3kVA. So τ≈1.14h, final rise≈41.1°C; new ratings (i)393kVA, (ii)454kVA. ## Answer #### Given - Rating $S$ = 350 kVA; maximum efficiency at 85 % 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: $$ \begin{aligned} P_i &= (0.85)^2 P_c = 0.7225\,P_c \\ \text{Full-load losses} &= P_i + P_c = 1.7225\,P_c \end{aligned} $$ 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 %: $$ \begin{aligned} 0.7225 + x^2 &= 1.15 \times 1.7225 = 1.9809 \\ x^2 &= 1.2584 \;\Rightarrow\; x = 1.1218 \\ \text{New rating} &= 1.1218 \times 350 = 392.6\ \text{kVA} \end{aligned} $$ #### (ii) Allowable temperature rise 50 °C Losses allowed are now in the ratio $1.15 \times \dfrac{50}{41.14}$: $$ \begin{aligned} 0.7225 + x^2 &= 1.15 \times \frac{50}{41.14} \times 1.7225 = 2.4073 \\ x^2 &= 1.6848 \;\Rightarrow\; x = 1.2980 \\ \text{New rating} &= 1.2980 \times 350 = 454.3\ \text{kVA} \end{aligned} $$ **Answer: time constant ≈ 1.14 h (68.5 min); final rise ≈ 41.1 °C; new rating (i) ≈ 393 kVA, (ii) ≈ 454 kVA.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. From rises 24 °C (1 h) and 34 °C (2 h), the value of a = e^(-1/τ) is: A. 0.4167 B. 0.6 C. 1.4167 D. 0.5 Answer: A. 1 + a = 34/24 = 1.4167, so a = 0.4167, giving τ = 1.14 h and θm = 41.14 °C. (Key point: Time constant method) 2. With maximum efficiency at 85 % of full load, the iron loss equals: A. 0.85 Pc B. Pc C. 0.7225 Pc D. 1.7225 Pc Answer: C. At maximum efficiency iron loss = copper loss at that load = 0.85² Pc. (Key point: Iron loss) 3. For the same final rise, a 15 % increase in λ allows: A. 15 % more copper loss only B. 15 % more total losses C. 15 % less losses D. 15 % more current exactly Answer: B. θm = losses/(Sλ), so allowed total losses scale with λ. (Key point: Loss limit with fan) 4. The new loading x satisfies: A. x = 1.15 B. 1.15 x^2 = 1.7225 C. 0.7225 + x^2 = 1.15 x 1.7225 D. x^2 = 1.15 x 1.7225 Answer: C. Iron loss is constant, so only copper loss changes: 0.7225 + x² = 1.9809, x = 1.1218, about 393 kVA. (Key point: New rating (i)) ## Close variants set in other exams - [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× ## 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 - [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 (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× - [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 - [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 - [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 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