--- title: "Quarter-load heat run and 20% overload time" subject: "Electric Machine Design" chapter: 2 key: "semester_5/electric_machine_design:quarter-load-overload-40-50" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/quarter-load-overload-40-50 site: BCT Engineering --- # Quarter-load heat run and 20% overload time Electric Machine Design · Chapter 2: Heating and cooling of electric machine - Asked in: 2075 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/quarter-load-overload-40-50 - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2 ## Question A transformer has a temperature rise of 40°C after 3 hrs and 50°C after 4 hours on quarter load. What is the final steady temperature rise at full load? If the transformer is working on 20% over load, how long will it take to attain the same temperature rise provided that maximum efficiency occurs at 65% of full load? ## Short answer Quarter-load data: 40°C (3h), 50°C (4h) → x=0.8689, τ=7.11h, θmq=116.26°C. Pi=0.4225Pc, full-load losses 1.4225Pc → θmf=341.0°C. 20% overload losses 1.8625Pc → final rise 446.5°C. Time to 341.0°C: 341=446.5(1-e^(-t/7.11)) → t=10.27h (616min). Final full-load rise 341.0°C, time ≈10.27h, τ=7.11h. ## Answer For heating from cold, the temperature rise after time $t$ is $\theta = \theta_m\left(1-e^{-t/\tau}\right)$, where $\theta_m$ is the final steady rise and $\tau$ the heating time constant. The final rise is proportional to the total losses ($\theta_m \propto P_i + x^2P_c$). #### Step 1: Time constant and final rise at quarter load Let $x = e^{-1/\tau}$ ($\tau$ in hours). Then $$ \begin{aligned} 40 &= \theta_{mq}(1-x^3), \qquad 50 = \theta_{mq}(1-x^4)\\ \frac{50}{40} &= \frac{1-x^4}{1-x^3} \;\Rightarrow\; x^4 - 1.25x^3 + 0.25 = 0\\ (x-1)(x^3 - 0.25x^2 - 0.25x - 0.25) &= 0 \end{aligned} $$ $x = 1$ is not acceptable, so solving the cubic: $x = 0.8689$. $$ \begin{aligned} \tau &= \frac{1}{\ln(1/0.8689)} = 7.11\ \text{h}\\ \theta_{mq} &= \frac{40}{1-0.8689^3} = \frac{40}{0.344} = 116.26\ ^\circ\text{C} \end{aligned} $$ Check: $116.26(1-0.8689^4) = 50\ ^\circ$C. #### Step 2: Losses at different loads Maximum efficiency at 65% load means iron loss = copper loss at that load: $$ P_i = (0.65)^2P_c = 0.4225\,P_c $$ | Load | Total loss | |---|---| | Quarter load | $0.4225P_c + P_c/16 = 0.4850P_c$ | | Full load | $0.4225P_c + P_c = 1.4225P_c$ | | 20% overload | $0.4225P_c + 1.44P_c = 1.8625P_c$ | #### Step 3: Final steady rise at full load $$ \theta_{mf} = 116.26 \times \frac{1.4225}{0.4850} = 341.0\ ^\circ\text{C} $$ #### Step 4: Time on 20% overload to reach the same rise Final rise on overload: $$ \theta_{mo} = 116.26 \times \frac{1.8625}{0.4850} = 446.5\ ^\circ\text{C} $$ Time to reach 341.0 °C (the full-load final rise) on overload, starting cold: $$ \begin{aligned} 341.0 &= 446.5\left(1-e^{-t/7.11}\right)\\ e^{-t/7.11} &= 1-\frac{341.0}{446.5} = 0.2363\\ t &= 7.11\ln\frac{1}{0.2363} = 10.27\ \text{h} \end{aligned} $$ **Answer: final steady rise at full load = 341.0 °C; time on 20% overload ≈ 10.27 h (about 616 min); $\tau$ = 7.11 h.** (The given data lead to a very high rise and long time constant; the method is what matters.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With maximum efficiency at 65 % load, the full-load total loss is: A. 1.4225 Pc B. 1.8625 Pc C. 0.485 Pc D. 1.65 Pc Answer: A. Pi = 0.65² Pc = 0.4225 Pc, and adding Pc gives 1.4225 Pc. (Key point: Losses at loads) 2. The final rise at full load is obtained using: A. Rise proportional to load current B. Rise proportional to total losses C. Rise proportional to time constant D. Rise equal for all loads Answer: B. θm is proportional to losses: 116.26 x 1.4225/0.485 = 341 °C. (Key point: Use of proportionality) 3. The time constant for the overload case is: A. Larger by 20 % B. Smaller by 20 % C. Zero D. Same, 7.11 h, since mass and cooling are unchanged Answer: D. τ = Gh/(Sλ) does not depend on load. (Key point: Time constant unchanged) 4. The time on 20 % overload to reach 341 °C comes from e^(-t/τ) equal to: A. 341/446.5 B. 1 - 341/446.5 = 0.2363 C. 446.5/341 D. 1 - 446.5/341 Answer: B. 341 = 446.5(1 - e^(-t/τ)) gives e^(-t/τ) = 0.2363 and t = 10.27 h. (Key point: Overload time) ## Close variants set in other exams - [Quarter-load heat run and 30% overload time](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aquarter-load-overload-30-40) — Electric Machine Design, chapter 2, asked 2× ## Related in Electric Machine Design - [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 - [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× - [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 - [Field winding final working temperature](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Adc-motor-field-winding-temperature) — Basic Electrical Engineering, chapter 1