---
title: "Quarter-load heat run and 30% overload time"
subject: "Electric Machine Design"
chapter: 2
key: "semester_5/electric_machine_design:quarter-load-overload-30-40"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/quarter-load-overload-30-40
site: BCT Engineering
---

# Quarter-load heat run and 30% overload time

Electric Machine Design · Chapter 2: Heating and cooling of electric machine

- Asked in: 2078 Bhadra, 6 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-30-40
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2

## Question

A transformer has a temperature rise of 30°C after 2 hrs and 40°C after 3 hrs on quarter load. What is the final steady temperature rise at full load? If the transformer is on 30% overload, how long will it take to attain the same temperature rise provided that maximum efficiency occurs at 70% of full load?

## Short answer

Quarter-load rises 30°C (2h) and 40°C (3h): solving yields a=0.7676, τ=3.78h, θq=73.03°C. Pi=0.49Pc, full-load losses ratio 1.49/0.5525 → θFL=196.9°C. 30% overload losses ratio 2.18/0.5525 → final rise 288.1°C. Time to reach 196.9°C: 196.9=288.1(1-e^(-t/3.78)) → t=4.35h (261min). Time constant 3.78h, full-load steady rise ≈197°C, time on 30% overload ≈4.35h.

## Answer

#### Given
- Heat run at **quarter load**: rise 30 °C after 2 h, 40 °C after 3 h (from cold)
- Maximum efficiency at 70 % of full load
- Find the final steady rise at full load, and the time on 30 % overload (1.3 × full load) to reach that rise

#### Step 1: time constant and final rise at quarter load
With $a = e^{-1/\tau}$:

$$
\begin{aligned}
30 &= \theta_q(1 - a^2), \qquad 40 = \theta_q(1 - a^3) \\
\frac{1 - a^3}{1 - a^2} &= \frac{1 + a + a^2}{1 + a} = \frac{40}{30} \\
3a^2 - a - 1 &= 0 \;\Rightarrow\; a = \frac{1 + \sqrt{13}}{6} = 0.7676 \\
\tau &= \frac{1}{\ln(1/0.7676)} = 3.781\ \text{h} \\
\theta_q &= \frac{30}{1 - 0.7676^2} = 73.03\ ^\circ\text{C}
\end{aligned}
$$

#### Step 2: losses
Maximum efficiency at 70 % load gives $P_i = 0.7^2 P_c = 0.49\,P_c$ ($P_c$ = full-load copper loss).

| Load | Losses (in units of $P_c$) |
|---|---|
| Quarter load | $0.49 + 0.25^2 = 0.5525$ |
| Full load | $0.49 + 1 = 1.49$ |
| 30 % overload | $0.49 + 1.3^2 = 2.18$ |

#### Step 3: final rise at full load
The final rise is proportional to losses (same cooling):

$$
\theta_{FL} = 73.03 \times \frac{1.49}{0.5525} = 196.9\ ^\circ\text{C}
$$

#### Step 4: time on 30 % overload
Final rise on overload: $\theta_{OL} = 73.03 \times \dfrac{2.18}{0.5525} = 288.1$ °C. Time (from cold, same $\tau$) to reach 196.9 °C:

$$
\begin{aligned}
196.9 &= 288.1\left(1 - e^{-t/3.781}\right) \\
e^{-t/3.781} &= 1 - 0.6835 = 0.3165 \\
t &= 3.781 \times \ln\frac{1}{0.3165} = 4.35\ \text{h} \approx 261\ \text{min}
\end{aligned}
$$

**Answer: heating time constant = 3.78 h; final steady rise at full load ≈ 197 °C; on 30 % overload the same rise is reached in ≈ 4.35 h (about 4 h 21 min).**

(The data give an unusually high full-load rise. This follows directly from the figures given; 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 70 % load, the quarter-load losses in units of Pc are:
   A. 0.5525
   B. 0.25
   C. 1.49
   D. 0.49
   Answer: A. Pi = 0.49 Pc, copper at quarter load = 0.0625 Pc, total 0.5525. (Key point: Losses at quarter load)

2. How is the full-load final rise found from the quarter-load final rise?
   A. Multiply by 4
   B. Multiply by the square of 4
   C. Multiply by the ratio of total losses 1.49/0.5525
   D. Multiply by 1.49 x 0.5525
   Answer: C. Final rise is proportional to total losses, so 73.03 x 1.49/0.5525 = 196.9 °C. (Key point: Scaling to full load)

3. To find the time on 30 % overload to reach the full-load rise, which relation is used?
   A. 196.9 = 288.1 e^(-t/τ)
   B. 196.9 = 288.1(1 - e^(-t/τ))
   C. 196.9 = 73.03(1 - e^(-t/τ))
   D. t = τ
   Answer: B. On overload the final rise is 288.1 °C and heating from cold follows 1 - e^(-t/τ), giving t = 4.35 h. (Key point: Overload time)

## Close variants set in other exams

- [Quarter-load heat run and 20% overload time](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aquarter-load-overload-40-50) — 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×