---
title: "Full-load heat run and 50% overload time"
subject: "Electric Machine Design"
chapter: 2
key: "semester_5/electric_machine_design:full-load-overload-20-32"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/full-load-overload-20-32
site: BCT Engineering
---

# Full-load heat run and 50% overload time

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

- Asked in: 2070 Asar, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/full-load-overload-20-32
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2

## Question

A transformer has a temperature rise of 20°C after one hour and 32°C after two hours on full load. What is the final steady state temperature rise at full load? If the transformer is working on 50% over load, how long will it take to attain the same temperature rise? Take the copper losses on full load equal to twice the iron loss.

## Short answer

Full-load rises 20°C (1h) and 32°C (2h): a=0.6, τ=1.958h, θm=50°C. Copper loss = 2× iron loss, so full-load losses = 3Pi, 50% overload losses = 5.5Pi, overload final rise 91.67°C. Time to 50°C: 50=91.67(1-e^(-t/1.958)) → t=1.543h (92.6min). Final steady rise 50°C, time on 50% overload ≈1.54h.

## Answer

Heating from cold follows $\theta = \theta_m\left(1-e^{-t/\tau}\right)$. Final rise is proportional to total losses.

#### Time constant and final steady rise at full load
Let $x = e^{-1/\tau}$ ($\tau$ in hours).

$$
\begin{aligned}
20 &= \theta_m(1-x), \qquad 32 = \theta_m(1-x^2) = \theta_m(1-x)(1+x)\\
\frac{32}{20} &= 1 + x \;\Rightarrow\; x = 0.6\\
\tau &= \frac{1}{\ln(1/0.6)} = 1.958\ \text{h}\\
\theta_m &= \frac{20}{1-0.6} = 50\ ^\circ\text{C}
\end{aligned}
$$

#### Losses
Full-load copper loss $P_c = 2P_i$.

- Full load: $P_i + 2P_i = 3P_i$
- 50% overload: $P_i + (1.5)^2(2P_i) = 5.5P_i$

Final rise on overload:

$$
\theta_{mo} = 50 \times \frac{5.5}{3} = 91.67\ ^\circ\text{C}
$$

#### Time to reach 50 °C on 50% overload

$$
\begin{aligned}
50 &= 91.67\left(1-e^{-t/1.958}\right)\\
e^{-t/1.958} &= 1-\frac{50}{91.67} = 0.4545\\
t &= 1.958\ln\frac{1}{0.4545} = 1.543\ \text{h}
\end{aligned}
$$

**Answer: final steady rise at full load = 50 °C ($\tau$ = 1.958 h); on 50% overload the transformer reaches 50 °C in about 1.54 h (≈ 92.6 min).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With rises 20 °C at 1 h and 32 °C at 2 h, the final steady rise at full load is:
   A. 32 °C
   B. 40 °C
   C. 50 °C
   D. 60 °C
   Answer: C. 1 + x = 1.6 so x = 0.6, and θm = 20/0.4 = 50 °C. (Key point: Final rise)

2. With Pc = 2Pi at full load, the losses at 50 % overload compared with full load are:
   A. 5.5 Pi against 3 Pi
   B. 4.5 Pi against 3 Pi
   C. 3 Pi against 5.5 Pi
   D. 2.25 Pi against 3 Pi
   Answer: A. Overload loss = Pi + 1.5² x 2Pi = 5.5 Pi, so the final rise becomes 50 x 5.5/3 = 91.67 °C. (Key point: Loss ratio)

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