---
title: "Heating constant and 1-hour rating (40 kVA)"
subject: "Electric Machine Design"
chapter: 2
key: "semester_5/electric_machine_design:oil-temperature-40kva-rating"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/oil-temperature-40kva-rating
site: BCT Engineering
---

# Heating constant and 1-hour rating (40 kVA)

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

- Asked in: 2082 Chaitra (new course), 4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/oil-temperature-40kva-rating
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2

## Question

A 40 kVA, 11000/400 V, 3-phase transformer is working in an ambient temperature of 35°C on full load. Its oil temperature is recorded as follows: 59°C after 1.5 hours; 71°C after 3 hours. Its full-load copper losses are 2 times the iron losses. Calculate: i) Heating time constant ii) Final steady temperature rise iii) 1-hour rating

## Short answer

Heating time constant τ = 1.5/ln 2 ≈ 2.164 h. Final steady temperature rise is 48 °C. For a 1‑hour rating the required final rise is 129.72 °C; using loss ratio (1+2k²)/3 = 129.72/48 gives k = 1.885, so rating = 75.4 kVA.

## Answer

Temperature rise above ambient (35 °C):

- after 1.5 h: $\theta_1 = 59 - 35 = 24\ ^\circ$C
- after 3 h: $\theta_2 = 71 - 35 = 36\ ^\circ$C

Heating from cold: $\theta = \theta_m(1-e^{-t/\tau})$.

#### i) Heating time constant
Let $x = e^{-1.5/\tau}$. Since 3 h = 2 × 1.5 h:

$$
\begin{aligned}
\frac{\theta_2}{\theta_1} &= \frac{1-x^2}{1-x} = 1 + x = \frac{36}{24} = 1.5 \;\Rightarrow\; x = 0.5\\
\tau &= \frac{1.5}{\ln 2} = 2.164\ \text{h}
\end{aligned}
$$

#### ii) Final steady temperature rise

$$
\theta_m = \frac{24}{1-0.5} = 48\ ^\circ\text{C}
$$

(Final oil temperature = 35 + 48 = 83 °C.)

#### iii) One-hour rating
The 1-hour rating is the load that, starting cold, brings the rise to 48 °C in exactly 1 hour. Let its final rise be $\theta_m'$:

$$
\begin{aligned}
48 &= \theta_m'\left(1-e^{-1/2.164}\right) = \theta_m'(1-0.630) = 0.370\,\theta_m'\\
\theta_m' &= 129.72\ ^\circ\text{C}
\end{aligned}
$$

Final rise ∝ losses. With $P_c = 2P_i$ at full load, losses at load fraction $k$ are $P_i(1+2k^2)$, and at full load $3P_i$:

$$
\begin{aligned}
\frac{1+2k^2}{3} &= \frac{129.72}{48} = 2.702\\
k^2 &= \frac{3(2.702)-1}{2} = 3.554 \;\Rightarrow\; k = 1.885
\end{aligned}
$$

One-hour rating $= 1.885 \times 40 = 75.4$ kVA.

**Answer: $\tau$ = 2.164 h, final steady rise = 48 °C, 1-hour rating ≈ 75.4 kVA (1.885 × full load).**

## Check yourself

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

1. Using rises 24 °C at 1.5 h and 36 °C at 3 h, the heating time constant is:
   A. 2.164 h
   B. 1.5 h
   C. 3 h
   D. 1.0 h
   Answer: A. 1 + x = 1.5 gives x = 0.5, so τ = 1.5/ln2 = 2.164 h. (Key point: Time constant)

2. The final steady temperature rise is:
   A. 36 °C
   B. 83 °C
   C. 60 °C
   D. 48 °C
   Answer: D. θm = 24/(1 - 0.5) = 48 °C, a final oil temperature of 83 °C. (Key point: Final rise)

3. With Pc = 2Pi at full load, losses at load fraction k are:
   A. 3 Pi k
   B. Pi(1 + 2k^2)
   C. Pi(1 + 2k)
   D. 3 Pi k^2
   Answer: B. Iron loss is constant and copper loss varies as k², giving Pi + 2Pi k². The 1-hour rating comes out as 1.885 x 40 = 75.4 kVA. (Key point: Loss expression)

## Close variants set in other exams

- [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×

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