---
title: "New kVA rating with fan cooling (150 kVA)"
subject: "Electric Machine Design"
chapter: 2
key: "semester_5/electric_machine_design:fan-cooling-150kva-rating"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/fan-cooling-150kva-rating
site: BCT Engineering
---

# New kVA rating with fan cooling (150 kVA)

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

- Asked in: 2070 Chaitra, 6 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-150kva-rating
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2

## Question

The temperature rise of a 150 kVA transformer is 25°C and 37.5°C after 1 and 2 hours respectively starting from cold condition. Calculate its heating time constant and final steady temperature rise. If the rate of heat dissipation is improved by 20% with help of external fan, find the new kVA rating for same steady temperature rise. The maximum efficiency occurs at 80% of full load.

## Short answer

Rises 25°C at 1h and 37.5°C at 2h give a=0.5, τ=1.443h (86.6min), θm=50°C. Max eff at 80% → Pi=0.64Pc, full-load losses 1.64Pc. 20% better dissipation allows 1.2×1.64=1.968 times Pc; solving 0.64+x²=1.968 → x=1.1524, new rating 172.9kVA. Heating time constant 1.443h, final steady rise 50°C, new kVA 172.9kVA.

## Answer

#### Given
- 150 kVA; temperature rise from cold: 25 °C after 1 h, 37.5 °C after 2 h
- Fan improves heat dissipation by 20 %; maximum efficiency at 80 % of full load

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

$$
\begin{aligned}
1 + a &= \frac{37.5}{25} = 1.5 \;\Rightarrow\; a = 0.5 \\
\tau &= \frac{1}{\ln 2} = 1.443\ \text{h} \approx 86.6\ \text{min} \\
\theta_m &= \frac{25}{1 - 0.5} = 50\ ^\circ\text{C}
\end{aligned}
$$

#### Step 2: losses
At maximum efficiency (80 % load), iron loss = copper loss at that load:

$$
\begin{aligned}
P_i &= 0.8^2 P_c = 0.64\,P_c \\
\text{Full-load losses} &= 0.64P_c + P_c = 1.64\,P_c
\end{aligned}
$$

#### Step 3: new rating for the same final rise
Since $\theta_m = \dfrac{\text{losses}}{S\lambda}$, a 20 % higher $\lambda$ allows 20 % more losses for the same $\theta_m$. At $x$ times the old full load:

$$
\begin{aligned}
0.64 + x^2 &= 1.2 \times 1.64 = 1.968 \\
x^2 &= 1.328 \;\Rightarrow\; x = 1.1524 \\
\text{New rating} &= 1.1524 \times 150 = 172.9\ \text{kVA}
\end{aligned}
$$

**Answer: heating time constant = 1.443 h (86.6 min); final steady rise = 50 °C; new rating with fan ≈ 172.9 kVA.**

## Check yourself

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

1. With rises 25 °C and 37.5 °C at 1 h and 2 h, a = e^(-1/τ) equals:
   A. 0.5
   B. 1.5
   C. 0.64
   D. 0.25
   Answer: A. 1 + a = 37.5/25 = 1.5, so a = 0.5, τ = 1/ln2 = 1.443 h and θm = 50 °C. (Key point: Time constant)

2. When load changes from full load to x times full load, which loss stays constant?
   A. Copper loss
   B. Total loss
   C. Iron loss
   D. Neither
   Answer: C. Iron loss depends on voltage and flux; copper loss varies with x². (Key point: Constant iron loss)

3. With 20 % better dissipation, the condition for the new load x is:
   A. x^2 = 1.2
   B. 0.64 + x^2 = 1.2 x 1.64
   C. x = 1.2 x 1.64
   D. 0.64 + x = 1.968
   Answer: B. Allowed losses become 1.2 x 1.64 Pc = 1.968 Pc, giving x = 1.152 and about 172.9 kVA. (Key point: New rating)

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