---
title: "Motor temperature after shutdown"
subject: "Electric Machine Design"
chapter: 2
key: "semester_5/electric_machine_design:motor-cooling-after-shutdown"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/motor-cooling-after-shutdown
site: BCT Engineering
---

# Motor temperature after shutdown

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

- Asked in: 2075 Asoj, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/motor-cooling-after-shutdown
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2

## Question

An induction motor is heated to a temperature of 60°C and is shut down. Calculate the temperature at a time 20 minutes after the shut down if the cooling time constant is 60 minutes. The ambient temperature is 30°C.

## Short answer

Initial rise is 30 °C. After 20 min the rise is 30 e⁻²⁰/⁶⁰ ≈ 21.5 °C, so the motor temperature ≈ 51.5 °C.

## Answer

After shut-down no heat is produced, so the machine cools exponentially towards ambient:

$$
\theta = \theta_i\,e^{-t/\tau_c}
$$

where $\theta$ is the rise above ambient, $\theta_i$ the rise at the moment of shut-down and $\tau_c$ the cooling time constant. (From $0 = Gh\,d\theta + S\lambda\theta\,dt$.) The cooling time constant is usually larger than the heating time constant because ventilation stops when the machine stops.

#### Data
- Temperature at shut-down = 60 °C, ambient = 30 °C, so $\theta_i = 30\ ^\circ$C
- $\tau_c = 60$ min, $t = 20$ min

#### Calculation

$$
\begin{aligned}
\theta &= 30\,e^{-20/60} = 30 \times 0.7165 = 21.50\ ^\circ\text{C}\\
\text{Temperature} &= 30 + 21.50 = 51.50\ ^\circ\text{C}
\end{aligned}
$$

```text
 temp
 60 |\
    | \
 51.5|- -\  (t = 20 min)
    |     `-._
    |         ``--.____
 30 |- - - - - - - - - -``---- ambient
    +-------------------------> t
```

**Answer: temperature 20 minutes after shut-down ≈ 51.5 °C.**

## Check yourself

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

1. After shut-down, the rise above ambient falls as:
   A. θ = θi(1 - e^(-t/τc))
   B. θ = θi e^(-t/τc)
   C. θ = θi - t
   D. θ = θi e^(t/τc)
   Answer: B. With no heat produced, 0 = Gh dθ + Sλθ dt gives an exponential decay. (Key point: Cooling law)

2. For a motor at 60 °C with ambient 30 °C, θi in the cooling equation is:
   A. 60 °C
   B. 90 °C
   C. 30 °C
   D. 0 °C
   Answer: C. The decay is of the rise above ambient: 60 - 30 = 30 °C. (Key point: Initial rise)

3. After 20 min with τc = 60 min, the motor temperature is:
   A. 21.5 °C
   B. 60 °C
   C. 30 °C
   D. 51.5 °C
   Answer: D. Rise = 30 e^(-1/3) = 21.5 °C, and adding the ambient gives 51.5 °C. (Key point: Final temperature)

4. Why is the cooling time constant of a self-ventilated motor usually larger than the heating one?
   A. The ventilation fan stops with the machine, lowering λ
   B. Mass increases when shut down
   C. Losses increase
   D. Specific heat rises
   Answer: A. τ = Gh/(Sλ) rises when λ falls, and the shaft fan no longer blows air. (Key point: Cooling vs heating constant)

## Related in Electric Machine Design

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