Motor temperature after shutdown

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

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

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

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