Temperature rise-time curve under heating

A machine heats up because its losses produce heat faster than it can be dissipated at first. The temperature rises exponentially towards a final steady value, at which all the heat produced is dissipated.

Assumptions

• The machine is a homogeneous body with uniform temperature.
• Heat produced $P$ (W) is constant; heat dissipation is proportional to the temperature rise (Newton's law of cooling).

Notation

$P$ = losses (W), $G$ = mass (kg), $h$ = specific heat (J/kg°C), $S$ = cooling surface (m²), $\lambda$ = specific heat dissipation (W/m²°C), $\theta$ = temperature rise above ambient at time $t$.

Derivation

In a small time $dt$ the temperature rises by $d\theta$. Energy balance:

$$\underbrace{P\,dt}_{\text{produced}} = \underbrace{G h\,d\theta}_{\text{stored}} + \underbrace{S\lambda\theta\,dt}_{\text{dissipated}}$$

Rearranging:

$$\begin{aligned} \frac{dt}{Gh} &= \frac{d\theta}{P - S\lambda\theta} \\ \frac{S\lambda}{Gh}\,dt &= \frac{d\theta}{\dfrac{P}{S\lambda} - \theta} \end{aligned}$$

Let $\theta_m = \dfrac{P}{S\lambda}$ and $\tau = \dfrac{Gh}{S\lambda}$. Integrating, with $\theta = \theta_i$ at $t = 0$:

$$\begin{aligned} \frac{t}{\tau} &= \ln\frac{\theta_m - \theta_i}{\theta_m - \theta} \\ \theta &= \theta_m\left(1 - e^{-t/\tau}\right) + \theta_i\,e^{-t/\tau} \end{aligned}$$

Starting cold ($\theta_i = 0$):

$$\theta = \theta_m\left(1 - e^{-t/\tau}\right)$$

As $t \to \infty$, $\theta \to \theta_m = P/(S\lambda)$, the final steady temperature rise: all heat produced is then dissipated.

 rise
  ^          theta_m ---------------------
  |               _..--~~~~~~
  |          _.-'
  |       .-'  <- 0.632 theta_m at t = tau
  |     /
  |   /  initial slope = theta_m / tau
  | /
  +---|------|------|------|------> t
     tau    2tau   3tau   4tau

Heating time constant

$$\tau_h = \frac{G h}{S\lambda}$$

It is defined as:

1. the time to reach the final rise $\theta_m$ if the initial rate of rise continued unchanged (no heat dissipated); or
2. the time to reach 63.2 % of the final rise, since $1 - e^{-1} = 0.632$.

In practice the machine is considered thermally steady after about $4\tau$–$5\tau$ (98–99 %). The time constant is large for large, heavy machines with poor cooling, and is reduced by better ventilation (larger $\lambda$).