Heating and cooling time constants

The heating time constant is the time a machine would take to reach its final steady temperature rise if heat were never lost to the surroundings, i.e. if the initial rate of rise continued. Equivalently, it is the time to reach 63.2 % of the final rise. The cooling time constant is the corresponding time during cooling: the time for the temperature rise to fall to 36.8 % of its initial value.

Heating

Let $P$ = 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. In time $dt$:

$$P\,dt = G h\,d\theta + S\lambda\theta\,dt$$

Solving with $\theta = \theta_i$ at $t=0$:

$$\theta = \theta_m\left(1 - e^{-t/\tau_h}\right) + \theta_i\, e^{-t/\tau_h}$$ $$\theta_m = \frac{P}{S\lambda}, \qquad \tau_h = \frac{G h}{S\lambda}$$

Starting from cold ($\theta_i = 0$): $\theta = \theta_m(1 - e^{-t/\tau_h})$. At $t = \tau_h$, $\theta = 0.632\,\theta_m$. In practice the final temperature is reached after about $4\tau_h$–$5\tau_h$.

Cooling

When the machine is switched off ($P = 0$) from a rise $\theta_i$:

$$0 = G h\,d\theta + S\lambda_c\theta\,dt \;\Rightarrow\; \theta = \theta_i\, e^{-t/\tau_c}, \qquad \tau_c = \frac{G h}{S\lambda_c}$$

At $t = \tau_c$, $\theta = 0.368\,\theta_i$.

Curves

 rise                       rise
  ^   theta_m ----------     ^ theta_i
  |        _.-~~~~~          |\
  |     .-'   heating        | \   cooling
  |   /                      |  '.
  |  / 0.632 theta_m at tau  |    '-._ 0.368 at tau_c
  | /                        |        ~~--.____
  +------------------> t     +------------------> t

Comparison