Natural vs artificial convection; heating curve

Convection is the transfer of heat from a hot surface by a moving fluid (air, hydrogen, oil). It is the main way heat leaves an electrical machine. Depending on what drives the fluid, it is called natural or artificial (forced) convection.

Natural vs artificial convection

The dissipation coefficient for forced air flow is often written $\lambda_c = \lambda_{c0}(1 + 0.1\,v_a)$, where $v_a$ is the air velocity in m/s. Doubling the cooling therefore lets a much larger loss be removed for the same rise.

Temperature rise–time curve of an electrical machine

Assumptions: the machine is a homogeneous body with uniform temperature; losses $P$ are constant; heat dissipated is proportional to the temperature rise.

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

In time $dt$, heat produced = heat stored + heat dissipated:

$$P\,dt = G h\,d\theta + S\lambda\theta\,dt$$ $$\begin{aligned} \frac{d\theta}{dt} + \frac{S\lambda}{Gh}\theta &= \frac{P}{Gh} \end{aligned}$$

Let $\theta_m = \dfrac{P}{S\lambda}$ (final steady rise) and $\tau = \dfrac{Gh}{S\lambda}$ (heating time constant). The equation becomes

$$\tau\frac{d\theta}{dt} + \theta = \theta_m$$

Separating variables and integrating from $\theta_i$ at $t = 0$:

$$\begin{aligned} \int_{\theta_i}^{\theta}\frac{d\theta}{\theta_m - \theta} &= \int_0^t\frac{dt}{\tau} \\ \ln\frac{\theta_m - \theta_i}{\theta_m - \theta} &= \frac{t}{\tau} \\ \theta &= \theta_m\left(1 - e^{-t/\tau}\right) + \theta_i\,e^{-t/\tau} \end{aligned}$$

From cold: $\theta = \theta_m(1 - e^{-t/\tau})$.

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

Cooling (losses removed): $0 = Gh\,d\theta + S\lambda_c\theta\,dt$, giving $\theta = \theta_i e^{-t/\tau_c}$ with $\tau_c = Gh/(S\lambda_c)$.

Link between convection and the curve

• Forced convection raises $\lambda$. This lowers both the final rise $\theta_m = P/(S\lambda)$ and the time constant $\tau = Gh/(S\lambda)$.
• For the same permitted rise, a forced-cooled machine can carry larger losses, i.e. a higher output from the same frame.
• In self-ventilated machines the fan stops at standstill, so $\lambda_c < \lambda$ and the cooling time constant is longer than the heating time constant.