0
$\begingroup$

There is a question that asks me to find the diffusion time assuming an initial temperature and given thermal conductivity $k$, specific heat capacity $C$ and density $\rho$. From the heat conduction equation (assuming diffusion in only one direction):

$$\frac{\rho C}{k}\frac{\partial T}{\partial t} = \frac{\partial^2T}{\partial x^2}$$

Then there is somehow a conclusion that given the length of the diffusion wall $L$, the diffusion time is $t_D = L^2/\alpha$. How does this come about? What additional assumptions must be made?

$\endgroup$

1 Answer 1

2
$\begingroup$

Useful scaling relations can often be obtained by replacing differentials with finite differences:

$$\frac{\rho C}{k}\frac{\partial T}{\partial t} = \frac{\partial^2T}{\partial x^2};$$

$$\frac{\rho C}{k}\frac{\Delta T}{\Delta t}\sim\frac{\Delta T}{(\Delta x)^2}.$$

(Note that $\Delta T$ doesn't get squared on the right because the operator is $\frac{\partial^2}{\partial x^2}$.)

From this we obtain $\Delta t=\frac{\rho C(\Delta x)^2}{k}=\frac{(\Delta x)^2}{\alpha}$, where $\alpha\equiv\frac{k}{\rho C}$ is the thermal diffusivity.

This relation isn't exact by any means; it just tells us that we can expect the diffusion process over distance $x$ to have progressed fairly far along by time $\frac{x^2}{\alpha}$. (For some configurations, we're at $1-e^{-1}=63\%$ of the asymptotic final solution; two time constants would give us $86\%$, and three $95\%$.) A better solution may require solution of the original differential equation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.