3
$\begingroup$

Suppose that there is a slab with thickness $2L$ and other dimensions of the slab are very bigger than $L$. Both sides of the slab are kept in the constant temperature $T_0$. In time $t=0$ a heat source starts to generate heat uniformly with constant $q$.

Its easy to see that steady state temperature is$$T_{s.s}=T_0+\frac{ql^2}{2k} \left(1-\frac{x^2}{L^2} \right),$$where $ -L\le x \le L$ and $k$ is the thermal conductivity of the slab.

Now, when we solve this mathematically and use , for example Fourier series, the time needed to reach steady state is infinite. But physically, it doesn't make any sense! One intuition for this is that with infinite time to reach steady state, generated heat, $qA_s L$, is infinite!

Now can we find a good approximation for the time to reach steady state? How it is related to Fourier number $\alpha t/L^2$? I think $\alpha t/L^2 \approx 1$ is not a good answer.

$\endgroup$
  • 1
    $\begingroup$ Sure. This is all worked out in many heat transfer books. The best encyclopedia of heat transfer solutions (which definitely has the solution to this problem) is Conduction of Heat in Solids by Carslaw and Jaeger. And you are correct about the Fourier number being the key dimensionless group for determining the practical amount of time to reach the steady state. $\endgroup$ – Chet Miller Jul 20 '16 at 23:41
  • $\begingroup$ Thanks Chester. The book you mentioned has the solution to this problem. But there is no discussion on time needed to reach steady state. I would be grateful if you could give more details, like pages numbers, if it is possible :) $\endgroup$ – Ghartal Jul 21 '16 at 3:48
  • 1
    $\begingroup$ All the exponential terms in the series expansion die out before the very first term in the expansion. You can see the Fourier number in the exponent of the first term. Just determine when this term has decayed to less than about 0.01% of its initial value (or any other percent you wish to specify), and that will tell you the "effective time" to reach steady state. $\endgroup$ – Chet Miller Jul 21 '16 at 12:02
  • $\begingroup$ @ChesterMiller sir please see my question here : physics.stackexchange.com/questions/370746/… $\endgroup$ – Ghartal Nov 25 '17 at 13:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.