2
$\begingroup$

I'm trying apply the Laplace transformation to solve the non-dimensional heat conduction PDE for a hollow cylinder with convection boundary conditions and a non-homogenous initial condition.

$$\frac{\partial^2 \theta(r,t)}{\partial r^2}+\frac{1}{r}\frac{\partial \theta}{\partial r}=\frac{\partial \theta}{\partial t}$$

Initial condition:

$$\theta(r,t=0)=1$$

Boundary conditions:

$$\frac{\partial \theta}{\partial r}\big\rvert_{r=\mathrm{a/R}}=\mathrm{Bi_1}\theta\big\rvert_{r=\mathrm{a/R}}$$

$$\frac{\partial \theta}{\partial r}\big\rvert_{r=\mathrm{b/R}}=\mathrm{Bi_2}\theta\big\rvert_{r=\mathrm{b/R}}$$

where $\mathrm{a, b, R, Bi_1, Bi_2}$ are physical constants, and $\theta, r, t$ are dimensionless temperature, space and time, respectively.

My attempt:

I first applied the laplace transformation,

$$\frac{\partial^2 \overline{\theta}}{\partial r^2}+\frac{1}{r}\frac{\partial \overline{\theta}}{\partial r}=s\overline{\theta}-1$$

IC:

$$\overline{\theta}(r,s=\infty)=1/s$$

BCs:

$$\frac{\partial \overline{\theta}}{\partial r}\big\rvert_{r=\mathrm{a/R}}=\mathrm{Bi_1}\overline{\theta}\big\rvert_{r=\mathrm{a/R}}$$

$$\frac{\partial \overline{\theta}}{\partial r}\big\rvert_{r=\mathrm{b/R}}=\mathrm{Bi_2}\overline{\theta}\big\rvert_{r=\mathrm{b/R}}$$

I solved for the homogeneous and particular solutions of the now ODE to get,

$$\overline{\theta}(r,s)=A I_0(s r)+B K_0(s r) + 1/s$$

Which doesn't look like it has enough constants to apply the initial and boundary conditions. What could I be doing incorrectly?

Edit- I just found this: https://math.stackexchange.com/a/243003/642640. It is a similar problem, and it looks like the solution starts off looking similar to mine, however the arguments of the modified bessel functions look different, and I'm not sure how to do that correctly. I'm also still unclear on how to treat the boundary conditions.

$\endgroup$
  • $\begingroup$ The solution should be $\bar\theta(r,s) = \frac{1}{s}+ A(s)I_0(\sqrt{s}r) + B(s)K_0(\sqrt{s}r)$. The solution needs to decay at $s\to\infty$, so $A(s)=0$. I think the problem may be overdetermined $\endgroup$ – Dylan Feb 8 at 17:36
  • $\begingroup$ @Dylan Agreed. However I don't understand how I can have two boundary conditions, and initial condition but only two constants I need to solve for, A and B. If I use the initial condition I find that A=0. Then I can use either of the boundary conditions to find B. $\endgroup$ – omar Feb 8 at 17:40
  • $\begingroup$ The IC is already included in your treatment of the Laplace transform of the time derivative $\partial\theta/\partial t$: it corresponds to the $-1$ in the r.h.s. of the transformed equation. You cannot write it as an additional (and oddly written) IC in the transformed space as you did. $\endgroup$ – Paul Enta Feb 8 at 21:19
  • $\begingroup$ The best I can say is, try using the 2 boundary conditions to solve for $A$ and $B$, and see if that solution decays by luck. $\endgroup$ – Dylan Feb 9 at 3:03

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.