I will try to describe briefly how I am modelling the problem. (Please bear with the length). The governing equation describing temperature for a block at steady state is

$$\nabla^2 T = 0$$ where $\nabla^2 T = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$ defined on $x \in [0,L] ,y \in [0,l], z \in [0,w]$ This is prescribed with the following boundary conditions

$\frac{\partial T(0,y,z)}{\partial x}=\frac{\partial T(L,y,z)}{\partial x}=0 \rightarrow Neumann$

$\frac{\partial T(x,0,z)}{\partial y}=\frac{\partial T(x,l,z)}{\partial y}=0\rightarrow Neumann$

$$\frac{\partial T(x,y,w)}{\partial z}=p_h\bigg(T_h - T(x,y,w)\bigg) \rightarrow Convection $$

$$\frac{\partial T(x,y,0)}{\partial z}=p_c\bigg(T_c -T(x,y,0) \bigg)\rightarrow Convection$$

The situation can be better understood from the attached figureenter image description here

Two fluids are in thermal contact with the separating wall and flow perpendicular to each other on either side of the wall. The inlet temperatures of both the fluids is known. Actually, there are two separate equations which govern them as

$\frac{\partial T_h}{\partial x} + \frac{b_h}{L} (T_h - T) = 0\rightarrow T_h=\frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T\mathrm{d}x$

Known that : $T_h(0,y,-w)=T_{h,i} \rightarrow $ a constant

$\frac{\partial T_c}{\partial y} + \frac{b_c}{l} (T_c - T) = 0 \rightarrow T_c=\frac{e^\frac{-b_c y}{l}b_c}{l}\int e^\frac{b_c y}{l}T\mathrm{d}y$

Known that: $T_c(x,0,0)=T_{c,i} \rightarrow $ a constant

So,the quantities $T_h$ and $T_c$ can now be substituted in the original last two boundary conditions. For example, the last bc would now look like

$\frac{\partial T(x,y,0)}{\partial z}=p_c\bigg(\frac{e^\frac{-b_c y}{l}b_c}{l}\int e^\frac{b_c y}{l}T\mathrm{d}y -T(x,y,0) \bigg)$

Hence, this becomes a Robin type condition where all terms on the LHS and RHS are functions of $T$ (which is not analogous to any example I have encountered in textbooks, where such b.c. usually have a free stream temperature defined)


I used:

$T(x,y,z)=\sum_{m,n=1}^{\infty}T_{nm}(z)\cos(\frac{n\pi x}{L})\cos(\frac{m\pi y}{l}).$

where $T_{nm}(z)$ is the undetermined $z$ function. Substituting this expression in $\nabla T^2 =0$, I obtain

$T_{nm}(z) = A_{nm}^{+}e^{\gamma z} + A_{nm}^{-}e^{-\gamma z} $ where $\gamma^2 = {(\frac{n\pi}{L})^2 + (\frac{m\pi}{l})^2 }$. Now, the undetermined coefficients are $A_{nm}^{+},A_{nm}^{-}$ which need to be determined using the $z$ boundary conditions.

Henceforth, the $z=0$ BC (using @Dyaln suggestion) , becomes

$$\frac{\partial T(x,y,0)}{\partial z} = p_c\bigg(e^{-b_cy/l}\left[T_{ci} + \frac{b_c}{l}\int_0^y e^{b_cs/l}T(x,s,z)ds\right] - T(x,y,0)\bigg) $$

On applying this boundary condition:

$$ \frac{1}{p_c}\sum_{n,m=1}^\infty \cos(\frac{n\pi x}{L})\cos(\frac{m\pi y}{l})\gamma ( A_{nm}^{+} - A_{nm}^{-}) = e^{-\frac{b_c y}{l}}T_{ci} + U + V - S - T $$


$U =\sum_{n,m=1}^\infty ( A_{nm}^{+} + A_{nm}^{-}) \frac{(b_c)^2}{(b_c)^2 + (m\pi)^2} \cos(\frac{n\pi x}{L})\cos(\frac{m\pi y}{l}) $

$V = \sum_{n,m=1}^\infty ( A_{nm}^{+} + A_{nm}^{-}) \frac{b_c m\pi}{(b_c)^2 + (m\pi)^2} \cos(\frac{n\pi x}{L})\sin(\frac{m\pi y}{l})$

$S = \sum_{n,m=1}^\infty ( A_{nm}^{+} + A_{nm}^{-}) \frac{(b_c)^2}{(b_c)^2 + (m\pi)^2} \cos(\frac{n\pi x}{L}) e^{\frac{-b_c y}{l}} $

$T = \sum_{n,m=1}^\infty ( A_{nm}^{+} + A_{nm}^{-})\cos(\frac{n\pi x}{L})\cos(\frac{m\pi y}{l})$

Subsequently, I would have to use the $z=w$ BC to get another equation in terms of $A_{nm}^{+}, A_{nm}^{-}$.

My question

(1) What I could not figure out till now is how to handle the exponential term while using orthogonality?

An approximation [Update]

From the physical problem at hand, one of the boundary conditions could be rewritten in an approximate from as

$$\frac{\partial T(x,y,w)}{\partial z}=p_h\bigg(\frac{T_{hi}+T_h(x=L)}{2} - T(x,y,w)\bigg)$$

$$\frac{\partial T(x,y,0)}{\partial z}=p_c\bigg(\frac{T_{ci}+T_c(x=l)}{2} - T(x,y,0)\bigg)$$

On using the suggestions of Dylan, the BC now takes the form

$$p_h^{-1}\frac{\partial T(x,y,w)}{\partial z} = \frac{1}{2}\bigg[T_{hi}(1+e^{-b_h}) + \frac{e^{-b_h}b_h}{L}\int_0^L e^{\frac{b_h s}{L}}T(s,y,z) \mathrm{d}s\bigg] - T(x,y,w)$$

$$p_c^{-1}\frac{\partial T(x,y,0)}{\partial z} = \frac{1}{2}\bigg[T_{ci}(1+e^{-b_c}) + \frac{e^{-b_c}b_c}{l}\int_0^l e^{\frac{b_c s}{l}}T(x,s,z) \mathrm{d}s\bigg] - T(x,y,0)$$

These BC(s) remove the dependence on the $e^{\frac{-b_c y}{l}}$ & $e^{\frac{-b_h x}{L}}$. Are these now solvable ?

  • $\begingroup$ @Christoph Any thoughts or suggestions would be very helpful. $\endgroup$ – Indrasis Mitra Feb 26 at 10:15

Again, not a full answer, but too long to put in a comment.

Since you know the initial values of $T_h$ and $T_c$, try writing them as

\begin{align} T_h(x,y,z) &= e^{-b_hx/L}\left[T_{hi} + \frac{b_h}{L}\int_0^x e^{b_hs/L}T(s,y,z)ds\right] \\ T_c(x,y,z) &= e^{-b_cy/l}\left[T_{ci} + \frac{b_c}{l}\int_0^y e^{b_cs/l}T(x,s,z)ds\right] \end{align}

This doesn't make the math easier, but now you know the B.C.s are inhomogeneous.

  • $\begingroup$ This was an excellent workaround. Thanks. I used your suggestion and for one sub-problem at $z=0$, I reached the equation where $A_{nm}$ is spposed to be determined. The problem of it being homogeneous is now resolved as you can see on my addition to the original question. The problem now stems from the fact that I have terms like $e^{\frac{-b_c y}{l}}$. I cannot figure out how to use orthogonality on them. $\endgroup$ – Indrasis Mitra Feb 19 at 13:08
  • $\begingroup$ I added another attempt to the original problem, this time without just guessing the $z$ function to be a $\cosh$ type. But i still face the problem of handling the exponential term $e^{-\frac{b_c y}{l}}$. Any suggestions would be helpful. $\endgroup$ – Indrasis Mitra Feb 23 at 5:26
  • $\begingroup$ Do you find something inherently wrong in my attempts to solve the problem ? Am I doing something extremely naive here. I am not trying intentionally to be abstract, but even after a lot of searching I do not find any examples analogous to this where such exponential terms crop up in the Fourier coefficient expression ? Are situations where such BC(s) appear fall out of the realm of separation of variables method ? $\endgroup$ – Indrasis Mitra Feb 26 at 13:33
  • $\begingroup$ @Dyaln I would have to admit that I am still stuck at the problem because of the exponential terms. I hope you find the time to suggest how should I use them to find out the Fourier coefficients. $\endgroup$ – Indrasis Mitra Feb 28 at 3:18
  • $\begingroup$ You should definitely decompose the exponential terms into Fourier series, but I can't say more as these B.C.s are way more complicated that $A_n = \text{something}$ or $A_n+B_n=\text{something}$. In my last attempt I ended up with something like $B_n = A_n + \text{something in terms of n} + \sum_{k=0}^\infty (\text{something in terms of k})A_k$. It's a real mess, so I'm not sure if my approach is the best one. But I don't know a better method. $\endgroup$ – Dylan Feb 28 at 5:09

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