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I need to solve the three-dimensional Laplace equation ($\nabla^2T = 0$) where $\nabla^2=\frac{\partial^2}{\partial x^2} +\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$ in the domain where $x\in[0,L];y\in[0,l]$ and $z\in[0,\mu]$. The boundary conditions are $$T\vert_{0,y,z} = T_{hi}, T{(L,y,z)} = 0 \tag A$$ $$T\vert_{x,0,z} = T_{ci}, T{(x,l,z)} = 0\tag B$$ $$\frac{\partial T}{\partial z}\vert_{x,y,0} = k_c(T\vert_{x,y,0}-T_{c,av}) \tag C$$ $$\frac{\partial T}{\partial z}\vert_{x,y,\mu} = k_h(T_{h,av} - T\vert_{x,y,\mu}) \tag D$$

Here, $$T_{c,av} = \frac{1}{2}\Bigg(T_{ci}+e^{-b_c}\Bigg[T_{ci}+\frac{b_c}{l}\int_0^l e^{b_c s/l}T(x,s,z)\mathrm{d}s\Bigg]\Bigg) \tag E$$

and, $$T_{h,av} = \frac{1}{2}\Bigg(T_{hi}+e^{-b_h}\Bigg[T_{hi}+\frac{b_h}{L}\int_0^L e^{b_h s/L}T(s,y,z)\mathrm{d}s\Bigg]\Bigg) \tag F$$

I have decided to sub-divide the problem into three parts and adding up the solution of each sub-problem finally. Can be better understood from the following schematic:

enter image description here

SP1,SP2 are identical problems. The last two figures at the end describe SP3 with each figure showing $z=0$ and $z=\mu$ face respectively.


SP3

The other B.C.(s) for SP3 are:

$$T(0,y,z) = T(L,y,z) = T(x,0,z) = T(x,l,z) = 0 \tag G$$

along with B.C.(s) $\bf\mathrm(C),\mathrm(D)$

Hence, SP3 has two non-homogeneous Robin type BC at $z=0$ and $z=\mu$ respectively. I assume a preliminary temp. distribution as follows:

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

$$T_{nm}(z) = A_{nm}e^{\gamma z} + B_{nm}e^{-\gamma z} \tag I$$

I plan on evaluating $A_{nm}$ and $B_{nm}$ using the two linear equations in two unknowns that would be generated on applying $(\mathrm{H})$ in the boundary conditions : $\bf\mathrm{(C)}$ and $\bf\mathrm{(D)}$ .


My questions are:

  1. Is this method of sub-dividing the problems logical or is there any conceptual flaw in this approach ?
  2. The physical situation that these equations describe should not allow the temperature on $z=\mu$ wall go above $T_{hi}$ and $z=0$ wall go below $T_{ci}$. My solution defies these constraints. Can anyone give this a try and help me out by verifying this ?

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