I have the 3D Laplace equation:

$$\nabla^{2} T_w = 0$$

where $\nabla^{2}=(\frac{\partial^{2}}{\partial x^2}+\frac{\partial^{2}}{\partial y^2}+\frac{\partial^{2}}{\partial z^2})$ defined on $x \in [0,L]$, $y \in [0,l]$ and $z \in [0,w]$.

with the following boundary conditions

$$\frac{\partial T_w(0,y,z)}{\partial x}=\frac{\partial T_w(L,y,z)}{\partial x}=0 \rightarrow Neumann \rightarrow \textbf{(1)}$$

$$\frac{\partial T_w(x,0,z)}{\partial y}=\frac{\partial T_w(x,l,z)}{\partial y}=0 \rightarrow Neumann \rightarrow \textbf{(2)}$$

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

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

Contextual information

How the third and fourth boundary conditions were arrived at

There are two additional equations

$$ \frac{\partial T_h}{\partial x} + \frac{b_h T_h}{L} = \frac{b_h T_w}{L} \rightarrow T_h = \frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T_w\mathrm{d}x $$ Known that $T_h(0,y,w)=T_{hi}$ (Inlet temperature of the hot fluid) So, we can write:

$$T_h(x,y,z) = e^{-b_hx/L}\left[T_{hi} + \frac{b_h}{L}\int_0^x e^{b_hs/L}T_w(s,y,z)ds\right]$$

So, the third BC becomes:

$$\frac{\partial T(x,y,w)}{\partial z} = p_h\bigg(e^{-b_hx/L}\left[T_{hi} + \frac{b_h}{l}\int_0^x e^{b_hs/l}T_w(s,y,z)ds\right] - T_w(x,y,w)\bigg) \rightarrow \textbf{(3)}$$

Similarly, $$ \frac{\partial T_c}{\partial y} + \frac{b_c T_c}{l} = \frac{b_c T_w}{l} \rightarrow T_c = \frac{e^\frac{-b_c y}{l}b_c}{l}\int e^\frac{b_c y}{l}T_w\mathrm{d}y$$

Known that $T_c(x,0,0) = T_{ci}$ (Inlet temperature of the cold fluid). Again, we can write:

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

Hence, the fourth BC becomes:

$$\frac{\partial T(x,y,0)}{\partial z} = p_c\bigg(e^{-b_cy/l}\left[T_{ci} + \frac{b_c}{l}\int_0^y e^{b_cs/l}T_w(x,s,z)ds\right] - T_w(x,y,0)\bigg)\rightarrow \textbf{(4)}$$

$l,L,b_h,b_c,p_h,p_c$ are all constants $>0$

  • $\begingroup$ What is the argument of $T_s$ in the integral? I am a bit confused since x and y are used as integration variables as well as fixed variable. $\endgroup$ Feb 10, 2019 at 14:30
  • $\begingroup$ @MarkusSprecher I have added a portion called Contextual Information to my original question to explain your query. $\endgroup$
    – Avrana
    Feb 10, 2019 at 16:26
  • $\begingroup$ @MarkusSprecher I have edited the question and added more background information to offer more explanation. Any advice on how to approach the problem would be helpful. $\endgroup$
    – Avrana
    Feb 11, 2019 at 12:55


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