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how can a multi wall heat transfer problem with internal heat generation be solved? E.g.

|| heat generation ($\lambda_1,l_1,A_1$) | wall1 ($\lambda_2,l_2A_2$) | wall2 ($\lambda_3,l_3,A_3$) | $T_\infty,\alpha_\infty$

Heat flow is assumed only from left to right (positive x-direction) because of symmetric construction. || denotes the plane of symmetry. I'm also only interested in the stationary solution. The wall where the heat generation is situated has length $l_1$ with a surface area of $A_1$ and a thermal conductivity of $\lambda_1$. The characteristic values for the other walls are denoted respectively. On the right end of the multi wall body, convective heat transfer with a heat transfer coefficient of $\alpha_\infty$ is assumed.

My idea is to start with the stationary one dimensional heat transfer equation $$0 = \lambda(x) \frac{\partial^2 \vartheta}{\partial x^2} + Q$$

Can I just integrate over the 3 different walls? I'm struggling on how to properly consider the Q then. I hope you can help me.

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    $\begingroup$ I love heat transfer problems but there's a big piece missing here: what's left to the left most wall? Also, walls with different surface areas but in touch with each other? How? $\endgroup$ – Gert May 27 '20 at 14:29
  • $\begingroup$ For my investigations it's most suitable to assume, that the walls have perfect contact with each other, hence, no heat transfer coefficient needs to be considered. To the left it's basically the same composition of walls. Therefore (because of symmetry) I intend only consider the right side and 'only' consider half the produced power loss. $\endgroup$ – Steradiant May 27 '20 at 14:49
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    $\begingroup$ So the assembly has a plane of symmetry that runs through the middle of the internally heated wall? $\endgroup$ – Gert May 27 '20 at 14:52
  • $\begingroup$ Exaktly, this is the reason why I only (want to) consider one side, i.e. to the right of the symmetry plane. I adapted my post accordingly. $\endgroup$ – Steradiant May 27 '20 at 15:00
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    $\begingroup$ I'm answering this but it'll take a while. Ta. $\endgroup$ – Gert May 27 '20 at 15:06
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Regards the central layer, for steady state of the quasi-$1D$ problem, apply Fourier's heat equation with uniform heat sink $Q$:

$$\lambda_1\frac{\text{d}^2T}{\text{d}x^2}+Q=0$$

(No partials are needed, because $1D$ and steady state)

$$\frac{\text{d}^2T}{\text{d}x^2}=-\frac{Q}{\lambda_1}$$

$$\Rightarrow \frac{\text{d}T}{\text{d}x}=-\frac{Q}{\lambda_1}x+c_1$$ $$\Rightarrow T(x)=-\frac{Q}{2\lambda_1}x^2+c_1x+c_2$$

We'll assume the temperature at $x=-l_1/2$ and $x=+l_1/2$ to be $T_1$ because of symmetry. This allows to determine the integration constants $c_1$ and $c_2$: $$T_1=-\frac{Ql_1^2}{8\lambda_1}-\frac{l_1c_1}{2}+c_2$$ and: $$T_1=-\frac{Ql_1^2}{8\lambda_1}+\frac{l_1c_1}{2}+c_2\tag{1}$$ Adding up these equations: $$2T_1=-\frac{Ql_1^2}{4\lambda_1}+2c_2$$ So that: $$c_2=T_1+\frac{Ql_1^2}{8\lambda_1}$$ Inserting this in $(1)$ shows that $c_1=0$, so that: $$\boxed{T(x)=-\frac{Q}{2\lambda_1}x^2+T_1+\frac{Ql_1^2}{8\lambda_1}}\tag{1'}$$


The worst in terms of tediousness is now behind us.

Because of symmetry we assume that half of the heat flux $Q$ now flows to the right (and the other half to the left, of course).

It's crucial to understand that the heat flux that flows through one layer of the composite wall is equal to the heat flux that flows through the layer to its left, as well as the heat flux that flows through the layer to its right.

Simply put, $\frac{Q}{2}$ is the heat flux that flows through each layer of the wall, due to the steady state condition. The heat flux $\frac{Q}{2}$ is imposed and in steady state conditions means all temperatures are constant in time.

For any layer of the wall, we have:

$$\frac{Q}{2}=-\lambda A\frac{\text{d}T}{\text{d}x}=-\lambda A\frac{T_l-T_r}{l}$$ where $T_l$ is the left hand temperature of the layer and $T_r$ is the right hand temperature of the layer and $l$ is the thickness of the layer.

For a series of layers we can write (excluding the central layer no. $1$):

$$\boxed{\lambda_2 A_2\frac{T_1-T_2}{l_2}=\lambda_3 A_3\frac{T_2-T_3}{l_3}=...=\lambda_i A_i\frac{T_{i-1}-T_{i}}{l_i}}\tag{2}$$


There remain two problems with how this problem was posed.

Firstly, it has to be defined how the outermost layers (left AND right) lose heat to the environment. E.g. if we assume heat loss by convection then the heat loss would be given by ($n$ being the outermost layer):

$$h_n A_n(T_n-T_{\infty})\tag{3}$$

where $T_{\infty}$ would be the environmental temperature.

Secondly: I don't see how in a quasi-$1D$ problem the condition:

$$A_1 \neq A_2 \neq ... \neq A_n$$

can be maintained without exposing parts of the layers to the environment.

It also prevents the Thermal Resistance concept being applied here.

But if we assume all $A$ to be identical then we can re-write $(2)$ with $(3)$ as:

$$\frac{Q}{2}=\lambda_2 A\frac{T_1-T_2}{l_2}=\lambda_3 A\frac{T_2-T_3}{l_3}=...=\lambda_i A\frac{T_{i-1}-T_{i}}{l_i}=h_n A(T_n-T_{\infty})$$

Assuming $Q$, $h_n$ and $T_{\infty}$ are known, each temperature point can now be recursively determined, if needed. E.g. the surface temperature of the outer (right) wall is given by:

$$\frac{Q}{2}=h_nA(T_n-T_{\infty})$$ where $h_n$ is the convection coefficient of the outer layer to the environment. Then from $T_n$, determine $T_{n-1}$, $T_{n-2}$, all the way to $T_1$ can be determined.

If needed, the central temperature $T(0)$ can also be determined from $(1')$ with $T_1$.


The temperature profile, here (NOT to scale) schematised for $4$ layers (RHS of the assembly):

Theta schematised:

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  • $\begingroup$ Thanks a lot for the great explanation. I indeed see the point with the different surface areas, I'll have to rethink this again. To the first note, I've defined convective heat transfer ($\alpha_\infty$ in my case) and due to symmetry, I assumed it on both sides. One thing bothers me though, how can I now really evaluate $T_1$ and $T_3$ where the latter one is the wall temperature (surface temperature at the very right end of the wall structure)? $\endgroup$ – Steradiant May 27 '20 at 16:42
  • $\begingroup$ Re your last question, I'm still editing to that effect. $\endgroup$ – Gert May 27 '20 at 16:45
  • $\begingroup$ Done. If you like it, upvote it! $\endgroup$ – Gert May 27 '20 at 18:24
  • $\begingroup$ Thanks a lot! This also means, that if I'm only interested in the wall temperatures rather than on the temperature development inside one wall, I don't need to start with Fourier's equation but I can directly use the thermal resistance concept right? $\endgroup$ – Steradiant May 29 '20 at 5:44
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    $\begingroup$ Yes, that is correct! $\endgroup$ – Gert May 29 '20 at 13:42

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