# Heat flow in response to a thinning barrier

I'm having a technical problem of heat transfer that I can't find the best approach to deal with. I'm grateful for any advice that can point me in the right direction!

Here is an idealised model:

I have a stack of $$n$$ metal sheets with different temperature on each side of the stack ($$T_h$$ and $$T_c$$). Heat is transferred from the hot side to the cold side, we can assume a 1D transfer. The thickness of each plate is $$h$$ so the total thickness of the stack is $$n \times h$$, assuming $$n$$ plates in the stack. $$T_h$$ and $$T_c$$ are fixed by circulating fluid.

I know the surface temperatures of both sides of the stack, $$T_h$$ and $$T_c$$, and the transfer is in steady state. I also know the thermal conductivity, the density and all properties of the metal. However, thermal conductivity depends on the actual temperature of the metal. I also assume that the sheets are in perfect thermal contact.

Now, I'm removing sheets, one by one, from the cold side of the stack. The surrounding temperatures and hence the surface temperatures $$T_h$$ and $$T_c$$, remain the same. We can assume that the surface of the newly exposed sheet will instantly have the temperature of the surrounding medium. The total thickness changes from $$n \times h$$ to $$(n-1) \times h$$.

The transfer can no longer be steady state, as the thermal gradient of the top sheets is steeper. Eventually, a new steady state will be reached and the gradient will be readjusted to the new thickness - until I remove another sheet.

I'd think that every removal of a sheet should induce a transient heat pulse.

How can I estimate the heat-flow pulse in response to removal of the sheets (=thinning of the barrier)?

This is not a homework, I could add additional details about the process, but I'm just after the conceptual solution. Any assumptions work.

• When you remove a sheet, the sheet below it will not instantly be at $T_c$, but rather will be a little above $T_c$, and will take some time to cool down. – probably_someone Oct 19 '18 at 3:28
• Thanks for a good comment, @probably_someone. The surface of the sheet below will instantly be at $T_c$, due to cooling fluid, but it takes some time for the sheet to cool down all through, due to the somehow limited thermal conductivity. I'll try to clarify the question. – Tactopoda Oct 19 '18 at 3:43
• "How can I estimate the heat-flow pulse in response to removal of the sheets?" What is the type of answer you are looking for? Do you want to know every detail about the transient, such as: 1) A plot of the Temperature vs. Distance vs. time, from the time of removal of a sheet for any arbitrary $(T,x,t)$, 2) Creation of a set of equations that will allow you to solve for any arbitrary $(T,x,t)$, or 3) Qualitatively understand the transient of Temperature at each time and distance after the time of the sheet removal? – Thomas Lee Abshier ND Oct 19 '18 at 4:32
• Thanks, @Thomas Lee Abshier ND. I'm looking for a way to approach the problem, from your alternatives, that would probably be (3) and maybe (2). I tried to made the question general as my real application contains a lot of additional parameters and assumptions. – Tactopoda Oct 19 '18 at 4:49
• In your description, you seemed to suggest that the thermal conductivity is a function of temperature. Is this correct? Once we get this settled, I can show you how to solve this problem. – Chet Miller Oct 19 '18 at 11:42

I have a feeling you've lost interest in this. Anyway, I'm going to present only the part of my solution addressing the initial and final steady states. If you're interested in the transient part of the solution, get back with me.

Consider a flat slab of material situated between x = 0 and x = L. The temperature at x = 0 is $$T_h$$ and the temperature at x = L is $$T_c$$. The material has a thermal conductivity k which varies linearly with temperature: $$k=k_c+(k_h-k_c)\frac{(T-T_c)}{(T_h-T_c)}$$ The solution to the steady state heat conduction equation for this situation is given by: $$\theta+\frac{(k_h-k_c)}{(k_h+k_c)}\theta (1-\theta)=1-\frac{x}{L}\tag{1}$$where $$\theta$$ is the dimensionless temperature: $$\theta=\frac{(T-T_c)}{(T_h-T_c)}$$ For purposes of the present analysis, Eqn. 1 is a little unwieldy to work with because of the quadratic in temperature on the left hand side. In the analysis of the transient problem, it would be much more convenient if the temperature were expressed explicitly in terms of this distance x. This can be done by obtaining the quadratic solution for $$\theta$$ and retaining only the linear term in the dimensionless thermal conductivity variation: $$\theta=\left(1-\frac{x}{L}\right)-\frac{(k_h-k_c)}{(k_h+k_c)}\frac{x}{L}\left(1-\frac{x}{L}\right)\tag{2}$$ You mentioned that the thermal conductivity can vary with temperature by up to 50% over the range of temperatures of interest. I have made a plot of equations 1 and 2 to see how they compare for a 50% variation of thermal conductivity, and have found that the two results virtually superimpose. Therefore, Eqn. 2 is valid for the present purposes.

Eqn. 2 describes the final temperature variation in a slab of initial length L+h after a layer of thickness h has been removed from the end at x = L (and the slab has been allowed to thermally equilibrate): $$\theta_f=\left(1-\frac{x}{L}\right)-\frac{(k_h-k_c)}{(k_h+k_c)}\frac{x}{L}\left(1-\frac{x}{L}\right)\tag{3}$$ The initial temperature variation in our slab before the layer was removed was: $$\theta_i=\left(1-\frac{x}{L+h}\right)-\frac{(k_h-k_c)}{(k_h+k_c)}\frac{x}{L+h}\left(1-\frac{x}{L+h}\right)\tag{4}$$If h is small compared to L, we can linearize Eqn. 4 with respect to h and obtain: $$\theta_i=\theta_f+\left(\frac{x}{L}+\frac{(k_h-k_c)}{(k_h+k_c)}\left[\frac{x}{L}\left(1-\frac{x}{L}\right)-\left(\frac{x}{L}\right)^2\right]\right)\frac{h}{L}\tag{5}$$

The actual difference between $$\theta_i$$ and $$\theta_f$$ will be relatively small, on the order of h/L, and the transient temperature variation will lie between these two nearly equal variations. Therefore, we can, with good accuracy, use their average in describing the variation of thermal conductivity with position x during the transient part of the response.

I think I'll stop here for now.

• Thank you Cheter! I've not lost interest, but just had some busy days. The site told me that there was too many comments on the initial question, and I tried to clarify the question. It's an interesting observation that the transient phase could be estimated by interpolation of the steady state conditions. I'll have a proper look soon. – Tactopoda Oct 23 '18 at 21:29
• Yes. It's my intention to show you how to obtain the transient transition between the initial and final steady states (even though these states are pretty close to one another). But for now, I'll give you more time to digest what has been done so far. Please let me know when you are ready to see the next installment that covers the analytic solution for the transient part of the response. – Chet Miller Oct 23 '18 at 23:11
• Took me some time to get back to the problem. Thanks to your effort, I have a working model now. Thanks! – Tactopoda Nov 13 '18 at 20:42

Let's initially assume that the thermal resistance due to the interfaces between the sheets is negligible.

The initial temperature across the sheets then satisfies the steady-state heat equation

$$\nabla\cdot\left[k(T)\nabla T\right]=0$$

with boundary conditions $$T(x=0)=T_\mathrm{c}$$ and $$T(x=nh)=T_\mathrm{h}$$, where $$k$$ is the thermal conductivity of the sheet material.

Since you say that the thermal conductivity is temperature dependent, it's not worth getting into analytical approaches to solve this equation. You'll almost certainly need to solve it numerically.

You now must numerically solve the transient heat equation

$$\rho c\frac{dT}{dt}=\nabla\cdot\left[k(T)\nabla T\right]$$

between $$x=h$$ and $$x=nh$$ with initial temperature $$T(x,t=0)$$ being the steady-state temperature distribution you just obtained and $$T(x=h,t)=T_\mathrm{c}$$ and $$T(x=nh,t)=T_\mathrm{h}$$. Here, $$\rho$$ is the sheet material density, and $$c$$ is its heat capacity, as discussed in the link.

You'll find, as you might intuit, that the newly exposed region will quickly cool down to the fixed cold temperature. This "information" will then propagate via diffusion through the stack, causing the final temperature profile to be steeper in the remaining material. This transition will be largely complete after several multiples of the characteristic time constant of the system, $$\tau=\left[(n-1)h\right]^2\rho c/k$$ (applying the general diffusive scaling relationship $$\tau=L^2/D$$, where $$L$$ is a characteristic distance and $$D$$ is the diffusivity, here the thermal diffusivity).

Now, if the thermal resistance due to the interfaces between the sheets isn't negligible (and indeed may have its own temperature dependence), then the numerical solution would get a little more complex because your solid material (the stack of sheets) has turned essentially into a composite. Heat transfer textbooks describe how to use finite-difference schemes, for example, to discretize the width of the sheets, incorporate the interface thermal resistances, and obtain a numerical solution of both the steady-state and time-dependent temperature profiles within the sheet stack.

• Thanks for explanation. I'll try to apply it on my problem. – Tactopoda Oct 21 '18 at 1:24