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What are we supposed to do when two bodies are in contact to solve heat transfer over time?

Consider this one dimensional problem:

Suppose I have Body $A$ in perfect contact with Body $B$ via a surface $S$.

  • $K_{a} \ne K_{b}$
  • $T(x<x_{s}, 0) = T_{a}$
  • $T(x>x_{s}, 0) = T_{b}$

What is the evolution of $T(x_{s}, t)$ over time, i.e. the evolution of the interface temperature ?

We can't use the Fourier's Law at $x_{s}$ as $K(x)$ is not continuous in $x_{s}$

This answer doesn't work as there is a discontinuity of temperature: Heat transfer between two surfaces


If anyone is interested in this problem, I wrote and explained a python program solving this type of transcient heat problem using the "ghost nodes" method. https://github.com/camillechaize/Composite_Wall_MCP

rob
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2 Answers2

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Since you don't mention any type of thermal contact resistance, I'm going to ignore that aspect and assume that the contact is thermally perfect.

As @Gert notes, the bodies must be initially separated to have different temperatures at their surfaces, as a single point or plane can have only one well-defined temperature. At the instant of contact, we find that conduction, convection and radiation across the quickly narrowing gap rapidly force the surfaces to assume the same interface temperature.

Immediately after contact and for a while, the interface temperature $T_\text{int}$ is

$$T_\text{int}=\frac{T_\text{a}(x,0)\sqrt{k_\text{a}\rho_\text{a}c_\text{a}}+T_\text{b}(x,0)\sqrt{k_\text{b}\rho_\text{b}c_\text{b}}}{\sqrt{k_\text{a}\rho_\text{a}c_\text{a}}+\sqrt{k_\text{b}\rho_\text{b}c_\text{b}}},$$

where $k_i$ is the thermal conductivity of material $i$, $\rho_i$ is the density, and $c_i$ is the constant-pressure specific heat capacity. Note that for identical materials, the interface temperature lies halfway between the two initial temperatures, which we'd expect by symmetry.

Interestingly, this interface temperature is a fixed value that depends only the material properties (here assumed to be temperature-independent) and the initial temperatures! At the interface, we predict a steadily decreasing heat flux—but constant temperature—as the initially sharp temperature gradient evens out:

Proof (adapted from Incropera & DeWitt's Fundamentals of Heat and Mass Transfer):

The temperature within each body satisfies the conduction heat equation

$$\alpha_i\frac{\partial ^2 T_i(x,t)}{\partial x^2}=\frac{\partial T_i(x,t)}{\partial t},$$

where $\alpha=\frac{k}{\rho c}$ is the thermal diffusivity. At the interface (taken as $x=0$), for time $t>0$, we have the boundary conditions $$T_\text{a}(0,t)=T_\text{b}(0,t)=T_\text{int};$$ $$k_\text{a}\frac{dT_\text{a}(0,t)}{dx}=k_\text{b}\frac{dT_\text{b}(0,t)}{dx}.$$ The first condition reflects continuity of temperature at a single point; the second reflects an energy balance using Fourier's law of conduction: the fluxes $q^{\prime\prime}=-k_i\frac{dT_i}{dx}$ at the interface must be equal.

For a while after contact, the interface doesn't yet "know" about the other ends of the bodies (and vice versa). In other words, at times less than approximately the rule-of-thumb diffusion time, $\frac{L^2}{\alpha}$ (where $L$ is the body length), the ends are only minimally affected by the joining process and in turn can have little thermal effect on the interface.

(As a specific example, if you were to heat the very end of a 1 cm and 2 cm steel sample, you'd see the same spatiotemporal temperature distribution for the first few seconds for both samples, until the heat wave substantially diffused 1 cm and then started to respond to the length difference.)

Thus, for a while, we can idealize the bodies as semi-infinite, which eases the solution process substantially. In boundary-condition terms, we express this as $T_\text{a}(x\to\infty,t)\to T_\text{a}(x,0)$, $T_\text{b}(x\to-\infty,t)\to T_\text{b}(x,0)$.

We introduce the similarity variable $\eta_i=\frac{x}{2\sqrt{\alpha t}}$, whereupon the heat equation becomes

$$\frac{\partial ^2 T_i(\eta)}{\partial \eta^2}=-2\eta\frac{\partial T_i(\eta)}{\partial \eta}.$$

Integrating twice, we obtain

$$T_i(\eta_i)=C_1\int_0^{\eta_i}\exp(-\eta^{\prime2})\,d\eta^\prime+C_2$$

(where $\eta^\prime$ is a dummy variable of integration) and ultimately

$$T_i(x,t)=T_i(x,0)+[T_\text{int}-T_i(x,0)]\,\text{erfc}\frac{|x|}{2\sqrt{\alpha_i t}},$$

where $\text{erfc}$ denotes the complementary error function. For small $x$ or large $t$, $T_i(x,t)\approx T_\text{int}$; for large $x$ or small $t$, $T_i(x,t)\approx T_i(x,0)$, with a smooth transition between the two extremes.

The corresponding flux is $$q^{\prime\prime}=-k_\text{a}\frac{dT_\text{a}}{dx}=\frac{k_\text{a}(T_\text{int}-T_\text{a}(x,0))}{\sqrt{\pi\alpha t}}.$$

Setting the fluxes equal as described above, we obtain the short-time interface temperature $T_\text{int}$ given above.

Starting around the approximate characteristic time $\frac{L^2}{\alpha}$ of thermal diffusion down the bodies, and over many multiples of this characteristic time, the interface temperature transitions to the final uniform temperature of the two bodies

$$T_\text{int}=\frac{T_\text{a}(x,0)C_\text{a}+T_\text{b}(x,0)C_\text{b}}{C_\text{a}+C_\text{b}},$$

where $C_i$ is the heat capacity of body $i$. Note that for identical bodies, the interface temperature remains halfway between the two initial temperatures, which we'd again expect by symmetry.

This long-time interface temperature is obtained simply from a steady-state energy balance of both bodies. The interpretation here is that the diffusion process has come to the physical end of the hotter body that supplies heat flux and the end of the colder body that absorbs it. The temperature gradients that persisted within the bodies now vanish, there being nothing to sustain them. The final temperature, at steady state with zero heat flux, depends on the bodies' heat capacities (but not their thermal conductivities, as no fluxes remain).

  • There is something I quite don't understand in your answer. You describe 2 different $T_{int}$: one "just after contact" and one "at the end", but in the final solution, there is no difference between those two $T_{int}$ as $\forall t, T_{i}(x=0,t)=T_{i}(x=0,0)+[T_{int}−T_{i}(x=0,0)]erfc(\frac{|x=0|}{\sqrt{2αit}}) = T_{int}$ (as I assume $T_{i}(x=0,0) = T_{int}$) But shouldn't the interface temperature change over time ? – Camille Chaize Jul 28 '22 at 21:09
  • The interface temperature doesn't change over time if you assume semi-infinite bodies (e.g., extending infinitely far to the left and the right of the point of contact). However, no real body has infinite length. At some point, you come to an end to the regions of the hotter body to drive heat flux and regions of the colder body to absorb this flux. At that point, the interface temperature does change toward a final value corresponding to complete thermal equilibrium. This value is independent of the thermal conductivity and depends only on the heat capacity. Does that clarify things? – Chemomechanics Jul 28 '22 at 21:15
  • Nice answer. But where does $u$ pop up from, please? – Gert Jul 28 '22 at 22:47
  • That's just a dummy variable for the integration; it could have any name. It might have been clearer to use $\eta^\prime$, although this can get confused with a derivative of $\eta$. – Chemomechanics Jul 28 '22 at 23:09
  • It clarifies a lot. But I'm precisely wondering what's happening between the initial dynamic steady state and the final steady state for the interface temperature. – Camille Chaize Jul 29 '22 at 08:00
  • When you give the corresponding flux at the end you chose the body A, why not B if the fluxes are equal? This is something I don't understand, how to apply this "equal flux" boundary condition at the interface, as for each of the bodies, you need the other's "temperature derivative at the interface". – Camille Chaize Jul 29 '22 at 08:30
  • @CamilleChaize The equal fluxes in $A$ and $B$ helps ensure $T_i(x,t)$ is a smooth function over all of its domain. Together with $T_\text{a}(0,t)=T_\text{b}(0,t)=T_\text{int}$. – Gert Jul 29 '22 at 12:28
  • @Chemomechanics You do have the same initial problem as I have though: you move from the unphysical two temperatures in one point to a common $T_\text{int}$ instantaneously! – Gert Jul 29 '22 at 12:57
  • @CamilleChaize "When you give the corresponding flux at the end you chose the body A, why not B if the fluxes are equal?" I chose A as an example; simply replace the $a$ subscript with $b$ to get the flux expressed in terms of B's properties. – Chemomechanics Jul 29 '22 at 16:03
  • @Gert I ignored the (very rapid, over the "instant of contact" as noted by Incropera & DeWitt) transient increase from the original temperature to the fixed interface temperature. I also ignored the finite speed of phonons, temperature-dependent material properties, anisotropy and inhomogeneity in the bodies, material diffusion, thermoelastic effects, etc. Of course one is free to extend this result to include those factors. – Chemomechanics Jul 29 '22 at 16:11
  • I've edited my answer to address these comments. – Chemomechanics Jul 29 '22 at 16:52
  • @Chemomechanics: Ok, that's great. +1 from me. – Gert Jul 29 '22 at 17:28
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The situation you describe as:

$$T(x<x_{s}, 0) = T_{a}$$ $$T(x>x_{s}, 0) = T_{b}$$

has no physical significance because it would mean that:

$$\nabla T=\left(\frac{\partial T}{\partial x}\right)_{x=x_s}=|\infty|$$ and: $$\mathrm{q}=-k_a\nabla T=-k_b\nabla T=-|\infty|$$

This extremely high heat flux $\mathrm{q}$ causes the temperatures immediately left and right of the 'theoretical' boundary to equalise very quickly.

The interface temperature will then continue to evolve until a constant (in time) heat flux $Q$ flows through the ensemble.

That principle is well illustrated by this paper on the "Thermal resistance for a composite wall", which shows how to (quite simply) calculate the interface temperatures for layers with different $k$.

The calculation is for steady state (all relevant temperatures, including interface, have become time-invariant).

However, calculating the transient, i.e. $T(x_s,t)$ is no mathematical sinecure, as this Q&A of mine on a composite rod clearly shows.

I hope this helps.

Gert
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    I don't see how this answers the question. We can always assume a very thin interface thickness with a very high thermal conductivity, which makes the initial temperature gradient high but not infinite. There is certainly physical significance to pushing two materials of different temperatures together! Carslaw and Jaeger provide an answer (in the form of erf functions) for two semi-infinite regions with the same thermal diffusivity, but the present question is somewhat different from that simple case. – Chemomechanics Jul 27 '22 at 22:22
  • @Chemomechanics "There is certainly physical significance to pushing two materials of different temperatures together!" Nice straw man. I was of course referring to OP's proposition of having two different $T$ in one single $x$ as having no physical significance, nothing else. – Gert Jul 27 '22 at 23:07
  • Thank you for the clarification. But the temperatures are not different for any finite length of time; they instantaneously converge to the same intermediate value. – Chemomechanics Jul 27 '22 at 23:49
  • I quote from Incropera & DeWitt: "An interesting permutation of case 1 occurs when two semi-infinite solids, initially at uniform temperatures TA,i and TB,i, are placed in contact at their free surfaces...If the contact resistance is negligible, the requirement of thermal equilibrium dictates that, at the instant of contact...both surfaces must assume the same temperature Ts, for which TB,i < Ts < TA,i." I've added an answer discussing the temperature at and around the interface. – Chemomechanics Jul 27 '22 at 23:49