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So I have this side-view drawing.

a drawing

Now I wonder, will such a multi-layer material have asymmetric heat conduction properties?

Namely, because of radiative conduction, reflective aluminum surface and absorbing paint surface, will heat conduction to the left of the drawing be higher than heat conduction to the right?

Will the effect be significant/noticeable at all?

Is it at all theoretically possible to construct material with such properties in the general case?

ulidtko
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3 Answers3

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The second law of thermodynamics forbids materials that conduct better in one (forward) direction than the reverse direction - such a material placed between two containers at thermal equilibrium would drive the temperature away from equilibrium, decreasing the entropy of the whole system and paving the way for a perpetuum mobile...

Reflectance and absorbance of material is the same (at a given wavelength) which is why left-right and right-left processes will happen at the same rate.

You drawing will have anisotropy in the vertical versus horizontal directions - but not left-right versus right-left asymmetry. Sorry.

Floris
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    Yeah. Excellent! I did have the feeling that something is fundamentally wrong with this concept. – ulidtko Sep 22 '14 at 13:36
  • It's a perfectly reasonable question... incidentally, multi layer insulators can be very interesting (especially to reduce radiative heat losses). I hesitate to use the word "cool"... See this recent answer – Floris Sep 22 '14 at 13:39
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    @Floris wrote "The second law of thermodynamics forbids materials that conduct better in one direction than another..." Unless I have misunderstood what you are saying here you seem to claim that conductivity is a scalar, when, in fact, crystals have tensor conductivities and this is known since the time Stokes has investigated the symmetry of these tensors. – hyportnex Sep 22 '14 at 14:32
  • @user31748 I mean the tensor is symmetrical : a->b should the same as b-> a – Floris Sep 22 '14 at 14:34
  • But the question "will such a multi-layer material have anisotropic heat conduction properties" of @ulidtko was about the anisotropy of that arrangement. The standard meaning of anisotropy is direction dependence and I believe in his arrangement the effective "vertical" and "horizontal" conductivities may be different. – hyportnex Sep 22 '14 at 14:49
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    @user31748 off course it was understood that vertical conductivity may well be different from horizontal. I just didn't found simple enough way to word that, and left it that way hoping it'd be clear from the context. "Anisotropic heat conduction along the horizontal line"? Sounds really silly, IMO. – ulidtko Sep 22 '14 at 15:05
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    Note I had already confirmed the horizontal vs vertical anisotropy in my last paragraph. I changed a couple of words in the offending phrase - hope it is less ambiguous now. – Floris Sep 22 '14 at 15:07
  • @ulidtko - I changed the wording in your question to "asymmetric"; I think that better represents what you were trying to say. If you agree, then I will update the wording in the answer to use the same phrase as well. If you disagree, you can "roll back" the edit - this will just undo what I did. – Floris Sep 22 '14 at 15:17
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    In that case it seems that @ulidtko is asking about what is normally called reciprocity. Although I do not believe that his example is such but non-reciprocity, that is asymmetry of the conductivity tensor is again not prohibited by any law, specifically not by the 2nd law. It is usually claimed that symmetry will follow from the so-called "Curie principle" and Onsager's "microscopic reversibility" but this is a very controversial subject. Truesdell in his "Rational Thermodynamics" says that this symmetry has not been verified for at least 11 crystal classes, but this was written in 1984. – hyportnex Sep 22 '14 at 15:36
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    @user31748 - you are right this is really asking about reciprocity. I would encourage you to to expand your thoughts into an answer of your own; I think it would add more than just comments (which have a tendency to disappear over time...) – Floris Sep 22 '14 at 15:48
  • I agree with the last comment; it seems that @user31748 is capable of writing an in-depth explanation, especially regarding the theoretical part. – ulidtko Sep 22 '14 at 16:35
  • Not sure I understand this argument - if two containers are at equilibrium with each other then they are at the same temperature - correct? So if you put this material between those containers then the temperature gradient in the material is zero and so the heat flux through this material is zero no matter what the heat conduction is - correct? Then why would this drive the system out of equilibrium? – Maxim Umansky Sep 22 '14 at 18:32
  • @MaximUmansky - the argument for the asymmetrical conductivity is that the reflectivity of the two surfaces is different - and that therefore there will be more heat flux in one direction than the other (different emissivity in both directions - net heat flow). If that were the case then the foil would become a heat pump. But the argument doesn't take into account reciprocity: more reflection == more absorption. That's what saves the day. – Floris Sep 22 '14 at 22:04
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    OK, so the issue is not the asymmetrical thermal conductivity, here it is asymmetrical radiation transport. The latter does violate thermodynamics. But I am not sure anything wrong would happen if diffusive thermal transport was asymmetrical. – Maxim Umansky Sep 23 '14 at 02:33
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If we assume that both the aluminum foil and the paint have scalar heat conductivities then their composite will also be scalar, and thus symmetrical. Being both material either poly-crystalline or amorphous this is probably reasonable assumption.

@Floris suggested that I expand on this, but not being my area I can only summarize a few ideas from Truesdell's Rational Thermodynamics, chapter 7.

It is an old problem going back to Stokes (1851) whether the heat conductivity tensor $\mathbf{K}$, that is the tensor in the linear Fourier law between temperature gradient and heat flux written as $ \vec q = \mathbf{K} \nabla T $ for crystalline materials is always symmetric or not. Assume the simplest case of $\mathbf{K}$ independent of the temperature and deformation, then one gets from the 1st law $\rho c \partial_t{T} = \mathbf{K_{+}} \nabla^2{T}$ and from the 2nd law that $\mathbf{K_+}$ is positive definite, where $\mathbf{K_{\pm}} = \frac {1}{2} (\mathbf {K}\pm \mathbf{K^T})$. Notice that in the 1st law only the symmetric part of the tensor shows up, hence the natural inclination to ignore the skew part.

Stokes proved that for heat conduction there are 13 types of crystals and for seven of those types $\mathbf{K_-} = 0$, $\mathbf{K}$ symmetrical. Out of the 32 optical crystal classes 19 are forced to have symmetric conductivity tensor. In only two triclinic classes the skew part $\mathbf{K_-}$ may seem to have any value, and in the other cases at least one component will always vanish. Stokes conjectured that $\mathbf{K}$ mus always be symmetrical but he could not prove so. In the ensuing decades may experiments were conducted to verify it either theoretically or experimentally (Voigt, Curie, Soret, etc.) Gurtin (1969) supposedly proved that a rigid heat conductor cannot be thermally stable unless its $\mathbf{K}$ is also symmetric but these are nontrivial results.

hyportnex
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    Right now your score is 911-1-1-6 including badges. I feel bad destroying the rotational symmetry but this deserves an upvote... – Floris Sep 23 '14 at 02:46
  • @Floris, aww, that ain't destroying — rather, spontaneous symmetry breaking ;] – ulidtko Sep 23 '14 at 15:13
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It's tempting to think that heat will flow more readily across a cavity from a high emissivity surface to a low emissivity surface than in the opposite direction when the surface temperatures are interchanged, but this is fallacious. The reason is that repeated inter-reflection between the surfaces restores symmetry.

Call the radiant flux across the cavity from left to right $W_{LR}$ and the radiant flux from right to left $W_{RL}$, after all inter-reflection has been accounted for. Then if the left and right surfaces have emissivities $\epsilon_L$ and $\epsilon_R$ and absolute temperatures $T_L$, and $T_R$, respectively, the radiant balances at the two surfaces can be expressed as

Left surface: $W_{LR} = \epsilon_L \sigma T_L^4 + (1- \epsilon_L) W_{RL}$

Right surface: $W_{RL} = \epsilon_R \sigma T_R^4 + (1- \epsilon_R) W_{LR}$

On solving for $W_{RL}$ and $W_{RL}$ the net radiant heat flux from left to right is found to be

$$ W_{LR} - W_{RL} = \frac{\sigma (T_L^4 - T_R^4)}{\frac{1}{\epsilon_L} + \frac{1}{\epsilon_R} -1 } $$ which simply reverses sign when $T_L$ and $T_R$ are interchanged.

MartinG
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