# Can a black coating increase the efficiency of a heat dissipator?

I have an aluminium profile that contains too many LEDs and heats up around 70°C, which is about +20°C above my expectation. Would an external black coating increase the efficiency of the heat dissipation by the aluminium?

Possibly. Implicit in your question is the assumption that radiative heat transfer is playing or could play an important role in your configuration (vs. convection). If so, then applying a "black" coating (and thus increasing the emissivity to essentially 1, with the caveat that we're talking about the maximum wavelengths of emission at 70°C) could benefit you. However, note that the coating itself may hinder heat transfer across various interfaces.

I would plug the relevant numbers into the various formulas for heat transfer ($$hA_\mathrm{surface}(T-T_\infty)$$ for convection, $$kA_\mathrm{cross\,section}\Delta T/\Delta x$$ for conduction, $$\sigma\epsilon A_\mathrm{surface\,facing\,surroundings}(T^4-T^4_\infty)$$ for radiation, as described in any introductory heat transfer textbook and at many locations online) to estimate the relative magnitude of convection vs. radiation before attempting to optimize a heat transfer mechanism that might be unimportant.

As an example, the natural convection coefficient $$h$$ can be very broadly estimated to around $$10\,\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{K}^{-1}$$ by order of magnitude (I'm sure exceptions exist in certain geometries). From the numbers you've given, one can estimate that changing the emissivity from 0 to 1 (by anodizing the aluminum, for instance) could potentially boost the outgoing heat flux by a detectable amount, but probably less than 100%. Furthermore, as other posters have noted, you may have other options to increase the outgoing heat flux much more substantially—using fins or a fan, perhaps.

• One thing to remember, for practical purposes, is that the black of interest for the temperature in question is most likely not the same black as human would perceive it with naked eye. Nice depiction is on wiki - see the trashbag example. Looks black but it isn't for body temperature thermal radiation. – luk32 Oct 3 '18 at 11:25
• @luk32 It's actually the opposite for paint. Except for metallic colors, a decently thick layer of paint has emissivity >0.8, regardless of color. The nanoparticles, like TiO2, that provide scattering in pigment are too small to work at thermal IR wavelengths. – user71659 Oct 3 '18 at 22:20
• @user71659 I am not sure what you mean by opposite. I didn't make any statement regarding paint. What I meant is that you can pick any eye-visible color because relevant part of EM spectrum is elsewhere. It needs to be black in IR and that's not what people usually mean by black. I think we agree. – luk32 Oct 4 '18 at 14:44
• @luk32 Opposite in that you point out a trash bag looks black but is transparent at thermal IR, whereas paint can look any color and it is still "black" in IR. (The issue with the trash bag is its too thin; if you get the material thick enough, it works as a good IR absorber... ) Note that by conservation of energy, transmissivity + absorptivity + reflectivity = 1, and the issue with the trash bag is a high T, whereas emissivity is 1-A. – user71659 Oct 4 '18 at 18:21

Yes, a black body will radiate more heat away than a white body. That much is correct.

However, this is true for each wavelength individually. A paint that reflects red but absorbs blue will radiate blue light effectively, but not red light. And the overall effect on heat radiation is the emission of a perfect black body times the absorption coefficient for each wavelength.

Now, what kind of wavelengths does a 70°C black body radiate most? Well, most definitely not visible light. It's somewhere in the infrared spectrum. And that is the problem: Black color says something about visible wavelengths only, not about infrared absorption. You can judge the absorption in the visible band of wavelengths, but you need a high absorption in a totally different band.

Thus, when you buy typical black color, you do not know whether it will increase or decrease the heat radiation of a 70°C object. Of course, you might guess that the absorption coefficient of the relevant infrared band is correlated with that of visible light, but that does not need to be the case. I mean, you can buy blue color, and you can buy red color, and each will absorb the light that the other reflects. Your black color will be white in some other band of wavelengths, but in which?

• Re, "A paint that...will radiate blue light." I don't believe the paint has been invented that could survive the temperatures needed for it to radiate significant amounts of blue light. – Solomon Slow Oct 3 '18 at 1:27
• I once had this very same discussion with an engineer who was measuring temperatures using a thermal camera. He told me that most non metallic surfaces are pretty good black bodies in the thermal-IR part of the spectrum. This would include black paint (and also white paint, BTW). – Edgar Bonet Oct 3 '18 at 7:38
• @EdgarBonet That sounds likely to me, unfortunately I don't know anything about it. So I didn't want to make any unfounded claims. If you have a reference for this, I would be happy to either add this to my answer or upvote your answer should you choose to turn it into an independent answer. – cmaster Oct 3 '18 at 7:50
• @SolomonSlow So true :-) Nevertheless, the red paint does radiate some blue light at room temperatures, it's just totally insufficient to be seen... – cmaster Oct 3 '18 at 7:53

You seem to be wanting to make use of black body radiation. The formula for energy dissipated through black body radiation is $$\sigma T^4$$, where $$\sigma=5.670367(13)\times10^{−8}\ \mathrm{W\ m^{-2}\ K^{-4}}$$. Plugging in $$T=343\ \mathrm K$$, we get $$785\ \mathrm{W\ m^{-2}}$$. However, it would also be absorbing heat from the surroundings. If we model the surroundings as a black body emitting at temperature $$300\ \mathrm K$$, we get $$459\ \mathrm{W\ m^{-2}}$$, for a net of $$326\ \mathrm{W\ m^{-2}}$$. This means that for every square centimeter, you'll get getting around $$33\ \mathrm{mW}$$ of cooling. This is probably going to be only a small fraction of the cooling you need. Most of your cooling is coming from conduction to the air, and then convection within the air, and as others have pointed out, a black coating will likely decrease the heat conductivity of the aluminum. It will likely be more productive to increase the surface area and air flow, and decrease the surrounding temperature.

The modelling of the surrounding as being a black body at room temperature is, of course, questionable, but even without it, you'll be getting only $$78\ \mathrm{mW\ cm^{-2}}$$. (And if it's in an enclosed space, you may be getting less than $$33\ \mathrm{mW}$$, as it will be heating up the surroundings, and that heat will just be radiated back.) Furthermore, it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile. Unless it's in a dark room, it will be absorbing heat from whatever lighting there is in the room. Thus, its net black body heat exchange may be positive, in which case making it darker would make things worse even without taking into account conductivity. Since we're not actually dealing with perfect black body radiation, there is a possibility that you can decrease its albedo in $$343\ \mathrm K$$ range without significantly increasing it in the visible range, but that's rather advanced engineering.

• (1) Where does the factor of $1/\pi$ come from? (2) You seem to say that the surroundings may effectively be at >70°C ("it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile"). How does this work, assuming this is a habitable space? – Chemomechanics Oct 2 '18 at 17:44
• (1) en.wikipedia.org/wiki/… (2) The effective black body temperature is the temperature a body would have to reach for the amount of heat it is emitting to equal the heat it is absorbing. If you have a light bulb at 500K, then you would have to reach 500K to be in thermodynamic equilibrium with it. But if you have ways of losing heat other than black body radiation, such as conduction with the air, then you can remain at a temperature below that. – Acccumulation Oct 2 '18 at 18:01
• (1) The factor of $1/\pi$ applies to the radiance, not the heat flux. (2) The filament of a nearby incandescent light bulb occupies a minuscule solid view angle. – Chemomechanics Oct 2 '18 at 18:05
• I edited the answer to remove the bogus factor 1/π. As pointed out by @Chemomechanics, this arises from a confusion between radiance (σT⁴/π) and radiant exitance (σT⁴). Pending approval. – Edgar Bonet Oct 3 '18 at 9:21
• @EdgarBonet I've rejected the edit because I don't think it's okay to fix such things since the OP explained that that is deliberate. The edit conflicts with the original intent; it isn't for editors to decide factual accuracy. If you believe it's incorrect, it's a better idea to leave comments and vote. – Chair Oct 3 '18 at 10:59

Your heat sink gains heat from its contact with the hot LEDs and looses it to the environment (air) around it. Since it uses convection to lose heat, any coating that has less thermal conductivity than aluminium will act as an insulator and will cause it to get hotter, which will work against you.

As one of the comments point out, you can increase the amount of heat lost to the environment by increasing the surface area of your heat sink.

You can also look for a material with higher thermal conductivity so heat is transported faster inside the material.

• You've immediately assumed that radiation is negligible, but the whole point of the question is about optimizing radiative heat transfer. So I think this assumption needs more detailed support. – Chemomechanics Oct 2 '18 at 17:14
• @Chemomechanics - Read the question again. "Would an external black coating increase the efficiency of the heat dissipation by the aluminium?" There is no restriction to radiative transfer, and in fact most of the heat loss of the heatsink will be by convection of the air. – MaxW Oct 2 '18 at 18:41

There are thermal dissipative coatings which can be used to improve heat removal performance. These have both good IR emissivity and low thermal resistance, and are very different from regular black paint (they are not necessarily black to begin with).

Typically dissipative coatings are only used at high temperatures (hundreds of degrees). At lower temperatures, even specialized coatings only harm heat dissipation by adding thermal resistance.

Temperatures around 70°C are completely normal for modern LEDs, so unless your LEDs came with a datasheet which specifies a lower temperature, or the profile in question should remain cool for other reasons (like contact with skin), I would advise to leave it as it is. If you have to make it cooler, either find a bigger profile, a profile optimized for convection cooling (bigger surface area) or increase the airflow.