Black Heat Sinks- Better?

[Whizzard9992]

Ok so this came up in another thread, so I wanted to post about it.

In a nutshell, it was said that painting a HS flat black would increase its thermal efficiency.

I'm an engineer, so my knowledge of physics is mostly limited to electricity. What I do know is that "color" is an indication of the ability of a material to absorb or reflect certain parts of the color spectrum when ambient light hits the material. I can't see how a pigment would make a material radiate heat any better.

Although I'm skeptical, I'm genuinely curious. A heat sink is all about convection and surface area, and I would think that "paining" a heat sink would at least reduce the thermal effectiveness of the sink, and at most have no benefit (Even if the paint was a better conductor than the material, you're still limited by the material's conductivity).



At any rate, can someone fill in the blanks for me?

(Original Thread - Posts by NovemberWind and SirHeck) *edit*

 

[ajfink]

If it was laser-pitted black, then maybe, but not paint. Maybe black thermal grease.

 

[Whizzard9992]

Ah, yeah, I remember reading something about that.

All metals can turn black - Article on ZDnet

Perhaps this is a rumor born from the above article.

 

[Whizzard9992]

(<-- Correcting myself)

Ah. Pitting is different than the article I read...

 

[jaguarskx]

I'm no engineer or physicist, but wouldn't the black paint act as a layer of insulation, thus slightly reducing the heatsink's ability to radiate out the heat absorbed from the CPU?

 

[pignoli]

Whilst black paint would radiate the heat better as a function of it's colour, yes I'm pretty sure that that effect would be negligible compared to the insulating properties of a layer of paint.

The laser pitting looks like a far more promising prospect: If I'm interpreting that article right, it could not only turn the metal black, giving it better radiative properties by way of colour, but also dramatically increase the surface area of the treated metal. If it were possible to get an appreciable amount of air circulation around these microstructures, it could increase the cooling capacity of the material by a large amount. The problem would be in getting the air to flow around such tiny structures: at this kind of scale the viscosity of the air itself would probably prevent the extra surface area being useful and could even trap a layer of air (think of fur) at the surface of the metal, insulating it immensely. These are problems very similar to those pointed out in another thread about immersion cooling.

 

[ajfink]

Whilst black paint would radiate the heat better as a function of it's colour, yes I'm pretty sure that that effect would be negligible compared to the insulating properties of a layer of paint.

The laser pitting looks like a far more promising prospect: If I'm interpreting that article right, it could not only turn the metal black, giving it better radiative properties by way of colour, but also dramatically increase the surface area of the treated metal. If it were possible to get an appreciable amount of air circulation around these microstructures, it could increase the cooling capacity of the material by a large amount. The problem would be in getting the air to flow around such tiny structures: at this kind of scale the viscosity of the air itself would probably prevent the extra surface area being useful and could even trap a layer of air (think of fur) at the surface of the metal, insulating it immensely. These are problems very similar to those pointed out in another thread about immersion cooling.

That was my thinking as well. I think manufacturers of high-end cooling should take a look at it.

 

[Nitro_mule]

The original poster is partially right: a blackened HS surface has higher emissivity, which is the degree to which an object produces far infrared radiation (=heat radiation) compared to a perfect blackbody.

However, the energy radiated from the HS is also determined by the temperature of the 'receiving' surface, which is the inner surface of the computer's case.

You can 'easily' compute the radiation by yourself, it's

Q = 5.67e-8 x A x e x (Th^4 - Tc^4), Q in Watt

with A (in m²), the 'exposed' surface of your HS, this does not include the surface between the actual cooling fins,
e, emissivity which is equal to 1 for a perfectly black HS (i.e. carbon black)
Th, heat sink temperature (in Kelvin)
Tc, case surface temperature (in Kelvin)

Taking some standard values for these variables, I only get a heat dissipation due to radiation of 1 to 5 Watts.

Which is fairly negligible, considering your HS has to pull > 100 Watt under full load.

 

[balister]

The only way a black heatsink would help is if you are totally depending on radiative heat transfer. Today, most heatsinks in computers are using convective heat transfer (fan blowing air over the heatsink). In the case of where you are using convective heat transfer, it wouldn't matter much if the heatsink was black, silvery, or puce.

 

[Paperdoc]

The only way a black heatsink would help is if you are totally depending on radiative heat transfer. Today, most heatsinks in computers are using convective heat transfer (fan blowing air over the heatsink). In the case of where you are using convective heat transfer, it wouldn't matter much if the heatsink was black, silvery, or puce.

This is EXACTLY right! And yes, the paint's thermal transfer rate VERY likely would make it act more like an insulator, completely negating any advantage for radiant heat dissipation (which is, as others have said, only a minor contributor to overall heat removal).

 

[dasickninja]

I think the reflective silver works best for now.

 

[chuckshissle]

Better off buying a small low rpm fan can a can of spray paint.

 

[Whizzard9992]

Thanks for the input.

I'm genuinely curious, but this is what I've found:

http://en.wikipedia.org/wiki/Emissivity

No flaming wiki, if you please.


A black body isn't something that's necessarily black; it's something that absorbs all electromagnetic radiation, and appears black as a result. Painting something black only means the surface will absorb more of the visible spectrum, and doesn't make the heat sink itself more of a "black body," and thus it shouldn't increase the amount of emissive radiation by any substancial amount.


My personal opinion is that this is a misinterpretation of physics rules, more specifically what defines an object as a "Black Body." Feel free to correct me if I'm wrong.

 

[kwalker]

The only way the black paint could aid in cooling would be if the sink was hollow with vent tubing exhausting outside of the case.
The heat transfer would cause an updraft effect such as in black solar chimney piping for natural solar venting.
This is dependent on the titanium content in the pigment for solid heat generating objects.
It has been used in the automotive industry for the past 20 years to aid in the thermal stability of the internal combustion engine helping to keep the temperatures at an optimum operating temperature of 223 to 227 deg F.
As jumping jack stated black absorbs heat energy and white reflects.
Under the right circumstances this theory can be applied.

 

[HotFoot]

Black will reabsorb radiative energy, what you want is to reflect that away --- so shiny and white is the best means of doing so....

This is incorrect. Black does absorb heat better, but it also radiates it better. It's all based on emissivity. Shiny white objects have low emissivity, and dull, more black object have high emissivity. Emissivity is to radiative heat transfer and conductivity is to conductive heat transfer. This is why the SR-71 Blackbird is black: because it flying at Mach 3.5 the air no longer cools the jet, but actually heats it (beginnings of hypersonic effects). Radiative heat transfer becomes the most effective way for the spy plane to keep "cool". The same goes for the X-15 and the heat shielding panels of the space shuttle.

As for a black heatsink, I don't think that it would be effective. Effective radiative heat transfer generally requires very large temperature differences between the object and ambient. You certainly don't want your heatsink to be that hot. Convective heat transfer is the norm for CPU cooling because it works very well.

Other posters noted about the laser pitting to make a metal black. This is a great idea, but I'm sure it would be expensive, at least for a while. The black colour might not give you a huge boost in cooling effectiveness, but it won't hurt. As for increasing surface area of the material with the nano-scale pits, the pits are on the order of the wavelength of visible light, which is too small for the fine structures of turbulence to fit into. Once an object is below the Kolmogorov scale, it will have no effect on the flow.

This brings me to a different idea. In wind tunnels, sometimes it is necessary to "trip" the flow from laminar to turbulent in a test. One way to do this is to create a region of small bumps, about the size of a grain of sand, over which the flow must go. This may be quite effective for CPU coolers, as turbulent flow has a much greater effectiveness for convective heat transfer.

 

[HotFoot]

Sigh... I can see where you're going on the wrong path here. Putting the objects in sunlight exposes them to an environment with a high radiation energy... I'll say density. Black does absorb sunlight to get hotter better than white (or better still a mirror-finished piece of aluminium.

In the case of the CPU, it is not sitting in sunlight. The temperature of the CPU is higher than the vast majority of objects within line-of-sight of the CPU cooler, so that the balance of radiative heat transfer will be away from the cooler. It's all about equilibrium (you don't need university for this, high school will do). While heat is absorbed better by a black object, it is also radiated away faster. The heat radiated by the body is proportional to:

eATobj^4

where e is the emissivity, A is the surface area, and Tb is the absolute temperature of the object.

The energy absorbed by radiative heat transfer is proportional to:

eATs^4

where Ts is an averaged value for the temperature of the bodies within view of the object.

Therefore, the energy balance is the difference between the two, which amounts to ea(Tobj^4-Ts^4). This is the equilibrium part. If you put something that's at room temperature in view of the Sun, the overwhelming radiation coming out of the Sun will increase the average Ts so that the balance is heat being absorbed by the object. If the object is shaded from the Sun and it is warmer than it's surroundings, then it will emit radiation heat. The latter case describes the situation in which you should find your CPU, so the higher the emissivity of the cooler, the better it will be for radiative heat transfer.

All that said, radiative heat transfer will be an order of magnitude weaker than convective heat transfer.

 

[Pepicek]

Its all about surface area. That's what laser pitting does. You can also get a very dramatic increase in heat transfer in cutting up your fins. The computers on the Space shuttle (the one responsible for engine - fuel management) have procupine fins.

Sorry this is not a direct answer, but one I think you will find more effective.

 

[HotFoot]

Neither of your experiments demonstrate how a black-coloured body doesn't emit more radiative heat energy than a white-coloured body of the same temperature. Here's an experiment to prove that it does.

Take two bars of the same material, preferably something that can absorb a lot of heat. Paint one white and one black. Heat both bars to the same temperature in an oven, and then stand them on a mat or towel in a dark room. Measure the temperature of both over time, and you will find the black one cools faster.

A counter-example is a good coffee thermos. What keeps the coffee cool is a piece that's generally metal or glass that sandwiches a vacuum chamber. The vacuum cuts out conductive heat transfer, which is great. But the tube is always polished in the good thermoses to a near-mirror finish to keep radiative heat transfer to a minimum as well.

I'm quite sure that the folks at NASA and Lockheed's Skunk Works didn't have their heads up their A**ses when they went through all the trouble to make the X-15 and SR-71 black instead of just buffing up the metal.

 

[HotFoot]

See my edit above for emissivity (the constant your Raleigh-Jeans equation) and you will see that color (black, white yellow or red) has nothing to do with the effectiveness of an object to emit.

From that Table, under Paints:

Black, e=0.96 (I imagine this differs between flat, glossy, etc.)
White, e=0.95 in one case

Lower down on the page...

Oxidized Aluminium, e=0.11. (The range for e is from 0 for perfect reflectivity to 1 for perfect absorption).

It seems to me that it might not really matter what colour paint you choose, anything will dramatically increase the emissivity of a cooler (mine's got aluminium fins). In light of this, don't use white paint for the experiment with the hot rods. Use something with a metallic shine.

 

[HotFoot]

You are still trying to treat the color black as a black body.... this is incorrect, a black body radiator is a hypothetical beast.... furthermore, your T^4 law is actually incorrect, as mentioned by the ultraviolet catastrophe above. It only works for temperatures above the threshold to emit in the black body range. Below 200 or 300 deg C, emission is not the predominant heat transfer mechanism.

I just thought of a good counter-example to the high-temperature requirement you state for radiative heat transfer: the Earth. Radiative heat transfer is the only method for the Earth to balance it's temperature, since there's no conduction and no convection worth mentioning between Earth and space. If the Earth weren't able to radiate heat (mean temperature something like 15C), then the heat gained from the sun would continually build until the Earth reached a much higher temperature. It would have to be 200 to 300C to emit heat.

 

[HotFoot]

So what you're saying is that an object of a certain colour, or material, whatever your determining factor is, will be more effective at either absorbing or emitting radiative heat, rather than have the same vale of emissivity for both cases. Also you're saying that radiative heat transfer is not in agreement with the ~eA(T^4) until a fairly high temperature is reached.

This is a little far-fetched for me to believe. Can you explain my previous counter example(s)?

 

[HotFoot]

I thought I should put up some numbers so people can see the effect of changing emissivity on something like a CPU cooler. This should demonstrate that while emissivity increases can improve thermal radiative heat transfer, the total cooling power of radiation is an order of magnitude below the cooling provided by convective heat transfer.

All of this work is done on the basis of what is taught in a university course on heat transfer. If people want to try to prove that what universities are teaching in classical physics courses is so inaccurate as to be unusable, you should take that up with the Engineering accreditation council or other authority in your area.

I will use two coolers as examples. Both coolers have an effective surface area of 0.08 square metres. One is aluminium, for which I've used an emissivity estimate of e=0.11. As Jack says, e is a function of temperature, but it takes a large change in temperature to create a small change in e, so I'll be using this linearising simplification. The other cooler is painted or in some way made black, or as it turns out pretty much any paint colour will do, and I'll appoints its emissivity to be 0.95. Effective surrounding temperature will be 300K.

For the first one (aluminium surface), consider the case where the cooler temperature is 50C (323K). The power of radiative heat transfer emitted from the object is then 5.43 Watts. The power of radiation absorbed by the object is 4.04 Watts, so the total radiation cooling power is 1.39 Watts. This is pretty insignificant compared to the TDP of a CPU. If the cooler's temperature is reduced to 40C, the total radiative cooling power is reduced to 0.75 Watts.

For the second cooler (black surface), at 50C the power out is 47.4 Watts and the power in is 35.3 Watts, for a total balance of 12.1 Watts of radiative cooling. At a cooler temperature of 40C, the total radiative cooling is 6.5 Watts. Again, even with very high emissivity, this cooling is inadequate for a CPU.

 

[misiu_mp]

Painting anything black will make it behave more like the theoretical black body. Black is black because it absorbs radiation and that is exactly the main attribute of a perfect black body.


Imagine one black and one shiny body. Put them in a room at say ... room temperature . After a while they will reach thermal equilibrium.
Now as we all agree the black body absorbs more radiation than the shiny body. Still it stays at the same temperature as the shiny body. The only explanation is that it EMITS more radiation then the shiny body.
Assuming emissivity is a property of the body and it is bound to it and constant, whatever the temperature, we can conclude that a black body will always emit more radiation than a shiny body at the same temperature.

JumpingJack, you say that radiation is not dominant at lower temperatures. Thats true but it is also valid for absorbtion. If not much is radiated, not much is there to absorb.

 

[HotFoot]

I never stated that radiative heat transfer would be adequate for CPU cooling in a PC. I stated the opposite. However, our disagreement was on the laws of physics. Yes, classical physics aren't exactly right. F does not exactly equal MA. But you have to be in pretty extreme circumstances to notice the difference between what's really happening and what Newton and his peers wrote.

I've merely been trying to defend the physics as I learned from my teachers and textbooks.

 

[HeavyF]

Absence of evidence is not evidence of absence.

Stop worrying about the bullsh*t color of your heatsink and get laid. For the love of Garrett Oliver, won't somebody think of the children.