paint heatsinks

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Why unbelievable? Shiny heatsink has to lose almost all heat by convection. Black heatsink can use radiation too. Radiation is clearly significant, as otherwise nobody would ever bother to blacken a heatsink but just use a slightly larger one.

I can remember a school lab experiment. Square metal beaker containing hot water. Different sides with different paint finishes. Some form of heat radiation detector - I forget what. Clear differences between sides.
 
Right-o:

From Conrad:

"Coating Material
Textured black polyester powder coating has been chosen as the standard finish on all coated Conrad heatsinks and provides:

a quality, durable and attractive finish capable of withstanding elevated temperatures,
increased thermal dissipation in the order of 5% to 8% (depending on the heatsink) under natural convection."



Fwiw, I am surprised that black polyester (powder coat no doubt) actually is an improvement and not a hindrance, but they must have tested it.
 
Remember that a standard heatsink with fins is not the same as a flat sheet of polished aluminium. The fins increase the radiation - not because of their surface area (which increases convection) but because radiation bounces around between the fins and the main surface. In effect the fins form a surface which to a limited extent behaves like acoustic foam wedges in an anechoic chamber. The linked test found 24% difference on a smooth cylinder, where there is no scope for this effect. A cylinder with a rough surface would show less difference. Fins are just a bad approximation to a rough surface.
 
Fascinating.

I am surprised still that the insulative properties of paint seem to not effect the results.

I read through the flashlight temperature test thread, and thought perhaps the insulation effect was skewing the results. But no, he had imbedded the thermocouples inside holes in the tubes! Good practice. So, those results seem legit. Someone made the case that the difference in temperature made about a 1% diff in light output for those LEDs. But that difference in temperature makes much more difference for hot running devices in audio in many cases.

DF96, my understanding is that the width between fins below a certain dimension limits natural convection. And, there is a limit to the rate of natural "free" convection. So too wide and you reduce the number of fins. What I recall is that the fins *do* increase the surface area, and that is the primary benefit. Radiation bouncing = bread toaster, mutual heating. What we want is a path to the air, and the greater the surface area per BTU of heat the lower the temp per surface area, or the higher the deg/W that a given heatsink in convection can cool. Fins are more than an approximation to a "rough suface", afaik they are a way to substantially increase the surface area which permits higher cooling per sq dimension - which is a practical consideration for real world equipment.

In fact some extrusions have ripples or other surface variations on the fins themselves - again to increase the effective surface area without making the fins longer.
 
Fins surface area increases convection (within reason), as there is more area in contact with air.

Much of the radiation from anywhere on the surface is likely to hit the heatsink again. A flat surface radiates in all direction, not just perpendicular. Hence much of the radiation is 'wasted' and just heats up the heatsink - although this slightly boosts the effective thermal conductance of the aluminium. Imagine the heatsink is all black. Then looking at it all you see is a black area. It is this area which matters for radiation, not the actual surface area. Hence the main effect of fins for radiation is to increase the radiation coefficient, not the area. (There is a small increase in area at the ends).
 
Just another comparable situation that is well known.
Under-hood of cars that use white hi-temp paint on headers keeps under-hood temperatures "cooler" than headers that are painted black....common knowledge.
Imagine if you will, infra-red energies traveling thru a metal mass encountering the mass barrier, ...if it is a shiny "reflective surface" or the next best thing, a bright white surface, will reflect back off. A flat black absorbing best, similar to a "Black body".
It's almost as if it was similar to transmission line effect.....the load absorbing the energy.

_________________________________________________Rick..........
 
Remember that a standard heatsink with fins is not the same as a flat sheet of polished aluminium. The fins increase the radiation - not because of their surface area (which increases convection) but because radiation bounces around between the fins and the main surface. In effect the fins form a surface which to a limited extent behaves like acoustic foam wedges in an anechoic chamber. The linked test found 24% difference on a smooth cylinder, where there is no scope for this effect. A cylinder with a rough surface would show less difference. Fins are just a bad approximation to a rough surface.
you either did not read my post from a couple of years ago or did not beleive it.

The radiation that is emitted sideways froma fin is substantially absorbed by the facing fin.

The net area for the radiation is NOT the fin area. The radiation area is the heatsink "box", i.e. the rectangular box that would enclose the reactangular heatsink.
For a fan cooled heatsink the dominant heat loss is via convection. This is forced convection where much of the flow is turbulent.

In a passive heatsink the convection is less dominant but still very much more than the radiation loss. Very short fins will rely more on radiation. No fins, i.e. flat plate will probably have similar convection & radiation losses.

Out in space, where there is no significant air to flow over the sink surface, the radiation loss is by far the dominant.

I still believe there is something that ABL has not shown us about their "test procedure" to give those unbelievable results.
 
AndrewT said:
The radiation that is emitted sideways froma fin is substantially absorbed by the facing fin.

The net area for the radiation is NOT the fin area. The radiation area is the heatsink "box", i.e. the rectangular box that would enclose the reactangular heatsink.
Actually, that is precisely what I was trying to say. I obviously didn't say it clearly enough for you.
 
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