Copper Heatsinks

Gepetto said:
An interesting math problem...one of those that your 8th grade math teachers gives you in a word problem.

Suppose each hole that you drill is 0.25" in diameter into your 1/8" thick heatsink. The surface area that you remove from both sides of the heatsink as a result equals 0.0981 square inches. The surface area that you add on the circumference of the hole you drilled is 0.0981 square inches. Net zero on radiant area and you have lost conductivity through the heat transfer path.

Next, suppose each hole that you drill is 1.00" in diameter into your 1/8" thick heatsink. The surface area that you remove from both sides of the heatsink as a result equals 1.57 square inches. The surface area that you add on the circumference of the hole you drilled is 0.393 square inches. Big net loss on radiant area and you have lost conductivity through the heat transfer path.

Next, suppose each hole that you drill is 0.125" in diameter into your 1/8" thick heatsink. The surface area that you remove from both sides of the heatsink as a result equals 0.024 square inches. The surface area that you add on the circumference of the hole you drilled is 0.049 square inches. Net gain on radiant area and but you have lost some conductivity through the heat transfer path.

The answer is the usual engineering answer, "that depends" :)

The rule is that if the hole radius being drilled is less than the thickness of the material being drilled through, you win on surface area. Else you lose.

Also remember that in still air, the topside half of a drilled hole is not as effective as the bottom half.
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Thanks for all of that, still kinda goes over my head as math was not something I excelled in. Hope to see more on the copper fins too. So " it depends" is the definitive answer as with many things. It will take me a day or two to get in my head.
 
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