0
$\begingroup$

I mean, it's the ratio of actual heat transfer to heat transfer if the material had infinite conductivity.

How is it of any practical, or even theoretical use? How does it help, in saying one fin is better over other?

Improve this question
$\endgroup$
2
$\begingroup$

If the material has infinite conductivity, the entire fin would be at the base temperature.

This represents an ideal scenario. If the temperature of the fin were the same as the base all the way through, that would mean the heat transfer between the fin and cooling medium would be a theoretical maximum. In reality, you will lose some of that efficiency, because the imperfect conduction means that the further from the base the fin is, the lower the temperature, and thus heat transfer is reduced.

There's some level of tradeoff between increased area of fins, and the decreased temperature due to non-infinite conduction; but the "ideal" fin the efficiency compares to does not have that tradeoff.

Improve this answer
$\endgroup$
  • $\begingroup$ So, if i get it correctly, "if my efficiency is low, parts of my fin are redundant"? $\endgroup$ – Rohit Oct 10 '19 at 17:29
  • $\begingroup$ @Rohit I don’t think redundant is the right word. They likely still contribute, just not near as much as they could if they were more conductive. You could probably increase the efficiency by using a more conductive material, or shorter fins; but that still may reduce the overall heat transfer from the fins. $\endgroup$ – JMac Oct 10 '19 at 17:34
  • $\begingroup$ so, "increasing the extension of the extended surface gives diminishing returns". and "instead of one long fin, it's a better idea to get more than one shorter fins, for the same use of the material"...? $\endgroup$ – Rohit Oct 10 '19 at 17:40
  • $\begingroup$ @Rohit Those generalizations likely depend on the size, temperatures, and surrounding fluid conditions. I would be wary to make any broad general rules like that. The first part may be true; but I'm not even sure if it is universal, or only applies after some critical length. $\endgroup$ – JMac Oct 10 '19 at 17:52
0
$\begingroup$

Well, think about a heat sink with one small fin. You could model the heat transfer to a convecting fluid with a bit of work.

Now think about a larger heatsink with multiple fins the same size as the first one. You'd expect it to transfer more heat to the convecting fluid, right?

So we can simplify things a bit by assuming that the material of the heat sink conducts perfectly. So now it is only the size and shape of the heatsink that governs how much heat is moved. Now we can compare two heatsinks and see which one moves more heat. That ratio is just a convenient way of doing this.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.