Air can contain heat. The amount of heat in a certain volume of air is given by (Linearized, which is "good enough" for engineering):
Q=c_V * V * T
c_V in this formula, is the c_V of a monoatomic (3/2), diatomic (5/2), etc coeficient at constant pressure used on basics ideal gasses formulas?
The value does not matter. It just needs to be reasonably constant.
In the formula you assume V is the V in litres of your case, isn't it?
Well... No, rather an arbitrary volume around the heatsink where the temperature is reasonably uniform
Also, if the air surrounding is hotter than the heatsink, by the equilibrium of temperatures theory between 2 objects, the surrounding air will transfer heat to the heatsink although the sorruonding air is moving. Am I correct? So, the lesser temperature (then less Q it has) the airflow through the heatsink has, the bigger amount of heat (Q) the airflow can carry out (evacuate) from the heatsink.
Whoa! If your intake temperature is hotter than the heatsink, please take the case out of the oven! I think it is safe to assume that the air is cooler than the heatsink
But actually, heat transfer works more efficiently/faster when the temperature difference is greater. Therefore you want a big temperature difference between the air and the heatsink, which you can by either providing cool intake air (thus the importance of ducts), or by letting the heatsink run comparatively hot (as the graphic card manufacturers do).
As mentioned before, once dynamic equilibrium is established, your fans must move the heat generated by the processor, thus Q/t is fixed
As far as I know, copper has better thermal conductivity than aluminium and actually, then most of modern heatsinks use heatpipes based on copper. The little I know about them is that when your heatsink reaches the maximum amount of heat he can conduce by itself, the extra heat is not transferred to the heatsink, thus it is retained by the CPU's capsule, growing the temperatures of the CPU's capsule.
So far you are correct.
So establishing an airflow through the heatsink will have the ability to carry out so much Q as the equilibrium between the 2 corpses can. I mean, if your heatsink is at 80ÂºC and you have airflow at 0ÂºC, then the equilibrium between these temperatures are 40ÂºC, so at a constant 80ÂºC by the heatsink, the 0ÂºC airflow will move out 40ÂºC. I don't know what the exact formulas are because I have not studied so much about thermodinamics, so I can assume it is not linear but I think it works someway as I have said, does it not?
You might be confusing static and dynamic equilibrium. If you would bring two exactly equivalent systems into contact, and one had 80Âº and one 0Âº, the static equilibrium would indeed result in both systems having 40Âº.
The dynamic equilibrium I meant is that the heat flow out of the system (the case) is equivalent to the heat per time generated within the system (due to the CPU). For the argument I made above, the exact manners of transport are irrelevant.
Regardless, I'll try to explain what I vaguely remember from lectures long past - if there are any true experts here, please chime in
In the neighbourhood of the heatsink, a heat transfer process takes place between the metal and the surrounding air. These processes are more efficient/faster the greater the temperature difference between air and the heatsink metal is. In the beginning, when we switch the CPU on, there is no temperature difference and thus no heat transfer. Thus, heat builds up in the CPU/heatsink, and raises its temperature. Finally, if we wait long enough, the heatsink will reach a certain "dynamic equilibrium" temperature so that the air can actually carry all the heat away (and if that temperature is too high your CPU fries).
What that temperature exactly is depends on a lot of boundary conditions, material constants and whatnot in complex ways. But it should scale at least approximately linear with the amount of air that comes into contact with the heatsink surface during a given time frame, which is essentially the argument given above.
Ok if I have not misunderstanded, the more airflow (at less temperature) is moving around the heatsink, the more Q it can evacuate from the heatsink. So less airflow => more T at the heatsink.
The last thing I want to ask you is the following: I've read that the nearest the fins of the heatsink are, the more difficult for the airflow is to move. So, if you have closer fins, you have to blow air at higher pressure in order to circulate it through the fins. More pressure usually means more RPMs from the fan so higher noise. Then, a combo of a heatsink with the 'proper distance' between its fins + a common fan, makes better scores in temperature than a heatsink with closer fins at same fans' speeds. Am I correct?
In principle, yes.
Yet.... Well, if you have closer (=more) fins, the contact area between the air and the heatsink is bigger. Thus, heat transfer from heatsink to air is more efficient, so you could theoretically get away with a lower equilibrium air current through the heatsink than with more sparse fins. Thus, more sparsely spaced fins are not always better, because at some point the contact area becomes too small for efficient heat transfer.
But then again, more fins make it harder for the air to move, you need more pressure for the same airflow, and if you missed the "optimal" spacing (which probably also depends on the current temperature of air/heatsink), you will not reach the equllibrium current with the same amount of fan RPMs.
Now, the problem is that the pressure and air current a fan can generate depend on the geometry of the fan (oh, and generated sound depends on the geometry, too...). Due to this, it is hard
to find the optimal fan/heatsink design. Which is why we are here, and the SPCR guys measure these things for us