(Ed.note: You’re going to see some lopsided pictures; those using smaller resolutions may have to scroll. Normally I resize them, but
if I did that in this case, you simply couldn’t read any of the graphs)
I’ve worked out some possibilities for cooling CPU’s.
I’m an master-engineer in metallurgy, so i have access to some good simulation
software for this problem. (FEMLAB, which works under MATLAB)
First, some assumptions:
degrees C don’t matter in the calculationes.)
Now: we are only left with two bottlenecks: The thermal conductivity of the
heatsink, and the thermal coefficient between the heatsink, and the air.
As we know, less airspeed on the fins requires more surface for
the air to flow over to remove the heat equally efficient. So I’m going to find
out if the thermal conductivity of the heatsink is big enough to spread the heat
out to a large enough surface.
Since the needed surface area (on the fan side) of the heatsink is 1/(relative
airspeed), I will only simulate the temperature differences in the heatsink itself.
As a start, and to confirm I’m using the program correctly, I did a very simple
A copper cube of 1cm^3, 100W flowing in from the top, and out at
the bottom. The sides are isolated, so this is a very easy calculation to
confirm by calculating by hand.
The thermal conductivity of copper is 402W/(cm*K) and should give about 0,25 C difference between the bottom and top.
The simulation looks like this:
This looks correct with a 0.249 C temperature difference (mesaures in meters and
To make further calculations easier for this computer (I only have 256MB ram,
and FAMLAB needs HUGE amouts of ram; wish I had a 1GB or so.); I’m going to
divide the calculation in two parts: the heatspreader-base, and one of the fins
connected to it.
OK, it’s just a sketch which is out of proportion.
Imagine the 1cm^3 “cube” of copper to be a part of the heatspreader and placed
over the CPU-core. Now the cube would conduct heat out from the sides (to the
surrounding material) Since the sides are bigger then the top, i would expect a
lower temperature difference between the CPU-core and the top of the base.
Surface Temperature at the bottom of the base
Surface Temperature at the top of the base
Temperature ISOs in the base
Heatlines in the base
Pretty? I love all these nice visualisations.
Note that the temperature differences in the base is only about 0.1C.
The fins? I assumed there is 20 of them on top
of the base, each removing 5W. The dimentiones are 1x40x40mm, with 1 mm space
between them. I also assumed equal heat removal (W/cm^2) all over the two sides. Here
is one of the fins:
Temperature of the Copper Fin
The temp-differance is 1.0C (I’m sorry i made it upside down, but the
heat-physics still apply)
If we sum it up, the total temperature difference in the heatsink would be 1.1C
between the hottest point and the coldest. That’s not much at all.
Even if we used aluminium, the temperatures wouldn’t rise much. Actually it proportional
with the heat conduction: deltaT* (K Copper/K Aluminium) = 1.1C * (401/237) =
From this we learn that the heat conduction in the heatsink is not a big
bottleneck in heat removal. It’s so small that i can assume the same
temperatures in the whole heatsink for further calculationes.
This shows that the major bottleneck in a cooler is the transport of heat from the fins
to the flowing air.
It’s not coincidental that an
upgrade of the fan can lower the temperatures considerably, or by closing
close the heatsink and letting water flow through it.
To optimise weight vs. cooling capability, I would
have made these changes:
If someone wants to make a totally fanless PC, airflow is a
must. While natural convection can work, it needs a huge surface area, and
thus a huge heat sink. Such huge heatsinks (30×30 cm or so) may work, if you
have ok air circulation around it.
The temperature drop in the heatsink it self
will not be a huge problem. A few degrees C DeltaT in the heatsink is ok, but
you need a huge surface area and good aircirculation. Altso, to make the
convection as good as possible, the fins shuuld be placed vertically.
I will do some calculations of a “new model” with new dimentions (6x6cm
and 30x30cm. If i can figure out the aerodynamics and
heatconductinon copling in FEMLAB (and get myself some more RAM) i might even
come up with some actual simulations on a full heatsink and fan-combo, and a
megaheatsink without a fan.