[bigben2k]

I've been purposely avoiding this thread, because I didn't have the time to post a coherent reply, but ya'll have had my attention.

As Bill pointed out 2 pages back, look at the temperature gradient.

The power that a CPU emits is what we want the waterblock to capture, and take away. Is it a 100% efficient process? no. Some heat is being dissipated by convection to air directly from the CPU (wether it's through the side of the core, of through the substrate. We recently went over this at OCAU, and even the block is only 97-98% efficient.

The actual heat source is imbedded within the silicone die and as such, is a layer all on its own, in the transition of the heat into the water.

Each transition layer will have a temperature range, or differential, from the hot side (CPU side) to the cold side (water side). If there were no temperature gradients, the materials would be perfect heat conductors, and I hope everyone can accept that there's no such thing.

This temperature gradient is measured or specified by its C/W rating, which is a measurement of the heat resistance of the material. In our cases, this unit applies to the whole block, where normally, one would use the same concept applied to a specific thickness of material.

So for a 70W source (feel free to deduct the 2-3% inefficiency), you can literaly calculate a temperature differential of 13.3 degrees C, in the case of the best known block: Cathar's White Water (according to Bill's testing, under a specific set of variables, where the thermal resistance measured is 0.19 C/w).

The other layers are composed of:
-the TIM joint (included in Bill's measurement, at 0.05)
-the water
-the radiator
-the ambient air (the ultimate destination of the heat)

By increasing the flow rate, you decrease the thermal resistance of the water (or more specifically, the thermal resistance between the waterblock and the water), as well as the resistance within the rad (i.e. water-to-rad resistance).

The total power (aka heat) travels through a series of resistances, to eventually be dissipated in the room. 70 Watts is 70 Watts, no matter how you look at it. (but myv65 will point out the ever so important additional heat source, provided by the pump).

If some of the individual resistances are lowered, the total resistance is lowered, since they are all in series (i.e. follow each other).

If an individual resistive layer has been improved, it will have a lower thermal gradient.

If you add all the thermal gradients, you get the temperature differential, between the CPU and the ambient air.

Since the ambient air temp is not going to change (for the sake of the discussion), you can simply add the thermal gradients to it, to find your result: your CPU temp.


What's interesting to note throughout all this, is what Tom observed, when he purposely burned an AMD processor (by removing the heatsink altogether): the die temp reached temperatures in excess of 350 deg F. That's your 70 Watts of power going through the die, and trying to make it into the ambient air, but with such a high resistance, that the gradient was too high for the CPU to handle.