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So the radiator really does have more energy to transfer. It must, as that is the only way to reduce the temperature of the CPU.
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And therein lies your problem.
That statement is WRONG.
The processor has a given output of heat. Lets say 70W, for the sake of argument. If you had an infinitely efficient radiator, and an infinitely efficient waterblock, the processor would be at ambient temperature, BUT THERE WOULD STILL ONLY BE 70W BEING TRANSFERRED INTO THE WATER AND ONLY 70W TRANSFERRED INTO THE AIR BY THE RAD!!!!!
This is the key principle to understand.
If you don't understand this then there is no point in me carrying on.
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Furthermore, the decrease in radiator temperature means that the radiator/fan relationship is not quite as good as it used to be.
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Again, I believe you may be missing the point some what. For a given CFM there will be a C/W ratio for the radiator. Rather than saying the radiator will be cooler, how can it be? This implies that heat is being transferred from the fins of the radiator to the air at greater than the efficiency rating for the rad.
The air temp is fixed - yes?
The C/W ratio at a given cfm is fixed - yes?
The heat load is fixed at 70W - yes?
Then the fins of the rad WILL be at ambient temp + 70 x C/W (at the given CFM)
For a 70W load this is the equilibrium position and will not change!
This will not be affected by a change in flow rate, however, a change in flow rate will change the C/W ratio for heat going from the water to the fins.
We have already established, that the fin temperature is dependent on the fixed heat load, the fixed ambient temp and the fixed efficiency, resulting from a fixed CFM. Lets call this Tf, for fin temperature.
Now lets start with a given C/W for a given flow rate. The average water temperature in the radiator should be such that the temperature difference divided by the C/W is equal to 70.
In other words, the average water temp in the rad needs to be;
Tf + C/W x 70
Now lets improve the efficiency. An improved efficiency is a lower C/W ratio, agreed? So lets call this (C/W - D). D for difference.
So now, the average water temp in the rad will have to be;
Tf + (C/W - D) x 70 = (Tf + C/W x 70) - (D x 70)
Can you now see that the average water temp from an improved efficiency is now lower than before by an amount (D x 70).
Granted, D is small, probably of the order of 0.01. However 0.01 x 70 is of the order of 1 degree difference in average water temp throughout the loop.
Now I will concede that the water temperature does not vary linearly as it passes through the rad, but I have taken the liberty of evaluating the temperature change removing 70W from a water flow at 6lpm. which is roughly 1.5 gpm.
It turns out to be 0.16 degrees.
We know therfore, that the average temperature of the water has a margin of error of 0.16 degrees.
Dropping this by 1 degree will reduce the temperature of the water going into the block, which will in turn reduce the temperature of the cpu.
Sorry the last bit's rushed, but I've run out of time.
Any further questions or points you would like to disagree with, carry on.
8-ball