They way I consider the efficiency of waterblocks/loops and so on is as follows.
As Cathar has stated, the CPU is giving out a heat load, lets say 90W. For a given flow rate, the rad will be able to dissipate 90W for a particular temperature differential between water and air. The water should stabilise at this temperature.
Now we can assume (roughly) that we have water at this temperature flowing through the block. The block has a C/W ratio. Lets say 0.2 for a good block. This implies that for every 0.2 of a degree difference in temperature, 1 watt of heat will be transferred to the water. So for 90 Watts, you would need a temp differential of 18 degrees between the water and the block.
A more efficient block would be able to transfer 90W from the die with a lower temp differential, lets say 0.17 C/W ratio.
There would now need to be 15.3 degrees difference between the water and the die.
However, you are still only getting 90W of heat transferred into the water, which if the flow rate is the same as before, will raise the temp of the water by the same amount, and for the same flow rate, the rad should still be performing the same.
I'm sure it then gets complicated because the cpu will most probably clock higher, introducing more heat into the water raising it's temperature. But you would have changed the conditions for comparison.
Now consider raising the flow rate. The rad will perform slightly more efficiently, (lower C/W ratio between water and air for rad), thus for the same heatload, the water temp should be slightly lower, but only very slightly, as the C/W ratio will not change much. The block should now also perform with a lower C/W ratio, thus decreasing the temp difference between the water and the block required to remove 90W.
It is my understanding that people often think of this the wrong way round.
If you want to transfer 90W from the cpu to the air (via water cooling), you need to consider the following stages, in this order!
(Consider a constant flow rate - constant C/W ratios in block and rad)
1. How much warmer wil the water have to be than the air blown through the rad, for 90W to be transferred.
90 x C/W for rad = dTaw
(dTaw is temp difference between air and water).
2. For the same flow rate and given C/W ratio for block, how much higher will the temp of the cpu have to be than the water in the block, for 90W of heat to be transferred to the water.
90 x C/W for block = dTwc
(dTwc is temp difference between water and cpu).
So if the ambient temperature is A, then for a given flow rate, the temp of the cpu should be A + dTaw + dTwc.
This is very approximate as it assumes no other sources of heat in the loop, and assumes that the temp of the water doesn't vary throughout the loop, which for a high flow system is not far off.
However, the general principle still stands, If you are working from a given ambient air temp with a given heatload, you need to work back fro the air to consider how much hotter each stage will have to be to allow the transfer of 90W, until you reach the cpu. Add these up and you can get a rough estimate of the expected CPU temp.
The main point of this however, aws to explain why a more efficient waterblock, does NOT raise the temps of the water, merely the temp difference between the water and the cpu. IE, lowers the CPU temp.
8-ball
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