[Dix Dogfight]

Alot of the dabate here has been around the issue of compairability of the different tests across the globe.
The problem with compairing person X results with Y's is the calibration of the measurement setup.
I've bee thinking about it for a while and this is what I've come up with as a part of a solution. Using a die-simulator and doing the measurements this way:

We want to measure how "good" the thermal interface between the cpu and WB is.
The problem with measurement of a WB is to know the exact amount of applied power and the effect of the pump/rest of system.
Therefore we must calibrate the setup.
First measure the exact amount of water in the system, se picture 1, we call this volume V.
Then run the pump without the cpu simulator on. For a half hour or so and measure the temps every 30-60 seconds.
The total time is depending on the amount of water and the C/W of the box/hoses. But the important part is that you measure untill the temp is stable.
This gives you a calibration curve/plot to correct our measurement with. The plot will/should look something like picture 2.
Now we can calculate how much power, P_pump, the pump adds to the water by using the slope, K_pump, of the curve in the beginning of the measurements and the measured amount of water V.

P_pump = K_pump * theta * rho * V (Watts)

(rho = 1kg/litre = desity of water)
(theta = 4180 J/(kg*C) = specific heatcapaciy of water)

A typical pump (eheim 1250) consumes 28W. With a setup with one litre this should give us a slope of K_pump=0.0067 C/s.
1/K_pump=150 s/C = it takes 150 seconds to heat one litre of water one degree celsius.

Now wait untill the water reaches ambient again and redo the temp measurement but with the cpu simulator on and measure the power it consumes.
P_cpu=U*I

Now we know the total applied power P_total=P_pump+P_cpu and the WB C/W of the system is easily calculated using the equilibrium temp at load.
If we take the slope in the beginning of the load measurement, K_load and subtract K_pump we get K_wb.
To calculate the C/W of the WB we need the temp at the cpu simulator an ambient temp:

(T_cpu - T_ambient)/((K_load - K_pump)*4180*rho*V)=C/W of the waterblock


If we want to compare different measurements from different places/persons we do:
EXAMPLE
Lets say person X measures a MAZE and a SPIRAL and person Y measures a MAZE and a SWIFTECH. Both persons has done the measurements the described way.

Person X:
MAZE_X=0.40C/W
SPIRAL_X=0.35C/W

Person Y:
MAZE_Y=0.55C/W
SWIFTECH=0.50C/W

Which is better SWIFTECH or SPIRAL?

SPIRAL_X/MAZE_X =0.40/0.35=1.14
SWIFTECH_Y/MAZE_Y=0.50/0.55=1.10

Since the SPIRAL har a relative difference that is bigger than the SWIFTECH the the SPIRAL is better.
(PLEACE NOTE THAT THESE NUMBERS ARE MADE UP AND IN NO WAY RELATE TO THESE WATERBLOCKS, i simply stole the names to make it easier to read)



The same setup can be used to measure different radiators. By simply exchanging the WB with a rad and heat up the water to, lets say, 60 degrees celsius.
Do the measurements like above, with/without fan on rad, to get the K values, the slopes in the beginning, and then calculate the C/W th same way.
(what about the fan some might think. Well if you use a monster fan that "saturates" the rad = any higer airflow won't make any difference,
then the rad is working under optimal/best conditions, then the airflow will not reflect on the measurement).



What do we need for this to work = conclusions :
To compare we need a WB/rad that both persons use as referense.
Since all comparisons are relative the difference between systems are irrelevant.
Since the calibration is made, the relative K_WB is indifferent from different setups.
We need good isolation around the reservoir to get a big delta temp.

Now I'm beginning to get tired and I've probably missed something somewhere but please look at the idea as a whole.