That's the bit I don't quite understand. If the water only gets a few degrees above ambient, it's not going to be efficiently transfering heat to the air. Hence surely it'll need a very large radiator? Or high air volume.
I'm sure someone can provide the maths to show relation to heat disippation, water temperature and surface area. I can imagine a system like PS3 with the radiator on the bottom of a big fan and a peristatic pump moving water around (I'm pretty sure when I used them in biochem that we worked with very watery fluids, but perhaps the amount of effort to squeeze the pipe increases) geared off the fan motor, but I don't know if that's actually a feasible design.
Sorry Shifty I missed this one. I believe the relationship you are looking for is version of:
Q = U A dT'
Where,
Q is the rate of heat transfer in J/s or W
A is the surface area available for heat transfer in say cm²
U is the overall heat transfer coefficient of the system in W/cm²K
dT' is the LMTD (log-mean temperature difference) in degrees K
U is a constant and a function of the heat transfer mechanism (i.e. radiative, conductive, convective or combination of them), the properties of the heating/cooling fluids (e.g. thermal conductivies, velocities etc) and the dimensional arrangment of the flow path of each heating/cooling fluid. U is given by the design of the heat exchanging unit, it's materials, and flowing fluids from which heat is transferred one to another.
The LMTD is a logarithmic mean temperature difference used because with this system (i.e. likely counter or cross-current flow of air/water) temperature change along the flow path of each fluid is not linear.
So if you have ambient air at temp. t1, and water into the radiator at temp. T1, say your outlet temperatures are t2 (air) and T2 (water), the temp. differences (use R) at the inlet and outlet are:
Rin = T1 - t2
Rout = T2 - t1
The dT' or LMTD is:
(Rin - Rout) / Ln(Rin / Rout)
For an air/water system as discussed the water is the "hot" fluid and the air the "cold" fluid. The outlet air temp. will never exceed the inlet water temperature as the air or cooling fluid is the poorer heat conductor.
Basically if the temperature difference between the air and water into the system is small, t2 will be very close to t1, thus Rin is low, and thus your LMTD value is also small. So for a given amount of heat that you want to eject from the system (i.e. the heat produced by the processor core) the area you would need becomes huge.
Shifty is correct, in that a cooling water circuit is not an efficient way to remove heat from a console/PC, regardless of your fancifulness in radiator design. It boils down ultimately to temperatures of the water and that of the air, in which case the difference will always be small and thus imply low heat transfer performance of heat ejecting into the air at the radiator.
You could on the other hand run the cooling water circuit hotter, using a lower water volume, thus allowing for a greater thermal driving force for heat transfer at the radiator/fans end of the circuit. The only problem is that the cores you're trying to cool also run hotter. In such a case you'd ask yourself whether it's worth it then, as you could simply use heatpipes, heat sinks and fans, and save all the added complexity.
Otherwise you run large water volumes in your circuit, keep your cores cooler, but suffer shitty heat transfer at the radiator end because your temperature approach to the ambient air temperature is small (thus requiring a large radiator with lots of surface area to get the required heat out of the system).
Hope that clarifies