sorry for doubleposting, but i thought i might be on to something.
Talked a bit with a mate yesterday about how much a radiator heats up the air that passes through. He said that it a function of CFM and Watt load.
So he gave me this formula:
Fixed air temp raise for given wattage is wattage * 60 seconds per minute / ( 1.293 kg per cubic meter * 1010 J per kg * 0.0283 (cubic meters per minute) per cfm * fan cfm) I think it is that now correctly..In other words 1,6 celsius cfm per watt. (That is 1,6 * wattage / cfm = raise in celsius.)
According to that formula, if i were to put a 200W load on the radiator, with 81 CFM (3x27, Nexus 12cm at 7V) the air delta would roughly be 4Â°C. Meaning that 25Â°C air ambient would become 29Â°C after the first rad. The Thermochill site states that it can handle a 200W load on those exact Nexus 7V fans with a 10Â°C air/coolant delta. (http://www.thermochill.com/PATesting/index.php#PA1203
). So at full load, the temperature of the coolant would be around 35Â°c.
Now if we consider the second rad, in series, thus stacked, in the following layout:
Nexus fans (push) --> ThermoChill (1) --> Shroud --> ThermoChill (2)--> Nexus fans (pull)
ThermoChill (2) will have to eat 29Â°c hot air, and might be loaded with 300W. According to the same ThermoChill Site linked as above, the fans can manage that load at 12V at a 10Â°C air/coolant delta. So the coolant would be around 39Â°C. Now even if those fans were to run at less than 12V, they'd still could manage the load, albeit at a higher coolant/air delta.
Naturally all of this makes no sense if we were talking just about one watercool loop. But i think it could make sense in two loops.. the first being the "cold" loop for the processor, the second "hot" one for the graphic cards.
Also i know that the fans will never be able to push/pull through as much CFM as stated in the fan databases, nor that the radiator has perfect thermal transfer. That's why i chose to use very high loads in my calculation, namely the power that the CPU and GPUs will draw from the PSU:
CPU : e6750 email@example.com will actually eat 135W of power (according to http://www.extreme.outervision.com/psucalculatorlite.jsp
. But i also want to keep enough reserves for overclocking a quadcore heavily. If i could find a jewel doing 4ghz on watercooling it'd ask for 245W at the same Vcore as the DualCore chip.
GPUs: I decided to rate a HD2900XT at around 150W in full load. Somehow difficult to evaluate, as there seems to be contradicting information on that. So that'd mean 300W for a Crossfire system.
I doubt that all power they drain will be transformed into heat. (Unfortunately i don't have stats that tell how much power is transformed into heat).
One problem with this kind of setup is that it could heat up a lot if the ambient runs high, let's say 35Â°c ambient in summer. Then the water temperature in the first rad would be 45Â°, and 49Â°c in the second. I don't know if that kind of water temperature could still manage to actually cool computer parts. I could ramp up all 6 Nexus fans to 12V, but then the setup wouldn't be really quiet anymore.
I understand that the optimal solution would be to allow each radiator to have its own access to fresh air. But unfortunately I'm a bit "limited" with my case. Concerning the layout i presented in the above post, i was told that such a setup would easily make a thermal circle, meaning that the hot air would be sucked directly into the the rads again... making proper cooling impossible.
So for now, i don't see any other possibility for me than stacking the rads; using only one rad wouldn't manage the heat load. This leaves me with a few questions:
- Would this setup work, considering it uses two watercooling loops, while fully knowing the second one will heat up more?
- Are my calculations concerning airflow heat-up actually correct?
- Is there any information about the actual amount of heat that chips dump related to the wattage they eat?
Thank you very much