Quote:
The equation is T = 1.76*(Watts)/CFM where T is in Celsius
so for 100Watts and 40 CFM the delta T will be 4.4 degrees C. If you double the airflow to 80CFM the delta T will drop to 2.2 degrees C, thus lowering the whole water cooling system temp by 2.2 C. If you increased airflow to 160CFM you would drop temps by 3.3 degrees instead. Note that if you can drop the air temperature going into the radiator by 3.3C (by keeping it away from hot air exhaust say) you get the same benefit as Quadrupling the airflow.
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Equations are not my strong point for sure but it seems to me this doesn't have any real world relevance. If you have a 40CFM fan on a rad at whatever pressure it puts out who's to say that replacing it with a 80CFM fan is going to double the pressure (or CFM for that matter) as well? It would seems to me the pressure of the fan needs to rise linearly with the CFM of the fan in order for this equation to work. Most fans are not built the same so I imagine one at 80CFM would have a different pressure rating than another at 80CFM. And being the pressure are not similar the over all CFM of the fan would be reduced by lack of pressure power of the fan. So just dropping a 80CFM fan may not in reality push an extra 40CFM through the rad and/or may not double the pressure completely throwing off your entire calculations making that equation useless.
Of course I would love unregistered or pH to set me strait here as this is just off the top of my head which may be in my ass.
