Pro/Forums - View Single Post - Custom WC setup

heatwave · 07-05-2004, 06:20 PM

Angry - obviously the temperature rise will have to be kept under acceptable conditions.

Cathar - if you are the same cathar that made the twin *big @$$* radiators in Overclockers.au

, yes I grossly overestimate the heat load for both CPU and GPU so this will be my watercooler for quite some time. My 9800 Pro only puts out 65W but the 6800 Pro does put out 110W under max. load. If I'l be running Vcores above 2V, I'll be producing more heat than the hottest CPUs today which some exceed 100W already - thats 210W already combined.

I am assuming the temperature of the CPU to be cooled from 75C to 45C - where the temperature of the air increases from 35C to 45C.

Using the heat transfer law Q = U*A*delta T, for a flowrate of 3L a minute, that is like removing 2000W of heat based on the specific heat of water and its flowrate.

Also only 4 square meters of area is required but this equation although useful for simplicity, doesn't show the properties of the metal being used as the heat exchanger, let alone what flowrate of air is required to cool it down. I'll look into more details for more equations.

07-05-2004, 06:20 PM	#4
heatwave Cooling Neophyte Join Date: Jul 2004 Location: Down Under Posts: 11	Angry - obviously the temperature rise will have to be kept under acceptable conditions. Cathar - if you are the same cathar that made the twin big @$$ radiators in Overclockers.au , yes I grossly overestimate the heat load for both CPU and GPU so this will be my watercooler for quite some time. My 9800 Pro only puts out 65W but the 6800 Pro does put out 110W under max. load. If I'l be running Vcores above 2V, I'll be producing more heat than the hottest CPUs today which some exceed 100W already - thats 210W already combined. I am assuming the temperature of the CPU to be cooled from 75C to 45C - where the temperature of the air increases from 35C to 45C. Using the heat transfer law Q = UAdelta T, for a flowrate of 3L a minute, that is like removing 2000W of heat based on the specific heat of water and its flowrate. Also only 4 square meters of area is required but this equation although useful for simplicity, doesn't show the properties of the metal being used as the heat exchanger, let alone what flowrate of air is required to cool it down. I'll look into more details for more equations.