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Unread 04-08-2004, 02:01 PM   #9
OvcA
Cooling Neophyte
 
Join Date: Dec 2001
Location: Slovenia
Posts: 13
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4067 W seems to be the right number (just W, meaning J/s, not W/min) although it is obviosly wrong. I realy do not see the poin in dividing by 60. As it stands we have:

P = (m/dt)*c*dT = (700 l / h) * (1000 kg / m^3) * (4186 J / kg K) * 5 K = (700 / 3600 l/s) * (1 kg/l) * (4186 J / kg K) * 5 K = 4070 J/s = 4070 W

If we acount for the reduced flow we get a result in the order of magnatude of 10^2 which is still quite considarably off. I can't imagine a CPU producing more than 200 W of heat while being only water cooled. We than have to take into account the losses from heat transfer and secondary heat paths which means that the flow through the radiator would have to be about 35 l/h netting a whooping 95 % flow drop. A bit to much I should think.
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