Quote:
|
Originally Posted by Cathar
|
Maybe I missed something, and I believe that radiator are more efficient with more flow... but the graph you put isn't an asnwer. The answer is given by the graph posted by lolito_fr.
All graphs on the HE 120 series test page are based on a 5°C coolant/air differential.
As input is equal to
coolant/air differential * specific heat capacity of coolant * mass flow rate when you modify the flow with a constant coolant/air differential you modify the input, so with more flow you introduce more heat in the rad, so the results is evident : rad dissipated more heat, but it doesnt mean than a rad with the same input but with more flow will disspated more heat (this appears to be right as shown by the graph dt vs flow at 71 W)
1 Watt = 1 J/s
cp = 4185 J/kg/°C (depends of temperature, but very few changes)
lpm/60 gives liters per second
1 liter of water is about 1 kg
so lpm/60 * cp * delta T gives kg/s * J/kg/°C * °C = J/s = W
On the graph you posted :
at 2 lpm, you have 2/60 * 4185 * 5 = 697.5 W in input
at 10 lpm, you have 10/60 * 4185 * 5 = 3487.5 W in input
(hum, it's amazing, maybe i missed something because they are great values, am I wrong ?)
Ok, I got it 
, I keep my wrong reasoning to show you how I am stupid
and to let you understand what I meant. After reading of the article, the delta T isn't between inlet air and inlet coolant but between inlet air and the average temp of coolant resulting of dissipation, so it doesn't mean the input is directly proportional to flow as I believed.
But I dont understand your quote : my sentence "We need same graph but with a constant power as a CPU will put the same power in water in a high or a flow rate. So the coolant/air differential will increase when flow decrease..." is true to my mind, and it's exactly the graph posted above. The answer is evident on this graph, not on your where we don't know nothing about input.