[pauldenton]

Originally Posted by mtbdrew

Thanks for the mental work out, you are making me dust the out corners I have not used in a while.

One point you are still missing though. You are making the assuption of the heat being applied at one time. However, it is being applied by five different sources which cause a 5:1 ration of heating to cooling.

err no - at equilibrium the cooling = the sum of all the heating (by definition - otherwise it's not at equilibrium.)

Originally Posted by mtbdrew

This of course is a simplified way to look at it. To do this accurately you would have to measure the time it take for a set amount of water to travel through the rad and each heat source.

that's the flow rate - in our example 4 litres/minute or 66.667 ml/second

Originally Posted by mtbdrew

But for this example lets just say it takes 1 second for a millileter of water to travel through each device. In this case for every second of cooling you get 5 seconds of heating.

ahh this is where you are misunderstanding i think... we are dealing with a continuous flow of water - in every minute we have 60 seconds of heating per heatsource and 60 seconds of cooling per rad - at no time is there ever less, or more water in a component.

Originally Posted by mtbdrew

The calculation is made moore complicated because each heat source is a different value but you get the general idea. To go back to my original state he will either need a monster rad or two seperate rads. Either methods should increase the amount of time the water gets cooled. Ideally you would want a 1:1 ratio to bring your calculation back into line.

what you have to take on board is that a rad's cooling ability is proportional to the temperature differential between water and air.
when you add additional heat sources then the equilibrium water temperature will rise until it's cooling ability matches the heat. to use the figures we used earlier:
CPU block = 70w
pump = 10w
total heat input 80w
say the rad/fans can cool this at a water temperature of 3C above the air flowing through it. if we add the gpu blocks (and imagine for a moment that the resulting flow drop has no effect on the rad's efficiency) then we have:
CPU block = 70w
pump = 10w
2xGPU blocks = 60W
total heat input 140W
now the equilibrium point would be +3Cx140/80 i.e 5.25C - so our water temperature has risen by 2.25C as a result of added heat input.
but at this (new) equilibrium the same calculations about temperature differentials apply-
the coolest water is directly after the rad, then we have the following rises after components:
pump 10/279= approx 0.035C
CPU block 70/279 = approx 0.25C
GPU bbocks 30/279 = approx 0.1075C each
total 0.5C...

so the position in the loop makes very little difference (unless you either have very low flow, or huge heat-eg TECs)

you could add a second identical rad, and (given the same flow airflow/air temp) then you'd halve the water:air differential (ie the +5.25C would drop to +2.625C) but the rises over the hot components wouldn't change (the ones over the rad would halve due to the temp. drop - each rad would now give an approx 0.25C drop not a 0.5C one) - the payoff from the extra/better rad(s) is lower water temps throughout: the rise over a component is unchanged(at teh same flow)