prove me wrong
As you wish.
More volume just means more energy is required to raise the temperature of the medium. Eventually it will reach equilibrium and the loop temp will stabilise, it just takes longer to do so. What does make a difference is the mass flow rate taken in a 2 dimensional cross section. Which gives you the heat transfer across any block or radiator. Its makes no difference to the block/rad how much fluid is behind or in front of it volume wise, all that matters is the mass passing through it for any given second. That flow rate is determined by the pressure head of the pump and the losses from resistances in the loop.
---------------
The mean heat transfer can be expressed as
dQ = m Cp dT
Heat transfer = mass x Cp x temperature difference
Cp, Specific heat capacity of pure water at 0.01 °C = 4210 J/kgK
For 2kg of water heated by a cpu at 200 J. Approx
Approx 200/8420 = 0.024C increase per second until equilibrium.
So if we doubled the mass, we halve the temperature increase. But it would still increase until heat in = heat out. Oh as an aside, the reservoir of water would also be radiating heat at a rate dependent on a lot of factors I wouldn't care to calculate, suffice to say it adds to the cooling effect. You could extend this to an extreme, a huge tank that takes hours to warm up and also radiates more from surface area than you are dumping in. But that still does not effect the flow rate in the loop, that's down to the pump, and the rate that heat can be taken out of the block or radiator. Moving on.
-------------
Assume as in heat exchangers (rads & blocks) the fluid flow is continuously heated and therefore a rate applies.
W = dQ/dt = M Cp dT
Power = mass flow rate x Cp x temperature difference
where
Q is heat energy entering the coolant, once equilibrium is reached its the heat leaving the radiator in watts (W).
M is mass flow rate (density x velocity x cross sectional area)
Cp specific heat capacity (water is 4186 J/(Kg°C))
T2-T1, delta T, difference in temperatures from inlet to outlet for any block or rad etc.
Or basically the heat transfer is directly proportional to the mass flow rate. So if you increase the flow rate you increase the heat transfer. Now if you assume the cpu is dumping a constant heat load Q (dQ/dt is Watts or joules/second), then increasing the mass flow rate decreases the water deltaT.
---------------
Example.
A computer WC rig with a flow of 2GPM & 7/16" bore. The water moves at 1.3m/s. A 250W heat load. Water at 22C has a density of 997 kg/m3 giving a mass flow rate of 0.126 kg/s, specific heat capacity for water is 4186 J/(Kg°C)
250 = 4186 x 0.126 x dT
Therefore dT = 0.473°C
If you doubled the flow to 4GPM? Fluid velocity increases to 2.6m/s, mdot (mass flow rate) increases to 0.251 kg/s.
Therefore dT = 0.237°C
So for the 250w going into the cpu block, say with a coolant inlet temp of 20C, the outlet is 20.47C. If you
seriously increased flow by doubling it you only get a marginal improvement of 20.23C.
With these kinds of flow rate, once the system has reached equilibrium heat in = heat out, then increasing flow even doubling it does little. It also means the temperature anywhere round the loop varies very little, and all more flow does is reduce the slight deltaT's towards the equilibrium temp.
---------------------
Best to have the least for the loop, less water goes round the loop quicker so quicker heat transfer out from what is in there
Is what i heard from an old pro
I believe my point still stands, that the volume of flow will outlay the volume of water
Let me explain. A lower total volume doesn't make the water go around quicker. For any given length of loop with or without a res (excluding losses from the res itself) the flow is the same and dependent on the pump head. Less volume means the same 'bit' of water goes round more often, but at the same speed. As the 'bit' of water passes the block it picks up heat and as it passes the rad it drops it. If you have a res, that 'bit' of water has to sit and wait in the res, but the queue of water 'bits' in front have been passing through the block at the same rate. The same amount of energy has been picked up and dropped. All that will have happened is that whilst waiting in the res the 'bit' of water will have dropped some heat to its neighbours and especially if its near the outside of the res. But generally the queue (res) isn't that big and the 'bit's of water don't hang around long enough (flow) to loose a lot.
Now maybe your old pro meant keep the loop short, which would indeed increase flow rates and improve heat transfer minutely. But, as I showed above the flow rates are so high generally as to make differences in deltas very small. Jon is correct, volume acts as a buffer to temperature variations, but considered over a reasonable period of time with a uniform heat dump and with such small reservoirs all things being equal the loop temps just level out.
Its like anything if you take it to extremes, a really long loop with a little pump will have low flow rates and therefore poor heat transfer. A giant reservoir will take an age to warm up and can absorb a huge amount of energy. But with these sorts of flows and res sizes the only way to improve cooling is to add more radiators/fans or improve the thermal resistance of the blocks.
