Cooling heat production

heh... you couldn't have said that more clearly, cheers

 

Sorry if i offended you Biff but i genuinely wasn't trying to be condicending. Its one of the problems of the written word as it can be interpreted in other ways then it was intended.

My understanding of this runs along the same lines as cheebs. Once the radiator reaches its limit it cant remove the heat to the air quicker then the water is adding heat to the rad. At best the rad effectively stops functioning at the worst it could start adding heat to the water.

 

fivecheebs, I can understand what you're saying but it doesn't wash - the efficiency of any heatsink is measured in *C/W - that is, broadly, how close the hot side can be kept to the cold side for a given thermal load. So if you have a 300W loop and a rad which is rated at .02*C/W, your coolant would sit at 6*C above ambient (300*.02). Increasing the load should have a linear effect on the temperature delta.

In your example increasing the load from 300W to 310 W would cause the coolant temp to rise slightly, as with the new thermal load the rad will be unable to maintain the same delta-T. However, as soon as the temp increases, the thermal transfer rate of the rad increases too, and it eventually levels off at (by my example) 6.2*C above ambient (310*.02). Now, obviously, the water doesn't immediately jump to the new temp, and the more water in the loop the longer it will take, but when it gets there its temp will level off.

So, even assuming you can make accurate assumptions about the parameters (fan speed, flow rate, water additives, coolant temperature tolerances) of the 'average' watercooling loop, I still don't understand how it can be said that a rad is limited to e.g. 400W without specific design parameters for the system in question.

The alternative is, of course, that I am missing something and there is a 'ceiling' on heat transfer through a rad, but I really can't envisage how that would work. Can anyone enlighten me?

 

A rad can only add heat to the water if the water is colder than ambient, in which case you'd need to be using some kind of active heatpump (phase-change / peltier) to keep the water cold, or else you've put some ice in the res. In either case, you wouldn't be using a rad to cool your water.

I think the max heat rating for each rad is more of a guideline than anything else, as to the maximum system they'd recommend using with it. It can't be anything else.

 

Hmmm...
I am not a theoretical expert in thermodynamics, but two statements here strike me as questionable.

The radiator does not "stop working" under these conditions, it continues to dissipate the 300w it always has.

What is described here is an exponentially increasing temp spiral.
Would the system not shut down in a matter of seconds (or blow the loop apart as phase change takes place) if this occurred?

It seems to me that the waterloop would reach a state of equilibrium (at a higher temp) instead.

For instance....
Imagine a system built of the finest components (>insert dream setup here<).
This arrangement results in a CPU temp of X.
Now, replace this loop with one of lesser capability (>think TT Aquarius<).
This results in a CPU temp of X+x, but the system continues to work (albeit at a higher temp)...it does not go crazy and accelerate into thermal shutdown (well, maybe with the Aquarius it would, but you get my point ).

Of course, I could be wrong....

 

I think we need someone like 8Ball on the case

 

Just to add to this. I remember reading somewhere that as the temp of the HS (rad) increases, the °C/W rating would slightly decrease. Assuming that your ambient air stays at the same temp so there is a higher deltaT. IIRC this has to do with greater amounts of turbulence at the surface of the HS. This is what I had in mind when I said the HS becomes more efficient the hotter it gets, maybe this could be said more accurately, but thats the idea.

 

That's quite possible but I've never heard it so can't comment. In any event I think the effect would be small. Are you saying that increased rad temp improves or diminishes its heat transfer capability? If the °C/W rating decreases, then deltaT decreases as well (assuming a fixed heat load).

 

I'm saying the °C/W gets lower, but I'm sure it is a very small change. For example say you had a rad that had a °C/W of 0.1, so with 100W you have a deltaT of 10°C. If you put in 150W the deltaT, assuming the °C/W remained the same, would be 15°C. But because there is a higher deltaT between the rad and the ambient, the °C/W is lower so with 150W the temp rise may only be say 14°C. Which works out to about 0.093°C/W. I don't know if this is true, I'm only saying that I think I remember reading it somewhere.

 

sounds quite plausible to me - higher temp = greater molecular movement = more collisions = quicker heat transfer = smaller deltaT

 

In short (and bypassing all the technical bits) the assumption made by HWLabs is that a 100CFM airstream is passing through the rad (which is not quite the same as slapping a 100CFM fan onto it, as there will be airflow resistance from the rad).

This means that max. Wattage is, as you summise, partly a function of the ammount of air passed through the rad. HWLabs just decided to set this at 100CFM for their comparison. The max. Watts value stated by HWLabs (or any other manufacturer, for that matter) is therefore purely under the condition of a 100CFM airflow. This was just some arbitrary condition they specified. In reality the CFM going through your PC's rad will, for the sake of your hearing and sanity, be much lower, but setting a condition of a 100CFM makes the capacity of their rads look better.

Think of car mileage: some manufacturer will specify, say: 60mpg (motorway), 36 (urban), 45 (mixed). What exactly is the average traveling speed at "motorway", "urban" or "mixed" may be open to interpretation, and usually the small print will specify some numbers there. It will also state: "Your mileage may vary".

 

It's quite clear that the heat transfer capability of a rad depends partly on the airflow through it - otherwise we'd all get rid of the fans with no detriment to the cooling. The question at hand is why the rad manufacturers are stating a maximum head load capability of, say, 400W, when you could quite happily put 600W of heat into the water and the rad would still dissipate it, albeit at a higher differential to the ambient air temperature.

So far no-one has come up with a reason why there would be an absolute ceiling on the power that a rad can dissipate, so can we conclude that the manufacturers' ratings are simply recommendations?

 

If it is recomendations, then a Black Ice Micro (rated for 275W) is pretty much good enough to cool my 226W peltier. Which I think is wishful thinking.

 

I did the say in my second post (which i missed out in the first) that the rad effectivley stops working.

Now i dont mean it switches itself off, as it is obvisouly still dissapating heat, but its no longer lowering the temperature of the water as the CPU is adding more heat then it can get rid off. That in my mind means it isn't working in the context of cooling my hardware.

 

but the amount of heat a rad dissipates varies linearly with deltaT, so as the water heats up and deltaT increases, the amount of heat dissipated by the rad will increase, until it reaches an equilibrium point where it is dissipating the same amount of heat as the components are dumping into the water. At that point the water will not get any hotter unless (a) the ambient air temp rises; (b) the fan is turned down / off or obstructed so that airflow is reduced; or (c) the amount of energy being dumped into the water increases again. In any of these three cases the water will heat up until the new value of deltaT is sufficient for the rad to once again be able to maintain the water temp where it is. Conversely, reducing the heat being dumped into the water, reducing ambient air temp or increasing airflow will mean that the rad can maintain a lower water temp, and the water temp will reduce accordingly.

 

We're not saying that this doesn't happen, we're just after why because it doesnt make sense to us. As far as I can see it a HS has a °C/W rating and I would think this should hold true no matter how hot until something happens like the coolant boils or solder joints melt.

 

If you continue to dump heat into the radiator, at some point it will pass the point at which it can lose that heat at least as fast as it is introduced. The thing will go white hot, explode, turn into a puddle of molten metal, whatever. The recommendations of the manufacturers state simply that:

"at this CFM it can dissipate this maximum amount of Watts while still maintaining an equilibrium. Surpass that Wattage and the whole ting will continue to heat up until either your CPU overheats or the rad melts down.

 

(just because I love keeping arguments going )
Don't forget that a radiator doesn't 'get rid' of heat, it just dumps it into the room. The manufacturer's specs are making the (fairly resonable) assumption that the room housing the rad is sufficiently large, and has a good enough movement of air, that the ambient temperature will not change.

(based on how warm my legs are getting next to my exaust fans, maybe not such a good assumption...)

 

So what you're saying then past a given heat load the thermal resistance (°C/W) of the rad will climb and approach infinity? It would have to in order for the temperature to endlessly continue to climb without a continuously rising heat load, until a failure occurs.

I'm not convinced.

 

it's well worth reading this for an overview of radiator performance...
http://thermal-management-testing.com/ThermoChill.htm

key point is that heat dissipated rises linearly with temperature differential.

when you turn the system on, the water temperature will rise until a steady state is reached at which the rad(s) shed(s) heat as fast as the components add it. the higher the heat added by the components cooled by the system (not forgetting the pump!) the higher the water temperature will rise to.

iirc hardware labs ratings for the black ice rads is based on a 40 C temperature differential! (which is the kind of figure used in rating domestic central heating rads...) which is why they seem so high when compared with thermochills quoted figures, which are based on a somewhat more liveable 10C difference.....

{so yes biff, a BIM will cool your tec - IF you are willing for the water temperature to be very high indeed: maybe high enough that your CPU won't be particularly cool...}