No, if the heat is, as you put it, pulled more efficiently from the processor, then a lower temperature differential will be required between the cpu and the water (the average water temp for a given heat load being determined by the radiator, the ambient air temp and the airflow, NOT the waterblock), thus the cpu can be cooler while still dissipating the same heatload into the water.

The efficiency of the waterblock determines how much hotter the cpu must be than the water, NOT the other way round.

8-ball

Man, mebbe I'm just confused (still hung over from last night ... dizzy and hurting).

DAMN YOU MARKETING BASTARDS FOR FSCKING WITH MY HEAD!

In the end, the coolant temperature will be slightly higher if the block is more efficient. Basically, if you have X amount of heat generated from the CPU, it all has to go somewhere. If the block is less efficient, that heat will be partially absorbed by other components than the water such as the block itself (i.e., having a hot block but cool coolant), the CPU packaging (god forbid) and the motherboard, etc. The heat has to go somewhere ... the chip doesn't get rid of it by staying hotter. If the chip temp goes down due to a more efficient block, the water temperature will go up ever so slightly.

The difference won't be that noticable, but the difference will be there. It HAS to be. Heat doesn't just disappear.

That make more sense?

Damn ... I need some meds. I think I have a migrane coming on. I'm starting to get the tunnel vision thing and these weird faded spots in my vision ... and damn does my stomach hurt. This has to be more than a damn hangover.

I guess the best way to explain that would be to take a heat source, like a soldering iron, and place it in a bucket of water. The bucket will cool via evaporation, simulating the radiator in this case. When you plug in the iron (increasing the heat load to simulate the increased block efficiency), the water will heat up, and the water will evaporate faster (simulating the greater efficiency of the radiator under greater dTs). Regardless of the fact that the water is evaporating faster, though, the water will be warmer than before. Now, the difference we are talking about isn't as large as turning a soldering iron on, but it is there. There will be an increase in coolant temperature if a waterblock is more efficient.

lol, stay with the slime

no, eh
read what 8 ball is saying
do NOT try to think up analogies

a given applied heat (load), will heat the coolant 'x' amount;expressed in Watts, and dependant on the flow rate will result in a FIXED temp rise
(really basic physics here, google 'heat capacity calculations')

the efficiency of a wb will be apparant in the temp gradient across it
i.e. how hot does it have to get to transfer that 'fixed' amount of heat into the coolant

back to the slime pit, boy

Airspirit- I can say I've felt how you do, this physics stuff can be so confusing. The thing to remember is no matter what the processor dissipates X watts of heat at equilibrium. It does seem like to bring the CPU temperature lower, we have to dissipate more heat, right? but thats wrong, if you "dissipate more heat" with some change, your processor is not at equilibrium anymore, and the temperature will fluctuate to a new temperature where it will dissipate X watts of heat again. The difference between all heatsinks and waterblocks, is the delta T required to dissipate X watts. Poor cooling needs a very high delta T between the Die and the coolant(air or water) to be able to dissipate X watts, while better cooling needs a lower delta T.

The heat load is dependant on the heat losses through paths other than the wb. The size of these secondary heat losses is dependant on the efficiency of the wb. For example:-
http://www.jr001b4751.pwp.blueyonder.co.uk/SecW.jpg

In other words I get it that all WBs absorb pretty much the same.

http://www.jr001b4751.pwp.blueyonder.co.uk/SecW.jpg at a fixed temperature........ right??

Yes.
A fixed thermal enviroment except for the wb.

So, in other words, the more efficient the block is, the more thermal energy that will be dumped into the water, correct? And this would heat up the water, correct? Am I missing something?

"applied heat (load)"
not so well stated, intent was to consider that specific portion of the applied heat seen as a load by the wb
obviously the secondary path 'losses' are irrelevant to an assessment of the wb's performance
(they are consequential)

and for this reason the quantification of "W" must include an assessment of the secondary losses
this can be done with a heat die - and I do so - but is about impossible using a CPU

better ?

no hara, not if they have different "C/W"s
that's what's being measured, eh ?

airsprit
no, again no

the whole cooling 'chain' is a cascade of gradients
more efficient = lower gradient
always same amount of heat being shuffled
(less 'losses' here and there, per Les' correction)

billa, i have this question bugging me for a long time:

imagine a cpu always dissipate 80w.

then why the diference between the idle and load?

CPUs don't produce a constant amount of heat; they use more power (and therefore generate more heat) when working than when idle.

aaaahhhh k thanks, because i have hearded that in this forums and got confused.

Fine.
Was not questioning the assessment of secondary losses.in your test results.
Sums for the Heat Die suggest any variation in load would be less than 1% in any wb's efficiency versus flow characterisation.
http://www.jr001b4751.pwp.blueyonder.co.uk/SecW1.jpg
Dunno whether this correlation exists or is detectable.

Possibly of interest is that the R(insulation) of 15.375c/w corresponding to the experimental 98.4% absorbtion(1.6% loss) would be that calculated(Kryotherm) for a 50x50x90mm lump surrounded by 22.4mm thick layer of Polyurethane foam. Not sure but think this is not too remote from reality.

8-ball... after reading your first post and disregarding all the fancy smanchy graphs and technical bibbel babbel (sorry #Rotor, BillA...) the point you made became quite clear...:D
My brain works better on simple explanations...:)

I should have caugth this one myself as I was quoting but I was kinda caugth up in the process of translating... I positively hate german... ie the language and the spelling...
Nothing against germans at all...:) They have 2-3 rules and 999 exceptions....:D

Basicly: It may be absorbing more heat(per 'unit'), but the water's cooler in the first place because of the increased 'cooling performance'?. Therefore it's cooler?...

But if you were putting sustained 20degC water through the two blocks(no rad, just 'once used' water), it would be warmer then?...


Is any of that right?...

Not quite sure what you're referring to. If it's why the cpu is at the temperature it is, then I'll try and explain in a little more detail.

A few points.

1. Heat flows DOWN a temperature gradient.
2. The amount of thermal energy that will flow down a unit thermal gradient (ie 1 degree cooler per metre) for a block of unit cross section (1mx1m) is the coefficient of thermal conductivity.

So, for a given heat exchange process, if we want to transfer MORE thermal energy, then we need to INCREASE the thermal gradient.

If we have a fixed coolant temp and a fixed heat load, the thermal gradient must be increased until the thermal energy which will flow down the gradient matches the heat dissipation of the source. The heat source will then stay at a constant temperature, as the thermal energy it is producing is ALL being removed. This doesn't mean it will cool down, as this would reduce the thermal gradient.

The whole heat exchange from cpu to air can be considered as a series of individual heat exchanges, each with an associated thermal resistance (reciprocal of the efficiency), where a greater resistance requires a greater thermal gradient (driving force) in order to transfer the same amount of thermal energy.

1. CPU > Waterblock base - (TIM C/W)
2. Waterblock base > waterblock fins - (block C/W)
3. Waterblock fins > water - (convective heat transfer coefficient)
4. Water > radiator wall - (convective heat transfer coefficient)
5. Radiator wall > radiator fins - (radiator C/W)
6. Radiator fins > air (convective heat transfer coefficient)

For a given heat load, each of these steps will have an associated temperature difference required to transfer that exact amount of heat. These can simply be added up to find the total temperature difference between the air and the cpu, where the air temp is the controlling variable.

Factors such as flow rate, fluid viscosity, thermal diffusivity of fluid and the geometry of heat exchanger surfaces will affect the resistances/efficiencies, thus affecting the temperature difference associated with each step.

As for the difference between idle and load.

If the thermal energy load is resuced, ie, cpu at idle, then the temperature difference required at each stage to transfer that amount of thermal energy will be reduced. Thus the overall temperature difference will be reduced, and since the ambient is still at the same temperature, the cpu will cool down.

Obviously, there will be a delay due to the time it takes for the water and the copper to drop in temperature, which will happen because the heat transferred into them from the cpu will drop, yet they are still at a temperature to transfer the initial load to the next medium, so more thermal energy will be transferred out than that which is coming in, so it will cool down.

That's why larger copper blocks and systems with large volumes of water will not react so quickly to heat loads.

I hope all of that makes sense.

8-ball

What was the topic again? Oh yea.
http://becooling.safeshopper.com/36/192.htm?780
" This is one of the highest (if not best) performing blocks currently on the market for AMD and P4 cpus. High flow, multiple, parallel 1/16" deep and wide micro channels. Channels are curved on the base to maximize turbulence. 1/2" Chrome barbs, 1/4NPT threads, black anodized aluminum cover prevents corrosion. Copper alloy 110 base is machined, lapped and polished.

O-ring design provides an excellent seal against leaks. Pretested prior to shipping."

This one seems a bit BS-ish to me.

peace.
unloaded

a stretch, but second only to WW
review up after I move this week (I guess)

8-Ball
correction noted re my inverted phraseology
(using gradient as a magnitude)

Huh? I'm used to tilt being an informal reference to proof by contradiction, but I know that's not what you mean. What do you mean?

pretty sure he was referring to the abuse of a pinball machine :)

He meant that Airsprit's comment is out-of-order (a pinball reference)

The heat being the same, the heat dissipation will also be the same, which means that the water temp will also be the same, so the only thermal gradient that changes, is somewhere between the inside of the core, and the top of the baseplate of the waterblock, just before the heat hits the water.

Hmmm, I agree with the more efficient waterblock leading to warmer water, because of the secondary heat paths. Les is plotting it. I thoerised about it in this thread:

http://forum.oc-forums.com/vb/showth...hreadid=148310

The differences are very minor though. Something that would be near impossible to pick up under normal operation.

pick your premises and stick to them

one can consider a wb's response to a heat load (applied heat warming the coolant)
or
a heat die->TIM joint->wb 'system' (minimal secondary losses, quantified)
or
a CPU->TIM joint->wb 'system' (large secondary losses, rather undefined)

and for each of the 3 there are different considerations, hence different 'answers'

yea, pinball (airspirit was making the same 'bad' moves)

I will try and explain with reference to your example.

You are quite correct that when you turn the soldering iron on, there will be an increase in the surrounding water temperature.

However, this is NOT the equilibrium temperature, as at this point, the thermal energy transferred to the air via evaporation of water will be LESS that the thermal energy transferred from the soldering iron into the water.

Because of this, the thermal energy of the water will continue to rise, (so the temperature will rise), and consequently more thermal energy will be transferred to the atmosphere vie evaporation. The temperature will continue to rise until the transfer of thermal energy via evaporation is the same as the transfer of thermal energy from the soldering iron is the same.

So now we have a situation where there is no NET change in thermal energy in the water. The same amount comes in as goes out. This is the equilibrium condition, which is what would happen if you ran a computer crunching for a couple of hours.

The difference in the temperature between the soldering iron and the water is determined by the convective heat transfer coefficient for the soldering iron surrounded by water. This coefficient gives a value of temperature difference required between the edge of the soldering iron and the water which will cause the transfer of 1W of thermal energy. So supposing it is a 70W soldering iron, the required temperature difference between the soldering iron and the water is the heat load divided by the convective heat transfer coefficient .

Note that regardless of the temperature of the water, 70W WILL be dissipated, PROVIDED that the soldering iron manitains this temperature difference above the water.

This is like the efficiency of the waterblock, but there are more stages to consider in the heat transfer, but it is still a coefficient of efficiency.

Improving the efficiency, (ie a lower C/W) will mean that the same 70W is transferred to the water, but a lower temperature difference is required.

AS SUCH, AT THE EQUILIBRIUM CONDITION, THE CPU WILL BE COOLER, AND NOT THE WATER WILL BE HOTTER.

However, as has been said, this does not take into account the secondary heat paths of a fixed efficiency. Essentially, the total heat output of the cpu is conducted away via a number fo routes, each with varying efficiency. By increasing the efficiency of one of them, you will increas the proportion transferred by this route.

However, I still believe that this was not your understanding of why the water will warm up, VERY slightly, as reinforced by your argument above.

So back to your example, once an equilibrium condition has been reached, changing the efficiency of heat transfer into the water WILL NOT change the water temp. The only way to change the water temp is to change the efficiency of heat transfer from the water to the air, as this would change the temperature difference between water and air, OR change the ambient air temperature.

Supposing we have a fixed ambient air temperature, the efficiency of heat transfer from water to air could be IMPROVED by directing a fan at the surface of the water. This will reduce the temperature difference between water and air, BUT the temperature difference between the water and the soldering will not have changed, though due to the water now being at a lower temperature, the soldering iron will also be at a lower temperature.

So I hope you can see from this that, FOR A GIVEN HEAT LOAD, changing the efficiency of a heat transfer stage will ONLY change the temperature delta of the heat producing component above the heat absorbing component. (by heat producing, I mean the component from which the heat is being transferred, so in a water block, the fins are the heat producing component and the water is the heat absorbing component.)

Does this make sense?

8-ball

Much easier to think of everything as an equalibrium problem. Theres waaay to much calc to be dealing with instantanous conditions when you can be using algebra.

Think of the CPU load as an average value, the radiator performance as an average and the flow rate as an average. That way you can just add componets together without any messy (and unsolvable) differential equations.

Here's a goody:
http://www.bluecooling.com/welcome.htm

Read this thread at sysopt. Some people are just so easily misinformed. :D

"a digital thermistor" :D

that just made my day.