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A Universal Formula To Rate The Performance of Any Cooling Solution

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Zerileous

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Jun 21, 2002
Hrm well I finally had a chance to read all of this. I see that you're controlling for die size, which can be a good thing, and this can either prove or disprove our gut feelings that these smaller die sizes are producing less heat energy but running higher temperatures because there is less surface area to remove the heat.

I don't think this formula gets us apples to apples comparisons of different cooling solutions that are tested under different circumstances. Yes ambient temp and die size are controlled, but as others have mentioned, IHS and TIM are not controlled. So yes we can look at the system as a whole, as you say, but cannot compare x-cooler to y-cooler just by crunching some numbers for our dissimilar systems.

You may find this article to be interesting: https://www.gamersnexus.net/guides/3561-cpu-cooler-testing-methodology-most-tests-are-flawed

So you want me to run a load of my choice (or is this assumed prime95 small FFT?) and use software reported wattage? For my CPU 3 different wattage values are reported in HWiNFO64. Currently folding on 4/6 cores, "CPU Core Power" is reported around 70W, "Core+SoC Power" is reported around 80W, and "CPU PPT" (which is the power draw of the entire socket) is reported around 85W. I also have two VRM readings reported in HWiNFO64, and adding them together gives me 89W. Which of these readings would you like me to use in your formula?

Can I also use the formula for my GPU and see if the rating is similar? It should be since it's the same loop.

I don't feel like Blaylock's question was conclusively answered. Rather there was, with what appeared to me to be a bit of a condescending tone, a statement that I interpreted as, "when you get it, you'll know the answer." So I will ask, does my water cooling loop need to be at an equilibrium temperature or can it be near ambient? This is a yes or no question for the purposes of participating in your other thread only. This is Not a theoretical question in the context of my post.

Finally is there a way to combine both my GPU and CPU into one score? Otherwise each score may suffer a slight penalty of the idle heat from the other component, as they share a cooling loop.
 

mackerel

Member
Joined
Mar 7, 2008
So something is wrong with the mackeral numbers.

Temps and power were as reported by the CPU via hwinfo64. I looked at the individual reported core temps and averaged manually. I also looked at the CPU reported power which was used.

The load I was using was Prime95 128k FFT per core. This is small enough to not be ram limited and I find representative of near worse case load for a CPU with good AVX performance (most Intel since Haswell, AMD only from Zen 2). On Intel side up to original Skylake I found this roughly tracks TDP. It was only after Skylake when Intel's process advancements became derailed that they started turbo-ing much above TDP.

I forgot to add, for the 8086k I'm running that turbo disabled. So it will be closer to TDP under Prime95 than if turbo were enabled.

For illustration attached is a screenshot of hwinfo64 to give example of the values I'm using. I'm using the same load, but didn't let it burn in so values will be lower than previously mentioned. Also ambient is probably lower as it is early morning right now.View attachment 209992
 

Zerileous

Senior Member
Joined
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So you're saying the utility of this equation is tuning fan curves? Measuring if or when a point of diminishing returns is reached in a specific system at a specific fan speed? Or is it just academic? I know you asked to keep discussion out of this thread, but I am replying to a discussion post by yours truly.

The problem with Zen2 architecture is that there are multiple dies on the CPU, but in the case of the 3700x which we've seen a few times in this thread, only one of two dies contains CPU dies cores. An IO die is placed on the package of the CPU but is produced using a larger 12nm process compared to the 7nm core die. The majority of the heat is coming from a very small surface area. When the equation is performed with only the core wattage and core surface area the HRQ is almost double! Anyway I point all if this out because you state,
If we were talking about the Thermodynamic concept of Heat Transfer, we only consider the heat generation portions. But since we are dealing with Heat Removal, we are constrained to the size of the actual contact block, which is, for all intents and purposes, the size of the die itself.
I am assuming you're presuming that the die itself would act as some sort of heat spreader, conducting heat uniformly into the IHS. Silicon is actually a poor thermal conductor so this is problematic (hence why Intel is now shaving down their CPU dies, in order to have less thickness of silicon before the heat gets to the STIM and IHS). Also the area rather than volume of silicon is used in the formula, so it doesn't account for thicker vs thinner dies. Finally this all becomes most problematic with a Zen2 CPU where there is literally air insulation between the core and IO dies.
 
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Dr_Emmett_Brown

Dr_Emmett_Brown

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I see that you're controlling for die size, which can be a good thing, and this can either prove or disprove our gut feelings that these smaller die sizes are producing less heat energy but running higher temperatures because there is less surface area to remove the heat.

Probably the most intelligent, succinctly stated thing I have read in this post thus far.

I don't think this formula gets us apples to apples comparisons of different cooling solutions that are tested under different circumstances. Yes ambient temp and die size are controlled, but as others have mentioned, IHS and TIM are not controlled. So yes we can look at the system as a whole, as you say, but cannot compare x-cooler to y-cooler just by crunching some numbers for our dissimilar systems.

I think this highlights the nomenclature misunderstanding that has been persistent all of this time.

This formula DOES NOT rate any individual SUBSET of the cooling solution, including the cooler itself.

IT RATES THE SYSTEM AS A WHOLE, including the water blocks/cold plates/heat pipes + the TIM plus every other heat transfer layer from start to finish.

The other components that you identified as "not controlled" are implicitly included because they play a role in the observed temperatures. A better TIM yields a lower temperature, for example, so that will show up in the equation.

You are right in that a "cooler" should not have its rating penalized because of some crappy TIM.

But, again, recall this equation just shows how hard the system as a whole is WORKING to remove heat.




So you want me to run a load of my choice ... Which of these readings would you like me to use in your formula?

The block is in contact with the entire chip, so the smaller heat contributions to the entire package are also heating it up. They must be included, even though we have no "average temperature" data for these components, nor a straightforward way to integrate them even if we did. Fortunately, they are small contributors.

Can I also use the formula for my GPU and see if the rating is similar? It should be since it's the same loop.

I've never given this any thought. I have my GPU cooled in a separate loop.

I don't feel like Blaylock's question was conclusively answered. Rather there was, with what appeared to me to be a bit of a condescending tone, a statement that I interpreted as, "when you get it, you'll know the answer." So I will ask, does my water cooling loop need to be at an equilibrium temperature or can it be near ambient? This is a yes or no question for the purposes of participating in your other thread only. This is Not a theoretical question in the context of my post.

A question posed smugly will usually merit the tinge demonstrated in my response. Any question asked with sincerity will be answered to the best of my ability without any such veil as aforementioned.

The short answer to your question is: No.

Again, don't think of the formula as a "High Score Rating." At lower startup temperatures, your cooling solution is not working as hard compared to a "heat soak interval" that has elapsed. The higher the number means the harder it is working AT THAT PARTICULAR INSTANT given the data points observed.

Finally is there a way to combine both my GPU and CPU into one score? Otherwise each score may suffer a slight penalty of the idle heat from the other component, as they share a cooling loop.

I will have to set up some experiments but I suspect it would need an expansion to the formula to account for additional factors.
 
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godevskii

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Apr 2, 2012
And maybe not so surprising at 67C under a 95W load the score of 10,157 indicates it is working harder than 73C under a 97W load with a score of 9,196.

I don't know if its me but this does not compute, how can it be working harder at a lower temp/load?
To me you formula isn't reaching the saturation point of the cooler but instead the value indicates something like an efficiency loss?!?!?!? The cooler is the same, room temp is the same. How can it be removing less heat with a higher temp/load? The 3rd scenario should at least return the same value as the 2nd scenario, it can't remove less heat if the cooler properties have not changed and the heat is there to dissipate.
 
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Dr_Emmett_Brown

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I don't know if its me but this does not compute, how can it be working harder at a lower temp/load?

Ask yourself this question: If "cooling solution A" removes more heat more than "cooling solution B", which is working harder?

In other words, the cooling solution is lowering the temperature more, therefore it is working harder.
 

EarthDog

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It isn't you. That's what I said in the chat thread too. :)

The cooling loop doesn't work any 'harder' at a given temperature. It's cooling properties are its properties no matter what (again all variables not the heat load the same). The temperatures are a byproduct of everything... including the die size... including the internal TIM/solder...including the IHS, including the TIM between the IHS and base of the heatsink/cold plate... etc.

EDIT:
In other words, the cooling solution is lowering the temperature more, therefore it is working harder.
Nope. They are both working the same... up to their potential. One has more capacity/cooling capability than the other, but, they aren't working harder. One solution is conducting more heat from the baseplate out, but they work the same no matter what.....the cooling properties are the cooling properties. Now, if you said more efficient or has more cooling capacity, sure. But so long as fans stay the same speed or pumps, the cooling solution works the same regardless of load.

PS - I moved the inevitable off topic posts back to this thread. Try to keep the discussion here.

Dr - next time you start a thread like the other and want users to come back here, consider linking the thread to make it easier. I've edited your first post and added a link to this thread.
 
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Dr_Emmett_Brown

Dr_Emmett_Brown

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The cooling loop doesn't work any 'harder' at a given temperature.

This cannot possibly be true.

1. Fan speeds can be set to vary to heat sensor loads. Spinning faster = more CFM = more cooling to radiator fins, etc.
2. If a cooling solution can remove 200W of heat and it ALWAYS removes this much heat, what happens when the CPU is generating 100W of heat? Are you saying it still removes 200W of heat?

No, no, no, you are certainly mistaken.

A cooling solution has a MAXIMUM capacity, and it DOES NOT WORK as hard when you first power up your computer and everything is at or near room temperature.

It works harder when the ambient temperature is higher.

It works harder under full load than it does under idle.
 

godevskii

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Apr 2, 2012
If a cooling solution can remove 200W of heat and it ALWAYS removes this much heat, what happens when the CPU is generating 100W of heat? Are you saying it still removes 200W of heat?
No one suggested that.

A cooling solution has a MAXIMUM capacity
No one is disputing this.





In example, if a cooling solution can remove 200W, at a load of 100W its removes X amount of heat, but at a load of say 150W its got to to remove X+Y until this reaches its maximum capacity.
Yet according to your formula, (the 3rd scenario) even with more load/temperature, SAME CONDITIONS/PROPERTIES ON THE COOLING APPARATUS, it manages to removes LESS heat then 2nd scenario, explain that?

I honestly think that formula needs some debugging.
 

EarthDog

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This cannot possibly be true.
1. You may have missed that I said all other things remaining the same. This includes fan speeds (also pump speeds/flow rates when talking water as well.)

2. The cooler is removing the heat load that it is given. That doesn't mean it is working any harder or less. If the cooler/mater/system is able to move 385 w/k it does so regardless if there is 100W load or 200W load. The difference is in the output... in this case temperature.

If I set an air cooler's fan at 700RPM static, it does the same 'work' at idle as it does when fully loaded. The rate of heat removal doesn't change because the properties of the cooling system remain the same. What does change is the heat load and with it going up, the temperature goes up. Eventually, the device gets saturated/equilibrium... and that is another part of the conversation...

I think I have a problem with the term 'works harder'. Because, to me, a cooler isn't doing work, per say. For all intents and purposes, it's a passive system outside of fans/pumps. It's working the same. There is more heat transfer, but that is a product of the heat source, not any 'work' the cold plate/cooling system has to do.
 
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Dr_Emmett_Brown

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1. You may have missed that I said all other things remaining the same.

Well then of course!

2. The cooler is removing the heat load that it is given. That doesn't mean it is working any harder or less. If the cooler/mater/system is able to move 385 w/k it does so regardless if there is 100W load or 200W load. The difference is in the output... in this case temperature.

Yes but now you are saying Watts per K, previously everyone was saying watts. Technically, by your statement, you are not disagreeing with me, and I am agreeing with this one.
Details are important!

If I set an air cooler's fan at 700RPM static, it does the same 'work' at idle as it does when fully loaded.

Agreed. But when "The cooling loop doesn't work any 'harder' at a given temperature" was said, this left itself open to the implication that this was true for "any given temperature."

I inferred you meant DIFFERENT temperatures.

"The cooling loop doesn't work any harder at a fixed, constant temperature" is the unambiguous statement I agree with.


The rate of heat removal doesn't change because the properties of the cooling system remain the same.

Once at a steady state, I agree. First turn your system on, the heat removal rate is still climbing to this level, otherwise the CPU temp would instantaneously drop to less than ambient the instant after throwing the "power on" switch. There would be no derivative functions in all of thermodynamics if things were constant.

What does change is the heat load and with it going up, the temperature goes up. Eventually, the device gets saturated/equilibrium...

Indeed.

I think I have a problem with the term 'works harder'. Because, to me, a cooler isn't doing work, per say. For all intents and purposes, it's a passive system outside of fans/pumps. It's working the same. There is more heat transfer, but that is a product of the heat source, not any 'work' the cold plate/cooling system has to do.

OK. I shall hunt for a more appropriate set of terms.
 
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EarthDog

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Agreed. But when "The cooling loop doesn't work any 'harder' at a given temperature" was said, this left itself open to the implication that this was true for "any given temperature."

I inferred you meant DIFFERENT temperatures.

"The cooling loop doesn't work any harder at a fixed, constant temperature" is the unambiguous statement I agree with.
Wouldn't it be true for any given temperature? Typically you see temperatures follow 1:1 with ambient. In that if my system yields 90C at 150W with 22C ambient, I'll hit 100c with 32 ambient. Same amount of heat transfer but different ambient temps yield different end temps. But is the cooler doing anything different at all?

Does this extrapolate to sub ambient cooling methods or would the equation have to change?

22C room temp, -150C cpu temp @ 150W, 149mm die area
22c room temp, -40C cpu temp @ 150W, 149mm die area
22c room temp, 90C cpu temp @ 150W, 149mm die area


Edit:
You are still thinking about the "High Score" mentality, and this is not the correct intended use of the formula.
You said this to blaylock.. but the first post says this is a rating. The higher the value (if you dont want to call it a score) the better it performs, right? What is the intended use of the contrived end value? We've heard rating, snapshot, etc and higher is better beat removal, but... I'm still left looking for more than an arbitrary value in the end.
 
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bmwbaxter

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Jun 9, 2010
Just curious as to what is trying to be decided here?

A cooler has set amount of cooling capacity. This is going to be dependent on size, design, type etc... The formula from the first post will have to be consistently applied across different types of CPU's and coolers to give any meaningful data. You would have to take into account the design of the cooler in your testing.

Taking the temps after 3 mins on an air cooler and water cooler will not give accurate numbers since the water will be absorbing excess hit faster than the radiator can dissipate it. The CPU will also change the outcome, soldered or TIM IHS we know makes a difference in obtainable performance. Doesn't help having 800W of cooling if the heat can't get off the die and into the external environment.

I feel like this equation would be better to compare results between the same cooler on different CPU's or different mounts. But you can already do the same thing with a normal temperature reading...
 

Blaylock

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Wouldn't it be true for any given temperature? Typically you see temperatures follow 1:1 with ambient. In that if my system yields 90C at 150W with 22C ambient, I'll hit 100c with 32 ambient. Same amount of heat transfer but different ambient temps yield different end temps. But is the cooler doing anything different at all?

Does this extrapolate to sub ambient cooling methods or would the equation have to change?

22C room temp, -150C cpu temp @ 150W, 149mm die area
22c room temp, -40C cpu temp @ 150W, 149mm die area
22c room temp, 90C cpu temp @ 150W, 149mm die area


Edit:You said this to blaylock.. but the first post says this is a rating. The higher the value (if you dont want to call it a score) the better it performs, right? What is the intended use of the contrived end value? We've heard rating, snapshot, etc and higher is better beat removal, but... I'm still left looking for more than an arbitrary value in the end.

This is what I've been trying to say since my first post on this topic. Doc has tried explain how at least 7 highly knowledgeable people just don't get it. Well, I'm here to say Doc you don't get it. Thank you for the thought experiment. I think it has merit but without the ability to except that someone else might know something or understand something that you don't is going to simply kill this concept. Good luck in the future cuz I'm out.

Peace be with you.
 

godevskii

Member
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Not that it matters much, but just to clarify, some people including the author have been calling this formula an equation a few times. This is not an equation.
 

bmwbaxter

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Not that it matters much, but just to clarify, some people including the author have been calling this formula an equation a few times. This is not an equation.

An equation generally involves a = sign which this does. A formula is just a statement of define outcome not always in regards to math. H20 is the formula for water as an example.
EDIT:I agree that it doesn't matter much especially in the context of this thread
 
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