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I think I can defend myself on this one.
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Very wrong!!! What you are stating is that the water will be cooled further if it is left in the radiator for a longer period of time. SO WHAT!! Let us suppose that we have a radiator with an internal volume of 1 litre. This is attached via a pump to a reservoir (no heat load), with a capacity of 9 litres (including the tubes between res, pump and rad. Lets say the water in this "system" is at 50 degrees C and we want to cool it down to ambient with the rad. There are TWO ways of doing this. The are extremes at either end of the scale of reality, though they are reasonable for the argument here. 1. Pump one litre of water into the radiator, then leave it until it is cooled down to ambient. Then pump the next liter in and so on ten times until all of the water has been cooled to ambient. 2. Now pump the water at such a speed that the turbulence in the radiator reduces the barrier thickness (slow moving/stagnant water at the interfaces between copper and water) effectivel to zero. Keep pumping the water around until the water reaches ambient. Now I ask you. Which one will reach ambient first. In both cases, we are removing EXACTLY the same amount of heat energy from the water. Answer - 2 Your approach to this is wrong. I am correct in saying that you need to consider the whole system to be able to understand the implications of what happens when you change something like the flow rate. In fact, rereading your post, you have already shot yourself in the foot. Your argument, be it already flawed, that the water leaving the block will be colder is preceded in the previous paragraph by a statement that water leaving the radiator will be warmer. These will cancel each other out. For the sake of argument, lets assume that the water temperature does not vary throughout the loop. This is a reasonable assumption as it actually doesn't vary by much more than a degree anyway. Now lets think in the opposite direction. Lets start from the air. Lets assume we have a cpu producing 70W. The radiator only has to dissipate 70W No more, because there IS no more. The performance of a radiator as we require it, is determined by the efficiency. C/W - the temperature difference between the coolant and the air required to transfer 1 Watt of heat. This value decreases with increasing flow, as the barrrier region is broken down at the surface of the water. This is a layer of water which "sticks" to the surface. This is not good, as heat must get through this barrier from the faster flow of water in the center of the tube to the copper tube. Water is not a good conductor. What breaks it down - turbulent flow from faster flowing water. So we have reduced the C/W value by increasing the flow. Can you now see, that to remove a fixed value of heat, we will need a lower temp difference. So the average coolant temp will be reduced. Equally, the C/W ratio for the waterblock will also be reduced by the increased flow. This means the temperature diffeence between the cpu and the water can be lower while still having the block remove 70W of heat. But the coolant is already cooler than before (on average) and the cpu will be cooler relative to the water than before. In essence we have reduced the temperature difference required at the two heat exchanging steps, and the temperature is fixed at one end, the ambient air temp. So now, can you see that you HAVE to consider the whole system otherwise you will fall into the trap of believing that slow flow in a rad is good. The only other thing to consider, is the "point of diminishing returns". This is the point where by additional heat produced by the larger pump, and friction in the tubing from increased flow, means that the overall heatload goes up. At first, the increased efficiency will more than compensate for the additional heatload, but there will come apoint where the heatload will become to much, and more flow will result in and increase in cpu temperature. However, this opint is beyond what most people's systems are capable of, particularly with the heatercores and high flow blocks which dominate the market at the moment. I hope this clears things up a little. 8-ball |
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8-ball: Nice explanation. |
So let me see if I understand 8-Ball correctly. Please pound me mercelessly if I'm incorrect :dome:
So, with the water moving too slowly, we get a barrier of water itself that keeps the heat from properly getting to and from the flowing water to transfer the heat out of the system at the radiator and into the system at the source (CPU). If we move the water too fast there is a point at which we will begin to add more heat than we are removing with the added efficiency. So the idea is to get that balance where we move the water just fast enough to break down that barrier of stagnant water with turbulance and cavitation, but not so fast that we are adding more heat with the need for a stronger pump that only contributes to the problem. Is that the basics of what you were saying 8-Ball? I feel this thread has taken a toll on my grey matter... :drool: |
That's probably the best summary of the principles behind watercooling I have seen. Short and to the point.
Though you might want to remove "cavitation" as this is where negative pressure either at pump inlets are around ship propellers causes the water to vapourise. This is not really a good thing as it will wreck efficiency. Though I knew what you meant. Edit that out and it will be a good summary to this thread which I now have linked in my sig for the benefit of anyone who believes otherwise. Glad I could help 8-ball |
Yeah, you're on the right track now!
But the balance point is so far up there, that you're not likely to reach it: if you ever get a pump that puts in enough heat as to make a significant impact, post a picture of it, OK? ;) BTW, the more "turbulent" the flow is, the easier it will be for the water to shed its heat (or pick it up). The measurement of turbulence is in Reynolds, where <2'000 is considered laminar, 2'000 to 4'000 is in the transition zone, and >4'000 is fully turbulent. If you can run the numbers, you'll find out that no one is achieving fully turbulent flow (jet inpingement excluded). |
and the toll would be far higher were facts used instead of 'theoritical' assumptions
8-Ball is wrong those rads that I have tested show a C/W 'peak' (a minima actually) at 1 to 1.5 gpm (will try to post this afternoon) now what ? as a friend of mine (expert in CFD modeling with a 30 teraflop computer) says: another beautiful theory destroyed by facts |
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I'm very much looking forward to your article, Bill. Your contribution to the watercooling world is absolutely amazing. I thought you were going to drop the whole radiator issue, where we still needed more info. Thank you. |
That would probably be one of the posts from last night. Imeant to read over those and check for wine related mistakes.
However, it is still the case that 1.5Gpm is still quite highfor a watercooling loop. Add to that the fact that most waterblocks will continue to gain efficiency beyond that flow level. BillA, can you tell me which post that was in as I haven't the time to reread them all. Too much revision to do. As for the minima at 1.5GPm, If you were recording the temperature of water as it entered the radiator, could friction within the rad contirbute to additional heat while not appearing on your measured temp? Just a thought as I'm trying to get my head round it all. 8-ball |
the coolant inlet temp (offset) is held at 5.0°C (±0.6) wrt the rad inlet air
coolant temp measurement resolution 0.01°C (±0.02 inst uncertainty) taken at inlet and outlet C/W calculated using mean temp between inlet and outlet as the difference between this and the "logarithmic mean temperature difference" (normally used for heat exchanger calculations) is close to nothing for these deltaTs note that the values are quite small I am just observing that 'more is not necessarily better', even though that is what the theory says and if you observe the graphs linked to it is seen thet some behave VERY differently lots of reasons for such I agree that few WCing systems achieve a 1.5gpm flow rate, but the serious can do so without much difficulty - just a LOT of attention to detail |
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You're example, where you suggests that pumping water around will cool faster, is indicative of the general problem of understanding this issue....you continue to take components out of the system to review their behavior. Pumping water around and around will NOT cool the water faster. Why? Because of every loop you have a waterblock adding 6 therms of energy to the water! You will never understand what is happening if you separate the components. The zero flow example was an extreme. Just like saying that at infinitely fast flow, the waterblock is the same temperature as the water and the water temperature never rises. It simply illustrates what the pertinent physics formulas will tell you. Finally, when you reduce the temperature of the water, you increase the difficulty in extracting the same amount of heat energy from it. The closer the water temperature gets to ambient, the harder it is to cool it. That's why the benefit from faster flow is countered in the radiator. The water from the block is at a lower temperature, so the radiator will have a more difficult time performing it's job. The easiest way to see this is to just create a simple hypothetical system and use the heat tranfer formula to see how transfer rates and loads are affected. |
It is true that you will NEVER actually reach a CPU temperature that matches the coldest air temperature measured going into your radiator.
As stated the closer you get to ambient temp the harder it is to get to that point... kind of like a half-life or half-step... if, with every step, you cover half the distance to your goal... you will never reach your goal. Practiacally speaking, you would get close enough that you wouldn't care about that last 2 to 5 degrees. The CPUs we are trying to cool have an EXPECTED operating temperature of what... 40° C to 50° C? Average room temperatures in a home are what? 21° C to 25° C? That's a pretty big difference... if you can get your CPU core temperatures down under 35° C you are doing well in my opinion. If you are under 30° C you are doing really well.. in fact I am more than happy with just being under 30° C. If you get under 25° in a room that is 21° C without a peltier or some chilled water setup... you are doing EXTREMELY well. Again, I was always under the impression that the goal was to keep our CPUs cool under load, especially when we overclock. If they look good and function quietly in addition... then so much the better. My goal is to have cool and quiet... not just a pressure head that could lift a car and cool a volcano. :p |
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do some (remedial for this site) reading then make a list of each thermal impedance between the CPU and the air then do some more research and ascribe a value to each then add them up -> then you will understand what tripe you just posted get serious |
I don't think I've ever been accused of posting "tripe" but thanks... :D
I'm just quoting what I know... my system temps are BOTH under 30° C... if this is NOT considered low... well I apologize to all the readers and will post no more comments on actual temperatures. :shrug: |
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I was merely describing two situations where we are attempting to cool the same 10 litres of water from the same temperature with the same radiator and no additional heat load. The only difference is that in one scenario, we have water moving, breaking down the barrier layer, and in the other, we have the water stationary with effectively a full barrier layer, ie the thermal energy from the water at the center of the tube/channel must conduct through the water rather than turbulent flow randomly carrying it into the copper wall. And yes, I agree that the water entering the rad will be cooler, but each "packet" of water is carrying less thermal energy. It is plain to see that the radiator will dissipate more heat if the water is warmer, but it doesn't have to dissipate more heat. The heat load is set by the processor. We know that the processor will rise to such a temperature that the thermal gradient across the WB baseplate transfers 70W of heat. However, this does not determine the water temperature directly. The water temperature will be set by the efficiency of the radiator and the resulting delta T required over the ambient air temp so that it can remove the 70W. If you suddenly slow the flow rate down in a watercooling loop, it is the reduction in the efficiency of the radiator which will cause the overall "average" water temperature to rise. This in turn will cause the cpu to heat up to a temperature where the delta T is sufficient to transfer 70W to the water. I'm not separating components as such, merely working back from the only fixed variable we have, the air temp, at each stage considering the efficiency of the heat exchanger and the corresponding delta T required to move 70W, which is all we are asking the system to do. Ask yourself, this, how is it that increasing the efficiency of both the waterblock and the radiator would result in an increase in the cpu temp, as this seems to be what you're implying. 8-ball |
Well, not having access to a "Lab quality" temperature meter... I have simply used several commercially available thermometers and averaged the results.
Once that was done at idle and under load I "adjusted" my MBM5 software to compensate. As stated before the sensors on most motherboards are neither accurate or placed in very good spots. I then remeasured my temps with the external thermometers (yes they had thermister probes) and verified that my MBM5 temps were within 1 degree C. I will see if I can obtain a temp meter from the lab and bring it home to verify. My ambient air temperatures in my apartment range from 21° C to 25° C this is also based on those same external thermometers. I have not directly monitored the water temperature, so I guess I can't give numbers for that. :shrug: If my methodology is flawed... please relate the proper proceedure for measuring these temps without a lab quality meter. :confused: |
BillA,
Am I being daft somewhere along the line because it appears that the theory which I am describing is sound, yet it is still causing trouble. I would definitely like to have a discussion about why the radiators have a max efficiency around 1.5GPm. The point I tried to make earlier was that no matter how accurate your measuring devices, you would not be able to detect any additional thermal load reducing from increased friction WITHIN the radiator itself as this is AFTER your initial temperature reading. I'll have a think about this this afternoon and try and explain my thinking a little better. That's provided you don't point out any obvious mistakes which I'm making and have not seen. 8-ball |
I take NO exception to the 'theory' of thermo or fluid mechanics
(even I am not so daft) but 'reality' (in the form of un-accounted for influences) often skews the results http://www.thermal-management-testing.com/6ddcube.gif can you describe what is occuring here ? (pop quiz, lol) note that max efficiency is unique for each rad re the frictional heating come on, more than offset by the internal pressure drop WAY too small and the readings are at the inlet and outlet, any such is included (as is the effect of the pressure drop - several threads on this) |
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Please correct my impression if it is wrong, but I get the impression that you think the radiator doesn't need to cool as much because the water temperature is lower. This, I'm afraid, is completely wrong. Yes, the water is cooler, but the rad has to cool more water for a given time period than with a slower flow rate. So the radiator really does have more energy to transfer. It must, as that is the only way to reduce the temperature of the CPU. Quote:
In a typical system, we have water, at a given temperature, going into the waterblock. The water is heated to a higher temperature and then goes to the radiator. There, it is cooled back to the starting temperature and the cycle begins again. When we increase flow, the starting water temperature is assumed to stay the same. This means that the increase to waterblock performance is based solely on increased flow. The performance of the radiator, however, will be affected by two factors. The fundamentals of heat transfer tell us that the water exiting the waterblock is now at a lower temperature. Since we are now starting with a lower temperature for the radiator, the performance is affected by increased flow, *and* decreased temperature. Since a temperature decrease will reduce performance, the benefit from increased flow will be countered. Yes, the radiator can now move heat faster, but the waterblock's performance increase is still bigger. The radiator will get more heat than its performance increase can account for. The bottom line is that the radiator is going to have to work harder. Furthermore, the decrease in radiator temperature means that the radiator/fan relationship is not quite as good as it used to be. This fact exacts another toll on the radiator. The result from these additional influences is that the radiator is walloped with lots of extra work. So, what happens to CPU temperature? If you have a good radiator that can rise to the challenge, and cool water down to the point it did before, then you will have a lower CPU temperature. However, if the radiator can't rise to the challenge and can't remove the heat, then it will leave the heat in the water. This will undo the flow benefit, and temperatures will rise to the point where there is once again balance between the heat taken from the CPU and the heat expelled by the radiator. This balance will occur at a higher CPU temperature. So we see that it all depends on the radiator. Anything you can do to help the radiator do its job better, will help you realize the benefit of increased flow. Pushing more air through, or changing the rad outright, may be required. If you had a good radiator to begin with, and you weren't using all of the radiator's cooling capacity, then increasing flow will drop your CPU temperature by putting that extra capacity to work. |
a minor clarification to Graystar's description
all rad 'capacity' is always utilized, its not as if there is ever unused capacity what is happening is that a rad has 2 efficiency curves the major one related to the air side flow rate the minor one being the liquid side flow rate (although this can be significantly manipulated) the two affect each other, so a shift in either to a 'point' of greater efficiency will result in increased cooling - but since the actual goal is the lowest possible coolant exit temp from the rad, which is in conflict with the optimization of the wb's performance needs (always max flow rate); the "theoretical" calculation of the optimum compromised flow rate is beyond anyone without actual performance curves for the relevant components |
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Well... now that I know my calibration method and measuring method were off... I retract my temp statements.
Until I've been able to modify a CPU and do proper bath calibration and later measurement, I will not quote any more temps. :cry: I feel I owe Jaydee and BillA an apology in this area too. :( I came up with the proceedures I use all on my own and it shows, thanks for the info guyz... :rolleyes: |
pHaestus is kinda 'out in front' with this stuff
although few wish to accept what he says |
Christ, I'm gone for a day and BillA manages to blow everyone's cerebral cortex. Let me get the paper towels.
Okay, BillA, I don't know exactly why you're getting the charts you're getting. Not saying you're violating any rules of physics, but your current explainations, if I'm reading them right, are. For your rad, the heat transfer coefficient of the air should not be adversely affected by improving the heat transfer coefficient of the water. They are completely independent phenomena. My first job when something goes against theory is to recheck everything. What sort of precision can you get by generating replicate data? What is the accuracy of your temperature measuring equipment? Your flow metering equipment? Your pressure metering equipment? What are your raw data so others can check your calculations? And, assuming you're correct on all this, what is really happening? Alchemy |
short answers (trying to get my main computer back up)
for sure my explanations are inept yes, air and water quite independent, and treated in the experiment's design as independant let me get my box back up and plot the new data (many problems with the old) but those bumps are quite real not so difficult really but in the other thread perhaps ? - Null-A Ben started another later |
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That statement is WRONG. The processor has a given output of heat. Lets say 70W, for the sake of argument. If you had an infinitely efficient radiator, and an infinitely efficient waterblock, the processor would be at ambient temperature, BUT THERE WOULD STILL ONLY BE 70W BEING TRANSFERRED INTO THE WATER AND ONLY 70W TRANSFERRED INTO THE AIR BY THE RAD!!!!! This is the key principle to understand. If you don't understand this then there is no point in me carrying on. Quote:
The air temp is fixed - yes? The C/W ratio at a given cfm is fixed - yes? The heat load is fixed at 70W - yes? Then the fins of the rad WILL be at ambient temp + 70 x C/W (at the given CFM) For a 70W load this is the equilibrium position and will not change! This will not be affected by a change in flow rate, however, a change in flow rate will change the C/W ratio for heat going from the water to the fins. We have already established, that the fin temperature is dependent on the fixed heat load, the fixed ambient temp and the fixed efficiency, resulting from a fixed CFM. Lets call this Tf, for fin temperature. Now lets start with a given C/W for a given flow rate. The average water temperature in the radiator should be such that the temperature difference divided by the C/W is equal to 70. In other words, the average water temp in the rad needs to be; Tf + C/W x 70 Now lets improve the efficiency. An improved efficiency is a lower C/W ratio, agreed? So lets call this (C/W - D). D for difference. So now, the average water temp in the rad will have to be; Tf + (C/W - D) x 70 = (Tf + C/W x 70) - (D x 70) Can you now see that the average water temp from an improved efficiency is now lower than before by an amount (D x 70). Granted, D is small, probably of the order of 0.01. However 0.01 x 70 is of the order of 1 degree difference in average water temp throughout the loop. Now I will concede that the water temperature does not vary linearly as it passes through the rad, but I have taken the liberty of evaluating the temperature change removing 70W from a water flow at 6lpm. which is roughly 1.5 gpm. It turns out to be 0.16 degrees. We know therfore, that the average temperature of the water has a margin of error of 0.16 degrees. Dropping this by 1 degree will reduce the temperature of the water going into the block, which will in turn reduce the temperature of the cpu. Sorry the last bit's rushed, but I've run out of time. Any further questions or points you would like to disagree with, carry on. 8-ball |
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Alchemy |
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