I think I said pretty much the same thing as alchemy, which reassures me somewhat. I always worry on its only two people carrying out a debate. It can turn into a shouting mach pretty quickly.
Alchemy, does my post state anything which contrevenes any fundamental laws. I'm pretty sure it doesn't, unless I forget ny advanced thermodynamics course I did a couple of years ago. 8-ball |
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WRONG
you putting suction on the board traces ? re-try that one |
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I will be the first to admit that there are effects which will alter the theoretical outcome, as with everything, but the theory still stands and is relatively accurate. As I say over and over, yes you do need to consider the whole system, but the only data we know is the flow rate, the efficiencies, the heat load and the ambient air temp. The only things we can directly control are the heat load, the ambient air temp and the flow rate. (air and water) The efficiencies are determined by the flow rates, which in turn determine the delta T required to shift the set heat load. THIS IS HOW IT WORKS. Consider this. There are equations for the rate at which the water cools in the rad, the rad temp relative to the ambient temp, the water temp relative to the rad temp, the waterblock temp relative to the water temp and the cpu temp relative to the waterblock temp. This is essentially a set of simultaneous equations which can be solved. However, the only temperature we know is the ambient air temp, so we MUST work backward from this. Is this clear. The thermodynamic principles I outlined earlier are correct. Quote:
By increasing the flow rate, have we mysteriously added more water? The only concept you need to grasp is that watercooling is all about different components/mediums reaching equilibrium temperatures relative to the only defined temp, the ambient air, such that the differences in temps with the corresponding heat transfer coefficients result in the treansfer of the defined heat load. Changing variables such as flow rate air or water, will increase the heat transfer coefficients, thus reducing the temperature differences. Note this doesn't take into account heat from the pump, though differnt pumps may produce the same flow with different heat input. This is one of those real world factors that is hard to control. but hey, lets get the basics sorted first. 8-ball |
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The last statement here is false. The cpu puts out a fixed heat load for a given frequency. This does not change! The cpu temp is reduced by increasing the efficiencies....... Oh god, someone please shoot me. I can't go on like this! Over at Bit-Tech, they have a nice head banging against a wall smilie. This would be appropriate. 8-ball |
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We have the "70W" going in and "70W" going out of the same volume of water. IE all of the water in the system.
I think much of your problem comes from considering each molecule of water. Overcomplicated and always likely to end in trouble. 8-ball |
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If not, grab a couple of books, put toghether a solid understanding of the most basic aspects of thermodynamics (fluid dynamics is rough), then reread this thread. Other people are trying to help you get an accurate understanding of what is happening, and you keep providing spurious counterexamples. |
I must recant
it is now very clear Graystar knows the words, but not the principles sorry all Graystar - stop posting 'till you sort it out (you will come out better so) |
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Hey what can I say, I'm a scientist. I'm not ashamed of that. Damnit, where's the blushing smilie gone?!? 8-ball PS, I tried to PM you earlier, but your mailbox was full. Do you not check them anymore? |
Smoking???? This guy's hooked up straight to the IV man. I want to know if he does tarrot cards and palm reading on the side:D Oh wait, I think I just reached equalibrium, now if I could only down more beer than I piss.....
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stop it, your making me laugh too loud, and everyon else in the flat is asleep.
8-ball |
no man, it chills your dick
got to hold it to prevent freeze up EDIT: 8 ball cleared out the box, or click the link to the site and e-mail me man, this thinking has given mve a head cold |
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Dammit Bill.
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Creepy, 3 posts at exactly the same time.
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Of course, the answer is that we are *not* removing all of the heat energy produced...only most of it. What we can't get remains to keep the core hot. Quote:
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If we didn't remove all of the heat that the CPU is producing, it would continue to rise in temperature until it killed itself. What you need to understand that heat and temperature are not the same things... and that the heat flow requires a temperature differential to drive it.
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I give up.
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Oh go on then, one more try.
In the thread re your tiny waterblock, you stated that you need a temperature gradient to move the heat (I almost thought you actually DID know what you were talking about when I saw that and that you'd been stringing us all along. Not to be I guess) It's still the same. The temperature difference sets up a temp gradient which will provide a driving force for the transfer of heat from the cpu to the waterblock. If for some reason the coolant rose in temp (like you'd suddenly plugged a hot tap/fawcet into your loop), the temp difference would be too low. Ie the temp gradient would be shallower. I believe from your other thread that you understand the implications of this. Less than the "70W" will be tranferred from the cpu to the waterblock. But the processor is still producing "70W", so, excess thermal energy will begin to build up in the processor. This will cause the temperature in the processor to rise. As this happens, the temp gradient will gradually steepen, driving more and more heat off the processor per second as it steepens, until a point comes when the gradient is the same as it was before, allowing the transfer of 70W of heat from the processor to the block. The processor will stabilise at this temperature, because it is producing 70W and the temp gradient is just right to transfer 70W. You could say it is in equilibrium. Now suppose the cpu begins to idle. The power output drops to 60W, yet there is still a gradient driving 70W of heat into the block. Because the block is now removing more heat than is being produced, the cpu will cool down, with the thermal gradient getting shallower as it does so, until such a point is reached where 60W is transferred to the waterblock. Again, the temp will stabilise, and the cpu and the block are "at equilibrium". In reality, this process happens very quickly, with the heat removed by the block matching the fluctuations of the cpu power output. I hope this makes sense, as it is a key principle to understand. If not, query it bt by bit and myself or others will try and help. We may come across as stubborn, but we would be less so if you stopped and listened to what we said just for a moment, to allow a coherent argument to be put together. What do you think? 8-ball PS, sorry if I came across as a little harsh earlier. |
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8-Ball, good luck. I give the hell up. And I'm drunk. But mostly, I give the hell up. Alchemy |
I should have given up last night when I was drunk, though it seemed to make me even more determined. (read stubborn)
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Such is the difficulty of trying to explain things in text in a fluid discussion. What I type may make sense to me (at the time) but not to others. So, maybe if we keep this up we might all end up on the same page at some point. :) Your description of the temperature changes is correct, but that's not what I was addressing. What I was saying is that you can't keep the core hot for free. The CPU is at a temperature above ambient and remains there so as long as the computer is on. Some thermal energy is used in maintaining that elevated temperature. So when a processor dissipates 70W of power in one hour, I'm saying that 238 BTUs are removed by water cooling, and 1 BTU is used keeping the core hot. (values are demonstrative only) So, when you increase flow, and increase the heat transfer rate of the system, you will now remove 238.1 BTUs during that hour, leaving .9 BTUs to heat the core. This give us a cooler core. So you really are removing more heat, but the amount of thermal energy produced and transferred remains the same (as it must.) I hope that's a better explanation. |
nope
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Sounds to me like Graystar is confusing secondary heat paths with the heat energy "needed to keep to CPU hot".
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I'd like to invite Graystar over for a while. You see, my wife wants this lily pond in the back yard and Graystar seems the perfect person to keep digging a hole deeper, and deeper, and deeper. . . . .
Speaking as someone with a BSME and ~15 years into an engineering career, I must say that your writings in this thread would not make sense no matter how much beer one had consumed. Please, please heed the advice of countless folks here and step back for a bit. Not only are your posts technically incorrect, but they're wandering about rather aimlessly, too. Sorry for such harsh words, but you have earned them. Getting back on track The bit about radiator effectiveness vs flow rate has puzzled me for a while. I profess that I can not offer an explanation for the information that Bill has measured. I would say there are a couple of things to consider. One is that all of our wonderful theory boils down to empirical observation. The equations for determining what happens with "an infinite flow field over a flat plate" are cumbersome enough to confuse most people, yet don't come close to describing what may happen inside a radiator. The fact is that flow regimes near a radiator's inner tube walls are beyond what simple equations can model. I can only surmise that the interior geometry and surface finish make them particularly amenable to specific flow velocities. The other thing has been mentioned here before, but is worth restating for clarity. This involves the energy from pumps and the true energy balance on the system. As 8-Ball has observed, the power from a CPU is practically a constant. What will differ, albeit minutely, is the percentage of CPU heat lost through secondary pathways. As the chip gets closer to ambient, less of its thermal energy will leave via secondary paths and a greater percentage will go into the fluid. I will not attempt to quantify this other than to say "it's minor". The pump energy remains one of the most misunderstood (and denied). When you look at the fluid you must consider all sources of heat input and exit. Even with inline pumps, most of the juice flowing from the wall socket ends up in the fluid. Because radiators are fairly restrictive and because pump power tends to rise approximately vs flowrate^2, you could reach a point where power dumped in via the pump starts to be a major player in the game. The radiator must get rid of both the CPU's energy as well as the pump's. Sure, speeding water up will generally increase the convective coefficient, but the total energy input to the water also goes up. This may not matter if you're using a pump that peaks at 12 watts, but start getting into 30, 50, or 200 watt pumps (yeah, I own a 200W pump, but it's for that lily pond I mentioned earlier) and the CPU power can become the insignificant factor. Given all of this, I would expect there to be a unique flowrate for each radiator where the increase due to added fluid velocity no longer outweighed the added power to make that fluid move faster. If one was to graph it, I suppose it would be sort of a wide, upside-down parabola. Bill's data seems to hint at this, but doesn't continue tailing off with added flow. Ah heck, maybe his data is just playing tricks with my mind. Then again, maybe there's something missing somewhere in his measurements. I don't think the latter is true, but I can't completely dismiss it either. One thing remains certain in my mind. When you're working with ambient air as your radiator cooling source, the added bang from really high flowrates simply is not worth the bother. Getting even a few °C requires higher air flow through the radiator and good fluid flow through the system. So you buy your loud 120mm fans and your Iwaki pump. Congrats. Now you have something darn near as loud as a really good air heatsink, but with nowhere near the portability or reliability. It may only be viewed as another hobby, destined to consume time and money for a payback that can only be measured in one's sense of satisfaction with doing something new/different. |
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