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-   -   Thoughts on water pressure & flow rate... (http://forums.procooling.com/vbb/showthread.php?t=6385)

8-Ball 04-18-2003 03:51 PM

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

Graystar 04-18-2003 03:55 PM

Quote:

Originally posted by Alchemy
I'll be sure to call up Newton, Fourier, Bouissinesq, Grashof, Brinkman, Nusselt, Prandtl, Reynolds, Biot, Sieder, and Tate and let them know their concepts of heat transfer are flawed. They will be most displeased.
:rolleyes:

Quote:

All temperatures depend on the heat transfer coefficients. Increasing one of them (or two, if you increase flow rate) will affect water temperature everywhere. So, no.
Lets say I have a radiator the size of Texas. The temperature of the water coming out of it is at ambient. Now lets say I add a couple processors into my water cooling loop. The water out of the radiator is still at ambient temperature because the radiator is just so damn big! The ability of the radiator is something you didn't consider. So, yes.

Quote:

You're pretending two mutually dependent events are independent. They are not. For constant heat rate, the heat transfer coefficient and the delta-T are inversely proportional. Delta-T (and thus water temperature) decreases *because* the heat transfer coefficient increases with increasing flow rate. They are cause and effect, not competing causes.
I'm not pretending anything. First, I can't even figure out what you're talking about. Second, my primary point has always been that you must consider the operation of the WC as a whole when tying to determine the effects of a change. Nothing happens independently. Also, the temperature of the water drops because you now have the same amount of heat absorbed over a larger mass.

Quote:

You're saying that increasing flow rate increases the power output of your CPU. Think about that for a moment.
I don't even see "CPU" mentioned in that quote. What are you smoking??

Quote:

If you do nothing to the airflow, then the temperature difference between the radiator tubes and the air will not change. Altering the water flowrate only changes the temperature between the water and the radiator tubes.
So you agree that you have water at a lower temperature entering the radiator, but somehow the temperature of the radiator stays the same...right. :rolleyes:

Quote:

Increasing flowrate increases heat transfer. Increasing flowrate causes adverse effects to cooling *only* when you are putting in enough energy by pumping that the improvement in heat transfer coupled with the increased heat load yields higher temperatures.
I'm sorry, but that is simply not correct.

Skulemate 04-18-2003 04:18 PM

Quote:

Originally posted by Graystar
I don't even see "CPU" mentioned in that quote. What are you smoking??
I think what he's getting at is the fact that we're dealing with an equilibrium situation when cooling a CPU... the heat transferred is always equal to the heat output of the chip. The radiator will never work any harder than this... that's all the heat there is. (Neglecting pump heat, of course.)

Graystar 04-18-2003 05:29 PM

Quote:

Originally posted by Skulemate
I think what he's getting at is the fact that we're dealing with an equilibrium situation when cooling a CPU... the heat transferred is always equal to the heat output of the chip. The radiator will never work any harder than this... that's all the heat there is. (Neglecting pump heat, of course.)
Yes, we have an equilibrium situation. but we can shift the states of equilibrium of all the components involved in our watercooling system. When we increase the performance of the WC system we do get a little more energy out of the system. That is why the CPU temperature drops...it has less heat energy within it.

BillA 04-18-2003 05:38 PM

WRONG

you putting suction on the board traces ?

re-try that one

8-Ball 04-18-2003 05:38 PM

Quote:

I'm not pretending anything. First, I can't even figure out what you're talking about. Second, my primary point has always been that you must consider the operation of the WC as a whole when tying to determine the effects of a change. Nothing happens independently.
I'm sorry to say, but your grasp on the principles involved is lacking.

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:

Also, the temperature of the water drops because you now have the same amount of heat absorbed over a larger mass.
Now who's smoking what?:shrug:

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

8-Ball 04-18-2003 05:43 PM

Quote:

Originally posted by Graystar
Yes, we have an equilibrium situation. but we can shift the states of equilibrium of all the components involved in our watercooling system. When we increase the performance of the WC system we do get a little more energy out of the system. That is why the CPU temperature drops...it has less heat energy within it.
My god, you really don't know what you're talking about.

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

Graystar 04-18-2003 05:44 PM

Quote:

Originally posted by 8-Ball
By increasing the flow rate, have we mysteriously added more water?
That is exactly what we have done. If you can't see that then I guess there's no point in discussing it further.

8-Ball 04-18-2003 05:47 PM

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

theetruscan 04-18-2003 05:52 PM

Quote:

Originally posted by Graystar
That is exactly what we have done. If you can't see that then I guess there's no point in discussing it further.
Graystar, please tell me you're being sarcastic, here especially, but throughout this thread as well . . . please!

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.

BillA 04-18-2003 06:40 PM

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)

8-Ball 04-18-2003 06:50 PM

Quote:

Originally posted by unregistered
I must recant
Had to go to dictionary.com on that one.

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?

gone_fishin 04-18-2003 06:54 PM

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.....

theetruscan 04-18-2003 06:56 PM

Quote:

Originally posted by gone_fishin
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.....
Remember, if you piss faster, your piss gets heavier.

8-Ball 04-18-2003 07:12 PM

stop it, your making me laugh too loud, and everyon else in the flat is asleep.

8-ball

BillA 04-18-2003 07:12 PM

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

Since87 04-18-2003 07:12 PM

Quote:

Originally posted by theetruscan
Remember, if you piss faster, your piss gets heavier.
But does your pee-pee get colder?

Since87 04-18-2003 07:13 PM

Dammit Bill.

8-Ball 04-18-2003 07:13 PM

Creepy, 3 posts at exactly the same time.

Graystar 04-18-2003 07:30 PM

Quote:

Originally posted by theetruscan
Graystar, please tell me you're being sarcastic, here especially, but throughout this thread as well . . . please!
Flow rate is volume over time. When you increase the flow rate you increase the volume over a given amount of time. So if you double the flow where you had a quart of water flowing by previously, you now have 2 quarts flowing by. It doesn't matter if it's from the same gallon of water that is circulating over and over. The fact is that for a given amount of time more water is being used for heat transfer.

Graystar 04-18-2003 07:38 PM

Quote:

Originally posted by 8-Ball
We have the "70W" going in and "70W" going out of the same volume of water. IE all of the water in the system.
All of it eh? Then how is it that the CPU is hot? I mean, if *all* of the heat energy is being removed, I'd expect the CPU to be at ambient temperature.

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:

I think much of your problem comes from considering each molecule of water. Overcomplicated and always likely to end in trouble.
Just as oversimplification will lead to trouble.

Skulemate 04-18-2003 07:44 PM

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.

8-Ball 04-18-2003 07:48 PM

I give up.

8-Ball 04-18-2003 08:10 PM

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.

Alchemy 04-18-2003 08:34 PM

Quote:

Originally posted by Graystar
All of it eh? Then how is it that the CPU is hot? I mean, if *all* of the heat energy is being removed, I'd expect the CPU to be at ambient temperature.

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.

Jesus Christ.

8-Ball, good luck. I give the hell up. And I'm drunk. But mostly, I give the hell up.

Alchemy

8-Ball 04-18-2003 08:46 PM

I should have given up last night when I was drunk, though it seemed to make me even more determined. (read stubborn)

Graystar 04-18-2003 09:23 PM

Quote:

Originally posted by 8-Ball
Oh go on then, one more try.
At my job we have an internal chat system. Sometimes one person is trying to explain something to another, and for some reason or another, the knowledge is just not getting through. So one person calls the other on the phone and 30 seconds later they realize they were saying the same thing and that everyone's understanding is in the right place.

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.

BillA 04-18-2003 09:37 PM

nope

Cathar 04-18-2003 10:18 PM

Sounds to me like Graystar is confusing secondary heat paths with the heat energy "needed to keep to CPU hot".

myv65 04-18-2003 10:48 PM

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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