Thoughts on water pressure & flow rate...

[bigben2k]

Originally posted by 8-Ball
I would definitely like to have a discussion about why the radiators have a max efficiency around 1.5GPm.

Let's do that, shall we?

[Graystar]

Originally posted by 8-Ball
If you reread my post, though it is possible I didn't make the point clear enough, there is no waterblock in the situation I have described.

Oh no, I got that clearly. That's why I said what I did. In the WC system you don't get multiple chances to cool the water. You only get one. That's an important difference.

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.

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

I'm glad you asked!

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.

[BillA]

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

[jaydee]

Originally posted by MMZ_TimeLord
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.

Your temp senario defy the laws of physics so maybe you can better explain why those laws do not apply with your computer. Then we can get into testing methods.... But for a start: http://forums.procooling.com/vbb/sho...&threadid=5765

[MMZ_TimeLord]

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.

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...

[BillA]

pHaestus is kinda 'out in front' with this stuff
although few wish to accept what he says

[Alchemy]

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

[BillA]

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

[8-Ball]

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.

And therein lies your problem.

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.

Furthermore, the decrease in radiator temperature means that the radiator/fan relationship is not quite as good as it used to be.

Again, I believe you may be missing the point some what. For a given CFM there will be a C/W ratio for the radiator. Rather than saying the radiator will be cooler, how can it be? This implies that heat is being transferred from the fins of the radiator to the air at greater than the efficiency rating for the rad.

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

[Alchemy]

Originally posted by Graystar
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.

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.

When we increase flow, the starting water temperature is assumed to stay the same.

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.

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.

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.

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.

You're saying that increasing flow rate increases the power output of your CPU. Think about that for a moment.

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.

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.

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.

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.

Alchemy

[8-Ball]

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]

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.

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.

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.

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??

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.

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]

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]

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]

WRONG

you putting suction on the board traces ?

re-try that one

[8-Ball]

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.

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]

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]

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]

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]

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]

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]

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]

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 Oh wait, I think I just reached equalibrium, now if I could only down more beer than I piss.....

[theetruscan]

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 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]

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

8-ball