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

But does your pee-pee get colder?

Dammit Bill.

Creepy, 3 posts at exactly the same time.

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.

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.

Just as oversimplification will lead to trouble.

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.

I give up.

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.

Jesus Christ.

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)

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.

nope

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

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.

15 years.....hmmm.

Hey 8-Ball and Alchemy, when you recover from your hangovers take another stab. If nothing else, you guys are helping me get a better handle on this stuff.

Although Graystar is phenomenally obstinate about holding onto his ignorance and misunderstanding, there is some evidence that he may actually be able to acknowledge (at least to himself) that he is wrong, from the fact that he hasn't posted in this thread in a few days. (Sorry, my thought processes seem to consist primarily of run on sentences, and parenthetical expressions)

Once again you are displaying fundamental misunderstanding.

Ever put coffee in a thermos and drink it a few hours later? Still hot? Hmmm. Where's the energy input that maintained the temperature of the coffee?

Unless there is a path for thermal energy to leave an object, it will maintain its temperature. If thermal energy is continuously added to an object with no 'heatsink' the object will continually increase in temperature.

You seem to have 'power' and 'energy' confused.

Power is a rate of energy usage/dissipation/flow/whatever. So it would make sense to say, "When a processor use 239 BTU's (70 Watt-hours) of energy at a constant rate for an hour, the rate of energy usage is 70 Watts."

Yes there is greater thermal energy in a processor that has been running for some time, than in one that has not. That elevated energy level is 'achieved' in the process of reaching thermal equilibrium. Once equilibrium is attained though, no further energy is "used" in maintaining that elevated energy level. (and corresponding temperature)

At equilibrium the flowrate of electrical energy that is dissipated in the CPU must be equal to the flowrate of heat energy leaving the CPU, otherwise the energy in the CPU would continually be climbing, and the temperature would be climbing with it.

To go back to your example:

If in one hour 239 BTU's of electrical energy were dissipated in the processor and only 238 BTU's of thermal energy left the processor then it is not at equilibrium yet. Are you suggesting that after the next hour there are 2 BTU's left, and then 3, 4...?

Graystar, I know that you understand that the coolant flowrate is the same through all components of a water cooling loop. Is it such a stretch to understand, that the energy flowrate is the same through all the components of a thermal system at equilibrium?

Think about it.

Thanks for reminding me.

Shortest kegger ever. Ah well, got my money's worth. Sucks to be sober before last call, though.

Anyway, I'm not sure what I can tell you. Seems like you got this stuff down.

Which reminds me. . . BillA, why exactly are you calculating disspation in BTUs instead of BTU/hr in the charts on page 2 of this thread? Or is that just a recurring typo you have? I was hoping to see your calculations for your charts, but you don't have any on your site. Presenting dissipation in terms of energy is nonsensical, though I think you already know that.

As for Graystar, I should have more patience, I guess. Hell, one of my teammates on our capstone chemical engineering design project got confused about the difference between heat and temperature. Maybe it's not as obvious a distinction as it seems to me.

Still pisses me off when people don't listen.

Alchemy

To quote from Bills article* :-
" “Btus” as used in this article are Btus/hr and can be converted to Watt*hours by multiplying by 0.2931"

* http://thermal-management-testing.co...esting%201.htm

:mad:

Since87... why did you have to remind him? I was tired of arguing :p

Graystar,

believe it or not, I think we might be making "progress".

I am now "refreshed" and ready for a long work filled day. Saturday I know, but finals are finals and it would be a shame to waste 20 years of education!

This is SO true, but you need to try and step back and get into the head of the author. Spend time making sure you really understand what has been said before picking through it.

Back to the subject in hand.

Would it help if I tried to explain it from a slightly different perspective.

Firstly, lets assume that we have a lump of metal on the table. This piece of metal is 1kg and has a specific heat capacity of 5 J/g/K.

If 5 KJ of thermal energy are "injected" into the block of metal, the temp rise would be defined as

(energy input) divided by (mass x specific heat capacity)

With the numbers I have suggested, the block would rise in temp by 1 degree. Now if we insulated this block such that none of the thermal energy can escape, it would stay at 1 degree above ambient.

So for no thermal energy going into or out of the block (perfect insulator) it will stay at the same temperature.

Now imagine we bury a resistive heater in the block, which has a power rating of 70W, ie 70 Joules of thermal energy per second.

At the same time we put a water jacket around the block inside the insulation which only removes 70W.

So we have 70W in and 70W out, IE, every second 70 Joules in and 70 Joules out. But we still have the original 5KJ, and that aint changing.

So the net change to the thermal enery of the block is zero. So, according to the specific heat capacity, the temperature will not change.

To raise the temp of the block some more, we need only add a finite amount of energy. Another 5KJ would raise the temp by 1 degree again.

So lets try and apply this to a processor. I'll do this from the other side as you seem insistent on considering what happens to the energy from the cpu, rather than starting at the radiator as I have suggested.

Suppose we have a processor at a temperature T. Doesn't matter what temp this is! This processor has a power output of 70W. So going back to our metal block example. If we don't want the temperature of the cpu to go up or down, then we need 70W to be removed. Most of this will be removed by the water, but some will also be conducted away by the pins, though this is minor and we will assume it is negligible.

So we want 70W of heat to be conducted into the water.

As we established before, thermal energy is conducted down a thermal gradient. The rate at which it is conducted is proportional to the value of the thermal gradient. So we need a thermal gradient which will allow the conduction of 70 Joules per second. We also need to get the heat into the water which is where the efficiency comes in. But back to the cpu.

We still want to keep the cpu at the original temperature T, so we need the water to be at a temperature T-x such that the gradient conducts 70W.

It doesn't matter what the value T is, provided that the temperature difference, x, will allow the conduction of 70W.

So again, we have no net change in the thermal energy in the processor, so according to the equation relating specific heat capacity and energy, thus no temperature change.

Again, x is the impotant value, not T, as it is the temperature DIFFERENCE which is necessary to conduct 70W.

Now lets say that we know what the water temperature is, because we do. It is determined by the air flow, the air temperature, the heat load - 70W - and the efficiency of the radiator.

So we know T-x, and we know x, we can work out the processor temperature.

So I have tried to explain what happens to the thermal energy from the moment it is converted from electrical energy in the processor, and at the last minute fixed the water temp.

So to recap,

1. 70W of thermal energy converted from electrical energy.
2. Temperature difference, x, drives the conduction of 70W of thermal energy away from the cpu.
3. We fix the water temperature by fixing the air temperature. Thus we know the cpu temperature.
4. The cpu stays the same temperature - SO THERE IS NO NET CHANGE IN ENERGY from the equation

deltaQ = m c deltaT

Does this make sense. I should have introduced this fundamental equation earlier.

By its definition, if the cpu is at a constant temperature, there must be no net change in thermal energy (delta Q).

In order for there to be no net change in thermal energy, the water/pins must be conducting away the same power as is porduced by the cpu.

Any questions, fire away.

Anyone else, does this make sense?

8-ball

I would like to thank ALL of you for this thread. It was a great excercise for fundamentals. Every time Greystar posted and I thought, 'that's not right', someone who's better at explaining (most of you) would chime in and try to correct him.

Thank you!

I've been purposely avoiding this thread, because I didn't have the time to post a coherent reply, but ya'll have had my attention.

As Bill pointed out 2 pages back, look at the temperature gradient.

The power that a CPU emits is what we want the waterblock to capture, and take away. Is it a 100% efficient process? no. Some heat is being dissipated by convection to air directly from the CPU (wether it's through the side of the core, of through the substrate. We recently went over this at OCAU, and even the block is only 97-98% efficient.

The actual heat source is imbedded within the silicone die and as such, is a layer all on its own, in the transition of the heat into the water.

Each transition layer will have a temperature range, or differential, from the hot side (CPU side) to the cold side (water side). If there were no temperature gradients, the materials would be perfect heat conductors, and I hope everyone can accept that there's no such thing.

This temperature gradient is measured or specified by its C/W rating, which is a measurement of the heat resistance of the material. In our cases, this unit applies to the whole block, where normally, one would use the same concept applied to a specific thickness of material.

So for a 70W source (feel free to deduct the 2-3% inefficiency), you can literaly calculate a temperature differential of 13.3 degrees C, in the case of the best known block: Cathar's White Water (according to Bill's testing, under a specific set of variables, where the thermal resistance measured is 0.19 C/w).

The other layers are composed of:
-the TIM joint (included in Bill's measurement, at 0.05)
-the water
-the radiator
-the ambient air (the ultimate destination of the heat)

By increasing the flow rate, you decrease the thermal resistance of the water (or more specifically, the thermal resistance between the waterblock and the water), as well as the resistance within the rad (i.e. water-to-rad resistance).

The total power (aka heat) travels through a series of resistances, to eventually be dissipated in the room. 70 Watts is 70 Watts, no matter how you look at it. (but myv65 will point out the ever so important additional heat source, provided by the pump).

If some of the individual resistances are lowered, the total resistance is lowered, since they are all in series (i.e. follow each other).

If an individual resistive layer has been improved, it will have a lower thermal gradient.

If you add all the thermal gradients, you get the temperature differential, between the CPU and the ambient air.

Since the ambient air temp is not going to change (for the sake of the discussion), you can simply add the thermal gradients to it, to find your result: your CPU temp.


What's interesting to note throughout all this, is what Tom observed, when he purposely burned an AMD processor (by removing the heatsink altogether): the die temp reached temperatures in excess of 350 deg F. That's your 70 Watts of power going through the die, and trying to make it into the ambient air, but with such a high resistance, that the gradient was too high for the CPU to handle.

Exactly what I have been saying all along.

8-ball