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

Let's hope that we can clear this up, once and for all.

I found that most people that have a hard time understanding the whole concept, or have a difficulty getting into such a discussion, have the difficulty of not being able to differentiate the starting conditions from the running (aka balanced) conditions: sure your water temp is 20 deg, when you start your PC, but it will rise.


No. Twice heated, twice cooled (if you double the flow rate).

The quantity of water is irrelevant: it will reach a specific temperature (and temperature gradient) regardless: it would just take a bit more time with a larger quantity of it.

By having a lower flow rate, the water temperature will be spread over a larger range. That's your increased thermal gradient. The average temp remains the same.

Yes, there is a temp difference, within the rad, but the flow rate is such that it would be hard to measure. The average flow rate of a rig is ~50 gph, that's roughly 1 gpm (gallon per minute): think about it.

hey guys, we have been feeding a troll
I just wasted 15min reading the OCers thread Since87 (?) linked to previously
and this Greystar is gameing the world

but eliciting some explanations of great clarity, over, and over, and over, and over

I would suggest just following his posts with the caption:

Caution: lots of words, no understanding, too stupid to learn; Greystar is struttin' his stuff

I've been starting to wonder about that.

Man I'll be fuming if I've wasted that much of my precious revision time for no good reason.:mad:

8-ball

OK, I've got a question which is kind of related to the original title of the thread.

Suppose you have flow vs head loss charts for various lengths of tubing, a radiator and a waterblock, and maybe a gpu block as well. How would you combine this to produce a head loss vs flow curve for the whole loop from the pump outlet round to the pump inlet for comparison with the pump p-q curve.

Also, can you do the same thing with flow vs efficiency charts to try and predict a systems sweetspot.

8-ball

Absolutely.

You must have three things:
the pressure drop curve for the heatercore/rad
the pressure drop curve for the block
the PQ curve for the pump.

As long as you keep the flow speed less than 2 or 3 feet per second within the tubing, you can basically ignore it (unless you want to be accurate).

You then add the pressure drops for the block and core, overlay this total over the pump curve, and see where the lines cross: that's your resulting flow rate and total pressure drop.

It's easier said than done, but that's the principle. I usually run the numbers at different flow rate increments.

I don't think you have to worry about blowing off your fittings with that pump; I have one, and the 4' head spec seems pretty generous...:evilaugh:

It does have a built-in valve to tailor flow rate, but I doubt you'll run it at anything but wide-open.

@8-ball,

In priciple, you can "add" the curves as Ben has suggested. Practically speaking, it doesn't work to perfection. As you likely know, the impact on flow from a given object doesn't begin and end at the object's physical boundaries. Fortunately, the very nature of a centrifugal's curve means that it minimizes the errors in your estimate. What I mean is that if you guess your resistance higher than it really is, the flow will not go up much because increasing the flow will increase the delta-P. Likewise, if you estimate your resistance lower than it really is, the corresponding drop in flowrate drops the system resistance a little. It's a sort of attenuated error, I guess one could say. Does that make sense?

By adding, do you mean that I can develop a curve by considering each discrete flow rate in turn and adding the corresponding pressure drops for all of the components.

How about efficiencies? Or is that a bit more complicated.

8-ball

Yes, you can simply add the dP of each component, for a given flow rate.

As myv65 pointed out, this method is not terribly accurate, and a small margin of error on the pressure, won't amount to a big difference on the flow rate.

It gets a bit tricky if you have an odd configuration, where you've got an extra component running in parallel, like a chipset block.

Some handy, "good enough" techniques:

Most devices have a PQ curve of the form:

dP= Q^2 * Rf

where

dP is the pressure differential across the device
Q is the the flowrate through the device
Rf is what I call the flow resistance

Based on Bill's data, waterblocks generally match the equation very well, radiators are more aberrant but 'close enough' for meaningful quicky checks.

You can calculate Rf for a device by picking a point towards the right end of a device's PQ curve and solving the equation. A quicky shortcut for determining Rf, is to just take the pressure drop at 10 lpm and divide it by 100.

So, find Rf for the devices in the loop. Then add them up for a combined Rf, or:

dP-system = Q^2 * ( Rf-block + Rf-rad + Rf-tubing... )

Plug the equation into Excel and graph. Voila! System PQ curve.

As myv65 pointed out, the nature of centrifugal pump PQ curves tends to minimize the effect of errors, so although this is a somewhat sloppy technique, you still can get results accurate enough to do meaningful comparisons fairly quickly.

Correct me if I'm wrong, but that equation only works if the curve is actually a straight line, no?

From what I've seen, for a typical range of flows, the curve is actually "curved"!:D

Add all the head loss charts together. That is, for each flow rate, make the pressure the sum of the pressure on each head loss chart at that flow rate.

As you probably know, you can then take this overall head loss chart and plot against the pump curve. The interesection point is the head loss and flow rate of the entire system.

Not sure what you mean by "flow vs efficiency charts."

Alchemy

Ben... Since87's equation is that of a parabola.

I thought it would be somewhere between parabolic and cubic. Doesn't the "flow resistance" Rf increase as Q increases? If so, it wouldn't be parabolic, would it? I could be totally off base here, but I am curious enough to ask.

Thanks, that's what I thought.

By efficiency, I mean C/W.

I guess what I'm asking is can you produce a C/W vs flow chart for each stage in the transfer of heat from the cpu to the air, then sum these to determine an overall C/W vs flow for the entire loop, including all of the waterblocks.

Factor in the variation in heat load for different flow rates and try and predict the sweet spot for the system.

Or have I just ridiculously oversimplified what people are trying to do in the thermal simulator forum.

8-ball

It's just some messy nomenclature on Since87's part. The dp is really the flow resistance for the component, as a function of the flow. The term he's calling the flow resistance, Rf is really a constant, needed to shape the parabola.

Hot damn, you're right!

Let's put it to the test, with some data I've extracted from Bill's roundup:

Using the "Surplus"'s curve, I extracted the following info:
dP @ 2.0 gpm: 1.6 psi
dP @ 1.5 gpm: 1.0 psi
dP @ 1.0 gpm: 0.3 psi
(a rough extraction, but let's take a look!)

Ignoring the units,
using the first set of numbers, Rf = 0.4
using the second set of numbers, Rf = 0.44
using the third set of numbers, Rf = 0.3

Hum...:shrug:

as has been pointed out, the individual resistances can be simply added
but this sum will be 'correct' only at that specific flow rate as each component's rate of variation is different

and consider the potential accuracy, 15% would be unusually good
this is due to the quite small 'individual' resistances being measured; the instruments' accuracy and the substantial effect of rounding errors

Null-A Ben
if you are going to calculate; use the Chevette hc data, it is much more accurate

what a pleasant thread without having to push a rope uphill

:confused:

You got it right. Last I heard, that's exactly what's supposed to be going on in the simulator. Except I don't think the variation in heat load will be considered, since it's hard to say with inline pumps how much heat is going into the water. And if you just consider the energy input to the water by pressure increase you get . . .

Hm. 1.5 gpm for an overall pressure drop of 5 psi (~10ft H2O) gives you an energy increase in the water of 3 W. So, maybe 5% of the CPU load.

If you have flowrate constant, flow resistances add up just like resistances in an electrical circuit.

[Edit - removed non sequitir. Thanks, Since87.]

And to nitpick on what someone else said, centrifugal pumps *are* analagous to voltage sources. But they are not analagous to *ideal* voltage sources. They induce a potential in the fluid (pressure), but that potential is a function of flow rate. I've only studied basic circuits so I don't know if there are any voltage sources that act this way. But it's still a decent analogy, in my opinion.

Alchemy