Maze3 vs MCW462

[Player0]

I've just finished my latest waterblock review here (http://www.liquidninjas.com/reviews.php?op=showcontent&id=17) for your viewing pleasure. This time, I pitted the DangerDen Maze3 waterblocks versus the SwiftTech MCW462-UH. I also had the opportunity to test the different Maze3 top options. Go take a look at the results!

[kms]

good review Tom, shame about the antec though :)

[lechumbl]

Hi Tom,

Looks good, run it...........

[masher]

Article Text:

> "The longer water molecules stay in the waterblock to
> absorb heat, the better the waterblock usually performs"

This is incorrect, and almost completely backwards from the real truth.

There are only 3 factors which affect the cooling performance of a circulating fluid.

1. The thermal conductivity of the fluid and contact medium (in this case, water & copper).
2. The physical geometry of the liquid/solid interface (primarily the amount of contact surface area)
3. The dT present at the contact medium. (temperature differential between the water and the copper).

The longer any single water molecule absorbs heat, the lower dT will be, and efficiency will decrease. This is why higher flow rates increase performance- the cooling fluid has less time to absorb heat, and is thus presented with a higher dT.

The tradeoff in designing watercoolers lies in #2. "Maze" type designs offer a very high surface area for heat exchange (both in actual physical surface area and in 'virtual' area due to fluid turbulence) but can restrict flow and thus decrease dT. A well designed exchanger does decrease flow somewhat, but more than balances this with a large exchange area.

In a single-exchanger system, the less time any single fluid molecule remains in the exchanger, the better. However, computer water cooling systems use 2 (or more) exchangers- one for heat input and one for output. If both exchangers are equally efficient, then the optimal linger time for the system would be when water exiting any block has 1/2 dT as water entering. Usually the cooling radiator for these systems is far more efficent, though, with larger surface area and and active cooling. The cpu block thus may be around 1/3 to 1/2 as efficient, and the flow rate should be adjusted accordingly.

[kms]

Originally posted by masher
Article Text:

> "The longer water molecules stay in the waterblock to
> absorb heat, the better the waterblock usually performs"

This is incorrect, and almost completely backwards from the real truth.

Hi Masher :) To low a flow and the coolant saturates with heat energy & can absorb no more so cooling is vastly inefficient, too high a flow & theres not enough time to transfer sufficient heat energy to the coolant (i'll dig an interesting link out for you in a second - biggest is not always best). So a mid point compromise is the truth not the complete opposite ?

footnote - i'm often wrong though :D

[kms]

http://www.amdmb.com/article-display.php?ArticleID=179

[Player0]

Masher, well put.

Indeed, the statement I made was a little 'blanket'. Yes, once any single water molecule absorbs heat, it instantly becomes less effective in absorbing heat. However, until the temperature of the water molecule is equal with that of the waterblock, it still absorbs heat, even if its much less efficiently.

Also remember that while water is in a waterblock, it will experience much turbulance. The more the better. This means that more molecules are coming in contact with the waterblock. But it also means that the water molecules are sharing the heat with other molecules around. This potentially decreases the temperature of each molecule, allowing it to absorb more heat.

Back to my blanket statement. I meant that in the context of which waterblocks work better, in general, waterblocks which keep water molecules the longest in the block are generally the waterblocks that work the best.

In any event, the amount of time any water molecule spends in a waterblock is tiny tiny tiny. What is important to realize is that the water entering any small CPU waterblocks is almost never heated to saturation point. This can be proven by measuring the temperature of the water going in versus the water going out. The temperature difference will be hardly anything. That is proof that the water molecules dont have enough time to reach saturation, and still have alot of water cooling potential.

In real world terms, waterblocks with the longest channels (or largest caveties like the Swifty) work the best out of the other designs, especially when powered by a high-flow pump (the more the better) because this improves turbulance.

There are many many factors at play, and I was probably being over general in my statements in that article.

[masher]

Player0,

Picture a heat exchanger in a state of stasis. It has a certain number of water molecules inside, each with its own kinetic energy, which translates to an average temperature somewhere between (ambient) and (exchanger surface temp). Now, roll time forward just enough so that a single molecule has exited the system. Since water is an incompressible fluid, this means the exiting molecule is replaced by a single, cooler molecule on the intake side. T(av. coolant) therefore has decreased, and the overall efficiency increases.

The real-world coolers don't perform better because they're holding the water longer...they perform better because --despite holding the water longer -- their longer channels and larger cavities present a larger exchange surface area. Take that same cooler and push water through it faster (keeping any one molecule in for less time) and it will perform better.

Anyway, I think we're probably saying the same thing in a different fashion by now. Otherwise, an excellent review. Best of luck.

[Player0]

There never seems to be a clear answer to anything these days hehe. Surface area is important. But then look at the MCW462. It has a hideously small internal surface area, but performs nearly as good as the Maze3. Why? I dont/cant say for sure. The large cavity allows water to stay in the block longer, and this means it takes more advantage of turbulance. You wouldnt think something with such a small surface area would work well. But it does.

In a real life cooling system, you want a waterflow somewhere in the middle. Yes, I agree and always have agreed that waterblocks benefit from faster waterflow. Unfortunately, radiators are just the opposite. What that means is you want a waterflow that isnt too fast, and isnt too slow.

Yeah, I suppose we are pretty much saying the same thing. I do understand your point. Im not sure I necessarily agree with the 'why'. On that same note, I can't really prove things either way :)

I do beleive that turbulance is more important that surface area though. The MCW462 backs that theory :)

[kms]

are waterblocks more efficient as flow rate is increased without limit ? Does heat conduction occur instantaneously ? or with a time constant dependent upon conductor material used ?

[masher]

Yes, any single waterblock is more efficient as flow is increased. Its a logarithmic curve, however. An infinitely-large, infinitely-fast flow rate equates to a cooling fluid that is always at ambient, and has an unlimited ability to sink heat. Still, the rate of heat exchange isn't infinite, its limited by the factors I gave above.

In practice, the cooling rate between a flow of, say, 20lpm and 2000lpm are minimal.

[Player0]

I think KMS's point, if i can be so bold, is that water molecules need a certain amount of time to absorb a specific amount of heat. If they are rushed too fast through the waterblock, they may not have any time to absorb any decent amount of heat energy.

The heat transfer from waterblock to water molecule certainly isnt spontanious. It takes time, like boiling water. There is a point where if the water moves to fast, its not picking up heat. Just where that point lays, I dont know.

Anyone have those equations?

[masher]

I understand the point, but its wrong. If you want a mathematical treatment, you need to compute the Enthalpy using partial differentials. You start with:

dH = [@H/@T)subP dT + [@H/@P)subT dP

Where H is the Enthalpy of the system, @H/@T is the partial derivative of H with respect to time, and @H/@p is the partial with respect to pressure. With water as a circulating fluid under relatively small dT changes, its PVT surface can be approximated as flat, and the second term is constant. The first partial can be defined as the specific heat of the system then, and you can solve for it. Ignoring the heat input from the quasistic work function of the circulation itself, you get:

c(p) = 1/N(@H/@T)subP

N being the molar amount of water within the heat exchanger.

This is a macroscopic analysis using classical physics. I can give you a quantum-statistical analysis using Bose-Einstein combinatorials, but it would take a few pages.

An easier way to understand it is this: yes, it takes a finite amount of time for a single molecule to absorb one quanta of heat energy. However, a) that time is far far shorter than you're going to get outside a laboratory. And b) even if you did manage it...so what? That molecule exited the system before it changed quantum states, but it was replaced by another molecule which did. For any incompressible fluid, the total number of molecules within the heat exchanger at any given time is a *constant* value...regardless of flow volumes or velocities.

If this wasn't true, then you could set up a "perfect insulator" simply by flowing one substance past another at a certain velocity, and heat would never flow across the boundary. If you manage this, your water cooling system would vanish within the event horizon of a black hole, about the time the ghosts of Maxwell, Fermi, and Einstein rose from their graves and strangled you.

[Player0]

I have not dealt with this type of physics since college quite a few years ago. Whether its right or wrong, would take a decent amount of research for me to determine. I don't *see* the answer in what you're showing us.

You need to better describe your formulas. I have no idea what things like subP or dP mean. You need to put this in terms someone without a degree in physics can readily understand. I don't take anyones word for anything, no matter how complicated it looks ;)

[masher]

> "I don't take anyones word for anything, no matter
> how complicated it looks..."

Player0,

A very good trait indeed...the sheeplike acceptance of authority always stands in the way of true discovery.

Let's start with the simplest possible equation, the one used to model the efficiency of a Carnot heat engine:

dT = K [ (T1-T2)/T1 ]

- dT is the time rate of change of Temperature (rate of heat flow).
- K is a fixed constant depending on the physical geometry of the system (size, shape, materials used etc)
- T1 is the source temperature (high side)
- T2 is the reservoir temperature (low side)

So, to increase heat flow without changing the geometry of the system, you can either:

- Increase T1
- Decrease T2

But in a cooling system, you're not trying to maximize heat flow...you're trying to minimize T1. Yes, increasing your CPU temperature would increase the amount of heat removed from it, but that defeats the purpose.

That only leaves T2 reduction as an option. You can do this by exhausting heat to a colder location (e.g. putting your radiator in a bucket of ice). But since we're dealing with a circulating fluid, we can also reduce T2 by moving the fluid faster. This means at any given slice of time, the portion of fluid actually in contact with the T1 boundary will be cooler.

As a macroscopic, simplistic analysis, this has flaws. Primarily we're treating "K" as a constant. For a solid-solid boundary, this is valid. For a circulating fluid, we have to deal with turbulence. And changes in flow rates can change turbulence, sometimes in a discontinous fashion. A tiny increase in flow can increase turbulence by a factor of 10...and a tiny increase beyond that can reduce it by the same amount. There are no simple equations to compute turbulence-- it can only be modeled (approximated) with some very complex calculations.

However, turbulence doesn't increase heat loss by 'trapping' molecules and allowing them to absorb heat longer. This in fact reduces heat flow. Turbulence changes "K", by bringing more molecules in contact with the exchange surface and creating a "virtual" contact surface much larger than the physical one.

Now on to the question itself. Is there any minimum time a particle (in this case, an h20 molecule) to remain in contact with the T1 boundary, such that the particle will not absorb heat? The answer is yes. Does this reduce heat flow? The answer is no.

To prove that rigorously requires a statistical treatment such as BE stats. Since most people *with* a physics degree have a tough time with it, its not really practical to explain in non mathematical terms. However, a gedanken experiment (a thought picture) can give you a pretty good feel for why this is true.

Picture a watercooler large enough only to only hold a single molecule of water. Water is incompressible. This means the reservoir can never hold two or zero molecules. When the molecule inside flows out, it is instantly replaced by one other.

"Heat" is the kinetic energy of particles, and it is transferred by the elastic collisions between particles. At the quantum level, this is essential random. It is theoretically possible to happen instantly, or not for an extremely long period-- you could put your hand on a hot stove, and not get burned until an hour later. With all the trillions of atoms in your hand, though, the chance of this happening is many millions of times longer than the age of the universe.

Still, its a stastistical chance, and we have to model heat flow on that basis. In our one-molecule exchanger, lets assume the chance of a particle-particle collision is one per second. Odds are, a water molecule remaining inside one second will experience one interaction, and thus be raised one quantum state. If we adjust our flow rate to one molecule per second, then on the *average*, every molecule leaving the system will be 1 state higher.

Any individual molecule may have experienced two state changes though, or more...or zero. The mechanics of heat exchange says that, once you've had one collision, the chances of another decrease. So once a molecule has changed state, we need to remove it as fast as possible, or our heat flow will decrease.

There's no way we can do this however. We can't observe molecules, and remove them once they change state. So what if we just increased the flow rate, to say 100 molecules/second. On the *average*, 99 of those would flow through without ever changing state, and only one would experience a collision. Some may experience more than one obviously, but the average number of collisions per second is the only thing we care about. Why? Because "average collisions per second" is the definition of heat flow.

Our flow is so high than almost all the water moving through the system never even absorbs a single quantum of heat. Now, what is our average rate of heat flow? Its easy to compute-- its just the collision probability: 1 per second (minus a tiny fraction to allow for the lowered chance of collision on the 1 particle in 100 that actually had a state change, before it manages to exit the system). If we up our flow to 1000/sec or 10,000/sec...our rate gets closer to 1. It can never quite reach 1, the theoretical maximum. But it can get so close you can't measure the difference.

So now it should be obvious. Even if the interaction time (flow rate) is far faster than the mean collision time, heat flow is not reduced. In fact, it is increased (up to a certain ceiling value) by increasing the chance of interactions between particles at ambient temperature, instead of ambient+1 or ambient+2, etc, etc.

[tripodal]

seems so deceptively simple, yet it was difficult to explain
nice job

[kms]

forgot about this thread :D followed some but openly admit not all you've typed but sorry am still not totally convinced your right mainly through personal observations / fact on a larger scale, that being at the power stations i work at we day in day out have to reduce flow though our preheater stage (where condensate is passed back through the highest / coolest part of the boiler before returning to the deareator. Reasons being a) for better dearation conditions & b)as if the flow is too high the stack temperatures rise i.e. less heat is being absorbed into the returning condensate so efficiency is down. This priniciple is common among the thousands of stations worldwide so why on a smaller scale, again with water in fluid a state, admitedly at lower pressure, temperatures and flow rates is it different? :)

[Player0]

First of, water IS compressable. It just takes alot of pressure to compress it.

Secondly, this thread has gotten to complex. When we make things overly complicated, we tend to make mistakes.

Regardless of quantum physics and thermodynamic math, it *does* take a measurable amount of time for water to heat up. It takes forever to boil a pot of water, because it is so good at absorbing heat.

I'd say that your math was incorrect, but you havent shown any mathematical proofs. Just a bunch of thrown together theories, which frankily dont add up.

If you take a cup of water, and throw it in to a pot on the stove for 10 seconds and then take it off the heat, the water will be slightly warmer. If its kept in the pot for 20 seconds, its remarkably warmer. The longer the water is in the pot, the more heat it absorbs until it's temperature rises to boiling (or whatever temperature the pot is at).

This is NOT unsimilar to what happens in the waterblock. The waterblock is at a certain temperature, say 40c degrees, and holds a couple ounces of water. This is a scaled down version of the stovepot experiment, but the physics are the same.

If you had an open system where water didnt circulate, it simply is pumped from one bucket, through the waterblock, in to the other bucket, you would see that as the speed of your pump increases, water starts getting warmer because theres more molecules interacting with the waterblock. But then at some point, as you pump faster and faster, the water doesnt have time to absorb as much heat, and eventually barely absorbs any measurable amount.

Your mathematical theories are great, but the pragmatic truth of it all is that:

Too low flow can reduce waterblock efficiency.
Too high flow can reduce waterblock efficiency.
Somewhere in the middle is where you want to be.

If youre equations are too complicated to see the real life truth, then you need to step back and simplify things down. Your trying to tell me that putting a cup of water in to a pot for 10 seconds will yeild more temperature increase than putting it in for 20 seconds, but i dont buy it. Sorry.

[Selenium]

Too low flow reduces efficiency because water loses its ability to absorb more heat easily.

Too high flow reduces efficiency because the radiator doesnt have enough time to liberate the heat that the water has gained.

What if you had seperate loops. A high flow for the water block and a low flow for the radiator. Both of them running out of a sump.

Hook up a rheostat or pot to both pumps and tweak em till they are just right

[masher]

KMS,

You're absolutely right. However the mechanics of compressible gases are quite different. Not only do you have the enthalpy (and temperature change) due to compression/expansion, but in most systems like that, most of the "work" is done from the state change from water @ 100C to gas @ 100C.

Player0,

Well, I have to laugh. I've tried to explain it mathematically and in layman's terms. Yes, water is technically compressible, but for the purpose of thermodynamics of any system running under a few million psi, it is not. Your example doesn't correlate to a waterblock for one simple reason, which I've tried to explain several times. You can take water off a stove, but you cannot remove water from a waterblock. It always holds the exact same volume of water, regardless of flow. I have a masters in physics, specializing in magnetohydrodynamics, so my "thrown together theories" are rather much the same as the rest of the educated world.

I won't bother to post here again, but good luck anyway.

[lechumbl]

Hi masher,

I was going to welcome you, but after reading all of this post you seem to have a rather interesting attitude.
I guess all of that education made you more superior than us low lifes.
Have a good day, and I hope that you find a bbs that caters to your level.

Take care.......

[Player0]

masher, you're not impressing anyone here with your credentials. I guess we are all too dumb to understand your holy knowledge of physics.

Why one molecule leaving and another entering has any bearing on this conversation, I dont know. The molecule is still inside the waterblock for a specific amount of time, and if that time is too short, it doesnt pick up as much heat energy, and if the time is too long, it becomes saturated.

The fact that the water only holds a certain amount of water is quite obvious. Thats sort of a given isnt it? I know my copper waterblocks aren't expanding ;)

[tripodal]

you know, i never really did get his point....

[Synthohol]

this post was like the nightmare you have as a kid, you only remember the fear not the substance certainly not the outcome.
ill try not to click here too often if at all geeeezzz!


fwiw, i do however believe everything has a liquid state, even ninjas!!!:D