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Discovery may lead to improved CPU cooling

One of the major issues that computer makers have to overcome as computer processors get faster is how to cool them effectively. Enthusiasts already know that the easy way to get the most performance out of a processor is to keep them very cool. Many of the record breaking machines that destroy benchmarks are cooled by exotic means like liquid nitrogen.

Many commercially available computers today use liquid cooling with radiators and a circulating fluid similar to the system used to cool vehicles today. As processors become ever faster, researchers are always looking for new and novel ways to cool the processor. A pair of researchers at the University of Rochester has discovered a new method that may one day lead to a very efficient and effective cooling system for processors.

Researchers Chunlei Guo and Anatoliy Vorobyev published their finding s in the journal Optics Express. The paper outlines the discovery made by the pair of a method to make liquid flow vertically along a silicon surface without needing pumps or mechanical devices to overcome gravity. The pair describes the phenomenon as similar to how water rises in a straw; only with their system, no outside force pushes the water.

The process works by using ultra-powerful lasers to create nanometer-scale structures on the surface of the silicon. The creation of these nanoscale structures on the silicon surface makes the silicon much more hydrophilic. Water placed on the silicon is much more attracted to the surface and the water molecules literally climb over each other to get closer to the surface, crating forward motion at a rate of 3.5cm per second. The team points out that the same process will work on metal, but the process is much more important on silicon where it has potential for cooling processors. 

Most computers today are cooled by air using a fan while some low power machines are passively cooled with heat sinks. With fan cooling, the air around the CPU absorbs the heat and the fan dissipates the heated air. Liquid is a much better method of cooling a CPU, but the catch is that most liquid cooling systems are bulky and not appropriate for most mainstream computers.

Michael Scott is a professor of computer science at the University of Rochester, and is not part of the research. However, Scott believes that the breakthrough that allows silicon to essentially pump its own water has the potential to become widely used in personal computers.

Scott said, "Heat is definitely the number one problem deterring the design of faster conventional processors."

Heat isn’t the only concern with creating faster processors though; getting more transistors onto a piece of silicon is a major part of increasing performance. Some predict that Moore's Law will be broken in a bit more than a decade. The reason for the prediction is that we will reach a point where transistors simply can't be made smaller.

To continue making faster processors, new technologies will need to be developed. Research is being conducted today into using other materials for making microprocessors such as graphene and even DNA.



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Wait . . . what?
By MrPeabody on 3/17/2010 10:32:36 AM , Rating: 5
If I understand this article correctly, these researchers have basically created a kind of hyperactive capillary action that draws water up against the pull of gravity.

Forgive an old man his deteriorating mental faculties, but . . . how is this not free energy? Not in the strictest sense, I suppose, but if we grouped enough of these little laser-etched silicon doohickeys together, could we run a tiny waterwheel?

Are my dreams of carbon-neutral municipal street lighting for my ant farm approaching fruition?




RE: Wait . . . what?
By Iridium130m on 3/17/2010 10:39:15 AM , Rating: 2
so the question I have is how much energy went into making those precise carvings with the high powered laser?

How long does the silicon last before degrading and the effect goes away?

What happens when the water reaches the top and has no more silicon to be attracted to?


RE: Wait . . . what?
By Drag0nFire on 3/17/2010 11:04:39 AM , Rating: 3
quote:
What happens when the water reaches the top and has no more silicon to be attracted to?


This is the answer to the infinite energy paradox. If the water is attracted to the silicon, then it can more to a lower potential energy by attaching itself to the silicon (like a ball rolling down hill). It's possible in certain circumstances for this potential energy to be enough to do work against gravity. However, once it is attached at low potential energy, there will be no way of removing it.


RE: Wait . . . what?
By Bravo16 on 3/18/2010 10:09:24 AM , Rating: 3
quote:
What happens when the water reaches the top and has no more silicon to be attracted to?


There are various things one could do when the water reaches the top.

One could for example make a non carving section where the water runs down again into reservoir. That is only one idea, it would depend on the specific application.

You could also create a device that looks the same on both ends and with a reservoir on both ends. Water runs up the silicon into the "top" reservoir and as it become full and heavy it could swing around on a its own axis due to gravity and repeat the process.

One could harvest energy from the turning motion and coupled with a few of these devices have a continually turning axle. Free energy =)

Just one idea, could be wrong could be right.


RE: Wait . . . what?
By Gungel on 3/17/10, Rating: 0
RE: Wait . . . what?
By Shig on 3/17/2010 11:02:09 AM , Rating: 4
Until recently we couldn't effectively manipulate things at the nano-scale. Now we can and biology is the master of building things from the 'bottom up'.

We're going to see so many things going into technology over the next 10-20 years that are simply copied directly from biological systems.


RE: Wait . . . what?
By 91TTZ on 3/17/2010 11:33:55 AM , Rating: 2
This doesn't overcome the basic laws of physics. All this does is create enhanced capillary action and doesn't "pump" any water. It merely attracts it until all the laser etched silicon is wet.


RE: Wait . . . what?
By dark matter on 3/17/2010 12:46:37 PM , Rating: 1
As you are well aware the basic laws of physics go to pot at the nano scale.


RE: Wait . . . what?
By porkpie on 3/17/2010 12:50:17 PM , Rating: 2
Eh? They do no such thing.


RE: Wait . . . what?
By jlips6 on 3/17/2010 1:16:25 PM , Rating: 3
Quote: as you are well aware, newtonian physics go to pot at the nano scale

fixed it.


RE: Wait . . . what?
By Jaybus on 3/17/2010 1:55:14 PM , Rating: 2
Nothing is going to pot. This is the capillary effect, or "wicking" effect. By causing the water to spread out all over the surface, evaporation is increased. Neither Newtonian nor Quantum mechanics are required to explain it, but rather thermodynamics. It is still an evaporative cooling method.


RE: Wait . . . what?
By porkpie on 3/17/2010 3:41:17 PM , Rating: 5
" Neither Newtonian nor Quantum mechanics are required to explain it, but rather thermodynamics."

Err, there isn't any physics other than Newtonian and Quantum mechanics. You can explain wicking (and thermodynamics in general) on either a Newtonian or a quantum level.


RE: Wait . . . what?
By Yaron on 3/17/2010 4:51:29 PM , Rating: 3
porkpie,

From time to time I see you comment on this website.
It looks like your background in science (especially physics) is very strong - way stronger than most or any of the readers here. To be honest, I am curious: what is your background man? What industry are you working in?

I hope I am not crossing the line here, but I am truly curious :)

Cheers.


RE: Wait . . . what?
By Spookster on 3/17/2010 6:29:48 PM , Rating: 4
He's a terrorist working at a Nuke power plant.


RE: Wait . . . what?
By porkpie on 3/17/2010 7:10:41 PM , Rating: 3
Actually, I'm just a cranky old engineer with the vice of correcting others.


RE: Wait . . . what?
By SilthDraeth on 3/17/2010 10:57:11 PM , Rating: 2
Cranky engineers are the best kind.


RE: Wait . . . what?
By Suntan on 3/18/2010 12:48:32 PM , Rating: 1
No they are not. The ones that accept that they do not, in fact, know everything about every subject, are the best kind.

The ones that constantly sit around armchair quarterbacking other’s work, and demanding that everyone accept their knowledge as law, are just the ones that never have the stones to get up and do something useful with their time. At least that has been my personal experience with them.

Signed - an engineer that has spent the majority of his career specializing in thermodynamics and fluid mechanics. And one that saw nothing of substance in this article, or the one linked to at the University of Rochester website, that would make me go out on a limb and claim these people have done anything more novel than find a new way to treat the material surface such as to promote capillary wicking…

--Oh, and no (good) engineer I know would spend time pissin-n-moaning about which came first, Newtonian Physics or Thermodynamics… They are both just explanations of the actions found in the universe, and neither (or both, if you want to look at them as the same thing) explains all the actions of the universe correctly.

-Suntan


RE: Wait . . . what?
By whiskerwill on 3/18/2010 12:57:24 PM , Rating: 2
Wow Suntan, that post sure sounds cranky to me.

Pot, meet kettle.


RE: Wait . . . what?
By PitViper007 on 3/18/2010 3:19:44 PM , Rating: 2
I've worked with several cranky old engineers over my various careers. They tend to be the ones you WANT to work with, assuming you can get past the crankiness. :)


RE: Wait . . . what?
By chemist1 on 3/18/2010 3:41:01 AM , Rating: 2
To the "cranky old engineer with the vice of correcting others":

The statement to which you object is in fact correct. Neither Newtonian Mechanics (NM) nor Quantum Mechanics (QM) are required -- classical thermodynamics, which is a phenomenological theory that (unlike NM and QM) is independent of the details of the system, is sufficient. Thus neither NM nor QM are required. Nor can thermodynamics be viewed as a mere subset of NM and QM, for two reasons: (1) one cannot derive the laws of thermo (via the science of statistical mechanics) using only NM and QM -- one also needs, for instance, the ergodic hypothesis. Further, NM and QM are not as broadly applicable as thermo. NM is inconsistent with special relativity, and QM has not yet been reconciled with general relativity. Thermodynamics, on the other hand, has been successfully applied to problems involving both special and general relativity (for the latter, see Hawking's work on black holes).

Or, to put it more simply: thermodynamics is not simply reducible to NM or QM. Thus there are "other physics" besides NM and QM (not to mention general relativity, but I know you weren't thinking of that).

Yours truly,

A theoretical chemist with the vice of correcting others :)


RE: Wait . . . what?
By chemist1 on 3/18/2010 3:50:50 AM , Rating: 2
correction: I shouldn't have said thermo is "more broadly applicable" than NM or QM, since there are places where classical thermo does not apply (small systems). I should have said that thermo works in areas where NM or QM (as currently configured) do not. But even with that, my point stands: that thermo is not reducible simply to NM or QM, and thus is indeed "other physics."


RE: Wait . . . what?
By chemist1 on 3/18/2010 5:35:03 AM , Rating: 2
On further reflection, I should acknowledge that the ergodic hypothesis can be considered equivalent to the assumption of equal a prior probability in quantum mechanics, so in that sense QM does contain an equivalent assumption. I should further acknowledge that Hawking's work on black hole thermodynamics involved the development of an analogous formulation of thermodynamics. Nevertheless, thermo is still "other physics."


RE: Wait . . . what?
By porkpie on 3/18/2010 8:26:47 AM , Rating: 2
"Nor can thermodynamics be viewed as a mere subset of NM and QM, for two reasons: ... (snip)"

Unfortunately, you are incorrect. Thermodynamics is a subset of mechanical statistics; it can be treated in a classical manner, or a quantum statistical one.

The entirety of physics is contained in the two subsets of classical (Newtonian) and modern physics. Now, in loose speech, we sometimes call modern physics quantum physics, though it also contains GR/SR. However, thermodynamics in general has *not* been sucesfully wedded to relativity...though its one of the hot theoretical topics of the last several decades, and success here would be a great leap forward to a grand 'theory of everything'.

There is some good information below about the problems with work on relativistic thermodynamics:

http://www.nature.com/nphys/journal/v5/n10/full/np...
http://www.physicsforums.com/showthread.php?t=1942...

" Thermodynamics [has] been successfully applied to problems involving both special and general relativity (for the latter, see Hawking's work on black holes)."

Incorrect again. Penrose's treatment of black holes is based on classical thermodynamics; Hawking expanded that to a semi-quantum statistical model:

http://iopscience.iop.org/0264-9381/17/2/701?ejred...

"...a theoretical chemist with the vice of correcting others :)"

We in physics call chemists failed molecular physicists. :)


RE: Wait . . . what?
By thepalinator on 3/18/2010 8:57:12 AM , Rating: 5
quote:
We in physics call chemists failed molecular physicists. :)
GEEK FIGHT!


RE: Wait . . . what?
By JonnyDough on 3/18/10, Rating: -1
RE: Wait . . . what?
By Yaron on 3/18/2010 11:49:15 AM , Rating: 2
"GEEK FIGHT!"

Incorrect. This is a super geek fight!

Yours truly,
A theoretical dude with the vice of correcting others :)


RE: Wait . . . what?
By chemist1 on 3/19/2010 5:14:37 AM , Rating: 2
To your last comment: Sigh. OK, we can go there if you insist: and physicists are just chemists without social skills :)

To the issue of black holes and thermodynamics: see http://en.wikipedia.org/wiki/Black_hole_thermodyna...
which shows that the laws of black hole thermodynamics are analogous to those of classical thermodynamics. But, better yet, never mind -- the issue of thermodynamics and GR was a distraction from my main point, so I should not have mentioned it.

And my main point remains: thermodynamics is a phenomenological theory whose laws and results are independent of the details of the system. NM and QM, by contrast, do depend on the details of the system. And because thermodynamics is independent of the details of the system, it can be applied without *knowledge* of such details. Therein lies its power -- a power that NM and QM lack. And when the original poster said that thermodynamics would suffice, that's exactly what he meant -- that we don't need to concern ourselves with the details of the system -- we don't need to perform an analysis using NM or QM -- but, rather, we can just apply these broad laws of thermodynamics, which do not require knowledge of such details. Hence I maintain that the original poster's statement was correct -- there's no need to do an analysis using NM or QM when a simpler, more direct (and thus more elegant) tool is available.

Also, could you please explain to me (and I'm not being cheeky here), how one can get the zeroth law of thermodynamics from NM or QM?

I conclude with Einstein's famous quote on classical thermodynamics:
"A theory is the more impressive the greater the simplicity of its premises, the more different kinds of things it relates, and the more extended its area of applicability. Therefore the deep impression that classical thermodynamics made upon me. It is the only physical theory of universal content which I am convinced will never be overthrown, within the framework of applicability of its basic concepts."


RE: Wait . . . what?
By porkpie on 3/19/2010 9:38:48 AM , Rating: 2
"ould you please explain to me (and I'm not being cheeky here), how one can get the zeroth law of thermodynamics from NM or QM?"

You can't derive ANY basic law of physics from anything else. That's just the point. They're axiomatic: postulated or observed (or usually both). You can't derive action-reaction (Newton's 3rd Law) or Einstein's invariance of light or conservation of momentum. You use them -- just as you do the laws of thermodynamics -- to derive higher order relationships.

" thermodynamics is a phenomenological theory whose laws and results are independent of the details of the system. NM and QM, by contrast, do depend on the details of the system"

This isn't true at all. As example, consider the gravitational force from an object. You don't need to know it's composition, or even its precise mathematical shape. You simply need to know total mass and the centroid. . . two objects of radically different shapes and makeup can generate the same force.

Thermodynamics is primarily a statistical science. So are many aspects of QM and (to a lesser extent) NM. It's an inherent property of statistics to mask internal details in favor of mean behavior.


RE: Wait . . . what?
By chemist1 on 3/20/2010 3:02:44 AM , Rating: 2
"You can't derive ANY basic law of physics from anything else. That's just the point."

No, that's precisely not the point. You seem to have a lot of technical knowledge, but what's needed at this stage is not technical knowledge but just clear thinking. Think it through: My question about the zeroth law was clearly a challenge to your statement that there is no new physics in thermo -- that it's all just (i.e., follows from) NM/QM. That is what you communicated, right? But, for that to be true, thermo has to be entirely obtainable from NM/QM. Yet if the zeroth law of thermo can't be entirely obtained from (i.e., seen as a reformulation of the ideas/assumptions already present in) NM/QM, that means that that there is indeed new physics in thermo, and that your statement is therefore wrong. Am I being clear enough here?

And note that, contrary to what you imply, there is no problem with fundamental laws in one field being seen as equivalent to fundamental laws in another. So, for instance, the ergodic hypothesis in statistical mechanics is equivalent to the theory of equal a prior probability, and thus does not contain new physics. So the question I raise is a legitimate one: are the physical ideas contained within the zeroth law of thermodynamics already present within QM/NM (the way conservation of energy is), or are they not? If you can't show that they are, then you can't support your position that there is no new physics in thermo (relative to NM/QM).

Regarding you second point about whether NM is phenomenological: you are correct, it is, but again, that's not relevant to the discussion. And here this is my fault: I was not complete enough in my argument. What I should have said is that, if one uses NM in an attempt to derive the laws of thermo, one does this through stat mech, which does require assumptions about the internal nature of the system (that it's composed of particles, for one). I.e., while NM by itself may be (as you correctly say) independent of the internal details of the system, you can't derive thermo from NM without using NM in a way that does depend on the internal details of the system.

And you still have failed to directly address the thing that started this: I contend that (1) the original poster's statement that you don't need NM/QM, only thermo, was a legitimate one; (2) it reflected a sophisticated recognition that one use the simplest, most direct tool (which certainly could be thermo, if you are talking about evaporative cooling); and (3) given this, your criticism of the poster (particularly starting with the condescending "Errr ...") was misplaced and gratuitous -- it can be argued that there is different physics in thermo and, more importantly, it sounded like you were more interested in showing off (at that poster's expense) than in trying to see the big-picture value in what he/she was saying (point no. 2). I will close humorously by recalling my rejoinder to your comment about chemists (like me) being failed molecular physicists: Having said all that, I shouldn't be too hard on you, after all, you are a physicist, and thus shouldn't be expected to have actual social skills :).


RE: Wait . . . what?
By porkpie on 3/20/2010 10:23:56 AM , Rating: 2
quote:
for that to be true, thermo has to be entirely obtainable from NM/QM. Yet if the zeroth law of thermo can't be entirely obtained from (i.e., seen as a reformulation of the ideas/assumptions already present in) NM/QM, that means that that there is indeed new physics in thermo, and that your statement is therefore wrong
Surely you can see the logical fallacy in that statement. I'll illustrate by counterexample. Is the law of conservation of momentum derivable from NM? No. Does that mean its not part of NM? Ludicrous! What about Newton's own three laws? They're not derivable either

Further, your focus on the Zeroth law is rather myopic. On any textbook on thermodynamics, this topic might garner a page or two. The other 400+ pages (in my book, at least) are concerned with the derivation and manipulation of further laws, relationships, and outcomes -- all of which are derived directly from mechanics. The Fundamental Equation of Thermodynamics? Mechanics. Kinetic Model of an Ideal Gas? Mechanics. Entropy and Equilibrium Conditions? Mechanics. Maxwell Relations? Mechanics. Boltzmann Distribution? Mechanics.

In fact, half the texts on statistical thermodynamics simply call it statistical mechanics . . . that's how closely related the topics are.

quote:
it reflected a sophisticated recognition that one use the simplest, most direct tool (which certainly could be thermo, if you are talking about evaporative cooling
Please explain how one can derive evaporative cooling without invoking kinetic theory, itself derived from basic mechanics.

quote:
after all, you are a physicist, and thus shouldn't be expected to have actual social skills :).
I'm not a physicist...I'm an engineer who works in the field. In other words, a failed physicist who actually draws a decent paycheck :)


RE: Wait . . . what?
By chemist1 on 3/20/2010 8:38:33 PM , Rating: 2
"Is the law of conservation of momentum derivable from NM? No. Does that mean its not part of NM? Ludicrous! What about Newton's own three laws? They're not derivable either."

Your statement is both correct and not relevant to the argument at hand. I.e., it does not respond to what I've written, since nowhere have I claimed that fundamental laws of a field are derivable using that field!. You're not disagreeing with me; rather you're not understanding me. Maybe it's my fault for not explaining it better, but I thought I was being clear. Let me try one more time: the first law of thermodynamics is conservation of energy. Yet the law of conservation of energy is also contained within QM. Thus, if thermodynamics were simply "conservation of energy," you could argue, correctly, that thermodynamics contains no new physics relative to what's in QM. Right? And note that, in making such an argument, nowhere am I saying that conservation of energy is "derivable." Right? But thermo is not just the first law. It also contains, for instance, the zeroth law. And I maintain that the zeroth law articulates a physical principal that is not present in QM or NM. Hence it is wrong to say that thermo contains no new physics beyond what is in QM or NM.. And nowhere, in making such an argument, am I talking about "deriving" fundamental physical laws. Rather, I'm simply saying that, in assessing whether theory A contains no physics not already present in theory B, one must ask if all the fundamental laws in theory A can really be obtained from (i.e., are already present in, or are reformulations of) what's already in theory B (as I did with the example of conservation of energy).

[I don't think this explanation is any clearer than what I gave before; so, if you still somehow think I am claiming that fundamental laws can be derived from within the field that is based upon them (which is of course circular), then let me ask you to do this: please print off just your last post and this reply, and show them to one of your colleagues. With a different pair of eyes, he or she will hopefully understand what I'm trying to say, and explain it to you in a way that I can't.]

Now, as to the zeroth law itself: while my eyes may be myopic, my logic is not :). The number of pages devoted to the zeroth law in textbooks is irrelevant. That is determined by pedagogical issues, not epistemological ones, and it is only the latter that concern us here. By asking how many pages are devoted to the zeroth law in textbooks, you are asking the wrong question. [Indeed, according to http://en.wikipedia.org/wiki/Zeroth_law_of_thermod... the zeroth law "is arguably the most fundamental of the four numbered laws of thermodynamics. It was called the zeroth law because the need to state it explicitly was not understood until after the First, Second, and Third Laws had been named and become commonplace." ] Instead, the questions you should be asking are: (1) Is the zeroth law an essential part of the formulation of the thermodynamics? Answer: It is. (2) Is the zeroth law merely definitional, or are there actual physical ideas in it? Answer: There are. (3) Are the physical ideas present in the zeroth law already contained within QM/NM? Answer: This is the crux of it; I contend they are not, from which it follows that it is incorrect to say that thermo contains no physics not already present in QM/NM.


RE: Wait . . . what?
By porkpie on 3/20/2010 9:58:01 PM , Rating: 2
quote:
But thermo is not just the first law. It also contains, for instance, the zeroth law. And I maintain that the zeroth law articulates a physical principal that is not present in QM or NM. Hence it is wrong to say that thermo contains no new physics beyond what is in QM or NM
But this chain of reasoning is a logical error even more glaring than the first. Your conclusion is that Zeroth Law lies outside of of NM...because it lies outside of NM. Circular reasoning at its finest!

You: "The Zeroth Law is not part of classical mechanics".
Me: "Why?"
You: "Because the Zeroth Law lies outside of classical mechanics"

But there is a far more devastating rebuttal to your argument. The Zeroth Law, in fact, follows automatically form the First and Second Laws (see link below for just one reference of this). Now, you've already admitted that the First Law is a consequence of classical mechanics. The Second Law is as well, it is a logical result of Liouville's Theorem (see second link) when applied to entropy of an ensemble of microstates.

Thus, 0LTD is simply a restatement of classical mechanics, and contains no "new physics". Quod erat demonstrandum.

http://adsabs.harvard.edu/abs/1961AmJPh..29...71T
http://www.physicsforums.com/showthread.php?t=1652...


RE: Wait . . . what?
By porkpie on 3/20/2010 10:03:22 PM , Rating: 2
And to add to the above, here is a reformulation of the Zeroth Law on the basis of quantum mechanics, rather than Newtonian:

http://www.sciencedirect.com/science?_ob=ArticleUR...


RE: Wait . . . what?
By thepalinator on 3/20/2010 10:26:45 PM , Rating: 3
Can someone please tell me whose winning?


RE: Wait . . . what?
By chemist1 on 3/20/2010 10:52:06 PM , Rating: 2
"But this chain of reasoning is a logical error even more glaring than the first. Your conclusion is that Zeroth Law lies outside of of NM...because it lies outside of NM. Circular reasoning at its finest!"

This is silly, and patently not at all what I was saying. All you've demonstrated here your inability to follow what wrote. Really, please run this by one of your colleagues -- hopefully he or she can explain it to you.

Regarding your references, I will look into them. But you should know that just because someone asserted back in 1961 that the zeroth law is merely a consequence of the first and second does not mean that it's true -- or even that this assertion has gained general acceptance in the field. Again, I will examine this assertion more closely.


RE: Wait . . . what?
By porkpie on 3/20/2010 11:32:38 PM , Rating: 2
"just because someone asserted back in 1961..."

The QM reformulation was from a paper in 1985. Here's a paper from 2001 that establishes 0LTD based on classical (Newtonian) statistical mechanics:

http://adsabs.harvard.edu/abs/2001nsma.conf..237A

"This [is] patently not at all what I was saying."

You concluded that thermodynamics lies outside NM, based on the assumption that 0LTD lies outside NM. But your assumption seems to be derived from nothing more than your belief in your conclusion.

If I've misunderstood, please let me know. I don't profess to infallibility, but I saw nothing in your post outside the above to justify your initial premise.


RE: Wait . . . what?
By chemist1 on 3/21/2010 12:34:53 AM , Rating: 1
You've misunderstood my argument but, honestly, I lack the energy to explicate it yet again. Again, please show it to a colleague. He or she should be able to see that I wasn't being circular.

And again, I will investigate those papers you referenced, and the whole zeroth law issue -- I think it's an interesting one. Two of my colleagues are world-class scientists who have done significant work in the fundamentals of thermodynamics (one published a highly-regarded textbook on the subject), and have thought deeply about such issues; I will consult with them. That may take some time, though -- check back in about a week!


RE: Wait . . . what?
By chemist1 on 3/22/2010 5:20:45 PM , Rating: 1
OK, spoke with one of our two resident thermodynamics experts. (This is a professor emeritus who is considered one of the world's experts on thermodynamics. As a colleague once told me: "Thermodynamics is a deep and subtle subject. If you want *an* answer, talk to anyone in the department. If you want the right answer, talk to [this guy].")

So, here's what he said: thermo cannot be obtained from NM/QM, because thermo requires the physical concept of temperature, and neither NM nor QM talk about temperature, nor can temperature be obtained from them. Hence thermo does indeed contain physics that is not in NM/QM. And where is the concept of temperature formally introduced into classical thermodynamics? That's right -- it's the zeroth law. [Alternately, if you go step-by-step through the derivation of statistical mechanics, temperature is typically introduced following the use of the Method of Lagrange Multipliers, at which point the unknown term "beta" is identified as being 1/kT.]

Regarding those papers that assert you can derive the zeroth law from the first and second: this has not been generally accepted. As evidence, if you could do this, that would be of significant fundamental importance, and would therefore be featured prominently when textbooks discuss the zeroth law. Yet no textbook (at least none that I know of) mentions this -- hence that assertion is still considered well, an assertion. [And lest you misconstrue this argument as circular, let me reconstruct it:
A -> B = !B->!A, where A=being able to derive zeroth law from 1st and 2nd, and B=this fact appearing in textbooks. So here I am saying, given the reasonable assumption of
A->B, the assertion that the zeroth law can be obtained from the 1st and 2nd is false.]


RE: Wait . . . what?
By porkpie on 3/23/2010 2:47:30 PM , Rating: 3
"(This is a professor emeritus who is considered one of the world's experts on thermodynamics"

I have to call shens on this. Temperature *does* have an exact analogue in Statistical Mechanics. I don't know anyone with a physics background ('emeritus expert' or not) who doesn't know this.

Further, I don't know you pulled the discussion of temperature from, but it's sadly mangled. Beta is not an "unknown term", it IS the SM definition of temperature...or the multiplicative inverse of temperature, in units of energy, if you prefer. It relates the microstate SM interpretation of energy to the macrostate interpretation of temperature.

Thermodynamic beta is the linkage between SM and thermodynamics that I described in my earlier posts.

The only bearing 'Lagrange Multipliers' has on the issue is that its a convenient way to maximize a function. There are plenty of other derivations, however, see below for details:

http://en.wikipedia.org/wiki/Canonical_ensemble

Further, 0LTD does not introduce temperature. We didn't even begin to discuss the concept of the zeroth law until the 1930s.


RE: Wait . . . what?
By chemist1 on 3/23/2010 10:13:35 PM , Rating: 2
From your own link: "comparing with thermodynamic formulae [classical thermodynamics], it can be shown that [beta] is related to the absolute temperature. SO IT IS AN UNKNOWN UNTIL, TO GET THE SAME FORMULAIC RESULTS IN SM AS THOSE YOU ALREADY HAVE IN CLASSICAL THERMO, YOU REALIZE BETA MUST BE EQUAL TO 1/KT. THAT'S WHERE TEMPERATURE IS INTRODUCED INTO SM. YOU CAN'T GET IT FROM MECHANICS ALONE!! YOU CAN ONLY DETERMINE THAT BETA IS EQUAL TO 1/KT BY USING THE RESULTS IN CLASSICAL THERMO [I.E., BY REQUIRING THE FORMULAS FOR THE STATE FUNCTIONS THAT YOU GET FROM SM TO BE IDENTICAL TO THOSE FROM CLASSICAL THERMO]!!!! UNTIL YOU DO THAT STEP BETA IS AN UNKNOWN!! HOW CAN YOU NOT SEE THIS???

AND IF YOU DISAGREE WITH ME, JUST SAY YOU DISAGREE. LEAVE ASIDE THE CHILDISH PUT-DOWNS, LIKE SAYING MY ARGUMENTS ARE "SADLY MANGLED" OR "LUDICROUS." HAVE YOU EVER STOPPED TO THING THE REASON YOU HAVE TROUBLE WITH MY ARGUMENTS MIGHT BE THAT YOU SIMPLY DON'T UNDERSTAND THEM (AS SEEMS APPARENT TO ME)?

AND THE ZEROTH LAW IS FORMALLY NEEDED TO DEFINE TEMPERATURE. THAT IS WHY THE ZEROTH LAW WAS INTRODUCED. BEFORE THAT TEMPERATURE WAS USED, BUT THE STRUCTURE OF THERMODYNAMICS WAS NOT FORMALLY CORRECT UNTIL THE INTRODUCTION OF THE ZEROTH LAW.


RE: Wait . . . what?
By chemist1 on 3/20/2010 3:23:21 AM , Rating: 1
Oh, and, by the way, maybe my understanding of the zeroth law and NM/QM is wrong here -- i.e.. maybe the physics of the zeroth law are indeed contained within the physics of NM/QM. I don't think they are, but I've not investigated this in depth (indeed, stimulated by our exchange, I will now do so). And if you could convincingly demonstrate how the physics of the zeroth law are merely reformulations of what's already in NM or QM, I would be very interested to hear of it. But, again, until you can do so, I do not believe you can support your claim that thermo contains no new physics relative to NM/QM.


RE: Wait . . . what?
By chemist1 on 3/20/2010 3:50:02 AM , Rating: 1
And ... thank you for a stimulating discussion!


RE: Wait . . . what?
By Yaron on 3/20/2010 6:29:11 AM , Rating: 2
chemist1 is threatening to dethrone porkpie as DT's top super geek!
:O


RE: Wait . . . what?
By chemist1 on 3/20/2010 7:57:32 PM , Rating: 1
I am not worthy of such an honor and, besides, I am just passing through while porkpie has, I understand, a history of posting contributions here.


RE: Wait . . . what?
By AnnihilatorX on 3/17/2010 11:12:32 AM , Rating: 2
Plant relies on capillary action to tranport water, I believe this is a bit different as it doesn't involve thin tubes.


RE: Wait . . . what?
By Iaiken on 3/17/2010 11:28:39 AM , Rating: 4
Capillary action can be achieved with ANY surface that water prefers to adhere to over other water, regardless of shape. In this case, it is a seemingly flat shape with grooves (zomg! That's half a tube! No wai!).


RE: Wait . . . what?
By roostitup on 3/17/2010 7:50:37 PM , Rating: 2
It's different because a tree in a sense is using vacuum like forces to draw water up (kind of like sucking through a straw). The tree opens many stomas on each leaf which allows water to transpire into the atmosphere, creating a vacuum. In other words, the water evaporating out of the leaves creates a vacuum that sucks water from the ground through the roots.


RE: Wait . . . what?
By porkpie on 3/17/2010 11:13:03 AM , Rating: 2
"Forgive an old man his deteriorating mental faculties, but . . . how is this not free energy? "

Despite a few other incorrect replies, the actual answer is that the water loses heat energy as it rises. In an energy-conservation sense, it's no different than water evaporating and rising into the sky to form clouds.


RE: Wait . . . what?
By dark matter on 3/17/2010 12:48:17 PM , Rating: 2
But he wouldn't need warm water to drive the waterwheel. Cold water will do.


RE: Wait . . . what?
By porkpie on 3/17/2010 12:54:45 PM , Rating: 2
Absolute temperature is meaningless; temperature differentials are what contain extractable energy. There is also potential energy in from hydrostatic pressure, osmotic pressure, matric effects, and other sources.


RE: Wait . . . what?
By HotFoot on 3/17/2010 1:46:27 PM , Rating: 2
If that's true, Carnot sure sold us a big heap of crap!


RE: Wait . . . what?
By porkpie on 3/17/2010 2:07:02 PM , Rating: 3
Huh? Carnot efficiency is defined by the ratio of input to output temperature. You can have a fluid at 1 million degrees...but unless you have a temperature differential, you can't extract work.


RE: Wait . . . what?
By HotFoot on 3/17/2010 2:49:52 PM , Rating: 1
Ratio of absolute temperatures.

If that weren't true, then you could plug Celsius into the equation instead of Kelvin.


RE: Wait . . . what?
By porkpie on 3/17/2010 3:25:09 PM , Rating: 2
You're still missing the point. The OP said cold water was being used, rather than hot, the implication being the water was 'too cold' to derive energy. In the context of thermodynamic work, the absolute temperature is meaningless; what matters is the differential:

Heat pump at 0C, exhausting at -136C = 50% efficient.
Heat pump at 273C, exhausting at 0C = 50% efficient.


RE: Wait . . . what?
By HotFoot on 3/17/2010 4:29:35 PM , Rating: 2
At the bottom of your post there, you have one differential of 136C and one differential of 273C. If all you're saying is that there has to be a temperature difference for a heat engine of any type to work, then agreed. But I consider thermal efficiency to be a significant part of the discussion of what heat engines do. So, from my perspective, absolute temperature is a pretty important concept.


RE: Wait . . . what?
By porkpie on 3/17/2010 4:32:56 PM , Rating: 2
I'm not sure why you're not getting it. Absolute temperature doesn't control efficiency. In the example I gave above, the two system have radically different temperatures . . . but the same efficiency. A very hot working fluid can provide low efficiency, and a cold working fluid a high efficiency. Or vice versa.

What matters is the reservoir ratio, not the absolute temperature.


RE: Wait . . . what?
By Keeir on 3/17/2010 6:28:10 PM , Rating: 3
Okay

Carnot Efficiency

A heat engine can be no more efficient than

1-Tc/Th

Where Tc is the absolute temperature of the exhaust/enviroment/etc and Th is the absolute temperature of the heat engine input.

In Porkpie's example

273K -> 136.5K into the equation results in .5
546K -> 273K into the equation also results in .5

The efficieny is determined by the relative drop in energy. The more the relative drop in energy, the greater the possible efficieny of the system. Keep in mind, efficieny is not absolute output energy.

Thus just because a water based system might start at 100C doesn't mean it will have a higher efficieny than one that starts at 50C. Even if the total output of the system is greater, it still possible that the 50C is more efficient based on output temperatures.


RE: Wait . . . what?
By HotFoot on 3/17/2010 1:44:02 PM , Rating: 2
I hope you're joking.


RE: Wait . . . what?
By Shining Arcanine on 3/17/2010 3:31:13 PM , Rating: 2
If the water loses heat as it rose, why are the clouds gaseous and not solid? Wouldn't losing heat mean that it would freeze?


RE: Wait . . . what?
By porkpie on 3/17/2010 3:38:38 PM , Rating: 2
When water vapor rises, it undergoes adiabatic cooling from expansion. It doesn't generally freeze (though it will if it rises far enough) because its gaining convective and radiative heat energy from the environment as it rises.

But that's a rather different process than what happens during capillary action.


RE: Wait . . . what?
By HotFoot on 3/17/2010 4:33:09 PM , Rating: 2
Of course the energy going in to making the warm water vapour rise comes from the potential energy of an equivalent volume of denser fluid descending. Heat doesn't actually push the vapour up any more than it pushes it down, or sideways.


RE: Wait . . . what?
By Fritzr on 3/17/2010 5:59:54 PM , Rating: 2
Many clouds are composed of ice crystals others of water droplets. Clouds are diffuse, but when in the form of a mist are not gaseous.


RE: Wait . . . what?
By XZerg on 3/17/2010 11:14:02 AM , Rating: 2
i am betting on heat being the necessary energy provider for this to work. If not that then there is some possibly undefined factor working here that behaves similar to how magnets do.


RE: Wait . . . what?
By AnnihilatorX on 3/17/2010 11:21:49 AM , Rating: 2
Not necessarily. Take an extreme example, superfluidity.

Superfluids literraly creeps up any container seemingly defying gravity until it coats a thin film of itself over the container.

Anyway, what I am trying to say is, energy is available everywhere. Temeprature gradient, internal energy of the liquid, be it thermal, potential or kinetic. This certainly does not violate conservation of energy if it works.


RE: Wait . . . what?
By Iaiken on 3/17/2010 11:26:13 AM , Rating: 2
You're forgiven, you forget that trees and plants have been doing this for millions of years. Though the pace of capillary action in plants is far slower than 3.5cm per second.

This entire process is made possible by the same mechanics that cause surface tension on water. If you place a droplet of water on a hydrophobic material, it forms a roughly spherical shape in spite of gravity.

What is essentially happening is that the water is trying to (with the aid of a hydrophilic material) increase it's surface area in spite of gravity. This water could not be used to do work so it doesn't count as free energy. It is simply to move water from point A to point B in similar fashion to how plants do it all day every day.


RE: Wait . . . what?
By porkpie on 3/17/2010 11:51:58 AM , Rating: 2
" This water could not be used to do work"

Actually it can (and is). It's how capillary based water pumps work (though the amount of energy extracted is tiny).

On a larger basis, a similar principle enables osmotic power stations. I believe they're building one somewhere in Scandinavia large enough to power an entire town.

In high school physics, you learn that water has gravitational potential energy. It actually has potential energy from a large number of other sources, such as solute potential, matrix potential (surface tension), pressure potential, etc.


RE: Wait . . . what?
By Myg on 3/17/2010 8:19:44 PM , Rating: 2
So, if we were to build large enough power plants all over the world convering heat energy to gravitational potential energy would that solve global warming? ;-D


RE: Wait . . . what?
By MrPeabody on 3/17/2010 12:00:02 PM , Rating: 4
I guess this is a cause of my mental hiccup:
quote:
crating forward motion at a rate of 3.5cm per second.

To me, this implies two things:
(a) A lapse in proofreading.
(b) A relatively significant velocity imparted to the water.

I don't really care about (a) , but would the velocity in (b) be enough for the water to exit from the top? If so, could the water not then be dropped through a hypothetical miniature water-wheel and return to the basin where it started?

I'm assuming that it has to be. Otherwise, it strikes me that this magical non-mechanical cooling technique would back up and stall quicker than my john after a visit from the mother-in-law. Please don't ask.


RE: Wait . . . what?
By stirfry213 on 3/17/2010 11:46:40 AM , Rating: 2
quote:
By creating nanometer-scale structures in silicon, Guo greatly increases the attraction that water molecules feel toward it. The attraction, or hydrophile, of the silicon becomes so great, in fact, that it overcomes the strong bond that water molecules feel for other water molecules. Thus, instead of sticking to each other, the water molecules climb over one another for a chance to be next to the silicon. (This might seem like getting energy for free, but even though the water rises, thus gaining potential energy, the chemical bonds holding the water to the silicon require a lower energy than the ones holding the water molecules to other water molecules.)


Taken from the link in the article: http://rochester.edu/news/show.php?id=3566

Its seems to be quite vague, but that may be due to the fact that I don't understand it.


RE: Wait . . . what?
By JediJeb on 3/17/2010 2:12:53 PM , Rating: 2
Think of a magnet lying on a table, then you pass a metal bar over it and it jumps up to the bar from the table. The attraction of the magnet to the metal bar is greater than the gravitational attraction between the magnet and the earth so it moves upward. Similar principal is happening here, where the hydrophilic attraction of the water to the silicon is greater than the hydrophilic attraction of water atoms to each other, which makes the water "jump up" to the silicon. I haven't read the whole article but it is probably coming from something like hydrogen bonding or surface tension modification.

Take a bowl of water, sprinkle some black pepper on the surface. Now take a drop of dish soap and let it run down one side of the bowl, and watch what happens once it comes in contact with the water. You will observe more or less an opposite phenomena where the water untouched by the soap is more attracted to itself than to the water that has been touched by the soap. (It is also a really neat experiment to show to kids)


RE: Wait . . . what?
By rs1 on 3/17/2010 3:47:00 PM , Rating: 5
It's not free energy because the same force that draws the water up through the silicon capillary system also makes it reluctant to exit the capillary system once it reaches the top/end. Getting the water back out requires an additional energy input. As counter-intuitive as it is, by filling the capillary system the water is actually creating a stable equilibrium state, and breaking that equilibrium to force the water back out requires additional energy.

A paper towel is a good comparison for this. Paper towels can draw water up into themselves through capillary action, even against the force of gravity. However, the vast majority of the water stays in the towel after this, until additional energy is applied (for instance, by wringing the towel) to force the water out.

So the infinite water-wheel idea won't work, unless the researchers have also devised a way to coax the water out of the capillary system that doesn't consume any energy itself.


RE: Wait . . . what?
By GourdFreeMan on 3/17/2010 4:06:47 PM , Rating: 2
As is mentioned in the other thread in this comment section, the laser-etched pipe/chamber will only continue to draw water until the entire etched surface is wet. To make a pump that keeps water in motion you need another agency, such as evaporation at one end of the tube, to remove water from the surface. In a sealed environment the rate of evaporation will decrease dramatically as you approach the saturation point of water vapor in the air.

This prevents such a device from being a perpetual motion machine of the first kind. You don't get unlimited "free energy"[1].

What I am going to say next is probably going to draw some flak, so if you are satisfied with the above explanation, do not read any further. The second law of thermodynamics is a statistical statement and can be violated. It is true almost all of the time because it is merely a statement that a system is more likely to be in a large group of indistinguishable states than in a smaller group of indistinguishable states, if all of the states represent regions of equal probability. Evaporation functions as a Maxwell's Demon of sorts because it allows molecules with higher kinetic energy to escape the potential well created by Van der Waals that hold molecules in a fluid together. In an open system, the free energy[2] of the system would cause evaporation until there was not enough free energy[2] left for a single molecule to escape the fluid. This is almost a perpetual motion machine of the third kind. However, in a closed system, the molecules that escape the system will eventually return to the fluid or adhere to the physical boundaries of the container (condense). Eventually a steady state will be reached where the rate of evaporation and condensation are equal (saturation point). That doesn't mean that the cycle of evaporation and condensation has stopped, only that the rates are equal (possibility of a perpetual motion machine of the second kind). It doesn't mean you can extract energy from the system and expect the rate of evaporation to remain the same. It also doesn't describe a practical physical method of doing so (practicality of a perpetual motion machine of the second kind).

---
[1]"free energy" - pseudoscience term meaning an unlimited or cheap supply of harnessable energy likely inspired by the schemes to make use of free energy[2]
[2]free energy - 19th century term that meant something similar the modern notion of thermal energy. Wikipedia suggests Helmholtz free energy or Gibbs free energy of the system as the historical meaning of this term, but my understanding from reading early works prior to the formalization of statistical mechanics suggests the term thermal energy would be a better fit.


RE: Wait . . . what?
By GourdFreeMan on 3/17/2010 4:14:54 PM , Rating: 2
Minor correction -- in the third paragraph "molecules that escape the system will eventually" should read "molecules that escape the fluid will eventually"


This is a misleading article title.
By 91TTZ on 3/17/2010 12:04:32 PM , Rating: 2
This invention does not "pump" any water at all. It merely draws is via capillary action.

There is a key difference:

If it pumped it, you could have water traveling continuously in a closed circuit the way your car's radiator system or a liquid cooled PC setup works.

If it draws it via capillary action, it would only draw the water onto the laser etched areas until they're all wet. Once they're all wet the water will not flow.

You can see this if you've ever used capillary pipettes. The pipette will draw the liquid into it until the pipette is full, after that it stops. It doesn't keep pumping the water through it continuously. That would be a violation of the basic laws of physics because you'd have work being done by a stationary object with no energy input.




RE: This is a misleading article title.
By porkpie on 3/17/2010 12:11:26 PM , Rating: 2
Untrue. Capillary action can and does pump fluids. See below link for a discussion on nanoscale capillary pumps:

http://www.rsc.org/publishing/journals/LC/article....


RE: This is a misleading article title.
By 91TTZ on 3/17/2010 1:42:40 PM , Rating: 3
I just read that article in its entirety and it supports what I said. Basically, there must be a capillary reservoir of at least the same capacity as the volume you're trying to move. In other words, this will not continuously pump fluid; instead, it will only continue to suck fluid via capillary action until the capillary surface area is saturated. At that point, the movement of fluid stops. Creating structures on a nanoscale definitely helps increase the surface area that the fluid can adhere to, but it doesn't negate the basic rules of physics which enable it to work.

"The simplest possible capillary pump is a microchannel having a sufficient volume to accommodate the volume of liquid that needs to be displaced"

They then go into detail about more complex designs which increase the amount of surface area.


RE: This is a misleading article title.
By porkpie on 3/17/2010 2:12:33 PM , Rating: 2
You're considering only hydrostatics. A capillary pump involves two components: the capillary action which draws the fluid inward/upward AND some mechanism (usually evaporative) for removing the fluid from the capillary chamber.

Consider a tree. Over its lifetime, it pumps many hundreds of tons of water from its roots to its leaves. Why does that pumping action never stop? Because water continually evaporates from the leaf pores.

Here's a similar, albeit mechanical capillary pump:

http://www.wired.com/wired/archive/11.11/start.htm...


By HotFoot on 3/17/2010 4:40:58 PM , Rating: 2
I think the misconception here is cleared up as follows:

The capillary action is not adding energy to the system. The pumping in this example is provided by the energy added from a heat source - like your CPU heating up the fluid in a heat pipe. Capillary action works on potential energy. In the heat pipe example, liquid moving back towards the heat source via capillary action is similar to water flowing down a river towards the ocean. I wouldn't call what the river or capillary does "pumping", because that implies playing an active, rather than passive, part.


RE: This is a misleading article title.
By rs1 on 3/17/2010 4:02:26 PM , Rating: 5
In this case, wouldn't the energy input be coming from the piece of silicon itself? I mean, it sounded like the intent was to set up CPU's so that they have this capillary system integrated into them. If that's the case, then whenever the CPU is running it is dumping heat into the system. Probably enough heat to force the liquid that is currently in the system out, thus causing more water to be drawn in. That's the way all cooling systems should operate, really, by using the energy of the thing being cooled to power the cooling operation itself.

The one thing I don't get is what would prevent the water from being forced out of the entrance portion of the capillary system rather than the exit. In theory if there's nothing to direct the flow of water, and the heat from the CPU is applied more-or-less in the center of the capillary network, then the liquid should show an equal flow in all directions. That could inhibit the intake of new water into the system. Maybe they also figured out how to build little nano-scale valves into their system, to ensure that the water can only flow in the desired direction?


By HotFoot on 3/17/2010 4:44:30 PM , Rating: 2
What you describe in your first paragraph is very much like the common heat pipe in most high-performance air CPU coolers. This technology could improve the wicking action over the standard sintered metal and make heat pipes more effective.


By porkpie on 3/17/2010 4:49:07 PM , Rating: 2
"Maybe they also figured out how to build little nano-scale valves into their system, to ensure that the water can only flow in the desired direction?"

Won't work...that's essentially just another Maxwell's Demon.


perpetual motion machine
By melgross on 3/17/2010 6:58:03 PM , Rating: 2
There is one way to get a perpetual motion machine, and it's been done.

I'm assuming that everyone here has heard of Brownian Motion?

For those that haven't, it's the motion of molecules in a liquid (for simplicities sake). This is everlasting, even at absolute zero. Nano machines have been made that take advantage of that fact. By building machines with moving parts constrained by one way ratchets, they have been able to make them move quickly. It's believed that with more research, they will have machines that are powered by this that can do some actual work.




RE: perpetual motion machine
By porkpie on 3/17/2010 7:08:44 PM , Rating: 2
"It's believed that with more research, they will have machines that are powered by this that can do some actual work."

Mathematically impossible. What you're describing is just another variety of the Feynman Ratchet...google it to see why it doesn't work.

Now, there has been discussion on powering a nanomachine with Brownian motion in a fluid with a temperature gradient in it...but that's an entirely different thing, as the work being done is coming from the energy applied externally to maintain that gradient.


RE: perpetual motion machine
By camylarde on 3/18/2010 8:04:13 AM , Rating: 2
Brownian motion stops at absolute zero. That is how they found the temperature value in the first hand.


RE: perpetual motion machine
By porkpie on 3/18/2010 8:33:35 AM , Rating: 2
Quantum Brownian Motion exists even at absolute zero. Classical Brownian Motion vanishes at 0K.


Why is this different?
By Iketh on 3/17/2010 6:52:13 PM , Rating: 2
Don't heat pipes already use this technique? When the fluids reach their boiling point, it travels up the tube to the fins where it condensates back into the "wick" and travels back to the hot spot again, no matter where gravity is pulling it.

Does this just allow for more efficient designs of these heatpipes?




RE: Why is this different?
By Vagisil on 3/17/2010 8:09:45 PM , Rating: 2
Have you ever seen a heatpipe sticking out of your core2duo?

This cooling takes place inside of the processor hopefully cooling the inners and transferring heat to the plate ontop.

Benefits may include:

Faster processors due to cooler inner temperatures (they'll probably crank up the clock speeds and fill the temperature gap)

More efficient transfer of heat to the plate allowing more efficient cooling.

This has nothing to do with heat pipes in the literal sense, maybe just the functionality.


I made a pump too
By Jellodyne on 3/17/2010 10:55:12 AM , Rating: 2
I also made a magic pump from a thin glass tube. Think I can gt a doctorate from the University of Rochester? I can use it to pump water, as long as I have an infinite pool of water, connected to an infinite amount of dry glass tube.




RE: I made a pump too
By Shining Arcanine on 3/17/2010 3:33:38 PM , Rating: 1
No you cannot. There is a limit to how high capillary action will take things, which is why ferns only grow to a certain height and no higher. At some point, the weight of the water will make it impossible for it to continue to rise, and that point is usually 2 to 3 meters with ferns. It is the same with your "infinite" tube.

Of course, gravity will weaken as the water rises, but not nearly as fast as the weight of the water will increase.


Old, old
By chrisld on 3/18/2010 1:01:06 PM , Rating: 2
Making liquids move uphill using a gradient of surface energy is decades old. Why not do some actual science and achieve something A. new and B. useful?




Wow, people are missing the point
By KIAman on 3/18/2010 7:00:04 PM , Rating: 2
The water will only spread enough to evenly distribute its volume over the surface. Once that's done, energy must be applied to remove it to repeat the cycle.

The point everyone is missing is the energy happens to be waste heat from a processor. So it "feels" like this movement of water is free because the heat from a processor is wasted anyways. We only care about transistors switching on a processor, not the heat. We aren't using it to cook eggs.

Just like a paper towel. Soak one corner and watch as water magically distributes throughout the towel. Once saturated, nothing happens until you use a blow-dryer to dry the water out. Then be amazed as the water distributes from the corner soaked in water throughout the towel again.




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