Graeff’s experiments and 2LoD: Replication and Implications

Posted: June 28, 2012 by tchannon in atmosphere, Gravity, methodology

This is part four of a four part guest post by ‘Lucy Skywalker’.

Graeff’s experiments and 2LoD: Replication and Implications

Lucy Skywalker recaps: In Part One I described my visit to Graeff’s seminar. In Part Two I described some of his experiments in detail. In Part Three I showed how he developed the backing theory. Finally in Part Four I now consider the implications of this work, and plans for replicating the experiments. Replication is of crucial importance both to Climate Science in particular, and Science in general; without it, no theory is sacrosanct.

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Here is a replication of one of Graeff’s experiments, assembled by him and ready to go. This particular experiment seems to be simpler in its results than the experiments we’ve looked at. But first, I want to think about implications of his work, to gauge where we want to pitch in.

IMPLICATIONS OF GRAEFF’S WORK

Graeff has demonstrated that a modification of the full statement for the Second Law is needed, and that this is possible without contravening the essence of the Second Law. One place where his modification is clearly of importance is Climate Science.

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Atmospheric temperature drops on average about 7°C for every kilometre altitude gained, with variations depending on humidity and other factors. This is the adiabatic lapse rate which is very familiar to meteorologists and airplane pilots. Yet there is no theoretical basis that includes the molecular effect of gravity in the way Graeff shows must be at work – as far as I am aware. Now we suddenly have a very beautiful, very simple and very satisfying explanation for the adiabatic lapse rate, namely gravity on individual molecules, OFFSET more or less by convection to produce a dynamic and somewhat unstable equilibrium.

The following is my fuller explanation, which I hope others can improve:

By gravity, gas molecules fall and gain kinetic energy which is warmth. But this increased kinetic energy causes them to repel each other more actively and the gas will either expand-and-rise, or warm. We have a continual balancing-act between what individual molecules do in microscopic response to gravity (fall and get warmer) and what groups of free molecules can do all together – ie convect (expand and rise as a parcel, gaining momentum as wind but cooling due to both expansion and loss of gravitational kinetic energy). The very existence of the adiabatic lapse rate (a.l.r.) strongly suggests the presence of a gravitational temperature gradient T(Gr). It appears that T(Gr) (-0.07K/m = 70°C/km) is about nine-tenths offset by convection in the free atmosphere to produced the familiar adiabatic lapse rate of around 7°C/km – and a habitable planet. This is unfamiliar, so it feels tricky at first. One has to imagine single molecules under gravity, and at the same time, parcels of molecules able to have a net collective action ie convection. In the free atmosphere, convection nearly overcomes T(Gr), but not quite, and the a.l.r. is the result. But in the far denser oceans, convection wins over T(Gr) so cool water being heavier sinks. If there were convection impedance, warmth would increase with depth – as happens in the solid earth.The convection needed to balance T(Gr) in air is scarcely noticed on this planet. But the sun shining through clear air to warm Earth’s surfaces creates noticeable convection currents which, again, undo most of the warming.The true greenhouse effect occurs in greenhouses where convection is impeded.

ls-7

I suspect that early meteorologists may have understood or intuited all this – it is not far-removed from common-sense. There is indeed sound evidence for the “greenhouse gas” properties of substances like CO2, O3 and CH4: we can see the evidence for ozone in the diagram above – but the ozone effect at least seems to be insitu, not projected down. Yet the IPCC claim a mere 33-degree greenhouse effect. This contradicts common-sense and the ozone effect, and when we look at the 70 degrees temperature difference between Earth’s tropopause and Earth’s surface; it is even further from explaining the 100-odd degrees temperature difference between the lunar surface temperature (measured now by Diviner) and the Earth surface temperature.

The gravity effect, as theorized by Graeff, can explain this large difference with no trouble at all (gravitational temp. gradient, minus local convection, equals local adiabatic lapse rate). It can explain the warmth in places below sea level, and down deep mines. Graeff’s theory also makes sense of the violent atmosphere on Jupiter and the very high temperatures of Venus’ surface and the interior of Jupiter. It starts to explain why our own planet is extremely hot at the core yet temperate at the surface. I have found nothing in the solar system that Graeff’s theory does not start to explain.

Graeff’s theory remedies a huge missing link.

HOW TO ALLOW FOR OUR HUMAN REACTIONS?

People have strong reactions to challenges to familiar ways of thinking. Some people will go to great lengths to stay in denial of evidence, rather than look at something that requires them to consider changing their thinking and their belief system, maybe risk losing grants, status, obsessions, or credibility. What an embarrassment if one’s lifelong “expertise” might not look so great any more. We’ve seen the whole of orthodox Climate Science close ranks and do this to skeptics, in spades.

In Graeff’s case, it is not only “warmists” who may have such a reaction. It is “climate skeptics” as well. Graeff’s theory not only challenges one of the most sacrosanct of all the laws of science, it also gives strong credence to other challengers like Nikolov and Zeller.

OK, we have to ask questions. Surely Graeff’s challenge would have been seen and accepted at the time of Maxwell, if it was good science?? Actually there was one good scientist, Loschmidt, who did dispute Maxwell. The really extraordinary thing is that until Graeff, nobody had checked Loschmidt’s challenge to Maxwell’s belief (and Gibbs’ mathematical theory) by practical experimentation.

REPLICATION OF EXPERIMENTS

Clearly we have to do very thorough checking, with replication of experiments. But we can save ourselves anxiety by remembering that only a few years back, Graeff’s experiments at laboratory scale would have yielded temperatures too small and too fluctuating to measure. We didn’t have suitable materials, or precision thermocouples and thermistors, or the computer power to record long sequences. But Graeff has developed sufficient methodology, and has shown what accuracy and consistency to expect, and how to wring statistically significant results out of fluctuating data, so that we can experiment and get valid results. The graphs below are a reminder from his water experiment. They show (1) how he wrung significance from the subtle effect of gradient fluctuations (thermocouples n 1-8, fine scale to LHS) lagging temperature rates of change (thermistors 9-14 – larger scale to RHS),

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and how he used those temperature gradient fluctuations (thermocouples) plotted against extermal temperature changes (thermistors) to obtain a very exact reading for the temperature gradient when the external conditions are not changing (point where change per hour = zero).

NB: the thermistor readings above, while clearly accurate in showing change (they clearly move in step) are not so accurate in absolute terms. Therefore it is neither clear nor necessary to know which lines represent which, between 9 and 14.

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We will still have to deal with challenges rooted in emotional denial. We can expect a lot of repetition and a great show of apparently relevant knowledge that may be valid as claimed, though it is more likely to be straw men etc. This can be tiring and depressing. Scientist have even committed suicide following such responses, even when their work has been correct and accepted years, decades or even centuries later.

But the other side of the challenges is eventually a very strong certainty. There is no way Science can escape dealing with emotional reactions. Rather, it is surely far better scientific practice to acknowledge human nature, and by acknowledgement, develop ways to deal with it. This is a big challenge to Science today – how to handle so-called “subjective” factors that Science rightly excluded early on, that are now kicking back so hard as to make corrupt nonsense of whole branches of Science.

At the simplest level, we can deal with this issue by simply becoming aware of what’s happening, without judgement, and preferably with compassion, whenever we see human frailty. Blogs can be brilliant places for developing this higher part of our human nature. Some awaken more quickly than others. But we are all on the journey.

At the end of all this, there is the excitement of real scientific discovery. Here is something important, right under our noses.

WHAT CAN WE DO NEXT?

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I want to see Graeff’s work replicated in ways people can trust. I believe this is crucial at all levels. Without basic replication, people will reasonably say that Graeff could be mistaken (though I personally have no doubts). Replication is thus the key to:

  • Put right a flaw in a very basic and very important law of physics
  • Accept that even the Second Law can still have amazingly “obvious” flaws
  • Open up possibilities regarding alternative energy production – possibilities that may have been prematurely excluded
  • Help restore Climate Science to a proper scientific footing – and stimulate reform therein
  • Open the way to a more holistic approach to Science altogether – we ignore “subjective” factors at our peril.

My latest news is that Graeff is actually working right now with a professor at an American University who is hoping to replicate his experiments. This is the first time anyone has actually chosen to replicate him, and collaborate. Professor Sheehan has certainly taken a warm and perceptive interest, but has not undertaken replication. A Japanese professor set up an experiment, but using a centrifuge without enough controls to be able to measure a true gravitational effect as Graeff has done. Lack of replication is not for want of Graeff trying to interest the universities that should be doing this work…

I regard this US development as of crucial importance. If Graeff’s work is now going to be replicated in proper fashion, and written up so as to be able to achieve at least the first two points above, I shall feel that my involvement in direct replication is no longer needed. However, we are not yet at this point. And I do not know how many of my other points the professor may be open to. So there may still be work for us to do, particularly to help this work to be grasped and honoured by Climate Science.

If nothing comes of the US development, then I will still need to start with replication – in which case, the questions are:

  • Building a team – who wants to become involved? How can we build a good team? How about a trip to Graeff for this? Etc… I am willing to be a first contact point.
  • Location – should we start where I live and where there are already many others interested in parts of science that orthodoxy will not examine? Does Tallbloke want to take on this project where he lives? Can we find a friendly college laboratory? Etc…
  • Details: basic materials, sensitive equipment, skills, premises? How to find somewhere that maintains a constant temperature like Graeff’s thermostatically-controlled cellar? Do we simply collect scrap insulation / metal sleeves / etc or do we have money to go for “the best”? How much hands-on knowledge do we need? Can someone handle datalogger setup, calibration, and conversion to Excel on a PC? What can Graeff advise out of a lifetime of engineering? Etc…
  • Communication: getting the word out to the professionals who should be doing this work – do we want to aim to produce a paper for peer-review? should we include Sheehan and Graeff and others? at what point do we step back with “mission accomplished” ?

I am hoping to visit Graeff again this year, perhaps with a small group of people who are interested like myself. I think that such a visit would make a huge difference. And it may be now or never, since he is 84 – and it would be a great loss to all of us if we miss this opportunity.


Prepared for the web from documents supplied by Lucy Skywalker.

TNC.

Comments
  1. […] Graeff’s experiments and 2LoD: Replication and Implications […]

  2. Richard111 says:

    Thank you for these postings. Very interesting and thought provoking for this layman. Now for the silly layman question. 🙂 CO2 is slightly more massive than N2 and O2 molecules yet the claim is it is well mixed in the atmosphere. Plants and water on the Earth’s surface are consuming CO2 almost as fast as we produce it. I have to assume gravity ensures the plants don’t run short at their fixed locations. Is there a gravity induced lapse rate of CO2 concentration in the atmosphere? I would also expect high concentration of CO2 down wind of major cities. You certainly get soot and stuff downwind.

  3. Bryan says:

    Lucy Skywalker says

    “The very existence of the adiabatic lapse rate (a.l.r.) strongly suggests the presence of a gravitational temperature gradient T(Gr).”

    Yes indeed!
    A metre square perfect absorber at Earth orbit facing the Sun will reach a temperature of about 120C at radiative equilibrium.
    Construct a very long adiabatic tube facing the Sun on top of the square placed at the equator.
    Fill it with an relatively inert gas like N2.
    A lapse rate will develop quite close to the Earths – 9.8K/km as Cp(air) is very close to Cp(N2).
    If we ignore the small radiative contribution of N2 the only way the tube can transfer heat will be by radiation from the base.

    This simple theory tells us that at a height of about 13km to temperature will drop to near 3K.

    So a greenhouse theory explanation of the lapse rate is unnecessary.

    Graeff’s experiment goes a bit beyond this.
    To replicate his experiment unfortunately is practical only for a university department (or similar) or a rich amateur with plenty of spare time.

    Thanks for your lucid ,very readable, posts.
    You certainly have our attention and I will follow developments with interest.

  4. tchannon says:

    Richard111, Bryan,

    I recommend not jumping the gun by extending the possibility of a tiny effect having significance at a large scale. First lets get to the first base of clear proof there is an effect.

    CO2 mixing? There is plenty of evidence showing nothing surprising, it is mixed to varying degree of goodness, with highs and lows. Some claim this is not so, mostly AGW people who rely on perfect mixing as part of their assertions (particular problem over ice cores), yet others know this is not the case and show this.

    Altitude effects seem to be minor and rare, eg. pools of volcano gas forming a CO2 lake, yet air movement will rapidly dissipate this under normal circumstances.

  5. ferdberple says:

    The inversion of the temperature gradient in the oceans seems likely a result of the incomprehensibility of water as compared to air. A cubic meter of sea water at the bottom of the ocean weighs very much the same as a cubic meter of sea water at the top of the ocean. Thus, there is very little resistance to moving water upwards.

    This is not the same case in air, where a cubic meter of air at the surface weighs considerably more than a cubic meter at altitude. Nature overcomes this problem by expanding the cubic meter of air as it rises, thus reducing its weight. However, this expansion also cools the cubic meter of air, increasing its density and counteracting the convection.

    Thus, the incomprehensibility of water aids the movement of water upwards as compared to air. Which is indeed fortunate for humans. We live at the boundary between air and water, and both of these have their maximum temperature at this boundary.

  6. tchannon says:


    http://www.simetric.co.uk/si_water.htm

    Go figure, it ain’t simple.

    There is a question re: Graef, so does the claimed gradient in water vary with water temperature?

  7. Trick says:

    Bryan 9:43am – “Construct a very long adiabatic tube facing the Sun on top of the square placed at the equator. Fill it with an relatively inert gas like N2.”

    Now that Bryan has constructed a “very long” Graeff’s dewar experiment, I’ll apply what V&G, B&A and Akmaev all tell us about the lapse rate.

    1st thing tidy up a little house keeping….Bryan continues: “.. the only way the tube can transfer heat will be by radiation from the base. “

    As we learn in V&G paper, “Exner (1925) pointed out that the confusion arose from defining the problem in an inconsistent way.” And then Emden 1928 published on ‘‘Periodically recurring errors, or misconceptions.”

    Here, Bryan 1st tells his the column is adiabatic then “..the tube can transfer heat.” Anyone else see a perfect application of Emden and Exner here? Hint: dictionary.com has a great definition for “adiabatic”, I will leave the investigation of that to the reader.

    Those following the bouncing ball can see we have an adiabatic column closed to work in/out at earth’s surface and for all intents and purposes the top adiabatic condition let’s specify is closed to work in/out also. Bryan now has an enclosed adiabatic column w/no work in/out like Graeff sans the powder. Let’s say Byan’s very long col. goes up to P(hmax) = 264.36 hPa (one of the dots on the standard atm. curve) and the column is left alone to LTE as V&G use for their air example of section 2b (I can do less work on my environment this way.)

    Now I have learned, posters here cannot see the cartoon for Bryan’s column in B&A; rest assured it shows an enclosed adiabatic column (just like I clarified for Bryan’s) which points us to Verkley 2b. (Rule out V2a isothermal soln. since that V&G Fig. 1 column is open to work in/out at top & bottom.)

    V&G 2b (details from B&A): T(z) = T0 * ((P(z)/P0)^k), where k = R/Cp & calculated for air (N2 is too expensive per fill of approx 3.5tons):

    R = 287, Cp = 1004, P0 = 1000 hPa, T0 = 302.04 K (28.89 C), P(z) = 264.36 hPa

    T(z) = 302.04 * ((264.36/1000)^ 287/1004) = 302.04 * (0.26436)^ 0.285856 = 302.04 * .683646

    T(z) = 206.49 K which by Mark 1 eyeball inspection lies ~on V&G dashed line in their Fig. 2 as it should (I have not made an egregious math error anyway; rule of typo’s is they are spotted right after posting still applies).

    T(z) = -66.66 C so V2b non-isothermal soln. lapsed 28.89 + 66.66 = 95.55 C.

    The standard atm. reaches this P ~ 264 hPA around a z of ~10km. The standard atm. lapse rate is 6.4 C/km so avg. standard atm. lapses ~64C in these 10km. The ideal air non-isothermal non-GHG lapses ~95C vs. real (GHG et. al.) atm. standard of ~64C give or take an eyeball.

    Bryan continues: “So a greenhouse theory explanation of the lapse rate is unnecessary.”

    Any discussion?

  8. Bryan says:

    Trick says:

    “Here, Bryan 1st tells his the column is adiabatic then “..the tube can transfer heat.” Anyone else see a perfect application of Emden and Exner here? Hint: dictionary.com has a great definition for “adiabatic”, I will leave the investigation of that to the reader.”

    Its a reasonable definition of adiabatic!

    Adiabatically restricted horizontally but without making that explicit as the context makes that unnecessary.
    The derivation of the DALR assumes you have a heated base to start with.
    For Earth conditions you also have ‘cooling’ by radiation to space at the TOA.

    My example was to emphasise that you don’t need to specify inclusion of a radiative gas to have a lapse rate.

    So no need (so far) for a greenhouse gas theory.

  9. dp says:

    How does this relate to a bottle of compressed gas heating during compression and freezing when that pressure is released? No gravity is involved, just good ol’ second law stuff.

  10. ferdberple says:

    A lapse rate will develop quite close to the Earths – 9.8K/km as Cp(air) is very close to Cp(N2).
    This simple theory tells us that at a height of about 13km to temperature will drop to near 3K.
    =======
    I get a different number. surf temp 290K / 9.8K/km = 30km for the height of the atmosphere as a function of KE at the surface. Which is petty close to observation.

    Actual results will differ because KE of air molecules has a probability distribution around the average, which suggests atmospheric density might show something similar.

    Is there any observational evidence that atmospheric density is modified by the M-B distribution, centered around 30km? With allowance for condensation of water.

    http://en.wikipedia.org/wiki/File:Comparison_US_standard_atmosphere_1962.svg
    http://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution

  11. ferdberple,

    We can’t let you go on being incomprehensible when you are so incompressible…!

  12. Bryan says:

    ferdberple says

    I get a different number. surf temp 290K / 9.8K/km = 30km- quite correct.
    I made the classic error of forgetting to change C to K

  13. ferdberple says:


    Wikipedia shows 30km as the height of the atmosphere.

    from above: 290K / 9.8K/km = 30km (surface temp / gravity = atmospheric height)

    This has to be more than simply a coincidence. It shows that the height of the atmosphere is linearly related to gravity and surface temperature.

    This suggest that the KE of molecules at the surface is what determines the atmospheric height, because at 30km the average KE of an air molecule falls to zero because of gravity.

    As a point of conjecture: What determines the relationship between pressure/density (non-linear) and the lapse-rate (linear). Can it be the M-B distribution? A product of chance resulting from elastic collisions, resulting in molecules having lower or higher energy according to a probability distribution? Thus limiting them to reaching different altitudes according to a probability?

  14. Richard111

    Thanks. Is CO2 proportionately lower at higher altitude? Simple answer, I suspect but don’t know. But hey, if water droplets can stay aloft, it can’t be too difficult for CO2.

    And yes, CO2 levels can vary hugely, eg high downwind of cities and industry, and low downwind of forest, and variations for night and day and seasons, as Beck’s collection of firstrate scientists’ tests shows.

  15. Bryan, thanks.

    You have to be careful not to confuse the temperature gradient due to gravity, with the effect of the Sun heating the earth’s surface which transmits that heat to adjacent air. Both effects, in my understanding, are heavily modulated by convection – which varies with geometry and confuses the issue further.

    But yes, the ghg effect seems to evaporate somewhat. Plus Nikolov and Zeller, with the corrected Stefan-boltzmann equation, showed the difference the atmosphere makes is more like 100 degrees, certainly not 33 degrees as IPCC asserts. However, I still think ghg (ozone) explains the temperature pattern through the stratosphere and mesosphere. But it’s a local effect, not a radiated one.

  16. ferdberple says: July 6, 2012 at 3:48 pm

    The inversion of the temperature gradient in the oceans seems likely a result of the incomprehensibility of water as compared to air….

    First Tim, yes water is densest at 4C, so the ocean floor is at this temperature and liquid, rather than cooler and frozen, thank goodness. I never cease to be amazed by the fine-tuning that makes life possible.

    Second, why does the atmosphere show an adiabatic lapse rate, but not the ocean? This is my hunch: Imagine a cube of water 1m x 1m x 1m, sitting in front of you at sea level. Now imagine that cube of water vaporized but taking up the same ground area. It will fill a column 1000m or so high. What this says to me is that gravity has a thousand times more room to make its influence felt in air than in water. Thus it is perhaps not so surprising that in the open air, convection nine-tenths overcomes the gravitational temperature gradient, but in a room, convection wins completely (ceiling warmer than floor), and with liquids, convection always wins unless it is suppressed.

  17. Tenuc says:

    Lucy Skywalker says:
    July 7, 2012 at 8:58 pm
    “…I never cease to be amazed by the fine-tuning that makes life possible…”

    Hi Lucy, I think you’re confusing chicken with egg… 🙂

    The type of life we observed here on Earth has, over a long period of time, evolved/adapted to the conditions that exist here. If our world had a different make-up, then some other form of life could maybe be in residence here???

  18. dp says: July 7, 2012 at 5:46 am

    How does this relate to a bottle of compressed gas heating during compression and freezing when that pressure is released? No gravity is involved, just good ol’ second law stuff.

    Yes indeed. In your situation, no gravity is involved, just the gas laws.

    However, if you want to understand how both are related, you have to think yourself into the molecular level of reality. Here, expansion means fewer molecules even though they are moving just as fast: hence cooling. But 2LoT will kick in to speed up those depressurised molecules so that their net temperature is the same as adjacent temperatures. With regard to the gravitational temperature gradient discussed here, you have to imagine the effect of gravity on each molecule. Think of the Newton’s Cradle balls but in the vertical plane.

  19. I want to replicate, or at least, to ensure replication. And I want to get the word out. I think Graeff is so important to Science, perhaps more than he realizes.

    Anybody interested in helping?

  20. Tim Folkerts says:

    Lucy says: “Now we suddenly have a very beautiful, very simple and very satisfying explanation for the adiabatic lapse rate, namely gravity on individual molecules, OFFSET more or less by convection to produce a dynamic and somewhat unstable equilibrium.”

    Why do you need a new, fuller explanation? The nature of the dry adiabatic lapse rate is will known, and available all over the internet.

    http://en.wikipedia.org/wiki/Lapse_rate#Dry_adiabatic_lapse_rate
    http://www.gps.caltech.edu/classes/ese148a/lecture11.pdf
    http://acmg.seas.harvard.edu/people/faculty/djj/book/bookchap4.html
    http://www.physics.umt.edu/borealis/Environmental%20Lapse%20Rate.pdf

    What SPECIFICALLY do you disagree with in these? What is wrong with any of these, which provide a clear, exact derivation of the lapse rate based on well-established thermodynamics? How is your explanation different & better? Could you quantify to what extent the equilibrium is unstable in your explanation?

  21. Tim Folkerts says:

    Lucy says: ‘First Tim, yes water is densest at 4C … ”

    Actually, fresh water is densest at 4 C. Sea water is densest at lower temperatures .. closer to -4 C due to the presence of salt. You need more subtle reasoning to determine the temperature and state at the bottom.

  22. Q. Daniels says:

    Tim Folkerts wrote:
    Could you quantify to what extent the equilibrium is unstable in your explanation?

    From the Wikipedia link:
    If the environmental lapse rate is larger than the dry adiabatic lapse rate, it has a superadiabatic lapse rate, the air is absolutely unstable — a parcel of air will gain buoyancy as it rises both below and above the lifting condensation level or convective condensation level.

    Graeff is measuring a superadiabatic lapse rate.

    A column of air with convection suppressed and perhaps some other unrecognized conditions achieves an equilibrium which is greatly in excess of the DALR.

    That seems pretty straight-forward to me. What’s not to understand?

  23. Trick says:

    Tim F 7/9 4:05am – “What is wrong with any of these, which provide a clear, exact derivation of the lapse rate based on well-established thermodynamics?”

    Probably nothing , though I have yet to read them all. Once again, in the very 1st link, Tim can learn the difference with Graeff’s experiment and the real atmosphere. The difference is, again, in the assumed constraints. Graeff’s very insulated & likely close to adiabatic closed dewar can do no work on the air around it and here, plainly, is the difference in Tim’s wiki clip and, for me at least, thoroughly answers Tim’s other question “What SPECIFICALLY do you disagree with in these?” – wiki:

    “ As the air parcel expands, it pushes on the air around it, doing work (thermodynamics). Since the parcel does work but gains no heat, it loses internal energy so that its temperature decreases.”

    The enthalpy in the ideal column (V&G 2a) setup loses internal energy (enthalpy) by pushing on the air around it in exactly the right amount to make the classic constraint solution isothermal – no gradient.

    When there is no internal energy lost when the enclosed (V&G 2b) ideal column does not push on the air around it, such as Graeff’s dewar experiment, there is 0 enthalpy lost externally (no push change term) and that changes the classic column constraint solution to non-isothermal gradient larger than the real atmosphere.

    V&G 2b sets an ideal limit on the highest real atm. lapse rate. The real atmosphere column being slower than ideal lapse gradient is due to many things: GHGs, pushing on the air around it, aerosols, sun heating at the 288K base, heating&cooling in/out and radiating at the top to 3K, et. al.

  24. Trick says:

    Q. Daniels 6:19am – “A column of air with convection suppressed and perhaps some other unrecognized conditions achieves an equilibrium which is greatly in excess of the DALR.”

    The classic ideal isothermal column LTE has less to no buoyancy therefore has less convection than the real atm. Suppressing convection would tend to lessen bouyancy, right? If right, suppressing convection should push the DALR down toward isothermal constant gradient not up and “greatly in excess of DALR.”

    Graeff should measure less convection with powder, therefore less gradient leading to a need for greater sensitivity in the thermocouples the more convection is suppressed. Graeff should get rid of the powder in B74 experiment.

  25. Tim Folkerts says:

    Q. Daniels say: “Graeff is measuring a superadiabatic lapse rate.”
    But Q. Daniels also quotes Wikipedia

    If the environmental lapse rate is larger than the dry adiabatic lapse rate, it has a superadiabatic lapse rate, the air is absolutely unstable — a parcel of air will gain buoyancy as it rises both below and above the lifting condensation level or convective condensation level.

    A “superadiabatic lapse rate” with “absolutely unstable” air means the air is actively moving. In Greaff’s case, the air would be actively moving WITH NO ENERGY input. This is a classic perpetual motion machine — you could put a miniature wind gauge in the tube and it would perpetually measure air rising and settling . This is clearly not the maximum entropy state.

    Once again, it is POSSIBLE that perpetual motion machines exist and Graeff as discovered one, but I am not holding my breath.

  26. Tim Folkerts says July 9, 2012 at 4:05 am

    You are simply ignoring Graeff’s results. Please explain Graeff’s results, both theoretically and regarding significance. Please keep focus on this.

    Graeff is getting seven times the Wikipedia / world / “appeal to authority” calculated lapse rate of ~10K/km, when he impedes convection. Measured DALR at ~7K/km tend to be lower than the Wiki calculated of ~10K/km. How do you explain that theoretically? To me, it doesn’t quite fit. Graeff fits better. And explains more. And explains more simply, and more logically, with a result that’s just as simple (see my previous article). What’s not to like?

  27. Tim Folkerts says:

    “A column of air with convection suppressed and perhaps some other unrecognized conditions achieves an equilibrium which is greatly in excess of the DALR.”

    Perhaps we need to agree on terminology. It is possible to achieve a “steady state” solution that is non-isothermal, but (according to classic thermodynamics) this not the “equilibrium” solution ( = after a long time, with no work or heat exchange either internally or externally). It is even quite possible to exceed the DALR. But in classical thermodynamics, this always requires some external work or external heat sources/sinks.

    “Thermals” are a standard example. The ground is heated, the lapse rate is exceeded, and convection starts. This can continue for quite some time as a “steady state” condition — but it is not the equilibrium condition.

    People seem to think that Graeff is achieving “thermals” (ie active convection) within his tube, but that he is doing it without the heat input.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    As a thought experiment, consider a “Graeff tube” (perfectly insulated on all sides) 1 km tall with a cross section of 1 m^2. Imaging putting a heating/cooling plate at the bottom (inside the tube) and another at the top that can be set to temperatures T_b and T_t respectively. Fill it with dry air. For the sake of argument, let T_b = 300 K

    According to classical thermodynamics…

    If T_t = T_b, then no energy input is required to either the top or bottom plate, and the column is in equilibrium. There is no conduction or convection.

    If (300 – 9.8 K ) < T_t < 300 K, then there will be conduction thru the column, but no convection. To maintain a steady state conduction, the top must be continually cooled, and the bottom must be continually heated. This only requires a very tiny heat source at the bottom and an equivalent heat sink at the top and (microwatts, I believe) because conduction thru the air is very poor.

    If T_t < (300 – 9.8 K ), then the air becomes unstable, and convection starts. Now we need considerably more heating power at the bottom and cooling power at the top for the "steady state condition" because convection is much more efficient at transporting energy. But this is NOT an "equilibrium" condition.

  28. ferdberple says: July 7, 2012 at 5:36 pm

    Take care with Wikipedia’s “height” of atmosphere. Look here for comparisons. Broadly speaking, the tropopause is at 10km height average / “normal” though that varies with latitude etc. At tropopause, pressure is about 1/10 that at sea level. But pressure keeps on falling steadily through the ozone-rich stratosphere and mesosphere. “Linear” ALR only extends up to the troposphere. And is it truly exactly linear? Wiki formula suggests yes but I doubt that explains enough.

  29. Tim Folkerts says:

    Lucy Skywalker says: “Please explain Graeff’s results, both theoretically and regarding significance. Please keep focus on this.”

    That is indeed the core of the discussion. And the answer is that classical thermodynamics CANNOT explain the results. There is no point in trying to explain this using classical physics (just like there is no point in trying to explain electron orbitals using classical physics).
    Either
    1) Graeff is wrong, and needs to improve his experiment
    2) Classical physics is wrong, and the scientific community needs to improve its theories.

    “Graeff is getting seven times the Wikipedia / world / “appeal to authority” calculated lapse rate
    I would say you have this argument backwards. Appealing to “Graeff” is an appeal to authority, since he and he alone is getting his result. I am making an appeal to “physics” as presented in numerous textbooks and webages. And that you can follow and derive yourself if you are so inclined.

    “Measured DALR at ~7K/km tend to be lower than the Wiki calculated of ~10K/km. How do you explain that theoretically?
    No … the measure ENVIRONMENTAL LAPSE RATE is ~ 7 K/km. There are several reasons the environmental lapse is not equal to the DALR. There is moisture in the air. The atmosphere is not “adiabatic” (a parcel or air can lose energy by conduction and radiation). The sun does not always heat the ground enough to maintain the moist ALR. There is no reason to expect that the actual atmosphere would follow the DALR. The DALR is simply a dividing line between two different types of behavior.

    “To me, it doesn’t quite fit. Graeff fits better. And explains more. And explains more simply, and more logically, with a result that’s just as simple (see my previous article). What’s not to like?”

    What’s “not to like” is that your explanation ignores a lot of well-established physics. Since you have readily admitted that you are not an expert in thermodynamics, you must (if you are honest to yourself) acknowledge that your opinion of what “seems to fit” may not the most reliable indicator of the correctness of your conclusions.

  30. Tenuc says: July 7, 2012 at 9:06 pm

    Lucy Skywalker says:I never cease to be amazed by the fine-tuning that makes life possible…”

    Hi Lucy, I think you’re confusing chicken with egg… 🙂

    Yes. Which came first??

  31. Tim Folkerts says: July 9, 2012 at 4:13 am
    Lucy says: ‘First Tim, yes water is densest at 4C … ”

    Actually, fresh water is densest at 4 C…

    Correction noted re. salt water max density. But overall, my point still holds. Seawater sinks in liquid form and the part that freezes, floats.

  32. Q Daniels, Tim F

    I would rather say that Graeff is measuring a temperature gradient that appears to be due to gravity when convection is impeded. Strictly speaking, it would be correct to still call it an adiabatic lapse rate.

    People seem to think that Graeff is achieving “thermals”(ie active convection)

    Tim F, there is no perpetual motion here, Graeff’s experiment is in static equilibrium due to convection impedance, and equilibrium is the correct word. The problem is with the theoreticians who thought equilibrium could mean isostasy in a vertical tube of air.

  33. br1 says:

    Trick:

    I really know I shouldn’t, but maybe one more post. Here goes:

    “When there is no internal energy lost when the enclosed (…) ideal column does not push on the air around it, such as Graeff’s dewar experiment, there is 0 enthalpy lost externally (no push change term) and that changes the classic column constraint solution to…”

    a constraint which has constant internal energy. A closed adiabatic system that can do no work on the surroundings has constant internal energy. This is V&G constraint 2 in the introduction, which gives an isothermal answer.

    “Graeff should measure less convection with powder, therefore less gradient leading to a need for greater sensitivity in the thermocouples the more convection is suppressed. Graeff should get rid of the powder in B74 experiment.”

    but he has measured both of these already. He measures a higher gradient with the convection suppressing powders and less gradient when convection is allowed. The gradient is not attributed to bulk convection, it is attributed to direct kinetic energy reduction due to conversion to gravitational potential energy.

  34. br1 says:

    Tim Folkerts:
    “And the answer is that classical thermodynamics CANNOT explain the results. There is no point in trying to explain this using classical physics (just like there is no point in trying to explain electron orbitals using classical physics).
    Either
    1) Graeff is wrong, and needs to improve his experiment
    2) Classical physics is wrong, and the scientific community needs to improve its theories.”

    I agree.

    My dabbling with classical physics has been to fully appreciate why there is no predicted temperature gradient. I can now see this from several different viewpoints. It has also been useful in appreciating why the laws are the way they are, what assumptions may have been involved and if any of them are invalid. All good stuff!

    Saying the experiment needs improving is also a tough one – where is the flaw? If Lucy goes ahead with the replication, what do you recomend as the most critical point to look out for? After all, we wouldn’t want to replicate the flaw by not knowing what it is.

  35. I said “Graeff is getting seven times the Wikipedia / world / “appeal to authority” calculated lapse rate.

    Tim F said: “I would say you have this argument backwards. Appealing to “Graeff” is an appeal to authority, since he and he alone is getting his result. I am making an appeal to “physics” as presented in numerous textbooks and webages. And that you can follow and derive yourself if you are so inclined.”

    Tim F it is you who have the meaning of “appeal to authority” backwards.

    Form: Authority A believes that P is true. Therefore, P is true.

    Bertrand Russell:

    It is not what the man of science believes that distinguishes him, but how and why he believes it. His beliefs are tentative, not dogmatic; they are based on evidence, not on authority or intuition.

    I am not “appealing to Graeff” but looking at his statistically significant data. “He and he alone” has done the experiments, simply because nobody else has so far replicated them, despite numerous invitations to universities. I am trying to get support for replication, in writing all these articles. That was my purpose!! Replication is ultimately needed so as NOT to appeal to authority.

    On the evidence so far encountered, I think that Graeff is right and that “the scientific community needs to improve its theories.” I don’t think classical physics is “wrong” here, it is simply a particular detail of the Second Law’s interpretation, that was NEVER tested – until Graeff.

  36. Trick says:

    Tim F 3:28pm – Terminology: “Thermals” are a standard example. The ground is heated, the lapse rate is exceeded, and convection starts.”

    This is forced convection starting; glider pilots use it. There is no forced convection in Graeff’s experiment, only free convection can exist in the dewar B74 (another terminology is Graeff includes gravity convection).

    br1 4:36pm – “This is V&G constraint 2 in the introduction, which gives an isothermal answer.”

    Conservation of energy constraint 2 also gives non-isothermal answer! Wow, huh? The V&G intro is a summary – less specific than V&G section 2b when read in conjunction with B&A (I will get page & paragraph in a few days). You apparently are having trouble seeing from the intro that constraint 2 (constant energy internal plus potential) HAS to be applied to energy accounting across the control volume in V2a AND V2b AND V2c due to 1st law. V&G just assumes the reader knows this in the intro. or the conclusion; specifically in their math you can see they enforce constant energy (gas enthalpy) accounting across the control volumes of sections 2a, 2b and 2c.

    Constant (internal & potential) energy is golden & in part leads to the real atm. gradient, the non-isothermal gradient AND the isothermal no gradient depending on ideal gas, entropy max. or not, closed or open energy accounting across the control volume and if needed solid state eqn. in the detail math. This is why constraints are important – they select the ideal isothermal or non-isothermal solution.

    Lucy 3:20pm – “Graeff is getting seven times the Wikipedia / world / “appeal to authority” calculated lapse rate of ~10K/km”

    Graeff’s tube B74 if actually completely full of powder no longer has only a free ideal gas eqn. of state like in wiki/world/authority, the gas is all trapped in the glass beads. A solid eqn. of state has to be invoked and no ref. you or Tim F cites include the solid eqn. of state lapse. Conduction in solids could be leading to the 7x lapse found in B74 dewar for all we know so far.

    Tim F and br1:

    “Either
    1) Graeff is wrong, and needs to improve his experiment
    2) Classical physics is wrong, and the scientific community needs to improve its theories.”
    I agree”

    Or 1) Graeff’s solid state & gaseous state B74 experiment is good & replicable, 2) classical physics is right, 3) scientific community HAS improved its theories to find V2b enclosed, adiabatic column non-isothermal and 4) WE just need to pay attention to the constraints defined.

    I (along with Emden and Exner) keep finding it is 4) WE that need to pay close attention to constraints used in theory or experiment. Control volumes really help eliminate confusion.

    There must be a CFD program that could be set up with exact B74 situation and run to find the analytical answer. Anyone have access to one?

  37. Tim Folkerts says:

    We seem to be running around in circles. Lets go back to the very basics.

    Convection is a type of bulk motion (ie not simply the random motion of molecules due to thermal energy).
    Perpetual convection is a type of perpetual motion.

    Therefore, if you say there is perpetual convection, you are saying there is perpetual motion.

    If you say there is perpetual motion, then either
    1) there must be a perpetual source of energy/heat/work or
    2) you have a perpetual motion machine requiring no energy input.
    (Tor example, the oceans are a “perpetual motion machine” because they are continuously getting energy from the sun and losing energy to space. If the sun stopped shining, the ocean convection would slowly and irreversibly diminish.)

    I could imagine that Graeff has discovered that somehow gravity can induce a temperature gradient in STATIC AIR (although I still disagree with that). I could not image that Graeff has discovered that air continuously convects as a perpetual motion machine to create a temperature gradient WITHOUT any energy input.

  38. Tim Folkerts says:

    I don’t understand the terminology you use, Trick: “There is no forced convection in Graeff’s experiment, only free convection”

    To me “forced convection” would be like a convection oven, where a fan blows the air around.
    “Free convection” would be thermals, driven by temperature gradients.

    There is no convection in a pan of water sitting on the counter-top. You could stir the water to get it moving, but eventually the water will come to rest. convection only starts when you HEAT the pan on the stove. There is also no convection in an isothermal column of air. Only when you exceed the adiabatic lapse rate will convection spontaneously develop.

    What sort of “free convection” do you imagine that exists in Graeff’s insulated columns of air ?

  39. Trick says:

    Lucy 4:56pm – The social critic Bertrand Russell quotes are great. My favorite at the top of this page:

    http://www.brainyquote.com/quotes/quotes/b/bertrandru101364.html

    and one from a guy wiki lists as his protégé:

    “Whereof that which one cannot speak, thereof one should remain silent.” – Ludwig Wittgenstein

    But of course in Godot Act 2 a good blog philo. really comes thru:

    “…let us try and converse calmly, since we are incapable of keeping silent.”

  40. Tim F: Since you have readily admitted that you are not an expert in thermodynamics, you must (if you are honest to yourself) acknowledge that your opinion of what “seems to fit” may not the most reliable indicator of the correctness of your conclusions.

    This is a tricky one to handle.

    Sometimes an outsider can see flaws more easily. Graeff describes in his book, how he refused to look up “expert” device patents because he found it easier to think things through for himself, first, and check later. I follow a similar pattern.

    Indicators that help me gauge the correctness of my conclusions here: Study, intensive study. I started with the reference to Graeff in Tallbloke’s article on Loschmidt – because this article looked as if it might hold material clarifying Nikolov and Zeller’s hypothesis where it “seems” to Anthony Watts to disregard the Second Law. Graeff’s paper on the water test passed all my preliminary “smell tests” developed from the time I gained high marks at high school in maths and physics. I was always strong on deductions and by comparison, weak on memorizing. This is a plus when it comes to proving new hypotheses, and to approaching a subject I know little of, and focussing on obtaining key relevant information.

    It was clear that Graeff;s presentation lacked some of the characteristic presentation details one might expect from peer-reviewed papers; but the substance was good. Abstract; history; experiment; data; suggested conclusions; suggestions for reframing 2LoT. All totally logical. This gave me the impetus to check Graeff’s website. Appallingly designed and difficult to track key info but his interest in alternative energy piqued my interest. Found links to his book and bought it through Amazon. Wrote to Prof Sheehan in the US who has run two conferences on current challenges to 2LoT, and who is very supportive of Graeff. He contacted Graeff for me, Graeff invited me to his seminar, I jumped at the chance for a THIRD level of checking this work with all my senses and BS meters.

    This all amounts to a high level of relevant expertise – which I have and you do not. Sure, I had difficulty following the Wikipedia proof for the ALR. But this does not invalidate Graeff’s results or my ability to appreciate their statistical significance and theoretical coherence. At this point, the WP proof is of marginal relevance. It does not take away Graeff’s inexplicably high figures. Still, I do now own Bohren and Albrecht, having been encouraged by studying Graeff’s own copy of this book. And I did get to the point, with Nikolov and Zeller, where I could understand the key maths there. Even Monckton the whizz mathematician did not reach that point – much to my surprise and sorrow.

    I might peruse B&A to understand the standard ADLR formula derivation – and to deepen my understanding of how this might relate to Graeff.

    But first, I think, replication!

  41. Trick says:

    Tim F 5:40pm and 5:49pm:

    Tim – I have not gone back and checked, but a debate formed around not allowing convection in V&G section 2b and allowing convection in V&G section 2b. Indeed the authors try to define their own terms as I’ve written about before (turbulent convective mixing…etc.).

    Most writers do not take the time to write they mean forced or free convection, they just write “convection” as above & leave the reader to figure the meaning from context (free or forced). In V2b section, the column is free to convect or not convect, no limitations are in the math. Convection terminology doesn’t matter.

    This terminology stuff is really a side show, the terms are NOT precise but the math and constraints ARE precise. You cannot see B&A math, constraints or cartoon figure yet for the non-isothermal soln.; in the coming days I will have a copy of their text again and try to communicate specifically the exact math and constraint implementation in a calm conversation if you want to participate.

    My view, can learn a lot about the real atm. gradient once it is seen being in between the isothermal and non-isothermal ideal gradients. Discussions of why that is – then would have a solid, precise math and constraint foundation.

    Also, the B74 dewar set up should be possible to better discuss w/folks that “get” the theory foundation. Lucy admits she does not. That is not a problem except in communication, prolific tester Tom Edison didn’t need all the theory either.

  42. Trick says:

    Lucy 6:09pm – “I might peruse B&A to understand the standard ADLR formula derivation…”

    Thank you for the effort. B&A text employs a writing style that allows insight into the theory w/o needing to grok ALL or even a lot of the theory they use. It is a great text, wish I had learned from it originally. Thermodynamics would not have been the pain it was at the time.

  43. Trick says:

    Tim F 5:40pm: “Therefore, if you say there is perpetual convection, you are saying there is perpetual motion.”

    In the ideal gas setup, there IS perpetual motion as neither the isothermal nor non-isothermal ideal gas loses any energy after LTE is achieved. The inelastic molecules bounce around forever within an adiabatic inelastic control volume. This is after all the reason no real adiabatic insulation exists AND the real collisions are a bit elastic.

    NB: this PM is ok w/2nd law as S is allowed to remain constant in ideal reversible process.

  44. Tim Folkerts says:

    Trick says: “In V2b section, the column is free to convect or not convect, no limitations are in the math.”

    No. V2b specifically is about a situation more gas is moving up and down (and not losing/gaining any heat in the process). This IS the math constraint that is imposed. This is why “Potential temperature” is constant, rather then plain old temperature. Conduction is specifically ignored in this section (so that the gas never comes to thermal equilibrium.

    ~~~~~~~~~~~~~~~~~~~~
    “Most writers do not take the time to write they mean forced or free convection”

    However you want to talk about it, there are basically two sources for convective motion. 1) Physical (fans, pumps, etc) and 2) thermal (hot & cold areas causing expansion and contraction).

    No matter how you slice it, Graeff’s experiments carefully try to eliminate both tot these, and there should be no bulk motion within the columns.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~

    [While we are at it, “adiabatic” is used in two different senses in this discussion. It can mean the entire column is insulated from the rest of the universe (but heat can flow within the gas column. It can also mean each subset of the gas is insulated from the rest of the gas (and from the rest of the universe). This is what is implied when talking abut the “DALR” and V2b. It is an ADDITIONAL constraint that is not REQUIRED. But IF this constraint is applied, then the gas will not be able to come to internal equilibrium, since heat cannot be transferred thru the gas by conduction. Then a gradient will naturally develop and “potential temperature” will be constant.]

  45. Tim Folkerts says:

    You just don’t seem to hear what I am saying, Trick. You say ‘The inelastic molecules bounce around forever”, which, of course, is true and obvious. But molecules bouncing around forever is NOT convection. Convection always involves organized bulk motion, not simply random molecular motion. Period. End of story.

  46. Trick says:

    Tim F 6;59pm – “No. V2b specifically is about a situation more gas is moving up and down (and not losing/gaining any heat in the process). This IS the math constraint that is imposed.”

    Which V2b formula specifically writes that out please?

    Tim F 7:11pm“You just don’t seem to hear what I am saying, Trick.”

    Well, I read. I am trying to understand. Once again you write – leaving off the qualifier free or forced which means I must try and understand which you mean from context – when you write: “Convection always involves organized bulk motion, not simply random molecular motion. Period. End of story.”

    From context I must assume you mean V2b. So if you mean forced convection, what force organizes the bulk motion away from being free random packets of molecules? There is no force other than gravity, PV=nRT, 1st and 2nd laws all of which are invoked in V2b and show no molecules move into “organized bulk motion”. There is random mixing that is termed “turbulent convective mixing” which implies free convection with no bulk organization (it is turbulent but again I have to assume that from context).

    In the real atm. there IS forced convection with sun heating say from the surface which DOES organize some molecular packets into bulk motion. No sun to do that in Graeff or V2b.

  47. Q. Daniels says:

    Tim Folkerts wrote:
    Convection is a type of bulk motion (ie not simply the random motion of molecules due to thermal energy).
    Perpetual convection is a type of perpetual motion.

    Therefore, if you say there is perpetual convection, you are saying there is perpetual motion.

    The term “Perpetual motion” is itself a problem. It has a several meanings, including three technical ones based on which Law of Thermodynamics is being violated.

    I think you’ve used the term to mean “something which keeps moving forever”.

    If you say there is perpetual motion, then either
    1) there must be a perpetual source of energy/heat/work or
    2) you have a perpetual motion machine requiring no energy input.

    Item 2 can be fairly easily demonstrated, so long as there is no energy output. Orbital Mechanics and Superconductors both seem to fall into this category.

    I don’t see any energy output from the convection itself. (Energy input) + (energy output) = 0. I don’t see a problem in terms of Conservation of Energy.

    Only the Second Law poses a potential problem.

    There’s an assumption that thermodynamic systems have a single, stable attractor.

    To my mind, the question is whether or not that is a valid assumption.

  48. Tim Folkerts says:

    Q. Daniels says: “Orbital Mechanics and Superconductors both seem to fall into this category [of perpetual motion machine].”

    That is a good point. An individual object (like a planet) that is not subject to any sort of friction can move for ever. There are even some collections of particles that defy classical thermodynamics (like superfluids and superconductivity) and can move in a collective state indefinitely without energy input.

    But no one, as far as i can tell, has ever suggested that common room temperature acts is such a way. If you stir some water, it will start spinning. As time passes, that organized motion will slowly dissipate until you are left with only molecular-level random motions. The water will never start spinning again by itself. If you stir the air in a column (eg by flipping over Graeff’s tube), the air will gain some organized, bulk motion. As time passes, that organized motion will slowly dissipate until you are left with only molecular-level random motions. The air will never start spinning (or convecting or sloshing back and forth) again by itself.

    Or to put it another way, to give the air some bulk motion, you have to give that air extra kinetic energy. That extra energy must come from somewhere. In Graeff’s experiment, that would have to come from some the gas spontaneously cooling to perform work other parts of the gas. But this, again, is a standard violation of the 2nd Law and would be rewriting standard physics if it could happen.

  49. Tim Folkerts says:

    Trick says: “Once again you write – leaving off the qualifier free or forced which means I must try and understand which you mean from context – when you write: “Convection always involves organized bulk motion, not simply random molecular motion. Period. End of story.”

    And once again it doesn’t matter if it is “free” convection or “forced” convection (whatever you mean by those terms). Convection always involves organized macroscopic motion, not simply random molecular microscopic motion. Period. End of story.

    “There is random mixing that is termed “turbulent convective mixing” which implies free convection with no bulk organization”

    Define “free convection” which does not involve macroscopic, organized motion.

  50. Trick says:

    Tim F 1:08am – Free convection has no external heat source causing heat transport like in forced convection where the sun is warming the ground, a fire is heating the smoke, a range is boiling the water.

    Get free convection from the internal energy within an enclosed system adiabatic control volume and any transient difference in local density due to random collisions creating one at constant S. There is no need to suppress free convection in Graeff’s B74 dewar that I can find, so far at least.

    Pretty sure B&A text will have more on the subject once I get it in a few days; it is still in transit.

  51. Q. Daniels says:

    Tim Folkerts wrote:
    Or to put it another way, to give the air some bulk motion, you have to give that air extra kinetic energy. That extra energy must come from somewhere. In Graeff’s experiment, that would have to come from some the gas spontaneously cooling to perform work other parts of the gas. But this, again, is a standard violation of the 2nd Law and would be rewriting standard physics if it could happen.

    I believe that is what is being suggested here.

    When the work decays by friction, it is again returned to heat. Energy is conserved. It is only the Second Law that is being challenged.

    It doesn’t take any great leap to say that there are fluctuations (thermal noise) within a volume of gas.

    That brings up the interesting question of whether or not the thermal noise dissipates with exactly the same probability as its creation, or if the pressure gradient has an effect. I haven’t explored this angle of math yet, so I don’t know.

    Graeff has created an environment where the modes of thermal noise are pretty restricted, but not eliminated.

    The energy involved is very small (~kbT), but the number of beads involved is huge and the time are short.

  52. ferd berple says:

    ferdberple says:
    July 7, 2012 at 5:36 pm
    Wikipedia shows 30km as the height of the atmosphere.
    from above: 290K / 9.8K/km = 30km (surface temp / gravity = atmospheric height)
    This has to be more than simply a coincidence.
    ==========
    Repeating this exercise with Venus (numbers from Wikipedia)

    735K / 8.9m/s/s = 82.5 km predicted height of atmosphere of Venus from KE due to gravity.

    Wikipedia shows atmospheric height of venus is about 80-90km. This shows that surface temperature is independent of the ghg composition of the atmosphere, because the relationship holds on both earth and venus, which have very different ghg compositions.

    Surface temperature is determined by gravity and atmospheric height. The effective kinetic energy of atmospheric molecules in free-fall from the top of atmosphere equals the surface temperature. The more solar energy, the more the atmosphere expands, the greater the KE of molecules at the surface, the greater the surface temperature.

    http://en.wikipedia.org/wiki/File:Venusatmosphere.svg

  53. ferd berple says:

    Tenuc says:
    July 7, 2012 at 9:06 pm
    The type of life we observed here on Earth has, over a long period of time, evolved/adapted to the conditions that exist here. If our world had a different make-up, then some other form of life could maybe be in residence here???
    ======
    One could equally argue that life has shaped conditions on the planet and continues to do so today. That life and conditions co-evolved.

    The problem arises when the changes made by humans are seen as unnatural, while the changes made by other life forms are seen as natural. This is an illogical belief because humans are a life-form and part of nature. It is this illogical belief that leads to illogical conclusions.

  54. ferd berple says:

    Tim Folkerts says:
    July 9, 2012 at 2:49 pm
    Once again, it is POSSIBLE that perpetual motion machines exist and Graeff as discovered one, but I am not holding my breath.
    =========
    For this to be a perpetual motion machine, there must be no external source of energy. However, in this case gravity is the external energy source.

    If you were to repeat the Graeff experiment in orbit around the earth it would show no temperature gradient, because the net acceleration due to gravity would be 0.

    The Graeff experiment shows a temperature gradient only when the frame of reference is accelerating, such as when standing on the surface of a planet. It is this 9.8 m/s/s force that provides the energy to create the gradient.

    What has misled so many is that we see ourselves as standing still while resting on the surface of the earth. However, due to gravity we are experience a continual force of 9.8 m/s/s, exactly as though we were in a rocket in open space with the engines on. If we were to release an object inside the rocket, it would “drop” to the “floor” opposite to the direction of acceleration.

    It is this acceleration that provides the energy source. For example, if one was to build a stirling engine with the hot side at sea level and the cold side on the top of a mountain, it would also appear to be a perpetual motion machine. However, it would simply be harnessing the difference in KE in air molecules at different elevations due to gravity.

  55. Tim Folkerts says:

    Q. Daniels says: “It doesn’t take any great leap to say that there are fluctuations (thermal noise) within a volume of gas.”

    I agree that there is always thermal noise. It certainly is present (and detectable) in electronics (http://en.wikipedia.org/wiki/Johnson%E2%80%93Nyquist_noise). That is actually the ONE line of thought presented so far that I think is worthy of pursuing (although I suspect it, too, will lead to a dead end).

    Microscopic fluctuations will cause some microscopic areas to be warmer or cooler, to be more compressed or less compressed. But the odds of even a small volume (say 1 mm^3) being significantly away from equilibrium (say 1 K warmer than average) for a short duration(say 1 second) will, I am quite sure be astronomically small. I don’t see this driving the “convection” needed to create an “adiabatic lapse rate”.

  56. ferd berple says:

    Trick says:
    July 9, 2012 at 6:09 pm
    prolific tester Tom Edison didn’t need all the theory either.
    =======
    Theory should only serve as a guide to where the right answer might be. However, many see theory as saying where the right answer cannot be.

    This is of course a wrong application of theory, because it assume that theory is 100% correct and without error. When in fact theory is only accurate within bounds. Those bounds, where theory is known to be accurate, are set by testing.

    If you haven’t tested a theory outside of the bounds, then it may still be correct, or it may not be. Graeff’s experiment appears to be testing theory outside the bounds of previous testing. People are arguing that Graeff’s results cannot be correct, based on theory.

    However, if the theory has never been tested outside the bounds, then the theory argument is at best weak. If Graeff is replicated, it will be the theory that will be modified, or at least the interpretation of the theory. If Graeff is replicated then most likely the interpretation of the 2nd Law will be seen to be incorrect, not the 2nd Law itself.

  57. ferd berple says:

    Tim Folkerts says:
    July 10, 2012 at 4:07 pm
    Q. Daniels says: “It doesn’t take any great leap to say that there are fluctuations (thermal noise) within a volume of gas.”
    =======
    The M-B distribution establishes the probability.

    It can be replicated by elastic collision (conduction). When two pool balls collide for example, depending on the relative direction of travel and angle of incidence, there is a probability function associated with the resulting momentum and KE for each of the balls.

    This distribution does not lead to all balls having the same energy, because it is possible for a slow moving ball to increase the speed of a fast moving ball, further reducing the speed of the slow moving ball. Otherwise, if a slow moving ball could not increase the speed of a fast moving ball, then over time all balls would have the same KE.

    It is this probability function that allows fluctuations. However, due to the large number of balls in even a small sample, these fluctuations have low probability of being large over large volumes. Another interpretation of the gradient could be to see gravity as a sorting mechanism for the fluctuations, though it seems more natural to view the trajectories of the pool balls as curved under the influence of gravity, which changes the M-B distribution along the direction of acceleration.

  58. ferd berple says:

    Thinking more on this problem, it would appear a possible solution to the Graeff experiment might be to calculate the M-B distribution in an accelerating frame of reference. It would be interesting to see if this has been done. If it shows a KE shift along the axis of acceleration, this would be consistent with the development of a gradient.

  59. br1 says:

    Trick:
    ” You apparently are having trouble seeing from the intro that constraint 2 (constant energy internal plus potential) HAS to be applied to energy accounting across the control volume in V2a AND V2b AND V2c due to 1st law.”

    It’s not that I have trouble with it, but rather disagree with it entirely because it’s wrong.

    Verkley agrees:
    “As we will show below, the same requirement is found if one relaxes 2 by allowing neighboring layers to do work on the layer under consideration; constraint 2 is then to be replaced by 2′, a constant enthalpy.”

    note the words “replaced by”.

    Then he says:
    “Bohren and Albrecht arrive at their constraint 3 by starting with a constraint similar to 2′, and then
    modify it in an approximate way, which in fact amounts to replacing 2′ by 3.”

    note the words “replacing 2′ by 3”. These are clearly not “and” statements, they are “replaced by” statements. Verkley says this even more explicitly with:

    “It is the *purpose of this article* to suggest a way of incorporating 3 in the maximization problem
    without sacrificing the constraint 2′, which after all stems from the first law of thermodynamics. We will, in other words, pose 3 as an additional constraint to 1 and 2′.”

    so having an “and” constraint is entirely novel in Verkley, and was not done in B&A who sacrificed the enthalpy constraint 2′. Showing how to do this is the purpose of Verkley 2c, and is not present in 2b or B&A.

    “specifically in their math you can see they enforce constant energy (gas enthalpy) accounting across the control volumes of sections 2a, 2b and 2c.”

    specifically not. Note Verkley Eqn(15), you can mathematically see that enthalpy is not conserved. Conservation of potential temperature does not mathematically imply conservation of enthalpy. This is why enthalpy has to be explicitly written into Eqn(19).

    “You cannot see B&A math, constraints or cartoon figure yet for the non-isothermal soln.; in the coming days I will have a copy of their text again and try to communicate specifically the exact math and constraint implementation in a calm conversation”

    It won’t do you any good – this is exactly the part that B&A screwed up. Correctly quoting something that is wrong doesn’t make the quote correct, see for example:

    https://tallbloke.wordpress.com/2012/05/29/lucy-skywalker-graeffs-second-law-seminar/#comment-26892

    As I don’t wish to clutter this thread, I might have enough energy to continue on the other thread for a while which is already so cluttered it can’t be harmed any further.

  60. br1 says:

    ferd berple:

    “Thinking more on this problem, it would appear a possible solution to the Graeff experiment might be to calculate the M-B distribution in an accelerating frame of reference. It would be interesting to see if this has been done.”

    Sure has, many times, I even did my own version. Check out:

    http://www.slideshare.net/brslides/analytic-velocity-distribution-under-gravity

    What do you reckon?

  61. Trick says:

    br1 4:42pm – “…this is exactly the part that B&A screwed up.”

    You haven’t even read B&A yet! Get a copy, you will trace thru & see they have not made a mistake. Give B&A a fair chance. No mistake in Verkley either, you are being run in circles with the sideshow terms & not examining the detail math steps just the summary in V&G. Akmaev confirms B&A and V&G, completely.

    Here is where some of your confusion seems to be: Constraint 2 is constant energy = constant enthalpy. Constraint 2’ is also constant energy = constant enthalpy. General by definition in the control volume:

    Gas Enthalpy = energy = 1/2mv^2 + mgh + work done on environment (i.e. pV) must be constant by Law 1.

    When V&G use the prime on constraint 2 (i.e. 2’ in V2b) they are just distinguishing the third term (pV) which has a non-zero value in V2a because work (pV) is allowed on the environment and in V2b no work is allowed on environment so third term is 0.

    Constraint 2 in V2a: control volume (cv) energy = ideal gas enthalpy = 1/2mv^2 + mgh + pV

    Constraint 2’ in V2b: control volume energy = ideal gas enthalpy = 1/2mv^2 + mgh + 0

    That last 0 in 2’ changes V2a to V2b allowing no work on environment, conserving enthalpy H and arriving at a V2b non-isothermal gradient larger but “remarkably” close to real atm. V2b same enclosed constraint as Graeff dewar B74.

    So yes, replace constraint 2 with 2’ in V2b b/c the cv is changed to enclosed so no work on environment (now pV=0) & energy = enthalpy is still conserved constant in V2b, no external heating allowed, cv is adiabatic. Constraint 2 and 2’ conserve energy, constant H must happen in V2a and V2b. By Law 1.

    All this is shown in detail in B&A – go thru it! It helps, I was confused with V&G too until I read the B&A derivation details cited by V&G.

    br1 continues for V2b: “Note Verkley Eqn(15), you can mathematically see that enthalpy is not conserved.”

    Geez. See your confusion? I can see V&G Eqn. 15 DOES conserve enthalpy in 2b as is derived from the 1st law and the 1st law requires conserve energy = conserve ideal gas enthalpy constant as I just showed (again!). Re-read this slow.

  62. Q. Daniels says:

    Tim Folkerts wrote:
    Microscopic fluctuations will cause some microscopic areas to be warmer or cooler, to be more compressed or less compressed. But the odds of even a small volume (say 1 mm^3) being significantly away from equilibrium (say 1 K warmer than average) for a short duration(say 1 second) will, I am quite sure be astronomically small. I don’t see this driving the “convection” needed to create an “adiabatic lapse rate”.

    That’s right. 1K is a huge amount, and 1 mm^3 is a much larger volume than Graeff is apparently using.

    I don’t have the technical data on the beads Graeff used. Various manufacturers offer hollow glass beads in the 1-30 um range.

    I get a 1 um^3 volume as having ~2.7×10^7 molecules, with an average z fluctuation of (1/2)*kbT, or 5×10^-8 of the volume’s thermal energy.

    We’re looking to explain a gradient of 7×10^-8 K/um, or ~2×10^-10 of the thermal energy per micron.

    Don’t trust my math on this.

    Also of note, 1 um is ~14x the mean free path, so at smaller volumes, an increasing fraction of molecules would cross the volume without colliding. For this reason, 1 um^3 is probably near the peak of this effect (if it’s real).

  63. br1 says:

    Trick:
    “See your confusion?”

    No, I see yours.

    That will take another post, (probably a whole series), so I’ll continue at

    https://tallbloke.wordpress.com/2012/05/29/lucy-skywalker-graeffs-second-law-seminar/#comment-28723

  64. ferd berple says:

    br1 says:
    July 10, 2012 at 5:20 pm
    http://www.slideshare.net/brslides/analytic-velocity-distribution-under-gravity
    What do you reckon?
    ====
    do you have a graphic for this showing the results? I did a quick and nasty simulation in excel but didn’t have time for a full simulation.

  65. ferd berple says:

    question: does the KE of the particles remain constant with altitude?

  66. pochas says:

    I’d like to know how much power you can extract by connecting a thermopile between the top and bottom of a Graeff tube, if any. With that information we can think about how to rewrite the laws of thermodynamics.

  67. br1 says:

    ferd berple:
    “do you have a graphic for this showing the results?”

    Have a look at the related presentation, which accompanies the analytic solution:
    http://www.slideshare.net/brslides/maxwellboltzmann-particle-throw

    in particular slide 8.

    “question: does the KE of the particles remain constant with altitude?”

    The KE of any given particle slows down with altitude, as shown on slide 1 of the throw presentation. However, the *average* KE of all the molecules that reach any altitude remains constant, as shown on slide 8.

    The reason for this is that in order for a set of particles to get to a high altitude they must start off hotter than average at low altitude. Then when they rise they cool. It just so happens to turn out that the temperature they cool to is the same as the temperature of the layer they started from! So the temperature of the particles which reach any height remains constant (according to this simple theory). The ‘cold’ particles at low altitude can’t reach high altitude, so they don’t contribute to the average KE at high altitude.

    Temperature is all about distributions – hence the maths in the analytic presentation is all about finding the distribution with height.

    Of course, real life is more complicated than that, which is one reason I remain fascinated by Graeff’s work.

  68. Trick says:

    ferd asks: “question: does the KE of the particles remain constant with altitude?”

    br1 – “The KE of any given particle slows down with altitude… ..the *average* KE of all the molecules that reach any altitude remains constant..”

    Take a careful look & there is usually 1 particle that reaches the highest altitude so if “any given particle slows down with altitude” how can it be possible “..the *average* KE of all the molecules that reach any altitude remains constant..” at the highest altitude where there is only 1 particle to avg?

    If that 1 particle has slowed under gravity as you say then avg. of 1 particle KE at that highest altitude must be lower but you say KE remains constant at any altitude. How can that be?

  69. br1 says:

    Trick:
    “If that 1 particle has slowed under gravity as you say then avg. of 1 particle KE at that highest altitude must be lower but you say KE remains constant at any altitude. How can that be?”

    If one takes an instantaneous snapshot of a finite number of molecules, then there will be fluctuations according to particle number. But if one takes another snapshot at a slightly later time, then the fluctuations will be different. Taking the average of all the fluctuations over an arbitrarily long period of time gives the expected value, which is isothermal.

    In the analytic maths of the probability distribution function, the value of the PDF goes to arbitrarily small values. However, the PDF is a smooth function, so it is already using the averaged out values. On the other hand, the throw simulation selects individual molecules and has temperature fluctuations depending on how many particles are counted – you can see this by comparing slides 4/5 with slides 8/9.

    Grok?

  70. Trick says:

    br1 2:24pm – There are no slides 4/5 or 8/9 I see in the 4:37am post cylinder which you say “the KE of any given particle slows down with altitude…but avg. KE remains constant at any altitude”. Do not grok how that can be. Show the theory formula please.

  71. ferd berple says:

    Trick says:
    July 11, 2012 at 1:16 pm

    ferd asks: “question: does the KE of the particles remain constant with altitude?”

    br1 – “The KE of any given particle slows down with altitude… ..the *average* KE of all the molecules that reach any altitude remains constant..”
    ============

    That is interesting because it wasn’t what I found when I did a crude ideal gas simulation without collisions, where all sides of the container, including the bottom were thermally neutral, and the top was sufficiently high as to be infinite.

    What I found was that when I started the simulation with all molecules evenly distributed, the average KE did not remain constant. Instead the averaged peaked some distance above the bottom, then fell off with altitude.

    The problem I have with your argument is this:

    “The KE of any given particle slows down with altitude”
    “the *average* KE of all the molecules that reach any altitude remains constant”

    This to me doesn’t make sense. For example, consider the highest layer, the top layer. The molecules at this layer are moving slowly, as they are about to fall back to earth. Lets call these the terminal molecules. They all have low KE.

    Now consider the layer just below this, top-1. It also contains the terminal molecules moving slowly just about to fall back to earth. These have the same KE as the terminal molecules in the layer above.

    However, this top-1 layer also contains the molecules moving to and from the top layer above. Lets call these the moving molecules. These moving molecules all have higher KE than the terminal molecules.

    However, the top layer only has terminal molecules, while the top-1 layer has both terminal and moving molecules. Thus the molecules in the top-1 layer must have a higher average KE than the molecules in the top layer.

    This argument can then be repeated for each of the layers, giving an energy distribution like this:

    T = terminal
    M = moving
    KE(M) > KE(T)

    top = KE(T) / T
    top -1 = (KE(T) + KE(M) ) / (T + M)
    top -2 = (KE(T) + KE(M) + KE(M2)) / (T + M + M2)

    where KE(M2) > KE(M) > KE(T)
    etc.

    This is the result I saw in the simulation and it argues strongly that the gas is not isothermal in the presence of gravity (acceleration)

  72. br1 says:

    Trick:
    “There are no slides 4/5 or 8/9 I see in the 4:37am post cylinder which you say “the KE of any given particle slows down with altitude…but avg. KE remains constant at any altitude”. Do not grok how that can be. Show the theory formula please.”

    The 4:37am post was something ferd berple posted, that is nothing to do with me personally.

    I was referring to my own work – slides 4/5 (few samples) and 8/9 (many samples) refer to this presentation:
    http://www.slideshare.net/brslides/maxwellboltzmann-particle-throw

    Theory formula is here:
    http://www.slideshare.net/brslides/analytic-velocity-distribution-under-gravity

  73. br1 says:

    ferd berple:
    “For example, consider the highest layer, the top layer.”

    if you have an infinitely tall column (as you were implying) then there is no top layer. The velocity probability distribution continues with finite value to infinitely high velocity. No matter how high you look, sooner or later you will find a molecule passing by. There is no such thing as a layer of terminal molecules – a faster one will come along soon enough.

    Saying that, I would still like to know more about your model – I think the problem is that you are using the microcanonical ensemble. If you can see this paper
    http://iopscience.iop.org/0143-0807/17/1/008
    then it is possible to get a temperature gradient for a small sample of molecules that interact with eachother but do not interact thermally with the walls. My simulation had thermal interaction with the ground floor, so the molecule velocity would get randomised each time it hit the floor. Hence my solution represents thermal equilibrium with a heat bath. If you have a reflective floor and walls, then the molecules preserve the initial information you have given them, and the velocity profiles will not represent thermal equilibrium.

  74. Trick says:

    br1 5:30pm – Ok I looked at your theory formula link and see again an assumption T is constant wrt dv in eqn. 4 to perform the integration. This is the same constant T assumption used in barometric formula derivation (your link says it reproduces barometric formula) shown in ferd’s 4:37am posted site.

    This is a close to reality assumption since variation in T wrt dv is small h << kT/mg. This variation in T is so small after all Graeff is having a difficult time measuring it with no powder.

    ferd 4:13pm – In your layer by layer reasoning, hold T assumed constant as in your 4:37am posted video since it varies so little, see what happens.

  75. br1 says:

    Trick:
    “Ok I looked at your theory formula link and see again an assumption T is constant wrt dv in eqn. 4 to perform the integration.”

    that is not an assumption, it comes from what a ‘probability density function’ is, which in this case is given in Eqn(1). If Eqn(1) is correct, then Eqn(4) is correct, exactly, no approximations or assumptions whatsoever. Eqn(1) represents thermal continuity between a heat bath and a gas at the plane interface between the two, the reference paper is given. To say that T has a single value in Eqn(1) is to say that the starting velocity distribution has a single temperature. This is the starting condition we are interested in when discussing what happens to this temperature later as a function of height.

    Furthermore, note Eqns(1) and (4) are at the ground plane – there is absolutely zero height involved at this stage of the derivation. It is not a case that h is ‘small’ giving a good approximation – there is no h whatsoever. There is no h in Eqn(1) or Eqn(4). Even if you try to argue about the width of an atom, this derivation lives in continuous-space theory land, and in that world this derivation is exact and approximation free.

    After looking at ground level, you can make h>>kT/mg to your heart’s content, the derivation of Eqn(11) holds without approximation or assumption and the distribution will be isothermal.

    By the way, this is meant to be a ‘replication and implications’ thread.

  76. Trick says:

    br1 7:49PM – Eqn. 1 is correct and so is eqn. 4 before integration is carried out. The link states velocities are considered normal to the wall. If the velocities were parallel to the wall I would agree, T is constant parallel to the bottom wall and is constant in the WALL but it can’t be in the ideal gas velocities normal to the wall right next to the wall due to PV=nRT.

    Here eqn. 1 is good as the velocity v, n and T vary in the phi1 distribution. Eqn. 4 is good for the same reason. Eqn. 4 cannot be integrated unless T is held constant (or a f(v,T) substituted) in dv over va to vb normal to wall velocities because the v between va and vb vary – the whole point of the article – & therefore gas T must vary va to vb in the gas dv with PV=nRT unless T is just assumed constant wrt dv to make things easy We all agree T varies with v AFAIK. If va = vb then the integration is ok also, then T really is constant in the normal to wall velocity distrib.

  77. ferd berple says:

    br1 says:
    July 11, 2012 at 5:40 pm
    if you have an infinitely tall column (as you were implying) then there is no top layer. The velocity probability distribution continues with finite value to infinitely high velocity. No matter how high you look, sooner or later you will find a molecule passing by. There is no such thing as a layer of terminal molecules – a faster one will come along soon enough.

    Saying that, I would still like to know more about your model –
    ===
    Pick a layer of arbitrary height. There are on average more molecule reaching terminal velocity in that layer than on average in any layer above this. These all average out to be the same for each KE for each layer.

    However, except at low altitude, there are more molecules passing through this layer on average to higher layers, than there are on average in any higher layer. Thus, by inequalities you can show that on average the current layer has greater energy than any layer above it.

    What is different in your simulation is that you allowed energy to enter the system from the base of the column.

    “My simulation had thermal interaction with the ground floor”

    I didn’t do this and the result was a gradient.

  78. ferd berple says:

    br1 says:
    July 11, 2012 at 5:40 pm
    if you have an infinitely tall column (as you were implying) then there is no top layer.
    ========
    The top layer exists at the limit. The layers are arbitrary in size, and shrink to zero at the limit. By computing the number and velocity of balls moving in each layer, you can see if there is a gradient.

  79. br1 says:

    Trick:
    “T must vary va to vb in the gas dv with PV=nRT unless T is just assumed constant wrt dv to make things easy”

    Can we at least agree that the *floor* has a temperature? Not a temperature as a function of velocity. Not several values of temperature. The *floor* has just one value of temperature?

    When a gas molecule is in contact with the floor, it equilibriates with it and picks up that *one* temperature, as the starting condition is that the gas has the temperature of the floor at the plane of the floor. It gets emitted from the floor with a velocity in the distribution according to that temperature. It can’t have any other temperature than the temperature of the floor, hence no other value of T can be used in the velocity distribution.

    This is not ‘to make things easy’, it is a consequence of saying that the gas is in thermal equilibrium with the floor at the plane of the floor.

    It doesn’t even apply to what velocities the molecules have when they hit the floor, technically it only applies to the starting velocities the molecules have at the instant they are emitted from the floor.

    If the gas is not in thermal equilibrium with the floor, then you have a different system. The solution linked only deals with thermal equilibrium.

  80. br1 says:

    ferd berple:

    I only skimmed your post before, I have read it a couple of times more now. You said:
    “That is interesting because it wasn’t what I found when I did a crude ideal gas simulation without collisions, where all sides of the container, including the bottom were thermally neutral, and the top was sufficiently high as to be infinite.

    What I found was that when I started the simulation with all molecules evenly distributed”

    so let’s see if I understand this. Most importantly for me, you are saying that the particles don’t interact with each other, and they only reflect off the walls. To me this says that the molecules will preserve the information you give them at the start. If the starting conditions don’t represent a thermal equilibrium solution, then the system has no way to attain equilibrium because there are no interactions.

    It sounds to me like you have a lot of non-interacting single particle systems. For a single particle that simply bounces off the ground, it is clear that KE drops with height. Unless you have an ensemble of particles which have a M-B type distribution of velocities, then you will indeed get a variaiton of average KE with height. The M-B distribution is, as far as I know, the *only* distribution which *doesn’t* have a temperature gradient with height. If you make up some arbitrary distribution, then it could have either a positive or a negative gradient with height (yes, positive gradients are possible mathematically), and this can even be ‘lumpy’ so that one can have a ALR type profile at one height, and an inversion type profile at another height, determined *only by the starting velocity distribution*. This sounds like what you have. How did you select the initial velocities?

    If you allowed particle-particle interactions (quite hard to implement properly) then at least the system could come to equilibrium within itself, or if you allowed particle-wall interactions (much easier to implement) then the system could come to equilibrium with an external heat bath. Then you would find that the gradient dissipates to nothing. As it stands, you seem to have no way to reach equilibrium.

    “What is different in your simulation is that you allowed energy to enter the system from the base of the column.”

    enter and exit. This is the mechanism by which my system comes to thermal equilibrium.

  81. Trick says:

    br1 11:06am – “Can we at least agree that the *floor* has a temperature?”

    Yes, this is a given in the link.

    “The *floor* has just one value of temperature?”

    Yes. T1. Have to assume floor molecules are solid state. An unseen heat source with a thermostat holds the floor surface constant at temperature T1 I guess.

    “When a gas molecule is in contact with the floor, it equilibriates with it and picks up that *one* temperature, as the starting condition is that the gas has the temperature of the floor at the plane of the floor.”

    No. The gas particle is not like one billiard ball that could do that. For the gas particle itself, the temperature is undefined – the particle is a system unto itself.

    The ideal gas particle in contact with the thermal wall will feel the temperature of the wall, the pressure above, the temperature above, the number of particles right around it, and the volume above. The particle’s vertical velocity (hence KE) will equilibrate with all those at LTE. PV=nRT.

    An alternative is to assume the wall is adiabatic and the gas itself has initial condition T1 at h=0 across the horizontal surface plane.

    “It can’t have any other temperature than the temperature of the floor, hence no other value of T can be used in the velocity distribution.”

    If no other value of gas T than T1 for these gas particles, then va=vb at T1 for those particles. Just look at that va=vb case – the answer of the integration says n=0 in that case. This is not satisfying, so obviously another unseen term from the integration applies since T varies and that term comes from proper integration with T set equal to some f(v). PV=nRT would be a good starting point. Actually T varies across particles, as each particle is affected from above not just from below in contact w/the floor.

    Or just assume gas T constant at T1 va to vb & proceed to integrate over dv isothermally as done in your link. This is not satisfying either since the particles lose kinetic energy climbing to a distribution at phi2, so they can’t be at the same T (isothermal) in an enclosed container as ferd is reasoning out.

    A satisfying alternative is V&G 2a and 2b which allow for PV=nRT and define constraints fully consistent with law 1 and law 2 and gravity. All this discussion gives me a deeper appreciation for their work & Graeff’s experiment.

    NB: The author of the link does not explicitly say assume isothermal T, one has to look at the formulas involved to trace nature’s pea under the magician’s thimble.

  82. br1 says:

    Trick:
    “The ideal gas particle in contact with the thermal wall will feel the temperature of the wall, the pressure above, the temperature above, the number of particles right around it, and the volume above. The particle’s vertical velocity (hence KE) will equilibrate with all those at LTE. PV=nRT.

    actually, in the simulation and theory collisions are not taken into account, so no your sentence here will not apply. For example, in the particle throw simulation there was only one particle, and this was thrown up and down successively and the average taken later. So there is no case of LTE with surrounding particles. In the analytic solution see ‘notes on assumptions’ at the end which also states no collisions.

    Of course a real system will have particle collisions, but that would be another level of description.

    Hence we conclude that the simulation result is exact, and for non-interacting particles under gravity which are in thermal contact with a floor, the temperature does not vary with height.

    This then also has implications for a ‘one step more real’ situation: as the non-interacting case has no temperature variation with height, then one can see that allowing LTE between adjacent vertical layers will not change the temperature, and the column will remain isothermal.

  83. ferd berple says:

    br1 says:
    July 12, 2012 at 11:31 am
    “What is different in your simulation is that you allowed energy to enter the system from the base of the column.”
    enter and exit. This is the mechanism by which my system comes to thermal equilibrium.
    ========
    This however is not the Graeff experiment, in that the column of air is specifically prevented from interacting with the outside environment, including the bottom. It would seem a simple thing for you to recreate the conditions of the initial problem if you already have a working model. Otherwise there is no way to know if the interaction with the bottom is what creates the iso-thermal result.

    The problem is of course that my model also does not recreate the Graeff experiment. It simply showed that the relationship between velocity and distance was not linear along the axis of acceleration for falling objects. The change in velocity along the acceleration axis was not matched by the change in hang time, which would in the absence of some other disturbance should lead to a gradient.

    That seemed to me to be the key, that hang time in each layer would have to offset the change in velocity due to gravity for the result to be iso-thermal. As particles accelerate/decelerate due to gravity, this changes the time they spend in each layer, and if this didn’t balance the change in velocity, then a gradient must result.

    For example, when an object is dropped from height, its KE increases, but the time it spends in any layer decreases. If these two exactly balance, for any arbitrary sized layer along the path of the falling object, then I would agree that gravity does not create a gradient. If they don’t balance, then this suggests to me that a gradient must result, unless something like collisions provides the restoring force.

    =====
    How did you select the initial velocities?
    ===
    The particles were evenly distributed, then I instantly accelerated the frame to 9.8 m/s/s, and measured the KE based on resulting velocities and hang time in each layer, similar to the falling object described above. The only interaction was with the bottom of the frame, which was thermally neutral.

    The model is crude. I specifically didn’t add collisions because of the difficulty in getting an accurate result. It struck me as similar to the 3 body problem, where small timing errors would creep in and render the simulation inaccurate.

    Is there user friendly simulation software that accurately handles collisions?

  84. Trick says:

    br1 3:52pm – “…we conclude that the simulation result is exact..”

    Simulation is exact for the simulation (the computer is a good calculator) but the simulation does not exactly simulate nature since PV=nRT is not part of the integration of eqn. 4 for the simulation.

    “…as the non-interacting case has no temperature variation with height, then one can see that allowing LTE between adjacent vertical layers will not change the temperature, and the column will remain isothermal.”

    Yes of course for the simulation b/c isothermality is assumed in simulation’s integration of eqn. 4.

    No assumption of isothermal in nature where nature integrates eqn. 4 consistent with ideal gas law & nature is always observed to obey PV=nRT within reasonable earth tropospheric P, n & T ranges.
    Laws of nature: PV=nRT in control volume, total energy conserved in control volume, max. entropy in control volume at LTE all satisfyingly operate in V&G and in Graeff’s B74 dewar (esp. with no powder).

    Think of this:

    You could numerically integrate eqn. 4. Find v, T, m at a point va get na. Take a small step dv to a v point between va and vb. Find n there from v,T, m there. Rinse and repeat with small enough steps up to vb and add all the n. You then have na-b. Note you had to vary T from PV=nRT, constant energy, max. entropy at LTE to obey natural laws.

    Or you could skip variance in T and just use T1 with v varying say by whatever rule you want, say gravity as shown in the link. Now you know na-b according to your gravity rule and get isothermal result (with ANY rule! Not just gravity). T1 could be assumed constant by whatever rule everywhere in va to vb – interesting huh?

  85. Trick says:

    ferd 3:57pm – “Is there user friendly simulation software that accurately handles collisions?”

    Here’s a few. Searching forward along these lines might find a reduced setup, free, easy to use pre-existing software. Seems like Lucy (or Graeff!) would want to do so in order to model up Graeff’s experiment without then knowing the exact theory involved. Just have to be very careful about the software and how it exactly models what you want to simulate. GIGO is best avoided in that way.

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

  86. br1 says:

    ferd berple:
    “This however is not the Graeff experiment, in that the column of air is specifically prevented from interacting with the outside environment, including the bottom.”

    while my simulation is not exactly the Graeff experiment, one should note that the air certainly does come in thermal contact with the bottom of the container, and that while heat can pass through the system (albeit slowly, no insulation is perfect), the main reason for all the insulation and thermal equalisers is to prevent thermal *gradients* in the immediate environment of the container.

    “The particles were evenly distributed”
    I presume then that they started with zero velocity? The simplest modification I would suggest, if you are willing, is to distribute the particles with a height spacing according to the barometric formula. Taking ground level as having height 0, place the initial particles at positions hn=exp(n/h0), where hn is the height position of particle n, n increments by 1 for each particle (n=1,2,3, …) and h0 is a scale height to give reasonable numbers. Note the exponential has a plus sign inside as we are dealing with position not density. Then when averaging over a layer, make sure the layer is thick enough to cover maybe at least 10 particles. I would like to hear what result this gives.

    The next simplest modification I can think of is that instead of the particles merely reflecting off the ground floor, re-initialise their vertical velocity using the equation:

    vz=sqrt((-2*kB*T/m)*log(1-rand));

    where rand is an evenly spaced random number between 0 and 1. This represents thermalisation with the floor.

    Particle-particle collisions are tough. Have a look at Physion (www.physion.net) – I think there may be some way to get actual relevant data from it, as it seems to allow some coding interactions with the main simulator. I have only used this for ‘look-see’ stuff, but even if ultimately it doesn’t give what you want, it is still a great piece of software!

  87. br1 says:

    Trick:

    “since PV=nRT is not part of the integration of eqn. 4 for the simulation”
    you are wrong here in so many ways I don’t even know where to start. Not only is PV=nRT totally consistent with the maths/simulation conditions, but it doesn’t come in to Eqn(4) because Eqn(4) represents thermal contact at a plane of temperature T. Volume is irrelevant – volume of what?

    “Yes of course for the simulation b/c isothermality is assumed in simulation’s integration of eqn. 4.”
    this is no way an ‘of course’ result, and note there is no isothermality assumed with *height*. Do you even realise that Eqn(4) applies at ground level *only*? That temperature in Eqn(4) is ‘constant with velocity’ is only the condition that the ground plane has a temperature. By itself it says nothing about what temperature one may find at different altitudes. The whole point is to find out whether the temperature drops with *altitude*. Eqn(9) is the first time the distribution at an altitude above ground level is written.

    “Find v, T, m at a point va get na.”
    and from previous post
    “If no other value of gas T than T1 for these gas particles, then va=vb at T1 for those particles.”

    oh dear, I don’t think I can take your mangling of both physics and maths any more. I just can’t bear to start an explanation of what a probability density function is. I think I’m done here (again).

  88. br1 says:

    I wrote:
    “place the initial particles at positions hn=exp(n/h0)”

    this is quite a simple simulation to write. I tried my own version and instead of isothermal or decreasing temperature profile, I got an *increasing* temperature profile. So this initial condition also doesn’t satisfy thermal equilibrium (or indeed the barometric formula! Reasons for this later). However, it does say that in such a simulation you can get pretty much any profile you like, depending on starting conditions! Because this type of simulation has no means to reach an equilibrium, one needs a proper justification of what the starting altitudes/velocities actually are, and one starting set is not equivalent to another starting set.

    I would still like to hear what ferd finds.

  89. ferd berple says:

    br1 says:
    July 13, 2012 at 10:29 am
    while my simulation is not exactly the Graeff experiment, one should note that the air certainly does come in thermal contact with the bottom of the container, and that while heat can pass through the system (albeit slowly, no insulation is perfect), the main reason for all the insulation and thermal equalisers is to prevent thermal *gradients* in the immediate environment of the container.
    ============

    I’m sorry but the difference between what you modeled and the Graeff experiment are sufficient to rule your results “not significant” with respect to any conclusions about Graeff’s experiment.

    Why not adjust your model so that it matches the experiment?

  90. br1 says:

    ferd berple:
    “Why not adjust your model so that it matches the experiment?”

    I have tried this in very many ways, and failed every time. I would love to be able to find a gradient.

    What do you recommend?

  91. Tim Folkerts says:

    You all might look at http://www.falstad.com/gas/ [ demands Java ]

    It seems to be a pretty good simulation. Plus the java code is there. So you could, presumably, modify it to give the KE in the top and bottom halves separately (I’m not a Java programmer).

    Just playing around with temperature, gravity, number of particles, you can “see” that the colors (= KE) have similar distributions at the top and bottom. With gravity turned up, this corresponds to a considerable altitude variation between top and bottom.

    PS “Kinetic theory” is a good term to search for, rather than “ideal gas”

  92. ferd berple says:

    br1 says:
    July 13, 2012 at 1:00 pm
    I would still like to hear what ferd finds.
    ====

    The starting point for my model was equivalent to a container of gas at absolute zero in weightless (free fall) conditions. This seemed to me as valid a starting point as any. The container did react with the gas.

    I then applied a constant acceleration of 9.8 m/s/s to the container. The gas molecules we sufficiently sparse that no two molecules overlapped each other spatially along the axis of acceleration. Thus, as they fell to the bottom of the container and rebounded there were no collisions. The bottom of the container did not randomize the rebounds in any fashion. The particles fell and rebounded vertically. Thus, they simply regained their original position within the container, then fell to the bottom once more.

    In this respect, the model was very similar to what one would obtain by dropping a perfectly elastic ball from altitude onto a perfect surface, excluding all air resistance.

    The model for this experiment was inspired by the vector analysis of falling objects, such as horizontally fired cannon balls and cannon balls dropped from the mast of a moving ship. That their acceleration due to gravity is independent of their original motion.

    Thus it seemed to me that the complex motion of molecules and the collisions can simply be ignored, and the KE of the molecules under the influence of gravity examined in much simpler fashion.

    I now believe that a single molecule is all that is required to study the question, which can be approximated by dropping a perfectly elastic ball from altitude onto a perfect surface, excluding all air resistance.

    If this model shows that the average KE of the ball per layer of altitude remains constant during free fall, then there is no way a gradient can develop in the container. However, if the change in KE is not balanced by the change in “hang time” through each layer, then there is the possibility of a gradient.

    This is a relatively simple experiment to perform on paper. Drop a perfectly elastic ball from 100 m altitude and calculate the average KE within each meter to the ground. If it remains constant then no gradient can form.

    However, my excel model showed (which may well be in error) that the average KE of each meter was not constant, that it increased towards the ground.

    This is not as obvious as it might first appear, because while the KE of the ball is always higher closer to the ground, it is also moving faster vertically, so the time it spends in each meter is reduced, which reduces the average energy in that meter. The purpose of my model was to explore if these two factors cancelled each other out, which if they did would in my mind be sufficient to conclude that no gradient was possible. However, what I found was that they did not cancel each other out.

  93. ferd berple says:

    Tim Folkerts says:
    July 13, 2012 at 2:26 pm
    You all might look at http://www.falstad.com/gas/
    ====
    Thanks, looks like quite a good simulation

  94. ferd berple says:

    Tim Folkerts says:
    July 13, 2012 at 2:26 pm
    You all might look at http://www.falstad.com/gas/
    ====
    If you turn the gravity all the way up and the molecule count all the way down, you have the single molecule experiment I described above. From there one can calculate the average KE at each level, to see if it remains constant.

    In other words, is a single molecule iso-thermal? Does the KE of each layer of altitude remain constant when there is only one molecule? My excel model showed a single molecule was not iso-thermal.

  95. Trick says:

    br1 11:02am – “Volume is irrelevant – volume of what?”

    Graeff’s enclosed container.

    br1 – “By itself it says nothing about what temperature one may find at different altitudes.”

    Yes, of course, since there is no way to say after eqn. 4 the simulation or link says anything about gas temperature at different altitudes – that info. is lost here in that temperature becomes always and everywhere T1 the value of the base wall in the simulation so find isothermal with height by definition after eqn. 4.

    br1 – “The whole point is to find out whether the temperature drops with *altitude*.”

    YES! However the link br1 supplies shows gas T irrelevant to altitude after eqn. 4 – the whole point of this thread.

    br1 – “Eqn(9) is the first time the distribution at an altitude above ground level is written.”

    And eqn. 9 has only T1 constant from the assumption of isothermality in previous eqn. 4 integration.

    br1 – “I don’t think I can take your mangling of both physics and maths any more.”

    Where? Be formula specific. I’m interested to look into any thread relevant mangling by any poster.

    I agree with ferd 2:04pm, br1 simulation should be adjusted to match Graeff’s experiment in order to have T relevance to the topic of this thread.

  96. ferd berple says:

    Tim Folkerts says:
    July 13, 2012 at 2:26 pm
    You all might look at http://www.falstad.com/gas/
    ====
    Off to work but will look at this again. I’ve been able to get it running locally and the code looks straight forward. With luck I may be able to adapt the energy distribution display to show average KE over time, as a function of altitude, which would in my mind settle this question.

  97. ferd berple says:

    ferd berple says:
    July 13, 2012 at 2:32 pm
    The starting point for my model was equivalent to a container of gas at absolute zero in weightless (free fall) conditions. This seemed to me as valid a starting point as any. The container did react with the gas.
    ===
    typo.

    The container did NOT react with the gas. Except for perfectly elastic collisions at the bottom.

  98. ferd berple says:

    Tim Folkerts says:
    July 13, 2012 at 2:26 pm
    You all might look at http://www.falstad.com/gas/
    Trick says:
    July 13, 2012 at 3:01 pm
    I agree with ferd 2:04pm, br1 simulation should be adjusted to match Graeff’s experiment in order to have T relevance to the topic of this thread.
    ====

    If you enable the “heater” check box on the simulation, this look to me like the br1 model. If you turn the heater low, almost all the way to the left, it seems to work as a neutral randomizer, which I expect is similar to br1’s model. All the way to the left and the heater appears to be a cooler.

  99. br1 says:

    ferd berple:
    “However, my excel model showed (which may well be in error) that the average KE of each meter was not constant, that it increased towards the ground.”

    yes, this is the standard result in such a situation, see http://iopscience.iop.org/0143-0807/17/1/008
    eqn(8) for example.

    However, there is a huge difference between one bouncing ball and an ensemble of bouncing balls, as the reference also points out. The distribution of velocities in the ensemble is absolutely vital. For example, if you re-run your simulation, but instead of having the particles initially equally spaced, have them spaced according to hn=exp(n/h0) (as suggested above), then I guarantee (almost! I’m not quite sure what is in your simulation) that you will get an ‘equilibrium’ temperature profile that increases with altitude. I would be very interested to hear what you actually get, it should be easy for you to try.

  100. Tim Folkerts says:

    Ferd asks: “In other words, is a single molecule iso-thermal? Does the KE of each layer of altitude remain constant when there is only one molecule? My excel model showed a single molecule was not iso-thermal.”

    Well, this depends on how you define your terminology.

    1) There is really no meaning to the “temperature” of a single particle at a single time. You could talk about the temperature of a collection of particles at a single time, or the temperature of a single particle averaged over time, or equivalently, talk about the distribution of KE over time (which is what I assume you mean)

    2) If the particle CANNOT exchange energy with the surroundings at all, then no, the distribution will not be the same at all altitudes. The particle will always have the same total energy, and will definitely have lower average KE the higher you go. (This is the “microcanonical ensemble” where the total energy is constant).

    2b) In in the limit of large numbers of particles, the distribution will approach a constant temperature with altitude. Since there are a HUGE number of particles even in Graeff’s column, this isothermal result would apply here.

    3) If the particle CAN exchange exchange energy with the surroundings, then the distribution SHOULD be isothermal with altitude. (This is the “canonical ensemble” where the total energy can vary around some average energy.) This would be like turning on the “heater” in the simulation (which should more properly be thought of as a constant temperature thermal reservoir.) Also, since no insulation is perfect, this would also apply to Graeff’s experiments.

    Those are the conclusions of the Velasco paper (https://tallbloke.files.wordpress.com/2012/01/s-velasco.pdf). If you want to challenge any of these, then you need to be able to follow Velasco’s math and explain what he got wrong (or how I misinterpreted his results).

  101. br1 says:

    Tim Folkerts:
    “Those are the conclusions of the Velasco paper (https://tallbloke.files.wordpress.com/2012/01/s-velasco.pdf).”

    ok, great, the paper has been uploaded for general access. This is also the paper I was referring to.

  102. Q. Daniels says:

    Thermal noise:

    What follows is a mental model, not a rigorous analysis. It needs to be fleshed out.

    First, consider the thermal noise of a large volume of gas without gravity.

    In a large volume, there will be local fluctuations of temperature, pressure and velocity. I’ll focus on local fluctuations of bulk velocity, because that’s easiest for me to visualize.

    Consider a series of parcels arranged in a vague loop, moving along the loop. Eventually it will be damped out, but it is probable that such loops will emerge spontaneously. The reason this happens is that the system is insufficiently damped to prevent noise from scaling.

    Such a system is never exactly in hydrostatic equilibrium. The difference scales with volume. Further, the differences have Entropy.

    Now, consider a large volume of gas with gravity.

    The same loop can occur horizontally. The question is, what conditions would be required for the loop to include a vertical component?

    The Isothermal Profile would clearly suppress vertical loops, because the gas would be out of thermal equilibrium at the top and bottom. In this case, the noise is no longer symetric in all axis. Gravity is acting to damp the vertical component of the noise.

    An Adiabatic Profile would, however, allow the loops to form vertically as well as horizontally. In other words, the vertical component of thermal noise retains the same distribution as the horizontal components.

    Reconciling with Graeff:

    My back of the envelope calculation is that Graeff’s results can be fully reconciled with this if the gas within the beads is heavy (ie Xe, SF6, Halons or heavy refrigerants).

    Incidentally, the text just before Verkley (1) assumes hydrostatic equilibrium, which prevents reaching this result.

  103. Tim Folkerts says:

    Q. Daniels says: “In a large volume, there will be local fluctuations of temperature, pressure and velocity. … Such a system is never exactly in hydrostatic equilibrium. The difference scales with volume. ”

    Typically those variations scale with 1/n^(0.5). So if a typical gas molecule has a speed of 400 m/s, then 4 molecule would have typically an average speed of ~ (400 m/s) / (4^0.5) = 100 m/s. If you look at 1E12 particles (which is still a very tiny volume; about 1 trillionth of a mole) ), a typical average speed would be 4E-4 m/s. So that TINY parcel of air (on the order of 0.1 mm on a side) is moving well under 1 mm/s (and the next second it could be moving a different direction).

    So, yes, the fluctuations scale with volume, but the LARGER the volume you look at, the SMALLER the fluctuations will be. And those fluctuations are pretty negligible for sizes of the order (1 mm)^3. I don’t see how you would possibly get any significant “loops” any larger than a fraction of a mm that have any significant net air flow.

    “The Isothermal Profile would clearly suppress vertical loops, because the gas would be out of thermal equilibrium at the top and bottom.”
    Only if you accept conclusions counter to traditional thermodynamics. “Thermal equilibrium” = “uniform temperate” — that is the 0th Law.

    “An Adiabatic Profile …”
    Could you define what you mean by “An Adiabatic Profile”? “Adiabatic” = no flow of heat. Do you mean the profile within an adiabatically sealed container? Or the profile where gas within the container cannot exchange heat with other regions within the container?

    Which ever you choose, it is not a priori obvious what the temperature profile would be for such a gas.

  104. tchannon says:

    1/f noise

    Very strange business.

  105. Q. Daniels says:

    Tim Folkerts wrote:
    I don’t see how you would possibly get any significant “loops” any larger than a fraction of a mm that have any significant net air flow.
    I’m considering relatively small flows within a much larger volume. The total volume is very close to average, but small parcels can differ.

    There are two different questions. Suppose we have a 1 meter cube, and consider it as a collection of non-isolated 1mm cubes. If you ask for the average velocity for a smaller cube, it will be very close to zero. If, instead, you ask for the velocity of the fastest cube, that’s a much higher number. That’s asking “Which of these 10^9 numbers is the largest?”

    “The Isothermal Profile would clearly suppress vertical loops, because the gas would be out of thermal equilibrium at the top and bottom.”
    Only if you accept conclusions counter to traditional thermodynamics. “Thermal equilibrium” = “uniform temperate” — that is the 0th Law.

    Perhaps my sentence wasn’t written clearly.

    Suppose you model the loop in an Isothermal Profile. Rotate the loop by 180 degrees. The top and bottom of the loop would be out of thermal equilibrium. This would represent a significant fluctuation, and is therefore very unlikely to happen.

    “An Adiabatic Profile …”
    Could you define what you mean by “An Adiabatic Profile”? “Adiabatic” = no flow of heat. Do you mean the profile within an adiabatically sealed container? Or the profile where gas within the container cannot exchange heat with other regions within the container?

    Which ever you choose, it is not a priori obvious what the temperature profile would be for such a gas.

    I meant an Adiabatic Lapse Rate.

    The question is, “What temperature profile allows parcels to drift without requiring an exchange of thermal energy between parcels?” That question has exactly 1 specific answer.

    The next question in series is, what density is required for thermal noise to be the dominant form of energy transport within a gas?

  106. Q. Daniels says:

    br1 wrote:
    For example, if you re-run your simulation, but instead of having the particles initially equally spaced, have them spaced according to hn=exp(n/h0) (as suggested above), then I guarantee (almost! I’m not quite sure what is in your simulation) that you will get an ‘equilibrium’ temperature profile that increases with altitude. I would be very interested to hear what you actually get, it should be easy for you to try.

    I also get a temperature profile that increases with altitude.

    The requirement for such appears to be a significant change in particle density across the Mean Free Path. It looks like what’s happening is that the slower particles are more likely to get knocked off path in the middle regions, while the faster particles tend to pass through. I’m not sure if this is real, or an artifact of simulation.

    Does the linked Velasco paper have a good enough answer to be worth $33?

  107. br1 says:

    Q. Daniels:
    “Does the linked Velasco paper have a good enough answer to be worth $33?”

    It has been uploaded by some kind soul here: https://tallbloke.files.wordpress.com/2012/01/s-velasco.pdf

    Well worth a read – it specifically addresses the ‘paradox’ that even though each particle loses KE with height, the ensemble at each height has the same temperature.

  108. Q. Daniels says:

    Ah, ok. I’ve got that paper, and read it. It was part of my basis for commentary on fluctuations.

    It doesn’t address the subject of increasing temperature with increasing z. I still don’t know if that’s a bug or real.

    If it’s real, it would suggest there are actually three regimes for the temperature profile of a gas. In order of increasing density, they would be:
    1) Increasing temperature with increasing z. This could be a result of a MFP that is long enough for a substantial pressure gradient. The fastest molecules preferentially escape the lower levels, distorting the distribution.
    2) Uniform temperature, as per the classical solution
    3) Decreasing temperature with increasing z, as per my comments above.

  109. Trick says:

    Q. Daniels 10:17am – “(Velasco) doesn’t address the subject of increasing temperature with increasing z. I still don’t know if that’s a bug or real.”

    Well, it CAN be real and physical since when I walk up from my basement T increases with z. To check if it is a program bug, you would have to rigorously check the control volume (cv) heat flows and accelerations of whatever 1mm cube or 1m cube you model.

    Your cubes will be in hydrostatic equilibrium when parcel vertical accelerations are much smaller than g and heat flows across cv are nil.

    For instance in the real atmosphere, a thumb rule is on scales > ~10km the parcels are in hydrostatic equilibrium. Of course, reality has exceptions to thumb rules as you reason.

    The Velasco paper is in agreement with B&A and V&G that Graeff’s container will be non-isothermal at equilibrium – the case for which the paper uses the term “microcanonical ensemble” and shows: “…one gets…for a finite adiabatically enclosed ideal gas in a gravitational field the average molecular kinetic energy decreases with height.” shown in Velasco eqn. 8.

    The paper corroborates the assumption “the average molecular kinetic energy” is proportional to temperature in eqn. 4.

    Interestingly Velasco paper also describes what happens for the real condition of Graeff’s experiment being in a larger heat bath (his basement) as N goes to infinity (the term then being “canonical ensemble”).

  110. Tim Folkerts says:

    Q Daniels says: “If, instead, you ask for the velocity of the fastest cube, that’s a much higher number. That’s asking “Which of these 10^9 numbers is the largest?” “

    Actually, it is not all that different. The distribution of speeds will follow the normal distribution. 1 out of a billion is only about z=6, ie the fastest block will only be about 6 standard deviations from the mean, or about 6 times faster than a typical block. Multiplying a negligibly small speed by 6 is still going to be very slow.

    “Suppose you model the loop in an Isothermal Profile. Rotate the loop by 180 degrees. The top and bottom of the loop would be out of thermal equilibrium. ”
    I agree there, because doing such a thing would require net work. Since the density drops as you go up, rotating a loop raises the center of mass. The part that had been on the bottom would rise, cooling. The top on the top would compress, warming. But I am not sure why this is a big deal. Such motion could only occur if a) outside work was done on the loop, or 2) some extraordinarily improbably event occurred whereby the rest of the gas did net work on the loop, meaning that the system was out of thermal equilibrium.

    (There is a constant challenge in classical thermodynamics — we have to look at large enough groups of particles that random fluctuations are insignificant, but look at small enough sections that interesting things can be measured. If you REALLY want to look at things on the level where the “random noise” is important, then you want to be looking at statistical mechanics.)

    “I meant an Adiabatic Lapse Rate.”
    OK. So you are assuming the parcels don’t exchange heat with surrounding parcels. But, of course, this assumes that the thermal conductivity is zero, which we know it false. For large, rapidly moving parcels, thermal conductivity can be ignored. For small parcels moving at a snail’s pace, conductivity is indeed important.

    “The question is, “What temperature profile allows parcels to drift without requiring an exchange of thermal energy between parcels?” That question has exactly 1 specific answer.
    This is indeed an interesting question, and the answer is the adiabatic lapse rate. But the verb could be changed from “requiring” to “allowing”. You have to move a parcel without ALLOWING any heat exchange. As mentioned above, this happens (approximately) for large, fast parcels of air where conduction is ignored.

    For small, slow parcels of air, you can’t stop conduction. In the limit of slow, small parcels of air, conduction would wipe out any changes due to raising and lowering particles.

    I suppose, in principle, the actual profile would be somewhere between isothermal and the adiabatic lapse rate. But in the limit of large numbers of particles (ie the thermodynamic limit), the profile would approach the isothermal result. At least I am pretty sure of that. (And that is the conclusion of Velasco, as I understand his paper).

    So if you really want to pursue this beyond hand-waving, is seem you will have ot study the distribution of speeds and temperatures of tiny parcels of air, and compare this to the ability of conduction (and other random fluctuations) to mitigate any temperature differences that occur.

  111. Tim Folkerts says:

    Some definitions are needed!

    “Microcanonical” = “perfectly insulated from the rest of the universe” = “constant total energy”
    “Canonical” = “can exchange heat with a thermal reservoir” = “energy can fluctuate around some average value”

    and for completeness …
    “Grand canonical” = “can exchange heat AND particles with a thermal reservoir” = “energy AND # of particles can fluctuate around some average value”

    Note that none of these deal with the SIZE of the ensemble. You could have 2 particles in any of these ensembles, or you could have 10^23 particles in any of these systems.

    “for a finite adiabatically enclosed ideal gas in a gravitational field the average molecular kinetic energy decreases with height.” shown in Velasco eqn. 8.”
    Before you get too excited about this, I would challenge you to calculate the profile for N2 at T = 300 K and g = 9.8 m/s^2 for a) 1 molecule of N2 b) 100 molecules of N2 and c) 1 mole of N2. I suspect you will be disappointed at the theoretical profile that Velasco predicts.

  112. ferd berple says:

    Tim Folkerts says:
    July 13, 2012 at 2:26 pm
    You all might look at http://www.falstad.com/gas/
    ====
    Ok, I’ve got an initial report on the code.

    1) The model exhibits gradual creep in the total KE when gravity is turned on. A single vertically bounding molecule that only contacts the bottom does not maintain the same KE between bounces.

    From a quick look the effect appears to be dependent on the number of molecules and simulation speed, which suggests it is a loss of precision error.

    2) The code run fine under the free NetBeans IDE Java SE download. I was able to import the package, make changes to the code, and observe the effects

    http://netbeans.org/downloads/index.html

    3) First change to try. You may have noticed that the color each molecule remains constant until it contacts a wall. However, this is wrong when gravity is turned on. To fix this search for this code

    m.dy += gravity*dt;
    m.x += m.dx*dt;
    m.y += m.dy*dt;

    This piece of code calculates the change is speed (dy) due to gravity and the change in distance (x,y) for each molecule (m).

    Change this to

    m.dy += gravity*dt;
    m.x += m.dx*dt;
    m.y += m.dy*dt;
    setColor(m);

    Now, as the KE of a molecule changes due to gravity the color will change as well.

    Looking at these equations they appear to double the change in distance due to gravity, but this should not cause the KE to creep.

    4) The sensitivity of the gravity slide bar can be adjusted on this line:

    if (e.getSource() == gravityBar)
    gravity = gravityBar.getValue() * (.001/20);

    Change the (.001/20) to (1) for example and the model will have 20k times greater gravity at the right most slider limit.

  113. Tim Folkerts says:

    BR1 says: ” … that you will get an ‘equilibrium’ temperature profile that increases with altitude.”

    I may have missed something, but aren’t you all doing models where the particles don’t interact with each other? In that case, the particles cannot come to equilibrium, since they cannot redistribute energy amongst themselves.

  114. ferd berple says:

    Ok, I can see what is wrong with the model. The code uses a very simple technique to “bounce” off the ground and top of the container. Code modified slightly to eliminate html problems.

    if (m.y LT upperBound+r || m.y GE winSize.height-r) {
    .. m.dy = -m.dy;

    This is not precise as it occurs after the collision has actually occurred and will change the KE of the molecule each time it bounces, when there is acceleration present. The technique is fine in the x direction, but not the y direction when gravity is turned on.

    The code corrects for this problem when molecules collide, so it would see that sort of correction needs to be added here.

  115. ferd berple says:

    Nope, looked at the correction for collisions. While the code corrects for t,x,y it does not correct for dy, which means that collisions will also change the KE when acceleration is present.

    double t = (-b-java.lang.Math.sqrt(b*b-4*a*c))/a;
    // backtrack m to where they collided.
    // (t is typically negative.)
    m.x += t*m.dx;
    m.y += t*m.dy;

    there should be a correction here for gravity to match the earlier code. Maybe something like:

    m.x += t*m.dx;
    m.y += t*m.dy;
    m.dy += t*gravity;

  116. ferd berple says:

    Looking further at this, the correction is even more complicated, as the solution for collision time

    double t = (-b-java.lang.Math.sqrt(b*b-4*a*c))/a;

    does not allow for changing m.dy due to gravity.

    a and b both depend on dy, which has not been corrected for gravity. and you cannot do this without knowing t. so you would have to iterate towards a solution for t, which has not been done in this model.

    This points to the difficulty in creating code that is a 100% accurate representation of the real world. Even in what seems a very simple model, the KE is not being correctly calculated in the presence of gravity.

  117. ferd berple says:

    ps: my analysis of the code is no reflection on the efforts of the author. There is nothing “wrong” with the code. It makes some simplifying assumptions to cut down on the amount of work (coding) required to produce something that looks reasonable realistic.

    The errors are small, but cumulative. Whether they are sufficient to change the result is unknown. Since we are concerned with KE and gravity, it would seem that the errors in KE due to gravity would need to be corrected before one could begin to put faith in the results.

    Even then, nothing says that the real world behaves exactly like these models, regardless of how accurate we make them. And Murphy says we can never make them completely accurate. Every non-trivial program has at least one undiscovered bug.

  118. Trick says: July 9, 2012 at 5:34 pm

    Lucy 3:20pm – “Graeff is getting seven times the Wikipedia / world / “appeal to authority” calculated lapse rate of ~10K/km”

    Graeff’s tube B74 if actually completely full of powder no longer has only a free ideal gas eqn. of state like in wiki/world/authority, the gas is all trapped in the glass beads.

    No, actually the spaces between the tiny spheres of glass are absolutely crucial here, because they effectively constitute a true column of air, just one that has had convection impeded.

  119. ferd berple says:

    having said that, the model appears worthy of further investigation because a large portion of the work is already done.

    I expect the “energy distribution” display can be readily adapted to show average KE per layer and the corrections required for gravity appear to be limited in scope. The MB distribution display provides a quick check on the health of the model itself, in case coding changes alter the distribution.

  120. Tim Folkerts says:

    Ferd,

    It sounds like you have done a great job digging into that code. Is there an easy way to sum up the average KE for say the top half and the bottom half separately (or perhaps 10 divisions at different altitudes)? With gravity turned way up high, then is should be easy to see the adiabatic lapse rate if it exists.

  121. ferd berple says:

    OK, here is a correction to fix one of the gravity errors. Otherwise if you turn up the gravity, the energy drains out of a single particle. This does not correct the boundary error, which adds PE.

    //m.dy += gravity*dt; //old
    m.x += m.dx*dt;
    m.y += m.dy*dt;
    m.dy += gravity*dt; //new
    m.y += gravity*dt*dt*.5; //new
    setColor(m); //new

  122. ferd berple says:

    Tim Folkerts says:
    July 14, 2012 at 5:50 pm
    Is there an easy way to sum up the average KE for say the top half and the bottom half separately (or perhaps 10 divisions at different altitudes)? With gravity turned way up high, then is should be easy to see the adiabatic lapse rate if it exists.
    ========
    that is my plan, once I correct the obvious errors in PE/KE. I’ll post back here as I progress.

    This is my first time with the NetBeans IDE. Works pretty well. You can change the code on the fly as the model is running which is always nice.

  123. ferd berple says:

    this corrects for PE at the boundary in the y direction

    if ((m.y < upperBound+r && m.dy < 0)|| (m.y >= winSize.height-r && m.dy > 0)) { wallF += Math.abs(m.dy)*m.mass; //if (m.y < upperBound+r) m.y = upperBound+r; //if (m.y >= winSize.height-r) //m.y = winSize.height-r-1;

  124. Trick says:

    Lucy 5:48pm: “No, actually the spaces between the tiny spheres of glass are absolutely crucial here, because they effectively constitute a true column of air.”

    Picture the simulation Tim F found and ferd is looking thru filled completely with powder consisting of hollow glass beads with entrapped air. Would it look anywhere near a true column of air? I think no. Ha, maybe ferd could program that for grins.

    You would have all those colorful gas-like bouncing balls all jiggly inside the beads & squirming between the HUGE glass beads, the state would look more like a solid than a gas!

    Set air molecules colorful and glass beads that are 10^10 mol more massive in white! The glass beads would just sit there with colorful dots trapped inside & around them.

    The ideal gas theory shows no powder is needed; powder only reduces the ideal gas gradient toward 0 like in a solid (where 0th law does apply) making LTE gas gradient much harder to measure.

  125. br1 says:

    Tim Folkerts:
    “I may have missed something, but aren’t you all doing models where the particles don’t interact with each other? In that case, the particles cannot come to equilibrium, since they cannot redistribute energy amongst themselves.”

    In the berple sims, yes that is the case, and was one of my first comments. However I still find it interesting that even though KE always drops with height, it is possible to have ‘steady state’ solutions that are hotter at higher altitude, no matter how long you leave them. As a berple sim can have decreasing or increasing gradients, it can obviously have an isothermal solution, depending on starting conditions.

    My own simulation had thermal interaction with the floor at ground plane only, and this is enough to reproduce the canonical ensemble which is isothermal with altitude.

  126. Trick says:

    Tim F 3:55pm: “Some definitions are needed! “Microcanonical” = “perfectly insulated from the rest of the universe” = “constant total energy””.

    More exact defn. microcanonical = “adj. Describing any closed system of constant volume which is thermally isolated from its surroundings, and whose total energy is constant and is known.” = Graeff’s B74 experimental dewar w/o the powder.

    http://www.wordnik.com/words/microcanonical

    Tim F continues: “I would challenge you to calculate the profile for N2 at T = 300 K and g = 9.8 m/s^2 for a) 1 molecule of N2 b) 100 molecules of N2 and c) 1 mole of N2.”

    Ok I wondered about that myself so – fair warning – a microcanonically long post & semi-calculating (conserving energy & work) & barring typo.s for:

    a)

    N=1 of an N2 molecule in f 1 dimension in z (easy but no easier f math) where total E = ideal gas enthalpy = sum over m of (1/2mv^2+mgz +0) for Velasco eqn. 8 becomes (m=mass of 1 N2 molecule, g=earth’s, m = 1 molecule N2):

    K(z) = (E)/(1 + 2*1 – 2) * (1- mgz/(E)) = E * (1-mgz)/(E) = E – mgz

    This says eqn. 8 reduces to and just restates what we already know microcanonically KE = TE – PE for 1 billiard ball-like ideal molecule right? Or restating Total energy = potential energy + kinetic energy + 0 work on environment enthalpy of the 1 ideal gas molecule under ideal enclosed adiabatic boundary constraints.

    (Sounds like ferd could see if this holds for 1 gigantic N,v in the kinetic theory of gas simulation Tim F. found.)

    Set z=0 and find E = K(z) + mgz = K(0) + 0 = K(0) proportional to T(0) = 300K for ref. initial condition set N2 Kb at 300K (eqn. 4), find E.

    Set z = any reasonable earth height up thru troposphere and find K(z) = E – mgz proportional T(z) < (T(0)=300K) showing ideal Velasco gradient non-isothermal in z.

    ——————————————

    b)

    N = 100, m = mass of 100 N2:

    K(z) = E/(100 +2*100 – 2) * (1 – mgz)/E) = E/298 – mgz

    Set z=0 to find the initial condition K(0) = E/298 proportional T(0) =300K initial condition w/N2 Kb, find E.

    Set z = any height and find K(z) = E/298 – mgz or proportional T(z) < (T(0)= 300K) for Velasco gradient non-isothermal in z.

    ——————————————–

    c)

    N = 1mole = ~6.022 × 1023, m = mass of mol N2.

    K(z) = E/(mol + 2*mol -2) * (1- mgz/E) ~ E/3*mol – mgz

    K(z) ~ E/(3 mol) – mgz

    Set z=0 to find the initial condition K(z) ~ E/3mol proportional T(0)=300K for initial condition set N2 Kb (eqn. 4), find E.

    Set z = any height and find K(z) ~ E/3 mol – mgz proportional T(z) < (T(0) = 300K) for Velasco gradient non-isothermal in z.

    ———————————–

    d) extend Velasco eqn. 8 to a tall in (z) & 1m in (x dim.) & 1m in (y dim.) for Velasco f 3 & GHG-free air column with earth’s initial conditions at surface:

    1 m^^2 tall z of N = earth ideal adiabatic enclosed non-GHG atmospheric column from surface z=0 at 1atm. (1000 hPa) up to ~10km (~225 hPa):

    Sum m air ~ 7,633kg., E ~ 2.01 * 10^9 Joules, T(0) ~ 302K, T(10km) ~ 215K (w/in an eyeball). Get non-isothermal profile drop from 302K to 215K all a,b,c T drops w/similar initial numbers or IOW in a) 1 gigantic billiard ball-like N2 molecule of 3.5tons or so going fast.

    Hello ferd? Does that 1 N ~3.5 ton T(0) 300K initial condition 1m^^2 ~ 10km tall work out arithmetic wise? What is the velocity variance? I’m good with Fortran but my entropy never increased up to Java in my LTE.

    ——————————–

    All a), b), c), d) per Velasco have non-isothermal ideal gas gradient as Velasco tells us in the clip after eqn. 8 (again: “…for a finite adiabatically enclosed ideal gas in a gravitational field the average molecular kinetic energy decreases with height.”). Even the real atm. has a non-isothermal avg. gradient DALR for more reasons than ideal reasons.

    NB: I bet there is a typo left in there somewhere (maybe even a material one), find it (them?) for an IOU of 1 thank you each. An excel spreadsheet will show the numbers exact to 12 or more digits in Velasco eqn. 8 non-isothermal lapse for any increasing N. I will leave that to any grad. students here to look up.

    —————————

    2nd NB: Now what happens as N approaches infinity (and beyond tm pixar) per Velasco, you know the edge of the universe heat bath? E/N is finite! Well, some like to think the universe is adiabatic and enclosed TE = PE + KE + 0. So get non-isothermal all the way to E/N (both at infinity) at finite ~3K right now as Velasco implies in the limit & V+G graph seem to imply. Then heat death arrives as 3K goes to 0K with expansion.

    Some think the universe allows no heat in/out but does work on its surroundings (pixar: beyond!) for E = KE + PE + pV. Well, maybe so, in which case switch the universe to isothermal and no heat death. Thus seems like we do have something to look forward to in real LTE for optimists among y’all that seem to like the isothermal soln. (with non-zero pV term and thus T(0)=288K finite constant to E/N at infinity) better. I see now why that is. No heat death looking forward.

  127. Trick says:

    1 correction thank you to myself:

    c) N = 1mole = ~6.022 × 10^23

  128. br1 says:

    Q. Daniels:
    “Ah, ok. I’ve got that paper, and read it. It was part of my basis for commentary on fluctuations.

    It doesn’t address the subject of increasing temperature with increasing z. I still don’t know if that’s a bug or real.”

    I think that paper includes particle-particle collisions. Otherwise Eqn(8) would be meaningless. With one particle in the system, the gradient in Eqn(8) is clear, but how do you add a second particle? You could add it exactly in step with the first particle, and the gradient would be identical. I’m pretty sure the solution is based on equipartition of available energy, so the two particles must be able to swap energies.

  129. Q. Daniels says:

    br1 wrote:

    With one particle in the system, the gradient in Eqn(8) is clear, but how do you add a second particle? You could add it exactly in step with the first particle, and the gradient would be identical. I’m pretty sure the solution is based on equipartition of available energy, so the two particles must be able to swap energies.

    I grabbed NetLogo, which has a model. “GasLab Gravity Box”. For collisions, the model computes the velocity of the center of mass, which leaves both particles coming in head on at the same speed. It then scatters the results at a random angle relative to the center of mass. The overall results look pretty good.

    I added instrumentation to allow averages over time. Avg. Temp = Total binned Energy / total binned count. Over the course of 5,000 ticks, with a factor of 1.25 density gradient, the temperature changes by about 5%. It looks like it’s some function of the square of (the density gradient over 1 MFP).

    I can run either Microcanonical or Canonical (bottom wall thermal). That doesn’t affect the results.

    Possible sources of error are:
    1) The probability of collision is unrealistic. This seems more likely.
    2) There’s a distortion in the binning.
    3) The scattering angle distorts the results. This seems unlikely.

  130. Tim Folkerts says:

    That’s a good start, Trick. I would have selected a 3 dimensional gas, but 1 D will work too.

    So,
    K = [ E / (3N – 2) ] * [ 1 – (mgz/E) ]

    But remember that the total energy, E = Ne, where e = average energy per particle. The first term becomes
    [ e / (3 – 2/N) ]
    For N = 1, this is “e”. For large N, this approaches e/3. So the first tern is somewhere between e/3 and e. Not terribly exciting.

    The second term is more interesting:
    [ 1 – (mgz)/(Ne) ].
    For large N, this term simply becomes “1”. In other words for large N, the value of z doesn’t matter. What ever the lapse rate is for 1 particle, divide that by N to get the lapse rate for N particles.

    Specifically, for N = 100, m = 28 u, e = 3/2 kT and T = 300 K, KE drops off by 0.07% from z = 0 to x = 10,000 m. That would a drop of 0.22 K in 10 km, or a lapse rate of 0.022 K / km. Increase the count to N = 1 mole and Excel gives a lapse rate of zero (to the accuracy that Excel can calculate). In other words, the predicted lapse rate is too small to measure for any macroscopic collection of gas molecules.

  131. Trick says:

    Tim F – Thanks. I wonder why the lapse of some 87 degrees then in V&G z=0 to z=10km? Will look into this, interesting.

  132. br1 says:

    Trick:

    While I wish to encourage anyone who is willing try some maths and post it in public, maybe your attempt here will give you pause for thought about your maths ability?

    “K(z) = E/(mol + 2*mol -2) * (1- mgz/E) ~ E/3*mol – mgz”

    In my country, children are generally ten or eleven when they learn how to multiply out brackets. Fortunately in this case there wasn’t anything hairy like an exponential in this equation, otherwise it might have taken another 300 posts to argue what is wrong with it.

  133. br1 says:

    Q. Daniels:
    “I grabbed NetLogo, which has a model. “GasLab Gravity Box”.”

    I’ll check that out.

  134. Q. Daniels says:

    Tim Folkerts wrote:
    “The question is, “What temperature profile allows parcels to drift without requiring an exchange of thermal energy between parcels?” That question has exactly 1 specific answer.”

    This is indeed an interesting question, and the answer is the adiabatic lapse rate. But the verb could be changed from “requiring” to “allowing”. You have to move a parcel without ALLOWING any heat exchange. As mentioned above, this happens (approximately) for large, fast parcels of air where conduction is ignored.

    I’m exploring a theoretical framework more than looking for specific values at this stage. Your comment is correct for physical systems. My choice of verb was deliberate for a speculative system.

    For small, slow parcels of air, you can’t stop conduction. In the limit of slow, small parcels of air, conduction would wipe out any changes due to raising and lowering particles.

    Agreed.

    I suppose, in principle, the actual profile would be somewhere between isothermal and the adiabatic lapse rate. But in the limit of large numbers of particles (ie the thermodynamic limit), the profile would approach the isothermal result. At least I am pretty sure of that. (And that is the conclusion of Velasco, as I understand his paper).

    So if you really want to pursue this beyond hand-waving, is seem you will have ot study the distribution of speeds and temperatures of tiny parcels of air, and compare this to the ability of conduction (and other random fluctuations) to mitigate any temperature differences that occur.

    Agreed, in principle. I would, however, expect the other random fluctuations to have the same average lapse rate as velocity.

    We have one questionable data point, which if valid, says that there is an error in one of the things we know to be true. Given that, it makes sense to me to look carefully at each of the things known to be true, and evaluate the question, “What if this isn’t true?”

    Your feedback has helped me sort through possibilities, and is appreciated.

    Right now, the most interesting (to me) thing “known to be true” is the concept of thermodynamic systems damping to nearly indistinguishable from equilibrium. This is a concept that predates our understanding of non-linear dynamics (aka Chaos), and has been found to hold true for only a narrow subset of problems. It is an assumption that, in other contexts at least, has been shown to unduly restrict both the data and solution spaces.

    As you say, if I want to move beyond hand-waving, I’ve got some work to do.

    I see a few ways forward:
    1) Brute force simulation of a reasonable number of particles (10^6-10^7). Getting a simulator to run fast enough to handle that could be entertaining.
    2) Hunt down someone else’s tools, either mathematical or simulation.
    3) Extract something from an unspecified location.

  135. Q. Daniels says:

    Tim Folkerts wrote:
    For small, slow parcels of air, you can’t stop conduction. In the limit of slow, small parcels of air, conduction would wipe out any changes due to raising and lowering particles.

    I wrote:
    Agreed.

    After further review…

    The DALR is 0.0098 K/m. The thermal conductivity of air is 0.025 W/m*K at 300K. The thermal conductivity is fairly flat from 1 to 10^-5 bar.

    Unless I’ve screwed my arithmetic again, we’re talking about a quarter milliwatt per square meter to push a column of gas from isothermal to adiabatic.

  136. Trick says:

    br1 7:41am – correctly points out another typo. in the 7/14 9:43pm post. Corrected Velasco eqn. 8 for Tim F’s parameters:

    b)

    K(z) = E/298 – mgz/298

    c)

    K(z) = E/(3 mol -2) – mgz/(3mol-2)

    No material change, Velasco eqn. 8 is still non-isothermal profile w/any reasonable earth z height for all Tim F Ns.

    Two thank you issued to br1. Double check though since br1 still might earn more IMHO.

  137. Trick says:

    Tim F 2:59am – “The second term is more interesting: [ 1 – (mgz)/(Ne) ]. For large N, this term simply becomes “1″.”

    Don’t think that is so because m AND N (numerator & denominator) will increase in proportion. Does that affect your computations?

    I played around with some Velasco numbers too and in doing so ran across another question. Nowhere in the paper does Velasco mention the word “entropy”. They do use the term “thermal equilibium” but only in the sense of asking the original question. Not in the answer w/gravity anywhere obvious.

    So far as I can see, Velasco 1996 does not mention the task of maximizing the entropy with g for LTE as detailed in Bohren&Albrecht 1998 text (I just received my copy again). It will be interesting to compare detail arithmetic to get a Velasco Eqn. 8 idealized non-isothermal temperature delta for earth surface up to ~10km and see how that compares to V&G’s ideal non-GHG non-isothermal profile surface 302K to ~215K up at ~10km altitude.

    I’m not yet finished plugging together all the Velasco numbers, I really need a grad. student as proper units and constants with such large & small numbers gives me instant headaches (like ferd is finding I suppose). Thank goodness V&G did the calculation from their theory with max. entropy to check results.

  138. br1 says:

    Trick:
    “K(z) = E/(3 mol -2) – mgz/(3mol-2)

    No material change, Velasco eqn. 8 is still non-isothermal profile w/any reasonable earth z height for all Tim F Ns.”

    Sigh.

    Next thing to note is that m is the mass of one molecule. As the first term has no z dependence, and the second term is the PE of one molecule divided by the total number of molecules in the system (on the order of 10^24 in this case), then the profile is isothermal as far as any measurement is concerned.

    The next thing to note is that the microcanonical ensemble not only doesn’t exist in nature, but I don’t think there is even a good approximation to it. One needs absolutely 100% reflective container walls and the tiniest thermal interaction, including swapping of radiation, wipes out any minute effect that one may predict in idealised theory. While it is certainly good to know about such things, Graeff’s system is certainly not a microcanonical system.

  139. Trick says:

    br1 – ““K(z) = E/(3 mol -2) – mgz/(3mol-2)…the second term is the PE of one molecule divided by the total number of molecules in the system…”

    No, not one molecule. Double check 7/14 9:43pm to find the second term mass m in C) calculates the PE of Tim F.’s challenge for one mol of molecules – see in the post that “m = mass of mol N2.”

    It is interesting to note that the first term K(z) is constant and second term is variable with z. There is some z where K(z) will hit zero for each total E in a,b,c. The altitude cannot then be larger in z and have molecules though the math just keeps calculating lower than absolute 0 for T at higher z. So the physical reason exists to limit z to reasonable earth atmosphere (troposphere) in which we are interested.

    br1 continues – “The next thing to note is that the microcanonical ensemble not only doesn’t exist in nature, but I don’t think there is even a good approximation to it.”

    Yes there is no perfect insulation, though my coffee stays hot in a microcanonical ™ thermos for long enough & idealized theory teaches us “remarkably” well about earth’s atmosphere per V&G. Even Graeff’s B74 results presented by Lucy show the insulation is effective for quite awhile – Graeff container initially establishes non-isothermal gradient early in those results.

    The theory of V&G and Velasco should enable a Graeff container initial non-isothermal T profile calculation before his basement heat bath reservoir arrives by s-l-o-w conduction. I’d like to apply Velasco theory to Graeff’s “early” results once my headaches are cleaned out; will take some time to get the arithmetic right & verified. Small differences of large numbers and all.

  140. Tim Folkerts says:

    Trick says: “Don’t think that is so because m AND N (numerator & denominator) will increase in proportion. Does that affect your computations?”

    “m” is the mass of each particle, not the total mass of all “N” particles. This is not spelled out explicitly in the paper, but it is most clear in equation 4. This equation gives the average MOLECULAR kinetic energy as basically 1/2 m v^2 (the “v^2” term is some ugly weighted average). Thus “m” is clearly the mass os a single molecule (which I assumed to be 28 u = 4.65E-26 kg.

    So this doesn’t impact my calculations at all. 🙂

    “There is some z where K(z) will hit zero for each total E in a,b,c. The altitude cannot then be larger in z and have molecules though the math just keeps calculating lower than absolute 0 for T at higher z. ”

    True.
    For a single N2 molecule @ T = 300 K, this is altitude is about 14 km
    For 100 N2 molecules @ T = 300 K, this is altitude is about 14,000 km,
    For 1 mole, that “altitude” would be way beyond the end of the Milky Way! (but of course, “g” would not be constant, so the whole analysis would be moot).

    See https://docs.google.com/spreadsheet/ccc?key=0AgM8XE4GABYQdGI3eG84eE1fQ0gyaDVRNU45Z0ZGX0E
    for the calculations and a handy graph.

  141. ferd berple says:

    br1 says:
    July 14, 2012 at 9:36 pm
    As a berple sim can have decreasing or increasing gradients, it can obviously have an isothermal solution, depending on starting conditions.
    =========
    what?? I never said that. you appear to be inventing conclusions. I never observed a reverse gradient.

  142. Tim Folkerts says:

    Q. Daniels says” “Unless I’ve screwed my arithmetic again, we’re talking about a quarter milliwatt per square meter to push a column of gas from isothermal to adiabatic.”

    Yes, I got similar results once long ago, so I believe you are indeed in the right ballpark.

    Which brings us back to my original concern about Graeff’s experiment. His cross-sectional area is well under 1 m^2 — closer to 0.01 m^2. That means that a energy flows on the order of microwatts will be enough to create his gradients (assuming that isothermal is the equilibrium profile). It would be tough to eliminate microwatt energy transfers thru wires, thru stray EM fields, etc. Thus means that to be sure the effect is real, any microwatt energy sources/sinks have ot be eliminated.

    A quick calculation shows that a 1 m long 30 AWG copper wire with temperature difference of 1 K between the ends will conduct ~ 20 microwatts. I don;t know what gauge or material of wire he uses for the lead to the thermocouples, but this could easily influence the experiment. The electronics used to record the data will surely warm up a few degrees during operation, so such thermal conduction thru the thermocouple leads would have to be very carefully eliminated.

    P.S. This is also why Graeff’s results — even if confirmed — are unimportant in the actual atmosphere. A few mW/m^2 could set up the Graeff effect in the atmosphere. A few mW/m^2 of cooling at the bottom could eliminate or reverse the Greaff effect. The atmosphere has energy flows on the order of 200 W/m^2 heading up thru it (sunlight heating the surface; IR to space cooling the top). The Graeff effect would be lost in the noise with these overwhelming energy flows.

  143. ferd berple says:

    Q. Daniels says:
    July 15, 2012 at 12:14 am
    For collisions, the model computes the velocity of the center of mass, which leaves both particles coming in head on at the same speed. It then scatters the results at a random angle relative to the center of mass. The overall results look pretty good.

    Possible sources of error are:
    3) The scattering angle distorts the results. This seems unlikely.
    =========

    I don’t agree. The energy of a collision will on average be increased in the downward direction by gravity. Thus a random scattering is guaranteed to introduce error, unless corrected.

  144. br1 says:

    Ferd berple:
    “what?? I never said that. you appear to be inventing conclusions. I never observed a reverse gradient.”

    Please read back over my comments to you. You will see that I requested you more than once to rerun your sim using different starting conditions. One suggested starting condition is hn=exp(n/h0) where n=1,2,3, etc. I wrote my own sim in what I think is your conditions, and this gives me (and Q. Daniels) an ‘equilibrium’ (that is, you can run the sim for as long as you like) temperature profile that increases with height. If I start my sim with an equal spacing in height then I get a temperature drop with altitude, as you do. So I would be very interested to hear what you get with an exp starting distribution in your own sim.

  145. Trick says:

    Tim F 7:25pm – “(m)is most clear in equation 4.”

    Funny, m just disappears in eqn. 4. As you write, m is NOT defined in Velasco & neither is T. Eqn. 4 caused my biggest headache trying to do the arithmetic – just what does f stand for in Velasco eqn. 4? T is undefined if this f means just 1-D molecule (my 1st assumption) but now I suspect Velasco using f here means some function of Kb * T where possibly T is Tavg. integrated over the column. So I propose neither of us is right, m stands not for 1 molecule mass nor total N*(1 mass).

    Maybe you are interested to figure this out while I do so too. I think I am wrong to just use PE(N*m) in B) and C) though it works in a) because at any z need the potential energy of the N(z) molecules since K(z) would want to use just the potential energy of just the N molecules at z since constant TE = sum (KE(z) + PE(z)).

    It is becoming possible that Velasco Eqn. 8 has no closed form solution like V&G does (and I suspect why that is but I don’t yet know for sure). Pulling the earlier Velasco paper to look for T and m defn.s seems like a good thing.

  146. Q. Daniels says:

    Tim Folkerts wrote:
    Yes, I got similar results once long ago, so I believe you are indeed in the right ballpark.

    Thanks.

    From what I’ve written above, the thermal noise would be best solved in the Grand Canonical Ensemble. Rough calculations say I’m looking for something on the order of 10^-12/meter. One region for investigation is a divergence between real and ideal gases, which probably means non-factorable, and usually ignored as ”. Straight into the heart of darkness.

    “It’s a bug hunt.”

    Once upon a time, I knew the Grand Canonical well enough to undertake such a task with confidence. Now, I have a high probability of stupid errors. That’s not enough to dissuade me, but should be taken as a caveat.

    Which brings us back to my original concern about Graeff’s experiment. His cross-sectional area is well under 1 m^2 — closer to 0.01 m^2. That means that a energy flows on the order of microwatts will be enough to create his gradients (assuming that isothermal is the equilibrium profile). It would be tough to eliminate microwatt energy transfers thru wires, thru stray EM fields, etc. Thus means that to be sure the effect is real, any microwatt energy sources/sinks have ot be eliminated.

    A quick calculation shows that a 1 m long 30 AWG copper wire with temperature difference of 1 K between the ends will conduct ~ 20 microwatts. I don;t know what gauge or material of wire he uses for the lead to the thermocouples, but this could easily influence the experiment. The electronics used to record the data will surely warm up a few degrees during operation, so such thermal conduction thru the thermocouple leads would have to be very carefully eliminated.

    Sure, and thermocouple wires have different thermal conductivities, of necessity. For example, Chromel and Alumel (K-type thermocouples) are 19 W/mK and 30 W/mK respectively.

    Graeff’s experiments B76 and B428 go part way to addressing this. In B76, he placed his apparatus in a drum, which could be rotated, reversing the vertical orientation. This does not eliminate persistent effects possibly aligned with gravity, including EMI.

  147. Tim Folkerts says:

    Trick says: ” … m is NOT defined in Velasco & neither is T.”

    And neither is “g” (gravitational field strength) or “k_B” (Boltzmann constant). Writers have to assume the read has some basic knowledge of the topic. Do you still doubt that “m” is the mass of a molecule?

    “Funny, m just disappears in eqn. 4.”
    Not funny at all! Totally expected! It is well-known that all gas molecules at the same temperature have the same average KE. So mass should not affect the average KE. If would be odd if “m” did NOT disappear!

    “just what does f stand for in Velasco eqn. 4?”
    As he clearly states, it means the number of dimensions of the gas. It is theoretically easy to imagine a 1 D gas (where the particles are only free to move in 1 direction) or 2 D gas, in addition to a more standard 3D gas. He is just being more general.

    Basically, the first half of the paper confirms the classical result: if you have a sysmte in contact with a thermal reservoir (a canonical ensemble) then the distribution is isothermal.

    The second half looks at the alternate –that we have a small number of particles that are thermally isolated (a microcanonical system). Here we do get a non-isothermal solution. But for Large N, the system approaches isothermal.

  148. Trick says:

    Tim F 9:47pm: “As (Velasco) clearly states, (f) means the number of dimensions of the gas.”

    I hear what you are saying.

    Tim F: “..all gas molecules at the same temperature have the same average KE.”

    Velasco: “..the temperature is proportional to the average molecular kinetic energy.”

    Text book 1: Temperature is just another name for the mean kinetic energy density of molecular motion. Define m = 3KbT in 3D. T = absolute T in Kelvin.

    Text book 2: Define = 3KbT in 3D. T = absolute T in Kelvin.

    Velasco f=3 in 3D; so ok Eqn. 4 = = 3KbT/2 checks out by definition (with only N2 molecules take m outside brkts.)

    Tim F continues: “So mass should not affect the average KE.”

    It is system complicated. The texts have m defined as particle mass so KE and absolute Temperature scale up with increasing N particles of mass m to tune of pV=NKbT as N scales up in the enclosed container, so does T and p. (NB: p = 1/3 (N/V) * )

    Since gas Total energy TE = KE + PE + pV in the “sysmte in contact with a thermal reservoir” then if I’m allowed to assume a rigid container for both Velasco large 1st half and small 2nd half, pV= 0 and

    TE – KE = PE and unless TE and KE scale up with N exactly the same, seems cannot keep PE constant at 1 particle m for N=1, 100, 1 mole.

    As m scales up with N, KE scales up, TE scales up, PE then should scale up somehow with N. PE remaining as one particle m constraining TE – KE = constant in a special case of 1 particle m is not physical that I can think of. Maybe you can fill me in.

    Until filled in that m=1 during PE scaling, the question I have remains in Velasco Eqn. 8: how does mgz scale up with N?

    An interesting question, but you seem to think mgz remains 1m particle of N2 for N = 1, 100, 1 mole, 10km tall column of gas and doesn’t scale. Seems to me PE (mgz) scales with N somehow in eqn. 8. And probably not the way I did it on 7/14 9:43pm (using N*m).

    So one material typo thank you is hereby issued to Tim F. Would not have seen that if I hadn’t tried the Tim F. challenge.

    Meantime I’ll be researching how PE is text book & paper defined with ideal gas of N identical particles.

  149. Trick says:

    Ha, more typos I see with the sideways v html tag hiding things. I’ll just repost below using braces for avg. instead. Maybe mod.s can delete my 1:17am post.

    ————————-

    Tim F 9:47pm: “As (Velasco) clearly states, (f) means the number of dimensions of the gas.”

    I hear what you are saying.

    Tim F: “..all gas molecules at the same temperature have the same average KE.”

    Velasco: “..the temperature is proportional to the average molecular kinetic energy.”

    Text book 1: Temperature is just another name for the mean kinetic energy density of molecular motion. Define m[v^2] = 3KbT in 3D. T = absolute T in Kelvin.

    Text book 2: Define [mv^2] = 3KbT in 3D. T = absolute T in Kelvin.

    Velasco f=3 in 3D; so ok Eqn. 4 [K(z)] = [1/2mv^2] = 3KbT/2 checks out by definition (with only N2 molecules take m outside brkts.)

    Tim F continues: “So mass should not affect the average KE.”

    It is system complicated. The texts have m defined as particle mass so KE and absolute Temperature scale up with increasing N particles of mass m to tune of pV=NKbT as N scales up in the enclosed container, so does T and p. (NB: p = 1/3 (N/V) * [mv^2])

    Since gas Total energy TE = KE + PE + pV in the “sysmte in contact with a thermal reservoir” then if I’m allowed to assume a rigid container for both Velasco large 1st half and small 2nd half, pV= 0 and

    TE – KE = PE and unless TE and KE scale up with N exactly the same, seems cannot keep PE constant at 1 particle m for N=1, 100, 1 mole.

    As m scales up with N, KE scales up, TE scales up, PE then should scale up somehow with N. PE remaining as one particle m constraining TE – KE = constant in a special case of 1 particle m is not physical that I can think of. Maybe you can fill me in.

    Until filled in that m=1 for PE scaling, the question I have remains in Velasco Eqn. 8: how does mgz scale up with N? An interesting question, but you seem to think mgz remains 1m particle of N2 for N = 1, 100, 1 mole, 10km tall column of gas and doesn’t scale. Seems to me PE (mgz) scales with N somehow in eqn. 8. And probably not the way I did it on 7/14 9:43pm (using N*m).

    So one material typo thank you is hereby issued to Tim F. Would not have seen that if I hadn’t tried the Tim F. challenge.

    Meantime I’ll be researching how PE is text book & paper defined with ideal gas of N identical particles.

  150. wayne says:

    After sitting through the first three hours of a refresher on statistical mechanics (http://www.youtube.com/watch?v=H1Zbp6__uNw) at Stanford on the derivation of Boltzmann distribution, there appears to me to be a disjoint between the Helmholtz free total energy (lecture #3 at ~57 minutes) which by definition should include the potential gravitational energy m.g.h, being all energy available to do work, and the per energy level E(j) within the sum of the partition function defining the Boltzmann distribution that is internal kinetic energy only. Anyone know where this view might be right or wrong?

    That seems to me to be where the missing natural lapse rate may lie, within that disjoint itself.

  151. Tim Folkerts says:

    Trick says: “Since gas Total energy TE = KE + PE + pV … ”

    PV is *work*, not *energy*. The energy is the KE + PE (+ rotational & vibrational energy, if they exist in the particular system). Q & W are *transfers* of energy that can change the energy of a system, but they do not belong TO the system.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    I think it is time to bow out of this discussion, for two reasons

    1) we can covered about as much as we can. Hashing out further details with any sort of efficiency would probably require sitting down in person with a large white board and lots of coffee.

    2) MORE IMPORTANTLY, it seems odd to be working out all these details when neither of the principles (Lucy or Graeff himself) are interested in participating. That suggests to me that, even if we do learn something, it is not going to go anywhere.

  152. ferd berple says:

    report on Gas.java http://www.falstad.com
    =======
    1) added a display for total energy
    2) with gravity turned off model conserves energy exactly, even with collisions.
    3) with gravity turned on, model does not conserve energy

    code changes
    1) fixed molecule color so it changes as KE changes
    2) fixed the boundaries so energy is conserved with gravity
    3) work in progress to conserve energy between collisions – small non-random leakage
    4) added quick and nasty energy average by altitude

    observations
    1) without gravity, model is isothermal
    2) model (now) conserves energy with gravity turned on – until there are collisions
    3) gradient when gravity is turned on and no collisions.
    4) consistent reverse gradient with collisions – energy leakage – no conclusion possible at present

  153. Brian H says:

    It seems to me that the 2nd Law is not challenged or modified, just contextualized. It applies in the absence of acceleration of the system. Under acceleration, the appropriate forces and energy flows must be accounted for.
    __
    Thought experiment for water: multiple layers in an insulated case, separated by thin flexible membranes (e.g., mylar) to prevent convection except within each layer. Start with water at consistent temp. throughout, then track changes.

  154. Trick says:

    Tim F 4:14am – “That suggests to me that, even if we do learn something, it is not going to go anywhere.”

    Lucy wrote she’s not up on the theory. Seems both Graeff and Lucy in replication could benefit from someone accurately computing the theoretical lapse in the B74 no-powder container (to x64 digits, ha). Then the experimentalists would know what to look for.

    Seems you agree the microcanonical container is non-isothermal and that is all that is needed. My thought is to compute the B74 lapse (w/o powder) from both Velasco and V&G paper.

    Right now I have proven I don’t know for sure how PE scales in V Eqn. 8. Need that clarified. I am running down the 1995 Roman et. al. paper to better trace thru the m in Eqn. 8 mgz. I expect to find it really is m(z) i.e the total mass in the K(z) volume element which would then scale with N.

    Tim F continues: “PV is *work*, not *energy*. The energy is the KE + PE (+ rotational & vibrational energy, if they exist in the particular system).”

    Work is energy (f * d) and that pV term in the definition of gas enthalpy is work done (energy expended) by the gas on the environment if the container is non-rigid (as in V&G 2a pV term .NE. 0). Gas enthalpy defined = internal energy + pV. That’s why need a rigid container for pV = 0 in V&G 2b and likely in Velasco 1st large half and 2nd small half.

    Yes, some of the particle energy is in rotational & vibrational, chemical, mass, quantum radiation levels et.al. All of which authors rule out as not important ideally. Some might just be important in the Graeff container as written earlier so any experimental difference from ideal theory could be run down.

    I’m surprised Graeff is first to try all this, seems like some student/researcher must have tried before.

  155. Trick says:

    ferd 4:49am:

    Reads like good progress. Interesting work. A reverse gradient with collisions seems to imply the collisions are somehow not ideal elastic in the java implementation. Collisions also need to obey PV=nRT, since nothing in the collision math says that – has to be implemented as a forced adjustment I imagine.

  156. ferd berple says:

    report on Gas.java http://www.falstad.com

    5) with the heater turned on (as a cooler), the leakage in total energy (KE + PE) due to incorrectly calculated collisions (with gravity) can be compensated so an approximately constant temperature and energy is achieved.

    6) In the absence of an error in calculating collisions, the heater should not be required to achieve equilibrium.

    7) my conclusion is that the need for an energy exchange with the bottom of the container to achieve equilibrium is a sign that the model is incorrect, that it is somehow gaining or losing energy and thus the conclusions from such a model are not reliable when applied to the Graeff experiment.

  157. ferd berple says:

    Tim Folkerts says:
    July 16, 2012 at 4:14 am
    I think it is time to bow out of this discussion, for two reasons
    ——–

    I will continue reporting here, as my progress on fixing the model code has been very good. I believe I’m withing spitting distance of having a model with theoretically correct collisions and no energy leakage.

    I would appreciate if anyone is able to come up with a quick way to solve the equation of collision between two molecules with gravity turned on. The code in questions follows these comments in the gas.java source:

    // handle a collision
    // first, find exact moment they collided by solving
    // a quadratic equation:
    // [(x1-x2)+t(dx1-dx2)]^2 + [(y1-y2)+…]^2 = mindist^2

    I can do the coding if someone can give me the solution to incorporate acceleration (gravity) into the above formula, or provide an alternative formula.

  158. br1 says:

    ferd berple:
    “The code in questions follows these comments in the gas.java source:”

    the good news is that the quadratic collision eqns for t and t2 don’t need to be changed. The acceleration terms in the relative distance cancel out, because both particles experience the same acceleration.

    The backtrack can then be modified to:

    // backtrack m to where they collided.
    // (t is typically negative.)
    m.x += t*m.dx;
    m.y += t*m.dy-0.5*gravity*t^2;

    The last terms is negative, as he seems to take t as negative for the collision time. However, I am confused by this section of code because he only seems to backtrack one of the particles. Surely the collision should be symmetric? In which case you should copy-paste the above two lines but change for particle 2

    m2.x += t*m2.dx;
    m2.y += t*m2.dy-0.5*gravity*t^2;

    I would also expect a backtrack of y velocity is needed:

    m.dy += gravity*t;
    m2.dy += gravity*t;

    before doing the COM stuff.

    Then under
    // send m on its way

    you’ll need to change m.y to
    m.y -= t*m.dy-0.5*gravity*t^2;

    (again, subtracted as t is negative here (yuck!)), and update the velocities
    m.dx -= gravity*t;
    m.dy -= gravity*t;

    I think the reason he doesn’t backtrack the second particle is because he takes its position as constant during the collision, and only changes its velocity in
    // adjust m2’s momentum so that total momentum
    // is conserved

    to conserve momentum. This is itself an approximation, even when there is no gravity. In which case you will also have to ‘send m2 on its way’ by copy-pasting the above lines and putting in for particle ‘2’.

    I haven’t tried the above mods, so they may need some adjustment, but I think there is some chance they are close to correct 🙂

    p.s. When you have time, I would still like to see your other sim re-run with a different initial condition (hint, hint).

  159. br1 says:

    Trick:
    “I am running down the 1995 Roman et. al. paper to better trace thru the m in Eqn. 8 mgz. I expect to find it really is m(z) i.e the total mass in the K(z) volume element which would then scale with N.”

    The Roman1995 paper is the one to go for alright, I downloaded this a year and a half ago from http://iopscience.iop.org/0143-0807/16/2/008

    Near the start of Section 2 they say

    “m is the mass of a particle”

    and it is clear that that is the meaning it has all the way through the paper. So yet again your instincts have lead you astray, I’m afraid to say.

    On the positive side, you are right to note that Velasco Eqn(8) makes no sense above some value of z – the authors effectively note this in Roman Eqn(4), where you are not allowed to consider a state with energy greater than the system energy.

  160. Trick says:

    br1 5:44pm – Near the start of Section 2 they say “m is the mass of a particle””

    Good, then in the volume element z + dz for which we want sum KE(z) there would be some N*m number of single particles for the integration over the volume element for K(z) so that in Eqn. 8 find m(z)gz is justified where m(z) = N*m in that volume element which is what I initially assumed. If m is for different particles say N2, O2, etc. then have to perform the integration wrt the mass of each particle * its N fraction in the volume element..

  161. Tim Folkerts says:

    Trick says: ” … so that in Eqn. 8 find m(z)gz is justified where m(z) = N*m in that volume element … ”

    Unless you can show the specific mathematical steps deriving Eqn 8 from the previous equations, I see no reason to accept your “assumption” that the author suddenly switched notation without the slightest warning to the readers.

  162. ferd berple says:

    ferd berple:
    “The code in questions follows these comments in the gas.java source:”
    the good news is that the quadratic collision eqns for t and t2 don’t need to be changed. The acceleration terms in the relative distance cancel out, because both particles experience the same acceleration.
    ========
    thanks, I found an equation for the solution and can see this is true:

    0 = |D_ACC|^2 * t^3 + 3 * dot(D_ACC, D_VEL) * t^2 + 2 * [ |D_VEL|^2 + dot(D_POS, D_ACC) ] * t + 2 * dot(D_POS, D_VEL)

    where:

    D_ACC = ob1.ACC-obj2.ACC = 0 !!!!

    Fingers crossed. I appreciate the comments. A second pair of eyes is always a big help.

  163. Trick says:

    Tim F 12:20am: “I see no reason to accept your “assumption” that the author suddenly switched notation without the slightest warning to the readers.”

    Just the author’s ref. to the 1995 paper and in eqn. (1) there is a hidden sum m in numerator. But I dunno yet, didn’t see the issue until trying to do real numbers. Here’s one reason found in a clue by putting numbers to Eqn. 8 in 3D as you say shows the issues. Kg-m-J units. Please try it yourself, take a couple minutes, look up own info. – compare to me independently.

    1 N2 m = 4.6 * 10^-26 kg

    Use V&G 2b column where microcanonical T declines 302K to ~215K at ~10,000m.

    Total E = 2 x 10^9 Joules

    Eqn. 8 avg. KE (0) = 2 x 10^9 * (1-0) = 2 * 10^9 J

    With 1 molecule in Eqn. 8 avg. KE (10,000) = 1.99999999999544 x 10^9 J

    This is not a avg. KE w/proportional drop in T of 215/302k ~ 0.712. Here T drops below 300 in 12th digit.

    Put your own numbers in Eqn. 8, headaches happen w/m=1 molecule and total E = huge.

    Now try that with sum m = 7.63 x 10^3 kg. Find 302K goes down to 188K at 10km. Much better result. As you wrote, the author leaves some things to the informed reader I guess.

    Now I would say – going in if right about Velasco – the Velasco 188K vs. Verkley 215K at 10km is that Velasco has not bothered to max. out entropy at LTE – the 188K is just the column’s unequilibrated starting point.

    From which we can learn something else….

  164. br1 says:

    ferd berple:
    “Fingers crossed. I appreciate the comments. A second pair of eyes is always a big help.”

    I’m going to install a Java SDK to have a go at this code myself. It may well have been easier to rewrite it myself, but it is good to have third party code for independence.

    Next thing I note is that the thermal interaction with the heater is all wrong. This is the section starting

    if (!bounce && nix >= heaterLeft && nix <= heaterRight &&

    The equation for y velocity should be
    m.dy=Math.sqrt((-2*kB*heaterTemp/m.mass)*Math.log(getrand(1000)/999.0))

    and for x velocity is
    m.dx=2*(getrand(1)-0.5)*Math.sqrt(2*kB*heaterTemp/m.mass)*erfinv(getrand(1000)/999.0)

    It's not clear to me whether Java has erfinv (which is the 'inverse error function'), maybe you can sort it out. If you can, let me know. The problem with falstad's version is that while it gives a random average energy to the particle, it does so with a non-thermal distribution.The correct equations above are got from Eqn(1) and Eqn(2) of http://polymer.chph.ras.ru/asavin/teplopr/ttk98pre.pdf

  165. ferd berple says:

    br1 says:
    July 17, 2012 at 3:09 pm
    I’m going to install a Java SDK to have a go at this code myself.
    =====
    I’ve got the code working with constant energy (to many decimal places). I’ll post a copy once it stabilizes.

    One problem I’m having is when there are lots of molecules and gravity. The ones on the bottom get trapped below the boundary, because of the weight of molecules above.

    The author’s solution was simply to bring them back up to the boundary, but when I adjust for PE, sometimes the molecule has such low KE, that this would create negative energy.

  166. Tim Folkerts says:

    Trick says: “Use V&G 2b column where microcanonical T declines 302K to ~215K at ~10,000m.”

    You can’t compare the two papers — they are completely different situations. Averaging over the microcanonical states (Velasco) gives the equilibrium values. This would correspond to Verkley 2a (in the limit of N –> infinity)

    Verkley 2b is NOT equilibrium since it includes convection (or it forbids heat transfer between different parts of the column, which also prevents equilibrium).

  167. Trick says:

    Tim F 4:21pm: Velasco 2nd half microcanonical is an enclosed container by paper’s definition. No work allowed outside container on environment, the enthalpy pV term is 0 in Velasco microcanonical (TE=PE+KE+0). Verkley 2a is an open to work on environment container – by Verkley 2a paper defn. enthalpy pV term is not 0 (TE=PE+KE+pV) so Velasco and V2a constraints are not comparable. Verkley 2a container is open and not the same as Graeff closed container.

    Velasco microcanonical 2nd half is Verkley 2b but no entropy maximized in Velasco – both are enclosed container just like Graeff. How many times do I need to write that? You can see this in the detail eqn.s for gas enthalpy. To compute the lapse in Graeff (your “Graeff effect”) use V2b and Velasco microcanonical soln.s. We could get to that once the papers are understood.

    Show me the eqn. in Verkley 2b derivation that forbids heat transfer between different parts of the column. There is no V2b eqn. forbidding internal heat transfer to equilibrium – if there was the entropy could not be maximized to LTE which is done plain as day in the V2b citation.

    I’ve tracked Roman 1995 paper to my local library a couple miles away – will look at that soon. I will get to the theory prediction of Graeff effect shortly, when I understand the Velasco paper ref.s.

  168. br1 says:

    ferd berple:
    “I’ve got the code working with constant energy (to many decimal places). I’ll post a copy once it stabilizes.”

    Just to let you know I also have it working under NetBeans, so whenever you have an update I’ll be ready to run.

    I made all the gravity mods mentioned above, but still get some energy oddities, so I have some work to do there, and I haven’t looked at the heater yet. The main thing I wanted to do for the moment was get temperature vs. height with gravity switched on. I now have that for three sections of the container, divided in thirds, to see if a trend is visible.

    Within the noise of the sim, there is no trend. Whenever you think you see a trend, all you have to do is wait a little while and the trend will reverse, or jumble up between the three sections. Saying that though, the noise is high, often greater than +-10%, so one can’t really say anything definitive from that yet, and the total energy also fluctuates (but less than 1%). The next step is to take averages over multiple updates to get a long term view and bring the noise down.

  169. tchannon says:

    Folks, I suspect you are thrashing things out, not content I can follow. if you think there is reason for an article, speak up. [co-mod]

  170. ferd berple says:

    br1 says:
    July 18, 2012 at 12:32 am
    I made all the gravity mods mentioned above, but still get some energy oddities.
    ======
    I added debug logic to check the energy levels after each manipulation until I got the signs and time right.

    I have 5 energy layers, no obvious leaks and a reasonable consistent negative gradient. The changes are quick and dirty but it appears to be stable. Haven’t done any cleanup.

    http://www.filedropper.com/gassource20120717

  171. ferd berple says:

    tchannon says:
    July 18, 2012 at 2:52 am
    if you think there is reason for an article, speak up. [co-mod]
    ====

    Yes, this source
    Tim Folkerts says:
    July 13, 2012 at 2:26 pm
    You all might look at http://www.falstad.com/gas/

    combined with help from
    br1 says:
    July 16, 2012 at 4:10 pm
    “The code in questions follows these comments in the gas.java source:”

    Has given me what I needed to throw together what appears to be a reasonably good model for investigating gas theory and may well eventually be the basis for an article along with source and a working model anyone can download.
    http://www.filedropper.com/gassource20120717

  172. ferd berple says:

    Thinking on the negative gradient I have one explanation. The average KE for a layer is only computed when a molecule is in the layer. When the layer is empty, the average remains unchanged.

    This seems wrong to me, as it would seem to me to make a layer appear warmer than it really is. There should be an average over time for it to be meaningful, which might change the gradient.

    any comments/suggestions?

  173. ferd berple says:

    For example, when I turn the gravity up far enough, the upper layers rarely see any molecules. Those that make it up there are high energy, but most of the time the layer is empty. It seems like those few high energy molecules skew the result because time is not taken into account.

  174. ferd berple says:

    ps: the zero button resets the e1 through e5 energy level readings.

  175. ferd berple says:

    ps: under some conditions I still see a bit of energy leakage. Likely the code to rescue the occasional low KE molecule that gets pushed out of bounds on the bottom when there are lots of molecules. Still a couple of bugs to be found.

  176. ferd berple says:

    TE on the bottom right of the display shows total energy for the model.

  177. ferd berple says:

    keep the speed down on the simulation and the leaks go away. as the simulation speeds up the time slice gets bigger, and likely less accurate as things get missed.

  178. ferd berple says:

    with two different sized molecules (setup 2) and the speed turned down it looks isothermal.

  179. br1 says:

    ferd berple:
    ” The average KE for a layer is only computed when a molecule is in the layer. When the layer is empty, the average remains unchanged. This seems wrong to me”

    This is ok, it is only the molecules present that contribute. All it means is that the fluctuations will be greater, so you have to wait for longer time for the averages to stabilise.

    I’ve downloaded your code – nice! I particularly like your ‘zero’ button and the long term energy averages. While I’m still playing with it, I have a few comments:

    1, Sometimes I get a gradient, but it is always hot on top and cold on bottom. Notice that E1 is for high altitude and E5 for low altitude. We know this from the code

    if(m.y < winSize.height*1/5){
    totalE1 += m.mass*(m.dx*m.dx+m.dy*m.dy)/2 ;
    countE1++;
    }

    because m.y is upside down, and also if you turn down the number of molecules, you can track which E gets updated according to which altitude the molecule is at.

    2, The gradient is very time step dependent – if you turn down the time step, the profile becomes closer to isothermal. Hence the numerics have not really converged yet, as the solution should not depend on time step. One can't believe the result unless the values have converged with time step.

    3, There are particle timing issues in the code. What I mean by this is that when one counts the position and KE of each molecule, it should be done at equal time steps. Hence the particles should always be at their proper positions at the end of a full time step before they are counted. As it stands at the moment, when the molecule bounces off the wall, you have placed it at the wall (conserving energy) for the next time step. This is not quite correct – one has to also 'send the molecule on its way' for the remainder of the time after impact, because otherwise its position at the next time step is not correct. While this doesn't change total energy, it changes the counting because the counting depends on position. Similarly for particle-particle collisions, you seem to have deleted the 'send m on its way' section. This needs to be there, as does an equivalent 'send m2 on its way', or again the counting is off because the positions are off. Because there are more collisions at low altitude, whatever effect this has will be greater at low altitude and less at high altitude, so may give a false gradient. This is an error which will depend on time step value – the smaller the time step, the less error there should be, which fits with comment 2 above. I might get a chance to deal with this, but feel free to beat me to it!

  180. ferd berple says:

    1, Sometimes I get a gradient, but it is always hot on top and cold on bottom. Notice that E1 is for high altitude and E5 for low altitude.
    ===========
    Yes, that is the reverse gradient I have been seeing.

    2, The gradient is very time step dependent – if you turn down the time step, the profile becomes closer to isotherma
    =========
    The simulation is less precise as you increase the speed. It isn’t the display that slows down, it is the time slice that changes. need faster computers to solve this one.

    3, There are particle timing issues in the code.
    =========
    I’ve moved the first 1/3 of updateGas (the move code) to the end of updateGas to solve this. This revealed a bug in the collision code. I was using r instead of m.r and m2.r to work out if the collision was forbidden (due to bounce within time t), which explains the energy leakage when using different sized molecules. Fingers crossed no more energy leaks. So far rock solid even with everything cranked up to the stops.

    4. I’ve added some code to properly scale the histogram in updateHistogram

    you need to replace
    double maxke = energy ? 70 : 15;

    with an actual value by inspecting all the molecules. Makes a big difference. Everything shows MB distribution.

    I’ve also changed some defaults to make more display available, less controls. I’ll let it run for awhile to see if it stable then publish.

  181. Trick says:

    Tim F 7/17 12:20am –

    “Trick says: ” … so that in Eqn. 8 find m(z)gz is justified where m(z) = N*m in that volume element … ”. Unless you can show the specific mathematical steps deriving Eqn 8 from the previous equations…”

    Ok, my local college librarian printed out Roman, Velasco et. al. 1995 for me. It shows the momentum of a particle with mass m is defined to be summed over the interval 0 to total E in 1995 eqn. (4). So m (mass of a particle) is to be multiplied by N “identical structureless” particles when N .GT. 1.

    Indeed, the paper even gives examples for N = 3 up to 6 using (N*m)*g*z (eqn. (56)).

    For consistent container constraints:

    1) Roman, Velasco 1995 paper in eqn. (1) sets the gas enthalpy term pV = 0 inside the “vessel”. By reference then, Velasco 1996 microcanonical is also a pV=0 rigid “vessel”.
    2) V&G 2b sets enthalpy term pV = 0 in B&A cartoon Fig. 4.1b showing a rigid “atmosphere” column.
    3) Graeff’s dewar B74 can be considered rigid relative to T&P fields of interest in dewar V.

    All constraints 1,2,3 above are consistent with pV=0 for no external work. This rules out V2a with pV not= 0 as the V2a air column does work on the air column above and below; it is this work that allows the classic soln. to be isothermal & which Verkley is generally showing us. Even with a specific numerical example.

    Can y’all move on to get an agreed theory solution for Graeff’s container (Tim F’s “the Graeff effect”) now after all this?

    Maybe even the computational work of br1 and ferd eventually can gain confidence to do so. Though IMO they will need a CFD program consistent with & implementing ideal gas law, Law 1 & Law 2 theory.

  182. br1 says:

    ferd berple:
    “new source”

    I downloaded this but still didn’t like the way the collision detection worked, so I put back in the ‘send m on its way’ lines, turned down the time step slider sensitivity, and added some extra output numbers: instantaneous gravitational potential energy (correcting for being ‘the right way up’ this time!!! (ferd will know what I mean)), and long term averages of how many particles were in each section.

    With 500 particles and gravity turned up so that there was a particle number ratio of about 15:1 between the lowest and highest altitude slices, I ran the prog for about ten minutes (after waiting for an initial time before pressing ‘zero’), and ended up with the following ‘temperature’ profile:

    high altitude
    105.3
    103.1
    105.6
    105.5
    105.6
    low altitude

    So approximately isothermal.

    We can compare this to Velasco. Note that the max PE for any particle was about 428 units (checked by turning particle number down to 1), and the total system energy was 100,000 in the run under discussion, then the temperature at the highest altitude should be about (1-428/100000)=0.996 of the temperature at ground level. This corresponds to about a 0.4 temperature drop for the above run, or just less than 0.1 per level. This is also consistent with the simulation which showed 0.3 difference top to bottom, and there was still some noise being settled out.

    Next to note is that a particle number ratio of 1:15 with altitude corresponds to an atmospheric height of 21.7 km, which by my eyeball corresponds to a 22% reduction in temperature. This is obviously very different than the sim result, and is incompatible with it. Roderich’s gradient is even higher than atmospheric, so that is also incompatible.

    Ferd: what do you get?

  183. ferd berple says:

    Ferd: what do you get?
    ========

    Isothermal so far, within the limits of the drift of the model. The smaller the molecules, the more varied the molecules the more isothermal.

    Which in some respects may be exciting, because if Graeff’s work can be replicated it would mean it likely represents new science.

    My surface layer is consistently slightly cooler than the other layers, which makes me think there is a bug or two. I’m a bit surprised as I show no energy gain/loss.

  184. br1 says:

    ferd berple:
    “My surface layer is consistently slightly cooler than the other layers, which makes me think there is a bug or two. I’m a bit surprised as I show no energy gain/loss”

    Well I was trying to explain that we not only need the correct energy but also correct position, as we are trying to estimate KE vs position (=altitude). When I put back in the ‘send m (and m2) on it’s way’ lines, my lower layer has the same temperature as the other layers. One also needs a small time step (I reduced the slider step), and also to wait a good length of time before pressing the ‘zero’ button. Note there is still a very minor glitch in the collision routine, as at small time step (and setting all the colours to white) one can definitely see rare position jumps in high density areas. You may also be seeing the glitch when the density is high and a particle is forced below the bottom boundary and can’t be normalised properly.

    But yes, the general result is as close to isothermal as makes no difference.

  185. br1 says:

    “But yes, the general result is as close to isothermal as makes no difference.”

    So now we’ve reproduced the classical result, there are lots of interesting experiments that can be set up.

    The first thing I not is that the isothermal solution doesn’t depend on initial conditions – you can start the sim far from equilibrium, yet after a while all the velocities get randomised and goes isothermal. This is important to keep in mind. The following are the three most interesting simulation experiments to look for gradients:

    1, Have a fan in the bottom right hand corner. This can be implemented by checking for collisions against the bottom right hand section of the floor (say the rightmost fifth of the floor). Instead of simply reflecting, as the sim does at the moment, have the particle reflect and then add a velocity boost +v0. This represents an impact with a fan blade. As this only occurs at the bottom right of the system, there should be a convection cell set up which rotates counter-clockwise. The convection should give a temperature gradient with altitude. As the velocity boost also represents an increase in KE, then there should be a slow temperature rise of the system as a whole.

    2, Instead of a fan, have a heater at the bottom right of the floor, and add a cooler at the bottom left edge of the floor. One heater by itself will preserve an isothermal profile, but a heater and cooler should give a temperature profile. Obviously this will be true right-to-left due to heater position, but here we are interested in a profile with altitude. As both heaters are placed on the floor it is not immediately obvious that there will be a temperature gradient with altitude. The heaters should set up convection, so from the convection one expects a gradient with altitude.

    3, Include a large particle in the sim for Brownian motion. While the sim already includes one, we should reduce its mass so that it is more like a balloon which floats quite easily. As this large object pushes the little particles out of the way, it may set up spontaneous convection. So one may get temporary convection currents which give temporary temperature gradients, but because each temporary convection current should have a negative temperature profile with height, then maybe one can get a long term negative gradient.

    Note that experiments 1 and 2 require energy supplies, so even if they produce gradients there are heat flows so do not represent thermal equilibrium. It will be interesting to see when those supplies are turned off that the system will revert to being isothermal.

    Experiment 3 is perhaps the most interesting in the sense that if a long term gradient can be established, then this could be what we are really interested in. The prediction therefore is that there will be no noticeable temperature gradient. I would love to be wrong on this one!!!

  186. br1 says:

    This should be fun! I can also see how the circulations work up to a certain altitude, so above that altitude one should get an isothermal result. Yes – we can reproduce the troposphere and thermosphere with this little model!

    Won’t get any time for this today, but I’m looking forward to the weekend 🙂

  187. Trick says:

    ferd, br1 – What is the gas pressure taken from the sim at the bottom, z=0?

  188. ferd berple says:

    Ok, I’ve uploaded a new source with the recommended changes plus a bunch of code cleanup.

    http://www.filedropper.com/gassource20120719

    the energy is stable. the negative energy boundary problem appears solved. collisions are sent on their way. time slice 10 times smaller. 6 layers of atmosphere. new “counts” button showing sampling counts/ratios.

    I’ll need to add more code to get pressures at various layers. The average pressure in my current run (short time, gravity turned to the max, 500 large molecules) is 2500. Temp is 1070. Volume 366. All layers are +- 30 degrees of this with no pattern apparent, except the bottom layer which today is running consistently hot (as opposed to cold yesterday).

    Ratio in counts between layers is about 1.9, which I take to mean there is 1.9^5 pressure difference top to bottom, but I could be wrong on this. checking counts top layer has 100k and the bottom layer has 2200k molecule samples in the current run, which is a difference of 22 in pressure.

    total energy is 1.7 million. isothermal as far as I can see. One proviso. I can see that the sim is very sensitive to small errors, but I haven’t yet seen anything to indicate a large positive gradient.

  189. Trick says:

    ferd 4:08pm: N=500, p=2500, T=1070, V=366 results in the sim having units w/Boltzmann’s constant at 1.71 where in standard units Planck and Boltzmann published 1.38 * 10^-23 Joules/Kelvin for the constant.

    Boltzmann’s constant is not a universal constant (like the speed of light) – it is simply a unit conversion factor. Can the sim work with standard units such that Boltzmann’s constant is obtained?

  190. ferd berple says:

    here are the formulas from the code
    m = molecule

    totalKE += m.mass*(m.dx*m.dx+m.dy*m.dy)*.5;
    temp = (2./3)*totalKE/molCount;

    wallF calculated when molecule impacts boundary:

    wallF += Math.abs(m.dx)*m.mass;
    wallF += Math.abs(m.dy)*m.mass;
    a = 2*(winSize.width+(winSize.height-upperBound)-4);
    pressure = 1e4 * wallF/(3*a);

    vadj = 4e-4;
    volume = (winSize.width-2)*(winSize.height-upperBound-2)*vadj;

  191. ferd berple says:

    I’ve uploaded an updated gas simulation anyone should be able to run.

    Unzip and click index.html. You may have to enable java in your browser.

    http://www.filedropper.com/gas20120720

    details: contains latest source plus an updated gas.jar with the compiled code. Edit index.html with wordpad to optimize the display for your screen.

    Edit the above line to change the height and width to best suit your display.

  192. br1 says:

    Hi ferd,

    Good stuff!

    I have more good news – I can now fix the heater (which is not a proper thermal source in the present implementation). I haven’t tried it in java yet, so will let you know once I have it debugged. Physically correct heaters will allow real temperature gradients to be measured under convection, but incorrect heaters can cause invalid gradients due to bad velocity distributions (which is the way it stands at the moment).

    I’m not sure the best way to get the updated code to you – a seven day trial of filedropper might work, but longer term I’ve set up an email account

    filetransferxyz
    at
    gmail
    dot
    com

    drop me an email and I’ll attach it back to you.

  193. Q. Daniels says:

    br1 wrote:
    high altitude
    105.3
    103.1
    105.6
    105.5
    105.6
    low altitude

    Is the 103.1 persistent?

    I won’t have the time to give this attention it deserves for a couple days. I’m curious about this, though. I can see a vague pattern, where the 103.1 would be real and persistent.

  194. br1 says:

    Q. Daniels:
    “Is the 103.1 persistent?”

    not really. The lower three numbers were quite stable, but the upper two had tendencies to drift, due to the smaller particle count at higher altitude. It was intriguing how even after ten minutes accumulating a long term average, the values could suddenly start to wander off again. So to be honest the high altitude value of 105.3 was prone to drift too, but there is nothing comparable to a DALR present. You can see this for yourself, to your heart’s content, now that ferd has posted an easily executable version.

  195. wayne says:

    ferd berple, I have your latest copy of the code installed, it does compile and runs fine. However, with gravity ‘on’ full there is a marked temperature gradient, about 32 at the bottom and 7 at top after a few minutes but all temperatures keeps growing at all levels very slowly, the apparent ratios level to level seem to stay constant. Were you aware of this? Know which area to start looking for it first? (I just don’t want to start at level zero looking for it myself if possible since you have spent some time decyphering the flow :))

  196. ferd berple says:

    ferd berple, I have your latest copy of the code installed, it does compile and runs fine. However, with gravity ‘on’ full there is a marked temperature gradient, about 32 at the bottom and 7
    ============
    Let the sim run for awhile, then press the “zero” or “reset”.

    The values T0-T5 and C0-C5 are accumulators that keep increasing since the last time you press “zero”. Reset presses zero for you, and reloads using current settings.

    This can cause misleading results if the sim is transitioning between states. If you change gravity, or anything else, press the zero button after the sim stabilizes.

    If you still get the gradient, that would be significant/ For my part I get hot at the top and cold at the bottom – the opposite of Graeff – which I’m calling a negative gradient.

  197. ferd berple says:

    br1 says:
    July 20, 2012 at 5:17 pm
    I’m not sure the best way to get the updated code to you – a seven day trial of filedropper
    ==
    filedrop works for me. I’ll monitor this post. I don’t monitor my email as it is a junk cutout.

    Something else left to correct is the timing at the boundaries. The code corrects the energy, but not the time. My thought was this simulates a rough boundary, but perhaps it does something to the temps as well. If you have a suggestion for a time correction at the boundary I’ll code it in.

    I’m still seeing a negative gradient.

    57.1
    55.5
    53.8
    51.7
    49.4
    47.0

    setup 1. Gravity 3/4, sim speed 1/6, 500 mols.

    It wanders but the reverse gradient is there. Anyone else getting this?

    Also, anyone tried the latest gas.jar via index.html?

  198. ferd berple says:

    ps: the results in my last post are from the gas.jar version running under IE.

  199. ferd berple says:

    my previous might have been confusing:

    This (the accumulators) can cause misleading results if the sim is transitioning between states.

    – because you could have a large amount of accumulated data without gravity skewing the gravity results, or vice-verse. the accumulators should probably zero automatically when you change gravity or the heater.

  200. Q. Daniels says:

    I ran ferd’s model, with some modifications:
    1) Increased the max # of molecules. This should be run with small molecules, or the MFP becomes too short.
    2) Reduced gravity by a factor of 10
    3) Increased the number of digits displayed to 6

    My times are not very accurate.

    Results for n=500 speed=min+2, gravity=1/10th ferd’s min, integrating from 1 to 15 minutes.
    T0=3.57
    T1=3.46
    T2=3.44
    T3=3.42
    T4=3.40
    T5=3.35
    A negative temperature gradient. The gradient is lower with more molecules.

    I reset the counters at 15 minutes. I observe that after 15 minutes, there are still thermal waves sloshing around, even with just 500 particles. I can’t see them directly, I can only watch the temperature profile swing. At 30 minutes from start of run, a slight negative gradient is observed.

    After ~45 minutes, profile is:
    T0=3.429
    T1=3.413
    T2=3.429
    T3=3.470
    T4=3.447
    T5=3.422

    I make that as a slight positive gradient, with some wall effects at the top and bottom.

    I’m considering turning the gravity down further, and letting it run substantially longer.

  201. Q. Daniels says:

    My previous comment was at least partially incorrect. I tried with 1% of ferd’s initial minimum gravity and n=500, and a negative gradient of about 5% established itself fairly quickly.

    With n=4999, I do get a slight positive gradient, on the order of 1%, which appears to be consistently higher than the density gradient.

    I’m contemplating instrumentation to analyze the patterns of movement.

    Of course, this matches my bias, so caveat emptor.

  202. Q. Daniels says:

    Ah ha!

    Gravity = 0.0001 ferds and n=2500 is exhibiting what I was looking for. I’ll instrument it, and send the updated code to br1.

    Good job, ferd and br1.

  203. Q. Daniels says:

    I made the following changes:
    1) Reduced the gravity by 1000x.
    2) Increased the max particles to 5000.
    3) Increased the number of digits displayed to 6
    4) Added instrumentation. A new graph shows a BFL temperature gradient, averaged over 20 cycles (cyc_len). The green line is 0 gradient.

    For best results, I run at minimum gravity (0.001 ferds), and n>= 2500.

    The new graph shows a nice chaotic oscillator, with some pretty significant swings in gradient. The amplitude varies over time, but doesn’t appear to stabilize. The overall average gradient is not isothermal.

  204. ferd berple says:

    Q. Daniels says:
    July 21, 2012 at 8:33 am
    Gravity = 0.0001 ferds and n=2500 is exhibiting what I was looking for. I’ll instrument it, and send the updated code to br1.
    =========
    I’ve added code to correct for time at the boundaries:

    t = (dist2-dist1)/(vel2-vel1)
    then used this to send the molecules on their way.

    If you’d like to file drop your source I’m keen to add your changes as well, especially any new instrumentation. Probably via a new checkbox then I’ll upload a new gas.jar with source so anyone can run.

    I think there is a huge advantage to circulate the source and verify with many eyes. The experience so far on this project points to closed, unverified source as a likely source of error in climate models. If I write this up, I think that will be the slant. The accuracy improvement due to collaborative open source.

  205. Q. Daniels says:

    I think I’ve dropped it at:
    http://www.bitupload.com/afdt33rl13eo

    After running it for ~8 hours, it’s still showing oscillations of over 5% of temperature.

  206. ferdberple says:

    New jar with updated source. Still seeing negative gradient. Which is interesting. One possible explantin is that GHG actually cools the atmosphere, not warms it. Which explains the hot thermosphere where GHG is rare.

    Added time correction at boundaries and increased molecules to 2000. Made the speed and gravity controls exponential to increase their range. Decreased the low range and increased the high. Changed the grid which appears to smooth the motion.

    http://www.filedropper.com/gas20120721

    Also fixed a few bugs. On occasion the code was throwing NAN’s where there were lots of collisions.

  207. ferdberple says:

    still getting a gradient

    2000 molecules, all same size, low gravity, low speed, 10 hours. no energy change. 56,699.

    top diff
    9.91
    9.77 0.14
    9.66 0.11
    9.53 0.13
    9.41 0.12
    9.3 0.11
    bottom

    This is a pretty consistent difference between the altitudes.

    Which is actually significant because is begs the question, why is the atmosphere cold at altitude?

  208. ferdberple says:

    Here is another simulation at higher gravity, same run time, same pattern

    top diff
    496.284
    487.18 9.104
    476.07 11.11
    461.73 14.34
    450.45 11.28
    439.46 10.99
    bottom

  209. Q. Daniels says:

    ferd,

    I think you’re looking at very high gravity for the scale. With a ~7% pressure gradient, I get a ~2% temperature gradient.

    For the model, at STP, we’ve got the a vertical scale equivalent of much less than a micron. Getting even a 7% pressure gradient across a micron takes a lot of gravity.

    With ~1% pressure gradient, it may be showing the opposite temperature gradient.

  210. Q. Daniels says:

    Another change I made, this time in the html: width=1200 height=1000

    That gives a gas canvas of 900^2, which helps a lot.

    For n=4999, slow simulation, low gravity, after 4 hours:
    Temp-n Cn/Cn-1
    0.9069 E=6942.7
    0.9080 1.0076
    0.9122 1.0002
    0.9129 1.0030
    0.9117 1.0058
    0.9104 1.0006* (Raw is 0.99389, but last layer is 149? tall instead of 150)

    I instrumented the temperature gradient on horizontal slices as well, and the variation appears slightly greater than vertical.

  211. br1 says:

    Hi guys,

    Before claiming a gradient can I make one request – please reduce the time step by at least a factor of two, and zero the sim again. The code has known issues with time step, and a result is only meaningful if it gives te same result at different time steps.

    On my PC, this makes the sim run at ‘watching paint dry’ speeds, (I also reduced the time step slider sensitivity) but the results are always isothermal!

    I’ve implemented and debuged the heater, added differential heating, and added a fan. These give interesting results but not what I expected! I nearly always get hot on top and cold on bottom (converged in time step), and I can see the very obvious reason why! Will have to write that up soon.

  212. Q. Daniels says:

    br1,

    Ah, yea. I turned down the sim speed by a factor of 10. Now the low end speed seems to work better. I’ll give a report after the paint finishes drying.

  213. ferd berple says:

    br1 says:
    July 22, 2012 at 9:01 am
    Hi guys,
    Before claiming a gradient can I make one request – please reduce the time step by at least a factor of two, and zero the sim again.
    ===========
    The current code is 100 times more sensitive on time slice over the original on the low end setting. Adjusting the grid size appears to have resolved the the “jumpiness” that was sometimes apparent in collisions.

    The previous code did not correct for time at the boundaries, which could be argued calls the previous model runs into question.

    Also, the collision code should probably also ensure that both momentum and energy is conserved. It appears to only solve for momentum. This also calls the results into question.

  214. ferd berple says:

    There is a fluctuation in total energy around 9 decimal places as a result of collisions.

  215. ferd berple says:

    correction. the fluctuation is there with a single molecule, no collision. unless there is an error in the time slice or motion calculations, this must reflect the precision of type double.

  216. ferd berple says:

    http://www.filedropper.com/gas20120722

    updated source – time correction was 2x. increased grid resolution. turned on gravity by default. increased number of decimals for energy reading to better detect drift.

  217. ferd berple says:

    I’m using the default sim on the latest 20120722 source, running in a browser from index.html, one sim per core. no changes to defaults/sliders. energy looks stable short term to 8-9 decimals.

  218. ferd berple says:

    I’m wondering if it might be possible to add conduction to the sim? By changing the radius of the molecules as their kinetic energy changes? While leaving the mass unchanged? Or some other technique?

  219. br1 says:

    ferd berple:
    “I’m wondering if it might be possible to add conduction to the sim?”

    It’s already there, the collisions of molecules do that. The only thing fundamentally missing is radiation (and the fact this is a 2D sim, not a 3D sim).

    I didn’t get as much time as I would have liked to explore convection, but I must say it is rather hard to get anything you would really call convection. It is more like diffusive behaviour. After playing with the model, I would expect convection is only really visible at much larger particle number. I’ll add a few instruments such as integrated x velocities (to show the sideways motion of convection at top and bottom of the loop), which I’m sure will give non-zero answers, but I’m not sure when the description crosses over from diffusive to convective.

    Anyway, when adding in two heaters of different temperature on the bottom plane of the sim, which gives a hot column and a cold column, I found the average temperature increases with height in a very noticeable manner. The reason is simple – it is given by the barometric formula. Density reduces as exp(-mgh/kT), and if there were only one heater then the temperature would be isothermal. If you take two adjacent columns at different temperature, then the colder one has much fewer particles at high altitude. Taking an average at each altitude, there are approximately equal numbers of ‘hot’ and ‘cold’ molecules at low altitude, but much more ‘hot’ molecules than ‘cold’ at higher altitudes. Even if each column remained isothermal, the mix would vary with height, and so give a gradient which is hot on top and cold on bottom.

    Because there are more ‘hot’ than ‘cold’ molecules at higher altitude, there is noticeable diffusion of particles from the hot column to the cold column at high altitude, and I guess a diffusion of cold particles from the cold column to the hot column at low altitude (though that is harder to see). This does set up a circulation of sorts, but a surprisingly weak one which is hardly observable in the sim. The thermal contact with the floor is surisingly good at randomising the flow – the next thing I’ll do is mix in reflective collisions with thermal collisions, maybe this will help ‘lubricate’ the system.

    To get a DALR I feel requires a heck of a lot more particles in the sim.

  220. br1 says:

    I should also add that if I switch off the heaters, the system reverts to isothermal fairly quickly.

  221. Bryan says:

    br1 says;

    “I didn’t get as much time as I would have liked to explore convection, but I must say it is rather hard to get anything you would really call convection.”

    For convection do we not require a certain ‘stickiness’ between the gas particles?
    This force at a distance between the molecules is ignored in basic kinetic theory.
    However it seems to me to be necessary to explain parcel theory.
    Perhaps a modification along Van Der Waals equation is necessary?
    This must be very difficult to implement!
    At higher density the effect would really kick in.

    All readers are in admiration of the work br1, fred,Q. Daniels and others are putting in.

  222. br1 says:

    Bryan:
    “For convection do we not require a certain ‘stickiness’ between the gas particles?”
    good question. My intuition says no we don’t need it, but these days I’ve learned to not trust my intuition until backed by a good simulation! So until I can simulate convection in a clear manner, I won’t say for sure. At the very least I expect to see something like http://en.wikipedia.org/wiki/Granular_convection (without the Brazil nuts!) which should apply to non-attractive objects.

    “This force at a distance between the molecules is ignored in basic kinetic theory.
    However it seems to me to be necessary to explain parcel theory.”
    Parcel theory is another subject I don’t really trust my intuition on. While it looks simple from a distance, the more you think about ‘what’s a parcel’, the less obvious it gets (to me anyway). I suspect that to have something called a parcel, one has to have a uniform property over a large enough volume (and number of molecules) that a molecule in the middle of the parcel cannot get to the edge of the parcel in the amount of time it takes for the parcel to do something. Even that is not clear to me. But for a mini-model such as we have, can we ever have a ‘parcel’ moving around? I don’t know, somehow I doubt it.

    “Perhaps a modification along Van Der Waals equation is necessary?
    This must be very difficult to implement!”
    I implemented a van der Waals model with both finite particle size and interparticle attraction in a 1D gas. Even this was tricky, as the model suffered definite ‘energy leaks’ that I never fully got rid of. As the attractive force is nonlinear with distance, and involves lots of molecules, modeling a particle-particle bounce in a finite time step is quite challenging. FWIW, the model came out isothermal, but it could definitely do with more attention.

  223. ferd berple says:

    http://www.filedropper.com/gas20120723

    new jar with source that initializes gravity to positive value. previous looked like it was setting gravity, but wasn’t.

  224. ferd berple says:

    br1 says:
    July 23, 2012 at 11:04 am
    It’s already there, the collisions of molecules do that. The only thing fundamentally missing is radiation (and the fact this is a 2D sim, not a 3D sim).
    ===========
    If this is a non-radiative gas such as N2/O2, then according to GHG theory radiation will not be present. Which if a column of air is isothermal, means that GHG must be cooling the atmosphere at altitude, not warming it.

    It seems to me there is another affect we are missing.

    A large rising/falling column of air, once it gets going, should be able to carry molecules within it, sort of like the flow of a river, only vertically. The molecules within this flow will be traveling faster vertically than would normally be the case.

    As such, since they will spend less time in the influence of G, they will not lose/grain as much PE as would be expected for the distance traveled. This would have the effect of increasing/decreasing temperature between the layers, which might cause a gradient.

    It could be that a heater at the bottom of the sim, hot at one end and cold at the other, sort of like the earth’s surface, so that there is no net change in energy might simulate this situation.

  225. br1 says:

    I wrote:
    “The thermal contact with the floor is surisingly good at randomising the flow”

    Indeed, this behaviour is basically that of a boundary layer: http://en.wikipedia.org/wiki/No-slip_boundary_condition

    What happens for a 100% thermal wall is that the molecule deposits all its KE at the wall, then picks up a new random velocity associated with the wall temperature. As the random thermal velocity has (by definition) equal probability of being in any direction, this is independent of any flow or stream there may be in the gas above the wall, and the net gas velocity travelling along the wall will be zero. As the gas molecules leave the surface, they will carry this ‘no-flow’ information into any stream that may be above, giving the gas some ‘viscosity’ in regions near the boundary, and impeding flows parallel to the boundary. Without some power supply to continually drive the flow, there will therefore be no net flow or circulation. Even if a power supply has established a flow, once the supply is turned off then the flow will go to zero, due to the no-flow thermal interaction with the walls.

    Hey, this sim is great!

    As I don’t want a stationary boundary layer, I’ll turn down the thermal interaction, and make most of the wall interaction reflective. This should provide for slip, and may make flows easier to come by. It won’t affect the no-flow equilibrium condition, but it should help the power sources establish a flow more easily.

  226. br1 says:

    This then answers a long-standing question I had – if you have given a gas (or liquid) a vortex motion, and according to the conservation of angular momentum this cannot be destroyed, then will it in fact continue or not?

    The clear answer here is that it will NOT continue, in spite of conservation of angular momentum. The reason is that the thermal interaction with the walls destroys the angular momentum internally in the container, as it is passed to an infinite external reservoir.

    Even if the vortex is well away from the walls, in the ‘middle’ of a very large container, there will always be some mixing propagating towards the edges, eventually the thermal boundaries will have their way and all net motion will be shut down. Given long enough time, every molecule will interact with the wall and pick up a ‘no-flow’ thermal velocity.

    So there can be no persistent macroscopic flows in a container using the canonical or grand-canonical ensemble. The situation is different for the micro-canonical ensemble, but they don’t really exist. The situation is also different when one has two or more heat source interactions, but then this involves heat flow which must eventually run out. Ho hum.

  227. Q. Daniels says:

    br1 wrote:
    “Perhaps a modification along Van Der Waals equation is necessary?
    This must be very difficult to implement!”
    I implemented a van der Waals model with both finite particle size and interparticle attraction in a 1D gas. Even this was tricky, as the model suffered definite ‘energy leaks’ that I never fully got rid of. As the attractive force is nonlinear with distance, and involves lots of molecules, modeling a particle-particle bounce in a finite time step is quite challenging. FWIW, the model came out isothermal, but it could definitely do with more attention.

    I have Lance Olav Eastgate’s excellent CS Thesis and source code, which implements this. I can pass the code along later.

    We can switch to this at some point. It’s a lot slower.

    For the current one, we should do a code merge. Could I get your source? I’ve got ferd’s already. Maybe after lunch and some day job I’ll do a merge.

  228. tjfolkerts says:

    Now that you all have done all these modifications/improvements to the original simulation, has any thought to contact the original author/programmer to suggest reading thru these comments and including some of these tweaks to the code?

  229. ferd berple says:

    I continue to see a negative gradient. It does not go away as the speed decreases.

    I’m seeing little if any speed sensitivity in the current code, so long as the speed is low enough that the energy E= remains constant.

    As you turn up the gravity and/or decrease the number of molecules the motion becomes so large that the current code cannot correct for it and the total energy becomes unstable. The next release will likely correct for this.

  230. Q. Daniels says:

    My latest code:
    http://www.bitupload.com/pt9pdhe6fcmf

    It adds tracking for horizontal fluctuations.

  231. Q. Daniels says:

    I merged my code with ferd berple’s 20120723.
    http://www.bitupload.com/a0uhelc2pvw7

    Two things I didn’t change:
    1) I left maxMolCount at 5000, instead of 2000
    2) I left a linear gravity, with a lower minimum value. The region I find interesting isn’t all that wide.

    Also, I’ve changed the html file to (almost) square the window.

  232. Frank de Jong says:

    Two questions / remarks for the modelers.

    1) The original falstad.com version has a closed top (as in Graeff’s experiments). To get true atmospheric behaviour, I guess the top should be open (and gravity turned up a lot so that you don’t have to make the container 100 miles high). Did anyone try this?

    2) Graeff uses a factor, related to the degrees of freedom of the gas molecules. As I understand it, you are using spherical particles in 2D only, so you have only 2 degrees of freedom for the X and Y movement. What would be the influence of introducing a 2-sphere model (so as to model N2 and O2) so that there is a third, rotational degree of freedom?. I can imagine the collisions become more complex to model, but the introduction of rotational energy might have an effect because it will be more pronounced for higher pressures (more collisions) than lower.

    Thanks for your time.

    Frank

  233. Q. Daniels says:

    Frank de Jong wrote:

    1) The original falstad.com version has a closed top (as in Graeff’s experiments). To get true atmospheric behaviour, I guess the top should be open (and gravity turned up a lot so that you don’t have to make the container 100 miles high). Did anyone try this?

    As best I can tell, the behavior in this case is very different from Graeff’s. The results ferd and I get are that it develops a strong temperature gradient, with the top being hotter. I suspect it relates to the change in density over the mean free path.

    2) Graeff uses a factor, related to the degrees of freedom of the gas molecules. As I understand it, you are using spherical particles in 2D only, so you have only 2 degrees of freedom for the X and Y movement. What would be the influence of introducing a 2-sphere model (so as to model N2 and O2) so that there is a third, rotational degree of freedom?. I can imagine the collisions become more complex to model, but the introduction of rotational energy might have an effect because it will be more pronounced for higher pressures (more collisions) than lower.

    I take this to mean two linked spheres, such that they can have rotation. This is an interesting suggestion. It shouldn’t have a qualitative effect, but definitely a quantitative one.

  234. br1 says:

    Q. Daniels:
    “I have Lance Olav Eastgate’s excellent CS Thesis and source code, which implements this. I can pass the code along later.”

    I found the thesis here: http://lanceolav.com/research/thesis_candscient.pdf

    A nice piece of work, including particle-particle attraction. I paid particular attention to his thermal walls discussion, and was relieved that I had done it correctly. I’ll get my code to you tomorrow.

  235. ferd berple says:

    http://www.filedropper.com/gas20120724

    I’ve updated further fixes. The previous code was not correcting for the second particle in a collision being out of bounds as a result of a collision. I’ve added a general case fix for this problem and extended it to other areas that were giving problems.

    I’ve also added code to automatically zero the accumulators and dial back the simulation speed in case the code detects an error in the simulation. This can happen as temperatures increase and the motion of the particles becomes large.

  236. ferd berple says:

    the new code also automatically zero’s the accumulators if you change the model parameters that affect energy.

  237. ferd berple says:

    Frank de Jong says:
    July 24, 2012 at 10:18 am
    1) To get true atmospheric behaviour, I guess the top should be open (and gravity turned up a lot so that you don’t have to make the container 100 miles high). Did anyone try this?
    ============
    Download

    http://www.filedropper.com/gas20120724

    unzip and double click index.html.

    Select small molecules and with low gravity, press reset. Wait for the molecules to fall such that most are in the lower half, then quickly turn the gravity to max. The molecules will rarely if ever reach the top of the container.

    Let the sim run for awhile and press zero. Please report back with any observed results.

  238. ferd berple says:

    http://www.filedropper.com/gas20120724_1

    Here is the same code with an option to remove the walls.

  239. ferd berple says:

    Q. Daniels says:
    July 24, 2012 at 10:35 am
    As best I can tell, the behavior in this case is very different from Graeff’s. The results ferd and I get are that it develops a strong temperature gradient, with the top being hotter
    ========
    I continue to get this gradient. I’ve searched the sim for errors and corrected both for energy and time, but still this gradient persists. I’m beginning to suspect it is not an artifact of the code.

    I’ve produced the version without any walls to investigate this question, to see if this has any effect.

  240. Q. Daniels says:

    ferd,

    Can you retrieve the merged code I made?

    With ~4k molecules, and gravity 1 step above minimum, I don’t get that gradient any more.

  241. ferd berple says:

    Q. Daniels says:
    July 24, 2012 at 7:26 pm
    ferd,
    Can you retrieve the merged code I made?
    With ~4k molecules, and gravity 1 step above minimum, I don’t get that gradient any more.
    +++++++++++
    Yes, I had a look. The version you sent, gravity is so low that it is likely noise will overwhelm any gradient.

    You can tell this by looking at the ratio’s on the right. C1/C0, etc. They should all be well in excess of 1. I normally run 1.5 and above.

    As I increase gravity, I see a gradient. Hot on top.

  242. Q. Daniels says:

    ferd berple wrote:
    The version you sent, gravity is so low that it is likely noise will overwhelm any gradient.

    Yes. That’s the point of my inquiry right now. How does the system behave when noise overwhelms the gradient, yet gravity is still present?

    You’re welcome to switch it back if you want to explore other regions. I called it out so you would be able to switch it easily.

    More important was that I merged the changes we’ve made.

  243. ferd berple says:

    http://www.filedropper.com/gas20120725

    new “no walls” version with “infinite” sky. Display only shows bottom portion of sky, but it extends upwards quite a distance. Collisions should still take place above the upper boundary, only we can’t see them. The 6 temp layers only cover the visible part of the display.

  244. Q. Daniels says:

    ferd,

    Could you drop me a line at q_daniels_t-2l@yahoo.com ?

  245. br1 says:

    Hi ferd,

    I downloaded your latest http://www.filedropper.com/gas20120725

    However, the ‘send it on its way’ at the wall collisions are not correct. The equations on lines 666, 691 and 722 are all of the form

    double t = (m.y-oy)/(m.dy-ody);

    but they should all be

    double t = (m.y-oy)/(m.dy);

    Note that ody is not needed, as otherwise the particle seems to go at twice the speed it is going at, and the particle covers the distance m.y-oy only once.

    Also note that lines 691 and 722 should ideally be quadratic equations, but I’m happy to ignore that for the moment. I solved for that in my 1D simulations, but I think the effect will be very minor. We can come back to it if the above improvements don’t make your wrong-way-up gradient go away. I can’t check it at the moment, but will later.

  246. ferd berple says:

    br1 says:
    July 25, 2012 at 3:11 pm
    double t = (m.y-oy)/(m.dy-ody);
    ========
    I’ve posted a new source which includes the noWalls option, plus a correction so that collisions take place off the display screen. This latest also corrects for a weakness in the grid logic.

    http://www.filedropper.com/gas20120725_2

    You appear to have an older source. “Send it on its way” was corrected for the doubling of time in a previous release.

    It reads as follows:

    (in x. direction:)

    double t = (m.x-ox)/((m.dx-odx)*.5);

    You are correct, this is redundant, as there is no acceleration in the x direction and this can be simplified (since m.x = -odx) as

    double t = (m.x-ox)/(m.dx); (change in above source)

    However, (in y. direction:)

    “Also note that lines 691 and 722 should ideally be quadratic equations,”

    I believe we can remove the quadratic by calculating m.y from ody, t, and g. Since delta m.y (g*t) is constant, averaging should work.

    double t = (m.y-oy)/((m.dy-ody)*.5);

  247. ferd berple says:

    You can tell if you have an older source, as you will be missing adjustXY(molecule).

    adjustXY(molecule) is where I take care of things off the screen, including noWalls. The latest code also implements a fix in gridAdd(Molecule m) to make sure nothing leaves the collision detection grid.

    Anything out of the visible area gets lumped into the limits of the grid, which means collision detection takes a bit longer for off screen items as more combinations must be looked at. However, the off screen items are “less dense” so it appears a reasonable trade off to avoid changing the grid detection approach.

  248. br1 says:

    ferd:

    Fair enough, that should be ok (though I’m not entirely convinced that the last point is exact, but I’m not too worried about it).

    The zip I downloaded from http://www.filedropper.com/gas20120725_2 may have an updated .jar file, but the .java file is from the 21st of July and doesn’t contain the code you mention. I deleted my cache to make sure I wasn’t looking at an old download. Could you check again?

    By the way, I’ve sent Q. Daniels the thermal walls implementation. There is a fan implemented also, but that will take another refinement before being robust. I only had a very brief play with parameters, but by changing the ratio between reflective and thermal wall interactions, and turning up the fan speed, I seemed to get fairly convincing convection. More later.

  249. ferd berple says:

    br1 says:
    July 25, 2012 at 3:11 pm
    don’t make your wrong-way-up gradient go away.
    ========
    So far nothing has made the gradient go away. Please try the latest source and see if you can find a combination of settings that gives a decent density gradient without a temperature gradient.

    I’m also surprised by the results. Next release I’ll lump the molecules above the display into the temp and counts for the upper layer. Very small code change.

    I could also separate walls and sky into separate check boxes. not sure if that would provide any new information, as the nowalls was intended to simulate gas surrounding a planet.

  250. ferd berple says:

    my mistake, I forgot to update the source when I updated the jar

    http://www.filedropper.com/gas20120725_3

    here is the latest source

  251. ferd berple says:

    the latest source includes this line:

    // System.out.println(“off screen collision “

  252. br1 says:

    ferd:
    “here is the latest source”

    Great! All wall collisions are now exactly correct, all times are exact according to Newton’s uniformly accelerated motion equations, and particles sent on their way correctly. Nicely done, no further improvement possible.

    That is for the general particle-wall impacts, but the next worry is when a particle is pushed out of bounds during a particle-particle collision. I fear as you notch up the density near the lower boundary that this is the next thing that could mess up the solution. Less clear what to do about that, I’ll give it a look.

  253. ferd berple says:

    I’ve made this change to the time slice.

    int inc = (int) (sysTime-lastTime);
    //dt *= inc/8.;

  254. ferd berple says:

    correction: the change to the time slice above caused the energy to drift slowly. change removed.

  255. ferd berple says:

    br1 says:
    July 25, 2012 at 5:30 pm
    I fear as you notch up the density near the lower boundary that this is the next thing that could mess up the solution. Less clear what to do about that, I’ll give it a look.
    ===========
    I also struggled with that. In the end I added code to check for any energy loss/gain and automatically reduce the time slice until it stops or the time slice cannot go any lower without hitting zero.

  256. br1 says:

    “By the way, I’ve sent Q. Daniels the thermal walls implementation. There is a fan implemented also, but that will take another refinement before being robust.”

    I’ve now sent him another one, hopefully he can merge all the files together and post the combined sim. My latest version has slider controls to allow:
    1, a temperature difference between the left and right hand half of the heater. This can be used to investigate convection due to a temperature difference, and also investigate temperature vs height.
    2, a ‘thermal wall factor’ slider which if set to the left gives a purely reflective floor, while if set to the right gives a purely thermal floor. If set in between it gives a mix between reflective/thermal.
    3, fan X and Y velocity which controls a blower on the right hand side of the floor, which you can control strength and direction.

    I also corrected for the heater collision detection routine, which was not done properly until now.

    Playing with these gives some nice looking convection, but the temperature profile with altitude is always hot on top. This is not a simulation error this time – I explained it a few posts up, but it shows it is actually quite difficult to get a temperature gradient which is cold on top!

  257. ferd berple says:

    br1 says:
    July 26, 2012 at 9:58 am
    This is not a simulation error this time – I explained it a few posts up, but it shows it is actually quite difficult to get a temperature gradient which is cold on top!
    ==========
    It is yet to be shown that the hot on top situation is a simulation error “this time”.

  258. ferd berple says:

    http://www.filedropper.com/gas20120726_1

    some small changes to get rid of quirky collisions.

  259. ferd berple says:

    br1 says:
    July 26, 2012 at 9:58 am
    it is actually quite difficult to get a temperature gradient which is cold on top!
    ========
    What I’m seeing in the sim is that the ratios on the right Cn/Cn-1, these are not equal between layers, especially when using the default simulation. They decrease from top to bottom.

    When I switch to Gas, Small the ratios between layers becomes much more equal as I would expect if pressure/density is exponential.

    I wonder if what we are seeing is an artifact of the bigger molecules and/or the number of molecules? The ratio’s also appear to be affected by the number of molecules. Too many molecules and too little space?

  260. br1 says:

    ferd berple:
    “I wonder if what we are seeing is an artifact of the bigger molecules and/or the number of molecules? The ratio’s also appear to be affected by the number of molecules. Too many molecules and too little space?”

    That will certainly affect the number ratio between layers – if you jam packed the whole box with molecules, turning it solid, then there would be equal density with height no matter how much the gravity.

    Whether that can affect temperature or not I’m not sure (somehow I doubt it, but it would be nice to find out). One thing we do know for sure is that it stresses the collision handling routine!

    I was thinking along similar lines for your gradient – if instead of having a factor of 2 counts between layers, have a factor of sqrt(2)=1.4. Then check whether the temperature gradient across three layers in the first case is the same as the gradient across six layers in the second case.

  261. ferd berple says:

    br1 says:
    July 26, 2012 at 4:33 pm
    One thing we do know for sure is that it stresses the collision handling routine!
    ============
    The collision handler may be the problem. It does two things. It builds a grid of molecule locations (called grid) and searches only those molecules in grid [x +/ -1][y +/- 1]. This limits the molecules searched, but always builds collisions left to right, top to bottom first (or maybe the reverse, you get the idea). This means the collisions are likely skewed preferentially in a direction.

    Secondly, it should be possible to only allow collisions within the current time slcie. however, the collision handler does not find some (low energy?) collisions unless you look back as much as 10-20 time slices. This is for sure creating problems in the sim, as it means some collisions are found sooner than others, possibly based on energy.

  262. ferd berple says:

    I’ve changed the logic back to match the original. first I move a ball, then I check for collisions.

    However, with just two mols, the time calculated for the collisions a fair percentage of the time exceeds the time slice. This doesn’t seem possible if the collision time is being calculated correctly, because neither mol moves farther than 1 time slice in time between collision checks.

    When two balls are overlapped, the overlap must have come about during the past time slice, if the logic is working correctly, so this indicates to me there must be an error, likely in the calculation of time.

    double t = (-b-java.lang.Math.sqrt(b*b-4*a*c))/a;
    double t2 = (-b+java.lang.Math.sqrt(b*b-4*a*c))/a;

  263. br1 says:

    ferd berple:
    “so this indicates to me there must be an error”
    for sure.

    Just to let you know I’ll be away for over a week, and can’t contribute much during that time. Q. Daniels has my latest code, so I hope he pitches in for me.

  264. ferd berple says:

    found the problem.

    should be:

    double t = (-b-java.lang.Math.sqrt(b*b-4*a*c))/(2*a);
    double t2 = (-b+java.lang.Math.sqrt(b*b-4*a*c))/(2*a);
    //if (java.lang.Math.abs(t) > java.lang.Math.abs(t2))
    // t = t2;

  265. ferd berple says:

    http://www.filedropper.com/gas20120728

    lots of testing to eliminate errors as potential cause of negative gradient.

    molecules increased to 5000, radius cut in half, collision time corrected, spurious collisions ignored, collision detection randomized, fixup for potential negative energy at lower boundary.

    Total Energy rock solid. Collisions appear rock solid. will run it at paint drying speed and report.

  266. ferd berple says:

    there is a weakness in the collision detection routine. two molecules are only considered to have collided when they are overlapped during a time slice. to be accurate this requires time slices to essentially be zero.

    still haven’t been able to answer why collisions take place more than 1 time slice in the past, when molecules are only able to move one time slice. when two molecules are overlapped, the condition that caused it should have been during the past time slice. yet we see much larger past times, which are correct, because they are giving the correct separation between the molecules (2* radius).

  267. wayne says:

    ferd, still following you. I broke off and wrote a c# version, much simpler in some respects to also try and get all interactions as perfect as possible. The conservation of total energy, ke+pe, on my version is now basically perfect. After three million iterations the only digit that has changed is in the 15th least sig digit. The interaction at the top and bottom took a bit to get this perfect… I’m using a converging Newton-Raphson here to steer away from any square roots for precision digits. Seems the Java is using a quadratic here. Mine also has no interaction at the walls, as a planetary column would have, perfectly homogenous vertically so when one molecule leaves to the left another identical molecule appears at the right to replace it, a simple if x .LT. zero add width type of logic to handle the horizontal conservation of energy. After a day of running mine settles into a small ~2 “degree” bias warmer at the bottom but I still don’t really trust it yet for I too seem to have the same collision problem and I’m going to try to carry mine to use an intersection of segments type of logic instead of only registering an intersection when they overlap after a position update. So, I am still here in the background.

    a) Update positions into new set of variables by dt
    b) Pre-calculate all new x1-x0, y1-y0 vector lengths
    c) Detect segment crosses, if not continue
    d) You now have t & s (0 to 1) ratios of intersection so see if by velocities if they hit
    e) Do the intersection momentum transfer logic

    Also i am now using the same type of pressure logic found in the US.StdAtm76 code, inversed actually, to initialize the column density under gravity so the molecules don’t bounce so bad at startup… it was taking hours for this oscillation-bouncing signature to die out, repeating from warm at top to warm at bottom over and over but dying out… very time consuming !!

    Where the Java version was quite lengthy mine is so far a very minimal ~300 lines. Roger and Tim have my e-mail so just ask them for it if you should want the code to compare. Java and c# are very close to the same base language being both c based, you can almost verbatim swap code segments, it’s just that this is where my tools lie over the years so for dev speed that is my preference.

  268. ferdberple says:

    wayne says:
    July 31, 2012 at 4:10 am
    =====

    Wayne I’d be very interested in a copy of your code. I wrote c,c++ commercially for years and have VS all in place for c#. pls email a copy to ferdberple(AT)yahoo.com and I’ll watch for it. I’d very much like to get two different sims running to eliminate code dependent problems.

    my latest is here

    http://www.filedropper.com/gas20120801
    a fixup of the collision detection routine.

  269. wayne says:

    Hi ferd, sorry, got caught up in Watts et al 2012. I’m half way through cleaning up a cleanup version three and I’ll see if I can finish it tonight before sending it to you tomorrow. Great, since you have VS I don’t have to make it all one unit for simplicity and include instructions how to use the console compiler!☺ (that’s how I was compiling the java, dos style… snail dev but VS editor will colorize java code and even does some limited bracket auto formatting, I had never tried that but it does work)

  270. br1 says:

    ferd berple:
    “I’d very much like to get two different sims running to eliminate code dependent problems.”

    I’m back from travels, and on the way I managed to solve the collision problem exactly. I’ve written my own version from scratch, and this time instead of using a predetermined time step, I calculate the time to the next collision between every molecule, and to the walls. Whatever the least time is, in the future of course, is used as the time step. As there are 15 significant figures calculated, this practically guarantees that only one collision is dealt with in any step, so this can be solved exactly. As it stands, it could be written in a less computationally intensive manner, but it looks like it works even at high packing density. I’ve also added a long term temperature record vs time, so that you can judge whether the temperatures have stabilised.

    I have written it in Matlab, which should run under Octave, I’ll revise it a bit more and send it on to you. I haven’t had time to go through all the parameter space yet, but I’ll keep going and let you know when I have some solid results.

  271. br1 says:

    I’ve been trying to make my latest code less CPU intensive, and just about have it. One result I have found is that Velasco1996 and Roman1995 make an unstated assumption, that their results are only exactly valid for zero diameter particles. As the particles also need to swap energy by collisions, which can’t happen if the particles have zero size, then this makes their model slightly inconsistent. At least, one should take their result as applying in the limit of small particle size.

    If one doesn’t have infinitesimal particles, then the temperature results differ both in magnitude and in gradient as compared to Velasco Eqn(8). The magnitude of the average KE is less than that given in Eqn(8), which can be explained in the limit of stuffing the box full of finite sized particles. In that case, there is no room for the particles to fall, so the gravitational potential energy is higher than it would be if the particles were infinitesimal. This leaves less KE per particle than the infinitesimal maths expects, and violates the result given in Roman where the expectation of KE equals expectation of GPE for a 2D gas. So that is fun! Might even be worth a paper in itself, though I’m aware of some lossless granular gas papers that I’ll have to check over again.

    It is less clear to me why the temperature gradient with height should differ – the gradient one gets for finite sized molecules seems to be greater than that given by Velasco Eqn(8). Not by much, and it still goes to isothermal as one increases particle number or reduces particle density, so it certainly doesn’t give anything near the gradient we are looking for, but interesting nevertheless.

  272. br1 says:

    Hehe! I think I have the collision routine perfect (running the model as an ‘event driven simulation’), and here’s another result: There is a stated assumption in Roman/Velasco which is that the container is infinitely tall. But when one puts a lid on the container, this prevents molecules from reaching infinite height (obviously!). What is less obvious about this is that it raises the expected temperature of the gas!

    The reason for this is that the expectation of KE is practically independent of height. However, the GPE varies linearly with height. By capping the column, one does not allow the GPE of any particle to increase beyond a certain amount, and the centre of mass of the gas is lower than it would be if the column were of infinite height. This makes the total GPE lower than expected, which leaves more energy to be shared amongst the KE. Hence the average KE is slightly higher than Velasco Eqn(8). Going from an infinite column to a capped column might be equivalent to an adiabatic compression – you have the same molecules but confined in a smaller volume, so the temperature increases.

    I have run different container heights, and one needs to go to surprisingly high columns in order for Eqn(8) to hold.

    ferd: I’ll send the program on to you shortly.

  273. ferdberple says:

    br1 says:
    August 4, 2012 at 7:15 pm
    I calculate the time to the next collision between every molecule, and to the walls. Whatever the least time is, in the future of course, is used as the time step.
    ===========
    This is pretty much the conclusion I’d come to as to the only way to make the simulation truly realistic, however I ran into a computational brick wall.

    For example, consider a molecule resting on the “ground” with vertical velocity of zero. It will always be the molecule that collides “next”.

    In effect, a molecule sitting on the ground with zero vertical motion is bouncing infinitely fast, with a bounce height of zero.

    This is only one example of what can happen. My concern was that in a situation where there were lots of molecules in contact, such as can happen with gravity, that the time to collision would be zero and the simulation would quickly grind to a halt.

    A secondary problem was that as the time slices became smaller, floating point precision might also start to become an issue, as you added up lots of very small time slices.

    Both these concerns, but the first one especially, seemed to indicate that computationally you needed to set a limit on how small the time slice could become.

    Looking forward to seeing the code.

  274. ferdberple says:

    br1 says:
    August 7, 2012 at 11:02 am
    infinitely tall. But when one puts a lid on the container, this prevents molecules from reaching infinite height (obviously!). What is less obvious about this is that it raises the expected temperature of the gas!
    =========

    Interesting!!

    I can confirm that if you move the top of the container up and down the temperature decreases and increases. However, if you move the walls in and out the temperature does not change.

    This I don’t understand, because it is my understanding that if you reduce the size of a container, the pressure must go up and this should lead to an increase in temperature.

    What am I missing?

  275. ferdberple says:

    br1 says:
    August 7, 2012 at 11:02 am
    Going from an infinite column to a capped column might be equivalent to an adiabatic compression – you have the same molecules but confined in a smaller volume, so the temperature increases.
    =============

    This is what I thought as well, but when I modified the sim to move the walls in/out, I didn’t see a temperature change.

    Which to my mind is exactly what should happen, as moving the walls does not change the KE of the molecules (except perhaps slightly due to the acceleration of the container walls). But this contradicts the observation that compressing a gas by reducing the volume of the container will heat the gas.

  276. ferdberple says:

    thinking more on this, moving the walls didn’t raise the temperature because the sim doesn’t add the movement of the wall to the motion of the molecules. in fact it is purposely designed to do this.

    however, this means that moving the top of the container up and down is cooling/heating the gas by a different mechanism than moving the walls.

    What it means is that (for example) a piston working vertically should have slightly different efficiency than a piston working horizontally.

    This may indeed be worthy of a paper if it has been written up previously.

  277. ferdberple says:

    Thinking more on this,the worst case design if you want compression to heat a gas is to use a vertical piston with the gas above the piston.

  278. br1 says:

    ferd berple:
    “however, this means that moving the top of the container up and down is cooling/heating the gas by a different mechanism than moving the walls.”

    yes indeed.

    At first it crossed my mind that you had periodic boundary conditions sideways, so the width of the container has a different sort of meaning than the height. But even with solid walls, there is a different mechanism in the two cases.

    If one were to simply step the walls sideways to make the container wider, there will be no cooling of an ideal gas. This corresponds to ‘free expansion’, and only leads to cooling for gases with a non-zero Joule-Thompson coefficient – which only happens when intermolecular attraction is present. There should be cooling if you give the walls a finite velocity outwards, as then the molecules bounce off a moving object and so will lose KE.

    If one were to step the top of the container vertically upwards, the situation is different as the average centre of mass of the entire gas also rises a little bit as it expands to fill the new space. As this is true for the *entire* volume of gas, the energy required to do so must come from the KE of the entire gas, and so it cools a little bit. Note that this is very different than the case of an individual molecule!!! For an individual molecule, its KE does of course reduce as it moves upwards in the gravitational field, but it is the energy of the *ensemble* of molecules at each height that determines temperature, and we have been finding that this remains close to isothermal!

  279. Bryan says:

    br1 says:

    “If one were to step the top of the container vertically upwards, the situation is different as the average centre of mass of the entire gas also rises a little bit as it expands to fill the new space. As this is true for the *entire* volume of gas, the energy required to do so must come from the KE of the entire gas, and so it cools a little bit.”

    This means that the gas has gained gravitational potential energy at the expense of thermal energy.

    Would there not be a apparent second law violation?

  280. br1 says:

    Bryan:
    “This means that the gas has gained gravitational potential energy at the expense of thermal energy.

    Would there not be a apparent second law violation?”

    This is the exact question I had which brought me to Graeff in the first place. I thought this might be a violation, googled what others had done, and came across Graeff. Unfortunately I’ve come to the opinion that while this particular effect is real, I can’t see how to use it. One may argue that removing the lid costs no energy, and yet I can find no way to undo this action without doing work – so it is not for ‘free’. How do you make a cycle out of this effect, extracting work as you do so? Everything I think of doesn’t work. For example, you can’t simply slip in the divider back in the original position, as then you lose the molecules that have reached the upper altitudes and the system is not the same. You can’t simply lower the upper molecules back into the lower part of the container, as you would have to use a piston and input mechanical energy. Etc, etc.

    I’d like to hear how you might propose to make use of it.

    p.s. I sent my simulation to ferd, let’s see what he makes of it.

  281. Trick says:

    Bryan, br1 –“..step the top of the container vertically upwards…extracting work as you do so?”

    First you have to check if any particles are reaching the top (i.e. the container’s Hmax or Zmax) against the g field. As br1 pointed out, Velasco eqn. 8 assumes an infinite high container like a column of earth’s atmosphere going up past p=0 space vacuum. After that it doesn’t matter if the column top is stepped up from moon to Pluto. If p>0 on top wall, it does. How?

    The particles constant total enthalpy = PE + KE + pV being forced constant by law 1 in a rigid, enclosed, adiabatic container means if you move the walls (or the top assuming p > 0 at top) have to account for the energy by/to surroundings in the pV term.

  282. ferd berple says:

    got the code and did a very quick read through. looks good. thanks

    1) do you consider the case where a collision early in your minimum time list changes the tail of the list? It seems likely that to be correct you need to recalculate the list after every collision, because that collision could insert a new event into your list, potentially ahead of the rest of the list.

    2) how do you handle the case of a molecule lying motionless (vertically) on the ground? It will collide with the ground with dt = 0, and continue to collide with dt = 0; As such, it will always be the minimum time event

  283. br1 says:

    Hi ferd,

    Just sent you another one, including differential floor heating and a vector plot of momentum to look for circulations. Another nice puzzle with respect to differential heating is raised which I’ll get to in a later post!

    In answer to your questions:

    1) yes you are right, but only the molecule that has actually collided needs to be recalculated (with respect to all the other molecules and walls). All the other molecule/wall collisions simply get stepped forward in time as they can’t possibly be affected. The new list is then thoroughly searched for the minimum next collision time, taking into account the precalculated list and the newly calculated list. To check this, I have run the simulation with a reduced number of molecules and with large molecule size – there are no ‘ghost’ molecules (where a collision might be overlooked and the molecules pass through eachother) and there are no ‘phantom collisions’ where a collision is taken out of order and the molecules bounce even though they haven’t reached each other yet. So it really looks like it is working.

    2) I do have an ‘if dt==0 break’ line in the code which will cause the program to quit, but this condition has never been met! So basically this situation does not arise in practice. Which is convenient!

  284. ferd berple says:

    I’m installing matlab and will give the latest a try.

  285. br1 says:

    Ok, here’s the puzzle.

    Take a look at this picture: http://s1257.photobucket.com/albums/ii503/brspics/?action=view&current=Gas2Dv5p0.png

    In the upper left quadrant is the molecule position picture – there are 400 molecules bouncing around inside a container, including gravity. The left half of the floor is at a ‘high’ temperature, the right half of the floor is at a ‘low’ temperature (you can tell this visually by noting the different molecule density).

    In the upper right quadrant of the picture is the temperature vs time of the gas at different altitudes to see whether the temperatures have stabilised – there is still some wobble, but they are fairly stable.

    In the lower left quadrant is a net momentum graph, showing the net flow of molecules at different subsections of the container (long term average). This shows a nice circulation.

    In the lower right quadrant is the temperature vs altitude (long term average).

    There are two puzzles – one is why the lower right quadrant graph temperature increases with altitude, but I’ve dealt with that in a couple of previous posts on this thread and am fairly happy with the reason.

    The second puzzle relates to the lower left figure (net momentum). There is a clear circulation where the gas above the hot part of the floor rises, and that above the cold part of the floor sinks. But the puzzle is the vector at the bottom of the graph, pointing left. This is saying that there is a direct flow of molecules from the cold part of the gas to the hot part. As the heat is only contained in the molecular motion (no radiation in this model), this is saying that heat (=molecules) is travelling from cold to hot. So… how come??? 🙂

    I’m pretty sure of the answer, and what is wrong with the question, but I still find it a nice question!

  286. br1 says:

    Here’s another nice little result: http://i1257.photobucket.com/albums/ii503/brspics/SpontaneousConvection.png

    This time, the floor is uniformly heated to 10 K, the ceiling uniformly cooled to 1 K, with gravity present. The 1200 molecules spontaneously arrange into a convection cell(s), despited the lateral symmetry (see bottom left figure)! The cell can sometimes go CW, sometimes CCW, depending on what the starting noise was, and sometimes you get double cells. So this model is really quite useful for ideal gas visualisation.

    Still no real idea how to get Graeff’s gradient, which is the ultimate goal.

  287. ferd berple says:

    this is saying that heat (=molecules) is travelling from cold to hot. So… how come???
    ==========
    wind flows from cold to hot at the surface and from hot to cold at altitude. the exception is the mid latitudes where the polar and tropical flows induce the westerlies to flow from hot to cold on the surface, making the high latitude temperate zone warmer than it would be otherwise.

    The model looks good in my quick check. the matlab plots are a bonus. fiddled around with different sizes for the container and different molecule counts and everything looked very stable. don’t know matlab so some aspects of the code are still a mystery.

  288. ferd berple says:

    Still no real idea how to get Graeff’s gradient, which is the ultimate goal.
    ===========
    Indeed. The dry air lapse rate is too close to G to be coincidental, yet the model results suggest that it cannot exist as a consequence of G. which suggests there is a mechanism at work that is not part of the model.

    it reminds me of the model not showing any heating when the walls moved inwards. some natural mechanism that is not properly included in the model,

    the model is telling us that KE/PE conversion during conduction is not the mechanism. But the atmosphere is telling us that some mechanism exists. my money is on the work done as air expands/contracts during convection, but how to model this?

  289. ferd berple says:

    and of course if the mechanism is related to convection, then this contradicts Graeff’s work to limit convection.

  290. ferd berple says:

    What about simply installing an elevator at both ends of the model. One end always goes up, the other always goes down. the next work being done with be zero as the upward work at one end should equal the downward work at the other.

  291. br1 says:

    ferd berple:
    “wind flows from cold to hot at the surface”

    sure, but the puzzle is why this doesn’t break 2LoT. After all, the molecules carry their coldness with them, so ‘heat’ is travelling from cold to hot.

    Or at least, that’s one confused way of describing it! 😉

  292. ferd berple says:

    Still no real idea how to get Graeff’s gradient, which is the ultimate goal.
    ===========

    Ok, reading wikipedia I can see what looks like a missing part to the sim.

    http://en.wikipedia.org/wiki/Lapse_rate
    As the air parcel expands, it pushes on the air around it, doing work (thermodynamics). Since the parcel does work but gains no heat, it loses internal energy so that its temperature decreases. The rate of temperature decrease is 9.8 °C per 1,000 m.

    This contradicts what we see when the sim walls are suddenly moved outwards. There is no change in observed sim temperature whatsoever. If the sim was matching reality, when the walls moved outwards at infinitely high speed, there would be a drop in temp.

  293. ferd berple says:

    more wikipedia
    http://en.wikipedia.org/wiki/Ideal_gas_law

    Isentropic process

    T2 = T1(V2/V1)^(1 − γ)

    The value used for γ is typically 1.4 for diatomic gases like nitrogen (N2) and oxygen (O2), (and air, which is 99% diatomic).

    This tells me that we should be seeing a temp change with the volume change, which the sim is not currently showing when the walls move. It appears that the sim models γ = 1.0

  294. br1 says:

    fer berple:
    “This tells me that we should be seeing a temp change with the volume change, which the sim is not currently showing when the walls move. It appears that the sim models γ = 1.0”

    I beg to differ. I fully expect the model to show γ = 1.66 http://en.wikipedia.org/wiki/Heat_capacity_ratio for a monatomic gas. The situation you simulated is not an isentropic process – in fact the entropy increases if you step the walls sideways, and this is not a reversible process (that is, you can’t simply step the wall back to its original position without losing molecules). To do the expansion isentropically, the wall must move (infinitesimally) slowly so that the pressure of the molecules is felt as it expands. The molecules will undoubtedly lose momentum as they bounce off a wall moving away from them, so the gas temperature will drop. This is a reversible process as you can then move the wall (infinitesimally) slowly inwards again to recover the initial configuration, the molecules gaining momentum as the wall moves back in. Doing this in the model is a bit tricky, but I might give it a go.

    I think the physics of the DALR is in the model, but the circumstances have not been set up to see it. I have some ideas on that, but it might require too large a particle number to be practical. Still I’ll definitely give this one a shot and let you know – this is where I’m heading with the circulation examples.

    p.s. I’ve nearly finished the GUI on my model – makes it a lot easier to set parameters!

  295. ferd berple says:

    br1 says:
    August 10, 2012 at 9:44 am
    To do the expansion isentropically, the wall must move (infinitesimally) slowly so that the pressure of the molecules is felt as it expands.
    ============
    Questions:

    1) When a piston compresses air in a cylinder, does the heating of the air occur only along the face of the piston, where the air molecules are being accelerated by the piston, or does it take place throughout the remaining volume of the cylinder.

    2) When a piston moves to expand a cylinder, does the cooling of the air only occur along the face of the piston, or does it take place throughout the remaining volume of the cylinder.

    3) When a piston moves a supersonic speed to expand a cylinder, such that few if any air molecules can contact the piston, does this prevent the cooling of the gas in the cylinder?

  296. ferd berple says:

    “The molecules will undoubtedly lose momentum as they bounce off a wall moving away from them, so the gas temperature will drop.”
    =========
    When the gas is expanding, the individual molecules are (on average) moving away from each other, similar to the wall moving away.

    So, if the molecules lose momentum as the wall moves away, they should also (on average) lose momentum as the gas expands and the molecules bounce off the molecules moving away from each other.

    This should be seen as drop in temperature as the walls move away, regardless of the wall’s speed. And this drop in temperature should occur throughout the gas, not just along the moving wall.

    Air released though a vent from a pressurized cylinder will still experience a cooling within the cylinder, without any movement of the walls. The released air will also cool, even though the gas itself may be accelerated to high speed as it escapes.

    I suspect there must be a net loss of momentum as the air molecules bounce off the molecules expanding away from them, which translates into a cooling of the gas as it expands.

    The sims are not capturing this effect.

  297. br1 says:

    ferd berple:

    Here are my answer to your questions. Other opinions may vary 🙂

    1) Heating occurs along the face of the piston only. But of course, there is conduction, there is the speed of sound (600 m.s-1) to propagate pressure waves, there is diffusion (molecular velocity 1000 m.s-1), there is turbulence… These all act to mix the temperature very quickly in practice, despite gases being poor conductors when compared to solids.

    2) likewise.

    3) This (and the similar example of air released through a nozzle) is governed by the Joule-Thomson coefficient http://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect . This effect is related to intermolecular attraction, which is one effect our models don’t have (yet!). However, this is not needed to explain the DALR! Also note in the figure on Wikipedia that gases can warm due to free expansion!

    I agree the sim does miss out on the Joule-Thomson effect, but the gas should very much have γ =1.66, and I believe if given the right conditions should have a DALR. I haven’t reproduced one yet, but I’m working on it 🙂

  298. br1 says:

    see also http://en.wikipedia.org/wiki/Joule_expansion which shows that free expansion for an ideal gas has Ti=Tf and is not an isentropic process for any gas.

  299. ferd berple says:

    What about the gas left inside the scuba tank? Why does it cool when the tank valve is opened?