This is a guest post from ‘Lucy Skywalker’ who has recently returned from a trip to Germany where she attended a seminar given by Roderich Graeff, the engineering concern owner who has been experimenting with equipment he has designed to test the Loschmidt gravito-thermal effect. This line of research is highly relevant to the theoretical work of Hans Jelbring, and also Nikolov and Zeller, who have proposed hypotheses to explain the thermal gradient found in the atmosphere causing the near surface air to be warm relative to higher altitudes.
SECOND LAW CHALLENGED? LOSCHMIDT VINDICATED?
Lucy Skywalker – May 28 2012
Few people know that in the last decade, there have been quite a number of serious challenges (as opposed to perpetual-motion challenges) to the hallowed Second Law of Thermodynamics. Dr Sheehan has organized conferences and written books about all this. He put me in touch with Dr Graeff, the one participant who has been running real experiments. These experiments appear to vindicate the theories of Loschmidt, who 150 years ago challenged his friend Maxwell’s belief that a vertical air column in equilibrium will be the same temperature top and bottom. Loschmidt maintained that gravity would cause the bottom molecules to be warmer than the top ones. But until Graeff, nobody had actually undertaken the experimental research needed to check these theories against measurements.
This work could be extremely important, not least for Climate Science, if it holds up to close scrutiny. After reading Graeff’s paper and his book, I went to Germany to join his seminar, and to examine for myself his apparatus that appears to measure vertical heat gradients in columns of air, water, and other substances in steady, non-convecting equilibrium, and appears to show that in isolation, they are warmer at the bottom than at the top.
On the journey, everything goes wrong. The plane fails to take off and we are abandoned in the airport. The lucky ones find a long queue waiting for help from two junior staff on broken-down laptops. I get new train tickets in Frankfurt, but cannot inform my host because I have the wrong number. The first train is late so it misses the second train. I swap for another ticket but the next train is also late so it misses the third train. Dr Graeff has been driving to the station and back for several hours. I tell him I know it is going to be special – it has to be, after all that.
And it is special indeed. But who am I coming to a seminar whose purpose is to question the hallowed Second Law of Thermodynamics and prove that heat can travel from cold to warm? That’s what my host asks me – and everyone else. Why am I here? and why are you here? and you? and you? Why do you think you can disprove the Second Law then? Who do you think you are to do this? And without a degree in physics too? And do you know what it says? Can you explain it to me please? Ah, but how many versions of the Second Law are there, and which one do you refer to?
I would have been thoroughly disconcerted by now. I never was out to disprove the Second Law – only to examine a modification to a common interpretation of it, that Graeff has spelled out as a result of thousands of hours of close experimental work. But there is a twinkle in mine host’s eye. At 84, Dr Roderich Graeff is as sprightly as the rest of us, listening intently and responding with a mixture of
sharp challenges and a wicked sense of fun. We’re not going to let him get away with any slipshod experiments – and neither is he going to let us get away with slipshod challenges. When I ask him, how can he be sure the measurements are as accurate as he claims, he turns on me and says, well, what do you think? How are you going to check how accurate they are?
Nullius In Verba.
Beautiful. I will have to work for my understanding, as Graeff had to do, but I love every minute of it.
The visit is a magnificent extension of Graeff’s book which I cannot recommend too highly, if you are open to the challenge. It tells you far more than I can tell here, about Graeff’s story, his life of engineering work, his scientific interests, his surprising discovery, his simple but well-fitting theoretical backing, and enough details to tempt you to replicate his work. He is a genius in classical style, who also comes from a deep place of caring about humankind. After surviving the firebombing of Hamburg (40,000 dead in one night) he wants to end all “Mutually Assured Destruction” systems, proactively build peace, and use his scientific and engineering talents to look for alternative energy sources.
Graeff’s whole house has become a laboratory – and yet it is still the warm home that he designed and built in the beautiful Black Forest many years ago. Empty Thermos flask interiors of all sizes stand around like Russian dolls, waiting to be stacked inside each other, to provide affordable insulation to the vital core experimental columns, whose temperature gradients between top and bottom are going to be measured with homemade thermocouples and fine thermistors, using the finest insulated wire you have ever seen. Sheafs of labelled pairs of wires hang out of large polystyrene boxes holding their precious cargoes of alternate layers of insulating and conducting sheaths that enclose the central mystery: Dewar bottles aka thermos flask interiors containing the core columns of solid, liquid or gas to be tested.
It all looks a mess, but in fact the experiments are well orchestrated. How many has Dr Graeff now done? About 852 1/2, this last decade or so, he tells me. Is this true? Certainly it may be, because he has carefully numbered each one. Why are there so many? Well, he has needed to check out the environment, vertical orientation, effects of size, effects of number of layers, the insulations, thermal equalizers, thermocouples, the instruments’ bias and sellotape fixings, the convectance impeders, substances actually tested, dataloggers, and software. You would think the universities would be fighting each other to have the priviledge of testing Graeff’s amazing work. But no. With a few notable exceptions like Prof Sheehan, the academics refuse to touch anything that challenges the Second Law, even if only to modify its interpretation. No, they cannot fault the experiments. And no, they cannot deny the results either. But still they fear… what? Leprosy? So Graeff goes back to his work, patiently testing over and over again, each tiny detail that might, just might upset the results. But they don’t. The results continue to hold. Tiny, but indisputable, like grit in the mouth.
Calibrate the thermocouples? You have to get the feel of the whole thing, Graeff says. And he is right. This is not 852 separate experiments, this is 852 facets of one basic experiment. Think the thermocouples might give a skewed result? Right, let’s reverse their connections and take the average to eliminate bias. Let’s also run a series of experiments in which the core is inverted. See how long the core needs to settle down… seems to take just under 2 days to invert its gradient if it’s convection-impeded gas or liquid (conductive metals are quicker to readjust), so let’s run a series of 4-day cycles and see how they average out. See how close the spots on the measurement lines are to each other? See what steady lines they make? All that is proof that the measuring devices are measuring true, not wild, and the fact that we can reverse the apparatus and get similar results suggests strongly that it is gravity that realigns the temperature of the inner core to a negative gradient, not a fault in the apparatus.
Still not certain about the thermocouple calibration? But that’s not the point, Graeff says. And he’s right again. The point of the experiments is that they are consistently showing a negative vertical temperature gradient at all in the core axis, even though that is surrounded by layers in which there is no gradient, which are surrounded by layers tending steadily towards the normal interior-of-room positive gradient, warmer at ceiling level than at floor level. Okay. So are all 852 experiments showing a negative temperature gradient at the core, surrounded by a positive gradient, that suggests gravity is, as Loschmidt suggested, affecting the thermodynamic equilibrium of all those core columns?
Well, perhaps a few experiments did not work. Perhaps a Dewar bottle broke. Perhaps a mouse got in and made a nest. Perhaps the cellar door was draughty and letting in the cold outside air. Perhaps the thermocouples were badly attached. Perhaps the Keithley datalogger was playing up. Perhaps the Excel program was faulty. Perhaps… but aren’t we forgetting Einstein’s words, to the effect that just one effective experiment was all that was needed to topple an accepted law? And aren’t 800 varieties of that experiment enough? What hoops has the man got to go through? Hasn’t he earned recognition?
Why is no university even bothering to check those experimental methods and results, let alone replicate them or study Graeff’s theory which fits the results like a silk glove? How many laboratories are going to kick themselves because they didn’t recognize the telltale evidence of a genius at work, I wonder?
Perhaps it’s easier for me as an outsider to spot genius
I can feel that frisson a mile off. I recognize the heavy, stuck shape of the doubts of the orthodox when “appeal to authority” is missing from the menu; I recognize the hoops and the highs; I have examined for myself what proofs and patience and precisions are really needed. Still not convinced? Read my Amazon review of Graeff’s book. Get the book itself and read it twice. Then come and meet Graeff and check out his tests for yourself, ask more questions. But by that time, you ought to have realized that he really is on to something important, and it’s something that we too can check and replicate and even publish.
James Clerk Maxwell would be proud.
Next part:
Gathering an Experimental Replication Team. I will describe the experiments in more detail, with background about the 2nd Law controversies. I will describe Graeff’s theoretical underpinning that seems to work and requires omitting a commonly-held assumption about the Second Law, and allows for the effect of gravity as an unavoidable outside influence. Certainly there is no reason to abandon the 2nd Law or even rewrite it in essence. And I’m putting out word to gather a little local team to replicate Graeff’s work.







Trick:
“There is no new maximization w/LM in part c”
You keep skipping Eqn(19) for some bizarre reason (I wonder why?). With your knowledge of what Lagrange multipliers look like, can you see any in Eqn(19)? Usually they have symbols such as lambda or mu.
“Or kindly show me a maximization process I missed anywhere after eqn. 21 (a restatement just combining 2a and 2b soln.s)”
There are only four eqns in section 2c, Eqn(21) is the third. It clearly comes from Eqn(20), not only because it is obvious but note that Verkley says “(Eqn(20) so that we should have Eqn(21)”. Pretty clear really. But Eqn(20) comes from Eqn(19), not only because it is obvious but as Verkley says “Eqn(19). That gives … Eqn(20)”. Unmissable once you see it. So Eqn(21) comes from Eqn(19), and is not a trivial copy-paste from earlier. But what is Eqn(19)?
That is the question.
I’ve clearly given my interpretation, what is yours?
It’s so rare to see science discussed in a sustained way at this depth where the protagonists manage to keep talking to each other. It’s a credit to both of you. I just want to see it continue – hence the gentle nudge.
Trick 6/11 12:54pm – “Look and ye shall find NO maximization of eqn. 19 in 2c.”
br1 6/12 7:56am– “You keep skipping Eqn(19) for some bizarre reason…? (I wonder why?)”
I can see the wonderment. Might come from: ““The historic development of thermodynamics has been…particularly susceptible to logical insecurity..”
br1 continues – “But what is Eqn(19)? That is the question.”
Trick 6/11 12:54pm: “Eqn. 19 is combining eqn. 9 from 2a and eqn. 15 from 2b.”
Truesdell 1980: “Thus I write of thermodynamics…a field accursed by…retreat.”
br1 continues: “I’ve clearly given my interpretation, what is yours?”
Retreating to Trick 6/11 8:13pm:
“…Verkley uses (Lagrange multiplier LM) in 2a and 2b for the maximization process (of iso and non-iso) then he copies them to 2c & combines. There is no new maximization w/LM in part c – there can’t be b/c Verkley is showing a 3rd profile soln. not at LTE/max. entropy in between 2a and 2b which in the limit converges to either iso or non-iso LTE/max. S. “
That 3rd Verkley profile in 2c (Akmaev’s monotonic) is the real atmosphere profile, shown as the standard atmosphere in Verkley Fig. 2.
There can be no further maximization in Verkley 2004 2c because here the paper seeks to tie the threads of ideal iso and non-iso max. S LTE profiles together to theorize the real standard atmosphere profile always operates somewhere in between these two ideal solutions below max. S.
The reality is the real atmosphere air column doesn’t ever achieve max. entropy S (except maybe in rare transient coincidence). This is a key point of the paper – placing iso and non-iso idealized profiles as the bookends for reality, the real advancement as the paper says.
Then Akmaev 2008 weighs in handing off the ball to fullback Chebyshev who smashes thru nature’s defense and carries the ball across the goal line w/o ever fumbling so far as I can see. Game over.
NB: Truesdell 1980 says thermodynamics didn’t get to be an adult science until somewhere around the early 1960’s. The caloric theory (heat being a physical entity fluid) not being formally dismissed so far as Truesdell can find before 1973.
Truesdell p.41: “…thermodynamics, from its beginnings, has been a sick science, its sores unprobed by conceptual analysis and uncleansed by logical criticism….caloric theory forbids an ideal gas to have constant specific heats will prove fatal to it.”
Trick:
“The reality is the real atmosphere air column doesn’t ever achieve max. entropy S (except maybe in rare transient coincidence). This is a key point of the paper”
Yippee! A statement we can agree upon!
Unfortunately I think the agreement will be short lived (as I disagree with most of your other statements), but common ground is always nice to have.
Staying on Eqn(19) a bit more, can we further agree that it *does* set up the problem to always find the max entropy achievable *under the constraints* of fixed M, H and L, but that the *maximised* entropy it finds does not in general equal the ‘absolute maximum’ entropy that is possible?
This is a slightly different formulation than your statement, but if we tread carefully we may make progress.
br1: “.. the *maximised* entropy (eqn. 19) finds does not in general equal the ‘absolute maximum’ entropy that is possible?”
The only way I can make sense of this is that some possible physics are left out of eqn. 19 that could lead to even higher ideal S entropy value if considered. Conceptually then, I can see that the ideal max. S computed for iso and non-iso “does not in general equal the ‘absolute maximum’ entropy that is possible”.
Now I point out either a) you choose to add a possible physics present in the standard atmosphere (GHG, aerosols, non-adiabatic walls) or b) you choose to add some possible physics NOT present in the real standard atmosphere. To eventually get to “the ‘absolute maximum’ entropy that is possible”
Remember to re-measure fairly:
a) standard atmosphere profile with only the possible physics you choose and
b) alien possible physics acting on the standard atmosphere profile.
Don’t see how this adds anything, but your turn here in our quiet little talk shop area that TB encourages. Effect is to change H/L curve and unique alpha I suppose.
Trick:
“The only way I can make sense of this is that some possible physics are left out of eqn. 19 that could lead to even higher ideal S entropy value if considered.”
In my mind it is simpler than that. I was agreeing with your statement “the real standard atmosphere profile always operates somewhere in between these two ideal solutions below max. S.”
My view revloves around the following two statements:
1, Atmosphere operates below max S.
2, Eqn(19) determines how to maximise S given constrained M, H and L.
It is the constraints which cap S from rising to a value of S that could be achieved if the constraints were different. That may sound tautological, but I think is an important point to consider in trying to answer the following questions: what are the limits on the constraints, and why do the constraints have the values they do?
So let’s look at the constraints again. We can justify a fixed M because there is only so much mass in the atmosphere. We can’t simply double the value of M in the maths without dealing with conservation of matter in practice. Similarly for H, if we take a column from ground level up to space then there is a fixed energy in that column, and we can’t arbitrarily change that energy without dealing with where that energy is coming from – there is after all conservation of energy to consider.
But what of L? That is where I see all the trouble and squirming in the papers. What does ‘potential enthalpy’ really mean? Why should L have a fixed value? Can I arbitrarily double the amount of L in a column of atmosphere – what law would it break?
The answer to that last question is that there is no *fundamental* law describing conservation of L. It is possible in a totally closed insulated container of gas that the value of L can change without any inputs and without breaking any laws. In this sense it is quite different to changing M or H which would have to be supplied from somewhere. But in the case where there is heavy turbulence, it so happens that L tends to a value that doesn’t depend much on what other parameters are doing. B&A tried to justify this, but according to Verkley and Akmaev they failed, Verkley tried a bit harder but admitted failure even in his own paper, whereas Akmaev may have got to some better justification (I haven’t had time to digest that yet).
The point being that it is the constraint on L which limits how close the entropy can get to its ‘absolute maximum’. The word ‘absolute’ is used here because M and H *have* to be constrained (at least for the atmosphere as a whole), you can’t get around them without breaking the fundamental laws of physics, or supplying heat or matter. But the value of L is much more contingent – how heavy is the turbulence? how strong is the external heating?
In a closed insulated container, if the gas starts off in a turbulent state, that turbulence will die down. The value of L therefore can spontaneously change in an isolated system. In such an isolated system, because L can change it will not be constrained, and so can no longer constrain S. S will therefore rise until it reaches a value constrained by H. This is a ‘hard’ constraint (conservation of energy), so S maxes out at *the* maximum value. Akmaev explicitly shows that the S of a system for a given H but unconstrained L is always greater than the S of a system for a given H and a constrained L. Therefore *any* constrained L system will have lower S than the equivalent isolated system with unconstrained L.
I hope you can agree that this is the case for Verkley 2c.
Lots of things above you might not like – have at it.
br1 2:36pm – “.. gas starts off in a turbulent state, that turbulence will die down..”
Not much free time today, or even thru the weekend, this is the big deal Verkley discusses in 2b & any reader has to come to grips with. B&A approximates. And Akmaev settles exactly with fullback Chebyshev. I come to grips with it & grok/write this as vigorous mixing of the molecules.
Sure, in the beginning there are pockets with more or less turbulence (or vigorous mixing call it what you will) in the closed control volume and then at LTE Akmaev shows us rigorously how the turbulence or vigorous mixing has become smoothly continuous at max. S which is non-isothermal eqn. of state PV=nRT and since it is the max. S point, isentropic. Exactly.
br1, either you or I are a bit confused. I keep reading Akmaev alongside you statement that “2, Eqn(19) determines how to maximise S given constrained M, H and L.” but H (enthalpy) *is* both P and L (see equation 1, 2, and first page and a quarter). The only place H is used is as scale height. It seems L is the enthalpy here, not H in this paper so you using H and L in a statement leaves me scratching my head. Are you sure you have the parameters in Akmaev’s paper correct? I agree with Trick that the way Akmaev wrote this paper has little to desire (it’s not easy to follow due to the phrases and re-definition of past used symbols) but I can follow his flow though.
From what I gather from these papers is that the enthalpy depends greatly on how you define the start point of the ‘system’, but I start it as a very tall, totally evacuated and insulated column with ‘x’ amount of gas (~10 Mg here) at the bottom in liquid or solid state and zero K. This is P & L in Akmaev’s paper. So you track the energy slowly added to the system until the bottom reaches and stabilizes at your target temperature, let’s say 288 K. At this point the column is either isothermal or has a lapse rate with the bottom warmer, hence, natural lapse rate.
The total energy added is ‘L’ (or P). If a lapse rate exists, it would take additional heat to bring the system to a constant 288 K with constant agitation so ‘L’ is at a minimum at equilibrium compared to an isothermal state.
Contrary, if it was found the stabilized column to be isothermal, a later lapse rate would be quite impossible without a constant movement of energy from the top to the bottom, which again is impossible internally or without external transfer of energy across the walls or the top and bottom. So one thing you seem to see is that a lapse rate result has lower enthalpy than an isothermal column.
Both Earth and Venus have mechanisms to decrease this natural DALR, water in Earth’s case, sulfuric acid rain in Venus’s case, some radiation from the surface in both cases, along with the accompanying evaporation and convection near the equators.
If you look at this from an isothermal viewpoint, isothermal is natural and the only reason any planet would have a lapse rate is that there is more energy constantly arriving at the surface and a constant loss of heat at the top, or movement of that energy back down to the surface, radiation here is assumed. In Earth’s case this seems feasible and that is what AGW proponents firmly believe is happening. But if you then turn and look at Venus with an assumed natural isothermal atmosphere, you would need a constant movement of energy from the upper atmosphere to the lower atmosphere, or solar input to the surface, to maintain this huge temperature differential. This would be a continuous cycle and some say this is due to “back-radiation” from all of the co2 in Venus’s atmosphere. But this continuous cycle of energy needed to maintain the ~8.5 K/km lapse rate over 94 equivalent Earth atmosphere thicknesses would be a perpetuum mobile and this where the true violation of the second law of thermodynamics creeps in. You cannot have such a cycle of energy from the atmosphere itself for little solar radiation is absorbed by Venus. In Earth’s atmosphere it is harder to see this same principle but it is there also.
So I keep looking where Akmaev may have made math errors in his derivations, it sure seems it is there somewhere. His explicit constraint equations and surface equation in his development of the Lagrange multiplier sure would have helped. You can assume what he used but it is an assumption.
That’s how I see it so far.
Hi wayne,
No time for a detailed reply at the moment, I’ll be back for your and Trick’s post later.
But a quick clarification, Eqn(19) is from Verkley. Sorry for swapping between papers, but Akmaev is more explicit on a few points (which I’ll return to next time).
P in Akmaev is H in eveybody else’s notation, this is enthalpy.
L is the same symbol across the papers, but this is NOT enthalpy. Akmaev calls it ‘potential enthalpy’, and this is quite a different beastie than enthalpy. It relates to ‘potential temperature’ which is quite a different thing than temperature. Of course there are relations, but examples of the differences are that potential enthalpy is not conserved in an isolated system whereas enthalpy is, and an isothermal column has increasing potential temperature with height.
So the behaviours of these quantities are not always intuitive.
More later.
“But this continuous cycle of energy needed to maintain the ~8.5 K/km lapse rate over 94 equivalent Earth atmosphere thicknesses would be a perpetuum mobile and this where the true violation of the second law of thermodynamics creeps in. You cannot have such a cycle of energy from the atmosphere itself ”
In light of that how would one explain the heating from adiabatic compression as air descends ?
And air MUST descend as much as it rises.
I’m coming round to the view that the atmosphere really doesn’t warm much from solar irradiation but rather from adiabatic compression as air is forced down in one place in response to rising air elsewhere.
Surface warming too would be a minor player because that simply supplies the energy for the initial uplift and the consequent decompression negates it in the rising column.
So we are back to what I suggested in another thread. Surface heating is cancelled out by subsequent decompression and at top of atmosphere solar energy in is cancelled by longwave out.
If surface heating and top of atmosphere heating are both cancelled out then the remaining heat from adiabatic compression is left over as a warming bias and all it is is the energy from the initial compression being constantly renewed and being locked into the system until such time as the sun goes out or the atmosphere dissipates to space.
Note that compression will only generate heat if heat from insolation has energised the gas molecules in the first place by lifting them off the surface. No insolation and gases just lie on the surface in solid form.
It isn’t a perpetuum mobile because rising air working against gravity is exactly balanced by falling air working with gravity and the movement (work) is driven by density differentials and uneven insolation which causes some portions of the atmosphere to be less dense and therefore lighter than other portions.
There is, therefore, no breach of the Laws of Thermodynamics.
“the only reason any planet would have a lapse rate is that there is more energy constantly arriving at the surface and a constant loss of heat at the top.”
That is exactly what does happen in any atmosphere with a circulation.
In the descending air column potential energy is converted to heat as compression increases towards the surface.
Quite simply the compression of the descending air transfers as much energy to the surface as is needed for radiation to space from the surface so as to create a radiative balance at top of atmosphere.
If there were no GHGs at all then ALL the energy that needs to get transferred to space from the surface will be delivered to that surface by adiabatic compression.
You would get a more turbulent circulation in a non GHG world but the lapse rate would be just the same and the top of atmosphere radiative balance will still be maintained.
The presence of GHGs means that the atmosphere needs to be less turbulent since some energy is radiated straight out to space by the GHGs and so does not need to be returned to the surface by adiabatic compression.
The concept of adiabatic compression needs to be incorporated into the planetary energy budget as a means of returning energy to the surface in place of so called downwelling infra red.
There is no such downwelling. What is actually being measured is the IR energy emanating from the air around the sensor which is being generated as a result of the compression process.
You don’t even need to propose any transfer of energy from air to surface. Instead what happens is that the temperature of the compressed air inhibits cooling of the surface until the surface acquires much the same temperature as the air.
However, over water surfaces any temperature variation that results from internal movement of the water will dominate for a time until the water cycle evens things out again back to the temperature dictated by surface pressure.
On Venus the sulphur dioxide cycle does the trick, On Mars the CO2 cycle. On other planets the methane cycle.
On any planet where any material can phase change at ambient temperatures that will help the equilibriation process. In the absence of such materials the planet simply adopts a more vigorous circulation to the same end.
Trick:
“Sure, in the beginning there are pockets with more or less turbulence (or vigorous mixing call it what you will) in the closed control volume and then at LTE Akmaev shows us rigorously how the turbulence or vigorous mixing has become smoothly continuous at max. S which is non-isothermal eqn. of state PV=nRT and since it is the max. S point, isentropic. Exactly.”
But it is not *the* max S point, it is only the max S point under the conditions of heating and energy dissipation, as he discusses in depth from Eqns(25) to (31). Hence the isentropic system (such as in Verkley 2b) is far from equilibrium, as he discusses in depth and states explicitly many times.
I have now re-derived all his equations. I appreciate a bit more the function w(p), and how this gives the whole of Verkley 2a, 2b and 2c and more, including unstable profiles with a lapse rate greater than the DALR. His conclusion across *all* these profiles is “the further a monotonic temperature profile is from the isotherm, the further it is from the thermodynamic equilibrium”, so again Verkley 2b is not in thermodynamic equilibrium.
His whole spiel from Eqn(25) to (31) is to explain how a stable lapse rate (which includes isothermal, Verkley 2a, high entropy) can get to a DALR (=Verkley 2b) which has lower entropy. He notes that “turbulence in a stable stratification (or forced convection) cannot be sustained without a continuous energy supply”. So energy needs to be supplied, and he introduces a *dissipation* term which will increase entropy in order to go from high entropy isothermal to low entropy DALR. The dissipation term takes up the slack in the entropy, but needs to be fed with energy to do so.
That the final isentropic state (=Verkley 2b) is thermodynamically unstable and will revert to an isothermal state once the continuous energy supply is turned off is stated in places such as “the further a monotonic profile deviates from the isotherm (note Verkley Fig. 2), the further it deviates from the thermodynamic equilibrium.” and “TL(p) (=Verkley 2b) cannot be in thermodynamic equilibrium.”
So all in all, Verkley 2b is lower entropy than Verkley 2a, is thermodynamically unstable and will revert to isothermal once the energy supply runs out.
A final Akmaev quotation:
“The classical isothermal solution by Gibbs is the *ultimate* state of *maximum possible* entropy under the conservation of energy … Contrary to what has been suggested in some previous studies, the presence of gravity does not change the outcome”.
“once the continuous energy supply is turned off”
The energy supply must be continuing as long as density differentials caused by uneven surface heating at the bottom of the column continue to give rising parcels of air with consequent decompression and exactly equal falling parcels of air with consequent compression.
In practice, the energy supply in an atmosphere open to space is the sun which causes density differentials in the first place by heating the surface and materials in the atmosphere differently at different heights and at different locations around the planet.
The presence of gravity is important because the same strength of gravitational field varies in its effect on air parcels of different densities, hence rising and falling at different locations.
Furthermore, if there were no gravity all the warmed molecules would drift off into space as soon as they changed from solid to gas.
So, turn off the sun and all the molecules congeal on the surface. Turn off gravity and they all drift into space.
Change anything else and all that happens is an energy redistribution within the atmosphere.
How does that sit with Verkley and Akmaev ?
Stephen Wilde:
“How does that sit with Verkley and Akmaev ?”
Sounds perfect to me.
Now for the argument – what is the vertical temperature profile of a gas in a closed insulated container at finite temperature under gravity?
br1 9:47pm – “…what is the vertical temperature profile of a gas in a closed insulated container at finite temperature under gravity?”
Reasonable earth gas T profile from surface ~1000 hPa up to ~250 hPa starts off with whatever monotonic profile initial conditions the weather may be at the time the theoretical insulated container closes. Verkley 2b/B&A prove the T profile will become non-isothermal, isentropic at LTE. Akmaev confirms Verkley/B&A and makes available some exact statements for relevant T distributions.
“what is the vertical temperature profile of a gas in a closed insulated container at finite temperature under gravity?”
“whatever monotonic profile initial conditions the weather may be at the time the theoretical insulated container closes.”
Sounds reasonable to me.
Trick:
“Verkley 2b/B&A prove the T profile will become non-isothermal, isentropic at LTE. Akmaev confirms Verkley/B&A and makes available some exact statements for relevant T distributions.”
You are incorrect about Verkley 2b, and Akmaev’s exact statements prove it.
I could repeat the exact statements, but I’ve done that already. You don’t address any of the points about them which flatly contradict your position, so I guess there is no point.
Stephen Wilde:
““whatever monotonic profile initial conditions the weather may be at the time the theoretical insulated container closes.”
Sounds reasonable to me.”
but what does the temperature profile become in the long term?
As it is closed and insulated, then according to Verkley and Akmaev, it will become isothermal no matter what profile it had when the container closed.
According to Trick the profile becomes the DALR no matter what profile it had when the container closed.
According to Graeff’s experiment it becomes about ten times the DALR, but only when convection is suppressed.
Do you have any suggestions as to how Graeff’s experiment could be the case?
For every million molecules of air there is an aerosol. Thermophoresis will induce a net force making the aerosol move to an area of lower temperature. The force of gravity is acting to concentrate aerosols and thermophoresis is inducing them upward to cooler areas where their mass, being greater than air molecules, returns them downward sooner but still retaining some heat from previous encounters with the warmer energetic molecules near the surface.
http://aerosol.ees.ufl.edu/Thermophoresis/section01.html
Don’t suppose aerosols might be causing the results found in these experiments?
br1
I’ll have to confess at trhis point that I don’t really concern myself with what might or might not happen once one closes off the container because at that point it ceases to become an adequate analogue for atmospheric behaviour.
The whole point about a planetary atrmosphere is that it is open to space and readily expands radially with virtually no resistance from external influences.
I know that the point of the Graeff experiments is to show that even in an enclosed container gravity will somehow create a lapse rate but to my mind that is unnecessary in order to account for the effect of gravity on gases under an open sky subjected to uneven warmig from external irradiation.
As I currently see it, one doesn’t need Graeff’s experiments to show any previously unobserved effect of gravity in order to explain the behaviour of planetary atmospheres.
Nonetheless if pressed on the point I would currently say that as soon as one encloses the container one removes the one feature that allows the circulation that could give rise to a lapse rate. That feature is the ability of the vertical column to expand radially with height.
The lapse rate develops because decompression towards the top causes cooling and compression towards the bottom causes warming and one gets neither decompression nor compression unless there can be radial expansion at the top so I would expect that in due course any lapse rate existing at the time of closing would disappear all other things remaining equal.
Reading more of the above somewhat belatedly I see that I’ve not said anything new in relation to Graff’s actual findings which appear to be different from what I and others would have expected.
To summarise for other newcomers to this thread:
i) Graeff’s experiments appear to show a lapse rate type temperature gradient in a closed container in the absence of convection within the container.
ii) That finding is contrary to ‘established’ science which requires convection to set up the lapse rate
iii) Graeff and others are doing all they can to exclude the possibility that the experimental result is flawed due to the influence of factors other than gravity.
iv) If gravity were to have such an effect it would need to be incorporated into current theories about how the actual lapse rate around planets arises.
However, I don’t see why we need Graeff’s experiments to explain observations whether he is right or wrong because the atmosphere is not a closed container. That feature makes all the difference since it allows convection and so far as I am aware all the numbers work well enough as per the gas laws and the standard atmosphere without needing an additional component from gravity at all.
What am I missing here ?
Has anyone considered that the process of getting to an isothermal state from the point where the container is closed might take a very long time and that Graeff might be taking his measurements before the process has been fully concluded ?
Or that external influences however well screened might actually prevent the final isothermal state from ever actually being achieved ?
Indeed, if the stabilisation period is long enough and the scope for disruption large enough even factors such as the rotation of the planet and the pull of the moon could prevent a completely isothermal outcome ever being achievable.
Everything around us is always in motion all the time. That could disrupt the outcome of the experiment whatever the level of screening.
Stephen 11:37am: “…one gets neither decompression nor compression unless there can be radial expansion at the top…lapse rate…disappear…”
Well that’s a new one. This top post has a closed air column pressure gradient caused by gravity – weight from above decreases with increasing height. There is a P(z) gradient.
PV=nRT. V,n,R constant so T(z) varies as P(z), z vertical dimension. No energy in/out. Can’t be isothermal. Move a thermometer up & down at LTE and check. Not sure why br1 can’t see that, gets lost in the context I guess.
Very easy really and 3 informed, critical recent authors prove non-isothermal obeys all the laws issuing tickets to some classic work (& 2 are even easy to read, ha). Yes, no radial expansion continuing work in/out but there you have it, lapse rate does not disappear.
Take away gravity from enclosure somehow, get no pressure gradient, P(z) then constant so T(z) = constant, becomes isothermal. Just that easy.
But yeah make V variable, allow energy in/out varying rotating big as earth, now need a supercomputer. Not easy.
br1 9:38am – “(Trick’s) incorrect about Verkley 2b, and Akmaev’s exact statements prove it.”
Understanding comes from a few small steps at a time. I’ll stick with it if you want. Clear out the context. Let’s start at the very beginning. Top post defines the analysis control volume. Again, reasonable earth assumptions.
1) Start by computing the total GHG-free air constant mass per unit area of a gas layer in a control volume between any two heights under gravity g.
Any problem with that starting point? I don’t grok latex, would put in the formula if I did. You can ref. B&A 1998 for the formula(s).
2) Add in the hydrostatic equilibrium pressure change with height in the gravity field.
Ok still?
3) Compute the total enthalpy per unit area of the layer realizing the layer possesses potential energy per unit area in earth’s gravity field.
This will look something like H = gas internal energy (or KE+PE) + (gas volume * boundary pressure w/environment). If you are still with me, or even not, I’ll continue.
4) From that, realize energy conservation (Law 1) imposes a constraint that total dry static energy is constant in the layer (within adiabatic control volume).
Following me? Anything incorrect? There are 9 more steps. If you want to see them let me know. They do get more – way more – difficult.
Trick:
In a container with a pressure reduction from bottom to top so that there can be expansion with height:
with circulation – lapse rate.
without circulation – isothermal.
because with no circulation the energy will travel through the material only by conduction which will equalise over time.
Graeff specifies no convection (at equilibrium) which means no circulation which should mean isothermal but his results show otherwise.
How does he ensure that there really is zero circulation ?
Is zero circulation possible in a container on a planet spinning through space ?
Gravity could be causing the non isothermal result not via its vertical pull on the materials in the container but by applying a centrifugal force inducing a circulation.
How could one screen that out from a surface bound experiment ?
“This top post has a closed air column pressure gradient caused by gravity – weight from above decreases with increasing height. There is a P(z) gradient.”
The pressure gradient alone has no necessary thermal consequence. Despite the different densities at top and bottom conduction alone would ensure an isothermal outcome eventually.
To get a lapse rate you have to have constant decompression and compression which requires a circulation between regions of differing pressures.
No energy is added or taken out but the energy that is present from the beginning gets redistributed with more going to the compressed regions to create a lapse rate.
Work is being done by the circulation and that work redistributes the available energy from the isothermal state or possibly prevents an isothermal state ever being achieved.
So if Graeff is getting any sort of lapse rate he has to have a circulation in there somewhere.
Stephen 12:18pm:
Of course you know I will say the real questions are: “..process of getting to a NON-isothermal state…will anything real ever prevent the final NON-isothermal state from ever actually being achieved?”..for the top post setup.
Either way you point out 2 tough questions driving experiments b/c thermo theory tells us precious little about time.
Stephen 12:31pm: “…with no circulation the energy will travel through the material only by conduction which will equalise over time.”
Not physical! This will violate conservation of total gas energy (enthalpy really) in Law 1 so theory says cannot happen. See step 4 above.
Why Graeff seeks to constrain convection escapes me. There is no need theoretically to do so in finding non-isothermal, isentropic point at LTE.
The time to LTE is not known b/c real heat flow overtakes theory at some unknown but experimentally determinable time. Graeff’s curves seem to indicate a short term and long term temperature trend difference; therein is a clue.
Stephen 1:12pm: “The pressure gradient alone has no necessary thermal consequence.”
Yes, there IS a necessary thermal consequence. To rephrase Lucy in 2nd post: “Alas if your theory is found to be against the ideal gas law eqn. of state PV=nRT, then I can give you no hope; there is nothing for it but to collapse.”
br1 9:38am – Continuing on to pinpoint just where br1 thinks Trick is incorrect:
5) From step 4 at 12:56pm, realize and compute the total entropy (S) of the layer over the height of the layer.
Need this eqn. because we want to eventually invoke 2nd law from which there is no escape. But we want to make sure we invoke the proper eqn. of state which doesn’t have height in it so…
6) Transform S integration from height to pressure by invoking hydrostatic eqn. above and ideal gas law eqn, of state: PV=nRT
7) Now just maximize this entropy S to find nature’s required ideal gas temperature profile over the height of the unit area layer from law 2, LTE is achieved at max. S.
8) To do that, confine the math maximization process to reasonable range of pressures found in earth’s atmosphere (roughly 1000mb down to 200mb ~80% of atmosphere) to make no serious errors.
With me still? Lost? Anything incorrect creeping in yet? 5 more steps to go. Increasingly difficult to grok.
Trick, I referred to the pressure gradient ALONE.
Thus referring me to the gas law is a non sequitur.Why do you think a column containing varying pressure cannot become isothermal in the absence of circulation.?
I don’t see how conduction leading to an isothermal state is not physical. Please explain.
As regards your 13 steps I must admit that my technical skills aren’t up to critiquing them but what I would say is that I generally find that if it gets that complex there is likely to be a flaw in there somewhere.
Nature invariably boils down to very simple concepts.
Stephen 2:15pm – “I don’t see how conduction leading to an isothermal state is not physical. Please explain.”
Ideal gas pressure can’t have a gradient ALONE, the eqn. of state controls all things.
The above IS a good question to ask, your expression is intuitive but incorrect. If conduction were able to do this, it would have to violate thermo Law 1. Conduction IS operating normally (as is convection) but neither can violate Law 1. You seem to understand how convection does not violate Law 1 – if something goes up, then something equal and opposite comes down. Ok, that means convection is cool with Law 1.
So what about conduction. Sort of the same thing: think if something heats up by conduction in the column something equal and opposite must cool down since no energy in/out.
This is proven formally & in step 4 where the gas enthalpy is held constant by invoking law 1. If the container is open, then H formula I posted at step 3) has the second term able to vary. However constraints change In this top post to a closed container, the H second term is constant so drops out to zero in the variation. Just left with varying H = gas internal energy (or PE+KE) term with height and hydrostatics tells us then is same as varying with pressure.
Stephen 2:15pm: “I generally find that if it gets that complex there is likely to be a flaw in there somewhere. Nature invariably boils down to very simple concepts.”
Slight mod.: “..there is MORE likely to be a flaw..”
Agreed, like the probe that missed Mars due to a simple unit conversion flaw.
Thermo nature boils down to the 0th, 1st and 2nd law, simple concepts, right? Why then can’t we all come to terms? All 3 simple concepts are invoked correctly in the 13 B&A steps along with the not so simple but really pretty easy eqn. of state ideal gas law b/c after all gas is with what we are dealing.
It is the math that gets hairy, especially in that inscrutable Akmaev paper.
“So what about conduction. Sort of the same thing: think if something heats up by conduction in the column something equal and opposite must cool down since no energy in/out.”
Well yes of course.
So in a sealed container, no convection, no energy added or removed, warmer at one end than the other energy will flow to and fro by conduction until it is all the same temperature.
Wouldn’t that occur whether or not there were a pressure gradient ?
How would it be a breach of the Laws of Thermo ?
Hi Trick,
Glad to take the discussion further. Nice approach to explain things – giving a list of the necessaries.
I got lost at step 7:
“7) Now just maximize this entropy S to find nature’s required ideal gas temperature profile over the height of the unit area layer from law 2, LTE is achieved at max. S.”
First question: what is ‘law 2′? You may say it later, but not clear yet.
Then a note: Akmaev explicitly states that the max. S attainable under certain constraint conditions is not the same as thermal equilibrium. One can have an entropy producing non-equilibrium dissipation term that allows the column S to be pinned at less than that of thermal equilibrium. However, let’s not get into that again at this stage of your presentation, so after dealing with the first question please continue.
“Wouldn’t that occur whether or not there were a pressure gradient ?”
To clarify, the temperature would be isothermal but because of the continuing pressure differential the molecules at the lower pressure would be carrying more potential energy than those at the higher pressure.
Now, that is the position with no circulation as proposed in the top post.
If one then introduces a circulation the molecules at lower pressure move to higher pressure so some of the potential energy converts to heat and the temperature rises as a result of compression.
Meanwhile the molecules at higher pressure move to lower pressure so some of their heat converts to potential energy and the temperature falls due to decompressio.
And then you have the lapse rate with denser molecules warmer than the less dense molecules.
The secret is that the changes in pressure swap potential energy for heat with the sign of the effect being opposite at each end of the pressure gradient so as to produce the lapse rate.
A simple redistribution of the available energy so no perpetuum mobile and no breach of the Laws of Thermodynamics.
Note that the whole process implies that with no circulation the equilibrium state must be or must be capable of becoming isothermal hence my suggestion that Gaeaff’s results are a consequence of a circulation of some sort remaining in his container despite his best efforts to exclude external influences.
Stephen 4:56pm – “So in a sealed container, no convection, no energy added or removed, warmer at one end than the other energy will flow to and fro by conduction until it is all the same temperature. Wouldn’t that occur whether or not there were a pressure gradient ?”
No – only get same temp. when no pressure gradient (i.e.no gravity). There IS convection in the sealed container under gravity since nothing in theory prevents some parcel of air moving up while an equal and opposite parcel moves down as you have earlier pointed out when discussing the real atmosphere. Gravity is a big game changer & only thing allowed to cross the closed control volume.
Why? More specifically with gravity field a closed, constant internal total gas energy (PE+KE) will flow to and fro by conduction, convection, turbulence, vigorous mixing, call it what you will – all physical and eventually given the adiabatic control volume settle down to non-isothermal gradient due to pressure gradient at LTE. Even at LTE, some perturbation might happen randomly or say nonrandom by a big heavy truck rumbling by Graeff’s house late in the night. System settles to non-isothermal again.
Stephen continues: “How would it be a breach of the Laws of Thermo ?”
With gravity & closed, 1st law of thermodynamics is violated when you say constant T. Energy can neither be created nor destroyed – simple, eh? Energy cannot be harmed but if you claim isothermal then energy will be destroyed as the gas particle climbs against gravity (isothermal temperature KE = same, PE goes up with h so non-constant total energy – against the law).
NB: this is the big deal about constraints. If change them to an open container, then get isothermal solution as the total energy is conserved by the air column doing work on the air column above and below (+/-PE = -/+ work on environment) & just the right amount of work to obtain isothermal at LTE. Very cool.
br1 5:31pm – “First question: what is ‘law 2′? You may say it later, but not clear yet.”
A: The 2nd law of thermodynamics, the one that humbles us all. But it is our friend to understand nature. Like friction is our friend, mostly.
br1 continues: “One can have an entropy producing non-equilibrium dissipation term that allows the column S to be pinned at less than that of thermal equilibrium.”
No. Get a ticket from the thermo police or a rap on the knuckles from my thermo prof. S cannot be pinned lower than max. allowed in any control volume, see the very humbling law right above. Not in the closed column of the top post or anywhere. Even in the universe control volume, a whole ‘nother can of worms.
Akmaev is showing how entropy can be produced by the ideal gas to climb up to S max. from a lower S. This is no easy matter, requires REALLY hairy math. You will see it when I reach the end of the B&A 13 steps. It is really step 14, but B&A merely announce it w/o proof. Akmaev formally proves how the entropy needed to get to max. S is produced mathematically.
Ok Trick.
The issue then boils down to whether the presence of gravity inevitably creates a circulation all else remaining equal.
The gravitational effect relies on density differentials arising first so that the gravity can act differentially on separate parcels of the medium thus allowing a circulation to develop.
In the atmosphere the density differentials arise from uneven insolation due to various factors but theoretically in a closed container with no energy added and none removed there should (after a while) be no such thermal unevenness, no significant density variations and so no circulation and the entire contents would become isothermal but the pressure difference induced by gravity at one end would remain.
The only difference would be that molecules at the less dense end would have more potential energy than, but the same sensible temperature as, those at the more dense end.
Thus my suggestion that a circulation is being induced some other way such as via centrifugal force from the spinning of the Earth beneath the experimental chamber.
I don’t see how Graeff could reliably exclude ALL circulation from his experiment and that circulation need be nothing to do with the vertical pull of gravity which for all intents and purposes would be even across the base of the container.Far more likely is a sideways pressure from the medium onto the walls of the container induced by the Earth’s rotation through the gravitational field.
Also possible is an unevenness in the energy distribution from molecule to molecule at the micro level which reacts with the gravitational field to create a circulation whatever Graeff might try to do to avoid it.
I also have in mind the possibility that molecules with different balances of heat and potential energy in a single container might set up a circulation betweeen themselves to destroy the isothermal balance that would otherwise occur and of course one could say that that is a direct effect of gravity as you suggest.
I think this is more a matter of getting the concepts right than doing the maths.
Then check out the concepts with data and the maths must follow.
This issue of constant compression and decompression in the atmosphere and in a closed container has been inadequately explored in the context of the effects of different gases on a planetary atmosphere.
To summarise:
I’m sure that br1 and myself and others are right to suggest that all other things being equal gravity alone does not give rise to a lapse rate.
However there is reason to think that you are right in suggesting that in the real world other things are never equal and the changes in other things inevitably interact with gravity to frustrate the isothermal outcome.
Basically, any circulation however small within the medium will prevent an isothermal outcome and introduce a lapse rate. A circulation is intimately connected to gravitational influence.
Simply proposing a ‘lack of convection’ as Graeff does is not enough. There must be no movement at all for the isothermal outcome to be achieved and I doubt that it is possible to construct a suitable experimental setup anywhere in the known universe.
Stephen 5:33pm: “…because of the continuing pressure differential the molecules at the lower pressure would be carrying more potential energy than those at the higher pressure.”
Yes. The light has clicked on. You are lawfully thinking through PV=nRT constrained by control volume unchanging total energy = KE + PE.
Stephen continues: “A simple redistribution of the available energy so no perpetuum mobile and no breach of the Laws of Thermodynamics.”
I am going to nudge your thinking along a little more now that you have puzzled it all pretty much correctly (give or take) thru to this point in 5:33pm post.
I am going to actually suggest what might seem to be some possible heresy: the theory of the closed, perfectly insulated ideal gas DOES allow a perpetuum mobile and no ticket need be issued by the thermo police. The 2nd law, immutable as it may be, is actually ok with that.
Q: Why? Puzzle it out or read below….
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A: 2nd law of thermo allows for constant entropy. It outlaws a decrease in entropy in the control volume LTE. Now granted, this constant entropy only occurs ideally. Graeff’s experiment cannot be ideal, as such Graeff cannot construct a perpetuum mobile, entropy increases in the cylinder unstoppably to LTE. For some unknown time constant, the entropy increase ought to follow thru to non-isothermal but eventually 2nd law says entropy of the Graeff cylinder must settle at LTE with its heat environment b/c no perfect insulation exists. This of course may take a long while. Like this thread…LOL.
“b/c no perfect insulation exists”
Like I said.
And without perfect insulation gravity will always induce a circulation and hey presto a lapse rate.
But that is not a perpetuum mobile in the sense that energy is coming from nowhere.
Instead it is a perpetual recycling of the energy already available within a gravitational field. It is a closed loop with no energy gain or energy loss.
No depletion of gravity either because the work taken up by air rising against gravity is offset by work in the opposite direction when air falls with gravity.
Going one step further, if there is top of atmosphere equilibrium between energy in and energy out then none of the energy in the system and involved in the compression/decompression cycle can ever escape until the sun gets weaker or the atmospheric mass declines.
Once a planet has gained energy in the atmosphere from the initial gravitational compression then it is never lost, merely constantly recycled up and down continually being maintained.
And that is why planets with atmospheres are always warmer than S-B would predict for a planet without an atmosphere.
Stephen 7:08pm: “…the entire contents would become isothermal but the pressure difference induced by gravity at one end would remain.”
You were making good PV=nRT progress but fall off the tracks here. The entire closed, insulated contents would become isentropic and NON-isothermal. Pressure gradient begets temperature gradient at LTE.
Stephen continues to summarize: “I’m sure that br1 and myself and others are right to suggest that all other things being equal gravity alone does not give rise to a lapse rate.”
Think of the enclosed, insulated column no gravity: LTE becomes isentropic, isothermal.
Think of the enclosed, insulated column with gravity: LTE becomes isentropic, NON-isothermal.
Gravity alone DOES give rise to a lapse rate. This has been known in the literature for some time – at least since the mid-1990s. It was unknown classically by the thermo grandmasters though they suspected it. They just couldn’t do the integrations required since integral calculus wasn’t as developed and/or they had not fully mastered it.
Truesdell 1980 points out their handicap really was lacking of the integral & differential math side. He has humorous instances of Fourier running in circles trying to prove stuff page after page and even just announcing he had worked some thermo things out where there is no surviving record of him doing so.
I mean – geez, thermo history & all the famous names just reminds me of blog handles in extremely slow motion. Decades to a whole century of threads to get to non-isothermal solution.
Meanwhile, back at the br1 ranch, where I stopped with step 8): wherein pretty much as usual integral calculus is made easier due to some simplifying assumptions that allow progress to understand nature.
Step 9) Transform variables to make the integral calculation much more simple (math. Dept. instruction will help us here – nothing new in therrno).
10) Obtain the function for which the layer’s entropy S is maximum subject to energy conservation constraint in the layer.
11) To simplify, constrain this maximization process to a linear set of temperature profiles (pretty good for earth’s atmosphere up to 200mb by observing the ICAO standard atmosphere.)
12) The mathematical condition for entropy S to be an extremum is that its derivative vanish, set the derivative of this function to zero.
13) Do the algebra, find the nature of the entropy S extremum IS a maximum i.e. isentropic point & only with non-isothermal temperature profile.
Voilà. QED.
14) Prove that the assumption in step 11) was actually not really needed generally. Akmaev paper shows the work. B&A offer discussion only.
“Think of the enclosed, insulated column with gravity: LTE becomes isentropic, NON-isothermal.”
Without a circulation ?
My position is that it can only become non isothermal with a circulation due to the need to compress and decompress so that density differentials lead to a higher temperature at the end nearest the gravitational field by switching kinetic and potential energy around.
With no circulation as proposed by Graeff for a perfectly insulated container (which I aver is not possible) then you could theoretically have an isothermal outcome. After all, if one prevents exchange of kinetic energy for potential energy nearest the gravitational field then it would become isothermal wouldn’t it ?
Since I don’t think that perfect insulation is possible I favour a non isothermal outcome in the real word as per Graeff’s observations and your contentions.
It may be a semantic issue as to whether the non isothermal outcome is due to gravity alone. I think it is gravity plus a circulation working with and against gravity as a reversible adiabatic process and so if the circulation could somehow be suppressed the outcome would only then be isothermal.
Either way the system is isentropic as you say because there is no change in entropy in a reversible adiabatic process and if circulation is suppressed within a perfectly insulated system then there is no change in entropy there either.
Stephen 8:26pm: Non-isothermal “Without a circulation ?”
Well, that depends on what you really mean. There is nothing in the theory for the closed, insulated container that imposes “circulation” or demands “no circulation”. No thermo law needs “circulation” explicitly. Else explain it better.
By “circulation” my explanation is you mean the ideal gas molecules are free to move & mix naturally from top to bottom w/o constraint – no laws are broken, no enthalpies or energies are harmed. If so, that is certainly ok with ideal gas law, 0th, 1st & 2nd thermo laws.
Graeff’s introduction of the powder in the B74 air experiment is bizarre; reducing convection is stunningly unneeded to find non-classic non-isothermal profile. At LTE all I can suggest happens is that the powder has simply wafted or dropped to the bottom of the cylinder and lies there inert. Like dust on my bookshelf here.
“you mean the ideal gas molecules are free to move & mix naturally from top to bottom w/o constraint”
Correct, with the proviso that such movement would form a pattern (circulation) according to the cause, nature and scale of the disequilibrium between the container contents and the environment outside the container.
Since there can be no perfect insulation there will always be something that will cause such movement in the container thereby preventiing an isothermal outcome.
To achieve an isothermal outcome Graeff needs to eliminate not only convection but also ANY movement of molecules up and down within the container. I don’t think he will ever manage that.
If there is ANY conversion of kinetic energy to potential energy and back again then there will be a lapse rate
The cause isn’t gravity alone but rather movement within a gravitational field and the engine for that is not gravity per se but density differentials that can arise for multiple reasons. Gravity then acts differentially on the density variations to induce movement up and down thereby converting energy back and forth between kinetic and potential in different locations.
The lapse rate is the physical manifestation of the differing proportions of kinetic and potential energy held by individual molecules or groups of molecules in different places.
Trick wrote:
Graeff’s introduction of the powder in the B74 air experiment is bizarre; reducing convection is stunningly unneeded to find non-classic non-isothermal profile.
Graeff wasn’t going for that. He was attempting to isolate the effects of gravity.
At LTE all I can suggest happens is that the powder has simply wafted or dropped to the bottom of the cylinder and lies there inert.
IIUC, he used sufficient volume of powder to fill the tube. This does bring into question the properties of the powder. As noted, they may be filled with air, and even sealed.
One of the models I used some years ago was a vertical column, open at the top, but sufficiently tall that the pressure at the top is negligible.
In the Micro-Canonical Ensemble, it’s pretty easy to calculate the required height for a ‘gravitational seal’. If you assign all of U to one particle as PE, then KE + U = U, or KE = 0. From this, it’s trivial to calculate the height z for which T = 0.
It’s a lot more difficult in the Canonical Ensemble, but I think a similar approach still works. The probability of having particles with extreme PE needs to match up with the probability of the system having the corresponding internal energy.
Thinking about that last phrasing, I might actually be able to crank through the math on that.
Trick:
“10) Obtain the function for which the layer’s entropy S is maximum subject to energy conservation constraint in the layer.”
Constraint of energy conservation is Verkley Constraint 2, which he says gives an isothermal profile. As this doesn’t seem to be what you are after, maybe you could clarify what you mean here?
“No. Get a ticket from the thermo police or a rap on the knuckles from my thermo prof. S cannot be pinned lower than max. allowed in any control volume, see the very humbling law right above. Not in the closed column of the top post or anywhere.”
Sure it can – just think of a refrigeration compartment. This is actually a very good analogy for Verkley 2b and all the Akmaev stuff. Now read Akmaev Eqn(25) and onwards. If one imagines the isentropic profile as analogous to a freezer compartment, one can ask the question – how can one go from a high entropy state (room temperature/isotherm) to a low entropy state (a cold freezer compartment/DALR)? Akmaev even says “the dry-adiabatic profile corresponds to a state of lower entropy than a stable profile with the same column enthalpy” how can you ignore such a statement? If you simply write Eqn(25) (by analogy, ‘the freezer cools down’), then all you get is a breaking of 2lot, which nobody is happy about. So Akmaev points out that one needs a continuous energy supply and a dissipation term, which he calls epsilon_t. This term basically acts (in the freezer analogy) as the plug in the wall, the motor and the compressor/decompressor – it represents the additional machinery at work which can drive the freezer compartment’s entropy lower. After a while, the freezer reaches its ‘equilibrium’ temperature, and this is a steady-state where the entropy of the freezer compartment no longer changes, and the Lagrange multipliers give as a stable solution. By analogy, this corresponds to the isentropic temperature profile, which is the steady-state of the atmosphere in the presence of turbulence and dissipation power supplies. But the freezer compartment is lower temperature than room temperature, so it has lower entropy than what the compartment would have if there wasn’t a motor attached sucking power from the wall socket and driving up the universe entropy at an equal or greater rate. That the freezer compartment is not at thermal equilibrium (despite the fact that the derivative of its S is not changing) is easily noted when one unplugs the power supply or the power supply runs out. In that case it will warm back to room temperature (the profile TL(p) will go back to being an isotherm). Akmaev says as much with “This also means that the dry-adiabatic profile TL(p) maximizing entropy under condition (2) always corresponds to a state with a lower column entropy SL than the maximum SP attainable with the same total energy: TL(p) cannot be in thermodynamic equilibrium.”
I’ve grokked the above, I hope you can grok it too. Could be a good chance for Stephen to do a bit of grokking too.
Stephen 9:38pm – “To achieve an isothermal outcome Graeff needs to eliminate not only convection but also ANY movement of molecules up and down within the container. I don’t think he will ever manage that.”
Yeah, can only manage that in gas by turning off gravity. But my understanding at 1st was Graeff is trying to find the non-isothermal ideal gas lapse rate different from the real atmosphere lapse rate. Of course, I’m second hand – all I know of his work is what others have written.
Q. Daniels 10:07pm – “…(Graeff) used sufficient volume of powder to fill the tube…”
Yes, Lucy was going to double check and in post 2 found: “The innermost Dewar (1) of 1/2 litre was filled with a fine powder in order to eliminate convection currents and radiation between the inner wall surfaces.”
If truly ½ litre of powder is in the ½ litre dewar, then I no longer have any idea what the B74 experiment is trying to accomplish (conduction in powder?). But it is interesting discussing the columns of free ideal air that have no powder – the classic setups.
br1 10:37pm: “Constraint of energy conservation is Verkley Constraint 2, which he says gives an isothermal profile.”
Energy conservation is 1st law of thermo, hence used as a constraint in law abiding Verkley 2a isothermal solution and 2b non-isothermal so to be consistent with thermo law.
What you are trying to grok I think is the gas enthalpy is different between an open air column in 2a and a closed air column in 2b.
2a open: control volume gas enthalpy = internal energy + work on environment = constant1
2b closed: control volume gas enthalpy = internal energy + 0 = constant2
B&A steps are for a closed container so gas enthalpy constant2 = internal energy + 0 = KE + PE + 0 = ½ mv^2 + mgh + 0 for this step:
“10) Obtain the function for which the layer’s entropy S is maximum subject to energy conservation constraint in the layer.”
The rest will take awhile to parse; maybe as long as LTE. Whew.
br1 10:37pm – “Akmaev even says “the dry-adiabatic profile corresponds to a state of lower entropy than a stable profile with the same column enthalpy” how can you ignore such a statement?”
I don’t. Translation from Akmaev-ese: “..the isentropic closed column corresponds to a state of lower entropy than observed standard atmosphere profile with the same column enthalpy”.
Translate into Table 1 Verkley-ese. Isentropic H = 2.0106. Observed H = 2.0032. They are not the same enthalpy because the isentropic is closed container (remember H = KE + PE + 0 = const. = 2.0106) and the observed is open container able to do work on the column environment above and below (H = KE = const. = 2.0032) – see less H.
Now Akmaev says set the SAME enthalpy H. Lower the isentropic enthalpy of 2.0106 down to 2.0032. Akmaev says find S is now lower for the isentropic than observed profile. That’s all, look at S = 4.3769 for isentropic – I suppose it could end up lower than observed S = 4.3764 at same H of 2.0032; I’ll let you do the math.
Ok, then why? Well the observed is higher entropy S than lower S isentropic at same enthalpy H because observed has more entropy S from say galactic rays, GHGs like water vapor clouds & CO2, aerosols, et.al. than ideal isentropic Any problem seeing that (barring typo’s that is)?
Using a refrigerator just moves the air columns to say the North Pole surface T < 288K avg. Doesn’t change anything.
br1 said:
“Could be a good chance for Stephen to do a bit of grokking too”
I’m not sure which point I need to grok.
I’ve agreed with br1 and Almaev that if one could prevent convection (actually ANY movement up and down) then in theory one should get an isothermal column despite a pressure difference at each end.
However I don’t think that such complete suppression of movement is possible at a molecular level in a gas. For example, even the movement of the container through the gravitational field as the Earth spins would produce forces within the column tending to induce movement.
If one could achieve an immobile and thus isothermal setup with pressure discrepancies at each end the kinetic energy would be the same throughout but the molecules at the lower pressure end would additionally have more potential energy.
Trick rightly points out that that would offend the gas laws and of course it would because such immobility is characteristic of solids and not gases.
As soon as molecules at the isothermal temperature start to move towards the low pressure end they will lose kinetic energy (cool) but gain potential energy and in the process force molecules at the other end to move in the opposite direction and gain kinetic energy (warm) at the expense of potential energy.
That would then create a non isothermal temperature gradient from high pressure to low pressure with highest temperature where pressure is highest.
So unless Graeff can somehow get his gases to behave in an immobile fashion as if the gas were a solid then he is never going to be able to create an isothermal outcome.
As far as I can see the introduction of movement within the gravitational field as the critical factor rather than the presence of gravity itself resolves the problem between br1 and Trick.
If a scenario could be created whereby there is no movement then br1 would be right. If movement is inevitable then Trick is right.
Trick:
“energy conservation constraint in the layer.”
no, there is no ‘energy conservation constraint in the layer’ in Verkley 2b!
Verkley himself refers to this in phrases in section 2b such as
“Even when on the scale of the turbulence the heating rate J would be zero, this now needs not to be the case. Indeed, (14) is expected to have a nonvanishing right-hand side”
So heating of the layer under consideration is explicitly allowed.
One can also see this simply by noting that H is unconstrained in Verkley 2b! H is allowed to vary to any value consistent with the constraint that L has the measured value of L. So again, conservation of energy is very much not assumed.
Verkley points out it is this that B&A make a mess of:
“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. This way of obtaining 3 can be criticized on the grounds that, had no approximation been made, one would have found an isothermal instead of an isentropic profile”
So if you strictly follow B&A you get an isothermal profile!
Maybe this has caused confusion.
“Now Akmaev says set the SAME enthalpy H. Lower the isentropic enthalpy of 2.0106 down to 2.0032. Akmaev says find S is now lower for the isentropic than observed profile. That’s all,”
no, he says it is lower than the isothermal profile, hence the isentropic profile is not in thermal equilibrium. Yet again I quote Akmaev p192:
“TL(p) cannot be in thermodynamic equilibrium” where TL(p) is the isentropic profile (=Verkley 2b). Check it out, Akmaev says this comes straight from his own Eqn(2) which is Verkley Eqn(13), it really can’t get any clearer.
“Using a refrigerator just moves the air columns to say the North Pole surface T < 288K avg. Doesn’t change anything."
The example of the refrigerator compartment is to point out a very clear example where S is pinned to a lower value of S than you would get if you didn't have "dynamical dissipative processes" (Akmaev p195) which can take up the entropy slack. The analogy goes as:
1, fridge compartment at room temperature vs fridge compartment at cold temperature
2, isothermal profile vs DALR profile
In each case one needs a power supply to reduce S, which becomes pinned at a value lower than what it would be in thermodynamic equilibrium. Once you turn off the power supply, the system reverts to 'true' equilibrium.
Stephen Wilde:
“I’m not sure which point I need to grok.”
don’t worry, I was just play around with the word ‘grok’, which is one of the better contributions Trick has made to this discussion
“If a scenario could be created whereby there is no movement then br1 would be right. If movement is inevitable then Trick is right.”
actually, we are both caught short in any case – Graeff’s experiment shows ten times the DALR!
Verkley Fig. 1: “The column is assumed to exchange no net heat with its surroundings…”. Verkley continues: “…the assumption of no net heat exchange, which is the essence of keeping H fixed…”. B&A figure for V2b: “adiabatic.. column boundary assumed to allow no heat in or out…”.
br1 9:56am – “no, there is no ‘energy conservation constraint in the layer’ in Verkley 2b!”
Emden 1926: ‘‘Periodisch wiederkehrende Irrtu¨mer’’. I agree with Emden (and Exner).
And building on Eddington in Lucy’s 2nd post, Trick: “…if your theory is found to be against the 1st law of thermodynamics I can give you no hope.”
br1 needs to better explain not being a law abiding poster.
Trick:
“br1 needs to better explain not being a law abiding poster.”
will do:
Note that your first two Verkley quotes lead up to Eqn(6), where he says:
“If in the variational process no net heat is to be transferred to or from the column, the enthalpy
Eqn(6) is to remain constant.”
which he uses in section 2a, Eqn(9).
But he doesn’t use this in 2b, instead he *only* uses Eqn(13) and Eqn(15). In these equations, constant H is not mentioned. That H is allowed to ‘float’ can be seen from Table 1 when comparing isothermal and isentropic profiles.
Furthermore in 2c he says:
“Keeping M and H fixed leads to a uniform absolute temperature (isothermal profile); keeping M and L fixed leads to a uniform potential temperature (isentropic profile).”
It’s the very fact that 2b doesn’t fix H that allows you to get a different profile.
Switching over to Akmaev, apart from all the quotes I’ve quoted numerous times, here is another one (Akmaev, p294, bottom right):
“Interestingly, it also appears impossible for a physical process described by Equation (31) to simultaneously conserve the column enthalpy (Equation (1)) as well, as suggested by Verkley and Gerkema (2004).”
It took me a few goes to grok that one, but it is important.
So one of the highlights of Akmaev’s paper, where he actually goes about pinpointing the physical processes which can reduce column entropy (by increasing it elsewhere via dissipation terms), does not conserve column enthalpy. Verkley noted this in his discussion on heating in section 2b.
Of course, there *is* conservation of energy, but it is just not conservation of energy *within the layer*. That B&A screwed up here is noted by both Verkley and Akmaev who say that B&A would end up with an isothermal profile if they followed their own instructions.
“actually, we are both caught short in any case – Graeff’s experiment shows ten times the DALR!”
Probably because the effect of the circulation (that he has been unable to eliminate) inside the container is proportionately ten times greater than the effect of the atmospheric circulation in the real world outside the container.
I don’t think he will ever be able to eliminate such circulation within any container.
Tim Folkerts says, June 7, 2012 at 5:34 pm: If there was something wrong with something as fundamental as either the 0th or 2nd Laws, it would have been damn near impossible to hide it for this long.
==============
Gravity is an energy source. There is nothing that violates thermodynamics if nature finds a way to make use of this energy source.
Molecules move. Under the influence of gravity their paths are modified from straight lines to curved lines. This curvature reflects a conversion between potential and kinetic energy.
When the sun evaporates water, and this later falls as rain and eventually powers a turbine, how is this any different? You are seeing the conversion between kinetic and potential energy by the sun, which we have learned to harness.
Gravity is energy. Harnessing gravity is not a violation of thermodynamics.
Ferd 2:51pm – “Gavity is energy.”
Strictly speaking gravity is a force field F = mg. When gravity is able to move mass a distance d then energy = F * d becomes part of the total energy in any control volume.
br1 1:01pm – “But (Verkley) doesn’t use (constant H) in 2b… “
Yes, Verkley does use constant H in 2b control volume. He must, it is the law. Even though no mention can you find. Some things (esp. in a condensed paper like this) are just assumed – the informed, critical reader might have to dig thru assumptions or cites, in this case the diggin’ leads to B&A cite.
br1 continues: “…constant H is not mentioned…”
You can’t see it explicitly in the condensed Verkley writing cited from B&A but in B&A which you can’t see, they do use constant H in 2b. See step 4 for 2b derivation of non-isothermal closed insulated column above 6/18 12:56pm: “…realize energy conservation (Law 1) imposes a constraint that total dry static energy is constant in the layer (within adiabatic control volume). “
br1 continues: “It’s the very fact that 2b doesn’t fix H that allows you to get a different profile.”
No. Maybe I’ve said a few times in different ways, but grok it please: the control volumes of 2a and 2b are defined different. So Verkley gets different H in each. But in each control volume, 2a & 2b, H must be conserved – it is the law. H (enthalpy) is the total gas energy in the control volume, energy cannot be harmed or created, energy H must be fixed. Account for energy.
Let me say it again, energy H is fixed in 2a and 2b control volume. It is the law. But H can be different if the control volume is different and I have shown br1 why above. (Hint: find the eqn. for enthalpy H.)
br1 continues: “Of course, there *is* conservation of energy, but it is just not conservation of energy *within the layer*.”
Let me say it a 3rd time, energy H is fixed or conserved or constant call it what you will & yes, even within the layer. Any control volume you name, I’ll respond to say total energy is conserved in there, just gotta’ do accounting right. Sometimes that’s tough, dang tough.
“That B&A screwed up here is noted by both Verkley and Akmaev who say that B&A would end up with an isothermal profile if they followed their own instructions.”
No. B&A conserves energy in the control volume. Always. B&A follow the law(s). Perfectly.
Trick:
“the informed, critical reader”
mmm hmmm…
“Let me say it a 3rd time, energy H is fixed or conserved or constant call it what you will”
do I detect a touch of exasperation?
“B&A conserves energy in the control volume. Always.”
there is a teensy weensy problem with this. This itsy bitsy problem has been pointed out before, but seeing as it hasn’t gone away, let’s look at it again.
Here is what Akmaev has to say about B&A (Akmaev p190):
“Bohren and Albrecht (1998), apparently anticipating that P cannot be conserved exactly under condition (2)…”
ooh, what’s this? B&A can see that enthalpy cannot be conserved during the L optimisation process?
Let’s continue the quotation:
“…explicitly assume that the conservation of L approximately implies the conservation of P”
what’s this? B&A are conserving enthalpy? But really they shouldn’t?
And there’s more:
“…as the ratio T/θ = Pi(p) does not change much for sufficiently close p1 and p2. However, this assumption is hardly defensible for any extended layer…”
B&A’s position is not defensible? What about that law we’ve been hearing about?
But what difference does it make? Read on:
“…because the maximum-entropy temperature profiles corresponding to the two constraints differ drastically (e.g. Verkley and Gerkema, 2004, their figure 2).”
so you get a completely different profile if you actually follow B&A than what they are claiming?
Well, let’s see what profile we would get if we followed B&A properly. I notice Verkley was quoted, what is his view on the subject?
Let’s see – Verkley discusses B&A at the top of p932:
“had no approximation been made, one would have found an isothermal instead of an isentropic profile”
Sounds clear to me. OK, so let’s grok that all together – B&A enforce conservation of enthalpy, but this is not correct for the problem they are trying to address and leads to the wrong solution.
But surely nobody could claim enthalpy is not conserved, what about the law?
Well, Akmaev has this to say (Akmaev, bottom of p194):
“Interestingly, it also appears impossible for a physical process described by Equation (31) to simultaneously conserve the column enthalpy (Equation (1)) as well”
seems like you blipped over this quotation in your last post. Akmaev has identified a process that doesn’t conserve enthalpy. But there’s more if one finishes that sentence:
“as suggested by Verkley and Gerkema (2004).”
in other words, if one reads Verkley properly, he also says this.
Maybe a trip to Specsavers might come in useful?
Watching Trick and br1 go around and around, I’m not sure either has the math completely right, and maybe neither.
Can either of you reconcile your understandings with Graeff’s results?
br1 11:10am – Trick blipped over: “Akmaev has identified a process that doesn’t conserve enthalpy….if one reads Verkley properly, he also says this.”
Is br1 really prepared to blip down the path purporting the Akmaev paper has found the 1st law of thermodynamics has been overturned with some identified process AND with support from Verkley paper?
If so, please go into more detail and better ‘splain the Akmaev identified process along with Verkley support precisely. Show your work and where I fall off the tracks. Until you do, I’ve already shown how both papers along w/B&A correctly support & invoke the 1st law of thermodynamics.
Q. Daniels 12:35pm: “Can either of you reconcile your understandings with Graeff’s results?”
If Graeff’s 1/2L inner dewar in B74 is really filled with 1/2L of powder as Lucy writes, then no. The ideal gas law is not applicable as PV=nRT is not the eqn. of state anymore.
Trick wrote:
If Graeff’s 1/2L inner dewar in B74 is really filled with 1/2L of powder as Lucy writes, then no. The ideal gas law is not applicable as PV=nRT is not the eqn. of state anymore.
I don’t understand this statement.
Within the beads, V and n are constant. Reading the Data Sheet, the beats account for 55 to 68% the total volume. The material of the beads accounts for roughly 5% of the volume, with the balance being air within the beads.
The space between the beads also has a constant, though irregularly shaped V. If the tubes are fully sealed, then n is constant as well. This volume also accounts for 32%-45% of the total volume.
Trick:
“If so, please go into more detail and better ‘splain the Akmaev identified process along with Verkley support precisely. ”
It’s not that 1lot is violated, it is simply a matter that the column can no longer be considered an isolated system. Energy passing into and out of the column changes the column enthalpy. At a smaller scale, energy passing into and out of a particular layer is absolutely necessary if one is to redistribute potential temperature in an optimised way. This is said so many times across Verkley and Akmaev that I won’t quote it all (and I’ve probably already done so), but here are two quotes to note:
Verkley 2b, p934,
“Even when on the scale of the turbulence the heating rate J would be zero, this now needs not to be the case. Indeed, (14) is expected to have a nonvanishing right-hand side”
Akmaev says a lot more about heating, but here is one, (Akmaev p188):
“It may be shown, for example, that for any initially stable stratification, maximization of entropy under condition (2) always requires additional energy, and so cannot occur spontaneously in an isolated layer.”
Note that ‘initially stable’ includes isothermal. ‘Additional energy’ clearly refers to energy input from outside the layer in cosideration, so this will not conserve the enthalpy of the layer. An ‘isolated layer’ is one that can have no energy transfer across it’s boundaries – such a layer clearly won’t do.
Another important one can be found on Akmaev p193
“Physically, turbulence in a stable stratification (or forced convection) cannot be sustained without a continuous energy supply”
So to get to the isentropic state, one needs a power supply. For the Earth’s atmosphere, this ultimately means the Sun’s input (note Akmaev’s discussion of radiation on p192), though at a local level it may manifest as turbulence due to, for example, “the reservoir of large-scale kinetic energy, for example by a background wind-shear instability.” (Akmaev p193). Such a process can dump energy into a layer, changing it’s enthalpy and sending it to the isentropic lapse rate. It is clear though that such a situation can hardly be described as ‘thermodynamic equilibrium’!
Grok or non-grok?
Q. Daniels:
“Can either of you reconcile your understandings with Graeff’s results?”
I certainly can’t.
But I also can’t find a flaw in Graeff’s experiment. I would love for the experimental results to stand, so I keep looking for possible ways to explain them. One reason I keep up the detailed discussion with Trick is that I want to learn all the little details about temperature gradients, and I have certainly learnt a lot since I came to this forum. For that: thanks!
br1 11:09am: “It’s not that 1lot is violated, it is simply a matter that the column can no longer be considered an isolated system.”
NO! Akmaev is tough to read but he is not telling us here what you write. The column in 2b CAN continue to be an isolated system non-isothermal, isentropic ideally. See the 13 steps above. They are cool with all laws so Akmaev is ok with that (he says in the conclusion).
br1 continues: “Akmaev says a lot more about heating, but here is one, (Akmaev p188): “It may be shown, for example, that for any initially stable stratification, maximization of entropy under condition (2) always requires additional energy, and so cannot occur spontaneously in an isolated layer.””
Watch the pea under Akmaev’s thimble very carefully. You are not doing so. Here Akmaev is writing about any initially stable column including one such as V2b (or one in the monotonic atmosphere at large) BEFORE reaching max. entropy point – see: “maximization of entropy under condition (2) always requires additional energy and so cannot occur spontaneously in an isolated layer.”
This is true – if additional energy cannot occur spontaneously in one isolated layer then another isolated layer has to give up energy. All authors agree on this point for 2b entropy rising up to max. entropy point at LTE. You even seem to explain this but I fear your context “outside the layer” means you think this is across the control volume – it is not crossing the cv. All this is happening inside the cv.
This (inside cv) is confirmed in the very NEXT Akmaev sentence which you didn’t clip (blipped by it I guess): “Conversely, energy has to be released by any initially unstable layer to maximize entropy and satisfy condition (2).” All inside cv.
br1: “So to get to the isentropic state, one needs a power supply.”
Yikes. See my fears are confirmed. No, here you are wrong; there is no power supply needed in Verkley 2b or in any step 1-13 above, or any need for one in theory as Akmaev JUST WROTE.
Then Akmaev extends all this to say generally: “…with the same column enthalpy, a temperature profile maximizing entropy under condition (2) possesses less entropy than any monotonic temperature profile that is closer to the isotherm (i.e. has a smaller lapse rate), and so cannot be in equilibrium.”
Here Akmaev is telling us if a column of real atmospheric air happens to attain non-isothermal, at the adiabatic max. entropy, this max. entropy is lower than what can be attained when let loose in the monotonic profile (real atm.) & therefore it cannot be in LTE anymore and so moves to higher entropy towards the “isotherm”. This is not surprising, Verkley Fig. 2 shows us that IS the case.
You are getting closer to the “Aha!” grok moment that when you encounter something puzzling in Akmaev – you can stop and eventually parse an understanding. If stop too long, expect fullback Chebyshev to run you right over though, ha.
Trick:
Just to let you know I’ll be travelling for the next four days, so won’t be able to correct all your errors for the next while. Apart from the string of errors in the first half of your post there was some comfort in your second last paragraph where you seem to say an isothermal profile has higher entropy than an adiabatic one at the same enthalpy.
I’ll have to come back to the rest of your post on Tuesday, but I can’t believe you said some of those things! Please try and think of the consequences before I have to point them out to you.
Back in a few days.
right, sorry for the delay, where were we… Ah yes:
Trick:
“NO! Akmaev is tough to read but he is not telling us here what you write.”
Oh yes he is! (this could take a while).
“Akmaev is writing about any initially stable column including one such as V2b (or one in the monotonic atmosphere at large) BEFORE reaching max. entropy”
yes, I know that, glad we agree.
“This is true – if additional energy cannot occur spontaneously in one isolated layer then another isolated layer has to give up energy.”
This is nonsense – isolated layers cannot exchange energy – by the *definition* of isolated. However, I may let you away with this one, as I will presume you know better and just mis-spoke. But worse, you sound like you are simply asserting Akmaev Eqn(25) which I hope we all agree is rubbish, as Akmaev is at pains to point out.
“All this is happening inside the cv.”
not only. Think about what happens for a whole column – if the column has a lapse rate less than the DALR and is therefore stable, how can the whole column get to the DALR without input from outside?
“This (inside cv) is confirmed in the very NEXT Akmaev sentence which you didn’t clip (blipped by it I guess): “Conversely, energy has to be released by any initially unstable layer to maximize entropy and satisfy condition (2).” ”
umm, this refers to a low entropy unstable system that is available to give up energy – did you ever consider how power supplies are defined in thermodynamics??? If you have a power supply in the column then fine, but the situation of an initially stable column is not like this, the power supply has to come from outside the cv. Otherwise one is just asserting Akmaev Eqn(25) which breaks 2LoT. His whole point in discussing Eqn(25) is to *identify the power supply*.
“Here Akmaev is telling us if a column of real atmospheric air happens to attain non-isothermal, at the adiabatic max. entropy, this max. entropy is lower than what can be attained when let loose in the monotonic profile (real atm.) & therefore it cannot be in LTE anymore and so moves to higher entropy towards the “isotherm”. ”
YES! Something we can agree on.
Which is also why a DALR column needs a power supply not only to reach DALR but also to *maintain* the lapse rate. It is that supply which prevents it from being ‘let loose’ from its non-equilibrium state. Note that when the column is at a steady-state DALR, the active power supply doesn’t actually heat up the column or increase enthalpy any more, but it is needed to replenish the dissipated energy of the ‘dynamical dissipative processes’ which by their nature are dissipative. The clue is in the phrase ‘dynamical dissipative processes’. Once the dissipation equals the supply then the column steady-states at the DALR and the net column enthalpy no longer changes. This is the whole discussion of Akmaev after Eqn(25), but it does not mean that because the enthalpy no longer changes that the system can be closed and still maintain the DALR! Maybe you could share what ‘dissipation’ means to you?
br1 – “This is nonsense – isolated layers cannot exchange energy – by the *definition* of isolated….you just mis-spoke.”
No. Here the term “isolated” means only separate; isolated does not mean insulated. Of course separate layers w/in an insulated column cv CAN exchange energy.
Trick: “All this is happening inside the cv.”
br1 – “not only”
Yes, only. Again, Verkley 2b control volume is perfectly insulated, enclosed & shown precisely in the B&A cartoon figure.
br1 – “..the power supply has to come from outside the cv.”
Nope. Read the line right above. The cv is perfectly insulated; no power supply allowed from outside cv in V2b. Akmaev is just telling us the power supply comes from inside cv, isolated layers exchanging energy (heat i.e. your power supply) INSIDE the cv up to max. entropy point of LTE.
br1 – asks: “Maybe you could share what ‘dissipation’ means to you?”
br1 answers correctly (& I have nothing to add except the column enthalpy is not net of anything): “Once the dissipation equals the supply then the column steady-states at the DALR and the net column enthalpy no longer changes.”
br1 –“umm, this refers to a low entropy unstable system that is available to give up energy – did you ever consider how power supplies are defined in thermodynamics??? If you have a power supply in the column then fine…”
Power supplies are important in electronics, aerospace, et. al. So YES, I have considered them in thermo also. Here though, br1 is getting closer to the real crux of the matter.
All authors of interest here discuss the problem of the V2b column being at lower than max. entropy & initially unstable then climbing to the max. entropy point. How does that happen? Well, the energy to do so comes from the isolated layers exchanging energy w/in cv to dissipate any excess energy (heat) imbalance over the stable DALR. When that V2b 13 step process ends, a zero & unstable DALR is voila converted at LTE to non-isothermal, isentropic (max. point) stable column in perpetuity (ideally anyway) & heat no longer flows. Any random perturbation is damped out quite quickly since there is no power supply across the cv. How that happens exactly is proved by Akmaev mathematically. Quite formally, actually.
Folks that get conduction in solids confused with enthalpy/entropy in ideal gases cannot see this without additional study.
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Trick:
“Here the term ìisolatedî means only separate”
oh dear, redefining the meaning of thermodynamics terms – please don’t.
” Verkley 2b control volume is perfectly insulated, enclosed & shown precisely in the B&A cartoon figure.”
no, Verkley 2b says that heating is allowed, Akmaev confirms it, and both say that B&A screwed up.
“Akmaev is just telling us the power supply comes from inside cv, isolated layers exchanging energy (heat i.e. your power supply) INSIDE the cv”
Not when you apply the statement “According to statement I, a stable temperature profile cannot be rearranged to maximize entropy subject to condition (2) with no additional energy input” to a whole column. If you have a *static* stable *column* and want it to have an isentropic temperature profile of the same L, you have to add energy to it to increase H – if you only swap energy between layers inside the column (conserving column H) then you reduce its entropy, thus violating 2LoT (Akmaev Eqn(25)).
“Power supplies are important in electronics, aerospace, et. al. So YES, I have considered them in thermo also.”
then I’d expect you to recognise them a bit better. Shame you don’t seem to.
“All authors of interest here discuss the problem of the V2b column being at lower than max. entropy & initially unstable ”
no that is not the case. Akmaev spends more time dealing with how a *stable* profile can reach isentropic, as that is where the problems lie. He says “the isothermal stratification is also statically stable in a gravity field”, so a *static* isothermal column can’t reach isentropic by itself. So one needs “a continuous energy supply from the reservoir of large-scale kinetic energy, for example by a background wind-shear instability”. Maybe you wish to include this power supply inside your cv, but what if it is not in the cv? Winds aren’t always present. And of course wind-shear instabilities don’t come for free! They themselves don’t arise without power supplies (the sun being the most obvious). If for example, one has a small open thermos flask in a closed room on a windless day and you close the lid, do you think wind-shear instabilities will spontaneously arise inside???
Finally, if one considers what happens if one has used a power supply to get from an isothermal column to a ‘steady-state’ isentropic column – we find that “TL(p) cannot be in thermodynamic equilibrium”. So the isentropic profile will *not* last ‘in perpetuity’ as you say, but can only be sustained with continued application of a power supply to continually replenish the dissipated energy of the wind turbulence, maintaining steady-state enthalpy that way. Once the power supply runs out, the system reverts to isothermal, “maximization of entropy in a thermally-isolated atmosphere yields an isothermal temperature distribution”.
This is now so obvious to me that I feel I’ve nothing more to learn here. While it’s actually been a pleasure discussing this with you for the last few weeks, maybe it’s time to move on.
br1 wrote:
p.s. For those interested in writing their own simulations for this, the phi1 distribution in the thermal walls paper is not actually a MB distribution. One can generate velocities in the phi1 distribution by using its inverse cumulative density function
V_phi1=sqrt((-2*kB*T/m)*log(1-rand))
where ‘rand’ is an evenly weighted random number selected between 0 and 1.
I finally got around to checking this. I get a different ICDF for phi1, ending in:
log(1-sqrt(rand))
It lines up better graphically, but I don’t trust it. Do you have a reference I could use for checking my math?
Also of note, whether I use (1-rand) or (1-sqrst(rand)) I get a very interesting artifact. I’m plotting the entire distribution, and then cutting out the lower half of the population, and subtracting out the minimum KE (sqrt(v^2-vmin^2)). It’s decidedly not the same temperature.
I did not expect that result.
Q. Daniels:
Quick reply:
The PDF is
phi1=abs(v).*(m/(k*T)).*exp(-m*v.^2/(2*k*T))
This can be integrated to give the CDF via
http://integrals.wolfram.com/index.jsp?expr=x*%28m%2F%28k*T%29%29*exp%28-mx%5E2%2F%282*k*T%29%29&random=false
Note that the integral goes from 0 to v, as the initial distribution is only defined over positive v (due to it representing a solid wall where the particles leave the wall in the positive direction).
This gives
CDF=1-exp(-m*v.^2/(2*k*T))
Solving for v goes as
exp(-m*v.^2/(2*k*T))=1-CDF
-m*v.^2/(2*k*T)=log(1-CDF)
v^2=(-2*k*T/m)*log(1-CDF)
v=sqrt(-2*k*T/m)*log(1-CDF)
Replace CDF with a random number and hey presto.
About cutting off the lower velocities – this will surely change the temperature, but I’m not sure what you are doing with that – can you elaborate a little?
ah, the problem is in your last step. The last formula should be:
v=sqrt((-2*k*T/m)*log(1-CDF))
The distribution of sqrt(log(1-CDF)) != log(1-CDF).
My solution wasn’t correct, either.
I’m working in Excel, using the CDF form. It’s an easy way to visualize the entire population.
What I was doing was selecting a delta-V such that half the population dropped out, and subtracting out that much energy from each line of the remaining population. Using the wrong distribution produced a positive temperature gradient.
Using the distribution above, I get the exact same curve with a reduced population. No change in temperature.
The assumption I used in doing that was that the population was not interacting. That is, all particles which left z=0 with sufficient velocity reached z(1/2) with energy E(1/2)=E0-mgz(1/2).
br1, drop me a line to q_daniels_t-2l
at
yahoo com
Q. Daniels:
“ah, the problem is in your last step. The last formula should be:
v=sqrt((-2*k*T/m)*log(1-CDF))”
Yes, apologies for the copy-paste error. That is the formula I used.
“Using the distribution above, I get the exact same curve with a reduced population. No change in temperature.”
Yes, that’s what it does alright. So if one starts with this distribution under gravity, one gets an isothermal temperature profile. I’ve tried including collisions and intermolecular forces but still get an isothermal profile.
One thing I’m toying with is NOT starting with this velocity distribution. After all, it is like an ideal blackbody emission, all very well in theory but real objects are not perfect blackbodies. Haven’t got a temperature profile so far, but haven’t exhausted the possibilities yet. We can continue via email if you like.
br1 wrote:
One thing I’m toying with is NOT starting with this velocity distribution.
I’ve been considering this. In particular, the beads interiors are Canonical Ensemble systems. The interiors of the beads appear to be about 1 micron in diameter.
At any given instant, the average velocity will not be zero. Over time, it averages to zero, but that’s not the same thing. Instead, the center of mass will be bouncing around the interior with an average energy of 25 meV. At its peak, the vertical component will be converted into potential energy and its temperature reduced by a few millionths of a degree. It will also deliver slightly more collisions to the top wall. Also of note, it spends more time near the peak than bouncing off the bottom.
That’s a really crude approximation of what’s going on.
The behavior of the center of mass (and the system as a whole) only differs from average because of gravity.
I haven’t modeled it out yet. This is just the math I can do in my head. Modeling is up soon.
This is a continuation of the discussion that started on
http://tallbloke.wordpress.com/2012/06/28/graeffs-experiments-and-2lod-replication-and-implications/#comment-28702
copy pasted here in full for convenience:
Trick said:
“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.”
Trick:
“You haven’t even read B&A yet! Get a copy, you will trace thru & see they have not made a mistake”
I’ve traced through Verkley who 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. This way of obtaining 3 can be criticized on the grounds that, had no approximation been made, one would have found an isothermal instead of an isentropic profile”
What do you make of that?
Akmaev says that B&A setting enthalpy constant in their equivalent of V2b is their mistake. Please re-read slowly:
“Bohren and Albrecht (1998), apparently anticipating that P cannot be conserved exactly under condition (2), explicitly assume that the conservation of L approximately implies the conservation of P , as the ratio T /θ = Pi(p) does not change much for sufficiently close p1 and p2. However, this assumption is hardly defensible for any extended layer, because the maximum-entropy temperature profiles corresponding to the two constraints differ drastically”
More details here:
http://tallbloke.wordpress.com/2012/05/29/lucy-skywalker-graeffs-second-law-seminar/#comment-26892
“Constraint 2 in V2a: control volume (cv) energy = ideal gas enthalpy = 1/2mv^2 + mgh + pV”
whoa, talk about confusion!
Internal plus potential energy is not conserved in V2a, because any work done by the layer will cool it, thus reducing 0.5.m.v^2. External work done is not counted in internal energy. V2a uses Constraint 2′ which has replaced Constraint 2. Constraint 2′ includes work done, Constraint 2 doesn’t. In a closed container, no pV work is done because the boundaries are fixed, hence Constraint 2 applies. What you have written is Constraint 2′.
“Constraint 2’ in V2b: control volume energy = ideal gas enthalpy = 1/2mv^2 + mgh + 0″
Not only is this constraint Constraint 2 (have you mistyped or do you really think this is 2′???), but this constraint is not used in V2b. V2b uses Constraint 3, which has replaced *both* Constraints 2 and 2′. “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*.”
This is made even more obvious by considering that V2c would be redundant if Constraint 2′ was already in V2b. Verkley has to *explicitly* write Constraint 2′ into V2c in Eqn(19) – confer with Eqn(15).
You also never dealt with Akmaev:
“Interestingly, it also appears impossible for a physical process described by Equation (31) to simultaneously conserve the column enthalpy (Equation (1)) as well” when he is describing what “appears to be a characteristic, if not the only possible, explicit representation of a process transporting heat vertically and generating entropy under constraint (2)”
Enthalpy is not conserved in the V2b variational process.
And I thought br1 6/30 10:04pm really meant maybe it was time br1 move on.
Anyway, since br1 will not go look at the detail math in B&A to resolve br1 confusion I find it interesting to try ending this confusion, one small step at a time. When I re-read V&A & Akmaev to do so I appreciate their work even more. Tracing thru V&A won’t end br1 confusion; br1must go get B&A. It is a must to write B&A detail math in this blog avoiding confusion – br1 not doing so gets confused….easily.
br1 11:13pm – “a constraint similar to 2′, and then modify it in an approximate way, ….What do you make of that?…”
Here br1 is confused because cannot see that a constraint similar to 2’ is not 2’ exactly, it is only similar. br1 needs to read B&A math eqn.s for why that is (will find the difference is open and enclosed containers).
Yes a constraint similar to 2’ (what do you know! the similar one is constraint 2) will lead to an isothermal solution not isentropic profile. A constraint exactly like 2’ which I wrote correctly clipped 10:25pm (“have you mistyped?” No, not here at least) will lead to isentropic non-isothermal profile as shown by B&A, V&G and confirmed exactly by Akmaev.
br1 continues: “ This way of obtaining 3 can be criticized on the grounds that, had no approximation been made, one would have found an isothermal instead of an isentropic profile…What do you make of that?”
Here is constraint 3): a constant integrated potential temperature. Ok, so B&A, V&G and Akmaev avoid the criticism of the approximation by doing the math w/3 showing the exact solution of an enclosed adiabatic column with 2’ enthalpy & it is exactly max. S isentropic non-isothermal at LTE. Akmaev adds clarity to the exactness of V2b non-isothermal isentropic and how that works w/o approximation. br1 needs to read the B&A math entirely & not out of context to reduce confusion.
br1 continues: “You also never dealt with Akmaev: “Interestingly, it also appears impossible for a physical process described by Equation (31) to simultaneously conserve the column enthalpy (Equation (1)) as well.””
Yes, br1 has a continuing thimble and pea problem as I dealt with before. Here Akmaev develops eqn. 31 to show that V2b entropy can indeed climb up to a max. LTE for V2b and that V2b will not have the same enthalpy as V2a. Just exactly as I wrote & br1 clipped 10:25pm. This answers a question I tried to explain to br1 even earlier. br1 was confused about this and here finds his own answer but cannot see it. Oh well.
br1 11:13pm ends with shocker: “Enthalpy is not conserved in the V2b variational process.”
No shocker exists. Enthalpy 2’ IS conserved in V2b as it must be. Read B&A to see this in the math. It is a good read; br1 will reduce confusion.
Next.
I fear the problem is that it is simply not possible to suppress a circulation developing in a column subjected to a gravitational field and spinning in space around the centre of gravity of a planet.
That is bound to cause differential pressures at different points on the column wall and will force a circulation even if all other factors could be countered.
Once a circulation forms then the theoretically anticipated isothermal profile will be destroyed and I think that Graeff is measuring that effect.
Nonetheless it is true that gravity causes the effect, just not directly. The spinning motion is required to combine with the force of gravity to set up the temperature gradient via the circulatory process.
My first visit to this site,wich is quite interesting,but slightly esoteric,in that,all of the various”effects”pale into insignficance when compared to the major effects.It`s a bit like arguing how many angels can fit onto a pin head.Firstly you have to suspend rational thought and belive in a god.
So lets consider the major factors.
The 2nd Law applies only to a closed system. The Earth is definitely not a closed system,It receives an enormous amount of energy from the sun daily.This is done via EMR(photons),also it can lose heat into space via the same mechanism depending on the well know albido(and other reflective agents). Due to the Earths rotation all heat absorbing components,land,sea, and air are differentialy heated. This naturally will cause turbulance,as the temperatures”try” to equalise.As this is a continuous process there will always be atmospheric turbulance.The Earths rotation also will add to this,thus we have high and low pressue cells rotating in opposite directions.Other atmospheric components,cloud formation,Hadely cells,jet streams etc add to the tubulance.
The oceans cover 2/3 rds of the earths surface,and has the highest heat absorbing capacity,is subject to all of the same differential heating/cooling,so naturally there will be turbulance in the oceans, Also there are tidal effects adding to the turbulance.
There is also precession of the earths axis (Milankovitch cycles,closely associated with ice ages) that means the Earth never returns to the same position relative to the Sun. So that what is now the northern winter, in 130k yrs will be the nortern summer, the seasons (climate) are continuosly changing around the globe.
Finally as a critique to the climate change scientologists, yourr beliefs are a kin to that of a religion.ie it is based on suspension of need of provable scientific evidence.
Regards to all. Terry.
I would like to get some feedback,if you have time.