Guest post by Robert G. Brown
Duke University Physics Department
The Problem
In 2003 a paper was published in Energy & Environment by Hans Jelbring that asserted that a gravitationally bound, adiabatically isolated shell of ideal gas would exhibit a thermodynamically stable adiabatic lapse rate. No plausible explanation was offered for this state being thermodynamically stable – indeed, the explanation involved a moving air parcel:
An adiabatically moving air parcel has no energy loss or gain to the surroundings. For example, when an air parcel ascends the temperature has to decrease because of internal energy exchange due to the work against the gravity field.
This argument was not unique to Jelbring (in spite of his assertion otherwise):
The theoretically deducible influence of gravity on GE has rarely been acknowledged by climate change scientists for unknown reasons.
The adiabatic lapse rate was and is a standard feature in nearly every textbook on physical climatology. It is equally well known there that it is a dynamical consequence of the atmosphere being an open system. Those same textbooks carefully demonstrate that there is no lapse rate in an ideal gas in a gravitational field in thermal equilibrium because, as is well known, thermal equilibrium is an isothermal state; nothing as simple as gravity can function like a “Maxwell’s Demon” to cause the spontaneous stable equilibrium separation of gas molecules into hotter and colder reservoirs.
Spontaneous separation of a reservoir of gas into stable sub-reservoirs at different temperatures violates the second law of thermodynamics. It is a direct, literal violation of the refrigerator statement of the second law of thermodynamics as it causes and maintains such a separation without the input of external work. As is usually the case, violation of the refrigeration statement allows heat engines to be constructed that do nothing but convert heat into work – violating the “no perfectly efficient heat engine” statement as well.
The proposed adiabatic thermal lapse rate in EEJ is:
where g is the gravitational acceleration (presumed approximately constant throughout the spherical shell) and cp is the heat capacity per kilogram of the particular “ideal” gas at constant pressure. The details of the arguments for an adiabatic lapse rate in open systems is unimportant, nor does it matter what cp is as long as it is not zero or infinity.
What matters is that EEJ asserts that 
in stable thermodynamic equilibrium.
The purpose of this short paper is to demonstrate that such a system is not, in fact, in thermal equilibrium and that the correct static equilibrium distribution of gas in the system is the usual isothermal distribution.
The Failure of Equilibrium
In figure 1 above, an adiabatically isolated column of an ideal gas is illustrated. According to EEJ, this gas spontaneously equilibrates into a state where the temperature at the bottom of the column Tb is strictly greater than the temperature Tt at the top of the column. The magnitude of the difference, and the mechanism proposed for this separation are irrelevant, save to note that the internal conductivity of the ideal gas is completely neglected. It is assumed that the only mechanism for achieving equilibrium is physical (adiabatic) mixing of the air, mixing that in some fundamental sense does not allow for the fact that even an ideal gas conducts heat.
Note well the implication of stability. If additional heat is added to or removed from this container, it will always distribute itself in such a way as to maintain the lapse rate, which is a constant independent of absolute temperature. If the distribution of energy in the container is changed, then gravity will cause a flow of heat that will return the distribution of energy to one with Tb > Tt . For an ideal gas in an adiabatic container in a gravitational field, one will always observe the gas in this state once equilibrium is established, and while the time required to achieve equilibrium is not given in EEJ, it is presumably commensurate with convective mixing times of ordinary gases within the container and hence not terribly long.
Now imagine that the bottom of the container and top of the container are connected with a solid conductive material, e.g. a silver wire (adiabatically insulated except where it is in good thermal contact with the gas at the top and bottom of the container) of length L . Such a wire admits the thermally driven conduction of heat according to Fourier’s Law:
where λ is the thermal conductivity of silver, A is the cross-sectional area of the wire, and ΔT=Tb–Tt . This is an empirical law, and in no way depends on whether or not the wire is oriented horizontally or vertically (although there is a small correction for the bends in the wire above if one actually solves the heat equation for the particular geometry – this correction is completely irrelevant to the argument, however).
As one can see in figure 2, there can be no question that heat will flow in this silver wire. Its two ends are maintained at different temperatures. It will therefore systematically transfer heat energy from the bottom of the air column to the top via thermal conduction through the silver as long as the temperature difference is maintained.
One now has a choice:
- If EEJ is correct, the heat added to the top will redistribute itself to maintain the adiabatic lapse rate. How rapidly it does so compared to the rate of heat flow through the silver is irrelevant. The inescapable point is that in order to do so, there has to be net heat transfer from the top of the gas column to the bottom whenever the temperature of the top and bottom deviate from the adiabatic lapse rate if it is indeed a thermal equilibrium state.
- Otherwise, heat will flow from the bottom to the top until they are at the same temperature. At this point the top and the bottom are indeed in thermal equilibrium.
It is hopefully clear that the first of these statements is impossible. Heat will flow in this system forever; it will never reach thermal equilibrium. Thermal equilibrium for the silver no longer means the same thing as thermal equilibrium for the gas – heat only fails to flow in the silver when it is isothermal, but heat only fails to flow in the gas when it exhibits an adiabatic lapse in temperature that leaves it explicitly not isothermal. The combined system can literally never reach thermal equilibrium.
Of course this is nonsense. Any such system would quickly reach thermal equilibrium – one where the top and bottom of the gas are at an equal temperature. Nor does one require a silver wire to accomplish this. The gas is perfectly capable of conducting heat from the bottom of the container to the top all by itself!
One is then left with an uncomfortable picture of the gas moving constantly – heat must be adiabatically convected downward to the bottom of the container in figure 1 in ongoing opposition to the upward directed flow of heat due to the fact that Fourier’s Law applies to the ideal gas in such a way that equilibrium is never reached!
Of course, this will not happen. The gas in the container will quickly reach equilibrium. What will that equilibrium look like? The answer is contained in almost any introductory physics textbook. Take an ideal gas in thermal equilibrium:
where N is the number of molecules in the volume V, k is Boltzmann’s constant, and T is the temperature in degrees Kelvin. n is the number of moles of gas in question and R is the ideal gas constant. If we assume a constant temperature in the adiabatically isolated container, one gets the following formula for the density of an ideal gas:
where M is the molar mass, the number of kilograms of the gas per mole.
The formula for that describes the static equilibrium of a fluid is unchanged by the compressibility (or lack thereof) of the fluid – for the fluid to be in force balance the variation of the pressure must be:
(so that the pressure decreases with height, assuming a non-negative density). If we multiply both sides by dz and integrate, now we get:
Exponentiating both sides of this expression, we get the usual exponential isothermal lapse in the pressure, and by extension the density:
where P0 is the pressure at z=0 (the bottom of the container).
This describes a gas that is manifestly:
- In static force equilibrium. There is no bulk transport of the gas as buoyancy and gravity are in perfect balance throughout.
- In thermal equilibrium. There is no thermal gradient in the gas to drive the conduction of heat.
If this system is perturbed away from equilibrium, it will quickly return to this combination of static and thermal equilibrium, as both are stable. Even in the case of a gas with an adiabatic lapse rate (e.g. the atmosphere) remarkably small deviations are observed from the predicted P(z) one gets treating the atmosphere as an ideal gas. An adiabatically isolated gas initially prepared in a state with an adiabatic lapse rate will thermally equilibrate due to the internal conduction of heat within the gas by all mechanisms and relax to precisely this state.
Conclusion
As we can see, it is an introductory physics textbook exercise to demonstrate that an adiabatically isolated column of gas in a gravitational field cannot have a thermal gradient maintained by gravity. The same can readily be demonstrated by correctly using thermodynamics at a higher level or by using statistical mechanics, but it is not really necessary. The elementary argument already suffices to show violation of both the zeroth and second laws of thermodynamics by the assertion itself.
In nature, the dry adiabatic lapse rate of air in the atmosphere is maintained because the system is differentially heated from below causing parcels of air to constantly move up and down. Reverse that to a cooling, like those observed during the winter in the air above Antarctica, and the lapse rate readily inverts. Follow the air column up above the troposphere and the lapse rate fails to be observed in the stratosphere, precisely where vertical convection stops dominating heat transport. The EEJ assertion, that the dry adiabatic lapse rate alone explains the bulk of so-called “greenhouse warming” of the atmosphere as a stable feature of a bulk equilibrium gas, is incorrect.
So your error – according to Velasco et al, as well as myself, DeWitt Payne and probably Robert Brown too – is in taking that marginal height dependence of average molecular kinetic energy for small N as a temperature lapse rate. It just isn’t.
Oh yeah, count me in. And that difference vanishes at high T in the correct treatment, exactly as it should. But small N is where naive/simple treatments of things get very difficult, even for physicists, partly because the concept of temperature starts to break down. That’s why I prefer (and offered) a straight up thermodynamic argument in the simple (and relevant) thermodynamic regime where DALR has been hypothesized to create a thermodynamically stable thermal gradient.
If it does, the entropy of the Universe can easily be made to spontaneously decrease by coupling the ends so that they convert the temperature difference reversibly into work, reducing the overall temperature of the system with a strictly negative overall change in entropy.
rgb
Robert Brown wrote:
Piffle. The equivalence principle suggests that one cannot tell the difference. I agree that in a rotating frame the correction doesn’t have the same form as gravity, there is a coriolis force as well as additional pseudogravity, but this makes no difference in the argument.
The interior of a centrifuge isn’t a true accelerated frame. Your car experiment is a true accelerated frame.
I’ve ridden a centrifuge, and will admit that it certainly feels like one. I could not distinguish it from gravity. That does not, however, mean that it’s not distinguishable.
Gas molecules in a centrifuge only appear to follow accelerated paths if you’re using a cylindrical coordinate system. If you’re using a rectangular reference coordinate system, they follow straight lines. That’s the key difference. There is no acceleration between collisions at the molecular level.
You! Cannot! Violate! The! Second! Law! Of! Thermodynamics!
In this particular case, I am only questioning the validity of the test. I’m trying to choose my points carefully. This was a very narrow point.
There is no “Centrifugal Force”. That got pounded in to my head by a physics professor decades ago.
Q. Daniels says: “There is no “Centrifugal Force”. That got pounded in to my head by a physics professor decades ago.”
Time for a little humor: http://xkcd.com/123/
In this case, the cartoon is indeed more accurate than your physics prof. It is all a question of what frame of reference you choose. (And choosing inertial frames is indeed simpler in most cases, but that does not make it the only way to look at things.) While there are some differences between rotation and gravity, that does not refute the observation that they both achieve the same ends as far as the discussion goes.
(As an aside, in relativity freefall is the only inertial reference frame. Standing on a planet — even one that is not rotating — is an accelerated reference frame.)
Robert Brown says at 1/26 4:24pm:
“No, he doesn’t. Read 2.17, especially the exercise at the end. Or look at the exam question and solution I posted.”
Yes he does, check it out.
Robert, again, as I posted above, Caballero in 2.17 is talking about the no gravity isotropic velocity field ideal gas he shows in Fig. 2.3.
Quoting from 2.17 “Consider again a gas in a pipe, as in Fig. 2.3.” Fig. 2.3 has NO gravity.
The homework exercise is for dT/dx eqn. 2.75 which is the horizontal temperature gradient in hydrostatic equilibrium. Yes, horizontally dT/dx =0. No doubt. Gravity acts only in the z height. Please stop and carefully check your work as you would ask any student to do.
Again, 2.17 is for the isothermal no gravity isotropic velocity case, you cannot use Caballero here to support dT/dz =0 isothermal gas column in a gravity field.
Because my statement is true, to restate, I added the ref. in parentheses:
“Caballero supports isotropic velocities for isothermal T only in the no gravity ideal gas column case (in 2.17). Add gravity & Caballero in Sec. 2.3 supports gas column T is non-isothermal and constant entropy.”
Please re-read Caballero ref. cites again. Grok them as you asked me to do. And please return.
You may have another text ref. to cite to support your work, so far Caballero only supports a non-isothermal gas column in the presence of gravity, Sec. 2.3.
I will look any ref. right up, I have placed on order Caballero’s recommended: “Bohren and Albrecht’s excellent Atmospheric Thermodynamics.”
Paul Birch:
Again, I appreciate the attention.
Now, it was to obtain more information on which to make an assessment of whether Velasco et al. are right that I brought their paper up on this site; they fill books with what I don’t know about thermodynamics. But, frankly, nothing anyone has told me so far is anything I didn’t already know–or already know was wrong; the discussion has either been what Velasco et al.’s verbal description means, or how people on this site with credentials view it.
As I’ve said, I’m no physicist, and I claim no expertise in this area. And I’m not as sure of anything as many of the contributors here seem to be about everything. But I do have quite a bit of experience assessing all manner of physical-science-based matters where not getting it right could have significant financial implications. And that experience has made me pretty sure it’s best to do two things. First, go with the equations, not the verbal descriptions. Second, don’t rely on the opinions of experts, no matter how exalted, if in my opinion their logic doesn’t stack up.
So, although I respect your opinion, I have to go with mine. And it’s my opinion that, for the purposes of a theoretical discussion–which, let’s not forget, is what Dr. Brown initiated–Velasco et al. do indeed say that an isolated vertical gas column exhibits a non-zero lapse rate at equilibrium.
Robert Brown says at 1/26 4:24pm:
“Look, nothing … is going to permit a stable thermodynamic temperature gradient, because for one to exist you must not be able to use it to do work, not even in principle.”
I agree, in Caballero 2.3 non-isothermal ideal gas column in 2.3 where the velocities are higher at the bottom than the top, no work can come out of it. This means a system of constant entropy allowed by 2nd law, the process is reversible. Gravity increases molecules speed on way down, decreases speed going up. Same exact amount ideally.
“Otherwise as you do work with the thermal difference, you are reducing the heat content of the system (cooling it) and turning it into reversible work.”
No. The energy content of the system in Caballero 2.3 is not cooling. It is remaining constant. Thus heat of the system is NOT being cooled; b/c as I have written over & over energy is constant. Conserved. 1st law compliant.
You are smart, Caballero is so far, right. You can grok.
Trick,
Wanna bet that if you ask Caballero about this statement in 2.3:
and then about section 2.17 he will say that 2.17 is wrong? Somehow I don’t think so. My bet is that he will correct 2.3.
DeWitt Payne says:
January 25, 2012 at 9:49 am
“In the troposphere, the atmosphere is more transparent to SW radiation and less transparent to LW radiation so the surface temperature is warmer than the slab. The opposite is true in the stratosphere.”
______________________________________________________________
This analogy is very flawed because at the base of the troposphere is a real “slab” (the Earth’s surface) that can be warmed itself by solar insolation and then transfer perhaps as much as 70% of that energy back to the atmosphere via evaporation and diffusion (ie thermodynamics not just radiation.) In contrast there is no such physical slab at the tropopause.
I am not saying that the explanation for the inversion in the stratosphere is incorrect – just that there is much more going on in the troposphere and that it is totally wrong to imply that radiation is the sole cause of the lapse rate in the tropopause.
Evidence suggests that thermodynamics prevails at least two-thirds, and a significant role in the thermodynamics is played by “latent heat” in evaporated water which leap frogs up into the cloud levels and is then released up there by phase change. This has to affect the lapse rate and may explain the observed reduction from the much higher (almost double) rate in the lower 14,000 feet compared with the layers above that. (See data in my earlier post.)
Once again I find Professor Claes Johnson promulgating the most plausible hypothesis http://claesjohnson.blogspot.com/2010/09/lapse-rate-vs-radiative-forcing.html
Yes, Johnson is the one who proved computationally that an atmospheric greenhouse effect is a physical impossibility because low frequency radiation is not converted to thermal energy by a surface which is warmer than the emitting layers of the atmosphere. It’s worth reading: http://www.csc.kth.se/~cgjoh/blackbodyslayer.pdf
Be the first to prove him wrong – and back it up with empirical evidence of backradiation warming something. Apparently it cannot even melt frost on the ground.
PS I’d still like from you the reference to that experiment you wrote about on SoD where the gas did not absorb until the emitter became warmer. Clearly this supports Johnson’s hypothesis.
George E. Smith; says:
January 26, 2012 at 12:47 pm
I don’t judge people or the papers they write by anything but the content thereof and my own experience, work, tutoring and further private study in the fields of maths and physics since my university studies in the 1960’s and 1970’s. But, just for the record, Johnson is a Professor of Applied Mathematics (very akin to physics) with many published papers. http://www.csc.kth.se/~cgjoh/
If you want a quick summary of his main contribution in this field (proving the greenhouse effect is impossible) read the Radiation page on my site http://climate-change-theory.com
DeWitt Payne says at 1/26 6:00pm:
“My bet is that (Caballero) will correct 2.3.”
LOL. I’ll take that bet! For the sum of one $tygian bragging right.
Agree to let Boren text be the source of the bet decision? Does DeWitt have a copy of Boren already? Have a text quote correction ready? I’ll wait for mine to come in a few days. Takes my chances. I’ll stand w/Caballero lines up with Boren.
If you want to let Velasco paper eqn. 8 settle it, then that’s ok too.
Bill Hunter says:
January 25, 2012 at 7:48 pm
The problem outlined by Dr. Brown is interesting but largely irrelevant to the issue of whether the surface would be warmer in the absence of greenhouse gases.
Here is my reasoning on that.
It’s the analogy to the passive solar water heating system. First lets get a few things clear on that.
A good passive solar system does not need a greenhouse for the collectors to operate well. Often they use just plain black piping. It works nearly as well as pipes in a greenhouse because the 1,000 plus watts of solar radiation far exceeds radiation losses without the greenhouse.
Shouldn’t that be 240?
Trick says:
January 26, 2012 at 5:51 pm
Trick, you can’t just assert that “no work can come out of it”. Out here in the real world, if there is a temperature difference you can extract work from it, either in reality or in principle. The non-ideal gas column of of Caballero is not magically exempt from that rule, and the rule certainly is not suspended because Trick claims it won’t happen.
So if you want to look like other than a dilettante, you need to show, not claim but actually establish through logic and argument, that no work can come from the column. I see absolutely no reason that a temperature difference anywhere anytime cannot produce work either in principle or in reality.
w.
Why, when we are living in the midst of an experiment, whose chemical distributions are being measured to the best of our ability, would we try to create a planet of imaginary chemicals to understand our situation.
We are in the experiment, just measure it.
Dr. Brown, if you get bored patiently explaining physics to the gravityheads, take a look at the link above for a red-hot possibility for your next project. This is the claim that energy cannot be radiated from a cooler body to a warmer body …
w.
u.k.(us) says:
January 26, 2012 at 7:16 pm
Thought experiments have a long and glorious history. Einstein used them to convince people that relativity was a correct theory. Why? Because you can cut through the complexities and the practicalities that beset us, and come to a deeper understanding of the reality from contemplating simpler situations.
So your complaint, that this is a thought experiment without the complexity of the real world, falls on deaf ears. Folks here understand the value of these kinds of experiments, and the fact that you do not just means you should rethink your opposition to a long-established part of scientific inquiry.
w.
Willis Eschenbach says:
January 26, 2012 at 7:21 pm
=============
Einstein ?, i have thoughts, none of which discounted thought experiments, and it wasn’t a complaint, just an observation from my reference frame, such as it is.
Robert Brown wrote:
You! Cannot! Violate! The! Second! Law! Of! Thermodynamics!
At no point have I made any claim or representation regarding my skill in this manner.
I have not even claimed the ability to tie my own shoe laces together, though I will say that there are reports that I once tied Melvin Calvin’s (Chemistry, 1961) shoe laces together.
Tim Folkerts wrote:
In this case, the cartoon is indeed more accurate than your physics prof. It is all a question of what frame of reference you choose. (And choosing inertial frames is indeed simpler in most cases, but that does not make it the only way to look at things.) While there are some differences between rotation and gravity, that does not refute the observation that they both achieve the same ends as far as the discussion goes.
I disagree. A rotating reference frame does not even qualify as accelerated. That’s the problem.
Consider a non-moving particle, planet, or whatever, sitting by itself near a reference point and far from everything else. If you have a non-rotating frame, the physics is simple and straight-forward. The object remains where it is. If you change to a rotating frame perpendicular to the axis, explaining the circular motion of the object becomes much more difficult.
In the case of the centrifuge, there is acceleration only in the rotating frame. Let’s put the centrifuge in deep space for clarity. If you use the inertial frame, there is no acceleration applied to the individual molecules except when they hit each other or the wall. A molecule which had been brought to a full stop by gas collisions would remain in place until disturbed by another collision. This is very different from the effect of gravity, wherein molecules can come to full stop by gravity, and then accelerate out again without collision.
Gravity is a force that acts on every particle in the system, independent of mechanical interaction. A centrifuge only acts mechanically.
There is no “Centrifugal Force” to play the equivalent of Gravity. There is only the wall of the centrifuge, which occurs only at the wall of the centrifuge.
While reading about Maxwell’s demon, I came across an amazing invention that appears to shed light on the “gravitationally induced” temperature gradient idea. The Ranque-Hilsch Vortex Tube generates a temperature differential within a tube by injecting a compressed gas stream which then generates a vortex. The temperature along the walls is higher than the incoming gas, while the temperature within the center is colder than the incoming gas. The two temperature streams can be physically separated.
Here are some links, including a patent on an embodiment of the invention, if you are interested in learning more about this phenomenon:
http://www.filtan.de/ENGLISH/VTS_A.htm
http://www.me.berkeley.edu/~gtdevera/notes/vortextube.pdf
http://www.google.com/patents?hl=en&lr=&vid=USPAT3546891&id=KppYAAAAEBAJ&oi=fnd&dq=%22vortex+tube%22+gravity+%22temperature+gradient%22&printsec=abstract#v=onepage&q&f=false
Gee Willis, I look up heat dissipation and centrifuges, all they talk about is high temperatures and problems removing the excess energy. Seems you 200 C would not be too far off at that amount of g’s. But, duh, of course it would be much hotter at the outside where g’s are higher. Seems you just proved yourself wrong.
one example: http://www.freepatentsonline.com/5433080.html , there are many more.
@Doug Cotton
Whilst I trust you know my position on all this, I do not like pushing the “CO2 saturation” concept. Firstly, bands only appear blank at TOA because CO2 scatters radiation and hence, when you point an instrument at some place on Earth, very little appears to come directly towards you from that point. Warmists will argue that the bands just get wider,
+++++
OK, another time perhaps. You are correct re the conduction and of course that is normally left out of any discussion of ‘CO2 thermal blanketing’.
There are a lot of posts sent in the meantime. I see our contributing Author has again appealed to his own and his selected authority and put down an opinion which which he does not agree solely by observing that the other source was not correctly qualified to have a valid opinion. This is the essence of every weak and losing argument. On that basis alone I suspect the validity of whatever the thesis states.
I asked my friend Kevin why academics put so much energy into postulated planets and he replied, “Because there is so little at stake.”
Question: Is this putative planet rotating? At what rate? How deep is atmosphere? What is the equitorial centrepital acceleration rate just below the TOA and what is the acceleration due to gravity? It may be far colder a the top than has been considered so far because the expansion may well exceed what thermals can achieve. Realising this, will we again constrain the physics on the planet to bias the answer towards our favourite explanation?
u.k.(us) says:
January 26, 2012 at 7:16 pm
Why, when we are living in the midst of an experiment, whose chemical distributions are being measured to the best of our ability, would we try to create a planet of imaginary chemicals to understand our situation.
We are in the experiment, just measure it.
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2871909/
Willis says at 1/26 6:59pm:
“So if you want to look like other than a dilettante, you need to show, not claim but actually establish through logic and argument, that no work can come from the column.”
Learned a new word for amateur! Good one.
Fair to ask for logic in any post from pro or amateur. The logic here is that the ideal gas column in ideal closed container in a gravity field (top post Fig. 1) is a reversible system in equilibrium. Molecules go up and down reversibly. Caballero terms this hydrostatic equilibrium in Sec 2.3, Fig. 2.4.
2nd thermo law tells us for a single reversible system, sometimes called a cyclic process, it is impossible to produce work into the surroundings.
In the top post, the equation of hydrostatic equilibrium is shown as eqn. 5, in Caballero it eqn. 2.27.
Is my claim that “no work can come from the column” actually established through logic and argument for you now? There is certainly more to grok but your argument for the elevator speech length being adequate is strong.
Willis continues:
“I see absolutely no reason that a temperature difference anywhere anytime cannot produce work either in principle or in reality.”
Think the thermo grand masters would agree. Except in the post that Joules Verne made at 1/25 8:57pm there is a terrific paper posted that even they were perplexed about the more or less philosophical thought experiment being discussed in Fig. 1 top post on down.
The best one can do is have a view whether the process is isothermal or non-isothermal it turns out. It is interesting learning while supporting one’s view consistent with texts, papers and thermo laws, other posters. As you write, a noble blog purpose.
Call me out (w/logic) when my post is not established thru logic and argument consistent with natural laws. (Try not to make the moderators or me cry though, lol).
Wayne,
When I see people state “of course” it often points to the flaw in the arguement, because people are simply assuming, rather than proving. You assume the problem is that the outside is hotter, but provide no evidence.
When I look for centrifuges and heat dissipation, I see two things — neither of which is related the outer parts being warm. Instead, they discuss:
1) Dissipating heat created by the equipment. There will always be friction and vibrations that generate thermal energy. At high speeds this could be quite a problem, so dissipating the heat from the bearings and the motor can be a problem.
2) Trying to cool the samples below ambient temperatures. This is what your link is actually talking about.
“Before centrifugation occurs with many samples, temperature thereof must be precisely controlled. In practice, classification of the sample in a rotor must occur at controlled temperature. An example of such a temperature is 2° centigrade for certain biological samples. The sample must be brought to the temperature and during centrifugation the sample must be maintained at the temperature. In both events cooling of the chamber is required. Due to their small size and weight, thermoelectric devices using the Peltier effect are utilized. ”
I’d be interested to see links you find that discuss the outer edge of the centrifuge getting hot compared to the axis, especially links where the difference is ~ 200 C as you seem to think it should be.
Trick says:
January 26, 2012 at 8:17 pm
====================
OK, now that has been established, can we move on to Physics 102 ?
I’m sick of “heat” flow.
Clearly any body emitting photons has no idea where they will end up, photons being somewhat bereft of ideas. Photons are in fact so stupid, they only know that they are emitted and absorbed simultaneously. They have no notion that they have ever spent time travelling between atoms!
Those photons being emitted by The Git’s adipose tissue are streaming out in almost every direction. Some in the direction of the ground, some sideways and some towards the sun. Those heading sideways towards the trees will be absorbed by the trees just because they are cooler than The Git’s adipose tissue. Similarly, those emitted toward the ground will be absorbed for the same reason.
Those photons emitted toward the sun however have a different fate. The sun being ever so much hotter than The Git’s adipose tissue is not allowed to absorb those photons. Perforce they must either pass straight through the sun, or else be diverted around the sun before being allowed to continue their straight line path beyond. There is only one possible answer to this: the Will of God/Krishna/Jehovah/Inanna/Allah [delete whichever is inapplicable] which transcends even the Laws of Physics as decreed by God/Krishna/Jehovah/Inanna/Allah [delete whichever is inapplicable] .
This is an elevatorsqueak account of why “energy cannot be radiated from a cooler body to a warmer body”.
It is not the intention of the writer to demean, degrade, humiliate, despise, or disparage any god, godess, offspring of god, demiurge, inamorata, fetish or other supernatural being.