At The Hockey Schitick MS has posted a brave article
“The Greenhouse Equation”
This seems to be a culmination of a series of articles.
[UPDATE: and another article showing a fit against Standard Atmosphere http://hockeyschtick.blogspot.com/2014/11/the-greenhouse-equation-predicts.html /UPDATE]
He concludes with
I welcome all suggestions and refutations of the “greenhouse equation.” There can be only one valid theory of the 33C greenhouse effect, since if both the gravitational and greenhouse gas radiative forcing theories had merit, the Earth would be at least 33C warmer than the present.
What does the Talkshop have to say?
Click graph or link is here
I could probably produce a downloadable file for WxMaxima which is cross platform and able to solve that form but extending this to plots is not something I’ve used it for. Maybe a reader can investigate.
Post by Tim
Tallbloke's Talkshop 
Hi from Oz. I’m amazed and shocked – where is the term for CO2? 😉
Otherwise, well done. Now all we need is for Politburo to take notice…
It seems only fair to point out that Nikolov & Zeller started the process of explaining surface temperature in terms of equations, something that James Hansen failed to do while attempting to scare scientific illiterates with his loony “Runaway Greenhouse Effect”.
N&K created equations that modeled the observed temperature at the surface of rocky bodies in our solar system using pressure as the main variable. Their opponents called this “Curve Fitting”, so I tested their equations at a constant pressure of one bar and found they worked well:
Then Robinson & Catling produced a model that worked even better than N&K’s:
Are the “Hockey Schitick” equations an improvement on R&C? I hope y’all will be able to tell me as I am still trying to reproduce R&C.
That should put off any future climatologists.
“Are the “Hockey Schitick” equations an improvement on R&C? I hope y’all will be able to tell me as I am still trying to reproduce R&C.”
Well…you know my opinion on that 😉
I corresponded with R&C right after their paper was published and they confirmed that “most” of the GHE is due to atmospheric mass and that convection dominates over radiative forcing in the troposphere. They do believe, as most climatologists do, that the ERL is somehow set by GHG forcing, but they were not sure why, but the point of my posts is that the ERL is set by the center of mass of the atmosphere, not GHG RF.
Well if we look at other planets the ERL goes higher the more mass in the atmosphere. Its very low on Mars, about 5 km on earth and even higher on Venus. So according to Arrhenius we now know that its the ” chief mass of the atmosphere” thats control the temperature. The IR radiation is the end result of the temperature produced by pressure/mass/gravity.
So all climate reaserch by the IPCC is conducted on a wrong atmospheric model!
What a waste of money!
I always have trouble posting at the Hockey Schtick site because I can’t read the text where you prove you are not a robot. (God Help Me; maybe I am a robot and don’t know it? — a Blade Runner will terminate me ….)
So, to MS, I would like to say that the post is real science in action. The series of posts were exciting and informative: and I happen to agree with you. I guess I agree because I was trained in science in the early 70s before the madness of “back radiation” destroyed the scientific method — so I might be biased. But even if I am biased, you have produced a testable equation! You have given us a starting point to investigate. I did not think I would live long enough to see this sort of thing again in climate science. Congratulations. And I mean congratulations even if a flaw is found in the equation — you are doing science.
I am disposed to believe you are on to something as one of the chief fools from WUWT has already made at least two fact-free complaints about your work while citing Dr. Brown of Duke as some sort of god-like authority in such matters. (don’t misunderstand, I like Dr. Brown but he is not Maxwell)
I would like to state that I think that both H2O (in all its states) and CO2 have warming effects and cooling effects but “mass/gravity/pressure” of the atmosphere is the key to the riddle.
There is a region in the upper troposphere and above where radiation becomes dominant, so I have a hard time believing that the picture is complete without line-by-line radiation calculations. I do agree that the lapse rate controls the temperature profile, and that convection dominates radiation in the lower troposphere, but asserting that the center-of-mass equals the equivalent emissions height is a stretch, even if it does turn out to be approximately true.
HS says:
Para 1: mass/gravity/pressure
Para 3: mass/pressure/gravity
You are seem confused about the assumptions of your theory
Your theory is gravity/mass/weight /pressure/ temp.
That is
1. Gravity acts on mass
2. This causes weight
3. Greater weight cause higher pressure Barometric theory, gas confined by gravity
4. Higher pressure causes higher temp IGL, gas confined by a container
My objection is that in stages 3 and 4 you use two opposite and incompatible models of pressure.
IGL ignores gravity; the gas is confined by solid walls
The Barometric equations apply Newton’s 2nd law of motion, F = ma, to motion in a gravity field.
The gas is confined by gravity and is not confined by walls.
Hockey Schtick,
Your ‘ERL/atm centre of mass’ hypothesis is interesting, but ultimately has no explanatory power. The ‘ballpark’ relation that we find on Earth and Titan seems to be basically a happenstance. The two atmospheres are of fairly equal weight (mg).
When you formulate an hypothesis like this, what you want is to have your basic postulate, the assumption of a dependent relationship between two variables (ERL and atm centre of mass), work as a universal principle. You haven’t accomplished that at all. Your premise only (sort of) works for Titan and Earth.
To me, something else entirely is going on:
The lighter the atmosphere, the lower the theoretical planetary ERL temperature will be found. Conversely, the heavier the atmosphere, the higher up it will be. The reason? The heavier the atmosphere, the hotter it will make the surface, even with equal solar input, and the higher its ‘heat capacity’, so the higher you need to go for all the energy coming in from the Sun to finally escape back out to space.
Earth and Titan simply appear to be at the point on the continuum between say Mars (sfc pressure: 6 mbar) at the one end and Venus (sfc pressure: 92 000 mbar) at the other, where the BB emission equivalent temp layer has moved a few kilometers up from the surface (about 5.1 km on Earth, about 11-15 km on Titan (effective emission temp 80.6-84K)).* On Mars this height is at the solid surface itself, far below the ‘atm centre of mass’. On Venus it’s at the tropopause, high above the ‘atm centre of mass’.
* Titan’s sfc pressure: 1467 mbar; Earth’s sfc pressure: 1013 mbar. ‘Atm centre of mass’ on Titan is located 13 km off the surface. On Earth, the same altitude is 5.7 km up.
I’ve pointed this out many times before, to the likes of Joe Postma, who has claimed the universality of some kind of ‘law of averages’, where the ERL temperature layer would naturally lie in the middle of the full tropospheric ensemble, from 288K at the surface to 210K at the tropopause. However, this only (sort of) works on Earth: (288+210)/2 = 249K.
This is not a universal principle.
What you need to understand, Hockey Schtick, is that there is NO connection whatsoever between the total, final radiative flux emitted by a planetary system to space and the physical temperature of some specific layer inside the system.
The final planetary LW flux to space is simply there to balance the absorbed incoming SW/LW flux from the Sun, nothing more, nothing less. It does NOT come from ONE single layer. It is accumulated from ALL layers of the system, from the surface itself and every layer of air up to the ToA. Radiation escapes to space whenever it can, from wherever it can. It tries all the time, regardless of temperature. Some escapes directly from the surface, some is unable to do so before being transported by convection all the way to the tropopause and (some times) beyond. And much escapes from layers in between. This is a continuous process. It enters and it exits. All that enters also exits, although not necessarily at the same time and from the same place. There is no overall piling up of energy in the atmosphere in a steady state (dynamic equilibrium) situation.
The TEMPERATURE is set by completely different mechanisms. (Yes, it’s a mass/gravity phenomenon, but not the way you think.)
Anyway, the SURFACE temperature is set first. The tropospheric temperature profile climbs up from this baseline. Nothing ‘temperature-inducing’ propagates DOWNWARDS in the troposphere (after the solar input has been absorbed). The HEAT flows UP from the surface. Until it’s finally all released back out to space.
Just look at temperature plots of surface vs. troposphere. They are impressively similar, only with larger amplitudes in the latter (as one would expect from its lower ‘heat capacity’). However, you will notice how the surface always changes first and the troposphere follows. The heat propagates UPWARDS.
The hight of ERL may be a result of how much the atmosphere absorbs of incomming SW.
Thick atmosphere like Venus absorbs high up and on Mars the thin transparent atmosphere absorbs close to nothing, hence the surface ERL.
Im sure an absorbsion parameter will make it work on any planet.
It may not be perfect, but a giant leap in the right direction.
Mr Pettersen,
Please read my comment just above yours. The ERL concept is a chimaera, an imaginary theoretical (mathematical) construct. No such thing exists in the real world. So any hypothesis basing its postulates and premises on it, ultimately fails.
Kristian. You and i have talked about this earlier if i remember correctly, on another blogg.
Of course it dosn’t mean that all energy escape from that singel hight.
The ERL is as you say, just a mathematical average.
Hockey Schtick is correct in every essential particular and confirms what I have been saying here and elsewhere for years.
Climate Prat of 2014 – Voting has started. Have your say.
Pointman
Why must adiabatic and radiative factors be exclusive? Ultimately temperature isostacy is set by radiation, but does that photon heading for space care where its energy derives? If two factors equally sufficient push atmospheric temperature, their cumulative efforts are still limited by the efficiency of radiation to space.
Stephen 3:55pm: I hadn’t noticed you were saying earth atm. is isothermal as HS claims in the derivation of the top post eqn. at the HS site (11/27 top post eqn. (12) & (13)). Interesting. This should be good for some fun. LOL.
In science practice, the only way to determine actual non-isothermal T(h) and surface Tmedian for any planet with atm. is radiative transfer line by line (LBLTRM) integration over dh (sometimes called dz, sometimes called s). I am sure HS will vociferously debate the point. Even though a planet is windy, the LBLRTM analysis process integrates earth actual T(h) over dh as measured by in situ thermometry on radiosondes to within precision instrument accuracy. There are some good approximations like –g/Cp (off ~20% to tropopause) once surface T is found and the Poisson eqn. (off only ~10% at tropopause).
******
pointman 3:57pm: Science is not a voting machine. Leave that process to politics.
Trick.
HS is reproducing the observed temperature profile with a lapse rate.
That is not an isothermal atmosphere.
I suspect that you are misreading or misinterpreting or misrepresenting the equations you refer to.
This whole discussion is based on an extremely simplified model – that of the Earth as a black body (or a gray body) with a perfect thermal conductivity – no temperature difference between day and night, no difference between summer and winter. Guess what – the model temperature is off by 33 degrees!
Don’t fall into a modeler’s favorite trap of believing that the model is correct. There are so many things wrong with it. Introduce a speed of rotation, a thermal capacity and a thermal conductivity of the surface, to name just a few. The atmosphere is important, but it should not be the first thing to improve.
Pointman: Don’t tempt me…
Curious George, I recommend holding off certainty; here is a reason why.
In 1984 Essex published three seminal papers. These upset the assumptions made by many up to that time. Today there is still deep reluctance to accept what is.
The pith is that the atmosphere is always in a transition state as you and I know but ignored by certain parties. The atmosphere cannot be modelled as a terminal state, static, disc world, etc.
Hockey Schick has chosen a mean. If it turns out that is how a transition state means out, fair enough.
I don’t know.
All models are wrong. Some are useful.
Stephen 4:45pm: Suspects.
Ok, Stephen – I know you don’t have the math pre-req.s to understand HS eqn. (12) and (13) do show an isothermal T assumption. But there they are – would you like a short course in integral calculus?
The -g/Cp derivation makes the same isothermal assumption over dh as does HS, so this top post eqn. has presented nothing new in derivation in that respect. To date, I haven’t seen a chart of HS T vs. s from s=0 to s=tropopause. Given that we can see whether it beats -g/Cp or Poisson eqn. (from the 1890s) in accuracy. It has already been shown LBLRTM is pretty much dead nuts to within instrument accuracy even on a windy planet.
When I see a hairy eqn. like top post, first I check units. I do not yet see how one can twist g*s/C into units of Kelvin but it may be possible. R*T^0.25/g*m I will leave up to others, as the 1000 comes from nowhere and I do not know if 1000 has units, my head hurts enough. Then I might input s=0 and s=infinity and look at the result. According to HS, s=0 is announced at T=288.486 (11/28), no intermediate steps are shown. Maybe so.
But there is a problem at s=infinity. From the eqn. after “results” in top post, looks like the HS analyzed T eqn.goes to negative infinity. This can’t be physical as absolute T has a floor at 0K, the HS T eqn. should tend to 0 at infinity. It doesn’t.
Re complaints about curve fitting.
Curve fitting IF SUBSEQUENT EVENTS SHOW IT TO BE PREDICTIVE is a fascinating situation. It means there has been a reproduction of prior AND existing conditions by proxy. Proxies are like analogies in the literary world, in that they tell you some things about the object you are studying but are not the object itself. Sometimes this is enough, but you must remember that as the analogy/proxy is not the thing itself, there will be differences.
A successful proxy determined by curve fitting undermines the unique solution syndrome, the bete noire of CAGW. If there is another way that successfully reproduces AND predicts the phenomenon, then there are hidden, connective factors. Not saying that the “real” causitive forces are false, but that the presence of usable proxies not understood as to connection, implies that one’s certainty of the unique solution is wrong.
Warmer temperatures can make plants grow faster. But as we know, so can increased CO2 and N2.
…. and before someone chimes in with the Pirates vs CO2 graph, let’s consider this. If something pirates did reduced CO2, we’d want to know about it. Or the sociologist was able to tie CO2 levels and the timing of pirate activity to reduced agricultural output, economic collapse and unemployment amongst the young, violent males …. that would be good, too. Speculative to begin with, but isn’t all scientific discovery?
Curve-fitting is useful. At least for determining that the unique solution is untenable. The IPCC in fact has done curve fitting. But they claim it is a description of a unique solution. Therein lies the absurdity: one curve-fitting exercise used to prove the foolishness of other curve-fitting exercises.
Gymnosperm says:
“Why must adiabatic and radiative factors be exclusive?”
Quite!
Curious how, in this digital age, there seems to be an aversion for believing that an effect can be a combination of more than one influence, isn’t it?
It is clear that the troposphere is heated uniformly from the surface of the Earth. Thus, the center of mass of the troposphere is logical.
The temperature in the zone of the ozone increases as a result of a chemical reaction.
ren says: November 29, 2014 at 8:32 pm
“It is clear that the troposphere is heated uniformly from the surface of the Earth. Thus, the center of mass of the troposphere is logical.”
The tropospheric temperature, and temperature gradient, is determined by the rate of conversion of WV latent heat to sensible heat, offset by the rate WV EMR can dispatch that entropy to space.
Your so called center of mass has no meaning! l;;;;;;;;;;;;;;;;;;;;;;;;It is but an attempt to support the IPCC claim that the the surface is 33 degrees Celsius higher, than a meaningless number. Find a fake 255 Kelvin
somewhere in mid troposphere, and call that, “Effective Radiating Level”. Pull numbers from toilet to calculate an increasing 33 C to surface. All is thus explained by snake oil, just as the ClimAstrologists do it. Can there be no integrity anywhere?
ren says: November 29, 2014 at 8:38 pm
“The temperature in the zone of the ozone increases as a result of a chemical reaction.”
The temperature in this zone is reduced by the increase in CO2 EMR to space, providing the correct level of O3, without overheating. What a wonderful planet!
catweazle666 says: November 29, 2014 at 7:55 pm
(Gymnosperm says: “Why must adiabatic and radiative factors be exclusive?”)
“Quite! Curious how, in this digital age, there seems to be an aversion for believing that an effect can be a combination of more than one influence, isn’t it?”
In this TV age, curious children are intentionally dumbed down to only a single thought at a time. That makes a better customer! Not like remembering, not to go in the direction of the threshing machine while chasing that squirrel.
The concept of an adiabatic barrier can be a useful concept to isolate a particular energy transfer. Adiabatic compression in an Diesel engine, simply means that there is no time for any heat or mass transfer, all energy is contained. There is no adiabatic anything in this atmosphere.
Everything is mixed up with everything else!
“Why must adiabatic and radiative factors be exclusive?”
Not a correct comparison. The proper comparison is between radiation and conduction/convection.
If radiation is converted to kinetic energy in mass which causes conduction and is then moved up against gravity so that it is converted to PE then that PE can no longer participate in the radiative exchange because PE is not heat and does not radiate.
The same parcel of energy cannot both radiate and form part of the PE reservoir at the same time.
“There is no adiabatic anything in this atmosphere.”
All work done with or against gravity is adiabatic by definition.
Stephen Wilde says: November 29, 2014 at 11:16 pm
(“Why must adiabatic and radiative factors be exclusive?”)
“Not a correct comparison. The proper comparison is between radiation and conduction/convection.”
This is not a comparison, it is a question properly poised, Barrister! Please reply to the question!
“If radiation is converted to kinetic energy in mass which causes conduction and is then moved up against gravity so that it is converted to PE then that PE can no longer participate in the radiative exchange because PE is not heat and does not radiate.”
Please provide any physical evidence of your insane fantasy?
Stephen Wilde says: November 29, 2014 at 11:17 pm
(“There is no adiabatic anything in this atmosphere.”)
“All work done with or against gravity is adiabatic by definition.”
All “work done” by definition is “entropic”, and can never be adiabatic!
To see just how powerful adiabatic forces are spend a little time in southern Alberta where it can go from -30 C to 5 C in a matter of an hour thanks to westerly winds flowing over the Rockies and being heated as they descend.
–I welcome all suggestions and refutations of the “greenhouse equation.” There can be only one valid theory of the 33C greenhouse effect, since if both the gravitational and greenhouse gas radiative forcing theories had merit, the Earth would be at least 33C warmer than the present.–
I think the Earth is exactly where it should be in terms of temperature.
It seems the problem is we have some ideas [wrong ideas] about what temperature it should be.
Earth is not going to get 33 C warmer, nor earth like planet.
And if Venus were in Earth’s orbital distance, Venus would be a lot cooler.
Though Venus is perfectly fine in terms of it’s current temperature, but humans have the conception that they consider the idea of living under oceanic pressure of a massive atmosphere- because that is where the ground is, but they feel they can’t do this because it’s too hot.
So this is the monkey’s problem not a planet’s problem.
But since on the topic, Earth was apparently a lot warmer than it is now. Since we currently in Earth’s ice box climate and Earth most common temperature in it’s history is about 10 C warmer than 15 C.
And somewhat accepted explanation [not including the highly unlikely Greenhouse theory] is that geological processes and plate tectonic movements and continent building is part of the explanation for the Ice box climate we are in.
So how is all that related to the “One equation for earth temperature”
And also another question, it seems to me that there is large degree of uncertainty in regards to Titan’s atmospheric temperature profile. And I wonder about what temperature Titan temperature was at Earth air density- so 1.2 kg per cubic meter. And/or I am under the impression that at this much lower density than on the titan surface, and that it’s warmer at such a elevation as compared to surface temperature. Or assume that 1.2 kg per cubic meter is somewhere in Titan troposphere. [somewhere with zero and 40 -50 km elevation] And also gather that like Venus, Titan above it’s troposphere has or could have high velocity wind [racing globally around the moon].
As general matter, I am not very comfortable with comparing Earth to Titan because it seems there a high degree of uncertainty in regards to Titan’s atmosphere [though we also don’t seem to know much about Venus either- though it seems to me, it’s slightly more detailed data than Titan].
Thanks ren for the link to that graphic, which was perfect to illustrate this post to refute Willis Eschenbach’s claims
give you multiple h/t 🙂
http://hockeyschtick.blogspot.com/2014/11/why-atmosphere-is-in-horizontal.html
If anyone wants to try to rip this other new post…
The Greenhouse Equation predicts temperatures within 0.28°C throughout entire troposphere without radiative forcing from greenhouse gases
http://hockeyschtick.blogspot.com/2014/11/the-greenhouse-equation-predicts.html
Alan Poirier says: November 30, 2014 at 1:16 am
“To see just how powerful adiabatic forces are spend a little time in southern Alberta where it can go from -30 C to 5 C in a matter of an hour thanks to westerly winds flowing over the Rockies and being heated as they descend.”
I have been to BC. :Please define what you may mean by “powerful adiabatic forces”! The entropic air mass movements can be measured complete with increasing temperature with increasing pressure. Please show how this is in any way adiabatic, rather than entropic?
Will / Alan
http://www.weatheronline.co.nz/reports/wind/The-Foehn-wind.htm
Hockey Schtick says: November 30, 2014 at 2:16 am
“Thanks ren for the link to that graphic, which was perfect to illustrate this post to refute Willis Eschenbach’s claims give you multiple h/t 🙂 ”
Please continue to try to defend the indefensible, many, many, times worse than the whole Climate Clown deliberate SCAM.
HS, don’t want to deluge articles at the moment so I’ve added a referencing update. Traffic is fine. Holding most popular article at the mo.
Anyone really looking will work out what where.
” The entropic air mass movements can be measured complete with increasing temperature with increasing pressure”
Um…that would be adiabatic, or gravitational in the absence of wind. It literally means without the devil, the devil being fire of course. No fire required, it warms all by itself with increasing pressure.
My point is that if you consider the planet a “black ball” from space and let all the wind and back radiation and entropic (if you must) effects run their course invisibly, you might be able to write two equations, one for adiabatic/gravitational/entropic effects, and another for radiational effects each which would be sufficient to explain the 33K, yet the two would not be additive because the black ball increases its radiating intensity to the fourth power of any increase.
Will, I was about to explain it and then saw that tchannon had already supplied a link. Thanks, t, I should have known the phenomenon was experienced elsewhere. To truly appreciate it, though, you need to be in the middle of deep freeze! Cheers.
HS 2:16am: “The Greenhouse Equation predicts temperatures within 0.28°C throughout entire troposphere…”
Sure, apparently your top post greenhouse equation predicts the US standard atmosphere adequately – you or Wolfram did the algebra correctly – but you did NOT find actual “temperatures throughout entire troposphere”. Here’s an equation that predicts US Standard Atmosphere temperatures surface to troposphere even better, within 0.0C:
T(s) = To – 6.5 * s meters/1000
For US readers:
T(s) = To – 1.98 * s ft/1000
Occam’s razor and all. You have to realize the US Standard Atm. linear decreasing temperature up to tropopause you are matching is not any sort of actual atmosphere. The US Standard Atmosphere is a composite atmosphere designed by a committee, not a column of air preserved in a long glass tube in say Boulder, Colorado standards lab.
In its work, committee used the same eqn.s as you (e.g. IGL, barometric) with the same assumptions (isothermal over dz) which turns out is good one evidenced by test. They knew by measurement global Tmedian at surface (~288K) and ~255K at ERL i.e they knew the 33K from observations and matched. They connected the dots found 6.5 avg. slope. You have ~duplicated their work (more arcanely), nice going, many can learn from your work.
Now go out in your back yard field far from thermal energy sources (e.g. BBQ, asphalt), draw a m^2 square in the dirt. Hang a thermometer about 1.5m agl or any agl up to tropopause on still air day. Do not look at the thermometer. First, predict its still air T trace over the next 24 hours from your mass, gravity, pressure theory (no radiative transfer allowed!). Write it down. THEN look at the thermometer trace. See if you came within 0.28C as clouds go by, as day turns to night. I challenge you won’t come at all close. Just look at any weather station: 24 hour pressure, density, T are not shown proportional at all as in IGL as you assumed.
You CAN find To on the thermometer or T(z) at any z in m^2 column over your square within instrument error using one of many specialty LBLRTM codes available, confirmed by radiosonde thermometry.
–“If radiation is converted to kinetic energy in mass which causes conduction and is then moved up against gravity so that it is converted to PE then that PE can no longer participate in the radiative exchange because PE is not heat and does not radiate.”
Please provide any physical evidence of your insane fantasy?–
LOL.
Well, a molecule normally doesn’t go up [a meter or a cm in troposphere air density]
Something like a balloon will go up- and molecules of air not like a air molecules in balloon.
Instead one has pattern of collisions, and a such patterns does not radiate energy.
Though I think the basic point is gases are colliding without any lost of energy [ideal gases]
and the energy of gases is kinetic, kinetic does lose kinetic energy by radiating.
Of course the greenhouse gases will interact with certain wavelengths of radiant energy. but the discussion about is about kinetic energy and patterns of motion cause by heat adding kinetic energy to molecules.
and the energy of gases is kinetic, kinetic does lose kinetic energy by radiating.
meant:
and the energy of gases is kinetic, kinetic gases does not lose it’s kinetic energy by radiating.
Circulation, like Venus, very quickly be able to compensate for temperature. However, this direct sunlight determines the temperature increase in the troposphere. Therefore, the current change in the circulation causes great temperature fluctuations in the US. This is proof that greenhouse gases do not work.
markstoval, November 29, 2014 at 11:49 am
“I always have trouble posting at the Hockey Schtick site because I can’t read the text where you prove you are not a robot. (God Help Me; maybe I am a robot and don’t know it? — a Blade Runner will terminate me ….)”
My (several) attempts to comment on the “Hockey Schtick” failed miserably. I was beginning to think it was my fault.
The low traffic at that site may be a result of the “Captchas”.
That is the reality.
It should be noted that the temperature of the tropopause is essentially constant. Only strong hurricanes or volcanic eruptions are able to locally raise the temperature of the tropopause for a short period of time.
Mr Pettersen, November 29, 2014 at 8:46 am
“So according to Arrhenius we now know that its the ” chief mass of the atmosphere” thats control the temperature. The IR radiation is the end result of the temperature produced by pressure/mass/gravity.
So all climate reaserch by the IPCC is conducted on a wrong atmospheric model!
What a waste of money!”
It would be hard to find a better summary of the fraudulent discipline known as “Climate Science”.
Sadly many excellent scientists such as Judith Curry, Richard Lindzen and John Spencer give credence to the false Arrhenius (1896) hypothesis by discussing a “Sensitivity Constant” in degrees Kelvin per doubling of atmospheric CO2 concentration.
In his award winning movie “An Inconvenient Truth”, Al Gore claimed that [CO2] correlated with Antarctic temperatures for the last 800,000 years. While Al was right about the correlation he was wrong about the cause. The correlation depends on Henry’s law rather than the Arrhenius hypothesis:
If you are not impressed by Arrhenius you may like this:
You may think that Arrhenius is dead and buried, yet the loony James Hansen published a paper in 2003 that included this chart:
This slide shows Earth’s surface temperature varying by 4. 5 K/doubling of CO2. Exactly what Arrhenius predicted in 1896. It also shows the surface temperature varying by 16 K/halving of CO2 which would explain the last seven glaciation cycles (800,000 years) in terms of CO2. He fails to address the fact that [CO2] lags temperature by ~600 year. In the real world cause precedes effect.
The “Tooth Fairy” and the “Easter Bunny” are real to James Hansen. If you take loony Jim seriously I have a fine bridge in Brooklyn that you may want to buy.
It does calculate as HS said and this might be just a little bit easier to understand what is there.
/* DESCRIPTION OF VARIABLES INVOLVED e = natural log base eps = epsilon ELR = env. lapse rate (K/km) IRR = equil. rad. level f = proportion const g = grav. accel kHS = hydrostatic const M = comp. mol. mass S0 = solar const SB = Stefan's const T = temperature(z) z = height (km) */ /* SET SOME VARIABLE VALUES */ SB = 5.6704E-08 g = 9.80665 M = 0.0289644 e = exp(1) S0 = 1362 eps = 1 ELR = 6.5 z = 11 /* CALCULATE 'T' */ IRR = (S0·(1-0.3)/(4·eps·SB))^(1/4) f = 1000·log10(e) / log10(1/2) kHS = g·M/R T = IRR - ELR · ( z + (IRR/f) / kHS ) /* DUMP VARIABLES e ··············· 2.718282 ELR ············· 6.5 eps ············· 1 f ··············· -1442.695 g ··············· 9.80665 IRR ············· 254.6246 kHS ············· 0.03416319 M ··············· 0.0289644 S0 ·············· 1362 SB ·············· 5.6704e-08 T ··············· 216.7046 z ··············· 11 */Hockey Schtick, November 29, 2014 at 7:59 am
“I corresponded with R&C right after their paper was published and they confirmed that “most” of the GHE is due to atmospheric mass and that convection dominates over radiative forcing in the troposphere.”
Some “Climate Scientists” will communicate with the general public. Given that taxpayers pay their salaries why should any of them refuse to respond to questions from the great unwashed?
Here is a list of “Climate Scientists” who corresponded with me even though they realized that I was skeptical about CAGW (Catastrophuc Anthropogenic Global Warming). These people are reacting as real scientists should. I call them “Good Sports”.
Richard Alley, Evan Pugh Professor of Geosciences at Pennsylvania State University
Barry Brook, Sir Hubert Wilkins Chair of Climate Change at Adelaide University
David Catling, University of Washinton, Seattle
Scott Denning, Montfort Professor of Cimate Science at Colorado State University
Several people at the Danish Meteorological Institute
Albert Klein Tank, KNMI
Tom Peterson, NCDC Asheville
Tyler Robinson, University of Washinton, Seattle
Here is a list of people who ignored my requests for information or “Lawyered Up”:
Michael Mann
Jay Severinghaus
Thomas F. Stocker
I regard such people as charlatans who can’t defend their theories.
“My point is that if you consider the planet a “black ball” from space and let all the wind and back radiation and entropic (if you must) effects run their course invisibly, you might be able to write two equations, one for adiabatic/gravitational/entropic effects, and another for radiational effects each which would be sufficient to explain the 33K, yet the two would not be additive because the black ball increases its radiating intensity to the fourth power of any increase.”
They have to be additive otherwise one could not match energy in with energy out in the long term.
They simply must add up to produce exactly enough radiation to space from both surface and from within the atmosphere to match radiation received from space.
In practice, convection ensures that the balance is maintained.
All very similar to this:
Trick
This derivation (link) of the Barometric Equations does not assume an isothermal atmosphere.
Page 8 on
Click to access 1003.1508.pdf
Stephen Wilde, November 30, 2014 at 10:47 am & 11:23,
The 33 K is based on good mathematics and false assumptions about the surface properies of our planet. Why not discuss 56 K or 79 K?
While I once shared your enthusiasm for Nikolov & Zeller’s pressure based equations, they only work at the surface of rocky bodies. This left N&K vulnerable to charges of “Curve Fitting”. One should expect pressure based equations to work at any arbitrary pressure:
Bryan 1:23pm: Thanks for pointing that out. P.8 G&T is not discussing isothermal barometric formulae as HS used; G&T are redeveloping therein non-isothermal Poisson eqn.s of the 1890s – their resultant eqn.s (36), (42). Which they attribute as “well-known from standard textbooks”. Three things:
1) If HS assuming isothermal atm. over ds wants to know the actual T(s) at tropopause height above the temperate calm backyard m^2 dirt square, to within ~20%, and knowing the temperature 1.5m agl by looking at the thermometer, then HS can use DALR.
2) If HS assuming non-isothermal atm. over ds wants to know the actual T(s) at tropopause height above the backyard m^2 dirt square, to within ~10%, and knowing the temperature 1.5m agl by looking at the thermometer, then HS can use G&T eqn. (42) well known in textbooks as the Poisson eqn. of the 1890s.
3) If HS assuming isothermal atm. over ds wants to know the standard atm. T(s) at tropopause height above the backyard m^2 dirt square, to within 0.28C, and knowing the temperature 1.5m agl by looking at the thermometer, then HS can apparently use top post eqn. Alternatively, HS could use this and be off 0.0K from standard atm.:
T(s) = To – 6.5 * s meters/1000
Note none of the above uses radiative transfer! Note the last eqn. is of form: y=mx+b.
The actual results are good approximation 10-20%; but HS needs to know the temperature at one height (b) to get the T( s) answer from approx. slope (m) – in this case from the thermometer at 0 height agl (HS s=0). To analyze that temperature dead nuts without looking on the thermometer, HS is going to have to use radiative transfer. As is routinely done & well known in text books.
NB: This won’t work in a hurricane. Not even close.
Trick, I never make any assumption that the atmosphere is isothermal, just the opposite in the vertical plane, but remarkably isothermal in the horizontal plane, because of the horizontal equilibrium set by -g/Cp, which is not controlled by GHG RF. The horizontal isotherms are quite remarkable as ren has shown in the graphics he has posted. I explain by RF can’t control the lapse rate, barometric formula, greenhouse equation, etc. here:
http://hockeyschtick.blogspot.com/2014/11/why-greenhouse-gases-dont-affect.html
The one and only temperature used by the greenhouse equation is the calculated equilibrium temperature with the Sun, which is a constant. That’s it. If you believe otherwise, then show me specifically where the equation makes any assumption whatsoever about any temperatures other than Te located at the center of mass. All of the other temperatures with height are calculated by the equation.
HS 3:27am: Your assumption of isothermal atmosphere over dh is, again, for at least the third time, in your 11/27 post eqn.s (12) and (13).
dP = -(MPg/RT)dh (12)
∫dP/P = -(Mg/RT) ∫dh (13)
Vertical T profile is assumed constant at that step when T(h) comes out from under the integral sign.
T(h) is the vertical variation of T with height (your entire point matching US Standard Atm) so T cannot come outside the integral unless T is assumed constant (isothermal) over dh. Isothermal is inherent in the barometric formula you use. Isothermal eqn. (13) IS a good approximation for actual atm. (off ~20% surface to troposphere from radiosonde thermometry). So isothermal works reasonably well in ISA or US Standard Atm. design.
Poisson 1890s eqn. derivation moved DALR ahead adding non-isothermal w/o radiative transfer but is still off ~10% in T profiles from radiosondes. The only way to get a dead nuts T profile answer that compares to radiosondes for the actual atmosphere profile is to integrate
∫dP/P = -(Mg/R) ∫T(h)dh
by invoking the line by line radiative transfer methods (LBLRTM) – they are routinely successful to within instrument accuracy in meteorology. This is how the global monthly anomaly can be found down to like a tenth degree K. Too, basic radiative transfer text books show one layer works reasonably well for earth pressures with all measured input.
A paper by Clough & Iacono 1995 JGR showing LBLRTM in action from surface 1 bar to .0001 bar was discussed around here a few months back also from your site and another one:
I pulled that original paper. The confusing 400 labeled blue box at upper right scale does not appear in the original. Somehow got added incorrectly along the way.
Trick what you are pointing out about the barometric formulae doesn’t matter to the calculation of the troposphere temperature profile. There is one and only one point at which these three factors converge to determine the h at that location:
1. height of the center of mass where P=Ps/2 and T=255
2. Te and ERL at that same location =255K (Te assumed constant)
3. P=Ps/2 after density adjustment
there’s one point only are all three converge and which can be calculated. Once that point is determined, then the lapse rate -g/Cp from that point down to the surface and up to the tropopause can be calculated.
Thus, entire std atmosphere in tropo calculated knowing only Te, g, atm mass, Ps
Trick, how many degrees warming of the 33C GHE do you believe are from pressure vs. radiative forcing, and how do you calculate or determine your answer?
HS,
The height of the atm centre of mass on Earth is at 5.7 km, not 5.1 km. The actual centre of mass altitude is at 251K, not 255K. That’s 4 degrees off.
This is why I’m saying your assumption of a dependent correlation between theoretical ERL temp and atm centre of mass only ‘sort of’ works on Earth (and Titan).
It doesn’t work at all, however, on Mars and Venus. On Mars, the atmosphere is much lighter than on Earth/Titan; on Venus it is much heavier. The lighter and more tenuous the atmosphere, the further down you will find the ERL equivalent temp layer, the heavier and denser the atmosphere, the higher up you will find it.
But, once again, there is NO connection whatsoever between the mathematical construct of a planetary BB emission temp in space (solely based on its average LWIR flux from the ToA) and the physical temperature of some layer inside the planetary system.
So basing your hypothesis on the acceptance of that rGHE/AGW-invented ERL concept I’m afraid disqualifies it already at the onset.
The CoM height varies with latitude. This is related to the variable length atmospheric path incoming but not outgoing. This also varies seasonally with earth movement.
Related are lots of effects which rarely get mentioned let alone discussed.
This doesn’t necessarily void the concept of a mean but does make validation more difficult, if this is possible at all.
HS 9:46am: “Trick what you are pointing out about the barometric formulae doesn’t matter to the calculation of the troposphere temperature profile.”
Then leave out that isothermal assumption integral step, find what happens or do the integration with T(h) under the integral. To calculate actual nature dead nuts to get anomaly close enough for gov. work not an approximation as is ISA or US Standard Atm., DALR and Poisson eqn.:
∫dP/P = -(Mg/R) ∫T(h)dh
adds needed precision. Also, the work you have done cannot construct the Clough & Iacono 1995 chart.
“how many degrees warming of the 33C GHE do you believe are from pressure vs. radiative forcing, and how do you calculate or determine your answer?”
Photons aren’t marked like number truant children can be found, identified and returned to class. However some reasonable estimates can be made from lab tests. The atm. optical depth process showing contribution from pressure and amount each species is the Robinson&Catling 2013 Letter. Another reasonable paper starting place for doing so is discussed here:
–HS,
The height of the atm centre of mass on Earth is at 5.7 km, not 5.1 km. The actual centre of mass altitude is at 251K, not 255K. That’s 4 degrees off.–
It depends upon the temperature.
Can we assume you using 15 C at sea level?
And it also depends on humidity
In terms of actual location and at a particular time, 15 C at sea level is not common.
I mean the tropics don’t get this cold- so it doesn’t happen at 40% of the Earth surface.
And outside the tropics one has seasons and larger night and day variation in temperatures.
So the point of half the atmosphere in terms a elevation is generally more constant in tropics- not wide seasonal differences- not large difference of night and day and less variation in humidity.
As guess at equator middle of Pacific the point my bounce around by less than 100 meters.
Though as guess it seems unlikely it’s either 5.7 km or 5.1 km.
Or:
“Mount McKinley’s summit air pressure rises from increasingly warmer temperatures as summer is approached. From May 1 to July 1 the change in pressure is, on average, equivalent to “descending” a full 743 feet.”
http://www.cohp.org/ak/notes/pressure_altitude_simplified_II.html
And half sea level pressure is the point of 1/2 the atmosphere.
So when it warmer the 1/2 point is going to rise higher and despite widely swinging temperatures
in the non tropics, it seems that the 1/2 point would rarely get above freezing or rarely have any liquid water.
Or it seems instead of half the atmosphere one could also say at the freezing point of water- as the freezing point of water might be more relevant than the pressure.
Hockey Schtick says:
December 1, 2014 at 4:46 am
1. height of the center of mass where P=Ps/2 and T=255
2. Te and ERL at that same location =255K (Te assumed constant)
3. P=Ps/2 after density adjustment
there’s one point only are all three converge and which can be calculated. Once that point is determined, then the lapse rate -g/Cp from that point down to the surface and up to the tropopause can be calculated.
Thus, entire std atmosphere in tropo calculated knowing only Te, g, atm mass, Ps
ERL (measured LR) is set in the equation, that is, it is constant at 6.5K/km
LR is not calculated, it is a known constant, 6.5K/km
T is linear: y = mx + c (pressure variation is exponential)
Knowing one x,y and gradient m the line and thus all x,y points can be found
The equation for T is graphed using one point (Te = 255K, height (Te) = 5.1Km) and gradient T/h = 6.5K/km
The Greenhouse equation is a mathematical description of the observed facts.
T is not calculated using atm mass and pressure
LR = T/h is not caused by mass and pressure
LR is caused directly by reduction in g further from the surface
The Greenhouse Equation goes from surface 288K – tropopause 218K = 70K
The Greenhouse effect It doesn’t stop at 255K.
gbaikie 3:11pm: “(atm. center of mass) depends upon the temperature.”
PE=mgh. Use Ma for total atm. mass, a constant over thousands of years:
Atm. center of mass avg. height = PE/(Ma*g). I’d like to see Kristian 9:46am calculate or a cite for the 5.7km asserted.
Nuts. GHGs (Greenhouse Gases) are defined as having some absorption bands in the thermal infrared. An atmosphere with no greenhouse effect whatsoever is free of such gases, therefore it is absolutely transparent in said frequency band. Now, the atmosphere (at least below a height of 50 km) is always in local thermodynamic equilibrium, so Kirchhoff’s radiation law can be applied. That is, in any frequency band emissivity is equal to absorptivity. Which means the atmosphere can not emit radiation in the thermal infrared either, so ERL is not at an intermediate level in the atmosphere, but right at the surface.
Berenyi Peter,
The point is that in the vertical plane an atmosphere is always in thermodynamic DISequilibrium due to work done against gravity in holding the mass of the atmosphere off the surface converting KE to PE to produce an inevitable lapse rate even in a GHG free atmosphere.
For a GHG free atmosphere all radiation to space would indeed have to be from the surface but the surface for Earth would still need to be at 288K to get 255K out to space due to conduction and convection enabling the mass of the atmosphere to obstruct the free flow of radiation to space.
The ERL would still be at the height where the temperature is 255K i.e. some distance off the surface and not at the surface.
You can only have 255K at the surface if there is no atmosphere at all.
Mass not radiative capability creates the greenhouse effect.
Stephen 7:12pm: “You can only have 255K at the surface if there is no atmosphere at all.”
Only in Stephen’s imagination, not in science. On Mars, the ERL is just barely off the surface since surface Tmedian ~ Teff. Yet Mars has shown us dust devils so demonstrates an atmosphere too. A good point for HS to consider also. On earth, the atmosphere is not nearly as optically thin as Mars, so the ERL rises off the surface considerably more than Mars.
The Mars ERL is off the surface due to the mass of the atmosphere as is that of Earth.
Stephen 7:54pm – The atm. mass radiates on all planets having a relevant atm. That’s why they call ERL the effective radiating level.
The EFFECTIVE radiating level is not the same as the ACTUAL radiating level.
For a radiatively inert atmosphere the actual level is the surface but the effective level is higher up where the temperature is sufficient to radiate out as much as comes in and as HS says it is close to the centre of mass of the atmosphere.
The energy difference between the two locations then exchanges between surface and atmosphere in a constant adiabatic convective loop.
Stephen 8:50pm: Changes the subject to ARL from ERL. Adds nothing.
“For a radiatively inert atmosphere…”
There are none in nature only in Stephen’s imagination. All mass radiates.
“…as HS says it is close to the centre of mass of the atmosphere.”
Not on Mars. Same physics apply as on Earth. ERL on both planets is one optical thickness unit below the TOA as on all planets with relevant atmosphere.
I get what you mean, Stephen. Like the way “effective temperature” of a body is just the temperature of a hypothetical blackbody of a certain size and distance from sun. It isn’t an “actual” temperature just a theoretical one. So the ERL is just whatever part of the atmos that happens to have a temp that corresponds with the blackbody temp. But the “actual” radiating level could be somewhere else as in at Earth’s surface in the case of an Earth with no GHGs but an atmosphere, and higher up in the case of Earth WITH GHGs. That height depends entirely on amount of GHGs in the atmosphere, however that doesn’t mean the surface temp is necessarily affected. Why would it? The energy all gets radiated out one way or another, not like it could ever be “trapped” or anything, especially not moving at the speed of light n all.
Graham W
The height of the ERL depends on mass and if GHGs try to change it then convection changes instead of the height of the ERL or the surface temperature changing.
Otherwise you’ve got the idea.
OK. I’ve just mentally substituted “level of atmosphere that equals 255K” for “ERL” in what you’ve just written. Makes more sense to me that way.
Berényi Péter says: December 1, 2014 at 6:57 pm
(“We have also determined the effective radiating height (average) or ERL [Effective Radiating Level] in the troposphere (where T = the equilibrium temperature of Earth with the Sun), and found the ERL to be located as expected at the center of mass of the atmosphere”)
“Nuts. GHGs (Greenhouse Gases) are defined as having some absorption bands in the thermal infrared. An atmosphere with no greenhouse effect whatsoever is free of such gases, therefore it is absolutely transparent in said frequency band. Now, the atmosphere (at least below a height of 50 km) is always in local thermodynamic equilibrium, so Kirchhoff’s radiation law can be applied. That is, in any frequency band emissivity is equal to absorptivity. Which means the atmosphere can not emit radiation in the thermal infrared either, so ERL is not at an intermediate level in the atmosphere, but right at the surface.”
The atmosphere is near thermodynamic equilibrium but is always at a higher temperature than radiative equilibrium temperature. Mass “only” absorbs thermal EMR flux when the temperature of that mass is below that of radiative equilibrium. Kirchhoff’s Laws of Radiation. At all locations, in addition to transmitting all flux from below, that location adds and accumulates exit flux by its own radiance. This accumulation of exit flux continues all the way to 220 km. There is no ERL, every part of the atmosphere contributes to the total exit flux.
@Stephen
“There is no adiabatic anything in this atmosphere.”
“All work done with or against gravity is adiabatic by definition.”
Boy, wish you joined me in battle on the Greenhouse Effect thread. Had a raging battle on adiabatic and diabatic processes.
–Trick says:
December 1, 2014 at 9:06 pm
Stephen 8:50pm: Changes the subject to ARL from ERL. Adds nothing.
“For a radiatively inert atmosphere…”
There are none in nature only in Stephen’s imagination. All mass radiates.–
I would say all mass radiates, but gas masses only radiate a part of blackbody spectrum.
So compared a piece of paper [which absorbs and radiates in a wide spectrum] the accumulation all gases in existence don’t absorb and radiate as much. Or one could fold a piece of paper in manner that it might actually can direct radiant energy- or the paper could warm something.
Or basically the way to control “room temperature energy is related to conduction of heat and convective of heat.
And what this is about is the point in the atmosphere where convection the heat starts to not work very well.
Problem is there is another way to control “room temperature” type heat, and it’s swamp cooler.
Or a hot driveway can cooled by spraying water on it- mainly due to evaporation.
Though I haven’t actually tried, one could probably pour boiling water on hot driveway and the eventual result would be to cool it.
Will Janoschka says:
December 2, 2014 at 12:30 am
/////////////////////////////////////////////////
”There is no ERL, every part of the atmosphere contributes to the total exit flux.”
I agree with this as it agrees with empirical measurement. The strength of emission and altitude of emission from the atmosphere is constantly changing, being largely dependant on cloud patterns. Clouds are the strongest source of IR emission from the atmosphere to space, but they only have an average IR emissivity of around 0.7. You simply can’t pick a centre of mass for the atmosphere and claim it is radiating as a near blackbody from this altitude.
Konrad. says: December 2, 2014 at 2:20 am
Will Janoschka says: December 2, 2014 at 12:30 am
(”There is no ERL, every part of the atmosphere contributes to the total exit flux.”)
“I agree with this as it agrees with empirical measurement. The strength of emission and altitude of emission from the atmosphere is constantly changing, being largely dependant on cloud patterns. Clouds are the strongest source of IR emission from the atmosphere to space, but they only have an average IR emissivity of around 0.7. You simply can’t pick a centre of mass for the atmosphere and claim it is radiating as a near blackbody from this altitude.”
Agreed. An optical depth of media with partial transmittance is the distance through that media, at each frequency when the transmittance (from behind of ‘modulation’) falls to (1/e), 37%. 63% is emittance or
exitance from the mass of that media in any direction toward lower field strength. For an isolated cloud, the radiative solid angle is not limited to PI steradians,like a surface, but closer to 5 steradians. 63% x 5/PI, go figure.
Regarding comments on Mars, as Robinson and Catling point out in their paper, the surface pressure on Mars is only 0.006 bar, far too low to sustain convection or a lapse rate, thus none of the above discussion applies.
So Will, you think it’s just a complete coincidence that the temperature at the center of mass of both Earth and Titan is located exactly at the equilibrium temperature with the Sun, right? And you think increased GHGs have changed that exact point over the past 39 years since the 1976 Standard Atmosphere database, correct? And thus you think my equation that calculates the 1976 Standard Atmosphere database at every meter from 0-12000 based only on the center of mass at Patm/2 pressure and Te exclusively, without any GHG forcing whatsoever, is just one big coincidence, right?
–The EFFECTIVE radiating level is not the same as the ACTUAL radiating level.–
Here is definition of ERL which is given:
“ERL: The lowest level in the atmosphere from infra red radiation is able on average,
to escape upward to outer space without being reabsorded”
http://www.aos.wisc.edu/~aos121br/radn/radn/sld012.htm
And:
“At some height most radiation emitted upwards makes it to outer space without being reabsorbed
on the way. This height (in practice around 8-10 km) is called the Effective Radiating Level.”
And assume this has nothing to do with what you mean and I think you have to define what is meant by EFFECTIVE radiating level.
I think it’s a bad and/or confusing term to use.
Because the 1/2 halfway is has nothing to radiating at this level.
One would a more related term is the effective convection level- though such a term would still
need to be defined.
And as said above I tend to think it has more to do evaporation and/or the height of liquid water in the atmosphere.
-For a radiatively inert atmosphere the actual level is the surface but the effective level is higher up where the temperature is sufficient to radiate out as much as comes in and as HS says it is close to the centre of mass of the atmosphere.-
I think the only thing doing a significant amount of radiating [lost of heat- rather than re-radiating radiant light is liquids or solids rather any gas.
–The energy difference between the two locations then exchanges between surface and atmosphere in a constant adiabatic convective loop.–
I think this is key aspect of this.
Another way to look at it is the air above the surface is sort of like the ground below the surface.
As both are storing heat and both must be connect to surface to lose heat. Or ground under the surface can’t radiate directly to space, it must conduct or convect heat to the surface to have energy radiated to space. And the same applies to the air below the surface.
And the surface itself has little heat content.
So ground beneath surface is effectively a few inches or tens of kg per square meter, whereas
the air above the ground to 5 km is more than 4 tons of mass.
Hockey Schtick says: December 2, 2014 at 7:15 am
“So Will, you think it’s just a complete coincidence that the temperature at the center of mass of both Earth and Titan is located exactly at the equilibrium temperature with the Sun, right? And you think increased GHGs have changed that exact point over the past 39 years since the 1976 Standard Atmosphere database, correct? And thus you think my equation that calculates the 1976 Standard Atmosphere database at every meter from 0-12000 based only on the center of mass at Patm/2 pressure and Te exclusively, without any GHG forcing whatsoever, is just one big coincidence, right?”
Michael,
First. I was not trying to compare you with D. Cotton or J. Shore!
I was reading; and considering:
1. Why yet another model equation for affirming the nonsense numbers of the Climate Clowns?
2. Do averages of anything about this planet have any meaning whatsoever?
3. Why would an alternate to back radiation and CO2 do any good now?
No coincidence at all!
The first part of your equation is the same erroneous Solar irradiance, modified by the same erroneous reduction via albedo, and that all remains is absorbed by the surface although this planet has 220km of dispersive gas above that surface. Then make the erroneous claim that all is evenly spread into 4 steradians, and emitted by that surface with an erroneous 100% emissive surface. This via formula gives an erroneous 255 Kelvin, straight from the toilet!
The second part is putting that 255 Kelvin at an altitude pressure of 50 kPa, calling that center of mass of the atmosphere. The center of mass of the atmosphere is co-located with the center if mass of the planet. Your equation then curve fits a 6.5 C/km laps rate to derive a Earth std atmosphere, the same way as is used to calculate the atmospheric profile for Earth and Titan by others.
Impressed I am not! Please derive the control method used by this planet’ to exactly determine WV content at every location in the atmosphere, and still come up with the same average lapse rate, temperature and profile?
BTW There is no Greenhouse Gas! All fantasy!
Hockey Schtick says, December 2, 2014 at 7:15 am:
“Regarding comments on Mars, as Robinson and Catling point out in their paper, the surface pressure on Mars is only 0.006 bar, far too low to sustain convection or a lapse rate, thus none of the above discussion applies.”
But this is simply not true. There is both vigorous and consistent convection and a defined environmental lapse rate on Mars. The atmosphere being so thin simply makes the diurnal fluctuations around the ALR much greater.
http://theoilconundrum.blogspot.no/2013/03/standard-atmosphere-model-and.html
http://onlinelibrary.wiley.com/doi/10.1256/qj.02.169/pdf
http://en.wikipedia.org/wiki/Climate_of_Mars
HS 7:15am: “..the surface pressure on Mars is only 0.006 bar, far too low to sustain convection or a lapse rate…”
Thanks for the charts Kristian*. The two long term NASA Mars rovers also disagree with HS as their horizontal solar panels have been repeatedly cleaned of dust by convection – winds strong enough to do so over many years. Their cameras have photographed dust devils moving across the surface. Convection and thus a lapse rate (~4.5K/km) demonstrably exist on Mars and the ERL is a little under 1 km while the Patm/2 is several km above the surface.
Many pictures of Mars surface show dunes and ripples caused by convection.
R&C: “Mars’ low surface pressure of 0.006 bar means that it does not fall within our scope of examiningcommonalities in thick atmospheres…” The physics is the same on Earth and Mars; Mars simply falls off the R&C chart due such an optically thin atm.
*Note the hook back 0-1km in night time, shows the high energy capacity surface cooling faster by radiating to space (big window) than the relatively lower Cp atm. is able.
******
HS: ”So Will, you think it’s just a complete coincidence that the temperature at the center of mass of both Earth and Titan is located exactly at the equilibrium temperature with the Sun, right?”
HS simply assumes this; HS has still not shown a calculation for Earth or Titan atm. center of mass correctly using column avg. PE (PE=mgh).
Re Effective Radiating Level
Perhaps a better term would be “Optimum Radiating Level”, since it is usually deemed to be the altitude where the air is thin enough for ease of radiation, but also where the IR active gases are thick enough for radiation to be strong. Thus it is a trade-off, or an “Optimum” situation. And yes, it is an average, or statistical, reality, not a physical one as others have pointed out.
Trick
The Barometric Formulas and indeed the Ideal Gas Laws require the heat capacity Cp to remain constant while the temperature varies.
This is valid for Earth and Titan (mainly N2 and O2)
However it is not valid for Mars and Venus (both with high CO2 fraction.
http://www.engineeringtoolbox.com/carbon-dioxide-d_974.html
Ron C. – The ERL is physical temperature – checks out as radiosondes with thermometers routinely measure the planetary Teff ~ 255K at the level 1 optical thickness unit below the top of Earth atmosphere (tau = tau infinity – 1 where tau = optical thickness, tau infinity = total optical thickness).
If an atm. is optically thick, radiation emitted by the surface will mostly be absorbed within the atm. Thus, for optically thick atm.s as say in the R&C paper, the radiation emitted to space will mostly originate at some level above the surface, the ERL. Although Mars atm. is relatively optically thin, the planet still lifts its ERL off the surface a little less than 1km.
IR radiative cooling of the Earth by so-called greenhouse gases is strongest from about 4000 to about 11000 meters altitude and with the temperature dropping throughout the troposphere with increasing altitude, radiative cooling becomes less and less efficient. But it is rapid compared to the cooling effects of the lower troposphere.
Data from the NIMBUS satellites of the Earth’s emission spectrum into space show that the dominant water vapor emission is mostly from altitudes from 2.5 km to 6 km, CO2 emission is from 3.5 km to 20 or more km with most of it in the beyond 10 km altitude, and methane and nitrous oxide radiate mostly from 2 to 4.5 km altitude. The methane and nitrous oxide tend not to build up, since they are quickly broken down by UV radiation.
Note that because CO2 reabsorbs its emissions at lower altitudes or often has those emissions reabsorbed by water vapor, it is only from the upper edge of the troposphere that CO2 emissions manage to reach space. The temperature at the top of the troposphere has fallen to a frigid 217K. As a result, the altitude with the temperature matching the thermal equilibrium seen from space of 255 K is found at the top of the water emission zone at about 5000 m.
Of course, in the previous comment the altitudes are averages, since it is well known that the tropopause varies greatly over the earth’s surface, being very near to the ground during polar winter, for example.
Bryan says, December 2, 2014 at 1:30 pm:
“The Barometric Formulas and indeed the Ideal Gas Laws require the heat capacity Cp to remain constant while the temperature varies.
This is valid for Earth and Titan (mainly N2 and O2)
However it is not valid for Mars and Venus (both with high CO2 fraction.”
This circumstance is nowhere near sufficient to explain why the ERL equivalent temperature is found at the solid surface on Mars and at the tropopause on Venus, now is it?
Kristian
Mars is in some ways at the opposite extreme of Venus in terms of temperature and pressure. The
surface pressure is of Mars is 6 mb and mean temperature is ~215 K.
I further don’t expect the Barometric Formula derived for Earth with a near constant Cp for dry air to be accurate in application to Mars.
Venus surface temperatures are somewhere between 700K and 800K .
The Venusian atmosphere is opaque to visible light unlike Earth.
However if one compares the temperature and pressure profiles of Venus and
Earth we will see that they are both very similar. An important difference is
the atmospheric pressure on the ground, which is approximately 100 times than on
the Earth.
At 50 km altitude the Venusian atmospheric pressure corresponds to the normal
pressure on the Earth with temperatures at approximately 310K
However things are extremely complex (volcanic activities, clouds of sulfuric acid)
So again I would not apply the Earths Barometric Equations and expect to get meaningful results.
What is remarkable about the recent posts of HS on the Barometric Equations is how close the article in Nature is to the peer reviewed article of Gerlich and Tscheuschner where the falsification of the CO2 Greenhouse Effect was highlighted.
Gone is the radiative slabs model and other such nonsense favoured by the IPCC.
Gerlich and Tscheuschner recently stated that the days of CO2 greenhouse catastrophe malarkey are coming to an end.
Not before time.
Congratulations to Hockey Schtick
Bryan says, December 2, 2014 at 3:53 pm:
“Congratulations to Hockey Schtick”
The problem, Bryan, and I hope you can see this, is that Hockey Schtick’s argument is invalid already from the get-go. Because it makes use of two flawed (in fact, ridiculous) rGHE/AGW concepts in conjunction:
1) The idea of the ‘effective radiating level’ of a planet, and
2) The idea of lapse rate warming of the surface DOWN from this somehow fixed level.
Sorry, but whatever he’s found or not found, it all falls through on these two points alone.
Kristian
HS was referring to R&C article in Nature.
Clive Best had a take on it
Tallbloke examined HS’s post
Climate Etc seemed to have a parallel discussion but not much on HS posts.
All I took out of it was a development of the Barometric Equations to produce a temperature.
Since the article was not printed in full (or I missed it) I made a quick note of the names and intended to do an in depth study when a full account was available.
The ERM level is something that you question and I have yet to form an opinion on.
The idea of a local vertical centre of mass is interesting and its motion up and down with temperature change and hence the height of the tropopause gives the possibility of another energy storage system for smoothing out the Earths climate.
I reserved comment again to await the full paper.
My posts here and on other sites were just pointing out some related facts which have not been challenged.
I am surprised that the article has not been attacked by the alarmists as it implies CO2 is not responsible in any way for the adiabatic lapse rate
–The problem, Bryan, and I hope you can see this, is that Hockey Schtick’s argument is invalid already from the get-go. Because it makes use of two flawed (in fact, ridiculous) rGHE/AGW concepts in conjunction:
1) The idea of the ‘effective radiating level’ of a planet, and
2) The idea of lapse rate warming of the surface DOWN from this somehow fixed level.
Sorry, but whatever he’s found or not found, it all falls through on these two points alone.–
I would say there are more points of failure.
But In terms of 1, I would say the lowest 1/2 of atmosphere kinetic energy is making the most effect upon daily surface temperatures. Or if view Earth as a blackbody, include the first 1/2 of the atmosphere as the “blackbody surface”. The surface and 1/2 the atmosphere are tied together in terms of the temperature of a single 24 hour period.
Or an ideal blackbody is an invalid concept is sense it’s fictional idea of a very thin surface [doing magical things], but if one can think of a blackbody as a fat surface, that fat surface extends upward to about 1/2 the atmosphere.
The rest of atmosphere certainly effects things but one can ignore it in terms of looking at main factors. As for 2, 1/2 atmosphere is warming the point we measuring temperature [the white box 5 feet above the ground], but what is more important is the surface is warming [mostly] this first 1/2 of the atmosphere. Or said differently if this first 1/2 of atmosphere is not warmed up, the lower surface air will not be warmed up.
As for what else is missing- the ocean which covers 70% of the surface area is missing. And the fact that tropics is the region which receives most to the energy of the sun. These aspects are related in of general idea of first 1/2 of the atmosphere, but are not included.
Kristian 3:05pm, 4:53pm – The 1976 US standard atm. does design in 5.1km at 255.027K. The top post eqn. apparently does work as it apparently overlays the standard. That eqn. which HS shows is just sort of announced on 11/28 from barometric formula and IGL on 11/27 and 11/23. The chain of calculations provenance is not exactly made clear. Nor is HS assertion supported at all: “Pressure = 0.50 atmospheres at the approximate center of mass of the atmosphere”.
You made the comment 5.7km is center of atm. mass for earth – can you support that assertion?
“…Hockey Schtick’s argument is invalid…Because it makes use of .. flawed…1) idea of the ‘effective radiating level’ of a planet…”
You have not shown ERL idea is flawed only asserted your opinion. It is true HS has not supported his ERL location assertion. BTW: Mars ERL is not at the solid surface, it is less than klick above.
******
Bryan 6:18pm: “…(HS) implies CO2 is not responsible in any way for the adiabatic lapse rate..”
The top post eqn. does not imply such. Or support this assertion.
Trick says, December 2, 2014 at 6:58 pm:
“You made the comment 5.7km is center of atm. mass for earth – can you support that assertion?”
I simply eyeballed a standard atmosphere altitude/pressure chart. I see now, after having checked several, that the correct altitude is more likely 5.6 km.
“You have not shown ERL idea is flawed only asserted your opinion.”
There’s no need to ‘show’ it at all. One can know it’s flawed from simlple deduction and logic alone. The ERL is simply a theoretical level somewhere in the atmosphere, mathematically derived solely from Earth’s global LW flux to space, averaged from satellite measurements. You put this mean flux (239 W/m^2) into the Stefan-Boltzmann equation, assuming a BB surface radiating directly to space. This surface would then be at a temperature of 255K. No such single BB surface exists within the Earth system, emitting the entire 239 W/m^2 flux to space. Hence, there is no such surface/layer to measure down from to determine any ‘extra’ surface warming by lapse rate. The idea is flawed.
“BTW: Mars ERL is not at the solid surface, it is less than klick above.”
No. Mars’ ERL equivalent temperature level is the solid surface (or somewhat below it).
http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html
Stating that the mean global temp of Mars is -55C or thereabouts (a ‘classic’ figure) has no empirical basis. It is but a guess, a convenient one, a few degrees above the planetary BB emission temp. The data from the various landers around the surface, mostly recording during summer and mostly located in the lower latitudes, suggests considerably lower global annual means.
Kristian 7:56pm: Not sure how you can eyeball an earth altitude/pressure chart for avg. PE and its height in the standard column. Mars ERL is not est. below ground; here using isothermal assumption same as HS for top post result and NASA est. fact pages (with which you can successfully argue if show your better) data:
Mars Teff = 210.18K.
Mars surface global Tmedian = 214K
DALR = 4.5K/km
Because we know est. temperatures, we roughly know Mars 1 optical thickness unit below TOA (ERL) occurs ~ 0.85km above surface (3.82K down in T at 4.5K/km), rounded.
Earth 1 optical thickness unit below TOA (ERL) is derived not “solely from Earth’s global LW flux to space”. Known T=255K is routinely measured by radiosonde for height and computed from total mass path of absorbers. Physically, the ERL corresponds to the optimal trade-off between high density (which gives high emissivity) and little overlying atmosphere to permit the emitted radiation to escape.
“HS has still not shown a calculation for Earth or Titan atm. center of mass correctly using column avg. PE (PE=mgh).”
Yes I did in the first 2 posts of the greenhouse eqn series.
Newton’s 2nd law of motion at surface: F= ma = mg = 1 atmosphere by definition (1.01325×10^5 N/m2) asl at latitude of Paris France
Gravitational forcing from mass of atmosphere pushing down from above center of mass of the atmosphere F = 1/2mg = 1/2 atmosphere units by definition
Thus, center of mass in atmospheres located where P = 1/2 atm, which can then be calculated with the barometric formula
HS – that is not correct. You need to use the integral over h again, find the eqn.for column PE. Compute PE of the column then Zavg=PE/Ma*g
HS:
1) Center of mass is NOT the median point (where 1/2 the mass is on one side and 1/2 the mass is on the other side). CoM is a weighted average, where the distance as well as the mass matters. As a very simple example, consider a uniform rod 1 m long with a mass of 1 kg. Place 1 kg at one end. The CoM is 0.25 m from the end of hte rod, with 1.25 kg on one side and 0.75 kg on the other.
So for the earth’s atmosphere, the CoM will be well above the median point. You could simply say “median height” instead of CoM to fix this.
2) “you think it’s just a complete coincidence that the temperature at the center of mass of both Earth and Titan is located exactly at the equilibrium temperature with the Sun, right?”
I for one also think that. In particular the median height for the atmosphere of Venus is WAY hotter than any equilibrium temperature determined by insolation. This show that the results for earth and Titan are not some universal law and are indeed a coincidence.
Stephen Wilde, December 1, 2014 at 7:12 pm
“You can only have 255K at the surface if there is no atmosphere at all.”
The 255 K airless Earth temperature is wrong because it is based on unrealistic assumptions concerning the surface properties of our planet.
Much of what you say makes sense but you reduce your credibility by repeating this canard over and over again.
Kristian, December 2, 2014 at 9:31 am
QUOTE
Regarding comments on Mars, as Robinson and Catling point out in their paper, the surface pressure on Mars is only 0.006 bar, far too low to sustain convection or a lapse rate, thus none of the above discussion applies.”
But this is simply not true. There is both vigorous and consistent convection and a defined environmental lapse rate on Mars. The atmosphere being so thin simply makes the diurnal fluctuations around the ALR much greater.
UNQUOTE
Analyzing Mars is way above my pay grade but I liked your comment. If there was no convection on Mars how can one explain the dust devils and the dust storms that sometimes cover more than a quarter of the planet?
I concur with your opinion that diurnal variations are greater when atmospheric pressure is low, which in turn implies larger temperature gradients to drive convection. Conversely, diurnal variations are lower when pressure is high. Venus with its 93 bar surface pressure has lower diurnal variations than Earth in spite of its low rate of rotation.
“1) Center of mass is NOT the median point (where 1/2 the mass is on one side and 1/2 the mass is on the other side). ”
Wrong. In a 1d case which is this one as as stated by HS, half above, half below and is also where the pressure is 1/2 the maximum. The horizontal dimensions here have no hand of influence. I don’t agree with his further use and logic but in this matter he has it correct. You are speaking more in relation to torque or inertia where the mass distribution over distance does matter.
The mass of earth’s atmosphere is 5.1 x 10^18 kg
And Mars atmosphere is ~2.5 x 10^16 kg
http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html
Earth’s atmosphere has 200 times more mass than Mars, but
since Mars is smaller world it only has about 1/100th of atmosphere
of Earth per square meter [Earth has 510 million, Mars has 144.8 million square
km, so it’s a bit more per square meter. But roughly we have 10 tons per square
meter and Mars has about .1 tons per square meter]
What Mars atmosphere “look like” if it had 100 times more atmosphere or if
it had 10 tonnes per square meter.
Because of mars gravity being about 1/3 of Earth’s, it would have 1/3 the pressure
as Earth. so somewhere around 5 psi rather than earth’s 14.7 psi.
Assuming the 100 times added is CO2, one would still have a much lower density
at surface even if Mars remained about cold as it is.
I could not find much about Titan’s atmosphere but here:
http://cseligman.com/text/moons/titan.htm
It says about Titan’s atmosphere:: “Atmosphere 10 times mass of Earth’s, 10 times depth of Earth’s
Atmospheric surface pressure 1.6 times Earth’s”
Or somewhere around ~5 x 10^19 kg.
Titan is 1600 km smaller than Mars, and is less dense.
If transplanted Titan atmosphere onto this Mars [added large amount nitrogen] which has
10 tons of CO2 per square meter, then we the 5 psi + 1.6 times 2 of Atm of pressure. Which
about 52 psi [3.5 atm] giving much more pressure than compared to Earth and much more air density. So we then have which atmosphere about 10% CO2 and 90% nitrogen.
So per square meter there is about 200 tonnes.
Assuming this new Mars doesn’t get a lot colder [greenhouse effect theory probably would guess
it becomes very hot- in terms of parts per million there is a lot methane [which suppose to far more potent than CO2]]. But assuming it’s the same temperature Mars has always been, at what elevation would be half the atmosphere [1.75 atm].
Wayne says December 3, 2014 at 5:10 am:
“Wrong. … You are speaking more in relation to torque or inertia where the mass distribution over distance does matter.”
But that IS what is meant by “center of mass”. If HS or you want to use common scientific terminology, you should use the common meaning of the term. Here, HS should invent some new name like “median position”.
If you still think my definition is flatly “Wrong”, then find even ONE source that uses Center Of Mass the way you claim it should be used.
Tim Folkerts says: December 3, 2014 at 12:06 pm
Wayne says December 3, 2014 at 5:10 am:
((“1) Center of mass is NOT the median point (where 1/2 the mass is on one side and 1/2 the mass is on the other side). ”))
(“Wrong. In a 1d case which is this one as as stated by HS, half above, half below and is also where the pressure is 1/2 the maximum. The horizontal dimensions here have no hand of influence. I don’t agree with his further use and logic but in this matter he has it correct. You are speaking more in relation to torque or inertia where the mass distribution over distance does matter.”)
Ref:Tim Folkerts says: December 3, 2014 at 3:05 am
(“1) Center of mass is NOT the median point (where 1/2 the mass is on one side and 1/2 the mass is on the other side). CoM is a weighted average, where the distance as well as the mass matters. As a very simple example, consider a uniform rod 1 m long with a mass of 1 kg. Place 1 kg at one end. The CoM is 0.25 m from the end of hte rod, with 1.25 kg on one side and 0.75 kg on the other. So for the earth’s atmosphere, the CoM will be well above the median point. You could simply say “median height” instead of CoM to fix this.”)
Folkerts,
Your COM is correct only for rotational inertia or angular momentum just as Wayne stated! Your language usage correction is correct and most helpful for understanding. Your reply to Wayne:
“But that IS what is meant by “center of mass”. If HS or you want to use common scientific terminology, you should use the common meaning of the term. Here, HS should invent some new name like “median position”. If you still think my definition is flatly “Wrong”, then find even ONE source that uses Center Of Mass the way you claim it should be used.”
Is meant only to confuse! You did not define, you deliberately used a non relevant context!!!
Tim, you never use a relevant context, if you think you can get away with it. “One source”??? genuine baiting, and the waste of another’s time and effort. Do your own research when lacking understanding.
For EMR, rotational inertia or angular momentum, are not applicable ever. Gravity is not relevant except when the gravitational mass has considerable relative motion to the vector of electromagnetic flux. For EMR, the atmospheric COM is co-located with the center of mass of the earth (concentric) and is non weighted.
Will says: “Tim, you never use a relevant context … Your COM is correct only for rotational inertia or angular momentum”
My “context” is “all of physics and engineering” (not just rotational motion)! Look in ANY engineering or physics text on your shelf. Look in ANY online source. Ask ANY physics or engineering professor. “Center of Mass” is ALWAYS the weighted average x(CoM) = Σx(i)m(i) / Σm(i).
> Tim’s context is “physics”
> Tim’s context is never relevant
>> Therefore physics is never relevant in this discussion. 🙂
“Do your own research when lacking understanding.”
Do your own research when accusing someone of not understanding. You simply bluffed your way through a rebuttal; you were wrong AND you didn’t support your conclusion.
I didn’t specify a specific source because 1) you might not have it and 2) it’s irrelevant anyway since they all use this definition.
“You did not define [center of mass]”
You are right — I did not define “center of mass” as x(CoM) = Σx(i)m(i) / Σm(i). I shouldn’t need to. Just like I shouldn’t need to define “median” or “uniform” or “insolation”. If someone does’t know the definitions, then they don’t belong in this discussion.
“Do your own research when lacking understanding.”
Do your own research when accusing someone of not understanding. You simply bluffed your way through; you were wrong AND you didn’t support your conclusion.
“Tim, you never use a relevant context … the atmospheric COM is co-located with the center of mass of the earth”
You provide a perfect example of not using the relevant context (I won’t say you did it deliberately, since I doubt you even noticed). Yes, the CoM of the entire atmosphere will be at the center of the earth. But in the context of this discussion, we are not concerned about the CoM of entire atmosphere. We are concerned about CoM (or again, more precisely, the “median height”) of a single column of gas somewhere.
Will 7:33pm: “For EMR, the atmospheric COM is co-located with the center of mass of the earth (concentric) and is non weighted.”
Will – That’s pedantic to the max., as Tim points out. Let me spell it out for Will slowly. The context for HS discussion is atmosphere per unit area.
Now, in context, the center of gravity of a m^2 column of atm. rises agl as the PE increases from heating. Hint: the height (Z) agl is important concept in context.
PE=mgh
So for HS to correctly find the center of gravity of a column of air which HS did not correctly calculate due to varying density, rho, ground to space which is designed for his US Standard:
Center of gravity = Zcg =
Zavg. = (integral 0 to infinity rho*g*z dz)/(integral 0 to infinity rho*g dz)
You may recognize (I doubt it) the bottom being the total mass air in the column Ma assuming g is close to constant hence:
Zavg. = PE/Ma*g
The real task ahead of HS is to find where rho( z) is the density at column z:
(integral 0 to infinity rho( z)*g*z dz)
for HS US standard atmosphere m^2 column. His thesis depends entirely on doing this correctly. Otherwise a whole series of posts fails.
Tim Folkerts says: December 3, 2014 at 8:25 pm
(Will says: “Tim, you never use a relevant context … Your COM is correct only for rotational inertia or angular momentum”)
“My “context” is “all of physics and engineering” (not just rotational motion)! Look in ANY engineering or physics text on your shelf. Look in ANY online source. Ask ANY physics or engineering professor. “Center of Mass” is ALWAYS the weighted average x(CoM) = Σx(i)m(i) / Σm(i).”
The subject here is thermal EMR. Please show where any thermal EMR is subject to any mass moment (torque) within or without a gravitational field. Please stop being a “statistical mechanic”,
half your parts must fall off
Trick says: December 3, 2014 at 8:45 pm
(Will 7:33pm: “For EMR, the atmospheric COM is co-located with the center of mass of the earth (concentric) and is non weighted.”)
“Will – That’s pedantic to the max., as Tim points out. Let me spell it out for Will slowly. The context for HS discussion is atmosphere per unit area.”
It is not, except for those that can only think of Newtonian kinetic energy. The subject here is thermal EMR. Please show where any thermal EMR is subject to any mass moment (torque) within or without a gravitational field. Please stop being a “statistical mechanic”, half your parts must fall off, always! Please try to learn something of the principles of Electricity and Magnetism, before demonstrating your complete ignorance of EMR.
Thermal EMR is absorbed, or emitted from any surface or cross-sectional area, with no required mass. Its magnatude is dependent on the difference in radiative potential between surfaces. That difference in radiative potential at each frequency,is a highly non-linear function of the difference in absolute temperature between “surfaces”.
Max Planck’s equation correctly describes the maximum possible electromagnetic ‘radiance” (potential) at each frequency, at each temperature, and in each direction. No where is indicated any energy flux. Any aggregation of the variables in Planck’s equation, in opposition to Maxwell’s equations, can lead only to deceit of self, and others.
HS may as well picked the stratopause, as ERL with the characteristics of 30% reflective, 70% transmissive, from 0.1 micron to 2.5 micron, and 100% emissive at longer wavelengths. A Willis Eschenbach. steel greenhouse. This would better fit the meteorological fantasy of thermal EMR. (Everything under the stratopause is “CONVECTION”, never radiation). You folk may amuse/abuse yourselves with any sort of fantasy. Please do not call that knowledge or understanding.
Will 10:33pm: “The subject here is thermal EMR.”
Not at all Will. The subject is center of gravity in the top post “One equation for earth temperature”. Please try to pay attention.
“Thermal EMR is absorbed, or emitted from any surface or cross-sectional area, with no required mass.”
Fantasy! After mastering the cg subject Will, next please try to become competent in thermal EMR for proper discussion. Do your own research.
Trick says:
December 3, 2014 at 10:50 pm
//////////////////////////////////////////////////
”Fantasy! After mastering the cg subject Will, next please try to become competent in thermal EMR for proper discussion. Do your own research”
Do you realise just how much of a fool you just made of yourself Trick?! Will’s extensive experience in radiative physics has be covered several times on threads at Talkshop. Military stuff that had to work, not climastrology. What did you imagine Modtran, Hitran and Lowtran were developed for? What problem did you think they were solving? Which software are climastrologists misusing?
Trick says:December 3, 2014 at 10:50 pm ********************************************************************
(Will 10:33pm: “The subject here is thermal EMR.”)
“Not at all Will. The subject is center of gravity in the top post “One equation for earth temperature”. Please try to pay attention.”
Where in the equation does center of gravity of anything exist? can you not read?
(“Thermal EMR is absorbed, or emitted from any surface or cross-sectional area, with no required mass. Its magnatude is dependent on the difference in radiative potential between surfaces. That difference in radiative potential at each frequency,is a highly non-linear function of the difference in absolute temperature between “surfaces”.”)
“Fantasy! After mastering the cg subject Will, next please try to become competent in thermal EMR for proper discussion. Do your own research.”:
Please identify “any” error in my statement? Where ever does thermal EMR require mass?
Does Solar irradiation involve mass? Please try to learn something of the principles of Electricity and Magnetism, before demonstrating your complete ignorance of EMR.***********************—
Konrad 11:16pm: Yes. Line by line radiative transfer. Problem solved using radiation (commonly known as radar and radio et. al.) for putting a BGM-109D 450kg hi-ex over a campfire in Afghanistan from 1500 klicks standoff on dark, stormy nights sufficiently close scatter the camels which didn’t sleep at night for awhile in the future. Be specific on the software.
Better, more helpful, to stay on topic.
******
Will 11:20pm: “Where in the equation does center of gravity of anything exist?”
The eqn. and thesis of the top post HS eqn. step 3 11/27 entirely depends on the cg of the US Standard Atmosphere column computing out to h=5100m as shown. This computation for column cg is incorrectly done as density, rho, varies from 0 agl up to space. I have pointed out the correct way to calculate the cg of a column of air. Will has not, was pedantic writing the atm. cg is at center of earth unhelpfully.
“Thermal EMR is absorbed, or emitted from any surface or cross-sectional area, with no required mass….Please identify “any” error in my statement?”
Delete the “no” to eliminate Will’s error. To absorb & emit EMR, the surface requires mass.
“Does Solar irradiation involve mass?”
Yes. Actually it is solar radiation; the solar EMR is from the mass of H, He, some other constituents. /pedantic
“Please try to learn something of the principles of Electricity and Magnetism, before demonstrating your complete ignorance of EMR”
Take your own advice Will. I recommend a good text on atmosphere radiation.
12:30am: ”Mass “only” absorbs thermal EMR flux when the temperature of that mass is below that of radiative equilibrium.”
This must be why we walk into invisible trees and lampposts out in the sun all the time, they are unseen by our cooler retinas not absorbing their EMR flux. Technology of Romulan origin.
So Trick thinks a cold pit viper can ‘see’ i.r. from some source that is colder than it’s physical eyes. Amazing! And this is coming from someone that claims he knows radiation physics!
wayne – Humans can see objects in the visible range just fine even if objects are warmer or colder than retinas. Any help on air column cg computation for top post eqn.?
Trick says: December 4, 2014 at 2:18 am
“wayne – Humans can see objects in the visible range just fine even if objects are warmer or colder than retinas. Any help on air column cg computation for top post eqn.?”
Radiative genius Trick thinks that a human retina is at or above radiative equilibrium for visible reflected solar illumination. That brightness “temperature” is way above that of your retna.
“Delete the “no” to eliminate Will’s error. To absorb & emit EMR, the surface requires mass.”
To convert EMR to/from other types of power “may” require some mass. Flux requires no mass.
To convert latent heat of evaporation to EMR requires a mass “value” of H2O, to determine total energy exited to space. All mass remains within the atmosphere. EMR flux never requires mass itself, only some properties of energy that mass may have. Latent heat energy is never conserved in this atmosphere.
(“Does Solar irradiation involve mass?”)
“Yes. Actually it is solar radiation; the solar EMR is from the mass of H, He, some other constituents.”
Please demonstrate where solar radiant flux requires anything but the “radiance” of the sun?
Please demonstrate that air column cg has anything to do with EMR exit flux?
Just more Trikster’s feeble intent to confuse!
Will – with such radiative genius, computing the cg of an atm. gas column should be easy for you to show HS the correct method, please apply yourself to the topic.
Trick says: December 4, 2014 at 12:13 am
“Konrad 11:16pm: Yes. Line by line radiative transfer. Problem solved using radiation (commonly known as radar and radio et. al.) for putting a BGM-109D 450kg hi-ex over a campfire in Afghanistan from 1500 klicks standoff on dark, stormy nights sufficiently close scatter the camels which didn’t sleep at night for awhile in the future. Be specific on the software.”
The BGM-109 Tomahawk is a long-range, all-weather, subsonic cruise missile. was introduced by General Dynamics in the 1970s. Even today it has no visual or IR seeing capability. It is guided by precision GPS, and uses limited radar for collision avoidance. None of any control is aided by any Trickster LBLRTC. Trickster has never used such code and has never even read the users manual. That code does nothing of Trickster’s fantasy.
Oh yes the terrain following using the mil. version of gps (more accurate). One of those eyebrow raising it works? No reason why not and highly resistant to non high tech space country countermeasures. Won’t work if there is a real war unless there is a covert system up there we don’t know about.
Not sure how far optical gyros have progressed, another possibility.
Given the Chinese and Russians have space based navigation I suspect there would be reluctance to kill satellites short of the nuclear button.
Jamming GPS is quite a subject, very real for civilian. Most worrying is the proven kill receiver hack: force a clock advance where plenty of systems have no reverse gear, assume they can’t lock and time will be later than now.
Will – Check out the capability of the D model. Be helpful, please show HS the way to properly compute cg of atmosphere gas column, move the topic forward.
Trick says: December 4, 2014 at 3:54 am
“Will – Check out the capability of the D model.”
That is the sub-munitions version from Raytheon. It’s a wonder the thing works at all, from them. The Terrain matching Fl radar and altimeter profile works, as does laser target designation from near to the target. Image matching has not proven successful. None of this uses any radiative transfer code whatsoever.
“Be helpful, please show HS the way to properly compute cg of atmosphere gas column, move the topic forward.”
Take your column CG and shove it! I want no part of your deliberate misdirection.
Will – Not helpful. Thanks for trying your best though.
Tim C – Interesting idea about the clock.
Eh, you can’t be serious. The reason the climate system is in thermal disequilibrium is it being enclosed between two heat reservoirs at vastly different temperatures. One is a small patch of the sky (0.00054% of it) at 5778 K, while the the temperature of the rest (99.99946%) is 2.72548±0.00057 K. Thermal coupling between the climate system and these heat reservoirs is purely radiative, as no other heat transfer mechanism is operative across vacuum. As soon as you make the atmosphere completely transparent to all kinds of EM radiation, it is no longer coupled to any heat reservoir outside the system, only to the surface, which in turn is coupled to both the hot and cold reservoirs. That makes the bulk of the atmosphere isothermal with no lapse rate whatsoever, with a thin thermal inversion layer over much of the planet (except the hottest region close to the equator at noon). That makes convection pretty much impossible, so heat exchange between the surface and atmosphere is only by conduction (but heat conductivity of gases, as you surely know, is extremely low). That’s an atmosphere very different from what we have.
If, on top of it, the gas your hypothetical atmosphere was made of had short wave absorption lines (all gases have some in the hard UV end of the spectrum), the temperature inversion would go all the way up to ToA (Top of Atmosphere) and the negative lapse rate were only bounded by the poor conductivity of said gas. That means the atmosphere, even if it were made of some heavy noble gas like Argon, would boil away to space in a couple of million years, leaving the surface in pristine vacuum.
The primary role of GHGs (gases having absorption and emission lines in the thermal infrared) is cooling the upper atmosphere by establishing a direct coupling between it and the ultimate cold reservoir (the CMBR background). That’s what maintains a steady positive lapse rate and makes convection possible below the tropopause (which does not exist in atmospheres dominated by gases with more internal degrees of freedom than mono- or diatomic gases — in such atmospheres convection goes all the way up to ToA).
A static gravitational field in itself is insufficient to establish and maintain a temperature gradient. In a steady state there is no work done against gravity whatsoever.
Sorry Berenyi but that cannot be right.
Even for an atmosphere with no GHGs uneven surface heating will cause differing densities in the horizontal plane and that will lead to lighter parcels rising up relative to heavier parcels.
Once that begins KE (heat) converts to PE (not heat) and convection continues adiabatically.
Descent follows automatically.
Onme cannot suppress convection and cooling with height for a gas overlying an unevenly heated surface within a gravitational field.
Check out the data in the 1976 US Standard Atmosphere here:
http://hockeyschtick.blogspot.co.uk/2014/12/why-us-standard-atmosphere-model.html
There is no denying those 241 pages of thoroughly worked out and observationally authenticated data.
Your scenario just doesn’t happen in the real world. Period.
GPS,
not the article I had in mind but nevertheless the stupidity continues … and they don’t want LORAN
http://eandt.theiet.org/magazine/2011/04/gps-vulnerabilities.cfm?origin=EtOtherStories
LORAN, small part here
http://www.extremetech.com/extreme/193571-our-terrifying-reliance-on-gps-and-the-need-to-develop-a-ground-based-alternative/2
My company name GPSL (probably up for sale but that is a different matter, GPS is deliberate)
Can’t find the article, was fairly tech anyway. About a proof of concept exploit on receivers and it worked. Into factory reset to get out of it. Out there in use. Best not have any bright kiddies around.
Given there are the Chinese and Russian systems, maybe Indian sometime, EU want a money sucking system too, there is a lot going to happen.
Loran was IIRC knifed by Obama but here in Europe it struggles on. Linkage feasible with the Russian eastern Arctic route.
Combined receivers are I think not yet on the market.
The good one for Loran is penetration, low frequency will even go inside buildings.
Bad one? Oh yes, the prayer wheels mess it up, get rid of them.
Berényi Péter says: December 1, 2014 at 6:57 pm
“Now, the atmosphere (at least below a height of 50 km) is always in local thermodynamic equilibrium, so Kirchhoff’s radiation law can be applied. That is, in any frequency band emissivity is equal to absorptivity. Which means the atmosphere can not emit radiation in the thermal infrared either, so ERL is not at an intermediate level in the atmosphere, but right at the surface.”
Sorry, The atmospheric temperature at all levels below 240 km are at a temperature above that for radiative equilibrium, so Kirchhoff’s radiation law can be applied. That partially emissive gas not only transmits without attenuation any EMR from a higher radiance, it also adds to the radiative exitance as allowed by Maxwell’s equations. Exitance accumulates outward all the way to 220 km. All levels of the atmosphere are effective radiating levels.
Berényi Péter says: December 5, 2014 at 8:49 pm
Stephen Wilde says: December 1, 2014 at 7:12 pm
(” Berenyi Peter, The point is that in the vertical plane an atmosphere is always in thermodynamic DISequilibrium due to work done against gravity in holding the mass of the atmosphere off the surface converting KE to PE to produce an inevitable lapse rate even in a GHG free atmosphere.”)
“Eh, you can’t be serious. The reason the climate system is in thermal disequilibrium is it being enclosed between two heat reservoirs at vastly different temperatures. One is a small patch of the sky (0.00054% of it) at 5778 K, while the the temperature of the rest (99.99946%) is 2.72548±0.00057 K. Thermal coupling between the climate system and these heat reservoirs is purely radiative, as no other heat transfer mechanism is operative across vacuum.”
Yes, mostly! Space has an average brightness temperature of 6.8 Kelvin, only the CMB is that low.
“As soon as you make the atmosphere completely transparent to all kinds of EM radiation, it is no longer coupled to any heat reservoir outside the system, only to the surface, which in turn is coupled to both the hot and cold reservoirs. That makes the bulk of the atmosphere isothermal with no lapse rate whatsoever, with a thin thermal inversion layer over much of the planet (except the hottest region close to the equator at noon).”
The earth’s atmosphere is not at any frequency completely transparent. The lapse rate, sometimes called DALR depends not on radiation, but only on the thermostatic gravitational compression in the lower atmosphere. All gas increases in both pressure and temperature when mechanically compressed as the atmosphere formed. This thermostatic state, is maintained as adiabatic and isentropic as the same gravitational field prevents the expected heat diffusion to the lower temperature above. Rudy Clausius never insisted that spontaneous need happen.
“This makes convection pretty much impossible, so heat exchange between the surface and atmosphere is only by conduction (but heat conductivity of gases, as you surely know, is extremely low). That’s an atmosphere very different from what we have.”
Natural convection is but the hydrodynamic buoyancy. This has nothing to heat or temperature. Sometimes sensible heat is moved, along with the mass. Any mass movement, however, is always entropic, never adiabatic. What is the rate of latent heat transfer from saturated WV, to unsaturated WV? How would anyone measure that chemical heat transfer? Is partially condensed WV made up of polymer/tri-mers, rather than monomer/tri-mers? What is the density?
“If, on top of it, the gas your hypothetical atmosphere was made of had short wave absorption lines (all gases have some in the hard UV end of the spectrum), the temperature inversion would go all the way up to ToA (Top of Atmosphere) and the negative lapse rate were only bounded by the poor conductivity of said gas. That means the atmosphere, even if it were made of some heavy noble gas like Argon, would boil away to space in a couple of million years, leaving the surface in pristine vacuum.”
What nonsense! The gas would increase in temperature until all energy received by any means is dispatched to space via EMR. How do you boil a gas? The Earth like the Sun, has gravitational attraction keeping gas in place independent of temperature.
“The primary role of GHGs (gases having absorption and emission lines in the thermal infrared) is cooling the upper atmosphere by establishing a direct coupling between it and the ultimate cold reservoir (the CMBR background).”
OK! But all atmospheric levels not just the upper.
“That’s what maintains a steady positive lapse rate and makes convection possible below the tropopause (which does not exist in atmospheres dominated by gases with more internal degrees of freedom than mono- or diatomic gases — in such atmospheres convection goes all the way up to ToA).”
What TOA? Again nonsense, with no gravity, only EMR, the lapse rate in this troposphere would be minus 14-17 degree Celsius/km. The DALR of -9.8 degree Celsius/km requires more EMR flux outward from every level (higher temperature). The conversion of latent heat of evaporation requires even higher atmospheric temperature and even higher outward EMR flux from every level.
“A static gravitational field in itself is insufficient to establish and maintain a temperature gradient. In a steady state there is no work done against gravity whatsoever.”
No work need ever be done to maintain a thermostatic state. A bale of hay on a shelf, stays there.
What a wonderful planet!!! Except for ClimAstrologists who have nary a clue!!! -will-
Berényi Péter says:
December 5, 2014 at 8:49 pm
//////////////////////////////////////////////
”The primary role of GHGs (gases having absorption and emission lines in the thermal infrared) is cooling the upper atmosphere by establishing a direct coupling between it and the ultimate cold reservoir (the CMBR background).”
You are correct, however I would use the term radiative gases not GHG’s. “GHG’s” is a propaganda term that presupposes a net atmospheric radiative GHE.
Steven’s claim that tropospheric convective circulation in the Hadley, Ferrel and Polar cells defies the results of empirical experiment. Strong vertical circulation in these cells would stall and the atmosphere would overheat were it not for radiative-subsidence at altitude.
I have checked this via empirical experiment with gas columns and again via CFD –
– energy entry and exit only at low altitude causes R-B circulation to stall and the gas column temperature to rise.
Stephen Wilde says:December 6, 2014 at 12:16 am
“Check out the data in the 1976 US Standard Atmosphere here:
http://hockeyschtick.blogspot.co.uk/2014/12/why-us-standard-atmosphere-model.html
There is no denying those 241 pages of thoroughly worked out and observationally authenticated data. Your scenario just doesn’t happen in the real world. Period.”
There is no denying those 241 pages of utter nonsense. That nonsense completely ignores radiative exitance to space from all levels of the atmosphere. All of the “heat energy to gravitational potential energy” is what never happens in this troposphere. Authenticated fantasy with only a equation involving altitude and temperature with absolutely no understanding whatsoever. All of your meteorological nonsense about has been trash since 1965. Any attempt to claim saturated WV changes Cp is also trash. W/V must actually produce sensible heat upon partial condensation.
To conform to Kepler’s laws and the Clausius virial theorem, only Newtonian kinetic energy, with momentum, maintains a virial with gravitational potential. That kinetic can only have one constraint on DOF, gravity. The no-momentum heat energy in the troposphere is constrained also by every nearby molecule.
”
–Berényi Péter says:
December 5, 2014 at 8:49 pm
//////////////////////////////////////////////
”The primary role of GHGs (gases having absorption and emission lines in the thermal infrared) is cooling the upper atmosphere by establishing a direct coupling between it and the ultimate cold reservoir (the CMBR background).”
You are correct, however I would use the term radiative gases not GHG’s. “GHG’s” is a propaganda term that presupposes a net atmospheric radiative GHE.–
Is there nothing in the atmospheric but gases which radiate?
Is possible that there is more than a trillion tonnes of stuff which is not in gaseous state which is in the atmosphere.
How many trillions of tonnes of non-gases in atmosphere would enough to be more than insignificant?
Stephen 12:16am: On US Standard Atm. 1976: “There is no denying those 241 pages of thoroughly worked out and observationally authenticated data.”
The paper itself denies this in the Foreward. Stephen should actually read this stuff before commenting.
Pg. xiv: “…this U.S. Standard Atmosphere…is as follows: “…A hypothetical vertical distribution of atmosphere temperature, pressure, and density…”
Sure, the eqn. agreed to and adopted by fiat “is roughly representative of year-round mid-latitude conditions.” One has to realize the surface (0m) temperature 288.15K was determined by a committee. As was the lapse rate. The HS eqn. in top post is for hypothetical atm. not the actual atmosphere.
Great.
Decades of successful application of the Standard Atmosphere in aeronautics and folk here consign it to the trash can.
Does anyone happen to know where I can find the complete detailed derivation of the dry adiabatic lapse rate (DALR=g/cp)? To my surprise I thought this was within Wikipedia under “Lapse Rate” but you get to the very last line and that doesn’t give the reason why g/cp appears as the answer but instead points to a problem in a book that doesn’t show the answer or the derivation at all!
I’ve realized this is the last step having my complete thermodynamic and barometric derivation chains but I get to this supposedly easily last step and can find nothing of it on the web so far.
I can find boo-koo definitions and loose descriptions but that is not good enough. Why exactly is DALR = g/cp = ≈9.8 K/km from first thermodynamic principles?
wayne asks
“Does anyone happen to know where I can find the complete detailed derivation of the dry adiabatic lapse rate (DALR=g/cp)? ”
Click to access 1003.1508.pdf
Page 11
HS and Stephen Wilde have a fundamental misunderstanding of the adiabatic processes in the atmosphere.
For instance HS in answer to a reply to a blogger gives a rising balloon as an example of gravity doing work which turns into gravitational potential energy of balloon plus gas content.
What they don’t see is the falling atmospheric gas filling in the void left by the rising balloon.
PE gained by balloon is exactly equal to the PE lost by falling atmospheric gas so there is no net work done by gravity.
The adiabatic lapse rate sets the slope of temperature with changing altitude .
It does not give the actual surface temperature.
Trick is correct when he says the surface value given (288.15K) was determined by a committee.
AS far as I can make out there is no known way to determine the actual surface temperature of a particular location other than an actual on the spot measurement with a thermometer.
Berenyi Peter: “In a steady state there is no work done against gravity whatsoever.”
I think what Stephen Wilde is driving at is that the solar energy throughput and 20% absorption in the atmosphere and heating from the sun warmed surface is what causes the inflation of the atmosphere against gravity. If the Sun stopped, then the atmosphere would cool, shrink, and congeal on the surface. The cold dense air at high altitude want to fall to the ground under gravity, and indeed it does subside, only to be replaced by less dense air from below. There can be no ‘steady state’, but there is a dynamic equilibrium which involves this vertical circulation.
I agree that the cooling of the high altitude air by radiation to space assists this circulation, but I don’t believe it is solely or even dominantly the cause of it. It balances the solar energy heating the surface which warms the atmosphere from below – the heating and cooling processes both cause circulation by convection and subsidence, and they balance in order to bring about a dynamic ‘steady state’. Underlying that is the temperature gradient created by the differential absorption of solar energy by denser near-surface and rarer high altitude layers of the atmosphere, which also contributes to the support of the lapse rate.
So it’s not a question of ‘work done against gravity’, it’s a question of the reasons why the distribution of energy in the atmosphere is what it is. The force of gravity plays a part by causing a pressure gradient to exist in the atmosphere which determines where the denser more energy absorbant layers are and where the less dense layers holding less energy are. That density gradient is the reason why energy can be radiated to space more easily from the top of the troposphere, making that air colder, denser and causing it to subside.
Bryan says:
December 6, 2014 at 9:08 am
wayne asks
“Does anyone happen to know where I can find the complete detailed derivation of the dry adiabatic lapse rate (DALR=g/cp)? ”
Click to access 1003.1508.pdf
The dry adiabatic lapse rate of the troposphere, DALR
I prefer the theory neutral term temperature gradient
Use only: the 1st Law of thermodynamics, conservation of energy
And the 2nd law of motion, force = mass x acceleration F = ma,
For vertical motion in Earth’s gravity field F =mg
Kinetic, heat, energy (K.E.) = mass x specific heat x temp. = mcT
Gravity energy = mass x gravity x height = mgh
Total energy constant so
mgh = mcT
temp. grad. = T/h = g/c
T and G use Ideal Gas Law ( IGL ), which does not apply to the atmos.
For IGL: pressure caused by walls and not by gravity
For barometic equations: pressure caused by gravity and not by walls
Can’t use both in a derivation from theory
Roger Clague thinks that;
A rising parcel of air loses internal energy and converts an increasing proportion of it from kinetic to potential energy.
He then goes on to show how this gives the formula for DALR
Does this correspond to reality?’
Theres an easy way to find out.
Your theory is the loss of internal energy on ascent is turned into potential energy.
Lets work through a problem where we are all agreed about the numbers used
So take a mole of air at STP.
Work out its internal energy from formula KE =Nk x2.5T
N = Avogadro’s Number
k = Boltzmann’s Constant
T =Standard Temperature in Kelvin units = 273K
Now do the same at a height of one Kilometre with the same formula but with T2 = T -9.8
The 9.8 comes from the drop in temperature due to adiabatic lapse rate drop for one kilometre
Subtract this value of internal energy from the STP value.
= 204J
Now if you are correct this value will be equal to the gravitational potential energy gained.
PE = mgh
m = mass of one mole of air in kilograms = 0.029Kg
g = 9.81
h = 1000
All with the correct units implied
PE = mg(delta)T = 284J
You will find that the energies will not equal one another.
To get the correct answer you need the barometric formulas for an adiabatic atmospheric expansion.
Work done by parcel in expanding against surroundings(atmosphere)
= nCv(T2 – T1)
n = number of moles = 1
Cv =20.8
T2 – T1 = 9.8
Work done = 204Joules which is the same as the value of lost internal energy given above.
Your theory does not give realistic answers so although by coincidence they look the same the theory behind it is false.
Try and work them out for yourself
check your answers against mine
Internal Energy lost = 204J
Potential Energy at 1000m = 284.J
Work done in adiabatic expansion using the barometric formula’s = 204J
The parcel on returning to the Earth surface would gain 204J
That why your method (although quite plausible) does not work
Not one Joule of lost internal energy turns into gravitational potential energy
Stephen 5:50am: “..folk here consign (U.S. Standard Atm.) to the trash can.”
No, Stephen. Just your non-reading of 241 pages you comment upon. If you were to read them, then find therein: “Typical usages are as a basis for pressure altimeter calibrations, aircraft performance calculations, aircraft and rocket design, ballistic tables, and meteorological diagrams.”
Bryan,
Way too many people, it seems, believe that it is somehow the lifting of the air itself that cools it. The lifting is not an adiabatic (thermodynamic) process at all. It is a Newtonian (mechanical) process. A thermodynamic process specifically deals with energy transfers leading to changes in temperature. A mechanical process merely deals with changes in motion through various forces being applied.
Changing mechanical KE into gravitational PE and back is real enough; it simply has no bearing whatsoever on temperatures. Exchange of energy across system boundaries in the form of heat or work does.
The expansion and compression of the lifting/sinking air is the (only) adiabatic process going on. The expansion/compression of air is in itself not dependent on the lifting/sinking of the air for them to cool/warm the air. In the atmosphere you simply need to move the air up or down to change the external pressure. If you could change the external pressure in any other way, you wouldn’t need to move the air at all.
Expanding air cools adiabatically by doing work on its surroundings (thus losing internal energy), against the external pressure. Air being compressed warms adiabatically from the surroundings doing work on it (it thus gains internal energy).
A lot of confusion could be avoided on this particular topic if everyone simply read a little bit about the adiabatic process. It’s not that mysterious.
Kristian
“Bryan,
Way too many people, it seems, believe that it is somehow the lifting of the air itself that cools it. ”
Well hopefully my worked example will convince some people.
I’m not too hopeful about Stephen though.
If an explanation relies on text only, then a lot will ‘get it’ however some won’t.
If we have paragraphs of a near random arrangement of words like adiabatic, second law,entropy,kinetic energy,isothermal,potential energy,lapse rate and so on then it might look ‘scientific’ but is in fact nonsense.
Kristian: “Changing mechanical KE into gravitational PE and back is real enough; it simply has no bearing whatsoever on temperatures.”
In the classical mechanical view, KE is related to temperature and PE is not. So if KE gets converted to PE, then temperature falls, because total energy remains the same. If you disagree with this, then tell me why.
Bryan: Not one Joule of lost internal energy turns into gravitational potential energy
So where does the additional PE come from when the altitude changes?
Thanks Rog.
I’m watching this stuff with increasing disbelief and an increasing unwillingness to waste more of my time on it.
Rog said:
“I agree that the cooling of the high altitude air by radiation to space assists this circulation, ”
Absolutely right but think what it means to ‘assist’.
To assist means that the circulation has to be LESS powerful and so LESS energy gets back to the surface during adiabatic descent than was removed from the surface in adiabatic ascent.
So the reduced energy getting back to the surface in adiabatic descent offsets any warning effect from GHGs.
Supremely simple but just not sinking in.
Too many people think that assisting the circulation means enhancing it. Not so.
The circulation has a task to perform, namely maintaining atmospheric height. Assisting it means that the circulation has to work less hard to maintain atmospheric height.
AGW theory (like many contributors here) has it backwards as Hockeyschtick is clearly demonstrating with the equations used in recent posts and the utility of the Standard Atmosphere proves it.
Just look at the straight snow line across a mountain range regardless of vigorous convection up and down and try to say that the Standard Atmosphere does not define an absolute truth.
It could not have been used successfully in aeronatics for decades unless it were so.
And there is no term for the radiative characteristics of atmospheric gases in in the Gas Laws or the Standard Atmosphere.
tallbloke says, December 6, 2014 at 2:35 pm:
“In the classical mechanical view, KE is related to temperature and PE is not. So if KE gets converted to PE, then temperature falls, because total energy remains the same. If you disagree with this, then tell me why.”
You need to separate between the translational KE of indivudal molecules (through the kinetic theory of gases proportional to IGL absolute temperature (and pressure-volume)), also known as the ‘thermal energy’ of an ideal gas (the part of its internal energy [U] that corresponds to its temperature), and the mechanical KE of an object (or a parcel of air for that matter), the amount of kinetic energy it contains thanks to its macroscopic motion through space. The kinetic energy of an object simply has to do with how much work we must apply to get it into a certain speed or how much work it in turn is able to do on its surroundings at this certain speed. When you lift an object, it doesn’t get any cooler from this mechanical KE being turned into gravitational PE. You simply ‘charge’ the object with energy to be expended on its way back down, were you finally to let it drop.
Again, you need to separate between Newtonian (mechanical) processes and adiabatic (thermodynamic) processes. Only the latter concerns changes in temperature. The former only concerns changes in motion/forces.
tallbloke says:
“So where does the additional PE come from when the altitude changes?”
Here is an excellent post from HS
equilibrim
http://hockeyschtick.blogspot.co.uk/2014/05/maxwell-established-that-gravity.html
Maxwell
“This result is important in the theory of thermodynamics, for it proves that gravity has no influence in altering the conditions of thermal equilibrium in any substance, whether gaseous or not. ”
So left to itself over time a gas will move towards thermal equilibrium
Maxwell again
”This result is by no means applicable to the case of our atmosphere. Setting aside the enormous direct effect of the sun’s radiation in disturbing thermal equilibrium, the effect of winds in carrying large masses of air from one height to another tends to produce a distribution of temperature of a quite different kind, the temperature at any height being such that a mass of air, brought from one height to another without gaining or losing heat, would always find itself at the temperature of the surrounding air.”
So the unequal heating of the globes surface produces thermal disequilibrium and winds constantly mixing up the atmosphere
The air is compressible so the air nearest the surface will be at a higher pressure hence higher density.
Gravity imposes hydrostatic equilibrium meaning that hot air will rise up and colder air fall
Hydrostatic equilibrium is established much faster than thermal equilibrium
Maxwell again
“The extreme slowness of the conduction of heat in air, compared with the rapidity with which large masses of air are carried from one height to another by the winds, causes the temperature of the different strata of the atmosphere to depend far more on this condition of convective equilibrium than on true thermal equilibrium.”
Kristian,
The flaw in your post is that you are trying to deal with gases as though they were solid objects.
Gases get hotter when they descend throiugh a gravitational field and colder when they rise through a gravitational field.
To deal with that difference between gases and solids the Gas Laws were defined.
The difference between solids and gases arises because the molecular bonds between gas molecules are very weak and the density of gases is very low so gases respond readily to gravitational influences.
It isn’t just the up and down movement that matters for gases but also the distance between molecules which is pressure sensitive.
Unlike solids gases rise spontaneously upwards against gravity when one increases their kinetic energy and descend spontaneously with gravity when one decreases their kinetic energy.
No internal energy is lost or gained, it just transforms between KE which affects temperature and PE which does not.
I have seen that you have many things right in your previous posts but they are sadly invalidated by your lack of appreciation of the difference between the behaviour of gases and solids within a gravitational field.
Kristian said:
“Again, you need to separate between Newtonian (mechanical) processes and adiabatic (thermodynamic) processes. Only the latter concerns changes in temperature. The former only concerns changes in motion/forces.”
Total confusion.
Adiabatic processes are mechanical.
Bryan,
Maxwell is pointing out that thermal equilibrium cannot occur in the case of our atmosphere. Do you agree that in light of that an isothermal atmosphere is not possible even without GHGs ?
“the effect of winds in carrying large masses of air from one height to another tends to produce a distribution of temperature of a quite different kind,”
@ Kristian December 6, 2014 at 1:39 pm
I like what you said. There are a few little things I might quibble with (eg there are plenty of thermodynamic processes that do not involve temperature change), but I agree with the thrust of the comments.
You mentioned the concept of ‘systems’ and I think that is worth expanding on in this heated debate. Any discussion of energy thermodynamics (or even Newtonian mechanics) requires a carefully defined ‘system’. ‘Potential energy’ deals with the configuration of a system, so the whole concept of ‘potential energy’ only makes sense when you know what system you are talking about.
As a very simple Newtonian example, consider a 2 kg rock 5 m above the ground.
* If the ‘system’ is only the rock, then there is no reference for ‘configuration’ and you can’t talk about potential energy. If I let go of the rock, it gains KE because a force from ‘outside the system’ (gravity of the earth) does WORK on it.
* If the system includes the rock and the earth, then there is no work from outside the system (ie gravity does no work), and the rock gains KE because the rock/earth system looses PE.
*****************************************************************
Everyone here seems to be using the PE viewpoint rather than the gravitational work viewpoint, which is good. One less thing to worry about.
******************************************************************
But we still have to decide what ‘object’ we want to consider as the ‘system’.
* We could look at an individual molecule. If it is moving upward, it is gaining PE and (since no other work is being down on the ‘system’) the KE will decrease to conserve energy. But you can’t talk about ‘the temperature’ of an individual molecule. Any talk of temperature involves and average for many particles. We could for example, look at many particles at a specific time, or look at one particle for a long time. Of course this is simply ‘kinetic theory’ and is well studied. It is also a pain in the ass doing all the averages, so I only recommend this approach for people who want to invest a LOT of time doing lots of calculations.
* We could look at a ‘parcel’ of air containing many molecules. If it is moving upward, it is gaining PE — but the surrounding atmosphere is doing work on the parcel since the pressure on the bottom is greater than the pressure on the top. KE+PE does NOT stay constant and the drop in KE will not equal the gain in PE. Depending on the details of the situation, the parcel might cool a little or might stay the same temperature.
TB wonders: “So if KE gets converted to PE, then temperature falls .. ”
Not necessarily. The answer is wrapped up in what I just wrote. As an easy example, throw a rock upward. The gain of PE results in a loss of KE of the center-of-mass of the rock, but NOT a loss of KE of the individual molecules within the rock. The rock does NOT cool as it goes up!
Stephen Wilde says, December 6, 2014 at 4:07 pm:
“Adiabatic processes are mechanical.”
*Sigh* No, Stephen.
The adiabatic is the epitome of a thermodynamic process. I don’t know why I bother, because you’ve been shown this way too many times now without revealing even the slightest hint of wanting to come out of your bubble. It’s not like it’s a secret how the adiabatic process works, how and why expanding gases cool and how and why compressed gases warm, how it is all intimately linked to the 1st Law of Thermodynamics. It’s right there in any textbook or on any webpage discussing it. It is just so basic that I don’t really know what else I can suggest for you to do but simply … read.
I’ve asked you this question several times before, Stephen. You’ve previously refused to answer.
What does this equation tell you?
ΔU = Q – W
What is U? What is W? What is Q? What do they represent? What does this equation describe? And how does it relate to the adiabatic process (dU = δQ – PdV)?
If you can’t or won’t answer these questions, I’m sorry, but then there is no way we could ever engage in a discussion on this particular subject. Any argument you try to bring forth would simply fall to the ground. Because no one would know what your answers to these fundamental questions are. So no one would know whether you have any understanding whatsoever of basic thermodynamic principles or not.
Could you please just answer these questions, Stephen? So we can all know. And move on.
Tim F: throw a rock upward. The gain of PE results in a loss of KE of the center-of-mass of the rock
No. The loss of KE here is due to a force being applied in the opposite direction to the motion of the rock – the force of gravity. The acquisition of PE is completely passive WRT to causation.
Your conflation of an adiabatically moving air parcel and a rock with a force applied to it is just completely obfuscatory.
December 6, 2014 at 9:08 am
Bryan says:
wayne asks
“Does anyone happen to know where I can find the complete detailed derivation of the dry adiabatic lapse rate (DALR=g/cp)? ”
Page 11
Thanks Bryan for that link, turns out I already had that Gerlich & Tscheuschner 2010 paper saved on my machine!
Bryan says:
HS and Stephen Wilde have a fundamental misunderstanding of the adiabatic processes in the atmosphere.
For instance HS in answer to a reply to a blogger gives a rising balloon as an example of gravity doing work which turns into gravitational potential energy of balloon plus gas content.
What they don’t see is the falling atmospheric gas filling in the void left by the rising balloon.
PE gained by balloon is exactly equal to the PE lost by falling atmospheric gas so there is no net work done by gravity.
Yes, exactly, that has been said to Stephen at least many times and he seems to refuse to see what is actually happening, he ignores various realities to form his unphysical fantasies. Seems same for Hockey Schtick. They both fail to separate the energy that caused the decrease in density from the isentropic and reversible nature of gravity’s actions of convection/subsidence of bulk masses. On well.
A little more comments on why I am searching in that barometric area later. Hockey Schtick’s example will not work universally.
On further reading down the comments it seems I may need to add Tim Folkerts to the PE camp of Stephen Wilde and Hockey Schtick’s views on convecting ‘packets’ of air. PE is ALWAYS globally minimized, specifically by gravity no matter what is occurring! Without that fact there would never be convection upward, ever!
I light a match. The energy from that burning phosphorous instantly expands increasing globally the PE! Nature jumps in action to re-minimize the global PE, get this, by pushing that now-less-dense area of mass upward. As Bryan said, it is the gravitational falling of the surrounding mass’s PE that forces the less-dense volume upward. Amazing how incredibly mixed up so many are in relation to gravity, what it is and how it acts in various situations!
tallbloke says: December 6, 2014 at 2:35 pm
(“Kristian: “Changing mechanical KE into gravitational PE and back is real enough; it simply has no bearing whatsoever on temperatures.”
“In the classical mechanical view, KE is related to temperature and PE is not. So if KE gets converted to PE, then temperature falls, because total energy remains the same. If you disagree with this, then tell me”
Roger,
I know you wish to keep a discussion “going” on your blog, but why bash the few folk that demonstrate that they can “think” rather than only, teach! Kristain has shown that in hydrodynamic buoyancy, no weight or mass changes position in this gravitational field. Just how does PE change?
Roger it is only your claim, like that of the misguided meteorologists, that energy is ever conserved in this atmosphere. The purpose of this atmosphere is not to support but to rid the planet of waste energy, called entropy to space, never to return. The hydrodynamic properties help birdies and aircraft to function.
tallbloke says: December 6, 2014 at 2:45 pm
(“Bryan: Not one Joule of lost internal energy turns into gravitational potential energy”)
“So where does the additional PE come from when the altitude changes?”
Please demonstrate any additional PE? Even with the the large atmospheric bulge in the direction of the Sun, all is equilibrated by differential pressure potential and differential gravitational potential. Non-Newtonian sensible heat is never included. The idea that sensible heat, or chemical heat, is anything like Newtonian kinetic energy is part of the meteorological FRAUD.
Will J: There may be a misunderstanding that’s arisen here between individual ‘packets’ of air and the whole situation of the atmosphere. I’ll wait for Kristian to clarify.
tallbloke says: December 6, 2014 at 7:43 pm
(“Tim F: throw a rock upward. The gain of PE results in a loss of KE of the center-of-mass of the rock”)
“No. The loss of KE here is due to a force being applied in the opposite direction to the motion of the rock – the force of gravity. The acquisition of PE is completely passive WRT to causation.”
This is correct gravitational force decelerates the mass while it is gaining altitude (PE), till it stops. then that same gravitational force accelerates that same center of mass, gaining Newtonian momentum and energy (KE). Sensible heat is not involved.
“Your conflation of an adiabatically moving air parcel and a rock with a force applied to it is just completely obfuscatory.”
Any induced, rather than inertial, “motion” of a mass, requires force times distance (work). Work is always entropic, never adiabatic. See the post modern version of the second law of thermodynamics involving entropy..
Are we having fun yet? Reviewing everything from Aristotle Copernicus Galileo Newton? About time the Climate Clowns were put in their place, cleaning the latrine!
wayne says: December 6, 2014 at 7:46 pm
December 6, 2014 at 9:08 am
(Bryan says: “PE gained by balloon is exactly equal to the PE lost by falling atmospheric gas so there is no net work done by gravity.”
“Yes, exactly, that has been said to Stephen at least many times and he seems to refuse to see what is actually happening, he ignores various realities to form his unphysical fantasies. Seems same for Hockey Schtick. They both fail to separate the energy that caused the decrease in density from the isentropic and reversible nature of gravity’s actions of convection/subsidence of bulk masses. On well.”
The energy from the oxidation of propane, causing the balloon to move (upward force of more dense gas, in a gravitational field). Is completely lost to entropy, and then EMR to space, else that balloon, like the water in the form of WV, can never return to the surface. That energy/entropy must be radiated to space. Even if it could happen, (EMR to a higher temperature), cannot ever decrease energy/entropy. Only EMR to a lower temperature can do that. Well maybe by other means, not yet discovered. The ClimAstrologists keep digging deeper, rather than even trying to climb out of!
On Will Janoschkas December 7, 2014 at 1:52 am:
Will, I wouldn’t go so far as calling it fraud, It more like that the more meteorological minded are trained in their learning a set of fast shortcuts to come up with close to the same answer fast. I imagine what a meteorologist must go through to come up quickly a forecast every six or less hours and I really don’t know how they manage all of that data so quickly. So, have a heart.
It’s like convection. I could do a set of experiments of a one cubic meter balloon. Fill it with methane and record it’s acceleration upward. Now do the same for helium then hydrogen. The acceleration upward of each is must faster respectively in a 10:4:2 ratio and could come up with an impression that the smaller the molecular mass the faster that gas is pulling the balloon upward. Same with various temperature differences, the warmer, the faster the acceleration. A quick look-up table or algorithm gives them the correct answer fast and that is all they need but this speed of knowledge can lead to the wrong conclusion of root causes.
But from the physics side you know this simple view is not really true but to go all of the way down to first principles of calculating the weight displaced and the weight of the balloon itself to get a differential force and knowing F=ma so a=F/m the smaller the m the faster the acceleration but that is still not down to the very bottom. As Christopher provided in that former thread you really need to do an integral of the pressure at every point on both sides of the surface separating the balloon from the surrounding fluid to even get the accurate pressure difference and so on…. so much slower and a meteorologist is never going to go to such pains.
However, to me, it is better for everyone one here regardless of their backgrounds to know what is literally happening so we don’t get hung up on trivial misconceptions.
On Will Janoschkas December 7, 2014 at 2:58 am
Will, I haven’t said much about the radiative losses from every point from the surface upward but you are absolutely right on those points you keep pointing out. That original energy influx that causes the initial drop in density all gets eventually radiated to space both on the way up and what is left at the top keeps the temperature at that level constant for that level is also always radiating. Something radiating is always having its temperature lowering if not replenished unless maybe we are also speaking of state changes occurring where radiation can occur without the temperature changing. Just so you will know I do understand all of this is occurring at all times while I may be speaking of the barometric aspects.
Oops Will, sorry, that would be a 16:4:2 acceleration ratio wouldn’t it, but hope you got the underlying point I was trying to make.
Will, a hot “air” balloon is a bad example. The envelope is filled with a very light gas, water vapour from combustion as well as being warm. (water runs down the inside of the enveolope), not a lot of air in there. For obvious reasons it won’t stay up when the gas source vanishes.
A sealed gas bag device is different.
Will Janoschka says: December 7, 2014 at 2:58 am
wayne says: December 6, 2014 at 7:46 pm
“They both fail to separate the energy that caused the decrease in density from the isentropic and reversible nature of gravity’s actions of convection/subsidence of bulk masses.”
Wayne,
Perhaps it is time for a new word to remind us of a gravitational field. Not being creative, I suggest “Wensity” to remind that it is weight per unit volume, not density in a gravitational field for any sort of hydrodynamics. OTOH we have way to many useless “words”. -will- 🙂
Tallbloke says, December 6, 2014 at 7:43 pm:
“Your conflation of an adiabatically moving air parcel and a rock with a force applied to it is just completely obfuscatory.”
Nothing here is ‘adiabatically moving’, Tallbloke. The moving of the air is not the adiabatic part. That is the mechanical (Newtonian) part. It has no bearing on temperature. The adiabatic part is ONLY the expansion/compression against/by external pressure.
wayne says: December 7, 2014 at 3:15 am
On Will Janoschkas December 7, 2014 at 1:52 am:
“Will, I wouldn’t go so far as calling it fraud, It more like that the more meteorological minded are trained in their learning a set of fast shortcuts to come up with close to the same answer fast. I imagine what a meteorologist must go through to come up quickly a forecast every six or less hours and I really don’t know how they manage all of that data so quickly. So, have a heart.”
OK! The FRAUD is from the academics that try to instruct others on the reasons for the observations, with no understanding whatsoever. I admire the worker bees. They, that try to pick out the deterministic, from the statistical. Why they give chances of the opposite, is dear to me!
I would get so frustrated doing that kind of work.
“It’s like convection. I could do a set of experiments of a one cubic meter balloon. Fill it with methane and record it’s acceleration upward. Now do the same for helium then hydrogen. The acceleration upward of each is must faster respectively in a 10:4:2 ratio and could come up with an impression that the smaller the molecular mass the faster that gas is pulling the balloon upward. Same with various temperature differences, the warmer, the faster the acceleration. A quick look-up table or algorithm gives them the correct answer fast and that is all they need but this speed of knowledge can lead to the wrong conclusion of root causes.”
The speed of look up tables using past observations, is critical to the promotion of the deterministic guesses without the need of astrology!.
“But from the physics side you know this simple view is not really true but to go all of the way down to first principles of calculating the weight displaced and the weight of the balloon itself to get a differential force and knowing F=ma so a=F/m the smaller the m the faster the acceleration but that is still not down to the very bottom. As Christopher provided in that former thread you really need to do an integral of the pressure at every point on both sides of the surface separating the balloon from the surrounding fluid to even get the accurate pressure difference and so on…. so much slower and a meteorologist is never going to go to such pains.”
Wayne, If you think that is difficult, please let me introduce you to the other 100 partial differential equations that only deal with how energy is dispatched from this local region of the universe.
“However, to me, it is better for everyone one here regardless of their backgrounds to know what is literally happening so we don’t get hung up on trivial misconceptions.”
I agree Wayne,
My observations are. If you ever get to understand this Earth! This Earth must change! That is the prime design parameter! God got pissed at being “all powerful and all knowledgeable, -boring-“.
Engineers! , build something that even I do not know, else I will get new engineers!
1. Bryan says:
December 6, 2014 at 12:04 pm
Work out its internal energy from formula KE =Nk x2.5T
Kinetic theory of gases is for a gas in a container, at one temperature and pressure
The atmos has only one solid surface, it is constrained by gravity not walls.
Roger Clague says:
“Kinetic theory of gases is for a gas in a container, at one temperature and pressure
The atmos has only one solid surface, it is constrained by gravity not walls.”
Its not clear what point you are trying to make here
Is it that formula KE =Nk x2.5T is incorrect ?
If so all the thermodynamics textbooks are going to have to be rewritten.
Bryan says:
December 7, 2014 at 2:01 pm
Is it that formula KE =Nk x2.5T is incorrect ?
It is useful if applied to a gas in a container at one temp and pressure.
It does not correctly calculate anything when applied to the atmos.
The atmos. is not in a container.
Also its pressure and temp vary with height.
Roger
Its quite clear to me that you have never taken a course in thermodynamics.
I’m afraid that you will have to study the topic if you are to make any progress.
There are no short cuts.
Indeed. It is a dynamic equilibrium called steady state in order not to confuse it with thermodynamic equilibrium, which is an entirely different beast.
Thanks. However, in a hypothetical atmosphere lacking ingredients able to radiate in the thermal infrared, that does not work, obviously. Higher layers of the atmosphere can only get cooled by a colder surface below, through heat conduction across intermediate atmospheric layers. However, heat conduction requires a temperature gradient, that is, each layer should be cooler than the one above it. Such an atmospheric situation is called temperature inversion, which inhibits convection.
Remember, we are talking about a hypothetical atmosphere with no “greenhouse effect” whatsoever. That means it is not “easy” to radiate energy to space from any layer of the atmosphere, it is impossible.
1. tallbloke says:
December 7, 2014 at 2:05 am
There may be a misunderstanding that’s arisen here between individual ‘packets’ of air and the whole situation of the atmosphere.
I would also like a clarification of adiabatic cooling of the atmos. Is it a theory of temperature gradient for the whole atmos. ? Or is it a theory only for special situations.
It should be for all the whole atmos but to avoid difficult questions it supporters limit its application to upward convection in storm clouds and downward convection on mountain slopes.
How can air packets do work against air if the packets are everywhere. If the packets are working against air pressure in other packets everywhere they will run out of other packets to do their work on.
Bryan says:
December 7, 2014 at 2:50 pm
http://en.wikipedia.org/wiki/Kinetic_theory
The rapidly moving particles constantly collide with each other and with the walls of the container.
Where or what are the walls of the container in the atmosphere?
Roger
Its quite clear to me that you have never taken a course in thermodynamics.
I’m afraid that you will have to study the topic if you are to make any progress.
There are no short cuts.
Its not my Job to go over the basics with you.
It would require you to do at least 2 years of hard study before you could begin to understand the answers you ask for
Berényi Péter 3:35pm: “Remember, we are talking about a hypothetical atmosphere with no “greenhouse effect” whatsoever. That means it is not “easy” to radiate energy to space from any layer of the atmosphere, it is impossible.”
With those hypothetical physics, it is also impossible for the solar atm. to radiate energy to space. Earth’s main surface energy source would be geothermal.
A better hypothetical is to just consider thinning the optical depth of earth atmosphere (pressure unchanged) so that the solar atm. radiates same energy to space i.e. unchanged. This stepwise reduces surface kinetic Tmedian over time to earth’s brightness temperature as measured from space.
The HS eqn. in top post would not capture this change as it would be due to the distinct sources of opacity (i.e. various absorbing gases) changing within the atm. thru their mass extinction coefficients’ less mass mixing ratio. The standard atm. would be thought closer to the equator than mid-latitude.
You did hit the strawman hard, did you not?
The physics above is not hypothetical at all, only radiative properties of the atmosphere are, if we assume, with the author, it lacks absorption / emission lines in the thermal infrared (which is well approximated by a dry Argon atmosphere). It has absolutely nothing to do with solar radiation, as only a tiny fraction of solar energy is emitted in the thermal infrared and no change in composition of the solar photosphere is assumed anyway.
Berényi Péter 7:34pm” “assume…lacks…lines..in the thermal infrared..”
Here you change hypothetical; from “no GHE whatsover” & “impossible” to radiate – to now simply lacks IR mass extinction coefficient lines. This change turns the sun back on. Good move.
Hence, in the case of a hypothetical Ar atm., the single specie mass extinction coefficient is extremely small but non-zero. So Earth surface kinetic Tmedian with such an optically thin Ar atm. would approach Earth brightness Tmedian as measured from space over time. The HS eqn. in the top post cannot know this change has occurred except for the changes in Cp and R. The hypothetical US standard atm. would be estimated to occur nearer the equator than mid-latitudes.
tallbloke says: December 7, 2014 at 2:05 am
“Will J: There may be a misunderstanding that’s arisen here between individual ‘packets’ of air and the whole situation of the atmosphere. I’ll wait for Kristian to clarify.”
Roger,
It is the illusion of individual “packets”, And the conservation of energy of each, that is the basis for meteorological FRAUD! Please distinguish your ‘packet’ from my ‘packet’, and why do they not exchange energy if at different temperatures/densities?
Observations and Consequences of Solar Cycle 24:
The Perspective from Earth’s Upper Atmosphere.
Click to access PressConfMlynczakFinal.pdf
Will J: Please distinguish your ‘packet’ from my ‘packet’, and why do they not exchange energy if at different temperatures/densities?
I expect given enough time and if they are close enough together, they will.
Bryan says:
December 7, 2014 at 4:15 pm
Roger
Its quite clear to me that you have never taken a course in thermodynamics.
I have a a B.Sc ( Hons ) in chemistry Bristol University 1971. I have used IGL and K.E theory in the lab and industry scale calculations.
It would require you to do at least 2 years of hard study before you could begin to understand the answers you ask for
You mean it takes 2 years of climate science to know not to ask or expect an answer to such a simple question as:
Can the gas laws and kinetic theory of gases be applied to the Earth’s atmosphere which is contained by gravity not a container with solid walls?
Roger
It would appear that you need to look over your thermodynamics notes.
You appear to have forgotten most of them
For instance right at the beginning of our correspondence on this thread you said that the best way to get the adiabatic lapse rate was to use the formula that;
Thermal or heat energy (now known as internal energy) can be turned into potential energy
internal energy = cm(deltaT)
Potential Energy = mgh
Rearranging
(delta )T/h = g/C
So for example 1000Joules of heat energy can be turned into 1000Joules of gravitational potential energy.
Now theres something wrong with this idea but I’m going to let you figure it out for yourself.
Look up your old textbook if required then come back and tell me why you now think that this is a faulty method.
“Can the gas laws and kinetic theory of gases be applied to the Earth’s atmosphere which is contained by gravity not a container with solid walls?”
Of course the gas laws and kinetic theory still apply. Well, as long as you can define pressure and temperature for the gas in question. So we can’t apply the IGL once to the atmosphere as a whole, since different parts have different temperatures and pressures. But If I take a specific cubic meter with a well defined pressure and temperature, then PV = nRT applies even if that specific cubic meter of gas is not in a container with rigid walls. Walls are simply handy ways to conceptualize what is happening.
Suppose I have a 3m x 3m box with rigid walls. Can I apply the IGL to the center 1m x 1m? That cubic has no walls — does that mean we can’t use the IGL for that cubic meter? Surely PV=nRT will still apply where V = 1m^3 and n = the moles in that cubic meter.
The first two absorption bands of CO 2 is approx. 2μm and 2,8μm (Figure 2). The radiation emitted by the sun. The earth does not emit waves in this regard. Absorbing agents reduce the amount of solar radiation reaching the Earth’s surface in these bands, which play a role similar to ozone and ultraviolet absorber that protects the earth from the excess of the issue. The amount of energy that can absorb CO 2 in these areas can be estimated at about 4% of the total capacity to absorb carbon dioxide.
Another extent of absorption of CO 2 4 – 4,5μm. Earth emits a minimal amount of radiation. These are the waves on the boundary frequency emitted by the surface of the planet. In part this goes beyond the scope of the issue of the Earth. The graphs showing the amount of radiation emitted by the Earth, we see an almost horizontal line by selecting the energy of the emission wavelength. In this band the carbon dioxide absorbs about 8% of the total amount of infrared which is able to absorb. Due to the fact that the absorption in this band is only a fraction of CO 2 absorption capacity, as well as itself is extremely small infrared emission at this frequency can not be regarded as meaning that the absorption climate because it is the interaction is too little, and maybe even trace.
Fourth clearly marked on the said plot (Figure 2), the absorption band, the scope of the largest infrared absorption by carbon dioxide. It includes a wavelength of about 14μm to 18μm, so it is very large, and in addition to a large amount of radiation – in this field is absorbed approx. 88% of the total absorbed by CO 2 radiation. Infrared absorption peak for CO 2 falls to about 15μm, as shown in the following figure (Figure 3). It was at this fairly large absorption band of environmentalists see their greatest source of influence on the climate. It does not surprise me personally, because they have to search some data in support of their theory. It seems that this is evidence – infrared absorption band of high, but again, this is evidence of perverse. What matters is not the amount of radiation in fact, but its quality, that is the real power of influence. We are dealing here with high radiation wavelength, and the greater the wavelength, the lower the energy. By means of such radiation will not change in any significant way the temperature of the reasons why these wavelengths are not applicable in heating. There frequencies used bordering with visible light (from 0,78μm) and slightly larger wavelength, but those larger than 10μm did not play the role here (although they are present there, because of infrared filaments emit large range of wavelengths). Commissioned by the atmosphere part of such radiation is certainly not lead to a significant increase in the Earth’s surface temperature and will not increase the average temperature of the atmosphere.
https://translate.google.pl/translate?sl=pl&tl=en&js=y&prev=_t&hl=pl&ie=UTF-8&u=http%3A%2F%2Fekotest.republika.pl%2FEfekt%2520cieplarniany%2520-%2520krytyka%2520i%2520dyskusja%2520strona%25201.html&edit-text=
Let us see sudden increases in temperature in the stratosphere during the winter. They occur after a few days of low solar activity. As a result of the GCR 14C is formed which immediately combines with oxygen. 14CO2 formed absorbs part of the solar infrared radiation.
Tim Folkerts says: December 8, 2014 at 2:57 pm
(“Can the gas laws and kinetic theory of gases be applied to the Earth’s atmosphere which is contained by gravity not a container with solid walls?””
“Of course the gas laws and kinetic theory still apply. Well, as long as you can define pressure and temperature for the gas in question. So we can’t apply the IGL once to the atmosphere as a whole, since different parts have different temperatures and pressures. But If I take a specific cubic meter with a well defined pressure and temperature, then PV = nRT applies even if that specific cubic meter of gas is not in a container with rigid walls. Walls are simply handy ways to conceptualize what is happening.”
Leave it to Dr. Folkerts to bring up trivial nonsense. Walls or a “parcel” are but a way to intentionally fool others! The IGL, conservation of energy, or gravitational potential energy, have no applicability to temperature anywhere in this troposphere.
“Suppose I have a 3m x 3m box with rigid walls. Can I apply the IGL to the center 1m x 1m? That cubic has no walls — does that mean we can’t use the IGL for that cubic meter? Surely PV=nRT will still apply where V = 1m^3 and n = the moles in that cubic meter.”
More trivial nonsense.
Consider a cubic atmospheric “parcel” 7 km on each edge, lower GPS coordinates near Waco Texas, The center surface sq km has 1/4 pasture, 1/4 plowed ground, and 1/2 open stock tank. That 750 cubic kilometer space, with no adiabatic or rigid barriers. exhibits a stratified pressure, and a temperature gradient at all times. It has only a fixed volume. This 750 km^3 is but a wee “parcel”, (7 x 10^-10) of the atmosphere. Does it ever move? Can the surface temperature and altitude temperature profile ever be determined buy IGL, conservation of energy, and gravitational PE? Does this not also “require” consideration of periodic; isolation, WV production/condensation, and.EMR exitance to space. This is just for the simplified sixth grade version. For the real answers, you must have the coefficients for the other 120 partial differential equations describing this situation in fluid dynamics! Tim, please go somewhere and buy a clue!
Tim Folkerts says: December 8, 2014 at 2:57 pm
Whoops, 350 cubic kilometers, not 750 sorry!
Will says …
“The IGL, conservation of energy, or gravitational potential energy, have no applicability … “
* The IGL is a very good approximation to the behavior of the gas in the atmosphere (other than condensation/evaporation of H2O). As such it can be applied very easily and very usefully to gas in the atmosphere.
* Conservation of energy is applicable always, everywhere. It is a fundamental principle of science that has never been show to be wrong. As such, it is very useful for understanding processes in the atmosphere.
* Gravitational potential energy is also useful, but much less so because of the buoyant forces of surrounding air and the tendency of the atmosphere to automatically rearrange to minimize GPE. And I never mentioned GPE, so this is rather a strawman argument.
“Consider a cubic atmospheric “parcel” 7 km on each edge … “
Sure, we could consider such a “parcel”. It’s an interesting subset of the atmosphere to consider. But it is certainly not the only sort of parcel to consider. And you need to realize that this is NOT the sort of ‘parcel’ everyone else is talking about. If you want to make cogent contributions, you should at least know the context that others are using. Otherwise you are again attacking strawmen of your own construction. Try googling “thermodynamic air parcel” to get up to speed on the terminology. You may not *like* the standard use, but you can’t simply wish it away.
“Can the surface temperature and altitude temperature profile ever be determined buy IGL, conservation of energy, and gravitational PE?”
No one that I know of ever made such a claim (yet another strawman tonight!). Certainly temperature profiles in the atmosphere need to be consistent with basic physics like this, so these principle are relevant. Beyond that, you additionally need other principles — like absorption of incoming sunlight and evaporation/condensation and IR radiative transfer to start to get a handle on the temperatures within the atmosphere.
Tim Folkerts says: December 9, 2014 at 3:43 am
(Will says .“The IGL, conservation of energy, or gravitational potential energy, have no applicability … “)
“* The IGL is a very good approximation to the behavior of the gas in the atmosphere (other than condensation/evaporation of H2O). As such it can be applied very easily and very usefully to gas in the atmosphere.”
In this troposphere the H2O effect is everywhere present, The IGL is always wrong for any gas mixture and can never be used unless P, V, or T is an iso. Such is never true in this troposphere. What is your definition of “behavior of the gas”?
“* Conservation of energy is applicable always, everywhere. It is a fundamental principle of science that has never been show to be wrong. As such, it is very useful for understanding processes in the atmosphere.”
Energy need not be conserved in a gravitational field, but often is, such as Keppler’s laws. E. Noether (1919-IV2) algebraic demonstration. A gravitational field is not energy, it is a “potential” for possible isentropic energy transfer. In this atmosphere such fanciful claim is but a distraction from understanding.
“* Gravitational potential energy is also useful, but much less so because of the buoyant forces of surrounding air and the tendency of the atmosphere to automatically rearrange to minimize GPE. And I never mentioned GPE, so this is rather a strawman argument.”
Indeed Tim, you did not mention GPE, sorry! All meteorology depends on this, as the only explanation for this atmospheric behaviour, unless EMR effects actually dominate.
(“Consider a cubic atmospheric “parcel” 7 km on each edge … “)
“Sure, we could consider such a “parcel”. It’s an interesting subset of the atmosphere to consider. But it is certainly not the only sort of parcel to consider. And you need to realize that this is NOT the sort of ‘parcel’ everyone else is talking about. If you want to make cogent contributions, you should at least know the context that others are using. Otherwise you are again attacking strawmen of your own construction. Try googling “thermodynamic air parcel” to get up to speed on the terminology. You may not *like* the standard use, but you can’t simply wish it away.”
Most that use the term “air parcel” consider that a physical object rather than a fantasy or concept to assist in understanding. You do the same, with intent to confuse!
(“Can the surface temperature and altitude temperature profile ever be determined buy IGL, conservation of energy, and gravitational PE?”)
“No one that I know of ever made such a claim (yet another strawman tonight!). Certainly temperature profiles in the atmosphere need to be consistent with basic physics like this, so these principle are relevant.”
Everyone here that supports Michael’s, formula for lapse rate, agree that radiative effects need not be included for understanding of this atmosphere, only those three. This is scientific FRAUD! The radiative in/out flux and the clever way this planet adjusts that, is crucial to any understanding of this atmosphere.
“Beyond that, you additionally need other principles — like absorption of incoming sunlight and evaporation/condensation and IR radiative transfer to start to get a handle on the temperatures within the atmosphere.”
Indeed for the sixth grade level of understanding. Can you even identify some of the other 120 partial differential equations needed for adult understanding? Konrad can!!
Tim Folkerts says:
December 8, 2014 at 2:57 pm
“Can the gas laws and kinetic theory of gases be applied to the Earth’s atmosphere which is contained by gravity not a container with solid walls?”
Of course the gas laws and kinetic theory still apply. Well, as long as you can define pressure and temperature for the gas in question.
For IGL and K.E.T. of gases pressure is defined and calculated as elastic rebound from a solid wall.
So we can’t apply the IGL once to the atmosphere as a whole, since different parts have different temperatures and pressures.
I agree
Walls are simply handy ways to conceptualize what is happening.
Don’t agree. Rebounding from a wall is a real world mechanism. It is what is happening. It is the basis of the maths of IGL and K.E.T.
Suppose I have a 3m x 3m box with rigid walls. Can I apply the IGL to the center 1m x 1m? That cubic has no walls — does that mean we can’t use the IGL for that cubic meter? Surely PV=nRT will still apply where V = 1m^3 and n = the moles in that cubic meter.
But the atmos is big, not homogeneous, ( temp, pressure, density, gravity ) and has WV which changes phase. We can only compare it to atmos. of other planets and moons. Not a 3m cube at STP.
Bryan says:
December 8, 2014 at 1:11 pm
Thermal or heat energy (now known as internal energy) can be turned into potential energyinternal energy = cm(deltaT)
Potential Energy = mgh Rearranging (delta )T/h = g/C
Yes that is how to calculate T/h for the atmos.
I call T/h a temperature gradient, not the adiabatic lapse rate.
Adiabatic lapse rate assumes T/h
Can be found using IGL (internal energy lost to work against pressure) and is negative
I don’t accept those assumptions.
So for example 1000Joules of heat energy can be turned into 1000Joules of gravitational potential energy.
Now theres something wrong with this idea but I’m going to let you figure it out for yourself.
Look up your old textbook if required then come back and tell me why you now think that this is a faulty method
As far as I know the 1st law of thermodynamics, conservation of energy, means each form of energy can be changed into any other form
So heat energy can be turned into gravitational potential energy. A hot-air balloon does it.
Heat is motion. A pendulum converts gravitational energy to motion and back
.
The sun radiation energy is converted to gravitational energy in our atmos. A must admit I don’t know how, although I do have some clues about how radiation and gravity interact .
Roger
Your textbook did not get as far as the Second Law of Thermodynamics it appears.
Go look up what the Second Law says about conversion of internal energy (cm(delta)T part) into mechanical energy (mgh part).
Or put in simpler language can 1000J of heat energy be transformed into 1000J of gravitational potential energy with 100% efficiency.
Tim Folkerts says:
December 9, 2014 at 3:43 am
/////////////////////////////////////////////
For once (maybe the first time) I have to agree with Tim-
”Certainly temperature profiles in the atmosphere need to be consistent with basic physics like this, so these principle are relevant. Beyond that, you additionally need other principles — like absorption of incoming sunlight and evaporation/condensation and IR radiative transfer to start to get a handle on the temperatures within the atmosphere.”
HS got it wrong. Direct linear equations will not solve the problem. They can, as in the determination of the US standard atmosphere, produce a reasonable model. But the equations for US-SA as with HS’s attempt disregard physical mechanisms. Atmospheric temperature profile is highly dependant of the the speed of vertical circulation across the pressure gradient of the atmosphere. Half of this is dependent on radiative subsidence. No radiative gases, no joy.
If you want to skip the complexity of CFD for our radiative atmosphere then simply calculate the surface temp without atmospheric cooling. 312K. Current surface temp? 288K. What’s the net effect of our radiatively cooled atmosphere on surface temps? Not too hard is it?
Will asks: “What is your definition of “behavior of the gas”?”
In this case, I am talking about local behavior. If I take a sample of air anywhere in the atmosphere and measure P, V, n, and T for that sample, I expect PV=nRT to work pretty darn well. It won’t work exactly because no gas is idea (eg molecules have finite volume; van der Waals forces exist), but it is still highly useful.
“Energy need not be conserved in a gravitational field, but often is, such as Keppler’s laws. E. Noether (1919-IV2) algebraic demonstration.”
I would love to hear something you consider to be an example of conservation of energy failing — whether gravity is involved or not. If you want to bring up Noether’s Theorem, then you will have to show how the fundamental laws of physics are changing over time to show that energy is not conserved. Again, I would love to hear something you consider to be an example.
“Most that use the term “air parcel” consider that a physical object rather than a fantasy or concept to assist in understanding. ”
Now you seem to be using a false dichotomy. “Most” consider an ‘air parcel’ to be a something in between these two extremes. One visualization is to take a sample of air and put it inside a thin, airtight plastic bag and then follow that bag of air and see what happens to it. Certainly this limits mixing more than would happen in the real atmosphere. On the other hand, you can watch the smoke coming out of a chimney rather long distances, suggesting that even ‘unbagged’ air often stays in relatively well-defined ‘packets’ over long distances.
“Everyone here that supports Michael’s, formula for lapse rate, agree that radiative effects need not be included for understanding of this atmosphere, only those three.”
If you mean “Michael’s formula for lapse rate AND one fixed point on that slope” then I agree. They are picking one plausible point (the temperature at the median altitude) and thinking they have solved the problem. The fact that using the median altitude of Venus’s atmosphere is an utter failure doesn’t seem to phase them!
“All “work done” by definition is “entropic”, and can never be adiabatic!”
Speaking of ‘adult understanding’ …
Read up on “adiabatic compression/expansion” — it is a basic part of pretty my every heat engine cycle that involves adiabatic work done! In classical thermodynamics, work does NOT change entropy. Work can easily be isentropic; work can easily be adiabatic! Your statement is pretty much 100% backwards!
“If you want to skip the complexity of CFD for our radiative atmosphere then simply calculate the surface temp without atmospheric cooling. 312K. “
Just out of curiosity, how did you “simply calculate” 312 K?
[…] ‘Gravito-Thermal Effect’ (GTE) at a few notable climate blogs, like The Hockey Schtick, Tallbloke’s Talkshop, Clive Best and even Judith Curry’s Climate Etc. (in fact, this is where the lengthiest […]
“Mass extinction coefficient” is simply the absorption coefficient divided by density. Therefore in a medium with positive density (like air) a zero mass extinction coefficient implies no absorption at all. GHGs are gases with some absorption in the thermal infrared, by definition. The proposition “an atmosphere does not absorb radiation in said EM band” is equivalent to “there is no GHE in it”. And, of course, if it does not absorb thermal IR radiation, it can’t emit it either (by Kirchhoff).
The US standard atmosphere is an extremely weak example in this respect, because it is a wet atmosphere with lots of water vapor in it, a potent greenhouse gas, with abundant absorption / emission in the thermal infrared.
There are no zero mass extinction coefficients, all even Ar are positive.
I was really hoping this thread or the last one would bring out someone who could explain why we can so easily and closely calculate the venus temp-pressure profile from that of earth despite numerous vast differences. Did I miss it, or is it still a coincidence?
Tim Folkerts says: December 9, 2014 at 2:22 pm
(Will asks: )
“In this case, I am talking about local behavior. If I take a sample of air anywhere in the atmosphere and measure P, V, n, and T for that sample, I expect PV=nRT to work pretty darn well. It won’t work exactly because no gas is idea (eg molecules have finite volume; van der Waals forces exist), but it is still highly useful.”
I ask for a definition of the word “behavior” Tim answers with the words local “behavior. OK “What is your definition of “local behavior” of the gas”?” You seem to be assigning A constant ratio of two gas variables, from a complex algebraic equation (IGL) with all variables,except those two held constant,,
as the behavior of an atmospheric gas mixture”, with both “n” and “R” unknown. If increasing T results in lifting “behavior”, do to hydrodynamic buoyancy, your IGL is completely useless and only serves your intent to confuse!
(“Energy need not be conserved in a gravitational field, but often is, such as Keppler’s laws. E. Noether (1919-IV2) algebraic demonstration.”)
“I would love to hear something you consider to be an example of conservation of energy failing — whether gravity is involved or not.”
The easiest is Einstein’s E = mc^2. A mass accelerated by force to velocity “c” has Newtronian kenetic energy (mc^2)/2. Where comes the other half? Did the “force” loose the other half keeping energy conserved? One other is EMR to space, This can be accounted for to determine efficiency, but that energy of a different gauge group than sensible heat no longer exists.
“If you want to bring up Noether’s Theorem, then you will have to show how the fundamental laws of physics are changing over time to show that energy is not conserved. Again, I would love to hear something you consider to be an example.”
I do not have to show! It is up to you to understand when the first law of thermodynamics has been falsified, and must be redefined, as all except arrogant academics have accepted.
(“Most that use the term “air parcel” consider that a physical object rather than a fantasy or concept to assist in understanding. ”)
“Now you seem to be using a false dichotomy. “Most” consider an ‘air parcel’ to be a something in between these two extremes.”
Please read the comments in this thread, especially those of the fake meteorologists!
“One visualization is to take a sample of air and put it inside a thin, airtight plastic bag and then follow that bag of air and see what happens to it. Certainly this limits mixing more than would happen in the real atmosphere. On the other hand, you can watch the smoke coming out of a chimney rather long distances, suggesting that even ‘unbagged’ air often stays in relatively well-defined ‘packets’ over long distances.”
That may be your visualization but not that of the meteorologists. They insist that all gas species in the atmosphere are well mixed, when that suites their fantasay, so no smoke. At the same time to fit their other fantasy that a parcel is enclosed within an adiabatic barrier allowing no transfer of energy to/from that parcel. Both concepts are insane.
(“Everyone here that supports Michael’s, formula for lapse rate, agree that radiative effects need not be included for understanding of this atmosphere, only those three.”)
“If you mean “Michael’s formula for lapse rate AND one fixed point on that slope” then I agree. They are picking one plausible point (the temperature at the median altitude) and thinking they have solved the problem. The fact that using the median altitude of Venus’s atmosphere is an utter failure doesn’t seem to phase them!”
Michael’s formula has two parts the first is a ClimAstrologist claim of a specific temperature on or about this planet due to some nonsense radiative formula without understanding. The second is an agebraic expression for planetary lapse rate with no radiative part at all. Both parts are insane
(“All “work done” by definition is “entropic”, and can never be adiabatic!”)
“Speaking of ‘adult understanding’ …Read up on “adiabatic compression/expansion” — it is a basic part of pretty my every heat engine cycle that involves adiabatic work done! In classical thermodynamics, work does NOT change entropy. Work can easily be isentropic; work can easily be adiabatic! Your statement is pretty much 100,% backwards!”
In the adiabatic compression/expansion cycle, “no” work is done. It is a reversable cyclic process that “only” changes the form of energy while doing no work. Please show even one physical example of “work” that is reversible or isentropic? Please a definition of work that distinguishes “work” from “energy” This is your fantasy. You screwed up entropy, instead of correcting, you arrogant academics, choose to dig deeper, and try to create internal energy! Still insane. Poor students!
wayne says: December 6, 2014 at 6:28 am
“To my surprise I thought this was within Wikipedia under “Lapse Rate” but you get to the very last line and that doesn’t give the reason why g/cp appears as the answer but instead points to a problem in a book that doesn’t show the answer or the derivation at all!”
If I understand your question correctly:
on Wikipedia after combining the two formulas the Specific Volume and Density cancel each other out to one, leaving g/Cp as the formula for the DALR.
Pressure can be eliminated because the rising parcel will always expand to have the same internal pressure as the surrounding static atmosphere (which is in hydrostatic equilibrium).
If the parcel rises or sinks very slowly the adiabatic assumption will no longer hold, and energy exchange with the static atmosphere will modulate the DALR.
I assume this is the case in eg. high pressure areas, where the subsiding air sinks with something like 50 meter/hour.
Will, says “The easiest [example of non-conservation of energy] is Einstein’s E = mc^2.
First of all, why bring up relativity? Are there any molecules moveclose to the speed of light in the atmosphere? (And you think I go off on irrelevant academic tangents!)
Furthermore, in relativity there is a mass-energy equivalence, so in relativity you have to talk about conservation of mass-energy, which is indeed still conserved.
“A mass accelerated by force to velocity “c” has Newtronian kenetic energy (mc^2)/2. Where comes the other half?”
1) You can’t accelerate a mass to velocity “c” with any force. It is impossible!
2) You seem to think you moving article has KE = 1/2 mc^2 and some other energy 1/2 mc^2 to get a total of mc^2. Instead, the particle at rest already has rest energy = mc^2. A moving particle has ADDITIONAL energy. At low speeds this additional energyis approximately the classical 1/2 mv^2, but at relativistic speeds, the additional energy becomes much greater than 1/2 mv^2 (and can even become much greater than the rest mass of mc^2).
“In the adiabatic compression/expansion cycle, “no” work is done.”
I can understand having trouble with relativity — it is pretty esoteric and doesn’t come into play in real-world enginering (other than GPS).
But this statement is mind-boggling! During the expansion cycle, the gas applies a force to the piston and the piston moves. Since force*distance is work, the gas is doing work! The expansion cycle is always where the work is done! That is stroke 3 in a 4 stroke engine — the power stroke!
To follow-up on Ben’s reply to Wayne about lapse rate, the wikipedia page DOES tell you where they got the answer — ie “Combining these two equations”.
They had Cp dT = α dP = (V/m) dP = (1/ρ) dP
and dP = ρg dz
Plug dP from the second equation into the first an rearrange to get dT/dz
tjfolkerts says: December 10, 2014 at 4:12 pm
(Will, says “The easiest [example of non-conservation of energy] is Einstein’s E = mc^2″).
“First of all, why bring up relativity? Are there any molecules move inetic theory of everything.close to the speed of light in the atmosphere? (And you think I go off on irrelevant academic tangents!)”
Professor Folkerts’
I do not accuse you of irrelevant academic tangents! I do accuse you of irrelevant fifth grade nonsense! As an educator with a PHd, How about do you claim that EMR flux from the Sun is non-relativistic? How about EMR exit flux to space? Still non-relativistic? Please drop your fantasy of kinetic theory of everything, especially the falsified “statistical mechanics”?
“Furthermore, in relativity there is a mass-energy equivalence, so in relativity you have to talk about conservation of mass-energy, which is indeed still conserved.”
Your post modern non-science claims such perversion of Einstein’s equation! Please show the Lorentz transform that illustrates symmetry, and the conservation of your fake mass-energy.
(“A mass accelerated by force to velocity “c” has Newtonian kinetic energy (mc^2)/2. Where comes the other half?”)
“1) You can’t accelerate a mass to velocity “c” with any force. It is impossible!”
Indeed I cannot, nor do I wish to! Although most adults consider “c” an asymptote. Please indicate the Laplace transforms that show “c” cannot be approached to “any” degree by “force”. tjf fantasy.
“2) You seem to think you moving article has KE = 1/2 mc^2 and some other energy 1/2 mc^2 to get a total of mc^2. Instead, the particle at rest already has rest energy = mc^2.”
Wow! another post modern fantasy. Please show your rest energy doing (mc^2 – delta entropy) amount of “work”!
“A moving particle has ADDITIONAL energy. At low speeds this additional energy is approximately the classical 1/2 mv^2, but at relativistic speeds, the additional energy becomes much greater than 1/2 mv^2 (and can even become much greater than the rest mass of mc^2).”
If you would explain this nonsense in a way acceptable to anyone with an engineering degree visiting “Talkshop”, I hereby promise to buy a set of your knives that never get dull!
(“In the adiabatic compression/expansion cycle, “no” work is done. It is a reversible cyclic process that “only” changes the form of energy while doing no work. Please show even one physical example of “work” that is reversible or isentropic? Please [show]a definition of work that distinguishes “work” from “energy” This is your fantasy. You screwed up entropy, instead of correcting, you arrogant academics, choose to dig deeper, and try to create internal energy! Still insane. Poor students!In the adiabatic compression/expansion cycle, “no” work is done.”)
“I can understand having trouble with relativity — it is pretty esoteric and doesn’t come into play in real-world engineering (other than GPS).”
As a sparky, I find relativity quite proper, and comfortable! It is your post modern, mechanical nonsense, that is quite esoteric, fanciful, and backward.
“But this statement is mind-boggling! During the expansion cycle, the gas applies a force to the piston and the piston moves. Since force*distance is work, the gas is doing work! The expansion cycle is always where the work is done! That is stroke 3 in a 4 stroke engine — the power stroke!
That expansion is provided by the inertia of the flywheel, no work is doneI
I clearly qualified, “adiabatic compression/expansion cycle”. In any whole cycle, show any work being done by the adiabat? Please show a definition of work that clearly distinguishes “work” from “energy”? You cannot do such, within your fantasy, hence you have no work, only “energy”.
How does your fantasy claim that “energy” is in any way useful!. Each algebraic symbols in your delta(U) = (E-W) have no meaning in this physical. Please explain the meaning of W = (E – delta(U))? Then please explain your claim of isentropic “work”?
Tim Folkerts says: December 10, 2014 at 11:26 pm
“To follow-up on Ben’s reply to Wayne about lapse rate, the wikipedia page DOES tell you where they got the answer — ie “Combining these two equations”. They had Cp dT = α dP = (V/m) dP = (1/ρ) dP and dP = ρg dz. Plug dP from the second equation into the first an rearrange to get dT/dz”
Again, what intentional grade school misdirection by Professor Folkerts!
According to the despised “wikipedia”, the fake 1LTD, also holds this atmosphere? (using y as gamma). That expression mCvdT-VdP/y = 0, has been shown over and over in this thread to be false. Nowhere in this atmosphere is energy ever conserved! Thermal energy in this atmosphere is highly cyclic, with various delays, and is mostly deterministic. There is no need for conservation in any “part” of a cycle. Statistical mechanics is only the noise riding upon the deterministic.
Thanks, Will. We have had about four people, it seems to me, who throw around inappropriate formulae and other claims. They waste a lot of time, when others have been trying to gain understanding over several years. I’ve been pondering this problem with my Uni Physics textbook beside me (Young, 8th edition, old but far younger than me, enough of the puns). However, you are doing a far better job than I could, so thanks again. Brett
Brett Keane says: December 11, 2014 at 9:55 am
“Thanks, Will. We have had about four people, it seems to me, who throw around inappropriate formulae and other claims”
As long as the clueless masses cannot or will not understand something simple as convection this kind of discussions will continue.
Even Wikipedia has this correct:
“The dry adiabatic lapse rate (DALR) is the rate of temperature decrease with altitude for a parcel of dry or unsaturated air rising under adiabatic conditions.”
Notice the word RISING.
The idea that the DALR has anything to say about the temperature profile of the entire troposphere / atmosphere is beyond stupid.
An Englishman writes: conversations in English amongst the English are riddled with misunderstandings arising from the ambiguity of language generally.
Brett and Ben are highlighting a subtle problem of meaning and I think about are writing about extending a meaning of a descriptive phrase beyond its applicability.
At some risk of clouding matters further I suggest that any effects peripheral to the process described are not within the process even though in reality they are present.
I often find this kind of thing at the root of technical confusion and is very hard to see.
None of this is nor should be taken as a criticism of anyone, just that right end of stick is devilish hard to grasp.
@Ben Wouters says:
December 11, 2014 at 11:58 am: Once again, I am accused of something I did not say. Strawman stuff, and name-calling too. But this is my last comment on point-scorers. We have better things to do. Brett
There is some ambiguity here.
As Tim pointed out and thanks Tim for the attempt, but that Cp·dT = (1/ρ)dP substituting dP = ρ·g·dz from Wikipedia to be Cp = g·dh/dT which are both linear if geopotential height ‘h’ in stead of ‘z’ is used and solves to cp = 9.80665 [J/m/K] * 11000 [m] / (288.15 -216.65)[K] that give an operational cp of precisely 1508.715 J/kg/K that exactly matching to seven digits what the some 27 international agencies used as the very exponent used in the hydrostatic polytropic formation of the 1976 Standard Atmosphere troposphere linear profile in the first place! The computed lapse above is 6.5 K/km, not 9.75 K/km. That is why I explicitly asked if anyone had a better and more detailed derivation from bottom up — from first principles that is.
I see most commenters here just trivially toss in 1.0035 kJ/kg/K and then claiming some 9.75 K/km for the lapse of our atmosphere but even those equations praised by Tim Folkerts tend to show this is not so simply true, if true at all. Even the units without redefining ‘g’ as I did above are not correct in this too simple lapse=g/cp equation.
So I ask again, where the h–l did that ‘DALR=g/cp’ come from originally? What is its history?
wayne,
Not sure where you got the impression that the DALR in ‘reality’ is 6.5K/km. The DALR is 9.75K/km. The SALR is (on average) ~5K/km. The mean global ALR, for all the tropospheric water content evened out, ends up on the continuum between the two: 6.5K/km. Which becomes the template for the ELR. Globally and annually it must stabilise at this value.
wayne 8:28pm – Although HS seems to think the top post eqn. is the “one and only one”, its components started from lab test mid-1800s – the differences between dry and moist lapse rates were 1st demonstrated by lab experiments of James Espy using his “nephelescope” (an expansion cloud chamber) published in 1841 “Philosophy of Storms” p. viii of the introduction here:
http://books.google.com/books/about/The_philosophy_of_storms.html?id=reG7AAAAIAAJ
“..(air) will grow colder by about 1 degree and a quarter for every hundred yards of its ascent…”
Courtesy a history by James McDonald in 2 historical papers, the dry & moist lapse theory in support was developed from the work of William Thompson (Lord Kelvin) 1862-5 based on Joule conceptualizing, exact theory 1864 from Reye was more clearly written, same as Peslin 1868 paper, Hann 1874, Hertz as a hobby 1884 inspired by his mentor Helmholtz, then von Bezold 1888 (carefully derived the “pseudoadiabatic” ascent followed by dry adiabatic descent & added description of atm. energetics associated with precipitation), Poisson extended it to ideal case in the 1890s. Success as usual has lotsa’ fathers. Some were better writers and scientists than others & are better remembered as is demonstrated on blogs even today.
Wayne, I think you are looking at the equations incorrectly.
The DALR was indeed derived in the wikipedia article. The derivation makes very specific theoretical assumption that were NOT chosen to match the actual properties of the atmosphere. Specifically, the DALR applies to an atmosphere that is in hydrostatic equilibrium, and that is adiabatic (no heat flow from one part of the atmosphere to another) with no condensing gases. For such an idealized atmosphere, the theoretical lapse rate would be g/cp. For dry air, this is approximately g/cp = (9.8 m/s^2) / (1.00 kJ/kg/K) = 9.8 K/km.
[NOTE: the units do indeed work out correctly. And note that g & cp will both be varying as height and temperature change.]
Let me reiterate — this is an idealization (like a perfect Carnot Cycle or a frictionless surface) and not any sort of empirical estimate of the actual atmosphere. The real atmosphere is not adiabatic. Nor is it dry. Nor is is in hydrostatic equilibrium. This equation will not predict the actual lapse rate in the atmosphere. The best it can do is provide a limiting case — a limiting case that tells us that convection WILL be occurring if the ELR (environmental lapse rate) is greater than this ~ 9.8 K/km. If someone *did* use this equation and thought it would give the actual lapse rate, they would be badly mistaken.
(One can also calculate the Saturated Adiabatic lapse rate, which turns out to be ~ 5 K/km. This gives a different limiting case).
Other the other hand, the US Standard Atmosphere is entirely empirical — a linear lapse rate chosen to approximately match the ELR. The 6.5 K/km is NOT ‘computed’, it is ‘assumed’ because it is a handy number that happens to be close to observed average lapse rates. So your “operational cp” is found by working backwards from this assumed lapse rate.
Wow — three post answering the same question (in slightly different ways) at the same time. 🙂
tchannon says: December 11, 2014 at 1:07 pm
“Brett and Ben are highlighting a subtle problem of meaning and I think about are writing about extending a meaning of a descriptive phrase beyond its applicability.”
TimC,
I agree with your identification of the mess this thread has descended into. However Brett made only one comment about my comments. Ben Wolters attacked Brett ,for his POV, with the phrase “clueless masses” Bryan has been in his attempt to show in science that details matter and that dry adiabatic lapse rate is never -g/Cp. Wayne wishes to know where -g/Cp came from, while expressing the answer “which are both linear if geopotential height ‘h’ in stead of ‘z’ is used”.
Geopotential height is a meteorological construct falsely invented to indicate sensible heat .converts to “potential energy” with altitude!
The mass, but not the sensible heat in the atmosphere may do that, but only if the mass of the air in the troposphere were free to accelerate under the force of gravity. Fluid dynamics and buoyancy within this atmosphere prevent that. The Meteorological Religion is falsified.
“I often find this kind of thing at the root of technical confusion and is very hard to see.”
I agree! What was the topic? 🙂
What I take from this thread is that the basic troposphere temperature structure is explained by the barometric equations.
Radiation plays a negligible part and is ignored in the derivation of these equations.
Where radiation is significant is in cooling to space.
Very few now try to describe the Earth surface temperature as being heated radiatively by slabs in the atmosphere.
Five years ago that explanation was much more common.
Surely that is progress!
Trick says: December 11, 2014 at 10:41 pm
Thanks very much for this history. Fascinating to see that some 150 years ago already a good understanding was developed about the dynamics of rising air and cloud forming.
The linked book can be downloaded as a PDF.
From the Introduction:
“SYNOPSIS. When the air near the surface of the earth becomes more heated or more highly charged with aqueous vapor, which is only five- eighths of the specific gravity of atmospheric air, its equilibrium is unstable, and up-moving columns or streams will be formed. As these columns rise, their upper parts will come under less pres sure, and the air will therefore expand ; as it expands, it will grow colder about one degree and a quarter for every hundred yards of its ascent, as is demonstrated by experiments on the nephelescope, (58 to 68.) The ascending columns will carry up with them the aqueous vapor which they contain, and, if they rise high enough, the cold produced by expansion from diminished pressure will con dense some of this vapor into cloud ; for it is known that cloud is formed in the receiver of an air pump when the air is suddenly withdrawn. The distance or height to which the air will have to ascend before it will become cold enough to begin to form cloud, is a variable quantity, depending on the number of degrees which the dew point is below the temperature of the air ; and this height may be known at any time by observing how many degrees a thin metallic tumbler of water must be cooled down below the tempera ture of the air before the vapor begins to condense on the outside. The highest temperature at which it will condense, which is varia ble accordingly as there is more or less vapor in the air, is called the ” dew point,” and the difference between the dew point and the temperature of the air in degrees, is called the complement of the dew point.’ (117, 118, 129.) It is manifest, that if the air at the surface of the earth should at any time be cooled down a little below the dew point, it would form a fog, by condensing a small portion of its transparent vapor into little fine particles of water ; and if it should be cooled twenty degrees below the dew point, it would condense about one half its vapor into water, and at forty degrees below, it would condense about three fourths of its vapor into water, &c. This, however, will not be exactly the case from the cold produced by expansion in the upmoving columns ; for the vapor itself grows thinner, and the dew point falls about one quarter of a degree for every hun dred yards of ascent.”
” von Bezold 1888 (carefully derived the “pseudoadiabatic” ascent followed by dry adiabatic descent & added description of atm. energetics associated with precipitation)”
Probably a description of the Fohneffect Tim linked to earlier, which N&Z mistakenly used as a poster child for their atmospheric effect.
Tim Folkerts says: December 11, 2014 at 11:17 pm
“The DALR was indeed derived in the wikipedia article. The derivation makes very specific theoretical assumption that were NOT chosen to match the actual properties of the atmosphere. Specifically, the DALR applies to an atmosphere that is in hydrostatic equilibrium, and that is adiabatic”
The assumption is NOT that the atmosphere is adiabatic.
The assumption is that the rising parcel does not exchange energy with the surrounding air, making it possible to calculate the temperature drop due to the expansion AS IT RISES through the static atmosphere, that is in hydrostatic equilibrium.
From the American Meteorological Society:
http://glossary.ametsoc.org/wiki/Lapse_rate
“The decrease of an atmospheric variable with height, the variable being temperature, unless otherwise specified.
The term applies ambiguously to the environmental lapse rate and the process lapse rate, and the meaning must often by ascertained from the context.”
http://glossary.ametsoc.org/wiki/Dry-adiabatic_lapse_rate
“.A process lapse rate of temperature, the rate of decrease of temperature with height of a parcel of dry air lifted by a reversible adiabatic process through an atmosphere in hydrostatic equilibrium.”
It seems almost everyone has fallen into the trap set by calling the temperature profile of the static atmosphere and the temperature drop of a rising parcel both “lapse rate”
They are totally different concepts.
The clearest explanation and derivation of the barometric formulas (that I have found) is given here.
Click to access 1003.1508.pdf
All the caveats are carefully set out.
Any radiative contribution must be included in the bulk thermodynamic quantities like Cp they are certainly not extracted and overlaid to get a final result
wayne says: December 11, 2014 at 8:28 pm
“cp = 9.80665 [J/m/K] * 11000 [m] / (288.15 -216.65)[K] that give an operational cp of precisely 1508.715 J/kg/K that exactly matching to seven digits what the some 27 international agencies used as the very exponent used in the hydrostatic polytropic formation of the 1976 Standard Atmosphere troposphere linear profile in the first place!”
The ISA lapse rate of 6,5K/km has no process on which it is based. It is the average of a great number of soundings, relevant for the mid latitudes. It is used ao to calibrate altimeters, so that aircraft flying at different altitudes do not crash into each due to different altimeter calibrations.
The operational Cp you show:
in the thought experiment I discussed earlier, with the atmosphere on the surface at 0K, you not only need to supply the energy to warm the column to the numbers you gave, but also the energy for the air to expand against gravity (increase potential energy of all molecules)
Suggest everyone interested to follow this “classroom”, especially the first 5 pages:
http://www.tornadochaser.net/capeclass.html
Notice on page 3 how the temperature of the rising parcel follows the SALR (moist adiabat)
The difference with the temperature of the static atmosphere for every level gives the buoyancy force.
Bryan says: December 12, 2014 at 10:06 am
“The clearest explanation and derivation of the barometric formulas (that I have found) is given here.
http://arxiv.org/pdf/1003.1508.pdf All the caveats are carefully set out.
Any radiative contribution must be included in the bulk thermodynamic quantities like Cp they are certainly not extracted and overlaid to get a final result”
That G-T paper clearly sets out the formula without any rising or falling parcels! This is true science much unlike the religious dogma with no science from the American Meteorological Society. Please notice there is no fake meteorological Cp for air with saturated WV.
Bryan says: December 12, 2014 at 8:56 am
“What I take from this thread is that the basic troposphere temperature structure is explained by the barometric equations. Radiation plays a negligible part and is ignored in the derivation of these equations.”
Indeed the equations need no radiation in or out to indicate the basic structure from gravitational effects only. No Hydrostatic equilibrium is necessary either.
“Where radiation is significant is in cooling to space.”
Radiation in and out, coupled with extremely energetic conversion of WV latent heat to sensible heat produces all of the fascinating and perhaps dangerous weather we have on this planet.
“Very few now try to describe the Earth surface temperature as being heated radiatively by slabs in the atmosphere.”
All atmospheric heat in or transferred from the surface, both sensible and latent is completely transferred to space via EMR flux in accordance with all Thermodynamic Laws, Maxwell’s equations, and your barometric formula. None is returned to the higher temperature surface by any means whatsoever. What the Sun provides the atmosphere dispatches to space, after the WV creates the fascinating weather.
BTW since WV has twice the Cp of air,, the fake meteorological Cp for air with saturated WV. is also doubled giving a fake lapse rate for saturated LR of 1/2 the dry. This is where the meteorologists do the snake oil without having to explain the actual process. The Cp for heating saturated is the same as dry. The effective Cp for lowering that air until all WV condenses is minus 24 times the Cp for dry. 1% WV by mass with 2400 J/gram latent heat would raise the temperature of that 100 grams of air by 24 degrees Celsius, before rain. Fortunately that same WV radiates all that energy to space with little or no increase in temperature!
Tim Folkerts says:
December 10, 2014 at 11:26 pm
Cp dT = α dP = (V/m) dP = (1/ρ) dP
and dP = ρg dz
Plug dP from the second equation into the first an rearrange to get dT/dz
Pressure ( P ) is eliminated. It is assumed that the concept of pressure is the same in the 2 equations. It is not.
In the first P is caused by T, kinetic energy, motion. not gravity. As in IGL
In the second P is caused by gravity, not motion. As in the hydrostatic equation.
This problem is overcome by using the 1st law of thermodynamics, conservation of energy directly to the atmos. Without reference to pressure.
Change in heat energy = change in gravity energy
Mgh = mcT
T/h = g/c
heat energy = kinetic energy.
gravity energy = potential energy.
Change in heat energy = changes in gravity energy but of opposite sign.
Most of you are coming closer and closer to what I’ve been telling you.
Will says: “That G-T paper clearly sets out the formula without any rising or falling parcels!”
No.
Equation 14 in the G-T paper is dp/dz = − ρg or dp = − ρg dz. In other words, this equation relates a parcel rising a distance dz with a change in pressure of -ρg dz. Rising parcels are built into the equations from the very start. Integration yields the effects of moving the parcel a finite distance, z.
PS the wikipedia derivation utilizes the exact same dp = − ρg dz to introduce the effect of moving a parcel. All the rest in both sources simply relies on the basic properties of ideal gases (G-T just go into way more detail). So either BOTH derivations are flawed, or BOTH are acceptable.
“BTW since WV has twice the Cp of air,, the fake meteorological Cp for air with saturated WV. is also doubled giving a fake lapse rate for saturated LR of 1/2 the dry.”
No.
That is not how the saturated lapse rate is determined. The specific heat of dry air is still used, along with other properties of water to get the lapse rate. No one (that I have seen) uses the approach you claim.
http://en.wikipedia.org/wiki/Lapse_rate#Saturated_adiabatic_lapse_rate
“It is assumed that the concept of pressure is the same in the 2 equations.”
The concept of pressure *is* the same in both! Pressure is pressure; its the force exerted on an surface divided by the area of the surfce. A pressure gauge will read the same no matter what causes the pressure. The molecules will be moving still be moving with the same average speeds.
I am speaking of the constant long-term temperature gradient as seen in this set of skew-t plots. Those red lines are the temperature per altitude and half are randomly from June 02 to June 11, 2014 and half are taken randomly from December 02 to December 11, 2014. See, it doesn’t matter in which time of day or the season you choose, the slope is always right at 6.5 K/km right up to the tropopause no matter whether day or night, whether summer or winter, the slope (or lapse) is always a constant, on the average over long periods of time. So please stop dragging my examples into immediate weather effects!
The slope of the bottom yellow line (just eyeballed) is right at -6.5 K/km. The temperature goes from about 27.5°C to about -52°C in 40000 feet.
My primary question is why Earth’s average is 6.5 K/km and why, as I showed on that previous thread: “a lone skeptic voice”, also Venus is 7.7 K/km and Jupiter is 2.2 K/km and Titan is 0.9 K/km as these figures are all given by one single equation and they all match the probes data from NASA? The equation is polytropic, not adiabatic, not isothermal, spontaneous that is, and takes the form T=T0·(P/P0)^(1/DoF) and P=P0·(T/T0)^(DoF). The DoFs to match the know profiles end up right at these values, Earth: 5.2559, Venus:6.0044, Jupiter: 3.0034, Titan: 5.001. Why?
If this does not perk the curiosity of supposedly scientific commenters here, that is a bigger question why? Bryan seems to see its relevance.
The only conclusion I can see in this investigation is that the primary temperature gradient slopes are governed by one single thing, the expressed degrees of freedoms of the constitute molecules present in each atmosphere which though this does not set the temperature of a surface it does mean imho if the surface were to warm, the temperature above the troposphere must also warm. If the surface were to cool, the lower stratosphere must also cool. Likewise if the stratosphere cools the surface *will* cool unlike some climatologists hell bent on insisting on some radiation-only effects governing our entire atmosphere.
Stephen Wilde says:December 12, 2014 at 2:41 pm
“heat energy = kinetic energy” No!. Heat is not Newtonian kinetic!
“gravity energy = potential energy” No!. Gravitational potential = potential, never energy!
A difference in potential allows, but does not demand, a transfer of energy, 2LTD!
“Change in heat energy = changes in gravity energy but of opposite sign”.
Total Horse Shit! see below!
Roger Clague says: December 12, 2014 at 1:45 pm
Tim Folkerts says: December 10, 2014 at 11:26 pm
(“Cp dT = α dP = (V/m) dP = (1/ρ) dP
and dP = ρg dz
Plug dP from the second equation into the first an rearrange to get dT/dz”)
“Pressure ( P ) is eliminated. It is assumed that the concept of pressure is the same in the 2 equations. It is not. In the first P is caused by T, kinetic energy, motion. not gravity. As in IGL In the second P is caused by gravity, not motion. As in the hydrostatic equation.”
Therein is the fraud. Sensible heat, although called kinetic energy, by those that subscribe to the “kinetic theory of everything”, such as Tim F., and is a Red-Herring. Sensible heat is not Newtonian kinetic (no momentum). The velocity is not a vector. Heat is a completely different form of energy. A rock sitting on a post above the ground, has no more energy, than that rock on the ground. Only if the post is removed and the force of gravity allowed to act with vector acceleration, is that energy ever manifest. A mountain above sea level has no potential energy.
“This problem is overcome by using the 1st law of thermodynamics, conservation of energy directly to the atmos. Without reference to pressure. Change in heat energy = change in gravity energy
Mgh = mcT T/h = g/c”
This problem is not overcome, it is exacerbated by a deliberate attempt to apply 1LTD, where it does not apply, This Earth’s atmosphere! This atmosphere is completely open. No restriction on energy in, out, or contained therein. No Lorentz transform applies, no symmetry, no conservation of anything, including mass!.
Try to learn and understand! It is the trying, and the “aw shit”, that leads to learning, never a book or a professor, (lecturer).
wayne says: December 12, 2014 at 9:44 pm
“I am speaking of the constant long-term temperature gradient as seen in this set of skew-t plots.”
OK! It “is” the long-term statistical mean that is the FRAUD!
“If this does not perk the curiosity of supposedly scientific commenters here, that is a bigger question why? Bryan seems to see its relevance.”
Indeed, go check the variance, 2 sigma, 3 sigma, 5 sigma, and complete outliers to gain any understanding of this physical. The average is “pablum” for the gullible masses!
“The only conclusion I can see in this investigation is that the primary temperature gradient slopes are governed by one single thing, the expressed degrees of freedoms of the constitute molecules present in each atmosphere which though this does not set the temperature of a surface it does mean imho if the surface were to warm, the temperature above the troposphere must also warm. If the surface were to cool, the lower stratosphere must also cool. Likewise if the stratosphere cools the surface *will* cool unlike some climatologists hell bent on insisting on some radiation-only effects governing our entire atmosphere”
Wayne,
You have correctly, thank you, presented the maximum information contained within the mean. If we get together and agree to shitcan all of that, we “perhaps” can start to learn! What was the topic? 🙂
Tim Folkerts says: December 12, 2014 at 8:50 pm
“It is assumed that the concept of pressure is the same in the 2 equations.”
The concept of pressure *is* the same in both! Pressure is pressure; its the force exerted on an surface divided by the area of the surfce. A pressure gauge will read the same no matter what causes the pressure. The molecules will be moving still be moving with the same average speeds”
Have you “no” personal integrity, or is it simply that you can not think? Please describe the difference between “pressure” and “density” especially as that may pertain to a gravitational field?
wayne 9:44pm: “…takes the form T=T0·(P/P0)^(1/DoF)…why?”
As I understand your question, you want to research 1st principle difference in DALR (–g/Cp) and ELR 6.5 as hypothesized in top post. As I understand your plot, these red lines are daily atm. soundings obtained at Norman, Okla. in June and Dec. The lower surface temperatures being Dec. Red lines include effects of liquid water & radiative transfer as they show surface & altitude intercepts as well as eyeballed slopes.
http://www-frd.fsl.noaa.gov/mab/soundings/reply-skewt.cgi?data_source=RUC2&lon=&lat=&airport=oun
Here’s some thoughts pursuant to your quest. Might open some research ideas for you. First though, I would highly recommend asking (email) the folks listed at the bottom of the page. Ask them how the data is produced exactly, i.e. what is the RUC exactly. Apparently it combines input from variety of sensors.
1) The eqn. you write was developed by Poisson in the 1890s for ideal gas. It leads to DALR and, including the observed environment, the hypothetical top post equation developed by a committee. Gas DOF and gamma are related for an ideal gas (to me in an uninteresting way). Undoubtedly these daily red lines contain fine liquid water droplets driving earth’s 9.8 toward 6.5 LR. Why different: Liquid water is not an ideal gas.
2) For a rising parcel, the DALR & top post eqn. can be developed in two different ways: a) From the Poisson relations you write for an ideal gas undergoing an assumed adiabatic process – the usual eqn.s around here AND b) appealing to entropy conservation in a reversible adiabatic process – unusual around here as entropy is a virus that has escaped the lab and infected many people who are not scientists esp. those with a literary bent like to write on blogs. I sometimes am amused by the disconnect between scientific and literary entropy (esp. Will & Stephen).
3) For 2a – this usual route is actually closed to your research into your “why?” simply because liquid water is not an ideal gas and certainly the fine water droplet effects exist in your red lines. So you are going to have to use 2b an entropy approach. It is very difficult, two derivations in fact may not even agree. The starting point is the parcel total entropy S = Mv*Sv+Md*Sd+Mw*Sw where v, w, and d denote vapor, liquid water and dry air (water vapor invisible, fine liquid water droplets visible cloud).
Now it gets rough going from dry static energy (top post eqn.) to moist static energy and theorists find an eqn. for ascent (or descent) that is isentropic for dT/dz = –g/Cp – 1/Cp * d/dz(liquid water effect) where Cp is now for liquid water rather than water vapor. This is just a start as the d/dz term can be found mathematically.
When an unsaturated parcel rises isentropically, saturation is achieved when temperature and dew point converge. Then for earth’s typical saturation mixing ratio and temperature in d/dz term, the lapse rate changes by roughly a factor of two, from ~10 to ~5 degrees C/km. So you can see abrupt changes in the red lines slope occur (clouds) which visually avg. around 6.5 (yellow line).
Need more? Just ask.
Definitely hairier than the hairy top post eqn. Note in the above, radiation is unimportant as above discussion is about slope not intercepts. Need radiative transfer added (LBLRTM) to pin down the T intercepts to small anomaly not off by 10-20%. The red lines are from observations so don’t need LBLRTM analysis for these charts, radiative transfer is inherent.
Trick says: December 13, 2014 at 12:09 am
wayne 9:44pm: “…takes the form T=T0·(P/P0)^(1/DoF)…why?”
(“As I understand your question, you want to research 1st principle difference in DALR (–g/Cp) and ELR 6.5 as hypothesized in top post.”)
“The observed environment, the hypothetical top post equation developed by a committee. Gas DOF and gamma are related for an ideal gas (to me in an uninteresting way). Undoubtedly these daily red lines contain fine liquid water droplets driving earth’s 9.8 toward 6.5 LR. Why different: Liquid water is not an ideal gas.”
Thank you Trick! Liquid water is not even a gas, but the liquid that can be supported easily at altitude, by hydrodynamic process. The exact process of condensation of WV in this atmosphere is unknown, but does involve both electro-motive and magneto-motive forces. It is time for all to admit “beats the shit out of me”, then go with twigs to the sand and attempt to draw pictures, that the other guy, shakes head up and down, rather than sideways. This is called “science”.
http://en.wikipedia.org/wiki/Supercooling
“Water normally freezes at 273.15 K (0 °C or 32 °F) but it can be “supercooled” at standard pressure down to its crystal homogeneous nucleation at almost 224.8 K (−48.3 °C/−55 °F).”
That -48.3C seems rather close to Wayne’s -52C,
Co-incidence, or something significant?
Tim Folkerts says: December 12, 2014 at 6:05 pm
(Will says: “That G-T paper clearly sets out the formula without any rising or falling parcels!”)
“No. Equation 14 in the G-T paper is dp/dz = − ρg or dp = − ρg dz. In other words, this equation relates a parcel rising a distance dz with a change in pressure of -ρg dz. Rising parcels are built into the equations from the very start. Integration yields the effects of moving the parcel a finite distance, z.”
Where in that G-T paper is anything moving? A difference at a distance implys a difference at a distance, never any movement.
“PS the wikipedia derivation utilizes the exact same dp = − ρg dz to introduce the effect of moving a parcel.”
What total horse shit. no movement of anything is implied only (dSomething)/dx? It is always change in location rather than any movement of any mass.
“All the rest in both sources simply relies on the basic properties of ideal gases (G-T just go into way more detail). So either BOTH derivations are flawed, or BOTH are acceptable.”
There is no reliance on anything IDEAL, only careful measurement of this physical..
(“BTW since WV has twice the Cp of air,, the fake meteorological Cp for air with saturated WV. is also doubled giving a fake lapse rate for saturated LR of 1/2 the dry.”)
“No. That is not how the saturated lapse rate is determined. The specific heat of dry air is still used, along with other properties of water to get the lapse rate. No one (that I have seen) uses the approach you claim. http://en.wikipedia.org/wiki/Lapse_rate#Saturated_adiabatic_lapse_rate.
Look at your reference!, there is no reference to WV condensing at that altitude or location providing gobs of sensible heat to be radiated away to space. Only a stupid and distracting equation that results, in fictitious saturated lapse rate of 1/2 dry lapse rate. A clear attempt at deliberate FRAUD. Tim, you are a party!!
Anything is possible says: December 13, 2014 at 2:01 am
http://en.wikipedia.org/wiki/Supercooling
(“Water normally freezes at 273.15 K (0 °C or 32 °F) but it can be “supercooled” at standard pressure down to its crystal homogeneous nucleation at almost 224.8 K (−48.3 °C/−55 °F).”)
“That -48.3C seems rather close to Wayne’s -52C, Co-incidence, or something significant?”
Look at the temperature of the stratopause! At that pressure/temperature water can be solid, liquid, gas, depending on the latent heat of that molecule. We have much to ponder!
Tim Folkerts says: December 12, 2014 at 8:50 pm
“A pressure gauge will read the same no matter what causes the pressure. The molecules will be moving still be moving with the same average speeds.”
No, that is part of the problem right there, that thought.
v_rms = √(3·Rs·T) but P = ρ·Rs·T
The pressure depends on density and the rms speed does not. You can have a certain pressure at low density, very hot, that is the same pressure if very cold and dense and the mean velocities are going to be miles apart in the two cases with the same molecular mass therefore identical specific heat.
wayne says: December 13, 2014 at 2:42 am
Tim Folkerts says: December 12, 2014 at 8:50 pm
(“A pressure gauge will read the same no matter what causes the pressure. The molecules will be moving still be moving with the same average speeds.””
“No, that is part of the problem right there, that thought.
v_rms = √(3·Rs·T) but P = ρ·Rs·T
The pressure depends on density and the rms speed does not. You can have a certain pressure at low density, very hot, that is the same pressure if very cold and dense and the mean velocities are going to be miles apart in the two cases with the same molecular mass therefore identical specific heat.”
SWEET!!!
Thank you Wayne! My turn for a round, if I can move!
Trick — December 13, 2014 at 12:09 am:
First, thanks for your thoughts though you seem to not quite my search yet.
You say: “As I understand your question, you want to research 1st principle difference in DALR (–g/Cp) and ELR 6.5 as hypothesized in top post.”
Not really Trick, I could care less of the DALR, it doesn’t exist at climate time and spatial scales of any atmosphere I have been able to get enough data on. None of these atmospheres use γ/(1-γ) of (γ-1)/γ (adiabatic, Poisson) to match the profiles as that exponent. You can use γ/(1-γ) but you end up with an adiabatic dry ERL profile of 9.75 K/km on Earth and the same wrong answer on the other three bodies. Try it. Mater of fact try 2/(1-γ) and all four body’s cases of profiles fall in line. That is what I am searching for, why? Down to the correct derivation as to why. Without a scientific explanation in equations this is just some grand curiosity.
“1) The eqn. you write was developed by Poisson in the 1890s for ideal gas. It leads to DALR and, including the observed environment, the hypothetical top post equation developed by a committee. Gas DOF and gamma are related for an ideal gas (to me in an uninteresting way). Undoubtedly these daily red lines contain fine liquid water droplets driving earth’s 9.8 toward 6.5 LR. Why different: Liquid water is not an ideal gas.”
It doesn’t lead to DALR, that depends on the exponent formulated that is necessary to precisely match the pressure/temperature profiles from the bottom to the top of the linear segment found in the lower atmospheres. This seems to just point out that atmospheres are a polytropic process and not some idealized adiabatic and reversible process.
You say “Undoubtedly these daily red lines contain fine liquid water droplets driving earth’s 9.8 toward 6.5 LR. ” and I don’t agree there at all except water vapor specifically might explain why Earth’s DoF is coming up 5.256 instead of 5.0 that N2 and O2 woukd seem to be dictate as found in Titan’s N2 atmosphere, its IS right at 5.0. Is it because the other three are dry and Earth is wet — maybe. Water is one great mover of energy upward so could it appear to be a partial DoF? Maybe.
Your further points are appreciated but seems to show you still don’t understand how multiple atmospheres are all operated — identical or close best I can tell. You go on about some perfect idealized concepts on down and I don’t find any operate in that manner. Any thin layer that is not changing its temperature is receiving and shedding exactly the same energy always no matter what individual processes are causing the reception and the shedding to space. This is true universally. Entropy enters from the sun and leaves by infrared waste heat in some manner or the other. But once again, I do not want to speak of momentary weather effects that are so short in duration they average out at climate scales.
Will, still trying to understand exactly where you seem to object to what I am doing, well, or attempting. I’ll get back soon and thanks for just trying to see into what I am trying to explain.
Trick says: December 13, 2014 at 12:09 am
wayne 9:44pm: “…takes the form T=T0·(P/P0)^(1/DoF)…why?”
“As I understand your question, you want to research 1st principle difference in DALR (–g/Cp) and ELR 6.5 as hypothesized in top post. As I understand your plot, these red lines are daily atm. soundings obtained at Norman, Okla. in June and Dec. The lower surface temperatures being Dec. Red lines include effects of liquid water & radiative transfer as they show surface & altitude intercepts as well as eyeballed slopes.”
Trick, please try to stop being a blithering idiot! You are smarter than that! The ALR and ELR are always functions of whatever this planet wishes to do, right cheer, right now, and even over yonder. Your statistical nonsense, impresses this Earth not at all.
wayne 4:14am: The red curves are each the daily weather above a location in Oklahoma. You seemed to ask why? the yellow line “is always” at 6.5 for avg. weather = climate. The reason can be shown mathematically. I gave you a hint for the main reason why. Has to do with liquid water in the red curves not acting like an ideal gas.
Sure, you can add in DOF for a diatomic molecule say 3 each for rotation and translation then 2 for vibration and find gamma = 10/8 = 1.25. But measurements show gamma = 1.4. So ignore the vibrational 2 and assume the molecule is a rigid rod for gamma = 7/5 = 1.4 close to observed. But this is not satisfying, and what is more: atoms have structures – electrons orbiting nucleus of proton/neutrons in motion and bound by nuclear forces. Ignoring those is tantamount to cheating, all the DOF modes have mouths to feed. So turn to quantum mechanics. Even that I guarantee will get you back to considering gamma for monatomic gases as point masses works well even though they are not. Diatomics only need 7/5.
If at the end of your quest, you find if all the internal modes of motion of molecules, including ALL their components, contributed to the specific heats of gases, the world would be a much different place. It is not. Your quest will be futile. I can explain some more but off topic; consider the top post hypothetical eqn. works well (can obtain T within 10-20% of the actual red curves up to tropopause) with only point mass considered and gamma = 1.4.
wayne says: December 13, 2014 at 5:29 am
Will, still trying to understand exactly where you seem to object to what I am doing, well, or attempting. I’ll get back soon and thanks for just trying to see into what I am trying to explain.
Wayne, your explanations are always outstanding!
Wayne, I have no objection to what you try to do, only to “why” you wish to do that! In the lumber yard, average meters^3, of wood shipped, says nothing about, how much your employees wish to off you! The averages have no information whatsoever. Whatever you wish to do, I am pleased to assist. Head wagging sideways, mostly!
Trick says: December 13, 2014 at 5:40 am
“Sure, you can add in DOF for a diatomic molecule say 3 each for rotation and translation then 2 for vibration and find gamma = 10/8 = 1.25. But measurements show gamma = 1.4. So ignore the vibrational 2 and assume the molecule is a rigid rod for gamma = 7/5 = 1.4 close to observed. But this is not satisfying, and what is more: atoms have structures – electrons orbiting nucleus of proton/neutrons in motion and bound by nuclear forces. Ignoring those is tantamount to cheating, all the DOF modes have mouths to feed. So turn to quantum mechanics. Even that I guarantee will get you back to considering gamma for monatomic gases as point masses works well even though they are not. Diatomics only need 7/5.”
A great explanation of a deliberate chaotic troposphere. That will fix the know it all God! Thank you for pointing out that no earthling or group of political earthlings has any clue whatsoever.
“If at the end of your quest, you find if all the internal modes of motion of molecules, including ALL their components, contributed to the specific heats of gases, the world would be a much different place. It is not. Your quest will be futile. I can explain some more but off topic; consider the top post hypothetical eqn. works well (can obtain T within 10-20% of the actual red curves up to tropopause) with only point mass considered and gamma = 1.4.”
All of that is easy from the barometric formula! What you cannot do is know, or predict the weather or temperature at any location on or above this planet, at any time. The thermo/hydro/dynamics of this planet is so fascinating, that no greenies, mafia, or government can fix the outcome. They all claim they can!
Will Janoschka says:
December 13, 2014 at 6:00 am
wayne says: December 13, 2014 at 5:29 am
Will, still trying to understand exactly where you seem to object to what I am doing, well, or attempting. I’ll get back soon and thanks for just trying to see into what I am trying to explain.
“Wayne, your explanations are always outstanding!
Wayne, I have no objection to what you try to do, only to “why” you wish to do that! In the lumber yard, average meters^3, of wood shipped, says nothing about, how much your employees wish to off you! The averages have no information whatsoever. Whatever you wish to do, I am pleased to assist. Head wagging sideways, mostly!”
You ask why I wish to do so? Hmm. Never thought of that. I guess it is my long term background and interest has been in astronomy, was hand picked as a lab assistant in college and my major was not even in that area, first major in chemistry then moved to physiology but also wanted to solo circumnavigate in my sailboat, had to learn celestial navigation first and that got me interested. Later got hooked deeply into solar system integration systems and wrote a close match for the JPL Horizon system from the ground up. When I find something curious I want to know it down to the very core.
So when I became curious of this blame of co2 and trillions of taxes because of this puny increase about eight years ago, from physiology of course it is harmless, I first wanted to know what everyone else is not speaking of, especially about atmospheres in general and that has been on my mind since but until I could get a good grasp on thermodynamics and general atmospheric physics, till then I has to put that question on hold until about two years ago. So it took me taking some twelve courses to get me back to speed in those areas.
Does that help you understand ‘why’? If I can first understand what is common of all atmospheres just being gases held by gravity with solar input, and that is important to me, then I can turn my attention to the bigger question of “ok, why is earth’s atmosphere so trivially assumed so different as if it is special?”. Is it really? There are so many questions never approached, never really allowed to be asked, shied away like the plague, as if the answers are going to pop the bubble and all of climate ‘science’ and climate blogs are going to go poof if ever correctly answered so best to never ask.
wayne says, December 12, 2014 at 9:44 pm:
“See, it doesn’t matter in which time of day or the season you choose, the slope is always right at 6.5 K/km right up to the tropopause no matter whether day or night, whether summer or winter, the slope (or lapse) is always a constant, on the average over long periods of time. So please stop dragging my examples into immediate weather effects!”
wayne,
What are you actually trying to say here? Are you saying that annually, the ALR/ELR is a steady 6.5K/km all over the world, over deserts, polar ice caps, mountain plateaus, rainforest basins and deep convecting tropical sea surfaces alike. Do you have plots similar to the single site you presented above from all of these different kinds of regions from all around the globe? I know that this is not the case. The 6.5K/km value is but a global mean on the continuum between the two ALR extremes: ~10K/km (DALR) at the one end, and ~5K/km (SALR) at the other.
We’re basically talking about the same thing, only from different perspectives. You say that water skews the average molecular DOF of Earth’s atmosphere away from the 5 of N2 and O2. We’re saying that water skews the ideal DALR into something much closer to the SALR (at the other end of the spectrum), because of the release of latent heat upon condensation in the air column.
Tim Folkerts says:
December 12, 2014 at 8:50 pm
Pressure is pressure; its the force exerted on an surface divided by the area of the surfce. A pressure gauge will read the same no matter what causes the pressure. The molecules will be moving still be moving with the same average speeds.
The IGL and the hydro-static equation are different theories/models of pressure.
In the IGL theory or model, the force is caused by molecules bouncing off a surface. Pressure is homogeneous.
In the hydro-static model, the force of caused by the Earth’s gravity. Pressure has a vertical gradient.
They are both theories of pressure. But of pressure under different conditions, so cannot be put equal.
Kinetic energy and gravity energy can be equated via the 1st law of thermodynamics, conservation of energy.
Roger and Stephen
One last attempt from me to correct your mistakes.
The kinetic energy of individual molecules of a gas is described as thermal energy or more correctly internal energy of the gas.
Mechanical kinetic energy of the same gas sample is the KE of the centre of mass of the gas sample.
They are two quite different things
Will Janoschka is correct when he says that the internal energy of the group has no momentum.
Do you know the difference between speed and velocity?
Velocity is a vector as is its momentum (MV)
Speed and energy are scalers – look it up.
The gas particles are moving randomly and their individual momentums cancel out leaving the momentum of the centre of mass as zero.
If a windmill is in such a gas sample it would not move even though the gas molecules are moving at average speeds of around 500m/s
.
On the other hand if the gas particles are in a wind of lets say 20m/s the centre of mass will have momentum and can do work such as making a windmill spin.
Direct lossless conversion of thermal energy into mechanical energy such as gravitational potential energy is impossible and is governed by the second law of thermodynamics.
Look up the second law as well.
Kristian says:
December 13, 2014 at 2:58 pm
The 6.5K/km value is but a global mean on the continuum between the two ALR extremes: ~10K/km (DALR) at the one end, and ~5K/km (SALR) at the other.
According to fig.1 of this paper, the range of LR is 5.5-6.5K/km. It never gets above 6.5K/km even over dry areas.
Click to access LapseRate-FAO06-IACP430.pdf
Bryan,
I tried to say the same thing (directed at Tallbloke) here:
But your above explanation is better in that it’s more easily understood. Thanks.
Ben Wouters says, December 12, 2014 at 9:44 am:
“The assumption is that the rising parcel does not exchange energy with the surrounding air, making it possible to calculate the temperature drop due to the expansion AS IT RISES through the static atmosphere, that is in hydrostatic equilibrium.”
*Sigh*
For the umpteenth time: The rising air parcel DOES EXCHANGE ENERGY WITH THE SURROUNDING AIR. It just doesn’t do it through a transfer of energy as HEAT, but rather through a transfer of energy as WORK. This is what the adiabatic process is all about, Ben:
ΔU = Q – W (1st Law of Thermodynamics)
U is the internal energy of the system (air parcel), corresponding to its temperature [T]; Q is the net transfer of energy as heat to/from the system; and W is the energy transferred to/from the system by work being done by/on the system.
For incremental steps in a so-called quasi-static (reversible) process:
dU = δQ – δW
dU = δQ – PdV
In an adiabatic process, the transfer of ‘heat’ [Q] across the system boundary is 0 by definition, so the entire change in system internal energy [dU], and thus temperature T, is due to the pressure-volume work [PdV] being done by/on the system:
dU = – PdV
The adiabatic process in one simple formula.
Kristian 6:05pm: What happened to the V*dP term in your dU differential of δW?
Roger Clague says, December 13, 2014 at 5:47 pm:
” http://ifaran.ru/old/ltk/Persona/Mokhov_pub/LapseRate-FAO06-IACP430.pdf “
Thanks, Roger. Wayne should check out this paper.
“According to fig.1 of this paper, the range of LR is 5.5-6.5K/km. It never gets above 6.5K/km even over dry areas.”
No, Roger. I’m looking at Fig. 1 too, you know. And it is plain to see that you’re wrong. All the dark areas have annual lapse rates steeper than 6.5. The upper limit is only the DALR. If you look at this plot, notice the Saharan lapse rate:
Likewise, if you look at higher latitudes in Fig. 1, you will see how annual lapse rates are lower than 5.5 (and even 5) K/km, confirmed by Fig. 2. This has to do with convective power vs. direct radiative cooling in these regions.
Bryan,
You are mistakenly regarding kinetic energy as movement in a single plane.
For gases the kinetic energy that matters is the speed of vibration of the molecules.
Lifting a vibrating molecule upwards against gravity so as to reduce pressure causes it to vibrate less and cool. That is the process whereby gravitational energy (not heat) replaces kinetic energy (heat). Internal energy is KE plus PE not just KE.
Kristian,
Regarding ΔU = Q – W
See here:
http://hockeyschtick.blogspot.co.uk/2014/12/how-gravity-continuously-does-work-on.html
For the portion of the work done that is done against gravity that work cannot simultaneously be done against surrounding molecules. There is some of the latter during adiabatic uplift only because no adiabatic process is perfect. There is always some leakage. Work done against gravity causes cooling by changing KE to PE. Temperature will decine with height even without GHGs.
Will Janoschka says: December 13, 2014 at 6:26 am
“All of that is easy from the barometric formula! What you cannot do is know, or predict the weather or temperature at any location on or above this planet, at any time. ”
Absolutely Will, you are correct, you cannot predict anything having to do with the chaotic weather or chaotic temperatures. I felt maybe that is what you were thinking I was attempting, by averaging, and that is not true at all, just that the entire tropospheres seems, by the data, to act as a quasi-static slope of a gradient at any one location.
When it gets colder, the entire profile shifts left (colder) as a unit and the lapse stays on the average constant. Just the opposite when it is warmer like in the summer but the gradient also stays constant and I think that says so much as to why AGW is completely incorrect. In order for the AGW theory to even exist it requires the slopes of the gradients to change as co2 is added — cooling the lower stratosphere and warming the lower troposphere and the data seems to show that is not possible without massive changes is the barometric equation’s exponent components, either g and operating cp and mankind changing g is of course impossible.
Right now I need to read Ferenc’s latest paper.
Kristian says:
December 13, 2014 at 6:25 pm
” http://ifaran.ru/old/ltk/Persona/Mokhov_pub/LapseRate-FAO06-IACP430.pdf “
All the dark areas have annual lapse rates steeper than 6.5.
So the range of LR is 5-7K/km. Still not near to the DALR at 10K/km.
The Saharan LR looks like 30K/ 5.5 km = 5.5K/km, again far from DALR at 10K/km
According to your theory LR of the dry poles should be high, near the DALR at 10K/km. Not lower than wet areas as it is.
Kristian says: December 13, 2014 at 6:05 pm
*Sigh*
Are you even aware that pressure drops with increasing altitude?
A rising parcel (or balloon etc.) experiences an ever decreasing pressure.
It equalizes this lowering pressure by expanding (= increasing volume).
So we have decreasing pressure and increasing volume for the parcel, containing the same amount of molecules all the time.
Guess what happens with the temperature: it drops.
Guess how fast: according the DALR = 9,8K/km., until condensation kicks in, releasing latent heat.
Kristian says: December 13, 2014 at 2:58 pm
“We’re saying that water skews the ideal DALR into something much closer to the SALR (at the other end of the spectrum), because of the release of latent heat upon condensation in the air column.”
Pse show where on this planet massive condensation in the air column (=cloud formation) happens,
(enough to skew your “ideal DALR” to the average 6,5K/km) and upward vertical movement of the air is NOT involved.
I’ll help with ONE example: fog forming near the surface when the temperature drops after sunset to below the dewpoint.
wayne says: December 12, 2014 at 9:44 pm
” So please stop dragging my examples into immediate weather effects!”
wayne says: December 11, 2014 at 8:28 pm
“So I ask again, where the h–l did that ‘DALR=g/cp’ come from originally? What is its history?”
By asking about the history of the DALR you yourself initiated a discussion about convection etc.,
since the DALR is closely related to convection.
The old text Trick linked to is very informative.
Stephen 7:33pm: “Kristian, Regarding ΔU = Q – W See here:”
Done. Change in control volume internal energy is ΔU. Stephen and HS: What are the text book assumptions for atm. thermo. you use after writing this comprehensive 1st law?
Oh, I see, Stephen misplaced his 1960s atmosphere text and can no longer look them up. I tried to get Kristian interested also at 6:22pm.
Note Stephen’s “see here” HS ref. along with Kristian arrives at dU = δQ – Pdv fairly mysteriously. No light of day on assumptions, so I’ll shine some light in 1), 2) and 3):
1) Why the d and why the δ? Oh, I see, Stephen’s understanding of differential calculus is limited. Somebody look that up for Stephen. I like the Truesdell ref. which says not all that important.
2) Why the minus sign? This working term – P*dv comes from work done (mgh) by a moving weight put on top of a movable cylinder compressing a volume of gas. If that weight is put on “quasi-statically” i.e. slowly, the pressure doesn’t change very fast and the v*dP term can then be considered 0 as dP/dt ~0, this is the answer I was looking for 6:22pm from Kristian. Convince yourself the minus sign is needed when a gas is compressed dV/dt is negative and the gas internal energy increases.
3) What are “quasi-static” assumptions? Calculations show means weight moves cylinder much less than speed of sound. OK. “Quasi-static” also implicitly means the radiation bath the cylinder is held within doesn’t change in any appreciable way affecting the increase of temperature from compression. I suppose room or lab temperature is held constant steady state with thermostat controlling a furnace.
Uh-oh.
After that “quasi-static” point, Q has its origins in interactions between systems of molecules with different temperatures and densities. Stephen’s dU = changes in KE + PE, always mistakenly leaving out the working P*V term, which Stephen now sees used by Kristian and the HS ref. in – P*dv as I’ve been pointing out.
Q also may have a component that results from interactions of molecules with radiation as molecules emit and absorb radiant energy from the surrounding photonic bath & changes in these processes are vital to understanding the precise workings of the atmosphere down to anomaly on order of 0.1K. The top post eqn. does come close (within 10% to 20% in T) by ignoring radiation in “quasi-static” Q to actual atm. troposphere T profiles wayne showed 9:44pm (in red). HS LR -6.5 K/km top post eqn. is the yellow line up to tropopause.
This means “quasi-static” assumption allows Stephen’s ref. to proceed w/o the radiative component of Q allowing HS conclusion to be the only remaining components are “How Gravity continuously does Work on the atmosphere to control pressure & temperature”. Because radiation and v*dp effects in actual atm. Q were assumed to be left out at the beginning assumption of “quasi-static”.
Similar to Stephen’s mass and gravity debates which don’t even include the p*dv term, just argue from PE+KE 1st law view: molecules as point mass without pressure. Lacking the 1960s text book Stephen “lost” is a real impediment to his complete understanding of changes in Q (from p*dv, v*dp and radiation) adding to meteorology the needed precision in 1st law.
Ben Wouters says, December 14, 2014 at 3:20 pm:
“Are you even aware that pressure drops with increasing altitude?
A rising parcel (or balloon etc.) experiences an ever decreasing pressure.
It equalizes this lowering pressure by expanding (= increasing volume).
So we have decreasing pressure and increasing volume for the parcel, containing the same amount of molecules all the time.
Guess what happens with the temperature: it drops.
Guess how fast: according the DALR = 9,8K/km., until condensation kicks in, releasing latent heat.”
Hehe, yup, it’s called ‘PV work’, Ben. That’s the work that expanding air does as it pushes into (and thus against the external pressure of) the surrounding air masses, thus transferring some of its internal energy to them, not as ‘heat’, but as ‘work’, PV work. The opposite happens when the air is compressed.
You really should read up. This is common knowledge. It’s called ‘adiabatic processes’.
Ben Wouters says: December 14, 2014 at 3:32 pm
“By asking about the history of the DALR you yourself initiated a discussion about convection etc., …”
Not really. I was asking the relation of DALR that everyone ELSE seems to tie to long-term climate and I can never find even a trace of it there. That is why I wanted the detailed derivation.
But it seems I have finally answered that question to myself.
Took sitting through the first few lectures on a t.d. class more than once to drive it home. You say when a packet or very loose and baggy balloon rises in the gradient atmosphere it expands and cools and that is exactly correct Ben.
But one point the professors kept driving home so no student would miss it is that the work done on the surroundings is dependent not on the pressure of the gasses in the balloon but critically on the pressure of the surrounding’s pressure! And what sets the pressure of the surroundings? The exponent of that polytropic barometric equation and it is different vertical-meter for vertical-meter if the local gradient is isothermal (expon.=3.5) or if the gradient is the environmental lapse of 6.5 K/km (expon.=5.256) or it could be somewhere between but I just took the two well known cases.
Now if you then take the time to write a numerically integration for an example of this scenario it becomes perfectly clear as you view the output. The isothermal lower atmosphere, like in the case of between the surface with thick low cloud cover, the pressure doesn’t decrease near as fast as in the environmental case so the gases in the balloon are doing more work faster as each meter goes by as it rises and therefore the temperature in the balloon drops faster, 9.75 K/km or inversely ≈103 m/K. The environmental lapse is inversely ≈154 m/K, it takes longer for the balloon to lose 1K by the work it is doing against the surrounding’s pressure. Walla! I am finally satisfied.
So the cooling rate can never be really known unless you have some intimate knowledge of exactly what the local (in the mixing layer) pressure is in the case of weather’s case. This is why I was saying earlier that it is strictly densities that dictates the rising/falling and it is the local external pressure gradient that determines how fast some parcel or balloon will have the temperature decreased/increased if it is moving vertically, it just took a while for me to prove to myself that my thinking was correct there.
Thanks for the comments that forced me to learn that great lesson.
wayne says: December 15, 2014 at 3:28 am
“So the cooling rate can never be really known unless you have some intimate knowledge of exactly what the local (in the mixing layer) pressure is in the case of weather’s case. This is why I was saying earlier that it is strictly densities that dictates the rising/falling and it is the local external pressure gradient that determines how fast some parcel or balloon will have the temperature decreased/increased if it is moving vertically, it just took a while for me to prove to myself that my thinking was correct there.Thanks for the comments that forced me to learn that great lesson.”
For this thread,and atmosphere, half the pressure is below 5.6 km. Half the mass, (weight) is below 565 meters. This atmosphere is a real bitch.
Stephen Wilde says, December 13, 2014 at 7:33 pm:
For gases the kinetic energy that matters is the speed of vibration of the molecules.
Lifting a vibrating molecule upwards against gravity so as to reduce pressure causes it to vibrate less and cool. That is the process whereby gravitational energy (not heat) replaces kinetic energy (heat). Internal energy is KE plus PE not just KE.
You don’t lift each individual molecule up, Stephen. Each individual molecule isn’t the object here. You lift the air mass in bulk, en masse. Each indivual molecule flies around in all directions – up, down and sideways – at all times. The air itself doesn’t. You are mixing up two different worlds, the microscopic and the macroscopic. You mix up the thermal energy and the mechanical energy of a gas sample. Thermal energy (the KE of each individual molecule) can never be turned into gravitational PE. The two do not share the same phase space. Gravitational PE transforms into mechanical KE, what Bryan described as the mechanical kinetic energy of the centre of gravity of the gas sample. This particular form of energy has to do with the momentum, the motion, of the bulk air, whether it moves horizontally or vertically. All the kinetic energies of all the individual molecules making up the gas sample adds to zero momentum, zero motion. When you ‘charge’ an object (or an air parcel) with gravitational PE upon lifting it, you don’t cool it (for the air, this ONLY happens due to expansion against a progressively lower external pressure, for a rigid object it only happens due to the progressive cooling of surrounding air as it rises), you rather prepare it for the descent – the higher you lift it, the higher the content of gravitational PE, and the greater the potential momentum on the way back down the gravitational well.
The lifting itself has no bearing on temperatures, Stephen. Only the processes of expansion/compression change the temperature.
Roger Clague says, December 13, 2014 at 8:32 pm:
“So the range of LR is 5-7K/km. Still not near to the DALR at 10K/km.”
This is hopeless. You appear to just see what you want to see, Roger. Where do you see an upper limit of 7K/km!? The dark areas simply represent ‘beyond 6.5K/km’. The upper limit is of course the DALR at ~10K/km. No area would match this at an annual basis, because it would require surface heating and thus convection around the clock. But the Saharan ELR for instance, with a non-condensing atmospheric column, would track the DALR pretty closely during the day, as confirmed by the plot above.
“The Saharan LR looks like 30K/ 5.5 km = 5.5K/km, again far from DALR at 10K/km”
Hehe, no Roger. Take another look. The temperature at 1000 mbar (ground level) is about +31/+32 degrees Celsius, the temperature at 500 mbar (about 5.6 km up) is about -23/-24 degrees Celsius. That’s a drop of ~55K over a vertical distance of 5.6 km = DALR.
“According to your theory LR of the dry poles should be high, near the DALR at 10K/km. Not lower than wet areas as it is.”
I don’t have a ‘theory’, Roger. I simply observe what’s going on in our troposphere. Like I said (not sure if you read it): “This has to do with convective power vs. direct radiative cooling in these regions.”
Convection in the highest latitudes of our globe is pretty sporadic and thus quite inconsequential annually, not like in the tropics/subtropics and the frontal zones. At the same time, surface solar heating is tiny compared to lower latitudes, especially evened out across the year. These regions are to a larger extent than elsewhere on the planet part of a net radiative cooling regime. Such a regime, with very little radiative surface heating and very little resulting convection, but lots of radiative cooling to space, especially in winter, will be utterly unable to sustain an annual mean lapse rate on a par with the adiatic ones (which specifically depends on the two former mechanisms).
Check out the plot from an earlier comment of mine on another thread:
wayne says: December 15, 2014 at 3:28 am
“So the cooling rate can never be really known unless you have some intimate knowledge of exactly what the local (in the mixing layer) pressure is in the case of weather’s case.”
One of the assumptions when using the DALR is that the surrounding static atmosphere is in hydrostatic equilibrium. For other situations all bets are off.
The 2 examples you gave (isothermal and 6,5K/km profiles) do not allow dry (non condensing) convection. The 6,5K/km profile DOES allow initially condensing convection ubtill most latent heat has been used up.
Did you see the link to the buoyancy / CAPE class?
http://www.tornadochaser.net/capeclass.html
Do you now agree that the DALR and SALR are ONLY relevant for the temperature WITHIN a rising or sinking parcel?
Nice to see Noor van Andel agree that convection is highly overrated 😉
Click to access KNMI_voordracht_VanAndel.pdf
page 29:
“Radiation plays little role 1km.”
Kristian says: December 15, 2014 at 1:32 pm
” Where do you see an upper limit of 7K/km!? The dark areas simply represent ‘beyond 6.5K/km’. The upper limit is of course the DALR at ~10K/km. No area would match this at an annual basis, because it would require surface heating and thus convection around the clock. But the Saharan ELR for instance, with a non-condensing atmospheric column, would track the DALR pretty closely during the day, as confirmed by the plot above.”
Brilliant observation of the Sahara profile more or less matching a DALR profile.
Only problem is that the Sahara is under the receiving end of the Hadley circulation, and the air is actually SINKING above the Sahara.
So iso massive convection we have massive subsidence over the Sahara.
It’s ok though, creates a DALR like profile anyway.
Trick says:
December 14, 2014 at 7:20 pm
This working term – P*dv comes from work done (mgh) by a moving weight put on top of a movable cylinder compressing a volume of gas.
I agree. The adiabatic process occurs in the cylinder of a heat engine, the atmos is not enclosed by walls as is a cylinder of a heat engine.
Kristian says:
December 15, 2014 at 12:57 pm
Each individual molecule isn’t the object here. You lift the air mass in bulk, en masse.
But the IGL and K.E. theory use molecules are the object. That is the point of reductionism, to explain bulk properties, such as LR of the atmosphere using concepts at a smaller and simpler scale.
We explain the absorption/emission properties of gases by considering molecular shapes.
Thermal energy (the KE of each individual molecule) can never be turned into gravitational PE
A hot air balloon converts heat to gravitation energy.
Kristian says:
December 15, 2014 at 1:32 pm
” http://ifaran.ru/old/ltk/Persona/Mokhov_pub/LapseRate-FAO06-IACP430.pdf “
The dark areas simply represent ‘beyond 6.5K/km’
The values increase in band of 0.5K/km. So the dark areas are 6.5-6.9K/km.
The temperature at 1000 mbar (ground level) is about +31/+32 degrees Celsius, the temperature at 500 mbar (about 5.6 km up) is about -23/-24 degrees Celsius.
Which temp are you talking about, the one on the left or the one along the bottom? Both are about 5K/km.
I don’t have a ‘theory’,
You mean your ideas are common and therefore facts, mine are unusual, at the moment, and so are theories.
Calling the atmos temp grad adiabatic is a theory. The theory is that the LR is caused by heat being converted to work against pressure. Accept you have a theory and let me have my theory that is heat converted in the atmos into gravitational energy.
Roger Clague says: December 16, 2014 at 12:37 pm
” A hot air balloon converts heat to gravitation energy.”
The heater warms the air inside the balloon until the “density” of the “system balloon” is equal to or lower than the density of the air the balloon is floating in.
Equal means float, lower density means rise, higher density means sink.
No conversion of heat into potential energy at all.
Roger Clague says, December 16, 2014 at 12:37 pm:
“But the IGL and K.E. theory use molecules are the object. That is the point of reductionism, to explain bulk properties, such as LR of the atmosphere using concepts at a smaller and simpler scale.”
The kinetic theory of gases doesn’t explain the momentum/motion of bulk air, Roger. It explains pressure, volume and temperature of the gas.
“A hot air balloon converts heat to gravitation energy.”
Er, no. The air inside the ballon is less dense than the air on the outside, because it’s warmer, hence it lifts from positive buoyancy in an attempt to restore hydrostatic equilibrium.
No thermal energy is ever converted into gravitational potential energy in this manner. Only mechanical energy is.
“The values increase in band of 0.5K/km. So the dark areas are 6.5-6.9K/km.”
Are you being willfully obtuse, Roger? The pointy ends of the first and last ‘band’ are directly equivalent to the mathematical symbols of (greater than). If these two bands were indeed, as you claim, just regular 0.5K/km bands like the rest, why aren’t they square just like these and marked by 4.5 and 7 at the end? These kinds of scale bars are extremely common and I fear they all say the same thing, Roger. The ends are pointy and without denomination for a reason.
“Which temp are you talking about, the one on the left or the one along the bottom? Both are about 5K/km.”
The straight, diagonal temperature lines moving steeply up towards the righthand side and crossed one by one, perpendicularly by the rising blue Saharan profile. At the bottom, slightly above 1000mbar, it crosses the 30 degree line. At the top it moves a bit beyond the -20 degree line. You’re looking at the slightly curved, quasi-horizontal pressure lines. You can observe how equal pressure levels are progressively higher on the righthand side of the diagram than on the lefthand side. Why is this? Because on the righthand side, the temps are always higher than on the lefthand side, so the atmosphere expands thermally (becomes less dense). At ground level for instance, the temperature at 1000mbar is +31/+32C to the right and only about -15C to the left. I’ll repost the diagram here:
http://s1172.photobucket.com/user/Keyell/media/tphi_zpsa5cf20ce.png.html
“Calling the atmos temp grad adiabatic is a theory. The theory is that the LR is caused by heat being converted to work against pressure.”
First of all, no one’s calling the ‘atmospheric temperature gradient’ adiabatic. The atmospheric temperature gradient or profile is the ELR which is caused by the tight interaction of radiation and convection. These are very much diabatic processes.
The only thing that’s adiabatic in our atmosphere is … the ‘adiabatic lapse rate’ (the ALR).
You need to read up on the difference between the two (ELR vs. ALR).
Second of all, the ‘theory’ is NOT that the LR is caused by heat being converted into work against pressure. No heat is converted into work. Both ‘heat’ [Q] and ‘work’ [W] are both energy TRANSFERS across thermodynamic system boundaries. ‘Internal energy’ [U] is expended (and resupplied) by thermodynamic ‘work’ being done against (and by) external pressure. No energy gained by the atmosphere as heat (from the Sun or from the surface) is ever spent by the adiabatic process. All of this heat is shed to space through radiation. This HEAT GAIN/HEAT LOSS process doesn’t really have anything to do with the internal adiabatic process.
Kristian says, December 16, 2014 at 3:50 pm:
“The pointy ends of the first and last ‘band’ are directly equivalent to the mathematical symbols of (greater than).”
Not sure what happened here. Seems WordPress ‘snipped’ some of my words. It should say:
“The pointy ends of the first and last ‘band’ are directly equivalent to the mathematical symbols of < (less than) and > (greater than).”
[Mod note] WP interprets anything inside chevrons as code. If it’s bad code, it bins it.
Kristian 3:50pm – “..no one’s calling the ‘atmospheric temperature (ELR) gradient’ adiabatic..”
The USSA committee & HS are!
The ELR = 6.5 K/km from the Standard Atm. hypothetical also has T(dh) considered constant (adiabatic and in less precise terms “quasi-static”) over dh to do the integral as HS shows 11/27:
dP = -(MPg/RT)dh (12)
∫dP/P = -(Mg/RT) ∫dh (13)
The USSA eqn. 5 is the same as HS (12). Not as easy to find the next step in math in the details of the USSA paper however they state p. 10 transformations are introduced so “the resulting differential equation has an exact integral” thus to evaluate HS integral (12) and USSA (5) into an exact integral result (USSA 33b) same form as HS (13) resulting:
P = e^-((Mgh/(RT)) (15)
So I learned from HS work that ERL is really a hypothetical lapse profile up to Earth mid-latitude tropopause (~0.18bar) consistent with barometric formulae, radiation, convection in mid-latitude summer & winter as eyeballed by committee assuming constant T (adiabatic!) over dh to bring T outside of the integral as shown by HS to find committee determined 6.5 K/km which is at least between ~5 K/km moist (liquid water is not an ideal gas) and ~10 K/km dry.
As wayne informatively & in very cool graph showed 9:44pm, the hypothetical 6.5 K/km (yellow line) adiabatic derived ELR lies in between actual atm. noisy red soundings (it was cloudy some days!) for summer and winter mid-latitudes even as recent as this last summer/early winter 2014.
Trick says: December 16, 2014 at 9:43 pm
“As Wayne informatively & in very cool graph showed 9:44pm, the hypothetical 6.5 K/km (yellow line) adiabatic derived ELR lies in between actual atm. noisy red soundings (it was cloudy some days!) for summer and winter mid-latitudes even as recent as this last summer/early winter 2014.”
There is much contention that the ELR can be considered “adiabatic”. Does an adiabatic barrier prevent energy transfers between different “masses”, or does it also prevent energy transfers that are “not” between sensible heat and pressure? You can find no scientific agreement, only religious belief.
Is the change in energy “form” between latent heat of evaporation (chemical heat) and the conversion to sensible heat within that same mass, “adiabatic” or not? It is this change from latent heat to sensible heat, entirely within the same mass (one religion), or within the same “air parcel” (other religion), that results in the only change in the lapse rate! Remember that the same conversion to sensible heat is no reduction in mass or weight but “requires” a reduction of approx. 4%in pressure/volume of the atmosphere, (near 100% reduction of H2O volume). This atmosphere is a real bitch!
Will Janoschka says: December 17, 2014 at 12:55 am
You didn’t mention the religious belief of clueless nitwits that babble about climate as if they have a clue, but come up with statements like:
“For this thread,and atmosphere, half the pressure is below 5.6 km. Half the mass, (weight) is below 565 meters. This atmosphere is a real bitch.”
Half the pressure is below 5,6 km????
Half the weight is below 565 meters????
Kristian says:
December 16, 2014 at 3:50 pm
The straight, diagonal temperature lines moving steeply up towards the righthand side and crossed one by one, perpendicularly by the rising blue Saharan profile. At the bottom, slightly above 1000mbar, it crosses the 30 degree line. At the top it moves a bit beyond the -20 degree line
http://s1172.photobucket.com/user/Keyell/media/tphi_zpsa5cf20ce.png.html
I don’t think it is correct to read along a diagonal axis on a Cartesian graph.
We should compare the vertical temp with the vertical height, T/h = lapse rate ( LR )
h = 5km T = -40C
h = 0km T = -15C
dT/dh = 25C/5km = 5C/km
Where did you get this graph from? What is the red line?
The atmospheric temperature gradient or profile is the ELR which is caused by the tight interaction of radiation and convection. These are very much diabatic processes.
The only thing that’s adiabatic in our atmosphere is … the ‘adiabatic lapse rate’ (the ALR).
But , according to you and the mainstream consensus,
ELR = DALR – Latent heat effect
So ELR is also an adiabatic process
Roger Clague says: December 18, 2014 at 10:56 am
See http://www.atmos.illinois.edu/~snesbitt/ATMS505/stuff/09%20Convective%20forecasting.pdf
for a simple introduction in thermodynamic diagrams.
The Sahara profile is close to and (almost) exactly parallel to the 30C potential temperature, so by definition its lapse rate is 9,8K/km.
Graph is from this page:
http://www.st-andrews.ac.uk/~dib2/climate/lapserates.html
Reasonable description of meteorological basics, lousy explanation below the graph.
“The Sahara line (blue) is parallel to a dry adiabat (line of equal potential temperature: this is characteristic of a dry atmopshere well mixed by convection. ”
This profile is very likely caused by descending air due to the Hadley circulation.
“The Irish curve closely follows a saturated adiabat through most of the atmosphere, characteristic of a well-mixed, cloudy atmosphere. The abrupt change in direction just below 300 mbar is the tropopause.”
Over Ireland with surface temp ~15C the tropopause is very unlikely to be below 300 mbar.
Most probably the temperature inversion is due to the polar front.
Ben Wouters says:
December 18, 2014 at 4:00 pm
The Sahara profile is close to and (almost) exactly parallel to the 30C potential temperature, so by definition its lapse rate is 9,8K/km
Graph from
http://www.st-andrews.ac.uk/~dib2/climate/lapserates.html
which says
potential temperature is denoted q (Greek theta), and can be defined as:
the temperature that an air mass would have if it were moved dry adiabatically to a level at which pressure is 1000 hPa.
and
This is known as the Dry Adiabatic Lapse Rate (DALR), and is equal to 9.8°K/km
The temperatures are adjusted to the “ potential temperature” as “ if it were moved dry adiabatically”. So you get the answer you wanted, as you say “by definition”. That is by making the temperatures what you want them to be.
However the reference does concentrate on the hydro static equation, pressure caused by gravity producing weight.
Roger Clague says: December 19, 2014 at 5:36 pm
” So you get the answer you wanted, as you say “by definition”. That is by making the temperatures what you want them to be.”
Suggest to (re)read the introduction to thermodynamic diagrams I linked to.
Every line of equal potential temperature has a slope of 9,8K/km BY DEFINITION, and they are all parallel to each other.
If you draw a profile of real temperatures like the Sahara one, and it is parallel to a line of equal potential temperature, I don’t have to calculate its lapse rate, it just is the same as all lines of equal potential temperatures: 9,8K/km.
Do the calculation yourself:
31C at -31m to -22C at 5430m.
Ben Wouters says:
December 19, 2014 at 8:04 pm
I don’t have to calculate its lapse rate, it just is the same as all lines of equal potential temperatures: 9,8K/km.
Potential temps are temps you expect if LR = 10K/km. That is you are adjusting the temps from what is measured to what you expect according to your theory. Don’t you see a problem with that?
Why use a diagonal axis, not vertical/ horizontal as is normal with a graph?
Why not use measured T and h as here?
http://oceanworld.tamu.edu/resources/environment-book/atmosphere.html
Section headed structure
I suspect it is so incorrect statements such as:
“Actual vertical temperature gradients in the atmosphere are thus highly variable”
Can be made by St Andrews and those who refer to them
Roger Clague says: December 21, 2014 at 4:57 pm
” Potential temps are temps you expect if LR = 10K/km.”
Read AND understand:
http://en.wikipedia.org/wiki/Potential_temperature
http://en.wikipedia.org/wiki/Equivalent_potential_temperature
” Don’t you see a problem with that?”
Only problem I see is that you do not understand the concepts used.
Again http://www.atmos.illinois.edu/~snesbitt/ATMS505/stuff/09%20Convective%20forecasting.pdf
“Why use a diagonal axis, not vertical/ horizontal as is normal with a graph?”
see page 12, especially the first item. (actually page 8-12 are all relevant)
” Why not use measured T and h as here?”
measured T IS used. Altitude makes no sense, pressure is the relevant variable in the atmosphere, so a logarithmic pressure scale is used.
” I suspect it is so incorrect statements such as:
“Actual vertical temperature gradients in the atmosphere are thus highly variable”
Can be made by St Andrews and those who refer to them”
Incorrect?????????????
Which planet are you thinking of?
see http://weather.uwyo.edu/upperair/sounding.html
and see some profiles at various times and places.
Ben Wouters says:
December 19, 2014 at 8:04 pm
Every line of equal potential temperature has a slope of 9,8K/km BY DEFINITION, and they are all parallel to each other.
If you draw a profile of real temperatures like the Sahara one, and it is parallel to a line of equal potential temperature, I don’t have to calculate its lapse rate, it just is the same as all lines of equal potential temperatures: 9,8K/km.
Do the calculation yourself:
31C at -31m to -22C at 5430m
http://www.st-andrews.ac.uk/~dib2/climate/lapserates.html
I think the gradient of T/h is found using:
Horizontal and vertical axes and measured temperatures.
Not a diagonal axis and a vertical axis and “potential temperatures”
30C-5C/5km = 6K/km not 30C-(-23C)/5km = 10K/km
” http://ifaran.ru/old/ltk/Persona/Mokhov_pub/LapseRate-FAO06-IACP430.pdf “
LR is not adiabatic and never reaches 10K/km. It ranges from 5-6.9K/km for most of the surface.
To understand LR we need to look at raw data of T/h.
You think it’s OK to change T adiabatically, according to your theory and you also say h doesn’t matter, only pressure., another assumption.
A photon from Sol can, at any given instant, be either in transit between surfaces or molecules while in the atmosphere, or converted to heat. Not both. Re-emission cools the emitter. Eventually there is no receiver, and said photon or its descendant escapes to space, the Final Sink, leaving the planet unheated by its passage, except to the extent its wavelength has increased.
The GHE is responsible for an increase in the “in transit” pool of photons, but they can be treated as heat only when they leave the pool by absorption without instant re-emission.
Sats find linear rise in OLR with surface temperature rise, contrary to all GCMs; there are evidently few thermally adsorbed photons in an unenclosed atmosphere. No Maxwell’s Demon.