This article is part II of “A new Lunar thermal model based on Finite Element Analysis of regolith physical properties“, written primarily by gallopingcamel (Peter Morcombe), edited and prepared for WordPress by Tim Channon.
Figure 1 (click full size)
Modeling the Moon
A few months ago an analysis of the Moon’s equatorial temperature was posted here using two different types of engineering software. Tim Channon used SPICE circuit analysis software originally developed at Berkeley while I used Quickfield, a finite element analysis program developed by Tor Cooperative, a Russian firm, marketed outside Russia by Tera Analysis. In addition, several detailed comments were received from “br” who used LTSPICE from Linear Technology Inc.
Two very different methods. The results were identical.
Both Quickfield (in Student edition) and LTSPICE are freely available for download for those interested in replication or for further investigation.
Most “Climate Science” predictions cannot be falsified or reproduced by independent analysis and are therefore not “Science” in the generally accepted sense. In sharp contrast, the above modeling efforts are real science as we were able to reproduce the results of Ashwin Vavasada’s one dimensional linear model which in turn agrees closely with the Diviner LRE observations. Furthermore the analysis cannot be dismissed as “Curve Fitting” because it is based on measurable parameters such as the thermal and radiative properties of regolith.
Three different models have been validated so it is reasonable to use any or all of them to estimate surface temperatures in situations where direct measurement is difficult or impossible. Let’s start by looking at the surface temperature of an airless Earth using Quickfield.
Modelling an airless Earth
Would an airless Earth have the same temperature as the Moon? The short answer is “No” because the Earth rotates more rapidly than the Moon while its surface is dominated by oceans that have thermal properties quite different from lunar regolith. For clarity these two effects have been separated.
| DIVINER | MOON | MOON1 | EARTH | |
|---|---|---|---|---|
| Period (hours) | 709 | 709 | 24 | 24 |
| Surface | Regolith | Regolith | Regolith | Ice |
| Global average (K) | 197 | 196 | 209 | 234 |
| Equatorial av. (K) | 212 | 212 | 226 | 247 |
| Absorbance | 0.4-0.9 | 0.4-0.9 | 0.4-0.9 | 0.5 |
The first column shows the temperatures found by NASA’s Diviner LRE (Lunar Radiation Experiment). The remaining columns are results obtained using a model based on Quickfield. The second column corresponds to the Moon “as is”. The third shows the effect of reducing the Moon’s period to 24 hours and the fourth shows the effect of replacing regolith with ice.
Figure 2
No simple assumption can be made for Moon’s absorbance (1 – a). The absorbance is strongly dependent on the angle of incidence.
For convenience repeating figure 1
Figure 1 (repeat)
Referring to figure 1, the green plot shows Diviner LRE data published by NASA while the yellow plot shows model output. The excellent correlation could be achieved by curve fitting but the FEA model assumes the lunar regolith to consist of 50 layers, each ten millimeters thick and each with different thermal properties as measured by NASA missions (Apollo). The model assumes that the Moon’s surface emissivity = 0.95 in the thermal IR region.
The dark blue plot, MOON 1, shows the effect of reducing the Moon’s rotation period to 24 hours. Daytime temperatures don’t change much but night time temperatures are significantly higher.
The red plot, EARTH, shows the effect of replacing the Moon’s regolith with ice. Ice has a thermal conductivity that is orders of magnitudes greater than the upper layers of regolith so the surface temperature variations are much smaller. At the equator temperatures could exceed the freezing point of water so local melting is a possibility.
Thermal conductivity is a critical factor, lunar regolith has a very fine dust top layer, gravity is low lessening grain packing and also critically there is a hard vacuum which breaks the dominant convective heat transfer present where there is gas above a few torr. The result is an excellent insulator, minimally in contact fine grains reducing the area and increasing the path length for conduction.
Discussion
The quality of physics textbooks used in K-12 education and even in some universities is lamentable as shown by a study led by John Hubisz of NCSU. One might expect the textbook publishers to take advantage of what amounted to an in depth review of their products. Sadly that did not happen and the same errors have been repeated in later editions of most textbooks.
Universities around the world teach that the Greenhouse effect is 33 Kelvin based on the assumption that Earth’s average temperature is 288 K while the temperature would be 255 K if there was no atmosphere. The 255 K figure is mathematically correct as shown in this slide by Scott Denning, Monfort Professor of Climate Science, Colorado State University. The problem is that the assumptions made are unrealistic given that it would take a perfect thermal conductor to ensure an uniform surface temperature. An Albedo of 0.3 is not realistic for either regolith or ice.
The average temperature of an airless Earth should be 209 K if its surface was covered with at least 0.5 meters of regolith or 234 K if the surface was primarily made of ice.
Will physics textbooks get updated?………………….
Don’t hold your breath!
“br”, who wishes to remain anonymous, contributed heavily for part I but was was not involved in part II.
We have not provided turnkey code for part II, is similar to part I. Ask in public or privately if you need this.
Posted by Tim
Tallbloke's Talkshop 
‘The average temperature of an airless Earth should be 209 K if its surface was covered with at least 0.5 meters of regolith or 234 K if the surface was primarily made of ice.’
I take it this conclusion would not go down well with the ‘orthodoxy’.
I like the analysis. I am not as much of a fan of the Discussion.
“The quality of physics textbooks used in K-12 education … “
While this is interesting and important, it is not germane to this specific discussion about much more advanced topics.
“Universities around the world teach that the Greenhouse effect is 33 Kelvin based on the assumption that Earth’s average temperature is 288 K while the temperature would be 255 K if there was no atmosphere.”
The assumptions are a bit simplified, but that is pretty standard practice in science (it has even become a standard “joke” among physicist — http://en.wikipedia.org/wiki/Spherical_cow) . Very few scientist would consider the 33 K value anything other than a rough estimate based on the simple calculation that you linked to.
Specifically, the 255 K is the “blackbody equivalent temperature”. It assumes the current albedo. It assumes an emissivity of 1 for thermal IR. It assumes a uniform temperature. It does NOT assume no atmosphere per se, only that the atmosphere is transparent to IR. It does NOT assume the surface is bare rock nor that it is ice.
Compared to the “spherical cow”, these are pretty reasonable assumptions. The albedo is about 0.3 ± 10%. The emissivity is 5-10% lower than 1.0. The temperature is (mostly) uniform to within 10% (± 25-30 K) (largely because of fluid oceans and a fluid atmosphere).
The message certainly could be made clearer — that 33 K is only a rough estimate based on simplifying assumptions; that this effect includes water vapor & clouds, not just CO2 as part of “the greenhouse effect”; that changing any one aspect will lead to feedbacks that change other aspects. But overall, I don’t see these assumptions any less useful or informative than your assumptions of no atmosphere and then rock or ice for the entire surface of the earth.
OT but worth keeping an eye on
http://flux.phys.uit.no/cgi-bin/plotgeodata.cgi?Last24&site=tro2a&
Haven’t seen this strong negative Z at day time, hope that all the shaky areas are stable for next few days.
The real GHE using the flat plate SW collector, spherical emitter model is ~11 K.
You get this by calculating the mean surface temperature assuming no cloud or ice albedo, but with the seas in place so their emissivity is unchanged, for 341 W/m^2 emission to Space.
The 33 K claim is plain wrong because it assumes a single -18 deg C OLR emitter at 5 to 6 km altitude: this does not exist; the -18 deg c is a virtual, flux-weighted mean temperature of emission from zones ranging in altitude from 0 to ~20 km.
The original paper which made the single OLR emitter assumption, with zero experimental evidence of its existence should never have passed peer review: 1981_Hansen_etal.pdf
[mod] link = http://pubs.giss.nasa.gov/abs/ha04600x.html
That paper, plus Sagan’s incorrect aerosol optical physics and incorrect assumption of black body surface IR emission, has led to 33 years of utter science failure by Hansen, Lacis and Trenberth. i forgive Ramanathan because he’s a good experimentalist.
@ AlecM: The reviewers should have smelt a rat when they read the phrase ‘fabled northwest passage’ in the abstract – not very ‘scientific’ 😉
Alec says: “… assuming no cloud or ice albedo, but with the seas in place so their emissivity is unchanged, for 341 W/m^2 emission to Space.”
That is ANOTHER way to model the situation. It is interesting and useful, but it is no more “real” or “correct” than the models that Peter or Tim C or I discussed. ALL of them make assumptions counter to the “real” Earth. All of them give answers that are based on their own assumptions.
“The 33 K claim is plain wrong because it assumes a single -18 deg C OLR emitter at 5 to 6 km altitude “
No, the “standard model more properly assumes a single emitter at 0 km altitude, ie that the atmosphere is completely transparent to IR so that all IR comes directly from the surface. It then calculates that such a surface would be -18 C. It is the real Earth that has a -18 C temperature (on average) coming from an altitude of 5-6 km (on average).
What would be the effect of having an oxygen atmosphere on the moon?
Would it warm via conduction and convection from the surface?
Would that then increase the surface temperature?
How would the oxygen lose its heat to space?
This debate is important but it produces a sort of conundrum where one element of the argument is always hypothetical. Like the spherical cow, it always leaves a lingering doubt.
What value of total earth radiance do satellites measure and is there any agreement with the theory?
Tim F,
The -18C assumes a non-dynamic body when it is dynamic. There is an oversimplification by omitting rotation,
Peter indirectly mentions this “a perfect thermal conductor”. I have been talking about dynamic etc. for a long time without much traction. There is in reality only in-transition, not steady state. Day and night have fundamental effects.
Perhaps you can extend the maths.
I have something important to say about -18C next
The -18C problem
Long term readers may recall the extensive arguments surrounding N&Z (Nikalov and Zeller) first started on the Talkshop shortly after the police raid on Tallbloke.
A useful reminder can be found attached to a later blog article, ie. when many have had time to think and compute.
The correctness of the old article, which I was asked to prepare and post, was hotly disputed in comments. This raises the question of what are the correct mathematics?
“Kelvin Vaughan says:
What would be the effect of having an oxygen atmosphere on the moon?
…”
An atmosphere destroys the extremely thermally insulative surface layer of the moon. Yes it would be warmer.
To be fair the simulation earth ice surface cannot exist. Neither case is possible.
I looked briefly at the article. I am not impressed. There are some major flaws.
1) He concludes that ε = 1-a. This is wrong. Albedo, is the reflectivity of the incoming light (UV, visible, near IR). Emissivity, ε, is for thermal IR. There is no need for these two number to be related — just like there is no need for the reflectivity of blue light to be related to the reflectivity of red light. For example, black paint and white paint have very different albedos, but almost the same emissivity, which could not work with his equation.
This leads to a very wrong equation 4. At some level, this is a case of “two wrongs make a right”. What he calls “emissivity” is actually the “visible light absorbance” which is indeed (1-a). What most people call “emissivity” (ie the emissivity for thermal IR) is left out of his analysis. However this means he is effectively assuming the emissivity is 1.0. which is what many people do.
2) The integration looks good.
3) I think he messed up a couple things with the T_night.
* I get a much smaller irradiance from the earth back to the moon — closer to 0.04 W/m^2 rather than 0.4 W/m^2. That would cut the temperature down from ~ 50 K to ~ 30 K. Furthermore, that is only for a “full earth” for a small part of the month, so only a few nights would get as warm as this — most would be MUCH colder (ignoring heat capacity up until this point).
* Heat capacity works both ways. If heat capacity helps keep the surface warm at night, it will also help keep it cool during the day.
4) I agree that what he calls the “IPCC approach” gives a number that is clearly much larger than the observed results. But everyone who looks into this in detail knows that this is a MAXIMUM possible temperature, which is only achieved for an object with an entirely uniform temperature. The less uniform the temperature, the lower the average will be. For the earth that is not a bad assumption — the rapid rotation and flows of fluids keep the temperatures within ~ 10% of the average value. But the moon — with slow rotation and no atmosphere or ocean — will be much cooler than the simple estimate of 270 K. This is not a big surprise.
Kelvin Vaughn asks:
“What would be the effect of having an oxygen atmosphere on the moon?”
We may be thinking along the same lines. My goal is to find models that explain observations with reasonable accuracy. IMHO there are now several models that explain average surface temperatures on the Moon complete with detailed diurnal variations to an rms accuracy of better than one degree Centigrade at any latitude and by extension other airless bodies.
The next step is to add atmospheres that may consist of oxygen, nitrogen, carbon dioxide and other gases in a wide range of compositions. The work of Ramanathan and Pierrehumbert has been taken to the next level by Robinson & Catling. They have created a model that explains the observed temperature gradients in all seven bodies in the Solar System that have significant atmospheres. Thus far I have only had time to look at Titan. R&C’s analysis improves on Nikolov and Zeller’s “Unified Theory of Climate” significantly:
Currently I am looking at R&C’s analysis of Venus and Earth. They have been most helpful so this could proceed quite quickly. Without their help it may take a while to independently create a model that is at least equal to theirs given the fact that I have no funding. While I was able to reproduce Vavasada’s lunar model without his help, this is a much tougher problem than modeling airless bodies. It seems likely that my FEA software can create models that include cloud layers and phase changes from “First Principles”; something that R&C have yet to tackle.
Robinson & Catling are trying to construct models of planetary atmospheres that allow them to predict temperatures from planetary stratospheres down to the surface for any arbitrary gas composition. Their goal is to predict temperatures on exo-planets.
Click to access Catling.pdf
Please look at “Slide 35” in the above presentation. In a single slide, James Hansen affirms the Arrhenius (1896) estimate of 4.5 K/doubling of CO2 and the last seven glaciations in terms of a “Sensitivity Constant” of 16 K/halving of CO2 concentration. It has been obvious for years that Hansen and his apostles at GISS are zealots rather than “Scientists”. Ice core evidence says temperature drives CO2 rather than the reverse hypothesis promoted by Hansen et al.
Tim F. 7:54pm reply to AlecM 6:04pm: +1
Earth’s surface real GHE as commonly defined is measured at approx. 33K. AlecM has to find a reason to cast doubt on the measurements not the simplified theory calculations which reasonably and justifiably match the measurements.
******
Tim C. 2:20am: The simplified calculations supporting the measured 33K do not use the day night rotation period. Energy in – energy out is unaffected by rotation period – i.e. is the same for all rotation periods so simple Tmean will be independent of rotation. The diurnal surface T max. and min. around the mean will change with rotation periods. Faster spins reducing the range. Indeed, further down in regolith the max. and min. converge to the mean.
Earth’s Tmean simplified calculations do not make nor need the assumption of a perfect thermal conductor at the surface.
******
Tim C. 2:20am & gc top post: N&Z make the same assumption as the top post: “The model assumes that the Moon’s surface emissivity = 0.95 in the thermal IR region.”
This key assumption is inherent curve fitting to get a result close to the Diviner answer.
Actual regolith samples have been measured on Earth in the lab for emissivity and found to be much less than 0.95 meaning regolith is much more reflective AND diffractive than natural earth materials. As the top post discusses – much of the regolith is powder of diameter close to wavelength of interest. Invoking the standard Planck distribution therefore is ruled out even by Max Planck’s original paper.
Proper modeling of regolith emissivity needs to be inclusive of an extensive existing literature search and discussion on the subject not a simple assumption of regolith epsilon = 0.95 based on Earth’s natural materials which haven’t been pummeled for eons as has the airless moon’s regolith.
******
gc 4:39am: ”R&C’s analysis improves on Nikolov and Zeller’s “Unified Theory of Climate” significantly.”
+1 R&C have a concise paper explaining all planets and a moon with thick atm.s T profile. Very useful IMO. You will go far building on their work. Next step: start by googling something like “regolith emissivity” (easy but lot of reading) and figure out how to modify Planck distribution for diffraction in moon’s powdery surface (very hard but there is a text book on the subject from the early 1980s).
AlecM says: August 27, 2014 at 6:04 pm
The real GHE using the flat plate SW collector, spherical emitter model is ~11 K.
You get this by calculating the mean surface temperature assuming no cloud or ice albedo, but with the seas in place so their emissivity is unchanged, for 341 W/m^2 emission to Space.
The 33 K claim is plain wrong because it assumes a single -18 deg C OLR emitter at 5 to 6 km altitude: this does not exist; the -18 deg c is a virtual, flux-weighted mean temperature of emission from zones ranging in altitude from 0 to ~20 km.
The original paper which made the single OLR emitter assumption, with zero experimental evidence of its existence should never have passed peer review: 1981_Hansen_etal.pdf
[mod] link = http://pubs.giss.nasa.gov/abs/ha04600x.html
That paper, plus Sagan’s incorrect aerosol optical physics and incorrect assumption of black body surface IR emission, has led to 33 years of utter science failure by Hansen, Lacis and Trenberth. i forgive Ramanathan because he’s a good experimentalist.
Alec please replace the name Lacis with Schmidt, Andrew as you know is innocent. GavIn, however has a permanent bulls eye on his face!
“No simple assumption can be made for Moon’s absorbance (1 – a). The absorbance is strongly dependent on the angle of incidence.”
Tim, If you would go measure, any where on this Earth at every wavelegth on this Earth at 45 degrees from normal, emissivity is 0.5 reflectivity is 0.5., never your stupid computer generated geometrical bull shit. We measure, because geometry always lies, except if you are being extremely careful, then geometry never lies! No sense in giving a fool the upper hand.
Tim, Do you have any idea of how to measure emissivity at any wavelength? The measurement is allways AW shit!
As (an almost) black body radiator the earth radiates IR photons. Consider such a photon approaching the earth, where would the surface of the earth be encountered?
Opacity, as far as the photon is concerned, starts when it encounters the greenhouse gases, of which the most abundant is water vapour. 95% of this gas is found up to about 5km from the ground.
If this is the effective surface for our theoretical incoming IR photon, then perhaps it is also the effective surface for an outgoing IR photon. A black body radiating from this height would be seen to have a temperature of -18 Celsius.
Schrodinger’s Cat says: August 28, 2014 at 7:55 am
“Consider such a photon approaching the earth, where would the surface of the earth be encountered?”
See http://en.wikipedia.org/wiki/Sunlight#mediaviewer/File:Solar_Spectrum.png
Solar UVB/C is intercepted by ozone (and breaks it apart) in the stratosphere
Solar IR is intercepted by H2O (and CO2) in the troposphere.
Reflection (albedo) mostly in UV and visible.
The point is that as far as IR BB radiation is concerned, the effective surface of the earth is at an altitude of 5km. The surface of the earth has our ambient temperature as a consequence of the lapse rate.
The GHG trick is not extra conservation of heat, it is that in the case of IR radiation the effective surface of the earth is 5km up. IR photons find water vapour as opaque as solid mountains or the oceans.
The GHG effect is the mistaken belief that the black body surface is at sea level.
The idea of GH warming is due to assuming wrongly that the BB radiation is from the physical earth’s surface when that is impossible if the atmosphere contains GHG and is therefore opaque to IR.
Finally, consider the earth if there were no GH gases. The atmosphere would be transparent to IR all the way down to the surface. The BB radiative surface would coincide with the actual surface and the BB temperature would show about 15 Celsius.
Tim Folkerts says: August 28, 2014 at 4:25 am
“3) I think he messed up a couple things with the T_night.
* I get a much smaller irradiance from the earth back to the moon — closer to 0.04 W/m^2 rather than 0.4 W/m^2. That would cut the temperature down from ~ 50 K to ~ 30 K. Furthermore, that is only for a “full earth” for a small part of the month, so only a few nights would get as warm as this — most would be MUCH colder (ignoring heat capacity up until this point).”
Earthshine can only have an effect on the side that continuously faces the earth.
Imo the effect overall is negligible
To me the surprisingly high night temps are caused by moons hot core and the resulting geothermal flux at the surface. Some deep craters near the poles never receive any sunshine (and no earthshine), yet their temps are ~25K or so.
This can be simply explained by a geothermal flux of 0,05 to 0,1 W/m^2. Same values as for our earth.
A planet / moon in outer space and a geothermal flux of eg 0,05 W/m^2 would have an all around temperature of ~30K. Drilling towards the core, the temperature would rise, same as on earth.
Schrodinger’s Cat says: “Finally, consider the earth if there were no GH gases. The atmosphere would be transparent to IR all the way down to the surface. The BB radiative surface would coincide with the actual surface and the BB temperature would show about 15 Celsius.”
You need one more thought to complete this argument. If the surface were emitting 15 C BB radiation through that IR transparent atmosphere, there would be MUCH more energy leaving the Earth’s surface than arriving at the Earth’s surface. This would mean that the surface would necessarily start to cool. It would cool until balance was once again restored, which would be somewhere around -18 C (assuming that other factors like albedo remained the same).
So even under your arguments, it is very clear that the atmosphere’s ability to absorb & radiate IR is what allows the surface to be ~ 15 C rather than ~ -18 C.
Tim, you might be right, but then we don’t have a GHG free atmosphere, so the outgoing radiation is not lost from the surface as BB radiation. It does get transported upwards by convection, collisions, etc, until it reaches an altitude where the atmosphere is transparent to IR. The lapse rate defines the temperature of the black body at that altitude.
I think this is a situation where hypothetical scenarios don’t help. The GHG hypothesis does the same by assuming that the surface of the earth can be a BB when the atmosphere above it is opaque.
Schrodinger’s Cat says: “The lapse rate defines the temperature of the black body at that altitude. ”
That’s pretty much backwards. Energy balance determines the temperature at the “effective radiating surface”. That temperature must be around 255 K for the IR out from the effective radiating surface to equal the sunlight absorbed.
The lapse rate determines the temperature at OTHER altitudes. If the effective radiating surface is 5-6 km up (due to clouds & GHGs), then that level would be ~ 255 K and the physical surface must be ~ 30 K warmer due to the lapse rate. If the effective radiating surface is the physical surface (ie the atmosphere can’t emit/absorb thermal IR), then the physical surface would be ~ 255 K.
What are the parameters for the simulations where there is a transparent atmosphere and the answer is -18C?
If the answer is -18 C for the average temperature, the “standard” parameters that work are
* uniform temperature everywhere
* albedo = 0.3 (ie an average of 240 W/m^2 absorbed sunlight)
* emissivity = 1 for thermal IR
Many other choices could also lead to this temperature.
“uniform temperature everywhere”
Invalid for a sphere and point source.
@ Ben,
I hadn’t even thought about geothermal energy (“lunothermal”???). In any case, the night time temperatures can have wide variations with very minor changes in power. As such, any “average temperature” is rather suspect when one side could be 2.7 K or 50 K or even100 K with only a small change in the assumptions.
tchannon says: August 28, 2014 at 6:50 pm
“Invalid for a sphere and point source.”
Yes. Also the earth is not a static object. It is spin spinning on it’s axis and is following an elliptical orbit which changes the distance to the sun throughout the year. A further complication is that most of the landmass is in the NH which alters the way the system heats and cools in a none-repeating pattern. Difficult to try to come up with a simple linear model for Earth temperature that matches reality.
I find even the measured temperature of the moon is a puzzle. We are told it has a maximum temperature of 100C, but with little or no atmosphere it should be radiating all the energy it receives instantaneously into space. Anyone have a good theory as to why this doesn’t happen?
Tim C,
Experimentally, the Earth IS close to a uniform temperature (give or take ~ 10%)
Experimentally, the surface of the Earth is close to a blackbody (according to most people, anyway).
As such, estimating the “average temperature” as if the temperature were uniform and the emissivity were 1 is a “pretty good approximation” rather than “invalid”. Sure, the result will be better if a more accurate set of temperatures were used, but as a starting point for discussions, 255 K is not a bad estimate for the earth without “back-radiation” (all else being equal).
I assume that “255 K” is a simple by handy estimate of the average temperature. Non-uniform sunlight & temperatures will lower this average (as your calculations above confirm). So 255 K is an UPPER LIMIT on the temperature without GHGs. The warming is AT LEAST 33 K from the IR properties of the atmosphere.
Tim Folkerts says: August 28, 2014 at 12:28 pm
Nothing pertaning to the Moon
“You need one more thought to complete this argument. If the surface were emitting 15 C BB radiation through that IR transparent atmosphere, there would be MUCH more energy leaving the Earth’s surface than arriving at the Earth’s surface.”
This Earth is at no wavelength ever close to black body surface. The Earth’s atmosphere (all of it) replace the surface with a more effective radiator. You and your ClimAstrology cohorts have never even measured incoming and or exiting EMR. Nor can you decribe a meathod for doing so. All claims are fradulent fantasy.
“This would mean that the surface would necessarily start to cool. It would cool until balance was once again restored, which would be somewhere around -18 C (assuming that other factors like albedo remained the same).”
Because the atmosphere is creating the EMR exitance to space the, surface is kept at a higher temperature via the measured lapse rate, partialy adiabatic, never any atmospheric CO2.
Tim Folkerts says: August 28, 2014 at 6:28 pm
More of his toy planet, nothing of the moon!
“If the answer is -18 C for the average temperature, the “standard” parameters that work are
* uniform temperature everywhere
* albedo = 0.3 (ie an average of 240 W/m^2 absorbed sunlight)
* emissivity = 1 for thermal IR”
That may be for your toy planet in a Playstation-64 somewhere. None of those parameters are anywhere close to that measured anywhere on or about this planet Earth! All fraud.
Tim Folkerts says: August 29, 2014 at 1:02 am
Still nothing of the moon!
“Tim C, Experimentally, the Earth IS close to a uniform temperature (give or take ~ 10%)”
So plus or minus 30 kelvins! That gives a 2.2:1 change in radiance and exitance in the near direction to the Sun. That can only be uniform for purpose of intentional fraud.
“Experimentally, the surface of the Earth is close to a blackbody (according to most people, anyway).”
Most people huh! Please show where even one location on Earth’s surface where the spectral UV, visible, and IR, BRDF has been surveyed. Most people realize the surface is nothing like a black body as you fraudulently claim!
“As such, estimating the “average temperature” as if the temperature were uniform and the emissivity were 1 is a “pretty good approximation” rather than “invalid”. ”
Such a claim is not only invalid, it is intentionally fraudulent as above, made only by those with PHD’s that “did know better” or” should have known better” by training.
“Sure, the result will be better if a more accurate set of temperatures were used, but as a starting point for discussions, 255 K is not a bad estimate for the earth without “back-radiation” (all else being equal).”
What “back radiation”? Your intentional misuse of the Schuster Schwarzschild two stream approximation? Please show even one paper in the last 35 years that correctly describes the generation and propagation of IR EMR flux, in this Earth’s atmosphere. Academic fantasy.
“I assume that “255 K” is a simple by handy estimate of the average temperature. Non-uniform sunlight & temperatures will lower this average (as your calculations above confirm). So 255 K is an UPPER LIMIT on the temperature without GHGs. The warming is AT LEAST 33 K from the IR properties of the atmosphere.”
Please show any evidence that the radiatively induced surface temperature of this Earth is affected by anything but measured lapse rate. Higher lapse rate, higher surface temperature everywhere. Gravitationally induced higher temperature.
Will, please tone down what you write. This is far from the first time you look like you are attacking.
tchannon says: August 29, 2014 at 4:28 am
Will, please tone down what you write. This is far from the first time you look like you are attacking.
Ok I’ll let others attack the post modern Arrogant Academics
Thanks (tchannon) for that Sweger link. Like Sweger I welcomed the entry of physicists into the climate debate. In my optimistic imagining, the mumbo jumbo would be driven out by calculations that explained observations in terms of equations based on physical constants.
When I applied for a job in the Duke university physics department in 1990 I was shocked to find how small the Physics building was. When I mentioned that in my interview I was told that the tiny building was also the home of Mathematics department. Given that I studied physics at the Cavendish Laboratory in Cambridge (UK) the scale was underwhelming.
Over time I came to realize that while the Duke physics department was (is) tiny, it had the highest “brains to weight ratio” of any organization I ever worked for. Two winners of the FEL prize (John Madey and Vladimir Litvinenko) and several people who have gone on to head large departments in other universities.
I retired in 2002 and now work part time teaching quantum electro-optics. While I don’t teach at Duke or UNC I run three courses a year at NCSU which gives me the opportinity to drink adult beverages with my ex-colleagues at Duke. It gives me great satisfaction to find that Duke physicists speak out eloquently against loony energy policies based on the phony science that used to be called CAGW. Here are some of the people I am talking about:
Edward Bilpuch: http://today.duke.edu/2012/09/bilpuch
Nicola Scafetta: http://people.duke.edu/~ns2002/
Robert G. Brown: http://www.phy.duke.edu/~rgb/
‘Tim, If you would go measure, any where on this Earth at every wavelegth on this Earth at 45 degrees from normal, emissivity is 0.5 reflectivity is 0.5., never your stupid computer generated geometrical bull shit. We measure, because geometry always lies, except if you are being extremely careful, then geometry never lies! No sense in giving a fool the upper hand.’
Will
Do you have any understanding of the ‘cosine rule’? Do you understand the principles of how an infrared thermometer works?
If you did, then you wouldn’t be making foolish statements. It undermines your credibility about other things you may be quite conversant in.
[mod: alex tone down attacking others, disagree politely… in public –Tim C]
@Tim Folkerts:
The radiative equilibrium mean temperature of the airless Earth with Space as the opposite radiative emitter would, because there are no clouds or ice, be for 341 W/m^2. The answer would be between -2 and +5 deg C depending on your assumptions about IR emissivities. Your claim that it would be -18 deg C, implying 238.5 w/m^2 warming, is one of the clever lies in the IPCC scam.
Will, let’s consider a single sentence you wrote:
“Because the atmosphere is creating the EMR exitance to space the, surface is kept at a higher temperature via the measured lapse rate, partialy adiabatic, never any atmospheric CO2.”
How does the atmosphere create that EMR exitance (ie thermal IR escaping to space)? The largest single source is clouds, which radiate pretty much like a blackbody. But the other major source is GHGs. Gaseous H2O and CO2 both radiate rather effectively in specific wavelength bands, contributing to the EMR exitance. All of this is easily measured by satellites orbiting the Earth.
Since CO2 is ALWAYS a source of EMR exitance (along with clouds and other GHGs), then by your own words, it is ALWAYS a cause (along with clouds and other GHGs) of the surface being kept at a higher temperature.
Tim Folkerts,
That’s right. This also means that surface temperature is determined by both the altitude of the effective radiating surface and by the lapse rate. Models assumes that radiative phenomena have no effect on the lapse rate (the concept of GHG forcing is its corollary).
Can anyone justify this assumption?
I think it’s impossible.
@Tim Folkerts: I will answer your post, specifically “Since CO2 is ALWAYS a source of EMR exitance (along with clouds and other GHGs), then by your own words, it is ALWAYS a cause (along with clouds and other GHGs) of the surface being kept at a higher temperature.”
The partial exitance for well-mixed CO2 15 micron IR in the OLR is set by the temperature of the tropopause, defined by where convection ceases to operate as the main atmospheric heat transfer mechanism. The physics is different for non well-mixed H2O.
Low level atmospheric processes provide sufficient negative feedback to reduce CO2 climate sensitivity to zero on average.
Tim Folkerts says: August 28, 2014 at 10:11 pm
“I hadn’t even thought about geothermal energy (“lunothermal”???). ”
I prefer lunAthermal.
Suggest you start thinking about geothermal heat, -fluxes and gradients, since I’m convinced that they hold the answer to the question why our surface temperatures are so much higher than the sun can account for.
See eg http://diviner.ucla.edu/science.html
On the moon the night temps converge towards ~50 or 60 K iso 2,77K, even after 2 earth weeks of cooling. So the moon seems to have a “base” surface temp. of around 50K, before the sun starts adding its energy.
Earths base temp.is much higher.
Tim Folkerts says: August 29, 2014 at 1:02 am
“So 255 K is an UPPER LIMIT on the temperature without GHGs. The warming is AT LEAST 33 K from the IR properties of the atmosphere.”
Actually the warming is AT LEAST 33K, but not from the atmosphere.
The atmosphere makes it possible that the average solar 240 W/m^2 can MAINTAIN the average 290K surface temp., since the atmosphere reduces the energy loss to space to the same value. It is not the REASON for the higher temps.
A couple quick comments …
@Ben August 29, 2014 at 4:16 pm
The coldest place on the moon appears to be ~ 26 K (http://sservi.nasa.gov/articles/lro-finds-coldest-place-on-the-moon/), which corresponds to ~ 0.026 W/m^2. This is ~ 1/2 of your “base temperature”. Furthermore, it would be better to think of this as a “base power”. So consider if we have some spot on the moon with 1000 W/m^2 of sunlight, so that it reaches ~ 360K. The lunathermal heat will not warm the surface by the 25-50 K above 360 K. Rather it will raise the power by 0.025 W/m^2, resulting in a warming of ~ 0.02 K above 360 K.
Since both geothermal & lunathermal heat flows are way below 1 W/m^2, they are nearly insignificant compared to the 10’s or 100’s of W/m^2 related convection, sunlight, IR, etc.
@ AlecM August 29, 2014 at 2:26 pm
Feedbacks are certainly important and an issue that many people are trying to figure out. I have seen estimated climate sensitivities of up to ~ 4 C and down to ~ 0.4 C. You are about the only person I have seen that claims the value is exactly 0.0 C. How are you so certain, when even skeptical scientists give numbers that disagree with your conclusion?
@ Will Janoschka August 29, 2014 at 3:40 am
Here are a couple IR spectra taken from satellites — some actual data!
It is clear that:
1) the envelop for the curves is pretty close to that of a BB at surface temperature. Hence the surface is pretty close to a BB.
2) the radiation from the GHG’s take “bites” out of the spectrum, and hence these GHGs are LESS effective at radiating to space than the ground, not more effective as you claim.
There are a number of reasons to assume lunar geologic heat is very slight, but neither should DIVINER data be taken as gospel. The designers did their best with instrumentation, will still be in error.
Given the lunar surface is highly insulative the Apollo findings which included an excess of radioactive upper material, is insufficient, will not show at the surface.
Radioactive decay might be unevenly distributed meaning there would be regional variations in this source.
People like Tim Folkerts seem to thrive on turning the whole ‘GHG’ issue on its head.
The so-called ‘GHGs’ (the IR-active gases) are there to enable the atmosphere to cool to space, from whatever altitude (Earth’s final/total radiative flux to space is a cumulative one, after all).
The mere presence of a massive atmosphere (a fluid in a gravity field) on top of a solar-heated surface makes it impossible for that surface to accomplish radiative balance with its heat source (the Sun), because a significant share of the absorbed energy is brought up into the atmosphere by convection. The presence of IR-active gases in that atmosphere in turn makes it possible for it to get rid of this energy efficiently to space. They’re not required for the surface energy to end up in the atmosphere in the first place. Remove them and you would force the surface to warm to maintain its convective loss to a warming (but not adequately simply ‘thermally’ expanding) atmosphere.
Kristian says: “The mere presence of a massive atmosphere (a fluid in a gravity field) on top of a solar-heated surface makes it impossible for that surface to accomplish radiative balance with its heat source (the Sun), because a significant share of the absorbed energy is brought up into the atmosphere by convection.”
You miss a key fact here. That “significant share of the absorbed energy” can ONLY be brought up into the atmosphere if the atmosphere has a means to get rid of said energy.
If a “significant share of the absorbed energy” kept going into the atmosphere with no way to lose energy from the atmosphere, the atmosphere would keep warming! Of course, what really would happen is that as the atmosphere will warm and the temperature difference between surface and atmosphere will diminish. The “significant share” will diminish, asymptotically approaching “zero share of the absorbed energy” as we approach a steady-state condition.
And what will the conditions be at that point? Well, now that no more energy is required to further warm the atmosphere, all of the incoming solar energy will again be emitted as thermal IR straight to space (for our hypothetical IR-transparent atmosphere). So the surface will average ~ 255 K (give or take a little depending on the exact scehario) and the atmosphere will be about the same.
The “mere presence” of a massive atmosphere is not enough! That massive atmosphere MUST have a way to shed energy (which BTW also blocks the surface from directly shedding energy to space) or the surface will never warm above~ 255 K.
Alex says: August 29, 2014 at 7:46 am
(‘Tim, If you would go measure, any where on this Earth at every wavelegth on this Earth at 45 degrees from normal, emissivity is 0.5 reflectivity is 0.5., never your stupid computer generated geometrical bull shit. We measure, because geometry always lies, except if you are being extremely careful, then geometry never lies! No sense in giving a fool the upper hand.’)
“Will
Do you have any understanding of the ‘cosine rule’? Do you understand the principles of how an infrared thermometer works?” “If you did, then you wouldn’t be making foolish statements. It undermines your credibility about other things you may be quite conversant in.”
What you call cosine rule is termed Lambert”s Law of opaque surfaces with constant
absorptivity/reflectivity at each direction from normal. Lambert’s Law relates absorptivity (a surface property) with TimC’s word absorbance (flux absorbed, Watts/m^2). With almost all natural physical surfaces “absorptivity” also drops with angle from normal, they become highly reflective at grazing angles. Lambert’s Law cannot be used in such cases!
Infrared thermometers work, “when pointed at a surface at higher temperature than themselves” measure flux in the direction toward the instrument. When pointed at a lower temperature surface, always measure flux in the direction of that lower temperature surface. That flux one way or the other is always a linear function of the difference in opposing radiance (normalized field strength) resulting in a single Poyinting flux at that frequency. If the instrument also measures its own temperature then from that one way flux, the temperature of the “pointed at” surface may be approximated. This exact information is always contained within the technical information regarding that specific instrument.
“If you did, then you wouldn’t be making foolish statements. It undermines your credibility about other things you may be quite conversant in.”
[mod: alex tone down attacking others, disagree politely… in public –Tim
Please point out my foolish statements.
Tim Folkerts says: August 29, 2014 at 11:02 am
Will, let’s consider a single sentence you wrote:
(“Because the atmosphere is creating the EMR exitance to space the, surface is kept at a higher temperature via the measured lapse rate, partialy adiabatic, never any atmospheric CO2.”)
“How does the atmosphere create that EMR exitance (ie thermal IR escaping to space)? The largest single source is clouds, which radiate pretty much like a blackbody. But the other major source is GHGs. Gaseous H2O and CO2 both radiate rather effectively in specific wavelength bands, contributing to the EMR exitance. All of this is easily measured by satellites orbiting the Earth.”
H2O radiates extensively from 5-8 Microns and 14- 200 Microns, more than enough to regulate the whole temperature if this Earth Increasing CO2 only increases exitance in the stratosphere where it becomes thr fourth dominant gas. Remember every part of the atmosphere is at a higher temperature than it would be via only EMR maintaining radiative equilibrium. This is do to convection and the change from latent heat to semsible heat of water vapor.
The now beyond radiative equilibrium atmosphere at any altitude must always add to the exitance from below, (Kirchhoff’s Laws of thermal radiation) that is passed through that mass without attenuation without work, or increase in entropy. The atmosphere accumulates exitance all the way to 220 km, where that exitance may exceed any Solar absorptance. Depending only on atmospheric WV.
“Since CO2 is ALWAYS a source of EMR exitance (along with clouds and other GHGs), then by your own words, it is ALWAYS a causealong with clouds and ot (her GHGs) of the surface being kept at a higher temperature.”
The atmosphere is never a cause for higher surface temperature! The atmosphere is a “more” effective emiter of EMR to space than the surface can be! The most obvious feature of the atmosphere and especially clouds is they have cross sectional area and thus are not restricted via Lambert’s Law to an effective solid angle of one PI steradian. It is only the Gas Laws and Gravatational attraction that creates an adiabatic temperature increase at lowdes in every body with sufficient atmosphere in this whole Solar system including this Earth. There is no Green House Effect. There are no Green House Gases! All a deliberate scam.
Will says:
“The atmosphere is never a cause for higher surface temperature!”
But Will says:
““Because the atmosphere is creating the EMR exitance to space the, surface is kept at a higher temperature … ”
Perhaps Will can explain this contradiction.
Will says: :With almost all natural physical surfaces “absorptivity” also drops with angle from normal, they become highly reflective at grazing angles.”
This is true for smooth and/or polished surfaces. Glass reflects very well at glancing angles. So does water. So does polished granite.
But weathered granite is not particularly reflective at any angle. Nor is wood. Nor is grass. Rough surfaces are less reflective. As they get really rough, they reflect about as well at any angle, ie they become much closer to “Lambertian”. Thus “real” (ie rough) surface DO rather closely follow Lambert’s Law/
Smooth water is quite reflective at grazing angles, but the reflectivity only reaches 50% at angles at least ~ 82 degrees from normal — not 45 degrees as you claimed. Furthermore, water on the oceans is rarely calm, which reduces the ability of the light to hit at a grazing angle. Once again, any rough surface — including rough water — is less reflective at glancing angles than similar smooth materials.
“The atmosphere is a “more” effective emiter of EMR to space than the surface can be! The most obvious feature of the atmosphere and especially clouds is they have cross sectional area and thus are not restricted via Lambert’s Law to an effective solid angle of one PI steradian. “
I doubt I will be able to explain this to you without pictures, but I’ll take a quick shot.
Imagining looking (from far away) perpendicularly at a 1mx1m square of a surface that follows Lambert’s Law. It will emit some amount of thermal IR. Now imagine looking at a 2mx1m rectangle, but tilt it 60 degrees. The cosine law say you will get half as much thermal IR from each square meter (cos(60) = 0.5), but there are twice as many square meters, so the total reaching you from those 2 tilted square meters is the same as had been reaching you from the 1 perpendicular square meter. That should be intuitive — from far away both subtend the same field of view.
Continuing this argument, both a disk seen perpendicularly and a sphere of the same radius cover the same field of view. Thus the thermal IR will look the same from both; the thermal IR does not “fade to black” as you look toward the rim of the sphere. Lambert’s Law does NOT “restrict” the output the way you seem to think.
Tim Folkerts says: August 29, 2014 at 8:28 pm
@ Will Janoschka August 29, 2014 at 3:40 am
“Here are a couple IR spectra taken from satellites — some actual data!”
-Skip Tim’ homoginized spectral “normal” radiance taken “perhaps”, from some unknown altitude with cloudless skies over the tropical ocean.-
Note: Radiance only, no radiation or flux, no radiance data for greater than 25 microns
Measured “nadir” emissivity for the ocean surface from 8-13.5 microns is 0.945! The measured emissivity at 45 degree from normal is 0.496 for a emittance to space of less than half of a black bodyn at that angle. It gets much worse with increasing angle from normal. In addition 66% of the sky is covered with clouds permitting no direct exitance to space from the surface. The spectrum from 14 – 200 microns provides 75% of the exitance to space with its two PI steradian effective solid angle.
“It is clear that:
1) the envelop for the curves is pretty close to that of a BB at surface temperature. Hence the surface is pretty close to a BB.
2) the radiation from the GHG’s take “bites” out of the spectrum, and hence these GHGs are LESS effective at radiating to space than the ground, not more effective as you claim.”
It is clear that:
1. Such charts are used only to confuse individuals about the operation of this atmosphere.
2. ClimAstrologists demonstrate no knowledge of the atmosphere nor any understanding of this atmosphere’s effect on electromagnetic radiation at any wavelength.
3. There are absolutely no valid claims of Green house effect or of any so called Green house gases, which cannot be identified.
4. ClimAstrologist have no viable evidence that EMR “in or out”, plays any important part in determining the surface temperature of this planet. It does create interesting weather from time to time. It does create a wonderful biosphere.
One last comment to Will, who said: “H2O radiates extensively from 5-8 Microns and 14- 200 Microns, more than enough to regulate the whole temperature if [I assume that is a typo for “of”] this Earth”
I will simply refer you to the experimental result shown in the 2nd satellite images above. In the atmospheric window, the thermal IR from the ground is strong — close to that calculated for a BB @ 295 K. The emissions in the ranges you name for H2O are LESS THAN the emissions expected from the 295 K surface. Hence H20 “regulates” the temperature by REDUCING the outgoing IR and thereby causing the whole system to warm up! Remove that H2O and MORE thermal IR would leave, cooling the planet.
Tim Folkerts says: August 30, 2014 at 3:05 am
(“Will says: “The atmosphere is never a cause for higher surface temperature!”)
In anwer to your rediculous claims of “back radiation” from the atmosphere.
But Will says: ““Because the atmosphere is creating the EMR exitance to space the, surface is kept at a higher temperature … ”
Perhaps Will can explain this contradiction.
Easy you have a “contradiction” because refuse to read.
Wiil, (full quote)
“Because the atmosphere is creating the EMR exitance to space the, surface is kept at a higher temperature via the measured lapse rate, partialy adiabatic, never any atmospheric CO2.”
Gravity, not atmosphere, increases surface temperature.
Why must you use any opportunity to sow confusion?
Tim Folkerts says: August 30, 2014 at 4:08 am
Will says: (“With almost all natural physical surfaces “absorptivity” also drops with angle from normal, they become highly reflective at grazing angles.”)
“This is true for smooth and/or polished surfaces. Glass reflects very well at glancing angles. So does water. So does polished granite. But weathered granite is not particularly reflective at any angle. Nor is wood. Nor is grass. Rough surfaces are less reflective. As they get really rough, they reflect about as well at any angle, ie they become much closer to “Lambertian”. Thus “real” (ie rough) surface DO rather closely follow Lambert’s Law”
You measured these materials at what wavelengths using what procedures?
Smooth water is quite reflective at grazing angles, but the reflectivity only reaches 50% at angles at least ~ 82 degrees from normal — not 45 degrees as you claimed. Furthermore, water on the oceans is rarely calm, which reduces the ability of the light to hit at a grazing angle. Once again, any rough surface — including rough water — is less reflective at glancing angles than similar smooth materials. come highly reflective at grazing angles.”
Same question You measured what when how! Your silly graphs of nothing are again an attempt to confuse. An amplitude modulated signal, at some wave length incident atn 45 degrees from normal detected at the complementery 45 degrees will have that modulation reduced by how much? Who witnessed this precise measurement?
Tim Folkerts says: August 30, 2014 at 4:08 am
“Continuing this argument, both a disk seen perpendicularly and a sphere of the same radius cover the same field of view. Thus the thermal IR will look the same from both; the thermal IR does not “fade to black” as you look toward the rim of the sphere. Lambert’s Law does NOT “restrict” the output the way you seem to think.”
A total misunderstanding of every precept of spherical geometry. The disk is flat, not spherical, normal to the disk its apparent area is equal to the cross sectional area of the sphere of the same radius. At all other directions the apparent area of the disk reduces by cos (theta) and becomes zero at 90 degrees. The sphere however retains the same cross sectional area, it is isotropic (same on all directions). The difference is that the disk, has for exitance for that area into a maximum of one PI steradians into each hemisphere, while the sphere has a cross-sectional exitance into two PI steradians into the same hemisphere. Grade school geometry.
Tim Folkerts says: August 30, 2014 at 4:20 am
“One last comment to Will, who said: (“H2O radiates extensively from 5-8 Microns and 14- 200 Microns, more than enough to regulate the whole temperature of this Earth”)”
“I will simply refer you to the experimental result shown in the 2nd satellite images above. In the atmospheric window, the thermal IR from the ground is strong — close to that calculated for a BB @ 295 K.”
That is not thermal IR from anything no is it an image. It is a homoginized plot if normal spectral “radiance” (field strength) from some altitude over a cloudless tropical ocean.
“The emissions in the ranges you name for H2O are LESS THAN the emissions expected from the 295 K surface.” There are no emissions only a lesser “radiance” in those ranges. Integrated over the whole 2 PI steradians of very low radiance space the total exitance is or can be higher than all solar EMR converted to sensible heat.
“Hence H20 “regulates” the temperature by REDUCING the outgoing IR and thereby causing the whole system to warm up!”
Another TF use of the undefined “warm up”. Can you never say what you mean? The H2O exitance is the thing that cools this planet. That same H2O prevents incomming solar EMR by forming clouds
“Remove that H2O and MORE thermal IR would leave, cooling the planet.”
Why remove anything? Only to continue your nonsense conjecture of surface exitance?
This planet has long had much water which completely determines amount EMR in and out. You simply cannot defend your post modern fantasy of electromagnetic radiation, which is not statistical. Do not remove H2O, CO2, or CH4, Burn coal untill sufficient Nuclear power becomes available. Coal is mearly a tree that has not yet caught on fire!!
Tim Folkerts says, August 29, 2014 at 11:34 pm:
“You miss a key fact here. That “significant share of the absorbed energy” can ONLY be brought up into the atmosphere if the atmosphere has a means to get rid of said energy.”
We’ve been through this before, Tim. This is not a ‘fact’. This is what would be the case in your fantasy bubble world where unicorns, pink dragons and magical gases live side by side.
You completely ignore how real heated surfaces respond to impeded convection and prefer to live in lala land to try and justify how the surface will need to COOL with an atmosphere on top that is not able to shed its energy to space!
“If a “significant share of the absorbed energy” kept going into the atmosphere with no way to lose energy from the atmosphere, the atmosphere would keep warming!”
Exactly, that’s what I’m saying. The atmosphere would just keep warming, forcing the surface to warm even more to maintain a gradient. The atmosphere would thus keep expanding until it started to erode into space.
“Of course, what really would happen is that as the atmosphere will warm and the temperature difference between surface and atmosphere will diminish. The “significant share” will diminish, asymptotically approaching “zero share of the absorbed energy” as we approach a steady-state condition.”
A situation where the atmosphere on top of a solar-heated surface could not cool radiatively to space would give an Earth system that could never reach a steady state, it could never reach an equilibrium between incoming and outgoing. It would overheat. Until the atmosphere itself were gone, blown into the vacuum of space.
“And what will the conditions be at that point? Well, now that no more energy is required to further warm the atmosphere, all of the incoming solar energy will again be emitted as thermal IR straight to space (for our hypothetical IR-transparent atmosphere). So the surface will average ~ 255 K (give or take a little depending on the exact scehario) and the atmosphere will be about the same.”
This ongoing nonsense about blocking for convection, the ultimate cooling mechanism of a heated surface surrounded by a fluid, and then get this surface to COOL (not heat) by just switching to radiation through the fluid instead, no more convection needed, is quite frankly sad to behold. It’s the perfect radiative delusion.
Here’s the gist: Since convection is THE cooling mechanism of a solar-heated surface beneath a massive atmosphere, somehow blocking for it will cause heating, not cooling. The IR active gases strengthens the convective cooling of the surface by radiating the surface energy at altitude to space. If you then take out these gases, efficient convective cooling of the (still) solar-heated surface would be not as effective at the same temperature as before, and solar energy would thus automatically accumulate, making the surface warmer to reestablish (maintain) the temperature gradient. It wouldn’t respond by simply switching to radiation. Even worse, by cooling below the original temperature (with convection working properly) as it did. No, the surface would naturally get hotter and hotter in order to strengthen convection so as to prevent the temp gradient from buckling, the atmosphere progressively expanding in the process. You see, the surface will automatically respond like this, desperately trying to cool convectively to the atmosphere directly on top and will therefore accumulate the solar energy coming in until it can.
Heat stops flowing (‘steady state’) between two systems when there is only the two systems involved, Tim. When there ISN’T energy constantly being added from a third system, an external heat/power source, to the one system.
You’re is practicing the classical circular AGW way of doing ‘science’ by postulating your own premise as fact in order to show yourself to be correct. Yup, there will be a steady state with zero convective loss (because I say so), and FROM THAT you can see that I’m right. Priceless!
It’s not like the Sun stops shining, Tim. It’s not like the surface will all of a sudden find itself in a stable, purely radiative (BB) situation if you simply put a stop to convection.
I’ve told you this before: You can’t just set convective losses to zero and be rid of them. The real world doesn’t work like a mathematical equation; just strike out convection to get where we wanna be – radiation only. If you suppress natural conductive>convective transfer of energy from a constantly heated surface surrounded by a fluid, you will get warming, not cooling. Why? Because the energy from the heat source will accumulate at that surface until such transfer is restored. Naturally. Automatically. It’s what happens in the real world.
“The “mere presence” of a massive atmosphere is not enough!”
It sure is! It is what makes the mean global surface of the Earth so much warmer than the mean global surface of the Moon. The MASS of the atmosphere …
The point is, in the REAL Earth system, the atmosphere IS able to cool adequately to space. Therefore the Earth’s surface is able to stay at a relatively constant (steady-state) temperature, because convection runs smoothly, cooling the surface from the absorbed solar energy. The IR-active gases aid in this cooling process. Since there IS convection bringing solar>surface energy into the atmosphere, the surface itself could never reach a purely RADIATIVE equilibrium with its heat source, the Sun. The radiation emitted back out would rather be spread out from the surface itself to the ToA.
But this spread is NOT what sets the mean surface temp of the Earth. What sets the mean surface temp of the Earth is the MASS of the atmosphere resting on top of it setting a limit – at a certain temp – to convective/evaporative energy loss from the surface. The mass of the atmosphere is what reduces the temp gradient away from the solar-heated surface (‘heat capacity’ – the atmosphere’s got it, space doesn’t). It also suppresses free convective uplift + free evaporation from the ocean surface (‘weight’ – the atmosphere’s got it, space doesn’t). Both effects force the steady-state temp up.
Will
The cosine rule has nothing to do with Lambert’s law. It’s a mathematical calculation, in this case, connected to the amount of energy received/emitted from a surface at an angle away from the normal.
An Ir thermometer will give the most accurate reading when the cone of the reading is at 90deg to the surface, ie a circle. When it is used at an angle it takes in an elliptical surface which is clearly a greater surface area and the resultant reading would, clearly, be lower.
tchannon says: August 29, 2014 at 9:52 pm
“There are a number of reasons to assume lunar geologic heat is very slight”
Fully agree, but my point is that this temperature is not zero.
A calculation that uses SB implicitly assumes the zero incoming radiation temperature to be 0K.
This is almost correct for the moon, certainly not for earth.
Lets assume somehow a planet like the young earth has entered into the same orbit as our earth.
It’s effective temperature is depending on the assumed albedo, will be similar to that of the moon or earth. Lets say 270K.
The surface of this planet consist of bubbling lava, temperature eg. 1000K.
I hope no one will try to defend that this temperature is due to a greenhouse or similar.
This planet will radiate away like crazy, and cool down eventually.
Our current earth also has a temperature, and I can pinpoint it around 277K.
So BB calculations are invalid for this situation.
I had an interest in the temperature profile below the surface of the Moon to see what depth you’d have to go to, to get a habitable year-round temperature. As well as providing protection for micro-meteorite bombardment, living underground, if the temperature of the surrounding was adequate, would obviate the need for energy sources for heating/cooling.
On Earth, the average surface temperature is found only a few feet underground, which is why we can dig cool cellars in New England for vegetables and ice cellars in Inuvik, Canada. Does anyone have a thermal profile of the top 20m of the lunar surface?
I tried to figure out what the profile was from Apollo 17 data and failed. Not enough depth data. Those who have done this work have clearly investigated available data: has someone the answer to my question?
The same applies to Mars. On Mars, however, Olympus Mons has a huge number of open lava tubes – you can seen the massive collapsed ones in satellite imagery, and using Hawaii as an example, we know that OM is riven with lava tubes. There the score would be to use the lava tubes for living quarters. Easy to seal the walls, good protection from micros again, excellent stability for Earth-atmosphere pressures. Probable ice-CO2 sources not just at the top of OM but inside lava tubes as well. The attractiveness depends on the temperature of the rock, however; putting insulation along the walls to maintain a decent temperature could be more trouble than it is worth. Again, I don’t know what the thermal profile of the Martian surface is.
Kristian says:
August 30, 2014 at 10:32 am
The mass of the atmosphere is what reduces the temp gradient away from the solar-heated surface
If you mean
The mass of the atmosphere is what reduces the temp ( that is creates the temperature gradient, the lapse rate ) away from the solar-heated surface. Then I agree.
However this seems in conflict with your belief, we have discussed on another thread, that the lapse rate is caused by adiabatic expansion.
— Doug Proctor says:
August 30, 2014 at 6:03 pm
I had an interest in the temperature profile below the surface of the Moon to see what depth you’d have to go to, to get a habitable year-round temperature. As well as providing protection for micro-meteorite bombardment, living underground, if the temperature of the surrounding was adequate, would obviate the need for energy sources for heating/cooling.
On Earth, the average surface temperature is found only a few feet underground, which is why we can dig cool cellars in New England for vegetables and ice cellars in Inuvik, Canada. Does anyone have a thermal profile of the top 20m of the lunar surface?–
A meter below the surface is around -30 C. And 1 meter to 20 meter would be about the same [around -30 C]. Something buried 1 meter depth to 20 meter depth would be well insulated- or with active refrigeration it’s a very good freezer [high efficiency due good insulation]. Or add one adds heat, it’s good furnace; or say a well insulated hot water tank.
This in general is not much different than say a desert or any dry area on Earth.
In terms of living on Moon, on Earth one generally does not live in completely sealed environments.
Or on the Moon one needs life support systems, which would scrub CO2 from the air and deal with moisture [and other volatiles] generated from living human beings.
So basically underground on the Moon does not cause cooling or warming- though without any input of cooling or warming it would eventually passively cool down until it was about -30 C, whereas on earth it could be instead passively cool to around 10 C [or Earth average temperature is 15 C- though the warmth of tropics increases this average- on underground in tropic might be around 25 C.
Now if one wants to have the underground on the Moon to be warmer than -30 C one could reduce
the insulative property of the Moon’s top 1 meter. Or if one’s “roof” was concrete or solid rock, the sunlight would warm it up more. Or said differently the entire Moon is covered by about 3″ of very insulative dust. If one removed this dust [by say digging a hole] then this would alter the temperature beneath the surface.
So say take a 1 square km area on the moon and level and flatten with road equipment. Doing just this would significantly alter the thermal property of the lunar dust- it would not be fluffy and instead it become more compacted, and no longer be this type of super insulation.
I guess doing this would cause something buried within the 1 square km area to increase it’s passive [or average temperature] by say 10 C. Or it would be around -20 C instead of around -30 C.
Now one get more elaborate, and use something which conducts better than this- which if compacted acts like dry sand. So one fuse the regolith so it becomes like rock- but still is not a very good conductor of heat. The best practical conductor of heat are metals, and copper is one of best metals. Not that it’s practical, but if one covered the surface with 1 meter thick copper, it would have very significant effect upon the ground beneath it- instead of compacted regolith of -20 C, it could closer to 0 C.
Now there no known large amount of easily available copper on the Moon, but there is a lot of iron which could harvested from surface [there is a lot of “loose chunks” somewhat pure iron which could attracted with magnets]. And there is million of tonnes of fairly easy to mine water.
So with black painted iron pipes and water,one could harvest thermal solar energy. And on the Moon solar energy can boil water at 1 atm of pressure [unlike what one can do on Earth].
Or in terms roof of underground, one could 1″ plate steel encasing 1 meter of water- and this would work as good as the solid meter of copper.
Doug Proctor says: August 30, 2014 at 6:03 pm
“Does anyone have a thermal profile of the top 20m of the lunar surface?”
This post https://tallbloke.wordpress.com/2012/04/09/ashwin-vavasada-lunar-equatorial-surface-temperatures-and-regolith-properties-from-the-diviner-lunar-radiometer-experiment/
has a temperature profile for the equator. Below ~30 cm no more solar signature.
Imo the moon still has geothermal heat, so the temperature should increase from 30 cm going down.
see eg. http://www.dailymail.co.uk/sciencetech/article-2719809/Is-moon-s-core-MOLTEN-Centre-satellite-wrapped-layer-soft-rock-claim-scientists.html
— Ben Wouters says:
August 30, 2014 at 9:22 pm
Doug Proctor says: August 30, 2014 at 6:03 pm
“Does anyone have a thermal profile of the top 20m of the lunar surface?”
This post https://tallbloke.wordpress.com/2012/04/09/ashwin-vavasada-lunar-equatorial-surface-temperatures-and-regolith-properties-from-the-diviner-lunar-radiometer-experiment/
has a temperature profile for the equator. Below ~30 cm no more solar signature.–
So in middle of lunar day when surface is about 120 C [a frying pan at good temperature for cooking pancakes] a foot under surface it’s about -30 C.
And with the ground deeper than 1 foot is not warming or cooling during the day and nite cycle- at least as far as what can be determined from lunar orbit. Or if variation in temperature was less than 1 C, probably one couldn’t accurately determine such a small temperature differences [from orbit].
— Imo the moon still has geothermal heat, so the temperature should increase from 30 cm going down.
see eg. http://www.dailymail.co.uk/sciencetech/article-2719809/Is-moon-s-core-MOLTEN-Centre-satellite-wrapped-layer-soft-rock-claim-scientists.html —
It doesn’t on Earth and Earth has far more geothermal heat.
Instead on Earth one needs hundreds of feet, and the Moon [or Mars] instead it would be miles under surface. Though Mars has more geothermal heat than Moon [Mars is less dead] and one might find regions on Mars where it’s warmer nearer the surface. Or the Earth has geothermal heat at the surfaces- such with Iceland, Yellowstone, ocean floors, and etc. Such warm spots have not yet been found on Mars, though it’s possible that somewhat warmer regions could be found on Mars.
Or for example one might find liquid water less than 1 km under the Mars surface [might even be at surface or might be a 100 meters below surface], but very unlikely one could find what is possible on Mars surface, on the lunar surface.
And I would guess with either Mars or the Moon that such hot spots could be related to more recent
impactor events [or not necessarily, much to do with planetary interior heat].
Ben Wouters said:
“On the moon the night temps converge towards ~50 or 60 K iso 2,77K, even after 2 earth weeks of cooling. So the moon seems to have a “base” surface temp. of around 50K, before the sun starts adding its energy.
My model assumes that the Moon is covered in a uniform layer of regolith 0.5 meters deep. It is also assumed that below the regolith there is basalt bedrock at a uniform temperature of 240 K. This simplification reduces the problem to something that can be solved on my puny laptop.
Thus the surface temperature on the night side of the Moon is not 2.7 K as it would be if the regolith were a perfect insulator. Enough heat is conducted from the basalt to maintain a relatively toasty 95 K. In polar craters that are never reached by solar radiation the Diviner LRE shows surface temperatures down to 25 K (Hermite crater) which is consistent with the model.
Thanks for the answers!
Interesting that 30cm maintains a thermal difference of -61C average (212K) and -30C subsurface. The geothermal power gradient of the moon is almost insignificant for human purposes. Heat conductivity would be a different issue, probably a real problem, as any camper will tell you bare rock cools down fast at night (and heats up quickly during the day). But a slightly elevated structure encased in regolith filled walls, once warmed, would probably be easily maintained at a comfortable living temperature.
Use an existing crater, put the modules on a platform, bulldoze surrounding regolith on top, pipe circulating fluid through the surrounding regolith, heat added by simple black cloth through which fluid moves (drained during the night, or deal with freezing expansion issues). Heating is probably not the problem during the night but cooling is the problem as human activity adds heat.
Despite the hard vacuum of space being a good insulator, the Apollos 13 crew were freezing as well as suffering from hypoxia. So a secondary source of stored heat would be probably necessary, i.e. the colony modules would cool anyway. Considering the amount of solar energy available at 1364 +/- 6.8% W/m2 sunup to sundown to perpendicular to incident mirror, collector reflectors could be concentrated enough to even melt rock as a heat storage medium, providing a thermally suitable environment with passive systems doesn’t sound like too difficult a problem.
Mars obviously has a different problem, not temperature wise (average of -60C at the equator? I forget) but in energy source, even if the atmosphere doesn’t absorb/refract/reflect as a clear Earth atmosphere does. Lava tubes still work structurally, but maybe not so much energetically, due to thermal loss through tube walls. Again have to suspend living quarters. Certainly do-able as any Canadian winter resident understands: insulation and a cheap! power source is what makes Canada habitable year-round.
The subsurface of the Moon surprises me. At a surface of 212K and a near-surface of 243K, it means there is a continuous depth-to-surface transfer of energy. Whatever that power is involves math that is the reason I dropped out of astronomy and went into geology. You guys would know how to figure out the power drain and the total energy loss on a planetary scale: would that be significant? tell you anything about the internal heat source of the moon?
Tim Folkerts said, August 27, 2014 at 4:56 pm
“I like the analysis. I am not as much of a fan of the Discussion.”
While I am interested in CAGW to the extent that it affects government policies that raise the cost of energy my real concern is the false science that our children are are exposed to. Our kids are being taught that CO2 is a “Pollutant” and that rising CO2 emissions will cause catastrophic “Global Warming”.
One dimensional atmospheric models (e.g. Robinson & Catlings) should help us understand the effect of gas composition on surface temperature. Is pressure the major variable as Nikolov & Zeller say? Or is CO2 dominant as James Hansen contends?
Alex says: August 30, 2014 at 12:55 pm
“Will
The cosine rule has nothing to do with Lambert’s law. It’s a mathematical calculation, in this case, connected to the amount of energy received/emitted from a surface at an angle away from the normal.
An Ir thermometer will give the most accurate reading when the cone of the reading is at 90deg to the surface, ie a circle. When it is used at an angle it takes in an elliptical surface which is clearly a greater surface area and the resultant reading would, clearly, be lower.”
Since the detector’s projected solid angle did not change, why do you claim ” the resultant reading would, clearly, be lower.”? Did you run out of projected solid angle? The reading would be lower “only” if the emissivity in the new direction were lower!
gallopingcamel says: August 31, 2014 at 4:40 am
“My model assumes that the Moon is covered in a uniform layer of regolith 0.5 meters deep. It is also assumed that below the regolith there is basalt bedrock at a uniform temperature of 240 K. This simplification reduces the problem to something that can be solved on my puny laptop”
Fully understood.
IF I’m correct about the moon having a hot core, than the temperature towards the surface reduces from the core outward. The flux would be minimal, BUT the temperature gradient would settle to have the AVERAGE surface temp. just below the depth where no more solar influence is experienced.
For the equator this would be (390+100)/2 = ~245K.
For higher latitudes this temp. will be lower, eg lat60 (330+70)/2 = ~200K.
What would be the effect of setting the temp. of the basalt to 0K, like a comet would have after having cooled down completely in outer space, before entering a solar orbit like our earth?
gbaikie says: August 30, 2014 at 10:26 pm
– Ben Wouters says: August 30, 2014 at 9:22 pm
— Imo the moon still has geothermal heat, so the temperature should increase from 30 cm going down.
see eg. http://www.dailymail.co.uk/sciencetech/article-2719809/Is-moon-s-core-MOLTEN-Centre-satellite-wrapped-layer-soft-rock-claim-scientists.html —
“It doesn’t on Earth and Earth has far more geothermal heat.”
On Earth the temperature increase is ~25K/km.
http://en.wikipedia.org/wiki/Geothermal_gradient
Afaik no one has been drilling to any depth on the moon yet. 😉
Basic physics of heat flow says that — everything else being equal — smaller objects will cool more quickly than larger objects. The moon being much smaller will cool much faster. Certainly the core would still be warm, but the thermal gradient as you go down from the surface would almost certainly be less than on the earth.
Tim Folkerts says: August 31, 2014 at 3:35 pm
” Certainly the core would still be warm, but the thermal gradient as you go down from the surface would almost certainly be less than on the earth.”
Extremely likely, see http://www.dailymail.co.uk/sciencetech/article-1344980/Moon-liquid-core-just-like-Earth-reveal-sensors-left-lunar-surface-astronauts-40-YEARS-ago.html
for some numbers.
BUT as long a FLUX exists from core to space, however small, the subsurface will settle at the average surface temp. as caused by the sun.
eg 240K at 40 cm below the equatorial surface, but only 26K in polar craters without any sunshine at all.
Roger Clague says, August 30, 2014 at 7:48 pm:
“If you mean
The mass of the atmosphere is what reduces the temp ( that is creates the temperature gradient, the lapse rate ) away from the solar-heated surface. Then I agree.”
I mean exactly what I wrote. The atmosphere has a mass and hence a ‘heat capacity’, meaning it is able to warm. As opposed to space. Hence, with a massive atmosphere in place – at the same mean surface temperature – the energy loss per unit of time from the surface would be smaller than without one, because the temp gradient/difference from the solar-heated surface to its direct surroundings would be much reduced. If the incoming heat from the Sun is then equal to what it was, this means that the surface will naturally have to warm in order to be able to balance it.
The presence of a massive atmosphere on top of a solar-heated surface likewise forces this surface to warm by restricting buoyant uplift and evaporation rates at a certain temperature. To maintain balance with equal incoming: heavier atmosphere > warmer surface; lighter atmosphere > cooler surface. If the atmospheric weight is kept relatively constant (like on Earth), a warmer surface creates greater buoyant uplift and higher evaporation rates. And vice versa. This is mostly connected with larger or smaller input from the Sun. Compare the tropics with the higher latitudes.
The warming (as in ‘insulating’) effect of a massive atmosphere on the solar-heated surface of the Earth is very real, very big and very obvious if you just take the time to think about it.
Earth vs. Moon.
The effect simply has got nothing to do with thermal radiation/’atmospheric back radiation’ and CO2. IR-active gases work forcefully towards cooling the Earth (and hence also the surface).
Kristian says: “Hence, with a massive atmosphere in place – at the same mean surface temperature – the energy loss per unit of time from the surface would be smaller than without one, because the temp gradient/difference from the solar-heated surface to its direct surroundings would be much reduced.”
The conduction and/or convection heat loss from the surface with no atmosphere would be zero — there is no conduction or convection through a vacuum. The conduction and/or convection heat loss from the surface to an atmosphere would be positive (ie greater than zero) (assuming the atmosphere is cooler than the surface). So the conduction and/or convection heat loss with an atmosphere is LARGER than without one, not SMALLER as you conclude.
Thus surrounding the planet with a cool IR-transparent atmosphere would COOL the planet, not warm it! (Surrounding it with a warmer atmosphere would warm the surface, but such things are not observed in any real situations).
There is, of course, another way to reduce the energy loss from the surface. With no atmosphere — or an atmosphere transparent to IR! — there will be a strong IR radiation loss from the surface to space. This could be reduced by having the surface radiate instead to an atmosphere that is warmer than space but cooler than the surface. The net EMR flux from the surface is reduced and the energy loss would be smaller and the the surface would warm.
Tim Folkerts says: August 31, 2014 at 9:16 pm
(Kristian says: “Hence, with a massive atmosphere in place – at the same mean surface temperature – the energy loss per unit of time from the surface would be smaller than without one, because the temp gradient/difference from the solar-heated surface to its direct surroundings would be much reduced.”)
“The conduction and/or convection heat loss from the surface with no atmosphere would be zero — there is no conduction or convection through a vacuum. The conduction and/or convection heat loss from the surface to an atmosphere would be positive (ie greater than zero) (assuming the atmosphere is cooler than the surface). So the conduction and/or convection heat loss with an atmosphere is LARGER than without one, not SMALLER as you conclude.
Thus surrounding the planet with a cool IR-transparent atmosphere would COOL the planet, not warm it! (Surrounding it with a warmer atmosphere would warm the surface, but such things are not observed in any real situations).
There is, of course, another way to reduce the energy loss from the surface. With no atmosphere — or an atmosphere transparent to IR! — there will be a strong IR radiation loss from the surface to space. This could be reduced by having the surface radiate instead to an atmosphere that is warmer than space but cooler than the surface. The net EMR flux from the surface is reduced and the energy loss would be smaller and the the surface would warm.”
That is quite good, but still with intent to confuse poor students. Tim!
Let me give you a thought problem much closer to “this is”, rather than pink dragons. This Earth best we know with a surface reflectivity from 2.5 microns to 200 microns of 85% (15% emissivity).
No other changes to surface emissivity, amospheric WV, clouds, or production of WV in any manner. Just what is your predicted surface temperature? And why is that temperature so much higher than what is measured. All your fraudsters picked “high” long-wave emissivity from the toilet, with absolutely no measured evidence, and no understanding of Earth’s atmosphere!
A complete intentional fraud by persons yet to be indicted! (Is that OK, TimC?)
Tim Folkerts says, August 31, 2014 at 9:16 pm:
“Thus surrounding the planet with a cool IR-transparent atmosphere would COOL the planet, not warm it! (Surrounding it with a warmer atmosphere would warm the surface, but such things are not observed in any real situations).”
Interesting. So now a planetary surface WITH a massive atmosphere on top, only one WITHOUT radiatively active gases in it (so no way to radiate its energy to space), will actually all of a sudden turn out COOLER than if the surface was surrounded by the vacuum of space! That’s a new one. Do tell more. So 255K without an atmosphere? How much cooler than this with an IR-transparent atmosphere, Tim?
“Kristian says: “Hence, with a massive atmosphere in place – at the same mean surface temperature – the energy loss per unit of time from the surface would be smaller than without one, because the temp gradient/difference from the solar-heated surface to its direct surroundings would be much reduced.”
The conduction and/or convection heat loss from the surface with no atmosphere would be zero — there is no conduction or convection through a vacuum. The conduction and/or convection heat loss from the surface to an atmosphere would be positive (ie greater than zero) (assuming the atmosphere is cooler than the surface). So the conduction and/or convection heat loss with an atmosphere is LARGER than without one, not SMALLER as you conclude.”
As readers will notice, I said “…the energy loss per unit of time from the surface would be smaller …” while Tim ‘misunderstands’ this so that he can ‘correct’ his carefully constructed straw man: “…the conduction and/or convection heat loss with an atmosphere is LARGER…”
In other words, I don’t conclude this. Tim concludes. To confuse.
It’s pretty clear he’s got nothing.
Sigh. Kristian, I can’t seem to bring you up to the level I am discussing.
1) Can we agree that — for fixed conditions — the average long-term heat flow rate into the atmosphere must approach the average long-term net heat flow rate out of the atmosphere? Otherwise there would be a continuing net long-term temperature increase (or decrease). Sure, there will be energy flowing into the atmosphere during the day and energy flowing out out during the night, but averaged over the whole planet over a while year (or decade or century), the atmosphere as a whole is not gaining or losing net heat. (And similarly the average long-term heat flow rate into the SURFACE must approach the average long-term net heat flow rate out of the SURFACE).
If we can’t agree on this, there is really no point in the rest of the discussion. Once a stable temperature pattern has been established, there will be no net long-term flow of heat into any part of the system.
2) If we agree on (1), then the real issue is what the long-term solutions will be for different situations.
* With no atmosphere, there is heat in to surface from the sun, and thermal EMR out from the surface to space, and these two will be balanced. This gives the oft-quoted ~ 255 K average surface temperature result (which of course depends on a variety of assumptions; the top post for example shows that the average would be much lower than this for a slowly rotating object).
* With completely transparent atmosphere, there is no heat in to the atmosphere from the sun and no heat out to space (neither conduction nor thermal EMR work from a transparent atmosphere to the vacuum of space). The only other possible source sink for heat is the surface, but since we just established that all other heat flows are zero, the net long-term surface-to-atmosphere heat flow must ALSO average to zero.
Well, if the net long-term surface to atmosphere heat flow is zero, then we are STILL left with the only possible heat loss for the surface being thermal EMR to space. So the surface will STILL radiate as much thermal IR out as the sunlight in, and will still settle in at ~ 255 K.
****************************************************************************************
Kristian asks: So now a planetary surface WITH a massive atmosphere on top, only one WITHOUT radiatively active gases in it (so no way to radiate its energy to space), will actually all of a sudden turn out COOLER than if the surface was surrounded by the vacuum of space! That’s a new one. Do tell more. So 255K without an atmosphere? How much cooler than this with an IR-transparent atmosphere, Tim?
Here we are talking about the SHORT-TERM situation (perhaps I should have made that clearer since it clearly escaped your notice). if you suddenly added a cool, completely transparent atmosphere, then there would be a short-term heat flow from the surface to the cooler atmosphere, leading to a short-term cooling of the surface. But as the temperature difference between surface & atmosphere decreased, the heat flow would decrease. At the same time, the surface would radiate less thermal EMR to space. Eventually there again be a net gain at the surface and then the temperature would creep back toward the original stable pattern.
IF you added a warmer atmosphere, the reverse would occur — heat would flow from the warmer atmosphere to the surface. The atmosphere would cool; the surface would warm, but would then radiate more thermal EMR to space.
Either way, the eventual answer is “255 K no matter what the initial temperature of the atmosphere is”.
*******************************************************************************************
There would be other secondary effects. For example, a completely transparent atmosphere would cool the day-side and warm the night-side, and this would lead to a slightly higher average temperature. Ie WITH an atmosphere, the average temperature would be CLOSER to 255 K (but still not able to get any high than that) (where I am using your presupposed 255 K surface temperature for the bare planet).
Will says: “This Earth best we know with a surface reflectivity from 2.5 microns to 200 microns of 85% (15% emissivity).
Once again, you are pulling numbers from thin air, while claiming every published scientific paper and every published engineering table is wrong, as if you and you alone are privileged with the right answers. The earth you know may have a surface emissivity of ~ 0.15, but the earth we (the entire scientific and engineering community) know has a surface emissivity of ~ 0.85.
Convince me otherwise. Convince me that generations of scientists and engineers who make their living understanding thermal IR have gotten it horrible wrong. Tell me how IR thermometers calibrated with “standard” emissivity values give the right answer even though those emissivities are off by an order of magnitude.
Regarding Tim Folkerts says:
September 1, 2014 at 12:33 pm
===========================================
Interesting discussion all. I have a couple of questions, and a comment. First is heat cumulative? By this I mean take two atmospheres composed of ten thousand non GHG molecules per sq foot, and have them, each molecule, vibrating with the same energy/heat. A mercury thermometer will register the same T correct?
Now, while keeping all the molecules vibrating with the same energy, change one of the atmospheres to only 100 molecules per sq M. Now, due to far fewer molecules striking the glass mercury thermometer surface, will the less dense atmosphere register a lower T then the more dense atmosphere? (This is what I mean by is heat cumulative?)
Next question. We have this 255 k average T correct? But is the average T not the important point in elevating a system to its maximum T? Is not the highest T what really matters?
(By this I mean place a pot of water on a surface with a small, flame size T of 100 degrees F. Now, if the large pot of water is surrounded by 32 degree air, and has no lid and thin copper sides, the water will still warm, but not to 100 F due to the lack of insulation. Now take the same flame size 100 degree F heat source, and place it directly under the center of a raised center very thick pot, only thin right at the raised center heat source, with a sealed thick lid. Now the T in the pot will rise far higher, but can never exceed 100 degrees F can it? So, if the above is correct, the average T is not what matters in a system, herein descried as the pot of water, but the highest T is what describes the limit of a system.)
So we have above heat capacity and hottest input source. Now, before I go further I would like to ask you all to think of heat capacity as a function of residence time. By this I mean the following…
David’s law. (-; “Only two things can affect the energy content of a system in a radiative balance; either a change in the input, or a change in the residence time of some aspect of energy within the system.” (Two PHD guest WUWT authors accepted this law, Ira Glickstein, after much discussion, and a Dr. Brown.)
The earth’s system is defined as the oceans, land, and atmosphere.
All non-input change theories on climate are a manifestation of the affect of “residence time.” For instance, the GHE is based on increasing the residence time of certain WL of LWIR energy via redirecting exiting LWIR energy back into the system, while input remains constant, thus more total energy is within the system. The greater the increase in residence time of the energy, the greater the potential energy accumulation.
Clouds are capable of both increasing the residence time of some LWIR radiation from the surface, and decreasing the residence time of SW insolation from the Sun. The net affect is dependent on both the amount of energy affected, and the residence time of the energy affected, which is dependent on both the WL of the energy, and the materials said energy encounters.
It is true that 100 watts per sq. M of SWR, has the same energy as 100 watts per sq. M of LWIR, however their affect on earth’s energy balance can be dramatically different. In this sense, not all watts are equal.
For instance lets us say 100 watts of LWIR back radiation strikes the ocean surface. That energy then accelerates evaporation where said energy is lifted to altitude, and then condenses, liberating some of that energy to radiate to space. Now lets us assume the same 100 watts per sq M strikes the ocean, but this time it is composed of SWR, penetrating up to 800 ‘ deep. Some of that energy may stay with in the ocean for 800 years. The SWR has far more long term energy, and even warming potential then the LWIR.
Now let us consider how the material encountered manifests according to “David’s Law”. Taking the same 100 watts per sq M of SWR, and let us say it strikes a cloud. The residence time of this is very short, as the reflected SWR leaves earth at light speed. The same SWR energy, sans clouds, again strikes the ocean, much of that energy staying within the earth’s system for years, some for decades, some for centuries.
So in truth all climate theories depend on this law, and the positive or negative affect on earth’s energy budget and long term warming or cooling affect, is a simple mathematical expression of residence time. A traffic analogy helps in understanding this. Cars symbolize packets of energy…
Take a 120 mile long highway. (This is the earth’s system; land, oceans and atmosphere.) Input into the system is 10 cars per hour. (This is insolation) Each car is on the highway for 120 minutes, or two hours. (This is residence time) After two hours of input there are twenty cars on the highway. This is the total energy within the system.
Ten cars input per hour, twenty cars within the system, ten cars exit per hour; the system is in a radiative balance.
Now let us increase the residence time of each car from two hours, to twenty hours. After twenty hours the system will now contain 200 hundred cars, instead of ten. The system will have ten times the energy it formerly had before again establishing a radiative balance at 200 cars. This 1000 percent increase was achieved with zero change in input.
Knowing the residence time is the difficult part. Assuming a small change in very long residence time energy cannot affect the climate is , in my view, simple minded.
Please keep the idea of residence time in your debate. One possible manifestation of this is that the greater the residence time of the input energy, the more larger and larger area of the system can move towards, not the averageT, but the maximum T.
Tim Folkerts says, September 1, 2014 at 12:33 pm:
“Sigh. Kristian, I can’t seem to bring you up to the level I am discussing.”
Hahaha! Nice one. Let’s hear it, then …
“1) Can we agree that — for fixed conditions — the average long-term heat flow rate into the atmosphere must approach the average long-term net heat flow rate out of the atmosphere? Otherwise there would be a continuing net long-term temperature increase (or decrease).”
If you remove the atmosphere’s ability to shed its energy to space, but at the same time let the Sun keep heating the surface, the atmosphere will keep warming and expand. There will be no steady state until that atmosphere is gone into space. You can’t just get rid of conductive/convective and achieve a purely radiative situation for the surface as long as the atmosphere is still there, just by setting them to zero, as if the world worked like a mathematical equation, Tim.
You apparently (?) didn’t read my previous comments specifically addressing this nonsense. Well, you never replied, and here you are continuing along the same course, as if we hadn’t been through this several times before … So here they are again:
It’s just upthread, Tim.
Here’s a pertinent quote:
“Heat stops flowing (‘steady state’) between two systems when there is only the two systems involved (…). When there ISN’T energy constantly being added from a third system, an external heat/power source, to the one system.
You’re (…) practicing the classical circular AGW way of doing ‘science’ by postulating your own premise as fact in order to show yourself to be correct. Yup, there will be a steady state with zero convective loss (because I say so), and FROM THAT you can see that I’m right.”
“If we can’t agree on this, there is really no point in the rest of the discussion.”
That’s when the discussion begins, Tim. If you don’t want to be part of it, suit yourself.
“Once a stable temperature pattern has been established, there will be no net long-term flow of heat into any part of the system.”
There won’t be a stable temperature pattern established without IR-active gases in the atmosphere, Tim. That’s why IR-active gases are present in (and a crucial part of) all atmospheres that we know of. They keep the system as a whole stable by radiating the atmospheric energy to space, enabling the solar energy to run smoothly through the system: Sun > surface > troposphere > space.
“2) If we agree on (1), then the real issue is what the long-term solutions will be for different situations.“
I hope you understand by now that NO, I will not agree to your postulated premise to get where you want. Because it won’t happen like you postulate. Reality doesn’t work that way. Read my comments above …
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“if you suddenly added a cool, completely transparent atmosphere, then there would be a short-term heat flow from the surface to the cooler atmosphere, leading to a short-term cooling of the surface. But as the temperature difference between surface & atmosphere decreased, the heat flow would decrease.”
And then the surface would warm as a result. To maintain a gradient steep enough so that the surface is able to shed its energy as fast as it comes in from the Sun. Exactly.
You see, as soon as you place an atmosphere on top of the solar-heated surface, then conduction>convection will start removing energy from the surface. This energy is then not available for radiative emission. The convective loss REPLACES radiative loss, it doesn’t come IN ADDITION TO it. The surface only has a certain amount of energy to give away, the energy it gets in from the Sun. There will automatically be less energy radiated from the surface when conduction>convection is also available as energy loss mechanisms. This is what your purely radiative equations fail to take into account. Because they simply were not made for such scenarios.
Suppress convective loss from a heated surface surrounded by a fluid, and you will get swift and significant WARMING, Tim, not cooling.
“If you remove the atmosphere’s ability to shed its energy to space, but at the same time let the Sun keep heating the surface, the atmosphere will keep warming and expand. “
That is one of the most obvious violations of the 2nd Law of Thermodynamics I have ever seen. I suspect you will have a hard time getting anyone to back you on this point. (Anyone care to speak up here who thinks the atmosphere will indeed keep warming??) And until this most basic of points is cleared up, the rest of what you wrote is built on an unsupported premise.
The surface will not get above 255 K average — potentially as high as ~ 390 K at noon near the equator on a black rock. Heat cannot flow from the surface to the atmosphere unless the atmosphere is cooler than the ground. The atmosphere simply cannot continue warming above the hottest surface temperature without having heat moving from cooler (ground) to warmer (Atmosphere).
Or stated another way … if the atmosphere near the surface was warmer on average than the 255 K average ground temperature, then the atmosphere would shed energy TO THE GROUND by conduction. There is no need to shed it to space, since heat will naturally flow from the warmer atmosphere to the cooler surface, preventing the atmosphere from overheating the way you fear.
PS Of course, all this is hypothetical. All atmospheres of all known planets and moons do have GHGs, so they can all shed energy via thermal EMR.
PPS To head off a potential objection, I suppose a volcano could make a few areas that are locally warmer than this, but volcanoes are not common enough to put significant amounts of energy into the atmosphere. The hot plume of rising air would spread out and cool; somewhere the air would have to settle, causing it to cool and lose energy when it reaches the ground. So we could add ~ 0.1 W/m^2 average geothermal power to the system, but adding less than 1 W/m^@ average power to the ~ 240 W/m^2 solar power is not really going to matter.
Kristian 3:54pm: Discussing all this should lead to a better understanding of the top post but you can’t ever get past just the basic Planck paper & 1st law.
“If you remove the atmosphere’s ability to shed its energy to space, but at the same time let the Sun keep heating the surface..”
Nonsense physics. In your thought experiment, the sun’s atm. will also have its ability to shed its energy to space “removed”. Solids could still radiate I presume in this strange place so geothermal energy will make the surface glow to space and conduction will enable the atm. to equilibrate at surface + lapse but atm. won’t glow.
“There won’t be a stable temperature pattern established without IR-active gases in the atmosphere, Tim.”
Sure there will be Kristian – by 1st law which shows N2,O2 earth atm. stable Tmean will be just above 255K in this thought experiment. Surface & N2, O2 will still radiate to space based on thin optical depth given by species amount, the absorption coefficient of each, and surface pressure.
“The surface only has a certain amount of energy to give away, the energy it gets in from the Sun.”
Nonsense. The surface also “gets in” energy from geothermal, atm., stars, moon, new moon, planets, brown dwarfs, CMB, forest fires, power plants, friction, et.al. – from anything that uses up a fuel & is incident. Some energy that “gets in” is negligible.
“This is what your purely radiative equations fail to take into account. Because they simply were not made for such scenarios.”
Nonsense. The basic text book surface balance 1st law accounts for radiative, convective, and conductive energy transfer. Geothermal is basically considered negligible for earth but which you can easily include if you want. Look it up in any atmosphere radiation text book; I like Bohren 2006 p. 33.
”Suppress convective loss from a heated surface surrounded by a fluid, and you will get swift and significant WARMING…”
Concur. By 1st law. See solar pond discussions. Optically thinning an atm. reduces the surface Tmean by 1st law. Understanding this is key to achieve understanding of the top post. And then move it forward.
–David A says:
September 1, 2014 at 3:32 pm
Regarding Tim Folkerts says:
September 1, 2014 at 12:33 pm
===========================================
Interesting discussion all. I have a couple of questions, and a comment. First is heat cumulative? By this I mean take two atmospheres composed of ten thousand non GHG molecules per sq foot, and have them, each molecule, vibrating with the same energy/heat. A mercury thermometer will register the same T correct? —
10,000 molecules per cubic foot, is very thin air.
Gas doesn’t vibrate. Though since there is so many within a volume of space, they don’t move far before they collide. But If only have 10,000 molecules per foot, then they would be able to move quite a far distance before colliding with each other.
Gases are only warmed [have they velocity increased] by warmed surfaces or by other warmer gases. Or basically, the box holding the air will be same temperature as the air it’s containing- regardless of amount of gas [density, type, or amount of gravity].
“Now, while keeping all the molecules vibrating with the same energy, change one of the atmospheres to only 100 molecules per sq M. Now, due to far fewer molecules striking the glass mercury thermometer surface, will the less dense atmosphere register a lower T then the more dense atmosphere? (This is what I mean by is heat cumulative?)”
If one reduces amount of gas in a box, the gases cool, but the box would warm the gases- though the box can be cooled by the gases- in any case, eventually the box and gases will be same temperature. Added or removing gases from a box, is same as adding or removing energy or heat
from the box. Or this is what a heat pump or refrigeration is.
“Next question. We have this 255 k average T correct? But is the average T not the important point in elevating a system to its maximum T? Is not the highest T what really matters?”
Yes it’s important. The Moon’s max temperature from the sun is about 120 C [400 K]. Or the lunar surface can only warm to 120 C, no “greenhouse effect” or whatever [other magnification of sunlight] can raise temperature higher than this. So direct sunlight on Moon is 120 C. With earth, the atmosphere weakens the direct energy of sunlight so instead 120 C, it’s about 80 C.
So summer sun above head and one has asphalt parking lot with surface temperature of around 80 C- no kind of greenhouse effect will increase this temperature. Since we live on planet mostly covered with ocean, the max possible temperature of water should be considered and max surface temperature of water [largely controlled by evaporation] is about 40 C. So dry land max is about 80 C and water is about 40 C. The highest air temperature ever recorded was slightly below sea level, and n Death Valley:
“The world’s highest recorded air temperature is officially recognized by the World Meteorological Organization as 134°F (57.6°C) recorded at Death Valley, California, USA on 10 July 1913.”
Or air temperature above land is limited to about 20 C cooler than surface ground temperature,
whereas over water there is less difference- or the evaporated water is a gas of the atmosphere.
This is how things work:
You put an atmosphere (ANY atmosphere) around a solar-heated planetary surface. What happens?
That surface would no longer be able to balance the incoming solar flux with its outgoing radiative flux, because some of the energy originally absorbed from the Sun will be removed from the surface by way of convection and so less will be available for radiative emission than before.
What will this planet have to do? It NEEDS to balance the incoming with the outgoing to reach a dynamic equilibrium, a steady state.
Well, it now has an atmosphere at its disposal, something it didn’t before it was put in place. So it can spread its emission to space through a three-dimensional space rather than just emitting solely from a two-dimensional surface as before.
So what it does is, it uses the upward bulk movement of air to get that part of the solar energy that escaped the surface through that very same mechanism rather than by radiation above the level where the bulk transfer is still able to ‘replace’ upward radiative loss.
Which means, the ‘convective energy’ (the ‘rest of’ the surface energy) will ultimately rather be radiated to space from aloft.
The only reason for the need of this kind of three-dimensional ‘spread’ is the convective loss replacing radiative loss from the actual surface, so that it can never itself reach a radiative balance between IN and OUT with its heat source (the Sun).
The point here is, if 240 W/m^2 come IN, then 240 W/m^2 need to go OUT. In the end. It doesn’t matter from where. It just needs to go, from whatever level. It’s all about the amount of energy that needs to be shed. NOT about temperatures.
The surface cannot shed all its energy absorbed from the Sun (165 W/m^2) as radiation directly to space. Only about 20 W/m^2 of it, as it is on Earth. The rest is transferred to the lowermost air layer of the atmosphere (33 W/m^2 by radiation, 112 W/m^2 by conduction and evaporation) from where it is brought up and into the troposphere by mass transfer (convection).
So a mean flux of 145 W/m^2 from the two-dimensional surface enters the three-dimensional volume of the troposphere, while at the same time a flux of 75 W/m^2 is absorbed (also through a volume depth) coming in from the Sun. ALL of this (220 W/m^2) is then finally also emitted to space from the three-dimensional volume of the troposphere to space.
165 W/m^2 are absorbed and then escape the surface.
240 W/m^2 enter and then exit the ToA, the Earth system as a whole.
None of this has anything to do with what temperatures the energy is being emitted from. It only has to do with the fact that what energy enters per unit of time must also exit.
It’s the convective processes from the surface that determine how much energy that is ultimately to be radiated to space. Because the atmosphere is there to radiate. Just provide it with the energy. It’s all a giant balancing act. If convection brings less energy up, then less IR will be emitted to space. And vice versa. It’s really that simple. We see this very clearly in the OLR data. For instance, warm ENSO events bring lots of deep moist convection, humidifying and warming the troposphere. And OLR through the ToA shoots up. Cool ENSO events hold convection back, the troposphere dries and cools. And the OLR drops.
What controls total OLR through the ToA on Earth is basically the water cycle’s response to surface processes, mainly ENSO. OLR is very much correlated to the ENSO process and responds specifically to 1) surface temps >> 2) tropospheric temps and humidity, and 3) cloud cover.
Tim Folkerts says, September 1, 2014 at 12:33 pm:
“if you suddenly added a cool, completely transparent atmosphere, then there would be a short-term heat flow from the surface to the cooler atmosphere, leading to a short-term cooling of the surface. But as the temperature difference between surface & atmosphere decreased, the heat flow would decrease.”
So you add a cool, completely transparent atmosphere to rest on top of the solar-heated surface. There would initially be a very steep temp gradient, but as the atmosphere gradually warmed up, this would eventually stabilise into a normal (environmental) lapse rate gradient (quite close to the steepness of a gravity-based dry adiabatic lapse rate), maintained by the direct interaction between solar surface heating and its convective response. If the atmosphere then continues to warm from this stage (which it will without the ability to cool to space by radiating away the convectively transferred energy from the surface), then the surface will naturally need to warm also, so as to be able to maintain this gradient.
On our Earth, the process of maintaining the established average lapse rate by the direct coupling between solar surface heating and its convective response ensures that the troposphere and hence the surface ISN’T constantly warming and the tropopause ISN’T constantly lifting (they’re in a dynamic steady state). Why? How? Because of the constant tropospheric radiative cooling to space performed by the IR-active gases countering the convective warming of the troposphere. Keeping the system as a whole stable …
–Tim Folkerts says:
September 1, 2014 at 8:16 pm
“If you remove the atmosphere’s ability to shed its energy to space, but at the same time let the Sun keep heating the surface, the atmosphere will keep warming and expand. “
That is one of the most obvious violations of the 2nd Law of Thermodynamics I have ever seen. I suspect you will have a hard time getting anyone to back you on this point. (Anyone care to speak up here who thinks the atmosphere will indeed keep warming??) And until this most basic of points is cleared up, the rest of what you wrote is built on an unsupported premise. —
General don’t agree. Though I will say it’s well known that during Solar Max, Earth atmosphere does expand- as it affect the orbit of objects in low orbit [including ISS].
So Earth’s atmosphere above 100 km, becomes denser during Solar Max. And one should keep in mind Earth atmosphere is generally regarded to extend to about 800 km above the surface.
We talking about a very thin atmosphere, so thin that objects traveling around 10 Km/sec are affected by this atmosphere, whereas in vacuum of space or even atmosphere of moon, if anything traveling around 10 km/sec is not really affected- by this much thinner atmosphere.
“As most Boy and Girl Scouts can testify, if you hold a marshmallow close to a roaring camp fire it puffs up. A well-roasted marshmallow can grow to nearly twice its normal size, doubling its allure to a voracious sweet tooth.
Something similar happens to Earth’s atmosphere every 11 years when the sunspot cycle nears maximum. As solar activity increases, extreme ultraviolet radiation (EUV) heats our planet’s gaseous envelope, causing it to swell and reach farther into space than normal. ”
http://science.nasa.gov/science-news/science-at-nasa/2000/ast30may_1m/
Now if anyone is worried about “global warming” causing Earth’s atmosphere to somehow boil away,
it’s not the case. Nor would say if Earth average temperature were 10 C warmer, that this would cause as much problem for satellites in low earth orbit, to the degree that Solar Max does.
Or different way to say this, is global warming has far more with reducing how cold Earth gets, than about how hot our atmosphere gets.
Kristian, you were good up to here:
” If the atmosphere then continues to warm from this stage (which it will without the ability to cool to space by radiating away the convectively transferred energy from the surface), then the surface will naturally need to warm also, so as to be able to maintain this gradient.”
Once the surface has warmed to it’s pre-atmopshere temperatures, it now is warm enough to once again radiate ALL of the incoming solar power away as thermal IR. There is no power left to heat the atmosphere beyond this point!
If the surface warmed above the pre-atmopshere temperatures, it would (using the numbers of this discussion) be receiving 240 W/m^2, but losing more than 240 W/m^2 to space AND losing some additional power to the atmosphere AND using some power on top of that to warm the ground! Now you are violating the 1st Law of thermodynamics!
*************************************************************************
“… the surface ISN’T constantly warming and the tropopause ISN’T constantly lifting (they’re in a dynamic steady state). Why? How? Because of the constant tropospheric radiative cooling to space performed by the IR-active gases … ”
Yes! With IR active gases, the surface @ 255 K is no longer radiating 240 W/m^2 to space; it is radiative less because it is radiating (at least in part) to the much warmer atmosphere, which inhibits the heat lost from the surface. THIS is what allows the surface to warm. The mere presence of an atmosphere is not enough, as shown by the violations of the 1st and/or 2nd Laws discussed above.
Tim Folkerts says: September 1, 2014 at 12:58 pm
Will says: “This Earth best we know with a surface reflectivity from 2.5 microns to 200 microns of 85% (15% emissivity).”
Actual Statement from Will Janoschka September 1, 2014 at 5:22 am
(“Let me give you a thought problem much closer to “this is”, rather than pink dragons. This Earth best we know with a surface reflectivity from 2.5 microns to 200 microns of 85% (15% emissivity).
No other changes to surface emissivity, amospheric WV, clouds, or production of WV in any manner. Just what is your predicted surface temperature? And why is that temperature so much higher than what is measured. All your fraudsters picked “high” long-wave emissivity from the toilet, with absolutely no measured evidence, and no understanding of Earth’s atmosphere!”)
“Once again, you are pulling numbers from thin air, while claiming every published scientific paper and every published engineering table is wrong, as if you and you alone are privileged with the right answers.”
Yes! to show that you are incapable of solving even the simplest situation. Can you not even ask someone capable? “Let me give you a thought problem much closer to “this is”, rather than pink dragons.”
If I propose a “thought problem” I get to set the parameters for thought! You do not get to replace them with nonsense quotes from the Bible of the antropogenic, written by Rev. James Hansen. Like your fantasy of 255 Kelvin and 0.95 emissivity in the IR.
If you cannot do the math to get a revelent temperature with a 0.15 IR emisivity, why should anyone believe your nonsense bible? Please show any relevent engineering documents showing an emissivity of 0.85 from 2.5 microns to 200 microns at an angle of 70 degrees from normal?
“The earth you know may have a surface emissivity of ~ 0.15, but the earth we (the entire scientific and engineering community) know has a surface emissivity of ~ 0.85.”
You Tim Folkert, are definitely not part of the “scientific and engineering community”, but only an arrogant academic with no experience of this physical. Please tell us of any measurement of emissivity of any part of the Earth’s surface that you have done, or even witnessed?
“Convince me otherwise. Convince me that generations of scientists and engineers who make their living understanding thermal IR have gotten it horrible wrong.”
I am part of the engineering community making their living understanding thermal IR. I do not get it wrong because I do not listen to what Arrogant Academics preach, I measure instead.
I have no desire to attempt to convince you of anything, AFAIK you are hopeless.
“Tell me how IR thermometers calibrated with “standard” emissivity values give the right answer even though those emissivities are off by an order of magnitude.”
When was the last time you calibrated one? What is “standard emissivity”, or “right answer”?
In a 20 Celsius room, and a thermometric measured surface temperature of mild steel of 100 Celsius, what did the IR thermometer indicate at normal incidence? What did it indicate at 60 degrees from normal? Do you understand the instruction manual where they carefully explain the difference between thermometric, and radiometric, temperature?
A complete intentional fraud by persons yet to be indicted! (Is that OK, TimC?)
Ben Wouters said, August 31, 2014 at 10:29 am
“Afaik no one has been drilling to any depth on the moon yet. ;-)”
How deep would one need to dig to measure the bedrock temperature of the Moon?
According to Ashwin Vasavada, the regolith is such a poor conductor of heat that the temperature is constant at depths lower than 0.4 meters. My model confirms this which is why I assumed the regolith to be 0.5 meters deep. Thus if you take the Moon’s temperature at a depth greater than 0.4 meters you are reading the bedrock temperature.
Take a look at Figure 7 here:
http://onlinelibrary.wiley.com/doi/10.1029/2011JE003987/pdf
Note that the bedrock temperature is 240 K so I adopted that figure for my model.
Will Janoschka takes issue with my simplistic assumption that the average emissivity of the Moon in the thermal IR spectrum is 0.95.
When I entered that constant into my model the night time variance corresponded to the Diviner measurements with an rms error of less than 0.5 Kelvin.
So I moved on to the daytime short wave absorbance based on a Bond Albedo of 0.11. The rms error was several Kelvin so I concluded that the absorbance was non-Lambertian. Vasavada, Tim Channon and “br” had already noticed this and you can see a comparison of our model assumptions in the chart “ABSORBANCE vs. ANGLE OF INCIDENCE”.
The Diviner LRE uses seven detectors with spectra that overlap to estimate the Moon’s surface temperature. The measurements cover the entire surface of the Moon for more than a year. Even the “Level 3” data caused my laptop to crash so I am indebted to Tim Channon for providing temperature profiles that my puny laptop can handle.
gallopingcamel says: September 2, 2014 at 7:28 am
“Will Janoschka takes issue with my simplistic assumption that the average emissivity of the Moon in the thermal IR spectrum is 0.95.”
Where, when? I have never been to the Moon to measure. All I mentioned was that the Lunar regolith increases in conductivty and emissivity when stepped upon. I did ask TimC about oblique emissivity on this earth.
“When I entered that constant into my model the night time variance corresponded to the Diviner measurements with an rms error of less than 0.5 Kelvin.”
OK try 1.00! The Diviner instruments, I think, use measurements made normal to the surface.
They are trying for the most accurate radiometric temperature the can get. Not measurement of how the exitance may vary with angle from normal.
“So I moved on to the daytime short wave absorbance based on a Bond Albedo of 0.11. The rms error was several Kelvin so I concluded that the absorbance was non-Lambertian. Vasavada, Tim Channon and “br” had already noticed this and you can see a comparison of our model assumptions in the chart “ABSORBANCE vs. ANGLE OF INCIDENCE”.”
I do not understand your use of the word “absorbance”. The word only applies to partialy transmissive media! I would consider the regolith to be opaque! Lambertian does not mean the same emissivity/reflectivity at all wavelengths, it means same emissivity/reflectivity at all directions from normal at each wavelength individually. The emissivity/reflectivity are surface properties.
Emittance/Absorptance/Reflectance are used for what happens at the surface at various angles.
All are subject to geometry and must decrease by cosine (theta) “angle from normal” or more, and all go to zero at 90 degrees, As that surface disappears! If emissivity decreases at angels from normal reflectivity increases. This is important if you think about the gross nonsense of “albedo”, which ignores all forward scattering of incident flux.
“The Diviner LRE uses seven detectors with spectra that overlap to estimate the Moon’s surface temperature. The measurements cover the entire surface of the Moon for more than a year. Even the “Level 3″ data caused my laptop to crash so I am indebted to Tim Channon for providing temperature profiles that my puny laptop can handle.”
Indeed The radiance from seven bands can be used to recreate both black-body temperature and emissivity by matching to the Planck integral curve iteratively, adjusting apparent emissivity for each band, and converging on a best fit for one temperature and seven different vertical emissivities (reflectivity below 2.5 microns on the sun side). If you wish to know the calculated emissivities at each waveband, find the guy (not manager) at JPL that did it! Avoid the PR folk. Before asking learn the precise engineering language to use! Practice with all here! Here you get many tries. With an engineer at JPL you get only one try, else you are known by all as a kook! The guys there love to brag og their own effort, let them! They save the BS for the PR folk.
@ Will Janoschka.
Given my limited computer resources I have to look for simplfications such as the 0.95 emissivity which turned out to “close enough for government work”.
As I am looking at a transient heat transfer problem it would be nice if the Moon was a “Black Body”. As that is not the case I use the concept of “Absorbance” to arrive at a constant that correponds to the energy absorbed by the Moon’s surface relative to a black body at the same temperature. It is not necessary to know the details of the absorption spectrum.
With absorbance = 0.9, the model was in good agreement with measured temperatures at noon but the accuracy declined for lower angles of incidence. In my model the absorbance = 0.5 @ grazing incidence. IMHO this demonstrates that the Moon is non-Lambertian when it comes to absorption of incoming solar radiation.
I just realized that the Vasavada link I quoted is pay-walled so here is a more convenient link:
Looking ahead, my next task is to reproduce Robinson & Catling’s work on bodies with atmospheres. While this is a much more complex problem there are some simplifications that will help. For example I propose to remov the time variable (static heat transfer rather than transient). I am hoping it will run on my puny laptop.
gbaikie says, September 1, 2014 at 9:34 pm:
“–Tim Folkerts says, September 1, 2014 at 8:16 pm:
“If you remove the atmosphere’s ability to shed its energy to space, but at the same time let the Sun keep heating the surface, the atmosphere will keep warming and expand. “
That is one of the most obvious violations of the 2nd Law of Thermodynamics I have ever seen. I suspect you will have a hard time getting anyone to back you on this point. (Anyone care to speak up here who thinks the atmosphere will indeed keep warming??) And until this most basic of points is cleared up, the rest of what you wrote is built on an unsupported premise. —”
That’s my quote, gbaikie, not Tim’s. It is based on our hypothetical world with no IR-active gases in the atmosphere.
How is that situation a violation of the 2nd Law of Thermodynamics? There will never be a heat flux from cooler to warmer. And how, if you keep heating an atmosphere (convectively) that isn’t in turn able to cool (radiatively to space), how will it NOT keep warming and expanding until it starts being physically shed into space?
This is how insulation works. Very simple. If you let energy from the (heated) hot body warm the insulating layer (surrounding cold body), then the (heated) hot body will have to warm also. The more you warm the insulation, the warmer the hot body will become also, to maintain the temp gradient and the heat flow. (Remember, the hot body is itself heated. There is heat IN (from its hot reservoir) and heat OUT (to its cold reservoir, the insulating layer).)
Folkerts here simply wants to postulate as a factual outcome (because it suits his ‘radiative world view’) a situation where the warm (very warm), massive atmosphere all of a sudden will no longer insulate the surface. Because convective transfer is apparently just shut off. And voilà! a purely radiative black body situation for the surface, at 255K, even with the atmosphere still in place and even with the Sun still shining down on it. Ordinary gas dynamics and processes are obviously just considered an inconvenience, in the way, a nuisance, in Tim’s Planck universe. So he simply chooses to strike them out, set them mathematically to zero and be rid of them. To get where he wants: Radiation only.
Sorry, but the real world doesn’t work that way. There IS and WILL BE convective energy transfer going on from surface to atmosphere. If the atmosphere somehow ‘tries’ to reduce the environmental lapse rate, the normal temp gradient established away from the solar-heated surface by the tight coupling of solar surface heating and its convective response, stabilised to fluctuate around the gravity-based adiabatic lapse rate, by warming up, then convection will tend towards weakening and as a result, normal convective cooling of the surface will be hampered, meaning energy will naturally and automatically accumulate at/below the surface, raising its temperature, so that the temp gradient (the normal lapse rate) can be restored. It’s all about tropospheric stability. An ELR out of sync with the ALR will always be an inherently unstable situation. It can work out locally and even regionally for a while, but never globally. Without something happening.
A heated surface surrounded by a fluid (like a gas) in a gravity field that does not absorb or emit EMR, has to become pretty hot for the radiative heat loss to grow so much as to render the parallel convective heat loss negligible, enabling us to in effect consider that surface a pure radiator to space. Think red-hot iron or a bonfire. The surface of the Earth couldn’t become that hot and still keep its IR-transparent atmosphere. And a surface heated by the Sun and overlain by a sea of air as heavy as our atmosphere could certainly not become a pure radiator like that at -18C. And so would need to warm.
Tim F starts his response with:
“Once the surface has warmed to it’s pre-atmopshere temperatures”
and that is precisely where the sleight of hand is…at what point did the surface cool? Kristian was saying the surface would warm following the addition of an IR-transparent atmosphere, not cool. So it won’t be “warming back up to its pre-atmosphere temperature”. It (the surface) will (according to Kristian) be warming from the moment the IR-transparent atmosphere is “added” to the bare Earth. Not cooling first then warming back up to the temperature it was before the atmosphere was added.
The way I see it, Kristian’s point is, in the bare Earth scenario you have 240 w/m2 reaching the surface from the sun on average and 240 w/m2 leaving it, setting a surface temp of 255K because it is an entirely radiative scenario. But from the very first moment the IR-transparent atmosphere is introduced to the bare Earth, you no longer have 240 w/m2 leaving, because some of that available energy is warming the atmosphere through conduction/convection. So you immediately have a “radiative forcing” present as less radiation is leaving the planet than coming in. So surface temp must increase IMMEDIATELY since the forcing is present IMMEDIATELY in this highly hypothetical situation. There is no “it cools first because its losing energy through conduction/convection”…the loss of energy through cond/conv is the REASON the radiative forcing exists.
If I understand correctly?
gallopingcamel says: September 2, 2014 at 3:33 pm
@ Will Janoschka.
“Given my limited computer resources I have to look for simplfications such as the 0.95 emissivity which turned out to “close enough for government work”.
As I am looking at a transient heat transfer problem it would be nice if the Moon was a “Black Body”. As that is not the case I use the concept of “Absorbance” to arrive at a constant that correponds to the energy absorbed by the Moon’s surface relative to a black body at the same temperature. It is not necessary to know the details of the absorption spectrum.
With absorbance = 0.9, the model was in good agreement with measured temperatures at noon but the accuracy declined for lower angles of incidence. In my model the absorbance = 0.5 @ grazing incidence. IMHO this demonstrates that the Moon is non-Lambertian when it comes to absorption of incoming solar radiation.
I just realized that the Vasavada link I quoted is pay-walled so here is a more convenient link:
Looking ahead, my next task is to reproduce Robinson & Catling’s work on bodies with atmospheres. While this is a much more complex problem there are some simplifications that will help. For example I propose to remov the time variable (static heat transfer rather than transient). I am hoping it will run on my puny laptop.”
Thanks for the link! that helps to understand what you are doing. Your curve of surface temperature seems to contain a thermal lag in surface temperature. Where is directly at the Sun in your temperature graph? The Moon cannot absorb solar flux at any terminator. No absorption no albedo all flux misses the Moon, and goes directly to the opposing hemisphere. Instead of midnight being 12 make it 10 or 11.
Actually at any solar incidence greater than 45 degrees the reflective is into the opposing hemisphere, and is not part of “albedo”. It is the rest, that also is not absorbed. The Moon is cold. It receives much less flux than the ClimAstrologists claim, as does the Earth.
@ Will Janoschka,
In my “Figure 2”, Θ = 0 means “Noon” when the sun is directly overhead.
While my absorbance at grazing incidence (0.5) is lower than Vavasada and tchannon assume (0.4), it does not make much difference as the models are insensitive to absorbance near the terminator because Cos Θ is approaching zero.
Given the low conductivity of lunar regolith, the thermal lag is small. If you look closely at the blue curve in “Figure 1” you will see it lags the green and yellow curves noticeably around dawn (hour 18). That lag is small considering the 29 times difference in the rate of rotation. I think it is fair to claim that the Moon’s temperature is weakly affected by the rate of rotation.
gallopingcamel says: September 3, 2014 at 4:41 am
“@ Will Janoschka,
In my “Figure 2″, Θ = 0 means “Noon” when the sun is directly overhead.”
OK but this must be computation rather than measurement. I do not wish to depreciate your effort. Your charts defy geometry.
“While my absorbance at grazing incidence (0.5) is lower than Vavasada and tchannon assume (0.4), it does not make much difference as the models are insensitive to absorbance near the terminator because Cos Θ is approaching zero.”
Models indeed. they are incorect. you can have no energy transfer with no area Cos(PI/2).
Please define “absorbance”?
“Given the low conductivity of lunar regolith, the thermal lag is small. If you look closely at the blue curve in “Figure 1″ you will see it lags the green and yellow curves noticeably around dawn (hour 18). That lag is small considering the 29 times difference in the rate of rotation. I think it is fair to claim that the Moon’s temperature is weakly affected by the rate of rotation.”
Low conductivty, and low emissivity will only increase the lag. Your graphs have less slope than
Cos(theta) between PI/6 and PI/2.
Will, the original intention of the work was merely building a simulation to see what happened at a fundamental level. I knew what ought to work not whether it would since it had not been done before. I matched the little actual data I had.
The ground data fitted much the same but unrelated terrestrial data.
None of this was a surprise.
There was a problem, a significant error. I extracted the error curve and gave it so some fancy software here (unpublished) which ate it for breakfast. The answer was unexpected, perplexing. The solution was raise the irradiation angle law to a fractional power. The remaining error is very small.
I asked for idea or theories, an explanation. No takers.
I also checked lunar images which seemed to confirm the empirical law.
People asked so I extended to a globe.
In public that was it. In private I did rather more, unpublished. Backcalculating to substance (ground characteristics) this gave a sensible answer.
As I originally did this work there is no assumption about reflectance, admittance, whatever, is irrelevant. There is no figure in there, no-where to put it. A surface acts flux to heat and heat to flux converter. Certain people didn’t like this so others introduced the concept. Trying to explain this point gets no-where, never mind, been here before (out on a limb unable to pass across what I mean).
Some time later Peter picked up the [modelling] concept [used] and tried to use what other data is known, properties of various things to see if the results made sense. He succeeded.
All those involved have looked a lot at the surface problem. This is not a focus of major interest.
I could probably figure out more if I working on the Diviner data but it is to me a mess are better uses of my time.
Above is probably illegible, too tired.
[edited as moderator after sleeping –tim]
I add, it seems quite likely the angular law is combined with in the material law and error law, perhaps convolving.
—Kristian says:
September 2, 2014 at 5:21 pm
gbaikie says, September 1, 2014 at 9:34 pm:
“–Tim Folkerts says, September 1, 2014 at 8:16 pm:
“If you remove the atmosphere’s ability to shed its energy to space, but at the same time let the Sun keep heating the surface, the atmosphere will keep warming and expand. “
That is one of the most obvious violations of the 2nd Law of Thermodynamics I have ever seen. I suspect you will have a hard time getting anyone to back you on this point. (Anyone care to speak up here who thinks the atmosphere will indeed keep warming??) And until this most basic of points is cleared up, the rest of what you wrote is built on an unsupported premise. —”
That’s my quote, gbaikie, not Tim’s. It is based on our hypothetical world with no IR-active gases in the atmosphere.—
Yes, I know. I would say a world with no IR-active gases will not keep warming and expanding.
Or a Venus replacing it’s CO2 with nitrogen, will not keep expanding.
Or it would be little different than what Venus currently is. Probably a bit colder.
Or Venus very hot atmosphere doesn’t make Venus atmosphere very large- though it’s massive atmosphere makes it a bit larger. Or an 1 earth atm it’s at about 50 km. Or roughly it’s atmosphere is plus 50 km as compared to Earth. If instead it had say 1000 times earth’s atmosphere- it would be much hotter at surface and is be something like {very wild guess} +100 km instead of +50 Km.
Or when Earth or Venus formed it was much, much hotter than it it now and it’s atmosphere would not have been much higher [unless it had much more atmosphere- say huge amount of helium and hydrogen [most common elements in solar system or universe]. Oh another element which makes a higher atmosphere is lower gravity- and proto-earth had considerably less gravity.
Plus if put an earth size atmosphere on Mars, Mars would have larger atmosphere than Earth [maybe higher than Venus]. And a very cold world- a lot dimmer and enough atmosphere to make it miserably cold for any human.
–How is that situation a violation of the 2nd Law of Thermodynamics? There will never be a heat flux from cooler to warmer. And how, if you keep heating an atmosphere (convectively) that isn’t in turn able to cool (radiatively to space), how will it NOT keep warming and expanding until it starts being physically shed into space?–
I would not say it violates 2nd Law of Thermodynamics. Rather it’s not good understanding scale and gravity. Or it’s going to have lapse rate, yes?
Let’s see:
“There will never be a heat flux from cooler to warmer.”
Maybe/maybe not- or let’s say that is correct. So what?
“And how, if you keep heating an atmosphere (convectively) that isn’t in turn able to cool (radiatively to space)”
Venus has very little [or none] convection at it’s surface. Or sunlight doesn’t really reach surface and sunlight certainly can not warm the surface- Venus convection is higher up in it’s in atmosphere.
Let’s “try” [and fail] to think about earth being as hot as venus surface. The sunlight lacks the power to warm such a hot surface.Maybe this what is meant by “violation of the 2nd Law of Thermodynamics”. In fact the sunlight lacks the power to warm it even if Earth was at Mercury distance. Or it’s impossible for sunlight to warm earth’s surface to this temperature.
Or the temperature of sunlight at earth distance is about 120 C- or around the temperature of lunar surface at noon.
Or unless one has more atmosphere than earth has, it’s impossible to reach a temperature much higher than 80 C from sunlight [unless the sunlight is magnified].
gbaikie says, September 3, 2014 at 7:24 am:
“Yes, I know. I would say a world with no IR-active gases will not keep warming and expanding.
Or a Venus replacing it’s CO2 with nitrogen, will not keep expanding.
Or it would be little different than what Venus currently is. Probably a bit colder.”
How do you know? Have you tried it? You’re just assuming and asserting stuff here. Just like Tim is doing. And, yes, just like I’m doing. Because this is an entirely hypothetical situation. So no one REALLY knows what would actually happen in the end. Maybe there are factors that none of us thought of that will turn out decisive.
What we DO know is that there exists no atmosphere that we know of that contains no radiative gases whatsoever. Your pure N2 atmosphere is nowhere to be found. Neither is our planet Earth without any IR-active gases in its atmosphere. I say it simply wouldn’t work. We COULDN’T have an atmosphere without IR-active gases in it.
Well, either way, that’s why such a discussion will always just lead to entrenched positions without any real means of resolving the issue. A pointless endeavour in the end. I will never convince Tim. And he will certainly never convince me. So what’s the point?
Much more interesting, then, to look at our world as it is. REAL Earth. What do we observe? And what do we NOT observe? What is REALLY going on?
Why will the surface cool more slowly with a humid atmosphere and/or clouds above it than with a dry atmosphere and/or no clouds above it? This topic deserves a post of its own. Because the way this question is answered is central to the entire ‘atmospheric radiative greenhouse effect’ (rGHE) debate.
You talk about Venus. Let’s mention Mars also. Mars has a BB temperature when seen from space of 210K. There’s a thin atmosphere containing 95% CO2 resting on top of its global surface, about 8 times (IIRC) the absolute atmospheric content on Earth within a volume about 2.5 times as small. There is both convection and radiation going on from the surface. There is also certainly radiation being emitted from the atmosphere to space. But, the many surface measurements made by the different landers and rovers spread around the planet’s surface, clearly indicate that Mars’ ACTUAL physical global mean surface temperature would be LOWER than 210K by a few degrees. At least no higher. Even the measurements made during the Martian summer and/or in the equatorial/tropical belt (which includes most of the measurements from Mars) have recorded no higher average temps than 215-225K, which seems to be where the official (most quoted) number is derived from (-55C). This is surely way to high for an all-year, fully global mean estimate.
–Or a Venus replacing it’s CO2 with nitrogen, will not keep expanding.
Or it would be little different than what Venus currently is. Probably a bit colder.”
How do you know? Have you tried it? You’re just assuming and asserting stuff here. Just like Tim is doing. And, yes, just like I’m doing. Because this is an entirely hypothetical situation. So no one REALLY knows what would actually happen in the end. Maybe there are factors that none of us thought of that will turn out decisive.–
That is true. What I know is based theories- that I am aware of.
But I know of no theory which would allow for such constant expansion of an atmosphere.
Hydrogen and helium are not greenhouse gases and they comprise most of known mass
in this universe.
And therefore what you are suggesting could fundamentally alter our entire view of the formation of this universe. Perhaps this would be exciting, but it seems one needs the provide some evidence and/or theory [in which one could then use to look for specific evidence which would then support it].
–What we DO know is that there exists no atmosphere that we know of that contains no radiative gases whatsoever. —
As said above one can an overwhelming amount of hydrogen and helium- and these gases are quite unlike greenhouse gases- and they will even be more dominant in upper atmosphere- because these molecule move very fast compared to other gases- so all of our gas giants would fit this requirement of not having greenhouse gases [in any quantity of significant- or you have to go deep [+100 km deep] into the atmosphere to find them. Or one detect them if have impactor hitting the atmosphere or very powerful storm which drives them upward [say, red spot of Jupiter, perhaps to some amount.].
Though if only need small amounts of greenhouse gases, one say all gases [including nitrogen] will radiate some kind radiation in very small amounts- or greenhouse gases are notable in sense that comparatively, they radiate a lot and a lot at a particular wavelength [temperatures found near the Earth surface]. Or other way to say this is if include thermosphere of Earth, one has very high temperatures involved in which non-greenhouse gases would radiate more. Wiki:
“Thermospheric temperatures increase with altitude due to absorption of highly energetic solar radiation. Temperatures are highly dependent on solar activity, and can rise to 2,000 °C (3,630 °F). ”
http://en.wikipedia.org/wiki/Thermosphere
And here, we circled back to issue of solar max affecting the much higher atmosphere on Earth.
tchannon says: September 3, 2014 at 7:09 am
“As I originally did this work there is no assumption about reflectance, admittance, whatever, is irrelevant. There is no figure in there, no-where to put it. A surface acts flux to heat and heat to flux converter. Certain people didn’t like this so others introduced the concept. Trying to explain this point gets no-where, never mind, been here before (out on a limb unable to pass across what I mean).”
Thanks Tim, “A surface acts flux to heat and heat to flux converter.” I agree but highly nonlinear with any temperature! (1-albedo) is still flux not temperature. No need that temperature obey some geometrical function.
“I add, it seems quite likely the angular law is combined with in the material law and error law, perhaps convolving.”
In the VASAVADA ET AL paper I cannot find the channel 1,2 numbers, so no info on albedo with angle. The reflected flux should follow cos(theta) as the radiometer has constant solid angle (nadir), while solar illumination is spread as 1/cos(theta), if the surface is reasonably regular.
Kristian says: September 3, 2014 at 5:08 pm
gbaikie says, September 3, 2014 at 7:24 am:
(“Yes, I know. I would say a world with no IR-active gases will not keep warming and expanding.
Or a Venus replacing it’s CO2 with nitrogen, will not keep expanding.
Or it would be little different than what Venus currently is. Probably a bit colder.”)
“How do you know? Have you tried it? You’re just assuming and asserting stuff here. Just like Tim is doing. And, yes, just like I’m doing. Because this is an entirely hypothetical situation. So no one REALLY knows what would actually happen in the end. Maybe there are factors that none of us thought of that will turn out decisive.”
I think you nailed it, “So no one REALLY knows what would actually happen in the end.”
The ClimAstrologists Have no clue, only fantasy conjecture, which they call theory.
gbaikie says, September 3, 2014 at 8:32 pm:
“That is true. What I know is based theories- that I am aware of.
But I know of no theory which would allow for such constant expansion of an atmosphere.”
Really? It is no stranger than inflating a balloon until it bursts. You could equally keep heating the air inside the balloon and eventually it would blow. Same principle. An atmosphere that has no way of ridding itself of added energy will warm and expand until it is blown off into space. Nothing mysterious about that … in principle. It is a strange situation, I agree. But that is how our hypothetical situation is set.
Kristian says, September 3, 2014 at 5:08 pm:
“Why will the surface cool more slowly with a humid atmosphere and/or clouds above it than with a dry atmosphere and/or no clouds above it?”
To add to this question: Why will there be less IR radiated to space per unit of time from humid air and clouds than from drier air with no clouds?
Is it a ‘radiative’ effect as taken for granted by ‘Climate ScienceTM’? Or is it really a simple ‘heat capacity’ effect?
Kristian says: September 3, 2014 at 11:05 pm
Kristian says, September 3, 2014 at 5:08 pm:
“Why will the surface cool more slowly with a humid atmosphere and/or clouds above it than with a dry atmosphere and/or no clouds above it.
Lower surface temperature, is a characteristic of higher humidity air.
“To add to this question: Why will there be less IR radiated to space per unit of time from humid air and clouds than from drier air with no clouds?”
It is not less it is more, as the latent heat in condensation converts to sensible heat and is dispatched to space as EMR. To 6.8 Kelvin space 288 vs 255 Kelvin is trivial.
“Is it a ‘radiative’ effect as taken for granted by ‘Climate ScienceTM’? Or is it really a simple ‘heat capacity’ effect?”
I think it is both, To dispach thermal EMR that power must come from sensible heat and temperature will drop, unless that energy is replaced.
AlecM says:
August 29, 2014 at 9:47 am (Edit)
@Tim Folkerts:
Tim F said: the “standard model more properly assumes a single emitter at 0 km altitude, ie that the atmosphere is completely transparent to IR so that all IR comes directly from the surface. It then calculates that such a surface would be -18 C.
The radiative equilibrium mean temperature of the airless Earth with Space as the opposite radiative emitter would, because there are no clouds or ice, be for 341 W/m^2. The answer would be between -2 and +5 deg C depending on your assumptions about IR emissivities. Your claim that it would be -18 deg C, implying 238.5 w/m^2 warming, is one of the clever lies in the IPCC scam.
Tim Folkerts always avoids answering this point. He (and the global warming theorists) likes to have his cake and eat it: they need to postulate a cloudless water vapour free atmosphere which nonetheless has a 0.3 albedo in order to make room for fantasy effects of co2.
Kristian says:
September 3, 2014 at 10:59 pm
gbaikie says, September 3, 2014 at 8:32 pm:
“That is true. What I know is based theories- that I am aware of.
But I know of no theory which would allow for such constant expansion of an atmosphere.”
Really? It is no stranger than inflating a balloon until it bursts. You could equally keep heating the air inside the balloon and eventually it would blow. Same principle.
Suppose one had a pipe which was 1000 feet long. And one caps one end.
You put the 1000 foot pipe in the water. And water would come in the open side of the pipe, until such time as water and the air inside the pipe would destablize the pipe, and cause the pipe to flip
vertical. One can see similar thing by throwing a beer bottle in a lake- though with beer bottle the open end would up [due to shape of beer bottle] and with the pipe the open end would be down.
So one could have the 1000 ft pipe with say most of it filled with water and less than 1/2 filled with air. Let’s say 600 ft filled with water and 400 ft filled with air. Allowing about 300 ft of pipe to be above the water. Now let’s say one drills a hole in top of pipe and put valve in it, and let air out until there is 200 ft air in the pipe [making pipe less than 200 feet above the water.
And say the air pressure inside the pipe is 10 psi.
So if this was the case, then the air pressure would more or less stay constant. So if add air so there is 300 ft of it, then air pressure remains around 10 psi [and pipe is higher above the water.
And reduce air so there 100 feet of air, then also it remains about at 10 psi [and the pipe is less high above the water.
So rather than add air to raise the pipe, one want s to heat the air.
So if you have 200 feet of air inside pipe, how much warmer would you need to heat the air
to cause there to be 400 ft of air instead of 200 ft?
Clue:
“The effect of heat on the expansion of gases is stated in Charles’ Law. A gas occupies 1/273.15 more space for each 1° C. rise in temperature and, conversely, 1/273.15 less space for each 1° C. drop in temperature. ”
http://science.howstuffworks.com/dictionary/physics-terms/expansion-info.htm
And once you added enough heat to warm the air so it’s hot enough so there is 400 feet of air. how much further heating would you need to make 800 ft of air?
Will Janoschka says, September 3, 2014 at 11:35 pm:
“Kristian says: September 3, 2014 at 11:05 pm
Kristian says, September 3, 2014 at 5:08 pm:
“Why will the surface cool more slowly with a humid atmosphere and/or clouds above it than with a dry atmosphere and/or no clouds above it.
Lower surface temperature, is a characteristic of higher humidity air.”
Yes, but that is because of the NET effect of high humidity air. Tropical rainforest areas are consistently cooler by several degrees than tropical/subtropical desert areas on an annual basis. And this is ONLY because of the much larger amounts of H2O in the tropospheric column above the former.
I’m talking now, however, only of the reducing effect on surface COOLING rates, not of the countering (and evidently larger all in all) reducing effect on surface HEATING rates.
““To add to this question: Why will there be less IR radiated to space per unit of time from humid air and clouds than from drier air with no clouds?”
It is not less it is more, as the latent heat in condensation converts to sensible heat and is dispatched to space as EMR. To 6.8 Kelvin space 288 vs 255 Kelvin is trivial.”
Yes, and this is kind of an ironic (and maybe subtle) point that seems to pass the ‘climate establishment’ unnoticed.
Humid air and clouds per se actually DO end up radiating less to space per unit of time than dry air and clear skies. On average, across the diurnal cycle. But this is only half the truth. What one tends to forget, is how much more energy the former can and do hold compared to the latter. And how much warmer the troposphere tends to become with more H2O being injected into it from the surface. The troposphere is actually mainly warmed from the surface by the condensation of water vapour into droplets, clouds and precipitation:
Notice here how esteemed climate researchers Norman Loeb and Joel Susskind miss this point completely:
“Atmospheric temperature and water vapor mixing ratio contributions to LW TOA flux variability are negatively correlated (…), with lower (drier) temperature (water vapor) changes during the cold phase of ENSO reducing (enhancing) LW TOA flux. Overall, atmospheric temperature variations have a greater impact on LW TOA flux variability than water vapor changes.”
Click to access Loeb_et_al_ISSI_Surv_Geophys_2012.pdf
“OLR is very sensitive to the concentration of mid-upper tropospheric water vapor in very moist (i.e., most tropical) areas, in the sense that increasing water vapor concentration increases atmospheric absorption in some spectral bands and therefore lowers OLR, everything else being equal. OLR also decreases with increasing cloud cover, especially for high clouds.”
“The paper (…) used anomaly time series of surface skin temperature, mid-tropospheric water vapor, and cloud amounts derived from analysis of AIRS sounder data over this time period to explain why global and tropical mean OLR anomaly time series are positively correlated with the El Niño Index, which like global and tropical mean OLR, has decreased on the average over this time period as a result of phases of El Niño/La Niña oscillations.”
Click to access 20120012822.pdf
How do they figure the troposphere (especially the tropical one) warms in the first place during a positive ENSO event (an El Niño)?
It is of course warmed by a pronounced increase in the release of ‘latent heat’ (diabatic heating) in the tropospheric column, originally brought up from the surface by deep moist convection. More water vapour is transferred to the troposphere than usual. What happens when this finally condenses? Warming.
So the substantial tropospheric warming leading to the increased OLR through the ToA during positive ENSO events specifically results from more H2O brought up into the troposphere by strengthened convection, a direct response (negative feedback) to increased surface temps.
In other words: More energy transferred from the surface to the air above and transported high up into the troposphere by enhanced convection will ultimately lead to MORE radiation leaving the Earth system to space.
More energy from surface > more tropospheric warming > more emission to space.
The opposite happens during negative ENSO events: Less H2O from the surface > cooler troposphere > reduced OLR.
From Trenberth et al. 2002:
“ENSO events contribute to coherent interannual and even decadal fluctuations in the global mean temperature and, as we have shown, the nature of the ENSO contribution is quite complex. Part of it involves the recharge and discharge of heat in the tropical Pacific Ocean. During and following El Niño events the heat from the ocean is redistributed within the tropical Pacific, and much of it is released to the atmosphere, creating local warming. However, a major part of the ocean heat loss is through evaporation, and the heat is realized in the atmosphere as latent heating in precipitation. This diabatic heating drives large-scale atmospheric overturning that influences the response throughout the tropics and subtropics as well as influencing other teleconnections within the extratropics.”
“OLR is often used as a proxy for precipitation owing to the dominant changes that occur in cloudiness and high cloud tops. However, the algorithms for translating OLR to precipitation do not account for the real effects of surface temperature changes of OLR or of small changes associated with heating of the atmosphere. High SSTs are associated with the atmospheric convergence, deep convection, and thus low OLR, but also with a flux of latent energy into the atmosphere, condensation and heating of the atmosphere, and transport of heat to higher latitudes where it can be radiated to space. This increase in OLR, seen especially in the subtropics, may be interpreted erroneously by precipitation algorithms as less precipitation.”
Click to access 2000JD000298.pdf
““Is it a ‘radiative’ effect as taken for granted by ‘Climate ScienceTM’? Or is it really a simple ‘heat capacity’ effect?”
I think it is both, To dispach thermal EMR that power must come from sensible heat and temperature will drop, unless that energy is replaced.”
Yes, it is both. Sort of. But the reason a surface under a humid atmosphere and/or a cloud cover cools more slowly than in dry and clear conditions is quite obviously because of the larger ‘heat capacity’ of humid air and clouds than that of dry air. Humid air and clouds simply hold more energy per degree of temperature and thus will cool more slowly to space even if releasing as much IR per unit of time as the dry air. It’s all a matter of temperature gradients. If the atmosphere cools more slowly to space, then the surface below will in turn cool more slowly to the atmosphere. Basic stuff. (Also, ‘latent heat’ is released during the night as the air cools, slowing the process even more.)
The OLR thing is a bit more complex. Especially in the tropics during the day, humid air and clouds will transmit a lot less of the IR through to space than dry air and clear skies. Simply because they are storing up on energy to a much greater extent – higher ‘heat capacity’ means higher energy storage. (At the same time, humid air and clouds will greatly reduce surface heating rates during the day compared to dry and clear conditions.)
During the night, there is no reason to assume that humid air and clouds will radiate less energy to space per unit of time than a dry/clear atmosphere. Depends a lot on starting temps. But the cloud fraction is most likely larger during daytime than during the night in the tropics. On average. Because of the tight coupling between surface heating and convection. Humidity probably doesn’t vary that much, at least not over the oceans.
The point is, the overall effect of more humid air and more clouds in the tropics/subtropics will be less IR to space. Because of the clear daytime effect, storing rather than releasing, and (most likely) generally more clouds during the bright hours of the day than during the dark.
The ‘slowed cooling’ effect during the night is not necessarily related to less IR to space, but rather more energy needing to be released for the same amount of cooling. Plus the release of ‘latent heat’.
Tallbloke says: “Tim Folkerts always avoids answering this point. He (and the global warming theorists) likes to have his cake and eat it: they need to postulate a cloudless water vapour free atmosphere which nonetheless has a 0.3 albedo in order to make room for fantasy effects of co2. ”
I hear what you are saying, but I don’t understand why you feel this is such an important point (for two separate reasons)
1) It is common to formulate a hypothesis where only one of many variables is changed, and then to examine what impact that one variable has. One interesting hypothesis deals with eliminating IR properties but keep everything else the same (ie ~ 33 K cooler due to IR).
If you want to hypothesize what happens when both the IR properties and the albedo effects of the atmosphere are removed but everything else is kept constant, that is also interesting (~ 33 cooler from IR and ~ 15 warmer due to albedo).
Of course, this hypothesis could also never “really” happen on Earth. You would “really” also change the albedo of the surface. You would remove the moderating effects of the oceans. Oh — and you would also have to do something like change the primordial composition of the solar system to eliminate water and CO2. OR maybe make the sun much hotter so it could evaporate away the atmosphere.
The point is that NEITHER hypothesis (same albedo or different albedo) is especially realistic since either would require massive other complex changes to make it “reasonable”.
2) IR does have effects on the Earth’s temperature. CO2 does have effects on IR. Neither of these is “fantasy”.
No one says that CO2 causes the entire warming effect from IR. No one says there are not feedbacks. But the way you say it, you seem to imply that ANY impact from CO2 is “fantasy”.
Pipe, double volume per 273 K, and how it compares to an atmosphere.
With the atmosphere, unlike a relatively short pipe length, must include lapse rate.
Take say 4 km of the troposphere, and add 273 K to the temperature at the surface, it
does quite double the volume.
So say there is lapse rate of 6.5 K per 1000 meter elevation. Say sea level temperature
is 20 C. 6.5 time 4 km is 26 K, or at 4000 meters the air is -6 C.
So one could add 20 + 273 at surface to get 293 C and at 4 km elevation with 6.5 lapse rate it is 267 C.
But if volume is increase by 2, the elevation would 8 km instead of 4 km- or add 4 km of height
one reducing temperature by another 26 C.
So to get the doubling one needs to increase temperature higher than adding 273 K at the surface,
or warming conditions may alter lapse rate [it may be less or more than 6.5 C per 1000 meters].
And as we all know, most of Earth atmosphere is in the troposphere- something like 18,000 meter elevation in tropics. [[Or one also say with higher troposphere in Tropics, most of atmosphere is in tropics and within that troposphere.]]
So in terms warming the entire atmosphere so that it’s atmospheric height increase, one can essentially ignore the atmosphere above the troposphere. Or because most of atmosphere is in troposphere it’s what is being doubled by this much warmer world.
So if one could add more than 273 K to surface air temperature, this add about 18 km to atmosphere’s height in the tropics.
Tim F: Thanks for your considered response.
You are right that “You would remove the moderating effects of the oceans.” Indeed I have made this very point to you in previous similar discussions when we’ve been discussing the effect of gravity acting on the mass of the atmosphere on temperature. Without gravity to retain the atmosphere, there would be no oceans because they would boil away to space. I’m glad to see the message is gradually getting through.
The key finding in this excellent post, if you can tear yourself away from your perennial talking points and examples, is that an ice covered airless Earth would be a lot warmer than the waterless Moon, due to ice’s conductivity (red line on fig1), and Earth’s rotation rate. Consider now how much warmer again an airless Earth illuminated from one side by a point source (the Sun) would be with a surface covered in a liquid substance that had a lowish ‘effective emissivity’ (say around 0.67) that could advect energy towards the poles efficiently (minimising Holder’s inequality). If this liquid were also to be capable of advecting energy downwards during the day and advecting it upwards at night such that the surface temperature of the liquid remained the same within a couple of degrees throughout the rotation period, what would the average surface temperature be?
Regarding your other points:
2) IR does have effects on the Earth’s temperature. CO2 does have effects on IR. Neither of these is “fantasy”.
True, but I contend that due to the inability of ‘back radiation’ from on high to reach the surface before convection whisks any sensible heat it generates back upwards, the effect on surface temperature is a lot smaller than naive theorists calculate, and is limited to the extent to which the slight warming of the upper air slows down convection from the surface.
No one says that CO2 causes the entire warming effect from IR. No one says there are not feedbacks. But the way you say it, you seem to imply that ANY impact from CO2 is “fantasy”.
Well I was being too elliptical then, because that’s not what I meant. The fantasy, as I see it, is the claim that 398ppm of co2 and 20,000ppm of water vapour causes the earth’s surface to be 33C warmer than 255K. The model presented in this post shows that an ice covered *airless* Earth would average around 250K. So the additional ~40K we are at has to be proportionally allocated to the following dominant factors:
1) warming effect of suppression of evaporation by the MASS of the atmosphere under gravity generating pressure at the surface, causing the ocean to warm to a temperature at which it can lose energy as fast as it gains it.
2) Warming effect of GHG’s – warming the upper air and thereby slowing convection from the surface.
3) Cooling effect of GHG’s – preventing solar energy from reaching the surface in the first place, and radiating energy to space.
4) Cooling net effect of cloud radiative forcing vs cloud albedo reflecting enegy back to space.
Ideas on expanding this list and quantifying effects welcome.
My current, very sketchy, estimate is:
4) -18K
3) -40K
2) +20K
1) +77K
tallbloke says, September 7, 2014 at 1:12 pm:
“2) Warming effect of GHG’s – warming the upper air and thereby slowing convection from the surface.”
Could you please expand on this one, tb? As I (and Konrad and more) see it, ‘GHGs’ don’t warm the upper air at all. They specifically cool it. Convective transfer brings surface energy up to the upper parts of the troposphere from where it can be radiated to space … by the ‘GHGs’.
It is the transport of warmer air into cooler air regions by convection, plus the release of ‘latent heat’ on the way up, plus the absorption of incoming solar radiation, that warms the tropospheric air column, convection being the process maintaining the distribution of energy (and hence temperatures) along the lapse rate from surface to tropopause.
If anything, the IR-active gases in the troposphere work towards strengthening convection, by warming low (absorption from the surface) and cooling high (emission to space).
tallbloke says: September 7, 2014 at 1:12 pm
” The model presented in this post shows that an ice covered airless Earth would average around 250K. So the additional ~40K we are at has to be proportionally allocated to the following dominant factors:”
Model shows 234K average, so the difference to explain is more like 56K.
” 1) warming effect of suppression of evaporation by the MASS of the atmosphere under gravity generating pressure at the surface, causing the ocean to warm to a temperature at which it can lose energy as fast as it gains it.”
Would you expand on how this mechanism caused the deep oceans to cool ~18K since the Cretaceous peek temperatures? Did the mass of the atmosphere reduce dramatically or did the oceans evaporate more, and if so why?
Also why the start of the current cycle of glacials / interglacials has nothing to do with the deep oceans temperature dropping below a certain threshold, ~ 3K above present temperatures?
gbaikie says, September 4, 2014 at 4:48 am
and
September 6, 2014 at 9:18 pm:
You seem to forget about convection. You’re only talking about pure thermal expansion. Your pipe and the atmosphere are far from analogous.
The key points in our hypothetical scenario are: the Sun continues to heat the surface, the atmosphere is expandable, but under the influence of gravity, the solar-heated surface and the atmosphere on top are naturally convectively coupled, there’s a lapse rate (a tropospheric temp gradient away from the solar-heated surface) to sustain, the atmosphere is NOT able to cool by radiation to space.
Tim claims this setup will end up in a steady – and purely surface>space radiative – state and hence make the surface COLD. I claim this setup will be unable to reach a steady state until the atmosphere is physically gone, because convective loss from a heated surface surrounded by a fluid in a gravity field is THE cooling mechanism of any such surface and so can never be simply ignored (unless the surface becomes very hot, much, much hotter than the current surface of the Earth), and hence ends up making the surface HOT (hotter and hotter) in the process.
I feel strongly that I have reality on my side on this one. Things and processes actually going on out there in the physical world. What Tim has is a mathematical model where convective loss is simply conveniently set to zero so that he can focus ONLY on the radiative balance between surface and space in his postulated steady state.
Graham W says, September 2, 2014 at 9:04 pm:
“The way I see it, Kristian’s point is, in the bare Earth scenario you have 240 w/m2 reaching the surface from the sun on average and 240 w/m2 leaving it, setting a surface temp of 255K because it is an entirely radiative scenario. But from the very first moment the IR-transparent atmosphere is introduced to the bare Earth, you no longer have 240 w/m2 leaving, because some of that available energy is warming the atmosphere through conduction/convection. So you immediately have a “radiative forcing” present as less radiation is leaving the planet than coming in. So surface temp must increase IMMEDIATELY since the forcing is present IMMEDIATELY in this highly hypothetical situation. There is no “it cools first because its losing energy through conduction/convection”…the loss of energy through cond/conv is the REASON the radiative forcing exists.
If I understand correctly?”
Well, yes. I would call it “radiative imbalance” rather than “radiative forcing”, but other than that, this seems to be a key point. The surface simply cannot reach a radiative equilibrium with its heat source (the Sun) until it becomes very hot indeed (or actually, until the atmosphere is gone altogether). Because it will naturally and automatically lose lots of the incoming energy from the Sun to the atmosphere by way of convective transfer.
Energy brought from the surface to the atmosphere is not easily brought back down. Some will during the night, but certainly not all of it. And hence energy will slowly accumulate in the expanding atmosphere. The surface, however, never stops functioning as the heat source of the atmosphere, as long as it’s still heated by the Sun and directly convectively coupled with the atmosphere. A passive insulating layer can never become the heat source of the body actively heating it. And there can be no steady state with no more heat being transferred as long as energy keeps entering at one end. If energy is exiting at the same rate at the other end, there will be a steady temperature state (though still a gradient from heating to cooling end), but there will be a dynamic, continuous transfer of energy as heat from the warmer to the cooler regions. If energy can and does NOT exit at the other end in such a situation, temperatures will just keep rising. If the receiving system is able to expand, some kind of temperature gradient will be maintained through it, away from the heating end. Countering the heating somewhat, meaning it would not heat as fast. If it is NOT able to expand, it will for all intents and purposes eventually become isothermal a short distance away from the heating end and hence heat up fast until its confines can hold it no longer; but it will still remain cooler than its heat source, which in both cases will also heat up (as a natural response to the tendency towards reduced temp gradients/differences), only at different rates.
Kristian: I agree with much of your analysis and I should have been more specific. The ‘upper air’ I referred to is below the cloud deck. Clouds absorb outgoing IR and re-radiate some of it downwards. Nearly all of his downwelling IR will be absorbed and thermalised well above the surface, and buoyantly convected upwards again. Nonetheless, this will reduce the temperature differential between the surface air and the air at altitude, thus slowing convection from the surface – I think. If I’m correct, this will be part of the reason why surface air temperatures on cloudy nights tend to be warmer than on clear nights – slower heat loss from the surface.
Ben: Thanks for the correction with the 234K figure. I suspect that through some cataclysmic near space event the Earth did indeed lose a large percentage of its atmosphere. The fossilized skeletal remains of pterosaurs indicate that the near surface air was considerably denser in their era, if they were true fliers (which I think they were) rather than merely gliders (which I’m pretty sure they were not).
The consequent surface pressure reduction would have allowed a greater rate of evaporation and a consequent loss of ocean heat content and lowering of surface temperature.
tallbloke says: September 7, 2014 at 3:44 pm
See
http://en.wikipedia.org/wiki/File:Phanerozoic_Climate_Change.pngSeems we had at least 4 cataclysmic events in this period then.
What changes when the deep oceans start warming up again, supposedly due to increasing atmospheric mass (pressure) reducing evaporation, and why is this warming relatively fast, and the cooling relatively slow.
And more interestingly, how does this reduced evaporation warm the oceans a couple of kilometers deep?
Ben: There are a lot of factors involved, and I don’t know the answers to your questions. What we do know is that reducing surface pressure would also cause a lot of seismic activity and eruptions, which will add atmospheric mass again. How all those factors play out we can only speculate about.
Tim C,
The Bond albedo of the moon is typically given as ~ 0.12. In other words, 1- 0.12 = 88% of all sunlight falling on the moon is absorbed. How do you reconcile this with your graph of absorbance vs angle? It seems your average would be well below 0.88. Do you feel the references that give the Bond albedo as ~ 0.12 are wrong?
–Kristian says:
September 7, 2014 at 2:46 pm
gbaikie says, September 4, 2014 at 4:48 am
and
September 6, 2014 at 9:18 pm:
You seem to forget about convection. You’re only talking about pure thermal expansion. Your pipe and the atmosphere are far from analogous.–
I would agree that there are differences between a pressure vessel and an atmosphere. And perhaps one can indicate some of the differences by replacing pipe with hot air balloon- one would have quite warm/hot air in the upper part of the balloon as compared to open part of balloon at the bottom of balloon. Or with the pipe the water pressure “seals” the bottom of pipe, whereas with the bottom hot air balloon, it is not sealed.
–The key points in our hypothetical scenario are: the Sun continues to heat the surface, the atmosphere is expandable, but under the influence of gravity, the solar-heated surface and the atmosphere on top are naturally convectively coupled, there’s a lapse rate (a tropospheric temp gradient away from the solar-heated surface) to sustain, the atmosphere is NOT able to cool by radiation to space.–
I would say generally, that no gases can cool by radiating into space, rather gases can radiate into space, but the individual molecules of gases are not losing a significant amount of the energy they “have” by radiating.
One can carry a pot of hot water [or carry block warmish ice] and the pot of water or block of ice
can lose nearly all it’s heat by radiating into space.
Or block of ice or pot water have heat capacity and level of this heat can be lowered.
Whereas the heat capacity of gases in related their motion.
So if you can stop a gas molecule then this result is that a gas molecule has little energy and that a gas molecule doesn’t stop or slow down by radiating.
A gas molecule changes the amount energy it has by hitting stuff. Or matter in the state of a gas are continuously exchanging energy to other gas molecules [also can transform in a solid or liquid and be warmed or cooled [gain/lose kinetic energy] by hitting matter which is liquid or solid which has a different amount of heat.
But let’s say I am wrong, and gases can slow down by radiating their energy into space, therefore one could transport fast moving gas molecules to higher elevation in atmosphere so as allowing them can lose their average velocity.
So quantifying it, one could look at a block of air, say 1 km cube, and ask how much energy is the volume gas molecules losing into space.
And I assume therefore, that if losing heat, it must get cold [quickly] or be gaining amount energy in relation to amount it loses- if lose x amount of heat, then x amount of heat must be being transported into the 1 km cube of air.
Or one could measure the air flow- warm air in and cold air out.
Now, in contrast the conventional climate view [as it seems to me] that instead of air moving, one has radiant energy moving to and/or thru this 1 cube km of atmosphere.
And also in my view, clouds which transporting water vapor which is converted to water droplets can be very powerful ways to transport heat within such 1 km cube. But neither of these aspects seem to be involved with what you specifically talking about [I would say the water droplets are cooling rather than the H2O gas radiating heat into space].
So my question if one transporting heat of air to higher atmosphere so the air radiates and lose it’s energy, at what elevation would the 1 km cube of air loss have greatest loss of this heat- and so how much per second?
And it seems to me, that if dealing with thin, cold, and dry air, and one losing and gain a lot of energy, it would be more violent [windy] as compared to a denser and warmer air losing same amount of energy.
Tallbloke says: “You are right that “You would remove the moderating effects of the oceans.” Indeed I have made this very point to you in previous similar discussions when we’ve been discussing the effect of gravity acting on the mass of the atmosphere on temperature. Without gravity to retain the atmosphere, there would be no oceans because they would boil away to space. I’m glad to see the message is gradually getting through.”
The primary moderating effect that I was talking about was the large heat capacity. The oceans cool slowly at night and warm slowly during the day. Anything that evens out the day/night temperature swings will raise the average temperature.
You rather confuse me when you go on to postulate “without gravity”. Are you imagining that we could turn off gravity? Are you thinking about a much smaller Earth where gravity iess?
As for “boiling away” …
1) if the temperature is well below 0C, there will not be any boiling anyway — just ice.
2) Even with some liquid water around (or even sublimating ice), the resulting water vapor would not escape to space. The ability of a planet to hold an atmosphere depends on gravity and temperature and the molecular mass of a gas. Here’s a typical diagram summarizing the results:
The Earth can hold on to water vapor, so any evaporation would quickly lead to a partial pressure of H2O that would suppress evaporation. The H2O gas would increase a bit during the day due to evaporation and would decrease a bit at night due to condensation, but it would NOT escape to space. Even with no N2 or O2 around, the H2O will not boil away to space.
“The key finding in this excellent post, if you can tear yourself away from your perennial talking points and examples, is that an ice covered airless Earth would be a lot warmer than the waterless Moon … “
As I said early on, I think the science here is interesting and mostly good.
But I think you are slightly mis-representing the “key finding”. To me, the two key findings are
1) using reasonable values, the model can recreate the observed temperatures on the actual moon.
2) If changes are made to the model, changes that reduce temperature swings lead to warmer average temperatures (as expected). Faster rotation reduces temperature swings. Increasing the conductivity and/or specific heat of the surface reduces temperature swings. It is simply the thermal conductivity & larger specific heat of the ice that makes the difference. (In fact, the model uses rather “dark” ice that absorbs 1/2 of the incoming sunlight. “Real” ice which is lighter would reflect more light and hence be cooler than they calculate. )
Neither of these findings in surprising, but they show that the model is pretty good at what it is trying to do.
tallbloke says: September 7, 2014 at 3:38 pm
“Kristian: I agree with much of your analysis and I should have been more specific. The ‘upper air’ I referred to is below the cloud deck. Clouds absorb outgoing IR and re-radiate some of it downwards. Nearly all of his downwelling IR will be absorbed and thermalised well above the surface, and buoyantly convected upwards again.”
Roger,
This is electromagneric radiative flux, no way heat, all thermal EMR clearly follow all of Maxwell’s equations and all laws of thermodynamics. Your claim of electromagnetic flux in opposing directions at the same frequencies is a direct contradiction to all of Maxwell’s equations and has never been observed. As AlecM keeps pointing out there is no Schwarzschild two stream approximation. All thermal electromagnetic flux is proportional to the difference in opposing “radiances’. This the only flux that is ever observed and is carefully measured.
“Nonetheless, this will reduce the temperature differential between the surface air and the air at altitude, thus slowing convection from the surface – I think. If I’m correct, this will be part of the reason why surface air temperatures on cloudy nights tend to be warmer than on clear nights – slower heat loss from the surface.”
Surface temperature is actually slightly lower with increasing atmospheric WV. This decreases convection. The big change is what your body can radiate, on a clear, night it radiates 40 w/m^2 directly to space through the 8-14 micron window, 30w/m^2 to high clouds, and only 6 W/m2 to low clouds or overcast skies. You “have a feeling of warmer” with no change in body temperature.
In no case may thermal electromagnetic flux powered only by sensible heat be dispatched in a direction of higher radiance/temperature, as that would always decrease the entropy associated with that energy.
Tim Folkerts says:
September 7, 2014 at 10:22 pm
…
Do you feel the references that give the Bond albedo as ~ 0.12 are wrong?
Sorry about slow, busy offline of recent.
With the work I did there is no figure or adjustment or allowance etc. for assumption about the surface. Flux impinges on matter which does whatever it does. Matter emits however it does.
I adjusted the surface thermal impedance to produce the values measured. Whatever is going on is encoded in that.
In darkness it matches.
In light it matches.
There is no change in parameters for either.
Tim Folkerts says: September 7, 2014 at 11:13 pm
Tallbloke says:( “You are right that “You would remove the moderating effects of the oceans.” Indeed I have made this very point to you in previous similar discussions when we’ve been discussing the effect of gravity acting on the mass of the atmosphere on temperature. Without gravity to retain the atmosphere, there would be no oceans because they would boil away to space. I’m glad to see the message is gradually getting through.”)
“The primary moderating effect that I was talking about was the large heat capacity. The oceans cool slowly at night and warm slowly during the day. Anything that evens out the day/night temperature swings will raise the average temperature.”
Please explain your claim that averaging increases the average?
“You rather confuse me when you go on to postulate “without gravity”. Are you imagining that we could turn off gravity? Are you thinking about a much smaller Earth where gravity iess?”
If you can attempt to eliminate atmospheric WV TimF, Roger, certainly can eliminate “gravity” in your incessant thought fantasy! Go measure something, anything!
“As for “boiling away” …
1) if the temperature is well below 0C, there will not be any boiling anyway — just ice.
2) Even with some liquid water around (or even sublimating ice), the resulting water vapor would not escape to space. The ability of a planet to hold an atmosphere depends on gravity and temperature and the molecular mass of a gas.”
Gee, you finally remembered sublimation, did you have to look that up? From your #2, you seem to agree that all atmospheric and surface temperatures, anywhere, are highly dependent on gravitational attraction, not on the concentration of some ppmv CO2!
tchannon says: September 8, 2014 at 3:16 am
Tim Folkerts says: September 7, 2014 at 10:22 pm
(“Do you feel the references that give the Bond albedo as ~ 0.12 are wrong?”)
Sorry about slow, busy offline of recent.
“With the work I did there is no figure or adjustment or allowance etc. for assumption about the surface. Flux impinges on matter which does whatever it does. Matter emits however it does.”
Interesting, and what determines the magnitude of the exiting or impinging flux?
“I adjusted the surface thermal impedance to produce the values measured. Whatever is going on is encoded in that. In darkness it matches. In light it matches.
There is no change in parameters for either.”
Have you considered that the electromagnetic impedence of free space can go from 377 real ohms where energy is transfered, to 377 imaginary ohms with no energy transfer, all strictly controled by opposing field strength, or “radiance” at that frequency?
“Please explain your claim that averaging increases the average?”
I’m surprised this has escaped you. Or maybe I’m not so surprised. Read up on Hölder’s inequality for the mathematical details. It is a bit subtle, but this is a well-know idea among those who discuss the physics of global temperatures.
Or here is a simplified version that might be easier for you to understand. Suppose you have a blackbody at a uniform 255 K, radiating 240 W/m^2. Now suppose you have a blackbody that radiates 340 W/m^2 from 1/2 the surface and 140 W/m^2 from the other half (for example the day side and the night side) so that the average is still 240 W/m^2 for the entire planet . The warm side will be 278 K and the cold side 223 K, for and average of 251 K. Bump that up to 440 W/m^2 & 40 W/m^2 and the average drops to 230 K.
ANY variation away from a uniform power radiating from the entire surface would result in a COOLER average temperature. The less uniform, the lower the average temperature would be.
Basically, this is one of the key points to this whole thread — that more rapid rotation leads to more uniform temperatures, which leads to a higher average.
“From your #2, you seem to agree that all atmospheric and surface temperatures, anywhere, are highly dependent on gravitational attraction”
No. Read more carefully. There was nothing about gravity affecting temperature. It was about gravity AND temperature affecting how well a planet or moon can hold an atmosphere.
I will say that the surface temperature depends on the lapse rate, and the lapse rate depends on gravity, so in that sense gravity is an important factor — BUT ONLY IF there is an atmosphere to begin with.
Tim Folkerts says: September 7, 2014 at 10:22 pm
“Tim C, The Bond albedo of the moon is typically given as ~ 0.12. In other words, 1- 0.12 = 88% of all sunlight falling on the moon is absorbed. How do you reconcile this with your graph of absorbance vs angle? It seems your average would be well below 0.88. Do you feel the references that give the Bond albedo as ~ 0.12 are wrong?”
TimF, it is you that are wrong. Albedo has nothing to do with what radiative flux is not absorbed.
Both astronomical and Bond albedo deal only with the visible 0.45- 0.7 micron flux reflected (scattered) into a hemisphere convex in the direction of the Sun It does not include the 0.8- 2.5 micron energy back scattered nor any energy forward scattered into the opposing hemisphere.
Just more of your scam!
Tim Folkerts says:
September 8, 2014 at 4:22 am
“Please explain your claim that averaging increases the average?”
“I’m surprised this has escaped you. Or maybe I’m not so surprised. Read up on Hölder’s inequality for the mathematical details. It is a bit subtle, but this is a well-know idea among those who discuss the physics of global temperatures.”
What nonsense, you try to show some inequality between flux somewhat proportanal to temperature rasied to the fourth power and that of temperature itself Show what happens to your averages when the environment is at a higher temperature than the planet. More of your scam.
ANY variation away from a uniform power radiating from the entire surface would result in a COOLER average temperature. The less uniform, the lower the average temperature would be.
Only in an envrionment of lower temperature. No science whatsoever!
“Basically, this is one of the key points to this whole thread — that more rapid rotation leads to more uniform temperatures, which leads to a higher average.”
Only in an environment of much lower temperature. You have no science, only fantasy!
(“From your #2, you seem to agree that all atmospheric and surface temperatures, anywhere, are highly dependent on gravitational attraction”)
“No. Read more carefully. There was nothing about gravity affecting temperature. It was about gravity AND temperature affecting how well a planet or moon can hold an atmosphere.”
What BS. With no science! If this earth had a helium atmosphere the low altitude temperature must be less than the temperature at altitude at lower pressure.
“I will say that the surface temperature depends on the lapse rate, and the lapse rate depends on gravity, so in that sense gravity is an important factor — BUT ONLY IF there is an atmosphere to begin with.”
Are you finally accepting an atmosphere in this physical, rather than your fantasy in your Playstation-64?
Look, Will, you seem to be a fairly bright guy, but you need to be at least willing to learn. These are just a few of the first hits from a Google search — none seems to agree with you. Albedo can be calculated in various ways. The “regular” albedo is typically averaged over the visible part of the spectrum, but the descriptions of the “Bond albedo” pretty much always includes 1) the entire spectrum 2) scattered in all directions. You are clearly wrong on both counts.
So, unless you can find reputable sources that agrees that bond albedo is 1) only visible light and/or 2) only reflected into the hemisphere toward the sun, you should:
1) admit you were wrong.
2) apologize for the “Just more of your scam!” remark.
Tim F: You rather confuse me when you go on to postulate “without gravity”. Are you imagining that we could turn off gravity? Are you thinking about a much smaller Earth where gravity less?
This goes back to discussion of Ned and & Karl’s work. Gravity acts on the MASS of the atmosphere and this causes surface pressure. They hold that this in itself accounts for the higher surface temperature (due to increased near surface air density – I think). My view is that the pressure acting to suppress the rate of evaporation, which necessarily forces the ocean to rise to a higher temperature in order to be able to emit as much energy as it absorbs fro the Sun, an that this is the biggest factor in raising surface temperature above that of the Moon. This is because, as you also know, the ocean has a much higher heat capacity than the air, and as you also know, this even’s out day/night fluctuation, raising average temperature. You only have to stand in the open in the middle of a continent under a clear sky to feel how quickly the air cools after dusk to recognise that higher near surface air density alone isn’t going to account for the temp difference between Earth and Moon, even with the higher rotation rate.
Thanks for the escape velocity plot, I stand corrected on the ‘boiling away’ point. However, my point about suppression of evaporation stands. In a thinner atmosphere exerting less pressure at the surface, water would shed energy more readily and the oceans would be at a lower temperature. I wonder if you’re ready to address this point.
Will: Roger,
sorry about all that! please delete any repeats
I’m less concerned about double posting comments than I about this sort of thing:
What nonsense, you try to show some inequality between flux somewhat proportanal to temperature rasied to the fourth power and that of temperature itself Show what happens to your averages when the environment is at a higher temperature than the planet. More of your scam.
Will, it’s lots of fun having you around, and I appreciate your insights, but before submitting, please re-read your comments and remove the unnecessary provocations, like this:
You try to show some inequality between flux somewhat proportanal to temperature rasied to the fourth power and that of temperature itself Show what happens to your averages when the environment is at a higher temperature than the planet.
Tim F can be exasperating at times, but he is also right about some stuff and he mostly steers clear of discussing motive, so let’s leave the climate politics out and concentrate on science.
Then you can tell me what you mean by “when the environment is at a higher temperature than the planet” 🙂
When is space ever at a higher temperature than the planet? (True vacuum can’t even have a temperature) Or do you mean something else?
–Thanks for the escape velocity plot, I stand corrected on the ‘boiling away’ point. However, my point about suppression of evaporation stands. In a thinner atmosphere exerting less pressure at the surface, water would shed energy more readily and the oceans would be at a lower temperature. I wonder if you’re ready to address this point.–
I would like more details regarding the plot. First, Mars has nitrogen in it’s atmosphere- 2.7 %.
or 2.7% of 25 trillion tons is 67.5 billion tonnes of N2 in the atmosphere.
Mars also has trace H20 gas and methane in it’s atmosphere.
And our moon has colder average temperature than Mars.
Some imagine Mars was much warmer and had liquid oceans something like more than couple billions years ago [and lasting for hundreds of millions of years- if not billions. And generally, lots curious things about plot.
In terms of evaporation, it’s controlled by partial pressure of H20 in the atmosphere.
So it seems to me, with less gravity, one needs higher amount water vapor in atmosphere to limit the rate of evaporation.
I can’t wait for Dawn to get to Ceres, as it’s thought there significant amount water vapor in it’s atmosphere [and has very low gravity].
Tim Folkerts says:
September 8, 2014 at 5:26 am
” Bond albedo, defined as the fraction of the total incident solar radiation reflected by a planet back to space, is a measure of the planet’s energy balance.”
” The Bond albedo, named after the American astronomer George Phillips Bond (1825–1865), who originally proposed it, is the fraction of power in the total electromagnetic radiation incident on an astronomical body that is scattered back out into space.”
” Bond albedo (named for the American astronomer George Bond), also known as spherical albedo, is the fraction of the total incident solar radiation – the radiation at all wavelengths – that is reflected or scattered by an object in all directions; this is an important measure of a planetary body’s energy balance. ”
What total nonsense! Albedo is only the percentage ability to observe a planetary body by its diffuse reflection of visible illumination from the primary. Astronomically it is precisely the diffuse reflection from a flat surface with area equal to the cross-sectional area of that body. Bond albedo is that difference of a hemispherical surface rather than a flat surface. Albedo is never the percentage electromagnetic energy not absorbed by a body as you claim. Quite independent of whatever Wikipedia or Google may say. The non absorbed energy is always greater than albedo.
“Look, Will, you seem to be a fairly bright guy, but you need to be at least willing to learn.”
I have learned of the FRAUD by the post modern fools that call themselves Scientists!
“only reflected into the hemisphere “convex” toward the sun, you should:
1) admit you were wrong.
2) apologize for the “Just more of your scam!” remark”
I refuse!! It is you doing the SCAM!.
tallbloke says: September 8, 2014 at 9:43 am
Will: Roger, sorry about all that! please delete any repeats
“I’m less concerned about double posting comments than I about this sort of thing:”
“What nonsense, you try to show some inequality between flux somewhat proportional to temperature raised to the fourth power and that of temperature itself Show what happens to your averages when the environment is at a higher temperature than the planet. More of your scam.”
Will, it’s lots of fun having you around, and I appreciate your insights, but before submitting, please re-read your comments and remove the unnecessary provocations, like this:
(“You try to show some inequality between flux somewhat proportional to temperature raised to the fourth power and that of temperature itself Show what happens to your averages when the environment is at a higher temperature than the planet.”)
“Tim F can be exasperating at times, but he is also right about some stuff and he mostly steers clear of discussing motive, so let’s leave the climate politics out and concentrate on science.”
OK!
“Then you can tell me what you mean by “when the environment is at a higher temperature than the planet” :When is space ever at a higher temperature than the planet? (True vacuum can’t even have a temperature) Or do you mean something else?”
If this planet were in an environment with a “radiance” equivalent of 600 Kelvin (oven) , all radiative flux would be in the opposite direction. All TimF’s claims would be nonsense. TimF refuses to describe the conditions under which his claims may be correct. -will-
Will Janoschka says:
September 8, 2014 at 4:12 am
…
Interesting, and what determines the magnitude of the exiting or impinging flux?”
1. Solar, rotation, angular modification
2. Surface characteristics
Only talking about what I did originally, no assumption is made about 2.
The intention was contact of the purely fictional Stephan-Boltzmann equation with real matter. As must happen all incoming energy ultimately leaves again. The entire result is based on thermalisation even though some might be a lossless “optical” effect.
Like most if not all this stuff the overall effect is cancellation to unity so even if selective effects take place there is still x/x
tallbloke says: September 7, 2014 at 5:10 pm
“Ben: There are a lot of factors involved, and I don’t know the answers to your questions.”
Obviously I DO know the answer to those questions. Reduced/increased evaporation isn’t included in them. See
“What we do know is that reducing surface pressure would also cause a lot of seismic activity and eruptions, which will add atmospheric mass again.”
Are you really serious? Surface pressure on the continental crust (25-70 km thick) is just one atmosphere (equal to the pressure of ~10 meter of water).
Pressure on the oceanic crust (7-10km thick) is more like 400 atmosphere.
Problem in these discussions is that both sides are partly correct.
The warmists are correct in stating that without atmosphere the average temperature would be much lower on earth.
This doesn’t mean that the atmosphere is capable of warming the surface and the deep oceans to their current values.
So the sceptics are correct in stating that the science does not support the “atmosphere warms surface plus oceans” meme.
Bring in deep oceans heated by geothermal heat, and the whole debate can be finished imo.
–Ben Wouters says:
September 8, 2014 at 1:21 pm
tallbloke says: September 7, 2014 at 5:10 pm
“Ben: There are a lot of factors involved, and I don’t know the answers to your questions.”
Obviously I DO know the answer to those questions. Reduced/increased evaporation isn’t included in them.–
Things like planetary rotation and gravity, I believe are factors in a climate model.
And the climate model is not just for earth. One should able to use Earth’s climate to model
all planetary temperature- otherwise, it’s simply not science
–Problem in these discussions is that both sides are partly correct.
The warmists are correct in stating that without atmosphere the average temperature would be much lower on earth.–
There is no doubt in my mind that without an atmosphere there would be larger swings in daily temperatures in surface temperature [as one can’t have air temperature- though on earth if measure surface rather than air, one has a larger swings in temperature- which something people might forget]
–This doesn’t mean that the atmosphere is capable of warming the surface and the deep oceans to their current values.
So the sceptics are correct in stating that the science does not support the “atmosphere warms surface plus oceans” meme.
Bring in deep oceans heated by geothermal heat, and the whole debate can be finished imo.–
It would agree that oceans could be significantly warmed by geothermal heat. I don’t know how much, and I don’t think it has much to do with current global temperature. I also seems possibly that geothermal heat- large eruption under oceans could cause mixing of ocean water [mixing of ocean cools the surface of the ocean]. Also impactor can mix the ocean- could be related to warming or cooling- but again, don’t think it has much to do with Earth’s current climate.
Tim Folkerts, September 7, 2014 at 11:13 pm,
That is a very important chart for folks who are into exo-planets. Here is slide presentation you may like:
Click to access Catling.pdf
Several years ago I used to post at “Science of Doom”. My field is quantum electro-optics so what little I know about climate science I learned from DeWitt Payne, Leonard Weinstein, Grant Petty, Ramanathan, Nicola Scafetta, Robert G. Brown and Rodrigo Caballero.
Back then my interest was in the effect of CO2 on planetary surface temperatures. Is [CO2] going to cause catastrophic global warming on Earth? Was [CO2] responsible for a “Runaway Greenhouse Effect” on Venus?
I am happy to inform y’all that the answer to both of the above questions is a resounding “NO”. I am convinced that the 1896 Arrhenius hypothesis that attributes the “Greenhouse Effect” to trace gases such as CO2 and water vapor is false. The “Greenhouse Effect” that raises the surface temperatures of bodies with atmospheres primarily defined by the “bulk of the atmoshere” rather than “trace gases”.
Four years ago I was convinced that if the CO2 in the Venusian atmosphere was magically replaced by nitrogen the surface temperature would change very little. Replacing the CO2 with helium would probably raise the surface temperature significantly:
Back then my conclusions were based on simplistic “back of the envelope” calculations. It did not occur to me to apply engineering software such as PSPICE or Quickfield to these questions.
Tim Channon, “br” and this camel have shown that engineering software can explain the temperature on airless bodies with great precision. I am a fan of Nikolov & Zeller who dared to suggest that total pressure is more important than the chemical composition of atmospheres. I am also a fan of Robinson & Catling who produced a more accurate pressure based model:
Using Quickfield it was relatively easy to confirm Ashwin Vavasada’s one dimensional model of the Moon’s surface temperature. Is it likely that Quickfield can confirm (or refute) Robinson & Catling’s model?
My next post will attempt to analyze the effect of gas composition on Venusian surface temperature using Quickfield.
Peter, I haven’t had time to look at Robinson and Catling. Commenter ‘trick’ gave me the impression they were confirming the standard model which promotes atmospheric composition as being of primary importance, whereas you are saying they confirm our preferred pressure model. I’ll need to get up to speed.
Please let us know when your next post is ready, so we can reblog it here for discussion.
gbaikie says: September 8, 2014 at 8:32 pm
“There is no doubt in my mind that without an atmosphere there would be larger swings in daily temperatures in surface temperature [as one can’t have air temperature- though on earth if measure surface rather than air, one has a larger swings in temperature- which something people might forget]”
Heat capacity of the total atmosphere is equal to that of ~3 meter of water. Sun warms the upper 100 -200 meter of ocean directly. So the largest regulator of temperature swings are the oceans.
If we could have gel like water ( all properties of water, except the capability to evaporate) even without atmosphere the temperature swing in the tropics would be very limited.
“It would agree that oceans could be significantly warmed by geothermal heat. I don’t know how much, and I don’t think it has much to do with current global temperature.”
The current temperature of the deep oceans is ~273K. which is ~76K above the average surface temp of our moon. Unless you have a better explanation, my position is that this 273K is FULLY caused by geothermal heat. Surface heated water does NOT sink into the cold, dense deep oceans.
— Ben Wouters says:
September 9, 2014 at 3:54 pm
gbaikie says: September 8, 2014 at 8:32 pm
“There is no doubt in my mind that without an atmosphere there would be larger swings in daily temperatures in surface temperature [as one can’t have air temperature- though on earth if measure surface rather than air, one has a larger swings in temperature- which something people might forget]”
Heat capacity of the total atmosphere is equal to that of ~3 meter of water. Sun warms the upper 100 -200 meter of ocean directly. So the largest regulator of temperature swings are the oceans.
If we could have gel like water ( all properties of water, except the capability to evaporate) even without atmosphere the temperature swing in the tropics would be very limited.–
Well, I would say it this way, the ocean absorb most of the sunlight which reaches Earth. Or the land absorbs only small fraction of sunlight.
This is obviously true, simply because there is more ocean area, but this would be true if the there was a equal amount of land and water area.
Another reason ocean absorb more sunlight, is that something 80% or more of tropical area of Earth is covered by oceans.
Earth average temperature is somewhere around 15 C, Earth average temperature has in Earth past been 25 C [or more]. That addition of 10 C to average global temperature is in my option solely/mostly due to a warmer ocean.
Or I would say the only way to significantly warm earth so it is warmer than 15 C, is to warm the ocean. And the ocean is warmed by direct sunlight, or as you say ” Sun warms the upper 100 -200 meter of ocean directly”
Everyone knows the ocean in the tropics heats the world- Gulf stream warms Europe by about 10 C
and etc.
So tropics warms rest of world [or equally one can say the rest of the world cools the tropics- there are vast rivers of cold water flowing from the polar regions which is biggest element of this].
But it also seems the atmosphere is significant part of why the nights are not colder, disentangling the heat from ocean from retention of heat of an atmosphere without this heat is a bit tricky, but one way to say the obvious is that one needs the atmosphere to transport this heat. But if want to isolate all the effects of the ocean, it seems the atmosphere is still limiting how cold the night gets.
So land of course can involve water evaporating- or ocean don’t have a complete monopoly on water vapor- and so water vapor over land area in the atmosphere is part of why atmosphere inhibits wide swings in temperature of night and day.
But even if eliminate both ocean warming and warming heat content of water vapor, just the atmosphere seems to maintain heat.
One say atmosphere is equal to 3 meters of ocean, but each day one has this “3 meters of water”
cooling and warming by around 20 C. Or heat is added and heat is lost in this tonnage of atmosphere.
Take Mars. It has 24 hour day with slightly large axis lit than Earth. It swing day and nite by about 100 K. And it seems if put Mars at Earth distance, it would have even larger swings in temperature- or the day would get a lot warmer, and the night not get much warmer.
So keeping Mars where is it, and adding shallow global tropic ocean to the planet, would seem to me to add a lot warming to Mars. In the tropics of mars with an ocean, the swings in daily temperature could be around 30 to 40 K instead of 100 K [we talking about surface air temperature- the shallow ocean would not have such a swing in temperature- or “average ocean temperature” would remain at constant temperature [as does Earth’s ocean]. Though the surface of Mars ocean would have some temperature swing during night and day.
So shallow ocean on Mars would lower the day and night swings in temperature- in the Mars tropics.
But in the temperate zone on Mars [far away from the oceans] it seems one only slightly modify the swings in Mars night and day temperature. It might reduce the swing by say 5 K, but basically still have the 100 K swing in temperature. The only way to stop this swing in temperate [or arctic] zones would be to add more atmosphere, or extend the tropic ocean poleward.
And if extend to ocean very far poleward, it would start to cool the tropical ocean. And add more atmosphere it also cools the tropical ocean. [or tropical ocean warms it].
But ocean or air in Mars temperate zone also being warmed directly by sunlight- or more air [or water] can be warmed during the daylight. Or not just the tropics which is warming the air or water in temperate zone.
tallbloke 8:02am and gallopingcamel (gc)7:11am – Per Robinson & Catling 2013 (R&C) link by gc, both atmosphere pressure AND composition are important to atmosphere total optical depth (tau) IR opacity; which one is primary or dominate depends on two existing conditions.
R&C eqn. S12 shows the grey IR differential optical depth across any particular atmosphere layer with N distinct sources of opacity is calculated by a function of the product of each of N constituent’s: 1) mass extinction coefficient (i.e. degree of each gas being IR active) & 2) density summed over the vertical thickness of the layer.
If mass extinction coefficient is low (i.e. H2, N2, O2), then high enough pressure (molecular collision induced opacity) can dominate e.g. gas giants & Titan. If mass extinction coefficient is high (H2O, CO2, CH4), density low, then composition can dominate optical depth over pressure e.g. planet Earth.
Details to determine which of the two is primary on a specific planet with thick atmosphere are well known:
Once hydrostatic equilibrium is added (this is experimentally shown close approx. for atm.s in terms of 8-10km high chunks) S12 turns into S16 which calculates the change in atm. grey optical depth as a sum of each of N constituents (or N species): 1) mass extinction coefficient, 2) mass mixing ratio, 3) delta pressure over the layer/g. The word grey (or gray) is used because natural substances like atm. gases always found to transmit and/or reflect some incident EM energy thus no object in nature is perfectly black absorber.
Posters who generally write atm. pressure is important are correct for some conditions; posters who generally write composition is important are also correct since for each specific planet or moon with thick atmosphere, R&C show how the relative importance of each can vary. Modtran (as rgb writes on gc blog) or similar solver is useful to explore the summed function optical depth for given conditions (i.e. the product of low/high extinction or high/low pressure). rgb mentions Grant Petty 2006 text; also Bohren 2006 and R&C ref. text Peixoto 1992 go deep into the details.
R&C conclude: “For Earth, water vapor is the primary opacity source in the lower troposphere…” as opposed to pressure in Titan-like atmospheres where: “Opacity is provided by collision-induced absorption from the following pairs: N2‐N2, N2‐CH4, CH4‐CH4, and N2‐H2” function of pressure, and pressure then is primary from “…collision‐induced opacity (e.g., H2 in gas giant atmospheres in the Solar System)”.
For Venus R&C: “..first derived a profile of the grey infrared optical depth using the global‐average aerosol properties and distributions from Crisp (1989).” One must go look that up to learn more about R&C’s relative importance of composition and pressure for Venus extreme conditions. Still Venus troposphere falls in line on their Fig. 1 chart fairly well.
Trick says: September 10, 2014 at 12:51 am
“tallbloke 8:02am and gallopingcamel (gc)7:11am – Per Robinson & Catling 2013 (R&C) link by gc, both atmosphere pressure AND composition are important to atmosphere total optical depth (tau) IR opacity; which one is primary or dominate depends on two existing conditions.”
“Once hydrostatic equilibrium is added (this is experimentally shown close approx. for atm.s in terms of 8-10km high chunks) S12 turns into S16 which calculates the change in atm. grey optical depth as a sum of each of N constituents (or N species): 1) mass extinction coefficient, 2) mass mixing ratio, 3) delta pressure over the layer/g. The word grey (or gray) is used because natural substances like atm. gases always found to transmit and/or reflect some incident EM energy thus no object in nature is perfectly black absorber.”
This is all correct, lukewarmers, especially Robert G Brown use the measurements made on this atmosphere in an manner inapplicable for the actual radiant flux of this planet. RGB gets it correct about line broadening with pressure and temperature at all wavelengths, but clearly fail to understand what was measured.
Almost all the ClimAstrologists use the measured extinction as an extinction of FLUX! It is not flux and never intended to be, it is a measurement on attinuation of amplitude modulation at a frequency high enough to prevent any effects of radiative equilibrium. At radiative equilibrium all upwelling flux to colder, is already and simultaneously exited in a yet colder direction, no delay! The upwelling flux only maintains the temperature of that air. Ferenc Miskolczi, and Noor Van Andel got it correct, Michael and Ronan Connolly got it mostly correct. Most all others have their minders wipe off the drool! The hydrostatic equilibrium temperature gradiant for earth, N2, H2O, is always lower than a radiative only gradiant would be.
“Posters who generally write atm. pressure is important are correct for some conditions; posters who generally write composition is important are also correct since for each specific planet or moon with thick atmosphere, R&C show how the relative importance of each can vary. Modtran (as rgb writes on gc blog) or similar solver is useful to explore the summed function optical depth for given conditions (i.e. the product of low/high extinction or high/low pressure). rgb mentions Grant Petty 2006 text; also Bohren 2006 and R&C ref. text Peixoto 1992 go deep into the details.”
This is the problem,, it is not a summed function of optical depth except for modulation. For exiting flux to colder, it is an accumulation of additional flux all the way to 220 km, space has little opposing radiance. Not only the upwelling radiation accumulates, all convective transfer, whether sensible heat or latent heat converted to sensible heat upon condensation also powers the increasing EMR flux to space. None is returned to the surface. The atmosphere with its cross sectional area can be a much better radiator to space than the surface “can be”.
This flux to space is continuously variable as determined by the amount of water vapor added to the atmosphere. Any atmospheric CO2 effect is always compensated by these WV adjustments. Your ClimAstrologists have never even attempted to discover the method for performing such adjustments.
Trick: Thanks for the synopsis. If R&C are working with constant ‘from all sides’ average insolation to Earth, they will be able to get a more or less correct average result considering atmosphere (mass-density and composition) only. However, without the ocean’s heat capacity their theory won’t work for the real Earth illuminated from one side and turning every 24 hours. That’s where my point about the ocean rising in temperature to the level where it can shed energy as fast as it gains it comes in. And that swings the relative importance of the atm. pressure vs atm. composition towards pressure for Earth because it is pressure which is the biggest restriction on the ocean’s ability to shed energy, due to the suppression of evaporation. And it is also because convection dominates radiation in the lower atmosphere. Convection rates are a function of pressure.
gbaikie says: September 9, 2014 at 8:01 pm
“Earth average temperature is somewhere around 15 C, Earth average temperature has in Earth past been 25 C [or more]. That addition of 10 C to average global temperature is in my option solely/mostly due to a warmer ocean.
Or I would say the only way to significantly warm earth so it is warmer than 15 C, is to warm the ocean. And the ocean is warmed by direct sunlight, or as you say ” Sun warms the upper 100 -200 meter of ocean directly””
Almost fully agree. The whole system of solar radiation being thermalized in the upper ~100m of ocean, this layer losing this energy to space again via the atmosphere (reduces speed of energy loss) is capable of increasing the surface temperature of the oceans ~15K relative to the temperature of the deep oceans and is energetically “in balance”
So the temperature of the deep oceans decides what the resulting surface temperature will be.
Deep oceans ~275K => surface temp. ~290K
Deep oceans ~290K => surface temp. ~305K
This is obviously a simplification, but the general idea should be clear.
So dumping an amount of magma into the oceans that equals ~10% of their volume has a major impact on the surface temps on earth.
tallbloke says: September 10, 2014 at 8:17 am
“Trick: Thanks for the synopsis. If R&C are working with constant ‘from all sides’ average insolation to Earth, they will be able to get a more or less correct average result considering atmosphere (mass-density and composition) only. However, without the ocean’s heat capacity their theory won’t work for the real Earth illuminated from one side and turning every 24 hours. That’s where my point about the ocean rising in temperature to the level where it can shed energy as fast as it gains it comes in. And that swings the relative importance of the atm. pressure vs atm. composition towards pressure for Earth because it is pressure which is the biggest restriction on the ocean’s ability to shed energy, due to the suppression of evaporation. And it is also because convection dominates radiation in the lower atmosphere. Convection rates are a function of pressure.”
Roger,
Convection rates are also a function of temperature (see partial WV pressure vs temperature). The hydrostatic lapse rate with both increasing temperature and pressure in the direction of the surface almost cancels any change in rate of production of atmospheric latent heat near the surface. For oceans the controlling parameter is lateral wind. Same as what makes your boat go!!
Trick says:
September 10, 2014 at 12:51 am
You quote R and C like this
“For Earth, water vapor is the primary opacity source in the lower troposphere…” as opposed to pressure in Titan-like atmospheres
But you did not quote the whole sentence which goes on
For Earth water vapor is the primary opacity source in the lower troposphere, and its vertical profile depends strongly on pressure due to condensation at the colder temperatures of the middle and lower troposphere.
You replaced
And its vertical profile depends strongly on pressure… by
As opposed to pressure in Titan –like atmospheres.
You reversed the meaning of the sentence by quoting part of a sentence and ending with your own words.
R and C are not discussing the relative importance of pressure and composition.
The composition and pressure are both important.
It is H2O that absorbs ( and emits) most, CO2 only a little.
The absorbing is according to pressure.
Roger 10:43pm: You are correct to read the context from the clips I point to. My alternative is clip the whole paper when it is readily accessible linked by gc; my experience is many readers prefer the brevity & won’t read long posts. This is a useful style for me.
“You reversed the meaning of the sentence by quoting part of a sentence and ending with your own words. R and C are not discussing the relative importance of pressure and composition.”
No. I respectfully disagree. The evidence is clear. R&C ARE discussing the relative importance of pressure and composition. For just one example, see Fig. S5: A) Titan like (collision‐induced absorption), B) Jupiter-like, C) Terrestrial cases (water vapor is the primary opacity source in the lower troposphere).
“But you did not quote the whole sentence which goes on For Earth water vapor is the primary opacity source in the lower troposphere, and its vertical profile depends strongly on pressure due to condensation at the colder temperatures of the middle and lower troposphere.”
Here you seem to be misled by my bolded word “its” which as I read it refers back to water vapor profile dependent on pressure (of course!) and not meaning Earth’s lower troposphere opacity being primarily dependent on pressure which would reverse all their contextual use elsewhere in the paper.
tallbloke 12:51am: “If R&C are working with constant ‘from all sides’ average insolation to Earth, they will be able to get a more or less correct average result considering atmosphere (mass-density and composition) only.”
R&C consider optical depth based on sum of each of N species: 1) mass extinction coefficient and 2) density (eqn. S12).
”However, without the ocean’s heat capacity their theory won’t work for the real Earth illuminated from one side and turning every 24 hours.”
You are correct to think about a theory’s limiting assumptions. I know well these are too often ignored at many poster’s peril. I went thru several texts on my “shelf” looking for optical depth calculation assumptions.
I found only two major. R&C explicitly mention 1) assumption of hydrostatic equilibrium which allows them to convert from density S12 to pressure in eqn. S16 which I mentioned. The only other major assumption I can find for calculating optical depth is 2) the assumption of an isotropic medium which is good for planetary atm.s & not so good for Titan but is thought serviceable.
Now there are a couple other details in the theory development like integration that is analytically impossible so is done numerically, the two-stream approximation, and in one ref. there is some discussion of a tidally locked planet or moon posing an issue for optical depth. But it sort of just trails off leaving the reader if not lost a bit bewildered (like Davey Crockett – never lost only occasionally a might bewildered for 3 days in the woods).
Also optical depth uses S-B which means positive radii and negligible surface powder. This is good assumption for all the R&C worlds in their Fig. 1.
Nowhere have I found the atm. optical depth theory depending on assumptions of an ocean’s heat capacity, planet rotating 24hours and so forth. Indeed R&C apply the same optical depth theory to planets with no ocean, different rotations even counter, Titan where water is like quartz on Earth.
Conclude: R&C optical depth theory use is reasonable good physics & will work for the real Earth.
Trick: Nowhere have I found the atm. optical depth theory depending on assumptions of an ocean’s heat capacity, planet rotating 24hours and so forth.
Yes, it looks like they’ve completely missed the implications, demonstrated in the top post here by the ~30K difference between monthly rotating Moon and 24hr rotating Moon, and even bigger difference between Ice covered Moon Vs Dust covered Moon.
Conclude: R&C optical depth theory use is reasonable good physics & will work for the real Earth.
Well that would appear to be an opposite (and illogical) conclusion to mine, GC’s, and Tim C’s by the look of it.
tallbloke says: September 11, 2014 at 6:32 am
“Trick: Nowhere have I found the atm. optical depth theory depending on assumptions of an ocean’s heat capacity, planet rotating 24hours and so forth.
Yes, it looks like they’ve completely missed the implications, demonstrated in the top post here by the ~30K difference between monthly rotating Moon and 24hr rotating Moon, and even bigger difference between Ice covered Moon Vs Dust covered Moon.
Conclude: R&C optical depth theory use is reasonable good physics & will work for the real Earth.
Well that would appear to be an opposite (and illogical) conclusion to mine, GC’s, and Tim C’s by the look of it.”
Roger,
Optical depth theory, is but a theory. Throw all such away. What “must” be measured, on or near this planet, that will give some indication of what “is” on or near this planet?
Will: Trick is up to his old tricks again.
tallbloke says:
September 11, 2014 at 7:50 am
Will: Trick is up to his old tricks again.
OK, What “must” be measured, on or near this planet, that will give some indication of what “is” on or near this planet?
I would get clobbered by my my claim that “temperature” supplies no information whatsoever!
Well, first of all, Trick, the climate establishment, and their favourite theorists need to acknowledge that what is being measured is 1361W/m^2 of TOA incoming energy from a point source 93m miles to one side of Earth, not a flux of 240W/m^2 from all directions 24hrs/day.
How about showing us altered maths which includes rotation?
1. Show that rotation does not affect a body, in effect that our maths is wrong.
2. Show standard body temperature maths which takes into account rotation.
One of the two.
tchannon says:
September 11, 2014 at 12:24 pm
Show that rotation does not affect a body, in effect that our maths is wrong.
The R and C (2013 ) paper is a model for calculating lapse rate ( LR = atmos. temp. gradient ) not surface temperature.
Clearly there is a day/night surface temperature difference on Earth . There is not a day/night LR difference.
Rotation does not affect the LR of an atmosphere. That is a fact regardless of any maths.
Roger C: Rotation does not affect the LR of an atmosphere. That is a fact regardless of any maths.
Air gets a lot colder at night. That’ll make a big difference to the amount of LR buzzing around in it. That’s another fact regardless of any maths.
–Clearly there is a day/night surface temperature difference on Earth . There is not a day/night LR difference.
Rotation does not affect the LR of an atmosphere. That is a fact regardless of any maths.–
There is obviously a difference in LR emitted between night and day. The amount difference measurable in regard to LR emitted from the atmosphere could be small. And the amount difference measurable of LR emitted from the ocean will be small.
Both atmosphere and ocean retain their heat, and have dynamic process which mask the apparent loss of heat. With a desert surface there is less heat stored and therefore would larger difference of measurable LR.
Another factor is that globally there is not a uniformity of LR, and measurable difference point to problem of measuring. One measure difference large areas because one lacks the capability
to measure at fine resolution. Or the blind may not be able to see, but they might be able to see bright light.
But generally due to mixing at the local scale, it is actually blurring a real signal you can measure, so it’s not just an illusion, but it also real.
tallbloke 6:32am: R&C are discussing optical depth for a moon with thick atmosphere -Titan. Of course R&C missed Earth’s moon – it has no atmosphere. R&C paper is n.a. earth’s moon.
“Well that would appear to be an opposite (and illogical) conclusion to mine, GC’s, and Tim C’s by the look of it.”
What is your et. al. conclusion on the applicability of R&C tested optical depth theory to their world’s in Fig. 1?
9:42am: “..Trick…need to acknowledge that what is being measured is 1361W/m^2 of TOA incoming energy from a point source 93m miles to one side of Earth, not a flux of 240W/m^2 from all directions 24hrs/day.”
Acknowledged. The sun point source is always on meaning the planet is never in sun shade.
Planet wide: energy in – energy out = 0 near long term balance with anomaly.
******
Will 7:33am: “Optical depth theory, is but a theory. Throw all such away.”
Optical depth is easily checked by test; shine a beam thru the medium of interest – count the photons missing from the exitance of the beam. If Will prefers, just change ref. to optical depth testing, then no need to throw away.
As the tested optical depth of the atmosphere increases, more of the test beam is absorbed.
Tim C. 12:24pm: “How about showing us altered maths which includes rotation?”
The maths are worked in detail here to find certain points of confusion for non-rotating & rotating planets ” with no infrared-absorbing atmosphere” with the statement: “In fact, the standard presentation in climatology textbooks is accurate in all material respects.”
Click to access 0802.4324v1.pdf
Note this was published before Diviner results became available so Table 1 is now out of date.
Note the NASA page showing moon Teff = 270K vs. Diviner ~197K measured by satellite radiometer.
That difference is interesting and applaud gallopingcamel’s efforts to explain it. AFAIK no one has to date. For me, I think any explanation will have to include the non-lambertian and non-Plankian non-S-B diffraction physics of the regolith where the reflection/diffraction is huge and regolith emissivity has been measured around 0.6 IIRC depending on view angle vs. gc and N&Z and NASA and text book assumed ~0.95 to 1.0.
Trick,
The Moon may have a regolith emissivity of 0.6 at some wavelength but averaged over the thermal IR region the emissivity appears to be close to 0.95. There is little evidence of non-Lambertian behavior in that part of the spectrum although Vasavada did recommend further study of this issue.
I looked at the equatorial Tmin (the temperature at the lunar dawn). With e = 0.95, my model says Tmin = 99 K in good agrement with observations.
With e = 0.6, Tmin = 111 K.
tallbloke says:
September 11, 2014 at 8:16 pm
Roger C: Rotation does not affect the LR of an atmosphere.
TB: Air gets a lot colder at night. That’ll make a big difference to the amount of LR buzzing around in it.
The effect of day/night on LR is a good way to test different theories of the cause of LR.
The best data I know of is this, which you first linked to.
Click to access angeo-19-1001-2001.pdf
Fig.4 Diurnal evolution of the vertical, 4km. to 20km., profile of temperature Kerala, India. Daily temp range 24-27C
I calculate from this:
time LR K/km
0700 4.9
1100 4.8
1500 4.8
1900 4.5
2300 4.9
0300 4.9
From 1700 temperature starts to fall.LR falls. As expected, reduced temperature difference from surface to space
normal > reduced LR.
From 1900 the LR returns to day-time value, 4.9.
The tropopause height has continues to increase.
The tropopause height is greatest from 2400 to 0300.
The tropopause height is greatest when the temperature is at its lowest. This is not expected from the ideal gas law.
The LR
1. is similar day and night
2. adjusts quickly to change from day to night
This suggests LR is caused by EMR, not mass transport of air by convection
Tallbloke opines: “Well, first of all, Trick, the climate establishment, and their favourite theorists need to acknowledge that what is being measured is 1361W/m^2 of TOA incoming energy from a point source 93m miles to one side of Earth, not a flux of 240W/m^2 from all directions 24hrs/day.”
It seems you are the one who needs to get up to speed.
Clearly these models DO include rotation. They DO include variations from day to night. They include convection and evaporation and many other factors. Here is a cool video of the output of a GCM. You can see the daily changes from night to day (especially in the Amazon River Basin). https://www.youtube.com/watch?v=qh011eAYjAA. They can’t get daily variations if they don’t use daily changes in solar input!
Here is a 30 min video that describes GCMs and explains (among other things) that the time scales for calculations are on the order of 10 minutes. When done with small grid size and short time steps, these models are used to predict the weather a few hours into the future. When done with larger grids and larger time steps, the models are used to project conditions weeks or decades into the future. https://www.youtube.com/watch?v=ZtK8oqQm7wQ
–Fig.4 Diurnal evolution of the vertical, 4km. to 20km., profile of temperature Kerala, India. Daily temp range C24-27–
What does Kerala, India have to do with it And a daily temp range of 24 to 27 C in terms of air temperature seems like a small range.
And since it’s low temperature I assume it’s the temperature at 4 km elevation- though 27 C at 4 km seems surprisingly warm.
According to wiki, average high temperature of Kerala India in August is 29 C and average low is 23 C. This is close to 24 to 27 C. But linked ref is for” tropical station Gadanki (13.5 N,
79.2 E)” Which somewhere around Tirupati: 13.65°N 79.42°E.
Kerala is “Lying between north latitudes 8°18′ and 12°48′ and east longitudes 74°52′ and 77°22′”
Wiki says Tirupati temperature in August has 34.8 C average high and average low of 25.5 C.
So somewhere in Kerala a daytime 27 C high at surface seems plausible, though seems on cool side for surface temperature at tropical station Gadanki.
Anyhow I am wondering why there is is this low range in daily air temperature- it suggest to me high humidity and other factors involved.
Roger Clague,
“This suggests LR is caused by EMR, not mass transport of air by convection”
A temperature gradient is always related to heat flux. As there is a radiative flux and a convective flux in our atmosphere, the lapse ratte is necessarily linked to these two phenomena.
Tim Folkerts,
“Clearly these models DO include rotation. They DO include variations from day to night. They include convection and evaporation and many other factors.”
But they do not include the effect of a change in the radiative structure of the atmosphere on the lapse rate :
“convection is what determines the temperature gradient of the atmosphere”
(Ramanathan et al. 1978)
TimF
“Clearly these models DO include rotation. They DO include variations from day to night.”
Did Tallbloke qualify what he meant?
The only accepted GCM is a weather GCM. These do attempt to including day and night. I have previously mentioned they make a dog’s breakfast of night, is widely acknowledged. (see discussions about nocturnal jets etc)
The performance in an absolute sense is zero to an upper limit of ~10 days of sensible results.
The second “GCM” tries to climb on the reputation of a weather GCM for validity, the climatic GCM which forecasts out 10 to 100 years. Rather solid evidence says these cannot manage 1 year.
Lots and lots of fudges, anything to try and avoid actual computation, the mushroom problem.
I could spell things out except I am very busy creating new stuff.
Can you find a climatic CGM coder, Mr Harry and ask them about say nocturnal jets?
Alternately tell us about the Met Office seasonal forecasts..
Tim Folkerts says: September 13, 2014 at 9:10 pm
(“Tallbloke opines: “Well, first of all, Trick, the climate establishment, and their favourite theorists need to acknowledge that what is being measured is 1361W/m^2 of TOA incoming energy from a point source 93m miles to one side of Earth, not a flux of 240W/m^2 from all directions 24hrs/day”.”)
“It seems you are the one who needs to get up to speed.”
““A general circulation model (GCM), a type of climate model, is a mathematical model of the general circulation of a planetary atmosphere or ocean and based on the Navier–Stokes equations on a rotating sphere with thermodynamic terms for various energy sources (radiation, latent heat). “”
Please show where any of the GCMs are ever based on the Navier–Stokes partial differential equations? None has a computer big enough or fast enough to do that within a lifetime. Let alone “how to verify”. All NASA GISS has is Hansen’s Playstation-64. The Navier–Stokes equations will have about 200 partials and only 7 orthogal variables. The Solar irradiance mentioned by Roger is one of those seven. Within this atmosphere all that goes to hell, because of the partials. There is no energy budget, only few measurements of temperature, with no observation of sensible heat.
What is completely absent is any energy transfer other than EMR, to or from this Earth.
Your ClimAstrologists have no clue as to what they are doing!!
phi says:
September 13, 2014 at 11:15 pm
A temperature gradient is always related to heat flux. As there is a radiative flux and a convective flux in our atmosphere, the lapse ratte is necessarily linked to these two phenomena.
The way I link them is:
Radiative flux causes the vertical temperature profile which causes convection
EMR > LR > Convection
The consensus view is
convection > LR > EMR
“convection is what determines the temperature gradient of the atmosphere”
(Ramanathan et al. 1978)
Can you explain how? The adiabatic expansion idea is nonsense.
1. The atmosphere is not adiabatic
2. Rising air is meeting air at lower pressure so cannot be doing work
–phi says:
September 13, 2014 at 11:15 pm
A temperature gradient is always related to heat flux. As there is a radiative flux and a convective flux in our atmosphere, the lapse ratte is necessarily linked to these two phenomena.
The way I link them is:
Radiative flux causes the vertical temperature profile which causes convection
EMR > LR > Convection–
If one had a million watts of lasers ]you choose the wavelength which is best] spread over
1 square km [a watt per square meter] how much could you change the lapse rate?
Roger Clague,
“Can you explain how [convection is what determines the temperature gradient of the atmosphere]?”
No, I can not. But it is a basis for modelisation.
That said, as convection operates heat transfer, it has an influence on the gradient, can be seen in this graph of Manabe : http://climatephys.files.wordpress.com/2012/06/manabe_strickler.jpg
Some tasty words from Science of Doom :
“As Ramanathan and Coakley pointed out in their 1978 paper, convection is what determines the temperature gradient of the atmosphere but solving the equations for convection is a significant problem – so the radiative convective approach is to use the known temperature profile in the lower atmosphere to solve the radiative transfer equations.”
(http://scienceofdoom.com/2010/04/)
Roger Clague says: September 14, 2014 at 10:24 am
“Radiative flux causes the vertical temperature profile which causes convection”
There is no physical way you can demonstrate that. In this atmosphere the radiative flux is determined by the vertical temperature profile, which is determined totally by gravitational potential and condensing WV in the atmosphere. The total radiative flux exiting the atmosphere to space is the sum of all energy transfers into the atmosphere. No surface EMR is required, little is ever detected.
“In this atmosphere the radiative flux is determined by the vertical temperature profile, which is determined totally by gravitational potential and condensing WV in the atmosphere.”
But why try to systematically eliminate one of the components of the lapse rate? There are two different types of flux which affect the gradient. It’s annoying and complicated but this is so and not otherwise.
gbaikie says: September 14, 2014 at 10:58 am
“Radiative flux causes the vertical temperature profile which causes convection
EMR > LR > Convection– If one had a million watts of lasers ]you choose the wavelength which is best] spread over 1 square km [a watt per square meter] how much could you change the lapse rate?”
By no measurable amount. With your megawatt laser and the 5 megawatts/T of entropy generated on your sq km, the temperature of the “whole vertical column” of troposphere may increase by 0.3 Celsius, but the lapse rate will be identical. The change in temperature of the “whole vertical column” changes by much more than that from noon to midnight, while the lapse rate only changes by the amount of WV in that column. With more WV, the surface is at a lower temperature, as that WV radiates more entropy to space than can the surface.
Of course, no one factor controls the lapse rate completely under all circumstances. But the one factor that plays the biggest role would clearly be convection.
* Conduction through air is very poor, and for the most part can be ignored. For example, when the thermal gradient is 6.5 K/km, the conduction through the atmosphere is on the order of 0.0002 W/m^2. Thus a parcel of air exchanges little heat via conduction.
* As Will has pointed out, IR only travels short distances within the atmosphere, so the IR exchange within the atmosphere is also rather small. Thus a parcel of air exchanges little heat via radiation.
Since close to zero heat is exchanged from a parcel of air to its surroundings, the parcels can be considered (quite close to) adiabatic.
Convection, however, is different. If the lapse rate is small (ie below ~ 10 K/km for unsaturated air or ~ 5 K/km for saturated air), there is no convection at all. If the lapse rate gets even a little larger, convection starts. Furthermore, convection can increase dramatically to carry away more air and more heat. The lapse rate cannot get significantly above the adiabatic lapse rate because that would simply cause more convection which would cool the surface.
For those who might like an electrical analogy, convection is like a diode; conduction and radiation are like resistors. Connect them in parallel. At small currents, all the current will go thru the resistors and the the voltage will be proportional current. Once the current gets larger, the proportion of current through the diode rises and the voltage starts to level out and never gets above ~ 0.7 V.
–Will Janoschka says:
September 14, 2014 at 12:12 pm
gbaikie says: September 14, 2014 at 10:58 am
“Radiative flux causes the vertical temperature profile which causes convection
EMR > LR > Convection– If one had a million watts of lasers ]you choose the wavelength which is best] spread over 1 square km [a watt per square meter] how much could you change the lapse rate?”
By no measurable amount. With your megawatt laser and the 5 megawatts/T of entropy generated on your sq km, the temperature of the “whole vertical column” of troposphere may increase by 0.3 Celsius, but the lapse rate will be identical. The change in temperature of the “whole vertical column” changes by much more than that from noon to midnight, while the lapse rate only changes by the amount of WV in that column. With more WV, the surface is at a lower temperature, as that WV radiates more entropy to space than can the surface.–
So if we change it a bit, say total of 5 million watts of lasers [again, whatever wavelength [or power level per each laser- and you are not limited to one type of laser, rather it’s just the total power is limited to 5 MW of output power] and put it in a smaller area of 100 meter square rather than 1000 meter square, thereby giving 500 watts per square meter, then could this have measurable effect upon the lapse rate?
As control or comparative scheme to the 5 MW laser. What if one had pool 1 meter deep by 100 meter by 100, in which one put near freezing water and later change it by putting in 95 C water and measured any difference. So the pool has less watts per square meter involved, and one can’t choose the wavelength, nor direct the the power output like a laser- and probably much cheaper to do [one might find such pools of such a size [or larger] and heated around this temperature by volcanic heat, somewhere in the world. But in any case, it would easier and cheaper to do.
Anyhow would 5 MW have much effect upon lapse? And how would it compare to water kept at 95 C
in terms of it’s possible effect upon the lapse rate. And with water one also have the water covered with plastic and also remove the plastic.
“Since close to zero heat is exchanged from a parcel of air to its surroundings, the parcels can be considered (quite close to) adiabatic.”
This is so adiabatic that the entire convective flux ends up being dissipated radiatively.
Superb example of unacceptable simplification.
Phi, I am not sure I understand your objection. Or perhaps I was not clear enough.
I was considering a parcel of air within the atmosphere — neither too close to the ground nor too close to the tropopause. Air nearby (say within 100 m) is going to be pretty close to the same temperature, so neither conduction nor radiation will be exchanging much heat with the surroundings.
Certainly there will be SOME conduction & radiation. However, since the lapse rate is observed to be rather close to the adiabatic lapse rate, then that is evidence that the adiabatic approximation is rather good.
Near the tropopause, the IR can travel farther, and here the lapse rate DOES start to diverge significantly from the theoretical adiabatic lapse rate. Then the convective heat flux (from the surface toward the upper troposphere) DOES get dissipated radiatively (from the upper troposphere toward space).
@Tim Folkerts,
On Earth, radiation causes a positive lapse rate in the stratosphere. The same goes for Jupiter, Saturn, Titan, Uranus and Neptune.
The solitary exception is Venus which has a negative lapse in the stratosphere owing to the high concentration of CO2:
Tim Folkerts,
Radiative dissipation begins at the surface and continues over the entire troposphere column.
“However, since the lapse rate is observed to be rather close to the adiabatic lapse rate, then that is evidence that the adiabatic approximation is rather good.”
Maybe. In the same spirit, we may as well ignore the effect of increased CO2 levels on surface temperatures.
Insofar as two different types of flux provide the heat transfer, neglecting the effect of one of the two on the gradient is a gross error.
gallopingcamel says: September 15, 2014 at 6:21 am
“@Tim Folkerts, On Earth, radiation causes a positive lapse rate in the stratosphere. The same goes for Jupiter, Saturn, Titan, Uranus and Neptune.”
What! Will someone please define Lapse rate positive or negative, of what is observed on various
bodies with gravitational potential. The gas giants with H2 and He have an increase in temperature with decreasing pressure. All carefully measured by engineers, not academic physics lecturers. Science by Physicists is dead.
“The solitary exception is Venus which has a negative lapse in the stratosphere owing to the high concentration of CO2:”
That CO2 has no measurable Cv or Cp. We know that specific heat of CO2 at the critical point has a Chaotic value, partially dependent on direction of approach to the critical point, especially in the presence of a gravitational field.
phi says: September 15, 2014 at 8:22 am
Tim Folkerts, (“Radiative dissipation begins at the surface and continues over the entire troposphere column.”)
That should be radiative “accumulation” of flux to space all the way to 220 km.
(“However, since the lapse rate is observed to be rather close to the adiabatic lapse rate, then that is evidence that the adiabatic approximation is rather good.”)
“Maybe. In the same spirit, we may as well ignore the effect of increased CO2 levels on surface temperatures..Insofar as two different types of flux provide the heat transfer, neglecting the effect of one of the two on the gradient is a gross error.
Please demonstrate any detectable difference in the types of your imaginary flux?
Will Janoschka,
In the troposphere there is at least heat transfer by convection (and conduction) and heat transfer by radiation. If there is an issue with that, it is well above my modest knowledge.
Have you a problem with this graph (http://climatephys.files.wordpress.com/2012/06/manabe_strickler.jpg) and if so which one?
Tim Folkerts says: September 14, 2014 at 3:08 pm
“Of course, no one factor controls the lapse rate completely under all circumstances. But the one factor that plays the biggest role would clearly be convection.”
On what is this assumption based?
Most convection never gets above ~2-3km. In high pressure areas large air masses are actually sinking towards the surface.
Only a few convective cells grow into cumulonimbi, and MAY reach all the way to the tropopause.
And for every m^3 of air that convects upward, somewhere else a m^3 of air has to sink downward.
Convection lasts usually 10-20 minutes (or shorter) before the convecting parcel reaches an altitude with the same density and the convection stops. DURING the convection negligible amounts of energy are exchanged with the surrounding air, and AFTER the convection the parcel is at an altitude where the surrounding air has roughly the same temperature.
So how is the lapse rate (atmospheric temperature profile) mostly set by convection????
phi says September 15, 2014 at 11:07 am
Will Janoschka,
“In the troposphere there is at least heat transfer by convection (and conduction) and heat transfer by radiation. If there is an issue with that, it is well above my modest knowledge”
There is no issue please note that all energy transfer is outward to cold space. No transfer of energy to the higher temperature surface, by any means whatsoever.
” Have you a problem with this graph
http://climatephys.files.wordpress.com/2012/06/manabe_strickler.jpg ” and if so which one?”
Thank you for that graph. I have been looking for our radiative lapse rate since 1976.
Notice the “thermal radiative” has the highest decrease in temperature with altitude. Meaning that radiative is a poor determanant for atmospheric temperature vs altitude. Note that temperature at any altitude would be higher for the DALR or the ELR, meaning that convection and transfer of latent heat converted to sensible heat increases any EMR radiative exitance to space. In complete opposition to the ClimAstrologists claim of stored heat in this atmosphere.
gallopingcamel 6:08pm: “..averaged over the thermal IR region the emissivity appears to be close to 0.95.”
To understand lunar IR emissivity properly, one needs to make a thorough search and reading of the papers in this subfield – regolith emissivity – a dielectric property of the lunar surface so both subfields are important to master.
In the handful of papers I’ve read – testing of the extremely reflective lunar regolith (from Apollo sample) shows its emissivity averages much, much less than the 0.95 generally found in natural earth material. Darshan Shrestha is an author that has studied regolith from a dielectric expert’s POV based on test and radiative transfer analysis. Find his published charts of lunar regolith brightness temperature & emissivity vs. incidence angle: emissivity shown averaging 0.68 down to 0.5 at 70 degrees angle.
In his paper there are many confirming references to follow up. Plus there are charts of lunar regolith emissivity vs. frequency, depth (emissivity 0.75 down to 0.35 at 5cm deep), bulk density, rock fraction, scatter size.
If your analysis work is to be fundamental & not tuned to a known experimental result, then this testing and analysis will need to be encompassed. Here’s another confirming paper on line to consider:
http://onlinelibrary.wiley.com/doi/10.1029/2011JE003874/pdf
phi says:
September 14, 2014 at 11:03 am
Some tasty words from Science of Doom :
“As Ramanathan and Coakley pointed out in their 1978 paper, convection is what determines the temperature gradient of the atmosphere but solving the equations for convection is a significant problem – so the radiative convective approach is to use the known temperature profile in the lower atmosphere to solve the radiative transfer equations.”
SOD says: Convection determines LR but we can’t solve the equations.So we will use radiation equations.
Tasty? Not science though.
You refer me to his explanation of the convection causes LR consensus theory.
The first law of thermodynamics – conservation of energy – says that if there is no change in energy then work done by a parcel of air in expanding must equal the change in heat.
Rising air is rising into lower pressure, it isn’t doing work against pressure. It is doing work against gravity.
m c T = m g h decrease in K.E.= increase in G.P.E. ( gravitational potential energy
LR = T/h =g/c
Consensus says c = cp air = 1J/K
DALR = 10m/s2 divided by 1J/K
= 10K/km
ELR = DALR – latent heat effect
THis latent heat effect is as big as the adiabatic effect. But is never calculated..
I say used cp H2O(g) = 2J/K
LR = 10/2
= 5K/km
Correct for moist air.If the air is very dry this can rise to 6.5K/km.
Tim Folkerts says:
September 14, 2014 at 3:08 pm
IR only travels short distances within the atmosphere,
Some IR reaches the surface from the sun and some IR from the surface reaches space, not short distances.
so the IR exchange within the atmosphere is also rather small.
I am suggesting that temperatures in the troposphere are caused by IR energized H2O(g) molecules colliding with N2 or O2 molecules.
Thus a parcel of air exchanges little heat via radiation.
A parcel of air is continuously being bombarded by energised H2O(g), so its temperature will be the same as all other air at the same height.
Roger Clague says: September 15, 2014 at 4:27 pm
“Rising air is rising into lower pressure, it isn’t doing work against pressure. It is doing work against gravity.”
Rising air IS NOT DOING WORK AGAINST GRAVITY, it is pushed up by the pressure gradient force. It rises “for free” as long as its density is lower than the density of the surrounding air.
Try to understand ‘hydrostatic balance’ and ‘pressure gradient force’.
“Rising air is rising into lower pressure, it isn’t doing work against pressure”
It is doing work to push the air it is moving into “out of the way” while expanding due to the lower pressure of the surrounding air.
Roger Clague says: September 15, 2014 at 4:47 pm
“Some IR reaches the surface from the sun”
Over 50% of the solar radiation that reaches the surface is IR. This is AFTER reflection and massive absorption by H2O and also CO2.
Will says: “Radiative [accumulation] begins at the surface and continues over the entire troposphere column.”
Well, sort of.
Near the absorption bands of CO2 or H2O (eg near 15 um), IR travels very limited distances in the tropsphere (ie the atmosphere is “optically thick”). Effectively no 15 um IR photons travel more than 100 m before getting absorbed by a CO2 molecule in the atmosphere.
At any given altitude within the troposphere, the upward flux is a function only of the ~ 100 meters below and the 100 m above. There is no “accumulation” from 200 m down or 2000 m down (well, maybe 0.0001% of the flux comes from more than 500 m down, but that is insignificant).
Near the top of the troposphere, there is “accumulation”. The net upward flux rises rapidly as you get closer to the tropopause. Looking down from space, what you see is flux from the top of the troposphere, not some accumulation of flux from all the way down to the surface.
Will says: ” Note that temperature at any altitude would be higher for the DALR or the ELR, meaning that convection and transfer of latent heat converted to sensible heat increases any EMR radiative exitance to space. In complete opposition to the ClimAstrologists claim of stored heat in this atmosphere.”
The first sentence is correct. If convection could somehow be suppressed but the atmosphere was still able to emit/absorb IR as it does now, the surface would indeed be even warmer than it is now.
However, if convection still existed but the atmosphere could be made transparent to IR, the surface would be much cooler than now.
The fact that convection has a cooling effect on the Earth does not negate the fact that IR has a warming effect. IR warms the surface; convection partially limits that warming. There is no contradiction between your first sentence and what scientists understand.
Tim Folkerts,
“Near the absorption bands of CO2 or H2O (eg near 15 um), IR travels very limited distances in the tropsphere (ie the atmosphere is “optically thick”).”
And what do you want to demonstrate ?
At 15 um, the addition of CO2 has a cooling effect. So what?
Phi says: “And what do you want to demonstrate? At 15 um, the addition of CO2 has a cooling effect. So what?”
At that point, I was demonstrating that IR in GHG bands does not particularly “accumulate”, but the IR upward at each level within the troposphere is (to a very good approximation) only a function of the atmosphere in a narrow shell above and the shell below because those shells very effectively block all IR from more distant shells . Only near the top of the troposphere does the IR start to “accumulate” because the shells above become more and more transparent.
BUT! If you want to look deeper, I was demonstrating that the addition of CO2 has a net WARMING effect on the surface!
* With more CO2, the level where the shells start to get transparent will be a little higher.
* Since the shells higher up are cooler, they will emit less thermal IR to space than the shells a little lower would have.
*Since less thermal IR is leaving, energy will build up within the layers below, warming them.
There are lots of feedbacks that impact the magnitude of this warming, but the net effect is clearly warming of the lower layers (and the surface) not cooling.
>> TIM: IR only travels short distances within the atmosphere,
> ROGER: Some IR reaches the surface from the sun and some IR from the surface reaches space, not short distances
In context, the original statement was about IR within absorption bands of the GHGs. For example, wavelengths near the 15 um band of CO2. So 15 um IR only travels short distances. IR within the “atmospheric window” certainly can travel from the sun to the surface or from the surfce to space.
>ROGER: I am suggesting that temperatures in the troposphere are caused by IR energized H2O(g) molecules colliding with N2 or O2 molecules.
That is certainly one mechanism for energy transfer. There will be some direct warming of H2O by solar IR.
>ROGER: A parcel of air is continuously being bombarded by energised H2O(g), so its temperature will be the same as all other air at the same height.
I was probably being a bit hasty in saying that radiation is not so important here. It would be an interesting back-of-the-envelop calculation to estimate the relative cooling rates from convection vs radiation (but I don’t have time right now).
Tim Folkert,
If I’m not mistaken, the CO2 transmittance is still significant for thicknesses far exceeding that of our atmosphere. Your reasoning is far too schematic.
“* Since the shells higher up are cooler, they will emit less thermal IR to space than the shells a little lower would have.”
This situation is very theoretical and, in fact, they will not emit less because the equilibrium is instantaneous through a local induced warming.
The question is to quantify the effect at the surface. The answer depends on the evolution of the temperature gradient. It is an open quesion as the adiabatic simplification is clearly abusive.
phi says: “If I’m not mistaken, the CO2 transmittance is still significant for thicknesses far exceeding that of our atmosphere. Your reasoning is far too schematic.”
This satellite IR spectrum looking down at the earth suggests that the 15 um radiation only comes from the cold upper troposphere, not significantly from the warmer layers farther down.
“in fact, they will not emit less because the equilibrium is instantaneous through a local induced warming.”
I disagree. 1) the warming may be quick, but not instantaneous. 2) If the new top layer (call it layer n+1) blocks some of the IR from below, the former top layer (layer n) will emit less IR to space and will warm. This will cause layer n-1 to emit less IR upward, so it will warm. And so will layer n-2. And on down to layer 1 (the surface).
” The answer depends on the evolution of the temperature gradient.”
That is indeed the crux of the matter. With no feedback and no change in the lapse rate, the effect is clear — warming all the way down. If the lapse rate changes and/or there are other feedbacks, the effect at the surface would be larger or smaller. I don’t have the expertise to say exactly what all those feedbacks might do. But there is no reason to think that the feedbacks are “smart enough” to exactly compensate and cause no warming at the surface.
Tim Folkert,
“This satellite IR spectrum looking down at the earth suggests that the 15 um radiation only comes from the cold upper troposphere, not significantly from the warmer layers farther down.”
Look closely, at 15 um, it comes from the stratosphere.
That said, if you integrate according to temperature, you will see that the part of the lower layers is far from negligible.
“1) the warming may be quick, but not instantaneous.”
In all cases, the warming is much faster than the CO2 voyage from exhaust pipes.
“If the new top layer (call it layer n+1) blocks some of the IR from below, the former top layer (layer n) will emit less IR to space and will warm.”
If the layer n+1 blocks IR, it also means it warms!
“But there is no reason to think that the feedbacks are “smart enough” to exactly compensate and cause no warming at the surface.”
It’s not a matter of feedback but of intial effect and I did not claim that there would be an exact compensation.
phi says: September 15, 2014 at 9:18 pm
Tim Folkert,( “This satellite IR spectrum looking down at the earth suggests that the 15 um radiation only comes from the cold upper troposphere, not significantly from the warmer layers farther down.”)
“Look closely, at 15 um, it comes from the stratosphere.
That said, if you integrate according to temperature, you will see that the part of the lower layers is far from negligible”
The important part is that is zenith radiance That gas cross-sectional area with all most no CO2 molecules radiates equally in every direction into a 2PI steradian hemisphere. This high altitude atmospheric cross-sectional area can indeed radiate to space twice the J/s to space as any surface can do. adjust your graphs accordingly. This mear geometry!
(“1 the warming may be quick, but not instantaneous.”)
“In all cases, the warming is much faster than the CO2 voyage from exhaust pipes.’
EMR can tranfer limited power at 3x 10^8 m/s
Gravitional force can transfer 10^4 that power at 80 m/s. Go figure.
(“If the new top layer (call it layer n+1) blocks some of the IR from below, the former top layer (layer n) will emit less IR to space and will warm.”)
Ah Tim and his undefined will “warm”, with no possible meaning.
“If the layer n+1 blocks IR, it also means it warms!”
There is no blocking of any EMR exit flux if that air is in radiative equibrilium!
(“But there is no reason to think that the feedbacks are “smart enough” to exactly compensate and cause no warming at the surface.”)
Timmy Timmy Timmy there are no feedbacks on or around this Planet Earth. None are needed!
Atmospheric WV controls all. Why don’t you arrogant Physics lectureres find out what controls the amount of atmospheric WV, at any location, at any time? You seem so intellegent!
Bryan says:
September 16, 2014 at 1:30 pm
To get the correct answer you need the barometric formulas for an adiabatic atmospheric expansion.
see:Page 8 of this link
Click to access 1003.1508.pdf
An important question is what laws can we apply to the atmosphere?
G and T say
By combining hydrodynamics, thermodynamics, and imposing the above listed assumptions for planetary atmospheres one can compute the temperature profiles of idealized atmospheres.
The air of the atmosphere obeys an equation of state of an ideal gas
For the ideal gas we use the reversible work form p dv
Adiabatic expansion and ideal gas law cannot be applied to the atmosphere
Adiabatic expansion
1. In an air packet the N2 and O2 molecules are getting energy from collisions with IR energized Water vapor, WV, H2O(g)*. The packet takes the temperature of the surrounding air.
2. Rising air is meeting lower pressure. It is not doing work against pressure
Barometric Formulas, require Ideal gas law conditions
1.Enclosed
2.Small volume
3.Temperature, pressure and density homogeneous
4 Temperature and pressure near STP.
4.Constant gravity
The atmosphere breaks these in spectacular fashion
1. Open at top and sides
2. 10^10 km^3
3.Temperature, pressure and density vary
4.Temperature and pressure go very far from STP
5.Gravity varies vertically
G and T take 58 equation to get to:
T/h = g/cp
Hydrostatic equilibrium and conservation of energy only
Gravitational potenial energy (GPE) = force x distance = f x d
Newtons 2nd law f=ma
in the atmosphere, d = height (h), a = g
GPE = m x g x h
Kinetic energy ( K.E.) = mcT
mgh = mcT ( thermo 1st law )
T/h = g/c
This derivation is shorter and uses less assumptions but is not taught.I suspect this is because if more people knew gravity alone causes the “greenhouse effect” then the belief that a trace gas CO2 is the cause will be less convincing.
Roger Clague says: September 16, 2014 at 10:04 pm
Bryan says:September 16, 2014 at 1:30 pm
(“To get the correct answer you need the barometric formulas for an adiabatic atmospheric expansion.see:Page 8 of this link”)
“G and T take 58 equation to get to:
T/h = g/cp
Hydrostatic equilibrium and conservation of energy only
Gravitational potenial energy (GPE) = force x distance = f x d
Newtons 2nd law f=ma
in the atmosphere, d = height (h), a = g
GPE = m x g x h
Kinetic energy ( K.E.) = mcT
mgh = mcT ( thermo 1st law )
T/h = g/c
This derivation is shorter and uses less assumptions but is not taught.I suspect this is because if more people knew gravity alone causes the “greenhouse effect” then the belief that a trace gas CO2 is the cause will be less convincing.
Gee Roger,
Thank you for your helpfull essay on how the value of “g” was carefully determined long ago on a planet with 78% N2 atmosphere!
Your treatment of the symbol “T” is pure horseshit! T depends on degrees of freedom of that gas. Try again with He in the troposphere (pressure greater than 10 KPa). Then explain!
@Will Janoschka, September 15, 2014 at 9:57 am
Your comment baffles me as none of it makes any sense. The gas giants all have negative lapse rates when the pressure exceeds 1 bar. When pressure exceeds one bar temperature rises at a constant rate of -g/Cp unless vapors are present. Only Earth and Titan have oceans of liquids which modify the lapse rate.
Then you say this:
“That CO2 has no measurable Cv or Cp. We know that specific heat of CO2 at the critical point has a Chaotic value, partially dependent on direction of approach to the critical point, especially in the presence of a gravitational field.”
The specific heat of CO2 has been measured. At constant pressure (Cp) the generally accepted value is 1.126 kJ/kgK at 700 K which happens to be very close to the Cp of Nitrogen (Cp = 1.098 at 700 K). Consequently in 1967 Carl Sagan estimated the surface temperature of Venus and concluded that it would make very little difference whether CO2 or nitrogen was the dominant component of the Venusian atmosphere.
The following year Sagan published a correction to his 1967 paper when he realized that the Cp of CO2 is more pressure dependent than the Cp of nitrogen. You can find Sagan’s correction here:
http://adsabs.harvard.edu/doi/10.1086/149625
gallopingcamel says: September 19, 2014 at 4:41 am
@Will Janoschka, September 15, 2014 at 9:57 am
gallopingcamel says: September 15, 2014 at 6:21 am
“@Tim Folkerts, On Earth, radiation causes a positive lapse rate in the stratosphere. The same goes for Jupiter, Saturn, Titan, Uranus and Neptune.”
(“What! Will someone please define Lapse rate positive or negative, of what is observed on various
bodies with gravitational potential. The gas giants with H2 and He have an increase in temperature with decreasing pressure. All carefully measured by engineers, not academic physics lecturers. Science by Physicists is dead.”)
“Your comment baffles me as none of it makes any sense. The gas giants all have negative lapse rates when the pressure exceeds 1 bar. When pressure exceeds one bar temperature rises at a constant rate of -g/Cp unless vapors are present. Only Earth and Titan have oceans of liquids which modify the lapse rate”
Peter,
I guess by your reply That negative lapse rate means precisely decreasing temperature with increasing altitude above some imaginary solid or liquid surface, without regard to cause, diabatic or adiabatic, maybe not, please explain? The Adiabatic Lapse rate (That caused only by lower compression by upper air mass because of gravatational attraction) never -g/Cp, but from the entropy, enthalpy, curves if each gas. (He) increases in temperatute with controlled expansion while above 48 Kelvin. The colder outer temperature of the gas giants including the Sun, is not determined by of gravitational attraction.
“Then you say this:”
(“That CO2 has no measurable Cv or Cp. We know that specific heat of CO2 at the critical point has a Chaotic value, partially dependent on direction of approach to the critical point, especially in the presence of a gravitational field.”)
“The specific heat of CO2 has been measured. At constant pressure (Cp) the generally accepted value is 1.126 kJ/kgK at 700 K which happens to be very close to the Cp of Nitrogen (Cp = 1.098 at 700 K).”
At the critical point of CO2 (304.25 K) and (7.39 MPa), or 217 Kelvin, 518kPa, It is a line not a point, any attempt to measure specific heat in the Earth’s gravitational field depend highly on direction of temperature, direction of pressure, and to orentations of both directions with respect to the gravitional field. Venusian atmosphere goes right through this line.
“Consequently in 1967 Carl Sagan estimated the surface temperature of Venus and concluded that it would make very little difference whether CO2 or nitrogen was the dominant component of the Venusian atmosphere.”
That is Carl’s second error about Venus. First was the 1964 phd thesis with the two stream approximation, that which AlecM bitches of!
“The following year Sagan published a correction to his 1967 paper when he realized that the Cp of CO2 is more pressure dependent than the Cp of nitrogen. You can find Sagan’s correction here:” http://adsabs.harvard.edu/doi/10.1086/149625
I guess that is the third one!
Bryan says:
September 18, 2014 at 7:17 pm
Up Thread I did a worked example for Stephan showing this.
I expect you mean
Bryan says:
September 17, 2014 at 11:14 am
Loss of internal energy for a mole of air at STP rising to a height of 1000m is 203J
Work done against surroundings is 203J (use of barometric equations)
So not one Joule of lost internal energy was turned into gravitational potential energy!
And yet the mole of air is now 1000m from the ground with this mass of air having apparently acquired 284J
I would like to see your calculations. However here are mine.
Work done against pressure = PdV
P1V1 = P2V2 IGL T constant
V2/V1 = P1/P2
= 1 bar/0.9bar
= 1.1
V2 = 1.1V1
dV = V2 -V1
= 1.1V1 – 1V1
= 0.1V1
V1 = 20litres
= 0.02m^3
dV = 1/10 x 2/100m^3
PdV = 100 000kg.m/s^2 x 1/10 x 2/100m^3
= 200kg.m^2/s^2
= 200J
Work done against gravity
mgh
m = V x density
= 0.02m^3 x 1kg/m^3
= 0.02kg
g = 10m/s^2
h = 1000m
mgh = 0.02kg x 10m/s^2 x 1000m
= 200kg.m^2/s^2
= 200J
No conclusions from this comparision
The G and T paper is about LR. So the real test is calculating LR.
Click to access LapseRate-FAO06-IACP430.pdf
fig.1
Why does the LR only reach a maximum of 6.5K/km. over deserts and not the 10K/km. dry adiabatic lapse rate DALR?
The problem with your approach is ignoring the 0.003% change in gravity in the atmosphere.
Imagine an object at rest falling from 1 000m
v^2 = u^2 +2as
v is final velocity
u is initial velocity = 0
v^2 = 0 + 2 x 10m/s^2 x 10 000m
v^2 = 200 000m^2/s^2
v = 450m/s
Change in gravity in the atmosphere cannot be ignored.
The barometric equation is similar to the gravitational LR only in the lower 10km of the troposphere. The altimeters on planes are calibrated by reference to local air traffic control. They may be all wrong , but by the same amount, so no collisions.
Roger Clague says:
“P1V1 = P2V2 IGL T constant”
So right at the start you say that the temperature is constant as the air parcel moves from ground level up to 1000m!
Everybody else would expect the temperature to drop by 9.8K
I think you had better start again
\\
Will says: “This high altitude atmospheric cross-sectional area can indeed radiate to space twice the J/s to space as any surface can do “
Nothing can absorb more IR than a blackbody. Similarly, nothing can emit more IR than a blackbody. Subsequently, any layer of gas can at best match what a BB can radiate. The gas certainly cannot radiate twice what this BB surface can radiate. Your understanding of geometry fails you.
“Ah Tim and his undefined will “warm”, with no possible meaning.”
ummm … “to warm” means to get warmer = to rise to a higher temperature. All I can conclude from your statement is it that you think temperature has no possible meaning (or that you simply enjoy being argumentative).
“There is no blocking of any EMR exit flux if that air is in radiative equibrilium!”
Sigh
1) The example given was specifically NOT in equilibrium! It was not even in a steady-state condition.
2) After spending so much time discussing how opposing radiance reduces EMR flux, I am astounded that you are now arguing against yourself. If there is flux from an object to the cold of outer space, you get a large flux. If the same object at the same temperature radiates to a layer of gas (cooler than the object but warmer than space), there will be less flux from the object because there will be some radiance from the gas back downward. We have “blocked” some of the exit flux from the original object with the layer above it.
“there are no feedbacks on or around this Planet Earth.”
This is so silly I almost don’t know where to begin! Pretty much the entire climate system is one huge feedback system, where any one change causes changes in other aspects of the climate. I don’t think you will find a SINGLE person in this discussion to support this dubious claim.
Bryan says:
September 19, 2014 at 11:32 am
RogC says P1V1 = P2V2 IGL T constant
So right at the start you say that the temperature is constant as the air parcel moves from ground level up to 1000m!
Everybody else would expect the temperature to drop by 9.8K
I am calculating the energy change using
1. your isothermal/adiabatic method
2. my K.E. to gravitional potential energy method
contant T is one of Gand T postulates. You refereed me to G and T. I am only doing the calculation using G and T postulates because you have not posted your calculation.You say 203J, 284J and I think 243J but don’t show your working! You must be a climate scientist.
Then when I do what you failed to do, show how you got 203J, you say I am at fault.
Also I notice that your posts of 17 09 11.14am and 18 09 7.17pm. have disappeared. Why is that?
How do you explain and calculate a temperature gradient with a theory that postulates T is constant ( isothermal atmosphere )?
The temperature profile is what we want to calculate, why have as an assumption of your model of the value of what you want to find out?
I have never come across a successful theory that does that.
Can you give me an example of one?
Roger Clague
You have got your threads and replies mixed up.
The replies are on the thread
“Hockey Schtick: CO2 does what exactly?”
I only noticed by chance that you were replying to that thread here instead of the original thread.
If you read my replies there and similar replies from Tim Folkerts and Trick it might help you out of your confusion.
Tim Folkerts says: September 19, 2014 at 12:07 pm
(Will says: “This high altitude atmospheric cross-sectional area can indeed radiate to space twice the J/s to space as any surface can do “)
“Nothing can absorb more IR than a blackbody. Similarly, nothing can emit more IR than a blackbody. Subsequently, any layer of gas can at best match what a BB can radiate. The gas certainly cannot radiate twice what this BB surface can radiate. Your understanding of geometry fails you”
Will you please go back to high school and try to learn solid geometry!!! Using Planck’s integral
On the left axis is a “per steradian) or (sr^-1). What do you think that means? An isotropic black body of cross sectional area one sq meter, with no opposing radiance will radiate just that flux at each wavelength uniformly into four PI steradians Never limited to the one PI steradian of a flat surface as in your fradulantly used S-B equation or law or however you are misusing correct science in your post modern nonsense. The high atmosphere may do the same with its cross sectional area. The lower hemisphere has higher radiance so no flux is detached in the direction of that lower temperature.
“Ah Tim and his undefined will “warm”, with no possible meaning.”
ummm … “to warm” means to get warmer = to rise to a higher temperature. All I can conclude from your statement is it that you think temperature has no possible meaning (or that you simply enjoy being argumentative.”
Scientifically and Lingustically “to warm” like “to heat” are “only” verbs both meaning to add heat energy to. No temperature need change i.e boiling water. Simarly a coat may make one feel more comfortable, warmer (colloquial never scientific) without increasing body temperatiue. To warm always requires an energy source, never just a reduction in exit flux. You again clearly intend to deceive.
(“There is no blocking of any EMR exit flux if that air is in radiative equibrilium!”)
“Sigh 1) The example given was specifically NOT in equilibrium! It was not even in a steady-state condition.”
What example given? The whole Earth atmosphere quickly regains thermodynamic and radiative equilibrium which means no spontanious “change” in flux or temperature, attempting to regain equilibrium. What part of the atmosphere is not in or above radiative equilibrium?
“2) After spending so much time discussing how opposing radiance reduces EMR flux, I am astounded that you are now arguing against yourself. If there is flux from an object to the cold of outer space, you get a large flux. If the same object at the same temperature radiates to a layer of gas (cooler than the object but warmer than space), there will be less flux from the object because there will be some radiance from the gas back downward. We have “blocked” some of the exit flux from the original object with the layer above it.”
If you would even try and do it correctly, you would notice radiative flux from the surface is almost eliminated, less than 1 watt/m^2. This is because of the near equal radiance fron the atmospheric WV within 10 meters. The use of the Schuster-Schwarzschild two stream approximation, for earths atmosphere, is the greatest fraud attemped by your post modern non-science.
That one W/m^2 goes straight through the entire atmosphere without attenuation. That flux has another 1/2 W/m^2 added for every 10 meters increase in altitude all the way to 300 meters altitude. Above that 16 W/m^2 gets added to greatly, as reduced pressure/density limits any radiance opposing new exit flux. The whole atmosphere all the way to 220 Km accumulates exit flux from whatever at its equilibrium radiance to low radiance space with increasing radiative solid angle. There is no radiating altitude! The whole atmosphere contributes exit flux to space.
(“there are no feedbacks on or around this Planet Earth.”)
“This is so silly I almost don’t know where to begin! Pretty much the entire climate system is one huge feedback system, where any one change causes changes in other aspects of the climate. I don’t think you will find a SINGLE person in this discussion to support this dubious claim.”
No problem! Your claim is that this planet is inately deterministic and subject to the whims of post modern non-scientists. Only the adjustable WV in the atmosphere controls radiative exitance of thermodynamic entropy collected in this cold atmosphere. This is but one of few ways this planet rids itself of useless energy no mater how aquired. Your post modern idiots, have no clue as to how energy enters or exits the Earth system, nor how the atmospheric WV is controled. They have never even looked. All $130 billion just to sell carbon credits. A SCAM.
@Will Janoschka, September 19, 2014 at 9:04 am
My field is electro-optics including lasers and relativistic accelerators. Thermodynamics gave me fits back in 1961 when I got my bachelors degree in physics. Even so I have great confidence in the many textbooks that derive the DALR in terms of thermodynamics.
When Rodgrigo Caballero was at University College, Dublin he published some course notes on the internet which included a strikingly lucid derivation of the DALR and also an explanation concerning the effect of vapors. Unfortunately the link no longer works but thanks to the miraculous “Wayback Machine” you can find it here:
Click to access PhysMetLectNotes.pdf
In “2.20 The dry adiabat” starting around page 40 you will find a clear explanation. Please note that the derivation does not specify whether energy is transferred by convection, conduction or radiation.
Thanks for motivating me to find the Caballero notes. I have been careful to store a copy this time!
Brian & Roger Clague,
You may like Chiefio’s take on PIVNURT:
–http://web.archive.org/web/20111107035900/http://mathsci.ucd.ie/met/msc/PhysMet/PhysMetLectNotes.pdf
In “2.20 The dry adiabat” starting around page 40 you will find a clear explanation. Please note that the derivation does not specify whether energy is transferred by convection, conduction or radiation. —
From reference above:
“Now consider an atmosphere whose vertical temperature structure is such that Θ is constant.
Such an atmosphere has the following interesting property: any parcel brought adiabatically
to the reference pressure p0 will have the same temperature on arrival as the ambient air. In
other words, the temperature structure is invariant with respect to adiabatic rearrangements.
However much you stir this atmosphere, the temperature structure does not change. In
an incompressible fluid (e.g. water), this is only possible if the temperature is uniform.
Since the atmosphere is compressible, however, the invariant profile actually has temperature
decreasing with height, at a constant rate known as the dry adiabatic lapse rate.”
I think part about stirring is interesting way to explain it.
If one stirs a gas or liquid, you stop convection [assuming one stirs it enough]. Or the stirring would balance the heat. And the convection of air molecule is doing the same thing- it’s balancing the heat. And/or such convection is way to stir [a gas or liquid].
So if have some water which has same temperature, it’s been stirred or mixed. And if atmosphere is dry and has dry adiabatic lapse rate [on Earth it’s 9.8 K per km of elevation] then that atmosphere has uniformly mixed [well stirred]. Or if it does not follow this dry adiabatic lapse rate
it has *not* been stirred. And if one changes the temperature of the air some portion of the air, then it’s not uniform temperature, and it will mix until it is uniform.
So if the ground surface is hotter than the air, it will heat the air near the ground and create an imbalance which will be corrected. And if ground is colder than air and it causes the air at surface to cool, then this also will be corrected.
How it’s corrected can only be explained by convection as it’s the only process which is able to balance heat. Or it’s what is mixing the gas.
gallopingcamel says: September 20, 2014 at 5:45 am
“@Will Janoschka, September 19, 2014 at 9:04 am
My field is electro-optics including lasers and relativistic accelerators. Thermodynamics gave me fits back in 1961 when I got my bachelors degree in physics. Even so I have great confidence in the many textbooks that derive the DALR in terms of thermodynamics.”
Hi Peter,
Some of that is fun! My electro-optics is mostly a wavelengths greater than 2.5 microns. Almost all trying to detect, or measure, something way over yonder, through the mess we call atmosphere.
“When Rodgrigo Caballero was at University College, Dublin he published some course notes on the internet which included a strikingly lucid derivation of the DALR and also an explanation concerning the effect of vapors. Unfortunately the link no longer works but thanks to the miraculous “Wayback Machine” you can find it here:”
http://web.archive.org/web/20111107035900/http://mathsci.ucd.ie/met/msc/PhysMet/PhysMetLectNotes.pdf “In “2.20 The dry adiabat” starting around page 40 you will find a clear explanation. Please note that the derivation does not specify whether energy is transferred by convection, conduction or radiation.” “Thanks for motivating me to find the Caballero notes. I have been careful to store a copy this time!”
Thanks for thr reference! I have also stored, however I now have now more than a terabyte across 6 machines., By tomorrow evening I will have no clue as to where it may be.
I remember from a co-worker the definition of an adiabatic barrier, no pressure, no energy across!
Consider some gas within an adiabatic barrier bag (non-rigid). Stand on some plywood on the bag. In the bag pressure will increase as you transfer all energy of getting on the plywood. That energy is distributed among pressure and temperature depending on the gas. Step off the bag, the plywood rises to a same level, as both internal pressure and temperature adiabatically return to the original. I guess something like charging a “good” battery! In this atmosphere the adiabat just is, get used to it.
The adiabat is storage for energy with no work being done whatsoever on the storage media. Any work is a reversible thermostatic extentensive property. Specific heat is intensive, temperature is extensive, as is mass pressure, and sensible heat.. YMMV especially with temperature!
Post modern thermodynamics is much worse today than 1961. Totally useless for engineering. Just what is the probility of adding sensible heat thus reducing temperature?
I much prefer “If it fell down?”, (don bild it dat way nomore).
@gbaikie, September 20, 2014 at 8:05 am,
I found Caballero’s quote about “Stirring” most helpful when trying to understand the Dry Adiabat given my difficulties with thermodynamics. It helped me understand some strange weather effects, at least in a qualitative way.
When the lower atmosphere is unusually still (no stirring) you can get “Temperature Inversion” that means that temperature rises with altitude. In North Carolina we used to get ice storms where rain falls through progressively colder air until it frozeon contact with trees or the ground. The ice build up could rip branches from trees and down power lines.
In the UK (and Florida) we don’t get such ice storms but the ground can cool very rapidly (by radiation) when skies are clear. The resulting temperature inversion makes ground fog commonplace. In deserts where the humidity is low ground fog can’t form so ground temperatures can plummet dramatically.
Without going into details this is exactly what I have tried to get across many times, inversions form nightside from radiative cooling which leads to atmospheric cooling failing, one of the reasons why spin is so critical to reality. Note that this applies with a radiatively inert atmosphere. (maybe why it doesn’t go down well)
Wind turns up, nothing is simple. Without it an inert gas planet will be hot.
Another way of putting this.
A mug of hot drink will cool down quickly without a lid.
Whereas an open top supermarket deep freeze has low loss.
@Will (Janoschka),
It is great to hear from another photon jockey! You are working in a spectral region I have limited experience of. In 1970 I built dye lasers operating from 380-800 nm. Next, fiber optics using semiconducor lasers from 850-1550 nm which I still teach on a part time basis including a class that starts next week in South Carolina.
http://www.bdidatalynk.com/PeterMorcombe.html
Much later I worked on John Madey’s accelerator driven “Mk3 FEL” operating from 2-10 microns. Later still I was project manager for the HIGS (High Intensity Gamma Source) designed by Vladimir Litvinenko (now at BNL) and Henry Weller.
http://www.tunl.duke.edu/web.tunl.2011a.higs.php
http://www.bnl.gov/newsroom/news.php?a=1224
When I retired in 2002 this machine operated up to 11 MeV but since then there have been some amazing upgrades under the direction of Ying Wu. The beam energy has increased by an order of magnitude but the brightness even more:
Click to access mopea077.pdf
tchannon says: September 21, 2014 at 3:33 am
“Without going into details this is exactly what I have tried to get across many times, inversions form nightside from radiative cooling which leads to atmospheric cooling failing, one of the reasons why spin is so critical to reality. Note that this applies with a radiatively inert atmosphere. (maybe why it doesn’t go down well)
Wind turns up, nothing is simple. Without it an inert gas planet will be hot.
Another way of putting this.
A mug of hot drink will cool down quickly without a lid.
Whereas an open top supermarket deep freeze has low loss.”
Is this only temperature/density affecting convection? Is the gravitational field having a more direct effect on conduction and/or radiative flux?
gallopingcamel says: September 21, 2014 at 3:58 am
“@Will (Janoschka),
It is great to hear from another photon jockey! You are working in a spectral region I have limited experience of. In 1970 I built dye lasers operating from 380-800 nm. Next, fiber optics using semiconducor lasers from 850-1550 nm which I still teach on a part time basis including a class that starts next week in South Carolina.”
Peter,
Thank you, I have little experiance in high energy anything. Trying to measure atmospheric attenuation of 8-14 micron IR radiative flux modulation, with an atmospheric path length of 400 km,
results in another AW SHIT. To this day I have little confidence in “what” I was measuring. By God, I was measuring “something”, best I could! “Do not fuck with my numbers”.
FYI: ‘Lunar volcanoes suggest the moon may still be warm’
http://www.newscientist.com/article/dn26371-lunar-volcanoes-suggest-the-moon-may-still-be-warm.html
“Young volcanism indicates possibly more magma, or magma at higher temperatures, or magma at shallower depths, or all of the above,” she [researcher] says. The heat powering this activity may come from gravitational tugs from Earth or the decay of radioactive elements beneath the moon’s surface.
[…] The paper will be of some interest to Talkshop readers given the work done on simulating Lunar temperature over the past few years. Work is ongoing in private involving others, with a four way confirmation. One article is here. […]