Debunking the Greenhouse Gas Theory with a Boiling Water Pot
New experiment proves the greenhouse gas theory is bogus and thus there is zero scientific proof of man-made global warming.
The radiative greenhouse gas effect theory (GHE) is what underpins the science of man-made global warming theory. It insists our planet radiates at a defined rate as a blackbody (BB) to create an ‘energy loop’ in the atmosphere.
Below we detail the experiment to prove there is no such BB energy for that to occur and it is therefore time to abandon the failed climate theory.
A believer in the GHE theory challenged me to perform an experiment that would show the error of my ways as a ‘non-believer.’ I duly performed the experiment and it duly confirmed my observations from my earlier modest experiment. The challenger summarized it nicely in the following exchange:
GHE Believer: “The participation of other means of heat loss such as conduction and evaporation does not alter the emissivity…”
Nick Schroeder: “Depends how emissivity is defined. Your definition is too narrow.”
GHE Believer: “…but it (emissivity) is not a property of the surroundings or situation…”
Nick Schroeder: “Call it what you want, the surroundings certainly matter, they prevent BB emission.”
GHE Believer: “…1) you have the strange notion that blackbody radiation is not possible if other means of heat transport are also present. 2) A solid or liquid can still radiate essentially as a blackbody while also conducting heat to a gas around it or losing heat by evaporation.”
Nick Schroeder: “I have presented two classical experiments, physical proof, evidence demonstrating that point 1) is and point 2) is not. You have presented nothing. Consensus doesn’t count. In the game of science, I’m up 2 to 0
So, here it is, a gallon of water, 8.33 lb, happily boiling away with a nameplate rated 1,500 W heater.
At this altitude thermo water properties has saturation temperature a little under 200 F and that’s about what is measured. The IR thermometer reads about 178 F. Not impressed. The type K T/C reads about 197 F.
The wattmeter reads about 1,180 W. Power flux calcs as 42,815 W/m^2. The only way energy gets out of the pot is through the water surface. If 97{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} this energy left surface as radiation the temperature would have to be 1,206 F. If this experiment were conducted in a vacuum, that would be the case as it was with the heating element of the modest experiment.
A surface at 200 F and 0.97 emissivity radiates 1,021 W/m^2.
Radiative + non-radiative = 42,815 W/m^2
Ratio of actual radiative power flux to total input measured power flux aka emissivity: 1,021/42,815 = 0.0238
Now what I discovered of interest is increasing the area. This is why heat exchanger manufacturers add fins, increasing the radiative surface area. There is an area factor on the attached spread sheet. If the area were tripled, radiation’s share/emissivity increases to 0.215. Six times the area and emissivity rises to 0.859.
Which brings us back to the prime issue.
For RGHE theory to function as advertised the surface of the earth must radiate as a BB, 16 C, 289 K, 396 W/m^2. This upwelling energy feeds the 333 W/m^2 GHG energy loop. No upwelling BB 396 means no energy for the 333 GHG energy loop. The GHG molecules receive zero energy to “trap” or absorb or re-radiate to simultaneously warm the atmosphere and surface.
But as demonstrated in both the modest experiment and the boiling pot the non-radiative heat transfer processes render such BB emission impossible. No BB upwelling + no GHG loop + no RGHE + no CO2/GHG = no man-caused climate change.
Below are the full details:
Modest Experiment in the Classical Style
Heat is energy in motion flowing from a hot/higher energy source to a cold/lower energy sink. A relatively hot surface transfers energy/heat to its surroundings through several processes: conduction, convection, enhanced convection or advection, latent and radiative processes. The greater the number of these modes and the more effective those modes the lower will be the surface’s operating temperature and share of radiation.
Emissivity is the ratio of the actual radiative heat emitted by a surface to the S-B BB ideal radiation based on the surface temperature or input to the system. As radiation’s share of the total heat transfer processes decreases so does its emissivity.
If half of the incoming radiation to a surface is reflected or passed through the maximum emissivity will be 0.5
If half of the outgoing radiation from a surface is handled by non-radiative processes emissivity will be 0.5.
Experimental procedure
A 125 W electric heating element is operated in open air and its surface temperature recorded.
A small biscuit fan blows air across the heating element and the temperature recorded.
A water spray bottle is used to wet the heating element and the temperature recorded.
The heating element is placed inside a one cubic foot steel box to inhibit convection and temperature recorded.
A vacuum is pulled on the steel box to remove molecules leaving radiation as the only heat transfer mode and the element’s surface temperature recorded. (elevation 6,300 feet, Baro P 24 “Hga)
Conclusion
This experiment demonstrates conclusively that the emissivity of radiative heat transfer is heavily dependent on the various modes of heat transfer. In a situation where a surface transfers heat into a contiguous participating media through the various modes, emissivity of the radiative heat transfer will reflect its respective and smaller contribution. For the K-T diagram 63/160 = 39.4{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117}.
Assuming an S-B BB emissivity of 1.0 for the earth’s surface and an average temperature of 16 C / 289 K to calculate an upwelling LWIR power flux of 396 W/m^2 is simply not supported by physical evidence. In line with the results demonstrated in this experiment and the values on the K-T power flux balance diagram, earth’s surface radiative emissivity is about 0.16, 63/396.
Without an up/down/”back” radiating loop the radiative greenhouse effect theory fails. When RGHE theory fails, so does the concept of man-caused climate change..
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Ed Bo
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Nick:
Even by your standards, this is one of your most ridiculous postings yet!
“Emissivity”, like most technical terms, has a very specific definition. The GHE believer you cite understands this. You obviously do not.
Whatever you are using here, it is most certainly NOT emissivity. Since you made up an entirely new concept, it needs an entirely new name, so let’s call it “kwyjibo”. (Google it…)
Emissivity, which is consistently defined as the ratio of thermal radiation emitted to the theoretical maximum (aka “blackbody”) thermal radiation that could be emitted by that substance at that temperature. Period.
Your experiments and your subsequent Kwyjibo calculations provide absolutely NO (as in nada, zilch, zip, zero) information about the actual radiative emissivity of water. You did not in any way demonstrate that “the non-radiative heat transfer processes render such BB emission possible.” Your results are entirely consistent with the arguments of your “GHE believer”.
You claim to have education in heat transfer. Did you sleep through the lectures on radiative heat transfer? Because you do not understand it at all!
P.S. In our last exchange, I pointed out that your heat transfer calculations were literally off by a factor of a million. Why did you go silent?
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Nick Schroeder
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By definition a BB perfectly absorbs 100 % and perfectly emits/radiates 100%. Emissivity is the ratio between what a surface actually radiates and what it would radiate if it could perfectly radiate all 100% of it. Because of the contiguous participating media, i.e. atmospheric molecules, 100% is not available for radiation ergo the surface cannot radiate as a BB.
Emissivity & the Heat Balance
Emissivity is defined as the amount of radiative heat leaving a surface to the theoretical maximum or BB radiation at the surface temperature. The heat balance defines what enters and leaves a system, i.e.
All of it entering/absorbed, W/m^2 = radiative + conductive + convective + latent
Emissivity = actual radiative / all of it, W/m^2 = radiative / (radiative + conductive + convective + latent)
In a vacuum (conductive + convective + latent) = 0 (as demonstrated in the modest experiment) and emissivity equals 1.0.
In open air full of molecules other transfer modes reduce radiation’s share and emissivity, e.g.:
conduction = 15%,
convection =35%,
latent = 30%,
radiation & emissivity = 20%
Per K-T diagram: convection 17 (10.6%) + 80 (50.0%) + 63 (39.4%) = 160 (All of it)
K-T emissivity = actual 63 / theoretical most possible 396 (288 K & 1 emissivity)
“….heat transfer calculations were literally off by a factor of a million.”
Run that by again. I don’t always get pinged on every comment.
I have other thangs to do than sit in front of the key board and repeat myself.
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Ed Bo
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Nick:
You remain as confused as ever. Emissivity is DEFINED AS the ratio of thermal radiative power emitted to the maximum possible thermal radiative power that could be emitted. The “100{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117}” you cite is this maximum possible thermal RADIATIVE power, which is called the “blackbody” radiaition. It is most certainly NOT the total heat transfer by all means.
Emissivity has NOTHING to do with other modes of heat transfer. A body with 0.80 emissivity will emit 80{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} of what a blackbody would emit at the same temperature, regardless of what else is going on.
This is how it is defined in every single science and engineering textbook or professional reference I have ever seen (and I have seen many of them). I challenge you to find a single source that matches your definition.
The actual definition is useful because it allows you to easily compute the radiative output given only the material and its temperature. Liquid water has a (fixed) emissivity of about 0.96, as any engineering reference will tell you. So for your case of boiling water, one can compute the radiative power flux density as:
Qout = 0.96 * 5.67E-08 * 373^4 = 1053 W/m^2
This gives the radiative output for that condition REGARDLESS of what else is going on.
0.96 is close enough to 1.0 that treating the emissivity as 1.0 gives an error of only a few percent, not the large errors you claim.
BTW, you forgot in all your calculations to include the ambient IR radiation impinging on the water. In a room at 20C (293K), with most solid substances having an emissivity of about 0.95, this comes to:
Qamb = 0.95 * 5.67E-08 * 293^4 = 397 W/m^2
So the net radiative balance of the water surface is 1053 – 397 = 656 W/m^2, even for a vacuum (and the inner walls of your vacuum chamber are likely to be significantly warmer than 20C). Neglecting the radiative input is horribly slopping, and if it were in a real engineering project, tantamount to negligence.
Oh, and the proper definition of emissivity, as opposed to your made-up one, is useful because absorptivity is equal to emissivity, so we know that the water will absorb about 96{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} of the incoming radiation. Your definition provides no such useful information.
But even accepting your made-up definition, nothing that you show does ANYTHING to disprove the radiative GHE.
And my earlier comment on a different thread, where I prove you were off by a factor of a million, can be found here:
https://principia-scientific.com/slayers-confirmed-pressure-not-co2-determines-planets-temps/#comment-21151
You say: “I have other thangs to do than sit in front of the key board and repeat myself.”
It’s probably for the best. You shouldn’t be repeating your silly errors.
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Nick Schroeder
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Posted a reply to the million factor at the other site.
The heater was delivering 1,180 W, 42,800 W/m^2. That’s ALL of the energy entering the system. A 1.0 emissivity BB has to emit ALL of that.
The actual radiative at 200 F is 1,053 W/m^2 or 1,053/42,800 = 2.38% = emissivity. 397 W/m^2 from the cold surroundings to the hot water is just flat impossible.
It is my contention demonstrated through experiment that since BB radiation from the surface is not possible the 396 W/m^2 upwelling LWIR is a theoretical “what if” calculation and does not in fact exist. (TFK_bams09)
With no source of energy the 333 W/m^2 up/down/”back” GHG energy loop also cannot and does not exist.
There is no energy for the GHG molecules to absorb/trap and re-emit nor “warm” the atmosphere, earth or water molecules.
There is no man-caused global warming or climate change.
Respond to that sequence.
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Ed Bo
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Nick: You go off the rails before you even get started. You say: “The heater was delivering 1,180 W, 42,800 W/m^2. That’s ALL of the energy entering the system. A 1.0 emissivity BB has to emit ALL of that.”
NOOOO! Completely wrong! This would get you flunked out of any introductory heat transfer class. You completely misunderstand the meaning of “emissivity”, as I keep telling you. I have challenged you to find a single source that uses your definition of emissivity, and you have not (I suspect because you can’t).
If radiation were the only mode of heat transfer to ambient, a body with emissivity of 0.5 would still need to emit 100% of the power to ambient radiatively. It would need to be a higher temperature than a body with an emissivity close to 1.0, but it could do it.
I see also that you do not understand the very basic concept of radiative exchange. This is how radiative heat transfer is introduced in EVERY heat transfer textbook I have ever seen. Here’s a link to the text than MIT’s ME department is using now:
http://www.mie.uth.gr/labs/ltte/grk/pubs/ahtt.pdf
It first introduces the concept of radiative exchange in the overview chapter, on page 40. In the chapter on radiative heat transfer, starting on page 487, the very first section is “The problem of radiative exchange”.
It is very clear that they teach that radiative heat transfer is a two way process, with the colder body transferring power to the hotter body (just less than the hotter body transfers to the colder body). The texts said the same thing when I studied ME at MIT many years ago.
You clearly do not understand the most basic concepts of heat transfer.
You say: “the 396 W/m^2 upwelling LWIR is a theoretical “what if” calculation and does not in fact exist.”
But we have actual measurements of this upwelling flux, knowing magnitude and spectrum. It is NOT a theoretical “what if” calculation — it is an EMPIRICAL FACT!
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jerry krause
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Hi Ed,
I write this to you for two reasons, One, I want to read your response to the idea I have had. Two, I want to make this idea a matter of record so I can conveniently refer it to others.
Robert Beatty, an Australian mining engineer, and I were discussing the volcanic activity that had occurred on my property in southern Oregon USA. And he wrote: “The top of the tube freezes quickly when lava is flowing.” And I wrote: This is a stupendous observation based upon a fundamental scientific’fact’ like the existence of gravity is a fundamental scientific fact. The interior of a ‘solid’ planetary body with volcanic activity must be cooled by radiation from its surface Hence, when I reasoned that the Venusian surface was so hot because the heating of its atmosphere above the its cloud deck by photochemical reactions, I was wrong. It was the principally the scattering of the upwelling IR radiation being emitted by it hot surface back toward the surface by the sulfuric acid cloud droplets which was severely limiting the transmission of this radiation to space. And the volcanic activity is irrefutable evidence of the nuclear fission occurring in its interior just as our volcanic activity is the result of such nuclear fission occurring in the earth’s interior. But the seasonal warming and cooling of surface soil layer at most land locations is irrefutable evidence that solar radiation is the principal cause of the earth’s surface temperature. And why the top of the tube freezes quickly when lava is flowing.
Ed, What do you think of this? Is it simple or is it simple?
Have a good day, Jerry
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Ed Bo
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Hi Jerry:
Interesting question. First, some background for perspective. The geothermal power flux density from the earth’s core, averaged over the surface, is a little less than 0.1 W/m^2. We have good measurements of this. Compared to the time-and-area-averaged 240 W/m^2 absorbed from the sun, this is trivial.
When this ~0.1 W/m^2 flux is traveling through solid rock, it creates a gradual temperature gradient. Very deep hard rock mines are hot, but most of us don’t really notice it.
Where there are gaps in the crustal rock (a VERY small fraction of the earth’s surface), it is possible for very hot molten rock from deep down to find its way to the surface. Once it reaches the surface, it is now exposed to a much colder environment, and it can dissipate its thermal energy must faster.
(Remember that molten rock is hot enough that its thermal radiation is of much shorter wavelengths than the longwave IR that is emitted at surface ambient temperatures, and the atmosphere is highly transparent to this SW radiation, so it can radiate directly to the almost-absolute-zero of space.)
You are correct that the high opacity of Venus’ atmosphere to IR, along with its great thickness, is the key reason that Venus is so hot. It effectively has many more layers of insulation against the cold of deep space than the earth does.
But the best estimates of Venus’ geothermal flux are that it is even less than Earth’s, so trivial compared to the solar input flux and the IR output flux.
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Herb Rose
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Hi Ed,
Your statement that the atmosphere of Venus insulates it from the cold of space is misleading. The thick atmosphere of Venus is absorbing the energy from the sun and distributing it around the planet. That is why the dark side of Venus, which is never exposed to sunlight, has the same temperature as the day side of the planet. On Mercury which is receiving more energy from the sun than Venus, but has no atmosphere, the day side reaches a temperature of 800 degrees while the night side of the planet is -290.
On Earth our atmosphere is also absorbing most of the heat from the sun and distributing to the surface and into space. Just because the atmosphere is transparent does not mean it doesn’t absorb energy. An undisturbed pool of water will have the surfac water heated by the sun while the deeper water remains cool even though the water is transparent. Our atmosphere is equivalent to a layer of water 33 feet deep at sea level.It is absorbing energy during the day and transmitting that energy into space and to the surface the Earth at night.
It is incorrect to speak of space as having a temperature because there is little matter to transfer energy. In the space between the Earth and the sun the temperature may be zero but the energy traveling from the sun to the Earth is greater than the energy striking the Earth. (Energy decreases with distance.)
Have a good day,
Herb
Ed Bo
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Hi Herb:
You say: “Your statement that the atmosphere of Venus insulates it from the cold of space is misleading.”
No, it is not at all misleading. It is THE KEY reason that Venus temperature is so high. And the atmosphere is far more opaque to longwave radiation than it is to shortwave solar radiation, so it gains energy primarily from the bottom and loses energy from the top, setting up a negative lapse rate.
But the fact that the optical thickness of the atmosphere to longwave IR is so great means that the thermal energy has time to distribute horizontally to the night side of the planet. (You are not correct that Venus has a permanent night side, but its day/night cycle is 116 earth day/night cycles.)
You are also incorrect that: “On Earth our atmosphere is also absorbng most of the heat from the sun and distributing to the surface and into space.” Two-thirds of the incoming solar radiation that is not reflected (by clouds) reaches the surface.
You say: ” Just because the atmosphere is transparent does not mean it doesn’t absorb energy.”
What??? The definition of transparency is non-absorbency. And your observations of a pool are not what I’ve ever seen!
You say: “It is incorrect to speak of space as having a temperature because there is little matter to transfer energy.”
For the purposes of conductive or convective transfer, the density is so low that its temperature is irrelevant. However, for the purposes of radiative heat transfer (what I am talking about, space behaves exactly like a blackbody at 2.725K (+/-0.001K) both in magnitude and spectrum. That is REALLY COLD!!!
Herb Rose
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Hi Ed,
Your comment that the atmosphere of Venus is horizontally transmitting heat is correct. This called absorbing heat. You are correct that only the shortest wavelength reach the surface of Venus. The atmosphere is so opaque that even during the day the sun is not visible so there is no day on Venus. To believe the very limited energy contained in the wavelengths shorter than ultraviolet are heating the surface of Venus is ludicrous when it represents such a small portion of the energy emitted by the sun.
Clouds do not reflect light. In order to reflect a surface must be smooth enough to return the light in the direction in which it came. Water droplets are round and absorb and scatter light. Light absorb is also emitted back towards the source but this is not reflection.The reason clouds appear white is the same reason frosted glass appears white. Light is being absorbed and emitted.
Transparency refers to the visible wavelength and dos not mean the material is not absorbing energy. A glass container is transparent but it can contain a lot of heat and burn you none the less.
On your trip I suggest you try some scuba diving since your swimming experience seems to be in indoor pools. You will need to wear a wet suit, even in the tropics, because the water temperature drops with depth (thermocline) and you should also take a flashlight because the gray corral you see can be quite colorful but you cannot tell because of the amount of light absorbed by the tranparent water.
Space has no temperature but that is not the same as absolute zero. Temperature is a measurement of the kinetic energy of molecules and until radiated energy (disturbance in the electric and magnetic fields) encounters matter it cannot be converted to kinetic energy. If you were in orbit above the atmosphere the side facing the sun would have a temperature of 250 degrees (Thats really hot) while the side receiving radiated energy from the Earth would be -250 degrees.
If you were on the surface of the Earth under the atmosphere the temperature would be 50 degrees. The energy difference for that 200 degrees is contained in the atmosphere so I don’t see how you can claim that 67% of the energy from the sun strikes the Earth’s surface.
Have a good trip,
Herb
Ed Bo
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Herb:
Back from my trip now.
You make so many mistakes in a single post! “Clouds do not reflect light.” What??? Of course they do! How else could you see clouds, either from space or from the surface? They don’t generate their own (visible) light. It is diffuse reflection, not specular reflection, but it most certainly IS reflection.
You completely misunderstand the difference between reflection and absorption/re-emission. Reflection is at the same wavelengths as the incoming EMR. Re-emission of absorbed radiation is at the wavelengths determined by the temperature of the absorbing body.
You say: “even during the day the sun is not visible so there is no day on Venus.” Wrong!!! Just look at the pictures from the Soviet Venera craft on the surface of Venus. They were taken in ambient sunlight.
And like Nick, you make up your own definitions for terms that have well defined precise meanings, and then you wonder why no one agrees with you.
Transmission of heat IS NOT absorbing heat.
Transparency refers to any wavelength of EMR, not just visible. If you mean that glass is highly transparent to visible wavelengths, but highly absorbent of longwave infrared, then SAY SO!
You need to devote yourself to a prolonged period of study of these basic principles and terms before you will have anything useful to contribute to these discussions.
jerry krause
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Hi Ed,
Thank you for your response. My idea (ideas) is (are) now on record and your ideas are also now on record.
Have a good day, Jerry
Reply
jerry krause
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Hi Ed,
I read that the temperature profile (gradient) of the Venusian atmosphere has been observed to be near the theoretical adiabatic lapse rate from a carbon dioxide atmosphere and the gravity of Venus.
At Wiki, about ‘adiabatic’ I read: ” In thermodynamics, an adiabatic process is one that occurs without transfer of heat or mass of substances between a thermodynamic system and its surroundings. In an adiabatic process, energy is transferred to the surroundings only as work.[1][2] The adiabatic process provides a rigorous conceptual basis for the theory used to expound the first law of thermodynamics, and as such it is a key concept in thermodynamics.
Some chemical and physical processes occur so rapidly that they may be conveniently described by the term “adiabatic approximation”, meaning that there is not enough time for the transfer of energy as heat to take place to or from the system.[3]
By way of example, the adiabatic flame temperature is an idealization that uses the “adiabatic approximation” so as to provide an upper limit calculation of temperatures produced by combustion of a fuel. The adiabatic flame temperature is the temperature that would be achieved by a flame if the process of combustion took place in the absence of heat loss to the surroundings.”
Does the absorption of your proposed mechanism to explain the observed temperature of the Venusian surface violate the adiabatic condition? Now I must clarify the Wiki “in the absence of heat loss to the surrounding” a bit by changing heat to energy so that all the energy being created by the combustion of hydrocarbons is being emitted as radiation.
It seems that in your proposed case the emission of absorbed radiation is only in the downward direction and no radiation is being emitted in the upward direction toward space.
Have a good day, Jerry
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Ed Bo
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Jerry: About to start a trip, so don’t have much time.
The opacity of Venus’ atmosphere (and other planets’ as well) to LWIR is sufficient to create a lapse rate greater than adiabatic. Physicists and meteorologists call this an “unstable lapse rate”. When you have a lapse rate this large, upward convection starts, driving the lapse rate back towards adiabatic.
The convection is fast enough, and the thermal conductivity of air is so low, that it is a very good approximation that an upward parcel of air is adiabatic.
I am not claiming that the atmosphere only radiates downward. Where on earth did you get that idea?
Nick Schroeder
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“…find a single source that uses your definition of emissivity…”
From a thread involving Trenberth:
“I do not know what the equation he (That’s me.) has is meant to depict but nothing on earth.
The emissivity of the oceans are about 0.97. Only in deserts is the emissivity notably lower than unity.”
Desert e lower because non-radiative conduction & convection & reflection reduces radiation’s share.
Trenberth’s and my definition are the same.
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Ed Bo
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Nick:
No!! Trenberth is using the standard definition of emissivity, not your bizarre imaginary one.
Oceans have tremendous evaporative losses, averaging over 100 W/m^2 globally. There is no way he could calculate 0.97 emissivity using your definition.
Of course, deserts do not have these evaporative losses.
Once again, you demonstrate you have no idea what you are talking about.
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Nick Schroeder
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“But we have actual measurements of this upwelling flux, knowing magnitude and spectrum. It is NOT a theoretical “what if” calculation — it is an EMPIRICAL FACT!”
The Instruments & Measurements
But wait, you say, upwelling LWIR power flux is actually measured.
Well, no it’s not.
IR instruments, e.g. pyrheliometers, radiometers, etc. don’t directly measure power flux. They measure a relative temperature compared to heated/chilled/reference thermistors or thermopiles and INFER a power flux using that comparative temperature and ASSUMING an emissivity of 1.0. (And 1.0 is WRONG!!!) The Apogee instrument instruction book actually warns the owner/operator about this potential error noting that ground/surface ε can be less than 1.0.
That this warning went unheeded explains why SURFRAD upwelling LWIR with an assumed and uncorrected ε of 1.0 measures TWICE as much upwelling LWIR as incoming ISR, a rather egregious breach of energy conservation.
This also explains why USCRN data shows that the IR (SUR_TEMP) parallels the 1.5 m air temperature, (T_HR_AVG) and not the actual ground (SOIL_TEMP_5). The actual ground is warmer than the air temperature with few exceptions, contradicting the RGHE notion that the air warms the ground.
https://www.linkedin.com/feed/update/urn:li:activity:6384689028054212608
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Ed Bo
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Nick:
Seriously? No scientist or engineer knows how to measure radiative flux for the last 150 years?
Have you ever looked at spectroradiometer data that shows the magnitude by wavelength of LWIR emissions, not just the overall magnitude?
You completely misunderstand the operation of these sensors. You say they “don’t directly measure power flux.” Well, thermometers don’t directly measure temperature, and barometers don’t directly measure pressure.
The sensor in these devices, which has a KNOWN emissivity/absorptivity and is NOT assumed to be 1.0, increases in temperature if the incoming flux is greater than outgoing, and decreases in temperature if the incoming flux is less than outgoing. There is a definite, well-defined relationship between the differential flux and the resulting sensor temperature.
Where you have to be careful is in using the derived value for incoming flux to determine the temperature of the source of the incoming flux, whose emissivity may not be known. My inexpensive little kitchen IR thermometer, which assumes an emissivity of about 0.95 when it does the conversion, works well on almost everything I use it on, hotter or colder than ambient, except for aluminum foil, which has a very low emissivity. Most solid and liquid substances we deal with have emissivities of about 0.95. (So even there, assuming 1.0 is not a large error.)
But since you don’t understand the most basic concepts of radiative heat transfer, including the definition of emissivity or the phenomenon of radiative exchange, you can’t even get started in a real analysis. I expect this of people with no formal education in the subject, but you claim this education. I don’t see any evidence of that.
But you keep dodging my major points. Why???
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Herb Rose
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Gentlemen,
In today’s science there is no reason for this disagreement.
A popular way to preserve food is to freeze the cooked food down to -30 to -40 degrees and convert the water content into ice. A strong vacuum is then applied that causes the solid water molecules to convert directly to gaseous water molecules, without going through a liquid stage (sublimation). By removing the water the food is preserved and by not adding more heat to the food, more of the flavor is preserved.
In the atmosphere where the temperature is -30 to -40 degrees and there is a strong vacuum gaseous water molecules condense into liquid water droplets. These droplets do not freeze but form clouds that produce the rain needed to grow the food.
It is possible today to have conditions in one location produce a set of results while the same conditions in a different location produce the opposite results.
This is due to the magical power of modern physics where the observer choses the reality.
You both can be right and there is no reason to argue.
Peace,
Herb
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Nick Schroeder
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Mark Twain observed, “The trouble with most of us is that we know too much that ain’t so.”
Adding to the “Δ33C without an atmosphere” (see other article) that completely ain’t so is the example of Venus.
Venus, we are told, has an atmosphere that is almost pure carbon dioxide and an extremely high surface temperature, 750 K, and this is allegedly due to the radiative greenhouse effect, RGHE. But the only apparent defense is, “Well, WHAT else could it BE?!”
Well, what follows is the else it could be. (Q = U * A * ΔT)
Venus is 70% of the distance to the sun so its average solar constant/irradiance is twice as intense as that of earth, 2,615 W/m^2 as opposed to 1,368 W/m^2.
But the albedo of Venus is 0.77 compared to 0.31 for the Earth – or – Venus 601.5 W/m^2 net ASR (absorbed solar radiation) compared to Earth 943.9 W/m^2 net ASR.
The Venusian atmosphere is 250 km thick as opposed to Earth’s at 100 km. Picture how hot you would get stacking 1.5 more blankets on your bed. RGHE’s got jack to do with it, it’s all Q = U * A * ΔT.
The thermal conductivity of carbon dioxide is about half that of air, 0.0146 W/m-K as opposed to 0.0240 W/m-K so it takes twice the ΔT/m to move the same kJ from surface to ToA.
Put the higher irradiance & albedo (lower Q = lower ΔT), thickness (greater thickness increases ΔT) and conductivity (lower conductivity raises ΔT) all together: 601.5/943.9 * 250/100 * 0.0240/0.0146 = 2.61.
So, Q = U * A * ΔT suggests that the Venusian ΔT would be 2.61 times greater than that of Earth. If the surface of the Earth is 15C/288K and ToA is effectively 0K then Earth ΔT = 288K. Venus ΔT would be 2.61 * 288 K = 748.8 K surface temperature.
All explained, no need for any S-B BB RGHE hocus pocus.
Simplest explanation for the observation.
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Ed Bo
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Nick:
So you are back to cutting and pasting your same tired spiel you keep using to spam unrelated threads.
You haven’t even tried to deal with my key criticisms:
1.) Your equation for conduction is just plain wrong! The units don’t even work out correctly. (Didn’t you learn in high school that if the units don’t come out correctly, the equation CANNOT POSSIBLY be correct?)
2.) When you use the correct form of that equation — replacing dT with dT/dz — your resulting value of Q is a factor of a million too low.
3.) Your claim that there is a heat transfer mechanism over decreasing temperature to high in the atmosphere where the energy can be radiated to space is the VERY DEFINITION of the greenhouse effect, not a refutation of it!
Why don’t you even try to deal with these? (I think I know…)
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Robert Beatty
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One important aspect Jerry and I did not discuss is the importance of having a high mountain chain to act as a cloud stirring device. Earth had a very high mountain range in its early history as a consequence of the Moon separation. Venus had no such history which means the blanketing effect of the high cloud bank is still in place, much as it was in primeval times.
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