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Total Emissivity of the Earth and Atmospheric Carbon Dioxide: A Note from Nasif S. Nahle

Introduction

Central to the theory of Anthropogenic Global Warming (AGW) is the assumption that the Earth and every one of its subsystems behaviors as if they were blackbodies, that is their “emissivity” potential is calculated as 1.0. [1]

But this is an erroneous assumption because the Earth and its subsystems are not blackbodies, but gray-bodies. The Earth and all of its subsystems are gray-bodies because they do not absorb the whole load of radiant energy that they receive from the Sun and they do not emit the whole load of radiant energy that they absorb. [8] [9] [10]

Furthermore the role of carbon dioxide is misunderstood.   According to AGW hypothesis, carbon dioxide is the second most significant driver of the Earth’s temperature, behind the water vapor, which is considered the most important driver of the Earth’s climate. [2] Other authors of AGW discharge absolutely the role of water vapor and focus their arguments on the carbon dioxide. [3]

What is the total emissivity of carbon dioxide?   I will consider this question with reference to the science of radiative heat transfer.

Total Emissivity of the Carbon Dioxide – The Partial Pressures Method

In 1954, Hoyt C. Hottel undertook an experiment for determining the total emissivity of the carbon dioxide and the water vapor [6]. He found that the total emissivity was linked to the temperature of the gas and its partial pressure. As the temperature increased above 277 K, the total emissivity of the carbon dioxide decreased, and as the partial pressure (p) of the carbon dioxide increased, its total emissivity also increased.

Hottel found also that the total emissivity of the carbon dioxide in a saturated state was very low (Ɛcd = 0.23 at 1.524 atm-m and Tcd = 1,116 °C). [6]

As Hottel diminished the partial pressure of the carbon dioxide, its total emissivity also decreased in such form that, below a partial pressure of 0.006096 atm-m and a temperature of 33 °C, the total emissivity of the carbon dioxide was not quantifiable because it was almost zero. [6] [7] [8]

After Hottel’s experiment, in 1972, Bo Leckner made the same experiment and corrected and error on the graphs plotted by Hottel. However, Leckner’s results placed the carbon dioxide in a lower stand than that found by Hottel. [6] [7]

The missing part, however, remained at the real partial pressure of the carbon dioxide in the Earth’s atmosphere and instantaneous temperatures. Contemporary authors, like Michael Modest, and Donald Pitts and Leighton Sissom made use of the following formula to know the total emissivity of the carbon dioxide considering the whole emissive spectrum, at any instantaneous tropospheric temperature and altitude [6] [7] [8]:

Ɛcd = [1 – (((a-1 * 1 –PE)/(a + b – (1 + PE)) * e (-c (Log10 ((paL)m / paL)^2))] * (Ɛcd)0 [8]

Introducing 7700 meters as the average altitude of the troposphere and the real partial pressure of the atmospheric carbon dioxide (0.00038 atm-m), the resulting total emissivity of the carbon dioxide is 0.0017 (0.002, rounding up the number).

Evidently, the carbon dioxide is not a blackbody, but a very inefficient emitter (a gray-body). For comparison, Acetylene has a total emissivity that is 485 times higher than the total emissivity of the carbon dioxide.

After getting this outstanding result, I proceeded to test my results by means of another methodology that is also based on experimental and observational data. The algorithm is outlined in the following section.

Total Emissivity of CO2 – Mean Free Path Length and Crossing Time Lapse of Quantum/Waves Method

The mean free path length is the distance traversed by quantum/waves through a given medium before it collides with a particle with gravitational mass. The crossing time lapse is the time spent by the quantum/waves on crossing a determined medium; in this case, the atmosphere is such medium.

As the carbon dioxide is an absorber of longwave IR, we will consider only the quantum/waves emitted by the surface towards the outer space.

The mean free path length of quantum/waves emitted by the surface, traversing the Earth’s troposphere, is l = 47 m, and the crossing time is t = 0.0042 s (4.2 milliseconds). [9] [10]

Considering l = 47 m to know the crossing time lapse of quantum/waves through the troposphere, I obtained the crossing time lapse t = 0.0042 s. By introducing t into the following equation, we obtain the real total emissivity of the atmospheric carbon dioxide:

Ɛcd = [1-(e (t * (- 1/s))] / √π [9] [10]

Ɛcd = [1-(e (0.0042 s * (1/s))] / √ 3.141592… = 0.0024

Therefore, the total emissivity of the atmospheric carbon dioxide obtained by considering the mean free path length and the crossing time lapse for the quantum/waves emitted from the surface coincides with the value obtained from the partial pressures method:

Ɛcd 1 = 0.0017 = 0.0017

Ɛcd 2 = 0.0024 = 0.0024

The difference is 0.0007, which is trivial in this kind of assessment.

Conclusions

In the introduction I asked: What is the total emissivity of carbon dioxide?

In this note I have calculated the real total emissivity of the atmospheric carbon dioxide at its current partial pressure and instantaneous temperature to be 0.002.

Clearly carbon dioxide is not a nearly blackbody system as suggested by the IPCC and does not have an emissivity of 1.0. Quite the opposite, given its total absorptivity, which is the same than its total emissivity, the carbon dioxide is a quite inefficient – on absorbing and emitting radiation – making it a gray-body.

Accepting that carbon dioxide is not a black body and that the potential of the carbon dioxide to absorb and emit radiant energy is negligible, I conclude that the AGW hypothesis is based on unreal magnitudes, unreal processes and unreal physics.

Acknowledgements

This blog post was inspired by Chapter 12 of the book ‘Slaying the Sky Dragon. 

“This first catechism will be referred to in a later figure as the ‘Cold Earth Fallacy’, and it is based on the erroneous assumption that the earth’s surface and all the other entities involved in its radiative losses to free space all have unit emissivity. The second catechism has already been discussed: the contention that Venus’ high surface temperature is caused by the ‘greenhouse effect’ of its CO2 atmosphere.”

-Dr. Martin Hertzberg. Slaying the Sky Dragon-Death of the Greenhouse Gas Theory. 2011. Chapter 12. Page 163. [11]

http://www.amazon.com/Slaying-Sky-Dragon-Greenhouse-ebook/dp/B004DNWJN6

References

[1.]  Hertzberg, Martin. Slaying the Sky Dragon-Death of the Greenhouse Gas Theory. 2011. Chapter 12. Page 163.

[2.]  http://www.bom.gov.au/info/GreenhouseEffectAndClimateChange.pdf (Page 6).

[3.]  http://www.aip.org/history/climate/co2.htm

[4.]  http://www.zypcoatings.com/ProductPages/BlackBody.htm

[5.]  http://www.ib.cnea.gov.ar/~experim2/Cosas/omega/emisivity.htm

[6.]  Hottel, H. C. Radiant Heat Transmission-3rd Edition. 1954. McGraw-Hill, NY.

[7.]  Leckner, B. The Spectral and Total Emissivity of Water Vapor and Carbon Dioxide. Combustion and Flame. Volume 17; Issue 1; August 1971, Pages 37-44.

[8.] Modest, Michael F. Radiative Heat Transfer-Second Edition. 2003. Elsevier Science, USA and Academic Press, UK.

[9.]   Lang, Kenneth. 2006. Astrophysical Formulae. Springer-Verlag Berlin Heidelberg. Vol. 1. Sections 1.11 and 1.12.

[10.]  Maoz, Dan. Astrophysics in a Nutshell. 2007. Princeton University Press. Pp. 36-41

 [11.]  Dr. Hertzberg is an internationally recognized expert on combustion, flames, explosions, and fire research with over 100 publications in those areas. He established and supervised the explosion testing laboratory at the U. S. Bureau of Mines facility in Pittsburgh (now NIOSH). Test equipment developed in that laboratory have been widely replicated and incorporated into ASTM standards. Published test results from that laboratory are used for the hazard evaluation of industrial dusts and gases. While with the Federal Government he served as a consultant for several Government Agencies (MSHA, DOE, NAS) and professional groups (such as EPRI). He is the author of two US patents: 1) Submicron Particulate Detectors, and 2) Multichannel Infrared Pyrometers.  http://www.explosionexpert.com/pages/1/index.htm

**************

Read more from Nasif by scrolling here: http://jennifermarohasy.com/blog/author/nasif-s-nahle/

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237 Responses to “Total Emissivity of the Earth and Atmospheric Carbon Dioxide: A Note from Nasif S. Nahle”

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  1. Comment from: Nasif Nahle


    @RWFOH…

    Thanks for the links to those blogs where I defeated a horde of trolls. I demonstrated that my calculations were right and AGW ideas were wrong.

    They, as you, never showed a single number, a single algorithm which contradicted my peer review. Just trolling, like this of yours.

    Thanks again for your deference. LOL!

    :D

  2. Comment from: Nasif Nahle


    I’m still waiting for a scientific demonstration from Gavin, RWFOHthat my calculations are wrong.

  3. Comment from: Nasif Nahle


    I’m still waiting for a scientific demonstration from Gavin, RWFOH and Luke that my calculations are wrong and that my paper is not peer reviewed.

    :D

  4. Comment from: Nasif Nahle


    I’d like to call your attention to this data from Hottel’s experiment:

    Hottel found also that the total emissivity of the carbon dioxide in a saturated state was very low (Ɛcd = 0.23 at 1.524 atm-m and Tcd = 1,116 °C).

    Notice that the partial pressure of the carbon dioxide was 1.524 atm, which is 4010 times higher than the atmospheric partial pressure of this gas, and that the temperature was 1116 °C, 66 times higher than the standard temperature of the atmosphere. Nevertheless, the total emissivity of the carbon dioxide was the insignificant 0.23.

    :)

  5. Comment from: gavin


    Nasif; I said early on in our exchanges fancy math don’t = science. I said recently I can’t follow your flow ie logic as most radiative stuff requires the use of black body physics for starters and it’s false engineering to go down the CO2 plant path.

    It’s a pity we can’t see the CO2 molecule in action but it either absorbes or emits an IR photon as a black body in math terms or it dosen’t work as a spectrum filter in practice. Yuo can’t wriggle out of that. Btw when did Hertzberg review your stuff?

    Also there is another furnace product SO2. What can you say about that re plants or climate change??

  6. Comment from: debbie


    Luke, Rwfoh, Gavin
    I thought the basic questions being asked here were:
    Do the current models assume CO2 has an emissivity of 1 in our atmosphere?
    Is that assumption correct or does the research we have been directed to by Nasif and others indicate that it is incorrect?

    If that assumption is correct (and if that is the figure in the radiative heat transfer models?) then you have nothing to worry about.
    If that asumption is incorrect then the models need updating don’t they?
    I wasn’t trying to get ahead of myself Luke, I would really like to know what the figure is and what figure is being used in the models.
    I also think it has to be relevant to our atmosphere doesn’t it?
    CO2 in a furnace is not the same as CO2 in our atmosphere is it?
    Plant absorption of CO2 is an extremely important part of the way CO2 operates in our atmosphere.
    Our atmosphere is definitely not a furnace.

    Gavin (at least I think it was the same Gavin) posted this elsewhere on Jen’s site

    “There is a simple way to tell the difference between propagandists and scientists. If scientists have a theory they search diligently for data that might actually contradict the theory so that they can fully test its validity or refine it. Propagandists, on the other hand, carefully select only the data that might agree with their theory and dutifully ignore any data that disagrees with it”

    It tends to apply here as well don’t you think?

  7. Comment from: Nasif Nahle


    @ Gavin…

    Thanks for your comment.

    You say:

    Nasif; I said early on in our exchanges fancy math don’t = science. I said recently I can’t follow your flow ie logic as most radiative stuff requires the use of black body physics for starters and it’s false engineering to go down the CO2 plant path.

    Maths is the language of science, therefore, science comes first and maths after science. Hottel and others carried out the experiments, observations, etc., scientific method; soon after, they and others derived the formulas -science language- that all scientists on thermal energy transfer apply.

    Maths permits us to make predictions and calculations over real conditions. Thus, we corroborate those calculations with observations and experimentation. That’s what Lapp conducted to present his PhD thesis. After him, many other scientists conducted experiments and completed and corrected the charts by Hottel. This is how science works.

    Calculations on black bodies, which do not exist in nature, are useful because we can have an idea about what is happening in a determined process, for example the Stefan-Boltzmann equation on black bodies. However, if we want to know what is actually occurring, we make use of gray bodies formulas; again, for example the Stefan-Boltzmann equation, but introducing the emissivity/absorptivity of the system that we are studying.

    You say:

    It’s a pity we can’t see the CO2 molecule in action but it either absorbes or emits an IR photon as a black body in math terms or it dosen’t work as a spectrum filter in practice. Yuo can’t wriggle out of that. Btw when did Hertzberg review your stuff?

    Yes, it’s a pity that we cannot see molecules in action. Nevertheless, if the molecule of CO2 would absorb or emit as a black body, it would be more efficient than the own Sun and it would be a sorce, not a simple absorber/emitter of IR. Starting from the fact that the wavelength emitted by the surface has not the same wavelength than the quantum/wave absorbed and the quantum/wave emitted by the carbon dioxide is not the same than the wavelength of the absorbed quantum/wave:

    http://jennifermarohasy.com/blog/2011/03/recycling-of-heat-in-the-atmosphere-is-impossible/

    From here, the problems start because this issue on the change of wavelengths has been extensively verified and it makes impossible that the carbon dioxide reabsorbs an emitted quantum/wave from another CO2 molecule. Consequently, CO2 does not absorb/emit as a black body just because it is not a black body.

    In a furnace, like that of Hottel where he heated the carbon dioxide up to 2505 °C to confirm the total emissivity of the carbon dioxide, you can make the air as a whole heats up at a given temperature. It does not happen in nature because the energy incoming from the Sun and the solar energy absorbed by the surface are limited amounts. If another Sun sudenly appeared, or we we lost all of our carbon dioxide (a super-coolant) we would be in serious problems. :)

    You say:

    Also there is another furnace product SO2. What can you say about that re plants or climate change??

    To be honest, I have not made calculations on SO2. But some experiments show it is a good absorber of thermal energy.

    NSN

  8. Comment from: Nasif Nahle


    It is worthy to notice that Hottel found that the total emissivity of the carbon dioxide at 2502 °C (147 times higher than the standard temperature of the Earth) and a partial pressure of 5 atm (five times the effective pressure of the atmosphere at sea level) was 0.09.

    We don’t need imagination to know that under the physical conditions of the Earth, with average standard temperature of 17 °C and a partial pressure of the atmospheric carbon dioxide of 0.00039 atm the carbon dioxide cannot have a total emissivity higher than 0.002.

    Is carbon dioxide a good absorber/emitter of heat? No, it’s a very weak one; it is a coolant, not a warmer.

    NSN

  9. Comment from: cohenite


    Well, now Roetc has left the building casting aspersions about my character and linking to a SoD thread where “someone named Mark really took him [Nasif] to task and the whole facade came crashing down.”

    I suspect Mark is Tamino but that is of no consequence; the exchange between Nasif and someone who I do respect, DeWitt Payne, is interesting: DeWitt says this [on June 21, 2010 at 5:22 pm]

    “The unit atm m in SI units has the dimension m only (a Dobson Unit is 0.00001 atm m so 300 Dobson units is 0.003 m), which is a little confusing I admit. It took me a while to figure that out. It’s the mass of gas per unit area divided by the density of the gas at 1 atm. So then the mass of gas/m2 is atm m times the density. One atm m of a gas with the density of dry air at STP (1 atm, 273.2 K) has a mass of 1.293 kg/m2. 1 atm in SI units has the dimensions kg m-1 sec-2. As you should be able to see, they’re really quite different. 3.04 atm m of CO2 then has a mass of 5.971 kg. 7988 atm m of dry air has a mass of 10,328 kg/m2 or the mass/m2 of the atmosphere (101,325 Pa/9.81 m/2(g)). So again, 3.04 atm m of CO2 is not 3,000,000 ppmv, it’s 3.04/7988 or 0.00038 ppmv.

    One can also have kPa m, which would then be the mass of gas/m2 divided by the density of the gas at 1 kPa pressure. Or in English units, atm foot, which would be the mass in pounds of gas/square foot divided by the density at STP in pounds/cubic foot. Once you understand that atm m is a length, it’s all very simple. And it is a length.

    For an optically thin emission source, doubling the path length doubles the emission. By saying that CO2 in the entire vertical column of the atmosphere is 0.00038 atm m, you’re really saying that the atmosphere is only 1 m thick.”

    Nasif replies:

    “1. You don’t know a reliable formula for obtaining the total emissivity of an absorbent gas in the atmosphere.

    2. You mix things pertaining to one concept with other unrelated things, for example, DU with carbon dioxide.

    3. You don’t know the real meaning of the units atm m because you say that 0.00038 atm m is equal to one meter thick. You’re absolutely wrong. 0.00034 atm m, not 0.00038 atm m, is the thickness that the atmosphere would have if it were composed exclusively by carbon dioxide at its current density, which would be 0.34 millimeters. It couldn’t be otherwise because the density of the carbon dioxide in the Earth’s atmosphere is 0.00062 Kg/m^3, and its mass fraction is barely 383 ppmV. But you say it is 3.04 atm m, which is equivalent to 3,000,000 ppmV, which is equivalent to 5.81300 kg / m^3.

    4. You don’t know how to obtain partial pressures because you said that 3.04 atm is the partial pressure of the carbon dioxide in the atmosphere. See what you said:

    DeWitt responds:

    “Dobson units are relevant because they are the same as atm m (@STP) except for a scale factor of 100,000.

    I don’t say 0.00038 atm m is 1 m. Quite the opposite. 0.00038 atm m = 0.00038 m (@ STP in SI units). and 1 atm m = 1 m (@STP in SI units). atm m, as I said has the dimension meters. What you continue to miss is that the vertical column of the atmosphere of the Earth at STP is 7988 atm m, not 1 atm m.”

    Nasif:

    “If the mass fraction of carbon dioxide is 383 ppmV per each cubic meter, at an altitude of one meter a P = 1 atm abs, and a temperature of 308 K, the density of that cubic meter of air is 0.00067 Kg/m^3.

    I am introducing real magnitudes, so it is what we observe in nature.

    Those 0.00067 Kg/m^3 of carbon dioxide represent 0.038% of the total mixture of gases comprising the air, which is equivalent to 0.038% of the absolute pressure at the sea level, which is the absolute pressure applied by the whole column of the mixture of air, i.e. 1 atm.

    Consequently, the partial pressure of the carbon dioxide in the atmosphere, at the sea level, is 0.038% / 100 % * 1 atm at every meter of the air column, assuming that the mixture of gases in the atmosphere is homogeneous in temperature and density at any altitude below 23500 m.

    We know this is not real because the temperature, the composition, the density and the pressure of the air decrease with altitude. You are dismissing this fact, assuming that the partial pressure of the carbon dioxide has to be multiplied by the height of the column of air. This practice gives always false data.”

    The point here is a conflict about how to best calculate the partial pressure and mean free path length of CO2; can it be done by extrapolating from the first meter at sea level over the whole atmospheric column, as I think DeWitt advocates, or at every level, as I think Nasif advocates.

  10. Comment from: gavin


    “Starting from the fact that the wavelength emitted by the surface has not the same wavelength than the quantum/wave absorbed and the quantum/wave emitted by the carbon dioxide is not the same than the wavelength of the absorbed quantum/wave:” is another English mess, let’s see the math hey

    Debbie; CO2 that comes out of a furnace with all it’s radience characteristics there is that extra CO2 in the atmosphere that AGW science follows. CO2 is Carbon Dioxide regardles of source. I claim we know how it transports radiation from combustion and other uses such as lasers where changing concentrations with other gases have an inpact on exchanger preformance.

    The process of combustion is not cooled by the production of CO2

  11. Comment from: debbie


    OK Gavin,
    I get that. “The process of combustion is not cooled by the production of CO2″
    Considering the whole point of creating combustion in a furnace is to create heat, the production of that much CO2 would create a huge problem if it was an efficient coolant in a furnace.
    Got that bit, no worries.
    BUT…is the process of combustion HEATED by CO2?
    In a furnace, isn’t CO2 a by product of the combustion process? Does it actually have amazing heating properties of its own in the furnace or indeed when its released into the atmosphere?
    I’m not arguing that we’re putting more of it out there. I don’t think anyone is arguing about that.
    BUT, the question about the way CO2 behaves in our atmosphere as opposed to a furnace is a valid question.
    The atmosphere is not an impenetrable cast- iron like object that lets no heat out or in.
    A furnace does not have humans and animals and oceans and plant life living inside it that absorb and process CO2 in many different ways. All species on Earth need CO2 as part of their survival. It is not a toxic gas. Is it an efficient heating gas?
    Isn’t that part of what Nasif is trying to explain?
    Are we correct to asume that CO2 has an emissivity of 1 in our atmosphere?
    Are the models in fact assuming that CO2 does have an emissivity of 1?

  12. Comment from: Nasif Nahle


    @Gavin…

    You say:

    The process of combustion is not cooled by the production of CO2

    Of course not because you have an external operator, you, and a primary source of heat, gas, electricity or coal; I don’t know your furnaces. You have to admit that if you inject carbon dioxide to the furnace and do not put to work the source of thermal energy, the carbon dioxide will NEVER heat up your furnace.

    :)

  13. Comment from: Nasif Nahle


    Inside most furnaces, the air heats up thanks to a source of thermal energy that heats it up.

    A furnace must have a good system to avoid the thermal energy scapes fastly to the environment. Reflective walls, refractive materials, etc., but most important, some way to avoid the warm air scapes to the cooler surroundings (convection).

    Debbie says the truth, a the Earth is not a furnace, evidently.

    NSN

  14. Comment from: Nasif Nahle


    @Cohenite…

    Heh! The best part came when I messed the brackets and DeWitt found the error. I corrected my mistake but… The results were the same!

    After verified that my numbers were correct and that he obtained the same numbers, DeWitt left the dialogue very happily because he had found a reliable way to calculate total emissivities.

    All the best,

    NSN

  15. Comment from: cohenite


    Nasif; where did the debate with DeWitt resolve itself with him being happy with your view, I can’t seem to find it, and could you clarify which system for measuring the CO2 variables you prefer; extrapolating from the first 1 metre to the rest of column or measuring each metre seperately?

    I have just reread your exchange with Mark/Tamino and he is persistent; the only escape clause I can find is when he says this:

    July 18, 2010 at 10:38 am “It also gives you 238K when calculating the temperature change of the earth when you use it with your figure of 181.64W/m^2 if you assume emissivity of 1 (tarmac, for example, would be pretty close)).”

    Did you in fact assume an emissivity of 1?

  16. Comment from: gavin


    Debbie; it seems I need to labor the point about furnaces, it’s the sudden large amount of CO2 after combustion that should be the target of any one fishing for CO2 emissivity at a pressure of one atmosphere minus a teensy bit = to your suck in between words or about 2cm negative in a u tube water manometer given air is the source of O2 necessary for combustion. Note too in that process, O2 is almost completely used in an efficient fuel conversion.

    Engineers need to calculate radiant heat transfers for all gases remaining in the air stream in addition to conduction and convection right down to the last flue gas heat exchanger that’s used for warming the fresh air coming in with fuel. Btw reflection is not much of an issue for heat exchangers since they are very black steel and mostly far away from red hot boiler tubes close to combustion. That pesky CO2 and unfortunately SO2 along with nitrogen are the medium for all radiant heat transfer after combustion.

    I made these observations while watching real furnaces in action, wood, coal, oil & gas fired often together and sometimes after fuel conversions. However I started as instrument tec in a team including several physicists building a pilot reactor for combusting black (organic) liquor and air in combination at extremely high pressures in order to salvage our caustic from a large wood chip pulping process.
    I reckon it’s worth pondering what happens to CO2 and it’s heat transfer characteristics in all conditions (2000 psi as described above) and say it’s not enough to hunt down the AGW science by it’s self from some arbitrary position outside our growing climate circles. Getting a proper handle in the right field is vital if our practice is to improve

  17. Comment from: debbie


    Thanks Gavin,
    That helps a lot.
    There’s no question that we release a lot of extra CO2 due to our production of power and also other processes that burn solid fuels like steel production.
    Most of that production of CO2 is because we furnace the solid fuels (particularly coal).
    I also agree that it’s important to understand what happens to CO2 after it is released into our atmosphere and we defnitely need to understand what its heat transfer characteristics are once it is released.
    I am still not convinced that its emissivity characteristics are the same as they would be inside a furnace.
    Once it is released into the atmosphere a whole new set of variables are influencing CO2 characteristics.
    One of them is that carbon as a singular molecule is heavier than air.
    Another is that our life cycle on planet Earth uses carbon and CO2 in many different ways.
    Another is that our atmosphere transfers energy in many different ways and it is not the same as a furnace.
    I would be happier if I was convinced that CO2, once released into the atmosphere, definitely had that emissivity factor of 1.
    I am still convinced that if it doesn’t have that emissivity figure of 1, then the AGW climate models need updating.
    I don’t think the way the calculations have been formulated are wrong.
    I am questioning the inputs.
    I know without doubt if the inputs are incorrect, then the results will be too.
    It doesn’t matter how complex the models are, they are not able to question the inputs.
    That is definitely up to the people who key in the inputs.
    Also…I’m assuming it’s a typo error (there is those pesky inputs again!) but I didn’t understand the latter part of your first paragraph:
    “minus a teensy bit = to your suck in between words or about 2cm negative in a u tube water manometer given air is the source of O2 necessary for combustion. ”
    ???
    Do you mind explaining what you meant to say?

  18. Comment from: Nasif Nahle


    @Gavin…

    I do not know too much on furnaces; however, there is a statement from you that is not congruent with your large post:

    I reckon it’s worth pondering what happens to CO2 and it’s heat transfer characteristics in all conditions (2000 psi as described above) and say it’s not enough to hunt down the AGW science by it’s self from some arbitrary position outside our growing climate circles. Getting a proper handle in the right field is vital if our practice is to improve

    There is not hunting of climate science, but only demosntrating that AGW is pseudoscience and antiscience.

    On the other hand, we have the charts on the total emissivity of the carbon dioxide which are based on experiments and observations. For example, you should know that Hottel, Leckner and Lapp carried out their experiments on a very wide range of temperatures that, perhaps, coincide with the temperatures inside your furnaces. They tested the carbon dioxide from 5 °C to 2600 °C.

    Those charts can be read from any book dealing with heat transfer. It’s quite strange that you, who works at furnaces, have never seen them.

    :)

  19. Comment from: Nasif Nahle


    @Cohenite…

    Perhaps science of doom erased the “inconvenient” posts. When I asked Dan to answer by me, it was because I had been banned to post on science of doom blog. So, it would not be exceptional that science of doom had erased those posts.

    I have not visited his/her blog to post there, but only to see or read his self-promoting articles.

    NSN

  20. Comment from: Nasif Nahle


    @Cohenite…

    At higher altitude, the partial pressure and temperature of the carbon dioxide decrease, you know. In consequence, the total emissivity of this gas decreases sensibly with altitude. Above the first kilometer, the total emissivity of the carbon dioxide is quite insignificant (paL = 0.00034 atm), therefore, if you calculate the total emissivity of the carbon dioxide introducing its partial pressure at 1 meter of altitude (0.00039 atm), then at 1000 meters (0.00034 atm), 2000 meters (0.0003 atm), 3000 m (0.00026 atm), etcetera, the results will be the same than at one meter of altitude because we cannot add a total emissivity at 1 m plus the total emissivity at 1000 meters, at 2000 meters…. etcetera. It would not be an addition but an average.

    Regarding your second question, of course not, I was looking for the real emissivity of the carbon dioxide, not the temperature by the energy absorbed by a black body.

    If the incident solar radiation on the surface of the Earth is 249 W/^2, and I said that the Earth’s surface absorbed only 181.64W/m^2, I was obviously considering an absorptivity of the Earth of 0.73, not of 1.0.

    That was a distracting tactic from Tamino/Mark because 249 W/^2 is the number given by K-T in their annual energy budget calculations.

    NSN

  21. Comment from: cohenite


    Thanks Nasif

  22. Comment from: Graeme M


    This may be tangentially relevant, but I make no claims regarding its accuracy. I found it on a recent Judith Curry blog comment and I expected some kind of response from other posters, but so far zilch. Perhaps I miss the point or I’ve not kept across the various debates well enough, but this does sound interesting. Any views?

    http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html

  23. Comment from: Mack


    Science of Doom :) :)
    What sort of an idiot would call himself “Science of Doom”?
    You have science, real science.
    And then you have science of doom!!!!
    Aaahahahahahahahahahahahaha.

  24. Comment from: Martin Mason


    This is very good stuff from Nasil, very thought provoking. The responses from the AGW side are as always hackneyed, predictable and in the main unconvincing in the extreme. Do they not understand why people don’t buy CAGW? Peer review is not the solution for making the debate on climate science more open and useful, it is the problem which stops it happening.

    I’m quite disappointed with the Blogs that refuse to publish Nasils theories.

  25. Comment from: cohenite


    Graeme M, this may be of interest:

    http://www.ilovemycarbondioxide.com/pdf/Rethinking_the_greenhouse_effect.pdf

    But you should also read this:

    http://scienceofdoom.com/2010/06/12/venusian-mysteries/

  26. Comment from: gavin


    Debbie; a brief note re ‘Do you mind explaining what you meant to say’

    The water manometer was a portable instrument I carried in boiler house commissioning usually after a shutdown for annual maintenance. It was never more than a coil of clear poly tubing, a 1/2 Lt drink bottle for tap water and a common rule or measuring tape

    Even the most basic steam generator required 3 instruments, No1 a drum water level indicator, 2 a steam pressure gauge, 3 a combustion chamber draft indicator to ensure that no gaseous fuel mixtures exceeded atmospheric pressure at any time. Since this is mostly about balancing the suck of the stack and blow of fans feeding the furnace under varying operating conditions i.e. hot and cold furnace I used to suck and blow their instrument and mine with a great deal of caution because thick furnace dust got into our tapings often.

    A fine tuned system could run safely with a minus 1/2″ water gauge differential with ambient P

  27. Comment from: cementafriend


    Gavin, it seems you are back in the rail steam engine era or the old B&W spreader stokers.
    Boilers are a bit more sophisticated these days. Further, fluid bed boilers are pressurised but maybe you have never heard of them.

  28. Comment from: Grant Petty


    Nasif,

    I have read the first part of your message, and I was able to immediately conclude that you have never taken a course in atmospheric radiative transfer or, at the very least, you didn’t grasp what you were taught.

    I say this on the basis of having written a textbook on atmospheric radiative transfer and having taught atmospheric physics at the university level for 20 years.

    It is of course your prerogative to dismiss my expertise on the subject you are writing about. In fact, I fully expect you to do so, because I have encountered many like you, and you will be quite convinced that your faulty understanding of radiative transfer is superior to that of someone who studies and teaches it for a living. And so will your supporters in this website. That gloomy prediction does not restrain me from pointing it out for the sake of other readers who might happen across my reply before you get around to deleting it.

    If you want to debate the role of CO2 in global warming, at least do yourself and your readers the favor of doing your homework first. Start by understanding that no gas has an intrinsic emissivity, and so your comparison of emissivity in a laboratory with that in the free atmosphere is meaningless, because it doesn’t take into account the different in the size of the sample.

    – Grant Petty

  29. Comment from: Mack


    Sciece of Doom,
    I wish to apologise for my ad hom attack,

  30. Comment from: Nasif Nahle


    Grant:

    I have read the first part of your message,

    Which one of my messages?

    and I was able to immediately conclude that you have never taken a course in atmospheric radiative transfer or, at the very least, you didn’t grasp what you were taught.

    You are wrong in every one of your assertions on the above paragraph.

    I say this on the basis of having written a textbook on atmospheric radiative transfer and having taught atmospheric physics at the university level for 20 years.

    If you wrote a wrong book on atmospheric radiative transfer, it’s not my problem, but yours.

    It is of course your prerogative to dismiss my expertise on the subject you are writing about. In fact, I fully expect you to do so, because I have encountered many like you, and you will be quite convinced that your faulty understanding of radiative transfer is superior to that of someone who studies and teaches it for a living. And so will your supporters in this website. That gloomy prediction does not restrain me from pointing it out for the sake of other readers who might happen across my reply before you get around to deleting it.

    I’m not dismissing your “expertise” on the subject and I don’t care about your “expertise. If you are wrong from a scientific standpoint, it is your problem, not mine.

    I only ask you to show me what the experiment on total emissivity of carbon dioxide you have conducted through your “20 years of expertise” and the numbers on the total emissivity of the carbon dioxide in the atmosphere, at current conditions, are. I don’t care your expertise, even if you were Albert Einstein or Newton.

    If you want to debate the role of CO2 in global warming, at least do yourself and your readers the favor of doing your homework first. Start by understanding that no gas has an intrinsic emissivity, and so your comparison of emissivity in a laboratory with that in the free atmosphere is meaningless, because it doesn’t take into account the different in the size of the sample.

    Show me your numbers for the total emissivity/absorptivity of the atmospheric carbon dioxide is. And please, don’t try to say that total emissivity/absorptivity of the atmospheric gases is out of this world or that it is meaningless to know the radiative thermal energy is in the atmosphere because you would be wasting your “20 years of expertise” on the matter. My work is based on observations and experimentation and I have proven it from different scientific viewpoints: the one provided by physical chemistry, the one provided by quantum physics, the one provided by astrophysics, and the one provided by quantum thermodynamics. If you consider that the climate viewpoint contradicts the scientific knowledge because the laws of physics don’t apply to climate, then you’re handling pseudoscience.

    :)

  31. Comment from: Neutrino


    Nasif,
    I have a few questions about how you came up with a mean free path of 47m for a photon through CO2.

    In your article “Mean Free Path of Photons…” you calculate the distance, quoted below:

    “l = m / (n σ) (References 1 and 2).
    Where l is for the mean free path length, m is for the mass of the gas, n is for the number of molecules of the substance per gram, and σ is for Thomson’s Cross Section (6.7 x 10^-25 cm).

    Introducing magnitudes:
    l = (6.894 x 10^-7 g) / (2.2 x 10^14 (molecules/g) * 6.7 x 10^-25 cm) = 4677.1 cm”

    a) Something that should jump right out at anyone is the mismatch in the units. You claim to be calculating a distance yet the equation as written has units of g^2/cm. As written this equation makes no sense.

    b) m, from the value used above is the mass of just one cubic cm of the gas. n, is simply wrong by many many many orders of magnitude. And σ should have units of cm^2.

    c) What are you calculating the Mean Free Path of? CO2, at atmospheric temperature and pressure, is not an ionized gas with free electrons and as such will not cause Thomson Scattering. The question should be what is the Mean Free Path before absorption not before scattering.

    d) Where did you get that equation from? I do not have access to your reference (1) but reference (2) does not contain it. It does have something similar(equ. 3.41) but that is not what you have used above.

    I look forward to your reply.

  32. Comment from: Nasif Nahle


    @Neutrino…

    To you question a) Obviously molecules is not a unit. It should have said. Many people like you do not know about this, so it is a good observation:

    Introducing magnitudes:

    l = (6.894 x 10^-7 g) / (2.2 x 10^14 /g * 6.7 x 10^-25 cm) = 4677.1 cm

    You say:

    b) m, from the value used above is the mass of just one cubic cm of the gas. n, is simply wrong by many many many orders of magnitude. And σ should have units of cm^2.

    Sorry, but who is wrong is you. I hope you know how to calculate molecular mass. If you know how to do it, you’ll find that “m” is correct.

    c) What are you calculating the Mean Free Path of? CO2, at atmospheric temperature and pressure, is not an ionized gas with free electrons and as such will not cause Thomson Scattering. The question should be what is the Mean Free Path before absorption not before scattering.

    I am not making a calculation for an ionized gas. Do you know with which particles of a molecule the quantum/wave interacts?

    You say:

    d) Where did you get that equation from? I do not have access to your reference (1) but reference (2) does not contain it. It does have something similar(equ. 3.41) but that is not what you have used above.

    A pity.

    :)

  33. Comment from: Nasif Nahle


    Errata:

    l = (6.894 x 10^-7 g) / (2.2 x 10^14 /g * 6.7 x 10^-25 cm) = 4677.1 cm

    Corrected:

    l = (6.894 x 10^-7 g) / (2.2 x 10^14 g * 6.7 x 10^-25 cm) = 4677.1 cm

    :)

  34. Comment from: Science of Doom


    You can see some examples of spectrometer measurements of atmospheric radiation from Grant Petty’s book in:

    Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Ten

    People who are interested in science will notice that there are some measurements around 15um that don’t fit in with Nahle’s theories.

    People interested in ways to ignore science – talk among youselves.

    Previously I asked Nahle about the inconvenient measurements in his many hundreds of comments on Lunar Madness and Physics Basics – for example, here.

    Nahle wasn’t interested in actual atmospheric measurements because they didn’t fit his theory. Why waste time with real measurements?

    A bit like not being interested in evidence that his absurd claim (“atmospheric physics assumes that the atmosphere radiates as a blackbody”) is false. Why bother reading what the textbooks say when you can pretend the complete opposite?

    You can see how censored the persecuted Nasif Nahle was on my blog by checking the hundreds of comments he made on that article. Actually it was his counter-commenter who was the reason I suspended all comments on that article for one day. One day. Persecution.

    I invite the few people who read this blog who are interested in real science to take a look at Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Ten and ask Nahle here to explain how it is that the atmosphere radiates so strongly around 15um.

    When you get the usual bluster perhaps it will help make sense of the puzzle.

  35. Comment from: Science of Doom


    cohenite on March 30th, 2011 at 10:43 am

    The most important point to understand is that differential absorption (dI/ds) is proportional to the number of absorbers in the path, ds.

    This is stated as the Beer-Lambert law.

    This is why any formula which calculates absorptivity or emissivity has to take account of the total number of absorbers.

    And, therefore, if the number of absorbers is expressed via density or number of molecules/unit volume you have to take account of the total path length. If you assume a 1m path length has the same number of molecules as a 1000m path length you will be making a big mistake.

    It’s so fundamental that when someone claims it isn’t important you just have to laugh. No point arguing with them. Or when they can’t see that bar.cm on a graph means you have to include path length.

  36. Comment from: Nasif Nahle


    @ScienceofDoom…

    Mine is not a model, in the first place, but reality. The formula that I applied is everywhere in good books of radiative heat transfer and the methodology is following the scientific method.

    As the authors of the experiments said, they took all the “strong” bands of the absorption spectrum of the carbon dioxide, so your arguments are absolutely false.

    It seem you write in your blog whatever comes to your mind. The backradiation measured is no more than 34 W/m2 during daylight, so you are saying, besides, lies.

    :D

  37. Comment from: Nasif Nahle


    @ScienceofDoom…

    You say:

    Nahle wasn’t interested in actual atmospheric measurements because they didn’t fit his theory. Why waste time with real measurements?

    This is another lie and it demonstrates you know nothing of physics. The formula applied on this assessment introduces actual atmospheric measurements, that is, instantaneous measurements.

    Stop saying lies about something that you ignore.

    :D

  38. Comment from: Nasif Nahle


    @scienceofdoom…

    I considered the total path length of 7700 m.

    Again, you’re not saying truth.

    :D

  39. Comment from: Nasif Nahle


    @ScienceofDoom…

    Whatever number you give in your website is false if it doesn’t coincide with reality, I mean with the results of experimentation made by Hottel, Leckner, Lapp, Ludwig, Sarofim, etcetera, etcetera.

    Cannot you make the difference about your fantasies and the reality demonstrated by so many scientists?

    :D

  40. Comment from: Neutrino


    Nasif,

    a) Even though you have changed your equation it still doesnt have the right units. As written it has the units of cm^-1 not cm.
    Also the Thomson cross section has units of cm^2 not cm, its an area not a length.

    b) Estimating the mass of a molecule of CO2 is simple, Avagadro’s Number or the concept of a mole does not even have to be used. One molecule of CO2 has 44 nucleons, given the mass of a nucleon is 1.67*10^-24g the mass of a CO2 molecule will be approximately 7.35*10^-23g. Given the above value the number of molecules per gram would be aproxximately 1.36*10^22. This is different from your number by 100million!

    c) So what are you calculating then? Electrons that interact with an incoming photon are bound inside the molecule. Using the cross section for a free unbound electron to represent the cross section of a CO2 molecule is wrong.

    d) A pity? Pity that I do not have access to a text you reference? Or pity that the equation you used is not referenced in the one I do?
    Where did you get that formula from? If its in the first reference could you please quote from it, or if you believe it is in the second tell me where. The equation as written makes no logical sense to me, what I am trying to do is understand what you believe it represents.

    n is what? the mass of a single CO2 molecule? the mass of 1 cm^3 volume of CO2? the mass of all CO2 in the atmosphere?
    m is what? does it have units of g^-1 or g? flipping that unit changes the meaning of the quantity.
    σ is what? the cross section of a free electron or the cross section of a molecule?
    If l is a length shouldnt the RHS of the equation also be a length?

    I am trying to understand what you have done, actual answers would help that.

  41. Comment from: Nasif Nahle


    @Neutrino…

    A pity that you only resort to the internet and don’t have good sources of scientific knowledge… Books. :)

    a) The units are as follows and I didn’t change anything, just eliminated the term “molecules”:

    l = g molecule / (molecule g * cm)

    Another pity is that you don’t know what the terms are and the formula.

    Besides, you are not eliminating units in a correct way. Do you express length in X/cm? :)

    b) Molecules of CO2 per gram (n) = (1.6 x 10^-8 mol * 6.02 × 10^23 molecules) / 44.01 (g/mol) = 2.2 x 10^14 molecules per g. You are deducing a number by a simple rule of three. It’s an incorrect methodology. Period.

    c) You have not answered my question.

    n is for the number of molecules of the substance per gram. Period.

    m is the mass of a molecule of CO = 26.894 x 10^-7 g

    σ is the cross section of a molecule of CO2 is 7.143^-25 cm and to obtain cm instead cm^2 you divide the square root of the cross section of the molecule by a correction factor given by the c, le and v. The method is explained in both references.

    Regarding the origin of the formulas, here they go again:

    Lang, Kenneth. 2006. Astrophysical Formulae. Springer-Verlag Berlin Heidelberg. Vol. 1. Sections 1.11 and 1.12.

    Maoz, Dan. Astrophysics in a Nutshell. 2007. Princeton University Press, Princeton, NJ. Pp. 36-41

    :)

  42. Comment from: Neutrino


    Nasif,

    I am just going to focus on one point putting aside all the other issues until this one can be resolved, your comment:
    m is the mass of a molecule of CO = 26.894 x 10^-7 g
    (assuming that it should read “CO2 = 6.894″ and above is a simple typo)

    Numbers from you:
    m(co2) = 6.894*10^-7g
    ρ(co2) = 6.894*10^-7g/cm^3
    N = 6.02*10^23mol^-1
    molar mass = 44.01g/mol

    Numbers from me:
    m(proton) =1.67*10^-24g

    Are you actually trying to say that 1 molecule of CO2 has a mass that is 100,000,000,000,000,000(10^17) times of a proton?

    Are you actually saying that there is only 1 molecule of CO2 in a cm^3 volume?

    Those are both direct results of you asserting that a single molecule of CO2 has a mass of 6.894*10^-7g. Both are also absolutely absurd.

    BTW. Using Avagadro’s Constant and the molar mass(both of which you supplied) its simple to calculate the mass of CO2 molecule.
    molar mass / Avagadro’s Constant = mass of 1 unit
    44.01g/mol / 6.02*10^23mol^-1= 7.31*10^-23g
    (which is pretty damn close to my approximation earlier of 7.35*10^-23g)

  43. Comment from: Nasif Nahle


    @Neutrino…

    I can see your highly confused. Here again the calculation of Molecules of CO2 per gram (n):

    Molecules of CO2 per gram (n) = (1.6 x 10^-8 mol * 6.02 × 10^23 molecules) / 44.01 (g/mol) = 2.2 x 10^14 molecules per g

    Am I saying that there is only 1 molecule of carbon dioxide per gram of air? The number 2.2 x 10e+14 is 220000000000000 molecules per gram.

    You say:

    Those are both direct results of you asserting that a single molecule of CO2 has a mass of 6.894*10^-7 g. Both are also absolutely absurd.

    No, it is not absurd. It is you who is deeply confused and your argument is absurd. :)

    First of all, you MUST obtain the density of carbon dioxide in the atmosphere. It is 6.894 x 10^-7 g/cm^3. REMEMBER that the following calculations will be made in grams and centimeters.

    Second, you MUST calculate the number of moles of that sample of carbon dioxide:

    Number of moles of CO2 = (mass) / (molar mass) = 6.894 x 10^-7 (g/cm^3 * cm^3) / 44.01 (g/mol) = 1.6 x 10^-8 mol

    In the third place you MUST to calculate the mass of one molecule of carbon dioxide taking into account the number of moles of CO2 in the sample and the molar mass of the carbon dioxide:

    Mass of CO2 = (# of moles) * (molar mass) = 1.6 x 10^-8 mol * (44.01 g/mol) = 6.894 x 10^-7 g

    Then you get the number of molecules of carbon dioxide per gram of the sample:

    Number of molecules of carbon dioxide per gram = moles * Avogadro’s number / molar mass = (1.6 x 10^-8 mol * 6.02 × 10^23 molecules/g) / 44.01 g/mol = 2.143 x 10^14 molecules * g

    GOT IT?

    Don’t you know what is the molar mass of the carbon dioxide? It’s not the mass of one proton:

    http://www.webqc.org/molecular-weight-of-CO2.html

    You can use the calculator on that page. And remember, one molecule of carbon dioxide is not a single proton.

    If it is not enough, read the following book:

    Boyer, Rodney. Concepts in Biochemistry. 1999. Brooks/Cole Publishing Company: Thomson Corporation; Stamford, CT.

    Now, this game of you is absurd. Please, take a good book on analytical chemistry, read it and corroborate each one of my formulas. Stop guessing.

    And remember, this is not a classroom and I’m not obtaining a single buck on teaching you basic chemistry.

    :D

  44. Comment from: Neutrino


    Nasif,

    Your entire post is a misdirection, well most of it is. My last post dealt solely with what the mass of 1 molecule of CO2 is. That was m from your equation. Any discussion of n doesnt have a bearing on the mass of 1 molecule of CO2 and is a distraction.

    All that is needed to calculate m is the molar mass(44.01g/mol) and Avagadro’s Constant(6.02*10^23mol^-1) both of which you provided.

    I never implied the molar mass of CO2 is equal to the mass of one proton, infact I used the molar mass that you provided. The point of comparing the mass of CO2 to that of a proton(or neutron) is that it shows your value for CO2 is not even in the balll park of possible values. A molecule of CO2 contains 22protons and 22neutrons yet you maintain the mass of 1 molecule of CO2 is 10^17 times the mass of the individual compents. Hint: it should be about 44 times the mass of the individual components. Yours is an unreasonable claim.

    Additionally the point of comparing the mass of 1 CO2 molecule to the density you provided was to show that given your mass and density you are claiming that a cube 1cm on a side contains just 1 lonely molecule. Your mass is simply the density multipled by volume of 1cm^3. Again this is an absurd proposition.

    The only two things anyone needs to find the mass of any molecule/atom are the molar mass and Avagadro’s constant, since we both agree on the value of those two things(heck im even using the values you provided) how we also dont agree on the mass of CO2 is baffling.

    Do you agree that the unit mass of a molecule is given by:
    molar mass / Avagrado’s constant = unit mass

    (Hint: try it for something other than CO2, say a Hydrogen atom.
    1.01g/mol / 6.02*10^23mol^-1 = 1.67*10-24g
    Not suprisingly thats the mass of 1 proton and exactly what we would expect the mass of a hydrogen atom to be just as its not suprising that CO2 is approximately 44 times the mass of a proton)

  45. Comment from: cohenite


    Hi SoD; I’ll repeat your whole comment:

    “Comment from: Science of Doom April 3rd, 2011 at 9:05 pm

    cohenite on March 30th, 2011 at 10:43 am

    The most important point to understand is that differential absorption (dI/ds) is proportional to the number of absorbers in the path, ds.

    This is stated as the Beer-Lambert law.

    This is why any formula which calculates absorptivity or emissivity has to take account of the total number of absorbers.

    And, therefore, if the number of absorbers is expressed via density or number of molecules/unit volume you have to take account of the total path length. If you assume a 1m path length has the same number of molecules as a 1000m path length you will be making a big mistake.

    It’s so fundamental that when someone claims it isn’t important you just have to laugh. No point arguing with them. Or when they can’t see that bar.cm on a graph means you have to include path length.”

    The point I was trying to understand was, when calculating the absorptivity of the CO2 concentration in the atmosphere, is it reasonable to extrapolate from the first 1 meter to the whole path length over the atmospheric column, or should every meter be seperately calculated.

    As you know, Geiger shows that maximum backradiation in the CO2 wavelength comes from under 100 meters above the surface;

    http://scienceofdoom.com/2010/08/15/height-of-emission-of-olr-and-dlr/

    This would tend to lend weight to the need to calculate CO2 absorption and emission at every level because clearly CO2 emissivity is less as you go higher; I can’t remember whether Nasif advocates that or not but Miskolczi has done that:

    http://www.friendsofscience.org/assets/documents/E&E_21_4_2010_08-miskolczi.pdf

    I am also interested in your reference to Beer-Lambert [BL] because Miskolczi’s calculations of the optical density [OD] are intrinsically connected to BL; if BL is the integrated form of dI/dx = -Iadx, I = Ioe-ax, where a is the linear absorption coefficient, then OD is the product of the density and the distance, d = ρx, expressed in g/cm2;where, if a’ = a/ρ, I = Ioe-a’d. The advantage is that a’ is not a function of the density and distribution of the absorber, only on the number of absorber molecules present. In other words what Miskolczi calculated in his line by line analysis of the atmospheric spectra is more accurate than attempting to average over the whole column.

    Miskolczi has of course found that OD has not moved, so even though CO2 concnetration has increased there must be a counterbalancing effect, probably from water vapour reduction, again probably at the crucial higher levels.

    I would really like you, Nasif and Grant Petty to comment on that and the new guy, Neutrino, if he can stop showing off the advantages a good education long enough.

  46. Comment from: Neutrino


    Nasif,

    I must apoligize, I really should have put more thought into what you have actually written.

    From your “Mean Free Path…”

    Mass of CO2 (m) = 6.894 x 10^-7 g

    Molecules of CO2 per gram (n) = (1.6 x 10^-8 mol * 6.02 × 10^23 molecules) / 44.01 (g/mol) = 2.2 x 10^14 molecules per g

    Both of the above are really just expressing a mass in a different way.
    The first is straight forward, 1 molecule = 6.894*10^-7g.
    The second is restating the mass but in another way, 2.2*10^14 molecules = 1g.

    And here is the problem(well not the problem but one of many):
    Those two numbers do not agree with each other!

    1) 1 molecule = 6.894*10^-7g
    2) 2.2*10^14 molecules = 1g

    Solve 2) for 1 molecule
    3) 1 molecule = 1g / 2.2*10^14
    4) 1 molecule = 4.55*10-15g

    Combine 1) & 4)
    5) 1 molecule = 1 molecule (stating the obvious)
    6) 4.55*10-15g ≠ 6.894*10^-7g (or not)

    The calculations that you are doing generate contradictory results, your maths are meaningless.

  47. Comment from: Nasif Nahle


    @Neutrino…

    Contradictory to those that don’t understand basic chemistry. The procedure is correct; the only problem is that you don’t undertand the procedure.

    There are many books on chemistry from which you can learn the correct procedure and dissipate your enormous MEANINGLESS confusion.

    :D

  48. Comment from: Nasif Nahle


    @Neutrino…

    Oh! I forgot to say:

    The rule of three that you use to “calculate” molar mass, mole, molecules per gram, mass of one molecule, etc. doesn’t work in chemistry and are ridiculous.

    There are well defined formulas to get reliable and correct results; I applied those formulas in the article that was published in other website, not here, and was peer reviewed.

    XD

  49. Comment from: Nasif Nahle


    @Cohenite…

    I’m still waiting their educated answers.

    NSN

  50. Comment from: Nasif Nahle


    @Cohenite…

    The problem with Neutrino is that those very clear formulas surpass his rule of three. :)

    All the best,

    NSN

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