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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: Science of Doom


    cohenite on April 4th, 2011 at 12:48 pm:

    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.

    If you want to calculate the absorptivity of CO2 in a path through the whole atmosphere you need to know the total amount of CO2 through that path.

    This is easily done with a well-mixed gas like CO2 because the pressure and therefore density of the atmosphere vs height is easily calculated. (Or you can look it up, for example, for the “US standard atmosphere”).

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

    ..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.

    Of course I agree. In fact, calculating the total absorptivity through the whole atmosphere is interesting but not quite the best approach. See note at end.

    .. 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..

    The calculations you reproduce of optical thickness and of the integrated form of the Beer-Lambert law are correct. And as you correctly state, you just need to know the total number of absorbers in the path (plus the capture cross-section) – which can be derived in a variety of ways to reach the same result.

    I’m sure that Miskolczi has done this because it is atmospheric physics 101.

    Note on calculating Absorptivity:

    To calculate radiative transfer through the atmosphere requires the absorption and emission at each height in the atmosphere (in practice this is done numerically using a number of layers in the atmosphere).

    That is, if you calculate absorptivity for the whole atmosphere, then separately calculate emission for the whole atmosphere you won’t get the correct result for the emission of radiation from the top of atmosphere.

    The equations are known as Schwarzchild’s equations.

    Subrahmanyan Chandrasekhar, who won the Nobel Prize for Physics in 1983, wrote what seems to be the original work on the subject in 1950 – “Radiative Transfer”. You can find it at Google Books.

    This is the standard approach for solving this problem, you can find it, for example, in Atmospheric Radiation: Theoretical Basis by R.M.Goody (1964) and in most textbooks on atmospheric physics. (You won’t find a different approach).

  2. Comment from: Nasif Nahle


    @ScienceofDoom…

    You say:

    To calculate radiative transfer through the atmosphere requires the absorption and emission at each height in the atmosphere (in practice this is done numerically using a number of layers in the atmosphere).

    I did it and the total emissivity of the carbon dioxide decreases very little every meter up to 7000 meters of altitude.

    Sorry, but if you don’t make your calculations, you will never be the light at the end of the tunnel.

    :D

  3. Comment from: Neutrino


    Nasif,

    Maybe you missed the question so I will repeat it.

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

    This really should not be a difficult question, yes or no?

  4. Comment from: Nasif Nahle


    @Neutrino…

    Your question is simple and unnecessary. The formula is l = m / (n * σ), and

    n = ρCO2molecule * Number of molecules of CO2 * 1.96 (FCO2)

    Your question is meaningless in the context of the calculation.

    :D

  5. Comment from: Neutrino


    Nasif,

    Yes I agree it is a very simple question, yes or no?

    Since the m in your equation is the mass of a CO2 molecule it is neither unnecessary nor meaningless.

    Just to connect this all back to original post:

    You use the transit time to calculate total emissivity.
    You use the mean free path to calculate the transit time.
    You use the mass to calculate the mean free path.

    If you can not defend your value for the mass then the rest of the calculations are useless as a result.

  6. Comment from: Nasif Nahle


    @Neutrino…

    Just answer your own question by determining the units of your result. You’ll see that your methodology is not correct.

    :)

  7. Comment from: Nasif Nahle


    @Neutrino…

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

    :D

  8. Comment from: Nasif Nahle


    @Neutrino…

    You won’t solve your doubt if you don’t solve the following algorithm:

    n = ρCO2molecule * Number of molecules of CO2 * 1.96 (FCO2)

    And the mass of CO2 in one cm^3 of air is 6.894 x 10^-7 g

    Could you wait few days ahead to read my article on this issue?

    :)

  9. Comment from: Neutrino


    Nasif,

    Are we even talking about the same thing???

    molar mass has units of g/mol
    Avagrado’s constant has units mol^-1

    Do you dispute the above?

    So… molar mass / Avagadro’s constant has units of:
    g/mol / mol^-1 = g*mol/mol = g

    Which is a mass and exactly what I thought we were looking for. I am at a loss to understand your logic.

    Then you repost these two confused statements:
    Mass of CO2 in the atmosphere (m) = 6.894 x 10^-7 g
    And the mass of CO2 in one cm^3 of air is 6.894 x 10^-7 g

    So, if m is your mass of 1 molecule of CO2 then you are asserting that there is only 1 molecule per cm^3!!!

    The above is an absolutely absurd statement but it follows directly from your two comments.

  10. Comment from: Nasif Nahle


    @Neutrino…

    Try with those units. Make the division:

    (g/mol) / (molecules/mol)

    Got it?

    Sometimes I am in a hurry and do not complete some sentences. m is the mass of CO2 in one cm^3.

    To calculate n you need the mass of the carbon dioxide in one cm^3.

    Please, wait for my next article is published and you will get up. I hope Dr. Jennifer finds it fits as to be published on her blog.

    :)

  11. Comment from: cohenite


    Neutrino, since you are obviously a guy who means business, what do you calculate as the emissivity of CO2?

  12. Comment from: Nasif Nahle


    @Cohenite…

    He or she cannot understand that the formula does not require the mass of one molecule, but the value of n for molecular carbon dioxide, which under vibrational dephasing, rotational relaxation, etc. is known as n = ρCO2molecule * Number of molecules per gram * 1.96 * (FCO2).

    Excelent question. I reproduce it here. Cohenite asks you:

    Neutrino, since you are obviously a guy who means business, what do you calculate as the emissivity of CO2?

    NSN

  13. Comment from: Neutrino


    Try with those units. Make the division:
    (g/mol) / (molecules/mol)
    Got it?

    Yes I think I got it, you do not demonstrate the ability to do simple arithmetic. The above is still g/molecule.
    Hint: a / b = a * (1 / b)

  14. Comment from: Neutrino


    Comment from: Nasif Nahle April 4th, 2011 at 5:46 am
    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

    Comment from: Nasif Nahle April 5th, 2011 at 3:34 pm
    m is the mass of CO2 in one cm^3.

    Comment from: Nasif Nahle April 5th, 2011 at 5:56 pm
    n = ρCO2molecule * Number of molecules per gram * 1.96 * (FCO2).

    I am trying to understand Nasif’s formula for calculating the mean free path of a photon. In that formula he has both m and n but seems to freely change what those two variables represent.

  15. Comment from: Neutrino


    cohenite,

    I have no idea what the total emissivity of CO2 is in the atmosphere. I have no qualms admitting that. Based on what Nasif has been writing he also has no idea what the total emissivity is.

    But as such it is a poorly formed question, terrestrial or longwave radiation ranges from 4μm to 40μm(roughly) so the only range of interest is that. It does not matter what the emissivity of CO2 is in any wavelength outside that range because the emission will be zero regardless of the emissivity.

    To actually find the emissivity within that range I agree with SoD’s earlier comment that an integration across all the pertinent wavelengths would have to be done.(ie: emissivity is wavelength dependant) I do not have the ability to do that so I would have to defer to someone else in this.

    What I can do though is read articles and papers put forward by people attempting to do that and make observations and critiques of them. Nasif’s articles(“Total Emissivity of the Earth…” and “Mean Free Path of Photons…” which his ultimate calculation of emissivity relies upon) are simply wrong. He does not consider wavelength and treats the cross section of a molecule as equal to that of a free electron to undergo scattering(the question is about absorption and emission not scattering). These two points would seem to invalidate any conclusion he would have about emissivity. Further confusion comes from his continually vague and shifting definitions of the terms of base equations. This renders any final conclusion of his equally vague and non descriptive.

    All I have tried to do here is show the underpinnings of Nasif’s claims are baseless. I don’t think it is very practical to talk about high level concepts when the underlying foundation is so corrupt.

    I ask Nasif one more time, and you could answer as well, what is the mass of one molecule of CO2? Since it is part of his equation to calculate mean free path which leads to transit time which leads to total emissivity it matters that he uses the correct value.

  16. Comment from: Nasif Nahle


    @Neutrino…

    Please, do not extend your ignorance over me. If you ignore something doesn’t mean that the whole world ignores it.

    I have calculated the total emissivity/absorptivity of carbon dioxide and other substances through two methodologies and I have found the same result.

    If you ignore what the total emissivity is, go to Hottel, Leckner, Lapp, Ludwig, Sarofim, etc., experiments and tables.

    The fact that something does not fit your imagination does not mean that the results of observations and experimentation are wrong.

    Sorry, but your attitude is not professional and I see you are learning too much from me.

    There are formulas that you can read from any book. Do not say that the investigators are wrong, or that my claims are baseless just because you don’t like the reality.

    Read the authors I have mentioned. All of them are wrong?

    Hah!

  17. Comment from: Nasif Nahle


    I’m writing an article on how to calculate the total emissivity as from the mean free path length.

    My formulas, variables, etc., are found on books and articles of astrophysics and chemistry. I provided the references, go over them and see that my calculations are completely adhered to the scientific methodology.

    If you think I am wrong, demonstrate it with your numbers. If you confess that you don’t know what the total emissivity is, ALL HAS BEEN SAID BY YOUR OWN MOUTH.

    :D

  18. Comment from: Nasif Nahle


    @Neutrino…

    First of all, by your own words, you don’t know that the total emissivity is, therefore your following statement is wrong:

    I ask Nasif one more time, and you could answer as well, what is the mass of one molecule of CO2? Since it is part of his equation to calculate mean free path which leads to transit time which leads to total emissivity it matters that he uses the correct value.

    No, I don’t need the mass of one molecule of CO2, but its mass in the atmosphere. Your deductions are absolutely wrong. You, the same as SoD and others only talk, but do not demonstrate that the books on astrophysics and radiative heat transfer are wrong.

    :D

  19. Comment from: Neutrino


    Nasif,

    I will try this one more time.
    I am critiquing how you calculate emissivity via the transit time of a photon. The transit time is conditional on the mean free path, therefore if the mean free path calculation is wrong then so is the transit time and therefore the resultant emissivity as well.

    The fact that I do not know the total emissivity has no bearing on whether or not my critique of your calculation for the mean free path is valid. And no I do not think texts on astrophysics and radiative heat transfer are wrong. In fact the one that you referenced, Dan Maoz: Astrophysics in a Nutshell, I believe is clear and concise on the topic. The problem is you are taking the equations out of context and twisting them.

    I believe you are trying to use eq. 3.37 and eq. 3.40 from the above text to calculate mean free path.
    eq. 3.37 l = 1/(n*σ)
    eq. 3.40 n ≈ ρ/m(h)
    where:
    l = mean free path
    n = particle density
    σ = particle cross section
    ρ = density
    m(h) = mass of hydrogen atom

    The text is calculating the mean free path before a photon interacts with an electron inside of a star. As an approximation of the particle density of free electrons the text uses a proxy of the number of hydrogen atoms.

    Rewriting eq. 3.40 for CO2 makes it:
    1) n = ρ(co2)/m(co2)
    Combining that with 3.37 yields:
    2) l = m(co2)/(ρ(co2)*σ)
    Your equation is:
    3) l = m/(n*σ)
    where:
    m = mass of CO2
    n = number of molecules CO2 per gram
    σ = Thomson electron cross section

    As written your 3) does not equal the text 2) which you cited (for starters n ≠ ρ(co2)). Also as I said before the units of your 3) do not produce a length. Several times you have changed your definition of terms so it’s a little hard to actually pin you down(ie: is m the mass of one molecule or the mass of one cm^3 volume?).

    To prevent confusion and foster actual dialogue could you write out your equation and what the terms are as I did above.

  20. Comment from: cohenite


    Hi Neutrino; “To actually find the emissivity within that range I agree with SoD’s earlier comment that an integration across all the pertinent wavelengths would have to be done.(ie: emissivity is wavelength dependant) I do not have the ability to do that so I would have to defer to someone else in this.”

    Defer away.

    http://miskolczi.webs.com/Context-and-background.htm

  21. Comment from: Neutrino


    Cohenite,

    Perhaps my comment was not quite precise. I do not have the capability to do the full blown analysis so I must defer to someone else to do that task. Not necessarily defer to their result.

    As for Miskolczi’s paper, I would have to read the full piece before I could comment on it. But with a cursory look at the abstract and your link it does not appear to provide a number for the total emissivity of the CO2 component of the atmosphere.

    Regardless of whether or not Miskolczi’s paper is solid does not change the point I have been trying to make in this thread:

    The calculations Nasif performs have no basis in fact.

  22. Comment from: cohenite


    Neutrino, I must say it is surprising that someone with your background in atmospheric physics has not read Miskolczi. The thing about Miskolczi is that you must defer to his “result” because, as Nindetharma notes, he has shown from OBSERVATION that the mean infra red opacity of the atmosphere has not changed significantly over the last 60 years or so. Given, that the infra-red opacity of the atmosphere must have increased because of additional CO2, the only logical conclusion that can be drawn from his OBSERVATIONS is that some other green-house must be compensating for the increased opacity due to CO2. It doesn’t take a rocket scientist to realize that the “other green-house gas” is water vapor.

    In fact, if you actually do a spectral analysis of the infra-red opacity of the atmosphere
    over the last 60 years, you find that the spectral signature is exactly the same as the spectral signature of water vapor changes in the atmosphere that are induced by the El Nino/La Nina/ENSO climate phenomena.

    The increase in infra-red opacity of the atmosphere due to CO2 is roughly being negated by a decrease in infra-red opacity due to the decrease in the specific humidity (water vapor mass per Kg of air) of the upper troposphere in the tropics.

    In this respect I suppose this makes the emissivity of CO2 moot but certainly not the Earth’s emissivity which, on the basis of Miskolczi’s observations, have not changed.

    In respect of Nasif, he is correct about the relative importance of the radiative properties of water vapor and CO2, with CO2 much less important, and therefore having lessor emissivity, than water. This is confirmed by the fact that the decline in the specific humidity (water vapor mass per Kg of air) of the upper troposphere in the tropics is much less than the increase in CO2 concentrations.

    I guess the issue of CO2 emissivity is a peripheral issue to Miskolczi’s OBSERVATIONS [Nasif may disagree] but still a crucial one and I look forward to the details of that emissivity being resolved.

  23. Comment from: gavin


    comment deleted. Jennifer

  24. Comment from: cohenite


    Well, go ahead gav, refute Miskolczi’s OBSERVATIONS about warming NOT being due to an increase in the GHG effect; perhaps a pithy homily, preferably involving pretty foreign backpackers, will do the trick.

  25. Comment from: Nasif Nahle


    @Neutrino…

    That’s why I was telling you that your calculations were wrong, considering the algorithm, but you never read my posts.

    Please, wait until the article is published.

    :)

  26. Comment from: Nasif Nahle


    @gavin…

    LOL!

    I aaaam heeeerrre… I doooo eeeexxiiiisssst…. LOL!

    Gavin, why you don’t read scientific books and the articles published by those investigators? You’ll be surprised to have been living in the fantasy world of AGW.

    :D

  27. Comment from: Nasif Nahle


    @Gavin…

    You say that Neutrino “nailed” my calculations on total emissivity of the carbon dioxide.

    WRONG! Neutrino has not shown a single number which demonstrates that the total emissivity of the carbon dioxide is different or higher than 0.002

    BESIDES… He/she spent the time on this blog on trying to demonstrate the mass of a molecule of carbon dioxide, when everybody knows it is 44.01 g.

    ADDITIONALLY… Neutrino told us that he/she ignores what the total emissivity is; consequently, he/she CANNOT understand the calculations and the remainder of the physics contained in my article

    :D

  28. Comment from: Nasif Nahle


    @Gavin…

    And… if you have in doubts, read what the molecular mass of the carbon dioxide and each one of the components of the aire are from here:

    http://www.engineeringtoolbox.com/molecular-mass-air-d_679.html

    Thanks! LOL!

    :D

  29. Comment from: cohenite


    Nasif; it is good that you [singular] are still in good spirits; I don’t know where gav gets these ideas from; it is true that the mass of the CO2 molecule is common knowledge but I’ll let you argue the toss with Neutrino and I’ll try and follow as best I can.

    I presume your next article is about saturation, line and pressure broadening etc; I know that is not directly connected with Miskolczi but what is your opinion of his observations and results?

  30. Comment from: Neutrino


    Wow,

    Yes, I have not shown a single number that would put total emissivity below or above your value of 0.002.
    But I have shown that your calculation is wrong, and I do not need to propose an alternate value of total emissivity to do that.

    Yes, I have spent a great deal of time on this blog in regards to the mass of one CO2 molecule. Sadly it has been to little success. Everybody most certainly does not know that one molecule has a mass of 44.01g. What everyone, including myself, who has even a cursory understanding of molar mass and the periodic table knows is that the molar mass of CO2 is 44.01g/mol. That together with the knowledge of what a mole is gets you the mass molecule.

    Then you misstate my position. I never said I ignore the total emissivity, I stated I did not know the value. Besides that, not knowing the ultimate value of a thing does not preclude me, or anyone else, in being able to critique the calculations you try to do to get to it.

    And for the record, I am a he.

  31. Comment from: Nasif Nahle


    @Neutrino…

    But I have shown that your calculation is wrong, and I do not need to propose an alternate value of total emissivity to do that.

    In your dreams.

    :D

  32. Comment from: Nasif Nahle


    @Neutrino…

    In my next article I will talk about the overlapping absorption spectral bands, and immediately after, I will send another article on the determination of the total emissivity considering the mean free path length and crossing time lapse of quantum waves.

    Be patient.

    :D

  33. Comment from: Martin Mason


    Can anybody help me with a question on radiation? I instinctively believe that a cold body can’t transfer heat to a warmer body but it can radiate towads the warmer body. If the radiated wave back from GHGs in the atmosphere can’t be readsorbed and re-emitted by the surface, what does it do?

  34. Comment from: Environmental Geography


    [...] Source of the above quote is: < http://jennifermarohasy.com/2011/03/total-emissivity-of-the-earth-and-atmospheric-carbon-dioxide/ [...]

  35. Comment from: Project Proposal [Final] « Environmental Geography


    [...] of the Earth and Atmospheric Carbon Dioxide: A Note from Nasif S. Nahle. Nasif S. Nahle. <http://jennifermarohasy.com/2011/03/total-emissivity-of-the-earth-and-atmospheric-carbon-dioxide/>  [September 25, [...]

  36. Comment from: Doug Cotton


    Comment from: Martin MasonApril 11th, 2011 at 1:31 pm

    Can anybody help me with a question on radiation? I instinctively believe that a cold body can’t transfer heat to a warmer body but it can radiate towards the warmer body. If the radiated wave back from GHGs in the atmosphere can’t be reabsorbed and re-emitted by the surface, what does it do?

    It resonates with the target molecule and is effectively re-emitted rather like being reflected at the speed of light. None of its energy is converted to thermal energy. I prefer to use the term “scattered” in order to avoid implying that it is either reflected (in the true sense of the word) or absorbed – which most people assume means it does some warming.

    Now, when and why does it resonate? Well, the frequency distribution of a blackbody has a peak which is proportional to absolute temperature. Study carefully the first plot here http://scienceworld.wolfram.com/physics/WiensDisplacementLaw.html and note that the plot for a warmer temperature always envelopes that for a cooler temperature. Hence radiation from a cooler source can only have frequencies which can resonate with those of a warmer body. So all such radiation never leaves thermal energy behind. In contrast, radiation from a warmer source will always have some frequencies (at the right) which cannot resonate with a cooler target. It is the energy in radiation with these frequencies which has to be retained and is thus converted to thermal energy. This is actually necessary for the Second Law of Thermodynamics to apply.

    Hence spontaneous radiation from a cooler atmosphere cannot add thermal energy to a warmer surface. Since it cannot add thermal energy it cannot either increase the rate of warming of the surface in the morning or slow the rate of cooling on the evening.

    Herein lies the collapse of the atmospheric radiative greenhouse conjecture.

  37. Comment from: B. Meth


    “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.”

    You seem to be missing the point that it’s not the partial pressure p that’s important, it’s p*L.
    To quote from literature: “In a Hottel’s chart, the total emissivity of a gas, averaged over all radiation wavelengths, is plotted against temperature for several values of the partial-pressure-beam-length product (PL).”
    Clearly, the value you mention “0.006096 atm-m” is not p but p*L, as indicated by the units used. The highest value in the chart I have is 155kPa.m, which corresponds to a partial pressure of 0.00038 atm and L= 4km, good enough for a rough estimation and corresponding to an emmisivity of about 0.2 within a wide temperature range.

    Regarding the use of astrophysical formulae, I fail to see how you can justify them in this context: CO2 molecules don’t re-emmit the absorbed heat immediately, and you don’t take into account conductive heat transfer to other air molecules. The idea that radiant heat from the surface would exit the atmosphere with a time constant of 4.2 milliseconds seems pretty wild.

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