Atmospheric CO2 radiation

Discussion in 'Environment' started by Wuwei, Sep 15, 2019.

  1. Wuwei
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    Wuwei Gold Member

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    Some people, including the infamous Dr. Happer say that CO2 cannot radiate due to a plethora of molecular collisions that abort almost all of the ability to radiate15 micron photons. Why are they wrong?

    The short answer:
    A molecule vibrating in the15 micron radiation mode can hold that energy for 0.335 seconds.
    The number of molecular collisions during that time is over a billion. However the vast majority of those collisions will simply bounce off without changing the molecular vibration state. Because of that, almost all CO2 molecules will keep their energy until they emit 15 micron radiation into the atmosphere.

    The long answer:

    These are the details of the short answer:

    https://apps.dtic.mil/dtic/tr/fulltext/u2/771554.pdf from Table V
    The average lifetime of a CO2 molecule in a bending vibration mode (15 micron radiation wavelength)
    is 0.335 seconds.

    Frequency of Molecular Collisions
    The average atmospheric molecular collision frequency is 3.25×10⁹ collisions/second.

    If just one of those billions of collisions were able to quench the CO2 vibrations, the probability of a CO2 radiating would be roughly 1 / 3.25×10⁹x0.335 which is 1 in 1.09 billion.

    At first sight it would seem that the collision rate is so high that, for practical purposes all radiation vanishes. However inelastic collisions that would quench a CO2 vibration state are rare. Elastic collisions simply bounce off without changing any internal energy and are predominant.

    https://pure.tue.nl/ws/files/3478579/109243.pdfOn Vibrational Relaxations in Carbon Dioxide”
    TABLE III Translational Transition Probabilities Page 33.
    This table gives the probabilities that a collision will quench a CO2 molecule in a 15 micron bending mode. The second column shows the probability is Pd² = 2.0 10⁻¹³ at 300 K. An extrapolation to the global average temperature of 288 K temperature gives a quenching collision probability = 1.7×10⁻¹³. That probability is very small.

    Number of CO2 molecules per m³ @ 400ppm = 1.01×10²².
    The number of atmospheric CO2 molecules in a 15 micron excitation state is
    1.01×10²² x 2/9 = 0.244×10²² (2/9 comes from Equipartition Principle)

    The probability that a single collision will quench that excited state = 1.7×10⁻¹³
    The probability that a single collision does not quench the excited state = (1- 1.7×10⁻¹³)
    The probability that none of the 1.09 billion collisions quenches the excited state =
    (1 − 1.7×10⁻¹³)^1.09×10⁶ ≅ 1 − 1.7×10⁻¹³ × 1.09×10⁶ = 1 − 1.853×10⁻⁷ = 0.999999815
    The approximation after the first term comes from eliminating very small polynomial higher order terms.

    The number of excited CO2 not quenched by collision is 0.244×10²²x0.999999815
    That is almost a certainty that all CO2 molecules in a “bending” vibration state will radiate 15 microns.

    The radiation density is the energy of a 15 micron photon times the number of radiating CO2 molecules. The energy of a 15 micron photon = 1.3 10⁻²⁰ J
    The total radiation energy of all 15 micron photons in a cubic meter
    = 0.244×10²² x 1.3 10⁻²⁰ Joules = 0.317 x 10² J
    The average radiation energy per cubic meter per second is
    31.7 J / 0.335 s = 95 Watts.

    Conclusion:
    The resulting radiation density at atmospheric pressure and 15℃ is around 95 Watts radiating isotropically within a cubic meter at 15℃ . Radiation is a far faster mode of energy transfer than conduction. The role of GHG radiation in the atmosphere energy transfer should not be underestimated.
     
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  2. mamooth
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    mamooth Gold Member

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    Ruh-roh. You just debunked one of SSDD's favorite pseudoscience rants.

    Not that it will have any effect on him.
     
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  3. Billy_Bob
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    Billy_Bob Platinum Member

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    BWHAAAAaaaaaaaa..

    Oh my gawd.... Not this crap again..

    Looks like Bill needs to whip out his crayons and deal with the BS... Lets tear this fallacy apart brick by brick...
     
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  4. Billy_Bob
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    Billy_Bob Platinum Member

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    HE debunked NOTHING.. He made an assumption that is not based in observed fact.
     
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  5. mamooth
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    mamooth Gold Member

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    It's not debatable that molecules almost never lose vibrational energy in collisions.

    Thus, your PSI cult physics is debunked.

    Can you find a single actual physicist who supports your kook fantasy physics? No, you can't. That would be because you just made everything up.
     
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  6. Billy_Bob
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    Billy_Bob Platinum Member

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    This is Item one.

    Your post tells us that the original temperature of the air mass is 288K (58.73 Deg F). This indicates that you are using DOWN-WELLING RADIATION as an input. You are using total energy input, through the air mass, of the sun to obtain your output.

    This negates any energy from the earth being shed as the AGW hypothesis dictates. So right on its face you deceive those your trying to win over. The 95W/m^2 is incorrect when you are looking at energy emitted from the black body of the earth. You improperly convert joules into watts..

    A Joule is a Watt Second (NOT Watts per second). A Joule is equal to one watt of power delivered over 1 second of time. That's the correct definition. However, this means you divide your wattage by the total number of seconds it is applied to derive Joules. If you are determining wattage output you multiply.

    This means your power output is 31.7 J *0.335 = 10.6195 Watts NOT 95 Watts..

    Your paper doesn't even know how to properly convert Joules... Its no wonder you folks think the world is going to burn up...
     
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    Last edited: Sep 15, 2019
  7. Billy_Bob
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    With that number of collisions the energy contained is removed KINETICALLY, not by emission.

    Its funny how you think that a molecule that collides some 30,000 times during the time energy can resides will some how emit rather than move it kinetically. So I take a look at this paper and its assumptions, again I find that they do not know how to properly find the proper formation of the bending cycle and where it is blunted.

    When I pull up your links I find this;

    upload_2019-9-15_20-48-41.png

    You make no attempt to even calculate the proper collision rate and bending blunt rate. You make a wild ass assumption and post it as fact.

    At this point your not worth even trying to debate this subject. You cant even get the basic math right and there is no hope of a complex problem being correct... You made assumptions on the gases pressure, temperature, and speed of motion, none of which bear any resemblance of reality.

    IF I have time tomorrow, I will do the math and show you how really wrong you are.
     
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  8. percysunshine
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    percysunshine Gold Member

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    It would probably help your thesis if the empirical data were treated with a consistent number of significant figures.
     
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  9. Billy_Bob
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    Billy_Bob Platinum Member

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    INCORRECT ASSUMPTION!

    Using the tables provided by Dr Jay Blauer you made a set of serious errors. The JPL (Air force Jet Propulsion Laboratory) indicted that they did not know CO2's bend blunting rate and were unable to determine it;

    upload_2019-9-15_21-23-24.png

    You cite this work and yet you misuse it even when they tell you they were unable to determine the blunt factor and the interaction factors.

    You are really reaching for something to say "gotcha" aren't you.
     
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  10. Billy_Bob
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    Billy_Bob Platinum Member

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    Worse still you are using rates found at the temperature of 1500K;

    upload_2019-9-15_21-30-39.png

    This is the temperature of rocket exhaust not the earths atmosphere...

    There is really no where to go with this pile of crap. All of your assumptions are done outside of reality. IF the atmosphere were 1500K it is quite possible your 1m^2 of atmosphere could be radiating at 95 Watts...
     
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