Determining the Total Emissivity of a Mixture of Gases Containing Overlapping Absorption Bands: A Note from Nasif S. Nahle

Abstract

According to anthropogenic global warming (AGW) theory, carbon dioxide increases the potential of water vapor to absorb and emit IR radiation as a consequence of the overlapping absorption/emission spectral bands. I have determined the total emissivity of a mixture of gases containing 5% of water vapor and 0.039% of carbon dioxide in all spectral bands where their absorptivities/emissivities overlap. The result of my calculations is that carbon dioxide reduces the total absorptivity/emissivity of the water vapor, working like a coolant, not a warmer of the atmosphere and the surface.

Update  April 8, 2011.   There was an error in calculating the overlapping bands, discovered thanks to criticism from ‘Neutrino’.   The errors are now shown with lines through them, the correct figures beside them.   The ‘adjusted’ calculations give a greater cooling effect from carbon dioxide . 

Introduction

Since the popularization of AGW theory in 1988, proponents have argued that carbon dioxide causes an increase in the total absorptivity of the atmosphere1, 2, 3.

For example, at Environmental Defense1 it is argued that:

As humans emit greenhouse gases like CO2, the air warms and holds more water vapor, which then traps more heat and accelerates warming.”

And at Science Daily2 that:

Climate warming causes many changes in the global carbon cycle, with the net effect generally considered to be an increase in atmospheric CO2 with increasing temperature — in other words, a positive feedback between temperature and CO2.”

Masato Sugi and Jun Yoshimura3 claim that:

By the overlap effect of CO2 and water vapor absorption bands, the existence of CO2 significantly reduces the cooling rate of water vapor…

These arguments suggest that by increasing the concentration of carbon dioxide in the atmosphere there will be warming of the atmosphere.

However, according to results from experimentation made by H. C. Hottel11, B. Leckner12, M. Lapp13, C. B. Ludwig14, A. F. Sarofim15 and their collaborators14, 15 on this matter, the combined effect of overlapping absorption bands causes a reduction on the total absorptivity of a mixture of gases4, 5, 6.

My assessment reinforces the argument made by H. C. Hottel11, B. Leckner12, M. Lapp13, C. B. Ludwig14, A. F. Sarofim15 and their collaborators14, 15 because my calculations coincide with the results obtained from the algorithms derived from their experiments.

In 1954, Hoyt C. Hottel conducted an experiment to determine the total emissivity/absorptivity of carbon dioxide and water vapor11. From his experiments, he found that the carbon dioxide has a total emissivity of almost zero below a temperature of 33 °C (306 K) in combination with a partial pressure of the carbon dioxide of 0.6096 atm cm.

Seventeen years later, B. Leckner repeated Hottel’s experiment and corrected the graphs12 plotted by Hottel. However, the results of Hottel were verified and Leckner found the same extremely insignificant emissivity of the carbon dioxide below 33 °C (306 K) of temperature and 0.6096 atm cm of partial pressure.

Hottel’s and Leckner’s graphs show a total emissivity of the carbon dioxide of zero under those conditions.

The results of Hottel and Leckner have been verified by other researchers, like Marshall Lapp13, C. B. Ludwig14, A. F. Sarofim15, who also found the same physical trend of the carbon dioxide.

On the other hand, in agreement with observations and experimentation carried out by the same investigators11, 12, 14, 15, 16, the atmospheric water vapor, in a proportion of 5% at 33 °C, has a total emissivity/absorptivity of 0.4.5, 6

The total emissivity/absorptivity of water vapor combined with its high specific heat capacity and its volumetric mass fraction makes water vapor the most efficient absorbent and emitter of Infrared Radiation among all gases forming the Earth’s atmosphere.

In contrast, the carbon dioxide has negligible total emissivities and partial pressures as a component of the atmosphere (the partial pressure of the carbon dioxide at the present atmosphere is 0.0051 atm cm). 

So what is the effect of a combination of water vapor and carbon dioxide at current conditions of partial pressure, temperature and mass densities in the atmosphere?

Methodology

The whole range of spectral absorption of both gases and an effective path length (Le) of 7000 m were considered for calculating the total emissivity of a mixture of water vapor and carbon dioxide in the atmosphere. I have made use of formulas on radiative heat transfer taken from the references numbered as 4, 5 and 6.  However, I made use of the main formula to calculate the total emissivity of a mixture of gases in the atmosphere, where their absorption bands overlap, that was derived by H. C. Hottel11, B. Leckner12, M. Lapp13, C. B. Ludwig14, A. F. Sarofim15 and their collaborators14, 15, and enhanced by contemporary authors as Michael Modest5, as from the results of observations, as from the results of experimentation.

The effective path length is the length of the radiation path through the atmosphere. It differs from the geometrical distance travelled because the radiation is scattered or absorbed on entering and leaving the atmosphere. In a vacuum there is no difference between the effective path length and the geometrical path length. As this assessment deals with the atmosphere, I considered the effective path length in my calculations.

The volumetric mass fraction of water vapor in the atmosphere fluctuates between 10000 ppmV and 50000 ppmV 10. This variability allows the water vapor to show a wide range of high total absorptivities and total emissivities which may vary according to the temperature of the molecule of water vapor. For this reason, I considered the maximum mass fraction of the water vapor in the atmosphere.

The water vapor potential to absorb shortwave infrared radiation from the solar photon stream makes of it the most efficient absorbent of Infrared Radiation. In quantum physics, a photon stream is a current of photons emitted by a source that behave as particles and waves and have a specific directionality i.e. from the source towards the surroundings.

After concluding my analysis, Dr. Charles R. Anderson called my attention to the observation that these calculations constituted further evidence for his theory about the cooling effect of carbon dioxide on the Earth’s surface. When Dr. Anderson and I further examined the calculations, we found that carbon dioxide not only has a cooling effect on the surface, but also on the molecules of other gases in the atmosphere.

The total emissivities of the atmospheric carbon dioxide, water vapor and oxygen were obtained by taking into account the mean free path length of the quantum/waves through those gases, taken individually, and the time lapse rate that a quantum/wave takes on leaving the troposphere after colliding with molecules of carbon dioxide, water vapor and oxygen. This set of calculations will be described in a future article.

Total Emissivity of a Mixture of Water Vapor and Carbon Dioxide in the Current Atmosphere of the Earth

On July 3, 2010, at 10:00 hr (UT), the proportion of water vapor in the atmosphere at the location situated at 25º 48´ N lat. and 100 º 19’ W long., at an altitude of 513 m ASL, in San Nicolas de los Garza, Nuevo Leon, Mexico, was 5%. The temperature of the air at an altitude of 1 m was 310.95 K and the temperature of the soil was 330 K. I chose this location, near my office, because it is an open field, far enough from the city and its urban effects.

From this data, I proceeded to calculate the following elements:

1.    The correction factor for the overlapping emissive bands of H2Og and CO2g.

 2.  The correction factor of the total emissivity of carbon dioxide where the radiative emission bands of both gases overlaps, considering that the partial pressure of the carbon dioxide is 0.00039 atm.

 3.  The total emissivity of the mixture of water vapor and carbon dioxide in the atmosphere.

 4.   The total normal intensity of the mixture of water vapor and carbon dioxide in the atmosphere.

 5.  The change of temperature caused by the mixture of water vapor and carbon dioxide in the atmosphere.

Obtaining the correction factor for the overlapping emissive bands of H2Og and CO2g

To obtain the total emissivity of the mixture of water vapor and carbon dioxide in the atmosphere, we need to know the equilibrium partial pressure of the mixture of water vapor and carbon dioxide. The formula for obtaining the equilibrium partial pressure (ζ) of the mixture is as follows:

ζ = pH2O / (pH2O + pCO2)         (Ref. 5)

Where pH2O is the partial pressure of water vapor in a proportion of 5% in the atmosphere –which is an instantaneous measurement of the water vapor, and pCO2 is the partial pressure of the carbon dioxide.

Known values:

pH2O = 0.05 atm

pCO2 = 0.00039 atm

Introducing magnitudes:

ζ = pH2O / (pH2O + pCO2) = 0.05 atm / (0.05 atm + 0.00039 atm) = 0.9923

Therefore, ζ = 0.9923

Obtaining the total emissivity of a mixture of water vapor and carbon dioxide in the atmosphere:

Now let us proceed to calculate the magnitude of the overlapped radiative emission bands of the water vapor and the carbon dioxide. To do this, we apply the following formula:

ΔE = [[ζ / (10.7 + 101 ζ)] – 0.0089 (ζ)^10.4] (log10 [(pH2O + pCO2) L] / (pabsL) 0)^2.76 [Ref. 5]

Known values:

ζ = 0.9923

pH2O = 0.05 atm

pCO2 = 0.00039 atm

(pabsL)0 (absolute pressure of the mixture of gases on the Earth’s surface) = 1 atm m

Le = (2.3026)) (Aas / μa) = 7000 m

Introducing magnitudes:

ΔE = [(0.992 / 110.892) – (0.0089 * (0.992)^10.4] * (log10 [(0.05 atm + 0.00039 atm) 7000 m] / (1 atm m)0)^2.76 (Ref. 2)

ΔE = [0.00076] * (2.55 atm m / 1 atm m) = 0.0019; rounding up the cipher, ΔE = 0.002

Therefore, the correction addend for the overlapping absorption bands is 0.002

Consequently, the total emissivity of the mixture water vapor and carbon dioxide is as follows:

E mixture = ECO2 + EH2O ΔE = 0.0017 + 0.4 – 0.002 = 0.3997

Total Normal Intensity of the energy radiated by the mixture of gases in the air:

Therefore, the total normal intensity (I) (or the spectral radiance over wavelength) caused by the mixture of water vapor and carbon dioxide in the atmosphere is:

I = Emix (σ) (T)^4 / π        (Ref. 5 and 6)

I = 0.3997 (5.6697 x 10^-8 W/m^2 K^4) (310.95)^4 / 3.1416 = 67.44 W/m^2 sr

By way of contrast, the spectral irradiance over wavelength caused by the surface (soil), with a total emissivity of 0.82 (Ref. 1 and 5), is as follows:

I = Esurface (σ) (T)^4 / π (Ref. 5 and 6)

I = 0.82 (5.6697 x 10-8 W/m^2 K^4) (330 K) / 3.1416 = 203 W/m^2 sr

Following Dr. Anderson’s recommendation (which I mentioned above in the abstract) I calculated the overlapping bands of a mixture of water vapor (4%), carbon dioxide (0.039%) and Oxygen (21%).

The calculation for a mixture of atmospheric Oxygen (O2), Water Vapor (H2O) and Carbon Dioxide (CO2) is as follows:

ζ = pO2 / (pO2 + pCO2) = 0.21 atm / (0.21 atm + 0.00039 atm) = 4.1675 0.9981

ζ = pO2+CO2 / (pHO2 + pO2+CO2) =  0.9981 4.1675 atm / ( 0.9981  4.1675 atm + 0.05 atm) = 0.9881 0.9522

Consequently, the equilibrium partial pressure of the mixture of oxygen, water vapor and carbon dioxide in the atmosphere is 0.9881 0.9522

And the change of the total emissivity of the mixture is:

ΔE = [[ζ / (10.7 + 101 ζ)] – 0.0089 (ζ)^10.4] (log10 [(pH2O + pCO2 + pO2) L] / (pabsL) 0)^2.76 [Ref. 5, 11,12,14 and 15]

ΔE = [[0.9881/ (10.7 + 101 (0.9881)^10.4)] – 0.0089 (0.9881)^10.4] (log10 [(0.26039 atm) 7000 m] / (1 atm)^2.76 = 0.00989

ΔE = [[0. 9522/ (10.7 + 101 (0.9522)^10.4)] – 0.0089 (0.9522)^10.4] (log10 [(0.26039 atm) 1 m] / (1 atm)^2.76 =  0.008 * 26.11 = 0.2086

And the total emissivity of the mixture of gases in the atmosphere is:

E mixture = ECO2 + EH2O ΔE = 0.0017 + 0.4 + 0.004 – 0.00989 0.2086 = 0.3958 0.1971; or 0.2 if we round up the number.

Evidently, the mixture of oxygen, carbon dioxide and water vapor, at current conditions of temperature and partial pressures, causes a sensible decrease of the total emissivity of the mixture of air.

The general conclusion is that by adding any gas with total emissivity/absorptivity lower than the total emissivity/absorptivity of the main absorber/emitter in the mixture of gases makes that the total emissivity/absorptivity of the mixture of gases decreases.

In consequence, the carbon dioxide and the oxygen at the overlapping absorption spectral bands act as mitigating factors of the warming of the atmosphere, not as intensifier factors of the total absorptivity/emissivity of the atmosphere.

Conclusions

My assessment demonstrates that there will be no increase in warming from an increase of absorptivity of IR by water vapor due to overlapping spectral bands with carbon dioxide. 

On the overlapping absorption spectral bands of carbon dioxide and water vapor, the carbon dioxide propitiates a decrease of the total emissivity/absorptivity of the mixture in the atmosphere, not an increase, as AGW proponents argue 1, 2, 3.

Applying the physics laws of atmospheric heat transfer, the carbon dioxide behaves as a coolant of the Earth’s surface and the Earth’s atmosphere by its effect of diminishing the total absorptivity and total emissivity of the mixture of atmospheric gases.

Dr. Anderson and I found that the coolant effect of the carbon dioxide is stronger when oxygen is included into the mixture, giving a value of ΔE = 0.3814, which is lower than the value of ΔE obtained by considering only the mixture of water vapor and carbon dioxide.

by Nasif S. Nahle, Director of Scientific Research Division at Biology Cabinet Mexico

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

Acknowledgments

I am very grateful to Dr. Charles R. Anderson, PhD, author of the Chapter 20 in the book Slaying the Sky Dragon-Death of the Greenhouse Gases Theory, especially page 313  for his valuable help on realizing the cooling role of the Oxygen in the atmosphere.

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

References

1.  http://www.edf.org/documents/5596_GlobalWarmingWaterVapor_onepager.pdf

2.  http://www.sciencedaily.com/releases/2010/01/100127134721.htm  

3.  http://journals.ametsoc.org/doi/pdf/10.1175/1520-0442(2004)017%3C0238%3AAMOTPC%3E2.0.CO%3B2

4.  Manrique, J. A. V. Transferencia de Calor. 2002. Oxford University Press. England.

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

6.  Pitts, Donald and Sissom, Leighton. Heat Transfer. 1998. McGraw-Hill, NY.

7.  Van Ness, H. C. Understanding Thermodynamics. 1969. General Publishing Company. Ltd. Ontario, Canada.

8.  Engel, Thomas and Reid, Philip. Thermodynamics, Statistical, Thermodynamics & Kinetics. 2006. Pearson Education, Inc.

9.  Anderson, Charles R. Slaying the Sky Dragon-Death of the Greenhouse Gas Theory. 2011. Chapter 20. Page 313.

10.  http://www.eoearth.org/article/Atmospheric_composition

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

12.  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.

13.  http://thesis.library.caltech.edu/2809/1/Lapp_m_1960.pdf

14.  Ludwig, C. B., Malkmus, W., Reardon, J. E., and Thomson, J. A. L. Handbook of Infrared Radiation from Combustion Gases. Technical Report SP-3080.NASA. 1973.

15.  Sarofim, A. F., Farag, I. H., Hottel, H. C. Radiative Heat Transmission from Nonluminous Gases. Computational Study of the Emissivities of Carbon Dioxide. ASME paper No. 78-HT-55.1978

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193 Responses to Determining the Total Emissivity of a Mixture of Gases Containing Overlapping Absorption Bands: A Note from Nasif S. Nahle

  1. Neutrino April 7, 2011 at 6:14 am #

    For starters,

    ζ = pH2O / (pH2O + pCO2) (Ref. 5)
    ΔE = [[ζ / (10.7 + 101 ζ)] – 0.0089 (ζ)^10.4] (log10 [(pH2O + pCO2) L] / (pabsL))^2.76 [Ref. 5]

    Where in the references do these two equations appear?
    I ask for several reasons.
    First, you have two listings as reference 5 so which one is it?
    Second, since the first listing is over 800 pages and the second being 237 it’s a lot of material to sift through.
    Third, so I can understand what the equations mean by looking at the terms and context of it.
    And finally just to see that you have transcribed them correctly.

    Oddly you have another equation:
    E mixture = ECO2 + EH2O – ΔE
    Which is unreferenced but it appears it is eq. 1.2-18 from the second 5. referenced.

    But it doesn’t really matter where those equations came from since you cannot even use them correctly as you yourself have written them down.

    ζ = pO2 / (pO2 + pCO2) = 0.21 atm / (0.21 atm + 0.00039 atm) = 4.1675
    ζ = pO2+CO2 / (pHO2 + pO2+CO2) = 4.1675 atm / (4.1675 atm + 0.05 atm) = 0.9881

    Really??? 4.1675??? It is just arithmetic at this point! Then to compound that error you use the calculated number ζ, which is wrong but not even the point for this part, as the partial pressure of O2 + CO2 which it is clearly not.

    ΔE = [(0.992 / 110.892) – (0.0089 * (0.992)^10.4] * (log10 [(0.05 atm + 0.00039 atm) 7000 m] / (1 atm m))^2.76 (Ref. 2)
    ΔE = [0.00076] * (2.55 atm m / 1 atm m) = 0.0019; rounding up the cipher, ΔE = 0.002

    Ignoring the fact that you now are citing a different source for this equation.
    You lose track of the brackets and attach the ^2.76 solely to the (1 atm m) term(or just drop it altogether) As written the equation clearly has the ^2.76 as a power to (log10 [(pH2O + pCO2) L] / (pabsL)) term. If the ^2.76 is applied as the brackets require(ie: as you wrote it!) then the result is 0.0097 not the 0.0019 you calculated.

    ΔE = [[ζ / (10.7 + 101 ζ)] – 0.0089 (ζ)^10.4] (log10 [(pH2O + pCO2 + pO2) L] / (pabsL) 0)^2.76 [Ref. 5]
    ΔE = [[0.9881/ (10.7 + 101 (0.9881)^10.4)] – 0.0089 (0.9881)^10.4] (log10 [(0.26039 atm) 7000 m] / (1 atm)^2.76 = 0.00989

    The first term changes when it is rewritten! [ζ / (10.7 + 101 ζ)] somehow becomes [0.9881/ (10.7 + 101 (0.9881)^10.4)] when values are introduced. Besides which even calculating with the edited formula and continuing the incorrect placement of the ^2.76 power from before I cannot seem to replicate your result.

    Is there really any point in discussing this article given the above errors?

  2. Nasif Nahle April 7, 2011 at 6:23 am #

    @Neutrino…

    Reference 5.

    :D

  3. Schiller Thurkettle April 7, 2011 at 6:35 am #

    Dr. Nasif,

    Many thanks for this cogent explanation of the role of OC2 and its relation to global warming. While I cannot follow the math, your plain-English descriptions match with what I and others have been suggesting for years.

    An additional sidelight to this work: it explains how in the past we were able to have an ice age with vastly more CO2 than today!

    It might be interesting to run your numbers again, but with CO2 at 4,500 ppm instead. (Late Ordovician ice age.) Relevant graphic at:
    http://www.geocraft.com/WVFossils/PageMill_Images/image277.gif

  4. Nasif Nahle April 7, 2011 at 6:49 am #

    @Neutrino…

    ζ = pO2 / (pO2 + pCO2) = 0.21 atm / (0.21 atm + 0.00039 atm) = 0.9981

    ζ = pO2+CO2 / (pHO2 + pO2+CO2) = 4.1675 atm / (0.9981atm + 0.05 atm) = 0.9525

    ΔE = [[ζ / (10.7 + 101 ζ)] – 0.0089 (ζ)^10.4] (log10 [(pH2O + pCO2 + pO2) L] / (pabsL) 0)^2.76
    ΔE = [[0. 9522/ (10.7 + 101 (0. 9522)^10.4)] – 0.0089 (0. 9522)^10.4] (log10 [(0.26039 atm) 7000 m] / (1 atm)^2.76 = 0.01334

    And the total emissivity of the mixture of gases in the atmosphere is:
    E mixture = ECO2 + EH2O – ΔE = 0.0017 + 0.4 + 0.004 – 0. 01334 = 0.3924

    Fixed! Thanks for the observation. Now the effect of cooling is higher than in my previous calculation. THANKS A LOT, NEUTRINO!!!

    You say:

    It is just arithmetic at this point! Then to compound that error you use the calculated number ζ, which is wrong but not even the point for this part, as the partial pressure of O2 + CO2 which it is clearly not.

    Read the reference. LOL!

    You are again wrong in your concept of partial pressure. If you have 1 atm and from this magnitude you have 21% of O2 and 0.039% carbon dioxide, what are the partial pressures of those gases?

    :D

  5. Nasif Nahle April 7, 2011 at 7:27 am #

    @Schiller Thurkettle…

    You’re welcome.

    Thanks for the graph and the suggestion. I have noticed that the graph is updated with recent paleodata.

    It would be interesting to calculate the total emissivity of a mixture of water vapor in a proportion of 1%, carbon dioxide at 4500 ppmV, and a surface temperature of about 16 °C.

    NSN

  6. Neutrino April 7, 2011 at 10:11 am #

    1. “ζ = pO2 / (pO2 + pCO2) = 0.21 atm / (0.21 atm + 0.00039 atm) = 0.9981

    Great you fixed the arithmetic, now update the article.

    2. “ζ = pO2+CO2 / (pHO2 + pO2+CO2) = 4.1675 atm / (0.9981atm + 0.05 atm)= 0.9525

    And back to confusion, 4.1675atm is not the partial pressure of anything I am aware of in this discussion. As well 0.9981 is not a partial pressure either. It is a dimensionless number you just calculated, just look up above at 1.. Adding atm to it doesn’t make it a pressure.

    3. “ΔE = [[ζ / (10.7 + 101 ζ)] – 0.0089 (ζ)^10.4] (log10 [(pH2O + pCO2 + pO2) L] / (pabsL) 0)^2.76
    4. ”ΔE = [[0. 9522/ (10.7 + 101 (0. 9522)^10.4)] – 0.0089 (0. 9522)^10.4] (log10 [(0.26039 atm) 7000 m] / (1 atm)^2.76 = 0.01334

    Again that mysterious ^10.4 appears in 4. when it’s not in 3. and still unclear where the ^2.76 applies. Exactly as written the result should be 0.02606. Curious as to why the 0.9525 from 2. changes to 0.9522 in 4. as well.

    Until the head article is updated to correct the gross mathematical mistakes it is meaningless to discuss its conclusions. These are not typos, they are errors. The “pHO2” in 2. above is a typo, calculating an erroneous result and then using that result in the next equation is not.

    My understanding of partial pressure is that if we have 1atm with 21% O2 and 0.039%CO2 then the partial pressure of O2 is 0.21atm and CO2 is 0.00039atm.

  7. jennifer April 7, 2011 at 11:00 am #

    Off topic questions and answers have been deleted. Jennifer

  8. Nasif Nahle April 7, 2011 at 12:51 pm #

    @Neutrino…

    And back to confusion, 4.1675atm is not the partial pressure of anything I am aware of in this discussion. As well 0.9981 is not a partial pressure either. It is a dimensionless number you just calculated, just look up above at 1.. Adding atm to it doesn’t make it a pressure.

    Strange observation. If you have 20 chickens and buy other 20 chickens, how many chickens you will have? 40 chickens:

    (0.05 atm + 0.00039 atm) = 0.05039 atm.

    If you have 20 chickens and you gift all your 20 chickens, how many chickens will you have?

    0.05 atm / 0.05039 atm = 0.9923 (without units)

    It is elemental arithmetics.

    Again that mysterious ^10.4 appears in 4. when it’s not in 3. and still unclear where the ^2.76 applies. Exactly as written the result should be 0.02606. Curious as to why the 0.9525 from 2. changes to 0.9522 in 4. as well.

    It seems mysterious to you because you don’t know the formula. Read the reference 5.

    :)

  9. debbie April 7, 2011 at 7:06 pm #

    While the maths is definitely way beyond me I would definitely like to thank Nasif and others who are staying with this topic.
    I do understand how statistical modelling works.
    What I can recognise is that Nasif is seriously questioning a ‘basic assumption’ in those AGW models.
    I agree it is absolutely necessary that we ALWAYS question basic assumptions in any statistical models.
    If their basic asumptions are not correct (which in this case would be the total emissivity of CO2 in the IPPC climate models) then the answers can’t possibly be correct.
    That’s how it works.
    That’s why they need constant checking and constant updating when new data and new variables become available.
    If you put garbage in, you’ll get garbage out.
    Stats can’t work any other way because those complex models can’t question the inputs.
    That has always been up to their human masters.
    Good for you Nasif!

  10. Luke April 7, 2011 at 9:26 pm #

    So Debbie you don’t understand the maths yet you’re thrilled for Nasif to tell you anyway ? And now you think AGW models are statistical models do you? Deary me.

    Debbie, Nasif could tell you anything. Debbie – the mark of a director of a science institute would be to publish these revelations. He hasn’t. So it’s junk.

  11. gavin April 7, 2011 at 9:48 pm #

    Debbie; this is not statistical modelling or “garbage in garbage out” and neither is it new it seems when I checked the following

    “Total Normal Intensity of the energy radiated by the mixture of gases in the air:

    Therefore, the total normal intensity (I) (or the spectral radiance over wavelength) caused by the mixture of water vapor and carbon dioxide in the atmosphere is:

    Emix (σ) (T)^4 / π (Ref. 5 and 6)

    I = 0.3997 (5.6697 x 10^-8 W/m^2 K^4) (310.95)^4 / 3.1416 = 67.44 W/m^2 sr”

    But elsewhere we find I = 669 W/m^2 sr after proper analysis with calculus in “Radiative Heat Transfer” – Michael F Modest (I like this whole text including section 6) from

    http://books.google.com.au/

    see Chapter10.9 “Total Emissivity and Mean Absorbition Coefficient” and many other things of great interest

  12. cohenite April 7, 2011 at 10:30 pm #

    Gav, how’s that saucer of night-time, back-radiation heated milk going?

  13. Luke April 7, 2011 at 10:51 pm #

    Well Cohenite – it’s not frozen ! THE POINT

  14. cohenite April 7, 2011 at 11:06 pm #

    THE POINT?! The point, Mr backradiation, chief groupie for the Philipona fan club, is that clear-sky, night-time LDR from CO2 wavelengths should warm the milk to the same extent as daytime insolation since they have similar w/m2; on reflection the LDR is greater than daytime insolation, 363w/m2 compared with 166w/m2:

    http://www.atmos.illinois.edu/colloquia/080430.htm

    Some warm milk for beddie byes please luke.

  15. Neutrino April 7, 2011 at 11:31 pm #

    From last pot:
    1. “ζ = pO2 / (pO2 + pCO2) = 0.21 atm / (0.21 atm + 0.00039 atm) = 0.9981”

    2. “ζ = pO2+CO2 / (pHO2 + pO2+CO2) = 4.1675 atm / (0.9981atm + 0.05 atm)= 0.9525

    What was the strange observation in relation to 2.?

    My observations were:
    a) The value you entered for pO2+pCO2 in the numerator was 4.1675atm and it is wrong. Where did it come from? We are talking about a standard atmosphere here so entering a value that is in excess of 4 times that is on its face simply incorrect. Where did you get the value of 4.1675atm from?
    b) The value you entered for pO2+pCO2 in the denominator was 0.9981atm and is also wrong. The unitless number of 0.9981 which was calculated from 1. is not a pressure, simply adding atm to it does not make it so. It is a ratio of pressure’s and not a pressure itself.

    Additional observation:
    c) pO2+pCO2 in both the numerator and denominator should be the same value since they are the same variable. The value of pO2+pCO2 is 0.21039atm.

    And to be parsimonious the answer to your question:
    If you have 20 chickens and you gift all your 20 chickens, how many chickens will you have?
    20 chickens – 20 chickens = 0 chickens. And I agree it is elemental arithmetic that you are continually getting wrong.

    3. “ΔE = [[ζ / (10.7 + 101 ζ)] – 0.0089 (ζ)^10.4] (log10 [(pH2O + pCO2 + pO2) L] / (pabsL) 0)^2.76
    4. ”ΔE = [[0. 9522/ (10.7 + 101 (0. 9522)^10.4)] – 0.0089 (0. 9522)^10.4] (log10 [(0.26039 atm) 7000 m] / (1 atm)^2.76 = 0.01334

    No Nasif. It seems mysterious to me because going from 3. to 4. the formula changes. So one of either 3. or 4. is wrong, they are not the same formula. Which one is correct? As written they both cannot be be.

    To make it simpler to see:
    1) a – b^c = x
    2) a^c – b^c = x
    a = 2
    b = 3
    c = 4
    1) 2 – 3^4 = -79
    2) 2^4 – 3^4 = -65
    The above 1) and 2) are not equal, and neither are your 3. and 4..

    Nasif, it is an 800+ page text, where exactly does this formula appear? (ie: what is the number of the formula in the text) As well I have not even begun to talk about the meaning of the formula as we are still on its form. I do not need to know the context of the formula to know that the form is incorrect, and neither should you. If you correctly reproduced it in 3. then you did not correctly reproduce it in 4..

    (Thank you to gavin for suggesting google books, I had been looking online for a copy with no luck)

  16. Neutrino April 7, 2011 at 11:42 pm #

    Hello debbie,

    Thank Nasif for what? For presenting an incoherent set of math?

    Yes I agree Nasif is seriously questioning basic assumptions. But he is doing so in a way as to completely invalid his own conclusions.

    You say:
    I agree it is absolutely necessary that we ALWAYS question basic assumptions in any statistical models.
    If their basic asumptions are not correct (which in this case would be the total emissivity of CO2 in the IPPC climate models) then the answers can’t possibly be correct.
    That’s how it works.

    And I agree wholeheartedly. Apply that same logic to Nasif.

    Just because an answer that someone gives you is what you expect or want does not mean that their assumptions and process to arrive at that number was correct or even valid.

  17. Nasif Nahle April 8, 2011 at 12:12 am #

    @Neutrino…

    So you don’t read correctly either. The formula is on page 342.

  18. Nasif Nahle April 8, 2011 at 12:22 am #

    @Gavin and Neutrino…

    Don’t digg deeper on an issue that you completely ignore or you will be ridiculed, like the arithmetics of Neutrino. The formula is also in other books and articles.

    Calculate by yourself and you’ll obtain -0.0016, which I rounded up to -0.002

    The number is negative, so it is discounted from the sum of the total emissivity of carbon dioxide plus the total emissivity of the water vapor. It’s simple, but it seems that you and Gavin cannot accept reality and cannot do simple basic arithmetics operations.

    :D

  19. Neutrino April 8, 2011 at 12:40 am #

    Thank you Nasif.
    For future ease of communication if you use a formula from a source could you please cite the formula number as well. As in this case it would be 10.146.

    Quick notes and a request.

    a) Looking at the formula the inclusion of the ^10.4 that I bolded in 4. is erroneous and should not be there.

    b) The log10 should apply to ((pH20+pCO2)L/paL) not just to what is in the numerator that you have. This is actually obvious to preserve units.
    (log10 [(pH2O + pCO2 + pO2) L] / (pabsL) )^2.76
    Should be:
    (Log10[ (pH20+pCO2+pO2)L/ pabsL ])^2.76

    c) It is questionable to use the same formula 3. for both CO2+H20 and CO2+H20+O2 as the parameters may not be exactly the same, 3. is specifically for H20+CO2. I would have to read further to be sure on this.

    And could you actually address the issues that I have raised.

  20. Nasif Nahle April 8, 2011 at 1:27 am #

    @Neutrino…

    I told you the formula is in other books on radiative heat transfer. Why to resort to only one source?

    Do the calculation by yourself and find the result. Now tell me, what your result is?

    :D

  21. Nasif Nahle April 8, 2011 at 1:39 am #

    @Neutrino and Gavin…

    First you were suggesting that I had invented the correction factors, coefficients and powers, I demonstrated that the formulas exist with the same constants and variables I applied. Now you’re trying to say that the formula cannot be applied to other mixtures, when the authors show the way to make the calculations of complex mixtures of gases. This is a meaningless childish game of you.

    To put a final point to your nonsensical arguments, I recommend you to read the following sources to find the origin of my formulas:

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

    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.

    http://thesis.library.caltech.edu/2809/1/Lapp_m_1960.pdf

    Ludwig, C. B., Malkmus, W., Reardon, J. E., and Thomson, J. A. L. Handbook of Infrared Radiation from Combustion Gases. Technical Report SP-3080.NASA. 1973.

    Sarofim, A. F., Farag, I. H., Hottel, H. C. Radiative Heat Transmission from Nonluminous Gases. Computational Study of the Emissivities of Carbon Dioxide. ASME paper No. 78-HT-55.1978

    I always acquire the literature on which an author of a book based his/her calculations. So go on and make your own calculations. You’ll found that the results are the same results I have obtained.

    By the way, the chart in Modest’s book is incomplete. To find the complete graph, corrected by Leckner, read it from:

    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.

    Bye

  22. Neutrino April 8, 2011 at 4:06 am #

    Why resort to only one source you ask.

    Well because that is the source you cite for the formula. If there is a difference between how you write the formula and how your source writes it I would have to go with the source being correct and you having made a transcription error. The formula that you cite as coming from “Modest, Michael F. Radiative Heat Transfer-Second Edition. 2003.” does not actually appear in that text. Their 10.146 is very similar to your equation but there appears to be copying errors. All I am doing is trying to get you to be accurate in your statements.

    ∆є = [(ζ/(10.7 + 101ζ)) - 0.0089ζ^10.4](log10[ph20 + pco2)L / p(a)L])^2.76

    That is the correct formula.

    It is not equal to what you wrote.

    I never said you invented correction factors and coefficients, and I suspected the mysterious power was a typo. After seeing the actual formula from the cited author I confirmed that indeed it was erroneous. The other errors in your formula all appeared to be transcription errors.
    The salient point is that you did not copy the formula you are using correctly from the source.

    I did not emphatically say the formula cannot be used for other gasses. The word I used was “questionable”. Besides, the source describes the above equation for CO2+H20. It would be logical to assume that with a different gas the overlap between emission bands will be different necessitating different coefficients. It is not that the formula is specific to H20+CO2 but that the coefficients are specific to those two gasses.

    As for calculating the ∆є myself:
    1) pco2 = 0.00039atm
    2) ph20 = 0.05atm
    3) L = 7000atm.m
    4) p(a)L = 1atm.m

    5) ζ = ph20/(ph20 +pco2)
    6) ζ = 0.05atm/(0.05atm + 0.00039atm)
    7) ζ = 0.99226

    8) ∆є = [(ζ/(10.7 + 101ζ)) - 0.0089ζ^10.4](log10[ph20 + pco2)L / p(a)L])^2.76
    9) ∆є = [(0.99226/(10.7 + 101*0.99226)) - 0.0089*0.99226^10.4](log10[0.05atm + 0.00039atm)7000m / 1atm.m])^2.76
    10) ∆є = 0.009731

    That’s compared to your 0.0019 using an incorrectly copied formula.

  23. Nasif Nahle April 8, 2011 at 5:11 am #

    @Neutrino…

    With your number, the effect of cooling is higher. Then you’re supporting my conclusion:

    E mixture = ECO2 + EH2O – ΔE = 0.0017 + 0.4 – 0.0.009731= 0.391969

    Which is the total emissivity of the mixture.

    I had obtained 0.3958, that is a total emissivity that is higher than the total emissivity obtained from your calculation.

    :D

  24. debbie April 8, 2011 at 8:11 am #

    Luke and Gavin,
    If it makes you happier I will call them ‘computer models’.
    Same difference…. they produce graphs and projections.
    They still put garbage out if you put garbage in.
    It doesn’t change the fact that Nasif is questioning a basic assumption in those models.
    The atmosphere is complex and so are those models. It is possible that the models are correct, but, it is just as likely that they are not correct because all the input data is not correct.
    Whether you like it or nor, Nasif is questioning something that is ‘assumed’ about the atmosphere. He is also questioning the same ‘assumptions’ in the publications that you keep referring to.
    I’m much more interestd in the discussion between Nasif and Neutrino than your huffing and puffing about publications.
    Neutrino is engaging about the actual theories.
    I still congratulate Nasif for qestioning the inputs and I now congratulate Neutrino for doing the same.
    If they continue to focus on the actual issue here, which is the total emissivity of CO2 and how that influences AGW theories, we might get some more answers.
    As I’ve said before, for me the jury is still out. I’m not a warmist or a sceptic or any other name you may want to tag me with.
    I DON’T believe ‘the science is settled’ one way or the other.
    I believe we have more to learn .

  25. Neutrino April 8, 2011 at 8:45 am #

    My main point has been that you use formulas improperly. As such your articles have no merit.
    If you can conceded that the calculation I have done above corrects what you originally posted in your article then sure we can move on and talk about the consequence of ∆є. Also if that is the case could you update the head article to correct the errors that I have pointed out? If you will not do that I will continue to point out each error that you make.

    As for ∆є,
    That ∆є reduces the overall emissivity of a mixed gas, compared to the sum of the individual component gasses, is both obvious and completely mainstream. (ie: if both gasses have an emissivity of 1 over a common band then the combined gas will not have an emissivity of 2 over that same band therefore the combined gas must have an emissivity that is less than the individual sums)

    What then are the consequences of the mixed gas having a lower emissivity than the sum of the individual gasses?

    It radiates less at the same temperature. To maintain the same flux at TOA, which is required for the earth’s energy balance to be maintained, the earth surface, the atmosphere or both must heat up.

  26. jennifer April 8, 2011 at 9:43 am #

    I’ve made changes to the calculations based on advise from Nasif. Nothing has been deleted, but the calculations in error are now shown with a line through them.

    We are grateful to Neutrino for finding the errors in the calcultions.

    The net result is that the cooling effect from carbon dioxide is greater than previously calculated.

  27. cohenite April 8, 2011 at 9:54 am #

    A question to both Nasif and Neutrino; if, as Nasif asserts, and appears to have proved, Neutrino’s finding of some calculation errors not withstanding, where CO2 and H2O overlap on the spectrum there is less emissivity than would be the case with CO2 and H2O absorbing seperately, is this due to the greater absorbing capacity of either CO2 or H2O at those overlapping wavelengths?

    If it is not due to a greater absorption capacity of either gas then what is causing it? As it stands this seems to be a major poke in the eye of AGW theory and, as Nasif claims, the idea that positive feedbacks from H2O accelerate the initial warming caused by CO2.

  28. cementafriend April 8, 2011 at 10:48 am #

    Cohenite, let me quote from Perry’s Chemical Engineering Handbook page 5-23 “Carbon Dioxide-Water Vapor Mixtures – When these gases are present together, the total radiation due to both is somewhat less than the sum of the separately calculated effects because each gas is somewhat opaque to radiation from the other in the wavelength rergions 2.7 and 15 micron”

    Please do not mention photons because they do not exist.
    keep well

  29. Nasif Nahle April 8, 2011 at 11:04 am #

    @Cohenite…

    Thanks for your comment.

    I must say that it is not my discovery, but it is the discovery of the researchers who made the experiments and observations from nature (Hoyt Hottel, Lapp… etc.).

    By following the formulas of the authors mentioned, I calculated the effect of cooling under current conditions, and found that the effect of less emissivity and absorptivity is increasing as carbon dioxide increases in the atmosphere.

    Nowadays, the explanation given by some authors is related to the number of available energy microstates at the molecular and atomic levels and the change the molecules with low absorptivity cause on the wavelength of the quantum/waves.

    My assessment, although contained some errors due to failures in my computer about the transcription of numbers and formulas, provides a qualified explanation on the Ordovician ice age, when the carbon dioxide mass fraction was 12 times higher than in present days. It happens because during a period of warming huge quantities of carbon dioxide in the oceans and desert’s sands are released into the atmosphere and it causes an effect of cooling. The higher the mass fraction of carbon dioxide in the atmosphere, the higher effect of cooling will be.

    It explains many other observed phenomena which I will reveal step by step. Of course, if Dr. Jennifer accepts to publish them in her blog, which is a trustworthy and equitable blog and if you, the readers adhered to the theory of truth, are prone to read them. :)

    NSN

  30. William Gray April 8, 2011 at 11:05 am #

    Whats all this mean:

    Harries et al

    Satellite data for the last thirty years shows that the amount of heat (long wave radiation) leaving the Earths atmosphere into space has decreased, exactly the wave length absorbed by CO2 so the argument is over. ”

  31. cohenite April 8, 2011 at 11:09 am #

    Ok, I won’t mention photons, or Zasa Gabor. What I will mention is that AGW relies on positive feedback from water to small CO2 heating; this effect quantified by Nasif and brought to the unwashed masses who hitherto were not aware of this seems to quash those feedback estimates.

    It would also seem to have an impact on lapse rates by not steepening them and therein showing why a tropical hot spot, the centre-piece of AGW, would not form. I think the effect would also impact on condensation rates and keep cloud formation lower and therefore create a negative feedback. These are all potential Miskolczi mechanisms for showing why the greehouse effect, optical density, has not increased over the period of post WW11 warming allegedly due to AGW.

    I woudl especially be interested in Science of Doom’s response since he has put up several posts showing CO2 domination over water in the overlapping spectrum and in terms of backradiation. And backradiation is the final point; it must be less than shown if this effect is right.

  32. cohenite April 8, 2011 at 11:18 am #

    For a critique of Harries see this:

    http://landshape.org/enm/interpretation-bias/

  33. Neutrino April 8, 2011 at 11:47 am #

    Thank you jennifer for correcting some of the errors in the main article.

    Remaining errors: (not just typos)
    ΔE = [(0.992 / 110.892) – (0.0089 * (0.992)^10.4] * (log10 [(0.05 atm + 0.00039 atm) 7000 m] / (1 atm m))^2.76< (Ref. 2)
    ΔE = [0.00076] * (2.55 atm m / 1 atm m) = 0.0019; rounding up the cipher, ΔE = 0.002

    The log10 and the ^2.76 are not applied in correct place, corrected answer should be 0.009687.
    All following uses of 0.002 are also errors.

    ζ = pO2 / (pO2 + pCO2) = 0.21 atm / (0.21 atm + 0.00039 atm) = 4.1675 0.9981
    ζ = pO2+CO2 / (pHO2 + pO2+CO2) = 0.9981 4.1675 atm / ( 0.9981 4.1675 atm + 0.05 atm) = 0.9881 0.9522

    Use of 0.9981 in the second line is incorrect. 0.9981 is not a pressure, and specifically it is not po2+pco2 which is 0.21039atm.

    Now for questions about the formulas.

    The above equation, what is the proper numerator? Seeing how the order of doing the calculations, po2+pco2 first versus ph2o+pco2 versus ph2o+po2 first would all seem to lead to different numerators with common denominators. Which one is correct?

    ΔE = [[ζ / (10.7 + 101 ζ)] – 0.0089 (ζ)^10.4] (log10 [(pH2O + pCO2 + pO2) L] / (pabsL))^2.76

    Similar comment on this, is this appropriate? The formula as cited is for CO2+H2O, just rewriting it include O2 seems problematic.

    For demonstration purposes consider the calculation with just O2 and H2O. The overlap between O2 and H2O should not be the same factor as the overlap between CO2 and H2O. Why? Because O2’s emission bands are dissimilar to CO2’s, therefore the overlap and resultant correction should be different. Yet if we calculated the correction with O2’s pressure set to CO2’s(just for the sake of demonstrating the point) we would arrive at the same correction. That seems on its face a contradictory result.

    To use this specific formula for something other than H2O+CO2 without addressing this point is questionable.
    Even the ζ calculation has an issue when doing gasses other than CO2+H20. Which gas goes in the numerator? For O2+H2O the result would be 0.8077 or 0.1923 respectively, which is correct?

    Please do not just respond “Read the references”. If you believe that the equations can be used this way cite where you believe the references support that point.

    As to cooling via lowered emissivity.

    The authors that Nasif are quoting did not find that lowered emissivity cooled the atmosphere, what they found was that emissivity changed due to mixing. (at least what he has shown of the references he has supplied do not conclude cooling, if they indeed do predict cooling please cite those specific examples)
    The cooling effect is a speculative response to that lowered emissivity.

    That speculation, to my understanding, is misguided as it contradicts basic physics.

    As I said before.
    If emissivity of an object is lowered then its emitted flux will reduce. For the object to stay in equilibrium it must heat up until it is emitting its former flux. If the atmosphere’s emissivity drops by adding more CO2 then either it, the earth, or both must heat up to compensate for the reduced emissions.

    Nowhere did Nasif calculate a cooling of the atmosphere, what he did is calculate a reduced emissivity, which has been standard physics already known for a long time, as proved by the many references he cites.

  34. Nasif Nahle April 8, 2011 at 12:15 pm #

    @Neutrino…

    You say:

    To use this specific formula for something other than H2O+CO2 without addressing this point is questionable.

    Even the ζ calculation has an issue when doing gasses other than CO2+H20. Which gas goes in the numerator? For O2+H2O the result would be 0.8077 or 0.1923 respectively, which is correct?

    Please do not just respond “Read the references”. If you believe that the equations can be used this way cite where you believe the references support that point.

    As always, you’re suggesting that I invented the formulas. GO TO THE SOURCES. I demonstrated once and once again that I didn’t invent the formulas and you have read them from a single source in the internet. So, there is no other way for you that to go to the sources.

    You say:

    The authors that Nasif are quoting did not find that lowered emissivity cooled the atmosphere, what they found was that emissivity changed due to mixing. (at least what he has shown of the references he has supplied do not conclude cooling, if they indeed do predict cooling please cite those specific examples)
    The cooling effect is a speculative response to that lowered emissivity.

    What demonstrates that you don’t read well or you have not read those authors.

    That speculation, to my understanding, is misguided as it contradicts basic physics.

    It does not contradicts basic physics. AGW contradicts basic physics and I am demonstrating it.

    I’m showing what basic and advanced physics say. You have not given a single number from YOUR basic physics. You only have found some errors of transcription, which were corrected.

    You say:

    Nowhere did Nasif calculate a cooling of the atmosphere, what he did is calculate a reduced emissivity, which has been standard physics already known for a long time, as proved by the many references he cites

    With which you are demonstrating that you don’t know Kirchhoff’s Law. If the effect of the carbon dioxide and the Oxygen is of lessening the total emissivity of the mixture of gases, it is doing it also with the total absorptivity. Kirchhoff’s Law, and it is a LAW, confirms what I am saying. The total emissivity equals to the total absorptivity. Consequently, the total absorptivity is also lowered, by the same effect.

    :D

  35. cementafriend April 8, 2011 at 12:15 pm #

    Neutrino above is using twisted logic (or rather no logic). The main point is that the radiant absorption of CO2 in the atmosphere is insignificant. Convective heat transfer is the main form of transfering heat from surfaces to the gases of the atmosphere. Over oceans there is also evaporation. There is mixing of the atmosphere through winds which result from pressure differentials (which in turn result from rising hot air and condensation of water vapor). As the air rises it cools which is due to expansion and is determined by the lapse rate. The dry adiabatic lapse rate is 9.8 K/1000m. The reduced environmental average lapse rate of 6.5 K/1000m comes from the release of latent heat from the condensation of water. Clouds, water vapour and CO2 radiate to space at the top of the atmosphere. Because radiation form CO2 has such a small (insignificant) effect increasing the CO2 level when doubling also has no measurable effect.
    I suggest again everone reads the following http://climategate.nl/wp-content/uploads/2011/02/CO2_and_climate_v7.pdf

    keep well

  36. Luke April 8, 2011 at 12:27 pm #

    Debbie – Nasif’s work is also a “model” as are all abstractions of reality.

    No peer reviewed serious publication means not worth investing in. The internet is full of stories.

  37. Nasif Nahle April 8, 2011 at 12:38 pm #

    @Luke…

    Wrong, my dear Watson. My calculations are based on observations and experimentation, not in guessings.

    Let me explain how this stuff on the cooling effect of the carbon dioxide and the oxygen works:

    A lessening of the total emissivity means also a lessening of the total absorptivity, according to Kirchhoff’s Law.

    The surface has a measured emissivity of 0.82; it means that the surface emits 82% of the energy it absorbs.

    The mixture of gases in the atmosphere, water vapor, oxygen and carbon dioxide, has an emissivity/absorptivity of around 0.2, which means that the mixture of gases, as they are in the atmosphere, is about four times lower than the emissivity/absorptivity of the surface.

    Therefore, the atmosphere has more available microstates that can be adopted by the energy and it takes (absorbs) energy from the surface. In other words, they are taking the heat crossing the boundary of the surface.

    By taking heat from the surface, because the atmosphere has a lower emissivity than the surface, it is cooling the surface.

    Regarding the cooling effect of the carbon dioxide and the oxygen in the molecules of the atmosphere, if water vapor was alone in the atmosphere, it would have an emissivity/absorptivity of 40%, which means that it would warm the atmosphere; nonetheless, with the presence of carbon dioxide and oxygen, water vapor emissivity/absorptivity is lowered to 31.28%, i.e. 8.72% less than if it was alone in the atmosphere. It is a cooling effect because the water vapor is absorbing and emitting a lower amount of energy.

    This is how the cooling effect works. The arguments of Neutrino are nonsense.

    :)

  38. cohenite April 8, 2011 at 12:47 pm #

    Neutrino has agreed that there is a reduction in the emissivity from a combination of CO2 and H2O in the overlapping spectrum when both gases are present when compared when only either one is present; in fact Neutrino calls this “standard physics “; I don’t recall any AGW literature or view referring to it; in fact the AGW literature says that CO2 increased warming causes increasing water vapor and further increased heating; the meaning there is that water is a positive feedback which amplifys the warming from CO2.

    What Nasif has shown is that water is not a positive feedback; think about it.

    If an amount of water vapor, w, receives radiation in the overlapping spectrum it will absorb and emit rw; if CO2, c, is by itself in the overlapping spectrum it will aborb and emit rc. When both water and CO2 are present the total radiation absorbed and emitted will not be rw+rc, it will be rw+rc-e.

    The AGW equivalent is rw+rc+e.

    One extra issue, apart from the ones about lapse rate and backradiation I mentioned earlier is:

    1 Does e vary with variations in CO2 and/or H2O? That is, if CO2 levels increase does e get bigger?

  39. jennifer April 8, 2011 at 1:08 pm #

    Luke, Go and read Thomas Kuhn to understand how peer review can stifle progress in science. Jen

  40. Mack April 8, 2011 at 1:28 pm #

    Cementafriend,
    You are right about the total dominance of convection in the cooling of the earth.
    I think a huge proportion of this occurs over the Pacific.
    If you fly over the Pacific during daylight you may see mile after mile, thousands upon thousands, of those column like cloud formations,some with flat tops.
    There must be horrendous uplift there and if some of those CO2 molecules get hauled up to higher altitudes, it may well act as a refrigerate. ???

  41. Johnathan Wilkes April 8, 2011 at 1:30 pm #

    Jennifer,
    I don’t think Luke takes notice and going to read and educate himself re. “peer review”
    The best we could manage so far by pointing out his obsession with peer review and publication, is that he keeps quiet about it for a few days.

    Small mercy but better than nothing!

  42. Mack April 8, 2011 at 2:05 pm #

    When I say refrigerate, I’m relying on the work input of convection to lift the molecule to altitude, it then lose heat, and then the weight of the cold molecule to bring it back down to pick up more heat etc etc.

  43. Mack April 8, 2011 at 2:23 pm #

    “cold molecule ?”
    Is it a cold molecule heavier than a warm molecule? Help the layman out please Nasif.
    I’m sorry for interrupting.

  44. Neutrino April 8, 2011 at 2:29 pm #

    Nasif,

    I believe that the equations are real. I have read them. But you need to establish that they are valid over the range you are using them. Instead of just telling me to go read the sources why don’t you actually engage in debate? Besides I have gone to the sources, not all of them of course, and what I have seen so far does not establish that that specific formula is for anything other than CO2+H20. Why not comment on what I actually critiqued? As well as what I wrote of the ζ formula?

    I agree with your comment about Kirchhoff, it is confusing to me to see absorptivity not equal to emissivity.
    Since you like to tell me go to read the reference, I will return the favour. Check your reference 5. leading up to eq. 10.148a & 10.148b. As best I understand it the effect is at least in part that total emissivity is modified by self absorption within the gas(ie: a molecule emitting from within the gas may be absorbed before the photon escapes). For absorption this doesn’t happen for obvious reasons.

    Also, from your comment:
    The surface has a measured emissivity of 0.82; it means that the surface emits 82% of the energy it absorbs.
    Is one of the classic misrepresentations of Kirchhoff’s.

    If a surface has emissivity of .82 it will absorb 82% of incident radiation, but it will always emit 100% of the energy it absorbs(assuming here bodies in isolation and only radiative transfers occurring, no convection/conduction). For that to happen the surface will have to be at a higher temperature than an ideal emitter would have to be to emit the equivalent flux. Having an emissivity of .82 means the object emits 82% of a black body at the same temperature.

  45. Neutrino April 8, 2011 at 2:40 pm #

    Hello cohenite,

    I say it is standard physics because the texts and papers that Nasif has used are up to 50+ years old. The fact the equations are in mainstream texts leads to the fact it is part of the established view on radiative transfer.

  46. debbie April 8, 2011 at 2:42 pm #

    Luke,
    Of course Nasif’s work relies on modelling.
    That’s why Neutrino was right to question it.
    Nasif also recognised that some of the questions were valid and corrected his calculations.
    I think they’re both doing an excellent job.
    Let’s see where it goes.
    Even if you don’t, they obviously think it is worth discussing.

  47. Mack April 8, 2011 at 3:03 pm #

    Problem solved,
    Just substitute the word “molecule” with “gas” :)

  48. cementafriend April 8, 2011 at 3:32 pm #

    neutrino, I suggest you look up a definition of Kirchhoff’s law, absorptivity and emissivity are equal at “a surface in its surroundings at its own temperature” see Perry’s Chemical Engineering Handbook Page 5-25. For a gas the absorptivity has to be determined for the temperature of the source eg the absorptivity for radiation from the sun is almost zero because it does not absorb SW radiation. At the earths surface the absorptivity and emissivity are the same but at the top of the atmosphere (say 210K) CO2 could be absorbing from the earth surface (say 288K) and radiating to space (2K). The absortivity and emissivity are in the latter case different. This, of course, is a simplification because there is mixing, and radiation in the particular wavelengths from the surface may have been already absorbed lower in the atmosphere.
    Nothing about climate is simple but if one can not understand the simplified overall concepts then one can not grasp the more complex detailed concepts.
    Consider this
    UV radiation from the sun is important at the very top of the atmosphere with regard to the chemical reactions with oxygen/ozone and some other gases.
    Visible light and SW IR from the sun are important in heat transfer to the earth surface (particulary oceans)
    Convection and evaporation are the main forms of heat transfer between the surface and the atmosphere.
    LW radiation occurs to space at the top of the atmosphere from clouds, water vapor and CO2 in that order and from the earths surface through the wavelength window.
    What happens in between is complex detail that involves thermodynamics and fluid dynamics but all the available information (eg CO2 concentration changes lagging temperature changes, lack of tropical hots spots, cyclical climate changes etc) shows that radiation to and from CO2 in the mid-atmosphere has no or insignificant influence. Dr Van Andel paper (link above) gives evidence for that.

  49. cohenite April 8, 2011 at 3:38 pm #

    Neutrino says: “I say it is standard physics because the texts and papers that Nasif has used are up to 50+ years old. ” That would explain why I was not aware of it since I am older still.

    But my ancient age aside the point I was making was this scientific fact contradicts AGW modelling which supposedly shows a positive feedback from water vapor to increases in CO2; do you agree?

  50. Mack April 8, 2011 at 3:54 pm #

    In fact the planet might be a hell of a lot cooler by this mechanism if all that “absorption” and “emission” of energy took place with CO2 that the AGWers will try to have us believe.
    Aahahahahahahahaha (I think)

  51. Nasif Nahle April 8, 2011 at 4:27 pm #

    @Neutrino…

    You say:

    If a surface has emissivity of .82 it will absorb 82% of incident radiation, but it will always emit 100% of the energy it absorbs(assuming here bodies in isolation and only radiative transfers occurring, no convection/conduction). For that to happen the surface will have to be at a higher temperature than an ideal emitter would have to be to emit the equivalent flux. Having an emissivity of .82 means the object emits 82% of a black body at the same temperature.

    Impossible, Neutrino. The surface is not a blackbody. Only the non-existen blackbodies emit 100% energy. I suggest you to read a bit more on blackbodies, subsurface materials, conduction, convection and radiation, sinks of thermal energy, gravity effect on quantum/waves, change of wavelength and frequency after absorption and emission, microstates, latent heat, momentum transfer, evaporation, and the scientific method. The latter because you think that the experiments that are 50 years old are invalid when you said:

    I say it is standard physics because the texts and papers that Nasif has used are up to 50+ years old. The fact the equations are in mainstream texts leads to the fact it is part of the established view on radiative transfer.

    Arrhenius ideas have more than 100 years, and you continue using them, though have been falsified.

    Another of your strange comments:

    Is one of the classic misrepresentations of Kirchhoff’s.

    I think this conclusion of yours is because you don’t know what the total emissivity is. Perhaps, you are thinking that Kirchhoff Law is only applicable to arbitrary bodies in thermal equilibrium. My question to you is, are molecules in thermal equilibrium with their surroundings or not?

    :D

  52. Neutrino April 8, 2011 at 5:47 pm #

    Wow, I knew we really didn’t have a lot of common ground but after that statement I think we are miles apart.

    Quick digression with a simple thought experiment so I can make sure I understand you correctly.

    Body in isolation, no means of convection/conduction only radiative transfers involved. Variables as follows:
    1. є = 0.9
    2. Q = 100W
    3. q = єQ
    4. p = єq
    Q is the incident power, q is the absorbed power and p is the emitted power.

    I think we both agree that q is 90W.
    Based on what you are saying, if I understand you correctly, that 4. above is correct, that p = 81W.

    Am I correct in describing your position?

  53. Neutrino April 8, 2011 at 6:12 pm #

    And to answer your question. Yes, in general molecules are in thermal equilibrium with their neighbors.

  54. Luke April 8, 2011 at 7:40 pm #

    Oh look Jen – the non-existent greenhouse effect (NEXT !)

    http://i165.photobucket.com/albums/u43/gplracerx/PettyFig8-2.jpg

  55. cohenite April 8, 2011 at 8:03 pm #

    luke, would you be a good chap and interpret your graphs for all us plebs while I try and figure out whether Neutrino has asserted the atmosphere is in a Zeroth condition by virtue of his answer to Nasif.

  56. gavin April 8, 2011 at 8:52 pm #

    Cohenite; for a layman’s Atmospheric “equilibrium” see this Radiative equilibrium and thermal equilibrium model

    http://adsabs.harvard.edu/abs/1964JAtS…21..361M

  57. Louis Hissink April 8, 2011 at 9:33 pm #

    Physical Equilibrium only pertains to close systems.

    The earth system isn’t, so what’s the argument about?

  58. gavin April 8, 2011 at 9:39 pm #

    Debbie; After noting the citations for the above work, let’s move on to a smooth flow with this illustrated outline for our advance model Atmosphere

    “Atmospheric Radiative Transfer and Climate” by David Straus

    Beginning with the basic heat transfer with a simple model

    “Radiative Equilibrium and the Lapse Rate”

    http://www.iges.org/straus/CLIM_710/Radiation_Climate.pdf

  59. cohenite April 8, 2011 at 9:54 pm #

    gav, your link is to a model which assumes isothermailty to test the temperature effect of various ghg’s; the study shows that a thermal equilibrium is unrealistic and a dynamic thermal [dis]equilibrium is more realistic!

  60. mkelly April 8, 2011 at 11:22 pm #

    http://www.scribd.com/doc/34962513/Elsasser1942

    Please everyone read the paper on radiative heat transfer in the atmosphere. Dr. Elsasser is good at starting with basics up through the difficult.

  61. mkelly April 8, 2011 at 11:27 pm #

    Comment from: Louis Hissink April 8th, 2011 at 9:33 pm

    Physical Equilibrium only pertains to close systems.

    The earth system isn’t, so what’s the argument about?

    A closed system allows no matter to cross the boundary. The volume of the atmosphere does not change so the earth-atmosphere system is closed. Unless you want to count photons as matter.

  62. Louis Hissink April 8, 2011 at 11:43 pm #

    mkelluy,

    So ?

  63. Louis Hissink April 8, 2011 at 11:46 pm #

    mkelly,

    In addition, it’s energy, not matter, that is at the core of the argument here.

  64. RWFOH April 8, 2011 at 11:51 pm #

    Nasif has calculations that “prove” that increasing CO2 concentrations in the atmosphere cause cooling? Now to reconcile the hottest decade recorded in recent history with elevated atmospheric CO2 concentrations that are unprecedented in recent geological times….

    Nasif also claims his “calculations are based on observations and experimentation”. What observations? What experimentation?

    The only observation relevant to Nasif is he is “100% baloney” and is either incompetent or an ideologically driven fraud. As soon as I fashion myself a hat from tinfoil I’ll join you in writing off 200 years of solid science and kneel before the Messiah who will deliver us from Agenda 21.

  65. Nasif Nahle April 9, 2011 at 1:08 am #

    @Neutrino…

    You’re contradicting yourself. In one post you say the emissivity of the surface is 1.0 and accept that its absorptivity is 0.82; nevertheless, if you have molecules in thermal equilibrium, the emissivity equals to the absorptivity, which means that the molecules would be absorbing 100% of the energy received and emitting 100% of the energy received by radiation. This is false in the REAL world.

    :D

  66. Nasif Nahle April 9, 2011 at 1:11 am #

    @RWFOH…

    Guau, Guau! Haven’t you read the references? Oh! You didn’t, so you continue living in fantasyland.

    Yours is ad hominem, not science.

    :D

  67. Serioso April 9, 2011 at 5:17 am #

    Can’t one accurately determine the effects of H2O and CO2 by using a line-by-line spectroscopic model such as MODTRAN?

  68. Neutrino April 9, 2011 at 6:39 am #

    Nasif,

    Can you actually answer a question put to you? Ill repeat:

    Body in isolation only radiative transfers. Variables as follows with one additional condition, that the body is in equilibrium(I had assumed that before but neglected to state it, I apologize for that):
    1. є = 0.9
    2. Q = 100W
    3. q = єQ
    4. p = єq
    Q is the incident power, q is the absorbed power and p is the emitted power.
    I think we both agree that 3. q = 90W is true.
    I think that you believe 4. p = 81W is true.

    Do you endorse that given 1, 2, and 3 that 4 is a truthful statement? I am asking so I can understand your position and not misrepresent you.

    Please give me a simple direct answer. I realize this is slightly off topic from the head article but it is important to actually understand your starting position accurately.

  69. Luke April 9, 2011 at 6:55 am #

    Well Coho – you might ponder why one graph is almost the inverse of the other.

  70. Mack April 9, 2011 at 7:47 am #

    “I am asking so I can understand your position and not misrepresent you”
    Aaahahahahahahahahahahahaha

  71. Nasif Nahle April 9, 2011 at 8:49 am #

    @Neutrino…

    Your reasoning is wrong and, consequently, your question is also wrong. I think it obeys to your ignorance on what the total emissivity/absorptivity is. Emissivity/absorptivity is the potential a thermodynamic system has to absorb and emit energy. If you say that a thermodynamic system has a total emissivity of 100% you’re applying a false concept because you’re talking about a blackbody, and blackbodies don’t exist in nature.

    If what you want to say is that at zero Kelvin the thermodynamic system has released the whole amount of its energy, then you perhaps would be correct, but you are not talking about total emissivity.

    To put the things more easier to you:

    Total emissivity/absorptivity refers to RADIATIVE HEAT TRANSFER, as well as thermal conductivity refers to CONDUCTIVE HEAT TRANSFER, and Prandtl Number, thermal diffusivity, Grashof Number, buoyancy, viscosity and Convective Heat Transfer Coefficient refer to CONVECTIVE HEAT TRANSFER. You are dismissing many other mechanisms of transfer of energy, work and sinks of energy.

    In RADIATIVE HEAT TRANSFER, the total emissivity equals to the total absorptivity:

    α = (ε/εb) = ε

    Pitts and Sissom. Heat Transfer. Pp. 295, 314.

    Modest. Radiative Heat Transfer. Pp. 75.

    http://en.wikipedia.org/wiki/Kirchhoff's_law_of_thermal_radiation (Third paragraph).

    Hidenobu Wakabayashi and Toshiro Makino. Experimental Verification of Kirchhoff’s Law on Thermal Radiation at an Electromagnetic Wave Level. ASME/JSME 2007 Thermal Engineering Heat Transfer Summer Conference, Volume 3.

    Please, do not argue if you don’t know what we really are talking on.

    :)

  72. cohenite April 9, 2011 at 9:37 am #

    Yes luke, I do ponder that, but I’m wondering what it has to do with the combined reduced emissivity of H2O and CO2.

    Neutrino, I’ll play; yes p=81W, according to your figures; carry on I want to see where you’re going with this.

  73. Boadicea April 9, 2011 at 9:42 am #

    “Oh look Jen – the non-existent greenhouse effect (NEXT !)

    http://i165.photobucket.com/albums/u43/gplracerx/PettyFig8-2.jpg

    1. The area under the 20km graph is far larger than the one looking up, ..meaning that there is far more energy going out at the the TOA, than was emmitted from the bottom.

    2. Water has a far bigger impact than Co2..still

  74. Neutrino April 9, 2011 at 9:43 am #

    It truly is amazing your inability to answer a direct question.

    1, 2, 3 & 4 in the previous post represent what I think your belief is. If it is not simply type “No”. If it is then simply type “Yes”.

    The reason i believe this is you statement:
    The surface has a measured emissivity of 0.82; it means that the surface emits 82% of the energy it absorbs.

    Did I accurately reflect you belief with 1, 2, 3 & 4 from my previous post?

  75. Nasif Nahle April 9, 2011 at 9:55 am #

    @Neutrino…

    Sorry, but I find your comments a bit aggressive. Here the answer to your question:

    Your question is:

    1. є = 0.9
    2. Q = 100W
    3. q = єQ
    4. p = єq

    Q is the incident power, q is the absorbed power and p is the emitted power.

    I think we both agree that 3. q = 90W is true.

    I think that you believe 4. p = 81W is true.

    No, we don’t agree.

    Let us assign an arbitrary temperature to the surface, let’s say it is 320 K and an arbitrary temperature to the air above the surface -let us say it is 310 K. Here we apply the Stefan-Boltzmann equation:

    q/A = ε (σ) (Tair^4 –Ts^4) = q/1m^2 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) ((310 K)^4 – (320 K)^4)

    q = -58.14 W/m^2 (1 m^2) = -58.14 J/s (1 Joule/s = 1 Watt).

    The minus sign means that the surface is emitting power towards the air, i.e. it is releasing energy due to the temperature gradient, from higher energy density to lower energy density.

    Now, let’s calculate the amount of power the surface absorbs. To do this, we apply the same formula, but without considering the air temperature, but only the surface temperature:

    q = A (ε) (σ) (Ts^4) = q/1m^2 = 1 m^2 (0.82) (5.6697 x 10^-8 W/m^2 K^4) (320 K)^4 = 487.5 J/s.

    And the surface is absorbing:

    q = A (ε) (σ) (Ts^4) = q/1m^2 = 1 m^2 (0.82) (5.6697 x 10^-8 W/m^2 K^4) (320 K)^4 = 487.5 J/s.

    Now the load of energy absorbed by the air at temperature 310 K:

    q = A (α) (σ) (Ts^4) = q/1m^2 = 1 m^2 (0.2) (5.6697 x 10^-8 W/m^2 K^4) (310 K)^4 = 104.72 J/s.

    And the load of energy emitted by the air at temperature 310 K:

    q = A (ε) (σ) (Ts^4) = q/1m^2 = 1 m^2 (0.2) (5.6697 x 10^-8 W/m^2 K^4) (310 K)^4 = -104.72 J/s.

    I hope this helps. I apologize.

    :)

  76. Nasif Nahle April 9, 2011 at 9:59 am #

    Typo:

    q = A (ε) (σ) (Ts^4) = q/1m^2 = 1 m^2 (0.2) (5.6697 x 10^-8 W/m^2 K^4) (310 K)^4 = - 104.72 J/s.

    It should have said:

    q = A (ε) (σ) (Ts^4) = q/1m^2 = 1 m^2 (0.2) (5.6697 x 10^-8 W/m^2 K^4) (310 K)^4 = 104.72 J/s.

    No minus sign.

    I highly appreciated person reccomended me to not write to fast.

    :)

  77. Luke April 9, 2011 at 10:11 am #

    It seems Cohenite that despite the maths games leading to who knows where and defintely not publication, that the physics is working as assumed. I’m sure Thomas Kuhn knows best. ROFL.

  78. Neutrino April 9, 2011 at 10:20 am #

    Thank you for answering. There is no aggression on my part. You simply stated something that I believe was incorrect so I am merely trying to narrow down the exact disagreement between us.

    You missed a vital point of the question though. The body is in isolation there is no air above it, consider it to be free floating in space and illuminated by a distant star. The point was to create a simplified environment, so the differences between our opinions would be clear.

    To recap:
    There is only one body and it is sitting in a vacuum, it has an incident power on it of 100W(Q), of which the body absorbs 90W(q)(which I still think we both agree on).

    When the body is in equilibrium what is the emitted power from the body(p)?.

  79. Nasif Nahle April 9, 2011 at 10:58 am #

    @Neutrino…

    No problem.

    If we are considering an isolated system, we cannot introduce total emissivity because it is isolated and there is not interchange of energy with the surroundings or other bodies.

    We should then consider it an ideal blackbody, with total absorptivity of 100% and total emissivity of 100%.

    We should assign also an blackbody temperature to the system to know how much energy it can absorb and emit. Let’s say it is 255.15 K:

    q = A (1) (σ) (Ts^4) = q/1m^2 = 1 m^2 (0.2) (5.6697 x 10^-8 W/m^2 K^4) (255 K)^4 = 239.73 J/s.

    And it will emit the same load of energy per second.

    It would not be heated up to 255 K by such star.

    :)

  80. Priest Admiral Asperamanka April 9, 2011 at 11:07 am #

    I hope Nasif Nahle makes this work more widely available. Measurements of IR absorption by gases, including water and carbon dioxide, have been used for decades to control industrial furnaces. We now see that the equipment made by
    – Signal USA,
    – Ecotech,
    – P.C.E. Instruments,
    – Sick,
    – Campbell Scientific,
    – T.S.I. Inc,
    – Horiba,
    – M.R.U. Instruments,
    – Asea Brown Boveri,
    – Yokogawa,
    – Serinus,
    – Barber-Colman
    – etc. etc. etc.
    are all wrong. Industries have been unwittingly using incorrect measurements all over the world, every day, for years and years. Mr. Nahle stands to make a fortune, if he can design control equipment that, at last, gives accurate figures.

  81. Neutrino April 9, 2011 at 11:20 am #

    Again thank you,
    But you are still overly complicating the situation. I am gaining more information on the difference between us, so again thank you for responding.

    Using a blackbody hides any differences in our methods, so keep the example with an actual emissivity. If assigning an emissivity troubles you in this scenario please treat it as a thought experiment and just assume the emissivity is 0.9. From this a close approximation would be quartz glass.

    I am not sure where the miscommunication is coming from here.
    The example has an Incident of 100W and 0.9 emissivity. Given those values how much of is absorbed and, after the body equilibrates, how much is emitted?

  82. Neutrino April 9, 2011 at 11:24 am #

    gah, spurious ” in there, see if this works

  83. Nasif Nahle April 9, 2011 at 11:54 am #

    @Neutrino…

    The star emitting 100 W has a temperature; what its temperature is?

    To know how much the planet of quartz is absorbing we must know what its T0 is. Otherwise, we could not apply the Stefan-Boltzmann equation.

    Remember that emissivity/absorptivity is temperature dependent because temperature determines the wavelength distribution of radiation.

    Could you provide these data?

    :)

  84. Neutrino April 9, 2011 at 1:15 pm #

    I am sorry this has been drawn out so much, was not my intention when the question was first posed it is basically a Yes/No question, all these complications are unnecessary.

    The temperature of the source is irrelevant; all that is required is that the object has an Incident of 100W.

    For example, the source could be a 20m diameter sphere at 1000K with our body at a distance of 238m.
    Or the source could be a star with radius of 700million meters at 3000K with our body at a distance of 150billion meters.
    Or the source could even be a 100W laser focused on our body(at any distance).

    Assuming I did all the calculations right, the first two examples will both have a flux of 100W/m^2. So if our object is 1m^2 in size then it will have an incident of 100W.

    The point being that what is important is that our body receives 100W not where the 100W came from.

    Initial temperature of the quartz body is also irrelevant. If the body starts with an initial temperature it will just take a different amount of time to equilibrate.

    The original question was rather simple. A object has an incident power of 100W, with emissivity 0.9 it will absorb 90W, how many Watts will it emit(after equilibrium)?

    Thank you again, your posts have greatly helped me to understand your position.

  85. Neutrino April 9, 2011 at 1:21 pm #

    I might suggest if we are going to continue this perhaps find a new thread and stop the hijacking of this thread. As I said before that was not my intention as I seriously did expect a simple Yes or No answer.

  86. Mack April 9, 2011 at 3:56 pm #

    It all seems all pretty logical to me Neutrino…..
    Yes, Water DOES water down the heat. :)

  87. Nasif Nahle April 9, 2011 at 4:34 pm #

    @Neutrino…

    Sorry for the delay. I think the following explanation will help you to understand my position:

    In this scenario, the star is 204.93 K and the quartz planet absorbs 93 J/s of power in one second; it means 93 J, or 93 W*s of energy if and only if the initial temperature of the planet was 1 K.

    This amount of energy would make the temperature of the planet increases up to 198.17 K.

    Now we can apply the Stefan-Boltzmann equation. The planet with temperature of 198.17 K will be absorbing:

    q = (0.93) (1 m^2) (5.6697 x 10^-8 W/m^2 K^4) ((204.93 K)^4 – (198.17 K)^4) = 11.68 W, or 11.68 J/s.

    If the quartz planet is isolated, it would emit 0 W. However, if we continue considering this scenario, we would find that the planet would be emitting towards the cold, three-dimensional, unbounded space:

    q = (0.93) (1 m^2) (5.6697 x 10^-8 W/m^2 K^4) ((198.17 K)^4 – (204.93 K)^4) = -11.68 W, or -11.68 J/s.

    Its T0 would be ~122 K, and q0 would be 81.3151 W, not 198.17 W.

    As you can see, the thermal equilibrium is not reached until both bodies acquire exactly equal temperature.

    :)

  88. Nasif Nahle April 9, 2011 at 4:39 pm #

    @Neutrino…

    I agree; I don’t like either hijacking these threads. Dr. Jennifer has been very tolerant with us all.

    However, this topic is quite interesting.

    Best,

    NSN

  89. Neutrino April 9, 2011 at 5:38 pm #

    cohenite & Mack,

    The point of my recent exchange with Nasif was to understand his position, I think that I now do. Never did I intend to derail the thread as much as I did, but it happened. If a thread opens where we can discuss absorption, emission and radiative exchanges then I will engage the topic there.

    That I do not agree with Nasif is rather obvious, but this is not the place for that discussion. If Dr. Marohasy gives us the opportunity to continue then I will but otherwise I will revert to the actual head article for this thread.

  90. Neutrino April 9, 2011 at 5:41 pm #

    Thank you again for responding to my questions Nasif.

    This will be me last post in this thread in relation to the current topic.

  91. Luke April 9, 2011 at 6:27 pm #

    Cripes Neutrino – don’t mind us – please continue. It’s not like the rest of us were adding anything to the point (whatever the point is ?).

    I think we got past 4000 comments once….

    Jen how about giving these guys an open thread? We beseech thee ! Anything for clarity (I think).

  92. cohenite April 9, 2011 at 6:28 pm #

    Well, so much for that; a case of emissive interruptus; and who is that masked stranger, Neutrino?

  93. william gray April 9, 2011 at 7:44 pm #

    Luke your very funny, and helpfull.
    I have to speculate, that you persistantly resist changing your position deliberately. And that you are paid by someone to do so. If true I,d be asking for my money back.
    Hows sceptical science going, I recall some years back our author posting there about the cooling effect of Co2.

  94. Priest Admiral Asperamanka April 9, 2011 at 8:36 pm #

    See http://ronney.usc.edu/ On the subject of radiative heat transfer in gases, Professor Ronney knows whereof he speaks.

    Professor Ronney provided a spreadsheet to calculate radiative heat transfer using the Hottel Leckner method. For an atmosphere containing 3% water and 390 ppm carbon dioxide at a temperature of 33 degrees C and an earth surface temperature of 15 degrees C, this spreadsheet shows:

    emissivity H2O 0.57
    emissivity CO2 0.14
    total emissivity 0.65
    (Mr. Nahle reports correctly that the emissivities are not additive: 0.65<0.57+0.14)
    absorptivity 0.68
    absorption coefficient 0.0002

    The absorption co-efficient is small but not zero. In other words, carbon dioxide absorbs heat that would otherwise radiate out to space and warms up.

    And that, dear people, is what climate scientists have been saying since
    Climatic Change: Are We on the Brink of a Pronounced Global Warming?
    Author(s): Wallace S. Broecker
    Science, New Series, Vol. 189, No. 4201 (Aug. 8, 1975), pp. 460-463

  95. Mack April 9, 2011 at 8:37 pm #

    I dunno Cohers but I gave him a yes answer that he wanted to hear and should’ve been said to him in the first place to keep him happy.( Of course with a little assistance from Nasif.)
    :) :) ;) :):)

  96. jennifer April 9, 2011 at 10:18 pm #

    I am happy to start a new thread if Neutrino would like to send me something to begin it with… one or two paragraphs so I have a title, and we have a beginning, would be sufficient.

    Otherwise, continue here, you both have an audience.

    PS I already have another blog post from Nasif – another in his series. But have been reluctant to post it until we got to the end/exhuasted ourselves here.

    PSS If Neutrino had a post on the more general topic, I could hold off further with Nasif’s next contribution. To give us all a chance to better understand where Neutrino is coming from etcetera?

    Let me know.

  97. Nasif Nahle April 10, 2011 at 12:24 am #

    Thank you very much, Dr. Jennifer!

    It’s fine to me.

    NSN

  98. Neutrino April 10, 2011 at 1:59 am #

    Getting back to the original topic
    (I will send Dr Marohasy an email later today in regard to the off topic conversation that got started here.)

    I have no objection to the equations that were used, I can read the references that I have access to and they all look like good solid science.

    The overlap equation(∆є) is calculation of the overlap between CO2 and H20. As cited it is specifically for those two gases. To reinforce that point read the line immediately preceding the formula in Modest:
    As noted before, in a mixture that contains both carbon dioxide and water vapor, the bands partially overlap and another correction factor must be introduced, which is found from
    This is not a general formula for calculating an overlap between any two substances; it is specific to CO2 and H2O.

    Thinking about this logically it should be an obvious conclusion. CO2 and H2O share emission bands so therefore they overlap, this formula accounts for that. For two gasses that do not share emission bands there would be no need for a correction(redundant obvious statement). Yet if we used this equation on the two non overlapping gasses we would still get an overlap(actually the same overlap if the partial pressures of these two new gases were the same as CO2 and H2O).

    O2 and CO2 barely overlap in emission bands, so as a matter of point their overlap correction should be smaller than that of CO2 and H2O. For a second pretend that the O2 concentration in the atmosphere dropped to 0.05atm. Apart from the fact we would all be dead rather shortly the overlap between CO2 and O2 would still exist and should still be less than that of CO2 and H2O. Changing the pressure doesn’t change where the bands are, although it may affect the shape to a degree. The point still holds though that O2 and CO2 barely overlap. Yet if we were to now calculate their correction using the above equation we would again find they have the same correction as CO2 and H20.

    We know that that should not be the case, therefore it is obvious that the above equation is specific to CO2 and H2O. (quickly raises O2 partial pressure back to 0.21atm so we don’t all die)

  99. Nasif Nahle April 10, 2011 at 4:05 am #

    @Neutrino…

    The overlapp formula is for any mixture of gases and the procedure I applied is correct. If you don’t like the results, it is your problem.

    I have provided the references from which you can read that the formula applies to any mixture of gases.

    It is not a matter of thinking logically, which would take you to erroneous conclusions, but of making the things according to the scientific methodology and the known data.

    In the case of the star emitting 100 W and a planet absorbing 0.93, the logics took you to an erroneous conceptualization on total emissivity/absorptivity, which are coefficients by which the variables must be multiplied. It is not so easy as to multiply the energy incomming by the absorptivity. You must know the temperature of the surface and the temperature of the emitter, in this instance; if you don’t know those variables, you cannot apply those coefficients.

    I hope you understand that logics is not always the truth.

    :D

  100. Nasif Nahle April 10, 2011 at 4:10 am #

    Neutrino,

    This conclusion of yours is absolutely erroneous:

    We know that that should not be the case, therefore it is obvious that the above equation is specific to CO2 and H2O. (quickly raises O2 partial pressure back to 0.21atm so we don’t all die)

    A calculation won’t make the oxygen depletes.

    :D

  101. Neutrino April 10, 2011 at 6:12 am #

    Nasif,

    Your reference to Modest. Radiative Heat Transfer. clearly states that the equation that you used is for CO2+H2O.

    If you believe that another of the references specifically endorse that formula for use with gasses other than CO2+H2O then please state exactly where(page and paragraph) in the reference this occurs. As yet I have not been able to find such a reference anywhere. Better yet just post a quote of the reference stating that the formula can be used for any gases.

    Logic is king, and logic says that the formula is specific for the overlap between the spectral bands of CO2 and H2O. The manner of the overlap is built into the formula as the only inputs to it are pressures. Overlap between O2 and CO2 would require a different formula as their bands overlap in a different manner.

    As for the example, let me restate it a different way,
    Two identical chambers that both contain 0.00039atm CO2. Chamber A also contains 0.05atm O2 while chamber B has 0.05atm H2O.
    What is the overlap of each chamber?

    Since the formula only uses partial pressures to calculate overlap the above two chambers would have the same calculated overlap. That result is patently false. O2 doesn’t overlap CO2 significantly whereas H2O does. ∆є(co2+h2o) > ∆є(co2+o2).

  102. Nasif Nahle April 10, 2011 at 8:27 am #

    @Neutrino…

    Read Leckner, Hottel, Pitts and Sissom, etc. By the way, also read about Dalton’s Law of partial pressures of gases.

    If you don’t know about the science and heat transfer, don’t argue.

    Logic without observations and experimentation is nothing.

    :D

  103. Nasif Nahle April 10, 2011 at 8:38 am #

    @Neutrino…

    You’re not looking for my posture. What you’re doing is what you said in your first pot in the previous thread, to discredit real science. I have demonstrated that you ignore many things about radiative heat transfer:

    You didn’t and don’t know what the total emissivity/absorptivity is. How is it that you argued against my article?

    You didn’t know, and don’t know what the partial pressure is. How is it that you are aguing that the formula doesn’t exist?

    You didn’t know that to calculate the energy absorbed or emitted by a substance we MUST to know temperatures. How is it that you argued that my calculations were wrong?

    You didn’t know, and don’t know what the Dalton’s Law is. How is it that you argue that the formula that I applied doesn’t exist?

    You said that I had invented the formulas to calculate the total emissivity as from partial pressures; I demonstrated that the formulas exist and gave you the reference where you could find them. You verified that the formulas exist. How is it that you continue saying that I invented formulas, like this one on Dalton’s Law?

    Gathering all the above things, I conclude that your purpose is not to “understand” my posture, which by the way is absolutely adhered to science; what you’re looking for is to discredit my work, which is based on academic knowledge.

    :D

  104. Neutrino April 10, 2011 at 8:39 am #

    An interesting result of the ∆є equation.

    Since the overlap correction is basically preventing double counting where CO2 and H2O both emit the overlap at its most extreme could be the entire addition to emissivity due to the lower of the two gasses.
    Since CO2 is the weaker of the two that limits the correction to be no more than the CO2’s contribution.

    The calculated correction is ∆є = 0.00973. So the minimum emissivity for CO2 is є = 0.00973.
    Nasif maintains that CO2 є = 0.0017 so we have a contradiction if my assertion that ∆є represents the minimum є for CO2. Well only if we accept Nasif’s number.

    Nasif cites Hottel often saying that the emissivity is that low. From the introduction to this article:
    In 1954, Hoyt C. Hottel conducted an experiment to determine the total emissivity/absorptivity of carbon dioxide and water vapor11. From his experiments, he found that the carbon dioxide has a total emissivity of almost zero below a temperature of 33 °C (306 K) in combination with a partial pressure of the carbon dioxide of 0.6096 atm cm.

    But what he is neglecting to mention is that our atmospheric concentration of CO2 is not 0.6096atm.cm.
    We have km’s of atmosphere above our head that contains 0.00039atm CO2. Hottel also found that the total emissivity of CO2 at 1atm.m is near 0.18. Which does everyone think is a more accurate representation of CO2 in our normal atmospheric environment, 0.6096atm.cm or 1atm.m? I vote for the later.

    Don’t take my word for this, trust Nasif’s own sources. This graph from a paper by Lapp clearly displays both Hottel’s results(dashed line) and three computed results based on different models. Page 22 in the pdf(page 14 of the paper).

  105. Nasif Nahle April 10, 2011 at 8:59 am #

    @Neutrino…

    Yes, and see that the carbon dioxide starts emitting at 0.06 feet-atm (0.0183 atm m) while the partial pressure of the carbon dioxide, at 390 ppmV is 0.00039 m atm.

    The line doesn’t start below 0.06 feet-atm because the emissivity of the carbon dioxide is almost zero below that partial pressure. I have mentioned it in previous posts and the graph confirms it.

    You’re wrong in your calculation because a partial pressure of the carbon dioxide of 0.00039 atm m is not 0.6096 atm cm, but 0.039 atm cm. You’re quite wrong in your calculation and this is another demonstration that you’re not seeking for knowledge, but to discredit my work with false arguments.

    Read this nonsense:

    We have km’s of atmosphere above our head that contains 0.00039atm CO2. Hottel also found that the total emissivity of CO2 at 1atm.m is near 0.18. Which does everyone think is a more accurate representation of CO2 in our normal atmospheric environment, 0.6096atm.cm or 1atm.m? I vote for the later.

    Sir, the partial pressure of the carbon dioxide is not 1 atm m. So you’re saying nonsense and misinterpreting the graphs.

    :D

  106. Nasif Nahle April 10, 2011 at 9:03 am #

    In adition, Hottel mentioned that the emissivity of the carbon dioxide stops at 33 °C and a partial pressure of the carbon dioxide of 0.02 atm feet, which is 0.6096 atm cm.

    Consequently, I’m right, you’re wrong.

    :D

  107. Nasif Nahle April 10, 2011 at 9:21 am #

    To all readers,

    For more clarity, read the data on page 11 of the pdf, or page 3 of Dr. Lapp’s thesis.

    The chart shows the emissivities of the carbon dioxide through scales of partial pressures and temperatures and at 1 atm of total pressure. Total pressure is the total pressure of the atmosphere, while partial pressure refers to the pressure exerted by any gas taken individually, in this case, the partial pressure of the carbon dioxide is 0.00039 atm.

    Notice that Lapp observed that the carbon dioxide starts emitting at a partial pressure of 0.04 atm feet, which is about 0.0122 atm m. The emissivity of the carbon dioxide is very low at that partial pressure, i.e. emissivity = 0.062 at 300 K of temperature.

    Below that partial pressure 0.04 atm feet of the carbon dioxide at 300 K, emissivity of carbon dioxide is not depicted because it is almost zero. AGW is a fantasy, and the carbon dioxide clearly is not a greenhouse gas. At least while the temperature of the atmosphere is not 327 degrees Celsius! (600 K).

    When one read a source, we must read it completely and, if one doesn’t know how to interpret graphs, we should ask for help from someone that is not prone to cherrypicking data.

    :D

  108. Mack April 10, 2011 at 9:44 am #

    Nasif,
    Your general conclusion was that by adding any gas with total emissivity/absorptivity lower than the total emissivity/absorptivity of the main absorber/ emitter in the mixture of gases makes that the total emissivity/absorptivity of the mixture of gases decreases.
    Well that makes common sense to me Nasif
    It’s like the speed of a stream of traffic is always lowered by the slow coaches :)
    How come a dumb-ass layman like myself can see this.

  109. gavin April 10, 2011 at 10:04 am #

    Guys; the pressure of any gas floating round my head any day is 30″ Hg give or take a little fpr elevation re SL. The gas in my handy man compressor it around 100 PSI at peak and all gases inc H2O and CO2 O2 etc are at 100 PSI and will stay at the same temp due to work by the piston until I inflate a tyre or two.

    Air is air for engineers regardless of how one dreams about its individual gas analysis that can only be done as a pure gas

  110. Neutrino April 10, 2011 at 10:30 am #

    Nasif,

    I admitted to not knowing what the value for total emissivity of CO2 is, not to not knowing what the concept of total emissivity is. I have to thank you on this point, because of your references I now have a better idea of what the numerical the value of total emissivity for CO2 is. And it is not 0.0017. Admitting when you do not know something is not a weakness.

    You have said this before that I do not know what a partial pressure is, can you quote something of mine that would demonstrate that belief?

    I defer the discussion as to absorption and emission for a later thread as I have already emailed Dr. Marohasy. As a start to that future discussion: energy absorbed is a function of emissivity and incident(q = εQ) and emission is a function of emissivity and temperature(p = εσT^4.A), it’s as simple as that.

    In reference to Dalton’s Law, see above. Where have I ever calculated partial pressure incorrectly?

    I never said you invented formulas, post a quote where I say that. What I did say was you transcribed them incorrectly, and then applied them over situations that they were not valid for.

    My purpose is and has always been sincere. I have read statements from you that I believe are wrong, I have tried to discuss them with you to further understand your point of view. The ultimate goal was to correct the errors that I have seen in your posts.

    As I was writing this your last post appeared, it deals with partial pressures and your belief I do not know what they are.

    In response to that:
    A partial pressure is a pressure. It has units of atm(or bar or Pascal or….).
    Units atm.m(or ft.atm, atm.cm) is not a partial pressure.

    Find a source that uses units of atm.m for partial pressures and back up your assertion. All of the formulas that you have used in this post that use partial pressures use units atm. Your own references contradict you on this.

    The graph I linked goes down to 0.04ft atm and up to 3ft atm(or use the table Nasif pointed to, they are the same data). To avoid confusion converting all units to atm.m.

    So from 0.012atm.m to 0.9atm.m is the range of the graph. At 0.012atm.m the emissivity is 0.062, this is already above Nasif’s claimed value of 0.0017. So it all hinges on what the pressure distance is for the atmosphere in bulk.
    Simplest calculation(which is wrong) would be to just multiply the atmospheric column by the partial pressure of CO2, that yields 2.73atm.m(using 7km as atmospheric column). Although this number is wrong(because as altitude increases pressure drops) it provides the upper limit of the value.
    Just looking at the first kilometer of the atmosphere we get a value of 0.39atm.m. This is well within the range where CO2 is emissive, taking a value from the chart we are in the range of 0.15.

    Nasif is confusing a partial pressure with a pressure distance. As such he is reading values off of a chart that are orders of magnitude form where he should be.

  111. Nasif Nahle April 10, 2011 at 10:57 am #

    @Neutrino…

    You say:

    So from 0.012atm.m to 0.9atm.m is the range of the graph. At 0.012atm.m the emissivity is 0.062, this is already above Nasif’s claimed value of 0.0017. So it all hinges on what the pressure distance is for the atmosphere in bulk.

    You’re wrong again. The partial pressure of carbon dioxide at 390 ppmV is 0.00039 atm m, not 0.062 atm m.

    You say:

    Just looking at the first kilometer of the atmosphere we get a value of 0.39atm.m. This is well within the range where CO2 is emissive, taking a value from the chart we are in the range of 0.15.

    Would you say that the partial pressure of the carbon dioxide at 7000 m is 2.73 atm m? The total pressure of the atmosphere on the surface is 1 atm. Are you suggesting that the partial pressure of the carbon dioxide at the surface is more than twice the sum of all the components of the atmosphere?

    You say:

    Nasif is confusing a partial pressure with a pressure distance. As such he is reading values off of a chart that are orders of magnitude form where he should be.

    It is you who is confused. In my calculations, to get the partial pressure of the carbon dioxide, I applied the following formula:

    (paL)m / (paL)0 = 0.225 * t^2

    You know it? See your confusion, not mine.

    :D

  112. Neutrino April 10, 2011 at 11:02 am #

    Just to point out to make Nasif’s stance perfectly clear. He says:

    The chart shows the emissivities of the carbon dioxide through scales of partial pressures and temperatures and at 1 atm of total pressure. Total pressure is the total pressure of the atmosphere, while partial pressure refers to the pressure exerted by any gas taken individually, in this case, the partial pressure of the carbon dioxide is 0.00039 atm.

    The chart he is referring to has temperature in K and the an optical depth in ft.atm. The chart(and graph) do not have partial pressure anywhere on it. They do state that observations were done at total pressure of 1atm though. Comparing 0.00039atm to 0.006096atm.m(or any other value in atm.m) is apples to oranges and is meaningless.

  113. Nasif Nahle April 10, 2011 at 11:12 am #

    @Neutrino…

    Well, now I see that you don’t know that simple formula to calculate the partial pressure of the carbon dioxide in the atmosphere. I won’t mention other methods because I doubt you know them.

    :D

  114. gavin April 10, 2011 at 11:15 am #

    Guys; you can’t have a “partial” pressure in the atmosphere so mind your choice of language please

  115. Nasif Nahle April 10, 2011 at 11:23 am #

    @Neutrino…

    You say:

    The chart he is referring to has temperature in K and the an optical depth in ft.atm. The chart(and graph) do not have partial pressure anywhere on it. They do state that observations were done at total pressure of 1atm though. Comparing 0.00039atm to 0.006096atm.m(or any other value in atm.m) is apples to oranges and is meaningless.

    See what your purpose is on your nonsense? Unfortunatelly for you, I detected your objective since your first post.

    If you don’t read the whole articles of Hottel, Leckner and Lapp, you’ll get more and more confused and you’ll see apples and oranges everywhere.

    Hottel’s and Leckner’s graphs clearly say they “determined the effective emissivity of a hemispherical gas system of radius L at a partial pressure p“.

    Please, if you don’t understand the above assertion, you are who are seeing oranges and apples everywhere.

    :D

  116. gavin April 10, 2011 at 11:24 am #

    Btw; in practice we can use the same “gauge” in PSI what ever for any gas, AIR, steam, CO2, 02, N2 and so on

  117. Nasif Nahle April 10, 2011 at 11:32 am #

    @Dr. Jennifer…

    I’m asking for your authorization to send you a copy of the graph by Hottel and Leckner, so you can see the erroneous arguments of Neutrino.

    Thank you,

    Nasif S. Nahle

  118. gavin April 10, 2011 at 11:47 am #

    Nutrino; RE “ASERTIONS” In the abstract for “Methane Gas Visualization Using Infrared Imaging System and Evaluation of Temperature Dependence of Methane Gas Emissivity” we find this statement

    “Gas emissivity is the main uncertainty in gas temperature measurement using thermal cameras”

    http://pubs.acs.org/doi/full/10.1021/ie901340g

    I suggest we consider that in this case, methane is also viewed as part of the air at normal presure

  119. gavin April 10, 2011 at 12:00 pm #

    “The term “emissivity” has a lot of qualifiers..”

    http://www.evitherm.org/default.asp?ID=601&mode=html

    I feel Nasif is not showing us any of the qualifiers

  120. Nasif Nahle April 10, 2011 at 12:07 pm #

    @Gavin…

    Gavin, Gavin… Cannot you read “TOTAL emissivity”?

    :D

  121. Neutrino April 10, 2011 at 12:18 pm #

    Wow!

    determined the effective emissivity of a hemispherical gas system of radius L at a partial pressure p“.

    And this is supposed to convey that the graph is plotted against partial pressure? It says it right there, you even bolded it, the graphs that I am looking at from Lapp are clearly plotted with optical depth not pressure.

    Distance X Pressure = Optical Depth. The charts and graphs are in relation to depth not pressure.

    What do you think the optical depth of the atmosphere is? (aka what number should we be reading off on the graph at to get the total emissivity for CO2)

    As for the graph you want to send to Dr Marohasy, is it significantly different from the one in Lapp?
    Since the one in Lapp contains a reproduction of Hottel’s results why not just address the one you have in the paper linked(since anyone who wants to can easily see it for themselves).

  122. cohenite April 10, 2011 at 12:19 pm #

    Well said Mack [Comment from: Mack April 10th, 2011 at 9:44 am]; and that is the point; Nasif’s calculations which as Neutrino admits are based on solid, albeit ancient science, establishes, amongst other things that water feedback is, if positive, constrained by extra CO2. That in itself should be frontpage news.

    A couple of other things before we get bogged down in detail again; the Lapp paper contradicts Neutrino’s point above, or at least deals with the complaint by Neutrino [Comment from: Neutrino April 10th, 2011 at 8:39 am] that Nasif’s calculations about partial pressure are based on unreal Eartly conditions; in the Abstract to the Lapp paper they say this:

    “We note that the success of these calculations does not depend upon a fit to Hottel’s data at 300K since (a) we are able to calculate the total emissivity at 300K with fair accuracy from spectroscopic data and (b) we are able to estimate the parameters required by our model directly from available spectroscopic information. The use of Hottel’s data is adopted only as a convenience for this test calculation in order to provide a consistent check on our method of calculating emissivities at ELEVATED TEMPERATURES [my bold]”

    2 Secondly the point not addressed yet [although I could have missed it in all this excitment] is, does CO2 increase continue to reduce the combined emissivity of H2O and CO2 in the overlapping spectrum; if it does is it a linear, exponential, or what, decrease?

  123. cohenite April 10, 2011 at 12:33 pm #

    I have to say upon reading further into Lapp’s thesis that it is just one big refutation of many of the official capacities for CO2 as the agent of AGW; for example on pages 20-21 of the Pdf, Happ discounts pressure broadening and wings expansion of CO2 emissivity in the context that at Earth conditions of optical depth with moderate overlapping [of the water and CO2 spectrums] they tend to cancel each other or at least cannot be counted twice.

    So, as well as having CO2 and H2O reduce each others greenhouse effect when both are present, the capacity of CO2 not to be saturated is also limited.

  124. gavin April 10, 2011 at 1:45 pm #

    I think we should leave cohenite and nasif Lapping round blogsphere

  125. Nasif Nahle April 10, 2011 at 1:58 pm #

    @Neutrino…

    No, not different, but expanded a lot, I mean more specific. Lapp confirmed the experiments of Hottel and Leckner.

    :D

  126. cohenite April 10, 2011 at 2:44 pm #

    gav, that’s the wittiest thing you’ve said; good to see the supplements are working.

  127. Nasif Nahle April 10, 2011 at 3:37 pm #

    I’m Hotteling, Lapping and Lecknering, etc. LOL!

  128. Nasif Nahle April 10, 2011 at 3:45 pm #

    @Neutrino…

    The graph on page 22 of the pdf is the graph plotted by Lapp. The graph on page 10 of the pdf, or page 2 of the thesis, shows the graph by Hottel.

    You’re cherrypicking and misinterpreting the graphs. The graph you refer to is not Hottel’s graph, but Lapp’s graph.

    Sorry, but again, you’re trying to confuse the readers.

    :D

  129. Neutrino April 10, 2011 at 4:37 pm #

    Again Nasif you are misrepresenting my words.

    I never attributed the graph on page 22 to Hottel, I did say:
    This graph from a paper by Lapp clearly displays both Hottel’s results(dashed line) and three computed results based on different models.

    The reason I referenced Lapp’s graph and not Hottel’s was purely on readability. Since you agree that Lapp reproduces Hottel I see no reason for you to say I cherrypicked Lapp.

    Again what do you think is the correct value of atm.m to use to read off the emisivity form Lapp’s(or Hottel’s) graph?

  130. gavin April 10, 2011 at 4:58 pm #

    Nasif should ask cohenite to look for a proof of concept test given all the literature above

  131. cohenite April 10, 2011 at 5:08 pm #

    gav, old stick, the proof of concept is well established; the concept is that when CO2 and H2O are together instead of there being an amplification of their emissivity and therefore absorptivity and therefore heating capacity there is a reduction.

    AGW is based on CO2 and H2O positively reinforcing each other’s heating capacity; the basic science tells us this is not physically true; AGW is based on a lie; Neutrino, smart person that he is, can quibble about Nasif’s adding up till the cows come but the fact is the cows should not have been left out in the first place.

    Do you get it gav, or are you like the Earth mother I and a mate met when leaving the Sydney carbon tax protest who berated us for wearing No Carbon Tax t-shirts and said nothing would convince her that humans were not destroying the planet.

  132. gavin April 10, 2011 at 5:40 pm #

    cohenite, for such a little guy you do try so hard to peep over the barrier to that temporary (1960’s) traffic jam on the other side.

    Let’s see your proof of concept starting from say the 1990’s. I suggest this be your first practical

    “The measurement, instrumentation, and sensors handbook” By John G. Webster on Google Books

    see chapter 32.97

  133. spangled drongo April 10, 2011 at 6:13 pm #

    It is ironic that gav wants an empirical proof of concept when he’s been happy to accept GCM projections in the past.

    Personally, I find this discussion between Nasif and Neutrino fascinating even though the math is way beyond me.

    I hope they can continue this to a mutually agreeable position.

  134. Debbie April 10, 2011 at 8:11 pm #

    So do I spangled.
    At this point, it is looking more likely that those AGW climate models have been telling a porky about the properties of CO2 in our atmosphere.
    Before anyone loses their cool, I said ‘likely’ not ‘definitely’ !

  135. Luke April 10, 2011 at 8:54 pm #

    Why Debbie – coz an unpublished blogger has appealed to your point of view. Pullease !
    So let’s get this right – 1000s of physics papers are “wrong” simply because you like like Nasif’s message…. ROFL and LMAO !

  136. gavin April 10, 2011 at 9:03 pm #

    Debbie; don’t you think it’s a bit strange that after 50 years there have been no “fresh” atmospheric CO2 emissivity experiments?

  137. cohenite April 10, 2011 at 10:00 pm #

    You’re a dope gav; ask your mate luke about his hero Philipona’s experiments with CO2 backradiation “emissivity experiments”; then go and look at Miskolczi’s which are confirmed by Hottel and Lapp in this post. You 2 idiots don’t get it; CO2 reduces H2O’s emissivity; even luke’s other mate SoD confirms that:

    http://scienceofdoom.files.wordpress.com/2010/07/spectra-h2o-co2-15um.png

    What would it take for you 2 to be convinced AGW is wrong? Nothing.

  138. Luke April 10, 2011 at 10:20 pm #

    What does it take to get Cohenite excited about unpublished sceptic chatter – anything will do !

    You really are a prince among thieves Coho.

  139. Nasif Nahle April 11, 2011 at 1:20 am #

    @Debbie…

    Sorry for the delay on answering your post.

    Yes, I’m modeling, Debbie. However, the difference is that I am basing my calculations on real numbers and emissivities/absorptivities obtained from experimentation and observations, i.e. from the real world.

    Conversely, AGWer modelers base their calculations on fictitious numbers, variables and constants, all of them produced by their hysterical imagination.

    That’s a great difference. :)

    NSN

  140. Nasif Nahle April 11, 2011 at 1:28 am #

    @Luke…

    The greatest stupidity than I have listened or read is that the science constructed through the centuries by hundreds of scientists and engineers is invalid only because a scientist who has not published in highly discriminatory magazines writes about it.

    For your knowledge, I am a member in good stand of the APS; they do not authorize subscriptions to people without a solid institutional and academic endorsement.

    :D

  141. Luke April 11, 2011 at 3:53 am #

    “Conversely, AGWer modelers base their calculations on fictitious numbers, variables and constants, all of them produced by their hysterical imagination.” What utter rot – this shows what a ratbag you are. And you are who again?

    Stop obfuscating Nasif – get published or be consigned to the scrap bin of history.

  142. Luke April 11, 2011 at 4:00 am #

    ” they do not authorize subscriptions to people without a solid institutional and academic endorsement” hahahahahahahahhaaaa – classic – hear that Cohers !

  143. Nasif Nahle April 11, 2011 at 4:58 am #

    @Luke…

    By your own mouth… LOL!

    :D

  144. gavin April 11, 2011 at 6:36 am #

    Debbie; when Nasif admits he is modeling you know this whole pile of “assessments” is on very shaky ground and I suggest more so than a few of us who use the black body concept in our physics which is a classic form that can be built in practice with real hardware good enough to run the calcs. See all the engineering literature.

    I say this partial pressure model for emissivity determination cannot be built by Nasif, Cohenite or any one else we know today though we can have a discrete volume of mixed gas molecules whizzing round any time at atmospheric pressure. It can be as large as you like too but that requires a bit of integration that should be shown in full as original work, not cobbled from other text at will. We should also know the difference between a PhD thesis and classic physics as we have with the gas laws and their appropriate application.

    Btw this was a field I worked and studied in through the 1960’s & 70’s as a technician. Fortunatly, one night while driving home late after a physics lecture I had a brain storm that left me floating in a universe of brilliant light for quite a few moments where I had all the models known also others and could get an answer for any question at will. So for months I tried to write it up but did not have either enough math or language skill to handle it.

    Meantime others noticed the out of body stuff that went with a huge hole in memory so I had a long holiday from studies till it all went away. Research in measurements and methods from then on became a quest for short cuts to standards and feet on ground concepts that could be sold to your average customer on a daily basis. Lets say disassociation with the grass roots in physics is a dangerous path to follow for a number of reasons, the first is being out on a limb with a hand saw while contemplating people toiling on the factory floor below.

  145. el gordo April 11, 2011 at 7:35 am #

    ‘Lets say disassociation with the grass roots in physics is a dangerous path to follow’.

    Which explains why Anthony won’t touch it.

  146. cohenite April 11, 2011 at 8:51 am #

    So gav had his “did he inhale or not” moment at college and concludes “disassociation with the grass roots in physics is a dangerous path to follow for a number of reasons”.

    Exactly, and this is what AGW has done, time after time, it has diverged with actual physical observations to the extent that AGW is now intent on adjustments of those observational records.

    But in respect of Nasif’s post; is it grass roots physics, or not, that CO2 and H2O in combination reduce the total emissivity?

    Simple question, so answer it and gav can go back to amazing us all with his tales from his hippie days.

  147. Nasif Nahle April 11, 2011 at 1:16 pm #

    @Gavin…

    Debbie; when Nasif admits he is modeling you know this whole pile of “assessments” is on very shaky ground and I suggest more so than a few of us who use the black body concept in our physics which is a classic form that can be built in practice with real hardware good enough to run the calcs. See all the engineering literature.

    Wrong, I’m not modeling climate, but using real data to assess something that can be modeled based of real, real variables and constants, not from imagination.

    :D

  148. Neutrino April 11, 2011 at 1:32 pm #

    Nasif,
    If you are using real data to get the emisivity of CO2 in the atmophere what value of atm.m were you using to represent the CO2 in the atmospheric column to get the emisivity?(since Hottel’s and Lapp’s measurements and calculations were plotted against atm.m it is important)

  149. Dave April 11, 2011 at 4:11 pm #

    Nasif and fellow commentators.

    I am a skeptic by nature, but admit I was once a believer in CAGW not because I was well informed, I wasn’t. Simply because I trusted and believed the headlines and honorable scientific body’s behind them. But the dire prediction kept coming at an alarming rate, to the point I realized I better bring myself up to speed on global warming issue. The evidence that really started to turn my belief system into one of caution was Al Gores an Inconvenience truth. I realized that if one piece of evidence was misrepresented then others could also be. It was the grand statement that Mnt Kilimanjaro was losing its glaciers due to man-made global warming? You see I lived in Tanzania for many years. I have hiked the mountain on 2 wonderful occasions. It is a well known local and scientific fact that deforestation, land use and population growth had changed the weather pattern around and on the mountain and little or no rain or snow falls there anymore, there fore the glaciers were no longer getting the snow build up to keep the glaciers building or stabilizing them. I realized Mr. Gore and his scientific team was blowing smoke on this one then why not more? Since then I realized they were and are blowing smoke on nearly everything else. It wasn’t the science after all but the money, big money and not the trickle of oil money the AGW alarmist always scream about, turns out there are trillions of tax payers dollars at stake in the way of carbon taxes and cap and trade this has exposed the dirty underbelly of the CAGW participants and everybody seem to have their price!

    99.99 % of skeptics including myself have never received a penny for or efforts to expose this criminal activity and the only oil I use is in my car and on my salad!

    I’m now solidly in the skeptics camp, till I can see some real un-blemished evidence and not second rate nonsense that the IPPC has brought forward, I could go into the myriad of other so called science and fear mongering that has made me an avid follower of all thing’s involving the earth’s environment and our solar system. For that I thank you Mr. Gore, but I have also seen yours, the environmentalist, the scientist and governments dark side, and for that you get no respect from me!!!!

    Back to the subject of H2O and CO2. I have followed the discussion with awe, this was the equivalent of a scientific tennis match, this is real science in action I must congratulate Nasif, you have defended your position well with solid science, calculations and evidence. I’m convinced you have acquainted yourself well. You have obviously gathered the necessary tools to go far in your field of endeavor.
    Thank you and keep on producing quality research.
    David.

  150. Dave April 11, 2011 at 4:13 pm #

    Nasif and fellow commentators.

    I am a skeptic by nature, but admit I was once a believer in CAGW not because I was well informed, I wasn’t. Simply because I trusted and believed the headlines and honorable scientific body’s behind them. But the dire prediction kept coming at an alarming rate, to the point I realized I better bring myself up to speed on global warming issue. The evidence that really started to turn my belief system into one of caution was Al Gores an Inconvenience truth. I realized that if one piece of evidence was misrepresented then others could also be. It was the grand statement that Mnt Kilimanjaro was losing its glaciers due to man-made global warming?

    You see I lived in Tanzania for many years. I have hiked the mountain on 2 wonderful occasions, It is a well known local and scientific fact that deforestation, land use and population growth had changed the weather pattern around and on the mountain and little or no rain or snow falls there anymore, there fore the glaciers were no longer getting the snow build up to keep the glaciers building or stabilizing them. I realized Mr. Gore and his scientific team was blowing smoke on this one then why not more? Since then I realized they were and are blowing smoke on nearly everything else. It wasn’t the science after all but the money, big money and not the trickle of oil money the AGW alarmist always scream about, turns out there are trillions of tax payers dollars at stake in the way of carbon taxes and cap and trade this has exposed the dirty underbelly of the CAGW participants and everybody seem to have their price!

    99.99 % of skeptics including myself have never received a penny for are efforts to expose this criminal activity and the only oil I use is in my car and on my salad!

    I’m now solidly in the skeptics camp, till I can see some real un-blemished evidence and not second rate nonsense that the IPPC has brought forward, I could go into the myriad of other so called science and fear mongering that has made me an avid follower of all thing’s involving the earth’s environment and our solar system. For that I thank you Mr. Gore, but I have also seen yours, the environmentalist, the scientist and governments dark side, and for that you get no respect from me!!!!

    Back to the subject of H2O and CO2. I have followed the discussion with awe, this was the equivalent of a scientific tennis match, this is real science in action I must congratulate Nasif, you have defended your position well with solid science, calculations and evidence. I’m convinced you have acquainted yourself well. You have obviously gathered the necessary tools to go far in your field of endeavor.
    Thank you and keep on producing quality research.
    David.

  151. J.Hansford April 12, 2011 at 1:20 am #

    Makes good reading guys…. Keep it up.

  152. Neutrino April 12, 2011 at 9:12 am #

    Still waiting on that value of atm.m that you used to get the atmospheric emissivity of CO2. Since the emissivity is central to your argument it is important how you came to that value. Since Hottel is your source for emissivity then you cannot claim to have an emissivity unless you have a corresponding atm.m.

  153. Nasif Nahle April 13, 2011 at 4:26 am #

    @Neutrino…

    You can still waiting the time you wish. The formulas are given, the charts are documented; it’s just a matter of you to see and read them well.

    :D

  154. Nasif Nahle April 13, 2011 at 4:30 am #

    @Dave…

    Back to the subject of H2O and CO2. I have followed the discussion with awe, this was the equivalent of a scientific tennis match, this is real science in action I must congratulate Nasif, you have defended your position well with solid science, calculations and evidence. I’m convinced you have acquainted yourself well. You have obviously gathered the necessary tools to go far in your field of endeavor.
    Thank you and keep on producing quality research.

    Thank you very much for your kind words! I’ll keep doing my work adhered to science.

    Best,

    NSN

  155. Nasif Nahle April 13, 2011 at 4:31 am #

    @J. Hansford…

    Thanks!

    NSN

  156. Neutrino April 13, 2011 at 9:12 am #

    To try and recap this for anyone interested.

    Nasif claims that the entire column of atmospheric CO2 has an emissivity of 0.0017.
    For this number he references Hottel.
    In 1954, Hoyt C. Hottel conducted an experiment to determine the total emissivity/absorptivity of carbon dioxide and water vapor. From his experiments, he found that the carbon dioxide has a total emissivity of almost zero below a temperature of 33 °C (306 K) in combination with a partial pressure of the carbon dioxide of 0.6096 atm cm.

    He then shows that many researches(B. Leckner, Marshall Lapp, C. B. Ludwig, and A. F. Sarofim) all confirm Hottel’s results. That is all fine and good, I accept that as should any reader. It is well established.

    But what did Hottel actually claim?
    Hottel’s results are a plot of Emissivity vs atm.m(ft.atm in his case but converted to SI here). atm.m is not a partial pressure. It is the combination of a partial pressure and a distance. Hottel claimed that for sufficiently small values of atm.m that the emissivity is almost zero, not that emissivity is almost zero for small partial pressures. Nasif’s above quote conflates the two.

    So to apply Hottel’s results to the atmosphere in bulk the partial pressure must be multiplied by the path distance through it. Nasif is using 7km as the value for the atmosphere so ill continue using that number to stay consistent. Over 7km the atmospheric pressure drops to approximately 60% of the surface value. Assume a linear decrease in partial pressure from 0-7km and integrate. That gives you a value of approximately 2.2atm.m. Nasif is claiming that the value should be 0.00039atm.m which has no basis in fact. Unless of course he wants to conclude that the atmosphere is only 1m thick.

    To put this another way, Nasif is claiming that 1m of atmosphere radiates the same as 2m and 10m and 100m and 1000m and 7000m. That simply put is wrong. Multiplying the surface partial pressure by 1m does not result in either a partial pressure or an accurate representation of the atmospheric column for use in obtaining emissivity from Hottel.

    As for what a partial pressure is, Nasif has repeatedly stated that I do not know Daltons Law. This law, in a nutshell, that states that the sum of individual partial pressures sums to find the total pressure of a gas. Nowhere is there any mention of a quantity in atm.m, that’s because atm.m is not a partial pressure.

  157. Nasif Nahle April 14, 2011 at 12:44 am #

    @Neutrino…

    By the way, I have sent another essay where I explain how to make correctly these calculations.

    You’ll take classes on physics from these short essays I have been posting here.

    :)

  158. Neutrino April 14, 2011 at 5:06 am #

    Until you stop conflating a partial pressure(atm) and a pressure.distance(atm.m) you will be unable to use Hottel, Farag, Lapp or any other similar source to acquire the emissivity of the atmospheric CO2.

    None of your sources found emissivity’s from a partial pressure, they all find emissivity’s from pressure.distance’s.

  159. Neutrino April 14, 2011 at 7:00 am #

    Thank you Nasif for posting your calculations,

    Your calculation of solar flux at solar surface:
    Q/m^2= 0.9875 x (1 m^2) (5.6697 x 10^-8 W/m^2 K^4) (5804.135 K)^4 = 6.354 x 10^7 W/m^2

    Completely agrees with my calculation:
    7) p = 3.84*10^26W
    Dividing my p by the surface area of the sun, 4π(6.955*10^8m)^2, produces a flux of 6.32*10^7W/m^2.

    So considering the slight difference in our initial conditions these numbers agree. So I ask again why do you think my numbers are wrong if when you calculate it you get the same result?

    Your calculation for flux at earth:
    I = 6.354 x 10^7 W/m^2 (6.08 x 10^18 m^2) / (2.83 x 10^23 m^2) = 1366.34 W/m^2
    So = 6. 354 x 10^7 W/m^2 * (6.96 x 10^8 m / 1.5 x 10^11 m)^2 = 1368 W/m^2
    (I assume the difference between these two is the rounding error as they are identical calculations)

    Also completely agrees with my calculation:
    11) Q = 1.74*10^17W
    Dividing my Q by the disc of the earth, π(6.37*10^06m)^2, produces a flux of 1366W/m^2.

    Again considering the slight difference in our initial conditions these numbers agree. So I ask yet again why do you think my numbers are wrong if when you calculate it you get the same result?

    Where we part ways is this:
    What is the reason that we must introduce 1 m^2? Because to apply the Stefan-Boltzmann equation we must find the smallest surface area that resembles a flat surface. Here, Neutrino failed on the correct procedure.
    That’s the solar energy emitted by the Sun in direction to the Earth.
    And the flux of power the Earth would absorb per unit area is:
    The average flux of power inciding on the surface of the the Earth obtained by measurements is 329 W/m^2.
    Such amount of power flux causes the standard temperature of the Earth of 290 K, or 17 °C, without torturing mathematics and without resorting to any unreal “greenhouse effect”, but the power of the Sun alone.

    First, calculating for 1/m^2 is just a flux whereas my calculation was for total power. Both produce the same overall results, and neither one is better. One is a flux the other is a power, they both are correct. This is demonstrated by the fact you can convert between them by either dividing the power by total surface area or multiplying the flux by total surface area.
    There is no need for the surface to be flat, just for the area to be known.

    Second, how did you arrive at 329W/m^2, what measurement?(KT(page 4) finds that surface direct solar intensity is only 184W/m^2 of which 23W/m^2 is reflected) Since this is a thought experiment with no atmosphere how would you calculate that number?

    Third, how does 329W/m^2 result in a temperature of 290K? If we were to use the 329W/m^2 and calculate the temperature of a body emitting that much we would be at 276K via the Stefan-Boltzmann equation(using emissivity of 1).

    Could you finish the calculations off with:
    How much is absorbed by the surface.
    How much is emitted by the surface.
    What temperature of that surface.

    Since those were the interesting numbers it’s important to the discussion.

  160. Neutrino April 14, 2011 at 7:01 am #

    Opps wrong thread, delete if you can please.

  161. Nasif Nahle April 15, 2011 at 4:05 am #

    @Neutrino…

    All of your questions were answered from a scientific standpoint in the correct thread.

    :)

  162. Neutrino April 21, 2011 at 10:27 am #

    A delayed response but…

    In regards to the formula Nasif has thrown out as a calculation of partial pressure:
    It is you who is confused. In my calculations, to get the partial pressure of the carbon dioxide, I applied the following formula:
    (paL)m / (paL)0 = 0.225 * t^2
    You know it? See your confusion, not mine.

    Only confusion on my part is why Nasif thinks the above formula calculates a partial pressure of CO2 in the atmosphere.
    This is not a formula to calculate partial pressure but one to calculate (paL)m which is itself simply a parameter for the emissivity calculations. It most certainly is not the partial pressure, or even the pressure.distance, of the CO2 in the atmosphere. The only input to the above formula is the temperature of the gas, (paL)o is a constant, and CO2 does not appear anywhere.

    Besides, as clearly described by the source of the above formula(Modest, p344) the correct formula to use is:
    (paL)m / (paL)o = 0.054 / t^2 for when t < 0.7
    (paL)m / (paL)o = 0.225 * t^2 is for when t > 0.7
    (t is defined as T/To, with To equal to 1000K).

    With the range of terrestrial temperatures t < 0.7 so the first formula should be used not the second which Nasif is quoting.

    Regardless of the improper use of the above Nasif doesn’t seem to be consistent in what he believes is the pressure.distance for atmospheric CO2:

    From lead article: (my bold)
    (the partial pressure of the carbon dioxide at the present atmosphere is 0.0051 atm cm).

    From post: (my bold)
    The partial pressure of carbon dioxide at 390 ppmV is 0.00039 atm m

    Putting aside the fact he erroneously calls a value of pressure.distance(atm.m) a partial pressure(atm) his claims are not consistent.

  163. Nasif Nahle April 24, 2011 at 7:19 am #

    @Neutrino…

    There is always time for you say nonsense:

    Only confusion on my part is why Nasif thinks the above formula calculates a partial pressure of CO2 in the atmosphere.
    This is not a formula to calculate partial pressure but one to calculate (paL)m which is itself simply a parameter for the emissivity calculations. It most certainly is not the partial pressure, or even the pressure.distance, of the CO2 in the atmosphere. The only input to the above formula is the temperature of the gas, (paL)o is a constant, and CO2 does not appear anywhere.

    Besides, as clearly described by the source of the above formula(Modest, p344) the correct formula to use is:
    (paL)m / (paL)o = 0.054 / t^2 for when t 0.7
    (t is defined as T/To, with To equal to 1000K).

    The formula, as it is in Modest’s book, is correctly applied in my calculations. Period.

    :D

  164. Nasif Nahle April 24, 2011 at 7:22 am #

    @Neutrino…

    From lead article: (my bold)
    “(the partial pressure of the carbon dioxide at the present atmosphere is 0.0051 atm cm). ”

    From post: (my bold)
    “The partial pressure of carbon dioxide at 390 ppmV is 0.00039 atm m”

    Putting aside the fact he erroneously calls a value of pressure.distance(atm.m) a partial pressure(atm) his claims are not consistent.

    LOOOOL!!! What’s the “lead” article? Where do you get that 0.0051 atm cm???

    Is it compulsory of what? I mean, your lies… Heh!

    :D

  165. Nasif Nahle April 24, 2011 at 7:24 am #

    @Neutrino…

    You say:

    Putting aside the fact he erroneously calls a value of pressure.distance(atm.m) a partial pressure(atm) his claims are not consistent.

    Why don’t tell this to the author of the book? Heh!

    :D

  166. Nasif Nahle April 24, 2011 at 7:26 am #

    Now there are two “unknowns” to Neutino… He doesn’t know what a hell is total emissivity; now he’s demonstrating that he doesn’t know what partial pressure is… :D

  167. Nasif Nahle April 24, 2011 at 7:38 am #

    Now learn real physics, Neutrino:

    Partial pressure of a gas in a mixture of gases is:

    ppmV / (1 x 10^6 ppmV * Ptot) = 0.00039 atm

    Effective partial pressure of a gas in a mixture of gases is:

    (paL)m / (paL)o = 0.054 / t^2 for when t 0.7

    After this, you took a line at the foot of a table:

    (t is defined as T/To, with To equal to 1000K).

    Why you don’t buy the book and read the whole chapter, Neutrino?

    If you divide T by 1000 K, you will obtain a partial pressure for carbon dioxide of 0.0000234, which will make the total emissivity of carbon dioxide disappears into a vacuum… :D

  168. Nasif Nahle April 24, 2011 at 7:56 am #

    Just for this guy, Neutrino, stop saying lies on me. Here the calculation goes:

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

    Introducing magnitudes:

    Ecd = [1- ((((2.97 – 1 * 1 – 0.987022 atm)) / ((2.97 + 0.1) – (1 + 0.987022 atm)))) * e (-1.47 (Log10 ((0.648 atm m / 0.3048)^2))] * (0.002) = 0.0019

    No way, Neutrino, YOU LOST!

    :D

  169. Nasif Nahle April 24, 2011 at 7:59 am #

    Ecd = [1- ((((2.97 – 1 * 1 – 0.987022 atm)) / ((2.97 + 0.1) – (1 + 0.987022 atm)))) * e (-1.47 (Log10 ((0.648 atm m / 0.3048 ATM M)^2))] * (0.002) = 0.0019

    It’s desperating how a person that doesn’t know arithmetics, what the total emissivity is, what the units W are for, is capable of questioning the work of REAL scientists, including me… I’m real… LOL!

    :D

  170. Neutrino April 24, 2011 at 10:38 am #

    Nasif,
    LOOOOL!!! What’s the “lead” article? Where do you get that 0.0051 atm cm???

    “lead” article would be the article that started this thread, which you wrote and which contains the line “(the partial pressure of the carbon dioxide at the present atmosphere is 0.0051 atm cm).

    You quote me then add:
    “Putting aside the fact he erroneously calls a value of pressure.distance(atm.m) a partial pressure(atm) his claims are not consistent.
    Why don’t tell this to the author of the book? Heh!

    The authors of Radiative Heat Transfer do not conflate a partial pressure and a pressure.distance.

    For anyone interested here is a link to the section of the text that contains these formulas(p 342) and tables(p 344).

    From the table 10.5 the formula for (paL)m can be found.
    Clearly stated is that t = T/To, To = 1000K and (paL)o = 1atm.cm.
    Therefore t ≈ 0.25 for terrestrial temperatures.

    The formula:
    (paL)m / (paL)o = 0.054 / t^2, t < 0.7
    Is not a partial pressure, it is a correlation constant. (paL)m is a pressure distance and has no connection to the partial pressure of CO2. This should be fairly obvious to anyone actually looking at the formula because the partial pressure of CO2 is not in the equation. If the partial pressure of CO2 is 0.1atm or 0.000001atm the value of (paL)m does not change.

    Looking at Nasif’s calculations his values of the correlation constants are:
    (paL)m = 0.648atm.m
    a = 2.97
    b = 0.1
    c = -1.47
    Pe = 0.987022atm

    Calculating each of these from the formulas listed in table 10.5 we get quite different numbers:
    (assuming T = 253K(-20C) as an average, t = 0.25)
    (paL)m = (0.054 / t^2) * (paL)o
    (paL)m = (0.054 / 0.25^2) * 1atm.cm
    (paL)m ≈ 0.00864atm.m

    a = 1 + 0.1/t^1.45
    a = 1 + 0.1/0.25^1.45
    a = 1.75

    b = 0.23

    c = -1.47

    Pe = (p +0.28pa)/po
    Pe = (1atm + 0.28(0.00039atm))/1atm
    Pe = 1.0001

    And paL is simply the partial pressure of CO2 multiplied by the path length.
    (as noted before there is a problem calculating this because the partial pressure is not constant throughout the atmospheric column)
    paL = 0.00039atm * 7000m
    paL ≈ 2.73atm.m

    Where Nasif is getting his numbers is a mystery to me. One of the values that is just read off of the chart(b) is wrong, he has units on a dimensionless number(Pe).

    As a general note it is entirely confusing why he is even bothering to do that particular calculation anyways. From the text it is clear that the formula is scaling factor for emissivity when total pressure is different to 1atm. Since we are talking about an atmosphere that is already at 1 atm then it is a pointless calculation. Just read off from the charts or graphs what the emissivity of CO2 is at the current pressure.distance. So the question comes back to:

    What do you think is the current pressure.distance for the atmospheric column?
    0.3048atm.m which you used in this last calculation
    0.0051atm.cm which you stated originally in your article
    0.00039atm.m from an earlier post

  171. Neutrino April 24, 2011 at 10:45 am #

    Just as a quick note:

    The definition of a partial pressure is simply:

    “The pressure that one component of a mixture of gases would exert if it were alone in a container.”

    It has units of a pressure, something that has units of pressure.distance is not a partial pressure.

  172. Neutrino April 24, 2011 at 12:29 pm #

    Quick correction:

    Both c’s should be just 1.47 without the ‘-‘

  173. Nasif Nahle April 24, 2011 at 3:13 pm #

    @Neutrino…

    Where Nasif is getting his numbers is a mystery to me. One of the values that is just read off of the chart(b) is wrong, he has units on a dimensionless number(Pe).

    Yes, because you don’t know what the total emissivity is…

    Calculating each of these from the formulas listed in table 10.5 we get quite different numbers:
    (assuming T = 253K(-20C) as an average, t = 0.25)
    (paL)m = (0.054 / t^2) * (paL)o
    (paL)m = (0.054 / 0.25^2) * 1atm.cm
    (paL)m ≈ 0.00864atm.m

    Yes, because you’re not calculating correctly. The emissivity of 0.002 is given from experimentation at 611 K, not 1000 K, so you have to correct the value 1000 K.

    Anyway, I did the calculation introducing 1000 K, and the result is the same: 0.0019

    Ecd = [1- ((((2.97 – 1 * 1 – 0.987022 atm)) / ((2.97 + 0.1) – (1 + 0.987022 atm)))) * e (-1.47 (Log10 ((0.648 atm m / 0.3048 ATM M)^2))] * (0.002) = 0.0019

    LOL!

    :D

  174. Nasif Nahle April 24, 2011 at 3:29 pm #

    @Neutrino…

    You say:

    Is not a partial pressure, it is a correlation constant. (paL)m is a pressure distance and has no connection to the partial pressure of CO2. This should be fairly obvious to anyone actually looking at the formula because the partial pressure of CO2 is not in the equation. If the partial pressure of CO2 is 0.1atm or 0.000001atm the value of (paL)m does not change.

    Hahahaha! Now you made me laugh… You see the consequences of not knowing physics?

    (paL)m and (paL)0

    Do you see those “a” after the p? It is for “absorber”… You say “It doesn’t change” Hahaha! What a crazy argument!!! Besides, when you have seen that the partial pressure of the carbon dioxide in the atmosphere is 1 atm-m??? You’re managing false numbers. However…

    Let’s introduce your numbers, just to enjoy it awhile:

    (paL)m ≈ 0.00864atm.m

    a = 1 + 0.1/t^1.45
    a = 1 + 0.1/0.25^1.45
    a = 1.75

    b = 0.23

    c = -1.47

    Pe = (p +0.28pa)/po
    Pe = (1atm + 0.28(0.00039atm))/1atm
    Pe = 1.0001

    Ecd = [1- ((((1.75 – 1 * 1 – 1.0001)) / ((1.75 + 0.23) – (1 + 1.0001)))) * e (-1.47 (Log10 0.00864 atm.m / 1 atm.m)^2))] * (0.002) = [1-(0.00357 * 0.00177)] * 0.002 = 0.9999937 * 0.002 = 0.00199998

    The same result, but confusing…

    :D

  175. Nasif Nahle April 24, 2011 at 4:19 pm #

    @All readers…

    The correct formula to calculate the total emissivity of an absorber gas is as follows:

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

    Notice that the bolded term is not (paL)0, but paL. That is why I introduced the partial pressure of the carbon dioxide, not as (paL)0, but as paL, which is 0.3048 atm m.

    Regarding the value of (Ecd)0 of 0.002, it was obtained by experimentation and it is the total emissivity of the carbon dioxide at a partial pressure of 0.00039 atm.

    Neutrino’s confusion is because the partial pressure of any gas in the atmosphere is given by its percentage multiplied by 1 atm and after divided by 100. The addition of all partial pressures of the atmosphere must give 1 atm.

    For atmospheric carbon dioxide, its partial pressure is (0.039% * 1 atm) / 100 = 0.00039 atm.

    For the atmospheric N2, its partial pressure is (78.09% * 1 atm) / 100 = 0.78 atm.

    The O2 has a partial pressure of (20.946% * 1 atm) / 100 = 0.20946 atm.

    The Argon is present in the atmosphere exerting a partial pressure of (0.9340% * 1 atm) / 100 = 0.00934 atm.

    Water vapor has a partial pressure of (5% * 1 atm) / 100 = 0.05 atm.

    These partial pressures of the main gases in the atmosphere are the pressures exerted by the whole column of air on the surface by a gas taken individually.

    To calculate the partial pressure by length of a given gas, we must to follow another procedure. The latter is what generated the confusion of Neutrino.

    On Neutrino’s confusion, he confusingly took the number of paL as 1 atm m; however, he didn’t read well the text and he didn’t noticed that paL was expressed in bar cm. 1 bar cm is 0.0098692 atm m. From this mistake, he calculated:

    paL = 0.00039atm * 7000m
    paL ≈ 2.73atm.m

    Which is an error because the partial pressure exerted on the surface is the pressure of the whole column. paL is not more than the measured paL on the surface.

    Anyway, if we introduce his erroneous paL, the result is the same:

    Ecd = [1- ((((1.75 – 1 * 1 – 1.0001)) / ((1.75 + 0.23) – (1 + 1.0001)))) * e (-1.47 (Log10 0.00864 atm.m / 2.73 atm.m)^2))] * (0.002) = [1-(0.00357 * 0.0001026)] * 0.002 = 0.99999634 * 0.002 = 0.0019

    As you can see, the calculation including the 7000 m of altitude gives the same result than the calculation considering only 1 m of altitude.

    Thanks for your patience.

    NSN

  176. Alasdair Fairbairn April 25, 2011 at 7:49 pm #

    May I butt in here? I do this with diffidence as I am but an engineer without benefit of access to the academic circles.

    I found the paper by Nasif S Nahle extremely interesting as it directly impinges upon my own thoughts on the GHG hypothesis, which, I consider, does NOT comply with the laws of thermodynamics. [The engineering perspective!]

    I had difficulty following the details of the paper and the ‘ding- dong’ battle on the question of units. This was mainly due to a confusion in my mind on the exact definitions being used; albeit obvious to those involved.
    This was initially triggered by the conclusion in the paper which stated:

    “On the overlapping absorption spectral bands of carbon dioxide and water vapor, the carbon dioxide propitiates a decrease of the total emissivity/absorptivity of the mixture in the atmosphere, not an increase, as AGW proponents argue 1, 2, 3.”

    Now my interpretation of the term – ‘emissivity/absorptivity’ is Emissivity divided by [1 – Albedo] which, if correct, and applied to the Stefan-Boltzmann equation would INCREASE temperature should the value of this term DECREASE. This, of course being totally contrary to the thrust of the paper!

    However I also noted that in the abstract the term was reversed to ‘absorptivity/ emissivity':

    “The result of my calculations is that carbon dioxide reduces the total absorptivity/emissivity of the water vapor, working like a coolant, not a warmer of the atmosphere and the surface.”

    Thus removing my query.
    Hence my confusion. Could Nasif S Nahle elucidate?

    ON A COMPLETELY SIDE ISSUE; but as we are talking about partial pressure; has thought been given to effect of Henry’s law on the solubility of CO2 in water?

    As the partial pressure increases then so does the solubility; thus in inaccurate and rough terms should the partial pressure of CO2 double then the capacity of the oceans to absorb it also doubles.

    This raises the question as to whether the puny contribution of fossil fuel CO2 emissions could EVER achieve a doubling of CO2 concentration in the atmosphere. Somehow I doubt it, as my ‘back of the envelope’ calculations indicate. – Any thoughts?

  177. Neutrino April 26, 2011 at 4:59 am #

    Nasif,
    What is the pressure.distance of the atmospheric column of CO2?

    You have stated three different values so far:
    0.3048atm.m, 0.0051atm.cm, and 0.00039atm.m

    Since it is simply defined as paL, partial pressure * beam length, why is this such a difficult term for you to nail down?

    If you want to find the emissivity for the entire volume of CO2 then why would you use a value of L as anything other than the entire atmospheric column?

    Using a L of 1m would just be finding the emissivity of 1m of CO2 at 0.00039atm.

  178. Nasif Nahle April 26, 2011 at 5:26 am #

    @Alasdair Fairbairn…

    However I also noted that in the abstract the term was reversed to ‘absorptivity/ emissivity’:

    “The result of my calculations is that carbon dioxide reduces the total absorptivity/emissivity of the water vapor, working like a coolant, not a warmer of the atmosphere and the surface.”

    Thus removing my query.

    Hence my confusion. Could Nasif S Nahle elucidate?

    Sorry for my way on writing it colloquially. I should have written “absorptivity and/or emissivity” instead “absorptivity/emissivity”, which would lead to confuse the “/” by the symbol of division. The same would happen if I had written “absorptivity-emissivity” instead “absorptivity and/or emissivity” because it would be confused with a subtraction. Sorry for that and thanks for the observation.

    NSN

  179. Nasif Nahle April 26, 2011 at 5:32 am #

    @Neutrino…

    Nasif,
    What is the pressure.distance of the atmospheric column of CO2?

    You have stated three different values so far:
    0.3048atm.m, 0.0051atm.cm, and 0.00039atm.m

    Since it is simply defined as paL, partial pressure * beam length, why is this such a difficult term for you to nail down?

    If you want to find the emissivity for the entire volume of CO2 then why would you use a value of L as anything other than the entire atmospheric column?

    Using a L of 1m would just be finding the emissivity of 1m of CO2 at 0.00039atm.

    In a single paragraph:

    You are wrong on introducing 1 atm m instead the conversion from bar cm to atm m. Period.

    I have done the calculations introducing the whole length of the column of air and the result is lower than the result I obtained even as I introduced your erroneous “paL” of 2.73 atm m, which is not 1 atm m but 0.00987 atm m:

    Ecd = [1- ((((1.75 – 1 * 1 – 1.0001)) / ((1.75 + 0.23) – (1 + 1.0001)))) * e (-1.47 (Log10 0.00864 atm.m / 2.73 atm.m)^2))] * (0.002) = [1-(0.00357 * 0.0001026)] * 0.002 = 0.99999634 * 0.002 = 0.0019

    That’s all… I’m right, you’re wrong.

    :D

  180. Neutrino April 26, 2011 at 6:02 am #

    Yes I have been using bar and atm interchangeably, they are not identical units but about 1% different. Doing as I did introduces a small error to the end result.

    Do you even understand what that formula calculates?
    That formula (eq 10.145) is to calculate the scaling of emissivity when the total pressure is not 1atm. Since we are talking about our atmosphere it is already at 1 atm(at least near the surface). So of course the result should equal 1 regardless of the L used.
    (pal)o which is 1bar.cm converts to 0.00987atm.m is a constant does not appear in that formula(except in part to calculate (paL)m, which if you look above I used 0.01atm.m as a value for (paL)o).

    paL is the thing we are calculating the emissivity from.
    The error that you continue to make is that εo is the emissivity with that pressure.distance. The value of 0.002 you are using is simply wrong.

  181. Nasif Nahle April 26, 2011 at 4:01 pm #

    @Neutrino…

    Yes I have been using bar and atm interchangeably, they are not identical units but about 1% different. Doing as I did introduces a small error to the end result.

    No, no, no, Neutrino… You used 1 atm m instead 1 bar cm. The book from which you took the value of paL is in bar cm, not in bar m; therefore, you must to convert 1 bar cm to atm m… Capici?

    1 bar cm = 0.00986923267 atm m

    Capici? Capici? (unederstood?)

    :D

  182. Nasif Nahle April 26, 2011 at 4:07 pm #

    @Netrino…

    Poor Neutrino:

    The error that you continue to make is that εo is the emissivity with that pressure.distance. The value of 0.002 you are using is simply wrong.

    Haven’t you read the papers by Hottel, Lapp, Sarofim, Ludwig, Leckner, etc.? You’re just prattling. You don’t know what you’re talking about.

    :D

  183. Neutrino April 27, 2011 at 3:11 am #

    Nasif,

    Are you referring to paL or (paL)o?

    For (paL)o yes I originally used 1atm.m because I assumed you had the correct constant in your article.
    From your article:
    (pabsL)0 (absolute pressure of the mixture of gases on the Earth’s surface) = 1 atm m

    Once I got access to a copy of Radiant Heat Transfer I found out that assumption was wrong.
    From Modest Table 10.5:
    (paL)o = 1 bar cm
    After that I began using 0.01atm.m for (paL)o.

    Yes I have been using bar and atm interchangeably, this introduces about a 1% error.

    If you mean paL then I am not sure what your point is. paL is simply the partial pressure of the gas of interest(pa) multiplied by the path length(L).

    Since the path length in question is 7000m and the partial pressure is 0.00039atm how can paL be anything other than 2.73atm.m?

    If you are not using 7000m then you are not calculating the emissivity of the entire column. For example, using 1m path length paL = 0.00039atm.m, this will just get you the emissivity of 1m of co2.

    Yes I have read Lapp and Farag and the relevant parts of the text by Modest. All of those are quite clear that the emissivity of a gas is a function of its temperature, partial pressure, path length, and total pressure.

    If you feel one of the other authors counters that please quote them.

    In specific, you keep claiming Hottel concluded that the emissivity of the atmospheric CO2 is negligible, where exactly did he come to that conclusion?

    Your quote from the article:
    In 1954, Hoyt C. Hottel conducted an experiment to determine the total emissivity/absorptivity of carbon dioxide and water vapor11. From his experiments, he found that the carbon dioxide has a total emissivity of almost zero below a temperature of 33 °C (306 K) in combination with a partial pressure of the carbon dioxide of 0.6096 atm cm.

    Does not say that the emissivity of the atmospheric column of co2 is almost zero, but rather that at 0.006096atm.m co2 has almost no emissivity. A pressure.distance of 0.006096atm.m, correlates to a path length of only 15.6m at co2’s partial pressure of 0.00039atm. That hardly can be considered the total atmospheric column.

  184. Nasif Nahle April 29, 2011 at 3:14 am #

    @Neutrino…

    Does not say that the emissivity of the atmospheric column of co2 is almost zero, but rather that at 0.006096atm.m co2 has almost no emissivity. A pressure.distance of 0.006096atm.m, correlates to a path length of only 15.6m at co2’s partial pressure of 0.00039atm. That hardly can be considered the total atmospheric column.

    You must read Hottel’s, Leckner’s, Sarofim’s and Ludwig’s articles and books. Otherwise, you will continue in the error.

    I have made the calculations, as the scientific procedure indicates, and the three methodologies give the same total emissivity reported by the researchers, at a partial pressure of CO2 = 0.00039 atm, and T = 300 K, the carbon dioxide has a total emissivity of 0.002.

    Three methodologies, observation and experimentation by SEVERAL researchers CANNOT be wrong.

    AGW idea is just that, a myth.

    :D

  185. Nasif Nahle April 29, 2011 at 3:16 am #

    @Neutrino…

    Nasif,

    Are you referring to paL or (paL)o?

    I’m not referring to anything. Read the formula and find your mistake by yourself.

    :D

  186. Neutrino April 29, 2011 at 1:02 pm #

    Nasif,

    You keep saying that Hottel et al say that at 0.00039atm of CO2 the emissivity is very low. The problem is that from everything I have read so far not one of them has made that claim.
    They all agree at very low pressure.distance CO2 has a very low emissivity. I have no reason to object to their findings.

    My objection is with your claim not any of theirs.
    Everywhere in the referenced material that I have read emissivity is plotted or calculated against a pressure.distance not just a pressure. So your claim that at 0.00039atm CO2 has an emissivity of 0.002 is unsupported by your references.

    Your article is trying to assert something about the emissivity of the atmospheric CO2, as such the distance used has to be comparable to the actual height of the atmosphere. Using a value of 0.00039atm.m does not represent the atmospheric column but rather just 1m of it.

    If you really do believe that one of those authors maintains that emissivity can be plotted from just pressure or that the atmospheric column of CO2 represents a very low pressure.distance then please cite that exact point.

  187. gavin May 4, 2011 at 6:42 pm #

    For the record, the issue of Nasif’s peculiar math and physics goes back quite a while, see

    “A debate between Hans Erren and Nasif Nahle” 2007 on “ukweatherworld”

    I have to note there is nothing published despite the Nasif campaign

  188. David Socrates May 22, 2011 at 8:10 pm #

    Nadir,

    Excuse my ignorance but until I read your article I did not know that there was an issue with the fact that most of CO2’s bands overlap those of water vapour. I thought that the debate was all about the fact that some of the bands bands do not overlap, thus providing additional absorption not already achieved by the water vapor. I thought the discussion had evolved as follows:

    1. WARMIST: CO2 enhances the greenhouse warming effect because it’s absorption bands do not wholly overlap with water vapour. Therefore it absorbs additional radiant energy, thus causing additional warming.

    2. SKEPTIC: Agreed, but this is such an insignificant increment percentage-wise that it can be ignored. Water reigns supreme!

    3. WARMIST: Agreed the CO2 warming effect is small in itself, but the small warming effect that it does have in turn results in a much larger increase in atmospheric water vapour and this does result in a significant temperature rise. (The “water vapour feedback multiplier”.)

    Now you seem to be claiming something completely different that has the opposite (negative) effect. Fair enough but then the important question arising is which of these two effects (band overlap and band non-overlap) predominates and in what proportion? Otherwise we are in danger of a spending all our time on qualitative arguments without reference to likely quantitative contributions from the two effects – which makes it impossible to make any real-world assessment of their respective importance.

  189. Nasif Nahle July 20, 2011 at 12:40 am #

    @David Socrates…

    I been absent from this blog because I have been a bit busy on experiments and other professional issues.

    The main error is to attribute the knowledge of overlapping bands to me, when it is basic scientific knowledge that every physicist knows. I, as many scientists also, thought that overlapping bands enhanced the absorptivity of the mixture of gases. Well, the real thing is the opposite, i.e. it makes the absorptivity and the emissivity of the mixture decreases.

    I have conducted an experiment that demonstrates that the “greenhouse effect” by retention of radiation does not exist, so the AGW idea is a complete fallacy.

    NSN

  190. Derek B July 20, 2011 at 12:09 pm #

    Neutrino, thanks for your valiant efforts to keep Nahle honest. But it doesn’t look like he’s scientist enough to own up to his blunder.

  191. Derek B July 22, 2011 at 1:01 pm #

    Had a closer look at the whole paper. There’s a possibly crucial error in the way it calculates the emissivity of a mixture. It assumes a fixed total pressure (one atmosphere). Thus, adding CO2 to the mixture effectively displaces some of the H2O. That could certainly lead to reduced total absorptivity, but it’s not how the atmosphere works.
    The partial pressure of the water vapour is dictated by the temperature and water’s SVP curve. Adding CO2 increases the total pressure – it does not displace any of the water vapour.
    It would be interesting to see whether the cooling effect persists when that is corrected.

  192. Doug Cotton May 3, 2012 at 2:21 pm #

    In my view, when adding CO2 to existing water vapour, you have to take into account which CO2 frequencies overlap existing WV frequencies. The end result cannot be that there is a higher intensity, because the intensity of radiation at any frequency is limited by the Planck curve. See this post.

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