Determining the Total Emissivity of a Mixture of Gases Containing Overlapping Absorption Bands: A Note from Nasif S. Nahle
Posted by Nasif S. Nahle, April 6th, 2011 - under News, Opinion.
Tags: Climate & Climate Change, Physics
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
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.
Makes good reading guys…. Keep it up.
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.
@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.
@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
@J. Hansford…
Thanks!
NSN
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.
@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.
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.
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.
Opps wrong thread, delete if you can please.
@Neutrino…
All of your questions were answered from a scientific standpoint in the correct thread.
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.
@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.
@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!
@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!
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…
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…
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!
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!
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
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.
Quick correction:
Both c’s should be just 1.47 without the ‘-’
@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!
@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…
@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
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?
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.
@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
@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.
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.
@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?)
@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.
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.
@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.
@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.
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.
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
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.
@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
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.
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.
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.