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01 April 2013

The Earth Surface Temperature without Greenhouse Gases: The Shade Effect of Infra-Red Active Gases



It is commonly said that because the radiative power from the Earth system into space is the same as that of a black body with a temperature of about 255K and the observed average temperature of the Earth’s surface is about 14.5ºC or 287.65K, the surface of the Earth is nearly 33K warmer than it would be if there were no greenhouse effect.  The greenhouse effect is said to be due to infra-red active gases such as water vapor, carbon dioxide, methane gas, and nitrous oxide.

The problem with this argument is that while the Earth system is in radiative equilibrium with space, the surface of the Earth is not.  Thus, there is a complex relationship between the equilibrium temperature of the Earth’s surface and its equivalent radiative temperature as a black body radiator.  Some of the radiation into space is direct from the Earth’s surface, but most of it is from the infra-red active gases of the many layers of the atmosphere.

I have previously shown that the infra-red gases probably produce more cooling than warming.  Those arguments were not simple enough for most people to follow and they did not show what the amount of cooling of the Earth’s surface actually was.  I am going to show a surprisingly simple proof in this article that the infra-red gases, commonly called greenhouse gases, cause the surface temperature of the Earth to be much cooler than it would be if our atmosphere had no infra-red active gases in it.

Consider the following NASA energy budget of the Earth:




As I have discussed in an article called The Unsettled Earth Energy Budget, one should add a small amount of energy absorbed by the Earth’s surface due to back-radiation from the atmosphere.  My estimate for that back-radiation is about 1 to 2% of the solar insolation radiation at the top of the atmosphere.  Since other NASA Earth energy budgets put the amount of energy absorbed from the direct solar insolation at 48 or 49%, let us only add the 1% back-radiation amount to the 51% absorbed solar insolation shown above.  In 2010, the average top of the atmosphere solar insolation was 1365.8 W/m2.  The average solar insolation over the day at a spot on the rotating Earth is one-quarter this amount.

At equilibrium, the power absorbed by the Earth’s surface equals the power emitted by the Earth’s surface.  If the Earth’s surface were an interface with vacuum, the total emitted power is given by the Stefan-Boltzmann Law and we have the equilibrium condition:

Pabs = ε σ T4,

Where ε is the Earth’s surface emissivity, σ = 5.6697 x 10-8 W/m2K4, and T is the average surface temperature of the Earth.  Thus we have:

Pabs = (1365.8/4)(0.52) = ε (5.6697 x 10-8 W/m2K4)(287.65K)4

Solving the equation for ε, we find:
ε = 0.457

This is an emissivity somewhat less than half what most commonly is claimed for the Earth’s surface.

It is common to claim that emissivity is nearly black body-like, with 0.95 < ε < 0.98.  The claim is also made that like a near black body, the Earth both absorbs and emits a continuum of infra-red radiation in the mid- and far-infra-red ranges characteristic of a black body radiator with a temperature somewhere between 220 and 360K.  It is claimed that the absorptivity and the emissivity are matched as they would be in a black body radiator.  Yet, infra-red spectroscopy in the laboratory on common laboratory FTIR spectrometers show that the absorption spectra of water, minerals, soil, and plant materials are not at all similar to that of a black body.  The very characteristic spectra of these materials are used to identify them or similar materials.  Yet, there are those who claim that the emissivity of water which covers 71% of the Earth’s surface is in this near black body range and that so are the emissivity and absorptivity of most organic materials, such as plant materials.  Such a claim is inconsistent with the NASA Earth energy budget shown in the diagram above.

Now, we will make the assumption that the Earth’s surface characteristics remain the same, so its emissivity remains 0.457, but that we eliminate all of the infra-red active gases, commonly called greenhouse gases.  Now in reality, the emissivity would change under such a condition, but that is a result that we can separate from having clouds and greenhouse gases in the atmosphere.  It is the presence of clouds and greenhouse gases that are said by careless scientists to cause the greenhouse effect and make the Earth’s surface nearly 33K warmer than it would otherwise be.  What would the surface temperature of the Earth really be without these gases?

To find out, we must ask how the solar insolation and any back-radiation from the atmosphere which are absorbed by the Earth’s surface change.  The back radiation issue is clear.  There is none without the infra-red active gases called greenhouse gases.  Now we examine the solar insolation absorbed by the surface.  The 20% reflected by clouds is no longer reflected by the vanished clouds.  The 3% of solar radiation absorbed by clouds is no longer absorbed.  Of the 16% of solar insolation absorbed by the atmosphere, let us suppose that 6% was absorbed by greenhouse gases as a low-ball estimate.  Now the NASA diagram above showed that 51% of the solar insolation was absorbed by the surface and 4% was reflected.  So, without greenhouse gases, the solar insolation incident upon the surface would be about 51% + 4% + 20% +3% + 6% = 84%.  If 4% was reflected from the surface out of 55% incident, then assuming the same reflectivity the amount reflected without so-called greenhouse gases would be (4/55)(84%) = 6%.  The solar insolation power density absorbed by the surface would then be (84 - 6)% = 78%.

The surface temperature can now be calculated for the surface of the Earth under the condition of no so-called greenhouse gases in the atmosphere and with the elimination of clouds.  We now see that this temperature is given by the following equation:

Pabs = (1365.8/4) (0.78) = (0.457) (5.6697 x 10-8 W/m2K4) T4

T = 318.41K

The effect of having no infra-red active gases in the atmosphere is then a surface temperature which is 30.76K higher than the observed average surface temperature of 287.65K!  In other words, the so-called greenhouse gases have a very strong cooling effect upon the Earth’s surface temperature, which is of nearly the same strength as the commonly claimed warming effect on the Earth system as a whole is.  Let us be clear that the claimed 33K warming is actually neither a surface warming of that amount or even an Earth system warming of that amount because the Earth system is not actually a black-body radiator.

The main part of this cooling effect is provided by clouds, but there is also some absorption of incoming solar insolation by infra-red active gases in the atmosphere.  Clouds and the infra-red active gases reduce the radiation absorbed by the surface from about 78% to about 52% and this provides a strong cooling effect.

If the Earth system as a whole has some kind of greenhouse warming based upon a posited black body radiator equivalent model that does not mean that the surface of the Earth is warmer.  In fact, I have just proven that it is actually about 31K cooler than it would be if there were no infra-red active gases in the atmosphere.  Because it is the surface temperature of the Earth that has much the greatest effect upon the fortunes of mankind, it is very deceptive to claim that the Earth is 33K warmer than it would otherwise be because of so-called greenhouse gases such as water vapor, carbon dioxide, methane gas, and nitrous oxide.

The reality is that the infra-red active gases act more like an umbrella providing the Earth’s surface with shade to keep it cool than like a greenhouse to keep it warmer.  It is a much more realistic description of the infra-red active gases to call them shade gases, rather than greenhouse gases.

1 comment:

david russell said...

The SB formula that leads to an emissivity of .47 seems flawed. If the Earth were interfaced directly to outer space, the absorption of solar radiation would be 100%, not 51% (ok, there'd be some reflection, but you can't use 51% because that includes the effect of non-exiting clouds and missing stratospheric ozone shading, etc).

You surely can't use 51%, right?