## Tuesday, October 26, 2010

### Thick Clouds

We have been using our computer program, TEP2 to estimate how much the Earth's surface will tend to warm up if we make certain permanent changes to its atmosphere. In With 660 ppm CO2 we found that doubling the atmosphere's CO2 concentration from 330 ppm to 660 ppm tends to warm the Earth's surface by 1.5°C. In High Clouds we found that the formation of high, thin clouds in a clear sky will tend to warm the Earth's surface by 38°C.

Today we consider the effect of thick clouds. We start with the Earth on a typical clear spring day at latitude 30°N, as represented by our EA1.txt absorption spectra. Our TEP2 program tells us that the total escaping power is 252.5 W/m2. We allow a thick cloud to form from altitude 3 km to 12 km. In Clouds we argued that a cloud this thick will reflect 90% of the Sun's radiation and absorb 100% of the Earth's radiation. We simulate such a cloud by setting the absorption of our 3-km, 6-km, and 9-km layers to 1.0 for wavelengths 5 μm to 60 μm to produce EA4. We apply TEP2 to EA4 and obtain the following output.
`------------------------------------------------      Name    Temp       BB     Layer   Escaping------------------------------------------------   Surface    290.0    401.1    388.8      0.0      0-km    280.0    348.5    262.6      0.0      3-km    270.0    301.3    292.0      0.0      6-km    250.0    221.5    214.0      0.0      9-km    230.0    158.7    152.3    120.8     12-km    220.0    132.8     22.2     10.1     15-km    220.0    132.8     14.4      5.5------------------------------------------------Total:                                   136.5------------------------------------------------`
This output is exactly the same as the one we obtained for high clouds. Both the high clouds and the thick clouds have the same effect upon the total escaping heat. The difference between high clouds and thick clouds is that thick clouds block 90% of the Sun's heat from reaching the Earth, while high clouds block only 10%.

As we did in High Clouds, let us suppose that the Earth radiates as much heat as it absorbs when its surface and atmosphere are as described by EA1. The total power escaping the earth is 252.5 W/m2, so the total power arriving from the Sun must be 252.5 W/m2 also. In the conditions described by EA4, 90% of the heat arriving from the Sun is reflected back into space, so only 25.2 W/m2 is absorbed by the Earth. But the total power escaping the Earth is 136.5 W/m2. When these large clouds form, the Earth will start to radiate 111 W/m2 more than it absorbs. It will cool down. In our TEP2 program, we decrease the temperature of the Earth and its atmosphere until we arrive at the following output.
`------------------------------------------------      Name    Temp       BB     Layer   Escaping------------------------------------------------   Surface    194.3     80.8     76.0      0.0      0-km    187.6     70.2     59.0      0.0      3-km    180.9     60.7     56.4      0.0      6-km    167.5     44.6     40.8      0.0      9-km    154.1     32.0     28.7     22.3     12-km    147.4     26.8      4.5      2.9     15-km    147.4     26.8      2.1      1.1------------------------------------------------Total:                                    26.3------------------------------------------------`
Here we see the total escaping heat is 26.3 W/m2, which is pretty close to the 25.2 W/m2 arriving from the Sun through the thick clouds. The Earth's surface has cooled by 96°C.

According to our simulation, the formation of a 9-km thick layer of cloud at altitude 3 km has a cooling effect sixty-four times stronger than the warming effect of doubling the atmosphere's CO2 concentration.

UPDATE: When we look at discussions like this one about clouds, we see that climatologists recognize the power of clouds to cool and warm the Earth. But we don't see them giving us values in Centigrade in the same way that they do for a hypothetical doubling of CO2 concentration.

1. One immediately feels that the matter is treated with the incisive common sense of an engineer. Chapeau!

Some observations, if you allow.

1) January 6, 2010.
“Carbon dioxide absorbs infra-red light. Infra-red photons are just right for exciting the atomic bonds within the carbon dioxide molecule. After absorbing an infra-red photon, the carbon dioxide molecule vibrates. The molecule is now hotter. That's what heat is: vibration of atoms.”
As far as I know, the kinetic theory of gases tell us that = 3kT/2, i.e. the average (translational) molecular kinetic energy is proportional to the absolute temperature; the rotational and vibrational energies increase heat capacity of a molecule without affect its temperature. Is it correct?

The mechanism of energy absorption/emission requires that there’s always equilibrium between EM that becomes Heat Energy (i.e. molecular KE) and vice versa. At room temperature that’s absolutely true for solids and liquids, but not for gases, wherein EM easily turns into HE which is exchanged in every collision as molecular translational KE and rotational energy, not as vibrational energy because only the translational and rotational degrees of freedom are unfrozen whereas the vibrational ones are still frozen. The energy transformation from KE to (vibrational EM) has a probability more and more lower than the transformation from (vibrational EM) to KE. That means that one can’t apply Plank’s and Kirchhoff's law to CO2 15 microns IR into Earth’s atmosphere (or of another planetary atmosphere) since the planetary temperature is too low to thermally excite the CO2 molecules at the bending resonance frequency needed to emit a 15 microns photon. That’s true for a star, not for a planet.
Hence for any gas at room temperature there wouldn’t be radiative symmetry for the vibrational or higher frequencies, whereas there would be radiative symmetry only for rotational and translational frequencies.
How is it possible to take in account this asymmetry?

3) Work by Convection
I think that potential energy cannot be neglected because of the great variation of altitude between the bottom and the top of troposphere. So both the work done by the air from (4) to (2) and the one done upon the air from (2) to (4) are overestimated of the gravity’s work that is zero in the whole cycle. In other words there works only the buoyancy, naturally if I’m not being wrong.

4) Second law of thermodynamics
The second law must be valid also for RTE.
Energy is a scalar physical quantity that acts in different forms, all equivalent between them. If not constricted (potential energy) it spreads in order to be shared with environment flowing according to the own density gradient. In other words the energy flows in a gradient field and it’s well known that the field lines never intersect, i.e. at any point of the space exists only one vector (the resultant of some presumable components) tangent to it and the effect caused at that point is only one: the vector, if not zero, solely operates in a direction or in the opposite direction.
The EMR energy density, given by Plank’s law, is a decreasing function of the temperature, then the EMR energy can flow only according to decreasing temperatures, as does the heat flow.
So it’s true to claim that the lower (hotter) layer emits towards the higher (colder) layer, but the opposite is wrong because contradicts the second law of thermodynamics.

5) I am waiting for your conclusions with great interest. I would like to suggest, if it’s allowed, to also extend your proof to Venus that seems very different, but has tropopause pressures and temperatures quite analogous to those of the Earth. I feel that everything will go OK.

Regards.

2. Dear Anonymous,

I am greatly honored at the attention you have paid to my posts.

"the rotational and vibrational energies increase heat capacity of a molecule without affect its temperature. Is it correct?"

That is not correct.

Rotation in the context of molecules and heat means one atom rotating with respect to another within the molecule. Vibration is one atom within the molecule moving towards and away from another atom. Translation is the entire molecule moving as one.

We can break up a road surface by dropping a ten-kilogram iron chisel upon it from on high (translational), swinging the chisel at the road on the end of a wood shaft (rotational), or vibrating the chisel tip with a sledge-hammer (vibrational).

Suppose two CO2 molecules collide. One is vibrating vigorously. The other is translating slowly. The vibrating molecule might kick the other with its vibration, sending it shooting off. In that case, the vibrating molecule will go shooting off in the other direction.

With a very large number of molecules and a very large number of collisions, an equilibrium evolves in which vibration, rotation, and translation are all occurring in particular ratios.

But translational movements are more effective at imparting momentum to other molecules . When two molecules collide, they must be moving towards one another, and they will always exchange momentum. But there need not always be an exchange of vibrational or rotational momentum. Translational momentum is very much more effective at imparting momentum to other molecules.

A molecule that moves twice as fast has twice the momentum to impart and bumps into things twice as often. It is four times more effective at imparting momentum. If it is twice as heavy, it has twice the momentum, so it is twice as effective at imparting momentum. Thus the ability of a molecule to impart momentum is proportional to its kinetic energy.

Consider two volumes of gas, one is helium one is CO2. They are at the same temperature and we mix them together. According to the Zeroth Law of Thermodynamics, there is no net heat exchange between them. The helium molecules impart momentum with the same vigor as the CO2 molecules. We try to heat up our mixture. We must increase the translational kinetic energy of both the He atoms and the CO2 molecules by the same amount, so that they will once again be equally vigorous in their ability to impart momentum. But as we increase the CO2 molecule's translational energy, we find that this energy gets re-distributed into its vibrational and rotational movements. We must put more energy into the CO2 molecule to raise its translational kinetic energy than we must put into the He atoms to achieve the same result.

The heat capacity of CO2 is greater than that of He. But energy is not "trapped" in vibrational or rotational modes, nor are these modes "frozen".

Yours, Kevan

3. Dear Anonymous,

There is no such thing as a "frozen degree of freedom". When you say, "there’s always equilibrium between EM that becomes Heat Energy," I don't think this statement has any physical meaning. How would I observe the truth of your statement?

The principle of radiative symmetry does not use "Kirchoff's Law" as I know it, nor "Plank's Law". It uses the Second Law of Thermodynamics, and involves a thought experiment proposed by Kirchoff. No knowledge of rotation or vibration is required.

When you say "the planetary temperature is too low to thermally excite the CO2 molecules at the bending resonance frequency needed to emit a 15 microns photon," you are incorrect. Objects at 20°C absolutely do emit radiation at 15 μm. We can take thermal images and watch them emit such radiation.

Although the average energy of molecules at 20°C is much smaller than the energy of a 15-μm photon, there are always some molecules with very much more energy than others, and when energy is concentrated by chance in a vibration, this vibration can emit a photon and in doing so come to a stop. Conversely, a vibration can start through the absorption of a photon.

Yours, Kevan

4. Dear Anonymous,

I have been thinking about your (3). To raise 1 kg of air 10,000 m we must do 100 kJ of work, which is a lot. If, at the same time, 1 kg of air sinks by 10,000 m, it does 100 kJ of work. So I think you're right: we can't neglect it, but it cancels. All convection cycles rely upon the fact that the downward-going air does the work of lifting the upward-going air.

Yours, Kevan

5. Dear Anonymous,

On your point (4), you say, "So it’s true to claim that the lower (hotter) layer emits towards the higher (colder) layer, but the opposite is wrong because contradicts the second law of thermodynamics."

The Second Law of Thermodynamics refers to the net heat flow between two objects. And "heat" is a form of "energy". Energy, meanwhile is a mathematical construction that has no physical existence. It is a useful concept, but when we start thinking of it as a physical thing, we are on the wrong track. Your energy fields do not, in fact, exist.

A cold layer radiates heat. A hot layer radiates heat. A cold layer can absorb heat radiated by a hot layer. A hot layer can absorb heat radiated by a cold layer.

The Second Law says that there cannot be more heat passing from the cold to the hot than there is passing from the hot to the cold, not unless you have some kind of engine that does work to force the heat to move in such a direction.

So, a cold cloud at 10 km will radiate heat downwards, which will be absorbed by the Earth. Meanwhile, the Earth is radiating heat that will be absorbed by the cloud. You can add up all the heat exchanges, and you'll know that you have done your sums wrong if you get a net heat flow from the cold cloud to the warm Earth.

Yours, Kevan

Yours, Kevan

6. Dear Anonymous,

On your point (5) about Venus: good idea. I will be sure to look at the Venus situation soon.

Yours, Kevan

7. Dear Kevan,

I am sorry. I forgot to sign my previous post because I was in a hurry. I will be careful not to forget again it.
Let me improve some statements.

1) Temperature and heat capacity for gases
I would like to point the topic “The relationship of temperature, motions, conduction, and heat energy” (here)
The empirical relationship between temperature and pressure of a gas, at constant volume, tells us that p = (nR/V)*T; the pressure, by kinetic theory of gases, is the rate of translational momentum exchanged between the gas molecules and a wall of unitary area perpendicularly to the wall, and this rate is proportional to mean translational KE of molecules. Hence also the thermodynamic temperature is only a function of molecules translational motion.
A photon absorbed excites a rotational/vibrational mode of the molecule and can be re-emitted without affecting the temperature of neighboring molecules, because the rotational/vibrational energy is a potential and internal energy. If that don’t occur, The photon is thermalized, i.e. its energy is shared with all other degree of freedom and with the other molecules because the natural tendency of a system is towards a state of equipartition of energy. The temperature raises because is raised the translational KE.

.... continue

8. 2) Frozen and Unfrozen
I try to better explain my thought, specifying that this terminology isn't a my personal definition.
Referring to hydrogen (here), at lowest temperatures, below 100K, its ability to store energy is due only to translational degrees of freedom and the other degrees don’t contribute (they are blocked as if they were “frozen”). Contrarily, above room temperature (for T greater than 300K), the rotational degrees of freedom are fully excited (they are unblocked as if they were “unfrozen”). Thus the rotational contribution to heat capacity approaches its equipartition value, i.e. R per mole.
Idem for vibrational degrees of freedom, but the fully excited temperature is at least about 3000K, for how much I can remember.

I agree with you that “objects at 20°C absolutely do emit radiation at 15 μm” but an object (liquid or solid) has all its degrees of freedom fully “unfrozen” and so there’s the same probability both for the thermalization of the radiative energy and for the opposite conversion of the thermal energy (translational KE), whereas, at the same temperature, the CO2 molecule has its vibrational degrees of freedom fully “frozen”.
You say “although the average” (thermal) “energy of molecules at 20°C is much smaller than the energy of a 15-μm photon, there are always some molecules with very much more energy than others, and when energy is concentrated BY CHANCE in a vibration, this vibration can emit a photon…”
That’s true but the fact that a molecule can emit a 15-μm thermal photon BY CHANCE don’t imply that there’s symmetry in absorption/emission, rather it really shows the contrary.

4) Second Law
The Second Law of Thermodynamics don’t refers only to the heat flow between two objects, as it have an ampler meaning that concerns every form of energy.
Let’s take back the relationship above p = const*T. Both the terms depict an energy density and have units of J/m^3. The “const” must have units of J/(m^3*K) i.e. it’s a volumetric heat capacity. The LHS term measures the energy density in mechanics units, while the RHS term measures it in thermal units.
For energy exchange there’s needed an energy flow and that occurs only if there’s a not-zero grad(p) or a not-zero grad(T) and that is directed along the decreasing p or T.
If you assume that the energy flows in the opposite direction, i.e. along increasing T (you say “a cold layer radiates heat … a hot layer can absorb heat radiated by a cold layer”) then the energy also can flow along increasing pressure. In other words it would be normal that an adrift boat sails upstream the river, or that the outlet flow of a water turbine spontaneously returns to the higher feeding reservoir, and again, that a gas escaped by a cylinder spontaneously re-enters it.
In my opinion all the later three fluid kinematics cases are obviously absurd as the previous thermal case as all they contradict the second law of thermodynamics.

Regards.

Michele

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12. Sorry, excuse me, my PC went crazy.

Michele

13. Dear Michele,

No problem about the extra comments you posted. It seems to me the final three were duplicates. I deleted them, but I will restore them if you ask me to. I hope I did the right thing. I look forward to reading your comments in more detail and answering them.

Yours, Kevan

14. Dear Michele,

This time I think I am understanding you better. I will point out the few places where I think we may disagree.

You say, "A photon absorbed excites a rotational/vibrational mode of the molecule and can be re-emitted." I suppose it must be possible for the photon to be re-emitted. If such a re-emission were common, we would see cold CO2 glowing with 15-μm photons when we illuminate it with 15-μm photons. But that is not the case. Instead, we see a 10-cm thickness of pure CO2 absorbing almost all 15-μm photons with no glow of 15-μm photons coming out the other side.

See the following post for the pure CO2 spectrum:

http://homeclimateanalysis.blogspot.com/2010/02/co2-spectrum.html

I refer to the source of the spectrum, and you can take a look at the apparatus they used to measure absorption length.

The re-emission that you speak of must be very rare indeed. Almost every photon absorbed remains absorbed, and its energy becomes heat.

I know what you mean when you say, "Referring to hydrogen (here), at lowest temperatures, below 100K, its ability to store energy is due only to translational degrees of freedom" It is not possible to put just any small amount of energy into a vibration. Only certain exact quantities of
energy will be accepted, and at 100 K the chance of an adequate quantum of energy entering the H2 vibration are vanishingly small. It is here that my description of molecules as little lumps connected by springs starts to break down.

So I think I know what you mean by "frozen". But if the H2 vibration was really "frozen", it would be unable to absorb a photon when it is at 100 K. So far as I know, however, H2 at 100 K will absorb a 161.4-nm photon just as easily as H2 at 1000 K.

Yours Kevan (more to follow)

15. Dear Michelle,

I see what you mean about air not going back into the cylinder, and I of course agree with you. But suppose we had a cylinder that was pushing air out at an average speed of 1 m/s. The pressure in the cylinder is higher, and the air is moving out. But the air molecules are moving at hundreds of meters per second. Many of them are indeed re-entering the cylinder. Many are leaving. On average, however, more are leaving than entering, and so your statement is true when you look at the entire volume of air. But it is not true for individual air molecules. These molecules are more likely to leave the cylinder, but not certain to leave.

In the case of the Earth and a cold cloud layer, there is more heat passing from the Earth to the cloud layer. But the clouds are radiating heat as well, and some of this heat is absorbed by the Earth.

In this case, therefore, it seems to me that we do not, in fact, disagree.

Yours, Kevan

16. Dear Kevan,

Only tow points.

“But the air molecules are moving at hundreds of meters per second. Many of them are indeed re-entering the cylinder. Many are leaving.”
A gas at rest is only macroscopically at rest as the molecules keep on their fluctuations around the center of mass of the particle that includes them. Indeed, the vectorial sum of the all velocities relative to the center of mass of the particle, is statistically zero because the velocities, also if chaotic, are really isotropic. The agitation motion of molecules only causes the local equipartition of macroscopic characteristics as pressure, density, temperature (in practice of energy density) within the particle, and it don’t cause any transfer of the energy or of the mass. The macroscopic transfer of mass (mechanic energy) occurs only if within two different parcel of the gas the pressures are different. With different temperatures there’s always a transfer of thermal energy and upon given conditions also of mass. This transfer is measurable (and indeed exists for real world) only at macroscopic scale.

“ … if the H2 vibration was really "frozen", it would be unable to absorb a photon when it is at 100 K”
Also at lowest temperatures a molecule is at its ground level of vibrational energy. So, if a photon interacts with the molecule, their EM fields don't have any difficulty to vectorially add each other bringing the molecule at its first energetic level, and that is always possible. Conversely, the excitation of the vibrational mode by collisions requires that the force due to rate of momentum change during the collision is enough strong for the axial stiffness of the molecule (i.e. that the temperature is sufficiently high).
I feel that the first way of excitation is very more probable than the second.

Regards, Michele