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Location: The Hague, The Netherlands | DeWitt and Madmartigan: please only talk physics, starting here. |
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Location: Tennessee, USA | Jan Pompe - 26/9/2007 17:47
A small point to both of you: if you want calculate the heating effect of surface radiation (ΔT)/m^2 for a column of air you'll need to work out the integral of the energy absorbed by the CO2 over the distance to extinction, since there will be a gradient. I think DeWitt is right that the mixing is (order of nano-seconds so almost) instantaneous and that the sensitivity calculation should include the heat capacity all the atmospheric components I don't think though that we need to nor can include the whole 200km column but just to the point of extinction. Have fun gentlemen.
Considering that forcings from ghg's are calculated at the tropopause, IIRC, all the action is in the lower ~15 km (less at the poles because the tropopause is at lower altitude there). The extinction from CO2 in the 15 micrometer band can be crudely approximated at current concentration by assuming a value of 1 (opaque) from 13.5 to 17 micrometers and zero everywhere else ( Petty, p275). That band, though, only represents about 10% of the total absorption. About 10% of the surface LW radiation is not absorbed at all and passes directly to space. The other 80% is absorbed by water vapor. Water vapor concentration falls off exponentially with altitude much more rapidly than CO2 (scale height ~2 km compared to about 8 for dry air), so emission from water vapor to space happens at lower, warmer altitudes.
Assume the tropopause is at a pressure of 200 mbar. So 80% of the atmosphere is below the tropopause or 8,000 kg m-2. Ignoring water vapor, the heat capacity of air is 1004 J kg-1 K-1. Using the K &T figure of 390 W m-2 emitted from the surface, then 39 W m2 is absorbed by CO2 in the troposphere. Multiplying that out gives 0.42 K/day heating. Of course you don't actually get the heating because the CO2 is also emitting in all directions, enough of which escapes to space to balance the energy absorbed. Even if you include water vapor, the heating in the absence of radiative cooling (which would be a violation of Kirchhoff's Law) would be 3.8 K/day. But if you include water vapor, the heat capacity is higher so the heating is less.
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Location: Tennessee, USA | Madmartigan - 25/9/2007 22:53
The assumption that we have 168 W/m^-2 of energy incoming from the Sun while we have a lost of energy of 492 W/m^-2 is a violation to the first law of thermodynamics.
The equilibrium is not possible because Earth is not an isolated system. How could it be that a colder surface (the atmosphere) is not warmed up by a warmer system (the surface) if there is a gradient of temperature? It would be a violation to the second law of thermodynamics. The atmosphere is always emitting radiation to other colder systems, so the atmosphere is always colder than the surface because of convective energy flow, how could you obtain the thermal balance so the heat emitted by the surface was not transmitted to the atmosphere and prevent the surface to be cooled by the air that descends from the upper layers of the troposphere? It is a violation of the second law. When there is a gradient of temperature, the heat flows from the warmer system to the colder system. The colder system acquires energy from the warmer system until the balance occurs. However, the atmosphere is an always moving system, it's not static, so the convection is always occurring and the surface is cooled by any colder system towards which the energy can be dispersed.
We have 168 W m-2 sunlight absorbed by the surface and 324 W m-2 absorbed by the surface from LW IR emitted downward by the atmosphere. The total, 492 W m-2, is equal to the energy lost from the surface.You can have a temperature gradient with no heat flow. I was confused by that for a while when I first started to look at all this also. But it's Energy that flows from higher to lower. At constant height and pressure, that's from warmer to colder. But the pressure of the atmosphere decreases with height and the gravitational potential energy increases. Lift a kg of air from the surface to 10 km. That's 98,000 J of work (g * z). That has to come from somewhere. Where it comes from is the kinetic energy of the air when it is adiabatically expanded. For dry air with a heat capacity of 1004 J kg-1 K-1, that's 98 K lower. Adiabatic expansion is a good approximation because the thermal conductivity of air is low. Conversely, if you lower a kg of air from 10 km to the surface, it is compressed and and the temperature goes up. But the gravitational potential energy is lowered by the same 98,000 J. If the atmosphere were isothermal, then the kg of air lifted to 10 km would be colder and denser than the air around it and it would fall back towards the ground. As far as convection, over large scales hydrostatic equilibrium can be assumed. Averaging over a horizontal area of, say, 100 km2, the up and down motions cancel out. Once again, on scales greater than ~ 10 km, the atmosphere remains in hydrostatic equilibrium. Overall, it can safely be stated that atmospheric motions with horizontal scales greater than 10 km are always in hydrostatic equilibrium.
All the math behind this is here.
Edited by DeWitt 26/9/2007 20:37
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Location: Sydney Australia | DeWitt
Considering that forcings from ghg's are calculated at the tropopause, IIRC, all the action is in the lower ~15 km
Why are you going off at a tangent? Aren't you and Madmartigan trying to calculate the amount of heating one can expect in a column of air above the surface due to radiation? In the time taken for a photon, that actually makes it to the point of extinction, to get there there won't be much convection going on so that can be ignored for the moment. Of course only that portion that GHGs can absorb only need to be considered and bear in mind the net radiation will only be from warmer to cooler and proportional to the difference To^4 - Ti^4. Once you get this sorted out you can start to think about convective effects and also what effect doubling CO2 would have at this level. |
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Location: The Hague, The Netherlands | Jan, IMHO Infrared radiation doesn't directly heat the atmosphere, because it are only the trace gases that absorb infrared .
Edited by coolhansnl 27/9/2007 01:58
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Location: Sydney Australia | Hans,
Jan, IMHO Infrared radiation doesn't directly heat the atmosphere, because it are only the trace gases that absorb infrared.
So are you saying this is correct?
? T = q/m (Cp) = 359.6 (J/s) / 0.00069 Kg (842 J/Kg*K) = 618.95 K/s
and there is some very hot CO2 mixed in with the cooler gases? How long do you think such a thermal in-equilibrium would last? |
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Location: The Hague, The Netherlands | Just one collision, there are sufficient cool O2 and N2 and Ar molecules around: approx 999620/380 |
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Location: Sydney Australia | Hans
So the heat absorbed by the trace gases is fairly quickly spread throughout the surrounding gases thus warming them. No? |
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Location: The Hague, The Netherlands | how much? |
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Location: Tennessee, USA | coolhansnl - 27/9/2007 01:58
Jan, IMHO Infrared radiation doesn't directly heat the atmosphere, because it are only the trace gases that absorb infrared .
I think you are correct in a global annual mean model, but not for the reason stated. You don't get net atmospheric heating because there is radiative balance at every altitude. As much energy is emitted as absorbed. Kirchhoff's Law again. Greenhouse warming happens because there is a temperature gradient causing gross emission/absorption from a non-condensing gas like CO2 to decrease with altitude and causing the partial pressure of water vapor to decrease faster than the total pressure as well as lowering the emissivity. That means each layer emits and absorbs less as altitude increases. That allows the effective brightness temperature of the Earth observed from space to be 255 K calculated from total LW energy emitted of 235 W m-2 while emitting 324 W m-2 downward at the surface. If the troposphere were isothermal, there would be no greenhouse warming because the top and bottom of the atmosphere would emit the same amount of radiation and the average surface temperature would be much lower. Probably even lower than 255 K because the oceans would freeze, drastically raising the albedo. The ice ball Earth scenario is supposed to have actually happened at one point. The exact extent of freezing of the oceans is in dispute.
Edited by DeWitt 27/9/2007 19:22
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Location: Tennessee, USA | Jan Pompe - 26/9/2007 20:50 DeWitt Considering that forcings from ghg's are calculated at the tropopause, IIRC, all the action is in the lower ~15 km
Why are you going off at a tangent? Aren't you and Madmartigan trying to calculate the amount of heating one can expect in a column of air above the surface due to radiation? In the time taken for a photon, that actually makes it to the point of extinction, to get there there won't be much convection going on so that can be ignored for the moment. Of course only that portion that GHGs can absorb only need to be considered and bear in mind the net radiation will only be from warmer to cooler and proportional to the difference To^4 - Ti^4. Once you get this sorted out you can start to think about convective effects and also what effect doubling CO2 would have at this level.
I don't think I'm going off on a tangent. I'm trying to get to the point where all parties agree on the terms used in the debate/discussion. There isn't a great deal of emission/absorption above 15 km. Using the Archer MODTRAN interface and the 1976 atmosphere locality with no clouds or rain LW IR upward is 261.625 W m-2 at 70 km and 258.799 W m-2 at 15 km. Downward IR is 15.1 W m2 at 15 km and 0.05 W m-2 at 70 km. Under these conditions the tropopause is at about 11.5 km, defined as the point where the temperature stops declining, and the temperature of CO2 emission upward in the 15 micrometer (667 cm-1) band at altitudes above the tropopause is slightly less than 220 K also corresponding to the temperature at the tropopause. Part of this is because the lower stratosphere is isothermal, so no greenhouse there. The rest is because the pressure is so low so there's not much there to emit or absorb. The upward emission is more than 235 W m-2 because there are no clouds. I do need to move the altitude up, though. According to the MODTRAN results, the tropopause is at 18 km in the tropics, on average, so 20 km instead of 15.
Edited by DeWitt 27/9/2007 19:35
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Location: Sydney Australia | Hans
Not much but 32 watts worth for CO2 and 93 watts worth for H2O (Charnock and Shine). To arrive at a temperature change we would have to divide that figure by the heat capacity of the atmosphere up to the point of extinction (Jack Barrett's estimate 100m).
Where does that leave this? The surface can warm only the CO2 and H2O molecules by radiation (as N2 and O2 don't absorb infrared), so direct radiation has a minimal effect on temperature change of the total atmosphere mass (CO2 and H2O being trace gases).
the N2 and O2 don't need to able to absorb infrared as the obtain rather quickly that energy (125 joules/m^2s) from those that can.
DeWitt
I don't think I'm going off on a tangent.
Maybe not but you are trying to build your house by erecting the roof before you lay the foundations. Before we can be worrying about heat radiating from the atmosphere we need to get the heat there in the first place. This does not occur at the tropopause but at the surface and most of that at the first few metres. Lets build this house brick by brick. What bothers me about the Keihl and Trenberth paper is that you have 235 watt incoming and 235 watt outgoing and 324 watts going the rounds between GHG and the surface - this looks like a perpetuum mobile to me. I'd like to find out if it is. |
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Location: Tennessee, USA | Maybe not but you are trying to build your house by erecting the roof before you lay the foundations. Before we can be worrying about heat radiating from the atmosphere we need to get the heat there in the first place. This does not occur at the tropopause but at the surface and most of that at the first few metres. Lets build this house brick by brick. What bothers me about the Keihl and Trenberth paper is that you have 235 watt incoming and 235 watt outgoing and 324 watts going the rounds between GHG and the surface - this looks like a perpetuum mobile to me. I'd like to find out if it is.
I didn't really intend this to be a complete explication of the radiative energy transfer part of enhanced greenhouse theory from first principles. I was replying to Nasif/Madmartigan's first point posted at Climate Audit and copied here by Hans in the opening post: "1. A colder system heats up a warmer one." For that, K & T is relevant. The heat comes from sunlight absorbed by the surface and the atmosphere totaling 235 W m-2. That energy can only leave by radiation to space. If it doesn't, it gets warmer. So radiation from the top of the atmosphere is fixed, more or less. It's the more or less that is important. In fact, current measurements have the difference between in and out at less than 1 W m-2. So, IMO, it's irrelevant whether you start at the top and work down or the bottom and work up.
I'm pretty sure that more than the first few meters are involved in the initial absorption. I've heard that, but never seen the original reference. Do you have it? I used to think that was reasonable until I started to play around with multilayer radiative models and found that for a gray atmosphere, the total absorption/emission of thermal IR in a 500 meter layer has to be quite small, less than 15%, for realistic values of temperature. Using Archer MODTRAN, at 0.5 km looking down, you can barely detect the beginnings of saturation of the 667 cm-1 band of CO2 (1976 std. atm., otherwise default). The rest of the emission spectrum is indistinguishable from the surface emission at 0 km height. The net loss of energy is only 1.5 W m-2 out of 360. I can post the graphs, but I recommend that you do it yourself. That helped me a lot. Particularly, watch the temperature of emission of the CO2 band as height increases. It gets colder and colder until you get to the tropopause and then doesn't change. That lost radiant energy warms the surface enough to emit sufficient energy in the rest of the spectrum to balance the equation. So CO2 is an effective ghg because it's main absorption band is saturated.
Edited by DeWitt 28/9/2007 01:02
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Location: Tennessee, USA | Everyone, One of my many character flaws is my tendency to be unaware of a shift into what someone referred to elsewhere as 'egotistical jerk' mode, or 'I'm smarter than you so I'm right and you're wrong'. Please kick me (rhetorically) if that happens again. Jan, What bothers me about the Keihl and Trenberth paper is that you have 235 watt incoming and 235 watt outgoing and 324 watts going the rounds between GHG and the surface - this looks like a perpetuum mobile to me. I'd like to find out if it is.
The net flow is in the right direction. The surface emits 390 W m-2 and the atmosphere only returns 324 W m-2. So the energy is there and the surface radiation plus absorbed incoming solar radiation heats the atmosphere, not the other way around. An analogy might be an equilibrium constant k = kAB/kBA for the reaction A ↔ B. (I really need the left and right arrow on top of each other to show that A goes to B and B goes to A.) The reaction rate constants, kAB and kBA and hence fluxes in each direction can be quite high, but there is no net change in concentration as long as the ratio of the concentrations of A and B is equal to the equilibrium constant. You can feed A into the system and remove B at the feed rate of A and nothing changes. But the conversion rates of A to B and B to A can be orders of magnitude higher than the feed rate of A. Does that help?
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| DeWitt - 28/9/2007 01:54
You don't get net atmospheric heating because there is radiative balance at every altitude. As much energy is emitted as absorbed. Kirchhoff's Law again.
That is not true. We have also convections moving energy, not just radiation.
If you think of a layer in atmosphere. It gets radiation from below and fom above. It also radiates according its temperature. But the radiation is not in balance. There are convections that move heat up and down. If the layer get more heat by convection than it looses, it heats up. And that means that it radiates more than it gets radiation. And, of course, if the layer gets less ...
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Location: Tennessee, USA | aha - 28/9/2007 17:02 DeWitt - 28/9/2007 01:54
You don't get net atmospheric heating because there is radiative balance at every altitude. As much energy is emitted as absorbed. Kirchhoff's Law again. That is not true. We have also convections moving energy, not just radiation. If you think of a layer in atmosphere. It gets radiation from below and fom above. It also radiates according its temperature. But the radiation is not in balance. There are convections that move heat up and down. If the layer get more heat by convection than it looses, it heats up. And that means that it radiates more than it gets radiation. And, of course, if the layer gets less ...
Kirchhoff's Law (Wikipedia is as good a source as any for this) requires that each layer of the atmosphere be in radiative balance. Sensible (convection) and latent (evaporation/condensation of water vapor) heat transfer affect the temperature which controls the absorption and emission of radiation.
Net vertical convective heat transfer is mainly in the boundary layer which is considered to be well mixed. The thickness of the boundary layer varies a lot. It can be as little as 10 m on a calm night and as thick as 2 km on a hot, windy day. It also varies with the surface roughness. On scales over 10 km, the atmosphere is considered to be in hydrostatic equilibrium with convection up and down balanced. Convection parallel to the surface does not violate hydrostatic equilibrium and in fact is required to move some of the excess heat absorbed in the tropics to higher latitudes where there is a deficit. The ocean currents move the rest. Also, the K&T energy balance includes convective heat transfer from the surface (24 W m-2). Most of the incoming 168 W m-2 absorbed by the surface is transferred to the atmosphere by latent heat transfer (evaporation and condensation of water vapor, 78 W m-2) and net radiation (66 W m-2).
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Location: Tennessee, USA | Jan,
Here's another way of looking at the 324 W m-2 moving back and forth between the surface and atmosphere. Construct a cube with the inside walls having an emissivity/absorptivity of 1. If the walls are at a temperature of 288 K, each wall emits 390 W m-2 (Stefan-Boltzman). Yet the walls won't cool because their emission is matched by absorption of an equal amount of radiation emitted from the other walls. The calculated global annual mean brightness temperature of the atmosphere looking up from the surface is 274 K. So when the sun goes down, the surface cools. On a calm night, sensible and latent heat transfer go away until the dew or frost point is reached. Then there is latent heat transfer to the surface.
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| DeWitt - 28/9/2007 08:01
I'm pretty sure that more than the first few meters are involved in the initial absorption.
It debends on the wave number. At about 670 1/cm a 100 m layer absorbs like 95% while at 1000 1/cm almost all surface radiation gets through.
Using Archer MODTRAN, at 0.5 km looking down, you can barely detect the beginnings of saturation of the 667 cm-1 band of CO2 (1976 std. atm., otherwise default). The rest of the emission spectrum is indistinguishable from the surface emission at 0 km height. The net loss of energy is only 1.5 W m-2 out of 360.
You can not calculate the difference like that. The radiation that you see looking down is not coming only from the surface, but also from the GHG in the 0.5 km layer.
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Location: Tennessee, USA | aha - 28/9/2007 17:51 DeWitt - 28/9/2007 08:01
I'm pretty sure that more than the first few meters are involved in the initial absorption. It debends on the wave number. At about 670 1/cm a 100 m layer absorbs like 95% while at 1000 1/cm almost all surface radiation gets through. Using Archer MODTRAN, at 0.5 km looking down, you can barely detect the beginnings of saturation of the 667 cm-1 band of CO2 (1976 std. atm., otherwise default). The rest of the emission spectrum is indistinguishable from the surface emission at 0 km height. The net loss of energy is only 1.5 W m-2 out of 360. You can not calculate the difference like that. The radiation that you see looking down is not coming only from the surface, but also from the GHG in the 0.5 km layer.
I know the energy difference number is not really indicative, but the shape of the spectrum should be. As you move up, the emission spectrum changes to reflect the temperature of the layer where saturation at each wavelength occurs. At 500 m you can see the dip from the 15 micrometer band, so it saturates at less than that. 95% absorption at 667 cm-1 for a path length of 100 m seems reasonable. But that represents no more than 10% of the total energy. Given that ~10% of the emitted radiation isn't absorbed at all (global annual mean, probably not true in the tropics). What's the optical depth where the other 90% of the emitted radiation from the surface falls to 1/e of it's initial value? I would be surprised if that actually turns out to be "a few meters" as stated above.
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| DeWitt - 29/9/2007 00:30
aha - 28/9/2007 17:02 DeWitt - 28/9/2007 01:54
You don't get net atmospheric heating because there is radiative balance at every altitude. As much energy is emitted as absorbed. Kirchhoff's Law again. That is not true. We have also convections moving energy, not just radiation. If you think of a layer in atmosphere. It gets radiation from below and fom above. It also radiates according its temperature. But the radiation is not in balance. There are convections that move heat up and down. If the layer get more heat by convection than it looses, it heats up. And that means that it radiates more than it gets radiation. And, of course, if the layer gets less ... (Wikipedia is as good a source as any for this )
Thank you, but I already know Kirchhoff's Law. Your link quite correctly states:
"At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity"
It does not say
Kirchhoff's Law requires that each layer of the atmosphere be in radiative balance.
which would not be true.
Sensible (convection) and latent (evaporation/condensation of water vapor) heat transfer affect the temperature which controls the absorption and emission of radiation.
(I am used to refer with convection to sensible and latent heat together, since water vapour moves with the other gases.)
Don't you see that the two quotations are inconsistent? How can convection (sensible and latent heat) affect the temperature, if radiation balance already defines it?
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A debate between Hans Erren and Nasif Nahle
