## Tuesday, April 21, 2015

### Eli and the Merry Elves

Some time ago, Eli and his merry elves  put together a lengthy comment on an even more lengthy paper (aka piece of trash) by Gerhard Gerlich and Ralf Tscheuschner, that being a paper so bad that it really was not worth the work, except the work the merry elves did was a piece of play.

Now the Rabett is quite happy with the project. It was maybe the first published blog generated reply to such nonsense (thus the grandfather of the 97% paper), and even happier about those who took part, some of whom blog, some of whom tweet and blog to this day, Chris Ho-Stuart, Chris Colose, Joel Shore, Arthur Smith and Joerg Zimmerman.

A major part of the comment was showing that absorbing layer models of the atmosphere lead to a warmer surface, in perfect agreement with the second law of thermodynamics.  What happens, of course, is that each absorbing layer re-emits IR radiation, a part of which is absorbed by the layer below.  This slows the rate at which the lower level cools by radiation.  If the lowest level is heated by an outside source (such as the sun) and an equilibrium is established so that the energy into the system matches that of radiation from the system, then the temperature of the lowest level at equilibrium is higher than it would be in the absence of absorbing layers.

Of course, this did not meet with understanding amongst the lard heads, and Eli ran into it again recently on Bishop Hill.  Curiously Chris Colose has been thinking about the problem too and has a couple of recent  posts on the subject.

Eli's introduction to thermal radiation shielding was building very high temperature ovens (> 1200K) with multiple levels of radiation shielding during his graduate research, so, on an experimental level the answer was clear, but today while searching the net he came across a book on radiative transfer by Robert Siegel which considers the problem in detail starting with parallel piles of heat shielding layers which emit diffusely (e.g. the same in all directions)

in really complete detail.  The model includes different emissivities for the inside and outside walls of each shielding level.  Eli is not going to go full SoD on the bunnies, but those interested can find a detailed derivation of the heat flow per unit area between two parallel plates in just about any book on thermal transfer, or you can corner John Abraham at the next AGU.  When a steady state is established the amount of heat flowing per unit area through each level q must be the same

$\begin{matrix}&space;q\left&space;(&space;\dfrac{1}{\epsilon&space;_{1}}&space;+&space;\dfrac{1}{\epsilon&space;_{11}}-1\right&space;)&&space;=&space;&&space;\sigma&space;\left&space;(&space;T_{1}^{4}-&space;T_{s1}^{4}\right&space;)\\&space;q\left&space;(&space;\dfrac{1}{\epsilon&space;_{12}}&space;+&space;\dfrac{1}{\epsilon&space;_{21}}-1\right&space;)&&space;=&space;&&space;\sigma&space;\left&space;(&space;T_{s1}^{4}-&space;T_{s2}^{4}\right&space;)\\&space;&&space;\begin{matrix}&space;\cdot&space;\\&space;\cdot&space;\\&space;\end{matrix}&space;&&space;\\&space;q\left&space;(&space;\dfrac{1}{\epsilon&space;_{\left&space;(N-1&space;\right&space;)}}&space;+&space;\dfrac{1}{\epsilon&space;_{N1}}-1\right&space;)&&space;=&space;&&space;\sigma&space;\left&space;(&space;T_{s\left&space;(&space;N-1&space;\right&space;)}^{4}-&space;T_{sN}^{4}\right&space;)\\&space;q\left&space;(&space;\dfrac{1}{\epsilon&space;_{N2}}&space;+&space;\dfrac{1}{\epsilon&space;_{2}}-1\right&space;)&&space;=&space;&&space;\sigma&space;\left&space;(&space;T_{sN}^{4}-&space;T_{2}^{4}\right&space;)\\&space;\end{matrix}$                   (1)

Following Siegel, if we add these equations up, the right hand side is σ(T14-T24).  Dividing by the co-factor of q on the left hand side yields

$q=\dfrac{\sigma&space;\left&space;(&space;T_{1}^{4}-T_{2}^{4}&space;\right&space;)}{1/\epsilon&space;_{1}+1/\epsilon&space;_{2}-1+\sum&space;\left&space;(&space;1/\epsilon&space;_{n1}+1/\epsilon&space;_{n2}-1&space;\right&space;)}$                     (2)

Heat transfer books usually stop there, because the MEs are interested in how to design shielding for thermal or cryogenic applications.

OTOH, Rabett and friends were looking at the case of a planet where the amount of incoming energy from the Sun or the star of your choice is q.  The emissivity of the surface is going to be something like 0.95, that of the atmosphere at different levels, well that depends on the pressure, concentration and spectra of greenhouse gases, and, of course the specific humidity and where the clouds are.  For CO2 the contribution is going to be between 0.19 and 0.12.  For water vapor higher, as high as water vapor goes before condensing out

However, we can gain insight by setting ε1 equal to 1 and letting all of the other levels have the same emissivity, both inside and outside each shielding level.  In that case

$q=\dfrac{\sigma&space;\left&space;(&space;T_{1}^{4}-T_{2}^{4}&space;\right&space;)}{1/\epsilon&space;_{2}+N\left&space;(&space;2/\epsilon&space;_{s}-1&space;\right&space;)}$                                                           (3)

At a steady state, the same amount of energy has to be radiated to space.  If there are no shielding levels, the amount of heat radiated per unit time is σT1o4.  Consider the case where there is only the outermost heat shield (N=0) then

$\sigma&space;T_{1o}^{4}=\dfrac{\sigma&space;T_{1}^{4}-\sigma&space;T_{2}^{4}}{1/\epsilon&space;_{2}}$                                                                      (4)

Canceling σ, multiplying both sides by 1/ε2 and bringing T1o4 to left hand side we get

$\left&space;(1/\epsilon&space;_{2}&space;\right&space;)&space;T_{1o}^{4}+&space;T_{2}^{4}&space;=&space;T_{1}^{4}$                                                                   (5)

All terms on the left hand side are positive, ε2 is less than or equal to 1, therefore T1, the temperature where there is one heat shielding level is greater than T1o, the temperature of the surface if there is no blocking.

If there are N equivalent heat shielding layers between the innermost and outmost layers, then similarly

$\left&space;(1/\epsilon&space;_{2}+N\left&space;(&space;2/\epsilon&space;_{s}-1&space;\right&space;)&space;\right&space;)&space;T_{1o}^{4}+&space;T_{2}^{4}&space;=&space;T_{1}^{4}$                                   (6)

The added term on the left hand side is again positive (if εs =1 then it is simply equal to N.  If εs  < 1 then (2/εs -1) > 1.  In either case, especially the latter, T>  T1o . The same can be done for  spherical geometries, but one has to consider geometric factors, the ratios of the areas of the various shells to each other.

Siegel and other heat transfer books do the derivation.

How important are the geometric factors?  They scale as An/Ao where A=4πR2 so at the risk of offending the punctilious the ratio is (Rn/Ro)2 The radius of the earth is 6371 km.  Using a 10 km high atmosphere basically the troposphere, or at least the effective level which radiates to space in the CO2 bands,  (Rn/Ro)2 = (6381/6371)2 = 1.003, so there will be a .3% difference from treating the system as a nest of sphere's or a series of parallel plates.  Close enough.

Fernando Leanme said...

Lets see if you get this:

You are on a high speed boat traveling between Key West and the nearest spot on the Cuban coast. The boat speed is 30 mph, wind speed is 10 mph in your face. You are holding a can of beer in your hand. The beer temperature is 4 degrees C. What's the beer temperature after 30 minutes, and which way is the wind blowing?

EliRabett said...

37 C. Eli drank the beer.

Next.

Pinko Punko said...

Haha. That was going to be my comment, that Eli drank the (carrot?) beer.

Oale said...

not going into maths, but would it be simplest to just say warmer water vapor expands more thus tropopause rises giving a thicker insulating layer, basically the same effect as in turkish sauna after releasing more steam in it.

EliRabett said...

Oale, it's a bit more complicated than that. The effect is driven by the raising of the level at which radiation to space is possible. Since water vapor does not rise very high because of condensation, the level at which it radiates to space is much lower and warmer than the level at which CO2 radiates to space. Take a look at the link.

John said...

Thanks, Eli.

Let me add my two cents. This is NOT a heat engine. Or even "like" a heat engine.

G&T (gin and tonic) think that it is, but they're wrong.

John Garland said...

Is there condensation on the can? What is the relative humidity?

Brian Dodge said...

Oale, there is also another effect due to the difference between CO2 and H20 behavior. Since CO2 is noncondensing (despite what you may have heard at WTFUWT), ~1/4 of the CO2 in the atmosphere is above the tropopause, whereas the H2O vapor concentration at the tropopause has fallen to << 1%(lapse rate + clausius-clapeyron). The back radiation from the mass of CO2 above the tropopause keeps the troposphere warm enough to hold the total column water vapor and actual average lapse rate between fully moist and dry adiabatic that is observed.

Here's a question - why is the temperature at the tropopause lower near the equator than at the poles?
see http://weather.uwyo.edu/cgi-bin/sounding?region=naconf&TYPE=GIF3ASTUVE10&YEAR=2015&MONTH=04&FROM=2212&TO=2212&STNM=04270
versus http://weather.uwyo.edu/cgi-bin/sounding?region=naconf&TYPE=GIF%3ASTUVE10&YEAR=2015&MONTH=04&FROM=2212&TO=2212&STNM=78762

Hank Roberts said...

hm, height of the tropopause decreases as you move toward the poles.

Russell Seitz said...

Warmer than Fernando thinks: he neglected to boost the beer can's Prandtl number to compensate for the velocity vector of the Gulf Stream.

lithophyte said...

John,

I think it is a heat engine, but not in the component that Eli has outlined here (the absorptive component). The radiative components are linearly additive. Once the energy gets into the ocean the coupled atmosphere-ocean system is working like a heat engine. Hence the high non-linearity of warming.

Russell Seitz said...

Lithophyte is invited to calculate the partition of the CO2 in the beer between Eli. the atmosphere, and the Gulf Stream

Extra credit for correcting for the corresponding differences in 18 O and 13C ratios.

Brian G Valentine said...

So why do you continue to "debunk" Gerlich and Tseuschner, Eli?

In your Weltanschauung, isn't this tantamount to repetitive debunking of ESP,Area 51, Chemtrails, Scientific Creationitism, and assorted other lunacy?

"When a steady state is established the amount of heat flowing per unit area through each level q must be the same ..."

What "steady state" do you refer to? Weren't G&T careful to explain that q = (sigma)T**4 applied uniquely to the equilibrium radiation of the Holraum?

Do you think God set forth the exponent 4 to apply to "thermal radiation"?

That, and your Sisyphus efforts to defeat an argument you have yet to accomplish, remain mythical.

Brian G Valentine said...

Hohlraum my error.

EliRabett said...

Brian, fwiw, this post came out of the usual @ Bishop Hill. Eli is a nice bunny and he won't send you over there, but there is a Bryan. . .

Anyhow, the point of this is eq. 5, which, AFAEK, is not a very common, but a very straightforward way of showing how external Hohlraumen raise the temperature of internal Hohlraumen

and no, G&T, well,

DaveScot said...

Kirchoff's law says good absorbers are good emitters. CO2 impedes cooling below emission altitude and enhances it above. Thus it still all boils (pun intended) down to hydrologic cycle and how that accelerates (or not) transfer of heat from ocean surface to the (now) more efficient emission altitude. Lapse rate feedback (negative and not well enough measured or modeled to say how large it is) negates to some extent the contention that the emission level is colder than it would be at lower CO2 ppmv. A smaller lapse rate means the air column is warmer at any given altitude.

The fact of matter is that climate science is trying to find a needle in a haystack. Signal to noise ratio is too small for reaching any credible conclusions.

Thanks for playing.

DaveScot said...

The wind is to our back at 20mph and and the beer temperature is of course 101F which is a bunny's normal body temperature.

Interesting that the beer begins at the average temperature of the global ocean, by the way. Nice touch there.

Fernando Leanme said...

Dave, you get A+. I had a bonus question similar to that in a heat transfer final. Most of the guys were so wiped out by the exam itself they failed to answer the question. Some of them bogged down trying to derive the heat transfer equation for a human hand holding a can of beer, and crazy stuff like that. But that exam had a really tricky question. The sob posed a heat transfer problem for triangular lead pipes.