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## A Challenge for Bryan

Bryan needs no introduction on this blog, but if we were to introduce him it would be as the fearless champion of Gerlich and Tscheuschner.

Bryan has been trying to teach me some basics on heat transfer from the Ladybird Book of Thermodynamics. In hilarious fashion we both already agree on that particular point.

So now here is a problem for Bryan to solve.

Of course, in Game of Thrones fashion, Bryan can nominate his own champion to solve the problem.

Case A

Spherical body, A, of radius ra, with an emissivity, εa =1. The sphere is in the vacuum of space.

It is internally heated by a mystery power source (let’s say nuclear, but it doesn’t matter), with power input = P.

The sphere radiates into deep space, let’s say the temperature of deep space = 0K to make the maths simpler.

1. What is the equation for the equilibrium surface temperature of the sphere, Ta?

Case B

The condition of case A, but now body A is surrounded by a slightly larger spherical shell, B, which of course is itself now surrounded by deep space at 0K.

B has a radius rb, with an emissivity, εb =1. This shell is highly conductive and very thin.

2a. What is the equation for the new equilibrium surface temperature, Ta’?

2b. What is the equation for the equilibrium temperature, Tb, of shell B?

Notes:

The reason for the “slightly larger shell” is to avoid “complex” view factor issues. Of course, I’m happy to relax the requirement for “slightly larger” and let Bryan provide the more general answer.

The reason for the “highly conductive” and “thin” outer shell, B, is to avoid any temperature difference between the inside and the outside surfaces of the shell. That is, we can assume the outside surface is at the same temperature as the inside surface – both at temperature, Tb.

For anyone who wants to visualize some numbers: ra=1m, P=1000W, rb=1.01m

This problem takes a couple of minutes to solve on a piece of paper. I suspect we will wait a decade for Bryan’s answer. But I love to be proved wrong!

### 105 Responses

1. The paper by Gerlich and Tscheuschener commented – as an independent paper http://www.ing-buero-ebel.de/Treib/Comment-G&T.pdf

On the basis of these and other accusations have Gerlich and Tscheuschner partially for properly corrected – but also written new nonsense http://www.skyfall.fr/wp-content/gerlich-reply-to-halpern.pdf

2. A much more catchy title would be

‘The Amazing Case of SoDs Inability to Read’

The little puzzle presented above was already answered early on in the G&T paper

” It is an interesting point that the thermal conductivity of CO2 is only one half of that of nitrogen or oxygen. In a 100 percent CO2 atmosphere a conventional light bulb shines brighter than in a nitrogen-oxygen atmosphere due to the lowered thermal conductivity of its environment.”

Page 12

So if you increase the insulation around a powered source the source equilibrium temperature will rise…..Duh

• SoD asked you “What is the equation for the equilibrium surface temperature of the sphere, Ta?” You haven’t answered that question, you’ve answered a different one. So accusing SoD of inability to read is somewhat ironic. Please answer the question he asked you.

• on August 7, 2014 at 1:15 pm DeWitt Payne

William,

Good luck with that. Bryan never answers inconvenient questions that might alter his world view. He always deflects to something different, as above. See for example his introduction of photosynthesis as a major flaw in energy balance calculations.

• The stationary heat conduction plays a negligible role in the atmosphere – so what should this mention? In the atmosphere is mainly convective thermal conduction.

But one of them is correct: Stronger insulation (at the resting heat conduction through lower quiescent thermal conductivity – in the atmosphere by more CO2) increases the temperature of the source: By the light bulb, the filament temperature – by the atmosphere, the surface temperature.

• In an oxygen atmosphere a conventional light bulb would burn the filament out in a couple of seconds. Eli has done the experiment;).

Which, among other things is why the Rabett would not trust one word from Gerlich and Geppetto.

• I believe G&T were referring to a conventional light bulb filled with argon in a chamber filled with CO2. Flashlight bulbs are filled with krypton instead of argon to reduce thermal conductivity, producing the same brightness using less power than argon-filled bulbs. The effect of a chamber filled with CO2 surround a convention light bulb appears complicated by convective heat loss, in addition to radiation and conduction. An example of such a problem (9-47) can be found at

Although thermal conductivity is not used directly, I suspect it contributes to one of the parameters (Prandtl number?)

3. on August 7, 2014 at 1:19 pm | Reply stevefitzpatrick

Willis’ steel shell example again. Of course the inner sphere warms due to the shell; double pane windows lose less heat in winter than single pane. How could it be otherwise?

Skydragon Slayer types are immune to rational argument, so nothing will change their minds.

4. on August 7, 2014 at 1:20 pm | Reply DeWitt Payne

Bryan,

In case the problem is too difficult for you, here’s a cheat sheet: http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/

The comments are sometimes amusing. Exactly one person who initially thought that Willis was wrong changed his mind.

• DeWitt

I find the comments by Luboš Motl and Tom Vonk are the most useful contributions.

• on August 7, 2014 at 6:08 pm DeWitt Payne

Bryan,

Tom Vonk’s comment is wrong about the S-B Law. The units do not include per steradian, sr-1. See for example here. The form of the Planck equation integrated over all wavelengths gives irradiance in energy per unit volume (see Planck Radiation Formula here). He’s also wrong about the equation only being valid for a hemisphere. It is valid for a unit area surface integrated over a half sphere defined by the orientation of the surface. His comment about the areas of the spheres not being equal is nitpicking, which he more or less admits.

the irradiance “conservation” is wrong but doesn’t impact much because you considered a case where r~R

. Motl doesn’t disagree with the fundamental calculation either, only that it is much simpler than the actual atmosphere. A point that Willis makes in the original post:

however, it that[sic]is not physically possible to model the Earth as a two-layer system.

For that matter, it’s not possible to model the actual atmosphere with multiple gray layers either.

There are versions of the Planck equation that give power per unit area per steradian per unit frequency or wavelength. But if you integrate those over all frequencies or wavelengths, the constant is not equal to 5.670E-08. It would be 5.670E-08/pi.

• DeWitt

I have bookmarked a number of the links comments and will work through them carefully.
When completed I will give a substantive reply

• on August 7, 2014 at 9:07 pm stevefitzpatrick

DeWitt,
You are a prince to go through this all (yet) again. Let’s see if your efforts have any effect. My personal bet: not an ice cube’s chance in Hades.

• SoDs challenge was intended to ultimately to show how insulation actually Heats the warmer surface.

You all know that!

Its tiresome to trail down that thread again.

Just to point out that no Physics textbook on the planet would support him.

However G&T (see above) noted that CO2 atmosphere made an electric bulb brighter.
No doubt that the greenhouse advocates could see some support there.
However G&T say its all down to the thermal conductivity of CO2 being half that of air.
Also G&T (in a less confrontational mode) advised climate modellers not to set atmospheric thermal conductivity at zero.

The scientific method would be to test by experiment but that seems to be avoided by climate science.

However the climate itself is one ongoing test for the ‘CO2 greenhouse effect’ and for nearly twenty years its not going to plan for the greenhouse advocates

Lots of people are very confident they can do heat transfer on the article you linked but don’t actually produce the equations.

That’s why I thought it was better to have Bryan produce the equations. Bryan is clearly quite an expert. So it’s going to be very interesting to learn from him.

5. Bryans `glory way:

6. The limit of an infinitely conductive or infinitely thin shell is no shell at all and the surface of the heated internal shell would be unchanged. Adding any thermal conductivity and thickness to the outer shell would necessarily create a resistance requiring an increase in the surface temperature of the heated shell to maintain outgoing power equal to the internal source. Perhaps a posting of the full equations that include a finite outer shell conductance and thickness, and a larger than negligible distance between the shells would eliminate some of the mystique surrounding backradiation.

For the record, I don’t see how a shell model enclosing a vacuum helps resolve how any increase in CO2 affects the surface temperature. The main question is whether increasing CO2 contributes to an increase or decrease in the net thermal conductance of the atmosphere. Of course this is off topic here.

• Chic,

As you say not so relevant for the problem at hand.

I’m looking forward to seeing Bryan’s equations and learning some basic heat transfer.

This problem is nothing to do with climate at all. We all have to start somewhere.

For interest and totally unrelated to this article, I did an analysis of a different but simple system in Do Trenberth and Kiehl understand the First Law of Thermodynamics?

As always, the comments are quite entertaining.

7. The way the question is stated with a “black” shell , the shell is going to be invisible to the inner sphere and will have no effect on its surface temperature since it will absorb and re-radiate all the energy impinging on it , It will be no different than a shell simply defined in the vacuum around the inner ball .

Given a total power of P watts , obviously the temperature of the surface of a ball of radius around it will be the temperature corresponding via SB to
P % 4 * pi * ra ^ 2

The temperature of the shell will follow the same function of its radius .

• Bob,

Let’s deal with case 1 first. Can you write down your equation for Ta once it reaches steady state. You have a % symbol in the equation and you use “corresponding”.

Just so there is no confusion:

Ta(case 1) = ?

• the notation is what Kx.com’s K uses . The ” % ” is used by both K and Iverson’s K with it’s original meaning of “divide” . If used “monadically” , that is without a value on the left , it means “reciprocal” . It’s use seems a natural to me .

The flux at the surface is the total power divided by the surface area of the sphere . Then applying the SB conversion from flux to temperature

( ( P % 4 * pi * ra ^ 2 ) % sb ) ^ % 4

where sb is the Stefan-Boltzmann constant . or rearranging and simplifying

( ( P % 4 * pi % sb ) ^ % 4 ) * ra ^ % 2

So the surface temperature decreases as the square root of the radius of the sphere . This is the same law demonstrated by the inner planets other than Venus : http://cosy.com/Science/PlanetTempPlotTa.gif .

Incidentally , I’ve gotten a ( cheap ) copy of Goody’s 1964 Atmospheric Radiation and am ordering a copy of Incropera and DeWitt ( 2006 , again , cheap ) which was mentioned here .

I have commented to a number of people , including Heartland Institute that this site is the best I know actually attacking the physics . I think a more neutral name which does not assume an outcome would be “Science of Destiny” .

• Bob,

Interesting.

And you are also going to put forward that:

Ta(case 2) = Ta(case 1)?

And what would you state for Tb(case 2)?

• As Chic also commented , if the outer shell has emissivity , and therefore absorptivity = 1 , ie , is black , then its temperature will be given by the same formula applied to its radius . And obviously , being black , won’t reflect anything .

• Interesting stuff. I can see a possible tiny problem with it, but let’s wait for Bryan.

Bryan, can you comment on Bob’s formula for us.

8. on August 8, 2014 at 1:18 am | Reply Climate Weenie

Can we add a medium that transports energy from the sphere to the shell?

9. Damn , I dislike not being able to edit .

That should be ” Iverson’s J ” . http://jsoftware.com/ .

10. We do not necessarily consider spherical shells, it to be enough parallel surfaces. In 1879, Stefan has described the heat transfer by radiation:

W = sigma (T1^4 – T2^4)

You are welcome to work Stefan read: http://www.ing-buero-ebel.de/strahlung/Original/Stefan1879.pdf p. 414 and agrees with the measurements of Dolung Petit 1800 match. Because of the non-linearity (T^4) changes, the heat flow between the plates with T1 and T2, when a further plate is interposed.

In the atmosphere but are given not the temperatures, but the heat flows – and therefore the temperatures change.

Bryan – read the commentary on the work of Gerlich and Tscheuschner http://www.ing-buero-ebel.de/Treib/Comment-G&T.pdf and you should open your eyes.

11. SoDs last comment on previous thread says

“it’s clear that, as with G&T, Bryan has not been able to conceive of a problem where body A is provided with a continual separate source of energy (e.g. the sun on the surface of the earth).

Once you have a continual source of energy “internally” heating body A, when you increase the temperature of body B that it is exchanging energy with A, the temperature of body A must increase.”

Back to Ladybird Thermodynamics once again

My reply above should have been enough but lets spell it out…………..

“The Amazing Case of SoDs Inability to Read

The little puzzle presented above was already answered early on in the G&T paper

” It is an interesting point that the thermal conductivity of CO2 is only one half of that of nitrogen or oxygen. In a 100 percent CO2 atmosphere a conventional light bulb shines brighter than in a nitrogen-oxygen atmosphere due to the lowered thermal conductivity of its environment.”

Page 12

So if you increase the insulation around a powered source the source equilibrium temperature will rise…..Duh”

…………………………………………………..

Body A = The continuously electrically heated lamp bulb
Body B = The insulating surroundings

Climate Science seems to rely on meme that a passively heated object having the newly found ability to heat the object that was heating it.

Physics on the other hand puts it like this;

Increasing the insulation(B) around(A) will restrict the heat loss from A thus forcing A to a higher equilibrium temperature.

Spontaneous Heat transfer is only from the higher temperature object to the lower temperature object.

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/seclaw.html#c3

Now the same link will give you all the equations you require to bring you back to reality
There is no point in reinventing the wheel

• Bryan, it is really a stronger insulation increases the temperature of the source – but the static thermal conductivity plays almost no role.

More CO2 means more insulation and thus higher surface temperature.

• > There is no point in reinventing the wheel

C’mon Bryan. Answer the original question: “What is the equation for the equilibrium surface temperature of the sphere, Ta?”

Its a single simple one-line equation that’s needed. Why are you wasting so many words – 30, 40, 50 lines or so – all not answering the question?

12. Bryan, the Stefan-Boltzmann law between two plates with T1 and T2 you should recognize. If not, then you is not going to help:

W = sigma (T1^4 – T2^4)

If now therebetween a further plate is placed, it will take an average temperature Tm – so that the heat flows from T1 to Tm of Tm to T2 and are equal. If this middle plate still has a heat resistance, the T1-facing side will be warmer than the T2-facing side. So that the temperature differences are small and therefore also the heat flow. For simplicity, we assume that both sides of the middle plate have the same temperature Tm. Of course, for the pairings T1 / Tm and Tm / T2 – but applies the Stefan-Boltzmann law with the same heat flows:

W‘ = sigma (T1^4 – Tm^4) = sigma (Tm^4 – T2^4)

This is a defining equation for Tm:

2 Tm^4 = T1^4 + T2^4 oder Tm^4 = (T1^4 + T2^4)/2

The inserted into W’ gives:

W‘ = sigma [T1^4 – (T1^4 + T2^4)/2] = sigma [2 T1^4 – T1^4 – T2^4]/2 = sigma [T1^4 – T2^4]/2

Compared with the first equation it follows W’ = W / 2

But if W ‘= W is supposed to be (as with the atmosphere – sun) must then T1 to a value T1′ to rise, as T2 is constant, that is:

W = sigma [T1‘^4 – T2^4]/2 = sigma (T1^4 – T2^4)

or

T1‘^4 – T2^4 = 2 (T1^4 – T2^4) = 2 T1^4 – 2T2^4

converted

T1‘^4 = 2 T1^4 – T2^4

Thus, T1’ is significantly greater than T1.

Bryan, when you do not have the Stefan-Boltzmann law dispute want, then you should understand the greenhouse effect.

13. It is kind of strange. Bryan has been explaining the basics to me from the “Ladybird Book of Thermodynamics”, and now says:

No point reinventing the wheel

Yes, there is. When I studied – old school – they created a scenario and we had to provide the equations – and solve them – for that scenario. We couldn’t just write “Textbook A gives the solution, no point me doing it again“.

I know, so yesterday!

Already I can see that there’s a lot for me to learn. First of all I don’t see how the thermal conductivity of different gases fits into this problem, given there are no gases and the sphere is in a vacuum – that’s why we are asking you to give us the equations.

Maybe there is an unexpected term in there beginning with “k” (thermal conductivity). My equations don’t have “k” in them.

Would you like me to provide my equations sealed to a trusted 3rd party? You provide your equations. The trusted 3rd party opens them and publishes them?

14. Only 0.03% of a decade so far..

• on August 8, 2014 at 11:10 am | Reply stevefitzpatrick

SoD,
He is not going to answer…… Ever.

Hard to not jump in with the correct calculation; reminds me of tutoring one of my kids in algebra when they were in junior high. It’s almost painful to watch the process. Of course, my kids actually wanted to learn algebra, which is not the case here.

– If you want to do it right, taking into account the solid-angle issue, the equation gets slightly cumbersome. I know you’re taking (rb – ra)/ra << 1, to avoid that.
– But another way to set up the geometry: Take a cylindrical tube of length L, made of perfectly reflecting (εa = 0) & perfectly insulating material. Cap the bottom (z = 0) with more of the same.
– Suspend the power source at z = L/3, not touching anything: Use magnets, whatever.
– Put a cap of normal stuff (εa =1) at z = 2L/3. This corresponds to your 1-shell problem, Case A.
– Add a second cap, also with εa =1, at z = L. This now corresponds to your 2-shell problem, Case B.

Because everything except the two top caps has εa = 0, everything is perfectly reflecting, and there should be no concern about geometrical solid-angle factors. The math looks exactly the same as your approximations – but now it's exact.

• on August 8, 2014 at 9:00 pm | Reply DeWitt Payne

Neal,

I’m not sure the solid angle for a large outer sphere matters. The radiation emitted by the inner surface of the outer sphere that doesn’t intersect the inner sphere will be exactly matched by the emission from the spot on the outer sphere where it’s absorbed, much like a hohlraum. The temperature of the outer sphere is determined only by it’s area and the total power it sees from the inner sphere. The surface of the inner sphere will see a sky at that temperature.

16. DeWitt:
– The equation I’m thinking about relates the power emitted by the inner shell to the power emitted internally by the outer shell, via the Stefan-Boltzmann constant. All the emissions from the inner shell impinge upon & therefore transfer heat to the outer shell; but only a fraction of emissions from the outer shell inward impinge upon the inner shell.
– If the heating element is turned on, the inner shell’s temperature is higher; and in the case that the heating element is turned off, the temperatures of the two shells must be the same.
– The calculation has to be done with consideration of Lambert’s law, which is applicable for blackbody radiation.

17. cont’d:

This reference might also be helpful:
http://en.wikipedia.org/wiki/View_factor#View_factors_of_differential_areas

• Neal,

The above answer of DeWitt Payne (August 8, 2014 at 9:00 pm) is correct and explains why the complex calculation that you propose is not needed. It’s surely possible to go through all those details, but that’s not necessary, when we know already, what the answer must be.

As DeWitt writes the temperature of the outer shell is determined by Stefan-Boltzmann law from the total power of internal heating and the surface area of the shell. The inner sphere is heated by radiation that corresponds to the temperature of the outer shell and by the internal source of heat. The first term is proportional to the surface area of the sphere, no need to use Lambert’s law. The second term is given by the internal source and is thus independent of the surface area. Thus the inner sphere gets the hotter the smaller its radius assuming that the internal heat source has a fixed power.

We can consider three alternatives for changing one parameter:

Increasing the radius of the outer shell lowers it’s temperature, and thus also the temperature of the inner sphere.

Reducing the radius of the inner sphere keeping internal heat source at constant power makes the inner sphere hotter, but does not change the temperature of the outer shell (as discussed above).

Reducing the radius of the inner sphere keeping the power per unit surface area fixed lowers the temperature of both the outer shell and the inner sphere.

• Pekka:

Yes, I agree with these qualitative points. But I suspect there is an important difference in approach.

Consider an extension of the problem: multiple spherical shells of absorptive stuff:
– The outside shell has radius Ro, area Ao = 4πRo^2
– The next shell inward has radius R1, area A1 = 4πR1^2
– …
– The innermost shell has radius Rn, area An = 4πRn^2
– The power generated in the center is P
– Ro > R1 > R2 >… > Rn
– Do NOT assume that (Ro – Rn)/Ro << 1

I don't think I'll be giving too much away to state:

P = σ * Ao * To^4

But the fun question is:
What is Tn^4 ?

I think there is a surprise.

18. Ebel

You have shown interest in G&Ts observation that the light bulb glows brighter in a CO2 atmosphere than in air.
G&T put this down to increased thermal conductivity.

This new paper discounts a radiative effect but instead opts for a mainly convective effect.
However I would not rule out increased thermal conductivity .
Its interesting to note that Argon is used as the filler gas in double glazing units for that reason.

http://hockeyschtick.blogspot.co.uk/2014/08/paper-proves-bill-nyes-faked-greenhouse.html

• Bryan, you are confusing something (though not surprising because of the work of Gerlich and Tscheuschner).

For incandescent and double window is about the stationary thermal conductivity – and which is the smaller, the higher the molecular weight. The greenhouse effect is about the radiative transfer resistance as a result of often absorption and emission. Also to overcome the radiation transport resistance is a temperature difference required (II. Law of thermodynamics).

Since oxygen and nitrogen be no absorptions and emissions, the radiative transfer resistance of oxygen and nitrogen is equal zero, but CO2 and other greenhouse gases to absorb and emit and thus have a radiative transfer resistance. The higher the concentration, the more absorptions / emissions and consequently the higher the radiation transport resistance.

One idea you could even here at G & T p. 50, Equ. (63) get where the optical thickness (radiative transfer resistance) is called. For oxygen and nitrogen kv is equal to 0, ie optical thickness 0 or no radiation transport resistance. In the greenhouse gases kv is not 0 and concentration-dependent.

19. Ebel a correction

increased thermal conductivity should be decreased thermal conductivity.

• Bryan,

I know we are only at 0.06% of a decade, but a couple of other metrics are also interesting.

a. Elapsed time so far by Bryan / time taken by SoD to write out the equations and the whole article = 100x

Or do you want the world to see that you can’t actually do basic heat transfer?

• SoD

Why waste time by reexamining the Willis Eschenbach type steel greenhouse.

I would much rather bring to your attention this peer reviewed paper.

It backs the findings of the famous R W Wood experiment.
It is an honest attempt to remove some bias from climate science.
In addition there are lots of equations for you to get stuck into.
If you find any difficulty I will of course be only to glad to help out.

• Bryan,

The paper shows that the TV demo has nothing to do with the greenhouse effect. After reading the paper, I basically agree. But once that’s been shown, the paper itself has nothing further to do or say about the greenhouse effect – and it states this. Grand.

Now weren’t we talking about the issues of radiative heat transfer, with an eye towards how they DO have to do with the greenhouse effect?

We’re having so much fun, it shouldn’t be legal! Come on and get your hands dirty!

• Bryan,

Indeed, why would you waste time in learning something as it’s so easy to continue writing irrelevancies, including giving references to “papers” that have nothing worthwhile to tell?

• Bryan,

It backs the findings of the famous R W Wood experiment.

Infamous would be a better adjective. It vanished into well deserved obscurity for nearly a century until it was mistakenly resurrected. I guess that’s appropriate with all the recent fascination with zombies.

The paper you cite in this comment has almost nothing whatsoever to do with the poorly documented R W Wood experiment. The containers in the paper aren’t insulated. They are open at the top. In other words, they do not resemble a greenhouse at all because there is nothing to block convection or radiation.

Bye.

20. “It backs the findings of the famous RW Wood experiment.”

In the experiment of Wood naturally occurs also the greenhouse effect – but in what value?

To measure the greenhouse effect by Wood 2 thermometers are used, with an accuracy of better than 1 mK.

These values ​​are obtained by using the same equations and values ​​as for the atmosphere.

• Ebel,

The problem is that R.W. Wood’s experiment has nothing to do with what we call the greenhouse effect. The atmosphere is NOT like a box with a glass lid over it. Now if you were to establish a temperature gradient over a distance, and pump in enough CO2 that this distance had optical depth of ~5, you could test the GHE. But not via R.W. Wood’s demonstration.

• In Experiment by Woods naturally occurs on the greenhouse effect – but the proportion of the greenhouse effect on the measured values ​​lost in the uncertainty of the measured values​​. Therefore, the values ​​obtained are not evidence against the greenhouse effect, but only a measure of the thermal resistance of the envelope.

• Ebel

G&T rely on the Wood experiment but also on the work of Schack as supporting evidence.

Page 71

Would you agree with Schack and if not why not.

• See http://www.ing-buero-ebel.de/Treib/Comment-G&T.pdf p. 70 – 72

Moreover Schack writes in the article, cite by G & T:

The absorption of a gas passing through the heat radiation in steady state exactly equal to the heat radiation of this gas. Because if this stocks deviations, would in this gas be formed in a cavity temperature differences, which is not possible according to the second law of thermodynamics.

21. Bryan: The problem SoD proposes for you would be one given as homework in an undergraduate engineering heat transfer course as soon as radiative transfer is introduced. (Nothing to do with climate science, but vital to understanding how these processes work in engineered systems.)

The professor would expect a student who understood the material to answer it in a few minutes on a couple of lines. He would also conclude that a student who could not get it at all after a few tries and having his errors explained to him is incapable of understanding the field as a whole.

As I look back at old threads at this site, you have been faced with this problem for four years now, and as far as I can tell, you haven’t even been able to make a stab at a solution.

• on August 9, 2014 at 6:24 pm | Reply Steve Fitzpatrick

I don’t think it is ability. It is that he doesn’t want to accept the basic concepts on whic GHG driven warming is built.

If I were SoD, he would be banned or ignored. Slayers arn’t worth it.

• Most likely Bryan doesn’t want to enter in discussion that’s based on logical succession of well established facts. He knows that such an approach would prove that all the comments he presents here are illogical and manifestly wrong. Declining from entering such a discussion he can pretend to believe the opposite.

In the first case, not much point giving him so much “airspace” with his confident proclamations on thermodynamics when he doesn’t know anything about the subject.

In the second case.. [blog etiquette doesn’t allow writing out the consequences]

So also sadly it seems I should limit Bryan’s comments at least when they are repetitions of previous statements. Perhaps as Bryan has a page devoted to his capabilities I might move them here.

That said, we do need a new champion who doesn’t mind the limelight.

Bob has put forward a solution. Bob (if you are still here) would you like to have your answer examined? I did notice a possible tiny problem with it.

23. i like (d) to read in this blog!

But this boring Desaster with Bryan is to stupid for SoD!

The Etikette:

it’s frustrating when no one answers your question the way you want it answered. Maybe people have just missed your point or ignored you or haven’t understood what you are really getting at.

The moderator reserves the right to just capriciously delete comments which use as their premise that standard textbook physics is plain wrong.

This is aimed to reduce the continual stream of unscientific rubbish that gets placed here as comments.

24. Are we still going to wait for Bryan, or is he due some time after Godot?

25. I vote for Godot

26. I think the straw man arguments of G&T will be sticky in the internet forever.

Science of Doom: I do want to congratulate you on your post. Thanks

27. I think Bryan is to excuse. He has read G & T and believes a professor must still write the truth. G & T have used their knowledge to write a work that evokes misconceptions in laymen – by omitting essential facts and adding immaterial items. G & T are smart enough that whole drawn so that a direct error is virtually impossible to prove them – but experts are sure to impress with such work hardly.

Two examples:

It is Hölder’s inequality called correctly, but claims that the effective radiation temperature (-18 ° C) would be wrong. The Hölder says something completely different. The effective radiation temperature is the maximum value of all possible average temperatures – each real average temperature must be lower. And this is confirmed by G & T with their bills.

Second example:

“These extremely different approaches show, that even the physically well-founded radiative transfer calculations are somewhat arbitrary.” P. 50 Of course, the experts know where and what shape is to be applied and again is the transition between the two forms. The Milne wrote in 1928 – http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1928MNRAS..88..493M&data_type=PDF_HIGH&whole_paper=YES&type=PRINTER&filetype=.pdf

28. Here I am

where is Brayn?

• I think he went looking for you.

He had a textbook on thermodynamics to sell; in good condition, barely used.

29. Dear SOD,

is it possible to get in contact with you via email.
If so I would appreciate if I could answer a couple of questions with respect to your radiation transfer code.

Best regards
Günter Heß

• Gunter, sure, email is:
scienceofdoom – you know what goes here – gmail.com

30. Dear Mr. Ebel,

exactly. G&T did a careful job in avoiding precise and meaningful thought experiments.
Per example, in their thought experiment involving Hölder’s inequality they used the example of a tide-locked planet for the earth, which is obviously wrong.

best regards
Günter Heß

31. We seem to have reached a state in which “Those who know do not speak, and those who speak … have stopped speaking.” I have a two-part brain-teaser; maybe it will be too easy. I will present the first part now:

Case I:
We have the perfectly absorbing shell of radius Ro; internally it is heated by another perfectly absorbing shell of radius Ro/10; the source of the inner shell’s heat is a small nuclear reactor producing a stable level of power, Po. This is being done while orbiting the Mars in free-fall, so everything is held together by magnets instead of struts. There is no gas, etc., in the shells;; and the 3-degree K background should be neglected.

As we know from having worked SoD’s first problem, in the steady-state, the outer shell has a definite temperature, and the inner shell has a different but also definite temperature. We denote these as To and Ts, respectively.

Case II:

Now we change one aspect: Instead of a spherical shell of radius Ro/10, we use a regular tetrahedral shell of edge-size 0.4*Ro. The steady-state temperatures are To’ and Tt, respectively.

Question 1:
What are To’/To and Tt/Ts ?

32. I don’t know (and frankly don’t care) whether this gets posted or not.

I think SoD has some very real problems with reading comprehension.

I gather he would like me to do an exercise similar to the Willis Steel Greenhouse.
Since there is already a posted article (with over 400 comments) it has been done to death.

There is nothing wrong with the Willis simple model.

As more shells are added and the power kept constant the core temperature will rise.
I see this as radiative insulation.

Now in my very first comment I pointed out that G&T noticed that a light bulb placed in CO2 will be brighter than in a N2 atmosphere.
This they attributed to reduced thermal conductivity
So obviously it causes the filament temperature to be now higher.

Did that have to be said!!!!

Conductive insulation will produce the same result as radiative insulation.
Now anyone can put the trivial equations down for the Willis model.

But what’s the ….ing point since we agree on the outcome?

I think several posters here find misrepresenting G&T easier than trying to understand their paper.
Fine if thats how they want to waste their time then good luck

• on August 10, 2014 at 2:46 pm | Reply stevefitzpatrick

Bryan,
Well then it seems you do believe in radiative transfer. Good. Now you need to consider not solid steel shells, but multiple wire mesh shells with mostly open area. This begins to better describe the influence of GHGs. Add more GHG (more open-mesh shells) and that also warms the surface. If you can wrap your head around that concept, then you will have made progress.

• “I think several posters here find misrepresenting G & T Easier than trying to understand Their paper.”

We have understood as professionals the work by G & T and recognize the shortcomings and superfluous in the work of G & T.

Therefore: where is your attempt Wood and Schack defend?

By the way, the back radiation is microphysically the Mehode that the net radiation is lower when a insulation is present.

Again to Schack: Schack notes that an absorbing gas to both sides radiates because it absorbs radiation from both sides and must radiate in the isothermal condition nch both sides so there is no heating or cooling by itself. The radiation may be refer as backradiation and thereby reduces the net radiation or a heated surface, where the net radiation is provided.

• Bryan, you say:

“Conductive insulation will produce the same result as radiative insulation.”

and:

“As more shells are added and the power kept constant the core temperature will rise.”

So do you agree that as you add more radiative insulation (i.e. greenhouse gases) to the earth’s atmosphere, the core (i.e. surface) temperature will rise?

For extra credit, in an earlier post you say:

“Climate Science seems to rely on meme that a passively heated object having the newly found ability to heat the object that was heating it.

Physics on the other hand puts it like this;

Increasing the insulation(B) around(A) will restrict the heat loss from A thus forcing A to a higher equilibrium temperature.”

Explain how these two explanations are really for the same phenomenon.

• Curt
Nobody (I hope) would say a cold object can spontaneously heat a warmer object.
Its very clear when you say that insulation reduces the heat loss of a warmer object
Climate science however might describe the same event as the colder object warms the warmer object.
CO2 in the Earths atmosphere can act in several ways
Nighttime insulating outbound IR – warming and so on

• All claims that climate science or some climate scientists would wrongly claim that a cold object warms a warmer one are misrepresentations and pure strawman fallacies.

In every case the issue is about reducing heat losses, which leads to a higher temperature that the same situation with a larger heat loss.

That applies equally to claims of G&T, Bryan and all others guilty of perpetuating the same logical fallacy.

• Bryan, you say:

“Climate science however might describe the same event as the colder object warms the warmer object.”

“Climate science describes this event has the warmer object having a higher temperature due to the presence of a colder object, compared to the case of the presence of even colder objects.”

• You lot need to stop feeding the troll(s).

• Curt
Perfect formulatrion

• Bryan:

Then you agree with the standard formulation of the radiative greenhouse effect — that the presence of colder radiatively insulating gases results in the earth’s surface having a higher temperature than in the case where the earth’s surface was in direct radiative contact with the even colder deep space.

Are we in agreement now?

• Yes Curt that fine…….but this part is often missed out

“compared to the case of the presence of even colder objects.”

Which then implies that a cold object will increase the temperature of a hotter object without qualification

33. Bryan,

off course you are a very clever guy, but

why are you not able to answer one of the simple questions asked by SoD?

34. “Climate science HOWEVER might also describe the same event as the colder object warms the warm object.”
The Earth’s surface is warmed by the sun, this forget the skeptics often again.

By impeding the outflow of heat from the surface due to CO2, the temperature of the surfaces increases as the more insulation is increased by more CO2.

35. Bryan,

G&T created the same perception that you have for the last 4 years. That you don’t believe more GHGs can increase the surface temperature of the planet because it violates the 2nd law of thermodynamics.

It seems you believe that more GHGs increasing the surface temperature of the planet is not a violation of the 2nd law of thermodynamics.

Readers of the many thousands of your statements on this blog would not know this simple fact.

Now, you haven’t answered the questions – what are the formulas for Ta(case 1), Ta(case 2), Tb(case 2)?

We need the formulas from you because you pass yourself off as someone who understands physics.

If you can’t work out the formulas then that explains why we get so many irrelevant and misleading statements from you – you don’t understand physics, at least you don’t understand anything about heat transfer, your choice of subject to comment on.

If you can work them out but don’t want to supply them then obviously you aren’t interested in dialogue.

And if you do supply them I will bring them out each time you make one of your misleading statements. Formulas are crystal clear, unlike 99% of your contributions.

• SoD

A physics degree requires 4 successive years of physics and two of mathematics.

You just make yourself ridiculous by harping on about trivial equations.

I’m certainly not going to jump through hoops on command.

The G&T paper is jam packed with equations yet apparently you are not to happy about that.

Why not ?

• Bryan,

the G&T Nonsense is your only reference to the Physics of the Atmosphere?

You do not answer simple questions, because you know that your stupid web party will end right than. You are a boring loser, nothing else.

• “The G & T paper is jam packed with equations yet apparently you are not to happy about that.”

Bryan, I have answered your questions about Schack and Wood, optical thickness and backradiation technically correct. This you mention, but refer back to the nonsense of G & T.

Read times “On the equilibrium of the Sun’s atmosphere” of Schwarzschild 1906 http://adsabs.harvard.edu/abs/1906WisGo.195…41S

Schwarzschild spoke therein of optical mass instead of optical thick, which is used today. Schwarzschild also referred to as the back-radiation inward migratory energy and shows the temperature rise due to absorption down – so all bases. Unfortunately, I have found by Schwarzschild only the abstract in English, the work is in German.

Gold has already in 1908 the rise of the tropopause by more CO2 suspected http://ia600701.us.archive.org/26/items/philtrans05311580/05311580.pdf and knew the work of Schwarzschild unlikely.

36. Bryan has elected not to supply these equations.

I’ll put up another post with the equations, and also some graphs.

37. […] « A Challenge for Bryan […]

38. One more thing: Fourier had already in 1824 the greenhouse effect correctly described – if he also used other words http://geosci.uchicago.edu/~rtp1/papers/Fourier1827Trans.pdf p. 12/13

Sunlight (light heat) is converted at the surface in the infrared (dark heat). The dark heat is obstructed on the way into space and therefore the temperature at the surface is higher. The onset of convection reduces the effect, but does not eliminate it.

“The mobility of the air, which is displaced rapidly in all directions and which rises when it is heated, and the irradiation by dark heat in the air diminishes the intensity of the effects which would take place in a transparent and solid atmosphere, but it does not completely eliminate these effects.

In effect, if all the layers of air of which the atmosphere is formed were to retain their density and transparency, but lose only the mobility which they in fact possess, this mass of air would become solid, and being exposed to the rays of the Sun, would produce an effect of the same type as that which we have just described. The heat, arriving in the form of light as far as the solid surface of the Earth, suddenly and almost entirely loses its ability to pass through transparent solids; it will accumulate in the lower layers of the atmosphere, which will therefore acquire elevated temperatures. One will observe at the same time a diminution of the degree of heat acquired as one moves away from the surface of the Earth.”

So the opposite of what G & T Fourier underneath. Also assume G & T a mistake in the year. The first release of 1824, was published in another journal the paper again in 1827.

Bryan has you not even the knowledge of Fourier in 1824?

39. […] by okulaer Science of Doom (SoD) has apparently issued a challenge of some sort to a commenter going by the name of ‘Bryan’. This is how SoD describes […]

40. Here is a very simple experiment Bryan can do at home. Take a light bulb. Measure the temperature of the surface when it is turned on (a clear bulb would be best). Wrap it tightly in aluminum foil. Measure the temperature of the foil surface. Try not to electrocute yourself.

• Eli,

Bryan doesn’t do math or experiments. In spite of the fact that both Vaughan Pratt and I have provided far more details on what we did than Wood, he won’t consider our results as being valid. Of course, he takes NN’s obviously flawed results as gospel because it agrees with his preconceptions.

A Wood-like experiment is not difficult to do. You can get all the parts you need at your local big box building supply. Then you need a clear sky for several hours and the patience to keep the boxes oriented correctly. I think this would be a guaranteed prize winning science fair project, but I haven’t been able to interest anyone yet. I say Wood-like because the experimental details of Wood’s note are seriously lacking in detail and rock salt plates aren’t easily available.

I’ve decided that the easier way to do it is to not use the sun. I have a 12″ square flat resistive heater to supply energy to the box and I’ll orient the insulated box with the heater on the top so it radiates downward to minimize convection, which is substantial when using the sun for energy. I think I’ll use a Variac rather than one of those switching dimmers to control the power. Then I’ll know how much power is going into the system, which will make it a lot easier to do the calculations.

41. Well this has been a funny funny funny read. Its been a while since I’ve done thermodynamics, but Pekka, Ebel, and SOD all seem to make sense when read. Bryan and Bob Armstrong’s posts all look like gobbledygook to me.

It’s nice to see SOD hauling these guys out into the light and demonstrating their lack of understanding.

I think Anders nailed it when he said “we need a better class of skeptic”;
http://andthentheresphysics.wordpress.com/2014/07/08/we-need-a-better-class-of-climate-skeptic/

42. Final answers to challenge: Ta1 = 193K; Tb = 192K; Ta2 = 231.43K

P = 1000W; Aa = 12.56m^2; Ab = 12.81m^2;

Ta = (P/Aao)^25 ; Tb = (P/Abo)^25; Ta2 = (.1576P/o)^.25

• Ta2 = 1.205Tb. Ta = Tb. Hence all the confusion, since you probably meant Ta2 when you were using variable Ta in your comments.

• Of course the answer to Ta2 is based on SoD’s assumption that Tb = (P/Ao)^.25. And that Ta = Tb. We could start with the assumption that Tb = (Ta^4 – P/Ao)^.25 and that Ta > Tb, which is always true. So the fact that Ta > Tb doesn’t necessarily mean Ta2 > Ta.

If we assume Ta and Tb are the same temperature, they could together become Ta2 and the zero Kelvin surroundings could become Tb2. Since Tb2 = 0.00K, the new equation would be Ta2 = (P/Ao)^25. In that case, Ta2 = Ta.

• mikejacksonauthor,

Of course the answer to Ta2 is based on SoD’s assumption that Tb = (P/Ao)^.25. And that Ta = Tb. We could start with the assumption that Tb = (Ta^4 – P/Ao)^.25 and that Ta > Tb, which is always true. So the fact that Ta > Tb doesn’t necessarily mean Ta2 > Ta.

And now it turns to rubbish so quickly. My hat is temporarily back on.

Don’t base on anything on my assumptions.

Please produce 2 equations with your working which are based on physics. As requested repeatedly in the comments on the page where this discussion started.

• I spotted anther problem: If Ta2 > Ta, then the mean temperature of a combined Ta2 and Tb would be greater than Ta and P would increase. You can’t have more P then you started with. Here’s the equations:

Tab = (Ta2 + Tb)/2

Ts = surrounding environment at 0.00K.

P = oAe(Tab^4 – Ts^4) > oAe(Ta^4 – Ts^4)

• Scienceofdoom, If you can get 2000W out of 1000W, again, what is stopping you from building a machine that produces unlimited energy? You also have a second problem: a violation of thermodynamics. Heat always flows from hot to cold, never from cold to hot. So shell B can’t heat up shell A beyond what 1000W will give.

You you start with a temperature for shell A that corresponds with your maximum power 1000W, then you add to that! As if you could! You start out with a system that is 100% efficient, then add more! You can start with less and add more.

For example, if the shell A was less than 193K and your heat source was suffering heat loss, and then you added shell B to insulate, so you have less heat loss, that could bring your temperature up to a maximum of 193K. 100% efficiency! You don’t get to have more than that! In reality, both shells lose heat to space, since hot always goes to cold, not vice versa.

I’ve worked it all out for you:

Your problem really has four variables: the mysterious power source P, Ta, Tb and space Ts.

Ta = 193.4K; Tb = 192.4K; Ts = 0.00K;

Aa =12.56m^2; Ab = 12.81m^2

P=1000W; P = Pab + Pbs (law of conservation)

This is the total power between Ta and Tb:

Pab = o((12.56m^2)193.3^4 – (12.81m^2)192.4^4) = 0.00W.

This is the total power between Tb and Ts:

Pbs = oAb(192.4^4 – 0) = 1000W.

Notice you have 1000W going in and out and no violations of heat transfer rules. So what has happened when shell B was added? It caused the area that was formerly occupied by space to warm up to 192.4K. When there was only shell A, the mean temperature was 193.4/2 = 96.7K. When shell B is added, it increased to (193.3 + 192.4)/2 = 192.85.

What happens in a greenhouse? Same thing. The ground warms the air when there is a glass shell around it, but the air does not warm the ground that is warmer than the air. Understand?

43. […] A Challenge for Bryan & A Challenge for Bryan – The Solution […]

44. Just to satisfy myself I worked out the real-world equilibrium temperatures of Ta and Tb. To do this, however, I had to give your magic power source a surface area of 12.31M^2.

When it’s the power source alone, 1000W (194.3K) is lost to space Ts (0.00K).

When you add shell A, it’s equilibrium temperature is 162.6K. Only 500W is lost to space and 500W is used to maintain Ta. 500W + 500W = 1000W.

When you add shell B, Ta increases to 174.7K. Tb = 146.14K. 333.3W is used to maintain Ta; 333.3W is used to maintain Tb; 333.4W is lost to Ts.
333.3 + 333.3 +333.4 = 1000W.

Note that Ta does increase like you said, but also note that its temperature never exceeds its power source, nor does the power ever exceed 1000W.

To solve this problem I simply divided the total power by the number of shells and Ts (space). That provides the value of each power drop. The sum of the power drops should equal 1000W. The power drops should be equal to each other so each shell receives the same power in as power out to maintain equilibrium.

45. on November 2, 2018 at 10:44 am | Reply Barry Jacobsona

Hi, i am late to the party here by about 4 years, so I don’t know if anybody will see my comment. Nevertheless, I deeply appreciate this physics-based discussion by people who advance rational arguments pro and con. Debate is very healthy, and hopefully will lead to the truth.

A major question I have is the following: The discussion mainly surrounds whether applying a heat source to the inner sphere will cause its temperature to be affected by the presence, properties or numbers of outer spheres. But this is not the situation with the earth, to begin with. The earth is not generating its own heat internally, except for a small geothermal component due to reactions at its core. But these are not usually considered in AGW discussions or models. Instead, the earth is being heated externally by a heat source far away, above the inner sphere of this discussion and its surrounding metal shell. The question is how much radiation gets through. At first glance, the better the outer shell insulates, the less the temperature of the inner sphere should be. So here is where it gets really complicated. On the way in, the frequency of the radiation is high (say 3000K), coming from the hot sun. It partially gets thru atmosphere, and heats the earth to a certain surface temp (say 300K). Now this lower freq radiation goes back up. Due to absorption properties of CO2, it is supposedly blocked, due to atomic resonance peaks at lower freq, which did not block it on way in at higher freqs. One question is whether such a one-way device can exist (like a rectifier in electronic circuits). More questions in a later post. Thanks.

46. Barry,

At a simple level the atmosphere is mostly transparent to solar radiation, therefore the sun heats the earth’s surface.

The atmosphere is opaque to terrestrial radiation, therefore emission of radiation from the climate system is from (on average) a few kilometers up in the troposphere.

Why is this? Absorption is wavelength dependent. Solar radiation is centered around 0.5μm, terrestrial radiation is centered around 10μm.

If the atmosphere absorbed terrestrial and solar radiation equally then the greenhouse effect wouldn’t exist.

Many articles on this site explaining the details. Maybe start with The Greenhouse Effect Explained in Simple Terms.