The “Steel” Greenhouse versus the “Real” Greenhouse

Written by Ross McLeod on July 22, 2013. Posted in Current News

The actions of the advocates of Climate Science have certainly caused significant public controversy.

Perhaps the most interesting result has been the public discussion itself.  If one takes the time to view the posts and comments on many websites that present climate science one is left with the inescapable viewpoint that everyone is an expert and that ridicule and sarcasm are actually cogent arguments that have become more important than the discussion at hand.

Perhaps, before we continue, we could humbly remember that Einstein said:

“All these fifty years of conscious brooding have brought me no nearer to the answer to the question, ‘What are light quanta?‘

Nowadays every Tom, Dick and Harry thinks he knows it, but he is mistaken.” (Albert Einstein, 1954)

Ok, let’s put aside the idea that any of us really understands quantum physics for now – I certainly don’t (and I’m simply waiting for some “clever” person to take the obvious cheap shot) and let’s consider the merits of the argument that are presented – both “for” and “against”.

This article began as an attempt to verify or dismiss the claims made in an “article” describing a “Steel Greenhouse” proposal which has been presented as scientific “fact” at least twice on the “Watts up with That” website.

I was simply amazed at the amount of certainty expressed by commenters and the sarcasm, scorn and invective inflicted upon anyone who dared disagree with the orthodoxy.

The interesting thing about this “thought” bubble is the inconsistency between real scientific results, which are often confusingly used to promote the greenhouse effect, and this model of the supposedly “same” thing.

I am writing this article as a “proof” that the “Steel Greenhouse” model is not a valid expression of science and must be dismissed.

I present several “proofs” of this claim in this article.

It is not sufficient to simply dismiss these with some inane, childish sarcasm and reference to “authority”.

So, in the words of one of the most underrated musical talents of the 1960s – The Small Faces –

“Are you all seated comfortable,
too square on your botty ?
Then I’ll begin”

Now, I say unequivocally that the “Steel Greenhouse” model as presented has no basis in real science.

It makes the claim that adding the “close fitting shell” must cause the radiative flux from the “planet” to double – that claim is simply, demonstrably wrong.

Initially we need to refresh our understanding of what was claimed.

So let’s re-examine WUWT’s Steel Greenhouse proposal – I hope they don’t mind that I borrowed their diagrams and some of the text.

“This planet is at equilibrium.

The natural reactor in the core of the planet is generating energy that at the planet’s surface amounts to 235 W/m2.

It is radiating the same amount, so it is neither warming nor cooling.

Now, imagine that without changing anything else, we put a steel shell around the planet. Figure 2 shows that situation, with one side of the shell temporarily removed so we can look inside.”

 Figure 1.

“Now, note what happens when we add a shell around the planet. The shell warms up and it begins to radiate as well … but it radiates the same amount inwards and outwards. The inwards radiation warms the surface of the planet, until it is radiating at 470 W/m2. At that point the system is back in equilibrium.

The planet is receiving 235 W/m2 from the interior, plus 235 W/m2 from the shell, and it is radiating the total amount, 470 W/m2.

Figure 2.

There are numerous reasons why that is wrong and these are listed below:

1. A mathematical analysis from first principles says that the ratio of Q’, the rate of energy loss after the shell is fitted to Q, the original rate of energy loss sans the shell is Q’/Q = R²/(R² + r²) where r is the radius of the “planet” and R is the radius of the shell.

   A mathematical proof of this is supplied as Addendum 1.

   For the trivial case in the proposal being discussed this result says that the rate of energy loss from the shell to space becomes half the original rate of energy loss once the shell is fitted.

   That is Q’ = Q x 1²/(1² + 1²)  That is Q’ is half of Q.

   This is in stark contrast to the claim that is made in the “Steel Greenhouse” proposal that Q’ = 2 Q.

   This claim is wrong – Q’ does not equal 2 x Q nor does the shell ever radiate – Q’’ – at a value equal to the initial Q – ever !

   The claimed result is demonstrably impossible.

2. The claim that the temperature of the shell must reach the same initial temperature as the sphere for thermodynamic equilibrium is wrong.

   The claim that this equilibrium requires that the rate of energy loss from the sphere therefore must double to maintain equilibrium in accordance with the laws of thermodynamics is wrong.

   The proponents of this assertion make a fundamental mistake in their interpretation of the law of conservation of energy.

   A mathematical proof of this is supplied as Addendum 2.

3. The claim that the effect claimed for the “Steel Greenhouse” – that is the shell causes the rate of energy loss from the planet to “double” – is “proof” of the “real greenhouse effect” is simply false.

   This claim is so fundamentally at odds with actual real scientific evidence I am amazed anyone continues to argue this fallacious concept as “proof” of the back radiative “greenhouse effect”.

   Real scientific data disproves the “Steel Greenhouse” proposal.

   A “proof” of this is supplied as Addendum 3.

4. The problem with Back Radiation.

   Back Radiation is claimed to be the driver of this phenomenon.

   A “proof” disputing this is supplied as Addendum 4.

Any person wishing to dispute these claims should control their immediate reaction to dismiss these because they have their own preconceived notions or understandings.

Rational discussion requires understanding the claims made sufficiently to provide a cogent argument as to why they are wrong.

Any argument to the contrary cannot stand without a “proof”.

If you read on you will see that I offer “proofs” of each claim made above.

You may disagree but if you cannot offer sound mathematical and scientific “proof” that clearly disproves my “proofs” then you will have failed to falsify my “proofs” and they stand unchallenged.

I await a successful challenge.

The curious should read on.

The truly intelligent will try to disprove the following 4 “proofs”.

The buffoons are free to troll and flame their absurdities as they choose – sarcasm without any rational intellectual argument to back it up is meaningless nonsense, demonstrates an inferior intellect and to be quite frank –

 I simply don’t care !

Addendum 1

Mathematical Proof of the general relationship between the rate of energy loss between a sphere and a “close fitting” shell as proposed in the “Steel Greenhouse” model.

This is a simple application of the Stefan-Boltzmann law – H_net = A ξ σ (T⁴ – T_o⁴).

Let Q be the rate of energy loss from the sphere alone.  Let T_Sphere be this equivalent temperature, T_o be the surrounding temperature.

Now fitting the shell undoubtedly changes the amount the sphere radiates.

Let Q’ be the rate of energy loss from the sphere and shell “combination”.  Let T_Shell be the temperature of the shell.  As of this point this is unknown.  (The text makes no arbitrary assumption about this value.)

Let Q’’ be the rate of energy loss from the outer surface of the shell to the surroundings.

Finally let r be the radius of the sphere and R be the radius of the shell.

The text solution writes 3 equations as follows:

Q = 4 π r² σ (T_Sphere⁴ – T_o⁴) as ξ is unity.

Q’ = 4 π r² σ (T_Sphere⁴ – T_Shell ⁴) as ξ is unity.

Q’’ = 4 π R² σ (T_Shell ⁴ – T_o⁴) as ξ is unity.

As Q’’ = Q’ the solution is.

4 π R² σ (T_Shell ⁴ – T_o⁴) = 4 π r² σ (T_Sphere⁴ – T_Shell ⁴)

4 π R² σ T_Shell ⁴ – 4 π R² σ T_o⁴ = 4 π r² σ T_Sphere⁴ – 4 π r² σ T_Shell ⁴.

4 π R² σ T_Shell ⁴ + 4 π r² σ T_Shell ⁴ = 4 π r² σ T_Sphere⁴ + 4 π R² σ T_o⁴.

(R² + r²) T_Shell ⁴ = (r² T_Sphere ⁴ + R²T_o⁴).

T _Shell ⁴ = (r² T_Sphere ⁴ + R²T_o⁴)/ (R² + r²).

Q’ = 4 π r² σ (T_Sphere⁴ – T_Shell ⁴) = 4 π r² σ (T_Sphere⁴ – (r² T_Sphere ⁴ + R²T_o⁴))

                                                                                —————————

                                                                                          (R² + r²).

Q’ = 4 π r² σ (R² T_Sphere⁴ + r² T_Sphere ⁴ – r² T_Sphere ⁴ – R²T_o⁴)

                    ———————————————————

                                             (R² + r²).

Q’ = 4 π r² σ R² (T_Sphere⁴ – T_o⁴)

                     ———————–

                            (R² + r²).

or Q’ = Q x R²/ (R² + r²) – because 4 π r² σ (T_Sphere⁴ – T_o⁴) = Q.

A word to the wise – you should read this carefully.

This concludes Addendum 1.

Addendum 2

The claim that the shell must reach the same temperature as the sphere in its initial state in accordance with the laws of thermodynamics is demonstrably wrong and arises out of a misinterpretation of the law of conservation of energy.

This claim has already been shown to be wrong by the “proof” offered in Addendum 1.

The remainder of this discussion is also general in development.

Let Q be the original Flux in Watts per square metre from the sphere initially.

Let Q’ be the flux radiated from the sphere to the shell once it is fitted.

Let Q’’ be the flux radiated from the shell in Watts per square metre once the shell is “fitted”.

It is indisputable that the sphere is radiating a total of 4 π r² x Q Watts, or joules per second.

It is also indisputable that this energy (per second) is absorbed by the shell which causes the heating of the shell and thus it commences radiating.

The shell reaches a certain temperature defined by the amount of energy absorbed and radiates over double the surface area of the sphere – an exterior surface and an interior surface.

The shell has no other source of energy – it all comes from the sphere.

Initially the shell receives 4 π r² x Q Watts and it is logically impossible for this amount of energy to cause a temperature beyond the temperature where the shell radiates at anything more than Q/2 because this, and only this, conserves energy.

Of course, the issue of back radiation is to be discussed later.

But this discussion dismisses absolutely any possibility that it is valid to assume that the flux radiated by the shell – Q’’ – is equal to Q initially.

If you are honest you must admit that the proponents of the claim that the flux from the sphere must double because the shell radiates over 2 surfaces have simply made a mistake.

I’ll say again – you cannot supply a total 4 π r² x Q Watts to a shell and obtain the result that it radiates at Q Watts per square metre over double the surface area and claim you are obeying any law of thermodynamics !

The mistake they have made is they are NOT conserving energy at all !

Their proposal conserves flux – not energy – and this is simply not correct.

Energy must be conserved – flux and temperature be damned – they will assume whatever value satisfies conservation of energy principles!

This mistake highlights why it is arrogant to assume you know it all and dismiss any alternate proposal out of hand.

A true scientist assesses a proposal and tries to falsify it without any preconceived assumptions.

Failing that they try to verify it.

This proves the claim that the shell must assume the same temperature as the initial temperature of the shell is in error.

This also falsifies the claim that the shell (radiation Q’’) will radiate at a rate equal to the initial rate of the sphere – Q.

Logic sound science and mathematics demands Q’’ be half Q at least initially.

Remember – I said the Back Radiation discussion is still to come.

A word to the wise – you should read this carefully.

This concludes Addendum 2.

Addendum 3

The claim that the effect claimed for the “Steel Greenhouse” – that is the “shell” causes the flux from the planet to “double” – is “proof” of the “real greenhouse effect” is simply false.

This claim is so fundamentally at odds with actual real scientific evidence I am amazed anyone continues to argue this fallacious concept as “proof” of the back radiative “greenhouse effect”.

This discussion is not mine and I state this right up front.  I have shamelessly copied this discussion text from Dr Roy Spencer’s website.

I do not agree with Dr Spencer on a lot of issues over the “greenhouse effect” but he offers a cogent argument in this text as to why the “Steel Greenhouse” concept is fundamentally wrong.

(By the way I simply hate all this I know better than you crap – can’t we discuss and argue like mature, civilised, rational adults ?)

http://www.drroyspencer.com/2012/03/slaying-the-slayers-with-the-alabama-two-step/

STEP 1:
Temperature is determined by rates of energy gain and energy loss. It does not matter whether we are talking about the human body, a car engine, a pot of water on the stove, or the climate system. The temperature (and whether it is rising or falling) is determined by the rates of energy gain and energy loss. In the case of the climate system, the Earth receives energy from the sun (primarily at visible wavelengths of light), and loses energy to outer space (primarily at infrared wavelengths). A temperature rise can occur either from (1) increasing the rate of energy gain, or (2) decreasing the rate of energy loss. The greenhouse effect has to do with the 2nd of these possibilities.

STEP 2:
Infrared absorbing gases reduce the rate at which the Earth loses infrared energy to space. Satellite measurements of the rate at which the Earth loses infrared energy to space have been made as early as the 1970′s, from the NASA Nimbus 4 spacecraft. The following plot shows the IR intensity (vertical axis) as a function of IR wavelength (horizontal axis). The area under the jagged curve is proportional to the rate of energy loss to space. Note that at the wavelengths where water vapor, carbon dioxide, and ozone absorb and emit IR energy, the rate of energy loss by the Earth is reduced.

http://www.drroyspencer.com/wp-content/uploads/Petty-fig6.6-modified.png” align=”left” border=”0″ height=”221″ hspace=”9″ width=”293″>

Now, let’s take Steps 1 and 2 together: If you add more of these “greenhouse gases” and nothing else changes then the rate at which the Earth, as a whole, loses energy to space is reduced. “

And there you have it. 

How can the flux be equal to the original, both up and down, as claimed for the “Steel Greenhouse” proposal and yet “the rate at which the Earth, as a whole, loses energy to space is reduced”?

That clearly is not possible.

I agree completely with Dr. Spencer’s conclusion that there is experimental proof that the “shell” reduces the rate of radiation to space – I reserve my right to disagree with his conclusions about the consequences.

Logic demands if a theory disagrees with experiment the theory is wrong.

“It doesn’t matter how beautiful your theory is, it doesn’t matter how smart you are. If it doesn’t agree with experiment, it’s wrong.” – Richard Feynman.

This concludes Addendum 3.

Addendum 4

The problem with Back Radiation.

I consider I have established beyond doubt that – at least initially – the sphere cannot radiate at any value higher than 117.5 Watts per square metre from each surface.

So is the following a fair representation of how the back radiation is supposed to work?

I warn you in advance this is tedious.

So let’s start at this point – the shell emits 117.5 back to the sphere which absorbs it and now it is at a temperature approaching ~281 Kelvin which is equivalent to a flux of 352.5 Watts per square metre = 235 + 117.5.

So now the sphere radiates at 352.5 Watts per square metre, the shell absorbs 352.5 Watts per square metre and thus radiates 176.25 Watts per square metre out to space and 176.25 Watts per square metre inwards at an equivalent temperature of ~236.12 Kelvin.

This 176.25 now combines with 352.5 and heats the sphere to ~310.75 Kelvin such that it radiates 528.75 Watts per square metre.

This must be true because the shell is not yet radiating the required 235 Watts per square metre out to space for the claimed equilibrium!

The shell absorbs this 528.75 Watts per square metre and it radiates 264.375 Watts per square metre out to space and 264.375 Watts per square metre inwards at an equivalent temperature of ~261.3 Kelvin.

Hmm – I wonder what went wrong – it missed the equilibrium point of 235 Watts per square metre – how careless of it.  Further, why does it stop at all?

Why does the back radiation simply not keep compounding endlessly?

Of course this must be a fictional scenario; surely anyone can see this is a runaway process creating energy out of nothing.

Is it possible that there is some other starting value that leads to “balance” – 470 from the sphere and 235 out and in from the shell?

I claim the answer is no.

It is not valid to have 235 emitted by the sphere every second and have different amounts radiating out and in such that the back radiation accumulates to 235 back to the sphere.

Show me the sequence that achieves this and why it is a valid consideration!

But there are several other sound scientific reasons why this “Back Radiation” concept is fundamentally flawed.

I will be tedious and quote from another text again:-

“While a body at absolute temperature is radiating, its surroundings at temperature T_s are also radiating, and the body absorbs some of this radiation.

If it is in equilibrium with its surroundings, T = T_s and the rates of radiation and absorption must be equal.

For this to be true, the rate of absorption must be given in general by H = A ξ σ T_s ⁴.  Then the net rate of radiation from a body at temperature T with its surroundings at temperature Ts is:-

H_net = A ξ σ (T⁴ – T_s⁴).”

Sears and Zemansky’s – University Physics with Modern Physics – Young and Freedman.

Surely I do not need to point out the obvious?

If T > T_s – that is the difference is positive – then net energy transfers from the object.  The equation says that absolutely.

If T = T_s – that is the difference is zero – then there is no net energy transfer from the object or the surroundings.  The equation says that absolutely.

If T < T_s – that is the difference is negative – then net energy transfers from the surroundings to the object.  The equation says that absolutely.

As the shell is always at a temperature lower than the sphere the net heat flow is always from the sphere to the shell – this is an absolutely fundamental result of Thermodynamics – indisputable!

The “Steel Greenhouse” proposal states that the shell temperature is always less than the sphere while it is increasing until it reaches the initial temperature of the sphere.

The Stefan-Boltzmann equation says categorically that this will not occur as the flow of heat is always from sphere to shell!

If a situation where the temperatures are equal means no net flow and therefore no heating of either – equilibrium – explain how it is possible for the shell to “heat” the sphere when there is a flow of “heat” away from the sphere !

The only way this process of “shell warming sphere” can occur is if the shell is transmitting more energy to the sphere than the sphere is transmitting to the shell.

If this is not true then the shell cannot be providing “net” energy to the sphere.

If the sphere is not gaining “net” energy or the “net” exchange is zero the sphere’s temperature simply cannot increase!

We have established that every Watt or joule per second provided by the internal power supply of 235 Watts per square metre radiative output is required to simply keep the shell at a temperature of ~213 Kelvin radiating 117.5 Watts per square metre over 2 surfaces – refer to the “proofs” contained in Addendum 1 and 2.

If the net flow of “heat” is always from the sphere to the shell unless the shell temperature exceeds the sphere temperature and the internal power supply is totally committed simply keeping the radiation out to space and back to the sphere at 117.5 Watts per square metre – the obvious question is where is the energy coming from to heat the sphere beyond the original ~255 Kelvin?

Of course everyone is going to sigh, claim this guy is too stupid for words and blindly claim it is the back radiation despite the overwhelming evidence that real science says this is not possible.

But how can that be if the net flow of heat is always from the sphere to the shell – always?

I have my own personal opinion as to how this phenomenon can occur.

So put aside your prejudices, your claim that you know how this stuff works and let’s at least discuss it.

I said in Addendum 2 –

“Energy must be conserved – flux and temperature be damned – they will assume whatever value satisfies conservation of energy principles!”

Everyone is assuming they know what those values are and they must be right because they have studied science.

Well I don’t think any of this is simple at all and many of the “tools” being applied are being applied incorrectly.

So, for what it is worth, here is my take on back radiation.  I claim it is consistent with reality.

I do not deny back radiation – at all.  To do so is stupid in the extreme.

I do, however, deny that it has a thermal effect on any object that is initially at a higher temperature.

Firstly, H_net = A ξ σ (T⁴ – T_s⁴) demands this. 

There is no possible reason to ignore this.

I have heard argument similar to – An object does not know the state of the object that emitted the photon it is absorbing.  It simply must absorb it and this means that while the “net” transfer of energy is from hot to cold this absorption must mean that back radiation can cause energy transfer and thus cold can warm hot.

The claim is analogous to claiming that all photons are created equal.

While that may be a noble ideal of human philosophy it is certainly not true of photons.

Anyone who doesn’t think a photon is emitted in exactly the circumstances that caused the emission and it has energy levels and other properties in accordance with that emission is denying all the science that lead to the development of Thermodynamics.

If you consider this diagram – how do you propose that any individual photons emitted by an object at 800 K – the light blue line in the graph – can ever impart sufficient energy to raise the elevation of the curve of the 800 K line so that it approaches the 900 K line?

That proposal is absurd – they clearly do not have sufficient energy.

All that happens is that these photons from the 800 K object simply replace photons of equivalent energy that are continually being radiated by the 900 K object – that is they simply fill in “hole” left when a photon at an energy level at or below the blue line section of the 900 K curve area is “pushed” out – they can never impact any above the blue line.

In all cases the total energy of a cooler object’s curve is entirely contained within a hotter object’s curve.

Back radiation is simply part of the energy exchange that is occurring all the time.

We all know energy transfer only has thermal impact when there is “net” energy exchange occurring and the Stefan-Boltzmann equation used throughout this article clearly says that energy exchange only has thermal implications when net energy flows from hot to cold.

I guess you could claim that somehow back radiation accumulates much like inflating a balloon.

I consider this to be analogous to a mechanical problem.

I am trying to move a 1000 kg object by myself.  I obviously am not capable of exerting sufficient force to do this on my own.

The “back radiation from cold objects causes heating of hot objects” analogy is similar to claiming that if I simply keep pushing long enough the energy will accumulate and I’ll move it.

I simply do not believe it.

This concludes Addendum 4.

References

Essentials of College Physics 7^th Edition Raymond A Serway, Chris Vuille

University Physics with Modern Physics 13^th Edition Young and Freedman

Physics for Scientists and Engineers Sixth Edition Paul A Tipler and Gene Mosca

Problems and Solutions on Thermodynamics and Statistical Mechanics (Major American Universities Ph.D. Qualifying Questions and Solutions) Volume 5 Edited by Young-Kuo Lim

Trackback from your site.