# Revisiting the Steel Greenhouse

Written by Joseph E Postma

## Solar Flux is not Q

This was touched on in the last post, but the steel greenhouse deserves a full logical physics treatment of heat flow to explain what actually happens and why the radiative greenhouse effect believers have it wrong.

First I just want to repeat something.  The radiative heat flow equation

Q = A*σ(Th4 – Tc4)

does not stand for conservation of energy where Q is the solar energy, and the hot term is the Earth’s surface and the cool one the atmosphere.  Conservation of solar energy does not occur between the surface and the atmosphere.  Why would solar energy be conserved as a difference in flux, as a heat flow, between the surface and atmosphere?  This reinterpretation of the equation has no logical or physical basis.  Solar energy is conserved to the outside of the system, not inside between the surface and atmosphere.  I just don’t know why this reinterpretation of the heat flow equation would have been invented.  The hot and cool term represent sources of energy, and Q is the local difference between them resulting in heat flow; the solar heating has to be one of the sources, on the right hand side of the equation, because it is a source, and the energy from the Sun is not dependent upon the difference between two other arbitrary and undefined sources.

You see, one of the ways to create science sophistry, is to simply misinterpret or worse lie about what the physics equations mean.  It is as simple as that is all they have to do, is lie about what an equation means, because most people don’t have the physics training to be able to know what the equation actually means and what the terms actually refer to.

If you want the heat flow equation with solar energy as a term, then you have to write

Q = A*(I*(1-α) – ε*σ*TS4)

where I*(1-α) is the local energy from the Sun impinging on a local surface with temperature TS.

## Steel “Greenhouse”

The diagram by Willis Eschenbach, as featured by Anthony Watts on WUWT, is shown above. The scenario is a heated sphere at temperature Tsp concentrically trapped inside a passive shell who’s temperature is denoted as Tsh.  The interior sphere is the only thing with power with constant output.  The interior sphere has radius rsp and the enclosing shell has radius rsh.

The most general statement of the radiative heat flow equation is

Q = A*(F1 – F2)

where F1 and F2 are a local flux from the two different sources.  That is, the fluxes correspond to a specific single location, not their own locations independently.  By convention, the location which the fluxes correspond to are that at the location represented by F2.  That way, if Q is positive, then it denotes positive heat flow into the location specified by F2, and so that location would rise in temperature.  If Q is negative, it denotes heat flow out of the location F2, and so if F2 is not a powered source of heat, then it will fall in temperature.  However, whether or not the heat flow out of F2 towards F1 can actually cause an increase in temperature at location F1, requires its own analysis with all of the terms switched to the new location F1.  We will work through this for the steel greenhouse.

For the heat flow at the interior of the shell, we have

Qsh = Ash*(Fsp – Fsh).

The flux from the sphere is reduced by its distance from the shell, while the shell has the local flux, and so

Qsh = Ash*σ*[Tsp4 * (rsp/rsh)2 – Tsh4].

For the heat flow at the interior sphere, we have

Qsp = Asp*(Fsh – Fsp).

The flux from the shell is isotropic and symmetric at any location inside the shell, and so it doesn’t reduce with distance from the shell to the sphere, and so

Qsp = Asp*σ*(Tsh4 – Tsp4).

Now since the interior sphere is the thing providing power, the passive shell will eventually come to thermal equilibrium with the sphere which means that, locally, at the location of the shell, the heat flow will be zero, i.e. Qsh = 0 and so

Tsh = Tsp * (rsp/rsh)1/2

Since rsp < rsh the temperature of the shell will be less than the sphere.

Now that we know the equilibrium temperature of the shell, we can calculate the heat flow at the sphere,

Qsp = Asp*σ*(Tsp4 * (rsp/rsh)2 – Tsp4)

= Asp*σ*Tsp4*((rsp/rsh)2 – 1).

Since rsp < rsh the heat flow will be negative, meaning away from the sphere, towards the shell.  Thus, the shell can not heat the sphere.

At equilibrium, at the shell, the interior heat flow is zero; however, the exterior of the shell also emits radiation and it emits according to its temperature:

Psh = Ash*σ*Tsh4

= Ash*σ*Tsp4*(rsp/rsh)2

For conservation of energy this has to be equal to the power supplied by the interior sphere, so

Ash*σ*Tsp4*(rsp/rsh)2 = Asp*σ*Tsp4

4*π*rsh2*σ*Tsp4*(rsp/rsh)2 = 4*π*rsp2*σ*Tsp4

which is 1 = 1 which means that energy is indeed conserved.

So you see what happening there.  The exterior shell is passive, it is a passive medium.  Nothing passive raises the temperature of the thing heating it.  The passive medium simply passes the energy through it, to its exterior.

Now this is where we immediately get the objection that a passive blanket makes you warmer.  This is a problem with the typical psychological profile of the majority of humans but scientists especially: the dominant sensing rather than intuitive mental profile.  Your physical senses do not determine truth; physical senses do not determine physics or tell you much at all about the underlying mathematics of a process.  Putting on a blanket is not what the radiative greenhouse effect is about.  A physical blanket stops convective heat loss, exactly like how a real greenhouse operates.  Just because being warmed by a coat is something you sense with your animal nature, does not mean that it logically or mathematically translates to the radiative redefinition of the greenhouse effect.  Logical thinking is associated with the intuitive psychological profile, not the sensing.  A blanket is doing the physical greenhouse effect, not the radiative greenhouse effect (as distinguished here).  If you want to use your animal physical sense perception to “empirically” test for the radiative redefinition of the greenhouse effect, then the correct way to do that has only ever been to arrange a situation which produces only the supposed radiative effect, not some mix-up with the physical convection-blocking effect.  Sensing personalities have a hard time appreciating that logic; intuitives do not.

That is, look at your face in a mirror, and feel the heat being returned.  Your face can detect temperature changes as small as 0.1 degrees centigrade, or about 1/5 of a degree Fahrenheit. You should feel a huge warming effect from all that facial infrared radiation being returned, but there is nothing sensed larger than 0.1 K.  You should feel it instantaneously given the speed of light with a mirror inches from your face, but you feel nothing.  If you want a sense-perception based experiment to test for the radiative version of the greenhouse effect of climate science, that is it, and it fails the test once again.

Just be logical!  What would happen if you shrunk the outer shell until it simply became the new surface for the interior sphere?  Then you just have a sphere with a slightly larger surface, depending on how thick the shell became as it reduced its radius while conserving its total mass.  A sphere with a slightly larger surface area than before simply has a slightly lower energy flux density at the surface, and hence lower temperature.

You see, what Willis Eschenbach and all these radiative greenhouse schemes are trying to do is to conserve temperature.  When Willis’ steel greenhouse is set up to conserve 240 W/m2, Willis is trying to conserve temperature.  If you don’t like putting energy flux density in terms of temperature, then just put it in terms of density: Willis was trying to conserve energy density.  Energy density / temperature does not need to be conserved, and is not the meaning of conservation of energy.  The energy density of 240 W/m2 can not be conserved when it radiates out to a larger surface area; by definition the density has to decrease, and hence the temperature associated with it as well, while energy is conserved.  Willis tried to get around this by shrinking the shell area to as close as possible to the sphere area, but again this just makes the shell the new surface for the sphere, and the original energy density still always has to decrease with any distance at all from the sphere.  If the shell is shrunk to zero thickness and sits on the sphere, then you just have the surface of the sphere.

If they would just apply logical physics and math to the problem it would be clear how to correctly conserve energy.  Of course it has always seemed that there is a predisposition to protect the belief in this alternative radiative reinterpretation of the greenhouse effect for climate science.

Consider what Willis’ steel greenhouse analysis contained: did it lay out the heat flow equation, describe the heat flow at the interior of the shell, and then at the surface of the sphere, make a point about how total energy needs to be conserved, and did it have all the physics math you see above?  No it didn’t have any of that.  It should have.  All it had was that Willis wanted 240 W/m2 to be emitted from the outside of the shell – that was thepredetermined result Willis desired to show; he didn’t work it out and find it, he wanted it to force it that way.  The interior sphere surface had a temperature of -180C, and Willis wanted the outside of the shell surface to be -180C also.  Is there any reason for desiring this?  Does it follow conservation of energy and the laws of heat flow?  Is temperature or density a conserved quantity in physics?  No, and Willis was able to get around that simply by not performing the mathematical analysis of the heat flow that should have been performed.  Of course, this was much like the later light bulb experiment by Curt Wilson and Anthony Watts, who simply invented a predetermined interpretation of their results specifically by avoiding performing the mathematical heat flow energy conservation analysis that should have been performed.  This is how easy it is to create science sophistry and belief in the climate science version of the greenhouse effect.

## Energy Conservation and Work and Heat

Recall that at equilibrium, no heat flows at the interior of the shell

Qsh = 0

while the heat at the sphere is given by

Qsp =  Asp*σ*Tsp4*((rsp/rsh)2 – 1).

In fact, at any location between the sphere and the shell the heat flow from the sphere is given by

Qsp =  Asp*σ*Tsp4*((r/rsh)2 – 1)

where rsp ≤ r ≤ rsh; when r = rsh then Qsp = 0.

On the outside of the shell, of course,

Psh = Ash*σ*Tsp4*(rsp/rsh)2

which conserves energy, and this would be equal to the heat flow from the exterior of the shell in an environment at 0K.  Note that the heat flux and temperature is not a conserved quantity – only total energy is.  Heat flow is just the thing we call Q, while the total energy is represented by the terms Q is dependent upon.  This is of course the distinction that flux/temperature and energy are different things – we do not conserve energy by conserving flux/temperature, in fact we conserve energy because flux and temperature are not conserved…that is the only way it can work.  It is in fact the only mathematical solution and the only way to avoid it is to not perform the mathematical solution, and just make up other things instead, not based on applying the heat flow equation, at least in the correct way.

Note that the interior of the shell readily radiates according to its temperature,

Psh = Ash*σ*Tsh4,

however, the heat flow this is capable of is equal to zero, because it is in equilibrium with the energy from the interior sphere.  The heat and energy from the interior sphere is obviously conserved on the outside of the shell, where the emission from the shell is translated as heat to the cooler exterior

Qsh = Ash*σ*Tsh4

if the exterior is 0K.  This is not a doubling of the available energy from the sphere emitted by the shell, because the shell is not capable of performing any work in its interior; only the sphere is.  Of course, if the exterior outside the shell is some other temperature, and lets assume it is a cooler one so that we don’t have to re-do the heat flow analysis with heat flowing from the exterior through the shell back inside(!), say it is like a cosmic background temperature, then

Qsh = Ash*σ*(Tsh4 – TCMB4).

Since the σ*TCMB4 basically comes from a source infinitely far away, then the shell heat flow eventually ends up as zero, i.e. heat death.  As a function of distance from the shell, its flux is given by σ*Tsh4*(rsh/r)2 where r is the distance from the shell, and rsh ≤ r ≤ ∞.  Then as a function of distance the heat flow from the shell is

Qsh = 4*π*r2*σ*(Tsh4*(rsh/r)2 – TCMB4)

and so the shell heat merges with the background when Qsh = 0, at

r = rsh*Tsh2/TCMB2.

For the Earth and its blackbody temperature of 255K and the real CMB of 2.7K, then the ‘r’ is 56.8 million kilometers; beyond that the Earth can’t provide heating above what the CMB would be providing (of course the Sun is still around and significant at this distance in any direction from the Earth).  The energy gets lost to the entropy of the universe, though of course it is conserved.

Read more from Joe Postma at climateofsophistry.com

## Comments (11)

• ### Pat Obar

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[quote name=”Leonard Weinstein”]This writeup is a typical example of why actual scientists that understand radiation Physics have no respect for many of the Principia Scientific Institute publications.[/quote]

Dr. Weinstein,
I have read som of your papers and agree with most of your your conclusions. This is outragious. Do you claim to be an actual scientist that understand radiation Physics?
If so you are the only “one”. Thousands of MEs understand thermodynamics but admit little undersatnding of radiation and Maxwell’s equations. Thousands of EEs understand radiation and Maxwell’s equations but admit little understanding of thermodynamics. You claim the understanding of both. [b]You claim the understanding of GOD[/b] Please fill the rest of us in on your understanding of “back radiation”.
Please include why it is that coherent radiation can deliver energy/second = power, while thermal radiation can only deliver entropy/second or entropy flux. The only difference wasw the introduction of “temperature” into “photons”. This may explain why photons “do not go to higher temperatures”. Thank you

• ### Pat Obar

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joeldshore says: 2013/12/10 at 8:07 PM
“Let’s take a look at all of this logic in Alan Siddons ’s article:”

[b]Joel[/b] The system as a whole is receiving a certain amount of energy from the sun and must radiate the same amount back into space. This determines the temperature of the shell. Then the temperature of the sphere is determined by the condition that the amount it radiates has to equal the amount it receives from the sun [or by its own thermal production if it is powered from within instead] and the amount it receives from the shell.

[b]Pat[/b] Reality: The sphere It receives no radiative power back from the shell all power is radiated only to the colder temperature outside the shell and the sphere. Any claimed radiative power from the lower temperature shell to the higher temperature sphere is a claim that violates 2LTD and such a claim is simply invalidated, and is never demonstrated. It is only claimed by (AAA) members that have not a clue.

[b]Joel[/b](And, if you don’t believe in back-radiation from the shell, it doesn’t matter because you can still say mathematically that the temperature of the sphere has to be such that the net power that it radiates to the shell, A*sigma*[(T_sp)^4 – T_sh)^4] has to equal the power that it receives from the sun or its own power source and you’ll get the exact same answer.

[b]Pat[/b] Reality: Since the results are the same Why violate 2LTD by claiming “back radiation”? In your fictional sphere, shell, space non-thought experiment, the fictional sphere temperature did increase by 1/2^(1/4) as now it is radiating not to the temperature of space. The the fictional sphere is now radiating to the temperature of your fictional shell.
The same result would occur if the emissivity of the sphere were reduced to 50%. Your (AAA) has no evidence that any increase in CO2 has produced such a reduction in emissivity of the Earth and its atmosphere, which are not radiatively coupled at all but only convectively coupled. Radiation between the surface and the atmosphere is non existent.
Just like your non existent “Greenhouse Gas Theory”. That is not a theory, It is a fantasy, that rises not even to the level of “conjecture”? Are we having fun yet?

• ### Pat Obar

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joeldshore says: 2013/12/10 at 8:07 PM
“Let’s take a look at all of this logic in Alan Siddons ’s article:”

[b]Allen[/b][i]In greenhouse physics, the sky responds to the surface by matching its temperature and emission. The surface then responds to the sky by getting warmer and radiating twice as much as it would without the influence of “greenhouse gases.” But then, why is it that these gases cease to respond to the surface’s higher temperature and emission? Facing a body that’s radiating 478 W/m², this infrared-absorbing layer should supposedly stay in character and radiate 478 W/m² back to the surface and another 478 W/m² out to space. Yet it stubbornly remains at -18° although the surface it responds to has reached 30°.[/i] [I]

[b]Joel[/b] Fallacy: Bodies in physics “stay in character” rather than satisfying the laws of physics.
[b]Joel[/b] ” I recommend you try to learn as much as you can about solving steady-state problems.”

[b]Pat[/b] Reality: No one cares of anything that Dr. (hic) Joel Shore may or may not say.

[b]Joel[/b] ” The steady-state solution is that the shell has to radiate as much back out into space as is being received by the system from the sun.
The reason the sphere in the steel greenhouse has to radiate twice as much as it did originally is not because this is in its character but rather because this is the steady-state solution that satisfies the equations for radiative balance:”

[b]Pat[/b] Reality: The shell radiates outward exactly the amount of real heat power as it actually receives from the sphere this means that the shell has no effect on the actual heat power radiated from the sphere (all the heat power it has received from the Sun or elsewhere). There is no more actual heat power to be radiated anywhere except outward from both the sphere and the shell. There is no heat power to ever be radiated by the so called “back radiation”

• ### Pat Obar

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joeldshore says: 2013/12/10 at 8:07 PM
“Let’s take a look at all of this logic in Alan Siddons ’s article:”

[b]Allen[/b] [i]It shouldn’t come as a surprise, moreover, that no one has ever demonstrated (let alone explained) how body X, which is as warm as body Y, can manage to raise the temperature of Y… while the now-warmer Y cannot manage to raise the temperature of X, even though thermal law dictates that it HAS TO because a temperature difference now exists.[/i]

[b]Joel[/b] Fallacy: Once again, the heat flow between two bodies interacting with the rest of the universe is sufficient to determine how their temperatures will change.

[b]Pat[/b] Reality: As long as nothing measurably changes, nothing measurably changes. Adding a Fake Greenhouse theory, changes nothing. Nothing in the Fake Greenhouse theory is measurable, all fantasy.!!

• ### Pat Obar

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joeldshore says: 2013/12/10 at 8:07 PM
“Let’s take a look at all of this logic in Alan Siddons ’s article:”

[b]Allen[/b] [i]This is confirmed by a European Space Agency chart.European Space Agency Chart
The section I’ve highlighted in blue describes the radiative response of a fully-absorptive flat plate. With a radiant barrier affixed to its backside, the 1-s plate can only radiate in one direction. Stimulated by a 1367 W/m² light beam, then, this one-sided surface responds with a 1367 W/m² infrared emission, a 100% return. The same plate without a radiant barrier (2-s) is able to radiate from both surfaces, however. So, having twice the area to radiate with, it yields only a 50% response from each side. As you see, it’s simply a matter of dilution: the same bundle of energy has been spread over a larger surface area. Please observe that a larger area of emission also prevents the two-sided plate from getting as warm as a single surface target. In short, the standard greenhouse diagram depicts a two-sided atmospheric layer that is impossibly warm and radiating an impossibly large amount of energy.[/i]

[b]Joel[/b]Fallacy: If I say 2 + 2 = 4 and you can show that the solution to a completely different problem is 8, that must mean that I am wrong.

[b]Pat[/b] Reality: A fixed amount of power radiating from a surface that doubles in area cuts the flux to 50% of its previous value. When the same environment the absolute temperature of the surface, will be reduced to 1/2^(1/4) its previous value, for all environments. The Earth has not increased its surface area or environment, space!
(Continued)

• ### Pat Obar

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joeldshore says: 2013/12/10 at 8:07 PM

“Let’s take a look at all of this logic in Alan Siddons ’s article:”

[b]Allen[/b][i] But here’s the snag: satellites report that the Earth emits to space the same amount of thermal energy as it receives from the sun. In other words, there is no evidence of radiative insulation, no physical sign of “greenhouse gases” acting like a heat-retaining blanket. [/i]

[b] Joel[/b] Fallacy: The steady-state solution is for the Earth’s emissions to be reduced if greenhouse gases act like a heat-retaining blanket.

[b]Pat[/b] Reality:the Earth emits to space the same amount of thermal energy as it receives from the sun. The exitance spectrum will look like this: http://www.barrettbellamyclimate.com/userimages/MODA.jpg ), and has not changed With CO2 concentrations changing from 270 ppmv to 400 ppmv at the 4 KM mauna loa observatory. That same chart shows that the Earth is nowhere near a black-body,m It does not even have the same temperature at various wavelengths. The S-B equation can never be used.
(Continued)

• ### Claudius Denk

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[quote name=”Leonard Weinstein”]This writeup is a typical example of why actual scientists that understand radiation Physics have no respect for many of the Principia Scientific Institute publications.[/quote]

I can’t even imagine how frustrating it must be to be so sure you are right and so completely unable to say how or why.

[quote name=”Leonard Weinstein”]
The article is basically wrong, and I have explained in detail on previous responses why.[/quote]

Put up or shut up.

[quote name=”Leonard Weinstein”]
The only comment I agree with is that the shell does not transfer “heat” to the sphere (by definition of heat transfer), but it does cause the sphere to heat up due to the transfer of back radiation energy (you can have energy transfer both ways, but heat transfer only refers to NET energy transfer), and this requires a higher sphere equilibrium temperature for a given energy net transfer for net energy balance.[/quote]

Scientific frauds always pretend they understand something and but are incapable of discussing/debating details. Just like a used car salesman.

[quote name=”Leonard Weinstein”]
This concept seems beyond the present authors ability to understand.[/quote]

Maybe you should find hobby that doesn’t involve complex things, like facts.

• ### JP

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Sorry Leonard but Willis’ greenhouse result is a simple violation of conservation of energy. He has 235 W/m^2 being emitted by the core, then 235 W/m^2 being emitted by the sphere. Therefore the ratio of the power emitted by the shell over that of the sphere is

Pshell/Psphere = 235/235 * (Rshell/Rsphere)^2 = (Rshell/Rsphere)^2

which is greater than 1 when Rshell > Rsphere. A ratio greater than 1 means that more energy is emitted than provided, thus violating conservation of energy.

On the other hand if you want to say that the sphere is emitting 470 W/m^2, then the ratio of the energy is

Pshell/Psphere = 235/470 * (Rshell/Rsphere)^2 = 0.5 * (Rshell/Rsphere)^2

and this ratio becomes larger than 1 when (Rshell/Rsphere)^2 becomes larger than 2. It is smaller than one when less than 2 of course. So this assessment of the energies don’t conserve either.

Finally, if you want to say that both the sphere and the shell emit a total power of 470 W/m^2, then the ration of power emitted to that supplied is, again

Pshell/Psphere = (Rshell/Rsphere)^2

which doesn’t conserve energy as soon as Rshell > Rsphere.

[quote]the shell does not transfer “heat” to the sphere (by definition of heat transfer), but it does cause the sphere to heat up due to the transfer of back radiation energy[/quote]

Actually that definition of heat transfer DOES mean that the shell can not heat up the sphere, and thus change its temperature…as per the definition.

The sphere heats the shell; the shell emits at a lower flux due to its larger radius.

Use the heat flow equation of thermal physics. Q = A*s*(T1 – T2). It’s quite straightforward.

• ### Leonard Weinstein

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This writeup is a typical example of why actual scientists that understand radiation Physics have no respect for many of the Principia Scientific Institute publications. The article is basically wrong, and I have explained in detail on previous responses why. The only comment I agree with is that the shell does not transfer “heat” to the sphere (by definition of heat transfer), but it does cause the sphere to heat up due to the transfer of back radiation energy (you can have energy transfer both ways, but heat transfer only refers to NET energy transfer), and this requires a higher sphere equilibrium temperature for a given energy net transfer for net energy balance. This concept seems beyond the present authors ability to understand.

• ### JP

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