## Entering the SkyDragon’s lair

Posted: March 10, 2013 by tallbloke in Analysis, Astrophysics, Energy

I’ll probably regret this, but I felt the need to place a comment on ‘SkyDragon’ Joe Postma’s site, on a thread where he has had a huge rant about Willis Eschenbach’s ‘Steel Greenhouse’ toy planet concept.

Figure 2: 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. The shell is receiving 470 W/m2 from the planet, and it is radiating the same amount, half inwards back to the planet and half outwards to outer space. Note also that despite the fact that the planetary surface ends up much warmer (radiating 470 W/m2), energy is conserved. The same 235 W/m2 of energy is emitted to space as in Figure 1. [Note]: Figure 2 is not to scale; the shell is very close to the planet surface, so areas are ~equal

Joe goes postal on this, saying in part:

Anyone who thinks that there is any actual modern physics or mathematics in this description of the greenhouse effect and who can’t immediately identify the absurd degree of pseudoscience and illogic is a complete moron.  These people are complete, unfettered idiots, and are a disgrace to mathematics….Willis just arbitrarily doubled the amount of energy available, so that he could add half of it back to the original 235 W/m2 in order to double it.  Just arbitrarily doubled out of nowhere.  Just made up bullshit….And then what is strange, is that Willis stops this energy doubling process for no reason!  If at the beginning, a 235 W/m2 output comes back to double itself to 470 W/m2, increasing its own temperature, then why doesn’t the 470 W/m2 output double again from itself coming back to increase itself yet again?

Here’s my  response:

tallbloke says:
2013/03/10 at 6:45 AM
Hi Joe.
Much as I have my differences with Willis Eschenbach, I have to say his steel greenhouse model is theoretically correct. Where he’s gone wrong  in the past is by misapplying it to the real Earth, which doesn’t have a vacuum between sky and ground, has lots of potential energy locked up in the hydrological cycle, and the dayside/nightside differential you have previously posted about.

The colder radiating to warmer thing is easy to understand. The key is to consider the net outward radiation. At equilibrium, the outer shell has to radiate 235 to space. We can all agree on that. The inner and outer surfaces of the outer shell will both radiate, not quite equally, but near enough that we can disregard the difference. We can all agree on that too I hope. Added up, the inner and outer surfaces have to be radiating at 470 in total in order for 235 to be going to space and maintaining equilibrium. Therefore the planet surface will reach equilibrium with the outer shell by heating up until it is radiating 470 too.

Your main complaint seems to be that the ‘back-radiation’ from the inner surface of the outer shell can’t possibly ‘heat’ the planet’s surface because that would violate the second law of thermodynamics. Quite right too… but that’s not what happens. What actually happens when the planet is first suddenly surrounded by the steel shell is this:

The planet radiates 235 as it was doing before, but it is absorbed by the outer shell, which then radiates 117.5 outwards and 117.5 inwards. Obviously this is only half what the planet is radiating outwards, so it isn’t going to make the planet surface hotter than it already is by itself. But what it will do is add to the total amount of radiation the planetary surface is receiving. Whereas before it got 235 from the decaying radiation, and nothing from space (disregarding the CMBR), it’s now getting 235 from below, and 117.5 from above for a total of 352.5.

In order to be at equilibrium, it’s going to rise in temperature until it’s radiating the same 352.5. But this isn’t the end of the story, because the outer shell is now receiving that 352.5 and will radiate half out and half in. So now the planets surface will still be getting 235 from within and 176.25 from above for a total of 411.25. In an iterative process, the surface will warm until it is radiating at a high enough rate to get the outer surface of the surrounding shell to radiate at 235 to space so the whole ensemble can reach equilibrium. That rate will be 470, because that is the level at which the outer shell can split its re-radiation such that 235 goes to space.

Now, I know you will object to this on the grounds that radiation from colder objects isn’t absorbed by warmer objects due to some mysterious ‘pseudo-scattering’ process Claes Johnson didn’t manage to explain to Jeff ID’s satisfaction, but ask yourself this:

If the ‘back-radiation’ incident on the surface is somehow ‘scattered’, where is it scattered to? if it isn’t absorbed by the surface, it has to be re-absorbed by the inner surface of the outer shell. But if that were the case, the outer shell would never rise above the temperature where it is radiating a total of 235. If that were the case only half of that would be radiated out to space. In which case equilibrium won’t be reached between the planet and space, as it was before the steel shell was added. Something, somewhere, is going to get very hot indeed if that carries on for any length of time. What would that something be? The planet’s core? Its surface? the steel shell? What else is left?

Best to you.

Rog TB.

1. Bryan says:

Correction should be

The only people impressed by your pseudoscience are the unfortunate bewildered uninformed who can be persuaded that heat spontaneously moves from lower to higher temperatures objects.

Tim’s pseudoscience must be infectious.

2. wayne says:

TJF: “If the shell was like a balloon that expanded (but was kept at 254 K by heaters), then the temperature of the planet would NOT change as the shell expanded or contracted. The planet would remain at the elevated temperature of 302 K for any size of shell. (And before anyone says I just contradicted something wrote earlier, please pay careful attention to the specific conditions for the size and/or heating of the shell.)”

Glad you finally see the light Tim, that is what I was saying to you on earlier posts. If the “effective emission level”, the EEL that you speak of often was to ever raise in altitude due to an increasing co2 it is going to have zero affect on the surface temperature all else remaining the same (temps), the environmental lapse would decrease in slope. But you are right, the temperature of the shell does matter and a higher EEL altitude with greater surface area is going to be cooler at a fixed input and the temperature of the surface would cool proportionally (unless the sun outputs more or albedo decreases). See, co2 is a net zero agent, one aspect says it will warm but the other aspect equally says it cools, result: flat temperatures for decades. The sun was warmer in the 80’s and 90’s no matter what NASA was able to capture from instruments… you could see it daily watching SOHO.

So as to spheres and their shells, all in all, altitudes do not matter, and the EEL can finally be put to bed, just the temperature of shells and the amount of IR leaving without any interaction matter. Agreed?

3. tallbloke says:

Tim F: Last time I went outside there was still a troposphere there. Willis’ vacuum packed toy planet has little to do with Earth, although radiation levels are a good temperature diagnostic within the system. Stand by for the results of some clever code work by Wayne.

4. gbaikie says:

“If Earth was creating a solar wind, this would disrupt and interact with the
Van Van belt and give Earth a mini-corona.”

Got to wondering what actually causing solar wind. So googled:
“The solar wind emanates from the Sun in all directions, but seems to emmanate most readily from the Sun’s coronal holes. Exactly what causes the solar wind to be accelerated, or “blown” into space is not well understood.”
Hmm:
“In fact, 1 million tons of particles come from the Sun every second!”
And:
“It is considered to be the continual expansion of the Sun’s atmosphere.”
http://www.windows2universe.org/glossary/solar_wind.html&edu=high

So a planet can out gas it’s atmosphere [lose it’s atmosphere]. So
the Sun is out gassing it’s atmosphere.
But the Sun has massive gravity well. I imagine, few would think of a gas giant at
a body which out gases. And our gas giants are tiny compared to the Sun.

And can star not out gas, or can star out gas at slower rate.
So can a star have atmosphere that doesn’t leak?

So what would such star need not to have as much out gassing?
It seems our sun is creating solar wind [and CMEs] relating to it’s magnetic
activity.
So perhaps if our sun didn’t rotate as quickly.
Also our gas giants [and bodies like Earth with it’s magnetosphere] magnetic
fields are to some extent interacting with Sun’s magnetosphere. So perhaps
a star with fewer or no planets with magnetospheres.

Hydrogen plasma is quite magnetic, what star burn through all it’s hydrogen-
that should cause less magnetic activity. For instance a white dwarf is just
a huge amount of hot carbon [a massive diamond]. So how could a white
dwarf make a solar wind?

5. gbaikie says:

“In fact my contention is that GHGs play no part in the transfer of energy up the atmospheric column (although they do play a vital role, obviously, in transforming incoming radiation from the Sun to kinetic energy and in transforming kinetic energy at the top of the atmosphere to radiation to space.”

It seems that if just talking about transferring energy up atmospheric column, vaporizing water
is doing 99.9% [or some number near all] of this.
On Mars where CO2 freezes and evaporates, CO2 would be fairly dominate in this regard [though there also some water on Mars which freezes and evaporates]

Interesting picture of Mars:

Description
English: “This false color photo of the surface of Mars was taken by Viking Lander 2 at its Utopia Planitia landing site on May 18, 1979, and relayed to Earth by Orbiter 1 on June 7. It shows a thin coating of water ice on the rocks and soil. The time the frost appeared corresponds almost exactly with the buildup of frost one Martian year (23 Earth months) ago. Then it remained on the surface for about 100 days. Scientists believe dust particles in the atmosphere pick up bits of solid water. That combination is not heavy enough to settle to the ground. But carbon dioxide, which makes up 95 percent of the Martian atmosphere, freezes and adheres to the particles and they become heavy enough to sink. ”
From wiki article here:
http://en.wikipedia.org/wiki/Water_on_Mars#Viking_program

6. Tim Folkerts says:

Bryan,

I see you completely side-stepped the glaring hole I pointed out in your model (actually two glaring holes now). Instead you “blather” (one of your favorite words) about “pseudo science” and “Maxwell’s Demon”.

Shall we assume that you agree with my objection since you couldn’t defend your own model?

So here is a simple challenge. Find ONE SPECIFIC THING that you think I got wrong — a specific equation or sentence that you think has some glaring error. Tell us specifically what you think the answer should be.

7. Second experiment
Isolated heated hot object
Distant (well 9cm) cold plate/warmplate
shows hot object temperature increase when external cold plate is replaced with identical warm plate.

Not the final experiment. but getting there.

8. Tim Folkerts says:

wayne says: March 15, 2013 at 10:47 pm …

Wayne, you make an interesting point (although it is not exactly the point I was making in the paragraph of mine you were quoting).

This particular “steel shell” model would behave much as you suggest. The “lapse rate” does adjust to whatever is needed to keep the shell at ~ 254 K in this model (minus a small amount due to the “radius cooling effect”). So whether the shell is 10 m or 1000 m or 10,000 m above the planet, the shell will be 254K and the planet will be 302 K.

I would suggest, however, that this is not necessarily the same way the earth with an atmosphere would behave. It is certainly not the way the steel planet would behave with an atmosphere.

Suppose we added a pure N2 atmosphere under the shell. So by definition, we have avoided latent heat of water and IR within the atmosphere. With similar gravity to earth, the dry adiabatic lapse rate will be ~ 10 K/km. So if the shell was 10 km above the surface, the “IR lapse rate” will be (302K-254K)/10km = 4.8 K/km. Since this is less than the DALR, then the atmosphere will be stable, there will be no convection, and the planet would be 302 K.

If the shell was 5 km up, we gets similar result. The “IR lapse rate” is 48K/5 km = 9.6 K/km, which is still a stable atmosphere so there will be no convection and the planet will be 302 K.

If, however, the shell was only 4 km above the surface, then the “IR lapse rate” = 48K/4km = 12 K/km is GREATER than the DALR. Convection will kick in, trying to lower the lapse rate back to 10 K/km. The surface will now only be ~ 254K + (4 km)*(10 K/km) = 294 K. There will now only be ~187 W/m^2 of IR going from planet to shell, so that leave ~ 235 – 187 = 48 W/m^2 of convection.

So …
* above ~ 4.8 km, the energy all goes as IR and none as convection, that the planet remains the same 302 K temperature.
* below ~ 4.8 km, more and more energy goes as convection, and the planet gets cooler and cooler (down eventually to 254 K with the shell right above the surface.

I could say more, but I think this is enough for starters. Basically, I think in our atmosphere, convection is always important and will always keep the lapse rate more or less the same, requiring the surface to change temperature. You believe otherwise.

9. Tim Folkerts says:

Roger,

The main things this model tells us about earth is that
* When viewed from space, we are seeing radiation coming from a “shell” above the earth. (Actually several shell that are semi-transparent and at different altitudes due to various GHGs and clouds).

* This shell (or these shells) will try to raise the surface temperature above what it would be without those shells to block IR created at the surface.

That’s it.

(To expand just a little, there are two very different “mass effects” that I recognize.
* The first is to “even out temperatures” which steals energy from the warm day-side and gives it back to the cool night-side.
* second is to “raise the shell”. More mass “inflates the shell to a higher altitude”. As outlined in the previous note to wayne, raising the shell will, in at least some circumstance, warm the planet. But of course, this effect only works when there is an IR blocking shell (like CO2) in the atmosphere that can be raised.)

10. suricat says:

Tim Folkerts says: March 16, 2013 at 3:48 am

“(To expand just a little, there are two very different “mass effects” that I recognize.
* The first is to “even out temperatures” which steals energy from the warm day-side and gives it back to the cool night-side.
* second is to “raise the shell”. More mass “inflates the shell to a higher altitude”. As outlined in the previous note to wayne, raising the shell will, in at least some circumstance, warm the planet. But of course, this effect only works when there is an IR blocking shell (like CO2) in the atmosphere that can be raised.)”

Hallelujah!!! A mention of the ‘latent energy’ in “The first” that ‘ISNT’ mentioned as part of the properties of ‘The Toy’!

Did I say that? What’s this new phenomenon? 😉

Shouldn’t we stay ‘on thread’ with our posts? 🙂

Best regards, Ray.

11. wayne says:

Tim, it is an interesting thought isn’t it, thank for bending my mind around onto the area and radius. You are right, distance doesn’t matter.

I immediately thought of the sun, make it the shell with the sphere centered within, the shell is the energy source this time, no internal energy from the sphere. That sets the shell to emit 340.6 W/m² both inward and outward, but that is the TSI/4 we are so familiar with in climate “science” and the internal of that shell and anything within it must equalize to 5.24°C no matter what the composition is of the sphere or no matter what it has about it (atmosphere).

That to me is a very, very interesting point and deserves some investigation, so, I might not be here ever moment for a while, got some work to do on this revelation. It’s all in the geometry isn’t it.

Just so no one has to calculate this figure, that sphere (5.24°C Earth) is missing 55.4 W/m² if it really has an average surface temperature of 289.1 K. Now where do you imagine that missing energy is coming from? CO2 does not manufacture energy.

12. Bryan says:

Tim F says

” Bryan, I see you completely side-stepped the glaring hole I pointed out in your model”

Tim’s so called glaring hole is that…….

“and the shell on the outside expands. ”

Not necessarily, there are metals with no expansion coefficient.

I do not tend to repeat myself just because you fail to understand when something is carefully explained to you.

If you still do not understand that the Willis model contradicts the second law and probably the first then why don’t you read a thermodynamics text book.

13. lgl says:

wayne
That ‘missing’ energy is coming from the atmosphere. The steel shell would reemit 50%, the atmosphere is reemitting 60%. Many (most?) wavelengths are absorbed and reemitted several times, meeting several shells if you like.

14. tallbloke says:

Tim F: The main things this model tells us about earth is that
* When viewed from space, we are seeing radiation coming from a “shell” above the earth. (Actually several shell that are semi-transparent and at different altitudes due to various GHGs and clouds).

One of the shells is on the earth’s surface, and emits a big percentage of OLR direct from surface to space (the ‘window’).

* This shell (or these shells) will try to raise the surface temperature above what it would be without those shells to block IR created at the surface.

You still haven’t given us a mechanism by which this can happen. The free path length is short, and downward directed radiation energy is soon absorbed in collisions with upward convecting air. How is the back-radiated energy going to ‘warm the surface’?

(To expand just a little, there are two very different “mass effects” that I recognize.
* The first is to “even out temperatures” which steals energy from the warm day-side and gives it back to the cool night-side.

Agreed, evening out differentials and reducing swings leads to higher average temperature

* second is to “raise the shell”. More mass “inflates the shell to a higher altitude”. As outlined in the previous note to wayne, raising the shell will, in at least some circumstance, warm the planet. But of course, this effect only works when there is an IR blocking shell (like CO2) in the atmosphere that can be raised.)

Water vapour is a much better example than co2 because it is 1) a much more powerful IR blocker and 2) It works to cool as well as warm in the lower troposphere, where co2 absorption/re-emission/collision does nothing at all to overall temperature.

By limiting yourself to two effects of mass, you don’t recognise the effect of mass with gravity acting on it to create a pressure/density gradient increasing the heat capacity of the nearer surface air.

And you don’t recognise the effect of the mass of the atmosphere limiting the rate of evaporation from the ocean, which must force it to a higher temperature than it would be at in a lower pressure environment. You are in denial of basic thermodynamic realities.

15. wayne says:

lgl, read that description again, you must have misinterpreted it somewhere as you read.

The sphere is totally inert, if it has an atmosphere it is also inert, just mass, neither is creating energy in this case so no, there is no re-radiating atmosphere and mysterious warming in this case. Heck, the sphere could be a any matter, a bowling ball, a Mars with nearly 100% CO2 atmosphere, an Earth minus any radioactive atoms in the interior, they all will equalize over time to that ~5°C temperature. So how is the Earth now at 16.1°C per TFK’s paper? (btw: I do believe that 16.1 is approximately correct, probably too high but close) Careful, it’s a rather hard question to answer if I didn’t make some mistake specifying it.

16. lgl says:

wayne
Yes I probably did misinterpret somewhere, but it’s not easy when you are mixing setups.
The way I interpret it
1. Shell heated by the Sun, 340 inward and 340 out and planet surface 5 C. Probably all good (I haven’t check your calcs)
2. Then you want the 340 to warm the surface to todays 16 C, but the two are totally different scenarios. The real planet has a heat source at the surface, the absorbed solar, which b t w is around 200 W/m2 (170 directly + 30 reemitted from the atmosphere), not 340.
The shell will radiate both inwards and outwards, the planet only outwards, so shell and planet being the source will give totally different results.

17. wayne says:

lgl, this printout from my program might save you a lot of time on the math, you know, where this thought even came from, the calcs are simple, just remember this is a powered BB shell of 1 AU radius, continuously warmed enough to have a radiance power of 340.6 W/m² (solar) both one the inside and on the ouside of the shell. The object in question is at the center with no energy production of any type. This is basically where climate ‘science’ kindly started me off a few years back and I should have seen this apparent paradox right off the bat.

σ = ‹kSB›
5.6704e-008 W/m²/K^4

AreaOfHalfSun = 2·π·(696E+6)² /*we only have use of power from 1 hemisphere*/
3.0437e+018 m²

AreaOf1AuShell = 4·π·(149.6E+9)²
2.8124e+023 m²

OutputOfSun = AreaOfHalfSun · σ·5772²²
1.9157e+026 W

BothSideRadiance = OutputOfSun / AreaOf1AuShell /*inside&outsize of shell*/
681.2 W/m²

340.6 W/m²

EffectiveTemp = √√(OneSideRadiance/σ) /*objects inside shell*/
278.4 K

EffectiveTempC = k2c(EffectiveTemp)
5.238 °C

Now you can just point if you see any mistakes, I can’t see any.

18. lgl says:

wayne
Like I said, that part is probably ok.
But then you bring in the 16C which is from a totally different setup, with heat generated at the planets surface (from absorbed solar radiation).

19. wayne says:

lgl: “The real planet has a heat source at the surface …”

Our planet’s surface is not an “energy source”! Our energy source comes from outside, so does this one, precisely the same as reality, see the TFK2009’s graphic energy budget.

This is no trick lgl. I’m not trying to convince you or anyone of anything. I’m trying to get everyone to think, analytically, help me answer this rather curious configuration of a steel shell encased planet. Willis was very slick in his starting conditions, I’m not, this should be an exact parallel case to the sun and Earth only being evenly warmed the way climatologists seem to always calculate energy flux (W/m²) and budgets. I see a problem, don’t you?

20. lgl says:

wayne
No I don’t see a problem (other than the silly 3x feedback amplification only existing in the minds of the CAGWers, but that’s another story).
Our energy source is the Sun, but the heat source is the surface where the radiation energy is converted into heat energy. Your shell heat source is a totally different thing.

21. gbaikie says:

“I immediately thought of the sun, make it the shell with the sphere centered within, the shell is the energy source this time, no internal energy from the sphere. That sets the shell to emit 340.6 W/m² both inward and outward, but that is the TSI/4 we are so familiar with in climate “science” and the internal of that shell and anything within it must equalize to 5.24°C no matter what the composition is of the sphere or no matter what it has about it (atmosphere).

That to me is a very, very interesting point and deserves some investigation, so, I might not be here ever moment for a while, got some work to do on this revelation. It’s all in the geometry isn’t it.

Just so no one has to calculate this figure, that sphere (5.24°C Earth) is missing 55.4 W/m² if it really has an average surface temperature of 289.1 K. Now where do you imagine that missing energy is coming from? CO2 does not manufacture energy.”

If the sun had shell at Earth distance, it would radiate our solar constant- 1363 watts per square meter.
It would radiate 1361 W/m-2 outward and 1361 W/m-2 inward and I would think have little effect
of warming the Sun- the flow of heat is outward.
Though if inner part of shell was reflective, then it might heat the sun.

If one had the inner part of sphere reflecting most of the sunlight nearly perfectly and one were at inner planet like say Venus. The entire sky is light, but the reflected light is not pointing at Venus- we could assume mostly directed at the Sun.

Venus is 108.2 million km from Sun. And Earth is 149.6 million km. So Venus is 41.4 million km from closest part of sphere and since the sphere is so large it’s essential very similar to a flat wall. And in this wall, the sun will reflected. You will see a smaller twin of the sun. It will emit far less energy
than the actual sun. It will be like being 149.6 + 41.4 million km from the Sun. So 191 million km
from the sun [closer than Mars]. So getting 2600 W/m-2 from sun and somewhere near 1000 W/m-2 from the reflection.
Therefore Venus probably wouldn’t get much hotter than it is presently from this reflected sunlight.

The next part is how warm is this sphere. How energy will it emit due to it being warm.

It seems with a reflected inner surface, one could design it to be quite cool, but it also seem one could design it to have high temperature. By which I mean one could surface temperature being around 400 K.
If the inner sphere is 400 K. Then it could reflect 1363 W/m-2 plus emit infrared inward and requiring some heat to emitted outward [no matter how well it’s radiantly inhibited. Though if designed so the inner surface was coolest one could make it, then one would have easier job or inhibiting the radiation emitted from outer part of shell.

So say designed so inner surface is around 400 K- [1451 W/-2 of infrared].
If assume IR is reflected like the sun’s light- coming from same spot in sphere that reflects the Sun.
Therefore the combined reflected sunlight plus infared light is approaching energy that Venus would receive directly from the Sun. And according, could make Venus a bit warmer.

The next part is about Venus ability to emit energy. The sky is not 2 K, but could as high as 400 K.

Now if assume the inner surface is 400 K, does it mean that from Venus distance it is 400 K.

Let’s start by imagining shell is 2 K. How much would our current Venus heat this shell per square meter? So take it at it’s closest of 41.4 million km, how does heat a square meter.
Or how much does the current Venus heat stuff at earth distance. It would seem Venus heats Earth by considerable less than our Moon heats Earth. So safe to say less than 1 watt/m-2.

So it might be reasonable to assume the radiant from Venus reaches the 400 K inner sphere and doesn’t warm it, but it’s not going to return to Venus, and if it did, it’s not going warm Venus.

Now could make argument that the heat will not flow to warmer surface. Which certainly reasonable.
but it get back to question how warm is it in relation to Venus. Yes, we know it’s 400 K- cause that is what we said it was. But only small part of sphere is radiating a significant amount radiation at Venus, so from Venus prospective most of sphere can not measured from Venus
as being hot. Or if looking perpendicular to inner shell surface it’s somewhere around 2000 W/m-2.

All the above could have nothing to do the post I quoted.
The ” 5.24°C” makes me think it’s referring to the temperature of ideal blackbody sphere
at Earth distance. Wiki says it’s 5.3 C: “If an ideal thermally conductive blackbody was the same distance from the Sun as the Earth is, it would have a temperature of about 5.3 °C”
http://en.wikipedia.org/wiki/Greenhouse_effect
And so if had blackbody shell above earth’s surface.
Now in regard to that, and this statement “It’s all in the geometry isn’t it.”
I would say an ideal blackbody sphere is 2 dimensional.
By which I mean referring to radiation outward from spherical surface. If 3 dimensional
it would not be divided by 4, but divided by 8 [outer and inner surface area].
So if you put ideal blackbody on surface one could ignore this- one can imagine the
surface beneath it is same temperature.
Now this part:
“that sphere (5.24°C Earth) is missing 55.4 W/m² if it really has an average surface temperature of 289.1 K.”

So, the 5.24°C is a uniform temperature- one can say it’s average temperature.

I would say that if Earth had uniform temperature of 5.24°C it would warmer
than our present Earth. Earth is mostly ocean and the entire ocean is around 3 C.
So Earth’s were 5 C, it would be a warmer world. If just looking the surface temperature
of Earth it’s much warmer than 5 C, but surface is tiny fraction of the ocean.
Or if our ocean were 5 C, we might consider ourselves as having left the ice house period we been been in for millions of years- and we would not have polar ice caps.

The basic problem is people assume ideal blackbody gives one warmest conditions possible,
whereas it’s mostly telling you how hot something can in general get at a certain distance
from the sun. Or how much the sun can heat something at a certain distance if one absorbs
all the heat and emits it as perfect blackbody spectrum.
If it absorbs more heat than emits, than you get higher temperatures. And to do this, one needs understand it from 3rd dimensional view. And including time- the 4th fourth dimension is also useful.

22. wayne says:

gbaikie, bless you for thinking but you went off like a shotgun. 😉 Whoa. It’s incredibly more simple than you are making it. I’ll try one last attempt and if not one person can say they do see exactly what I am describing and the problem, I give.

First, there is no longer a sun, no other planets, no solar system, just two objects, a clump of mass (planet), no poles, it doesn’t matter if it spins or not, in the center… and a shell of 1 AU radius about it, look at Willis’s picture above. The differences to this and the Earth/Sun is totally in the geometry, think geometry. In the real Earth/Sun case 1/2 of a hemisphere of the sun shines parallel rays to 1/2 hemisphere of the Earth every second, always, at any moment. You are right, the intensity is ~1362 W/m². But by the spherical curve of the Earth you have to divide by two, 1362/2 W/m². But wait, that is one side only, divide by two again, 1362/4 W/m². I’m sure you have hear this a million times. So, on the average the Earth receives ~340.5 J/s/m². Look at my math, that is what the sun outputs if spread over that shell unless I made some stupid mistake.

Now, make the shell to have the power to radiate outward at that ~340.5 J/s/m², the same SB power you will find on the inside, just like in the Earth/Sun case but now the sphere is actually able to absorb equally at any point on its surface, a maximum of that ~340.5 W/m² radiance and will if the planet in the center starts out at absolute zero. But over time that clump of matter called the planet will absorb and warm to what temperature? About 5.3°C. Now how is it ever going to get any warmer? It is now at equilibrium, like a cup of coffee in a room (shell). The shell is constantly radiating an equal ~340.5 outward into the universe. Simple, it can’t and never will it seems which should cause someone else to ask further questions but it seems no one has gotten that far yet. I’m still waiting.

So, that’s my question. Keep it so simple. Just geometry, energy, flux and such (distances don’t seem to matter as TimF pointed out, atmospheres shouldn’t matter either in this case).

One question comes to my mind, if the Earth were in a 0K space environment (I know, ignoring the 2.7K) could it be feasible that over time the real Earth would still have an inherent internal energy density to have a stable temperature of just 0.54 K as a starting point (within limits of time)? That should do it, or, am I wrong and why.

23. wayne says:

Well, on that last paragraph, answered my own question. It would require ~176K as a starting point, not 0.54K. Forget that thought. 😉 Must lie in that a great area of Earth does receives an intensity of ~1000+ W/m2 a great amount of the time and able due to that instantaneous large temperature differential to bury that much energy under the soil and in the top few feet of the oceans for the pass over the night. Probably just too tired, need a break.

24. Tim Folkerts says:

Bryan says: ” Tim’s so called glaring hole is that…….“and the shell on the outside expands. ” “

Good try, but of course that still doesn’t save your model.
1) Even metals with a small coefficient of thermal expansion like Invar still have SOME expansion.
2) You have also stipulated a large coefficient of expansion for your planet so that it can expand outward to touch the shell. So a small amount of cooling will make it contract and loose contact.

Basically, you are postulating situation where the two surfaces get closer as the planet warms, but don’t get farther apart as they cool. (Among other problems.)

*******************************************************
“If you still do not understand that the Willis model contradicts the second law and probably the first …

Willis’ model DOES NOT contradict the 2nd Law (nor does your fancy expanding/contracting model, but let’s stick to your claim about Willis’ model).

Where & when specifically does …
* Σ( P_i ln(1/P-i) ) decrease?
* heat (the net flow of thermal energy) go spontaneously from cold to warm?
* mechanical work get done with no heat rejected to a cold reservoir?

These are standard statements of the 2nd Law. None of these happen. Go ahead … pick one and explain where this happens in Willis’ model. Heck, find your OWN statement of the 2nd Law and show how it is violated.

25. Tim Folkerts says:

Wayne,

I think I understand your model after you re-explained it in your post to gbaikie. It is an interesting scenario.

I do think that lgl has a very legitimate point about the location of the heater. Heating the outer shell is conceptually very different from heating the inner planet. It is the same sort of difference you would get if you heated the outside of your house’s walls vs heating the inside of the walls. If you heat the outside to 20 C (when the air temperature is say 0 C), then the interior will also be 20 C. If you move those same heaters at the same settings to the interior, then the exterior of the walls will still be 20 C, but the interior will be much warmer.

Sunlight is such an “interior heater”. The photons travel freely (or at least fairly freely) through the atmosphere and deposit their energy at the surface, generating thermal energy there.

26. A C Osborn says:

What I find most amazing about this thread is that a few people can keep on repeating the same thing over and over again, although in different formats, but when “apparent” Physicists say that they are wrong it is completely ignored, they say that the physicists are wrong and then just carry on as if nothing has happened.

27. wayne says:

“Sunlight is such an “interior heater”. The photons travel freely (or at least fairly freely) through the atmosphere and deposit their energy at the surface, generating thermal energy there.”

But that is one area I don’t understand why that phrasse keeps being said over and over again. Speaking of radation, on the solar side 340 tries to get to the surface 161 makes it, 47%. Not such a great heater. On the IR side 396 strives to get out of the system and 238 succeeds, 60%. So which one passes energy through our atmosphere system with ease? See, those seem to just be cherry picked words to me.

28. Tim Folkerts says:

Tallbloke, I agree mostly with what you say March 16, 2013 at 9:09 am

1) I was starting with the two most obvious (and I believe the two most important effects) of mass. I was not specifically trying to deny that any other effects are possible.

2) At some level, the “mechanism” is simply conservation of energy. Less energy is observed leaving the ToA in the 15 um band that would be emitted by a blackbody at 255K. Therefore some part of the earth MUST emit more energy than a 255 K blackbody.

You could also say the mechanism is “the lapse rate” (determined primarily by convection * adiabatic compression/expansion). Air that is cooled at ~ 12 km up by radiating energy to space from ~ 220 K CO2 molecules will warm as it descends. By the time it gets to the surface, it has warmed by compression (presumably to ~ 288 K).

29. tallbloke says:

Tim F: The vast majority of energy radiated to space is radiated to space by water vapour molecules and the ground, not co2 molecules.
True or false?

30. Roger Clague says:

David Socrates says

GHGs play no part in the transfer of energy up the atmpospheric column (although they do play a vital role, obviously, in transforming incoming radiation from the Sun to kinetic energy and in transforming kinetic energy at the top of the atmosphere to radiation to space.

It is the mass of the earth, land and seas and air that converts radiation into kinetic energy.

http://clivebest.com/blog/?p=4475

According to this post fig.1GHGs ( other than H2O ) provide 15% of outgoing IR. If not there then IR will still escape by other means.

Land, sea, air ( mainly N2 and O2 )and water determine the weather and climate. GHGs, other than water, are not vital.

31. Tim Folkerts says:

Roger, I would say that is true. I don’t have specific numbers handy, but my intuition says
* clouds radiate the most IR (they are close to black bodies and cover the majority of the earth)
* land and water vapor are next
* CO2 is 4th

In defense of the importance of CO2
1) it is clearly increasing
2) it is higher altitude than the others, so it “trumps” the others in the bands where it absorbs.

Clouds and water vapor can change as a result of other factors (both natural and anthropogenic), so they can increase and decrease — hence the tricky aspect of feedbacks. But CO2 is always increasing. Thus its impact should also always grow (due to both broadening of the band and raising of the emission height). (As Clive pointed out earlier, the changes are not always as simple as an across-the-band decrease in emitted radiation.)

So while the other factors are larger MAGNITUDE, CO2 could well have a larger CHANGE. (The devil is in the details).

32. Bryan says:

Tim F says

“Bryan says: ” Tim’s so called glaring hole is that…….“and the shell on the outside expands. ” “

Good try, but of course that still doesn’t save your model.
1) Even metals with a small coefficient of thermal expansion like Invar still have SOME expansion.”

Tim you will really have to try harder to avoid making stupid mistakes

http://www.caltech.edu/content/caltech-scientists-use-high-pressure-alchemy-create-nonexpanding-metals/

So now for the second time we have established that ‘your big glaring hole’ does not exist.

Remember that you specified a planet and a thin shell.
Its quite obvious that the planet will be in contact with the shell for a considerable time given its large heat capacity and the small heat capacity of the shell.

Still we are making progress if you realise that the shell radiating for a considerable time at 470W/m2 is against the second law

Some people fail to realise that for instance a cold cup of coffee spontaneously heating up never happens.
This would be permitted by the first law but not the second.

Likewise a passive shell cannot transfer energy such that an object at 254W/m2 can spontaneously transfer heat to an object at 302K.

I think that in the dim and distant past you might have attended a thermodynamic class but have forgotten most of it.
But you surely remember Maxwell’s Demon and how it it failed due to second law considerations.
If you can take the Willis model energy stream and alternately allow the alternating 235 and 470 W/m2 radiation to heat a 2 adjacent gas containers such that one is hotter than the other then mechanical work is easily arranges.

Of course only in fairyland is it possible that such an arrangement works.

33. Kristian says:

Seriously, people. You’re not seeing the forest for the trees, here. Use your logic and your intuition. I especially implore Tallbloke himself to think this through, for I consider him to have a naturally open mind on matters such as this. And open minds can be changed.

Willis’ planet/sphere model is based (among other things) on the two following premises:

1) Conduction is ‘efficient’ – if not perfect, then at least expeditious
enough to be disregarded as a significant hindrance to heat flow from
inside to outside.

2) Both bodies (core planet & surrounding shell) should radiatively be
treated as black bodies – absorbing and emitting in exact accordance to
absolute temperatures.

His ONLY point is that the shell has two surfaces, while a solid sphere only has one – therefore there will always be a temperature difference between the shell and the planet.

Er … wrong.

How will a spherical shell in space supplied with a certain amount of energy balance this gain with loss to keep a steady state at a fixed temperature?

I wholeheartedly agree with cementafriend on this matter. Thermal radiation is a way for a body to lose heat. To cool. Period. It is not some inherent property completely independent from and unaffected by the surroundings of the body. As all experience tells us, the body cannot and will not lose heat by radiating in the direction of warmer. It can only lose heat by radiating in the direction of cooler. And it does so adequately.

Remember, this is the quantum world we’re dealing with here, not our everyday macroscale life. It’s not like a linear exercise where we follow and track each little photon on its rollercoaster ride through the Cosmos. This is about continuous flow (‘waves’) of energy across temperature gradients. There is a certain order of things.

Look,

We have two different bodies in space, both spheres with an outside surface area of 1 square metre. The one sphere, however, is solid and the other one is hollow, i.e. it is a spherical shell enclosing an internal void. Ignoring any minor differences in area between the inner and the outer surface of the shell, we can conclude that its total surface area is 2 square metres.

Now, the solid sphere has a nuclear power source within. This source provides a constant wattage to keep the sphere at a steady temperature of 300K. This sphere is equivalent/comparable to Willis’ core planet and remains at a state of dynamic equilibrium – output to the surroundings (heat loss) balances input from the heat source (heat gain). In other words, we can deduce from this that the power input is 459 W.

The hollow sphere (shell) also has a nuclear power source within, only spread out evenly across the inside of the sphere, like a paper-thin intradermal layer, so to say. And this source provides the same power input as the source in the solid sphere. The shell is also aspiring to reach a state of dynamic equilibrium, like its solid counterpart.

But now the question arises: Will the shell hold the same temperature as the solid sphere at equilibrium? Can it ever balance loss and gain of heat at all? Attaining a fixed temperature?

The solid sphere will act like a perfect black body and remain at balance (fixed temperature, surface heat loss = nuclear heat gain). No problem. But what about the hollow sphere, the shell? If it acted anything like Willis’ shell, then half of the outgoing surface flux (heat loss) would be radiated off the outer surface and the other half off the inner surface (459 W / 2 m^2 = 229.5 W/m^2 emitted each way).

But if this were the case, we would have a problem on our hands. The shell would become progressively hotter and hotter and hotter and hotter. It would be unable to ever balance its surface heat loss and its nuclear heat gain and hence keep a steady temperature.

Why?

Because the inner surface can only ever radiate upon itself. The shell can not lose heat to the vacuum within. The internal space is nothing like the external space in this regard. It can not be defined as ‘surroundings’. It can not be defined as a ‘cold reservoir’. It is part of the system. It is internal. Every hypothetical photon emitted from the inner surface would eventually hit a surface of equal quality (temperature) to its origin. There is nothing to provoke a heat loss flux, the transfer of heat away from the inner surface of the shell, since there is no temperature gradient. Inward heat loss is simply unrealisable.

But hey, it’s ridiculous, right, to assume then that the heat loss from this one half of the surface area of the shell is simply zero, ‘forcing’ the other half to compensate and emit a flux rather twice as large as expected? Because that would be … just wrong, eh? So, no, the outgoing flux from the outer surface of the shell HAVE TO be 229.5 W/m^2, in accordance with Willis’ logical model. The only problem then is that only half the power provided by the nuclear energy source is effectively expelled at any one time from the system – to space. The rest somehow ends up piling up in the interior, forever unable to escape. So the system will heat up. Indefinitely. Unstoppably. The system heat loss forever doomed to be half of the nuclear heat gain.

I hope you can all see how such a situation would be absurd, and that a body’s heat loss to its COOLER SURROUNDINGS is all that matters, the one thing that will balance the input energy (heat gain) and keep the body at a constant temperature.

I can find no reference anywhere where both inner AND outer surface is taken into account when considering heat loss from a hollow sphere. I’ve rather found the opposite, like here:

(Notice the following statement and equation: “The energy loss per unit time by the shell is
Q” = 4πR^2 σ(T1^4 – T0^4).” No ‘times two’.)

In Willis’ planet/shell model there is no way the shell would be able to lose heat inwards towards the planet. That is where the heat GAIN is coming from. The shell’s only heat LOSS would go out to space, from its outer surface to a cooler region (an infinite cold reservoir). At dynamic equilibrium, this flux would balance the incoming flux from the core planet to the inner surface, maintaining the shell’s equilibrium temperature – at this point dynamically equal to that of the core planet.

34. Kristian says:

“Radiation is heat transfer by the emission of electromagnetic waves which carry energy away from the emitting object.”

How is energy being carried away from the shell by the shell radiating inwards upon itself?

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

35. Tim Folkerts says:

Bryan, You are getting closer to understanding all this … but still not there!

“Still we are making progress if you realise that the shell radiating for a considerable time at 470W/m2 is against the second law

No, it is not.

Where & when specifically does …
* Σ( P_i ln(1/P-i) ) decrease?
* heat (the net flow of thermal energy) go spontaneously from cold to warm?
* mechanical work get done with no heat rejected to a cold reservoir?
* any other statement of the 2nd law get violated?
You never seem to actually get to the math. You never even say which version of the 2nds Law you are basing your conclusion on. Instead you are always resorting to what I have heard referred to as “proof by bold assertion”. 🙂

Perhaps this it too obvious to everyone else (and too subtle for Bryan), but the shell is not “spontaneously” getting hotter. There is a heater in the system!
* The heater supplies heat to the planet.
* the planet supplies heat to the shell
* the shell supplies heat to space.

All flows are ALWAYS from hotter regions to colder regions.

To use your coffee cup analogy, I could take a cup of coffee and insert a 235 W electric heater. I will alternately wrap the cup with insulation and then unwrap the insulation. When the insulation is in place, the heat flow from the coffee to the room will be less than 235 W (and the water will warm up). When the insulation is removed, the heat flow from the coffee to the room will be more than 235 W (and the coffee will cool off).

Energy is conserved. Heat always flows from hot to cold. No problems with thermodynamic laws once again.

PS Since you are so keen on specific numbers and specific metals, how about you propose specific values for EVERY important parameter.
* thermal expansion of shell and planet
* heat capacity of shell and planet
* radii of shell and planet (at some specific temperature)
* (emissivity is set to 1 for all surfaces)
* (heaters are set to 235 W/m^2 (although you are welcome to change this if you want)).
Find the temperatures and radii and IR radiations as a function of time. Then show us the calculations where either energy is not conserved or the 2nd law is violated. You “know” the models don;t work, so you should have no problem showing real numbers. You can’t just say “for a long time” or “cools slowly”.

36. Tim Folkerts says:

Roger C says: “It is the mass of the earth, land and seas and air that converts radiation into kinetic energy. “
Yes, but it is the land and seas and CLOUDS and GHG’s that convert the KE back into the radiation that leaves the earth.

Both halves of the process are important.

37. tallbloke says:

Tim F: In defense of the importance of CO2
1) it is clearly increasing
2) it is higher altitude than the others, so it “trumps” the others in the bands where it absorbs

We’re straying off topic, this really belongs in the Wiki/IPCC greenhouse thread. High altitude water vapour has been decreasing.

Maybe Miskolczi is right.

38. Bryan says:

Tim Folkerts, if TB offered a prize for Brass Neck of the year you would be an outright winner.

After I corrected your latest post for glaring mistakes you write this……..

“Bryan, You are getting closer to understanding all this … but still not there!”

Unbelievable!

39. Tim Folkerts says:

Kristian says March 16, 2013 at 6:44 pm ….

The book you linked to is a slightly different scenario — with the “planet’s” temperature being held fixed, not the power. Still, it is quite similar. Basically, this problem is finding how much you can turn down the power of the heaters when you add the shell around the planet.

Kristian: “I can find no reference anywhere where both inner AND outer surface is taken into account.”
It’s there.

The equations in the book’s solutions are for the NET flows of energy (ie “heat”).

Q’ represents the heat leaving the surface AND the heat arriving at the shell. The ” -T1″ term is the radiation from the inner surface of the shell back to the inner sphere.

40. Tim Folkerts says:

Bryan … once again it is time to retire from the conversation with you. It’s just not worth the time.

You steadfastly refuse to back up any of your claims with numbers or equations, so it is impossible to move ahead. Good luck re-writing all the textbooks on thermodynamics.

41. Bryan says:

Tim F

You appear to say on the one hand that there is no reason why the Willis planet shell should not radiate at 470W/m2
Yet a number of your posts try to prove it hardly happens.
In the intersts of clarity could you indicate at what point would you see a second law problem

If the shell radiates at 470W/m2

A. 100% of the time
B 50% of the time
C 25% of the time
D 10% of the time

42. Tim Folkerts says:

I shouldn’t bite … but the 1st law limits the outgoing power to and average of 235 W/m^2. Anything above 470 W/m^2 for 50% of the time would violate the 1st Law.

In either Willis’ scenario or your variation, the only (semi-plausible) way I can think that the 2nd law would be violated would be if the shell were warmer than the planet. (The shell has no heat source itself and is only heated by the planet; having the shell warmer would require heat moving from the “cooler planet” to the “warmer shell”.)

43. Tim Folkerts says:

hmmm … I realized after posting that having the shell radiate 470 W/m^2 for very long would also violate the 2nd law — basically for the reasons above. To radiate at 470W/m^2, the temperature would necessarily be at 302 K. The planet would have to transmit a net 470 W/m^2 W/m^2 to the shell to keep it warm, but the planet is only getting 235 W/m^2, so the planet is necessarily cooling to provide the “extra 235 W/m^2”. But a planet that has cooled below 302 K cannot provide heat to a shell at 302 K.

So the new revised answer says that the radiation from the shell could only “be” 470 W/m^2 “for an instant”. After than, it MUST start cooling and start radiating less — both to satisfy the 1st Law and to satisfy the 2nd Law. It could in principle stay “close” to 470 W/m^2 for a while (but I suspect that would take an even more clever scenario than has been proposed so far).

44. Westy says:

Later today I will do this simple kitchen experiment.

Turn the small ceramic hot plate onto low, place a candy thermometer on it, give it 15 minutes and take a reading.

Place a stainless steal mixing bowl over the plate and thermometer, give it 15 minutes and take a second reading.

Will this second reading be A: Unchanged. B: Off the scale at around 50% hotter. C: Other.

I’ll report back after picking up a candy thermometer and doing it.

45. lgl says:

Westy
It will be hotter, but mostly beause the mixing bowl will prevent convection, assuming you are planning to place it upside down. To get closer to the planet/shell model you should place two bowls upside down on the plate, the inner representing the planet and the second, a little larger, representing the shell.

46. lgl says:

hmm.. maybe buttom-up would be better wording…

47. lgl says:

.. or even bottom-up 🙂

48. Roger Clague says:

In another post on this blog Richard Lindzen describes the CAGW scam and is rightly praised for his brave opposition to it.

It is interesting that his model of the atmosphere is the similar to Willis’s steel greenhouse.

tp://texmex.mit.edu/pub/emanuel/PAPERS/greenhouse.pdf

In fig.1 his atmosphere

Has no mass and heat capacity
Is thin and has large conductivity
Is separated from Earth by a vaccuum

The model gets the wrong answer, 303K, for surface temperature.

His fig.2 is more interesting as it refers to the lapse rate. But it is also wrong. It is the raising effective emission height theory.
How can a small increase in trace gas increase the height of the troposphere and where is the evidence of it happening?

The lapse rate should be called the enhancement rate. It increases from a cool TOA downwards because of gravity. It does not reduce (lapse ) from a surface heated by radiation.

49. Bryan says:

Tim F says

“So the new revised answer says that the radiation from the shell could only “be” 470 W/m^2 “for an instant”.

I think we are converging in ideas in that we recognise that if the planet and shell were to radiate for long periods in the region of 470W/m2 there would be violation of the second law.
The thin shell has a very small heat capacity and the planet a very large heat capacity .
It seams reasonable to conclude that the shell would be radiating at around 470W/m2 for longer than an instant.

You ask for equations and there are the usual ones in this paragraph on the second law.
However the copy function on my computer will not reproduce them.
Likewise I don’t have the facility to type mathematical symbols.
However I have a large number of physics and thermodynamic text books and can direct you to the relevant pages if that helps.
While agreeing with much of what Postma said I thought his personal comments on you were well over the top.
You are obviously a smart guy and quite entitled to trenchantly defend your opinions.

http://thermopedia.com/content/1195/

Second Law

The microscopic disorder of a system is described by a system property called Entropy. The Second Law of Thermodynamics states that whenever a process occurs, the entropy of all systems involved in the process must either increase or, if the process is reversible, remain constant. “As a result, it is impossible to construct a machine operating in a cyclic manner which is able to convey heat from one reservoir at a lower temperature to one at higher temperature and produce no other effect on any other part of the surrounding.” This formulation of the Second Law has been attributed to Clausius.

50. gbaikie says:

“gbaikie, bless you for thinking but you went off like a shotgun. 😉 Whoa. It’s incredibly more simple than you are making it. I’ll try one last attempt and if not one person can say they do see exactly what I am describing and the problem, I give.

First, there is no longer a sun, no other planets, no solar system, just two objects, a clump of mass (planet), no poles, it doesn’t matter if it spins or not, in the center… and a shell of 1 AU radius about it, look at Willis’s picture above. The differences to this and the Earth/Sun is totally in the geometry, think geometry. In the real Earth/Sun case 1/2 of a hemisphere of the sun shines parallel rays to 1/2 hemisphere of the Earth every second, always, at any moment. You are right, the intensity is ~1362 W/m². But by the spherical curve of the Earth you have to divide by two, 1362/2 W/m². But wait, that is one side only, divide by two again, 1362/4 W/m². I’m sure you have hear this a million times. So, on the average the Earth receives ~340.5 J/s/m². Look at my math, that is what the sun outputs if spread over that shell unless I made some stupid mistake.”

“Now, make the shell to have the power to radiate outward at that ~340.5 J/s/m², the same SB power you will find on the inside, just like in the Earth/Sun case but now the sphere is actually able to absorb equally at any point on its surface, a maximum of that ~340.5 W/m² radiance and will if the planet in the center starts out at absolute zero. But over time that clump of matter called the planet will absorb and warm to what temperature? About 5.3°C. Now how is it ever going to get any warmer? It is now at equilibrium, like a cup of coffee in a room (shell). The shell is constantly radiating an equal ~340.5 outward into the universe. Simple, it can’t and never will it seems which should cause someone else to ask further questions but it seems no one has gotten that far yet. I’m still waiting.”

Sorry, and still not sure I get it. Probably missing some information in earlier posts.

So you saying the shell has radius of 1 AU [earth orbit]. And you seem to be putting this “clump of mass” where the Sun is.
But also having this shell warmed as though the sun was at the center.

So I will make easier and say on clump of mass, you measure energy received on one square meter of surface, and you find this energy is 340.5 J/s/m². And every square meter of the planet is getting this same amount and continuously.
Rather worry around the temperature of the sphere 1 AU distance.
So we have ~340.5 W/m² coming to surface and surface is radiating ~340.5 W/m² and the surface temperature is about 5.3°C.
And question is: “Now how is it ever going to get any warmer?”

Let’s assess the situation. There is no visible light. So it’s dark. No mention of volcanic activity.
So no light and no volcanic activity should mean no life could develop- so no reason for there being any oxygen in atmosphere.
So generally it would seem quite hard to do this.
But let’s see. So there some kind of atmosphere? And I assume the atmosphere is warmer near the surface and there is a lapse rate?
So if I dig a big hole, it should warmer at bottom of hole than top of hole, due to lapse rate.

If there is cooler air at higher elevation, I could float things up [with balloons] in the air and warm the cooler air. And if the amount energy is more intense at high elevation that would help, but not needed.

If there isn’t any water in the air, and people are running around without a spacesuits, then if one were to cause there to be some water in the air, people will not evaporate as much water and feel warmer.
If there is water in the air, if I were cause it to condense at higher elevation it would be another means of warming colder air at higher elevation.

You seem to think that shell and surface of clump of mass are radiating ~340.5 W/m² if true
there is no way to magnifying the energy from the shell.

I think the radiation at the shell would diminish over a AU distance and also think that if 340.5 W/m² reach TOA and one had 1 atm quantity of atmosphere [any type of gas] that it reduce
by time it got to the surface.

But I don’t think greenhouse gases involving some kind of radiant effects will warm the clump of mass’s surface temperature.

51. Max™ says:

Uh, not getting into the whole back and forth, but this was a pretty significant error that I felt should be corrected:

“So, on the average the Earth receives ~340.5 J/s/m². Look at my math, that is what the sun outputs if spread over that shell unless I made some stupid mistake.”

A horizontal surface at 1 AU from the Sun will receive 1366~ W/m², the interior of a Dyson Sphere with a radius of 1 AU will receive–before albedo–that amount, as would a hypothetical 1 AU radius steel sphere (well, that sphere would be massive and probably have unaccounted for effects on the Sun, but we’re ignoring that), the 340 W/m² value is when you average a disc worth of radiation across the front and back of a sphere at 1 AU.

52. Tim Folkerts says:

Bryan, as I said once before, the process will look something like this (if the planet makes contact @ 302 K and loses contact @ 290 K as you postulated):

* the planet at 302 K and the shell at 254 K, making contact
** the shell warming rapidly to 302 K and emitting 470 W/m^2
*** the shell cooling slowly to 290 K and emitting 401 W/m^2
**** the shell losing contact with the planet
***** the shell cooling quickly to 244 K and emitting 200 W/m^2
****** the shell warming slowly to 254 K and emitting 235 W/m^2 (and the planet warming slowly from 290 K to 302 K)
* the planet at 302 K and the shell at 254 K, making contact

“Rapidly” and “slowly” will be relative, and will depend on the relative heat capacities. If
__ (heat capacity of the shell) << (heat capacity of planet)
then "rapidly" approaches "instantaneously"

Yes, the time that it takes to cool from 302 K to 290 K will be quite long, so the shell will radiate between 400-470 W/m^2 for "a long time". But then the contact will be broken and it will take a "long time" for the planet to warm back up from 290 K to 302K. During this time, the shell radiates only in the range 200-235 W/m^2 to space. (It is receiving 400-470 W/m^2 from the planet, but radiating 1/2 up and 1/2 down.)

But you are underestimating the time it takes to warm back up. During the cooling, the planet is losing 400-470 W/m^2 (conducted to the shell which radiates it away) and gaining 235 W/m^2 from the heater. So the average loss is around 200 W/m^2 while cooling. However, during the warming phase, the planet is radiating 400-470 W/m^2 to the shell, getting 200-235 W/m^2 back as radiation, and getting 235 W/m^2 from the heater. Thus it is gaining between 35 W/m^2 (when cool) and 0 W/m^2 when warm (ie it is asymptotically approaching 302 K). So the average gain will be somewhere around 15 W/m^2 while warming. Thus it is warming much slower than it cooled and it spends much longer rewarming.

Roughly, it looks like ~ 1/12 of the time radiating ~ 435 W/m^2 from the shell and ~ 11/12 of the time radiating ~ 220 W/m^2 from the shell, for a weighted average of 238 W/m^2. Given the crudeness of the estimates, this is remarkably close to the “correct” answer of 235 W/m^2 for the average!

53. Westy says:

The simple kitchen experiment.

Unfortunately I couldn’t find the thermometer I had in mind that would have laid flat on the hot plate but decided to go ahead with the only one available. The one used was a metal spike with a circular needle gauge on top, bent in an effort to get more of the spike closer to the hot plate. 220 F being the high of the gauge. After 15 minutes on the hot plate on its lowest possible setting this thermometer showed a temp of 160 F so a 50% increase would send it off the dial.

After 15 minutes with the stainless mixing bowel covering both the hot plate and thermometer the reading was unchanged, still showing 160. However, after 30 minutes it showed an increased by about 4, up to 164. No 50% increase.

54. mkelly says:

Kristian says: March 16, 2013 at 6:44 pm

Sir, you are correct. I have tried to tell Mr. Folkerts that same thing for days. No times two! Mr. Folkerts proved the steel shell problem false with his ratio, but for some reason he won’t accept his own success. His ratio is exactly what that link you provided shows. Once you use the ratio you find the inner shell will be lower in temperature than the steel planet surface therefore the shell cannot provide heat to the planet.

Another way to look at the pictured problem is convert the 470 and the 235 watts per meter sqrd to temperatures and see that 255 K is not going to heat 302 K then or any of the time divisions from the moment the shell is put in place.

55. mkelly says:

Westy, sir, your slight increase in temperature being measured is probably from the air inside the bowl. The air can attain a temperature via conduction of the hot plate. So some rise should be expected.

56. Bryan says:

Tim F

You hit the nail on the head when you said previously

“So the new revised answer says that the radiation from the shell could only “be” 470 W/m^2 “for an instant”.

Any reasonable person would say that the contact time would be considerably more than an instant given the high thermal capacity of the planet and the very small thermal capacity of the shell.

If such a device ( a vacuum plus a passive shell) could extract heat from a low temperature source then split the heat into a high temperature stream and a low temperature stream
We could then expect the following;

1. Dump the low temperature component to the surroundings

2. Use the high temperature stream to do mechanical work between high temperature reservoir and surroundings.

This would enable us to utilise almost limitless free energy but unfortunately is science fiction.

In case you think that this is exactly what a ground source heat pump does think again.

A heat pump requires a separate power supply.

I think its time to stop this dialog for the moment as we tend to be repeating ourselves .
i would recommend that you read up on the following;

The Maxwell demon .
The impossibility of spontaneously converting low frequency radiation to high frequency radiation without significant loss.

57. Tim Folkerts says:

Bryan, ending here is probably good. Let me leave you with a few brief points

1) Try putting numbers with your ideas. At least estimate orders of magnitude. This will give you an idea of what sort of results you are looking at.

2) Time scales should be expressed with respect to the time for one “cycle” of the system. For an object like earth (if my back-of-the-envelop calculation are right), it would take months for the earth to warm or cool by 1 K. So one cycle would take years or decades (which is not at all surprising). So any phrase like “almost immediately” could mean “several days” and “a long time” needs to be at least a few months if we are talking about real planets. The times would, of course, be much shorter for smaller “models”.

3) For any reasonable coefficients of thermal expansion, when the warm planet and the cool shell touch, the shell will warm and expand, while the planet will cool and contract. This would move then “quickly” apart. This is not a “fatal” flaw for your thought experiment, since there are lots of OTHER ways we could imagine the two objects being put into thermal contact at any times we wanted

4) Assuming we make the contact work as you want, then roughly
~ 1/12 of the cycle would have the shell ~ 300 K radiating over 400 W/m^2
~ 11/12 of the cycle would have the shell ~ 245 K (not 254K) radiating less than 235 W/m^2
~ 0/12 for the transitions of the shell from warm-to-cool and cool-to-warm.
These will change as we adjust the scenarios, but we will always get energy conserved.

5) Most importantly, you are still misinterpreting the 2nd Law.
“We could then expect the following;
1. Dump the low temperature component to the surroundings
2. Use the high temperature stream to do mechanical work between high temperature reservoir and surroundings.”

Yes — it would look something like this Stirling engine: http://www.youtube.com/watch?v=SyszkssxVD4 . Such things exist and are easy to make. They do not violate the 2nd Law. We could put them all over the planet and the exterior of the shell and have them run quite nicely.

58. gbaikie says:

A fundamental question about climate science is how do you warm anything above blackbody
temperature.
So at earth distance this is a uniform temperature of about 5 C and a maximum temperature
of around 120 C [400 K].

First, I will define a uniform temperature of 5 C as sea level surface temperature AND the surface
below it as being always 5 C or higher temperature.
And as this has been defined Earth’s uniform temperature currently is below a uniform 5 C.

Earth is an example of how a planet can have uniform temperature below 5 C and average
temperature of around 15 C.

With climate science a focus is upon the average global temperature and could ask why is Earth’s the average temperature higher than it’s uniform temperature?

rather than it’s average global temperature. So the globalized 240 W/2-m energy input and output
is mostly about Earth’s uniform temperature.

So getting back to question why is Earth’s average temperature above it’s uniform temperature?
The basic mechanisms which allows this to occur is that Earth systems- atmosphere and ocean and lands- retain heat. This can be seen in moderate swings in temperature between day and night- this can be seen regionally where warm days remain warm at night and cold days do not get much cooler at night. Or Globalize in terms of average global temperature comparison between day and night has not much of a difference.
One can also say that large portions of Earth surface area do not reach below freezing during the night and for most of the year- when talking large portion of Earth surface area one has a mostly uniform temperature of above freezing [0 C].

One should say that global or regional climate is about uniform temperature or how cool does it get. And how warm it gets is about weather. Average global surface temperature is weather.

I was going to say more, but it happens I think I said enough.

59. Arfur Bryant says: March 13, 2013 at 12:34 am

tallbloke says: March 12, 2013 at 11:38 pm
[“Arfur: Go take a look at TheFordPrefect’s experimental work, that should help answer it for you.”]

“RADIATION FROM A COOL BODY SLOWS THE COOLING OF A HOTTER BODY”.

This is not the same as radiation from a cool body WARMS a warmer body, which is what the toy planet suggests.

———————————————–
In case you missed it I tried another experiment (an electrically warmed body – using a stable power input – in an insulated box i.e. at a fixed temperature when input power = power out then placing a room temp object or warm object in view of an IR transparent window)

The result:
the warm object caused the object with stabilized power input’s temperature to warm 0.3degC

A Cool Object Reduces Energy loss from a Hot Object
http://climateandstuff.blogspot.co.uk/2013/03/a-cool-object-reduces-energy-loss-from.html

Whilst I can still improve this experiment I believe the result is significant. Perhaps next week I will try the same box but an electrically heated external block.

60. Bryan says:

Tim Folkerts

I was hoping to end this exchange but you repeated something that is evidently wrong after I had given you a direct link

You said

“Even metals with a small coefficient of thermal expansion like Invar still have SOME expansion.”

Whereas in reality we have metals that have no expansion coefficient

http://www.caltech.edu/content/caltech-scientists-use-high-pressure-alchemy-create-nonexpanding-metals/

Now careful readers will note that you are desperate to avoid the conclusion that the Willis planet and shell will be radiating to space at 470W/m2 for considerable time.
If the contact time did not matter you would not have made the quote below.

Your recent post shows you backtracking from your last nights post where you said

“So the new revised answer says that the radiation from the shell could only “be” 470 W/m^2 “for an instant”.

It is not edifying to see a grown man squirm and be forced into contradicting himself within hours.

How can anyone take seriously any further comment from you on this topic?

Stop being a propagandist and follow logic to conclude that the Willis planet is an impossible construct.

61. Roger Clague says:

Bryan says:
March 17, 2013 at 5:06 pm

Willis planet is an impossible construct.

I agree. It is also has no water which is very unlike Earth.

The oceans has a high specific heat and absorb lots of radiation. Water vapour has a big latent heat and phase change at near atmosphere temperature so transports a lot of energy.
H2O has 3 atoms,is IR active and emits radiation to space.

It absorbs radiation ,transports heat and then emits radiation. Water rules the atmosphere.

The abundance and properties of water make it a big factor in climate

62. A C Osborn says:

thefordprefect says: March 17, 2013 at 5:00 pm well done with the experiment.
I am a bit surprised that you consider an increase of 0.45% as “Significant”.
Now that you have the experiment set up can you show what TF has been saying by leaving it running until the temperature of the Cold Plate reaches equilibrium, by which time the Hot plate should be showing nearer a 50% increase, 100 times what you have currently seen?

63. Tim Folkerts says:

And Bryan, you keep worrying about what is basically “trivia” while avoiding the “guts” of the issue.

I will stipulate that there are materials with small coefficients of thermal expansion. I will even do you one better and point to materials with NEGATIVE coefficients of thermal expansion over certain temperature ranges — water being the most famous ( http://en.wikipedia.org/wiki/Negative_thermal_expansion ). That is actually what you would want, so that the shell tightens around the planet when the planet touches it, and stays in contact for a while.

None of that really matters. That will change the temperature at which the planet and shell lose contact and will change the overall length of the cycle, but will not change the relative times. All this talk of coefficients of thermal expansion is “trivia” in the sense that there are numerous way that we could, in our thought experiments, cause the planet and/or shell to make and break thermal contact at any time and any temperature we want.

The “cooling phase” for the planet from T_upper = ~ 302 K down to (some number we choose between T_lower = 254 K and 302 K) will take less time than the “warming phase” for the planet from T_lower back to T_upper.

The real questions are”
* “what will the temperatures be, and what will the radiation from the various surfaces be?”
* “how long will the parts of the system be at various temperatures?”
* “is the 1st Law of Thermodynamics ever broken during our process”
* “is the 2nd Law of Thermodynamics ever broken during our process”

“Now careful readers will note that you are desperate to avoid the conclusion that the Willis planet and shell will be radiating to space at 470W/m2 for considerable time.
And careful readers will note that you are desperate to avoid calculating pretty much anything — times; temperatures; entropy; heat. When it comes to the actual “guts” of the scenario, you only talk about “considerable times” or the “quality of radiation” without ever providing numbers.

As Lord Kelvin once said

“In physical science the first essential step in the direction of learning any subject is to find principles of numerical reckoning and practicable methods for measuring some quality connected with it. I often say that when you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meagre and unsatisfactory kind; it may be the beginning of knowledge, but you have scarcely in your thoughts advanced to the state of Science, whatever the matter may be.”

So far, I am just about the only one providing any equations or numbers (lgl is one exception I recall — and of course Willis in the top post). You ask me for a hard number for the time that the shell can radiate 470 W/m^2, but you provide no answer yourself.
I tell you that
1) conservation of energy limits the radiation to 50% @ 470 W/m^2
2) since the planet will immediately start to cool from 302K after making contact, the shell must also immediately start to cool, so the shell will immediately start dropping from 470 W/m^2

We can move forward, Bryan, when you tell us how long you think the shell can remain EXACTLY at 302K radiating EXACTLY 470 W/m^2 — perhaps you can provide the differential equation for us. We can move forward when you provide numbers that show when you think the laws of thermodynamics are violated. We can move forward when you can predict at what temperatures the planet and shell will make and break contact with any coefficients of thermal expansion and any radii you want to postulate. We can move forward when you can give the equation (or even a link to some webpage) for calculating the “quality of radiation” that you want me to read up on.

I have given numbers (or at least relative numbers) for most of these. Until you “step up your game” to a more mathematical level, our discussion will most likely continue to be “of a meagre and unsatisfactory kind”.

64. A C Osborn says: March 17, 2013 at 6:57 pm
I am a bit surprised that you consider an increase of 0.45% as “Significant”.
Now that you have the experiment set up can you show what TF has been saying by leaving it running until the temperature of the Cold Plate reaches equilibrium, by which time the Hot plate should be showing nearer a 50% increase, 100 times what you have currently seen?
————————-
Unfortunately the experiment requires a thermal imaging camera (£30,000) which others use. I must do this for hours after work. so no easy repeat.

However if you look at the plots on my page you will see that the external warm object rapidly cools to room temperature so there will be no stable final temperature.

The next version to try, is an external source with constant power input (giving a near stable temperature) and to allow temperatures with and without the warm plate to fully stabilize.

The experiment is not to test the iron greenhouse of this post but simply to show that a cold object will change the temperature of a hot object. The first experiment showed that heat loss was reduced. This experiment is to show that the temperature of a hot plate with constant power input is increased by a the presence of a cooler plate (but hotter than background).
The significance I talk about is that there is visible and continuous and significant (above noise level)heating starting from the time the warm plate was introduced.until recording stopped.

Simulating a steel greenhouse is a completely different kettle of fish!

65. tallbloke says:

Thanks
Rog TB

66. lgl says:

Westy
My kitchen experiment was a bit more successful. 30W soldering iron heating a 0.12 m2 ‘planet’. Temp without shell, 55.7C Temp with shell, 67.4C, which means 12.4 Watts was mysteriously ‘created’ by the shell, who would have thought.
12.4 W is of course far from 30 W, but not bad considering this was not done in vacuum.

67. lgl says:

Tim
Equations or numbers wouldn’t help. You see, they are able to dig up Q” = 4πR^2 σ(T1^4 – T0^4), but unable to see that Q” = 0 if T1=T0. Like I said, not much hope.

68. tallbloke says: March 17, 2013 at 8:37 pm

————-
I would prefer to get to the end of the experimentation before I do a conversion of text and pictures for reposting as you suggested.

If you want you can certainly copy and paste from my blog. However, I do add the odd bit of info as I think of it!
Cheers
Mike

69. Bryan says:

Previously I said

“Now careful readers will note that you are desperate to avoid the conclusion that the Willis planet and shell will be radiating to space at 470W/m2 for considerable time.”

Tim Folkerts now says

“And careful readers will note that you are desperate to avoid calculating pretty much anything — times; ”

Easy to do,…… why did you not ask earlier?

Zero time.

You said pretty much the same thing “for an instant”, only I dispute even the instant

Why does it not require a differential equation to prove that?

The second law of thermodynamics

Likewise;

Does it require a differential equation to calculate the spontaneous increase in temperature of a warm cup of coffee placed in a colder room.

Does it require a differential equation to prove that water does not spontaneously flow uphill?

Does it require a differential equation to prove that an exploding bomb will not spontaneously reassemble?

You can if you like use a differential equation to find the number of angels on a pinhead.
Will the number convince you that it is real.

Now back in the real world the Stirling Engine does not violate the second law as you implied.
Page two of this PDF shows that when used as a heat pump it must be supplied with energy to do this work.

It has this in common with all other heat engines.

Further on there are several differential equation describing the theory behind the engine.

I would recommend that you study them carefully.
If you find the theory too difficult I will happily explain any of the awkward mathematics.

70. mkelly says:

LGL, you are exactly right when T1=T0 then q/a is zero. Which means the shell can never pass heat to the steel planet as the shell is always at a lower temperature than the planet by virtue is has a larger surface area.

71. lgl says:

mkelly
And nobody is claiming there is a net heat transfer from the shell to the planet. The point is if T1=T0 (or close) there is little or no net transfer from the planet to the shell, and there must be a transfer of 235 W/m2, any less and the planet will heat up.
The impossibility of spontaneously converting low frequency radiation to high frequency radiation without significant loss.” It’s unphysical nonsense.

72. Kristian says, March 16, 2013 at 6:44 pm: I can find no reference anywhere where both inner AND outer surface is taken into account when considering heat loss from a hollow sphere.

Kristian,

I don’t want to enter into a long discussion about this but I was struck by your comment above. I assume we are talking about a solid black body sphere of surface area A and a hollow black body sphere of outer surface A and inner surface A (assumed the same for simplicity).

1. In the case of a solid black body sphere that is heated to a fixed steady state temperature by a power source of X watts, it will radiate from its surface at a power intensity of X/A Wm-2. Using the S-B law, let’s say this corresponds to a temperature T.

2. In the case of a hollow sphere that is heated to a fixed steady state temperature by a power source of X watts, it will initially radiate from its inner and outer surfaces at a power intensity of X/2A Wm-2. But in this case the inner surface will also absorb radiation at a power intensity of X/2A Wm-2 as you suggest. So to ensure steady-state radiative balance this additional radiation back into the inner surface must be emitted from the outer surface. So the outer surface will therefore emit X/A + X/A = X Wm-2, exactly as does the solid sphere. Again using the S-B law, this corresponds to the same temperature T as for the solid sphere.

Prevost’s Theory of Exchanges (1791).

73. I notice that in all the hundreds of comments, few have discussed the SkyDragon proposition. I am not sure I understand it myself but wonder whether we could re-start this conversation with a simple energy flow diagram as follows:

Energy Flow

If I understand the SkyDragon/Claes Johnson radiation model correctly, there is 235Wm-2 ‘back radiation’ from the under surface of the shell (as there must be by the definition of a black body) but that radiation is not absorbed by the core. Instead, it is deflected back from the core surface. So it then gets absorbed back into the shell, as shown in my right hand diagram. The consequence of this is that the energy flow rate through shell AND core is 235Wm-2. So both shell and core are at a temperature of 254K (simple application of S-B law).

In contrast, the IPCC ‘back radiation’ model allows the 235Wm-2 of ‘back radiation’ from the under surface of the shell to be absorbed by the core. The consequence of this is that the energy flow rate through the core is 470Wm-2 and the energy flow rate through the shell is 235Wm-2. So the core is at 334K 302K and the shell is at 254K.

First we have to be certain that I have represented the Claes Johnson Model correctly.

Then we can have a rational discussion about the two models without indulging in all the irrelevancies of comparative surface areas, size of gaps between shell and core, and other mind-numbing sophistry that passes for serious scientific discussion…but isn’t.

DC

74. wayne says:

The trouble is no one has cut open the sphere to see what is inside it. I envision upon doing that you have a multitude of possibilities having to do with the thermal flow. On possibility is that hidden inside that sphere you have an exact copy of what you are looking in the diagram above, yet another sphere and shell inside that sphere.

Willis didn’t mention that? … what a dirty trick! So many words wasted.

So really, in that case, the 470 W/m-2 of radiative energy density has been hidden inside the sphere and blocked all of this time, just not able to be expressed… cooled 235 W/m-2 by the 0K void beyond … no “back radiation necessary” in that case, right? Can you think of some other case hidden inside that sphere that would not be yet another similar case of flow? Hmm…

75. Arfur Bryant says:

thefordprefect says:
March 17, 2013 at 5:00 pm

TFP,

I congratulate you on performing these experiments.

The difference between your experiment and the toy planet is that you are introducing a separate energy source into the arena. My response to tallbloke was in the context of the toy planet system, where the only energy available came from the source itself, as I believe I quoted.

I am inclined to suspect that what you have achieved is to reduce the rate of heat loss from the hot plate by placing a radiating object in the IR window of the case. I am not a scientist and therefore will graciously concede if you can somehow show that the source is heated by its own reflected/re-emitted radiation. I am also very interested in how that would work on the molecular level with lower energy photons increasing the energy level of the hot plate! But, like I said, I’m not a scientist…

Again, seriously well done on carrying out the experiment. The debate needs more data and less theorising, especially by me 🙂

76. lgl says:

David
Thank you for the flow diagram. Now everybody should be able to see how ridiculous the SkyDragon/Claes Johnson model is. Why on earth would the ‘U-bend’ be there? Between two object at the same temperature. And it it’s there the green arrow can’t be there, because there is no net flow between the objects at the same temperature. But you should not call the other the IPCC model, it’s basic well known physics.

77. lgl says:

typo1. And if it’s there …
typo2. no net flow between two objects …

78. Tim Folkerts says:

Bryan says: “Page two of this PDF shows that when used as a heat pump it must be supplied with energy to do this work.

But there is no “heat pump” in Willis’ model (nor in yours either)!

When do you think that heat (the net flow of thermal energy due to a temperature gradient) moves from a cooler area to a warmer area? Where is the “heat pump”? I only see heat flowing from warm to cool exactly as it naturally should — the diagram posted recently shows this plainly. (And even your modified version still only has heat flowing from warm to cool.)

******************************************************

And your answer to ‘how long will the shell radiate at 470 W/m^2’ has mysteriously switched from “the shell-planet combination will radiate at 470W/m2 for quite sometime” to”zero time” and you think I am inconsistent???

******************************************************

Now I really am out of this sub-thread. It is REALLY not worth it any more.

79. Tim Folkerts says:

David,

Thanks for bringing sanity and focus back to this thread!

Your diagram accurately reflect the “standard” interpretation of back radiation — the interpretation that led Willis to his temperatures.

I still have no idea what Johnson/Postma are thinking. I suspect that other diagram represents their thinking, but I have given up trying to follow their convoluted logic.

80. Kristian says:

David,

Yes, thanks for backing up my position.

The next step is seeing how this situation is no different to Willis’ planet/shell model. In my scenario, the shell’s heat input comes from inside the shell ‘structure’. In Willis’ scenario, it comes from the shell’s interior void (the planet as such doesn’t really have to ‘be there’). The heat input to the shell, however, warms it in similar fashion in both scenarios. The shell then needs to balance this heat gain with equal heat loss. You agree with me (and other quoted sources) that this can only happen from the shell’s outer surface, to space.

Therefore we have,

1) My scenario: 459 W/m^2 energy input (heat gain) – shell temperature 300K – 459 W/m^2 heat loss (from outer surface)

2) Willis’ scenario: 235 W/m^2 energy input (heat gain) – shell temperature 254K – 235 W/m^2 heat loss (from outer surface)

Both after having reached steady states.

The bottom line here is: As long as the system at equilibrium is losing energy as fast as it’s gaining it (as is the case in Willis’ model), there is no need to invent an amplification of the heat source temperature. There will be no difference in temperature between planet and shell at equilibrium. The planet will simply warm the shell until it’s reached (basically) the same temperature as the planet itself, in order for the shell to finally be able to rid the system of an equal amount of energy per unit time as the planet’s nuclear source is providing to the system.

(Just remember that this will always be a dynamic, one could almost say ‘precarious’, equilibrium – if the planet didn’t put out its constant heat, the shell would cool at once, and if the shell didn’t adequately eject its heat continuously out to space, the planet would warm. Insulate the planet completely by making the inside of the shell a ‘perfect’ mirror, and you would have runaway warming – the energy would have nowhere to go.)

What is insulation? What does it actually do? It creates and maintains a temperature gradient across the insulating layer, like a thermally resistant steel shell or the wall of a house.

If you have a room with a heat source and well-insulated walls, the heat from the heat source warms the inner surface of the wall faster and more, while the outer wall is warming slower and
less – the heat generated by the heat source takes longer to escape the system (the house). How does this happen? Less of the heat received and absorbed by the inner surface of the wall is effectively transported through to the outer surface of the wall, and more of the heat is rather piling up (KE) at the inner wall itself. The propagation of heat from inside to outside is simply slowed down. With less insulation, more of the heat will be conducted effectively through the wall, the inside will not warm as quickly and the outside will warm more quickly. The heat from the heat source escapes the system more readily.

So the inner wall (and the room) warms from insulation. The heat source? Not so much. Some, though. And when it’s switched off, it will cool more slowly WITH insulation in place than WITHOUT.

But this is all about conductive insulation. It does not pertain to Willis’ model. Because that aims to show how the radiative GHE works (you know, the AGW one), not the convective one.

Radiative insulation works in a very distinct way – by reflection.

Reflection is quite a different thing from absorption and reemission. When radiation is reflected, it is sent back from where it came from or deflected/scattered. It is specifically NOT absorbed by the surface it’s directed at. Hence, it cannot increase the temperature of the surface either.

This is common knowledge to engineers. But apparently not so to ‘climate scientists’. The atmosphere does not radiatively insulate the surface of the Earth. Because it doesn’t reflect thermal radiation from the surface back down again. And neither does Willis’ shell. Earth’s reflectivity (albedo) affects the incoming solar radiation. That’s it.

No, the atmosphere rather insulates the surface of the Earth ‘conductively’. By way of mass.

This involves convective processes. Not radiative.

81. Arfur Bryant says: March 17, 2013 at 10:23 pm

The difference between your experiment and the toy planet is that you are introducing a separate energy source into the arena….
I am inclined to suspect that what you have achieved is to reduce the rate of heat loss from the hot plate by placing a radiating object in the IR window of the case.

————————-
The experiment is not to simulate the iron green house it is simply to show that energy from a cool source will add to the energy in a hot source.

The warm outside plate is placed 9cms from the window any “hot radiation” from the internal heated plate will have to travel 2 times this distance to get back to the hot plate. (and pass twice through a window which attenuates IR)
The cool plate radiation passes through the window hits the hot plate and the hot plate cools more slowly or seemingly warms due to the added radiation from the cool plate.

These 2 experiments have been done at different time with different set-ups They both seem to show that cold radiation is absorbed by a hot plate which is so often not believed.
I would like to try the last experiment with a heated external plate. This should then show an internal temperature which stabilises at a higher temperature with a warmer than ambient plate outside the box.

82. Kristian says:

Also worth noting, the energy reaches the surface of Willis’ core planet by way of conduction, not radiation, as is the case with solar radiation to Earth.

83. Kristian says:

David Socrates says, March 17, 2013 at 10:00 pm:

Heh, no, you completely missed it, David. Remove the green u-turn arrow and then you have it. Heat loss. Heat gain. Remember? Dynamic equilibrium. The continuous ‘pull’ from the outer loss to space is at balance with the equally continuous ‘push’ at the surface of the core planet. A constant system flow-through (throughput, if you will) of energy/heat from nucleus to space. Maintaining steady state.

84. Max™ says:

Wouldn’t it be:

…….. ^ …………… “Skydragons”
…….. |
…… 235 W/m²
————————————- 254 K [235 in, 235 out]
..^..
235 (470-235) W/m²
————————————- 334 K [235 in, 235 out]
……..^
……..|
……235 W/m²

vs

…….. ^ ……………. IPCC
…….. |
…… 235 W/m²
————————————- 254 K [470 in, 470 out]
..^ …………….. 235 W/m²
470 W/m² ………v
————————————- 334 K [470 in, 470 out]
……..^
……..|
……235 W/m²

Not sure why you compared a (254, 254) example with a (254, 334) example.

85. Tim Folkerts says:

Kristian, if we “remove the green u-turn arrow” then there is 235 W/m^2 flowing from one surface at 254K to a second surface at 254 K. How is that supposed to work? How does the energy know which of the 254 K surfaces it should be leaving and which one it should be entering? The temperatures are the same, so there cannot be any heat flowing!

86. wayne says:

David, by Cleas Johnson’s example, how would it ever work in reality? There is no difference between the temperature of the two surfaces so the left 235 Wm-2 upward arrow could not possibly exist. However, I do agree just a little with Claes example for I see that right U-shaped flow more like a circular cycle, those 235 Wm-2 do nothing, you might as wll see them as cancelling each other but it does leave the surface of the sphere at 302K (470 Wm-2). It’s just I hate the term “back radiation” as if it is somehow “special” thermal radiation. I’ll continue to call it downwelling radiation for there is always equal or more upwelling radiation. In the Earth case we are only dealing with 23 Wm-2 at best (TFK). Miskolczi has it even lower. The proper SB works perfectly in every case I have every come across so why won’t everyone just speak in ‘net’ figures? Much easier and always correct to boot. 😉

87. gbaikie says:

“Kristian, if we “remove the green u-turn arrow” then there is 235 W/m^2 flowing from one surface at 254K to a second surface at 254 K. How is that supposed to work? How does the energy know which of the 254 K surfaces it should be leaving and which one it should be entering? The temperatures are the same, so there cannot be any heat flowing!”

If one has a pot of water on the stove the bottom of pot which in contact with the water will not be
above 100 C. So if you have boiling water one has flow of heat from the heating element, despite there not being difference in temperature.

The vacuum of space has no temperature but one could assume it has potential to cool to about 2 K. So a surface which has temperature of say 5 K will have a flow heat into the the vacuum of space.

There will be a delay in transfer of heat from the inner part of the shell to outer part of the shell,
the amount of the delay is based upon the type material [steel] and the steel shell thickness, and the difference [2 K and temperature and of inner surface temperature of shell.
Since one does one does not need many watts per square meter of heat conducted the delay should not be very significant- particularly if the steel is not very thick.

Or another way to look at it is that for the outer surface of shell to radiate 235 W/m^2 it needs no more or less than 235 W/m^2 being added from the inner surface.

One can’t have a flow of heat from a colder object to warmer object. The IPCC model is wrong.

Now, a shell could slow the rate of radiated energy that can radiate from core surface, and the result of this would be the core interior becoming warmer.
Or a cooler object can inhibit the flow of heat from a warmer object- can reduce the rate of cooling
by some amount [if cold surface is reflective to radiant energy it can be quite significant].
In Steel greenhouse you not dealing the core surface cooling, but rather the surface is constantly being heated, and so if amount radiation if prevented from radiating from the core surface the core interior temperature will rise.

88. gbaikie says:

“Radiative insulation works in a very distinct way – by reflection.

Reflection is quite a different thing from absorption and reemission. When radiation is reflected, it is sent back from where it came from or deflected/scattered. It is specifically NOT absorbed by the surface it’s directed at. Hence, it cannot increase the temperature of the surface either.”

I believe reflective surfaces also have low emissivity of blackbody heat. So reflective surface
can reflect a very high percentage of radiant energy, but they will absorb some of this radiant energy and not emit much of the heat [absorbed- or gained].
So for Radiative insulation one should have shiny surface on both side- one side reflects radiant energy and other side doesn’t emit much radiant energy.
And so many layers shiny material on both side can be used- and the first layer will warm up over time [so the additional layer inhibit this heat from getting thru].

89. Rosco says:

There is a simple method to debunk the original assertion – which is inserting a shell around the original “planet” causes the planet to emit double the amount of radiation it is capable of generating itself.

Let’s ignore all of the errors associated with geometry and agree that the outer shell is only an infintiesimal amount larger – so small we ignore it completely.

Also let’s assume Willis’ proposition is correct and the shell causes the effect he describes.

Now I do not believe his assertion for one minute but I will not argue that point – I will demonstrate the absurdity by assuming he is right !

Let us consider the new arrangement of a “planet” inside the steel shell which is emitting 235 to space. The shell is also emitting 235 backradiation to the planet and the planet is emitting its own 235 plus the 235 backradiation.

Now this whole ensemble looks just like the original planet – a surface emitting 235 to space.

So lets place another shell around this ensemble !

According to the original proposition this second shell will heat up to be emitting 235 to space.

According to the original proposition this second shell will also now be backradiating 235 back to the original shell.

Working our way inwards then we have the new outer shell radiating 235 out and 235 in for a total of 470.

But now the original inner shell now has to radiate 470 out and the same in for 470 + 470 = 940.

And the inner has to radiate 940 out.

Add another shell and another and … and pretty soon the whole thing self destructs.

This is a perpetuem mobile and is clearly impossible.

Everything I have said for adding a second shell agrees with the original proposition in terms of the mathematics claimed yet the result is clearly absurd – energy out of nothing !

90. gbaikie says:

If I understand the SkyDragon/Claes Johnson radiation model correctly, there is 235Wm-2 ‘back radiation’ from the under surface of the shell (as there must be by the definition of a black body) but that radiation is not absorbed by the core. Instead, it is deflected back from the core surface. So it then gets absorbed back into the shell, as shown in my right hand diagram. The consequence of this is that the energy flow rate through shell AND core is 235Wm-2. So both shell and core are at a temperature of 254K (simple application of S-B law). ”

If there was any surface which was cool enough, the back radiation could absorbed.
So say put tower on core surface and and thermally insolated, then put block of ice on top of it- and the back radiation from shell could heat it.

91. donald penman says:

The model that is proposed by the ipcc assumes the outcome that back radiation is cycling between the inner surface of the shell and the outer surface of the planet but I don’t believe that the inner surface of the shell can send back as much radiation as the outer surface sends to the inner surface of the shell.The outer surface of the planet will cool until it is in equilibrium with the inner surface of the outer shell ,this seems more likely to me .

92. Kristian says:

Tim and all,

Apparently I didn’t explain myself clearly enough.

You must not forget how this setup works. Willis’ shell (like the core planet) is a black body absorber/emitter. Imagine we ‘slowed down’ the transfer of heat from planet to shell to such a degree that we could see only one single wave or pulse of energy being propagated each second (235 J/m^2).

What would we see?

One pulse would leave the surface of the planet, the energy being replaced by conductive transfer from the nuclear power source to the surface. It would thus be at 254K, ready to send off the next pulse.

The first pulse meanwhile would hit the inner surface of the shell. It would be absorbed immediately, raising the temperature of the shell from 0 to 254K (same temperature as the planet). The shell would then accordingly emit to its cooler surroundings (space) as much energy as it received – 235 J/m^2.

Upon doing so, however, the temperature of the shell would drop back to 0 K again. The energy pulse received and absorbed would have been ejected again from the body.

So the shell would have to wait for the next pulse to raise its temperature back to 254K.

Now speed up the transfer rate again. You would not see the constant rise and fall in the shell’s temperature, because it would happen in a near-continuous flow. But in principle it would occur with every wave of energy sent from planet to shell.

This is the ‘push’/’pull’ process. DYNAMIC equilibrium. No APPARENT heat transfer. But still a very real one at work.

93. Rosco says:

I hadn’t read much of this before commenting above but I cannot believe lgl’s comment nor do I understand what he means.

“This shouldn’t be more difficult to grasp than if not for all the energy received from your cooler than body temp surroundings, you would freeze to death in a few minutes. At rest your body will generate around 100 Watts and if 2 m2 in area that’s 50 W/m2. If moved to outer space those 100 Watts will be able to sustain a body temp of 176K (disregarding the 2,7K background radiation)

Now, moved back to your living room your body temp will (according to well known physics) be 310K. The only change is the energy received from the surroundings at 293K, which is lower than your body temp, but hey, according to Postma-physics that is impossible. Your body temp can’t go above 293K, cold can’t warm warmer, remember?”

I am sorry but Joe is completely right – you just make stuff up !!

About the only thing you got right is the 310 K body temperature !!

My body temperature is determined by chemical reactions designed to sustain the processes of life and part of that is to generate sufficient thermal energy to maintain the ability of catalytic reactions chatacterized by ensyme reactions – eg the Krebs citric acid cycle etc. In this manner we accomplish chemical reactions that sustain life that otherwise would require a higher thermal threshold.

Normal core body temperature is maintained by a variety of mechanisms at approximately 37 degrees C – metabolism increases when the body senses it is cooling including the shivering reaction and sweating commences when the body needs to shed excess heat.

“Well known physics” has nothing to do with core body temperature at all.

Our skin temperature can be below that but our metabolism is tuned to keep it within a few degrees.

So lets analyse what you claim.

“At rest your body will generate around 100 Watts and if 2 m2 in area that’s 50 W/m2. If moved to outer space those 100 Watts will be able to sustain a body temp of 176K (disregarding the 2,7K background radiation).”

Everything in this is complete rubbish – total BS !

Firstly, my body is at 310 K and radiates about 523 W/sq metre accordingly. My surface area has absolutely nothing to do with this at all – it is the temperature/radiation the Stefan-Boltzmann equation gives in W /sq metre already.

Secondly, what space are you talking about ? You make the same absurd claim that space is “cold”.

Cold or hot are physical concepts associated with kinetic energy of molecules – space is a near vacuum, has no substance and it is absurd to apply physical concepts.

I note you mention the 2.7 K background radiation that is the minimum detected to date.

But the space around the Earth is literally awash with solar radiation of the order of 1370 W/sq metre.

There is no way this can be considered cold and your statement about freezing to death in a few minutes is nonsense.

In fact the exact opposite is the problem.

Exposed to the full power of solar radiation in space objects like the international Space Station have keeping cool as the problem not trying to avoid freezing. When Skylab was launched its parasol and solar panels failed to deploy properly. When astronauts arrived to fix the problem they found the internal temperature was over 70 degrees C or 343 K.

That people, some with PhDs, can assume that because there is very little radiation in deep space it is necessary that our atmosphere acts like a blanket keeping us warm and protected from the cold of space is an amazing display of not thinking before speaking.

And that a PhD does not guarantee you are right.

Data from the Moon indicate that the Moon, a heated surface radiating effectively to space, cools at a rate of no more than 1 K per hour – as a lunar day is about 708 Earth hours.

The Moon’s surface cools from 390 K to 100 K – 290 K in 354 Earth hours.

So vacuum space, even in the shadow of the Moon does not cause cooling as you claim – there is real evidence that your statements are just wrong !

In high humidity and say 30 degrees C most people sweat fairly profusely.

I’ve seen comments that this is caused by the radiation from a few percent of the atmosphere.

This is of course complete nonsense !

Ask any biologist and they’ll tell you our body’s prime mechanism for removal of heat, evaporation at the skin surfaces is impeded by high humidity as the air already contains as much moisture as it can contain.

Faced with failure of its cooling system the body simply tries harder and sweat pours out.

Simple science not some imaginery effect that people simply make up.

94. Rosco says:

Another way to consider any possible radiation from a cooler body is to consider the so-called blackbody radiation curves of three objects at different temperatures – say 400 K, 350 K and 300 K.

The curves of both of the cooler ones are wholly contained within the curve of the hotter one, and the curve of the 300K one is also whooly contained within the 350 K curve.

Now, let us consider the 300 K and 400 K ones in thermal isolation. At thermal equilibrium they both reach 350 K – this is reasonable.

So the 300 K body heats up to 350 K while the 400 K one cools.

The curves prove there is no justification for the claim radiation from the cooler body can induce a higher temperature in the warmer one.

The total energy avaiable is the area under the curve and the cold one has insufficient energy to induce heating in the warmer one.

The claims of the rules of thermodynamics being satisfied because the “net” flow is still from hot to cold are straw men arguments. If the “net” flow is from hot to cold and the result is always the hot one cools while the cool one warms then it is equally valid to state there is no flow of energy from cool to hot.

There is, remarkably, experimental proof that energy does not flow from cold to hot and it is over 213 years old.

It is Pictet’s experiment on the apparent reflection and radiation of cold.

I could not believe it when I first encountered it but in a nutshell it is this.

Two concave metal mirrors are placed 16 feet apart, facing each other. At the focal point of one mirror a temperature measuring device was placed.

After allowing time for equilibration to occur a glass of iced water was placed at the focal point of the other mirror.

Immediately the temperature of the measuring device dropped dramatically.

If the iced water was placed away from the focal point the effect was either less dramatic or non-existant.

16 feet apart is sufficient to remove any conduction effects.

This conclusively establishes that the heating of warmer objects by cold objects does not occur !

Google Pictet and satisfy yourself the reports are genuine. I have to believe they are and at least the result accords with thermodynamics.

95. Bryan says:

Tim Folkerts brings up the Stirling engine ‘to prove something’

I supply a pdf file with all the theory showing that despite Tim’s claim, the Stirling Engine obeys all the laws of thermodynamics.
I even offer to help him with the difficult mathematics if required

Now Tim says (rather than accept reality)……

“But there is no “heat pump” in Willis’ model (nor in yours either)!”

Unbelievable !!!

My observation that the Willis model breaks the second law by the fact that a metal planet with the usual metal expansion on heating, coupled with a shell with no coefficient of expansion would touch.
This would result in the planet and shell radiating to space at 470W/m2 for quite some time.

An impossible scenario.

Tim then asked me how long I thought this could happen for
I said zero time because it defies the second law and the planet surface does not radiate at 470W/m2 at any time.

Tim now takes these two quotes out of context

“470W/m2 for quite some time”……..the impossible Willis model

“I said zero time” ………..the reality model

Then says I changed my mind.

Unbelievable !!!

Now what is the reason for Tim’s irrational behavior.

1. Tim is thick and cannot read
2. Tim is smart but thinks other readers are dumb and cannot spot his tricks
3. Tim is a propagandist who is blinded by some cause

96. tallbloke says:

Bryan, Tim
Have you both agreed on the heat capacity of the planet?
It’s easy to get into misunderstandings with incomplete specs, and tempting to think the worst of people who disagree with you, or selectively quote them to make a point. Please step back, draw breath and go through it from the beginning with a fully specified and agreed set of parameters.

It seems to me that a large planetary heat capacity would soak up energy for a long time before the surface reached the full 470. Likewise the planet/shell combo could radiate more than 235 to space for some time before it cooled back to 235. I think understanding hysteresis is the answer to this issue.

97. Kristian, March 18, 2013 at 12:58 am says: David Socrates says, March 17, 2013 at 10:00 pm: Heh, no, you completely missed it, David. Remove the green u-turn arrow and then you have it. Heat loss. Heat gain. Remember? Dynamic equilibrium. The continuous ‘pull’ from the outer loss to space is at balance with the equally continuous ‘push’ at the surface of the core planet. A constant system flow-through (throughput, if you will) of energy/heat from nucleus to space. Maintaining steady state.

Kristian, I haven’t “missed” anything. I have simply presented two options for discussion.

One (my left hand diagram) is the conventional Willis/IPCC/standard radiation physics scenario.

The other (right hand diagram) is what I believe to be the Claes Johnson/Postma/Sky Dragon scenario.

If you want to take away the green u-turn arrow from my right hand diagram so that there is just a straight flow of 235Wm-2 from the core through the shell and out to space then by all means draw your own (third) diagram to show that.

Remember, this thread started because of a article by Joe Postma on his own blog that was distinctly uncomplimentary towards people like me who believe my left hand diagram is correct. I think (but like Tim Folkerts I can’t be sure) that Joe promotes the scenario represented in my right hand diagram.

I believe we need to check politely with Joe to see whether my right hand diagrammatic representation is correct. Because, if it is, then either (i) the photons leaving the inner surface of the shell, or (ii) the surface of the core, must be endowed with telepathic knowledge of a kind not known to us mortals.

Why?

Because individual photons can be of any wavelength – it is only the aggregated average of all the wavelengths of photons that defines the temperature of the emitting surface (i.e. the peak of the Wien curve). So how does the receiving core surface “know” whether or not to absorb an individual arriving photon? It might be coming from a hotter body, or it might be coming from a cooler body.

I think we should look at this issue seriously because it is a stumbling block to further progress, as I tried to demonstrate in my ATE Part I and ATE Part II articles.

If you (or Postma!) are correct, it knocks on the head the whole ‘radiative transfer’ mantra. Great! But if, as I suspect, you (and Postma!) are both wrong, then skeptics (and I am one) are simply wasting our time over an issue where we are simply wrong.

In this latter case, this does not, as some nervous skeptics assume, bring the house down and pass ‘game set and match’ to the warmists. Far from it, it enables skeptics to address the real flaws in the warmist mantra – as I have tried strenuously to do in my two articles.

98. Rosco says, March 18, 2013 at 8:39 am: Two concave metal mirrors are placed 16 feet apart, facing each other. At the focal point of one mirror a temperature measuring device was placed. After allowing time for equilibration to occur a glass of iced water was placed at the focal point of the other mirror. Immediately the temperature of the measuring device dropped dramatically … This conclusively establishes that the heating of warmer objects by cold objects does not occur !

Thanks for that excellently articulated summary of Pictet’s experiment. Good, so we can all be confident that heating of warmer objects by colder objects does not occur. But that is not what is under discussion here.

We are discussing two scenarios:

(1) Whether the exchange of radiation between a warmer object and a colder object can slow down the rate at which the warmer object cools compared with the rate at which it would cool if it were just radiating to empty space. Let’s call this the non-steady-state scenario.

(2) Whether, if a system consisting of two objects, one of which is receiving a fixed continual flow of energy, and the other of which is free to radiate to space, will settle down to a point where they are at different temperatures. Let’s call this the steady-state scenario.

The first scenario is actually supported by Pictet’s experiment. What is happening is that radiation from the warm thermometer is being received by the glass of iced water which warms up (towards ambient). Radiation from the iced water is received by the thermometer, which cools down (below ambient). Of course over time the two bodies (thermometer and glass of iced water) reach the same ambient temperature.

The second scenario is not falsified by the Pictet experiment. And it is this second scenario that is the one relevant to the earth-atmosphere system where a heated earth radiates to the atmosphere which then radiates to space.

99. Kristian says:

David Socrates says, March 18, 2013 at 10:32 am:

You apparently do not read what other people are writing, David. Your mind simply appears to be made up. Therefore, discussing this issue with you sort of becomes an exercise in futility.

I’m telling you that your rendition of Postma’s view on the planet/shell model is not correct, but to you that seems to be like water off a duck’s back. You’re building a strawman to be torn down, David.

So why would I (or any other commenter) want to discuss your strawman with you?

100. tallbloke says:

Kristian: The first pulse meanwhile would hit the inner surface of the shell. It would be absorbed immediately, raising the temperature of the shell from 0 to 254K (same temperature as the planet). The shell would then accordingly emit to its cooler surroundings (space) as much energy as it received – 235 J/m^2.

The vacuum between the planet and the shell is no warmer than space, so why would the shell preferentially absorb all the 235 on the inside, transfer it to the outside and emit it all to space? 117.5 will go each way (assumng low mass and blackbody characteristics).

101. Tim Folkerts says:

Tallbloke,

Yes, hysteresis is quite important. The heat capacities of the shell and sphere will affect the times for different parts of the cycle. But of course they can’t affect the overall conservation of energy. While the planet is “soaking up energy” to warm to 302 K = 470 W/m^2, the planet will be radiating less than 470 W/m^2 and the shell will be radiating less than 235 W/m^2.

In Willis’ model, that is the end of the story. The planet and shell asymptotically approach 302 K & 254 K.

In Bryan’s model, the various parts of the system oscillate in temperature as the two make & break contact. After the two make contact, quite a bit more than 235 W/m^2 is radiated from the shell for a relatively short time (ie less than half the total time for the cycle). After contact is broken, a little less than 235 W/m^2 is radiated from the shell for a relatively long time (ie more than half the total time for the cycle).
* Energy is conserved.
* heat always flows from warmer to cooler.
* the heat flow could be used to run a heat engine.
* there is no heat pump that requires external work to be done on his system.

Bryan has a fun little scenario, but in the end it simply reconfirms the “standard” interpretation of thermodynamics for this system.

102. Max™ says:

David, I posted this earlier, your diagram does not represent the “skydragon” explanation at all.

…….. ^ …………… I’m fairly sure this is the version Postma would endorse.
…….. |
…… 235 W/m²
————————————- 254 K [235 in, 235 out]
..^..
235 (470-235) W/m²
————————————- 334 K [235 in, 235 out]
……..^
……..|
……235 W/m²

vs

…….. ^ ……………. IPCC
…….. |
…… 235 W/m²
————————————- 254 K [470 in, 470 out]
..^ …………….. 235 W/m²
470 W/m² ………v
————————————- 334 K [470 in, 470 out]
……..^
……..|
……235 W/m²

The difference is only in the amount of energy said to be absorbed by the various surfaces, if radiation from a cooler surface is subtracted from the power of the radiation leaving the warmer surface, then the skydragons are right.

If radiation from a coolder surface is added to the power of the radiation leaving the warmer surface, then the IPCC model is right.

___________

I’m not sure why your diagram has a 255 shell/255 planet and a 255 shell/334 planet being compared, so I cleaned it up to more accurately demonstrate the differences between the two explanations.

103. Tim Folkerts says:

Kristian,

Your “pulse heater” is an interesting variation, but ultimately it will not “save the day”. If we look in “slow motion” there are ALWAYS photons traveling both ways. The shell NEVER cools to 0 K.

Or perhaps a better way to think about it is that your shell is completely transparent! The energy “goes in one side” and then an equal amount immediately “goes out the other side” — exactly as you would expect for a transparent shell. The energy is never absorbed by the shell. In this case, the planet would indeed be 254 K (and the shell could actually have any temperature). The planet would emit 235 W/m^2, the inner surface of the shell would receive 235 W/m^2 and the outer surface of the shell would emit 235 W/m^2

So perhaps Joe Postma is simply confusing “transparent” with “black body”! They both reflect no energy and they both will emit all the energy they receive (once a steady-state condition has been achieved).

Hmmm … interestingly, a 1/2 silvered mirror should have the same net result as a blackbody (the energy just never gets thermalized by the shell).

104. lgl says:

jeez… three typos in one sentence… let me try again.
“Why on earth would the ‘U-bend’ be there? Between two object at the same temperature. And if it’s there the red arrow can’t be there, because there is no net flow between two objects at the same temperature.”

105. wayne says:

“Radiation from the iced water is received by the thermometer, which cools down (below ambient). ”

Absolutely worded wrong. The ice water was warmed by the radiation from the thermometer! That loss of energy by the thermometer lowered the reading.

106. lgl says:

Rosco
There’s nothing wrong with what I said except I should have said ‘moved outside the solar system’ instead of ‘moved to outer space’. You are of course right about the solar heating so sorry about that confusion.
“Firstly, my body is at 310 K and radiates about 523 W/sq metre accordingly”
Correct, and if your body is 2 m2 you would lose 1046 Watts. Even Lance Armstrong on drugs is not capable of more than 500 W over time so he too would be dead pretty soon outside the solar system (vacuum is another story). We survive because we receive nearly as much energy from the cooler than body surroundings as we are emitting, net loss just around 100 W.
And you can’t compare the huge heat capacity of the Moon to that of a human body.

107. lgl says:

Kristian
“Apparently I didn’t explain myself clearly enough”

Yes you did, clearly as wrong as ever.
“The shell would then accordingly emit to its cooler surroundings (space) as much energy as it received – 235 J/m^2.”
No, if it’s emitting 235 J/m^2 to space it’s also emitting 235 J/m^2 to the planet.

108. Kristian says, March 18, 2013 at 1:09 pm: David Socrates says, March 18, 2013 at 10:32 am: I’m telling you that your rendition of Postma’s view on the planet/shell model is not correct, but to you that seems to be like water off a duck’s back.

Well then, since you didn’t actually tell me what was wrong with my diagram as a representation of Postma’s position, I shall simply ask Postma and report back. 🙂

109. Max™ says, March 18, 2013 at 2:13 pm: David, I posted this earlier, your diagram does not represent the “skydragon” explanation at all.

Max, Postma in his own blog article says that Radiation trapped inside a cavity does not increase its own temperature spectrum and nor, commensurately, does it increase the vibration spectrum and hence temperature its own material source. Trapped radiation combines in superposition in constructive and destructive interference and it was in solving this problem that led to the blackbody spectrum and quantum mechanics via Planck. Radiation trapped inside a cavity simply does not change its own temperature/radiation spectrum nor does it induce higher temperature than the temperature spectrum that it is, and that is a premise which underlies quantum mechanics.

Whether or not Postma is right, that is what he says. And I have chosen in my right hand diagram to represent that “trapped radiation” as a non-absorbed u-turn. Your suggested ‘modification’ to my diagram removes any sign of it. So, I would suggest that it is a less accurate way of representing Postma’s position than mine.

110. lgl says:

David
You are of course right.
Joe replied to me: “The other 117 gets scattered right back to the shell, warming it up until it gets to equilibrium with the planet, with both at 235 and equal temperature.”
Which is what your diagram is showing. We are wasting our time on nonsense.

111. tallbloke says:

Whether or not it is a waste of time depends on the value you place on trying to help move understanding forward for others as well. Tough going though. 😉

112. Tim Folkerts says:

The thing about Postma is that he knows just enough physics to REALLY screw things up! That quote @ March 18, 2013 at 5:25 pm is a perfect example. Cavity radiation DOES underlie the origins of QM, but the cavity in that case was defined as being at a specific temperature. He is somehow assuming that all the walls of a cavity must be at a single temperate, which is, of course, nonsense.

The only reason to continue this discussion is to point out the errors and “inoculate” people against this sort of nonsense!

113. Kristian says:

tallbloke says, March 18, 2013 at 1:37 pm:

Sorry, but this is getting downright silly. I thought this was already explained.

‘Point taken’ was part of your reply, tallbloke, to cementafriend upthread when he explained to you the ‘preferred’ direction of energy waves (‘photons’) across temperature gradients, you know, like water doesn’t spontaneously flow uphill. Read again what he wrote. Because apparently the point wasn’t taken after all … (my apologies, no condescension intended).

A heated shell will lose ALL of its heat to its surroundings, tallbloke. Not half of it. The interior space is NOT its surroundings. It is not a cold reservoir. It is its interior. It is part of the system. Every possible facing surface is and always will be exactly the same temperature (KE). There is no escape.

You all keep insisting that conduction/resistivity and heat capacity (insulating properties) of the shell be brought into the mix. The whole point of Willis’ model is that they couldn’t. If they did, he wouldn’t have a point. It would render his model and his argument moot. Because that would be describing something else completely than the radiative greenhouse mechanism.

The shell can not conductively insulate the planet. That would not prove or have any bearing on the hypothesized atmospheric radiative effect. It also can not radiatively insulate the planet (by reflecting the incoming radiation), for the same reason. Willis’ ONLY point is that the shell has two sides. Well, the one side (the interior one) quite literally (as I’ve shown and linked to) does not enter the equation on heat loss. Whatever energy/heat the shell absorbs, it HAS TO and WILL emit in total to the surrounding cold reservoir – space.

Read again: “Radiation is heat transfer by the emission of electromagnetic waves which carry energy away from the emitting object.”

How is energy being carried away from the shell by the shell radiating inwards upon itself?

114. Kristian says:

David, you say: “Whether or not Postma is right, that is what he says. And I have chosen in my right hand diagram to represent that “trapped radiation” as a non-absorbed u-turn. Your suggested ‘modification’ to my diagram removes any sign of it. So, I would suggest that it is a less accurate way of representing Postma’s position than mine.”

For Heaven’s sake! The cavity is a way to approximate a black body’s ability to absorb all the radiation that falls on it. It’s got nothing to do with Willis’ model. There is no need to postulate ‘trapped radiation’ there. The shell radiates freely to space all its absorbed energy. Why wouldn’t it?

Do you picture the shell as somehow equivalent to the cavity, or …? What is going on? FYI, the ‘trapped radiation’ in the cavity case is analogous to black body ‘absorbed radiation’, the cavity representing in a way ‘the surface’ of a black body (before reemission).

115. Max™ says:

Whether or not Postma is right, that is what he says. And I have chosen in my right hand diagram to represent that “trapped radiation” as a non-absorbed u-turn. Your suggested ‘modification’ to my diagram removes any sign of it. So, I would suggest that it is a less accurate way of representing Postma’s position than mine.“~ David

My modification is because you’re comparing two different things, the one on the left would be a sphere heated internally with a spherical shell at some distance from it, the one on the right with the U-bend would be a black body cavity heated from the outside evenly, they are in no way the same thing, and directly comparing them as though it fairly represents a counterargument is a strawman.

Postma and others claim that radiation from a colder surface will reduce the power of radiation coming from a warmer body.

There is no “photons know where they are going”, there is no “U-turn”, there is no “internally heated sphere+outer shell would be the same temperature”, those are all easily knocked down counterarguments or “strawman” arguments.

116. donald penman says:

I have no idea how the photons interact with the planet or shell I have a simple visualisation that energy flows out from the centre of the planet until it can no further when it reaches the outside surface of the planet then this energy is transported outwards by a photon until it reaches the inside of the shell where it can flow through the shell until it reaches the outside of the shell then a photon has to be emitted for energy to keep flowing outwards.The flow of energy has to be away from the energy source which is at the centre of the planet there is no flow of energy coming in from space mentioned .

117. Tim Folkerts says:

Kristian says: “Read again: “Radiation is heat transfer by the emission of electromagnetic waves which carry energy away from the emitting object.”

Kristian, i would suggest that you are succumbing to what I like to call “sound-bite science” — basing a scientific opinion on one particular phrase or sentence as if it were authoritative. If you remain at that level, then is is easy to be mislead by one poorly worded definition.

For example, that same Hyperphysics site says this under the entry for the Stefan-Boltzmann Law

The energy radiated by a blackbody radiator per second per unit area is proportional to the fourth power of the absolute temperature and is given by
P/A = σT^4

If the hot object is radiating energy to its cooler surroundings at temperature Tc, the net radiation loss rate takes the form
P = ε σ(T^4 – T_c^4)
http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/thermo/stefan.html#c1

These statements are slightly different from the page you quoted, and now support the “standard” view instead of your view. (In fact, I just sent the author a suggestion to update the pages to be more consistent).

“Hyperphysics” (like wikipedia) is a good starting point, but neither is fool-proof nor aithoritiative.

118. Rosco says:

lgl repeated one of the classic mistakes of misinterpretation of facts.

“Westy
My kitchen experiment was a bit more successful. 30W soldering iron heating a 0.12 m2 ‘planet’. Temp without shell, 55.7C Temp with shell, 67.4C, which means 12.4 Watts was mysteriously ‘created’ by the shell, who would have thought.
12.4 W is of course far from 30 W, but not bad considering this was not done in vacuum.”

What has happened here is a well established fact known for centuries which has little to do with radiation.

Heat is lost to the environment in an atmosphere primarily by conduction to and convection in the atmosphere.

Without a “lid” there is nothing to prevent this phenomenum.

Put a “lid” on it and you either stop or slow this process.

This has been known for centuries – it is why kettles are a more or less sealed unit, it is why we put the lid on saucepans, it is why boilers in power stations are enclosed.

It is why people built greenhouses – to prevent convective cooling by trapping the air heated by the sun.

It has little to do with trapping radiation – most probably nothing at all.

If the steel greenhouse is real as described then what is now accepted as real is a surface radiating 235 to space with the planet inside it.

Everytime you add another shell you double the energy if the original concept is valid and everytime you do this you end up with a surface radiating 235 to space.

So from the original to a second shell you have 940 required in the planet to balance the arithmetic.

Outer shell is always 235 out + 235 backradiation. The original inner shell now has to radiate 470 to the new outer one and thus now backradiates 470 as well. And the original core now has to radiate 940.

Everytime you add another shell you double the energy.

If it is valid for the original shell it is equally valid for each new one !!

Surely everyone can see this is completely impossible !

If one shell causes an increase in energy for the core from 235 to 470 then every extra shell doubles the energy.

This is exactly what is being stated for the first one and everybody seems to think that is OK.

By this logic you can have in excess of 3.85 MW simply by adding only 14 shells around the original 235 Watt source.

This reality means Engineers are the dumbest people on Earth and we should turn over all engineering to climate scientists.

Simply build a few vacuum shielded shells around the boilers of power stations and reduce the consumption of coal by a factor of 2 for each shell.

Engineers deserve to be prosecuted for negligence for missing this no brainer !!!

They have cost us all a fortune in unnecessary fuel costs – maybe they’re in the employ of big oil.

If you still want to believe this nonsense explain why the original scenario is OK but adding extra shells in exactly the same way isn’t !

Or maybe you really believe you can double energy tthis way and have endless amounts from tiny sources ?

119. Rosco says:

If Willis’ scenario is correct then Planck, Einstein, Wien, Stefan and Boltzmann were all wrong !

They never postulated doubling energy by trapping it with outer shells !

I’ll say again

If it is valid for one shell then it is valid for any number of extra shells you care to add and the result is infinite energy from tiny core amounts.

That is not science – it’s magic !!

You need to come up with a valid explanation why it is valid for the first shell but not additional ones and there simply isn’t one – if so please tell me why – I am open minded and if you can convince me I’ll believe.

120. lgl says:

Rosco
I used a plastic bag to prevent convection. How else could I get 29 W emitted from a 30 W heater?

No, adding more shells will add 235 W/m2 each time, 470, 705, 940, 1175 …
and remember this is not creating energy, you can’t tap any of it.

121. tallbloke says:

Hi Rosco.

It’s a fun idea, but as soon as you take any power out of the system centre, the amount left to heat up the shells diminishes, and so will the feedback. The total amount you’d be able to draw will never exceed the 235 the nuclear core is generating, and if you took 235, the shells wouldn’t heat at all. So if you take 200, the outer shell will only emit 70 once the system reaches steady state , 35 in, 35 out. Total out = 200 ‘tapped off’ +35 emitted from outer shell to space = 235

I recommend you do the calcs to prove this to yourself. Just use two shells, and start at the steady state with the planet emitting 900. Take 200 from the planet, and watch the shells and planet cool until the outer one emits 35 to space, at which point the system reaches a steady state again.

No free lunch, no magic energy creation, Planck et al can rest easy in their graves, because there’s no spare power to make them spin. 😉

By the way, everyone gets the chance to use the ‘magic’ flourish once here.

122. Rosco says:

There is absolute proof that a surface radiating to space loses energy at a slow rate – much slower than conductive and convective cooling.

The observed data from the Moon shows that the surface cools at a rate of no more than 1 K per Earth hour, and substantially less than this once it has cooled to terrestrial temperature levels.

You can put hot water in your freezer and have ice in less than eight hours at a mere minus 15 C.

Say the water cools from 60 to zero then freezes you have a cooling rate of several orders of magnitude greater than a surface radiating to space – greater than 8 K per hour.

Remember the water has a higher thermal capacity and also the latent heat of fusion.

All of our engines and other machinery rely on convective cooling. A car’s “radiatior” is really a convective device – it is the air flowing over the fins in the “radiator” that removes the heat – not radiation.

Central heating is really a convective device not the misnamed “radiator”.

A car engine would simply explode if it relied on radiation to cool it.

Have you ever observed the “shimmering” in the air just above the ground’s surface on a hot day ?

This is due to refraction of light due to the reduced density of the heated air convecting away from the surface.

This simple observation confirms what real scientists have known for centuries – an atmosphere acts to reduce the temperature of a heated surface !

There is absolute proof that our atmosphere and the most common and “powerful” greenhouse gas reduce the heating power of the sun.

Earth’s albedo is generally taken to be 0.3 whilst the Moon is generally regarded as 3 times less.

A large proportion is due to cloud tops in the atmosphere.

Clearly there is no possible argument against this – our atmosphere and the most powerful GHG reduce the Sun’s heating power.

And, surprise surprise a non GHG – oxygen ionized to ozone – removes substantial amounts of the most energetic component of sunlight – UV.

There is really compelling evidence that radiation from cold objects cannot increase the temperature of hot objects.

First there is the “blackbody” curves themselves where the cold curve is already whooly contained within the warmer one. If the cold one “heats” the warmer one why do they equilibrate to a curve above the cold one but less than the hot one. The hot curve must increase if energy is coming from the cold one.

Also, common sense says this is not true – you don’t put ice into you drink to warm it up.

And it has been experimentally demonstrated to be false – the only “evidence” for the unrealistic belief of energy from cold heating hot is rwisted thought bubbles that make no sense unless you want to believe them.

The greenhouse effect is an unproven hypothesis and most people cite evidence that simply isn’t true to support it.

If the Sun can heat the Moon’s surfaces to 390 K why is that not observed on Earth ?

1. – the atmosphere and clouds reduce the incoming energy by 30 % and approximately 20 % is absorbed by the atmosphere on its way in thus reducing the heating potential at the surface; and,
2. – at the Earth’s surface where the power of the sun is greatest the majority of the surface is ocean. Water evaporates without increasing in temperature and this substantial amount of energy is transported to the upper atmosphere hwere this energy is released during precipitation without ever having been involved in heating the surface.

Even ice can sublimate and disappear without ever melting (which is what happens on Kilimanjaro where the temperature never rises above freezing – clearance of the tropical forest reduced humidity on the plains below reducing snowfall to recharge the glaciers).

You can easily observe ice sublimating – make some cubes and leave them in your freezer – after some time they will simply vanish.

123. lgl says:

Here’s a two-shell model http://virakkraft.com/Shell%20planet.png

124. Arfur Bryant says:

thefordprefect says:
March 18, 2013 at 12:33 am

TFP.

Best of luck with your continuing experiment. It still could be that you are adding to the insulation by placing the warm plate near the IR window. If the IR window was not there (ie a solid wall), would not the hot plate equilibrium temperature be slightly higher?

Maybe you should try moving the same warm plate closer to (say 3cm) the window and further away (say 15 cm) and see what happens to the hot plate temp.

Similarly, instead of a warm plate, use a cold plate (below ambient) instead. The cold plate will still; radiate and should therefore – according to your implication – add energy to the hot plate.

I’m inclined to suspect that Rosco and Kristian are correct. To me, it all boils down (probably an unfortunate choice of words) to the question “Is the ‘low energy’ radiation from a cooler object absorbed for net gain by a warmer surface?” Or, another way, how does the energy level of a hot surface become increased through or by the addition of lower energy radiation? I do not think that you can just ‘add’ radiation if the original source of the radiation is the only source.

Again apologies for my non-scientific wording. I hope you understand my intended meaning.

125. tallbloke says:

Rosco, maybe you’ll have missed my reply to you here, while you were busy typing…

I agree with nearly everything you said in your last comment by the way. However, the subject of this thread is Willi’s vacuum-packed toy planet, not Earth.

126. Rosco says:

tallbloke

my belief is exactly what you state – you can’t take the energy out else the arithmetic fails BUT it is equally impossiblr that the energy at the core doubles simply by adding a shell or 2 or 14.

Again – do the arithmetic as Willis proposes – one shell causes 470 at the core – the ” new” arrangement is just like the original “planet” – it may not be a solid BUT it is a spherical surface radiating 235 out.

Thus adding another shell is exactly the same as adding the original – except that this one now requires 940 W at the core to make the arithmetic add up.

Adding each additional shell is as valid as the original – I stopped at 14 for 3.85 MW.

Continue this series on and you have 2^n x 235 W required at the core to balance the arithmetic.

Another 6 shells – a total of 20 requires 246 MW ate the core if Willis’ original hypothesis is correct – 30 requires 252 GW.

You can see where this is going ?

I agree completely with free and open discussion.

I have called lgl out on what he has said – I didn’t use insulting terms.

I believe he has the wrong viewpoint on the 2 quotes I highlighted.

The space around Earth is not cold – it has the 1370 W/sq metre from the sun constantly flowing throught it. the only way for an object to cool in that scenario is to be shielded from this by a substantial object such as a planet. The Moon loses energy at a maximum of 1 K per Earth hour so why would a person freeze in a few minutes especially when our metabolism keeps our core at a constant temperature while it has energy left ?

That was a silly statement and deserved highlighting – even if only to give people the opportunity to disprove my assertion – without insult of course.

I see he has corrected the statement about space by acknowledging the fact of near Earth space.

The other one about heating a plate and observing an increase in temperature when a lid is put on it is also a mistake.

Trapping radiation may play some miniscule part in this observation but it is a well known and proven fact that preventing free convective cooling by trapping the heated air – heated by conduction by the way as air is transparent to IR so cannot be heated by radiation supposedly.

This is provable by noting that while gteenhouses heat up they NEVER heat up beyond the power of the solar radiation at the locality they are situate – in fact they don’t make it to that because the glass eventually conducts heat away to the atmosphere outside.

Glass doesn’t trap radiation – it is simply a poor conductor.

You guys in the UK really know this because you know double glazing is a more effective insulator than single glazing and you know it is the trapping of the air between the 2 glass layers that reduces heat loss – any product will achieve this effect including IR transparent materials if they have sufficient rigidity to trap the air.

A vehicle with a sun roof left partially open will never heat up as much as one with thw windows down an equal amount and both will always remain cooler than one with the windows up – even if the other variables remain the same.

So I am not trying to insult but simply state why I think something said as fact is incorrect.

Many people would remember” 2001 – A Space Oddysey” where HAL locks Dave out of the spaceship without any protective apparatus at all.

Lots of “experts” said Arthur C Clarke wrote BS by having Dave survive this ordeal but real experts – NASA scientists – agree he got it completely right.

Dave would not have frozen on exposure to a vacuum even deep space. It would take many hours for that to occur.

And there is insufficient air pressure inside our bodies for that to have been a major problem exposed to zero pressure.

Obviously lack of air is the major problem but a few minutes is no problem for a well trained astronaut.

The main threat of sudden exposure to a vacuum is the release of gases normally in solution in the bloodstream. Whilst there is no hard experimental data on this – the subject could die – the medical opinion is that a short exposure – perhaps up to a minute, maybe more – is probably survivable.

Of course in near Earth space the UV would cause considerable tissue damage to the skin.

Anyone cal call me out on my proposals as I have done – I love a good argument and will always take the full half hour over the five minute one.

127. bwdave says:

I’ve spent many days trying to catch up with all of the comments back and forth, and see that little progress has been made convincing any of the “standard view” physics believers that energy only flows from hot to cold, even when believers offer “If the hot object is radiating energy to its cooler surroundings at temperature Tc, the net radiation loss rate takes the form P = ε σ(T^4 – T_c^4)” to support their belief. It’s about physics, not semantics. The term “net” doesn’t imply that T_c is heating T.

I only wish to offer a related thought. If CO2 in the atmosphere causes surface warming because it is radiating heat, why isn’t it used between double pane windows?

128. Rosco says:

bwdave says

“I only wish to offer a related thought. If CO2 in the atmosphere causes surface warming because it is radiating heat, why isn’t it used between double pane windows?”

Actually it is a slightly better insulator than normal air. It also heats up more than air with the same energy input.

This is why those CO2 “radiation traps” in bottle demos work – CO2 requires less energy to heat up and has a lower thermal conductivity so loses the heat slower – but that only explains normal thermal data already well known.

To prove radiative “trapping” the test needs to be in a vacuum chamber and show that CO2 doesn’t cool at all – which it obviously does.

The answer to bwdave is simply the benefit is not significant enough to justify the extra costs.

and tallbloke

yes I agree we are discussing Willis’ toy planet.

I simply believe if his assertion is right then adding another layer represents his original assertion and his claimed result must also apply.

If this is a valid hypothesis then simply adding to it should not lead to the unbelievable result of runaway energy generation by simply increasing the number of shells.

That it must lead to this result invalidates the whoel hypothesis in my opion.

I could always be wrong – but I will take some convincing that the original proposal is sound while my extrapolations are just wrong somehow.

129. tallbloke says:

Rosco: my belief is exactly what you state – you can’t take the energy out else the arithmetic fails

The arithmetic doesn’t fail. It gives you the correct answer for the amount the outer shell will radiate to space for a given amount of ‘tapping off’ of the energy at the core.

The space around Earth is not cold – it has the 1370 W/sq metre from the sun constantly flowing throught it

Please discuss the Earth/Sun on the IPCC/Wikipedia greenhouse theory thread. This thread isn’t about the Earth, it is about Willis’ vacuum-packed toy planet. It isn’t near a star.

130. Tim Folkerts says:

Rosco,

There is a lot that is correct in what you say, but unfortunately a lot that is also wrong.

Yes, the “kitchen experiment” and many every-day situations rely on convection and conduction, so they typically don’t do a good job of demonstrating the role of radiation. But when there is no atmosphere, then radiation is “the only game in town” — radiation must be responsible for the results in the “shell model” or in the earth’s heat loss to space.

You are also close on your analysis of multiple shells. The temperature of the core WOULD keep getting higher as you added more and more shells. Each shell would actually add 235 W/m^2 to next shell down (not double it as you concluded). But yes, you could in principle get to 3.85 MW/m^2 = 2870 K radiating from the inner most sphere with ~ 16000 shells.

Here is the tricky part — this is not illogical! The math gets a bit tricky, but as all the shells are warming up toward their steady-state conditions, the outer-most shell will be radiating less than 235 W/m^2 (and the next shell will be radiating less than 2×235 W/m^2, etc). Once the inner shell was close to 2870 K, you could remove the shells and — for a brief time! — get this bright radiation of 8.85 MW/m^2. But the planet (and all the shells) would cool quickly and the radiation would drop. The shells could be put back into place and the system would slowly warm back up — all the while radiating less than 235 W/m^2. You would find that — on average over one cycle — the total power radiated to space would be 235 W/m ^2.

As Tallbloke suggests, you should do the math. You should find that:
* Energy is conserved.
* Heat always moves from warmer to cooler.
* No laws of physics are violated.

131. Arfur Bryant says:

All,

This is not meant to be mischievous…

If the toy planet roles were reversed, and the shell was heated by the same constant power source with a much colder (but above 0K) planet inside, would the shell still get warmer according to Willis’ theory?

132. Tim Folkerts says:

Arfur … No, with the heater on the outside the interior would not get hotter. The whole interior would be 254 K once steady-state conditions were achieved.

133. Tim Folkerts says:

Rosco says: “The Moon loses energy at a maximum of 1 K per Earth hour so why would a person freeze in a few minutes especially when our metabolism keeps our core at a constant temperature while it has energy left ?

There are two big reasons.
1) During the “lunar day”, the incoming sunlight changes very slowly, so the temperature will pretty well track the temperature dictated by the sunlight. By “nightfall” the surface will already be quite cold – below 200 K it would appear from graphs on-line. At these temperatures there is little radiation emitted and little cooling. The person will be ~ 310K — so it will be losing 10x or more energy from each m^2

2) The moon has a much smaller surface-to-volume ratio. It will naturally change temperature much more slowly. It has a much bigger “reservoir” of energy to draw upon.

(Sunlight will have a huge impact on this. A person in sunlight will be getting a huge input of solar energy, while a person shadowed from the sun would have no such input.)

134. gbaikie says:

“Tim Folkerts says:
March 19, 2013 at 12:00 am

Rosco says: “The Moon loses energy at a maximum of 1 K per Earth hour so why would a person freeze in a few minutes especially when our metabolism keeps our core at a constant temperature while it has energy left ?

There are two big reasons.
1) During the “lunar day”, the incoming sunlight changes very slowly, so the temperature will pretty well track the temperature dictated by the sunlight. By “nightfall” the surface will already be quite cold – below 200 K it would appear from graphs on-line. At these temperatures there is little radiation emitted and little cooling. The person will be ~ 310K — so it will be losing 10x or more energy from each m^2

2) The moon has a much smaller surface-to-volume ratio. It will naturally change temperature much more slowly. It has a much bigger “reservoir” of energy to draw upon.

(Sunlight will have a huge impact on this. A person in sunlight will be getting a huge input of solar energy, while a person shadowed from the sun would have no such input.)”

To be in environment such as higher than 45,000′ on Earth, Mars, or Moon a person needs a pressure suit or will be dead within 2 couple minutes [and unconscious within 30 seconds unless person has some training for emergency situation of being in a vacuum].
The person will die from temperature, the person will die because he can’t breathe without a pressure suit – a oxygen face mask will not prevent this death.
If you have pressure suit and have a hand exposed- the hand will freeze, very quickly. But the hand will not freeze because space is cold, rather it will freeze because water boils in vacuum well below body temperature. The hand will freeze in direct sunlight- even if you were at Mercury distance from the sun.

No spacesuits have heaters:
http://science.howstuffworks.com/space-suit4.htm
Though I have heard warmed gloves for spacesuits. And one might need warmed boots if walking
around in a dark lunar crater which was 30 K.
So due to contacting cold surfaces. If not contacting things which are cold, you should not need heaters.

I would guess that if you were in very cold conditions of Earth, say -100 C, you would be cold
in a space suit [even with warmed gloves and boots].
And of course if in water near 0 C, one would need a dry suit and there time limits due to such cold conditions.

Other then warmed boots for walking around in dark lunar crater, one probably need to adjust how how cooling the space suit does. And don’t roll around in the regolith.

135. gbaikie says:

“Rosco says:
March 18, 2013 at 10:14 pm

There is absolute proof that a surface radiating to space loses energy at a slow rate – much slower than conductive and convective cooling.

The observed data from the Moon shows that the surface cools at a rate of no more than 1 K per Earth hour, and substantially less than this once it has cooled to terrestrial temperature levels.

You can put hot water in your freezer and have ice in less than eight hours at a mere minus 15 C.

Say the water cools from 60 to zero then freezes you have a cooling rate of several orders of magnitude greater than a surface radiating to space – greater than 8 K per hour.”

I am interested in the reference for “Moon shows that the surface cools at a rate of no more than 1 K per Earth hour”

The only reference I have seen is diviner and lunar eclipse:
“The orbit path progressed from east to west (right to left), with each orbit separated by roughly two hours. The first two data swaths were taken before the eclipse, the three center swaths were taken during the eclipse, and the last two swaths were taken after the Moon had reemerged from Earth’s shadow.”
http://www.diviner.ucla.edu/blog/
So it’s simple to say that in 2 hours temperatures cooled far more than 2 K:
http://www.diviner.ucla.edu/blog/
The orange appears to be 200 to 220 K and yellow is about 180 K with blue about 120 K.
So roughly more than 10 K per hour and less than 30 K per hour
Though note differences of cooling rate, small areas still orange in mostly yellow 3rd pass.
And coldest pass [4th pass- four hours start of total eclipse] still small areas 140 K.
Or as they say:
“The data show an average decrease in surface temperature during the eclipse of around 100K, with some locations remaining warmer than others.”

136. Tim Folkerts says:

bwdave,

It is important to be careful about words .. they have very specific meanings in science.

*Energy* flows both ways. *Heat* (=net energy) flows from warm to cold.

P = σ A T^4 is the energy that flows from a blackbody surface at temperature T
P = σ A (T^4 – T_s^4) is the heat (= net energy) that flows from a blackbody surface at temperature T surrounded by blackbody surfaces at T_s.
(Note that if T_s > T, then the number is negative and heat flows INTO the first object, exactly ass expected).

This is the “standard” physics. Unfortunately people often get sloppy about the word “heat”.

So for example, when you say “The term “net” doesn’t imply that T_c is heating T.” that would be a correct phrasing. The cooler surroundings at T_c are supplying energy to the warmer object, but the net flow = “heat” is the other way — T_c is NOT heating the warmer object (exactly as you said). The hotter object is “heating” the cooler surroundings. The cooler object slows the cooling of the warmer object by reducing the heat leaving the warmer object.

137. Max™ says:

Add a shell 1 meter above the surface, the power output from the sphere is distributed to give 230 W/m^2 at that distance, as radiation follows an inverse square law.

The total output of that shell is then 29,522,041.8 Watts, add another shell 1 meter above the first, the power reaching it from the first shell is reduced to 225.8 W/m^2 at that distance, giving a total of 29,521,204.9 Watts from that shell, keep adding shells and as expected the power radiated by the outer shell is reduced upon being distributed across the larger area.

4*pi*100^2 =
125,663.7*235 = 29,530,969.5/128,189.5 = 230.3

4*pi*101^2 =
128,189.5*230.3 = 29,522,041.8/130,740.5 = 225.8

4*pi*102^2
=130,740.5*225.8 = 29,521,204.9/133,316.6 = 221.4

4*pi*103^2
=133,316.6*221.4 = 29,516,295.2/135,917.8 = 217.1

This result is the same even if you skip the inner three shells and just have the 100 m radius sphere and a 104 m radius shell, for example.

138. gbaikie says:

” lgl says:
March 18, 2013 at 10:16 pm

Here’s a two-shell model http://virakkraft.com/Shell%20planet.png

At what depth would the rock be molten. Less than 1 meter?
More than 1 meter?

139. wayne says:

“.. exactly ass expected).

This is the “standard” physics. Unfortunately people often get sloppy about the word …”

No Tim, I perfectly understand what you are saying about physical “standard” male tendencies but I don’t know if I would call it physics, maybe so. 😉

140. tchannon says:

Perhaps it is asinine wayne, not that I had noticed asses are especially that way.

141. tchannon says:

gbaikie,
it might be an idea to look at the simulation I posted on the Talkshop. Agrees with diviner.

This is one of the reasons I wonder how many really understand SB when it is connected to a real object. My absence from these (to me) amusing long threads might be related.

The lunar surface is an excellent thermal insulator, fine dust. Thermally the whole thing is about a flux and thermal impedance, thermal capacity. You can’t ignore the dynamic effects on the earth or moon yet that is done.

You mention slow cooling. Keep in the mind the thermal capacitance.

Too late, I am probably writing twaddle.

142. Arfur Bryant says: March 18, 2013 at 10:21 pm
. It still could be that you are adding to the insulation by placing the warm plate near the IR window. If the IR window was not there (ie a solid wall), would not the hot plate equilibrium temperature be slightly higher?

Maybe you should try moving the same warm plate closer to (say 3cm) the window and further away (say 15 cm) and see what happens to the hot plate temp.

Similarly, instead of a warm plate, use a cold plate (below ambient) instead. The cold plate will still; radiate and should therefore – according to your implication – add energy to the hot plate.
———————-

The external warm plate makes very little difference to the heat loss through the window it is too far away. IR once it leaves the hot plate will need to trave 2×9 cm and pass 2 times through the IR (lossy) window and be reflected from the warm plate so will have little effect on the hot plate temperature.
Moving the warm plate will cause different amounts of radiation to pass throght the window and be absorbed by the hot plate (inverse sqaure law). so 3cm will cause the hot plate to be warmer and 15cm will be cooler than the temperature at 9cm.

The set up last used had a thermocouple embedded in the hot plate. With 2 windows open and ambient temperature plate in position at 9cm and power input at 1.88W the stable hot plate temperature was 71.1degC Covering both windows with thermal insulation as used in the box gave a temperature of 75.4degC after 30mins (and still rising).
This increase would be due to 2 reasons – IR being thermalised by the insulation and improved conductive insulation (the double glazed IR windows would not be perfect conductive insulators)

In all my experimental descriptions I have stressed that the warm plate is ABOVE the ambient temperature (or at ambient temperature). Above ambient there is more energy travelling from warm to hot than if there is no plate or a plate at ambient. If the external plate is taken below ambient then obviously there will be less energy travelling from cold to hot plates than if the external plate is at ambient. This will cause the hot plate to stabilise to a temperature lower than if exposed to ambient.
If the ambient is space at 2.7K then placing a 200K (-73degC) plate will again cause the hot plate to stabilise to a warmer temperature than with 2.7K.

It is all about balance of power!

power into hot plate is
electrical 1.88W fixed
through camera window IR from camera at approx 27degC + some background
through warm plate window IR from warm plate + some background IR from 21degC black body
back radiation from internal surfaces of box

Power out of hot plate
IR loss through windows (depends on source temperature and IR transparency of windows. Once outside the box the IR loss can no longer affect the hot plate temperature unless reflected back in)
conductive losses to wall/windows of box (set by insulation properties of box/window)
convective loss to air in box (set by insulation properties of box/window)
IR loss to wall of box.
Conductive losses throughelectrical connections to plate

The ambient temperature was 21C throughout
Power loss is substantially constant (difference caused by 65.3degC to 65.6degC black body radiation/conduction will be small)
Power input is 1.88 watts constant
camera IR unknown but constant
background IR constant
back radiation from box (substatially) constant
Radiation from location of warm plate – the only variable of note!

The temperature of the external hot plate in this setup is the only variable affecting he hot plate temperature.
1.88 watts + IR from ambient temp warm plate + constant background gives 65.3degC hotplate temp
1.88 watts +IR from 45degC warm plate + constant background gives 65.6degC hotplate temp

143. bwdave says:

Tim Folkerts, note that in your original quote T > T_c.

144. Kristian says:

Tim,

I am not interested in discussing this issue with you, since you are clearly not willing to even try to get to grips with my position. You keep picking out little snippets here and there to misunderstand and misrepresent rather than address the full content of what I write.

Frankly, I can’t be bothered dealing with such debating tactics, which you seem to apply on a regular basis.

Only one thing, Tim. If the inside surface of the shell acted like a black body surface, then it would absorb ALL the energy received from the planet (235 W/m^2) and immediately reemit it in the same direction. The outside surface would not receive an energy flux of any kind and would accordingly not emit one either.

You cannot have both surfaces emit the same flux. And especially not half the flux received on one side only.

So clearly this is not what happens. The shell is not an insulating layer.

The other option is that the energy flux from the planet is absorbed and warms the shell ‘as a unit’. The shell has no heat capacity, no conductive resistance. So it will instantly upon absorption of the energy flux from the planet acquire the temperature of 254K. The shell as a whole acts like a black body. And accordingly, it will instantly need to emit an equal amount of energy (because more is coming in from the planet continuously) to its surroundings. It doesn’t matter if the energy is coming from the inside or the outside or within the shell itself. It cannot get rid of the energy emitting it inwards upon itself. It needs to leave the system. It can only get rid of it emitting it outwards, to the infinite cold reservoir of space.

The surroundings of Willis’ planet/shell system will always be space, Tim. It never changes its temperature. It will always (hypothetically) be 0 K. The flux from shell to space would be 235 W/m^2 because that is the flux the shell needs to to get rid of, and because the temperature difference between the shell and space is 254K.

That’s it.

145. Gabriel van den Bergh (GabrielHBay) says:

I have been following this debate with fascination. As a non-physicist, I am intrigued by how obviously knowledeable people can emphatically disagree on the interpretation of what each one in his own right interprets as unassailable laws of physics.

This can surely only mean one thing: invalid application of the laws, or that some or all of the laws are subject to some as yet unidentified or misunderstood prior constraints or caveats. (Kindly be patient with me if the way I express my thoughts are not to the standard of the company here!)

So, what am I as a lay person to make of his? Eager to learn, I find that neither the learned posters nor the laws and esoteric explanations and calculations convince me from either side.
The simple and reduced question appears to be: Does the presence of the shell cause the temperature of the sphere to be higher than it would be in the absence of same?

Having mulled over this question for some days now, I would venture the following. Just for fun, in defiance of all the high physics in evidence here, I shall call it Gabriel’s Law.. (to raucous laughter):

“The only way the temperature of the surface of the sphere would be elevated by the presence of the shell would be if the shell offered resistance, or an impediment, to the flow of energy from the sphere to the cold space beyond the shell”.

This is for me the bottom line to the entire argument.

As I see the original thought experiment, it seems that the shell would in fact not offer resistance (black bodies and perfectly thin and all that) but I may be wrong in my understanding of all the implications. Yet, as far as my law is concerned, I believe that to be valid.

146. tallbloke says:

Hi Gabriel and welcome. Glad we’ve got you thinking. The shell would have to be thick enough to reduce the chance of radiation passing unimpeded between its constituent atoms to pretty much zero. Because it takes time for heat to be conducted across a thickness, and black body surfaces re-emit what they absorb pretty much immediately, heat absorbed from the planet on the inward side will lead to the inner surface of the shell getting hotter than the outside. There has to be a differential for heat to flow by conduction from the inner to outer surface. That would mean a bit more than half of the re-emitted radiation would be back towards the planet. That would mean the planet would end up slightly hotter than 470 before steady state is reached and the outer shell surface attains the temperature where it emits 235 to space.

Gabriel’s law is only going to work for see through steel. 😉

147. tallbloke says:

Kristian: If the inside surface of the shell acted like a black body surface, then it would absorb ALL the energy received from the planet (235 W/m^2) and immediately reemit it in the same direction. The outside surface would not receive an energy flux of any kind and would accordingly not emit one either.

What would stop heat passing from the inner surface to the outer surface of the shell by conduction?

You cannot have both surfaces emit the same flux. And especially not half the flux received on one side only.

See my response to Gabriel above. The inner surface would emit slightly more than the outer.

So clearly this is not what happens. The shell is not an insulating layer.

It doesn’t need to be an insulator. Your statement is a non-sequiteur.

It needs to leave the system. It can only get rid of it emitting it outwards, to the infinite cold reservoir of space.

The vacuum between the shell and the planet is just as cold as space. It’s a vacuum, it has no temperature, by definition. So there is no ‘preference’ on the part of the shell to emit in one direction rather than another.

148. Gabriel van den Bergh (GabrielHBay) says, March 19, 2013 at 6:05 am: The only way the temperature of the surface of the sphere would be elevated by the presence of the shell would be if the shell offered resistance, or an impediment, to the flow of energy from the sphere to the cold space beyond the shell.

I think that is a good commonsense argument. But whether or not it is true is the matter of debate here. I suspect the answer is not going to be resolved to everyone’s satisfaction except by experiment.

Beware of experts and do keep plugging on. On the subject of ‘resistance’ to energy flow, you might want to look at my ATE Part I and ATE Part II articles.

Nobody could agree on much there either. 🙂

DC

149. Kristian says:

Ok, tallbloke.

I’m done with this thread. You just keep reiterating the same points over and over even after I’ve offered countering arguments, which you have not addressed … and multiple times at that. What you’re saying here, I’ve answered already. This has become a ‘YES! – NO! YES! – NO!’ kind of squabble.

That is not constructive debate as I see it.

Thanks for having let me participate …

Final words from me: Conduction or conductive resistance have no part in the radiative greenhouse mechanism. Introduce that and you move into the realm of convection. You’re completely blinded by the notion that Willis’ shell is a real stell sheel with real solid material properties. If it were, Willis’ model would prove something else entirely: The CONVECTIVE ‘greenhouse’ effect mechanism – the real one, the ATE.

Willis’ planet/shell model, on the other hand, is supposed to be a conceptual description of the radiative mechanism. So bringing conduction into the mix would render his whole argument moot, purely an academic irrelevance.

Gabriel is of course entirely right. Plainly considering what Willis’ model is actually about, there is no resistance to the energy flow from planet to space. Simply asserting there is, just won’t do.

Cheers!

150. gbaikie says:

Never give up! Never surrender!

I think it’s possible to make the Willis Greenhouse- to make the core surface which gets
235 W/m2 and becomes a temperature which radiates 470 W/m2.

Under the Earth surface one has less Watts per meters second added and higher
As someones else have mentioned there could problem of how it’s described.

So first off a vacuum makes perfect insulation in terms convection and the conduction
of heat. And most of us know how insulate against radiant energy in a vacuum- multiple
layers of reflective material.

So I will give a design of a Willis Greenhouse which can generate power which will use
geothermal energy.
A problem with Geothermal energy is the lack high temperature, and limited locations
which there is enough hot material at not very deep under the ground.

So the use of this Willis Greenhouse is one could use more areas less powerful geothermal energy and it would bring the heat up closer to the surface.
A down size is it has to be fairly large. It could easily be too expensive.
So need a big box and the box needs to be in a vacuum. This is fairly daunting when one has
a large volume in a 14.7 psi atmosphere. One needs a structure which can withstand the massive weight of the atmosphere. But one could use a technology which is about 2000 year old- a domed ceiling. The largest stone dome is said to be 85 meter in diameter: Global Pagoda. Which is suppose to use no pillars and just using stone- and suppose to last 1000 years.
So the limit might be 50 to 100 meter in diameter and one doesn’t to limit it to being made from can use whatever material is sufficient to do the job. But point is one uses compressive strength of
materials.
So build this over solid rock and one going heat up this solid rock. The reason you a large area
is the heat will conduct sideways and so a larger area reduces this. And you want a large thermal mass- meters deep and hundreds square meter area. So say in total one has 1000 cubic meters
of hot rock in which has piping draw out the heat.
So 2,000,000 kgs of rock. And rock has specific heat of about 0.75 kJ/kg K.
Assume say 100 K as range of useful energy: 75 kJ/kg.
Which is 150 million kilojoules- 150 billion watt seconds
41.6 million kilowatt hours.
Or a 1000 MW generator could run for 41.6 hours [minus the loss of energy
converting thermal to electrical energy].
The idea is energy storage not a constant supply- one needs time to build up
the heat needed. And would have optimal temperature to operate most efficiently
but could deliver power when not reached optimal temperature- if is greater
need of power.
In terms of economics one calculate total energy available in say one year.
So say area 300 square meter and usable thermal energy is 1 watt per square.
So 300 watts per second. Hmm that’s only 9.4 billion watt seconds a year.
Which is not economical, 15.9 years to get the 150 billion watts seconds.
Need more powerful geothermal energy and/or more area.

So 40 meter diameter is 1256 square meters and if got
10 watts per square meter, it might closer to being economical.

Anyways as far astopic of post, one stacks 100 to 1000 layers reflective
layer in this vacuum- and would cause the rock underneath it, to heat
up?

151. Kristian says:

David,

We appear to agree very much on where the ATE is coming from and what actually sets Earth’s surface temperature. It is a convective/conductive effect, based on the fact that the atmosphere has a mass.

So please discard the idea that a conductive version of Willis planet/shell model is anything like the radiative version of it. It will get you nowhere.

152. tallbloke says:

Kristian: Final words from me: Conduction or conductive resistance have no part in the radiative greenhouse mechanism. Introduce that and you move into the realm of convection.

No. Both sides of the shell are surround by a vacuum. There is no convection in vacuums. There is conduction within metals though, whether they are surrounded by vacuum, air, jello, water, oxtail soup, sand or whatever.

You’re completely blinded by the notion that Willis’ shell is a real stell sheel with real solid material properties. If it were, Willis’ model would prove something else entirely

That is what it is specified to be in this gedanken experiment. I said many times in this thread we are discussing that gedanken experiment, not the real Earth. That’s why I posted the other article on the IPCC/Wikipedia greenhouse effect. If you want a discussion about that, place you comments there, not here.

On this thread, you have made many assertions which I have rebutted with simple well known physics. You haven’t responded, except by trying to shift definitions.

153. Kristian says, March 19, 2013 at 11:15 am: Ok, tallbloke. I’m done with this thread.

Kristian is all over the place. On March 19, 2013 at 6:00 am he said: If the inside surface of the shell acted like a black body surface, then it would absorb ALL the energy received from the planet (235 W/m^2) and immediately reemit it in the same direction. The outside surface would not receive an energy flux of any kind and would accordingly not emit one either. You cannot have both surfaces emit the same flux. And especially not half the flux received on one side only.

This shows that he simply doesn’t understand the definition of a black body which has BOTH absorptivity/emissivity of 1.0 at all wavelengths AND perfect conductivity. It’s pointless to have a discussion about the Willis shell model except in the context of a true blackbody shell and core.

154. Gabriel van den Bergh (GabrielHBay) says:

Well TB, thank you for your response. Much appreciated. I do notice that Gabriel’s Law has not been falsified by your explanation, since it does imply that the globe surface will heat up if there is impediment to the energy flow. What you have clarified is that there will indeed be such impediment, due to the shell needing to have a certain thickness, giving rise to a temperature gradient across the thickness, presumably due to the steel not being a perfect conductor of heat.

Question 1: Were in our thought experiment the shell to be a perfect conductor and still a black body, what then? In my own thought experiment (hey, we are all entitled to play the game!) I would content that, since there would then seemingly be no impediment to energy flow from sphere to outer cold space, there is no way that anything can heat up in the inner space between sphere and shell, and that would include the surface of the sphere. Seems to me that heating up the surface of the sphere to a temperature higher than it would have been without the shell would only be possible if some energy was retained, at least in the start-up stage. How, if there is no resistance to it escaping? (Yeah, I know, I am supposed to be looking at the radiation laws… but no-one seems to be able to agree on the application of those)

Question 2: If the shell displays a temperature gradient, which implies that it is an imperfect conductor, then surely it is, sort of by definition, an insulator? What if it were a perfect insulator? Then, in this extreme case the outer surface of the shell would never heat up and therefore never emit and the whole thing would soon come to a fiery end.

So, by reading these questions together, are we saying that the degree of heating up of the sphere is somewhere in between and (perhaps only) dependant on the insulating qualities of the shell? I am such a heretic, I know (heh-heh). Nevertheless, I would like to think that my maverick questions do raise issues if looked at carefully. What I do vaguely remember from my largely long forgotten engineering studies (don’t ask) is that one can learn a lot by looking at what happens at the extreme limits of a parameter…

(BTW thank you for a wonderful blog and the tolerance you always display with regard to out-of-the-box thinking. I always come here when I get bored with the more mainstream stuff. Never mind that things often go way over my head… I still find it stimulating)

155. gbaikie says:

tallbloke says:
March 19, 2013 at 9:12 am

Hi Gabriel and welcome. Glad we’ve got you thinking. The shell would have to be thick enough to reduce the chance of radiation passing unimpeded between its constituent atoms to pretty much zero. Because it takes time for heat to be conducted across a thickness, and black body surfaces re-emit what they absorb pretty much immediately, heat absorbed from the planet on the inward side will lead to the inner surface of the shell getting hotter than the outside. There has to be a differential for heat to flow by conduction from the inner to outer surface. That would mean a bit more than half of the re-emitted radiation would be back towards the planet. That would mean the planet would end up slightly hotter than 470 before steady state is reached and the outer shell surface attains the temperature where it emits 235 to space.

Gabriel’s law is only going to work for see through steel. 😉

But planet has stone which conducts less than steel.
As I said earlier having this much energy go thru rock is a bit unrealistic- less
unrealistic if one fair amount volcanic eruption and it’s some average
amount.

But I can agree that steel would slow conduction of heat- but question is
how much and would be significant- and 235 W/m-2 not a lot heat for
metals to conduct. Rock is poor conductor and one going thru meters
[or km] and steel isn’t great as compare other metals but imagine we
talking about less than meter of it and per inch it’s conducts about
10 times better than an inch of rock
http://www.engineeringtoolbox.com/thermal-conductivity-metals-d_858.html
http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html

“A plane wall constructed of solid iron with thermal conductivity 70 W/moC, thickness 50 mm and with surface area 1 m by 1 m, temperature 150 oC on one side and 80 oC on the other.

Conductive heat transfer can be calculated as:

q = 70 (W/moC) 1 (m) 1 (m) (150 (oC) – 80 (oC)) / 0.05 (m)

= 98,000

156. tallbloke says:

Gabriel: Question 1: Were in our thought experiment the shell to be a perfect conductor and still a black body, what then? In my own thought experiment (hey, we are all entitled to play the game!) I would content that, since there would then seemingly be no impediment to energy flow from sphere to outer cold space, there is no way that anything can heat up in the inner space between sphere and shell, and that would include the surface of the sphere. Seems to me that heating up the surface of the sphere to a temperature higher than it would have been without the shell would only be possible if some energy was retained, at least in the start-up stage. How, if there is no resistance to it escaping? (Yeah, I know, I am supposed to be looking at the radiation laws… but no-one seems to be able to agree on the application of those)

Quick lunchbreak reply: The discussion should be framed in terms of the rate at which energy is being produced, and the temperatures the components of the system have to be at in order for a steady state to be reached whereby the same amount of energy is lost to space as is generated by the nuclear core. The shell can absorb and reach a temperature consonant with the energy it absorbs, same for the planet surface layer, which absorbs energy from both above and below. Vacuums are ‘perfect insulators’ and there is nothing in them to get hot. By definition they have no temperature. Only Matter can be at a temperature. If the steel is a high conductance alloy, then both sides will radiate around the same amount because the energy it is receiving on the inside from the planet is absorbed and heat spreads through the material rapidly. The ‘startup stage’ when the shell is introduced around the planet leads to a situation whereby the planet’s radiation is absorbed by the shell, and it emits half to space and half back towards the planet. The planet absorbs this ‘back-radiation’ and then re-emits it along with the energy coming up from the nuclear core.

Question 2: If the shell displays a temperature gradient, which implies that it is an imperfect conductor, then surely it is, sort of by definition, an insulator? What if it were a perfect insulator? Then, in this extreme case the outer surface of the shell would never heat up and therefore never emit and the whole thing would soon come to a fiery end.

Yes, But see caveats below. Remember that the reason roof lining polystyrene has silver foil glued to one side of it is because polystyrene is better at slowing conduction than radiation. The foil is to reflect the radiation back inwards.

So, by reading these questions together, are we saying that the degree of heating up of the sphere is somewhere in between and (perhaps only) dependant on the insulating qualities of the shell?

No. The insulating quality of the steel shell will have a small effect on the setup, making the planet/shell system get very slightly hotter internally in order to reach steady state where 235 is being emitted to space, but most of the reason the surface gets hotter is due to the effect of the shell radiating back towards it while at the same time the nuclear core is pumping heat into the surface layer from below. That’s the currently accepted radiation theory. Ford’s second experiment backs it up to a certain extent.

I am such a heretic, I know (heh-heh). Nevertheless, I would like to think that my maverick questions do raise issues if looked at carefully. What I do vaguely remember from my largely long forgotten engineering studies (don’t ask) is that one can learn a lot by looking at what happens at the extreme limits of a parameter…

Sure. When considering this stuff, we must always remember that perfect blackbodies, perfect insulation etc don’t exist, and at the limit, paradoxes can be generated by assuming them to be real.

157. Tim Folkerts says:

bwdave says: “Tim Folkerts, note that in your original quote T > T_c.

Yes, and the main point of that particular post was the dangers of relying on individual sentences (or equations) without really understanding the physics. In point of fact, the equation still works with T_c > T, you just get a negative answer meaning the heat is flowing into the object from the warmer surroundings.

158. Tim Folkerts says:

Kristian, I focus on “snippets” in order to try to address one issue and to respond to what people say (so I don’t misquote them or address a point they didn’t make).

For for instance .. you make one key point: “Only one thing, Tim. If the inside surface of the shell acted like a black body surface, then it would absorb ALL the energy received from the planet (235 W/m^2) and immediately reemit it in the same direction.”

As tallbloke also pointed out, this is half wrong. A black body will absorb everything as you say, but the energy is then thermal energy of the object. The object will re-emit from its surface according to its own temperature — NOT according to how much light it absorbed in the last instant.

“You cannot have both surfaces emit the same flux
They emit according to their own temperature. If the two sides are the same temperature, then they will emit the same radiation.

The model supposes a thin shell with good thermal conductivity, in which case the temperatures will be close to the same. The inside will be some amount ΔT warmer than the outside, but ΔT can be as small as we want in our thought experiment by making the shell thin enough.

“The other option is that the energy flux from the planet is absorbed and warms the shell ‘as a unit’. The shell has no heat capacity, no conductive resistance. So it will instantly upon absorption of the energy flux from the planet acquire the temperature of 254K. The shell as a whole acts like a black body. And accordingly, it will instantly need to emit an equal amount of energy (because more is coming in from the planet continuously) to its surroundings. It doesn’t matter if the energy is coming from the inside or the outside or within the shell itself. It cannot get rid of the energy emitting it inwards upon itself. It needs to leave the system. It can only get rid of it emitting it outwards, to the infinite cold reservoir of space.

This is close, but not quite the correct “other option”. Upon absorption, the shell will acquire 235 W/m^2 of flux and emit 235 W/m^2 of flux — we must conserve energy, not temperature. In order to emit 235 W/m^2 from an area twice as large as the planet, it must emit 235/2 W/m^2 and have a temperature of 213 K.

The energy does not need to immediately leave the system — it could also warm the system. And in fact, that is exactly what it will do.

*******************************************

The most obvious error I see in your conclusion is that you have 235 W/m^2 of thermal IR flowing from one surface at 254 K to another surface at exactly the same 254 K. How does heat know which way to go? The Zeroth Law of Thermodynamics concludes that two objects at the same temperature cannot exchange heat (ie cannot have a net energy flow from one to the other).

159. Kristian says:

Tallbloke,

So exactly what bearing does conduction have on the radiative GHE mechanism?

You still don’t see that Willis’ model portrays the conductive/convective (relating to Earth – no convection in vacuum. DUH!) ‘GHE’ (the real ATE), NOT the radiative? Good luck with that.

160. Kristian says, March 19, 2013 at 11:19 am: David, We appear to agree very much on where the ATE is coming from and what actually sets Earth’s surface temperature. It is a convective/conductive effect, based on the fact that the atmosphere has a mass. So please discard the idea that a conductive version of Willis planet/shell model is anything like the radiative version of it. It will get you nowhere.

Kristin,

I really don’t know where you get your misconceptions from. The Willis planet/shell model is a MODEL designed to challenge (successfully or unsuccessfully) those who do not believe in back radiation. So I really don’t have any idea about a ‘conductive’ version to discard. 🙂

The importance of the Willis model is that it makes us all think and argue our position more carefully with the aim of establishing some fundamental principles of radiation that enable us to move on.

161. Tim Folkerts says:

Gabriel, I like your “Law”. It sums up the situation quite well.

But you say “it seems that the shell would in fact not offer resistance”, while I conclude exactly the opposite.

The ONLY energy leaving the planet is IR radiation and ALL of that is COMPLETELY blocked by the shell (by definition it is an opaque blackbody). That blocked energy will -accordingto your Law, warm the shell and the planet, until eventually …
* the shell is hot enough to transfer a net 235 W/^2 to space (ie it is 254 K)
* the planet is hot enough to transfer a net 235 W/^2 to the shell (ie it is 302 K)

162. tallbloke says:

Kristian: You still don’t see that Willis’ model portrays the conductive/convective (relating to Earth – no convection in vacuum. DUH!) ‘GHE’ (the real ATE), NOT the radiative? Good luck with that.

I’m sorry but I can’t understand what you’ve written.

163. Roger Clague says:

Kristian says:
March 19, 2013 at 6:00 am

If the inside surface of the shell acted like a black body surface, then it would absorb ALL the energy received from the planet (235 W/m^2) and immediately reemit it in the same direction. The outside surface would not receive an energy flux of any kind and would accordingly not emit one either.

Kristian makes a good point.. A black-body radiator can have only one surface. The surface that receives the radiation also radiates all of it.

If the black body is the inner surface of the shell it will absorb 235W/m2 and radiate 235W/m2. The outer surface will not get hot so will not radiate. The planet surface is now getting 235W/m2 from inside + 235W/m2 from outside. It will emit 470Wm2. The shell will now get 470m2 and emit 470W/m2.
The planet and shell will get hotter until one melts or explodes.

If the outer surface is the black body it will radiate 235W/m2 to space.

164. tallbloke says:

Roger C:
A black-body radiator can have only one surface

Where is this stated then? Link?

165. Gabriel van den Bergh (GabrielHBay) says:

Hi Tim. Thanks for the comment. My understanding was that the shell, while a black body, is also a perfect, or near perfect, conductor of heat. If that were true, any heating up of the inner surface would sort of instantaneously be present on the outer surface as well, and enable immediate radiative loss to the cold outer space. Thus the shell would really effectively be transparent to the energy received at the inner surface and offer no resistance. I must have misunderstood.. 🙂

Seems to me that if not a perfect conductor of heat the shell must be an (at least partial) insulator, and does the whole game then not change? LOL So confusing…

Happy that you like my simple little practical Law… Thank you.

166. Gabriel van den Bergh (GabrielHBay) says, March 19, 2013 at 11:47 am: Were in our thought experiment the shell to be a perfect conductor and still a black body, what then?

Gabriel,

You have mis-thought!! A black body IS a perfect conductor by definition. It is also a perfect absorber and a perfect emitter of radiation of all wavelengths.

In the original article, Willis Eschenbach explicitly says that his core and shell are black bodies. Therefore it is absolutely pointless worrying about real materials such a real steel that fall somewhat short of the ideal.

The point of such imaginary models, as I am sure you can see, is to remove as much of the irrelevant clutter of variables so that the true didactic (i.e. educational) point can be explored in isolation. And the true point in this case surrounds the following divergence of opinions:

(1) There is NO back-radiation from a cooler body towards a warmer body. So in the case of the Willis core-shell model, core and shell will attain the same temperature (eventually, when at steady state).

(2) There IS back radiation from a cooler towards a warmer body but it is deflected straight back from the surface of the warmer body towards the cooler body where it is re-absorbed and does not contribute additional energy to the warmer object. So in the case of the Willis core-shell model, core and shell will attain the same temperature (eventually, when at steady state).

(3) There IS back radiation from a cooler towards a warmer body, and it IS absorbed by the surface of the warmer body. So in the case of the Willis core-shell model, the core will therefore be at a higher temperature than the shell.

For what it is worth, let me give you my opinion on these three alternative scenarios.

Scenario (1) means that the cooler body must somehow “know” not to emit photons in the direction of a warmer body. This seems to me to be highly implausible, if not downright ridiculous. Especially when you consider that the warmer receiving body might be at the other end of the universe from the cooler body. Personally, I think this option is a non-starter.

Scenario (2) has been championed by Joe Postma, Professor Claes Johnson and the so-called ‘SkyDragon group of scientists. But for this option to work, the warmer body would have to somehow “know” that an arriving photon had been emitted by a cooler body than itself. But individual photons don’t have a ‘temperature’. They are just packets of energy and can be of any strength, low or high. Therefore there is no information contained in the photon to tell the warmer body about the temperature of the sending body. [‘Temperature’ is a macroscopic statistical phenomenon determined by the average strength of the photon stream emanating from a particular body.] I still await clarification from the proponents of this option on how this mechanism could possibly work, and in the meantime remain very skeptical.

Scenario (3) has been championed by Willis Eschenbach (a climate change skeptic) and by warmists generally. It has neither of the drawbacks of options (1) and (2) and so is the one I (also a climate change skeptic) currently favour.

167. Gabriel van den Bergh (GabrielHBay) says, March 19, 2013 at 4:03 pm: Hi Tim. Thanks for the comment. My understanding was that the shell, while a black body, is also a perfect, or near perfect, conductor of heat. If that were true, any heating up of the inner surface would sort of instantaneously be present on the outer surface as well, and enable immediate radiative loss to the cold outer space.

Gabriel,

Perfect, NOT near perfect, conductor of heat, as I explained earlier. But you also have to remember that when radiation is absorbed by a surface it ceases to exist . The energy is converted to heat (kinetic energy) in the black body. The total amount of heat in the black body defines its temperature. And its temperature then defines the level of outgoing radiation from its surface(s), which, at steady-state, must of course equal the level of incoming radiation.

Also, in my opinion Clive Best is competely wrong to suggest that a black body can have only one radiating surface. The shell in the Willis thought experiment is a perfectly conducting, perfectly absorbing, perfectly emitting, two-surfaced, black body. Keep saying that to yourself over and over. It will avoid a lot of confusion. 🙂

168. Tim Folkerts says:

We do have some interesting dilemmas regarding “perfect” materials. For instance, a metal could be a perfect thermal conductor, but then it would also be a perfect electrical conductor, and perfect electrical conductors would be perfect reflectors (the opposite of a perfect blackbody).

Of course nothing is “perfect”. But for a thought experiment we can usually either
1) treat things as ideal (as long as two idealizations are not mutually exclusive)
2) Or we can recognize that we can get “close enough” to ideal (eg a thin enough shell that is a good enough thermal conductor painted with a thin enough layer of black paint) that the error between “real” and “model” is “close enough” to ideal and the differences between the model and reality is no more than δT and δP.

Gabriel says: ” If that were true, any heating up of the inner surface would sort of instantaneously be present on the outer surface as well, and enable immediate radiative loss to the cold outer space.
I agree so far …

“Thus the shell would really effectively be transparent to the energy received at the inner surface and offer no resistance.”
I would say that HALF is “available” at the inside and HALF at the outside — the energy is absorbed into the material and then “spread out” randomly throughout the shell so it could be anywhere. There is no reason for the heat to only go to one side — there is no reason that only the atoms ion the outer shell start vibrating, but the atoms on the inside don’t!

Only a transparent shell would be “effectively transparent”. 🙂

169. Roger Clague says:

tallbloke says:
March 19, 2013 at 3:57 pm
Roger C:
A black-body radiator can have only one surface

Where is this stated then? Link?

It follows from the definition, perfect absorber and perfect emitter.

A perfect emitter emits all the radiation energy it receives. None is converted to K.E. It is a perfect insulator. For a shell only the inner surface radiates. The outer surface does not get hot and does not radiate.

Davis Socrates says

A black body IS a perfect conductor by definition. It is also a perfect absorber and a perfect emitter of radiation of all wavelengths.

Not a perfect conductor, a perfect insulator.

170. Gabriel van den Bergh (GabrielHBay) says:

Hi David. Must confess, I actually did-do understand Willis’ black body shell perfectly. The reason I sounded uncertain is really because I was deliberately trying not to come across as too assertive (as a layman) in view of TB’s reply to my original post, in which he expressed the thought about the inner surface of the shell being at a higher temp than the outer shell, in turn implying less than perfect conductivity. I have generally been gentle in the expression of my questions etc. in order to elicit responses focussed on the issues rather than get embroiled in the rather explicitly expressed frustrations that had developed on this thread. I apologise if my understated tone led to misunderstanding. Thank you for your very extensive reply. Much appreciated. I am here to learn.

I did see on Claes Johnson’s site (I think it was) some brief mention of the idea of a standing wave becoming established between the cooler and hotter body, which would prevent backradiation. Forgive me if I misunderstood this, but it did seem as if this may be in the realm of some still undefined factors at work. Perhaps physics that still need to be explored? I have no idea what the characteristics of such a thing would be, and whether or not it may exist as a mechanism separate to the radiative transfer from the hot to cold body. I do however strongly suspect, on the basis of no more than a practical intuition, that there is physics still to be discover pertaining to this entire issue. The energetic differences of opinion almost demand that one accepts that the radiative science is far from settled.

For the record, I am quite a hard core sceptic in many ways and normally require really good argument to sway me.. but sway I will if the argument passes my rather strongly developed instincts. I am old enough to have learnt to rely on those… more often right than wrong 😉 There.. now I have blown my cover… (damn)

Very interesting thread, this. Goes to the heart of important issues, even if only a thought experiment.

171. A C Osborn says:

Gabriel van den Bergh (GabrielHBay) says: March 19, 2013 at 6:05 pm says “Very interesting thread, this. Goes to the heart of important issues”
Where most popular Forums regurgitate “stories” this site tries to do actual Science, which is why I find it so interesting. I sadly follow 20 forums everyday, so I know what most of them talk about.

172. Kristian says:

David Socrates says, March 19, 2013 at 11:27 am:

Sorry, I just couldn’t leave this unanswered:

“This shows that he simply doesn’t understand the definition of a black body which has BOTH absorptivity/emissivity of 1.0 at all wavelengths AND perfect conductivity. It’s pointless to have a discussion about the Willis shell model except in the context of a true blackbody shell and core.”

So with perfect conductivity, David, where is the resistance to energy flow in Willis’ model? The shell just happens to let through half the flux to the other side and keeps the other half at the inner surface?

Listen, either there is 235 W/m^2 emitted straight back from the inner surface and nothing from the outer. Or the entire flux is instantly conducted through the shell, emitted from the far side – as black body radiation (heat loss to surroundings).

Or maybe tallbloke can rebut my ‘assertion’ with ‘well known physics’ that a shell heated by a nuclear source within the shell itself would not warm to infinity (half the energy supplied emitted outwards, half inwards) …?

173. Kristian says:

Shite, that last one should read:

“… a shell heated by a nuclear source within the shell itself WOULD warm to infinity (half the energy supplied emitted outwards, half inwards) …?”

174. Kristian says:

Roger Clague says, March 19, 2013 at 6:02 pm:

Exactly, Roger C. They simply do not want to get the obvious point that a black body emit all absorbed energy TO ITS SURROUNDINGS. Not half to its interior.

175. Tim Folkerts says:

Roger C mistakenly thinks: “A perfect emitter emits all the radiation energy it receives. None is converted to K.E. “

Don’t fall victim to “sound-bite science”. The link you provide is only a superficial introduction to blackbodies.

A blackbody absorbs all radiation that hits it. By absorbing all the energy, the energy gets change from EM energy of the photon to thermal energy of the object. It is now indistinguishable from all the other thermal energy of the object. How the object deals with its thermal energy is a different question — it might emit some as radiation, it might conduct some to other parts of the object, it might undergo a phase transition, it might …

A blackbody is also a “perfect emitter” in the sense that it emits according to Planck’s law. This is the largest possible radiation from the surface. A non-blackbody would emit LESS than this. A blackbody does NOT instantly convert all of its thermal energy into thermal EM radiation!

176. Kristian says, March 19, 2013 at 6:18 pm: So with perfect conductivity, David, where is the resistance to energy flow in Willis’ model?

The 470Wm-2 from the core is absorbed by the shell. “Absorbed” means that the radiation is anihilated and replaced by an equivalent amount of Kinetic Energy in the shell, AKA heat.

But a hot body continuously radiates energy. In this case, we are assuming steady state, so the shell must also be emitting 470Wm-2. It has two equal area surfaces so it emits 235Wm-2 from each surface.

There is nothing complex or sophisticated about this. It follows from the definition of a black body (instantaneous distribution of heat to all parts) and the S-B law (governing the amount of energy per square metre emitted by a black body at a given uniform temperature).

I guess you could say that the ‘resistance to energy flow’ through the system is a consequence of the fact that the black body shell has two radiating surfaces so it MUST radiate half its energy flow back towards the core.

As I know you already appreciate, this is a highly theoretical thought experiment, not an earth-atmosphere model.

177. donald penman says:

The way i see things happening ,when a photon is absorbed by the inner surface of the shell it is turned into thermal energy but the outside of the shell is colder than the inside surface so energy will flow from hot to cold and no photon will be emitted by the inner surface but a photon will be emitted by the outer surface once enough energy has accumulated on the outside surface.

178. mkelly says:

David Socrates says:

March 19, 2013 at 4:20 pm

(3) There IS back radiation from a cooler towards a warmer body, and it IS absorbed by the surface of the warmer body. So in the case of the Willis core-shell model, the core will therefore be at a higher temperature than the shell.
Scenario (3) has been championed by Willis Eschenbach (a climate change skeptic) and by warmists generally. It has neither of the drawbacks of options (1) and (2) and so is the one I (also a climate change skeptic) currently favour.

Sir, please provide a radiative heat transfer equation showing this feed back loop for raising the temperture of the core.

Also does your acceptance of #3 not go against your earlier stated opinion on back radiation that it existed but did no work? If Q= U + W how can Q increase without increase of W since U is established.

179. Kristian says, March 19, 2013 at 6:27 pm: Roger Clague says, March 19, 2013 at 6:02 pm. Exactly, Roger C. They simply do not want to get the obvious point that a black body emit all absorbed energy TO ITS SURROUNDINGS. Not half to its interior.

Kristian, You are jumping around again. Which model are you on about now? I never am quite sure.

MODEL A:
If you are talking about the Willis shell+core model then the power, 235Wm-2, originates inside the core. The shell radiates 235Wm-2 upwards to space and 235wm-2 downwards towards the core (NOT just to its interior void). The core emits 235+235 = 470Wm-2 upwards towards the shell.

MODEL B: But if you are talking about a shell-only model where the 235Wm-2 (nuclear?) power source is inside the shell and there is only a void in the shell’s interior (no core), then the shell radiates 235Wm-2 upwards to space and 235wm-2 downwards into its ‘interior’ void, the latter radiation being re-absorbed not by a core (because there isn’t one) but by the shell itself. So arithmetically, the radiation through the internal void has a zero effect on the temperature of he shell: it isn’t thermalised as KE because there is no core to thermalise it in – it just returns repeatedly to the shell to balance the energy flow budget, doing no work.

180. donald penman says, March 19, 2013 at 7:07 pm: The way i see things happening ,when a photon is absorbed by the inner surface of the shell it is turned into thermal energy but the outside of the shell is colder than the inside surface…

Donald,

For the umteenth time…a black body conducts heat instantaneously to both sides. The outside and the inside surfaces are, by definition, at exactly the same temperature, as is every granule of the interior of the shell. So both sides radiate equally, determined only by the S-B law.

181. Tim Folkerts says:

Kristian,

I do agree with you that a perfect blackbody would not be a perfect conductor as David was claiming (and would in fact tend to be a good insulator).

But we can still make something that is CLOSE to the ideal (eg a thin sheet of copper with a thin coat of black on both sides.

But I can’t agree with “They simply do not want to get the obvious point that a black body emit all absorbed energy TO ITS SURROUNDINGS.”
Consider a block of ice (very close to a blackbody for thermal IR) in a 0C room. Place an IR heater above the ice. The ice will ABSORB the energy and MELT. If it “emitted all the absorbed energy”, the IR light could never melt nor warm the ice!

If the shell emitted all the absorbed photons, it could never change temperature.

“So with perfect conductivity, David, where is the resistance to energy flow in Willis’ model?”
The “resistance” is at the surface of the blackbody! As I pointed out a couple times already, it there were no “resistance” at the surface to IR photons, then the photons would go through = TRANSPARENT!

*The photons stop (the “resistance”).
*The photons’ energy turns into thermal energy.
*The thermal energy gets turned back into photons by atoms at the surface
* There are two identical surfaces with identical atoms with identical thermal energy that will both emit identical spectra of photons.

182. mkelly says, March 19, 2013 at 7:25 pm [quoting my post of March 19, 2013 at 4:20 pm to Gabriel]: (3) There IS back radiation from a cooler towards a warmer body, and it IS absorbed by the surface of the warmer body. So in the case of the Willis core-shell model, the core will therefore be at a higher temperature than the shell…Sir, please provide a radiative heat transfer equation showing this feed back loop for raising the temperature of the core.

This is a trivial request. The Willis model is not a multi-layer atmosphere or anything like that. It is just two theoretical black bodies, one enclosed within the other. The inner black body (the core) has an internal power source of 235Wm-2 and it also receives back radiation power from the enclosing shell of 235Wm-2. Consequently (from S-B law) its temperature is 334K. As a body at 334K it emits 470Wm-2 towards the shell.

What more is there to say? The core is in steady-state balance at 334K, whilst the shell is in steady-state balance at 254K.

Now how about something that you might positively contribute: why not give us your opinion on the impossibility (or otherwise) of each of the alternative options (1) and (2) that I outlined for Gabriel?

183. Tim Folkerts says:

Gabriel suggests: “The energetic differences of opinion almost demand that one accepts that the radiative science is far from settled.”

Or the other interpretation is that the science is settled, but many people have not studied enough physics to recognize the right answer. 😉

184. lgl says:

Kristian
What is it with you and numbers?

I keep asking myself, are they just joking?

185. wayne says:

Seems this has nothing mentioned of black bodies except by the commenters. Does it not have to follow these equations for this approximation, assuming linear temperature gradient:

F_in = F_out + ((F_out+F_on)/2/σ)^0.25 · shell_thickness / Cs
(recurse to precision)

F is for flux.
Cs is the thermal conductivity.

and the spheres temperature follows to always have 235 W/m² flowing:
F_sphere = F_out + F_in

so

F_sphere = 2·F_out + ((F_out+F_on)/2/σ)^0.25 · shell_thickness / Cs

I get:
Thickness 1 cm, Cs=0.001 W/m/K, F_sphere = 5068 Wm-2, T_sphere = 546.8 K
Thickness 1 cm, Cs=0.1, F_sphere = 497.5 Wm-2, T_sphere = 305.8 K
Thickness 1 cm, Cs=10, F_sphere = 470.25 Wm-2, T_sphere = 301.8 K
Thickness 1 cm, Cs=100000, F_sphere = 470 Wm-2, T_sphere = 301.73 K **close

Thickness 1 m, Cs=0.001, F_sphere = 2066620 Wm-2, T_sphere = 2457 K
Thickness 1 m, Cs=0.1, F_sphere = 5067.8 Wm-2, T_sphere = 547 K
Thickness 1 m, Cs=10, F_sphere = 495.7 Wm-2, T_sphere = 305.8 K

Approximation for Fukushima: (swag) 😉
Thickness 0.3 m, Cs=0.12131, T_outside = 1000 K, F_sphere = 116489 Wm-2, T_sphere = 1197 K, melt yet?

My lesson: don’t even try to make sense of Willis’s thought experiment without at least allowing for the shell’s thickness and its conductivity.

What is surprising is when you open up some small “windows” through the shell to let a portion directly out… you end up basically with Miskolczi’s numbers for the Earth/atmosphere, very close.

186. Kristian says:

David, let me ask you a simple question.

A black body receiving and absorbing a continuous incoming energy flux and accordingly reemitting (as required by definition) an equal energy flux (from whichever side of the body, you insist it’s all the same) of 470 W/m^2, would that not have an emission temperature (as per the S-B equation) of 302K?

Well, Willis’ black body shell receives, absorbs and reemits an energy flux of 470 W/m^2. But its temperature? 254K.

Remember, the absorption always occurs before the reemission. There would be no emission without the absorption first. So why doesn’t the incoming flux of 470 W/m^2 raise the temperature of the shell to 302K? It doesn’t matter if the incoming flux is coming from the outside (a sun), the inside (emitting planet) or within (‘intrastructural’ nuclear source), the flux is still absorbed completely and then reemitted to balance. Why doesn’t the S-B equation work on Willis’ shell?

Your ‘IPCC model’ seems to claim the following:

# CORE absorbs 235+235= 470 W/m^2, emits 470 W/m^2, S-B temperature 334K (?!, really 302K)

# SHELL absorbs 470 W/m^2, emits 235+235= 470 W/m^2, S-B temperature 254K.

What’s wrong with this picture?

187. Tim Folkerts says:

David, you better check your numbers

* 235 W/m^2 = 254.73 K
* 470 W/m^2 = 301.73 K (not 334 K as you wrote)

Also, the model here may assume perfect thermal conduction, but that is not a required feature of all blackbodies. Indeed, it is in general NOT what you would expect from a blackbody.

On the other hand, we can still use the “sheet of copper with a thin coat of black paint” approach. A 1 m thick sheet of copper would only be ~ 0.6K difference from inside to outside. Make that 1 cm thick and we are down to less than 0.01 K difference across the shell, which is “close enough” for any practical purposes in this discussion. With a thin sheet of copper, the interior might be 254.00 K and the exterior 253.99 K — hopefully no one will object to such an approximation being called “the same temperature”.

188. lgl says:

mkelly
The equation for a positive feedback loop is, Gain=1/(1-B) when there like in this case is no forward gain.
B is the loopback gain, in this case 0.5 because the shell has two sides, so G=2.

189. Kristian says:

If you want to argue ‘Ah, but the NET incoming flux to the shell is 470-235= 235 W/m^2!’, even if the actually absorbed flux IS 470 W/m^2, there’s no getting around that, then consider this:

The NET outgoing flux from the core planet is also 470-235= 235 W/m^2. Why, then, is it OK for the planet to have a temperature of 302K, but not for the shell?

The temperature is set by the absorption of the INCOMING flux, then the emission of the OUTGOING flux is set accordingly.

190. Kristian says:

Aarghh, I promised to leave this thread alone. Hard work it seems …

191. Tim Folkerts says:

Kristian claims: “Remember, the absorption always occurs before the reemission.”

NO! Don’t remember that. The two are basically independent.

Photons are EMITTED based on the object’s own thermal energy. The atoms are constantly vibrating –gaining and losing energy. When an atom at the surface has some energy, it might emit some of that energy as a photon. The hotter the object, the more energy the atoms will have and the more photons they will emit. The atom does NOT have to first absorb a photon in order to emit a photon.

When a photon gets ABSORBED, it makes that atom vibrate a little extra — it has slightly higher than average thermal energy. As that atom interacts with other nearby atoms, it exchanges energy, and the energy of that photon is added to the general “fog” of thermal energy within the object. It will NOT stay only with the atom that absorbed the photon, waiting to be emitted.

192. tallbloke says:

Has anyone here heard of the double slit experiment? I think there probably are open questions around ‘photons’ and ‘E/M waves’ and how energy can manifest itself as either…

193. lgl says:

Kristian
# CORE absorbs 235+235= 470 W/m^2, emits 470 W/m^2, S-B temperature 334K (?!, really 302K)
Those two 235 are the same square meters

# SHELL absorbs 470 W/m^2, emits 235+235= 470 W/m^2, S-B temperature 254K.
These two 235 are different square meters so it’s 235, not 470. The surface area of the shell is two times that of the planet, remember? This is like stating if the temp of the southern hemisphere is 15 C and the temp of the northern is 16, then the global temp is 15+16=31C.

194. lgl says:

Kristian
“Why, then, is it OK for the planet to have a temperature of 302K, but not for the shell?”

Because then the shell would emit 470 out and 470 in and it would have cooled because there is not that much energy available.

195. Max™ says:

Scenario (2) has been championed by Joe Postma, Professor Claes Johnson and the so-called ‘SkyDragon group of scientists. But for this option to work, the warmer body would have to somehow “know” that an arriving photon had been emitted by a cooler body than itself. But individual photons don’t have a ‘temperature’. They are just packets of energy and can be of any strength, low or high. Therefore there is no information contained in the photon to tell the warmer body about the temperature of the sending body. [‘Temperature’ is a macroscopic statistical phenomenon determined by the average strength of the photon stream emanating from a particular body.] I still await clarification from the proponents of this option on how this mechanism could possibly work, and in the meantime remain very skeptical.“~ David Socrates

Again, this is a misrepresentation of their position as far as I can tell.

So rather than trying to explain why, let me propose an alternative you left out:

Scenario (4) the radiation reaching a warm surface from a cold surface reduces the power of the radiation leaving the warm surface as represented by the following equation: P = ε σ A T⁴ – Tc⁴ and as such the radiation from the cold surface is not able to increase the internal energy of the warm surface, only reduce the rate at which the warm surface loses energy.

196. Being contradicted by Roger Claque and Tim Folkerts is a rare experience 🙂 so it has given me pause for thought. I have done some thinking about whether or not the definition of a black body does or does not include the requirement that it be a perfect conductor.

On reflection, I think that Roger and Tim are both right that black bodies are by definition insulators in the electromagnetic sense, because they instantly, and 100% effectively, anihilate EMR when it enters their surfaces. Also a bit of googling has not produced much confirmation that black bodies are also required to be perfect thermal conductors, although I did find this:

Fortunately, it is possible to construct a nearly-perfect blackbody. Construct a box made of a thermally conductive material, such as metal. The box should be completely closed on all sides, so that the inside forms a cavity that does not receive light from the surroundings. Then, make a small hole somewhere on the box. The light coming out of this hole will almost perfectly resemble the light from an ideal blackbody, for the temperature of the air inside the box.
[http://docs.kde.org/stable/en/kdeedu/kstars/ai-blackbody.html ]

So maybe when considering the Willis shell-core model, we should just agree to assume perfect thermal conductivity in order to have a rational debate without introducing the complications of differential temperatures between the inner and outer surfaces of the shell, irrespective of whether perfect conductivity is a formal requirement of a black body. Just agreeing a very thin shell thickness achieves our objective of not gettig bound up in unnecessary complications.

After writing this I now see it is in contradiction to Wayne’s latest suggestion which is: My lesson: don’t even try to make sense of Willis’s thought experiment without at least allowing for the shell’s thickness and its conductivity..

No, no, no, Wayne – don’t even think about it!

197. Arfur Bryant says:

thefordprefect says:
March 19, 2013 at 4:57 am

TFP,

Thanks for the info. Best of luck with your next experiment. I understand your figured but I still think the 0.3C difference could be due to the small reduction in heat loss. Either way, I can’t prove that so I continue to applaud your experimentation and diligence.

I would, however, still like to see a link showing how the lower energy radiation numerically adds to the higher energy surface – if anyone has such a link?

Regards,

198. Arfur Bryant says:

Tim Folkerts says:
March 18, 2013 at 11:41 pm
[“Arfur … No, with the heater on the outside the interior would not get hotter. The whole interior would be 254 K once steady-state conditions were achieved.”]

Tim, thanks very much but what I actually asked was “…would the shell get warmer?” In Willis’ toy, the reflected energy from the shell adds to the emitted energy of the planet. So I ask would any emitted radiation from the ‘role-reversed’ planet (above 0K) add to the energy level of the ‘role-reversed’ shell?

I agree with Gabriel… very interesting thread.

199. Kristian says:

lgl,

What are you trying to pull here?

Consider the NET heat gain and loss for the shell vs. the planet at steady state in David’s ‘IPCC model’.

SHELL – NET gain (inner surface): 470-235= 235 W/m^2; NET loss (outer surface): 235-0= 235 W/m^2; balance – 254K.

PLANET – NET gain (from nucleus): 235-0= 235 W/m^2; NET loss (from surface): 470-235= 235 W/m^2; balance – 254K … no, wait!

200. Tim Folkerts says:

David,

The blackbody you are describing is the origin of the closely related term “cavity radiation”. Look at this image: http://www.instructables.com/image/F80NCK4FABDYVU0/Grab-a-soda-can.jpg
Notice how the opening to the coke can looks jet black, even though we know the interior is shine aluminum.

To implement something like this on willis’ shell, you would make something like this this heat sink, with fins on both sides. Notice how the space between is fairly dark, even though the metal is shiny.

I think it is easier just to use a thin coat of soot or a high-emissivity paint.

201. Arfur Bryant says:

Tim,

Sorry, I wrote ‘reflected energy’ when what I meant was re-emitted energy.

202. Kristian says, March 19, 2013 at 8:23 pm: David, let me ask you a simple question…

Tim Folkerts (March 19, 2013 at 8:50 pm) and Igl (March 19, 2013 at 8:59 pm) have between them hopefully put you right on this. But if you still can’t understand simple arithmetic – then I give up. 🙂

I would only add that the temperature of a black body dictates the outgoing radiation from its surface – not the incoming radiation into its surface. Of course these may be the same at steady-state but not necessarily. For example, the core has 235Wm-2 of non-radiation input from its thermal nuclear reactor and only 235Wm-2 of radiation from the shell. But it has 470Wm-2 of outgoing radiation, hence its enhanced temperature.

[Yes, apologies, my figure of 334K was incorrect – it should indeed have been 302K. It, of course, makes no difference to the conclusions.]

203. Tim Folkerts says:

Wayne,

I tried an exact solution to the question of the temperature gradient of the shell. I got a little larger numbers than you, but I agree with spirit of what you found. 🙂

For a good thermal insulator with thermal conductivity < 1, the gradient across even a 1 cm shell the shell can be large — many degrees. This raises the temperature of the planet by many degrees as well. But this should not be surprising — a styrofoam cooler can be several degrees different in temperature from one side to the other.

For a good conductor, the difference across that 1 cm shell will be minimal (on the order of 0.01 K) and the effect on the planet will be minimal.

So, yes, we need to consider the shell, but we can easily design a "real" shell that is close to the ideal. (Actually, the fact that the emissivity of the surfaces is not 1.00 is probably a much bigger "real" concern.)

204. Tim Folkerts says, March 19, 2013 at 10:06 pm: I think it is easier just to use a thin coat of soot or a high-emissivity paint.

Ever the practical enginer, I’m right with you there. 🙂

[Thanks also for pointing out the wrong core temperature in my diagram]

205. Tim Folkerts says:

Kristian ponders: ‘Consider the NET heat gain and loss for the shell vs. the planet at steady state in David’s ‘IPCC model’.

ANY object at steady temperature will have a net energy flow (ie “heat flow”) of 0 W/m^2 (barring something like a phase transition). Having zero net energy flow tells you nothing about the temperature. Both the shell and the planet have a net energy flow of zero so they are not changing temperature.

You CAN tell the temperature using an IR thermometer that measures the outgoing radiation from the surface. The planet has 470 W/m^2 outgoing IR = 302 K. Then re-aim the IR thermometer and find that the shell emits 235 W/m^2 of IR = 254 K.

206. Kristian says:

David Socrates says, March 19, 2013 at 9:40 pm:

“Being contradicted by Roger Claque and Tim Folkerts is a rare experience so it has given me pause for thought. I have done some thinking about whether or not the definition of a black body does or does not include the requirement that it be a perfect conductor.

On reflection, I think that Roger and Tim are both right that black bodies are by definition insulators in the electromagnetic sense, because they instantly, and 100% effectively, anihilate EMR when it enters their surfaces.”

Well, Roger contradicted you by backing me specifically when I said:

“If the inside surface of the shell acted like a black body surface, then it would absorb ALL the energy received from the planet (235 W/m^2) and immediately reemit it in the same direction. The outside surface would not receive an energy flux of any kind and would accordingly not emit one either. You cannot have both surfaces emit the same flux. And especially not half the flux received on one side only.”

This is how you responded to this at the time:

“Kristian is all over the place. … This shows that he simply doesn’t understand the definition of a black body which has BOTH absorptivity/emissivity of 1.0 at all wavelengths AND perfect conductivity. It’s pointless to have a discussion about the Willis shell model except in the context of a true blackbody shell and core.”

Some people are apparently more equal than others …

That aside, as far as I know, a real black body has zero heat capacity and also does not have any depth/thickness (that is, no potential for heat conduction from surface down). The only time I’ve heard ‘perfect conduction’ mentioned as a prerequisite in relation to black bodies is for objects that are not illuminated from all sides at once. In such case, the perfect conduction would ‘operate’ only laterally at the actual surface of the black body.

207. Tim Folkerts says:

Arfur,

Sorry if I was unclear. I meant EVERYTHING inside the outer surface of the shell would be 254 K when the heaters were mounted on the exterior of the shell. That includes the whole shell and the whole planet.

208. Roger Clague says:

wayne says:
March 19, 2013 at 8:20 pm

What is surprising is when you open up some small “windows” through the shell to let a portion directly out… you end up basically with Miskolczi’s numbers for the Earth/atmosphere, very close.

Not so surprising.
The atmosphere is a gas and radiation penetrates and interacts with molecules at all levels. The atmosphere is a black body at the top surface.
Wiilis’s greenhouse needed a little improvement, some thickness, conductivity and windows.

Please post details of the shell with windows calculations and the Miskolczi’s numbers.

Kristian

you are right, according to Willis’s analysis, the shell receives 470W/m2 and emits 470W/m2 so its temperature must be 301K.

# CORE absorbs 235+235= 470 W/m^2, emits 470 W/m^2, S-B temperature 302K

# SHELL absorbs 470 W/m^2, emits 235+235= 470 W/m^2, S-B temperature 302K

Where’s the greenhouse effect ?

Tim says

Also, the model here may assume perfect thermal conduction, but that is not a required feature of all blackbodies. Indeed, it is in general NOT what you would expect from a blackbody.

The model does assume perfect thermal conduction and the opposite is required for a black body.

209. Max™ says, March 19, 2013 at 9:35 pm

1. You say that Scenario (2) is a misrepresenation of the Postma, etc. position. I would not be at all surprised – I have never been fully able to understand it. If you know for sure what it is, do help me out. 🙂

2. As far as I can see your proposed Scenario 4 is very much the same as my Scenario 3. You actually talk about radiation going both ways and netting off. You say: the radiation from the cold surface is not able to increase the internal energy of the warm surface, only reduce the rate at which the warm surface loses energy. You also provide the correct mathematical calculation. But the critical question is how the netting off mechanism works physically and this you do not explain.

210. Roger Clague says, March 19, 2013 at 10:37 pm:

# CORE absorbs 235+235= 470 W/m^2, emits 470 W/m^2, S-B temperature 302K
# SHELL absorbs 470 W/m^2, emits 235+235= 470 W/m^2, S-B temperature 302K

Wrong! The core emits 470Wm-2 from ONE surface. The shell emits 235Wm-2 from TWO SURFACES.

The temperature of a black body is related through the SB law to the outgoing power intensity in watts per square metre, irrespective of the area of its surface(s)!!!!!!!

The power intensity out of the core is 470Wm-2.

The power intensity out of the shell is 235Wm-2.

Hence the difference in temperatures.

Please now accept that you have made an elementary logical error because otherwise this could go on for ever and ever. 🙂

211. Kristian says:

David Socrates says, March 19, 2013 at 10:17 pm:

“Tim Folkerts (March 19, 2013 at 8:50 pm) and Igl (March 19, 2013 at 8:59 pm) have between them hopefully put you right on this. But if you still can’t understand simple arithmetic – then I give up.”

Are you being serious? You’re the ones that cannot understand simple arithmetic. The numbers are right there in front of you. And still you deny them. Flatly. Or try your utmost to explain them away. Tim’s desperate attempts at obfuscating the simple facts here appears to impress you greatly.

A black body absorbs 100% of incident radiation. That energy is turned into KE. Which sets the temperature of the body. Which consequently emits a flux according to this temperature (level of KE). The emission is a consequence of the absorption. And they are equal.

How hard is this? Black body emission is the result, not the cause of the black body’s temperature. Its cause is the absorption … of the incoming flux.

You can compare this with the core planet say before the shell is placed around it. The energy flux to the surface from the nucleus is 235 W/m^2. THIS IS WHAT SETS THE TEMPERATURE. The outgoing flux simply needs to balance the incoming. When we ‘read’ the outgoing flux, we can TELL the temperature of the body. But it’s the incoming flux that is actually causing that specific temperature.

So, the core planet receives an incoming flux (from the nucleus) of 235 W/m^2. This doesn’t change. The shell in the end (according to your ‘IPCC model’) receives an incoming flux (at the inner surface) of 470 W/m^2. And still the planet ends up with a temperature of 302K while the shell ends up at 254K.

Go figure. This is pure AGW accounting. Upside-down. Magical.

212. Max™ says:

But the critical question is how the netting off mechanism works physically and this you do not explain.” ~David Socrates

Uh, radiation is a vector, two opposing vectors added together is the same as subtracting the cold surface emissions from the warm surface emissions.

213. Westy says:

If Gabriel can have a law so can I and it’s simple – “Your cold stuff can’t heat up your hot stuff.” Wait, what about Willis’s shell? My law must be a dud.

214. Bryan says:

Tallbloke asked for contact times given values for heat capacities and temperatures.

Taking Earth size as the planet and the specific heat capacity of steel(490J/kgK) we can make crude approximations.
Planet and shell touch and radiate to space with initial temperature of 302K and contracting Planet separates at 290K(say)
Assuming that the planet-shell combo radiates at maximum 470W/m2 to space it could maintain this for 5390 years.

This is a considerable underestimate as the maximum radiation (470W/m2 ) drops to 401W/m2 at 290K.
This is also a considerable underestimate for we also have ignored any contribution from the Nuclear core for this period.

The cycle spontaneously repeats itself endlessly.
This separates thermal energy spontaneously into two streams at higher and lower temperatures over time.
Thereby creating a capacity of producing mechanical work .
This is a classic problem and has been often discussed over the last 120years.

There seems no problem with scaling the Willis model up or down.
If a vacuum plus passive shell could increase temperatures at will its remarkable that such devices have never been commercially produced!

The conclusion is that this would violate the second law for reasons I have previously given.

215. tallbloke says:

Thanks Bryan, I’ll give this some thought. Nearly midnight here so I won’t try to formulate it now.

216. Kristian,

One more try.

Yes, I agree the core planet receives 235Wm-2 from its nuclear power source. And, yes, I agree if there is no shell, the core will radiate 235Wm-2 to space. And, yes, I agree under those circumstances its temperature will be 254K. All agreed. OK?

But when we (magically and instantaneously) introduce the shell, things change.

To make the discussion much easier, let’s introduce the shell instantaneously after it has been warmed up to a temperature of 254K. This means that, at the instant it is introduced, it will already be radiating at the rate of 235Wm-2 from BOTH its surfaces (simple application of the SB law). So the shell will now be emitting 235Wm-2 towards the core and 235Wm-2 towards space.

What happens next?

Well the power intensity into the core has now doubled from the original 235Wm-2 nuclear source to 470Wm-2. So for balance it will radiate out of its single surface at 470Wm-2 towards the shell.

But ha! 470Wm-2 is EXACTLY the power intensity into ONE of the shell’s surfaces that the shell needs to receive in order to be able to sustain an output power intensity of 254Wm-e across BOTH of its TWO surfaces.

Meanwhile, the core temperature will have been raised to 302K (simple application of the SB law to the increased input flow of 470Wm-2).

The system is now in steady state with its core at 302K and its shell at 254K.

Rationalise it this way
The core has only one surface whereas the shell has two. So pushing a total of 470W/m2 into a body that has only ONE outward radiating surface (the core) is obviously going to raise it to a higher temperature than pushing 470Wm-2 into a body that has TWO outward radiating surfaces (the shell).

217. Tim Folkerts says:

Bryan,

I get a similar time to cool by 12 K for a solid steel planet in contact with the shell (depending on just what mass I assume).

I agree that the time should be a little more than doubled to account for the cooling and the added power from the heaters. (ie the real power loss is between 470-235=235 W/m^2 and 401-235= 166 W/m^2.

So the cooling time would be 12,000 – 15,000 years or so.

What do you get for the rewarming phase — as the planet rewarms from 290 K to 302 K?
At first, the planet is 290 K radiating 401 W/m^2. The shell will radiate 1/2 in and 1/2 out = 200 W/m^2 back to the planet. The heaters add 235. So the net power into the planet is 235+200-401= 34 W/m^2. As the temperature of the planet approaches 302 K, the net power approaches 235+235-470= 0 W/m^2 (ie the temperature asymptotically approaches 302 K).
At best the net power is 35 W/m^2; at worst it is 0 W/m^2.

This is less than 1/10 the rate of cooling, so it would take more than 10 times longer to rewarm from 290K to 302K as it took to cool from 302K to 290K.

218. Tim Folkerts says:

Bryan, I am not sure what your concern is when you say

The cycle spontaneously repeats itself endlessly.
This separates thermal energy spontaneously into two streams at higher and lower temperatures over time.
Thereby creating a capacity of producing mechanical work .
This is a classic problem and has been often discussed over the last 120years.

* We have a cold reservoir (space) at 0 K.
* We have a continuous source of power to keep the hot reservoir warm.

1)There should be no problem using this to extract mechanical work.

2) This is not the typical “Maxwell’s Demon” problem where things spontaneously change from one temperature into a higher and lower temperature. The typical setting involves “stuff” at one temperature and a “demon” that sorts them out — putting the warmer atoms (or photons) on one side and the cooler atoms (or photons) on the other. Here we have very hot nuclear heaters — they heat some stuff to a warm temperature, and then later heat stuff to a slightly warmer temperature. We never have a single bunch of stuff that gets sorted into two different temperature piles. We only have a changing amount of “insulation” around the heaters.

219. wayne says:

Roger Clague, I had to dig back into Ferenc’s papers, brother, hard to do every time. Here’s how what I said seems to tie to his works.

Source: Greenhouse effect in semi-transparent planetary atmospheres
Ferenc M. Miskolczi
IDŐJÁRÁS Vol. 111, No. 1, January–March 2007, pp. 1–40

From the ERBE (2004) data product we estimated the five year average planetary equilibrium temperature as tE = 253.8 K, which resulted in a global average OLRA = 235.2 W m-2. From the same data product the global average clear-sky OLR is 266.4 W m-2.

and

For the Earth obviously the TA ≈ 0 condition apply and the A/ A 3/5 OLR SU = equation gives an optimal global average surface upward flux of A 392 SU = W m-2 and a global average surface temperature of
288.3 K.

Well since the 253.8 K is the 235.2 W/m² OLR, and surface upward flux of 392 by this you can extract the window flux:

By the equation: WinowFlux = 2·OLR – SurfFlux

235.2 · 2 – 392 = 78.4 W/m² window radiation

So, you end up with 238.2 – 78.4 or ~159.8 W/m² that goes both up and down at the ~”ToA”. i know the 80 is not exactly 78.4 but it sure is close.

In reality that 159.8 comes from somewhere (really smeared vertically near the ToA) but all ends up isotropic so for the 159.8 up and outward there is also an equal 159.8 downward (the inside of the shell) that may actually be just resistance to not radiation but convection, conduction, and latent heat transfer by temperature differentials… it has to, only a small amount of net radiation is coming from the surface except the window radiation and that doesn’t react at all, zip, out to space. That is the resistance that raises our surface temperature above with no atmosphere, the 159.8 or so it seems. What I am saying is it doesn’t have to be radiation at all, they are all energy transports, slow any and you get a rise in temperature at the base.

And sure enough that is what I remember of Miskolczi’s work, he rarely states things explicitly, just the relationships, you have to do the actual math (though quite simple), that his window radiation is about 10 W/m² above TFK’s 40 from the surface and 30 from the cloud tops.

That’s about all I can hand you easily, if you haven’t already read his works you really should.

Some of these thoughts have been developed while this thread has been running so some of it is new to me too, don’t know why I didn’t see this years ago.

220. gbaikie says:

“Clague says:
March 19, 2013 at 6:02 pm

tallbloke says:
March 19, 2013 at 3:57 pm
Roger C:
A black-body radiator can have only one surface

Where is this stated then? Link?

It follows from the definition, perfect absorber and perfect emitter.

A perfect emitter emits all the radiation energy it receives. None is converted to K.E. It is a perfect insulator. For a shell only the inner surface radiates. The outer surface does not get hot and does not radiate.”

First a blackbody must convert the radiant energy into KE. Or blackbody spectrum “expresses”
the KE in solids and liquids, so ideal blackbody must at least mimetic what matter radiates when it has certain amount of KE.

With climate models the ideal blackbody is only considered to radiate from one surface [outward from a spherical surface- rather than inward]- hence disk diameter divided by 4. And this disk area
is also consider one side.
And as said somewhere earlier this a problem with climate models- there are 2 dimensional rather
than 3 dimension.

But consider blackbody has having one side or two sides or one side of surface of a sphere or six side of a cube or whatever geometry you want.

With an idea blackbody without having insulation and it’s a flat panet [a thin material so ignore the edge area] at Earth distance it will 1/2 the blackbody temperature- side facing sun radiates 1363 W/m-2 divide 2 and side facing from sun radiate the other half.

But suppose this same panel was so close to the Sun that “half the sky was the sun” or side facing the sun can only radiate at the sun. in that situation one side insulated- one can have the side facing sun radiating energy at sun cause it’s the same or higher temperature [which what any insulation does [whether one talking insulation for radiant energy, conduction or convection- heat can’t flow it it’s the same or higher temperature than the surface].

When standing on Earth, 180 degree of 360 degrees [roughly] is the sky with other half the ground. And if 10 km up, you high enough to begin to see Earth curvature more clearly but still largely dominated 180 degrees facing earth
Since the Sun is so much bigger than Earth, being a million km above the 1.39 million km diameter
would have sun dominating the 180 degrees half of sky.

Try again from Earth the sun is 32 arcminutes. And 60 arcminute is 1 degree:
“The table shows that the angular diameter of Sun, when seen from Earth is approximately 32 arcminutes (1920 arcseconds or 0.53 degrees)”
http://en.wikipedia.org/wiki/Angular_diameter

Say had sphere which was 360 cm in circumference [one cm for each of 360 degrees] so [1.14 meters diameter or 57 32 cm in radius].
And it has surface area of 41,267 square cm. Half this for hemisphere: 20,633 square cm and the
sun would fill in .220 of square cm of area of the 20,633 square cm of surface of the hemisphere.
Or at Earth distance, the sun’s area is about 1/80,000th of the area in the 180 degree side facing the sun.
But if much, much closer, the sun could occupy more than 1/2 of this area.

With our steel greenhouse it’s similar to being near the sun- only side radiates as other side is insulated by warmth of planet.

221. wayne says:

Roger Clague, everywhere you see 159.8, replace it with 156.8. If you notice I carried a wrong figure forward… 238.2 instead of the correct 235.2. I should check my math before hitting that “Post Comment” button. 😉

I goes like this all together:

OLR, 235.2
Surf 392
Window, 235.2*2 – 392 = 78.4
Surface – Window is 392 – 78.4 = 313.6 that is equivalent to the planets surface, the window radiation leaves without interaction. So divide by two, 313.6/2 is the 156.8 which is equivalent to the inside of the shell which combined back with the window that slipped through the “holes” gives the 156.8 + 78.4 = 235.2 leaving the outside of the shell even though the outside of the shell is also at 156.8. In an atmosphere there is no real thermal conductivity right at the ToA.

Maybe that will make more sense in a nutshell.

222. Tim Folkerts says:

“Clarke’s Third Law” states:

Any sufficiently advanced technology is indistinguishable from magic.

Perhaps we need to postulate a corollary:

Any sufficiently advanced physics is indistinguishable from obfuscation.

🙂

223. bwdave says:

I think the big problem with the Willis Model is the apparent doubling of the black body’s surface area between energy receipt and energy transmission.

Tim Folkerts, I agree, warmer surroundings transfer heat to a cooler object, but like I said : “The term “net” doesn’t imply that T_c is heating T.”, where T_c is less than T; because a cooler object won’t heat a warmer one without help.

224. bwdave says:

The problem with the shell as a black body inside and out and also as a conductor, is that its surface area doubles between being irradiated and radiating.

Tim Folkerts, I agree, warmer surroundings transfer heat to a cooler object, but like I said : “The term “net” doesn’t imply that T_c is heating T.”, where T_c is less than T; because a cooler object won’t heat a warmer one without help.

225. bwdave says:

226. Bryan says:

Tim Folkerts says

“We have a cold reservoir (space) at 0 K.”

Could you give a practical method of dumping heat to space in the given circumstances?
Electromagnetic radiation emitted must represent the movement of the molecules of the steel shell.
In the classic problem I posed the higher and lower temperatures represent the movement of molecules of steel exactly like the demon example.

227. David says:

A physics law for thought….”Only two things can effect the energy content of any system in a radiative balance. Either a change in the input, or a change in the “residence time” of some aspect of those energies within the system.”

228. David says:

Food for thought…”Only two things can effect the energy content of any system in a radiative balance. Either a change in the input, or a change in the “residence time” of some aspect of those energies within the system.”

229. lgl says:

Kristian
“SHELL – NET gain (inner surface): 470-235= 235 W/m^2; NET loss (outer surface): 235-0= 235 W/m^2; balance – 254K.

PLANET – NET gain (from nucleus): 235-0= 235 W/m^2; NET loss (from surface): 470-235= 235 W/m^2; balance – 254K … no, wait!”

No, both the gain and loss of both the planet and the shell is 470 W/m^2, both in balance.
Set the area of the planet to1 m^2, then the area of the shell is 2 m^2 and the radiation from the planet is 470W/1m^2 and from the shell 470W/2m^2=235W/m^2, temps 302K and 254K

230. Kristian says:

David, you say:

“To make the discussion much easier, let’s introduce the shell instantaneously after it has been warmed up to a temperature of 254K. This means that, at the instant it is introduced, it will already be radiating at the rate of 235Wm-2 from BOTH its surfaces (simple application of the SB law). So the shell will now be emitting 235Wm-2 towards the core and 235Wm-2 towards space.”

No! You simply don’t get it, do you?

This shell (254K) will of course emit 235 W/m^2 to its surroundings (space). But that’s it. That’s its black body emission. This is one system. Consequently it has one effective surface. One boundary with space. Everything else, from this outer surface down to the core is internal to the black body shell system. The two opposite surfaces of the shell can simply not be considered equals as emitting black body surfaces. Please think about this.

The inward flux is 0. We’ve been through this before. Any hypothetical flux from the inner surface of the shell would always eventually hit an opposing surface of the exact same quality (temperature) and hence there will be no heat transfer in this direction. It doesn’t matter where the input energy to the shell is originally coming from (in this particular case it seemingly has no supply) – from the outside (a sun), the inside (core planet) or within the shell itself (‘intrastructural’ nuclear source). The absorbed heat will move directly and in its entirety to the outer surface. By necessity. The shell’s heat loss inwards upon itself will be 0. The entire heat loss (emission) goes out to its surroundings. Balance.

You even agreed to this upthread, remember? These are your numbers: Power input to shell, 470 W/m^2; inner surface output: 235-235= 0 W/m^2; outer surface output: 235+235= 470 W/m^2. So, incoming energy absorbed by black body system – 470 W/m^2; emitted energy from black body system to surroundings – 470 W/m^2; black body temperature – 302K.

Read my previous comment to you one more time. It is the incoming energy being absorbed by the black body that sets its temperature. And from this temperature it emits a corresponding flux to its surroundings. If the incoming flux is 470 W/m^2, the black body temperature will be 302K. Done deal. If the incoming flux is 235 W/m^2, then the black body’s temperature will be 254K.

You are simply needlessly inventing the inward 235 W/m^2 flux for the shell, David. The heat flux received and absorbed by the shell from the core planet in your ‘IPCC model’ diagram is … 470 W/m^2. Look at it. There is no way around it. Consequently, the shell needs to emit an equal flux … to its surroundings, as heat loss. To balance this. It has no business ejecting absorbed heat into its own interior.

470 IN from core, 470 OUT to space. Temperature: 302K. NOT 254K. You simply asserting that you need to split the incoming flux in two so that the corresponding emission temperature measured/detected from the outside of the shell will be 254K (from 235 rather than 235+235= 470 W/m^2, which is the actual input) is nothing short of bizarre. I cannot fathom why you keep on defending such a confoundingly illogical and obviously untenable position?

231. Roger Clague says:

David Socrates says:
March 19, 2013 at 10:52 pm

The shell emits 235Wm-2 from TWO SURFACES.
The power intensity out of the shell is 235Wm-2

I cannot agree. The shell is in a steady state. The shell is gaining 470Wm-2 so it must be losing 470Wm-2.

It is losing a total of 235Wm-2 to planet + 235Wm-2 to space = 470Wm-2

The model planet and shell are the same temperature 302K.

This model does not predict thermal enhancement (the greenhouse effect ).

232. gbaikie says:

“David Socrates says:
March 19, 2013 at 4:20 pm

Gabriel van den Bergh (GabrielHBay) says, March 19, 2013 at 11:47 am: Were in our thought experiment the shell to be a perfect conductor and still a black body, what then?

Gabriel,

You have mis-thought!! A black body IS a perfect conductor by definition. It is also a perfect absorber and a perfect emitter of radiation of all wavelengths. ”

I agree an ideal blackbody is this.

“In the original article, Willis Eschenbach explicitly says that his core and shell are black bodies. Therefore it is absolutely pointless worrying about real materials such a real steel that fall somewhat short of the ideal.”

But I as I recall Willis Eschenbach did not explicitly say it was an ideal blackbody and of course he called it a steel greenhouse, rather than a blackbody greenhouse.
Therefore it is reasonable to assume he meant steel coated with a material
which acted similar an ideal blackbody. Paint the steel flat black.
One do more just paint it, one also have some kind of texture to surface- or basically left it to engineers to design the surface that would absorb the all the incident radiant energy from the warmed core surface.
It should be noted that one can design a surface to have more surface area, so having the outer surface of steel greenhouse have twice the surface area as smooth surface is a possible option. Eg, have the surface have spikes or golf ball dimples.
For an engineer, the question is, is would the added cost be required to do something adequately- what is cheapest way to do something.

“The point of such imaginary models, as I am sure you can see, is to remove as much of the irrelevant clutter of variables so that the true didactic (i.e. educational) point can be explored in isolation. And the true point in this case surrounds the following divergence of opinions:

(1) There is NO back-radiation from a cooler body towards a warmer body. So in the case of the Willis core-shell model, core and shell will attain the same temperature (eventually, when at steady state).”

The universe is full of stuff radiating energy- cosmic background source is infinite heat/temperature, but it’s diffused throughout the universe. In terms of heating anything, it only matters if want something colder than 2 K or something is this cold.

So it’s not a issue of it’s existence, it’s a issue of whether it cause something to become warmer- or can matter use this energy to increase it’s kinetic energy- can matter having certain level of energy absorb this energy.
So it’s like a cold object and warmer air. The gas has kinetic energy, the question is not existence
of the object or the gas, the question is can warm air increase it’s energy when it hits the cooler object. So a gas is said to have a average velocity- and a molecule can have zero velocity, so one gas molecule could get velocity added from anything above absolute freezing. But we concerned about the average velocity of the gas- so a colder object does NOT warm the warmer gas.

“(2) There IS back radiation from a cooler towards a warmer body but it is deflected straight back from the surface of the warmer body towards the cooler body where it is re-absorbed and does not contribute additional energy to the warmer object. So in the case of the Willis core-shell model, core and shell will attain the same temperature (eventually, when at steady state). ”

Energy is conserved, so it is deflected or not absorbed- meaning converted into kinetic energy [heat] is correct.

The sun is shining and 1000 watts per square meter is hitting some surface, even if surface is black, one should not assume it’s absorbing 1000 watts per second. With an ideal blackbody we do make this assumption, UNLESS this blackbody isn’t already hot- though we seem to sort of assume a ideal blackbody doesn’t have a temperature [doesn’t hold energy- doesn’t get warm- rather it’s conducting all the heat away to be radiated and it’s the total surface area which is radiated is the limiting factor].

So sunlight at Earth distance can only heat something to a certain temperature, which loosely
speaking we can say it has a temperature at earth distance. This temperature is ideal blackbody surface which heats something and all the heat radiated only from one surface. Meaning flat panel one side insulated and side facing the sun is not insulated. Or the only thing stopping the sunlight from continuing to warm surface is the temperature can be increased further by the energy of sunlight [and sunlight [solar flux] is NOT magnified].
An alternative, is take all the energy emitted by the sun, total joules per second, and have a shell the size of Earth orbit emitting this energy outward, giving this shell a temperature.

So sunlight at Earth distance can said to have temperature. What similarly is the temperature of this back radiation?

(3) There IS back radiation from a cooler towards a warmer body, and it IS absorbed by the surface of the warmer body. So in the case of the Willis core-shell model, the core will therefore be at a higher temperature than the shell.”

With Willis core-shell model the interior is warmer than surface. An inch below the surface is warmer than 1/2″ below the surface. If you slow down the heat radiating from the surface, you prevent the 1/2″ below surface from losing heat, with nuclear reaction having no effective limit in temperature, the core would gain heat- become hotter.

To best way inhibit the radiation in vacuum from leaving the surface is to insulate it by reflecting it back, so you have the inner part of shell reflective to wavelengths of blackbody spectrum of 235 W/m2.
It also seems one doesn’t increase such insulation by making the reflective surface further away.
So having reflective surface only one foot away from surface should be improvement rather a any kind of reduction in terms of insulation ability.

So suppose one one ask how much improvement does one get from using a surface which reflective rather than surface which absorbs the radiation?

233. Kristian says:

David Socrates,

You were going to ask Postma about your righthand model diagram, if it’s a correct rendition of his view on this matter. I know you’ve asked him. And I (and I assume you also) know he’s given you his answer.

Why aren’t you presenting it here …?

234. Tim Folkerts says:

Things seem to be winding down .. so a few last comments.

A physics law for thought….”Only two things can effect the energy content of any system in a radiative balance. Either a change in the input, or a change in the “residence time” of some aspect of those energies within the system.”
–David

Why not just go with “change the input or change the output”? If people want to introduce a non-standard term like “residence time”, they better have a definition. How would you calculate the “residence time” for the energy in the shell or the energy between the planet and the shell?

******************************************

Tim Folkerts, I agree, warmer surroundings transfer heat to a cooler object, but like I said : “The term “net” doesn’t imply that T_c is heating T.”, where T_c is less than T; because a cooler object won’t heat a warmer one without help.
–bwdave

Perhaps that last clause is the key to many of the misunderstandings people have. In this scenario (and for the earth) there is a heater (nuclear or solar). The temperature that the planet reaches depends on the “help” that the heater receives. There could be “help” from insulation or “help” from ‘backradiation’. Neither of these “heats” the planet (in the technical sense of net flow of energy), but both of them “warm” the planet (in the colloquial sense of “higher temperature than it would have been otherwise”).

****************************************************

Could you give a practical method of dumping heat to space in the given circumstances?
Electromagnetic radiation emitted must represent the movement of the molecules of the steel shell.
In the classic problem I posed the higher and lower temperatures represent the movement of molecules of steel exactly like the demon example.
–Bryan

I suppose “practical” is a matter of perspective. You can “dump heat” from a container of water to space with a parabolic reflector, freezing the water on a night when the temperature is well above freezing. The whole shell is dumping heat to space via radiation.

Maxwell’s demon takes one set of things at one temperature and sorts them into “hot” and “cold”. There is no sorting. The whole set of atoms in the shell are warmed and cooled as a group.

235. Kristian says, March 20, 2013 at 4:40 pm: You are simply needlessly inventing the inward 235 W/m^2 flux for the shell, David. The heat flux received and absorbed by the shell from the core planet in your ‘IPCC model’ diagram is … 470 W/m^2. Look at it. There is no way around it. Consequently, the shell needs to emit an equal flux … to its surroundings, as heat loss. To balance this. It has no business ejecting absorbed heat into its own interior. [my bold]

OK we are narrowing down the area of our disagreement.

You seem to be implying that the black body shell “knows” which of its two surfaces is facing “out to space” and which is facing “towards the interior” and so it is therefore in the comfortable position of making sure it radiates all of its received energy flow to space.

So how does it “know” which direction is “towards space” and which direction is “towards the interior”?

My assumption is that it cannot possibly “know” which surface is which. To a dumb creature like a black body they are just surfaces. So my conclusion is that it must radiate from both surfaces equally because it just darned well doesn’t know which is which. Then, in the case of the flow “towards the interior”, the energy flow is either (i) absorbed by the core (IPCC Back Radiation Model); or (ii) deflected at the core surface without absorption and returned back to the shell (Claes Johnson/SkyDragon Model).

But whether one subscribes to the IPCC or Claes Johnson model, I am convinced that the shell’s inner surface sure as heck must also radiate downwards.

Take another look at my two diagrams and just make sure that you are not a Dragon Slayer after all. 🙂

236. Bryan says:

Tim Folkerts sys

“Maxwell’s demon takes one set of things at one temperature and sorts them into “hot” and “cold”. There is no sorting. The whole set of atoms in the shell are warmed and cooled as a group.”

The Willis model starts off at 254K and reaches 302K at the surface.
The expanding model has temperatures between 254K and 302 K at the shell
The arrangement is automatic and is as spontaneous as the shell model
We can store the energy at two extremes of temperature.
Remember this is a thought experiment and we can do this with ‘perfect’ insulation
This will enable us to do mechanical work between these two temperatures.
According to the second law we should not be able to do that.

You have avoided giving an example of doing work from the shell at a particular temperature.

Could you give a practical method of dumping heat to space in the GIVEN CIRCUMSTANCES?
This is to see if its possible to do mechanical work by that method.

Your parabolic reflector would be at the same temperature as the shell.
There is a further slight problem in that the radiation leaving the parabolic reflector will be less diffuse than otherwise.
Negative entropy change indicated.

Electromagnetic radiation emitted must represent the movement of the molecules of the steel shell.
In the classic problem I posed the higher and lower temperatures represent the movement of molecules of steel exactly like the demon example.

237. Roger Clague says, March 20, 2013 at 4:48 pm: David Socrates says, March 19, 2013 at 10:52 pm: “The shell emits 235Wm-2 from TWO SURFACES. The power intensity out of the shell is 235Wm-2.” I cannot agree. The shell is in a steady state. The shell is gaining 470Wm-2 so it must be losing 470Wm-2. It is losing a total of 235Wm-2 to planet + 235Wm-2 to space = 470Wm-2. The model planet and shell are the same temperature 302K.

Roger, take another look at my two diagrams. The left hand “IPCC Back Radiation Model” diagram is the one we are discussing.

When you say 235Wm-2 to planet + 235Wm-2 to space = 470Wm-2, you are simply confusing your ‘watts per square metre’ with your ‘watts’ (easily done!). But as you will clearly see from the diagram:

(a) 470Wm-2 is absorbed by ONE SURFACE.
(b) The 235Wm-2 balancing outward radiation is from TWO SURFACES.

If you still don’t get it, let me elaborate:

Suppose that the surfaces are each a billion square metres.

Then the power flowing from core to shell is 470,000,000,000 W.

And the power flowing from shell to space is 235,000,000,000 W.

And the power flowing from shell to core is 235,000,000,000 W.

And last time I looked 235,000,000,000 W + 235,000,000,000 W = 470,000,000,000 W.

A perfect energy flow balance!

So the power intensity is 470,000,000,000 W/ (2,000,000,000 m2) = 235Wm-2.

And this means that both surfaces are at a temperature of 254K (straightforward application of the S-B law).

Are you sure you are not a SkyDragon (right hand diagram) after all? 🙂

238. donald penman says:

Replace the glass in light bulb with metal and then measure the temperature of the metal shell it should be possible to determine if the shell radiates more from either side or if it radiates the same both sides.
http://home.howstuffworks.com/light-bulb.htm

239. Kristian says, March 20, 2013 at 4:58 pm: David Socrates, You were going to ask Postma about your right hand model diagram, if it’s a correct rendition of his view on this matter. I know you’ve asked him. And I (and I assume you also) know he’s given you his answer. Why aren’t you presenting it here …?

Because there was nothing of significance in his initial reply that we did not already know about the SkyDragon postion. I have responded again to him, asking him explicitly to conform that my right hand diagram + accompanying explanation is acceptable to him as a good representation of his position – or, alternatively, to suggest modifications. Unfortunately my response is now stuck in moderation because Joe is away on business in India.

I suggest you keep a lookout on the Postma site over the next few days if you want to keep up-to-date with developments.

240. Arfur Bryant says:

Tim Folkerts says:
March 19, 2013 at 10:34 pm

Tim,

Thank you for clarifying your earlier remarks. I agree with your last sentence.

But this means that Willis’ model cannot possibly be correct.

Why?

Because in the role-reversed model, with the heated shell and cool planet, in the time to reach equilibrium, no radiation from the cool planet could be absorbed by the warm shell because it would then warm above 254K. And yet that is what we are expected to believe with the Willis model. In Willis’ model, the 470W/m^2 arrow is therefore impossible (as I said right at the start of my comments) because the warm planet could not absorb (for net gain) the radiation from the cool shell.

Maybe I am being really dull but I have to repeat that the key to this model is not emission but absorption. How is it possible that low energy radiation can be absorbed (for net gain) by a higher energy surface?

If you have a link to such a possibility, please show me. I would be very grateful.

The downward 235W/m^2 arrow, IMO, does not get absorbed (for net gain). Therefore there is no 470 arrow.

I am not being obtuse. I really want to know how I can research absorption from a lower energy source. This has significant consequences in the climate debate and, so far, only Bryan has answered me. I agree with him but my objectivity demands a second opinion.

Regards,

241. wayne says:

David Socrates, here is how I see it, I disagree with both examples in your diagram:
(and I think you do too)

But there is a third way.

242. gbaikie says:

“Suppose that the surfaces are each a billion square metres.

Then the power flowing from core to shell is 470,000,000,000 W.

And the power flowing from shell to space is 235,000,000,000 W.

And the power flowing from shell to core is 235,000,000,000 W.

And last time I looked 235,000,000,000 W + 235,000,000,000 W = 470,000,000,000 W.

A perfect energy flow balance!”

This: “Then the power flowing from core to shell is 470,000,000,000 W.”
Should be: “Then the power flowing from the surface of core to shell is 470,000,000,000 W”

And you have the surface core warmer than would be just from nuclear decay,
SO, IF surface of core is warmer, it should be warming beneath the surface.

So you missing an arrow of the core surface heating the interior.

You missing it, if are assume the shell is increase the core surface temperature [which you do].

243. gbaikie says:

“wayne says:
March 20, 2013 at 8:18 pm

David Socrates, here is how I see it, I disagree with both examples in your diagram:
(and I think you do too)

But there is a third way.”

And energy density would mean nothing within the shell could cool below a certain temperature.

So without the shell, in vacuum with planetary core radiating 235 W/m-2, 10 km above the surface something could cool.
Say you have a pillar 1 km high made of stone. And on top of pillar you have a rock- the rock could
be quite cold. With the shell and your energy density you have a “room temperature” within the shell- so that rock on pillar could not be cooler than this.
Or a vacuum can have a temperature. In “normal” space vacuum you can passively cool to near 2 K- but this not available within the shell world [unless one is using active refrigeration {using energy}].
So one call it balance heat- or thermal equilibrium.

[And If put a column of air which it’s pressure was solely due to gravity [not pressurized so container with open top] one have lapse rate in temperature, the denser air near surface would have
higher temperature.]

244. Max™ says:

David, I imagine I sound like a broken record here…

Suppose that the surfaces are each a billion square metres.

Ok, then the radius is 8920.620585 m.

Then the power flowing from core to shell is 470,000,000,000 W.

Ok.

And the power flowing from shell to space is 235,000,000,000 W.

Hmmm, if the shell was 1 meter above the core, it would have an area of 1,000,224,213 m^2, and the power reaching it from the core would be 469.89 W/m^2 if the shell was at 0 K.

And the power flowing from shell to core is 235,000,000,000 W.

Well, assuming the inner surface was emitting half the total amount inwards and outwards, then at most it could emit 234,997,677,723 W back towards the core, reducing the net power reaching the shell to 234.89 W/m^2., so I would accordingly expect no more than that amount to be emitted to space.

And last time I looked 235,000,000,000 W + 235,000,000,000 W = 470,000,000,000 W.

A perfect energy flow balance!

So the power intensity is 470,000,000,000 W/ (2,000,000,000 m2) = 235Wm-2.

And this means that both surfaces are at a temperature of 254K (straightforward application of the S-B law).

Well, no, that’s not a straightforward application, a straightforward application would be to say “the shell surfaces must be at or above a temperature which would result in emission of 234.99 W/m^2 in and 234.89 W/m^2 out”, you can’t even begin to calculate a temperature from the SB law without knowing the emissivity, area, and ambient temperature first.

Strictly speaking you also need to know the geometry of the emitting surface.

_______________

Now, all of that can be ignored because none of it matters as much as this one point which I really wish to get across: energy must be conserved, not power.

Power tells you how rapidly energy arrives, but the total amount is what must be conserved, if this were not true then everything here would glow at the same temperature as the Sun. As it stands, both the geometrical decrease of the inverse-square law and the different energy densities in the absorbing surface guarantee that though the planet receives up to 1000 W/m^2 at times, it only has to emit at a quarter that power to balance the input.

The Earth receives 1.22×10^17 Watts on the day side but only has to emit 5.11×10^16 Watts from the day side to balance that input, as it can also emit 5.11×10^16 Watts from the night side.

Similarly the shell does not have to (and in fact can not) emit the same W/m^2 as the core emits, only the total energy is conserved, attempting to conserve power necessarily violates conservation of energy unless exact geometrical symmetry is preserved.

245. Roger Clague says:

David Socrates says

Suppose that the surfaces are each a billion square metres.
Then the power flowing from core to shell is 470,000,000,000 W.
And the power flowing from shell to space is 235,000,000,000 W.
And the power flowing from shell to core is 235,000,000,000 W.
And last time I looked 235,000,000,000 W + 235,000,000,000 W = 470,000,000,000 W.
A perfect energy flow balance!
So the power intensity is 470,000,000,000 W/ (2,000,000,000 m2) = 235Wm-2.

Applying the same proceedure ( Kirchoff’s Radiation Law ) at planet surface A m2

power flowing to surface from core = 235 A W
power flowing to surface from shell = 235 A W
total power to surface 470 A W
total area of surface 2 A m2 ( inside and out, same as shell )
power intensity =470 A W/2 A m2
= 235Wm-2 from surface
temperature of surface by S-B = 254K

246. Tim Folkerts says:

But Roger C, there is no “inner surface” that is radiating of the inner planet. And even if the planet were hollow, the inner surface is would be radiating to other parts of the inner surfaces at the same temperature. No eat flow; not temperture gradient.

I know this is subtle, but the heat flows OUT from the heaters toward space. The only gradient will be OUTSIDE the location of the heater. The same applies to insulation. Suppose you have a box with many layers of insulation. If you put uniform heaters on the outside, the entire interior will be the temperature of the outer walls. If you move the heaters in one layer, then the exterior will be the same temperature as before, there will be a gradient to to a higher temperature to where the heaters are now, and anything inside that location will be the same higher temperature.

247. suricat says:

gbaikie says: March 20, 2013 at 9:14 pm

“Suppose that the surfaces are each a billion square metres.

Then the power flowing from core to shell is 470,000,000,000 W.

And the power flowing from shell to space is 235,000,000,000 W.

And the power flowing from shell to core is 235,000,000,000 W.

And last time I looked 235,000,000,000 W + 235,000,000,000 W = 470,000,000,000 W.”

TBH, this is a geometric impossibility. The ‘shell’ must have a greater surface area than the ‘core’, thus, the ‘shell’s’ energy density is LESS than that of the ‘core’. If the core has an energy density of ‘ed’W/m^2, the shell must have an energy density of ‘ed x (core area / shell area)’W/m^2.

It’s the same for the back radiation. The ‘origin’ of the back radiation from the ‘shell’ provides LESS than the power density of the ‘core’ on emission from the shell, but when the back radiation ‘arrives’ at the core its intensity is equal to the ‘core’ intensity by way of the intervening geometry.

‘Core’ and ‘shell’ can’t possess the SAME ‘radiation density’ (W/m^2), as shown in the diagram, without refrigerating the ‘atomic power source’ within the core!

I hope that makes sense. 🙂

Best regards, Ray.

248. wayne says:

gbaikie, that’s a good way to view it. Without the shell, radiation decreases at the inverse squared and an object further from the lone sphere would therefore be ever cooler at equilibrium, being in space. And right, everything in a room (like in an afternoon, the temperature is constant) or the gap are at the same temperature. That being how you both have the 235 energy density everywhere while there is still also a 235 W/m² flowing ‘through’ that same space, the sphere surface has a bit higher temperature.

But yet that may be off a bit, I’m really not sure, need to read more. It seems there would be a linear gradient this time of that energy density in the gap since the sphere is warmer than the shell but again looking at it another way I can see it being exactly even. Hmm. Would an object, say a clump of coal, have a different temperature depending on where it is located, near the sphere or near the shell? I’ve never delved in ‘radiation energy density’ that far to say I know. In fact it does make sense that it would take on the same density as the warmer surface dictates, not the density that the cooler dictates, that higher state being always ‘transfered’, the other 235 W/m², that passes through tthe shell to space.

There are many very picky details hidden within this example. 🙂

249. suricat says:

wayne says: March 21, 2013 at 12:58 am

“There are many very picky details hidden within this example.”

I concur. If a ‘fixed’ value of energy transport is transported through a ‘fixed’ area, we tend to relate to this as a ‘W/m^2’ value.

This reminds me of some of the Feynman lectures and the question of ‘why’ when he’d reply with something like, ‘I told you ‘how’! Now you tell me ‘why”. Questions, questions, questions.

I’m ‘all ears’ wayne (these musings are rhetorical). 🙂

Best regards, Ray.

250. Max™ says:

The inverse-square law means energy density is lower as you move further from the core, it can’t be the same everywhere

251. Max, Suricat, Wayne, all…

Groan…I now put you all on notice that I am talking about a thought experiment in which all the surfaces are vanishingly close to being the same – not because they need to be but simply for simplicity of explanation of the energy flows involved.

Of course that is not true of real physial concentric shells, but you are all grown up enough to know that the purpose of a simplified model is to direct focus on to one physical principle at a time.

If you don’t like the concept of the surfaces actually touching (because if they did, yes, the physics would indeed change) then please consider that I am talking about a limiting case where they are an ‘indefinitely’ small distance apart. Ever heard of calculus? ‘dz–>0’, right?

If you still object to this approach then you will have to demonstrate clearly why, when the surfaces are a small finite distance apart, that real gap entirely demolishes the simple conceptual points about energy flows that I and others are attempting to make here.

Otherwise you are indulging in pure sophistry (finding an irrelevant or insignificant objection to the proposition so as to cast unreasonable doubt upon it).

252. wayne says:

” I’m ‘all ears’ wayne (these musings are rhetorical). 🙂 ”

Yeah Ray but I’m really asking for it if I go any further. 😆

Like I just read some words that are exactly the view I have long held but it cuts cross-grain to some here, or I get that impression:

“In quantum electrodynamics, disturbances in the electromagnetic fields are called photons, and the energy of photons is quantized.”

Fields? Ah, ha! Waves! Now what happened to the tiny BBs of energy? 🙂

I think better said… In quantum electrodynamics, disturbances in the electromagnetic fields are called photons, and the energy of these disturbances in the electromagnetic fields is quantized (complete 2π wave nodes implied by E=h·v). I’ve always thought you could never have an energy transfer of, let’s say, 1 3/8 Hertz worth of energy at a given frequency, only integer multiples, and I think not multiples but a singular cycle per interaction. Tim will say here multiple absorptions can occur and I agree in particular cases, depending on the temperature in question.

But I think I could be way off base on my comment to gbaikie. This info is hard to come by, every site speaks the same script… Maxwell equations, Poynting vectors, long derivations, coaxial cable example, and maybe a plane wave example… they are all clones! Hyperphysics speaks of ‘radiation energy density’ as being specifically energy in transit per volume, moving through space much the same logic as pressure in the kinetic theory of gases, using a 3d box.

What it doesn’t say is whether to calculate this density are just ‘net’ transfers used in the math, or, do you also need to add up even the weaker waves coming in the opposite direction (“back radiation”). Assuming the ‘net’ case you could even have a zero density if all temperatures are equal. The other way you would add powers and get up to double the density if at equilibrium, and it would never be zero except at 0K. Sometimes it is easier to compute such problems taking on a much simpler but incorrect viewpoint, but that is just to make the math simpler, allowing cancellations in the equations, but don’t necessarily reflect reality.

But you know, when speaking of continuous waves with continuous transfers of energy in one direction, like waves coming ashore or waves in the the wave machine at a water park, I’ve never seen smaller opposing waves pass untouched through the bigger waves of more power, I don’t think ever. I have seen smaller opposing waves take some of the power out of the bigger waves, waves have the ability to partially or totally cancel. As if nature already knew the end result and correctly also knew that the rest of that particular combination of events did not even need to complete. Seems that is called waves and also called the least action principle.

Still looking for a more exact set of definitions. Surely it is somewhere.

253. Kristian says:

Tim,

What Roger C is showing you, is how ridiculous your notion of the actual influx to the shell having to be split in two and thereby magically somehow reducing the temperature of the shell, really is. The (one) planetary surface gains its energy (according to your version) from two ‘sides’, from the outside (the shell to planet) AND from the inside (nucleus (via perfect conduction, one should assume) to surface. But as we know, only the latter is a real ‘heat transfer flux’. Logically, there is no real difference between the planet and the shell in this regard. For just as much as there can be no ‘counter-conduction’ (in the meaning ‘negative’ heat transfer) from planet surface to nucleus, there can be no ‘counter-radiation’ from shell to planet. The shell is ONE black body system. It has ONE effective black body surface, its boundary with its surroundings, its OUTER surface. It absorbs a specific power density flux, gains a corresponding emission temperature (235 W/m^2 – 254K; 470 W/m^2 – 302K) and emits an equal power density flux (as heat loss to surroundings/space) to balance.

It is really simple and basic. The radiative flux from planet to shell is transmitted just as swiftly and completely as it would be had the vacuum been filled with a solid material having perfect conduction. At steady state (dynamic equilibrium) there is a constant flux leaving the outer surface of the shell of 235 W/m^2. This has to be replaced right away. And it is, because at the same time there is an energy flux of 235 W/m^2 entering the same system from the planetary nucleus. This ‘vulnerable’ balance is kept in place by the radiative field between the planet and the shell. There is no apparent heat transfer, but energy is still constantly flowing from core through shell to space to maintain the balance – hence, ‘dynamic’.

If you bring conductive resistance to energy flow into this, or posit thermal mass (heat capacity) as a factor, then Willis’ model ‘proves’ the conductive/convective ‘GHE’ (refer to D. Socrates’ recent threads on the subject) – the REAL atmospheric effect, NOT the radiative (AGW) GHE. The radiative GHE does not exist. It is purely a theoretical construct.

254. Kristian says, March 21, 2013 at 11:35 am: Tim, What Roger C is showing you, is how ridiculous your notion of the actual influx to the shell having to be split in two and thereby magically somehow reducing the temperature of the shell, really is. The (one) planetary surface gains its energy (according to your version) from two ‘sides’, from the outside (the shell to planet) AND from the inside (nucleus (via perfect conduction, one should assume) to surface. But as we know, only the latter is a real ‘heat transfer flux’.

Kristian all you are demonstrating is that you are a SkyDragon whereas Tim is a warmist. whereas, for some strange reason, you think you are in a third category of not beleiving in back radiation at all. Take a look at my revised diagrams that I have today proposed to Joe Postma.

255. Max™ says:

If you still object to this approach then you will have to demonstrate clearly why, when the surfaces are a small finite distance apart, that real gap entirely demolishes the simple conceptual points about energy flows that I and others are attempting to make here.” ~David Socrates

Ok, in the thought experiment you have a power source which is capable of raising the temperature of a sphere until the entire sphere emits radiation at a power of 235 W/m^2, right?

If that is the case, then adding a layer in contact with the sphere will increase the surface area, right?

Whatever the internal power source supplies is sufficient to raise the temperature of the body until it emits 235 W/m^2 times the surface area.

If the area over which emissions occur is increased, the power of said emissions must decrease.

Discussions of infinitestimally thin layers are irrelevant, if you limit the discussion to that case, then nothing interesting happens.

________________

On the other hand, if you add a layer with some non-zero thickness, then you accordingly increase the area of the emitting surface, and the power emitted must go down.

The total emissions can remain the same, but the density of those emissions can not.

________________

If adding a solid layer to the sphere reduces the temperature of the emitting surface, due to necessarily reduced power, then adding a shell and gap with the same outer surface area as the solid layer will also reduce the density of the emissions, and thus it will have a lower temperature due to purely geometrical considerations before anything else is accounted for.

____________

Re: your diagrams, the skydragon side would have a cooler shell than the interior layer, it wouldn’t and indeed couldn’t be 254 and 254 unless both surfaces had a power supply and were emitting towards the other surface.

Could you explain how you would get from 254 planet and 254 shell to any other state?

256. Tim Folkerts says:

Let me leave everyone with a few final thoughts, as this thread has run its course, I think. No one is changing their minds anymore; no one is presenting anything really new.

**********************************************************************

1) Any thought experiment is an idealization. It is rarely worth arguing about details like exactly what the heat capacity of the planet is, or what the thermal conductivity of the shell is, or exactly how high the shell is above the planet. For the most part, these can be made “small enough” that the “kernel of truth” is still there, providing an answer that is only wrong by some small amount, δ.
* I can imagine the shell only 1 cm (or less) thick
* I can imagine the shell only 1 m (or less) from the planet
* I can imagine the shell has an emissivity of 0.99 (or better)
None of these will change the general answer more than a few percent, and hence they are not worth arguing about UNTIL the original idealized case has been worked out.

The only exception is when you can argue that there is a logical inconsistency in the set-up. We came close with “the shell is a perfect thermal conductor and a perfect blackbody” since no single material can have both properties simultaneously. But a combination of two materials (eg thin shell of copper painted black) can be “close enough”.

*****************************************************************************

2) The one simplest argument I see against the “same temperature” conclusion is that we would have 235 W/m^2 of heat going from one surface at 254 K to a different surface at 254 K. The materials are the same temperature and have all the other same properties, yet one is emitting 235 W/m^2 and one is absorbing 235 W/m^2.

That simply makes no sense to me — and it contradict the Zeroth Law of Thermodynamics and the idea of “same temperature”. And no one has provided any convincing argument why the heat behaves that way (other than that they had already assumed the two were the same temperature, and so the heat must flow that direction with that magnitude.)

*****************************************************************************

3) This whole discussion seems a lot like a fellow who watches a lot of shows about lawyers on TV and then thinks he is ready to represent himself in court.

A bunch of enthusiastic amateurs makes for a great discussion, but very few people in this discussion have derived Planck’s Law of radiation for a blackbody, or calculated the change in entropy for a process like this, or experimented with radiative cooling. Science is hard, and it takes a lot of time to develop skills and intuition.

Most of those arguing against the planet being warmer than the shell are either
1) nit-picking the details (see Point 1 above).
2) claiming the answer doesn’t “seem right” to them.

*************************************************************************

4) There have been two possible conclusions presented:
I) the planet will be ~ 2^(0.25) times warmer than the shell.
II) the planet will be ~ same temperature as the shell.

People on both sides have given what they think are good reasons to support their positions. Then we go to the webpages and textbooks, and we find that (other than a few exceptions like Joe Postma) they all conclude that the planet will be warmer than the shell.

It is an “exceptional claim” that generations of physicists and engineers and climate scientists got this wrong (while getting the bulk of thermodynamics right). It would take “exceptional evidence” to overturn their conclusions.

I ask you to honestly consider — do you have a better understanding of thermodynamics than the professors who teach the courses and write the textbooks? Professors in climate science, engineering, and physics?

*************************************************************************

The planet WILL be warmer than the shell by a factor of 2^(0.25) ±δ 🙂

257. Tim Folkerts says:

Max says: ” Ok, in the thought experiment you have a power source which is capable of raising the temperature of a sphere until the entire sphere emits radiation at a power of 235 W/m^2, right?”

No! The power source provides 235 W/m^2 IN ADDITION TO any other power supplied to the planet by any other means. The planet will (once a steady-state condition is reached) emit 235 W/m^2 + (any other power received by other means). The temperature in this case will rise until the energy emitted = energy absorbed.

The planet get 235 W/m^2 from the heaters, and 235 W/m^2 from the shell.
The planet emits 470 W/m^2 to the shell.
NET = 0 W/m^2

The shell gets 470 W/m^2 from the planet and 0 W/m^2 from space
The shell emits 235 W/m^2 to the planet and 235 W/m^2 to space.
NET = 0 W/m^2

* Energy is conserved.
* Heat always moves from warmer to cooler.

People talk about simple addition — well, it doesn’t get much simpler than this!

258. Max™ says, March 21, 2013 at 1:30 pm

Max,

1. If you don’t appreciate the virtue of simple plane-parallel representations of energy flows then there is nothing that I or anybody else can do to help you. Of course with real spheres and real shells there are finite gaps and finite thicknesses involved. And of course, even if these gaps and thicknesses are tiny, the Wm-2 figures will indeed go down by tiny fraction of the plan-parallel results as the energy passes through them. But it is pointless to argue any more about this if you reject the necessary approximations that go along with what I consider to be the major didactic advantages of the plane-parallel approach. We will just have to agree to differ.

2. Re. the SkyDragon diagram, using my simplified plane-parallel model, both shell and core are at the same temperature of 254K simply because both shell and core are emitting net radiation of 235Wm-2. (S-B law for an ideal black body).

3. You say: Could you explain how you would get from 254 planet and 254 shell to any other state? Er…for example I could get from a 254K planet and a 254K shell to a 354K planet and a 354K shell by increasing the power source from 235Wm-2 to 890Wm-2. (S-B law for an ideal black body). But surely you didn’t need me to tell you that? 🙂

259. Roger Clague says:

The steel greenhouse model does not apply procedures ( laws ) consistently.

For the shell out put is divided into 2 parts, up and down.

For planet output is not divided into 2 parts, up and down.

Why does the planet surface not send 1/2 the output down
into the planet? You may reply that it a 1 surface body not a 2 surface body. How does it know that? It does not. It will also radiate into the planet.

If we apply the same procedure to the planet, the temperature is a result of 235Wm-2, up and own, not 470W-2 .That is 254K.

The temperature of the planet remains the same. It is not heated by the shell.

260. mkelly says:

Tim Folkerts says:

March 21, 2013 at 1:54 pm
Most of those arguing against the planet being warmer than the shell are either…”

This is a large straw man. Nobody, at least not I, was arguing that the planet was not warmer than the shell. The planet most be warmer than the shell as that is where the thermal energy is originated.

The shell cannot supply heat (thermal energy) to the planet as it is at a distance from the planet and the r squared rule applies which makes it cooler.

If the planet temperature goes to 302 K then the shell must go to something less and that shows that the 235 W/m^2 cannot be true. I showed this to you with your own ratio using your own numbers.

By appling standard radiative heat transfer equations the steel shell problem as presented is shown to be wrong. Your declaration not with standing.

261. Tim Folkerts says:

Roger C — the “law” being applied is conservations of energy:

“The change in energy within a region = (energy added to the region) – (energy that leaves the region)”

In this case, think of the “regions” as concentric shells. And once a steady-state condition is achieved, then the change within any region is zero.

For regions INSIDE the heaters (ie shells in the interior of the planet), no heat flows into the regions from the inside, and none flows out from the outside. The temperature will be the same at the inside of the shell as the outside of the shell.

For regions BETWEEN the heaters and the planet’s surface (if the heaters are not right at the surface), 235 W/m^2 flow into the inside of the shell and 235 W/m^2 flows out the outside of the shell, With a flow of heat through a material, there will be a thermal gradient given by
dQ/dt = k A dT/dx.
So the temperature inside the planet will rise — but ONLY until you get to the layer where the heaters are. The remainder of the interior farther in will be a constant temperature.

For the regions containing the SURFACE of the planet, 235 W/m^2 is conducted in from below, and 235 is radiated in from above, and 470 is radiated up from the surface (there IS only one surface — facing outward).

For regions IN THE VACUUM, we have 470 in from the bottom and 235 in from the top; 470 out from the top and 235 out from the bottom. The net is still zero.

You get the picture .. every shell has zero change inside, because there is always as much heading in as heading out.

THAT is the rule in play here.

262. wayne says:

David, I’m not sure the right side of your revised diagrams is what is being said by the SkyDragons, really have no idea, haven’t followed his thread but a quick skim and I did notice he never seemed to say what the temperatures were at all. Left me blank as to his point. However that is definitely not what I am saying, I did forget to put in the temperatures in my third option so here’s my revised diagram as I see it, third option, this is the way the numeric integration settled:

One note: if the sphere and the shell *are* at the same temperature it is positively impossible for ANY energy to flow from the sphere to the shell, vice-versa, ever. It takes a temperature differential. Is this what you were trying to highlighting above as we were “put on notice”?

BTW: the gap in my case is very small, on the scale of meters or so. Seems you mentioned that being a problem if too large to others, I am ignoring the radius factor.

263. Kristian says:

Tim, lgl, David, Tallbloke and others,

Like Willis Escenbach and the IPCC, it appears you’ve all been bamboozled by the false supposition that the planet/shell case is somehow equivalent to lgl’s paper-thin heaters being 1 x 1 m on both sides. The quite fundamental difference, though, is that the former is a power DENSITY issue, the latter merely a power issue. If the two-sided heater receives a wattage of 470, this will produce a power density on EACH side of 470/2= 235 W/m^2, and the hypothetical black body emission temperature of the heater would be 254K. The larger the total surface area of the heater, the cooler it would get and the smaller the emitted power density flux. The thing to remember here, though, is that both sides emit their flux to the heater’s SURROUNDINGS, i.e. they carry energy/heat AWAY FROM THE SYSTEM. They are thermodynamically equal.

This is NOT (!!) the case, however, with the black body shell in Willis’ model. It receives a power density flux of 235 W/m^2 (or 470, depending on the scenario), and that’s it. Being a black body, it will HAVE TO emit an equal power density flux from its surface to stay in balance. In other words, the only reasonable ‘solution’ to this case, is the previously settled upon (with David Socrates) 235-235= 0 W/m^2 from the inner surface, 235+235= 470 W/m^2 from the outer.

Why is this? Shouldn’t a black body emit an equal power density flux from its ENTIRE surface? Yes. But the inner surface is NOT part of the surface of this particular black body system. It is internal. The shell’s only boundary surface is its outer one facing space (the surroundings of the black body).

Realise this, and everything will all of a sudden look very logical and physically right.

264. gbaikie says:

“Kristian all you are demonstrating is that you are a SkyDragon whereas Tim is a warmist. whereas, for some strange reason, you think you are in a third category of not believing in back radiation at all. Take a look at my revised diagrams that I have today proposed to Joe Postma.”

What a SkyDragon or warmist model be if instead of the shell absorbing IR, it reflected IR?

In terms of third category, I don’t think gases act like solids.

But suppose one stopped with the heated core, and used sunlight as source of energy. So the 235 W/m-2 entering from the top. Are the diagrams the same. Meaning for warmist the outer surface of
shell radiates 235 W/m-2, it’s inner part of shell radiates 470 W/m-2. And for SkyDragon it’s 235 W/m-2 on all surfaces?

265. Tim Folkerts says:

Kristian says: “Tim, lgl, David, Tallbloke and others,
Like Willis Escenbach and the IPCC …

AND every textbook on climate and thermal physics .. we have a solution agrees with all the laws of physics (which have been tested innumerable times).

Either
* All of the physics, engineering, and climate professors and textbooks are wrong about how thermodynamics works
* Kristian, Max, and Bryan (and Joe Postma) are wrong.

There certainly have been times when new science has been discovered, but that rarely happens on blogs by those with an undergraduate education dealing with well-studied fundamental postulates that form the backbone of innumerable practical applications.

Can you not at least consider the possibility that we (those on the “planet gets warmer” side) really have studied thermodynamics more than you and that you might be mis-applying the laws of thermo in the ways we suggest? Perhaps you could go to a local university and have a prof there give an opinion on this discussion to help you understand if you are right or if we are.

266. gbaikie says:

“2) The one simplest argument I see against the “same temperature” conclusion is that we would have 235 W/m^2 of heat going from one surface at 254 K to a different surface at 254 K. The materials are the same temperature and have all the other same properties, yet one is emitting 235 W/m^2 and one is absorbing 235 W/m^2.

That simply makes no sense to me — and it contradict the Zeroth Law of Thermodynamics and the idea of “same temperature”. And no one has provided any convincing argument why the heat behaves that way (other than that they had already assumed the two were the same temperature, and so the heat must flow that direction with that magnitude.) ”

Well, we dealing with a metal, which conduct heat fairly well.
If an outer surface radiated 235 W/m^2, it needs a input of 235 W/m^2.
If only given 234 W/m^2 it can’t radiate 235 W/m^2.
You claiming it must absorb more 235 W/m^2 in order to radiate 235 W/m^2.
The shell has heat budget which must balance [the input must balance output].

267. Roger Clague says:

Tim Folkerts says:
March 21, 2013 at 4:35 pm
the “law” being applied is conservations of energy

No, the law applied is Kirchoff’s Law of Radiation ( KLR ).
power flux in ( Wm-2 ) = power flux out ( Wm-2 )
Also Stefan- Bolzmann Law ( S-BL ) to convert power flux to temperature.

For the regions containing the SURFACE of the planet, 235 W/m^2 is conducted in from below, and 235 is radiated in from above, and 470 is radiated up from the surface (there IS only one surface — facing outward).

In from below is not the same as in from above. The top layer of the planet is getting energy from 2 directions. It has 2 faces. So it can lose energy in 2 direction, from 2 faces. Our planet seems to have a surface with one face because we are standing on it.

235Wm-2 is radiated to above and 235Wm-2 is conducted to below.
The temperature is 254K.

Wayne

The temperature of the planet in your diagram is 254K, not 302K. As explained above.

268. Tim Folkerts says:

mkelly says: ” Nobody, at least not I, was arguing that the planet was not warmer than the shell.

So the question seems to be “HOW MUCH warmer is the planet than the shell?”. The top post suggests 320K, 254K and 254K for the temperatures of the planet, the inner side of the shell and the outer side of the shell (to 3 sig figs).

For now, let’s assume the surfaces really are black bodies and only worry about the size and thermal conductivity of the shell is up for discussion.

* The temperature of the OUTER shell will be lower than 254 K by a factor that I have quoted before:
T_outer = (254 K) * (r_planet/r_outer)^(1/2).
As I have also stated, this less than 0.1% error even with the outer shell 10 km above the ground. Put the shell 10 m above the planet and the error is less than 1 ppm!
Let’s call this the “radius cooling effect”.

* The temperature of the INNER shell could be either warmer of cooler than 254 K. To conduct 235 W/m^2 of heat through the shell, there must be some temperature difference from the inside to the outside. Depending on the thickness and thermal conductivity and the extend of the “radius cooling effect”, this could leave the inner shell either above 254 K or below 254 K.
Let’s call this the “thermal conductivity warming effect”.

* The temperature of the planet could be either above or below 302 K, depending on the combined effect of the “radius cooling effect”. and the “thermal conductivity warming effect” leaves the inner shell above 254 K or below 254 K. The planet’s surface will receive 235 W/m^2 from the nuclear heater and σ(T_inner)^4. It will radiate enough to emit this same energy
Let’s call this the “radiative warming effect”

CONCLUSION
IF T_inner 254 K due to the combined “radius cooling effect”. and the “thermal conductivity warming effect”
THEN the planet will be cooler than 302 K due to the “radiative warming effect”

Note that the temperate planet can only get close to 254 K if the radius of the shell goes to infinity.

269. gbaikie says:

“wayne says:
March 21, 2013 at 6:08 pm

David, I’m not sure the right side of your revised diagrams is what is being said by the SkyDragons, really have no idea, haven’t followed his thread but a quick skim and I did notice he never seemed to say what the temperatures were at all. Left me blank as to his point. However that is definitely not what I am saying, I did forget to put in the temperatures in my third option so here’s my revised diagram as I see it, third option, this is the way the numeric integration settled:

One note: if the sphere and the shell *are* at the same temperature it is positively impossible for ANY energy to flow from the sphere to the shell, vice-versa, ever. It takes a temperature differential. Is this what you were trying to highlighting above as we were “put on notice”?”

So core surface is 302 K and shell is 254 K.
Does the shell have same temperature on inner side as it’s exterior?

“One note: if the sphere and the shell *are* at the same temperature it is positively impossible for ANY energy to flow from the sphere to the shell”

If one had a furnace the gas temperate will be same has interior wall temperature, and there certainly would be a flow of energy involved.
A significant difference of a furnace and our model is furnace has a lot energy per square meter which flow of heat can be greatly affected by materials which do not conduct heat well.
Rock stops a significant amount of heat from being conducted, whereas rock is not very good
insulation if one has lower temperatures- when less energy is conducted. Iron does inhibit the flow of heat from furnace by small amount, and if one deal cooler conditions, it inhibits it less.
Or copper would make a big difference in furnace condition as compared to iron- but for a house
the difference would be fairly insignificant.
For example a cooper roof vs steel/iron roof, wouldn’t have much difference in conducting heat, nor would matter much how thick either of them were.

So with our steel shell we could assume the inner and exterior temperature are the same, but this does not mean there can not be flow of heat, as the exterior is radiating energy, which being replaced from heated interior [has to be replaced]. And interior of shell need energy input, or it cools and can’t deliver the 235 W/m-2 to radiated into space.

Only because the shell has heat capacity, that it’s not like switch when energy is stoped being input into it. And the amount heat stored as heat capacity is not related amount energy something can received- rather it is like a battery in a circuit [takes a some time to charge and some time to discharge].

270. Kristian says:

Tim Folkerts says, March 21, 2013 at 7:11 pm:

So not even an attempt to actually counter what I write in the comment you’re replying to?

Just plainly resorting to argument by authority and numbers …

You don’t get the difference between the two-sided open air heaters and the ‘two-sided’ closed shell? Between power and power density? There is no ‘new physics’ here, Tim. Nothing revolutionary.

271. Roger Clague says:

Tim Folkerts says:
March 21, 2013 at 7:11 pm

you [ non-warmers] might be mis-applying the laws of thermo

No, you are mis-applying Kirchoff’s Law of Radiation ( KLR ). The shell losses energy from top and bottom. The planet surface only lose energy from the top
An important rule of theorizing is consistency. Apply laws in the same way unless you have a reason not to.

Why does the top layer of the planet not lose energy from top surface and bottom surface?

Applying your rules consistently does not make the planet warmer than the shell.

It is no use referring those who don’t agree with to a prof or a book

272. Tim Folkerts says:

Kristian , I have been countering what you say for days — you just are not listening.

Roger, It i not “warmers” who counter you – it is all of thermodynamics!

Every surface radiates according to its temperature — the planet (470 W/m^2 ± δ), the inside of the shell (235 W/m^2 ± δ), the outside of the shell (235 W/m^2 ± δ) . Where is the inconsistency? There is no “inside surface” of the planet to radiate. (And no, a hollow planet will not contractmy position.)

“It is no use referring those who don’t agree with you to a prof or a book
Now THERE is something we can both agree on! 😉

273. Arfur Bryant says:

Tim Folkerts says:
March 21, 2013 at 11:11 pm

[“Where is the inconsistency?”]

Tim,

I have pointed out the inconsistency!

Why will you not address my point that the planet has to absorb the downward 235 W/m^2 in order for it to radiate 470 W/m^2? Can you provide evidence to show that the radiation downward from the shell is actually absorbed by the planet’s surface?

Of course the planet will be warmer than the shell. I said this very early on. It is the only source of energy into the system! The shell cannot possibly emit 235 W/m^2 because it has a larger surface area and so the planet cannot emit 470 W/m^2 especially if there is no absorption for net gain at the planet’s surface. As the planet is warmer than the shell, Willis’s model is predicated upon the assertion that downward radiation is absorbed. For this to be true, there must be a mechanism for lower energy radiation to be absorbed for net gain by a higher energy surface.

Is this possible?

274. gbaikie says:

So let’s assume the Skydragons correct about the radient model. So one has constant 235 W/m-2
which is core surface and shell outer surface of 235 W/m-2 which is suppose to be 254 K [-19 C].

Now let’s add an atmosphere which has no greenhouse gases and we make it a small atmosphere
which is near a vacuum before reaching elevation level of the shell.
So that means air near surface air could warm to around 254 K [-19 C].
And some people might argue there would or would not be a lapse rate. I think the air temperature
would reduce per 1000 meters elevation. But doesn’t matter, and it will effect temperature of core surface or shell.

Now we going to add a greenhouse gas. This will be H20 ice. So we going to a bunch piles of ice which is 254 K. Ice conducts heat about the same as stone:
Ice (0oC, 32oF) [2.18 – k – W/(m.K)]
Concrete, stone [1.7 – k – W/(m.K)]
And also keep in mind water: Water [0.58 – k – W/(m.K)]
http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html

Water of course is liquid and IF there is difference in heat can transport
heat via convection- whereas stone and ice are solids and are just conducting
heat.
Now we going make neat stacks of ice- so 10 Km square and 1 km thick and
put them on the core surface. Then surround this ice with neatly stacked cubes
made of the material as core- or stone blocks. So make perimeter 1 km wide.

So we have cold ice, but cold ice will evaporate at this temperature- hence the addition
of greenhouse gases. So this means we will have 254 K air and some water vapor in this
air. Not much but some.
Some people may want to focus some attention on the possible warming effect of this added
greenhouse gas, but I want to first look at another factor.

On earth we don’t get as much thermal energy as this steel greenhouse does, yet if digs
1 km down anywhere on Earth, one could expect to find warmer conditions. Or there

So if one were to dig 1 km below the surface of the core, one would also expect higher temperature
at 1 km depth than at the surface of the core.
Or if you pile 1 km core stuff above the surface, the new top of surface may not be warmer than lower elevation surfaces [so might think it colder] but the old surface which has 1 km material
on top of it, the old surface should be considerable warmer.

So with the ice, you could have better conductor than the stone, but twice as much depth of ice
should would conduct heat worse than the stone. 1 km of ice should not conduct as much heat as 1/2 km of stone.

So I think any planet radiating 235 W/m-2 of thermal energy has molten rock fairly close to
the surface.
“In some of Yellowstone’s thermal areas, heat flow is over 100 watts per square meter, about 50 times that of Yellowstone’s average and ~2000 times that of average North American terrain.”
http://volcanoes.usgs.gov/volcanoes/yellowstone/yellowstone_sub_page_53.html

So it seems quite possible for ice at the old surface to become warm enough to become
liquid water. And if there isn’t much difference temperature of water, one has poorer
conduction of heat with the liquid water as compared to ice- water *could* insulate heat
better than stone- though if there is density difference [from warmer water being lighter
water have convect and conduct far more heat than stone. Though this convection mixes the temperature of water bringing a about uniform temperature.

Anyhow, one could ice at -18 C at surface and liquid water at some depth below the surface.

So, other add a bit H20 greenhouse gas, this shouldn’t changing the heat budget for Warmists. And for Skydragons the water vapor nor steel shell and also make no difference to surface temperatures.
Though the liquid water would seep into the ground, and cause all kinds of complications.

275. Max™ says:

1. If you don’t appreciate the virtue of simple plane-parallel representations of energy flows then there is nothing that I or anybody else can do to help you. Of course with real spheres and real shells there are finite gaps and finite thicknesses involved. And of course, even if these gaps and thicknesses are tiny, the Wm-2 figures will indeed go down by tiny fraction of the plan-parallel results as the energy passes through them. But it is pointless to argue any more about this if you reject the necessary approximations that go along with what I consider to be the major didactic advantages of the plane-parallel approach. We will just have to agree to differ.” ~David

There is no virtue to that representation, it misleads and obfuscates the actual processes.

2. Re. the SkyDragon diagram, using my simplified plane-parallel model, both shell and core are at the same temperature of 254K simply because both shell and core are emitting net radiation of 235Wm-2. (S-B law for an ideal black body).” ~David

_________

This is another example where the reasons why we disagree are not always obvious.

I completely agree, the core will always be warmer than the shell.

I completely disagree that the core will be warmer with the shell than it would be without it.

I would state this difference most clearly as: the shell will always be cooler than the core.

____________________________________________________

Tim F.

No! The power source provides 235 W/m^2 IN ADDITION TO any other power supplied to the planet by any other means. The planet will (once a steady-state condition is reached) emit 235 W/m^2 + (any other power received by other means). The temperature in this case will rise until the energy emitted = energy absorbed.

The planet get 235 W/m^2 from the heaters, and 235 W/m^2 from the shell.
The planet emits 470 W/m^2 to the shell.
NET = 0 W/m^2

The shell gets 470 W/m^2 from the planet and 0 W/m^2 from space
The shell emits 235 W/m^2 to the planet and 235 W/m^2 to space.
NET = 0 W/m^2

* Energy is conserved.
* Heat always moves from warmer to cooler.

People talk about simple addition — well, it doesn’t get much simpler than this!

The initial thought experiment specified that the planet was internally heated, there was no other power source for the planet.

The power source does not supply “235 W/m^2 IN ADDITION TO” whatever else you want to tack on.

The power source supplies enough energy to raise the temperature of the planet/core/body/whatever until it emits 235 W/m^2, the wording matters to someone who knows their physics, as I do.

I say this because, again, the geometry matters.

One can try to force it into a discussion of flat infinite planes, but that is not what Willis proposed, is it?

__________________________________

Let’s assume the power source has a radius of 4000 m and the planet has a radius of 8920 m as described above, the planet starts out emitting 235 W/m^2, and totals to 235,000,000,000 W.

The power source also puts out 235,000,000,000 W, which means it supplies 1168 W/m^2 to the interior of the planet.

In order for the planet to emit 470 W/m^2 the radius has to shrink to 6300~ m or the power source needs to double its output.

Adding a shell 1 m above the surface will reduce the power density to 234.99 W/m^2 or so, which may not seem significant, but it means the shell will always be cooler than the planet, and that means the shell can not supply power to the planet.

It is literally a physical impossibility to have the shell and planet at the same temperature, so mayhaps if you wish to discuss such a situation, you should clarify that you’re not defending the model Willis proposed?

276. Tim Folkerts says:

Arfur — the planet is postulated to be a blackbody, which (by definition) means it absorbs all incoming radiation. This is independent of the wavelength, so photons from any temperature of emitter will get absorbed. It doesn’t matter if the emitter is hotter or colder that the surface of the BB.

You seem to to be focusing mostly on the “radius cooling effect”. This will make the outside of the shell less than 254 K … but only a “tad” cooler. But the planet will still be WARMER than 254 K .. and not just a “tad” but a lot. (I just went through that a couple posts ago https://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/comment-page-3/#comment-47575)

277. Kristian says:

Tim Folkerts says, March 21, 2013 at 11:11 pm:

“Kristian , I have been countering what you say for days — you just are not listening.”

You’re funny. Still it’s rather sad to see a man clinging so desperately to pure dogma the way you do. There is no actual independent thinking in your reiterations.

Tim, you have not been addressing AT ALL during this thread the gist of what I’ve been saying, my MAIN points. You’ve rather meticulously made sure you AVOIDED addressing them, restating instead your obviously confused assumptions and baseless assertions over and over and over again, even after it’s been repeatedly pointed out how wrong they really are and why.

So how can you have been countering what I have been saying ‘for days’…? That’s exactly what you haven’t done. But of course, simply CLAIMING that you have is probably considered a valid rhetorical strategy in some circles.

I am asking you again, for the last time: “You don’t get the difference between the two-sided open air heaters and the ‘two-sided’ closed shell? Between power and power density?”

Or that a black body will have to emit (that is, get rid of) as much energy as it receives, not half of it. The shell in David’s ‘IPCC model’ diagram clearly receives an incoming flux of 470 W/m^2. It will of course absorb it entirely and acquire a corresponding emission temperature of 302K. And it will emit, according to its acquired emission temperature, a flux of 470 W/m^2, to balance this incoming flux. A black body does not rid itself of absorbed heat inwards upon itself. The entire flux will have to be emitted from the black body’s only real surface, its boundary with its surroundings – its outer surface.

None of these points have been addressed and certainly not countered ‘for days’ by you, Tim.

[Reply] I countered it by pointing out that the vacuum on both sides of the shell has no temperature, and the shell will emit more or less equally from both surfaces. “In on itself” is a meaningless phrase so far as I can see. TB

But you are right in one regard. This discussion is over. There is apparently nothing to be said to convince anyone that they’re wrong or that they should rethink their position. I’ve been repeating points the last couple of days – elementary at that – without anyone willing to take them on board at all. So I guess there is no point in proceeding.

[Moderation note] This comment is full of the rhetoric you accuse others of. Dial it back and do science please.

278. Kristian,

there is no point in getting angry just because your (sincere) opponents in this debate have a different point of view from yours, whether they are right or wrong to do so. In this forum you must accept that, like you, they are acting in good faith.

Tim Folkerts is a consistent advocate of the warmist position. I am a hard line skeptic. Yet we find common ground here on this very narrow issue. We could both be wrong. Or perhaps you and Roger Claque could both be wrong. That’s why we are debating it here. To get closer to the truth by learning a bit from one another.

We must remember we are discussing a thought experiment, designed to test our understanding of thermodynamcs and radiation. Whichever diagram is correct, will not in itself settle the warmist/skeptic debate over the real earth-atmsphere system and whether CO2 concentration does or does not affect the earth’s surface temperature.

This very narrow, but scientifially crucial issue we are discussing (IPCC versus SkyDragon energy flow thought experiment diagram) seems to have boiled down to an even narrower question:

Does a solid black body sphere, alone in the vacuum of space (i.e. without any surrounding shell) and with a constant power source inside it strong enough to produce an emission of I [Wm-2] from its surface towards space, at the same time emit I [Wm-2] inwards from that same surface towards its centre?

To me (and I think, from what he says, to Tim) this seems an utterly bizarre attempt to counter the correct mathematical point I made to Roger Claque earlier about the power intensities on BOTH sides of a surrounding shell so that the math still works out as you desire (i.e. in favour of the Skydragon diagram as opposed to the IPCC diagram).

Please let us instead accept that a solid black body sphere with a fixed internal power source that raises it to a temperature T [K] only radiates outwards at a power intensity I [Wm-2] related to T by the 134 year-old S-B equation. We need to look elsewhere for whether the IPCC or the SkyDragon diagram is the correct representation.

In other words, accept that point and then think more widely, as Postma and others have apparently done, about the physics involved. Surely we must all take the Postma viewpoint as a serious alternative until we find reasonable, logical, mathematical and physical grounds to refute it.

Otherwise we are going nowhere with this debate.

We are doing no useful work.

Just like the radiation between two bodies at the same temperature that goes round and round in a circle, separated by a vacuum. 🙂

279. Kristian says:

“I countered it by pointing out that the vacuum on both sides of the shell has no temperature, and the shell will emit more or less equally from both surfaces. “In on itself” is a meaningless phrase so far as I can see. TB”

Yes (you’re not Tim F), which I countered (multiple times) with the fact that the inside of the shell is not and will never be a cold reservoir for the shell. There will not be a net flux going in. This flux will be zero. Because the shell would be radiating upon itself (meaningless to you, apparently). It is the interior of the shell, of the ‘system’. ‘Emission’ brings energy OUT OF the system. The outside of the shell, its surroundings, space, is its cold reservoir.

This very simple fact is not sinking in.

You seriously seem to think that a black body shell receiving and absorbing an incoming flux for instance by radiation from the outside of say 470 W/m^2 will not reach an emission temperature of 302K, but rather 254K, because half of the reemitted flux will have to go inwards, from the inside of the shell to its core … Because it’s somehow the outgoing flux that dictates the temperature of the shell.

I’ve realised that this is a point we simply can’t get past. So I will not just dial back. I will withdraw completely (and this time for real :P).

280. A C Osborn says:

Kristian, I am amazed at your persistence, it was obvious days ago that agreement would never be met.
But well done for trying.

281. tallbloke says:

I’m just bemused by the idea that if you heat up a metal plate or sphere, it’ll only radiate from one side. I ask Kristian to try heating a frying pan or wok good and hot and then spitting on both sides of it to see what happens.

282. mkelly says:

Tim Folkerts says:

March 21, 2013 at 7:54 pm
* The temperature of the OUTER shell will be lower than 254 K by a factor that I have quoted before:
T_outer = (254 K) * (r_planet/r_outer)^(1/2).

This is absolutely wrong. There is no path for the planet to radiate to the outer shell. The outer shell is only connected to the inner shell via conduction. The planet is radiating only to the inner shell and that is the only other surface you can use in radiative heat transfer equations. You cannot violate the requirements of heat transfer to suit yourself.

Your ratio which I showed you several times shows that the inner shell will be a function of the radii of the planet and inner shell. We have been through this before and using your numbers the inner shell will be 301.7K if the planet is 302K.
T1(planet)= 302 K
R1(planet) = 6370 m
R2(inner shell) = 6380m

.9992* 302 = 301.758 K

T2(inner shell) = T1(planet surface) * sqrt(R1(planet/R2(inner shell) this is your ratio

This is where the discrepancy is. 301.7 K is not 235 W/m^2. You cannot get around that. Max, Kristan, I, and others keep pointing out the flaw and you ignore it.

283. Westy says:

Kristian, I can understand your frustrations and would like to thank you, and all who have stayed in the debate.

284. Tim Folkerts says:

One last try, Mkelly …

1) The planet has some total power P_pl generated somewhere inside it.
2) The planet has some total area A_pl
3) the outer surface of the shell has some area (A_outer) > (A_pl)

4) Once a steady-state condition as been established, there must be a net outward flow of power from the planet’s surface equal to P_pl (or the planet would be warming up)
5) Once a steady-state condition as been established, there must ALSO be a net outward flow of power from the shell’s outer surface ALSO equal to P_pl (conservation of energy)
6) Since (A_outer) > (A_pl), then (P_pl)/(A_shell) < (P_pl)/(A_pl)
7) Since (P_pl)/(A_pl) = 235 W/m^2, (P_pl)/(A_shell) < 235 W/m^2
8) Since 235 W/m^2 corresponds to 254 K, then the OUTER surface of the shell must be less then 254 K.

Too bad we can't all get together in a room with a good chalkboard. Since that is not possible, I think we will have to realize that we are quite possibly never going to get any farther in this discussion. Which is too bad, because there is some interesting physics here

285. Tim Folkerts says:

mkelly, there is one other effect at work that you might be thinking about.

Of the energy emitted as radiation from the planet, ALL of it hits the inner shell.
Of the energy emitted as radiation from the inner shell, only PART hits the planet.
(The rest of the energy emitted by the inner shell hits other parts of the inner shell).

I haven’t done all the math, but I am pretty convinced that (in the limit of a thin shell with good thermal conductivity), this will be the same “radius cooling effect” as I have calculated already.

This leaves the SHELL radiating just under 235 W/m^2 (both in and out) ie just under 254 K.

This means the planet will be absorbing just under 235 W/m^2 from the inner shell, absorbing exactly 235 W/m^2 from the nuclear heaters, and emitting just under 470 W/m^2 to the inner shell. This means the planet is just under 302 K.

***********************************
I was expanding on this to include the insulating effects of the shell. In this case the OUTER shell will still be jsut under 254 K, but the thermal gradient across the shell could easily cause the inner surface of the shell to be well above 254 K. This could lead to MORE than 235 radiating back down and more than 470 radiating back up = hotter than 302 K at the surface.

286. Tim,

For what it’s worth (which is probably not a lot now) your explanation makes perfect sense and doesn’t need a chalkboard. I am baffled as to what this is all about. Why anybody would be bothered to debate the obvious fact that, because of a miniscule increase in radii between core, shell inner surface, and shell outer surface, there is a corresponding miniscule fall-off in power per unit area is quite beyond me. Of course there is. So what?

The real issue is between the two models (IPCC and SkyDragon) with respective core surface temperatures of 302K and 254K – a huge difference of 48K and one very well worth debating. How grown men could be reduced to quibbling over a miniscule fraction of one degree Kelvin error in these numbers due to my using a simplified plane-parallel diagram is quite beyond my comprehension.

I suspect it is a diversionary tactic because the numbers in the IPCC model didn’t add up the way they wanted them to. When I pointed out their arithmetic flaw with respect to the power intensities at the double surfaced shell, they offered a ridiculous counter-argument that the core also has two surfaces. Because this was roundly rebuffed, they have simply diverted back onto a non sequiteur about miniscule radii differences.

I repeat what I said in my previous posting. If Postma is right, he is right for much more fundamental physical reasons than some minor mathematical discrepancy over radii.

I am more than happy to go on debating with you here what these reasons might be and whether they stand up, as I am with anybody else who is open minded enought to wish to continue the substantive debate.

287. lgl says:

mkelly
What will the temperature of the outer side of the shell be if the inner side is 301.7K?

288. Excellent question Igl. But you won’t get a straight answer!

289. Westy says:

Tallbloke, It is bemusing. When you heated the wok did the heat source get hotter because of the woks presence, from some back radiation. Westy’s law remember “Your cold stuff can’t heat up your hot stuff”. Doesn’t this apply to the shell heating the core of the Willis model?

290. Roger Clague says:

David Socrates says:
March 22, 2013 at 7:06 pm

counter-argument that the core also has two surfaces. Because this was roundly rebuffed

I made the objection about shell having 2 surfaces and the planet only 1.

Roger Clague says:
March 21, 2013 at 7:51 pm

The top layer of the planet is getting energy from 2 directions. It has 2 faces. So it can lose energy in 2 direction, from 2 faces.

I did not say the core has two surfaces

However your diagram shows the core with 2 surfaces.

Willis puts arrows outside the planet but,inconsistently, avoids putting an arrow inside the planet.

Let me put my objection another way.
The 235wm-2 ‘back radiation’ get emitted at the planet surface and is not absorbed and conducted away by the planet.
The 470Wm-2 from the planet is absorbed by the inner surface of the shell and is conducted to the other surface of the shell
This is inconsistent behavior by radiation at 2 single surfaces.

291. tallbloke says:

When you heated the wok did the heat source get hotter because of the woks presence, from some back radiation.

Of course.

292. gbaikie says:

“I repeat what I said in my previous posting. If Postma is right, he is right for much more fundamental physical reasons than some minor mathematical discrepancy over radii.”

Well Postma is right, if outer shell is radiating 235 W/m-2 and the source of heat [core surface] having same total area which providing 235 W/m-2 of energy.

It’s the same logic of determining how much energy is leaving earth- if Earth radiates 240 W/m-2 it must have a net gain 240 W/m-2. Any difference means a planet is gaining or losing heat.

[As we know the temperature of a planet is not related to this flow of energy- Venus has lower
incoming and outgoing heat flow than Earth and it is hotter than Earth. Or Earth is warmer
than 240 W/m-2 radiated from surface. But that’s another topic]

Postma is assuming 235 W/m-2 is being adsorbed by inner surface and heat is being conducted to outer surface of shell with little resistance. And since one is dealing metal and outer shell could cool to 2 K, this not an unreasonable assumption. If you said the inner part of shell was reflective
rather than absorbing the energy, Postma might give different numbers.

Now in terms of temperature, any body which has radioactive decay giving 235 W/m-2 of energy
could be made create very high temperatures.
Geothermal energy would be the only game in town. On Earth on average gets less than 1/1000th of this energy- and unless lava is erupting in the area one doesn’t even get small areas giving this much energy at the surface. But all this kind of activity which involves increasing temperature so it can used, doesn’t change the energy budget of the planet.

Now, a technical aspect regarding this, is question of whether or not the inner surface of the shell
can absorb 235 W/m-2 of energy. Or what does the incident radiation have to be for surface to absorb 235 watts per second.
Postma gives the obvious answer: 235 W/m-2 [if one assume the premise that this surface absorbs all of the incident radiation- if surface only converted 1/2 or 25% of incident radiation, one will have a different result].
One might ask the question, how can one absorb all the incident radiation. Well with solar thermal
heat generation. Which, on simple terms, can be garden hose filled with water lying in sunlight, you need something which can hold heat [have something which can be heated, store the energy, and put it somewhere. You need something cold to warm up. Or you can use something like oil or salt and concentrate the solar energy [get it much hotter than water boils]. If solar thermal energy’s task
was to melt ice [for some darn reason] one could get large efficiency in terms absorbing
all the incident radiation. If trying heat water for a practical purpose of hot water for home use- 50 to 60 C water- then one going to have efficiency of about 50% of total incident radiation being absorbed.
Now, with space, one in a sense has something very cold to heat- you have this purpose of melting ice.
So getting something near 100% efficiency should not be difficult.
If one has a lot of incident radiation it becomes more difficult- one might have use aluminum or silver. Or use radiating fins to shed more heat. Or one wouldn’t want steel on outer part of shell polished mirror like as it emits lower amount of radiation.
So if just limit to making it out of steel, one has numerous ways change the results. But if one trying
to make the steel surface behave like a blackbody [absorbing all incident radiation, and conducting the heat to outer surface which emits similar to blackbody] then any engineer should be able have a design so the inner surface of shell absorb most of the incident radiation.

293. Max™ says:

“The real issue is between the two models (IPCC and SkyDragon) with respective core surface temperatures of 302K and 254K – a huge difference of 48K and one very well worth debating. How grown men could be reduced to quibbling over a miniscule fraction of one degree Kelvin error in these numbers due to my using a simplified plane-parallel diagram is quite beyond my comprehension. “ ~David Socrates

Again, your second model is not an accurate representation of the skydragon position.

The skydragons would probably argue that the shell should always be colder than the core, and that the presence of the shell would not raise the temperature of the core.

Plane-parallel diagrams have nothing to do with Willis’ model, the geometrical effects you eliminate when moving to plane-parallel formats are rather important for understanding the disagreement between the two positions.

I myself fell into making the same error.

I would correct that in this manner:

^^^^ ______________ ^^^^
235 W/m^2 ________ 234.99 W/m^2
IPCC ____________ Skydragon
254 K ____________ 253.7 K
^^^^ ______________ ^^^^
470 W/m^2 ________ 235 W/m^2
302 K ____________ 254 K

The reason I feel this is the case is rather simple: if the gap were filled with the same material as the internally heated core, then the temperature would be determined by the surface area and internal power supply, if you compared two internally heated cores, one of the original size and one of the larger size, I am confident this would be the case.

As such it seems odd to claim the original size layer inside the larger core could be made warmer simply by removing the material and restoring the gap.

Indeed, in the case of the IPCC model, I would expect if the core were emitting 470 W/m^2, the shell should be at a temperature of around 301~ K, not 254 or 302.

In this case, the power output of the core has been doubled somehow.

Would the original surface layer still be 302 K if the gap was filled with the same material as the core, i.e. if you compared two IPCC models, both solid, one the original size, and one the size of the added shell, would the larger core have a higher temperature at the depth of the original surface?

8920 m + no shell = 254 K, 235 W/m^2 emissions
8920 m + 1 m high shell = 302 K, 470 W/m^2 emissions
8921 m + no shell = 254 K, 235 W/m^2 emissions

Would the temperature 1 m deep in the 8921 m object still be 302 k?

294. Tim Folkerts says:

gbaikie supposes: “Postma is assuming 235 W/m-2 is being adsorbed by inner surface and heat is being conducted to outer surface of shell with little resistance. And since one is dealing metal and outer shell could cool to 2 K, this not an unreasonable assumption.”

There are two HUGE flaws here.

1) If heat conducts well through the shell, then the temperature difference should be SMALL from the inside to outside.
2) if the outside is 2 K, it will be emitting 0 W/m^2 to space! Those 235 W/m^2 are getting absorbed at the inside but not leaving EITHER side!

This highlights the sorts of errors that the “254 K planet” side keep making. And frankly, Postma himself makes the biggest errors

295. Arfur Bryant says:

Tim Folkerts says:
March 22, 2013 at 2:46 am

Tim,

But I specifically pointed out that the 235 W/m^2 downward energy needed to be absorbed for net gain by the planet. In order for the planet to radiate at 470 W/m^2 you (and Willis) would have to show that it gained energy from its own ‘re-radiated’ 235 W/m^2 arrow. Where is the absorption for net gain in order to raise the energy level? If high energy radiation contacts a low energy surface the surface energy level increases and emits radiation at a higher level. What Willis’ model is postulating is that the two similar energy levels can be added numerically.

I am not focused on the radii problem but it cannot be ignored or dismissed because it means that the energy level of the downward arrow is actually (by however small a margin) lower than the planet surface. Therefore you are arguing that you can add a lower energy to a surface and yet apparently raise the surface energy level to twice its original!

Isn’t that a bit like saying you can add half a cup of water at 5 deg C to another half a cup of water at slightly less than 5 deg C and getting a full cup of 10 deg C?

All I’m asking is for you or anyone (other than Bryan) to show me a link proving this is possible.

296. gbaikie says:

“baikie supposes: “Postma is assuming 235 W/m-2 is being adsorbed by inner surface and heat is being conducted to outer surface of shell with little resistance. And since one is dealing metal and outer shell could cool to 2 K, this not an unreasonable assumption.”

There are two HUGE flaws here.

1) If heat conducts well through the shell, then the temperature difference should be SMALL from the inside to outside.
2) if the outside is 2 K, it will be emitting 0 W/m^2 to space! Those 235 W/m^2 are getting absorbed at the inside but not leaving EITHER side! ”

1] The temperature difference should be small.
2] The outside can’t be 2 K because it’s being heated, but it can cool to 2 K if there isn’t any heat added. It will lose heat until it’s 2 K.

2 K space environment is like cooling to room temperature, except with air temperature difference you have additional air convection losses [which aren’t generally significant unless the temperature difference is more than a few degrees- and there is not loss of heat from evaporation].

1] The temperature difference between interior and exterior of shell should similar to difference of temperature of sheet steel on the ground- on ground at night or during noon time sunlight.
Or on this 235 W/m-2 heated world, less than difference between the top inch of soil and 1 inch below the top surface- as soil conducts far less heat than steel does. Also same goes for solid rock.

2] Take a frying pan, have element turn up so frying pan reaches a fixed temperature- say 100 C,
drops water boil if put on the surface. Put pennies on frying pan, and put disk metal in top of frying
pan [on pennies]. This should not have much effect on temperature of frying pan. the pennies or
top and bottom side of disk. And you can ceramic spacer instead of pennies. And you put it in vacuum there still will not much differences. If the disk has reflective surfaces this will make make significant difference if in vacuum- but less effect if not in a vacuum.

297. Max™ says:

“I am not focused on the radii problem but it cannot be ignored or dismissed because it means that the energy level of the downward arrow is actually (by however small a margin) lower than the planet surface. Therefore you are arguing that you can add a lower energy to a surface and yet apparently raise the surface energy level to twice its original!

Isn’t that a bit like saying you can add half a cup of water at 5 deg C to another half a cup of water at slightly less than 5 deg C and getting a full cup of 10 deg C?” ~Arfur

Indeed.

It is also like saying if you take a sphere with the previously mentioned 1 million square meter surface area (8920~ m radius) and it has an internal power supply which raises the temperature until the surface emits 235 W/m^2 (total 235 megawatts) then you could do the following:

1. Add a shell 1 meter above the surface (so the radius of the shell is 8921~ m) and the shell will then emit 235 W/m^2 (total more than 235 gigawatts) and the original surface will warm until it emits 470 W/m^2 (total 470 megawatts) with no other changes made to the system.

2. Fill the gap between the shell and the surface with the same material as the shell and surface are made of, maintain the 235 W/m^2 emission from the shell surface and the original surface–now at a depth of 1 meter–will be at or near 302 K.

_______________

If this is the case, it seems to me that if we start with the original sphere+shell, assuming the sphere surface is emitting 470 W/m^2, we should be able to remove 1 meter of material inside the sphere and ask what the temperature would be afterwards.

If it is true that adding a 1 meter gap + shell above the surface raised the surface temperature until it doubled the emissions, then adding another gap within, such that the surface is a new shell, would also double emissions, right?

So the new surface would emit 470 and receive 470, totalling to 940 W/m^2.

Do that again, now with 1 “outer” and 2 “inner” gaps we have a total of 1880 W/m^2 leaving the inner surface, right?

But above I suggested that an internal power source, spherical, and 4000 m across could supply 1138 W/m^2 which would give 235 W/m^2 through a naive approximation of conduction following an inverse-square law.

Now we find that simply removing layers of material beneath the original surface+shell would produce more energy?

Why isn’t this being engineered and utilized?

[Reply] If you ‘utilised’ any of the energy, it wouldn’t be available to be radiated to the shell above, and so the ‘back radiation’ effect would diminish, and you’d find that the amount you ‘utilised’ plus the amount radiated to spce from the outermost shell would still only total the amount being produced by the nuclear core.

298. suricat says:

Guys, I won’t pretend that I’ve kept up with all the posts in this thread, but the failings of Willis’s ‘Steel Greenhouse’ are ‘legendary’ to all critics (sceptic, or warmist alike).

1. The “Planetary Nuclear Core” supplies energy to a ‘black body surface’.
2. The “steel shell” exhibits a ‘grey body’ behaviour (as Tim Folkerts has tried to remind us).
3. Most responses here wrongly ‘assume’ that the ‘shell’ also exhibits ‘black body’ behaviour.
4. A ‘black body shell’ (that ‘most posters’ assume) can’t cause “back radiation” because ‘the shell’ has no ‘reflective albedo’ (a ‘grey body’ property), it only ‘absorbs’ and ’emits’ ALL radiation, thus, a BB scenario CAN’T resolve this hypothesis.
5. The ‘energy’ (energy in total calories / second) generated by the Planet’s ‘core’ in the OP is the only amount of ‘energy’ that the ‘outer surface of the shell’ can radiate (irrespective of what goes on below this level).

Thus, no back radiation from a BB returned to the ‘source’ for wavelengths that should’ve been included for ‘escape from the system’, but the ‘shell’ isn’t ‘black body’! Is it? Surely, the ‘shell’ only ‘permits’ and ‘prohibits’ ‘energy transmission’ of wave lengths within the ‘shell’s’ ‘admittance / reflectance’ parameters.

This needs to be addressed, but I suppose that may well be ‘another subject’ (unless a ‘grey body’ is accepted for the ‘shell’ here)! 🙂

Arfur Bryant says: March 23, 2013 at 1:00 am

“Isn’t that a bit like saying you can add half a cup of water at 5 deg C to another half a cup of water at slightly less than 5 deg C and getting a full cup of 10 deg C?”

No. It would be ‘partly’ true for a BB scenario, but a real scenario is entirely different. 🙂

Best regards, Ray.

299. gbaikie says:

“Now we find that simply removing layers of material beneath the original surface+shell would produce more energy?

Why isn’t this being engineered and utilized?”

This can be done, if using a reflective surfaces for layers and it’s in a vacuum

So it’s not done because we don’t live in a vacuum.
If the Moon had enough “geothermal heat” [which it doesn’t] then it could done on the Moon.

But getting furnace type temperatures using sunlight on the Moon would be possible.
Let’s try google: solar furnace + lunar:

http://www.tvu.com/PSolarFurn.html
Excerpt:
“The Innovation. The innovation proposed in this program is to develop a high efficiency solar furnace core. This new furnace core will allow much higher furnace temperatures for the same solar energy collection or greatly reduced collection mirror area. Greatly increased efficiency is achieved by coupling vacuum insulation with an inside surface that is a nearly perfect reflector. Specific furnace technology that is available and has been under development at Thoughtventions (TvU) include 1) Low power furnaces; high efficiency heat containment furnaces, 2) High temperature fully and partially transparent furnaces (1200°C), 3) Improved heat radiation shields, 4) Very high temperature furnace technology (2500°C), 5) Vacuum insulation technology, 6) Monitoring technology including viewports and thermocouples, and 7) Furnace design technology.”

But you need a vacuum.
People don’t understand the value of a vacuum.
Having no air on the Moon is big plus.

300. lgl says:

gbaikie

“So it’s not done because we don’t live in a vacuum.”