## 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. Max™ says:

“But what it will do is add to the total amount of radiation the planetary surface is receiving.” ~tallbloke

Shouldn’t that be “it will reduce the total amount of radiation the surface loses”?

The outer shell is not an energy source for the surface, it is heated by the surface.

_______

Quick note:

Say the internally powered body has a surface area of uh, 1000 m^2, and is emitting 235 W/m^2, so the total output of the surface is 23,500 W/m^2, right?

Suppose the outer shell has a surface area of 2000 m^2, and is receiving 235 W/m^2 from the surface of the heated body.

If the outer shell emits 235 W/m^2 that is a total output of 47,000 W/m^2, right?

If the outer shell had twice the surface area of the internally powered body inside, wouldn’t it only need to emit 117.5 W/m^2 * 2000 m^2 to balance the 235 W/m^2 * 1000 m^2 supplied by the interior body?

2. tallbloke says:

Max:

Given the height difference between surface and TOA on Earth is ~15km compared to it’s diameter of 12700km, I think we can disregard area differences to keep the discussion simple. Fig2 is ‘not to scale’. Willis has pointed this out in previous versions of the idea.

The outer shell is not an energy source for the surface

This doesn’t alter the fact that it is radiating towards it. The radioactive decay is the energy source, the other components of the system simply determine how it gets redistributed and bounced around before finally departing for space.

“But what it will do is add to the total amount of radiation the planetary surface is receiving.” ~tallbloke

Shouldn’t that be “it will reduce the total amount of radiation the surface loses”?

Absolutely not.

3. michael hart says:

I also generally avoid these arguments, for a variety of reasons.
If I take a solid sphere that is painted black, and apply a new layer of ‘Stephan-Boltzman Acme Black Paint’, will the surface temperature change? No, I don’t think so.

I think the GHG argument is that one is applying an extra layer of grey paint to a grey surface, so in this limited case the answer might (might) be ‘Yes’-below the new surface.

Of course this model takes no account of movement, convective transport, state changes (latent heat) and clouds which exist below the new and old radiating surfaces-which makes it a model that is neither right nor very useful.

Further, I am confident that Modtran/Hitran models take no account of the fact that the “non-condensing greenhouse gas” carbon dioxide is actually very soluble in the predominant condensing greenhouse gas-water. All the King’s horses and all the King’s men, couldn’t frickin’ model water again.

4. A C Osborn says:

Suddenly Tallbloke admits to Back Radiation from a cooler surface heating a warmer one. Well, Well, my ghast is flabbered.

5. tallbloke says:

Michael: the toy model doesn’t need to take account of “movement, convective transport, state changes (latent heat) and clouds” because in the toy model, there aren’t any. There is only a vacuum.

ACO: Suddenly Tallbloke admits to Back Radiation from a cooler surface heating a warmer one. Well, Well, my ghast is flabbered.

No. The hotter object absorbs the radiative energy from the cooler object. But, energy is not heat. Energy is energy. Heat is a measure of the work done by energy. Black bodies radiate as much energy as they receive. If the black body is radiating more than it was before because it is receiving more energy than it was before, it’ll be doing it at a higher temperature, because radiation is proportional to temperature.

In this case the back-radiation isn’t making the surface hotter, it’s making it cool down slower (As Max nearly said earlier, but we need to go the long way round to make the point clear). Because the rate it can cool at is being restricted by the presence of the cooler radiating body surrounding it, but the internal heat it needs to dissipate is still arriving at the same rate, it accumulates heat faster than it can lose it, until it has risen in temperature to the point where it can once again lose energy as fast as it gains it.

6. graphicconception says:

Shouldn’t fig 2 have another 235 W/m2 coming in from outside?

In which case doesn’t that mean that the red arrow should be reduced to 235 W/m2 to make things add up?

[Reply] No. in the toy model, the nuclear heated planet is in free space, far from any star.

7. tallbloke says:

Things are warming up on Joe’s thread:

Joseph E Postma says:
2013/03/10 at 11:31 AM
Roger, none of what you said makes any sense. It is just making things up and not following any logic.

🙂

8. Anthony Watts says:

I say, apply for a grant and just do it. Solve the issue once and for all. /sarc

9. tallbloke says:

Excellent plan Anthony, call the Magratheans and get Slartybartfast over here right away. 🙂

10. michael hart says:

“Excellent plan Anthony, call the Magratheans and get Slartybartfast over here right away. 🙂 ”

Time for a quick skiing holiday before global warming sets in. 🙂

11. tallbloke says:

Michael: Time for a quick skiing holiday before global warming sets in. 🙂

Then we can do the wall of death on the inner shell surface FREEEEEEEOOOWWWWWWWWWWWWWW! 🙂

This one has been in moderation a while:

tallbloke says:
2013/03/10 at 11:55 AM
“The shell first sends back 1/2, then it sends back 1/4 of 235”

No. It sends back 1/2 of 235. Than after the next iteration it sends back 1/2 plus 1/4. Then after the next iteration it sends back 1/2+1/4+1/8. It is reducing each time because the extra amount of outgoing energy from the surface is reducing each time too. And that’s because half of the extra is being lost to space.

12. Tim Folkerts says:

Joe said: “These people are complete, unfettered idiots, and are a disgrace to mathematics….
Oh! The irony!

Tallbloke, I agree with everything you said. (Well, maybe not “Heat is a measure of the work done by energy.”, but that is not central to this discussion).

13. wayne says:

Maybe we should all view that shell as being just two meters off of the surface so we can don our spacesuits and our instruments and be able to walk about measuring the flux and temperatures of both the surface and the shell, if you are a tall bloke that is. 😉

Ok, let’s assume for a minute that this is correct, that flux does double if the outside shell is 100% opaque. If that shell is our atmosphere, ignoring any thickness, that shell is basically that opaque membrane to all IR frequencies ever emitted by the surface.

Let the W/m² unit be Fx for brevity. That would then imply that since our planet outputs ~238Fx, then there must be 476Fx coming from the surface and would require a surface temperature of 303K, but we are much cooler than than that on the average, what gives?

Well first, part goes directly to space right through our not-so-totally-opaque “shell”, some say 40Fx (TFK, +30Fx more from cloud tops), Miskolczi says more like 80Fx and it appears using this very simple analysis that Ferenc would be correct, using this assumption that would reduce the 476Fx to 396Fx emitting from the surface and there is the planetary atmospheric enhancement pecisely, ~288.5K.

This all gets back to whether a gas, specifically just any one gas’s percentage of composition, can “trap” energy over what the mass itself dictates, no gas absorbs in all of the lines, the whole spectrum, there is redistribution from equipartition and Boltzmann distribution, for now we have assumed (for analysis purposes) that normal atmospheric matter, with it’s mass, can absorb radiation. Duh, but everyone here knows there is absorption in all frequencies but the “window” frequencies, mostly very close to the surface.

To me this type of analysis once again says the Ferenc Miskolczi was right on the spot. The mean height of our atmosphere does not matter, though it is gravity bound, the total mass does matter. The window flux (optical thickness of IR radiation) doesn’t change with co2’s concentration one iota, and thank goodness it doesn’t. That is what killed the AGW conjecture, total absorption is completely mass governed in mixed GHG atmospheres and Venus data does seem to agree.

I have to agree with Willis and Rog on this one, any absorption, up to 100% opacity, the “shell”, does affects the temperature of surfaces below with constant energy inputs, it has to, but this does not mean it helps in any way the AGW “side”, that side has already been disproven by the science, many, the slower ones, just haven’t realized it yet.

14. Clive Best says:

There is nothing new in Willis’s model. Introducing a “steel shell” is irrelevant. He is just repeating an old argument found in several text books. A hypothetical 100% opaque atmospheric shell absorbs all IR frorm the surface and reaches a fixed temperature Teff needed to reach energy balance. The surface then ends up as 303K which is wrong.

see here

15. Bryan says:

There are flaws in Willis Eschenbach’s set up.
He says the surface of the steel sphere is very nearly equal to the planet surface, so we can ignore any surface area distribution of power difference.
Fair enough.
However at final equilibrium the steel sphere is forced to be at the original temperature of the planet to radiate to space at 235W/m2.
Since the planet and steel sphere are coupled in a radiative exchange, equilibrium between them will be reached only when the radiative fluxes are equal .
In other words the planets final temperature would be exactly the same as original.
Now in the real world there will be a difference between in temperature between the shell and the planet, only because of the difference in surface area which Willis choose to ignore.
The mass of the planet gives it immense thermal inertia .
So to raise its temperature by a smidgen would take a long,…. long time .

So any radiative insulating effect would be very small.

16. Curious George says:

As much as like Willis, his model has nothing to do with a greenhouse effect. If you like to sleep under stars, you know that a night under stars is colder than a night under cloudy skies. That is the greenhouse effect – and CO2 undoubtedly acts like a cloud, while not blocking the incoming solar energy. That said, the Earth is an incredibly complicated system, and our (modelers would like us to belive, actually their) models don’t reach the accuracy needed to predict anything.

17. tallbloke says:

My last word on the matter:

Hi Joe,

You said:
If the “what” is the current value at each iteration, which is now what you seem to be implying, this is an entirely different series and has a completely different convergence.

What I originally said was:
“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.”

117.5 is 1/2 of the original 235. And the difference between 117.5 and 176.25 is 1/4 of 235.
Therefore my series hasn’t changed, the progression is logical (since half of the excess is lost to space at each iteration), and I’m not “making stuff up”. My thanks to lgl for backing me up.

I do however completely agree with you that none of this applies accurately to the Earth, it’s just a toy model, though theoretically correct for the limited definitions it is based on. Indeed I said just that in my first comment:

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.

I’ll get out of the way and let others have a say.

Best to you.

Rog TB.

18. donald penman says:

This is the one million and sixth time we have seen this model of reality and i know all the steps know but just because I understand this model does not mean that I accept that it is a good model of reality.The idea seems to be that if someone does not agree with what you say about “the greenhouse effect” just keep putting out this stupid model with its limited assumptions of reality until they do agree with you.I don’t think this steel model can even be built so criticising the setup would be a waste of time

19. mkelly says:

Willis’ shell post had a 2000 or so meter distance between the surface and the shell. The surface area of he inner shell is larger than the sphere and the 235 must be divided over a larger area. Thus the inner shell will be at a lower temperature than the surface. Heat will flow from the surface to the inner shell and out. You cannot ignore the difference in area. All heat transfer stops when Tsurf is equal to Tshell, but that will not happen. And the radiation does not have to be absorbed it can be reflected. Reflection is a perfectly good thing for radiation to do.

20. tallbloke says:

Joe’s response:

Joseph E Postma says:
2013/03/10 at 2:37 PM

Cheers Rog…,

There’s no justification that half of “the excess” is lost to space/returned to the interior at each iteration. There is no law or anything that can “know” to return only half of “the excess”, especially when the basis of the original return was ostensibly to conserve energy. There’s no physical justification for the subsequent sequence and neither even is there for the initial return of 1/2.

The series is the same but reinterpreting it a few times doesn’t change the arbitrariness of it, and my analysis is exactly the same. There is no reason that 235 should be reduced by a factor of 2 each iteration – NOR does this entire discussion of “iterations” and sequences have anything to do with how heat flow actually functions or the behaviour of radiation trapped inside a cavity. We’re discussing unicorns. A meaningless simulacra of physics.

Also “excess” is just a very convenient term to use because it makes it sound like there is actually some excess somewhere that can be returned. There’s no excess, there’s a deficit under this setup, but the deficit is just re-worded as an “excess” because an excess sounds useful. Rather, the deficit of what exists on the outside is what is being turned into an addition on the inside, and only because we’ve already started with the assumption that the energy should be doubled, and then the geometric sequence of turning a deficit into an excess starting with a factor of 1/2 conveniently leads to a factor of two.

I think Joe has missed the fact that since the shell will radiate equally from both sides, this is where the factor of 1/2 comes from for the back-radiation. The rest follows from that.

21. Tim Folkerts says:

Bryan,

Your interpretation would work if the heater was moved from the planet to the shell. Then the shell would be heated to 255K by the 235 W/m^2 heater.
* The shell would get 235 W/m^2 from the heater and 235 W/m^2 from the surface’s IR radiation.
* The shell would loose 235 W/m^2 upward and downward by IR radiation.
* NET = 0 W/m^2; T = 235 W/m^2

* The surface would get 235 W/m^2 from the shell’s IR radiation
* The surface would lose 235 W/m^2 upward by IR radiation
* NET = 0 W/m^2; T = 235 W/m^2

But with the heater down on the planet’s surface, then Willis & Tallbloke are indeed correct.

It is analogous to heating the INSIDE of your house’s walls vs. the OUTSIDE. If you heat the OUTSIDE surface of the walls to 10 C (perhaps with 100 W/m^2 on a winter’s night), then the INSIDE of the house will also be 10 C. But if you move the same 100 W/m^2 heaters to the INSIDE surface, then the outside will still be 10C, but the inside will be much warmer.

************
Also, your claim that the surface will be warmer “only because of the difference in surface area which Willis choose to ignore” is wrong. The larger surface area of the shell would DIMINISH the effects of the warming due to the shell. Mathematically, we can make the shell as close as we want to the surface, making the difference in area arbitrarily small.

22. I have tried an experiment to see if a hot object cools slower in the presence of a warm object (warmer than background) the details are here:

Does Thermal Radiation Travel From Cool To Hot Bodies

Please feel free to criticise (by email if the comments do not work)

23. Tim Folkerts says:

Curious George says: “As much as like Willis, his model has nothing to do with a greenhouse effect. If you like to sleep under stars, you know that a night under stars is colder than a night under cloudy skies. That is the greenhouse effect

But that IS Willis’s model! The shell is the equivalent of the clouds. Both radiate energy downward, so both serve to keep the surface warmer than it wold be otherwise. (Now, he changes the input from solar energy to nuclear energy, but that does change the principles by which a radiating layer above the surface keeps the surface warmer than it would be without that radiating layer.)

24. Tim Folkerts says:

mkelley says: “Willis’ shell post had a 2000 or so meter distance between the surface and the shell.”

To 3 significant digits
6.37 x10^6 m = radius of earth
6.37 x10^6 m = radius of shell

To 3 significant digits
5.10 x10^14 m^2 = area of earth
5.10 x10^14 m^2 = area of shell

To 4 significant digits
253.7K = temperature of “bare planet”
253.7K = temperature of shell around planet

Why do you insist on distracting the discussion by focusing on a correction that is less than 0.1% ?

25. tallbloke says:

TFP: Well done for doing the empirical work! I’d like to repost that if you feel like bundling it up as a text file and set of images numbered in order with figure numbers in the text where they belong.

26. tallbloke says:

Tim F: The shell is the equivalent of the clouds. Both radiate energy downward, so both serve to keep the surface warmer than it wold be otherwise. (Now, he changes the input from solar energy to nuclear energy, but that does[n’t] change the principles by which a radiating layer above the surface keeps the surface warmer than it would be without that radiating layer.)

Steady now Tim F. It does change the principles because clouds are a negative feedback, due to the way they block more incoming solar radiation than they provide downward longwave radiation that actually reaches the surface. There are limited circumstances in which cloud cover makes the surface warmer (night-time in higher latitudes), but this doesn’t account for the generally high surface temperature by a long stretch.

27. Truthseeker says:

Tallbloke, I thing Willis and yourself are making a false assumption about the equality of the radiation from the outer sphere.

I do not see why the radiation would be equal in all directions. Surely the potential difference between the environments inside the sphere (235K) and outside the sphere (0K) make a difference. I would expect that all of the 235K radiation received by the sphere would radiate outwards as this is the greater potential difference. The Universe is always trying to reach equilibrium (all other things being equal) and so trying to get equilibrium between the 0K outside vaccum and the 235K outer sphere is where all of the radiated energy would go. Why would the radiation received by the outer sphere act like salmon swimming against the current of the radiation coming from the inner sphere and not go towards the 0K vacuum that provides no such impediment?

But hey, I am just trying to use practical logic here. After all I am not a scientist …

28. tallbloke says:

Truthseeker: good question. I don’t think all the radiation will go outwards, but I don’t think it’ll be an even split either. I don’t know the equations for this. Anyone?

29. Bryan says:

Tim Folkerts says

Bryan,

“Your interpretation would work if the heater was moved from the planet to the shell. Then the shell would be heated to 255K by the 235 W/m^2 heater.”

The central point made by Willis is that the radiation from a cold body must increase the temperature of the hotter source compared with the absence of the colder body.

If we reverse the Willis Eschenbach planet and hollow steel sphere as you suggest quite an unexpected thing happens (as far as IPCC theory goes).

Initially the heated sphere is at a higher temperature than the colder planet as you suggest.
The planet radiates to the steel sphere as usual .
The steel sphere absorbs the planets radiation as usual.

However all with a grain of common sense realise that the hollow steel sphere would be hotter without the presence of the internal colder planet than with it.

Postma is correct in his note on cavity radiation.

30. Max™ says:

Again, the radiation coming back from the shell is not an energy source for the surface, it can only ever reduce the energy lost by the surface, not increase it.

Additionally, let’s assume for some reason that the outer shell surface area is the same as the inner body, if the outer shell is radiating 235 out and 235 in, then it is still radiating twice what it absorbs.

31. Max™ says:

Whoops, posted too fast, “if it is radiating 235 out and 235 in, and the 235 in adds to the energy coming from the surface then the shell is producing energy magically”, is what I meant to say.

32. Tim Folkerts says:

Truthseeker.

Radiation from s surface is a function of the temperature and emissivity of the surface:
P = ε σ A T^4
The temperature other places do not matter (other than that they might influence the temperature of the original surface). So the radiation from a ~ 253 K surface will be 235 W/m^2 whether it faces a 0K surface or a 303 K surface or a 1000 K surface.

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

There is a separate issue that the two surfaces of the shell don’t a priori have to be the same temperature. Willis’ model assumes a thin shell with a good thermal conductivity so that the shell will have a single, well-defined temperature that applies to both surfaces. If the shell has some significant insulation properties, then the outer surface of the shell would still be 235 W/m^2 = 253K. The inner surface would be warmer than 253K and would radiate MORE than 235 W/m^2 and the surface would be warmer than 303 K.

33. Sparks says:

It’s a Dyson sphere, a hypothetical mega-structure originally described by Freeman Dyson.

34. Curious George says:

Tallbloke – Planck’s law of a black body radiation states that the radiation depends only on a temperature, not on anything else – of course, an ideal black body does not exist, but it is usually a good approximation.

Tim – I read Willis’s model as simply telling us that when you add an insulating layer between a heat source and a heat sink, the temperature at the heat source side would go up. Hardly surprising. You may see it as a greenhouse effect, but I feel that it is oversimplified for that purpose.

35. Tim Folkerts says:

Bryan says: “However all with a grain of common sense realise that the hollow steel sphere would be hotter without the presence of the internal colder planet than with it.”
All of these models assume that we wait “a long time” until equilibrium is achieved. Yes, a cold planet inside the shell will cool the shell for a while. But since we have billions of years to wait, the eventual equilibrium condition will be for the entire interior to reach the temperature of the shell (235 W/m^2 radiating from the surface @ ~ 253 K).

‘Postma is correct in his note on cavity radiation.”
Postsma’s descriptions of cavity radiation is approaching the level of ‘not even wrong’. It adds more confusion, rather than clarifying anything!

36. gbaikie says:

If a planet size of earth is radiating 235 W/m2 due to radioactive decay, how much heat is generated?
So Earth has 511 million square km which is 5.11 x 10^14 square meters. So
1.2 x 10^17 watts.
Actual Earth:
“The Earth’s internal thermal energy flows to the surface by conduction at a rate of 44.2 terawatts (TW”
http://en.wikipedia.org/wiki/Geothermal_energy
So 4.42 x 10^13 watts
2500 times more than Earth.

In this model Earth which is emitting 235 W/m2, we can assume there is very hot core in which there are miles of solid rock thru this energy is radiating thru. Though we could also assume there
is thin crust of rock. Which is it?

Hmm. And it may be impossible.
One could say have Earth’s core 2500 times hotter, but that seems like that leads to there being no solid rock.
Or one could simply have earth generate more radioactive decay until one has about meter solid rock floating hot lava.
A problem seems to be that the surface can radiate more energy than rock can conduct heat to the surface. I can imagine 1 meter less of solid rock on surface of lava but I think it’s a not a stable condition- it’s either heating or cooling.

If one starts molten ball lava and have it cool, it can stay at that level of heat in which one get about
235 W/m2 fairly long periods of time.
But steel conducts heat much faster than rock, so steel will radiate more energy than it receives.
Suppose the planet surface is molten steel which cools with hot rock beneath it adding some energy.
So a mile of steel in which the surface cools so it’s radiating 235 W/m2 [that would take a very long time to cool.
Now have meter thick steel above it separated by vacuum. If this meter layer of steel starts
5 C hotter and cools. You get a net flow back to the surface, but as the outer surface cools
you will get heat a heat gradient in the metal- the outer surface will be cooler.
Now, if the temperature in inner part of shell is the same temperature as planet surface, you would
not have transfer of heat. Therefore the planet surface cools slower.

So thermal property of steel is. Specific heat :
Carbon Steel 0.49 kJ/kg K

A meter cube of steel is 7.8 tonnes
7800 times 0.49 kJ is 3822 kJ

So an hour at 235 W/m2 is 846 kJ so 4.5 hours to lose one K
So started 5 C warmer and takes about one day for outer shell
to cool to surface’s temperature and would add an insignificant
amount of energy to surface during this day period.
And as said there will be heat gradient in the 1 meter of steel-
with outer surface cooler than inner part.
So order to cool the surface the steel shell inner part must be
cooler, making outer even more colder.
A meter length of vacuum is good insulation as is 1 mm or 1 km
length of vacuum.

I would say the basic problem is a planet can not be stable
with radioactive decay of 235 W/m2.
Or one will have times where entire surface will re-surface
with hot lava. So the surface is either in process of heating up
or cooling down.

37. mkelly says:

Tim Folkerts says: March 10, 2013 at 9:34 pm

Sir the larger surface area of the inner shell will have less w/m2 thus a lower temperature. Heat will flow from the sphere to inner shell to outer shell to space. And because the inner shell is lower in temperature it cannot transfer heat back to the sphere. If I am wrong please write a radiative heat transfer equation showing me. Tsurf is higher than Tinnershell so heat only travels in one direction. Engineering heat transfer book tells me not to ignore things like a difference in area. No matter how many significant digits.

38. michael hart says:

“Michael: Time for a quick skiing holiday before global warming sets in. 🙂

Then we can do the wall of death on the inner shell surface FREEEEEEEOOOWWWWWWWWWWWWWW! 🙂 ”

Is the Wall-of-Death you were referring to, TB
http://t3.gstatic.com/images?q=tbn:ANd9GcQFRYTOt1bANLUV8ebhEy9SQFxS620Ue6SheU0vuu3m47pXeQ_51Q

I used to walk past it regularly. The construction work in the background looks like one more wannabe IPCC climate models, only more succesful.

[Reply] I’ve seen these guys (and girl) in action. Awesome: http://www.wall-of-death.co.uk/index06.html

39. Tim Folkerts says:

Tallbloke — Certainly albedo is important, and clouds do indeed change that. The discussion was about cloud cover at night, so in that setting, the back-radiation from the clouds and the back-radiation from the metal shell both serve the same purpose — to provide additional energy to the surface to help keep it warm.

Curious George — this is indeed a simplification. But it is an important simplification. You consider it “hardly surprising”, but in fact many people like Postma seem to think it is completely wrong. The point of such a simple model is to highlight one simple fact — that any sort of shell that blocks (and therefore also radiates) IR radiation (whether it is a steel shell or a glass shell or clouds or GHGs) will have the effect of making the surface warmer than it would have been without the “back-radiation” from that shell.

Max — the shell cannot “reduce the energy lost by the surface”. Once a steady-state condition is reached, the shell will by definition neither gain nor lose energy independent of the presence or absence of a shell around it.

“if the outer shell is radiating 235 out and 235 in, then it is still radiating twice what it absorbs.
No.

SURFACE:
+ 235 W/m^2 in from heater
+ 235 W/m^2 in from shell’s IR
-470 W/m^2 out as IR (toward the shell)
NET = 0 W/m^2

SHELL
+ 470 W/m^2 in from surface’s IR
-235 W/m^2 from outer surface (toward space)
-235 W.m^2 from inner surface (toward planet)
NET = 0 W/m^2

Both the planet and the shell are receiving 470 W/m^2 and losing 470 W/m^2.
** Energy is conserved.
** The 2nd Law of Thermodynamics is obeyed.

40. donald penman says:

thefordprefect
With regard to the experiment if it is to model the Earth should it not be a cooler gas that is used to keep the warm object from cooling.Your experiment has far too much density and radiation to give an accurate picture of the Earth.

41. Tim Folkerts says:

mkelly says ” Engineering heat transfer book tells me not to ignore things like a difference in area. No matter how many significant digits.
But engineers also know not to waste time designing and machining a part to +/- 0.01 mm when +/- 1 mm will do. An engineer obsessing about too many significant digits will be a failure.

” And because the inner shell is lower in temperature it cannot transfer heat back to the sphere.”
And no “heat” is being transferred is such a way. “Energy” is transferred both ways — 470 W/m^2 from the warm planet to the cool shell, and 235 W/m^2 from the cool shell to the warm planet. But the net flow of energy = “heat” is 235 W/m^2 from the warm planet to the cool surface.

The 2nd Law of Thermodynamics is safe! (Please take this paragraph to your local university and ask a physics professor to confirm this if you don’t want to take my word for it.)

42. tallbloke says:

Joseph E Postma says:
2013/03/10 at 6:19 PM
Willis’ system is defined exactly like a pocket handwarmer in a shell. A pocket handwarmer is continuously putting out chemical energy (and radiative thermal energy), and this energy can’t auto-generate higher temperatures within itself and do more work than it does in the first place.

tallbloke says:
2013/03/10 at 6:24 PM
Joe: Thank you, so you agree my handwarmer analogy is apropos after all. Good, this is progress.

Now. If you insulate the handwarmer while it continues to produce the same amount of energy, of course it will get hotter. Why wouldn’t it?

Get a handwarmer and try it for yourself. 😉

43. Chris M says:

Willarse Effenbotch

Sorry Roger, it just had to be said 😉 By all means delete if you wish …

44. Sparks says:

Rog, a theoretical black body does exist in nature, If you scale mass up until light can not escape from it, therefor IR can not escape, and if you add this steel shell around it there will be absolutely no feedback whatsoever. If you begin to decrees the mass there will begin to be a weak feed back, but not enough to increase the energy of the source until a point is reached where the energy from the source begins reaching the shell, It does not increase the potential energy output of the source but there will be a point where an equilibrium is reached with the shell. I think you, Willis and Joe are correct, the argument is around different understanding of this “black body”. 🙂

45. Max™ says:

Say the sphere has a 100 m radius and emits 235 W/m^2 as a result of the internal power supply.
Let’s put a shell around it which is 1 m above the surface, and let’s suppose that the outer surface of the shell emits 235 W/m^2.

(4*pi*100^2)*235=29,530,970.94
(4*pi*101^2)*235=30,124,543.46

That’s an increase of 593,572.52 Watts.

If the outer surface of the shell emits the same total power as the inner sphere, it could only emit 230.36 W/m^2.

_______

It is worse for the diagram Willis used, where the shell could easily have twice the radius of the inner sphere, 100 m for the sphere, 200 m for the shell.

4*pi*100^2=125,663 m^2, A*235=29,530,970.94
4*pi*200^2=502,654 m^2, A*235=118,123,883.77

In order for the outer surface of the shell to emit the same total energy as the sphere it could only emit 58.75 W/m^2!

_______

With a sphere the size of the Earth, 4*pi*6371^2 = 510,064,471 km^2 surface area, an 11 km high shell has a surface area of 511,827,320 km^2, a difference of 1,762,848 km^2, if the shell does not create energy then it can only emit 234.19 W/m^2.

So right off the bat Willis is creating energy by setting the emissions from the shell and the surface to be the same value.

If the shell emits at full power inwards AND outwards, and the emissions add energy to the surface, that doubles the total energy emitted by the shell, doesn’t it?

_________________

“Radiation from s surface is a function of the temperature and emissivity of the surface:
P = ε σ A T^4″
~Tim Folkerts

Hmmm, see that letter I bolded?

46. Truthseeker says:

Tim,

I am not sure I agree with you that the environment does not make a difference.

Let me try an analogy.

We have a negatively charged pool of electrons. Say 1000 units worth of them. Nothing is connected to it and it is perfectly insulated, so they cannot go anywhere. We then connect two seperate electron difficient objects by a low resistance conductor, one to the left which is 500 units positive and one to the right which is 5000 units positive. The two positive objects are not connected to each other.

Are you saying that we are going to get 500 units of electrons going in each direction?

Let me know if my analogy is not valid.

47. suricat says:

tallbloke says: March 10, 2013 at 10:34 pm

“I don’t think all the radiation will go outwards, but I don’t think it’ll be an even split either. I don’t know the equations for this. Anyone?”

That’s a ‘no brainer’ really TB. The ‘thermal capacity’ property of the ‘outer shell’ is left to the imagination of the ‘reader’.

Let’s get one thing straight. ‘Radiation’ is what you get when there is no ‘mass’ that can interact with it. ‘Heat’ is what you get (well, ‘often’) when ‘mass’ gets in the path of ‘radiation’ that it can ‘interact’ with.

If the ‘thermal capacity’ of the ‘interacting mass’ (shell) is ‘zero’ (impossible, I know), then there is ‘as good as’ no interaction. Thus, radiation just continues in (probably) another direction.

If the ‘thermal capacity’ of the ‘mass’ (shell) is ‘greater than zero’ (probable), then the ‘shell’ will act as an ‘averaging device’ between the ‘source’ (sphere’) and ‘sink’ (space). Thus, generates a ‘time lag’ between ‘the source and the sink’ (this is beginning to sound like a ‘semiconductor’ property). The ‘thermal capacity’ of the ‘shell’ is essential data to enable this scenario’s reconciliation.

More on the ‘semiconductor’ like properties. No temporal value is assigned here! There ‘must’ be a temporal value assigned for the ‘changing temperature values’. This would be ‘along the lines of’ “natural decay rate of energy transfer”, between ‘sink, source and the “shell” grid (vacuum tube [valve] technology)’. Iteration just doesn’t cut the mustard (perhaps this is too hi-tec for this thread).

The ‘reactance’ is really important. If I tried to sell an ‘OP-Amp’ on the ‘iteration’ rate of the overall programme it would be a doddle. No, what the ‘customer’ wants to know is the ‘slew rate’ of the OP-Amp that can prove its performance for its ‘duty cycle’ (the ‘duty cycle’ is usually a lot shorter time scale than the ‘iteration’ rate).

I’ll leave it there. 🙂

Best regards, Ray.

48. Greg House says:

tallbloke says, (2013/03/10 at 6:45 AM): “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.”
=============================================================

Hi Roger,

by saying “so the whole ensemble can reach equilibrium” you can only mean that there must be equilibrium “at the end of the day” also meaning that exactly 235 goes to space then, not any more.

Unfortunately, the process you described will not lead to this “equilibrium”, because as this radiation and re-radiation goes on again and again, the shell at a certain point will radiate to the outer space more energy, then the “inner engine” of the planet produces. And that is apparently impossible. Which means, that the process you described can not exist. The model is inherently wrong, this is the only logical conclusion. Some people call it “reductio ad absurdum”.

Since you repeatedly made references to the discussion on the same topic on the Joseph E Postma blog, I think it would be appropriate if I posted the explanation I gave there in my next comment here.

49. gbaikie says:

“tallbloke says:
March 11, 2013 at 12:38 am

Joseph E Postma says:
2013/03/10 at 6:19 PM
Willis’ system is defined exactly like a pocket handwarmer in a shell. A pocket handwarmer is continuously putting out chemical energy (and radiative thermal energy), and this energy can’t auto-generate higher temperatures within itself and do more work than it does in the first place.

tallbloke says:
2013/03/10 at 6:24 PM
Joe: Thank you, so you agree my handwarmer analogy is apropos after all. Good, this is progress.

Now. If you insulate the handwarmer while it continues to produce the same amount of energy, of course it will get hotter. Why wouldn’t it?

Get a handwarmer and try it for yourself. 😉 ”

Let’s see, apparently:
“Iron Powder
Iron + oxygen + water = rust. The chemical reaction that turns iron into rust creates heat and is the secret behind hand-warmers.”

So if wanted to make hand warmer hot, I would start by using pure oxygen.
But seem to designed to be insulated.
I imagine unless added something like pure oxygen they are designed not
to get very hot.
Perhaps if you rub your hand together, it might get higher
temperatures.

Another kind of handwarmer uses Sodium Acetate [and it’s reusable].
it’s starts to form crystals at around 50 C so that kind couldn’t get
very warm as process which makes heat would stop if it gets
hotter.
I making some Sodium Acetate right now [something to play with].
It’s made from baking soda and vinegar.

[Reply] I was thinking of the old fashioned dangerous one I used to have with the glowing carbon rod you set fire to and put in the asbestos fibre lined box. You can’t buy them any more… 🙂

50. donald penman says:

The direction of the arrows from the inner core to the inside of the outer shell and from the inside of the outer shell and the inner core are straight towards each other but any point on the inside of the outer shell could radiate anywhere within the radius of the the outer shell I could draw another line between one point on the inside of the outer sphere and another point on the inside of the outer sphere which would then mean that the outer sphere will get more radiation then inner core.

[Reply] Hans Jelbring tried that one to get a perpetual motion machine with the S-B equation here:
https://tallbloke.wordpress.com/2012/02/25/hans-jelbring-stefan-boltzmann-law-and-the-construction-of-a-perpetuum-mobile/

51. Tim Folkerts says:

It calculates all the numbers for the simple 1-shell model – including the effects of the heat capacity of the planet.

The columns are
A) time (1 second steps)
B) temperature of the planet’s surface = previous temperature + (NetQ)/C
C) nuclear power in (set to a constant 235 W/m^2)
(ie using Q = mc ΔT for each square meter each second)
D) IR out from the planet using SB equation (used to find Column I)
E) IR in from the shell using SB (from Column K)
F) Net power into the planet. If this is positive, the planet warms; if it is negative, the planet cools/

H) temperature of the shell = previous temperature + (NetQ)/C
I) IR in from the planet (the same as column D)
J) IR up from the shell to space using SB
K) IR down from shell using SB (used to find column E)
L) Net power into the shell.
At the top you can set the initial conditions.

Built into the calculations are conservation of energy and all the heat flows based on current temperatures using SB. In other words, it is simple fundamental physics that anyone with an understanding of physics could verify (or try to refute). It is a simple yet effective numerical solution to the question at hand.

This spreadsheet DOES assume that the shell is “close” to the planet. To the extent that this is not correct, there will be some correction that would cool the shell slightly. Roughly, I suspect that each 1% increase in the radius above the surface will result in ~ 2% decrease in power/m^2 and ~ 0.5% drop in the shell temperature (but that is only a rough estimate).

************************************************************************
The heater at the surface is set to 235 W/m^2

Using C(planet) = 200 and C(shell) = 20 works pretty well to illustrate what is happening and lets the system get pretty close to steady-state within 500 seconds. (Of course, these would be small systems, but the principle could be scaled up.)

The initial temperatures are set to 254K for the planet and 211K for the shell. You can make them pretty much anything you want (I would suggest staying in the range 0-350 K). You can even start with the shell warmer than the planet. In every case, the planet approaches 2*(heater power) as the upward IR (ie ~ 302 K) and the shell approaches (heater power) as the IR upward and the IR downward (ie ~253 K).

52. Tim Folkerts says:

Truthseeker.

I don’t think that is a good analogy. Electrons have charge that affects other elections. Thus the presence of other charges nearby DOES affect other charges. Photons do not have this effect — other photons nearby do NOT inhibit the creation or movement of photons.

53. I have not read all the comments but it appears many who are commenting are mentioning supposed models or thought bubbles without having any real experience. The only thing that counts is actual measurement and then relate that to equations, determined from real experience, which could explain the process. This comparison process tells you if your measurements are reasonably accurate or the equations your are attempting to verify are wrong.
I have had experience with measurement in furnaces. My experience is as follows
a) with a poorly insulated furnace there will be heat loss from the walls which results in a flame temperature less than the theoretical and lower energy in the exhaust gases.
b) as the insulation is improved there is less heat loss from the walls, slightly higher flame temperatures, and higher energy in the exhaust gases
c) even with perfect insulation it is not possible that the flame temperature exceeds that of the theoretical temperature ie energy can not be created to exceed the energy input

My experience and reading of text books and journal articles based on real measurements tells me that there is no back radiation which can increase temperatures. Insulation has the effect of reducing heat loss. A furnace needs to be insulated to protect the outer metal shell. In a large scale combustion process the insulation has only a small effect on flame temperatures. Very good insulation such as a thick layer of ceramic fibre allows the combustion process to come more rapidly to equilibrium.
I find Prof Claes views on radiation and cut-off frequencies interesting.He could be right. My own view was that heat transfer occurs in waves at particular frequencies depending on the source and that these waves can be cancelled like that for sound or actual experience making objects invisible to radiation eg the stealth bomber. It is possible to neutralise the IR of heat seeking or homing instruments on missiles.

I am sorry Roger, you lost me with this post, I have always thought this was a thoughtful blog with input of engineers who have some experience. I hope it gets back to real science based on ethical engineering principles.

[Reply] Ouch. I am covering in this post what theory says will happen with a vacuum between the shell and surface. That doesn’t pertain in your furnace, where convection will likely prevent re-radiation from adding to the output from the heating element due to re-absorption in the air. However, follow the link to ‘TheFordPrefects’ blog and study the empirical test he conducted, then report back.

54. Sera says:

Why don’t you just explain to Joe how a capacitor works?

[Reply] Maybe a resistor would be more apropos? Then I could explain why computers have fans. 🙂

55. I meant Prof Claes Johnson http://claesjohnson.blogspot.com.au/
Also, I should have mentioned that in a well insulated furnace the wall temperature is close to that of the flame (it is possible to estimate both). By experience and definition of equations if the wall temperature and flame temperature (or any two objects) are the same, there will be no heat transfer (2nd law of thermodynamics). I challenge anyone to prove with actual measurements that this is wrong. Thought bubbles are not a proof.

56. Greg House says:

Here is my explanation from the Joseph E Postma blog I promised to post here:
=========================

Hi Roger,

let me demonstrate how absurd the Willis’ construction is in a slightly different way.

You will probably agree that if the planet in his fictional story is the only source of energy, then it is impossible that more energy is radiated to the outer space than the planet produces (please, tell me, if you disagree with that).

Now, if the shell is of almost the same radius, it is also of the same area as the surface of the planet, so we can neglect the difference and consider them to have the same area. And let us take another number to make the calculation more understandable for everyone.

So, let us say, the planet’s “inner engine” produces constantly the same amount of energy so that in absence of any shell the planet’s surface radiates according to it’s temperature 800 to the outer space. Now comes the shell and works as you wish, that is (1) the planet warms the shell by radiation, the shell radiates then both to the planet and to the outer space, the a half to the planet and the other half to the outer space and (2) what the shell radiates to the planet is fully absorbed by the planet’s surface and then re-radiated back to the shell in addition to what the planet’s “inner engine” constantly produces. Let us focus now just on what the shell would sooner or later radiate to the outer space.

It goes like that, step by step.

1. Initially the shell radiates nothing, then it receives 800 from the planet and radiates a half of it to the outer space, so, the outer space receives 400.

2. The shell radiates the same 400 back to the planet and the planet thankfully absorbs it (and gets there warmer), but not being egoistic at all the planet re-radiates it to the shell in addition to what the planet’s “inner engine” constantly produces, so, now the planet radiates 400+800=1200. Note, 1200 goes to the shell now.

2. Well, the shell is doing what it is supposed to do and radiates a half of this 1200 back to the planet and another half to the outer space, that is the outer space receives 600 now.

3. The same 600 goes back to the planet’s surface, is absorbed there, makes the planet warmer again and as a result the planet radiates 600+1200=1800 to the shell.

4. The shell radiates a half of this 1800 to the outer space as usual, that means 900 goes to the outer space now.

Well, that’s it, actually. Given that the area of the shell is practically the same as the area of the planet and that the radiation power of what is radiated to the outer space has increased from initial 800 to 900, the result is that the outer space receives more energy from the system than the system produces. I hope you understand that it is impossible in the real world. Which means that you construction is absurd, despite it looking good on paper.

[Reply] Hi Greg. Your maths went wrong at step 3. The planet radiates the constant 800 you specified was generated within, plus the back-radiation value of half what the shell is emitting, not a cumulative value. So it is 1400 emitted from the planet surface towards the shell at the end of step 3, not 1800. This is a dynamic system not a static one. No radiation is ‘trapped’ forever. So using your initial value of 800 from the planet surface, the back-radiation increases like this:

(1) 400 +800=1200 –> 600 to space and 600 back-radiated =(2)
(2) 400+200=600 +800=1400 –> 700 to space and 700 back-radiated =(3)
(3) 400+200+100=700 +800=1500 –> 750 to space and 750 back-radiated =(4)
(4) 400+200+100+50=750 +800=1550 –> 775 to space and 775 back-radiated =(5)
(5) 400+200+100+50+25=775 +800=1575 –> 787.5 to space and 787.5 back-radiated =(6)
(6) 400+200+100+50+25+12.5=787.5 +800=1587.5 –> 793.75 to space and 793.75 back-radiated =(7)
(7) 400+200+100+50+25+12.5+6.25=793.75 +800=1593.75 –> 796.875 to space and 796.875 back-radiated =(8)
(8) …

Hopefully, you can see this converges towards 800 but will never exceed that. So the shell will never emit more to space than the planet core is producing. No perpetuum mobile, no free lunch, energy conserved.

57. tallbloke says:

To save time and space I’ve made some bold inline replies at the bottom of overnight (UK time) comments.

Just to repeat my reply to Greg house:

Hi Greg. Your maths went wrong at step 3. The planet radiates the constant 800 you specified was generated within, plus the back-radiation value of half what the shell is emitting, not a cumulative value. So it is 1400 emitted from the planet surface towards the shell at the end of step 3, not 1800. This is a dynamic system not a static one. No radiation is ‘trapped’ forever. So using your initial value of 800 from the planet surface, the back-radiation increases like this:

(1) 400 +800=1200 –> 600 to space and 600 back-radiated =(2)
(2) 400+200=600 +800=1400 –> 700 to space and 700 back-radiated =(3)
(3) 400+200+100=700 +800=1500 –> 750 to space and 750 back-radiated =(4)
(4) 400+200+100+50=750 +800=1550 –> 775 to space and 775 back-radiated =(5)
(5) 400+200+100+50+25=775 +800=1575 –> 787.5 to space and 787.5 back-radiated =(6)
(6) 400+200+100+50+25+12.5=787.5 +800=1587.5 –> 793.75 to space and 793.75 back-radiated =(7)
(7) 400+200+100+50+25+12.5+6.25=793.75 +800=1593.75 –> 796.875 to space and 796.875 back-radiated =(8)
(8) …

Hopefully, you can see this converges towards 800 but will never exceed that. So the shell will never emit more to space than the planet core is producing. No perpetuum mobile, no free lunch, energy conserved

58. paulinuk says:

Anyone who’s cooked a roast in an oven knows it cooks faster if the oven door is closed. In other words it’s hotter in the oven with the door closed, ie with extra insulation. The greenhouse effect is real and tallbloke and Willis are right.

59. Greg House says:

[Reply] Hi Greg. Your maths went wrong at step 3. The planet radiates the constant 800 generated within plus the back-radiation value of half what the shell is emitting. not a cumulative value.
========================================================

Roger, the only math error I admit is my false numeration of the steps, I made step2 twice. To avoid confusion I will call them 2a and 2b if I need to mention them.

OK, let us look at the step 3. You need to take into consideration that the planet has already become warmer after receiving the first portion of back radiation from the shell, namely 400 (let us ignore the issue if it physically proven to work this way for debate’s sake), otherwise it would still radiate the same 800 regardless back radiation and this not your hypothesis. The planet radiates therefore 1200 to the shell now. The next portion of back radiation from the shell would be therefore 600.

Now, your adding this 600 to 800 instead of to 1200 would only be justified, if after the planet radiated 1200 and before it received this 600 the planet’s temperature dropped to the initial one (800 corresponds to the initial temperature). So, if you go on like that, it would require a constant increase in temperature of the planet, then decrease, then increase again, then decrease again and so on, always a drop to the initial temperature. Because we know that the surface radiates according to its temperature. To radiate every time the initial 800 plus an amount equal to back radiation from the shell requires the temperature to drop to the initial level every time, there is no way around it.

I would say, if it works this way, it is certainly a new word in physics.

However, if it does not, my reductio ad absurdum is correct and your hypothesis incorrect.

60. Greg House says:

paulinuk says (March 11, 2013 at 8:23 am): “Anyone who’s cooked a roast in an oven knows it cooks faster if the oven door is closed. In other words it’s hotter in the oven with the door closed, ie with extra insulation. The greenhouse effect is real and tallbloke and Willis are right.”
========================================================

Is it possible for you to imagine that trapping hot air does prevent a decrease in temperature (but only if the outside air is colder), but trapping radiation has no effect on temperature?

I mean, these a 2 different things.

61. tallbloke says:

Also, your reply to Paulinuk is incorrect, no radiation is ‘trapped’. It’s a dynamic system. This error is why you get a cumulative number instead of the correct instantaneous number. The planet surface only ever emits what it receives. And it is only ever receiving 800 from within plus half the emission of the shell. It doesn’t ‘store’ anything else to add to it. You are the one proposing a free energy concept, not me.

62. Bryan says:

Willis choose an unfortunate material for the shell.

From the engineers handbook

Steel Oxidized 0.79
Steel Polished 0.07
Stainless Steel, weathered 0.85
Stainless Steel, polished 0.075
Stainless Steel, type 301 0.54 – 0.63
Steel Galvanized Old 0.88
Steel Galvanized New 0.23

Also since P = ε σ A T^4

Now emissivity ε is a lot less than one (1) then the temperature of the sphere must be higher than the temperature of the surface to emit 235W/m2 !!!

How does the second law of thermodynamics go again?

63. tallbloke says:

Bryan, quite so. We have to imagine the steel is blathered with buoyant buckets of Boltzmann Blackbody bullshit. 😉

64. Greg House says:

tallbloke says, (March 11, 2013 at 8:41 am): “Greg, … your reply to Paulinuk is incorrect, no radiation is ‘trapped’. It’s a dynamic system. This error is why you get a cumulative number instead of the correct instantaneous number. The planet surface only ever emits what it receives.
==============================================

I did not mean the planet/shell thing, it was a general answer to Paulinuk’s analogy.

65. tallbloke says:

Greg: The toy planet surface only ever emits what it receives. And it is only ever receiving 800 from within plus half the emission of the shell. It doesn’t ‘store’ anything ‘trapped’ to add to it. You are the one proposing a cumulative free energy concept, not me. Remember we are dealing with a toy planet concept. Perfect blackbodies have no heat capacity and emit everything they receive. It’s a million miles from the real Earth and this is why the arguments arise over the applicability of the greenhouse theory to the real planet we live on. Nonetheless, so far as the theoretical situation goes, Willis’ model is correct.

66. Bryan says:

In the model that Willis proposes the returning radiation from the shell is instantly added to the the radiation the planet emits to derive a new planet surface temperature.

This is wrong.

In reality the returning radiation from a thin shell will only be fully absorbed by a black body.
Lets accept that the massive planet (that’s what planets are) is such a body.

The absorbed radiation will potentially have to raise the temperature of the full mass of the planet..
Its like emptying a bucket of water in the Ocean and thinking that the water level will rise significantly.

Thermodynamics in deriving its laws makes use of the concept of a heat reservoir.
That is objects that have such a huge heat capacity that the addition (or subtraction) of small amounts of heat makes no difference to their temperature.
This is a reality check.

67. Greg House says:

tallbloke says, (March 11, 2013 at 9:15 am): “Greg: The planet surface only ever emits what it receives.”
==================================================

Roger, let us try a step by step approach, maybe we could come to agreement easier this way.

Do you agree that bodies radiate depending on their temperature?

If yes, do you agree that to radiate more the body has to get warmer?

If yes, do you agree that in your hypothetical process back radiation has been absorbed by the constantly warming planet thus increasing it’s temperature further?

If yes, do you agree that in that process the “new” temperature only depends on the amount of back radiation and the actual temperature the moment before that amount of back radiation is absorbed?

If yes, then it is incorrect to add back radiation constantly to the same 800 like you did, as if the temperature of the surface has not changed at all in the process of absorbing back radiation, and my reductio ad absurdum is correct.

The key point it, as you can possibly see, that you can not ignore changes in temperature inherent in your hypothetical process.

68. tallbloke says:

Greg: If yes, do you agree that in that process the “new” temperature only depends on the amount of back radiation and the actual temperature the moment before that amount of back radiation is absorbed?

I agree that at each iteration the planet will be radiating more from a higher temperature, and if you look at the figures I wrote out, you can see that it is at each step; plus a diminishing increase is added at each step.

If yes, then it is incorrect to add back radiation constantly to the same 800 like you did

No. The planet gets a constant 800 from within. It gets an increasing amount of back-radiation from without as the temperature of the suddenly added shell increases due to radiation from the toy planet surface. But this backradiation doesn’t accumulate, because the planet surface is a perfect blackbody with no heat capacity. So at any moment it is only radiating what it receives at that moment, which is 800 from below, plus half of the shell emission. But as the system moves towards equilibrium whereby the emission to the heat sink equals the production of heat within, the differential reduces and the rate of increase diminishes until it stops at equilibrium.

I’ll see if adding extra annotation to my sequence makes it any clearer.

69. Westy says:

So why not add as many more shells as needed to get all the energy we need from the core of this toy?

70. Ronaldo says:

The determination of temperature distribution can be made simply by applying the SB equation, using emissivities of 1 (this is after all a thought experiment).
The outer surface temperature of the shell Tsh is defined by the radiant loss of (heat) energy to deep space say zero K for simplicity.. At equilibrium this is j= sigTsh^4, where sig is Boltzmann’s constant. and j is the radiated energy.
The temperature of the surface of the sphere is determined by the flow of energy from its surface to the inner surface of the shell (assumed equal to the shell’s outer surface in this scenario).
Thus the temperature of the sphere’s surface at equilibrium is defined as Tsp where the flow of energy from it to the shell is calculated by j= Tsp^4-Tsh^4. and this supports Willis’ shell model in this scenario.
Note that temperatures are set by the flow of energy through the system and and are dependent only on energy flow and the temperature of the heat sink – in this case outer space.
This demonstrates the relatively low efficiency of radiation as a means of heat transfer which is why forced cooling via conduction and convection is needed to remove heat from eg. internal combustion engines.
Given that this model is a simple thought experiment it is ok for that purpose. Once the real world inserts an atmosphere and water vapour into the gap between the sphere and the shell and complicates the heat generation in the sphere by imposing radiation from outside and involves the massive thermal inertia of the world’s oceans and our old friend Coriolis, beware of simple conclusions!

71. Greg House says:

tallbloke says, (March 11, 2013 at 9:56 am):

Greg: If yes, do you agree that in that process the “new” temperature only depends on the amount of back radiation and the actual temperature the moment before that amount of back radiation is absorbed?

I agree that at each iteration the planet will be radiating more from a higher temperature, and if you look at the figures I wrote out, you can see that it is at each step; plus a diminishing increase is added at each step.

=============================================================

Roger, it is important that you give a clear answer to that particular question to come forward, because that is a crucial point. If your answer is “no”, that is, if you do not think that in that process the “new” temperature only depends on the amount of back radiation and the actual temperature the moment before that amount of back radiation is absorbed, then please say so, just to avoid misunderstanding. I mean, there can only be either yes or no, right?

OK, “I do not know” would be an answer, too, no problem, in this case we will find a way to precede. But if it is “no”, then I would like to ask you to explain it.

By the way, if you find this question-answer approach annoying, just tell me, I will try to find another way then.

72. Tanner says:

Tallbloke: There seems to be a great deal of emotion and technical discussion on several blogs over the Green House effect and radiation. I would “simply” like to know if the overall nett effect is negative or positive? Simplified as follows:
On a hot sunny day when the clouds come in, some of the heat is reflected by the clouds and the air temperature near the earths surface cools down. The clouds effectively provide “shade”. Then at night the clouds “trap” some of the heat near the surface and the air temperature is warmer than if there were no clouds.
The reflective effect seems to be greater than the green house effect in the day and vice versa at night (since there is no sunshine to reflect). An absence of cloud cover generally leads to a higher day time temperature and lower night time temperature and cloud cover leads to a lower day time temperature and higher night time temperature but is the average temperature higher or lower with cloud cover?
Does the overall nett green house effect versus reflective effect lead to higher or lower air temperatures near the surface?
Or does the nett effect depend on the time of day or night the cloud cover comes in, the extent of clouds, how long the clouds remain, summer, winter, winds altitude etc.?

Or are we going to just keep debating whether the effect exists or not?

73. Ronaldo says:

oops. I should have included sig (Boltzmann’s constant) into the second equation ie.

j=sig(Tsp^4-Tsh^4).

my apologies

74. paulinuk says:

Geg House says:
“Is it possible for you to imagine that trapping hot air does prevent a decrease in temperature (but only if the outside air is colder), but trapping radiation has no effect on temperature?

I mean, these a 2 different things.”

Well yes, remove all the air and see what happens.

75. tallbloke says:

Greg, the problem is that you are double counting. You are trying to have the 800+400 and then add the 400 a second time along with the extra 200. You can’t have it both ways, because in a blackbody system, it’s an instantaneous – speed of light across a vacuum – response.

Tanner, you are talking about cloud feedback, not the greenhouse effect. The models say it is positive, i.e. warming. The empirical data from satellites says it’s negative overall. At different times and places, it can be one or the other, as you correctly summarized.

76. Tanner says:

Tallbloke: Thanks – much appreciated.

77. wayne says:

TB & GH, you are both making a critical mistake in your calculations it seems. You never accumulate any excess energy on any surface, probably because Willis’s mind twister is using a super-conducting shell, infinite conductivity, so it’s in all essence is invisible to radiation and has no defined temperature.

I view the examples, both of yours in 1/100th of a second step, not every second. When you say it sends 800 w/m² I see 8 going out, the shell sending 4 back and to space instead of 400, faster time steps. That way at least the temperature of the sphere’s surface doesn’t bobble to 0K every tick. When a surface sends some energy it cools, the other warms… but it sends it right back cooling… so the other can warm… back and forth and NOTHING IS CHANGING, at all! The temperatures are never changing.

Maybe a rethink is in order.

Tallbloke, thanks for posting this topic one more time, just like the SSB this is a long hanging unanswered question and maybe we can finally get to the bottom of this one too. Some will get pie in the face, myself for sure… I’m still straddling in the line of fire no matter what ends up being correct! 😉

78. Greg House says:

Roger, so, no answer, neither “yes” nor “no”. OK, no problem, I will not insist.

[Reply] Greg, just trying to get the groundwork right. If you can’t take it on board, then we can’t get further. To try to play it your way, yes, the temperature instantly drops when emission occurs from a blackbody with no heat capacity. You can’t have the cake and eat it. You can’t lose energy and keep the temperature

79. wayne says:

How about deviating from Willis’s (once again) totally unreal example and give the shell 1cm thickness with 100 W, or something, for the heat capacity of one sq. meter, then at least the shell can have a calculable temperature and then able to know what it really should radiate at as a BB. Also, every time the shell emits its temperature would not immediately hit 0K once again.

80. Greg House says:

tallbloke says, (March 11, 2013 at 9:56 am): “The planet gets a constant 800 from within. It gets an increasing amount of back-radiation from without as the temperature of the suddenly added shell increases due to radiation from the toy planet surface. But this backradiation doesn’t accumulate, because the planet surface is a perfect blackbody with no heat capacity. So at any moment it is only radiating what it receives at that moment, which is 800 from below, plus half of the shell emission. But as the system moves towards equilibrium whereby the emission to the heat sink equals the production of heat within, the differential reduces and the rate of increase diminishes until it stops at equilibrium.
==========================================================

To your “as the system moves towards equilibrium”, the problem is that your hypothetical system does not, if my calculations are correct. You actually need to prove that your hypothetical system “moves towards equilibrium”. Until you have proved it in frame of your hypothesis it is not a fact.

To “but this backradiation doesn’t accumulate”, I am not saying it does. What I am saying is that you can not ignore changes in temperature inherent in your hypothetical cyclical process. And you see, if we start considering changes in temperature instead of just making some radiation arithmetic ignoring the changes in temperature, the whole hypothetical cyclical process leads to more energy received by the outer space than is produced by the system planet-shall.

I fully understand that your arithmetic is OK if changes in temperature are ignored, so if I point out to changes in temperature and their consequences like increase in radiation accordingly, but you ignore it, you can never come to the right conclusion.

In other words, you ignore an inherent part in your hypothetical cyclical process that makes this process physically impossible, therefore this process still looks possible to you. This is the key point.

81. donald penman says:

Tallbloke it is not quite the same there no no planetery nuclear core in this example.
https://tallbloke.wordpress.com/2012/02/25/hans-jelbring-stefan-boltzmann-law-and-the-construction-of-a-perpetuum-mobile/
I see the energy from the nuclear core driving the enhanced warming of the outer shell and hence increasing heat loss from the outer shell,if the inner core was not warmer than the outside shell this would not happen and when they are the same temperature there is equilibrium.

82. Greg House says:

Greg House says, (March 11, 2013 at 10:38 am): “yes, the temperature instantly drops when emission occurs from a blackbody with no heat capacity.”
=====================================================

Thank you, Roger. So, the temperature drops to what and depending on what, in your understanding, if I may ask?

83. Tim Folkerts says:

Several of the comments relate directly or indirectly to the heat capacities of the planet and the shell. This gets tricky because …
* the heat capacity of the shell is assumed to be small so that it very quickly comes to a steady-state condition.
* the heat capacity of the planet is assumed to be very large but not infinite so that the planet slowly changes temperature.

In the limit that Cp(shell) goes to zero, then we get Tallbloke’s infinite series.
In the limit that Cp(planet) goes to infinity, we get Bryan’s “thermal reservoir” where the temperature of the planet never changes no matter what we do.

This also addresses Greg’s step: “If yes, do you agree that in that process the “new” temperature only depends on the amount of back radiation and the actual temperature the moment before that amount of back radiation is absorbed?”
The new temperature depends on the heat capacity as well.
* In the limit that Cp(planet) goes to zero, then the previous temperature doesn’t matter! The planet instantly changes to the new steady-state temperature regardless of any previosu temperature.
* In the limit that Cp(planet) goes to infinity, then the previous temperature matters, BUT the back radiation doesn’t because the temperature can’t change!
* Only when you can conceptually treat Cp as large but not infinite will we get the proper behavior.

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

My earlier spreadsheet tries to explicitly deal with those heat capacities. You can change the values for the heat capacities and see where the numbers lead. (Don’t go below about “10” units of Cp or the crude numerical integration can become unstable.) With a small Cp(shell) the result approach the numbers tallbloke has for the power radiating in various directions. ie, that the radiation down from the shell is 0.5*the radiation up from the surface throughout the process, and the radiation from the surface approached twice the power of the heater by it self.

84. Greg House says:

paulinuk says, (March 11, 2013 at 10:29 am): “Well yes, remove all the air and see what happens.”
==================================================

To find out if trapped radiation has an effect on temperature or not it is not necessary to remove air, it goes by comparison between 2 states, radiation trapped versus radiation not trapped, other conditions being equal.

85. tallbloke says:

Greg, too much too fast. I’m at work now, so it’ll have to wait, sorry.
Tim Folkerts can probably answer you better than I can anyway.

86. tallbloke says:

While this thread is hot I’ll just mention that voting for the bloggies is still open until March 17.

The ‘Finalist’ graphic top left is linked straight through for your convenience if you’d like to support us. 🙂

87. Ronaldo says:

The first law – energy can neither be created nor destroyed – applies!. The description of the behaviour of the simple sphere and shell with a constant energy generation (note generation not creation – the process of heating the sphere is one of energy conversion) can be expanded to allow for emissivities less than unity, thermal capacity and thermal conduction within the sphere and shell and as many shells as you like. The temperature of the sphere surface will simply reflect the various barriers to heat flow viz conduction within the sphere (for a given uniform heat generation rate the centre temperature of the sphere will be greater than that at ita surface), radiation across the sphere/ shell surface, conduction through the shell and radiation from the shell surface to outer space. Add another shell and the same principles apply.
Thus for a given energy flow from the sphere, its surface temperature is governed by the heat transfer processes between it and the final heat sink.
Thermal conduction will determine the temperature rise from outer shell surface to inner shell surface and radiatiative transfer will determine the shell outer surface temperature to space and the sphere surface temperature needed to drive this energy flow through space to the inner surface of the shell.

88. A C Osborn says:

Tallbloke,in your responses to Greg about the back Radiation adding to the Ball’s Radiation, how does that work exactly, I thought Radiation was dependent on temperature, ie excitement of atomic particles. To change from 800 to 400 +800=1200 what happens to the radiation, does the frequency change, does the photon flow change?
If so how does it do this without changing the original temperature of the ball?
After all the warmists insist that the Back Radiation warms the SURFACE of the earth, so willis’s example doesn’t do that, but still has a higher photon flow or higher frequency radiation.
This is extremely confusing.

89. Max™ says:

Willis stated that the error introduced by the disparity in surface area can be ignored.

If the sphere is the size of the Earth and the shell is 11 km above it, having the sphere emit 235 W/m^2 and the shell emit 235 W/m^2 means the shell is emitting 1.003 times as much as the sphere.

This becomes worse the greater the radius of the shell is in comparison to the sphere, but at no point is it ever just a trivial “oh that’s unimportant” sort of situation.

1.1986515×10^14 Watts is not “basically 1.2027942×10^14 Watts”, we’re talking about 400 Gigawatts of “free energy” being created here just from the initial conditions proposed.

That alone invalidates the thought experiment before anything else is even considered.

There can be no radiative equilibrium where a smaller inner sphere emits 235 W/m^2 and a larger shell absorbs that and then re-emits 235 W/m^2 too, that is producing energy from nowhere.
__________

Radiation from s surface is a function of the temperature and emissivity of the surface:
P = ε σ A T^4
″ ~Tim Folkerts

The A isn’t something that can be ignored.

90. Tallbloke, nothing new in Fordperfects link. Engineers found, just after the Stefan-Boltzman equation was formulated that for radiation heat transfer it is necessary to include emissivity factors and temperature of the receiver as well as the temperature of the source. Further in certain surrounds and heat exchange equipment it is necessary to have shape factors.

The experiment at that link indicates similar situation what I put in my post ie insulation reduces heat loss but can not increase the original energy (or original temperature) of the source.

The experiment says nothing about the supposed back radiation which does not exist.
Prof Claes Johnson’s proposals and calculations about threshold and cut-off frequencies can still be correct. Also, my view of energy wave cancellation.is not negated.
In the past I have linked to articles (including some from a Nobel prize Physicist) which have denied the existence mass-less photons. I am much more inclined to the concept of energy waves (x-rays, light waves, microwaves, sound waves etc). Prof Claes Johnson has been developing calculations, (eg http://claesjohnson.blogspot.com.au/2010/07/mathematics-of-blackbody-radiation.html) along those lines, which make a lot of sense. A specific photon at 14.8 micron wavelength or any other specific wavelength makes no sense.

91. tallbloke says:

Max: 400 Gigawatts of “free energy” being created here just from the initial conditions proposed.

Sounds like a big number. There again, the US generated 1.6×10^19 Watts of electricity supply in 2011.

What is it as a percentage of the total number of petawatts involved?

Earlier Max said: Say the sphere has a 100 m radius and emits 235 W/m^2 as a result of the internal power supply.
Let’s put a shell around it which is 1 m above the surface,

This turned out to make a difference of around 2%, but the ratio of surface to shell compared to sphere radius was 1:100. On the real Earth the ratio of surface to tropopause and planetary radius is more like 1:650, with a correspondingly much smaller percentage difference in temperature.

92. mkelly says:

Tim Folkerts says:

March 11, 2013 at 12:17 am
(Please take this paragraph to your local university and ask a physics professor to confirm this if you don’t want to take my word for it.)

Mr. Folkerts since you failed to present a requested radiative heat transfer equation to show that I was incorrect I will accept that I was correct and you are incorrect. When Tsurf = Tinnershell all heat transfer goes to zero and no heat is transferred back to the surface, but as the area of the inner shell is more than the surface the temperatures will not ever be equal.

As for your snide comment I find it offensive. I addressed you with respect and expect the same in return. [snip]

As others have commented the same ideas as I have as to AREA etc matters in this steel shell post it may be it is you who is not understanding.

[Reply] Tim accurately calculated the area issue to 4 sig figs earlier in the thread. Maybe it is you who is not keeping up. If you want respect, offer it too.

93. Roger Clague says:

Willis’s model does not define the mass of the planet or the shell.
If an object has no mass it cannot radiate. If its mass is large then the temperature will not change.

This is a general problem with Energy Budget Theory. Its all about radiation. Radiation is equated with heat. Mass is left out

http://earthobservatory.nasa.gov/Features/EnergyBalance/

Radiation from the sun is absorbed by matter. It becomes heat ( motion of molecules of matter ) in the atmosphere, then radiation again and then leaves.

The study of heat is thermodynamics.

94. tallbloke says:

Roger C: I agree Willis did a slack job of defining the setup, that’s how cowboy builders are. Tim Folkerts has offered a spreadsheet where you can play with parameters.

95. Max™ says:

Sounds like a big number. There again, the US generated 1.6×10^19 Watts of electricity supply in 2011.

What is it as a percentage of the total number of petawatts involved?” ~tallbloke

It is around a 0.3% difference… but whether it is a large value or not, one has to ask a rather important question: where did that energy come from?

Shouldn’t it be less than or equal to the power supplied by the sphere?

96. Bryan says:

Lets give Willis the benefit of the doubt and do a more realistic calculation.
Instead of heating up the whole planet, let the mass heated up be restricted to just the top 5 metres.

P.t = C.m.deltaT

P = power of back radiation = 235W/m2
t = time taken to increase surface temperature by one degree
C = average specific heat of crust material = 2000J/KgK
m = mass of 5cubic metres of crust = 3000×5 Kg
T = change in temperature = one Kelvin

235.t = 3000.5.2000.1

=> t = 36 hours.

I will leave it as an exercise to calculate the time taken for a thin metal shell to drop by one degree kelvin.

So in reality there is no instant doubling of temperature of the planet surface.

The thin shell loses heat faster than the planet surface rise of even a mere one degree rise in planet temperature.

97. Max™ says:

I’ll add an explanation which I guess I forgot to include: the same amount of power emitted by a sphere with a given surface area will necessarily produce a lower temperature when it is absorbed by a shell with a greater surface area than the sphere.

P = ε σ Area (T_hot^4 – T_cold^4) will give the right answer [snip]

[Moderation note] Innuendo removed.

98. Max™ says:

If the surface emits Area*235 W/m^2 to the shell which also emits Area*235 W/m^2, the shell has a greater surface area, so it is emitting more energy than the sphere did.

Where does the extra 0.3% for an Earth sized sphere+11 km high shell come from?

That isn’t a trivial error, is it?

[Reply] Not sure yet. Tim F has it under 0.1% so we have some figures to reconcile. In any case, it doesn’t make a huge difference to the basic theoretical argument.

99. oldbrew says:

cementafriend says: ‘The experiment at that link indicates similar situation what I put in my post ie insulation reduces heat loss but can not increase the original energy (or original temperature) of the source.’

100. tallbloke says:

CF: The experiment says nothing about the supposed back radiation which does not exist.

It’s ‘back-radiation’ which is slowing down the cooling of the primary heat source in the experiment isn’t it?

101. wayne says:

Tallbloke, sorry I questioned you. I wrote a small integration program before hitting the sack, updating all variables at one single instance, so time could no longer even be in question, and I thought it would still stall, not so, as Tim said:

“In the limit that Cp(shell) goes to zero, then we get Tallbloke’s infinite series.”

that is exactly what you get, in the end after many 1/100th second time slices the sphere ends up at 1600, the shell is 1600, and to space is 800 as it was without the shell. It’s going to be even harder now for someone to prove to me that this is not what does happen. And that is starting the shell at zero energy, 0K.

So, seems you were precisely correct to begin with.

(code available, 43 lines, if anyone wants to see it in action for themselves… like show me, I’m much that way)

102. Tim Folkerts says:

Suppose two different spheres emit the same total power as thermal IR. What will be the temperature ratios of the two spheres?

P_1 = P_2
ε_1 σ A_1 T_1^4 = ε_2 σ A_2 T_2^4 (use SB)
ε_1 σ (4π(r_1)^2) T_1^4 = ε_2 σ (4π(r_2)^2) T_2^4 (area of a sphere)
ε_1 (r_1)^2 T_1^4 = ε_2 (r_2)^2 T_2^4 (cancel)

(T_2/T_1)^4 = [ε_1 (r_1)^2 ] / [ε_2 (r_2)^2 ] (rearrange)

(T_2/T_1) = (ε_1/ε_2)^(1/4) * (r_1/r_2)^(1/2) (simplify)

We have been assuming blackbodies, so ε_1 = ε_2 = 1. Then
(T_2/T_1) = (r_1/r_2)^(1/2)

For “realistic” numbers, r_1 = 6370 km and r_2 = 6380 km, so
T_2/T_1 = (6370/6380)^0.5 = 0.9992
Or T_2 = 0.9992*255K = 254.8 K

So the “radius effect” provides well under 1K of cooling in this model. The “IR radiation effect” provides about 303K-255K = 48 K of warming. To first order, we can ignore the “radius effect”.. (Furthermore, we can model the shell as getting closer and closer to the surface, in which case the “radius effect” goes to zero, but the “IR radiation effect” remains 48K.)

If you want to have a “large shell” around the planet, then the effect becomes larger of course. If r_2 = 2 r_1, then the ratio would be 0.707, and the shell would cool to 255K *0.707 = 180 K. Now the “radius effect is bigger than the “IR radiation effect”.

103. tallbloke says:

Wayne: Your coding ability and thoroughness never fail to impress me. Thanks for checking it through. As I’ve been saying all along, this is how it is for the toy model. It has very little to do with the real Earth, for many and several reasons.

104. Greg House says:

tallbloke says, (March 11, 2013 at 4:16 pm): “It’s ‘back-radiation’ which is slowing down the cooling of the primary heat source in the experiment isn’t it?”
=================================================================

Roger, do you agree with this statement: “There has never been any scientific experimental proof that back radiation affects the temperature of the source to any not negligible extent”?

If you do not agree, please, name that experiment, except this one from 10 days ago.

Because, if there has been none despite all the claims, the only reasonable explanation would be that it is apparently impossible to prove. Therefore this new one does not seem credible. Everyone can simply make such a thing up.

105. mkelly says:

mkelly says:

March 11, 2013 at 2:17 pm
[Reply] Tim accurately calculated the area issue to 4 sig figs earlier in the thread. Maybe it is you who is not keeping up. If you want respect, offer it too.

Mod my post that he had the snide comment on started with “Sir…” since when is Sir a disrepectful ten. My request was stated politely and he failed to fill it. Before you choose sides maybe you should read the original post.

[Reply] I’ll tell which side I’m on Mr Kelly. I’m on the side of polite well reasoned debate free from snark snide innuendo and other time wasting asides. It’s not an easy job policing it, because everyone wants to slip one under the wire. It requires goodwill and generosity of spirit on all sides, so please show some. I have a lot of interesting stuff on my plate, and I’d rather not have to deal with inconsequential trivia as well. Thank you. TB.

106. wayne says:

“As I’ve been saying all along, this is how it is for the toy model. It has very little to do with the real Earth, for many and several reasons.”

TB, how very true. I did not perform detailed accounting of the physical joules in play to make sure there is not actually more physical energy flying about than has even been created by the nuclear source in the first place. That might be a further check to perform just to verify that all is ok.

Also… still not 100% sure that thermalization of this “back-radiation” physically occurs yet, but, I do agree with the slowing of radiation from the warmer surface. It seems that this fine point carries its own implications along with it. For if, at the warmer surface, this absorption and thermalization does occur, that energy just become part of the heat capacity and further radiation of that particular exact energy bundle from the back-radiation then exits not at the same frequency, but spread across the entire gray body spectrum once again. It just seems to me that this picky point does matter. I could swear I have seen spectrums looking down on the surface that do already show a very small divot in the co2 lines (a major component of the b.r.) and a reduction in the h2o lines and that implies that literal absorption and thermalization is not occurring in those lines and are just being suppressed (the slowing of the cooling). That might be the point that Joe Postma is pressing on.

107. mkelly says:

Tim Folkerts says:

March 11, 2013 at 4:28 pm

For “realistic” numbers, r_1 = 6370 km and r_2 = 6380 km, so
T_2/T_1 = (6370/6380)^0.5 = 0.9992
Or T_2 = 0.9992*255K = 254.8 K

So you now admit that the shell has a lower temperature than the surface of the sphere. Further, since heat can only flow from higher to lower temperature there is no way for the sphere to go to a higher temperature via standard radiation heat transfer equations.

Thank you for that.

Please show the equations that use the “IR radiation effect” you speak of the raise the temperature of the sphere.

108. tallbloke says:

Wayne said: still not 100% sure that thermalization of this “back-radiation” physically occurs yet, but, I do agree with the slowing of radiation from the warmer surface. It seems that this fine point carries its own implications along with it.

Yeah. Theory is one thing, and reality another. My magnifying glass won’t give me the answer to this one, and IR experts seem to be thin on the ground, and play the wise monkey more often than not. I guess we’ll figure it out eventually. See if you can find those plots you remember.

109. tallbloke says:

Greg House says: Roger, do you agree with this statement: “There has never been any scientific experimental proof that back radiation affects the temperature of the source to any not negligible extent”?

If you do not agree, please, name that experiment, except this one from 10 days ago.

Still trying the cross examination technique Greg? You may or may not be a lawyer, but I know for sure I’m not in the dock or the witness stand. 🙂

Because, if there has been none despite all the claims, the only reasonable explanation would be that it is apparently impossible to prove. Therefore this new one does not seem credible. Everyone can simply make such a thing up.

Do you know, it’s a funny thing, but Joe Postma accused me of ‘making stuff up’ on his blog last night too. Since then, a coder I trust has replicated my theoretical finding and confirmed it. In my opinion, before making such statements, the onus is on the person readying themselves to make the accusation to attempt replication of the experiment or theoretical method themselves and make the results available, like the person they are casting aspersions on has.

110. A C Osborn says:

Greg House says:
March 11, 2013 at 5:09 pm Therefore this new one does not seem credible
Greg, I am sure that it is credible, the only thing that adding the lower temperature Plate does is increase the Ambient Temperature between the 2 plates with it’s added warmth, note the 6 degree higher temperature of the back wall.
So the heat loss to the surrounding air has been slowed down because the temp differential has changed. Thus the whole area is slower to cool.
TFP says that if the second plate is at room temperature the heat loss will be the same as the no second plate version.

111. Roger Clague says:

Tim Folkerts says:
March 11, 2013 at 4:28 pm

You show that the radius effect is less than 1%. This has been done already. So you do not move the debate forward.

You move the debate back by asserting

The “IR radiation effect” provides about 303K-255K = 48 K of warming.

I and others do not accept that warming is an IR radiation effect. And that the warming is 48K.

Your model like that of Willis, Trenbirth and Nasa has no mass in it.

What is thermal IR?

IR is not thermal. Heat is thermal, radiation is not heat.
Radiation changes to heat when it is absorbed by matter.
Hot matter emits radiation, not heat

112. tallbloke says:

Roger C: I and others do not accept that warming is an IR radiation effect. And that the warming is 48K.

Quite right too. But please try to remember that what is under discussion here is a toy model with a vacuum between the ‘planet’ surface and a thin blackbody rigid shell a small distance above it, not the real Earth and it’s atmosphere. I would prefer it if Tim F used initial values entirely different to Earth’s to help prevent conflation of the toy planet with Earth.

113. Tim Folkerts says:

mkelly says: “So you now admit that the shell has a lower temperature than the surface of the sphere.
Of course. The surface of the shell is cooler than the surface of the planet. The surface of a big shell is (slightly) cooler than the surface of a small shell. I don’t think I ever argued against such things.

Much of the “art” of being a scientist or engineer is knowing which factors are important and which are not. The “radius cooling effect” is 1) small and 2) can be made arbitrarily small by bring the shell closer and closer to the surface. OTOH, the “radiative warming effect” exists for any radius and is much larger for any “reasonable” radii we would want to consider.

” Further, since heat can only flow from higher to lower temperature there is no way for the sphere to go to a higher temperature via standard radiation heat transfer equations.
Higher temperature than what?
The answer is that there is no way for the sphere to go to a higher temperature than the nuclear heater (or what ever other heater is involved) that is warming the surface. But the nuclear heater is postulated to provide 235 W/m^2 independent of the current temperature, so it can reach any arbitrarily high temperature (which is an approximation that would break down at high temperatures, but is fine for “low” temperatures around 300 K — there is that “art of knowing what is important” again).

Sunlight can be considered to be @ T = ~ 5780 K, so sunlight cannot — even in principle — heat anything above that temperature. Fortunately, the earth never gets anywhere near this warm, so we never have to worry about this limit.

114. J Martin says:

Whilst Anthony Watts comment had a ‘sarc’ tag at the end, it is in fact a perfectly feasible experiment that could be built and put into space and tested / measured.

115. Arfur Bryant says:

I hope no-one minds if I ask two genuine questions:

Can radiation from a cooler body/surface be absorbed (for net gain, as opposed to being instantly re-emitted) by a warmer body/surface)?

Am I correct in thinking that radiation needs to be absorbed in order for temperature to be increased?

Thanks in anticipation.

116. Max™ says:

[Reply] Not sure yet. Tim F has it under 0.1% so we have some figures to reconcile. In any case, it doesn’t make a huge difference to the basic theoretical argument.” ~tallbloke

Ok, I’m not sure why this doesn’t make a huge difference to the basic theoretical argument: it is still emitting more energy than it receives, and this is just from considering the outer surface emissions alone.

When you propose that the radiation emitted inwards is added to the energy leaving the surface (rather than subtracted from it, since it is a vector in a different direction), you now have doubled the discrepancy, and the presence of the shell means it is emitting twice as much energy as the entire surface of the sphere and then some.

This is not just an issue with math or error bars, this is encoded in the basic assumptions of the thought experiment. It is an issue of geometry.

1-D model: surface and shell have the same area
2-D model: shell has at least twice the area of the surface (as it emits up and down)
3-D model: shell always has more than twice the area of the surface (as it emits up and down, and must be larger than the surface to absorb all radiation)

Tim F:
P_1 = P_2
ε_1 σ A_1 T_1^4 = ε_2 σ A_2 T_2^4 (use SB)
ε_1 σ (4π(r_1)^2) T_1^4 = ε_2 σ (4π(r_2)^2) T_2^4 (area of a sphere)
ε_1 (r_1)^2 T_1^4 = ε_2 (r_2)^2 T_2^4 (cancel)

Ok, see what you did there?

You’re canceling the area down to a difference of radius.

The ratio between the area is 0.9965(55775552), the ratio between the radius is 0.9982(76402382), using 6371 and 6382 respectively.

(T_2/T_1)^4 = [ε_1 (r_1)^2 ] / [ε_2 (r_2)^2 ] (rearrange)

(T_2/T_1) = (ε_1/ε_2)^(1/4) * (r_1/r_2)^(1/2) (simplify)

We have been assuming blackbodies, so ε_1 = ε_2 = 1. Then
(T_2/T_1) = (r_1/r_2)^(1/2)

For “realistic” numbers, r_1 = 6370 km and r_2 = 6380 km, so
T_2/T_1 = (6370/6380)^0.5 = 0.9992
Or T_2 = 0.9992*255K = 254.8 K

Again, if you use the ratio of the area you get 255 K and 254.1 K, the black body temperatures for those values are 239.764 W/m^2 and 236.397 W/m^2.

In the case the sphere emits a total of 122,295,098,043 Watts while the shell would emit 120,994,442,985 Watts, the ratio of those is 0.9893(64618213).

________

None of that really matters though, I just felt it should be pointed out what was wrong with the manipulations Tim did with those values.

Canceling the 4*pi from the black body calculations produces an incorrect value, but to his credit, it is incorrect in the right direction, so at least Tim’s shell isn’t emitting more radiation than the sphere, as it is in the original thought experiment.

117. Bryan says:

As soon as real values of emissivity,mass and specific heat capacity are used it becomes clear that the back radiation from a thin shell cannot even raise the temperature of the surface by one degree kelvin .

It should also be noted that a radiative flux of 235W/m2 is indicative of a spectrum well into long wave infrared.
This area is well inside the radiative limits where the classical formula of Raleigh – Jeans still gives accurate results.

So no photons or their backradiation need be specified for accurate tracking of heat transfer.
You can of course still use a quantum mechanical approach with photons but the heat transfer results must be identical to those of Raleigh – Jeans.

The problem arises when misuse of the Stephan Boltzmann equation and black body radiative physics is misapplied.
This was clearly shown by the results of the Mariner mission to the Moon.

118. gbaikie says:

“Roger Clague says:
March 11, 2013 at 2:31 pm

Willis’s model does not define the mass of the planet or the shell.
If an object has no mass it cannot radiate. If its mass is large then the temperature will not change.

This is a general problem with Energy Budget Theory. Its all about radiation. Radiation is equated with heat. Mass is left out

http://earthobservatory.nasa.gov/Features/EnergyBalance/

Radiation from the sun is absorbed by matter. It becomes heat ( motion of molecules of matter ) in the atmosphere, then radiation again and then leaves.

The study of heat is thermodynamics.”

The temperature of large mass does not change much. And this due relatively small amount
surface area compared to large amount of heat stored in a larger volume.

And what is important is the speed that a large mass store heat- which is generally the conductivity
of heat through matter.

One way to look at it, is that to convert radiant energy into motion of matter, one needs some kind machine or mechanism. With Life this done primarily by converting sunlight into chemical energy
which can then used it to power movement, e.g. fins, legs, or wings use chemical energy to move.

I maintain that gases do not gain kinetic energy directly from radiant energy, but rather that matter with a structure [solids and liquids] convert radiant energy into energy which can cause gas molecules to move.
And that therefore atmospheres are also large volumes with low surface area which are capable of transferring heat out of the gaseous mass. Solids and liquids can increase the kinetic energy of gases and gases can increase the heat of liquids and solids with their kinetic energy.
Surfaces transfer energy in form of kinetic to gas molecules when warmer than the gas [and add
gas by evaporating] and gases convert their kinetic motion into heating liquids and solids when they are warmer than these surfaces.

The power or energy of radiant energy is it’s intensity [how much going in one direction per square meter] and how energetic the photons are [shorter wavelength photons have more energy].
A surface can emit more photons in “one direction” [a gas is not a surface].

A significant aspect of climate science should be the understanding that water is transparent
to the Sun’s radiant energy allowing this energy to be quickly conducted below the surface and that this energy converted into transfer back to the surface at a far slower rate.

Another aspect is a warmer surface conducts heat thru the solid at faster rate than a less warm
surface. So water [because it’s transparent to sunlight] and higher temperature surfaces such rock or sand transfer heat quicker into the substance than the heat leaves the warmed substance.

119. Kristian says:

Tallbloke,

I tend to agree with Joe Postma. There seems to be a fundamental misunderstanding here regarding how heat is transferred from the warm object to the cold.

Willis’ nuclear planet and its surrounding shell is a thermodynamic system. The nuclear planet is a black body. It will always emit exactly the same flux as it receives. It cannot emit less. It cannot emit more.

What is this thermodynamic system trying to achieve? Thermodynamic equilibrium. Maximum entropy. How is this to be accomplished? By equalizing the temperature of the core and the shell.

At the onset, the black body surface of the nuclear planet emits a net flux of energy of 235 W/m^2 towards the surrounding shell, holding a temperature of ~254K. That means, the heat transfer rate between the core planet and the shell is 235 W/m^2. At the beginning, the shell has a temperature of 0 K. This, however, starts increasing fast as soon as the heat from the core planet arrives.

Now, if there were no constant supply of energy to the surface of the core planet, the equilibrium temperature of the planet/shell system would end up somewhere around the midpoint between 254 and 0 K. The core planet would cool and the shell would warm, until they reached the same temperature. In our scenario, though, the core planet will maintain its original temperature because of its continuous supply of energy from within. That means the shell, completely dependent on the heat it receives from the core planet, is the only one of the two bodies that will have to change its temperature. It will need to raise it all the way from 0 to 254K. Only then the system will have reached thermodynamic equilibrium. The closer the shell gets to the core planet temperature, the slower the heat transfer between the two bodies. Why? The core planet will always emit the same outgoing flux – 235 W/m^2. But the warmer the shell gets, the greater its ‘counter-flux’ and the smaller the net flux (heat transfer –> lower warming rate). Until the flux ‘back’ from the shell to the planet is an equal 235 W/m^2. Then we have reached a state of thermal equilibrium and the heat transfer will end (it has reached zero). At this point, the surrounding shell has attained the same temperature as the core planet (~254K) and will accordingly radiate a flux of 235 W/m^2 out to space.

The misunderstanding seems to arise from the idea that the flux from the shell needs to be split in half, going two separate ways. This might seem intuitive and logical – ‘How can it give off a flux twice the size of the one it receives?’ The thing is, this has to do with the built up heat content of the shell over time, not with the specific rate of heat transfer at any one time, which is getting smaller and smaller the closer you get to equilibrium. When thermal equilibrium is reached, both the inside and the outside of the surrounding steel shell will have ended up with a temperature equal to that of the core planet. That’s part of the definition. The energy/heat supplied from the core during all this time (albeit at a constantly slower rate) will finally have ‘filled’ the entire 3D lattice of the shell with the adequate amount of KE. And it will radiate accordingly. Both surfaces will emit a flux of 235 W/m^2. So at this point there is a net flux within the system of 0 W/m^2, but a net flux leaving the system of 235 W/m^2, corresponding exactly to the continuosly generated power flux inside of the core planet.

The outside of the surrounding shell is now the steadily (equilibrated) radiating system surface to space, like the surface of the core planet used to be before the shell was placed around it.

120. tallbloke says:

Bryan: The problem arises when misuse of the Stephan Boltzmann equation and black body radiative physics is misapplied.
This was clearly shown by the results of the Mariner mission to the Moon

Is there a write up on the Mariner results problem anywhere Bryan? It sounds interesting.

121. tallbloke says:

Kristian, the way I’m reading what you’ve written, you seem to agree with me more than Joe. 😉

122. Bryan says:

tallbloke says “Is there a write up on the Mariner results problem anywhere Bryan? It sounds interesting.”

123. gbaikie says:

“Arfur Bryant says:
March 11, 2013 at 9:51 pm

I hope no-one minds if I ask two genuine questions:

Can radiation from a cooler body/surface be absorbed (for net gain, as opposed to being instantly re-emitted) by a warmer body/surface)? ”

A mirror can reflect energy and the mirror’s temperature has nothing to do with this energy.
So that is obvious.
But question is can cooler object prevent a warmer object from cooling as quickly and/or
can cooler object add heat to warmer object.
So a colder object can reflect the radiant energy from warmer object- this could cause warmer object to cool slower [the heat it was losing via radiation is returned to it].
The other part of it is colder object absorbs the energy from warmer object. So with the addition
of this heat and the heat a cold object already has [if above absolute freezing it has some heat]
can it add heat to warmer object. And answer is not by any significant amount- and generally speaking, no. As heat flows from warmer to cooler.

“Am I correct in thinking that radiation needs to be absorbed in order for temperature to be increased?”

Generally.
Temperature is a means of measuring- it’s a yard stick.
So it matters what you trying to measure.
Temperature generally refers to measuring how much energy in form of heat
something as. With gases it refers to the kinetic energy of the gases. Gases
is motion of molecules, with liquids and solids these are molecular structures
which bonded together- which if given enough energy in form of heat will
become gases.

124. Kristian says:

tallbloke says, March 11, 2013 at 11:19 pm:

“Kristian, the way I’m reading what you’ve written, you seem to agree with me more than Joe.”

Well, except you’re ending up with a core temperature of 302K (470 W/m^2) and a shell temperature of 254K (235 W/m^2), while I (and Joe) end up with a core temperature of 254K (235 W/m^2) and a shell temperature of 254K (235 W/m^2). I think the discrepancy is rather significant.

125. Tim Folkerts says:

Max says: “Ok, see what you did there? … You’re canceling the area down to a difference of radius….

Canceling the 4*pi from the black body calculations produces an incorrect value.”

Yes, I see exactly what I did there. I divided both sides of an equation by the same amount (4 pi). That is perfectly legitimate algebra. There is absolutely nothing wrong.

“Again, if you use the ratio of the area you get 255 K and 254.1 K, the black body temperatures for those values are 239.764 W/m^2 and 236.397 W/m^2.

In the case the sphere emits a total of 122,295,098,043 Watts while the shell would emit 120,994,442,985 Watts, the ratio of those is 0.9893(64618213).

Why would I want to switch from my correct results to your wrong results that produce imbalanced power outputs?

It is some much more powerful to simply use algebra & proportional thinking, but we can use numbers too.

If I use my values (r_1 = 6370km and r_2 = 6380km) and set the temperature to T_1 = 255 K for , then the other temperature is 0.9992159936 * 255K = 254.8000783699 (throwing in lots of digits).

The radiation flux from the smaller sphere (ie the bare planet or a shell that is infinitesimally above the surface of the planet) would be
F_1 = 5.6704E-8*255^4 = 239.75872344 W/m^2
The radiation flux from the bigger sphere (the outside of the shell 10 km above the planet) would be
F_2 = 5.6704E-8*254.8000783699^4 = 239.0077177198 W/m^2

The total powers would be
P_1 = F_1*A_1 = 239.75872344*4*PI()*6.37e6)^2 =1.2225E+017 W
P_2 = F_2*A_2 = 239.0077177198*4*PI()*6.38e6)^2 = 1.2225E+017 W

So I have predicted exactly what the ratio of temperatures should be (255K/254.800K) to balance power out for two different sized outer surfaces (6370 km vs 6380 km). Moving the shell up 10 km cools the outer sphere by 0.200 K. This is a real amount that ought to be taken into account, but it is much smaller than the heating effect of a shell around the planet.

” … so at least Tim’s shell isn’t emitting more radiation than the sphere “
Yep. It is emitting EXACTLY as much energy to space as the bare planet would have been! Exactly as the equations were designed to do.

126. gbaikie says:

“Kristian says:
March 11, 2013 at 10:57 pm

Tallbloke,

I tend to agree with Joe Postma. There seems to be a fundamental misunderstanding here regarding how heat is transferred from the warm object to the cold.

Willis’ nuclear planet and its surrounding shell is a thermodynamic system. The nuclear planet is a black body. It will always emit exactly the same flux as it receives. It cannot emit less. It cannot emit more.”

A problem is the sky [whether vacuum or earth atmosphere] radiates more energy than can transferred thru solid rock.
When one is talking about nuclear energy it doesn’t have a practical limit to how hot it gets- hence
one gets things like meltdowns with nuclear reactors.
Or quite simply to have surface of planet constantly radiating 235 W/m2 from radioactive decay
is “unrealistic”.
If the question was having a reactor with material which separated by a vacuum with a shell designed to radiating 235 W/m2 which then had additional shell- that would make it easier- or it could be something possible.
And if outer shell reflected little or none of the radiation and was same as inner shell. Then the inner shell and exterior shell would be around the same temperature. It be about same as making inner shell thicker.
Or heat would more or less flow at same rate- the outer surface radiate almost the same as
inner surface. But if energy is reflected back from outer shell, then the inner shell will increase
in temperature [the core is constantly added heat and and if inhibited in losing the heat it will become hotter and make inner shell hotter.

The difference between reactor and planet is [first planet is impossible] and core of planet would
take a long time to heat up.

127. Tim Folkerts says:

Kristian says”Well, except you’re ending up with a core temperature of 302K (470 W/m^2) and a shell temperature of 254K (235 W/m^2), while I (and Joe) end up with a core temperature of 254K (235 W/m^2) and a shell temperature of 254K (235 W/m^2). I think the discrepancy is rather significant.

Let’s think about that. The inner planet and the shell around it are at the same temperature in your model, yet the inner planet is radiating 235 W/m^2 to the shell. You have a heat flow with no temperature difference! How do you decide which 254K object receives heat, and which 254 K emits heat?

Your model could work if the heating elements were on the outer shell, but that is a completely different geometry. Then you have no heat flowing any direction,

128. Kristian says:

Tim Folkerts says, March 12, 2013 at 12:55 am:

Huh? So you didn’t read my March 11, 2013 at 10:57 pm comment. Which is the one Tallbloke gave his non-answer to. If you’d read it first you wouldn’t have had to misrepresent me as grossly as you do here.

Because you do of course know the difference between ‘radiative energy flow’ and ‘heat flow’ …?

129. Tim Folkerts says:

gbaikie says: “The nuclear planet is a black body. It will always emit exactly the same flux as it receives. It cannot emit less. It cannot emit more.”
No, a black body will always absorb ALL the radiation that arrives.
It will always emit P = σ A T^4 based on its own current temperature.
There is NO requirement that these two be the same.

Or perhaps you mean “a black body in a steady-state condition will always emit exactly the same flux as the f̶l̶u̶x̶ total energy it receives (ie incoming flux+ internal heaters)”. That I could agree with

But then condition Tallbloke and I (and the rest of the rational scientific community) suggest fits the bill.

The planet receives 235 W/m^2 of internal heating + 235 W/m^2 of IR flux, and emits exactly the same 470 W/m^2 flux.

130. Tim Folkerts says:

Kristian,

1) It seems that I attributed to gbaikie the quote in the previous post .. but he was only quoting you. So apologies to both of you.

But it doesn’t really change my opinion. You claim there that “The core planet will always emit the same outgoing flux – 235 W/m^2” but then a couple sentences late you claim “Then we have reached a state of thermal equilibrium and the heat transfer will end.”

Now there seems to be a NEW paradox to deal with. Rather than clarifying anything, we simply have to deal with a DIFFERENT impossible statement — is the heat flow 235 W/m^2 outward, or is it 0 W/m^2?

“Because you do of course know the difference between ‘radiative energy flow’ and ‘heat flow’ …?
With both regions at 254K, there would be a ‘radiative energy flow’ of 235 W/m^2 outward and a separate ‘radiative energy flow’ of 235 W/m^2 inward, and a ‘net radiative energy flow’ of (+235 – 235) W/m^2 = 0 W/m^2 which would be the ‘heat flow’ (because there is no other energy flow between the two regions). This is as it should be, since we would expect the ‘heat flow’ to be zero between different regions at the same temperature (that is pretty much the DEFINITION of “same temperature”)

131. suricat says:

TB & Guys. On a quiet day, I just read through this thread twice since my last post and its increased in length each time I got to what I thought was ‘the end’. Can’t digest it all. 😦

Look at the ‘toy’ again! The ‘shell’ is radiating 235W/m^2! The ‘same’ 235W/m^2 that the ‘core’ radiates! Geometry requires that a ‘shell’ around a ‘core’, however close, MUST posses a ‘greater’ area than the ‘core’ area! Thus, the total area of the ‘shell’ is emitting ‘more Jules per second’ than the ‘core’ can supply.

I don’t care WHAT is going on between the ‘shell’ and the ‘core’, but it certainly ‘looks’ as though the ‘core’ is being ‘freeze-dried’ by the ‘shell’! I know a vacuum is supposed to exist between the ‘core’ and the ‘shell’, but this reminds me, so much, of the atmospheric hydrological cycle.

Were the aims of this ‘construct’ to ‘elucidate’ the properties of ‘latent heat’, or was this just a ‘cock-up’? 🙂

Best regards, Ray.

132. gbaikie says:

“Let’s think about that. The inner planet and the shell around it are at the same temperature in your model, yet the inner planet is radiating 235 W/m^2 to the shell. You have a heat flow with no temperature difference! How do you decide which 254K object receives heat, and which 254 K emits heat?”
235 W/m^2 is a rate per second. It’s the same as 2.35 W/m^2 per 1/100th of second.
Heat flow is occurring almost instantaneously- nanoseconds.

So one has 235 W/m^2 radiating from the outer surface [2.35 W/m^2 per 1/100th of second].
The question is the heat flow from inner part shell to outer part facing vacuum and it must
be 235 W/m^2 or 235 W/m^2 can’t be radiated.
So rate watts per square which radiate depends on temperature and temperature depends
upon how fast heat can flow thru the material. Which depend on type of material it’s thickness and the temperature difference.

A chunk of steel in space will eventually cool to 2 K. A sheet of steel which so thin it’s somewhat transparent will reach 2 K much quicker as compare 1 cm thick plate steel.
A 1 cm thick meter square sheet has a certain amount heat depending on temperature.
With room temperature steel has specific heat of “Carbon Steel 0.49 kJ/kg K”
http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html
Unless someone can provide better numbers, I will assume 0.49 kJ/kg K regardless of
temperature.
So 1 cm thick and meter square steel weighs about 78 kg. So it has 38.22 kJ per K.
So if around -18 C and is radiating 235 W/m^2 it takes 162.6 seconds to cool
by 1 K. If instead it’s 1 mm thick it takes 16.3 seconds. And if a micron thick it
takes .016 seconds per K [60 K in one second]
“How thick must a foil be before its opaque?

Most metals are opaque at 0.25-0.3µ thick. Metals 0.2µ thick have an attenuation of about 8-10 orders of magnitude. ”

So steel foil to be somewhat transparent it needs to be about 1/2-1/4 of a micron and would cool
in vacuum of space to 2 K in fairly quickly [though as gets colder it doesn’t radiate anywhere near as much energy- and given a small amount heat it stay warmer than 2 K {distant body such Earth or the Moon can give such small amount of heat}]
So the top surface of the steel can cool very quickly- but normal steel doesn’t cool this
quickly because it’s thicker than micron and it’s heat is transferred to the surface so prevent it getting so cold.

As rough guess one can say difference is the 2 K, and even though steel conducts heat poorly
as compared to other metals. Example here:
http://www.engineeringtoolbox.com/thermal-conductivity-metals-d_858.html
Where depending on the type steel it’s 31- 21 as compared to Aluminum of 124 [Btu/(hr oF ft]
at 68 F, it seems a significant factor is thickness of steel according to formulas given here:
http://www.engineeringtoolbox.com/conductive-heat-transfer-d_428.html

133. wayne says:

Some interesting work on thermal radiation:

ROLE OF FLUCTUATIONAL ELECTRODYNAMICS IN NEAR-FIELD

but probably heed the words of the authors: “propagation of radiative energy is properly represented by electromagnetic wave approach, which requires the solution of the Maxwell equations. The fluctuational electrodynamics provides a model for thermal emission when solving a near-field radiation heat transfer problem, and the fluctuation-dissipation theorem provides the bridge between the strength of the fluctuations of the charges inside a body and its local temperature.”

Says to me… drop that “photonic” viewpoint, radiation is actually via em wave interactions, especially at surface interfaces.

Another I was just reading because it addresses heat transfer between two surfaces with a vacuum in-between, hmm…
A Green’s function formalism of energy and momentum transfer in fluctuational electrodynamics
http://arxiv.org/pdf/1302.0545v1
“Radiative energy and momentum transfer due to fluctuations of electromagnetic fields arising due to temperature difference between objects is described in terms of the cross-spectral densities of the electromagnetic fields.”

Neither may directly apply here… but they are interesting.

TB, you might wonder why I said I was not 100% sure, such papers (and many more I have read in the past) tell me to progress slowing into this area before stating absolutes.

134. philr1992 says:

What is this? Radiative units (W/m^2) are analogous to velocity…not material flux. You don’t add 5W/m^2 to 15W/m^2, microkinetic firing frequency is saturated at 15W/m^2. Nothing is going to heat this hypothetical nuclear core beyond it’s initial temperature. The universe would implode, otherwise.

I hope I just misunderstood the proposed argument. I’m exhausted, so that’s a possibility..

135. Max™ says:

I’m curious if my post above isn’t appropriate?

I don’t see why this wouldn’t be:

“Yes, I see exactly what I did there. I divided both sides of an equation by the same amount (4 pi). That is perfectly legitimate algebra. There is absolutely nothing wrong. ” ~Tim Folkerts

Ok, you’re reducing a calculation of the power radiated by the area of a 2 dimensional outer surface of a sphere to a calculation of the power radiated by the single point on the outside surface of a sphere.

You’re taking a calculated area which is needed for the equation and replacing it with something which is not an area.

Some other problems with this thought experiment: what’s the inverse-square law, why does the surface radiate twice the power it receives (inwards and outwards).

[Reply] Hi Max, nothing wrong, but we don’t have a moderator in every time zone like WUWT, and we need sleep occasionally.

136. tallbloke says:

Kristian: my March 11, 2013 at 10:57 pm comment. Which is the one Tallbloke gave his non-answer to.

Apologies; it was late – I was tired. I’ll re-read it.

Kristian said:The energy/heat supplied from the core during all this time (albeit at a constantly slower rate) will finally have ‘filled’ the entire 3D lattice of the shell with the adequate amount of KE. And it will radiate accordingly. Both surfaces will emit a flux of 235 W/m^2. So at this point there is a net flux within the system of 0 W/m^2, but a net flux leaving the system of 235 W/m^2,

So how long can the shell continue to radiate a total of 470W/m^2 whilst receiving only half that from the planet without cooling down? The same amount of time it too to heat up? what then? It’s a continuous process, what will happen is it will settle down to radiating 117.5 from each surface (assuming perfect conductivity) and then the planet will heat up because it is kicking out 235 but the shell is only losing half that to space. But when the planet heats up, the shell will receive more than 235…

Eventually, as the theory goes, the planet will kick out 470 and so will the shell. 235 to space for eqilibrium, and 235 back towards the planet, which will absorb and re-emit it along with the 235 from the core. No free lunch, no cake and eat it, energy conserved.

It is not a closed system. Energy is entering the surface from the core, and the same amount of energy is dissipating to space at the the end of the system. The temperature’s reached by the components in between depend only on the rate they can lose energy at. The outer shell surface has no problem losing 235 to space, because space is very cold, there is a big differential. The planet and the inner shell surface on the other hand are both trying to lose heat in proximity to another warm object, and in a theoretically perfect black body system they will get twice as much energy to deal with.

The real Earth of course is not a perfect blackbody system, and the theory does not apply, for many and several reasons.

137. Kristian says:

Tim Folkerts says, March 12, 2013 at 1:48 am:

Tim, you very much seem not to get what I’m saying. So I’m not sure there’s even any point trying to respond to your comment. But I’ll give it a go anyway.

I’m saying that your fundamental approach to the thermodynamic system at hand (core planet and surrounding shell) and its course towards thermodynamic equilibrium is mistaken. It is all a matter of regular heat transfer WITHIN one thermodynamic system. This is key to understanding the concept we’re dealing with. You have to consider Willis’ system as a whole.

The surface of the hot object at the core already has its temperature set by its constant internal energy supply. This will not change until this supply itself is turned up. The cold object around will at no point be able to cause a warming of the hot one beyond this temperature. It will ONLY ever gradually reduce the heat transfer between the two.

The point is this: As soon as the surrounding shell is put in place, the new thermodynamic system as a whole is not able to balance the power flux generated within the core with release to its surroundings (space) until it has reached thermodynamic equilibrium, i.e. a state of uniform temperature – the temperature of the core needs to spread across the entire system. This process starts as soon as the hot core begins to warm the cold surrounding shell.

Remember, the (cold) shell will warm only from the heat transferred to it from the (hot) core, fast at first, more and more slowly as the temperature of the shell approaches that of the core. This is simply because the heat transfer rate from hot to cold slows down (diminishes) as the temperature difference is gradually reduced. This is basic thermodynamics.

In the end there is a power flux of 235 W/m^2 being generated within the core planet, balanced by an equal flux of 235 W/m^2 leaving the outer surface of the surrounding shell to space, core planet and shell both maintaining a dynamic equilibrium temperature of 254K. Note the ‘dynamic’ term, because the equilibrium is dependent on the two fluxes (incoming and outgoing) balancing each other at all times. On a microscopic (quantum) level, this will never be fully attained. There will always be a ‘pull’ from core to outer wall. And remember – radiative (and conductive) heat transfer like this is in effect a matter of quantum physics. These processes are not always as intuitive as the more macro-scale processes (like mass movements – convection).

Without shell: a) 235–> b) 235–> (equilibrium)
( a) constant nuclear energy supply, b) surface of the core planet)

With shell (at first): 235–> 235–>
(internal heat transfer rate: 235 W/m^2 from core to shell)

Mid-way towards equilibrium: 235–> 235–>
(internal heat transfer rate: 117.5 W/m^2 from core to shell)

At equilibrium: 235–> 235–>
(internal heat transfer rate: ~0 W/m^2)

138. Kristian says:

tallbloke says, March 12, 2013 at 6:48 am:

“So how long can the shell continue to radiate a total of 470W/m^2 whilst receiving only half that from the planet without cooling down?”

Look, it’s very simple. Forget about the 470 W/m^2. At equilibrium, the system loses a flux of 235 W/m^2 through the outside surface of the outer shell. At the same time, the inner core produces an equal power flux of 235 W/m^2, providing the system with the same constant ‘incoming’ amount of energy/heat as the continuously lost/outgoing amount. This is a state of thermodynamic equilibrium, the temperature is constant and uniform throughout the system. The quasi-constant KE content is maintained by a fragile balance, though, between the ever shedded heat flux from the outer shell and the ever generated heat in the nucleus. As soon as the power from the core would start diminishing, the outgoing flux would be too large and the system would be pushed out of equilibrium again, until a new constant (and lower) temperature were found.

139. Kristian says:

Tallbloke,

I was just until very recently thinking like you when it came to this issue. But I’m quite sure now that this is the more correct way of seeing it. Just think about it …

140. Tim Folkerts says:

Kristian says says: “This is a state of thermodynamic equilibrium, the temperature is constant and uniform throughout the system.

No, this is a state of thermodynamic steady-state. In true thermal equilibrium, the system cannot have heat flowing through it.

Kristian, consider the analogous situation of heat flowing (by conduction rather than radiation) through the walls of a house on a cold winter’s night. The house has a furnace (rather than nuclear heater) that provides a fixed power (which we could express in W/m^2 of wall area). By your logic, there could be heat flowing the walls with no temperature gradient between the inside of the wall and the outside of the wall. The outside of the wall would be at the same same temperature as the inside of the wall. (It is not a perfect analogy, but it is pretty good.)

141. tallbloke says:

Tim F says: No, this is a state of thermodynamic steady-state. In true thermal equilibrium, the system cannot have heat flowing through it.

Good point Tim, I’ve been misusing the word equilibrium here. I was most of the way there though. I did point out that it’s not a closed system and that energy coming in at one end has to traverse various situations before dissipating at the other.

It seems our problem is to get people to understand that the emission and emission temperature can be different in different parts of the system (in this case they necessarily are) and still have the steady state as the overall result. I left this on Joe’s site earlier this morning, maybe it will help Kristian see our point:

tallbloke says:
2013/03/12 at 4:11 AM

Joseph E Postma says:
2013/03/11 at 6:57 PM

And, note, that force doesn’t increase itself due to a differential; voltage doesn’t increase itself due to a differential; and temperature doesn’t increase itself due to a differential!

However, a resistor with a current running through it will get very hot if you put it in a vacuum flask…
….and it will stop getting hotter once the outside of the vacuum flask is radiating as much energy as the resistor was in open air.

142. Max™ says:

The outside has to radiate the same total energy, but not at the same density.

If it were to radiate with the same power density as the inner surface, it would be giving off more energy than it receives from the surface.

This is why I disagree with Tim F. above, using the radius instead of the area for the SB calculation may be algebraically valid, but it is physically incorrect.

143. tallbloke says:

Max, we’ve already accepted that there will be a difference. Do you accept that it’s small (less than 0.5%) when considering the areas of spheres with a 10km/7000km radius difference?

using the radius instead of the area for the SB calculation may be algebraically valid, but it is physically incorrect.

Does it come up with the wrong figures for power density? How far out are they?

144. Max™ says:

It should be zero or less than zero, not positive, if such a critical flaw in the initial assumptions exists, how can the conclusion be trusted?

145. A C Osborn says:

Tim Folkerts says:
March 12, 2013 at 11:48 am
consider the analogous situation of heat flowing (by conduction rather than radiation) through the walls of a house on a cold winter’s night. The house has a furnace (rather than nuclear heater) that provides a fixed power (which we could express in W/m^2 of wall area). By your logic, there could be heat flowing the walls with no temperature gradient between the inside of the wall and the outside of the wall. The outside of the wall would be at the same same temperature as the inside of the wall. (It is not a perfect analogy, but it is pretty good.)

What rubbish, the use of “a cold winter’s night” provides the temperature differential of the wall with the house warmth on one side and “a cold winter’s night” on the other. Heat must “flow” through the walls until the outside equals the inside, which can’t happen due to the cold air outside continuously cooling the outside.

Surely that is the whole point of the Universe trying to reach equilibrium.
Or are you saying that is not what happens.

The extra shell added to the planet, even with a vacuum just extends the ball by the extra thickness added by the vacuum and the shell, the vacuum makes no difference, because as Willis admitted if there was any contact between the shell and the planet there would be “No Back Radiation”.
“If they could not do so, if for example the shell were held up by huge thick pillars that efficiently conducted the heat from the surface to the shell, then the two would always be at the same temperature, and that temperature would be such that the system radiated at 235 W/m2. There would be no differential heating of the surface, and there would be no greenhouse effect.”

Why not, you would still have “Radiating Surfaces” where they were not in contact.

146. tallbloke says:

Max: if such a critical flaw in the initial assumptions exists, how can the conclusion be trusted?

So no answer to my request for the magnitude of the error. I’m not surprised, because if you revealed it, you’d be exposed for making a mountain out of a molehill.

147. tallbloke says:

148. John in France says:

Please stop calling them “Skydragons” Roger. I thought that at least had been cleared up.

149. tallbloke says:
March 11, 2013 at 4:16 pm

CF: The experiment says nothing about the supposed back radiation which does not exist.

It’s ‘back-radiation’ which is slowing down the cooling of the primary heat source in the experiment isn’t it?
No! you may think it is semantics but my experience still says there is no backradiation.
The source of radiation (a high temperature object) does not receive/absorb any radiation from a lower temperature object which is the receiver of heat transfer. The amount of heat loss from a object depends on the surrounding configuration and temperature. One can measure the heat loss from a blackened thin wall copper pipe which has warm water flowing through it. If the surrounds are the same temperature as the water there will be no heat loss. I have carried out lots of measurements of heat loss from pipes carrying various fluids and gases. under various conditions.
As I said thought bubbles and related calculations which may or may not be correct (probably incorrect due to wrong application of S-B equation) are no replacement for direct verified measurements.
This post by Prof Claes Johnson, http://claesjohnson.blogspot.com.au/2013/02/big-bluff-of-pyrgeometer-dlr-as.html, makes a lot of sense. Please note my comment at the bottom of the page. Alarmists try and twist laws and relations, deliberately or because they have no experience and do not understand. It is saddening that there many here who dismiss the large amount of research by engineers and do not want to understand.

150. Tim Folkerts says:

AC Osborn says: “The extra shell added to the planet, even with a vacuum just extends the ball by the extra thickness added by the vacuum and the shell, the vacuum makes no difference, because as Willis admitted if there was any contact between the shell and the planet there would be “No Back Radiation”.”

You misunderstand. The vacuum makes a BIG difference! The vacuum is “insulation” (in fact you could consider it “perfect insulation” since there is absolutely no conduction allowed). The ONLY way to cool the planet is by radiation to cooler surroundings.
* If the planet is surrounded by outer space @ T=0K, then the planet will warm up to 254 K. At this point it can radiate enough energy to the surroundings to balance the energy from the internal heater
* If the planet is surrounded by a shell @ 254 K (and everyone seems to agree the shell will be this temperature (or 254K – δ if you want to include the “radius cooling effect”)), then the planet must be WARMER than 255 K to radiate energy to the shell. The planet will be 302K in fact, radiating 2 * 235 W/m^2 .

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

Willis was saying that if you replaced the “perfect insulation” with a “perfect conductor” then the temperature difference would go away. In other words, as more and more energy travels from planet to shell by mean OTHER than radiation, then the temperature difference will get smaller and smaller.

(As a prime example, suppose the shell was only 1 km above the surface. We would then have a lapse rate of 302K – 254K = 48 K/km. If you put in an atmosphere of pure N2, this would be highly unstable, and convection would ensue. Convection would limit the lapse rate to ~ 10 K/km. The shell would be 254 K (still set my radiation to outer space), but the surface would only be ~ 264K (probably a little higher). There would be ~ 40 W/m^2 of radiation, leaving ~ 195 W/m^2 carried by convection.)

151. tallbloke says:

CF: if you responded to the bolded part of what I wrote I missed it: did you?

“I am sorry Roger, you lost me with this post, I have always thought this was a thoughtful blog with input of engineers who have some experience. I hope it gets back to real science based on ethical engineering principles.”

[Reply] Ouch. I am covering in this post what theory says will happen with a vacuum between the shell and surface. That doesn’t pertain in your furnace, where convection will likely prevent re-radiation from adding to the output from the heating element due to re-absorption in the air. However, follow the link to ‘TheFordPrefects’ blog and study the empirical test he conducted, then report back.

As I keep emphasising, there is a world of difference between the theory of a contrived toy model planet and our real planet, and I agree with you that the real world understanding of engineers shouldn’t be ignored.

my experience still says there is no backradiation.
The source of radiation (a high temperature object) does not receive/absorb any radiation from a lower temperature object which is the receiver of heat transfer. The amount of heat loss from a object depends on the surrounding configuration and temperature.

Well, the presence (or not) of other radiating objects in the ‘surrounding configuration’ are going to affect the temperature of the surrounding aren’t they?

152. Tim Folkerts says:

CF says: ” One can measure the heat loss from a blackened thin wall copper pipe which has warm water flowing through it. If the surrounds are the same temperature as the water there will be no heat loss.

And there lies the crux. This correct as far as it goes, but it is changing the circumstances of the original thought experiment.
* Here you have a "heater" that is set to a fixed TEMPERATURE. When the pipe is at the same temperature as the surroundings there is "no heat loss" as you say, but there is also no need for any heat INPUT. The heater can and will shut down.

* In the original experiment, the heater is a fixed POWER. Again, as you say, IF the planet and the shell were at the same temperature, there would be no 'heat flow' = no energy leaving the planet to the shell (or to anywhere else for that matter). BUT we still have 235 W/m^2 flowing INTO the planet. So there is a NET increase of 235 J in to every square meter of surface every second. This WILL cause the temperature to rise on the planet. (Just like the temperature of the water in the pipe will rise if it is 1) already at the temperature of the surroundings AND 2) you supply energy from some heater.)

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

I have seen this sort flaw in several of the attempted rebuttals, where people effectively treat the heaters in the planet as being set to 254 K, rather than being set to 235 W/m^2.

153. Kristian says:

Tim Folkerts says, March 12, 2013 at 11:48 am:

“No, this is a state of thermodynamic steady-state. In true thermal equilibrium, the system cannot have heat flowing through it.”

I agree.

“Kristian, consider the analogous situation of heat flowing (by conduction rather than radiation) through the walls of a house on a cold winter’s night. The house has a furnace (rather than nuclear heater) that provides a fixed power (which we could express in W/m^2 of wall area). By your logic, there could be heat flowing the walls with no temperature gradient between the inside of the wall and the outside of the wall. The outside of the wall would be at the same same temperature as the inside of the wall. (It is not a perfect analogy, but it is pretty good.)”

It is not ‘pretty good’.

What is analogous to what here, Tim? The inside of the furnace is the nuclear power source and the outside surface of the furnace is the surface of the core planet? Then the air-filled space in between the surface of the furnace and the inside of the wall of the house is the vacuum between the core planet and the inside of the surrounding steel shell? The wall, including layers of air and insulating material, is the steel shell itself? And the ‘cold winter’s night’ outside is the vacuum of space surrounding the outer shell?

Or do you consider the inside wall to be analogous to the surface of the core planet and the whole stretch through the wall to the outside, the path from the core planet to the outside of the steel shell?

In either case, colour me unimpressed by your analogy, Tim. Insulation concerns and affects convection. It restrains convection. Like our atmosphere does. That’s why the surface of our planet is as nice and warm as it is. Not because of thermal radiation.

Let me give you a counter-analogy: Consider Willis’ nuclear planet and shell suspended in a vacuum like before, only this time around there is nothing but massive steel all the way from the surface of the core planet to the outside of the surrounding shell. The constant power flux from the core planet surface to all sides is 235 W/m^2. Disregarding pressure gradients (and difference in inner and outer surface area), how do you imagine this steel ball will manage to deliver a steady flux from its outer surface of an equal 235 W/m^2, keeping the balance between generated and emitted energy? Will the temperature be uniform throughout? Or will there be a temperature gradient from the core to the outer surface? What would this temperature gradient be like? How and why would this gradient be maintained at a steady state? Remember, no pressure involved, only conduction.

Once you turn the internal power source on, straight to full force, the outside surface is 0 K, emitting 0 W/m^2. At the point where balance is reached between produced energy in the nucleus and emitted radiative energy from the interface between steel ball and space, the outside surface is 254K and emitting 235 W/m^2. What has happened in between those two stages? What is the temperature at the inner (core planet) ‘surface’ at the point of balance?

154. Kristian says:

tallbloke says, March 12, 2013 at 11:59 am:

“… maybe it will help Kristian see our point.”

Tallbloke, I already see your point. It was my ‘point’ also until very recently. Now I’ve reconsidered.

What I wonder is, can YOU see OUR point …?

155. mkelly says:

Mr. Folkerts says:

For “realistic” numbers, r_1 = 6370 km and r_2 = 6380 km, so
T_2/T_1 = (6370/6380)^0.5 = 0.9992
Or T_2 = 0.9992*255K = 254.8 K

and also “The planet will be 302K in fact,…”

Mr. Folkert has with his ratio of T to r falsified the steel shell idea as presented. His ratio points out that given his numbers that T2 will be .9992 of T1. He further says that T1 will get to 302K which using his ratio makes T2 = 301.758K which cannot be true if the inner shell is radiating at 235 W/m^2. This is a serious inequality thereby falsifing the steel shell idea as presented.

Good work Mr. Folkerts. As we used to say in the Navy BZ or SH which ever one you like best.

156. tallbloke says:

So do you agree the resistor will get hotter inside the vacuum flask and will stop getting hotter when the outside of the flask loses as much energy as the resistor did in open air?

What I wonder is, can YOU see OUR point

Unfortunately not. I rechecked your earlier comment and I still don’t see how 235+235 = 235. According to your comment, it is something to do with
” The energy/heat supplied from the core during all this time (albeit at a constantly slower rate) will finally have ‘filled’ the entire 3D lattice of the shell with the adequate amount of KE. And it will radiate accordingly. Both surfaces will emit a flux of 235 W/m^2.”

Perhaps if you quantified ‘adequate’ and ‘all this time’ I might be able to start getting a handle on the maths underlying your proposition.

157. Roger Clague says:

Kristian says
The surface of the hot object at the core already has its temperature set by its constant internal energy supply.

TB says

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.

I see no reason why the hot object cannot increase in temperature.
TB is right and Kristian is wrong.

Constant energy flux ( J/sm2 )does not produce constant temperature until equilibrium is reached.

158. tallbloke says:

mkelly: This is a serious inequality thereby falsifing the steel shell idea as presented.

On the contrary, the area inequality is a trivial inequality which doesn’t do any harm to the basic theoretical physics principles demonstrated by the toy planet model.

Max and you need a new nitpick, this one’s getting all wore out. 😉

159. Roger Clague says:

Tim Folkerts says:
March 12, 2013 at 11:48 am
Kristian says says: “This is a state of thermodynamic equilibrium, the temperature is constant and uniform throughout the system.”

No, this is a state of thermodynamic steady-state. In true thermal equilibrium, the system cannot have heat flowing through it.

Thermodynamic equilibrium is a steady state, a dynamic steady state. Heat entering and leaving at the same rate.

Thermal equilibrium is two things in contact at the same temperature, so no heat flow. A completely different concept.

160. tallbloke says:

Roger C: thanks for the further clarification on this. I knew there was some reason I’d been using the word equilibrium.

161. Kristian says:

Tim Folkerts says, March 12, 2013 at 2:26 pm:

“The vacuum makes a BIG difference! The vacuum is “insulation” (in fact you could consider it “perfect insulation” since there is absolutely no conduction allowed). The ONLY way to cool the planet is by radiation to cooler surroundings.”

Tim, you’ve outdone yourself this time!

In what way [snip] does the vacuum in between the surface of the core planet and the inner surface of the surrounding steel shell constitute ‘a perfect insulator’??!!

Is it in the way the entire flux from the former is transferred instantly and without loss to the latter? Conduction would be hard-pressed to ever match, much less surpass, such an extraordinary transferring rate. [snip]

[Moderation note] Condescension removed

162. Tim Folkerts says:

Roger C says “Thermodynamic equilibrium is a steady state, a dynamic steady state. Heat entering and leaving at the same rate.”

No, that is NOT the standard definition! Google it!

Thermodynamic equilibrium typically means thermal equilibrium PLUS mechanical equilibrium + chemical equilibrium.

Thermodynamic equilibrium is MORE restrictive than thermal equilibrium, not LESS restrictive.

163. Tim Folkerts says:

Kristian ponders: “In what way [snip] does the vacuum in between the surface of the core planet and the inner surface of the surrounding steel shell constitute ‘a perfect insulator’??!!”

A perfect insulator is a material that prevent the conduction of thermal energy. A vacuum cannot — by definition — conduct thermal energy. Therefore vacuum is a “perfect insulator” regarding the conduction of thermal energy.

Now, vacuum is a very good medium for the transfer of photons, but that is a different question. In context, my statement is perfectly clear and perfectly correct.

164. tallbloke says:

Joes reply to my earlier post:

Joseph E Postma says:
2013/03/12 at 8:06 AM

“a resistor with a current running through it will get very hot if you put it in a vacuum flask…”

Well sure, but that’s not the GHE either.

“….and it will stop getting hotter once the outside of the vacuum flask is radiating as much energy as the resistor was in open air.”

The temperature it gets to is dependant upon the power running through it. If the flask absorbs the thermal radiation, then it can eventually get to the same temperature as the source.

And this guy has a postgraduate degree in astrophysics? Scary.

165. Tim Folkerts says:

Mkelly,

The calculations presented apply to the exterior of the outermost shell, given a fixed power on the inner shell. The results give the temperature of the outer shell as a function of the radius of the outer shell. You are making an apples-to-oranges comparison when you compare the outer shell to the inner planet.

If the bare planet was 255 K without a shell, and we then add a shell, then …

In the limit that r_shell = r_planet, then T_shell = 255 K as well (and the planet would be ~ 303 K)
In the limit that r_shell = infinity, then T=shell goes to zero (and the planet would be 255 K)
Anywhere in between, the temperature of the shell would be between 255K and 0 K; the temperature of the planet would be between 303K and 255 K.

For r_shell = 6380 km and r_planet = 6370 km, then T_shell = 254.8 K ie 0.2 K cooler than with r_shell = 6370 km. And the surface of the planet would be ~ 0.2 K cooler than the ~ 303 K it was with r_shell = 6370 km.

166. Tim Folkerts says:

Kristian proposes: “Let me give you a counter-analogy …nothing but massive steel all the way from the surface of the core planet to the outside of the surrounding shell.”

OK, we have 10 km of steel between the surface of the planet and the shell. This steel has 235 W/m^2 of power being conducted up through it (but no IR radiation now).

ΔQ/Δt = k A ΔT/Δx

where
ΔQ/AΔt = 235 W/m^2
Δx = 10,000 m
k = 40 W/(m*K) for steel

so ΔT = (235 W/m^2) * (10,000 m) / (40 W/(m*K) ) = 60,000 K!

(Clearly impossible — the steel would melt and start convecting, so the gradient would be much less than this. Only a thin layer (on the order of 100 m thick) near the shell would be solid.)

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

We could also postulate a “perfect conductor” as Willis did, in which case k = infinity, and then ΔT = 0 K. All of the power would move via conduction, with no net power moving via radiation. Of course, this is physically impossible, but makes for an interesting thought experiment.

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

Alternately, we could postulate the heaters on the OUTSIDE surface up on the shell. This would also mean no power flowing through the “gap” and the same temperature at the planet as at the shell. But that is a pretty boring and expected result.

167. philr1992 says:

I am a plasma physicist..The way in which I was educated, I think this should be very basic.

Here’s a better way to look at it perhaps…we have a marble sized “nuclear core”, and a stadium sized “outer shell”.

Looking at it that way, maybe the problem will become will become more apparent.

The interactive process is in equilibrium long before this “shell” somehow attains the same “temperature” as the small core. If the shell was emitting at the same intensity as the core, the system wouldn’t be in equilibrium

168. Bryan says:

Tim Folkerts

How far apart do the planet and the shell have to be for the planet to be twice as warm as the shell?

one kilometre
one metre
one millimetre
one micrometre
one nanometre

When does the magic begin?
Is it an instantaneous transformation?

169. Kristian says:

Tim Folkerts says, March 12, 2013 at 4:01 pm:

“A perfect insulator is a material that prevent the conduction of thermal energy. A vacuum cannot — by definition — conduct thermal energy. Therefore vacuum is a “perfect insulator” regarding the conduction of thermal energy.

Now, vacuum is a very good medium for the transfer of photons, but that is a different question. In context, my statement is perfectly clear and perfectly correct.”

No matter how you twist and turn, Tim, your statement simply does NOT support what you’re trying to argue. The energy/heat flux from the surface of the core planet is transferred just as quickly (in reality, even quicker and more completely) across a vacuum to the inner surface of the surrounding shell by way of radiation than it would have been by way of (super-)conduction through a hypothetical steel-like solid material filling that same space. How does this square with your ‘conduction would even out the temperatures from inside to outside, but not radiation’?

I don’t know why this is so hard to grasp. The core planet with no surrounding shell emits its flux of 235 W/m^2 directly to space at an emission temperature of 245K. Then we surround it with the steel shell. What needs to happen? The outer surface of the shell needs to attain the same emission temperature as the surface of the core planet. How is this accomplished? By transferring heat from the (warmer) core planet to the (colder) surrounding shell until it’s done and balance is reached. The core does not need to become gradually warmer in order for this heat to be tranferred. As long as it remains warmer than the colder shell, the transfer of heat will continue, albeit at a steadily slower pace. The heat from the core is absorbed by the shell and turned into KE, and this process will continue until it has reached the temperature at which it will emit 235 W/m^2 to space. This happens when its temperature (determined by its content of KE) is 254K.

Think of it this way: If you’ve got one radiator at 254K emitting a radiative flux of 235 W/m^2 and then place another one just next to it, also at 254K and emitting 235 W/m^2, do you consider the total flux to the surroundings from these two put together to be twice as large as the flux from just the single one? And if so, would the total temperature of the two in your opinion also be accordingly higher (302K)?

170. Bryan says:

I have sent a copy to the patent office of a wonderful new device utilising the latest IPCC radiative physics.

It consists of a planet with a shell one metre above it with a vacuum between.

Because a metal expands when heated the higher temperature planet will touch the shell and immediately send out a double intensity radiative pulse.
The planet will then cool and contract recreating the vacuum gap and get twice as warm.
This process will then repeat endlessly.

To compare it to the Willis Eschenbach model above.

The shell will never drop below 235W/m2 and quite often radiate at 470W/m2.

Some sceptics might find this hard to accept.

The usual complains about violation of the first and second law will no doubt be made.

However the science is settled and the device must work.

171. Kristian says:

Another thing. Don’t forget that as the heat transfer from the planet to the shell grows steadily SMALLER with time, the heat loss from the system as a whole (the outer shell surface to space) parallelly grows steadily LARGER, until the internal heat transfer is ‘dynamically’ 0 and the system heat loss is 235 W/m^2. So the 235 W/m^2 flux from the nuclear power source is ‘refound’ partly in the heat transfer flux between the two internal levels/bodies and partly in the system heat loss flux to space. As the one grows smaller, the other one grows larger. Until we have dynamic equilibrium.

172. donald penman says:

A C Osborn
I agree with you that the inner core will attain thermal equilibrium with the outer shell and 235 w/m2 will still be radiated to space because the inner surface of the shell can radiate to many other points on the inner surface of the of the shell , the shape of the inner surface of the shell is convex,rather than the inner core.The outer shell will not send back 235 w/m2 as back radition to the shell and will send back none when the inner core and outer shell are in thermal equilibrium.

[Reply] With radii of 6770 and 6780km (similar to Earth geometry) the angle required for direct re-absorption by the inward shell emission means a tiny fraction of the emission will be directly re-absorbed.

173. mkelly says:

Tim Folkerts says:

March 12, 2013 at 4:13 pm

Mkelly,

The calculations presented apply to the exterior of the outermost shell, given a fixed power on the inner shell. The results give the temperature of the outer shell as a function of the radius of the outer shell. You are making an apples-to-oranges comparison when you compare the outer shell to the inner planet.

Mr. Folkerts says:”If you want to have a “large shell” around the planet, then the effect becomes larger of course.”

Sir, I am not making any comparison these are your numbers not mine. You were talking about the planet and a shell so no mixing of apples and oranges. The pictured shell clearly shows 235 W/m^2 from inner and outer shell and clearly shows 470 from the planet or 302K. These are your numbers. Your ratio then says T2 must be 301.7K and that cannot have 235 W/m^2 as the proper output.

You have falsified the shell and I say congrats.

[Reply] Fig2 shows 470 from shell. 235 in and 235 out to space.

174. Greg House says:

Dear friends of back radiation’s :),

I know from my experience on blogs that it is sometimes hard to understand, let us say, that it is at least possible that back radiation won’t warm the source.

[Reply] Back-radiation doesn’t warm the source. It slows its rate of cooling. That is well understood by the friends of science here. I look forward to Claes Johnson’s followers conducting experiments and publishing their results as TFP did.

The counterargument is sometimes like “where does the back/trapped radiation go then?”

[Reply] Not a counterargument used here. I’ve told you three times no radiation is ‘trapped’. The only person here talking about ‘trapped radiation’ is YOU.

meaning that radiation must find a sort of a host to settle down. This seems to be a serious obstacle to accept the possibility that back radiation won’t warm the source.

Now, coming back the hypothetical planet-shell mutual warming process, how come nobody including the concept creators asks “where does the radiation go after leaving the shell in the direction outer space”? Because we have nothing there, so wished the creator Willis.

Well, what I would like to suggest is this. Since we have no problem with the radiation leaving the shell in the direction outer space doomed to never finding a new home, let us think that it is exactly the same way possible to have this desperate homeless back radiation in the inner space between the shell and the planet forever, still unable to warm. Please, give it a thought.

[Reply] Not in a vacuum. However, in the real atmosphere, plenty of LW radiation is buzzing around doing nothing in the way of warming, since it is the outcome, not the cause of the heat in the mass of the atmosphere.

Then, once there is no problem of unclear back radiation’s destiny, we can all look at this back radiation warming thing and ask ourselves “WTF??”. Then we do not need such, you know, explanations as “yes, the temperature instantly drops when emission occurs from a blackbody with no heat capacity” (must be to the absolute zero, OMG!). Then we can recall the R.W.Wood experiment and dismiss the “greenhouse effect” as contradicting reality. Please, give it a thought.

[Reply] Greg, you are Long on rhetoric, but short on understanding. And you quoted me out of context, which demerits you further in my view. The context was your own flawed view of the process, and I was pointing out the absurdity of the inevitable outcome. Further comment containing deceit will be binned.

175. Arfur Bryant says:

@gbaikie says:
March 11, 2013 at 11:36 pm

gbalkie,

Thank you for taking the time to answer my questions. They pretty much agree with my understanding.

All,

I will try to explain why I (a non-scientist) have a problem with the ‘toy’ planet:

1. Why would the 235W/m^2 emitted from the inner shell be absorbed by the planet? The planet is warmer than the shell. The shell cannot emit 235 to the planet as it is also emitting radiation to space. (I agree with the figure of 117.5 each way.)

2. Therefore, the arrow with ‘470’ leaving the planet is wrong.

3. If the shell is insulating the planet (which seems likely), it is reducing the rate of heat loss but it is not causing the planet to increase in temperature. If I lay outside on a cold night I would lose heat. If I wrap myself in a sleeping bag I lose heat less quickly, but I don’t increase my body temperature. (Granted, I don’t have a nuclear power station inside me…)

4. The text alongside the diagram states “The inwards radiation warms the surface of the planet, until it is radiating at 470 W/m2.” I believe this to be the cause of the problem. If the shell is emitting lower energy radiation toward the higher energy planet, why would the planet absorb any of this energy?

5. The shell will always have a greater area than the planet, so it can never emit 235W/m^2, even if it was all going back to the planet. This seems obvious to me. Are you sure you’re not seeing the forest for the trees?

Sorry if I’m being dull, but it seems to me that this debate centres around how much energy is absorbed, not radiated.

Regards,

176. Sparks says:

Willis didn’t give a timescale for this thought experiment, all theoretical additions are moot.

177. Sparks says:

A value of 470 Watts per meter squared without limits is simply that. there are conservation laws to adhere to, climate or not.

178. donald penman says:

The real earth does not have a steel shell around it as in this “thought experiment” so I don’t think what i propose would apply to the earth anyway.I don’t know what size this thing is but I dont think that what I propose would be insignifacant if the outer shell radiates less to the inner core than the inner core radiates to the outer shell then they must reach thermal equilibrium however long that takes in my oppinion.

179. tallbloke says:

Hi Arfur:
1)Look back through the thread at the matrix of figures I provided. The addition of the shell raises the amount of radiation being emitted by the planet, and so being absorbed and then emitted by the shell.
2)See 1)
3)The shell radiates half of what it receives back towards the planet
4)The planet has no trouble absorbing energy from the shell. I disagree with Willis that this ‘warms’ the planet. It doesn’t. It reduces the planets ability to shed energy so quickly. Because it is being heated from within by th nuclear core at a constant rate, it warms up. Having warmed up, it emits more radiation towards the shell which in turn re-emits half to space and half back to the planet. Round a set of iterative loops this repeating feedback eventually tails off and converges on the rate of emission whereby the outside of the shell loses energy to space at the same rate the nuclear core produces it.
5) For a planet the size of Earth, and a shell 10km above its surface, the difference in area is around 0.2%. It will make a difference to the simplified set of figures below, but not by much.

This set assumes the nuclear core produces an emission of 800W/m^2 at the surface, so step zero produces 400 back from the shell:
(0) 800 (planet) -> 400 to space and 400 back-radiated =(1)
(1) 400 +800=1200 –> 600 to space and 600 back-radiated =(2)
(2) 400+200=600 +800=1400 –> 700 to space and 700 back-radiated =(3)
(3) 400+200+100=700 +800=1500 –> 750 to space and 750 back-radiated =(4)
(4) 400+200+100+50=750 +800=1550 –> 775 to space and 775 back-radiated =(5)
(5) 400+200+100+50+25=775 +800=1575 –> 787.5 to space and 787.5 back-radiated =(6)
(6) 400+200+100+50+25+12.5=787.5 +800=1587.5 –> 793.75 to space and 793.75 back-radiated =(7)
(7) 400+200+100+50+25+12.5+6.25=793.75 +800=1593.75 –> 796.875 to space and 796.875 back-radiated =(8)
(8) …

Hopefully, you can see this converges towards 800 but will never exceed that. So the shell will never emit more to space than the planet core is producing. No perpetuum mobile, no free lunch, energy conserved, thermodynamic steady state achieved.

180. Sparks says:

It’s interesting don’t you think how the “green house effect” has evolved from being a basic idea, into a theory that causes a runaway thermal reaction, and then tries to survive by making loopholes in understood physics.

[Reply] 1) It isn’t a runaway reaction, it’s a positive feedback which slows down and converges at the steady state value. See the figures I just provided to Arfur above.
2) Argument by assertion doesn’t fly here, so back up what you claim with clear argumentation, or expect to see a strikethrough. Be very aware that we are discussing a toy planet thought experiment, not the real Earth’s atmosphere.

181. Sparks says:

Feedback’s exist, they are also weak, even waterfalls are supposed to produce a positive feed back, they do not produce enough energy to warm up the water they fall into.

[Reply] Feedbacks are weak once they are near a steady state. Here we are looking at a conceptual model where an entire planet is suddenly surrounded by a steel shell. Feedback is initially strong, but weakens as it approaches a thermodynamically steady state, as you can see from the figures I provided to Arfur above.

182. Sparks says:

Okay cool!

183. suricat says:

tallbloke says: March 12, 2013 at 12:07 pm

“Max, we’ve already accepted that there will be a difference. Do you accept that it’s small (less than 0.5%) when considering the areas of spheres with a 10km/7000km radius difference?

using the radius instead of the area for the SB calculation may be algebraically valid, but it is physically incorrect.

Does it come up with the wrong figures for power density? How far out are they?”

OK, One last try. 🙂 I wouldn’t dispute the difference between the radii of the ‘shell’ against the ‘core’, only that the ‘shell’ radius must always be greater than that of the ‘core’!

Having established the above, it now follows that ‘the radiating surface area of the shell’ must be GREATER than ‘the radiating surface area of the core’.

If we assume your “(less than 0.5%)” difference in surface area to be only ‘0.2%’ more than the core’s surface area, the scenario for a steady state then becomes:

Total core energy irradiated from the core surface = 100%

Total core energy irradiated from the shell surface = 100%

Energy intensity irradiated from the core surface = 235W/m^2

Energy intensity irradiated from the shell surface = 235 – 235 x 0.002 = 235 – 0.47 = 234.53W/m^2.

Did you notice that this is different and doesn’t ’emulate’ the ‘Toy’? The ‘Toy’ shows the ‘energy intensity’ per ‘unit area’ to be ‘the same’ for both the ‘core’ and the ‘shell’. This is a ‘physical impossibility’ and shows that the ‘thought experiment’ is ‘invalid’!

For the ‘shell’ to irradiate the same ‘energy intensity per unit area’ as the ‘core’, another source is required to make up the total energy deficiency of the ‘core’s’ output. 🙂

I just hope that helps. 😦

Best regards, Ray.

184. tallbloke says:

Ray: You say invalid, I say slight inaccuracy of simplified presentation justifiable in order to communicate the concept more easily.
It’s a nitpickers paradise this place. 😦

185. philr1992 says:

Tallbloke says:

“Look back through the thread at the matrix of figures I provided. The addition of the shell raises the amount of radiation being emitted by the planet”

Uh, what? How do you figure thermalization of photons emitted from the shell is taking place on the planet? The frequency emitted from the planet will not change. This has been laboratory tested. It sounds like you’re trying to count photons.

186. tallbloke says:

Philr: No, much cruder than that. I’m counting power in Watts. It’s an engineer thing. Anyway, I think of radiation as waves rather than photons.

The frequency emitted from the planet will not change. This has been laboratory tested

Tell us more, I’m always interested to learn.

187. Bryan says:

Thanks to Willis Eschenbach and his model .
It sums up in a very simple way what the Greenhouse Effect is, according to the theory.

It has been widely endorsed by various commentators and given a blessing by Skeptical Science.

The vacuum gap between planet and shell is said to be of no importance to the theory.

Unfortunately although beguiling, it turns out to be false physics.

If the metal planet has a large coefficient of expansion and the shell a smaller one it can be possible for the higher temperature planet will touch the shell.
The shell will now emitt 470W/m2
The planet will now cool and re-establish the vacuum and then heat up again.
The process will endlessly repeat.
The shell will never emitt less than 235W/m2 and will quite often emitt 470W/m2.

A passive shell has multiplied the output of the original nuclear reaction making the shell average output > original reactor input.

This is clearly contradicting the first and second laws of thermodynamics

188. tallbloke says:

Bad boy Bryan sticks the boot in 🙂

What about the loss of heat from the planet conducting through the shell to space? I expect it’ll be putting out quite a bit less from the surface while it builds up again.

189. Sparks says:

Where’s the “green house effect”?

[Reply] Well on planet toytown, it’s in the raising it’s surface temperature by around 45C, doubling the energy emitted by the surface of the planet. On Earth, Tim F will claim it’s around 33C. I say around 8C

190. Bryan says:

tallbloke says, “I expect it’ll be putting out quite a bit less from the surface while it builds up again.”

Yes but never less than 235W/m2

When the hypothetical planet cools and the vacuum re-established ;

the planet will still be radiating >235W/m2…….. or so the theory goes.

191. Westy says:

Talbloke,

Just wanted to say thanks for this post and the blog, thoroughly enjoyable, you deserve a win in this years Bloggies. That said, after much contemplation I’m with Kristian and Joe on this one. Cheers.

192. Arfur Bryant says:

tallbloke says:
March 12, 2013 at 9:14 pm

Rog,

“4)The planet has no trouble absorbing energy from the shell.” You then go on to say that the planet does eventually warm up (even though you disagree with Willis) because of this absorption.

I asked in my post of [March 11, 2013 at 9:51 pm] if anyone could provide evidence for cooler surfaces being able to absorb radiation from warmer surfaces. Could you provide that please, as it would go along way to sorting out my confusion?

To me, the issue is pretty simple. If a hotter body emits radiation which is then absorbed by a cooler body, then does the radiation being emitted by the cooler body get absorbed by the hotter? If yes, the ‘toy’ planet almost makes sense. If no, then it doesn’t come close to making sense.

The shell will always be cooler than the planet, by however small a margin, as it will never emit 235W/m^2. Your matrix proves it.

So I reckon this entire thought experiment rests on the ability (or not) of radiation from a cooler surface being absorbed by a warmer one. I’d really appreciate this being sorted!

To be fair, you did say at the start [“I’ll probably regret this…”]! 🙂

193. Arfur Bryant says:

Rog,

Sorry, I meant to say the shell will never emit 235 if the downward radiation is not absorbed, so halving the 235 then running your matrix.

194. tallbloke says:

Bryan: I’ll sleep on it. Basically, as soon as there is good conduction between planet and shell, the ‘greenhouse effect’ is short circuited.

195. Arfur Bryant says:

Oh bugger. Forget the 235 part. Just stick to the “shell will always be cooler than the planet” bit. I’m going to bed.

196. tallbloke says:

Arfur: Go take a look at TheFordPrefect’s experimental work, that should help answer it for you.

197. suricat says:

tallbloke says: March 12, 2013 at 10:38 pm

“Ray: You say invalid, I say slight inaccuracy of simplified presentation justifiable in order to communicate the concept more easily.”

If you suggest an ‘educational’ content, I must strongly disagree. The ‘subject matter’ of an ‘educational product’ must adhere to the ‘current truths’ of our ‘current knowledge database’. The ‘current subject matter’ in this thread is an ‘anomaly’ and not the educational product that one would present ‘without’ a ‘curriculum’ that led up to it.

Best regards, Ray.

198. Tim Folkerts says:

mkelly says: “You have falsified the shell and I say congrats.

No, I have improved the model.

First, I confirmed the basic model:
```If the shell is close enough to the surface that they can be considered to have the same radius, and if there is an internal power supply providing a heat flux of F_0 = P/A = 235 W/m^2 at the surface of the planet, then the planet's surface will radiate 2*F_0 = 470 W/m^2, and the shell will radiate F_0 = 235 W/m^2 upward and F_0 = 235 downward producing temperatures of ~ 302 K and 254 K respectively.```

I then extended the model to deal with cases where the radius of the shell was explicitly assumed to be larger than the planet below. Willis’ model is simply a special case of my results.

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

Showing that the IR radiation from the planet is (470 – δ) W/m^2 instead of exactly 470 W/m^2 does NOT falsify the model. It simply shows that this model, like all models is a simplification of the real world.

[Reply] It’s a simplification of the theoretical physics of a body surrounded by a vacuum and a steel shell, and nothing like the real world, if by ‘real world’ you mean our planet Earth.

199. graphicconception says:

May I respectfully suggest that those who have a problem with the differing radii and hence differing surface areas please consider an alternative formulation of what I think should be the same problem. (If it is, in fact different, I suspect someone might mention it!)

Consider a hollow cube of perfect insulating material made into a lidless box. Place a “core” sheet that has the energy source at the bottom of the box and the “shell” sheet at the top where the lid of the box sould be.

Both areas identical. The shell sheet emits 235 W/m2 into space. The core sheet was initially set up to radiate 235 W/m2 into the box.

200. gbaikie says:

“However, a resistor with a current running through it will get very hot if you put it in a vacuum flask…
….and it will stop getting hotter once the outside of the vacuum flask is radiating as much energy as the resistor was in open air.”

Because open air loses heat by convection.
If resistor is in room temperature and is not warmer than the air, putting it in a vacuum flask
should not cause it to get much hotter- if vacuum flask is not reflecting the heat, not warmer
at all.

A resistor is good comparison to nuclear heat- both can achieve relatively high temperatures-
unlike the energy of sunlight at Earth distance.

Now take resistor which isn’t warmer then air temperature, and not warmer in vacuum flask
and encase same resistor in ceramic sphere- the resistor could then get warmer, though
outside of ceramic sphere will not be warmer.
This would quite similar to this core of steel greenhouse.
If this ceramic sphere has outer shell of steel, the surface of ceramic will increase in temperature
by a small amount and if steel shell is reflective than a greater amount it will it be warmed.
Having surface of ceramic sphere slightly warmer will in turn cause entire sphere to warm and
will then cause resistor to be warmer.

What is occurring is net flow of heat is reduced by steel shell by fairly small amount.
What important is the resistance of the ceramic which caused to the
resistor to be warmer in the first place. By having a slightly warmer
surface you reducing the amount heat which passes thru the ceramic-
you making the ceramic a better insulator.
Perhaps a small amount add warmth at the surface will make a greater
temperature difference to the resistor.
This is what I see as the problem of the model- the factors causing
the 235 W/m2 are unknown and ignored.
Which is essentially the history of climate science- ignore the major
components and focus on some small insignificant aspect [and add to this,
the insignificant aspect is hopelessly badly defined].

201. Tim Folkerts says:

Bryan asks: “How far apart do the planet and the shell have to be for the planet to be twice as warm as the shell?” [NOTE: I assume you mean “to radiate twice as much power = to be 2^(0.25) times warmer.]

Any distance will do. As long as there is vacuum separating the two, the shell will be~ 254 W/m^2 and the planet will be ~ 302 K.

(To avoid various potential quantum mechanical effect, the two should probably be at least several wavelengths apart — 1 mm should be safe.)

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

Bryan also cleverly suggests:“If the metal planet has a large coefficient of expansion … The shell will never emitt less than 235W/m2 and will quite often emit 470W/m2.”

Yes, the shell WILL sometimes emit less than 235 W/m^2 to space, but you miss a couple important steps.

For the sake of simplicity, lets assume that we can inflate and deflate the planet at will, so we can expand it and contract it whenever we want.
Basically you are saying …
* start with no contact — planet @ 302 K & 470 W/m^2; shell @ 254 K & 235 W/m^2 (both up and down)
* inflate the planet to touch the shell; shell briefly emits 470 W/m^2 when it rises to the 302 K temperature of the planet, but both eventually cool to emit 235 W/m^2 @ 254 K
* deflate the planet to break contact …. this is where you go wrong.

The planet @ 254 K emitting 235 W/m^2 will break contact with the shell and will radiate 235 W/m^2 to the shell. The shell, radiating 235 W/m^2 up AND 235 W/m^2 down will cool quickly until it is 214 K radiating 117.5 W/m^2 up and 117.5 W/m^2 down.
The planet will start to warm, slowly approaching 302 K & 470 K, while the shell slowly warms toward 235 W/m^2 @ 254 K.

So the shell sometimes radiates MORE than 235 W/m^2 and sometimes radiates LESS than 235 W/m^2. I guarantee that the net effect will be to balance out to 235 W/m^2 on average. There are no violations of any laws of physics.

Although Bryan’s scenario is beguiling, it turns out to be false physics.

PS. All of the specific numbers are assuming no “radius cooling affect”, ie the shell is always “close” to the same radius as the planet.

202. Arfur Bryant says:

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. 470W/m^2 from the planet is wrong. Even if it could absorb the ‘cooler’ radiation, it would never receive 235 from the shell because the shell never received 235W/m^2 (even though the area difference is small). The planet will always be warmer than the shell. I’m not sure that TFP’s experiment shows absorption from the cooler body. In his experiment, the hot plate does NOT get hotter.

From your matrix, you assert: [“Hopefully, you can see this converges towards 800 but will never exceed that.”]. Well, not only will it never exceed 800, it will never even reach 800.

Rog, Kristian, anyone, am I making sense? I’ll get back in my box if I’m not…

Defo going to bed now.

203. Greg House says:

Greg House says, (March 12, 2013 at 7:43 pm): “Dear friends of back radiation’s 🙂 ,

I know from my experience on blogs that it is sometimes hard to understand, let us say, that it is at least possible that back radiation won’t warm the source.

Roger says: [Reply] Back-radiation doesn’t warm the source. It slows its rate of cooling.
=============================================================

Right, this is what I meant. I should have said “I know from my experience on blogs that it is sometimes hard to understand, let us say, that it is at least possible that back radiation won’t worm/slow down rate of cooling of the source” to avoid ambiguity. Because in certain cases there is no cooling, like in our particular hypothetical case where there was no cooling initially (because of the planet’s own source of energy), this was a precondition.

Anyway, if we assume that back radiation from a colder object slows down cooling of a warmer source as the source is in the process of cooling (but is still warmer than the colder object), it is hard to explain why would back radiation suddenly stop affecting the temperature of the source any longer if the source suddenly stops cooling. So, in this case we should logically expect an actual increase in temperature of the source, which would be literally warming.

Just for the record, it is not that I believe that this worming/slowing down rate of cooling by back radiation is a real physical process. This hypothetical Willis’ universe we have been discussing actually confirms this opinion of mine, because it leads to an impossible outcome, as I demonstrated earlier.

The only way to avoid this outcome in my demonstration is to assume that the black body planet radiates more and more retaining the same temperature all the time, but I do not see how it can be possible. So, it comes to contradiction one way or another and this disproves the initial assumption.

[Reply] In my opinion the only thing you “demonstrated earlier” was your misunderstanding. Maybe you should link back to what ever you thought was right, or restate it, rather than expect people to hunt through a 120 comment thread.

204. Tim Folkerts says:

Kristian says …
“I don’t know why this is so hard to grasp.
Neither do I. 🙂
Let’s figure it out.

“The core planet with no surrounding shell emits its flux of 235 W/m^2 directly to space at an emission temperature of 254K. “
OK

“Then we surround it with the steel shell. What needs to happen? The outer surface of the shell needs to attain the same emission temperature as the surface of the core planet.”
OK

“How is this accomplished? By transferring heat from the (warmer) core planet to the (colder) surrounding shell until it’s done and balance is reached. “
I’m still with you.

“The core does not need to become gradually warmer in order for this heat to be tranferred. As long as it remains warmer than the colder shell, the transfer of heat will continue, albeit at a steadily slower pace.”
Now it starts to get tricky. If you want to keep the planet at constant temperature then you would have to steadily decrease the power to the planet. 235 W/m^2 keeps the planet @ 254 K when surrounded by cold outer space. When surrounded by a warmer shell, less power would be needed to keep the planet warm so we could turn down the nuclear reactors and still stay at 254 K.

However the scenario postulates not constant temperature but constant power. In this case the temperature would continue to rise. Even after a steady-state is achieved, there sill must be a NET 235 W/m^2 from the planet to the shell.

“The heat from the core is absorbed by the shell and turned into KE, and this process will continue until it has reached the temperature at which it will emit 235 W/m^2 to space. This happens when its temperature (determined by its content of KE) is 254K.
Yes but, the INNER surface is ALSO radiating 235 W/m^2. So the shell has to be receiving 2×235 W/m^2 from a planet @ 302 K. Fortunately, the planet is receiving 2×235 W/m^2 (half from IR radiation and half from nuclear heaters).

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

PS For those who worry about the “radius cooling effect”, just add “-δ” to all the appropriate numbers above.

205. suricat says:

Tim Folkerts says: March 13, 2013 at 1:09 am

“Kristian says …
“I don’t know why this is so hard to grasp.
Neither do I.
Let’s figure it out.

“The core planet with no surrounding shell emits its flux of 235 W/m^2 directly to space at an emission temperature of 254K. “
OK

“Then we surround it with the steel shell. What needs to happen? The outer surface of the shell needs to attain the same emission temperature as the surface of the core planet.”
OK

“How is this accomplished? By transferring heat from the (warmer) core planet to the (colder) surrounding shell until it’s done and balance is reached. “
I’m still with you.

“The core does not need to become gradually warmer in order for this heat to be tranferred. As long as it remains warmer than the colder shell, the transfer of heat will continue, albeit at a steadily slower pace.”
Now it starts to get tricky. If you want to keep the planet at constant temperature then you would have to steadily decrease the power to the planet. 235 W/m^2 keeps the planet @ 254 K when surrounded by cold outer space. When surrounded by a warmer shell, less power would be needed to keep the planet warm so we could turn down the nuclear reactors and still stay at 254 K.

However the scenario postulates not constant temperature but constant power. In this case the temperature would continue to rise. Even after a steady-state is achieved, there sill must be a NET 235 W/m^2 from the planet to the shell.

“The heat from the core is absorbed by the shell and turned into KE, and this process will continue until it has reached the temperature at which it will emit 235 W/m^2 to space. This happens when its temperature (determined by its content of KE) is 254K.”
Yes but, the INNER surface is ALSO radiating 235 W/m^2. So the shell has to be receiving 2×235 W/m^2 from a planet @ 302 K. Fortunately, the planet is receiving 2×235 W/m^2 (half from IR radiation and half from nuclear heaters).

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

PS For those who worry about the “radius cooling effect”, just add “-δ” to all the appropriate numbers above.”

OK, so that’s a ‘big’ quote that I made TF! However, it just reflects the ‘adjustments’ that need to be made to the ‘OP’ (original post).

TBH, I don’t think the ‘Toy’ can be redeemed by ‘insertions’ of logic. It’s just ‘wrong’ on too many aspects. 😦

Best regards, Ray.

206. gbaikie says:

“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. 470W/m^2 from the planet is wrong. Even if it could absorb the ‘cooler’ radiation, it would never receive 235 from the shell because the shell never received 235W/m^2 (even though the area difference is small). The planet will always be warmer than the shell. I’m not sure that TFP’s experiment shows absorption from the cooler body. In his experiment, the hot plate does NOT get hotter.”

I say radiation from cooler body can slow the cooling of hotter body. And the most significant
aspect would be the reflection of the cooler body of energy from warmer body.

If something is making 1000 units of heat and it’s radiating 1000 units of heat, it has a certain
temperature.
Now if something which makes 1000 units of heat and can be caused to radiate 999 units of heat, the substance has temperature and the temperature is increasing. And continues to increase in temperature until it’s radiating 1000 units of heat.

So if you can cause something which is generating heat to then loss less heat it will increase
in temperature. And we will call this the net gain of heat of a energy budget.

It is this net gain of heat which makes James Hansen think Earth will resemble Venus.
It is the sum total of CAGW or AGW.
Though there is another factor one could call, compound interest- the more gain temperature
means more added heat. Or a warmed world, causes an even warmer world.
Or the significant difference between CAGW and AGW is the amount of compound interest
that is thought to be possibly involved.

Now the amount of net gain and compound interest is imagined to be involved is tiny. Or a watt
per second per square meter is large in comparison. Or there is no doubt that *if* one could add
1 watt [or joule of heat] per second, that after 1 year of time, one gets a lot of joules of heat:
3600 second in an hour 24 per day, and 356 days year: 31.5 million joules per square meter.
So for these idiots the question is what fraction of 1 watt per square meter on average is being added per second. And the raving idiots are wonder how much of the added heat is increasing the amount of net gain.

Btw, as for a sense of scale. 31.5 million joules per square meter.
One square meter of water 1 cm deep is 10 kg of water which has 4210 joules of heat per one degree of temperature [C or K] per kg. So 42100 joules per 10 kg, 31.5 million joules adds 749 C.
Or a meter deep is 7.49 C. And km deep is .00749 C. Or no one thinks there is as much as 1 watt per square meter being added, as this amount could actually be easily measured.

But Trenberth, Fasullo & Kiehl in their heat budget chart imagine .9 watt per square meter is the net gain:

207. gbaikie says:

-“The core does not need to become gradually warmer in order for this heat to be tranferred. As long as it remains warmer than the colder shell, the transfer of heat will continue, albeit at a steadily slower pace.”

Now it starts to get tricky. If you want to keep the planet at constant temperature then you would have to steadily decrease the power to the planet. 235 W/m^2 keeps the planet @ 254 K when surrounded by cold outer space. When surrounded by a warmer shell, less power would be needed to keep the planet warm so we could turn down the nuclear reactors and still stay at 254 K.

I say: How much would reduce this power would depend on, how much “steadily slower pace” is.
if shell was the mythical ideal blackbody then it would by definition of this body have a zero reduction in pace.
So we need to know the difference steel would conduct the heat as compared to blackbody which conduction heat perfectly.
Thin steel will conduct heat quicker than thick steel. With same thickness of Copper conducting better than steel. Diamond conducts better than copper and ideal blackbody conducts better than diamond.

However the scenario postulates not constant temperature but constant power. In this case the temperature would continue to rise. Even after a steady-state is achieved, there sill must be a NET 235 W/m^2 from the planet to the shell.-

Yes, one, could take a wild guess that something like 1 watt per square meter per second is the the rate of “steadily slower pace”, which cause core surface temperature to increase, which would cause the core temperature to increase. It would take a very long time to balance the net flow of heat. Probably get volcanic eruptions in this time period, add some complications.

208. Tim Folkerts says:

Arfur says (and many others echo the sentiment) …

“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.

I wish I knew the best way to say that the “toy planet” is not violating any laws of physics. Let me try a “reverse” approach.

Consider a nichrome wire with 10 Ω of resistance. Suppose I run 10 A through the wire, generating 1000W of power. This happens to be enough to warm the wire to 1000 K (glowing a nice red color) when put in a 300 K (room temperature) room.

Scenario 1) Suppose I move the wire to the interior of a furnace @ 1000 K, keeping the TEMPERATURE of the wire constant.
Q: How much current will I have to run thought the wire to keep it at 1000 K?
A: 0 Amps generating 0 W of power, since the furnace will keep it at 1000 K.

Scenario 2) Suppose I move the wire to the interior of a furnace @ 1000 K, keeping the POWER of the wire constant.
Q: What will the new temperature of the wire be?
A: Somewhere above 1000K (I would guestimate somewhere between 1150-1200K, but the number is immaterial.)

Scenario 3) Suppose I move the wire to the interior of a furnace @ ~1150 K, keeping the POWER of the wire constant.
Q: What will the new temperature of the wire be?
A: Somewhere above 1150K (I would guesstimate somewhere between 1250-1300K, but the number is immaterial.)

The Point of scenario 1 is that the surroundings can warm the object to their temperature without any separate power to the wire.

The Point of scenarios 2 & 3 is that the 1000 W of electrical power can warm the wire TO ANY TEMPERATURE as long as the surroundings are the right bit cooler than that temperature.

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

It is always the COMBINATION of the surroundings and the heater that causes the final temperature. Start with a T=0K planet.

* A planet with 235 W/m^2 steady heat and 0K surroundings will reach 254 K.
I think we all would would say the heater “warmed the planet to 254 K”.

* A planet with 235 W/m^2 steady heat and 254 K surroundings will reach 302 K.
I would tend to say that BOTH the surroundings and the heater together “warmed the planet to 302 K”.
If you prefer, you could say “the surroundings warmed the planet from 0K to 254K, and the heater warmed the planet from 254 K to 302 K”. Then you avoid having to say the “cool” 254 K surroundings “warmed” (or even “helped to warm”) the planet from 254 K to 302 K. This is semantics (and questionable physics), but it might make sense to some.

209. PhysicistPhil says:

” If you want to keep the planet at constant temperature then you would have to steadily decrease the power to the planet. 235 W/m^2 keeps the planet @ 254 K when surrounded by cold outer space. When surrounded by a warmer shell, less power would be needed to keep the planet warm so we could turn down the nuclear reactors and still stay at 254 K. ”

Again, radiative units are analogous to velocity only, not material flux or direct transport. The net frequency, or rate of “loss” from this hypothetical nuclear planet will always exceed the return rate from the shell..you’re not going to thermalize this backflux from the shell.

To put this in laymans terms…you’re NOT going to warm this hypothetical nuclear planet any further because it is already losing more energy to the shell than the shell is delivering to it…what is the thermalization mechanism?

You cannot “add” 5W/m^2 to 15W/m^2 to get 20W/m^2.

[Reply] Please start by explaining how radiation from a warmer body thermalises matter in a cooler body.

210. tallbloke says:
March 12, 2013 at 2:28 pm

CF: if you responded to the bolded part of what I wrote I missed it: did you?

“I am sorry Roger, you lost me with this post, I have always thought this was a thoughtful blog with input of engineers who have some experience. I hope it gets back to real science based on ethical engineering principles.”

[Reply] Ouch. I am covering in this post what theory says will happen with a vacuum between the shell and surface. That doesn’t pertain in your furnace, where convection will likely prevent re-radiation from adding to the output from the heating element due to re-absorption in the air. However, follow the link to ‘TheFordPrefects’ blog and study the empirical test he conducted, then report back.

Tallbloke, You should know when temperatures are very high (some flame temperatures I have measured exceed 3000C) radiation is a very important means of heat transfer, after all the heat transfer equation has Tf^4 in the calculation. The amount of convection depends on the position of the flame and the aerodynamics of the flame (dimensions, momentum and stability) Most refractories can withstand temperatures over 1500C and some much higher to over 3000C. There can be reasons to have a combustion chamber where no heat is transferred.eg in co-current driers where part of the product is likely to melt or is heat sensitive to a reaction. In such a combustion chamber the flame temperature maybe 1200-1500C and the wall temperatures the same. It is possible to measure the temperature, composition and flow rate of the gases leaving the chamber. It is possible to measure the temperature and flow rate of combustion air and the energy input from the fuel. The difference between input and output is the heat loss which can also be determined from the outer wall temperature of the furnace, the surface area and the surrounding conditions. It is of course impossible to obtain ideal conditions or zero heat loss.but the ideal situation can be approached. That is what engineering is about. Economics (taking into account safety) will determine how much heat loss can be allowed or is necessary in any process.

As I said in my experience I can not accept that there is any back-radiation which provides for new energy or increases the source temperature. There is nothing special in the atmosphere of planet earth that provides different conditions. Gases which absorb radiation at particularly wavelengths only do so from a source of higher temperature (the sun or some parts of the earth surface). These gases only radiate to receivers of lower temperature. It is false to assume that radiation occurs in all directions. Even if you accept the concept of photons (which I do not) then photons only travel in the direction of lower temperature which ultimately is space (assumed to be infinite and at zero K)

Roger, sorry for delays in replies. Also, do not know how to bold or highlight.
I have put in links to some of Prof. Claes Johnson’s mathematical posts. I hope you read some of them. I have been intrigued by his posts of mathematical computation on fluid dynamics (flight, sailing etc). His posts on radiation make some sense. I have yet to see any proof that he is wrong. I think those who point to supposed higher authority a) do not understand his maths b) do not understand the subject of the posts and c) have had no actual experience.(I am pleased he has accepted some of my comments) I have suggested to Prof Johnson and Dr A Makarieva that they should get together. Both have something to contribute in the understanding of atmospheric conditions.
You have put up some great thought provoking posts such as the recent one on the Fibonacci series and golden numbers. There is no doubt the sun and planets have an effect on our planet.

[Reply] Thanks for the response. Something you said confused me:
“photons only travel in the direction of lower temperature”
How do they know to avoid traveling towards warmer bodies?
I do however take on board what you’ve said elsewhere about the lack of thermalisation in low temperature situations, and that’s part of the reason I’m always open to being corrected. Also, the wikipedia page has very little to say:

http://en.wikipedia.org/wiki/Thermalisation

211. donald penman says:

If the diference between what the inner core produces and what the outer surface of the inner core radiates is thought of as the amount of energy needed to heat the outer shell then radiation emitted by the core will be large when temperature difference is large but will get less when the outer surface of inner core gets closer to the temperature of the outside shell.The outer shell will become the same temperature as the inner core because the inside of the outer shell is concave and will radiate to itself as well as the inner core which must radiate all of its energy to the outer shell.The mathematical model presented does not prove anything it just reflects the assumptions you are making in this steel shell world.

212. Kristian says:

(from core planet) 235 235 0 (to space)

(from core planet) 235 0 235 (to space)

[Reply] Hi Kristian. WordPress interprets anything between a left and right arrow as code. If it isn’t recogniseable as code, it bins it. I suggest you resubmit your main comment without using arrows, or for a left arrow use the ampersand symbol ‘&’ followed by lt (for ‘less than’). For a right arrow use & followed by gt for ‘greater than’.

213. tallbloke says:

Thanks everyone for the interesting overnight (UK time) comments. To save time I’ve added inline responses, so please check back.

214. Kristian says:

Tim Folkerts says, March 13, 2013 at 1:09 am:

We seem to be talking past each other, Tim. There are a couple of things I would like for you to consider.

1) If you’ve got one radiator at 254K emitting a radiative flux of 235 W/m^2 and then place another one just next to it, also at 254K and emitting 235 W/m^2, do you consider the total flux to the surroundings from these two put together to be twice as large as the flux from just the single one? And if so, would the total temperature of the two in your opinion also be accordingly higher (302K)?

and (more significantly)

2) At the very onset of the ‘new’ situation (shell placed around the planet), the inner surface of the shell would instantly gain 235 unrestricted W/m^2 of heat from the planet, while its outer surface towards space would gain zero flux. If these two surfaces were both perfect black bodies, they would instantly attain an emission temperature according to the absorbed flux and accordingly emit the same flux back – Inner: 254K, 235 W/m^2; Outer: 0 K, 0 W/m^2. Between the two surfaces, however, there’s a conductive layer of an unknown thickness and conductivity/resistivity. This is a definite weakness in Willis’ model, I feel, because it complicates the heat transfer business immensely. There is not just ideal thermal radiation to be considered. Determining the thickness and thermal conductivity/resistivity of the steel shell actually becomes decisive.

If the shell were a perfect insulator, both of its sides/surfaces would as mentioned act like perfect black bodies. But this would be an intolerable situation. No energy would be transferred from the inner surface to the outer, because the entire received flux from the planet would always just instantly be absorbed and reemitted from the inner shell surface, no matter how large, no matter how steep the temperature gradient across the shell. And hence, no energy would ever be able to leave the outer surface to space, it would forever be 0. In such a scenario, we would have a chain reaction and a meltdown.

235-> <-235 0->

If, on the other hand, the shell were a perfect conductor, the received radiative heat flux from the planet at the inner surface of the shell would instantly be absorbed and turned into KE, directly conducted away, tranferred through the shell without delay and out to the outer surface, from where the KE would just as promptly be turned back into thermal radiation and emitted without loss to space.

235-> <-0 235->

Clearly, reality would land somewhere in between these to extreme scenarios. But, and this is important, Willis’ planet/shell setup was NEVER meant to be anything but a conceptual model, a strictly hypothetical rendition of a ‘real’ situation, its main purpose to show how heat transfer by radiation differ fundamentally from that by conduction. And still we end up realising that the matter of conduction of heat through the steel shell is exactly what will decide the outcome of it all.

So my question to you, Tim, is: Do you consider the two surfaces of the surrounding shell both to be perfect black body surfaces, instantly absorbing and reemitting the entire radiative flux received and gaining a temperature in direct accordance with this flux? If so, you’ve made the steel shell a perfect insulator. Or do you consider we should disregard the complicating issue of conduction through the shell (its thickness and/or conductivity/resistivity) by making it perfect and thus simply letting us focus on what Willis appeared to be wanting us to focus on – the special properties of heat transfer by thermal radiation? Or, third option, would you like to suggest a specific thickness and/or a conductive value for the shell?

Remember, Willis’ basic premise seems to be that the heat transfer from the planet to the shell is very different in nature if radiated across a vacuum than if conducted through a solid (his connecting ‘pillars’). Here is what he says:

“Now, this magical system works because there is a vacuum between the planet and the shell. As a result, the planet and the shell can take up very different temperatures. If they could not do so, if for example the shell were held up by huge thick pillars that efficiently conducted the heat from the surface to the shell, then the two would always be at the same temperature, and that temperature would be such that the system radiated at 235 W/m2. There would be no differential heating of the surface, and there would be no greenhouse effect.

Let me summarize. In order for the greenhouse effect to function, the shell has to be thermally isolated from the surface so that the temperatures of the two can differ substantially. If the atmosphere [What atmosphere, Willis? There’s a vacuum!] or other means efficiently transfers surface heat to the shell there will be very little difference in temperature between the two.”

This does not take into account the thickness or the conductivity of the shell. It only focuses on the transfer BETWEEN the planet and the shell. Willis seems to be favouring my second option – ‘efficient’ conduction.

Problem is, the heat in Willis’ model is transferred just as efficiently from the planet to the shell by radiation across his vacuum as it would be through his conducting pillars. The energy and temperature would spread just as effectively. So why the postulated temperature differential?

215. Kristian says:

Thanks, TB

216. tallbloke says:

The wikipedia page on thermalisation is astonishingly uninformative I have to say:
http://en.wikipedia.org/wiki/Thermalisation

217. Roger says? [Reply] Thanks for the response.
“photons only travel in the direction of lower temperature” How do they know to avoid traveling towards warmer bodies?

Why does time go in one direction? why does electricity flow in the direction of higher to lower voltage? My explanation is that experience tells you. I have seen no better explanation. One can represent Fourier’s Law of conduction by an electrical circuit. Infra-red radiation is part of the electromagnetic energy spectrum. Even when energy is transferred with the help of work input such as refrigeration the 2nd law of thermodynamics is obeyed.
Anyway, I have stated I do not agree with the concept of billions/quadtrillions or what ever photons each at some specific wavelength such as 14.800000 or 14.800001 micron. I have said that the concept of energy waves made up of a series of wavelengths and amplitudes at different wavelengths (something like sound oscillations or music) makes more sense. The energy emitted is determined by the surrounds of the emitter. Prof. Claes Johnson has elaborated some of this concept in posts. Claes and many others have shown that energy waves can explain anything photons are supposed to do.
I could equally ask can you prove from experiments that photons exist and are unique (ie there is no other explanation). Then can you prove from experiments that a photon goes in any other direction than hot (higher enthalpy/entropy) to cold (lower enthalpy/entropy)? Now if you want to change words (as AGW morphed into climate change) and interchange photons and wave energy (I have seen at Lubos Motl the term -photons/waves) that is a different picture. Then there may be semantics and we may then be talking about similar concepts.

218. Bryan says:

Tim Folkerts says of my previous post (March 12, 2013 at 10:56 pm )

“Basically you are saying …
* start with no contact — planet @ 302 K & 470 W/m^2; shell @ 254 K & 235 W/m^2 (both up and down)
* inflate the planet to touch the shell; shell briefly emits 470 W/m^2 when it rises to the 302 K temperature of the planet, but both eventually cool to emit 235 W/m^2 @ 254 K
* deflate the planet to break contact …. this is where you go wrong.”

Tim, distorting my post will not let you escape from the inevitable conclusion that your ideas on the Willis model are completely wrong .

The metal planet with a high coefficient of expansion will when heated automatically touch the outer shell if the outer shell has a much smaller or no coefficient of expansion.
These metals exist.

No question of having to inflate the planet as you suggest!

The expanding planet at 302K touches the shell at 254K

The shell heats up to 302K and emits at 470W/m2 to space.

The planet cools down but not necessarily to 254K, only by enough to break contact lets say 290K.

The shell also cools down to 290K but will still be emitting more at more than 235 W/m2.

As cooling shell is still emitting more than the usual 235W/m2 towards the planet the planet will quickly heat up.
This process repeats itself endlessly with the result that the shell never emits less than 235 W/m2 and quite often 470W/m2

The physics I have used describes a fully spontaneous system.
The laws of thermodynamics fully apply and the first and second laws are clearly violated.

Improve your reputation here at TB and admit that you were wrong.

219. […] Bryan on Entering the SkyDragon’s… […]

220. oldbrew says:

‘Why does time go in one direction? why does electricity flow in the direction of higher to lower voltage?’

Why does water flow downhill? Line of least resistance.

221. tallbloke says:

CF: Point taken. I’ve put up a new post asking for links to the physics papers which tell us how ‘thermalising radiation’ works.

222. Tim Folkerts says:

Bryan says: “The shell also cools down to 290K but will still be emitting more at more than 235 W/m2.”

No. Once again, that is where you are wrong.

The shell is now receiving less than 470 W/m^2 — if the planet is 290K as you hypothesized, the planet will be radiating 401 W/m^2 and the shell will be receiving 401 W/m^2. In order for the shell to radiate 401 W/m^2, the shell radiates 200 W/m^2 down toward the planet from the inner surface and 200 W/m^2 out to space from the outer surface. This requires a temperature of 244 K.

Let me repeat — the shell radiates only 200 W/m^2 to space, which is indeed less than 235 W/m^2. Your claim that it will be more than 235 W/m^2 at this time is wrong.

Not until the planet warms to 302K and emits 2×235 W/m^2 can the shell radiate 2×235 W/m^2 total (half from the inner surface, and half from the outer surface.)

RECAP
When they first make contact, the radiation from the shell will be >235 W/m^2
When they first break contact, the radiation from the shell will be < 235 W/m^2

PS. All of this assumes that:
1) the shell in your model has negligible heat capacity compared to the planet, so that it "quickly" reaches the new steady-state conditions. The analysis would still work (but would get much messier) if you included heat capacity in your model.
2) the shell in your model is close to the same radius as the planet so the "small but real radius-cooling-effect" can be ignored.

PPS. The 'inflatable planet' was simply an effort to be able to easily control when the two touched and when they didn't touch without having to deal with the vagueness of how specific coefficients of expansion would make and break contact with shells of specific radii at specific temperatures. Sorry if that confused you.

223. tallbloke says:

Tim F: Once the planet is in contact with the shell, conduction will obviate the ‘inward radiation’ won’t it? If not, won’t we have to work out the same deal for all the ‘shell layers’ of the planet down to its core?

224. Greg House says:

cementafriend says:
March 13, 2013 at 8:58 am:

(Roger says? [Reply] Thanks for the response.
“photons only travel in the direction of lower temperature” How do they know to avoid traveling towards warmer bodies?)

Why does time go in one direction? why does electricity flow in the direction of higher to lower voltage? My explanation is that experience tells you. I have seen no better explanation.
=========================================================

I like this point very much and would like to elaborate a little bit.

I would rather say “experiments should tell you, not imagination”. Imagination, analogies etc. are an intrinsic part of the scientific process, but a real fact is only something that is proven to really work in reality. First come practical observations and experiment, then things are called somehow (like “photons”), then assumptions are made and further experiments follow. You can not just start thinking of photons as traveling particles of matter and do some arithmetic with them, because you do not know if this counting by analogy is correct. All this radiation arithmetic has no basis in reality until it is experimentally proven to work this alleged way in the real world. Even hypothetically it is not the only possibility that “photons” travel and do not care about direction, they can also already be everywhere and just get a sort of active if something makes them. But again, it is all just fantasy until experimentally proven correct. You can not just stop checking and proclaim a product of imagination to be a fact.

That back radiation warming/slowing cooling thing is very old, like 150 years old. And look what a professor for experimental physics did back in 1909: he conducted an experiment, instead of speculating “it must work this way, because…”. A very simple experiment, and back radiation effect was proven to be absent or negligible. This is how science works.

Sometimes it is also possible to refute something without experiments simply by demonstrating that the assumption inevitably leads to something impossible, so maybe we all should thank Willis for his brilliant idea, because it simply leads to an impossible outcome.

225. A C Osborn says:

Tallbloke, I have stated this once before, but nobody took any notice that time.
TFP’s experiment Did Not Prove Back radiation.
All it did was add extra energy to the Surroundings, thus changing the original heat flow pattern.
It also temporarily blocked the flow of radiation which added to the slowing down of the rate of cooling due to the lower Temperature Differential.
This is clear from the higher temperature of the back wall when the second plate is added.
TFP himself said that if the second plate is at Room temperature the heat loss graph would be the same as the no plate version, but he has not carried out the experiment to prove it. There must be a delay while the second plate is warmed up to allow sufficient energy to flow out, the air in the gap will just stay warmer for slightly longer.

In your last comment, it picks up a point I made earlier, if the shell is warmed by Conduction, why is it not also still radiating energy back to the Core heating it up?

Why does the energy stop radiating Back when temperature equilibrium is attained, as photons are still apparently flowing back?

226. Bryan says:

Tim Folkerts says

” shell @ 254 K & 235 W/m^2 (both up and down)”

Then you contradict yourself by saying that an shell at a higher temperature (290K) would radiate less than it did at 254K !

If the shell is at 290K as hypothesized, the shell will be radiating 401 W/m^2

Objects radiate according to their temperature as the Stephan – Boltzmann equation states.

Do you really think that the same object at a higher temperature radiates less than at a lower temperature?

Incredible !!

Do you really think that the shell has some kind of vested interest in the Greenhouse Theory that it will defy the laws of physics to keep Tim Folkerts happy?

Admit that the Willis model produces nonsense results and rejoin the rational world.

227. Bryan says:

Tim Folkerts

Which equation do you use to establish that an object at 290K radiates at 200 W/m2 ?

Do you think that a shell which was at 302K cannot cool down to 290K ?

Your fairy tale physics is just made up as you go along.

228. Tim Folkerts says:

Bryan says: “If the shell is at 290K as hypothesized, the shell will be radiating 401 W/m^2 …

You can hypothesisze that if you want, but it is still wrong!

The expanding planet at 302K touches the shell at 254K

The shell heats up to 302K and emits at 470W/m2 to space.

The planet cools down but not necessarily to 254K, only by enough to break contact lets say 290K.

That much is good.

The shell also cools down to 290K but will still be emitting more at more than 235 W/m2.

This is true ONLY AT THE MOMENT THE TWO BREAK CONTACT.

At that moment, the shell is indeed 290 K and is indeed emitting 401 W/m^2 outward and 401 W/m^2 inward. It is, however, only receiving 401 W/m^2 from the planet. That leaves a deficit of 401 W/m^2 for the shell. Since the shell is losing a net 401 W/m^2, it will necessarily cool rapidly from 290 K (remember we are assuming the shell is “thin” and has a much lower heat capacity than the planet). It must cool until it is only emitting the same power it receives. To emit a total of 401 W/m^2, it emits 200 W/m^2 from each surface, which is 244 K.

THE SHELL CANNOT REMAIN AT 290 K.

That’s worth saying one more time!
THE SHELL CANNOT REMAIN AT 290 K AS YOU ARE HYPOTHESIZING.

At this point, the planet will slowly start to warm from 290 K back toward 302 K. The shell will start to warm from 244 K to 254 K (always emitting less than 235 W/m^2).

Then the process will start over, with
* 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

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

It is only your false hypothesis about the shell temperature that is leading the to violation of conservation of energy.

229. Bryan says:

Tim Folkerts says in angry big character letters which shows he is under stress.

“THE SHELL CANNOT REMAIN AT 290 K.”

Who says the shell remains at 290K !

Certainly not me.

The shell is cooling from 302K down to 254 K and as it does so it is emitting more than the 235W/m2 that the Willis model emits.
This can only hasten the recovery of the planet to 302K and the cycle starts again.

Its good however that Tim now admits that the shell at 290K will radiate according to Stephan – Boltzmann equation so he must be well on his way back to sanity.

Tim says

“remember we are assuming the shell is “thin” and has a much lower heat capacity than the planet).”

I don’t remember making this additional condition however lets accept what you propose.

This means that the massive planet will hardly have lost any energy to the shell.
Most of the energy transmitted to space between 302K and 290K will have been supplied by the planet part of the combination.
Once separated the shell is now radiating at a higher value than the Willis model contact will resume rather quickly.
This means that once again the planet plus shell will radiate to space at a much greater intensity than the planet on its own.
This model on which the greenhouse theory depends clearly violates the first and second law.

230. Tim Folkerts says:

Bryan says: “The shell is cooling from 302K down to 254 K …

As I just explained, the shell cools from 302 K DOWN TO 244 K (not 254 K) in your scenario.

It will — at least for a while — be radiating at less than 235 W/m^2. This will “fix” the conservation of energy problem that you think exists.

231. Bryan says:

Tim Folkerts.

Despite your claims it would appear that the shell will be radiating to space at values well in excess of the Willis figure of 135W/m2 for considerable periods.

The exact excessive amount radiated to space will depend on actual values for the heat capacities of planet and shell.

Next we look at the second law.
The net result of the Willis model means that long wavelength back radiation with a Planck spectrum characteristic of 255K will be spontaneously upgraded to a shorter wavelength spectrum characteristic of 301K.

This process has never been achieved spontaneously.

Indeed this process you should know is impossible and contradicts the second law.

232. mkelly says:

Mr. Folkerts says: (T_2/T_1) = (r_1/r_2)^(1/2)

Sir this is your ratio for two spheres. Nowhere is there a mutilplication factor of 2. The T2 of the inner sphere (shell) is a function of T1(surface) and the square root of the ratio of the radii. Plug in the numbers and it proves the steel shell is false.

If T1 surface is 302 then T2 will be, using what you provided above of .9992, 301.758K. That is not 255 K nor is it 235 W/m^2.

Do you disavow your own math? If r1=r2 then T1 = T2. No double Watts nor increase in temperature.

233. Tim Folkerts says:

Bryan says: Despite your claims it would appear that the shell will be radiating to space at values well in excess of the Willis figure of 135W/m2 for considerable periods.

I think you meant “In agreement with your claims …. ”
I have clearly said that the power radiating from the shell is at times up to 470 W/m^2.

Bryan says: ” The net result of the Willis model means that long wavelength back radiation with a Planck spectrum characteristic of 255K will be spontaneously upgraded to a shorter wavelength spectrum characteristic of 301K. …

I don’t follow your objection. What do you mean by “spontaneously upgraded”? You seem to be suggesting that a 235 W/m^2 heater can’t warm something to 302 K?? Or perhaps that a 254 K object placed in contact with a warmer object can’t warm up ??

Which SPECIFIC statement of the Second Law are you referring to? How specifically is it being contradicted?

234. Bryan says:

Tim Folkerts.

Just at the moment of separation at a temperature of 290K the Greenhouse Effect resumes.

The planet will get a radiative contribution from the shell of 401W/m2 in addition to its own 401W/m2 according to the Willis – Folkerts theory.
This means that the planet surface temperature will jump to 345K from the 290K at separation.

Can you not see that this is so much nonsense?

Since the Willis numbers are just plucked out at random with a little care you could choose values that have a physical significance such as 373K the boiling point of water.

Produce steam during a temperature high and insulate it.

Then during a temperature low get a steam engine to produce work like lift objects higher than the planet surface.

Of course the whole thing is impossible as you should know.

235. lgl says:

tb

I finally managed to stop the kids ranting after suggesting this setup:
We have two equally sized radiators, and for the argument they are thin as a paper i.e only radiating from two sides. We supply enough power to keep one of them at 300K and the other at 400K when they are far apart. Then we place them very close to each other, sealing off the narrow gap between them completly so that no energy escapes from the gap. We supply the same amount of power to the radiators as before. Now, will the temperature of the radiators change? One, both, up or down?

Joe now refuses to show the calcs how much the radiators will emit (if areas say 1m2 each).
Hopefully he has found that his physics is impossible, but he will never admit it of course.

236. Bryan says:

Tim Folkerts says

“Which SPECIFIC statement of the Second Law are you referring to? How specifically is it being contradicted? ”

The specific claim was made in the post.

The net result of the process is that instead of radiating to space with a Planck spectrum characteristic of a 254K temperature it will at times be radiating to space with a Planck spectrum characteristic of 302K.

As you should know energy has quality as well as quantity.

100 Joules say of higher temperature radiation has a higher quality than 100 Joules of lower temperature radiation .

There is no process possible that will, without loss, will spontaneously convert lower frequency radiation into higher frequency radiation.

237. tallbloke says:

lgl: Joe now refuses to show the calcs how much the radiators will emit (if areas say 1m2 each).
Hopefully he has found that his physics is impossible, but he will never admit it of course.

Well he did kinda nail his colours to the mast in his opening salvo… 🙂
Thanks for your backup over there. I found it too hostile to continue
(As if I didn’t have my hands full here anyway).

238. A C Osborn says:

lgl says: March 13, 2013 at 5:35 pm “I finally managed to stop the kids ranting”

I am not sure how Joe and the other Kids on his site put up with your inane questions asking the same thing dozens of different ways for so long.
Why don’t you do exactly what you have proposed and come back on here with the results of your actual Experiment rather than playing thought games?

239. Tim Folkerts says:

Bryan, you are getting harder and harder to follow!

“The specific claim was made in the post.

Here are some specific ways the 2nd Law is often expressed.

The entropy of an isolated system never decreases.

Heat cannot spontaneously flow from cold regions to hot regions without external work being performed on the system.

it is impossible to devise a cyclically operating device, the sole effect of which is to absorb energy in the form of heat from a singlethermal reservoir and to deliver an equivalent amount of work.[

In every neighborhood of any state S of an adiabatically isolated system there are states inaccessible from S.

No heat engine can be more efficient than a Carnot Cycle.”

Please state which of these (or which other version you prefer that is not on the list) that you had used — a quote from your posts would be most helpful.

Based on that statement of the 2nd Law, calculate (or at least explain) how it is violated by the expanding/contracting shell model.

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

“There is no process possible that will, without loss, will spontaneously convert lower frequency radiation into higher frequency radiation.

Perhaps this is your previously-unstated version of the 2nd Law. I little restrictive, but it could work as a subset of the 2nd Law.

The shell does not “spontaneously get hotter” and thereby “spontaneously convert from low frequency radiation into higher frequency radiation” The shell touches something else that is warmer than it is and therefore the shell gets warmer than it was! THAT is the process that occurs. The “loss” is heat lost by the warm planet and added to the cooler shell. So this process is not “without lose” and does not contradict your statement.

Or does your version of the 2nd Law say things cannot get warmer by touching a warmer object???

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

“Then during a temperature low get a steam engine to produce work like lift objects higher than the planet surface.

Of course the whole thing is impossible as you should know.”

I know no such thing! We have the makings of a heat engine here:
** energy input at a high temperature (the nuclear heaters)
** energy output to a colder region (space)

Why would you think we could not run a heat engine with 1) a hot reservoir, 2) a cold reservoir, and 3 ) a continuous source of power? I’d be disappointed if someone couldn’t design a heat engine in such circumstances!

240. lgl says:

Osborn,

Too much work only to find what I am experiencing daily, that the cool surroundings is preventing my warm body being transformed into a lump of ice in minutes.

241. Bryan says:

Tim Folkerts says….. Bryan, you are getting harder and harder to follow!

Please state which of these (or which other version ..(of the second law)…you prefer that is not on the list) that you had used — a quote from your posts would be most helpful.

Then Tim answers himself reading my post where I say after a long post about the second law

“There is no process possible that will, without loss, will spontaneously convert lower frequency radiation into higher frequency radiation.”

Tim now gets the message and says

“Perhaps this is your previously-unstated version of the 2nd Law. I little restrictive, but it could work as a subset of the 2nd Law.2”

Do I need to spell out the ABC of thermodynamics?

This oscillating system will without any external source of energy purports to transform low frequency back radiation into high frequency radiation thus improving its quality.

The Nuclear Reactor which is an internal part of the system was only specified to give the planet a constant output of 235W/m2 so don’t change its specification now.

From the Kelvin statement of the second law

it is impossible to devise a cyclically operating device, the sole effect of which is to absorb energy in the form of heat from a single thermal reservoir and to deliver an equivalent amount of work.

The single resoirvior is the shell at 254K

100Joules of low frequency backradiation cannot be transformed into 100 Joules of higher frequency radiation such that this total energy can do work in the given situation.

242. A C Osborn says:

Bryan, for goodness sake stop confusing them with SCIENCE, they can’t take it.

243. physicistphil says:

Tallbloke…Regarding thermalization in isolated multi-body systems:

Obviously the systems-approach here is not a mirror to what we’re discussing but the same principles apply.

The argument that the inner core must “heat up” given initial IR impedance by the outer shell is utter nonsense *unless* the nuclear planet is emitting radiation at higher spectral frequencies (visible, UV, gamma, etc), which would be converted to thermal energy once intercepted by the shell. If the shell is emitting solely thermal/low frequency radiation, you have a saturated microquantum/vibrational firing frequency.

244. A C Osborn says:

physicistphil says: March 13, 2013 at 8:03 pm

Now you are it it as well, their heads will explode at this rate. /sarc off

245. mkelly says:

Bryan says:

March 13, 2013 at 7:42 pm

Bryan, Mr. Folkerts does not even admit his own math proves the steel shell false. I doubt he will agree with your valid point about quality of the radiation.

246. Arfur Bryant says:

@gbaikie says:
March 13, 2013 at 1:55 am

“I say radiation from cooler body can slow the cooling of hotter body.”

Thanks gbaikie. That is also what the Ford Prefect’s experiment showed. I have no problem with that. What I am trying to establish is seemingly fundamental (to me, anyway): Is the radiation from a cooler body absorbed by a warmer body? The answer to this question would go a long way to explaining exactly how the toy planet is supposed to work as advertised. i agree with your sentiments regarding cAGW.

**

@Tim Folkerts says:
March 13, 2013 at 2:42 am

“I wish I knew the best way to say that the “toy planet” is not violating any laws of physics. “

Tim, I’m not fussed about the analogy. I would really appreciate you answering (or having a try, anyway) my question to gbaikie above. Is the radiation from a cooler body absorbed by a warmer body? An answer would be much more appreciated than an analogy…

I think I understand what happens (on a molecular level) when radiation from a warmer body is absorbed by a cooler one. I am a bit cloudy on what would happen in the reverse case – on a molecular level.

Thanks,

247. Bryan says:

Arfur Bryant says:

I think I understand what happens (on a molecular level) when radiation from a warmer body is absorbed by a cooler one. I am a bit cloudy on what would happen in the reverse case – on a molecular level.

.
Now if one increases in temperature it will radiate more intensively, while the other radiates as before.
If the radiation is all thermal (longer wavelengths than say 3um) the molecules of the colder will increase in velocity and hence kinetic energy.
The colder object to hotter object radiation on the other hand although absorbed will not increase the molecular motion of the hotter any more than when they were at the same temperature.

248. Bryan says:

mkelly

The greenhouse effect illustrated by the Willis planet cannot add energy to the system by virtue of a shell producing a vacuum cavity.
As soon as real numbers and processes are included the theory falls apart.

249. Kristian says:

lgl says, March 13, 2013 at 5:35 pm:

Let me try to explain this to you, lgl.

We place two objects in an airless box. The one object has a temperature of 300K, the other 400K. The two freely emit according to their specific temperatures. If we were to let the thermal exchange between the two objects move all the way to equilibrium, they would both ideally end up with a temperature of 360K (not 350K). At this point, there would no longer be any heat transfer between the two objects. Up until then, though, there WOULD have been heat transfer from the hot to the cooler object. And if we plotted the RATE of this heat transfer through time, we would see that it followed a curve of exponential decay – it would start off very high but end up infinitesimally low. The hot (400K) object would cool all the way to equilibrium, while the cooler (300K) object would warm. The cooler object at no point made the hot object hotter than it originally was, even though it radiated a higher and higher flux towards it as it grew hotter itself. All it did was steadily slowing down the heat transfer rate, its own warming rate and the hot object’s cooling rate, until they were both effectively zero.

So far, so good. I hope.

Then, we change the setup a bit. This time around the two perfectly emitting objects in the airless box are different. One object is directly connected to a constant energy source supplying a power flux sufficient to heat the object’s surface to a uniform temperature of 254K. The other object enjoys no such constant supply of energy. It hypothetically holds a starting temperature of 0 K.

At the get-go, the warm object emits a flux of 235 W/m^2 towards the cold object and the cold object in return emits a flux of 0 W/m^2 towards the warm object.

What happens next? Where will the ensuing exchange process take us? What will the equilibrium state be like in this scenario?

Upon commencement, the radiative heat transfer from the warm to the cold object starts off at a prodigious rate, warming the cold object fast. But this warming once again slows down as time passes. This time the warmer object does however not cool in accordance with the warming of the colder object. The two objects do NOT meet in the middle at equilibrium. The heat transfer rate between the warm and the colder object follows the same general path as the one in the first scenario through time. But now equilibrium will not be achieved until the colder object has reached the temperature level of the warm object. At this time, heat transfer between the objects would effectively once again have come to an end.

Why, then, isn’t the warm object cooling to meet the cold object ‘half-way’ (213K) at equilibrium in this scenario?

Because it’s temperature/surface radiative flux is being maintained all along by its constant power supply. It will not drop in temperature as long as this source is active. It will not rise either as long as the source provides its constant power flux. This has to do with the internal (molecular) vibration (the level of KE) which relates to the intensity (frequency/wavelength) of the radiative flux received/emitted.

A key to understanding this outcome lies in realising that in the second scenario above (WITH power supply), as the heat transfer between the two objects WITHIN the thermodynamic system (the box) grows ever SMALLER towards equilibrium, the total system energy transmission to its SURROUNDINGS (the space around) parallelly grows ever LARGER towards that same equilibrium. In the first scenario above (WITHOUT any power supplies), this total outward system flux remains the same throughout – no extra heat generated. (In reality it would actually drop; this is after all only intended to be an idealised case cleared of any such real-world clutter. It was simply meant to prove a point.)

The heat generated and originally supplied to the warm object in other words goes into warming the COLD object (and therefore, by extension, the system as a whole) until it’s reached the constant temperature of the warm object. It does NOT (by way of ‘back radiation’ from the cold object) go into making the warm object itself warmer than it already is.

250. tallbloke says:

Bryan: The greenhouse effect illustrated by the Willis planet cannot add energy to the system by virtue of a shell producing a vacuum cavity.

I don’t think energy is added either. The conversion of the nuclear energy into kinetic energy provides a regular quantity of power into the system. The addition of the shell just makes the energy do some extra work in the system before it exits to the heat sink of space. Whether that extra work causes the planet to get hotter is the point at issue. The back-radiated energy has to go somewhere. If it is absorbed by the planet and re-emitted, then whether or not it caused the planet to vibrate more, it is emission additional to the regular steady quantity from the nuclear reaction. Theory seems to be that more emission means higher emission temperature. But maybe if some kind of reflection is occurring it ‘doesn’t count’. However, at the molecular level, reflection is a funny process in itself.

I’m not a radiation physicist and I don’t know the secrets of molecular level interactions between waves and matter. I just scratch my head over the numbers and use logic to say ‘it’s got to go somewhere’.

251. wayne says:

MaxTM, in answer to your question over at Anthony’s site that he so suddenly and rudely closed, my answer is b). Opposing powers subtract, never add. However, that does not mean that the warmer surface with an infinite energy source backing it up will not end up at a higher temperature, such an energy source does not depend on temperature differences. I read you already know that, this is for other readers. It will end that the two powers subtracted end up at the original 239 W/m^2.

This is where Joe Postma goes awry. His 235 W/m² radiative field trapped between the two BB surfaces at the same temperature has no power at all, the power from every point cancels the opposing and equal power from the opposite direction. So that field, even though there, and has a radiative density, has no power to move energy anywhere radiatively, he keeps saying that is this radiative field that keeps feeding the shell and keeping it a a temperature, not so. That is where his example falls apart. There still has to be a 235 differential in power in some direction (Poynting vectors) between the two points to do something.

Darn, I was just getting wound up to counter his views in that matter.

252. Kristian says:

Tallbloke says: “I just scratch my head over the numbers and use logic to say ‘it’s got to go somewhere’.”

It does go somewhere. It goes out to space. It goes into heating the SYSTEM as a whole (that is, ultimately its outer surface towards its surroundings).

253. Martin Hodgkins says:

I am just saying thanks to lgl for the following…

“We have two equally sized radiators, and for the argument they are thin as a paper i.e only radiating from two sides. We supply enough power to keep one of them at 300K and the other at 400K when they are far apart. Then we place them very close to each other, sealing off the narrow gap between them completly so that no energy escapes from the gap. We supply the same amount of power to the radiators as before. Now, will the temperature of the radiators change? One, both, up or down?”

I have been wondering about all this for days and it is now obvious to me that Joe is right. The radiators are going to be about 350K each. They certainly aint going to get any hotter with the same power input.

At last my mind is made up.

[Reply] So I wonder why Joe wouldn’t publish the calcs. 🙂

254. tallbloke says:

Whatever the outcome I’m declaring this a win for the quality and civility of debate here.

255. Tim Folkerts says:

mkelly says

“Mr. Folkerts says: (T_2/T_1) = (r_1/r_2)^(1/2)
Sir this is your ratio for two spheres.”
and later
“Bryan, Mr. Folkerts does not even admit his own math proves the steel shell false.”

Once again … that is the ratio for TWO DIFFERENT PLANETS of different radius with the same total power (or equivalently, two different shells around the same inner planet). See? Right there at the top of the post you where you got the equations?

Suppose two different spheres emit the same total power as thermal IR.

This is not the ratio for an inner sphere surrounded by an outer sphere. T_2 is the outside of a separate sphere, not “The T2 of the inner sphere “ as you suppose. The sole purpose of that equation is to find the temperature ratio of two spheres with the same TOTAL power (rather than the same power per square meter).

It is not, as you supposed “The T2 of the inner sphere “.

You were wrong about your reading of “(T_2/T_1) = (r_1/r_2)^(1/2)”. I will leave it to others to decide if you were any better about your thoughts on my discussion with Bryan.

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

And now it is time to retire from this sub-thread with Bryan. We have both explained our positions and neither seems to be able to convince the other of the “correct” understanding. Informed readers will be able to draw their own conclusions

256. mkelly says:

Tim Folkerts says:

March 14, 2013 at 2:00 am

Sir, I’m sorry you are not able to understand that the ratio you built applies to the steel shell problem. The math is exactly the same.

257. Tim Folkerts says:

mkelly,

Yes, the calculations apply to the steel shell problem. But they apply ONLY to the outermost shell radiating to space, not to the surface of a planet enclosed by an outer shell.

258. lgl says:

Kristian

Joe is not allowing me to answer you and Martin so hope tallbloke will.

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?
Of course, again, that’s not what is happening either (even if Joe has repeated that misconception a hundred times), it is still the internal power generation that is increasing the temperature.

259. lgl says:

Martin

Perhaps you would be so kind then to tell us how much power the two radiators at 350K would emit? And also verify my 3827 Watts for the total emission in the initial setup when the radiators were placed apart?

260. Tim Folkerts says:

Martin Hodgkins says
“The radiators are going to be about 350K each. They certainly aint going to get any hotter with the same power input.
At last my mind is made up.At last my mind is made up.”

with regard to lgl’s thought experiment …

We supply enough power to keep one of them at 300K and the other at 400K when they are far apart. Then we place them very close to each other, sealing off the narrow gap between them completly so that no energy escapes from the gap. We supply the same amount of power to the radiators as before. Now, will the temperature of the radiators change? One, both, up or down?

Let me see if I can change your mind!

The math is really pretty simple. For the sake of simplicity, lets assume these heaters are in space where the surroundings are ~ 0 K — this avoids any concerns about conduction or IR radiation from the surroundings. The radiators are also assumed to be black bodies. (The general results I will show are the same as long as the temperature of the surroundings is less than 300 K, but the math gets more complicated).

The 1m x 1m heater at 300 K requires 918.5 W (429.3 W of IR radiating from each side).
The 1m x 1m heater at 400K requires 2903.0 W (1451.5 W of IR radiating from each side).

When they are put together, they are a single sheet with an area of 1m x 1m and a heat input of 918.5 W + 2903 W = 3821.6 W, or 1910.8 W/m^2 of outgoing IR. To produce 1910.8 W/m^2. the surface will have to be 428.5 K.

Yes, BOTH sides are now significantly warmer than any of the sides of the heaters before.

(If the surroundings are @ 300K initially, then the two sides of the combined heater will be @ 400K. Basically, the 300 K heater will need no power, so the power for the 400 K heater can keep two sides warm, whether those are its own two sides, or the two sides of the combined heater.)

(As the temperature of the surroundings goes from from 0K to 300 K, the temperature of the combined heater goes from 428.5 K down to 400K)

(I better not sidetrack the discussion by contemplating what will happen if the surroundings are above 400 K)

261. Tim Folkerts says:

Dang — I have a numerical typo (caught when I looked at lgl’s comment).

The middle of my previous post should read.

When they are put together, they are a single sheet with an area of 1m x 1m and a heat input of 918.5 W + 2903 W = 3821.6 W, or 1910.8 W/m^2 of outgoing IR. To produce 1910.8 W/m^2. the surface will have to be 428.5 K.

Other than some round-off differences, lgl and I get the same answer (3826 W vs 3822 W). And i suspect we will get the same answer for the 428.5 K final temperature.

262. lgl says:

Yes Tim
I didn’t bother with accurate round-offs.
Thanks for verifying this, no wonder Joe refused to do so.

263. Kristian says:

lgl says, March 14, 2013 at 5:04 pm:

Firstly: Sorry, but you’re not addressing the content of my explanation to you. Not at all. Go back and reread what I’m actually writing.

Secondly: Go read David Cosserat’s thread on atmospheric throttling mechanisms. The atmosphere is making temperatures rise relative to the vacuum of space by restraining convection from the surface, not thermal radiation from the surface to the ToA. Insulation is all about suppressing convection, lgl. You’re warmer on Earth than in space, not because the thermal radiation you give off is more restricted from leaving your body (it’s not), but because you’re surrounded by an atmosphere with a mass (pressure/density/heat capacity). You’re mixing up.

264. Tim Folkerts says:

lgl,

The more I read from Joe Postma, the less impressed I am. I have pointed out specific errors (like when he was integrating over a cylinder when he though he was integrating over a sphere), but he still didn’t even consider that he might have been incorrect. This particular mistake inflated the average temperature of the earth (by basically treating the whole earth as being at the equator). This allowed him to downplay the importance of GHGs once again.

Tim

265. lgl says:

Tim
I haven’t read what you are referring to but I am not surprised. What surprise me more is that so many supposedly thinking creatures actually believe his nonsense.

266. TB, being away on holiday I missed this article until today.

You have my sincerest sympathies and also congratulations on the strong and clear responses you have generated to all the complete and utter blather you have received from most respondees, ranging from many who pretend to be specialists, and so should know better, to others who are obviously innocent of physics (no criticism in the latter case: everyone has to learn, constantly). And then there is always our information-free resident threadbomber to contend with, whom I know only too well from my Atmosphereic Thermal Enhancement (ATE) articles. 🙂 Compliments also to Tim Folkerts for his incredible patience in the face of such nonsense. Also to Wayne for being honest enough to see the light.

Over at Part I of ATE, I had to contend time after time with people who couldn’t emotionally handle the abstraction of a “thought experiment” – which is always highly limited in scope and simply intended as a didactic aid to teach people a fundamental point that they hopefully will never forget, thereby moving them on to their next stage of learning. But instead of discussing the thought experiment, such people impertinently suggest how you have got the experiment completely wrong or that “the atmosphere-earth system is not like that”, etc., etc.

Pathetic.

Here’s a simple thought experiment I offer as a challenge to anybody who does not believe in what some people care to call ‘back radiation’ (meaning, apparently, ordinary radiation that happens to be travelling in a direction they don’t approve of ): 🙂

(1) Inside a perfectly insulated evacuated tube we place two black bodies, one at either end of the tube with a vacuum gap in between them.

(2) Both bodies are a tight fit in the tube meaning they both have exactly the same cross-section as the inner cross section of the container. The consequence of this is that their only radiating surfaces are the ones facing one another across the vacuum.

(3) We initialise the system with both the black bodies at, say, 288K. They therefore initially radiate towards each other at 390Wwm-2 (deduced from simple application of S-B law).

The question to ask those who deby the existence of ‘back radiation’ is this. Will the bodies remain at exactly the same temperature for ever? If they will, the ‘back radiation’ in this example not only exists but (amazingly) flows in both directions and, moreover, does no work.

267. Kristian says:

Tim Folkerts says, March 14, 2013 at 5:13 pm:

“The 1m x 1m heater at 300 K requires 918.5 W (429.3 W of IR radiating from each side).
The 1m x 1m heater at 400K requires 2903.0 W (1451.5 W of IR radiating from each side).

When they are put together, they are a single sheet with an area of 1m x 1m and a heat input of 918.5 W + 2903 W = 3221.6 W, or 1910.8 W/m^2 of outgoing IR. To produce 1910.8 W/m^2. the surface will have to be 428.5 K.

Yes, BOTH sides are now significantly warmer than any of the sides of the heaters before.”

Whoa, hold your horses! What on Earth did you expect, Tim?

You’ve turned the two heaters into ONE! This one compound heater has the same surface area as any one of the two original heaters. Yet is has a power input much larger. Of course it’s gonna get hotter!

It’s like turning up the output from the nuclear energy source in Willis core planet from say 235 to 470 W/m^2 and saying: ‘Look, it’s warming up!’

Are you sure lgl meant to say that the two heaters were to be merged into one?

268. lgl says:

Kristian

The two heaters are not merged into one but they are put very close to each other and the narrow gap is perfectly insulated so that the total radiating area of both heaters to free space is only half of the initial. And the power input is the same as before, 3827 watts. Using Postma-physics the warm heater can’t warm up in this setting because the other heater is colder.

269. lgl says:

Kristian
“but because you’re surrounded by an atmosphere with a mass (pressure/density/heat capacity).”

Doesn’t matter. According to Postma-physics the colder surroundings can’t lead to a body temp higher than the surroundings, regardless type of energy transfer, be it radiation, direct contact or whatever.

270. oldbrew says:

Kristian says: ‘Insulation is all about suppressing convection’

And that’s the real greenhouse effect – which is why greenhouses work.

271. lgl says:

Kristian

Actually it doesn’t matter whether you merge the two heaters. Even if you do, according to Joe energy can’t flow from the 300K part to the 400K part, so the 400K side can’t get warmer.

272. Roger Clague says:

The steel greenhouse model is not a very realistic model. However it is useful if we can agree on what laws and logic to apply in a simple situation, in a relaxed manner, a self-educational exercise. Then we can look at a more real model with more confidence.

Bolding and/or capitalizing text does not help. Neither do insults and suggesting new models.

Kristian says:
March 13, 2013 at 8:07 am

This [ model ] does not take into account the thickness or the conductivity of the shell

The inner surface or the shell cannot, at the same time both radiate at 235W/m2 and conduct heat to the outer surface so it also radiates at 235W/m2.

273. Tim Folkerts says:

lgl — I thought of what seems a clever variation on your thought experiment.

Since at least some people think the result of your thought experiment is obvious, lets see if this is also “obvious” to them.

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

I start with one of lgls 1m x 1m heaters with a fixed power input — say 2900 W/m^2 of electrical power. This plate will be 400 K when isolated.

The other plate is a little different — it has a thermostat that keeps the temperature at 300 K. This plate will require 920 W of electrical power when they are far apart.

I contend that — as the two are brought together —
1) the first heater will warm up a bit (since it is naturally receiving some energy from the other heater)
2) the second will require less power input. (It would warm up otherwise, so we turn down the power input as we get closer).

Now here is the critical idea. AT SOME POINT WE CAN COMPLETELY TURN OFF THE POWER TO THE 2nd HEATER! If we brought the second heater all the way to the first, it would continue to warm all the way to 400 K when the distance went to zero (even with an infinitesimal vacuum gap between the two). We would have to back off a ways from the 2900 W/m^2 heater prevent the second “heater” get above 300 K. At this point the radiation from 2900 W/m^2 heater is sufficient to keep the 2nd “heater” at 300 K without the input of any external electrical power.

If people are with me so far, then we have agreed that:
1) the original 2900 W/m^2 heater is now warmer than 400 K.
2) the second “heater” needs no external electrical power — only the radiation from the 1st heater.

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

That is exactly what Willis’ steel shell is doing!

1) The planet has a fixed POWER of 235 W/m^2 (similar to the 1st heater).
2) The shell (cut up into 8 octants for simplicity and pulled far from the planet) has a fixed TEMPERATURE of 254 K (similar to the 2nd heater).

When the shell is far away, this would take 235 W/m^2 of power to the pieces of the shell. And at this point, the planet would be 254K as well.

As the pieces of the shell ear brought closer
1) the temperature of the planet slowly increase.
2) the external electric power needed for the shell pieces slowly decreases.

In the limit as the shell gets close the planet, the solution will be
1) The planet warms from 235 K toward 303 K — WITH A FIXED POWER OF 235 W/m^2 the whole time.
2) The external electrical power to the shell pieces decreases from 235 W/m^2 UNTIL NO EXTERNAL ELECTRICAL POWER IS NEEDED. All the power needed is supplied instead by the radiation from the planet.

274. Arfur Bryant says:

Bryan says:
March 13, 2013 at 9:50 pm

“The colder object to hotter object radiation on the other hand although absorbed will not increase the molecular motion of the hotter any more than when they were at the same temperature.”

Bryan,

Thank you very much for that straight answer. I have found your posts informative and clear. I am trying to get the basic processes clear and apply them to the toy planet. So…

The radiation from the cooler body is absorbed without net gain (of molecular motion). I would therefore surmise that the toy planet can not then emit at 470 W/m^2 (which is one of my problems with the toy), as to do so would require absorption for net energy gain. I also appreciate Kristian’s explanation to lgl above.

However, does that mean (reference tallbloke’s comment above) that the lower energy radiation is just lost (can that happen?) inside the planet, or does it get re-emitted without gain or loss and therefore just bounce about inside the gap between the planet and inner shell because it now possesses less energy than both the planet and the shell (which has received extra energy from the planet’s nuclear furnace in the time taken for the lower energy radiation to travel between the shell and the planet)? Given that the shell emits less than 235 (due to radius effect) but is always being ‘topped up’ by the planet this ‘may’ be feasible, but it could also be that the energy of the shell-emitted radiation remains the same and therefore the real destination for the lower energy radiation is space…?

The bottom line is that the planet does not heat up.

Sorry if I’m not making sense. To me, the debate re ‘backradiation’ centres around absorption, not emission.

@tallbloke:
Rog, this is easily the most civil and stimulating blog around. My sincere congratulations to you and all who contribute.:)

275. Westy says:

Can someone help me out here please. The toy planet with shell doesn’t have two power separate power supplies. If the shell is made larger or smaller will the temperature of the core change? If half or all of the shell is removed with the cores temp change?

276. gbaikie says:

“The question to ask those who defy the existence of ‘back radiation’ is this. Will the bodies remain at exactly the same temperature for ever? If they will, the ‘back radiation’ in this example not only exists but (amazingly) flows in both directions and, moreover, does no work.”

Yes the bodies will remain same temperature forever.
It goes in both both directions but I wouldn’t say flows in flows both directions. It flow if there is difference of heat.

It’s no different than if both ends were warmed by room temperature [even if “room temperature” was a furnace].

What happens if they radiating difference wavelengths but same watts per square meter? Then I think you would get a heat flow.
Or one blue and other end red, and shine light at each so they reflect same amount watts per square meter at each other blue each different color.

277. gbaikie says:

Or one blue and other end red, and shine light at each so they reflect same amount watts per square meter at each other [but] each different color.”

I mean shine blue light at blue color, shine red light at red color. And same per watts square meter of the color.

If shine on say white paper, the reflected light will turn both purple

278. Tim Folkerts says:

Westy askes: “If the shell is made larger or smaller will the temperature of the core change?”

Yes. The temperature of the shell will decrease as it gets bigger. I derived the equation earlier:
(T_2/T_1) = (r_1/r_2)^(1/2)

As the temperature of the shell decreases, the back radiation will decrease, which in turn will supply less energy back to the planet, which makes the planet cooler.

However, for “reasonable” sizes, this effect will be small — well under a 1%. So with a “large” shell, the planet might only warm to 301 K instead of 302 K.

If you remove half of the shell, that will cool the planet’s surface significantly. I did the math once but don;t feel inspired now to either look up the answer or to re-derive it. The temperature would be somewhere in between 254 K and 302 K — but not close to either end.

279. Kristian says:

David Socrates says, March 14, 2013 at 6:51 pm:

David, about your ‘simple thought experiment’. This is purely a case of heat transfer. The two bodies are at equal temperatures. The tube is perfectly insulated. In other words, the two bodies neither gain nor lose any heat. They cannot. Hence, they will both remain at the same temperature ‘forever’.

This is ‘back radiation’ to you? These are two independent bodies, David. Neither is heating the other. If they started out at different temperatures, heat would have flown from the warm to the cold until they’d reached equal temperatures, somewhere in between their two initial temperatures. And then the process would have stopped.

280. philr1992 says:

“lgl”

I stopped reading when you claimed the human body is kept “warm” by radiation from the atmosphere we live in..that is wrong.

We physicists may not agree on everything..but I assure you, we know that to be false. You’re technically bathing in the atmosphere..it insulates your body as warm bathwater would..no radiation required. When the gases surrounding your body are “colder” than your skin/extremity temperature, your body will cool until equilibrium between your skin temperature and the atmosphere is achieved..visa versa when the atmosphere is warmer.

Outer space has no temperature..it is a vacuum…your body would not lose much in the way of heat in the vacuum of space because there’d be nothing to advect heat from your body (hypothetically assuming you could survive).

281. Bryan says:

David Socrates

You arrive on this thread and immediately switch into extreme condescension mode!

“the complete and utter blather you have received from most respondees”

You go on to say;

“And then there is always our information-free resident threadbomber to contend with, whom I know only too well ”

Name names, rather than smear the general posters on this thread.

You go on to say

“Compliments also to Tim Folkerts for his incredible patience in the face of such nonsense.”

Its clear that you think the Sun shines from Tim’s rear orifice but do you really have to be so servile.

David, stop to consider the possibility that many (who do not believe that the Greenhouse Effect increases the Earths surface temperature by 33K) think you are a supreme disseminated of ‘blather’.

However they are too polite to tell you to your face.

Stick to posting science content that’s what most of this interesting thread was about.

282. philr1992 says:

To make sure I had not lost my mind, I asked three of my colleagues (plasma physicists) about this hypothetical system. As I suspected they all agreed with the earlier premise I had drawn.

Radiative units are analogous to velocity only, not material flux..the frequency wavelet (analogous to the total rate of loss) emitted by the nuclear planet is greater than the backflux emitted by the shell. The planet is losing energy at a given rate determined by it’s temperature, as is the shell.

You cannot “add” radiative units of the same spectral realm without creating energy..in fact you cannot add radiative units, period. In this case, the microkinetic firing frequency is saturated at the temperature of the planet.

283. Westy says:

Tim, You think the planet will get hotter as the shell shrinks. I still think the planet will be the same temp with a big shell, a small shell, or no shell. If the planet is getting hotter as the shell shrinks what happens to your heat when the shell gets small enough to touch the planet? Does this extra heat burst out into space?

284. Tim Folkerts says:

Westy,

Let me clarify one thing.

The shell I was talking about in my comment @ March 14, 2013 at 11:09 pm is a fixed area — just infinitesimally larger than the surface area of the planet so that it can fit around the planet. The shell must be sliced (for example, into two hemispheres) so that it can be pulled back from the planet. In this case, the planet should warm as I outlined.

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.)

When the shell gets close enough to touch, then yes, the shell would rapidly warm from 254 K to 302 K. But this is only temporary. The planet and shell would then start to cool back toward 254 K. If you pulled the shell back from the surface, then either
* it would COOL back down from 254 K (if you don’t add extra heaters).
* you would have to heat it to keep it at 254 K)

285. Tim Folkerts says:

Philr1992, could you explain this term? Google can not find a single reference to such a thing!

“We physicists may not agree on everything..but I assure you, we know that to be false.
And I can assure you that “we physicists” do not agree on this thing you assure us is false!
An object warms and cools by conduction, convection and radiation. All are important. The temperature of the air is certainly important, but radiation plays a key role as well.

286. mkelly says:

Tim Folkerts says:

March 15, 2013 at 4:43 am

Yes. The temperature of the shell will decrease as it gets bigger. I derived the equation earlier:
(T_2/T_1) = (r_1/r_2)^(1/2)

Now that you agree with me again, let’s plug numbers in. You say per the picture the 470 w/m^2 is 302 K. That is T1. Now let’s use the numbers of 6370 for planet (R1) and 6380 for outer shell (R2). It should be for the inner but let’s your position of the outer.

Your derived equation now says that T2= .9992 T1. Which is 301.753K. That temperarture cannot be reached if the outer shell is at 235 W/m^2 which is 253.7 K.

That is a mathematical failure in the posited problem of the steel shell. You falsified the problem and for the last time I say congrats.

287. lgl says:

Tim

Your clever variation is all good but I don’t it will help much. They base this nonsense on the absurd notion that there is a net 235 W/m2 transfer between two bodies having the same temperature. To achieve this the radiation in one direction is absorbed but the radiation in the other direction is ‘scattered back’, whatever that means, and is not being absorbed. When people get that stupid there really isn’t much hope.

288. Tim Folkerts says:

mkelly!

Do I really have to say this one more time? The equation derived is for the outermost shell. It applies to two different outermost shells of two different systems (or the same outermost shell at two different times after the radius has been changed). The shells always have the same total power radiating from them.

It does NOT apply to a planet and a shell around it, as you keep trying to do!

*IF* the outermost shell had a radius of 6370 km and a temperature of 302 K, then the total power inside the shell would have to be (470W/m^2) * (4*pi*(6.37e6)^2) = P(0). (NOTE: This is quite different from the original model that Willis created, where his shell was 254 K and his planet had 1/2 as much total power!)

*THEN IF* we increased the radius of the outermost shell but kept the total power inside constant at P(0), the temperature would drop to 301.753 K. Same power, but bigger surface = cooler surface.

What we have here is a conceptual failure on your part, not a failure of the model!

289. lgl says:

philr1992

I didn’t say “human body is kept “warm” by radiation from the atmosphere we live in”. I said “The only change is the energy received from the surroundings at 293K” and like Tim told you “The temperature of the air is certainly important, but radiation plays a key role as well.”
Anyway it’s irrelevant because the Postma-physics does not allow the body to warm by conduction either, from a cooler than body atmosphere.
And “your body would not lose much in the way of heat in the vacuum of space because there’d be nothing to advect heat from your body”
But it would in the way of radiation.

290. lgl says:

Kristian
“Firstly: Sorry, but you’re not addressing the content of my explanation to you. Not at all. Go back and reread what I’m actually writing.”

The reason I didn’t was my comment here was a copy of that at Joes blog, where I started with:
“This is still wrong for the reasons Joe will not allow me to repeat” but now having more time I will.

In your ‘explanation’ you are still doing the old Postma-blunder. As soon as the cold object starts heating it will send some energy back to the warm object and the net transfer from warm to cold will drop below 235 W/m2. Because there is a constant supply of 235 from the core which the planet can no longer get rid of it must heat up. As simple as that.

291. Bryan says:

The fact that most metals expand when heated can be used to show that the Willis model and Greenhouse Theory is unphysical and therefore pseudoscience.

The Willis planet has a constant supply of 235W/m2 and is initially at equilibrium with deep spacei at 254K.
Its temperature reaches 302K by the method of backraadiation when the passive shell is added it is proposed.
The shell is can be made of a metal with a small coefficient of thermal expansion and
the planet on the other hand with a much larger coefficient of thermal expansion.

The expanding planet at 302K will touch the shell at 254K

So the planet at 302K will touch the shell and the shell-planet combination will radiate at 470W/m2 for quite sometime before the planet cools enough to separate by contraction .
The process would endlessly repeat automatically without any external input of energy.

Now most people would see that this set up is outputting more energy than is being supplied.
This would be a violation of the first law (conservation of energy).

We can of course pick numbers where this is avoided (whether these numbers represent reality is doubtful).

But lets for the sake of the discussion accept some numbers where this is the case.

Some people think that as long as the first law is OK then that’s the end of the matter.

Not quite….
There is the second law.

In the above example lets say that 100 Joules initially at 254K is transformed into 80Joules at 302K and 20 Joules at 200K

This is quite acceptable as far as the first law is concerned but what about the second law?

This fictitious system would claim it is possible to spontaneously pump energy at a low temperature into energy at a high temperature endlessly without the input of energy external to the system.

Clausius and every physics book in the universe says this is impossible the according to the second law of thermodynamics.

292. Roger Clague says:

The steel greenhouse only considers temperatures and power (W/m2) of radiation. There is no thermal energy transfer in mass by conduction and convection.

The atmosphere, a gas, is modeled as a vaccum next to a thin solid with no mass. Not at all realistic.

Willis, David, Tim and IgI argue correctly for their radiation only model. But they should realize that the real atmosphere has mass, which is its most important property concerning temperature and energy transfers.

293. Tim Folkerts says:

Bryan says …The expanding planet at 302K will touch the shell at 254K …

We have already been through this — your reasoning is faulty on many fronts. Here is yet one more …

“The expanding planet at 302K will touch the shell at 254K …
OK, at this point the planet will COOL due to the contact with the shell and CONTRACT a little.
The shell will WARM due to contact with the planet and EXPAND a little.

Let me repeat — the planet on the inside contracts and the shell on the outside expands. The two will almost immediately loose contact (no matter what the relative coefficients of expansion are). The shell will never get much above 254 K and the planet will never get much below 302 K. The shell will — for a brief moment (not “for quite some time”) radiate a BIT more than 235 W/m^2 (not “470 W/m^2”). The shell will then cool a bit below 254 K and radiate a bit less than 235 W/m^2.

(I tried to rescue your thought experiment before by having an “inflatable planet” so you could prolong how long the two were in contact, but you were not interested. Of course, this doesn’t really fix your problems, but it at least would allow the system to “radiate at 470W/m2 for quite sometime”.)

So who is presenting an unphysical, pseudoscientific model?

294. Tim Folkerts says:

Roger C,

I agree that the atmosphere has mass and that is very important in the actual climate of earth (and I am pretty sure the others you named agree).

I would disagree a little that mass is “the most important property”. The “mass effect” and the “radiative effect” are intimately intertwined.

* Without the added mass of N2 & O2, the CO2 in the earth’s atmosphere would have little warming effect.
* Without the added radiative effect of CO2, the N2 & O2 in earth’s atmosphere would have little warming effect.

The whole point of Willis’s model is to highlight the mere existence of the radiative effect — an effect that many people reject.

295. Roger Clague says:

I studied physics before the satellite age. I remember the pop tune ‘Telstar’ I heard as a teenager. Now physics education includes more emphasis on radiation, especially as measured from in space. Probably mass has been considered ‘ the most important property’ for too long.

I think that in future we will need new ideas about the how matter and radiation interact. We are already discussing here posts about the way the sun and planets behave and influence each other in other ways than through mass alone.

296. tallbloke says:

Tim F:
* Without the added radiative effect of CO2, the N2 & O2 in earth’s atmosphere would have little warming effect.

Having just admitted the effect of atmospheric mass on surface temperature, Tim immediately goes back into denial. 😉

297. Tim Folkerts says:

And Tallbloke, having just posted and defended Willis’s model, now seems to think it makes absolutely no difference! 😉

298. gbaikie says:

The Earth has Van Allen belts which accelerate matter
like mass collider does. And mass colliders cause high temperature
collision.
The sun has a bigger and more powerful matter accellerator.

For any matter to leave the Sun it needs to have Sun’s escape
velocity: 617.5 km/s

For anything to leave Earth it needs Earth’s escape velocity:
11.2 km/s.
The particles in the Van Allen Belt are going much faster than Earth’s
escape velocity and are held due to magnetic force of Earth and
it’s interaction which solar wind.

So Sun has much stronger magnetic field than Earth and is the source
of high energy particles and solar wind.
Without this solar wind and energetic particles we wouldn’t have the Van Allen Belt-
Earth providing the magnetic field, but sun’s giving the other components.

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

The reason the Sun’s corona is hot is because the enough of vacuum
to allow these high particles to persist.
Or we would not have Van Allen Belt if there enough atmosphere where the van Allen
belts are [we need much bigger and hotter atmosphere to get a atmosphere so high].

299. Roger Clague says, March 15, 2013 at 5:50 pm: Willis, David, Tim and IgI argue correctly for their radiation only model. But they should realize that the real atmosphere has mass, which is its most important property concerning temperature and energy transfers.

Roger, one more time, will you please understand that (at least in my case) when I offer a radiation-only “thought experiment” on a blog that is specifically discussing ‘back radiation’, the purpose is didactic – to try to explain by simplified example how I believe radiation between two bodies at different temperatures works.

That does not allow you to accuse me (of all people) of ignoring the effects of atmospheric mass and the role of conduction and convection in providing a path for energy to space. This is what my two articles on Atmospheric Thermal Enhancement were all about. In fact my contention is that 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.

300. Bryan says:

Tim F says

“So who is presenting an unphysical, pseudoscientific model?”

You are and its pretty obvious to anyone who knows any physics.

Maxwell had a famous conjecture about how, if he could with a demon, separate a gas into hot and cold fractions he could contradict the second law.
Separated hot and cold regions can produce mechanical work.

This is a realisation of the conjecture as the shell becomes 302K and 256 K or lower for extended periods.
The separated fractions could give unlimited usable energy.

Of course this is fairyland where nonsense rules.

Anyone with any pretensions of a physics background knows that Maxwell on this occasion was wrong.
The second law will not be contradicted.
Not by Maxwell or Willis and certainly not by you.

So cut out the condescending remarks they just make you look ridiculous.

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