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.
Tallbloke's Talkshop 
What are the criteria of a good model? They are law, logic and correct predictions. This model has 2 bodies interacting and reaching thermal equilibrium through emitting, absorbing and conducting.
The criticisms of the physics
Relative area of surfaces
It is less than 1%.. Also Willis assumes the difference is not important.
Back radiation
The criticism is that energy flows from hot to cold in the model and this breaks the 2nd law of thermodynamics.
Number of surfaces
The shell is treated as a body which conducts from one surface to the other and radiates from 2 surfaces. The cores surface is treated as a body which transfers energy out from only 1 surface.
Logic
My criticism is that the rules of arithmetic used in the model are not consistent, when they are used consistently the model does not predict warming of the core surface.
The model does not lead to correct or uncontroversial predictions because it does not include mass and energy transfer by conduction or convection.
[Reply] The second law does not preclude energy from a colder object being absorbed by a warmer object, so far as I’m aware. Can you find a formulation of the law in any literature which does so?
Tallbloke,
Try spitting on both surfaces of a hollow sphere.
[Reply] OK, you hold it open
An update to my thermal experimentation:
The copper greenhouse!
http://climateandstuff.blogspot.co.uk/2013/03/the-copper-greenhouse.html
Please feel free to criticise, but I think this experiment shows that an iron greenhouse acts as Willis suggests!
Just don’t tell the skydragons!!!
“Now we find that simply removing layers of material beneath the original surface+shell would produce more energy?
Why isn’t this being engineered and utilized?”
Produce more energy? No.
Because the energy you are thinking was ‘excess’ energy was not excess in reality, this is the same decrease in energy found as you added layer that was not radiated from the inner surface but instead accumulated as internal energy to raise the temperature until equilibrium was reached. Furthermore, that energy originally raising the inner surfaces came not from “back radiation”, or better termed as ‘always equal or smaller but opposing radiation’, but came from the innermost nuclear reactions supplying this entire system with energy. Opposing radiation can oppose but can never warm. You must subtract opposing powers. A Watt unit is power and ‘J/s’ units are energy flux and even though they can be reduced to the same you must be very careful to keep them separated in reality or you will make incorrect assumptions of what is possible if seperated and what is actual. A difference in powers, a subtraction of opposing powers, is the flux, the work performed, it can either be a flux or an increase in temperature depending on the surrounding states.
[Reply] The second law does not preclude energy from a colder object being absorbed by a warmer object, so far as I’m aware. Can you find a formulation of the law in any literature which does so?
I was summarizing the arguments that have been made against the model.
My argument is whatever rules are used in a model they must be used in the same way in each similar situation. They are not in this model.
Thefordprefect says on his blog
I believe that this [ experiment ] shows that willis’s iron greenhouse model is likely to be valid.
I applaud the work you have done to make a practical test of the model prediction.
However in your experiment you observed a 0.6C increase from 75C.
That is 0.6K increase from 348K. 0.6/348 x 100 = 0.2%
Which is within the amount of error you can expect in such an experiment.
Willis predicts 48K
Why does the IR sensor measure 23-24C not 75C?
Roger C: You stated “The criticism is that energy flows from hot to cold in the model and this breaks the 2nd law of thermodynamics.”
Now, I’m assuming you meant that the other way round, because everyone accepts that energy flows from hot to cold, and that does not break the 2nd law. So you are claiming energy can’t flow from cold to hot.
However, I haven’t found any formulation of the 2nd law which says energy from a colder object can’t be absorbed by a warmer object, only that heat can’t flow from colder to warmer.
Do you accept that?
( core 1169 W/m^2 supply) <~ 4000 m radius, 235 Gigawatts
( original 235 W/m^2 planet emissions ) <~ 8920 m radius, 235 Gigawatts
( shell 235 W/m^2 outer surface emissions ) <~ 8921 m radius, 235 Gigawatts
Right?
Ok.
So now we willis says we have:
( core 1169 W/m^2 supply) <~ 4000 m radius, 235 Gigawatts
( post-shell 470 W/m^2 planet emissions ) <~ 8920 m radius, 470 Gigawatts
( shell 235 W/m^2 outer surface emissions ) <~ 8921 m radius, 235 Gigawatts
Now add a gap below the original planet surface:
( core 1169 W/m^2 supply) <~ 4000 m radius, 235 Gigawatts
( gap in planet 940 W/m^2 emissions ) <~ 8919 m radius, 940 Gigawatts
( post-shell 470 W/m^2 planet emissions ) <~ 8920 m radius, 470 Gigawatts
( shell 235 W/m^2 outer surface emissions ) <~ 8921 m radius, 235 Gigawatts
Hmmm, do that again:
( core 1169 W/m^2 supply) <~ 4000 m radius, 235 Gigawatts
( 2nd gap 1880 W/m^2 emissions ) <~ 8918 m radius, 1878.8~ Gigawatts
( 1st inner gap 940 W/m^2 emissions ) <~ 8919 m radius, 939.6~ Gigawatts
( post-shell 470 W/m^2 planet emissions ) <~ 8920 m radius, 470 Gigawatts
( shell 235 W/m^2 outer surface emissions ) <~ 8921 m radius, 235 Gigawatts
Hmmm, 2^n where n=number of gaps? Seems problematic if we try to conserve power rather than energy.
________________________
Myself, I think this is more realistic:
( core 1169 W/m^2 supply) <~ 4000 m radius, 235 Gigawatts
( 2nd gap 235.1 W/m^2 emissions ) <~ 8918 m radius, 235 Gigawatts
( 1st inner gap 235.08 W/m^2 emissions) <~ 8919 m radius, 235 Gigawatts
( post-shell 235 W/m^2 planet emissions ) <~ 8920 m radius, 235 Gigawatts
( shell 234.99 W/m^2 outer surface emissions) <~ 8921 m radius, 235 Gigawatts
Conserving energy rather than power seems right to me.
I have just responded to TFP on his latest Experiment thread as follows.
Sorry, your results appear to completely disprove the iron ball/shell theory unless you can show by calculation that 50% of the radiation from the Heat Source equals a rise in the temperature of the heat source of only approximately 1.25 degC.
Whereas the Calculation should at the same time be able to show that the Copper plate’s Temperature rose 51.5 degC using 150% of the radiation.
Do you really think that the 1.25 degC increase in the heat source represents the 50% increase in the iron ball temperature from the Willis theory?
Like Roger I believe that the TFP Experiment proves the case For the Dragon’s not disprove it.
ACO: If the Dragonslayers are right, why did the temperature rise at all?
thefordprefect
why didn’t you use bigger plates, all the way to the walls so that hot air couldn’t leak from the warm side to the cold side?
Max
No, adding more shells has already been discussed. Each shell only adds 235 W/m^2, not double.
The temperature rose because the interior conditions of the box have been changed.
For instance the GHGs (air) between the heat source and the plate could have been heated more due to Reflected radiation from the plate, which is not a perfect black body and not re-emitted radiation, which in turn would heat up the source slightly.
But note it does not even come close the sort of increases expected from “Back Radiation”.
A C O
It does not come close because most of the energy is transferred by air conduction.
Anyway the source shouldn’t warm at all since the air is colder than the source, and cooler can’t warm warmer, remember?
thefordprefect says: I believe that this shows that willis’s iron greenhouse model is likely to be valid.
This also means that the GHG theory is valid.
http://climateandstuff.blogspot.co.uk/2013/03/the-copper-greenhouse.html
Right up until the last line I was with you. 😉
If by “the GHG theory” you mean the ‘enhanced greenhouse effect’ in the open atmosphere, then I don’t think your experiment addresses this.
It doesn’t provide sufficient evidence for a 33C lift in the surface temperature either IMO.
It does show that back-radiation has an effect though, so well done with the experimental work.
lgl says:
March 23, 2013 at 5:17 pm Anyway the source shouldn’t warm at all since the air is colder than the source, and cooler can’t warm warmer, remember?
That was only an example of what could have happened, like I said the internal conditions have changed, as you yourself say the heat is lost/transferred due to Convection, which is now changed/restricted by the addition of the plate.
So come on with Willis Back Radiation what temperature should the Heat Source display with the Plate?
Same Question to Tim F.
tallbloke says: March 23, 2013 at 5:23 pm “It does show that back-radiation has an effect though, so well done with the experimental work”.
I agree with the second part “well done with the experimental work”, it is very good.
But I do not agree with the first part that “It does show that back-radiation has an effect”, in fact as Roger Clague says: March 23, 2013 at 3:59 pm “Which is within the amount of error you can expect in such an experiment.”, which I believe is due to the condition change, not back radiation.
ACO: Given that the results effectively replicate his earlier experiment too, I think the onus is now on those who don’t believe in currently accepted radiative physics to do some experiments of their own.
lgl says: March 23, 2013 at 4:42 pm
thefordprefect
why didn’t you use bigger plates, all the way to the walls so that hot air couldn’t leak from the warm side to the cold side?
————-
TFP: this would have changed the radiating area, I felt it best to keep the plate close to the heater and for the area radiating to be constant so the ir window restrictive size would have similar effect.—————————————
A C Osborn says: March 23, 2013 at 4:16 pm
Sorry, your results appear to completely disprove the iron ball/shell theory unless you can show by calculation that 50% of the radiation from the Heat Source equals a rise in the temperature of the heat source of only approximately 1.25 degC.
Do you really think that the 1.25 degC increase in the heat source represents the 50% increase in the iron ball temperature from the Willis theory?
—————————–
TFP: This sort of real world kitchen table top experiment in no way can EXACTLY replicate the iron greenhouse thought experiment.
As I said in the write up, perhaps the most important thing is the the 2 single plate runs the internal stucture of the warm box is the same (a restrictive plate changes the convection in the box in a similar way, so what explains the 0,6degC rise in temperature when the copper plate is present?
I was not looking for exact energy flows (for example I knew that the IR windows are not 100% transmissive, I knew the box is loosing heat through its walls, I do not know what the thermal capacity of the heater is, etc. All this is experiment does (and was expected to do) is show a warming where there shousd according to the slayers be none
——————
A C Osborn says: March 23, 2013 at 5:01 pm
The temperature rose because the interior conditions of the box have been changed.
For instance the GHGs (air) between the heat source and the plate could have been heated more due to Reflected radiation from the plate, which is not a perfect black body and not re-emitted radiation, which in turn would heat up the source slightly.
———————————
TFP yes the internal conditions have change but that is why I tried a IR “invisible” plate in place of the copper. The IR loss in this plate did cause a slight warming of the heater but no where near as much as the single copper plate.
but also remember that the plate will be cooler than the heater and slayer theory says energy cannot travel from cooler to hotter!
——————————-
tallbloke says: March 23, 2013 at 5:23 pm
Right up until the last line I was with you.
————————
TFP mmmmm! I suppose you are right I’ll modify that!
A C O
“what temperature should the Heat Source display with the Plate?”
It’s impossible to tell with all that mixed air. This is not a test of Willis model.
“No, adding more shells has already been discussed. Each shell only adds 235 W/m^2, not double.” ~lgl
Ok.
( core 1169 W/m^2 supply) <~ 4000 m radius, 235 Gigawatts
( 3rd gap 1175 W/m^2 emissions) <~ 8917 m radius, 1174~ Gigawatts
( 2nd gap 940 W/m^2 emissions~ ) <~ 8918 m radius, 939.4~ Gigawatts
( 1st inner gap 705 W/m^2 emissions ) <~ 8919 m radius, 704.7~ Gigawatts
( post-shell 470 W/m^2 planet emissions ) <~ 8920 m radius, 470 Gigawatts
( shell 235 W/m^2 outer surface emissions ) <~ 8921 m radius, 235 Gigawatts
Again, if the body in question has a surface area of 1 billion square meters and emits 235 W/m^2 from the entire surface, that's 235 Gigawatts, and it has a radius of about 8920 m.
From that we could work out that if the power source had a radius of 4000 m and was supply power across a vacuum, naively we would expect the surface to emit 235 W/m^2 when the power source was giving off 1169 W/m^2, as both values work out to 235 Gigawatts.
Adding gaps can not add 235 W/m^2 because that raises the output beyond 235 Gigawatts.
Conservation of energy is a thing. Conservation of power is not.
This is not a minor error, though it is also part of the issue I have with ignoring the inverse-square effects, comparitively that is indeed a small effect when placed next to a statement like “each shell only adds 235 W/m^2”, isn’t it?
Max: Conservation of energy is a thing. Conservation of power is not.
Have a think about what you mean by ‘output’ and how you would measure it.
“However, I haven’t found any formulation of the 2nd law which says energy from a colder object can’t be absorbed by a warmer object, only that heat can’t flow from colder to warmer.”~ tb
“Entropy is a measure of the disorder of a system. That disorder can be represented in terms of energy that is not available to be used. Natural processes will always proceed in the direction that increases the disorder of a system. When two objects are at different temperatures, the combined systems represent a higher sense of order than when they are in equilibrium with each other. The sense of order is associated with the atoms of system A and the atoms of system B being separated by average energy per atom – those of A being the higher energy atoms if system A is at a higher temperature. When they are put in thermal contact, energy flows from the higher average energy system to the lower average energy system to make the energy of the combined system more uniformly distributed – ie, less ordered. So the disorder of the system has increased – and we say the entropy has increased. But the process of increasing the disorder has removed the possibility that the energy that was transferred from A to B can be used for any other purpose – for example, work cannot be extracted from the energy by operating a heat engine between the two reservoirs of different temperatures. So although energy was conserved in the transfer (the first law), the entropy of the universe has increased in becoming more disordered (the second law) and consequently the availability of energy for doing work has decreased.”
http://www.calpoly.edu/~rbrown/entropy.html
Energy from a cold reservoir being absorbed in a warm reservoir is a decrease of entropy.
” Second Law of Thermodynamics: It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object. This precludes a perfect refrigerator. The statements about refrigerators apply to air conditioners and heat pumps, which embody the same principles.”
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/seclaw.html#c3
“Heat may be defined as energy in transit from a high temperature object to a lower temperature object. An object does not possess “heat”; the appropriate term for the microscopic energy in an object is internal energy. The internal energy may be increased by transferring energy to the object from a higher temperature (hotter) object – this is properly called heating.”
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heat.html#c1
Energy from a cold object being transferred to a warm object with no work performed to facilitate this transfer and/or no waste heat being rejected is specifically forbidden by the second law, entropy must remain the same or increase, energy from a cold object being transferred to a warm object is heat, and decreases entropy.
Well tallbloke, I would measure energy output in Joules, if I then consider Joules per second I get Watts for power, and if I then take into account the size of the surface or body or volume in question I get W/m^2 which is the intensity in the case of radiation.
Intensity is a vector, hence my insistence that you must subtract radiation from other sources, opposing vectors do not add. If you spray a garden hose at a fire hose the velocity and kinetic energy of the water from the garden hose doesn’t increase the velocity and kinetic energy of the water from the fire hose, does it?
[Reply] Opposing photon streams interpenetrate, they do not cancel or cause ‘subtraction’. If I shine a torch at a lightbulb does the room go dark?
Max: Energy from a cold object being transferred to a warm object with no work performed to facilitate this transfer and/or no waste heat being rejected is specifically forbidden by the second law
Point 1. Work is being performed by the nuclear core to generate heat. It has to go somewhere, and quite a lot of it goes into heating various elements in the system to temperatures in accordance with their ability to retain and radiate it.
Point 2. The laws refer to closed systems. Willis’ gedanken experiment is not a closed system.
Have you got that Max?
not a closed system
not a closed system
not a closed system
not a closed system
not a closed system
Once again:
not a closed system
Uh, tb, the shell is not performing work to transfer heat to the interior surface, and the laws are universally valid in closed systems, that does not mean you can simply ignore them otherwise.
Willis’ experiment is still an example of energy from a cold reservoir being added to the internal energy of a warm reservoir with no work being performed, and no heat being rejected in the process. It is beyond 100% Carnot efficiency, it is verboten by thermodynamics.
[Reply] Uh, Max, in Willis’ experiment the warm reservoir is the nuclear core, heat is rejected to the cold reservoir (space) and the temperature of the components in the system is what we are interested in. Carnot engines using radiation as the working fluid are NOT THE SAME as Carnot engines using gases as the working fluid, google the difference. Carnot engines have pistons. Where is the moving component of Willis’ system?
Max
I don’t know where your numbers come from but in his WUWT post Willis writes
“For an earth-sized planet with a shell two kilometers above the surface everywhere, the difference in area is only six hundredths of one percent.)”
If six hundredths of one percent is not small enough for you we can make the planet the size of the galaxy and totally ignore the radius difference.
-thefordprefect says:
March 23, 2013 at 1:29 pm
An update to my thermal experimentation:
The copper greenhouse!
http://climateandstuff.blogspot.co.uk/2013/03/the-copper-greenhouse.html
Please feel free to criticise, but I think this experiment shows that an iron greenhouse acts as Willis suggests!
Just don’t tell the skydragons!!!-
Cool.
How you know the copper isn’t reflecting IR?
What is results if replace copper plate with plate which is highly reflective to IR?
“However, I haven’t found any formulation of the 2nd law which says energy from a colder object can’t be absorbed by a warmer object, only that heat can’t flow from colder to warmer.
Do you accept that?”
What is meant by absorbed.
With radiant energy would absorbed mean converting radiant energy into heat [adding thermal energy to the object]?
[reply] No. That would violate the second law. I don’t see any prohibition against energy being absorbed and then re-emitted though.
The problem is in ‘time’. Everyone is saying things like… first it does this, then it does that… one emits and then it goes over there and is absorbed and such talk in itself is not even real in Mother Nature’s world of large numbers, huge numbers. On way to view her is seeing her dealing in simultaneous pairs of actions (integration).
So, if a surface at exactly the same moment both emits and absorbs an identical sized quanta, zero time between the two, what has happened to that surface? Has the temperature changed? Did the two events even occur? Are you sure? Can you prove it? How?
“gbaikie says:
March 23, 2013 at 7:47 pm
“However, I haven’t found any formulation of the 2nd law which says energy from a colder object can’t be absorbed by a warmer object, only that heat can’t flow from colder to warmer.
Do you accept that?”
What is meant by absorbed.
With radiant energy would absorbed mean converting radiant energy into heat [adding thermal energy to the object]?
[reply] No. That would violate the second law. I don’t see any prohibition against energy being absorbed and then re-emitted though.”
Ok. I can agree with that. But I think important to have different terms and lack of different terms
causes endless unnecessary confusion.
I think this process of not adding thermal energy, or as you say absorbed and re-emitted- is a key aspect related to the energy of sunlight in regards to Earth.
I have no problem with atmosphere directly absorbing and re-emitting sunlight, I have problem with
idea of atmosphere gas gaining directly from sunlight [or any radiant energy] thermal energy.
Or it is surfaces which are liquid or solids which can gain thermal energy from radiant energy.
And it’s only the gain and loss of thermal energy which is involved in Earth’s energy budget.
wayne says, “So, if a surface at exactly the same moment both emits and absorbs an identical sized quanta, zero time between the two, what has happened to that surface? Has the temperature changed? Did the two events even occur? Are you sure? Can you prove it? How?”
Excellent summary!
gbaikie says: March 23, 2013 at 7:07 pm
How you know the copper isn’t reflecting IR?
What is results if replace copper plate with plate which is highly reflective to IR
—-
The copper is spray painted with grey primer (from previous tests at work this is a reasonable IR absorber.
———
I have polished a copper plate (same dimensions) and will run the test again. I have also doubled the insulation on the sensor.
Max: I can set a control volume which encloses the planet and shell easily enough, accounting for the losses are easy enough, and one can show that the second law must hold within that volume.
I think this is where you are going wrong. Your control volume encloses the heat source (which means you need to account for the nuclear energy to heat conversion potential, and the rate it is released at) but doesn’t enclose the heat sink. The maths can’t work in this situation.
You don’t get to simply say “it isn’t a closed system” and then baselessly assert that the second law doesn’t apply in this case.
Be very careful about putting words in my mouth.
I’m not putting words in your mouth, you said it is not a closed system, and asserted that the second law does not apply.
“Point 2. The laws refer to closed systems. Willis’ gedanken experiment is not a closed system.
[Reply] Pointing out that Willis’ system is not closed is not “asserting that the second law does not apply”. What I’m getting at is the fact that the second law does not apply in a dynamic system with energy flowing through it in the same way it does in a closed system.
Myself, I am proposing that there are two possible outcomes from adding the shell, and I am saying one of them entails a violation of the laws of thermodynamics while the other does not.
1. Adding the shell raises the intensity of radiation at the surface of the planet to 470 W/m^2, but for some reason does not raise the intensity of the radiation emitted from the shell by the same amount,
[Reply] Both the planetary surface and the shell will emit ~470W/m^2, but the shell will only emit 235 of it to space.
though it still results in more than 235 Gigawatts being emitted.
[Reply] Emitted from where to where?
2. Adding the shell does not change the intensity of radiation at the surface of the planet, the radiation reaching the shell is reduced in intensity according to the inverse-square law, and the total emissions remain 235 Gigawatts.
[Reply] Since 235 has to go to space for the system to be in a steady state, am I to take it you are one of the people who believes a heated steel object will only radiate from one side?
In my IPCC and SkyDragon diagramsyou will note that both contain energy flows that balance perfectly. The issue of which diagram is correct therefore remains one of physics, not mathematics. And it certainly has nothing to do with marginal radius differences.
i suggest that the only realistic way forward now is to build on TFP’s experimental work to produce more robust results that will decide the issue one way or the other.
All else is just hand waving and assertion.
While I applaud the enthusiasm here, it is never going to be very effective having people who don’t understand thermodynamics trying to explain thermodynamics to other people who don’t understand thermodynamics.
Especially not in a system that involves numerous subtle theoretical and practical points.
Especially on a blog where errors can compound with errors and conversations are jumbled together.
If you have not at a bare minimum read through AND worked through the problems of an undergraduate thermal physics text like An Introduction to Thermal Physics by Daniel V. Schroeder or Thermal Physics by Kittel & Kroemer, then you have no business trying to teach others about thermal physics.
Tim Folkerts says the usual elitist utterances that we come to expect.
The poor schmucks who pay inflated energy bills and watch the world economy grind to a halt had better keep their mouths shut.
They may observe that CO2 concentration and troposphere temperature have never been correlated.
The temperature stopped rising 16 years ago.
Every prediction of IPCC science has proved false.
Yet Tim ignores the results of the giant ongoing climate experiment that we observe.
Speak up sceptics you cannot do much worse than Tim who is prepared to believe in fairy tales.
So says one brought up on Zemanskys ‘Heat and Thermodynamics’ and Adkins ‘equilibrium Thermodynamics’
” thefordprefect says:
March 23, 2013 at 10:04 pm
gbaikie says: March 23, 2013 at 7:07 pm
How you know the copper isn’t reflecting IR?
What is results if replace copper plate with plate which is highly reflective to IR
—-
The copper is spray painted with grey primer (from previous tests at work this is a reasonable IR absorber.
———
I have polished a copper plate (same dimensions) and will run the test again. I have also doubled the insulation on the sensor.”
My prediction is: If second plate is reflective, the heat source should increase in temperature.
Also first plate may increase in temperature and may increase the amount of radiant energy being emitted between first and second plate. But outer part of reflective plate should not have increase
in temperature or amount radiated [could decrease].
Well said Bryan.
I could point Tim F to statements he made on WUWT in the past which would make him cringe in the light of things he has learned at this blog since. However, it’s not productive, or conducive to collaborative learning. Perhaps he might have a think about that.
[Reply] Since 235 has to go to space for the system to be in a steady state, am I to take it you are one of the people who believes a heated steel object will only radiate from one side? ~tb
No, I am one of the (apparently few) people who understands that the actual flux between the inner surface of the shell and the planet would be very low as they would be nearly the same temperature, so only the outer surface will have a significant differential into which it could emit at full power.
The inner surface could not emit radiation at greater intensity than it receives, so it must always emit less than the planet surface does, due to geometrical constraints on the system.
As such there must always be a negative heat flow from the shell to the planet, which means the shell can not add energy to the planet, as that would be a spontaneous reduction of entropy.
_____________
My refrigerator is not a closed system, but it still lacks the ability to transfer energy to a warmer reservoir without work being performed.
If this thought experiment were true there would be some way to capitalize on it, and accordingly I should be able to say, put a thermos wall around my refrigerator, unplug it, and it would stay cold forever, or leave it plugged in and it would generate energy for free, either way it is impossible.
_____________
Once again I will point out that 235 W/m^2 do not have to go to space for the system to stabilize, so long as the total emitted is the same 235 Gigawatts supplied by the power source the temperatures will be stable. Intensity of radiation is not conserved, power is not conserved, energy is.
Power is the rate at which energy is supplied or work is performed, intensity is the density of power in a given direction.
The intensity of radiation from a cold body to a warm body is negative, that doesn’t mean there is no radiation, nor does it mean the radiation is magically deflected, I disagree with any skydragons who put forth that idea just as much as I disagree with anyone who claims all radiation increases the internal energy in the body it strikes, even if the body is warmer than the source of said radiation, and thus gives off more intense radiation.
So to answer your earlier semi-rhetorical question, tb, if you shine a flashlight at a light bulb, no it doesn’t make the room suddenly go dark, but it doesn’t make it brighter either, the vectors in that case are subtracted from each other with the dimmer source reducing the brighter source accordingly.
I should note that the flashlight casts a shadow though, but we can ignore that.
____________
As for thefordprefect, what would be the outcome if you performed the experiment with a mirror in front of the window instead of the copper? I’m interested in the outcome but do not have the same equipment available or I would try a few arrangements myself.
I had an idea for a way to handle the lack of vacuum.
Two hemispheres separated by a gap with a ring of material blocking direct view.
_—_ <- hemisphere edge gap+radiative shield
(( )) <~ hemispheres with heat source between
It should let convection take place while trapping the radiation from the heat source.
Max: So to answer your earlier semi-rhetorical question, tb, if you shine a flashlight at a light bulb, no it doesn’t make the room suddenly go dark, but it doesn’t make it brighter either, the vectors in that case are subtracted from each other with the dimmer source reducing the brighter source accordingly.
So presumably if we shine two equally bright torches at each other, the vectors will cancel and there will be no illumination of the dust particles in the air between the beams then? Or will the shadows cast by the dust particles prevent the canceling and allow the hemispheric illumination of the particles from each torch? What about penumbral effects? 🙂
Max: the actual flux between the inner surface of the shell and the planet would be very low as they would be nearly the same temperature, so only the outer surface will have a significant differential into which it could emit at full power. The inner surface could not emit radiation at greater intensity than it receives, so it must always emit less than the planet surface does, due to geometrical constraints on the system.
But since the outer shell surface must emit 235 to space for the steady state, there in zero left over for the inner shell surface to emit, according to your scheme, which is that the shell only gets hot enough to emit 235.
There is no such thing as cold radiation. So how does an uncooled microbolometer array at 50C in a thermal imaging camera take photographs of objects at -40C?
if there are say 2 objects in view one at -40C and one at -39C what enables the camera to record a difference?
Surely this can only be that the -40C object is allowing the bolometer to cool to a lower temperature than the -39C object. This cannot be true if cool objects radiation is rejected by a hot bolometer.
http://climateandstuff.blogspot.co.uk/2012/05/cool-body-can-transfer-measurable-heat.html
tallbloke: First part: pointing a flashlight at another flashlight will not make them brighter than they would normally be. The relative brightness will be greatest in a dark environment, if what is being claimed here were true then pointing a flashlight at the sun would be dangerous, one would think, what with it heating up to several thousand Kelvin and all. *pwoomf!*
[Reply] You said they would cancel. They won’t. You’re wrong. Photon streams and E/M waves interpenetrate. Live with it.
Second part: The outer surface must emit a bit less than 235 to space for steady state, the inward emissions are canceled and then some by the emissions from the planet, the intensity of radiation between the shell and planet is small but positive in the direction of the shell.
[Reply] You were talking in power terms remember? The shell must emit the same 235GW to space as is generated by the nuclear core for steady state. The fact that due to geometry this means a fractionally smaller power density is neither here nor there. Further bullshit from you will be binned, I’ve had enough of it.
thefordprefect: infrared radiation is not heat, it is only heat if it is coming from a warmer object. Microwaves are “colder” than infrared, being less energetic and only emitted spontaneously by very cold objects.
An object which is around 1 K will emit (microwave) radiation with wavelengths measured in millimeters, not micrometers/microns/µm like most of the infrared spectrum.
Bryan says“Tim Folkerts says the usual elitist utterances that we come to expect.”
Is it “elitist” to go to a doctor for medical advice instead of someone who took a health class in college?
Is it “elitist” to go to hire an experienced contractor to remodel my house, rather than a kid who took high school shop class?
Is it “elitist” to call a lawyer if I get arrested, rather than calling my friend who likes ot watch legal shows?
When there is something complicated or tricky to be done, why would you NOT want an elite practitioner?
I would look in a physics textbook or talk to a physics prof about a thermodynamics question, not some guy who makes freshman-level physics mistakes.
[Reply] Problem is, when I email learned prof’s to ask why they believe back radiation causes the surface to warm like Willis’ toy planet when there is no vacuum between the TOA and ground, but an upward convecting troposphere, they don’t answer. The reason is that the ’33C greenhouse effect’ on Earth is bullshit from start to finish, and they know it. It’s about time you admitted it is a fairy story.
***********************************************
Tallbloke says: “I could point Tim F to statements he made on WUWT in the past which would make him cringe … “
I would be interested to know which ones you might think would make me cringe. I am sure there would be a few.
But in any case, that only highlights my point. I suspect that I have as much formal education as anyone here when it comes to thermodynamics. But I had little experience with “climate science” until a few years ago. So I listened when others introduced me to the concepts like “top of atmosphere” and “dry adiabatic lapse rate” and “effective blackbody temperature”. I researched and applied my knowledge until I felt comfortable with the topics. Only then did I feel comfortable “teaching” about the topics. (And I try to limit my contributions to physics, since i am not at all an expert in climate models or politics or economics.)
[Reply] Nobody is an expert in climate science, certainly not the climate scientists. Theoretical physicists extrapolating belljar experiments to the real atmosphere is what caused has all the trouble in the first place.
*************************************************
As I said, it is cool to have a place like this to discuss and ponder and argue. But don’t expect to come to any resolution (as has been shown so clearly by over 500 posts which are not getting any closer to agreement). Recognize that maybe there really are incredibly bright professors who figured this out and wrote textbooks about it and who maybe, just maybe know more than you or I do.
[Reply] It has taken a while to reach the point where Max is cornered, but we get there in the end…
*************************************************
Various effects will make small differences, but in the idealized limit, the planet in Willis’ model really WILL be ~ 2^(0.25) times hotter with the shell than without.
[Reply] You should ponder why it is you felt the need to put ‘idealised’ in the above sentence, and then include the word ‘really’ and then capitalise the word WILL. Who are you trying to convince? Yourself? 😉
Experimentum summas judex -Albert Einstein-
Yes, the shell has to emit 235 Gigawatts
[snip irrelevance]
[Reply] Thank you. Now onto the substantive point. Under your proposed scheme, the shell would only receive 235 Gigawatts from the planet. So all of the energy would have be radiated from one side of the shell to space in order for the system to be in a steady state. But this is impossible, because the heated shell would radiate from both it’s inner and outer surfaces almost equally into the vacuum on either side of it, so your proposal (and the dragonslayer proposal) fails.
thefordprefect, I have no problem with the colder pixel if you will explain first how you see the 50°C camera ever recording any pixel in the scene that is also at 50°C. I think you agree that no net energy passes between these two pieces of matter at the same temperature even though there is a lens between performing scaling. How does that one case record and then we can delve into the differentials of hotter-than and colder-than. 😉
Kind of joking with you there… don’t you see there being a zero-level current at the ambient temperature of the camera? Or do you see the only way to record any pixel is it receiving a net flow of energy into the camera and not out of the camera thereby warming the scene at that colder pixel?
Climate and Stuff says about a 20C camera recording something much colder: “The bolometer is now receiving more energy than when -273C is focused. The bolometer therefore heats a fraction to say 20.1C Hopefully you can see where this is going. ….”
Yes I do…. in the trash !!
Another site bite the dust that I will never consider they know the least of physics:
Climate and StuffMy, my, and after a pretty good article explained to them that “Modern microbolometers measure temperature changes …” and detect radiant heat, not photons.
So instead they should have said 19.9C, not 20.1C. Another stuck in the mud about absolute zero, blackbodies, photons everywhere always warming never cooling, and “back radiation”. Brother!
thefordprefect, it seems I took your statement wrong and should have been responding to Max, again. Yes, colder radiation can and does modulate warmer radiation, even through a camaera lens. It is just that more energy goes out than comes in and the pixel in the camera cools and the scene that pixel sees warms (but that would be a very tiny warming wouldn’t it 😉 ) But I sure hope you see the flaws in that article you pointed at.
wayne says:
March 23, 2013 at 8:36 pm
Wayne and tallbloke…
[“So, if a surface at exactly the same moment both emits and absorbs an identical sized quanta, zero time between the two, what has happened to that surface?”]
Wayne, thanks, this is getting close to addressing the problem I have with the model.
tallbloke, it is not just that lower energy can be absorbed by a warmer surface, it is that the absorption has to be for net gain in order for the surface to then emit at 470 W/m^2.
Or am I wrong in thinking that a photon of the 235 downward arrow has a different amplitude/wavelength/frequency to a photon of the 470 upward arrow? If so, then the 235 photon must have been converted at or near the planet’s surface, which will involve heating the planet. This therefore means that the lower (cooler) energy radiation has helped to warm the planet. It is this that seems to contradict the 2nd law. And it happens apparently instantaneously.
How can the 235 photon be changed to a 470 photon without the 2nd law being broken? Can two lots of low energy simply be numerically added, especially when the underlying source is the same?
suricat says:
March 23, 2013 at 3:30 am
[“No. It would be ‘partly’ true for a BB scenario, but a real scenario is entirely different.”]
Thanks, Ray. It wasn’t meant to be a proper analogy, just to show that a numerical addition may not always apply.
I’m still trying to learn what happens in the real scenario! 🙂
The science doesn’t seem very settled, does it?
Regards,
Max
“the intensity of radiation between the shell and planet is small but positive in the direction of the shell.”
How does the planet get rid of the 235 generated then?
Tim Folkerts says, March 23, 2013 at 11:58 pm: While I applaud the enthusiasm here, it is never going to be very effective having people who don’t understand thermodynamics trying to explain thermodynamics to other people who don’t understand thermodynamics.
We are all learning, Tim. Including you.
[snip]
[Reply] I’ve emailed the rest back to you for re-use elsewhere. There are plenty of live threads for political debate on the talkshop right now. This one is for basic radiation theory. Thank you for your co-operation.
Tim Folkerts says in effect that only experts should comment on thermodynamics questions like whether or not there is a greenhouse effect.
Yes experts like Micheal Mann and Phil Jones will keep us right.
Other experts like Gerhard Gerlich, Ralf Tscheuschner, Miskolczi ,Gerhard Kramm, R W Wood you would not approve of.
There is fraud and malpractice behind much of climate science .
Second rate scientists hide and falsify results.
A cabal exclude any contrary views
Money is made and it curiously resembles the great bank fraud that has crippled the world economies.
Esoteric financial products were invented that only the ‘experts’ could understand.
Millions siphoned off and greedy bankers bloated with criminal gains.
The working and middle class are left much poorer as a result.
In Britain at the moment we are suffering the coldest March for 60 years.
This is now not unexpected because we have has a recent run of colder than average winters.
The deaths due to hypothermia are a shocking rebuke for the shameless ‘experts’ who persuaded the politicians to go along this path.
No prediction of climate ‘science’ stacks up
Yes sceptics are right to question much of ‘climate science’ , complacent ,deluded self appointed ‘experts’ like Tim should be ignored.
wayne says March 24, 2013 at 3:29 am
thefordprefect, I have no problem with the colder pixel if you will explain first how you see the 50°C camera ever recording any pixel in the scene that is also at 50°C.
—————————–
for example a bolometer at temperature of 50C when an object has these temperatures (heat sources) focussed:
object = -273C bolometer = 40C the temperature with NO external heat added to the internal 40C
object = -40C bolometer = 48.5C (say)
object = 50C bolometer = 50C (say)
object = 1500C bolometer = 95C (say)
All these temperatures are at equilibrium i.e. radiation in= radiation out (stability is reached within milliseconds in modern microbolometer arrays)
There are no cold rays!
the pixel microbolometer is in radiative balance with the object at the other end of its focus. at 50C bolo temp and 50 C object temp the radiation is balanced the same as at other temps
Arfur Bryant says:
March 24, 2013 at 7:14 am
How can the 235 photon be changed to a 470 photon
Good question. The surface it meets is at 254K .
At one time I had a 2 bar electric fire.
I turn on one bar, nice red glow.
I turn on the another bar. The colour of each of the 2 bars is the same as the one bar alone was.
That is the temperature of the first bar is not increased by the radiation from the second bar.
The radiation from the first bar is not changed by the radiation from second bar.
In a similar manner the radiation from the shell cannot increase the temperature of the core
That is evidence from my observations. But back to the the theory.
Why does the shell radiate from 2 surfaces but the core surface from only one?
Brian, I couldn’t agree more, have a look at the reaction to Richard Betts over on Bishop Hill.
http://bishophill.squarespace.com/blog/2013/3/24/bringing-politicians-to-booker.html
Arfur Bryant, I believe I see exactly what you are questioning and just might help you take on a new perspective. You keep speaking of 235 photons (254K) and 470 photons (302K), I guess meaning the total powers (the quantity and quality) of each, but I try to never look at them as such. I see a set quantity of 5 μm photons, a quantity of 10 μm photons, a quantity of 15 μm photons and such for both temperatures. Photons are different by the quanta of energy they carry and I am a bit confused whether you see quantities of photons at 235 and 470 or the power that each carries, or just the total energy. Maybe you see all at once, which would be great.
I have taken the time before writing back to your comment a spreadsheet of the BB Planck curves for three items, (470) 302K, (235) 254K, and the difference of the total energy carried between the 302K and 254K curves (delta of the two curves) and this might give you a new look. Only the quantity of 235 photons (and there is even more of these in the 470 group at each frequency) can be resonating between the sphere and the shell, that much energy staying constant, swapping back and forth, no T change. they must match in quantity and quality. But there is a remnant of the 470 group and that is what carries the other 235 W/m² from the energy source, through the shell and to space. These remnant photons are actually of higher “quality” in numbers than either the 235 or 470 groups. The remnants peak at a higher frequency when you look at the total quantity of energy they carry (surprised me!). You need to look at what the BB spectrums are doing might be a better way to word that..
I feel I’m not explaining this well, give me a moment to pick better words. Will be back to you. You are right, this might be one of the picky details where everyone keeps slipping right by each other, never able to see the same picture. Please don’t let TF sidetrack or confuse you too soon, at least give me a chance.
thefordprefect, I never said there were “cold rays”, don’t put words in my mouth. There is radiation coming from the cold scene but there is more radiation coming from the tiny pixel in the bolometer toward the scene than is coming from the scene to that pixel well. You are the one creating energy out of nothing, sorry, you have fallen line, hook and sinker to that articles wrong viewpoint. Read that article referenced within the article you pointed at and you will get a better picture of how they work. They measure differences in temperature to the ambient camera temperature, both warmer and colder. No “cold rays”, just more 50C radiation leaving than -40C radiation coming in and that tiny pixel well cools below the ambient T and the circuitry senses that difference in resistance.
thefordprefect, whoa, I must apologize. I though you were saying the camera was 40C so I took that wrong. You are saying IF you were to point that 50C camera at an absolute zero scene the bolometer pixels would cool to, say, 40C, the maximum possible cooling. Then the -40C scene would be somewhere between 40C and 50C. !! Jeepers! We are saying exactly the same thing but just using different words and trying to convince each other !! How about us dropping this OT sidetrack. Ok? I now see what you are saying and I’ll re-read that article again, may have made the wrong assumption there too. If so, they make it extremely easy to get the wrong impression. The absolute zero may have set me off.
Bryan says:
March 24, 2013 at 12:46 pm
OT but I was with you till this
Tim should be ignored.
I think Tim does well on a blog where he is in a minority. He could have an easier time commenting on the Guardian or similar warmist sites. He takes our arguments seriously and makes his defense of the consensus view with thermodynamic concepts.
I find that discussing the Energy Budget model with its well informed supporters helps me to understand better its faults and how eventually it will be possible to correct them.
Bryan what is your 1 sentence criticism of the Willis steel greenhouse model?
TB says
the heated shell would radiate from both it’s inner and outer surface
In the model, the other black- body, the top layer of the core, does not radiate from both it’s inner and outer surface.
Roger C: In the model, the other black- body, the top layer of the core, does not radiate from both it’s inner and outer surface.
Although it isn’t discussed or specified in the original thought experiment, it can be considered this way. Radiation from a surface into a vacuum is unimpeded. Radiation from the underside of the surface won’t get far into the core, because:
1.The material is dense and will absorb it within a very short distance.
2.There mus be a thermal gradient running from the core to the surface.
So during the time from when the shell is put around the planet and the planet surface starts to heat, there will be a phase where the surface is warmer than the material beneath. Heat will conduct downwards towards the nuclear core. The core will get hotter, since it is unable to get rid of heat as quickly as it used to, due to the reduction in the gradient. By the time the system reaches steady state, the core will be hotter than the equivalent of 470 at the surface, the gradient will be re-established, and the surface will be emitting 470 towards the shell.
[Tim F] takes our arguments seriously and makes his defense of the consensus view with thermodynamic concepts.
And appeals to the authority of the writers of textbooks on the radiative greenhouse effect on Earth.
tallbloke says:
March 24, 2013 at 6:22 pm
The material is dense and will absorb it
We agree. The surface absorbs at least some back radiation
The diagram does not say what form the 235Wm-2 is. But as all other energy transfers are by radiation we should assume it is also radiation. Energy can leave the planet so energy can enter.
There must be a thermal gradient running from core to the surface
According to the diagram the core and surface have only one temperature.
If the core is hotter than the surface, then why and how hot? It does not matter as we are discussing radiant energy not thermal ( Kinetic ) energy.
So during the time from when the shell is put around the planet and the planet surface starts to heat, there will be a phase where the surface is warmer than the material beneath. Heat will conduct downwards towards the nuclear core. The core will get hotter, since it is unable to get rid of heat as quickly as it used to, due to the reduction in the gradient.
We are discussing the claim of a steady-state as described in the diagram. We don’t know and it does not matter how it got to that state.
the core will be hotter than the equivalent of 470
See comment about thermal gradient
at the surface, the gradient
There cannot be a gradient at a surface
will be re-established
We are discussing a steady-state
the surface will be emitting 470 towards the shell
you assert what you claim to have proved.
“tallbloke says:
March 24, 2013 at 6:22 pm
Roger C: In the model, the other black- body, the top layer of the core, does not radiate from both it’s inner and outer surface.
Although it isn’t discussed or specified in the original thought experiment, it can be considered this way. Radiation from a surface into a vacuum is unimpeded. Radiation from the underside of the surface won’t get far into the core, because:
1.The material is dense and will absorb it within a very short distance.
2.There mus be a thermal gradient running from the core to the surface.
So during the time from when the shell is put around the planet and the planet surface starts to heat, there will be a phase where the surface is warmer than the material beneath. Heat will conduct downwards towards the nuclear core. The core will get hotter, since it is unable to get rid of heat as quickly as it used to, due to the reduction in the gradient. By the time the system reaches steady state, the core will be hotter than the equivalent of 470 at the surface, the gradient will be re-established, and the surface will be emitting 470 towards the shell.”
The core can increase in temperature due to flow of heat being insulated.
If Skydragons are claiming radiant energy in a vacuum can’t be insulated by reflective surfaces
then they are wrong. Or one can block the radiant energy from leaving the core surface by using
substances which reflect IR. With enough reflective insulation one make the core surface into hot lava. Or can one turn a surface below 0 C into a surface over 1000 C.
But the only effective way to do this, the only way to insulate radiant energy in a vacuum [as far as I know] is by using a reflective surface.
But the issue regarding the steel greenhouse is that we are apparently *not* using a reflective surface to insulate the core surface. But we do seem to talking about something *similar* to reflection which is the absorption of energy and emission of energy without heating. Or re-radiation
of radiant energy.
Both reflection and re-radiation has nothing to do with the temperature of whatever is doing the reflection and/or re-radiation. A mirror’s temperature has nothing to do the radiant energy it’s reflecting- the same amount of light is reflect from a mirror at 200 K as one at 400 K.
So re-radiation and reflection are similar, but not the same. One aspect about a mirror is a smooth
surface and for something to re-radiate one doesn’t need a smooth surface.
So droplets of water are reflective, but are unlike a flat surface- they are diffracting and reflecting light in many directions. So in some aspect a cloud or snow is more like re-radiating than a mirror
is.
It seems we need to explore the difference reflected radiation and re-radiated radiation and realize
the temperatures of such surfaces are perhaps an interesting aspect but not a controlling aspect.
Another important aspect in regards to reflective insulation is the shell is reflective on both sides
and it’s not clear to me which reflective side [inner or exterior side] has greatest effect.
Or if one has one side black and the other side reflective- what effect this has and does matter
which side is facing the heat source?
Ok Arfur, here are those pertinent BB ‘spectrum’ of the steel shelled planet.
At first I don’t want to lead you as to the way I view this (learned a bit myself in creating this) but would rather have you look at what these three spectrums are saying and then see if we are parallel in what this means.
But I will explain what the pieces of this graph are. The blue curve is the spectrum of BB radiation from the planet surface upward toward the inner surface of the shell. The magenta curve is the spectrum of the BB radiation from the inner shell downward toward the sphere, also it is the spectrum emitted from the outer surface of the shell to space, and, also (to me) it is that portion of the blue spectrum that resonates in the vacuum gap. The yellow is that portion of the blue spectrum that does not exactly match the 253.5K spectrum from the inner of the shell.
The arrows are what has to occur as the energy pass “through” the shell itself. Maybe view the downward arrow and that hysteresis shaped area containing it as the power to warm the shell itself (its frequencies are higher (see the gray max ticks), better “quality” energy with the ability to do work).
Well, see what you can gather from these “spectrums” or curves.
Roger C: “at the surface, the gradient”
There cannot be a gradient at a surface
If you misquote people this way, don’t be surprised if they ignore you from there on.
One more note on those curves. The area under the blue curve of course sums to total 470 W/m². But it might not be apparent that the area under both the yellow and the magenta curves are both exactly 235 W/m² even though they do not even coincide. (apparent from the math, that simple differential to create the yellow curve from the other two)
So if resonance swapping maintains the energy density in the gap the up and down radiation must exactly match both in frequency and number of photons for each frequency line. Hope you see what that implies right there (it means the yellow curve is also actual)
“Ok Arfur, here are those pertinent BB ‘spectrum’ of the steel shelled planet.
At first I don’t want to lead you as to the way I view this (learned a bit myself in creating this) but would rather have you look at what these three spectrums are saying and then see if we are parallel in what this means.”
There things I don’t understand.
the bottom of graph goes: 3.00E + 13, 2.00E + 13, 1.00E + 13, and then 0.00E + 00.
Corresponding to this is 5 microns space then 10 to 30 microns, and if guessing
40 or 50 microns would pretty close in distance on graph to 30 microns, as 50 microns
is closer to 30 microns than to zero.
It seems me the yellow line is indicating a warmer body which more distance than whatever
the blue line is indicating. Because it seems to me where a Planck curve peaks indicates temperature. So distance would only way for me to understand why the blue line has more intensity.
Now, go on:
“But I will explain what the pieces of this graph are. The blue curve is the spectrum of BB radiation from the planet surface upward toward the inner surface of the shell. The magenta curve is the spectrum of the BB radiation from the inner shell downward toward the sphere, also it is the spectrum emitted from the outer surface of the shell to space, and, also (to me) it is that portion of the blue spectrum that resonates in the vacuum gap. The yellow is that portion of the blue spectrum that does not exactly match the 253.5K spectrum from the inner of the shell.”
So, accordingly, blue line is intensified radiation reaching inner surface of the shell. By reflecting radiant energy one can intensify [or magnify or in a sense shorten the distance of source of radiant energy].]
So, check. Works fine, if radiation is so how intensified- which reflection can do. Whether re-radiating radiant energy can achieve similar things I will ignore at the moment. As mentioned in post before last one.
Next, magenta.
So magenta line [like blue line] also indicates lower temperature at further distance as yellow line.
So works fine if one dealing with reflection. Or if re-radiating is same as reflection.
wayne says: March 24, 2013 at 8:31 pm
Note to your ‘TinyPic’ graphic: You show the ‘peak’ displacement point of the curve, which is a detail of the ‘Wien Displacement Law’. If you were looking for the ‘Stephan-Boltzmann Law’ equivalent, it’s the ‘area under the curve’ (of which, the centre of area, is the ‘power factor’ point). 🙂
Best regards, Ray.
Roger Clague says:
March 24, 2013 at 2:04 pm
[“At one time I had a 2 bar electric fire. I turn on one bar, nice red glow. I turn on the another bar. The colour of each of the 2 bars is the same as the one bar alone was. That is the temperature of the first bar is not increased by the radiation from the second bar. The radiation from the first bar is not changed by the radiation from second bar.”]
Roger, thanks and I agree with your sentiment (I mostly do anyway as you talk sense). However I would have said that the WIllis planet differs from your two electric bar fire in that he asserts that the first bar heats the second bar (without switching the second bar on) and then the second bar, using the absorbed radiation from the first bar, then provides radiative energy to heat the first bar to twice its original energy!
Regards,
wayne says:
March 24, 2013 at 4:08 pm
Wayne,
First of all thank you very much for taking the time to try to explain this to me. It is most appreciated.
In answer to your confusion as to how I ‘see’ the photons, I will try to put it into words but please accept any non-scientific terms…
I suppose I see the photons as packets of wave-segments making up an energy flow from an emitting source to a receiving surface. The photons will be possibly a mix of amplitudes/wavelength/frequencies but there would be a ‘maximum’ power photon of a certain frequency which is not exceeded and which would give rise to the ‘235’ or ‘470’ energy level value. For a ‘470’ photon to be emitted from the surface would require more energy in the surface molecules than existed when the ‘235’ photons arrived.
Ok, to your BB graph…
What I notice first is that both the 235 and 470 curves share the same range of frequencies. That should at least make me re-think my visualisation!
However the graph obviously assumes that there is a 470 emission from Willis’ planet, which I am highly dubious of. The way I see it, in order for the 470 arrow to exist, the planet would have had to absorb all of the 235 arrow from the shell. My problem is that because I viewed that there would be a ‘maximum’ energy packet from the shell, that maximum would have to be increased at the surface in order for the planet to emit at 470. Your graph indicates that my assumption may be wrong, in that the range of frequencies is the same, so the reason for the energy level (irradiance?) of the 470 being higher than 235 is that there are simply either more photons or that they individually have more power. The fact that the highest frequency of the 235 curve is greater than that of the 470 curve does indeed cause me to ponder, as does the fact that the greatest delta occurs at 14 micrometres – which I would not expect if the photons were just numerically added to each other!
So, what happens at the surface in Willis’ model? If the 470 arrow exists, then the 235 photons are absorbed by the surface and seem to be added to the existing surface energy. This is not what I understood to happen when hotter thermal radiation is absorbed by a cooler surface. I thought the electrons rose to a higher orbit and thus the molecule could emit at a higher frequency. Obviously I was wrong.
Having said that, if the 470 arrow does not exist because there is no absorption of the 235 photons, then I may possibly still have a point.
There is no time allowance on your graph; it assumes that the 470 just exists. I’m not sure how significant that is. In reality, the planet will initially ‘heat’ the shell which, being cooler, will absorb. However, until equilibrium is reached the shell will emit at a lower energy level. But the energy level from the shell will always be lower because although the radius difference may be small, the fact is that the units of radiation are W/m^2 and therefore the radiation from the shell has to be relatively lower than that being emitted from the planet. In that aspect alone, the Willis model cannot be 100% accurate. But it may be accurate enough for generalisation.
My overall point is that I still can’t see how emitted radiation from one source can be absorbed by a second source and then re-absorbed by the initial source for net gain! There is no time line on your graph.
I seem to have rambled. If so, I apologise.
To get back to your hysteresis area; if the planet did not absorb any of the re-radiated 235 from the shell, there would be no such area. You would have two magenta curves. Actually the one original and the second slightly lower due to the radius effect.
Just out of interest, if you plotted one source at 235 and a second source (curve) at, say, 233, what would the graph look like? The crunch is can you add two lower energy curves and get one higher energy curve?
I am more than happy to continue my learning process if you see any potential for me to do so! But equally I’ll understand if you prefer not to… 🙂
Thank you again, Wayne.
Wayne and others,
Just read your last post about the areas…
One thing occurs: If the shell downward radiation were to be exactly the same as the planet initial radiation, would they not cancel out due to ‘phase cancellation’?
ps, in my job I’m told there are no stupid questions! 🙂
[snip]
The shell would only be receiving radiation from one side, [snip] and it would only be able to cool radiatively from one side.
[snip]
[Reply] One argument at once thanks Max. Then it would have to radiate from the opposite side, to space. Which means the shell conducts heat, which means it would happily radiate 117.5 from each side after receiving 235 from the planet. But this isn’t enough radiation to space to balance the production of heat from the planet and so the system will get hotter internally.
I made a mistake. I did not read the sentence carefully. By splitting it up, I misrepresented what I now understand you to mean. I undertake to read and quote you better.
The quote was copied exactly from your post. So anyone could see the mistake. I ignored a comma. I have some excuse for this as the next comma in the sentence need not be there. That the gradient is inside the core and not at the surface does affect my argument. Let me try again. Here is the whole sentence, a better way to quote.
By the time the system reaches steady state, the core will be hotter than the equivalent of 470 at the surface, the gradient will be re-established, and the surface will be emitting 470 towards the shell
If the core is hotter than the surface then the surface will also become the same temperature. That is hotter than before. It will emit at a greater flux density. This will affect the shell and its emissions and so on. So the model is not in a steady state.
Hi Roger C: Thank you for requoting me. There will be a gradient between the core and the surface, but it doesn’t matter for the purpose of the toy model, apart from answering your question about inwardly directed radiation. The point is, whatever happens with the underside of the surface, the surface ends up radiating 470 to the shell, while the shell ends up radiating 235 to the planet and 235 to space.
If there is anything that needs to be clarified, go ahead.
AB, right, agreed, no questions are stupid questions. 😉
“One thing occurs: If the shell downward radiation were to be exactly the same as the planet initial radiation, would they not cancel out due to ‘phase cancellation’?”
You know, that “phase cancellation” is exactly how I have always viewed it, but so many don’t seem to like that viewpoint at all that I shy away from even bringing it up anymore, and you know, it really doesn’t matter, if there are equal emissions and simultaneously absorptions, nothing happens there either, same end result, nada. The real answer is no one really knows, they have yet to actually capture a photon, it’s a concept.
If you are asking what would happen to the temperatures, that is different. At first, before reaching equilibrium, they would exactly cancel as you say, less cooling for the sphere, but the nuclear source is still pumping out the same energy and it is that energy which raises the sphere’s temperature. As the sphere gets even warmer that yellow line rises with the magenta and it is that energy that begins to radiate to space. Its really a bit more complex than that, you have three surface in question and at first they are all changing their temperatures constantly until equilibrium is established.
Look at that graph and see if you can visualize what happens to all three spectrums at each step after that cold shell is snapped about the sphere, the blue starting where the magenta is, the yellow starts identical to the blue, the magenta laying flat at zero radiance. As the magenta raises the yellow does not rise as fast as the blue… maybe you can see that and more importantly see that what appears to be what is actually occurring.
The big surprise to me was that the yellow ends up with more energy at higher frequencies and I had never realized that aspect before. That helps answers one big question.
Arfur Bryant says: March 24, 2013 at 10:59 pm
“One thing occurs: If the shell downward radiation were to be exactly the same as the planet initial radiation, would they not cancel out due to ‘phase cancellation’?
ps, in my job I’m told there are no stupid questions! :)”
I see a little light bulb switching on Arfur. Surfaces ‘exchange’ photons (molecules emit at the wavelength that they also absorb). 🙂
However, a BB surface/molecule (The Planet) emits and absorbs just about ‘all’ radiation, which leads to the question of ‘what are the absorption spectra of the steel shell’?
Just an ‘exercise’. 🙂
Best regards, Ray.
“Under your proposed scheme, the shell would only receive 235 Gigawatts from the planet. ~tb
Under your proposed scheme the shell would receive 470 Gigawatts from the planet, which is 235 Gigawatts more than the internal power supply makes available.
[Reply] We can move on to your critique of my theory after we’ve dealt with my critique of yours. So instead of partially quoting what I said and avoiding the issue, answer it.
wayne says: March 25, 2013 at 12:36 am
“You know, that “phase cancellation” is exactly how I have always viewed it, but so many don’t seem to like that viewpoint at all that I shy away from even bringing it up anymore, and you know, it really doesn’t matter, if there are equal emissions and simultaneously absorptions, nothing happens there either, same end result, nada. The real answer is no one really knows, they have yet to actually capture a photon, it’s a concept.”
Have more conviction wayne. It isn’t the ‘interception’ during ‘flight’, it’s the ‘exchange’ between ‘molecules’!
“If you are asking what would happen to the temperatures, that is different. At first, before reaching equilibrium, they would exactly cancel as you say, less cooling for the sphere, but the nuclear source is still pumping out the same energy and it is that energy which raises the sphere’s temperature. As the sphere gets even warmer that yellow line rises with the magenta and it is that energy that begins to radiate to space. Its really a bit more complex than that, you have three surface in question and at first they are all changing their temperatures constantly until equilibrium is established.”
Just like Earth, ‘Willis Eschenbach’s “Steel Greenhouse” model’ isn’t at a ‘thermodynamic equilibrium’, it’s in a ‘dynamic stasis’ (or it would be if he got it right)! It represents ‘his’ concept of an ‘analogy’ that can represent the machinations of a system that may be comparable to an ‘Earth’ scenario for the properties best exhibited by that ‘Earth scenario’. That’s to say, Earth is in a dynamic system that is ‘open’ for the purpose of ‘thermodynamic equilibrium’, but, for the most part, ‘closed’ for the purpose of a ‘thermodynamic response’. Enough. 🙂
“Look at that graph and see if you can visualize what happens to all three spectrums at each step after that cold shell is snapped about the sphere, the blue starting where the magenta is, the yellow starts identical to the blue, the magenta laying flat at zero radiance. As the magenta raises the yellow does not rise as fast as the blue… maybe you can see that and more importantly see that what appears to be what is actually occurring.”
Your ‘graph’ doesn’t originate at ‘bottom left’ as I’d expect, it originates at ‘bottom right’! Are your comments correct for this ‘bottom right’ orientation?
“The big surprise to me was that the yellow ends up with more energy at higher frequencies and I had never realized that aspect before. That helps answers one big question.”
What’s that question? Are you looking at ‘Peak’, or ‘RMS’ (there’s more to the life of a photon than just ‘radiance’)! 🙂
Best regards, Ray.
Thanks Ray, I actually searched for the correct word, I did, equilibrium was a poor choice, but the dictionary came up with “firm” (‘cease changing’ or ‘stop changing’) and I thought no one would understand what I was saying using that word ‘firm’ instead of just using equilibrium. You right, it’s an open system and will never be in strict thermodynamic equilibrium. I see your suggestion ‘dynamic stasis’. Better. I’m definitely no linguistic expert, quite the contrary, and realize I never will be! 😉
As to the X axis, that is the Planck blackbody relation plotted in frequencies instead of wave numbers or wavelengths as we are so used to seeing the great majority of the time, that’s all. I left it that way because it expands the high frequency portion that I was trying to highlight (so it wouldn’t be so squashed to the left side).
What would happen if [snip]
[Reply] What will happen is you will help keep discussion on track by dealing with one thing at once rather than tossing curve balls in three at a time and using the resulting smoke screen to avoid answering my points regarding your claims.
Ray, now to the guts of your comment, I greatly appreciate you spending the time thinking through this.
You mention, correctly, that blackbodies absorb all radiation falling on them, I agree. But that is the entire reason I made that graph! If the inner shell is radiating at 253.7K (235 W/m²), its spectrum (magenta) is shining (yes, I know it is IR, allow me some leeway with the words 😉 ) back onto the sphere’s surface below and that defines what downwelling radiation will be in addition to the energy coming from the nuclear core. So if the sphere shines upward with a spectrum of 302.7K (470 W/m²) this means that blue spectrum must also include all of the lines that will be shined downward in next infinitesimally small delta t. Getting the recursion there? So, that yellow curve is the main point of me even making that set of curves, it is what frequencies and radiance (really equivalent to counts of photons) that exceeds that portion in resonance. Can you look a bit deeper and see what I am saying?
If radiation, in order to do work (actually raise temperatures), must contain frequencies and amounts in excess of the spectrum that is already present at the receiver’s current temperature, then I could never figure out how that would ever occur… but I was evidently wrong… that is exactly what the yellow curve is showing, and I’ve never seen it so clearly separated out and plotted. Remember, that yellow curve is just blue radiance minus magenta radiance at each and every frequency. You would normally never come across that curve, in a way it is hidden within the blue spectrum.
See anything new there?
And Ray, I was not speaking of raising the sphere’s temperature, no, that is by the endless energy source remaining after that cancellation, I was speaking of the what energy raises the shell’s temperature as energy is also flowing though and to space. Didn’t want to leave a wrong impression.
I wasn’t partially quoting you, you partially quoted me by snipping most of my post and requesting I stick with one point at a time. That is fine, except you took it upon yourself to choose which point would be discussed by me, I did not feel it was as critical a point as the one I would have chosen.
Would you like to dictate the rest of my argument for me?
For the record my point which you snipped was in fact a reply to you:
“[Reply] One argument at once thanks Max. Then it would have to radiate from the opposite side, to space. Which means the shell conducts heat, which means it would happily radiate 117.5 from each side after receiving 235 from the planet. But this isn’t enough radiation to space to balance the production of heat from the planet and so the system will get hotter internally.” ~tb
My point was that you can see why your position is flawed most clearly when considering the results of filling the gap in the core/shell model.
In that case the outer surface will still emit 235 Gigawatts total, and if measured at any point beneath the shell to the original core layer I am confident that there would only be 235 Gigawatts flowing through that location.
If adding a solid layer of material does not require an increase in temperature, there is no reason to think the vacuum would do so.
[Reply] If you follow the link to Joe’s post and read Willis’ specification of the setup, you’ll find out he states that a conductive layer between the core and shell negates the heating. This is obvious when you think about it. Vacuums are good insulators. That’s why Mr Thermos got a patent.
Wayne
“its frequencies are higher (see the gray max ticks), better “quality” energy with the ability to do work”
How can the very low frequency, and presumably very poor “quality” radiation of a microwave oven in the GHz band heat your food to the THz band?
lgl, on your attempt to bring in some parallel to microwave ovens, I really can’t see that rotational dipole (heating) near-field effects apply here. Microwaves don’t rely on full spectrum blackbody radiation.
But lgl, I owe you a one big thanks without you even realizing it. You comment here clearly gave me what I sorely needed to answer a far-removed effect directly tying Venus’s atmospheric temperature profile to Earth’s atmosphere temperature profile and proving there is no difference between those two atmospheres. Also explains why the profile recorded by the Galileo probe into Jupiter’s atmosphere also follows this effect through as many as 16 Earth-equivalent atmospheres. I kept trying to look in the realm of attenuation (extinction) and the linearity found in your multiple-shell example of Willis’s shell game is precisely what I needed, explains it all. Thanks.
wayne
But wasn’t the idea that low-freq photons could not be “converted” to high-freq photons?
“But wasn’t the idea that low-freq photons could not be “converted” to high-freq photons?”
True, not when only speaking of diffuse Planck BB spectrums as the energy. Kirchhoff’s law, LBL, is involved there in the gap. Or do you know of some way to do so, could have some valuable applications! mind clueing me in? 🙂
The whole point that I see, and could be wrong, that it takes matter at some point to not convert low to high frequencies but higher frequencies to lower, repartition, that yellow curve portion does come out on the far outside of the shell also as a normal 235 W/m² (253.7K) BB spectrum. Maybe you can better explain that.
lgl says:
“How can the very low frequency, and presumably very poor “quality” radiation of a microwave oven in the GHz band heat your food to the THz band?”
Did you notice that the microwave oven is a machine?
Do you realise that the second law applies to spontaneous changes?
Another machine that apparently defies the second law is a refrigerator.
However if you study the Carnot Cycle and the second law you will see that the refrigerator like the microwave over is required to ‘do work’ to comply with the second law.
tallbloke says:
March 24, 2013 at 11:20 pm
There will be a gradient between the core and the surface, but it doesn’t matter for the purpose of the toy model, apart from answering your question about inwardly directed radiation. The point is, whatever happens with the underside of the surface, the surface ends up radiating 470 to the shell, while the shell ends up radiating 235 to the planet and 235 to space.
Willis in his original article
Is proud to admit his explanation of the greenhouse effect is a trick of geometry. A 2 faced shell balances a 1 faced sphere. . Although he calls the theory a 2 layer theory, the trick is to ignore what happens at the inner face of the layer at the surface of the planet. He ignores the 235Wm-2 inwardly directed ( travelling and absorbed ) radiation.
It is interesting that Willis’s diagrams does not show the 235Wm-2 from the core as an arrow striking the inner surface of the planet top layer. This might have alerted more people to flaw in the model.
Half the radiation from the planet layer must be into the the planet. Only 235Wm-2 is radiated to space. The temperature of the planet remains as it was.
How hot is the core? How does it get hot? How does a gradient answer my question about inward radiation? Willis does not mention a temperature gradient. He does no analysis within the planet. The planet has only one temperature. You should defend Willis’s model as he left it.
When the top layer of the planet is treated the same as the shell no temperature increase is predicted.
On Earth much radiation is absorbed by the oceans below the surface and leaves again as latent heat not as radiation.
Roger C: two points.
1. Do you accept that radiation is absorbed within a very short distance within solids, but isn’t absorbed in vacuums?
2. This thread is about Willis’ toy model, not the real planet we live on.
wayne, suricat,
I’m hanging in there…just. 🙂
“Look at that graph and see if you can visualize what happens to all three spectrums at each step after that cold shell is snapped about the sphere…”
But the blue curve would not exist as it is for 470 W/m^2 and when the shell is put in place the maximum radiation is 235 into the shell. So we would only have a magenta curve initially.
As the ‘upwards’ 235 hits the (steel) shell, the shell absorbs for net gain but it will never reach an energy level (heat) high enough to emit 235 W/m^2. At best it will re-emit those photon frequencies which are already ‘under’ the magenta curve.
At this point I see three options (maybe more):
A. The re-emitted radiation from the shell is cancelled by the radiation (same frequencies) still being emitted by the planet. As such, the planet does NOT absorb any shell radiation and does NOT heat up. This continues (at 50% reduction in shell absorption each cycle) until the system almost reaches ‘equilibrium/dynamic stasis’ at which point we just have a larger ball in space emitting slightly less than 235 W/m^2. ie. No 470 arrow ever existed. Hence the model as described is impossible.
B. The re-emitted radiation from the shell IS absorbed by the planet ‘for net gain’ which heats up. However it is generally accepted that something cool cannot ‘heat’ something hot. Hence the model is unlikely.
C. So maybe the shell’s radiation is absorbed by the BB planet for no net gain and we end up with a constant flux of emitted and re-emitted radiation in between the planet core and the shell. The model works (in numbers, kind of) but no heating takes place. Not sure I’ve sorted this.
Something else to throw into the mix is what exactly is the nuclear core doing? The initial Postma thread states the core is “…generating energy that at the planet’s surface amounts to 235 W/m2. It is radiating the same amount, so it is neither warming nor cooling.”
What it doesn’t state is whether or not the core is ‘governed’ at 235 W/m^2. If it is a black body it is not limited to the frequency of radiation it can emit, yet it is somehow limited to 235 W/m^2, presumably according to the heat is generates. If the planet is warmed by another (separate) source, maybe it can emit higher frequency (or at least more powerful) photons if that is possible. However, there is no other source here; there is only re-emitted radiation from itself.
Although I can have a go at understanding the curves on the BB Planck graph, I still maintain the graph assumes that 470 exists only because the model says it exists. How can the area under the magenta curve be added to by frequencies it didn’t have enough of to double up even if we added the 235 with another 235? E.g, the yellow line at about 7.5 micrometres is far higher than it would be if the magenta line was raised by twice its level. See what I mean? To me, the 470 blue curve is an addition to the model, not part of it. We need to look at the source of the blue curve.
Maybe I’m just getting confused. But I’m very happy to keep bashing this out as I can honestly say this is the most informative thread I’ve ever seen (and I’ve seen a LOT!)
wayne, if you could help me out by telling me your ‘big question’ which was answered, that might give me a clue. Thanks.
suricat, in reply to your ‘exercise’… “What are the absorption spectra of the steel shell?”… I’m not that advanced yet but I would suggest that the absorption spectra of the shell is a slightly smaller curve (but similarly shaped) than the magenta, as the photon frequencies are essentially the same. However I caveat my suggestion with a request to see what a curve of, say, 232 W/m^2 would look like – and what a curve of 116 W/m^2 would look like (50% of 232).
Both, please accept my sincere thanks for all your help. You are both extremely patient.
“ [Reply] If you follow the link to Joe’s post and read Willis’ specification of the setup, you’ll find out he states that a conductive layer between the core and shell negates the heating. This is obvious when you think about it. Vacuums are good insulators. That’s why Mr Thermos got a patent. ~tb
Mr. Dewar didn’t get the patent on it, the Thermos company won it, but he wouldn’t have gotten just a patent if his invention could make one side heat up without increasing the input, he would have gotten a Nobel, an army of followers, the keys to the kingdom, and a fancy hat. He damn near got the Nobel actually, and that is without it being able to generate energy like Willis’ thought experiment does.
A Dewar flask combines a vacuum with mirrored sides, mirrors are good insulators, vacuums prevent convection/conduction, limiting the forms of heat transfer to radiative.
Making a vacuum flask with black body walls will not be very effective.
Joe phrased it well just earlier actually: “Radiation is simply conduction at a distance, mediated in the exact same manner – just with real photons as opposed to virtual photons.” ~Postma
[Reply] That’s it! A lithium battery and a resistor. Boiling coffee in your thermos. We’re rich!
How can the very low frequency, and presumably very poor “quality” radiation of a microwave oven in the GHz band heat your food to the THz band?
Microwave oven cam increase water’s temperature above 100 C. As:
“The temperature of the water put to boil in the microwave, continues to increase and goes beyond the boiling point of water. ”
http://www.health.food-recipe-cooking.com/boiling-water-in-a-microwave-01.htm
I wonder how water could could get in microwave, assuming one had pressure vessel
or ambient pressure was higher- you could microwave oven inside a pressure container.
So water boils at 200 C at 225 psi [15.5 bars]:
http://www.engineeringtoolbox.com/boiling-point-water-d_926.html
So if pressure was 225 psi, could microwave oven cause the water to boil?
I tend to think it would. But also think there would some limit which may be
around 200 C.
But suppose one had water vapor at say 40 psi and no liquid water- so steam
say above 100 C, can a microwave oven increase the steam’s temperature?
So tend to think the microwave is causing molecules within molecular structure
[water or ice] to move. In other words with liquid water the molecules only move
if in a gas state.
Or another way to say this is one is adding heat which is like latent heat- or uniformly
evaporating water. Causing water to break it’s structural bond- which is what latent
heat is. So therefore as long as their is structural bonds to break, one can increase
the water’s temperature.
Water can evaporate at very low temperatures, so don’t require high temperature.
Now, I would say this could have something to do with it, but I also know that people
are convinced microwaves can used to heat substance other than water- microwaves
used to heat lunar regolith to high temperature, for instance. And of course if you use
metals in a microwave oven one get fireworks.
But as recall metal in microwave oven has something to do transferring voltage rather than
radiation.
“[Reply] That’s it! A lithium battery and a resistor. Boiling coffee in your thermos. We’re rich!” ~tb
Oh dear that sounds dangerous as hell to me for some reason, I can’t recall if lithium batteries like to get all ‘splodey or not, but yeah, if the planet and shell were silvered I would expect the planet to heat up a lot, probably until the mirror lining began breaking down.
The black body surface and black body shell should have the exact opposite effect, and be the coolest possible arrangement.
Wayne says:“So instead they should have said 19.9C, not 20.1C.”
When focused at -273C object, the sensor in the bolometer is 20 C.
According to Wayne, when focused instead at a -20 C object, the sensor will get colder.
hmmmm …. The sensor gets colder as the object it is looking at gets warmer. At what point, Wayne, will the sensor stop getting colder and start getting warmer? When the object it is looking at is 0C? 20C? 40?
Max Says:“The black body surface and black body shell should have the exact opposite effect … “
There are two “opposites” to a perfectly reflecting surface
1) a perfectly absorbing surface.
2) a perfectly transparent surface.
Unfortunately, your intuition was wrong. The transparent object has the “opposite effect”, ie no extra warming of the planet’s surface. The black surface has some effect — much less than the silvered surface, but still enough to warm the surface by a factor of a “mere” 2^(0.25).
Max says: “Mr. Dewar didn’t get the patent on it, the Thermos company won it, but he wouldn’t have gotten just a patent if his invention could make one side heat up without increasing the input, he would have gotten a Nobel, an army of followers, the keys to the kingdom, and a fancy hat.”
If you replace the concept “back radiation” with “insulation” then you will see how ridiculous this claim is. It is EASY to make things warmer or cooler with the same energy input, simply by changing how easy it is for heat to escape. Try this experiment — take the heat sink off your CPU in you computer. With the same energy input, I will bet you that you will make “one side heat up without increasing the input”.
Same idea with a Thermos bottle. Put a quart of water into a thermos bottle. Put a quart of water into a metal bottle. Put a 10 W heater in to both. The water in the thermos will get warmer and we will see “one side heat up without increasing the input”.
You seem to be confusing the idea of “one side heat up with a STEADY input” and the idea of “one side heat up without ANY input”.
[Reply] “Try this experiment — take the heat sink off your CPU in you computer.” Don’t try this at home kids! 😉
Bryan
Ah, so water knows when it is being radiated by a machine and then chooses to absorb the radiation, opposed to when it’s radiated by a blackbody, then it “scatters back” the radiation like Joe is imaginating, I see. Hard to keep up with all this groundbreaking new physics.
Arfur
“Maybe I’m just getting confused”
Guess you are, and I blame that on wayne of course 🙂
I can’t see any physical meaning of his yellow curve. Adding and subtracting curves like that does not make sense. But you can add the integrals of two curves to find how much energy has been absorbed, use that to calculate the resulting temperature and then find the emission spectrum.
The following is a general radiation heat transfer equation for black bodies radiating at each other.
(A1* F(1-2)* Gbb1) – (A2 * F(2-1) * Gbb2) = Q(1-2)
bb = black body
A = area
F = shape factor or view factor
G = SB* T^4
View factor for a concentric spheres is F(1-2) = 1. F(2-1) = (r1/r2)^2 Looks like Mr. Folkerts’ ratio.
There is no where to accommodate the feedback of the shell until the shell gets very large then it can see itself and F(2-2) = 1 – (r1/r2)^2.
lgl says
“Ah, so water knows when it is being radiated by a machine and then chooses to absorb the radiation”
No more so than butter inside a refrigerator realises its a good idea to lose some heat.
Arfur, seems the warmists are back and chiming in, picking statements out of context to marginalize anything I may say. I will no long be addressing them, just to those willing to just listen and consider.
Probably best to let you decide what that subtraction tells you, if anything, but you are correct, the sphere begins also as the magenta spectrum and ends up at steady state as the blue curve. You’re close. The inner shell is radiating down then as the magenta spectrum which means the sphere is then receiving the magenta spectrum from the inner shell and you might think Kirchhoff’s law comes in play here and the sphere is then in resonance with the inner shell, also at the magenta spectrum by his law. It sounds like some here don’t see it that way. But there is undeniably a difference there, the yellow difference, it’s a just mathematical subtraction? What of it? Help me decide.
Wayne says:
>>… but you are correct, the sphere begins also as the magenta spectrum and
>> ends up at steady state as the blue curve.”
Good … The “bare planet” would start at 254K, but would end up at 302 K with the shell placed around it. You got it right!
>> … The inner shell is radiating down then as the magenta spectrum which means
>> the sphere is then receiving the magenta spectrum from the inner shell
Exactly. The planet is receiving 235 W/m^2 from the inner surface of the shell. And it is receiving 235 W/m^2 from the nuclear heaters, for a total of 470 W/m^2.
>> … you might think Kirchhoff’s law comes in play here and the sphere
>> is then in resonance with the inner shell…
Could you define this mathematically? What specifically do you mean by “in resonance”? Where is “resonance” included in Kirchhoff’s laws?
>> … It sounds like some here don’t see it that way.
It would help if you could define “resonance”. It would help even more if you could explain how the planet can receive 235 W/m^2 from the shell, and 235 W/m^2 from the nuclear heaters yet only get rid of 235 W/m^2.
Or am I misunderstanding you? Do you mean what you said in the first sentence about the sphere ending up at “the blue curve” and emitting 470 W/m^2 at a temperature of 302K? Are you agreeing with Willis now that the planet will be 302 K?
suricat,
Been thinking about your exercise since getting up this morning. Don’t normally post this time of day but I need to amend my answer!
I do not know what the absorption spectra of steel is. I suspect it would not be as smooth as the magenta curve. I suspect it would have some peaks and troughs, as it is not a true black body, but not too many. Either way, I think the peaks would all be contained within the area under the magenta curve.
ps, if my first answer was correct… use that one! 🙂
lgl says:
March 26, 2013 at 1:20 pm
lgl, thanks for the input. My confusion would stem from not having had any advanced scientific training (I have basic training in many scientific disciplines), rather than just being thick. For example, I’m not thick enough to believe that a trace gas can exert a significant warming effect on the atmosphere! I see wayne’s curves being used as a teaching tool, whether they are accurate or not, so I’m happy to continue my discussion with him! 🙂
wayne says:
March 26, 2013 at 4:02 pm
Wayne,
The only area I’m slightly unsure about in your post is the “…sphere ends up as the blue curve”. If, indeed, the 470 W/m^2 is correct, then I can see that the yellow curve is valid (as a means of showing difference on a graph). My issue is how do we know that the 470 is actually present. I mean, if we accept Willis’ model as true, then I understand. But if we expect Willis’ model to be realistic, then I doubt the 470 exists at all. Does that make sense?
Ok, so I would decide that the 470 is unlikely, if not impossible. To be exact, even if the sphere were able to absorb the downward (slightly less than) 235 ‘for net gain’, it would still not emit 470 because of the radius effect (I cannot deny that effect, however small).
I am still puzzled by the different peaks of the yellow, magenta and blue curves. It appears that, at the far left of the graph, the higher energy blue curve has more radiance for the relevant frequency where the magenta has little or none. This alone might make the shape difference between the curves as the area under the yellow is the same as the magenta (I think you said). An increase in yellow area at the 5 micrometre level would necessitate a lowering of the yellow curve further to the right.
I’ve waffled enough…:)
tallbloke says:
March 25, 2013 at 9:43 pm
Do you accept that radiation is absorbed within a very short distance within solids, but isn’t absorbed in vacuums?
Yes. So the model is for a solid planet that absorbed all the back radiation in a short distance. Which is then all emitted again at a higher power.
Another problem with the model is that the explanation starts at the shell. Then the condition of the planet is worked out. The explanation should start where the energy starts from. That is
planet 235Wm-2, to shell at 235Wm-2 and then to space at 235Wm-2.
Roger
“Another problem with the model is that the explanation starts at the shell”
No, there is no problem. I have told you how you can start at the planet. Half the radiation is returned so Gain=1/(1-0.5)=2. 235*2=470.
It’s your ‘local skydragons’ made up physics that leads to problems;
1. There will be no energy transfer away from the planet if the temperatures are the same.
2. The shell only radiates from one side, alternatively the inward radiation is not absorbed by the planet. None of these are possible in the real world. Unscientific nonsense.
Arfur, I see what you are questioning. That is what I questioned for the first couple of years trying to IPCC’s script. But I have come to realize that this ATE is real, there is downwelling radiation that does not warm the surface per se, cold objects cannot raise the temperature of warmer objects, but that downwelling radiation can cancel the effect partially of what the warmer object is radiating upward that cools the surface. I am now trying to mathematically with proper physics explain how N-Z’s ~154K atmosphere-less Earth get’s to a steady state of 289.1K and I get much from this exercise (helping to answer my big question). So I now assume the boost in temperature is literal.
Over the last few years and quarter of a million comments I have read certain statements from others that know certain areas of physics much better than myself, and the mention of the “quality” of radiation keeps raring its head every now and then. Of course they are always speaking of the frequencies of such radiation, high frequency radiation that can perform certain effects (work) that low-grade thermal radiation cannot.
And Ray is correct, if using the spectrum from actual steel, not a BB ε=1 type of ‘steel’, the magenta curve would be far from what I plotted, the blue spectrum too. I also assume the shell and the sphere are both constructed from the same type of ‘steel’. But still you end up with two spectrums, you still have a warmer blue spectrum and a cooler magenta spectrum and subtracting the red on a line-by-line basis from the blue you still would have a yellow curve showing you the difference. Add the yellow to the magenta and you exactly have the blue.
Now that is where light bulbs began to go off in my mind. Ok, Kirchhoff’s law, on a line-by-line basis says what can be absorbed can equally be emitted at the same temperature and vice-versa. Well, that is what is occurring, L-B-L, between the warmer sphere and the cooler shell. Both the up and down of just that portion of the blue lines must exactly match. Once again, what of the yellow?
It immediately became apparent to me something had been left out of all of the IPCC, AGW and CO2 explanations over the years on IR radiation. The yellow. Look at Earth now, that was where Willis’s example was meant to end up since we are normally speaking of climate. The magenta (real now, ε temp –> 289.1K
Venus’s has a column mass that is 104.3 that of an Earth atm. column (per NASA) so lay in 104.3 ‘shells’ of Earth’s atmosphere and lets see what happens if we create a Venus-like planet with 104.3 shells of Earth’s atmospheres:
So allowing for the original window leakage, (238 – 80 = 158)
238(OLR) + 158(each added shell) * 104.3 = 16717 W/m² –> temp –> 737.1K
Wait! There is no mention of GHGs at all in that equation. If Venus actually has 0.95 portion of CO2 in the atmosphere and Earth has 0.0004 portion of CO2, then how the heck can this so-simple example of shells on shells on shells exactly portray what an actual Venus’s profile says? The two results should be miles apart but they are not, they are basically right on the spot.
I tried that on Jupiter’s atmosphere but that data has been massaged and modeled, there is an offset, but the lines are also close to parallel if you do the same layer by layer of equal masses to an Earth atmosphere there compared to the recorded Galileo profile. This deserves a closer look. Why is this so since the atmospheres are of such different composition? Well, seems to mean composition matters little as long as there is adequate opacity to be opaque in a portion of IR spectrum.
Don’t know about anyone else but that stopped me dead in my tracks! I shall never forget what I have learned by Willis’s cartoon steel planet example and I owe Willis for this one (even Willis and I definitely clash sometimes, usually on how people are treated)
So see Arfur, just because there is downwelling radiation (1/2) it still disproves all IPCC and the AGWers have been cramming into our minds, or that is the way I now see it. I also think it explains N&Z but still working on that.
Of course to take that viewpoint relies on many core physics principles. Like this shell example half does go up and half bears down and that is the isotropic nature of radiation within an opaque environment in IR. Like Postma, and I totally agree here, that radiation in an opaque environment is nothing more that fast long-reaching conduction. That even conduction has a “back-conduction” occurring (Newton’s 3rd law). I can even see that in thermals and latent heat do to the changes in temperature and temperature differences rule what radiation can possibly flow up and down. It’s all basically the same and guarantees that the shell on shell on shell nature combined with 1/2 up, 1/2 down forms atmosphere profiles in all but the diffuse upper fraction of a layer where solar wind and atomic and chemical effects take hold.
Said enough. That is just my view of what this article really points out and I’ll spend the next month or two checking and double checking, checking ever aspect, its correctness. If it’s wrong in parts, I’ll adjust as usual.
Wayne says: ” Well, seems to mean composition matters little as long as there is adequate opacity to be opaque in a portion of IR spectrum.”
A very good insight!
The lapse rate depends on the specific heat, which is quite similar for the gases typically found in the atmospheres of planets (around 1000 J/(kg*K). So it will not be surprising to find that Earth and Venus have similar lapse rates (especially since they also have similar gravity).
And it only takes a little bit of CO2 or H2O to create “adequate opacity” to IR. This is why it matters little that Venus is 96% CO2 and earth is 0.04%. But will have a similar lapse rate.
What DOES matter (for radiative balance) is the “diffuse upper fraction of a layer” that you mention — the layer where the opacity “ceases to be adequate”. At this layer, radiation to space becomes important.
And the reason that a 50% increase in CO2 on Earth matters is that this makes a difference in that “diffuse upper layer”. There would be 50% more CO2 in that layer, pushing the “adequate” layer just a little higher. Making that top layer just a little higher. Making that top layer just a little cooler. Making that top layer radiate just a little less energy to space. Making the surface just a little warmer to compensate and maintain energy balance.
[Of course, this is the “simple” version without feedback. Any number of different feedback options could come into play.]
[FWIW, I think N&Z have the “lapse rate” part correct, but miss the “adequate opacity” side.]
wayne says:
March 26, 2013 at 8:12 pm
Wayne, I take my hat off to you. That is great stuff. I can see where you are coming from. It all makes sense (or at least more sense than some current AGW ideas) and I hadn’t noted those Venus figures and the correlation.
I have no major problems with the ATE effect, and agree with you although my naturally sceptical mind will continue to question just about everything. All I can say is that you and Ray (and some others on here) have helped me a lot. I’ve learnt more on this thread than I have arguing about data and trends with pro-cAGW commenters on other blogs. I really like the way the discussion is handled on this blog – kudos to tallbloke!
There is still a lingering nagging at the back of my head regarding the blue curve. If some of the radiaition cancels each other out between the planet and shell, I’m not sure the blue curve would exist at all but I can see the argument as laid out with the 104 shells…
If you don’t mind, would you chat about the results of your investigation into this article and let me know? I give permission for tallbloke to give you my email address if that helps, or you could just wait for me to comment some time in the future.
Thanks again to you and Ray. 🙂
Tim Folkerts says:
March 26, 2013 at 8:48 pm
Tim, I agree Wayne’s comment was insightful.
However I don’t think you can justify your remarks regarding CO2 as some sort of ‘sequitur’ to his coment…
[“And the reason that a 50% increase in CO2 on Earth matters…”]
The problem is that there is no evidence to support the assertion that increasing CO2 actually does matter. I’ve queried this with you before.
I think you need to appreciate that, if you support the CO2-GHE-AGW theory, there was a GHE before 1850. Since that IPCC-chosen date, CO2 has increased by 40% and the global temp has increased by 0.8C. The two are not necessarily correlated but are contemporaneous. What nobody can confirm is how much of the 0.8C is due to CO2. If “increasing CO2 by 50% mattered…” then the 0.028% of CO2 existing in 1850 must have been a significant factor in causing the 32.2C GHE at that time. Unfortunately that significance has not manifested itself in observed data. You talk of ‘feedbacks’ but remember the 0.8C includes all feedbacks, all forcings and all natural factors. There is a big picture here. The earlier warming in the 20th century (1910-1945) effectively counters the cAGW “CO2 caused the warming in late 20th century” mantra, as does the flattening since 1998 when the CO2 effect should theoretically be accelerating.
I would say that a (nearly) 50% increase demonstrably has not mattered. It appears that the GHE can exist certainly to a significant extent without the benefit of CO2.
This is the best blog under the sun. Thanks all.
*wayne says: March 25, 2013 at 6:13 am
“Thanks Ray, I actually searched for the correct word”
That would be a thermodynamic “steady state”, but the term doesn’t mean a lot to many readers. Thus, the use of the ‘medical’ term “stasis” linked with “dynamic” to make readers think. 😉
*wayne says: March 25, 2013 at 6:43 am
“See anything new there?”
Only that, because you subtract magenta radiance from blue radiance at identical frequency, the yellow curve denotes the ‘extra “radiant” energy’ included in the blue curve that’s ‘over and above’ the radiant energy in the magenta curve that represents a ‘lower Planck level’ (phase shifted magenta curve) than in the blue curve. It may prove to be more productive if the magenta and blue curve ‘peaks’ were made to be coincident before production of the yellow curve.
This is becoming confused because the ‘two curves’ (magenta and blue curves) can’t exist together at the same time and ‘place’ (surface) for a ‘black body’. If you are looking for the difference in energy levels for the two emissions you’ll need to compare the two ‘areas’ enclosed below the relevant curves IMHO. This would also negate the need to account for any ‘phase shift’ to the ‘maximum radiance frequency’.
Sorry for the delayed response. 😦
Best regards, Ray.
Oh yes Arfur always skeptical. 🙂 I’m even skeptical of everything I just wrote, really. Did I make a mistake? Is it always half up and half down, like Newton’s “for each action there is always an equal and opposite reaction”? No matter the process? I know that is not really correct for radiation anyway, this is speaking of spheres and you have this factor called colloquially as the “dip of the horizon” effect in that you only have an exact split when precisely at a sphere’s surface for there is curvature present. Could there be even other factors?
On the blue curve, you question that, that is just the Planck blackbody curve or ‘spectrum’ of anything with emissivity of one at 301.7K. That is all that is. Nothing more. So that is one of those curves (magenta too) that will never change unless the steady-state temperature itself changes. It’s value, the height of the curve at every ‘x’, has no tie to the window radiation that is escaping except that it is where that window radiation originates.
I’m assuming you already know that (said for other reading that may not), so I really gather you are questioning the steady-state temperature in that particular configuration of energy and matter. That is highly idealized. If you add mass and heat capacities to these structures that does change, no doubt. Then the outer shell’s temperature is lower than the inner shell’s surface temperature, no longer equal, which is lower than the sphere’s. But that is dealing with a solid and gases have no stark ‘surfaces’. Radiation passes great distances right through gases if ever absorbed at all and in solid’s case it is always stopped in something like a micrometer so temperature (energy) gathers there. In an absorbing atmosphere there will always be a smooth monotonic gradient between so that factor doesn’t really enter into it when you make the mental jump to atmospheres like I did above. Really that is the lapse as Tim pointed out.
And yes, I’ll try to keep you posted, here if possible, if not directly.
TimF — you always seem to be looking down and warmer, never upward and cooler, forgetting its mate. 😉 Thanks of the kind comment but even though you see co2 having some great hand in current climate, I don’t, that is if albedo, solar irradiance, and the fraction of window radiation escaping without interaction never changes significantly. But we know clouds are rather fickle, some days the earth has more than others, the temperature does bobble about the mean a bit at every place and even when averaged. But overall, I think that shell example greatly questions of what we have been told of the 0.0004 concentration of carbon dioxide, in fact, it puts it in its proper insignificant place as far as I am now sure.
Don’t you see, you too proved part of this yourself above, a shell’s distance or its altitude matters not,even above a sphere, having to do with the 1/2 down portion, therefore a small increase in concentration may raise the ‘shell’ a tiny amount, still half up/half down always occurs, but that will affect nothing, the half up/half down still remains constant overall no matter the mode of trasport.
This has been one VERY informative thread, one of the best ever, even looking all of the way back to when got involved in this ‘physics of atmospheres’ years ago.
wayne says, March 26, 2013 at 8:12 pmSo see Arfur, just because there is downwelling radiation (1/2) it still disproves all IPCC and the AGWers have been cramming into our minds…
Wayne, exactly.
That was the very point I was trying to ram home in my ATE Part I and ATE Part II articles.
Of course back radiation exists. All bodies above absolute zero temperature (including atmospheres) radiate. And how would a body know not to radiate in a certain direction just because a million light years away (or 1 metre away) there happens to be a hotter body that would intercept it? My point all along has been that those skeptics who still don’t believe in back radiation (presumably because they think to do so would be giving in to warmism) simply have to get over it. Otherwise we cannot move on.
That is why I am interested in debating with Joe Postma. He, the arch sceptic, accepts that back radiation exists – heck, it’s just radiation, how could he do otherwise! But he and his associates claim that the physics is such that the back radiation simply offsets (partly cancels) the stronger forward radiation. This point of view has to be taken very seriously because it might well be true. If so, the skeptical cause is won. So why on earth would ANY skeptic think that was a bad thing?
Tim Folkerts says, March 26, 2013 at 8:48 pm:What DOES matter (for radiative balance) is the “diffuse upper fraction of a layer” that you mention — the layer where the opacity “ceases to be adequate”. At this layer, radiation to space becomes important.
Yes agreed. And this was the focus of ATE Part II was it not? It was a major step forward from havig to discuss earlier barmy warmist theories about back radiation in the bulk of the atmosphere causing warming (which it clearly does not). Sadly, we have now strayed a long way from the OUTPUT THROTTLING versus THROUGHPUT THROTTLING discussion. Until you can provide quantitative proof of OUTPUT THROTTLING, what you suggest is just a hypothesis to lay alongside the alternative hypothesis that the ATE is due to a THROUGHPUT THROTTLING EFFECT, a natural resistance to flow due to slow convection up the atmospheric column.
As I asked before, can you substantiate quantitatively your qualitative claims for OUTPUT THROTTLING?
David, sometimes you have to say something many different ways before everyone starts to get the same visualization, really understanding and all matching in what is even being spoken of. Sometimes I am one of the laggers, thinking of some other aspect and never quite getting in sych. Hope this helps you in your third segment. That is the one I have been waiting for… what exactlly happens at the ToA as the view from radiation’s standpoint opens up horizontally to the void at every increasing increment of altitude, the dumping of the energy to space. Wish some of these spectrums on the web were taken horizontally at various high altitudes, seems that would be very instructive.
“Of course back radiation exists.”
David, still wondering why most that you are addressing fight the gagging that term causes? So… stop calling it that! Call anything but, call it opposing radiation, downwelling radiation, down portion of the isotropic radiation, just downward radiation, anything but IPCC’s “back radiation” with magic warming only powers attached (lacking proper cooling half of the effect) . I too have developed an allergy to that particular term… causes instant blood pressure spikes. 😉
Formatting messed up, can you post this version instead of that one?
“That is why I am interested in debating with Joe Postma. He, the arch sceptic, accepts that back radiation exists – heck, it’s just radiation, how could he do otherwise! But he and his associates claim that the physics is such that the back radiation simply offsets (partly cancels) the stronger forward radiation. This point of view has to be taken very seriously because it might well be true. If so, the skeptical cause is won. So why on earth would ANY skeptic think that was a bad thing?” ~David Socrates
Rather than offsets or cancels, I find “subtracts from” more closely describes what is taking place.
Radiation is a vector, vectors in different directions can be manipulated mathematically, opposing vectors with similar magnitudes will produce a new vector after subtracting the smaller of the two from the larger one.
235 > — ~ 1 | 1 ~, net 234 to shell, 1 to space, 235 total
235 > — ~ 100 | 100 ~, net 135 to shell, 100 to space, 235 total
235 > — ~ 234.99 | 234.99 ~, net 0.01 to shell, 234.99 to space, 235 total
Doing it another way:
235 > — ~ 1 | 1 ~, 236 to planet, 1 to space, 237 total
235 > — ~ 100 | 100 ~, 435 to planet, 100 to space, 535 total
235 > — ~ 235 | 235 ~, 470 to planet, 235 to space, 695 total
Alternatively:
235 > — ~ 1 | 1 ~, 236 to shell, 1 to planet, 1 to space, 238 total
470 > — ~ 235 | 235 ~, 470 to shell, 235 to shell, 235 to space, 940 total
Regarding the heating of the shell;
1, The inside of the shell has to be heated by radiation from the inner core if that radiation stopped than the inside of the inner shell will instantly cool.
2, the outside of the shell is only heated by conduction from the inside of the shell if this conduction stops then the outside shell instantly cools.
3,For the inside of the shell to conduct heat to the outside of the shell then the inside of the shell must be warmer than the outside.if they reached the same temperature then conduction would stop and the outside would cool.
The shell cannot be at a constant temperature inside and out .
donald penman says, March 27, 2013 at 6:31 am.
As far as I can see your points 1 and 2 are not controversial. 🙂 They add nothing to the debate since we are talking about a steady state MODEL with a continuous power source. So neither of them apply.
Your third point would be true for a real shell but so what? Are you seriously suggesting that the tiny temperature drop across a very thin REAL shell in some strange way collapses the whole argument?
This is pure sophistry, in just the same way that people on this thread are still arguing that small different radius lengths of the core, inner shell surface, and outer shell surface, make a difference. Yes of course they would in the case of a real system but it is obviously completely marginal in its effect providing one stipulates that the surfaces are arbitrarily close to one another. And even with very small real gaps, one would still be left with the still very large temperature discrepancies of the Willis model paradox to explain away.
Why not concentrate on the actual paradox itself rather than fiddling at the margins? Postma and followers say it is simply due to a misunderstanding about the physics involved – not about some absurdly marginal heat conductivity or geometrical issue. They claim that so-called ‘back radiation’ exists but simply offsets the corresponding ‘forward radiation’. So why not concentrate on that substantive argument rather than fiddling at the margins of the debate?
That there would be large discrepancies even with small gaps is an assertion which needs to be justified, I don’t need to justify it if I can provide a plausible alternative without that discrepancy.
_______________
A slightly different tack: is it possible for the shell to cool by emitting radiation inwards towards the warmer planet?
Думаю, какую полезную информацию можно извлечь из этого материала
This thread is for discussion of Willis’ steel greenhouse gedanken experiment. Anybody tempted to wander off into talking about ‘back radiation’ on the Earth from it’s non-vacuum atmosphere should place their comment here:
Thank you for your co-operation – TB
David Socrates says:
March 22, 2013 at 7:06 pm
counter-argument that the core also has two surfaces.
The top layer of the planet radiates from its top surface upwards.
It gets energy, at 235Wm-2 from below into the lower surface.
The top layer of the planet has 2 surfaces.
Willis says
“This is called a “two-layer” model, with the two layers being the surface and the atmospheric shell.”
A layer has 2 surfaces.
The top layer of the planet radiates like the shell, up and down. 245Wm-2 up and 245Wm-2 down. Even into to a solid. The surface remains at its original temperature.
How is this criticism roundly rebuffed ?
donald penman — March 27, 2013 at 6:31 am
Donald, your exactly correct, on point 1, 2, and 3. No pure sophistry as David implies. What is pure sophistry is Willis’ steel greenhouse gedanken experiment, it’s totally impossible, unrealistic, can only be thought true by breaking a physics law along the way and you put your finger right on the point, point #3, where it slips by most. What people tend to forget is the cooling that equally and oppositely is always occurring if you have any warming occurring and at every point in the system it is only temperature differences that can possibly move energy anywhere, and if energy moves there will be equal amount of cooling as there is warming. Doesn’t matter if it is radiation, conduction, convection, thermals, state-change (latent heat transport), they all apply equally to both cooling and warming, always equal and opposite.
So you are right, for the shell to having a foot in reality that shell must have real mass and thermal conductivity and the inner of the shell has to be warmer than the outer or no energy flows. I also integrated that days ago and the sphere’s temperature does not settle at 301.7K, it settles around 333K depending on how thick you make the steel shell (how much mass) and the steel’s thermal conductivity.
Good insight there!
David’s mistake is him thinking that the very thin steel shell makes just an ignorable difference, no, it makes a large difference, 235 W/m² worth, you basically have two shells in play, not one… there is already a shell inside a shell built into Willis’s example. The only real difference is one is via radiation through a vacuum and the other is via conduction through steel, no difference between the two cases but the speed of transport. So, you really have the sphere at 333.9K, the inner shell at 301.7K and the outshell at 253.7 K. That is the real answer to Willis’ steel greenhouse gedanken experiment. If you make the shell even thicker with lower thermal conductivity both the inner shell and the sphere both just get even warmer from there.
I think this was already covered further up in this thread by other commenters.
ACO: Used reasoned argument please, your last comment was binned.
Wayne says: “Don’t you see, you too proved part of this yourself above, a shell’s distance or its altitude matters not,even above a sphere, having to do with the 1/2 down portion, therefore a small increase in concentration may raise the ‘shell’ a tiny amount, still half up/half down always occurs, but that will affect nothing, the half up/half down still remains constant overall no matter the mode of transport.
Tallbloke wants to stay focused this model, not the earth's real atmosphere. Let me just say very briefly that we both agree that raising the shell make (almost) no difference in the planet's temperature in Willis' model. In earth's real atmosphere, adding more CO2 is akin to adding additional shells, not akin to raising the one shell. Since the physics is very different, it should not be surprising to find a very different answer.
Roger Clague says, March 27, 2013 at 1:52 pm:The top layer of the planet radiates like the shell, up and down. 245Wm-2 up and 245Wm-2 down. Even into to a solid. The surface remains at its original temperature.
I’m sorry, Roger but you are really grasping at straws here. The CORE is a body with one surface. When I last looked at a thermodynamics text book it told me that such a body would radiate from its one surface (and that means radiate outwards!) at a power intensity dependant on its temperature (calculated according to the S-B law).
You only invented the magical inward radiation from the core surface when I pointed out the error in your earlier criticism where you hadn’t appreciated that the shell had two surface areas so the power into the shell was correctly balanced by the power out.
Why not concentrate on the Postma argument, which is that the physics of the interactions between cooler shell and warmer core is wrong. In fact what IS it about the Postma argument that you don’t like?
Roger Clague
Do you know what a surface is? It’s a boundary. Describe what’s above and what’s below the boundary, your ridiculous inner ‘surface’ of the planet.
wayne says, March 27, 2013 at 2:04 pm: What people tend to forget is the cooling that equally and oppositely is always occurring if you have any warming occurring and at every point in the system it is only temperature differences that can possibly move energy anywhere, and if energy moves there will be equal amount of cooling as there is warming…So you are right, for the shell to having a foot in reality that shell must have real mass and thermal conductivity and the inner of the shell has to be warmer than the outer or no energy flows.
Wayne, This is not worthy of your usual careful analysis.
First I think that the use of the words ‘warming’ and ‘cooling’ is extremely misleading. Whether you like it or not we are talking about Willis’ THOUGHT EXPERIMENT which is defined as being in steady-state. That means that there is NO warming and NO cooling occurring in any component in Willis’ model at all. And, although you are absolutely correct to say that no heat would be transferred from core to shell without a temperature difference, in Willis’ model there is a temperature difference (a big one!) so his THOUGHT EXPERIMENT doesn’t fall down in that regard.
Secondly, in my comment to donald penman I agreed with him that if the shell was not perfectly conductive there would indeed be a temperature drop from inner to outer shell surface but that, if the shell were sufficiently thin, the drop would be trivial compared with the huge difference claimed in the Willis model between core and shell.
AS long as people keep focussing on minor issues (thickness of shell; differences in the radii of core, inner shell surface, and outer shell surface; lack of perfect conductivity in the shell material) all we are doing is talking around the margins of the real issue which is why people think the physics of the Willis’ thought experiment is wrong.
Are you with Postma or not?
” if the shell were sufficiently thin, the drop would be trivial compared with the huge difference claimed in the Willis model between core and shell. “
To put some hard numbers here, the temperature difference due to conduction through the shell will be
ΔT = (235 W/m^2) * Δx / k = 235 * 1m / 40 = 6 K
for a steel shell 1 m thick with a thermal conductivity of 40 W/(m*K) for steel.
If the shell is 10 cm thick, the difference is 0.6 K
If the shell is 10 cm thick copper, the difference is 0.06 K.
The temperature difference would be similar on the planet’s surface. So the “shell insulation effect” will warm the inside of the shell and the planet by anything from ~ 0 K (for a thin conducting shell) to “a lot” for a thick insulating shell.
Clearly the original post assumes a thin conducting shell.
lgl says:
March 27, 2013 at 4:51 pm
above and what’s below the boundary
You make my point. A surface is a boundry which has above and below.
A boundry has 2 faces.
David Socrates says
The CORE is a body with one surface.
Energy enters the planet surface from below. Energy enters from below so energy can leave that way.
This is an educational exercise. We ignore many factors to get simplicity and clarity.We cannot ignore consistency. The analysis starts by asserting that the shell radiates up and down.In further analysis the planet surface should also be allowed to lose energy up and down.
Willis’s model creates energy from faulty geometry.
Are you with Postma or not?
We are discussing all objections to the model. Wayne has interesting questions about how quality of radiation. Radiation temperature come with a spectrum of frequencies.
We should only present arguments. We don’t need to form parties.
Roger,
Enjoy the luxury of not focussing on the one real issue.
All the best. 🙂
Its interesting to note that even in a solid object (like the steel planet ) it is mostly empty space.
If the biggest stadium in the world represented an atom, its nucleus would be the size of a pea in the middle.
So its quite reasonable to assume that the thermal emission of photon from an atom can go downward as well as any other direction.
Why then does the planet not heat itself up?
Roger C says: “Energy enters the planet surface from below. Energy enters from below so energy can leave that way.”
In principle, that is true. But not for the scenario proposed here.
The “nuclear heaters” are distributed uniformly under the surface of the planet. Vary simple arguments based on symmetry guarantee that no energy goes down from such heaters. Suppose that some of the energy from one square meter of one of the nuclear heaters went down — perhaps 230 W went up and 5 W went down. By conservation of energy, that extra 5 W/m^2 would have to come back up somewhere else. So some places would have LESS than 235 W/m^2 and some would have MORE than 235 W/m^2. But the entire planet is symmetric. There is absolutely no way to decide WHICH spot are MORE and which are LESS.
The only conclusion is that the power at every identical bit of the surface is identical = 235 W/m^2. All of the 235 W generated below each square meter moves radially outward to the square meter directly above the heater..
(Certainly if the heaters were not distributed uniformly (say all in one hemisphere and none in the other), then heat would flow from both sides = “leaves from the bottom of the heaters”. But Willis’s
model is symmetric. The heaters are distributed uniformly around the planet at uniform depths below the surface, all at uniform temperatures. No heat can possibly go “down” in such a scenario.)
Roger Clague
No, above the real surface or boundary is vacuum, below it is some solid matter. Now tell us what is above and below your imaginary boundary?
Tim F says
“But Willis’s model is symmetric. The heaters are distributed uniformly around the planet at uniform depths below the surface, all at uniform temperatures. No heat can possibly go “down” in such a scenario.)”
Did Willis specify this condition.?
It seems even more physically ridiculous than the steel planet GHE.
Or did you just add another ad hoc ‘get out of jail card’?
Bryan says, March 27, 2013 at 8:11 pm: Or did you just add another ad hoc ‘get out of jail card’?
Brian, the whole point of a model is simplification. Of course Willis was intending that the whole surface of the core was at the same temperature, implying isothermal heating. Otherwise you would have an incredibly over-complex scenario to analyse. What a silly argument. Why can’t you just go with the flow and address the overall problem instead of doing all this nitpicking.
Tim Folkerts doesn’t need a ‘get out of jail’ card. He (a warmist) is going with the flow of Willis’ MODEL. I (a skeptic) am going with the flow of Willis’ MODEL. How is it possible for him and me to disagree about the cause of atmospheric temperature enhancement but at the same time agree strongly on how to tackle the Will’s thought experiment? I guess it’s because from long professional experience we know how to play the ‘thought experiment’ game. You and others apparently don’t and so are reduced to irrelevantnit picking.
Get real guys or we are all wasting our time here. 🙂
Wow, a planetary scale nuclear core that outputs a constant 235 W/m^2, more than our averaged solar input per square meters and a shell of 1 meter thick of solid steel only warms the surface 6K?? Are you absolutely sure of that figure Tim?
When you say: The temperature difference would be similar on the planet’s surface. So the “shell insulation effect” will warm the inside of the shell and the planet by anything from ~ 0 K (for a thin conducting shell) to “a lot” for a thick insulating shell.
Clearly the original post assumes a thin conducting shell.
You are not following David Socrates. He says it must be viewed as a “perfect conductor” up against the 0K of space. A super conductor so to speak. So the shell basically does nothing, ~0K influence, but the reflected radiation, just because it is ‘there’, warms the surface ~48K. Have I read you right?
Oh, Tim, I am also assuming your 6K is plus that of the refected. Just so you won’t have to immediately jump onto that. 😉
Seriously, Bryan? Do we need to specify EVERY LAST DETAIL to make you happy? It’s a thought experiment not an engineering exercise!
We can’t say ANYTHING about the results with out making some glaringly obviously assumptions about the conditions. Without assuming the power is uniform, we cannot say what “the temperature” of the planet is since it will vary! Even though I disagree with your results, you have stated your opinions about “the temperature” of the planet and “the temperature” of the shell, so you yourself make this assumption that you now question.
PS. Yes, Willis DID specify. “For our thought experiment, imagine a planet the size of the Earth, a perfect blackbody, heated from the interior at 235 watts per square metre of surface area.” Every square meter of surface area has 235 W of power.
Wayne, the ΔT is the temperature difference across the shell.
The outside of the shell is 254 K
* With 1 m of steel, the inside of the shell is 254 K + 6 K = 260 K
* With 0.1 m of steel, the inside of the shell is 254 K + 0.6 K = 254.6 K
* With 0.1 m of copper, the inside of the shell is 254 K + 0.06 K = 254.06 K
* With 0.0001 m copper foil, the inside of the shell is 254 K + 0.00006 K = 254.00006 K
********************************************************
“You are not following David Socrates. He says it must be viewed as a “perfect conductor” up against the 0K of space. “
No — he only expects it to be “close enough to perfect” that the errors can be ignored.
“So the shell basically does nothing …
… other than stop all photons that reach it.
” … ~0K influence …
.. on the temperature difference between the inside of the shell and the outside of the shell.
“but the reflected radiation, just because it is ‘there’, warms the surface ~48K. …
There is no reflected radiation. By definition, a blackbody absorbs all and reflects none. But the radiation EMITTED by the shell (added to the power from the heaters) warms the planet by ~ 48 K compared to no radiation from the shell. This warming happens just because it (ie the energy of the radiation from the shell) gets absorbed by (ie added to) the planet.
” Have I read you right?”
Apparently not. 😦
Say one has 1 cubic meter of radioactive material which radiate 240 W/m-2. Have it encased in about 1 foot of steel and within it, radioactive material with long half life. So 40 W/m-2 per it’s 6 sides in a vacuum. And we could assume if covering planet at 1 meter thick it radiates 240 W/m-2
into a vacuum. And if instead 1 meter thick, if add another meter of the stuff it radiate 480 W/m-2.
So pave a planet with 1 meter of it , and it radiates 240 W/m-2, and if you put another 1 meter layer on top of it, it radiates 480 W/m-2.
So if second layer is not contacting the surface, but instead separated by vacuum, does this change anything?
So first layer radiates 240 W/m-2 and if separated with vacuum the second layer radiates 120 W/m-2 upwards and downwards.
So 1 km square by 1 meter thick slab of this material radiates 120 W/m-2 on both sides [since it’s large ignores the amount radiate at edge- or have edge well insulated]. And +1 million km away planetary surface that what it’s doing and when you bring it closer to planet it changes.
When put the 1 meter material on planetary surface it warm underneath it until get it warm enough
that insulate the heat. So putting it on cold planetary surface it starts off heating downward and upward by about 120 W/m-2, but if given enough time, it stops heating the surface and radiate 240 W/m-2 away from the planet. And if put another 1 meter layer on top of it, again it will begin warming the ground underneath it, but eventually the ground will warm up and cause 480 W/m-2 to radiate away from the planet. So this what this stuff is designed to do.
The question will it change design requirement or give different results if separated by say 1 meter
of vacuum space between two 1 meter thick surfaces?
So you have waited long enough so the planet with 1 meter shell is radiating 240 W/m-2, then bring the next layer so 1 meter above. And wait until the ground ground it to warm as much as it will
warm.
If we assume it does make any difference whether second layer is in contact or separated by 1 meter of vacuum, then the outer surface of shell will radiate 480 W/m-2.
And one could diagram it as 480 W/m-2 of heat being blocked from heating below the surface due
this surface being warm, and 480 W/m-2 radiating into space.
What we have obsessing on is the watts in between these shells.
The skydragon version is 480 W/m-2 “circulating” and IPCC version would be 960 W/m-2 of backradiation.
And the other part of argument is what happen if the second layer does not generate 240 W/m-2.
So instead being radioactive it is just plain old steel.
So question isn’t if put the two radioactive 1 meter layer together will get 480 W/m-2 radiating into space, but rather if put some distance of nothing between them whether this changes anything.
Or can design so it changes anything.
And I believe one can design it so it changes something, but I want to give others a chance
to explain how having vacuum between to two layer can change it.
David S says
” What a silly argument. Why can’t you just go with the flow and address the overall problem instead of doing all this nitpicking. ”
What is silly?
Apart from your absurd tendency to give condescending advice.
I along with I would imagine most people thought that the planet had a radioactive core at its centre which would give a uniform temperature at the surface.
The evenly distributed radioactive planet suggested is physically absurd.
It seems to have been invented to cope with Roger Clague’s perfectly rational question about why radiation cannot go down from the surface.
Why do you feel you need to reply for Tim Folkerts, its very odd ?
In fact a planet with a radioactive core producing a uniform surface radiation to space of 235W/m2 is really an infra red Sun.
Now the radiation from such an entity is the basic study of Astrophysics and I guess Postma would have studied this often.
So, to sum up these 700+ comments …
Tallbloke was right that Willis was right about this thought experiment.
🙂
wayne says:
March 26, 2013 at 11:22 pm
Wayne,
All noted. Thanks once again. I look forward to seeing the results of your investigation.
I agree this thread has been most informative.
All the very best,
Arfur
“So you have waited long enough so the planet with 1 meter shell is radiating 240 W/m-2, then bring the next layer so 1 meter above. And wait until the ground ground it to warm as much as it will
warm.
If we assume it does make any difference whether second layer is in contact or separated by 1 meter of vacuum, then the outer surface of shell will radiate 480 W/m-2.”
Should be:
So you have waited long enough so the planet with 1 meter shell is radiating 240 W/m-2, then bring the next layer so it’s 1 meter above it. And wait until the ground underneath it to warm as much as it will warm.
If we assume it does not make any difference whether second layer is in contact or separated by 1 meter of vacuum, then the outer surface of shell will radiate 480 W/m-2.
Note: it’s a bit redundant because as far as I know, there is no way not to have outer surface radiate 480 W/m-2.
Though other things can be changed.
The temperatures in various “locations”, including the outer surface can be changed.
But I am ignoring various possible temperatures which can be involved and strictly confining it
to energy budget [a heat energy budget].
A next step could be to look at temperatures.
But seems quite realistic that material can designed to give a certain amount of heat energy- that cubes of material can give same amount heat under varying conditions, Ie, whether it’s under water or in vacuum of space or piled on top of each other- kind of like pebble technology for nuclear reactors.
“Its interesting to note that even in a solid object (like the steel planet ) it is mostly empty space.
If the biggest stadium in the world represented an atom, its nucleus would be the size of a pea in the middle.
So its quite reasonable to assume that the thermal emission of photon from an atom can go downward as well as any other direction.
Why then does the planet not heat itself up?” ~Bryan
Interesting point, but photons interact with electron clouds, not the nucleus, unless we’re speaking of high energy x-ray lasers or gamma radiation, generally.
____________
As I asked earlier though, the shell can cool radiatively, right?
It can not cool by radiating towards a warmer surface, it is receiving more radiation than it loses in that direction.
It doesn’t “know” that the surface in that direction is warmer, but across distances this short it takes a couple of nanoseconds per meter, so any photons leaving are more than replaced immediately.
As such, no, the surface can not be treated as existing in isolation and radiating as though there are no other sources of radiation nearby, that is the first mistake Willis made, and it is commonly repeated.
___________
In fact this is what will happen when you add the shell:
Initially the shell will be cold, it will absorb close to 234.99 W/m^2, it will emit a small fraction of that amount inwards and outwards.
Then it will warm until it is emitting say, 100 W/m^2, then it will absorb 134.99~ W/m^2 from the planet, and emit 100 W/m^2 outwards.
Finally when it and the planet are in radiative equilibrium it will be emitting 234.99~ W/m^2, it will be absorbing a net 0.01 W/m^2 from the planet, and it will emit 234.99~ W/m^2 outwards.
Thus the emissions outwards will balance perfectly with those from the planet with no temperature increases from unphysical greenhouse effects.
David, to answer your question of Postma, I have so far read his thread down to about the 1/3rd mark, long sucker, and he finally stated what I was looking for, he said point-blank that to him both the shell and the sphere are and would stay at the same temperature at steady state, so I can now safely say to you that I do disagree with him at all. Many things he says around the edges are correct in physics, that is what it makes hard to find out exactly what he is thinking, precisely.
He further says that if the gap were filled with matter (I thought of sand, Chernobyl) this would make no difference, still 235 fx (254K), clearly wrong, not with a nuclear core inside. That showed me conclusively he reads this thought experiment completely wrong, as if the energy source is throttled to an equivalent steady temperature and not a constant energy output And, others that clearly state they see no possible way for it to ever get over 254K no matter how well it is insulated, and then he praises their brightness, shows he is off the mark, way off. That clenched my answer to you.
But you know, that may be the only difference in his explanation right there, and so his huge slip in logic. If the core was throttled to stay at a constant temperature (mentally converted to 235 W/m²) would his solution that he states over and over be correct? Seems it would, exactly. Isothermal from core outward. Since I have made such dumb, dumb mistakes before hope everyone doesn’t bear down on him too far, for that. But he should be thumped royally for saying everyone who didn’t agree with him was a dummy, ignorant, knows nothing – math – physics, … on that he deserves it.
Arfur Bryant says: March 26, 2013 at 9:54 pm
(wayne said)
“There is still a lingering nagging at the back of my head regarding the blue curve. If some of the radiaition cancels each other out between the planet and shell, I’m not sure the blue curve would exist at all but I can see the argument as laid out with the 104 shells…”
Well, wayne & Arfur. Don’t get ‘too’ ahead of yourselves. I wouldn’t want you to go off ‘half-cocked’ and a bit ‘miffed’ about something that I’ve said.
The ‘multiple shell’ with ‘central atomic energy source’ is best realised by a ‘top down’ calculation to find the true temperatures within the ‘system’. The ‘bottom up’ ‘shell on shell’ calculation needs an ‘iterative’ approach that increases the complexity of the calculation. Why? Let’s discuss.
For simplicity, let’s assume that both the planet and the shell in Willis’s “Steel Greenhouse” are BB structures and have zero thermal capacity.
1) The BB planet is ‘cold’ (~4k), also the shell to begin with, and the ‘atomic engine’ starts to add energy at a uniform rate of 235 W/m^2 at the planet’s surface area (this may also be back-calculated to a ‘surface temperature’ (Planck level) due to the W/m^2 emitted and the ‘total energy’ generated by the ‘atomic engine’ is easily realised by summing the energy available for each m^2 of the planet’s surface). This is an essential observation for both ‘bottom-up and ‘top-down’ methods of calculation.
2) The ‘multiple shell’ method observes what happens within the ‘first’ shell. What we observe is that the planet radiates photons at a Planck intensity equivalent to the temperature level of 235 W/m^2. Thus, the shell’s temperature and radiant intensity is increased by the amount of energy that the shell has received, but at what point in time is this interaction ‘completed’? This method necessitates iteration, after iteration, after iteration to arrive at a final conclusion.
3) The ‘belief’ that ‘back radiation’ can ‘warm’ (add temperature) to the ‘source of the energy’ is blatantly ‘false’. Any ‘back radiation’ is related to its ‘Planck relationship’ to the source that it came from. However, the scenario in Willis’s hypothesis involves a ‘source’ (planet) that outputs joules irrespective of infinitely variable temperatures that are ‘back-radiated’ towards ‘the planet’. The BB relationship may well permit an ‘interaction’ between the planet and the shell. Let’s look at this. No, let’s not. It’s too time consuming!
Rather, let’s look at a ‘top down’ calculation!
From ‘1’) (above), we know the ‘total energy input’ from the ‘atomic engine’ provides the ‘sole energy input’ to the ‘entire system’. Thus, the ‘entire system’ can only radiate this bulk quantity of energy whatever the ‘area of interaction’ (radiation density) in W/m^2 from the extremity of the outer surface of the ‘shell’ in a ‘steady state’ scenario.
Can anyone remember this ‘bulk’ energy quantum? My initiative for reading back the posts in this thread have ‘flat-lined’. 😦
We need to work ‘inwards’ from the outer shell surface to gain insight on any temperatures between the ‘heat source’ and its final dissipation when observing/investigating a ‘steady-state’ system.
Best regards, Ray.
The total energy output depends on the area, it’s 235 Gigawatts if the area is 1 billion square meters, in which case the radius would be 8920~ m. That 235 Gigawatt value can not change.
If there was some way that you could increase the energy density to 470 Gigawatts for the inner sphere, you can claim that it is ok because the outer sphere is only putting out 235 Gigawatts as it should.
…except, what happens if you open the shell after it reaches that steady state condition?
You have a power source which was only able to generate 235 Gigawatts, you put a shell around it, part of the system now has 470 Gigawatts flowing through it, and then you open the shell… does the extra 235 Gigawatts disappear?
No, that makes no sense.
[Reply] You are confusing energy with power. Again.
______
Again, the state where the shell and planet are in equilibrium is when the planet emits 235 W/m^2 towards the shell, the shell emits 234.99 W/m^2 (assuming 1 m gap) back towards the planet, net of 0.01 W/m^2 towards the shell, and the shell emits 234.99 W/m^2 to space.
The “arrrow” doesn’t recirculate uselessly, it is just canceled out by another “arrow”.
[Reply] You are confusing radiation physics with arithmetic. Again.
The shell can not cool by emitting radiation towards the planet,
[Reply] Yes it can, it just doesn’t end up cooler by doing so, because IT IS NOT A CLOSED SYSTEM AND ENERGY IS BEING GENERATED THAT IT CONSTANTLY RECEIVES.
therefore the shell can not warm the planet. The radiation from the shell is not added to that emitted by the planet.If you took this shell and planet, let it reach equilibrium, and opened the shell, you would only find 235 Gigawatts coming out, not 470 Gigawatts as Willis proposes.
[Reply] Willis doesn’t propose that 470 gigawatts “comes out”. Read it again.
Ray, Arfur, my response to you would be far OT and TB asked us politely not to here. How about later, different thread. I’m out most of tomorrow so I’ll drop a note as to where I can elaborate. This line dates back further than you seem to be aware of.
There is a huge problem with the Willis model that even given the most generous avoidance of physical reality its still fundamentally wrong.
There is an operational definition of temperature.
This is in line with the Ernst Mach, Karl Popper and Heisenberg school of Natural Philosophy.
The temperature of an object is that measured by a reliable thermometer calibrated against a
known and reliable source.
If a very thin piece of steel (say) is placed midway between planet and shell with a thermometer what would such a thermometer read.
Willis would say;
Pointed at the planet it would read 302K
Pointed at the shell it would read 254K
Absurd, the same thin piece of steel cannot have two radically different temperatures.
Postma would say;
Pointed at the planet it would read 254K
Pointed at the shell it would read 254K
Physically realistic.
So the Postma view of cavity radiation is correct and the Willis model false.
wayne says, March 28, 2013 at 3:14 am: David, to answer your question of Postma, I have so far read his thread down to about the 1/3rd mark, long sucker, and he finally stated what I was looking for, he said point-blank that to him both the shell and the sphere are and would stay at the same temperature at steady state, so I can now safely say to you that I do disagree with him at all.
Wayne,
Thanks for your fullsome and considered response. This is the way we should debate this matter, always moving forward.
However, before I can answer you I need to ask:
1. Postma has many threads. Which one were you reading? Then I can read it too and comment more meaningfully.
2. When you say I can now safely say to you that I do disagree with him at all er…um…do you mean that you disagree with him or that you agree with him? 🙂
Cheers,
David
Wayne. Ah…um…you obviously mean the Postma thread mentioned on the very first line of this top article. 🙂 Sorry, my mistake. I will check it out against your comments and reply back.
Bryan
“Willis would say;
Pointed at the planet it would read 302K
Pointed at the shell it would read 254K”
No, he would say close to 281 on both.
470 W/m^2 from inside + 235 from outside = 705, which is 352.5 reemitted from both sides.
Wayne,
First of all, in the above article TB refers to Postma’s article as a “huge rant”. It is and for me (and no doubt you as well) it makes very uncomfortable reading in places and is not my style of debate.
However, as scientists we must rise above that and look behind Postma’s anger to the physics that the SkyDragons espouse. Because the basic SkyDragon argument is actually very simple:
1. All bodies above 0K radiate. (this is uncontroversial I hope :-))
2. If a cooler body radiates towards a warmer body, that inflowing radiation to the warmer body doesn’t add to the warmer body’s fund of energy but instead reduces (subtracts from) the outgoing radiation from the warmer body that is flowing towards the cooler body.
That is the essence (as I understand it) of the SkyDragon position. It is not exclusive to Postma. One of its original protagonists was Prof Claes Johnson but unfortunately his quantum physical explanations may not be comprehensible to many readers. It has its roots in Prevost’s Theory of Exchanges (1791).
Now let me respond to your comments on Postma’s position:
(1) …he said point-blank that to him both the shell and the sphere are and would stay at the same temperature at steady state…
He is simply playing by the rules of the Willis thought experiment, just as I and Tim Folkerts are but, sadly, you and some others are not. I really don’t want to get involved in more debate with you and others about this but just to say that, in making his comments, Postma is assuming a very thin shell at a very short distance from the core surface with both shell and core having radiating black body surfaces and zero internal thermal resistance.
The onus is then on you or others to demonstrate (if you can) that those simplifications cause more than marginal arithmetical discrepancies (in both IPCC and Skydragon models).
(2) He further says that if the gap were filled with matter…this would make no difference, still 235 fx (254K), clearly wrong, not with a nuclear core inside.
Because in Willis’ model the core has zero thermal resistance it is isothermal. Once again, like Tim and me, Postma is just playing by the professional rules of a thought experiment discussion. Inside a real physical core, a real power source would have to be constructed such that it delivered a uniform heat flow to the whole of the surface but the Willis thought experiment excludes discussion of such details.
Once again the onus is on you and others to demonstrate (if you can) that this simplification causes more than a marginal</i arithmetical discrepancy (in both IPCC and Skydragon models).
(3)That showed me conclusively he reads this thought experiment completely wrong, as if the energy source is throttled to an equivalent steady temperature and not a constant energy output.
No, it shows me exactly the opposite – that you haven’t stuck to the rules of the thought experiment. Postma does NOT assume anything other than what the thought experiment specifies: (i) a fixed 235Wm-2 flux out of the surface; (ii) the shell is negligibly thick; and (iii) the void between core and shell is negligibly wide. So filling the void with matter would just construct a new composite solid core all of which would be isothermal (by virtue of zero thermal resistance throughout) and which would be radiating at 235Wm-2 (by virtue of the negligible radii differences).
Once again, the onus is on you or others to demonstrate (if you can) that those simplifications cause more than marginal arithmetical discrepancies (in both IPCC and Skydragon models).
In conclusion here for reference are my two diagrams (IPCC on left, SkyDragon on right):
[…] Second, they make their radiation arithmetic, assuming that the planet always retains the same temperature, but emits more and more, which “allows” them to always add the same initial 800 (initial power) to the increasing back radiation. The bodies, however, are known to emit according to their temperatures, hence their hypothetical cyclical process would lead to an endless mutual warming, so the whole thing falls apart at this stage too. There is, however, an abstract possibility to resolve this contradiction, namely by assuming that after emitting and before absorbing the planet’s temperature always drops to the initial one and then rises again and always to a higher and higher value. I offered it, hoping it would additionally demonstrate the absurdity of the initial assumption, but guess what? Roger took it indeed as a part of their process! Unbelievable. This is their physics. (https://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/comment-page-1/#comment-46160) […]
Format messed up again, this is the fixed version.
“[Reply] You are confusing energy with power. Again.” ~tb
No, I am not, power is the rate at which energy is delivered, it is measured in Watts, intensity is rate and density of energy delivery, it is measured in Watts per meter squared.
235 Gigawatts means every second the surface emits enough energy to perform a certain amount of work.
______
“[Reply] You are confusing radiation physics with arithmetic. Again.” ~tb
No, this is actually what Willis does when he naively adds the shell emissions to the surface input.
I am doing it the correct way, according to the physics of radiation and thermodynamics and everything else.
______
“[Reply] Yes it can, it just doesn’t end up cooler by doing so, because IT IS NOT A CLOSED SYSTEM AND ENERGY IS BEING GENERATED THAT IT CONSTANTLY RECEIVES.” ~tb
You did not read what I was saying.
I said the shell can not cool by radiating towards the planet because the shell is being warmed by radiation from the planet.
A surface warmed by a given energy source can not warm said energy source, shouting that it is not a closed system does not change this.
________
Willis proposes, and it has been argued by many in this thread, that the planet would output twice what it did without the shell present.
If the shell had louvres all over it and you first brought the system to what Willis and most here think is the equilibrium state, the shell would emit 235 W/m^2 times the area of the shell in m^2, and the planet would emit 470 W/m^2 times the area of the planet in m^2, right?
What would happen at that point if you opened the louvres?
_____
I’ll note the interesting twist that you said I was confusing arithmetic for radiation physics when this entire thought experiment is an example of confusing arithmetic for radiation physics.
Radiation physics involves adding and subtracting vectors.
A 2N left vector plus a 4N right vector equals a 2N right vector.
Also, David, I think this better represents the skydragon version: http://i341.photobucket.com/albums/o396/maxarutaru/fixedskydragonmodel_zps50023ce6.png
lgl says:
“No, he would say close to 281 on both.
470 W/m^2 from inside + 235 from outside = 705, which is 352.5 reemitted from both sides.”
There are several things wrong with this.
1. You are assuming that there is a isothermal temperature field in the gap of 281K
2. If there was a temperature specific event like a chemical reaction requiring 300K it could happen on the planet facing side but not on the gap facing side – unphysical
David S says
What a silly argument. Why can’t you just go with the flow and address the overall problem instead of doing all this nitpicking. ”
You can’t to decide which questions are asked. You insult my argument because you cannot answer it.
It is because the outward radiation is split by the shell and not by the planet surface that leads to different temperatures being predicted. Energy appears to be created by not constructing the model consistently.
Your diagram shows this. Radiation leaves the shell from up and down. Radiation from the core leaves only from the top surface.
This flaw is hidden by the incomplete definition and diagram of the starting conditions given by Willis, especially inside the planet.
The model can only be made consistent by further defining starting conditions. The model is theoretically incomplete.It cannot lead to uncontroversial predictions.
I am discussing TB’s claim that the Willis model is ‘theoretically correct’ . I am not discussing Postma’s opinions. They can be discussed on his blog.
The system as a whole loses energy only from the top, outer surface of the shell. This is 235W/m2.
Total energy lost is 235A J per second
A = area of surface ( shell and planet almost the same )
J = unit of energy, joule
The planet is radiating 470AJ per second
What happens to the 235AJ of energy per second that the planet radiates but is not lost to space?
Either it disappears contrary to 1st law or goes into core, as I say it should. The core heats up more, contrary to prediction of steady state.
Well I suppose it could be the back radiation that turns around and heads of up again.
Then it is the Willis model that is described in the diagram with the circulating back radiation labelled ‘Skydragons’ .
Bryan
1. No, I’m not assuming any temperature field at all because there is none, it’s vacuum.
2. No, I’m saying temp would be 281 on both sides.
Roger C
“What happens to the 235AJ of energy per second that the planet radiates but is not lost to space?”
Jeez.. can’t you see from the diagram that 235+235=470 enters the planet and 470 leaves the planet. Then how much net radiation is there to heat the planet?
lgl,
I admire your persistance at keeping on plodding away at it!
lgl says:
“No, I’m saying temp would be 281 on both sides.”
1.
Do you accept that the Planck spectrum of 302K
is different to the Planck spectrum of 254K
2.
Do you accept that the Planck spectrum of 302K is absorbed on planet facing side
and the Planck spectrum of 254K is absorbed on the shell facing side?
Then to deny that one face is at 302K and the other is at 256K is to deny that surfaces are black bodies despite being explicitly informed that they were in the Willis model
i
Bryan
1. Yes
2. Yes, but the radiation on both sides is absorbed and converted to kinetic energy, and your very thin piece of steel is a near perfect heat conductor so we can ignore the little temp difference between the sides.
I think what Bryan is getting at, lgl, is that if you inserted an arbitrarily thin piece of metal directly between the shell and planet, one side will receive radiation from the planet, one side will receive radiation from the shell.
What would the state of this thin piece of metal be after a given period of time?
I contend that the metal would approach equilibrium with the other two surfaces, planet and shell, respectively.
Do you think that sounds reasonable?
…
If so, how can it be in equilbrium with two surfaces at different temperatures simultaneously?
Simple proofs that back-radiation is not real
Proof by the fact that spurious absorption occurs.
In case of back-radiation one follows the hypothesis of the two-way heat flow by Prevost from back in 1971:
A surface emits heat only according to its own temperature, and absorbs eventually heat from another surface, depending on the temperature of the other surface.
In the one-way heat propagation formulation, the emission depends not only on the temperature of the surface which emits, but also on the temperature of the surface which absorbs that emitted heat:
for T1>T2 : q(1→2) = σ(T1^4 -T2^4) and q(2→1) = 0
In figure 1 are compared the two formulations.
Figure 1 Heat radiation between two plates
according to two implementations.
Formulation A Formulation B
one-way heat flow two-way heat flow
Prevost hypothesis,
avoid it
outer space outer space
↑σT2^4 = q ↑ σT2^4 = q
plate 2 ────── plate 2 ──────
↓ σT2^4
↑σ(T1^4 – T2^4) ↑σT1^4
plate 1 ────── plate 1 ──────
↑ q input ↑q input
In both formulations the input to the ground plates 1 is q, in steady-state conditions the output from the plates 2 is also q.
The temperature distributions turn out to be the same for the two formulations:
T1 = (2q/σ)^0.25 T2 = (q/σ)^0.25
Everybody seem to agree with this statement.
We could conclude that the two formulations are equivalent!
However, when we look to the absorption of plate 2, we see that in the one-way formulation plate 2 absorbs and emits a quantity of heat: q =σT2^4.
But according to the two-way formulation , plate 2 absorbs and emits a quantity of heat twice as much : 2q = 2σT2^4
We see that although the calculated temperature distributions are the same, the two-way formulation gives a non-physical spurious absorption by a factor 2.
If one takes a stack of N plates, plate 2 absorbs Nq for the two-way formulation , while in the one-way formulation plate 2 and all other plates absorb one single q.
Click to access IR-absorption_updated.pdf
Proof by taking plates with different emission coefficients.
The one-way heat propagation in fact has been the practice in engineering applications since the formulation by Stefan-Boltzmann.
In engineering applications pairs of surfaces are identified.
In case the emissivity of both surfaces ε = 1, the heat exchange by radiation is q= σ(T1^4 – T2^4) for the one-way formulation and q= σT1^4 – σT2^4 for the two-way formulation.
We have seen that the same temperature distribution will be obtained by the two formulations, only the absorption in the two-way formulation is not correct. But in engineering applications nobody cares.
However , in case the emissivities of the two plates are not equal but σ1 = ε1σ and σ2 = ε2σ only the one-way formulation can be applied!
We adopt the notation:
q(1→ 2) = q1,2 and q(2→1) = q2,1
we can write the one-way heat flow formulation:
For T1>T2: q1,2 = σ1,2(T1^4 – T2^4) q2,1 = 0
Surface 1 emits q1,2 and surface 2 absorbs it.
For T2>T1: q2,1 = σ2,1(T2^4 – T1^4) q1,2 = 0
Surface 2 emits q2,1 and surface 1 absorbs it.
The quantity σ1,2 = σ2,1 is calculated from (see Wikipedia, emissivity):
σ1,2 = ε1,2σ = σ2,1 = ε2,1σ
1/ε1,2 = 1/ε2,1 = 1/ε1 + 1/ε2 – 1
The relation σ1,2 = σ2,1 in fact confirms the one-way heat flow formulation.!
In case ε1 = ε2 :
1/ε1,2 = 1/ε2,1 =1 and σ1,2 = σ2,1 = σ
In case only one of the emission factors is 1, say ε1 =1
1/ε1,2 = 1/ε2,1 =1/ε2 and σ1,2 = σ2,1 = ε2 σ
This is important: even with ε1 =1, the black surface 1 only emits ε2 σ (T1^4- T2^4) to a surface 2 with T2T1, surface 2 emits ε2 σ (T2^4- T1^4) and the black surface 1 absorbs that heat.
Those readers who are not convinced yet, try to write down the exchange of heat between two plates with different emission coefficients, using the two-way heat flow formulation with σ1 = ε1σ and σ2 = ε2σ.
The relation from Wikipedia how to combine emissions for two different values is old and was published around 1860 in Germany in a Chemical Journal.
It turns out that the one-way heat flow formulation in the atmosphere gives coherent results for the heat balance of the planet.
The huge absorption by the atmosphere and the back-radiation as heralded by IPCC do not occur.
Click to access PROM_REYNEN_Finite_Element.pdf
Bryan says: “Then to deny that one face is at 302K and the other is at 256K is to deny that surfaces are black bodies despite being explicitly informed that they were in the Willis model.”
Bryan, you clearly do not understand the idea of a “blackbody”. A black body absorbs all EM energy that falls on it. That energy is added to the general thermal energy of the BB object. That energy might
1) warm the BB object
2) get radiated away by the BB surface
3) get conducted away from the BB surface
4) cause evaporation at the surface
5) ….
or any combination of these.
In this case, we have postulated that the sheet is at some steady temperature, so (1) can be ignored. Similarly (4) can be ignored. So we are left with radiation & conduction.
At the planet-facing-surface, 2*235 W/m^2 gets absorbed. 1.5 * 235W/m^2 gets emitted as IR radiation and 0.5 * 235 W/m^2 gets conducted away to the other side.
At the planet-facing-surface, 1 * 235 W/m^2 gets absorbed. 1.5 * 235W/m^2 gets emitted as IR radiation and 0.5 * 235 W/m^2 gets conducted in from the other side.
Both sides are balanced. That is what the idealized model predicts.
*******************************************************************
NIT_PICKY DETAILS:
This assumes perfect thermal conductivity. If the thermal conductivity is only “good” then the planet-facing-side will emit (1.5+δ) * 235W.m^2; the shell-facing-side will emit (1.5-δ) * 235W.m^2
and (0.5-δ) * 235W.m^2 gets conducted through. Then the shell-facing-side will be a little above 281 K and the other sode will be a little below 281 K.
Bryan’s conclusion only holds in the limit that the thermal conductivity goes to zero (eg we slip a thick layer of styrofoam in the middle of the steel sheet). Then δ = 0.5; no heat gets conducted thru the shell, and the temperatures are as he predicted. But that is very much NOT in the spirit of the situation Bryan proposed.
JWR, that is one heck of a paper. not that I am any kind of scientist. However because you are associated with the “Dragons” I fear that it will not get the credence it deserves.
JWR says: “T1 = (2q/σ)^0.25 T2 = (q/σ)^0.25
Everybody seem to agree with this statement.”
You would think so, but Joe Postma and many of his disciples in this discussion are sure that both surfaces are at T2 = (q/σ)^0.25.
JWR says: “the two-way formulation gives a non-physical spurious absorption by a factor 2”
I don’t think this is non-physical or spurious. For one thing, the two are mathematically equivalent, so at some level they are the exact same thing. There are all sorts of analogies I could make with money or circuits or water flow. All of those (and the model here by Willis) obey appropriate conservation laws and appropriate thermodynamics laws.
WHAT SPECIFIC PHYSICS LAW do you think is being violated by the steel shell model to make it non-physical?
JWR, you start the second argument by seemingly asserting your conclusion:
“Proof by taking plates with different emission coefficients….
For T1>T2: q1,2 = σ1,2(T1^4 – T2^4) q2,1 = 0 ”
If I read this correctly, you are asserting that q2,1 = 0. Why?
Going back to wikipedia, there is a nice little section at http://en.wikipedia.org/wiki/Thermal_radiation#Radiative_heat_transfer that discusses this topic. In particular, the equation for heat flow = net thermal energy transferred via IR radiation (the last equation in the section) is q1,2 in your notation. By swapping the “1” and the “2” we get q2,1. But that swap gives the same result — except with a minus sign.
So q1,2 = -q2,1, exactly as expected. NOT q2,1 = 0 as you wrote.
JWR
“write down the exchange of heat between two plates with different emission coefficients”
Yes lets do that, or better, modify Davids diagram with the shell now having an emissivity of 0.5
Basic physics of course also work with different emissivities.
JWR 4:43pm – “..confirms the one-way heat flow formulation.”
Yes, the spontaneous flow of macro heat is always one way however in the top post the planet temperature is being forced by the central heat source. Citing 1st & 2nd law is all that is needed which means the cooler shell in top post does not heat the planet, only the core heat source heats up the planet. Adding the shell, slows the forced planet surface radiative cooling which allows the planet to be forced warmer by the central heat source than it would be without the shell, all in accordance with 1st and 2nd law.
ACO 5:09pm “…one heck of a paper.”
The annotations are complex but are ok so long as they support the paper’s conclusion in this clear statement clipped & which seems to be the point of the paper’s formulas “Back-radiation of heat is a violation of the Second Law of Thermodynamics.” which is physically correct if the back radiation comes from a cooler source object than the warmer receiving object.
The cooler shell in top post cannot heat up the warmer surface of the planet in steady state or transiently, only the central heat source can do so. The cooler shell slows the forced planet surface radiative cooling which allows the surface to be forced warmer (heated to higher T) only by the central heat source.
Really have to closely watch the formulas and the statement language specifics for clarity on top post subject (pea under the thimble concept).
Also note, the planet core heat source is not creating energy, it is converting mass to energy. Eventually the mass will be used up. No PMM.
JWR, after looking at your paper a little, I see that you fall into a seductive misinterpretation of the 2nd Law of Thermidynamics.
“In the case of Ti > Tj the surface j would emit a quantity of heat from cold to warm: a violation of the Second Law of Thermodynamics.”
The 2nd Law in its modern form is based on this definition of entropy:
S = -k Σ P(i) ln(P(i))
The second law says that this quantity must increase (or stay the same) .
One classical corollary is that heat (the NET flow of energy) must go from hot to cold. This does not require that NO energy go from cold to hot, as long as the energy the other way is larger. As long as the NET energy flow is “the right way”, then S = -k Σ P(i) ln(P(i)) will increase.
For a simple example, consider a block of metal in contact with a slightly cooler block of metal. “Heat” is the net result of the collisions of the atoms on the surfaces of the two blocks. At least occasionally, an unusually fast atom in the cooler block will hit an unusually slow atom in the warmer block. Such collisions will send energy from the cooler block to the warmer block. This is NOT a violation of the 2nd Law because MORE collisions transfer MORE energy from the warmer to the cooler block.
This is “back-conduction”! It is very real. And “back-radiation” is just as real. Neither violates the 2nd Law.
Trick says: March 28, 2013 at 6:32 pm “Adding the shell, slows the forced planet surface radiative cooling which allows the planet to be forced warmer”
So you are saying that for maybe 1 microsecond the Planet gets slightly hotter than it was without the shell while the shell starts shedding the heat, but not much hotter?
Does it stay hotter?
Would you care to quantify as Tim F has done?
“slows the planet’s surface radiative cooling”
That’s the catch all. Hard to see. There doesn’t seem to be a single word to identify or describe that ‘slowing’ process. But ACO, think of the temperature differences and how fast they are changing. The shell starts out cold. It takes very little energy to raise its temperature markedly at first. So the temperature difference between the shell and planet decreases very rapidly at first and by SB the planet for a while cannot shed all of the 235 every second upward until a new equilibrium is reached. The planet warms by its own energy source as Trick stated.
Tim Folkerts SAYS
“In this case, we have postulated that the sheet is at some steady temperature”
Another unilateral postulate!
Willis did not deal with a temperature probe between the planet and the shell being irradiated by both.
Black bodies first of all have to absorb, hence surface effects and significantly different temperatures will exist on each face.
Your previous postulate about the planet being heated by a huge number of evenly distributed micro radioactive sources seems to be forgotten by David S who is back posting about the radioactive core.
ACO 6:53pm: “So you are saying that for maybe 1 microsecond…”
No I didn’t say that. As you clip, proper physics says both transiently and at equilibrium under the 1st law “Adding the shell, slows the forced planet surface radiative cooling which allows the planet to be forced warmer” only by the central heat source. That is what I’m saying.
“…care to quantify…”
1st law in this next form is very useful for energy accounting analysis of top post system, same form as correctly used even by Postma in past TB post, 2nd law formulates the direction of heat flow:
energy in – energy out = m*Cp*DT/dt W/m^2
Examination of the above 1st law heat eqn. (m=mass of object, Cp heat capacity) shows:
• energy in greater than energy out then object is warming DT/dt is positive
• energy in less than energy out then object is cooling DT/dt is negative
• energy in = energy out then object is in equilibrium Dt/dt = 0
Properly account energy in and energy out of a control volume thru conductive, convective and radiative processes find if object warming or cooling or at equilibrium.
Quantify top post: 235 W/m^2 in – 235 out W/m^2 = 0 at equilibrium or energy in = energy out, no convection process in the top post set up.
Transient case a little more work but ends in finding 235 = 470 – 235 is one possible solution to top post as is shown up there. Nature’s exact solution depends on global top post emissivity of inner shell emitting upon itself, this is NOT easy.
******
wayne 7:15pm says it well also. I can work thru the heat eqn. transient math but makes David think I am not lucid doing so, may not be worth the effort; wayne and Tim F. doing it well enough in their own lucid way.
My gracious Bryan, that was YOUR unilateral postulate!
Bryan said: “If a very thin piece of steel (say) is placed midway between planet and shell with a thermometer what would such a thermometer read…..
Then to deny that one face is at 302K and the other is at 256K .. ”
YOU assumed that the sheet was at a steady temperature! Your thermometer reads a steady temperature so clearly you were assuming a steady-state solution where the temperature had stabilized at a specific temperature (or a steady pair of temperature at two different locations).
I was simply correcting your incorrect conclusion about what the steady-state result would be!
Trick is correct to point out that the everlasting source of power in the Willis model is unphysical .
Mass is converted to heat and at some point will run out.
This point was made earlier by Cementafriend (sp?) where pointing out his experiences with furnaces.
The ultimate temperature was limited by the thermochemical properties of the fuel used.
A maximum temperature was attained and could not be exceeded by any amount of backradiation.
So the only way for the internal energy of the planet to increase depends on the reaction of the power source to any increase temperature of the planet
“The 2nd Law in its modern form is based on this definition of entropy” ~Tim F
Entropy is a way to quantify the amount of order/disorder present in a system, or the availability of energy which can be used to perform work.
If both surfaces are the same temperature then there isn’t there more disorder than the case where one surface is warmer than the other? Similarly, if both surfaces are the same temperature there is no energy gradient, so how would you extract work?
_______
If one surface is warmer than the other, and if this is the state which maximizes entropy, then I have a question:
What exactly prevents one from extracting work from this energy gradient?
If you put your hand just outside where the shell is you are supposed to feel the same warmth if it is there or if it is not and this seems unreasonable to me.That the planet ,the entire mass of the planet,heats up because the energy is slightly delayed in this system by the shell also seems to me unreasonable. I am sure that the shell does delay energy leaving the system but there is nowhere for extra heat to reside other than in the shell,there is a hypothetical (could not exist)vacuum inside the shell which cannot heat up.Any radiation that is absorbed by the surface of the planet from the shell cannot be conducted towards the centre of the planet it would have to reverse the temperature gradient of the planet in order to do that and if it did that all energy flowing from the surface outwards would stop.
Max™ says, March 28, 2013 at 12:29 pm: Also, David, I think this better represents the skydragon version…
Max my version of the diagram has already been agreed by Joe Postma. You are entitled to your own opinion but you should not have used my diagrammatic form with the same sub-heading because this may easily confuse other readers.
It is up to Joe whether he accepts your revised version as better than mine at representing the SkyDragon position. But my view is that your diagram is not appropriate nor indeed clear in its intent. The acid test is whether or not it confuses people trying to understand the SkyDragon position. I think it will confuse them.
Sorry but that’s my opinion.
Max
“If so, how can it be in equilbrium with two surfaces at different temperatures simultaneously?”
It’s the energies that need to be in equilibrium, not the temperatures. Energy into and out of all objects must balance. The temperatures are determined by the total energy supplied to the objects.
And your version of Postmas junk is wrong. He claims the radiation from the shell towards the planet is not absorbed but ‘scattered back’. Anyway your version don’t work either. The planet must emit net 235 W/m^2 not to heat up so your 0.01 W/m^2 will not do.
Dear lgl
Thank you for you answer with the drawing.
You confirm with that drawing that you do not use in your hand calculation what Prevost said and what is used in the software of IPCC.
You have the correct temperatures, certainly, but you did not use the hypothesis of Prevost!
Prevost says that the ground plate at T1 is emitting as if it were looking to outer space.
That is your red arrow of 940= sigma*T1^4.=4q
And the lower face of plate 2 is emitting e*sigma*T2^4=235=q , your green arrow, also as if the face were looking to outer space.
Prevost and IPCC say that the flux from 1 to 2 , q1,2=4q-q= 3q= (940-235)
So you did not show with the blue arrow what Prevost and IPCC are claiming.
With the blue arrow you end up with the correct figures. I see your ad hoc reasoning with the blue curve, but that does not correspond to what Prevost and IPCC are telling.
I propose you to follow my one-way heat formulation!
From 2 to outer space, e*sigma*T2^4 =235
From 1 to 2 with sigma1,2= e*sigma and q1,2= e*sigma*(T1^4-T2^4) =q=235
That is all.
From this follows e*sigma*T1^4= 2q and sigma*T1^4= 4q=940.
The same answer as in your picture where you correct the red and green arrows with your blue arrow , which is an ad hoc reasoning which you cannot use in a computer program.
Max wonders“If one surface is warmer than the other, and if this is the state which maximizes entropy … ”
Ah, but this is not the state that maximizes entropy. The maximum entropy would be when the nuclear energy that was “highly ordered: within the radioactive nuclei is spread throughout the universe and the planet and shell have both cooled to 2.7 K.
In the meantime, the nuclei are occasionally releasing energy and that energy is getting spread out among the atoms of the planet (thereby increasing entropy in the universe). The energy of the warm atoms in the planet is getting more spread out by spreading from the warm planet to the cool shell (thereby increasing entropy in the universe). The energy of he cool atoms in the shell set further spread out by leaving the cool atoms of the shell and moving out into cold space (thereby increasing entropy in the universe).
The system is TENDING TOWARD maximum entropy — but never actually getting there because more energy from the nuclei is continuously being added to the system so that it cannot reach equilibrium. As long as there is energy being input from the nuclear heater, the system is never at equilibrium. As long as this is true, then there is no problem extracting work from the temperature differences that exist.
Tim Folkerts says ;
“YOU assumed that the sheet was at a steady temperature! Your thermometer reads a steady temperature so clearly you were assuming a steady-state solution where the temperature had stabilized at a specific temperature (or a steady pair of temperature at two different locations). ”
No my point was that the whole arrangement was unphysical
JWR
“Prevost says that the ground plate at T1 is emitting as if it were looking to outer space.”
You seem unaware that we are discussing a toy model here. A planet with a steel shell around it, so I don’t care what Prevost and IPCC are telling, I’ve never seen them discussing this toy model.
And I did no ad hoc reasoning, just used the basic laws of thermodynamics.
Strange b t w that you agree with the 940, I doubt Postma would.
JWR says: “Prevost and IPCC say that the flux from 1 to 2 , q1,2=4q-q= 3q= (940-235)”
No, “Prevost and IPCC” — and all qualified practitioners of radiative heat transfer — would get the answer that lgl got.
If you don’t like his simple but correct calculations, you could use the equation I referred to earlier at:
http://en.wikipedia.org/wiki/Thermal_radiation#Radiative_heat_transfer
This rather intimidating equation correctly gives 235 W/m^2 from the planet to the shell for the numbers lgl gave. Exactly as lgl predicted based on much simpler (but also correct) calculations.
It is you, JWR, who is incorrectly calculating the answer that “textbook science” would predict.
***********************************************
This seems to be a very common problem — projecting ones own misunderstanding onto others. Physics and engineering professors for the last 100+ years have not been idiots and really do know a thing or two about how to determine radiative heat flow between surfaces.
lgl,
Why is it strange that I agree with the 940?
But when you are so clever with drawings, suppose the outer face of the shell has emissivity 1, and
only the inner face has emissivity 0.5.
In fact that is what I intended to ask in the first place, because both faces with emissivity 0.5 gives rise to a symmetric situation and thereby ad hoc reasoning.
I do not think that we discuss here only a toy.
The back-radiation is a non-physical phenomenon, a kind of auxiliary variable which gives for some cases the same temperature distribution as the one-way heat flow formulation.
But it gives non-physical heat absorption, and IOCC and others use that spurious absorption in the atmosphere to claim that the planet is heating up.
“Max my version of the diagram has already been agreed by Joe Postma. You are entitled to your own opinion but you should not have used my diagrammatic form with the same sub-heading because this may easily confuse other readers.” ~David S
Well, mine is less confusing than yours, and this is his response to mine:
“Well the one thing is that those could be walls, in which case the surface area differential problem of the shells game doesn’t exist. Say the core is an infinitely deep plane heat source and so we don’t care about what’s beneath it – it just holds a steady temperature and produces an output of 235 W/m2 forever – it is an infinite energy source. The “shell” is another plane then, a wall or whatever, just beside the infinite plane heat sink. It therefore gets the full 235 W/m2, and comes to the exact same temperature. What then? Then there is net 0 backwards, because q ~ (T2^4 – T1^4), and T2 is equal to T1.
Now here’s a seemingly paradoxical situation that confuses the GHE people. If there is net zero backwards, then isn’t there also net zero “forwards”? Don’t you need a temperature differential to transmit bonafide net heat energy, and if the “shell/wall” is equal to the infinite source/sink, then how can the source heat the shell?
But this is precisely it. The present occurs continuously and instantaneously. Heat transfer doesn’t occur in steps of one-second, but continuously instantaneously. The shell/wall is heated until its outward energy loss matches that of the source/sink. There is never any net q from the shell/wall back to the source/sink. As soon as energy is lost outwardly by the shell, it is instantaneously replaced by heat energy from the source underneath.
A radiation field that doesn’t transmit net energy is JUST like a force that does no work. We know all too well what a “force that does no work” means – we accept that intuitively, physically and mathematically. We all accept that forces can exist, in which real energy is expended, but where no work actually gets done. I.e., push against a wall. So, it is totally legitimate to write 235 W/m^2 back from the shell, and, that it has ZERO net transfer of “temperature” and heating power. The only net transfer which can exist is that on the outside of the shell/wall, and the only quantity this value can have is the quantity which comes in to it from the bottom.” ~Postma
I’m confident that he mentally corrected the circling arrows to one like mine, and I am confident that he does not actually think part of the energy circles around while the other part doesn’t.
He noted that it would be simpler to just think of it as being infinite walls, I like working with the same setup as the initial experiment, hence my using the values for a spherical arrangement.
_______
“And your version of Postmas junk is wrong. He claims the radiation from the shell towards the planet is not absorbed but ‘scattered back’. Anyway your version don’t work either. The planet must emit net 235 W/m^2 not to heat up so your 0.01 W/m^2 will not do.” ~lgl
The planet and shell equalize, the shell is emitting the necessary amount for the system to not heat up, the idea that the planet has to heat up and won’t heat the shell up accordingly is odd.
JWR asks: “But when you are so clever with drawings, suppose the outer face of the shell has emissivity 1, and only the inner face has emissivity 0.5.”
This is really not that tough …
The shell emits (1.0)*σ*(254^4) = 235 W/m^2 upward to space (required by Conservation of Energy)
The shell emits (1.0)*σ*(254^4) = 235 W/m^2 downward to the planet (since both sides are the same temperature), of which
>> 117.5 W/m^2 gets absorbed and
>> 117.5 W/m^2 gets reflected upward to the shell.
So the planet is getting 235 W/m^2 from the heaters, + 117.5 W/m^2 from shell radiation.
So the planet must radiate 1.5 * 235 W/m^2. Since the emissivity is only 0.5, the temperature must be
T = (1.5 * 235 / (σ*0.5) )^0.25 = 334 K
************************************************
Not too surprisingly, the general equation for IR heat transfer turns out to be the same whether the planet or the shell has the emissivity of 0.5:
>> P/A = 0.5 σ (T_p^4 – T_s^4)
which indeed works for this case or the previous case.
Funny how “textbook physics” is pretty good at these 100 year old problems!
“JWR
“write down the exchange of heat between two plates with different emission coefficients”
Yes lets do that, or better, modify Davids diagram with the shell now having an emissivity of 0.5
Basic physics of course also work with different emissivities.”
So with this modification, does shell have emissivity of 0.5 on both side?
It seems if shell had emissivity of .5 on it’s outer surface, this would raise the temperature of the shell and core. If instead it had only had emissivity of .5 on inner surface of shell, it would only increase temperature of core.
It isn’t +235 W/m^2 from the heater and +235 (or 117.5) W/m^2 from the shell.
It is +235 W/m^2 from the heater, +235 W/m^2 emitted by the planet, -235 (or -117.5) W/m^2 emitted by the shell.
________________
Where does the idea come from that you add radiation from a cold surface–one heated by the surface it is radiating towards–to the energy input of the warm surface rather than subtracting it from the output?
The planet cools by emitting radiation towards the shell, and in the process it warms the shell.
The shell can not cool by emitting radiation towards the planet, any losses in that direction are immediately replaced and more. The shell can only cool by emitting radiation to space.
“1. All bodies above 0K radiate. (this is uncontroversial I hope 🙂 )
2. If a cooler body radiates towards a warmer body, that inflowing radiation to the warmer body doesn’t add to the warmer body’s fund of energy but instead reduces (subtracts from) the outgoing radiation from the warmer body that is flowing towards the cooler body.
That is the essence (as I understand it) of the SkyDragon position. It is not exclusive to Postma. One of its original protagonists was Prof Claes Johnson but unfortunately his quantum physical explanations may not be comprehensible to many readers. It has its roots in Prevost’s Theory of Exchanges (1791).”
It seems to me if follow Willis rules for the model, the Sky Dragon view is correct.
It assume the blackbody surfaces are converting all radiant energy into heat which radiate [IR] at it’s blackbody temperature.
It seems to me that the skydragon position is merely following the same rules for blackbody that determine that if Earth were an ideal blackbody it’s uniform temperature would be around 5 C.
A problem seems to occur when one enter in different emissivities other than 1.
I imagine Postma would have different model if instead of blackbody, it involved
reflective surfaces.
In addition aspect is a blackbody would not re-radiate energy rather a blackbody
converts all incident radiation into heat then and radiates according to entire surface area [In accordance to the geometry of the blackbody].
And it seems from what I can understand regarding Prof Claes Johnson, is, he addresses issue
of re-radiation. And I would say in terms climate in general, greenhouse gas are about re-radiation and are not same in this regard to black bodies.
gbaikie says: March 29, 2013 at 1:57 am
This is true, but the Willis Model shows an impossible scenario where the shell radiates the same energy intensity to ‘space’ as the planet radiates to the ‘shell’ (the ‘doubled intensity’ is wrong. This should be shown as ‘temperature difference’ for the different Planck definition).
There ‘MUST’ be a difference in radiating area between the planet and the shell, thus, the ‘energy intensity’ (W/m^2) is inappropriately addressed in the ‘model’.
The ‘model’ is just, sooo WRONG!
That’s not to say that we can’t come up with one, a ‘model’, that’s more accurate as a ‘tutorial’!
Best regards, Ray.
Tim F says
“Physics and engineering professors for the last 100+ years have not been idiots and really do know a thing or two about how to determine radiative heat flow between surfaces.”
Here we can agree and that is why you will not find a mention of the ‘greenhouse effect’ in any serious textbook.
Max™ says, March 29, 2013 at 12:41 am
Max,
This is beyond parody.
Relating to your modified version of my diagram of the SkyDragon energy flows, you first say:
Well, mine is less confusing than yours, and this is his response to mine
You then include a long quotation from Joe Postma, which clearly and succinctly and correctly sets out the SkyDragon position but doesn’t mention the diagram other than pointing out politely to you that a plane-parallel representation is perfectly acceptable (as I use in my diagram, but unlike your proposed bastardised version which uses my plane-parallel format yet has power intensity values which imply a spherical model where the concentric surfaces have different areas).
You then say:
I’m confident that he mentally corrected the circling arrows to one like mine, and I am confident that he does not actually think part of the energy circles around while the other part doesn’t.
Well what arrogance. Didn’t you think to ask him? I agreed my diagram with Joe after a to-and-fro discussion on his blog. 🙂
TB, I just can’t help it, tried. I must hit the “Post Commit” button even though this may speak OT about reality. 😉
This is true, but the Willis Model shows an impossible scenario where the shell radiates the same energy intensity to ‘space’ as the planet radiates to the ‘shell’ (the ‘doubled intensity’ is wrong. This should be shown as ‘temperature difference’ for the different Planck definition).
There ‘MUST’ be a difference in radiating area between the planet and the shell, thus, the ‘energy intensity’ (W/m^2) is inappropriately addressed in the ‘model’.
The ‘model’ is just, sooo WRONG!
Ray’s right, but here’s another impossible scenario that does affect the surface temperature. The same for the sphere to the inner shell. Those intensities are never going to be the same either, due to geometry, outgoing is always greater than incoming even if you have perfect reflection at the shell, sphere outward > shell inward, remembering that half is always going through the shell. Couple that with what TimF says (probably regretting shining a light on it now, even though I already knew it was true), that looking at intensities inward from a shell now to an enclosed sphere, the distance between doesn’t matter at all, in that one case the geometry cancels the 1/r² natural spread of radiation outward with the 1/r² difference in area intercepted inward is the opposite direction.
Right Ray, let’s do get it real. Hope TB will agree. There can be no constant and perfectly back and forth of 235 between the two. This is not a perfectly elastic situation, never will be, and the entropy is always increasing even when not considering what is also happening at the outer shell outward that Ray points out. So even if down radiation were to thermalize it is always lower than what went outward cooling even if you try to imagine they are equal. IIRC, that is the physical reason actual thermalization can never occurs at all by radiation in such a BB cavity, there is but a resonance in the e.m. field.
Wikipedia say photons are disturbances in the underlying e.m. field. This time I do agree with them. But resonances in a cavity have no “disturbances”. You figure it out.
( under Electric Field )
( under Electromagnet Field )
Now that is what I learned while at my university in physics and astronomy.
This is what I hate the most of climate “science”… all of the physical effects they insist on saying “just ignore this small difference, just ignore that, this is too small to matter and that is too small to matter either”, … on and on, always with a thumb on the scales leaning to the warming side — I call that dishonesty. I can easily find enough ignored difference to completely zero out the 0.9°C they say will be due to CO2 concentration.
Max’s complaints about the area differences should not have been just brushed aside. They are real effects even in a hypothesized thought experiment if you are not going to skew and lie about it to make a false point (this is not a real point). This was a setup, although I did learn through it why there is no GHE to ever harm us (unless the either the mass of the atmosphere, albedo, or TSI alters)
To sum this up, the “back radiation” warming concept is null and void yet there IS an e.m. field resonance (< 235) present in the gap half caused by the inner shell, half caused by the warm sphere. Overlying this resonance are actual disturbances (= 235) always emanating when in steady state from the nuclear energy emanating from the sphere outward, the real ‘photons’, always outward. And yes Ray, this will always be less than 235 W/m². If willis wanted to prove IPCC is correct he should have made this entire example as an unrealistic set of infinite flat planes. (another place their thumb is on the warming scales)
Why don’t I just integrate this using real parameters and lets see just how big is this real difference really is, that makes it actually cooler than just imagined. Oh, that reality may be OT again.
Best regards right back Ray, good point made. Maybe TB will start another short post that allows us to look at a more real “steel shelled planet”, even where the “steel” can be partially transparent, may even have an emissivity not equal to one. You know, the actual science on the World’s Greatest Science Blog since WUWT has little real science discussions like this anymore.
(hope all of that fomatting made it through wp ok)
JWR
“Why is it strange that I agree with the 940?”
Because it’s silly, stupid, idiotic, insane and so on, that radiation from a cooler object can be absorbed by a warmer object, according to Postma, and you seemed to be one of his followers, glad if I was wrong.
Since I’m so clever with Davids drawings 🙂 here is the version for your requested emissivities
and again no ad hoc reasoning. The shell must emit 235 out and with e=1 that sets the temp to 254. Because of half emissivity on the inside there will only be 117.5 towards the planet. 235+117.5=352.5 which is what the shell must absorb from the planet and since e=0.5 (and absorptivity=0.5) the planet must emit 352.5*2=705.
In fact there is an even simpler way to do this using the positive feedback equation Gain=1/(1-B)
The inside emissivity is 0.5 so we know half of the radiation from the planet is returned from the shell. We also know that 1/3 of the remaining is returned because there is twice as much sent outwards than inwards from the shell. That is B = 1/2 + 1/6 = 4/6 and the equation becomes Gain=1(1-0.67)=3, and 235*3=705 from the planet.
Since Postma does not understand feedback this method is of course also insane nonsense to him.
I agree we are discussing more than a toy, but we are not discussing the climate system of the earth, which is what the IPCC is doing.
“Well what arrogance. Didn’t you think to ask him? I agreed my diagram with Joe after a to-and-fro discussion on his blog. 🙂 ” ~David S
It’s not arrogance, Joe knows his physics, if he agreed with your presentation it was because it was vague and he fixed the error.
Kristian confirmed this by doing a better job explaining things than I did here in this comment, which Joe said was a “great post”.
With a diagram as well: http://i1172.photobucket.com/albums/r565/Keyell/ShellPerfCond_zps0db643b0.png
To many,
I see that most people come up with the correct answers.
However the funny point is, that they indeed do not use the Prevost concept, which is also in the IPCC software, and which gives spurious absorption.
And IPCC uses that spurious absorption to alarm the general public that the planet is heating up.
Prevost said that the emission of a surface only depends on its own temperature, (and on sigma and on its own emission, but that came 100 year later).
The emitting surface can also receive heat from another surface, according to Prevost.
People subtract the two quantities to give what they call the net emission from the warmer to the colder surface. That is the two-way heat flow formulation.
The part in that net emission with the negative sign is given a physical interpretation and is called back-radiation.
***
In the one-way heat flow formulation , one starts immediately by considering a pair of surfaces, not necessarily with the same emission coefficient e.
For the pair a combined emission coefficient is defined: 1/e1+1/e2-1=1/e12=1/e21 (see Wikipedia, emissivity)
***
.
One-way heat radiation between the plates is defined by
heat flow= e12*sigma*(T1^4-T2^4) from 1 towards 2 if T1 bigger than T2 , from warm to cold and the other way around if T2 bigger than T1.
***
Prevost and IPCC claim surface 1 emits e1*sigma*T1^4
and surface 2 emits e2*sigma*T2^4
And in IPCC software: surface 1 absorbs the emission of surface 2, and
surface 2 absorbs the emission of surface 1
According to Prevost and IPCC software: quantity = e1*sigma*T1^4 – e2*sigma*T2^4
Still according to Prevost and IPCC
if quantity positive, then net-flow of heat from 1 to 2,
and they call e2*sigma*T2^4 the back-radiation from 2 which is absorbed by 1,according to IPCC,
and
if quantity negative, then net-flow of heat from 2 to 1,
and they call e1*sigma*T1^4 the back-radiation from 1 which is absorbed by 2, according to IPCC
Tim
Thank you for answering while I was asleep. Good we can provide 24/7 service here 🙂
(your scenario was not exactly what JWR suggested though)
David
“This is beyond parody”
Like about 400 of the comments.
Max
“the idea that the planet has to heat up and won’t heat the shell up accordingly is odd.”
I’m afraid mother earth does not care about your intuition.
gbaikie
See my other comment.
Max says: ” Joe knows his physics ..”
Thanks for that laugh. I needed a smile to start my day. 🙂
JWR says: “According to Prevost and IPCC software: quantity = e1*sigma*T1^4 – e2*sigma*T2^4”
1) There is no such thing as “IPCC” software. The IPCC does not carry out original research.
2) I find that it is usually dangerous to put words into someone else’s mouth, and then attack that position — that is the start of many a strawman argument! I suspect that the code used in climate models is a little more sophisticated than you might think. Do you have a reference to back up your claim that they use this equation?
3) Even your equations are simplifications. There are rather sophisticated software packages out there that do these sort of radiative transfer problems for the atmosphere. Check out MODTRAN.
lgl says: “your scenario was not exactly what JWR suggested though”
I was assuming that BOTH sides of the shell had an emissivity of 0.5.
You have only the INSIDE with an emissivity of 0.5.
It looks like we each did our our scenario correctly. It is not worth looking back to decide exactly which one JWR meant.
The main point, of course, is that both the “net flow model” or the “two flow model” work identically when done correctly.
@Tim Folkerts
You are right, when I speak about IPCC software, I am referring to K&T and similar diagrams who come up with 333 w/m^2 back-radiation. And that is coming from the huge absorption
Also Science of Doom uses the two-way heat flow description.
In the stack model which I use, I do not have back-radiation.
The LW radiation from the surface into the atmosphere is 6 W/m^2 + the 53 though the window.
Click to access PROM_REYNEN_Finite_Element.pdf
JWR
So you have found that the global average surface temperature is 179K (or 184K with emissivity 0.9), congrats, and good luck with trying to get this junk published.
@lgl
JWR
So you have found that the global average surface temperature is 179K (or 184K with emissivity 0.9), congrats, and good luck with trying to get this junk published
I do not know where you got that statement from.
JWR
You claim there is 6+53 W/m^2 emitted from the surface, which is what an object at 179K will emit.
lgl
There you go, with your concept of back-radiation.
You are good in graphics, and you know sigma*T^4 but not the physics.
Read the paper, as well as an earlier one, reference 4, where everything is explained.
JWR
You say “Also Science of Doom uses the two-way heat flow description.” There’s a good reason for that, it’s real, it’s being measured every day. Take a look at fig.2 here for instance, http://scienceofdoom.com/2010/07/31/the-amazing-case-of-back-radiation-part-three/
or go here for more measurements, http://www.esrl.noaa.gov/gmd/grad/surfrad/pick.html
You see, your mumbo jumbo can’t explain away the real world.
lgl, JWR’s figures are for the net upward IR, so his numbers are at roughly the same order of magnitude as the various measured results. His numbers do not imply a cold earth, as you were concluding.
JWR,
Your works is certainly a good start. The fact that your numbers are in the same ballpark as other estimates suggest that your model has something going for it.
A couple suggestions/questions
1) I don;t see clouds modeled in your work (other than their effect of albedo). Clouds are responsible for a large part of the downward IR to the ground and a large part of the upward IR to space (using the “two directional” terminology).
2) Your “wire grid” seems to absorb some of the light at all frequencies. But GHGs absorb all of the light at some frequencies.
As I understand it, by stacking enough wire grid layers, your model will absorb pretty much all of he surface IR. But that is not how GHGs work. They absorb specific wavelengths in short distances, but other wavelengths will get transmitted thought any number of layers.
Tim F 8:09pm: “JWR’s figures are for the net upward IR…”
Net of what?
The shell in top post physically emits from both inner and outer surface as we know since their T>0K but JWR’s paper ref. explains & shows somehow the shell would not emit from inner surface. So there is no “net upward IR” in JWR’s numbers. LW IR emitted from surface is all upward in the paper ref.d. Neither JWR nor the paper ref.d explains even though JWR says “everything is explained” by reading that paper. Not.
A reason explained in the paper is heat can’t flow from cooler shell to warmer surface which is true. But this isn’t an explanation for the correct warmer than no shell planet surface shown in top post.
What the paper doesn’t explain is that we do know the emission from cooler inner shell can slow the cooling of the warmer planet surface allowing the planet core heat source to have forced planet surface to higher temperature than without the shell. Have to watch the peas under the shells, LOL.
Tim
No, there is no ‘net’ in their model, there’s only upwards. One-way heat flow, remember?
Sorry Trick, your comment wasn’t there when I wrote mine.
suricat says:
March 28, 2013 at 3:20 am
[“3) The ‘belief’ that ‘back radiation’ can ‘warm’ (add temperature) to the ‘source of the energy’ is blatantly ‘false’. Any ‘back radiation’ is related to its ‘Planck relationship’ to the source that it came from. However, the scenario in Willis’s hypothesis involves a ‘source’ (planet) that outputs joules irrespective of infinitely variable temperatures that are ‘back-radiated’ towards ‘the planet’. The BB relationship may well permit an ‘interaction’ between the planet and the shell. Let’s look at this. No, let’s not. It’s too time consuming!”]
Ray,
Sorry, I’ve been offline fora day or so.
Thanks. I think that is the crux of the debate here. Its been fascinating. Thank you very much for all your help on the subject. I look forward to more discussion at a later date.
Regards,
wayne says:
March 28, 2013 at 8:15 am
Wayne,
Agreed! Thank you and Ray once again. All the best.
Regards
lgl,
Part of the challenge is that there are a large number of wrong ways to try to explain things, but only a small number of correct ways.
So in an equation like σ*T1^4 – σ*T2^4, we could say that “σ*T1^4” is the upward radiation and “σ*T2^4” is the “back-radiation”. Or we could say that “σ*T1^4 – σ*T2^4” is the net radiation heating “one way” from warmer to cooler. I am comfortable with either nomenclature, since they are mathematically equivalent.
I *think* that JWR is effectively trying to calculate the “net” IR, whether he wants to call it “net” or “one way”. (My main disagreement with him seems to be that he feels that “standard science” does not properly take into account the emissivities of the two surfaces or their “view factors” where I feel that these are easily accounted for in both the “two way” and the “one way” models.)
I *am sure* that I have no idea what Postma or the other “slayers” are saying.
Trick says: “Net of what? The shell in top post …”
Sorry — I was referring there to the paper that JWR wrote (not Willis’ model) that tries to calculate energy flow in as somewhat more “realistic” model.
Click to access PROM_REYNEN_Finite_Element.pdf
His “53 W/m^2” corresponds the the “40 W/m^2” atmospheric window in K&T
His “6 W/m^2” corresponds to the 356-333 = 23 W/m^2 net flow from surface to atmosphere in K&T
So his numbers are in the “right ballpark”.
lgl says as if to criticise
“One-way heat flow, remember?”
How many times do you need to be told?
Heat only flows one way.
From higher to lower temperatures.
Only in IPCC fairyland can heat flow spontaneously from cold to hot.
Every physics department and textbook agrees with Carnot, Clausius and Kelvin if you care to read or ask them.
Some like Trick and lgl seem surprised that a one way heat flow model gives all the correct answers without backradiation.
Perhaps they should look at this graph
http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html#c4
They will find that the Raleigh – Jeans – classical (no backradiation) formula gives correct answers for all radiation > 3um.
This includes all of Earths thermal radiation calculations and even covers the Willis model.
Only for wavelengths < 3um is the quantum theory essential.
And Bryan, how many times do YOU need to be told that nowhere in “IPCC fairyland” does heat flow from cold to hot?
Energy flows both ways; photons move both ways. ,But heat = net flow of energy is always from hot to cold.
Bryan says: “They will find that the Raleigh – Jeans – classical (no backradiation) formula gives correct answers for all radiation > 3um.
That is not even close to what the Raleigh -Jeans Equation says.
The formula is close to correct above 3um FOR THE SUN. The wavelengths where it is correct would be dependent on the temperature of the object. Only thermal radiation WAY above the peak in the curve is correct.
For cooler objects like the earth, it would only be correct above ~ 50 um.
“Max says: ” Joe knows his physics ..”
Thanks for that laugh. I needed a smile to start my day. 🙂 ” ~Tim F
How silly of me, thinking astrophysics was in any way related to physics, I’m sure he just managed to guess his way through a masters degree in the field. *eyeroll*
“Max
“the idea that the planet has to heat up and won’t heat the shell up accordingly is odd.”
I’m afraid mother earth does not care about your intuition.” ~lgl
1. I’m not talking about the real world, remember the point of the thread?
2. I’m not talking about my intuition, rather I am talking about that of those who think the shell will stabilize at a lower temperature than the planet.
Tim F. 10:28pm – “So his numbers are in the “right ballpark”.
Play ball for top post not earth system, the main difference being the location of the sun.
JWR referenced paper physics are not in the right ballpark for the top post set-up as the paper(s) assign 0 emissivity to the inner shell surface when the inner shell T of top post clearly is greater than 0K. This was Willis’ main point IIRC.
If not, my point is again:
JWR & the paper (s) don’t explain how we do know the non-zero radiation emitted from cooler inner shell can slow the cooling of the warmer planet surface allowing the top post planet core heat source to force planet surface to higher temperature than without the shell.
We can know this from everyday experience once the furnace is moved inside the “shell.” Don’t need any sigma’s or T^4’s for that, LOL.
Trick says: March 30, 2013 at 1:17 am
“JWR & the paper (s) don’t explain how we do know the non-zero radiation emitted from cooler inner shell can slow the cooling of the warmer planet surface allowing the top post planet core heat source to force planet surface to higher temperature than without the shell.”
That’s an easy one for BB stuff Trick. Radiation to 0k is what is expected of the Boltzmann equation, but when radiation is directed to a ‘warmer body’, the ‘potential’ of ‘radiative pressure’ is reduced (‘temperature difference’ enables energy transport by radiation). 🙂
Best regards, Ray.
I see Max still doesn’t understand thermal gradients across systems with heat sources and sinks.
A thermal gradient set up by radiation through a vacuum.
[Reply] Yes, fill a thermos flask with boiling water and check it out.
Is the shell a sink, or is the space outside of it a sink?
[Reply] Both, and you are a time sink.
suricat 1:43am: Not sure what you mean. Objects with T>0K enable radiative, conductive, and/or convective energy transport both ways. One object higher temperature than the other object temperature enables heat transport one way.
I’ll add that the only gradient which would exist is the one due to geometry, otherwise there is no mechanism capable of producing this effect… or should I say, no realistic physical mechanism.
[Reply] Sounds sciency.
Tim Folkerts says
“The formula is close to correct above 3um FOR THE SUN”
Read again
http://hyperphysics.phy-astr.gsu.edu/hbase/mod7.html#c1
Tim Folkerts says
“And Bryan, how many times do YOU need to be told that nowhere in “IPCC fairyland” does heat flow from cold to hot?2
lndeed it does not but several supporters of IPCC ‘science’ on here have said that it does.
Tim Folkerts says
“Energy flows both ways; photons move both ways. ,But heat = net flow of energy is always from hot to cold.”
Why state the bleedin obvious?
Do you not remember that you were forced to make a groveling apology for saying I said something different?
Tim
“So his numbers are in the “right ballpark”
How can you say that when half of the numbers are missing (close to zero that is)?
From his ref.4 http://www.tech-know-group.com/papers/IR-absorption_updated.pdf
“We see from table 2:
– Back-radiation of 324, 330, 315 by the IPCC authors with the two-way
heat propagation formulation and 0.0 for the multilayer one-way stack
model.
– Atmospheric absorption of 350, 355, 315 by the IPCC authors with the
two-way heat propagation formulation and 23 for the multilayer oneway heat propagation formulation.”
Can’t believe I’m spending time on this utter nonsense.
lgl,
Like you, I prefer the “two way model” since it represents the energy of the photons moving each way. There are photons created by the cooler shell that fly through space and that get absorbed by the warmer planet.
But there is nothing per se incorrect about doing subtraction and talking about the “net” flow”. The only problem comes when people start to think that somehow the photons moving from the cooler to the warmer are somehow “destroyed” or “pushed back” by the more numerous photons moving the other direction.
Bryan says: “Read again”
( http://hyperphysics.phy-astr.gsu.edu/hbase/mod7.html#c1 )
It is never a good sign when someone is apparently learning physics on the fly from a webpage!
Bryan, did you even read or understand the page you were linking to? The page specifically says the two formulas are approximately equal when
f << kT/h
Notice that temperature is right there in the formula. In other words, if you double the temperature, then you double the frequency at which the two formulas give similar results. Or if you cut the temperature by ~ 20 (say from 5800 K for the sun to 290 K for the earth) then you cut the frequency where the formula works by a factor of 20 and increase the wavelength where the formula works by a factor of 20.
For the sun, the two agree pretty well above ~ 3 um.
For the earth, the two agree pretty well above ~ 60 um.
*****************************************************
…. and THAT is why I state "the bleedin obvious".
* because it is "bleedin obvious" that the temperature matters in these equations.
* because it is "bleedin obvious" that many people DO think the photons only go one way.
* because it is "bleedin obvious" that your contracting shell still doesn't violate the 2nd Law of Thermodynamics.
* because it is "bleedin obvious" that the planet will be ~ 2^(0.25) times warmer than the shell, as Willis and Tallbloke and I and others have been saying from the beginning!
Lets ask him then.
JWR, if you are still around, how much back-radiation is there? and how much is the surface emitting? The actual numbers of course, not net numbers.
lgl
Of course there is no back-radiation of heat!
It is said explicitly, in the two papers.
You can see it in the figures;
ORL(in red) = mechanisms other than LW radiation +surfaceLW (in green)
surfaceLW(in green) = qwindow (in cyane) +LW into atm.
In fact the whole discussion is about the fact that back-radiation of heat does not exist.
In the December paper I found 16W/m^2 for qatmLW-in, in the march paper where I introduced the viewingfactors and the more correct distribution of CO2 I find 6W/m^2.
Of course there is an electromagnetic signal from the atmosphere telling the surface: do not send me the 390W/m^2( sigma*Ts^4), because I am not at zero K.
That is a signal, some people call it energy which is not very correct, but certainly not a heat flux.
In table 2 to which you refer, back-radiation for IPCC authors is 324,330,315 (and zero for me),
and the absorption is 350,355,315 (and 16 (December) resp 6 (march) for me.
The difference is 26,25, 0 and (16 or 6 for me).
In the K&T diagrams and table 2 you see the same values for mechanisms other than radiation.
Nasa has now also K&T diagrams without back-radiation of heat. I can’t find now the link, I will send it later.
As concerns measurement of back-radiation of heat, that is impossible, because it does not exist.
You can measure remotely the temperature of the atmosphere, and you can put that signal into Stefan -Boltzmann sigma*temperature^4. But claiming that it is a heat flux is not correct.
On those gadgets there is a switch: degrees? or W/m^2.
Since you have been looking in the December paper, you see a stack of N completely opaque slabs,
with the results, with the one-way and with the two-way formulation. The results for temperature are the same for the two formulations.
If you take one slab, then you have the results for an iron shell around the globe!
That example was given in that paper to show that the two-way formulation give spurious absorption.
lgl
This is a K&T diagram of NASA, without the non-existing back-radiation
JWR 5:17pm – The earliest provenance I can find of the NASA chart you link is from a high school teaching document NASA prepared in 2003 of problem based modules. The chart was intentionally set up with the downwelling atmosphere IR radiation (DWIR) removed to teach the HS students how to think of DWIR:
“Convection, evaporation, and radiation from the surface exceed the total amount of energy that was absorbed by the surface to begin with! This is impossible unless there is a missing element of the energy budget. In fact, there is: the Earth’s atmosphere contains water vapor, carbon dioxide, and other greenhouse gases, which absorb energy radiated toward space and then emit some to space and some back to the Earth’s surface.”
Unfortunately it seems this incomplete NASA HS teaching diagram has been given a life far beyond its original & incomplete intentions.
Here is the complete NASA Energy Budget diagram citing TFK09 correctly showing the DWIR from Earth’s atm. – this is the same as the DWIR that is emitted from the shell inner surface in the top post except of course Willis moved the forcing (sun) into planet core to make a well founded physical point about DWIR being non-zero when the shell inner surface T>0K:
http://earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
JWR 5:17pm – Here is the June 2003 NASA teaching document I clipped the quote from – about a 1MB download and where your linked incomplete Energy Budget chart comes from AFAIK:
Click to access 62319main_ICS_Energy.pdf
Tim Folkerts trashes the well regarded physics resource hyperphysics
( http://hyperphysics.phy-astr.gsu.edu/hbase/mod7.html#c1 )
“It is never a good sign when someone is apparently learning physics on the fly from a webpage! ”
Whats wrong with this perfectly orthodox physics link?
Tim thinks that only 10um radiation from the Sun qualifies for near equality for classical and QM treatments.
If the 10um radiation comes from say 256K does it not qualify?
Where does it say this in hyperphysics?
Look at the graph again it shows clearly that the two formulas converge around 3um
Can Tim find a source from anywhere to back his position?
Rather more amusing is Tim’s hope that no one advocating IPCC science states that heat can move spontaneously from a colder to a hotter object.
But lgl and Trick would disagree with Tim.
Its all happening here at tallblokes
Tim
I told you…
“Of course there is an electromagnetic signal from the atmosphere telling the surface: do not send me the 390W/m^2( sigma*Ts^4), because I am not at zero K.”
“As concerns measurement of back-radiation of heat, that is impossible, because it does not exist.”
Utter nonsense, but thank you for the confirmation JWR.
Bryan
Like I have said before, nobody is claiming there is a net heat flow from colder to warmer, but there is thermal radiation in both directions. Since you are unable to understand the difference, don’t bother.
JWR, Bryan et al
Don’t bother opening this link. http://scienceofdoom.com/2010/07/24/the-amazing-case-of-back-radiation-part-two/
These spectra do not exist. The instruments used do not exist. The measurements have never been done. All just fabricated by mad scientists.
lgl
Backradiation is not heat.
Dont take my word for it listen to Tim Folkerts.
On this point he is absolutely correct.
Bryan
But when it is absorbed it is converted to kinetic energy, the object gets warmer than it would have been without the absorbed radiation, get it?
lgl says
“But when it is absorbed it is converted to kinetic energy, the object gets warmer than it would have been without the absorbed radiation, get it?”
When a hot object emits radiation it loses internal energy and temperature drops.
If absorbed by a colder object the colder gains internal energy and temperature rises.
If the hotter object has no additional power supply the absorbed radiation from the colder object CANNOT INCREASE the temperature of the hotter object.
Get it?
If however the hotter object is supplied with an additional source of energy its temperature may increase.
This will happen if the new energy supplied by the power source is greater than the power loss of the warmer object.
Bryan says: :”Tim Folkerts trashes the well regarded physics resource hyperphysics”
No, I only trashed your completely confused interpretation of the perfectly fine physics presented there. Heck, I even used that same Hyperphysics link to show specifically what they did say!
Bryan further says: “Look at the graph again it shows clearly that the two formulas converge around 3um”
What the graph ‘clearly shows’ is that the two converge about 6x above the peak — the intensity peaks around 0.5 um for sunlight; the two converge around 3 um. For earth around 300 K, the peak is around 10 um and the two would converge around 60 um.
It is really not tough if you understand the math.
Bryan
You’re almost there. Just change that last sentence to “This will happen if the total energy supplied by the power source and the back radiation is greater than the power loss of the warmer object.
Since we seem to mostly be going around in circles, I think it really is time to bow out with a few last comments.
* Even though JWR has a poor verbal understanding of thermal radiation, his equation for the NET heat from thermal radiation seems to be correct. Even thought JWR thinks that his equations are new/different/superior, they really just are the equations from “standard physics”.
* Heat is the NET flow of thermal energy. “Forward-radiation” is “energy”, but it is not “heat”. “Back-radiation” is “energy”, but it is not “heat”. The difference between the two is “heat”.
* Any energy that gets absorbed by an object will tend to raise the temperature of the object. Any energy that leaves will tend to cool the object. The temperate at steady-state will depend on ALL energy inputs and outputs. Since “back-radiation” is energy, it will be PART of what determines the temperature.
I’m the one called a time sink while the above goes on?
“[Reply] Yes, fill a thermos flask with boiling water and check it out.“~ tb
I assume that the interior surface of the vacuum chamber in the dewar flask will begin radiating towards the outer surface of the vacuum chamber and somehow raise the temperature of the interior surface?
Note that initially before thermal equilibrium has been reached the interior surface will be warmer than the outer surface, though neither will ever be warmer than they started.
[Reply] The ability to miss the point is strong in some. The ThermosTM flask example was to demonstrate a thermal gradient, not to create a replication of Willis’ toy planet setup. You’d need to add a heating element for that. Then you’d find the system would indeed internally increase to a higher temperature due to the vacuum than it would without it.
I’m not sure if everyone here understands why there is a vacuum in a dewar/thermos/vacuum flask.
The presence of a vacuum or near vacuum prevents heat transfer through conduction and convection (except where the container lip makes contact) and the mirrored surfaces prevent radiative heat transfer.
It is the mirrors (preventing radiative losses) in conjunction with the vacuum (preventing conductive/convective losses) which would be responsible for the system heating up, not the radiative insulation ability of the vacuum by itself.
__________
“ The amount of heat that a spacecraft radiates into space and receives from the Sun can be controlled by the makeup of its surface. And this is the second secret of the vacuum bottle (or thermos): while the vacuum suppresses heat exchanges by conduction and air convection, exchange by radiation is suppressed by the shiny metallic coating of the bottle. This shiny coating reflects the heat radiation like a mirror and keeps it either inside the bottle (if the content is hot) or outside (if the content is cold).
A thermos bottle gets rid of convective heating or cooling with a vacuum, and only a small amount of diffusion happens through the top and the glass wall. The silvering of the glass helps limit radiative heating or cooling. But the thermal radiation is ALWAYS there, and that is what a spacecraft uses. To get rid of heat, you can point thermal radiators at the dark sky, and to warm up you can point at the Sun or Earth. The Sun warms the Earth through radiation, not convection or diffusion.” ~NASA FAQ
__________
“It is usually necessary to maximize the heat transferred by only a single mode in order to obtain
adequately low thermal resistances within equipment. Even though a complete cooling system
may include three modes of heat transfer, each particular heat path will usually emphasize a single
mode. Where a single mode dominates, other modes can often be ignored. For example, with
conduction as the predominant mode for parts operated in a vacuum, the conductive thermal
resistance can be made low by the use of thermal shunts. The heat transferred by radiation and
convection is almost negligible. That is, in the electro-thermal analogue,
the shunt thermal resistances due to radiation and convection are so large
that they are insignificant for design purposes” ~NASA
__________
More fun reading @Project Rho, as well as this bit on TCS design.
__________
Oh, and just for fun, if one layer receiving a constant input with a vacuum and another layer causes the layer receiving constant input to heat up, and adding other layers above the outer layer increases this effect, why does the Webb sunshield have so many layers?
I assume the silvered surfaces have nothing to do with it.
I do have to admit I was mistaken though. I incorrectly stated that a solid block the size of the gap would result in it being the same temperature at the top/outer layer.
“Why does the sunshield have five layers instead of just a single thick one? Each successive layer of the sunshield is cooler than the one below. The heat radiates out from between the layers, and the vacuum between the layers is a very good insulator. One big thick sunshield would conduct the heat from the bottom to the top more than 5 layers separated by vacuum. ” ~NASA
The solid block would be warmer, as the gaps would make it cooler at the top.
Though again, these are silvered layers, but hopefully that will help make my point clearer.
[Reply] Silvering creating mirrored surfaces on Dewar flasks suppresses radiative losses as NASA said. But it doesn’t ‘prevent’ them as you said. Mirrors get hot in the sun. Hotter than the air around them once the thermal mass of the glass has reached a stable temperature. Some useful research you’ve done here Max, Tim C and I were discussing whether introducing stuff about MLI on spacecraft would be a help or further confuse things on the phone yesterday.
I’m sorry, I was wrong.
“The principle behind MLI is radiation balance. To see why it works, start with a concrete example – imagine a square meter of a surface in outer space, at 300 K, with an emissivity of 1, facing away from the sun or other heat sources. From the Stefan-Boltzmann law, this surface will radiate 460 watts. Now imagine we place a thin (but opaque) layer 1 cm away from the plate, thermally insulated from it, and also with an emissivity of 1. This new layer will cool until it is radiating 230 watts from each side, at which point everything is in balance. The new layer receives 460 watts from the original plate. 230 watts is radiated back to the original plate, and 230 watts to space. The original surface still radiates 460 watts, but gets 230 back from the new layers, for a net loss of 230 watts. So overall, the radiation losses have been reduced by half by adding the additional layer.
MLI covering the heat shield of the Huygens probe
More layers can be added to reduce the loss further. The blanket can be further improved by making the outside surfaces highly reflective to thermal radiation, which reduces both absorption and emission.” ~the Wiki
We were all wrong in various ways though.
If the initial surface gave off 235 Watts, the addition of a thin but opaque layer 1 cm away and thermally insulated with an emissivity of 1 would allow the new layer to cool until it was radiating 117.5 Watts from each side, the original surface still radiates 235 Watts but gets back 117.5 from the added layer for a net loss of 117.5 watts to the layer and to space.
[Reply] Hurrah! You are three-quarters of the way to enlightenment. Now, tell me, if the original surface continues to radiate 235 Watts while only 117.5 is lost to space, do you think the inner surface will get hotter or stay at the same temperature?
If we take another look at the graphic that wayne pressented on March 24, 2013 at 8:31 pm:
We can see/observe that, thanks to wayne’s offering, the individual radiances at each frequency/wavelength supply differing ‘radiant potential’ (energy at that frequency/wavelength) for each temperature. OK?
The ‘yellow curve’ (remainder) represents the ‘blue curve’ (higher temperature) after the ‘magenta curve’ (lower temperature) has been subtracted from this ‘BB’ (Black Body) model/graphic. This shows that the ‘higher temperature’ (doubled Boltzmann energy transmission) contains a greater energy density than the ‘lower temperature’ (Boltzmann energy transmission). Herein lays an anomaly.
The anomaly is that more energy will always be transmitted to the lower energy/temperature body than can be transmitted to the higher energy/temperature body when ‘photons’ are *exchanged* between them.
However, should there be NO ‘lower temperature body’, the ‘higher temperature body’ would radiate ALL its energy to space/vacuum because ‘there is NO photon *exchange*’ at all.
Thus, it should be realised that “back radiation” is nothing ‘more’ or ‘less’ than “insulation for a radiation type of energy transfer”! However, it really can be confusing when the ‘insulation factor’ is included for an “energy transport” type budget!!!
I can only hope that this post provides ‘some’ understanding of the Willis Model. 🙂
Best regards, Ray.
So, Willis steel greenhouse doesn’t doesn’t increase the temperature of core surface.
Or adding multiple blackbody shells does not increase the temperature, one could say
that ideal blackbodies “were made” with the idea that they don’t increase the temperature.
And that sandwiching steel in between them does make much difference [as long as one
dealing with modest amounts of energy {235 W/m-2]}] as steel or any metal is pretty good
conductor of heat.
Next one might ask what could increase the temperature, and the answer *could be* that if one
used just about anything other than ideal blackbody surface for the shell [not something designed to perfectly absorb and emit radiant energy].
Now, can anything decrease the core surface temperature? One answer is one can increase the core surface’s surface area, but if you have more surface area at core and still had the blackbody shell which hasn’t had to increase surface area [it had a smooth outer surface] then increasing the surface area of surface of core, doesn’t make the core surface much cooler.
But say one had checker board of 1 km squares, with the black [or red] squares has emission of 1 and white squares had emissivity of .02. The temperature of the white square would higher temperature than black squares and all the squares would emit about 235 W/m-2. Or these black squares lower the core surface temperature in comparison to the white squares.
A planet warmed by sunlight, the white square would absorb less radiant energy, but it’s about amount energy absorbed over time, or the white squares do not heat up as fast as black squares.
But if you stop heat loss from convection of air, a shiny object can reach around the same temperature.
So a shiny surface on planet with a sun, has to do with have energy is absorbed rather relating to
it’s temperature, and thermal heated planet of Steel Greenhouse a shiny surface increases the temperature because it reduces the loss of heat from the core [and so core heats up].
So a surface which has emissivity of .02 radiating 235 W/m-2 is hotter than surface with emissivity
of 1 radiating 235 W/m-2. So with Steel greenhouse it’s pretty easy to make the surface have a higher temperature as compared to it’s temperature if it were merely a blackbody surface.
But can you this if instead of planetary thermal energy, the energy is coming from the Sun?
And one could also ask how warm can Earth get from Sunlight.
Or the averaged amount energy received from sun is 240 W/m-2 and emits this much. This how
much energy is absorbed and radiated. So to increase the temperature one could absorb more energy or emit less energy. If increase amount absorbed by same amount one emits, you are not getting much warmer, or you absorb less and emit less by the same amount, you aren’t getting warmer.
Or a with blackbody and no atmosphere you adsorb 340 W/m-2 and emit 340 W/m-2 and one has uniform temperature of 5 C if the surface has emissivity of 1.
If the blackbody wasn’t a blackbody in terms of it’s emissivity- say .5 rather than 1 it would much warmer than 5 C.
And if blackbody absorbed less than 1 and emitted 1 it would be cooler. Which what described in Greenhouse effect theory: absorbs .7 emits 1 and gives -18 C uniform global temperature.
Or if a blackbody received 1360 times .7 which is 952 and it was emitted over surface of sphere
[divide by 4 = 238] it’s 238 W/m-2 and with emissivity of 1 it indicates a temperature of around -18 C. But if the emissivity was lower than 1, it would higher uniform temperature than -18 C.
Now one could argue that Earth has emissivity of around 1. Water is:
Water 0.993 – 0.998
Ice 0.98
Snow 0.969 – 0.997
Sand 0.949 – 0.962
Granite 0.898
Green Grass 0.975 – 0.986
http://www.engineeringtoolbox.com/radiation-heat-emissivity-d_432.html
And earth surface is covered by 70% water.
But also can argue that Earth surface doesn’t absorb on average 70% of
sunlight and addition water absorb energy in a different way than any theorized
blackbody surface.
Let’s return to the blackbody world in a vacuum which has uniform temperature of 5 C.
Such world would be warmer than Earth and be radiating more energy than Earth: 340
vs 240 and it would be refrigerating the tropics at mad rate and dumping huge amount of energy
into the poles during it’s winter. Without getting into details, such warmth makes our entire ocean at 5 C and we have no polar ice.
So reasonable to say it would be warmer world than Earth.
Though Humans would feel fairly cold [as they evolved as a tropical beast], but plants would be happy as they would never freeze. So tropical plants could do ok in polar regions, and tundra and other plants with it’s adaption to cold weather would go extinct.
So I would argue it’s possible this world has never seen an average temperature as cold
as 5 C, but a uniform temperature of 5 C is warmer then we got.
So one should not confuse, uniform temperature and average temperature
The point is, that if blackbody world in a vacuum wasn’t doing such through in job of keeping the entire planet warm, it could have some regions which are warmer. It could mimic earth, warmer tropics and other half of the world be cooler and in this way lower it’s emissivity- have higher average temperature.
Now is there any parts of the worlds which absorb more per square meter than the average of 70% of the energy of the Sun. So 24 hours absorb 240 W/m-2 per second. So 20.7 million joules
heat created per square meter.
To increase a cubic meter of water by 1 C requires. At 5 C:
4.204 kJ/kgK). So 1000 kg by 1 C is 4.204 million joules. So about 5 C increase.
So air temperature 10 C, water 5 C, sunlight for one day with evaporation inhibited,
can it warm by 5 C? In kilowatts hours 5.76 Kilowatt hours of thermal heat.
Hmm, it’s possible, but solar water heaters get about 60% efficient and therefore
require about 10 kilowatts hour per day to get more than this. So someplace can
get this much per day during summer. Not Germany- the solar capital of the world:)
Not in summertime not ever..
And in a ocean the top 1 meter of water is not going to warm by 5 C in one day, anyplace
in the world. But if this is average than there should areas which exceed it.
Let’s look at a sidewalk, say starts at 20 C and warms to 50 C.
Concrete, stone: 0.75 kJ/kg K
http://www.engineeringtoolbox.com/specific-heat-solids-d_154.html
So 10 cm thick: 200 kg. So 150 kJ per degree. 30 times 150
is 4,500 KJ. 4.5 million joules. So sidewalks going to at best warm up
by 1/5 or 1/4 of the amount needed.
So if think of surfaces they are not absorbing this much energy.
So average ideal blackbody surface which had 30% sunlight blocked
would absorb 20.7 million joules.
The top one meter of ocean is not absorbing
this much nor are any land surfaces- water going to be better than land.
Now the top 10 meter or 100 meter of ocean may be adsorbing this much
energy or more. Mainly because they could be more efficient at storing heat,
than compared to 60% efficient solar water heaters.
Solar ponds:
“This means that the temperature at the bottom of the pond will rise to over 90 °C while the temperature at the top of the pond is usually around 30 °C”
http://en.wikipedia.org/wiki/Solar_pond
Blackbodies can’t do this.
Blackbodies in greenhouses can’t do this.
And keep in mind this temperature is under about 1 meter of water, so in order for
one to get such high temperatures, one has to have a high intensity
of sunlight passing thru the first meter of water.
Asphalt in the sun doesn’t get this hot, nor inside parked cars,
nor in insulated boxes.
This requires water and salt- something common in nature
and in terms of storing heat per square meter of incoming sunlight,
I think it’s reasonable that oceans out perform solar ponds in terms
total heat stored. So Earth is warmer covered with oceans, than it
would be covered by solar ponds.
Trick says: March 30, 2013 at 2:44 am
“suricat 1:43am: Not sure what you mean. Objects with T>0K enable radiative, conductive, and/or convective energy transport both ways. One object higher temperature than the other object temperature enables heat transport one way.”
It’s hard to get near to “0K”. We need a ‘protected environment’ to get below the ‘CMB’ (Cosmic Microwave Background) temperature of ~4K. However, the ‘toy’ we speak of only permits the “radiative” components of energy transport.
I concur with “heat transport one way” as being ‘energy transport’ only (discussion/elucidation on this definition seems prohibited within this thread. 🙂
Best regards, Ray.
“JWR says:
March 30, 2013 at 5:17 pm
lgl
This is a K&T diagram of NASA, without the non-existing back-radiation
http://principia-scientific.org/images/NASA_Energy_Budget_Fig_2.jpg”
I would say more than 3% is absorbed by water droplets of clouds, and
the atmosphere doesn’t absorb the sunlight’s energy.
So 16 % of sunlight is prevented from reaching a sensor detecting the sunlight
at the surface [probably more than this as there isn’t 1000 watts per square
during most of a clear day in most places on Earth on average].
It does reach the sensor because it’s mostly blocked by H20 in various forms- gas,
water, and ice [and water and ice can absorb this energy- evaporate mostly].
But much of it absorbed and re-emitted, and/or scattered and diffused. So
it’s blocked from reaching sensor but not warming gas.
“Herein lays an anomaly.
…
Thus, it should be realised that “back radiation” is nothing ‘more’ or ‘less’ than “insulation for a radiation type of energy transfer”! However, it really can be confusing when the ‘insulation factor’ is included for an “energy transport” type budget!!!”
Ah Ray, I think you may now be seeing what I kept trying to point out a few times above. I have no problem with your description as “insulation for a radiation type of energy transfer”, not at all, clear as a bell… that is the underlying harmonic oscillators I spoke of, my way to describe it. It’s just the lower frequencies of a portion of the blue spectrum that has to perfectly match the magenta to be, guess what, harmonic oscillators. How can something be harmonic and oscillate unless frequencies AND intensities match exactly? Yep, because that is present (radiative insulation… I like it) from the inner shell the spheres internal energy (235) will add on top and to the left (h.f.) portion (yellow) to form the blue spectrum over that lower-grade portion that is oscillating. Is that pretty close to what you are saying too?
lgl
You think that absorption of backradiation must always increase the temperature of the absorbing hotter surface.
This is despite no Physics textbook in the world agreeing with you.
A version of the Willis model can show that you are wrong,
This time its the shell that is warmer than space (lets say 350K) and the planet is not there but the vacuum is.
Now insert an object at 250K at the centre of the hollow sphere.
The 250K colder object will radiate towards the warmer shell.
The radiation will be absorbed by the shell
But no one thinks (I hope) that as a result, the temperature of the shell will increase.
Bryan,
I posed this exact problem/question some time ago on this thread (March 18, 2013 at 11:33 pm)…
No-one provided answer. (TF attempted but misinterpreted the question as seen in the next comment…)
I think it is the crux of the model debate.
“Bryan,
I posed this exact problem/question some time ago on this thread (March 18, 2013 at 11:33 pm)…
No-one provided answer. (TF attempted but misinterpreted the question as seen in the next comment…)
I think it is the crux of the model debate.
If the toy planet roles were reversed, and the shell was heated by the same constant power source with a much colder (but above 0K) planet inside, would the shell still get warmer according to Willis’ theory?”
According to his theory but not according to the Model as he defined it.
So heated shell with colder planet, would start radiating 235 divide by 2 in either direction, but after
cold planet warmed up, it would radiate the heat outward.
But a cold planet would take millions of years to warm up from the 117 1/2 W/m-2 which within a few decades may amount to less than 10 W/m-2 [so because of heat gradient and low power which it’s heating a mile underground- a million years wouldn’t enough time for planet to stop gaining some heat.
Though if one heats the shell so radiating 235 W/m-2 on both sides when planet is cold, it’s more power [twice] than one side of spherical surface radiating 235 W/m-2
In terms of the temperature of shell, when one has cold core, shell would be cooler, but as planet surface warmed so will the shell temperature increase, until it’s about the same temperature as if instead the core were providing 235 W/m-2 [about -18 C].
Bryan
“The 250K colder object will radiate towards the warmer shell.”
Almost there again.
Yes it will, but the shell will radiate even more towards the planet and warm it. There will be a net energy transfer from the shell to the planet so the shell will cool.
lgl says, March 30, 2013 at 7:36 pm: JWR, Bryan et al…Don’t bother opening this link. http://scienceofdoom.com/2010/07/24/the-amazing-case-of-back-radiation-part-two/
These spectra do not exist. The instruments used do not exist. The measurements have never been done. All just fabricated by mad scientists.
Well said 🙂
If you haven’t already come across it, there is a recent TB article devoted to the measurement of downwelling atmospheric radiation using pyrgeometers. [snip]
[Reply] and you can go there to discuss them.
Bryan says, March 31, 2013 at 8:32 am: lgl, You think that absorption of backradiation must always increase the temperature of the absorbing hotter surface.
Bryan,
How many times do you have to be corrected? lgl and others have said nothing of the sort.
Nobody in this or any other recent TB thread that I have seen suggests that radiation from a cooler radiating body directed towards a warmer radiating body, will cause that warmer body to increase in temperature. What is does do is to offset the cooling rate of the warmer radiating body.
Prevost’s Theory of Exchanges (1791).
Just to complete David’s point: The effect of slowing the rate of cooling in a closed system would never lead to an increase in temperature of the warmer body, however in a dynamic system such as Willis’ model, it can. This is because newly generated heat is being radiated from the core’s surface at the same time back-radiation from the shell is being absorbed and re-emitted.
In a static system entropy operates to equalize temperatures between the system components.
In a dynamic system, entropy still operates, but the system now includes the universe, which is cold. The nuclear core will end up cold too, but not for a very long time. In the meantime, the system will configure the temperatures of its components in the way it must for steady state to be reached. In the case of Willis’ toy planet, this will be when the core surface and shell are in radiative balance, both emitting the same amount of energy, such that the shell loses to space what the core produces.
The steady state will be reached when the iterative process described here a fortnight ago reaches equilibrium, which is when the core surface emits 470, and the shell emits 470, half out to space, balancing the production from the core, and half in towards the core.
However, this is the situation with a vacuum between the shell and the core surface, and doesn’t resemble the situation on Earth, which has a freely convecting atmosphere.
David
an international meteorological conspiracy – right …
No I haven’t read that thread but thanks for the summary, then I don’t have to torture myself reading it.
“The new layer receives 460 watts from the original plate. 230 watts is radiated back to the original plate, and 230 watts to space. The original surface still radiates 460 watts, but gets 230 back from the new layers, for a net loss of 230 watts.” From Max’s NASA example
Please note the 230watts is not added back on top of the 460 watts from the original surface. Per this NASA example Max found Willis’ steel planet is wrong as presented. This is a heat shield.
Theres are no places in any radiative heat transfer equation to add back in the radiation from the shield or other surface.
A typical equation for radiation shield is
q/a = 1/2* SB* (T1^4-T2^4) / (1/e1 + 1/e2 – 1)
So that q/a(1-3) =q/a (3-2) = q/a
As Max said the correct answer is the shell puts out half on each side of the original 235 if you wish to treat it as a heat shield.
If you find an equation that adds back in the radiation please show me.
I thought this problem seemed familiar for some reason.
“Consider a black sphere of radius R at temperature T which radiates to distant black surroundings at T = 0K.
(a) Surround the sphere with a nearby heat shield in the form of a black shell whose temperature temperature is determined by radiative equilibrium. What is the temperature of the shell and what is the effect of the shell on the total power radiatied to the surroundings?
(b) How is the total power radiatied affected by additional heat shields?
(Note that this is a crude model of a star surrounded by a dust cloud.)” UC Berkley
“Solution:
(a) At radiative equilibrium, J – J₁ = J₁ or J₁ = J/2. Therefore T₁⁴ = T⁴/2, or T₁
(b) The shield reduces the the total power radiated to half the initial value. This is because the shield radiates a part of the energy it absorbs back to the black sphere.“
Max, that fig 1.8 seems to be badly labelled. ‘Black sphere’ should be on one side, and ‘Shield’ should be on the other. Anyway, good to see an ‘official’ publication which vindicates the position taken by myself, lgl, David Socrates and some others.
Are we all happy that Willis’ toy planet theory is correct now?
Uh, how do you figure that it vindicates Willis’ take on things?
Black sphere ||J-> | surroundings
——————||<-J₁|–
————————-^ shield
At radiative equilibrium J-J₁=J₁ or J₁ = J/2
Max, the way I’m reading it, at radiative equilibrium, j is radiated by the sphere, j/2 is radiated back towards the sphere by the shield and j/2 is radiated outwards to surroundings by the other side of the shield. This is exactly what Willis’ model proposes is it not?
Yes, you have interpreted it correctly, Tallbloke.
The model that Max presented is slightly different than Willis’ model. This new model keeps the TEMPERATURE of the inner sphere constant (with half as much power required after the shell is in place). Willis’ model keeps the POWER the same for the sphere (with the temperature increasing as a result of teh shell). But the net result is the same: the sphere is warmer than the shell by a factor of 2^(0.25).
KInda cool how 100 year old physics still applies. 🙂
MaxTM, face it, you have approached this every way but upside down! 😉
But Willis’s example is not really just a planet with a steel shell, Willis’s designed this example for all intents and purposes to represent a spherical nuclear reactor, rods always fully out, it cannot be shut down, with and without a steel vacuum containment in place close about it. I just don’t understand why you cannot draw the parallels and know exactly what happens. If you add more and more layers of steel with vacuum gaps between, don’t you see the eventual melt down at some given added layer ‘n’ (unless radius differences are very large)? And don’t you draw from that with just that one shell that the sphere (reactor) is going to have a higher temperature than with no containment?
Also, this is not a parallel to the Berkeley example. Their example is flat and has constant temperature on the hot side, not constant convertion of matter to energy, big difference.
The question posed by UC Berkeley is not ‘official’. It is solved similar to Willis’s model because they teach a model similar to his.
I notice the black lines on the inside surface of the top layer of the sphere. They are there to indicate no radiation enters here. But why not?
Like Willis. Shell radiates up and down. Sphere only up. Not consistent.
There are other problems with the physics
“Max, the way I’m reading it, at radiative equilibrium, j is radiated by the sphere, j/2 is radiated back towards the sphere by the shield and j/2 is radiated outwards to surroundings by the other side of the shield. This is exactly what Willis’ model proposes is it not?” ~tb
Yeah, at radiative equilibrium 235 is radiated by the sphere, 117.5 back towards the sphere, 117.5 outwards.
The example made of a star surrounded by a cloud of dust is quite enlightening, actually.
Stars don’t heat up their surrounding dust clouds until they glow like stars themselves, why would anything else?
They do heat them up, and when viewed from the right direction they can be very dramatic, but a large part of the impressive appearance is the juxtaposition of starlight against shadows and clouds.
_____________
Willis proposes that 2J is radiated by the sphere, J is radiated back, and J is radiated outwards, or so I thought.
[Reply] The key phrase is ‘at equilibrium’. On further reflection, I think the problem posed is ambiguous, and inconclusive for our purposes.
Whoops, I just noticed the link was using the initial page I went to from the google search, not the one I intended to link to: problem 1023.
Can anyone give us an example of a planet with a nuclear core surrounded by a steel shell , I am sure there are many examples of stars being surrounded by dust clouds in the universe
“The steady state will be reached when the iterative process described here a fortnight ago reaches equilibrium, which is when the core surface emits 470, and the shell emits 470, half out to space, balancing the production from the core, and half in towards the core.”
One can have situation where one insulates the core surface so it’s input is 235 W/m-2 and it’s
heated to temperature in which it would radiate 470 W/m-2.
This can be done using reflective insulation in the vacuum.
So say core surface has average emissivity of .9, and it gets covered by a material which has emissivity of .45, which requires the surface to heat up so the core surface has twice as much being emitted so that 235 W/m-2 is emitted from this .45 emissivity surface.
It doesn’t matter much where this .45 emissivity surface is. It could lying on the surface, it could a 1000 feet above the core surface or it could a foot below the shell. The material is say .2 mm thick is designed to have one side of it have emissivity of .45 or half the emissivity of core surface.
One could make the other side of be quite reflective so it heat up core surface quicker- though
whatever kind of surface it is, main point is to have the one side have 1/2 the emissivity core surface.
The question is, will it make any difference and what kind of difference would it make if the insulation is removed after the point in which a equilibrium in temperature of core surface
and shell temperature is reached?
gbaikie
Using a shell with emissivity 0.5 will make the core radiate 940 W/m^2. Already forgot?
… with emissivity 0.5 the feedback factor becomes 0.75 and Gain=1/(1-0.75)=4
“[Reply] The key phrase is ‘at equilibrium’. On further reflection, I think the problem posed is ambiguous, and inconclusive for our purposes.” ~tb
The part that makes it unambiguous in my mind is: “(b) How is the total power radiated affected by additional heat shields?
(Note that this is a crude model of a star surrounded by a dust cloud.)
(b) The shield reduces the the total power radiated to half the initial value. This is because the shield radiates a part of the energy it absorbs back to the black sphere.“
Note that this is also a reply to your earlier comment: ” [Reply] Hurrah! You are three-quarters of the way to enlightenment. Now, tell me, if the original surface continues to radiate 235 Watts while only 117.5 is lost to space, do you think the inner surface will get hotter or stay at the same temperature?” ~tb
If the situation described would cause the core to heat up until it was emitting 470 and the shield 235 in/235 out, why would the inward radiation not cause it to heat up again, and again, and again?
” lgl says:
March 31, 2013 at 11:00 pm
gbaikie
Using a shell with emissivity 0.5 will make the core radiate 940 W/m^2. Already forgot?
http://virakkraft.com/half%20emissivity.png”
You making the shell the insulation.
I am saying put the insulation between core and shell and halving the amount emitted as compared to the emissivity of core surface.
It seems to me that if one has set amount such as 235 W and one halves it’s emissivity
it’s going lower the amount watts emitted by half. And since one has to have 235 W emitted, due nuclear decay energy generation, the temperature will increase until the lower emissivity surface
is emitting 235 W/m-2.
“… with emissivity 0.5 the feedback factor becomes 0.75 and Gain=1/(1-0.75)=4”
You talking about reflecting the energy from core surface. It’s not what I am talking about, as not interested too much about how reflective the insulation is which facing the core surface, rather it’s the outer surface of this insulation which has emissivity which I wanted to be 1/2 of core surface’s emissivity.
But if you interested in reflectivity, .5 emissivity isn’t very reflective, one could consider a more mirror like surfaces. With visible light, a common mirror can reflect more than 99% of this radiant energy, and could also similar effects similar materials with IR radiant energy. Like polished silver
is 0.02 – 0.03 at 300 K:
http://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html
So basically such insulation I am talking about would be like painting the core surface with material which results in emissivity of .45 with my assumption the original surface was about .9 emissivity.
Though it could have began with anything. Granite is suppose to be 0.45. and concrete is 0.85. But with natural terrain I thought .9 was reasonable.
But after looking above chart, if it started with limestone 0.90 – 0.93 and one re-paved with Granite: 0.45 that would be the same effect I am talking about.
Tim F. 6:53pm:
Thank you (and wayne 7:14pm).
Yes, “black sphere” T is held constant in Max’ linked berzerkley quiz answer, an important difference.
Is inconsistent w/Willis’ top post thought experiment which allows non-constant T of “planet” to rise when “shell” constructed. I was puzzling thru the maths until “getting” your comment, a cool 100 yr. old physical insight.
Now, maybe it’s just me, but I can’t see where it says that the sphere temperature is held constant.
In problem 1026 it goes over the rate of cooling as modified by the shield.
In 1025 it goes over infinite parallel plates at 300K and 4.2K.
1024 goes over two infinite plate black bodies radiating towards each other with an inserted reflective plate between them.
1023 goes over a black body radiating into a 0K background at temperature T, a shield is added, it is mentioned that this is actually a crude model of a star and a dust cloud, it doesn’t say it is held constant or forced that way, it just says it is at temperature T and the shield is introduced, and asks what the temperature of the shield will be at radiative equilibrium.
If you can show that the temperature of the sphere will be changed by the temperature of the shield at said radiative equilbrium, I’d love to see it, but so far I’ve seen numerous arguments of the form: “what you said happens, Max, but then this other thing happens too”, winding up with the conclusion we were examining.
_______
If the 235 Watts was somehow suppressed by the 117.5 coming back then there wouldn’t be 117.5 coming back in the first place. The full emission reaches the shield, it heats it up, but the shield has a larger surface area and “loses” half of the sphere emissions to space.
If there were no losses on the outside it would just be a description of a black body cavity with an isotropic 235 Watt radiation field inside.
_______
What if there was another layer to the model and the original sphere with a hollow shell with a power source inside radiating at it?
The hollow sphere is emitting 235 to the shield which emits 117.5 to space, right?
So what if the “other thing happens too” here?
The inner source is giving off what 470 to the now hollow sphere which gives off 235 to the shield which emits 117.5 to space, but now inward emissions cause the hollow sphere to heat up until it emits 470 and the shield emits 235 to space?
Wait does that mean the hollow sphere would be emitting more back towards the power source?
tallbloke says:
March 31, 2013 at 5:50 pm
[“Are we all happy that Willis’ toy planet theory is correct now?”]
Rog, David, lgl and others who think Willis’ model works…
I’m happy to play ‘Thread Idiot’ on this. No, I don’t think Willis’ model will work.
David’s statement A:
[“Nobody in this or any other recent TB thread that I have seen suggests that radiation from a cooler radiating body directed towards a warmer radiating body, will cause that warmer body to increase in temperature.”]
Is mutually exclusive with Tim’s statement B:
[“Willis’ model keeps the POWER the same for the sphere (with the temperature increasing as a result of teh shell).”]
In Willis’ model, the planet heats up because it receives radiation from the cooler shell! How is this consistent with David’s statement?
” Arfur Bryant says:
April 1, 2013 at 9:12 am
tallbloke says:
March 31, 2013 at 5:50 pm
[“Are we all happy that Willis’ toy planet theory is correct now?”]
Rog, David, lgl and others who think Willis’ model works…
I’m happy to play ‘Thread Idiot’ on this. No, I don’t think Willis’ model will work.
David’s statement A:
[“Nobody in this or any other recent TB thread that I have seen suggests that radiation from a cooler radiating body directed towards a warmer radiating body, will cause that warmer body to increase in temperature.”]
Is mutually exclusive with Tim’s statement B:
[“Willis’ model keeps the POWER the same for the sphere (with the temperature increasing as a result of teh shell).”]
In Willis’ model, the planet heats up because it receives radiation from the cooler shell! How is this consistent with David’s statement?”
I guess it depends upon how you think an ideal blackbody is suppose to work.
If you take a a meter square chunk of this steel greenhouse, and have facing the sun at Earth distance it would radiate from both sides. So 1360 divide by 2 giving 680 W/m-2 on side facing the sun and on side facing away from Sun.
And this because the sun is only a small portion of the sky.
If you were to get closer to sun, so the sun occupying most of sky, then I suppose some would agree that less energy would radiate sunward as compared to away from the Sun.
And this would the situation with shell encasing a planet. In the inward direction is mostly this warm core planet.
Or put mirror facing the core, no matter what angle you looked at the mirror, you would mostly see
the core planet.
So if right in front of of mirror [with back to planet] you see yourself framed by the planet, and if sideways and/or up and down and look at the mirror you will mostly see the planet in the reflection.
Arfur says: In Willis’ model, the planet heats up because it receives radiation from the cooler shell! How is this consistent with David’s statement?
Did you read my followup to David’s statement? If so, which part of it is there a problem with?
The key point is that the back radiation slows down the rate at which the planet surface can lose heat. Do you get that? It doesn’t heat it, it slows down the rate at which it can lose heat. Because the rate of output of the nuclear core does not slow down, heat builds up at the surface and so the temperature rises. It is the nuclear core doing the heating of the whole system, not the shell. The crucial difference between this situation and the way radiation problems are classically posed is that we are not dealing with a closed system with a fixed amount of heat in it. New heat is being generated, and lost, so the internal dynamics have to be considered, in order to work out how the heat is lost, and what temperature the various components have to be at in order for the system to be in a steady state.
“ Or put mirror facing the core, no matter what angle you looked at the mirror, you would mostly see
the core planet.” ~gbalkie
Uh, the view factor for the core guarantees that all radiation it emits will reach the shell, the shell can–for large enough gaps or shallow enough angles–radiate towards itself, and it can radiate towards the environment.
Max
We are still waiting for your answer to TBs question:
“Now, tell me, if the original surface continues to radiate 235 Watts while only 117.5 is lost to space, do you think the inner surface will get hotter or stay at the same temperature?” ~tb
tallbloke says, April 1, 2013 at 9:57 am: Arfur says: “In Willis’ model, the planet heats up because it receives radiation from the cooler shell! How is this consistent with David’s statement?” Did you read my followup to David’s statement? If so, which part of it is there a problem with? The key point is that the back radiation slows down the rate at which the planet surface can lose heat. Do you get that? It doesn’t heat it, it slows down the rate at which it can lose heat…
TB, Thanks. Saved me replying. We are cooking on gas!
Max 5:33am: “I can’t see where it says that the sphere temperature is held constant.”
It is confusing. UCB 1st sentence sets up black sphere at T radiating to distant surroundings also at T=0K.
They would have been more clear to write something like Ts=0 for the surroundings heat sink.
It does imply sphere T greater than surroundings T=0K or nothing ever emits. At radiative equilibrium with heat shield in place, the sphere is again at temperature T with shield T1 less than T but neither being =0K since they emit J and J1.
Implies UCB sphere with no shield emits, say J2, less than J where Willis’ model has the J held constant both with and without shell, not the T. So the equilibrium answers will be different once real numbers are plugged in per Tim F. and wayne’s explanation.
******
Max 10:47am:
Yes, Willis’ inner shell (and UCB shield) surface emits to itself (making exact simple S-B inapplicable) therefore the inner shell emissivity is not easy to know so nature’s exact solution is not easy to find meaning the top post numbers are only a close guess but not the exact answer. Probably good enough for conservative engineering work but not a science paper looking for 0.5C/century with confidence intervals significantly less than 0.5C.
lgl says:
Max
We are still waiting for your answer to TBs question:
“Now, tell me, if the original surface continues to radiate 235 Watts while only 117.5 is lost to space, do you think the inner surface will get hotter or stay at the same temperature?” ~tb
I find Max’s evasion of questions and introduction of new obfuscations very frustrating. So frustrating that a few times I’ve tried to force him to reply on threat of not publishing further comment until he does. I do my best to be fair to everyone and let them develop their argument as they wish, but I AM NEAR THE END OF MY TETHER with this sort of behavior. I will give Max a timeout from this thread unless he observes the rules of scientific discourse and answers fairly put questions properly.
Chicken. Egg. Both exist.
[Reply] So do cabbages and kings. Your point?
Arfur Bryant asks
David’s statement A:
[“Nobody in this or any other recent TB thread that I have seen suggests that radiation from a cooler radiating body directed towards a warmer radiating body, will cause that warmer body to increase in temperature.”]
Is mutually exclusive with Tim’s statement B:
[“Willis’ model keeps the POWER the same for the sphere (with the temperature increasing as a result of teh shell).”]
In Willis’ model, the planet heats up because it receives radiation from the cooler shell! How is this consistent with David’s statement?
………………………………………………..
Arfur is still waiting for a meaningful answer as must other readers?
…………………………………………………
My comment is how many times have we heard this dog eared refrain
“A cooler body will keep a higher temperature body warmer than it would otherwise be.”
Its almost always included in any IPCC advocates statement.
I have given an example where this is completely wrong.
As I said in my post to lgl
“A version of the Willis model can show that you are wrong,
This time its the shell that is warmer than space (lets say 350K) and the planet is not there but the vacuum is.
Now insert an object at 250K at the centre of the hollow sphere.
The 250K colder object will radiate towards the warmer shell.
The radiation will be absorbed by the shell
But no one thinks (I hope) that as a result, the temperature of the shell will increase.”
The presence of cooler body does not ‘make the shell warmer than if it was not there’ quite the reverse.
tallbloke
This has been a long and interesting thread perhaps a minimum consensus can be found around.
1.
Heat can never be transferred spontaneously from a lower to a higher temperature object.
2.
A lower temperature object can appear at times to make a higher temperature body ‘warmer’.
But this is just a coincidence.
In all such cases the alternative to the colder body is usually an even colder body.
So quite clearly this is a relative term.
The colder temperature usually suggested is that of empty space.