## Friday, October 20, 2017

### Why the Green Plate Effect Has Had an Effect

The Green Plate Effect (GPE) post is a gedanken experiment posed by your friendly bunny, which uses simplifications to think through the consequences of a proposition.  The proposition is that as several  have claimed, that the Greenhouse Effect (GHE) violates the second law of thermodynamics.

Discussion of the GPE has occupied more that a few places, including Rabett Run, Roy Spencer's bodega, and the Dragons Lair (be sure to wear protection when going there or better yet do not), but contrary to the Weasel, there does seem to have been an effect.  (BTW he has been tossed out of his condo and retreated to the original hovel)

The GPE post drove home two ideas:

1.  The GHE is not a statement about two bodies, a hot and a cold one, e.g. the surface and the atmosphere, but a statement about three bodies, a heat source, the absorber of energy from the heat source and the thermal shield, In the GPE these are the illumination source, the blue plate and the green plate.

This idealization can be extended to the sun, surface atmosphere system at the cost of mathematical complication involving things such as geometry, viewing angles, emissivity, thermal conductivity, diurnal cycles, etc.

Were bunnies to go acaveating it might be mentioned that the blue plate would cool more quickly in the absence of the green plate when the light was turned off.  This, happens in real life.  Night time temperatures fall much more quickly in the desert than in Mississippi and yes, Betty, water vapor does absorb IR emissions from the surface and yes, something is needed to heat the surface first.

2.  Simple analysis shows that in the GPE the presence of the green plate makes the blue plate hotter.

The myriad attempts, some here, some there, involve changing the problem to something else or they break down into first or second law contradictions or they tie themselves up into algebraic knots.  Mathocists are invited to look at the comments at RR, Dr. Roys, or Postma's Pablum Palace.

Christian Anders said...

Eli,

please allow me a little bit of critic.

You are right when you state that downgrading the complexity of a thought experiment is an often practiced method in physics.
But when you are dealing with people who have a strong unwillingness to understand, too much simplification might backfire, because it will be the very reason those people will not even try to understand what you want to show, since they can find enough to chew on in the experimental setup.

Would it have been too complex to assume not infinitely thin, but real sized plates which are simply extremely thin? You can stick to perfect conduction as an assumption anyway.

- It is easy to show that emission from the edges is of negligible power.

- Nobody would have tried to discuss if there is possibly no 2nd surface which can radiate.

Furthermore, assuming the plates to be infinitely large does cause a real problem in my opinion.

If that's assumed, one can argument that a radiative exchange like between the two plates has as well to take place between the first plate and the heat source, thus bringing up the temperature of the source with exactly the same mechanism.
What else is a constant heat source than a surface that constantly emits energy? And this has to have infinite dimension also, if the plates are assumed to be that way.

Because of this it might have been better to take a real star as a source an thin real shaped plates of something like 1m^2 or 10m^2 or so.

The problem is in my opinion, that those people who easily follow your abstraction, do so because they already know the correct physics even for other setups and so it comes easily to them. The others are just finding an additional barrier to jump over, or to run against eternally.

Of course you will never satisfy those who know you are wrong, they will simply ignore the existence of disturbing facts like the existence of a heating source, the defintion of adiabatic, known laws of radiatve energy transfer, or even the fact that a thought experiment does not need to be applicable to the real world.

That I have learned...

Additionally I share the opinion that came up in the comments of the first GPE article, that "equilibrium" is simply a wrong term for the stationary state of this setup.

Apart from that I agree with your conclusions.

Eli Rabrtt said...

Like many of the Gedenkenversuche this one can actually be done if the dimensions are relaxed but at the cost of complexity in analysis. Best example of why this is not the best way to go is maybe the difficulty of closing the last holes in the Aspect realizations of the EPR gedenkenversuch but Bell's theoretical exploration of the idealized experiment is accepted.

Christian Anders said...

You're right, but in this case I personally think the trade-off between complexity in analysis and avoiding a certain problem speaks for a little more complexity.

The problem with infinite size of the plates does create an additional effect not shown by your experiment, namely if the source is of matter and in non-infinite distance.

For this will not change the underlying principle you wanted to show, one could argue it does not really matter, but it exhibits your arguments for side-critics anyway.

Russell Seitz / Bright Water said...

"BTW he has been tossed out of his condo and retreated to the original hovel)"

The mustelids are not the only species on the move

It's the Megalopygids that have me worried !

Timothy Folkerts said...

I would argue that the greenhouse effect is a statement about FOUR bodies.
1) a heat source
2) the absorber of energy from the heat source
3) the thermal shield
4) a low-temperature heat sink.
The thermal shield must be interposed between the absorber and cold surroundings (or it is not a "shield"!).

The heat sink is rather abstract in many cases -- the cold 3 K universe. But I think this is worth an explicit mention. For many simple experiments, the heat sink is the 300 K walls and ceiling of a room. This leads to vastly different conclusions (mathematically, intuitively and practically) for 300 K background than 3 K background.

Christian Anders said...

Timothy,

according to your definition Eli's GPE is a four body problem too.

I suggest to simplify it further:

A constant working power source with a plain surface of infinite size and an infinitely large plate of a given thickness. Both surfaces assumed to be black bodies. Both parallel to each other without direct contact.

That reduces the setup to a three body problem, if the envoirenment is counted as a heat sink.

It will become more easy to see, what is to be shown: That the surface of the power source will have higher temperature with the plate in proximity than without.

This would be as well in 0K vacuum as in 300K of any atmosphere. The latter case is simply much harder to calculate. But the latter case also would be a very good analogy to a blanket that covers you at night.

Timothy Folkerts said...

Christian,

Yes -- whether the GHE or the GPE or any other experiment (actual or gedanken) related to this topic -- I am saying the heat sink is an integral part.

I have seen many variations of "room temperature CO2 doesn't warm my room temperature room, so the GHE is false". (Conversely, there are experiments where the 'shield' is at the temperature of the surroundings, yet gets credit for warming via the GHE.)

Christian Anders said...

Timothy,

Not sure what you mean:

"I have seen many variations of "room temperature CO2 doesn't warm my room temperature room, so the GHE is false"."

The reason for this is not a missing heat sink, since the walls of a room are mostly colder than the contained air.

The reason is, that in a room you have nearly no lapse rate and thus the vertical heat transport is too insignificant to be measurably influenced by CO2. A very small effect on a very small process is unobservable.

-----

Now that I thought a bit about how to visualize the principle of GPE qualitatively even easier than Eli did, what do you think of this:

Imagine the sun being enclosed by a sphere of a hypothetic material that will not melt, is a perfect black body and has finite heat conductivity. The sphere is of a given thickness and has a given inner diameter.
Will the surface of the sun in this constellation get hotter over time?

Something whispers in my ear that it won't, for the cold sphere can never warm up the surface of a star, not even with a fusion reactor helping...

Timothy Folkerts said...

Christian, it is very clear that the sun *would* warm up in your scenario. (There is a bit of a thermostat with the sun, but that would only partly compensate. Let's assume the nuclear reactions are constant for the sake of argument.)

If the sun starts radiating to ANY temperature above the background 3K, then it will be radiating less total power. If it is radiating less, it is keeping more. If it is keeping more, it must warm.

I can't see how this would be the least bit controversial.

John Garland said...

Question: Do the plates really have to be infinite in size to make this work, or effectively work?

How about if the plates are arbitrarily--but nowhere near infinitely--large and arbitrarily close--in a vacuum, of course--such that the radiation from the edges of each plate still had to hit the other plate? Doesn't that work? Or effectively work? That's how I've been visualizing it, anyway.

Eli Rabrtt said...

No but having to deal with edge effects over complicates the example. Timothy and Christian are over complicating also

Timothy Folkerts said...

Eli, I personally like your green plate example and the simple assumptions. It gets cleanly at the crux of the matter. I was responding to the situation Christian proposed.

I do, however, still think that specifically mentioning a low-temperature heat sink is valuable. We are talking a steady-state solution, which means a steady flow in and a steady flow out.

Christian Anders said...

Timothy,

not controversial? Well, not by people with a basic understanding of physics.

The likes of Betty on the other hand claim, that no surface heated by an power source ( like the surface of the sun with fusion reaction going on beneath ) is ever to get warmer with a colder body in proximity, like the sphere.

But maybe they just don,t see the simplified analogy GPE represents for that setup.

Russell Seitz / Bright Water said...

Since the sun's optical depth makes for a mighty greenhouse effect, readers are invited to calculate how much clolder it would be without all that 10 micron earthshine warming it up.

On another front, his supreme nebulosity,the President of Venezuela has exhorted the populace to take a bite out of the continuing food shortage and increase the carrot supply by eating their pet bunnies .

Unknown said...

I made the mistake of going over to Postma's site!
I posted a correction to his equation as follows;
“We now introduce the green plate, which can be approximated as infinite plane parallel with the blue plate, and for this geometry the heat flow equation between the blue and green plate is simply
Q = sigma * (Tb^4 – Tg^4)
The green plate stops rising in temperature when Q = 0, i.e. when the heat flow to it goes to zero which is thermal equilibrium. Therefore when Q = 0 the green plate has a constant temperature of
0 = Tb^4 – Tg^4
Tb = Tg”
This is flawed because you have omitted the heat flow from the back of the green plate.
The heat balance for the green plate is :
Q = sigma * (Tb^4 – Tg^4) – Tg^4
Thus the temperature stops rising when Q=0
Tb^4 = 2Tg^4
Tb = 1.19*Tg
Your solution would only be correct if the surroundings were at Tg, but the problem as posed is that the surroundings are ‘space’ so Ts is 3K. Even so an additional term of +Tg^4 would have to be added to the blue plate balance in that case.

He seized on the obvious typo in the Q equation (missing pair of brackets, I should have put the extra term inside them). The Tb:Tg equation is correct though. He proceeded to swear and abuse me. When I tried to post the correction to the first equation it appears that I am banned from the site, no loss!

Eli Rabrtt said...

I have really enjoyed these exchanges.

barry said...

Me too (assuming no sarcasm, Eli). The GPE is quite penetrable to this non-physics grad. Tackling the challenges to it has been a good exercise for the mind, and brought a bit more understanding of thermodynamics for the effort.

barrymoose

Betty Pound said...

Eli is incapable of understanding science.

He still believes 266 W/m2 can't heat 133 W/m2 to an equilibrium of 200 W/m2.

He still believes an object can recycle 50% of its energy off of an other object in order to warm itself up some more.

Eli is a crank, and his gedanken experiment has no relation to reality.

Betty Pound said...

There is no heat flow from green to blue.

You are 1.5 counting blue's energy to green.

You get a situation where blue emits 267 heat to space, and 266-133=133 heat to blue.

Blue can't have a uniform temperature of 262K, of it's emitting 266 W/m2 of HEAT to space and 133 W/m2 of HEAT to green. That is an object not in equilibrium, and definitely not 262K.

John Garland said...

No, Betty...

It's emitting 267 degrees to both the Sun and towards the green plate. You seem to want to make the nonilluminated side of the black body plate some kind of weird Maxwell's Demon that siphons off 1/2 of the energy on that side of the plate and sends it elsewhere entirely outside the system. That way lies the perpetual energy you are so fond of.

In any case, it's pretty straightforward to show that if the blue plate was indeed only radiating 133 W/m^2 towards the green plate, the green plate would only heat to 185K if I pressed the calculator keys right.

Why do you so greatly need to deny the greenhouse effect? Forget carbon dioxide's controlling role--deny it if you must--but your very life depends on the much more fundamental greenhouse effect from all greenhouse effect producing gases. This is 19th century science.

barry said...

Posting to see if I've learned anything. Feel free to correct me.

Flux density (energy per second per unit energy) cannot be summed.

I think some erred in the previous thread, in confusing flux density (energy per second per unit area) for flux (energy per second). In the latter case summing surface area flux density to get total flux is appropriate.

With that in mind...

Blue plate is emitting at 400 Watts total in equilibrium with energy received from the sun.

If its surface area is 1 meter square only, it is emitting at a rate of 400 w/m2.

If its surface area is 2 meters squared it is emitting at 200 w/m2.

Size of surfaces don't matter for this because one side receives 400 w/m2 from sun across its entire face. What matters is that there are two surfaces, not one.

The latter is the condition for the GPE.

Green plate receives at a rate of 200 w/m2 from the right side of the blue plate.

Warms up until it emits at a rate of 200 Watts, in equilibrium with what it receives from blue plate.

If its surface area is 1 meter square only, it is emitting at a rate of 200 w/m2.

If its surface area is 2 meters squared, it is emitting at 100 w/m2.

Again, size is not an issue. It does matter that there is vacuum between all three bodies, and that radiation is the only way in which energy is transferred.

On a planetary surface with no atmosphere constantly facing the sun, it doesn't matter how long you stand under the shade of black-painted tinfoil. The shade will never heat up enough to emit down to you at the same flux density as you'd get standing in the direct heat of the sun. Some portion of the energy received by the tin foil will always be going skyward.

You'll always be a little cooler in the shade.

The green plate will never be as warm as the blue plate. Not with vacuum between them.

barry said...

If I've got the metrics for blue/green plate right (total radiant flux in watts: flux density in w/m2), the 2-plate system is not in thermal equilibrium with input, and the system must warm to become so.

In a constant process over time...

Blue plate absorbs energy from green, making its rate of energy loss slower. Blue plate heats up as a result, and emits more both ways. Green now receives more energy, making its energy loss less efficient, and it heats up. This continues until the outflux from the 2-plate system is in equilibrium with the input from the sun. At all times the NET flow of heat between the 2 plates is from blue to green. Both plates heat up. The green plate cannot become as warm as the blue plate because it receives less energy than the blue plate.

Christian Anders said...

barry,

your comments are essentially correct. I have only three things to mention:

1) It's not that you stated something that would not happen, but this very sentence "blue heats up as a result [...]" I would change to "blue is heated up further by the sun as a result [...]".
The reason for this is that it could be misunderstood as if anything is happening by itself instead of being driven by the sun. There are people out there who can only very hard distinguish between driven by a heat source and itself.

2) It was discussed - and in my personal opinion it is more than pure semantic - that "equilibrium" is a false characterization of the steady state the system gets finally in. A real equilibrium would be a state where the heat transfer is zero for everywhere in a system - impossible with a heat source like the sun present.
You stated the phrase [...] thermal equilibrium with input [...]" and such a state is called a "steady state". As a matter of fact you got it right.

3) On this certain point you are wrong: "Green now receives more energy, making its energy loss less efficient [...]"
No. Energy loss always equals energy gain, for green as well as for blue. But only 50% at each plate can radiate directly to space if two plates are there. For both plates the mere presence of the other plate reduces cooling efficiency from 100% to 50%. But it is constantly 50%, not getting more or less efficient with more or less energy received.

-----

If you have idealised bodies like the plates in the GPE and you are not going for certain numbers, but only look to efficiencies of heat transfer, the problem can even be solved by a simple sequence.
For every plate you add on the right the right surface of the formerly last plate loses 50% of its cooling efficiency.
For energy is not to be destroyed, but the system goes to steady state anyway, this energy must be radiate away from the very left side of the system. Therefore at least the plate on the very left must radiate more energy directed left. 50% if alone. 75% if second plate present. 87,5% if third plate present. 93,75% if fourth plate present. And so on.
Hence all plates are energetically connected by radiation, this will also lead to a higher steady state temperature of all other plates.
That would converge at 400 W/m^2 for the utmost left plate and 0 W/m^2 for the utmost right.

It is like the story of the guy who needs eternally to reach his doorstep, for he is not allowed to make a step wider then half the space to his door, but must additionally make every step half so wide as the step before.

( edit: Just wanted to show that efficiency of cooling is a matter of number of plates, not energy input into a plate. Maybe too much text for this... )

barry said...

Thanks Christian,

Just one point to query.

"3) On this certain point you are wrong: "Green now receives more energy, making its energy loss less efficient [...]" "

But,

"For both plates the mere presence of the other plate reduces cooling efficiency from 100% to 50%"

Maybe it's my lack of familiarity with the precise language, but I don't see a lot of difference between "energy loss less efficient," and "cooling efficiency is reduced." Best I can do to figure it out, is to think of different materials conducting heat less or more 'efficiently,' and wonder if my phrasing inadvertently brings that concept into this one.

Unknown said...

Update:
Unknown said...
I made the mistake of going over to Postma's site!

As I outlined above I posted a correction to an equation in one of his posts and got abused and banned from his site. Since then he has deleted my post and his abusive reply! Definitely not a good site to visit, I can only assume that he knew he'd made a mistake and wanted to cover it up.

Christian Anders said...

barry,

ah I see, that needs to be clarified. Your sentence

"Green now receives more energy, making its energy loss less efficient [...]"

is not wrong in its consequence ( less efficient energy loss ), but in its reasoning ( because more energy is received ).

The relative reduce of energy loss is independent from the absolute value of received energy, since in relation it is always going down to 50%. That is due to the very existence of the second plate and would be the same with any amount of energy received, because that is how the system is designed.

It's not changing while blue is heated up, it's 50% constantly.
It could as well be that I readsomething into your words you did not meant, then I apologize.

John Garland said...

barry...

What matters is that there is 400 in because 1 m^2 is illuminated and 200 out per side as there are 2 sides only one of which receives any illumination. I think you meant that, but it seems a bit unclear as stated.

John Garland said...

When there is only the blue plate alone, of course.

Timothy Folkerts said...

"Flux density (energy per second per unit energy) cannot be summed."
First, I assume you meant "energy per second per unit AREA".

You certainly cannot have a cube that emits 300 W/m^2 from each side and try to sum that to get 1800 W/m^2 total from the cube.

On the other hand, on a sunny day, sunlight might provide 1000 W/m^2 of flux density to a patch of ground. If I got a mirror (or some other bright light) that could provide 500 W/m^2 to that same patch of ground, then the two can be summed to give 1500 W/m^2 of incoming flux density. Similarly, the earth's surface absorbs an average of ~ 160 W/m^2 of sunlight and ~ 330 W/m^2 of thermal IR from the atmosphere. The total incoming intensity is properly 490 W/m^2

"Blue plate heats up as a result..."
I would avoid "heats up" and "heated up" and any variation thereof.
* Use "warm up" if you mean "the temperature rises" (measured in K)
* Use "heat" if you mean "transfer of thermal energy" (measured in J).
The two idea are related but not the same. When people say "heats up " if blurs the distinction.

"Therefore at least the plate on the very left must radiate more energy directed left. 50% if alone. 75% if second plate present. 87,5% if third plate present. 93,75% if fourth plate present. And so on. "
I think you will find it works out to:
* 1 plates: 50% left, 50% right (ie 1 blue plate)
* 2 plates: 67% left, 33% right
* 3 plates: 75% left, 25% right
* n plates: (n)/(n+1) left, 1/(n+1) right (ie 1 blue plate, n-1 green plates)

With that in mind...
Let I(in) be the intensity of incoming sunlight (ie I(in)=400 W/m^2 here).
Let I(B,L) = intensity OUT from the left side of the
Let A be the area of one side of the plates. Let P=power emitted

Blue plate is emitting at P=I(in)*A total in equilibrium with energy received from the sun, which is P=400W if the (cross-sectional) area of Blue is 1 m^2.

If its surface area is 1 meter square only, it is emitting at a total rate of P=400 W, or I = 200 W/m^2 from each side.

No matter what the (cross-sectional) surface area, it is emitting at 200 w/m2 from each side (with no Green). The TOTAL power absorbed and emitted will change in proportion to A, but not the intensity (flux density) = P/A.

Christian Anders said...

Timothy,

you are of course right about the series.

barry said...

Hey Timothy,

I assume you meant "energy per second per unit AREA".

Yes I did.

I've been using the mirror idea, too.

The two ideas are related but not the same. When people say "heats up " it blurs the distinction.

Maybe if I'm talking to a physics grad, but I see what you mean. Either term gives skeptics conniptions.

No matter what the (cross-sectional) surface area, it is emitting at 200 w/m2 from each side (with no Green). The TOTAL power absorbed and emitted will change in proportion to A, but not the intensity (flux density) = P/A.

Understood. I came across this interesting tidbit while looking things up.

The Sun has a higher surface temperature, so it must radiate more energy per unit surface area than alpha Ori. In spite of this, alpha Ori has a far greater luminosity than the Sun! There is only one way that alpha Ori could radiate more total energy it must have a larger total surface area.

http://astronomy.nmsu.edu/geas/lectures/lecture23/slide02.html

John Garland said...

Re. alpha Ori, etc.: I've often wondered why greenhouse effect deniers don't realize that were they correct the greater of astronomical knowledge beyond the Solar System gained since the 19th century would be immediately falsified as well. The fundamental spectroscopic underpinnings involved in taking stellar spectra and inferring anything from them involve the very same underlying math and physical laws--in particular, the Stefan-Boltzmann Law.

Betty Pound said...

You can't state the problem correctly. You can't calculate correctly. You believe the sun prevents equilibrium on objects it shines on perpetually. Eli didn't prove the GH effect. He actually disproved it.

John Garland said...

When the Sun and the plate are at equilibrium, they will be at the same temps, true. However, this state is also known as the heat death of at least the Solar System and probably the universe as well!

winston said...

I'm confused by all of the pointless mystification going on. Eli's original post seemed perfectly clear and correct, although there seems to be a tacit assumption that 100% of the radiation incident on a plate is absorbed. It's just a case of equilibrium energy balance, and I can't see why anyone needs to drag in the 2nd law. With regard to Timothy's comment, why not just say that the blue plate emits a fraction n/(n+1) of the incident radiation to both left and right, and absorbs a fraction (n-1)/(n+1) emitted by the plate immediately to the right. In general, with n plates, the i'th plate emits a fraction (n + 1 - i)/(n + 1) on both sides, and absorbs the radiation emitted from the plates to its left and right.

barry said...

You believe the sun prevents equilibrium on objects it shines on perpetually.

Thermal equilibrium is when temperature is uniform across a system/s, right?

I believe that polar orbiting satellites or heliocentric orbit satellites (constantly in the sun) are not the same temperature as the sun.

Gator said...

I believe your green plate effect is a simplification of what we called "superinsulation" back in my cryogenics days. The insulation around a dewar is basically vacuum + layers of aluminized mylar. The vaccum to prevent heat transfer by conduction, the many layers of mylar to prevent heat transfer by radiation.

EliRabett said...

Gator: Not quite, because the aluminized mylar reflects the IR rather than absorbing it and then re-emitting thermal radiation.

Tom Dayton said...

Yes, Eli, but the key to Gator's note's relevance is the "multilayer" part. We use multiple layers to insulate spacecraft. IR reflection as you responded is a contributor, but the multiple layers are important contributors as well. https://en.wikipedia.org/wiki/Multi-layer_insulation

EliRabett said...

So soon you forget

http://rabett.blogspot.com/2008/09/light-dawns-there-are-styles-in-science.html

Also reflective IR coatings on halogen lamps

http://www.smithmillermoore.com/Pdfs/technicalarticles/3-6-13_DSI_LTB_The_Rebirth_of_theIncandescentLightBulb.pdf

Gator said...

Eli -- maybe, but the layers will equilibrate to an intermediate temperature and radiate. The main thing is that the layers should not be transparent. I guess the main thing is that this is a real world, very practical example of the engineering of radiative heat transfer. It works.

Tom Dayton said...

Eli, unlike your foil-wrapped light bulb example, the multiple layers around spacecraft do not touch each other and are separated by vacuum. So conduction and convection don't operate. I assume Gator's vacuum container similarly has vacuum between its layers.

EliRabett said...

Tom,

While what you say is true, glass is a pretty good insulator and wrapping the bulb w. tinfoil pretty much kills convection between the glass envelope and the bulb. Convection btw the outside should be the same for glass or Al wrapped glass. Same with the IR coating on the halogen lamps.

The multiple layers, of course, make a difference in the numbers, but not the principle of a single plate. Multiple layer atmospheric models are pretty much what GCMs use (~20 ??)

Suwannee Dave said...

Jeez this is getting complicated. I am a retired Landscape Architect, so no scientist but exposed to a little. I totally understand that something hot (the sun) can heat something smaller and cooler (the earth) and that that cooler thing can conserve its heat by having a thermal radiative blanket (the atmosphere) until it reaches equilibrium. When the thermal blanket is made more effective (more CO2 (see old science)) the equilibrium must increase the temperature on the cooler thing.

Eli Rabrtt said...

Yep but there are many out there trying to sow confusion and simple examples are needed to deal with them.

Part of the value of the GPE is the obvious thrashing it gas inspired at rr Dr roys and elsewhere makes the point of how empty the confusioinsts arguments are.

One of the things those of us on the other side need to learn is to keep things simple.

Wrt blankets there is a long history of objection that they only cut off convection, that they cannot warm a hotter thing etc. Even there the GPE has value bcs it shows how a source of constant heat energy plays an important role

Russell Seitz / Bright Water said...

"You believe the sun prevents equilibrium on objects it shines on perpetually. "

Betty, It's not the sun's fault if people paint white objects black, or darken atmospheres with infrared inks like CH4 or CO2.

David B Benson said...

Those interested in a deeper intellectual exercise about climate may wish to discuss
on whether we are headed for the climate of the mid-Pliocene.

Russell Seitz / Bright Water said...

I hear we may be in for a spell of the Apocalyptocene

Timothy Folkerts said...

I had a thought that might help some people -- basically turn the problem around and calculate how much you have to actively heat or cool the two plates to achieve a specific goal. I say this because I *think* everyone agrees that a flat blackbody surface radiating to space emits
P/A = sigma T^4,
and that the power from a hot flat blackbody to a nearby cold flat blackbody is
P/A = sigma (T(h)^4 - T(c)^4)

So assume two flat blackbody plates that are each 1m x 1m (called "Blue" and "Green"). Assume you want to actively keep Green at 220K when blue is actively held at various temperatures.

Hold Blue @ 0K
Actively heat Green @ 267W so it can radiate 133W to space (from its right side) and radiate 133W to Blue (from its left side)
Actively refrigerate Blue @ 133W so it can absorb 133W from Green (from its right side) and radiate 0W to space (from its left side)

Hold Blue @ 220K
Actively heat Green @ 133W so it can radiate 133W to space and 0W to Blue
Actively heat Blue @ 133W so it can absorb 0W from Green and radiate 133W to space.

Hold Blue @ 244K
Actively heat Green @ 67W so it can radiate 133W to space and absorb 67W from Blue
Actively heat Blue @ 267W so it can radiate 67W to Green and radiate 200W to space.

Hold Blue @ 262K
Actively heat Green @ 0W so it can radiate 133W to space and absorb 133W from Blue
Actively heat Blue @ 400W so it can radiate 133W to Green and radiate 267W to space.

Hold Blue @ 290K
Actively refrigerate Green @ 133W so it can radiate 133W to space and absorb 267W from Blue
Actively heat Blue @ 667W so it can radiate 267W to Green and radiate 400W to space.

It is easy to confirm that all these
1) agree with simply SB power calculations
2) conserve energy.

Also, it is also intuitive that if blue is cold, energy with flow from Green to Blue and Green will need an active heater to stay at 220K. Conversely, if Blue is quite hot, energy will flow from Blue to Green and Green will need active cooling to stay at 220K. Thus there must be a sweet spot where green requires neither heating nor cooling to stay @ 220K, and that happens to be when Blue is 262 K (ie when blue is getting 400W input power).

John Garland said...

izen discusses and provides animated gifs of a related version of the blue- green plate example here:

Not that deniers will accept it. The will to deny is strong in some.

Paul B said...

Every time I run across this topic I see the same process....

A fundamental simple model is proposed and before 3 commments have been made the thread is off to the races with ice cubes and wooden blocks, Plancks, Boltzmans, etc. Pretty soon the snark kicks off followed by name calling and mind numbing math when all that is called for is profoundly simple logic built on agreement with some basic principles before diving into the weeds. I propose a trivial exercise and if we can't get every single person here to agree to the concept we don't proceed so here goes (this is mostly aimed at Betty but YMCV).....

Define a system boundary around a blue plate being heated with a constant of S Joules/sec. The plate is very thin so ignore edge effects. When the plate equilibrates it will radiate energy from its two sides. The boundary then has one input, S and two outputs, A and B.

Write an equation for A and B in terms of S

Paul B said...

Give up? Here is the secret answer...

A=S/2, B=S/2. And the corollary is of course... A=B

Let's call it Robert's law and it States that a very thin equlibrated plate with input energy S has two output energies equal and opposite each other normal to the plate's plane.

Paul B said...

If you agree with Robert's law let's add another plate to the system in proximity to the first and aligned so that they share a perpendicular center line.

Plate 1, let's call it blue, is to the left of plate 2, we'll call it green.

Let's define a system boundary around the whole thing and feed an energy S to blue. Blue has two outputs that obey Robert's law. We can call them A. The A on the right strikes green and it has two outputs that also obey Robert's law. We'll call them B.

Write a system of equations that describe this system....

Paul B said...

A=S/2, B=A/2 from Robert's law
For the system S=A+B from first law
The blue plate is S+B=2A

So a little algebra gets you S+A/2=2A,
And then simply A=2S/3

That means B=S/3

So blue plate is spewing 2S/3 to space.
Green plate is spewing S/3 to space. System is in balance, blue plate is in balance, green plate is in balance.

And BTW blue plate 4S/3 energy out where it only had S without the other plate present.

Solution requires nothing more than 1st law and 1st year algebra. You never have to go down a temperature rat hole. Temperature is a silly thing to dwell on. It's all about energy