Entering the SkyDragon’s lair

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

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

Figure 2: Now, note what happens when we add a shell around the planet. The shell warms up and it begins to radiate as well … but it radiates the same amount inwards and outwards. The inwards radiation warms the surface of the planet, until it is radiating at 470 W/m2. At that point the system is back in equilibrium. The planet is receiving 235 W/m2 from the interior, plus 235 W/m2 from the shell, and it is radiating the total amount, 470 W/m2. The shell is receiving 470 W/m2 from the planet, and it is radiating the same amount, half inwards back to the planet and half outwards to outer space. Note also that despite the fact that the planetary surface ends up much warmer (radiating 470 W/m2), energy is conserved. The same 235 W/m2 of energy is emitted to space as in Figure 1. [Note]: Figure 2 is not to scale; the shell is very close to the planet surface, so areas are ~equal

Joe goes postal on this, saying in part:

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

Here’s my  response:

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

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

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

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

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

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

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

Best to you.

Rog TB.

Comments
  1. lgl says:

    Bryan

    No you haven’t given any “example where this is completely wrong.”
    I explained your error
    “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.”
    What of this do you not understand?

  2. Bryan says:

    David Socrates

    You were well off topic when you posted above;

    [snip]

    [Reply] Agreed, so take it to the pyrgeometer thread to pursue it. I’ve snipped David’s comment back to the link so your reason for having to make this comment has been removed. We are now so close to finalising this thread I’m not allowing side arguments. Sorry if this appears draconian, but I want this sewn up.

  3. tallbloke says:

    THIS THREAD IS NOT ABOUT EARTH, PYRGEOMETERS, OR THE IPPC. FURTHER COMMENTS INCLUDING REFERENCES TO ANYTHING ELSE BUT THE SUBJECT OF THIS THREAD ARE GOING IN THE BIN, WHOEVER POSTS THEM.

  4. Bryan says:

    lgl

    Its great that you now accept that heat is not transferred spontaneously from a lower temperature to a higher temperature body

    Its also great that you now accept that the presence of a lower temperature body will not always appear to make a higher temperature body ‘warmer than it would otherwise be’.
    Granted that sometimes this appears to be the case.
    Someone without a grounding in thermodynamics can easily be fooled into thinking that absorbed backradiation will have this effect.

  5. lgl says:

    Bryan
    I have never said there can be a net transfer from colder to warmer, so nothing new.
    Nor have I claimed putting a colder object inside a warmer object will warm the warmer object.
    Your example has nothing to do with the model discussed. It’s like putting an icecube in a glass of water. When has the IPCC or Willis or anybody else ever claimed the icecube would warm the water?
    To get this back on topic, if you put a shell of ice around that glass of water and place it outside the solar system, the shell of ice will slow down the cooling of the water.

  6. tallbloke says:

    Bryan says:
    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.

    I don’t think anyone is in disagreement with 1.

    I don’t think 2. helps much. What sort of coincidence? If the warmer body (the planet) didn’t have a heat source supplying more heat on an ongoing basis, the adjacent presence of a cooler body (the shell) rather than an even cooler body (space) wouldn’t make it warmer than it already was. It would just slow down it’s rate of cooling.

    And it would slow down the planet’s rate of cooling while the nuclear core was supplying more heat to the planet surface too. Which is why the planet surface would get hotter than it was when it was just radiating to space.

    Its not because the shell transfers heat to the planet. It’s because the planet has a harder job getting rid of the heat being generated by the nuclear core to space with the shell in the way, re-radiating some of the energy it is trying to get rid of to space back at it, slowing down its rate of cooling.

    It has to get hotter until it gets the shell up to the temperature where its outer surface can radiate as much to space as quickly as the planet’s surface did before the shell got in the way. Until it gets to that temperature, the shell will emit less to space than is being produced by the nuclear core and the planet will continue getting hotter, because steady state hasn’t been reached.

    Are we in agreement now?

  7. Bryan says:

    tallbloke says

    “I don’t think 2. helps much. What sort of coincidence? ”

    You will be aware of Roy Spencer’s “Yes Virginia ” thread and several other similar posts from other people.
    The format seems to confound most peoples common sense by suggesting that colder objects make higher temperature objects ‘warmer than they would otherwise be’

    They then go on to prove their point by calculations confirming that this seems to be the case.
    This they then say supports the idea of a ‘greenhouse effect ‘.

    All examples of this kind have the colder object compared to an even colder object.
    Most often the comparison is empty space.
    This had led to some thinking that this comparison is always true.

    Apart from hot object A and cold object B there is the background object C not often commented on.

    For temperatures where

    B > C A loses less heat than if C > B where in fact A would lose more heat than if B did not exist.

    So backradiation from B does not always slow down heat loss from A

  8. Max™ says:

    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

    The temperature of the inner surface is not determined by the radiation it emits, it should stay the same temperature as it reached without the shell.

    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. “ ~Trick

    UCB sphere with no shield emits J at temperature T to a background at 0K.
    UCB sphere with shield emits J at temperature T to the shield, which emits J₁ to the sphere and to the background.

    Inserting the numbers from the thought experiment:

    Sphere_noShield and sphere temperature of 254K emits 235 W to the background which is at 0K so it emits 0 W back.

    Sphere_yesShield and sphere temperature of 254K emits 235 W to the shield which is at 213.36K so it emits 117.5 W back and 117.5 W to the background at 0K.

    [Reply] See you around on another thread. If you say something worth publishing.

  9. Trick says:

    Bryan 4:44pm: “For temperatures where B > C, A loses less heat than if C > B where in fact A would lose more heat than if B did not exist. So backradiation from B does not always slow down heat loss from A.”

    Oh my, Bryan. Yes.

    When top post shell B is put in place with starting temperature greater than deep space C ~2.7K CMB, planet A always loses less heat (cools slower). This is Willis’ presumed set up and the top post does specify: “..add a shell around the planet. The shell warms up and it begins to radiate as well..”

    OTOH as Bryan says, when top post shell B temperature is put in place with a temperature less than deep space C ~2.7K and before shell warms up, planet A loses more heat (cools faster) with shell B in place until shell B temperature becomes higher than 2.7K CMB heated up from the planet core heat source.

    We can count on Bryan to set us straight when Willis or anyone else sets up a top post example with shell beginning say at T=1K and wants to know what happens to delta planet energy flux while shell is less than 2.7K, LOL.

    Homework: compute the energy to cool an equilibrated planet sized shell around the planet to 1K. State any difficulty; explain why “…object C not often commented on.”

  10. tallbloke says:

    Reluctantly, I’ve retired Max from the thread. Hopefully he use the spare time to try a simple experiment to test his hypothesis that adding a steel shell around a heat source will make no difference to the surface temperature of the heat source.

  11. Bryan says, April 1, 2013 at 2:53 pm: 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.”

    No. Radiation directed from a cooler body towards a warmer body will slow down the rate at which the warmer body cools.

    “Slowing down the rate at which a warmer body cools” is NOT the same as “keeping a higher temperature body warmer”. The first statement is about a cooling rate over time. The second implies that the cooler body is in some way able to heat the body, which is incorrect.

  12. Tim Folkerts says:

    David says: ” “Slowing down the rate at which a warmer body cools” is NOT the same as “keeping a higher temperature body warmer”. ”

    And we are back to semantics. 😦

    The physics is clear.

    The math is clear.

    * The planet would be ~ 254 K = ~ (235/sigma)^0.25 with the nuclear heaters and no shell
    * The planet would be ~ 302 K = ~ (2×235/sigma)^0.25 with the nuclear heaters and a shell
    * The presence/absence of the shell is responsible for this difference of ~ +48 K.

    It is only the words like “slows the cooling” or “warmer than it would have been” or “heats up ” or “warms up” that seem to be causing problems now. Those who know the physics and math don’t need to fret over the semantics. The calculations for the two cases are straightforward and have been presented numerous times in this thread.

    The words — especially “heat” — are a quagmire waiting to suck in the unsuspecting. Use the symbols!

  13. gbaikie says:

    “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.”

    1. mirror can cooler than the energy it is reflecting towards a warmer object.

    2. If put warm object in space, it can radiate in any direction and the difference in heat is it’s temperature and the 2 K of space. If put object warmer than 2 K next to it, you reducing the area
    that it can radiate into with this temperature difference.
    Per the way I explained it, the Willis steel greenhouse is NOT limiting the area in which heat can flow. Because with such an object in space, one does consider the inward direction as option for this warmed object to radiate into space. Rather it’s outward at 360 degrees into a 3 dimensional space.

    If Willis steel greenhouse did not involve a ideal blackbody surface- something that absorbs emits all radiant energy. Then in this case it could warm the core surface, by inhibiting the flow of the 235 W/m-2 radiant energy leave the core.
    And if using a surface which reflects nearly all the radiant energy of the core surface back to itself
    one will cause the core surface to warm at fastest rate. And if couple this with the surface which is facing outward away core surface which has very low emissivity. Then core surface can become very hot.

    So core surface temperature can be made to be very hot by using radiant insulation- but blackbody
    surfaces are NOT insulation.

  14. gbaikie says:

    “tallbloke says:
    April 1, 2013 at 6:55 pm

    Reluctantly, I’ve retired Max from the thread. Hopefully he use the spare time to try a simple experiment to test his hypothesis that adding a steel shell around a heat source will make no difference to the surface temperature of the heat source.”

    Using Steel would as important as the surface of the Steel.
    And if one dealing with relatively large areas and higher temperatures than room temperature
    the convection of heat will be a dominate aspect.

    The surface of steel can be about it’s smoothness:
    Stainless Steel, weathered 0.85
    Stainless Steel, polished 0.075
    Stainless Steel, type 301 0.54 – 0.63
    Steel Galvanized Old 0.88
    Steel Galvanized New 0.23
    http://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html

    [reply] Choose one. Doesn’t matter which. They’ll all affect the temperature of the heat source. Even with convection

  15. gbaikie says:

    “Bryan says:
    April 1, 2013 at 4:44 pm

    tallbloke says

    “I don’t think 2. helps much. What sort of coincidence? ”

    You will be aware of Roy Spencer’s “Yes Virginia ” thread and several other similar posts from other people.
    The format seems to confound most peoples common sense by suggesting that colder objects make higher temperature objects ‘warmer than they would otherwise be’”

    Or said differently, people think earth can become much hotter than compare to Earth which was in a vacuum and had an ideal blackbody surface. That Earth can become warmer than uniform temperature of 5 C.
    They imagine that today’s global average temperature is already warmer than a uniform temperature of 5 C- and it is not.
    Their problem is they listen to Hansen and become believers of his nonsense, who believe Earth can or could have become almost as warm as Venus.

    I believe that Earth could become warmer as compared to uniform temperature of C, but it can’t
    warmer than this within a thousand years, as it takes thousands of years to warm the entire ocean by 1 C.

    And I believe that Earth in the past may have been warmer than uniform temperature of 5 C perhaps even as warm as uniform temperature of 10 C, but I am more skeptical of Earth becoming warmer than uniform temperature of 10 C unless it somehow involves “cosmic events”, though I suppose some kind of freak massive volcanic events on Earth could be somehow unrelated to “cosmic events”.

    To think Earth could become warmer than uniform temperature of 10 C because of human activities
    involving the emission of CO2, basically requires the brainlessness of a NASA bureaucrat.

  16. donald penman says:

    The mass of the planet is greater than the mass of the shell the radius of the planet is greater than the thickness of the shell ,the planet would not have to warm very much to cope with the backradiation from the shell.You can’t just deal with flows of energy without taking into account the size of the objects radiating towards each other.

  17. Arfur Bryant says:

    tallbloke says:
    April 1, 2013 at 9:57 am

    [“Did you read my followup to David’s statement? If so, which part of it is there a problem with?”]

    Rog, with due respect, the part where you think the back radiation is absorbed for net gain, i.e, the planet warms up. Losing heat less quickly is fine. Warming up because it is losing heat less quickly is very different.

    If you and David (and lgl and Tim) are correct, you have to accept this scenario:

    Place an electrically powered oil-filled radiator in the centre of a bare room. The electricity powers the radiator which heats up, thus heating the room. Equilibrium is eventually reached. According to you, the mere presence of a second (but not connected to the power supply) radiator introduced into the room in proximity to the powered radiator will be sufficient for the ‘ON’ radiator to heat up, because it is receiving re-emitted radiation from the nearby ‘OFF’ radiator (and, according to you, it is losing heat less quickly).

    Amazing. SImilarly, the heat produced by a one-bar electric fire could be increased by simply adding another bar next to the first one but not even switching the second bar on. Want the first bar to get even hotter? SImple, just add more adjacent bars but don’t connect them to the electrical supply.

    Willis’ model adds the 235 back-radiation to the initial (and ongoing) 235 radiation. But the back-radiation cannot be added because that would require cooler radiation being added for net gain to the initial warmer source.

    I understand the argument that losing heat less quickly is possible, but the toy planet is not losing heat, so that argument is invalid. The back radiation is always ‘cooler’ than the planet’s emission… so it will not be absorbed for net gain… so the surface will not heat up.

    Its not a question of me not getting your argument, I just don’t agree with it.

    By the way, your very first sentence: “I’ll probably regret this.” at the very start of the thread is probably very true if you’re reaching the end of your tether! 🙂

  18. tallbloke says:

    Bryan says:
    Apart from hot object A and cold object B there is the background object C not often commented on.
    For temperatures where B > C, A loses less heat than if C > B where in fact A would lose more heat than if B did not exist.
    So backradiation from B does not always slow down heat loss from A

    Completely agree. In the case of the Willis toy planet setup we have a constantly radiating core surface A which is heating shell B by radiative transfer. C is space at 3K.

    What I’d like to know is whether we are now in agreement about the dynamic situation I outlined at the end of my previous comment, reproduced below:

    “[The toy planet’s surface] has to get hotter until it gets the shell up to the temperature where its outer surface can radiate as much to space as quickly as the toy planet’s surface did before the shell got in the way. Until it gets to that temperature, the shell will emit less to space than is being produced by the nuclear core and the planet will continue getting hotter, because steady state hasn’t been reached.

  19. tallbloke says:

    Arfur says:
    According to you, the mere presence of a second (but not connected to the power supply) radiator introduced into the room in proximity to the powered radiator will be sufficient for the ‘ON’ radiator to heat up, because it is receiving re-emitted radiation from the nearby ‘OFF’ radiator

    I think this is a poor analogy, because the second radiator would only get marginally warmer than the ambient temperature of the room, due to its proximity. Try removing all the walls on a cold night, then standing for extra unplugged radiators all around the powered one and putting one on top of the four as well. Now the powered radiator will get hotter than it was fully exposed to the cold night air.

  20. tallbloke says:

    Donald says:
    The mass of the planet is greater than the mass of the shell the radius of the planet is greater than the thickness of the shell ,the planet would not have to warm very much to cope with the backradiation from the shell.You can’t just deal with flows of energy without taking into account the size of the objects radiating towards each other.

    No time limit was specified for the system to reach steady state after the introduction of the shell. We could work out how long it would take for any mass of planet with a given specific heat capacity and conductivity though.

  21. Tim Folkerts says:

    Arfur says: “If you and David (and lgl and Tim) are correct, you have to accept this scenario:
    … which heats up, thus heating the room … losing heat less quickly … heat produced by a one-bar electric fire … the toy planet is not losing heat … the surface will not heat up…

    … re-emitted radiation … back-radiation .. initial radiation … ”

    Semantics. “Heat” is being used in at least two different ways — probably three (its hard to know). “Radiation” is also used in a vague way, with imprecise statements about “adding” and “net”.

    This makes it very difficult to respond. It be much better to use well-defined symbols. Like “let W = the constant work done by electricity on the oil-filled radiator; let Q = the heat from the radiator to the surroundings; let Φ(planet) = εσT(planet)^4 be the thermal IR power flux from the planet” . (And yes, electrical energy input is considered “work” in thermodynamics since it is not “heat”)

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

    PS. I have no problem with the spirit of your example. Putting insulation and/or reflectors and/or black bodies around the radiator will make it “warm up” (= “heat up” = “get hotter” = “have a larger temperature”) by reducing the heat to the surroundings for a while. ΔU(heater) = W(to the heater) – Q(from the heater). At steady-state, Q=W and ΔU(heater)=0, which means ΔT(heater)=0. Since we are reducing Q when we put a large panel nearby, ΔU will be positive until the heater heats up enough for Q to increase to once again match W.

    Now if you want to object to something I said, you can point to a specific equation or statement. Maybe you disagree with “Since we are reducing Q when we put a large panel nearby” and people can hash that out.

  22. Westy says:

    Apologies for my last post. Beer + blog = silly comment.

    Even so, because the argument keeps going around and around, what comes first? The shell can’t warm planet unless shell is / gets warmer then the planet. Or planet heats because the shell inhibits planets cooling. Correct me if I’m wrong, and I’ve been wrong many times, but Willis intended a non insulative shell so I’m still in the non heating planet camp.

  23. gbaikie says:

    [reply] Choose one. Doesn’t matter which. They’ll all affect the temperature of the heat source. Even with convection.

    The emissivity and absorption of radiant energy of materials as little to do with the heating requirements of a furnace [a heat source] as compared to conduction of heat and convection of heat. But if we lived in vacuum it would largely have to do the emissivity and absorption of radiant energy.

    To test affect of this steel sphere, one needs to mimic conditions of a vacuum which is surrounding by 2 K space. And we unable to mimic this this precisely- unless one goes into space. But we can
    make conditions fairly close to this.

    Say take steel sphere weighed so it will sink water, and have heater in a hollow part of sphere, have vacuum [does need to be vacuum like space as this would difficult to do]. Have heater insulated from conduction of it’s heat- so a heater is solid sphere suspended non-metallic string
    within sphere and has batteries within solid sphere and something to measure it’s temperature.

    In air with temperature of 5 C one has solid sphere heated to 10 C. This internal temperature can
    varying depending type surfaces of the steel [interior and exterior surfaces].
    If put sphere in water which was 5 C one would effectively turning exterior surface into a blackbody surface as it would behave in 5 C air. So the emissivity of surface it say .02, would not matter if placed in 5 C water, it would behave as though it were an ideal blackbody.

    So the solid sphere within the steel sphere would be warmer with exterior surface having low emissivity in air which was 5 C as compared to surface which behaved as a ideal blackbody.
    I hope other than nitpicking about how much temperature difference and ability to measure
    this difference there is no dispute, so far.

    But your question involves can we make the inner sphere cooler than 10 C.
    Instead water at 5 C, we could water at 0 C, or colder ice, or something like liquid nitrogen- all of which are warmer than 2 K space.
    Is 0 C water colder than space. Put astronaut in spacesuit in 0 C water and he will be cold unless wearing warm clothes plus one would not want to circulate cooled water which normally done with spacesuit in space to keep astronauts cool.
    In simple terms if one can keep warmer in space if the vessel is designed to keep warm whereas with submarine one can’t alter the emissivity of it’s outer surface to affect how much heat is lost- it’s
    conduction and convection of heat- which can be insulated but not by 1 mm of material, unless the “material” is vacuum.
    But our sphere has some vacuum, and because a vacuum or simply less air, the 0 C water could be warmer than space. But with liquid nitrogen one could getting colder than space.
    So put the sphere in liquid nitrogen and the heated solid sphere could become colder than 10 C, and if the emissivity of inner part of steel shell is similar to blackbody surface it will become solid sphere will become colder as compare surface which absorbs less radiant energy as a blackbody surface would.

  24. gbaikie says:

    “I understand the argument that losing heat less quickly is possible, but the toy planet is not losing heat, so that argument is invalid.”

    Net heat loss isn’t being changed.
    The same area core surface radiating 235 W/m-2 is close to total area of shell sphere radiating
    235 W/m-2. But don’t see what you mean by toy planet not losing heat. The toy planet is losing a staggering amount of heat. I suppose you mean It’s not losing or gaining in it’s loss of heat.
    Though it’s generation of heat is not changing would an easier way to say it.

    But net heat loss is simply the amount energy generate or captured [sunlight warming earth].
    We can agree that energy production of core planet is not increasing, and so one must radiate the same amount generated.
    But this is unrelated to temperature of various spot/location on or within the sphere.
    The interior of core could be as hot as interior of planet Earth [consider the vast amount geothermal energy- it should much hotter interior than Earth’s interior- or different crustal material as compared to Earth. So it could same high temperature as Earth’s interior temperature, though it could much higher, or even much lower].
    But we seem interested in temperature of surface. Which btw seems to be calculated with assumption that it radiates like a black body- that it’s surface emits near 1.
    If one assume the core surface emits near 1, then it’s temperature will be around -18 C.
    So a very cold planet at skin surface emitting a vast amount thermal energy is what we beginning with before steel shell is added.
    The Universe may have such a strange planet, but only because the Universe could have many strange things. It certainly alien world to the Earthlings, any reasonable scientists upon seeing could assume it’s not natural- evidence of some kind of alien construction perhaps.
    No doubt the number of stacked impossibilities regarding it, is one most interesting aspects of it.

    Though considering the amount of confusion regarding this, I wonder whether the geologist have correctly measured the amount of Earth geological heat. I don’t doubt Earth isn’t generating much
    energy that heats Earth, but rather if had 235 W/m-2 of geothermal heat, it wouldn’t heat Earth by much either. The big difference would much shallower crust, and very frequent volcanic eruptions- or we have oceans of lavas rather than oceans of water. [They lava oceans would heat the place up, but I mean not the ground which heats up to -18 C]. Or the geothermal heat is about 1/10 of watt on average per square meter, but has it always been [last couple billion years]. It could increase by 100 times this amount and not make a huge difference [other than shallower crust and more volcanic activity. I suppose keeping around the same amount global land mass indicates to some degree of constant geothermal heat.

  25. wayne says:

    “Willis’ model adds the 235 back-radiation to the initial (and ongoing) 235 radiation. But the back-radiation cannot be added because that would require cooler radiation being added for net gain to the initial warmer source.”

    AB, almost like the Abbott and Costello skit on “Who is on first base” isn’t it? LOL. Hard to come away after looking at the 470 W/m² and wondering where IS “that extra energy” actually coming from?

    Where does the addtional 235 come from? It comes from the sphere. But the sphere only has 235. No, it has 470. Well where did that additional 235 come from. It came from the shell. I think I already asked that but, so where does that 235 the sphere holds come from. The shell. Where did the shell get the 235. The sphere. But I’m sure the 235 from the sphere went through the shell and to space. Ok, so you have 235 left over. So take 235 from the shell and add it to the 235 that was left over. But there is now no left over. Yes there is. No there isn’t. That went to space. Now wait, you’ve got this all wrong….

    Can you just see a great skit on “Ok, who’s got the Back Radiation?”. 😆

  26. Tim Folkerts says:

    Wayne, say it with math and you will be SO much closer to understanding!

    * The sphere “has” internal energy “U”, which is NOT (235 W/m^2) * A.
    * The sphere continuously gains energy from the heaters at a rate of dW/dt = (235 W/m^2)*A
    * The sphere continuously looses heat,Q, to the surroundings (at varying rates depending on the recent circumstances — rates that could be MORE than 235 W/m^2 or LESS than 235 W/m^2).
    * Q is composed of thermal IR from the planet and thermal IR into the planet from the surroundings, as determined by the S-B Law.

    Your “Who’s on first” routine only sounds vaguely plausible because you are treating U, dQ/dt, dW/dt, and IR fluxes as if they were the same thing. That would be kind of like treating deposits, withdrawals, and net balance as the same thing and wondering why you can’t balance your checkbook.

  27. donald penman says:

    Tb
    The reply you gave to me assumes certain simplifications of the problem which i don’t accept.If there is back radiation from the shell then given the fact that as far as i know in this situation I can’t have this radiation not being absorbed by the planet then the surface must get warmer but I don’t accept that this heat travels from the surface into the centre of planet against a temperature gradient.I am looking at if the planet heats up itself with its own heat because of the restriction of energy loss from the surface and i think that the adjustment in the temperature gradient of the planet given its size relative to the surface would only need to be very small to cope with the warming of the surface. I think that some of us are determined to make us accept the unreal outcomes implied by this unreal model but I will just have to hold to my own view on what might happen here despite what everyone else assumes here will happen.

  28. gbaikie says:

    “Where does the additional 235 come from?”

    It can only come from the core.
    To have a surface which radiates 235 W/m-2 to have that same surface increase to 470 W/m-2 *requires* more heat.
    If the core does not have the power or intensity to warm this surface from 235 W/2-2 to 470 W/m-2
    and there no other source heat [and there isn’t] which is able to provide such intensity so as to get hot enough to radiate this intensity of watts per square meter, then it will not ever get this warm.

    It’s like expecting a thermos bottle making it’s contents hotter. Or if thermos bottle is meant keeping something colder than environment- making it cooler.

    Since interior of core is much hotter than surface of core and if you insulate the surface core this allows surface to eventually become hotter.
    But ideal blackbody doesn’t insulate, therefore shell with it’s surfaces designed to act like an ideal
    blackbody won’t insulate.
    So if follow the instruction given by Willis, his model does not increase
    the core surface temperature.

    There is the other aspect, one could think the shell is radiating energy at the surface but not warming the surface. Which means one thinks the core surface is reflecting this energy [or re-radiating it].
    And this is sky dragons diagram in which you have 235 W/m-2
    “re-circulating”.
    But I think, if there was actually an ideal blackbody, that it would not do this- if one assuming the ideal blackbody conducted the heat to the steel as well as say, silver would conduct the heat.
    Of course there isn’t actually a thing which is an ideal blackbody surface.
    Nor does anything convert any kind energy to a different kind of energy with100% efficiency.

  29. wayne says:

    No Tim Folkerts, no, won’t say it as math. IT WAS A JOKE. Ok?
    See the laughing face? That was a hint if someone had never heard of Abbott & Costello.
    Look it up. Wait, I’ll just look it up for you…

  30. wayne says:

    “Where does the additional 235 come from?”
    Sure hope you took my last comment as a joke! It was meant that way.

    gbaikie: It can only come from the core.
    Right, I agree with what you are saying, and it must have the power to do so.

    Along those lines, since that 235 power is created between the time a shell is added and a steady state is reached (and by integration it is exactly just that, the 235 wm-2 in this shell example), I guess that puts the age of our real “back radiation” quantity at, what, a billion years old? Older? Much older? That is how long our “back radiation” has been hanging around. By the way some others here think, it’s actually been bouncing back and forth, back and forth, not even needing to be spontaneous, nearly the same quantity, a little deviance here and there, for eons. It’s an old, old friend. Don’t think life would have even been possible and created without “back radiation” so long ago. 😉

    Two thoughts to consider from all of this discovery of shells:
    1) Since our atmosphere is already totally opaque to non-window lines, many times over, how does that play into the concept of a shell.
    2) If the answer to one is it moves the shell further away, I thank TimF for proving that this makes zero effect on the spheres temperature. It is invariant to distance. But might keep an eye on the total mass of the atmosphere… this discussion has shown so clearly that it does matter, besides the sun itself.

  31. tallbloke says:

    Donald: No-one can be forced to change their mind, and it’s not my intention to do so. I’m just trying to show through many different ways of stating the theoretical situation that the logical outcome is that the nuclear heated planet surface with no atmosphere will end up warmer with a steel shell round it than without. Simple experiments show some warming too.

  32. Arfur Bryant says:

    tallbloke says:
    April 1, 2013 at 11:45 pm

    [“I think this is a poor analogy, because the second radiator would only get marginally warmer than the ambient temperature of the room, due to its proximity. Try removing all the walls on a cold night, then standing for extra unplugged radiators all around the powered one and putting one on top of the four as well. Now the powered radiator will get hotter than it was fully exposed to the cold night air.”]

    Will it?

    The analogy was never meant to be exactly representative; it was meant to demonstrate that the ‘concept’ of your ‘back-radiation-causes-a-powered-object-to-heat-up-by-losing-heat-less-quickly…’ is wrong. Even if the unplugged radiator does not surround the initial ‘ON’ radiator, it will nevertheless re-emit some radiation toward the initial object. Herein lies the crux of the debate. The re-emitted radiation from the ‘OFF’ radiator/s does not ‘add’ to the temperature of the ‘ON’ radiator. You and David are arguing that the Willis model planet heats up because the back radiation from the shell is added to the initial 235 W/m^2, in the guise of ‘losing heat less quickly’.

    If you don’t like the radiator in the room analogy, try this one:
    The boiler in your house heats the water which is pumped through (e.g.) copper pipes. The boiler is the constant power source. According to you (and David/Tim/lgl), you could increase the temperature of the water running through the output copper pipe simply by placing a larger-diameter copper pipe around the initial pipe, so the back-radiation from the larger pipe would ‘add’ to the temperature of the narrow pipe simply by making it lose heat less quickly.

    The trouble is, the only way the water can heat up is by absorbing additional energy for net gain, not by absorbing its own emitted radiation at albeit a slightly different wavelength. The shell radiation will always be ‘cooler’ than the planet.

    All,

    Willis’ model is wrong on at least three counts:

    a. The 470 arrow ‘up’ is impossible without the core absorbing extra energy for net gain, and thus heating up (which is what Willis states). gbaikie is correct in his assessment here:
    [“gbaikie says:
    April 2, 2013 at 7:27 am
    “Where does the additional 235 come from?”
    It can only come from the core.
To have a surface which radiates 235 W/m-2 to have that same surface increase to 470 W/m-2 *requires* more heat.
If the core does not have the power or intensity to warm this surface from 235 W/2-2 to 470 W/m-2 
and there no other source heat [and there isn’t] which is able to provide such intensity so as to get hot enough to radiate this intensity of watts per square meter, then it will not ever get this warm.”]

    b. The 235 arrow to space is inaccurate because the shell surface area is greater than the planet surface area. I know (Tim) that the difference is small, but it serves to demonstrate that the shell will always be ‘cooler’ than the planet. Everyone seems to agree on that.

    c. The 235 arrow from the shell to the planet is inaccurate for two reasons: firstly, the sphere cannot emit at 235W/m^2 because of surface area difference and, secondly, not all of the radiation emitted from the inner surface of the sphere will strike the planet. Some will miss the planet and strike other parts of the shell. Either way, the planet does not receive 235 from above.

    This thread demonstrates that the partisan approach to the cAGW debate (warmist/sceptic) in no way absolves any single person from pursuing reason or logic. I will accept that I am wrong if my logic can be shown to be faulty. My reasoning tells me that Willis’ model is wrong.

  33. Arfur Bryant says:

    Wayne,

    Got the joke. I was thinking more along the lines of the Parrot sketch:

    “It’s not dead, it’s just resting…” 🙂

  34. Arfur Bryant says:

    gbaikie says:
    April 2, 2013 at 2:43 am

    [“I suppose you mean It’s not losing or gaining in it’s loss of heat.
    Though it’s generation of heat is not changing would an easier way to say it.”]

    Yes. Agreed.

  35. Arfur Bryant says:

    Tim Folkerts says:
    April 2, 2013 at 12:08 am

    Tim, please see my response to tallbloke above. It should explain what I mean. I think I am making my case pretty clearly.

    [“* Q is composed of thermal IR from the planet and thermal IR into the planet from the surroundings, as determined by the S-B Law.”]

    Define Q again? “* The sphere continuously looses heat,Q,…” If Q is the rate of heat loss then fine, the planet’s ability to lose heat is reduced. This does not make the planet warmer. If Q is the temperature of the planet, then it is not increased by the cooler radiation from the shell.

    2 bunches each of 235 bananas = 470 bananas
    2 cups of water each at 5 deg C does not equal 10 deg C.

    The radiation which was initially emitted from the powered core cannot be added to itself for net gain of energy or heat.

    I might easily reach the limit of my scientific knowledge here, but I will not reach the limit of logic. At its most basic level, science should make sense.

  36. tallbloke says:

    Arfur quotes tallbloke:
    tallbloke says:
    April 1, 2013 at 11:45 pm

    [“I think this is a poor analogy, because the second radiator would only get marginally warmer than the ambient temperature of the room, due to its proximity. Try removing all the walls on a cold night, then standing for extra unplugged radiators all around the powered one and putting one on top of the four as well. Now the powered radiator will get hotter than it was fully exposed to the cold night air.”]

    And responds:
    Will it?

    Yes. Try it.

    The analogy was never meant to be exactly representative; it was meant to demonstrate that the ‘concept’ of your ‘back-radiation-causes-a-powered-object-to-heat-up-by-losing-heat-less-quickly…’ is wrong.

    Well in my opinion my setup is much more like Willis’ proposition, because the differentials are closer to his gedanken experiment, and the powered radiator is fully surrounded, like his shell surrounding his internally heated toy planet

    Arfur offers a third setup:
    pipes, boiler, tubing, hot water, big plumbing bill.

    I’ll stick to the cheaper, more representative and easier to organise experiment I think. 😉

    On to the categorical claims:

    a. The 470 arrow ‘up’ is impossible without the core absorbing extra energy for net gain, and thus heating up (which is what Willis states)

    1) This is not what Willis states. This is what you state to be what he states necessarily entails. But….
    …2) There is no “extra energy” just the same RATE OF ENERGY SUPPLY being lost more slowly from the toy planet surface/shell system until the surface gets hot enough to cause the shell outer surface to lose energy to space at the same rate as the surface did did before it was surrounded by the shell.

    b. The 235 arrow to space is inaccurate because the shell surface area is greater than the planet surface area. I know (Tim) that the difference is small, but it serves to demonstrate that the shell will always be ‘cooler’ than the planet. Everyone seems to agree on that.

    Yes. It will be ~234.9W/m^2 but because there are more m^2, the total number of joules per second (watts) lost to space will be the same as was lost from the planet before it was surrounded by the shell, once the system reaches steady state.

    c. The 235 arrow from the shell to the planet is inaccurate for two reasons: firstly, the sphere cannot emit at 235W/m^2 because of surface area difference and, secondly, not all of the radiation emitted from the inner surface of the sphere will strike the planet. Some will miss the planet and strike other parts of the shell. Either way, the planet does not receive 235 from above.

    Yes. The the energy emitted by the shell and re-absorbed by the shell without striking the planet will be part of what maintains its temperature.

    This thread demonstrates that the partisan approach to the cAGW debate (warmist/sceptic) in no way absolves any single person from pursuing reason or logic. I will accept that I am wrong if my logic can be shown to be faulty. My reasoning tells me that Willis’ model is wrong.

    Sure, we are both doing our best to be logical within our own frameworks of understanding. I respectfully suggest that what we should be trying to understand here is the why the temperatures of the surfaces of the components of the system are what they are due to the need for the number of joules per second lost to space to be the same as the number of joules per second supplied to the toy planet surface by the nuclear core in order for the system to be in a steady state. Working out what is happening during the period of time between surrounding the toy planet with the steel shell and the composite system reaching a steady state is the key to this understanding IMO. Perhaps we could at least agree on that as a methodological approach and take it from there.

    If so, here’s my first suggested step.

    1) We agree that before one of us can point at the other and say “told you so”, the system has to reach as steady state after the shell is placed round the toy planet so we can get reliable unchanging temperature readings from the surfaces.

    Agree or disagree?

    If you agree, here my suggested second step:

    2) We agree that to be in a steady state, the composite toy planet/shell system will have to be losing energy to space at the same rate it is being produced by the nuclear core.

    Agree or disagree?

    If you agree, I’ll move on to my suggested third step in my next reply to you.

    Cheers

    TB.

  37. Tim Folkerts says:

    Wayne,

    Sorry if I missed the attempt at humor. Even with smileys it is difficult to recognized whether the humor was directed at those who AGREE with Willis & standard science, or directed at those who DISAGREE with Willis and standard science.

    We need scorecards to keep track who concludes what!

  38. tallbloke says:

    I think the DISAGREE’s still have it on a majority vote. But science isn’t about majorities or consensus, is it Tim? 😉

  39. mkelly says:

    From Willis’ “Steel Greenhouse”: “The planet is now being heated by 235 W/m2 of energy from the interior, and 235 W/m2 from the shell. This will warm the planetary surface until it reaches a temperature of 470 watts per square metre.”

    Tallbloke says: “1) This is not what Willis states.”

    I am afraid it is what Willis states Tallbloke. He uses the word “heated”,”and” and “warm”. So Arfur has it right as to the criteria of Willis’ steel greenhouse.

    If Willis said that the surface put out 235 and the shell from both sides put out 117.5 I would say nothing as would others I presume. This would comply with the example Max had from NASA and close to the equation I gave for a radiation/heat shield. But he does not. So on the heat shield idea the presented picture is wrong.

    Willis’ gave an equation from his book he used or would use and it is like the one I gave with the shape or view factor. Q = A1 * F(1-2) * G1 – A2 * F(2-1) * G2. He used it is a comment on the latest discussion of the planet/shell idea. Again the shape factor and area are determining factors.

  40. Tim Folkerts says:

    mkelly says: ” He uses the word “heated”,”and” and “warm”.”

    Again, this is semantics. “Heat” is perhaps the single most abused and misused word in thermodynamics. Basically. any time you see the word “heat” you have to ask yourself “now what did the author really mean there?”.

    In some cases, even the author has no idea what was really intended. In this case, Willis seems to be using heat = “some energy that contributes to the thermal energy of the planet” rather than the more technical definition “the net flow of thermal energy”.

    Words are a poor substitute for equations. If people want to understand, they MUST be able to express the ideas using equations.

  41. Tim Folkerts says:

    mkelly says: Q = A1 * F(1-2) * G1 – A2 * F(2-1) * G2

    Now we have something specific to discuss. This looks like the equation at http://en.wikipedia.org/wiki/Thermal_radiation#Radiative_heat_transfer.
    A = area
    F = view factor
    G = E = sigma T^4
    Q should really be “Q-dot” = dQ/dt

    For closely concentric shells, then A1 ≈ A2 and F(1-2) ≈ F(2-1) ≈ 1. Following wikipedia, this becomes
    dQ/dt ≈ A sigma (T1^4 – T2^4)

    Yes, the shape factors and areas are “determining factors” but in the model presented with closely concentric shells, these factors “determine” that
    dQ/dt ≈ A sigma (T1^4 – T2^4)
    is a good equation for the heat transfer in this situation. Only when the shell is “significantly” above the surface will this equation be a “bad” approximation.

  42. wayne says:

    Tim, so you don’t like being more accurate and you don’t like the way everyone today uses the word “heat”.

    I guess that includes Wikipedia as they say “radiative heat flow“, they speak of “The radiative heat transfer…”, Maxwell’s book speaks of it the same. I speak of it the same. MKelly speaks of it the same. Same with Willis. But not you. Do you not notice one is a noun and the other a very confusing use as a verb? You say this is just semantics and I agree, but why do you always waste so many words? Hmm, there must be some reason.

    Personally, I think it great others here are keeping track of every “pea” and “Q”.

  43. mkelly says:

    Tim Folkerts says:

    April 2, 2013 at 3:21 pm

    Actually, Mr. Folkerts the equation is from Willis’ book, but is fundamentally the same as the one I used in an earlier comment. I posted the same equation earlier with an explanation of all the variables. And since Willis’ steel shell is “several thousand meters” from the planet your “closely concentric” does not apply. Nice try though.

    Also I give Willis credit for knowing exactly what he wants to say. I will not think him a fool and use words incorrectly. He said what he said and I take him at his word. So the planet and shell as presented is wrong per him and the heat transfer equations.

  44. Tim Folkerts says:

    Arfur says:
    > Define Q again? “* The sphere continuously looses heat,Q,…”
    Q is the common symbol for heat, the net transfer of thermal energy due to a temperature difference.

    >If Q is the rate of heat loss then fine, the planet’s ability to lose heat is reduced.
    dQ/dt would be the rate of heat loss (often indicated by “Q” with a dot over it).

    >This does not make the planet warmer.
    ΔU = Q + W (using the convention where heat *into* and work *on* the system are positive)
    This is the First Law of Thermodynamics.

    This could also be expressed as rates of change by differentiating with respect to time.
    dU/dt = dQ/dt + dW/dt

    In steady-state, dQ/dt = -dW/dT and dU/dt is zero (and consequently dT/dt = rate of change of temperature = 0).
    In our example dW/dT = (235 W/m^2)*A = constant.
    If “the planet’s ability to lose heat is reduced”, while keeping a constant dWdt = 235 W/m^2 input, then necessarily dU/dt is no longer zero and the temperature will change.

    Decreasing the magnitude of dQ/dt WILL cause the planet’s temperature to increase.

    >2 cups of water each at 5 deg C does not equal 10 deg C.
    The analogy would be more like “a bucket of water with a 10 W electric heater reaches a temperature of 35 C (in a room at 20 C) — so that dW/dt = +10 W and dQ/dt = -10 W. If the heat loss is decreased from 10 W to 8 W (perhaps by adding a little more insulation), the temperature of the water WILL immediately start to increase — and it will continue to increase until the water is hot enough to loss 10 W through the better insulation.”

    >The radiation which was initially emitted from the powered core cannot be added to itself…
    This is a common misunderstanding. It is like saying that if I pay someone $100, he cannot later use that same money to pay me at some later date. The money I paid him yesterday can easily be added to the money I get from my work to increase my net balance.

    Similarly, the energy that went to the shell earlier is now the shell’s energy.

    > At its most basic level, science should make sense.
    Sort of. But not many people think relativity or quantum mechanics “make sense”. Even topics from basic mechanics like centripetal force don’t make sense.

    I have heard education described as the process of improving your intuition. It takes a while until this sort of stuff becomes intuitive and makes sense.

  45. Tim Folkerts says:

    mkelly says: ” And since Willis’ steel shell is “several thousand meters” from the planet your “closely concentric” does not apply. Nice try though.

    Sigh.

    It is not the ABSOLUTE difference that matters, but the RELATIVE difference.

    1 km difference out of 6370 km is “small” — it is 0.016%. Even “several thousand meters” will be less than 0.1%. Only if we are worrying about errors on the order 0.1% does this correction matter.

    So if we are being especially careful, the shell might be ~ 0.1 K cooler than the “simple” approximation and the planet in turn might be ~ 0.1 K cooler than the “simple” approximation. Since we have already been rounding to +/-1 K and +/- 1 W/m^2 anyway, the “more accurate correction” is basically “lost in the noise”.

    Yes — the “radius cooling effect” exists and is real
    No — it is not a large or important correction to the basic conclusion that the shell will be ~ 254 K and the planet will be ~ 302 K.

    Nice try though. 🙂

  46. Arfur Bryant says:

    mkelly says:
    April 2, 2013 at 2:39 pm

    [“I am afraid it is what Willis states Tallbloke. He uses the word “heated”,”and” and “warm”. So Arfur has it right as to the criteria of Willis’ steel greenhouse.”]

    mkelly,

    Thank you. That quote from Willis is exactly what I would have used in my reply to tallbloke.

    **

    tallbloke says:
    April 2, 2013 at 11:10 am

    tallbloke,

    There is no intention on my part to indulge in any “I told you so…” stuff. I hope you know me (digitally) better than that. I’m just trying to get to the truth behind the Willis model.

    As to the analogies, I understand if you don’t like them. Fine, I won’t bother, although I think the hot water pipe analogy is good. IMO the ‘key to understanding’ the model (because that is what this thread is about) is one simple area: Is it possible for radiation re-emitted from a cooler object to be absorbed by a warmer (and constantly powered) object for net gain (i.e, resulting in an increase in thermal energy)?

    Most here (and i think this includes those of you who agree with Willis) are quite happy to accept that heat (energy) cannot transfer from cold to hot. You and David (at least) have decided that re-emitted radiation from the cooler shell (you agree) will cause the temperature of the surface of the (toy) planet to heat up because the back radiation slows down the rate of heat loss. [“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.”]. The phrase “…and so the temperature rises.” is where the error lies. A rise in temperature is the same as ‘heating up’, which contradicts the words used earlier “It doesn’t heat it…”. You can’t have it both ways. Either the re-emitted radiation from the cooler shell causes the surface to heat up or it doesn’t. Your argument that it heats up ‘by proxy’ is invalid, as it would mean that heat really can flow from a cooler to a warmer object. Remember the radiation from the shell comes from the 235 arrow initially from the planet. There is no other source. Therefore you have to address where the 470 W/m^2 comes from. According to Willis’ model, the only way the planet can emit that is by heating up; the only way it can heat up is by (somehow) the radiation from the cooler shell. It doesn’t matter how you or David spin it, by your argument the cooler shell is responsible for the planet heating up. That is the logical fallacy IMO.

    Ok, to address your suggestion: “Perhaps we could at least agree on that as a methodological approach and take it from there.”

    Yes, as I am still happy to play the ‘Thread Idiot’, I’ll go along with that. 🙂

    1) Agreed.
    2) Agreed.

    Cheers,

    Arfur

  47. Tim Folkerts says:

    Wayne says: “Tim, so you don’t like being more accurate and you don’t like the way everyone today uses the word “heat”.

    More semantics.

    1) It is correct that I don’t like the way that “everyone uses the word ‘heat’ “. That is because “everyone” includes the myriad of people who use heat to mean ..
    * temperature (the heat is unbearable today)
    * thermal energy (boiling water has a lot of heat)
    * to do work (he heated his hands by rubbing them together)
    None of these matches the formal thermodynamical definition of “heat”.

    2) I do like using words accurately, which is why I have tried to only use “heat” in its correct thermodynamic sense throughout this discussion. But even more, I like to understand concepts and understand people. Do is wish Wills had been a little more careful in his choice of words? Of course! Am I going to try to derail a discussion because even scientists sometimes make mistakes or uses words colloquially? Nope! (although I might well suggest that they be more careful in the future.)

    3) To be pedantic, “heat” is only officially a noun in thermodynamics. Look at that article you referenced from wikipedia and I don’t think you will find the word “heat” used as a verb. Thus ANY time that ANYONE uses “heat” as a verb, they are being informal. Willis was using an informal meaning of “heat”, so we have to accept that it is not clear without context what he might have actually meant.

    What I *do* like is ΔU = Q + W. 🙂

  48. Arfur Bryant says:

    Tim Folkerts says:
    April 2, 2013 at 9:12 pm

    Tim,

    I am perfectly happy to admit to you and everyone else that I am not a mathematician and have very little education that topic. I am unable to follow your equations so I will gracefully refrain from indulging in such a debate. If you consider that to be a failure or an admission of such by me, then so be it.

    I am a ‘big picture’ guy. If you think the equations you have shown somehow prove that adding insulation can heat the core, then I must concede on the equation side of things. I can only repeat that ‘reducing rate of heat loss’ is not the same as ‘increasing temperature’. The core has heated the shell by radiation. If, in doing so, the rate of heat loss decreases, that makes sense to me. But the core heating up doesn’t make sense. Why would it? It has no reason to heat up; it is just losing energy at a reduced rate. The shell is also losing heat to space. Even accepting the 235 out to space is accurate (which I don’t), 235 is being lost from the system while 235 is delivered to the system. There is no need for a 470 arrow. There is no need to invent an increase in energy output from a source which at all times has been either warmer or equal to the rest of the system.

    You can change the analogy of the water cups if you want, but that analogy was to show the concept of not always being able to use numerical addition to achieve the desired result.

    That’s how I see it… non equationally! 🙂

  49. tallbloke says:

    Arfur: If you think the equations you have shown somehow prove that adding insulation can heat the core

    When you wear a duvet jacket outside on a winters day, rather than a T-shirt, the reason you are warmer is that your ‘core’ energy generating warmth at your skin isn’t lost so easily to the environment. Now, that sort of insulation doesn’t work in the same way as the radiative shell in Willis toy planet model, but the result is similar.

    The key point is that the effect isn’t anything to do with the duvet ‘creating extra energy’. It is to do with the rate at which the heat from your core escapes to the environment.

    Thank you for agreeing with my first two propositions.

    To recap:

    1) We agree that before one of us can point at the other and say “told you so”, the system has to reach as steady state after the shell is placed round the toy planet so we can get reliable unchanging temperature readings from the surfaces.

    2) We agree that to be in a steady state, the composite toy planet/shell system will have to be losing energy to space at the same rate it is being produced by the nuclear core.

    So, on to proposition 3:

    3) We agree that the thin steel shell will radiate from both sides equally, since both the inside and outside surfaces will be at the same temperature.

    If we agree on this, then as a logical deduction:

    4) We agree that the shell is radiating the same amount back towards the planet as the planet was originally radiating directly to space.

    Agree or disagree?

  50. Tim Folkerts says:

    Arfur,

    No analogy is perfect, but consider the following as a way to appeal to “common sense” and limit the need for equations …

    The planet will be represented by a large air tank, and energy will represented by the air. The heater is like an air pump that pumps air into the tank at a fixed rate — say 235 g/s. This will cause the pressure (analogous to temperature) to to build up in the tank (the total amount of of air inside is analogous to total thermal energy, U). This tank has various leaks — the seams were not welded very well. It turns out that when the pressure has built up to 254 PSI, the air leaks out at 235 g/s — the same rate is is being pumped in (analogous to heat “leaking from the planet as fast as it is being added by the heaters).

    Conservation of energy is analogous to conservation of air molecules.
    The “Second Law” says air spontaneously moves from higher pressure to lower pressure.

    Now, suppose I surround the tank with another wall identical to the first (except a fraction larger so it will fit around the other without touching). This wall has the same poorly welded joints.

    What will happen? Well, at the first instant, there is no pressure between the walls, so no air leaks out from the outer wall, and air starts to build up between the walls. As the pressure starts to build up, the air starts to leak faster and faster out from the outer wall (but still less than the 235 g/s that are being added. This means that the total air inside is building up. This ALSO means that the air will start to leak more slowly through the inner wall (because the pressure difference is decreasing between the original tank and the the space around it.

    Eventually the pressure will build up until the pressure BETWEEN the walls is up to 254 PSI and the air is leaking at 235 g/s from the outer wall. The air in the inner tank would be 2×254 PSI so that air would leak through the inner wall at 235 g/s. (This is one place the analogy breaks down — the pressure would presumably increase by a factor 2, while the temperature only increases by 2^(0.25) ).

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

    * I never had to “create” air molecules to make the pressure in the tank increase from 254 PSI to 508 PSI. This occurs naturally if I limit the escape of air.
    * Air never flowed from low pressure to high pressure.
    * Adding the shell temporarily reduced the flow to the surroundings, allowing the air to build up inside until a new steady-state is achieved.
    * If i suddenly removed the outer wall, air would leak much faster than 235 g/s — again without any need to violate common sense or the laws of physics.
    * if the air pump is moved so that it adds air between the walls, then at steady-state, the tank and the space between the walls will all be 254 PSI, with no flow into our out of the main tank.

  51. gbaikie says:

    “To recap:

    1) We agree that before one of us can point at the other and say “told you so”, the system has to reach as steady state after the shell is placed round the toy planet so we can get reliable unchanging temperature readings from the surfaces.

    2) We agree that to be in a steady state, the composite toy planet/shell system will have to be losing energy to space at the same rate it is being produced by the nuclear core.

    So, on to proposition 3:

    3) We agree that the thin steel shell will radiate from both sides equally, since both the inside and outside surfaces will be at the same temperature.”

    For one surface which radiates as blackbody to radiate 235 W/m-2 it’s temperature
    is about 255 K [-18 C]. For blackbody to radiate 2 sides at 235 W/m-2 it needs a temperature
    equal blackbody radiating 470 W/m-2 from one side which about 302 K.

    One can make the shell have temperature of 302 K, one simply halves the outer shell emissivity as compared to the core surface emissivity, but one can’t do this if one has stated the outer surface of shell has the emissivity of a blackbody.

  52. Tim Folkerts says:

    Gbaikie says: “For one surface which radiates as blackbody to radiate 235 W/m-2 it’s temperature
    is about 255 K [-18 C]. For blackbody to radiate 2 sides at 235 W/m-2 it needs a temperature
    equal blackbody radiating 470 W/m-2 from one side which about 302 K.”

    Nope!

    The first sentence is right — any surface that radiates as a blackbody at 254 K will radiate 235 W/m^2 of photons from that surface: P/A = sigma T^4.

    But it makes no difference what shape the surface is – sphere, cube, disk, spiral. Now, it will certainly take more total power to maintain that 254 K temperature with more total area, but that is a different issue.

  53. wayne says:

    Tim Folkerts:

    The planet will be represented by a large air tank, and energy will represented by the air. The heater is like an air pump that pumps air into the tank at a fixed rate — say 235 g/s. This will cause the pressure (analogous to temperature) to to build up in the tank (the total amount of of air inside is analogous to total thermal energy, U). This tank has various leaks — the seams were not welded very well. It turns out that when the pressure has built up to 254 PSI, the air leaks out at 235 g/s — the same rate is is being pumped in (analogous to heat “leaking from the planet as fast as it is being added by the heaters).
    Conservation of energy is analogous to conservation of air molecules.

    The “Second Law” says air spontaneously moves from higher pressure to lower pressure.

    Now, suppose I surround the tank with another wall identical to the first (except a fraction larger so it will fit around the other without touching). This wall has the same poorly welded joints.

    What will happen? Well, at the first instant, there is no pressure between the walls, so no air leaks out from the outer wall, and air starts to build up between the walls. As the pressure starts to build up, the air starts to leak faster and faster out from the outer wall (but still less than the 235 g/s that are being added. This means that the total air inside is building up. This ALSO means that the air will start to leak more slowly through the inner wall (because the pressure difference is decreasing between the original tank and the the space around it.

    Eventually the pressure will build up until the pressure BETWEEN the walls is up to 254 PSI and the air is leaking at 235 g/s from the outer wall. The air in the inner tank would be 2×254 PSI so that air would leak through the inner wall at 235 g/s. (This is one place the analogy breaks down — the pressure would presumably increase by a factor 2, while the temperature only increases by 2^(0.25) ).

    Tim, I’ve got to hand it to you, that is one fine analogy !! Hand claps. No kidding this time. My mind immediately made but one small alteration to bring it perfectly in line with what I see. Instead of leaky welds make that a semi-porous rigid surface, like hydrogen in a rigid rubber balloon under pressure, hydrogen atoms leaking at every infinitesimal point through the pores every moment, those pores are just the size of the atoms. Perfect. Make their size so the 254 PSI does leak precisely 235 g/s.

    In fact, to me that is even closer than I bet you see in it yourself. The hydrogen atoms are exactly as I see the photons, “disturbances in the e.m. field” as Wikipedia put it flowing from warmer to cooler surfaces in that bb cavity, even down to the way they interact when close to a surface in the near-field.

    I’m sure you see those atoms able to flow backwards too against the gradient flow, kind of “back-flow”, sorry, I don’t, I’ve read and learned enough in that area of physics over the decades that I simply cannot return to that simplistic view.

    So I call that an exact photon-to-matter conversion of Willis’s steel shelled planet for all intents and purposes! Can’t see a single conflict as of yet. One of the things over those decades that I have learned above all of physics, it is the incredible parallels, the symmetries, the equivalences all across the board and how nature makes sure there are none at the very finest level.

    Well done Mr. Folkerts, well done.

  54. gbaikie says:

    “What will happen? Well, at the first instant, there is no pressure between the walls, so no air leaks out from the outer wall, and air starts to build up between the walls. As the pressure starts to build up, the air starts to leak faster and faster out from the outer wall (but still less than the 235 g/s that are being added. This means that the total air inside is building up. This ALSO means that the air will start to leak more slowly through the inner wall (because the pressure difference is decreasing between the original tank and the the space around it.

    Eventually the pressure will build up until the pressure BETWEEN the walls is up to 254 PSI and the air is leaking at 235 g/s from the outer wall. The air in the inner tank would be 2×254 PSI so that air would leak through the inner wall at 235 g/s. (This is one place the analogy breaks down — the pressure would presumably increase by a factor 2, while the temperature only increases by 2^(0.25) ).”

    The air pressure should be the same in both tanks.

    But leaks in tanks don’t normally work this way.

    What one could do is have small opening in tank such as small pipe or pressure release valve.
    If inlet pipe is bigger than outlet pipe, it builds up pressure. So larger pipe will put in more
    volume, and the smaller pipe must expel more air at higher pressure to equal same volume.

    And with the second tank with same small size pipe, the result will be the same pressure
    in both containers. If second tank had smaller pipe, the pressure in both tanks would be the same
    but at both would be a higher pressure. And if pipe was bigger on second tank. the first tank would have same pressure, and outer tank would have lower pressure.

    Now with leaky pressure tank you assuming it could hold pressure and start to leak at some pressure [tanks don’t do this, they blow up] But having the outer have pressure does make the inner stronger or one does need any structural strength if the outer tank is taking the load.
    Or suppose you needed 1/2″ thick tank to withstand 200 psi. If outer tank was 1/2″ thick the inner tank doesn’t need to be 1/2″ [it could be 1/16th of inch] as outer is providing the structural strenght.
    And in addition one could make the inner hold say 50 psi with weaker walls, so it could hold 250 psi
    while outer tank withstood 200 psi.

  55. tallbloke says:

    Gbaikie: The air pressure should be the same in both tanks.

    No. The air pressure in the two tanks would only equalize as the pressure dropped to zero after the air pump (nuclear core) had failed and all the air had leaked to space through the holes in the outer shell.

    Once again, you need to consider the dynamic situation, not a static one. If we simultaneously stopped the air pump and plugged all the holes in the outer shell, then the pressure between the two reservoirs would equalize. That would be a static situation. But in the dynamic situation there will be a gradient of pressure, with big steps between the inside and outside of the inner and outer tank walls.

    This has been the biggest problem in this thread. People hear the word dynamic, but conflate the static concepts with the actual situation.

  56. Tim,

    Excellent analogy, well done. Now watch the [snip] pick it apart. 🙂

    [Reply] Stop it please.

  57. tallbloke says:

    I had drafted up a similar analogy with water, but wasn’t happy with it. Tim’s air pressure version works better.

  58. gbaikie says:

    “tallbloke says:
    April 3, 2013 at 8:15 am

    Gbaikie: The air pressure should be the same in both tanks.

    No. The air pressure in the two tanks would only equalize as the pressure dropped to zero after the air pump (nuclear core) had failed and all the air had leaked to space through the holes in the outer shell.

    Once again, you need to consider the dynamic situation, not a static one. If we simultaneously stopped the air pump and plugged all the holes in the outer shell, then the pressure between the two reservoirs would equalize. That would be a static situation. But in the dynamic situation there will be a gradient of pressure, with big steps between the inside and outside of the inner and outer tank walls.

    This has been the biggest problem in this thread. People hear the word dynamic, but conflate the static concepts with the actual situation.”

    Say one had large reserve tank of air and it has pressure of 150 psi.

    Take two spheres connect them with 1/2 pipe. And have 1/2 pipe which get air from reserve tank
    and at other end second have 1/2 pipe with valve. And also have valve from reserve tank.
    Assume reserve tank can release for 30 mins through 1/2 line at around 150 psi before lowering in pressure. And can compressor kick in at that point providing 150 psi thru another 1/2″ diameter pipe.

    So first, have valve turned of at end of pipe of second sphere and turn valve from reserve tank.
    All tanks pressurize to 150 psi. And then open valve at end of second sphere you get 150 psi thru
    1/2 pipe.
    Keep this valve open and turn of valve to reserve. Spheres go to 0 psi. Then open valve to reserve tank. The first sphere will reach 150 psi before the second sphere, and may take a few minutes for both spheres to reach 150 psi. Depends how big spheres are. But if given enough time
    regardless of how big the sphere are, they will both reach 150 psi.

    Now, cut a sphere in half, put cylinder same diameter as sphere, so hemisphere welded to one end.
    Weld flat plate other end. Put 1/2 pipe in plate and weld other hemisphere to this end.
    Repeat.
    Same results.

    [Reply] You’ll have to draw diagrams and upload and link them. It is not possible for me to visualise the system you are trying to describe from your notes.

  59. gbaikie says:

    “Tim Folkerts says:
    April 3, 2013 at 4:19 am

    Gbaikie says: “For one surface which radiates as blackbody to radiate 235 W/m-2 it’s temperature
    is about 255 K [-18 C]. For blackbody to radiate 2 sides at 235 W/m-2 it needs a temperature
    equal blackbody radiating 470 W/m-2 from one side which about 302 K.”

    Nope!

    The first sentence is right — any surface that radiates as a blackbody at 254 K will radiate 235 W/m^2 of photons from that surface: P/A = sigma T^4.

    But it makes no difference what shape the surface is – sphere, cube, disk, spiral. Now, it will certainly take more total power to maintain that 254 K temperature with more total area, but that is a different issue.”

    The power is 235 W/m^2 or 470 W/m-2 per second.

    So for blackbody radiating on two sides of 1 meter square it requires 470 W/m-2 per second
    to maintain a temperature of -18 C.
    And power to radiate 1 meter square on one side is 235 W/m^2 per second to maintain -18 C.

    This also true if a disk. And sphere same diameter of 1 meter square area requires 4 times 235 W/m^2 per second of power to maintain -18 C. But is assuming one radiating outward from the sphere, if also radiating inwards from sphere instead of 4, it’s 8 times 235 W/m^2 per second of power to maintain -18 C.

    [Reply] Watts = Joules per second. The shell has twice the surface area of the toy planet (neglecting the small error due to diameter difference) Because it has an outside and an inside. To maintain the shell at -18C the toy planet needs to radiate it with 2 x 235W/m^2, not 8 x 235W/m^2. Please acknowledge this before moving on to further points.

  60. Tim Folkerts says:

    Wayne, The idea of a semi-porous wall works well too. In fact, I would say it is better since makes the leaks more uniform over the whole surface.

    As for back flow …
    1) Photons can pass right through each other, while molecules cannot, so there is one huge difference.
    2) Even in the air, molecules will be moving both ways. At typical room temperatures, the molecules in the air are moving randomly with thermal speeds of several 100 m/s. So even as the air leaks from the tanks (at say 10 m/s), there are individual molecules moving “upstream” at a few 100 m/s.The pressure gradient makes sure the net flow of molecules is “downstream” (just like the temperature gradient makes sure the net flow of energy is from hot to cold).

  61. tallbloke says:

    Tim F: Photons can pass right through each other

    I’d like to see a video of that. 😉
    E/M waves interpenetrate with no measurable loss of amplitude too. So all talk of ‘cancelling’ of back radiation is experimentally proven wrong. Net radiation figures are arithmetical accounting, not physical limitations.

  62. Roger Clague says:

    gbaikie : For blackbody to radiate 2 sides at 235 W/m-2 it needs a temperature
    equal blackbody radiating 470 W/m-2 from one side which about 302 K.

    The shell absorbs 470W/m-2 and emits 470Wm-2. What is its temperature?

    Tim Folkerts :

    The planet will be represented by a large air tank, and energy will represented by the air

    Radiant energy does not behave like a gas. We are applying Kirchoff’s Law of Radiation and the S-B Law. We do not need a model of the model.

  63. Westy says:

    Tim, I can see what you are saying with the two tanks. If the tanks are elastic I think you would probably need more pressure, like blowing up two balloons instead of one. For two solid tanks with the same size outlets isn’t the pressure likely to equalize after the second is added so you shouldn’t need more pressure in the first to expel the same amount of air from both?

  64. tallbloke says:

    Roger C: You are incorrect. The shell is emitting (just under) 235W/m^2 (Joules/sec per square metre) from each of two areas totalling (just over) twice that of the toy planet surface. It’s temperature is therefore ~254K.

  65. tallbloke says:

    Westy: The pressure will not equalize. This is because we are looking at a dynamic system which has high pressure at the middle going to no pressure in space via two sets of restrictions.

  66. Westy says:

    Talbloke, it’s the same size restriction to space if we use one tank of two.

  67. tallbloke says:

    Westy: But two lots of restrictions, so higher pressure needed at the centre to get the original rate of loss to space once the second set is added. Try putting a restrictor in a length of a garden hose and see what happens to the flow rate. Then try putting two in and see how much it further reduces. You’ll find you need to open the tap further and so increase the pressure to get the original flow rate back again. In fact, you can cut out the messing around with the hose and just consider what a tap does….

  68. Bryan says:

    Whatever the Willis model is suposed to represent ,it has nothing to do with a ‘greenhouse effect’.

    Remember what this effect claims to be, not that I endorse any of it.

    1. Incident short wave solar radiation is mostly unaffected by troposphere.
    2. Much of the short wavelength radiation is absorbed by Earth surface.
    3. The resulting Earth surface temperature emitts long wavlength IR
    4. This IR is strongly absorbed and 50% re-radiated back to Earth, similar to some primitive ideas of how a glasshouse works.

    The variable emissivity values of Earth surface for short and long wave radiation is at the heart of any greenhouse effect theory.

  69. tallbloke says:

    Bryan, I agree with you 100% and this is what I’ve stressed over and over again in this thread. Willis’ toy planet and steel shell have nothing to do with the Earth and it’s atmosphere. The reason I want to get it straight that in terms of physics theory his toy model is correct, is we can then go on to show how the ‘greenhouse effect’ on Earth cannot possibly work in the way it is claimed to.

    So please, for the love of correct science, respond to my question to you, which was to ask if we are now clear that Willis’ model is correct.

  70. Arfur Bryant says:

    All,

    I don’t have much time this evening but I will try to respond to each of my co-debaters although some responses may be shorter than I would usually like…

    tallbloke says:
    April 2, 2013 at 10:41 pm

    [“When you wear a duvet jacket outside on a winters day, rather than a T-shirt, the reason you are warmer is that your ‘core’ energy generating warmth at your skin isn’t lost so easily to the environment. Now, that sort of insulation doesn’t work in the same way as the radiative shell in Willis toy planet model, but the result is similar.
    The key point is that the effect isn’t anything to do with the duvet ‘creating extra energy’. It is to do with the rate at which the heat from your core escapes to the environment.”]

    Rog, you feel warmer but the duvet jacket does NOT increase the core temperature. No amount of insulation will do that. In the same way no amount of shells will make the planet warmer. Please try to understand that reducing heat loss is NOT the same as increasing temperature. Willis’ model is predicated on the planet getting warmer because of the presence of the shell. It would not happen. Even if you ate constantly to maintain your body core output (like the planet), your core would not be warmer because of insulation. The clue is in your statement “isn’t lost so easily…”. Losing heat less quickly is still losing heat! You cannot add it to the core temperature. Just like you cannot add re-emitted radiation to the planet’s original 235.

    The only source of energy in this system is the source at the core which is producing 235 W/m^2. The planet cannot emit 470 when it only has enough power to emit 235 from the same surface. To do so, the planet would have to warm significantly enough for the energy level to increase to the point where it can emit 470. This cannot be done without the addition of a separate source of energy. Re-emitted radiation from the shell will always be ‘cooler’ than the original planet. Hence, it will not be absorbed for net gain by the planet. Hence the planet will not heat up. No amount of insulation will change that. Otherwise we would cook in a really togged up sleeping bag when camping!

    **

    [“So, on to proposition 3:
    3) We agree that the thin steel shell will radiate from both sides equally, since both the inside and outside surfaces will be at the same temperature.
    If we agree on this, then as a logical deduction:
    4) We agree that the shell is radiating the same amount back towards the planet as the planet was originally radiating directly to space.
    Agree or disagree?”]

    3) Guardedly agree. I am not completely sure that the radiation will be equal. There may be some imbalance as the temperature differential out to space is greater than the differential inward to the planet. I am not sure this has a bearing, hence the guardedness.

    4) Disagree. This is where your erroneous assumption kicks in. You have assumed the shell can emit 235 both ways as the end product of a circular argument. In order for it to do so, it will have to have received greater than 470 (because of the radius effect you want to ignore). The shell will not even receive 235 because of this effect. There is no way a power unit which can produce energy at 235 can be responsible for producing 470 without another, separate, source. Sorry, it is not a logical deduction. The logical deduction would be that the shell is back-radiating half of what it receives from the planet (caveated with my guardedness on point 3).

    Regards,

  71. wayne says:

    So Tim and TB, please explain your alternate explanation that the double-slit experiment should have no have places in space where no photons even exist. You both seem to view that photons can not possibly affect each other, just always pass through each other unimpeded, no cancellation. Please show me your sources, I’ll accept anything as further input on this subject.

    So I took a laser and the best mirror I could find. Shined that laser directly into the mirror from about 1″ away. You a can see a very weak pattern at the glass-air interface but is very weak. Now rotate that laser slowly to where the beam is no longer shining directly onto itself. At about 1/16″ you get an extremely bright light where there was basically nothing before. Rotate another 1/16″, it basically disappears again, another 1/16″ super bright. Wish I had a very sensitive multi-meter to measure the current pull on the battery as that laser is diametrically shining directly onto itself. Does the current draw actually decrease? Are “photons” actually cancelling? Everything I have read and studied from good sources says yes. See a good book on quantum electrodynamics.

    Blackbodies by definition have every frequency that the temperature allows in all directions and every polarization at all times in such a cavity. So if cancellation occurs in the double slit experiment then why not there. Susskind says, in so many words, that it is because all waves in that idealized bb case are in resonance ‘at the nodes. of the vibrations, there is NO transfer of energy at the nodes. If it were not precisely at the nodes there would be photons. It’s not just from him, I have read or listened to countless others on this subject, same thing.

    I guess we just disagree on that picky point. Or, maybe I have just listened to the ‘wrong’ lectures, read the ‘wrong’ physics books over all of these years.

  72. wayne says:

    TB, oh no, I now already what you are going to say. You are going to think I mean there is NO radiation flowing in the bb gap of Willis’s example, not true. What I said above only applies to the 235 that matches exactly in every aspect from the shell… there is still 235 of radiation (photons) flowing from the sphere outward. Just had to add that so this didn’t get misinterpreted.

  73. Arfur Bryant says:

    Tim Folkerts says:
    April 3, 2013 at 1:11 am

    The pressure analogy…

    Sorry Tim, your analogy is wrong. I’m surprised that so many seem to have accepted it, let alone lauded it. You have just made the same error that you make with the defence of the Willis’ model. Namely, that you have just doubled the pressure/radiation from the core (inner tank)! Why?

    [“The air in the inner tank would be 2×254 PSI so that air would leak through the inner wall at 235 g/s.”]

    Huh? You had already said that

    [“It turns out that when the pressure has built up to 254 PSI, the air leaks out at 235 g/s…”].

    So why does it now need 2×254 in order to leak the air? The system will reach an equilibrium with the same 254 in the inner tank. Once the middle space has been filled, the air will leak out of the outer tank at 235g/s. No need to invent a doubling of pressure. This is what Willis did. 235 W/m^2 will warm the shell and when equilibrium is reached the shell will be emitting 235 (well, not exactly due to radius effect but let’s go along with it for now) to space. There is no need for 470. it can’t happen. It doesn’t need to happen. It is an irrelevance.

    As you seem to want to play ‘Analogy tennis’, try this one:

    You place two electric fires facing each other in a room. One fire has one bar and the other fire has two bars. They both radiate toward each other. According to Willis, the two bar fire will get warmer (even if it is by losing heat less quickly!). Do you agree with Willis? If so, please provide me with a link which proves that a low energy source can heat a higher energy source.

    Regards,

  74. lgl says:

    “So why does it now need 2×254 in order to leak the air?”

    Arfur, think.

  75. tallbloke says:

    Arfur says: The planet cannot emit 470 when it only has enough power to emit 235 from the same surface

    This is a massive misunderstanding. The planet without a shell emits 235W/m^2 IN A STEADY STATE because that is the rate at which it is IN DYNAMIC BALANCE with its environment. Change the environment and the temperature will change, and since things emit according to their temperature, so will the emission. Wrap the planet in a duvet jacket, and the surface will get hotter, just like your skin does when you put a duvet jacket on when its cold weather. And a high tog sleeping bag will make your core hotter, which is why you start to sweat, then wake up and unzip a bit.

    3) Guardedly agree. I am not completely sure that the radiation will be equal. There may be some imbalance as the temperature differential out to space is greater than the differential inward to the planet.

    The shell has a vacuum on both sides of it. Vacuum’s don’t have a temperature, because there’s nothing in them to get hot, and since radiation doesn’t have eyes, it doesn’t know or care whether it is setting off in the direction of a warm planet or outer space. So the radiation will be equal in both directions, and even if it were different, I don’t think you are trying to claim it would be much different, just maybe a little bit different.

    Either way, you have now agreed from proposition 2) that at steady state, the shell’s outer surface has to lose as much energy to space as the planet surface did before, and from proposition 3) that the inner shell surface will be emitting more or less as much as the outer shell surface. So now we have 235W/m^2 being lost from a total area twice the surface area (just over) of the planet. This is where the logic has led us, and now we must work out how the planet, which is the only source of energy for millions of miles around, comes to be supplying the shell with sufficient energy to be radiating at that rate from that doubled up area.

    To make the shell radiate 235W/m^2 to space, despite the fact it is radiating as much inwards as outwards, the planet must be radiating towards the shell at 470W/m^2, since it has half the area of the shell’s inner and outer surfaces totaled together. So the multi-billion dollar question is, how did it get to be hot enough to be doing that? We know it must be doing that, because that is where the logic has led us. there is no other option. So in the next comment, I’m going to start at proposition 5), since 4) was a superfluous logical deduction rather than a proposition, and we’ll get to grips with the thorny issue which has caused all the trouble on this thread:

    How does radiation from a cooler body contribute to a warmer body’s temperature in a dynamic situation?

    Thanks for hanging in there, and thanks for agreeing, however guardedly, to proposition 3).

  76. tallbloke says:

    Wayne: The Shell is at 254K and the planet is at 302K. Won’t the radiation be at a different wavelength in each direction? Wouldn’t that avoid any ‘cancelling’ issues?

  77. tallbloke says:

    Arfur: So why does it now need 2×254 in order to leak the air?

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

    It’s a bit like Back_radiation 😉

  78. wayne says:

    Arfur, you are making this far too complicated. It takes a drop in pressure across a boundary for anything to be forced to cross that boundary. So both the sphere membrane and the shell membrane must have a drop in pressure across both of them for the air (or hydrogen) to cross each. That by itself dictates that the inner sphere must have the highest pressure, the gap lower than that, the void beyond at the least pressure. If the pressure in the sphere and the gap were identical nothing would flow. Tim has more being constantly pumped into the sphere. Walla, a two pressure gradient.

    I have finally proved beyond a doubt, to me anyway, that what IPCC is feeding us is baloney, but examples such as Tim’s and Willis’s is pure proper physics, you don’t need to try to disprove classic physics that has already been proven over hundreds of years to reach the same goal.

    Likewise radiation ‘pressure’ or power in an opposite direction from a cold object to a warmer object is real, just don’t call it “back radiation”, super bad term because of what power IPCC has implied wrongly to it.

  79. Bryan says:

    Tallbloke
    I still cannot accept several aspects of the Willis model.
    I stand by the points I made in posts about the expanding planet causing 470W/m2 to space for thousands of years being against the second law.
    Cementafriend made an earlier comment about the ability of a source to keep upping the quality as well as maintaining the quantity of output as the temperature increases.
    That the quality (in the second law sense) of the output must increase as the planet temperature increases is unphysical.
    Roger Clague was onto a similar point when he asked about the surface back radiating to the source at the centre.
    I felt that Tim Folkerts was unsettled by the question and then came up with what I felt was an addition to the Willis model.
    He specified that the planet was not heated from a central core but by billions of independent micro radiators evenly spread over the planet volume.
    Think of a power plant with cooling towers trying to cope if the surrounding atmospheric temperature reached 110C.
    I dont accept 200W/m2 and 300W/m2 being automatically equivalent to 500W/m2
    JWR made a contribution which Tim accepts give the right sort of numbers without backradiation.
    For all these reasons and more I remain sceptical.
    Like you I regret the vitriol of the exchanges between some of PSI and other more mainstream sceptics.
    Perhaps they are reacting to attacks made on them.

    Tim Folkerts is far from being a idiot and Anthony Watts is not an agent of the IPCC.
    On the other hand PSI contains good people too.
    Piers Corbyn, JWR, Tim Bell and others can make a contribution without spitting the sceptics into quarrelsome factions.
    Your own site offers a bridge where the science can be examined and the more loopy ideas abandoned .

  80. tallbloke says:

    Wayne: examples such as Tim’s and Willis’s is pure proper physics, you don’t need to try to disprove classic physics that has already been proven over hundreds of years to reach the same goal.

    Exactly. Furthermore, undertsanding and accepting the classical physics is going to help us disprove the IPCC Earth-greenhouse theory in a much better pragmatic and ‘acceptable’ way than trying to disprove it by attacking it with dragonslayer style nonsense avante-garde radiative physics.

    There will be a major post from David Cosserat tomorrow that will put another building block in place to achieve that from solid foundations. Mainstream climate ‘science’ is about to change forever.

  81. tallbloke says:

    Thanks Bryan,
    That’s a big comment which covers issues far wider than this thread, but it’s said, and must be responded to

    I’ll let Tim Folkerts respond to the physics of Mr Reynen. Tim Ball is not part of PSI, although PSI claim he is.
    I’m surprised to see Piers Corbyn’s name down as a PSI member. I’ll ask him.

    Roger Clague needs to think about how much of a role conduction has in solid objects.Perhaps you can get him to think about it. He doesn’t often respond to my critique of his comments. Today he tried to tell us the shell must be at 302K in order to radiate a total of 470W/m^2 and that’s just bollocks.

    I stand by the points I made in posts about the expanding planet causing 470W/m2 to space for thousands of years being against the second law.

    Well if the planet spends thousands of years getting cooler, it’ll also spend thousands of years getting hotter. During which time it’ll be radiating less than 235W/m^2 to space via the shell because it cooled to space from the outside to some depth. Don’t forget it takes less energy to lift something by nK from a cooler temperature than from a higher one. It shouldn’t be too difficult to specify a model which could be programmed to test this. I encourage you to do that. I think you’ll find energy is successfully conserved.

  82. Arfur Bryant says:

    All,

    Ok, Ok. I accept defeat graciously. I have misunderstood some basic concepts.

    tallbloke – I now accept your duvet analogy. I must have mis-read the definition of insulation as “reducing heat loss” to mean heat gain was impossible.

    lgl – you told me to think. I’ve done a fair bit on this thread (thank you all). As to the 2×254 psi, my thinking tells me that the constant pump from the core will increase the pressure in the inner tank until 254 is reached, a which point the leakage starts into the outer tank. As the leaks occur, the pressure would drop but for the constant pumping, so the pressure starts to build up in the outer tank. When this pressure reaches 254 the outer tank starts leaking to space. IMO the constant pumping is still happening but has not needed to exceed 254 psi because only 254 is needed to cause the leaking in both tanks. The pressure in the inner tank and the outer tank can still be 254 each and (Wayne) flow can still be occurring. The pressure at two points along a pipe can be the same even though flow is present. What is true is that the pressure difference exists between the pump output and outer space. Once the inner tank is at 254, it may as well not exist. I am, however, prepared to be wrong again!

    tallbloke & Tim,

    I still have a problem with your 235 back to the planet from the shell requiring 470 from the planet. Why? I am trying to figure it out but I think it comes down to a ‘net flow’ of radiation. Initially, the planet emits 235. With the shell (once at equilibrium) in place, 235 emits to space from the shell. Any 235 arrow back to the planet is compensated for by the 235 outbound. It would be as if the inner void did not exist (like the pressure analogy see above to lgl). No net flow from shell to planet means the 235 to space is essentially the same 235 that initially left the planet.

    I am hanging in, just, and appreciate your patience. I still have doubts over Willis’ model but I am unable to effectively (technically) argue my case, and therefore I concede your points and look forward to you finishing me off with the next phase (5)! 🙂

    Either way, thanks to all for the polite and helpful discussion.

  83. gbaikie says:

    “[Reply] Watts = Joules per second. The shell has twice the surface area of the toy planet (neglecting the small error due to diameter difference) Because it has an outside and an inside. To maintain the shell at -18C the toy planet needs to radiate it with 2 x 235W/m^2, not 8 x 235W/m^2. Please acknowledge this before moving on to further points.”

    Yes, watts equal joules per second. Or watts is a unit of power.
    Since watts could be watt hours [a different amount of power or 3600 times the power of a watt second] saying a watt second does add some clarification- though it can rightfully be seen as unnecessary redundancy.

    Well I meant converting a disk area to spherical area. With core surface is starting from a spherical area already, it’s 2 x 235W/m^2, and not 8 x 235W/m^2.
    Or it can treated as a flat surface to another same size flat surface.

  84. Westy says:

    If you wanted to build Tim’s two tank model with the both tanks having the same pressure you would have the hole in the second tank half the size of the hole in the first. If you wanted the first tank to have more pressure you would have the hole in the second tank twice the size of the hole in the first. To get even pressure in each tank the hole in each would be very close to the same size. To be fair the holes would need to line up or it would be like pumping through a longer restriction for the second tank, remembering the outside tank is only a fraction bigger then the inside the same as our Willis toy.

  85. Westy says:

    If the tank analogy is to hold won’t it get down to the Willis shell having a smaller, greater or equal resistance to heat transfer?

  86. wayne says:

    Wayne: The Shell is at 254K and the planet is at 302K. Won’t the radiation be at a different wavelength in each direction? Wouldn’t that avoid any ‘cancelling’ issues?

    Different wavelength? Yes tallbloke, of course they are… well, kind of. They have different spectrums but portions of each line do overlap if we are not speaking of, lets say, the 5800K sun and 288K IR spectrums.

    Each individual line in two spectrums, when both are not at the same temperature, will vary in the radiance at each frequency line. They have different radiant power.

    But, a portion of matching frequencies do actually overlap, that is if the temperatures are close together — that overlap portions that do match in radiance is what I am speaking of, just that portion. The portions that do not match (subtract the two radiances, one will be larger) can never resonate between those two objects, and that remaining portion, without a doubt, that is always the actual photons transferring energy in an attempt to equalize the two surface’s temperatures, no resonance there, there is nothing there (no opposing power) to resonate with.

    Did you see my set of spectrums earlier? The area under the yellow curve IS that remnant. It is the portion where the actual blue and magenta spectrums do not overlap. The volume under the yellow curve represent the actual photons, to me. That is the “disturbance in the e.m. field”, waves that have nothing to match with, they travel, called photons, at ever line contained within that area. The area under both the blue and magenta, the intersection of those two areas, has lines that do resonate, everything matches, no photon there, just vibrations in the e.m. field. That is the cancellation (probably a really bad word to describe that, it’s not really cancellation, definitely not the cancellation of any energy, but it is a cancellation of the effect of the opposing e.m. radiation power at each line)

  87. tallbloke says:

    Arfur: Thanks for your spirit of participation, we’re all learners here, and all conclusions are contingent on being overturned.

    You said: The pressure at two points along a pipe can be the same even though flow is present.

    Try it! My dad was a chartered civil water engineer all his working life and he used to delight in telling me that with an infinitely long pipe, you could apply infinite pressure at one end and have no flow at the other. More prosaically, if you double the length of a water pipe and keep the pressure constant at the inlet end, the flow at the outlet will be less than it was. Now, much of that is due to turbulence in the flow which causes frictional loss, so it doesn’t really apply to our radiative physics problem, to which I’ll now turn.

    To recap the argument succinctly:

    1) At steady state, the shell’s outer surface has to lose as much heat to space as the nuclear core generated at the planet surface before the shell was emplaced.
    2) The shell will re-radiate equally from both sides, which total fractionally over twice the area of the planet.
    3) To get the shell hot enough to satisfy 1) and 2), the planet has to be radiating towards the shell at just over twice the power density it formerly did direct to space.

    So how does the planet’s surface end up hot enough to do that?

    Lets have a bit of preamble before we formulate the next proposition. Consider the net radiation in the case with no shell and the case with the shell.
    No shell:
    Planet radiates 235W/m^2 into the vacuum of space. Space radiates 0W/m^2 back at the planet: Net figure outwards 235W/m^2

    With added shell and time enough for steady state to be reached (could be a looooong time 🙂 )
    Planet radiates 470W/m^2 towards the shell. Shell absorbs 469.8W/m^2 and reradiates 234.9W/m^2 to space and 234.9W/m^2 back to the planet: net figure outwards from planet 235W/m^2 because there are more square metres on the shell radiating back towards it.

    So the planet surface has the heat from the core, and is also absorbing and re-emitting the radiation that got re-radiated from the shell in its direction, but in net terms, it’s only getting rid of as much energy as it was before.

    But! (I hear several people shouting) the surface is at 302K and the shell is at 254K. So how can the radiation from the shell make the planet surface warmer?

    I reply: it isn’t making it warmer. It’s at a nice steady 302K and has been since it reached the steady state condition and the radiation coming from the shell and being absorbed and re-radiated by the planet along with the radiation due to the heat it acquires from the core is going outwards and being absorbed by the shell, maintaining it at a nice steady 254K as it loses 234.9W/m^2 to space, thus cooling itself at the same rate it recieves net energy from the planet. Everything is peachy and the numbers add up.

    OK you sneaky bastard (I hear several people cry) but how did the planet’s surface get to 302K from 254K without being heated by the cooler shell???!!!

    And the answer to that will have to wait for the next thrilling installment of how I single-handedly defeated the dragonslayers. (The other hand is currently holding a glass of 15 yr old Glenmorangie 🙂 )

  88. Trick says:

    tb 12:50am: “The other hand is currently holding a glass of 15 yr old Glenmorangie”

    Give or take a W/m^2 (or degree K), I’ll toast that up with a Macallan 15 otr.

  89. wayne says:

    TB, this is just because it shows you what I see maybe clearer.

    I was watching Prof. Lewin (MIT) and a student shaking a rope between themselves. Ok close. Just take that light rope and make it a heavy steel cable. You and I could get that cable oscillating harmonically between us, each of us just moving our arms slightly up and down, up and down, till both of us are at nodes. The only energy we need to exert to keep this going is just what it takes to overcome the air resistance and friction in the cable. It becomes very easy to keep that wave in motion. But nature, in the e.m. case, has no ‘air resistance’ or friction. These can oscillate with zero enery transfer though the two ends are now interlinked. If one of us stops moving suddenly, or makes an irregular movement, at the speed of the wave’s speed, you would get a huge kick for then all of that energy in that “cable field’ would transfer to you. Not instantly, but after the wave travel time. That is what I see as photons, energy transfer, and e.m. fields.

    See, it is not the viration itself that is the energy transfer, it is the waves themselves that hold the power. The two ends are just moving, vibrating, to minimize the energy at each point. As long as the lossless hormonic oscillations continue, nothing happens energy wise, it is when they fall out of sync or one end gets blocked, then “photons” naturally jump into action. Can you see it that way? You and Tim seem to see maximum energy transfers all of the time and I see mimimum energy transfers only when they have to.

  90. tallbloke says:

    Wayne: Hopefully, the ‘net’ figures I introduced in my last comment will allay your concerns. I’m a ‘principle of least action’ man at heart. Which is why my other half had to kick my butt to get some joinery done in the kitchen today. 😉

  91. wayne says:

    Joinery, hehe… read you there. It’s always easy to find some way to do nothing. 😉

    I won’t say much more on that subject of the intricacies of electromagnetism, photons and matter. As dear Dr. Feynman said, you’ll probably never be able to correctly visualize it anyway. Said he always tried and yet failed, but nevertheless, that is the way it is, nature that is. As weird as it seems all of the experiments and results that so precisely match say what really is there, it’s just that we humans can’t even seem to imagine it to the tee.

    Yeah, that a good place to stop on that side-track.

  92. gbaikie says:

    “gbaikie : For blackbody to radiate 2 sides at 235 W/m-2 it needs a temperature
    equal blackbody radiating 470 W/m-2 from one side which about 302 K.

    The shell absorbs 470W/m-2 and emits 470Wm-2. What is its temperature? ”

    It’s temperature is related to how much it emits per square meter.
    If it radiates as blackbody [a full spectrum of electromagnetic radiation] it’s the lowest
    temperature, as compared a surface emitting less of this full spectrum but emitting
    same watts per square meter.
    If one side of meter square radiates 470 W/m-2 and it’s a blackbody surface, then
    it’s temperature is about 302 K. If radiating two sides of panel, and so radiating 235 W/m-2
    on both side which totals 470 W/m-2. Then spectrum from both sides if re-combined should equal a blackbody spectrum of 302 K.

    So if blackbody spectrum flux reduces by 1/2 due to distance traveled [inverse-square law] that should same as spectrum as splitting the energy to go to either side of a panel. And indicates a temperature of 302 K

    So, I would say splitting blackbodies by 2 or 4 [as climatic models do] is problematic in regards to
    temperature. If add two blackbody temperature curves of 254 K, you don’t get a 302 K blackbody curve.

  93. Tim Folkerts says:

    Gbaikie says: ” Then spectrum from both sides if re-combined should equal a blackbody spectrum of 302 K.”

    No.

    Do the math and calculate the spectra. Or look at the spreadsheet that someone (I think Wayne) posted showing the two spectra and their difference. Adding 2 spectra @ 254 K is NOT the same as one spectrum @ 302 K. The most obvious difference is that the two will peak at different wavelengths.

  94. lgl says:

    Arfur
    The reason we needed the air pump in the first place was to get a pressure difference of 254 psi between the tank and the outside to get 235 g/s flowing. Any less than 254 resulting in less than 235 flow. Since the pump is supplying a constant flow of 235 g/s the pressure difference must remain at 254 psi so when pressure starts building up between the tanks the difference drops and that can’t happen because then we do not get rid of the 235 g/s which is a must. Just like the planet and the shell, the temperature of the planet must always be high enough to drive a 235 W/m^2 net radiation from the planet to the shell. Any less than 302K will give less than 235 W/m^2, the system is not in equlibrium and the planet temp will rise.

  95. Roger Clague says:

    TB says Wrap the planet in a duvet jacket, and the surface will get hotter, just like your skin does when you put a duvet jacket on when its cold weather.

    A jacket warms mainly by reducing convection and conduction not by reducing radiation. Willis’s model has no mass, conduction or convection in it. So the analogy is not helpful.

    The Willis model is not consistent in how it assigns temperatures.

    The planet receives 470Wm-2 and is said to be at 302K.
    The shell receives 470Wm-2 and said to be at 254K.

    Tim Folkerts says

    Adding 2 spectra @ 254 K is NOT the same as one spectrum @ 302 K.

    So we cannot add the 235Wm-2 from within to the 235W-2 back radiation to get a body at 302 K.

  96. tallbloke says:

    Roger C: Just because you don’t need analogies to understand this problem, doesn’t mean they are not helpful for others.

    The Willis model is not consistent in how it assigns temperatures.
    The planet receives 470Wm-2 and is said to be at 302K.
    The shell receives 470Wm-2 and said to be at 254K.

    That’s because the 470W/m^2 radiated from the planet towards the shell is re-emitted by the shell from twice the area of the planet’s surface which means half the power density, which means ~235W/m^2, which means ~254K.

  97. lgl says:

    Roger C
    Put on a thin plastic jacket and take trip to Greenland and see how long the reduced convection keeps you varm.

  98. wayne says:

    Arfur,

    “When this pressure reaches 254 the outer tank starts leaking to space. IMO the constant pumping is still happening but has not needed to exceed 254 psi because only 254 is needed to cause the leaking in both tanks. The pressure in the inner tank and the outer tank can still be 254 each and (Wayne) flow can still be occurring.”

    To me you’re still thinking a little off. It doesn’t take 254 before flow to space begins, just 1 psi works too, it’s just the flow to space would be very small. And on flowing with equal pressure, that does never happen. Okay, think of a mile long pipe. Does it take more pressure at one end to get a given flow at the exit end than if the pipe was just one foot long? Of course. Why? The pipe has surface friction, intermolecular friction, all along the pipe, kind’a like a continuous shell that restricts flow just a tiny bit at every foot along that mile. Taking any interval along that pipe the pressure is always going to be higher on the source side that on the exit side.

    The shell is just a a very strong restriction to the flow, pressure will build up on the source side until it reaches the flow necessary to exit what every is continually being injected at the source. Now look at the gap in the steel shell case, it does the same. You will end up with 2×pressure(T) on both sides of the sphere surface, 1×pressure on both sides of the shell, 0×pressure is empty space.

    Now just make that 10 shells, 100 shells, 1000 shells but let their resistance be very small (huge holes along the way) and you basically have the one mile pipe example but it looks more like 1×pressure, 0.999×pressure, 0.998×pressure, 0.997×pressure …. all along that pipe. See, there is always a drop in pressure if there is flow unless there is zero resistance, and there is never zero resistance, even with e.m. field gradients. That is to say there is only zero radiation flow if the temperatures on both sides are equal. Same with pipes, or tanks, or really anything you can dream up. That is why physics is so wonderful and interesting… just how the heck does nature do this so flawlessly everywhere and with everything?

  99. Tim Folkerts says:

    Roger C

    * We CAN add 235 W/m^2 and 235 W/m^2 and get 470 W/m^2. This is required by conservation of energy.
    * We CANNOT add the spectrum from a 254 K object and the spectrum from another 254 K object and get the spectrum from a 302 K object.

    The closest I can come to making gbaikie’s train of thought work would be

    * An object that receives 235 W and radiates with ε=1 from 1 m^2 will be 254 K.
    * An object that receives 470 W and radiates with ε=1 from 1 m^2 will be 302 K.
    * An object that receives 470 W and radiates with ε=0.5 from 2 m^2 will be 302 K (radiating 235 W from each m^2).

    We could then “combine the radiation” from the two sides of the 2 m^2 graybody (perhaps using a mirror or two) to recreate the same spectrum as we would get from a 1 m^2 blackbody at 302 K.

  100. gbaikie says:

    “lgl says:
    April 4, 2013 at 5:10 pm

    Roger C
    Put on a thin plastic jacket and take trip to Greenland and see how long the reduced convection keeps you warm.”

    A thin plastic jacket [or thick one] has a lot of convection losses.
    If use a “double pane” jacket of two thin “plastic panes” separated by air and you will lower
    convection loss. If you could somehow have the two panes always separated by a vacuum, you would have no convection loses.
    The main problem keeping warm in Greenland, is heat loss from conduction- have warm feet and
    that is a major factor in keeping warm.

  101. Arfur Bryant says:

    tallbloke and wayne…

    A couple of points for you both and then I think I can see where my mistake may have occurred:

    First, tallbloke says: ‘You said: The pressure at two points along a pipe can be the same even though flow is present.

    Try it! My dad was a chartered civil water engineer all his working life and he used to delight in telling me that with an infinitely long pipe, you could apply infinite pressure at one end and have no flow at the other.”

    Wellllll… My statement is true for a steady streamlined flow according to Bernoulli’s principle. If the flow speed is constant between the two points, then the pressure is the same. No disrespect to your Dad but if it is an infinitely long pipe, how can it have “…the other end”? I would say as it is infinitely long, then there will be a point where there is no flow, and at that point the pressure will be zero. I think that is what your Dad meant.

    Doesn’t that make sense?

    **

    wayne,

    I was assuming that the leaks only started when the pressure reaches 254psi (see Tim’s post: “This tank has various leaks — the seams were not welded very well. It turns out that when the pressure has built up to 254 PSI, the air leaks out at 235 g/s…”

    I can see now that I may have misinterpreted but I hope you can see that the way it is written could mean that the leaks only start when the pressure reaches 254. See? If that is the case, then your statement “It doesn’t take 254 before flow to space begins, just 1 psi works too…” does not necessarily apply – as the tanks were built to identical specs.

    However, given that I may have misinterpreted Tim’s sentence, then yes, 1 psi would be enough if the leaks could happen at 1 psi…

    By the way, as to the mile long pipe thingy, I go back to Bernoulli. In your example it would be not streamlined (steady), as friction may incur turbulence. in which case Bernoulli doesn’t apply. If it is a steady streamlined flow, then the pressure can be the same at two points.

    Anyway, that was my reasoning.

    **

    tallbloke

    Thanks for your kind words. I look forward to your next instalment after the Glenmorangie! 🙂

    Regards,

    Arfur

  102. Arfur Bryant says:

    Ooh Ooh! Do I get a prize for the one-thousandth post? 🙂

  103. Arfur Bryant says:

    lgl says:
    April 4, 2013 at 4:22 pm

    [“Arfur
    The reason we needed the air pump in the first place was to get a pressure difference of 254 psi between the tank and the outside to get 235 g/s flowing. Any less than 254 resulting in less than 235 flow. Since the pump is supplying a constant flow of 235 g/s the pressure difference must remain at 254 psi so when pressure starts building up between the tanks the difference drops and that can’t happen because then we do not get rid of the 235 g/s which is a must.”]

    Not sure about that lgl. To me, when the inner tank reaches 254 psi, the air is leaking out at 235 g/s into the outer tank. This is exactly the same as when the inner tank started to fill – at 235 g/s. Therefore the pressure will build up in the outer tank at that point (when the leaks start into the outer tank). When the outer tank reaches 254 psi, air will leak out at 235 g/s. The pump is constant; the pressure does not need to be twice 254. It just needs to be 254 in the inner tank for the outer tank to start filling. If the pressure drops as soon as the leaks start and the pump is constant, how can it get to 2 x 254? It only ever needs to be 254 to fill the outer tank. The inner tank need not exist then.

    I am thinking, honest! Maybe I need more quality and less quantity? 🙂

  104. gbaikie says:

    “Tim Folkerts says:
    April 4, 2013 at 7:20 pm

    Roger C

    * We CAN add 235 W/m^2 and 235 W/m^2 and get 470 W/m^2. This is required by conservation of energy.
    * We CANNOT add the spectrum from a 254 K object and the spectrum from another 254 K object and get the spectrum from a 302 K object.

    The closest I can come to making gbaikie’s train of thought work would be

    * An object that receives 235 W and radiates with ε=1 from 1 m^2 will be 254 K.
    * An object that receives 470 W and radiates with ε=1 from 1 m^2 will be 302 K.
    * An object that receives 470 W and radiates with ε=0.5 from 2 m^2 will be 302 K (radiating 235 W from each m^2). ”

    I would agree with above. If you say it had 0.5 emission [and it could be true] is easy way to indicate how it’s temperature could work out be 302 K

    And would add if had two source of 235 W/m-2, so 235 W/m-2 + 235 W/m-2 it will increase
    temperature of surface receiving it, it’s just different blackbody curve than compared to 470 W/m-2 source.
    With lot’s mirrors reflecting sunlight one cause high temperatures. It’s the same as magnifying
    sunlight- one is increasing the the height of the curve. And magnification is acting as though one
    were getting closer to the source of heat.
    Or one can look at it as “undoing” the diminishment of the source of heat
    due to distance [inverse square law].
    Though this unrelated to Willis steel greenhouse [the distance is considered too small to include as a factor- or distance is already considered close to zero. Which btw, one couldn’t assume at all IF this if one simply had 1 square meter as the source of heat, but with large surface area of the planet it is close to zero if distance is about 10 km (or 50 km) ].

    And the issue in regards to Willis steel greenhouse is what would be the insulation effects of the
    shell.
    So if steel shell with both it’s surfaces being a blackbody surface has ε=0.5. Is it because
    steel involved, or if it was simply, a blackbody surface with two sides emitting would it also have ε=0.5?

  105. Westy says:

    It’s been fun trying to keep up with the learned people here and I’m still willing to be convinced that the toy planet will heat with shell added but just can’t get there. Before the shell is added the nuclear powered core is working as hard as it can to heat and maintain the temp of the planets surface. When the shell is added the power source must now heat and maintain the temp of shell as well. There is no extra energy from an already maxed out power source to further heat the planets surface that is now being drained because it must share this energy to heat and maintain an always cooler shell. Is it as simple as this?

    [Reply] No

  106. wayne says:

    Arfur,

    You are a gentleman. Sorry, I did word something wrong, made a mistake — but nothing new.

    I like to flap my mouth, typing away at comments, and then spend the next hour(s) trying to prove myself wrong… and I’m successful quite often… gives me a real perk… for if I was never wrong I would then never be successful… right? In like manner, trying to mimic all great physicists I respect, like Feynman adding wrong, spend a whole lecture off by half on his main topic, Susskind regularly has students correcting him all of the time, I love it, I’ll be, they are human and so am I, real camaraderie. I get a real good feeling as long as I can be wrong just as often as they are! 😆

    Ok, I spent the time to roughly simulate what I told you, it was wrong. The pressure just inside the sphere would be 2×pressure+2×epsilon, but just outside that sphere’s surface the pressure would be 1×pressure+2×epsilon, epsilon being pressure proportional to whatever viscosity and friction you attribute to the liquid (or gas), the difference is 1 pressure. Just inside the shell it would be 1×pressure+epsilon it takes to reach the flow of the influx in the sphere, but outside the shell it would be just epsilon and quickly diminishing to zero as the liquid or gas floats away into the void, all assuming the gap width/radius is near zero and no gravity (not very realistic is it?). Once again, the difference across the membranes or surfaces are both 1×pressure. So I misled you as to what I was trying to portray, but I’ll can find nothing else wrong about the flow. It still takes the proportion of epsilon in pressure to move the volume and mass across the gap no matter what the absolute pressure is.

    Think of what pressure is. Pressure pushes in all directions, the net force, vector-wise, sums to zero, null vector. It takes a difference in pressure to be a directional force, and that force can do work to move matter, or do you see even that being wrong the way you see it?

    -w

  107. Westy says:

    It’s also probably not fair to define the pump in the tank analogy as constant flow and not by the pressure it can deliver and sustain because it would require incrementally more energy and the planet hasn’t got that.

  108. Roger Clague says:

    tallbloke says:
    April 4, 2013 at 4:38 pm

    That’s because the 470W/m^2 radiated from the planet towards the shell is re-emitted by the shell from twice the area of the planet’s surface which means half the power density, which means ~235W/m^2, which means ~254K

    You calculate the temperature of the shell from power density emitted.
    The shell must receive energy before it can emit energy.
    The temperature of the shell should be calculated using the power density it receives, 470Wm-2

    Tim Folkerts says:
    April 4, 2013 at 7:20 pm

    * We CAN add 235 W/m^2 and 235 W/m^2 and get 470 W/m^2. This is required by conservation of energy.
    * We CANNOT add the spectrum from a 254 K object and the spectrum from another 254 K object and get the spectrum from a 302 K object.

    Willis’s model says you can add the spectrum of a 254K object, the nuclear reactor, and another 254K object, the shell, and get the spectrum of a 302K object, the planet surface.

  109. Tim Folkerts says:

    Roger C says: “The temperature of the shell should be calculated using the power density it receives, 470Wm-2

    To be consistent, you would have to calculate the average power density it receives on ALL of the surfaces. Half of the square meters get 470 W/m^2 of IR radiation (the inside of the shell) and half get 0 W/m^2 (the outside). The average is, of course, 235 W/m^2, so it emits on overage 235 W/m^2.

    If the thermal conduction through the shell is good (which would be assumed in the top post), then both sides will be the same temperature and radiate 235 W/m^2 and both sides will be 254 K.

    Roger also says“Willis’s model says you can add the spectrum of a 254K object, the nuclear reactor, and another 254K object”
    The reactor is NOT a 254 K object — it is in effect a very hot object. (It is also modeled as heating the surface via conduction, not radiation, but doesn’t really matter here). With sufficient insulation, the nuclear material could heat the planet to 1,000’s of kelvins (maybe even 1,000,000’s). So we can combine the energy provide by the shell with the energy provided by the heater and get above the temperature of the shell.

    Think of it this way — evacuate a room where the walls are 20 C. Then put an “unheated” sheet of metal painted black on the floor (emissivity = 1) (let’s make the sheet 1m x 1m) (with some styrofoam between it and the floor to prevent conduction). Radiation will bring the sheet to 20 C. Now put a heater under the sheet that provides some ADDITIONAL energy (say 100 W). The sheet will warm above the temperature of the surroundings.

    The energy from BOTH sources combines to determine the final temperature of sheet.
    The energy from BOTH sources combines to determine the final temperature of planet.

  110. mkelly says:

    gbaikie says:

    April 5, 2013 at 12:42 am
    Though this unrelated to Willis steel greenhouse [the distance is considered too small to include as a factor- or distance is already considered close to zero.

    Which is why the model as presented is wrong.

    The sphere surface emits 470 W/m^2 and that must fall on the inner surface of the shell. No if ands or buts.

    So the two totals must be equal. Area * emissivity * SB * T^4 of sphere must be equal on inner surface of shell.

    Mr. Folkerts did the math earlier so I won’t go though it all, but we can factor out all common factors like pi, 4, SB, emissivity factor and will be left with
    r^2(sphere) * T^4(sphere) = r^2(inner shell) * T^4(inner shell)

    We know or are told 302 K for sphere and 254 K for inner shell. So lets pick a number for the radius of the sphere. Say 7000 m. What then must the radius of the shell be to satisfy the equation?

    I get 9895.6 m. See there is the problem the shell cannot be close and have a temperature of 254 K or if it is at 254 K then it must have almost half again as much radius.

    The problem as presented is not correct.

  111. lgl says:

    Arfur
    The pressure has to be 254 psi in the outer tank relative to outside to get a flow of 235 g/s
    The pressure has to be 254 psi in the inner tank relative to the outer tank to get a flow of 235 g/s
    as simple as that.
    Why wasn’t there a flow of 235 from the tank from the beginning, without the air pump, if no pressure difference is needed to get an airflow?

    Westy
    “it would require incrementally more energy”
    No, the air intake is from the inside of the outer tank so the pump never has to lift the pressure more than 254 psi, no incrementally more energy.

    Roger C
    Like I have said earlier you can add the integrals of two 235 spectra and the sum equals the integral of a 470 spectrum, but you can’t add values for each wavelength and say that the sum is the emission on that wavelength of a 470 spectrum. You can add the integrals which represents the energy but not the ‘shapes’ if that makes sense.

  112. lgl says:

    Westy
    Oops! I take that back, didn’t do much thinking there, sorry.

  113. Tim Folkerts says:

    mkelly,

    You are STILL misinterpreting the equation I derived:
    r^2(sphere) * T^4(sphere) = r^2(inner shell) * T^4(inner shell)

    This equation says that if the bare planet with radius r(sphere) would have a temperature T(sphere) then if a shell is added, the temperature of the outside of the shell will be T(shell). (And since the shell is assumed to be thin with good thermal conductivity, then the inside of the shell will be the same temperature.)

    This equation does NOT compare the temperature of a shell to the temperature of the planet inside that shell! This equation ONLY compares the temperatures of the outermost layers in the various circumstances.

    For your 7000 m planet & 9900 m shell, the planet would have been T(planet) = 254 K without a shell. With a shell, big shell will be T(shell) ~ 214 K, radiating 1/2 of 235 W/m^2 from twice as many m^2. The planet itself will be a little warmer than 254 K in this case, but not as warm as 302 K (I am pretty sure it would be 281 K, but that would take more time to re-confirm).

  114. Tim Folkerts says:

    Westy, try this …

    Before the s̶h̶e̶l̶l̶ insulation is added, the n̶u̶c̶l̶e̶a̶r̶ ̶p̶o̶w̶e̶r̶e̶d̶ ̶c̶o̶r̶e̶ furnace in your house is working as hard as it can to heat and maintain the temp of the p̶l̶a̶n̶e̶t̶s̶ interior walls’ surface on a cold winter’s night. When the s̶h̶e̶l̶l̶ insulation around your walls is added, the p̶o̶w̶e̶r̶ ̶s̶o̶u̶r̶c̶e̶ furnace in your house must now heat and maintain the temp of s̶h̶e̶l̶l̶ insulationas well. There is no extra energy from an already maxed out p̶o̶w̶e̶r̶ ̶s̶o̶u̶r̶c̶e̶ furnace in your house to further heat the p̶l̶a̶n̶e̶t̶s̶ interior walls’surface that is now being drained because it must share this energy to heat and maintain an always cooler s̶h̶e̶l̶l̶ insulation.

    Now I think it sounds painfully obvious that the maxed out furnace will indeed make the house warmer when insulation is added around the outside of the house.

    While the analogy is not exact, it is very close. The shell is “IR insulation” for the planet.

  115. donald penman says:

    Does the whole moon heat up when radiation is absorbed by the moons surface? What is different here?I think that the nuclear core might take a long time to respond to the shell materializing in empty space and only the surface will interact with the shell during this time. The shell is warming by retaining radiation that the planet has sent out to space and is reradiating this to the planets surface which can then radiate more because the surface is heated by radiation just like the moons surface.The flow of energy from the heat supplied by the core will be radiated faster because the surface has been heated by backradiation.

  116. gbaikie says:

    “Now I think it sounds painfully obvious that the maxed out furnace will indeed make the house warmer when insulation is added around the outside of the house.

    While the analogy is not exact, it is very close. The shell is “IR insulation” for the planet.”

    So what makes the best “IR insulation”?

    What is best insulated, is something that maintains it’s heat for longest period of time- a thermos
    which keep the coffee hot for a week, would be a very good thermos. It would have to have the best “IR insulation”.

    The worst “IR insulation” would something transparent to IR. So vacuum is worst IR insulation- as nothing is as transparent to IR as compared to a vacuum.
    So a shell made of material transparent to IR would be more insulation than vacuum, but it would the least insulation for a shell.
    We could say this transparent shell isn’t providing enough insulation to be considered important.

    Instead of being transparent can a material absorb the IR and be poor insulation?
    If the material is very cold then it’s similar to being transparent. So something absorbing
    IR which is cold is also poor IR insulation.

    The best insulation of IR is something that reflects the IR- and the temperature of the reflective material does have much to do with this insulation.

    And then there matter of the emissivity of material such as anywhere from 1 to .01 [lower emissivity are generally reflective.

  117. gbaikie says:

    “Does the whole moon heat up when radiation is absorbed by the moons surface? What is different here?”

    The moon is not covered with solid rock.
    The moon is covered with about 3 inches of fine dust. The fine dust in a vacuum is very good insulation. Better than anything we have, though aerogel would be similar:
    “Aerogel is a synthetic porous ultralight material derived from a gel, in which the liquid component of the gel has been replaced with a gas. The result is a solid with extremely low density and thermal conductivity.”
    http://en.wikipedia.org/wiki/Aerogel
    continuing:
    “Aerogels are good thermal insulators because they almost nullify two of the three methods of heat transfer (convection, conduction, and radiation). They are good conductive insulators because they are composed almost entirely from a gas, and gases are very poor heat conductors. Silica aerogel is especially good because silica is also a poor conductor of heat (a metallic aerogel, on the other hand, would be less effective). They are good convective inhibitors because air cannot circulate through the lattice. Aerogels are poor radiative insulators because infrared radiation (which transfers heat) passes right through silica aerogel.”

    The fine dust on the Moon is in very good vacuum and it’s basically silica and would block radiant
    energy, so because of the extreme vacuum it’s better than Aerogels.

    “A flower is on a piece of aerogel which is suspended over a Bunsen burner. Aerogel has excellent insulating properties, and the flower is protected from the flame.”

    So in picture [above] that looks like about 1/2 inch of aerogel stopping a Bunsen burner from heating the rose, so blast furnace type heat with 3″ in lunar dust would do similar job of protecting a rose from heating up, or better job than say couple feet of brick.

    So couple feet of brick heating up has much higher heat capacity than 3 inches of dust- a very fluffy material, that if compressed would conduct more heat thru it.
    So it stops heat from penetrating thru it and has less than 1/20th of heat capacity as
    2 feet of brick would have. So if the Moon had solid rock [or brick] it would store more- say 20 times the heat as would a lunar surface.

    Or sand on Earth is only vaguely similar to the dust on the Moon. And if the Moon were internally heated and scraped away the dust, it would very hot beneath it.

  118. Westy says:

    “Now I think it sounds painfully obvious that the maxed out furnace will indeed make the house warmer when insulation is added around the outside of the house.

    While the analogy is not exact, it is very close. The shell is “IR insulation” for the planet.”

    Can’t argue with what you say Tim and analogies and thought experiments aren’t perfect. I’m uncertain if Willis intended the shell to be insulation, plus my take is that the planets surface is the maxed out furnace that’s supposed to get so much warmer after his cold shell is added. The inside of the shell is more like your house and I still can’t think it’ll heat up enough to further warm the furnace, it can’t get warmer then it so only slow its cooling. Hey, as already mentioned, I’ve been wrong plenty of times and will be happy to learn something.

  119. Arfur Bryant says:

    lgl says:
    April 5, 2013 at 3:52 pm

    [“Arfur
    The pressure has to be 254 psi in the outer tank relative to outside to get a flow of 235 g/s
    The pressure has to be 254 psi in the inner tank relative to the outer tank to get a flow of 235 g/s
    as simple as that.”]

    No, lgl, it isn’t!

    Your line 1 is correct.
    Your line 2 is wrong if there is a flow between inner tank and the exit to space. Tim’s analogy states that, at 254 psi, the output flow rate matches the input flow rate. (This alone means that the pressure in the inner tank cannot increase above 254 – why would it?)

    The only pressure differential required in a ‘series’ of tanks is the differential between the pump and the exit. At the boundary between the inner and outer tanks, at every instant the air being lost to the outer tank is being replaced by the air being pumped into the inner tank. Hence no requirement for the pressure to increase above 254 – and certainly not double to 508!

    Sorry Tim and lgl, but the analogy falls down here. There is no doubling of pressure.

    Now it’s your turn to think! 🙂

  120. Arfur Bryant says:

    wayne says:
    April 5, 2013 at 1:46 am

    Wayne,

    Thanks for the kind words! I have yet to meet anyone who is right all the time!

    Please see my response to lgl above. I hope it explains why the pressure will not double in the inner tank. If the tanks are in series and exit flow rate is the same as the input flow rate, there is no requirement to double (or even increase) the pressure.

    Regards,

  121. lgl says:

    Arfur
    Try this, http://www.pipeflowcalculations.com/airflow/
    Doubling the pipe length is the same as adding the second tank.

    “The only pressure differential required in a ‘series’ of tanks is the differential between the pump and the exit.”
    No, a series of tanks is a series of resistors, and a series of resistors reduce the flow. You have to increase the driving potential to maintain the flow.

  122. lgl says:

    Arfur
    On second thought, since air is compressible you are probably right about the “certainly not double to 508!”, but it must be much higher than 254. Using a fluid would get us closer to 508.

  123. lgl says, April 6, 2013 at 9:57 am: …a series of tanks is a series of resistors, and a series of resistors reduce the flow. You have to increase the driving potential to maintain the flow.

    I love it. You must be an electrical engineer like me. Everywhere I look in this controversy I see electrical circuit analogies. Kirchoff sure was a clever guy.

  124. Tim Folkerts says:

    You know the conversation has been going on too long when we are discussing analogies of analogies of models of the greenhouse effect. LOL

  125. gbaikie says:

    “David Socrates says:
    April 6, 2013 at 2:14 pm

    lgl says, April 6, 2013 at 9:57 am: …a series of tanks is a series of resistors, and a series of resistors reduce the flow. You have to increase the driving potential to maintain the flow.

    I love it. You must be an electrical engineer like me. Everywhere I look in this controversy I see electrical circuit analogies. Kirchoff sure was a clever guy.”

    Speaking of resistors, I wonder if maybe this has to do with the earth having a pressure
    of 14.7 psi.
    Does it matter how much the atmospheric pressure is?
    If vacuum or Venus 92 atm of atmosphere, does it make a difference?

    Or perhaps gravity is involved?

  126. tallbloke says:

    Gbaikie: Yes. See many posts on gravity on this blog.

  127. wayne says:

    “The only pressure differential required in a ‘series’ of tanks is the differential between the pump and the exit.”

    Arfur, on more time, I’ll keep on until you see your mistake. I used the think the same way with the S-B relationship, the flow (W/m²) only depending on the temperature at the two endpoints. But that is not correct if mass is present, watch out, it will drive you a bit crazy ! 😉 I got a better understanding by turning to Venus because it is such an exaggerated example, a great tool. You have ~17000 W/m(740K) at one end and ~0 W/m(3K) space at the other end so naively 17000 W/m² should escape every second using your way of thinking, straight S-B, but is is not, not even close, only ~65 W/m² escapes even though this extremely high ‘pressure’ is present. The only answer is there must be ‘constrictions’ every ‘layer or shell’ due to the matter between the surface and the void of space.

    Take a 1″ pipe with ten joints along the way. If the flow is laminar and the joints have no constriction then your example is correct, it only depends on the pressure at the two ends assuming the length is a constant one by this relation.

    The relationships of pressure, pipe diameter and flow rates in laminar flow:
    FR = k · π·R^4 · (P−P0) / (8·N·L)
    FR is the flow rate, k is the constant of proportionality (units dependent), R is radius, P0 is the pressure at one end, P is the pressure at the other end, L is the length of the pipe, N is the viscosity.

    But if at each of those joints you constrict the flow to 1 cm radius, or 1 mm radius, or just pinholes, surely you cannot think that it does not take much more pressure at the influx end to maintain a constant effluent flow at the exit end, doesn’t it? if you have pressure gauges at every segment along the way between these constrictions are they not going to monotonically decrease segment by segment? In the shell example, these shells are constrictions that at least double the pressure needed across them if they are totally opaque.

    Back to Venus being a good teaching tool, take the 17000 and half it, half it again, and again, until you get to the actual W/m² exiting flow. Well, it is about 8 times. So you now have 2^8=256 shells segments. Take 17000 W/m² and divide by 256 and you get 66.4 W/m². Multiply by 4 (from 4π sphere) and divide by (1-.9) Bond albedo and you get a TSI for Venus of 2658 W/m². So Close. Why not exact 2614 W/m²? Because ln(17000/65.5)/ln(2) is not exactly 8, it is 8.0198… you really have about 259 shells but I think you should now get my drift. Try it, it works exactly the same for Earth’s atmosphere, the ‘constriction drop’ for both planets even though they have such drastically different atmosphere composition is just right at -159 W/m² per Earth atmosphere 1 m² column masses. That is where I knew what IPCC ‘believes’ is totally incorrect. The ‘why’ is that all non-window radiation lines are already totally opaque, you cannot get more opaque than opaque, and as TimF showed distance (altitude) does not matter looking downward when dealing with spheres and shells (imaginary layers).

    For Earth you don’t even get to divide by 2 once, just a fraction of 2.

    The only difference between Earth’s atmosphere and Venus’s atmosphere is our has a bit of leakage directly from the surface to space by-passing the shell(s), this is the window frequencies, directly from surface to space. Now I see why Miskolczi jumped right on that parameter, it is all that matters and it isn’t changing for some reason, see his papers. So when I say CO2 has near zero effect at these concentrations (it’s already totally opaque), that is why.

  128. Arfur Bryant says:

    Guys…

    Please bear with me here. I understand what you are saying but I am taking Tim’s analogy at its face value.

    Tim – LOL. Yes, I agree that the thread has moved away but you introduced this analogy!

    lgl, wayne and David: thank you for your continuing patience, but…

    lgl says [“No, a series of tanks is a series of resistors, and a series of resistors reduce the flow.”]

    But the tanks are NOT resistors! There is no resistance to the flow once the pressure is at 254 psi because Tim says that the exit flow rate is equal to the input flow rate of 235 g/s. See? The tank is just a cavity which has to be filled. When it is filled, the pressure inside the tank will be 254 psi and the flow will then continue unabated. No resistance, no resistor, no reduction in speed of flow, which means no reduction in pressure. To introduce another analogy – Tim please note 🙂 lol – fill a bottle with water from your kitchen tap but keep the tap running. When the bottle is full the excess water runs down the sides of the bottle and into the sink at the same rate at which it was flowing out of the tap. No pressure increase needed from the tap. Put another bottle under the first one (and catch all the water) and the same thing will happen. When the second bottle fills the water will still flow into the sink with no reduction to flow so no pressure change from the tap + gravity. A slightly better analogy would be to put a relief valve in the bottom of each bottle which opens at 254 psi but allows the full flow to exit.

    This is the key to Tim’s analogy: at 254 psi the output (exit to space) flow equals the input (pump) flow. There is now no resistance to this flow. No doubling of pressure required.

    Are you guys still sure that I’m wrong?

    Aside: in turbulent flow, there may be a pressure drop with a very long pipe but it would only need enough of a pump pressure increase to compensate for this.

  129. lgl says:

    Arfur
    Of course there is resistance. Without there would have been no pressure difference.
    So yes, I am still sure you are wrong.

  130. Arfur Bryant says:

    lgl,

    That is why the analogy is wrong.

    The pressure is between the pump outlet and space. Put a tank (shell) around the pump and there will of course be pressure initially but when the pressure reaches 254 psi according to Tim’s analogy the leakage commences at an exactly similar rate to that coming out of the pump! Now (as I said at the start) the inner tank need not exist. Air flows into the outer tank through the now useless inner tank. The pressure builds up in the outer tank until it reaches 254 psi, at which point the outer tank leaks at the same rate as the pump. Voila! The air in the outer tank leaks out to space and it only ever needed 254 psi to do so because 254 psi is needed to provide the leakage!

    The analogy (much lauded) is wrong, not me.

    Is there a hydraulic engineer out there somewhere?

  131. Tim Folkerts says:

    Arrfur says: ” … but when the pressure reaches 254 psi according to Tim’s analogy the leakage commences … “

    No — that was not my intent. The leak starts immediately. Even with 1 PSI in the tank, SOME air will leak through the cracks. But that leak will be ~ 1 g/s, so there is a net ~ 234 g/s added to the tank. The pressure builds up .. and the leak increases .. the pressure builds up more slowly .. the leak increases more slowly. It just so happens that when the pressure reaches 254 PSI, the leak reaches 235 g/s (or more specifically, when the pressure DIFFERENCE between inside and out is 254 PSI, the leak will be 235 g/s) A steady-state is achieved.

    To me, this was intuitive and obvious. If there is a hole in a tank, air (or water) will leak out faster as the pressure is increased.

  132. suricat says:

    Sorry I’ve been absent guys, I’ve had family problems. 😦

    I’ve tried to follow this thread and the hydraulic analogy is inconsistent, but ‘vaguely’ analogous. In ‘electronic terms’, the tanks are capacitors and the connecting pipe-work mimics the resistors. In ‘radiative terminology’, the ‘tanks and pipe-work’ are solely mass per se.

    There IS NO photon-photon, or wave-wave, interaction (though the ‘energy potential’ in the EM field can be observed, there is no interaction here). The electrical field generated by the electrons that surround a ‘mass entity’ is what generates an ‘energy interaction event’ between the EM field and the mass. Thus, photons/waves are just ‘ships in the night’, but a ‘mass object’ is where the ‘ships’ come to grief (if their ‘frequencies’ are at ‘near resonance’). 🙂

    That’s all I’ve time for just now.

    Best regards, Ray.

  133. tallbloke says:

    Arfur: I’m a qualified engineer and studied fluid dynamics to degree level. So if you want to appeal to the judgment of a hydraulic engineer that’s fine.

    You are wrong. You already admit you can’t do the maths. Nor have you built such systems or made measurements of their behaviour. Yet you state your opinion about them in categorical terms without blushing.

    I have designed and built such systems, and tested and measured their behaviour. At its simplest, the more restricting orifices you place in the way of a flow, the greater the pressure you need at the inlet for a given rate of flow at the outlet.

  134. Simulation of heat source and sink systems
    Best report seen so far is trick to find for free but it can be done:
    powersys_thermmod_gb_fhg.pdf
    Thermal Modeling of Power-electronic Systems
    Dr. Martin März, Fraunhofer Institute for Integrated Circuits IIS-B, Erlangen
    Paul Nance, Infineon Technologies AG, Munich

    Search
    Find a Chinese? site “milim”
    Select page 3
    select the thermal modelling text
    click the download

    According to the equations (1) and (3) heat conduction processes can therefore be
    modeled by a transmission line equivalent circuit diagram which, as shown in figure
    1, consists of R/C elements only. In addition the equivalences listed in table 1 exist
    between the electrical and thermal variables.
    Thermal ⇔ Electrical
    Temperature T in K ⇔Voltage U in V
    Heat flow P in W ⇔Current I in A
    Thermal resistance Rth in K/W⇔ Resistance R in V/A
    Thermal capacitance Cth in Ws/K ⇔Capacitance C in As/V
    Table 1: Corresponding physical variables

    Not that heat flow corresponds to current
    so a heater dissipating any power with a defined thermal capacitance but totally insulated will be equivalent to a capacitor across a current source – a straight line starting at 0,0 and finishing at infinity, infinity!

    These set ups are very easily modelled in any spice programme e.g. LTSPICE free – do a search.

    By using lossless transmission lines you can even simulate the time it takes for radiation to reach a shell and for the shell to pass half back (you get a staircase voltage ⇔ temperature). It is probably better to use [bi] sources [I=delay(voltage,time)] but you then need to think a bit.

  135. donald penman says:

    The reason why the whole moon does not heat up when radiation is absorbed by its surface has got nothing to do with moon dust because the moon has bare rock at the surface at many places and yet it is still true that only the surface heats up.The problem is one of scale and none of the analogies that have been put forward apply on astronomical scales only on engineering scale where you could possibly build something that works as you theorise.

  136. tallbloke says:

    Donald: the moon has heated up to 197K avg surf temp in the glare of the Sun. What temp do you think it would be at with no Sun? The equatorial temp stays at a steady 230K or so just a foot below surface under regolith.

  137. gbaikie says:

    “The reason why the whole moon does not heat up when radiation is absorbed by its surface has got nothing to do with moon dust because the moon has bare rock at the surface at many places and yet it is still true that only the surface heats up.”

    Old article, discusses:
    http://adsabs.harvard.edu/full/1953AuJPh…6…10J
    And another one:
    http://adsabs.harvard.edu/full/1958ApJ…127..751G

    Newer source, The Once and Future Moon, by Paul Spudis, Chapter 4:
    “We have known for a long time that the surface of the Moon is covered fine dust. If
    it’s surface were bare rock, we would see a bright reflection at the point on the Moon
    directly under the Sun as it rotates on it’s axis.
    Such an effect is similar to bright glare one sees when looking towards the Sun in late
    afternoon on lake, ocean or smooth body of water. Bare rock would also show this effect
    although not as mirror-like as water because a rock surface would be much rougher
    at fine scales.”
    Page 83
    Continuing:
    “The surface of the Moon does not display this type of bright, specular reflection. In fact by carefully studying the exact way light is reflected from the surface, scientist determined that the Moon was covered everywhere by dust- very fine dust. What was not known in the pre-Apollo days
    was how deep such dust layer might be.”

    There are some areas in which there isn’t meters of depth of compacted lunar regolith, and one has steep walls of bare rock and regions where it’s rocky [I.e, a field of boulders] so areas which can be accurately be called rocky. And fresh impact zones will favor a less dusty surface conditions.
    But I think a description that the moon looks similar region on earth which recently had a few inches
    of snowfall [lunar dust] as fair analogous description of vast majority of the surface of the Moon.
    And there are hundreds of Apollo of close up picture of the lunar surface and I don’t recall any with any horizon surface which did not have the surface covered in dust.

    Which is unlike like the surface of Mars, where one can see some areas of bare rocks [Mars has wind and it’s surface is shaped in large part by it’s wind].

  138. gbaikie says:

    “I have designed and built such systems, and tested and measured their behaviour. At its simplest, the more restricting orifices you place in the way of a flow, the greater the pressure you need at the inlet for a given rate of flow at the outlet.”

    No doubt old household plumbing with corroded pipes will deliver lower pressure. And if you
    taking shower and someone turns on sink faucet this affects the flow of a shower. If replace
    the old pipe with new one the same diameter, one see dramatic improvement, and still better
    if you replace with slighter larger diameter pipe.
    But though never had experience, I doubt if you put couple hot water heaters in a series you would get a significant drop in pressure.

    Nor would you expect one, since the cross sectional areas of radiator matrix are designed to be compatible with the pipework. Please stop the irrelevant banter.

    Now suppose one had leak in submarine, and some reason you had to abandon a section of this submarine. It seems eventually a flooded section would reach same pressure as outside the sub.

    Reply: Of course it would. But this is irrelevant to the discussion, which is about continuous flow, not a static equilibrium.

  139. Arfur Bryant says:

    Tim and tallbloke,

    You are missing the point that the problem is the analogy, not qualifications.

    Tim says: [“It just so happens that when the pressure reaches 254 PSI, the leak reaches 235 g/s (or more specifically, when the pressure DIFFERENCE between inside and out is 254 PSI, the leak will be 235 g/s) A steady-state is achieved.”]

    Tim, it makes no difference whether your analogy works like that or whether there is a pressure relief valve set at 254 psi. The problem with the analogy is manifest in your statement quoted above. I.e, when the pressure difference is 254 psi, the leakage rate is 235 g/s. This is exactly the same rate as the output of the pump! Hence there is no requirement (or even an ability) for the pressure to build up in the inner tank. Pressure would only build up if the outflow was less than the inflow! The steady state is achieved at 254. Once the leakage flows into the outer tank, the pressure in the outer tank will build up without any further pressure in the inner tank because the leakage is already happening at just the same rate as the initial pump outflow. Therefore the outer tank pressure will build up until it reaches 254 psi, at which point the outer tank leaks at 235 g/s again; the inner tank just becomes a filled cavity with flow through it at exactly the same rate. The pressure differential remains 254 between the pump and ‘outer space’.

    tallbloke,

    This is nothing to do with maths or qualifications. It is logic.

    [“At its simplest, the more restricting orifices you place in the way of a flow, the greater the pressure you need at the inlet for a given rate of flow at the outlet.”]

    What has that got to do with the analogy? The analogy has already stated that 254 psi gives a flow rate of exactly the same as the input There is no more restriction to flow! 235 goes into the inner tank and 235 goes out with a pressure of 254 psi. So now we start again in the outer tank; the pressure (which is being provided by an inflow of 235 g/s) builds in the outer tank until 254 psi, at which point the outflow is the same 235 g/s as goes in! No more restriction. No more pressure.

    The problem is with the analogy, not basic fluid dynamics. If the outflow at 254 psi is the same as the inflow, why on earth would you need to increase the pressure? Actually, how could you increase the pressure? You would need to introduce a restriction which would reduce the flow and falsify the analogy.

    The only other factor that could be here is if the flow was turbulent and then we could agree that the pressure reduces due to friction effects. That was not mentioned and actually should be ignored as it does not apply in the Wilis model.

    Remember my argument here is with the applicability of Tim’s analogy in trying to show a doubling of pressure is needed.

  140. Arfur Bryant says:

    Tim and tallbloke,

    My line 6 above should read:

    “Hence there is no requirement (or even an ability) for the pressure to further build up in the inner tank…”

  141. gbaikie says:

    “Reply: Of course it would. But this is irrelevant to the discussion, which is about continuous flow, not a static equilibrium.”

    Ok, so try again.
    “The planet will be represented by a large air tank, and energy will represented by the air. The heater is like an air pump that pumps air into the tank at a fixed rate — say 235 g/s. ”

    So planet or reservoir provides 235 g/s to a container constantly.
    We could say reservior has 1 million psi, and so could not matter what the pressure is in the first
    container being filled.
    I see no clarification of psi of reservoir, but a very high pressure should be not able to to changed by a comparatively low pressure. Or instead of a high pressure reservoir pressure, it could be a smart valve which exactly meters out a constant 235 g/s.
    Either way what we getting is a constant amount. Though gallon per second would depend upon it’s pressure in terms keeping same molecules of gas per second.
    So the g/s could the reservior has constant pressure and piston cylinder is filled with consistent
    amounts of gas molecules and each stroke of piston is this 235 gallons and and you have stoke per second. So fixed amount goes in.
    next part:
    “This tank has various leaks — the seams were not welded very well. It turns out that when the pressure has built up to 254 PSI, the air leaks out at 235 g/s — the same rate is is being pumped in (analogous to heat “leaking from the planet as fast as it is being added by the heaters).”

    I don’t like leaks. So replace with a pipe diameter that deliver 235 g/s into a vacuum when it has pressure of 254 PSI.
    And really the whole issue hinges on this, does something at 254 PSI which delivers 235 g/s into a vacuum, deliver significantly less molecules or significant less than 235 g/s if goes into container which has 254 PSI.

    And the claim is that to deliver this 235 g/s into some container with 254 PSI that it needs to have twice the pressure:
    “Eventually the pressure will build up until the pressure BETWEEN the walls is up to 254 PSI and the air is leaking at 235 g/s from the outer wall. The air in the inner tank would be 2×254 PSI so that air would leak through the inner wall at 235 g/s.”

    Now a problem with air is it gets denser under more pressure. So at same speed of going thru a hole one has more molecules of gas passing through the hole. Whereas with water, you would have same number of molecules [pretty much regardless of pressure of the water] if at same
    speed.
    So from this aspect it seems water is more likely than gas to need a higher pressure.
    But we keep with the idea of gas for the moment, it’s well known that air pressure will slow down
    in sense the rocket engine has poorer performance at sea level as compared to a vacuum.
    But rocket engine throwing out very fast molecules of gas, and the difference isn’t huge, and it
    also is affected rocket nozzle design. And purpose of a nozzle is the have the least least amount of
    pressure. One might claim bringing up rocket engine is irrelevant:)
    But as I said, I think water would better [and hopefully simpler] to make the case of needing twice the pressure.
    Now with water under pressure, one doesn’t get a flow of water if two bodies of water are at same
    pressure, but neither there isn’t much delay involved if side suddenly lacks a volume of water.
    And since dealing high pressure water it seems if anything there will be less delay.
    Another related topic to water is what called water hammer or hydraulic shock. Basically if running fire hose and turn it off too quickly the momentum of the water can destroy your pipes [and kill people].
    So it seems the lower the pressure involved the more possible effect of needing twice the Psi.
    So if took a basin of water. Put watertight barrier between them, and put a hole in the barrier
    which was an inch in diameter. And fill side with a garden hose, the side being filled should not become twice as filled as other side.And your garden hose has advantage of having momentum
    and pressure.
    Or the hole doesn’t get the advantage of momentum or pressure.
    And forgot put hole to outside of the second compartment.
    And water doubles it’s pressure every 10 meters in height.

    So let’s back basin. Seal hole in barrier. Turn garden hose so fills of basin halfway up and doesn’t get any higher. Unseal hole in barrier, fill first “container” with same flow as using before.
    And don’t you will get more few inches higher in first container.

  142. Tim Folkerts says:

    Arfur,

    if you want to take about a pressure relief valve, then you have a pressure relief valve through the outer wall that will open up when the pressure DIFFERENCE is 254 PSI — ie when the pressure between the walls in 254 PSI more than the pressure outside the outer wall = gauge pressure of 254 PSI. The relief valve through the inner wall will ALSO open up when the pressure DIFFERENCE is 254 PSI — ie when the pressure inside the tank is 254 PSI more than the pressure between the walls = gauge pressure of 2x 254 PSI inside the tank.

    That is how such valves are designed to work. That is how leaks through holes work. The pressure DIFFERENCE from one side to the other drives the flow, not the ABSOLUTE pressure. Air will not flow from one region @ 254 PSI to another region at 254 PSI.

    There is no problem with either fluid dynamics or the analogy (well there are ALWAYS problems with analogies, but this is not one of the problems here).

  143. lgl says:

    Arfur
    With only the inner tank a pressure differnce between inside and outside of 254 psi is needed to get a flow of 235 g/s. How can you imagine you will get the same flow when the pressure difference is close to zero? it’s just ridiculous.

    “You would need to introduce a restriction which would reduce the flow and falsify the analogy”
    We are introducing a restriction, the second tank with small leaks, but it’s the equivalent to the shell so no falsification.

  144. tallbloke says:

    Arfur: tallbloke, This is nothing to do with maths or qualifications. It is logic.

    Logic doesn’t help when you don’t understand the physical reality of pressures and flows.

    “Hence there is no requirement (or even an ability) for the pressure to further build up in the inner tank…”

    The pump is very strong and if you blocked all the holes in tank 1. the pressure would just keep increasing. The reason the pressure gets no higher than 254 psi is that’s the pressure that is sufficient for the loss through the holes to equal the rate the pump is pumping at, with no outer tank. Once there is an outer tank the pressure builds up in it until it is at 254psi and the air is pushed through the holes in the outer tank at the same rate as the pump. But now the outer tank is resisting air coming through from the inner tank with a pressure of 254psi, and so the pressure in the inner tank has to rise in order to re-establish the flow rate of 235gpm into the outer tank.

    Experience and knowledge of actual physical systems is something I can offer, but if you’re not going to take what I’m telling you as having any value, I can’t help you further.

    As lgl says, and as I told you two days ago, there will be a pressure drop at each restriction in the system, just like there will be a temperature drop at successive shells in a radiative system. And since the outer shell still has to lose as much to space as was originally lost from the toy planet surface to space directly, the whole system will develop a bigger temperature gradient across all the components until the outer surface does lose that much energy to space. You agreed the logic of this three days ago.

    It’s still true.

  145. Kristian says:

    tallbloke says, April 7, 2013 at 6:34 pm:

    “(…) since the outer shell still has to lose as much to space as was originally lost from the toy planet surface to space directly (…)”

    No, this is the fundamental misunderstanding. The planet only needs to lose from its surface the same amount of heat as is being provided from its nucleus. The shell doesn’t need to lose as much heat to space as the planet gains in order for there to be energy balance. Because the shell itself ‘takes hold of’ some of the heat from the planet.

    Look, in a closed system (this is indeed a thermodynamically closed system, Tallbloke, only not an isolated one), the heat input must equal the heat output. If not, the temperature will rise. So, planet without shell, if the surface didn’t manage to shed as much heat to space as it gained from within, then its temperature would rise indefinitely. But with the shell in place, the heat delivered to one object goes into sustaining the temperature of two. This means more of the output energy from the nuclear source will have to stay within the system at any one time than before the shell arrived. The steady state temperatures will be: Planet, 254K. Shell, 213K.

    Before the shell was emplaced, the planet’s surface rate of heat loss (Q) was 235 W/m^2 (matching the input). The input sustained its surface temperature of 254K.

    After the shell was emplaced, the planet’s surface rate of heat loss (Q) is exactly the same, 235 W/m^2 (still matching the input). But now the input sustains both the planet’s surface temperature of 254K and the shell’s temperature of 213K. Extra thermodynamic work is now continuously being done within the system. No piling up of energy.

    Half the heat flux from the planet’s surface goes into keeping up and keeping steady the temperature of the shell. The other half goes out to space from the outer surface of the shell. This is how it works:

    Q = Q’ + Q”
    Q: system input and planet surface rate of heat loss
    Q’: heat transfer rate planet –> shell
    Q”: heat transfer rate shell –> space

    At steady state (dynamic equilibrium), the situation looks like this:

    235 = (235 – 117.5) + 117.5 = 235 W/m^2

    The -117.5 is the ‘back radiation’ from the shell towards the planet. It reduces Q’, the heat transfer rate between planet and shell. But at the same time, the shell cools to space with an equal and opposite flux, +117.5. As Q’ shrinks, Q” grows. And Q remains the same.

  146. tallbloke says:

    Hi Kristian
    I see you guys still don’t understand that the toy planet’s surface will get hotter if the shell is only losing half as much to space as is being produced by the nuclear core.

    Half the heat flux from the planet’s surface goes into keeping up and keeping steady the temperature of the shell.

    Why is there any more ‘keeping up’ to do, where is the shell losing energy to, aside from the 117.5 going to space? You seem to think it’s getting ‘cancelled out’ on it’s way back to the planet, but since I know you understand energy is neither created nor destroyed, that can’t be so. Can it?

  147. Kristian says:

    Tallbloke,

    Did you even read my post? It appears to me you didn’t.

    You ask: “Where is it [the planet] losing energy to, aside from the 117.5 going to space?”

    To the shell, Tallbloke. You know, the 235 – 117.5 = 117.5.

  148. mkelly says:

    Kristian says: April 7, 2013at 7:58 pm

    Kristian what you describe is a radiation/heat shield. That could in fact be correct but it is not what Willis shows. Max had a NASA example of a heat shield and I gave the heat transfer equation for a that. It is as you say and I said up thread 235 from planet and 117.5 from shell.

    This problem is exactly the same as the sun-earth linkage. Sun equal planet earth orbit is the shell. Now all we need to do is apply the same equations.

    I agree with you. As presented this is not correct.

  149. tallbloke says:

    Hi Kristian, please re-read and respond to my comment, I clarified that by ‘it’ I’m referring to the shell, not the planet. Thanks.

    By the way, I did read your comment in full, and chose to ask you to clarify an aspect which doesn’t make sense to me. Lets take it one bit at a time. I agree that at first, 117.5 will be radiated to space. However, after a period of time consistent with the heat capacity of the planet, heat will build up, because only half the amount being generated is escaping to space.

  150. tallbloke says:

    MKelly: I think the NASA heat sheilds are protect tanks from temperature extremes aren’t they? There’s no power source in the tanks within them them which would build up heat, correct?

  151. lgl says:

    Kristian
    So now the shell is destroying half of the energy, great. Suppose we heated the shell to radiate 235 W/m2 towards the planet, then all the 235 from the planet would be destroyed. All the generated 235 W/m2 just disappearing, wow.

  152. Arfur Bryant says:

    Tim, lgl and tallbloke,

    You guys just aren’t getting it. What part of the output flow is the same as the input flow don’t you get?

    Why does the pressure build up in the inner tank? Because there is a restriction in that the leakage is less than the pump output. What happens when the pressure gets to 254 psi? The leakage into the outer tank is now the same as the input. So now what is providing the pressure difference? The pump, the same as it was in the beginning. All you have done with Tim’s analogy is to make the inner tank a cavity (a filled cavity now). It is now no longer a restriction! the pressure differential is now between the pump and the outer tank. The system is in a steady flow state.

    Now to the outer tank; the flow into this tank is 235 g/s which is exactly the same situation as we had between the pump and the inner tank. So what happens? The pump is still providing the pressure to the system and the analogy states that the flow into the outer tank is the same as that from the pump. Therefore, the outer tank will fill with this same flow rate and so the outer tank becomes just like the inner tank. When the pressure reaches 254 psi the output to space is the same as the input from the inner tank which is the same as the output from the pump!

    [” …there will be a pressure drop at each restriction in the system]”.

    There are no restrictions to flow when the inner tank is filled (254 psi) because the input and output flows are identical. Therefore the only pressure required is that from the pump.

    [“We are introducing a restriction, the second tank with small leaks…”].

    See above. ‘Once the inner tank is at 254 psi’, there is no restriction and the tank behaves as a filled cavity. If the outlet flow is the same as the inlet flow, how can there be a restriction? Go and fill a glass of water from the tap. When the glass is full, the water will overflow the glass at the same rate as it exits the tap. No additional pressure required.

    Air will not flow from one region @ 254 PSI to another region at 254 PSI.

    If the inner tank has an identical flow rate input and output, then there is no restriction to flow and therefore no discrete pressure! The tank is just a filled cavity and the pressure differential now is between the pump and the outer tank. This is where your analogy fails. You use the ‘leaks’ as a means of developing a pressure in the inner tank but then you remove the restriction causing the pressure by stating that the output flow is the same as the input flow! So the outer tank gets pressurised the same as the inner tank did until the pressure is 254 when the flow rates are identical and so any restriction is withdrawn. No restriction = no pressure. But the overall pressure difference is between the pump and outer space which is why the flow continues.

    The analogy is flawed. The “…pressure reaches 254 psi…” is only possible if there is a restriction. If you then effectively remove the restriction by making the flow rates the same, then you have effectively ruined your attempt at analogy.

    Go on, fill a glass from a tap, then fill another one below the first one and see the flow continue once the second tap is full with NO additional pressure from the tap.

    Finally, how can the pump cause a pressure of 508 psi in the inner tank if the pump output is only 235 g/s and the exit flow is 235 g/s? Its like asking a tap at medium (235 g/s) strength flow to produce a high pressure in a cake tin when the leaks from the cake tin are 235 g/s. Crap.

    That’s it.

  153. lgl says:

    Arfur
    When the pressure gets to 254 psi in the outer tank according to you it equals the pressure in the inner tank and then there will be no flow from the inner to the outer tank. If you are unable to understand this very basic logic it’s no use discussing any further.

  154. tallbloke says:

    Arfur: how can the pump cause a pressure of 508 psi in the inner tank if the pump output is only 235 g/s?

    Because pressure and flow rate aren’t the same thing. Block half the holes in the inner tank. The pump keeps pumping 235g/s. The pressure must rise until it is high enough to force 235g/s out of half the number of holes – yes?
    HTH.

  155. gbaikie says:

    “Tim Folkerts says:
    April 7, 2013 at 3:17 pm

    Arfur,

    if you want to take about a pressure relief valve, then you have a pressure relief valve through the outer wall that will open up when the pressure DIFFERENCE is 254 PSI — ie when the pressure between the walls in 254 PSI more than the pressure outside the outer wall = gauge pressure of 254 PSI. The relief valve through the inner wall will ALSO open up when the pressure DIFFERENCE is 254 PSI — ie when the pressure inside the tank is 254 PSI more than the pressure between the walls = gauge pressure of 2x 254 PSI inside the tank.

    That is how such valves are designed to work. That is how leaks through holes work. ”

    Using pressure relief values one can certainly make there be difference in pressure between the two containers. But the difference does not need to be 254 psi. It can be any difference you want it to be. Because one is assuming one does have mechanism which responds to pressure. But pressure relief valve are used to control the maximum pressure of something, and designing something to have leaks at certain pressure wouldn’t a replacement for a pressure relief valve. One could say a purpose of a pressure relief valve is to ensure a pressure vessel doesn’t develop leaks.

    But to state the obvious one can not cause more pressure than the maximum pressure which is
    pressurizing the containers. Which is the same with core and shell, one can insulate the loss of heat leaving the core and thereby increase it’s temperature [which has no practical limit- it can cause the rocks reach a temperature where rocks melt or even vaporize- which quite different temperature as compared to -18 C].
    So if source of pressure is as “unlimited” using pressure release values, one also have as unlimited amount of pressure and differences of pressures. Assuming these pressure containers
    have enough structure strength for such pressures.

    Normally pressure release valve is designed work from whatever the pressure is inside the container, but there no reason one couldn’t design one to work from the difference in pressure-
    gauge pressure. Now you design it so reservoir provides constant amount, but let’s say one put this same type release value on the reservoir, but instead of it being 254 psi, make it 300 psi.
    So supply cuts off from reservior if first container exceeds 300 psi.
    And first container valve to second opens when is sense it’s pressure difference is greater than 254 psi. And second opens to outside when it senses it’s pressure difference is greater than 254 psi.

    And you run it and watch what happens. So first contain builds up to 254 psi, and releases into
    second container. But second container won’t reach 254 psi. You get 300 psi in first container and
    46 psi in second container, and reservoir release valve trips.

    You have 3 engineer. One engineer says first container can not exceed 300 psi and be safe.
    Second engineer says, first tank pressure is no where near the 300 psi rating because it’s
    being strengthen by the 46 psi of exterior pressure of second container pressure. Second engineer says just remove the reservoir safety release valve and everything will be fine. The first engineer says, better to replace release safety valve on first container going into second container, instead if being based on gauge pressure, have based on absolute pressure, and have open when pressure of first container is 260 psi and therefore first tank will never exceed rating pressure of 300 psi.
    Third engineer says, let’s drill some small holes in first container in wall separating it second container, and have this modification not reduce the pressure vessel rating below, say 290 psi- using small holes and/or reinforce the strength of structure to compensate for the structural loss of the holes.
    And the first and second engineer response to third is, “We want the first container to hold more pressure, you idiot.”

    But the third engineer is married to daughter of the owner of the project, so they end up doing the idiot’s suggestion.
    So we got some small holes in wall of first container. So that means “eventually” the second container’s pressure will build up until it’s 254 psi. And the first container release valve never open again. Nor does the first container pressure exceed 300 psi, nor does second container
    exceed 254 psi.
    And the flow depends of how many holes and the maximum possible of 300 psi.
    So density of air at 300 psi and 46 psi of pressure difference and size of holes.
    One could have 290 psi in first container and 254 psi in second container, and all the release valves being useless- first never trips, the second one never opens and last one always open
    and always flowing same amount molecules, as coming from reservoir and being allowed thru the holes in container.
    New ownership. Problem: we want more pressure in first container.
    Solution remove useless control valve from reservoir. And fill holes.
    So first tank gets to 254 psi times 2. Or 508 psi. And apparently this tank doesn’t explode [the second engineer was correct]. And second tank is at 254 psi. Again same amount molecules are flow from reservoir, and first tank
    valve and out the second tank valve.
    The difference of pressure from first tank to second and from second to outside is 254 psi.
    But the density of air involved is different. If it was water instead of air the density would be the
    about the same.

  156. mkelly says:

    tallbloke says: April 7, 2013at 9:04 pm

    Reading the quote from NASA they maintained the same w/m^2 output from the source after the shield was in place. Further, my heat transfer book shows this 1/2 idea as an example with no change indicated from the source. The book says that heat shields add no heat to the system. I suggest this steel shell problem is nothing more than the sun and the orbit of the earth as the shell. The equations are the same for both.

    Understand I am only saying that Willis’ steel shell problem is presented wrong. Insulation will retard the rate of heat loss. That is why I say “as presented” in many of my comments.

  157. gbaikie says:

    “Kristian says:
    April 7, 2013 at 8:29 pm

    Tallbloke,

    Did you even read my post? It appears to me you didn’t.

    You ask: “Where is it [the planet] losing energy to, aside from the 117.5 going to space?”

    To the shell, Tallbloke. You know, the 235 – 117.5 = 117.5.”

    Not sure I read the post in question, but do you mean, shell transfer 117.5 from planet to space.
    And steel absorbs 117.5 W.
    And I suppose once you get warmed steel, the steel radiates at this temperature to
    space?

    Not agreeing or disagreeing.
    But it’s obvious if/when shell were cold, people would not consider it could be warming planet
    in anyway [unless it was a reflective surface]..
    But once it’s warmed it’s reasonable to assume one adding a new source of heat.
    So though warmed a bit, it’s not preventing energy from passing thru it, and adding it’s
    own source of heat which it is also radiating.
    So if a transparent shell, you could see the planet thru it.
    But it’s not transparent, but one probably could detect that it has warm planet under the shell.
    Something like, detect shell having a certain temperature also detecting it has some kind of internal source of heat.

    Dr Spock: “Captain we are detecting a dark object, it appears to be a warm planetoid with shell”

    Now it’s possibly it could be detecting by active sensors, but would it be beyond Spock’s capably
    to detect this merely by using passive sensors?

  158. suricat says:

    I don’t think an analogy is helping TB. Let’s return to the ‘toy’, but with a BB planet and shell (we can always get to a ‘grey body’ shell later).

    At the risk of repeating myself, I quote my post of March 31, 2013 at 4:20 am:

    “If we take another look at the graphic that wayne presented 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.”

    NB. There is NO energy exchange ‘BETWEEN photons’! Only energy exchange ‘between masses’ (BBs [Black Bodies]) can evoke photon emission, or absorption. For ‘EM radiation’, the ‘photon/EM wave’ is just the ‘delivery system’ for ‘radiant energy transport’ between ‘BB masses’. It always gets from A to B on the vector of its original departure (‘Relativity’ aside).

    To add to that post. A cooler body, with a ‘lower frequency Planck peak’, sends ‘photons/EM waves’ to a warmer body, with a ‘higher frequency Planck peak’, which results in the ‘warmer body’ overly filling the ‘Planck spectra’ that it can ‘radiate’ (emit) from the lower wavelengths/frequencies of its ‘curve’. This process is one of ‘mutual absorption’ between bodies.

    The ‘end result’ is that, when the ‘warmer body’ is supplied with an infinite and constant energy source, the ‘warmer body’ CAN’T emit enough of the ‘lower wavelengths/frequencies’ to maintain its ‘total power output’ for that ‘Planck temperature level’, so its Planck temperature ‘increases’.

    How is this possible? Not being able to emit energy at the lower wavelengths/frequencies, the body’s ‘internal energy’ is increased, thus, increasing its ‘temperature’ and ‘Planck curve peak’ to balance/reflect/disclose the increase in internal energy.

    I may be absent (again) for a while following this post, as Mum’s Funeral is on Thursday and I’ve a lot to arrange.

    Best regards, Ray.

    [Reply] Ray: Our thoughts and condolences are with you at this difficult time. All the best from everybody here at the talkshop.

  159. donald penman says:

    This seems to me where The willis model came from
    http://noconsensus.wordpress.com/2012/07/20/why-back-radiation-is-not-a-source-of-surface-heating/
    I have no problem with this when we are talkng about heated balls but when this logic is applied to planets it no longer makes sense to me .Some of us wish to see the Earth as an engineering problem and the the shell has the same place as the nose in the monty python sketch about wiches in that it makes the Earth less Earthlike and more like a machine.I do not see this as a thought experiment because a thought experiment deals with reality but you are allowed to make an observation where you can’t possibly do so it is not about dreaming up steel shells around planets which don’t exist.

    I am sorry to hear about your mum Ray you have my condolences.

  160. tallbloke says:

    Hi Donald. It seems to me that the value of considering gedanken experiments such as Willis’ steel greenhouse is that it allows us to isolate effects within the frame of physics, and examine them in the context of a system of sufficient simplicity for clear understanding to be reached. (Hoho). We don’t expect to be able to work out whether or not extra co2 would cause an increase in surface temperature on Earth from consideration of this model. Principally because it posits a vacuum between the shell and the planet surface.

    Leonard’s article goes a lot further, by considering three different models all in one session. He does it well too in my opinion. He shows that the ‘effective radiation level’ is not only a product of the action of IR active gases in the atmosphere, but also the other modes of energy transfer such as convection and latent heat. The upshot of that is that the hydrological cycle is utterly dominant and extra co2 is a fart in the hurricane.

    But we ARE NOT GOING THERE in this thread, which is solely to help people understand how the impedance of radiation leads to higher internal temperature without heat being transferred from cooler objects to warmer ones. The dragonslayers do not need to come up with ‘new physics’ for us to be able to slay the skydragon. The physics we already have (which is well supported by observation and experiment) can do the job just fine.

  161. Arfur Bryant says:

    lgl says:
    April 7, 2013 at 10:29 pm

    [“Arfur
    When the pressure gets to 254 psi in the outer tank according to you it equals the pressure in the inner tank and then there will be no flow from the inner to the outer tank. If you are unable to understand this very basic logic it’s no use discussing any further.”]

    lgl,

    You are grasping at straws and being silly. Of course there can be 254 psi in each tank and still be a flow between the pump output and outer space! The flow throughout is 235 g/s and the inner and outer tanks can be considered one tank.

    Read up on Bernoulli. From memory, his principle goes something like: “In a steady streamlined flow, the sum of energies remains a constant”. In a flowing system the pressure might change due to friction losses but in general if the flow maintains a constant velocity, so does the pressure. This was discussed a long time ago.

  162. Arfur Bryant says:

    tallbloke says:
    April 7, 2013 at 10:30 pm

    [“Because pressure and flow rate aren’t the same thing. Block half the holes in the inner tank. The pump keeps pumping 235g/s. The pressure must rise until it is high enough to force 235g/s out of half the number of holes – yes?
    HTH.”]

    Rog,

    Firstly, I have no idea what HTH means!

    Secondly, you are 100% correct. Pressure and flow rate aren’t the same thing! When have I said they were?

    Thirdly, yes, the pump keeps pumping 235 g/s. Ever wondered why? If you are correct about pressure and flow rate being different (and you are), then – if Tim’s analogy is correct and the inner tank pressure doubles to 508 psi and the flow rate remains constant – the pump must have a governor on it! This means the pressure inside the pump would have to increase in order to maintain the output flow of 235 g/s against the increased ambient pressure of the inner tank. Yes?

    Well, now we have a whole new ball game – but that’s not good news for Tim’s analogy and its supporters.

    If the analogy is meant to represent the Willis model so wonderfully (see all the plaudits earlier in the thread), then the presence of a governor changes lots of things. In the gas tank analogy the pump itself (the core) heats up (because pressure is analogous to temperature in Tim’s analogy) in order to maintain the energy (flow rate of 235 g/s), which is analogous to radiant energy in Willis’ model. But if Willis’ model where the 235 suddenly becomes 470 W/m^2 really was represented by Tim’s analogy, the governed core will have to reduce in temperature in order for the radiant energy to be maintained at 235 W/m^2! Therefore the addition of the shell has actually caused the core of the planet to cool! So much for analogies…

    Rog, eventually the pressure between the tanks will equalise (and can be considered one tank) and the flow rate will be a constant 235 g/s. No change in flow rate but the pressure differential will be maintained between the pump and space.

    The analogy is flawed.

  163. lgl says:

    Arfur
    Yes lets use Bernoulli to show how flow relates to pressure difference, this for instance:
    http://www.engineeringtoolbox.com/bernouilli-equation-d_183.html
    Now set P1=P2 and report back how large the flow will be.

  164. tallbloke says:

    Arfur: Rog, eventually the pressure between the tanks will equalise (and can be considered one tank) and the flow rate will be a constant 235 g/s. No change in flow rate but the pressure differential will be maintained between the pump and space.

    Here’s the logic:
    To reach a steady state, the outer tank has to lose as much air to space as the pump puts out (235g/s)
    To do that the pressure in outer tank has to reach 254psi. Once it does reach 254psi it is pushing back on the air trying to get out of the inner tank with that amount of pressure.

    If the inner tank was only at the same pressure, THERE WOULD BE NO FLOW FROM THE INNER TO THE OUTER TANK. But we know that at steady state, the outer tank must lose 235g/s to space. Therefore the same 235g/s has to pass from the inner to outer tank through the restriction. We know that it took 254psi to get the flow rate up to 235g/s when there was no pressure pushing back. So we know the pressure has to be 254psi higher than whatever the pressure in the space it is pushing air into is. Therefore the pressure in the inner tank will have to rise to twice the pressure in the outer tank to get the requisite flow rate from the outer tank into the void beyond.

    Arfur, your ‘logic’ tells you that you are right. My Welding set, pressure gauges, theoretical knowledge and practical experience tell me you are wrong. The tanks cannot be considered as one tank because of the restriction between them, and the pressure will never equalise unless you block up the leaks in the outer tank and stop pumping. You would then have a static situation. BUT THIS IS NOT A STATIC SITUATION.

  165. Arfur Bryant says:

    Ah, debating at its best.

    lgl, Why would I make P1 = P2? P1 is 254 and P2 is 0. Read your link again.

    Tallbloke, Smileys… really? I’m glad you’ve found a new friend. You can play fantasy engineering together.

    I am going to try to explain this to you in really simple steps:

    1. At Tim’s ‘end steady state’, you have the outer tank at 254 psi and the flow rate of 235 g/s.

    2. At Tim’s ‘end steady state’, you have 508 psi in the inner tank and 254 in the outer.

    3. At Tim’s ‘end steady state’, you have a fixed flow rate into the system of 235 g/s

    Why wouldn’t the two tanks equalise in pressure? Any excess pressure in the inner tank will flow into the outer tank. Even an inner tank pressure of 255 psi would start a flow on top of the already existing flow. There is no suggestion that the flow rate is a maximum of 235 g/s between the tanks. So as the excess (508-254) starts to flow into the outer tank, the outer tank would increase in pressure but for the fact that any increase in pressure above 254 in the outer tank will increase the flow rate to outer space. There is no suggestion that the exit flow is limited to 235 g/s. Tim has already admitted that you don’t need a differential of 254 psi to start the leakage. This process will continue until the tanks equalise at 254 psi.

    OF COURSE they can be the same pressure when there is a flow in a system! The pressure differential is between the pump and outer space. For goodness sake, what do you think would happen? You think the flow would stop in the middle of a system because the two tanks were the same pressure? Really? What would happen to the air in the outer tank? It would be evacuated to space, thus inviting more air from the inner tank. This air is constantly being re-supplied at 235 g/s! That’s what is providing the pressure for the system. OF COURSE it’s not a static system.

    The two tanks CAN be considered as one if the pressure is the same AND there is flow between them at the same rate as the input and output.

    [“Therefore the pressure in the inner tank will have to rise to twice the pressure in the outer tank to get the requisite flow rate from the outer tank into the void beyond.”]

    Nonsense. How does the inner tank get to 508? If there was any tendency for any increase in the inner tank the flow rate between the tanks would increase but then instantly the pressure in the outer tank would increase which would result in the exit flow increasing. No, once the tanks equalise, the flow rate will be constant because sufficient pressure is being provided by the pump to maintain 235 g/s into the system. Friction losses ignored.

    [“If the inner tank was only at the same pressure, THERE WOULD BE NO FLOW FROM THE INNER TO THE OUTER TANK.”]

    Nonsense. Yes there would be flow because there is a fixed flow into the system, and the system has an exit. The pump will have to work harder but you want to ignore that because it screws up the analogy relating to Willis’ model.

    Rog, I like you and what you have done with this blog. I admit when I am wrong. I’m not wrong. The analogy is flawed. If you want me to leave, just ask, but don’t try to tell me the tanks won’t equalise.

  166. Tim Folkerts says:

    Arfur. I can’t believe I actually just spent 30 min doing this but ….

    Go find two buckets (or plastic tubs from the grocery store) — one that will fit loosely inside the other. Poke similar holes in sides of each right near the bottom.

    EXPERIMENT 1: Put the smaller bucket in the sink and run water into it. It should fill to some level (depending on the water flow and the size of the hole.) For me, this was ~ 4″ deep.when it reached a steady-state.

    EXPERIMENT 2: Put the larger bucket in the sink and run water into it (without adjusting the flow rate from Expt 1). It should fill to some level (depending on the water flow and the size of the hole.) If the hole is the same size, this should be about the same depth as in Experiment 1 (and it was for me).

    EXPERIMENT 3: Put the smaller bucket in larger bucket and run water into it. The water very clearly filled to ~ 4″ in the outer bucket and ~ 8″ in the inner bucket.

    CONCLUSION. A pressure DIFFERENCE is required to make the water flow through the leak. The inner “tank” can and will and must rise to a greater pressure than the outer “tank”, which can and will and must rise to a greater pressure than the (literally in this case) “sink”.

  167. Tim Folkerts says:

    Just to be extra clear, that should be …

    EXPERIMENT 3: Put the smaller bucket in larger bucket and run water into the smaller bucket (still at the same flow rate).

  168. wayne says:

    Arfur: try http://hyperphysics.phy-astr.gsu.edu/hbase/ppois.html#poi
    I think you must have misinterpreted something basic far up this thread, like missed that there are two real orifices constricting the flow, or something. Don’t you feel everyone is saying the opposite to you… usually a good time to re-evaluate your own understanding of the example.

  169. tallbloke says:

    Tim F: Bravo. Simple, cheap, illustrative.

  170. tallbloke says:

    Hi Arfur, I know you are prepared to admit when you are wrong. I wouldn’t spend this much time if you weren’t.

    Do Tim’s experiment for yourself and report back. Make sure you make the holes in the sides of the buckets near the bottom, and the flow rate is high enough to get the level in the outer bucket high enough to cover the holes in the inner bucket. You’ll need sufficient difference in bucket diameter to allow the reduced volume in the outer bucket to generate a head of pressure once the inner bucket is introduced. PV=Constant so don’t be thinking that the changed volume affects the outcome. You could run water between the buckets before making holes in the smaller one to be sure of this. You’ll need to weight the smaller bucket to prevent it floating.

    Experimentum summas judex
    – Albert Einstein –

  171. From Postma’s blog
    thefordprefect says: Your comment is awaiting moderation.

    2013/04/10 at 5:45 AM
    OK.
    First what I hope is an uncontrovesial graph
    Temperature 0K to 10K – Quanta emitted LINEAR (because it fits the plot better but 10^4 works too) with respect to the temperature
    The core A is actively stabilized from 10K to 0K and emitts 100 to 0 quanta
    The shell B is actively stabilised from 0K to 10K and emits 0 to 100 quanta
    Apart from the 10^4 quanta vs temperature hopefully there is nothing “wrong” with this

    Now the other plots. This assumes the slayer premise that quanta are not utilised when they reach a hotter body than what they were emitted from. i.e. when A temperature is below B temperature A quanta have no effect on B (graph shows this as null)

    There is also the plot showing the warmist view that all quanta reach and add to the energy in the target
    You will note that when looking at net quanta from A to B (slayer mode) at the point A temperature = B Temperature the net quanta in B goes from 0 to -50 so it will begin cooling rapidly.
    The warmist mode shows no discontinuity!

    Perhaps you could show some plots which better represent your thoughts?

    [reply] This: “when the reach a colder body than what they were emitted from” contradicts this: “when A temperature is below B temperature A quanta have no effect on B” Please clarify.

  172. Oh dear! That’ll keep them arguing for days!

    “Now the other plots. This assumes the slayer premise that quanta are not utilised when the reach a colder body than what they were emitted from. i.e. when A temperature is below B temperature A quanta have no effect on B (graph shows this as null)”

    should be

    Now the other plots. This assumes the slayer premise that quanta are not utilised when the reach a hotter body than what they were emitted from. i.e. when A temperature is below B temperature A quanta have no effect on B (graph shows this as null)

  173. lgl says:

    Arfur
    “Why would I make P1 = P2? P1 is 254 and P2 is 0. Read your link again.”

    Because that is your claim. We are talking about the inner and outer tank where you claim P1(inner)=254 and P2(outer)=254 (not 254 and 0). This according to you will still give a 235 g/s flow, which is absurd.

  174. Arfur Bryant says:

    tallbloke, Tim, wayne, lgl

    This will probably be my last post on this subject. I believe I am in danger of practising what I complain about others doing – that is putting my understanding (assumption) first and ‘making the pieces fit’ in order to get my way instead of being objective. Being objective is something I hold dear and, as I may be becoming less than objective on this topic, I should – in all conscience – refrain. I appear to be up against a consensus, and we all know how difficult that is to combat. (This statement acknowledges that the consensus may well be correct in this case!)

    In all sincerity, I concede defeat. I obviously need to take some time to do some more research and learning. I accept my level of knowledge is apparently insufficient to compete.

    I am also embarrassed that I feel it necessary to concede twice in the same thread. So be it. We all learn. What doesn’t kill me will make me stronger.

    Thank you all for your patience and time.

    Rog, I hope you won’t mind me making comments on other threads if I feel able to do so.

    I wish you all the best, whichever side of the cAGW fence you stand.

    Kind regards,

    Arfur

  175. tallbloke says:

    Arfur: you’re welcome to comment on other threads. This one was not about the Earth and its convective atmosphere. That’s where the consensus between the people here from both sides of the (c?)AGW debate breaks down. We all know the same basic physics but when latent heat and convection join the party in the real atmosphere, things are no longer basic physics. So, not much of a consensus really. 😉

  176. Ilma630 says:

    All I can say is that folks should read http://climateofsophistry.com/2013/05/13/slayers-putting-up-not-shutting-up/ and Joe Postma’s other articles again, and closely. He clearly identifies that you cannot treat the interference of two radiating sources in the same way as two (conductive/convective) heat flows. Two objects radiating 235W/m2 at each other do NOT make a total of 470W/m2. His point is that radiation just doesn’t work that way which is why the ‘additive’ maths being used is nonsensical.

    He also clearly sets out a time-dependent model for Earth’s energy budget which fully accounts for the observed (real-world) temperatures with no need for any greenhouse/back-radiation effect. His point is that you cannot ‘average out’ incoming solar radiation to a single value applied constantly for 24 hours per day, the real world doesn’t behave that way.

    Just try shining two equal brightness torches at each other, or one into a mirror so the reflection is directly back on itself. Does the output of a torch magically increase, i.e. the torch get brighter? No, of course it doesn’t.