Let's finally move on. Could we finally take the very first step in this calculation? Please?
I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.
Show your calculations where we can see them. I'm not doing this just for me, I want to show other people just how much a clown you actually are. I am not going to install Sage today just to check your math, and probably neither is anybody else who sees this.... I have reasons for wanting it public-readable, and I will accept nothing else. [Jane Q. Public, 2014-09-02]
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches Jane's "steady-state".
Could we finally take the very first step in this calculation? Please?
I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.
Show your calculations where we can see them. I'm not doing this just for me, I want to show other people just how much a clown you actually are. I am not going to install Sage today just to check your math, and probably neither is anybody else who sees this.... I have reasons for wanting it public-readable, and I will accept nothing else. [Jane Q. Public, 2014-09-02]
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches Jane's "steady-state".
Then it should be easy to show how badly this approximation screws up the calculation, right? So why don't we check to see wrong these approximations are, by actually doing some calculations? Finally? Please?
I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.
Show your calculations where we can see them. I'm not doing this just for me, I want to show other people just how much a clown you actually are. I am not going to install Sage today just to check your math, and probably neither is anybody else who sees this.... I have reasons for wanting it public-readable, and I will accept nothing else. [Jane Q. Public, 2014-09-02]
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches Jane's "steady-state".
... You then (at ANY time, I don't care when) claim that a larger surface is at the same temperature, which requires the same amount of W/m^2. But you have more m! So the total power output is greater than your input. There is no way to weasel out of this, man. You're trying to output more power than you're putting in. This isn't even 11th-grade physics. Let's try it at something more like your level: You have 200 beans equally distributed among 10 squares. If you now take those beans, and divide them equally among 25 squares of the same size, how many beans do you now have per square? Show your work. [Jane Q. Public, 2014-09-04]
Cute. I've repeatedly explained ad nauseum that neglecting area ratios is an approximation. I've already shown how tiny the effects are for Earth's area ratio. Does this mean you don't intend to perform even the simplest calculation to confirm this? Why don't we check to see wrong these approximations are, by actually doing some calculations? Finally? Please?
I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.
Show your calculations where we can see them. I'm not doing this just for me, I want to show other people just how much a clown you actually are. I am not going to install Sage today just to check your math, and probably neither is anybody else who sees this.... I have reasons for wanting it public-readable, and I will accept nothing else. [Jane Q. Public, 2014-09-02]
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches Jane's "steady-state".
You were throwing around the term "equilibrium" rather loosely, and at one point you mentioned that "at equilibrium" the outer surface of the enclosing passive plate must be at the same temperature as the surface at the heat source:
Electric input of 509 W/m^2 is constant and the walls are held at 0F (255K). Therefore, the second plate has to radiate the same power out as the heated plate did before it was enclosed. So energy conservation at equilibrium requires that the second plate be at 150F (339K).
Again, Jane might not agree with the fact that the heat source warms after it's enclosed. But again, Jane could at least acknowledge that this is what I'm saying rather than trying to pretend that I somehow said all temperatures are the same. Please?
Will you acknowledge that no matter what you meant, this is still wrong? If the enclosing passive plate must radiate out the same power as the enclosed heat source, it cannot be at the same temperature, because radiated power is measured in W/m^2, and there are more m^2 in the enclosing passive plate. Therefore (SIMPLE MULTIPLICATION), because there is greater area they could not be at the same temperature and radiate outward the same power. No matter how you try to bullshit your way around this, it is still WRONG. [Jane Q. Public, 2014-09-04]
I've repeatedly explained ad nauseum that neglecting area ratios is an approximation. I've already shown how tiny the effects are for Earth's area ratio. For weeks you've refused to perform even the simplest calculation to confirm this. Why don't we check to see wrong these approximations are, by actually doing some calculations? Finally? Please?
... You refuse to use real materials with measured absorptivities and emissivities, and insist on using gray bodies and Kirchhoff's law, neither of which actually represent Spencer's experiment in anything like the real world, even though it was intended to be a model of the real world. I get that. But I want to make sure everyone else gets it too. I offered to use real materials with measured emissivities in the approximate temperature range we are talking about (though you refuse to acknowledge that), but you refuse to use them. Just so we know where we stand. I have already explained to you that there is no need to resort to gray bodies, and that we have plenty of information to calculate more realistic, real-world results. But whatever. You refuse to do anything but what you want to do, so let's just go with it for now. But I reserve the right to re-visit this issue. [Jane Q. Public, 2014-09-04]
... you insist on using gray bodies, because... you don't have the spare 15 minutes or so it might take to do approximate, more real-world calculations, and want to use Kirchhoff's law (although it really isn't necessary) to make your life easy....
I've been explaining for over a month that the heated plate warms after it's enclosed. I realize you don't agree, which is why I'm trying in vain to get you to finally perform a single, solitary calculation of your own.
I never said I disagree with this. Please find where I said that. On the contrary; I definitely agree that it warms. In fact it must: Spencer stipulated that it was to be inserted when it was colder than the heat source.... [Jane Q. Public, 2014-09-04]
I've been explaining for over a month that the heated plate (aka Jane's "source") warms after it's enclosed. I've only been wasting my final days because Jane's repeatedly disagreed by supporting Dr. Latour's ridiculous Sky Dragon Slayer claim that the heated plate (aka Jane's "source") simply remains at 150F after it's enclosed:
... the heat source does not become hotter. This is, and has been, the whole of Latour's argument, and it is valid. It is not crazy speculation by some nitwit, it is straightforward application of Stefan-Boltzmann law. Q.E.D., indeed. If the above inequalities hold (and they do), Latour's conclusion is the only one that is mathematically valid. [Jane Q. Public, 2014-08-02]
If Jane agrees that the heated plate (aka Jane's "source") warms after it's enclosed, then that's great news! In that case, we can all agree that the mainstream physics describing the greenhouse effect is accurate, obeys the laws of thermodynamics, and proves that the Sky Dragon Slayers are wrong.
... you had calculated this number for thermal equilibrium using Kircchoff's radiation law. But as I've explained many times now, there is no thermal equilibrium so Kircchoff's law does not apply.... as you state yourself that equation is derived from Kircchoff's radiation law, which does not apply here.... [Jane Q. Public, 2014-09-04]
... You pointed out to MIT's derivation of energy transfer between infinite gray bodies. It does not apply here because... that derivation makes use of Kircchoff's law which does not apply in Spencer's challenge.... [Jane Q. Public, 2014-09-04]
... you'll still have to modify your equation if that is based on the one you borrowed from MIT. I repeat that Kircchoff's law does not apply here.... [Jane Q. Public, 2014-09-04]
No, I calculated this number for a system which doesn't change with time. From now on I'll call this condition "steady-state" but that doesn't change the fact that my equations are based on conservation of energy in a system that doesn't change with time. Again, I only mentioned Kirchhoff's law to explain MIT's gray body approximation. Since emissivity isn't a function of wavelength, all surfaces aren't required to be at the same temperature.
... You pointed out to MIT's derivation of energy transfer between infinite gray bodies. It does not apply here because (a) we have specifically defined areas, they are not infinite... [Jane Q. Public, 2014-09-04]
Notice that the first example MIT applies their final equation to is a thermos bottle which doesn't have infinite walls. That's because a thermos bottle has no edges (just like our fully enclosed plate!) so the infinite plate approximation applies. If not, why did MIT use their equation to model a thermos? Were they talking about a thermos with infinite walls?
... any temperature of the source that is higher than the initial radiative equilibrium (150 deg. F) represents higher power output from that same source, any such higher temperature would violate conservation of energy.[Jane Q. Public, 2014-09-04]
No, Jane. If power in != power out in steady-state, that would violate conservation of energy. Because my equations are based on the principle that in steady-state power in = power, their solutions satisfy conservation of energy.
Your figure of 233 deg. F radiant temperature at what you called "equilibrium" represents a constant radiative power output from the heat source greater than its initial power output at 150 deg. F. Where is this additional power coming from? [Jane Q. Public, 2014-09-04]
No. The radiative power output is exactly the same as before the heat source was enclosed. It's hotter because radiative power output is proportional to T_h^4 - T_c^4. Before the heat source was enclosed, it was radiating to the chamber walls at T_c = 0F. After it's enclosed, it's radiating to the inside surface of the enclosing plate which is at T_c > 0F.
But as you said, it's pretty damned hard to prove anything without calculating it all the way through. So let's finally take the very first s
Earlier, when I saw your mentions of equilibrium, I thought you were referring to the steady-state that would eventually be achieved.... [Jane Q. Public, 2014-09-02]
As I said: "Energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At equilibrium, that rate is zero because the system doesn't change. So at equilibrium, power in = power out."
I explicitly said a system in "equilibrium" doesn't change, which Jane calls "steady-state". I repeatedly asked Jane if we could agree on that, but a month later Jane objected:
... Kircchoff's law does not apply to this experiment, and no situation arises in which the temperatures are the same everywhere, or the emissivities vs absorptivities. There is a steady-state arising from active (but constant) exchange. But there is no equilibrium. [Jane Q. Public, 2014-09-02]
Earlier, when I saw your mentions of equilibrium, I thought you were referring to the steady-state that would eventually be achieved. But even though you mentioned Kircchoff's law, it didn't sink in to my brain that you were referring to actual, literal equilibrium. Uh-uh. As they say in my neck of the woods: it ain't happenin'. [Jane Q. Public, 2014-09-02]
... A steady-state is NOT the same thing as equilibrium.... You do not get to re-define equilibrium any way you choose. Just no.... [Jane Q. Public, 2014-09-03]
Note that my definition of equilibrium is identical to this one: "Class 6- Equilibrium Temperature: Equilibrium means no change with time.... In equilibrium, we expect ENERGY IN = ENERGY OUT..."
Also note that this definition of equilibrium doesn't require a planet's south pole to be at the same temperature as its equator, or its surface to be at the same temperature as its tropopause (for planets with an atmosphere).
But from now on I'll call the system in "steady state" when its temperatures don't change with time, in the naive hope that we might actually be able to finally take the very first step in this calculation.
... THERE IS NO RADIATIVE EQUILIBRIUM HERE. THERE IS NO THERMAL EQUILIBRIUM HERE. None. You may not assume them.... In Spencer's challenge, thermodynamic equilibrium does not exist.... in Spencer's challenge there very definitely is no equilibrium. [Jane Q. Public, 2014-09-03]
... Kircchoff's law (and MIT's example) both assume no bodies involved are storing thermal energy, and there is thermal equilibrium. In fact that is how Kircchoff's law is derived: technically Kircchoff's law only applies at thermal equilibrium. MIT was free to apply it in their example because thermal equilibrium was assumed. However in Spencer's challenge there very definitely is no equilibrium. It is not appropriate to assume it or try to apply it here: the whole point is that we are trying t
Once again, no. I never said that all surfaces were at the same temperature. I've already explained that the final outer temperature of the enclosing shell doesn't happen at the same time as the initial temperature of the heated plate. Initially, the heated plate is at 150F and the enclosing shell is cooler than 100F. But because power in > power out, the plates slowly warm to a new steady-state. By the time the outer temperature of the enclosing shell is ~149.6F (accounting for area differences), the heated plate is ~233.8F. This doesn't change even if we neglect area differences: the enclosing shell and the heated plate are never at the same temperature. Again, that's why I called them T_c and T_h.
You did say it, quite clearly. I quoted you twice and linked to your web page. LATER you changed your tune. I can accept that you changed it later, but you did say it. [Jane Q. Public, 2014-09-04]
Once again, no. I never said that all surfaces were at the same temperature. I've already explained that the final outer temperature of the enclosing shell doesn't happen at the same time as the initial temperature of the heated plate.
By the time the outer temperature of the enclosing shell is ~149.6F (accounting for area differences), the heated plate is ~233.8F.
But you do not give any justification for this answer, you just throw it out there. [Jane Q. Public, 2014-09-04]
I've been explaining for over a month that the heated plate warms after it's enclosed. I realize you don't agree, which is why I'm trying in vain to get you to finally perform a single, solitary calculation of your own. But even if you don't agree with my statement that the heated plate warms after it's enclosed, can't you at least acknowledge that this is what I'm saying rather than trying to pretend that I somehow said all temperatures are the same?
... It may not be assumed that the temperatures are the same! As you have done at least once.... [Jane Q. Public, 2014-09-03]
Once again, I never said that all surfaces were at the same temperature.
You USED this before to ASSUME all surfaces were at the same temperature! I quoted you saying it in a post above, and you referenced that passage just the other day. In fact this was the source of much of the misunderstanding here.... [Jane Q. Public, 2014-09-03]
Once again, I never said that. In reality, I said that both sides of a thermal superconductor are at the same temperature. This was the source of much of the misunderstanding here, and you strongly objected to the notion of a thermal superconductor. Again, that's why I calculated the small temperature difference across an aluminum shell with finite conductivity. [Dumb Scientist]
Electric input of 509 W/m^2 is constant and the walls are held at 0F (255K). Therefore, the second plate has to radiate the same power out as the heated plate did before it was enclosed. So energy conservation at equilibrium requires that the second plate be at 150F (339K).
You were referring to "the second plate", as opposed to the "heated plate". That corresponds to what I have been calling the "passive" or "enclosing" plate.... [Jane Q. Public, 2014-09-03]
Once again, no. I never said that all surfaces were at the same temperature. I've already explained that the final outer temperature of the enclosing shell doesn't happen at the same time as the initial temperature of the heated plate. Initially, the heated plate is at 150F and the enclosing shell is cooler than 100F. But because power in > power out, the plates slowly warm to a new steady-state. By the time the outer temperature of the enclosing shell is ~149.6F (accounting for area differences), the heated plate is ~233.8F. This doesn't change even if we neglect area differences: the enclosing shell and the heated plate are never at the same temperature. Again, that's why I called them T_c and T_h.
So once again, I never said that all surfaces were at the same temperature.
If you don't particularly mind, could we finally take the very first step in this calculation? Please?
Yes, I mind very much. There is no point in doing any calculations at all until we rid you of the false assumptions you have been making about this experiment (as I have been trying to do). They have been leading to incorrect results, and moving on would be a waste of everybody's time.... There is no equilibrium in this experiment, either thermal or radiative. Period. You may not assume them, or use formulas that are only appropriate for equilibrium. Get past that and move on, or stay stuck here. That's up to you. But unless and until you do, there is simply no need for me to go any further. Your refutation to this point has been demonstrated to be invalid. [Jane Q. Public, 2014-09-03]
I'm very sorry. I take full responsibility. Can we please move on?
... you can take out the epsilons, since I thought we had already agreed we don't need them. (If they represent emissivity.)...
Nonsense. They are figures at at incident radiation of 1367 W/m^2, which is sunlight at 1 AU, for the very reason that it is an approximation of Earth insolation. So in fact it would make a good representative example of what Spencer's model is supposed to be all about.... ESA gives observed values for integrated emissivity and absorptivity for aluminum. This is a good approximation and it is used in the real world for aluminum in a vacuum.[Jane Q. Public, 2014-09-03]
Those ESA absorptivities are for absorption of sunlight. Consider the first diagramhere which shows that 6000K sunlight has much shorter wavelengths than the radiation from objects at the temperatures we're considering. In fact they hardly overlap. But the emissivities are for radiation emitted by much cooler objects. That's one reason why those ESA emissivities aren't equal to their absorptivities.
Here's a good explanation of this problem: "... white paint is quoted as having an absorptivity of 0.16, while having an emissivity of 0.93.[9] This is because the absorptivity is averaged with weighting for the solar spectrum, while the emissivity is weighted for the emission of the paint itself at normal ambient temperatures...."
If you really insist on gray bodies that's up to you; but I do not acknowledge that there is any legitimate reason to NOT use reasonable approximations of integrated absorptivity and emissivity. [Jane Q. Public, 2014-09-03]
Since the absorption values you indirectly cited are for absorption from the 6000K radiation from the Sun, that seems like a legitimate reason not to use those values in a thought experiment where nothing is at 6000K. Again, another reason is that we'd have to recreate MODTRAN to derive heat transfer between non-gray bodies where emissivity and absorptivity are arbitrary functions of wavelength.
And once we debugged that new MODTRAN clone, we'd have to test it in a simple case, like the case of gray bodies where emissivity and absorptivity don't depend on wavelength. So we might as well solve the simple problem first.
We might be talking past each other. What you're calling steady-state is what I'm calling equilibrium. Radiative thermodynamic equilibrium doesn't require all surfaces to be at the same temperature, it simply means that temperatures don't change with time. At radiative equilibrium, power in = power out, which also means irradiance in = irradiance out.
You USED this before to ASSUME all surfaces were at the same temperature! I quoted you saying it in a post above, and you referenced that passage just the other day. In fact this was the source of much of the misunderstanding here.... [Jane Q. Public, 2014-09-03]
Once again, I never said that. In reality, I said that both sides of a thermal superconductor are at the same temperature. This was the source of much of the misunderstanding here, and you strongly objected to the notion of a thermal superconductor. Again, that's why I calculated the small temperature difference across an aluminum shell with finite conductivity.
As you said, the best we can realistically do is graybodies where emissivity = absorptivity. If you'd like to use a different emissivity just let me know, and we can both independently calculate the required electricity to check each other's answers.
After considering the situation I changed my mind. Since we are discussion what is supposed to be a real model of a real situation, we can use real emissivity and absorptivity.... [Jane Q. Public, 2014-09-02]
You were right when you said the best we can realistically do is graybodies where emissivity = absorptivity. Otherwise we'd need to derive a new equation where heat transfer is an integral over wavelengths. In other words, we'd have to recreate MODTRAN. I simply don't have time for that.
After considering the situation I changed my mind. Since we are discussion what is supposed to be a real model of a real situation, we can use real emissivity and absorptivity. And the emissivity of aluminum (as you pointed out yourself some time ago) is different from the absorptivity by a factor of about 3. The ESA figures are observed figures for aluminum plates in near-vacuum, so those figures would appear to be perfect. [Jane Q. Public, 2014-09-02]
I've never pointed that out. I've repeatedly shown youGoodman 1957 where Table 1 lists aluminum's emissivity as 0.113 from 100C to 300C.
In contrast, you're citing ESA figures from page 32 which are at 0K (-273C). But nothing in this experiment is anywhere near that cold.
Also note that Goodman 1957 specifically tests the gray body approximation and concludes that "Pure aluminum appears to act like a gray body when its radiating surfaces are at temperatures lower than 400C."
Again, if you'd like to use a different emissivity just let me know, and we can both independently calculate the required electricity to check each other's answers.
THERE IS NO THERMODYNAMIC EQUILIBRIUM IN THIS EXPERIMENT. There is a steady-state, but no actual equilibrium. That is not possible, because we are actively pumping heat in at one "end", and pumping it out of the other. Since one of the requirements of thermodynamic equilibrium is that all surfaces be at the same temperature... [Jane Q. Public, 2014-09-02]
We might be talking past each other. What you're calling steady-state is what I'm calling equilibrium. Radiative thermodynamic equilibrium doesn't require all surfaces to be at the same temperature, it simply means that temperatures don't change with time. At radiative equilibrium, power in = power out, which also means irradiance in = irradiance out.
... Kircchoff's law does not apply to this experiment, and no situation arises in which the temperatures are the same everywhere, or the emissivities vs absorptivities. There is a steady-state arising from active (but constant) exchange. But there is no equilibrium. [Jane Q. Public, 2014-09-02]
Earlier, when I saw your mentions of equilibrium, I thought you were referring to the
... What do you want to use for material? We might as well use the same material throughout. So if you want to use aluminum for source, passive plate, and walls that is fine with me. We know then, from ESA that the emissivity of aluminum in vacuum is approximately 0.15, and absorptivity 0.05.[Jane Q. Public, 2014-09-02]
Again, the materials are oxidized aluminum with emissivity = 0.11 for these temperatures. As you said, the best we can realistically do is graybodies where emissivity = absorptivity. If you'd like to use a different emissivity just let me know, and we can both independently calculate the required electricity to check each other's answers.
I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.
Show your calculations where we can see them. I'm not doing this just for me, I want to show other people just how much a clown you actually are. I am not going to install Sage today just to check your math, and probably neither is anybody else who sees this.... I have reasons for wanting it public-readable, and I will accept nothing else. [Jane Q. Public, 2014-09-02]
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches equilibrium. I promise to provide public-readable versions of my Sage worksheet from now on.
I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.
Show your calculations where we can see them. I'm not doing this just for me, I want to show other people just how much a clown you actually are. I am not going to install Sage today just to check your math, and probably neither is anybody else who sees this.... I have reasons for wanting it public-readable, and I will accept nothing else. [Jane Q. Public, 2014-09-02]
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches equilibrium. I promise to provide public-readable versions of my Sage worksheet from now on.
Obviously we'll have to agree to disagree. But I thought you wanted to do some actual calculations? As you say, it's pretty damned hard to prove anything without calculating it all the way through. So why don't we take the first step?
... it's pretty damned hard to prove anything without calculating it all the way through.... So why don't you draw a diagram, and simply perform all the calculations?... I'll even go with your own example of the passive plate enclosing the heat source, for now.... Then an enclosing plate is introduced, at a temperature (initially) less than that of the source. We can, if you wish, assume it is a hollow sphere, of some reasonable thickness, so the interior and exterior areas differ, and of a smaller external radius than the outside wall, so again they don't touch. Vacuum in between. And we begin our analysis. The starting point and equilibrium are both relevant points that should be calculated.... I don't insist, but to avoid ambiguity and to make things expressible on a standard keyboard, this is how *I* would label things: S for heat source, so radiative temperature T of S would be T(s). Passive plate (or shell) P. Outside enclosure or wall W. Absorptivity A so absorptivity of P would be A(p). Emissivity E.... [Jane Q. Public, 2014-09-02]
Note that it reduces to my simpler blackbody Eq. 1 if E(s) = E(w) = 1.
If you'd like me to clarify what my variable names for a particular equation would be in your terminology, just ask.
I've specified the dimensions. The heated plate is a sphere with radius 6371 mm and surface area A_h. The enclosing plate is a 1 mm thick concentric shell with an inner radius of 6378 mm, surface area A_c1 on the inside, and A_c2 on the outside. The chamber is also a concentric sphere with inner radius 6386 mm, so there's a 7 mm gap on both sides of the enclosing shell. Again, the plates and walls are oxidized aluminum.
At equilibrium, net heat flow out (in W/m^2) equals "electricity". The first step is to calculate that constant variable "electricity" which describes electrical power per square meter heating the sphere to 150F without an enclosing shell. I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches equilibrium.
Electric input of 509 W/m2 is constant and the walls are held at 0F (255K). Therefore, the second plate has to radiate the same power out as the heated plate did before it was enclosed. So energy conservation at equilibrium requires that the second plate be at 150F (339K).
Utter nonsense. The temperature of the outside of your enclosing sphere is determined entirely by its absorption minus its emission, with absorptivity and emissivity factored in. If your interior heat source were emitting at (your figure) 509W/m^2, and that is being absorbed by the interior surface of your enclosing sphere (which MUST have larger radius than the source, since they can't contact), then your outside surface, being of even larger area, must therefore be colder.... So you're INVENTING ENERGY OUT OF THIN AIR.... [Jane Q. Public, 2014-09-02]
I've already showed you that the outer surface of an enclosing shell with an area ratio similar to Earth's warms to ~149.6F. I've explained that neglecting area ratios is a tricycle: a simple approximation that helps us learn. It's like the "frictionless pulley" or "massless rope" or "blackbody" approximations. Again, in this case the tricycle isn't too inaccurate compared to the bicycle, it's much easier to learn, and it provides a sanity check on the more complicated calculation. As the area ratio approaches "1.0" the bicycle should give the same answer as the simpler tricycle. And it does.
... your prior analysis was still wrong. It is VERY easy to show this. Presume you have an initial source at T = 150 deg. F. It has a surface area of 1 m**2. Therefore (let's just assume your figure for power output here, it doesn't really matter and it's good enough for this illustration): it's emission is 509W/m**2. Let's say the EXTERIOR of your enclosing shell has an area of 2 m**2. However, your words (though in a slightly different context): power in = power out. Since the total power (W/m**2 times X m**2) must be the same in as out, the exterior of your shell cannot have the same irradiance. The same must be true if this were just one solid sphere, rather than a hollow sphere enclosing another sphere. Solving for the Stefan-Boltsmann relation at 509W/m**2 times 1 m**2 is total number of watts. If you try to multiply the same emission rate over 2 m**2 you get a DIFFERENT answer. That's just a fact. By assuming an external temperature of 150 deg. F, you have just created tangible energy from the vacuum. Congratulations. [Jane Q. Public, 2014-09-02]
When the area ratio departs far from 1.0, the tricycle becomes very inaccurate, so one should use the more complicated bicycle. But again, the Earth's area ratio is roughly 1.0025, so in that case the tricycle isn't too inaccurate.
Once again, I've already accounted for the area ratio to obtain the more complicated and more accurate solution.
But the second plate also radiates the same power in, toward the enclosed heated plate. Just like the cold chamber walls do. Now consider conservation of energy just inside the second plate (but outside the first) at equilibrium. We can solve for the insulated heated plate's temperature using Eq. 1 by setting Tc = 150F (339K). That yields an insulated heated plate temperature of 235F (386K).
No, it doesn't! The irradiation is total for the entire hollow sphere, not for each surface. You have to divide the total irradiance b
... It is the engineering textbook answer. Claiming it is nonsense does not make it so. It was your own model that violated conservation of energy. But to see why, it's easiest to solve the general case first, then look at a specific case. I told you I had reasons to solve the general case first.... Well, then, I guess you do admit defeat. It doesn't take much time to obtain a textbook on the subject (you were given references 2 years ago and it's not that hard to find others)... [Jane Q. Public, 2014-09-01]
No, the PSI Sky Dragon Slayers told you it's the engineering textbook answer. I showed you MIT's final expression which reduces to my Eq. 1 for blackbodies, and is consistent with theseequations and Eq. 1 in Goodman 1957. Physicists and engineers have been using thermodynamics for decades in the real world that contradicts Dr. Latour's Slayer nonsense.
... I am disputing that given reasonable chosen dimensions it is anywhere near an intractable problem.... [Jane Q. Public, 2014-09-01]
I never said the problem is intractable. Just that it's more complicated than the spherically symmetric problem. Again, do you dispute that equilibrium temperatures for a non-enclosing plate would vary across the plate surfaces rather than being simple numbers like with a spherically symmetric fully enclosing plate?
Maybe I should explain that. Consider Dr. Spencer's first illustration. Presumably the heated plate at "150F" has finite conductivity, so its lack of spherical symmetry means that its corners will be cooler than the plate's side's midpoints. That's because the corners are closer to the cold chamber walls than those midpoints.
An integral over the heated plate's surface might average to "150F" but (unlike a spherically symmetric plate) it can't have that temperature everywhere as long as it has finite conductivity. But at least the single heated plate has bilateral symmetry; the left and right hand side midpoints have the same temperature.
Adding a cool plate removes even that bilateral symmetry. The left hand side's midpoint warms the least because it's still radiating to the 0F chamber walls. The right hand side's midpoint warms the most because it's now radiating to the (initially) 100F cold plate.
Since enclosing a spherically symmetric plate warms it from 150F to ~233.8F for area ratios similar to Earth's, the right hand side's midpoint won't warm past ~233.8F. But it has to warm to conserve energy because at equilibrium power in = power out.
I can't be more specific without programming a finite element model. But Dr. Latour never even allowed for the heated plate's temperature to be different on each side. As long as we're only considering materials with finite conductivity, this would only be poss
Maybe I should explain what I meant by saying that equilibrium temperatures for a non-enclosing plate would vary across the plate surfaces. Consider Dr. Spencer's first illustration. Presumably the heated plate at "150F" has finite conductivity, so its lack of spherical symmetry means that its corners will be cooler than the plate's side's midpoints. That's because the corners are closer to the cold chamber walls than those midpoints.
An integral over the heated plate's surface might average to "150F" but (unlike a spherically symmetric plate) it can't have that temperature everywhere as long as it has finite conductivity. But at least the single heated plate has bilateral symmetry; the left and right hand side midpoints have the same temperature.
Adding a cool plate removes even that bilateral symmetry. The left hand side's midpoint warms the least because it's still radiating to the 0F chamber walls. The right hand side's midpoint warms the most because it's now radiating to the (initially) 100F cold plate.
Since enclosing a spherically symmetric plate warms it from 150F to ~233.8F for area ratios similar to Earth's, the right hand side's midpoint won't warm past ~233.8F. But it has to warm to conserve energy because at equilibrium power in = power out.
I can't be more specific without programming a finite element model. But Dr. Latour never even allowed for the heated plate's temperature to be different on each side. As long as we're only considering materials with finite conductivity, this would only be possible for a spherically symmetric enclosing plate.
Dr. Latour's answer wasn't "reasonably precise". He claimed that the heated plate wouldn't warm at all when the cold plate was added, even if it completely enclosed the heated plate such that K = 1. This is a specific prediction of "0.0000...F" warming. Since energy conservation means that adding a cold plate has to warm the heated plate, he's only off by a factor of infinity.
I've already explained why this is BS excuse. Latour didn't need finite element modeling to come up with a reasonably precise answer, and neither would you. [Jane Q. Public, 2014-09-01]
Latour's answer is ridiculous Sky Dragon Slayer nonsense which violates conservation of energy, as I've shown.
Once again, solving a problem without spherical symmetry means you'll have to solve for equilibrium temperatures which aren't constant across the heated and passive plates. Those equilibrium temperatures wouldn't be simple numbers. They'd be complicated functions that would vary across the plate surfaces. Contrast that with a spherically symmetric enclosing plate, where equilibrium temperatures are just simple numbers.
Are you disputing that equilibrium temperatures for a non-enclosing plate would vary across the plate surfaces rather than being simple numbers like with a spherically symmetric fully enclosing plate?
I simply asked why you refuse to show where Latour was wrong in Spencer's original challenge, not the "enclosing" variant of it. [Jane Q. Public, 2014-09-01]
Because, unless you dispute the above facts, that would require a complicated finite element model due to its lack of spherical symmetry. I simply don't have that much time left. And again, we'd have to test that complicated model in a case where an analytic solution is available anyway...
I'm not saying we should solve simpler problems before moving on to more complex problems. [Jane Q. Public, 2014-09-01]
Okay, then we disagree. It's always helpful to solve simpler problems before moving on to more complex problems. The simpler problem is easier to learn, and often serves as a sanity check on the more complex problem.
That got a minor mention later in his article, is not included in his diagrams, and is NOT the problem I originally presented to you. As I have said many times before, AFTER you refute Latour's calculations regarding Spencer's original challenge, which did not have the passive body enclosing the heat source, I would be happy to move on to the other issue... with no additional stipulations or additions to the problem Spencer describes. But you haven't gotten there yet. Cart before the horse, with a straw-man riding the cart. [Jane Q. Public, 2014-09-01]
Again, Latour's calculations allowed for K = 1: "K is the fraction of radiation from the first bar absorbed by the second colder bar, 0 < K <=1."
The only way K = 1 is if the cold plate completely encloses the first heated source. Otherwise, radiation from the side of the source opposite the cold plate couldn't possibly be absorbed by the cold plate, which would force K < 1. So once again, the fact that Dr. Latour included the possibility that K = 1 means that his claim applies to all geometries.
If not, why doesn't he deal with edge effects? The only ways to eliminate edge effects are if the plates are infinite, or if the cold plate completely encloses the heated source.
Why don't you just shut up and do it? Why have you been so mightily struggling, like a fish on a hook, to avoid it? [Jane Q. Public, 2014-09-01]
Again, I don't have enough time to program a finite element model to account for the fact that a non-fully-enclosing plate would cause plate temperatures to vary across their surfaces. But even if I did, the first thing I'd do after debugging it would be to check the finite element solution in a case where a simple analytic solution can be obtained. Namely, a fully-enclosing passive plate, where the plate temperatures are simple numbers.
By the way, since you keep insisting that only a particular geometry could refute Dr. Latour's treatment, could you please show where he specified the dimensions of the plates? Or where Dr. Spencer did? Otherwise, even if I had enough time to do so, how could I possibly program this complicated finite element model with the specific geometry that would finally convince you the Slayers are wrong?
... The problem is that there is no such thing as a thermal superconductor of this kind, and you aren't seeing that it leads to contradictions. The only way it could exist would be if it had NO thermal effect on its surroundings whatever. So it's the ultimate straw-man argument. There is no way it can be legitimately used to demonstrate anything. [Jane Q. Public, 2014-09-01]
Again, we'll have to agree to disagree about thermal superconductors. That's why I've repeatedly pointed out that I've already solved this problem with an aluminum enclosing shell, and it also warms the heated plate (aka Jane's "source") to ~233.8F.
No, they didn't, because it's a different problem, being given a theoretical treatment. You keep doing that, but I'm not buying. Two infinite plates, neither of which is heated, is not even remotely the same situation, and it's also theoretical only. They're not taking into account certain real-world factors pertaining to Spencer's experiment. Latour does. Not that they're doing anything wrong... given the context of their situation: infinite non-heated grey bodies. This is not Spencer's experiment. [Jane Q. Public, 2014-09-01]
No, it's exactly the same problem. The same infinite sum of absorption and reflection. The plates are only "infinite" to avoid having to model fringing field effects around the plate edges. And note that Dr. Latour doesn't model edge effects either, so his plates are either infinite or the passive plate completely encloses the "source". Either way, there would be no edges.
Notice that the first example MIT applies their final equation to is a thermos bottle where the inside wall is heated by hot fluid.
You did not point to a calculation he performed on Spencer's situation and prove it wrong. You took what you incorrectly called an analogous situation and called that wrong. Which has been my whole point here. You keep claiming something else represents Spencer's experiment, but you won't tackle Spencer's actual, original experiment. You have consistently refused, for over 2 years.... You continue to refuse to actually do what you said you'd done: refute Latour's treatment of Spencer's challenge. [Jane Q. Public, 2014-09-01]
Again, Dr. Spencer's actual, original experiment included the possibility of a fully-enclosing passive plate. And so did Dr. Latour's treatment of it. If you don't agree, please show where Dr. Latour specifies the dimensions of the plates before wrongly concluding that T remains 150.
In fact, as far as I can tell nobody's specified the plate dimensions except for me. Since the argument I'm refuting never specified the plate dimensions, why would the plate dimensions matter?
... I repeat: get the experiment with the two separate plates (actively heated plate and passive plate) right first. Then you can move on to a fully-enclosing plate. You say it's simpler but in a way it's not; you're trying to ride a bicycle when you haven't even managed to ride your tricycle without falling off....
At equilibrium, the enclosing shell radiates the same power out as the heated plate did before it was enclosed. But its area is 1.0025 times larger, so its outer temperature is 149.6F (338.5K) instead of 150.0F (338.7K).
In order for what you say to be correct, then the "enclosing shell" you refer to is not the heated plate enclosing the source. Which would mean you were talking about a completely different experiment, not even the one Spencer mentioned with the heated plate enclosing the source. [Jane Q. Public, 2014-08-31]
We might be talking past each other. What you're calling the "source" is what I've been calling the "heated plate" with temperature "T_h" in all my equations. I've called the other enclosing plate the "cold plate" with temperature "T_c". As I've repeatedly and consistently stressed, "T_c" is only identical on both sides of the enclosing cold plate if it's a thermal superconductor.
I'm sorry for any confusion this caused, but as you can tell I really am talking about the experiment Dr. Spencer mentioned. We're just using different words, and again I'm sorry for not noticing this miscommunication earlier. I take full responsibility.
... But your hypothetical thermal superconductor could not store heat like a black body and remain a superconductor. That's a contradiction. So it's a different creature, from your imagination. This is why I say: leave it out. There is no way you can try to demonstrate anything else with it, either, without leading to a contradiction. And it's not part of the original experiment anyway; it's nothing but misdirection. [Jane Q. Public, 2014-08-31]
We'll have to agree to disagree about thermal superconductors. I'm sorry for trying to simplify the problem in a way that ultimately just caused us to waste so much time. Again, I take full responsibility.
But again, I've already solved this problem with an aluminum enclosing shell, and it also warms the heated plate (aka Jane's "source") to ~233.8F.
... I'm not interested. Original experiment. Latour's treatment of it. Show where he was wrong. Period. Stop prevaricating. [Jane Q. Public, 2014-08-31]
That was Dr. Spencer's original challenge. He included the possibility of a fully-enclosing passive plate. And so did Dr. Latour's treatment of it. If you don't agree, please show where Dr. Latour specifies the dimensions of the plates before wrongly concluding that T remains 150. Also, why did Dr. Latour explicitly allow for K = 1 and k = 1, which describes a fully-enclosing blackbody passive plate?
Dr. Latour really did wrongly claim that a fully-enclosing passive plate wouldn't warm the heated plate (aka Jane's "source"). I've shown that his claim violates conservation of energy. As long as the shell is warmer than the chamber walls (which it is), the net radiative heat loss from the heated plate (aka Jane's "source") is reduced. So power in > power out, which means the heated plate either warms or energy isn't conserved. Just like how a bathtub fills up.
"Stop prevaricating"? Really? I've showed that Dr. Latour was wrong because his claim violates conservation of energy. Again, in physics that's a really big mistake.
Don't you see that you threw in this whole "thermal superconductor" schtick without considering what properties a thermal superconductor must actually have? In order to superconduct, it must be the same temperature everywhere, always. The only way this would be even remotely possible were if it were a perfect radiator... [Jane Q. Public, 2014-08-30]
Superconductors are distinguished from aluminum by internal properties, not radiative surface properties. That's because conduction happens inside materials, whereas radiation is emitted and absorbed on surfaces.
... The only way this would be even remotely possible were if it were a perfect radiator, with emissivity of 1. It would also be a perfect absorber, absorptivity of 1. Regardless of wavelength. So while this might not technically be true, for all practical purposes it is: a thermal superconductor would be completely transparent to all radiation... [Jane Q. Public, 2014-08-30]
No. As I've explained, emissivity = 1 and absorptivity = 1 is the definition of a blackbody. A completely transparent material would have transmittance = 1 and absorptivity = 0. Blackbodies can't be transparent.
... a thermal superconductor... has no "thermal mass". So it would have absolutely no effect on anything in this experiment. For practical purposes, it would not exist. Your idea that you can get around this by placing some kind of thin lining on its interior doesn't work. It's still as though it weren't there at all... all you have left for practical purposes is the thin shell, nothing else.... [Jane Q. Public, 2014-08-30]
I've already solved this problem with an aluminum enclosing shell rather than a thermal superconductor shell. Both shells warm the heated plate to ~233.8F.
... That's why I say: no more prevarication. No more beating about the bush. Take Spencer's original challenge, apply Latour's thermodynamic treatment of it, and show where it is wrong. Anything else constitutes failure to back up your claim that Latour is wrong and -- as you have said more than once -- some kind of nutcase. You've had more than 2 years. That is plenty. [Jane Q. Public, 2014-08-30]
Dr. Spencer's original challenge included the possibility of a fully-enclosing passive plate. And so did Dr. Latour. Note that Dr. Latour never specifies the dimensions of the plates (as Jane began to) before wrongly concluding that T remains 150. This means his incorrect conclusion must apply to all geometries, including a fully-enclosing passive plate. In fact, notice that Dr. Latour explicitly allows for K = 1 and k = 1, which describes a fully-enclosing blackbody passive plate.
So Dr. Latour wrongly claimed that a fully-enclosing passive plate wouldn't warm the heated plate. I've shown that his claim violates conservation of energy. As long as the shell is warmer than the chamber walls (which it is), the net radiative heat loss from the heated plate is reduced. So power in > power out, which means the heated plate either warms or energy isn't conserved. Just like how a bath
No, I'm not wrong. You calculated the outside temperature from the inside temperature, saying it's LOWER because of its greater area. This much is correct. THEN you try to say that with a thermal superconductor, the inner temperature would be the same as outside. Except you just calculated that outside temperature from a WARMER interior. You quite literally can't have it both ways. EITHER you're claiming a superconductor has a different temperature on both sides, or you're claiming that the inside has 2 different temperatures simultaneously. [Jane Q. Public, 2014-08-30]
Remember that the inner surface of the enclosing shell is different than the surface of the heated plate. The inner and outer surfaces of the enclosing shell are at exactly the same temperature because it's a thermal superconductor. That's what I've always been saying, despite your attempts to pretend otherwise.
The surface of the heated plate at equilibrium, however, is warmer than the inner surface of the enclosing shell. It has to be.
Here is an excellent example of this (19.3.2), which illustrates why it is a straw-man argument that is not relevant to the problem at hand. In this case the walls are warmer, not cooler, and the radiation shield is blocking the thermocouple from the radiation inward from the chamber walls, so that it can get an accurate temperature reading of the air without interference from the walls. In your case, it is the opposite: the walls are cooler than the thermocouple. But in neither case is the situation a representation of equilibrium (for example in this case, air is convecting away some of the heat of the thermocouple). The shield is absorbing and emitting radiation, too, it's just that it is isolated from the chamber walls, and so is closer to the ambient temperature of the medium being measured. This is in no way related to our experiment at all. It is in a vacuum. There is no "medium" to measure, with an ambient temperature. Not even remotely. [Jane Q. Public, 2014-08-30]
I've repeatedly linked to that excellent example. Despite your incoherent protests, it's a relevant example where a passive plate reduces radiative heat loss from a warmer source, warming it to a higher equilibrium temperature. It's a real world example which shows Jane and the Sky Dragon Slayers are wrong.
See? Same shit different day. You won't sit down and do the calculations start-to-finish, instead you do one small part, then start indulging in your hallmark game of out-of-context he-said, she-said, toss in a straw-man, then claim it's all proved.... It's simply another illustration of the depths of hand-waving you will go to, rather than actually doing all the calculations on the actual experiment from start to finish. All you're doing is tossing in more straw-men and irrelevancies. You won't do the actual experiment. The only reasonable conclusion to be drawn here is that you won't do it because you know you're wrong. [Jane Q. Public, 2014-08-30]
Don't you see the irony here? I've repeatedly done the calculations "start-to-finish" by deriving and solving equations describing the final equilibrium temperature of the enclosed plate using increasingly realistic scenarios. I've repeatedlytold you that you'd only be able to understand this thought
Slashdot Mirror
Let's finally move on. Could we finally take the very first step in this calculation? Please?
#Calculate constant electrical power/area heating 1st plate.
var('sigma T_c T_h electricity epsilon_h epsilon_c')
eq1 = electricity == sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1)
soln1 = solve(eq1.subs(T_c=255.372,T_h=338.706,sigma=5.670373E-8,epsilon_h=0.11,epsilon_c=0.11),electricity)
soln1[0].rhs().n()
ANSWER: 29.3986743761843
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches Jane's "steady-state".
I already said yes.
Could we finally take the very first step in this calculation? Please?
#Calculate constant electrical power/area heating 1st plate.
var('sigma T_c T_h electricity epsilon_h epsilon_c')
eq1 = electricity == sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1)
soln1 = solve(eq1.subs(T_c=255.372,T_h=338.706,sigma=5.670373E-8,epsilon_h=0.11,epsilon_c=0.11),electricity)
soln1[0].rhs().n()
ANSWER: 29.3986743761843
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches Jane's "steady-state".
Then it should be easy to show how badly this approximation screws up the calculation, right? So why don't we check to see wrong these approximations are, by actually doing some calculations? Finally? Please?
#Calculate constant electrical power/area heating 1st plate.
var('sigma T_c T_h electricity epsilon_h epsilon_c')
eq1 = electricity == sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1)
soln1 = solve(eq1.subs(T_c=255.372,T_h=338.706,sigma=5.670373E-8,epsilon_h=0.11,epsilon_c=0.11),electricity)
soln1[0].rhs().n()
ANSWER: 29.3986743761843
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches Jane's "steady-state".
Cute. I've repeatedly explained ad nauseum that neglecting area ratios is an approximation. I've already shown how tiny the effects are for Earth's area ratio. Does this mean you don't intend to perform even the simplest calculation to confirm this? Why don't we check to see wrong these approximations are, by actually doing some calculations? Finally? Please?
#Calculate constant electrical power/area heating 1st plate.
var('sigma T_c T_h electricity epsilon_h epsilon_c')
eq1 = electricity == sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1)
soln1 = solve(eq1.subs(T_c=255.372,T_h=338.706,sigma=5.670373E-8,epsilon_h=0.11,epsilon_c=0.11),electricity)
soln1[0].rhs().n()
ANSWER: 29.3986743761843
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches Jane's "steady-state".
Once again, no. I've repeatedly explained that the outer surface of the enclosing passive plate is never at the same temperature as the heat source.
Again, Jane might not agree with the fact that the heat source warms after it's enclosed. But again, Jane could at least acknowledge that this is what I'm saying rather than trying to pretend that I somehow said all temperatures are the same. Please?
I've repeatedly explained ad nauseum that neglecting area ratios is an approximation. I've already shown how tiny the effects are for Earth's area ratio. For weeks you've refused to perform even the simplest calculation to confirm this. Why don't we check to see wrong these approximations are, by actually doing some calculations? Finally? Please?
I've been explaining for over a month that the heated plate (aka Jane's "source") warms after it's enclosed. I've only been wasting my final days because Jane's repeatedly disagreed by supporting Dr. Latour's ridiculous Sky Dragon Slayer claim that the heated plate (aka Jane's "source") simply remains at 150F after it's enclosed:
If Jane agrees that the heated plate (aka Jane's "source") warms after it's enclosed, then that's great news! In that case, we can all agree that the mainstream physics describing the greenhouse effect is accurate, obeys the laws of thermodynamics, and proves that the Sky Dragon Slayers are wrong.
No, I calculated this number for a system which doesn't change with time. From now on I'll call this condition "steady-state" but that doesn't change the fact that my equations are based on conservation of energy in a system that doesn't change with time. Again, I only mentioned Kirchhoff's law to explain MIT's gray body approximation. Since emissivity isn't a function of wavelength, all surfaces aren't required to be at the same temperature.
Notice that the first example MIT applies their final equation to is a thermos bottle which doesn't have infinite walls. That's because a thermos bottle has no edges (just like our fully enclosed plate!) so the infinite plate approximation applies. If not, why did MIT use their equation to model a thermos? Were they talking about a thermos with infinite walls?
No, Jane. If power in != power out in steady-state, that would violate conservation of energy. Because my equations are based on the principle that in steady-state power in = power, their solutions satisfy conservation of energy.
No. The radiative power output is exactly the same as before the heat source was enclosed. It's hotter because radiative power output is proportional to T_h^4 - T_c^4. Before the heat source was enclosed, it was radiating to the chamber walls at T_c = 0F. After it's enclosed, it's radiating to the inside surface of the enclosing plate which is at T_c > 0F.
But as you said, it's pretty damned hard to prove anything without calculating it all the way through. So let's finally take the very first s
As I said: "Energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At equilibrium, that rate is zero because the system doesn't change. So at equilibrium, power in = power out."
I explicitly said a system in "equilibrium" doesn't change, which Jane calls "steady-state". I repeatedly asked Jane if we could agree on that, but a month later Jane objected:
Note that my definition of equilibrium is identical to this one: "Class 6- Equilibrium Temperature: Equilibrium means no change with time. ... In equilibrium, we expect ENERGY IN = ENERGY OUT ..."
Also note that this definition of equilibrium doesn't require a planet's south pole to be at the same temperature as its equator, or its surface to be at the same temperature as its tropopause (for planets with an atmosphere).
But from now on I'll call the system in "steady state" when its temperatures don't change with time, in the naive hope that we might actually be able to finally take the very first step in this calculation.
Once again, no. I never said that all surfaces were at the same temperature. I've already explained that the final outer temperature of the enclosing shell doesn't happen at the same time as the initial temperature of the heated plate.
I've been explaining for over a month that the heated plate warms after it's enclosed. I realize you don't agree, which is why I'm trying in vain to get you to finally perform a single, solitary calculation of your own. But even if you don't agree with my statement that the heated plate warms after it's enclosed, can't you at least acknowledge that this is what I'm saying rather than trying to pretend that I somehow said all temperatures are the same?
Once again, I never said that all surfaces were at the same temperature.
Once again, no. I never said that all surfaces were at the same temperature. I've already explained that the final outer temperature of the enclosing shell doesn't happen at the same time as the initial temperature of the heated plate. Initially, the heated plate is at 150F and the enclosing shell is cooler than 100F. But because power in > power out, the plates slowly warm to a new steady-state. By the time the outer temperature of the enclosing shell is ~149.6F (accounting for area differences), the heated plate is ~233.8F. This doesn't change even if we neglect area differences: the enclosing shell and the heated plate are never at the same temperature. Again, that's why I called them T_c and T_h.
So once again, I never said that all surfaces were at the same temperature.
I'm very sorry. I take full responsibility. Can we please move on?
Those ESA absorptivities are for absorption of sunlight. Consider the first diagram here which shows that 6000K sunlight has much shorter wavelengths than the radiation from objects at the temperatures we're considering. In fact they hardly overlap. But the emissivities are for radiation emitted by much cooler objects. That's one reason why those ESA emissivities aren't equal to their absorptivities.
Here's a good explanation of this problem: "... white paint is quoted as having an absorptivity of 0.16, while having an emissivity of 0.93.[9] This is because the absorptivity is averaged with weighting for the solar spectrum, while the emissivity is weighted for the emission of the paint itself at normal ambient temperatures. ..."
Since the absorption values you indirectly cited are for absorption from the 6000K radiation from the Sun, that seems like a legitimate reason not to use those values in a thought experiment where nothing is at 6000K. Again, another reason is that we'd have to recreate MODTRAN to derive heat transfer between non-gray bodies where emissivity and absorptivity are arbitrary functions of wavelength.
And once we debugged that new MODTRAN clone, we'd have to test it in a simple case, like the case of gray bodies where emissivity and absorptivity don't depend on wavelength. So we might as well solve the simple problem first.
Once again, I never said that. In reality, I said that both sides of a thermal superconductor are at the same temperature. This was the source of much of the misunderstanding here, and you strongly objected to the notion of a thermal superconductor. Again, that's why I calculated the small temperature difference across an aluminum shell with finite conductivity.
That's also
You were right when you said the best we can realistically do is graybodies where emissivity = absorptivity. Otherwise we'd need to derive a new equation where heat transfer is an integral over wavelengths. In other words, we'd have to recreate MODTRAN. I simply don't have time for that.
I've never pointed that out. I've repeatedly shown you Goodman 1957 where Table 1 lists aluminum's emissivity as 0.113 from 100C to 300C.
In contrast, you're citing ESA figures from page 32 which are at 0K (-273C). But nothing in this experiment is anywhere near that cold.
Also note that Goodman 1957 specifically tests the gray body approximation and concludes that "Pure aluminum appears to act like a gray body when its radiating surfaces are at temperatures lower than 400C."
Again, if you'd like to use a different emissivity just let me know, and we can both independently calculate the required electricity to check each other's answers.
We might be talking past each other. What you're calling steady-state is what I'm calling equilibrium. Radiative thermodynamic equilibrium doesn't require all surfaces to be at the same temperature, it simply means that temperatures don't change with time. At radiative equilibrium, power in = power out, which also means irradiance in = irradiance out.
Again, the materials are oxidized aluminum with emissivity = 0.11 for these temperatures. As you said, the best we can realistically do is graybodies where emissivity = absorptivity. If you'd like to use a different emissivity just let me know, and we can both independently calculate the required electricity to check each other's answers.
#Calculate constant electrical power/area heating 1st plate.
var('sigma T_c T_h electricity epsilon_h epsilon_c')
eq1 = electricity == sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1)
soln1 = solve(eq1.subs(T_c=255.372,T_h=338.706,sigma=5.670373E-8,epsilon_h=0.11,epsilon_c=0.11),electricity)
soln1[0].rhs().n()
ANSWER: 29.3986743761843
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches equilibrium. I promise to provide public-readable versions of my Sage worksheet from now on.
HTML characters "& gt;" for "greater than" and "& lt;" for "less than" (without the spaces).
#Calculate constant electrical power/area heating 1st plate.
var('sigma T_c T_h electricity epsilon_h epsilon_c')
eq1 = electricity == sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1)
soln1 = solve(eq1.subs(T_c=255.372,T_h=338.706,sigma=5.670373E-8,epsilon_h=0.11,epsilon_c=0.11),electricity)
soln1[0].rhs().n()
ANSWER: 29.3986743761843
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches equilibrium. I promise to provide public-readable versions of my Sage worksheet from now on.
Obviously we'll have to agree to disagree. But I thought you wanted to do some actual calculations? As you say, it's pretty damned hard to prove anything without calculating it all the way through. So why don't we take the first step?
Here's my Eq. 2 using your variable names:
net heat flow = sigma*(T(s)^4 - T(w)^4)/(1/E(s) + 1/E(w) - 1) (Eq. 2J)
Note that it reduces to my simpler blackbody Eq. 1 if E(s) = E(w) = 1.
If you'd like me to clarify what my variable names for a particular equation would be in your terminology, just ask.
I've specified the dimensions. The heated plate is a sphere with radius 6371 mm and surface area A_h. The enclosing plate is a 1 mm thick concentric shell with an inner radius of 6378 mm, surface area A_c1 on the inside, and A_c2 on the outside. The chamber is also a concentric sphere with inner radius 6386 mm, so there's a 7 mm gap on both sides of the enclosing shell. Again, the plates and walls are oxidized aluminum.
At equilibrium, net heat flow out (in W/m^2) equals "electricity". The first step is to calculate that constant variable "electricity" which describes electrical power per square meter heating the sphere to 150F without an enclosing shell. I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches equilibrium.
I've already showed you that the outer surface of an enclosing shell with an area ratio similar to Earth's warms to ~149.6F. I've explained that neglecting area ratios is a tricycle: a simple approximation that helps us learn. It's like the "frictionless pulley" or "massless rope" or "blackbody" approximations. Again, in this case the tricycle isn't too inaccurate compared to the bicycle, it's much easier to learn, and it provides a sanity check on the more complicated calculation. As the area ratio approaches "1.0" the bicycle should give the same answer as the simpler tricycle. And it does.
When the area ratio departs far from 1.0, the tricycle becomes very inaccurate, so one should use the more complicated bicycle. But again, the Earth's area ratio is roughly 1.0025, so in that case the tricycle isn't too inaccurate.
Once again, I've already accounted for the area ratio to obtain the more complicated and more accurate solution.
No, the PSI Sky Dragon Slayers told you it's the engineering textbook answer. I showed you MIT's final expression which reduces to my Eq. 1 for blackbodies, and is consistent with these equations and Eq. 1 in Goodman 1957. Physicists and engineers have been using thermodynamics for decades in the real world that contradicts Dr. Latour's Slayer nonsense.
That's why Jane, Dr. Latour and the rest of the Slayers disagree with the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society.
I never said the problem is intractable. Just that it's more complicated than the spherically symmetric problem. Again, do you dispute that equilibrium temperatures for a non-enclosing plate would vary across the plate surfaces rather than being simple numbers like with a spherically symmetric fully enclosing plate?
Maybe I should explain that. Consider Dr. Spencer's first illustration. Presumably the heated plate at "150F" has finite conductivity, so its lack of spherical symmetry means that its corners will be cooler than the plate's side's midpoints. That's because the corners are closer to the cold chamber walls than those midpoints.
An integral over the heated plate's surface might average to "150F" but (unlike a spherically symmetric plate) it can't have that temperature everywhere as long as it has finite conductivity. But at least the single heated plate has bilateral symmetry; the left and right hand side midpoints have the same temperature.
Adding a cool plate removes even that bilateral symmetry. The left hand side's midpoint warms the least because it's still radiating to the 0F chamber walls. The right hand side's midpoint warms the most because it's now radiating to the (initially) 100F cold plate.
Since enclosing a spherically symmetric plate warms it from 150F to ~233.8F for area ratios similar to Earth's, the right hand side's midpoint won't warm past ~233.8F. But it has to warm to conserve energy because at equilibrium power in = power out.
I can't be more specific without programming a finite element model. But Dr. Latour never even allowed for the heated plate's temperature to be different on each side. As long as we're only considering materials with finite conductivity, this would only be poss
Maybe I should explain what I meant by saying that equilibrium temperatures for a non-enclosing plate would vary across the plate surfaces. Consider Dr. Spencer's first illustration. Presumably the heated plate at "150F" has finite conductivity, so its lack of spherical symmetry means that its corners will be cooler than the plate's side's midpoints. That's because the corners are closer to the cold chamber walls than those midpoints.
An integral over the heated plate's surface might average to "150F" but (unlike a spherically symmetric plate) it can't have that temperature everywhere as long as it has finite conductivity. But at least the single heated plate has bilateral symmetry; the left and right hand side midpoints have the same temperature.
Adding a cool plate removes even that bilateral symmetry. The left hand side's midpoint warms the least because it's still radiating to the 0F chamber walls. The right hand side's midpoint warms the most because it's now radiating to the (initially) 100F cold plate.
Since enclosing a spherically symmetric plate warms it from 150F to ~233.8F for area ratios similar to Earth's, the right hand side's midpoint won't warm past ~233.8F. But it has to warm to conserve energy because at equilibrium power in = power out.
I can't be more specific without programming a finite element model. But Dr. Latour never even allowed for the heated plate's temperature to be different on each side. As long as we're only considering materials with finite conductivity, this would only be possible for a spherically symmetric enclosing plate.
Dr. Latour's answer wasn't "reasonably precise". He claimed that the heated plate wouldn't warm at all when the cold plate was added, even if it completely enclosed the heated plate such that K = 1. This is a specific prediction of "0.0000...F" warming. Since energy conservation means that adding a cold plate has to warm the heated plate, he's only off by a factor of infinity.
Latour's answer is ridiculous Sky Dragon Slayer nonsense which violates conservation of energy, as I've shown.
Once again, solving a problem without spherical symmetry means you'll have to solve for equilibrium temperatures which aren't constant across the heated and passive plates. Those equilibrium temperatures wouldn't be simple numbers. They'd be complicated functions that would vary across the plate surfaces. Contrast that with a spherically symmetric enclosing plate, where equilibrium temperatures are just simple numbers.
Are you disputing that equilibrium temperatures for a non-enclosing plate would vary across the plate surfaces rather than being simple numbers like with a spherically symmetric fully enclosing plate?
Because, unless you dispute the above facts, that would require a complicated finite element model due to its lack of spherical symmetry. I simply don't have that much time left. And again, we'd have to test that complicated model in a case where an analytic solution is available anyway...
Okay, then we disagree. It's always helpful to solve simpler problems before moving on to more complex problems. The simpler problem is easier to learn, and often serves as a sanity check on the more complex problem.
Again, Latour's calculations allowed for K = 1: "K is the fraction of radiation from the first bar absorbed by the second colder bar, 0 < K <=1."
The only way K = 1 is if the cold plate completely encloses the first heated source. Otherwise, radiation from the side of the source opposite the cold plate couldn't possibly be absorbed by the cold plate, which would force K < 1. So once again, the fact that Dr. Latour included the possibility that K = 1 means that his claim applies to all geometries.
If not, why doesn't he deal with edge effects? The only ways to eliminate edge effects are if the plates are infinite, or if the cold plate completely encloses the heated source.
Again, I don't have enough time to program a finite element model to account for the fact that a non-fully-enclosing plate would cause plate temperatures to vary across their surfaces. But even if I did, the first thing I'd do after debugging it would be to check the finite element solution in a case where a simple analytic solution can be obtained. Namely, a fully-enclosing passive plate, where the plate temperatures are simple numbers.
By the way, since you keep insisting that only a particular geometry could refute Dr. Latour's treatment, could you please show where he specified the dimensions of the plates? Or where Dr. Spencer did? Otherwise, even if I had enough time to do so, how could I possibly program this complicated finite element model with the specific geometry that would finally convince you the Slayers are wrong?
Again, we'll have to agree to disagree about thermal superconductors. That's why I've repeatedly pointed out that I've already solved this problem with an aluminum enclosing shell, and it also warms the heated plate (aka Jane's "source") to ~233.8F.
No, it's exactly the same problem. The same infinite sum of absorption and reflection. The plates are only "infinite" to avoid having to model fringing field effects around the plate edges. And note that Dr. Latour doesn't model edge effects either, so his plates are either infinite or the passive plate completely encloses the "source". Either way, there would be no edges.
Notice that the first example MIT applies their final equation to is a thermos bottle where the inside wall is heated by hot fluid.
Again, Dr. Spencer's actual, original experiment included the possibility of a fully-enclosing passive plate. And so did Dr. Latour's treatment of it. If you don't agree, please show where Dr. Latour specifies the dimensions of the plates before wrongly concluding that T remains 150.
In fact, as far as I can tell nobody's specified the plate dimensions except for me. Since the argument I'm refuting never specified the plate dimensions, why would the plate dimensions matter?
We might be talking past each other. What you're calling the "source" is what I've been calling the "heated plate" with temperature "T_h" in all my equations. I've called the other enclosing plate the "cold plate" with temperature "T_c". As I've repeatedly and consistently stressed, "T_c" is only identical on both sides of the enclosing cold plate if it's a thermal superconductor.
I'm sorry for any confusion this caused, but as you can tell I really am talking about the experiment Dr. Spencer mentioned. We're just using different words, and again I'm sorry for not noticing this miscommunication earlier. I take full responsibility.
We'll have to agree to disagree about thermal superconductors. I'm sorry for trying to simplify the problem in a way that ultimately just caused us to waste so much time. Again, I take full responsibility.
But again, I've already solved this problem with an aluminum enclosing shell, and it also warms the heated plate (aka Jane's "source") to ~233.8F.
That was Dr. Spencer's original challenge. He included the possibility of a fully-enclosing passive plate. And so did Dr. Latour's treatment of it. If you don't agree, please show where Dr. Latour specifies the dimensions of the plates before wrongly concluding that T remains 150. Also, why did Dr. Latour explicitly allow for K = 1 and k = 1, which describes a fully-enclosing blackbody passive plate?
Dr. Latour really did wrongly claim that a fully-enclosing passive plate wouldn't warm the heated plate (aka Jane's "source"). I've shown that his claim violates conservation of energy. As long as the shell is warmer than the chamber walls (which it is), the net radiative heat loss from the heated plate (aka Jane's "source") is reduced. So power in > power out, which means the heated plate either warms or energy isn't conserved. Just like how a bathtub fills up.
"Stop prevaricating"? Really? I've showed that Dr. Latour was wrong because his claim violates conservation of energy. Again, in physics that's a really big mistake.
Since you just linked to this
Superconductors are distinguished from aluminum by internal properties, not radiative surface properties. That's because conduction happens inside materials, whereas radiation is emitted and absorbed on surfaces.
No. As I've explained, emissivity = 1 and absorptivity = 1 is the definition of a blackbody. A completely transparent material would have transmittance = 1 and absorptivity = 0. Blackbodies can't be transparent.
I've already solved this problem with an aluminum enclosing shell rather than a thermal superconductor shell. Both shells warm the heated plate to ~233.8F.
Dr. Spencer's original challenge included the possibility of a fully-enclosing passive plate. And so did Dr. Latour. Note that Dr. Latour never specifies the dimensions of the plates (as Jane began to) before wrongly concluding that T remains 150. This means his incorrect conclusion must apply to all geometries, including a fully-enclosing passive plate. In fact, notice that Dr. Latour explicitly allows for K = 1 and k = 1, which describes a fully-enclosing blackbody passive plate.
So Dr. Latour wrongly claimed that a fully-enclosing passive plate wouldn't warm the heated plate. I've shown that his claim violates conservation of energy. As long as the shell is warmer than the chamber walls (which it is), the net radiative heat loss from the heated plate is reduced. So power in > power out, which means the heated plate either warms or energy isn't conserved. Just like how a bath
Remember that the inner surface of the enclosing shell is different than the surface of the heated plate. The inner and outer surfaces of the enclosing shell are at exactly the same temperature because it's a thermal superconductor. That's what I've always been saying, despite your attempts to pretend otherwise.
The surface of the heated plate at equilibrium, however, is warmer than the inner surface of the enclosing shell. It has to be.
I've repeatedly linked to that excellent example. Despite your incoherent protests, it's a relevant example where a passive plate reduces radiative heat loss from a warmer source, warming it to a higher equilibrium temperature. It's a real world example which shows Jane and the Sky Dragon Slayers are wrong.
Don't you see the irony here? I've repeatedly done the calculations "start-to-finish" by deriving and solving equations describing the final equilibrium temperature of the enclosed plate using increasingly realistic scenarios. I've repeatedly told you that you'd only be able to understand this thought