Instead of holding electrical heating power constant, Jane held the source's radiative power output constant. That held source temperature constant and forced electrical heating power to change.
No, that is not correct. You made assumptions that are, to be blunt, bullshit nonsense.
Since the emissivity for every object in our system is the same, power output is proportional to the T^4. Period. End of story.
Draw your boundary around the heat source. Power in = power out (your own principle). Therefore the power in is 41886.54 Watts, which is the power initially being radiated out.
SPENCER stipulated that this power is held constant. It wasn't my idea. It's a condition of the experiment.
By the Stefan-Boltzmann law, since the power in remains constant, then UNLESS power is taken up from some other source, the temperature will remain constant. This follows directly from the S-B radiation law, which you seem to be disputing.
Another requirement of the S-B law, and also of thermodynamics: since EVERY other object in the system is at a lower temperature than the heat source, NET heat transfer is in ONLY one direction: from hotter to colder.
Therefore, no energy is flowing "backward" to boost the output of the heat source.
Yet another fact that follows directly from the S-B law, is that nearby cooler bodies have zero effect on the output of the heat source. They don't "suck" power from it, nor (see above) do they "lend" power to it.
The only logical conclusion -- the only physically possible conclusion, unless you dispute the Stefan-Boltzmann radiation law, is that the heat source does not change temperature. Power out = power in, and is constant. Everything else is cooler, so it remains a constant. There is no further energy or power flowing "backward" the heat source.
The Stefan-Boltzmann law clearly shows that no NET radiation from cooler objects is absorbed; it is either transmitted, reflected, or scattered. Since these are diffuse gray bodies, they do not transmit. That leaves reflection and scattering. For our purposes, the net effect is that it is all reflected.
You are imagining some kind of power input to the heat source that doesn't exist. Further, if the heat source became even hotter, as you assert, it would require even MORE power, because as you say, power in = power out. That was YOUR assertion. Draw your boundary around the heat source itself. There is no net radiation absorbed from outside, and the supplied power remains constant.
It this whole "proof" of yours, I have shown where you have contradicted yourself at least 3 different ways.
Jane might wonder why he can't derive a single equation which works for all these cases.
I don't know where you get this idea, because I did. I used the S-B equation to find my solution. I used the textbook equations for heat transfer. Yes, I ignored area because the areas were so similar. But it was still a reasonably accurate approximation. I checked my work, and it wasn't off by more than a fraction of a percent.
But Jane can't even admit there's a difference between holding electrical heating power constant and holding the source's radiative power output constant.
Because there isn't any. Your own "boundary" principle says so. This isn't a matter of differential equations at this point. Do you think we're all idiots? Power in = power out. Your Newmann and Dirichlet boundary conditions are just more straw men. We don't need them to find the answer to this. Plain old algebra works just fine, because everything is at steady-state. So knock off the bullshit, because I see right through it, and so will the others I show this to.
Again, warming the chamber walls is like partially closing the drain on a bathtub where water is flowing in at a constant rate
Which is not only false (the S-B relation again, which says it only relies on its radiant temperature, not the temperature of cooler bodies nearby), but another straw man, because the chamber walls aren't warmed. They are held at a constant 255.37K.
Hopefully these are just more badly-worded sentences because they all require absorptivity = 0.
No, they don't. Gray body radiant power vs. temperature is expressed by S-B equation, and we already know that gray body absorptivity = emissivity. I was using the proper equation, and you were using it too (if improperly). Are you trying to tell me that the equation YOU have been using is invalid?
Yet again, you have contradicted yourself. You're a great bullshitter but I've caught you out and you've already been proved wrong. All this trying to twist out from under the obvious any way you can only confirms that you were bullshitting all along. Be a man and admit the truth, because people ARE going to see this. Why do you want to look more foolish than you do already?
But for gray bodies it's just an approximation because it ignores reflections. After obviously failing to explain that we need to account for reflections, I decided to agree to disagree. For two gray bodies interacting with small view factors (e.g. Earth's tiny view factor of the Sun) reflections can be safely neglected. But the chamber wall completely encloses the source, so its view factor is 1. That's why MIT's equation is more accurate here: it accounts for reflections.
Complete bullshit again. We were assuming diffuse gray bodies. Further:
But the chamber wall completely encloses the source, so its view factor is 1.
No. If the surfaces are numbered 1, 2, 3, 4 as I did in my solution, F12 = F34 = 1. In the other direction (as you already know, and so do I) it is R1/R2, where R1 is the smaller diameter. F21 = F43 = 0.9989.
But in this context it is already "dirt simple", as I pointed out before. These are diffuse gray bodies. (1 - emissivity) is assumed to be the "reflection", which in this context also includes scattering but no transmission. This is already accounted for in the equations, such as the heat transfer equation you borrowed from Wikipedia.
If you like, you can use the preferred method (according to Wikipedia) for calculating the respective radiant output of the surfaces: the Radiosity Method. That method explicitly accounts for reflection (1 - emissivity). And I already know that it confirms my solution. So go ahead. I simply didn't show it in my brief write-up because I intended it to be a brief write-up. I do intend to show it in the fuller version.
Since Jane's proposed equation is missing the "(s)*T4^4" term, it doesn't reduce to this simpler Eq. 1J.2 for blackbodies where (e) = 1. So it's wrong.
More nonsense. The S-B relation says that the radiative power out of a body is P = (epsilon)*(sigma)*T^4. It is not wrong. It is a simple equation that is well-known to physicists. You claim to be a physicist, so why don't you know it?
The equation you are trying to use there is a partial equation for heat transfer, not radiant power output. They're not the same things. The proper equation for power out given radiant temperature is right there in the above paragraph. It can be found in any heat transfer textbook and many physics books.
Didn't you notice that MIT's equation is essentially the SAME equation as Wikipedia's heat transfer equation, except for areas? I sure did. Why didn't you notice that?
I repeat: I checked my solution using Wikipedia's equation, including the areas AND the view factors AND the reflections. It checked out just fine, thank you very much. Why don't you try it yourself and see?
But after using Jane's equation in pointless attempts to illustrate more fundamental problems in Jane's analysis, I wanted to stress once again that MIT's equation is more appropriate for enclosing chamber walls because it accounts for reflections.
It doesn't matter. It still checks out. Although I'd say that Wikipedia's equation is more correct because it includes area and view factor, which MIT's equation does not.
Other than your mention of the equations in the latter part of your comment, it is easy to show that EVERYTHING ELSE is just plain nonsense. You are trying to dispute the Stefan-Boltzmann radiation law and its corollaries. Excuse me, but that didn't work in the beginning, and it still isn't working. You've added nothing worthwhile to the conversation since.
You've been owned, man. BE enough of a man to admit it. Because everybody's going to know it anyway.