But Jane's equation is different:

electrical power per square meter = (s)*(e)*Ta^4

**YES!!! This is a different equation!** It's not an equation for *heat transfer*! It's the Stefan-Botlzmann RELATION between radiative power out and temperature for gray bodies. **It is used for calculating RADIATIVE POWER OUT versus TEMPERATURE and vice versa. ***It is not for heat transfer and I'm not using it for heat transfer*. YOU are the one who is getting them confused, not me. This *other* equation shows that radiative power is dependent ONLY on emissivity and temperature. It does not depend on other bodies. For the third time (today): it's a temperature vs. power equation, not a heat transfer equation. Further, "electrical" is your own addition. The equation is for power. It doesn't specify "electrical". * *** [Jane Q. Public, 2014-09-22] **

My equation for electrical power is different than the equation for radiative power out, which is why it's bizarre that Jane keeps using the equation for radiative power out to determine electrical power. That's what I've been trying to tell Jane: we don't disagree about the equation for radiative power out. The equation for radiative power out is simply a **part** of the equation for conservation of energy: power in = power out through a boundary where nothing inside is changing. That's why we need to use a heat transfer equation to determine electrical heating power, not just an equation for radiative power out.

... it is not necessary to account for cooler bodies in the **temperature versus power out** equation. ... The second equation you cited above is the STANDARD equation for calculating radiative power out of a gray body. I showed you where it was in Wikipedia. It also just happens to be in my heat transfer textbooks. The answer is 82.12 W/m^2. It is the textbook answer. It isn't going to change. Why don't you look it up in a textbook and discover that for yourself? ... Radiative power out of the warmer body is dependent ONLY on emissivity and thermodynamic temperature. Anything else violates the second law of thermodynamics. It isn't controlled or mitigated by nearby cooler bodies. ... * *** [Jane Q. Public, 2014-09-22] **

I've already agreed that it's not necessary to account for cooler bodies in the **temperature versus power out** equation. Again, we're not disputing the equation for radiative power out. We're disputing the equation describing conservation of energy around a boundary drawn around the heat source:

power in = electrical heating power + radiative power in from the chamber walls

power out = radiative power out from the heat source

Since power in = power out:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

This doesn't violate the equation for radiative power out. It simply uses that equation to account for the power flowing out of the boundary, and uses that same equation for radiative power to describe radiative power flowing into the boundary.

... I will repeat: I did not and do not claim that no radiation is absorbed. Just no net radiative power. Any that does get absorbed is just re-transmitted... * *** [Jane Q. Public, 2014-09-22] **

Jane's been calculating the required electrical heating power, which requires using a net heat transfer equation to describe power in = power out through a boundary around the source. Because Jane's equation doesn't even include a term for "radiative power in", Jane's equation **does** claim that no radiation is absorbed **at all**.

If Jane would reconsider conservation of energy and include a term for "radiative power in", **then** Jane could honestly say he was only claiming that no net radiative power is absorbed by the source. Until then, Jane's equation claims that no radiation is absorbed by the source **at all**.

But Jane's equation is nonsense, because absorption is controlled by absorptivity. So we could only ignore the power radiated from the chamber walls if the source's absorptivity = 0. But then its emissivity = 0, so it also couldn't emit any radiation, so it couldn't be a heat source.

Look at your S-B equation above. What does it say? No net radiative power is absorbed by warmer bodies from cooler bodies. You said so yourself. But NOW, you're claiming that it is. You contradict yourself. ... * *** [Jane Q. Public, 2014-09-22] **

No, I said the source has to absorb some radiation as long as it has absorptivity > 0. I **never** said the source would absorb more radiation than it emitted. In fact I said the opposite "happens automatically".

... The chamber walls neither transfer any of their net radiative power to the heat source, nor do they cause the net radiative power of the heat source to be any less. They have NO EFFECT. Net energy flows only FROM the heat source to the walls, and the temperature of the walls effects heat **transfer** only, not radiative power of the heat source. ... * *** [Jane Q. Public, 2014-09-22] **

If the temperature of the walls affects heat transfer, they also affect how much electrical heating power is required to keep the source at 150F. Note that I said "electrical heating power" and not "radiative power out" because these are two very different things. Calculating "radiative power out" just requires writing down the Stefan-Boltzmann law. Calculating "electrical heating power" requires drawing a boundary around the heat source at steady-state, and setting power in = power out.

... 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. ... * *** [Jane Q. Public, 2014-09-19] **

**Of course** boundary conditions are needed to find the answer, because they determine what "plain old algebra" is used, even at steady-state. We're all applying Dirichlet boundary conditions to the chamber walls, but Jane mistakenly applied them to the source as well, instead of the correct Neumann boundary conditions. Jane also continues to wrongly insist that Jane held electrical heating power constant **as well as** holding source temperature constant. So apparently in Janeland there's no difference between Neumann and Dirichlet boundary conditions. If that's true, why do physicists and engineers use different names for Neumann and Dirichlet boundary conditions?

... The areas in his equation were unnecessary ... Therefore the areas were irrelevant and about all he accomplished with his large equation was to further confuse the issue. ... * *** [Jane Q. Public, 2014-09-10] **

After I originally solved a simple equation without areas, Jane objected that neglecting areas was a "fucking logical error!". That's why I had to solve the more accurate large equation with areas, even though I warned Jane that it wouldn't substantially change the answer.

The large equation with areas was also necessary because:

... 788.01 W != 721.44 W (!!!) Power is not conserved. ... the inner surface of the cavity has twice as much area, so the **total power radiated** is twice as much. Power is not conserved. ... * *** [Jane Q. Public, 2014-09-07] **

Jane confused himself about areas so badly that he claimed "power is not conserved". So I explained that Wikipedia’s equation correctly takes into account areas and view factor.

Wikipedia's equation conserves energy because the view factor from chamber walls to enclosed source equals the **area ratio**. If the view factor didn't vary exactly like that, energy really **wouldn't** be conserved.

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. * *** [Jane Q. Public, 2014-09-19] **

As I said, the view factor from enclosed source to chamber walls is 1. If Jane wants to calculate the view factor in the other direction, the link I've repeatedly given Jane shows that for smaller radius R1, F21 = (R1/R2)^2 = 0.9978.

If the view factor varied as the **radius ratio** like Jane claims, energy really **wouldn't** be conserved. The view factor has to vary as the **area ratio**, which is the square of the radius ratio.

... I'd say that Wikipedia's equation is more correct because it includes area and view factor, which MIT's equation does not. ... * *** [Jane Q. Public, 2014-09-19] **

If only I'd mentioned that repeatedly.

... 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? ... * *** [Jane Q. Public, 2014-09-19] **

Of course I noticed that they're both net heat transfer equations, which is why I used them both in the same way to get nearly identical answers. I'm using MIT's and Wikipedia's equations because they yield radiative "power out minus power in". These net heat transfer equations **are** the proper equations for applying conservation of energy to a boundary around the source.

In contrast, Jane's clinging to an equation for "power out" and incoherently trying to justify ignoring "power in" through that boundary.

... I will make use of only ONE of your assumptions: that the enclosing plate (hollow sphere) is, due to thermal conductivity, **approximately** the same temperature on both sides. It's only 1mm thick after all, and the thermal conductivity of aluminum was a stipulation of yours so it will be the same to a couple of decimal places, give or take. So the answer won't be exact, but it will be reasonably accurate. Certainly close enough to demonstrate the concept. ... * *** [Jane Q. Public, 2014-09-10] **

When I approximated the enclosing shell as a thermal superconductor, Jane insisted that there's no way to demonstrate anything with it without leading to a contradiction, and that it was nothing but misdirection and a fantasy ultimate straw-man argument.

When Jane approximates the enclosing shell as a thermal superconductor, it's reasonably accurate and certainly close enough to demonstrate the concept.

A cynic might suspect double standards.

... I have already explained how your "boundary" **assumed** that all the power was output from the **outside** of the enclosing sphere. ... you neglected to account for the fact that the hollow sphere has TWO surfaces it is radiating from. You left out half the m^2 in A, so your figure for W/m^2 was off by very nearly 100%. Q.E.D. * *** [Jane Q. Public, 2014-09-11] **

Once again, Jane's completely wrong. When I held the source temperature constant, I reproduced Jane's result. So we're actually disagreeing about what to hold constant. If Jane's hilarious "Q.E.D." were correct, I wouldn't have been able to reproduce Jane's result simply by changing what variable I held constant.

... Add them together for the total heat transfer: 27.7832 + 27.7813 = 55.5645 total heat transfer. This checks against our initial calculation which was 55.5913. The difference is only 0.0268, or about 0.1%. Close enough for what we're doing. ... * *** [Jane Q. Public, 2014-09-10] **

Ironically, Jane's off by ~100%. Again, Jane's total heat transfer dropped to 27.8 W/m^2 after the shell was added, so Jane's meaningless 55.6 W/m^2 value is ~100% higher than the actual value.

... Factor out (e*s) from both sides. (Despite khayman80's assertion that we cannot do this, yes we can. It is the same scalar and the same constant on both sides.) ... * *** [Jane Q. Public, 2014-09-10] **

Once again, I never asserted that. In fact, I repeatedly showed Jane an equation derived by factoring out the sigmas and epsilons from both sides. Only Jane/Lonny Eachus could repeatedly quote that equation and even agree with it, then accuse me of asserting the opposite.