... I used the proper equation for radiative power, which at steady-state doesn't depend on other bodies. So there is no "difference" term. Just temperature. That's simple physics. You are trying to use a heat transfer equation to calculate power out of a single body at known temperature. That's just plain WRONG. ... [Jane Q. Public, 2014-09-24]
No, Jane tried to use an equation that only calculates radiative "power out" when Jane needs to use an equation for heat transfer that calculates radiative "power out minus power in".
If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out".
Just no. If radiation goes in and comes right back out, we do not need to account for it, because then the NET amount of that particular radiation crossing your boundary is ZERO. A = A. You do know how to add and subtract, right? You know what a zero is, right? [Jane Q. Public, 2014-09-24]
Jane's accounting for "power out" without including a term for "power in". That's not A = A, it's A = 0 because one of the terms has been ignored. It's led Jane to the absurd conclusion that electrical heating power doesn't depend on the cooler chamber wall temperature. If that's the case, then how did we detect the 2.7K cosmic microwave background radiation with warmer detectors? How do uncooled IR detectors see cooler objects? Again, why is Venus hotter than Mercury?
... All the radiation going IN from the cooler body just goes right back OUT again, making the NET radiation crossing your boundary from the cooler body zero. If that were not so, then you'd have net energy being transferred from a cooler body to a hotter one, which is a violation of the second law of thermodynamics. As I've explained to you many times now. You're just plain wrong. ... [Jane Q. Public, 2014-09-24]
This is complete gibberish, Jane. Power radiated in from the chamber walls needs to be accounted for using one term. Power radiated out from the source needs to be accounted using another. Once again, accounting for power flowing in doesn't violate the second law of thermodynamics or somehow imply net energy transfer from cool to hot, no matter how many times Jane wants to assert that nonsense. However, failing to account for power flowing in does violate conservation of energy, because power in = power out through any boundary where nothing inside is changing.
So Jane refuses to retract his absurd claim that view factors vary as the radius ratio, which violates conservation of energy. A cynic might have expected as much, given how Jane flagrantly violates conservation of energy by incoherently ignoring radiative power passing in through a boundary around the heat source.
I made no such claim, you liar. As you well know, the view factor from the surface of the inner sphere to the inner surface of the outer sphere is 1. The calculated view factor from the outer sphere to the inner was 0.9998... [Jane Q. Public, 2014-09-24]
Jane made no such claim? Jane keeps making that absurd claim! Again, 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.
Jane's campaign of educating ignorant, stupid physicists about physics has only just begun. Jane still needs to educate Prof. Brown and Lonny Eachus still needs to educate Dr. Joel Shore.
No, I don't need to educate either one. They can both pick up a textbook on heat transfer and see that I am correct. I'm not arguing with them. [Jane Q. Public, 2014-09-24]
Of course Jane argued with Prof. Brown and wasn't able to "educate" him. Of course Lonny Eachus argued with Dr. Joel Shore and wasn't able to "educate" him. Why not, Jane? Do those physicists not have heat transfer textbooks, or are they just ignorant and stupid?
... I did NOT make broad claims in this recent exchange about "greenhouse gas" or any such thing. So I'm not arguing with those other people. I simply showed YOU to be wrong. ... [Jane Q. Public, 2014-09-24]
But Jane does make broad claims:
.. the CO2-warming model rely on the concept of "back radiation", which physicists (not climate scientists) have proved to be impossible. I'm happy to leave actual climate science to climate scientists. But when THEIR models rely on a fundamental misunderstanding of physics, I'll take the physicists' word for it, thank you very much. .. [Jane Q. Public, 2012-07-05]
Jane/Lonny Eachus insists that an enclosed source doesn't warm, which means CO2 emissions couldn't cause warming. That's why Jane/Lonny Eachus needs to educate the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society.
Jane/Lonny Eachus wins a silver medal in psychological projection for telling me to "be a man for a change" but Slayer CEO John O'Sullivan still takes the gold.