From: David Jonsson
What I mean is that regardless of how efficient the
thermophotovoltaic is there is no other way for heat-energy to escape the
enclosure except as IR-light converted to electricity. That is naïve. IR light will escape - whether some small fraction of it can be converted to electricity, or not. You cannot keep it from leaving - nor can you force more of it to be converted than a matching band gap will permit. You cannot even reflect very much of it. You seem to be pulling the donkey with the cart. As a practical matter, even in a perfect vacuum - if you are removing electrical current, you must have conductive wires to do that. But the main problem is that IR will radiate from any surface - whether or not the means exists to convert part of it to electricity – which in your example depends on a succession of overlapping band gaps. You simply cannot “force” heat to be converted when there is an easier path - and since conversion is anti-entropic - the easy path is blackbody radiation without conversion. Whatever heat flux is not converted usually 95% of it - will radiate. Most of it will be missed - since IR conversion is inefficient. Having said that – cough, cough … you can follow up on the mysterious Qu-tube. It is one of those “holy grails” of alternative energy that has been around for years. http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20080009660_2008009120. pdf … said to be a superconductor of heat. But NASA did not confirm the claim. If Dr. Qu’s claim were to be true (and let me add that there are people who I respect who will tell you that they have seen it) - then it would potentially do most of what you suggest - to the extent that a superconductor of heat becomes a superconductor of electricity. But even so, this kind of tube would not qualify as thermophotovoltaic, at least not as defined in the article. It would simply be a superconductor of heat – where the “easy path” is a vector that is also anti-entopic. Jones
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