From: Steve High
At 1:31 of part 1 of his July 21 demonstration Mills tells us that in order for the Sun Cell to produce 10 megawatts of electricity it will need to create 25 megawatts of light energy, as the PV cells have a 40% conversion ratio. I presume that will leave 15 megawatts to be dissipated as heat (it has to go somewhere, right?) With that much heat wouldn't the reactor itself nearly glow with the intensity of the sun? I'm just a simple country doctor so probably I'm missing something. Steve – You probably do not take this too seriously, so you are not missing the fact that it is not science – it is hucksterism with a dose of sophistication. The “40%” conversion ratio should make it clear to anyone who follows solar cells that Mills is blowing smoke. Affordable cells for use in mass production are below 20% efficiency. The Boeing/Spectrolab triple junction cells which have produced 40% on occasion are not available at reasonable cost. And even if they become available - they are actually rated for 32% continuous and are now extraordinarily expensive – 500 times more than silicon per watt. NASA can pay that but can you? So instead of dissipating 15 megawatts to get 10, less the reprocessing overhead (assuming that it works at all for more than a few days) it would be closer to dissipating 40 MW to get 5… which isn’t bad if it were true. But given Mills’ track record, do not sign a check just yet. If history is an indicator - you will never see it being sold. Mills has raked in a few million more from a few more suckers, in this round of investment, and in 2015 there will be something else. SunCell what? Maybe in a museum some day.
