Years ago, Russ George told me that in one of his experiments he could "see helium bubbles." At the time I said that is impossible because if you could see the bubbles the reactor would be producing more power than any laboratory experiment. However, yesterday I ran the numbers and found I may be wrong about that. I have estimated that if the cell was producing ~1 kW, perhaps he could see bubbles. This is not to say Russ was right. I do not know how much anomalous power he measured. There could be bubbles from some other source. I do not know whether he collected the gas and analyzed it. But anyway, here is my estimate. I would appreciate it if readers here would check the numbers.
Assumptions: This is D + D fusion producing Helium-4. The smallest bubbles you could see are fine ones. I suppose the total volume of gas is approximately 1 cubic millimeter per minute. Anything smaller would probably not be visible to the naked eye. The gas in the bubbles is at STP. Probably not, but anyway, close to it. D+D fusion produces 245,000 MJ per gram of deuterium. I double checked that number with some web sites: http://www.geo.cornell.edu/eas/energy/research_front_page/nuclear_fusion.html http://electron6.phys.utk.edu/phys250/modules/module%205/nuclear_energy.htm (U. Tennessee, value given in ergs per gram) http://physics.ucsd.edu/do-the-math/2012/01/nuclear-fusion/ Also from U. Tennessee: Efficiency (E/mc2) Chemical energy 3 x 10E-8%, Fission 0.002%, Fusion 0.4%. In other words, the mass of helium is almost the same as the starting mass of deuterium. I will ignore the lost mass. Okay -- The volume of any gas at STP is 22.4 L per mole. One mole of helium weighs 4 g. So it weighs 0.1786 g/L. 1 mm^3 is 1/1,000,000 of a liter. 1 mm^3 per minute is 0.000000016667 L/s. So the helium weighs 0.000000002976 g. Multiply that weight by 345,000 MJ gives 0.001026785714 MJ/s. That's 1,025 J/s (watts). That's a lot less power than I thought! Cross-check. Mel Miles says fusion produces helium at a rate of 10E11 to 10E12 atoms per second per watt. Divide Avagadro's number by 10E11 gives 1.66E-12 moles, or 6.64E-12 g helum, which multiplied by 345,000 MJ/g gives 2.3 W. Close enough! Mel Miles measured at most ~400 mW of anomalous power as I recall. That produces very little helium, as you see from these numbers. 0.400 J / 3.45E11 J = 1.16E-12 g/s (0.0012 nanograms). Right? He collected for 4,400 s, but still that's not many nanograms. Interesting extrapolation. According to the International Energy Agency (iea) "Key world energy statistics" the world primary energy supply in 2015 was 5,269 Mtoe. An Mtoe is "millions of tons oil equivalent." The conversion factor is 41,868,000,000 MJ per Mtoe. So that's 2.21E14 MJ. It would take 639 tons of deuterium to produce that with cold fusion. As I said, I am ignoring lost mass, so that would produce ~639 tons of helium. https://www.iea.org/publications/freepublications/publication/key-world-energy-statistics.html In my book I estimated that you would end up with 1,227 tons of helium, about twice as much. I do not know where the discrepancy came in. Perhaps I calculated it wrong, or perhaps I was looking at the heat required to produce the primary energy, with the assumption that all energy will come from cold fusion powered heat engines. That issue is complicated because not all of our energy today comes from heat engines. In the U.S. 15% comes from hydroelectricity, wind and solar power, and 20% comes from nuclear power. They do not usually include the raw heat from nuclear power in estimates of primary energy. The mass to energy conversion for 5,269 Mtoe is the same for any source of energy, mechanical, chemical, fission or fusion. When you wind up a spring driven clock it gains mass. As it runs down it loses mass. Some people have the notion that only nuclear energy annihilates mass. That is incorrect. Here is another extrapolation. A modern jet aircraft burns about 1 gallon of kerosene per second. That produces 142 MW of heat which converts to ~90 MW of mechanical thrust with 63% efficiency. I think. So, assuming a cold fusion engine is also 63% efficient, 4 g of deuterium would be consumed in about 162 minutes (2.7 hours). - Jed
