If you split the time between activator and ecat as he suggests you get slightly different numbers:
0.91 * 35% of the time = 0.3185 kWh/h 1 * 65% of the time = 0.65 kWh/h total output = 0.9685 kWh/h input = 0.9 * 35% of the time = 0.315 kWh/h COP = 3.07 On Tue, May 14, 2013 at 4:48 AM, Alan Fletcher <a...@well.com> wrote: > I can't figure out how Rossi claims a COP of 100-200 > > May 12th, 2013 at 9:59 PM > http://www.journal-of-nuclear-physics.com/?p=802&cpage=9#comment-694786 > > Dear Dr Joseph Fine: > Please don’t go too far: just, for now, let’s limit to what I wrote about > the Activator/E-Cat cycle. Please read carefully what I wrote. More than > that is not possible to get, so far. Our basic module is made by an > apparatus in which we have 2 components: an activator, which consumes abour > 900 Wh/h and produces about 910 Wh/h of heat. This heat activates the E-Cat > and then goes to the utilization by the Customer, so that its cost is paid > back by itself. This activator stays in function for the 35% of the > operational time of the syspem of the apparatus. The E-Cat, activated by > the heat of the Activator, works for about the 65% of the operational time, > producing about 1 kWh/h without consuming any Wh/h from the grid. Combining > these modules we can make E-Cats of 1 kW , 10 kW, 100 kW, 1 MW , > respectively, of power. > Warm Regards, > A.R. > > - - - > > kWh/h > mouse output 0.91 > ecat output 1.00 > > total output 1.91 > total input 0.90 > > Total COP = total output/input 2.12 > > > > -- Patrick www.tRacePerfect.com The daily puzzle everyone can finish but not everyone can perfect! The quickest puzzle ever!
