Here’s an important correction, I think, that pertains to the discussion we have been having about the cat and mouse.
On the Journal of Nuclear Physics, Andrea Rossi wrote to Joseph Fine:
Our basic module is made by an apparatus in which we have 2 components: an activator, which consumes about 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.
On vortex-l, Patrick Ellul wrote in reference to this post:
I thought he might have done a typo, and mean 10 kWh/h so I emailed him to ask.
He confirmed that it was a typo, and he meant 10kWh/h for the ecat, just like it has always been.
If that is the case:
0.91 * 35% of the time = 0.3185 kWh/h
10 * 65% of the time = 6.5 kWh/h
total output = 6.8185 kWh/h
input = 0.9 * 35% of the time = 0.315 kWh/h
COP = 21.65
So, taking into account this correction and rounding the numbers, the way I see it is that the output of the Cat is about 10 times the output of the mouse, and it is turned on about twice as long as the mouse is on, so you are going to get approximately 20 times as much energy out as you get in.
Regarding the activator ‘paying for itself’, I now think that Rossi means that the customer is paying for the input energy anyway (electricity or gas) and it ends up being used in the output mix, and with the E-cat in place that input energy is multiplied by about 20.
That’s my current thinking on the matter — could change with more information.