How Does the E-Cat Pay For Itself?

There has been a flurry of Q & As on the Journal of Nuclear Physics since Andrea Rossi started to describe some more details about the ‘cat and mouse’ configuration of the new hot cat. Rossi has given similar results a number of times, and yet there still seems to be some confusion about how he gets to the ‘100-200 COP’ calculation. Here are three different attempts to explain from Rossi:

It’s simple: if the COP of the activator is >1, whatever the number, it gives heat that repays for itself, besides activating the E-Cat. The economy of the system is: the Activator consumes 1, yields 1.x as heat to the Customer AND activates the E-Cat: the E-Cat consumes nothing from the grid , gives its energy to the Customer.

while the E-Cat is turned on, no other source of energy comes to the system. When the Mouse is turned on, the E-Cat is turned off and in this phase the Activator draws energy from the heat source. When the E-Cat is turned on ( about 65% of the operational time) the denominator is zero, no energy comes from any source to heat the Activator and the E-Cat, while the E-Cat is turned off ( about 35% of the operational time) the activator draws energy from the heat source, but at the same time produces for the Customer an amount of heat that is equal or more than the energy consumed, so that it is pays the energy that consumes by itself.

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.

The key question here is how does the activator ‘pay for itself’, when it is has a COP of only just above 1? Is the energy required to control the ‘Cat’ so small as to be almost insignificant — and the rest of the heat is passed through to the output of the system?

To avoid off-thread discussion in the previous post, let’s keep our discussion about this configuration here. Ok — discuss!

UPDATE: Thanks to reader Glenn for providing this image to go along with our conversation. He is asking for critiques, so please feel free to comment!

proposed e-cat setup