So Rossi is saying the volume of the reactor assembly minus the fins and
water tank is 0.318 m^3 / 106 E-Cats = 3,000 cm^3 per E-Cat module.
Based on the reactor assembly being 5 cm thick, as Lewan stated, I get
the other dimensions as 24.5 x 24.5 cm which agrees with other
statements and would fit inside the finned heat dissipation assembly we
have seen.
On 12/24/2011 1:28 PM, Aussie Guy E-Cat wrote:
http://www.journal-of-nuclear-physics.com/?p=510&cpage=39#comment-155427
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Andrea Rossi
December 23rd, 2011 at 11:37 AM
<http://www.journal-of-nuclear-physics.com/?p=510&cpage=39#comment-155427>
Dear Alby:
I never banned you. Probably your comment has been spammed by the
robot, as it happens, unfortunately, sometimes to our Readers. When
it happens, just repeat the send, as you correctly made, but be sure
that in the address from where you send the comment or in the same
comment there are not advertising signals.
About your question:
The 1 MW plant that has been tested on October 28th 2011 by the
Consultant of the Customer, he has calculated what follows:
1- the total volume of the 106 reactors (net of the heat exchangers
and the water tanks): 0.318 cm (cubic meters zero point three
hundred eighteen)
2- the total amount of kWh of energy produced in 5.5 hours has been
2,585.00 kWh (two thousand five hundred eighty five kWh)
3- to produce 2,585.00 kWh with the best existing batteries you need
a battery volume of at least 53 cm (fifty three cubic meters)
Conclusion: the energy produced has been 53/0.318 = 166 times more
that the maximum amount of energy you could have produced with the
most energy-dense batteries existing in the market.
I want also to add that:
a- if you read the Focardi-Rossi paper published on this same
journal, you will see that an E-Cat has worked for 180 days, 24
hours/day
b- we made and make regularly tests which last more than 20-30 hours
c- presently we are heating a factory of ours with our E-Cats.
Warm Regards,
A.R.
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