COP eCat between 125 and 143….(?)

Here’s a question sent by John Schut:

“Did I understood correctly that Anrea Rossi has indirectly stated that the COP of the eCat is between 125 and 143? This because he stated that the eCat uses 7 à 8 KWatt/hour while producing 1 MWatt/hour.”

John’s question is based on a statement made by Andrea Rossi yesterday on the Journal of Nuclear Physics when asked by a reader what the input power was when the 1 MW plant was in self-sustain mode, including consumption of the air conditioner and ‘entertaining gadgets.’

Rossi replied:

Andrea Rossi
August 1st, 2015 at 7:00 PM
Alexvs:
Rethinking, what you ask for can be answered: the comsume of current when the 1MW E-Cat is in ssm, producing 1 MWh/h is betwen 7 and 8 kWh/h. The air conditioner is not included, because powered from a source independent from the E-Cat, as well as the light. Entertaining gadgets are Physics books, they do not consume current. The current is consumed by the control panels, triacs, transformers, safety systems.
Warm Regards,
A.R.

Someone responded by asking Rossi if this meant that the COP would be between 125 and 142 (as John asked)

Rossi responded:

“You must make a distinction between the ssm and the non ssm cycles. We will give the data related to the performance of the 1 MW E-Cat after the end of the tests on course.
For example now, at 7.54 a.m. of Sunday August 2, She is not in ssm; all stable, all good, no problems by now, and the control system has put Her in not ssm mode.”

So at this point it really is not possible to know the average COP of the plant, because we don’t know the duration of the SSM periods in comparison to the non-SSM periods. We don’t have any data to go by, except the spot-checking we get when Rossi is asked on the JONP what the plant is doing at a particular moment. I have tried to compile those responses on this thread:

http://www.e-catworld.com/rossis-1mw-plant-performance-updates/

From the responses compiled, Rossi says that out of 14 entries, 10 times he mentions that the plant is in SSM, 4 times the plant is “stable”. A reading once in a while is random, however, and I’m not sure we can determine a statistically significant pattern from these data points.


  • Well, Rossi certainly seems to be the leader today. As it appears the Solar Hydrogen Trends device is just a battery, not an energy source at all, their COP has to be less than 1. Defkalion claimed a COP far above 20, but where is their product now? They have been silent so long it appears they hit a roadblock or were bought up by a secretive larger corporation. Brillouin has interesting ideas, but has never proven a high COP. The simplified hot fusion companies are years away from having a usable product, but long term they could be competition.

  • Albert D. Kallal

    But due to so many small reactors, they not all going to be powered at the same time – so in a merry go round approach, you have the units staggered.

    In other words you don’t put power to all 100 units – you put power to a few
    and once they SSM, you use that power for other units. Once you get them all staggered, then you only powering 20% of the units at a given time (assuming 80% SAM). So you would never “sync” them all at the same time to require power.

    In fact this issue MUCH shows how using many smaller
    devices is better than one big unit that would require huge amounts of power
    when “on”. And this also applies to startup – you only need power to start a few, and once they SSM, you use that power to startup other units.

    Rossi has stated the max input is about 250,000 watts,
    but he also stated above that the power consummation was only about 8000 watts
    to get that 1 million watts output. This is only possible by staggering a bunch
    of small units like a conductor managing a musical symphony.

    Regards,
    Albert D. Kallal
    Edmonton, Alberta Canada

  • Albert D. Kallal

    Actually, would not 80% ssm at a COP cop of say 8 give you

    Think of 10 units of energy.

    Input: 1 1 0 0 0 0 0 0 0 0 = 2 units of energy supplied
    1 2 3 4 5 6 7 8 9 10
    Output:8 8 8 8 8 8 8 8 8 8 = 80 units out

    So your COP will be 40.

    At 90% ssm (9 units), then you get 80/1 = 80

    If the static COP is only 3, then

    Input: 1 1 0 0 0 0 0 0 0 0 = 2 units in
    1 2 3 4 5 6 7 8 9 10
    Output:3 3 3 3 3 3 3 3 3 3 = 30 units out

    30/2 then cop = 15

    Of course the above assumes a “constant” output when in
    SSM, and I suspect it more of a “downward” curve that the controller “kicks” on
    the power before the curve drops below the tempature in which the LENR effect stops.

    So perhaps another analogy is those guys that spin plates – you run around and give each plate a spin like this (think of EACH plate as a LENR reactor core!). Just needs a “spin” once in awhile to keep going!

    https://www.youtube.com/watch?v=Zhoos1oY404

    Regards,
    Albert D. Kallal
    Edmonton, Alberta Canada

    • John Schut

      I don’t really understand your remark.
      I get the impression that you mixup COP and Output(?).
      You speak of 90% ssm (9 units), but the input is still 8×0 (which must be 9×0).
      Furthermore you set the output differently, while it should be always the same (1 mW/h because that’s the output of the E-Cat).The COP is nothing more or less then the outputed 1 mW/h divided by the totally inputted power.

      • Albert D. Kallal

        thanks – that is a type-o – yes I mean 9 units for the 2nd example of same cop, but THEN I go back to original 8 units with a different COP – I edited the post to make this more clear.