The failure of one module of the Rossi 1 MW reactor will not cause the entire 1 MW reactor to fail. Its performance will only degrade gracefully.

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When the core of the module overheats or melts, the surface of the nickel nanopowder will fail before the nanopowder enclosure will fail since the enclosure will be cooled by low temperature steam or water which would remove heat, effectively cool the enclosure, and support its structural strength. The failure of the nanopowder will cause the individual module to cool and be ineffective at generating thermal power. It would be analogous to a failure of one pixel of your computer screen; if one such pixel grows dark, your screen will not fail but its performance would degrade. You would still be able to use the screen, just the picture would not be as sharp. So too with the Rossi reactor; it would still generate heat, but not so much as before. Its capacity would be reduced until its performance would eventually degrade below a certain predefined lower threshold. When this low bound threshold is reached, the entire reactor is considered to have failed. On Thu, Sep 29, 2011 at 3:34 PM, Horace Heffner <hheff...@mtaonline.net>wrote: > > On Sep 29, 2011, at 4:02 AM, Man on Bridges wrote: > > Hi, >> >> On 29-9-2011 8:27, Horace Heffner wrote: >> >>> Looking at the other side of the coin, the probability of catastrophic >>> failure, suppose there is a 0.1% chance per hour one of the E-cats can blow >>> up spreading steam throughout the container. There is thus a 0.999 >>> probability of success, i.e. no explosion for one E-cat, operating for one >>> hour. The probability that all 52 E-cats perform successfully for a 24 >>> hour test period is then 0.999^(52*24) = .287. That means there is a 71.3% >>> chance of an explosion during a 24 hour test. >>> >> >> Me thinks you are wrong. Your statistical probability calculation is based >> upon the fact that the chance of a single Ecat exploding is influenced by >> it's behaviour earlier, >> > > This is false. The probability in each time increment is assumed to be > independent. For there to be success there must be no failures for any time > increment. If there are T time increments, and the probability of failure > in any time increment is p, the probability of success q=1-p in each time > increment is independent of the other time increments, and the probability > of success in all time increments is q^T (only possible if what happens in > each time increment is independent event), and the probability of any > failure having occurred is thus 1-(q^T). > > > > which of course is not true. Statistically each Ecat has it's own >> independent chance of explosion at any given moment which does not change >> over time. >> > > The instantaneous probability of failure is zero. Zero time results in zero > probability because lim t->0 q^t = 1 for for all 0=<q<=1 and positive t. > Therefore lim t->0 1-(q^t) = 0. Note that I provided an assumption of > 0.001 percent probability of failure *per hour*. > > > > With your probability of 0,1% chance per hour this would result for the >> whole of 52 Ecats then in a chance of explosion at any given moment of 1 - >> (0.999^52) = .05 or 5%. >> > > No. The probability of at least one E-cat failure in the 52 E-cat system, > based on the assumption of 0.001 probability of failure of an individual > E-cat in an hour is 1-(0.999)^52 = 0.506958 = 5%. Your number 5% is right, > but your interpretation of it representing an instantaneous moment is wrong. > > > > > >> Looking even a bit more closer again this would mean that if the chance of >> explosion is 0.1% per hour then the chance of explosion is 2,77e-7 per >> second at any given moment for a single Ecat, which would result for 52 >> Ecats into 1-((2,77e-7)^52) = 0,00001444434 or 0,00144% at any time. >> > > The phrase "at any time" makes the above statement nonsensical. > > An hour represents 3600 seconds, which are 3600 independent events of 1 > second duration. Let a be the probability of failure in 1 second, and > b=(1-a) be the probability of success in 1 second. We have the given > probability p of failure for 3600 seconds being 0.001, and the probability > of success of one E-cat for one hour being q = 0.999. The probability of > success (no failures) for the 3600 1 second independent time increments is > > q = 0.999 = b^3600 > > b = q^(1/3600) = 0.999^(1/3600) > > a = 1 - 0.999^(1/3600) = 2.779x10^-7 > > Note that a is the probability of failure in one second, not "at any time". > This is totally consistent with the probability of failure in one E-cat in > one hour being 5%. In other words, going backwards: > > p = 1-(1-a)^3600 = 1-(1-2.779x10^-7)^3600 = 1-0.999 = 0.001 > > My calculations are therefore self consistent. The time intervals are all > treated as independent events. Your interpretation of "moment" is perhaps a > conceptual problem. > > > >> Kind regards, >> >> MoB >> >> > Best regards, > > Horace Heffner > http://www.mtaonline.net/~**hheffner/<http://www.mtaonline.net/~hheffner/> > > > > >