This last post is a wonderful one. The way the E-Cat produces and then radiates energy is a complete unknown and there is a absolute and uncompromising need in this unique situation to calibrate the temperature sensor used in this particular kind of test in a complete and fined grained detail if the true COP of this reactor is to be determined reliably.
On Wed, Oct 29, 2014 at 12:36 PM, David Roberson <[email protected]> wrote: > All indications are that the visible spectrum contains very little of the > energy being radiated so what we see can not be used to figure the radiated > power. Many other variables appear to get into the fray which forces us > to rely upon calibration if we are to achieve accurate accounting of the > radiated and convected power. It is unfortunate that the input power was > not the same during both the dummy run and the active one since the > increased apparent temperature would have clearly demonstrated excess power > if any was present. > > I am left with believing that excess power was generated due to the rapid > increase in calculated output power when a small increase in input power > was applied. This is a characteristic of an ECAT system with positive > thermal feedback. A passive system would not display this behavior. > > Dave > > > > -----Original Message----- > From: Alan Fletcher <[email protected]> > To: vortex-l <[email protected]> > Sent: Wed, Oct 29, 2014 11:35 am > Subject: Re: [Vo]:MFMP interviews spokesman from WILLIAMSON > > *From: *"Eric Walker" <[email protected]> > *Sent: *Tuesday, October 28, 2014 10:20:52 PM > > On Tue, Oct 28, 2014 at 12:36 PM, Alan Fletcher <[email protected]> wrote: > > Basically what happens is that as the temperature changes the peak of >> the blackbody spectrum moves through different parts of the >> emissivity/wavelength curve. > > > Are you assuming a standard Boltzmann curve that just shifts its peak > according to emittance? Is it possible that the frequency and > heat-dependant combination of emittance, transmissivity and reflection make > it so that there is a distribution other than a Boltzmann distribution for > the alumina shell? > > Eric > > Yes, that's how Planck's formula/integration works. It TRIES to send a > Boltzmann curve, but this is modulated by the emissivity spectrum. > As the temperature increases the spectral peak get higher and shifts to > shorter wavelengths. If the emissivity is higher then the total power will > increase, otherwise (as in this case) it decreases. > > Per Manara the transmission looks negligible outside the visible range, > where there's practically no blackbody power anyway up to 1400C. (It moves > to the visible at much higher temperatures -- 4000 to 6000C). > > >
