For comparison I checked the underside of my electric kettle and the power rating is approx. 1500 W
Harry ----- Original Message ---- > From: Daniel Rocha <[email protected]> > To: [email protected] > Sent: Tue, June 21, 2011 12:01:14 PM > Subject: Re: [Vo]:E-Cat proven to be a hoax? > > No problem, that's ok :). BTW, I made some more calculations, after > GoatGuy told me something: > > > > ************************* > > Without the thermal conductivity worked out, the "blackbody" > approximation is in error. Use 50° C instead. (i.e. a > hot-but-can-be-held temperature). Also, don't forget that > net-radiative has to come from the temperature differential, not the > temperature absolute. A 50°C blackbody in a 50°C room will neither > gain, nor lose, heat by SBL. > > P = AεσT4 where A = area, ε = emissivity, σ = 5.67×10-8 > > ( 3 m × 3.14 × 0.02 m × 0.8 emissivity × 5.67×10-8 × ((50 + 273)4 - > (25 + 273)4 )) = 25.6 W > > That's more like it. I'd believe that 25 watts would maybe be > radiated as the differential. I chose 80% as the greybody emissivity > (which is in line with black rubber). I'm not sure what you were > doing the extra multiplications by the Stefan-Boltzman constant for, > but … the above works out well, directly. > > And again, I don't think any of this materially is significant > compared to Rossi's stated “running at about 2,500 watts, stably”. > You know? > > G O A T G U Y > > **************** > > And I answered: > > ********************** > > Sure, I used the temperature differential. The emission of the hose > emiting to the evironment at 100C, and the environment heating the > hose ar 30C. Well, that's the difference anyway... Well, eventually, > there must be conservation of energy, so I guess if I used the > internal part of the hose and the environment, that would be more > correct so, 110*(1.25/2)=69W. Per meter, it gives 23W. I made several > approximations during all the way, but 23W is a bit close to 25.6. > >
