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.
