On Thu, Jun 23, 2011 at 3:37 PM, Abd ul-Rahman Lomax <a...@lomaxdesign.com>wrote:
> At 11:56 AM 6/23/2011, Jed Rothwell wrote: > >> Abd ul-Rahman Lomax wrote: >> >> My sense, from the weak steam coming out of the end, is that what seems >>> to be marginal at the end is an indication that more power is being >>> generated than the input electrical power, but I'd not want to claim that >>> this demo shows that, it's way too shaky. >>> >> >> No, it isn't shaky. The water would be 60°C or less in most of these tests >> if there was no anomalous heat. There would be no trace of steam. The only >> question is: Is there a lot of anomalous heat, or only a little? Who >> cares?!? >> > > This appears to be taken from a probable error in the Kullander/Essen > report. They claimed that the temperature would have not exceeded 60 degrees > without excess heat. In fact, I think that what they intended to say was > that the rate of change would not have exceeded the rate up to 60 degrees if > not for excess heat. > > My guess is that it would still reach boiling point, at roughly the time > predicting by extrapolation of the rate of temperature rise, because the > device is insulated and most heat will not leave unless the water starts > boiling. It's a 600 watt steam kettle, I'd expect such to boil water. Just > not as quickly as seen. Nooooooo. Not you, too. Now I find myself defending the Rossi crew against a LENR advocate. It's flowing water, not a kettle. So the input power can only heat it so much. Power in = mass-flow-rate * specific heat * temperature difference So, temperature difference = 300W / (1.73 g/s * 4.2 J/K g) = 41 K If the input temperature is 20C, then the maximum output is 61C. If you accept the numbers as given. It's not like a kettle. The reason the graph shows a gradual increase in the temperature when the power is first applied, is that the reactor has to heat up first, and that absorbs some of the power. When it reaches equilibrium temperature, all the power goes in to heating the water.
