crisply summarizes a lot of the critical evaluation by Cude, Heffner, and Murray... elementary over estimation of excess heat by Rossi in all his demos... thanks to Mary Yugo and the original source 123star --
https://mail.google.com/mail/u/0/?tab=nm#inbox/133dbc70ac37ba45 123star November 23, 2011 - 7:24 pm | Permalink References: http://en.wikipedia.org/wiki/Heat_capacity http://en.wikipedia.org/wiki/Latent_heat From: http://ecatnews.com/?p=1392&cpage=1#comment-9761 123star November 23, 2011 - 7:24 pm | Permalink On Sat, Nov 26, 2011 at 5:29 AM, Peter Heckert <peter.heck...@arcor.de> wrote: > The question is, how did they measure the energy input? > This is not documented. > > This colonel engineer confuses kg and g. > He measures a hydrogen consumtion of 1.7000 kg and dont write down all > significant digits. > Then he subtracts this from a value that means gramm. but is mistakenly > labeled as "kg". > How can we win a war where precise decisions must be made in seconds? ;-) > He makes many handwritten corrections and erasures to ensure he can read his > own writing. > He has two different ways to write a one: "1" and "|" in one and the same > document. > He uses decimal point and decimal "," alternating in one and the same > document. > How can we believe he measured or calculated the electrical energy or the > diesel consumption correctly? > Possibly he has confused more than that? > > This is not a Nato colonel engineer with 30 year of experience in a > multi-language military organisation. > > Peter > > > Am 26.11.2011 13:51, schrieb Berke Durak: >> >> On Fri, Nov 25, 2011 at 12:30 PM, Mary Yugo<maryyu...@gmail.com> wrote: >>> >>> I didn't originate this. I reprint it with minor changes from >>> ecatnews.com. >>> ... >> >> Interesting! Let's run the figures for the 1 MW demo. >> >> Energy input : 66 kWh -> 238 MJ >> Water claimed to be vaporized : 3716 l >> Average output temperature : 104.5 C >> Average input temperature : 18.3 C >> Energy required to heat 3716 kg of water from 18.3 C to 104.5 C : >> (104.5 - 18.3) * 4.181e3 * 3716 = 1.34 GJ >> >> So: >> >> COP = 1.34 GJ / 238 MJ = 5.63 >> >> So if he and Fioravanti mistook very wet steam for steam, he only >> has a cold fusion reactor with a COP of only 5.63, instead of a COP >> of 9.49 GJ / 238 MJ = 39.9. What a scammer!!!! >> >> I mean, when I pay $2,000,000 for a cold fusion reactor with a COP >> of 40, I don't expect to be given a cold fusion reactor with a COP >> of 5.63. Jeeez!!! That's what you get when you go for cheap Italian >> knock-off "University of Baloney" cold fusion reactors.
