On Sat, Jun 25, 2011 at 8:58 PM, <[email protected]> wrote: > In reply to Joshua Cude's message of Fri, 24 Jun 2011 16:20:48 -0500: > Hi, > [snip] >>I was talking about running it above boiling, but way below the level needed >>to boil it all. Different thing. And it's easy. The power can range within a >>factor of 7. In this case, anywhere between 600W and about 5 kW. > > BTW (the latent heat of steam) / (the heat energy required to bring water to > the > boil) is a factor of about 6.7 (depending on starting temperature of water), > and > curiously close to the COP Rossi claims to be aiming for. > In short, if virtually none of the water were converted to steam, and he was > assuming that it all was, then it would neatly explain the conversion factor > he > is claiming. >
You might be thinking of another scenario - but if I'm guessing what you are saying then the best anyone could do is about 1.86 to 1 ratio. But this assumes that any liquid hot water needed to cool water vapor in a heat exchanger is included in the calculation (otherwise the ratio would be worse, less than 1.86 to 1). I did this calculation, shown below, weeks ago. Basically in this fraudulent set up, the Ecat would do the following: 1. Create 1 kg of 99.9 C water from 10 C water which requires (99.9 - 10) x 4.18 kJ/kg/C = 376 kJ 2. Using same water from step 1, make 1 kg of water *vapor* requiring 2257 kJ. Total input to Ecat required at this point is 376 + 2257 = 2633 kJ 3. Condense water vapor into micro droplets (i.e. fog) deep *inside* the Ecat using a heat exchanger and use this heat to heat 6.00 kg of cold liquid water from 10 C to 99.9 C. This is because 2257 kJ /376 kJ/kg = 6.00 kg (note that the units are correct). Also, note that at this point the total input energy is still 2633 kJ. The actual/real end result is 6.00 kg of 99.9 C water and 1 kg of micro liquid water *droplets* (fog or steam with 0% quality). A gullible observer would think that the Ecat just produced 6 kg of hot water and 1 kg of water *vapor* when it really made 6 kg of hot water and 1 kg of hot *liquid* water droplets. The gullible observer would think that the energy normally needed to create this is 4890 kJ because: (6 kg + 1 kg) x (99.9 - 10) x 4.18 kJ/kg + (1 kg) x 2257 kJ/kg = 4890 kJ While in *reality* it took the following amount of electrical energy: (6 kg + 1 kg) x (99.9 -10) x 4.18 kJ/kg = 2633 kJ So, the gullible observer would see 2633 kJ of electrical energy go into the Ecat and 4890 kJ of thermal energy leave the Ecat. This is a ratio of 4890/2633 = 1.86 I can't think of any way of increasing this ratio using any other similar method. Jeff > Regards, > > Robin van Spaandonk > > http://rvanspaa.freehostia.com/project.html > >
