On Tue, Jul 5, 2011 at 8:37 PM, OrionWorks - Steven V Johnson < svj.orionwo...@gmail.com>wrote:
>From Josh, > > For brevity sake I'm just going to focus on the following: > > > I don't think the quality of the video is good enough to judge that. > > Fair enough. > > > Take a look at figure 2.2.3 on the site Iverson just linked to. > > Follow the constant pressure path ABCD. It indicates clearly that at > > constant pressure, as soon as you get dry steam, it can be heated > > above the boiling point, but that if the steam is wet, it can't be. > > Dry "steam" will most certainly increase in temperature *IF* it can > hang around long enough in a super heated environment to absorb > additional heat. > That is the question. > It's simpler than that. It's a matter of conservation of energy. If the power increases slightly above what is necessary to boil all the water to produce dry steam, then unless the additional heat finds its way through the insulation, the only way to remove that heat is by heating the steam to a higher temperature. At equilibrium, power in must equal power out. Probably some combination happens, but removing the heat by heating the vapor is certainly more efficient than dissipating the heat through the insulation. If you think it doesn't hang around long enough when the power just exceeds dry steam, then the fluid is not removing heat fast enough, and so the ecat will get hotter. That means the water will boil a little earlier, and the steam has to pass by more of the heated walls of the ecat, and at a higher temperature. Very soon, the water boils away early enough, and the ecat is hot enough so that the gas *is* able to remove the additional power, and a new equilibrium is established with the steam coming out at a higher temperature. Then the power in once again equals power out. > > However, I repeat. A key point in all of this conjecture is the fact > for the 100% "dry" H2O gas to increase in temperature much above 100 > C, it has to hang around long enough within a super heated environment > in order to absorb additional heat energy. > Yes, but it must get hotter if power in is to equal power out. > > I am under the impression that the water that was being heated in > Rossi's demo e-cat was NOT under any pressure, meaning it is not > maintained within a contained environment. > Once again, pressure is not necessary. > Therefore, the newly > > converted gas, which BTW is constantly expanding, quickly exits the > heated reactor chamber. > Yes, and indoor air quickly leaves a furnace element, but it still gets hotter. > Once the H2O leaves the confines of heated > > reactor chamber (which it will quickly do) it no longer has a chance > to increase much above the 100 C temp it had initially acquired. > Obviously not. It must get heated before it leaves the cell. Therefore how can this newly formed H2O gas be expected to be much > above 100 C if it doesn't have a chance to hang around long enough to > absorb additional heat energy. > How can it not? If the power-in exceeds the power needed for dry steam, as it probably would occasionally if there were ordinary fluctuations in the ecat output of the sort described in the secret 18-hour experiment, then the power-out must also increase. And the only way for the power out to increase, if the steam is already dry, is for it to get hotter.
