I think one has to take into consideration the specific heat... think of the water as a big 'heat capacitor', and although it is at the boiling point (BP) at point 'B' on the graph, it absorbs a shitload more heat energy by the time it gets to point 'C' on the graph, imparting more and more kinetic energy to both the liquid and vapor water molecules, which reduces the likelihood of finding any liquid water in the steam. Joshua's insistance that the temperature of the steam MUST be well above BP is ASSUMING that the capacitor is full, and there's no where else for the heat to go but into the vapor molecules. What if the E-Cat is operating with a 98% 'full charge' on the heat-capacitor? It would still have considerable capacity left to absorb heat fluctuations without significantly changing steam temperature. >From the measurements and statements made by the Rossi camp, I would bet that >the E-Cat is operating at SLIGHTLY to the LEFT of point 'C'... where there's enough kinetic energy in the steam to maintain it at less than 1.5% liquid water (by mass) in the chimney, but still able to absorb some modest heat fluctuations from the reactor without significantly changing the temperature of the steam in the chimney -- i.e., what fluctuations that do occur in the reactor heat output are dampened by the heat storage capacity of the water / steam capacitor. Thus, ***IF*** the reactor's heat output is stable enough, it could achieve what they are saying... Anybody have any insights as to how stable the heat output of the reactor is? Does it fluctuate by 10 watts, 100 watts or 1000 watts? And over what time period?
-Mark _____ From: Joshua Cude [mailto:joshua.c...@gmail.com] Sent: Tuesday, July 05, 2011 7:30 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Analysis of e-Cat test by E. Storms 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.
