On Sat, Jun 25, 2011 at 7:52 AM, Jouni Valkonen <[email protected]>wrote:
> 2011/6/25 Joshua Cude <[email protected]>: > > > > > > On Sat, Jun 25, 2011 at 1:02 AM, Jouni Valkonen <[email protected] > > > > wrote: > >> > >> 2011/6/25 Joshua Cude <[email protected]>: > >> > Well it might be if the reactor were at the bottom of a tea pot, and > the > >> > output at the top of the pot. But the input and output to the reactor > >> > are > >> > both horizontal at the same level. > >> > >> here was your misunderstanding. This is not true, because water input > >> is at the same level as reactor core, but water output or boiling > >> surface is above the reactor core somewhere in the chimney. > > > > The output of the entire device is at the end of a hose in another room. > The > > output from the *reactor* is horizontal, and level with the input. Then > it > > flows through a pipe into a chimney. > > This is NOT TRUE because chimney is filled with water. You seem to > have really hard time to understand this. > How does a chimney filled with water contradict the statement that the fluid exits the reactor cell horizontally? > > > No because the chimney is not situated vertically above the reactor. > > According fluid dynamics this is completely irrelevant, if it is first > horizontal and then bends into vertical.It does not change a > thing. Well it does not change some things. But it changes the things you are hanging your hat on. The reason there is regulation, you claim is because the heating element is at the bottom of a reservoir, like in a teapot. That way, only liquid is heated directly. And you claim it would overheat if it dried out. But if the reactor is horizontal, before the chimney, then the input side of the reactor is always being wetted with cold water, and if the power is high enough to vaporize all the water, then at the exit side, the reactor would no longer contact liquid, The steam would have to be formed in the reactor, because there is no further heating provided in the chimney, and therefore after the initial wate got pushed out by the steam, the chimney would not have water in it. > > > The heating does not happen in the chimney. > > It is completely irrelevant where the heating happens, because chimney > is filled with water. No. If the steam is formed before the chimney, then it will push the water out of the chimney. > In the tea pot heating does not happen in the > water surface.Chimneys diameter is large enough, so that chimney > effectively prevents all the sputtering and mist formation. That is > the whole point of having a chimney. > In a teapot, the element is below the water, the area is at least 20 times larger, and the power is a few times lower than claimed for the ecat. The ratio of gas pressure per unit area is probably 100 times lower. Try 5 kW of power boiling water into a 1 inch pipe, even if it opens into a 2 or 3 inch pipe, and see how fast the water will get pushed out. > Try to understand this that chimney is filled with water. And this > undermines ALL your arguments against E-Cat. > I know you want all arguments against the ecat to be undermined so you can continue to believe in this fantasy, but wishing for it won't make it so. ====== I think we've gotten a little side-tracked with all the discussion of what happens inside the ecat. All we really know is what comes out. The output is at a very flat temperature within a degree or two of boiling, and there is some indication of steam, and in Lewan's run, very clear indication of flowing liquid water. I don't see how that is evidence of dry steam, even if you argue that it might be consistent with it. What we need is evidence. Volume flow rates, or condensation and heat measurement. That sort of thing. But I'm curious about one thing from all the people that are satisfied that the steam must be dry: What would you expect to see at the output hose and on the temperature reading if the input power was well above the power necessary to reach the boiling point, but well below the power necessary to vaporize all of it? In the Krivit demo, to be specific, what would you expect to see if the input power was say 2 kW? Remember, in this run, if you believe Rossi's flow rate, 600 W would bring all the water to the bp, and 5 kW would vaporize it all. So, what comes out if the ecat transfers 2 kW to the water, by whatever method? Show your work.
