Some consolidated and clarified text follows.
... I'm heading out for a couple of days, so I don't have time to check your method or numbers
(assuming a 94 °C boiling point) of:
Isn't that a bit low?
You still need to take into account the fact that the output in mode 1 is VD (Vapour+Drops), and in modes 2 and 3 it's FVD(Fluid+Vapour+Drops)
In mode 3 it will take longer for the copper device to fill with
water, i.e. come to equilibrium. However, since mode 3 creates about
50% more steam, the percolator effect, vs a simple overflow effect,
should be more apparent.
The "percolator" effect is a bit confusing, because it depends on a) a build-up of water and b) enough steam to blow the accumulated water up the tube (in the flow regimes I'd call it a "slug" of water.)
From your separate post, I'd put in a transparent outlet and collect fluid water at two points.
Schematically (you need a non-proportional font here)
hose (sloping)
Chimney ==#=====================#====== Gas (Vapor+Drops) ---> sparge
|| T || T
|| ||
|| ||
| || | | || |
|W||W| |W||W|
|W||W| |W||W|
|W||W| |W||W|
|W||W| |W||W|
|W||W| |W||W|
|WWWW| |WWWW|
*----* *----*
Overflow Fluid Condensed Fluid
The hose should slope, with NO valleys/kinks that could collect water.
"T" is a T-junction (possibly with a tap). Would be good to have a thermometer here, too.
Catch the Overflow fluid and condensed fluid separately.
You could sparge the final output. (With Jed's method sparge into a tank with a lower water height than the traps I've shown.)
>Joe Catania
> Mon, 22 Aug 2011 18:37:21 -0700
>I think that tube diameter in the horizontal section is probably significant for this type of experiment. >Cantwell's copper tube diameter may not be much higher than his heater diameter thus it may not be a good >comparison with this aspect of Rossi's device.
What happens in the horizontal tube is only of academic interest. The vertival tube is the most important for considering flow regimes.
> Jouni Valkonen
> Mon, 22 Aug 2011 21:52:49 -0700
> Alan, you should explain more. I fail completely see your point, how does it > confirm your hypothesis?
See my reply to Jed. > Or at least you cannot make any calculations from > your idea of "dryout", *I* can't calculate the dryout points at the moment ... but others HAVE for different situations. > therefore it seems to be rather irrelevant concept. If everything had been done right it wouldn't be necessary. My aim here is to "eliminate" the left-half of NASA's Temp-Enthalpy diagram. > You also are misusing physical concepts. E-Cat produces always high quality steam and hot water, but never low quality steam. You can only make low quality steam by cooling rapidly high pressure steam and these conditions might happen in steam turbines. But E-Cat is working roughly at constant pressure (3.3kPa pressure difference is insignificant), therefore the steam quality is there always ca. 98% (±0.015). That 98% is true for a kettle boiler, but I don't think it IS for a TUBE boiler. You can't NOT boil a kettle, but you can cut off a tube at any suitable point. (In nukes, to AVOID dryout = burnout).
