I have understood. I have even worked with many types of pumps, including those with constant, fixed flow, as in this case- the peristaltic pump.. Very probably the system works in this way- you have a core, very hot in the center of the device- Ni and H reacting in a metallic tube.. By external heating - with the input current a resistor heats the core to a temperature (350 C?) and the reaction is started. It releases heat and if this heat is not removed fast enough the core overheats and stops working. With sufficient cooling as in this case it works - probably the inner temperature oscillates in some limits. The entire quantity of cooling water is evaporated and a bit overheated. We don't know how the Ecat is controlled, probably the heat furnished by the resistor is automatically adjusted to the cooling load. Remember that Rossi told that he heat after death regime, zero input (after the start of reaction) can be dangerous.
You can calculate the flow of water in non-evaporative hot water regime for this case approx 10-15 kW. The experiment had to be adjusted to what the Bologna University has, including practical experience with. Laboratory size peristaltic pumps yes greater gear pumps no. So they have decided to use the Ecat as a steam generator and not as a water heater. Now there are too many unknown unknowns- e.g. is the description of the device in the patent can be false in part, we don't know the effective section for the water flowing through the generator. And steam is high currency energy (it can be transformed in electrical energy) while hot water is considered low currency energy, ergo, it is better to work with steam. Plus the heat transfer surface is very hot. OK we can speculate a lot, we can try to demonstrate that the cell is not working, or it is not overunity with absolute certainty, or is collection of big bad IFs- but this seems contraproductive. Better wait and get more information. Waiting with empathy and expectation or with hostility and denial- personal choices. On Wed, Feb 9, 2011 at 6:48 PM, Stephen A. Lawrence <[email protected]> wrote: > I'm getting really tired of this. > > Peter, you didn't read, or didn't understand, what I wrote. > > You don't seem to understand the fundamental point, which is that the rate > of boil-off is being determined by the pump, with no feedback from the > reactor. The flow rate is fixed and 100% of the water is boiled to steam. > > If the reactor were generating 10% more power than needed to exactly boil > off the water, just where do you think that excess power would go? > > > > On 02/09/2011 11:02 AM, Peter Gluck wrote: > > Jed is right, it is an open system and even if the surface of heating is at > > 300 C, the time of contact is short and the steam cannot be overheated > much. > > On Wed, Feb 9, 2011 at 5:50 PM, Stephen A. Lawrence <[email protected]>wrote: > >> >> >> On 02/09/2011 10:22 AM, Jed Rothwell wrote: >> >> Stephen A. Lawrence <[email protected]> wrote: >> >> >>> The energy produced was apparently *exactly* what was needed to boil away >>> the input water -- no more, no less. >>> >>> And *that* is strange. >>> >> >> Nope. That's steam at 1 atm. It never gets any hotter than just above >> boiling. >> >> >> NO. Jed, I can't believe you're making this mistake! >> >> That's *exactly* like saying oxygen can't get any hotter than -183C (its >> boiling point) unless you raise the pressure above 1 atmosphere! >> >> There is nothing magic about water vapor -- it's just another gas, and it >> can exist at 1 atmosphere at any temperature above its boiling point. >> Increase its temperature while holding the pressure steady, and its density >> drops, that's all. >> >> Now, if you boil water in an *open* boiler with a *submerged* heating >> element, the temperature of the steam will never go above 100C (give or take >> a degree). The temperature of the steam in that case is pegged to the >> temperature of the water through which it must pass, and the temperature of >> the water is fixed at boiling, unless you close the boiler and raise the >> pressure. >> >> But in this case the heating element (the walls of the tube) is only >> submerged until the water boils. After that, the steam is in direct contact >> with the heating element, and no longer in close contact with liquid water, >> and there is nothing to keep its temperature from rising well above boiling. >> >> The geometry of the water jacket may be more complex than a simple tube >> but the same argument applies: Once the water has boiled away and the inner >> wall of the water jacket is in direct contact with the steam, the steam >> temperature is no longer fixed at boiling. >> >> >> >> >>> It comes out faster with more enthalpy if the pump adds more energy to >>> it. >>> >>> >>> THAT'S THE POINT! >>> >>> If the reactor produced even a few hundred watts more than what was >>> needed to vaporize the water, the temperature of the steam would have been >>> substantially higher than boiling. >>> >> >> Nope. It would just move faster out of the end of the hose, as I said. >> You have to raise the pressure to make the temperature go up. >> >> >> Sorry, that is completely wrong. >> >> Look, if it's moving faster out of the end of the hose, but it's the same >> number of moles of steam (which it *must* be, because the pumping rate is >> fixed), then the steam must be more "spread out", right? It must be taking >> up more volume per mole. Volume coming out is the integral of the flow >> rate, flow rate is the speed of the steam times the area of the hose >> opening; ergo, if it's going faster, you've got a larger volume coming out. >> >> Pressure is fixed, number of moles are fixed, and the volume has >> increased. What's that tell us? >> >> PV = nRT; let's solve for T. >> >> T = PV/nR >> >> 'n' is fixed, 'R' is a constant, 'P' is fixed, 'V' has increased -- so the >> temperature has also increased. >> >> QED. >> >> >> >> >> - Jed >> >> >
