Robert Leguillon <robert.leguil...@hotmail.com> wrote:
> Alright, I'll go over this again: Please, listen. > The E-Cat is not an 8 gallon pot that you take off a heater. The heater is > INSIDE. > When the E-Cat is turned off, there is still a hot iron core at 600 - > 1000C, depending on who you listen to. It has lead shielding. It is not possible to heat the iron to this temperature with these heaters. We know that most of the metal was nowhere near that hot, because the reactor was not incandescent. Most materials become incandescent above 525 deg C. However, I said add two gallons of water if you like. That has more thermal energy in it than hot iron at 1000C, because the specific heat of water is 10 times larger than iron. > If there are adjacent cats, they will also heat up, storing additional > heat. Depending on the description, this may or may not be in the bottom; it > may or may not be connected to the heat exchanger. > The heat exchanger plays no role in how quickly the reactor will cool. > If it slowly releases heat, the temperature in the E-Cat will be pegged at > boiling, which varies based on back pressure. If it is releasing enough > energy to boil .5 g/s at 117C at 1.5 ATM of pressure, and some obstruction > causes the pressure to increase to 1.51 ATM of pressure, the boiling > temperature could increase to 118C, but the amount of water boiled would > decrease to .48 g/s without violating any law of physics. You are missing the point. Whether the rate changes or not, after 4 hours IT WOULD BE COOL. The metal would no longer be so hot it burns you. The surface of the reactor would not be as hot as it was in the beginning. If it is boiling away less water than the paristaltic pump is putting in, > water could overflow. That has no bearing on the fact that IT WOULD BE COOL. It must cool down. > Overflowing water would necessarily be at an increased g/s of H2O, causing > a temperature spike in the thermocouple. The thermocouples have nothing to do with this. Please forget about them. Concentrate on what happens when you leave an insulated metal container of water in room temperature air, and you gradually replace all the water with tap water, twice. IT COOLS DOWN. It must cool down. > Sloshing or percolating water would make these spikes more temporary and > erratic. The thermocouples could be taken out of the picture. They could be taken to Mars. That would not affect the behavior. It would not prevent the thing from cooling down. > This explains bumps in both sets of thermocouples without violating laws of > physics. > But it does not explain how you can add 60 L of tap water to a 30 L container and still have it be as hot as when you started. > Your 8 gallon pot doesn't have an internal source of heat. Your proposal of > "adding tap water" doesn't work, because we have no idea what the input > water flow actually was. Yes, we do know. We can hear the pump running. But for the sake of argument, suppose the flow rate was zero. No water was added. The reactor was just left there on the table. The flow rate as measured by Lewan magically emptied it out and yet there was water left in it. Okay, even if all you do is leave it alone, IT HAS TO COOL DOWN. The surface is hot; it is radiating heat. It cannot get hotter. It cannot stay that hot. An object of this size made of insulated metal filled with water must cool down. You can confirm with the pot of water too; just don't replace 4 gallons per hour. Furthermore, if there is no water coming in, then the only way it could reach the heat exchanger would be by boiling. The whole 30 kg of water was vaporized in that case. Do you seriously think that hot iron can hold 67,800 kJ of heat? How much iron? What temperature would it be? How do you get it that hot with ordinary 3 kW electric heaters? It would be incandescent and it would destroy the heaters. Also, it would cool down according to Newton's law, not at an even pace. Frankly, your hypotheses are full of holes. You should have thought of some of the objections I raised here on your own. Please remember that "skeptical" assertions that call a result into question do not get a free pass. Your arguments must meet the same standard of rigor as anyone else's. You cannot blithely declare the flow rate might have been zero when people observed the pump working and condensed water flowing out of the primary loop at a rate that would have drained the box before the test ended, and when we know there was water left in it at the end. That does not add up. - Jed
