On Sat, Dec 3, 2011 at 2:08 PM, Jed Rothwell <jedrothw...@gmail.com> wrote:
> Mary Yugo <maryyu...@gmail.com> wrote: > >> >> > The total amount supplied during the warm up phase is easily measured. >> It is >> > the total amount of electricity supplied. >> >> Maybe it's easily measured but in fact it wasn't continuously recorded >> in most if not all of Rossi's tests. > > > It has been recorded and it is stable. > You sound like Rossi in the infamous video. And about as convincing. > > > > We know >> > how much heat a body of this size and surface temperature radiates. That >> > alone was much more than the total that went in. >> >> I don't think we have the necessary measurements to calculate heat >> loss by radiation. > > > All you need is the surface area and temperature. See: Stefan-Boltzmann > Law. Since you have some expertise in calorimetry I am surprised you did > not realize this. > Well, no, you need a little more. You also need the emissivity of the aluminum foil, which can vary anywhere from a few per cent to a few times 10%. Presumably, since it was being used for insulation, it was closer to the low end. But we didn't even have the temperature, except from Lewan's guess. How do we know he can tell the difference between aluminum foil at 60C or 40C? Even at 60C, in a room at 30C, that would give blackbody emission of about 500W. For 10% emissivity -- a high estimate -- that's only 50W by radiation. The losses by conduction are going to be larger. 20 kg of fire brick at 500C, or 10 kg at 1000C will provide a kW for 3 hours. So, you're not even in the ball park. And anyway, if you're suggesting something like a kW of losses (by any means) through that insulation, then, inside Rossi's megacat, there would have been some 50 kW thermal energy. Even with the doors open, that would have been one hot sauna. > > >> >> With the obvious insulation, I doubt it was >> anything like the total that went in. >> > > Insulation cannot prevent heat from coming out of the reactor. It can only > slow it down. Since the outer surface was nearly as hot as the inside, this > insulation hardly delayed it at all. You can see that it would cool down in > 40 min. > How do you know the temperature inside? Rossi claims hundreds of degrees in the core. The input energy could have heated some thermal mass to that level. You can clearly see that it would not cool down in 40 min. As matter of fact, after the experiment, when the inside could have been much closer to the boiling point of water, and Rossi more than doubles the coolant rate, it still took 50 minutes to reduce by only about 10C in temperature. > > > >> It is obvious that the thing would cool down to >> > room temperature in ~40 min. >> > >> >> I don't know about that. Again, Rossi refused the methods needed to >> confirm your assertion as he did any definitive testing. >> > > You do know about that. The decay curve is clearly shown in the graph, > several times. That was definitive. The method is simple: cut the power and > watch the temperature decline. > You keep using that word "definitive". I don't think you know what it means. When the power was cut, and the coolant flow more than doubled, it dropped 10C in 50 minutes. How is that cooling to room temperature in 40 minutes? And anyway, you yourself said he doesn't want to convince everyone. Something that's definitive is convincing to everyone. > We know the upper limit for the thermocouple error. It was 0.1 deg C by >> the >> > only serious analysis -- but even if it was more it would not be enough >> to >> > negate this conclusion. >> >> If by thermocouple error you include thermocouple placement, it could >> be a lot more than that. > > > No, it would not. See: > > > http://lenr-canr.org/RossiData/Houkes%20Oct%206%20Calculation%20of%20influence%20of%20Tin%20on%20Tout.xlsx > > Unless you can show a problem here, Houkes wins this debate. Anyway, as I > pointed out, two different methods were used to measure the heat output, > and they were in reasonable agreement, so there is no large problem with > the thermocouples. > Who appointed you judge in this debate? Rossi could have easily measured the temperature of the water by putting a probe in the water and avoided any dispute. That he didn't means he loses this debate. The other method suffers from the usual problem of not knowing the steam quality. The best the evidence shows is about a kW output, and that is miles from evidence of a nuclear reaction. If a demonstration claiming nuclear energy is so pitiful that losses due to radiation or storage in fire bricks are in the same order of magnitude, then it's a pretty good bet it's not nuclear energy. He's starting out with a factor of a million or so in energy density.That should be easy to demonstrate. Quibbling about losses due to radiation, or energy stored in fire in a demonstration of *nuclear* energy is a sure sign
