It's also worth noting that after the "hiccup" in the curve at about 80C, the curve seems to go up more steeply. Since the water's being heated more at that point, that indicates a significant increase in power generated versus the amount generated at "ignition" (at 60C).
That, in turn, suggests that the generated power continues to increase after ignition. I haven't made any attempt at teasing out any additional numbers from the slopes of the curve. On 04/08/2011 12:43 PM, Stephen A. Lawrence wrote: > > On 04/07/2011 02:10 PM, Jed Rothwell wrote: > >> Essen, H. and S. Kullander, Experimental test of a mini-Rossi device >> at the Leonardocorp, Bologna 29 March 2011., in NyTeknik. 2011. >> >> http://lenr-canr.org/acrobat/EssenHexperiment.pdf >> > Interesting paper; thanks, Jed! > > The temperature graph is particularly interesting, I think. If we > assume the thermal mass of the system remains reasonably constant > throughout the initial heating phase, then, by looking at the curve > slopes, we can form an estimate of how much heat was being generated by > the device up to the point where the temperature levels off at 100C. > > At the start, the water isn't being heated, so all the heater's output > is going into warming the device. So, the slope of the curve at the > beginning corresponds to the warming rate when 300 watts is applied to > *just* heating the device. I've drawn that line in, using Gimp, in red. > > At the point where the effluent is at 60 degrees, it's taking 300 watts > just to heat the flowing water. However, at that point "ignition" takes > place, and another heat source begins to warm the device. Since the > joule heater is taking care of warming the water (at 60C), all the > *additional* heat will go into warming the device. So, the slope of the > curve at that point corresponds to however much heat is being generated > by the reaction, applied solely to warming the device (as, again, the > heater's output is sufficient to warm the water to that temperature). > I've drawn a tangent line to the curve at 60 degrees in blue. > > The ratio of rise/run of the blue line is proportional to the power > heating the device, and by comparing that ratio to the ratio for the red > (300 watt) line, we can figure out the power generated inside the device > at the moment of "ignition". To do that easily, I copied the blue line > and moved it to share an end point with the red line, dropped a "plumb > line" to the axis, and used Gimp's measuring tool to get the relative > tangents of the red and blue line's angles. > > The power generated at "ignition" can thus be seen to be > > (349 / 149) * 300 = 703 watts. > > If the water were *just* heated to 100C and not vaporized, the heat > produced by the device, by the calculations in the paper, would be (95 * > 6.47) - 300 = 315 watts. Thus, according to the slopes in the graph, > there seems to be enough heat generated to heat all the water to 100C, > with about 400 watts left over to boil some amount of it. OTOH the > energy required to vaporize *all* the water isn't evident in the graph, > for whatever reason. > > As I said, interesting. > >
