Jed, if you review your figure 8 for the Oct 21 test that I am concentrating upon you will see a plot of the reactor internal temperature. It appears pretty obvious that the internal temperature is quite high for a short period of time and not during anywhere near the complete cycle. Also, the reactor wall temperatures are very restricted in time when compared to the drive pulse dead time period.
I have a difficult time accepting the premise that the power is constantly being generated during the complete period from this figure. It is much more likely to be restricted to .3 hours maximum. Have you given this figure much thought? I do not expect the anonymous heat to be proportional to input power in any linear fashion. Also, the time domain emission of that heat will not match the input. My model does not really care about the exact shape of the input pulse at this point, only the number of joules emitted. Thanks for smoothing out the data for me. What I see looks fairly clean. I realize that there remains a major difference in opinion between you and Gigi concerning the pump heating. I want to remain out of that argument but need the best proven information to use for my model. He has done extensive curve fitting and I have asked him to prove his case better. Jed, the system time constant is a bit less than 6 hours. That means that it takes several of these periods before an input no longer effects the final temperature. What do you suppose will happen if the ambient takes a step upwards? It will be many hours before the transient finally settles out. The same will happen for a step in pump power as well. I agree completely with you that if the ambient is completely stable and the pump power constant then any change in the coolant temperature is directly determined by the joules added by the drive power pulse and the excess power waveform. Unfortunately, that condition has not been met since the ambient is changing constantly. Also, since there is thermal leakage from the system, the coolant temperature will slowly fall with the system time constant determining the rate. In this case the time constant is 5.84 hours according to Gigi's excellent analysis. We obviously do not agree in several important ways, but that should not be a reason to prevent us from working as a team in order to prove that the Mizuno system is generating excess power. So far the indications are very positive, but I aim for the best possible proof. We are getting close to that goal. You will appreciate the end product of this exercise. Dave -----Original Message----- From: Jed Rothwell <jedrothw...@gmail.com> To: vortex-l <vortex-l@eskimo.com> Sent: Fri, Jan 16, 2015 8:34 pm Subject: Re: [Vo]:Jed's Results Look Good So Far David Roberson <dlrober...@aol.com> wrote: If you have a method of determining the actual shape in time of the energy being released by the active LENR wire please give me that information. I can not imagine it being only 3 watts instantaneous as you seem to be implying. As far as I can tell it is. It is hard to separate out the spike from the pulse and the anomalous heat but I do not see any indication that the anomalous heat is much higher at one point than another. It gradually fades away at the end. As I said in the report, you cannot tell when exactly it stops; you can only see that losses exceed anomalous heat sometime in the evening. When you download the spreadsheets you have as much information as I do. That would suggest that the same number of joules in each second are constantly being generated throughout the pulse repetition period. I would certainly think that as the structure cools down, less heat is generated. There is no anomalous heat effect proportional to input power. If it were proportional, I would suspect an instrument artifact. Cold fusion does not work like that. It is triggered by some sort of stimulation -- such as heat or laser light -- and then it goes off on its own, at its own pace, at a power level completely unrelated to the stimulus power. And, if past experience is a guide then the rate of production falls off very quickly as the temperature drops. I do not think so. Pd-D systems producing heat after death continue for a long time as they cool down. To recap, I am currently using your data from Oct. 21, 2014 to model the amount of heat generated by Mizuno's device during that day. The spreadsheet shows temperature changes every 24.4 seconds. There is a large lag between the water temperature and the reactor body temperature, and on the scale of individual readings there is a lot of noise. Despite these problems, you can derive power in watts by multiplying the temperature difference from one reading to the previous one, and multiplying that number by 4.12 (conversion factor) * 4000 g (of water) / 24.4 s. This results is extremely noisy so I suggest you data smooth it with the spreadsheet function, making it an average over several minutes. You will see that it never gets anywhere near 40 W, and most of the time it is far lower than 4 W -- which you think is the pump input power. As I said, I cannot imagine where you got that number. As I have often pointed out, the heat from the pump has already driven the system up by 0.6 deg C after the first 90 minutes, and can drive it no higher. So you will not see any sign of the pump in the minute-to-minute changes after that. When there is only the pump running and ambient is stable, the Delta T change over 24 s is zero, and the Delta T change over 24 minutes is zero, and if ambient were stable then you would find that over 24 hours, days or weeks it is . . . zero, exactly. This is dead obvious from the data. Why Gigi does not understand it I cannot say. He clearly does not understand calorimetry. If you think you are detecting 4 W from the pump -- or 4 W from anything for that matter -- you do not understand this either. If you do not data smooth the segments you will find huge power changes, up 100 W in one reading and down 80 W in the next. This is noise. It is caused by the instrument electronics and by water swirling around in the tank. You can simulate it at home with a thermocouple in a pot of water over a stove, for example. I averaged out power over 10 segments of 24 s each, graphed that, and then computed total energy from those instantaneous power measurements. The answer is about the same as taking energy from the total change in temperature for the whole day, which is the method I use. The second answer, from the 10-segment average power, is slightly higher. It is probably more accurate, because it captures heat before the heat "leaks" from the reactor. The difference is minor because the insulation is good and the temperature difference from ambient is small. - Jed
