>From the report:"The impression was that the loss of heating power was minor. >Consequently the heat produced by the E-cat in self sustained mode should have >been clearly larger than the heat from the power that was lost when the >electric resistance was switched off. " What a crock! A minor loss of heating >power is exactly what one expects from thermal inertia. There is no anomalous >heat.
Also since 1.8 grams were collected as overflow and only ~3 grams flowed in we have 1!.2 grams at most converted to steam. This means about 2700W. That's close enough for me to the 2600W input. ----- Original Message ----- From: Finlay MacNab To: [email protected] Sent: Wednesday, September 14, 2011 8:49 PM Subject: RE: [Vo]:E-cat news at Nyteknik Excellent observation! If this was a closed system with no FLOWING WATER EXITING THE SYSTEM you would have a point. As it is you have only discredited your argument about thermal inertia. Congratulations! I find your hand waving arguments completely unconvincing. Please describe in detail the geometry of the system you propose could account for the observed changes in temperature taking into account the well known rate of heat exchange between water and metals/other materials and the heat capacities of the various materials. Also, please account for the energy inputs and outputs to the device during its operation. 5 minutes with a text book will convince anyone with half a brain that what you describe is more improbable than cold fusion itself! Please do everyone here a favor and give a rigorous explanation of how "thermal inertia" can explain the rossi device. Please use equations and data to back up your claims. If you don't want to do this please stop spamming this message board and distracting from more interesting discussion. ------------------------------------------------------------------------------ Well, at a setting of 9 you have the same temp rise in 35 minutes as temperature fall in 35 minutes after power-off. ----- Original Message ----- From: Mark Iverson-ZeroPoint To: [email protected] Sent: Wednesday, September 14, 2011 4:55 PM Subject: RE: [Vo]:E-cat news at Nyteknik JC stated: “(and note that this takes considerable time in the ramp up)” Where he is referring to the long time it takes to ramp up the E-Cat’s internal temperature on startup… Mr. Catania, do you realize that the electrical power into the E-Cat’s resistance heater was NOT started at 100%, it was started at a setting of ‘5’ and RAMPED UP slowly over 40 minutes! Here is the time progression for resistance heater power… Timestamp PLC Setting DeltaTime (minutes) --------- ----------- ---------- 18:59 5 0 19:10 6 11 19:20 7 10 19:30 8 10 19:40 9 10 We know that the ‘Setting’ is referring to the duty cycle, but we do not know exactly what the relationship is… since 9 is the MAXimum setting, and Lewan states ‘power was at this point constantly switched on’, then a setting of ‘9’ is presumably a 100% duty cycle. (?) Since the PLC’s are programmable, we cannot assume that a setting of ‘5’ is 50% or 60%; it could even be programmed to be 10% duty cycle. So no useful calculations OR conclusions can be made during this ramp-up phase. -Mark From: Joe Catania [mailto:[email protected]] Sent: Wednesday, September 14, 2011 11:58 AM To: [email protected] Subject: Re: [Vo]:E-cat news at Nyteknik I think it caused a rise. There is no rise. Its your imagination. The temperature at power off is too low and must be discarded. If I bring a piece of metal the size of an E-Cat to some temperature (and note that this takes considerable time in the ramp up) and then I cut the power, the temperature will not instantaneously drop. It will stay at the same temperature and decline slowly. There is much too much mass for what your talking about to happen. I have to laugh at the fact that if you saw the temp drop even a hundredth of a degree at power down you would have declared the thermal inertia regime over and the CF regime to have begun.
