Gigi, I have another issue that you might be able to discuss. You made two independent simulations of the behavior of the Dewar temperature and reactor body over time. In one case the pump was working and in the other if had failed. I took the values that you calculated for the thermal components of your model and get very different results for the combined time constant than what you have shown for the individual ones.
The thermal masses should be added in parallel directly which you seem to have done. I combined the thermal impedances in what I consider the proper manner. When a final calculation of the time constant is computed by using the two others, the number does not come very close to matching what you are using for the first case. Please take time to perform that combination on this forum for us to view and analyze. Also, please turn that answer into an actual hour figure for us. This will be very important as we attempt to understand the impact of residual drive signal and any additional due to LENR activity. Begin with the time constant you calculate in hours. Regards, Dave -----Original Message----- From: David Roberson <[email protected]> To: vortex-l <[email protected]> Sent: Tue, Jan 13, 2015 3:27 pm Subject: Re: [Vo]:"Report on Mizuno's Adiabatic Calorimetry" revised I may have answered my own question below. The drop in ambient acts much like a negative signal as I have proposed before. Eventually the delta will become zero for both signals. Forget the first question and concentrate upon the next one. I notice that in both curves that you use to determine the pump power leakage actual true signal power due to the 20 watt drive and any device LENR power was contributing to the total. The shape of the temperature curves with time clearly show an initial rise during the first few hours that affect your final answer. What have you done to subtract this effect from your determination of the constant average power presumed to be leaking from the pump? It would be useful if you include this heat input into your model and see how that would modify the average power that is coming form the pump. Since you know the shape of this heat signal and you assume that no additional power is added by the LENR effect, it should be easy to model it as an additional heat source. Spice would handle this nicely. Dave -----Original Message----- From: David Roberson <[email protected]> To: vortex-l <[email protected]> Sent: Tue, Jan 13, 2015 2:49 pm Subject: Re: [Vo]:"Report on Mizuno's Adiabatic Calorimetry" revised Dear Gigi, I have begun to analyze your report and find something that does not seem logical according to my understanding of heat flow. On your figure A2 I see that you have overlaid your simulation results upon Jed's figure. The correspondence between the curves is remarkable and you should be commended for your work. The issue that I need to resolve is that the delta temperature between the Dewar and ambient is actually increasing during this time. Also, the delta for the reactor is becoming less with time as I was expecting. In order for the temperature delta to increase you would have to supply some form of heat power to that device. The model that you are using is extremely simple and certainly does not suggest that anything more complex would be happening. How do you explain that the delta is increasing? Is there some process that is supplying extra power into the Dewar once the pump is turned off? Regards, Dave -----Original Message----- From: Jed Rothwell <[email protected]> To: vortex-l <[email protected]> Sent: Tue, Jan 13, 2015 2:06 pm Subject: Re: [Vo]:"Report on Mizuno's Adiabatic Calorimetry" revised Gigi DiMarco <[email protected]> wrote: The refrigerator example is quite evident, but is unfit to our situation, by various causes. The main one is that there you have an abrupt change of air temperature, while in the 18h test the air temperature is falling at a modest rate of 0,36 °C/h that is very simple to follow for the calorimeter. No, it isn't. That is why a gap opens between the room temperature and the calorimeter, and the gap persists until early morning. If from now on the losses are equal to the pump power, since you have Loss = K * deltaT and PumpPower = loss = constant and since K is valid over a broad range of deltaT you should have a constant deltaT. No, it isn't. See Newton's law of cooling. So going back to the plots in the missing file you considered only the first 1.5 hour only because just after the ambient temperature starts decreasing. What have it happened if the ambient did not change for 5-6 hours? Can you answer this question? Yes, I can. If ambient stays stable, the reactor and water temperature will remain stable at 0.6 deg C above room Where in the data do you see that 0.6 °C is the maximum? It goes no higher after 1.4 hours. You can see this in other data sets as well, such as early in the morning with this data set. Whenever ambient remains stable for a few hours or more, the reactor temperature always settles 0.6 deg C warmer. Please don't be contemptuous and dismissive; it is not the case. If someone does't understand calorimetry it is not me. You do not understand Newton's law of cooling and you cannot tell the difference between ambient cooling and heat generation in a cell. In my opinion, you are terribly confused and totally unqualified to do calorimetry. - Jed
