Dave, I've passed your message to the colleague of mine who is playing with the simulation. Actually we do not have just one unknown parameter (the overall transmittance) but also the behaviour of the heat coming from the motor pump.
As Bob clearly points out As Gigi says some of this energy goes into the water as heat and some to the ambient. However as the ambient gets cooler a larger fraction goes to the ambient, because it becomes a better heat sink relative to the water pathway. Bob is perfectly right. Another complication is the fact that on the reactor vessel the temperature is taken in a few spots so that the thermal capacity concept is not easy to be applied when the reactor is fed bu the power pulses. But we are working on it. So now we are here a few persons reasoning on the thermal behaviour of the calorimeter. It's a pity that Jed is not willing to interact on this matter and preferred to insult me (of course I replied, I'm elderly aged and I got a lot of experience from my life). By the way Jed made a HUGE MISTAKE in the missing file and in his report when using the Newton's law of cooling, the same law he says I don't know. *Actually*, *Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e. the temperature of its surroundings). http://en.wikipedia.org/wiki/Convective_heat_transfer <http://en.wikipedia.org/wiki/Convective_heat_transfer>* You need two conditions to apply it 1) The ambient temperature is stable 2) The cooling body has no internal source of heat Jed meets the first requirement by suitably choosing a time interval in which the ambient is stable (it is said in the missing file) but he's not aware at all of the second condition that is not fulfilled since an internal heat source is delivering heat to the calorimeter [the pump always on]. This fact can be understood quite simply: the law states that if the requirement are satisfied the temperature difference [absolute value] will get smaller and smaller with time [in the real world it vanishes]. If a heat (cool) sorce is present this never happens. A ball with a heater inside will never reach the ambient temperature. The pump is a heater. This is the reason why Jed fails in estimating the pump power: his derived constants are completely wrong. Best regards 2015-01-13 23:59 GMT+01:00 David Roberson <[email protected]>: > Gigi, > > I just recalculated the combined thermal time constants and now I believe > you have them right. I must have performed that calculation 5 times and > kept getting a different answer! The thermal K's that you used are > inverted from the normal R's(resistors) that I always use when calculating > time constants. Since one value is so much larger than the other the > inverses are vastly different. > > Please check yourself to ensure that my latest figure is correct. I > finally get a thermal time constant of 5.84 hours. There is little doubt > that the power pulses and any resulting LENR power will influence the > output with that large of a time constant. This is particularly true since > you begin your analysis only about 2 hours after the last pulse. > > In the case of the dead pump, the problem is multiplied by the extreme > time constant of the Dewar which is 36.11 hours according to my latest > calculation using your values. When the pump stops, most of the energy > contained within the Dewar is locked in place. The same is not true for > the body of the power vessel which only has a time constant of 4.2 hours. > > Dave > > > > -----Original Message----- > From: David Roberson <[email protected]> > To: vortex-l <[email protected]> > Sent: Tue, Jan 13, 2015 4:46 pm > Subject: Re: [Vo]:"Report on Mizuno's Adiabatic Calorimetry" revised > > Take your time Gigi, we want to get to the facts. I am very impressed > by the simulations that you have shown and how well they match the curves > made by Jed. Now, we need to verify that things add up as they should by > combining thermal resistance and capacities of two parts to get to the > whole. > > They must combine according to normal physical laws. My first attempt > using your models did not seem to match properly. That is what I want you > to show. > > And, if the thermal time constant is large, then average power due to the > test itself will show up. As you know, you are assuming that there is > nothing except for the input drive signal of 20 watts and the pump leakage > power. I want to see how those parameters impact your results. Then, we > need to determine whether or not we can match the curves of Jed with a > actual input signal due to LENR using your model. > > I also still believe that the pump power is less than you are > considering. Nevertheless, I will keep an open mind and give your model an > opportunity to show its strength. > > Thanks, > > Dave > > > > -----Original Message----- > From: Gigi DiMarco <[email protected]> > To: vortex-l <[email protected]> > Sent: Tue, Jan 13, 2015 4:35 pm > Subject: Re: [Vo]:"Report on Mizuno's Adiabatic Calorimetry" revised > > Ok, we need some time to perform the full set of simulations. But please > do not take for sure that the pump only test had exactly the same > configuration than the test run had. We will present them as soon as we are > confident that all the problems have been settled. > > 2015-01-13 22:18 GMT+01:00 David Roberson <[email protected]>: > >> 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 >> >> >
