Re: [Vo]:Alternate Calculation and Calibration Method for Mizuno Report
This is really good news Jed. Now the calorimeter seems to be working in the right way [stable ambient and enough time, as we said at the beginning of the discussion]. From your short data presented I suppose it took 7 hours for the cooling liquid to reach a final stable average temperature. From now on, the power transferred from the pump to the water exactly compensates the losses to the ambient. Some data are, however, missing in your announcement: what is the value of the ambient temperature in the last 28 hours? Moreover, is it possible to get data for the transient part, I mean the first 7 hours. 2015-01-30 20:32 GMT+01:00 Jed Rothwell jedrothw...@gmail.com: And . . . Here is definitive proof of what I say. It happens that Mizuno sent me a large data set which includes about 35 hours where nothing was happening. He left the computer running, and he left the heating on so the ambient temperature was reasonably stable. I see it was especially stable during the last 28 hours. The pump was also running all this time. Here are the cooling water temperatures for the first 14 hours and the second 14 hours: First 14: Average 17.85°C, min 17.64°C, max 18.08°C Last 14: Average 17.81°C, min 17.65°C, max 18.10°C There is no difference. The pump did not change the temperature. Using the adiabatic method of calorimetry we see *zero heat* in this data set. There is NO INCREASE in the water temperature even though the pump is adding heat the whole time. It has reached the terminal temperature for the pump input. Really, people should stop debating this. The pump cannot possibly affect this method of calorimetry. - Jed
Re: [Vo]:Jed's Results Look Good So Far
Great Your results look very meaningful, Dave. We are still thinking about the problem of not uniform temperature across the reactor vessel just after the power pulse are applied, but it seems to me that you have solved them. I hope to read the full report very soon. GG 2015-01-16 5:29 GMT+01:00 David Roberson dlrober...@aol.com: I ran out of patience waiting for an input from Gigi and decided to construct a simple numerical model of the calorimeter used with the Mizuno test. After playing around with it for a while using the thermal values derived by Gigi and his team, I have come to my first conclusion. It is a bit early and I might find a glaring error somewhere with further analysis but felt it was reasonable to offer an interim report. It is consistent with the model for the device to be generating an output power of 100 watts. I have rounded the value off at this time until further research can pin point it more accurately. Even if 4 watts of leakage is present due to the pump action, the calorimeter appears to be fairly immune to that input provided the ambient does not change more than 2 degrees during the test. Again, this is a first pass result and subject to revision. Dave
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Jed, you continue to be wrong. If you have constant ambient and constant heat source the temperature difference will stay constant. No exponential decrease. Sorry. 2015-01-14 21:16 GMT+01:00 Jed Rothwell jedrothw...@gmail.com: I wrote: 2) The cooling body has no internal source of heat That is not true. As long as the body is hotter than the surroundings and the heat source within it is at constant power the law applies. The coefficient changes. . . . And the intercept is not zero, obviously. - Jed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Jed, all that you say has nothing to do with the Newton's law of cooling. It is by far more complex and it is what we are trying to simulate. With good and promising results I must say. You need the complete Fourier equation. It is time to go to sleep. Best regards. 2015-01-14 22:12 GMT+01:00 Jed Rothwell jedrothw...@gmail.com: Gigi DiMarco gdmgdms...@gmail.com wrote: you continue to be wrong. If you have constant ambient and constant heat source the temperature difference will stay constant. No exponential decrease. Sorry. If the power remains the same for the entire test, that is correct. It reaches the terminal temperature, and it does not not fall. If ambient remains steady, so does the reactor temperature. If ambient rises or falls, the reactor temperature follows with a long lag. On the other hand, if you reduce power, the temperature declines. That is what you see in Mizuno's data after the heat pulses and after anomalous heat fades away. It falls exponentially. The temperature gradually falls back down to within ~0.6 deg C of ambient (which is a moving target when the room is cooling off). It always reaches that temperature by the next morning. That is convenient for Mizuno, because it lets him start a new test every day. If the insulation were better, he would have to have active cooling to bring the reactor back down to the starting point. Or he would have to start at an elevated temperature, which would make comparing tests complicated. - Jed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
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 dlrober...@aol.com: 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 dlrober...@aol.com To: vortex-l vortex-l@eskimo.com 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 gdmgdms...@gmail.com To: vortex-l
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Excuse me Jed, but I think that is very simple for you to say that I do not understand calorimetry if you reply to a question that I did not ask. 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. But the main question is that in your report you made a statement that is not true. This is you: *The temperature rose for 1.5 hours until it stabilized 0.6°C above room temperature (Fig. 19.) It stabilized because heat losses equal the power from the pump.* If from now on the losses are equal to the pump power, since you have Loss = K * deltaTand PumpPower = loss = constant and since K is valid over a broad range of deltaT you should have a constant deltaT. Instead, deltaT keeps increasing for a few hours when considering the missing file. I fully agree with your words, howewer. So in the case we both do not understand anything of calorimetry. 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? Where in the data do you see that 0.6 °C is the maximum? The true is that you simply stop there and are happy with your data, but there is no theoretycal reason. Please don't be contemptuous and dismissive; it is not the case. If someone does't understand calorimetry it is not me. By the way we performed a full simulation of this calibration: we got exactly the same curve but at a much higher pump power. We shall show you and Mizuno the results, hopefully at ICCF-19. Will you be there? Regards 2015-01-13 16:52 GMT+01:00 Jed Rothwell jedrothw...@gmail.com: Gigi DiMarco gdmgdms...@gmail.com wrote: I could say that this is false but I will be fair and I will say that this is not true. From the missing file (Mizuno's data) we get the following situation for the difference between water and ambient temperature (4h 2.5°C) (5h 2.9°C) (6h 3.1°C) (7h 3.2°C) (8h 3.2°C) (9h 3.2°C) (10h 3.1°C) (11h 3.1°C) (12h 3.1°C) (13h 3.0°C) (14h 2.9°C) (15h 2.8°C) (16h 2.7°C) (17h 2.6°C) (18h 2.5°C) were the temperatures are taken at the beginning of the hour. How do you explain this? The ambient temperature is falling. The reactor is well insulated so it takes longer to cool off than the room does. That is all there is to it. You can simulate this easily with the following steps: 1. Fill a glass with warm water. 2. Measure the temperature difference between the water and air. 3. Move the glass to the refrigerator, and measure the difference between the water and the air in the refrigerator. 4. The second temperature difference will be much larger, because the water does not instantly cool off. Does that mean there is a source of heat in the water? No. If you do not understand this, you are not qualified to do calorimetry. I pointed this out before. I will not point it out again, and I will not discuss this with you again. - Jed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Dear Jed, in your report you write: *The temperature rose for 1.5 hours until it stabilized 0.6°Cabove room temperature (Fig. 19.) It stabilized because heat losses equal the power from thepump. In other words, with low input power after 1.5 hours, this system acts as an isoperiboliccalorimeter. Based on the cooling curve after the ambient temperature fell during the night, this0. 6°C difference indicates that the pump delivers only ~0.4 W of heat to the circulating water.More to the point, with this method of adiabatic calorimetry the 0.6°C temperature increaseover ambient is not included in the calculation of excess heat, because the pump is left on all thetime, and it always does the same amount of work, so the temperature is always 0.6°C aboveambient. To be specific, with this method, the starting water temperature is subtracted from theending water temperature, and the starting temperature is already 0.6°C warmer than ambient.With other methods of calorimetry, heat is measured by comparing the reactor temperature toambient. With these methods, heat from the pump has to be subtracted from the total, or it willbe mistaken for excess heat.* I could say that this is false but I will be fair and I will say that this is not true. From the missing file (Mizuno's data) we get the following situation for the difference between water and ambient temperature (4h 2.5°C) (5h 2.9°C) (6h 3.1°C) (7h 3.2°C) (8h 3.2°C) (9h 3.2°C) (10h 3.1°C) (11h 3.1°C) (12h 3.1°C) (13h 3.0°C) (14h 2.9°C) (15h 2.8°C) (16h 2.7°C) (17h 2.6°C) (18h 2.5°C) were the temperatures are taken at the beginning of the hour. How do you explain this? Thanks 2015-01-10 17:50 GMT+01:00 Jed Rothwell jedrothw...@gmail.com: I wrote: Where do you see that? At what hour? At hour 2.2 it reaches the peak. The water temperature is 23.3°C and ambient is 22.8°C. I meant to say: At hour 1.4 it reaches the peak. Taking the value at 2.2 hours, the water temperature is 23.3°C and ambient is 22.8°C. At 2.2 hours the numbers are stable. I refer to Fig. 19 on p. 25 here: http://lenr-canr.org/acrobat/RothwellJreportonmi.pdf - Jed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
*I missed the simulation for some reason. Where can I find that? Sorry if I overlooked it.* In a previous message I gave you a couple of links. In the second link, in the *APPENDIX *you will find the simulation *The fact that you measure 4.5 watts versus a specification of 3 watts maximum suggests that something is wrong with your procedure. How do you explain that difference?* Instead, I think it suggests that you do not read carefully what we write since you seem able to understand what we write The pump absorbs from the grid a given amount of electrical power: for the sake of simplicity let's say 12 W. According to the data sheet 3 W are transformed into mechanical work and, eventually, transformed into heat inside the water. The other 9 W are directly dissipated into heat: part of this heat, as we measured is transferred to the water. It would be hard to say that if the pump wall near the water chamber is at 50°C the heat is not transferred to the water. I do not see any kind of thermal isolator in the disassembled pump. If 1.5 W is transferred to the water everything is OK. *I hope now you can remove the confusion in your mind.* 2015-01-13 0:07 GMT+01:00 David Roberson dlrober...@aol.com: I missed the simulation for some reason. Where can I find that? Sorry if I overlooked it. Do you have data that shows the mass flow rate when a 10 mm tube is attached to the pump output? I assume that a large pipe is on the suction port. You need to attach a full length 10 mm tube to the pump and measure the flow rate and heating as a main step. There are far too many variables associated with operation of the pump with the 5 mm pipe. I have pointed out several problems that need to be addressed. If you do this and also measure the AC power into the pump and then clean up the pump bearings so that the frictional losses are low then that will go a long way toward proving your position. Do you have any method of verifying that the frictional losses are as low as those of the pump used by Mizuno? The fact that you measure 4.5 watts versus a specification of 3 watts maximum suggests that something is wrong with your procedure. How do you explain that difference? Also, the difference between what you measure and what Mizuno and Jed measures may be nothing more than those associated with operation in a different pump pressure range and a damaged pump. These types of questions remain unanswered. Dave -Original Message- From: Gigi DiMarco gdmgdms...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Mon, Jan 12, 2015 5:34 pm Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised Dave, you said nothing about simulations that should be a confirmation of our experiments. But I think that we can do something more: what will convince you that we are right and Mizuno is wrong? Regards 2015-01-12 23:17 GMT+01:00 David Roberson dlrober...@aol.com: Dear Giancarlo, Thanks for publishing your report in English so that many of us that do not speak Italian can understand it. There is no disagreement between the method that I used to calculate the kinetic transport power and what you would have calculated with the same numbers since we used the same basic principles. I relied upon the information from Jed about the mass flow rate of the pump where he stated that Mizuno had told him that it was 8 liters per second. If you match that rate with your 5 mm pipe as you have stated as a plan for replication of Mizuno's experiment then you will obtain my results. I do not have a pump and 16 meters of 10 mm inside diameter tubing before me to determine exactly what flow rate is obtained. It is going to be necessary for you to either obtain a matching pipe or for us to verify exactly what flow rate is being measured by Mizuno before a final answer can be established. Jed apparently believes that the friction within the 16 meter tubing is not sufficient to reduce the unloaded pump fluid flow rate to a value that is anywhere close to the 2.31 liters per minute that you are proposing. In your report, you state that you are matching the performance seen by Mizuno as far as fluid flow rate is concerned but I strongly doubt that this is occurring. If you make additional calculations you will see that the pressure required at the pump output is (10 mm/5 mm)^4 or 16 times as large when achieving the same flow rate for a 5 mm tube as compared to a 10 mm tube. This is a dramatic difference and you find that you quickly run out of head room when using the 5 mm tube for your test. Just this reason alone should be sufficient for you to realize that your replication attempt is failed. And, as further supporting evidence, the pumping power needed to reach the 8 liters per minute flow rate when using a 10 mm tube is only .192 watts which is well within the operational range of the MD-6. We can approach the power required
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
According to Jed the pump is never turned off. So this is the real fact: there is no excess heat, only the pump. Plus the calorimeter external and internal time constant (capacity+resistance) We can overlaid the experimental figures only by using the calorimeter parameter and an estimated pump power [in line with our measurements]. That's all Sorry about that. 2015-01-13 20:49 GMT+01:00 David Roberson dlrober...@aol.com: 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 jedrothw...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Tue, Jan 13, 2015 2:06 pm Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised Gigi DiMarco gdmgdms...@gmail.com 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 * deltaTand 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
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Jed, I think you should study heat transfer. I suggest you the book by Incropera et al. In one comment you say that the loss is equal to the pump power and the system stay constant; in the following comment you do not remember this and start speaking about the Newton's law of cooling. You present your arguments as a priest would present his Bible readings. Jumping from one chapter to the other. Could you tell us why the equation loss = supplied power doesn't hold during the steady state? Have you a good argument or only fatwas? The calorimeter loss equation doesn't imply that the room temperature stay constant. Why do you insist on that? Ask Dave... Maybe you can trust him better than me. Regards 2015-01-13 20:05 GMT+01:00 Jed Rothwell jedrothw...@gmail.com: Gigi DiMarco gdmgdms...@gmail.com 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 * deltaTand 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
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 dlrober...@aol.com: 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 dlrober...@aol.com To: vortex-l vortex-l@eskimo.com 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 dlrober...@aol.com To: vortex-l vortex-l@eskimo.com 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 jedrothw...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Tue, Jan 13, 2015 2:06 pm Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised Gigi DiMarco gdmgdms...@gmail.com 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 * deltaTand 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
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Dave, as promised and while you still insist saying that we were deeply wrong, we have put on-line two different updates 1) https://gsvit.wordpress.com/2015/01/12/further-measurements-on-the-md-6k-n-pump-used-by-tadahiko-mizuno/ 2) https://gsvit.wordpress.com/2014/12/10/analysis-of-jed-rothwells-report-about-his-calorimetry-performed-on-mizunos-cell/ The first one shows how you are terribly wrong with your calculations based on the kinetic energy only. We show that your assumption are completely wrong just referring to usual pump working diagram. In the pump under test you can not have simultaneously maximum head and maximum flow rate; the working point we chose was such that we had almost the same working conditions Mizuno had. Please take your time to read our post before commenting. The major result is that we measured 43°C in the pump body very close to the water so it is really easy to understand that, despite what Jed says, the pump motor delivers a lot of heat to the water; it is this the power we measure and it is by far much more that the mechanical power (3 W maximum from the data sheet). But, let me say that the second link is even more interesting [you have to go to the end of the article, the Appendix]: we set up a software simulation tools and were able to replicate by simulation the Mizuno's measurement. It was enough to evaluate the overall thermal transmittance of the system that is constant at least for the considered temperature range. If we simulate the Mizuno's curve starting from a time instant when the reactor is no more generating excess heat, it is possible to evaluate the only source of heat: the pump. We have to use only the room temperature as provided by Mizuno's data and the system starting temperature. The pump power turns out to be about 4 W. So we get comparable results by using very different methods 1) Pump theory and data sheet 2) Experiment 3) Simulations All the rest are only free words. We are going to apply the simulation to all the Mizuno's experiments to see if we can get those curves without any excess heat. Regards and take it easy. Please, consider to read all the articles in our site concerning the Mizuno's experiment. Gigi aka Giancarlo 2015-01-12 19:09 GMT+01:00 David Roberson dlrober...@aol.com: Bob, You have uncovered a pump specification that proves that the replication work by Gigi and allies is not accurate. They report to have determined that approximately 4.5 watts of thermal power is being absorbed by the circulating water under their test condition. This amount of reported power is clearly more than the pump should add and they need to explain why we should accept their data as accurate. Also, I have performs extensive calculations within a spreadsheet that is based upon the lift head versus fluid flow rate of this model pump. It is capable of delivering less than 1 watt of fluid power into the water coolant under the best of conditions. My actual calculation is .75 watts at 6 liters per minute which I rounded off for convenience to 1 watt. I included both potential as well as kinetic energy related powers. Any additional power imparted to the water must come from pump friction and thermal leakage through the construction materials. Without further careful measurements we or Gigi can not assume that the pump used by Mizuno is operating at its specification limit of 3 watts. Of course the measurement of 4.5 watts by Gigi is certainly not representative of a pump that is in good condition. The pump manual has several warnings about how easy it is to damage it and that strongly suggests that Gigi and his team has done just that in order to obtain their non representative performance. No one but Mizuno knows the status of his pump during those tests so the only conclusion that can conservatively be drawn is that the skeptical report by Gigi and team should not be considered valid. The pump manual states that the water reservoir must be at least 1 foot above the pump input port in order to prevent possible air intake along with the coolant water. Operation under conditions that do not meet this requirement can damage the pump according to the manual. Unfortunately, in both of the cases being discussed this was not done. The setup used by Gigi very clearly shows the pump mounted above the Dewar by several inches. The same appears true for Mizuno's experiment. Dave -Original Message- From: Bob Cook frobertc...@hotmail.com To: vortex-l vortex-l@eskimo.com Sent: Mon, Jan 12, 2015 12:15 pm Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised Jed-- I have researched the pump characteristics further and find that this pump has a low efficiency and would use at most about 3 watts of power in heating the circulating water. This is consistent with what you have stated. I am not sure how Mizuno measured the 10.8 Watts of power used by the pump. I think
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Jeff, I could agree entirely with you. I've have some problems with the internal and external calorimeter time constants that are too short. But let's go on and assume that what you say is completely right. Now can you tell me where in the Mizuno's results (excel files and figures) you see this behaviour? I do not see it, so if you tell me which is the right curve we can discuss about it. 2015-01-12 22:58 GMT+01:00 Jeff Driscoll jef...@gmail.com: Jed is correct, when the pump is turned on and everything reaches steady state, (using his example) the pump is putting in 4 watts of power to the tubing, the reservoir and the LENR chamber and all these tubes and the LENR chamber emit 4 watts of thermal power to the ambient at steady state. Then when the LENR experiment is turned on, any delta T can be attributed to the LENR device, not the pump (assuming the pump doesn't change speed). On Mon, Jan 12, 2015 at 4:10 PM, Jed Rothwell jedrothw...@gmail.com wrote: Gigi DiMarco gdmgdms...@gmail.com wrote: The major result is that we measured 43°C in the pump body very close to the water so it is really easy to understand that, despite what Jed says, the pump motor delivers a lot of heat to the water . . . You are wrong. This is not what I say. This is what Fig. 19 proves. If your graphs show something else, your experiment is different. Perhaps you are using a different kind of pump, or more pressure in the tubes, or perhaps you have confused the effects of falling ambient temperature with rising water temperature, as you did before. In the second paper you wrote: GSVIT-1) We do not agree at all. The pump was not stopped during the test and, as Rothwell says, we are speaking about a differential temperature increase equal to +2.5°C. . . . No one said the pump is stopped during the test. It runs all the time. If it were stopped, the test would fail because the heat from the reactor would no longer be collected. The pump power turns out to be about 4 W. Suppose, for the sake of argument, that is true. And suppose that raises the temperature by about 6°C. (Obviously that cannot be true because nowhere do we see a 6°C elevation above ambient, but let us pretend it is true.) In that case, all of the excess heat calculations must begin at a baseline 6°C above ambient, because the pump is always left on. Therefore this has absolutely no impact on the excess heat measurement. - Jed -- Jeff Driscoll 617-290-1998
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Dave, you said nothing about simulations that should be a confirmation of our experiments. But I think that we can do something more: what will convince you that we are right and Mizuno is wrong? Regards 2015-01-12 23:17 GMT+01:00 David Roberson dlrober...@aol.com: Dear Giancarlo, Thanks for publishing your report in English so that many of us that do not speak Italian can understand it. There is no disagreement between the method that I used to calculate the kinetic transport power and what you would have calculated with the same numbers since we used the same basic principles. I relied upon the information from Jed about the mass flow rate of the pump where he stated that Mizuno had told him that it was 8 liters per second. If you match that rate with your 5 mm pipe as you have stated as a plan for replication of Mizuno's experiment then you will obtain my results. I do not have a pump and 16 meters of 10 mm inside diameter tubing before me to determine exactly what flow rate is obtained. It is going to be necessary for you to either obtain a matching pipe or for us to verify exactly what flow rate is being measured by Mizuno before a final answer can be established. Jed apparently believes that the friction within the 16 meter tubing is not sufficient to reduce the unloaded pump fluid flow rate to a value that is anywhere close to the 2.31 liters per minute that you are proposing. In your report, you state that you are matching the performance seen by Mizuno as far as fluid flow rate is concerned but I strongly doubt that this is occurring. If you make additional calculations you will see that the pressure required at the pump output is (10 mm/5 mm)^4 or 16 times as large when achieving the same flow rate for a 5 mm tube as compared to a 10 mm tube. This is a dramatic difference and you find that you quickly run out of head room when using the 5 mm tube for your test. Just this reason alone should be sufficient for you to realize that your replication attempt is failed. And, as further supporting evidence, the pumping power needed to reach the 8 liters per minute flow rate when using a 10 mm tube is only .192 watts which is well within the operational range of the MD-6. We can approach the power required to match Mizuno's flow rate from another direction if you wish. The mathematics implies that the power required to drive a certain ratio of flow rates varies as that ratio to the third power. In your case that means (8/2.31)^3 or 41.53 times less than to reach 8 liters per minute. To take your example: 41.53 * .074 watts = 3.07 watts. (your numbers). So again, you would need to have 3.07 watts of pumping power delivered to the water stream in order to reach 8 liters per minute of mass flow rate just as I have shown. Giancarlo, you are the one that must defend your procedure to show that it truly replicates the experiment conducted by Mizuno. I am merely demonstrating why you have failed to do so. Unless you can prove that you are not damaging the operation of the pump in some manner by your technique then you can not expect me or anyone else to take seriously your claim that you have proven that there is no additional power being generated by Mizuno's device. Why are we expected to accept the notion that a pump that is being driven into overload by high pressure operation per your demonstration is not adding significant additional power into the water stream? The forces acting upon the pump are very much increased by your choice of pipe diameter and it does not take much imagination to expect the internal bearings to overload in a manner that generates significant heating as a consequence. I can not say with certainly that your technique is completely without merit, but you are also left with many issue to resolve before you can claim a good reproduction of the cooling system used by Mizuno. And, since you see powers that fail to match those derived from the experiment, it suggests that you are making some major error. If we continue to discuss this subject in additional dept, I believe that we will eventually come to a mutual understanding with respect to your effort. I remain neutral in my acceptance of whether or not excess power is being generated by the Mizuno experiment and I hope that you remain flexible. I await your response to this posting and perhaps we should begin considering additional tests that you can perform to help verify the facts. I like the horizontal flow demonstration that you used to measure the mass flow rate for the 5 mm tubing. Can you do the same with 10 mm as a beginning step? Best Regards, Dave -Original Message- From: Gigi DiMarco gdmgdms...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Mon, Jan 12, 2015 3:44 pm Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised Dave, as promised and while you still insist saying
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
No, never. I'm only an amateur that follows LENR from the outside. My job is different: I run a company working in the railway field (power converters): nothing to do with LENR. 2015-01-12 22:02 GMT+01:00 Daniel Rocha danieldi...@gmail.com: Gigi, were you part of Defkalion Europe? 2015-01-12 18:52 GMT-02:00 Jeff Driscoll jef...@gmail.com: I have not followed this debate closely, but I assume Jed is correct. So Dave, how do you address this statement: The steady state baseline includes the heat from the pump, any diversion from the baseline indicates excess heat. On Mon, Jan 12, 2015 at 3:44 PM, Gigi DiMarco gdmgdms...@gmail.com wrote: Dave, as promised and while you still insist saying that we were deeply wrong, we have put on-line two different updates 1) https://gsvit.wordpress.com/2015/01/12/further-measurements-on-the-md-6k-n-pump-used-by-tadahiko-mizuno/ 2) https://gsvit.wordpress.com/2014/12/10/analysis-of-jed-rothwells-report-about-his-calorimetry-performed-on-mizunos-cell/ The first one shows how you are terribly wrong with your calculations based on the kinetic energy only. We show that your assumption are completely wrong just referring to usual pump working diagram. In the pump under test you can not have simultaneously maximum head and maximum flow rate; the working point we chose was such that we had almost the same working conditions Mizuno had. Please take your time to read our post before commenting. The major result is that we measured 43°C in the pump body very close to the water so it is really easy to understand that, despite what Jed says, the pump motor delivers a lot of heat to the water; it is this the power we measure and it is by far much more that the mechanical power (3 W maximum from the data sheet). But, let me say that the second link is even more interesting [you have to go to the end of the article, the Appendix]: we set up a software simulation tools and were able to replicate by simulation the Mizuno's measurement. It was enough to evaluate the overall thermal transmittance of the system that is constant at least for the considered temperature range. If we simulate the Mizuno's curve starting from a time instant when the reactor is no more generating excess heat, it is possible to evaluate the only source of heat: the pump. We have to use only the room temperature as provided by Mizuno's data and the system starting temperature. The pump power turns out to be about 4 W. So we get comparable results by using very different methods 1) Pump theory and data sheet 2) Experiment 3) Simulations All the rest are only free words. We are going to apply the simulation to all the Mizuno's experiments to see if we can get those curves without any excess heat. Regards and take it easy. Please, consider to read all the articles in our site concerning the Mizuno's experiment. Gigi aka Giancarlo 2015-01-12 19:09 GMT+01:00 David Roberson dlrober...@aol.com: Bob, You have uncovered a pump specification that proves that the replication work by Gigi and allies is not accurate. They report to have determined that approximately 4.5 watts of thermal power is being absorbed by the circulating water under their test condition. This amount of reported power is clearly more than the pump should add and they need to explain why we should accept their data as accurate. Also, I have performs extensive calculations within a spreadsheet that is based upon the lift head versus fluid flow rate of this model pump. It is capable of delivering less than 1 watt of fluid power into the water coolant under the best of conditions. My actual calculation is .75 watts at 6 liters per minute which I rounded off for convenience to 1 watt. I included both potential as well as kinetic energy related powers. Any additional power imparted to the water must come from pump friction and thermal leakage through the construction materials. Without further careful measurements we or Gigi can not assume that the pump used by Mizuno is operating at its specification limit of 3 watts. Of course the measurement of 4.5 watts by Gigi is certainly not representative of a pump that is in good condition. The pump manual has several warnings about how easy it is to damage it and that strongly suggests that Gigi and his team has done just that in order to obtain their non representative performance. No one but Mizuno knows the status of his pump during those tests so the only conclusion that can conservatively be drawn is that the skeptical report by Gigi and team should not be considered valid. The pump manual states that the water reservoir must be at least 1 foot above the pump input port in order to prevent possible air intake along with the coolant water. Operation under conditions that do not meet this requirement can damage the pump according to the manual. Unfortunately, in both of the cases being discussed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Jed, I think you did not catch the importance of time constants in your calorimeter. I do not know how to explain it in more details. You will continue to say no forever. Do you think that simulation are a valid tools as far as they reproduce exactly the experiments? 2015-01-12 22:10 GMT+01:00 Jed Rothwell jedrothw...@gmail.com: Gigi DiMarco gdmgdms...@gmail.com wrote: The major result is that we measured 43°C in the pump body very close to the water so it is really easy to understand that, despite what Jed says, the pump motor delivers a lot of heat to the water . . . You are wrong. This is not what I say. This is what Fig. 19 proves. If your graphs show something else, your experiment is different. Perhaps you are using a different kind of pump, or more pressure in the tubes, or perhaps you have confused the effects of falling ambient temperature with rising water temperature, as you did before. In the second paper you wrote: GSVIT-1) We do not agree at all. The pump was not stopped during the test and, as Rothwell says, we are speaking about a differential temperature increase equal to +2.5°C. . . . No one said the pump is stopped during the test. It runs all the time. If it were stopped, the test would fail because the heat from the reactor would no longer be collected. The pump power turns out to be about 4 W. Suppose, for the sake of argument, that is true. And suppose that raises the temperature by about 6°C. (Obviously that cannot be true because nowhere do we see a 6°C elevation above ambient, but let us pretend it is true.) In that case, all of the excess heat calculations must begin at a baseline 6°C above ambient, because the pump is always left on. Therefore this has absolutely no impact on the excess heat measurement. - Jed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Alain, I'm not accusing anyone of having hidden an excel file; I'm just saying that Jed removed that file from his archives where I found it several weeks ago. I don't know why he removed it, maybe he could explain... Jed says it is of no importance to the present discussion; I find it of paramount importance. I call this, scientific democracy. There is no cospiracy around, but only measurement data: Mizuno's data and ours. Full stop. If you are not able to follow a scientific discussion please feel free to be silent. When we proved that Celani was right with his electrochemical compression at 80 bars, I do not remember you speaking of conspiracy against skeptics... 2015-01-10 10:45 GMT+01:00 Alain Sepeda alain.sep...@gmail.com: it move to an accusation of having hidden an excel file... conspiracy... now my tactic is to force the people denying LENR to be clear out the conspiracy theory they support so witness see it is huge and impossible. conspiracy is the easy answer to things one cannot accept... not only in science 8( 2015-01-09 21:48 GMT+01:00 Jed Rothwell jedrothw...@gmail.com: Bob Cook frobertc...@hotmail.com wrote: I see that we are not communicating accurately. To quote you in response to Alain message regarding this subject several days ago, I will not bother with further communications. I meant I would not discuss the matter over at the Italian web site. - Jed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
critical I observe it, can propose critics. this is a great problem of todays conspiracy theories supported by mainstream, that make real critics hard to separate from pure denial. Edmund storms in his book was clear that deniers are toxic to the discovery of reality as they prevent sane analysis and push circle the wagon or I am tired. this is the symptom of the dead clock right twice a day, which give no information. nothing personal (I always kind when I am personal), this is what I could say to any denier like Lewis or Hansen (take it for them), or our beloved mindguards,... all that is in fact coldly, rationally, a very common situation. Conspiracy theories grow in our societies, and there is nothing exceptional in cold fusion story. There is also conspiracy theories in LENR constellation, that make me laugh, and I would understand than the accused lose some of their flegme. I blame incompetence with high ego, as the main source of that tragedy. People like Lewis, quite competent, but much more vexed and egotic than their great competence, pretended to be sure on what they should not. and less competent people with even more ego and influence amplified that error, until nothing could be step back. and it is tragic that less and more important people became mindguard for that individual failure, transforming the tragedy of 5-10 egotic people into a western academic tragedy. maybe Mizuno made an error, but your demo, the conspiracy theory behind, is even worse than Rossi's demo, and charge of evidence is on your side given the mass of other evidence. sorry to compare Rossi with you. (Rossi will survive that comparison.) 2015-01-10 12:26 GMT+01:00 Gigi DiMarco gdmgdms...@gmail.com: Alain, I'm not accusing anyone of having hidden an excel file; I'm just saying that Jed removed that file from his archives where I found it several weeks ago. I don't know why he removed it, maybe he could explain... Jed says it is of no importance to the present discussion; I find it of paramount importance. I call this, scientific democracy. There is no cospiracy around, but only measurement data: Mizuno's data and ours. Full stop. If you are not able to follow a scientific discussion please feel free to be silent. When we proved that Celani was right with his electrochemical compression at 80 bars, I do not remember you speaking of conspiracy against skeptics... 2015-01-10 10:45 GMT+01:00 Alain Sepeda alain.sep...@gmail.com: it move to an accusation of having hidden an excel file... conspiracy... now my tactic is to force the people denying LENR to be clear out the conspiracy theory they support so witness see it is huge and impossible. conspiracy is the easy answer to things one cannot accept... not only in science 8( 2015-01-09 21:48 GMT+01:00 Jed Rothwell jedrothw...@gmail.com: Bob Cook frobertc...@hotmail.com wrote: I see that we are not communicating accurately. To quote you in response to Alain message regarding this subject several days ago, I will not bother with further communications. I meant I would not discuss the matter over at the Italian web site. - Jed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Jed, I'm sorry but if you take the 18 hour experiment file and draw the water temperature against the room temperature you will find a temperature rise at the equilibrium higher than 2.5 °C. This is a huge amount which, is incompatible with what you and Mizuno say. Your mistake is to think to have a good adiabatic calorimeter whereas you don't. To be good you should have a stable ambient temperature or appropriate time constants. Unfortunately, these concepts do not seem clear to you and you do not care that the external time constant of the test system is less than 6 hours, or about one quarter of the period of variation of the ambient temperature while it should always be considerably higher; even more in your case as the fluctuation of the ambient temperature is very high. I think that anybody here familiar with calorimetry can judge what I'm saying. Take all the Mizuno's measurement and consider the excess temperature of the water against the ambient. If the test run (including power pulses) and the pump run have similar values Mizuno is wrong. Remember that power dissipation is linear with that temperature difference so the ambient is the real baseline. To convince you: start again the experiment with the alleged reaction, but in the same time decrease the room temperature by at least 10 degrees opening the window (it's wintertime); do you really think that you are going to find an increase in the water temperature? If the water temperature decreases shall we have a negative excess heat? Think about it. Truth is the best for anybody. Regards 2015-01-10 16:16 GMT+01:00 Jed Rothwell jedrothw...@gmail.com: Gigi DiMarco gdmgdms...@gmail.com wrote: . . . for example what about the heat transferred from the motor to the water? Jed says it is negligible: we'll show that this is not true, you will see a photo of the pump gear and you will decide yourself. I did not *say* it is negligible; Mizuno *proved* it is negligible, by doing an 18-hour calibration. This is not the kind of issue decide yourself. It is not decided by debate or by appeal to theory. This is the kind of thing you measure and prove by experiment. Once Mizuno proves his point, there is no point to arguing. You could do a million dollar project lasting a year, but you are still wrong. If you find more than a fraction of a watt of heat in the water in your test, that proves your setup -- or your pump -- is not the same as Mizuno's. Questions relating to experimental science must be settled by experiment. Once they are settled, they must be considered closed. We have to move on to other questions. Otherwise no issue will ever be settled; no debate ended; and no progress will be made. It was reasonable to wonder how much heat the pump adds to the water, even though this heat cannot affect the calorimetry or change the conclusion. It was reasonable to wonder, and to ask Mizuno to check. Once he did check, that should have settled the question. The skeptics love to move the goalposts to keep all arguments alive forever. In essence, they are still debating whether hydrogen in palladium can produce 100,000 eV per atom. They move the goalposts down the field, out of the stadium, into the next county. - Jed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Jed, just as an example, in the missing file, in the row 989 which corresponds to 24131.191 seconds the room temperature is 18.78 °C and the water temperature is 21.90. Doing some mathematics we get that the temperature difference is 3.13 °C that appears to be higher than what you say. Why did you choose 1.4 hours? Giancarlo 2015-01-10 17:50 GMT+01:00 Jed Rothwell jedrothw...@gmail.com: I wrote: Where do you see that? At what hour? At hour 2.2 it reaches the peak. The water temperature is 23.3°C and ambient is 22.8°C. I meant to say: At hour 1.4 it reaches the peak. Taking the value at 2.2 hours, the water temperature is 23.3°C and ambient is 22.8°C. At 2.2 hours the numbers are stable. I refer to Fig. 19 on p. 25 here: http://lenr-canr.org/acrobat/RothwellJreportonmi.pdf - Jed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
that you linked do not directly take that into consideration since it becomes a portion of the hydraulic load from what I interpret. The added pressure required to accelerate the fluid is not handled as different than normal frictional loading. I contend that it is in fact a different mechanism and is actually very measurable in this particular case where there is no intentional hydraulic loading. Unfortunately the power lost due to friction inside the pipes is merged with this kinetic energy term. The one thing that is certain is that the heat transported in this manner will be 16 times as much as that transported by the experiment of Mizuno if pipe is used that is 1/2 the diameter and the fluid flow rate and treatment remains equal for both cases. If instead your test system does not treat the circulating water in the same manner then you are not performing a valid comparison for replication. Can we begin a collaboration by agreeing that you are confident that no heat power is not transported by means of kinetic energy of the fluid within this system? We must start somewhere if we are to use physical theory to guide our hand. This seems like a logical way to begin since I have derived equations that suggest you are wrong in this belief. Are you willing to make such a stand? So far I have asked many questions but have received few answers. Theory is important, at least that is what physicists state when they attack cold fusion claims! Forgive me for assuming that you were hiding behind obscure generalities. I was not aware that you were associated with the CSVIT group. I find it odd that you fail to support any theoretical understanding of this system since that would appear the most likely method of getting to the truth. I am willing to offer many theoretical stands that you or anyone among your party are welcome to prove erroneous. So far I have not seen a rebuttal to my equations. I am an electrical engineer as well and have retired from the normal working world in most respects. I hope we can use your experience with radar cooling systems to our advantage as we seek the truth about this issue. Unfortunately, I suspect that the systems that you encounter are of a continuous nature where this particular cause is hidden from view. The cooling fluid will likely deliver heat into the fluid sink tank that originates as a result of acceleration of the coolant by your pumps. Perhaps you have seen where the coolant appears to be hotter than can be accurately attributed to the expected pipe friction when the power amplifiers are shut down? Of course it is possible that all excess heating in an environment of this type is attributed to frictional losses when it is actually more complicated than many realize. Take care and lets uncover the real facts, Dave -Original Message- From: Gigi DiMarco gdmgdms...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Thu, Jan 8, 2015 4:05 pm Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised Dear Dave, I do not think we need so much calculation; better to perform a new measurement on a 10 mm pipe to test you hypotesis. I hate to say that we did it and the power dissipation increases a little bit, as any engineer would have expected: you will find soon the results here https://gsvit.wordpress.com/ I advise you to read the full article as well, so you can find all the theory you need. Please feel fre to ask any questions you like. In case you would like to take a look of the Mizuno 18 hour pump calibration you find here the file that Jed can not find anymore https://dl.dropboxusercontent.com/u/66642475/Mizuno2014-11-20.xlsx in the very first sheet (mio) you can find the water temperature increase against the room temperature coming from Mizuno's data. Take your time to think about it. Jed can confirm that the data are the original ones. By the way regarding your statement *I consider it poor form to hide behind obscure generalities * my name is Giancarlo De Marchis and I belong to the *GSVIT Group;* I thought it was clear, sorry. I'm an electronic engineer and I design water cooling systems [with pumps] for RADARs and high power converters. Normally they works fine. Regards 2015-01-08 19:40 GMT+01:00 David Roberson dlrober...@aol.com: The flow rate is going to be reasonably close to the 9 liters per minute specification from the manufacturer. I have a graph from Iwaki America that shows the expected rate as a function of the lift head facing the pump. At zero meters of head which corresponds to atmospheric pressure the rate is 9 liters per minute. At approximately .6 meters of lift the rate is still about 7 liters per minute. How much do you calculate as the effective head due to friction within the pipe? The experiment that claimed around 4 watts of pump induced power uses a pipe that is 5 mm
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Dear Dave, you still insist on your calculation neglecting what I wrote to you in an earlier message regarding the fact that increasing the pipe the power goes to zero when calculated according to your mathematics. We have just published the new experiment with the theory and diagrams behind it. https://gsvit.wordpress.com/2015/01/10/ulteriori-misure-sulla-pompa-md-6k-n-utilizzata-da-tadahiko-mizuno/ Unfortunately it is only in Italian; you have to wait a bit to have the official English translation I'm not sure to finish it by tomorrow. However, google translate makes a good job. Feel free to make all your comments; I'd rather like on our site so that is very easy for us to reply. 2015-01-10 21:06 GMT+01:00 David Roberson dlrober...@aol.com: Thanks Jed. If the water alone recovers 1.3 watts with average drive drive, and more resides within the vessel, then you are in great shape. If you have the chance, I would greatly appreciate it if you could ask Dr. Mizuno about the measured flow rate. My earlier calculation using 9 liters per minute clearly suggests that the skeptics made a major error by using the 5 mm pipe. As the calculations show, they will find that kinetic energy and thus power transport will be 16 times as much as seen had they used 10 mm pipe assuming the flow rate is constant. As you know I am discussing this aspect of their report and hope to resolve the issue soon. I am confident in my analysis. I have approached the problem from a couple of different directions and keep getting the same result. Dave -Original Message- From: Jed Rothwell jedrothw...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Sat, Jan 10, 2015 2:42 pm Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised David Roberson dlrober...@aol.com wrote: Jed, looking at figure 6, the Oct 21 data I calculate that the average power is 1.3888 watts. That is 20 watts * 500 seconds / 7200 seconds = 1.3888 watts. Yes, that is the answer I got, in Table 1. However, bear in mind that is for the water alone. Not for the reactor, which has a slightly larger thermal mass than the water, and much worse insulation. Estimating that, I get 3.4 W total, on average. Based on a very rough estimate of unaccounted for heat losses and Newton's law of cooling I guess the actual average power is about 7 W. In other words, the reactor metal plus the water are recovering about half of the heat. If Mizuno applies that amount of power continuously what would you expect the temperature to do? With 1.3 W input I expect to see nothing, as I said in the paper on p. 9. That is, in fact, what I saw when I did a similar test. There is too much noise, and the water recovers only about one-fourth of the heat, as I said. So I figure you would have to input ~7 W continuously to see this temperature rise. Mizuno hopes to do that kind of simulation but I do not know when. Actually, now that ambient fluctuations are reduced, you might see 1.3 W in the reactor. That would put ~0.5 W into the water I guess, about twice as much as the pump. It might raise the water temperature by ~1 deg C after an hour or two. It is hard to say. The only way to find out is to do a test and measure it. My gut feeling is that the temperature would increase along a constant slope once the transients are settled down. Well, it increases for a while, but at low power it then soon stops rising as the calorimeter goes from being adiabatic to isoperibolic. That takes 1.4 hours at ~0.2 W. I do not know how long it takes at 0.5 W or 3 W. At any power level it must eventually stop heating, when losses equal input power. Losses increase with the rising temperature, per Newton's law. Also, can you verify that the water flow rate is actually nominally 8 liters per minute? That's what Mizuno said. I suppose he measured it when dumping out the cooling water. He had to change out the Dewar reservoir a couple of times. I think that is what the pump spec. sheet says. There is hardly any resistance, and no grade, so I guess it should be close to maximum performance. - Jed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
*Basically, after 2.4 hours you should either ignore the rest of the data, or use a much more complex modeling method which takes into account the lag.* This is exactly what we have already done. You will read it soon. So please publish the original file so that Dave and other can check our results. We could discuss in detail the matter at ICCF-19. *Many thanks* 2015-01-10 18:28 GMT+01:00 Jed Rothwell jedrothw...@gmail.com: Gigi DiMarco gdmgdms...@gmail.com wrote: just as an example, in the missing file, in the row 989 which corresponds to 24131.191 seconds the room temperature is 18.78 °C and the water temperature is 21.90. Doing some mathematics we get that the temperature difference is 3.13 °C that appears to be higher than what you say. That is because the ambient temperature is falling rapidly. The reactor + Dewar are well insulated so they retain heat for a long time. It takes several hours for them to catch up to ambient. They finally do catch up early in the morning, except they remain slightly warmer because of the heat from the pump. That is why I do not want to use this data. The rapid, large fall in ambient greatly confuses the issue. It has confused you. You have confused a temperature difference from heat with a temperature difference caused by falling ambient. This is why it is important in calorimetry to keep ambient stable and not to try to use data when ambient is changing rapidly and with a big temperature change, either up, or down. That is the main reason I deleted this data. I plan to put back the first several hours only, to prevent this kind of confusion. Actually, I hope to replace the whole spreadsheet with new data when Mizuno improves the heating and airconditioning to eliminate these ambient fluctuations. Why did you choose 1.4 hours? I did not choose 1.4 hours. I meant to say that the temperature rise from the pump heat stabilizes at 1.4 hours (as shown in Fig. 19). I also meant to say that the ambient was fairly stable for the first ~2.4 hours that day, and we should only be looking at those first 2.4 hours. That is why I cut off the graph. When ambient starts to fall, the reactor and Dewar lag behind, and it becomes impossible to see the effect of the pump heat or to derive the constant for Newton's law of cooling. Basically, after 2.4 hours you should either ignore the rest of the data, or use a much more complex modeling method which takes into account the lag.
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
*What did Hoistadt say in response to your letter regarding this question? * That they do not have time to discuss on the blogsphere about their paper that has appeared only in the blogsphere. So I asked him to whistle to me when they get the paper published on Nature or Science. Giancarlo PS I agree with most of your technical consideration but it is time to go to bed here now. Regards 2015-01-10 19:07 GMT+01:00 Bob Cook frobertc...@hotmail.com: Gigi, Dave and Alain-- You, Gigi, wrote to Alain: I, personally, do not think that LENR are real but we are speaking about some specific experiments: it took 15 minute to me to understand that electrical power measurements were wrong in the TPR2. It's my job, I design and build power electonics and usually I use smart methods to measure power. In the TPR2 there was a hidden wattmeter; I simply found it and I wrote to Hoistadt. Rossi was very aggressive against me, whereas normally he lets people say whatever they want; so I'm sure I was right. What is the evidence for the hidden wattmeter you, Gigi, say you found? Were there data that it the hidden wattmeter provided that you know about, and if so, what were those data? Your professed understanding of this should be laid out. What did Hoistadt say in response to your letter regarding this question? Incidentally, I have a similar question to yours regarding the Mizuno test as you have implied in the following comment: It is written in the Mizuno's data, our demo is only a further proof. If you take a look of the data when the pump fails you will see that immediately both water and reactor wall temperatures start to decrease: in the presence of a reaction the wall temperature should have increased. Jed and Mizuno perform an experiment without hydrogen: the result is the same they got with hydrogen. The conclusion is that there is still some residual hydrogen in the reactor. My conclusion to these reported conditions was that the pump was on and supplying energy to the system, and was not the absence of a poor vacuum with residual hydrogen in the reaction chamber. A simple measure of pump power usage during testing could resolve this issue. In fact using the old test setup such a test should be run to measure this power usage at various temperatures. This would help resolve the issue of how much energy is introduced into the water bath. Insulating the pump to reduce the heat loss to the ambient in a run would further allow determination of the pump efficiency as a function of flow. A separate measure of differential pressure drop across the pump would establish the constancy of the flow during the reaction period and the base lining operation Jed had identified. Maybe MUMP should barrow the Mizuno test setup and run the same test with their own monitoring and ambient condition controls. The radiation monitors should not be necessary to barrow. Even the recording computer would not have to be barrowed. Bob Cook - Original Message - *From:* Gigi DiMarco gdmgdms...@gmail.com *To:* vortex-l@eskimo.com *Sent:* Saturday, January 10, 2015 7:25 AM *Subject:* Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised Alain, I must confess that I've some problems to follow your statements. You should stick to the facts not to general theories or books. I, normally, run a company and at the end of the month I provide the food for a few dozens families, including mine. I've no time for cospiracies. I, personally, do not think that LENR are real but we are speaking about some specific experiments: it took 15 minute to me to understand that electrical power measurements were wrong in the TPR2. It's my job, I design and build power electonics and usually I use smart methods to measure power. In the TPR2 there was a hidden wattmeter; I simply found it and I wrote to Hoistadt. Rossi was very aggressive against me, whereas normally he lets people say whatever they want; so I'm sure I was right. Two of us measured and discovered the Defkalion trick in the water flow measurement (or do you think it was really Gamberale?); if you like I can send you the proofs privately. It seems to me that only you and Peter Gluck are still confident that the hyperion works [in the meantime Defkalion disappeared almost]. Mario Massa was a good friend of Sergio Focardi and he built and tested a calorimeter for the Piantelli-Focardi cell. While the Piantelli measurement was showing an excess heat, the calorimeter showed a little less than 100% that translates into no excess heat [I hope you understand that it is very difficult to build a fake calorimeter that gives a COP=1 exactly]. He was not anymore allowed to stay in the vicinity of the cell. You of course can continue to call this conspiracy, but your idea will not turn LENR into real. By the way Mario is a good friend of Bill Collis as well: you can ask him if he considers
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Dear Eric, I was not present at the Defkalion test, but at GSVIT we are four people [?] Gamberale was present, of course. He did repeat later the measurement (after a few weeks), alone. In the Lugano report they perform a dummy measurement: it is enough to calculate the resistances in the load. In the run test, they provide the power dissipated in the wires before the load; this is the hidden wattmeter, since inverting the power values you can get the rms currents in the load. These currents are in the range 40-50 A rms, as it is written in some part of the report. With these currents the power dissipation is three times the power measured by the authors. It's linear and clear... I wrote it to Hoistad and just after Rossi was very aggressive against me, inviting me to read Electricity for dummies, saying that his resistance drops by a factor of three passing from 450°C to 1200°C and then stays constant up to 1450°C Of course of this VERY special inconel wire there is no trace in the TPR2. I think it is really VERY VERY VERY unprobable to get such a resistor respecting all the other constraints. That's all. 2015-01-10 20:42 GMT+01:00 Eric Walker eric.wal...@gmail.com: On Sat, Jan 10, 2015 at 7:25 AM, Gigi DiMarco gdmgdms...@gmail.com wrote: Two of us measured and discovered the Defkalion trick in the water flow measurement (or do you think it was really Gamberale?); if you like I can send you the proofs privately. Hi Giancarlo, Thank you for the careful analysis of Tadahiko Mizuno's and Jed's calorimetry (I am not in a position to weigh your claims and Jed's responses and take no position). As I was reviewing this thread, I noticed some interesting details that were mentioned. Am I correct in understanding the following? - You and another person were the ones who did the investigation of the flowmeter used in DGT's demo sometime back and concluded that it was malfunctioning, leading to an incorrect report of the amount of steam flowing through the exiting tube and an incorrect derivation of the power that was being produced. Gamberale was relying on your investigation when he went public with his claims about DGT. - You looked at the Lugano report by Levi et al. and concluded there was a hidden wattmeter. Can you say more about this wattmeter and what you believe its effect to be in the conclusions of the Lugano report? Have I gotten any of this wrong? Regards, Eric
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Peter, I fully agree with you: the time will tell the truth. The cause was a water wave bouncing forth and back so that to provide extra pulses in the flowmeter. There is an oscilloscope photo taken by connecting it to the flowmeter. What you simply miss is that it was very simple to obtain a high COP with Argon filling. It should be impossible, shouldn't it? But I realize that it is completely useless to try to discuss with believers: you will have always an excuse to use. In the meantime 20+ plus years have already passed and the average age of the followers is more than 70. Who is the CTO at Defkalion now? How many researchers do they employ? Regards 2015-01-10 21:53 GMT+01:00 Peter Gluck peter.gl...@gmail.com: The process is not manageable in this way, it is completely chaotic. At the demo Mats Lewan has helped at the testing of the flowmeter. The Gamberale report was accepted immediately by you and many of our colleagues- do you think it cn be really used such a trick to obtain consistent results as in the demo? I a sure you know the reason for which DGT has uninvited you. However we can discuss this a long time...let's wait to see waht happens with the Hyperion, this is what really counts, OK? peter On Sat, Jan 10, 2015 at 10:36 PM, Jed Rothwell jedrothw...@gmail.com wrote: Peter Gluck peter.gl...@gmail.com wrote: Because you refered to me as some last faithful of Defkalion, if you have discovered the Defkalion flowmeter trick( define it exactly please!) can you explain how can be obtained results as in the demos of July 22 and 23 by manipulating two valves? Can you reoeat the trick and in which limits of precision and extension can you adjust the flow? This is described in detail here: http://lenr-canr.org/acrobat/GamberaleLfinaltechn.pdf The limits are described in the report. When you set the valves in a certain way, you can reduce the flow down to zero while you make the flow rate look high, by having the same same water slosh back and forth triggering the flowmeter with each slosh. You have to adjust the valves in a way the flow meter installation manual warns against, and you have to use a mixture of water and steam this flowmeter was not designed to measure. I do not think you can adjust precision. The result is bound to be chaotic, as the flowmeter regurgitates the sloshing water. This chaotic response can be seen in the report Fig. 2. Gamberale clearly describes how he triggered the problem deliberately: The test was performed by replicating as closely as possible the thermal variations observed during the tests carried out by the DGT technicians (see for example Figure 4). The goal of the test was to verify the behavior of the flowmeter during the strong boiling of the water inside the coil that surrounds the reactor. We verified that, by suitably selecting the adjustment of the valves upstream and downstream of the flowmeter, the production of steam at low flow regime produces turbulence and induces a regurgitation of the inlet water able to temporarily reverse the direction of flow within the flowmeter itself. Below is a description of the sequential processes that lead to an erroneous measurement of the flow of water passing through the cooling coil of the reactor: 1 - At very reduced water flow the water remains in the coil for a time sufficient to reach the boiling inside the coil 2 - The volume increase resulting by the production of steam produces a pressure increase which tends to push the fluid upstream . . . Whether this was done by Dekalion by accident or on purpose I cannot say. Gamberale suspects it was on purpose. I can say two things: 1. If it was not dishonest it was incompetent. The problem should have been revealed by testing the flowmeter, by dumping the water into a bucket over a measured time. Any flowmeter manual tells you to do this. Defkalion refused to do this for months. Gamberale finally did it, and the problem was instantly revealed. 2. Defkalion originally invited me to visit. I told them I would bring instruments and do this test. They seemed to get cold feet. I told them I would not go if I could not do this test. They then uninvited me, three times, and eventually cancelled the trip. Then they spread rumors about me that I am engaged in some kind of unnamed nefarious business, which they said was the reason they cancelled. I suspect they uninvited me because I insisted on measuring the temperatures and flow rates myself, with my own instruments, and they knew damn well these numbers were bogus. I had the same experience with Patterson, who relented, and with Rossi, who said no measurements! and uninvited me immediately, the day we first discussed it. He invited Krivit instead. Krivit no only made no measurements, he made practically no observation, and he did not even write down or report Rossi's measurements. Then he attacked Rossi. Rossi would have come out
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
*Kinetic energy of the water carries the power into the storage medium so it can be calculated by the reliable formula E=1/2*M*V^2.* This is completely wrong: the pump power is not transformed into kinetic enegy of the water, otherwise you will get after a while an infinite velocity, not only for the water inside the tube but for cars on motorways as well. Only at the beginning, for a few seconds, the pump energy becomes kinetic energy. After this starting phase the pump power compensates the losses induced by hydraulic resistance in the pipe. Where is the hydraulic resistance in you analysis? Nowhere, so the analysis is wrong. 2015-01-08 13:07 GMT+01:00 Roarty, Francis X francis.x.roa...@lmco.com: Nicely done Dave! A skeptic has unwittingly provided positive evidence and reproduced Jed’s results in one fell swoop! *From:* David Roberson [mailto:dlrober...@aol.com] *Sent:* Wednesday, January 07, 2015 6:00 PM *To:* vortex-l@eskimo.com *Subject:* EXTERNAL: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised Guys, I believe that I have an explanation for the variation in measurements performed by the latest critic and Jed. I have long wondered about the physics of that pump system so I felt like it was time to do a bit of math. Unless I made a major error in calculations, both results make complete sense. The author of the negative report states that he is using pipe that is 1/2 the diameter of the one used by Mizuno. This is the key to the mystery. Consider the following derivation: The pump is rated at 9 liters per minute when the net lifting head is zero. A calculation of the flow rate yields 150 grams/second. i.e. 9 liters/min * 1000 cm^3/liter * 1min/60 seconds=150 cm^3/second. And, 1 gram/cm^3 is understood. The area of the 1 cm inside diameter pipe is pi * r ^2, which in this case is 3.14159 * (.5 cm)^2 = .7854 cm^2. The velocity of the water inside the pipe is 150 cm^3/sec / .7854 cm = 191.02 cm/second. Kinetic energy of the water carries the power into the storage medium so it can be calculated by the reliable formula E=1/2*M*V^2. To get power, you use the amount of water brought up to speed in 1 second. So we have E=1/2 * 150 grams * (191.02 cm/second)^2 = 2.738 x 10^6 gram-cm^2/sec^2 imparted upon the water in each second. These units are in ergs, so to get to joules you multiply it by 10^-7 which yields .2738 joules in each second. This is the definition of .2738 watts. Jed has measured numbers that fall into this range and has confidence in his results. Now our favorite skeptic claims that he is using .5 cm pipe instead of the 1 cm pipe used by Mizuno and does not realize that he is making a major error. But, the area of that pipe is reduced by a factor of 4 since it is exactly 1/2 the inner diameter of the original. With a factor of 4 reduction in area comes an increase in the velocity of the water flowing through it by that factor 4 in order to achieve the same mass flow rate. Every thing else being equal you find that the energy imparted upon the water that is sped up from rest to a velocity that is 4 times that from the first case yields the square of that factor. In which case it is 4^2 which is 16 times. Guess what? .2738 watts * 16 = 4.38 watts. So, the skeptic has verified the measurement performed by Jed! I love it when the math holds up so well. Congratulations Jed, you got it right. Dave -Original Message- From: Alain Sepeda alain.sep...@gmail.com To: Vortex List vortex-l@eskimo.com Sent: Wed, Jan 7, 2015 3:51 pm Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised done, question is if it will be moderated. They won't dare. anyway question now is not to convince, but to deliver to the industry. 2015-01-07 20:39 GMT+01:00 Jed Rothwell jedrothw...@gmail.com: Alain Sepeda alain.sep...@gmail.com wrote: your last sentence is enough to understand they screw up somewhere. If you would like to send them the last sentence please do so. I do not have the time or the inclination to deal with such people. - Jed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Sorry Dave but I do not agree at all with your DIY physics about pumps. 1) We actually don't know the actual power flow: you assumed 9 l/m : who told you? any flow meter around? 2) The physics of pumps is well known, there is no need to re-invent it see for example the first equation in the box here http://www.thermexcel.com/english/ressourc/pumps.htm as you can see the mechanical power depends not only on the flow rate (that we do not know) but also on the pressure loss, that we do not know either. I think we have to wait for the excel file from Jed; there we can find the way to solve our problem. Gigi 2015-01-08 17:22 GMT+01:00 David Roberson dlrober...@aol.com: Gigi, While Jed is locating that information for you may I request that you make a calculation of the kinetic energy contained within the moving water exiting the pump? Then, do the same thing for the kinetic energy of water that is about to enter the intake pipe of the pump. Do you agree that the difference in heat must be deposited within the standing liquid? Dave -Original Message- From: Gigi DiMarco gdmgdms...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Thu, Jan 8, 2015 10:54 am Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised *Mizuno measured the heat added to the system by the pump. There is no point to appealing to a theory or hypothesis about how much heat there may be when it has actually been measured for 18 hours by running the pump only. * dear Jed, I could not find anymore the excel file of this 18 hour measurement [it used to be http://LENR-CANR.org/Mizuno/Mizuno2014-11-20.xlsx] In that file it was clearly shown that the water temperature, with no excess heat, rised by 2.5 °C in a stable way against the room temperature. Is not it too much for 0,24 W? Could you post the file again? Many thanks 2015-01-08 16:39 GMT+01:00 Jed Rothwell jedrothw...@gmail.com: Gigi DiMarco gdmgdms...@gmail.com wrote: This is completely wrong: the pump power is not transformed into kinetic enegy of the water, otherwise you will get after a while an infinite velocity, not only for the water inside the tube but for cars on motorways as well. Let me point out again that this entire discussion is irrelevant for two reasons, which I clearly explained in the paper, starting on p. 24: 1. Mizuno measured the heat added to the system by the pump. There is no point to appealing to a theory or hypothesis about how much heat there may be when it has actually been measured for 18 hours by running the pump only. 2. It makes *no difference* how much heat is added to the system by the pump. Whether the temperature goes up 0.6°C, or 6°C or 10°C, and whether this temperature represents a half watt, or 5 W, or 10 Watts is completely irrelevant. The pump is left running all the time. Therefore all of the heat from the pump is in the baseline temperature of the system. Mizuno measures from the baseline to the terminal high temperature at the end of the test, just as the temperature begins to fall. He does not measure from the ambient temperature. I wish the people writing these critiques would spend a few moments reading the paper, but they never do. I am not even going to bother adding these remarks to the latest paper. I am busy. If someone here would like to, feel free to add these points. It is a waste of time, I think. - Jed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
*Mizuno measured the heat added to the system by the pump. There is no point to appealing to a theory or hypothesis about how much heat there may be when it has actually been measured for 18 hours by running the pump only.* dear Jed, I could not find anymore the excel file of this 18 hour measurement [it used to be http://LENR-CANR.org/Mizuno/Mizuno2014-11-20.xlsx] In that file it was clearly shown that the water temperature, with no excess heat, rised by 2.5 °C in a stable way against the room temperature. Is not it too much for 0,24 W? Could you post the file again? Many thanks 2015-01-08 16:39 GMT+01:00 Jed Rothwell jedrothw...@gmail.com: Gigi DiMarco gdmgdms...@gmail.com wrote: This is completely wrong: the pump power is not transformed into kinetic enegy of the water, otherwise you will get after a while an infinite velocity, not only for the water inside the tube but for cars on motorways as well. Let me point out again that this entire discussion is irrelevant for two reasons, which I clearly explained in the paper, starting on p. 24: 1. Mizuno measured the heat added to the system by the pump. There is no point to appealing to a theory or hypothesis about how much heat there may be when it has actually been measured for 18 hours by running the pump only. 2. It makes *no difference* how much heat is added to the system by the pump. Whether the temperature goes up 0.6°C, or 6°C or 10°C, and whether this temperature represents a half watt, or 5 W, or 10 Watts is completely irrelevant. The pump is left running all the time. Therefore all of the heat from the pump is in the baseline temperature of the system. Mizuno measures from the baseline to the terminal high temperature at the end of the test, just as the temperature begins to fall. He does not measure from the ambient temperature. I wish the people writing these critiques would spend a few moments reading the paper, but they never do. I am not even going to bother adding these remarks to the latest paper. I am busy. If someone here would like to, feel free to add these points. It is a waste of time, I think. - Jed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Dear Dave, I do not think we need so much calculation; better to perform a new measurement on a 10 mm pipe to test you hypotesis. I hate to say that we did it and the power dissipation increases a little bit, as any engineer would have expected: you will find soon the results here https://gsvit.wordpress.com/ I advise you to read the full article as well, so you can find all the theory you need. Please feel fre to ask any questions you like. In case you would like to take a look of the Mizuno 18 hour pump calibration you find here the file that Jed can not find anymore https://dl.dropboxusercontent.com/u/66642475/Mizuno2014-11-20.xlsx in the very first sheet (mio) you can find the water temperature increase against the room temperature coming from Mizuno's data. Take your time to think about it. Jed can confirm that the data are the original ones. By the way regarding your statement *I consider it poor form to hide behind obscure generalities* my name is Giancarlo De Marchis and I belong to the *GSVIT Group;* I thought it was clear, sorry. I'm an electronic engineer and I design water cooling systems [with pumps] for RADARs and high power converters. Normally they works fine. Regards 2015-01-08 19:40 GMT+01:00 David Roberson dlrober...@aol.com: The flow rate is going to be reasonably close to the 9 liters per minute specification from the manufacturer. I have a graph from Iwaki America that shows the expected rate as a function of the lift head facing the pump. At zero meters of head which corresponds to atmospheric pressure the rate is 9 liters per minute. At approximately .6 meters of lift the rate is still about 7 liters per minute. How much do you calculate as the effective head due to friction within the pipe? The experiment that claimed around 4 watts of pump induced power uses a pipe that is 5 mm diameter and about .5 meter in length. Please do the math if you have the equations to determine exactly what flow rate should be expected. The author of that report completely failed to take into account pump power being transported by means of the fluid acceleration. And, it is obvious that he was not aware that the faster the fluid moves, the more power it transfers. This is an obvious mistake and I am pointing it out. As I asked you before, take the time and use whatever equations you can locate in the literature to calculate the amount of kinetic energy that is imparted upon a liquid by the acceleration due to pump action. There apparently is no need to reinvent the physics of pumps to perform this calculation. If you do this one task, you will find that the heat power comes close to that which is measured by the two independent experimenters. Also, you will find that the amount of power due to this process depends greatly upon the area of the pipe carrying a constant amount of fluid mass per unit of time. That power will come out 16 times as much for a pipe that is 5 mm compared to one that is 10 mm in diameter. Do the math! If you counter that the flow rates do not match due to changes in size of the pipe, then it becomes apparent that the test performed by the skeptic does not agree with the one he is attempting to replicate which negates his results. How can you possibly believe that it is a coincidence that my calculations yield a result that is close to what is being measured? It is quite simple to figure out the kinetic energy imparted upon a mass of water that is accelerated by some means. Just read my derivation and tell me where an error is located other than just stating that no flow meter was present to prove the rate. I will be happy to review any evidence that you present to support your position. I am as amazed as you are that the calculations came out that well. Your earlier contention was that there is no energy transport due to acceleration of the liquid by pump action which ends up in a holding tank for the active liquid. You pointed out several terrible consequences if that were true. None of those are seen in real life so I assume that you now do not hold that position. Is this true? Before you continue to shoot down my proposal I expect you to show some mathematical support for your contentions. So far that has not happened. Take the time to add support to your position or you should back away from taking such a negative stance. I consider it poor form to hide behind obscure generalities. Dave -Original Message- From: Gigi DiMarco gdmgdms...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Thu, Jan 8, 2015 12:52 pm Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised Sorry Dave but I do not agree at all with your DIY physics about pumps. 1) We actually don't know the actual power flow: you assumed 9 l/m : who told you? any flow meter around? 2) The physics of pumps is well known, there is no need to re-invent it see for example
[Vo]:WLT Disproof
The following paper: Low Energy Neutron Production by Inverse-beta decay in Metallic Hydride Surfaces S. Ciuchi http://arxiv.org/find/nucl-th/1/au:+Ciuchi_S/0/1/0/all/0/1, L. Maiani http://arxiv.org/find/nucl-th/1/au:+Maiani_L/0/1/0/all/0/1, A. D. Polosa http://arxiv.org/find/nucl-th/1/au:+Polosa_A/0/1/0/all/0/1, V. Riquer http://arxiv.org/find/nucl-th/1/au:+Riquer_V/0/1/0/all/0/1, G. Ruocco http://arxiv.org/find/nucl-th/1/au:+Ruocco_G/0/1/0/all/0/1, M. Vignati http://arxiv.org/find/nucl-th/1/au:+Vignati_M/0/1/0/all/0/1 has just been uploaded to ArXiv http://arxiv.org/abs/1209.6501 The paper addresses the WLT as presented by Widom, Larsen and Srivastava in peer-reviewed journals and presents very strong objections to the theory that can be summarized as: It has been recently argued that inverse-beta nuclear transmutations might occur at an impressively high rate in a thin layer at the metallic hydride surface under specific conditions. In this note we present a calculation of the transmutation rate which shows that there is little room for such a remarkable effect. It is worthwhile to say that some of the authors are preminent scientists. Luciano Maiani is a San Marino citizen physicist best known for his prediction of the charm quark with Sheldon Lee Glashow and John Iliopoulos. He became Director General of CERN, serving from 1 January 1999 through the end of 2003. Moreover, he is the past-President of the Italian National Research Council (CNR, 2008-2011). Giancarlo Ruocco is deputy Dean at Rome University La Sapienza and Director of the Physics Department. He is in charge for research coordination activities at La Sapienza. The paper will be (or has been, actually I do not know) submitted to peer-reviewed journals to be published; probably in the same journal in which the WLT has been proposed. Since WLT forms the basis of a number of experimental approaches to LENR's (including Brillouin and NASA) maybe it's wise to read and try to understand the paper. Cheers GG
Re: [Vo]:Do you think Rossi will still be too busy?
I think Rossi is adding to English more words, coming in some sense from Italian, than in the last 5 centuries. Clownerie has been translated by Akira into travesty. Clownerie is another Rossi's invention, if I can imagine what take places into his brain I think the right path is clown == pagliaccio == pagliacciata (the act of being a clown) == clownerie; I think it should be translated into buffoonery (clowning, silly behaviour) 2012/2/14 Chemical Engineer cheme...@gmail.com http://blog.newenergytimes.com/author/sbkrivit/ *Smith Offers $1 Million Prize for Successful E-Cat Demo*http://blog.newenergytimes.com/2012/02/14/smith-offers-1-million-prize-for-successful-e-cat-demo/ by *Steven B. Krivit* http://blog.newenergytimes.com/author/sbkrivit/ *From:* Margot Egan [on behalf of Dick Smith] *Sent:* Tuesday, 14 February 2012 12:51 PM *To:* Andrea Rossi Re. E-CAT *Subject:* from Dick Smith in Australia Re. U.S. One Million Dollars for Successful Re-Testing of E-CAT *To: Andrea Rossi** From: Dick Smith* Dear Mr Rossi *Re: USD1,000,000 for Successful Repeat of E-CAT Demonstration* Dick Smith is my name. I am writing to you from Sydney, Australia. Possibly the best information in relation to my background is on Wikipedia - see *HERE* http://en.wikipedia.org/wiki/Dick_Smith_%28entrepreneur%29. Some time ago I was contacted by Mr Sol Millin of the Byron New Energy Trust. Mr Millin has had extensive communication with you. Mr Millin communicated the advantages of your ECAT unit in relation to energy and how it had the potential to solve the world’s energy problems. Mr Millin said that he had your authority to act on your behalf in relation to an agreement for the Australian “rights” to your invention. After some discussion, I agreed that I would invest AUD200,000 provided that evidence could be shown that the unit actually worked as claimed. There has been a lot of to’ing-and-fro’ing since then, with Mr Millin claiming that he has provided me with the evidence and with my insistence that this is not the case. At one stage Mr Millin even sent me an email (attached) threatening to sue me for one-hundred-million-dollars if I did not proceed with sending him my AUD200,000. As Mr Millin and I do not seem to be getting anywhere on this issue, I have determined a way that we could possibly break this nexus, i.e. I would like to offer you USD1,000,000 for a successful repeat of the March 29, 2011 demonstration. One million US dollars will be made out to you as a Bank cheque or will be held in an escrow account if you desire. I do not want to know how the unit operates, nor to have a share in the profits from any sales. My satisfaction will come from knowing that if the unit is successful, then some of the world’s greatest problems – especially in relation to climate change – will be solved. I point out that over the last few decades my wife and I have donated many millions of dollars to scientific research, much of it without any immediate results. We have not complained about this. My offer is very simple, which I will restate: I ask you to repeat the March 29, 2011 demonstration purported to show that your E-CAT unit had an output power of many times the input power through LENR (low energy nuclear reactions). As the sole judges as to whether this can be repeated correctly, I suggest we use the two Swedish scientists, Kullander and Essen, as they attended the March 2011 demonstration and wrote a report. I would be happy to cover any reasonable cost of having them flying to Italy to attend the repeat of the demonstration. They can then check the wires (because, as you know, there have been claims that the wiring may have been misconnected) and also the power output of the unit in relation to both the heated water and the steam. I would be happy, with Kullander and Essen as the sole judges as to whether the unit has the power output you have claimed, to hand you an irrevocable Bank cheque for USD1,000,000 made out in your name if the demonstration is successfully completed. If Kullander and Essen are not available, I am happy to agree with you on two other individuals of similar expertise to attend the new demonstration as the judges. I am sure we can come to an agreement as to who would be suitable and independent. It would also be necessary to have a third person - who you and I can agree on - to assist with the type of measuring equipment and its accuracy so there are no doubts that the scientific community will accept the results. I understand the 29 March 2011 demonstration took place over a period of more than six hours and showed a power multiplication of approximately ten times. To make the demonstration test even fairer, I would be happy if the demonstration to qualify for the assignment of the USD1,000,000 were reduced to a five-hour period and with a power multiplication ratio of at least eight times.
[Vo]:Re: WL
Dear all, I succeded in fixing the issue during the night, while sleeping... From the point of view of congruence of figures, a term is missing in Eq. (23), that is the inverse of 4*pi*epsilon-zero (the dielectric constant). In you put this term, then the figure in Eq.(25) turns out to be exact. a is the Bohr radius as stated in Eq. (24). I don't know which unit system the authors are using (maybe atomic units) but since from time to time they switch to MKSA, I think it would have been savier to clearly state it, otherwise the poor reader has to figure out a lot of things. Maybe the strange equal sign has a different meaning from what I could imagine. I've still some concerns about the assumptions, but this is another issue, I'll keep you informed if I found other discrepancies. Thank you for your attention Grazie dell'interesse, Giovanni. 2012/1/31 Gigi DiMarco gdmgdms...@gmail.com I've a problem with the WL theory. I read carefully their published paper http://newenergytimes.com/v2/library/2006/2006Widom-UltraLowMomentumNeutronCatalyzed.pdf and I found what seems to me to be a major flaw. I'm sure I'm totally wrong but I would ask you to check. It is only arithmetics, no advanced physics. My attention was catched by Eq. (25), where an electric field around one million of millions V/m appears. Too much, I told myself. As a comparison the proton induced electrical field at a Bohr distance is only about 10 to minus 7 V/m, that is 18 orders of magnitude less. So I checked the calculations starting from Eq. (23) where the electric field is 4 times proton charge divided by 3 times Bohr distance to the third power, all multiplied by a term, under square root, that represents the proton displacement during its oscillatory motion. In Eq. (25) a term equal to the Bohr distance is transported under the square root. So the term to be evaluated reads: 4 |e| / 3 a^2 This term provides us with a numerical value equal to 7.63 V/m, that is 11 orders of magnitude less than the value appearing in the paper. That turns out to be a huge problem for the authors, since the threshold criteria for electron capture Eq. (6) and Eq. (27) are no more satisfied by a large amount and the ultra low momentum neutron plus neutrino pair can not be produced. Is anybody here that can confirm or disproof my calculations? Best regards GDM
[Vo]:WL
I've a problem with the WL theory. I read carefully their published paper http://newenergytimes.com/v2/library/2006/2006Widom-UltraLowMomentumNeutronCatalyzed.pdf and I found what seems to me to be a major flaw. I'm sure I'm totally wrong but I would ask you to check. It is only arithmetics, no advanced physics. My attention was catched by Eq. (25), where an electric field around one million of millions V/m appears. Too much, I told myself. As a comparison the proton induced electrical field at a Bohr distance is only about 10 to minus 7 V/m, that is 18 orders of magnitude less. So I checked the calculations starting from Eq. (23) where the electric field is 4 times proton charge divided by 3 times Bohr distance to the third power, all multiplied by a term, under square root, that represents the proton displacement during its oscillatory motion. In Eq. (25) a term equal to the Bohr distance is transported under the square root. So the term to be evaluated reads: 4 |e| / 3 a^2 This term provides us with a numerical value equal to 7.63 V/m, that is 11 orders of magnitude less than the value appearing in the paper. That turns out to be a huge problem for the authors, since the threshold criteria for electron capture Eq. (6) and Eq. (27) are no more satisfied by a large amount and the ultra low momentum neutron plus neutrino pair can not be produced. Is anybody here that can confirm or disproof my calculations? Best regards GDM
[Vo]:hydrogen refill
I don't know if it was already reported but I found some oddies in the hydrogen consumption October 6th test, from Lewan's report Weight hydrogen bottle: - before filling: 13606.4 grams - after filling: 13604.9 grams Total loaded: 1.5 grams Pressure H2 Bottle: 55 bar Reduced: 15 bar October 28th test, from Fioravanti/Rossi's report - before filling: 13604.5 grams - after filling: 13602.8 grams Total loaded: 1.7 grams We can derive the following conclusions: - no intermediate experiments/tests have been made since 0.4 grams is likely to be a leakage, unless a different vessel was used - almost same quantity of Hydrogen is needed for either 1 or 107 fat-cat systems OR - Fioravanti Rossi report misleading figures Since Rossi in his blog says that 18 kg are needed in 6 months to charge the 1MW plant, 100 grams are needed in 1 day. Giancarlo
