Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
2015-01-09 0:00 GMT+01:00 David Roberson dlrober...@aol.com: Many of the cold fusion skeptics conclude that LENR is not possible because there is no theory to support it. An article describe that https://www.fightaging.org/archives/2015/01/the-scientific-institution-is-biased-against-shortcuts-to-the-production-of-practical-technology.php it match kuhn vision too. anomalies are ignored or rationalized until there is a perfect theory to explain all. reality is not a problems , it can be denied easily.
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
I haven't read Mizuno's report - so I might be mistaken in my comments but if Mizuno is at steady state with the pump on for many many hours, then when he turns on the LENR experiment, he will only see a delta T that is due to the LENR experiment and the pump heat doesn't matter at all. On Thu, Jan 8, 2015 at 6:48 PM, Alain Sepeda alain.sep...@gmail.com wrote: 2015-01-09 0:00 GMT+01:00 David Roberson dlrober...@aol.com: Many of the cold fusion skeptics conclude that LENR is not possible because there is no theory to support it. An article describe that https://www.fightaging.org/archives/2015/01/the-scientific-institution-is-biased-against-shortcuts-to-the-production-of-practical-technology.php it match kuhn vision too. anomalies are ignored or rationalized until there is a perfect theory to explain all. reality is not a problems , it can be denied easily. -- Jeff Driscoll 617-290-1998
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Mixuno would see a temperature differential as you say, however what fraction of energy introduced by the reaction is above the input energy of the electrical pump and or other electrical inputs?To get a COP you need the steady state in-put energy to determine this. Thus, the problem becomes one of determining the relationship between between energy and the system temperature relative to ambient at a steady state condition. If the reaction energy is introduced totally as heat, the determination should be pretty good assuming the calibration of the pumps input energy is well known. That calibration is the question that is being debated I believe. In Mizuno's test I believe the differential pressure that the pump put out did not change much; hence, the energy used should follow the specification for the pump in the pump head curve accurately. However, if the reaction caused a significant change in the differential pressure and, hence, the flow, such information would be necessary to accurately extrapolate the total energy, pump plus reaction to temperatures above that produced by the pump alone. I agree with Dave's analysis of the energy related to the flow changes in the system. However, they would only represent a portion of the pumps energy output--frictional losses in the piping, pressure drops at nozzles etc, and heat losses from the pump must also be added to confirm the pump head curve vs power is accurate. It seems that the Mizuno team should have accomplished such an in-house calibration to confirm the vendor's specs. While they were at calibration, I would have used a known dummy electrical heater to determine the temperature/energy input curve. Bob - Original Message - From: Jeff Driscoll To: vortex-l@eskimo.com Sent: Thursday, January 08, 2015 5:11 PM Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised I haven't read Mizuno's report - so I might be mistaken in my comments but if Mizuno is at steady state with the pump on for many many hours, then when he turns on the LENR experiment, he will only see a delta T that is due to the LENR experiment and the pump heat doesn't matter at all. On Thu, Jan 8, 2015 at 6:48 PM, Alain Sepeda alain.sep...@gmail.com wrote: 2015-01-09 0:00 GMT+01:00 David Roberson dlrober...@aol.com: Many of the cold fusion skeptics conclude that LENR is not possible because there is no theory to support it. An article describe that https://www.fightaging.org/archives/2015/01/the-scientific-institution-is-biased-against-shortcuts-to-the-production-of-practical-technology.php it match kuhn vision too. anomalies are ignored or rationalized until there is a perfect theory to explain all. reality is not a problems , it can be denied easily. -- Jeff Driscoll 617-290-1998
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Dear Gigi, I look forward to seeing the results of the new test with 10 mm pipe. Could you include some form of drawing that shows the location of the relative pump pipe input and output locations. I am confident that you realize that my calculations are based upon a system where the pipe entering the Dewar from the pump does not continue directly into the pipe that makes up the return path for the coolant. A continuous pipe would not deliver its load of kinetic energy into the tank. That type of system would behave like a heat exchanger. In most hydraulic systems the power generated by the pump is intended to drive some form of hydraulic load such as a cylinder. In these cases the amount of power lost due to kinetic energy transport is negligible. If you consider a typical log splitter for example, most of the time the hydraulic fluid is directed to bypass the cylinder by a control valve. Of course energy is imparted upon the fluid by the same process that I calculated in this case due to it being accelerated by the pump action. The typical pump for a log splitter is a constant displacement device where a fixed flow rate is generated. Even when the cylinder is bypassed you will find that heat power finds it way into the excess oil storage tank. Some of that heat is due to kinetic energy transport among other reasons. Why avoid calculations in this particular case? Many of the cold fusion skeptics conclude that LENR is not possible because there is no theory to support it. Here is a simple example of kinetic physics that most high school students would be capable of understanding. You have a mass of water that is initially at rest. It is acted upon by a pump which sets it into forward motion along a pipe at a certain velocity that can be established by knowing the flow rate and the cross section area of the pipe. And, the water slows down inside a Dewar which must accept the kinetic energy from the flow. I find it interesting that you and the other skeptics are reluctant to make that simple calculation. Why? And, where is there a problem with my procedure? Did I make an error in calculating the kinetic energy of the water traveling within the pump output pipe? If you can show me my error I will gladly concede the point. I honestly would like to see a scientific reason that your team finds that 4+ watts of excess heat power while at the same time Mizuno measures less than 1 watt. Here we both have an opportunity to seek the truth by using scientific principles and so far you have avoided that offer. Unless I missed something, you are still harbor the belief that there is no transport of heat from a pump into a relatively large load region in the form of lost kinetic energy of the fluid. The equations 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.
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
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.commailto:alain.sep...@gmail.com To: Vortex List vortex-l@eskimo.commailto: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.commailto:jedrothw...@gmail.com: Alain Sepeda alain.sep...@gmail.commailto: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
*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
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
[Vo]:news, bad for Europe, good for LENR
Dear Friends, I advice you to skip the introduction but read the rest of: http://egooutpeters.blogspot.ro/2015/01/lenr-news-january-8-2015.html Thank you, Peter -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Actually you are mistaken. Most of the kinetic energy imparted to the water will be deposited into the storage tank. This is due to the fact that the water is not immediately redirected into the return pipe. The only way that what you say could be close to correct is if the pipe is a continuous loop without any break. Consider what happens to moving water once it leaves the end of the pump drive pipe and enters the hydraulic tank. All the kinetic energy is released into that tank except for the very tiny amount that by good fortune happens to be directed into the return pipe at the velocity it enters. The amount must be less than a few percent at most. Perhaps you can explain where the incoming kinetic energy is deposited? If you transform the experiment such that a very large sink is used it becomes obvious that what I propose is true. In that possible system, the incoming water mixes with the much larger water that is at rest inside the tank. Water at rest has no kinetic energy due to motion and therefore any that happens to arrive through the input pipe by logic has to add to the static potential energy within the tank. Then, it is necessary for you to explain why the amount of kinetic energy as calculated yields the correct measurements as seen by two independent observers. Do you consider this a coincidence? I performed this calculation as an attempt to prove to myself that the idea would not yield enough kinetic transport power to impact the reported results. I was surprised just as you to realize that it does a pretty good job of matching the measurements. From a physics point of view the idea is sound. Of course I did not include frictional losses within the pipe since I hoped for a quick answer to my question. The guy that conducted the skeptical test system used a much shorter pipe as compared to the one used by Mizuno. He apparently was attempting to use a shorter half diameter pipe while maintaining the same mass flow rate. This was a good idea except for the flaw that I pointed out. So ask yourself why would a skeptic not be concerned when his experiment yielded results that are so far off from what Jed measured? Jed used reasonable practice by monitoring the water temperature rise and fall rates to estimate the amount of energy being deposited by the pump into the hydraulic tank. Can you show how his methods are wrong? My calculations show that he was right on target and so was the measurement performed by the skeptic. I have an interesting question that perhaps you have considered. Since the kinetic energy from the incoming water finds it way into the Dewar it eventually must show up as heat. That incoming water also has plenty of momentum that must be deposited into the Dewar. I would like very much to see the water behavior within that tank. It does not take much visualization to expect the water to begin swirling around rapidly until frictional effects dissipate the incoming momentum. We might have a pretty good vortex generator. I noticed that the warm up method used by the skeptic appears to have some form of baffle inside the small holding dish. I suspect that this was included in an attempt to prevent excess sloshing of the water inside that vessel. Dave -Original Message- From: Gigi DiMarco gdmgdms...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Thu, Jan 8, 2015 7:25 am Subject: 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
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Jed, as you have pointed out, Mizuno calibrated out the heat due to the pump operation. That should be enough to end the skeptical responses unless they contend that it is not possible to calibrate in that manner. I have shown quite simply that the original measurements of yours as well as theirs are close to what should be expected. It is left to the skeptics to explain where that kinetic energy ends up if not within the holding tank. Also, if they believe that my calculation of the kinetic energy of the accelerated water is inaccurate they should show why. This is fairly simple physics. Dave -Original Message- From: Jed Rothwell jedrothw...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Thu, Jan 8, 2015 10:40 am Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised 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
David Roberson dlrober...@aol.com wrote: I have shown quite simply that the original measurements of yours as well as theirs are close to what should be expected. It is left to the skeptics to explain where that kinetic energy ends up if not within the holding tank. More to the point, they should show why this makes a difference, given that the method measures from the baseline to the final temperature. - Jed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
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
[Vo]:Eta Carina is in the Scinece News again
Eta Carina (aka ate a car) is arguably the greatest intergalactic threat to life Earth, considering that it could already be responsible for known solar phenomena in our sun (including possibly the 11/5.5 year) cycle going back billions of years, and it has a mysterious connection to our sun. It is fairly close to us in cosmological terms. Here is the most recent story. In the past, Eta Car has been the brightest star in the night sky (in the Southern hemisphere) and may become so, once again. It is a binary, and one of the two could be a quark star. http://www.csmonitor.com/Science/2015/0108/NASA-unravels-mysteries-of-huge-s pace-explosion About three years ago this came up again on Vortex in this thread https://www.mail-archive.com/vortex-l%40eskimo.com/msg63181.html And this is not the first time we have given notice to the death star. It has a long ominous history, even on vortex. Curiously - the last poster in the thread above seems to think that there could be Rossi connection.
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
Gigi DiMarco gdmgdms...@gmail.com wrote: 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] . . . Could you post the file again? I had to remove this for complicated reasons beyond the scope of the discussion. I will put back a partial version of it. - Jed
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
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 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
[Vo]:Re: CMNS: news, bad for Europe, good for LENR
Dear Chino, The author DAVID HAMBURG (Science) cannot be found You have to write to David Rowan (editor) at in...@wired.co.uk and/or da...@davidrowan.com please let me know it it is a solved problem Warm greetings, Peter On Thu, Jan 8, 2015 at 6:38 PM, Peter Gluck peter.gl...@gmail.com wrote: Dear Friends, I advice you to skip the introduction but read the rest of: http://egooutpeters.blogspot.ro/2015/01/lenr-news-january-8-2015.html Thank you, Peter -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com -- You received this message because you are subscribed to the Google Groups CMNS group. To unsubscribe from this group and stop receiving emails from it, send an email to cmns+unsubscr...@googlegroups.com. To post to this group, send email to c...@googlegroups.com. Visit this group at http://groups.google.com/group/cmns. For more options, visit https://groups.google.com/d/optout. -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
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 the
Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised
Gigi DiMarco gdmgdms...@gmail.com wrote: I think we have to wait for the excel file from Jed; there we can find the way to solve our problem. I cannot provide that today, but there is a graph from it in the paper, on p. 25. It shows the first 2.8 hours. As I said, the pump is usually left on all the time. For the purpose of this test it was turned off for a day and the reactor and Dewar were left to cool down. As you see the water temperature climbed after the pump was turned on. After 1.5 hours the terminal temperature was reached, 0.6°C above ambient. It never goes any higher. Other tests have confirmed this. This is the baseline temperature including the pump heat. Mizuno uses this starting temperature and measures to the terminal temperature at the end of the test just before the cell and water temperature begin to decline. If you were to come in first thing in the morning, turn on the pump and immediately turn on the experiment, the pump heat would be included in the excess heat. That would be a problem. However, Mizuno never does that. If the equipment is off he turns it on an hour or two before beginning an experiment. As I said, in most cases he leaves the pump, power supplies and data collection running overnight, as you see in the graphs. - Jed
