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 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
