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 <[email protected]> To: vortex-l <[email protected]> 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 <[email protected]>: Nicely done Dave! A skeptic has unwittingly provided positive evidence and reproduced Jed’s results in one fell swoop! From: David Roberson [mailto:[email protected]] Sent: Wednesday, January 07, 2015 6:00 PM To: [email protected] 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 <[email protected]> To: Vortex List <[email protected]> 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 <[email protected]>: Alain Sepeda <[email protected]> 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
