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<mailto:alain.sep...@gmail.com>> To: Vortex List <vortex-l@eskimo.com<mailto: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<mailto:jedrothw...@gmail.com>>: Alain Sepeda <alain.sep...@gmail.com<mailto: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
