Now, I'm going to take a que from Horace (where are you Horace?), and do some basic algebra... Those who have been beating the measurement issue beyond dead-horse status do remember what algebra is????
Water in = Water out (water in whatever form) For the e-Cat demos/tests, we have: Liquid_Water_in = Water_Vapor_out + Water_Liquid_out (by MASS, not volume of course) Now, from your middle school days of elementary algebra, rearrange terms to get: (mass of liquid Water out) = (mass of Liquid Water in) - (mass of Water vapor out) What we are looking for (mass of liquid Water out), is what's left over when we subtract the mass of the water vapor in the output steam from the total water into the unit. If ABD's research into what exactly is being measured by the instrument (the mass of water vapor?), then we have BOTH terms on the right side of the equation and we can solve for the left side... There is also a reason why Rossi and Galantini ***always make a point of saying*** that the pressure inside the 'chimney' can be considered to be pretty much always at atmospheric pressure. Let's see if you can sit back, think, and figure out why that's important. -Mark
