It's bed time here; so, I'll have to check it out tomorrow. Meanwhile three phase power is calculated by:
Watt's Law: W = V avg. x A avg x p.f. x 1.732 Where: W = wattage (watts) Vavg = average voltage of the three separate phases (volts) Aavg = average current of the three separate phases current (amps) p.f. = average power factor or the three separate phases 1.732 = a constant necessary with 3 phase. Meanwhile again, here's another "revelation" occurring late today: https://drive.google.com/file/d/0B8mt4mJOTGvBei1sbThCMzJybm8/view?usp=sharing Sorry, I have a lot going on at the moment. On Tue, Oct 28, 2014 at 9:41 PM, Jones Beene <[email protected]> wrote: > This guy makes an interesting point that is not clear to me. > > Terry, Dave, Bob et al - what do you EEs who have looked at the input power > think about this approach? > > “The currents in the three C1 wires are all equal and they are measured by > the true RMS > > instrument PCE-830. The three heating resistors are also equal and therefore > they will > > all be heated by equal currents, I2. The authors of the report have assumed > that I2 is half > > of the current in the C1 wires. That turns out to be not true. Instead the > full current I1 is alternating between the two wires in the C2 wire pairs, > so the voltage drop will be the same as for a single wire. For calculation > of the resistance Re in the wire system, see paragraph E1in the spreadsheet > and reference. > > From: Brad Lowe > > http://lenr.fysik.org/eCat/COP=1_or_3.pdf > > Sent from my iPhone
