Well Dave, you have made a good and convincing argument. My hat is off to you and I need to eat it with a big public helping of crow.
It seems like if we go back to basics, the average power is integral((I dot V)dt)/integral(dt). If I is a constant, then you can pull I outside the integral and you get the average power as I x integral(Vdt)/integral(dt), which means the average power is I times the average voltage. Thank you for taking the challenge, making me rethink, and putting me straight! Regards, Bob On Mon, Oct 27, 2014 at 5:53 PM, David Roberson <[email protected]> wrote: > Bob, I take that as a challenge. I am not offended my friend, but find > this a great opportunity to prove what I am saying is correct. I predict > that you will agree with me once you have an opportunity to dig deeper into > the subject. > > It is not clear to me what you are showing in your example, perhaps due to > a problem with my display. Let me choose an example for you to consider. > Again, we can assume that the current being delivered into the load is > exactly 1 amp. If we further assume that the load resistance is 1 ohm, > then under DC conditions we will measure precisely 1 volt across the load > resistor. > > I and I assume you would calculate the power as being 1 watt delivered to > the load resistor under this static condition. Now, suppose that the > resistance changes to .5 ohms. In that case the voltage becomes exactly .5 > volts. This results in a power being delivered to the resistor of .5 > watts. For the other half of the AC square waveform the resistor becomes > 1.5 ohms. In that case the power delivered becomes 1.5 watts since 1 amp x > 1.5 volts = 1.5 watts. Since we are assuming a symmetrical AC waveform, > this is a pretty good example of that with numerous harmonics that also get > into the act. The assumed waveform is therefore a 1 volt peak to peak > square wave that is riding upon a 1 volt DC average. > > So the total power average becomes (.5 watts + 1.5 watts) / 2 = 1 watt. > Each half of the waveform makes its contribution and they balance each > other out about the normal DC average of 1.0 watt. This is true for all AC > waveforms, regardless of the harmonic content provided that the current > retains a constant DC value. > > I have stated this on numerous occasions and it is a general concept. > Power can only be extracted from a source current that flows at the same > frequency as the source voltage. In this case the current is at a DC > frequency, so no power can be extracted from the source except into a DC(0 > Hertz) voltage related load. > > Dr. McKubre essentially made the same statement with respect to his > experimental setup. Another feature of a constant current environment is > that the power delivered into the load varies directly with the load > voltage and not proportional to the square of the voltage as is normally > encountered. That is what allows the average to be used in this case > instead of having to deal with the messy RMS waveform additions. > > If you have reservations about what I have stated I strongly suggest that > you put together a Spice model. That will prove that what I am saying is > right on target. > > Dave > > > > > -----Original Message----- > From: Bob Higgins <[email protected]> > To: vortex-l <[email protected]> > Sent: Mon, Oct 27, 2014 3:11 pm > Subject: Re: [Vo]:questions on McKubre cells and AC component > > I hate to say this, but what you say is absolutely wrong. You could > only do as you describe if the "voltage" being averaged is the RMS > voltage. You cannot take the average voltage and multiply it by the > average current to get average power. For example, suppose that the > voltage was V=1_0.5sin(wt). The average of this voltage is 1. Lets say we > have a constant current of 1A. By your method, the power would be 1 watt. > However, the actual power is: > > P = (1A) sqrt(mean(1+0.5sin(wt))^2)) = (1A) sqrt(mean(1 + sin(wt) + > 0.25sin(wt)^2)) = (1A) sqrt(1+.25 (mean(.5 - .5cos(2wt)))) > > P = (1A) sqrt(1.125) = 1.0607 Watts > > On Mon, Oct 27, 2014 at 11:58 AM, David Roberson <[email protected]> > wrote: > >> The instantaneous power being delivered by the source is equal to the >> product of the current and voltage. When the current is constant, only DC >> voltage loads can accept power and thus energy from the source. All of the >> AC voltages that appear across the source terminals integrate to zero >> during a full cycle and do not enter into the input power equation. This >> understanding seems to escape most people until they review the theories >> carefully. I had to prove roughly the same issue to several skeptics that >> thought that DC due to load rectification of the AC power source could be >> used to sneak extra power into the earlier ECAT. They thought this was >> possible since the input power meter did not monitor DC directly. >> >>
