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AVERAGE power revisited



Original poster: "Gary Johnson by way of Terry Fritz <twftesla-at-uswest-dot-net>" <gjohnson-at-ksu.edu>

John, Terry, Harvey:

I did not want to start a long thread on power, but rather to encourage
those of us who should know the definitions to use them with precision.

Terry, you are right that my definition of average power implies steady
state operation.  It also implies single frequency operation.  You would not
believe all the papers I have reviewed for the IEEE Transactions on
Instrumentation and Measurement that were attempting to define real,
reactive, and apparent power for non sinusoidal waveforms. I finally came to
the conclusion that further efforts to develop more mathematics were just
nonproductive.  But as long as we stick with single frequency, steady state
operation, the math works well.

John, you ask about power gain. I think you understand it correctly, but I
would probably use the term "power processor" instead.  Power gain might be
interpreted as implying over unity operation, or "free energy", which I
don't think you want to say.  As you know, the primary capacitor accepts
power from the power company relatively slowly (a quarter cycle of the
60(50) Hz wave) and then discharges it rapidly. The input current waveform
is not sinusoidal, but it can be considered to consist of a 60(50) Hz
fundamental plus harmonics.  The watthour meter acts as a low pass filter
and basically just responds to the fundamental.  If the reading on the
watthour meter increases by 1 kWh in 1 hour of operation, then we say you
have used an average power of 1 kW. This idea of energy divided by time
works all the way down to one cycle, so we could talk about the average
power for 1/60 of a second and still be mathematically correct. Of course,
if your coil does not fire every cycle, the value you arrive at for a given
cycle may not be representative.

As far as I know, the energy in an hour's worth of sparks will be less than
the energy recorded on your watthour meter. However, the sparks are not
continuous, but exist only a small fraction of the time. When you take the
energy in a spark divided by the time of existence, the power will be much
higher. I hate to use the term average power for the spark since it is
intermittant and variable, preferring peak power or instantaneous power
instead. You asked about 3 kW in (average) and 1700 kW out (peak). I
certainly have no problem with those numbers.

Harvey, you asked about the difference between average and rms power.
Consider a capacitor (ideal) connected to household power. It absorbs power
(the integral of instantaneous voltage times instantaneous current over
time) until it reaches peak voltage. It then delivers power back to the
utility the next quarter cycle. It absorbs power again the next quarter
cycle as voltage goes in the negative direction, and so on. The plot of
power as a function of time is symmetric, positive half the time and
negative the other half. When we take the *average* of this power, we get
zero. The capacitor does not get hot, nor does the watthour meter turn. On
the other hand, if we calculate the *rms* power, we take the square root of
the integral of the square of the instantaneous power, and get a number the
same as if we had a resistor drawing the same current. Like I say, this
computation has no meaning that I know of.

Also, Harvey, you claimed that average power for a resonant circuit should
be computed without using power factor. It is true that power factor is
unity at resonance, but this does not change the definition. Average power
into a passive circuit (R, L, and C), single frequency, steady state, is
given by the rms voltage across the terminals times the rms current into one
terminal times the power factor. If you can reference a circuit theory book
that says otherwise, I would be interested in seeing it.

Gary Johnson