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Re: AVERAGE Power/My Stupendous Error



Original poster: "harvey norris by way of Terry Fritz <twftesla-at-uswest-dot-net>" <harvich-at-yahoo-dot-com>


--- Tesla list <tesla-at-pupman-dot-com> wrote:
> Original poster: "by way of Terry Fritz
> <twftesla-at-uswest-dot-net>" <free0076-at-flinders.edu.au>
> 
> 
> 
> On Mon, 8 Jan 2001, Tesla list wrote:
> 
> > Original poster: "by way of Terry Fritz
> <twftesla-at-uswest-dot-net>"
> <Parpp807-at-aol-dot-com>
> > 
> > Is it really so confoosin? The average power is
> the product of the
> > RMS voltage and current.
> > 
> > Cheers,
> > Ralph Zekelman 
> > 
> 
> And the power factor.
> 
> 
> Darren Freeman
> 
Sorry for raisin such a ruckus as I was the original
poster for this,and got upset and made derogatory
comments about the status quo of electrical
engineering  and also against those that propound it.
My Sincere Apologies for this. My misunderstanding was
that I had assumed the reactive power and the actual
I^2R  power expended on the inductor would be
approximately the same values. I now understand this
is only true at unity power factor. The
misunderstanding I had was that I became confused and
thought that what was being said was that the power
computed by Ohms law on the inductor or I^2R had to be
further reduced by multiplying in the power factor,
which of course is false and due to my
misunderstanding of the problem. Why I thought that VI
or the apparent power would equal I^2R, or the true
inputed power is explained after several extracts from
the following; 
     Here are some useful references from Herbert W
Jackson's "Introduction to Electric Circuits"
concerning  Power in Alternating Circuits. It may be
possible by reading these references  to see where my
misunderstanding came in.

   "Since a pure inductance in an ac circuit
alternately takes from and returns to the source equal
amounts of energy, THE AVERAGE OR TRUE POWER IN A PURE
INDUCTOR IS ZERO."[Intoduction to Electric Circuits,
Herbert W Jackson, pg 387]

    As can be imagined a PURE capacitance is defined
in the same way as a PURE inductance. Because of the
reactance there is no  true (average)power input. This
is because during one half cycle energy is being
stored in either magnetic or electric fields and in
the next half cycle it is being returned. That is the
reason and necessity for the term average. At any
particular moment in time there MAY be a power input,
but the total AVERAGE power input over cycle time is
zero. Now concerning a pure resistance; from pg 385,
"Since the voltage drop across a pure resistance must
be exactly in phase with the current through it, the
instantaneous voltage and current reach their peak
values simultaneously.  Therefore whenever i is a
positive quantity, e is also a positive quantity; and
whenever i is a negative quantity, e is also a
negative quantity. Since the product of two negative
quantities is a positive quantity, the instantaneous
power graph for a pure resistance is always positive."

     Now what happens when we are making the power
curve for the pure inductance example is that the
impressed amperage lags behind the impressed voltage
by 90 degrees. Since the instantaneous power curve is
the voltage times the amperage charted out over time,
this product will be a sine wave of twice the
frequency of the input frequency. This is because at
every quarter cycle of the input frequency either the
voltage or amperage quantities have reached zero in
their polarity change. Again the average power is
zero, since power in = power out. Now let us see what
Jackson says about adding a "true" resistance into the
reactive circuit. From pg 390;
    "The instantaneous  power graph for an ac circuit
containing both resistance and reactance is again a
sine wave of twice the frequency of the voltage and
current. But the instantaneous  power graph is neither
all positive, as in the case of a pure resistance, nor
equally positive and negative, as in the case of a
pure reactance. Since the instantaneous power graph is
more positive than negative, there is an average or
true-power component, which represents the true-power
input to the resistance portion of the circuit. A
portion of the positive instantaneous power is offset
by the negative instantaneous power TO ACCOUNT FOR THE
REACTIVE POWER OF THE REACTANCE OF THE CIRCUIT."

    Jackson now describes the condition of power in a
circuit containing resistance and reactance.
    "If a circuit consists of equal resistance and
inductive reactance in series, the current lags the
the applied emf by 45 degrees. Therefore when we plot
the instantaneous power it is a positive quantity
between 0 and 135 degrees  and becomes a negative
quantity only for the short period of time between 130
and 180 degrees." Glancing at this graph in the text,
it is also noted that the area under the negative
portion is also very small compared to the positive
portion. So at least here we can see that there is not
quite a linear relationship of the amount of returned
energy by varying the phase angle conditions from 90
degrees to 45. 
    Now here is how I made my monumental error of
assuming that the apparent power was equal to the true
I^2R value. I had  noticed that in resonance that only
one of the equations for power would be valid.
Ordinarily from Ohms Law V=IR. The Power requirement
ordinarily means that multiplying both sides of the
equation by I yeilds VI=I^2R. But these quantities are
not equal in resonance, or the way I understood
things,(which has already been shown to be wrong!) To
give an example the 1000 ohm 56 henry coil will
experience a voltage rise during series resonance, so
a meter measurement of the voltage times the amperage
will give a VI that does not equal I^2R, but a much
higher value. Therefore for series resonance I^2R must
be used for the true power input. By this
understanding and the fact that I thought that I
understood the meaning of VI being the apparent power
of a reactive curent; I simply deduced that the
reactive power measurement method yielded a result far
higher than the actual true power input. Therefore I
THOUGHT I knew what this meant, but evidently I did
not. Now for the inverse condition of parallel
resonance because of the parallel resonant rise of
amperage if the power were measured with the amperage
term I^2R inside the circuit it would yeild another
false reading of the true input power,being far
greater than the true input power; thus to derive it
in that case the true power would be derived by VI as
measured in the input. So in all this confusion I
assumed that in the inductor case of reactive current,
since there was no resonant rise of voltage, that VI
WOULD EQUAL I^2R. This is where I made MY STUPENDOUS
ERROR! I knew that in the non resonant condition the
amount of current in the  new phase angle condition
had already been reduced  from the resonant current
conditions by whatever phase angle that inductance
made in the new phase angle. This is why I reasoned
that it was okay to call VI the apparent power,and
that this should equal the true power because the
amount of amperage in that circuit had already been
reduced by the new larger phase angle. This is then
why I then considered it heresy to say that to find
the true power input the cos of the phase angle must
be multiplied by that result. I now FINALLY understand
that when this is done it yeilds the I^2R quantity,
which is the true power expended on the inductor. So
sorry for all the Hoi Poloi here, I have seen the
error of my ways. At least I have finally done so, and
have actually been convinced of the truth.

Apologies Again, HDN 


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