W5DXP wrote:
wrote:
W5DXP wrote:
I forgot to add. At the "zero-power points", either voltage or current is
zero. All that means is that all the energy is contained in the opposite
field.
Not quite all. It also means that there is NO power since P = V x I.
It means there is no NET power transfer.
Do not be afraid to admit that you have changed the definition of P = V
x I
and therefore do not accept the standard definition.
When I learned Pinst = Vinst x Iinst there were no caveats about how
Pinst meant Pnet. Instantaneous energy is flowing or it is not. When
Pinst
is 0 for all time, then there is no energy flowing.
To satisfy your theory (and minimize double think), you have had to
change
this to Pnet is zero to allow these cancelling powers to flow. So be it.
There are power flow vectors
in both directions that are equal in magnitude. Reference: _Fields_
and_Waves_in_Communications_Electronics_ by Ramo, Whinnery, & Van Duzer,
section 6.10, page 350, where they describe Pz-, the reflected wave
Poynting vector and Pz+, the forward wave Poynting vector.
There can be lots of energy present but none of it is flowing past
the zero voltage or zero current point; hence no power.
There is a forward power flow vector and a reflected power flow vector.
There is no net power flowing past any point.
True, sort of. At the quarter wave points where voltage or current are
always
zero, there is no energy flowing. Period.
At other points, energy flows in one direction for a quarter cycle and
then in the other direction for the next quarter cycle, producing a net
of zero. A true instantenouse power meter (one which measures V and I
and
displays V x I) will easily demonstrate this. As a thought experiment,
move such a true power meter along a shorted or open line and think if
its indications in the time domain, then do the averages. It will be
quite instructive. Repeat for a line terminated in other than its
characteristic impedance.
By the way, since energy flows forward for a quarter cycle and backwards
for the next, the maximum distance travelled by this energy is one
quarter
wavelength on the line. It is not flowing all the way to the end of the
line and then back. There is not enough time for this to happen (on a
multi-wavelength line) since it changes direction every quarter cycle.
There are, however, equal
magnitude component constant power flow vectors flowing in both directions.
To believe that energy is flowing across a zero voltage or zero current
point requires the rejection of the view that instantaneous power is
equal to instantaneous voltage multiplied by instantaneous current.
No, it doesn't. It only requires acceptance of Ramo, Whinnery, & Van Duzer.
However, to reject energy flow across a zero voltage or zero current
point requires a confusion of cause and effect. Energy flow in both
directions is the *CAUSE* of the standing waves. You simply cannot
turn around and say that standing waves eliminate their own cause
but continue to exist anyway.
Not quite. Standing voltage and current waves (which are not waves in
the normal sense) can be observed on the line. They can be measured with
real voltage and current instruments; as can real energy flows with
a real (V x I) power meter (but not a 'Bird watt' meter which
is doing something quite different). It happens that if you assume the
existence of forward and reverse voltage and current waves, mathematical
functions can be derived that will produce the same distribution of
voltage and current as observed on the line. This is extraordinarily
convenient some analysis but does not mean that these assumed waves are
real.
A mechanical analogue would be to look at a guy wire on a pole. You can
analyze the forces as two vectors at 90 degrees (or any other angle of
convenience!), but never make the mistake of assuming that there are
actually two guy wires present. Just because it is mathematically
convenient to assume the existence of two vectors does not mean they
exist.
Rejection of
P = V x I would have wide impacts on our understanding of electrical
power and energy flows.
Nobody is rejecting it. If the lossless stub is one second long, it takes
two seconds of *POWER* to bring it to steady-state. If the stub contains
no moving energy, where did all those joules go?
This energy is indeed stored in the stub. None of it moves across zero
voltage
or current boundaries. Between these boundaries, it is indeed moving as
it changes from being stored in the capacitance and inductance of the
line.
This time variation of the location of the stored energy produces the
observed
voltages and currents on the line.
They cannot disappear
Absolutely. They do not disappear.
or stand still.
With AC excitation they do not stand still, but when similar analysis is
done
for a line excited with a DC source, the energy does indeed stand still.
An
open line stores the energy in the capacitance and a shorted line stores
it
in the inductance.
If they are moving, power exists.
Yes, but it never moves more than a quarter wavelength.
....Keith