Standing-Wave Current vs Traveling-Wave Current
There's been a lot of confusion between average and instantaneous power.
Let me try to clarify a little.
Keith Dysart wrote:
. . .
Now power is really interesting. Recall that
P(x,t) = V(x,t) * I(x,t)
. . .
This is correct. Cecil and others have often muddled things by
considering only average power, and by doing this, important information
is lost. (As was the case of the statistician who drowned crossing a
creek whose average depth was only two feet.)
Roger wrote:
I would suggest that you add a caveat here. The power equation is true
if the measurements are across a resistance. If we are also measuring
reactive power (or reflected power), then we need to account for that.
And here's where one of the common errors occurs. The fundamental
equation given by Keith does "account for" reactive power. If I(t) and
V(t) are sinusoidal in quadrature, for example, then V(t) * I(t) is a
sinusoidal function (at twice the frequency of V or I), with zero
offset. This tells us that for half the time, energy is moving in one
direction, and for the other half the time, energy is moving in the
opposite direction. The average power is zero, so there is no net
movement of energy over an integral number of cycles. This is what is
called "reactive power".
On the other hand, if V(t) and I(t) are in phase, the product is again a
sinusoid with twice the frequency of V or I, but this time with an
offset equal to half the peak value of V times the peak value of I. What
this tells us is that energy is always moving in the same direction,
although its rate (the power) increases and decreases -- clear to zero,
in fact, for an instant -- over a cycle. The average power equals this
offset. So here the "reactive power" is zero. I've described this as
energy "sloshing back and forth", which Cecil has taken great delight in
disparaging. But that's exactly what it does, as the fundamental
equations clearly show.
When V(t) and I(t) are at other relative angles, the result will still
be the sinusoid at twice the frequency of V and I, but with an amount of
"DC" offset corresponding to the net or average power and therefore the
average rate of energy flow. The periodic up and down cycles represent
energy moving back and forth around that net value.
No additional equation or correction is needed to fully describe the
power or the energy flow, or to calculate "real" (average) power or
"reactive power".
A careful look at what the power and energy are doing on an
instantaneous basis is essential to understanding what the energy flow
actually is in a transmission line. Attempting to ignore this cyclic
movement and looking only at average power can lead to some incorrect
conclusions and the necessity to invent non-existent phenomena (such as
waves bouncing off each other) in order to hold the flawed theory together.
Roy Lewallen, W7EL
|