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

Roy Lewallen wrote:

...
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.)
...


This:
2. Microwave ovens use standing waves to cook food. This means that
nodes, where the amplitude is zero (where the wave crosses the x-axis),
remain at nearly fixed locations in the oven, and cooking won't occur at
those locations.

From he
http://faculty.fortlewis.edu/tyler_c..._microwave.htm

Now, you can argue that any damn way you wish, but "standing waves of no
power" is a myth for idiots!

Regards,
JS

John Smith wrote:
Roy Lewallen wrote:

...
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.)
...


This:
2. Microwave ovens use standing waves to cook food. This means that
nodes, where the amplitude is zero (where the wave crosses the x-axis),
remain at nearly fixed locations in the oven, and cooking won't occur at
those locations.

From he
http://faculty.fortlewis.edu/tyler_c..._microwave.htm

Now, you can argue that any damn way you wish, but "standing waves of no
power" is a myth for idiots!

Oddly, "John", another source tells us:

Another hazard is the resonance of the magnetron tube itself. If the
microwave is run without an object to absorb the radiation, a standing
wave will form. The energy is reflected back and forth between the tube
and the cooking chamber.

http://en.wikipedia.org/wiki/Microwave_oven

Dave K8MN

John Smith wrote:
Roy Lewallen wrote:

...
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.)
...


This:
2. Microwave ovens use standing waves to cook food. This means that
nodes, where the amplitude is zero (where the wave crosses the x-axis),
remain at nearly fixed locations in the oven, and cooking won't occur at
those locations.

From he
http://faculty.fortlewis.edu/tyler_c..._microwave.htm

Now, you can argue that any damn way you wish, but "standing waves of no
power" is a myth for idiots!

Another source, "John", says:

Why is food cooked in a microwave oven sometimes not cooked uniformly?

Inside the microwave oven, the microwaves bounce off the metal internal
walls and set up complex 'standing wave' patterns. As with any wave,
microwaves have peaks and troughs and the intensity of the microwaves is
greatest in the peaks and troughs and lowest at points in between.

So if some food is near one of the peaks it will absorb lots of
microwaves and get really hot, while if it is midway between peaks and
troughs it may receive hardly any microwaves and so not get very hot at
all.

http://www.bbc.co.uk/food/tv_and_rad...icrowave.shtml

So you're telling us that what cooks food in a microwave oven is the
standing waves and that it isn't because the food itself is the load for
the output of the magnetron? You'd have us believe that standing waves
which result from operating a microwave oven without such a load are
present and actually cooking food?

Dave K8MN

I'm surprised that standing waves seem so hard for people to understand.
They're simply a spatial pattern formed by the interference between
forward and reflected waves or, if you prefer, from the solution to a
general transmission line problem with boundary values applied. I'd hope
that even a brief look at a text would make it clear what standing waves
are, and what they are not.

Roy Lewallen, W7EL

Roy Lewallen wrote:
I'm surprised that standing waves seem so hard for people to understand.
They're simply a spatial pattern formed by the interference between
forward and reflected waves or, if you prefer, from the solution to a
general transmission line problem with boundary values applied. I'd hope
that even a brief look at a text would make it clear what standing waves
are, and what they are not.

Roy, if you would take a "brief look at a text", you
would know NOT to try to use standing wave current phase
to measure the delay through a 75m loading coil. At any
point in time, the standing wave current phase is essentially
the same value all up and down a 1/2WL dipole including
through any loading coils. Phase does NOT equate to delay
in a standing wave environment.
--
73, Cecil http://www.w5dxp.com

Dave Heil wrote:

...
Inside the microwave oven, the microwaves bounce off the metal internal
walls and set up complex 'standing wave' patterns. As with any wave,
microwaves have peaks and troughs and the intensity of the microwaves is
greatest in the peaks and troughs and lowest at points in between.
...

Actually, a bit more than that, even ...

Water molecules are slightly magnetic, they are spinning like hell on
those "humps of the standing wave"--friction cooking, you will excuse my
"artistic authors' license" ... or not, in ALL of this ... ;-)

However, the standing wave is much more appreciated, by me--at this
point, thanks to Cecil. And, my attention is always drawn towards
"little oddities" which can serve as diversion. And, the standing wave
IS cooking the turkey--in my humble opinion ... you could say, "I
believe in standing waves."

But then, tomorrow night, I'll be up very late with cookies and milk.
Yanno, I've never seen 'em--yet? ;-)

Regards,
JS

John Smith wrote:
Dave Heil wrote:
[...]

forgot ...

Merry Xmas Heil,
JS

John Smith wrote:
And, the standing wave
IS cooking the turkey--in my humble opinion ... you could say, "I
believe in standing waves."

If one thinks about it, one will realize that an unchanging
steady-state standing wave cannot cook the turkey. Cooking
the turkey would require the standing wave to give up energy
and if it does, it is no longer a steady-state standing wave.
All of the joules/sec delivered to the load during steady-
state is from traveling-wave energy whether it be in a
microwave oven or in a transmission line.
--
73, Cecil http://www.w5dxp.com