Re: rotating magnetic field

On Wed, 20 May 2009 23:33:32 -0700 Don Kelly <dhky@xxxxxxx> wrote:

|> | This is a misconception of the maximum power transfer theorem.
|> | For a given source impedance (including the lines) the power delivered
|> to
|> | the load (DC case, AC is a touch messier) will be a maximum when the
|> load
|> | matches the source impedance. The efficiency at this point will be 50%
|> which
|> | is lousy.
|>
|> I think we're using different terminology to say the same thing.
|
| We are Definitely NOT saying the same thing.

Then you would have to be saying something totally wrong about very basic
electrical theory, and I'm sure you know that volts/amps -> ohms, so what
else can it be?


|>
|> Sounds like Harry Potter engineering to me.
|>
|> If the source has a voltage of 10 volts and there is a current of 0.5
|> amps,
|> then the source system impedance is 20 ohms.
| ----
|
| And that would mean that the load is 0 so the power delivered to the load
| and the efficiency will be 0. Please note that I also specified the load
| resistance. That is , I specified the system ( quoting you: "the system
| (source and load) impedance"). "System" impedance is NOT "a description of
| the work-transfer relationship". If it is, then the output power of a step
| up transformer should be higher than the input power because the impedance
| V/I is higher. is that what you believe?

If the voltage at the source is X and the current is Y, then the source's
system impedance is effectively X/Y. But that doesn't tell you what the
series impedance is. If you know the actual series impedance, you can then
figure out the transfer relationships for all external series impedances.
Or you can experimentally observe the changes in current with varying series
impedances and interpolate the series impedance internal to the source.

AC would have to consider reactive effects (and do the impedance calculations
with complex numbers). DC doesn't need to consider that once the current is
at a steady state. So for DC the series impedance is just the wire resistance
and any effective resistances that might exist due to voltage drop at contacts.


|> | For a 10 ohm source and 11 ohm load, the current will be 0.48A and the
|> load
|> | power will be 2.49Watts at an efficiency of 43%
|>
|> Sounds to me like you're talking about series impedance/resistance, not
|> system impedance.
| ----
| But I am considering the "system impedance" by your definition. Note that
| I specified both source and load impedances.
|
| Think Thevenin. The whole of the interconnected power grid behind your 120V
| outlet appears as a single equivalent voltage source (120V nominal) behind
| an impedance. You then plug in your kettle which is a resistive load and
| this load is in series with the source and its impedance as seen at the
| outlet.

The AC power grid is much more complex. Correct calculations need to know too
many things. But that level of accuracy is generally not needed. For example,
we don't need to calculate the exact fault current; we only need to calculate
a practical ceiling for the fault. That's done by assuming an infinite fault
current behind a reasonable level of components (usually the last transformer)
and the wiring resistance.


| Lets see, given the whole NW power grid as seen from your outlet, the open
| circuit voltage is 120V and the impedance looking back into the grid is
| dominated by the local distribution transformer (ignoring the LV
| lines)-which for a 5KVA transformer is of the order of 0.1-0.2 ohms so that
| the system consists of a 120V source in series with a 0.2 ohm impedance as
| seen from the terminals of the outlet. The kettle is an impedance of the
| order of 10 ohms connected in series with the source and its impedance.
| Congratulations, you got it right .

When you plug in a 1440 watt kettle, you are NOT (by a longshot) coming anywhere
close to efficiently transferring all the power from the generating plants. This
situation doesn't even need to consider power transfer efficiency. Power loss in
transmission is all that is of concern (losses in the generator are, too, but
that is something the power company has to deal with).

If you run a generator at a reduced capacity (which means a reduced excitation
field for AC since it has to stay the same speed), you are changing its system
impedance (of the generator, not of the the whole grid).


|> A low SERIES impedance is, of course, very much desired. But that is not
|> what I was talking about.
| ------
| Then what hat are you talking through? Elucidate.
|
| You have a generator of some sort and this generator has internal impedance,
| and presents 2 terminals (DC or single phase) to the world. Multiple
| windings in the generator will increase the voltage as well as increase its
| impedance *in series with the internal voltage source* You connect a load
| at the terminals and this load IS in series with the source and internal
| impedance (same current= series).

It has the internal resistance (DC at steady state, so disregard reactance).
That is the series impedance. This is like a generator with zero internal
resistance with a big resistor in series with it. That resistor dissipates
power due to the voltage drop it has, which is proportional to its resistance
times the square of the current. The source system impedance is measured
without that resistor. The effective system impedance is measured with that
resistor (because you really can't avoid the resistor, since it is part of
the winding in a real world).

The higher the output voltage, the higher the source impedance. From that
you can then figure the effect of the transmission series impedance. The
higher the source impedance, the lower the effect of the transmission series
impedance. The load will also have a higher impedance. But since its effect
is the desired effect (for example, if it is a 1440 watt kettle, you desire
the dissipated heat there). At 120 volts, that's 12 amps, so the impedance
of the load is 10 ohms. That assumes no voltage drop in transmission so all
120 volts is applied to the load. If we look at a scenario with absolutely
zero series impedance everywhere (except in the heating elements of the kettle
itself, of course), including the generator winding, then we have a pure
100% power transfer (all power generated is dissipated in the load and not
anywhere else). Then the system impedance is unimportant. We could make it
be lower (12 volts, 120 amps) and it would still be a 100% transfer. But in
reality, there is series impedance, so we want to have the system impedance
of the source and load be higher to overcome the series impedance (all of it,
including the series component in the generator itself).


|> Your box seems to be locked tight, not letting you outside.
| ----------
| My box is open but it is true that it doesn't admit nonsense which is what
| you are peddling. You are hand waving and your what- iffing ignores reality.
| As long as it does, it is pipe dreaming. It does show that, in spite of
| your experience with commercial wiring practice, it appears that you have
| nearly no real knowledge of either electromagnetic energy conversion nor
| circuit analysis and (from previous posts) power systems.
| Why not deal with these deficiencies? you obviously have the ability to
| think and could do well, if you stopped to fill in the gaps. Why waste this
| capability? It is difficult to see the forest if you ignore the trees. This
| is what you are doing.

It's not nonsense. It's just not sense to you.

You're still convinced that the magnetic field from a rotating disk, where the
axis of rotation is in line with the magnetic polarity orientation, is just
the same as a non-rotating disk? You're still convinced that a metallic
conductor oriented radially and rotating around the same axis (such as being
carried by a non-conductive non-ferrous disk) will have a potential due to
the field and the motion, whether the magnet is rotating with the conductor
or not? You're still convinced that a conductor in a magnetic field will have
a potential due to the field and the motion, whether the magnet is in motion
with it or not (over a uniform field with no changes in orientation or density
or stength within the area of motion)?

So what we need is a real life device that can show the induced potential in
a conductor (wire segment) without depending on current (so we don't have to
complete a circuit loop) so we can really see where the magnetic field is
inducing a potential, and where it is not.

Note that I still use the term "induced" for expressive purposes, even though
the reality is that magnetic fields, electric potentials and current, and
mechanical force and movement, taking place in time, are merely simultaneous
components of the one and same nature of matter, time, and space.

--
|WARNING: Due to extreme spam, googlegroups.com is blocked. Due to ignorance |
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| Usenet from these places, find another Usenet provider ASAP. |
| Phil Howard KA9WGN (email for humans: first name in lower case at ipal.net) |
.

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