Re: rotating magnetic field

On Mon, 25 May 2009 16:57:42 -0700 Don Kelly <dhky@xxxxxxx> wrote:
|
|
| <phil-news-nospam@xxxxxxxx> wrote in message
| news:gv75k001j84@xxxxxxxxxxxxxxxxxxxx
|> 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?
|
| -----------
| I pointed out your misunderstanding of the maximum power transfer theorem
| and I also pointed out the inadvsability of applying it in practice. Then
| you prattled on about the need for high "system impedance " (undefined
| term)which is nonsense. Finally you pull Ohm's law to show ???
| Please note that Ohm's Law is not a power relationship but if you use it and
| power =EI. please note that the resultant expression for power is inversly
| proportional to resistance.
| The weakness in circuit theory is yours.

You didn't point to anything specific. It appears to be more of a case
that we are using different terminology.

A system delivering X watts with voltage P and current Q, where X=P*Q, is
a system with a higher impedance than one delivering X watts with voltage
P/10 and current Q*10.



|> |> 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.
| -----
| The impedance in this case it the total effective impedance of the "system"
| not the source's internal impedance. So what is it that you are talking
| about.
| -------------

Your appear to be referring to the series impedance (or resistance) by
your reference to "source's internal impedance". I have been talking
about the system impedance of the source, not its series impedance.


|> 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.
| -------
| OK except that the source has resistance as well. For both AC and DC, a
| Thevenin model can be used to represent a complex system >

I assume you mean the Thevenin theorem. I didn't see a need to use it at
this point since it doesn't distinguish the kinds of impedances you might
see inside the source.


|> 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.
| ----------
| One can represent the complex power grid by a simple Thevenin model- it is
| done all the time. A more generalized form of this is the Zbus matrix which
| is commonly used. As for a practical ceiling for the fault- it is apparent
| that you didn't read what I said (below).
| Nobody assumes an infinite fault current upstream of the last transformer-
| what is assumed is an "infinite bus" which is an ideal, finite, voltage
| source (with, it is true, an infinite current on short circuit because an
| ideal source has 0 series impedance). Somehow, working with an infinite
| current

And how does this diversion even relate to the original point? Are you
trying to drag this over to a discussion of how to model a regional AC
power grid? I don't have the interest in that at this time.



|> | 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).
| --------
| You are going off on a tangent again- with a mix of truth and fiction.
| Certainly a kettle is a miniscule part of the system and one is not so
| stupid that it is going to have a measurable effect on the grid and the
| total loss. However, all losses, including the transmission system),
| upstream of the customer's meter, are the power company's problem and the
| power company like to keep losses ( real and reactive) as low as
| possible-this is not conducive to having resistances and reactances any
| higher than the minimum possible.

It's not a miniscule part if your dealing with a Faraday-like generator
than can produce perhaps 3000 watts.

I suggest we not talk about power grids. This was, afterall, originally a
discussion of a generator and its efficiency (and presuambly also the
efficiency of the connected transmission system, which might be a mere one
meter long, or hundreds of kilometers).


| ------------
|>
|> 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).
| --------
| The above statment is pure nonsense. You have no idea of what is involved in
| the control of a generator in a grid.
| Changing the excitation of a generator changes its reactive output, not the
| real power. To change real power, you tweak the governor to try to change
| its speed to change the phase of the internal voltage.
| You are also using "impedance" The actual impedance of a generator or motor
| doesn't change -Ohm's Law doesn't apply to active sources. Noone in the
| utility business looks at a generator or motor as a "system impedance">

A phase change along does not increase the power.

And I'm not interested in an AC generator right now, or the phase issues
it has in an interconnected network. I'm more interested at the moment
in a DC generator, or perhaps at some point, more than one DC generator.


|> 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.
| OK
| But in
|> reality, there is series impedance,
| OK
| 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).
| NOT OK In effect the generator is modelled as an ideal voltage source
| behind a series resistance (line + generator internal resistance) So you
| have something like:
| E=(Ri+Rl +RL) I where Ri is the generator internal resistance, Rl is the
| line resistance and RL is the load resistance. Now the generator resistance
| is Ri. The source is a constant (or externally controlled) ideal source
| which has no resistance (by definition). If we consider the generator alone
| {E-RI} we can forget about Ohm's Law as it doesn't apply. If you wan't to
| call E/I a "system" impedance, fine, but it is a meaningless term.

This is a series model. It's telling you how the circuit will behave if you
have a fixed voltage and known series resistances. This has nothing to do
with the system impedance. The system impedance matters when the source is
NOT the ideal voltage source (which is a real life scenario).

I don't see any point in trying to apply this model.

--
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| Phil Howard KA9WGN (email for humans: first name in lower case at ipal.net) |
.

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