Re: rotating magnetic field



<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.
>

|>
|> 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.
-------------

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.
-------
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 >


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


| 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.
------------

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 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).
----------
Yes, the DC generator can be considered as an ideal source with a resistance in series. This series resistance is as low as it can practically be made (superconducting if you want to spend the money). It is not a "big" resistance but just the internal winding resistance of the machine- low losses are desirable.
>
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.

--
Don Kelly
dhky@xxxxxxxxxxxx
remove the x to reply

.

Relevant Pages

  • Re: rotating magnetic field
    ... I pointed out that the concept of maximum power transfer does not assume that one matches the source impedance to the load- but the load to the source. ... So for DC the series impedance is just the wire ... When you look at a generator again you commonly use a Thevenin equivalent unless you reallly want to write a set of simultaneous. ...
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  • Re: rotating magnetic field
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