Determination of direction in AC Power Flow

On 11/23/06 9:47 PM, in article LHv9h.356076$1T2.43031@pd7urf2no, "Don


<snip>
I fully agree. Maybe Don can be more more persuasive in explaining the carrier charge is irrelevant when it comes to what is known as circuit theory. I sure have failed.
Bill -- Fermez le Bush
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Slight slight slight is accurate.

Noone said they just move chaotically. Noone said they are just "joy-bouncing" around.

Well if the typical speed is 10^6 m/s and the drift velocity is 10^-4 m/s then one part in 10^10 of the velocity ends up averaging forward motion. The net current may be as little as one part in 10^10 of the moving charge. Even if this is out by a few orders of magnitude it still looks like your opinion is incorrect.
and then you might get a

OK, go ahead and clue me in.
as to the fact that this isn't age old electrostatics, this is

You think the way we spell words influences the movement of electrons in a conductor? By the way, electrostatics also starts with ELECTR, since you didn't notice.
j
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<snip>
Agreed. In another life, we studied neutron diffusion. Neutrons born from fission have very high velocities and slow by colliding with other materials. Once they reach equilibrium thermal energies, they continue colliding with materials until absorbed. The total distance a neutron travels is important in reactor design. The distance is determined in two parts, the 'slowing down length' and the 'diffusion length'.
Both electron movement and neutron movement are akin to Brownian motion. Many collisions, resulting in many steps in random directions, ending with a net movement.
The thermal energy of an electron is almost (if not all) in the form of simple kinetic energy, so solving KE=1/2 m V^2 can give the velocity of an electron that has 'average' thermal energies.
The rate of 'drift' due to Brownian motion is influenced by the application of an electric field.

You two may just be argueing about symantics. The electron at any instant may have a speed of 10^6 m/s, but it doesn't make a net progress through the lattice at that speed. Yet, with an electric field applied, it does make progress through the lattice.
daestrom
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On Sun, 19 Nov 2006 19:04:29 GMT, "daestrom"

That, sir, is exactly what it does. Thank you. sic itur ad astra
What does one refer to the quantified measure of that progress as?
Could it be The COULOMB.
This is the reason why the same current passed through smaller and smaller gauge wires yields higher and higher heat in the wire. It isn't that the wire has a higher resistance, it is because one is passing the same numbers of electron past a smaller and smaller cross sectional bottleneck of lattice.
A light bulb should go off in one's head. et sic de similibus
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Gave us:

---------- Definitely not. Trying to use the coulomb as a measure of progress is like trying to use cubic feet as a measure of flow of water. Something important is left out. Think about it. ----------

------ Which appears as an increase in resistance to anyone who is trying to observe it.
Note that resistance is expressed in terms of resistivity, length and cross section. You are trying to explain on the basis of one of the terms whereas the resistivity is the term related to the properties of the material -part of which is the lattice structure.
The wire does have a higher resistance and what you say is one way of explaining part of the reason for it ( which is not the whole truth but maybe a reasonable "lie to children" -other such "lies" work as well and can apply to conductors without a lattice structure). However what is observed is that there is a "voltage =function of current" relationship which is given the name "resistance".
In circuit analysis, one of the basic "circuit elements" is one where there is this relationship between current and voltage. It is a model - a mathematical model if you will- which correctly, within limits, represents observed reality. Circuit analysis involves modelling of the external behaviour of systems and components of these systems and is independent of models of the internal physics. It really is not concerned with what the actual charge carriers happen to be, the lattice structure, etc. What is involved is trying to get a model which correctly predicts the current, voltage and power relationships which are all externally observed.
The internal physics is very important but there is a tendency to use incomplete and sometimes inaccurate models- based on a search for a simple answer based on simple models such as the Bohr atom and a lattice structure that looks like a Tinkertoy construction -useful but still models and actually inadequate models on the basis of what we know now. These models do change as we learn more of the physics involved and trying to use these to explain electrical phenomena is a useful but possibly incorrect approach to the physics BUT leads to confusion when one tries to apply them to circuit theory and analysis where they aren't necessary. In other words, don't substitute an explanation of "why" there is resistance, for the observed relationship expressed by v =Ri (which is NOT Ohm's Law ).
Circuit theory does not depend on the physics of particular charge carriers, and lattice structure- treat them as separate matters. Evolution in the understanding of physics can change your lattice and electron ideas but has no effect on circuit theory.
--

Don Kelly snipped-for-privacy@shawcross.ca
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| How does one determine the direction of AC power flow? | | I understand how you can measure voltage and current with simple | instrumentation, but this is AC and the average value of those | parameters is zero. How can you tell in which direction the power is | going?
Don't average them. Measure voltage and current at common instants letting each instant figure its own power level and direction. Then average those to get real power and direction. If that averages out to zero, then you have no power flow (power factor is 0).
| Let's put this in the form of a puzzle: There are three adjacent | soundproof rooms A, B, and C. You are told that only one of the | following conditions is true: | | 1. There is an AC generator in A feeding power through open buss bars | in room B to a resistive load in room C. | | or | | 2. There is an AC generator in room C feeding power through open bus | bars in room B to a resistive load in room A. | | You are shut inside room B and are to determine whether condition 1 or | 2 described above is true. Remember, the rooms are soundproof so you | can't tell from sound leakage whether room A or C has the generator. | Also since the load is 100% resistive, assume that the power factor is | 1.
I think being shut in a room with open buss bars violates safety rules :-)
Of course you mean, the challenge is to figure it out entirely from inside that room with only the intrument(s) you figure in advance (the topic of the question) you will need brought in with you.
| Questions | | 1. Can you determine the direction of power flow just from | measurements to the AC buss bars in room B ?
Sure.
| 2. What sort of instrumentation would you need?
Something that measures voltage and current at an instant, gives you a power reading from that instant, and averages the power readings from many such instants across a cycle.
| 3. Do you need to break the circuit to make the measurements?
No. A clamp on ampmeter would do. But for direction alone, I don't think you even need that, as you can measure the direction of the magnetic field between the conductors to determine polarity.
I suspect there is some simpler answer. A coil of wire sufficient to handle the voltage applied (maybe with its own resistance to restrict current) that can fit between the buss bars could show field alignment by its orientation.
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| Phil Howard KA9WGN (ka9wgn.ham.org) / Do not send to the address below |
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On 11/11/06 10:44 AM, in article snipped-for-privacy@news3.newsguy.com,

You really need a fast multiplying device and averager--a wattmeter. But there still is an ambiguity that must be overcome. If you measure the magnetic field alone, you still don't know the direction of the Poynting vector for ac unless you can measure the electric field simultaneously. Bill -- Fermez le Bush
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| On 11/11/06 10:44 AM, in article snipped-for-privacy@news3.newsguy.com,
| |> No. A clamp on ampmeter would do. But for direction alone, I don't |> think you even need that, as you can measure the direction of the |> magnetic field between the conductors to determine polarity. | | You really need a fast multiplying device and averager--a wattmeter. But | there still is an ambiguity that must be overcome. If you measure the | magnetic field alone, you still don't know the direction of the Poynting | vector for ac unless you can measure the electric field simultaneously. | Bill
However if you have a field developed from voltage (a small current between the bars that is introduced) interacting with a field developed from current through the bars, this should get you a polarity equivalent to the electric field. You just need to know the orientation you will be getting out of the coil that is attached between the bars (I don't remember that rule at the moment but I could look it up if I needed it).
I believe that is exactly what a wattmeter is doing.
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Beachcomber wrote:

Simple. Measure the voltage from busbar to busbar at the entry/exit points at each side of the room. The side with the higher voltage is the power generation side, sending power to the one with the lower voltage - the power using side.
You need an AC voltmeter capable of detecting and displaying whatever theoretical voltage delta you name. Same procedure for DC with a DC meter.
No need to break the circuit.
Ed
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| Simple. | Measure the voltage from busbar to busbar | at the entry/exit points at each side of | the room. The side with the higher voltage | is the power generation side, sending power | to the one with the lower voltage - the power | using side. | | You need an AC voltmeter capable of detecting | and displaying whatever theoretical voltage | delta you name. | Same procedure for DC with a DC meter.
Suppose the buss bar is 10cm by 10cm solid copper cross section, and the current is not more than 100 milliamps, and you have on the order of 600 or more volts. Is there a meter good enough for that?
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snipped-for-privacy@ipal.net wrote:

In thought experiments there is always a meter good enough.
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On 12 Nov 2006 01:18:52 -0800 snipped-for-privacy@gmail.com wrote: | | snipped-for-privacy@ipal.net wrote: | |> |> Suppose the buss bar is 10cm by 10cm solid copper cross section, and the |> current is not more than 100 milliamps, and you have on the order of 600 |> or more volts. Is there a meter good enough for that? |> | | In thought experiments there is always a meter good enough.
*LOL* Great answer!
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You assume that AC power has a specific direction in each line - It does not - rather, it has direction in TWO lines (from the generator to the load).
Power is always positive into the load, and in AC it's magnitude varies in time and magnitude.
I.e., there is a voltage difference across the load for half the cycle, and then there is the same voltage difference but of different polarity across the load for the other half cycle. Both portions of the waveform deliver power to the load (equal if the waveform is symmetrical. (dP^2/R at any given instant)
Neutral is a reference point for voltage only (usually the same as earth ground in residential systems)
(continued below)

You could measure the field from the generator through the walls with a directional field meter, since only the generator will put out a magnetic field.
Other than that, I don't think it can be done using only the buss bars, for the previous reasons and because.
1) You have E-field, H-field (B), which vary with frequency (i.e., they reverse according to frequency along the time scale), and TTBOMK give no indication of direction.
2) Kirchoff requires current in=current out at any given instant in your loop. Thus, current on either bar is the same
3) Gauss requires any point along the no-resistance bussbars to be the same voltage with respect to any point - i.e., you have two perfect conducting busbars, thus voltage on a bar on one side of the room with reference to the other bar is the same
Resistance is the same, as long as the lines are intact.
Given the constraints and looking at it from s physics, engineering and a technical standpoint, I don't think it can be done as ststaed.
But, as Einstein said - it only takes one....

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that would better read "varies instantaneously with time"

and
same
the
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On 11/13/06 11:32 PM, in article
wrote:

How do you distinguish a load from a source? What is if your load is a negative resistance? Is circular reasoning showing up here?
Bill -- Fermez le Bush
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