mivey
Senior Member
So even at 0.7c the impact is only 0.01% of the copper weight.
3 Phase Current Flow
- Thread starter kody916
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A few fallacies in your analysis:
1. The electrons in a conductor or within the conduction band of a semiconductor (energies above the band gap) are NOT moving in a jump from one place to another. They are perfectly content to be anywhere in between.
2. If you go quantum on it, they are actually in a superposition state of several still not discrete locations, and the relative probabilities can change smoothly from one toward another.
3. The total energy an electron gains by moving across any sized gap (since it depends only on the voltage difference) is only one electron volt. That energy gives us, for relativistic purposes, nothing near the speed of light. If it did, relativistic effects would have been seen long before they actually were.
4. I think that your acceleration number is just plain wrong, considering the electric field produced by 100 volts over a one foot gap, but I have not actually done the calculations myself to confirm that. But in any case, when you consider the time over which that acceleration continues, it would still get nowhere near light speed.
(For purposes of discussion, I will consider 1/10th of c to be nowhere near light speed.)
Enuf alredy, back to the original question:
Enuf alredy, back to the original question:
In general the neutral current in a wye is,
In = Ia + Ib + Ic
For a delta,
Ia + Ib + Ic = 0
Provided that the sense of all currents is defined consistently.
These equations hold for instantaneous or phasor analysis.
FWIW, I agree with Bes, that if two sinusoids are not in phase, they are out of phase--no restriction to 180 degrees.
Enuf alredy, back to the original question:
In general the neutral current in a wye is,
In = Ia + Ib + Ic
For a delta,
Ia + Ib + Ic = 0
Provided that the sense of all currents is defined consistently.
These equations hold for instantaneous or phasor analysis.
FWIW, I agree with Bes, that if two sinusoids are not in phase, they are out of phase--no restriction to 180 degrees.
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mivey
Senior Member
If the charge carriers do not move from their place of rest there is no modification of the fields and no gradient created. We know they move and we know the fields change and we know a gradient is created. The movement of these charges and the change in the fields, at whatever the individual particle speeds and individual distances actually are, takes place at near light speed. We know the feedback propagates across the circuit at light speed and the resulting movement of charges takes place within a few light cycles or less.
The actual response starts as soon as the first feedback wave crosses the circuit. The movement is completed shortly thereafter and if the low energy state is satisfied without further feedback, the move will be completed at near light speed. As I stated before it is more of a sloshing action as the charge carriers receive additional feedback from changes in the circuit.
The electrons are not perfectly content to be anywhere because equilibrium has been disturbed. I did not use "jump" as discrete step function but just to say they moved from one place to another due to the feedback they got. I could have just said the move from one place to another but when you move in less time than it takes light to cross the circuit a few times.
I was not referring to discrete locations but the probabilities do change.
Any gap and one eV? We know that an electron beam in a TV tube gets the electrons moving pretty fast. In fact, at 25 kV they are about 1/3 light speed: electron speed_m/s = SQRT[ (-2 * (25000_V) * (-1.6E-19_C) / 9.11E-31_kg) ] = 93,709,971_m/s = 0.31c
The changes in the electric field propagate across the circuit at light speed. The movements of the charges are completed at near light speed.
The velocity of individual carriers is a different matter. But even at electron velocities near light speed, the change in relativistic mass is still a minute fraction of the conductor mass and would not cause the conductor to weigh as much as an elephant.
mivey
Senior Member
Hey rattus, where you been? Good to hear from you.
If the conductors of a DC supply turn into insulators at the moment of switching on, the electromagnetic signal originated at the supply side on switching still travels to the load end, The point is an electromagnetic signal once created will travel from one end to another end irrespective whether there is an intervening conductor or not. In other words, it has an independent existence. So the charge carriers moving from their place of rest, modification of the fields, gradient creation etc are all secondary effects due to the passage of electromagnetic signal through the conductor.
ggunn
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Not necessarily. In the Bohr model the electrons are already moving at a pretty good clip in their orbits. Yeah, I know; which atom an electron is associated with is merely probaballistic (misspelling intentional).
kwired
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I just wanted to remind everyone, myself included, what this thread is about.
Once some of these guys get on a topic like this has evolved to there is no going back.
mivey
Senior Member
Now consider the time for the secondary responses. The signal travels across the circuit at the speed of light. The secondary responses begin when they see the signal. As the secondary responses occur, they also send a signal through the circuit and cause additional secondary responses. This is the feedback I spoke of and is like a wave sloshing over the circuit.
In these secondary responses are the movements of the surface charges. The velocity of the individual particles will depend on the intensity of the electric field and how far they travel. They may move slow for short distance or fast for a long distance or who knows but as they move they modify the fields and send additional signals through the circuit.
This relocating of the surface charges can settle down pretty quick and essentially be over in less time than it takes light to cross the circuit a few times. Since the movement is a secondary response to the light speed signal, and takes a few light cycles or less to finish, this movement of the charges to a new equilibrium point takes place not at light speed but at near light speed.
ggunn
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Why in the world would we want to go back? The original boring question was answered eons ago.
mivey
Senior Member
Once the original question is answered (and it has been, more than once) the only fun left is in related spawned discussions.
mivey
Senior Member
Hear! Hear!Why in the world would we want to go back? The original boring question was answered eons ago.
I do not understand why this should be the case. What contradiction arises if the relocating of the surface charges can settle down pretty slow as the initial electromagnetic signal for the D.C supply is no way affected by it?
Just by way of information old chap, DC isn't usually what most people would take to be three phase.
mivey
Senior Member
Either it is a conductor or not. If it is a conductor with surface charges they will move and modify the field.
At a speed independent of the speed of the field.
mivey
Senior Member
No. The feedback delay is part of the total time. For electrical circuits with a large time constant, Te, compared to the light crossing time, Tc, the feedback delay can be insignificant. You can have:
Te>>Tc: insignificant feedback delay
Te>Tc or Te~Tc or Te<Tc: feedback is a significant portion of the time required
Te<<Tc: feedback delay is the majority of the time required.
Just to get the scale of things, here are some approximate relaxation times:
copper: 1.53E-19s
aluminum: 2.5E-19s
low doped Si: 3.5E-11s
water: 3E-9s
distilled water: 3.6E-6s
glass: seconds...hours
fused quartz: 1.7 months
Total time of what?
mivey
Senior Member
For the charges to re-locate.
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