Single Phase vs Three Phase flow of electrons with no neutral/grounded path?

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Besoeker

Senior Member
Location
UK
The theory is that the house ground went bad, and the coax (which had it's shield connected to house ground) became the ground.
The ground should not carry current in normal operation. Losing it should not cause current to flow in an alternative ground unless there was a fault.

Now an electrician speculated that losing the ground caused all the 120V circuits to be 240V.
Losing the neutral on a center tapped 120-0-120 system would result is some circuits getting more than their rated voltage and some getting less. Neutral and ground are not the same thing.

The only way I can think this would happen is if there was a short from neutral to one of the hot legs.
That ought to trip the overcurrent protection device for that circuit.
 

charlie b

Moderator
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Location
Lockport, IL
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Retired Electrical Engineer
Electron drift - any chance we can get an explanation how it works..................
What I said in Post #15 (i.e., the bit about electrons making the round trip from source to load to source and around and around again a bazillion times in one direction during each half cycle, and doing it in the opposite direction during the other half cycle) is not, as I had confessed in advance, strictly true. Any given single electron will generally move (at something close to the speed of light) no further than from the vicinity of one atom to the vicinity of the next atom in line. It will be some other electron that will make the leap from that second atom to the vicinity of the third atom along the line. Then some other electron will continue the process by making the leap from atom 3 to atom 4. And so it goes. Eventually, that first electron I mentioned might make another leap, and will thereby continue its progress down the wire. The speed at which any single electron moves down the wire is on the order of molasses flowing in winter time. That is what is meant by ?electron drift.?

However, from the point of view of a person standing outside the wire, there is no way to tell the difference between the first electron (moving from atom 1 to atom 2) and the second electron that continues the process by making the leap from atom 2 to atom 3. All electrons look the same to the outside observer. So as far as the observer can discern, it is the same electron that is moving along the wire from the source to the load, and back, and around and around the circle it goes. That is why my description in Post #15 is essentially true, for practical purposes.

Now, there I?ve gone and done it! So when I get home, I will boil one strand of spaghetti, and use it to give myself 300 lashes. :ashamed1:
 

Besoeker

Senior Member
Location
UK
I had asked that we not go there. :happysad: That will cost you 300 lashes with a wet noodle!
Shame on you!
You can't construe wiggle as drift.
:p
Slightly more seriously, I was making the point that trying to envisage the circuits in terms of electron flow isn't particularly helpful or relevant in AC circuits. Nor in DC for that matter. Current is the quantity that most of us can deal with in units we are familiar with.
 

EEC

Senior Member
Location
Maryland
Another stab at understanding

Another stab at understanding

Again the question is about no neutral in the 230V circuit.

Single phase panel circuit breaker #1 the current at peak voltage in the upper portion of the sine wave (i won't use the word positive) in the first half of a cycle travels through load to Single phase panel breaker #3

At the same time Single phase panel circuit breaker # 3 the current at peak voltage in the lower portion of the sine wave (won't use the word negative) in the first half of a cycle travels through the load to Single phase panel breaker #1

I believe that at the same period of time the sine wave cannot be exactly the same because you could not get 230V by measuring voltage between the two breakers. Therefore the two sine waves need to have different instantaneous values. I believe they are 180 degree difference in single phase circumstances.

Therefore the current is in opposite direction at the same instant in time. This does not seem possible that this could happen but I am not understanding it any other way yet. Please continue to break it down so the picture is clear. Remember I am not talking about any current in any neutral.
 

Besoeker

Senior Member
Location
UK
Again the question is about no neutral in the 230V circuit.

Single phase panel circuit breaker #1 the current at peak voltage in the upper portion of the sine wave (i won't use the word positive) in the first half of a cycle travels through load to Single phase panel breaker #3

At the same time Single phase panel circuit breaker # 3 the current at peak voltage in the lower portion of the sine wave (won't use the word negative) in the first half of a cycle travels through the load to Single phase panel breaker #1

I believe that at the same period of time the sine wave cannot be exactly the same because you could not get 230V by measuring voltage between the two breakers. Therefore the two sine waves need to have different instantaneous values. I believe they are 180 degree difference in single phase circumstances.

Therefore the current is in opposite direction at the same instant in time. This does not seem possible that this could happen but I am not understanding it any other way yet. Please continue to break it down so the picture is clear. Remember I am not talking about any current in any neutral.

No neutral so you have just a single phase 230V supply. And just one sinewave. Not two.
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
Single phase panel circuit breaker #1 the current at peak voltage in the upper portion of the sine wave (i won't use the word positive) in the first half of a cycle travels through load to Single phase panel breaker #3
I?m OK with that description.

At the same time Single phase panel circuit breaker # 3 the current at peak voltage in the lower portion of the sine wave (won't use the word negative) in the first half of a cycle travels through the load to Single phase panel breaker #1
No. It?s the same current. I think the part you are missing is that when current leaves breaker position #1, passes through the load, and returns to breaker position 3, that is all that there is. You do not have a ?second current? that departs from position 3.

During the half cycle in which position 1 voltage (with respect to ground) is ?in the upper portion? (using your terminology), the voltage (with respect to ground) on position 3 is in ?the lower portion.? Thus, the voltage at position 1 is higher than that of position 3, and current is therefore driven in the direction of 1 towards 3. During the other half cycle, the voltage at position 3 is higher than that of position 1, and current is therefore driven in the direction from 3 to 1.

Keep in mind that voltage must be measured between two points. There is no such thing as ?voltage at a single point.? If you measure voltage between breaker position #1 and ground, you will read 115. If you measure voltage between breaker position #3 and ground, you will read 115. But the two are opposite in sense (i.e., when one is in the upper portion, the other is in the lower portion). As a result, the voltage between positions 1 and 3 is 230.
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
Again the question is about no neutral in the 230V circuit.

Single phase panel circuit breaker #1 the current at peak voltage in the upper portion of the sine wave (i won't use the word positive) in the first half of a cycle travels through load to Single phase panel breaker #3

At the same time Single phase panel circuit breaker # 3 the current at peak voltage in the lower portion of the sine wave (won't use the word negative) in the first half of a cycle travels through the load to Single phase panel breaker #1

I think you can solve your dilemma if you look closely at what you are saying. At the same instant, you described current "leaving" breaker 1 in the upper portion of the sine wave. You also describe current "leaving" breaker 3 in the lower portion of the sine wave.

What happens to the current at breaker 1 when the sine wave gets to the lower portion of the sine wave? The current is no longer "leaving" the breaker, but since it is alternating current, it is "entering" the breaker.

So if the current is "leaving" breaker 1 (at the upper portion of the sine wave) then it is "entering" breaker 3 (at the lower portion of the sine wave) at the same time. The current is flowing in the same direction in the circuit at the same instant in time. "Leaving" one breaker, and "entering" the other.
 

Smart $

Esteemed Member
Location
Ohio
I believe that at the same period of time the sine wave cannot be exactly the same because you could not get 230V by measuring voltage between the two breakers. Therefore the two sine waves need to have different instantaneous values. I believe they are 180 degree difference in single phase circumstances.

Therefore the current is in opposite direction at the same instant in time. This does not seem possible that this could happen but I am not understanding it any other way yet. Please continue to break it down so the picture is clear. Remember I am not talking about any current in any neutral.
I believe the problem you are having is related to how we measure the voltages. For this discussion, refer to the following diagram.

splitphasevoltagewaveforms.gif


The reference point for voltage measurements is the negative or common lead of the measurement equipment.

When you measure voltage at circuit breaker #1, you are placing your negative or common lead on the neutral and the positive lead on the breaker terminal. You measure 115V RMS and the black line represents the waveform, as if displayed on an oscilliscope.

When you measure voltage at circuit breaker #3, you are placing your negative or common lead on the neutral and the positive lead on the breaker terminal. You measure 115V RMS and the red line represents the waveform, as if displayed on an oscilliscope.

However, if you were reverse your lead and place your negative or common lead on the terminal of circuit breaker #3 and your positive lead on the neutral, you would still measure 115V RMS, but the black line represents the waveform, as if displayed on an oscilliscope.

Furthermore, if you were to place your negative or common lead on the terminal of circuit breaker #3 and your positive lead on the terminal of circuit breaker #1, you would measure 230V RMS, and the magenta line represents the waveform, as if displayed on an oscilliscope.
 

Rick Christopherson

Senior Member
Any given single electron will generally move (at something close to the speed of light) no further than from the vicinity of one atom to the vicinity of the next atom in line.
Electrons have mass, and therefore, they do not approach the speed of light. "Electricity" does approach the speed of light, but not the electrons themselves. The speed of the electrons are relatively slow, but the propagation of the flow is very fast.

A good analogy to this is to consider bumper-to-bumper traffic with the bumpers literally touching. The first car in the lane lurches forward at 1 mile per hour, but because all of the bumpers are already touching, the last car in the lane also moves forward at the same instant, but only at 1 mile per hour. Each car moves at only one mile per hour, but the effect from the beginning to the end of the chain is almost instant.

Electricity is the same. The electrons move fairly slowly, but their effects down the length of a circuit are almost instantaneous.
That is what is meant by ?electron drift.?
The above analogy does NOT describe Electron Drift (In hindsight, it does represent Electron Drift Speed). Electron drift is the effective movement of electrons taking into account (or actually, ignoring) the "meandering" of an electron on its route. Electrons do not travel in straight lines or move in coherent directions.

A good analogy for describing electron drift would be a pinball machine. As the ball comes down from the launcher ramp at the top of the machine, it will bounce all over the place. Sometimes moving sideways; sometimes even moving backward (uphill). However, after some given time, all of the balls will eventually find their way to the bottom of the machine. The ball may travel hundreds of feet just to make the 3-foot progression to the bottom of the machine.

Electron drift speed is the effective speed of the electron (or pinball) to travel the relative distance (3 feet for a pinball machine) regardless how far it actually traveled in the process.

 

__dan

Senior Member
Are there differences in the flow of electrons in a single plase 230V circuit to a 208V 3-phase circuit uses only two of the 3 phases?
1. Take the single phase 115V part of the 230V circuit. (Does the current flow at a different time then the other 115v part of the 230V circuit?)
2. What is the path that the electrons flow on for the 115V part of the 230V circuit [since there is no return path for each leg of the 230V circuit, no neutral/grounded path]?
3. Whats the difference in a 208V circuit between two phases for 208V feeding a load. There does seem to be a time difference between phases.
4. In both of the above cases do the electrons flow back to the source or what happens to them.
5. Can you provide a reference source to back up your conclusions, if so what are they?

1. no difference

2. The sum of the voltages around any closed loop = zero (Kirchoff's law). 115v source through 115v load through the neutral, or if no neutral, 230v source through two 115v loads in series.

3. Three phase originates in and is ideally suited for rotating machinery, motors and generators. In a two pole rotating machine, the armature is a North to South constant bar magnet and it sweeps past the A, B, C field coil pairs, generating a sinewave voltage in the field coils. The A, B, C field coils are physically seperated by place and so also seperated by time, as the rotating armature magnetic field induces voltage sequentially, in the A field coil, then the B field coil, then C, then A again.

Connecting a load to A and C in series at 115v source volts each, if A and C were matching sinewaves in time, the voltage would add in series to 230v (0 degree phase difference). Since A and C sinewaves are generated at different times, for part of the cycle C is negative when A is positive and the sum sometimes is a subtraction of voltage rather than addition. The sum of two sinewaves is another sinewave, but at 208v rms because of the part of the cycle where the voltages subtract rather than add. If A and C were seperated by 180 deg, the sum in series would be 0 volts.

4. Voltage travels from source to load around the circuit. The individual electron mass barely moves. You can walk faster than the electron moves.

"Numerical example
Electricity is most commonly conducted in a copper wire. Copper has a density of 8.94 g/cm?, and an atomic weight of 63.546 g/mol, so there are 140685.5 mol/m?. In 1 mole of any element there are 6.02?1023 atoms (Avogadro's constant). Therefore in 1m? of copper there are about 8.5?1028 atoms (6.02?1023 ? 140685.5 mol/m?). Copper has one free electron per atom, so n is equal to 8.5?1028 electrons per m?.
Assume a current I=3 amperes, and a wire of 1 mm diameter (radius in meters = 0.0005m). This wire has a cross sectional area of 7.85?10-7 m2 (A= π?0.00052). The charge of 1 electron is q=1.6?10−19 Coulombs. The drift velocity therefore can be calculated:
bdd2b1a60d0e13256dcc15164858165a.png

a0c2947b4fd6d6707b72979beea58c8b.png

12a67b127c91044834b2a6ee866dcdcb.png

Analysed dimensionally:
[V] = [Amps] / [electron/m3] ? [m2] ? [Coulombs/electron]
= [coulombs] / [seconds] ? [electron/m3] ? [m2] ? [Coulombs/electron]= [meters] / [second]Therefore in this wire the electrons are flowing at the rate of 0.00029 m/s, or very nearly 1.0 m/hour"

http://en.wikipedia.org/wiki/Drift_velocity

The question you're asking, to understand it and to work it, the essential information is the transformer turns ratio, V primary /V secondary = turns primary / turns secondary, and the kVA rating.

In the single phase example of one iron core, one primary winding, and two secondary windings. The primary is energized and current flow induces a dense magnetic field in the iron core. A changing magnetic field is necessary to induce voltage in the secondary winding (Faraday or Lenz). Run DC through the primary and secondary voltage is zero, no matter what else.

There is a right hand rule for current flow through coils or turns of wire and magnetic field direction. The iron core's magnetic field is changing direction at the same frequency as the sinewave voltage driving the primary current.

On the secondary side, it can be simple. Every coil or turn of wire occupies the same place and the same time, so every turn of wire sees the same magnetic core field and same induced voltage, in synch. Turns connected in series, the voltages add. Turns connected in series but wound in the opposite diirection, CCW adding to CW turns, the voltages subtract. Add the turns matching polarity in parallel and the current rating adds. Add turns in parallel but opposite polarity (opposite winding direction) and the secondary is shorted.

Single phase, two secondary 115v windings. Added in series wound in the same direction, the turns ratio from primary to secondary changes by a factor of 2 and the voltage doubles.

Single phase, two secondary 115v windings. Added in series wound in opposite directions and you have a secondary with zero turns and zero volts. +X turns + -X turns = zero turns. This is how an auotransformer secondary at 24 volts can either add 24 v or subtract 24 v depending on reversing two wires.

Single phase, two secondary 115v windings. Added in parallel wound in the same direction, the turns ratio stays the same, voltage is same, and the current rating doubles. There is zero voltage difference between each winding with matching sinewaves, so no short through the other winding. Add in parallel, wound in opposite directions and one winding voltage peaks when the other is low and the reverse. The sum of the voltages is 2x the winding voltage and short circuit current flows because the winding is a low impedance path with a voltage difference imposed on it.
 
Location
Ohio
The ground should not carry current in normal operation. Losing it should not cause current to flow in an alternative ground unless there was a fault.


Losing the neutral on a center tapped 120-0-120 system would result is some circuits getting more than their rated voltage and some getting less. Neutral and ground are not the same thing.


That ought to trip the overcurrent protection device for that circuit.

Thanks for the response.

I was assuming some sort of fault.

Neutral is bonded to ground, correct? If we lose the ground at "ground" but it's still bonded, and there's a fault, do they become one and the same?

Would it properly trip the over-current device if we had no ground? Doesn't the breaker require ground to function properly?
 
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Smart $

Esteemed Member
Location
Ohio
Thanks for the response.

I was assuming some sort of fault.

Neutral is bonded to ground, correct? If we lose the ground at "ground" but it's still bonded, and there's a fault, do they become one and the same?

Would it properly trip the over-current device if we had no ground? Doesn't the breaker require ground to function properly?
Hard to interpret your questions using only the word "ground" to mean several aspects of a properly wired system, but...

Neutral is typically bonded the grounding system at or near the power source. The grounding system has equipment grounding, non-current-carrying metallic parts grounding, and also earth grounding (the grounding electrode system, GES). Losing any part of this grounding system will still allow overcurrent devices to trip as long as the remaining paths provide sufficiently low enough impedance for the fault current to return to the source.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
One of the hardest things for me to get through my head in considering three phase power is the effect of phase angle. Phase angle is only a factor if all three phases are involved or if a zero reference point is used. If you put an oscilloscope across two phases of a three phase system you will not see some sort of weird waveform, you will see a simple 60Hz sine wave. In fact if you could vary the phase angle at will to any angle, the only thing that would change would be the amplitude of the wave form you see; the shape of the wave would not change. I don't know why that was so hard for me to visualize.
 
Location
Ohio
Hard to interpret your questions using only the word "ground" to mean several aspects of a properly wired system, but...

Neutral is typically bonded the grounding system at or near the power source. The grounding system has equipment grounding, non-current-carrying metallic parts grounding, and also earth grounding (the grounding electrode system, GES). Losing any part of this grounding system will still allow overcurrent devices to trip as long as the remaining paths provide sufficiently low enough impedance for the fault current to return to the source.

Where I say ground, I'm referring to the center tap ground at the transformer. The "lost ground" would be at a single ground rod or water pipe (old house). So the GES.

So any other ideas as to why Cable Guy cutting his co-ax would cause household appliances to smoke if it's not a fault?
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Where I say ground, I'm referring to the center tap ground at the transformer. The "lost ground" would be at a single ground rod or water pipe (old house). So the GES.

So any other ideas as to why Cable Guy cutting his co-ax would cause household appliances to smoke if it's not a fault?
So the coax was providing the neutral, and when that was lost, the appliances on the more lightly loaded leg of the 240 saw a voltage rise, which smoked some of them, which caused loads to drop off, which caused the voltage to go up even more...

He's lucky he didn't electrocute himself when he cut the coax, or burn the house down.
 
Location
Ohio
So the coax was providing the neutral, and when that was lost, the appliances on the more lightly loaded leg of the 240 saw a voltage rise, which smoked some of them, which caused loads to drop off, which caused the voltage to go up even more...

He's lucky he didn't electrocute himself when he cut the coax, or burn the house down.

I'm guessing he was careful, as he saw it was melted and assumed current flow.

He did say he got shocked on the metal frame door when he tried to enter the house.
 
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