MWBC= more heat or Less heat?

[rt66electric]

GREAT REPLY !!! thanks

GREAT REPLY !!! thanks

Sir, I accept your challenge! :grin: I can't explain everything first, so I'll start with how electricity flows through a conductor, using your favorite analogy, a pipe (sort of), but instead of air or water, marbles.

Imagine a garden hose filled from end to end with marbles. What happens if you force another one in one end? A marble pops out of the other end. Not the same marble, but a similar marble.



Electrons orbit the nuclei in layers (aka valence shells.) The first layer can contain one or two electrons; if the atom happens to have a third electron, it will be the start of the second layer (up to eight), etc.

If the outermost layer is filled, or nearly filled, the electrons are bound tightly to the atom, and cannot be easily bumped away by another electron. These materials are electrically insulative.

Materials that are electrically conductive happen to have only one (or sometimes two) electrons in the outermost layer, and these 'free electrons' are easily bumped away and replaced by another one.

Gold, silver, copper, aluminum, and most other metals all happen to have only one electron in their outermost layer (aka a valence shell), so that electron can easily be displaced by a 'rogue' from outside.

A new electron introduced into one end of a conductor takes the place of the free electron of the 'first' atom, which bumps into the free electron of the next atom, and so forth. There is one requirement:

The free electron from the 'last' electron must have a place to go, which is why an electric circuit must be a loop, to provide a pathway for what we often call current 'returning to the source.'



Now, back to pipe: imagine a closed loop of 1" pipe with a circulating pump in the loop. The 'current' of water will flow through the loop as fast as the pump can pump it. Let's add that speed heats the water.

Now, in order for anything useful to be done, we add a hydraulic motor halfway around the loop, to spin, say, a fan. The harder the fan is to spin, the harder the pump is to spin, so will need a longer handle.



Electrical analogy: The pump is a generator, the pipe is a conductor, the hydraulic motor is a load, the water pressure is the voltage, and the resultant flow (based on pressure and resistance) is the current.

If we decrease the diameter of the pipe (smaller wire gauge), friction (resistance) adds to the difficulty of moving the water; likewise, a larger pipe reduces the friction. We also have to move the load.



Wow! I expected this to be a short analogy. Oh, well. I think you get the idea. Let me know if this is either too basic or too complex.

I like your exlpaination better than chemistry 101. I would still like an analogy(or visual) that would explain how a tic tracer operates ??
What is the magnetic field surrounds a wire.
In reference to a plain XFRM,- How do electrons move through the insulation from one wire to another?? -- Or-- Are there free electrons that are outside the insulation and are magnetically attracted to the inner pa rt of insulation and ride along the outside.


Sound is a regular "vibration" of minute air pressure changes as detected by a eardrum.

Is insulation similar to a drumhead? Vibrations(or movement) over the drumhead is transfered to vibrations under the drumhead, the "sound" is insulated(drumhead) but motion is transfered.

How is emf related to air pressure-or-sound??

Can EMF transmission be similar to the operation of "sympathetic strings"?? through a membrane, Instead of air/sound pressures--- magnetic attaction is moving the valence electrons sympathetically on both sides of the insulation.


I'm making this up as I go along

Sound waves cannot travel in a vacum. So much for my analogy. Is there anything that EMF cannot travel through??

Thank you for your avenues of Ideas.


'now for something a completly different subject" the weather is !! HOT !!!


DENNIS

 

[hardworkingstiff]

What is the magnetic field surrounds a wire.

You're not going to like this, it's a magnetic field.

In reference to a plain XFRM,- How do electrons move through the insulation from one wire to another??

They don't. The magnetic field created by the electron movements in primary windings move the electrons (induce the voltage) in the secondary windings.

How is emf related to air pressure-or-sound??

I don't think it is.

I'm making this up as I go along

hummmmm

'now for something a completly different subject" the weather is !! HOT !!!

Very hot in Wilmington, NC. The extra weight I put on last year is wearing me out right now.

 

[ActionDave]

I can see the relationship, wife acts like a magnet, you got stuck to her, and been makeing sparks ever since lol

You have me figured out darn near to perfection.

What is the magnetic field surrounds a wire.

Seriously- nobody knows.

 

[mivey]

I still can't see electrons flowing one direction at point A (let's say, straight down to Point B) while at the same time, electrons are flowing the other direction at Point B (traveling toward Point A).

The following is the answer:

They are flowing in the same direction looking from the outside, but looking from the neutral and looking toward point A you will see the electrons coming at you and turning and looking at point B you will see they going away from you.

But how about a hands-on metering example: The +/- voltage measurement has been covered by explaining that the black lead is kept at the center point while the red lead is placed at points A & B for a voltage measurement. Easy enough.

The current direction can also be relative. Let's use an ammeter so the front of the meter faces the same direction as I am facing. Let's call current flowing from the front of the meter to the back positive, or toward me. All of the measurements from this point on will be made at the same time.

If I stand at point B, facing A, and clamp on the wire just above point B, the current is shown to be flowing toward me. If I reach further up and clamp near point A, the current is also shown to be flowing toward me. This is the way to look at it when we are using the driving voltage as a reference. The current is positive and so are the voltages as measured from point B.

Now stand at the center point and use it as a reference. Face A and clamp on the wire near A and the current will read as flowing toward you. The voltage and current both read positive. Now turn and face point B. Clamp on the wire near B and the current will read as being negative. So will the voltage.

This is why we say the measurements are relative. It is the same as saying you are traveling in a straight line like the electrons (or holes). You are deemed to be going North until you cross the North pole, then you are deemed to be going South. You did not turn around, and neither did the electrons. There is a relative reference point that makes the difference.

If a car is traveling, one person may say it is traveling toward them, but another may say it is traveling away from them. It kind of depends on where you are standing.

For AC, we assume the current direction to be from + to - during the positive half cycle of the voltage sine wave.

FWIW, if you are really concerned about the electrons instead of the assumed current direction, we normally say current flow direction is opposite of the direction of electron flow (i.e., we say hole flow is in the same direction as current flow).

 

[mivey]

this discussion is mixing two different aspects, that of the physics and the measurement thereof

Bingo.

We can't see the electrons or apply our meter at the exact point of a sine wave so we assume polarities and current directions when we set up our measurements. The signs of our results, relative to the assumed signs, will determine if we made the correct assumptions for the steady-state, instantaneous, etc. conditions and will allow us to tie back to the physical world.

There is no penalty for using the correct or incorrect sign at the beginning as you will correct the signs of the results. Some of the sign assumptions at the beginning may actually make your calculations easier. However, if you do not correct the signs at the end (or at least account for them), then there can be significant penalties in the physical world.

 

[mivey]

What is the magnetic field surrounds a wire.

Think of it as peer pressure. We can't physically see it, but we know it exists. Some things are more sensitive to peer pressure (magnetic fields) than others (like iron).

The transfer of peer pressure is influenced by the medium of transfer. Peer pressure in a newspaper article is not as strong as that applied by a friend standing right in front of you. The force exerted by a magnetic field is also transfered more efficiently through different mediums.

You can be shielded from peer pressure and magnetic fields. I guess you could go on and on but you should get my analogy.

It is an invisible field which exerts force. There is no magic, we just don't have the ability to actually see the field, only the results. The results prove that there is actually something there, even if we don't have the means to see it. We know our senses are limited. Some animals have some senses that are better than ours. Who knows, maybe one day we will actually gain the sense to see a field.

FWIW, some of the greatest things in life are invisible: love, joy, peace, etc. We can't actually physically measure these things, but we can see the results and know they are real things. Contrary to a familiar phrase, seeing is not always believing.

Neither is tasting, touching, hearing, or any combination of what we are restricted to by our physical senses. A magnetic field is proof of that. So is infrared light, sounds beyond our hearing range, etc. We have developed tools that allow us to look beyond our senses to a world previously unkown to our ancestors. Who knows what else is out there that we have not been able to detect yet?

 

[rt66electric]

I love these discussions !!

I love these discussions !!

I love these discussions !!!!! Thanks for your patient replies.. Anybody else have obscure explainations?? DENNIS

 

[Dennis Alwon]

I love these discussions !!!!! Thanks for your patient replies.. Anybody else have obscure explainations?? DENNIS

Heck I am lost-- I gave up a long time ago. :grin: I still keep thinking how can we create more heat with the same amount of amps on the MWBC vs individual neutrals. We should market this somehow....:grin:

 

[SAC]

I have worked with a fair amount of factory made equipment that would run two 120 volt lamps in series supplied from 208, even though each lamp will have some differences it worked just fine. :smile:

And there are probably about a billion of those x-mass tree lights out that that work just fine on the same principal. Until one goes out, that is... :grin:

 

[Smart $]

... However, if you do not correct the signs at the end (or at least account for them), then there can be significant penalties in the physical world.

.
L:grin:L

 

[hurk27]

And there are probably about a billion of those x-mass tree lights out that that work just fine on the same principal. Until one goes out, that is... :grin:

if they are "out" don't think thay are working fine

just kidding, you ment "out there"

 

[david luchini]

Bypassing the neutral

Bypassing the neutral

I have to disagree with Iwire that the neutral is bypassed in a balanced MWBC. I think everyone would agree that current flows in the path of least resistance. If you connect two 100 watt light bulbs (A-N & B-N) with a common neutral, current will flow from Phase A through the light where it will then reach the common neutral point where the current can then follow a zero resistance path back to the neutral (I'm ignoring the impedance of the branch wiring) or a high resistance path through the second light bulb.

The current will take the zero resistance path back to the neutral. The current flowing from Phase B through the second lamp will also follow the zero resistance path along the neutral. Following Kirchoff's Current Law, the current flowing out of the common node (along the neutral) will be the sum of the current flowing into the node - which equals the sum of the current from A-N and from B-N.

The current flowing on the neutral would be 0.83<0 Amps + 0.83<180 Amps. The sum of these two phase current equals 0.0<0 Amps. The measured amps on the neutral in this balanced system is 0, NOT because the neutral is bypassed, but because BOTH currents are following the neutral path and because the currents added together equal zero.

 

[mivey]

I have to disagree with Iwire that the neutral is bypassed in a balanced MWBC. I think everyone would agree that current flows in the path of least resistance. If you connect two 100 watt light bulbs (A-N & B-N) with a common neutral, current will flow from Phase A through the light where it will then reach the common neutral point where the current can then follow a zero resistance path back to the neutral (I'm ignoring the impedance of the branch wiring) or a high resistance path through the second light bulb.

The current will take the zero resistance path back to the neutral. The current flowing from Phase B through the second lamp will also follow the zero resistance path along the neutral. Following Kirchoff's Current Law, the current flowing out of the common node (along the neutral) will be the sum of the current flowing into the node - which equals the sum of the current from A-N and from B-N.

The current flowing on the neutral would be 0.83<0 Amps + 0.83<180 Amps. The sum of these two phase current equals 0.0<0 Amps. The measured amps on the neutral in this balanced system is 0, NOT because the neutral is bypassed, but because BOTH currents are following the neutral path and because the currents added together equal zero.

I do not, and neither will most here, agree that current flows in the path of least resistance: it flows in all paths.

As for the rest, it depends on how you define bypassed. In a perfectly balanced circuit, there is no flow of electrons down one side of the neutral wire and back up the other that are canceling each other out. It is just that there is no potential difference available to push current down the neutral wire. You can't have the current without the voltage.

 

[david luchini]

Perhaps I oversimplified the statement that current follows the path of least resistance - will you accept current flows in inverse proportion to the impedance of the paths. The lower the impedance, the greater the current flow.

I cannot agree with you that there is "there is no potential difference available to push current down the neutral wire." Where did the potential difference go? It exists from A-N, or from B-N, or if the two phases are slightly out of balance, but suddenly disappears if the two phases are perfectly in balance.

I did not infer that electrons are flowing down one side of the neutral wire and back up the other to the other side to cancel each other out. I simply said that by KCL, the neutral is not "bypassed." The current flowing out the the common node is equal to the sum of the currents flowing into the node. That is, the current on the neutral (In) equals the sum of the current flowing from phase A to neutral (Ia) plus the current flowing from phase B to neutral (Ib.) In=Ia+Ib. In the balanced case, if Ia=1.0<0 and Ib=1.0<180 then In=0.0 amps. The neutral wire hasn't been bypassed, but as Kirchoff tells us, is carrying the sum of the two phase currents, which happens to be zero amps.

 

[mivey]

The lower the impedance, the greater the current flow.

If the voltage is the same across the impedances, that will be true.

I cannot agree with you that there is "there is no potential difference available to push current down the neutral wire." Where did the potential difference go? It exists from A-N, or from B-N, or if the two phases are slightly out of balance, but suddenly disappears if the two phases are perfectly in balance.

Sketch it out.

When the load is balanced, the neutral point of the loads is coincident with grounded conductor connection point. And there is no difference in the voltage drops across the loads (no potential difference).

Consider the unbalanced load with no grounded conductor. When the load is unbalanced, the neutral point of the loads is not coincident with desired grounded conductor connection point (the intersection of the loads) but is located somewhere inside one of the loads. The voltage drops across the loads are different (potential difference).

If you tie the grounded conductor to a point inside the load with the bigger voltage drop, such that the potential on either side of the grounded conductor is the same, no current will flow in the grounded conductor.

Now tie the grounded conductor to the intersection of the unbalanced loads (where there is a potential difference on either side of the connection). The connection point voltage will be forced up to a level where the potential drop on either side is the same.

Because we tapped in at a point where there was a potential difference, the voltage drops across the loads have been changed, resulting in a net current in the grounded conductor.

We have forced this circuit away from its equilibrium point, and it will draw more power. Remove the grounded conductor and the circuit will return to equilibrium and draw less power.

I did not infer that electrons are flowing down one side of the neutral wire and back up the other to the other side to cancel each other out. I simply said that by KCL, the neutral is not "bypassed." The current flowing out the the common node is equal to the sum of the currents flowing into the node. That is, the current on the neutral (In) equals the sum of the current flowing from phase A to neutral (Ia) plus the current flowing from phase B to neutral (Ib.) In=Ia+Ib. In the balanced case, if Ia=1.0<0 and Ib=1.0<180 then In=0.0 amps. The neutral wire hasn't been bypassed, but as Kirchoff tells us, is carrying the sum of the two phase currents, which happens to be zero amps.

The grounded conductor is not used (iwire calls this being bypassed) in the balanced case so carries nothing. Remove the grounded conductor and what carries the sum then, the air? There is nothing to carry.

True enough, it is not bypassed in the sense that a switch has been flipped. That is why I said it depends on the definition of bypassed. Iwire could have said nullified or something else instead, but he did not.

How about: "The neutral has been rendered useless for carrying current", or "The neutral is not needed for carrying current anymore"? If you bypass an alarm switch, didn't you essentially counter-balance the magnetic field? It is just a word.

We know what he meant. I can't believe I have typed this much stuff about it. I guess I was bored.

 

[david luchini]

You say that the grounded conductor is not being used, but Kirchoff's laws tell us that it is. It is carrying the sum of the two phase currents. Just because the two phase currents sum out to zero, does not take the neutral conductor out of the circuit. Consider the node between the two loads where the phase A load conductor, neutral conductor, and phase B load conductor meet - that node has 3 lines connected. By KCL the sum of the currents into the node must equal zero so that In + Ia + Ib = 0. If you consider that the current out of the node on the neutral is negative of the current in on the neutral (ie, - In) then the current flowing out of the node on the neutral is -In = Ia + Ib. If Ia and Ib are balanced and 180 degrees opposite phase, then when Ia is 1, Ib is -1 and the sum is zero. The neutral has been rendered useless for carrying current, it is simply carrying a sum of two current which always equals zero. It is still capable of carrying current and still an active part of the circuit.

You ask "if you remove the grounded conductor what will carry the sum." The answer is nothing. If you remove the grounded conductor, you no longer have the same circuit, but instead a completely new circuit. There is no sum to carry. In the new circuit, the node between the two loads has only two lines entering the node so that under KCL Ia + Ib =0 or Ia = - Ib.

Or another way to think of having the neutral conductor in the circuit and then removing it, suppose you had a slightly unbalanced load, say 90 watts on one phase and 100 watts on the other with a common neutral. In this case, the neutral will carry a slight current ( around 0.08 amps) but if we remove the neutral where does this current go - into the air? No, clearly by removing the neutral we have created a completely new circuit, the same way we would with a balanced circuit. The neutral in the first case is still an active part of the circuit, and carries the sum of two currents which happen to add up to zero.

 

[LarryFine]

The measured amps on the neutral in this balanced system is 0, NOT because the neutral is bypassed, but because BOTH currents are following the neutral path and because the currents added together equal zero.

It sounds like you're agreeing that the math says there's zero neutral current, but that physics says that two opposing currents actually are flowing in the neutral.

The opposing point of view is that, since opening the neutral will have absolutely no effect on the circuit's current, your theory is inaccurate; there's no current.

We've had this discussion before. An AC ammeter, or even two DC ammeters and current-steering diodes, would show no current at all. Even DC analysis agrees.

 

[Flex]

After 117 posts, whats the verdict? More or less heat?

 

[LarryFine]

Consider the node between the two loads where the phase A load conductor, neutral conductor, and phase B load conductor meet - that node has 3 lines connected.

If Ia and Ib are balanced and 180 degrees opposite phase, then when Ia is 1, Ib is -1 and the sum is zero.

It is still capable of carrying current and still an active part of the circuit.

If you remove the grounded conductor, you no longer have the same circuit, but instead a completely new circuit. There is no sum to carry.

The neutral in the first case is still an active part of the circuit, and carries the sum of two currents which happen to add up to zero.

I still disagree. If there is zero voltage difference between two points, a conductor between them will carry zero current. Zero current is zero current. No different than an open circuit.

 

[LarryFine]

After 117 posts, whats the verdict? More or less heat?

Less, with the usual all-other-things-being-equal caveat.