Understanding theory, 300.3(B)

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chris kennedy

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Miami Fla.
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60 yr old tool twisting electrician
The 08 NECHB commentary following 300.3(B);

This general rule remains consistent with electrical theory; that is, to reduce inductive heating and to avoid increases in overall circuit impedance, all circuit conductors of an individual circuit must be grouped. Similar requirements are found in 300.5(I).
How does increased distance between conductors increase overall circuit impedance?

Thanks
 

ohmhead

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ORLANDO FLA
Well hey Chris hows the weather in south florida ?

Iam working in Jacksonville its a little cool need a jacket in the morning .The question i think is all about inductive reactance between conductors it makes ac ohms go up ac power reactance .
 

chris kennedy

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Location
Miami Fla.
Occupation
60 yr old tool twisting electrician
As conductors of the same circuit are moved further apart, the impedance of the circuit will increase. As per 250.4, we want an effective ground fault current path, with as low an impedance as is possible.

How is the impedance increased? The resistance added by the effects of heating? Other factors? Should I have payed attention in school?
 

ohmhead

Senior Member
Location
ORLANDO FLA
Well Chris its the new Duval county court house were just starting underground on monday so no pictures yet .

Ive been ther 4 months planning and auto cadding the distribution lots of fun but id still like to be back in Orlando its warmer its only 2 hours each way nice drive !!!:D
 

ohmhead

Senior Member
Location
ORLANDO FLA
Well Chris this induction effect is increased its harder to cancel out by distance between your conductors . When conductors are far apart in a raceway or cable tray induced eddy currents are present now in the center of a conductor this pushes the normal current flow to the outer shell of the conductor so resistance is higher less copper to flow thur . Which is ac impedance added and yes also you are correct heat makes resistance higher this is more heat in circuit .

I never payed attention in school thats why iam a electrician
 
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Besoeker

Senior Member
Location
UK
The 08 NECHB commentary following 300.3(B);

How does increased distance between conductors increase overall circuit impedance?

Thanks
It increases the inductance.
If I recal correctly, the factor is ln(d/r) where d is the distance between the conductors. Bigger d, bigger inductance, bigger impedance.
 

charlie b

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Lockport, IL
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I couldn’t address this question until I got back to work, since that is where I keep my power engineering textbooks. I think the author(s) of that commentary have it wrong.

Electrical theory tells us that between any two wires there is a value of mutual inductance. Besoeker had the right idea:
If I recall correctly, the factor is ln(d/r) where d is the distance between the conductors. Bigger d, bigger inductance, bigger impedance.
Close. I had to look this one up. The inductance between two parallel circular wires of equal radii is equal to 4 x 10*(-7) ln(D/r’). But as you say, if the distance between the wires is increased, then the value of inductance is increased, and so too is the value of inductive reactance.


What the author(s) of the NECH commentary missed is that that is a good thing! The inductive reactance calculated using this formula would tell you the amount of leakage current that would flow between the wires. This is not “circuit impedance.” This impedance is not in series with the load. Current does not flow from the source, via one circuit conductor, through this “impedance,” then through the load, and finally back to the source via the other circuit conductor. Thus, increasing this impedance will not adversely affect the circuit. This is a leakage path, and is in parallel with the load. That makes it a bad thing. That means that it would be to our advantage to make the impedance larger, so as to make the leakage current smaller. And in order to make that happen, we need to route our circuit conductors as far away from each other as possible.

But we don’t do that. More to the point, the rule in 300.3(B) tells us to route the conductors in the same raceway. That puts the conductors close to each other, reducing the value of “D” in the formula above. That will reduce the mutual inductance, reduce the impedance, and allow for more leakage current. This turns out to be a small effect, one that has no influence on the circuit performance. So as I see it, it was mere pomposity on the part of the commentary author(s) to make the claim that electrical theory proves the rule to be a good idea. They knew a little about theory, but only enough to make it seem impressive.

Good question, Chris. And good catch.
 
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Besoeker

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Location
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Besoeker had the right idea:Close. I had to look this one up. The inductance between two parallel circular wires of equal radii is equal to 4 x 10*(-7) ln(D/r?).
What I gave was just the factor which shows why the inductance increases, not the means of calculating inductance.
 

ELA

Senior Member
Occupation
Electrical Test Engineer
What the author(s) of the NECH commentary missed is that that is a good thing! The inductive reactance calculated using this formula would tell you the amount of leakage current that would flow between the wires. This is not ?circuit impedance.? This impedance is not in series with the load. Current does not flow from the source, via one circuit conductor, through this ?impedance,? then through the load, and finally back to the source via the other circuit conductor. Thus, increasing this impedance will not adversely affect the circuit. This is a leakage path, and is in parallel with the load. That makes it a bad thing. That means that it would be to our advantage to make the impedance larger, so as to make the leakage current smaller. And in order to make that happen, we need to route our circuit conductors as far away from each other as possible.



But we don?t do that. More to the point, the rule in 300.3(B) tells us to route the conductors in the same raceway. That puts the conductors close to each other, reducing the value of ?D? in the formula above. That will reduce the mutual inductance, reduce the impedance, and allow for more leakage current. This turns out to be a small effect, one that has no influence on the circuit performance. So as I see it, it was mere pomposity on the part of the commentary author(s) to make the claim that electrical theory proves the rule to be a good idea. They knew a little about theory, but only enough to make it seem impressive.

Good question, Chris. And good catch.


The concern is addressing the inductance of the loop formed by the hot wire and the ground return if a fault occurs.

The inductance of the loop increases as the distance between the conductors increases. This is indeed a series impedance that would tend to resist current flow during a fault. It is not about leakage current.
 

Razzap

Member
I am no engineer, but here is what I was taught about Art. 300.3(B). It is similar to Art. 300.20(A) and (B) where it talks about grouping conductors together to avoid induced currents in a metal enclosure. When single A-C wires are run thru individual holes in a ferrous (magnetic) material, it causes the material to act like a single-turn secondary of a transformer. Induced currents will heat up the material. This is how the hand-held transformer-type soldering "guns" work - a single secondary turn is shorted out (the tip), and induced current from the primary flows thru it - high current, low voltage - heating the tip. Back to Art. 300.3(B), if individual phases are run in separate, parallel ferrous (magnetic) conduits, each conduit becomes like the "secondary" of a transformer that is shorted out. If all three phases are run in a single conduit (grouped together), the individual phase inductances cancel each other out, reducing the inductive heating effect. Hope this helps, it is hard to explain without pictures.
 

crossman

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Location
Southeast Texas
Am I thinking correctly here?

The issue at hand is "self inductance." The magnetic field around a single conductor carrying a current is such that it produces a countervoltage in the wire that opposes any change in the current flow in that wire... and is seen as an impedance.

Now, when we have two wires of a circuit in close proximity, the magnetic fields are equal but opposite. This tends to cancel out the magnetic fields and does away with the self inductance.

Another way of looking at it... the original field of the single conductor creates a counter emf in that wire. But bringing an equal and opposite magnetic field into the picture, and it creates an aiding emf in that original conductor via xfmr action, so thereby the counter voltage and the aiding voltage cancel... the magnetic field no longer has an effect on the current in the wire.
 

jghrist

Senior Member
Am I thinking correctly here?

The issue at hand is "self inductance." The magnetic field around a single conductor carrying a current is such that it produces a countervoltage in the wire that opposes any change in the current flow in that wire... and is seen as an impedance.

Now, when we have two wires of a circuit in close proximity, the magnetic fields are equal but opposite. This tends to cancel out the magnetic fields and does away with the self inductance.

Another way of looking at it... the original field of the single conductor creates a counter emf in that wire. But bringing an equal and opposite magnetic field into the picture, and it creates an aiding emf in that original conductor via xfmr action, so thereby the counter voltage and the aiding voltage cancel... the magnetic field no longer has an effect on the current in the wire.
Yes, you are thinking correctly. Except that the magnetic fields are not equal, so there is a net positive inductance. If the wires are close together, the magnetic fields are nearly the same; they almost cancel and the total inductance is small. If the wires are far apart, there isn't much cancellation and the total inductance is large.
 

chris kennedy

Senior Member
Location
Miami Fla.
Occupation
60 yr old tool twisting electrician
Mind if I bump this?

Mind if I bump this?

Another thread running today got me thinking about this again and I can't seem to wrap my pea brain around this. If this question was answered then I didn't understand and I apologize. If someone could post an example using math I sure that would do it.

Thanks and enjoy the day!
 

winnie

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Location
Springfield, MA, USA
Occupation
Electric motor research
Charlie,

I am having trouble understanding your analysis.

I think that the formula that you give is for the inductance of two long straight wires which are electrically in parallel. It is analogous to the self inductance of a single conductor, just for the special case where the single conductor is composed of two wires spaced apart. This is _not_ the formula for mutual inductance.

I don't see how mutual inductance between two wires would cause the leakage that you describe. As I understand mutual inductance, a change in the current flowing through one circuit element induces a voltage in another circuit element; this means that for to parallel wires the change of current flowing through one wire will induce a voltage in the other wire.

I don't see how the induced voltage represents leakage from one conductor to the other; I guess the analogy would be that mutual inductance in a transformer does not mean leakage from the primary to the secondary.

I think that for the conductors of a circuit, the mutual inductance of one wire to the other will tend to balance out the self inductance of those wires. The current in the two wires is the same, but flowing in opposite directions. So the polarity of the mutual induced voltage will be the opposite of the polarity of the self induced voltage. As you spread the wires apart, the self inductance stays the same, but the mutual inductance drops, so the net result is greater over-all circuit inductance.

-Jon
 

Smart $

Esteemed Member
Location
Ohio
Valid or not, I grabbed the formula below from this website:

Inductance of Straight Parallel Wires carrying equal and opposite currents is calculated from:

L = 0.1 + ( 0.4 * loge ( D / r ) ) microhenries/metre

where D is the separation, and r is the wire radius, and D >> r.
 
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