When is a neutral wire considered a conductor

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james_mcquade

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Hi everyone,

I have run into a situation at a plant and i would like your thoughts.

We have a 208 volt, 3 phase system with neutral and ground.
The power circuit is approximately 270, 250, 260 amps and has a neutral current of 50 amps.

My question is, when do you count the neutral as a current carrying wire?
what percentage of phase current warrents the neutral to be considered a current carrying wire? I am running into more and more neutral currents out here.

thanks in advance,
james
 
The neutral will always be a current carrying conductor as its purpose is to carry the unbalanced current from the ungrounded conductors.

But for the purpose of ampacity adjustments for more than 3 current carrying conductors a neutral that only carries the unbalanced current of a multiwire circuit are not required to be counted as CCC's.

Take a look at 315.(B)(4).

Chris
 
james_mcquade said:
The power circuit is approximately 270, 250, 260 amps and has a neutral current of 50 amps.

My question is, when do you count the neutral as a current carrying wire?
Click to expand...
You don't need to. Each amp on the neutral conductor is an amp that one of the line conductors isn't carrying, so its contribution is compensated for.

what percentage of phase current warrents the neutral to be considered a current carrying wire? I am running into more and more neutral currents out here.
Click to expand...
Unless you've got a lot of high-harmonic-content loads, you needn't worry about neutral current. It will show up any time your phase loads don't match.

The neutral is doing its job (keeping the service neutral point as close to the same potential as the utility neutral as possible) by conducting that current.
 
james_mcquade said:
Hi everyone,

I have run into a situation at a plant and i would like your thoughts.

We have a 208 volt, 3 phase system with neutral and ground.
The power circuit is approximately 270, 250, 260 amps and has a neutral current of 50 amps.

My question is, when do you count the neutral as a current carrying wire?
what percentage of phase current warrents the neutral to be considered a current carrying wire? I am running into more and more neutral currents out here.

thanks in advance,
james
Click to expand...

I neutral = ( A^2 + B^2 + C^2 - AB - AC - BC ) ^0.5

This formula is used for nuetral current calculation. According this, neutral current comes 17.32 Ampere and you have mentioned 50 amps. Are the other32.68 ampere neutral current is due to harmonic or what ?

My question is, when do you count the neutral as a current carrying wire?
Click to expand...

only when the load is not balanced.
 
james_mcquade said:
Hi everyone,

I have run into a situation at a plant and i would like your thoughts.

We have a 208 volt, 3 phase system with neutral and ground.
The power circuit is approximately 270, 250, 260 amps and has a neutral current of 50 amps.

My question is, when do you count the neutral as a current carrying wire?
what percentage of phase current warrents the neutral to be considered a current carrying wire? I am running into more and more neutral currents out here.

thanks in advance,
james
Click to expand...
Put a clamp on the bond wire. If you see current there then you would have an issue to look into.
 
LarryFine said:
You don't need to. Each amp on the neutral conductor is an amp that one of the line conductors isn't carrying, so its contribution is compensated for.
Click to expand...

Larry... one time I recall you gave a method to find the current on the neutral that was much simpler than anything I had ever seen... would you share that one again, please?

It was plain old common sense... which is why I can't remember it I guess.


thanks
 
Hameedulla-Ekhlas said:
I neutral = ( A^2 + B^2 + C^2 - AB - AC - BC ) ^0.5

This formula is used for nuetral current calculation. According this, neutral current comes 17.32 Ampere and you have mentioned 50 amps. Are the other32.68 ampere neutral current is due to harmonic or what ?
Click to expand...

Ham, I believe in the formula that you list, the values of A, B & C are the values of current flowing from A, B & C to neutral (or Ian, Ibn, Icn.)

The values listed in the post are the currents flowing on the phases, and cannot be plugged directly into that formula. For instance, if 270A is flowing on A, only 100A may be flowing from A to neutral.

There isn't enough information in the post to "calculate" the neutral current.
 
realolman said:
Larry... one time I recall you gave a method to find the current on the neutral that was much simpler than anything I had ever seen... would you share that one again, please?
Click to expand...
It's not so much a calculation as it is an explanation, in response to a question about how a shared neutral and two of three lines of a wye supply can carry as much current as the two phases.

Start with a 4-wire feeder, with all three phases evenly loaded; no neutral current, right? Now, subtract one amp from any one phase; the neutral current is has now become that same one amp.

Keep going until you've eliminated that one phase's load entirely, as if it never existed. Now, the neutral current is equal to the two remaining lines' current, but, that only works if they're still equal.

Note that two phases and a neutral has the same CCC count as three phases and a neutral, so there's no derating penalty for running all three phases. Conductor fill requirements still apply, of course.
 
raider1 said:
The neutral will always be a current carrying conductor as its purpose is to carry the unbalanced current from the ungrounded conductors.

But for the purpose of ampacity adjustments for more than 3 current carrying conductors a neutral that only carries the unbalanced current of a multiwire circuit are not required to be counted as CCC's.

Take a look at 315.(B)(4).

Chris
Click to expand...

As Chris pointed out, this section of the NEC is where the requirements for determining whether or not a neutral conductor is a ccc.


A neutral conductor can carry current and not be considered a ccc.
 
david luchini said:
Ham, I believe in the formula that you list, the values of A, B & C are the values of current flowing from A, B & C to neutral (or Ian, Ibn, Icn.)

The values listed in the post are the currents flowing on the phases, and cannot be plugged directly into that formula. For instance, if 270A is flowing on A, only 100A may be flowing from A to neutral.

There isn't enough information in the post to "calculate" the neutral current.
Click to expand...

The line current is the same as the phase current in a wye connection because each phase winding is connected in series with its corresponding line wire.

Remember that current in series circuit is the same throughout all parts of the circuit.

I(line) = I(phase) (Io)

To calculate the current in the common conductor of a three phase four-wire wye system, we must examine how the currents combine from all three phase wires. To analyse a balanced condition, consider that the common conductor will carry the imbalance of the three phase or hot conductor. If the three phase are equal as in supplying a three phase motor, the neutral current is the imbalance of the three phase conductors. In this case there is no imbalance and the neutral current is zero.

The formula used to find the neutral current in a three phase wye system is

I neutral = ( A^2 + B^2 + C^2 - AB - AC - BC ) ^0.5 ---1

You can quickly tell what the neutral current is if all the phase current are equal

For instance if they are all 10 A then

According to eq-1 the I neutral equals to zero.

to find the neutral current in a three phase system when the phases are not equal, we use same formula and perform a calculation. If the phase currents are different such as

A phase = 5 A
B phase = 10 A
C phase = 12 A

According to equation- 1 formula the I neutral becomes to 6.25 A

So, now compare this example with your own example.
 
Hameedulla-Ekhlas said:
The line current is the same as the phase current in a wye connection...
Click to expand...
Ham,

What David means was that the OP did not explicitly state whether or not the given amperages included three phase and/or single phase line-to-line loads. These loads do not contribute to neutral current or the balancing of line-to-neutral loads. Additionally, single phase line-to-line loads may contribute to unbalanced line amperage... and though three phase loads are assumed to be balanced, in the real world their line amperages may be 'lightly' unbalanced under normal operating conditions.
 
Smart $ said:
Ham,

What David means was that the OP did not explicitly state whether or not the given amperages included three phase and/or single phase line-to-line loads. .
Click to expand...

Smart$: Incase he has taken a approximately or it is not clear for him than ok. In wye connection, single phase load amperage is equal to three phase amperage.
 
Hameedulla-Ekhlas said:
Smart$: Incase he has taken a approximately or it is not clear for him than ok. In wye connection, single phase load amperage is equal to three phase amperage.
Click to expand...

Your still not understanding my point. Let's say on a 208/120 Wye system, we measure the currents of a feeder going to a panel. We measure 270A on A, 260A on B, 250A on C, and 50A on Neutral. We don't know how much of the current on the feeder to the panel is being connected to phase to neutral loads, how much is going to phase-to-phase (single phase) loads, and how much is going to phase-to-phase-to-phase (three phase) loads.

So let's imagine that there are two large e-phase motors connected to the panel, which draw 100A ea. That gives us 200A of load that has no neutral connection. Let us also imagine that the remaining load from each is all neutral connected load. So we have 70A from A-N, 60A from B-N, and 50A from C-N. In your Ineutral formula, A would be 70, B would be 60, and C would be 50. Without knowing how much of the current measured on the feeder is phase-to-neutral connected, you cannot apply your formula.

How this makes it more clear.
 
Hameedulla-Ekhlas said:
The formula used to find the neutral current in a three phase wye system is

I neutral = ( A^2 + B^2 + C^2 - AB - AC - BC ) ^0.5 ---1

You can quickly tell what the neutral current is if all the phase current are equal

For instance if they are all 10 A then

According to eq-1 the I neutral equals to zero.

to find the neutral current in a three phase system when the phases are not equal, we use same formula and perform a calculation. If the phase currents are different such as

A phase = 5 A
B phase = 10 A
C phase = 12 A

According to equation- 1 formula the I neutral becomes to 6.25 A

So, now compare this example with your own example.
Click to expand...

Ham,

If you had a three phase wye system with phase currents of 10<0, 10<-130, and 10<90, then by your formula, the neutral current will be zero.

However, if these phase currents are all supplying phase to neutral loads (that is to say from phase A-to-N, there is 10A<0, B-to-N is 10A<-130 and C is 10A<90) then from KCL, In=Ia+Ib+Ic.

Adding Ia+Ib+Ic ((10+j0) + (-6.43-j7.66) + (0+j10)) gives a current flowing on the neutral of 4.27Amps. Why the discrepancy?
 
Hameedulla-Ekhlas said:
I neutral = ( A^2 + B^2 + C^2 - AB - AC - BC ) ^0.5

This formula is used for nuetral current calculation. According this, neutral current comes 17.32 Ampere and you have mentioned 50 amps. Are the other32.68 ampere neutral current is due to harmonic or what ?
Click to expand...

The above formula is an approximation. It is only correct if the three currents A, B, and C are exactly 120 degrees out of phase, and contain no 'triplen' harmonics.

The difference between the measured 50A and the predicted 17A could be caused by harmonics, but it could also be caused by power factor differences on the three legs. It can even be caused by single phase L-L loads.

Consider a very simplified system that has a single 120V 100A unity power factor load connected between phase A and neutral, and a single 208V 100A unity power factor load connected between phases B and C. Such a system has 100A on phases A, B, and C, all the loads have the same power factor, and no harmonics, yet the neutral current is 100A. The prediction equation fails because the phase angle difference on the three line currents is not 120 degrees.

-Jon
 
winnie said:
The above formula is an approximation. It is only correct if the three currents A, B, and C are exactly 120 degrees out of phase, and contain no 'triplen' harmonics.

The difference between the measured 50A and the predicted 17A could be caused by harmonics,


-Jon
Click to expand...

Yes, due to harmonics I am agree with you.
 
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