'neutral' heresy...

[ToolHound]

Please suggest how to reconcile the following two Code passages, that seem to differ as to current on a service entrance neutral:


From Handbook Commentary following 240.4, quote:
"The increased ratings given in Table 310.15(B)(6) are based on the significant diversity inherent to most dwelling unit loads and the fact that only the two ungrounded service or feeder conductors are considered to be current carrying."
(underlining added)

in contrast to:

From 220.61, quote:
"The feeder or service neutral load shall be the maximum unbalance of the load determined by this article. "

Thanks , --ToolHound
​

 

[A/A Fuel GTX]

In a properly installed system, the maximum unbalanced load on the neutral will be significantly less than the load on the ungrounded conductors. I don't see a contradiction in the two statements.

 

[ToolHound]

In a properly installed system, the maximum unbalanced load on the neutral will be significantly less than the load on the ungrounded conductors. I don't see a contradiction in the two statements.

A/A Fuel GTX. Thanks. Well, ok. The statement about the ungrounded conductors threw me a little (i.e., "only the two ungrounded service or feeder conductors are considered to be current carrying." )

--ToolHound

 

[GoldDigger]

Neutral "there-esy"

Neutral "there-esy"

A/A Fuel GTX. Thanks. Well, ok. The statement about the ungrounded conductors threw me a little (i.e., "only the two ungrounded service or feeder conductors are considered to be current carrying." )

--ToolHound

Two different problems are being addressed. If an unbalanced load is created, the neutral will have to be capable of safely carrying the full current of one of the phase leads. No relief here from that requirement.
But for the purpose of derating bundled conductors, you have to consider the heating produced in each of the wires. In the case of a completely balanced load, there will be current in all of the phase wires but not in the neutral. In the case of an unbalanced load, there will be current in the neutral only to the extent that there is a reduced current in one or more of the phase lines so the total heating remains the same.

There is an important design element that modifies this: For highly non-linear loads, there may be equal currents in all of the phases, but still a current, possibly as large as the sum of the phase currents, in the neutral. In this case the neutral needs to be oversized and also should be considered as current carrying for derating purposes.

 

[al hildenbrand]

The statement about the ungrounded conductors threw me a little (i.e., "only the two ungrounded service or feeder conductors are considered to be current carrying." )

Here's a thought experiment:

Consider a balanced load on an existing 240 / 120 Volt single phase electrical service, where all the load is 120 Volt, and the total running load at the start of this thought experiment totals 100 Amps on each of the two energized service conductors.

L1 = L2 = 100 Amps. What is the current on the grounded service (or feeder) conductor? Answer: Zero Amps

Now, turn off all the load connected to L1. Leave the load on L2 running.

What happens. L1 goes to zero Amps, and the current on L2 now returns to the source through the grounded conductor.

You can redo this experiment with any combination of resistive load, whether 120 Volt or 240 Volt.

The heat created in the three conductors of a single phase 240 / 120 Volt service (feeder) is always equal to the heat of just two conductors.

 

[GoldDigger]

The heat created in the three conductors of a single phase 240 / 120 Volt service (feeder) is always equal to the heat of just two conductors.

Except in the case of non-linear loads, which I guess you would not be able to call balanced in a true sense even though the power on each phase would be identical.
Similarly, if only one phase had a low power factor there could still be substantial neutral current. The heating in the wire does not depend on the phase relationship between the applied voltage and the current. It is just I²R, independent of V_circuit.

 

[al hildenbrand]

Except in the case of non-linear loads, . . .

Absolutely. That's why I edited in the word "resistive" in the sentence before your quote.

My thought experiment is intended to start at the level of DC models.

The effects of harmonic currents are for the next class.

 

[ToolHound]

Here's a thought experiment:

Consider a balanced load on an existing....

al hildenbrand. Thanks. That is an instructional thought experiment. Makes me think of 'conservation of mass and energy' as well as, of course, principles of phase and neutral currents in a split phase system. --ToolHound

 

[al hildenbrand]

Thanks.

:thumbsup: You're welcome.

 

[jumper]

I predict 100+ posts for this thread.

 

[don_resqcapt19]

Except in the case of non-linear loads, which I guess you would not be able to call balanced in a true sense even though the power on each phase would be identical.
Similarly, if only one phase had a low power factor there could still be substantial neutral current. The heating in the wire does not depend on the phase relationship between the applied voltage and the current. It is just I²R, independent of V_circuit.

I didn't think that the neutral currents on a single phase system with non-linear loads are additive like they are on a 3 phase system.

 

[infinity]

I didn't think that the neutral currents on a single phase system with non-linear loads are additive like they are on a 3 phase system.

I agree.

 

[jumper]

I didn't think that the neutral currents on a single phase system with non-linear loads are additive like they are on a 3 phase system.

I agree.

AFAIK, They are not.

 

[charlie b]

In the case of an unbalanced load, there will be current in the neutral only to the extent that there is a reduced current in one or more of the phase lines so the total heating remains the same.

Not quite true. To make the statement correct, change the phrase "remains the same" to the phrase "remains the same or will be lower."

The heat created in the three conductors of a single phase 240 / 120 Volt service (feeder) is always equal to the heat of just two conductors.

Also not quite true. To make the statement correct, change the phrase "equal to" to the phrase "equal to or lower than."

The heat generated in a wire varies with current passing through it, but the relation is not linear. Heat varies with (I^2)R. Here is a simple math experiment. Let's consider R to be 1 ohm, for simplicity. Let's say the balanced load would be 10 amps. Here are three circumstances.

1. Total balance. Phase A = 10 amps, Phase B = 10 amps, Neutral = 0 amps.

• Heat in Phase A is (10 x 10) x 1, for a total of 100 units.
• Heat in Phase B is (10 x 10) x 1, for a total of 100 units.
• Heat in Neutral is (0 x 0) x 1, for a total of 0 units.
• Total Heat is 100 + 100 + 0, for a total of 200 amps.

2. Total imbalance. Phase A = 10 amps, Phase B = 0 amps, Neutral = 10 amps.

• Heat in Phase A is (10 x 10) x 1, for a total of 100 units.
• Heat in Phase B is (0 x 0) x 1, for a total of 0 units.
• Heat in Neutral is (10 x 10) x 1, for a total of 100 units.
• Total Heat is 100 + 0 + 100, for a total of 200 amps, (SAME AS BEFORE).

3. Somewhere in the middle of the other two. Phase A = 5 amps, Phase B = 10 amps, Neutral = 5 amps.

• Heat in Phase A is (5 x 5) x 1, for a total of 25 units.
• Heat in Phase B is (10 x 10) x 1, for a total of 100 units.
• Heat in Neutral is (5 x 5) x 1, for a total of 25units.
• Total Heat is 25 + 100 + 25, for a total of 150 amps (LOWER THAN BEFORE).

Conclusion: If there is an imbalance, the total heat generated by the phase wires and the neutral wire will be equal to, or lower than, the heat that would have been generated in a balanced loading (i.e., no neutral current) condition.

 

[ggunn]

I didn't think that the neutral currents on a single phase system with non-linear loads are additive like they are on a 3 phase system.

Why wouldn't they be? I'm not challenging you; I really don't know.

 

[Smart $]

Why wouldn't they be? I'm not challenging you; I really don't know.

Well, technically they can be... just not in the same sense as three phase. In three phase, it is triplen harmonics (multiples of three times the base frequency) that are additive. I'm not certain of terminology for split phase, perhaps "duplen" harmonics, i.e. multiples of two times the base frequency, would be additive (Actually I'm not certain about the multiples, but two times would be).

 

[Smart $]

Not quite true. To make the statement correct, change the phrase "remains the same" to the phrase "remains the same or will be lower."

Also not quite true. To make the statement correct, change the phrase "equal to" to the phrase "equal to or lower than."

The heat generated in a wire varies with current passing through it, but the relation is not linear. Heat varies with (I^2)R. Here is a simple math experiment. Let's consider R to be 1 ohm, for simplicity. Let's say the balanced load would be 10 amps. Here are three circumstances.

1. Total balance. Phase A = 10 amps, Phase B = 10 amps, Neutral = 0 amps.

• Heat in Phase A is (10 x 10) x 1, for a total of 100 units.
• Heat in Phase B is (10 x 10) x 1, for a total of 100 units.
• Heat in Neutral is (0 x 0) x 1, for a total of 0 units.
• Total Heat is 100 + 100 + 0, for a total of 200 amps.

2. Total imbalance. Phase A = 10 amps, Phase B = 0 amps, Neutral = 10 amps.

• Heat in Phase A is (10 x 10) x 1, for a total of 100 units.
• Heat in Phase B is (0 x 0) x 1, for a total of 0 units.
• Heat in Neutral is (10 x 10) x 1, for a total of 100 units.
• Total Heat is 100 + 0 + 100, for a total of 200 amps, (SAME AS BEFORE).

3. Somewhere in the middle of the other two. Phase A = 5 amps, Phase B = 10 amps, Neutral = 5 amps.

• Heat in Phase A is (5 x 5) x 1, for a total of 25 units.
• Heat in Phase B is (10 x 10) x 1, for a total of 100 units.
• Heat in Neutral is (5 x 5) x 1, for a total of 25units.
• Total Heat is 25 + 100 + 25, for a total of 150 amps (LOWER THAN BEFORE).

Conclusion: If there is an imbalance, the total heat generated by the phase wires and the neutral wire will be equal to, or lower than, the heat that would have been generated in a balanced loading (i.e., no neutral current) condition.

Also note in case #2, the utilized power is half that of case #1, yet the generated heat in the conductors is equal.

PS: You stated the totals as amps when it should be units (of heat).

 

[jumper]

Well, technically they can be... just not in the same sense as three phase. In three phase, it is triplen harmonics (multiples of three times the base frequency) that are additive. I'm not certain of terminology for split phase, perhaps "duplen" harmonics, i.e. multiples of two times the base frequency, would be additive (Actually I'm not certain about the multiples, but two times would be).

Smart, I gotta disagree about your "duplen"- even numbered harmonics being additive.

Harmonics can cause overloading of conductors and transformers and overheating of utilization equipment, such as motors. Triplen harmonics can especially caus e overheating of neutral conductors on 3-phase, 4-wire systems. While the fundamental frequency and even harmonics cancel out in the neutral conductor, odd-order harmonics are additive. Even in a balanced load condition, neutral currents can reach magnitudes as high as 1.73 times the average phase current.

http://ecmweb.com/content/fundamentals-harmonics

 

[Smart $]

Smart, I gotta disagree about your "duplen"- even numbered harmonics being additive.



http://ecmweb.com/content/fundamentals-harmonics

That statement is regarding three phase, four wire systems???

 

[Smart $]

Smart, I gotta disagree about your "duplen"- even numbered harmonics being additive...

Okay, after a little more research and theorizing, it is not full-wave harmonics which are additive, but rather current of half-wave switch-mode power supplies.