e57 said:
In the NEC it is a single circuit that "shall be permitted to be considered as multiple circuits".
Think about the circuit itself (loads) without the neutral, it is simply a leg to leg or phase to phase series circuit.
Roger
Derating
- Thread starter Ken 6789
- Start date
- Status
In the NEC it is a single circuit that "shall be permitted to be considered as multiple circuits".
Think about the circuit itself (loads) without the neutral, it is simply a leg to leg or phase to phase series circuit.
Roger
- Location
- New Jersey
- Occupation
- Journeyman Electrician (retired)
A 3-wire linear MWBC is either two or three CCC's depending on the system it's supplied from. From a 120/240 volt system it would be considered 2 CCC's. From a 208Y/120 volt system it would be 3 CCC's. This information is in 310.15(B)(4)(a) as Roger has posted.
I got a little ahead of myself with my first post. In my mind this was a 120/240 volt system hence my answer that it would be only 6 CCC's. It could very well be (he's using EMT after all) a 208Y/120 volt system in which case the number of CCC's would be 9. Either way #12 conductors with either a 70 % or 80% derating factor would still be permitted to supply loads protected by a 20 amp OCPD.
I got a little ahead of myself with my first post. In my mind this was a 120/240 volt system hence my answer that it would be only 6 CCC's. It could very well be (he's using EMT after all) a 208Y/120 volt system in which case the number of CCC's would be 9. Either way #12 conductors with either a 70 % or 80% derating factor would still be permitted to supply loads protected by a 20 amp OCPD.
e57
Senior Member
- Location
- San Francisco, CA
Still doesn't answer the question on how it is black in one section of the code yet white in another. Or is it just because 'I say so' type of thing?
SiddMartin
Senior Member
- Location
- PA
infinity said:
thank you for info
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
This question seemed to get passed over, so I'll chime in.Ken 6789 said:
It's either use #10 wire or 15a breakers, not both. When designing a circuit, you start with serving the load. If you need 20a circuits, then you need to use #10 wire. If you want to keep the #12, then you need 15a breakers.
It can sometimes be more economical to run two smaller conduits than one larger one, considering conductor re-sizing. Also, for your 12 circuits, it could be more economical to run a feeder to a sub-panel closer to the loads.
Bob NH
Senior Member
- Location
- The "Live Free or Die" State
You could run 3 MWBCs in the conduit, with 6 CCCs. Then connect the extra neutral where they branch to separate circuits and install GFCI receptacles as the first receptacle on each circuit.
- Location
- New Jersey
- Occupation
- Journeyman Electrician (retired)
Bob NH said:
Good point, but the 6 CCC's would be dependent on the supply system used. For 208Y/120 it would be 9 CCC's.
e57
Senior Member
- Location
- San Francisco, CA
roger said:
These are not 240 circuits where the neutral would not matter and carry no current - they are 120 individually utilized circuits and the neutral carries current - IMO once you 'consider' and use them as multiple circuits that is exactly what they are - right?
Please explain to me how they could be a 'multiple' individually utilized circuit and if not loaded evenly they are carrying current - they could be considered the same and not current carrying at the same time?
Enlighten me - because I can't see how they can be both at the same time.
e57 said:
Mark, all MWBC's are a single circuit, they are "allowed" to be considered multiple circuits.
In the example of one ungrounded conductor carrying 10 amps, the other carrying 5 amps, and the neutral carrying 5 amps shows this is one circuit, 5 amps of the 10 amp ungrounded conductor is being canceled by the 5 amps flowing in the other ungrounded conductor and the unbalanced 5 amps is be carried by the neutral.
All three conductors are performing together and are one circuit regardless of what terminology a code may allow or not allow.
Roger
Bob NH
Senior Member
- Location
- The "Live Free or Die" State
e57 said:
The basis for the code permitting the "unbalanced load only neutral" to be neglected when using a multiwire branch circuit, or any other case where the neutral carries only the unbalanced load, is that the sum of the (current squared) values for all of the conductors in the circuit (and therefore the heating) never exceeds the (current squared) value for the ungrounded conductors with balanced load.
For example, it you have a 20 Amp MWBC with balanced load the sum of the (current squared) = 20x20 + 20x20 + 0x0 = 800 Amps squared.
If there is an unbalanced load, such as 20 Amps on one conductor and 5 Amps on the other, then the sum of the (current squared) = 20x20 + 5x5 +15x15 = 650 Amps Sqaured.
- Location
- New Jersey
- Occupation
- Journeyman Electrician (retired)
Bob NH said:
Bob what type of system are you referring too? And I'm not sure I'm understanding your math. Could you please explain your calculations?
e57
Senior Member
- Location
- San Francisco, CA
Bob NH said:
You had me until the parts in red....
e57
Senior Member
- Location
- San Francisco, CA
So are you saying if I had (For the sake of easy numbers) 10a and 20a (30 total) it would be the same as 15a and 15a?
One would have 10a on the neutral the other zero....
10a A, 10a N, 20a B = 40a total*
vs
15a A, 0 N, 15a B = 30a total
*From what I gather from Roger's explaination is that the 10a (of heat) on the neutral is cancelled somehow to be only 30a total. (Unless covered by 310.15B4 b or c)
One would have 10a on the neutral the other zero....
10a A, 10a N, 20a B = 40a total*
vs
15a A, 0 N, 15a B = 30a total
*From what I gather from Roger's explaination is that the 10a (of heat) on the neutral is cancelled somehow to be only 30a total. (Unless covered by 310.15B4 b or c)
Last edited:
e57
Senior Member
- Location
- San Francisco, CA
roger said:
You gonna make me pull out a meter on a sunday?!?!?:roll:
Besides it being a very inacurate method of measuring current - what's your point - where does this current (and heat created by it for derating purposes) on the neutral vanish to? And why does it not count as current any longer? Even though it is still there and on a conductor.
Give me an explaination that does sound like 'just because' - work with me here....:roll:
e57 said:
It is a very accurate method of measuing current and is the way a GFCI works, it is also a great way to see if there are net currents flowing somewhere besides where they should be flowing.
e57 said:
It doesn't vanish unless the loads on both sides of the neutral are perfectly balanced, but the total of all three conductors added together would still be the same as if one side were an equal percentage higher than the other side were low.
e57 said:
Mark, I honestly don't know how to explain it to you if you can't see how all the conductors work together as a single circuit.
Why do you think on imbalanced MWBC's there is damage to equipment if the neutral is lost? When you understand this you will see how it is a single circuit. Look up "Edison" circuit
Something to ponder, how many circuits are going to an average 120/240 volt range?
Roger
Bob NH
Senior Member
- Location
- The "Live Free or Die" State
Originally Posted by Bob NH
The basis for the code permitting the "unbalanced load only neutral" to be neglected when using a multiwire branch circuit, or any other case where the neutral carries only the unbalanced load, is that the sum of the (current squared) values for all of the conductors in the circuit (and therefore the heating) never exceeds the (current squared) value for the ungrounded conductors with balanced load.
For example, it you have a 20 Amp MWBC with balanced load the sum of the (current squared) = 20x20 + 20x20 + 0x0 = 800 Amps squared.
If there is an unbalanced load, such as 20 Amps on one conductor and 5 Amps on the other, then the sum of the (current squared) = 20x20 + 5x5 +15x15 = 650 Amps Sqaured.
The basis for limiting the number of current carrying conductors in a raceway is to limit the heating, which is a result of the power dissipated in the conductors.
The power dissipated in a conductor is: Power = (current sqaured) x Resistance.
Now consider a MWBC of #12 with 20 Amp breakers, and let's calculate the power dissipated in the three conductors for various current conditions. Because the resistance of each conductor is the same and the length is the same we can disregard the actual resistance and calculate the (current squared) = (Amps x Amps) for each of the 3 conductors in the MWBC and add them together. The largest number is the one that provides the greatest heating.
Case 1: First let's take the CODE case of maximum current in each of the ungrounded conductors; which results in zero current in the neutral.
Case 1: Total Amps Squared = 20 x 20 + 20 x 20 + 0 x 0 = 400 + 400 + 0 = 800 Amps squared.
Case 2: 20 Amps in one ungrounded and therefore 20 Amps in the neutral
20 x 20 + 0 x 0 + 20 x 20 = 800 Amps squared.
Case 3: 20 Amps in one ungrounded conductor and10 Amps in the other ungrounded conductor, and therefore 10 Amps in the neutral.
20 x 20 + 10 x 10 + 10 x 10 = 400 + 100 + 100 = 600 Amps squuared.
This is the case with minimum heating when ONE of the ungrounded conductors is carrying 20 Amps. I won't try to show the differential calculus here.
Case 4: 20 Amps in one ungrounded conductor and 15 Amps in the other ungrounded conductor, and therefore 5 Amps in the neutral.
20 x 20 + 15 x 15 + 5 x 5 = 400 + 225 + 25 = 650 Amps squared.
CONCLUSION:
The code permits conductors that carry only the unbalanced portion of the load to be neglected because all cases with an unbalanced load resulting in current in the neutral dissipate LESS ENERGY IN THE CIRCUIT CONDUCTORS than the code case of maximum load in the ungrounded conductors.
The basis for the code permitting the "unbalanced load only neutral" to be neglected when using a multiwire branch circuit, or any other case where the neutral carries only the unbalanced load, is that the sum of the (current squared) values for all of the conductors in the circuit (and therefore the heating) never exceeds the (current squared) value for the ungrounded conductors with balanced load.
For example, it you have a 20 Amp MWBC with balanced load the sum of the (current squared) = 20x20 + 20x20 + 0x0 = 800 Amps squared.
If there is an unbalanced load, such as 20 Amps on one conductor and 5 Amps on the other, then the sum of the (current squared) = 20x20 + 5x5 +15x15 = 650 Amps Sqaured.
e57 said:
The basis for limiting the number of current carrying conductors in a raceway is to limit the heating, which is a result of the power dissipated in the conductors.
The power dissipated in a conductor is: Power = (current sqaured) x Resistance.
Now consider a MWBC of #12 with 20 Amp breakers, and let's calculate the power dissipated in the three conductors for various current conditions. Because the resistance of each conductor is the same and the length is the same we can disregard the actual resistance and calculate the (current squared) = (Amps x Amps) for each of the 3 conductors in the MWBC and add them together. The largest number is the one that provides the greatest heating.
Case 1: First let's take the CODE case of maximum current in each of the ungrounded conductors; which results in zero current in the neutral.
Case 1: Total Amps Squared = 20 x 20 + 20 x 20 + 0 x 0 = 400 + 400 + 0 = 800 Amps squared.
Case 2: 20 Amps in one ungrounded and therefore 20 Amps in the neutral
20 x 20 + 0 x 0 + 20 x 20 = 800 Amps squared.
Case 3: 20 Amps in one ungrounded conductor and10 Amps in the other ungrounded conductor, and therefore 10 Amps in the neutral.
20 x 20 + 10 x 10 + 10 x 10 = 400 + 100 + 100 = 600 Amps squuared.
This is the case with minimum heating when ONE of the ungrounded conductors is carrying 20 Amps. I won't try to show the differential calculus here.
Case 4: 20 Amps in one ungrounded conductor and 15 Amps in the other ungrounded conductor, and therefore 5 Amps in the neutral.
20 x 20 + 15 x 15 + 5 x 5 = 400 + 225 + 25 = 650 Amps squared.
CONCLUSION:
The code permits conductors that carry only the unbalanced portion of the load to be neglected because all cases with an unbalanced load resulting in current in the neutral dissipate LESS ENERGY IN THE CIRCUIT CONDUCTORS than the code case of maximum load in the ungrounded conductors.
- Location
- New Jersey
- Occupation
- Journeyman Electrician (retired)
Bob NH said:
Bob, what system do your calculations correspond too? And do you have a formula to go along with the numbers?
Bob NH
Senior Member
- Location
- The "Live Free or Die" State
infinity said:
I was responding to the several posts earlier in the thread where e57 appeared to be inquiring about the rationale supporting the provision of 310.15(B)(4)(a) that permits the neutral carrying only the unbalanced load to be omitted from the count when applying the provisions of 310.15(B)(2)(a). A 240/120 Volt multiwire branch circuit with two ungrounded conductors and one neutral is an example of how that provision may be applied.
My calculation shows the rationale for that "omit the neutral from the count" provision; which is that the power dissipated in the conductors that must be counted is ALWAYS greater than the power that will be dissipated in those two conductors plus the neutral for all cases where the neutral is carrying the unbalanced current of the two ungrounded conductors.
The only formula required is Power = I^2 * Resistance. That formula is a result of the application of Ohm's Law as shown below.
The power dissipated in a resistive load is equal to the voltage drop across the load x current, based on the following:
Voltage drop = Current x resistance
Therefore, Voltage drop x current = Current x current x resistance = current squared x resistance.
Since the resistance of the conductors is constant, the only varible that must be considered to prove the point is the sum of the values of Amps Squared for the three conductors of the MWBC.
As shown in my previous posts, the maximum value of the sum of Amps Squared, and therefore the maximum heating in the MWBC, is for the condition where the current is balanced with the maximum value in each of the ungrounded conductors. That maximum value is also equal to the case where one of the ungrounded conductors carries the maximum load and the other carries zero, while the neutral carries the unbalanced current equal to the current in the ungrounded conductor.
- Location
- New Jersey
- Occupation
- Journeyman Electrician (retired)
Bob NH said:
So you were referencing a 120/240 volt system. I ask because 310.15(B)(4)(b) tells us that in a Wye system a 3-wire MWBC would have 3 CCC's not 2 as in the 120/240 volt system.
- Status