sparkycoog
Member
- Location
- Texas
This is a 3 part question so to help make my questions clearer, I put the main question in bold and tried to separate the question to make things easier. If it did the opposite, I apologize in advance.
Art. 334.80 allows only 2 wires in a bored hole where thermal insulation, firecaulk, etc will be used-before we have to apply the adjustment factors in table 310.15(B)3a . That part is easy, but at what point, are other NM wires bundled together required to have adjustments? Is it when the bundle essentially becomes a raceway - ie. exceeds 24 inches without air circulation?
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When calculating the adjustments, is there a time when the neutral conductors not counted? The rule on Neutral conductors states that a the neutral may or may not be a current-carrying conductor. In accordance with 310.15(B)(4)(a), a neutral conductor that carries only the unbalanced current from other conductors of the same circuit shall not be required to be counted when applying the provisions of 310.15(B)(2)(a). But, if the neutral is part of a three-wire circuit consisting of two phase conductors and the neutral conductor and it is supplied from a four-wire, three-phase, wye-connected system, the neutral must be counted as a current-carrying conductor.
The first part of that statement confuses me because unless we have a perfectly balanced circuit, there is never a time when a neutral conductor is NOT carrying current. What am I missing here?
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The actual calculation confuses me a little too. In particular, what the role of ambient temperature is here. I've read that Romex would be treated as having a rating of 90 degrees C. How exactly is that role used? Is it to use Ambient Temperature Corrections from table 310.15(B)(2)(a) the way it would be for a regular conduit, or is used to treat the conductors based on the 90 degree C ratings in table 310.104(A)? ie. each current carrying conductor in 12/2 Romex is now rated for 30 amps?
In other words, suppose you had 20 12/2 romex wires bundled in an airtight situation over 24 inches long (if that is in fact the trigger to apply adjustments), would each conductor now be 30 amps x 40% = 12 amps?
Art. 334.80 allows only 2 wires in a bored hole where thermal insulation, firecaulk, etc will be used-before we have to apply the adjustment factors in table 310.15(B)3a . That part is easy, but at what point, are other NM wires bundled together required to have adjustments? Is it when the bundle essentially becomes a raceway - ie. exceeds 24 inches without air circulation?
-----
When calculating the adjustments, is there a time when the neutral conductors not counted? The rule on Neutral conductors states that a the neutral may or may not be a current-carrying conductor. In accordance with 310.15(B)(4)(a), a neutral conductor that carries only the unbalanced current from other conductors of the same circuit shall not be required to be counted when applying the provisions of 310.15(B)(2)(a). But, if the neutral is part of a three-wire circuit consisting of two phase conductors and the neutral conductor and it is supplied from a four-wire, three-phase, wye-connected system, the neutral must be counted as a current-carrying conductor.
The first part of that statement confuses me because unless we have a perfectly balanced circuit, there is never a time when a neutral conductor is NOT carrying current. What am I missing here?
----
The actual calculation confuses me a little too. In particular, what the role of ambient temperature is here. I've read that Romex would be treated as having a rating of 90 degrees C. How exactly is that role used? Is it to use Ambient Temperature Corrections from table 310.15(B)(2)(a) the way it would be for a regular conduit, or is used to treat the conductors based on the 90 degree C ratings in table 310.104(A)? ie. each current carrying conductor in 12/2 Romex is now rated for 30 amps?
In other words, suppose you had 20 12/2 romex wires bundled in an airtight situation over 24 inches long (if that is in fact the trigger to apply adjustments), would each conductor now be 30 amps x 40% = 12 amps?