Announcement

Collapse
No announcement yet.

Annex D Example D3(a)

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

    #31
    Originally posted by Xptpcrewx View Post
    Tortuga, We are getting warmer!

    If you look at the Neher-McGrath formula I is the "ampacity" which, as you have stated, is the load a conductor can carry "continuously", so why are we taking 125% of I for continuous loads if I is already accounts for this?
    From the UL Molded Case Circuit Breakers Marking and Application Guide page 13, 2016 Edition
    38.
    100 Percent Continuous Rated — Unless otherwise marked for continuous use at 100
    percent of its current rating, a circuit breaker is intended for use at no more than 80 percent of its
    rated current where in normal operation the load will continue for three hours or more. A breaker
    with a frame size of 250 A or more, or a multi-pole breaker of any current rating greater than 250 V,
    may be marked to indicate it is suitable for continuous use at 100 percent of its current rating. The
    marking is “Suitable for continuous operation at 100 percent of rating only if used in a circuit breaker
    enclosure Type ____or in a cubicle space______by_____ by _____ inches” or an equivalent
    statement. This type of breaker may also be marked to indicate it is to be used with wire sized for a
    75°C conductor with 90°C insulation and used with 90°C wire connectors.
    https://www.ul.com/wp-content/upload...Breaker_MG.pdf
    Comments based on 2017 NEC unless otherwise noted.

    Comment


      #32
      Originally posted by Xptpcrewx View Post
      If you look at the Neher-McGrath formula I is the "ampacity" which, as you have stated, is the load a conductor can carry "continuously", so why are we taking 125% of I for continuous loads if I is already accounts for this?
      I think you've already answered this.

      Originally posted by Xptpcrewx View Post
      Do you have a reference I could read up on about the purpose of the additional 25% and how its related to the mass of the termination
      The load in the example D3(a) is 119A. A #1awg conductor would carry a 119A load, but the Code makes you use a minimum #1/0awg conductor, which forces the terminations to be larger.

      But the conductor only has to have an ampacity to supply the load. The additional 25% really doesn't relate to the conductor sizing, other than as a means to increase the termination sizing.

      Comment


        #33
        Originally posted by david luchini View Post
        I think you've already answered this.
        Not sure I follow what you mean here...

        Comment


          #34
          One other question. Under the feeder neutral conductor section of this example, it states that 210.11(B) does not apply to these buildings. I’m looking at 210.11(B) and I don’t see anything that excludes industrial buildings. Am I missing something?

          Comment

          Working...
          X