UF Cable - Conductor fill question

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bjp_ne_elec

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When using say 12/2 UF cable, where do you derive the inch^2 area. I'm finding the individual conductors - obvisouly in Table 5 - but I'm getting a brain cramp on "multi-conductor" cable - which I would consider UF.

I know I'm safe at 3/4" - but looking to see if 1/2" will meet NEC for again, 12/2. Not just looking for a Yes or No - want this to sink in this time. I have another similar scenario coming up, but nailing this will cover a my other situation.

Thanks

Brett
 
Generally, an oval cable is considered to have a circular area equal to a round cable of the same diameter as the larger dimension of the oval cable. (now, inhale)
 
So Larry - I just rough measured a chunk of 12/2 UF - and it's approimately 3/8 (0.375) x 7/16 (0.4375) - so you're saying take 0.4375 x 0.4375 = 0.1903.

This would say that using 40% fill - Schedule 40 - you would need 0.1903/0.4 = 0.475 in^2. This would translate to 1" Sch 40 PVC. Now if this is correct, how do you get a piece of 12/2 UF in to a granite lamp post that has a 1-1/4" hole in it - as I'm wanting to transition in with Sch 40 PVC, as opposed to letting the UF go unprotected at the post's edge - where it enters at the bottom.

One thing I'm struggling with, now that I'm researching it, is Note 5 to the tables in Chapter 9, which states - "For conductors not included in Chapter 9 (and that is true for UF), such as multiconductor cables, the actual dimension shall be used. What I'm not sure, is how this ties to % fill - as I say 12/2 UF with ground is "More Than 2 Conductors"

Brett
 
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Brett, You need to treat this as one cable according to note 9 table 1

(9) A multiconductor cable of two or more conductors shall be treated as a single conductor for calculating percentage conduit fill area. For cables that have elliptical cross sections, the cross-sectional area calculation shall be based on using the major diameter of the ellipse as a circle diameter.

so if your cable is .4375 at the widest point , then use that for figuiring conduit size at 53% one conductor.
Rick
 
I have sleeved 12-2 uf in 3/4 pvc and will never ever try that again.The pvc was for protection because digging 18 inches was not possable and while i did not meet code this was the best i could do for a friend.
 
bjp_ne_elec said:
So Larry - I just rough measured a chunk of 12/2 UF - and it's approimately 3/8 (0.375) x 7/16 (0.4375) - so you're saying take 0.4375 x 0.4375 = 0.1903.
Click to expand...

No, the area of a circle is pi times radius squared = 3.14 x .2187 x .2187 = 0.1501
 
Eprice - I had calculated it based on it's rectangular square inches - not picking up on the Note 1 and Larry's mention of circular area (duh).

Not sure the reasoning about bumping it up by figuring circular area - but who am I to argue withe the code. Kinda of like a "square peg in a round hole". UF cable is more rectangular than oval - but again, heh.

Brett
 
bjp_ne_elec said:
So Larry - I just rough measured a chunk of 12/2 UF - and it's approimately 3/8 (0.375) x 7/16 (0.4375) - so you're saying take 0.4375 x 0.4375 = 0.1903.


Brett
Click to expand...


This calculation would give you the area of a square. You would need to multiply .1903 by 79% to find the area of the circle when each side of the square measured .4375.
 
Couldn't one of the CMPs just add cables to the tables so we don't have to argue about how to calculate the radial bilateral monolithic corinthian cross section of an elliptical cable? :D
 
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"radial bilateral monolithic corinthian"- sounds like something I want to sneak out in the backyard and light one up!

Sniff, Sniff - must be Brett's burning rope again, and he's always claiming he's burning copper.

Brett
 
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