How many 12-2 cables can I put in a short run of conduit?

Question

I am putting in a few new outlets in an enclosed patio, running a new circuit through the attic using 12-2. I would like to go from the panel to the first outlet, then to the second and then the 3rd. The run would be via the attic and the cable would just be exposed on the vertical run from the attic down to the outlet.

I would like to run this exposed section, about a 10 ft vertical run per outlet, in conduit. I would prefer to not have to run multiple conduits per box and was hoping I could use 3/4" LT or PVC - again just for the vertical section from the attic to the outlet. It isn't a wet or damp location, I just want it protected from any possible accidents.

Would this be acceptable to pull two 12-2 cable per 3/4" conduit at least for the first 2 outlets or do I need to put in a junction box along the way and then just run a 1/2" down to the outlet? I'd really prefer not to do this.

I know I shouldn't put cable in conduit for the full run, which I am not doing. This is just for exposure protection and I am not clear what the best practices are.

Answer 1

Liquidtight

If by "LT" you mean Liquidtight flexible conduit (metallic or nonmetallic), and you're looking for protection from physical damage. That's not an option, since it's not permitted to be used where subject to physical damage.

PVC

You'll have to use schedule 80 PVC, to provide protection from physical damage.

Calculate Fill

To calculate conduit fill for cables, you treat each cable as a single conductor and calculate the fill based on the major diameter of the cable (looked up from manufacturer documentation). For Southwire® Romex® SIMpull® 12/2 with ground cable, the major diameter is listed at 410 mils.

A = πr²
A = π * ((410/2)/1000)²
A = π * (0.205)²
A = π * 0.042025
A = 0.13202543126711106084639258818232 in.²

Since you're using the conduit as a sleeve, and not a complete conduit system, you should be able to fill the conduit to 100%. However, filling to 100% will make it impossible to pull the cables, so you'll want to stay well below that.

3/4" Schedule 80 PVC has a total area of 0.409 in.², of which two 12/2 with ground NM-B cables will fill 65% of. Depending on how long the sleeve is, this might make it quite difficult to pull the cables through.

Fill = Conductor Area / Total Area
Fill = (0.132025 * 2) / 0.409
Fill = 0.26405 / 0.409
Fill = 0.6456 = 65%

To put this in perspective. If this was two wires in a complete conduit system, you'd only be able to fill to 31% of the total area of the conduit (0.127 in.²).

Wet and Damp Locations

If this area can be considered damp or wet, you'll have to use cable rated for wet/damp locations. In this case, you'll want to use UF-B cable (463 mil major diameter). Two of these cables will fill 82% of the conduit, making it nearly impossible to pull by hand (depending on the length).

Fill = (((π * ((463 / 1000) / 2)²) * 2) / 0.409) * 100
Fill = 82.33%

Answer 2

I just recently tore out an installation where someone chicken-choked two 12/3 NM into a 3/4" EMT conduit. I just tore the conduit out wires and all; I cannot get the NM out of the conduit, it's in there so tight. I can't imagine how it was ever accomplished! It carries no more current than 4x THHN wires, which go effortlessly through 3/4" EMT.

For two 12/2 NM cable, you need at least 1" conduit (by calculations below) but it'll still be a difficult pull. For two 12/2 UF, you'll need at least 1-1/4" conduit.

Get pi out of the picture: Think in circular inches

When you get into large wires, the copper dimension starts to be expressed in a unit called K C Mil, or M C M.

• K or M = 1000 of ...
• C means circular area, instead of square area.
• Mil or M means "Mil", or 1/1000 (of an inch).

So a circular mil means the area of a circle 1/1000" across. KCMil or MCM mean 1000 of those circular mil units. A circular inch is 1 million CMil or 1000 KCMil.

So a "circular inch" is not a weird unit, and is in fact used in electrical already in the fashion of the KCMil.

Figuring out cross sections the easy way

Suppose your wire is W inches width in the widest dimension. The area of the wire is

W * W

and our unit is circular inches (or circular mils if we started with mils). Wow, that was easy. OK, so say your conduit is I width in interior cross section. The conduit is *I * I* obviously.

With 1 wire in a conduit, the one wire can't fill more than 53% of the conduit. That means the max wire area can be

I * I * .53

That means to be legal, this must be true:

I * I * .53 >= W * W

Algebra algebra algebra, and we get

I >= W * 1.38

Assuming 1 wire in conduit, the conduit ID must be 138% of the wire width.

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Now, this extends for 2 wires in conduit, where only 31% conduit fill is allowed (because of the strong tendency for 2 wires to bind, especially when they are wide cables). So if the other cable is Y width, it's literally

I * I * .53 >= W * W + X * X

And if both wires are the same size W, then minimum I evaluates to 2.54 * W for 2 wires of W width in conduit.

Now since 12-2NM is 0.40" wide (400 mil), that means 2 in conduit requires 1.016" conduit ID. Fortunately most 1" trade size conduit is somewhat larger than 1" actual. That's your answer.

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For 3 same-sized cables in conduit (40% fill allowed), it evals to 274% of cable max width.

For 4 same-sized cables in conduit (ditto), it evals to 317% of cable max width.

Answer 3

The rule of thumb that I've seen in researching this is to not fill the conduit more than 40% full, however, I cannot find a definitive citation for this. The general idea is found in Chapter 9 of the NEC.