Multiconductor cable conduit fill

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Cow

Senior Member
Location
Eastern Oregon
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Electrician
I'm having a little trouble calculating conduit fill for a muliticonductor cable without an overall outer sheath. I see three possible options:

1. Calculate it as a single cable. Take the overall O.D. of the cable to determine area. Conduit fill is limited to 53%.

2. Calculate it as individual conductors as usual. Conduit fill is limited to 40%.

3. Per the wire rep, calculate it like #2, but conduit fill is limited to 53%.

What brought this on, is I'm trying to stuff 3 phase 200 amp wire with a ground into an 1.5" pvc conduit. I haven't had any luck finding a wire manufacturer that makes compact copper yet either.
 
Cow said:
I'm having a little trouble calculating conduit fill for a muliticonductor cable without an overall outer sheath. I see three possible options:

1. Calculate it as a single cable. Take the overall O.D. of the cable to determine area. Conduit fill is limited to 53%.

2. Calculate it as individual conductors as usual. Conduit fill is limited to 40%.

3. Per the wire rep, calculate it like #2, but conduit fill is limited to 53%.

What brought this on, is I'm trying to stuff 3 phase 200 amp wire with a ground into an 1.5" pvc conduit. I haven't had any luck finding a wire manufacturer that makes compact copper yet either.
Click to expand...

You added this after I posted.

My answer is now #2.
 
Cow said:
I'm having a little trouble calculating conduit fill for a muliticonductor cable without an overall outer sheath. I see three possible options:

1. Calculate it as a single cable. Take the overall O.D. of the cable to determine area. Conduit fill is limited to 53%.

2. Calculate it as individual conductors as usual. Conduit fill is limited to 40%.

3. Per the wire rep, calculate it like #2, but conduit fill is limited to 53%.

What brought this on, is I'm trying to stuff 3 phase 200 amp wire with a ground into an 1.5" pvc conduit. I haven't had any luck finding a wire manufacturer that makes compact copper yet either.
Click to expand...

You'll probably find very similar answers, whether you go by 1 or 2. And I'd say both would be correct. The deal is that the fill percentages generally represent about 75% of the inner diameter to be used for wiring. A round wire in a round pipe simply fills this a lot more efficiently than numerous wires. Two wires are a special case, because no matter how you arrange them, you are left with a lot of empty space. Fill the pipe with the two wires wall-to-wall, and you fill the pipe to 50%.
 
If conductors are a multiplexed assembly it is one assembly and pulls like one assembly, I say 53% fill is allowed based on diameter of the entire assembly.
 
kwired said:
If conductors are a multiplexed assembly it is one assembly and pulls like one assembly, I say 53% fill is allowed based on diameter of the entire assembly.
Click to expand...

Where does Chapter 9 Table allow this? Note 9 says:

(9) A multiconductor cable, optical fiber cable, or flexible
cord of two or more conductors shall be treated as a
single conductor for calculating percentage conduit fill
area.

Nothing about an "assembly"?
 
jumper said:
Where does Chapter 9 Table allow this? Note 9 says:

(9) A multiconductor cable, optical fiber cable, or flexible
cord of two or more conductors shall be treated as a
single conductor for calculating percentage conduit fill
area.

Nothing about an "assembly"?
Click to expand...

A cable is a cable with or without a sheath. IMO
 
jumper said:
Where does Chapter 9 Table allow this? Note 9 says:(9) A multiconductor cable, optical fiber cable, or flexiblecord of two or more conductors shall be treated as asingle conductor for calculating percentage conduit fillarea.Nothing about an "assembly"?
Click to expand...

I believe it is covered by what you quoted.

I also agree with:
luckylerado said:
A cable is a cable with or without a sheath. IMO
Click to expand...
 
IMHO both 1 and 2 plausible interpretations of the rules for this case.

The code explicitly describes option 1 for multi-conductor cables.

Since the 'cable' doesn't have an outer sheath, then it is arguably simply a pre-made bundle of individual conductors, covered under option 2.

Option 3 is IMHO not code compliant. It ignores the fact that there will be spaces between the individual conductors.

-Jon
 
winnie said:
IMHO both 1 and 2 plausible interpretations of the rules for this case.

The code explicitly describes option 1 for multi-conductor cables.

Since the 'cable' doesn't have an outer sheath, then it is arguably simply a pre-made bundle of individual conductors, covered under option 2.

Option 3 is IMHO not code compliant. It ignores the fact that there will be spaces between the individual conductors.

-Jon
Click to expand...
But if they are twisted together or otherwise bound in some other way - they will behave like one cable when being installed.
 
FWIW, in the specific case of two conductors I would expect that two conductors in an overall sheath would be far less likely to jam while being pulled than the same two conductors without a sheath, provided that they are wide enough to rid up the raceway wall. So the calculation for one conductor should not be used, but rather two conductors even if they are supplied as a unit, as long as there is no sheath.
 
electrofelon said:
Ditto....?

I assume this discussion is about "URD", "mobile home feeder", or perhaps commonly called by other names but are just several USE-2/RHH-2 or XHHW that are factory twisted together? What requires us to call this a cable vs X single conductors?
Click to expand...

You ever try to pull that kind of product through a conduit body without untwisting the conductors first? They act like one assembly if you don't.
 
kwired said:
But if they are twisted together or otherwise bound in some other way - they will behave like one cable when being installed.
Click to expand...

Absolutely! This was Cow's option 1.

If they were bound very loosely then they would act more like individual wires, Cow's option 2.

Option 3 what the one I was rejecting as a possibility. This was 'treat like an individual wire (permitting 53% fill for 1 cable in a conduit) but calculate the area of this cable by simply summing the area of the individual wires making up the cable. This ignores any spaces between the wires, or any irregularities on the OD.

-Jon
 
It would of been spun into the cable. I was planning on XHHW.

It seems the overall diameter of the entire cable assembly is about 1.38", I believe. I won't even try to attempt pulling that into an 1.5" pvc conduit! We went with Plan B for the quote. We'll run conduit and wire from a nearby 400 amp disconnect and abandon the existing undersized 1.5" conduit/wire feeding the equipment.

Thanks for the replies.
 
Cow said:
It would of been spun into the cable. I was planning on XHHW.

It seems the overall diameter of the entire cable assembly is about 1.38", I believe. I won't even try to attempt pulling that into an 1.5" pvc conduit! We went with Plan B for the quote. We'll run conduit and wire from a nearby 400 amp disconnect and abandon the existing undersized 1.5" conduit/wire feeding the equipment.

Thanks for the replies.
Click to expand...
Step up voltage so you can use smaller conductors;)

Unless you are already at 480 volts, then you get into needing over 600 volt conductors and equipment.
 
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