raceway fill with romex

[jb_04]

I had a question on my journeyman exam that went something like this.

20" pvc with 14-2 dimension of .108 by .052, 10-3 with dimension of .4151, and so on...

basically gave two numbers for the 2 wire cables and one number for the 3 wire cables.

In the numbers above, do I multiply .108*.052, then add .4151 to it? Does this give me total cross sectional area? Does this make sense?

Thanks a lot!

 

[infinity]

Here's the applicable code section:

Chapter 9 Tables
Notes to Tables:
(9) A multiconductor cable or flexible cord 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.

 

[Dennis Alwon]

Since the flat cable has 2 dimension I will assume it is the length and width. The area would be L x W.

I don't know what the number is for 10/3 cable. If it is diamber than no you would not just use it.

 

[jb_04]

Here's the applicable code section:

Yes I was aware of this. I just have never seen the diameters of cable before.

 

[Stopmoving]

You would have to find the area of each cable using the formula for the area of a circle. For the 2 cable romex you would use the long dimension as the diameter. For the 3 cable you only have one measurement, that would be it's diameter. Add the two areas together and that would be your total. Do not forget to use the appropriate column (1 conductor; 2 conductor; or 2 or more conductors) for the correct percentage fill.

In case you've forgotten the area of a circle is pi (3.14) x the square of the radius.

 

[jb_04]

Since the flat cable has 2 dimension I will assume it is the length and width. The area would be L x W.

I don't know what the number is for 10/3 cable. If it is diamber than no you would not just use it.

So, do you think I did it correctly? It only gave 1 number for the 3 wire cable.

 

[jb_04]

You would have to find the area of each cable using the formula for the area of a circle. For the 2 cable romex you would use the long dimension as the diameter. For the 3 cable you only have one measurement, that would be it's diameter. Add the two areas together and that would be your total. Do not forget to use the appropriate column (1 conductor; 2 conductor; or 2 or more conductors) for the correct percentage fill.

In case you've forgotten the area of a circle is pi (3.14) x the square of the radius.

Wow, so i guess I did it wrong.

 

[Stopmoving]

Better now, when you have people willing to help, then at test time when you're on your own. I wish I could get half my students to visit this site.

 

[infinity]

You would have to find the area of each cable using the formula for the area of a circle. For the 2 cable romex you would use the long dimension as the diameter. For the 3 cable you only have one measurement, that would be it's diameter. Add the two areas together and that would be your total. Do not forget to use the appropriate column (1 conductor; 2 conductor; or 2 or more conductors) for the correct percentage fill.

In case you've forgotten the area of a circle is pi (3.14) x the square of the radius.

I agree with Stop's method. For a conduit of 20" your fill capacity would be 60%.

 

[jb_04]

Cool, thanks for the help!