Conduit fill for multiconductor cable

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philly

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If I am given the diamater for a 3/c #10 and 3/c #2 multiconductor cable do I then take the radius of each of these given diamaters and use pi*r^2 to find the area of each of these cables. Can I then add these two areas and compare this calculated total against the values given in Chapter 9 Table 4 for conduit fill %?

If two cable diameters are. .47" and 1.015" then the radius for these are .235" and .5075". The two areas respectively for these two cables are then .173"in sqrd and .809"in sqrd. Adding these two areas together it comes out to be .982 in_sqrd.

Now if I take this area of .982 and compare it to the Article 344 section of table 4 in chapter 9 for 2" conduit I see that the 40% fill for 2" is 1.363" sqrd. So since my calculated area of .982 is less then the 40% value of 1.363" then it is o.k to pull these cables in this 2" size conduit?

Anybody see any problems with my calculations?
 
If you are pulling two cables, you can't use the 40% fill, you are limited to 31% per Chapter 9, Table 1. In this case you are still good to go as the 31% fill is 1.056.
 
Cables: Qty Cable Outside Diameter
Cable #1 1 0.470
Cable #2 1 1.015

Cross Section/Square Inches 0.548

RMC (minimum conduit size) 1 1/2
Conduit: Table 4 Total Area In Sq 2.071
Table 1 Percent of Cross Section 31%
Available Inch Square 0.642 (Greater than the Cross Section/Square Inches) Correct Conduit Size

Note:
1) Chapter 9 Table 1, Note (5): "For conductors not included in Chapter 9, such as multiconductor cables, the actual dimensions shall be used."
2) Formula: Square Inches = cable outside diameter squared x (3.1416/4) x Qty of cables
 
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tryinghard said:
Cables: Qty Cable Outside Diameter
Cable #1 1 0.470
Cable #2 1 1.015

Cross Section/Square Inches 0.548

RMC (minimum conduit size) 1 1/2
Conduit: Table 4 Total Area In Sq 2.071
Table 1 Percent of Cross Section 31%
Available Inch Square 0.642 (Greater than the Cross Section/Square Inches) Correct Conduit Size

Note:
1) Chapter 9 Table 1, Note (5): "For conductors not included in Chapter 9, such as multiconductor cables, the actual dimensions shall be used."
2) Formula: Square Inches = cable outside diameter squared x (3.1416/4) x Qty of cables
Click to expand...
Cable 1 with an OD of 0.47" has an area of 0.173 square inches.
Cable 2 with an OD of 1.015" has an area of 0.809 square inches.
The total area of the two cables is 1.063 square inches and well over the permitted fill of 0.642 square inches for two conductors in 1 1/2" rigid conduit.
 
don_resqcapt19 said:
Cable 1 with an OD of 0.47" has an area of 0.173 square inches.
Cable 2 with an OD of 1.015" has an area of 0.809 square inches.
The total area of the two cables is 1.063 square inches and well over the permitted fill of 0.642 square inches for two conductors in 1 1/2" rigid conduit.
Click to expand...

Oops spreadsheet mistake! :-? :cool:

Cables: Qty Cable Outside Diameter
Cable #1 1 0.470
Cable #2 1 1.015

Cross Section/Square Inches 0.983

RMC (minimum conduit size) 2
Conduit: Table 4 Total Area In Sq 3.408
Table 1 Percent of Cross Section 31%
Available Inch Square 1.056 (Greater than the Cross Section/Square Inches) = Correct Conduit Size For Cables
 
tryinghard said:
Oops spreadsheet mistake! :-? :cool:

Cables: Qty Cable Outside Diameter
Cable #1 1 0.470
Cable #2 1 1.015

Cross Section/Square Inches 0.983

RMC (minimum conduit size) 2
Conduit: Table 4 Total Area In Sq 3.408
Table 1 Percent of Cross Section 31%
Available Inch Square 1.056 (Greater than the Cross Section/Square Inches) = Correct Conduit Size For Cables
Click to expand...
Why is your total area for the two cables 0.08 square inches less than mine?

I see that I added wrong...your total is correct...
 
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