There are times when we need to add conductors to an existing raceway. But do you know how do so without damaging either the existing or new conductors, or both? You can ensure a quality and Code-compliant installation by first determining the available spare space in the raceway and then calculating the number of conductors allowed in this spare space. Here’s a five-step process that will keep you on the straight and narrow.
Step 1: Determine the raceway’s cross-sectional area for conductor fill [Table 1 and Table 4 of Chapter 9].
Step 2: Determine the area of the existing conductors [Table 5 of Chapter 9].
Step 3: Subtract the cross-sectional area of the existing conductors (Step 2) from the area of permitted conductor fill (Step 1).
Step 4: Determine the cross-sectional area of the conductors to be added [Table 5 of Chapter 9 for insulated conductors and Table 8 of Chapter 9 for bare conductors].
Step 5: Divide the spare space area (Step 3) by the cross-sectional area of the conductors to be added (Step 4).
Now you should be ready to tackle a sample calculation.
Q. An existing 1-inch EMT contains two No. 12 THHN conductors, two No. 10 THHN conductors, and one No. 12 bare (stranded) conductor. How many additional No. 8 THHN conductors can be added to this raceway?
Step 1: Cross-sectional area permitted for conductor fill
Raceway = 0.864 x 0.4 = 0.3456 sq in.
Nipple = 0.864 x 0.6 = 0.5184 sq in.
Step 2: Cross-sectional area of existing conductors
No. 10 THHN [0.0211 sq in. x 2 = 0.0422 sq in.]
No. 12 THHN [0.0133 sq in. x 2 = 0.0266 sq in.]
No. 12 bare [0.0060 sq in. x 1 = 0.0060 sq in.]
Total cross-sectional area of existing conductors = 0.0748 sq in.
Note: Ground wires must be counted for raceway fill.
Step 3: Subtract the area of the existing conductors from the permitted area of conductor fill.
Raceway more than 24 inches long: 0.3456 sq in. – 0.0748 sq in. = 0.2708 sq in.
Nipple (less than 24 inches long): 0.5184 sq in. – 0.0748 sq in. = 0.4436 sq in.
Step 4: Cross-sectional area of the conductors to be installed.
No. 8 THHN = 0.0366 sq in.
Step 5: Divide the spare space area (Step 3) by the conductor area.
Raceway = 0.2708 sq in. ÷ 0.0366 sq in. = 7.4 or 7 conductors
Nipple = 0.4436 sq in. ÷ 0.0366 sq in. = 12.1 or 12 conductors
We must round down to 18 conductors because Note 7 to Table 1 only applies if all of the conductors are the same size and same insulation.