According to the National Electrical Code, the number of conductors in a raceway shall not be more than what can be installed or withdrawn without damage to the conductors or to the conductor insulation. This can be achieved by limiting the total area of the conductors in the raceway so that they don't exceed the percentage of cross-sectional area of the raceway as listed in Table 1 of Chapter 9 (Table 1).
The cross-sectional area of a raceway is dependent upon the raceway's diameter, which is different for each type of raceway. See Table 4 of Chapter 9. For example, the total cross-sectional area of various types of 1-inch raceway is shown in Table 2
The NEC doesn't list a requirement that there be a maximum distance between junction or pull points. However, it does stipulate there shall not be more than 360° of total bends between raceway terminations (Chapter 9, Table 1, Fine Print Note 1) (Fig. 1).
Sizing raceway using Annex C. Annex C in the NEC identifies the number of conductors permitted in a raceway in accordance with the raceway fill limitations of Table 1 of Chapter 9 and the cross-sectional area of the given raceway. However, this Annex Table can only be used if all of the conductors are the same size and are of the same insulation type.
Let's look at a few examples to see how this works.
Example No. 1: How many 1/0 AWG THHN conductors can be installed in 2-inch electrical metallic tubing? (Fig. 2)
(a) 2 conductors
(b) 3 conductors
(c) 4 conductors
(d) 7 conductors
The answer is (d), 7 conductors. You get this answer directly from Table C1 in Annex C.
Example No. 2: How many compact 6 AWG XHHW conductors can be installed in 1.25-inch electrical nonmetallic tubing?
(a) 10 conductors
(b) 6 conductors
(c) 16 conductors
(d) 13 conductors
The answer is (a), 10 conductors. This answer comes directly from Table C2A in Annex C.
Example No. 3: How many 18 AWG TFFN fixture wires can be installed in 0.75-inch liquidtight flexible metallic conduit?
(a) 40 conductors
(b) 26 conductors
(c) 30 conductors
(d) 39 conductors
The answer is (d), 39. This answer comes directly from Table C7 in Annex C.
Sizing a raceway using Tables 4 and 5When the conductors to be installed aren't the same size and/or they don't have the same insulation type, use these steps to properly size the raceway.
Step 1: Determine the cross-sectional area of each conductor.
Refer to Table 5, Chapter 9 for insulated conductors.
Refer to Table 8, Chapter 9 for bare conductors (Note 3, Table 1).
Step 2: Determine the total cross-sectional area for all of the conductors.
Step 3: Size the raceway using Table 4, Chapter 9.
40% for raceways that contain three or more conductors [Table 1]
60% for raceways 24 inches or less in length (i.e. nipples) [Note 4, Table 1]
Let's take a look at another example calculation to help clarify this procedure.
Example No. 4: Feeder conductors are installed in Schedule 40 rigid nonmetallic conduit. The raceway is more than 200 feet long, and it contains three 500 kcmil THHN conductors, one 250 kcmil THHN conductor, and one 3 AWG THHN conductor. What size RNC raceway is required for these conductors?
(a) 2 inches
(b) 2.5 inches
(c) 3 inches
(d) 3.5 inches
Step 1: Determine the cross-sectional area of each conductor by referring to Table 5, Chapter 9.
500 kcmil THHN=0.7073 sq. in.
250 kcmil THHN=0.3970 sq. in.
3 AWG THHN=0.0973 sq. in.
Step 2: Determine the total cross-sectional area for all conductors.
500 kcmil THHN=0.7073 sq. in.×3 wires=2.1219 sq. in.
250 kcmil THHN=0.3970 sq. in.×1 wire=0.3970 sq. in.
3 AWG THHN=0.0973 sq. in.×1 wire=0.0973 sq. in.
Total=2.1219+0.3970+0.0973=2.6162 sq. in.
Step 3: Size the RNC raceway at 40% fill per Table 4 of Chapter 9.
2.5 in.=1.878 sq. in. (too small)
3 in.=2.907 sq. in. (just right)
3.5 in.=3.895 sq. in. (larger than required)
The correct answer is (c), 3 inches.
Sizing raceways for low-voltage cablesThe raceway conductor fill limitations contained in 300.17 apply to the following signaling systems:
Control and signaling cables [725.3(A) and 725.28]
Fire alarm nonpower-limited cables (760.28)
Optical fiber cables (installed with power conductors) (770.6)
Sound system (audio) cables (640.23)
Except for optical fiber cable, the NEC prohibits the mixing of technology cables with power, Class 1, or nonpower-limited fire alarm circuit conductors in the same raceway.
Use the actual diameter of the cable to determine the area for raceway fill [Chapter 9, Table 1, Note 5]. The formula for determining the area is:
Area=3.14×(0.5×diameter)2
A multiconductor cable is considered a single conductor for calculating conduit fill [Chapter 9, Table 1, Note 9].
Let's work through another example to make sure you understand the calculations involved.
Example No. 5: What size electrical metallic tubing is required for two nonconductive optical fiber cables having a diameter of 0.25 inch and eight 12 AWG THHN power conductors?
(a) 0.5 inch
(b) 0.75 inch
(c) 1 inch
(d) 1.25 inches
Step 1: Determine the cross-sectional area of each cable and conductor.
The area of a cable is determined by the following formula:
Area=3.14×(0.5×diameter)2
Optical fiber=[3.14×(0.5×0.250 in.)2]=0.0491 sq. in.
12 AWG THHN [Chapter 9, Table 5]=0.0133 sq. in.
Step 2: Determine the total cross-sectional area for all conductors.
Optical fiber=0.0491 sq. in.×2 cables=0.0982 sq. in.
12 AWG THHN50.0133 sq. in.×8 conductors=0.1064 sq. in.
Total=0.0982+0.1064=0.2046 sq. in.
Step 3: Size the EMT raceway at 40% fill, per Table 4, Chapter 9 of the NEC.
A 0.75-inch EMT=0.213 sq. in.
Therefore, the correct answer is (b), 0.75 inch.
Raceway fill recommendationsThe following technology cables aren't required by the NEC to be installed in a raceway:
CATV (coaxial) cable (Art. 820)
Class 2 or Class 3 cables (725.52)
Optical fiber cables (770.3)
Power-limited fire alarm cables (760.3)
Network-powered broadband communications cables (Art. 830)
Radio and television cables (Art. 810)
Communications (twisted pair) cables (Art. 800)
However, if any of these cables are installed in a raceway, they should be installed in accordance with the installation recommendations contained in the BICSI Cabling Installation Manual. This installation guideline recommends that raceway runs be limited to 100 feet, have no more than two 90° bends, and have a maximum pull force of 25 pounds for Cat. 5 cable, and have a maximum pull force of 100 pounds for optical fiber cable.
Since most installers have no idea how to limit the pulling tension on signal or communication cables, the generally accepted practice is to size the raceway so that the cables don't exceed the percentage fill listed in Table 1, Chapter 9 of the NEC.
Let's look at another example.
Example No. 6: What size electrical metallic tubing is required for four Cat. 5 plenum cables (diameter 0.167 inch) three 12-strand nonconductive optical fiber cables (diameter 0.250 inch), two 24-strand nonconductive optical fiber cables (diameter 0.438 inch) (Fig. 3)?
(a) 0.5 inch
(b) 0.75 inch
(c) 1 inch
(d) 1.25 inches
Step 1: Determine the cross-sectional area of each cable.
Cat. 5 (4-pair)=3.14×(0.5×0.167 in.)2=0.0219 sq. in.
Optical fiber (12-strand)=3.14×(0.5 × 0.250 in.)2=0.0491 sq. in.
Optical fiber (24-strand)=3.14×(0.5×0.438 in.)2=0.1439 sq. in.
Step 2: Determine the total cross-sectional area for all conductors.
Cat. 5 (4-pair)=0.0219 sq. in.×4 cables=0.0876 sq. in.
Optical fiber (12-strand)=0.0491 sq. in.×3 cables=0.1473 sq. in.
Optical fiber (24-strand)=0.1439 sq. in.×2 cables=0.2878 sq. in.
Total=0.0876+0.1473+0.2878=0.5227 sq. in.
Step 3: Size the EMT raceway at 40% fill per Table 4, Chapter 9.
A 1.25-inch EMT=0.5980 sq. in. Therefore, the correct answer is (d), 1.25 inches.
Raceway sizing for “elliptical technology cables.”Note 9 of Table 1 in Chapter 9 states that for cables having an elliptical cross section, the area calculation shall be based on the major diameter of the ellipse.
Example No. 7: What size electrical nonmetallic tubing is required for one hybrid fiber/data cable? The minor diameter of the ellipse is 0.188 inch and the major diameter of the ellipse as a circle diameter is 0.5 inch (Fig. 4).
(a) 0.5 inch
(b) 0.75 inch
(c) 1 inch
(d) 1.25 inches
Step 1: Determine the cross-sectional area of the cable, based on the major diameter of the ellipse as a circle diameter.
Hybrid cable=3.14×(0.5×0.50 in.)2=0.1960 sq. in.
Step 2: Determine the total cross-sectional area of the cable.
Hybrid cable=0.1960 sq. in.×1=0.1960 sq. in.
Step 3: Size the ENT at 53% fill (one conductor fill), per Table 4, Chapter 9.
0.5 in. ENT=0.131 sq. in. (too small)
0.75 in. ENT=0.240 sq. in. (just right)
1 in. ENT=0.416 sq. in. (larger than required)
Therefore, the correct answer is (b), 0.75 inch.