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Quizzes on the Code

Jul 1, 2000 12:00 PM, By James Stallcup, Jr., NEC and OSHA Consultant

Sizing Conduits—Chapter 9

Choose the best answer:

1. What size rigid metal conduit (RMC) is required to enclose 18 No. 10 THWN copper conductors?

(a) 3/4 in.
(b) 1 in.
(c) 1 1/4 in.
(d) 1½ in.

2. What size electrical metallic tubing (EMT) is required to enclose two No. 6 THWN copper conductors, two No. 8 THWN copper conductors, and four No. 10 THWN copper conductors?

(a) 1 in.
(b) 1 1/4 in.
(c) 1 1/2 in.
(d) 2 in.

3. What size EMT nipple does the Code require to enclose 28 No. 12 THWN copper conductors that are installed between a panelboard and junction box?

(a) 3/4 in.
(b) 1 in.
(c) 1 1/4 in.
(d) 1 1/2 in.

4. What size RMC is required to enclose a 1 5/8 in. (1.625 diameter) multiconductor cable?

(a) 1 1/2 in.
(b) 2 in.
(c) 2 1/2 in.
(d) 3 in.

5. If you try to pull three 350kcmil THWN copper conductors through a 2½-in. EMT, would the conductors become jammed?

(a) Yes
(b) No

Answers and Discussion

1. (c). When working with the same size conductors, Note (1) in Chapter 9 refers you to Appendix C for determining appropriate conduit or tubing size. Table C8 lists the maximum number of conductors and fixture wires you can install in RMC. Since a 1-in. conduit is approved to hold 17 No. 10 THWN copper conductors, you must jump to the next size up: a 1 1/4-in. RMC.

Stallcup’s Code Loop: Note (1), Table 1, Chapter 9; Table C8, Appendix C

2. (a). When sizing a conduit enclosing different size conductors, Note (6) in Chapter 9 refers you to Table 5 and 5A for dimensions of conductors and to Table 4 for conduit or tubing dimensions.

Note the cross-sectional area (in square inches) of each type of conductor in the bundle by referring to Table 5. Then, compute a total area of conductor fill by adding each conductor area within the bundle. Use this total to determine the required conduit or tubing size, referencing Table 4.

Solution: Per Table 5, Chapter 9, the cross-sectional area of each conductor is:

No. 6 Cu = .0507 sq in.

No. 8 Cu = .0366 sq in.

No. 10 Cu = .0211 sq in.

Therefore, the total conductor fill is calculated as:
[2 x .0507 sq in.]+ [2 x .0366 sq in.] + [4 x .0211 sq in.] = .259 sq in.

Table 4, Chapter 9 requires a 1-in. EMT. We used the “Over 2 Wires 40%” column, since we worked with six conductors.

Stallcup’s Code Loop: Note (6), Table 1, Chapter 9; Table 5, Chapter 9; Table 4, Chapter 9

3. (b). Calculate the total cross-sectional area of the 28 No. 12 THWN copper conductors. Per Table 5, Chapter 9, the cross-sectional area of each is .0133 sq in. Therefore, the total conductor fill is:

28 x .0133 sq in. = .3724 sq in.

Select a conduit, look up its “Total Area 100%” value in Table 4, and multiply this by 60%. If this number is greater than the total conductor fill value, the conduit is acceptable. If not, select the next larger size and recalculate.

Solution: A 1-in. conduit has a 100% Total Area of .864 sq in. Therefore, .864 sq in. x 60% = .5184 sq in. Since .5184 sq in. is greater than .3724 sq in., a 1-in. EMT nipple is acceptable for use in this case.

Stallcup’s Code Loop: Note (4), Table 1, Chapter 9; Table 5, Chapter 9; Table 4, Chapter 9

4. (c). Note (9), Table 1, Chapter 9 says “a multiconductor cable of two or more conductors are to be treated as a single conductor for calculating percentage conduit fill area. For cables that have elliptical cross-sections, the cross-sectional area calculation must be based on using the major diameter of the ellipse as a circle diameter.”

Solution: First, calculate the cross-sectional area of the conductor.

Area = pi x 2[(diameter)²/4] = 3.14159 x [(1.625)²/4] = 2.074 sq in.

Then, based on the “1 Wire 53%” column in Table 4, a 2 1/2-in. RMC is required for this application.

Stallcup’s Code Loop: Note (9), Table 1, Chapter 9; Table 4, Chapter 9

5. (b). FPN No. 2 in Table 1, Chapter 9 states, “when pulling three conductors or cables into a raceway, if the ratio of the raceway (inside diameter) to the conductor or cable (outside diameter) is between 2.8 or 3.2, jamming can occur.” To determine this, look up the conductor diameter in Table 5 and the tubing diameter in Table 4.

The approximate diameter of a 350kcmil THWN copper conductor is 0.817 in. The internal diameter of a 2 1/2-in. EMT is 2.731 in. Then, compare the ratio of the EMT to the conductor: 2.731 in./0.817 in. = 3.342.

Since the ratio falls outside the range of 2.8 to 3.2, jamming should not occur.

Stallcup’s Code Loop: FPN No. 2, Table 1, Chapter 9; Table 4, Chapter 9; Table 5, Chapter 9

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