If Chapter 9 is the last place you think to look when designing a new electrical system, you’re really missing the mark, because the data in this section of the NEC is the key to determining proper raceway size. The tables in this chapter also help you avoid conductor overheating due to excessive conduit fill and conductor damage during installation from jamming or excessive tension. A great way to show you how best to use the tables in Chapter 9 is to teach by example — so here we go.
Let’s assume you need to run the following copper conductors in a single raceway of undetermined length: three 12 AWG THHN, three 14 AWG THW, three 10 AWG RHH, three 6 AWG XHHW, and one 8 AWG bare equipment grounding conductor. What are the steps necessary to come up with the proper size conduit for this situation?
Note 6 of Table 1 provides you with instructions when working with combinations of conductors of different sizes, and refers you to Table 5 of Chapter 9 for the approximate cross-sectional area of each individual conductor. A check of Table 5 reveals the following information:
12 AWG THHN conductor = 0.0133 sq in.
14 AWG THW conductor = 0.0139 sq in.
10 AWG RHH conductor = 0.0437 sq in.
6 AWG XHHW conductor = 0.0590 sq in.
To find the cross-sectional area of the bare equipment grounding conductor, you must refer to Table 8. Note (8) to Table 1 allows you to use Table 8, which yields the following:
8AWG bare conductor = 0.013 sq in.
Summing the cross-sectional area of these individual conductors yields the following total:
3 × 0.0133 = 0.0399 sq in.
3 × 0.0139 = 0.0417 sq in.
3 × 0.0437 = 0.1311 sq in.
3 × 0.0590 = 0.1770 sq in.
1 × 0.013 = 0.0130 sq in.
0.0399 + 0.0417 + 0.1311 + 0.1770 + 0.0130 = 0.4027 sq in.
As noted in Table 1, conduit and tubing fill shall be limited to 40% when working with more than two conductors. Table 4 features data for 12 different types of raceway. For our example, we’ll select EMT. A quick look at the “Over 2 Wires 40%” column reveals that a 1¼ in. EMT (with a total inside area of 0.598 sq in.) will accommodate these conductors. This leaves a remaining space of 0.1953 sq in. for future conductors, which could be pulled in at a later date. It’s also important to point out that Note 4 of Chapter 9, Table 1 allows a nipple (maximum length of 24 in.) between boxes, cabinets, and similar enclosures to be filled to 60% of its cross-sectional area.
Table 2 in Chapter 9, which covers both manual and machine bending of the raceway, provides field bend requirements for conduit and tubing. A field bend is defined as any bend or offset performed by the installer using proper tools and equipment to achieve the correct radius of the curve of the centerline of the raceway. The values shown in these columns are minimum bending radii. However, it’s important to note that over-bending can result in damage to the tubing or conduit, which subsequently could result in conductor damage. Improper or over-bending conduit or tubing can flatten the curvature of the bend and lead to difficulty in pulling in the conductors.
How do you read this table? If you’re looking to install a 90° bend stub-up for a ½-in. piece of EMT, then your minimum bend radius would be 4 in. Based on my field experience, however, this value is typically 5 in. to 6 in., depending on the brand of the bender you’re working with.
Table 8 in Chapter 9 provides a lot of data you can use in determining the resistance of copper and aluminum conductors and calculating voltage drop on your circuits. Some of you may find it odd that one of the two main headers in this table is shown as “Direct Current Resistance at 75°C.” The following example will help you understand what it means and how to use it.
For a random selection of size 8 AWG copper stranded and uncoated conductor, Table 8 shows a resistance of 0.778 ohms per 1,000 ft at 75°C. But what if you need to design the system to 90°C? In this case, the resistance would have to be increased by 15°C (75°C + 15°C = 90°C). To account for this change in temperature, you must apply the formula for temperature change as outlined in Note 2 of this table.
R2 = R1 × [1 + a(T2 -75)]
R2 = 0.778 × [1 + 0.00323(90 – 75)]
R2 = 0.8157 (i.e., the revised resistance for an 8 AWG copper conductor at 90°C)
So what’s the key takeaway point for this short and sweet article? Chapter 9 may be tagged as the forgotten chapter due to its position in the book; however, the proper use of its requirements should never be dismissed or taken for granted.
Werning is a freelance writer based in Forney, Texas, who formerly owned an electrical contracting firm and currently teaches electrical Code classes. He is also a member of the Texas chapter of the IAEI. He can be reached at firstname.lastname@example.org.