Sizing Auxiliary Gutter

Nov 23 / James Stallcup
Stallcup Electrical Education is committed to helping electricians, apprentices, and industry professionals sharpen their skills with a weekly calculation problem based on the National Electrical Code (NEC). Designed to reinforce critical thinking and real-world application, these problems challenge learners to apply NEC principles to electrical design, load calculations, and compliance scenarios. Whether you're preparing for an exam or looking to refine your expertise, Stallcup’s weekly calculation problems provide a valuable learning tool to keep you engaged and up to date with industry standards. Stay ahead of the curve and enhance your electrical knowledge—one calculation at a time! 
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Sizing Auxiliary Gutters
(Based on the 2023 NEC)

Example Problem: What size metallic auxiliary gutter is required to house 3 - 350 KCMIL THWN copper conductors (feeder) that have 3 - 1/0 AWG, 3 - 1 AWG, and 3 - 2 AWG THWN copper conductors spliced to them?


Step 1: Finding sq. in. area

Table 5, Ch. 9

350 KCMIL THWN = .5242 sq. in.
1/0 AWG THWN = .1855 sq. in.
1 AWG THWN = .1562 sq. in.
2 AWG THWN = .1158 sq. in.


Step 2: Calculating sq. in. area
Table 5, Ch. 9

.5242 sq. in. x 3 = 1.5726 sq. in.
.1855 sq. in. x 3 .5565 sq. in.
.1562 sq. in. x 3 .4686 sq. in.
.1158 sq. in. x 3  .3474 sq. in.
Total = 2.9451 sq. in.


Step 3: Sizing gutter

366.22(A)

sq. in. area divided by 20% = total fill
2.9451 sq. in. ÷ 20% = 14.7255 sq. in.


Step 4: Selecting gutter Chart

4" x 4" = 16 sq. in.


Step 5: Applying 75% fill for splices
366.56(A)

sq. in. of gutter x 75% = fill area
16 sq. in. x 75% = 12 sq. in.


Solution: A 4 in. x 4 in. metallic auxiliary gutter is required with only 12 in. of the gutter space used for splicing conductors.

Solution: