Published on *EC Mag* (http://www.ecmag.com)

The National Electrical Code (NEC) contains an introduction, nine chapters and eight annexes. Article 90 is the introduction to the NEC. This article contains specifications that are essential to all chapters and sections in the Code.

The National Electrical Code states its purpose in 90.1(A). “The purpose of the Code is the practical safeguarding of persons and property from hazards arising from the use of electricity.” Section 90.1 continues by covering the Code’s adequacy, its intention and its relation to other international standards. While most articles have a scope that describes what the article covers, Article 90 explains what the entire Code covers. While 90.3 explains the arrangement of the Code, 90.4 specifies its enforcement. Article 90.5 explains mandatory rules, permissive rules and fine print notes.

This section also explains the use of brackets in the NEC. Brackets containing section references to another NFPA document are for informational purposes only and are provided as a guide to indicate the source of the extracted text. These bracketed references immediately follow the extracted text. Formal interpretation procedures are discussed in 90.6. Section 90.7 covers examination of equipment for safety. Future expansion and convenience is mentioned in 90.8(A) and 90.8(B). These sections state that limiting the number of circuits in a single enclosure minimizes the effects from short circuit or ground fault in one circuit. The last section in Article 90 is 90.9. This section explains the units of measurement in the Code.

Article 220 contains requirements for calculating branch-circuit, feeder and service loads. Last month’s Code in Focus covered electric cooking equipment in 220.55. This month, the discussion continues with calculating loads for electric ranges and other cooking appliances in dwelling units.

The second note under Table 220.55 provides instructions for finding the maximum demand for ranges of unequal rating over 8¾ kW through 27 kW. In accordance with Note 2, find an average value of rating by adding together the ratings of all ranges to obtain the total connected load, and then divide by the number of ranges. After finding the average value, the maximum demand in Column C must be increased 5 percent for each additional kilowatt of rating or major fraction thereof by which the rating of individual ranges exceeds 12 kW. For example, what is the service demand load for five 13-kW, five 15-kW and five 17-kW household electric ranges? Start by adding together the ratings of all ranges to obtain the total connected load [(5 × 13) + (5 × 15) + (5 × 17) = 65 + 75 + 85 = 225]. Next, find the average value by dividing the total connected load by the total number of ranges (225 ÷ 15 = 15 kW). The average value for these 15 ranges is 15 kW (see Figure 1).

Because we have found the average value for the 15 ranges, it is as if this is a new question: What is the service demand load for 15 15-kW household electric ranges? Find the percentage by which Column C must be increased. A 15-kW range exceeds 12 kW by 3 kW (15 – 12 = 3). Since Column C must be increased 5 percent for each additional kilowatt of rating above 12, the maximum demand listed in Column C for 15 ranges must be increased by 15 percent (3 × 5% = 15%). Find the demand in Column C for 15 ranges, and then multiply by 15 percent (Column C for 15 ranges is 30 kW). The increased amount is 4.5 kW (30 × 15% = 4.5 kW). This increased amount must be added to the Column C demand load for 15 ranges (30 + 4.5 = 34.5 kW). The service demand load for five 13-kW, five 15-kW and five 17-kW household electric ranges is 34.5 kW (see Figure 2).

When applying Note 2, the range rating must not include a fraction of a kilowatt. The fraction must either be dropped or rounded up to the next whole kilowatt rating. For example, what is the service demand load for five 14-kW, five 16-kW and five 17-kW household electric ranges? Find an average value of rating by adding together the ratings of all ranges to obtain the total connected load. The total connected load is 235 kW (5 × 14) + (5 × 16) + (5 × 17) = 70 + 80 + 85 = 235. Now divide the total connected load by the number of ranges to find the average value of rating (235 ÷ 15 = 15.67 kW). The average rating of all 15 ranges is 15.67 kW. Notes 1 and 2 specify that the range rating must be increased for each kilowatt of rating or major fraction thereof by which the rating of the individual ranges exceeds 12 kW. A major fraction is .5 and larger. Since the .67 is a major fraction, round the average rating of 15.67 up to 16 kW (see Figure 3).

Now find the service demand load for 15 16-kW ranges. Because Column C is based on 12-kW ranges, subtract 12 from 16 (16 – 12 = 4). Since 16 kW exceeds 12 kW by 4, multiply 4 by 5 percent to find the amount Column C must be increased (4 × 5% = 20%). The maximum demand listed in Column C for 15 ranges must be increased by 20 percent. The increased amount is 6 kW (30 × 20% = 6 kW). This increased amount must be added to the Column C demand load for 15 ranges (30 + 6 = 36 kW). The service demand load for five 14-kW, five 16-kW and five 17-kW household electric ranges is 36 kW (see Figure 4).

Dropping the fraction or rounding the fraction up to the next whole kilowatt rating should be done only once: after finding the average range rating, just before finding the percent of increase. It is not necessary to round up or drop the fraction of each individual range. For example, what is the service demand load for three 13.6-kW, three 14.9-kW and four 16.6-kW household electric ranges? Although these ranges have fractions of kilowatt ratings, do not round the rating up or drop the fraction at this time. First, find an average value of rating by adding together the ratings of all ranges to obtain the total connected load: (3 × 13.6) + (3 × 14.9) + (4 × 16.6) = 40.8 + 44.7 + 66.4 = 151.9. Next, find the average value by dividing the total connected load by the total number of ranges (151.9 ÷ 10 = 15.19 kW). Since the .19 is not a major fraction, drop it (see Figure 5).

Now, find the service demand load for 10 15-kW ranges. Subtract 12 from 15 (15 – 12 = 3). The maximum demand listed in Column C for 10 ranges must be increased by 15 percent (3 × 5% = 15%). Find the demand in Column C for 10 ranges, and then multiply by 15 percent (Column C for 10 ranges is 25 kW). The increased amount is 3.75 kW (25 × 15% = 3.75 kW). This increased amount must be added to the Column C demand load for 10 ranges (25 + 3.75 = 28.75 kW). The service demand load for three 13.6-kW, three 14.9-kW and four 16.6-kW household electric ranges is 28.75 kW (see Figure 6).

Next month’s Code in Focus will continue the discussion of feeder and service load calculations.

**MILLER**, owner of Lighthouse Educational Services, teaches classes and seminars on the electrical industry. He is the author of “Illustrated Guide to the National Electrical Code” and “The Electrician’s Exam Prep Manual.” He can be reached at 615.333.3336, charles@charlesRmiller.com or www.charlesRmiller.com.