Branch-Circuit, Feeder and Service Calculations, Part XXVIII

by Charles R. Miller
Published: June 2008

Article 220 – Load Calculations

220.55 Electric Ranges and Other Cooking Appliances—Dwelling Unit

At 3 pm on September. 4, 1882, the Pearl Street Station started providing customers in New York with electric power. It was the first central-station electric generating plant, and Thomas A. Edison developed it. Although this first generating plant produced direct current, alternating current soon followed. Fifty-nine customers reluctantly consented to have their houses wired on the promise of three free months of electric power. If the service proved to be unsatisfactory, they had the option to discontinue the electric service. This new way of lighting was far from unsatisfactory. It was sensational.

But, with this new frontier came a new danger: electrical fires. There was such an outbreak of fires caused by electrical problems that something had to be done. For more than a decade, there was no standardized electrical code. By 1895, there were five different electric installation codes in use throughout this country, but no single set of codes was accepted by all. In March 1896, 23 people met to develop a national set of rules for electrical construction and operation. The National Board of Fire Underwriters, which is now the American Insurance Association, published the first National Electrical Code (NEC) in 1897. In 1911, the National Fire Protection Association (NFPA) assumed sponsorship and control of the NEC.

The first sentence of Article 90 reads, “The purpose of this Code is the practical safeguarding of persons and property from hazards arising from the use of electricity.” This objective has remained constant throughout the Code’s existence.

One contributing factor that led to electrical fires was overloaded circuits, and without load calculation requirements, it still would be a problem today. The NEC contains detailed requirements for calculating loads in branch circuits, feeders and services. These load calculation requirements are in Chapter 2, Article 220. Last month’s article 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.

Without reading the notes under Table 220.55, it would appear that Column C is good only for ranges rated 12 kW or less. But, the first two notes under Table 220.55 provide instructions for finding the maximum demand for ranges over 12 kW through 27 kW. Note 1 applies when one or more ranges have the same kilowatt rating. For ranges individually rated more than 12 kW but not more than 27 kW, the maximum demand in Column C must be increased by 5 percent for each additional kilowatt of rating or major fraction, which is .5 and larger (Note 1 under Table 220.55).

The demand loads in Column C must be increased when the ranges are rated more than 12 kW. The amount of increase is 5 percent for each additional kilowatt over 12 kW. To find the percent of increase for more than one range, subtract 12 from the average kilowatt rating. Then multiply that number by 5 percent. Next, multiply the percentage of increase by the Column C demand. This number is then added to the maximum demand for the given number of ranges in Column C. For example, what is the service demand load for 15 15-kW household electric ranges? First, 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 by 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 15 15-kW ranges is 34.5 kW (see Figure 1).

 
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When applying Note 1, the range rating (before finding the percent of increase) must not include a fraction of a kilowatt. The fraction must either be dropped or rounded up to the next whole kilowatt rating. This fraction is a fraction of a kilowatt, not a fraction of a watt. An example of this is a range with a 14.5-kW rating.

Although the range is 14,500 watts, the .5 is a fraction of a kilowatt. Notes 1 and 2 specify that the range rating must be increased for each kilowatt or major fraction of rating. Therefore, when the fraction is .5 or more, the fraction of kilowatt rating must be rounded up to the next whole kilowatt rating. For example, what is the feeder demand load for one 14.5-kW range? Since the .5 is a major fraction, round the 14.5 up to a 15-kW range and find the demand load. Because Column C is based on 12-kW ranges, subtract 12 from 15 (15 – 12 = 3). Since 15 kW exceeds 12 kW by 3, multiply 3 by 5 percent to find the amount Column C must be increased (3 × 5% = 15%). The maximum demand listed in Column C for one range must be increased by 15 percent. The increased amount is 1.2 kW (8 × 15% = 1.2 kW). This increased amount must be added to the Column C demand load for one range (8 + 1.2 = 9.2 kW). The feeder demand load for one 14.5-kW range is 9.2 kW (see Figure 2).

When the fraction is less than .5, the fraction of kilowatt rating can be dropped. For example, what is the feeder demand load for one 14.4-kW range? Since the .4 is not a major fraction, drop the .4, and find the demand load for one 14-kW range. Subtract 12 from 14 (14 – 12 = 2). Since 14 kW exceeds 12 kW by 2, multiply 2 by 5 percent to find the amount Column C must be increased (2 × 5% = 10%). The maximum demand listed in Column C for one range must be increased by 10 percent. The increased amount is .8 kW (8 × 10% = .8 kW). This increased amount must be added to the Column C demand load for one range (8 + .8 = 8.8 kW). The feeder demand load for one 14.4-kW range is 8.8 kW (see Figure 3).

 
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Though Note 1, under Table 220.55, specifies over 12 kW through 27 kW, the note applies only to ranges rated 12.5 kW through 27 kW. If the range is rated less than 12.5, the fraction can be dropped. For example, what is the service demand load for one 12.49-kW range? Since the .49 is not a major fraction, drop the .49 and find the demand load for one 12-kW range. After dropping the fraction, the range rating is only 12 kW, and Column C is based on ranges rated not more than 12 kW. In accordance with Table 220.55 and the notes under the table, a 12.49-kW range has a maximum demand of 8 kW (see Figure 4).

 
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Next month’s column continues 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 NFPA’s “Electrical Reference.” He can be reached at 615.333.3336, charles@charlesRmiller.com or www.charlesRmiller.com.

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