Cooking Equipment Calculations, Part IV

220.19 Electric Ranges and Other Cooking Appliances

Provisions for calculating branch-circuit, feeder and service loads are in Article 220 of the National Electrical Code. The four parts of this article include: I. General, II. Feeders and Services, III. Optional Calculations for Computing Feeder and Service Loads and IV. Method for Computing Farm Loads. Although the first part contains some general requirements, most of the provisions pertain to branch-circuit calculations. The second part includes various demand factors for computing feeder and service loads. All branch-circuit demand loads must be added to the demand loads in Parts II, III or IV to determine feeders and services. The calculated load of a feeder or service must not be less than the sum of the loads on the branch circuits supplied, as determined by Part I of Article 220, after any applicable demand factors permitted by Parts II, III or IV have been applied (220.10).

Performing load calculations is easier when the requirements can be broken down into a series of understandable steps. One step included in most dwelling calculations is cooking equipment. Specifications for household electric ranges, wall-mounted ovens, counter-mounted cooking units, and other household cooking appliances individually rated in excess of 1 3/4kW are covered in 220.19 and Table 220.19. Last month’s In Focus continued discussing household cooking equipment calculations. The discussion continues this month with a detailed explanation of cooking-equipment demand factors.

When ranges over 12kW through 27kW are all rated the same, the ranges must be calculated in accordance with the first note under Table 220.19. When they are not rated the same, use the second note. For ranges individually rated more than 8-3/4kW and of different ratings, but none exceeding 27kW, an average value of rating shall be computed by adding together the ratings of all ranges to obtain the total connected load (using 12kW for any range rated less than 12kW) and dividing by the total number of ranges (Note 2 under Table 220.19). This note only applies when there is more than one range.

The first step is to find the average value of rating for all ranges. This step is not necessary in the first note because the ranges are rated the same. To find the average, first total the kilowatt ratings of all the ranges. Next, divide the total kilowatt rating by the number of ranges. For example, what is the average kilowatt rating for two 13kW ranges, two 15kW ranges, and two 17kW ranges? These ranges have a combined total of 90kW [(2 ¥ 13) + (2 ¥ 15) + (2 ¥ 17) = 26 + 30 + 34 = 90]. Next, divide 90kW by six ranges. The average rating for these six ranges is 15kW (See Figure 1).

After finding the average kilowatt rating, increase the maximum demand in Column C by 5 percent for each kilowatt or major fraction thereof by which the average value exceeds 12kW. This last step is calculated the same as Note 1 under Table 220.19. The demand loads in Column C are for ranges not over 12kW rating. When the average is 12.5kW or more, the demand must be increased. The amount of increase is 5 percent for each additional kilowatt over 12kW. Find the demand in Column C for the total number of ranges and increase that demand by the appropriate percentage. For example, what is the service demand load for two 13kW ranges, two 15kW ranges, and two 17kW ranges? As discussed earlier, the average rating for these six ranges is 15kW. Next, find the percentage by which Column C must be increased. A 15kW range exceeds 12kW by 3kW (15 – 12 = 3). The demand in Column C must be increased by 15 percent (3 ¥ .05 = .15). The demand load for six ranges is 21kW. The increased amount is 3.15kW (21 ¥ .15 = 3.15kW). Add the increased amount to the original demand (21 + 3.15 = 24.15kW). The service load for these ranges is 24.15kW (See Figure 2).

When applying Note 2, the average range rating (before applying Column C) must not include a fraction of a kilowatt. The fraction must either be dropped, or rounded up to the next whole kilowatt rating. After finding the average kilowatt rating, if it is less than .5, drop the fraction. If the average kilowatt rating is .5 or more, round the rating up to the next whole number. For example, what is the service demand load for nine 13kW ranges, eight 14kW ranges, and eight 17kW ranges? These 25 ranges have a combined total of 365kW [(9 ¥ 13) + (8 ¥ 14) + (8 ¥ 17) = 117 + 112 + 136 = 365]. Next, divide 365kW by 25. The average rating for these 25 ranges is 14.6kW. Since the .6 is a major fraction, round the 14.6 up to a 15kW range and find the demand. A 15kW range exceeds 12kW by 3kW (15 – 12 = 3). The demand in Column C must be increased by 15 percent (3 x .05 = .15). The demand in Column C for 25 ranges is 40kW. Increase the demand in Column C by 15 percent (40 ¥ .15 = 6kW). Add the increased amount to the original demand (40 + 6 = 46kW). The service demand load for these ranges is 46kW (See Figure 3).

Round the fraction of a kilowatt up or down only after finding the average value. Include all fractions when finding the total kilowatt rating of the ranges. For example, what is the service demand load for five 13.1kW ranges, five 14.6kW ranges and five 15.7kW ranges? These 15 ranges have a combined total of 217kW [(5 ¥ 13.1) + (5 ¥ 14.6) + (5 ¥ 15.7) = 65.5 + 73 + 78.5 = 217]. Next, divide 217kW by 15 ranges. The average rating for these 15 ranges is 14.47kW. Since the .47 is not a major fraction, drop the fraction and find the demand for 15 14kW ranges. A 14kW range exceeds 12kW by 2kW (14 – 12 = 2). The demand in Column C must be increased by 10 percent (2 ¥ .05 = .10). The demand in Column C for 15 ranges is 30kW. Increase the demand in Column C by 10 percent (30 ¥ .10 = 3kW). Add the increased amount to the original demand (30 + 3 = 33kW). The service demand load for these ranges is 33kW (See Figure 4).

An important part of this provision that must not be overlooked is the wording inside the parentheses, “using 12kW for any range rated less than 12kW.” When calculating by the second note under Table 220.19, any range rated less than 12kW must be changed to 12kW before finding the average rating. For example, what is the service demand load for six 14kW ranges, six 15kW ranges, and eight 9kW ranges? Since these ranges are of unequal ratings and over 12.5kW, ranges rated less than 12kW must be changed to 12kW. The 9kW ranges must be changed to 12kW before finding the average rating. Using 12kW instead of 9kW, the combined total is 270kW [(6 ¥ 14) + (6 ¥ 15) + (8 ¥ 12) = 84 + 90 + 96 = 270]. Next, divide 270kW by 20. The average rating for these 20 ranges is 13.5kW.

Note that if the 9kW ranges were not increased to 12kW, the average rating would have been only 12.3kW. Since the .5 is a major fraction, round the 13.5 up to a 14kW range and find the demand. A 14kW range exceeds 12kW by 2kW (14 – 12 = 2). The demand in Column C must be increased by 10 percent (2 x .05 = .10). The demand in Column C for 20 ranges is 35kW. Increase the demand in Column C by 10 percent (35 ¥ .10 = 3.5kW). Add the increased amount to the original demand (35 + 3.5 = 38.5kW). The service demand load for these ranges is 38.5kW (See Figure 5).

Next month’s In Focus continues the discussion of household cooking equipment computations in 220.19. EC

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