Branch-Circuit, Feeder and Service Calculations, Part XXVII

by Charles R. Miller
Published: May 2008

Article 220 – Load Calculations

220.55 Electric Ranges and Other Cooking Appliances—Dwelling Unit

Load calculation requirements are in Article 220 of the National Electrical Code (NEC). This article provides requirements for calculating branch-circuit, feeder and service loads. Requirements in Article 220 are divided into five parts.

Part I covers general requirements for calculation procedures. Part II provides calculation provisions for branch circuits. Feeder and service calculation requirements (sometimes referred to as the standard method) are in Part III. Part IV specifies optional feeder and service load calculations, and it contains optional load-calculation procedures for a single dwelling unit, an existing dwelling unit, a multifamily dwelling, two dwelling units, a school, an existing installation and a new restaurant. Farm loads must be calculated in accordance with Part V of Article 220.

Loads calculated in accordance with the provisions in Article 220 are used with requirements from other articles to find conductor sizes and ampere ratings for overcurrent protection. For example, results from Part II in Article 220 are used with the provisions in 210.19 to size branch-circuit conductors and 210.20 to determine ratings for overcurrent protective devices (fuses and breakers).

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.

Appliances can have different ratings within Column A and different ratings within Column B. For example, what is the service demand load for 10 2½-kW ovens, 10 3-kW ovens, 10 5-kW cooktops and 10 6-kW cooktops? As specified in the third note under Table 220.55, calculate each column separately, and then add the results together.

There are 20 units that are within the limits of Column A. The 2½-kW ovens have a rating of 25 kW (10 × 2½ = 25). The 3-kW ovens have a rating of 30 kW (10 × 3 = 30). The ovens have a total rating of 55 kW (25 + 30 = 55). Because 20 ovens fall within the limits of Column A, the demand factor is 35 percent. The calculated demand for the ovens is 19¼ kW (55 × 35% = 19.25).

There are 20 units that are within the limits of Column B. The 5-kW cooktops have a rating of 50 kW (10 × 5 = 50). The 6-kW cooktops have a rating of 60 kW (10 × 6 = 60). The cooktops have a total rating of 110 kW (50 + 60 = 110). Therefore, the demand factor is 28 percent. The calculated demand for the cooktops is 30.8 kW (110 × 28% = 30.8). Now add the results to find the total demand load (19.25 + 30.8 = 50.05). The demand for these household cooking appliances from Columns A and B is 50.05 kW (50,050 watts). Since the total number of units is 40 (10 + 10 + 10 + 10 = 40), find the demand load in Column C for 40 units. Fifteen added to 40 appliances is 55 (15 + 40 = 55). The demand in Column C for 40 units is 55 kW. Finally, compare the loads, and select the lower. With this example, the lower number is from Columns A and B. The service demand for these household cooking appliances is 50.05 kW (see Figure 1).

 
CIF_May_08_01.jpg
 
   

Up until this point in the series, there has been no discussion of household cooking appliances with ratings above 12 kW. Ranges over 12 kW through 27 kW must be calculated in accordance with either the first or second note under Table 220.55. Use the first note when there is one range or when all of the ranges are rated the same. When all the ranges do not have the same kilowatt ratings, use the second note under Table 220.55. The demand loads in Column C are only applicable if the appliances are rated 12 kW or less. For ranges individually rated more than 12 kW but not more than 27 kW, 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 (Note 1 under Table 220.55).

Because ranges have such a wide variety of ratings, one column cannot be used to provide demand factors for every size. But, with the instructions specified in the notes under Table 220.55, one column can be used for all types of range calculations. When ranges are rated more than 12 kW, the demand load must be increased to compensate for the increased rating. The table is not applicable for ranges rated more than 27 kW because ranges rated more than 27 kW would not be considered household ranges. A 28-kW range rated at 240 volts draws 117 amperes (28,000 ÷ 240 = 116.6667 = 117).

Note 1 applies when one range has a kilowatt rating that is more than 12 kW. This note also applies when there is more than one range, and all the ranges have the same kilowatt rating. The demand loads in Column C, whether one or more ranges, must be increased when the individual ranges are rated more than 12 kW. The amount of increase is only 5 percent for each additional kilowatt over 12 kW. To find the percent of increase, subtract 12 from the kilowatt rating of the range and 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 one 13-kW household electric range? First, find the percentage by which Column C must be increased. A 13-kW range exceeds 12 kW by 1 kW (13 – 12 = 1). Since Column C must be increased 5 percent for each additional kilowatt of rating above 12, the maximum demand listed in Column C for one range must be increased by 5 percent (1 × 5% = 5%). The increased amount is 0.4 kW (8 × 5% = 0.4 kW). This increased amount must be added to the Column C demand load for one range (8 + 0.4 = 8.4 kW). The service demand load for one 13-kW range is 8.4 kW (see Figure 2).

With each increased kilowatt of rating, the demand load increases 5 percent. For example, what is the service demand load for one 14-kW household electric range? Because Column C is based on 12-kW ranges, 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 0.8 kW (8 × 10% = 0.8 kW). This increased amount must be added to the Column C demand load for one range (8 + 0.8 = 8.8 kW). The service demand load for one 14-kW range is 8.8 kW (see Figure 3).

 
CIF_May_08_02_and_03.jpg
 
   

The last two examples showed the calculation method for finding the demand load for ranges rated more than 12 kW. The calculation method is the same for ranges through 27 kW. For example, what is the service demand load for one 22-kW household electric range? Because Column C is based on 12-kW ranges, subtract 12 from 22 (22 – 12 = 10). Since 22 kW exceeds 12 kW by 10, multiply 10 by 5 percent to find the amount Column C must be increased (10 × 5% = 50%). The maximum demand listed in Column C for one range must be increased by 50 percent. The increased amount is 4 kW (8 × 50% = 4 kW). This increased amount must be added to the Column C demand load for one range (8 + 4 = 12 kW). The service demand load for one 22-kW range is 12 kW (see Figure 4).

 
CIF_May_08_04.jpg
 
   

Although this range is rated 22,000 watts, the maximum demand required by Table 220.55 is 12,000 watts.

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.

 
Featured Books November-26-2008
forumbutton_.gif

bannertabs4.png

Featured Books:



NFPAs_Electrical_References.png

NFPAs_pocket_electrical_references.png
Read more...