220.19 Electric Ranges and Other Cooking Appliances

Branch-circuit, feeder and service-load calculation provisions are in Article 220 of the National Electrical Code. While Part I is titled “General,” most of the first part pertains to branch-circuit calculations. Part II contains provisions and demand factors for calculating feeder and service loads. The provisions in Part I can be employed to size branch circuits, or can be added to other demand loads in Parts II, III or IV to determine feeders and services. The computed load of a feeder or service shall not be less than the total 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].

Computing feeder and service load calculations can sometimes be overwhelming. Breaking down load calculations into a series of steps can make a difficult task much easier. Different types of occupancies will require different steps. For example, a one-family dwelling will not have all of the same steps as a commercial building. One step included in most dwelling calculations, and some non-dwelling calculations is cooking equipment. Demand loads for electric ranges and other cooking appliances are in 220.19 and Table 220.19. Last month’s In Focus concluded with a discussion of the second note under Table 220.19. This month, the discussion begins with the third note.

In lieu of the method provided in Column C, it shall be permissible to add the nameplate rating of all household cooking equipment rated more than 1 3/4kW but not more than 8 3/4kW and multiply the sum by the demand factors specified in Column A or B for the given number of appliances, as per Note 3 under Table 220.19. If the cooking equipment is more than 1 3/4kW but less than 3 1/2kW, the demand can come from either Column A or Column C. If the equipment is 3 1/2kW to 8 3/4kW, the demand can come from either Column B or C. For example, what is the service demand load for two 4kW ovens and two 5kW cooktops? The four units have a combined rating of 18kW. The demand factor percent for four appliances in Column B is 50 percent. After applying the demand factor percent from Column B, the demand load is 9kW (18 ¥ .50 = 9). The demand load in Column C for four appliances is 17kW. Since it is permissible to use either column, select the lowest demand. The service demand for two 4kW ovens and two 5kW cooktops is 9kW (See Figure 1).

When calculating demand loads for household cooking equipment, if the kilowatt ratings fall in Columns A or B, compare them to Column C and select the lowest demand. Start by calculating the demand load from Column A or B. Then, compare that demand with the one from Column C. Finally, select the lowest demand load. For example, what is the service demand for six 8kW ranges? The demand load from Column B is 20.64kW (6 ¥ 8 ¥ .43 = 20.64). The demand from Column C for six appliances is 21kW. The lowest demand for six 8kW ranges comes from Column B (See Figure 2).

The demand loads from Columns A and B will not always be lower than the ones found in Column C. By adding one more 8kW range to the previous example, the column having the lowest demand changes. What is the service demand load for seven 8kW ranges? The demand load from Column B is 22.4kW (7 ¥ 8 ¥ .40 = 22.4). The demand from Column C for seven appliances is 22kW. While the lowest demand for six 8kW ranges comes from column B, the lowest demand for seven 8kW ranges comes from Column C (See Figure 3).

The second half of the third note provides instructions if the ratings of cooking appliances fall under both Column A and Column B. First, calculate the demand loads for the appliances that fall under Column A. Next, calculate the demand for appliances that fall under Column B. Finally, add the results together. For example, what is the service demand load for 30 3kW ovens and 15 5kW cooktops? The ovens are calculated from demand factors in Column A. The demand load from Column A is 27kW (30 ¥ 3 ¥ .30 = 27). The cooktops are calculated from demand factors in Column B. The demand load from Column B is 24kW (15 ¥ 5 ¥ .32 = 24). Add the results together to find the demand load. The service demand load for 30 3kW ovens and 15 5kW cooktops is 51kW (27 + 24 = 51). To make sure this is the lowest demand, find the kilowatt demand from Column C. Since all the appliances are rated less than 12kW, find the demand for all 45 appliances. Multiply 45 appliances by .75 and add 25. The demand in Column C for 45 appliances is 58.75kW (45 ¥ .75 = 33.75; 33.75 + 25 = 58.75). In this example, the lowest demand load is from Columns A and B (See Figure 4).

Up to this point, the purpose of all cooking equipment calculations was to size feeders and services. Nothing has been mentioned about sizing branch-circuits that feed cooking equipment. The fourth note under Table 220.19 specifies how to calculate branch-circuit loads. This note discusses several types of appliance installations. The first sentence stipulates how to calculate the branch-circuit load for one range. It shall be permissible to compute the branch-circuit load for one range in accordance with Table 220.19. For example, what size 75 C copper conductors are required to feed a 14kW, 240V range? First, calculate the branch-circuit demand load. Since this range exceeds 12kW, the demand in Column C must be increased 5 percent for each additional kilowatt of rating by which the rating exceeds 12kW. A 14kW range exceeds 12kW by 2kW (14 - 12 = 2). The maximum demand listed in Column C for one range must be increased by 10 percent (2 ¥ .5 = .10). The increased amount is .8kW (8 ¥ .10 = .8). This increased amount must be added to the original demand load (8 + .8 = 8.8). The branch-circuit demand for one 14kW range is 8.8kW or 8,800W. To determine the minimum conductor size, the branch-circuit ampacity must be known (watts/volts = amperes). The conductor ampacity must be rated at least 37A (8,800/240 = 36.6 = 37). The minimum size 75 C, copper conductors required for a 14kW, 240V range is 8 AWG (See Figure 5).

What size 75 C, copper conductors are required to feed an 8kW, 230V range? Calculate the demand load from Column B by multiplying 8kW by 80 percent (8 ¥ .80 = 6.4kW). Use the demand load of 6.4kW (or 6,400W), because it is lower than the Column C demand of 8kW. The conductor ampacity must be rated for at least 28A (6,400/230 = 27.8 = 28). The minimum size 75 C, copper conductors required for an 8kW, 230V range is 10 AWG (See Figure 6).

Next month’s In Focus continues the discussion of calculating branch-circuit loads for household cooking equipment. 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.