Branch-Circuit, Feeder and Service Calculations, Part XXV

by Charles R. Miller
Published: March 2008

Article 220 – Load Calculations

220.55 Electric Ranges and Other Cooking Appliances—Dwelling Unit

Understanding how to perform load calculations in accordance with the National Electrical Code (NEC) is essential for the professional electrician. Loads must be calculated before installing branch-circuits, feeders and services. Chapter 2 in the NEC contains 10 articles. Article 220 contains requirements for calculating branch-circuit, feeder and service loads. Results from calculations in Article 220 are used with requirements from other articles to find conductor sizes and ampere ratings for overcurrent protective devices. For example, results from calculations in Parts III, IV and V of Article 220 are used with the provision in 215.2(A)(1) to find the minimum feeder-circuit conductor size. Feeder conductors shall have an ampacity not less than required to supply the load as calculated in Parts III, IV and V of Article 220 [215.2(A)(1)]. This same section continues by specifying that the minimum feeder-circuit conductor size, before the application of any adjustment or correction factors, shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load. Likewise, results (from calculations in Article 220) are used with provisions in 215.3 to find the minimum size fuse or breaker permitted for feeders.

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.

One example last month calculated the demand for 25 5-kW counter-mounted cooking units (cooktops). The demand factor for those 5-kW cooktops was found in Column B, which is for household cooking appliances rated at least 3½ kW but not more than 8¾ kW. The demand factor percent then was multiplied by the total kilowatts of the cooktops to give the service demand load. Appliances can have different ratings but still be within the limits of one column. For example, what is the service demand load for five 3½-kW wall-mounted ovens, five 5-kW counter-mounted cooking units and 10 8-kW ranges? Because none of the appliances are rated less than 3½ kW or more than 8¾ kW, the demand factor will come from Column B. The first step is to find the total number of units (5 + 5 + 10 = 20). Now, find the demand factor percent across from 20 units (28 percent). Next, find the total kilowatts of all the appliances. The ovens have a total rating of 17½ kW (5 × 3.5 = 17.5). The cooktops have total rating of 25 kW (5 × 5 = 25). The ranges have a total rating of 80 kW (10 × 8 = 80). The combined rating of all the appliances is 122.5 kW (17.5 + 25 + 80 = 122.5). Finally, multiply the total kilowatt load by the demand factor percent (122.5 × 28% = 34.3). The service demand load for five 3½-kW wall-mounted ovens, five 5-kW -counter-mounted cooking units and 10 8-kW ranges is 34.3 kW (see Figure 1).

Look back at the first example, and compare the calculated load from Column B with the maximum demand from Column C. The total number of units from the first example was 20. The demand load from Column C for 20 units is 35 kW. The calculated load from Column B for five 3½-kW wall-mounted ovens, five 5-kW counter-mounted cooking units and 10 8-kW ranges was 34.3 kW. In the first example, the lower of the two demand loads was from Column B (see Figure 3).

A number of electricians and contractors will find this series on branch-circuit, feeder and service calculations very helpful when preparing for a journeyman or master electrician’s examination. One type of exam question pertaining to household cooking equipment that often is misunderstood and, therefore, answered incorrectly includes or uses the word minimum. This type of question can be tricky because of the wording. Below is an example of this type of question.

What is the minimum demand load in kilowatts for 10 8-kW ranges?

A. 20 kW     B. 25 kW     C. 27.2 kW      D. 80 kW

The ranges have a total rating of 80 kW (10 × 8 = 80). The demand factor percent across from 10 units in Column B is 34 percent. The calculated load from Column B is 27.2 kW (80 × 34% = 27.2). The demand load in Column C for 10 units is 25 kW. Because the word minimum is in this question, it looks like the correct answer should come from Column B, especially since the word maximum is in the heading of Column C. Because it can lead to confusion, disregard the word minimum in this question. In accordance with Table 220.55, when comparing Columns B and C, it is permissible to select the lower of the two. The demand load for 10 8-kW ranges is 25 kW. The correct answer is B (see Figure 4).

Another type of exam question pertaining to household cooking equipment that often is answered incorrectly includes or uses the word maximum. Below is an example of this type of question.

What is the maximum demand load in kilowatts for five 8-kW ranges?

A. 18 kW     B. 20 kW     C. 25 kW     D. 40 kW

The ranges have a total rating of 40 kW (5 × 8 = 40). The demand factor percent across from five units in Column B is 45 percent. The calculated load from Column B is 18 kW (40 × 45% = 18). The demand load in Column C for five units is 20 kW. With this question, it looks like the correct answer must come from Column C because the question is asking for the maximum demand. The maximum demand load required by Table 220.55 is the lower of the two columns. Disregard the word maximum in this question. The demand load for five 8-kW ranges is 18 kW. The correct answer is A (see Figure 5).

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.