BranchCircuit, Feeder and Service Calculations, Part XXIX
by Charles R. Miller
Published: July 2008
Article 220 – Load Calculations
220.55 Electric Ranges and Other Cooking Appliances—Dwelling Unit
The National Electrical Code (NEC) contains an
introduction, nine chapters and eight annexes. Article 90 is the
introduction to the NEC. This article contains specifications that are
essential to all chapters and sections in the Code.
The
National Electrical Code states its purpose in 90.1(A). “The purpose of
the Code is the practical safeguarding of persons and property from
hazards arising from the use of electricity.” Section 90.1 continues by
covering the Code’s adequacy, its intention and its relation to other
international standards. While most articles have a scope that
describes what the article covers, Article 90 explains what the entire
Code covers. While 90.3 explains the arrangement of the Code, 90.4
specifies its enforcement. Article 90.5 explains mandatory rules,
permissive rules and fine print notes.
This
section also explains the use of brackets in the NEC. Brackets
containing section references to another NFPA document are for
informational purposes only and are provided as a guide to indicate the
source of the extracted text. These bracketed references immediately
follow the extracted text. Formal interpretation procedures are
discussed in 90.6. Section 90.7 covers examination of equipment for
safety. Future expansion and convenience is mentioned in 90.8(A) and
90.8(B). These sections state that limiting the number of circuits in a
single enclosure minimizes the effects from short circuit or ground
fault in one circuit. The last section in Article 90 is 90.9. This
section explains the units of measurement in the Code.
Article
220 contains requirements for calculating branchcircuit, feeder and
service loads. 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.
The second note under
Table 220.55 provides instructions for finding the maximum demand for
ranges of unequal rating over 8¾ kW through 27 kW. In accordance with
Note 2, find an average value of rating by adding together the ratings
of all ranges to obtain the total connected load, and then divide by
the number of ranges. After finding the average value, 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. For example, what is the service
demand load for five 13kW, five 15kW and five 17kW household
electric ranges? Start by adding together the ratings of all ranges to
obtain the total connected load [(5 × 13) + (5 × 15) + (5 × 17) = 65 +
75 + 85 = 225]. Next, find the average value by dividing the total
connected load by the total number of ranges (225 ÷ 15 = 15 kW). The
average value for these 15 ranges is 15 kW (see Figure 1).
Because we have found the average value for the 15
ranges, it is as if this is a new question: What is the service demand
load for 15 15kW household electric ranges? Find the percentage by
which Column C must be increased. A 15kW range exceeds 12 kW by 3 kW
(15 – 12 = 3). Since Column C must be increased 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 five 13kW, five 15kW and five 17kW household
electric ranges is 34.5 kW (see Figure 2).
When applying Note 2, the range rating must not include
a fraction of a kilowatt. The fraction must either be dropped or
rounded up to the next whole kilowatt rating. For example, what is the
service demand load for five 14kW, five 16kW and five 17kW household
electric ranges? Find an average value of rating by adding together the
ratings of all ranges to obtain the total connected load. The total
connected load is 235 kW (5 × 14) + (5 × 16) + (5 × 17) = 70 + 80 + 85
= 235. Now divide the total connected load by the number of ranges to
find the average value of rating (235 ÷ 15 = 15.67 kW). The average
rating of all 15 ranges is 15.67 kW. Notes 1 and 2 specify that the
range rating must be increased for each kilowatt of rating or major
fraction thereof by which the rating of the individual ranges exceeds
12 kW. A major fraction is .5 and larger. Since the .67 is a major
fraction, round the average rating of 15.67 up to 16 kW (see Figure 3).
Now find the service demand load for 15 16kW ranges.
Because Column C is based on 12kW ranges, subtract 12 from 16 (16 – 12
= 4). Since 16 kW exceeds 12 kW by 4, multiply 4 by 5 percent to find
the amount Column C must be increased (4 × 5% = 20%). The maximum
demand listed in Column C for 15 ranges must be increased by 20
percent. The increased amount is 6 kW (30 × 20% = 6 kW). This increased
amount must be added to the Column C demand load for 15 ranges (30 + 6
= 36 kW). The service demand load for five 14kW, five 16kW and five
17kW household electric ranges is 36 kW (see Figure 4).
Dropping the fraction or rounding the fraction up to the
next whole kilowatt rating should be done only once: after finding the
average range rating, just before finding the percent of increase. It
is not necessary to round up or drop the fraction of each individual
range. For example, what is the service demand load for three 13.6kW,
three 14.9kW and four 16.6kW household electric ranges? Although
these ranges have fractions of kilowatt ratings, do not round the
rating up or drop the fraction at this time. First, find an average
value of rating by adding together the ratings of all ranges to obtain
the total connected load: (3 × 13.6) + (3 × 14.9) + (4 × 16.6) = 40.8 +
44.7 + 66.4 = 151.9. Next, find the average value by dividing the total
connected load by the total number of ranges (151.9 ÷ 10 = 15.19 kW).
Since the .19 is not a major fraction, drop it (see Figure 5).
Now, find the service demand load for 10 15kW
ranges. Subtract 12 from 15 (15 – 12 = 3). The maximum demand listed in
Column C for 10 ranges must be increased by 15 percent (3 × 5% = 15%).
Find the demand in Column C for 10 ranges, and then multiply by 15
percent (Column C for 10 ranges is 25 kW). The increased amount is 3.75
kW (25 × 15% = 3.75 kW). This increased amount must be added to the
Column C demand load for 10 ranges (25 + 3.75 = 28.75 kW). The service
demand load for three 13.6kW, three 14.9kW and four 16.6kW household
electric ranges is 28.75 kW (see Figure 6).
Next month’s Code in Focus will continue 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.
