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.
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).
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.
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.
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).
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).
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).
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
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.