Branch-Circuit, Feeder and Service Calculations, Part XXVIII
by Charles R. Miller
Published: June 2008
Article 220 – Load Calculations
220.55 Electric Ranges and Other Cooking Appliances—Dwelling Unit
At 3 pm on September. 4, 1882, the Pearl Street
Station started providing customers in New York with electric power. It
was the first central-station electric generating plant, and Thomas A.
Edison developed it. Although this first generating plant produced
direct current, alternating current soon followed. Fifty-nine customers
reluctantly consented to have their houses wired on the promise of
three free months of electric power. If the service proved to be
unsatisfactory, they had the option to discontinue the electric
service. This new way of lighting was far from unsatisfactory. It was
But, with this new frontier
came a new danger: electrical fires. There was such an outbreak of
fires caused by electrical problems that something had to be done. For
more than a decade, there was no standardized electrical code. By 1895,
there were five different electric installation codes in use throughout
this country, but no single set of codes was accepted by all. In March
1896, 23 people met to develop a national set of rules for electrical
construction and operation. The National Board of Fire Underwriters,
which is now the American Insurance Association, published the first
National Electrical Code (NEC) in 1897. In 1911, the National Fire
Protection Association (NFPA) assumed sponsorship and control of the
The first sentence of Article 90
reads, “The purpose of this Code is the practical safeguarding of
persons and property from hazards arising from the use of electricity.”
This objective has remained constant throughout the Code’s existence.
contributing factor that led to electrical fires was overloaded
circuits, and without load calculation requirements, it still would be
a problem today. The NEC contains detailed requirements for calculating
loads in branch circuits, feeders and services. These load calculation
requirements are in Chapter 2, Article 220. Last month’s article
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.
reading the notes under Table 220.55, it would appear that Column C is
good only for ranges rated 12 kW or less. But, the first two notes
under Table 220.55 provide instructions for finding the maximum demand
for ranges over 12 kW through 27 kW. Note 1 applies when one or more
ranges have the same kilowatt rating. For ranges individually rated
more than 12 kW but not more than 27 kW, the maximum demand in Column C
must be increased by 5 percent for each additional kilowatt of rating
or major fraction, which is .5 and larger (Note 1 under Table 220.55).
demand loads in Column C must be increased when the ranges are rated
more than 12 kW. The amount of increase is 5 percent for each
additional kilowatt over 12 kW. To find the percent of increase for
more than one range, subtract 12 from the average kilowatt rating. 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 15 15-kW household electric ranges?
First, find the percentage by which Column C must be increased. A 15-kW
range exceeds 12 kW by 3 kW (15 – 12 = 3). Since Column C must be
increased by 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 15 15-kW ranges is 34.5 kW
(see Figure 1).
When applying Note 1, the range rating (before
finding the percent of increase) must not include a fraction of a
kilowatt. The fraction must either be dropped or rounded up to the next
whole kilowatt rating. This fraction is a fraction of a kilowatt, not a
fraction of a watt. An example of this is a range with a 14.5-kW rating.
the range is 14,500 watts, the .5 is a fraction of a kilowatt. Notes 1
and 2 specify that the range rating must be increased for each kilowatt
or major fraction of rating. Therefore, when the fraction is .5 or
more, the fraction of kilowatt rating must be rounded up to the next
whole kilowatt rating. For example, what is the feeder demand load for
one 14.5-kW range? Since the .5 is a major fraction, round the 14.5 up
to a 15-kW range and find the demand load. Because Column C is based on
12-kW ranges, subtract 12 from 15 (15 – 12 = 3). Since 15 kW exceeds 12
kW by 3, multiply 3 by 5 percent to find the amount Column C must be
increased (3 × 5% = 15%). The maximum demand listed in Column C for one
range must be increased by 15 percent. The increased amount is 1.2 kW
(8 × 15% = 1.2 kW). This increased amount must be added to the Column C
demand load for one range (8 + 1.2 = 9.2 kW). The feeder demand load
for one 14.5-kW range is 9.2 kW (see Figure 2).
When the fraction is less than .5, the fraction of
kilowatt rating can be dropped. For example, what is the feeder demand
load for one 14.4-kW range? Since the .4 is not a major fraction, drop
the .4, and find the demand load for one 14-kW range. 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 .8 kW (8 × 10% = .8 kW). This
increased amount must be added to the Column C demand load for one
range (8 + .8 = 8.8 kW). The feeder demand load for one 14.4-kW range
is 8.8 kW (see Figure 3).
Though Note 1, under Table 220.55, specifies over 12 kW
through 27 kW, the note applies only to ranges rated 12.5 kW through 27
kW. If the range is rated less than 12.5, the fraction can be dropped.
For example, what is the service demand load for one 12.49-kW range?
Since the .49 is not a major fraction, drop the .49 and find the demand
load for one 12-kW range. After dropping the fraction, the range rating
is only 12 kW, and Column C is based on ranges rated not more than 12
kW. In accordance with Table 220.55 and the notes under the table, a
12.49-kW range has a maximum demand of 8 kW (see Figure 4).
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.