Electrical System Design
Chapter 01: Load Characteristics
A study defines the NEC based on:
Electrical Systems Design/ Theodore R. Bosela
ISBN: 0-13-975475-X
Pearson Education, Inc., 2003, USA
Chapter 01
Load Characteristics:
General Lighting Load:
The unit load for various occupancies (volt.Amp/sq. Ft) are shown in table 220.3A in the
NEC.
Example: Determine the estimated lighting load for an office area that measures 50 ft x
100 ft. The office contains 20 fluorescent lighting fixture, each having a
ballast input of 277V, 0.7Amp.
Solution: S1 = (50 ft x 100 ft) x 3 VA/sq. Ft = 17,500 VA
Also, S2 = (20 fixture) x 277V x 0.7Amp = 3,878 VA
Note: Although the actual connected lighting load S2 is 3,878VA, the estimated
demand load will be 17,500VA based on the NEC minimum requirements.
The number of branch circuits required for general lighting in this area must
be based on an estimated load of 17,500VA, not 3,878VA. Again, the greater
of the two values applies.
The above table requires some explanation as follows: First, although table 220.3A applies to
general lighting loads, the load from general purpose receptacles outlets in habitable areas of
dwelling units, hotels, and motels is permitted to be included with the general lighting load
for the purpose of load estimation. (Since the designer may not know at the time of design
what the final layout will be, this extra provision for load must be included).
You can always exceed NEC minimum requirements. Use common sense and judgment
when estimating loads on circuits where you suspect large lads might be connected.
General & Special Purpose Load:
For non-motor-operated: The volt-amperes rating as calculated from the actual current
and voltage ratings of the device shall be used as the load.
For Motors: Motors loads are also calculated based on the full load current of the motor
with 25% added to allow for slight overload conditions on the motor.
For Multi-motors circuits: The main feeder or branch circuit supplying multiple motor
loads, 25% of the load of the largest motor in the group must be
added to the total motor load.
Example: Determine the estimated demand load expressed in amperes and in VA for the
following individual motors.
Solution: 1/3 hp, 115V, single phase S = 115V x 7.2Amp
2 hp, 230V, single phase S = 230V x 12Amp
40 hp, 460V, three phase S = 3 x 460V x 52Amp 20 hp, 220V, three phase S = 3 x 200V x 62.1Amp
Motor full load currents used for the above example can be found in the NEC in the
following tables: 430.148, and 430-150.
Example: for multiple motors (Motor full load currents listed in NEC table 430-150)
Solution: 20 hp, 460V 27Amp
40 hp, 460V 52Amp
60 hp, 460V 77Amp + 77Amp x 25%
175.3Amp and S = 3 x 460V x 175.3Amp
Additional Notes about Motors
Design Grade of AC Motors A, B, C & D:
NEMA standard motor designs with a range of torque-versus-speed characteristics that are
graded from low to high to meet specific applications requirements. The most common B
design has normal starting torque, starting current, breakdown torque, and full load slip.
As can be expressed, the starting torque exerted by an induction motor started on full voltage
exceeds the starting torque of one started on reduced voltage.
In fact, that torque is proportional to the square of the applied voltage. For example, if the
starting voltage reduced to 80% of its rated value, starting torque will be only 64% of the full
voltage starting value. Reduced starting voltage lower starting current, but it also increases
the time needed for the motor to accelerate because of reduced starting torque.
Motor Duty Cycle:
The ratio of a motors operating time to its rest time. A motor that can operate continuously within the temperature limits of its insulation system after it has reached normal operating
temperature (equilibrium) is rated as continuous duty. A motor that never reaches
equilibrium temperature and must be shut down to cool periodically is rated as intermittent
duty.
Motor Frame Size:
NEMA has standardized motor frame sizes and other motor dimension. Fractional horse
power AC motor frame sizes are 42, 48, and 56 and integrated horse power AC motor frame
sizes are 143 to 449.
Motor Horsepower
A measure of the rate of work equal to lifting 33,000 lb to a height of 1 ft per 1 minute. One
horsepower equals 746W. For motors it is a function of torque and speed. Horsepower in
motors is given by: HP = T x RPM / 5250, where T is torque in pounds-feet (lb-ft) and RPM
is revolutions per minute.
Ambient Temperature:
The temperature of the air surrounding the motor. The NEMA standard maximum ambient
temperature is 40 degree C.
Household Electric Cooking Equipment: (NEC Table 220-19)
For the purpose of load estimation, electric ranges are considered individually. A counter
mounted cooktop and not more than two wall mounted ovens all supplied from the same
branch circuit may be considered as a single unit for the purpose of load estimation based on
the following rule.
THE RULE: for household electric ranges (or equivalent cooktop/ oven) individually rated
at not over 12kW, the estimated demand may be determined from column A of table 220-19.
If the range (or equivalent cooktop/ oven) has a rating over 12kW, but not exceeding 27kW,
the demand from column A must be increased by 5% for each kW in excess of 12kW. The
following example illustrate the procedure:
Example: Determine the estimated branch circuit demand load for the following
household electric cooking equipment. Each of these units is supplied from an
individual branch circuit.
10kW range 15kW range 19kW range
6kW cooktop 8kW oven
Solution:
10kW range since this range has a rating less than 12kW, the load is 8kW as
read from table 220-19, even though the range has rating equals
to 10kW. 8kW should be used here.
15kW range there are 3kW excess of 12kW. The percentage increase is
therefore equals to 3kW x 5% per kW = 15%, so the load shall
equal 8kW x 115%.
19kW range there are 7kW excess of 12kW. The percentage increase is
therefore equals to 7kW x 5% per kW = 35%, so the load shall
equal 8kW x 135%.
6kW cooktop the branch circuit demand load is equal to the rating of the
cooktop. The branch circuit supplying the cooktop must have a
rating sufficient to supply cooktop demand load 6kW.
8kW oven the branch circuit demand load is equal to the rating of the
oven. The branch circuit supplying the oven must have a rating
sufficient to supply oven demand load 8kW.
The 8kW is the base for load calculations.
Demand Factor:
The actual demand load for a particular occupancy will be less than the sum of connected
loads due to the diversity of equipment use. Not all equipment will be running at any given
time. Sizing a feeder or service equal to the sum of the connected loads would result in
excessive sizing of these elements. The oversized services and feeders would, of course, be
able to carry the connected load, but would not be economical.
Demand factors for general lighting loads are give in NEC tables 220.11 and 220.13.
Example 1: Of how to use NEC table 220.11.
Solution:
General lighting 200 sq. ft x 3VA/sq. ft 6,000VA
Small appliance branch circuit 4 @ 1,500VA per circuit 6,000VA
Laundry 1 @ 1,500VA per circuit 1,500VA
Miscellaneous general use receptacles 20 @ 180VA per circuit 3,600VA
Total 17,100VA
Based on table 220.11 of the NEC
First 3,000VA @ 100% demand 3,000VA x 100%
Next 3,001VA to 120,000VA @ 35% demand (17,100VA-3,000VA) x 35%
Demand load = 7,935VA
If the total load consists of loads that are prevented from operating at the same time
(simultaneously), only the larger of the two loads shall be considered in the total load
calculations. A good example of this is electric heating and central air conditioning
equipment. Obviously, the electric heating element will not operate at the same time as the air
conditioning unit. Another example is duplex pumping systems where only one pump motor
is permitted to run at the same time.
If four or more fastened in place appliances are present, Section 210.17 of the NEC permits a
demand factor of 75% to be applied to the sum of the full load current ratings of these
appliances. This demand factor shall not apply to electric ranges, clothes dryers, space
heating, or air conditioning equipment. Example of fastened in place appliances that may be
subjected to this demand factor are sump pumps, garage door operator, garbage disposals,
refrigerators, microwave ovens, deep freezers, dishwashers, etc. Even though many of these
appliances may be cord-and-plug connected, they can be considered fastened in place not
easily removable.
Example 2: The resident in Example 1 has the following fastened in place appliances.
Determine the estimated demand load for these appliances.
Solution:
Footer sump pump 1/3 hp, 115V, 7.2A = 828 VA
Laundry sump pump 1/3 hp, 115V, 7.2A = 828 VA
Overhead door operator 1/2 hp, 115V, 9.8A = 1,127 VA
Refrigerator 7.0A, 115V = 805 VA
Deep freezer 6.0A, 115V = 690 VA
Microwave hood 6.5A, 115V = 748 VA
Garbage disposal 1/2 hp, 115V, 9.8A = 1,127 VA
Dishwasher 900W = 900 VA (consider PF is 1.00)
Total = 7,053 VA
Applying the 75% demand factor to the total connected load for these appliances results in:
Estimated demand load = 7,053 VA x 75% = 5,290 VA
Example: Determine the total estimated demand load for the residence discussed in
Example 1 and Example 2. In addition to the loads specified in these
examples, the residence has an electric hot water tank rated at 5kW, 230V, and
air conditioning unit rated at 230V, 24A, and an electric range rated at 10kW.
Use this estimated demand load to determine the required service entrance
ampicity.
Solution: From Example 1 the subtotal for general illumination was 7,935 VA. From
Example 2 the subtotal for the appliance demand load was 5,290 VA. The
remaining items are electric range, air conditioning unit, and electric hot water
tank heater. In addition, since the demand load on the service is being
determined, it is necessary to add 25% of the largest motor load to the
calculation. The largest motor load will most likely be the air conditioning
compressor unit. Therefore, 25% of the full load current rating of the A/C unit
will be added to the calculations. All loads are tabulated below:
General illumination 7,935 VA
Fastened in place appliances 5,290 VA
Electric range (based on NEC rule) 8,000 VA
Hot water tank (actual rating) 5,000 VA
Air conditioning unit (volt x amp) 5,520 VA
25% of largest load 1,380 VA
Total = 33,125 VA
This demand load of 33,125 VA corresponds to a load current of 138A @ 240V. Therefore,
the service entrance rating must be based on an estimated demand of 138A. In this example, a
200A service would be appropriate to supply the load and allow for load growth.
Enclosure Type NEMA & NEMA Definitions:
NEMA 1 Enclosure designed for indoor use.
NEMA 3R Enclosure designed for GENERAL outdoor use.
NEMA 4 Enclosure protects against physical contact with internal energized switch
parts, falling dirt, dust, and wash-down with noncorrosive elements. Usually
has gasketed doors. May be used for indoor or outdoor applications.
NEMA 4X Same protection as NEMA 4 but also protects against corrosive elements.
Usually has gasketed doors. May be used for indoor or outdoor applications.
NEMA 12 Enclosure protects against physical contact with internal energized switch
parts, falling dirt, dust, and oil seepage. Usually has gasketed doors.
Generally used for indoor applications.
NEMA 13 Enclosure protects against physical contact with internal energized switch
parts, falling dirt, dust, oil seepage, and oil spray. Usually has gasketed
doors. Generally used for indoor application.
Appendix
Ground Fault Circuit Interruption Receptacle Operation
Equipment Ground
Grounded Conductor
Ungrounded Conductor
Push to Test
Resisto
r
GFS
GFS Ground Fault Sensor