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