 General lighting.
 Showwindow lighting.
 Track lighting.
 Sign and outline lighting.
 Other lighting.
Today I will explain the design calculations for the general lighting branch circuits as follows.
You can review the following previous articles for more information:
1 General lighting branch circuit
1.1 Definition:
General lighting outlets are those Outlets intended for general use for fixedinplace luminaires (lighting fixtures). They are only used for lighting for the normal use of the occupants and Its intensity should be adequate for any type of work performed in the area.
1.2 Lighting fixtures not included in this category:
 Specialized task lighting (Showwindow lighting, Track lighting, accent, specialty, or display lighting).
 Any special lighting for workshops, photography labs, or studios that may be located in the dwelling.
1.3 Calculation Method
Determining the general lighting load as per NEC will be based on the load per area method as follows:
 The NEC cod introduce minimum general lighting loads (in VA/ft2) for various types of buildings in Table 220.12.
 Within the same building, there are normally several different types of areas like storage, office, hallways, and cafeterias, these areas must be considered separately if their (VA/ft2) values are available in table 220.12.
 The general lighting load is calculated by multiplying the floor area (in ft2) of a building by its unit load (in VA/ft2) derived from the above table.
 If the load is continuous (as in Most commercial structures), the calculated load is multiplied by 1.25 (the inverse of 80%) to determine the circuit requirements. (Please review the definition for the term “continuous loads” in Article)
How to calculate the Floor area?
 The floor area for each floor shall be calculated from the outside dimensions of the building, dwelling unit, or other area involved.
 For dwelling units, the calculated floor area shall not include open porches, garages, or unused or unfinished spaces not adaptable for future use (like some attics, cellars, and crawl spaces).
A 25,000 ft2 office building is being designed. What is the general lighting load and what load does the circuit need to supply?
Solution:
From Table 220.12, the unit load for an office building is 3.5 VA/ft2.
The general lighting load is determined by multiplying this value by the square footage of the building:
The general lighting load = 3.5 VA/ft2 x 25,000 ft2 = 87,500 VA
So, the general lighting load is 87,500 voltamperes.
However, the load is continuous and can only be 80% of the load supplied by the circuit. This value must be multiplied by 1.25 to determine the circuit requirements:
the circuit rating = 87,500 VA x 1.25 = 109,375 VA
so, The circuit is designed to supply 109.375 KVA
1.4 Notes for table 220.12:
Important!!!
Don’t
apply the values of table 220.12 before reviewing the following notes.

 The unit values herein are based on minimum load conditions and 100 percent power factor (i.e. Load in VA = Load in Watt) and may not provide sufficient lighting for the installation contemplated. So, the designer can choose a higher value based on the existing design conditions.
 Under any conditions, don’t use values less than that specified in table 220.12, there are no exceptions.
 The general lighting load unit values specified in table 220.12 includes the following loads:
 All generaluse receptacle outlets of 20ampere rating or less, including receptacles connected to Bathroom Branch Circuits,
 The outdoor receptacle outlets,
 generaluse receptacle Outlets used in Basements, Garages, and Accessory Buildings.
 Wall lighting outlet used in Habitable Rooms,
 Wall lighting outlets used in hallways, stairways, attached garages, and detached garages,
 Wall lighting outlet used in Storage or Equipment Spaces (like attics, underfloor spaces, utility rooms, and basements),
 Wall lighting outlet used in Guest Rooms or Guest Suites In hotels, motels, or similar occupancies.
 The NEC method and table 220.12 are applied for any Additions to Existing Installations for both dwelling and nondwelling installations.
 Energy saving–type calculations (which used to reduce the connected lighting load and actual power consumption) are not permitted to be used to determine the minimum calculated lighting load if they produce loads less than the load calculated according to 220.12.
Example#2:
A (2) floors (basement and main) dwelling unit have dimensions as show in below image. Calculate the total general lighting load for this unit and the Minimum Number of General Lighting Branch Circuits.
Solution:
Step#1: Calculate the total area of the dwelling unit as follows:
a Basement area:
area “A” = 17 FT 9 IN. X (4 FT 4 IN. + 18 FT 6 IN. + 6 FT 0 IN.) = 17 FT 9 IN. X 28 FT 10 IN. = 512.0 FT2
Note that as mentioned in notes for table 220.12, the crawl space and garage areas are not considered as they are included in the NEC method calculation.
b Main floor area:
AREA “A”= 6 FT 0 IN. X (11 FT 0 IN. + 17 FT 1 IN.) = 6 FT 0 IN. X 28 FT 1 IN. = 168.5 FT2
AREA “B”= 18 FT 6 IN. X (6 FT 3 IN. + 3 FT 8 IN. + 11 FT 0 IN. + 17 FT 1 IN.) = 18 FT 6 IN. X 38 FT 0 IN. = 730.0 FT2
AREA “C”= 4 FT 4 IN. X (3 FT 8 IN. + 11 FT 0 IN. + 17 FT 1 IN.) = 4 FT 4 IN. X 31 FT 9 IN. = 137.5 FT2
AREA “D”= 20 FT 2 IN. X (3 FT 8 IN. + 11 FT 0 IN.) = 20 FT 2 IN. X 14 FT 8 IN. = 295.5 FT2
Total main floor area (IN FT2) = 1304.5 FT2
So, Total area of the dwelling unit = 512.0 FT2 + 1304.5 FT2 = 1816.5 ft2
Step#2: The unit load listed in Table 220.12 shows 3 voltamperes per ft2 for dwelling units.
The general lighting load for the dwelling unit = 1816.5 ft2 x 3 watts per ft2 = 5449.5 watts
1.5 Notes for NEC method for calculation of lighting branch circuit load
Important!!!
The
NEC doesn’t introduce a procedure for calculating the actual full load for
the individual lighting fixtures in a general lighting branch circuit.

 If the required information for calculating the actual full load for every individual lighting fixture in the circuit is available, in this time, you can calculate the actual load for the lighting branch circuit by summing of actual full load for its individual lighting fixtures.
 In this case, you need to compare the values obtained from NEC method with that obtained from actual load method and select the greater load value to be used in the design.
 But actually, The NEC method for calculation of lighting load is not required if the actual full load for every individual lighting fixture in the circuit is determined.
 Methods for determining the actual full load for every individual lighting fixture in the circuit is explained in our course " Advanced Course for Lighting Design  Level I " and I recommend reviewing these methods very well.
 Section 220.18 (b) states that For circuits supplying lighting units that have ballasts, transformers, autotransformers, or LED drivers, the calculated load shall be based on the total ampere ratings of such units and not on the total watts of the lamps. This means that we must take into account the losses in light fixture switchgear (ballast, internal wiring, etc.) so, you must use the current rating of the ballast, not the tube wattage.
Example#3:
A fluorescent lighting fixtures with 4 numbers 2 feet lamps, 18 watt/ lamp. Calculate the actual load for this lighting fixture.
Solution:
The actual total load of fixture = 4 lamps x 18 watt/lamp + losses
So, we can’t know the actual losses, we will use the same equation in another form
The actual total load of fixture = 4 ballast x watt/ballast = 4 x 20 w = 80 watt
1.6 Determining the Minimum Number of General Lighting Branch Circuits
Example#4:
For the dwelling unit in example# 2 above, Calculate the Minimum Number of General Lighting Branch Circuits.
Solution:
From example#2, the general lighting load for the dwelling unit = 5449.5 watts
However, the load is continuous and can only be 80% of the total load. This value must be multiplied by 1.25 to determine the circuits requirements:
the Load Rating = 5449.5 watts x 1.25 = 6811.9 watts
The total ampere = 6811.9 watts / 120 V = 56.77 A (note that , NEC method assumed that PF=1)
 If we design 15ampere circuits:
 If we design 20ampere circuits:
Because it is not possible to have a partially powered electrical circuit that operates properly, and because these are minimum units, the total number of branch circuits with use of 15ampere circuits is (4) circuits (3.78 circuits rounded up to 4 circuits) and (3) circuits with use of 20ampere branch circuits (2.84 circuits rounded up to 3 circuits).
1.7 Determination of maximum Permissible number of lighting fixtures on a general lighting branch circuit
You need to review the definition for branch circuit rating explained in the previous article " Branch Circuit Design Calculations – Part Two ".
Rules to be applied are as follows:
 Section 210.23 states that:
 In no case shall the load exceed the branchcircuit ampere rating.
 An individual branch circuit shall be permitted to supply any load (any number of outlets) for which it is rated.
 Section 210.23(A)(2) permits a 15 or 20ampere branch circuit supplying lighting outlets to also supply utilization equipment fastened in place, such as appliances or an air conditioner. Under the conditions specified in this requirement, the utilization equipment load must not exceed 50 % of the branchcircuit ampere rating (7.5 amperes on a 15ampere circuit and 10 amperes on a 20ampere circuit).
Note: according to 210.52(B), such fastenedinplace equipment is not permitted on the smallappliance branch circuits required in a kitchen, dining room, and so on.
Example#5:
A 4 feet long, twolamp fluorescent fixture ballast draws 0.7 amps at 120 volts. How many of these fixtures can be connected on a 20amp circuit?
Solution:
This is a continuous load, so the current used by the lights can only be 80% of the circuit current rating:
Allowable current = 20 A x 0.80 = 16 A
By dividing the allowable load by the load of each lamp, the total number of Fixtures is determined:
the total number of Fixtures = 16 A /0.7 A = 22.8 fixtures
So, the maximum number of fixtures on the circuit is 22.
Example#6:
Allowable current = 20 A x 0.80 = 16 A
By dividing the allowable load by the load of each lamp, the total number of Fixtures is determined:
the total number of Fixtures = 16 A /0.7 A = 22.8 fixtures
So, the maximum number of fixtures on the circuit is 22.
Example#6:
In example#5, If some utilization equipment fastened in place will be added to the circuit How many of these fixtures can be connected to the circuit?
Solution:
This is a continuous load, so the current used by the lights can only be 80% of the circuit current rating:
Allowable current = 20 A x 0.80 = 16 A
By following section 210.23(A)(2), the Allowable current = 16 A x 50% = 8 A
By dividing the allowable load by the load of each lamp, the total number of lamps is determined:
the total number of Fixtures = 8 A /0.7 A = 11.4 fixtures
So, the maximum number of fixtures on the circuit is 11.
Special Rules In nondwelling buildings:
In the next article, I will explain design calculations for other categories of Lighting Branch circuits. so, please keep following.
Allowable current = 20 A x 0.80 = 16 A
By following section 210.23(A)(2), the Allowable current = 16 A x 50% = 8 A
By dividing the allowable load by the load of each lamp, the total number of lamps is determined:
the total number of Fixtures = 8 A /0.7 A = 11.4 fixtures
So, the maximum number of fixtures on the circuit is 11.
Important!!
For
good design, usually assume that the general lighting branch circuits will
have some utilization
equipment
fastened in place and must be derated to 50% of its rating.

Special Rules In nondwelling buildings:
 Lighting branch circuits that supply fixed lighting units with heavyduty Lampholders can have 30 A or 40 A ratings.
 Lighting Outlets for heavyduty Lampholders shall be calculated at a minimum of 600 voltamperes.
 Use The above rules when it is applicable.
In the next article, I will explain design calculations for other categories of Lighting Branch circuits. so, please keep following.
Hello Eng. Ali,
ReplyDeleteAn industrial information has been conveyed with every minute detail. Contents are mind blowing for any practicing professional. Heads up.
Lastly, not trying to point out the mistake here but there was confusion in the mismatch of data provided in the Example2. Hoping this would clear for the other readers, this is being pointed out:
1) Dimension in the pictorial representation has 10 FT 6 IN in the place of 18 FT 6 IN.
2) In Main Floor area calculation for AreaB, computed total area would be 703 FT2 in the place of 730 FT2 (Total Area has no mismatch though).
Once again, thanks for the priceless contribution of information.
Naveen
Hi Naveen,
Deletethanks for your cooperation for highlighting any mistakes, and i think that your note#1 is not accurate the two lengths are identical in the figure of example#2. but your note#2 is accurate , the area "B" must equal 703 ft2, i think it was a typing error. thanks again.
I downloaded the image & viewed in PICASA. I may be wrong though.
DeleteThanks for responding.
nice
ReplyDelete