Artificial Lighting DesignArtificial Lighting Design

Task lighting for general purpose rooms involves Task lighting for general purpose rooms involves the installation of light sources that will provide the installation of light sources that will provide the the optimum amount of lightoptimum amount of light, and , and distributed as distributed as evenly as possibleevenly as possible in the work areas. in the work areas.

General purpose such as classrooms that will be General purpose such as classrooms that will be used for relatively short duration, that do not used for relatively short duration, that do not require a variety of quantity nor quality, differ from require a variety of quantity nor quality, differ from commercial work places that may require a commercial work places that may require a serious consideration of the four aspects of serious consideration of the four aspects of quantity and quality.quantity and quality.

Two important elements of lighting design Two important elements of lighting design include units of measurement:include units of measurement:

FOOTCANDLE LEVELFOOTCANDLE LEVEL of light on a work plane of light on a work plane – the designer must choose the amount of – the designer must choose the amount of general illumination light desired for the general illumination light desired for the particular task. Recommendations of light particular task. Recommendations of light quantity in footcandles are given in the text and quantity in footcandles are given in the text and other publications that deal with lighting design.other publications that deal with lighting design.

LUMEN OUTPUT LUMEN OUTPUT from a lighting source. from a lighting source. Lighting fixtures contain units of illumination. A Lighting fixtures contain units of illumination. A standard 100 watt incandescent bulb produces standard 100 watt incandescent bulb produces 1700 lumens. A 48” 40 watt T12 fluorescent tube 1700 lumens. A 48” 40 watt T12 fluorescent tube produces approximately 2800 lumens. Other produces approximately 2800 lumens. Other output data is included in the text. output data is included in the text.

The Illumination Engineering SocietyThe Illumination Engineering Society has has established standards for general purpose lighting established standards for general purpose lighting for classrooms and offices. Several factors for classrooms and offices. Several factors govern:govern:

– Size and shape of the room; Size and shape of the room; room cavity ratioroom cavity ratio– Light reflectance values of floors, walls, & Light reflectance values of floors, walls, &

ceilingsceilings– Maintenance of the system Maintenance of the system (LLF)(LLF)– Characteristics of the light sourceCharacteristics of the light source

Lumen outputLumen outputCoefficient of UtilizationCoefficient of Utilization

– Control of artificial lighting devicesControl of artificial lighting devices– Desired lighting level in footcandlesDesired lighting level in footcandles– Height of light source above work planeHeight of light source above work plane

The IES formula for determining the number of The IES formula for determining the number of lighting fixtures required for certain conditions is:lighting fixtures required for certain conditions is:

Number required = Number required = FC level x room areaFC level x room area

lumens x C.U. x L.L.Flumens x C.U. x L.L.F

where:where:

FC = lighting level desired in footcandlesFC = lighting level desired in footcandles

room area is in square feetroom area is in square feet

lumens is total amount per fixturelumens is total amount per fixture

C.U. = coefficient of utilization of fixtureC.U. = coefficient of utilization of fixture

L.L.F. = light loss factor in maintenanceL.L.F. = light loss factor in maintenance

Procedure for using the formulaProcedure for using the formula

11 The designer first determines the level of light The designer first determines the level of light desired.desired.

22 Calculate the area of the room in square feet.Calculate the area of the room in square feet.

33 Select a suitable lighting fixture, and determine Select a suitable lighting fixture, and determine the the total amount of lumens the fixture will produce.total amount of lumens the fixture will produce.

44 Determine the light loss factor. A discussion of Determine the light loss factor. A discussion of this this process follows.process follows.

55 From the manufacturer’s data for the lighting From the manufacturer’s data for the lighting fixture fixture selected, a coefficient of utilization will be selected, a coefficient of utilization will be determined.determined.

In order to do this, the reflectance values of the In order to do this, the reflectance values of the room room ceiling and walls must be known. Floor ceiling and walls must be known. Floor reflectance is reflectance is usually assumed at 20%.usually assumed at 20%.

The The room cavity ratioroom cavity ratio must then be determined, must then be determined, which which involves the length and width of the room, involves the length and width of the room, and the and the height the lighting fixtures will be mounted height the lighting fixtures will be mounted above the above the work plane. work plane.

Room Cavity Ratio is determined by the formula:Room Cavity Ratio is determined by the formula:

l = length of the rooml = length of the room w = width of roomw = width of room

and 5 is a constant,and 5 is a constant,

RCR will be a number between 0 and 10, to be used in RCR will be a number between 0 and 10, to be used in conjunction with the conjunction with the illumination data of the lighting illumination data of the lighting source chosensource chosen..

Where: h = height of light above work planeWhere: h = height of light above work plane

RCR = 5h x (l + w) / (l x w)

As an example problem, choose As an example problem, choose Room 102 in the Room 102 in the Architecture BuildingArchitecture Building

The room is 32’ x 31’The room is 32’ x 31’

Ceiling height = 12’Ceiling height = 12’

And h = 9.5 feetAnd h = 9.5 feet

(desktops are 30” (desktops are 30”

high)high)

RCR =RCR =

5 x 9.5 x (32 +31) /5 x 9.5 x (32 +31) /

(32 x 32) = (32 x 32) = 3.0163.016

IES recommends from IES recommends from 5858 to to 7878 footcandles of footcandles of illumination for a school classroom. Choose then, illumination for a school classroom. Choose then, a design level of a design level of 70 footcandles70 footcandles..

Fluorescent lighting is desirable for this task since Fluorescent lighting is desirable for this task since the lumen output and color of light is more the lumen output and color of light is more acceptable for general illumination.acceptable for general illumination.

An acceptable fluorescent lighting unit would be a An acceptable fluorescent lighting unit would be a “Williams” 440CW“Williams” 440CW, which contains , which contains four tubesfour tubes, , each with a lumen output of each with a lumen output of 28002800, for a total of , for a total of

4 x 2800 = 4 x 2800 = 11,200 lumens11,200 lumens - per fixture. Each tube - per fixture. Each tube is 40 watts, for a total of 200 watts per fixture, is 40 watts, for a total of 200 watts per fixture, including allowance for the ballast.including allowance for the ballast.

Two Points of Clarification in the formula:Two Points of Clarification in the formula:

The text refers to calculating the “number of The text refers to calculating the “number of luminaires” with the formula. The meaning here is the luminaires” with the formula. The meaning here is the number of units that produce light. But consider that a unit number of units that produce light. But consider that a unit may have one or more light sources, such as a chandelier may have one or more light sources, such as a chandelier with many light bulbs, or a fluorescent fixture that may with many light bulbs, or a fluorescent fixture that may have from 2 to 6 light tubes.have from 2 to 6 light tubes.

Replace the word “luminaire” with “light fixture”, Replace the word “luminaire” with “light fixture”, which would contain the number of light sources for which which would contain the number of light sources for which it is designed, and the number of lumens equals the lumen it is designed, and the number of lumens equals the lumen output “per light source” times the number of light output “per light source” times the number of light sources. sources.

Consider also that you require either candlepower or Consider also that you require either candlepower or lux as the quantity of light desired. If lux is what you lux as the quantity of light desired. If lux is what you calculate, the room area must be in square meters. If you calculate, the room area must be in square meters. If you calculate for candlepower, the room area is in square feet. calculate for candlepower, the room area is in square feet.

Reflectance Values of the RoomReflectance Values of the Room

Since Room 102 has windows, we will for the time Since Room 102 has windows, we will for the time being ignore the free daylight and assume the being ignore the free daylight and assume the calculation will be suitable for night time calculation will be suitable for night time classroom use.classroom use.

The reflectance values for the room are:The reflectance values for the room are:– CeilingCeiling = 80 %= 80 %– FloorFloor = 20 %= 20 %– WallsWalls = 50%= 50%

Calculate the room cavity ratio for the room:Calculate the room cavity ratio for the room:

RCR = 5h x ( l + w / l x w )RCR = 5h x ( l + w / l x w )

= 5 x 9.5 ( 31+32 / 31x32)= 5 x 9.5 ( 31+32 / 31x32)

= 47.5 ( 63 / 992 )= 47.5 ( 63 / 992 )

= 47.5 x .0635= 47.5 x .0635

= 3.016= 3.016

Determine the Coefficient of Utilization (CU), Determine the Coefficient of Utilization (CU), using the fixture manufacturer’s data:using the fixture manufacturer’s data:

From the CU chart:From the CU chart:In the column for ceiling reflectance, choose 80.In the column for ceiling reflectance, choose 80.Then under the wall reflectance of 50, read the Then under the wall reflectance of 50, read the column of numbers directly below. Find the column of numbers directly below. Find the number nearest the RCR calculated (3.016), and number nearest the RCR calculated (3.016), and see the Coefficient of Utilization Value of see the Coefficient of Utilization Value of

0.620.62At this point, from the IES formula, we have:At this point, from the IES formula, we have:

FC = FC = 7070Room Area = 32x31= Room Area = 32x31= 992992Lumens per fixture = Lumens per fixture = 11,20011,200C.U. = C.U. = 0.620.62

Next, find the Light Loss Factor, which is a Next, find the Light Loss Factor, which is a measure of the level of maintenance you will measure of the level of maintenance you will assume will be done.assume will be done.

Light loss factor is a percentage number, or how Light loss factor is a percentage number, or how efficient each item will be, using a scale of zero to efficient each item will be, using a scale of zero to one, where one is perfect, and zero is no one, where one is perfect, and zero is no maintenance at all.maintenance at all.

All the numbers are allowances All the numbers are allowances you will makeyou will make, , based on expected performance. Refer to the based on expected performance. Refer to the chart:chart:

1

1

1

..

.9

11

1

1

.9

.9

.9

The Light loss factor is then,The Light loss factor is then,

11 x x 11 x x 11 x x .9.9 x x 11 x x .9.9 x x .9.9 x x .9.9 = = .6561.6561

In other words, the allowance for cleaning In other words, the allowance for cleaning the room and fixtures, and changing the room and fixtures, and changing burnouts, is only burnouts, is only 65% efficient65% efficient..

To complete the calculation, insert all the numbers into To complete the calculation, insert all the numbers into the formula, andthe formula, and

Number of fixtures requiredNumber of fixtures required = =

(70 x 992) / (11,200 x .62 x .6561)(70 x 992) / (11,200 x .62 x .6561)

= = 15.24 fixtures15.24 fixtures = say sixteen = say sixteen

Then arrangement of the fixtures in a symmetrical Then arrangement of the fixtures in a symmetrical manner to distribute light evenly must be considered. manner to distribute light evenly must be considered. One could consider 4 rows of 4 fixtures eachOne could consider 4 rows of 4 fixtures each, which , which certainly would be an even distribution in a nearly certainly would be an even distribution in a nearly square room.square room.

Total watts for the fixtures = 16 x 200 = Total watts for the fixtures = 16 x 200 = 3200 watts3200 watts

Consider an Incandescent SourceConsider an Incandescent Source

Now consider a comparison of a different type of Now consider a comparison of a different type of light source: light source: Say use incandescent down-lights in Say use incandescent down-lights in a recessed housing flush with the ceiling.a recessed housing flush with the ceiling.

A “Lightolier” RC150 fixture would contain one A “Lightolier” RC150 fixture would contain one recessed flood light, of recessed flood light, of 150 watts and 3500 lumen 150 watts and 3500 lumen output.output.

Assume that LLF is the same as beforeAssume that LLF is the same as before, and , and compare the two types. Consider the chart for the compare the two types. Consider the chart for the Coefficient of Utilization:Coefficient of Utilization:

In the column of numbers for 80% ceiling and 50% In the column of numbers for 80% ceiling and 50% walls, note the walls, note the C.U. numberC.U. number closest to the RCR of closest to the RCR of 3.16 = 3.16 = .78.78

Now insert the new figures into the formula and Now insert the new figures into the formula and find:find:

Number of fixtures required =Number of fixtures required =

( 70 x 992 ) / ( 3500 x .78 x .6561) =( 70 x 992 ) / ( 3500 x .78 x .6561) =

36.67 fixtures = say 36,36.67 fixtures = say 36,

or arranged in a square roomor arranged in a square room6 rows of 6 lights each.6 rows of 6 lights each.

Consider a High Intensity Discharge Consider a High Intensity Discharge SourceSource

Now consider a high intensity discharge lamp Now consider a high intensity discharge lamp such as a Metal Halide that could be used inside. such as a Metal Halide that could be used inside. Considering a comparison with the incandescent, Considering a comparison with the incandescent, with a C.U. of .78 and a LLF of .6516:with a C.U. of .78 and a LLF of .6516:

The smallest Metal Halide is 175 watts with a lumen The smallest Metal Halide is 175 watts with a lumen output of 14,000. output of 14,000.

Plugged into the formula, the number of light units Plugged into the formula, the number of light units would be ( 70 x 992 ) / ( 14,000 x .78 x .6516 ) = 9.76 would be ( 70 x 992 ) / ( 14,000 x .78 x .6516 ) = 9.76 fixturesfixtures

For comparison say 5 fixtures in two rows.For comparison say 5 fixtures in two rows.

10 fixtures x 200 watts (including ballast) = 2,000 10 fixtures x 200 watts (including ballast) = 2,000 wattswatts

All fixtures give approximately the same amount of All fixtures give approximately the same amount of light, but consider a comparison of power use:light, but consider a comparison of power use:

FluorescentFluorescent: 16 x 200 = : 16 x 200 = 3,200 watts3,200 watts

IncandescentIncandescent: 36 x 150 = : 36 x 150 = 5,400 watts5,400 watts

Metal HalideMetal Halide: 10 x 200 =: 10 x 200 = 2,000 watts2,000 watts

Metal halide units would be by far the most Metal halide units would be by far the most efficient solution in terms of power, but an even efficient solution in terms of power, but an even distribution of light in the room would probably distribution of light in the room would probably suffer, unless an acceptable arrangement and suffer, unless an acceptable arrangement and diffusion system could be utilized.diffusion system could be utilized.

H.I.D lamps cannot be operated with dimmers.H.I.D lamps cannot be operated with dimmers.

PRACTICE PROBLEMPRACTICE PROBLEM

PRACTICE PROBLEMPRACTICE PROBLEM

A room is 40’ x 34’ with a ceiling height of 10 A room is 40’ x 34’ with a ceiling height of 10 feet. The work plane is 30” above the floor. The feet. The work plane is 30” above the floor. The ceiling and wall reflectance are 80 and 30 percent. ceiling and wall reflectance are 80 and 30 percent. The light loss factor is 0.70. Total lumens per The light loss factor is 0.70. Total lumens per fluorescent fixture = 11,200. Desired illumination fluorescent fixture = 11,200. Desired illumination at the work plane is 75 footcandles. Find:at the work plane is 75 footcandles. Find:

Room area:Room area:

Room cavity ratio:Room cavity ratio:

The coefficient of utilization:The coefficient of utilization:

Number of fixtures required for 75 Number of fixtures required for 75 footcandlesfootcandles

Room area = 40 x 34 = 1360 s.f.Room area = 40 x 34 = 1360 s.f.

Room cavity ratio: 5h= 5x 7.5 = 37.5; l+w = 74;Room cavity ratio: 5h= 5x 7.5 = 37.5; l+w = 74;

RCR = 37.5 x 74 / 1360 = 2.04RCR = 37.5 x 74 / 1360 = 2.04

The coefficient of utilization: from chart, 80, 30 % The coefficient of utilization: from chart, 80, 30 % = = 0.65 0.65

Number of fixtures required for 75 footcandlesNumber of fixtures required for 75 footcandles = =

(75 x 1360) / (11,200 x .65 x .62) =(75 x 1360) / (11,200 x .65 x .62) =

102,000 / 4,513.6 = 22.59 = 23, and for 102,000 / 4,513.6 = 22.59 = 23, and for distribution, probably equals 24 fixtures.distribution, probably equals 24 fixtures.