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

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## Artificial Lighting Design

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**Artificial Lighting Design**• Task lighting for general purpose rooms involves the installation of light sources that will provide the optimum amount of light, and distributed as evenly as possible in the work areas. • General purpose such as classrooms that will be used for relatively short duration, that do not require a variety of quantity nor quality, differ from commercial work places that may require a serious consideration of the four aspects of quantity and quality.**Two important elements of lighting design include units of**measurement: FOOTCANDLE LEVEL of light on a work plane – the designer must choose the amount of general illumination light desired for the particular task. Recommendations of light quantity in footcandles are given in the text and other publications that deal with lighting design. LUMEN OUTPUT from a lighting source. Lighting fixtures contain units of illumination. A standard 100 watt incandescent bulb produces 1700 lumens. A 48” 40 watt T12 fluorescent tube produces approximately 2800 lumens. Other output data is included in the text.**The Illumination Engineering Society has established**standards for general purpose lighting for classrooms and offices. Several factors govern: • Size and shape of the room; room cavity ratio • Light reflectance values of floors, walls, & ceilings • Maintenance of the system (LLF) • Characteristics of the light source • Lumen output • Coefficient of Utilization • Control of artificial lighting devices • Desired lighting level in footcandles • Height of light source above work plane**The IES formula for determining the number of lighting**fixtures required for certain conditions is: • Number required = FC level x room area lumens x C.U. x L.L.F where: FC = lighting level desired in footcandles room area is in square feet lumens is total amount per fixture C.U. = coefficient of utilization of fixture L.L.F. = light loss factor in maintenance**Procedure for using the formula**1The designer first determines the level of light desired. 2 Calculate the area of the room in square feet. 3 Select a suitable lighting fixture, and determine the total amount of lumens the fixture will produce. 4 Determine the light loss factor. A discussion of this process follows. 5 From the manufacturer’s data for the lighting fixture selected, a coefficient of utilization will be determined. In order to do this, the reflectance values of the room ceiling and walls must be known. Floor reflectance is usually assumed at 20%. The room cavity ratio must then be determined, which involves the length and width of the room, and the height the lighting fixtures will be mounted above the work plane.**Room Cavity Ratio is determined by the formula:**l = length of the room w = width of room and 5 is a constant, RCR will be a number between 0 and 10, to be used in conjunction with the illumination data of the lighting source chosen. Where: h = height of light above work plane RCR = 5h x (l + w) / (l x w)**As an example problem, choose Room 102 in the Architecture**Building • The room is 32’ x 31’ • Ceiling height = 12’ • And h = 9.5 feet (desktops are 30” high) • RCR = • 5 x 9.5 x (32 +31) / (32 x 32) = 3.016**IES recommends from 58 to 78 footcandles of illumination for**a school classroom. Choose then, a design level of 70 footcandles. • Fluorescent lighting is desirable for this task since the lumen output and color of light is more acceptable for general illumination. • An acceptable fluorescent lighting unit would be a “Williams” 440CW, which contains four tubes, each with a lumen output of 2800, for a total of 4 x 2800 = 11,200 lumens - per fixture. Each tube is 40 watts, for a total of 200 watts per fixture, including allowance for the ballast.**Two Points of Clarification in the formula:**The text refers to calculating the “number of luminaires” with the formula. The meaning here is the number of units that produce light. But consider that a unit may have one or more light sources, such as a chandelier with many light bulbs, or a fluorescent fixture that may have from 2 to 6 light tubes. Replace the word “luminaire” with “light fixture”, which would contain the number of light sources for which it is designed, and the number of lumens equals the lumen output “per light source” times the number of light sources. Consider also that you require either candlepower or lux as the quantity of light desired. If lux is what you calculate, the room area must be in square meters. If you calculate for candlepower, the room area is in square feet.**Reflectance Values of the Room**• Since Room 102 has windows, we will for the time being ignore the free daylight and assume the calculation will be suitable for night time classroom use. • The reflectance values for the room are: • Ceiling = 80 % • Floor = 20 % • Walls = 50%**Calculate the room cavity ratio for the room:**RCR = 5h x ( l + w / l x w ) = 5 x 9.5 ( 31+32 / 31x32) = 47.5 ( 63 / 992 ) = 47.5 x .0635 = 3.016**Determine the Coefficient of Utilization (CU), using the**fixture manufacturer’s data:**From the CU chart:**• In the column for ceiling reflectance, choose 80. • Then under the wall reflectance of 50, read the column of numbers directly below. Find the number nearest the RCR calculated (3.016), and see the Coefficient of Utilization Value of 0.62 At this point, from the IES formula, we have: FC = 70 Room Area = 32x31= 992 Lumens per fixture = 11,200 C.U. = 0.62**Next, find the Light Loss Factor, which is a measure of the**level of maintenance you will assume will be done. • Light loss factor is a percentage number, or how efficient each item will be, using a scale of zero to one, where one is perfect, and zero is no maintenance at all. • All the numbers are allowances you will make, based on expected performance. Refer to the chart:**11**1 1 1 .9 1 1 .9 .9 .9 . .**The Light loss factor is then,**• 1 x 1 x 1 x .9 x 1 x .9 x .9 x .9 = .6561 • In other words, the allowance for cleaning the room and fixtures, and changing burnouts, is only 65% efficient.**To complete the calculation, insert all the numbers into the**formula, and • Number of fixtures required = (70 x 992) / (11,200 x .62 x .6561) = 15.24 fixtures = say sixteen Then arrangement of the fixtures in a symmetrical manner to distribute light evenly must be considered. One could consider 4 rows of 4 fixtures each, which certainly would be an even distribution in a nearly square room. Total watts for the fixtures = 16 x 200 = 3200 watts**Now consider a comparison of a different type of light**source: Say use incandescent down-lights in a recessed housing flush with the ceiling. • A “Lightolier” RC150 fixture would contain one recessed flood light, of 150 watts and 3500 lumen output. • Assume that LLF is the same as before, and compare the two types. Consider the chart for the Coefficient of Utilization:**In the column of numbers for 80% ceiling and 50% walls, note**the C.U. number closest to the RCR of 3.16 = .78**Now insert the new figures into the formula and find:**• Number of fixtures required = ( 70 x 992 ) / ( 3500 x .78 x .6561) = 36.67 fixtures = say 36, or arranged in a square room 6 rows of 6 lights each.**Now consider a high intensity discharge lamp such as a Metal**Halide that could be used inside. Considering a comparison with the incandescent, with a C.U. of .78 and a LLF of .6516: • The smallest Metal Halide is 175 watts with a lumen output of 14,000. • Plugged into the formula, the number of light units would be ( 70 x 992 ) / ( 14,000 x .78 x .6516 ) = 9.76 fixtures • For comparison say 5 fixtures in two rows. • 10 fixtures x 200 watts (including ballast) = 2,000 watts**All fixtures give approximately the same amount of light,**but consider a comparison of power use: • Fluorescent: 16 x 200 = 3,200 watts • Incandescent: 36 x 150 = 5,400 watts • Metal Halide: 10 x 200 = 2,000 watts • Metal halide units would be by far the most efficient solution in terms of power, but an even distribution of light in the room would probably suffer, unless an acceptable arrangement and diffusion system could be utilized. • H.I.D lamps cannot be operated with dimmers.**PRACTICE PROBLEM**A room is 40’ x 34’ with a ceiling height of 10 feet. The work plane is 30” above the floor. The ceiling and wall reflectance are 80 and 30 percent. The light loss factor is 0.70. Total lumens per fluorescent fixture = 11,200. Desired illumination at the work plane is 75 footcandles. Find: Room area: Room cavity ratio: The coefficient of utilization: Number of fixtures required for 75 footcandles**Room area = 40 x 34 = 1360 s.f.**Room cavity ratio: 5h= 5x 7.5 = 37.5; l+w = 74; RCR = 37.5 x 74 / 1360 = 2.04 The coefficient of utilization: from chart, 80, 30 % = 0.65 Number of fixtures required for 75 footcandles = (75 x 1360) / (11,200 x .65 x .62) = 102,000 / 4,513.6 = 22.59 = 23, and for distribution, probably equals 24 fixtures.

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