Artificial Lighting

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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
1   The 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:



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


         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
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:
     1
     1
     1
1        1

1        .9

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
Consider an Incandescent Source
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.
Consider a High Intensity Discharge
              Source
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
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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