Lighting architecture notes lecture

1. LUMEN METHOD OF LIGHTING DESIGN:
Lumen method of design is the most widely used for the determination
of lighting layout for providing illumination on the working plane from
overhead lamps arranged in a regular pattern.
•This method is based on the concept of Utilisation factor.
Utilisation factor = ratio of the total flux received on the working plane
(Fr) to the total flux emitted by all the lamps (Fl).
For example , if all lamps together emit 10 000 lumen and a plane 0.8 m
high over the whole of the room receives 5000 lumen.
Utilisation factor = Fr = 5 000 = 0.5
Fl 10 000
UF
•For downward direct lighting 0.4 to 0.9
•For diffusing fittings 0.2 to 0.5
•For indirect lighting 0.05 to 0.2

2. LUMEN METHOD OF LIGHTING DESIGN:
Maintenance factor: a further allowance should be made for dirt on
the fitting or deterioration of lamp output. UF should be multiplied by
maintenance factor (MF) . It is usually taken as 0.8.
Illumination is the flux received divided by the area (A). If the room is
50 m², then illumination is :
E = 5000 = 100 lux (lumen/m²)
50
1. If we know the lamps’ output, we can calculate the illumination:
E = F X UF X MF where, F = lamps’ output
A E = Illumination
UF= Utilization factor
MF = Maintenance factor

3. LUMEN METHOD OF LIGHTING DESIGN:
2. If we know what illumination we want to get, we can find the lamp
output necessary to achieve this:
F = A X E
UF X MF
So this method can be used either as a checking tool or directly as a
design tool.

4. LUMEN METHOD OF LIGHTING DESIGN:
Problem 1: An office 10m X 5m requires an illumination level of 300
lux on the working plane. It is proposed to use 40 watt fluorescent
fittings having a rated output of 2440 lumens each. Design the lighting
scheme.
E = Lux on working plane = 300 lux
A = Floor area = 10m X 5m
F = total lamps’ output = n X 2440 lumens
where, n = number of lamps
Now, using the formula,
F = A X E
UF X MF
n X 2440 = 10 X 5 X 300 , n = 15.39
0.5 X 0.8
So we may use 15 /16 fittings of 40 watt fluorescent lamps.

5. LUMEN METHOD OF LIGHTING DESIGN:
Problem 2: Consider a private office of 150 sq. ft. area. We wish to
use a specific lamp that has 2440 lumens to achieve 40 fc.
E = on working plane = 40 foot candle
A = Floor area = 150 sq. ft.
F = total lamps’ output = n X 2440
where, n = number of lamps
Now, using the formula,
F = A X E
UF X MF
n X 2440 = 150 X 40 = 6.14
0.5 X 0.8
So we may use 6 fittings of 40 watt fluorescent lamps.
We need, 6 nos. Single lamp luminaire
3 nos. 2 lamp luminaire
2 nos 3 lamp luminaire
1 no. 6 lamp luminaire

6. WATTS-PER-SQUARE FOOT METHOD:
•Multiply room area in sq.ft. by watts/sq.ft. (refer table).
•How many watts of either fluorescent or halogen sources you need to
achieve.
Sl.
No.
Average light level desired
& typical application
Watt per square foot
fluorescent/ compact
fluorescent /HID lights
Watt per square foot of
incandescent / halogen
lamps
1 2.5 – 5 fc (foot-candle)
Hotel corridors, stair
towers
0.1 – 0.2 0.3 – 0.7
2 5 – 10 fc
Office corridors, parking
garages, theatres(house
lights)
0.2 – 0.4 0.7 – 1.0
3 10 – 20 fc
Building lobbies, waiting
areas, elevator lobbies,
malls, hotel function
spaces, school corridors
0.4 – 0.8 1.0 – 2.0

7. WATTS-PER-SQUARE FOOT METHOD:
* These levels are for general lighting only. it is not good to produce high
light levels of general light using halogen sources. However we can use
halogen sources for accent lighting in these space types.
Sl.
No.
Average light level desired
& typical application
Watt per square foot
fluorescent, compact
fluorescent or HID lights
Watt per square foot of
incandescent or halogen
lamps
4 20 – 50 fc
Office areas, classrooms,
lecture halls, conference
rooms, ambient retail
lighting, industrial
workshops, gyms
0.8 – 1.2 Not recommended*
5 50 – 100 fc
Grocery stores,
laboratories, work areas,
sports courts
1.2 – 2.0 Not recommended*

8. WATTS-PER-SQUARE FOOT METHOD:
PROBLEM 1:
•Average overall proper light level for a class room is about 50 foot-
candle.
•Using table – approximate amount of light needed = 1.2 watts/sq.ft. of
fluorescent lighting.
•Area of classroom = 800 sq.ft.
•Total lighting power needed for this room = (800 X 1.2) watt = 960 watt.
•If we plan on using luminaires with 2-40 watt lamps:
We will need = 960 = 12 luminaires
40 X 2
•If we plan on using luminaires with two 32-watt lamps each:
We will need = 960 = 15 luminaires
32 X 2

9. WATTS-PER-SQUARE FOOT METHOD:
PROBLEM 2:
•Consider the house lighting for a motion picture theater:
Recommended light level = 10 fc
•Lighting needs to dim over a full range – hence choose halogen lighting.
•Using table – approximate amount of light needed = 1.0 watts/sq.ft.
•Area of theater hall = 3000 sq.ft.
•Total lighting power needed for this room=(3000X1.0) watt = 3000 watt.
•If we plan on using luminaires with 60 watt downlights:
We will need = 3000 = 50 luminaires
60
•If we plan on using luminaires with 100 watt downlights:
We will need = 3000 = 30 luminaires
100