4.4 Illumination design& calculation
By Lumen design method
Lumen design method for artificial
Lighting :
• Normal design process
• For satisfactory illumination on
working plane
Decide :
• Level of Illumination (Lux)
• Type of luminaire/lamp
• Total no of lamps/ luminaires
• It’s lumen out put
• Layout design spacing
• Mounting height
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3.
4.4 Illumination design& calculation
• No. of lamps calculated from total
lumens needed on working plane
• By formula,
• N= Ex Ax V/FxUFxMF
Where,
• N = No of lamps required
• E = Require Illuminance of a
working plane (Lux)
• A = Area of working plane (sq.
m.)
• V = planning factor ( 1-1.25)
• F = Lumen output of lamp
• UF = Utilization factor
• MF = Maintenance factor
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4.
4.4 Illumination design& calculation
Utilization factor (UF):
• Simply ratio of total flux on working
plane (Fp) to total flux emitted by
total lamps (Fl )
• UF = Fp/Fl
• Measure efficiency of lighting
scheme
• Provided by manufacturer
Maintenance factor (MF):
• Due to deposit of dust on lamp,
Lumen output may less than
designed value
• Normally slightly over design for that
effect
• Normally 0.8 or 0.9, if lamps are
cleaned regularly
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5.
Calculation
Q. A roomof 5m (L)x 3m (B)x 3m (H) is lit
uniformly with 40 W fluorescent lamps.
Determine the no of lamps required for
supplementary of day light if Illumination
is 150 Lux on the working plane 0.75 m
above floor level & the mounting height
is 1.5m above the working plane. The
value of Utilization is 0.5 & MF is
0.8.Draw plan & section with lamps.
Solution:
Given,
• Area, A = 5mx 3m = 15 sq. m.
• Number of lamps, n = ?
• Illumination, E = 150 Lux
• 40 W F Lamp output, F = 2400 Lumen
• Maintenance factor, MF = 0.8
• Utilization factor, UF = 0.7
• Assume planning factor, V= 1 (1-1.25)
We know, according to Lumen
design method,
n = Ex AxV/Fx MFx UF
Or, n = 150x 15x1/ 2400x 0.8x 0.7
Or, n = 2250/1344
Or, n = 1.67
Say, n = 2
Layout of lamps for uniform distribution:
length of a room= 5 m
L1 = Length/ no of lamps
L1= 5m/2= 2.5 m (between lamps)
L2 = 2.5m/2 = 1.25m (between lamp & wall)
Breadth of a room B = 3 m
B1 = B/2
B1 = 3/2
B1 =1.5 m
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6.
Calculation
Q. A studioof 6mx 8m is lit uniformly with
12 nos of 1.2m long 40 W fluorescent
lamps. Total flux received on the
working plane 0.8 m above the floor
level is 19000 Lumen. Each lamp has
the output of 2800 lumen. Calculate a.)
the value of Utilization b.)Illumination
on the working plane.
Solution:
Given,
• Area , A= 6mx 8m = 48 sq. m.
• Number of lamps, n = 12
• Lengh of lamp = 1.2 m
• Total flux on plane = 19000 Lumen
• 40 W Lamp output = 2800 Lumen
• Assume maintenance factor,
MF = 0.8
• Assume planning factor = 1
We know,
E = Total flux on working
plane/Total area
E = 19000 Lumen/ 48 sq.m.
E = 396 Lux
We know, according to Lumen
design method,
n = Ex AxV/Fx MFx UF
Or, 12 = 396x 48x1/ 2800x 0.8x UF
Or, UF = 19000/12x 2800x 0.8
Or, UF = 0.71 ( i.e. 71 %)
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7.
Calculation
Q. A hallof 10mx 8mx 4m is illuminated by
24 nos of 40 W fluorescent lamps
uniformly distributed on the ceiling.
Calculate Illumination level in a hall if
the value of Utilization is 0.5 & MF is
0.8.Draw plan & section with lamps.
Solution:
Given,
• Area, A = 10mx 8m = 80 sq. m.
• Number of lamps, n = 24
• Illumination, E =?
• We know, 40 W F Lamp output,
F = 2400 Lumen
• Maintenance factor, MF = 0.8
• Utilization factor, UF = 0.5
• Assume planning factor, V = 1
We know, according to Lumen
design method,
n = E x AxV/Fx MFx UF
E x A x V = n x Fx MF x UF
E = n x Fx MF x UF/ A x V
Or, E = 24x 2400x0.5X 0.8/ 80x 1
Or, E = 23040/ 80
Or, E =288 Lux
Number of lamps, n = 24
• LCM 24= 6 x 4 Or, LCM 24= 8 x 3
Length, No= 6 lamps
• Spacing, L= 10m/6= 1.66 m
• Distance between lamp to lamp
• L1= 1.66m/2= 0.83 m
• Distance between lamp to wall
Breath, No= 4 lamps
• Spacing, B = 8m/4= 2 m
• Distance between lamp to lamp
• B1= 2m/2= 1 m
• Distance between lamp to wall
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8.
INVERSE SQUARE LAW
INVERSESQUARE LAW FORMULA:
• E = 4 ^ I/4 ^ d2
• E = I/d2
Where,
• E = Illumination
• I = Luminous Intensity of light (Cd)
• d = distance (m)
• According to the law, Illumination
from a point source reduces with
square of the distance.
• A source I Candela emits a total flux
of 4^I lumens at a distance (d) this
flux will be distributed over a sphere
of radius i.e a surface 4^d2.
• E = I/d2 applicable when direction
of light perpendicular to surface ( i.e
β = 0 angle)
• If plane is tilted, same light distributed
over larger area
• Hence illumination level reduced or less
by cos β
• Then E = I/d2 X cos β (Lux)
• E β = En X cos B
• Where, En = Illumination on a normal
plane
• Eβ = Illumination on a normal plane tilted
by angle β degree
• Angle β = Angle of incidence
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9.
CALCULATION – BYINVERSE SQUARE LAW
Q. If the angle of incidence is 45 degree,
intensity of light is 900 Cd and
distance of a point receiving light is
1.35 m, calculate the illumination level
at this point.
Given :
• I = 900 Cd
• d = 1.35 m
• Angle β = 45
• E = ?
Now, we know,
• E β = En X Cos β
• En = I/d2
• En = 900/ 1.35 x1.35 = 493.8 Lux
• Then
• E β = En X cos β
• E β = 493.8 (Lux)X cos 45
• E β = 493.8 (Lux)X0.71
• E β = 350 Lux
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10.
INTEGRATION OF DAY& ARTIFICIAL LIGHTING
• In case of unilateral opening
• & depth of a room > 6 m
• Then opposite wall area be darker
with less natural lighting level
• In large room, store, hall, corridor,
auditorium, etc.
• Then we need to lit other side by
artificial lamps
• Day light curve (1)
• Artificial light curve (2)
• Combination of natural & artificial
light curve (3)
• This is integration of Day & artificial
lighting for general illumination of
a room
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11.
Solar tube
• Solartube – modern technology
• used to lit room inside
• Source - Sun light
• By collector, reflected pipe &
diffuser
• Through roof
• As shown in figure
• Passive method of day lighting
• by active device
• Free of cost for energy source
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b. Tutorial :
Casestudy of Lighting & Acoustics (RT):
• Selection : theatre, cinema hall, shopping mall,
conference hall, radio, tv, music studio, art
gallery, church, hotel, auditorium etc
• 6 students / group
• Selection of case from Kathmandu i.e.
Rastriya Naachghar/ UWTC/City centre
• Selection of case from internet i.e. Sydney
opera house/ Kimbell art gallery/Ronchamp
church
• Artificial lighting, Day lighting & acoustic
design
• Eye survey, Photo, drawing, details,
• Opening design, %, type – unilateral,…..
• Luminaire, type, direct, indirect, CFL, special
lamp
• Acoustic design – shape & size of hall, wall,
ceiling, flooring material,
• report 13
14.
b. Tutorial :
Casestudy of Lighting & Acoustics (RT):
Day light analysis:
• Area of openings = Min 25 %
• Opening design, %,
• type – unilateral, Bilateral, …..
• Unilateral :
• D = 2.5 H
• D = ?
• In a long corridor of hotel, hostel, hospital,
school, …?
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