ILLUMINATION ENGINEERING 305

three attributes frequency, wave length and velocity and the three quantities are related y
the equation,
V=y (6.1)
where is the wave length and ythe freguency of the wave. The wave
lengths used by
iluminating engineers are very short and the units used are micron (10 metre) or Angstrom
A(10metre). The complete spectrum of radiant energy extends over a tremendous range ol
frequency as shown in Fig. 6.1. The visualregion extends from about 0.4 micron to 0.75 micron.
The illuminating engineer is interested in the
conversion of electrical energy into radta
energy of such frequencies as to be visible to the human eve and also in the conversion o
radiant energy intoelectrical energy by photocells or by
thermocouples.
100 0.03 A
19
3x 10 0.1 A

Ultraviolet region

7.5 x 100.41
4.3 x 10 -0.75 Visible region
14

Infrared region
300 cm

8
Hz 1 3x 10 m

Fig. 6.1 Spectrum of Radiant Energy

6.3 DEFINITIONS

Plane angle
Aplane angle is subtended at a point and is enclosed by two straight lines lying in the
same plane, A plane angle is expressed in terms of degrees or radian. Aradian is the angle
subtended by an arc of a circle whose length equals the radius of the circle.
Solid angle
A
concept which frequently is used for illumination caleulation is the solid angle and is
explained as follows:
Consider an area Arelative toa point P (Fig. 6.2). If all points on the boundary of the
area Aare joined toP, a cone-like shape is formed at P and the angle subtended by the area A
at Pis known as the solid angle. Let P represent the centre of asphere as shown in the Fig. 6.2.
There will be a boundary of intersection where the solid angle subtended by area A passes
 306 GENERATION, DISTRIBUTION AND UTILIZATION OF ELECTRICAL ENERGY

through the sphere. This area on the sphere surface and area Aare subtending the same solid
angle at P. Therefore, solid angle is defined as:

Flg. 6.2 Concept of Solit Angle

Solid angle subtended by area A = Area of intersection at sphere surface
(Radius of sphere)
Solid angle is expressed in sterdians.
Solid angle subtended by a sphere at its centre
Area of sphere 4 T'
2
=4 n steradians.
Relationship between o and 0. Refer to Fig. 6.3. Let the segment ABC subtend a solid
angle ofo and the lines 0A and 0Csubtend a plane angle 0 at the centre '0 of the circle.

Fig. 6.3 Relationship between Solid and Plane Angles

Solid angle Surface area of segment ABC

2

2n rh 2 h

Also from Fig. 6.3

r-h
= COs 0/2 Or h=r(1 - cos /2)
Substituting in the relation for o above we have
)=
2Th 2 nr (1- cos 0/2)
= 2n (1- cos /2) (6.2)
 ILLUMINATION ENGINEEAING
307
Luminous flux
It is the rate of energy radiation in the form of
light waves and is denoted by = Q/
where Q is the radiant energy. It's unit is lumen,.
Lumen
One lumen is defined as the luminous flux emitted by a
unit solid angle, i.e., sOurce of one candle power in a
Lumen = candle power of source x solid angle.
Candle power
Hence candle power of a source is defined as the no. of lumens
per unit solid angle in a given direction. The term candle power is emitted by that source
intensity. used interchangeably with
Luminous intensity
The mean luminous intensity over a particular range of directions or
contained per unit solid angle in that zone. If the solid angle is infinitely smallzones
the
is the flux
no longer a mean value but the value in a specific direction. intensity 18
Herñce luminous intensity 18
mathematically defined as:
ÖF
I= Lt (6.3)
where &F is the differential luminous flux in a differential solidangle So. The unit for
intensity is candela or lumens/steradian. luminous
QBrightness or luminance
It is defined as the intensity of asource in a given direction dividedby the orthogonally
projected area of the source in that direction. The orthogonal projection of any element of area
of a surface S Ais given by Acos where is the angle between the normal to element
the direction of view. The element SA may have any size provided that this is small as
and
compared
with the distance at which the intensity measurement is made. The unit for
brightness is
lambert and is denoted by the letter L.
Uniform diffuse source
A uniform diffuse source is one in which the intensity per unit projected area is the
same from all directions of view.
lumination
Illumination of a surface is defined as the luminous flux received by the surface per unit
area. Its unit is lux or foot-candle and is denoted by E. lumination of one lux means one
lumen persq. metre whereas one foot-candle means one lumen per sg.ft. Astill bigger
illumination is Phot and is equal to lumen per sq.cm, ie.,1 Phot = 10,000 lux.
Relationship between I, L and E
Consider a uniform diffuse spherical source with radius r metres and luminous intensity
Icandela.
Then
-
I
L=
 OF ELECTRICAL ENERGY
GENERATION, DISTRIBUTION AND UTILIZATION
308

E= 4n =
and 4n r
2 2

Therefore E=IL (6.4)

6.4 POLARCURVE

The luminous flux emitted by a source can be determined from the intensity distribution curve
The luminous intensity or candle power of any practical lamp is not uniform in all directions
ne to its unsymmetrical shape. The distribution of light is given by polar curves. Aradial
ordinate pointihg în any particular direction on apolar curve represents the luminous intensity
of the source when viewed from that direction. If theJuminous intensity is measured in a
horizontal plane aBbeut a vertieal axis and a curve is plotted between candle power and the
angular position a horizontal polar curve is obtained. Similarly,vertical polar curve gives the
relationship of candle power in vertical plane passing through the lamp at various angles.
Typical vertical and horizontal polar curves for a filament lamp are shown in Fig. 6.4.
180 90 G0
150 120
150 30
120

180
90
8
60

30

(a)Vertical Polar Curve (b) Horizontal Polar curve

Fig. 6.4 Polar Curves

Slight Depression at 180° in case of vertical polar curve is -r sin 0
due to the position of lamp holder whereas at 0° in the horizontal
polar curve is due to coiled coil filament occupying an arc
subtending an angle lessthan 360°.
The fluxX emitted by a source can be obtained by a polar do
solid which is related to the vertical polar curve. The polar solid is
assumed to be concentric with a sphere (Fig. 6.5) and the flux
emerging from the sphere through zones defined by the angular
limits 0, and ., (in vertical plane) is found as follows:
From equation (6.3), the luminous intensity I, along (as Fig. 6.5
measured with respect to vertical axis) is given as:
dF
do
 ILLUMINATION ENGINEERING
309

dF = 1,do

(6.5)

From Fig. 6.5 it is clear that the diferential solid angle do

strip area is given as subtended by the differential
do = rde. 2rr sin = 2 sin 0 do
2
Substituting in equation (6.5), we have

Fo; SL, 2n sin de = 2n sin de (6.6)

,
The total flux emitted is given when varies
from 0° to n radians

i.e., F= Jl, sin d9 (6.7)

0

Since I, is not a sinmple function of 0, direct evaluation of integral is normally not

and hence numerical or graphical methods are used. The graphical method normally possible
is the Rousseau diagram and will be dealt in the next employed
section.
From equation (6.6) F is the total flux emitted by the source and if I, is considered to be
mean spherical candle power then
Total flux from source
Aamenn = MSCP =
4T

Also if e, =0and 0, =2, the integration eovers the luminous flux emitted below the
horizontal
T/2
Fog0 = 2n I, Jsin 9 dÝ = 2 n I,
Therefore

Flux of lower hemisphere

< lower mean hemispherical candle power = 2r
Similarly
Flux of upper hemisphere
the upper mean hemispherical candle power =
2r

6.4.1 The Rousseau Diagram

The Rousseau diagram is ageometric means by which the sine function of equation (6.6) can
be graphically portrayed together with the intensity I which is a function of 6. The diagram
gives an area proportional to the flux emitted by the source and from it the total flux and zonal
îlux can be determined. The Rousseaudiagram is constructed as follows:
 ENERQy
|310 GENERATION, DISTAIBUTION AND UTILIZATION OF ELECTRICAL

do
r do

K
0,
dy

Fig.6.6 Rousseau construclion

The polar curve of thegiven source is enclosed in a circle of radius r as shown in Fig. 6E
Draw a line MN Parallel to the vertical axis of the polar curve ata suitable distance. The circle
is divided intoangular zones and the intercepts of the zonal limits and the circle is projected
horizontally cutting the line MN at right angles.The intensity value related to each radial line
of the polar curve is plotted along the corresponding horizontal 1ine, the line MN being used as
base line e.g. CD is made equal to OK. Sufficient number of intensity values are plotted so that
asmooth curve can be drawn. It can now be proved that the area of the Rousseau diagram is
directly proportional to the flux emitted. Let the intensity at an angle be I, and CD is equal
to L,. The area of the differential strip = dy x I:
where dy=r de sin 0
Therefore the differential area =I,r de sin
Therefore area between the angles ,, and 6, will be

=r ,sin de (6.8)

But from equation (6.6)we have seen that the flux contained in any angular zone with
limits 8,and 6, is given by
l'e, 6, = J I, sin 0 de
Therefore, the area of the Rousseau diagram between 0, and 0, should be multiplied by
2r/r to obtain the flux in this zone. The total flux emitted will be obtained when we multiply
the total area of the diagram by 2/r. The units of the area will be in terms of unitsof intensity
x unit of length. The following example illustrates the application of Rousseau diagram for
evaluating total flux emitted by a uniform diffusing plane source.
As defined earlier a uniform diffusing source is one in which the intensity per unit
projected area is the same from all directions of view which means
L =I, cos (6.9)
and the polar curve willbe given as in Fig. 6.7.
 JLLUMINATION ENGINEERING
311

I, , cos 0

Fig. 6.7 Rousseau Diagram For a Uoilorm Dilfusing Sourco

The Rousseau diagram is a straight line graph since
each ordinate from point m
downward is proportional to cos and each ordinate is itself proportional to cos 0. The area of
Rousseau diagram is that of a right angled triangle. Therefore
The area =r
2

Therefore flux of the source = 2 X area of diagram

F=
2r Imr
Or =T I, Lumens (6.10)

65 LAW OF ILLUMINATION
Considera source of light and a receiving surface as shown in Fig. 6.8 placed at a distance D.
The receiving surface is oriented so that it is normal to the direction of flow of radiant power at
that point.

Receiving
surface

Fig. 6.8 Law of lllumination

The illumination E, is measured using suitable meter. It is found that for diferent
increasing values of distances the illumination decreases. With Dnear the source the values of
ED² vary in a marked manner but as D increases the values all approach a constant. This
limiting values of ED² is used as a measure of the strength of the source and is called the
luminous intensity and is expressed mathematically as:
l= Lt (ED') (6.11)
D ’ oo

In fact when D is large as compared to the size of the source, the above relation holds
Rood and from this relation we define the law of illumination, i.e.,
 312 GENERATION, DISTRIBUTION AND UTILIZATION OF ELECTRICAL ENERGY

Radiant energy
E= (6.12)
D²
and is an inverse square tawi.e..the illumination at a point
normal to the direction of flow of radiantpower from the
souree is inversely proportional to the square of the distance.
In case the normal to receiving surface subtends an angle 9
with the radiant energv as shown in Fig. 6.9.
Surface receiving
E = COs (6.13) Fig. 6.9 Effect of Incident Angle
D²
This is known as cosine law of illumination.
Suppose that values of illumination are to be calculated for different points on a horizontal
plane due toa single source with intensitydistribution symmetrical about the vertical axis as
shown in Fig. 6.10 it is given as:
I(0) I(0)
E=
D²
cos or E= (6.14)

h
as = COs Source
D

Equation 6.14) is known as cosine cube law andis useful

for evaluating illumination in street lighting etc. Equation (6.14)
holds good when the size of the source is small compared with the
distances involved where illumination is required to be calculated.
Such sources are considered as point sources. In actual practice,
however. point sources do not exist and inverse square law used
for evaluating illumination introduces lot of error. We will now
proceed to evaluate illumination due to surface sources when they Fig. 6.10 Cos Law
are prefectly diffusing.
For such a source I (o) = I, cos o where I = the candle power of the surface (or any
portion of surface) normal to itself (o = 0), I(0) = the candle power at an angle o from the
norrnal. Consider Fig. 6.11. The objective is tofind an expression for illumination at point P
due to a surface source,S.

Normal

dI (0)

Fig. 6.11 lumination Due to a Surface Source

 LUMINATION ENGINEERING
313

Consider a perfectly
diffusing
source
emitted by d4 is LdA where dA is the having area
differential of thebrightness
constant 8Ource. The total flux
and the normal
of L.intensity to
thesurface element dA is |From relation (6.10)|.
LdA

Therefore, intensity in the direction of P = d I cos ¢

dI)= dl,, cos = LdA COs (6.15)
Since dA isvery small as compared to r the inverse
dA is square Jaw holds and the illuminauou
at P due to
dI(6) dl,m COs 0 LdA cos
dE,=2 COs = COs = cos

Therefore, Total illumination at Pfrom the surface S is obtained by integration:

E,-4 cos ) cos dA (6.16)

where the integral is taken over the entire surface S. Equation (6.16) is a general expression
Cor the illumination at any point due to a perfectly diffusing source of any form. In case L 1S
Tariable a separate integration should be carried out.
6.5.1 lumination Due to a Strip
The strip is assumed to be of constant luminosity L, the width of the strip is Wand length H.
The value of illumination is desired at a point P on the floor, on a line making an angle of 90°
with the wall. Consider a differential element of width Wand length dh. Therefore, the intensity
of this element normal to the strip will be (Fig. 6.12).

h
90°
I(0)
D P

Fig. 6.12 llumination Due to a Strip

LW dh
dl, =
LW dh
Therefore dlo = dI, cos ) =

dl, =
LW dh
COS cOs
Therefore dE, = 2 COs
2
 GENERATION, DISTRIBUTION AND UTILIZATION OF ELECTRICAL ENERGY
314

From Fig. 6.12 h= D tan or dh = Dsec' do

0- 90-, r= Dsec
also
LW sin o cos
Therefore. E, = D sec² dÙ

LW sin cos
D sec² dÙ
D' sec

LW
sin cos dÙ
To cover the total length of the strip ¢ varies from 0° (when r is along D) to
1 H
sin
VH +D2
H
sin'
LW
WH² +D²
Therefore, E, = 2TD sin 2 dÙ
0

-LW cos 20 -LW

2TD 2 4TD (cos? sin? )
LW
4TD (sin' - cos² ) (6.17)
D H
From Fig. 6.12 cOs 0, = and sin , =
VD² +H2 D² +H²
Substituting in equation (6.17), we get
LW H?
B, = 2rDH +D² (6.18)

hen H<<D

LW H? LW
E, = 2rD D 2rD3 (6.19)
In case the illumination desired is at point P' as shown in Fig. 6.12, it is
given as
LW H'
E,= E, cos C= COS O, (6.20)
2rD H'+D²
In case H>> D

LW
E COS
p
2rD (6.21)
 IMINATION ENGINEERING
315
6.5.2 lluminatio Due to Circular Disc
Consider illumination at point P in the
cireular disc of radius R as shown in Fig. horizontal plane due to a x dx do
6.13. The disc is assumed
he of uniform luminosity and perfectly
nbe determined at any point P along thediffusing. The illumination
axis of the disc as shown
The area of the differential source = xd ¢ dx D

Therefore, Lx dodx
dl, =

Therefore, dl, = dl, cos = Lx dÙdx Cos P

D Fig. 6.13 Ilumination

where cos = Due to a Circular Disc
+ D2

Lx dodx D²
dE, = n(D² +x) (x²+ D2)
LD2 R 2r xdo da LD²
R
xdx
0 0 (a + D²) (x+ D²)²
Let x² + D² = r²or xdx = rdr
VR²+ D² VR²+ D2
rdr dr
E, = 2LD2 -=2LD2
D D

=-LD2
1 1 LR?
(6.22)
R +D² D² R'+ D²
when D >> R,equation (6.22) reduces to
LR2
(6.23)
E, D² D²
and the inverse square law holds goods. On the other hand if
R>> D, E =L (6.24)
overcast sky, the illumination is the
This means, for a large diffusing source, such as an
and is equal to the luminosity
same every where, independent of the distance from the source,
of the sOurce.

6.5.3 Ilumination Due to a Rectangular Source

source with uniform luminosity L and the
Consider a more practical source, a rectangular perpendicular
which is on a
llumination willbe determined on the opposite wall at a point P
6.14.
erected at one corner of the source as shown in Fig.
 316 GENERATION, DISTRIBUTION AND UTILIZATION OF ELECTRICAL ENERGY

90 P
Z=D

Fig. 6.14 llumination from a Rectangular Source

Consider adifferential area dxdy, the intensity due to this area along r will be
Ldx dy
1, = cOs 9

Therefore, the differential illumination at P due to the differential element

dE, = cos
H W
Therefore,
dx dy
D
Form the Fig. 6.14 COs =

where 2= D² +2+y2
Therefore, substituting these values we have
H W H W
LD? dx dy
0 0
T (D +r2
+x+y)2
First of all we perform intergration with respect tox
dx W 1 W
tan -1
2(D +y' [(D² +y²)+ W²] 2D² +y2,32 VD? + y²
H
LD² dy 1 -1 W
tan
2T +y')(D'+ w')+ y²jJD
(D' + y² j3/2 0 D' + y'
dy
The first integral in the parentheses is
given as
H H
dy 1
dy dy 1r
WJp+ ((D' +w') +y²] W D +y' W(D +w) +y']
y
0
The second term in the
or the parantheses can be integrated by parts and the final
illumination is given as: expression
 ILLUMINATION ENGINEERING
317

LD?
E, = 2n Lan
W W
dy

H
W
2n p² +H2 R
8in
D' +w? R

where R= / D' + W2 + '

xample 6.1. Determine the
that the illumination shall bedistance
for a 25 c.p. lamp from a normally
(a) 5 lux (b) 15 lux (c) 8 f.c. placed screen in 0rde
Solutio. Since the screen placed is normal to the lamp, the
illumination is given by
E
Therefore,
(a) r= JI/E = =2.24 m. Ans,
(6)
25
= 1.29 metre. Ans.
|
V15
25
(c) r=
=1.76 ft. Ans.
8

Exámple 6.2. A lamp gives 1500 c.p. in every direction below the horizontal and no illumination
above the horizontal. Determine the total radiation sent vertically downward.
Solution. Since the flux required corresponds to the one that lies in a plane passing through
the lamp vertically downwards, the angle of the plane is 180° or n radians, therefore, the flux
willbe 1500 Tlumens = 4710 lumens.
Example 6.3. (a) Alamp emits a total flux of light of 1500 lumens. What is its mean spherical
candle power ? (b) A plane surface is placed 3 metres from a 200 c.p. uniform source of light.
Calculate the intensity of illumination on the surface when it is () normal (i) inclined at 60
(iü)parallel to the rays.
Total flux
Solution. (a) The mean spherical c.p. = = 119.42 c.p. Ans.
4T

60

(b)(i) E= 200 =22.22 lux. Ans.

(a)
 GENERATION, DISTRIBUTION AND UTILIZATION OF
318

(0 Therefore, F- 22 22 sin 60 =
19.24 lux. Ans. ELECTRIGAL ENERGY

(b)
Flg. E. 6.3 (a) anxd ()

Ahi) Since surface is parallel to the rays, the illumination is zero. Ans.
Example 6.4. A200 c,p. lamp is hung 4 metres above the centre of a circular area of 5
diameter. Determine the illumination at the (i) Centre of area (ii) of Periphery
Average illumination. Also determine the average illumination if reflector of 80% the area metre
(iü)
used.
Solution. ef iciency is

4,717
4m

2.5 m

Fig. E. 64

ylumination at the centre of the

area =
200
=
42 12.5 lux. Ans.
Sis Assuming the source to have
edge will be uniform intensity in all directions,
illumination at the
200 200
E= COs0 4
4.7172 4.7172
X
4.717 =7.624. Ans.
If given candle
power is only vertically
200
downwards then
4
4.7172 COs x = 6.46 lux.
Ans.
4.717
 ILLUMINATION ENGINEERING
intensity
the possibilities when
(iii) To find average illumination we again consider both
is uniform and when it is otherwise. sub
intensity: The flux on tbe surfoce will be Io where o is the solid angle
(a) Uniform 0.9546 steradian.
tended by the area at point Oi.e., o = 2r(1 - 0.848) =
Therefore, flux= 200 >x 0.9546 = 190.9 lumens.
190.9
Therefore, illumination average E = =9.73 1lux
T (2.5)2

Average illumination with reflector :9.73 x 0.8 = 7.78 lux

(6) Intensity = l cos

dx

Fig. E. 6.5

rde
The width of circular ring dx = cos

rde
The differential area = 2r r sin 0
cos
Therefore, solid angle subtended by the circular ring at '0

2Tr sin 0de 1 2T sin

dw = de
cos r cos2 cos

Therefore, flux in this ring = I cos dw

sin
Totalflux =2r I | COs0
2
de =2rI tan sec de
0

1
= 2rIsec0 = 2rI
Jo Lcos
 320 GENERATION, DISTRIBUTION AND UTILIZATION OF ELECTRICAL

2n x 200x 0.717
ENERGY
4

2n x 200 x 0.717
Therefore, average illumination =
4 x n (2.5)2
= 11, 47 lux. Ans.
Average illumination with reflector
= 11.47 x 0.8 = 9.177 lux
Eyomnle65. Determine the height at which a light
should be placed over a floor in order that the source of having uniform spherical distribution
distance from its vertical line may be greatest.
intensity horizontal illumination at a iv
Solution, Let I be the intensity of the source. The
illumination at a distance of l meter will be
I cos Ih
E=
h²+12 (h'+12)3/2
For E to be maximum
(h2 +2)32, 1h 3/2 (h2 +12)12, 2h =0
Or h= Ans.
V2
Example 6.6. A lamp of 500 c.p. is placed 2 mnetres below a plane
the light falling on it. Determine mirror which reflects 80% of
illumination
lamp which is hung 5 metres above ground.
at a point 5 metres away from the foot of the

Solution. Let O be the position of the main source with 500 c.p. 0'
Itsimage 0' is 2 m behind the mirror and its candle power = 500
x 0.8 = 400 c.p. 2m

Illumination at A due to
2m
500 500 1
O= COs =
(5/2) (5V2)² /2
J106 m
=7.07 lux
Illumination at A due to 5m

400 400 9
52
O'= cos =
(V106)2 106 106 5m A

=3.29 lux or 3.3 lux. Ans. Fig. E.6.6

Total illumination at pointA =7.07 +3.3 = 10. 37 lux. Ans.
Example 6.7. Alamp has a uniform candle power of 300 in all directions, and is provided with
areflector which directs 60% of the total emitted light
uniformly on to a flat circular disc. of 6
metres diameter placed 6 m. vertically below the lamp. Determine the illumination (a) at
the
 ILLUMINATION ENGINEERING 327

the eye piece. The souree is, therefore, nCen asauniformly bright dise having the luminance of
murrounded
the source (multiplied by the percentage loss in the optical ayatem). Thim dise in seen
thin ring is
by a ring of light reflected by the white comparison murface. The lumninance of yraduatec
matched against the object by adjusting the range and main control knoha which are
compartment.
The white screen is illuminated by a standard Iamp placed in the whitened
instrument nee
The lamp is moved with the help ofa calibration ACrew only when the
recalibration. Light renches the white comparison surface afer passingthrough the opnl wina
of the main
nd the amount of this light is ndjusted by adiusting the gector disc by rmeans
control knob.

6.8 LUMENORLOX MTHOD O CALCULATORS

The lumen or fluxmethod of calculation for interior lighting is most widelyused. The caleulaton
is based on average illumination required on the working plane. The method is Useful wnere
the symmetry in the layout of lighting fittingsensures that the illurnination at any point aoen
not differ much from the average value. The average illumination is given as:
Average E Flux received on the working plane
Area of the working plane
FxNxUx M (6.27)
A
where P is flux output of each lamp, N the number of lamps, U the utilization factor, M the
maintenance factor and A the area of working plane. Here U is defined as:
U=
Flux received by the working plane
Total lamp flux
The procedure, therefore, consists of the following steps:
1. Select a suitable type of lamp and lighting fitting.
2. Choose a suitable mounting height above the working plane (h).
3. Layout of the lighting points is obtained using a suitable spacing to height ratio for
the fitting.
4. Choose a suitable value of E for a particular job.
5. Select a suitable maintenance factor (0.6 for dirty situation and 0.8 for a clean
situation).
lxw
6. Calculate the room index Ri = and determine the utilization factor from
h(l+ w)
tables.
7. Determinethe required lamp flux and select a suitable size of lamp and no. of lamps
for each fitting.
 328 GENERATION, DISTRIBUTION AND UTILIZATION OF ELECTRICAL ENEDA

Example 8:8. An illumination on the working plane of 32 lux is required in a room 80 mx 18

metres. The lamps are reguiredto be hung 4.5 mabove the work bench. Assume a utilization
kector of 0.5, lamp efficacy of 14 lumens per watt, and candle power depreciation of 0.2, estimate
the number rating and disposition of the lamps. Assume a suitable value of space ratio.
Solution. Assume a maximum permissible value of spacing to height ratio of 1.5
Total lumens required - 32 x 80 x 15 lumens
32 x 80 x 15
Lamp lumens required = 0.5 x 0.8_ M= 96000 lumens.
Assume 200watt lamp. The no. of lamps required are
96000
200 x 14
= 34.28 lamps
12
36 lamnps
i.e., three rows of 12 lamps each.
80
The maximum spacing would be m and hence maximum spacing to height ratio
12
80
would be = 1.48 which is close to the maximum permissible value of 1.5.
12 x 4.5

As an alternative let the no. of lamps be 42. This will have three rows of 14 each. The
80
maximum permissible spacing to height ratio will be = 1.27 which looks quite
14 x 4.5
reasonable and will provide proper illumination.
-80m
5.7k

15 m

Fig. E. 6.8

6.9 THE ELECTRIC LAMP

Light sources are usually divided into two classes: incandescent sources_ and luminescent
sQurces. An incandescent source is one which emits light solely because of its' high temperature
e.g. a tungsten lamp whereas, a luminescent source emits light due to ionisation and excitation
independent of temperature e.g. mercury vapour lamp, neon lamp, sodium vapour lamp and
firefly. These sources usually radiate only at certain wave lengths.
6/.1 Tungsten Lamp
It has been observed that a black body is the most
efficient radiator as it emits energy at every
wave length. A black body when heated to 6250°C it emits its
maximum energy within the
 ILLUMINATION ENGINEERING
329

ieible spectrum. Even though it is not possible to realise a perfect black body, the materials
sGed for lamp filaments approach it very closely. Tungsten with its meltingpoint at 3500° C is
rniversally adopted as a material for incandescent lamps. The safe
working temperature
of such a filament is about 3006°C. The cold resistance of the filament is approximately one
isteenth of its hot resistance, The filament resistance increases almost instantaneously (about
02 sec.), An inert gas, usually acombination of nitrogen and argon is filled in the bulbswhich
results in possibility of working the filament at higher temperature without any fear of oxidation
and evaporation of the filament on account of the pressure exerted by the gas. Usually bulbs
with rating40 Wand above are gas filled wherea_ below 40 watts they are vacuum type. Since
the use of gas filled tungsten lamp, slight improvements have been made in the design of the
flament such as the use of a coiled or helical filament instead of a straight wire and finally a
coiled coil filament in which the single helix is coiled again on itself as shown in Fig. 6.23. The
coiled coil filament can be worked at a relatively higher temperature as a result the output per
watt increases and the loss due to convection decreases. The efficacy in lumens per watt increases
as the size of wattage of the lamp increases. If the working voltage is marginallygreater than
the rated voltage, more light output is obtained but results in shorter life of thelamp. Operation
of the filament at a voltage lower than the rated results in loss of light.

Fig. 6.23 Coiled Coil Filament

two
Electric discharge in a gas at low pressure: When an electric current flows between
the
electrodes in a gas at approximately 1 mm pressure, cathode glow is produced very close to the
called
cathode, followed by the cathode dark space and a bright region to some extent
column (Fig.
negative glow. Beyond thisare the Faraday dark space and the luminous positive
expand until it can be made to
6.24). A decrease in pressure causes the cathode dark space to
expand. In many of
fillthe whole tube. An increase in pressure causes the positive column to
whole tube.
the commercial lamps the positive column fills almost the
Cathode glow
Cathode Anode
Faraday dark space

Negative glow Positive column

Cathode dark space

Fig. 6.24 Electrical Discharge in a Gas at Low Pressure

the voltage drop
If the cathode is heated so that it gives an adequate supply of electrons,
sodium lamp and the mercury
at the cathode is low. Thus hot cathode lamps such as the
voltage devices. In cold cathode lamps
vapour lamp using amercury pool are essentially low
perhaps ten times the ionization
the voltage drop at the cathode must be great.ly increased to
 330 GENERATION, DISTRIBUTION AND UTILIZATION OF ELECTRICAL ENERGY

potential. This high drop accelerates the positive ions towards Supply
the cathode and gives them sufficient kinetic energy so that
when they bombard the cathode they liberate electrons. In
order to reduce power loss at the cathode, a long lamp is
Capacitor
required which in turn necessitates a higher operating voltage
for these lamps. The bombardment of the cathode also causes Choke
its gradualdisintegration with consequent blackening of the 000
tube, and there is a tendency for "clean up" of the gas which Transformer
results in reduction in the life of the lamp.
Sodhum Vapour lamp
A sodium vapour lamp consists of an inner bulb of
special glass containing the sodium and the inert gas either Cathode
neon oT argon (1.5 mm pressure) and is fitted with two
filaments. This is enclosed in a larger bulb (Fig. 6.25)which is
evacuated to prevent the escape of heat as much as possible.
The efficacy of the bulb is high as the radiations fromit are Outer tube
the yellow sodiua ines which are near the maximum of the Discharge tube
visibility curve. The normal operating temperature of the tube
is around 300°C. Asmall transformer is included in the circuit
for heating the cathode and a chake for stabilizing the
discharge. To begin with, the discharge starts in the inert gas
and the temperature of the lamp increases gradually until Fig. 6.25 Sodium Vapour Lamp
sufficient metallic vapour is present to conduct the current.
Since the ionization potential of the metal vapour is low, it Capacitor
willcarry nearly all the current and the spectrum of the inert
gas will practically disappear soon. Thus the bulb starts as a
neon lamp and then gradually changes from red to yellow as 000
Choke
the tempekature rises. Becauseof the monochromatic yellow
light which makes the objectsappear as grey, these lamps are
used for street and high way lighting.
Mercury Vapour Lamp
The mercury vapour lamp is similar in construction to
the sodium vapour lamp and consists of a double glass bulb Outer tube
(Fig.6.26). Since the cathode is maintained in an incandescent
state by ionic bombardment, no heating circuít is required. Discharge tube
The operation of the mercury vapour lamp at low mercury
vapour pressure gives blue light and a high proportion of the
ultra violet rays and hence are unsatisfactory. A more recent
developmnent is a high pressure (1 to 2 atm) mercury lamp in Electrode
which a small but carefully measured drop of mercury is
introduced in the bulb. The vapour pressure rises until all the
mercury is vaporized when the light given out is with a bluish
tinge. Here also an inert gas in introduced for initial discharge
of the mercury. When the mercury vapour lamp is in use and Fig. 6.26 Mercury Vapour Lamp
 ILLUMINATION ENGINEERING 331

sAY it goes out, it will not restart until it has cooled down and the vapour pressure fallen to a
value sufhiciently low to allow restriking of the discharge in the inert gas. The overallefhcacy
of the bulb is approxinmately 35 lumen per watt. and the p.f. 0.65. Astill more recent development
in mercury lamp 1S an ultra pressure lamp operating at about 40 atmosphere and nas a
efticacy comparable with the sodium vapour lamps. These lamps are used for outdoor and
industrial lighting.
Fluorescent Lamp
The mercury vapour lamps operating at low pressure emit radiation about 000
while at high pressure they emit radiation of about3650 AU, Since both the emissions are m
the ultraviolet region, are of no use as source oflight, However, these radiations can be used
excite certain materials. When the excited molecules of these materials return to normal, toe
emit a radiation at a frequency different from that which caused the excitation and thusae
radiation emitted by the material may be within the visible zone sothat the ultraviolet radiations
are converted to light. Materials which possess this property are called fluorescent.
The fluorescent lamp is in the form of atube 3to 5cms. in diameter and 0.5m. to L
meter long with an electrode at each end which are in the form of coiled filaments coated wun
an electron emitting material. The inside of the tube is coated with fluorescent powder and as
the operating temperature of the tube is approximately 50° C, no outer tube is required. A
series choke for stabilizing the discharge and ashunt capacitor across supply terminals for
improvement of p.f. are used with the tube. The lagging p.f. is due to the choke only. Sometimes
a smallcapacitor of 0.05u Fis connected across the tube as shown in Fig. 6.27 to suppress radio
interference as the tube draws non-sinusoidal current from the supply on account of the
characteristics of the discharge.
Capacitor
(Radio Suppression)

Choke

Capacitor
(p.f.)
4-Supply

Fig. 6.27 Fluorescent Larmp Circuit

 ELECTRICAL ENERGY
332 GENERATION, DISTRIBUTION AND UTILIZATION OF

electrodes and
When the supply is switched on the starter provides a path through the
automatically thereby the
the choke to provide preheating. The starter switch then opens instantaneously in the
Currentis interrupted. Since this heating current reduces to zero value
and thus produces a
choke circuit also, the choke field collapses which releases stored energy
high voltage between the electrodes and causes the preionized tube to strike.
Mainly there are two types of starter switches (i) glow starter switch and (ii) thermal
starter switch.
bulb
The glow starter switch consists of a pair ofbimetal contacts sealed in a smallglass
Ulled with argon gas. When the supply is switched on the total supply voltage appears across
the open contacts, thereby an arc discharge takes place between the contacts. The heat from
the discharge closes the bË-netal contacts causing the preheat current to flow. The closure of
the contacts extinguishes the arc, thereby the bimetal contacts cool and open and the lamp
strikes. Asmall capacitor to suppress radio interference is fitted between the contact connections
outside the glass bulb. The glass bulb is usually mounted on aplastic base and inserted into a
small cylinderical metal canister. The base has metal studs or pins coming out from it which
are connected to the bi-metal contacts. An insulated socket is mounted in the control gear
housing to receive the metal studs of the starter switch so that it may be connected into the
controlgear circuit.
A thermal starter switch has almost similar appearance to a glow starter switch except
thát t employs a relatively larger size of the canister. This also has a pair of bimetal strips
which are initially closed rather than open as in case of glow starter. The contacts alongwith
the heater coilare enclosed in a glass bulb. The bulb is filled with a gas to improve the thermal
link between the heater coil and the contacts.
When the supply is switched on, the current flows through the choke, the starter heater
and the electrodes. The heater coil raises the temperature of the bi-metal
contacts and they
separate, thereby the current through the choke is interrupted and the consequent voltage
pulse of approximately 1000 volts causes the tube to strike. Once the tube strikes the tube
current flows through the starter heater and the bimetal contacts remain open.
Flourescent discharge lamps may be operated on a d.c. supply also. This requires the
use of a resistive balla_t for arc stability in addition to the choke
for producing the voltage
pulse. Theresistive ballast introduces very high power loss inthe
of positively charged mercury ions towards the
ballast. Also the migration
cathode results in low light output from the
anode end of the tube. For this reason whenever the
tube is to be used on d.c., a reversing
switch is normally used in the circuit to change the
every few hours.
direction of the current through the tube
It is usually preferred to use an
inverter circuit
frequency isusually 10 KHZ. With the advent of power whenever d.c. supply is available. The
ubes is extensively being used in transistors, this method of operation of
trains, buses and airerafts.
 ILLUMINATION ENGINEERING
333

610 FLOOD LIGHTING ANDCALCULATIONS

An important application of illuminating engincering is the flood lighting of large and open
areas. Buildings very ollen have to be lighted by flood lights close to their base owing to Spae
limitations. In order to achieve reasonable uniformity from side toside, the flood lights snoud
have awide lateral distribution and to achieve a reasonable brightness high up the builain
the peak intensity should be high. Also, in order that the base of the building is not too bright
the beam should be narrow in elevation. Larger buildings with sethack features and towers
require non uniform illumination. The flood ligits shoutd be placed inside the parapets and
enough above the floor level of the setback to avoid rain water falling on them but at such
elevation as to"shield the units from external view. For this, usually long sources such as
fluorescent, tungsten iodine and sodium are used when their light is controlled by a trougn
shaped paraboie reflector which results in good side ways spread.
Large areas such as football fields or railway yards are illuminated using anumber of
towers as high as 50 m, carrying the flood light projectors so that a largearea can be lit from
one tower and the glare is avoided.
When flood lighting is used on aerodromes very accurate control of light is required
especially on the aprons so as to avoid distracting the pilots. A tungsten iodine lamnp alongwith
an accurate glass parabolic reflector achieves this goal. A baffle is used to stop direct light from
the flood light.
The number of floodlights required to obtain a requisite level of illumination can be
calculated by the lumen method which is similar to lumen method for indoor fittings. The
number of lighting units required is obtained by dividing the total flux required by the flux
from one lighting unit.
:
The following terms are used in connection with the illumination calculations
Beam angle: The angle within which the diversity of ill umination produced on a surface
will be
lights there
at right angles to the beam does not exceed 10 to 1. For symmetric flood
two beam angles, one in elevation and another in azimuth.
angle.
Beam lumens: The luminous flux contained within the beam as defined in beam
It is usually taken as 25 to 30 % of the lamp lumens.
Average illumination (1m/m)xArea in nm?
No. of flood lights required = Beam lumens
the floodlight
The beam lumens as evaluated above are the theoretical output whichmade to obtain
correction must be
must have. but in actual practice due to various losses
actual lumeps.
is bound
Waste light fuctor: When several flood lights are illuminating a surface, therethe area to
beyond the edges of
to be some amount of overlap and also some of the light willfall
waste light factor and has a
be illuminated. These two factors taken together are known as monuments or statues.
as
value of 1.2for rectanBar areas and 1.5 for iIregular areas such
effectiveness of the
Depreciation factor: Due to dirt and dust on projector surface, the
provided than is theoretically
projector reduces from 50% to 100% and hence more light should be
adequate.
required so that the illumination should be
 334 GENERATION, DISTHIBUTION AND UTILIZATIN OF ELECTAAL
xple 6.9. Abuilding meaAuring 30 m 20 misto he floodlit on the front wide with brighte
L25 lumen/sq. metre. Co efficient of reflection of building wurfae i# 0.25. 1amps of B
having lumens output of B000 each are used. Ansuming beam factur 9s 0.6, waste
1.2and maintenance factor an ).76, determine the numnber of larnp# required. light farttr
Solution. The illumination required 25
100 Jux
).25

Total lumen required 100 30 20 L2

0.75
Lumen provided by one lamp 8000 x0.6
Therefore,
Number of lamps 100x 30 20x12
0.75 x8000 x 0.6 =20 larnps. An8.
Example 6.10. A lamp having
above street level. What will be uniform Juminous intensity of 300 C.p. is suspended 6 rnetres
the illumination on the ground (a)
and (b) 6 metres away from (a). vertically beneath the larnn
Solution. (a) The illumination required = 300 = 8.33 1ux.
36
300 300
(6) E = COs =
(6 V2)2 = 2.95 lux. Ans.
72x V2
Example 6.11. Aphotocell has a sensitivity of 12 u
The projected area of the Alumen and operates with a load of 1.5 MQ.
cell is illuminated by (a) a cathode is 5.08 x 3.75 cm².
60c.p. lamp at a Determine the output voltage when the
a100 Wlarmp
having an efficacy of 20 distance of 1.8 m. (b)a 6 c.p. lamp at 0.5
m. and (c)
lumens/watt at 2 m.

6/2
6m

6m

Fig. E. 6.10
Solution. (a) Solid angle
subtended = 5.08 xL82
3.75 x 10-4

Flux = 60 x5.08x3.75 x 104

(L8)²
 ILLUMINATION ENGINEERING 335

Therefore,
60 x5.08x 3.75 x 101
Voltage (L8)2
x 12 x 1.5 = 0.635 volts.

(6) 6x5.08 x3.75 x 10

Voltage= 0.5
x 12 x 1.5 = 0.823 volts.

(c) 5.08 x 3.75 x 104 2000

Voltage= 22 4T
x 12 x 15 = 1365 volts.

The major purposes of street lighting are

) to promote safety and convenience in the streetsat night through adequate visiblity,
(ii)to increase the community value of a street,
(iii)to increase the attractiveness of the street.
In awell lighted street objects are shown up in silhouette by the background of the road
surface. An object is shown up in silhouette when the general level of brightness of all or a
substantial part of it is lower than the brightness of its background. The method of discernment
predominates in the observation of distant objects on lighted streets and highways where the
object itself may possess relatively low average brightness in the direction of the observer
whereas the street posseses a relatively high background brightness.
For proper illumination of streets the main requirements for the light distribution from
the lamps are:
1. The lamps should give a peak of intensity at a sufficiently high angle to make best
use of the reflecting properties of the surface of the road (Specular reflection principle).
2. The intensities below the peak should gradually become smaller in order that the
luminance of the road close tothe lamp-post should not be so great as to cause a
patchy appearance.
3. The reduction in intensity above the peak should be sharp in order to reduce glare
but not s0 sh¡rp asto limit flexibility in the spacing of the lantern. Provision should
be made for adequate lighting of side walks where there is suficient pedestrian traffic.
Where ever two streets meet, the level of illumination at the intersections should at
least equal the sum of illumination recommended for the two streets.

O2 0BSTGN OF OOKE AND CAPACITOR

With arc discharge lamps of all kinds additional control gear is necessary. This is because the
effective resistance of the lamp decreases as the are current increases and an external impedance
io reguired tolimit this current. The requirements of such an impedance are as follows: