# Chapter6, Illumination Engineering¶

## Ex1 : Page 317¶

In :
from __future__ import division
from math import sqrt
# To determine the distance of a 25 cp lamp for various illumination

#Candle power of the lamp
I=25

# Various illumination levels
E1=5        #Case1
E2=15       #Case2
E3=8        #Case3

# According to the law of illumination  E = I/(r**2)
# Using the above equation we find the distances for the above three illuminations

r1= sqrt(I/E1)
r2= sqrt(I/E2)
r3= sqrt(I/E3)

print 'a) The distance for %g flux illumination from the normally placed screen is %0.2f m'%(E1,r1)
print 'b) The distance for %g flux illumination from the normally placed screen is %0.2f m'%(E2,r2)
print 'c) The distance for %g f.c illumination from the normally placed screen is %0.2f ft'%(E3,r3)

a) The distance for 5 flux illumination from the normally placed screen is 2.24 m
b) The distance for 15 flux illumination from the normally placed screen is 1.29 m
c) The distance for 8 f.c illumination from the normally placed screen is 1.77 ft


## Ex2 : Page 317¶

In :
from math import pi
#To determine the total radiation sent vertically downward from a lamp of 1500 cp

#Candle power of the lamp
P=1500

#Since the flux required corresponds to the one that lies in a plane passing through the lamp vertically downwards the angle is given by
Angle= pi  # Angle in radians

Flux=P*Angle

print 'The total radiation sent vertically downward is %g lumens '%Flux
# Answer in the textbook are not accurate.

The total radiation sent vertically downward is 4712.39 lumens


## Ex3 : Page 317¶

In :
from math import sin
#To calculate Spherical Candle Power and intensity of illumination

Tf= 1500                                    #Total flux
D=3                                         #Distance from the source
CP=200                                      #Candle power of the Uniform Source
E=CP/(D**2)                                  #Maximum Illumination Intensity
IntIllumi = lambda x : E*sin(x*pi/180)        #Function to determine the intensity of iluumination for various conditions

MSCP=Tf/(4*pi)                             #The Mean spherical candle power

#Various Angles at which the rays falls on the surface
t1=90
t2=60
t3=0

#Illumination Intensities at various angles
E1=IntIllumi(t1)
E2=IntIllumi(t2)
E3=IntIllumi(t3)

print 'a)The Mean Spherical Candle Power is %0.2f c.p'%(MSCP)
print 'b The illumination intensities for the following cases are: '
print 'i)   Normal : %0.2f lux'%E1
print 'ii)  Inclined at 60 degrees : %0.2f lux '%E2
print 'iii) Parallel : %g lux'%E3

a)The Mean Spherical Candle Power is 119.37 c.p
b The illumination intensities for the following cases are:
i)   Normal : 22.22 lux
ii)  Inclined at 60 degrees : 19.25 lux
iii) Parallel : 0 lux


## Ex4 : Page 318¶

In :
from __future__ import division
from math import acos, degrees, cos
#To determine average illumination and illumination at various points on the area

CP=200                  #Candle power of the lamp
D=4                     #Vertical Distance from the centre of the area to the lamp
R=5/2                   #Radius of the given area
De=sqrt((D**2)+((R)**2))    #Distance from the edge of the area
Reff=80/100             #Reflector Efficiency

Ec= CP/(D**2)            #Illumination at the centre of the area
cos_theta = (D/De)     #Angle made between perpendicular and edge distances
w=2*pi*(1-cos_theta) #Solid angle subtended by the area

#Assuming uniform intensity in all directions

#Illumination at the edge is
Eue=(CP/(De**2))*cos_theta
#Flux is given by Iw
flux=CP*w
#Average illumination
Euavg= flux/(pi*(R**2))
#Average illumination with reflector efficiency
Euavgr= Euavg*Reff

#When candle power is only vertically downwards
Eve=Eue*cos_theta     #Illumination at the edge of the area

#Total flux equation for non uniform intensity is derived and given by 2*pi*I*(sec(theta)-1)
fluxn=2*pi*CP*(1/cos_theta-1)
#Average illumination
Enuavg= fluxn/(pi*(R**2))
#Average illumination with reflector efficiency
Enuavgr= Enuavg*Reff

print 'The illumination at the'
print 'i) Centre of area = %g lux'%Ec
print 'ii)Periphery of the area '
print 'a) Uniform illumination = %0.3f lux'%Eue
print 'b) Candle power is vertically downward = %0.2f lux '%Eve
print 'iii) Average illumination and Average illumination with reflector efficiency'
print 'a) Uniform intensity = %0.2f lux and %0.2f lux respectively.'%(Euavg,Euavgr)
print 'b) Non Uniform intensity = %0.2f lux and %0.3f lux respectively. '%(Enuavg,Enuavgr)

The illumination at the
i) Centre of area = 12.5 lux
ii)Periphery of the area
a) Uniform illumination = 7.622 lux
b) Candle power is vertically downward = 6.46 lux
iii) Average illumination and Average illumination with reflector efficiency
a) Uniform intensity = 9.73 lux and 7.78 lux respectively.
b) Non Uniform intensity = 11.47 lux and 9.177 lux respectively.


## Ex6 : Page 328¶

In :
from math import atan
#To determine illumination at a point when a mirror is used

CP=500                   #Candle power of the lamp
Vd=5                     #Vertical distance from the point to the source
Hd=5                     #Horizontal distance fron the point to the source
D=sqrt((Vd**2)+(Hd**2))    #Distance between the source and the point
Md=2                     #Distance of the source from the mirror
theta=degrees(atan(Hd/Vd))       #Angle made between the source to point and the horizontal

#Lets consider the mirror reflection
R=80/100                 #Reflected rays ratio
Vdm=Vd+(2*Md)            #Vertical Distance from the point to the source reflection
Hdm=Hd                   #Horizontal distance from the point to the source reflection
Dm=sqrt((Vdm**2)+(Hdm**2)) #Distance between the source reflection and the point
thetam=degrees(atan(Hdm/Vdm))    #Angle made between the source reflection to point and the horizontal

#Illumination at A due to source
Es=(CP/(D**2))*cos(theta*pi/180)

#Illumination at A due to source refelction
Er=(R*CP/(Dm**2))*cos(thetam*pi/180)
print "Illumination at point A due to O' is %0.1f lux."%Er
#Total Illumination at point A
E=Er+Es
print "The total illumination at point A is %0.2f lux."%E

Illumination at point A due to O' is 3.3 lux.
The total illumination at point A is 10.37 lux.


## Ex7 : Page 320¶

In :
#To determine the illumination over a disc

MSCP=300                          #Uniform intensity of the lamp or Mean Spherical candle power
Vd=6                              #Vertical distance of the lamp from the disc
R=6/2                             #Radius of the disc
Reff=60/100                       #Reflector efficiency
D=sqrt((Vd**2)+(R**2))              #Distance from the source to the edge of the disc
theta=degrees(atan(R/Vd))                 #Angle made between D and Vd

Ecwr=MSCP/(Vd**2)                  #Illumination at the centre without reflector
Eewr=(MSCP/(D**2))*cos(theta*pi/180)     #Illumination at the edge without reflector

#Illumination at the centre is equal to the illumination at the egde with a reflector
Eer=MSCP*Reff*(4*pi/(pi*(R**2)))
Ecr=Eer

w=2*pi*(1-cos(theta*pi/180))           #Solid angle made by the surface
flux=MSCP*w                       #Total flux produced by the source
Eavg=flux/(pi*(R**2))             #Average illumination

print 'The illumination at: '
print 'a) At the centre: '
print 'i) With reflector : %g lux'%Ecr
print 'ii)Without reflector : %0.2f lux '%Ecwr
print 'b) At the edge of the surface:'
print 'i) With reflector : %g lux'%Eer
print 'ii)Without reflector : %0.2f lux '%Eewr
print 'c) Average illumination over the disc without the reflector : %0.2f lux'%Eavg

The illumination at:
a) At the centre:
i) With reflector : 80 lux
ii)Without reflector : 8.33 lux
b) At the edge of the surface:
i) With reflector : 80 lux
ii)Without reflector : 5.96 lux
c) Average illumination over the disc without the reflector : 7.04 lux


## Ex8 : Page 328¶

In :
from math import ceil
#To estimate the number rating and disposition of the lamps

E=32                    #Illumination required for the working plane
A=80*15                 #Area of the work bench
UF=0.5                  #Utilization Factor
MF=0.8                  #Maintenance Factor
SHR=1.5                 #Maximum permissible value of spacing to height ratio
h=4.5                   #Required height for the lamps to be hung above the work bench
Leff=14                 #Lamp efficacy
Tlumen=E*A              #Total lumens required
Llumen=Tlumen/(UF*MF)   #Lamp lumens required
l=(A/15)                #Length of the workspace

Nc=ceil(l/(1.5*4.5))     #Minimum number of lamps in a single row (Number of columns)

W=200                   #Assumed wattage of the bulb

NoL=Llumen/(W*Leff)     #Number of bulbs required, Calculated value

Nr=ceil(NoL/12)         #Number of rows calculated for the required criteria

N=Nc*Nr                 #Number of lamps necessary.

Sp=l/Nc                 #Length wise spacing between the lamps
print '''Assuming 200W bulbs,
in a rectangular workspace of 80m*15m, we require %g bulbs arranged in %g rows
& %g columns having a spacing of %0.2fm between them.'''%(N,Nr,Nc,Sp)

Assuming 200W bulbs,
in a rectangular workspace of 80m*15m, we require 36 bulbs arranged in 3 rows
& 12 columns having a spacing of 6.67m between them.


## Ex9 : Page 334¶

In :
#To determine the number of lamps required

A=30*20                 #Area of the building
B=25                    #Brightness in lumen/sq. metre
CoR=0.25                #Co - efficient of reflection
Lwatt=500               #Lamp Wattage
Ll=8000                 #Lamp lumens output
E=B/CoR                 #Illumination required
BF=0.6                  #Beam factor
WF=1.2                  #Waste light factor
MF=0.75                 #Mainteanance factor

Tlumen= E*A*WF/MF       #Total lumens required

Llumen=Ll*BF            #Lumens provided by one lamp

NoL=Tlumen/Llumen       #Number of lamps

print 'The number of lamps required are %g. '%(round(NoL))

The number of lamps required are 20.


## Ex10 : Page 334¶

In :
from math import cos, degrees, sqrt, atan, pi
#To determine the illumination on the ground

CP=300                  #Uniform luminous intensity of a lamp
Vd=6                    #Vertical distance from the ground to the lamp
Hd=6                    #Horizontal distance from the ground to the lamp
D=sqrt((Vd**2)+(Hd**2))   #Distance from the lamp to point 6 metres away from the vertical line
theta=degrees(atan(Hd/Vd))      #Angle made between D and Vd.

# Illumination
Ec=CP/(Vd**2)            #Vertically beneath the lamp
E6=CP*cos(theta*pi/180)/(D**2) #**=6 metres away from the centre

print 'The illumination on the ground :'
print 'a)Vertically beneath the lamp : %0.2f lux.'%Ec
print 'b)6 metres away from condition a) : %0.2f lux.'%E6

The illumination on the ground :
a)Vertically beneath the lamp : 8.33 lux.
b)6 metres away from condition a) : 2.95 lux.


## Ex11 - page 334¶

In :
#Determine the output voltage

Area=(5.08*3.75)*(10**-4)        #Area projected by the cathode
Is=12*(10**-6)                   #Sensitivity
Rl=1.5*(10**6)                   # Load of operation

#Function to calculate the output voltage for each case
volt = lambda a,b:Area*a*Is*Rl/(b**2)

#Case 1
CP=60                           #Lamp intensity
D=1.8                           #Vertical distance of the lamp from the cell
V1=volt(CP,D)
#Case 2
CP=6                            #Lamp intensity
D=0.5                           #Vertical distance of the lamp from the cell
V2=volt(CP,D)
#Case 3
W=100                           #Wattage of the lamp
eff=20                          #eficacy of the lamp
CP=W*eff/(4*pi)                #Lamp intensity
D=2                             #Vertical distance of the lamp from the cell
V3=volt(CP,D)

print 'The voltage output of the cells are :'
print 'a) 60 CP lamp at 1.8m : %g V'%V1
print 'b) 6 CP lamp at 0.5m : %0.3f V'%V2
print 'c) A 100W lamp having a efficacy of 20 lumens/watt at 2m : %0.3f V'%V3

The voltage output of the cells are :
a) 60 CP lamp at 1.8m : 0.635 V
b) 6 CP lamp at 0.5m : 0.823 V
c) A 100W lamp having a efficacy of 20 lumens/watt at 2m : 1.364 V