Chapter 4: Illumination

Example 4.1, Page 105

In [32]:
import math

#Variable declaration
F=1000;#intensity in lumens

#Calculations
MSCP=F/(4*math.pi);# MSCP of the lamps

#Result
print "MSCP of the lamp is %.f"%MSCP

MSCP of the lamp is 80

Example 4.2, Page 105

In [33]:
import math

#Variable declaration
V=250;#in volts
I=0.8;#in amperes
F=3000;#intensity in lumens

#Calculations
wl=V*I;#wattage of lapms ins watts
lpw=F/wl;# lumens per watts is
MSCP=F/(4*math.pi);# MSCP of the lamps
MW=MSCP/wl;#MSCP per watts

#Results
print "lumens per watt is %.f"%lpw
print "MSCP per watt of the lamp is %.1f"%MW

lumens per watt is 15
MSCP per watt of the lamp is 1.2

Example 4.3, Page 105

In [34]:
import math

#Variable declaration
d=0.4;#diamter in meter
p=0.20;#in percentage absorption
F=4850;# lumens

#Calculations
Fe=(1-p)*F;# flux emitted by the globe in lumens
sa=4*math.pi*(d/2)**2;#surface area in m^2
als=Fe/sa;#average lumninance of sphere in lumens/m^2

#Result
print "average lumninance of sphere in lumens/m^2 = %.f"%als

average lumninance of sphere in lumens/m^2 = 7719

Example 4.4, Page 105

In [35]:
import math

#Variable declaration
P=20;#filament power in watts
h=5;#height in meters
d=4;#diamter in meter
p=0.50;#in percentage absorption
ef=0.89;#efficiency in watts

#Calculations
cpl=P/ef;#candle power of lamp in CP
Lop=4*math.pi*cpl;#lu,inous output in lumens
Fe=(1-p)*Lop;# flux emitted by the globe in lumens
sa=math.pi*(d/2)**2;#surface area in m^2
als=Fe/sa;#average lumninance of sphere in lumens/m^2

#Result
print "average lumninance of sphere in lumens/m^2 =%.3f"%als

average lumninance of sphere in lumens/m^2 =11.236

Example 4.5, Page 106

In [43]:
import math

#Variable declaration
cpl=100.;#
h=5;#in meter
th=60;#in degree
F=1000;#intensity in lumens

#Calculations
MSCP=F/(4*math.pi);# MSCP of the lamps
ai=((cpl/h**2)*math.cos((90-th)*math.pi/180));#average intensity of illumination

#Results
print "MSCP of a lamp is= %.f"%MSCP
print "average intensity of illumination is %.3f lux "%ai

MSCP of a lamp is= 80
average intensity of illumination is 3.464 lux

Example 4.7, Page 106

In [45]:
import math

#Variable declaration
p=500;#lamp power in watts
mscp=1250;#
h=2.7;#in meters

#Calculations
ea=(mscp)/(h)**2;#illumination directly below lamp in lux
le=(4*math.pi*mscp)/p;#lamp efficiency in lumens/watts
h1=3;#meters
eb=((mscp)/(h**2)*(2.7**3/(h1**2+h**2)**(3./2)));
#illumnination at a point 3 meters away on the horizontal plane vertically below the lamp in lux

#Results
print "illumination directly below lamp in lux = %.2f"%ea
print "lamp efficiency in lumens/W = %.2f"%le
print "illumnination at a point 3 meters away on the horizontal plane vertically below the lamp in lux = %.2f"%eb

illumination directly below lamp in lux = 171.47
lamp efficiency in lumens/W = 31.42
illumnination at a point 3 meters away on the horizontal plane vertically below the lamp in lux = 51.33

Example 4.8, Page 107

In [47]:
import math

#Variable declaration
l=100;#illumination at a point directly below the lamp in lumens/m^3
cp=256.;#
h1=1.2;#in meters

#Calculations
h=math.sqrt(cp/l);#height in meters
x=math.sqrt(h**2+h1**2);#
x1=h/x;#
eb=((cp)/(h**2))*(x1)**3;#illumnination at a point 1.2 meters away on the horizontal plane vertically below the lamp in lux

#Results
print "height in meters is =%.1f"%h
print "illumnination at a point 1.2 meters away on the horizontal plane vertically below the lamp in lux =%.1f"%eb

height in meters is =1.6
illumnination at a point 1.2 meters away on the horizontal plane vertically below the lamp in lux =51.2

Example 4.9, Page 107

In [49]:
import math

#Variable declaration
L1=500.;#candle power 
h1=9.;#in meters
d=2;#distance in meters
I2=20;#illumination in Lux

#Calculations
x=math.sqrt(h1**2+d**2);#from pythagoras theoram
Cpx=((I2-(L1/h1**2))*h1**2)/((h1/x)**3);#candle power 

#Result
print "candle power of lamp two in CP =%.f"%Cpx

candle power of lamp two in CP =1204

Example 4.10, Page 107

In [50]:
import math

#Variable declaration
h1=10;#in meters
eL=1;#ASSUME
Ea=1./(10)**2;#

#Calculations
X=(((10**3)*eL)/10**2)*10*(1./Ea);
x=(X)**(2./3);#
y=math.sqrt(x-100);#

#Result
print "distance in meters is %.1f"%y

distance in meters is 19.1

Example 4.11, Page 108

In [54]:
import math

#Variable declaration
th=15.;#in degree
l=400;#candela
d=8;# meter
p=0.80;#in percentage absorption

#Calculations
Fe=p*4*math.pi*l;# flux emitted by the globe in lumens
dA=d*math.tan((th/2)*math.pi/180);#diameter in degree
sa=math.pi*(dA)**2;#surface area in m^2
als=Fe/sa;#average lumninance of sphere in lux

#Result
print "total flux in lumens = %.f"%Fe
print "average lumninance of sphere is %.f lux"%als

total flux in lumens = 4021
average lumninance of sphere is 1154 lux

Example 4.12, Page 108

In [56]:
import math

#Variable declaration
CP=1000.;#
h=12;#in meter
d=24;#diamter in meter

#Calculations
mil=CP/(h)**2;#maximum illumination in lux
mal=mil*(12/math.sqrt(12**2+12**2))**3;#minimum illumination in lux

#Results
print "maximum illumination is %.2f lux"%mil
print "minimum illumination is %.2f lux"%mal

maximum illumination is 6.94 lux
minimum illumination is 2.46 lux

Example 4.13, Page 108

In [58]:
import math

#Variable declaration
p=60.;#
CP=200.;#
h=6;#in meter
d=10;#diamter in meter

#Calculations&Results
mil=CP/(h)**2;#maximum illumination in lux
print "part (a). "
print "illumination at the centre of the area without reflector is %.2f lux"%mil
mal=mil*(h/math.sqrt(h**2+(d/2)**2))**3;#minimum illumination in lux
tl=4*math.pi*CP;#total lumens 
ts=(p/100)*tl;#total lumens reaching the surface
A=math.pi*(d/2)**2;#total surface area in m**2
alf=ts/A;#average illumination with reflector
x=math.sqrt(h**2+(d/2)**2);#
y=h/x;#
om=2*math.pi*(1-y);# in steradians
tfr=CP*om;#total flux reaching the surface
alwr=tfr/A;#average illumination without reflector
print "\npart (b). "
print "illumination at the edge of the area without reflector is %.2f lux"%mal
print "average illumination with reflector is %.1f lux"%alf
print "average illumination without reflelctor is %.1f lux"%alwr
#with the reflector the illumintaion at the edge and at the end will be the same since the reflection directs the 
#light uniformity on the surface

part (a). 
illumination at the centre of the area without reflector is 5.56 lux

part (b). 
illumination at the edge of the area without reflector is 2.52 lux
average illumination with reflector is 19.2 lux
average illumination without reflelctor is 3.7 lux

Example 4.14, Page 109

In [59]:
import math

#Variable declaration
CP=100.;#
h=6.;#in meter
d=16;# meter

#Calculations
x=math.sqrt(h**2+d**2);#
em=2*((CP/h**2)*(h/(d-h))**3);#illumination in the middle in lux
ee=((CP/h**2)*(1+(h/x)**3));#illumination iunder each lamp in lux

#Results
print "illumination under each lamp is %.1f lux"%ee
print "illumination in the middle is %.1f lux"%em

illumination under each lamp is 2.9 lux
illumination in the middle is 1.2 lux

Example 4.15, Page 109

In [60]:
import math

#Variable declaration
CP=800;#
h=10;#in meter
d=12;# meter

#Calculations
x=math.sqrt(h**2+d**2);#
x1=math.sqrt(h**2+(d/2)**2);#
em=((CP/h**2)*(1+(h/x)**3+(h/x)**3));#illumination iunder each lamp in lux
ee=2*((CP/h**2)*(h/x1)**3);#illumination at the centrelamp in lux

#Results
print "illumination under each lamp is %.1f lux"%em
print "illumination in the middle is %.3f lux"%ee

illumination under each lamp is 12.2 lux
illumination in the middle is 10.088 lux

Example 4.16, Page 110

In [61]:
import math

#Variable declaration
CP=400;#
h=10;#in meter
d=20;# meter

#Calculations
x=math.sqrt(d**2-h**2);#
ee=4*((CP/h**2)*(h/x)**3);#illumination at the centrelamp in lux

#Result
print "illumination in the middle is %.2f lux"%ee

illumination in the middle is 3.08 lux

Example 4.17, Page 111

In [62]:
import math

#Variable declaration
cp=500.;#cp
h=4.;#in meter

#Calculations
x=((2*cp*h**3)/h**2);#
y=((cp*h**3)/h**2);#
y1=cp/h**2;#
y2=y/2;#
y21=y1/2;#
d=math.sqrt((((x-y2)/y21)**(2./3))-h**2)*2.29;#

#Result
print "distance is,(m)= %.2f"%d

distance is,(m)= 9.52

Example 4.18, Page 111

In [63]:
import math

#Variable declaration
d=6;#in meter
h=4;#in meter
ef=20;#lumens per watt
uc=0.5;#utilization coefficient
il=750;# in lux

#Calculations
a=(math.pi/4)*(d)**2;#
F=a*il;#in lumens 
tf=F/uc;#total flux emitted by the lamp
watt=tf/ef;#wattage of lamp

#Result
print "wattage of lamp is %.f watts"%watt

wattage of lamp is 2121 watts

Example 4.19, Page 130

In [66]:
import math

#Variable declaration
vl=220.;#voltage of lamp
wl=60.;#wattage of lamp
wl1=75.;#in watts
v2=440.;# in volts

#Calculations
r1=((vl**2)/wl);# in ohms
r2=((vl**2)/wl1);# in ohms
i=(v2/(r1+r2));#in amperes
v1=i*r1;# volts 
v12=i*r2;#in volts
cp6=(math.ceil(v1)/vl)**4 *(100);#candle power 
cp7=(v12/vl)**4*(100);#candle power

#Results
print "potential drop across 60 watt lamps is %.f volts"%cp6
print "potential drop across 75 watt lamps is %.f volts"%v12
print "candle power of 60 watts lampe in percentage = %.f"%cp6
print "candle power of 75 watts lampe in percentage %.f"%cp7
#answer is wrong in the book

potential drop across 60 watt lamps is 154 volts
potential drop across 75 watt lamps is 196 volts
candle power of 60 watts lampe in percentage = 154
candle power of 75 watts lampe in percentage 62

Example 4.20, Page 131

In [67]:
import math

#Variable declaration
w=84;#watts
pf=0.7;#power factor
v=240;#in volts

#Calculations
i=(w)/(v*pf);# in amperes
rva=v*i*math.sqrt(1-pf**2);#relative volt-amperes 
cpf=1;#corrected power factor
rvas=v*i*math.sqrt(1-cpf**2);#
f=50;# in hertz
c=((rva)/(2*math.pi*f*(v)**2));#in farads

#Result
print "capacitance in (micro-F) is %.2f"%(c*10**6)

capacitance in (micro-F) is 4.74

Example 4.21, Page 131

In [68]:
#Variable declaration
v1=110;#in volts
cp1=16.;#in cp
cp2=25;#in cp
v2=220.;#in volts

#Calculations
ri=((cp1/cp2)*(v2/v1));#ratio of curents
dr=(ri)**(2./3);#ratio of diameters
di=(cp1/cp2)*(1./dr);#ratio of lengths

#Results
print "ratio of diameter is %.2f"%dr
print "ratio of length is %.3f"%di

ratio of diameter is 1.18
ratio of length is 0.543

Example 4.22, Page132

In [73]:
import math

#Variable declaration
c1=71.5;#candel power
v1=260.;#in volts
c2=50.;#candel power
v2=240;#in volts

#Calculations&Results
b=math.log(c1/c2)/(math.log(v1/v2));#
a=c2/(v2)**(4.5);#
print "part (a). "
print "constants are %.2e and %.1f"%(a,b)
v=250;# in volts
p=4.;#change in percentage
dvc=a*b*((v)**(b-1));#in candle per volts
dc=(1+(p/100))**b;#when voltage increase by 4%
pcp=((dc-1))*100;#percentage change in candle power
dc1=(1-(p/100))**b;#when voltage falls by 4%
pcp1=((dc1-1))*100;#percentage change in candle power
print "part (b)."
print "change of candle power is %.2f per volts"%dvc

#chage in candle power per volt is calculated wrong in the book
print "percentage change in candle power  when voltage increase by 4%% is %.1f"%pcp
print "percentage change in candle power  when voltage falls by 4%% is %.1f"%pcp1

part (a). 
constants are 9.73e-10 and 4.5
part (b).
change of candle power is 0.90 per volts
percentage change in candle power  when voltage increase by 4% is 19.2
percentage change in candle power  when voltage falls by 4% is -16.7

Example 4.23, Page 139

In [74]:
import math

#Variable declaration
dp=1.2;#depreciation factor
uf=0.6;#utiliazation factor
l=15;# in meters
b=6;# in meters
n=20;# no. of lamps
lw=250;# mscp in watts

#Calculations
a=l*b;#arean in m^2
tl=n*lw*4*math.pi;#/total lumens
lwp=((tl*uf)/dp);#lumens reaching on the working plane
e=lwp/a;#illumination on working plane in lux

#Result
print "illumination on working plane is %.f lux"%e

illumination on working plane is 349 lux

Example 4.24, Page 139

In [75]:
#Variable declaration
ef=40;#efficiency in lumens/watt
mil=80;# minimum illumination in lumens/m^2
dp=0.8;#depreciation factor
uf=0.4;#utiliazation factor
l=100;# in meters
b=10;# in meters

#Calculations
a=l*b;#arean in m^2
tl=a*mil;#/total lumens
glr=tl/(uf*dp);#gross illumination required
twr=glr/ef;#total wattage required

#Results
print "number of lamps of 150watt rating in 2 rows are 42" 
print "total wattage is %.f watts"%twr

number of lamps of 150watt rating in 2 rows are 42
total wattage is 6250 watts

Example 4.25, Page 140

In [76]:
#Variable declaration
h=4;#in meters
wp=75;#in lux
ef=14;#efficiency in lumens/watt
dp=0.2;#depreciation factor
uf=0.5;#utiliazation factor
l=72;# in meters
b=15;# in meters

#Calculations
a=l*b;#arean in m^2
mf=1-dp;#maintenance factor
glr=(a*wp)/(uf*mf);#gross illumination required
twr=glr/ef;#total wattage required
wec=twr/80;#wattage of each lamps

#Results
print "number of lamps of 150watt rating in 2 rows are 80"
print " wattage of each lamp ",round(wec,1)," watts equivalent to 200  watts"

number of lamps of 150watt rating in 2 rows are 80
 wattage of each lamp  180.8  watts equivalent to 200  watts

Example 4.26, Page 140

In [77]:
#Variable declaration
a=30*30;#
e=75;#
uf=0.5;#
df=1-0.2;#
le=15;#efficiency

#Calculations
ph=(a*e)/(uf*df);#
W=ph/le;#
ew=300;#W
N=W/ew;#

#Results
print "total number of lamps is %.1f (say 42)"%N
print "wattage of lamps is %.f W"%W

total number of lamps is 37.5 (say 42)
wattage of lamps is 11250 W

Example 4.27, Page 141

In [78]:
#Variable declaration
h=5;# in meters
el=100;#in lux
ef=16;#efficiency in lumens/watt
dp=0.2;#depreciation factor
uf=0.4;#utiliazation factor
l=60;# in meters
b=15;# in meters

#Calculations
a=l*b;#arean in m^2
glr=(a*el)/(uf*(1-dp));#gross illumination required
n=12*3;#total no. of 
twr=glr/ef;#total wattage required
wec=twr/n;#wattage of each lamp

#Results
print "number of lamps of 150watt rating in 2 rows are %.f"%n
print "wattage of each lamp ",round(wec,1)," watts equivalent to 500  watts"

number of lamps of 150watt rating in 2 rows are 36
wattage of each lamp  488.3  watts equivalent to 500  watts

Example 4.28, Page 141

In [79]:
#Variable declaration
h=5;# in meters
el=120;#in lux
ef=40;#efficiency in lumens/watt
tw=80;#in watts
df=1.4;#depreciation factor
uf=0.5;#utiliazation factor
l=30;# in meters
b=15;# in meters

#Calculations
a=l*b;#arean in m^2
glr=(a*el*df)/(uf);#gross lumens required
twr=glr/ef;#total wattage required
nt=twr/tw;#no. of tubes required

#Results
print "total wattage required is %.f watts"%twr
print " number of tubes required is ",nt,"  equivalent to 48 tubes"

total wattage required is 3780 watts
 number of tubes required is  47.25   equivalent to 48 tubes

Example 4.29, Page 142

In [88]:
#Variable declaration
el=50;#in lux
df=1.3;#depreciation factor
uf=0.5;#utiliazation factor
l=30;# in meters
b=12;# in meters

#Calculations
a=l*b;#arean in m^2
glr=(a*el*df)/(uf);#gross lumens required
watt=[100,200,300,500,1000];
lum=[1615,3650,4700,9950,21500];#
for i in range(0,5):
    n=glr/(lum[i]);#
    print "if ",(watt[i])," watt lamps are used then number of lamps required is %.f"%n

if  100  watt lamps are used then number of lamps required is 29
if  200  watt lamps are used then number of lamps required is 13
if  300  watt lamps are used then number of lamps required is 10
if  500  watt lamps are used then number of lamps required is 5
if  1000  watt lamps are used then number of lamps required is 2

Example 4.30, Page 146

In [89]:
#Variable declaration
ef=17.4;#in mumens/watt
dp=1.2;#depreciation factor
wlf=1.3;#waste light factor
uf=0.4;#utiliazation factor
l=50;# in meters
b=16;# in meters
n=16;# no. of lamps
lw=1000;# mscp in watts

#Calculations
a=l*b;#arean in m^2
tl=n*lw*ef;#/total lumens
lwp=((tl*uf)/(wlf*dp));#lumens reaching on the working plane
e=lwp/a;#illumination on the surface in lumens/m^2

#Result
print "illumination on the surface is %.2f lumens/m^2"%e

illumination on the surface is 89.23 lumens/m^2

Example 4.31, Page 146

In [106]:
import math

#Variable declaration
watt=[300,500,1000,1500];
lum=[5000,9000,18000,27000];#
el=50;# in lux
dp=0.8;#depreciation factor
wlf=0.5;#waste light factor
uf=1.2;#utiliazation factor
l=60;# in meters
b=15;# in meters
lw=1000;# mscp in watts

#Calculations
a=l*b;#area in m^2
tl=a*el#total lumens
lwp=((tl*uf)/(wlf*dp));#lumens reaching on the working plane
n = lwp/lum[1];#number of projector required
ang=2*math.degrees(math.atan(4.5/8));#size

#Results
print "number of projectors are,= %.f"%(n+1)
print "wattage is,(W)= %.d"%watt[1]
print "beam angle is %.f (degree)"%(ang+1)
print (round(n)+1), " projectors of ",(watt[1])," watts each with beam angle of ",(round(ang+1))," degree will be required"

number of projectors are,= 16
wattage is,(W)= 500
beam angle is 60 (degree)
16.0  projectors of  500  watts each with beam angle of  60.0  degree will be required