# Chapter8 Harmonic and PowerFactor with the converter system¶

## Example8.1,Pg.no.22¶

In :
import math
from math import sqrt
I5=0.2                                      #amplitude of 5th harmonic current in Kilo Amperes
Vp= 11/(sqrt(3))                            #Input supply phase voltage in Kilo Volts
P=5                                         #supply power per phase of filter in MVAR
Pc=P+((Vp**2*I5**2)/(5*P))                  #AC Converter power per phase in MVAR
Pc=round(Pc,2)
print 'value of AC converter power=Pc=',Pc,'MVAR'
C=(Pc*10**3*3) /(11**2*314)                 #capacitance of the ShuntFilter in milliFarad
C=round(C,2)
print 'value of the capacitance of shunt filter=C =',C,'millifarads'
L=(106*10**6)/(400*4*25*250*3.14**2)        #inductance of filter in mHenry
L=round(L,2)
print 'Inductance of filter=L=',L,'milliHenry'
Q=50                                        #value of Q
W5=2*3.14*5*50                              #angular frequency of 5th harmonic
R=(W5*L)/Q                                  #Resistance of filter in milliOhms
R=round(R,2)
print 'Resistance of filter=R=',R,'milliOhms'

value of AC converter power=Pc= 5.06 MVAR
value of the capacitance of shunt filter=C = 0.4 millifarads
Inductance of filter=L= 1.08 milliHenry
Resistance of filter=R= 33.91 milliOhms


## Example8.2,Pg.no.23¶

In :
import math
print 'For six pulse converter most effective harmonic is 6th and for worst case a=90 degree'
Wv=24.1                          #voltage ripple in percentage
print 'voltage ripple=Wv=',Wv,'percent'
Id=200.0
I6=(5*Id)/100                    #Harmonic current for 6th harmonic in amp
I6=round(I6,1)
print 'Harmonic current for 6th harmonic=I6=',I6,'amp'
Edc=460.0                          #dc voltage in volts
W=2*3.14*50
La=1.0                             #inductance already present in the circuit in milliHenry
L=((Wv*Edc*10)/(I6*6*W))-La      #additional inductance required in milliHenry
L=5.93-1.0
L=round(L,2)

For six pulse converter most effective harmonic is 6th and for worst case a=90 degree
voltage ripple=Wv= 24.1 percent
Harmonic current for 6th harmonic=I6= 10.0 amp


## Example8.3,Pg.no.24¶

In :
import math
from math import sqrt,pi
Id=200         #rated dc current in amperes
I2=0.817*Id    #AC line current in amperes
print 'AC line current of the thyristor=I2=',I2,'amperes'
E2=415         #AC line voltage in volts
Edc=400        #dc terminal voltage in volts
Xt=0.04*E2/I2   #effective reactance of the thyristor in ohms
Xt=round(Xt,2)
print 'effective reactance of the thyristor=Xt=',Xt,'ohms'
C=1-((Id*Xt)/(E2*sqrt(3)))                   #cosine value of the commutational angle
C=round(C,2)
print 'cosine value of the commutational angle=C=',C
CA=math.acos(C)*180/pi
CA=round(CA,2)
print 'commutation angle=CA=',CA,'degrees'
F=Edc/(1.35*E2*(1+C)/2)                #cosine value of the firing angle
F=round(F,2)
print 'cosine value of the firing value=F=',F
FA=math.acos(F)*180/pi
FA=round(FA,2)
print 'firing angle=FA=',FA,'degrees'
I2=0.817*Id                            #AC line current in amps
print 'AC line current=I2=',I2,'amps'
Ied=0.58*Id                            #current through each device in amps
print 'Current through each device=Ied=',Ied,'amps'
PF=F*(1+C)/2                           #power factor
PF=round(PF,2)
print 'power factor=PF=',PF
AP=sqrt(3)*E2*I2*PF                    #active power drawn from the mains in Watts
AP=round(AP,2)
print  'active power drawn from the mains=AP=',AP,'Watts'
RP=sqrt(3)*E2*I2*sqrt(1-PF**2)           #reactive power in VAR
RP=round(RP,2)
print 'Reactive power drawn=RP=',RP,'VAR'

AC line current of the thyristor=I2= 163.4 amperes
effective reactance of the thyristor=Xt= 0.1 ohms
cosine value of the commutational angle=C= 0.97
commutation angle=CA= 14.07 degrees
cosine value of the firing value=F= 0.72
firing angle=FA= 43.95 degrees
AC line current=I2= 163.4 amps
Current through each device=Ied= 116.0 amps
power factor=PF= 0.71
active power drawn from the mains=AP= 83390.99 Watts
Reactive power drawn=RP= 82709.96 VAR


## Example8.4,Pg.no.25¶

In :
import math
from math import pi,sqrt
Id=100           #rated dc current in amperes
I2=0.817*Id      #AC line current in amperes
I2=round(I2,2)
print 'AC line current of the thyristor=I2=',I2,'amperes'
E2=230           #AC line voltage in volts
Edc=200          #dc terminal voltage in volts
PF=math.cos(pi/4)*(1+math.cos(pi/10))/2        #power factor
PF=round(PF,2)
print 'power factor=PF=',PF
RP=sqrt(3)*E2*I2*sqrt(1-PF**2)           #reactive power to be supplied by shunt compensator in VAR
RP=round(RP,2)
print  'Reactive power to be supplied by shunt compensator=RP=',RP,'VAR'

AC line current of the thyristor=I2= 81.7 amperes
power factor=PF= 0.69
Reactive power to be supplied by shunt compensator=RP= 23557.8 VAR


## Example8.5,Pg.no.25¶

In :
import math
from math import sqrt
I11=400/11              #amplitude of 11th harmonic current in Amperes
V1= 11/(sqrt(3))        #Input supply phase voltage in Kilo Volts
P=7                      #supply power per phase of f i l t e r in MVAR/
Pc=P+((V1**2*I11**2*10**-3)/(11*P))         #AC Converter MVAR rating of the capacitor
Pc=round(Pc,2)
print 'value of MVAR rating of the capacitor=Pc=',Pc,'MVAR'
W=2*3.14*50
C=(Pc*10**6)/(V1**2*W)                    #capacitance of the ShuntFilter in microFarad
C=round(C,2)
print 'value of the capacitance of shunt filter=C =',C,'microfarads'
W11=11*W
L=10**8/(C*W11**2)                        #inductance of filter in mHenry
L=round(L,2)
print 'Inductance of filter=L=',L,'milliHenry'
Q=35                                      #value of Q
R=(W11*L)/Q                               #Resistance of filter in milliOhms
R=round(R,2)
print 'Resistance of filter=R=',R,'milliOhms'

value of MVAR rating of the capacitor=Pc= 7.68 MVAR
value of the capacitance of shunt filter=C = 606.41 microfarads
Inductance of filter=L= 0.01 milliHenry
Resistance of filter=R= 0.99 milliOhms


## Example8.6,Pg.no.26¶

In :
import math
print 'For six pulse converter most effective harmonic is 6th and for worst case a=90 degree'
h=6.0
Wv=24.1            #voltage ripple in percentage
print 'voltage ripple=Wv=',Wv,'percent'
Edc=460.0            #dc voltage in volts
W=2*3.14*50
Ldc=6.0              #total dc circuit inductance in milliHenry
I6=Wv*Edc*10/(Ldc*h*W)             #Harmonic current for 6th harmonic in amp
I6=round(I6,2)
print 'Harmonic current for 6th harmonic=I6=',I6,'amp'
Id=300.0
Wi=100*I6/Id             #maximum value of current ripple in percentage
Wi=round(Wi,2)
print 'max. value of current ripple=Wi=',Wi,'percent'

For six pulse converter most effective harmonic is 6th and for worst case a=90 degree
voltage ripple=Wv= 24.1 percent
Harmonic current for 6th harmonic=I6= 9.81 amp
max. value of current ripple=Wi= 3.27 percent


## Example8.7,Pg.no.27¶

In :
import math
from math import pi,sqrt
#given
A=pi/4
h=6
#calculations
Wv=sqrt(2)*sqrt(h**2-math.cos(A)**2*(h**2-1))*100/(h**2-1)
Wv=round(Wv,2)
print 'voltage ripple of the 6th harmonic=Wv=',Wv,'percent'
print 'For six pulse converter most effective harmonic is 6th and for worst case A=90degrees'
A=pi/2
Wv6=sqrt(2)*sqrt(h**2-math.cos(A)**2*(h**2-1))*100/(h**2-1)   #maximum voltage ripple in percentage
Wv6=round(Wv6,2)
print 'maximum voltage ripple=Wv6=',Wv6,'percent'
A=pi/4
h=12
Wv=sqrt(2)*sqrt(h**2-math.cos(A)**2*(h**2-1))*100/(h**2-1)
Wv=round(Wv,2)
print 'voltage ripple of the 12th harmonic=Wv=',Wv,'percent'
A=pi/2
Wv12=sqrt(2)*sqrt(h**2-math.cos(A)**2*(h**2-1))*100/(h**2-1) #maximum voltage ripple in percentage
Wv12=round(Wv12,2)
print 'maximum voltage ripple=Wv12=',Wv12,'percent'
PR=(Wv6-Wv12)*100/Wv6                                  #percentage reduction in max. voltage ripple
PR=round(PR,2)
print 'percentage reduction in max. voltage ripple=PR=',PR,'percent'

voltage ripple of the 6th harmonic=Wv= 17.38 percent
For six pulse converter most effective harmonic is 6th and for worst case A=90degrees
maximum voltage ripple=Wv6= 24.24 percent
voltage ripple of the 12th harmonic=Wv= 8.42 percent
maximum voltage ripple=Wv12= 11.87 percent
percentage reduction in max. voltage ripple=PR= 51.03 percent


## Example8.8,Pg.no.28¶

In :
import math
from math import sqrt
#givem
Wv=18.6
h=6
#calculations
C=sqrt(14.68/35)    #cosine of triggering angle
C=round(C,2)
print 'cosine of triggering angle=C=',C
A=math.acos(C)*180/pi
A=round(A,2)
print 'triggering angle of the device=A=',A,'degrees'

cosine of triggering angle=C= 0.65
triggering angle of the device=A= 49.46 degrees


## Example8.9,Pg.no.29¶

In :
import math
from math import pi,sqrt
E2=415              #AC line voltage in volts
Edc=380             #dc terminal voltage in volts
C=1.1*Edc/(1.35*E2)
C=round(C,2)
print 'cosine of the triggering angle=C=',C
A=math.acos(C)*180/pi
A=round(A,2)
print 'triggering angle of the device=A=',A,'degrees'
PF=C*(1+math.cos(pi/12))/2            #power factor
PF=round(PF,2)
print 'power factor=PF=',PF
Id=200
I2=0.817*Id
RP=sqrt(3)*E2*I2*sqrt(1-PF**2)/1000         #reactive power to be supplied by shunt compensator in KVAR
RP=round(RP,2)
print 'Reactive power to be supplied by shuntcompensator=RP=',RP,'KVAR'

cosine of the triggering angle=C= 0.75
triggering angle of the device=A= 41.41 degrees
power factor=PF= 0.74
Reactive power to be supplied by shuntcompensator=RP= 79.0 KVAR