# Chapter11 Control of DC Motors¶

## Example11.1,Pg.no.30¶

In :
import math
from math import pi
Idc=(200.0/460.0)*(1000.0)       #Rated motor current in amps
Idc=round(Idc,1)
print 'Rated motor current=Idc=',Idc,'amps' #Thus the main armature converter will be having dc side rating of 500Amps and 460 volts
Ac=552.0/1.35
Ac=round(Ac,2)
print 'Hence AC voltage for six pulse configuration=',Ac,'volts' #Hence a 3phase ,415v AC supply will be adequate for armature control
AC=230.0/1.35
AC=round(AC,2)
print 'AC rating of field converter =',AC,'V'
DCp=230.0*10.0
print 'DC power=',DCp,'Watts'
ACp=1.05*2300
print 'AC power=',ACp,'Watts'
Edca=(170+170/10)*1.35    #available voltage in volts
Edc=1.35*230
A=math.acos(Edca/Edc)*180/pi
A=round(A,2)
print 'Field converter shall be locked at an angle of A=',A,'degrees'

Rated motor current=Idc= 434.8 amps
Hence AC voltage for six pulse configuration= 408.89 volts
AC rating of field converter = 170.37 V
DC power= 2300.0 Watts
AC power= 2415.0 Watts
Field converter shall be locked at an angle of A= 35.61 degrees


## Example11.2,Pg.no.31¶

In :
import math
from math import pi
Vdc=440.0          #Rated dc voltage in volts
Edca=Vdc+Vdc/10.0  #Required voltage after allowing 10 % drop
print 'Required voltage after allowing 10 percent drop=Edca=',Edca,'volts'
Edc=1.35*415.0
C=Edca/Edc
C=round(C,2)
print 'Cosine of the locked angle=C=',C
A=math.acos(C)*180/pi      #locked angle in degrees
A=round(A,2)
print 'Converter shall be locked at an angle of A =',A,'degrees'

Required voltage after allowing 10 percent drop=Edca= 484.0 volts
Cosine of the locked angle=C= 0.86
Converter shall be locked at an angle of A = 30.68 degrees


## Example11.3,Pg.no.32¶

In :
import math
from math import pi
Edca1=230.0
N1=1000.0
N2=500.0
Eb1=210.0
Eb1=Edca1-20.0
print 'Eb1=',Eb1,'volts'
Eb2=Eb1*N2/N1
print 'Eb2=',Eb2,'volts'
V=40.0      #motor armature drop at rated load in volts
Edca2=Eb2+V
print 'Edca2=',Edca2,'volts'
C1=1.0     #cosine of the firing angle corresponding to 1000 rpm load
C2=C1*Edca2/Edca1           #cosine of the firing angle corresponding to 500 rpm load
C2=round(C2,2)
print 'Cosine of the firing angle corresponding to 500 rpm load=C2=',C2
A=math.acos(C2)*180/pi   #firing angle corresponding to 500 rpm load in degrees
A=round(A,2)
print 'firing angle corresponding to 500 rpm load A=',A,'degrees'

Eb1= 210.0 volts
Eb2= 105.0 volts
Edca2= 145.0 volts
Cosine of the firing angle corresponding to 500 rpm load=C2= 0.63
firing angle corresponding to 500 rpm load A= 50.95 degrees


## Example11.4,Pg.np.32¶

In :
import math
from math import pi,sqrt
Edca1=1.15*440.0    #Rated output voltage from the converter for rated speed of750rpm
print 'Rated output voltage from the converter= Edca1=',Edca1,'volts'
N1=750.0
N2=500.0
Edca2=Edca1*N2/N1
E2=415.0
C2=Edca2/(1.35*E2)
C2=round(C2,2)
print 'Cosine of the triggering angle=C2=',C2,
A2=C2*180/pi
A2=round(A2,2)
print 'Triggering angle=A2=',A2,'degrees'
PF2=C2*(1+math.cos(15.0*pi/180.0))/2
PF2=round(PF2,2)
print 'Powerfactor=PF2=',PF2
Id=200           #dc current in amps
I2=0.75*0.817*Id       #Current at 75 percent load in amps
RP2=sqrt(3)*E2*I2*sqrt(1-PF2**2)/1000    #Reactive power drawn at 75% load
RP2=round(RP2,2)
print 'Reactive power at 75 percent load=RP2=',RP2,'KVAR'
h=6
Wv=24.17        #maximum voltage ripple in percent
Wi=8             #maximum permissible current ripple in percent
I6=Wi*Id/100
print 'Sixth harmonic ripple current=I6=',I6,'Amps'
W=314
L=(Wv*Edca1*10)/(I6*h*W)
L=round(L,2)
print 'Inductance required in dc circuit=L=',L,'mH'
C1=Edca1/(1.35*E2)
C1=round(C1,2)
print 'Cosine of the triggering angle=C1=',C1
A1=C1*180/pi
A1=round(A1,2)
print 'Triggering angle=A1=',A1,'degrees'
PF1=C1*(1+math.cos(15*pi/180))/2
PF1=round(PF1,1)
print 'Powerfactor=PF1=',PF1
I1=0.817*Id      # Current at 75 percent load in amps
RP1=sqrt(3)*E2*I1*sqrt(1-PF1**2)/1000.0       #Reactive power drawn at 75% load
RP1=round(RP1,1)
print 'Reactive power at 75 percent load=RP1=',RP1,'KVAR'

Rated output voltage from the converter= Edca1= 506.0 volts
Cosine of the triggering angle=C2= 0.6 Triggering angle=A2= 34.38 degrees
Powerfactor=PF2= 0.59
Reactive power at 75 percent load=RP2= 71.12 KVAR
Sixth harmonic ripple current=I6= 16 Amps
Inductance required in dc circuit=L= 4.06 mH
Cosine of the triggering angle=C1= 0.9
Triggering angle=A1= 51.57 degrees
Powerfactor=PF1= 0.9
Reactive power at 75 percent load=RP1= 51.2 KVAR


## Example11.5,Pg.no.34¶

In :
import math
from math import pi,sqrt
Edca=460
E2=415
C=Edca/(1.35*E2)
C=round(C,2)
print 'Cosine of the triggering angle=C=',C
A=C*180/pi
A=round(A,1)
print 'Triggering angle=A=',A,'degrees'
Edca10=0.1*460
C10=Edca10/(1.35*E2)
C10=round(C10,1)
print 'Cosine of the triggering angle=C10=',C10
A10=C10*180/pi
A10=round(A10,1)
print 'Triggering angle=A10=',A10,'degrees'
Id=10**5/Edca        #dc current in amps
I2=0.817*Id          #Current at rated speed in amps
AP=sqrt(3)*E2*I2*C/1000
AP=round(AP,1)
print 'Active power drawn from the system at rated speed=AP=',AP,'KW'
RP=sqrt(3)*E2*I2*sqrt(1-C**2)/1000    #Reactive power drawn from the system
RP=round(RP,1)
print 'Reactive power drawn from the system=RP=',RP,'KVAR'
AP10=sqrt(3)*E2*I2*C10/1000
AP10=round(AP10,1)
print 'Activepower drawn from thesystem at 10 percentrated speed=AP10=',AP10,'KW'
RP10=sqrt(3)*E2*I2*sqrt(1-C10**2)/1000       #Reactive power drawn from the system
RP10=round(RP10,1)
print 'Reactive power drawn from the system=RP10=',RP10,'KVAR'
P=RP10/RP
P=round(P,1)
print 'P=',P
print 'Thus reactive power has increased by 74.589 percent due to reduction in motor speed'

Cosine of the triggering angle=C= 0.82
Triggering angle=A= 47.0 degrees
Cosine of the triggering angle=C10= 0.1
Triggering angle=A10= 5.7 degrees
Active power drawn from the system at rated speed=AP= 104.5 KW
Reactive power drawn from the system=RP= 72.9 KVAR
Activepower drawn from thesystem at 10 percentrated speed=AP10= 12.7 KW
Reactive power drawn from the system=RP10= 126.8 KVAR
P= 1.7
Thus reactive power has increased by 74.589 percent due to reduction in motor speed


## Example11.6,Pg.no.35¶

In :
import math
#given
Rp=72.79
Rp1=0.1*72.79     #Reactive power at rated speed and 10 percent load
Rp2=0.1*127.08    #reactive power at 10 percent speed and 10 percent load
print 'Reactive power at rated speed and rated load =72.79KVAR'
print 'Reactive power at rated speed and 10 percent load=',Rp1,'KVAR'
print 'reactive power at 10 percent speed and 10 percent load=',Rp2,'KVAR'

Reactive power at rated speed and rated load =72.79KVAR
Reactive power at rated speed and 10 percent load= 7.279 KVAR
reactive power at 10 percent speed and 10 percent load= 12.708 KVAR


## Example11.7,Pg.no.35¶

In :
import math
from math import pi
N1=500
N2=400
Eb1=410
Eb2=Eb1*N2/N1
print 'Eb2=',Eb2,'volts'
V=440             #operating voltage of dc motor in volts
P=100             #input power of dc motor in KW
Ia=P*1000/V
Ia=round(Ia,1)
print 'Ia=',Ia,'Amps'
Ra=(V-Eb1)/Ia;
Ra=round(Ra,1)
print 'Ra=',Ra,'ohms'
E2=415
Edca=Eb2+(0.75*Ia*Ra)       #terminal voltage of dc motor at 500 rpm and 75% load
Edca=round(Edca,1)
print 'Terminal voltage of dc motor at 500 rpm and 75 percent load=Edca=',Edca,'volts'
C=Edca/(1.35*E2)         #cosine of the triggering angle of the converter
C=round(C,1)
print 'Cosine of the triggering angle of the converter=C2=',C
A=math.acos(C)*180/pi            #triggering angle of the converter in degrees
A=round(A,1)
print 'triggering angle of the converter A=',A,'degrees'

Eb2= 328 volts
Ia= 227.0 Amps
Ra= 0.1 ohms
Terminal voltage of dc motor at 500 rpm and 75 percent load=Edca= 345.0 volts
Cosine of the triggering angle of the converter=C2= 0.6
triggering angle of the converter A= 53.1 degrees