Chapter 7: DC Generators

Example 7.1: Page 114:

In [1]:
from __future__ import division
import math

#  given  data:
p=8;  #  number  of  poles
a1=p;  #  in  lap  winding
a2=2;  #  in  wave  winding
fi=15*10**-3;#  in  wb
N=500;#  rev/min
Z=800;#  number  of  conductors  on  armature

#calculations:
emf1=(fi*Z*N*p)/(60*a1)#  when  the  armature  is  lap  wound
emf2=(fi*Z*N*p)/(60*a2)#  when  the  armature  is  wave  wound

#Results
print "when  the  armature  is  lap  wound, emf(V)  =  ",emf1
print "when  the  armature  is  wave wound, emf(V)  =  ",emf2
when  the  armature  is  lap  wound, emf(V)  =   100.0
when  the  armature  is  wave wound, emf(V)  =   400.0

Example 7.2: Page 119:

In [2]:
from __future__ import division
import math

#given  data:
Vt=200#  terminal  voltage  in  volts
Rsh=100;#shunt  fieldresistance    in  ohm
Ra=0.1;#  armature  resistance  in  ohm
l=60;#  number  of  lamps
w=40  #  in  watt
N=4;  #  number  of  poles

#calculations:
total_l=l*w#  in  watt
Il=total_l/Vt#  load  current
Ish=Vt/Rsh#  shunt  field  current
Ia=Il+Ish;
I=Ia/N;
Va=Ia*Ra#armature  voltage  drop  
Vb=1+1;#  brush  contact  drop  for  2  pair  of  poles
E=Vt+Va+Vb;

#Results
print  "(a)armature  current,Ia(A)  =  ",Ia
print  "(b)current  per  path  in  a  armature,I(A)  =",I
print  "(c)emf,E(Volts)  =  ",E
(a)armature  current,Ia(A)  =   14.0
(b)current  per  path  in  a  armature,I(A)  = 3.5
(c)emf,E(Volts)  =   203.4

Example 7.3: Page 119:

In [3]:
from __future__ import division
import math

#  given  data:
W=10  #  output  of  the  generator  in  k-w
V=250;#  voltage  in  volts
R=0.07;#  in  ohm
Rsh=63.2;#  shunt  resistance  in  ohm
Ra=0.05;#  armature  resistance  in  ohm
Vb=2;#  brush  contact  drop

#calculations:
Il=(W*1000)/V#  load  current  in  A
Vf=Il*R#  voltage  drop  in  feeder
Vt=V+Vf;
Ish=Vt/Rsh;
Ia=Il+Ish;
Vd=Ia*Ra#  voltage  drop  in  the  armature
E=Vt+Vd+Vb;
#Results
print  "(a)terminal  voltage,Vt(V)  = ",Vt 
print  "(b)emf,E(V)  = ", E
(a)terminal  voltage,Vt(V)  =  252.8
(b)emf,E(V)  =  257.0

Example 7.4: page 129:

In [4]:
from __future__ import division
import math

# given  data:
W=20000#  in  watt
V=200;#  in  volts
R=0.08;#  in  ohm
Rs=0.02;#  series  field  resistance  in  ohm
Rsh=42;#  shunt  ield  resistance  in  ohm
Ra=0.04;#    armature  resistance  in  ohm
iron_losses=309.5;#  iron  and  friction  losses

#calculations:
I=W/V;#  in  A
Vf=I*R;
Vs=I*Rs;
V1=Vf+Vs;#  voltage  drop  of  feeder  and  series  field
Vg=V+V1;
Ish=Vg/Rsh#  shunt  field  current
Ia=I+Ish;
Vd=Ia*Ra;
emf=Vg+Vd;
Ed=emf*Ia#  in  watt
copper_losses=Ed-W;
mech_in=W+copper_losses+iron_losses;
Bhp=mech_in/735.5;
efficiency=(W/mech_in)*100;

#Results
print  "(a)terminal  voltage,Vg(V)  = ",Vg
print  "(b)emf(V)  =",emf
print  "(c)copper  losses(Watt)  =  ",copper_losses
print  "(d)bhp  metric  of  the  primemover,Bhp  =  ",Bhp 
print  "(e)efficiency(%)  =  ",round(efficiency,1)
(a)terminal  voltage,Vg(V)  =  210.0
(b)emf(V)  = 214.2
(c)copper  losses(Watt)  =   2491.0
(d)bhp  metric  of  the  primemover,Bhp  =   31.0
(e)efficiency(%)  =   87.7

Example 7.5: page 129:

In [5]:
from __future__ import division
import math

#  given  data:
n=3    #  number  of  motors
n1=4  #  number  of  parallel  path  in  winding
i=30;#current  in  A
Bhp=65#  in  hp
Rsh=44;#  shunt  field  resistance
Ra=0.08;#  armature  resistance  in  ohm
V=440;#  voltage  in  V
Vb=2  #  we  know  ,  brush  contact  drops

#calculations:
I=i*n#  current  taken  by  three  motors
Ish=V/Rsh#  shunt  field  current
Ia=I+Ish;
I1=Ia/n1#  current  in  each  path
Va=Ia*Ra;#  armature  drop
E=V+Va+Vb;
E_power=E*Ia;
W=V*I#  in  watt
M_power=Bhp*746#  assume  Bhp=746  W
Copper_losses=E_power-W;
S_loses=M_power-E_power;
eta_e=(W/E_power)*100;
eta_c=(W/M_power)*100;
eta_m=(E_power/M_power)*100;

#Results
print  "(a)total  armature  current,Ia(A)  =",Ia
print  "(b)current  in  each  path,I1(A)  = ",I1
print  "(c)emf,E(V)  =  ",E  #  answer  is  wrong  in  a  book 
print  "(d)electrical  power  developed  in  watt  =  ",E_power  #  answer  is  wrong  in  a  book 
print  "(e)copper  losses  (W)  =  ",Copper_losses
print  "(f)stray  losses(W)  =  ",S_loses
print  "(g1)electrical  efficiency,eta_e(%)  =  ",eta_e
print  "(g2)commercial  efficiency,eta_c(%)  =  ",round(eta_c,2)
print  "(g3)mechanical  efficiency,eta_m(%)  =  ",round(eta_m,1)
(a)total  armature  current,Ia(A)  = 100.0
(b)current  in  each  path,I1(A)  =  25.0
(c)emf,E(V)  =   450.0
(d)electrical  power  developed  in  watt  =   45000.0
(e)copper  losses  (W)  =   5400.0
(f)stray  losses(W)  =   3490.0
(g1)electrical  efficiency,eta_e(%)  =   88.0
(g2)commercial  efficiency,eta_c(%)  =   81.67
(g3)mechanical  efficiency,eta_m(%)  =   92.8