# CHAPTER 7 - PRINCIPLE AND CONSTRUCTION OF DC MACHINES¶

## Example E3 - Pg 130¶

In [1]:
#Caption:Find effect of change in connection on voltage,current and output
#Exa:7.3
P=50000.#Power of generator(in watt)
V_b1=230.#Voltage of generator(in volts)
p=4.#Number of poles
a=4.#Number of parallel paths for lap winding
b=2.#Number of parallel paths for wave winding
C=268.#Number of conductors with LAP winding
t=2.#Two turns coils are used
c=t*2.#Conductors per slot
n=C/c
I_1=P/(V_b1)
V_b2=V_b1*b
I_2=P/(V_b2)
print '%s %.f %.1f' %('voltage(in volts) and Current(in A) for LAP winding=',V_b1,I_1)
print '%s %.f %.1f' %('voltage(in volts) and Current(in A) for WAVE winding=',V_b2,I_2)
print '%s %.f' %('Output is same for both connections(in watts)=',P)

voltage(in volts) and Current(in A) for LAP winding= 230 217.4
voltage(in volts) and Current(in A) for WAVE winding= 460 108.7
Output is same for both connections(in watts)= 50000


## Example E6 - Pg 132¶

In [2]:
#Caption: Select a two circuit armature winding for a d.c machine
#Exa:7.6
import math
p=4.#Number of poles
n=1000.#Speed of d.c. machine(in r.p.m)
V=400.#Voltage of d.cmachine(in volts)
B=0.04#Flux per pole(in weber)
s_1=41.#Slot 1
s_2=45.#Slot 2
s_3=51.#Slot 3
a=2.#Number of parallel paths
Z=(V*60.*a)/(B*n*p)
Z_c=Z/a
Y=(s_3+1.)/(p/2.)
t=3.#turns per coil
c=t*a
z=s_3*c
print '%s %.f' %('slots=',z)
print '%s %.f' %('turn coils=',c)
print '%s %.f' %('coils sides per slot=',t)
print '%s %.f' %('total number of conductors=',s_3)

slots= 306
turn coils= 6
coils sides per slot= 3
total number of conductors= 51


## Example E7 - Pg 132¶

In [1]:
#Caption: Find (a)e.m.f generated at 750r.p.m for lap wound (b)e.m.f generated at 600r.p.m for wave wound (c)Speed to be driven for 400V for same flux per pole
#Exa:7.7
import math
p=4.#Number of poles
B=0.04#Flux per pole(in weber)
c=740.#Number of conductors for lap connection
n_1=750.#Speed of machine(in r.p.m)
n_2=600.#Speed of machine(in r.p.m)
V=400.#Voltage of machine(in volts)
a=4.#Number of parallel paths for lap winding
b=2.#Number of parallel paths for wave winding
E=(B*c*n_1*p)/(60.*a)
print '%s %.f' %('(a)E.M.F generated at 750r.p.m for lap wound(in volts)=',E)
E_1=(B*c*n_2*p)/(60.*b)
print '%s %.f' %('(b)E.M.F generated at 600r.p.m for wavewound(in volts)=',E_1)
n=(V*60.*b)/(B*c*p)
print '%s %.1f' %('(c)Speed of machine(in r.p.m)=',n)

(a)E.M.F generated at 750r.p.m for lap wound(in volts)= 370
(b)E.M.F generated at 600r.p.m for wavewound(in volts)= 592
(c)Speed of machine(in r.p.m)= 405.4


## Example E8 - Pg 139¶

In [2]:
#Caption: Calculate (a)Total armature current (b)Current per armature path (c)Generated e.m.f
#Exa:7.8
import math
p=4.#Number of poles
P=4000.#Power of generator(in watts)
V=230.#Voltage of generator(in volts)
r_f=115.#Field resistance(in ohms)
r_a=0.1#Armature resistance(in ohms)
a=p#number of parallel paths
i_f=V/r_f
i_l=P/V
I_a=i_l+i_f
print '%s %.1f' %('(a)Armature current(in A)=',I_a)
i=I_a/p
print '%s %.2f' %('(b)Current per armature path(in A)=',i)
E=V+(I_a*r_a)
print '%s %.2f' %('(c)E.M.F generated(in volts)=',E)

(a)Armature current(in A)= 19.4
(b)Current per armature path(in A)= 4.85
(c)E.M.F generated(in volts)= 231.94


## Example E9 - Pg 139¶

In [3]:
#Caption:Find the speed at which it will run as a motor
#Exa:7.9
import math
P_g=110000.#Power of generator(in watts)
n=402.#Speed of generator(in r.p.m)
V=220.#Voltage of busbars(in volts)
P_m=10900.#Power of motor(in watt)
r_a=0.025#Armature resistance(in ohms)
r_f=55.#Field resistance(in ohms)
v_b=1.#Voltage drop at each brush(in volt)
i_l=P_g/V
i_f=V/r_f
I_a=i_l+i_f
V_a=I_a*r_a
E=V+V_a+(2*v_b)
I_1=P_m/V
i_a=I_1-i_f
v_a=i_a*r_a
E_b=V-(i_a*r_a)-(2.*v_b)
N_m=(n*E_b)/E
print '%s %.f' %('Speed at which generator will run as motor is(in r.p.m)=',N_m)

Speed at which generator will run as motor is(in r.p.m)= 372


## Example E10 - Pg 140¶

In [4]:
#Caption: Calculate the speed of the motor when it is loaded and takes 60A from the mains
#Exa:7.10
V=230.#Voltage of motor(in volts)
n=800.#Speedof motor(in r.p.m)
i=5.#Current taken by motor(in A)
r_a=0.25#Armature resistance(in ohms)
r_f=230.#field resistance(in ohms)
i_f=V/r_f
i_a=i-i_f
E_b1=V-(i_a*r_a)
i_al=i_l-i_f
E_b2=V-(i_al*r_a)
N=(n*E_b2)/E_b1
print '%s %.f' %('Required speed of motor(in r.p.m) is=',N)

Required speed of motor(in r.p.m) is= 752


## Example E11 - Pg 141¶

In [5]:
#Caption: Calculate Power and torque developed
#Exa:7.11
import math
p=4.#Number of poles
d=20.#Diameter of armature(in cm)
l=25.#Core length(in cm)
c=300.#Number of conductors
i_a=50.#Armature current(in A)
B=0.3#Average flux density(in weber/m**2)
n=1000.#Speedofmotor(in r.p.m)
T=(B*(l/100.)*(i_a/p)*c*(d/100.)*(1./2.))
s=(2.*math.pi*n)/(60.)
P=(T*s)/1000.
print '%s %.3f %s %.2f' %('Torque(in Nm) developed is=',T,'\nPower(in KW)=',P)

Torque(in Nm) developed is= 28.125
Power(in KW)= 2.95


## Example E12 - Pg 145¶

In [6]:
#Caption: Determineper pole (a)Number of cross magnetising ampereturns,and (b)Demagnetising ampereturns
#Exa:7.12
I=100.#Current(in A)
c=500.#Armature conductors
p=6.#Poles
a=2.#Wave wound
e=(10.*p)/2.
F_d=(c*I*2.*e)/(2.*a*p*180.)
print '%s %.f' %('(a)Number of cross magnetising ampereturns=',F_d)
F_c=(c*I)*(1.-((2.*e)/180.))/(2.*a*p)
print '%s %.f' %('(b)Demagnetising ampereturns=',F_c)

(a)Number of cross magnetising ampereturns= 694
(b)Demagnetising ampereturns= 1389


## Example E13 - Pg 147¶

In [7]:
#Caption:Find the time of Commutation
#Exa:7.13
import math
p=4.#Number of poles
n=600.#Speed of generator(in r.p.m)
d=0.4#Diameter of commutator(in m)
c=243.#Number ofcommutator segments
c_s=3.#Coil sides per layer
w=12.5#Width of brush(in mm)
W=0.6#Width of mica between commutator segments
W_c=(math.pi*d*1000.)/(c)
D=w-W+(2.*W_c)
V_c=(math.pi*d*n)/60.
T=D/V_c*(10.**(-3.))
print '%s %.5f' %('Time of commutation(in sec)=',T)

Time of commutation(in sec)= 0.00177


## Example E14 - Pg 150¶

In [8]:
#Caption: Find average reactance voltage produce due to commutation
#Exa:7.14
p=4.#Number of poles
I=300.#Current delievered by generator on full load(in A)
L=0.02*(10.**(-3.))#Inductance of each coil(in mH)
a=2.#Wavw wound
i=I/2.#Current in conductors in each path(in A)
T_c=0.00174#Time of commutation(in sec)
E_r=(2.*L*i)/T_c
print '%s %.2f' %('Average reactance voltage(in volts)=',E_r)

Average reactance voltage(in volts)= 3.45


## Example E15 - Pg 150¶

In [9]:
#Caption: Calculate the number of turns needed on each commutating pole
#Exa:7.15
import math
p=4.#Number of poles
P=125000.#Power delievered by generator(in watts)
V=230.#Voltage of generator(in volts)
z=240.#Armature conductors
B=0.3#Flux density under the interpolar gap(in weber/m**2)
g=0.01#Interpolar airgap(in m)
a=p#LAP connection
I_a=P/V
F_a=(z*I_a)/(2.*a*p)
A=(B*g)/(4.*math.pi*(10.**(-7.)))
A_t=A+F_a
T=A_t/I_a
print '%s %.2f' %('The number of turns on each commutating pole=',T)

The number of turns on each commutating pole= 11.89