# Chapter 5:Dc Motor Drives¶

## Example No:5.1,Page No:63¶

In [10]:
import math
from __future__ import division
from sympy import Symbol

#variable declaration
#motor ratings
V1=200     #rated voltage
Ia1=10.5   #rated current
N1=2000    #speed in rpm
Ra=0.5     #armature resistance
Rs=400     #field resistance
V2=175     #drop in source voltage

#calculation
flux1 = Symbol('flux1')
flux2=V2/V1*flux1
E1=V1-Ia1*Ra
E2=V2-Ia2*Ra
N2=(E2/E1)*(flux1/flux2)*N1

#results
#answer in the book is wrong due to accuracy
print"\nmotor speed is:N2=",round(N2,1),"rpm"

motor speed is:N2= 1983.5 rpm


## Example No:5.2,Page No:63¶

In [11]:
import math
from __future__ import division
from sympy import Symbol

#variable declaration
V1=220      #rated voltage
Ia1=100     #rated current
N1=1000     #rated speed in rpm clockwise
Ra=0.05     #armature resistance
Rs=0.05     #field resistance

#calculation
#turns is reduced to 80% then flux is also reduced by the same value and hence current is also reduced
Ke = Symbol('Ke')
Ia2 = Symbol('Ia2')
T1=Ke*Ia1**2      #flux is directly proportional to current Ia
T2=Ke*0.8*Ia2**2  #flux is directly proportional to current Ia
Ia2=-Ia1/math.sqrt(0.8)   #since T1=T2 and the direction is opposite

E1=V1-Ia1*(Ra+Rs)

Rs=.8*Rs       #Rs=80% of the field resistance 0.05ohm since the flux is reduced to 80%
E2=-(V1+Ia2*(Ra+Rs))

N2=(E2/E1)*(Ia1/Ia2)*(N1/0.8)   #since E=Kn*flux*N

#results
print"\nmotor speed is:N2=",round(N2,1),"rpm"

motor speed is:N2= 1117.7 rpm


## Example No:5.3,Page No:70¶

In [161]:
import math
from __future__ import division

#variable declaration
#motor ratings
V1=220     #rated voltage
Ia1=200    #rated current
Ra=0.06    #armature resistance
Rb=0.04    #internal resistance of the variable source
N1=800     #speed in rpm
N2=600     #speed when motor is operatingin regenerative braking

#Calculation
Ia2=0.8*Ia1        #motor is opereting in regenerative braking at 80% of Ia1
E1=V1-Ia1*Ra       #back emf at rated operation
E2=(N2/N1)*E1      #back emf at the given speed N2
V2=E2-Ia2*(Ra+Rb)  #internal voltage of thevariable source

#results
print"\n internal voltage of thevariable source:",round(V2),"V"

 internal voltage of thevariable source: 140.0 V


## Example No:5.4,Page No:70¶

In [12]:
import math
from __future__ import division
from sympy import Symbol

#variable declaration
#The ratings of the motor is same as that of Ex-5.2
V1=220      #rated voltage
Ia1=100     #rated current
N1=1000     #speed in rpm clockwise
N2=800      #given speed during the dynamic braking
Ra=0.05     #armature resistance
Rs=0.05     #field resistance

#calculation
T1 = Symbol('T1')
T2 = 2*T1   #dynamic torque is twice the rated torque
Ia2=Ia1*math.sqrt(T2/T1)   #since T=Kf*Ia**2
E1=V1-Ia1*(Ra+Rs)
E2=(Ia2/Ia1)*(N2/N1)*E1    #since E=Ke*Ia*N
Rb=E2/Ia2-(Ra+Rs)          #since E2=Ia2(Rb+Ra+Rs)  during braking

#results
print"\n braking current Ia2:",round(Ia2,1),"A"
print"\n required braking resistance Rb:",round(Rb,2),"ohm"

 braking current Ia2: 141.4 A

required braking resistance Rb: 1.58 ohm


## Example No:5.5,Page No:70¶

In [4]:
import math
from __future__ import division
from array import array
import numpy
import matplotlib.pyplot as plt
%matplotlib inline

#variable declaration
#ratings of the DC shunt motor which operated under dynamic braking
Rb=1      #braking resisance
Ra=0.04   #armature resistance
Rf=10     #field resistance

#magnetisation curve at N1
N1=600    #speed in rpm
If=[2.5,5,7.5,10,12.5,15,17.5,20,22.5,25]    #field current
E =[25,50,73.5,90,102.5,110,116,121,125,129] #back emf

#calculation
print"Field current   If:",If,"A"
x=(Rb+Rf)/Rb
Ia = [If * x for If in If]    #armature current
Wm=2*math.pi*N1/60
Ke_flux=[E / Wm for E in E]    #Ke*flux=constant
Ke_flux=[round(Ke_flux,3) for Ke_flux in Ke_flux]

Ke_flux=numpy.array(Ke_flux)
Ia=numpy.array(Ia)
T=numpy.array(Ke_flux)*numpy.array(Ia)   #torque
print"\nKe_flux :",Ke_flux
T=[round(T,1) for T in T]
print"\nTorque  :",T,"N-m"

#results
#plotting the values of Ke*flux vs If
If=[2.5,5,7.5,10,12.5,15,17.5,20,22.5,25]    #field current
plt.subplot(2,1,1)
plt.plot(If,Ke_flux,'y')
plt.xlabel('field current $I_f$')
plt.ylabel('$Ke*flux$')
plt.title('$If vs Ke*flux$')
plt.grid(True)

#plotting the values of  T vs If
If=[2.5,5,7.5,10,12.5,15,17.5,20,22.5,25]    #field current
plt.subplot(2,1,2)
plt.plot(T,If)
plt.xlabel('Torque $T$')
plt.ylabel('field current $I_f$')
plt.title('$T vs If$')
plt.grid()
plt.tight_layout()
plt.show()

print"\nFrom the plot we can see that when the torque is 400 N-m, "
print"the field current is If=19.3 A, and Ke*flux=1.898 when If=19.3 A"
T=400          # braking torque
If=19.13       # field current
Ke_flux=1.898  # Ke*flux
Ia=x*If
E=If*Rf+Ia*Ra    #since E=V+Ia*Ra
N2=(E/Ke_flux)*(60/(2*math.pi))  #required speed
print"Hence the required speed in is :",round(N2),"rpm"

Field current   If: [2.5, 5, 7.5, 10, 12.5, 15, 17.5, 20, 22.5, 25] A

Ke_flux : [ 0.398  0.796  1.17   1.432  1.631  1.751  1.846  1.926  1.989  2.053]

Torque  : [10.9, 43.8, 96.5, 157.5, 224.3, 288.9, 355.4, 423.7, 492.3, 564.6] N-m

From the plot we can see that when the torque is 400 N-m,
the field current is If=19.3 A, and Ke*flux=1.898 when If=19.3 A
Hence the required speed in is : 1005.0 rpm


## Example No:5.6,Page No:71¶

In [1]:
import math
from __future__ import division
from array import array
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline

#variable declaration
#the motor rating is same as that of Ex-5.5
N=600      #value of the speed given from the magnetization curve in Ex-5.5

Ra=0.04    #armature resistance
Rf=10      #field resistance
N1=1200    #given speed in rpm to hold the overhauling torque

#calculation
Wm=2*math.pi*N1/60    #angular speed at the given speed N1

#magnetisation curve at N=600rpm
If=[2.5,5,7.5,10,12.5,15,17.5,20,22.5,25]       #field current
E =[25,50,73.5,90,102.5,110,116,121,125,129]    #value of the back emf as given in Ex-5.5 for the speed N

#magnetisation curve at N=1200rpm
If=[2.5,5,7.5,10,12.5,15,17.5,20,22.5,25]       #field current
E1=[N1/N*E for E in E]                          #back emf at the speed N1
print"Hence the magnetization curve at 1200rpm is"
print"Field current   If:",If,"A"
print"Back emf is     E1:",E1,"V"

Pd=round(T*Wm,2)                               #power developed
x=round(Pd*Ra,1)
V=[(E1-Pd*Ra/E1) for E1 in E1]       #terminal voltage
V=[round(V,2) for V in V]
print"Terminal voltage V:",V,"V"

#results
#plotting the values of V vs If
plt.subplot(2,1,1)
plt.plot(V,If)
plt.xlabel('Terminal voltage $V$')
plt.ylabel('Field current $I_f$')
plt.title('$V vs If$')
plt.grid()

#plotting the values of E vs If
If=[2.5,5,7.5,10,12.5,15,17.5,20,22.5,25]       #field current
E =[25,50,73.5,90,102.5,110,116,121,125,129]    #value of the back emf as given in Ex-5.5 for the speed N
E1=[N1/N*E for E in E]                          #back emf at the speed N1

plt.subplot(2,1,2)
plt.plot(E1,If,'y')
plt.xlabel('$E$')
plt.ylabel('Field current $I_f$')
plt.title('$E vs If$')
plt.grid()
plt.tight_layout()
plt.show()
print"\nFrom the plot we can see that when the current If=25 A the terminal voltage is V=250 V with the back emf E=258V"

E=258           #value of the back emf in V at from the plot
V=250           #value of terminal voltage in V from the plot at E=258 V
If=25           #value of If in A from the plot at E=258 V
Ia=(E-V)/Ra     #armature current
If=V/Rf         #field current
Ir=Ia-If
Rb=V/Ir         #braking resistance

print"Hence the rquired braking resistance is ",round(Rb,3),"ohm"

Hence the magnetization curve at 1200rpm is
Field current   If: [2.5, 5, 7.5, 10, 12.5, 15, 17.5, 20, 22.5, 25] A
Back emf is     E1: [50.0, 100.0, 147.0, 180.0, 205.0, 220.0, 232.0, 242.0, 250.0, 258.0] V
Terminal voltage V: [9.79, 79.89, 133.32, 168.83, 195.19, 210.86, 223.33, 233.69, 241.96, 250.21] V

From the plot we can see that when the current If=25 A the terminal voltage is V=250 V with the back emf E=258V
Hence the rquired braking resistance is  1.429 ohm


## Example No:5.7,Page No:72¶

In [170]:
import math
from __future__ import division
from array import array
import numpy
import matplotlib.pyplot as plt
%matplotlib inline

#variable declaration
#ratings of the DC series motor which operated under dynamic braking
Ra=0.5         #total resistance of armature and field windings
Rf=10          #field resistance
T=500          #overhauling load torque in N-m
N=600          #speed at the overhauling torque T

#magnetisation curve at a speed of 500 rpm
N1=500    #speed in rpm
Ia=[20, 30, 40, 50, 60, 70, 80]       #armature current
E =[215,310,381,437,482,519,550]     #back emf

#calculation
Wm1=2*math.pi*N1/60
print"\nArmature current :",Ia,"A"
Ke_flux=[E / Wm1 for E in E]    #Ke*flux=constant
Ke_flux=[round(Ke_flux,3) for Ke_flux in Ke_flux]
print"\nKe_flux :",Ke_flux
Ke_flux=numpy.array(Ke_flux)
Ia=numpy.array(Ia)
T=numpy.array(Ke_flux)*numpy.array(Ia)   #torque
T=[round(T,1) for T in T]
print"\nTorque  :",T,"N-m"

#results
#plotting the values of Ke*flux vs Ia and T vs Ia
plt.subplot(2,1,1)
plt.plot(Ia,Ke_flux,'y')
plt.xlabel('Armature current $I_a$')
plt.ylabel('$Ke*flux$')
plt.title('$Ke*flux vs Ia$')
plt.grid()

plt.subplot(2,1,2)
plt.plot(T,Ia)
plt.xlabel('Torque $T$')
plt.ylabel('Armature current $I_a$')
plt.title('$T vs Ia$')
plt.grid(True)
plt.tight_layout()
plt.show()

print"\nFrom the plot we can see that at the given torque T=500 N-m the current Ia is 56 A, and Ke*flux is 8.9 at Ia=56 A"
Ke_flux=8.9    #value of Ke*flux at T=500 N-m from the plot
Ia=56          #value of Ia at at T=500 N-m from the plot
Wm=2*math.pi*N/60
E=Ke_flux*Wm   #required emf
x=E/Ia         #x=Ra+Rb
Rb=x-Ra        #required braking resistance
print"Hence the rquired braking resistance is ",round(Rb,3),"ohm"

Armature current : [20, 30, 40, 50, 60, 70, 80] A

Ke_flux : [4.106, 5.921, 7.277, 8.346, 9.206, 9.912, 10.504]

Torque  : [82.1, 177.6, 291.1, 417.3, 552.4, 693.8, 840.3] N-m

From the plot we can see that at the given torque T=500 N-m the current Ia is 56 A, and Ke*flux is 8.9 at Ia=56 A
Hence the rquired braking resistance is  9.486 ohm


## Example No:5.8,Page No:74¶

In [171]:
import math
from __future__ import division

#variable declaration
#ratings of the separately excited motor
V=220     # rated voltage
N=970     # rated speed
Ia=100    # rated current
Ra=0.05   # armature resistance
N1=1000   # initial speed of the motor in rpm

#calculation
E=V-Ia*Ra
E1=N1/N*E   #value of back emf at the speed N1
#(a)the resistance to be placed
Ia1=2*Ia         #value of the braking current is twice the rated current
Rb=(E1+V)/Ia1-Ra  #required resistance

#(b)The braking torque
Wm=(2*math.pi*N1)/60
T=E1*Ia1/Wm

#(c)when the speed has fallen to zero the back emf is zero
E2=0
Ia2=V/(Ra+Rb)
T2=Ia2/Ia1*T   #since the torque is directly proportional to the current

#results
print"(a)Hence required resistance is :",round(Rb,2),"ohm"
#answer for the resistance in the book is wrong due to accuracy
print"\n(b)Hence the required braking torque is :",round(T,1),"N-m"
print"\n(c)Hence the required torque is :",round(T2,1),"N-m"

(a)Hence required resistance is : 2.16 ohm

(b)Hence the required braking torque is : 423.3 N-m

(c)Hence the required torque is : 210.9 N-m


## Example No:5.9,Page No:84¶

In [151]:
import math
from __future__ import division
import cmath

#variable declaration
#ratings of the separately excited motor which operates under rheostatic braking
V=220     # rated voltage
N=1000    # rated speed
Ia=175    # rated current
Ra=0.08   # armature resistance
N1=1050   # initial speed of the motor in rpm
J=8       # moment of inertia of the motor load system kg-m2
La=0.12   # armature curcuit inductance in H

#calculation
E=V-Ia*Ra

#(a)when  the braking current is twice the rated current
Ia1=2*Ia
E1=N1/N*E
x=E1/Ia1   #x=Rb+Ra
Rb=x-Ra    #required braking resistance

#(b)to obtain the expression for the transient value of speed and current including the effect of armature inductance
Ra=x                   #total armature current
K=E/Wm
B=0
ta=La/Ra                    #time constant in sec
Trated=E*Ia/Wm              #rated torque
Tl=0.15*Trated              #load torque is 15% of the rated torque
tm1= float('inf')           #tm1=J/B and B=0 which is equal to infinity
tm2=J*Ra/(B*Ra+K**2)

a = ta
b = -(1+ta/tm1)
c = 1/tm2
# calculate the discriminant
d = (b**2) - (4*a*c)
# find two solutions
alpha1 = (-b-cmath.sqrt(d))/(2*a)
alpha2 = (-b+cmath.sqrt(d))/(2*a)

K3=tm2*Tl/J
K4=tm2*K*Tl/J/Ra

#transient value for speed
x1=((J*alpha2-B)*K1-(Tl-J*alpha2*K3))/(J*(alpha2-alpha1))
y1=((J*alpha1-B)*K1-(Tl-J*alpha1*K3))/(J*(alpha1-alpha2))

#transient value for the current
x2=(K*K1+alpha2*La*K4)/(La*(alpha2-alpha1))
y2=(K*K1+alpha1*La*K4)/(La*(alpha1-alpha2))

#(c) to calculate the time taken by braking operation and the maximum value of the armature current
#now Wm=0 for the braking operation and hence 151.5 exp(-0.963*t1)- 8.247 = 0 from the previous answer in (b)
a=K3/x1    #a=exp(-0.963*t1)
t1=-alpha1*math.log(a.real)        #take log base e on both sides
#now d/dt(ia)=0 for themaximum current and hence d/dt(26.25-593.1exp(-0.963*t)+566.8exp(-4.19*t) = 0 from the previous answer in (b)
b=abs(alpha2*y2)/abs(alpha1*x2)    #b=exp(-0.963*t)/exp(-4.19*t)
t2=math.log(b)/(-alpha1+alpha2)    #take log base e on both sides
t2=abs(t2)
ia=K4-x2.real*math.exp(-alpha1.real*t2)-y2.real*math.exp(-alpha2.real*t2)

#results
print"(a)Hence the braking resistance is :",round(Rb,3),"ohm"
print"\nb)The value of alpha1 :",round(alpha1.real,3),"and alpha2 :",round(alpha2.real,3)
print"\nHence the expression for the transient value for the speed is"
print"Wm=",round(x1.real,1),"exp(",-round(alpha1.real,3),"*t)",round(y1.real,1),"exp(",-round(alpha2.real,2),"*t)","-",round(K3,3)
print"\nHence the expression for the transient value for the current is"
print"ia=",round(K4,2),"-",round(x2.real,1),"exp(",-round(alpha1.real,3),"*t) +",-round(y2.real,1),"exp(",-round(alpha2.real,2),"*t)"
print"\n(c)Hence the time taken is :",round(t2,2),"sec"
print"   Hence the maximum current is: ",round(ia,2),"A"
print"\n Note : There is a slight difference in the answers due to more number of the decimal place "

(a)Hence the braking resistance is : 0.538 ohm

b)The value of alpha1 : 0.963 and alpha2 : 4.187

Hence the expression for the transient value for the speed is
Wm= 151.5 exp( -0.963 *t) -33.3 exp( -4.19 *t) - 8.247

Hence the expression for the transient value for the current is
ia= 26.25 - 593.1 exp( -0.963 *t) + 566.8 exp( -4.19 *t)

(c)Hence the time taken is : 0.44 sec
Hence the maximum current is:  -272.23 A

Note : There is a slight difference in the answers due to more number of the decimal place


## Example No:5.10,Page No:86¶

In [50]:
import math
from __future__ import division
import cmath
import numpy as np

#variable declaration
#ratings of the separately excited motor of Ex-5.9 which operates plugging
V=220     # rated voltage
N=1000    # rated speed
Ia=175    # rated current
Ra=0.08   # armature resistance
N1=1050   # initial speed of the motor in rpm
J=8       # moment of inertia of the motor load system kg-m2
La=0.12   # armature curcuit inductance in H

#calculation
E=V-Ia*Ra
#(a)when  the braking current is twice the rated current
Ia1=2*Ia
E1=N1/N*E
x=(V+E1)/Ia1   #x=Rb+Ra
Rb=x-Ra        #required braking resistance

#(b)to obtain the expression for the transient value of speed and current including the effect of armature inductance
#the values given below are taken from Ex-5.9
ta=0.194                    #time constant in sec
B=0
tm1= float('inf')           #tm1=J/B and B=0 which is equal to infinity
tm2=1.274
K=1.967
Trated=E*Ia/Wm             #rated torque
Tl=0.5*Trated              #load torque is 50% of the rated torque
Ra=Rb
#values of the coefficient of the quadratic equation for Wm
x1=(1+ta/tm1)/ta
x2=1/tm2/ta
x3=-(K*V+Ra*Tl)/J/Ra/ta
#values of the coefficient of the quadratic equation ia
y1=(1+ta/tm1)/ta
y2=1/tm2/ta
y3=-B*V/J/Ra/ta+K*Tl/J/Ra/ta

a = 1
b = x1
c = x2
# calculate the discriminant
d = (b**2) - (4*a*c)
# find two solutions
alpha1 = (-b+cmath.sqrt(d))/(2*a)
alpha2 = (-b-cmath.sqrt(d))/(2*a)

K3=x3/x2
K4=y3/y2

Wm_0=K1                       ;ia_0=0
d_Wm_dt_0=(K*ia_0-B*Wm-Tl)/J  ;d_ia_dt_0=(-V-Ra*ia_0-K*K1)/La    #Wm=K1 at t=0 and during braking rated voltage V is equal to -V

A = np.array([[1,1],[alpha1.real,alpha2.real]])
B = np.array([Wm_0,d_Wm_dt_0])
x = np.linalg.solve(A,B)
C = np.array([[1,1],[alpha1.real,alpha2.real]])
D = np.array([-K4,d_ia_dt_0])
y = np.linalg.solve(C,D)

#(c)to calculate the time taken for the speed to fall to zero value
a=-K3/x[0]                      #a=exp(-0.966*t1)
t1=alpha1*math.log(a)           #take log base e on both sides

#results
print"(a)Hence the braking resistance is :",round(Rb,3),"ohm"
print"\n(b)The solution for alpha are ",round(alpha1.real,3),"and",round(alpha2.real,3)
print"   Wm=",round(K3,2),"+ A*exp(",round(alpha1.real,3),"*t) +","+ B*exp(",round(alpha2.real,2),"*t)"
print"   ia=",round(K4,2),"+ C*exp(",round(alpha1.real,3),"*t) +","+ D*exp(",round(alpha2.real,2),"*t)"
print"   We have to find the value of A,B,C and D in the linear equation using the initial condition"
print"   A=",round(x[0],2),"B=",round(x[1],2), "C=",round(y[0],2),"D=",round(y[1],2)
print"\nHence the expression for the transient value for the speed is"
print"   Wm=",round(K3,2),"+",round(x[0],2),"*exp(",round(alpha1.real,3),"*t)",round(x[1],2),"*exp(",round(alpha2.real,2),"*t)"
print"\nHence the expression for the transient value for the current is"
print"   ia=",round(K4,2),round(y[0],2),"*exp(",round(alpha1.real,3),"*t) +",round(y[1],2),"*exp(",round(alpha2.real,2),"*t)"
print"\n(c)Hence the time taken is :",round(t1.real,2),"sec"
print"\n Note :There is slight difference in the answers due to accuracy i.e more number of decimal place"

(a)Hence the braking resistance is : 1.167 ohm

(b)The solution for alpha are  -0.966 and -4.189
Wm= -86.48 + A*exp( -0.966 *t) + + B*exp( -4.19 *t)
ia= 46.22 + C*exp( -0.966 *t) + + D*exp( -4.19 *t)
We have to find the value of A,B,C and D in the linear equation using the initial condition
A= 136.24 B= -26.28 C= -1188.2 D= 1141.98

Hence the expression for the transient value for the speed is
Wm= -86.48 + 136.24 *exp( -0.966 *t) -26.28 *exp( -4.19 *t)

Hence the expression for the transient value for the current is
ia= 46.22 -1188.2 *exp( -0.966 *t) + 1141.98 *exp( -4.19 *t)

(c)Hence the time taken is : 0.44 sec

Note :There is slight difference in the answers due to accuracy i.e more number of decimal place


## Example No:5.11,Page No:89¶

In [67]:
import math
from __future__ import division
import cmath

#variable declaration
#ratings of the separately excited motor
V=220     # rated voltage
N=600     # rated speed
Ia=500   # rated current
Ra=0.02   # armature resistance
Rf=10     # field resistance

#calculation
E1=V-Ia*Ra            #rated back emf at rated operation
Wm1=2*math.pi*N/60    #angular speed
Trated=E1*Ia1/Wm1     #rated torque
#(i) when the speed of the motor is 450rpm
N1=450             #given speed in rpm
Tl=2000-2*N1       #load torque is a function of the speed as given
Ia2=Tl/Trated*Ia1  #for a torque of  Tl as a function of current
E2=N1/N*E1         #for a given speed of 450rpm
V2=E2+Ia2*Ra       #terminal voltage for a given speed of 450 rpm

#(ii) when the speed of the motor is 750rpm
N1=750          #given speed in rpm
Tl=2000-2*N1    #load torque is a function of the speed as given
Wm_=2*math.pi*N1/60
Ke_phi1=E1/Wm1

#Since we know that V=Ke*phi*Wm+Ia*Ra  by solving we get that  0.02*(Ia_)**2 -220*Ia_ + 39270 = 0"
a = 0.02
b = -220
c = 39270
# calculate the discriminant
d = (b**2) - (4*a*c)
# find two solutions
Ia_1 = (-b-cmath.sqrt(d))/(2*a)
Ia_2 = (-b+cmath.sqrt(d))/(2*a)

Ke_phi=Tl/abs(Ia_1)
V1=V*Ke_phi/Ke_phi1   #required field voltage

#results
print"(i)Hence motor terminal voltage is :",round(V2),"V"
print"   And the armature current is :",round(Ia2),"A"
print"\n(ii)The solution for Ia_ are ",round(abs(Ia_1),1),"A and",round(abs(Ia_2)),"A"
print"    We ignore ",round(abs(Ia_2)),"A since it is unfeasible,\n    Hence armature current is :",round(abs(Ia_1),1),"A"
print"    Hence the required field voltage is :",round(V1,1),"V"

(i)Hence motor terminal voltage is : 164.0 V
And the armature current is : 329.0 A

(ii)The solution for Ia_ are  181.5 A and 10819.0 A
We ignore  10819.0 A since it is unfeasible,
Hence armature current is : 181.5 A
Hence the required field voltage is : 181.3 V


## Example No:5.12,Page No:91¶

In [173]:
import cmath
from __future__ import division
import numpy

#variable declaration
#ratings of the 2-pole separately excited DC motor with the fields coils connected in parallel
V=220      # rated voltage
N=750      # rated speed
Ia1=100    # rated current
Ra=0.1     # armature resistance

#calculation
E1=V-Ia1*Ra            #rated back emf at rated operation
Wm1=2*math.pi*N/60     #angular speed
Trated=E1*Ia1/Wm1      #rated torque
Ke_phi1=E1/Wm1
#(i) when the armature voltage is reduced to 110V
Wm2=2*math.pi*N2/60    #angular speed
E2=Ke_phi1*Wm2
#Now there are two linear equations...that we have to solve
#They are given by 0.3*N2+2.674*Ia2=500 and 0.28*N2+0.1*Ia2=110
a = np.array([[0.3,2.674], [0.28,0.1]])
b = np.array([500,110])
x = np.linalg.solve(a, b)
N2=round(x[0],1)      #let the motor speed be N2
Ia2=round(x[1],1)     #let the motor current be Ia2

#(ii)when the field coils are connected in series
K=Ke_phi1/2
Wm3=2*math.pi*N3/60    #angular speed
E3=K*Wm3
#Now there are two linear equations...that we have to solve"
#They are given by 0.3*N3+1.337*Ia3=500 and 0.14*N3+0.1*Ia3=220"
a = np.array([[0.3,1.337], [0.14,0.1]])
b = np.array([500,220])
x = np.linalg.solve(a, b)
N3=round(x[0],1)            #let the motor speed be N3
Ia3=round(x[1],2)           #let the motor current be Ia3

#results
print"(i)Hence the motor armature current is Ia2 :",Ia2,"A"
print"   And the required speed is N2 :",N2,"rpm"
print"\n(ii)Hence the motor armature current is Ia3 :",Ia3,"A"
print"    And the required speed is N3 :",N3,"rpm"


(i)Hence the motor armature current is Ia2 : 148.9 A
And the required speed is N2 : 339.7 rpm

(ii)Hence the motor armature current is Ia3 : 25.45 A
And the required speed is N3 : 1553.3 rpm


## Example No:5.13,Page No:102¶

In [81]:
import math
from __future__ import division

#variable declaration
#ratings of the separately excited motor
V=200    # rated voltage
N=875    # rated speed
Ia=150   # rated current
Ra=0.06  # armature resistance
Vs=220   # source voltage
f=50     # frequency of the source voltage

#calculation
E=V-Ia*Ra            #back emf
Vm=math.sqrt(2)*Vs   #peak voltage

#(i)when the speed is 750 rpm and at rated torque
N1=750         #given speed in rpm
E1=N1/N*E      #back emf at the given speed N1
Va=E1+Ia*Ra    #terminal voltage
cos_alpha=Va*math.pi/2/Vm
alpha=math.acos(cos_alpha)   #required firing angle in radian
alpha1=math.degrees(alpha)    #required firing angle in degrees

#(ii)when the speed is -500rpm and at rated torque
N1=-500         #given speed in rpm
E1=N1/N*E       #back emf at the given speed N1
Va=E1+Ia*Ra     #terminal voltage
cos_alpha=Va*math.pi/2/Vm
alpha=math.acos(cos_alpha)   #required firing angle in radian
alpha2=math.degrees(alpha)    #required firing angle in degrees

#(iii)when the firing angle is 160 degrees
alpha=160    #firing angle in degrees
E1=Va-Ia*Ra    #since Va=E1+Ia*Ra
N1=E1/E*N      #the required speed at the given firing angle

#results
print"(i)Hence the required firing angle is :",round(alpha1,1),"°"
print"\n(ii)Hence the required firing angle is :",round(alpha2),"°"
print"\n(iii)Hence the required speed is :",round(N1,1),"rpm"

(i)Hence the required firing angle is : 29.3 °

(ii)Hence the required firing angle is : 120.0 °

(iii)Hence the required speed is : -893.9 rpm


## Example No:5.14,Page No:103¶

In [179]:
import math
from __future__ import division

#variable declaration
#ratings of the separately excited motor is same as that of Ex-5.13
V=200    # rated voltage
N=875    # rated speed
Ia=150   # rated current
Ra=0.06  # armature resistance
Vs=220   # source voltage
f=50     #frequency of the source voltage
La=0.85e-3   # armature curcuit inductance in H

#calculation
E=V-Ia*Ra            #back emf
Vm=math.sqrt(2)*Vs   #peak voltage
Wm=2*math.pi*N/60    #synchronous angular speed

#(i)when the speed is 400 rpm and firing angle is 60 degrees
N1=400         #given speed in rpm
alpha=60       #firing angle in degrees
W=2*math.pi*f
x=W*La/Ra
phi=math.atan(x)
cot_phi=1/math.tan(phi)
Z=math.sqrt(Ra**2+(W*La)**2)
K=E/Wm

y=Ra*Vm/Z/K
a=(1+math.exp(-(math.pi*cot_phi)))/(math.exp(-(math.pi*cot_phi))-1)
Wmc=y*math.sin(math.radians(alpha)-phi)*a   #required angular speed in rps
Nmc=Wmc*60/2/math.pi     #required angular speed in rpm

E1=N1/N*E

#The equation Vm/Z*sin(beta-phi)-E/Ra+(E/Ra-Vm/Z*sin(alpha-phi))*exp(-(beta-alpha)*cot_phi)=0
#can be solved using trial method such that beta=230 degrees
beta=230                   #in degrees

Va=(Vm*(math.cos(alpha)-math.cos(beta))+(math.pi+alpha-beta)*E1)/math.pi
Ia=(Va-E1)/Ra
T1=K*Ia

#(ii)when the speed is -400 rpm and firing angle is 120 degrees
Le=2e-3        #external inductance added to the armature
L=La+Le
N2=-400         #given speed in rpm
alpha=120       #firing angle in degrees
x=W*L/Ra
phi=math.atan(x)
cot_phi=1/math.tan(phi)
Z=math.sqrt(Ra**2+(W*L)**2)
K=E/Wm

y=Ra*Vm/Z/K
a=(1+math.exp(-(math.pi*cot_phi)))/(math.exp(-(math.pi*cot_phi))-1)
Wmc=y*math.sin(math.radians(alpha)-phi)*a   #required angular speed in rps
Nmc1=Wmc*60/2/math.pi     #required angular speed in rpm
#The motor is operating under discontinous condition"
E2=N2/N*E

#The equation Vm/Z*sin(beta-phi)-E/Ra+(E/Ra-Vm/Z*sin(alpha-phi))*exp(-(beta-alpha)*cot_phi)=0
#can be solved using trial method such that beta=281 degrees
beta=281                   #in degrees

Va=(Vm*(math.cos(alpha)-math.cos(beta))+(math.pi+alpha-beta)*E2)/math.pi
Ia=(Va-E2)/Ra
T2=K*Ia

#(iii)when the speed is -600 rpm and firing angle is 120 degrees
N3=-600        #speed in rpm
alpha=120      #firing angle in degrees
E3=N3/N*E    #since Va=E1+Ia*Ra
Ia=(Va-E3)/Ra
T3=K*Ia

#results
print"(i)Hence the required torque is :",round(T1),"N-m "
print"\n(ii)Hence the required torque is :",round(T2,1),"N-m"
print"\n(iii)Hence the required torque is :",round(T3),"N-m"
print"\nNote : There is a slight difference in the answers because of accuracy i.e more number of decimal place"

(i)Hence the required torque is : 1067.0 N-m

(ii)Hence the required torque is : 336.4 N-m

(iii)Hence the required torque is : 1110.0 N-m

Note : There is a slight difference in the answers because of accuracy i.e more number of decimal place


## Example No:5.15,Page No:105¶

In [181]:
import math
from __future__ import division

#variable declaration
#ratings of the separately excited motor is same as that of Ex-5.13
V=200    # rated voltage
N=875    # rated speed
Ia=150   # rated current
Ra=0.06  # armature resistance
Vs=220   # source voltage
f=50     #frequency of the source voltage
La=2.85e-3   # armature curcuit inductance in H

#calculation
E=V-Ia*Ra            #back emf
Vm=math.sqrt(2)*Vs   #peak voltage
Wm=2*math.pi*N/60    #angular speed
W=2*math.pi*f

alpha=120       #firing angle in degrees
x=W*La/Ra
phi=math.atan(x)
cot_phi=1/math.tan(phi)
Z=math.sqrt(Ra**2+(W*La)**2)
K=E/Wm

y=Ra*Vm/Z/K
a=(1+math.exp(-(math.pi*cot_phi)))/(math.exp(-(math.pi*cot_phi))-1)
Wmc=round(y,3)*math.sin(math.radians(alpha)-phi)*a   #required angular speed in rps
Nmc=Wmc*60/2/math.pi     #required angular speed in rpm

E1=Nmc/N*E     #value of back emf at the critical speed of Nmc
Ia=(Va-E1)/Ra
Tc=K*Ia

#(i)when the torque is 1200 N-m and firing angle is 120 degrees
T2=1200         #given torque in N-m
Ia2=T2/K        #given terminal current for the given torque and the answer in the book is wrong
E2=Va-Ia*Ra
N2=E2/E*N

#(ii)when the torque is 300 N-m and firing angle is 120 degrees
T=300    #required torque in N-m
beta=233.492   #required angle in degrees
x=beta-alpha
E1=(Vm*(math.cos(alpha)-math.cos(beta)))/x-(math.pi*Ra*T)/(K*x)
N1=E1/E*N   #required speed

#results
print"The motor is operating under continuous condition"
print"The torque Tc is :",round(Tc),"N-m"
print"The answer for torque Tc in the book is wrong  due to accuracy in the decimal place  which leads to subsequent "
print"\n(i)Hence the required speed is :",round(N2),"rpm"
print"\n(ii)The equation Vm/Z*sin(beta-phi)-sin(alpha-phi))*exp(-(beta-alpha)*cot_phi)="
print"    (Vm*(cos(alpha)-cos(beta))/Ra/(beta-alpha)-pi*T/K/(beta-alpha) )*(1-exp(-(beta-alpha)*cot_phi)"
print"    can be solved using trial method such that beta=233.492 degrees"
print"\n  Hence the required speed is :",round(N1,1),"rpm"

The motor is operating under continuous condition
The torque Tc is : 396.0 N-m
The answer for torque Tc in the book is wrong  due to accuracy in the decimal place  which leads to subsequent

(i)Hence the required speed is : -506.0 rpm

(ii)The equation Vm/Z*sin(beta-phi)-sin(alpha-phi))*exp(-(beta-alpha)*cot_phi)=
(Vm*(cos(alpha)-cos(beta))/Ra/(beta-alpha)-pi*T/K/(beta-alpha) )*(1-exp(-(beta-alpha)*cot_phi)
can be solved using trial method such that beta=233.492 degrees

Hence the required speed is : 5.6 rpm


## Example No:5.16,Page No:110¶

In [182]:
import math
from __future__ import division

#variable declaration
#ratings of the separately excited motor
V=220       # rated voltage
N=960       # rated speed
Ia=12.8     # rated current
Ra=2        # armature resistance
Vs=230      # source voltage
f=50        #frequency of the source voltage
La=150e-3   # armature curcuit inductance in H

#calculation
E=V-Ia*Ra            #back emf
Vm=math.sqrt(2)*Vs   #peak voltage
Wm=2*math.pi*N/60    #angular speed
W=2*math.pi*f

#(i)when speed is 600rpm and the firing angle is 60 degrees
alpha=60       #firing angle in degrees
N1=600         #motor speed in rpm
x=W*La/Ra
phi=math.atan(x)
cot_phi=1/math.tan(phi)
Z=math.sqrt(Ra**2+(W*La)**2)
K=E/Wm

y=Ra*Vm/Z/K
a=1-math.exp(-(math.pi*cot_phi))
Wmc=round(y,3)*(b-c)/a   #required angular speed in rps
Nmc=Wmc*60/2/math.pi     #required angular speed in rpm

E1=N1/N*E     #value of back emf at the speed of N1
Ia=(Va-E1)/Ra
T=K*Ia

#(ii)when the torque is 20 N-m and firing angle is 60 degrees
T1=20       #required torque in N-m
alpha=60   #required firing angle in degrees
Ec=Nmc/N*E #motor back emf at critical speed of Nmc
Tc=K*(Va-Ec)/Ra    #torque at the critical speed

Ia=T1/K
E1=Va-Ia*Ra
N1=E1/E*N   #required speed

#results
if N1<Nmc :
print"(i)The motor is operating under continuous condition"
print"   Hence the required torque is :",round(T,2),"N-m"
if Tc<T1 :
print"\n(ii)The motor is operating under continuous condition"
print"    Hence the required speed is :",round(N1,1),"rpm"

(i)The motor is operating under continuous condition
Hence the required torque is : 32.68 N-m

(ii)The motor is operating under continuous condition
Hence the required speed is : 664.8 rpm


## Example No:5.17,Page No:113¶

In [183]:
import math
from __future__ import division

#variable declaration
#ratings of the separately excited motor
V=220    # rated voltage
N=1500   # rated speed
Ia=50    # rated current
Ra=0.5   # armature resistance
Vl=440   # line  voltage with 3-phase ac supply
f=50     #frequency of the source voltage

#calculation
#(i) tranformer ratio
alpha=0    #firing angle
Va=V       #motor terminal voltage is equal to the rated voltage when the firing angle is 0 degrees
Vm=math.pi/3*Va/math.cos(alpha)
Vrms=Vm/math.sqrt(2)        #rms value of the converter input voltage
a=(Vl/math.sqrt(3))/Vrms    #required transformer ratio

#(ii)value of the firing angle
E=V-Ia*Ra      #back emf at the rated speed

#(a)when the speed of the motor is 1200 rpm and rated torque
N1=1200        #speed of the motor
E1=N1/N*E      #back emf at the given speed N1
Va=E1+Ia*Ra    #terminal voltage at the given speed N1
alpha=math.acos(math.pi/3*Va/Vm)  #required firing angle in radians
alpha1=math.degrees(alpha)         #required firing angle in degrees

#(b)when the speed of the motor is -800 rpm and twice the rated torque
N1=-800        #speed of the motor
E1=N1/N*E      #back emf at the given speed N1
Ia=2*Ia        #torque is directly proportional to the current hence twice the rated current
Va=E1+Ia*Ra    #terminal voltage at the given speed N1
alpha=math.acos(math.pi/3*Va/Vm)  #required firing angle in radians
alpha2=math.degrees(alpha)         #required firing angle in degrees

#results
print"(i)Hence the required transformer ratio is :",round(a,3)
print"\n(ii)(a)Hence the required firing angle is :",round(alpha1,2),"°"
print"\n    (b)Hence the required firing angle is :",round(alpha2,2),"°"

(i)Hence the required transformer ratio is : 1.559

(ii)(a)Hence the required firing angle is : 34.64 °

(b)Hence the required firing angle is : 104.21 °


## Example No:5.18,Page No:117¶

In [184]:
import math
from __future__ import division

#variable declaration
#ratings of the separately excited motor is same as that of Ex-5.17 but is fed from a circulating dual converter
V=220    # rated voltage
N=1500   # rated speed
Ia=50    # rated current
Ra=0.5   # armature resistance
Vl=165   # line  voltage
f=50     # frequency of the source voltage

#calculation
E=V-Ia*Ra            #back emf at the rated speed
Vm=Vl*math.sqrt(2)   #peak voltage

#(i)during motoring operation when the speed is 1000 rpm and at rated torque
N1=1000        #speed of the motor
E1=N1/N*E      #back emf at the given speed N1
Va=E1+Ia*Ra    #terminal voltage at the given speed N1
alpha_A=math.acos(math.pi/3*Va/Vm)
alpha_B=180-math.degrees(alpha_A)         #required converter firing angle in degrees

#(ii)during braking operation when the speed is 1000 rpm and at rated torque
N1=1000        #speed of the motor in the book it is given as 100 rpm which is wrong
E1=N1/N*E      #back emf at the given speed N1
Va=E1-Ia*Ra    #terminal voltage at the given speed N1
alpha_A1=math.acos(math.pi/3*Va/Vm)
alpha_B1=180-math.degrees(alpha_A1)         #required converter firing angle in degrees

#(iii)during motoring operation when the speed is -1000 rpm and at rated torque
N1=-1000        #speed of the motor
E1=N1/N*E      #back emf at the given speed N1
Va=E1-Ia*Ra    #terminal voltage at the given speed N1
alpha_A2=math.acos(math.pi/3*Va/Vm)
alpha_B2=180-math.degrees(alpha_A2)         #required converter firing angle in degrees

#(iv)during braking operation when the speed is -1000 rpm and at rated torque
N1=-1000       #speed of the motor in the book it is given as 100 rpm which is wrong
E1=N1/N*E      #back emf at the given speed N1
Va=E1+Ia*Ra    #terminal voltage at the given speed N1
alpha_A3=math.acos(math.pi/3*Va/Vm)
alpha_B3=180-math.degrees(alpha_A3)         #required converter firing angle in degrees

#results
print"\n(i)Hence the required firing angle is :",round(alpha_B,1),"°"
print"\n(ii)Hence the required firing angle is :",round(alpha_B1,1),"°"
print"\n(iii)Hence for negative speed during motoring operation the required firing angle are :"
print"     alpha_A :",round(math.degrees(alpha_A2),1),"° and alpha_B :",round(alpha_B2,1),"°"
print"\n(iv)Hence for negative speed during braking operation the required firing angle are :"
print"    alpha_A :",round(math.degrees(alpha_A3),1),"° and alpha_B :",round(alpha_B3,1),"°"

(i)Hence the required firing angle is : 134.1 °

(ii)Hence the required firing angle is : 118.1 °

(iii)Hence for negative speed during motoring operation the required firing angle are :
alpha_A : 134.1 ° and alpha_B : 45.9 °

(iv)Hence for negative speed during braking operation the required firing angle are :
alpha_A : 118.1 ° and alpha_B : 61.9 °


## Example No:5.19,Page No:126¶

In [185]:
import math
from __future__ import division

#variable declaration
#ratings of the separately excited motor
V=230    # rated voltage
N=960    # rated speed
Ia=200   # rated current
Ra=0.02  # armature resistance
Vs=230   # source voltage

#calculation
E=V-Ia*Ra    #back emf

#(i) When the speed of motor is 350 rpm with the rated torque during motoring operation
N1=350      #given speed in rpm
E1=N1/N*E   #given back emf at N1
Va=E1+Ia*Ra  #motor terminal voltage
delta=Va/V  #duty ratio

#(ii) When the speed of motor is 350 rpm with the rated torque during braking operation
Va=E1-Ia*Ra  #motor terminal voltage
delta1=Va/V   #duty ratio

#(iii)maximum duty ratio is 0.95
delta2=0.95   #maximum duty ratio
Va=delta2*V   #terminal voltage
Ia1=2*Ia      #maximum permissable current
E1=Va+Ia1*Ra  #back emf
N1=E1/E*N     #maximum permissible speed
Pa=Va*Ia1     #power fed to the source

#(iv) if the speed of the motor is 1200 rpm and the field of the motor is also controlled
N2=1200      #given speed
#now the field current is directly proportional to the speed of the motor
If=N/N2     #field current as a ratio of the rated current

#results
print"(i) Duty ratio is :",round(delta,3)
print"\n(ii)Duty ratio is :",round(delta1,2)
print"\n(iii)Maximum permissible speed is :",round(N1),"rpm"
print"     Power fed to the source is :",round(Pa/1000,1),"kW"
print"\n(iv)Field current as a ratio of the rated current is :",If

(i) Duty ratio is : 0.376

(ii)Duty ratio is : 0.34

(iii)Maximum permissible speed is : 962.0 rpm
Power fed to the source is : 87.4 kW

(iv)Field current as a ratio of the rated current is : 0.8


## Example No:5.20,Page No:127¶

In [140]:
import math
from __future__ import division

#variable declaration
#ratings of the separately excited motor when it is operated in dynamic breaking
V=230    # rated voltage
N=960    # rated speed
Ia=200   # rated current
Ra=0.02  # armature resistance
Vs=230   # source voltage
Rb=2     # braking resistance in ohm

#calculation
#when the motor speed is 600 rpm and the braking torque is twice the rated value
Ia1=2*Ia     #torque is directly proportional to current
N1=600       #speed of the motor in rpm
E=V-Ia*Ra    #back emf
E1=N1/N*E
x=E1/Ia1-Ra  #x=(1-delta)*Rb
y=x/Rb       #y=1-delta
delta=1-y    #duty ratio

#(ii)if the duty ratio is 0.6 and and the braking torque is twice the rated value
delta1=0.6  #duty ratio
Ia1=2*Ia   #torque is directly proportional to current
E1=Ia1*((1-delta1)*Rb+Ra)   #back emf
N1=E1/E*N

#results
print"(i)Duty ratio is :",round(delta,2)
print"\n(ii)Hence the motor speed is :",round(N1,1),"rpm"

(i)Duty ratio is : 0.83

(ii)Hence the motor speed is : 1393.3 rpm


## Example No:5.21,Page No:128¶

In [186]:
import math
from __future__ import division
from array import array
import numpy as np

#variable declaration
#ratings of the series motor
N=600     #speed in rpm
Vs=220    #source voltage
Ra_Rf=0.12    #combine armature resistance field resistance
#magnetisation curve at N
If=[10, 20,30, 40, 50,  60, 70, 80]    #field current
E =[64,118,150,170,184,194,202,210]    #terminal voltage

#calculation
#(i)when the duty ratio is 0.6 and motor current is 60 A
delta=0.6    #duty ratio
Ia1=60       #motor current
Va1=delta*Vs #terminal voltage for the given duty ratio
E1=Va1-Ia1*Ra_Rf      #back emf for the given duty ratio

#for Ia1=60 A the terminal voltage is 194 V as given in the magnetization curve
N1=E1/E[5]*N          #motor speed for the given duty ratio

#(ii)when the speed is 400rpm and the duty ratio is 0.65
delta=0.65    #duty ratio
N2=400        #speed in rpm
Va1=delta*Vs #terminal voltage for the given duty ratio

#from the magnetization characteristic for the speed of 400rpm the current Ia=70 A
E1=Va1-If[6]*Ra_Rf      #back emf for the given duty ratio
T=(E1*If[6])/N2/(2*math.pi/60)    #required torque

#results
print"(i)Hence the motor speed is :",round(N1),"rpm"
print"\n(ii)Hence the required torque is :",round(T),"N-m"

(i)Hence the motor speed is : 386.0 rpm

(ii)Hence the required torque is : 225.0 N-m


## Example No:5.22,Page No:129¶

In [187]:
import math
from __future__ import division
from array import array
import numpy as np

#variable declaration
#ratings of the series motor which is the same as that of Ex-6.21
#The motor is operated using regenarative braking method
N=600     #speed in rpm
Vs=220    #source voltage
Ra_Rf=0.12    #combine armature resistance field resistance
#magnetisation curve at N
If=[10, 20,30, 40, 50,  60, 70, 80]    #field current
E =[64,118,150,170,184,194,202,210]    #terminal voltage

#calculation
#(i)when  the duty ratio is 0.5 and the braking torque is equal to the motor torque
delta=0.5       #duty ratio
Va1=delta*Vs    #terminal voltage
Ia1=If[6]       #current at rated motor torque
E1=Va1+Ia1*Ra_Rf      #back emf for the given duty ratio
N1=E1/E[6]*N          #for a current of 70 A E=202 V from the magnetization curve

#(ii)when maximum permisssible duty ratio is 0.95 and current is 70A
delta_max=0.95      #maximum duty ratio
Va1=delta_max*Vs    #terminal voltage
Ia1=70              #maximum permissible current
E2=Va1+Ia1*Ra_Rf    #back emf for the given duty ratio
N2=E2/E[6]*N        #for a current of 70 A E=202 V

#(iii)when the motor speed is 1000rpm and maximum current is 70A with duty ratio in the range of 0.05 to 0.95
Ia1=70              #maximum permissible current
N3=1000             #given speed in rpm
delta_max=0.95      #maximum duty ratio
E3=N3/N*E[6]        #terminal voltage
x=(E3-delta_max*Vs)/Ia1    #x=R+Ra_Rf   where R is the required external resistance
R=x-Ra_Rf    #external resistance

#(iv)when the motor is running at 1000rpm with current at 70
Ia1=70              #maximum permissible current
N4=1000             #given speed in rpm
Ra=Ra_Rf            #total value of armature resistance is assumed to be the same
E4=Va1+Ia1*Ra       #back emf for the given speed N4
E_=N/N4*E4
ratio=E_/E[6]       #fraction of the requuired number of turns to be reduced

#results
print"(i)Hence the motor speed is :",round(N1,1),"rpm"
print"\n(ii)Hence the motor speed is :",round(N2,1),"rpm"
print"\n(iii)Hence the required external resistance is :",round(R,1),"ohm"
print"\n(iv)Hence fraction of the  number of turns to be reduced is :",round(ratio,3)

(i)Hence the motor speed is : 351.7 rpm

(ii)Hence the motor speed is : 645.7 rpm

(iii)Hence the required external resistance is : 1.7 ohm

(iv)Hence fraction of the  number of turns to be reduced is : 0.646


## Example No:5.23,Page No:130¶

In [188]:
import math
from __future__ import division
from array import array
import numpy as np

#variable declaration
#ratings of the series motor which is the same as that of Ex-6.21
#The motor is operated using dynamic braking method
N=600     #speed in rpm
Vs=220    #source voltage
Ra=0.12   # armature resistance
delta_min=0.1   #manimum value of duty ratio
delta_max=0.9   #maximum value of duty ratio
#magnetisation curve at N
If=[10, 20,30, 40, 50,  60, 70, 80]    #field current
E =[64,118,150,170,184,194,202,210]    #terminal voltage

#calculation
#(i) maximum braking speed is 800rpm with armature current of 70 A
N1=800   #maximum braking speed
Ia=70    #armature current
E1=N1/N*E[6]   #at 70A motor back emf is 202V
Rbe=E1/Ia-Ra   #effective value of braking resistance
Rb=Rbe/(1-delta_min)   #required braking resistance

#(ii)when the speed of the motor is 87 rpm
#now torque is maximum when the duty ratio is maximum
N1=87   #speed in rpm
R=Rb*(1-delta_max)+Ra

Ia=If[4]   #value of armature current for the given value of E=184V
Ke_phi=E[4]/(2*math.pi*N)*60
T=Ke_phi*Ia   #required torque

#results
print"(i)Hence braking resistance is:",round(Rb,2),"ohm"
print"\n(ii)Hence the required torque is :",round(T,1),"N-m"

(i)Hence braking resistance is: 4.14 ohm

(ii)Hence the required torque is : 146.4 N-m