Chapter 6: Induction Motor Drives

Example No:6.1,Page No:147

In [197]:
import math
from __future__ import division
import cmath

#variable declaration
#ratings of the Y-connected induction motor 
f=50           # frequency in HZ
Vl=440         #line voltage in V
P=6            # number of poles
N=950          #speed in rpm

#parameters referred to the stator
Xr_=1.2        # rotor winding reactance in ohm
Rr_=0.4        # resistance of the rotor windings in ohm
Rs=0.5         # resistance of the stator windings in ohm
Xs=Xr_         # stator winding reactance in ohm
Xm=50          # no load reactance
 
#calculation
Ns=120*f/P     #synchronous speed in rpm
s=(Ns-N)/Ns    #full load slip
x=math.sqrt((Rs+Rr_/s)**2+(Xs+Xr_)**2)   #total impedance
Ir_=(Vl/math.sqrt(3))/x                  #full load rotor current
angle=-math.atan((Xs+Xr_)/(Rs+Rr_/s))    #angle in radian
Ir_=cmath.rect(Ir_,angle)                #full load rotor current in rectangular form
Im=Vl/math.sqrt(3)/Xm*(-1j)              #magnetizing current
Is=Ir_+Im                                #full load current

Zf=Rs+Xs*1j+1j*Xm*(Rr_/s+1j*Xr_)/(Rr_/s+1j*(Xr_+Xm))
Zb=Rs+Xs*1j+1j*Xm*(Rr_/(2-s)+1j*Xr_)/(Rr_/(2-s)+1j*(Xr_+Xm))
Z=Zf+Zb
I=(Vl/math.sqrt(3))/abs(Z)             #motor current
Wms=2*math.pi*Ns/60

#torque due to positive sequence
Tp=(1/Wms)*(3*I**2*Xm**2*Rr_/s)/((Rr_/s)**2+(Xr_+Xm)**2)
#torque due to negative sequence
Tn=-(1/Wms)*(3*I**2*Xm**2*Rr_/(2-s))/((Rr_/(2-s))**2+(Xr_+Xm)**2)
T=Tp+Tn          #net torque
Wm=Wms*(1-s)     #rated speed in in rad/sec
Tl=0.0123*Wm**2 #required torque of the load

#results
print"Full load motor current Is:",round(abs(Is),1),round(math.degrees(cmath.phase(Is))),"°","A"
print"Tp:",round(Tp,2),"N-m"
print"Tn:",round(Tn,3),"N-m"
print"\nSince I:",round(I,2),"A and N:",N,"rpm"
print"And I:",round(I,2),"A <","Is:",round(abs(Is),2),"A, the motor will run safely"

Full load motor current Is: 30.5 -25.0 ° A
Tp: 125.61 N-m
Tn: -3.299 N-m

Since I: 24.26 A and N: 950 rpm
And I: 24.26 A < Is: 30.54 A, the motor will run safely

Example No:6.2,Page No:156

In [199]:
import math
from __future__ import division
import cmath

#variable declaration
#ratings of the Delta connected Induction motor
f=50           #frequency in HZ
Vl=2200        #line voltage in V
P=8            #number of poles
N=735          #rated speed in rpm
#parameters referred to the stator
Xr_=0.55        # rotor winding reactance in ohm
Xs=0.45         # stator winding reactance in ohm
Rr_=0.1         # resistance of the rotor windings in ohm
Rs=0.075        # resistance of the stator windings in ohm

#calculation
Ns=120*f/P     #synchronous speed in rpm
s=(Ns-N)/Ns    #full load slip
x=math.sqrt((Rs+Rr_/s)**2+(Xs+Xr_)**2)   #total impedance
Ip=(Vl)/x                  #full load phase current
Il=math.sqrt(3)*Ip         #full load line current
Wms=2*math.pi*Ns/60        
Tl=(1/Wms)*(3*Ip**2*Rr_/s) #full load torque

#(i)if the motor is started by star-delta switching
y=math.sqrt((Rs+Rr_)**2+(Xs+Xr_)**2)
Ist=(Vl/math.sqrt(3))/y                #Maximum line current during starting
Tst=(1/Wms)*(3*Ist**2*Rr_)             #Starting torque
ratio1=Tst/Tl                          #ratio of starting torque to full load torque
z=Rs+math.sqrt(Rs**2+(Xs+Xr_)**2)
Tmax=3/(2*Wms)*(Vl/math.sqrt(3))**2/z  #maximum torque
ratio2=Tmax/Tl                         #ratio of maximum torque to full load torque

#(ii) If the motor is started using auto transformer
y=math.sqrt((Rs+Rr_)**2+(Xs+Xr_)**2)
Ist1=Vl*math.sqrt(3)/y         #starting current direct online
aT=math.sqrt(2*Il/Ist1)        #transofrmation ratio
Ilst=2*Il/aT                   #starting motor line current
Ipst=Ilst/math.sqrt(3)         #starting motor phase current
Tst1=(1/Wms)*(3*Ipst**2*Rr_)    #starting torque

#(iii) If motor is started using part winding method
Rs_=2*Rs
Xs_=2*Xs
y=math.sqrt((Rs_+Rr_)**2+(Xs_+Xr_)**2)
Ist2=(Vl*math.sqrt(3))/y      #starting line current 
Ip=Ist2/math.sqrt(3)          #starting phase current
Tst2=(1/Wms)*(3*Ip**2*Rr_)    #starting torque

#(iv) motor is started using series reactors in line
Rs_=Rs/3  ;  Rr_=Rr_/3
Xs_=Xs/3  ;  Xr_=Xr_/3
Il=2*Il        #line current at start
x=(Vl/math.sqrt(3))**2/(Il**2)    #x=(Rs_+Rr_)**2+(Xs_+Xr_+Xe)**2
y=x-(Rs_+Rr_)**2                  #y=(Xs_+Xr_+Xe)**2
z=math.sqrt(y)                    #z=(Xs_+Xr_+Xe)
Xe=z-Xs_-Xr_


#results
print"(i)Maximum value of line current during starting Ist:",round(Ist),"A"
print"   Ratio of starting torque to full load torque :",round(ratio1,3)
print"   Ratio of maximum torque to full load torque :",round(ratio2,2)
print"\n(ii)transofrmation ratio aT:",round(aT,3)
print"    starting torque :",round(Tst1),"N-m"
#answer for the starting torque in the book is wrong due to accuracy
print"\n(iii)maixmum line current during starting :",round(Ist2),"A"
print"     Starting torque :",round(Tst2),"N-m"
#answer for the starting torque in the book is wrong due to accuracy
print"\n(iv)value of the reactor Xe:",round(Xe,3),"ohm"

(i)Maximum value of line current during starting Ist: 1251.0 A
   Ratio of starting torque to full load torque : 0.173
   Ratio of maximum torque to full load torque : 0.83

(ii)transofrmation ratio aT: 0.627
    starting torque : 7041.0 N-m

(iii)maixmum line current during starting : 2590.0 A
     Starting torque : 8539.0 N-m

(iv)value of the reactor Xe: 0.527 ohm

Example No:6.3,Page No:159

In [202]:
import math
from __future__ import division
import cmath

#variable declaration
#ratings of the star connected Induction motor
f=50           # frequency in HZ
Vl=400         # line voltage in V
P=6            # number of poles
#parameters referred to the stator
Xr_=2         # rotor winding reactance in ohm
Xs=Xr_        # stator winding reactance in ohm
Rr_=1         # resistance of the rotor windings in ohm
Rs=Rr_        # resistance of the stator windings in ohm

#calculation
Ns=120*f/P     #synchronous speed in rpm
Wms=2*math.pi*Ns/60        

#(i)
Sm=-Rr_/math.sqrt(Rs**2+(Xs+Xr_)**2)   #slip at maximum torque
x=math.sqrt((Rs+Rr_/Sm)**2+(Xs+Xr_)**2)
Ir_=(Vl/math.sqrt(3))/x        #current at maximum torque 
Tmax=(1/Wms)*3*Ir_**2*Rr_/Sm   #maximum torque
N=(1-Sm)*Ns                    #range of speed

#(ii)an overhauling torque of 100Nm
Tl=100   #overhauling torque in Nm
# Tl=(3/Wms)*(Vl**2*Rr_/s)/y
# where y=(Rs+Rr_/s)**2+(Xs+Xr_)**2
a=(1/Wms)*(Vl**2*Rr_)/(-Tl)    #a=s*(Rs+Rr_/s)**2+(Xs+Xr_)**2
a = 17
b = 17.3
c = 1
# calculate the discriminant
d = (b**2) - (4*a*c)
# find two solutions
s1 = (-b-cmath.sqrt(d))/(2*a)
s2 = (-b+cmath.sqrt(d))/(2*a)

N2=(1-s2)*Ns           #motor speed and we neglect s1 

#slight difference in the answer due to accuracy

#(iii)when a capacitive reactance of 2 ohm is inserted in each phase stator
Xc=2      #reactance of the capacitor
Sm=-Rr_/math.sqrt(Rs**2+(Xs+Xr_-Xc)**2)   #slip at maximum torque 
x=math.sqrt((Rs+Rr_/Sm)**2+(Xs+Xr_-Xc)**2)
Ir_=(Vl/math.sqrt(3))/x        #current at maximum torque 
Tmax_=(1/Wms)*3*Ir_**2*Rr_/Sm   #maximum overhauling torque with capacitor
ratio=Tmax_/Tmax


#results
print"(i)Maximum overhauling torque that the motor can hold is:",round(abs(Tmax),1),"N-m"
print"   Range of speed is from ",Ns,"to",round(N),"rpm"
print"\n(ii)Now s*(1+1/s)**2+16s=",a
print"    Or 17s**s+17.3s+1=0"
print"The solution for s are ",round(s1.real,3),"and",round(s2.real,3)
print"\nTherefore Motor speed is: ",round(abs(N2)),"rpm"
print"Note :There is a slight difference in the answer due to the decimal place"
print"\n(iii) Ratio of maximum torque with capacitor and to maximum torque without capacitor is:",round(ratio,2)

(i)Maximum overhauling torque that the motor can hold is: 244.6 N-m
   Range of speed is from  1000.0 to 1243.0 rpm

(ii)Now s*(1+1/s)**2+16s= 17
    Or 17s**s+17.3s+1=0
The solution for s are  -0.956 and -0.062

Therefore Motor speed is:  1062.0 rpm
Note :There is a slight difference in the answer due to the decimal place

(iii) Ratio of maximum torque with capacitor and to maximum torque without capacitor is: 2.53

Example No:6.4,Page No:162

In [203]:
import math
from __future__ import division
import cmath

#variable declaration
#ratings of the motor is same as that of Ex-6.3
f=50           # frequency in HZ
Vl=400         #line voltage in V
P=6            # number of poles
#parameters referred to the stator
Xr_=2         # rotor winding reactance in ohm
Xs=Xr_        # stator winding reactance in ohm
Rr_=1         # resistance of the rotor windings in ohm
Rs=Rr_        # resistance of the stator windings in ohm
N=950         #full load speed in rpm
SR=2.3        #stator to rotor turns ratio

#calculation
Ns=120*f/P     #synchronous speed in rpm
Wms=2*math.pi*Ns/60        
s=(Ns-N)/Ns     #full load slip
x=math.sqrt((Rs+Rr_/s)**2+(Xs+Xr_)**2)
Irf_=(Vl/math.sqrt(3))/x         #full load current
Tf=(1/Wms)*3*Irf_**2*Rr_/s       #full load  torque

#(i)initial braking current and torque
S=2-s     #during plugging at 950rpm
y=math.sqrt((Rs+Rr_/S)**2+(Xs+Xr_)**2)
Ir_=(Vl/math.sqrt(3))/y         #initial braking current
ratio1=Ir_/Irf_
T=(1/Wms)*3*Ir_**2*Rr_/S       #initial braking  torque
ratio2=T/Tf

#(ii)when an external resistance is connected such 
#that maximum braking current is 1.5 times the full load current
Ir_=1.5*Irf_
x=(Vl/math.sqrt(3))/Ir_    #x=math.sqrt((Rs+(Rr_+Re_)/S)**2+(Xs+Xr_)**2)
y=x**2                     #y=(Rs+(Rr_+Re_)/S)**2+(Xs+Xr_)**2
z=y-(Xs+Xr_)**2            #z=(Rs+(Rr_+Re_)/S)**2
a=math.sqrt(z)             #a=(Rs+(Rr_+Re_)/S)
b=(a-Rs)*S                 #b=(Rr_+Re_)
Re_=b-Rs
Re=Re_/SR**2
T=(1/Wms)*3*Ir_**2*(Rr_+Re_)/S       #initial braking  torque
ratio=T/Tf


#results
print"(i)Ratio of initial braking current to full load current is:",round(ratio1)
print"    Ratio of initial braking torque to full load torque is:",round(ratio2,2)
print"\n(ii)Ratio of initial braking torque to full load torque when the resistance is added is:",round(ratio,3)

(i)Ratio of initial braking current to full load current is: 5.0
    Ratio of initial braking torque to full load torque is: 0.64

(ii)Ratio of initial braking torque to full load torque when the resistance is added is: 1.426

Example No:6.5,Page No:165

In [204]:
import math
from __future__ import division
import cmath

#variable declaration
#ratings of the star connected Induction motor
f=50           # frequency in HZ
Vl=440         # line voltage in V
P=6            # number of poles
Vp=Vl/math.sqrt(3)  #phase voltage
#parameters referred to the stator
Xr_=1.2       # rotor winding reactance in ohm
Xs=Xr_        # stator winding reactance in ohm
Rr_=0.4       # resistance of the rotor windings in ohm
Rs=0.5        # resistance of the stator windings in ohm
Xm=50         # no load reactance
N=950         # full load speed
Sm=2          # slip at maximum torque

#calculation
Rr_=Sm*math.sqrt(Rs**2+(Xs+Xr_)**2)    #Since Sm=Rr_/math.sqrt(Rs**2+(Xs+Xr_)**2)
Ns=120*f/P      #synchronous speed in rpm
Wms=2*math.pi*Ns/60        
s=(Ns-N)/Ns            #slip at 950 rpm

x=1j*Xm*(Rr_/s+1j*Xr_)
y=Rr_/s+1j*(Xr_+Xm)
Zp=Rs+1j*Xs+x/y
Ip=Vp/math.sqrt(3)/Zp    #In the book the value of Ip is wrong which leads to other subsequent wrong answers in the book
Irp_=Ip*(1j*Xm)/(Rr_/s+1j*(Xr_+Xm))
Tp=(1/Wms)*3*abs(Irp_)**2*Rr_/s
x=1j*Xm*(Rr_/(2-s)+1j*Xr_)
y=Rr_/(2-s)+1j*(Xr_+Xm)
Zn=Rs+1j*Xs+x/y
In=Vp/math.sqrt(3)/Zn
Irn_=In*(1j*Xm)/(Rr_/(2-s)+1j*(Xr_+Xm))
Tn=-(1/Wms)*3*abs(Irn_)**2*Rr_/(2-s)    #In the book the value of In is wrong

T=Tp-Tn
I=abs(Ip)+abs(In)
Rr_=0.4    # from the parameters of the motor referred to the stator
x=math.sqrt((Rs+Rr_/s)**2+(Xs+Xr_)**2)
If=(Vl/math.sqrt(3))/x       #full load current
Tf=(1/Wms)*3*If**2*Rr_/s     #full load  torque

ratio1=I/If
ratio2=abs(T)/Tf

#results
print"Ratio of braking current to full load current is:",round(ratio1,3)
print"Ratio of braking torque to full load torque is:",round(ratio2,3)
#answer in the book is wrong

Ratio of braking current to full load current is: 1.448
Ratio of braking torque to full load torque is: 0.565

Example No:6.6,Page No:171

In [8]:
import math
from __future__ import division
from array import array
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cbook as cbook
%matplotlib inline

#variable declaration
#ratings of the star connected Induction motor which operates under dynamic braking
f=50           # frequency in HZ
P=4            # number of poles
#parameters referred to the stator
Xr_=3.01       # rotor winding reactance in ohm
Xs=Xr_         # stator winding reactance in ohm
Rr_=4.575      # resistance of the rotor windings in ohm
Rs=1.9         # resistance of the stator windings in ohm
J=0.1          # moment of inertia of the motor load system in kg-m2
Id=12          # given DC current

N=1500         #given asynchronous speed
#magnetization chacrateristic at the given asynchronous speed
Im=[0.13,0.37,0.6,0.9,1.2,1.7,2.24,2.9,3.9,4.9,6,8,9,9.5]       #magnetization current
E=[12.8,32,53.8,80,106,142,173,200,227,246,260,280,288,292]     #back emf

#calculation
torque=[]
speed=[]
temp=[]
Is=math.sqrt(2/3)*Id   #value of stator current for two lead connection
Wms=2*math.pi*N/60
for i in range (1,14,1):
    x=(Is**2-Im[i]**2)/(1+2*Xr_*Im[i]/E[i])   #x=Ir_**2
    Ir_=math.sqrt(x)       #required rotor current
    y=(E[i]/Ir_)**2-Xr_**2
    S=Rr_/math.sqrt(y)        #required slip
    N=S*Ns                    #required speed
    T=(3/Wms)*(Ir_)**2*Rr_/S  #required torque
    speed.append(round(N))
    torque.append(round(T,1))
    temp.append(round(T,1))
print"Hence the magnetizatioin curve is"
print"Speed             :",speed,"rpm"
for i in range(0,13,1):
    torque[i]=-1*torque[i]
print"Braking torque is :",torque,"N-m"

#results
#plot of of torque vs speed
plt.figure(1)
plt.plot(torque,speed)
plt.xlabel('Torque, N-m') 
plt.ylabel('Speed, rpm')
plt.title('Torque vs Speed')
plt.grid(True)
plt.tight_layout()
plt.show()

#plot of Wm vs J/T
inertia_over_torque=[]
for i in range(2,13,1):
    J_T=1000*J/temp[i]
    inertia_over_torque.append(round(J_T,4))
print"J/t :",inertia_over_torque 

Wm=[1,4,8,12,16,20,25,55,95,125,160]  
#the values of Wm are taken for the angular frequency with maximum value of Wms=50*pi rad/s
plt.figure(2)
plt.plot(Wm,inertia_over_torque)
plt.xlabel(r'Angular speed $\omega_m$') 
plt.ylabel(' $J/T,1*10^-2$')
plt.title(r'$J/T   vs   \omega_m$')
plt.grid(True)
x=[6.5,6.5]
y=[2,4.5]
plt.plot(x,y,'blue')
plt.annotate('A',xy=(8,2), xytext=(8,2.1), arrowprops=dict(facecolor='black', shrink=0),)
plt.annotate('B',xy=(8,4.5), xytext=(8,4.6), arrowprops=dict(facecolor='black', shrink=0),)
plt.annotate('C',xy=(80,3.4), xytext=(80,3.5), arrowprops=dict(facecolor='black', shrink=0),)
plt.annotate('D',xy=(157,8.3), xytext=(157,8.4), arrowprops=dict(facecolor='black', shrink=0),)
plt.annotate('E',xy=(157,2), xytext=(157,2.1), arrowprops=dict(facecolor='black', shrink=0),)
plt.show()

print"Hence from the plot the area ABCDEA between the curve and the speed axis for speed change "
print"for synchronous to 0.02 times synchrnous speed is the stopping time is which equal to: 9.36 sec"

Hence the magnetizatioin curve is
Speed             : [2975.0, 949.0, 578.0, 421.0, 306.0, 246.0, 207.0, 174.0, 150.0, 128.0, 86.0, 56.0, 34.0] rpm
Braking torque is : [-2.6, -8.3, -13.5, -18.4, -24.8, -30.0, -34.0, -36.9, -37.5, -35.9, -27.9, -19.5, -12.2] N-m

J/t : [7.4074, 5.4348, 4.0323, 3.3333, 2.9412, 2.71, 2.6667, 2.7855, 3.5842, 5.1282, 8.1967]

Hence from the plot the area ABCDEA between the curve and the speed axis for speed change 
for synchronous to 0.02 times synchrnous speed is the stopping time is which equal to: 9.36 sec

Example No:6.7,Page No:176

In [206]:
import math
from __future__ import division

#variable declaration
#ratings of the star connected Induction motor
f=50           # frequency in HZ
Vl=2200        # line voltage in V
P=6            # number of poles
#parameters referred to the stator
Xr_=0.5        # rotor winding reactance in ohm
Xs=Xr_         # stator winding reactance in ohm
Rr_=0.12       # resistance of the rotor windings in ohm
Rs=0.075       # resistance of the stator windings in ohm
J=100          # combine inertia of motor and load in kg-m2

#calculation
#(i) during starting of the motor
Sm=Rr_/math.sqrt(Rs**2+(Xs+Xr_)**2)   #slip at maximum torque
Wms=4*math.pi*f/P                     #angular frequency
x=Rs+math.sqrt(Rs**2+(Xs+Xr_)**2)
Tmax=(3/2/Wms)*(Vl/math.sqrt(3))**2/x  #maximum torque
tm=J*Wms/Tmax                          #mechanical time constant of the motor
ts=tm*(1/4/Sm+1.5*Sm)                  #time taken during starting
Es=1/2*J*Wms**2*(1+Rs/Rr_)             #energy disspated during starting

#(ii) when the motor is stopped by plugging method
tb=tm*(0.345*Sm+0.75/Sm)        #time required to stop by plugging
Eb=3/2*J*Wms**2*(1+Rs/Rr_)      #energy disspated during braking 

#(iii)required resistance to be inserted during plugging
tb1=1.027*tm       #minimum value of stopping time during braking
x=1.47*(Xs+Xr_)    #x=Rr_+Re
Re=x-Rr_           #Re is the required external resistance to be connected
Ee=3/2*J*Wms**2*(Re/(Re+Rr_))      #energy disspated in the external resistor
Eb1=Eb-Ee                           #total energy disspated during braking 


#results
print"(i)Time taken during starting is ts:",round(ts,4),"s"
print"   Energy disspated during starting is Es:",round(Es/1000),"kilo-watt-sec"
print"\n(ii)Time taken to stop by plugging is tb:",round(tb,2),"s"
print"    Energy disspated during braking is Eb:",round(Eb/1000),"kilo-watt-sec"
print"\n(iii)Minimum Time taken to stop by plugging is tb:",round(tb1,2),"s"
print"     Required external resistance to be connected is Re:",round(Re,2),"ohm"
print"     Total Energy dissipated during braking is Eb:",round(Eb1/1000,2),"kilo-watt-sec"

(i)Time taken during starting is ts: 1.108 s
   Energy disspated during starting is Es: 891.0 kilo-watt-sec

(ii)Time taken to stop by plugging is tb: 3.08 s
    Energy disspated during braking is Eb: 2673.0 kilo-watt-sec

(iii)Minimum Time taken to stop by plugging is tb: 0.5 s
     Required external resistance to be connected is Re: 1.35 ohm
     Total Energy dissipated during braking is Eb: 1162.36 kilo-watt-sec

Example No:6.8,Page No:184

In [4]:
import math
from __future__ import division
import cmath
from sympy import Symbol
from array import array
import numpy as np

#variable declaration
#ratings of the delta connected Induction motor
f=50           # frequency in HZ
Vl=400         # line voltage in V
P=4            # number of poles
Pm=2.8*1000    # rated mechanical power developed
N=1370         # rated speed in rpm
#parameters referred to the stator
Xr_=5          # rotor winding reactance in ohm
Xs=Xr_         # stator winding reactance in ohm
Rr_=5          # resistance of the rotor windings in ohm
Rs=2           # resistance of the stator windings in ohm
Xm=80          # no load reactance

#calculation
Ns=120*f/P                    #synchronous speed in rpm
Wms=2*math.pi*Ns/60           #synchronous speed in rad/s
s=(Ns-N)/Ns                   #full load slip
x=(Rs+Rr_/s)**2+(Xs+Xr_)**2   #total impedance
T=(3/Wms)*(Vl**2*Rr_/s)/x     #full load torque
Tl=T
K=Tl/(1-s)**2                 #since Tl=K*(1-s)**2

#(i) when the motor is running at 1200 rpm
N1=1200            #speed in rpm
s1=(Ns-N1)/Ns      #slip at the given speed N1
Tl=K*(1-s1)**2     #torque at the given speed N1

y=(Rs+Rr_/s1)**2+(Xs+Xr_)**2   #total impedance
a=Tl*(Wms/3)*y*(s1/Rr_)        #Since T=(3/Wms)*(Vl**2*Rr_/s)/x      and a=V**2
V=math.sqrt(a)                 #required voltage at the given speed N1
Ir_=V/((Rs+Rr_/s1)+1j*(Xs+Xr_))#rotor current
Im=V/(1j*Xm)                   #magnetizing current
Is=Ir_+Im                      #total current
Il=abs(Is)*math.sqrt(3)        #line current

#(ii)when the terminal voltage is 300 V
V1=300  #terminal voltage in V
s = Symbol('s')
x=(Rs+Rr_/s)**2+(Xs+Xr_)**2   
T=(3/Wms)*(V1**2*Rr_/s)/x

#Now we have to solve for the value of slip 's' from the given equation 104s**4- 188s**3 + 89s**2 - 179s + 25=0"
coeff = [104,-188,89,-179,25]    #coeffcient of the polynomial equation 
s=np.roots(coeff)                #roots of the polynomial equation

T=K*(1-s[3].real)**2     #torque at the given terminal voltage of 300 V
N=Ns*(1-s[3].real)       #speed at the given terminal voltage of 300 V
Ir_=V1/((Rs+Rr_/s[3].real)+1j*(Xs+Xr_))#rotor current
Im=V1/(1j*Xm)                   #magnetizing current
Is=Ir_+Im                       #total current
Il1=abs(Is)*math.sqrt(3)        #line current


#results
print"(i)Required torque is Tl:",round(Tl,1),"N-m"
print"   Required motor terminal voltage is V:",round(V,1),"V"
print"   Required line current is Il:",round(Il,2),"A"
print"\n(ii)The roots of the polynomial equation are "
print"  s1:",round(s[0].real,3),"s2:",round(s[1].real,3),"s3:",round(s[2].real,3),"s4:",round(s[3].real,3)
print"  Hence Only s4:",round(s[3].real,3),"is valid"
print"\nRequired torque is Tl:",round(T,2),"N-m"
print"Required speed is N:",round(N),"rpm"
print"Required line current is Il:",round(Il1,2),"A"

(i)Required torque is Tl: 36.9 N-m
   Required motor terminal voltage is V: 253.2 V
   Required line current is Il: 17.89 A

(ii)The roots of the polynomial equation are 
  s1: 1.818 s2: -0.079 s3: -0.079 s4: 0.147
  Hence Only s4: 0.147 is valid

Required torque is Tl: 41.94 N-m
Required speed is N: 1279.0 rpm
Required line current is Il: 16.88 A

Example No:6.9,Page No:199

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
#ratings of the star connected squirrel Induction motor
f=50           # frequency in HZ
Vl=400         # line voltage in V
P=4            # number of poles
N=1370         # rated speed

#the frequency variation is from 10 Hz to 50 Hz
fmin=10     
fmax=50
#parameters referred to the stator
Xr_=3.5        # rotor winding reactance in ohm
Xs=Xr_         # stator winding reactance in ohm
Rr_=3          # resistance of the rotor windings in ohm
Rs=2           # resistance of the stator windings in ohm

#calculation
Ns=120*f/P     #synchronous speed
N1=Ns-N        #increase in speed from no load to full torque rpm
Wms=2*math.pi*Ns/60
s=(Ns-N)/Ns    #full load slip
#(i)to obtain the plot between the breakdown torque and the frequency
K=0.1
k=[]
frequency=[]
torque=[]
for i in range (0,9):
    K=K+0.1
    f1=K*f
    x=Rs/K+math.sqrt((Rs/K)**2+(Xs+Xr_)**2)
    Tmax=(3/2/Wms)*(Vl/math.sqrt(3))**2/x
    k.append(round(K,1))
    frequency.append(round(f1))
    torque.append(round(Tmax,2))
print"K    :",k
print"f    :",frequency,"Hz"
print"Tmax :",torque,"N-m"

#plotting the values of Tmax vs f 
plt.figure(1)
plt.plot(frequency,torque,'y')
plt.xlabel('$f$,Hz')
plt.ylabel('$Tmax$,N-m')
plt.grid(True)
plt.title('Torque  vs frequency characteristic')
plt.show()

#(ii) to obtain the starting torque and current at rated frequency and voltage
x=(Rs+Rr_)**2+(Xs+Xr_)**2
Tst=(3/Wms)*(Vl/math.sqrt(3))**2*Rr_/x     #starting torque at 50 Hz frequency
Ist=(Vl/math.sqrt(3))/math.sqrt(x)         #starting current at 50 Hz frequency

K=fmin/fmax     #minimum is available at 10 Hz
y=((Rs+Rr_)/K)**2+(Xs+Xr_)**2
Tst_=(3/Wms)*(Vl/math.sqrt(3))**2*Rr_/K/y     #starting torque at 10 Hz frequency
Ist_=(Vl/math.sqrt(3))/math.sqrt(y)           #starting current at 10 Hz frequency

ratio1=Tst_/Tst    #ratio of starting torque to the rated starting torque
ratio2=Ist_/Ist    #ratio of starting current to the rated starting current


#results
print"\n(i)Hence from the plot we can see that for a constant (V/f) ratio breakdown torque decreases with frequency"
print"\n(ii)Hence the required ratio of starting torque to the rated starting torque is :",round(ratio1,3)
print"    Hence the required ratio of starting current to the rated starting current is :",round(ratio2,2)

K    : [0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]
f    : [10.0, 15.0, 20.0, 25.0, 30.0, 35.0, 40.0, 45.0, 50.0] Hz
Tmax : [22.93, 31.18, 37.44, 42.22, 45.94, 48.89, 51.27, 53.24, 54.88] N-m

(i)Hence from the plot we can see that for a constant (V/f) ratio breakdown torque decreases with frequency

(ii)Hence the required ratio of starting torque to the rated starting torque is : 0.549
    Hence the required ratio of starting current to the rated starting current is : 0.33

Example No:6.10,Page No:201

In [209]:
import math
from __future__ import division
from array import array
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline

#variable declaration
#ratings of the star connected squirrel Induction motor is same as that of Ex-6.9
f=50           # frequency in HZ
Vl=400         # line voltage in V
P=4            # number of poles
N=1370         # rated speed

#the frequency variation is from 5 Hz to 50 Hz
fmin=5     
fmax=50
#parameters referred to the stator
Xr_=3.5        # rotor winding reactance in ohm
Xs=Xr_         # stator winding reactance in ohm
Rr_=3          # resistance of the rotor windings in ohm
Rs=2           # resistance of the stator windings in ohm

#calculation
Ns=120*f/P     #synchronous speed
N1=Ns-N        #increase in speed from no load to full torque rpm
Wms=2*math.pi*Ns/60
s=(Ns-N)/Ns    #full load slip
Tmax=torque[8]   #maximum torque as obtain from Ex-6.9
#to obtain the plot between the voltage and the frequency
K=0
k=[]
frequency=[]
line_voltage=[]
for i in range (0,10):
    K=K+0.1
    f1=K*f
    x=2*K*Wms*Tmax/3
    y=Rs+math.sqrt((Rs)**2+(K*(Xs+Xr_))**2)
    Vl_square=3*x*y
    Vl=math.sqrt(Vl_square)
    k.append(round(K,1))
    frequency.append(round(f1))
    line_voltage.append(round(Vl,1))
print"K    :",k
print"f    :",frequency,"Hz"
print"Vl   :",line_voltage,"V"
#plotting the values of line voltage Vl vs f 
plt.figure(1)
plt.plot(frequency,line_voltage,'b')
plt.xlabel('$f$,Hz')
plt.ylabel('Line voltae,volts')
plt.grid(True)
plt.title('Line voltage  vs frequency characteristic')
#for constant V/f ratio
x=[0,10,20,30,40,50]
y=[0,80,160,240,320,400]
plt.plot(x,y,'--')
plt.annotate('Constant V/f ratiuo', xy=(21, 160), xytext=(30, 160),arrowprops=dict(facecolor='black', shrink=0),)
plt.show()

print"\nHence for a constant breakdown torque at all frequencies,"
print"V/f ratio has to be progressively increased with increase in frequency"

K    : [0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]
f    : [5.0, 10.0, 15.0, 20.0, 25.0, 30.0, 35.0, 40.0, 45.0, 50.0] Hz
Vl   : [84.3, 123.8, 159.2, 193.7, 228.0, 262.3, 296.7, 331.1, 365.5, 400.0] V

Hence for a constant breakdown torque at all frequencies,
V/f ratio has to be progressively increased with increase in frequency

Example No:6.11,Page No:202

In [210]:
import math
from __future__ import division

#variable declaration
#ratings of the star connected squirrel Induction motor is same as that of Ex-6.9
f=50           # frequency in HZ
Vl=400         # line voltage in V
P=4            # number of poles
N=1370         # rated speed
#parameters referred to the stator
Xr_=3.5        # rotor winding reactance in ohm
Xs=Xr_         # stator winding reactance in ohm
Rr_=3          # resistance of the rotor windings in ohm
Rs=2           # resistance of the stator windings in ohm

#calculation
Ns=120*f/P     #synchronous speed
N1=Ns-N        #increase in speed from no load to full torque rpm
Wms=2*math.pi*Ns/60 #synchronous speed
s=(Ns-N)/Ns    #full load slip
D=Ns-N        #drop in speed from no load to full load torque at 50 Hz

#(i)when the frequency is 30 Hz and 80% of full load torque
f1=30          #given frequency in Hz
d=D*0.8        #drop in speed from no load to 80% full load torque
Ns1=120*f1/P   #synchronous speed at the given frequency f1=30 Hz
N1=Ns1-d       #required motor speed 

#(ii)when the speed is 1000 rpm for a full load torque
N2=1000       #given speed in rpm
Ns2=N2+D      #synchronous speed
f2=P*Ns2/120   #required frequency

#when the speed is 1100 rpm and the frequency is 40 Hz
N3=1100       #given speed in rpm
f3=40         #given frequency in Hz
Ns3=120*f3/P  #synchronous speed at the given frequency f1=40 Hz
D1=Ns3-N3     #drop in speed from no load to N1=1100 rpm
x=(Rs+Rr_/s)**2+(Xs+Xr_)**2
Tf=(3/Wms)*(Vl/math.sqrt(3))**2*(Rr_/s)/x      #full load  torque
T1=D1/D*Tf    #required torque 


#results
print"(i)Hence the required motor speed is :",round(N1),"rpm"
print"\n(ii)Hence the required frequency is :",round(f2,2),"Hz"
print"\n(iii)Hence the required torque is :",round(T1,2),"N-m"

(i)Hence the required motor speed is : 796.0 rpm

(ii)Hence the required frequency is : 37.67 Hz

(iii)Hence the required torque is : 19.52 N-m

Example No:6.12,Page No:204

In [211]:
import math
from __future__ import division

#variable declaration
#ratings of the star connected Induction motor is same as that of Ex-6.9
f=50           # frequency in HZ
Vl=400         #line voltage in V
P=4            # number of poles
N=1370         #rated speed

#parameters referred to the stator
Xr_=3.5        # rotor winding reactance in ohm
Xs=Xr_         # stator winding reactance in ohm
Rr_=3          # resistance of the rotor windings in ohm
Rs=2           # resistance of the stator windings in ohm

#calculation
Ns=120*f/P     #synchronous speed
N1=Ns-N        #increase in speed from no load to full torque rpm
#(i)when f1=30Hz and 80% of full load
Ns=120*f/P
f1=30          #frequency 
N2=0.8*N1      #increase in speed from no load to 80% of full torque rpm
Ns1=f1/f*Ns
N=Ns1+N2       #machine speed

#(ii)at a speed of 1000rpm
N2=1000         #given speed in rpm
N3=N2-N1        #synchronous speed
f3=P*N3/120     #required frequency

#(iii)when frequency is 40Hz and speed is 1300 rpm
f4=40          #frequency in hz
N2=1300        #speed in rpm
Ns=120*f4/P    #required synchronous speed in rpm
N4=N2-Ns       #increase in speed from no load speed in rpm
Tf=25.37       #full load torque as calculated in Ex-6.11
Tm=-N4/N1*Tf   #motor torque

#(iv) when the motor is under dynamic braking


#results
print"(i)Required speed is :",N,"rpm"
print"\n(ii)required frequency is:",f3,"Hz"
print"\n(iii)Required motor torque :",round(Tm,2),"N-m"
print"\n(iv)The value of the frequency,speed and motor torque calculated in (i),(ii) and(iii)"
print"    will be the same  when the motor is operated under dynamic braking"

(i)Required speed is : 1004.0 rpm

(ii)required frequency is: 29.0 Hz

(iii)Required motor torque : -19.52 N-m

(iv)The value of the frequency,speed and motor torque calculated in (i),(ii) and(iii)
    will be the same  when the motor is operated under dynamic braking

Example No:6.13,Page No:204

In [212]:
import math
from __future__ import division

#variable declaration
#ratings of the star connected Induction motor is same as that of Ex-6.9
f=50           # frequency in HZ
Vl=400         #line voltage in V
P=4            # number of poles
N=1370         #rated speed
#parameters referred to the stator
Xr_=3.5        # rotor winding reactance in ohm
Xs=Xr_         # stator winding reactance in ohm
Rr_=3          # resistance of the rotor windings in ohm
Rs=2           # resistance of the stator windings in ohm

#calculation
Wms=4*math.pi*f/P
f1=60          #frequency in Hz during speed control of the motor
K=f1/f         #the value of K at 60Hz 
x=Rs+math.sqrt(Rs**2+K**2*(Xs+Xr_)**2)
Tmax_=3/(2*K*Wms)*(Vl/math.sqrt(3))**2/x       #torque at 60 Hz frequency
z=Rs+math.sqrt(Rs**2+(Xs+Xr_)**2)
Tmax=3/(2*Wms)*(Vl/math.sqrt(3))**2/z          #maximum torque
ratio=Tmax_/Tmax  #ratio

#results
print"Ratio of Motor breakdown torque at 60Hz to rated torque at 50Hz is:",round(ratio,3)

Ratio of Motor breakdown torque at 60Hz to rated torque at 50Hz is: 0.727

Example No:6.14,Page No:209

In [213]:
import math
from __future__ import division

print"When operating at a frequency K times rated frequency f then"
print"Im**2=[((Rr_/Ksf)**2+(2*pi*Lr_)**2)/((Rr_/Ksf)**2+(2*pi*Lm+2*Pi*Lr_)**2)]*Is**2----(1)"
print"Since Im is constant for constant flux therefore"
print"K*s*f=constant--------(2)"
print"K*Wms*s=constant-------(3) which is the slip speed"
print"s*K=constant----------(4)"
print"Thereofre for a frequency K*f"
print"T=(3/K/Wms)*[(Is*K*Xm)**2*(Rr_/s)/((Rr_/s)**2+K**2*(Xm+Xr_)**2]"
print"T=(3/K/Wms*s)*[(Is*Xm)**2*(Rr_)/((Rr_/s/K)**2+(Xm+Xr_)**2]-------(5)"
print"\nHence for a given slip speed (K*Wms*s),K*s is constant and from (1) for a given K*s*f and constant flux"
print"operation Is is fixed. Now from (5) T is also fixed. Thus, motor develps a constant torque and draws a"
print"constant current from the inverter at all frequencies for a given slip speed"

When operating at a frequency K times rated frequency f then
Im**2=[((Rr_/Ksf)**2+(2*pi*Lr_)**2)/((Rr_/Ksf)**2+(2*pi*Lm+2*Pi*Lr_)**2)]*Is**2----(1)
Since Im is constant for constant flux therefore
K*s*f=constant--------(2)
K*Wms*s=constant-------(3) which is the slip speed
s*K=constant----------(4)
Thereofre for a frequency K*f
T=(3/K/Wms)*[(Is*K*Xm)**2*(Rr_/s)/((Rr_/s)**2+K**2*(Xm+Xr_)**2]
T=(3/K/Wms*s)*[(Is*Xm)**2*(Rr_)/((Rr_/s/K)**2+(Xm+Xr_)**2]-------(5)

Hence for a given slip speed (K*Wms*s),K*s is constant and from (1) for a given K*s*f and constant flux
operation Is is fixed. Now from (5) T is also fixed. Thus, motor develps a constant torque and draws a
constant current from the inverter at all frequencies for a given slip speed

Example No:6.15,Page No:210

In [214]:
import math
from __future__ import division

#variable declaration
#ratings of the star connected Induction motor 
f=50           # frequency in HZ
Vl=400         #line voltage in V
P=4            # number of poles
N=1370         #rated speed
#parameters referred to the stator
Xr_=3.5        # rotor winding reactance in ohm
Xs=Xr_         # stator winding reactance in ohm
Rr_=3          # resistance of the rotor windings in ohm
Rs=2           # resistance of the stator windings in ohm
Xm=55          # magnetizing reactance in ohm

#calculation
Ns=120*f/P     #synchronous speed in rpm
s=(Ns-N)/Ns    #full load slip
x=1j*Xm*(Rr_/s+1j*Xr_)
y=Rr_/s+1j*(Xr_+Xm)
Z=Rs+1j*Xs+x/y        #total motor impedance
Isf=(Vl/math.sqrt(3))/abs(Z)                  #full load stator current
Irf_=Isf*(1j*Xm)/(Rr_/s+1j*(Xr_+Xm))          #full load rotor current
Tf=(3/Wms)*abs(Irf_)**2*Rr_/s                 #full load  torque
N1=Ns-N        #full load slip speed
#(i) when the motor is operating at 30Hz
f1=30          #given frequency in Hz
#at rated slep speedvalue of torque and stator current is same as the rated value
T=Tf 
Is=abs(Isf)    
Ns1=f1/f*Ns    #synchronous at f1=30Hz
N2=Ns1-N1      #required motor speed at 30Hz

#(ii)at a speed of 1200 rpm
N3=1200       #speed in rpm
Ns1=N3+N1     #required synchronous speed
f1=Ns1/Ns*f   #required frequency at N2=1200rpm

#(iii)when speed-torque curves are assumed to be straight lines at 30Hz at half the rated motor torque
f2=30          #frequency in Hz
N1_=N1/2       #slip at half the rated torque
Ns1=f2/f*Ns    #synchronous at f1=30Hz
N4=Ns1-N1_     #required motor speed

#(iv) When the motor is operating at 45hz  and braking torque is equal to rated torque
f3=45          #given frequency in Hz
N1_=-N1        #slip speed when braking at rated torque
Ns1=f3/f*Ns    #synchronous at f1=45Hz
N5=Ns1-N1_     #required motor speed


#results
print"(i)At 30Hz the required value of Torque is T:",round(T,2),"N-m "
print"   Stator current is Is:",round(Is,4),"A"
print"   Motor speed is :",N2,"rpm"
print"\n(ii)Required inverter frequency is :",round(f1,2),"Hz"
print"\n(iii)Required motor speed at 30Hz is:",N4,"rpm"
print"\n(iv)Required motor speed at 45Hz is:",N5,"rpm"

(i)At 30Hz the required value of Torque is T: 22.71 N-m 
   Stator current is Is: 7.2442 A
   Motor speed is : 770.0 rpm

(ii)Required inverter frequency is : 44.33 Hz

(iii)Required motor speed at 30Hz is: 835.0 rpm

(iv)Required motor speed at 45Hz is: 1480.0 rpm

Example No:6.16,Page No:215

In [217]:
import math
import cmath
from __future__ import division

#variable declaration
#ratings of the Delta connected slipring Induction motor
f=50           # frequency in HZ
Vl=400         #line voltage in V
P=6            # number of poles
SR=2.2         #ratio of stator to rotor
#parameters referred to the stator
Xr_=1          # rotor winding reactance in ohm 
Rr_=0.2        # resistance of the rotor windings in ohm
s=0.04         # given slip when motor runs at full load

#calculation
Ns=120*f/P     #synchronous speed
Wms=2*math.pi*Ns/60
x=(Rr_/s)**2+Xr_**2
Tf=(3/Wms)*(Vl)**2*(Rr_/s)/x      #full load  torque
K=Tf/(Ns*(1-s))**2
N=850                  #speed of the motor in rpm
Tl=K*N**2              #torque at the given speed N
s=(Ns-N)/Ns            #slip at the given speed N
y=Tl*(Wms/3)/Vl**2     #y=X/(X**2+Xr_**2) and X=(Re+Rr_)/s
print"The torque at the given speed of 850rpm is:",round(Tl),"N-m"
print"With a slip of s:",s
print"To find the external resistance connected the given quadratic equation is X**2-6.633X+1=0"
print"With X=(Re-Rr_)/s where Re is the required external resistance" 
a = 1
b = -1/y
c = 1
# calculate the discriminant
d = (b**2) - (4*a*c)
# find two solutions
X1 = (-b-cmath.sqrt(d))/(2*a)
X2 = (-b+cmath.sqrt(d))/(2*a)

#results
print"The solution for X are ",round(X1.real,4),"and",round(X2.real,4)
Re1=X1*s-Rr_
Re2=X2*s-Rr_

if (Re1.real>0) :
    print"\nThe number Re1:",round(abs(Re1),3),"ohm is feasible:"
    R=Re1.real/SR**2
    print"Rotor referred value of the external resistance is:",round(R,3),"ohm"

if (Re2.real>0) :
    print"\nThen  Re2:",round(abs(Re2),3),"ohm is feasible" 
    R=Re2.real/SR**2
    print"Hence Rotor referred value of the external resistance is:",round(R,3),"ohm" 

The torque at the given speed of 850rpm is: 691.0 N-m
With a slip of s: 0.15
To find the external resistance connected the given quadratic equation is X**2-6.633X+1=0
With X=(Re-Rr_)/s where Re is the required external resistance
The solution for X are  0.1544 and 6.4786

Then  Re2: 0.772 ohm is feasible
Hence Rotor referred value of the external resistance is: 0.159 ohm

Example No:6.17,Page No:217

In [7]:
import math
from __future__ import division
from sympy import Symbol

#variable declaration
#ratings of the star connected Induction motor 
f=50           # frequency in HZ
Vl=440         #line voltage in V
P=6            # number of poles
Ns=120*f/P     #synchronous speed
#parameters referred to the stator
Xr_=1.2        # rotor winding reactance in ohm
Xs=Xr_         # stator winding reactance in ohm
Rr_=0.4        # resistance of the rotor windings in ohm
Rs=0.5         # resistance of the stator windings in ohm
Xm=50          # magnetizing reatance
a=3.5          # stator to rotor turns ratio
delta=0        # duty ratio  at the given breakdown torque
Sm=1           # slip at standstill
#calculation
#slip at maximum torque without an external resistance is Sm=Rr_/math.sqrt(Rs**2+(Xs+Xr_)**2)   
#when an external resistanc Re referred to the stator is connected 
x=Sm*math.sqrt(Rs**2+(Xs+Xr_)**2)     #x=Re+Rr_
Re=x-Rr_
y=0.5*a**2*(1-delta)    # y=0.5*a**2*R*(1-delta)   #y=Re
R=Re/y

N = Symbol('N')
Sm=(Ns-N)/Ns
c=(x*Sm-Rr_)/(0.5*a**2*R)  #c=(1-delta)
delta=1-c                  #given duty ratio

#results
print"variation of the duty ratio is",round(delta*1000/N,3),"*N*10**(-3)"
print"Hence the duty ratio must change linearly with speed N"

variation of the duty ratio is 1.195 *N*10**(-3)
Hence the duty ratio must change linearly with speed N

Example No:6.18,Page No:223

In [219]:
import math
from __future__ import division
import cmath

#variable declaration
#ratings of the star connected Induction motor 
f=50         # frequency in HZ
Vl=440       # line voltage in V
P=6          # number of poles
N=970        # rated speed
n=2          # ratio of stator to rotor
Sm=0.25      # it is given the speed range is 25% below the synchronous speed which is proportional to the Slip
#parameters referred to the stator
Xr_=0.4      # rotor winding reactance in ohm
Xs=0.3       # stator winding reactance in ohm
Rr_=0.08     # resistance of the rotor windings in ohm
Rs=0.1       # resistance of the stator windings in ohm
alpha=165    # maximum value of the firing angle in degress

#calculation
Ns=120*f/P   # synchronous speed
Wms=2*math.pi*Ns/60
#(i) transformer turns ratio
a=-Sm/math.cos(math.radians(alpha))  #since  Sm=a*math.cos(alpha)   
m=n/a        #since a=n/m where m is the transformer ratio

#(ii)When speed is 780 rpm and firing angle is 140 degrees
N1=780           #given speed
alpha1=140       #given firing angle
s1=(Ns-N1)/Ns    #slip at the given speed N1
Vd1=(3*math.sqrt(6)/math.pi)*s1*(Vl/math.sqrt(3))/n
Vd2=(3*math.sqrt(6)/math.pi)*(Vl/math.sqrt(3))/m*math.cos(math.radians(alpha1))
Rs_=Rs*(1/n)**2   #stator resistance referred to the rotor
Rr=Rr_*(1/n)**2   #rotor resistance referred to the rotor
Rd=0.01           #equivalent resistance of the DC link inductor
Id=(Vd1+Vd2)/(2*(s1*Rs_+Rr)+Rd)
T1=abs(Vd2)*Id/s1/Wms     #required torque

#(iii)when speed is 800rpm and firing angle is half the rated motor torque
N1=800           #given speed
s=(Ns-N)/Ns      #rated slip
x=(Rs+Rr_/s)**2+(Xs+Xr_)**2
Trated=(3/Wms)*(Vl/math.sqrt(3))**2*(Rr_/s)/x          #rated torque
T_half=Trated/2        #half rated torque
s1=(Ns-N1)/Ns          #given slip at speed N1=800rpm
Vd1=(3*math.sqrt(6)/math.pi)*s1*(Vl/math.sqrt(3))/n
cos_alpha1 = Symbol('cos_alpha1')
Vd2=(3*math.sqrt(6)/math.pi)*(Vl/math.sqrt(3))/m*cos_alpha1
Id=(Vd1+Vd2)/(2*(s1*Rs_+Rr)+Rd)
T=abs(Vd2)*Id/s1/Wms     #required torque
#let cos_alpha1=-X
X=-cos_alpha1
#since the given torque is half of the rated value
#To find the find the firing angle we assumed cos(alpha1)=-X
#The given quadratic equation is X**2-0.772X+0.06425=0
a = 1
b = -0.772
c = 0.06425
# calculate the discriminant
d = (b**2) - (4*a*c)
# find two solutions
X1 = (-b-cmath.sqrt(d))/(2*a)
X2 = (-b+cmath.sqrt(d))/(2*a)
alpha1=-math.acos(X2.real)   #since cos(alpha1)=-X where alpha1 is radians
alpha1=math.degrees(alpha1)  #angle in degrees
alpha1=180+alpha1            #required firing angle


#results
print"(i)Transformer ratio is: ",round(m,3)
print"\n(ii)Required torque is :",round(T1),"N-m"
print"    There is a slight difference in the answer for the torque due to accuracy i.e more number of decimal place"
print"\n(iii)The half rated torque at the given speed of",N1,"rpm is:",round(T_half,2),"N-m"
print"      With a slip of s:",s1
print"The solution for X are ",round(X1.real,4),"and",round(X2.real,4)
print"For X1:",round(X1.real,4)," the motor is unstable hence we discard this value and we use X2:",round(X2.real,4)
print"\nHence the required firing angle is :",round(alpha1,1),"°"

(i)Transformer ratio is:  7.727

(ii)Required torque is : 271.0 N-m
    There is a slight difference in the answer for the torque due to accuracy i.e more number of decimal place

(iii)The half rated torque at the given speed of 800 rpm is: 302.66 N-m
      With a slip of s: 0.2
The solution for X are  0.0949 and 0.6771
For X1: 0.0949  the motor is unstable hence we discard this value and we use X2: 0.6771

Hence the required firing angle is : 132.6 °

Example No:6.19,Page No:225

In [220]:
import math
from __future__ import division
import cmath

#variable declaration
#ratings of the star connected Induction motor is same as that of Ex-6.17
f=50           # frequency in HZ
Vs=440         # line voltage in V
P=4            # number of poles
#parameters referred to the stator
Xr_=1.2        # rotor winding reactance in ohm
Xs=Xr_         # stator winding reactance in ohm
Rr_=0.4        # resistance of the rotor windings in ohm
Rs=0.5         # resistance of the stator windings in ohm
Xm=50          # magnetizing reatance
a=3.5          # stator to rotor turns ratio

#calculation
Ns=120*f/P              # synchronous speed in rpm
Wms=2*math.pi*Ns/60     # synchronous speed in rad/s
#(i)when motor speed is 1200rpm with a voltage of 15+0j V
V=cmath.rect(15,0)
N1=1200  #speed in rpm
Vr_=a*V          #rotor voltage
s1=(Ns-N1)/Ns    #slip at the given speed N1=1200 rpm
Z=Rs+Rr_/s1+1j*(Xs+Xr_)                   #total impedance
Ir_=(Vs/math.sqrt(3)-Vr_/s1)/Z            #rotor current
phi_r=cmath.phase(Vr_)-cmath.phase(Ir_)   #angle between Vr_ and Ir_
Pr=3*(abs(Ir_))**2*Rr_                    #rotor copper loss
P1=3*abs(Vr_)*abs(Ir_)*math.cos(phi_r)    #power absorbed by Vr_
Pg=(Pr+P1)/s1              #gross power 
T=Pg/Wms                   #required motor torque

#(ii)when motor speed is 1200rpm with a unity power factor 
N1=1200  #speed in rpm
Ir_=cmath.rect(abs(Ir_),0)   #machine is operating at unity power factor
x=Ir_*Z                      #x=(Vs-Vr_/s1)*phi_r   where phi_r is the angle between Vr_ and Ir_

#x=a+b
d=(Vs/math.sqrt(3)-Vr_/s1*math.cos(phi_r))**2
e=(Vr_/s1*math.sin(phi_r))**2
f=x/(d+e)
theta=cmath.phase(f)        #required angle in radian

#Now we should solve for the quadratice equation for the rotor current
#  0.9*Ir_**2 + 50.8*Ir_ + 90.12 = 0
a1 = 0.9
b1 = 50.8
c1 = 90.12
# calculate the discriminant
d = (b1**2) - (4*a1*c1)
# find two solutions
Ir_1 = (-b1-cmath.sqrt(d))/(2*a1)
Ir_2 = (-b1+cmath.sqrt(d))/(2*a1)

Ir_=Ir_2           #Ir_2 is chosen because for Ir_1 the motor is unstable
Vr_sin_phi_r=abs(Ir_)/2.083
Vr_cos_phi_r=s1*(Vs/math.sqrt(3)+2.5*Vr_sin_phi_r)
Vr_=Vr_cos_phi_r+1j*Vr_sin_phi_r   #total rotor voltage referred to the stator
Vr_=Vr_/a                          #total  rotor voltage referred to the rotor
phase=round(math.degrees(cmath.phase(Vr_)))   #required phase of the voltage in degrees

#results
print"(i)The torque is :",round(T,2),"N-m and since it is negative the motor is operating in regenerative braking "
print"\n(ii)Now theta θ:",round(math.degrees(theta),2),"◦"
print"    The solution for Ir_ are ",round(Ir_1.real,3),"and",round(Ir_2.real,3)
print"    We choose Ir_:",round(Ir_2.real,3),"A since higher value corresponds to unstable region"
print"\n    Hence the required voltage magnitude is Vr:",round(abs(Vr_),2),"V",";phase:",phase,"◦"
print"\n    Note :There is a slight difference in the answers due to accuracy i.e more number of decimal place"

(i)The torque is : -8.61 N-m and since it is negative the motor is operating in regenerative braking 

(ii)Now theta θ: 43.83 ◦
    The solution for Ir_ are  -54.611 and -1.834
    We choose Ir_: -1.834 A since higher value corresponds to unstable region

    Hence the required voltage magnitude is Vr: 14.64 V ;phase: 1.0 ◦

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

Example No:6.20,Page No:234

In [221]:
import math
from __future__ import division
import cmath

#variable declaration
#ratings of the single phase Induction motor 
f=50           # frequency in HZ
Vs=220         # supply voltage in V
P=4            # number of poles
N=1425         # rated speed in rpm
#parameters referred to the stator
Xr_=6          # rotor winding reactance in ohm
Xs=Xr_         # stator winding reactance in ohm
Rr_=5          # resistance of the rotor windings in ohm
Rs=2           # resistance of the stator windings in ohm
Xm=60          # magnetizing reatance

#calculation
N1=1200      #when the motor is operating at the given speed in rpm
Ns=120*f/P   # synchronous speed
Wms=2*math.pi*Ns/60
s=(Ns-N)/Ns   #rated slip

Zf=1j*(Xm)*(Rr_/s+1j*Xr_)/2/(Rr_/s+1j*(Xr_+Xm))
Rf=Zf.real  ;    Xf=Zf.imag
Zb=1j*(Xm)*(Rr_/(2-s)+1j*Xr_)/2/(Rr_/(2-s)+1j*(Xr_+Xm))
Rb=Zb.real  ;    Xb=Zb.imag
Zs=Rs+1j*Xs
Z=Zs+Zf+Zb
Is=(Vs)/Z
T=(abs(Is))**2/Wms*(Rf-Rb)
Tl=T
K=Tl/N**2

#therefore for a speed of  of N1=1200 rpm we get 
Tl=K*N1**2      #required load torque for the given speed N1
s1=(Ns-N1)/Ns   # slip for the given speed N1

Zf=1j*(Xm)*(Rr_/s1+1j*Xr_)/2/(Rr_/s1+1j*(Xm))
Rf=Zf.real  ;    Xf=Zf.imag
Zb=1j*(Xm)*(Rr_/(2-s1)+1j*Xr_)/2/(Rr_/(2-s1)+1j*(Xr_+Xm))
Rb=Zb.real  ;    Xb=Zb.imag
x=(Wms*Tl)/(Rf-Rb)    #since Tl=(abs(Is))**2/Wms*(Rf-Rb)   and x=Is**2
Is=math.sqrt(x)
Z=Zs+Zf+Zb
V=Is*abs(Z)

#results
print"Hence the motor terminal voltage at the speed of",N1,"rpm is : ",round(V,1),"V"

Hence the motor terminal voltage at the speed of 1200 rpm is :  127.9 V