# Chapter4, Electrical Drives and Industrial Applications¶

## Ex1 : Page 191¶

In [1]:
from __future__ import division
from sympy import symbols, solve
#To Determine Motor Parameters of Series Motor rated at 220V

V=220  #Rated Voltage
N=1000  #Rated Speed
Ish=20  #Current without Armature being shunted
Ra=0.15  #Armature Resistance
Rf=0.2  #Field Resistance
Eb=V-(Ish*(Ra+Rf))  #Back EMF
K=Eb/(Ish*N)  #Motor Constant
Rsh=20  #Shunt Resistance

#After Armature is shunted
x=symbols('x')  #Field Current Variable Value
Ia=(Ish**2)/x  #Armature Current #Since Torque is assumed constant
I20=(V-(x*Rf))/Rsh  #Current Through the Shunt resistance

X=Ia+I20-x  #Polynomial to find field Current

x=solve(X, x)  #Numerical Value of Field Current

#Conditon to find the Positive Root
if abs(x[0]+x[1])==0:
x=x[1]
else:
x=x[0]

Ia=(Ish**2)/x  #Armature when shunted
Ebsh=V-(Ia*Ra)-(x*Rf)  #Back Emf when shunted
Nsh=Ebsh/(K*x)  #Speed when armature is shunted

#When Load is not there, Ia = 0
If=V/(Rsh+Rf)  #Field Current at No Load
Ebnl=V-(If*Rf)  #Back Emf at No Load

Rex=5  #External Resistance
#At Starting when Eb = 0
Is=V/(Rex+Rf+(Ra*Rsh/(Ra+Rsh)))  #Supply Current
Ifs=Is  #Field Current
Ias=Ifs*Rsh/(Rsh+Ra)  #Armature Current At the Start

#Torque is directly proportional to square of Current
T20 = (Ish**2)   #Torque at shunted condition
Tas= Ias*Ifs  #Torque at start

RT=Tas/T20  #Ratio of Torques

print 'a) The Speed of Motor if the Load Torque remains constant & magnetic circuit is assumed unsaturated is %g RPM'%abs(Nsh )
print 'b) The No Load Speed of the Motor is %0.f RPM'%Nnl
print 'c) The Starting Torque represented in terms of torque corresponding 20A (without armature shunted) is %0.1f times'%RT
# Answer is not accurate for part 1 in the book.

a) The Speed of Motor if the Load Torque remains constant & magnetic circuit is assumed unsaturated is 1403.42 RPM
b) The No Load Speed of the Motor is 1878 RPM
c) The Starting Torque represented in terms of torque corresponding 20A (without armature shunted) is 4.2 times


## Ex2 : Page 193¶

In [2]:
#To Determine the resistance to be inserted for various cases

N1=1500  #Rated Speed
N2=1200  #Reduced Speed
rs=N2/N1  #Ratio of final speed to original speed
eff=80/100  #Efficiency
Pr=10*(10**3)  #Power Rating
V=250  #Voltage supply
Rf=110  #Field Resistance
Ra=0.25  #Armature Resistance
If=V/Rf  #Field Current
Eb=V-(Ia1*Ra)  #Back EMF
Eb1200=Eb*rs  #Back EMF for 1200 rpm

Res =lambda y:((V-Eb1200)/y) #Function to calculate the Resistance of various cases
#Torque directly proportional to current
Ia2=(rs)*Ia1  # Torque directly propostional to speed
Ia3=(rs**2)*Ia1  #Torque directly proportional to square of the speed

#Resistances for Various Cases
R1=Res(Ia1)-Ra
R2=Res(Ia2)-Ra
R3=Res(Ia3)-Ra

Pow = lambda y:((V-Eb1200)/y)*(y**2)/1000  #Function to Find Power Loss
#Power Lost in kW for various cases
P1=Pow(Ia1)
P2=Pow(Ia2)
P3=Pow(Ia3)

print 'The External Resistances to be connected and the power loss for '
print 'a) The Load Torque is independent of the speed : %0.f ohm and %0.2f kW respectively '%(R1,P1)
print 'b) The Load Torque is directly proportional to the speed : %0.2f ohm and %0.3f kW respectively '%(R2,P2)
print 'c) The Load Torque is directly proportional to square of the speed : %0.2f ohm and %0.3f kW respectively '%(R3,P3)

The External Resistances to be connected and the power loss for
a) The Load Torque is independent of the speed : 1 ohm and 2.84 kW respectively
b) The Load Torque is directly proportional to the speed : 1.31 ohm and 2.274 kW respectively
c) The Load Torque is directly proportional to square of the speed : 1.70 ohm and 1.819 kW respectively


## Ex 3 : Page 194¶

In [3]:
from sympy import symbols, diff, solve
from fractions import Fraction
#To determine the speed for which the torque is maximum

#In The Figure, The author has taken the resistance in series with the motor as 0.2 ohm, but in the figure it is given as 1 ohm

#This Doesn't Affect the calculation of the speed but it does affect the Maximum Torque

#So if we consider 1 ohm we get 0.8333 instead of 2.5

#The Equation for the Torque is found out to be 2.5*K2*(V**2)*(1.1-K1*w)/((76.1-(50*K1*w)))

w=symbols('w')  #Variable Value of w

#Lets assume the value of 2.5*K2*(V**2) = 1 and K1 = 1 to particularly to calculate the co-effcients

T=2.5*(1.1-w)/((76.1-(50*w))**2)  #Torque

#Maximum Torque is derivative of the above equation
X=diff(T, w)  #Polynomial to find the value co - effcient of w
w=solve(X, w)
w=w[0]  #We Choose value less than 1 to suit the differenciation process

Tmax=2.5*(1.1-w)/((76.1-(50*w))**2)  #Maximum Torque Co-Efficient

print 'The Speed at which the torque is Maximum is (%g/K1) rad/sec'%w
print 'The Maximum Torque is (K2*(V**2)*%g*10**-4) Nm '%(Tmax/(10**-4))

The Speed at which the torque is Maximum is (0.678/K1) rad/sec
The Maximum Torque is (K2*(V**2)*5.92417*10**-4) Nm


## Ex 4 Page 195¶

In [4]:
from sympy import symbols, solve
#To Determine the slip at maximum torque and at full load
from math import floor
from __future__ import division
Tl=1  #Assume that the Full Load torque is unity
Tst=125*Tl/100  #Starting Torque
Tmax=275*Tl/100  #Maximum Torque
si=1  #At the Start
RT=Tst/Tmax  #Ratio of Starting Torque and Maximum Torque
#R1 is negligible, elminating the terms with R1 and hence formulating the starting torque and Maximum Torque we get

# Tst/Tmax = 2*sTmax*s/((sTmax**2)+(s**2))
#Where sTmax is the Slip at Maximum Torque and s is the slip at the start

sTmx=symbols('sTmx')  #Variable Value of slip at Maximum Torque

X=RT-(2*sTmx*si/((sTmx**2)+(si**2)))  #Polynomial to find sTmax

sTmx = solve(X, sTmx)

#Condition to find the value of sTmax less than 1
if floor(sTmx[0])==0:
sTmx=sTmx[0]
else:
sTmx=sTmx[1]

s=symbols('s')  #Variable value of slip for Starting torque at full load

Y=(Tl/Tmax)-(2*sTmx*s/((sTmx**2)+(s**2)))  #Polynomial to find s

s=solve(Y, s)  #Numerical Value of s

#Condition to find the value of s less than 1
if floor(s[0])==0 :
s=s[0]
else:
s=s[1]

print 'i) The Slip at Maximum Torque is %0.2f'%(sTmx)
print 'ii) The Slip at Full Load is %0.3f'%(s)

i) The Slip at Maximum Torque is 0.24
ii) The Slip at Full Load is 0.045


## Ex 5 : Page 195¶

In [5]:
#To Determine the Maximum Torque as ratio of nominal value of voltage
from math import sqrt

#Resistances and Reactance for Nomrmal Value of Volttage and Frequency
w=50  #Frequency of supply
V=1  #Assumed Voltage
R1=0.05
R2=0.05
x1=0.3
x2=0.3

#Resistance and reactance at half voltage and half frequency
wh=50/2  #Frequency of supply
Vh=1/2  #Assumed Voltage
R1h=0.05
R2h=0.05
x1h=0.3/2
x2h=0.3/2

#Function to Determine the Maximum Torque
Max = lambda a,b,c,d,e:(3*(a**2))/(2*b*(c+(sqrt((c**2)+((d+e)**2)))))
#Function to Determine to the Starting Torque
Start =lambda a,b,c,d,e,f:(3*(a**2)*f)/(b*(((c+f)**2)+((d+e)**2)))
#Maximum Torques and Starting Torques at
#Normal Voltage And Frequency
Tm=Max(V,w,R1,x1,x2)
Tst=Start(V,w,R1,x1,x2,R2)
#Half Voltage and Half Frequency
Tmh=Max(Vh,wh,R1h,x1h,x2h)
Tsth=Start(Vh,wh,R1h,x1h,x2h,R2h)

#Ratio of torques as fraction of its normal value
R1=Tmh/Tm
R2=Tsth/Tst

print 'i)The Maximum Torque at the reduced value of supply as a fraction of its normal value is %0.4f'%R1
print 'ii)The Starting Torque at the reduced value of supply as a fraction of its normal value is %0.2f'%R2

i)The Maximum Torque at the reduced value of supply as a fraction of its normal value is 0.9207
ii)The Starting Torque at the reduced value of supply as a fraction of its normal value is 1.85


## Ex 6 : Page 196¶

In [6]:
#To Determine the starting torque and starting current for various starters
from math import pi
V=400 #Rated Voltage #Phase Voltage #Delta Connected
Pr=50*735.5  #Rated Power
N=750  #Rated Speed
Z=2.5  #Impedance per phase
w=2*pi*N/60  #Angular Frequency
Ist=V/Z  #Starting Current
Isl=sqrt(3)*Ist  #Line Current

#DOL Starter
Tstd=((Isl/Il)**2)*s*Tl  #Starting Torque

#Star - Delta Starter
Tsts=Tl*((1/sqrt(3))**2)  #Starting Torque

#Autotransformer
tap=70/100  #Tapping
Tsta=(tap**2)*Tl  #Starting Torque

print 'The Starting Current for all the three starter are the same that is %0.2f A'%Isl
print 'The Starting torque for: '
print 'i) D.O.L Starter : %0.2f Nm '%(Tstd)
print 'ii) Star -Delta Starter : %0.2f Nm '%(Tsts)
print 'ii) Auto Transformer with 70 percent tapping : %0.2f Nm '%(Tsta)

The Starting Current for all the three starter are the same that is 277.13 A
The Starting torque for:
i) D.O.L Starter : 647.29 Nm
ii) Star -Delta Starter : 156.08 Nm
ii) Auto Transformer with 70 percent tapping : 229.43 Nm


## Ex 7 : Page 197¶

In [7]:
#To Detemine Minimum Starting Current drawn from the supply
from math import exp, pi, acos, degrees, sin
from sympy import symbols, solve
TR=0.5  #Torque Ratio, Tst:Tl
V=400  #Supply Voltage
Pr=30*735  #Power Rating
N=500  #Rated Speed
eff=0.85  #Efficieny
pf=0.88  #Power Factor
Is=150  #Short Circuit Current
pfs=0.25  #Power Factor of the Short Cicuit Current
#Case 1
x=symbols('x')  #Tapping Percent
X=TR-(((Is/Il)**2)*(x**2)*s)  #Polynomial to determine the 'x'
x=solve(X, x)
x=x[0]  #Taking the Postive Root
Ist1=Is*x*x  #Starting Current numerical Value
#Case 2
Ist2=sqrt((TR*(Il**2)/s))  #Starting Current for Full Voltage
Z1=V/(sqrt(3)*Ist2)  #impedance for the starting current at full voltage
Z2=complex((V/(sqrt(3)*Is))*pfs,(V/(sqrt(3)*Is))*sin(acos(pfs)))  #Phasor form of the Impedance of for the short cicuit current
R=sqrt((Z1**2)-(Z2.imag**2))
Rs=R-Z2.real  #Resistance to be connected the stator circuit

print 'i) The Tapping of the Transformer is %0.1f percent and the starting Current is %0.2f A\n'%(abs(x*100),Ist1)
print 'ii) The Starting Current is %0.2f A and the Resistance to be added to the stator circuit is %0.4f ohm\n'%(Ist2,Rs)
# Answer in the textbook is not accurate.

i) The Tapping of the Transformer is 89.7 percent and the starting Current is 120.69 A

ii) The Starting Current is 134.55 A and the Resistance to be added to the stator circuit is 0.4658 ohm



## Ex 8 : Page 198¶

In [8]:
#To Determine the value of resistor connected in series to run the fan at a particular speed

Pf=100*(10**3)  #Power rating of the fan
f=50  #Supply Frequency
V=400  #Supply Line Voltage
V1=V/sqrt(3)  #Phase Voltage
SR=1.4  #Ratio of stator to rotor turns
Rr=0.03  #Rotor Resistance per phase
R2=SR*SR*Rr  #Rotor Resistance referred to the stator
N=240  #Rated Speed
w=2*pi*N/60  #Angular Frequency
P=24  #number of poles
Ns=120*f/P  #Synchronous Speed
ws=2*pi*Ns/60  #Synchronous angular Frequency
s=(Ns-N)/Ns  #Slip
X2=symbols('X2')  #Reactance of Rotor referred to stator, Variable Value

X=T-(3*(V1**2)*R2/(ws*s*(((R2/s)**2)+(X2**2))))  #Polynomial Expression to find X2
X2=solve(X, X2)  #Numerical Value of Rotor Resistance
X2=X2[0]
N1=180  #Speed of the fan
s1=(Ns-N1)/Ns  #Slip
T1=T*((N1/N)**2)  #Torque
R21=symbols('R21')  #Variable Value
Y=T1-(3*(V1**2)*R21/(ws*s1*(((R21/s1)**2)+(X2**2))))  #Polynomial Expression to find R21
R21=solve(Y, R21)  #Numerical Value
#R21 = 0.7545054 and 0.0100813
#We have to choose the value of R21 such that it is greater than R2

R21=R21[1]
Rex=(R21-R2)/(SR*SR)  #External Resistance

print 'The Value of Resistance to be connected in series so that teh fan runs at 180 rpm is %0.3f ohm'%Rex

The Value of Resistance to be connected in series so that teh fan runs at 180 rpm is 0.355 ohm


## Ex 9 : Page 212¶

In [9]:
from math import pi, sqrt
#To Determine the time taken to start the motor
HP=746  #According to the TextBook
P=150*HP  #Power Rating
Lim=1.5  #Limited Current Factor
Eshp=5000  #energy stored per hp
N=750  #Rated Speed
w=N*2*pi/60  #Angular Frequency
Es=Eshp*P/HP  #Total Energy Stored
Ts=Tfl*Lim  #Starting Torque
Ta=Ts-Tfl  #Torque Available for acceleration
I=2*Es/(w**2)  #Moment of Inertia
a=Ta/I  #Angular Accleration
#Intergrating wrt angular frequency
f = lambda x : 1/a # function for integration
t=quad(f,[0,w])  #Time Taken to start #Angular Acceleration is constant

print '''The time taken to start the motor if the load torque is equal to full load torque,
during the starting period,
& the current is limited to 1.5 times the full load current is %0.2f seconds '''%(t)

The time taken to start the motor if the load torque is equal to full load torque,
during the starting period,
& the current is limited to 1.5 times the full load current is 26.81 seconds


## Ex 10 : Page 213¶

In [10]:
from __future__ import division
#To time taken and the number of revolutions made before the motor is stopped

P=50*735.5  #Power Rating
V=400  #Rated Voltage
N=750  #Rated Speed
w=N*pi*2/60  #Angular Frequency
I=20  #Moment Inertia
eff=90/100  #Overall Efficiency
Rb=2  #Braking Resistor

Ilb=V/(sqrt(3)*Rb)  #The Line Current at the start of braking
Te=Tm*Ilb/Ifl  #Torque for electric braking

Tb=Te+Tm  #Total Braking torque

K=Te/w  #Motor constant
B=Tb/I  #Retardation

f1 = lambda x:(I/(Tm+(K*x))) # function for integration
ts=quad(f1,[0,w]) #Time Taken for the motor to stop

f2 = lambda t:(((Tm+(K*w))*exp(-1*K*t/I))-Tm) # function for integration

#Note The answer in the textbook is wrong, Calculation error

print 'The Time Taken for the Motor to stop is %0.2f seconds\n'%ts
print 'The Number of revolutions undergone before it stops is %0.2f revolutions which is approx %g revolutions\n'%(NoR,round(NoR))

The Time Taken for the Motor to stop is 1.90 seconds

The Number of revolutions undergone before it stops is 9.81 revolutions which is approx 10 revolutions



## Ex 11 : Page 219¶

In [11]:
#To Determine the 30 mins rating of motor
from math import exp

Pr=100  #Power Rating in horse power
Tc=90*3600  #Time Constant
Tr=30*3600  #Time to determine the rating
theta=50  #Temperature Rise
theta1=50  #Temperature rise for the new case

P=symbols('P')  #Power rating for 30min variable value

#Loss Corresponding to 30 mins

P30=(P/Pr)**2  # Times the Power loss corresponding to the 100 hp motor

theta2=theta*P30  #Final Temperature rise if run  continuously

X=theta1-(theta2*(1-exp(-1*Tr/Tc)))  #Polynominal to find the 30min rating

P=solve(X, P)  #Numerical Value
#We Consider the Positive Real Value of the root

P=P[0]  #Power Rating of the 1/2 Hr

print 'The 1/2 Hr Power Rating is %0.1f hp'%abs(P )

The 1/2 Hr Power Rating is 187.8 hp


## Ex12 : Page 219¶

In [12]:
#To Determine the Continous Rating of the Motor for Suitable Conditions

T=120  #Total Time
t1=0  #Power Rise to 1050
t2=15  #Constant Power of 600
t3=85  #Regenerative Breaking at 200
t4=95  #Motor At Rest
t5=120  #End
#Note Power Rating are in Horse Power
HP1=1050  #First Rise
T1=(t2-t1)  #Time for Rise of the Horse Power
HP2=600  #Constant Power
T2=(t3-t2)  #Time For which the power is constant
HP3=200  #Braking Initial Point
T3=(t4-t3)  #Time Period of Breaking

#Different Intergration

f1 = lambda t:((HP1/T1)*t)**2 # function for integration
f2 = lambda t:(HP2**2) # function for integration
f3 = lambda t:((HP3/T3)*t)**2 # function for integration

HPrms=sqrt((I1+I2+I3)/120)

HPav=(((HP1*T1/2)+(HP2*T2)+(HP3*T3/2))/T)

print 'i) The H.P Rating according to the RMS Value of loading is %g hp\n'%HPrms
print 'ii) The H.P Rating according to the Average Value of Loading is %0.f hp\n'%HPav

i) The H.P Rating according to the RMS Value of loading is 507 hp

ii) The H.P Rating according to the Average Value of Loading is 424 hp



## Ex 13 : Page 220¶

In [13]:
#To estimate the final temperature of the machine and its time constant

Eff=90/100  #Efficiency
P=15*735.5  #Power Rating in W
D=75*(10**-2)  #Cylinder Diameter
L=120*(10**-2)  #Length of the Cylinder
OSHD=12  #Outer Surface Heat Dissipation
A=pi*L*D  #Curved Surface Area
M=450  #Motor Weight
SH=700  #Specific Heat

HTC=M*SH/(A*OSHD*3600)  #The Heating Time Constant
Loss=(P/Eff)-P  #Loss in Motor
print 'Loss in Motor = %0.1f W'%Loss
FT=Loss/(A*OSHD)  #Final  Temperature
OSHD2=20  #Semi Closed Cooling Capacity
LossSC=FT*A*OSHD2  #Loss in Semi Closed System
print 'Final  Temperature = %0.2f degree C'%FT
X=(L2/(L2+LossSC)-Eff)  #Polynomial to Calculate L2
L2=solve(X, L2)  #Numerical Value of The Load

print 'The Final Temperature of the Machine & its Heating Time Constant is %0.2f degree Celsius and %0.2f Hrs respectively'%(FT,HTC)
print 'The Load which could be delivered is %0.f W '%(L2[0])
Rating = L2[0]/736 # hp
print 'Rating = %0.f h.p.'%Rating

Loss in Motor = 1225.8 W
Final  Temperature = 36.13 degree C
The Final Temperature of the Machine & its Heating Time Constant is 36.13 degree Celsius and 2.58 Hrs respectively
The Load which could be delivered is 18388 W
Rating = 25 h.p.


## Ex 14 : Page 223¶

In [14]:
#To Determine the Moment of intertia for a slip of 8 percent
from math import log
To=0
s=8/100  #Slip
t=15 #time period for which the load torque was applied
Tm=675 # Limited motor Torque
Wo=2*pi*N/60  #Angular Frequency
K=s/Tm  #Motor Constant
I=(t/(Wo*K*log((Tl-To)/(Tl-Tm))))  #Moment of Inertia

print 'The Moment of inertia of the flywheel is %g Kg m**2'%I
#Please Note the Calculation Mistake in the book. (Power of e)

The Moment of inertia of the flywheel is 1453.01 Kg m**2


## Ex15 : Page 224¶

In [15]:
#To determine the Moment of intertia of the flywheel

Pi=100*(10**3)  #Power Rating
P=6  #Poles of the Machine
N=950  #Rated Speed
Wo=2*pi*N/60  #Angular Frequency
Tm=Pi*60/(2*pi*N)  #Rated Torque
Tmax=2*Tm  #Maximum Torque
Tl=Tmin+Ta  #Total Torque
tp=15  #Time for which the additional torque is applied
#Assuming Slip Characteristic to Be a Straight line
s=0.05

I=tp/(Wo*K*log((Tl-Tmin)/(Tl-Tmax)))  #Moment Of intertia

Tm1=1500  #Trasistion Torque

t=I*Wo*K*log((Tmax-Tmin)/(Tm1-Tmin))  #Time Taken to become 1500 Nm

print 'a) The Moment of intertia of the flywheel for Maximum Torque to be twice the rated torque is %0.1f Nm'%I
print 'b) The Time Taken after the removal of additional load before the motor torque becomes 1500 Nm is %0.2f seconds '%t

a) The Moment of intertia of the flywheel for Maximum Torque to be twice the rated torque is 4772.7 Nm
b) The Time Taken after the removal of additional load before the motor torque becomes 1500 Nm is 10.61 seconds


## Ex16 : Page 231¶

In [16]:
#To determine the speed and torque of a single phase full converter
from math import cos
P=5*735.5  #Power Rating
Vr=220  #Rated Voltage
N=1500  #Rated Speed

Ra=0.25  #Armature Resistance
Ia=20  #Rated Armature Current
Vs=250  #Supply Voltage
Kaphi=0.0278  #Motor Voltage Constant
a=30  #Firing Angle

Va=2*sqrt(2)*Vs*cos(a*pi/180)/pi  #Armature Voltage
Eb=Va-(Ia*Ra)  #Back EMF
w=Eb/(Kaphi*60)  #Angular Frequency in Radians per second
Ta=Kaphi*Ia*60  #Torque

print 'The Speed and Torque of the Single Phase Full Converter are %0.1f rad/s and %g Nm respectively '%(w,Ta)

The Speed and Torque of the Single Phase Full Converter are 113.9 rad/s and 33.36 Nm respectively


## Ex17 : Page 232¶

In [17]:
#To Determine the motor current and torque for a particular firing angle

P=15*735.5  #Power Rating of the Motor
Raw=0.2  #Combined Armature and Field winding resistance
N=1000  #Speed
K=0.03  #Motor Constant
a=30  #Firing Angle
E=250  #Supply Voltage

#Semi Converter
w=2*pi*N/60  #Angular Frequency
Vas=sqrt(2)*E*(1+cos(a*pi/180))/pi
Ias=Vas/(Raw+(K*w))  #Armature Current
Tas=K*(Ias**2)  #Motor Torque

#FullConverter
Vaf=2*sqrt(2)*E*cos(a*pi/180)/pi
Iaf=Vaf/(Raw+(K*w))  #Armature Current
Taf=K*(Iaf**2)  #Motor Torque

#Accurate Caclculation, Rounding Off not done.
print 'The Motor Current And Motor Torque for:'
print 'i) Semi Converter : %0.2f A and %0.2f Nm'%(Ias,Tas)
print 'ii) Full Converter : %0.2f A and %0.2f Nm'%(Iaf,Taf)

The Motor Current And Motor Torque for:
i) Semi Converter : 62.84 A and 118.48 Nm
ii) Full Converter : 58.33 A and 102.08 Nm


## Ex18 : Page 234¶

In [18]:
#To determine the pulse width for a particular average value of back EMF

E=220 #DC Supply
Raw=0.2  #Combined Armature and Field Winding Resistance
Ia=25 #Average Current
f=200 #Chopper Frequency
Eb=100  #Average Value of Back EMF

Ton=Eav*1000/(E*f)  # Pulse Width in m sec

print 'The Pulse Width for the 100V back EMF is %0.2f m seconds'%Ton

The Pulse Width for the 100V back EMF is 2.39 m seconds


## Ex 19 : Page 235¶

In [19]:
#To determine the range of speed control and duty cycle
E=220  #DC supply voltage
Ra=0.2
Kaphi=0.08  # Motor Constant
Ia=25  #Average Current
Eav=Ia*Ra  #Voltage Drop

#Eav = lamda * E
#For Maximum Speed, lamba =1

lamda=Eav/E

lamda1=1  #Maximum Speed
Eb= (E*lamda1)-(Ia*Ra)  #Back EMF

Speed = Eb/Kaphi  # Maximum Speed

print 'i) The Range of speed control is 0<N<%0.f '%(Speed)
print 'ii) The Range of duty cycle is %g<lamda<%g'%(lamda,lamda1)

i) The Range of speed control is 0<N<2688
ii) The Range of duty cycle is 0.0227273<lamda<1