# Chapter7, Electric Traction¶

## Ex1 : Page 352¶

In [1]:
from __future__ import division
from sympy import symbols, solve
from numpy import array
#To determine the Acceleration and Coasting Period

D=1.92
Ts=20  #Duration of stops
Vsch=40
V1=60.8
#Retardation
B=3.2
Bc=0.16

SchTime=D*3600/Vsch  # Schedule time in seconds

T=SchTime-Ts  #Time of Travel

a=symbols('a')  # Acceleration Unknown Value

V2=(V1-Bc*(T-(V1/a)))/(1-(Bc/B))  # From the Speed Time Curve

X=((V1+V2)*T-(V1*V2*((1/B)+(1/a))))-(7200*D)  #Polynomial Equation to find a
a=solve(X, a)  #Roots of the Characteristic Equation
#To Determine the positive roots
Y=array([abs(a[0])+a[0], abs(a[1])+a[1] ])
if Y[0]==0:
a=a[1]
else:
a=a[0]

Ta=V1/a  #Duration of acceleration
V2=(V1-Bc*(T-(V1/a)))/(1-(Bc/B))  # From the Speed Time Curve
Tcs=(V1-V2)/Bc  #Coasting Time Period

print 'The Acceleration of the Train is %0.2f Kmphp '%a
print 'The Coasting Period is %0.2f seconds'%Tcs

The Acceleration of the Train is 2.13 Kmphp
The Coasting Period is 110.80 seconds


## Ex2 page 353¶

In [2]:
#To determine the duration of acceleration and others
D=1.6
Vav=40
Bc=0.16
B=3.2
a=2  #Acceleration

T=D*3600/Vav  #Time of travel

V1=symbols('V1')  #Variable Value

V2=(V1-Bc*(T-(V1/a)))/(1-(Bc/B))  # From the Speed Time Curve

#Formula #7200*D = (V1+V2)*T-V1*V2(1/B+1/alfa)
X=((V1+V2)*T-(V1*V2*((1/B)+(1/a))))-(7200*D)  #Polynomial Equation to find V1
V1=solve(X, V1)  #Numerical Values for V1
V1=V1[0]  #Train Considered have speed close to this value
V2=(V1-Bc*(T-(V1/a)))/(1-(Bc/B))  # From the Speed Time Curve
Ta=V1/a  #Acceleration Time
Tc=(V1-V2)/Bc  #Coasting Time
Tb=V2/B  #Braking time
#Distance are calculated according to the area under their respective curves
Da=(Ta*V1/2)/3600  #Distance during acceleration
Dc=(((V1-V2)*Tc/2)+(V2*Tc))/3600  #Distance during coasting
Db=(Tb*V2/2)/3600  #Distance during braking

print 'The Time periods and the distance covered for:'
print 'Acceleration : %.2f seconds and %0.3f km '%(Ta,Da)
print 'Coasting : %0.2f seconds and %0.3f km '%(Tc,Dc)
print 'Braking : %0.2f seconds and %0.3f km '%(Tb,Db)

The Time periods and the distance covered for:
Acceleration : 27.06 seconds and 0.203 km
Coasting : 105.30 seconds and 1.336 km
Braking : 11.65 seconds and 0.060 km


## Ex3: page 354¶

In [3]:
#To determine the acceleration required to run the service

D=1
Ts=20  #Stopping Time
B=3
Vsh=30
ShT=D*3600/Vsh  #Schedule time
T=ShT-Ts  #Actual Run Time
Vav=D*3600/T  #Average Speed
Vm=1.25*Vav
a=symbols('a')  #Acceleration Variable

X=((2*Vm*T)-((Vm**2)*((1/a)+(1/B))))-(7200*D)  #Polynomial Equation to find 'a'

a=solve(X, a)[0]  #Numerical Value of the Acceleration

print 'The Accleration required to run the service is %g Kmphps'%a

The Accleration required to run the service is 1.8 Kmphps


## Ex4 : Page 356¶

In [4]:
#To determine the time taken by the train to attain a particular speed

M=200  #Mass of train
g=9.81  #Acceleration due to gravity
W=M*g
RI=10/100  #Rotational Inertia
We=W*(1+RI)
motor=4  #Number of Motors
Tm=5000  #Torque per motor
T=Tm*motor
V=40  #Speed to be Attained
N=0.9  #Gear Efficiency
Y=3.6/1  #Gear Ratio
R=91.5*(10**-2)/2
Ft=N*T*Y/R
r=40
a=symbols('a')  #Acceleration

X=((28.3*We*a)+(10*W*G)+(M*r))-Ft  #Polynomial Equation to find acceleration

a=solve(X)[0]  #Numerical Value of Acceleration
Time=V/a  #Time Taken to attain the required the necessary speed

print 'The Time Taken by the Train to attain 40 Kmphs is %0.2f seconds'%Time

The Time Taken by the Train to attain 40 Kmphs is 19.73 seconds


## Ex7.5 :Page 363¶

In [5]:
#To determine the Factors affeting the mechanics of the Train

D=1.92
Ts=20  #Duration of stops
Vsch=40
V1=60.8
#Retardation
B=3.2
Bc=0.16
SchTime=D*3600/Vsch  # Schedule time in seconds
T=SchTime-Ts  #Time of Travel
a=symbols('a')  # Acceleration Unknown Value
V2=(V1-Bc*(T-(V1/a)))/(1-(Bc/B))  # From the Speed Time Curve
X=((V1+V2)*T-(V1*V2*((1/B)+(1/a))))-(7200*D)  #Polynomial Equation to find a
a=solve(X,a)[1] #taking +ve root

Ta=V1/a  #Duration of acceleration
V2=(V1-Bc*(T-(V1/a)))/(1-(Bc/B))  # From the Speed Time Curve
Tc=(V1-V2)/Bc  #Coasting Time
Tb=V2/B  #Braking time
#Distance are calculated according to the area under their respective curves
Da=(Ta*V1/2)/3600  #Distance during acceleration
Dc=(((V1-V2)*Tc/2)+(V2*Tc))/3600  #Distance during coasting
Db=(Tb*V2/2)/3600  #Distance during braking

r=4.53*9.81  #Train Resistance in N per tonne
M=200  #Mass of Train
AWF=1.1  #Accelerating Weight Factor
SEOA=0.010726*(V1**2)*(AWF)/D  #Specific Energy Output during acceleration
SEOAr=0.2778*r*Da/D  #Specific Energy Output during acceleration against train resistance
SEO=SEOA+SEOAr  #Specific Energy for the run
SEOB=0.010726*(V2**2)*(AWF)/D  #Specific Energy Output during braking
SEOC=SEOA-SEOB  #energy utilized during coasting
rc=SEOC*D/(0.2778*Dc)#Mean Train Resistance during coasting

print '\ni)The Specific Energy Output for the Run is %0.3f Whr/Tonne-km\n'%SEO
print 'ii) The Energy Dissipated by the brakes is %0.3f Whr/Tonne-km\n'%SEOB
print 'iii)The Energy Utilized during coasting is %0.3f Whr/Tonne-km\n'%SEOC
print 'iv) The Mean Train Resistance during coasting is %0.2f N/tonne\n'%rc

i)The Specific Energy Output for the Run is 24.266 Whr/Tonne-km

ii) The Energy Dissipated by the brakes is 11.400 Whr/Tonne-km

iii)The Energy Utilized during coasting is 11.316 Whr/Tonne-km

iv) The Mean Train Resistance during coasting is 48.93 N/tonne



## Ex6 : Page 364¶

In [6]:
#To determine the Specific Energy Output Runs for Both the Cases
from sympy import symbols, solve
G=(1/80)*100
W=210
We=1.1*W
Vm=symbols('Vm')  #Variable Value of The Max Velocity
D=1.6  #Distance between Stations
Ftg=10*W*G  #Tractive Effort due to Gradient
Fta=13050  #Total Tractive Effort for Acceleration
Ftb=20445  #Total Tractive Effort for Retardation
ron=5.43  #Train Resistance when power on
rc=6.525  #Train Resistance when coasting
#Level Track
al=2
Bl=3.2

Vavu=40  #Average Speed
Tu=D*3600/Vavu  #Total Time Taken
au=(Fta-Ftg)/(28.3*We)  #Acceleration
Bu=(Ftb+Ftg)/(28.3*We)  #Retardation
Xu=((2*Vm*Tu)-((Vm**2)*((1/au)+(1/Bu))))-(7200*D)  #Polynomial Equation to Find Vm
Vmu=solve(Xu, Vm)[0]  #Numerical Value of Vm
Tua=Vmu/au  #Accelerating Period
Tub=Vmu/Bu  #Braking Period
Tucs=Tu-Tua-Tub  #Constant Speed Duration
Dua=Vmu*Tua/(2*3600)  #Distance Travelled During Accleration in km
Dub=Vmu*Tub/(2*3600)  #Distance Travelled During Braking in km
Duon=D-Dub  #Distance run with power on
SEOua=0.010726*(Vmu**2)*(We/W)/D  #Specific Energy Output during Acceleration
SEOug=27.25*G*Duon/D  #Specific Energy Output for Gradient
SEOur=0.2778*ron*9.81*Duon/D  #Specific Energy Output for resitance
TSEOu=SEOua+SEOug+SEOur  #Total Specific Output

Vavd=44  #Average Speed
Td=D*3600/Vavd  #Total Time Taken
Bd=(Ftb-Ftg)/(28.3*We)  #Retardation
Xd=((2*Vm*Td)-((Vm**2)*((1/ad)+(1/Bd))))-(7200*D)  #Polynomial Equation to Find Vm
Vmd=solve(Xd, Vm)[0]  #Numerical Value of Vm
Tdb=Vmd/Bd  #Braking Period
Tdcs=Td-Tda-Tdb  #Constant Speed Duration
Dda=Vmd*Tda/(2*3600)  #Distance Travelled During Accleration in km
Ddb=Vmd*Tdb/(2*3600)  #Distance Travelled During Braking in km
Ddon=D-Ddb  #Distance run with power on
SEOda=0.010726*(Vmd**2)*(We/W)/D  #Specific Energy Output during Acceleration
#Net Force Acting Downward due to gradient and resistance
Fnet=W*((10*G)-rc)
#Since Fnet is Postive, To run the train at constant speed brakes will have to applied therefore they cannot supply electric energy
TSEOd=SEOda  #Total Sepcific Enegy Consumption

print 'The Total Specific Energy:'
print 'Up Gradient : %0.2f Whr/Tonne km'%TSEOu
print 'Down Gradient : %0.2f Whr/Tonne km '%TSEOd

The Total Specific Energy:
Up Gradient : 62.48 Whr/Tonne km
Down Gradient : 20.05 Whr/Tonne km


## Ex 7: Page 366¶

In [7]:
#To determine the specific energy consumption
from __future__ import division
W=400
G=1
a=1.5  #Acceleration
Ta=30  #Acceleration Time
Tf=36  #Free running Period
Tc=25  #Coasting Period
B=2.6
r=45
RI=10/100  #Rotational Inertia Effect
Eff=75/100  # Overall Efficieny
g=9.81  #Accleration due to gravity

Vm=Ta*a
We=W*(1+RI)

#Distance Covered is equal to the area under the speed time curve
Da=Vm*Ta/(2*3600)  #Acceleration
Df=Vm*Tf/3600  #Free Run

Ftf=W*(r+(10*g*G))  #Tractive effort during free run period

Fta=We*277.8  #Tractive effort due to acceleration

#During Coasting, The Accelrating force is equal to Tractive effort during free run
#Retardation due to coasting is
Bc=Ftf/Fta

V2=Vm-(Bc*Tc)  #Speed of train after the coasting period
Tb=V2/B  # Braking period

Dc=((((Vm-V2)*Tc)/2)+(V2*Tc))/3600  #Distance covered during coasting
Db=V2*Tb/(2*3600)  #Distance covered during braking

D=Da+Df+Dc+Db  #Total Distance

D1=Da+Df  #Distance for which the energy is being spent

SEO=(0.010726*(Vm**2)*We/(W*D))+(27.25*G*D1/D)+(0.2778*r*D1/D)  #Specific Energy Output
#Note the Calculated Specific Energy Output during accleration is wrong.

SEOa=SEO/Eff  #Actual Specific Energy Output with 75% efficieny

print 'The Specific Energy Consumption is %0.2f Whr/Tonne Km'%SEOa

The Specific Energy Consumption is 67.78 Whr/Tonne Km


## Ex 8 : Page 368¶

In [8]:
#To determine the weight of the locomotive
from math import ceil
Wg=300  #Weight of the train to be hauled
RI=10/100  #Rotation inertia
Wa=20  #Permissible Weight of axle load
r=45
G=2
a=1   #Acceleration
Wl=symbols('Wl')  #Variable  Weight of locmotive
W=Wg+Wl  #Total Weight of the train
We=W*(1+RI)
Ft=((277.8*a*We/W)+(98.1*G)+r)*W

Fmax=(9.81*1000*Ma*Wl)

X=Ft-Fmax  #Polynomial Equation to find Wl

Wl=solve(X, Wl)[0]  #Numerical Value of the Weight of the locomotive

NoA=Wl/Wa  #Number of axles

print 'The weight of the locomotive and the number of axles is %0.1f tonnes and %g axles respectively\n'%(Wl,ceil(NoA))

The weight of the locomotive and the number of axles is 115.9 tonnes and 6 axles respectively



## Ex 9 : Page 370¶

In [9]:
#To deduce the speed current characteristic of the motor
from numpy import array
Ia=array([10,20,30,40,50,60,70] ) #Current
T=array([45,130,230,350,470,610,765])  #Torque
V=440  #operating voltage
r=0.5  #Armature Circuit Resistance

N=9.55*(V-(Ia*r))/(T/Ia)  #Speed

print 'For the Given Current: '
for ia in Ia:
print "%0.f "%ia,
print '\nThe Speeds tabulated are:'
for n in N:
print "%0.f "%n,
# Answer in the textbook are not accurate.

For the Given Current:
10  20  30  40  50  60  70
The Speeds tabulated are:
923  632  529  458  422  385  354


## Ex10: Page371¶

In [10]:
#To determine the speed torque curve for the series motor
from numpy import linalg
I=array([50,100,150,200,250,300])  #Field Current
Eb1=array([230,360,440,500,530,580])  #Armature Volts
Ra=0.07  #Armature Winding Resistance
Rf=0.05  #Field Resistance
Rt=Rf+Ra  #Total Resistance
P=4  #Poles of the machine
N1=600  #Series Motor Speed
Vc=600  #Constant Operating Voltage

# Note that the Suffix 1 and 2 have given according the question  1 stands for the case where the magnetisation curve has been given and 2 stands for the case where we have to find the speed torque curve

Eb2=Vc-(I*(Rt))
N2=N1*Eb2/Eb1  #Speed

T=9.55*Eb2*I/N2  #Torque

print 'The Tabulated Speeds for 600V are:\n'
for n2 in N2:
print "%0.f "%n2,
print '\n\nAnd their corresponding torques are:\n'
for t in T:
print "%0.f "%t,
# Answer in the textbook are not accurate.

The Tabulated Speeds for 600V are:

1550  980  794  691  645  583

And their corresponding torques are:

183  573  1050  1592  2109  2770


## Ex11 : Page 378¶

In [11]:
#To determine the horse power delivered by the locomotive
from math import sqrt

Ftl=35000  #Tractive Effort on a level surface
Ftg=55000  #Tractive Effort on a gradient
V=50  #Speed of the train
HP=735.5  #One Horse Power
Pl=Ftl*V*1000/(3600*HP)  #Power Output(in HP) of the Locomotive on the level track

#CASE 1:
#D.C Series Motor, Power is directly proportional to the root of the Trative Effort
HP1=Pl*sqrt(Ftg/Ftl)

#CASE 2:
#3 Phase Induction Motor, Power is directly proportional to the Trative Effort
HP2=Pl*(Ftg/Ftl)

print 'The Horse Power delivered by the locmotive when the motors used are:'
print 'i) D.C Series Motor is %0.1f H.P '%HP1
print 'ii) 3 Phase Induction Motor is %0.1f H.P '%HP2
#Note the Calculation Mistake in TextBook for case 2

The Horse Power delivered by the locmotive when the motors used are:
i) D.C Series Motor is 828.5 H.P
ii) 3 Phase Induction Motor is 1038.6 H.P


## Ex12 : Page 378¶

In [12]:
#To Determine the tractive effort shared by the two locomotives

Da=1.27  #Loco A wheel diameter
Db=1.244  #Loco B wheel diameter
S=5  #Slip of both the Locomotives in Case 1
Sa=S  #Slip of Loco A in Case 2
Sb=4  #Slip of Loco B in Case 2
Ftta=11325  #Total Tractive Effort in Case 1
Fttb=1309  #Total Tractive Effort in Case 2
Ft=5227  #Full Load Tractive Effort of Loco

x=symbols('x')  #Variable for Tractive Effort Exerted by Loco A

#Slips are same
#For 11325 kg
#% Speed
NA=100-(S*x/Ft)
NB=100-(S*(Ftta-x)/Ft)
Rna=NB/NA  #Speed Ratio

Y1=Rna-(Da/Db)  #Polynomial to find out 'x'
x=solve(Y1, x)[0]  #Numerical Value of Tractive Effort shared by A

#Tractive Efforts Shared by A Loco and B Loco
FtAas=x
FtBas=Ftta-x

#For 1309 kg
#% Speed
x=symbols('x')  #Variable for Tractive Effort Exerted by Loco A
NA=100-(S*x/Ft)
NB=100-(S*(Fttb-x)/Ft)
Rn1=NB/NA  #Speed Ratio

Y2=Rn1-(Da/Db)  #Polynomial to find out 'x'
x=solve(Y2, x)[0]  #Numerical Value of Tractive Effort shared by A

#Tractive Efforts Shared by A Loco and B Loco
FtAbs=x
FtBbs=Fttb-x

#Different Slips
#For 11325 kg
#% Speed
x=symbols('x')  #Variable for Tractive Effort Exerted by Loco A
NA=100-(Sa*x/Ft)
NB=100-(Sb*(Ftta-x)/Ft)
Rna=NB/NA  #Speed Ratio

Y3=Rna-(Da/Db)  #Polynomial to find out 'x'
x=solve(Y3,x)[0]  #Numerical Value of Tractive Effort shared by A

#Tractive Efforts Shared by A Loco and B Loco

#For 1309 kg
#% Speed
x=symbols('x')  #Variable for Tractive Effort Exerted by Loco A
NA=100-(Sa*x/Ft)
NB=100-(Sb*(Fttb-x)/Ft)
Rn2=NB/NA  #Speed Ratio

Y4=Rn2-(Da/Db)  #Polynomial to find out 'x'
x=solve(Y4, x)[0]  #Numerical Value of Tractive Effort shared by A

#Tractive Efforts Shared by A Loco and B Loco
FtAbd=x
FtBbd=Fttb-x

print 'The Tractive Effort shared A and B for:'
print 'i) Slips are %d percent for both the Locomotives for a Tractive effort of:'%S
print 'a) %d kg : %d kg by A and %d kg by B respectively'%(Ftta,FtAas,FtBas)
print 'b) %d kg : %d kg by A and %d kg by B respectively'%(Fttb,FtAbs,FtBbs)
print 'i) Slips are %d percent for A and %d percent for B for a Tractive effort of:'%(Sa,Sb)
print 'a) %d kg : %d kg by A and %d kg by B respectively'%(Ftta,FtAad,FtBad)
print 'b) %d kg : %d kg by A and %d kg by B respectively'%(Fttb,FtAbd,FtBbd)

#Please Note there is caluculation mistake calculation in the TextBook for the First Case

The Tractive Effort shared A and B for:
i) Slips are 5 percent for both the Locomotives for a Tractive effort of:
a) 11325 kg : 6685 kg by A and 4639 kg by B respectively
b) 1309 kg : 1728 kg by A and -419 kg by B respectively
i) Slips are 5 percent for A and 4 percent for B for a Tractive effort of:
a) 11325 kg : 6175 kg by A and 5149 kg by B respectively
b) 1309 kg : 1775 kg by A and -466 kg by B respectively


## Ex13 : Page 386¶

In [13]:
#To determine the traction control of 2 motors rated at 1500V

NoM=2  #Number of motors
V=1500  #Rated Voltage
I=500  #Starting Current
R=0.15  #Armature Resistance
r=50  #Specific Resistance
W=120
We=140
Ft=38000  # Tractive Effort Per motor

Vs=40  #Speed at the end of starting period
#Note the question is asked for 50 Kmph, But the answer is calculated for 40Kmph
#Therefore it is calculated for 40 Kmph
#By Changing the value of Vs to 50 Kmph, The Specific Parameters can also be obtained

a=symbols('a')  #Accleration Variable

X=(2*Ft)-((277.8*We*a)+(W*r))  #Polynomial To find 'a'

a=solve(X)[0]  #To find the Numerical Value of the Acceleration

t=Vs/a  #Starting Time

Vd=I*R  #Resistance drop per motor

ts=t*(V-(2*Vd))/(2*(V-Vd))  #Starting Series Time Period
tp=t-ts  #Starting Parallel Time Period

Vt=a*ts  #Speed at Transition
#Loss can be found out by the computing the area under the of the given figure in the text book for different period
#In series, the Voltage reduces, Hence 675
Rhel=((((V/2)-Vd)*I*ts)+((V/2)*I*tp))/(2*3600*1000)  #Loss per motor in kWhr
Rhe=NoM*Rhel  #Total Rheostatic Loss

print 'i) The Duration of starting period is %0.1f seconds'%t
print 'ii)The Speed of Train at Transition is %0.2f Kmph'%Vt
print 'iii) The Rheostatic Loss is %0.3f kWhr'%Rhe

i) The Duration of starting period is 22.2 seconds
ii)The Speed of Train at Transition is 18.95 Kmph
iii) The Rheostatic Loss is 2.205 kWhr


## Ex14 : Page 387¶

In [14]:
#To determine the speed of motors when connected in series

V=650  #Rating of motor and Supply Voltage when connected in parallel
IR=10*650/100  #Armature Drop
N1p=1000  #Speed of the first motor in parallel operation
D1=88  #Motor 1 wheels diameter
D2=86  #Motor 2 wheels diameter

#Current Remains Constant during the start

r=D1/D2  #Ratio of the first motor wheel diameter to the second motor wheel diameter

#N1/N2 = 1/r

V1=((V-IR)/(1+(r)))+((IR)/(1+(1/r)))  #Voltage of Motor 1 in series operation
V2=V-V1  #Voltage of Motor 2 in series operation

N1=N1p*(V1-IR)/(V-IR)  #Speed of motor 1 in series configuration
N2=N1*r  #Speed of motor 2 in series configuration

print 'The Speed of Motor 1 in series Configuration is %d rpm'%N1
print 'The Speed of Motor 2 in series Configuration is %d rpm'%N2

The Speed of Motor 1 in series Configuration is 439 rpm
The Speed of Motor 2 in series Configuration is 449 rpm


## Ex15 : Page 387¶

In [15]:
from __future__ import division
from numpy import array, nditer
#To determine the characteristics gear ratio and wheels are replaced

D1=86  #Diameter of wheels in Case 1
Y1=71/21  #Gear Ratio in Case 1
D2=88  # Diameter of wheels in Case 2
Y2=74/19  #Gear Ratio in case 2
I=array([50,100,150,200,250,300] ) #Current
L1=array([80,50,45,40,36,32] ) #Speed in Kmph
Ft1=array([2000,6000,12000,16000,18000,22000])  #Tractive Effort

# V = pi*D*N*60/(100*100*Y)
#According to the above equation, V is directly proportional to D*N/Y

#Angular Frequency And Torque remains the same irrespective of the gear ratio or change in diameter

#Hence the V is directly proportional to D/Y

#V id directly proportional to N
# N directly proportional to Linear speed
#Hence relating the above terms
#We get linear speed directly proportional to D/Y

def fun1(L1):
it = nditer([L1, None])
for x,y in it:
y[...] = x*D2*Y1/(D1*Y2)
return it.operands[1]

L2 = fun1(L1) #Linear Speed in Case two

#Similarly T = Ft * D/(200)
#Dividing by Y   (T/Y) = Ft*D/(200*Y)
#Taking two cases where we find Torque per gear ratio of one case
# And multiply the Torque to Gear ratio to the other gear ratio to get Torque
# From this process we T1*Y2/Y1 =T2
# We Get Ft is directly proportional to Y/D

def fun2(Ft1):
it = nditer([Ft1, None])
for x,y in it:
y[...] = x*D1*Y2/(D2*Y1)
return it.operands[1]

Ft2=fun2(Ft1)  #Tractive Effort in Second Case

print 'The New Characteristics are - \nCurrent in Amperes'
for i in I:
print "%d "%i,
print '\n\nSpeed in Kmph'
for l2 in L2:
print "%0.2f "%l2,
print '\n\nTractive Effort in N'
for ft2 in Ft2:
print "%0.f "%ft2,

# Answers are not accurate in the textbook.

The New Characteristics are -
Current in Amperes
50  100  150  200  250  300

Speed in Kmph
71.00  44.00  39.00  35.00  31.00  28.00

Tractive Effort in N
2251  6754  13509  18012  20264  24767


## Ex16 : Page 388¶

In [16]:
from __future__ import division
#To Determine the Resistance of the Various Steps in a Series Motor

P=20*735.5  #Watt # Power Rating
V=500  #Rated Voltage
Eff=80/100  #Efficiency
Raw=1  #Resistance of Armature and Windings
Iflux=10/100  #Increase in Flux

#VAriation in Current
Imin=1.5*I
Imax=2*I

fi_maxBYfi_min = 1+Iflux
FluxRatio=1+(Iflux)  #Flux raises by 10% for every step
# eqn : fimax/fi_min = (V-Imax*R2)/(V-*Imin*R1)
from sympy import symbols, solve
Is=V/Raw
M=Imax-Is
for i in range(1,5):

if((abs(M)+M)==0):
c=i  #Number of steps + 1
R=symbols('R')  #Variable Resitance
X=((V-(Imax*Raw))/(V-(Imin*R)))-FluxRatio  #Polynomial To Find The Next Resistance
Rs =solve(X, R)[0]  #Total Next Resistance
Rg=Rs-Raw  #Resistance of the ith Step
print "Resistance of %d step = %0.2f ohm"%(i,Rg)
Raw=Rs
Is=V/Raw
M=Imax-Is
print "Reqired steps = %d"    %i

Resistance of 1 step = 1.04 ohm
Resistance of 2 step = 1.26 ohm
Resistance of 3 step = 1.52 ohm
Resistance of 4 step = 1.85 ohm
Reqired steps = 4


## Ex17 : Page 390¶

In [17]:
from numpy import array
#To detemine the characteristics for tapping of 30 percent of the turns

I=array([50,100,150,200,250])  #Current
N1=array([73.5,48,41,37,35])  #Speed
Ft1=array([131,457,810,1163,1525])  #Tractive Effort in Kg
Fl=70/100  #Effective Tapping of the Turns (30% reduction)

#Consider flux corresponding to 250A as 100%, that is for 35 Kmph

Flux=(35*100)/N1  #Flux Percent for Current Value

#Reduction by 30% of turns is same as reducing the Ampere Turns by 30% which is same as reducing the Current by 30%

RI=Fl*I  #Reduced Current
RFlux=array([32,57,74,83,90])  #Reduced flux determined From the Graph of %Flux and Current in the TextBook

N2=(N1*Flux)/RFlux  #Speed for the new case

#Tractive Effort is directly proportional to the product of flux and current

Ft2=RFlux*Ft1/Flux  #Tractive Effort of the Second Case

print 'The Current in Amperes:'
for i in I:
print i,'\t',
print ''
print '\nThe Speed in Kmph:'
for n2 in N2:
print '%0.2f'%n2,'\t',
print ''
print '\nThe Tractive Effort in Kg :'
for ft2 in Ft2:
print '%0.1f'%ft2,'\t',

The Current in Amperes:
50 	100 	150 	200 	250

The Speed in Kmph:
109.38 	61.40 	47.30 	42.17 	38.89

The Tractive Effort in Kg :
88.0 	357.2 	702.2 	1020.4 	1372.5 	

## Ex18 : Page 400¶

In [18]:
#To determine the power fed into the supply system

W=200
RI=10/100  #Rotational Intertia
Eff=75/100  #Effeciency of Conversion
G=2
V1=60
V2=20
D=4000  #Distance covered during the speed change in m
r=40
We=(1+RI)*W

Ftb=W*((98.1*G)-r)  #Tractive Effort during Retardation

KE=0.01072*(We/W)*200*((V1**2)-(V2**2))/1000  #K.E(in kWhr) Available due to change in speed

EquiEE=Ftb*D/(3600*1000)  #Equivalent Electrical Energy

TE=KE+EquiEE  #Total Energy Regenarated

ERS=Eff*TE  #Energy Returned to the Supply

B=((V1**2)-(V2**2))/(2*D*3600)  #Retardation
Tb=V1-V2/B  #time taken during retardation

PRS=ERS*3600/Tb  #Power Returned due to Retardation

#If there is no change in speed, Net Reactive force is Tractive Effort due to retardation

Vm=V1  #Maximum Speed

#Power that Can be fed into the system
PFS=0.278*Ftb*Vm/1000  #In Kilowatts

APFS=Eff*PFS  #Actual Power Fed Back into the supply system

print 'The Power Fed Back into the supply system is %0.1f kW'%APFS

The Power Fed Back into the supply system is 390.8 kW


## Ex19 : Page 401¶

In [19]:
from __future__ import division
from math import pi
from numpy import array, arange
#To determine the value of resistance to be connected in the motor circuit
#Page 411
N1=600; #Intial Speed
E1=array([252,522,750,900,951]); # Voltages At 600 RPM
Ia=array([20,40,60,80,100]); # Field Current
N2=400; #Changed Speed
E2=E1*(N2/N1); # Voltages at 400 RPM
T=350;
Raw = 0.05; #Armature and Winding Resistance
print 'Field Current in A:\n'
for ia in Ia:
print ia,'\t',
print '\n'
print 'The Respective Voltages at 400 rpm (V):\n'

for e2 in E2:
print e2,'\t',
print ''

P=E2*Ia #To find the value of E and Ia for the Required Speed

PI=2*pi*N2*T/60 #Power Input
i=0
for p in P:
i+=1
X = p-PI

if (X)<0:
L = i-1
H = i
else:
continue

Ip=Ia[H]-Ia[L] #To find the Current Period
Im=Ia[L] #Starting Value of Current in the Particular Portion of the Curve

I=Im+(((PI-P[L])/(P[H]-P[L]))*Ip) #Current Required for 400 rpm
R=PI/(I**2) #Total Resistance required in the circuit
Re=R-Raw  #External Resistance

print '\n\nThe External Circuit that needs to connected in the motor circuit to limit the speed to 400 rpm is %0.1f Ohm\n'%Re

Field Current in A:

20 	40 	60 	80 	100

The Respective Voltages at 400 rpm (V):

168.0 	348.0 	500.0 	600.0 	634.0

The External Circuit that needs to connected in the motor circuit to limit the speed to 400 rpm is 8.7 Ohm