Chapter 10:Traction Drives¶

Example No:10.1,Page No:320¶

In :
import math
from __future__ import division

#variable declaration
Ma=480             #mass of each motor armature in kg  0.48tonne=480kg
Da=0.5             #average diameter of each motor in m
m=450              #mass of each wheel in kg
M=40               #combine wight of one motor and one trailer coach in ton
alpha=5            #accelaration
N=4                #number of DC motors
a=0.4              #gear ratio
r=20               #train resistance

#calculation
Jw=1/2*m*R**2      #inertia of the each wheel
nw=2*(N*2)         #total number of wheels
J1=nw*Jw           #total inertia of all the wheels

Jm=N*(1/2*Ma*(Da/2)**2)  #approximate inertia of all the motors
J2=Jm/a**2               #approximate innertia of the motor referred to the wheels

Fa2=(J1+J2)*alpha*1000/3600/R    #Tractive efforts for driving rorating parts
Fa1=277.8*M*alpha                #tractive efforts to accelerate the train mass horizontally
Fr=r*M                           #Tractive efforts required to overcome train resistance
Ft=Fa1+Fa2+Fr                    #Tractive efforts required to move the train
Tm=a*R*Ft/N                      #torque per motor

#results
print"\nTorque per motor:Tm=",round(Tm,1),"N-m"
Torque per motor:Tm= 3241.3 N-m

Example No:10.2,Page No:321¶

In :
import math
from __future__ import division

#variable declaration
M=100              #mass of each motor armature in tonne
Me=100
Tm=5000            #torque of each motor in N-m
Da=0.5             #average diameter of each motor in m
m=450              #mass of each wheel in kg
N=4                #number of DC motors
r=25               #train resistance N/tonne
a=0.25             #gear ratio
nt=0.98            #gear transmission efficiency
Vm=100             #speed in kmph

#calculation
Ft=nt*Tm*N/a/R     #Tractive efforts required to move the train
alpha=(Ft-(9.81*M*G+M*r))/(277.8*1.1*Me)  #accelaration
t=Vm/alpha         #time taken to attain speed of Vm

#results
print"\n time taken to reach a speed of 100kmph is :t=",round(t,1),"sec"
time taken to reach a speed of 100kmph is :t= 32.6 sec

Example No:10.3,Page No:321¶

In :
import math
from __future__ import division

#variable declaration
r=25              #train resistance N/tonne
M=500             #mass of the electric train in tonne
n=0.8             #combine effiency of transmission and motor
#speed time curve characteristics
t1=60       #characteristic for uniform accelaration at v1 in sec
alpha=2.5   #given accelaration in km/hr/sec at t1
t2=5*60     #characteristic for constant speed in sec
t3=3*60     #characteristic for coasting in sec
B=3         #dynamic braking deceleration in km/hr/sec

#calculation
Vm=alpha*t1  #peak voltage
Fg=9.81*M*G  #tractive effort required to overcome the force of gravity
Fr=M*r       #tractive effort required to overcome the train resistance
F_bc=Fg+Fr   #retarding force during coasting

Me=1.1*M
B_c=F_bc/(277.8*Me)   #deceleration during coasting
V=Vm-B_c*t3           #speed after coasting
t4=V/B                #characteristic for a dynamic braking of 3km/hr/sec

d1=1/2*Vm*t1/3600     #distance covered during accelaration
d2=Vm*t2/3600         #distance covered during constant speed
d3=1/2*(Vm+V)*t3/3600 #distance covered coasting
d4=1/2*V*t4/3600      #distance covered during braking
D=d1+d2+d3+d4         #distance during stops
D1=d1+d2
x=D1/D
y=1-x
E=(0.01072*Vm**2/D)*(Me/M)+2.725*G*x+0.2778*r*x   #specific energy output
Eo=E/n              #specific energy consumption

#results
print"\n Specific energy consumption is :Eo=",round(Eo,1),"Whptpkm"
Specific energy consumption is :Eo= 41.1 Whptpkm

Example No:10.4,Page No:323¶

In :
import math
from __future__ import division

#variable declaration
r=25              #train resistance N/tonne
M=500             #mass of the electric train in tonne
n=0.8             #combine effiency of transmission and motor
#speed time curve characteristics
t1=50       #characteristic for uniform accelaration at v1 in sec
alpha=3     #given accelaration in km/hr/sec at t1
t2=10*60    #characteristic for constant speed in sec
B=2.5       #uniform braking deceleration in kmphs to rest

#calculation
Vm=alpha*t1  #peak voltage
t3=Vm/B      #characteristic for a uniform braking of B=2.5 kmphs

#(i)during accelaration total tractive effort
Me=1.1*M
Fta=277.8*Me*alpha-9.81*M*G+M*r   #total tractive effort during accelaration
Da=1/2*Vm*t1/3600     #distance covered during accelaration ,and t1 is in seconds
Ea=Fta*Da*1000/3600   #energy output during accleration in Wh

#(ii)during uniform speed net tractive effort
Ftu=-9.81*M*G+M*r     #total tractive effort during uniform speed
#the answer for Ftu in the book is -105220 N  which is wrong which leads to the other incorrect answers in the book

Du=Vm*t2/3600         #distance traveled,and t2 is in seconds
Eu=Ftu*Du*1000/3600   #energy output in Wh

#(iii)during braking net tractive effort
Ftb=-277.8*Me*B-9.81*M*G+M*r  #total tractive effort for the net braking
Db=1/2*Vm*t3/3600     #distance covered during braking
Eb=Ftb*Db*1000/3600   #energy output during braking in Wh

E=Ea/n+n*(Eu+Eb)      #net energy consumption
D=Da+Du+Db            #total distance traveled
Eo=E/(M*D)            #specific energy consumption

#results
print"(i)Energy consumption during accelaration is :Ea :",round(Ea),"Wh"
print"   There is a slight difference in the answer due to the number of decimal place"
print"\n(ii)Energy consumption during uniform speed is :Eu :",round(Eu),"Wh"
print"\n(iii)Energy consumption during braking is :Eb :",round(Eb,1),"Wh"
print"\n Net Energy consumption  is E :",round(E,1),"Wh"
print"\n Total Distance traveled is D :",round(D,4),"km"
print"\n Specific Energy consumption  is Eo :",round(Eo,2),"Whptpkm"
#answers in the book are incorrect
(i)Energy consumption during accelaration is :Ea : 107862.0 Wh
There is a slight difference in the answer due to the number of decimal place

(ii)Energy consumption during uniform speed is :Eu : -594444.0 Wh

(iii)Energy consumption during braking is :Eb : -162352.4 Wh

Net Energy consumption  is E : -470610.4 Wh

Total Distance traveled is D : 27.2917 km

Specific Energy consumption  is Eo : -34.49 Whptpkm

Example No:10.5,Page No:325¶

In :
import math
from __future__ import division

#variable declaration
Mm=40       #weight of the motor coach in tonne
Mt=35       #weight of the trailer in tonne
r=30        #train resistance N/tonne

#calculation
#(a)when the motor to trailer ratio is 1:2
M=Mm+2*Mt     #weight of one unit
Me=1.1*M
Md=40         #weight on the driving wheels
Fm=9810*u*Md  #total tractive effort without the wheel
Ft=Fm         #at maximum accelaration
alpha=(Ft-M*r)/(277.8*Me) #required accelaration since Ft=277.8*u*alpha*M*r

#(b)when the motor to trailer ratio is 1:1
M=Mm+Mt       #weight of one unit
Me=1.1*M
Md=40         #weight on the driving wheels
Fm=9810*u*Md  #total tractive effort wihout the wheel
Ft=Fm         #at maximum accelaration
alpha1=(Ft-M*r)/(277.8*Me) #required accelaration since Ft=277.8*u*alpha*M*r

#results
print"(a)maximum accelaration on a level track  is alpha :",round(alpha,4),"kmphps"
print"\n(b)maximum accelaration when motor to trailer coaches ratio is 1:1 is alpha :",round(alpha1,3),"kmphps"
(a)maximum accelaration on a level track  is alpha : 2.2366 kmphps

(b)maximum accelaration when motor to trailer coaches ratio is 1:1 is alpha : 3.326 kmphps

Example No:10.6,Page No:326¶

In :
import math
from __future__ import division

#variable declaration
G=10        #up gradient of the locomotive
Ml=110      #weight of the locomotive coach in tonne
Mt=500      #weight of the train in tonne
r=35        #train resistance N/tonne
n=0.8       #80% of the locomotive weight is carried by the driving wheels
alpha=1     #acelaration in kmphps

#calculation
#when only the 110 tonne locomotive is present
Md=Ml*n     #weight of the motor
M=Mt+Ml     #total mass of the train
Me=1.1*M
Ft=277.8*Me*alpha+9.81*M*G+M*r  #total tractive effort required to move the train
Fm=Ft
u=Fm/(9810*Md)   #co-efficient of adhesion ,since Fm=9810*u*Md

#when another locomotive of 70 is added together
Md=Ml*n+70      # mass of the motor
M_=Mt+Ml+70     # mass of the train
Fm=9810*u*Md
Ft=Fm
M=Ft/(277.8*1.1*alpha+9.81*G+r)  #total mass of the train, since Ft=277.8*Me*alpha+9.81*M*G+M*r
W=M-M_          #weight of additional bogies to be attached

#results
print"\n Given co-efficient of adhesion is:",round(u,2)
print"\n Weight of additional bogies to be attached is:",round(W,1),"T"
Given co-efficient of adhesion is: 0.31

Weight of additional bogies to be attached is: 415.2 T

Example No:10.7,Page No:327¶

In :
import math
from sympy import Symbol
from __future__ import division

#variable declaration
Ml=1000      #weight of the empty train in tonne
Mt=5000      #weight of the fully loaded train in tonne
V=30         #maximum speed of the train
r=40         #train resistance in N/tonne
alpha=0.3    #acelaration in kmphps

n = Symbol('n')  #number of locomotive required
W=100            #weight of each locomotive

#calculation
Md=W*n
Fm=9810*u*Md
Fb=9.81*(Mt+W*n)*G-(Mt+W*n)*r
print"\nFm=",Fm
print"\nFb=",Fb
print"\nequating Fb and Fm we get"
n=535750/(245250-10715)
if (n>2) :
n=3
print"\nThe number of locomotives is n:",n
Md=W*n
M=Ml+W*n
Ft=277.8*1.1*M*alpha+9.81*M*G+M*r
Fm=9810*0.3*Md
if (Fm>Ft) :
print"\nThe train can be accelarated with ",n,"locomotives"
Fm= 245250.0*n

Fb= 10715.0*n + 535750.0

equating Fb and Fm we get

The number of locomotives is n: 3

The train can be accelarated with  3 locomotives