Chapter 11: Power Supply for Electric Traction

Example 11.1, Page 321

In [12]:
#Variable declaration
l=20;# in m
w=0.5;# weight per meter in kg
T=500;# Tension applied in kg

#Calculations
dell=(w*l**2)/(2*T);
two_S=2*(l+(2./3)*(dell**2/l));

#Result
print "Total Length(m) = %.5f"%two_S

Total Length(m) = 40.00267

Example 11.2, Page 321

In [13]:
import math

#Variable declaration
l=30;# in meter
w=0.72;# weight per meter in kg
E=640;# in kg/cm^2
d=1;# diameter in cm

#Calculations
T=E*(math.pi/4)*d**2;
dell=((w*l**2)/(2*T))*100;

#Result
print "sag(cm) = %.1f"%dell

sag(cm) = 64.5

Example 11.3, Page 322

In [14]:
#Variable declaration
l=30;# in meter
w1=0.9;# average weight of catenary wire in kg/m
w2=1.2#average weight of trolley wire in kg/m

#Calculations
w3=(20./100)*w2#average weight of dropper and fittings in kg/m
w=w1+w2+w3;
T=1000;#in kg
dell=(w*l**2)/(2*T);

#Result
print "sag(m) = %.3f"%dell

sag(m) = 1.053

Example 11.4, Page 340

In [15]:
#Variable declaration
I=300;# in A
R=0.08;# in ohm
Vd=6;# voltage drop in volts

#Calculations
I_dash=((R*(I/2))-Vd)/R;

#Result
print "Current(A) = %.f"%I_dash

Current(A) = 75

Example 11.5, Page 340

In [16]:
#Variable declaration
a=7;#far end voltage in volts
i=125;# in A
r=0.02;# in ohm
l=3;# in km

#Calculations
p=(i*r*l**2)/2;
I=((p-a)/(r*l));#

#Results
print "potential of the track at tha far end of the section in volts is %.2f"%p
print "Current carried by -ve feeder,I(A) = %.3f"%I

potential of the track at tha far end of the section in volts is 11.25
Current carried by -ve feeder,I(A) = 70.833

Example 11.6, Page 341

In [1]:
import sympy
from sympy.solvers import solve
from sympy import Symbol

#Variable declaration
ix=200;#amperes
r=0.02;#in ohms

#Calculations
x = Symbol('x')
y = solve(0*x**2 +12*x + 19, x)
ipx=ix*(3-(-1*y[0]));#in amperes
inx=2*ix;#in amperes
it=ipx+inx;#in amperes

#Result
print "current through negetive booster in amperes is %.2f"%it

current through negetive booster in amperes is 683.33

Example 11.7, Page 341

In [2]:
import sympy
from sympy.solvers import solve
from sympy import Symbol

#Variable declaration
ix=250;#amperes
vb=2;#in volts
r=0.02;#in ohms

#Calculations
x = Symbol('x')
y = solve(0*x**2 +16*x - 27.6, x)
pc=vb+(ix*r*(1.6)**2)/2;#in volts
pd=((ix*r*(y[0]**2))/2);#in volts
tcurr= (1.6*ix)+((ix*(3.2-y[0])));#in amperes
vnf=r*tcurr;#in volts
bnb=vnf-vb;#in volts
cb=((bnb*tcurr)/1000);#in kw

#Result
print "maximum potential drop on any two points on the rails in volts is %.1f"%pc
print "capacity of booster in kW is %.2f"%cb

maximum potential drop on any two points on the rails in volts is 8.4
capacity of booster in kW is 10.28

Example 11.8, Page 342

In [3]:
import sympy
from sympy.solvers import solve
from sympy import Symbol

#Variable declaration
i=200;# A/km
r=0.01;#in ohms/km

#Calculations
x = Symbol('x')
y = solve(0*x**2 +8*x - 20, x)
i1=400;#in amperes
i2=(4-y[0])*i#in amperes
tc=i1+i2;#in amperes
vcn=r*tc;#in volts
nb=vcn-4;#in volts
rb=(tc*10)/1000;#

#Result
print "rating of the booster in kW is %.f"%rb

rating of the booster in kW is 7

Example 11.9, Page 343

In [26]:
#Variable declaration
vw=60;#in volts
vt=12;#in volts
vs=600;#in volts
vr=578;#volts
vn=10;#in volts

#Calculations
tv=vw+vt;#in volts
va=vs-tv;#in volts
vtn=tv-vn;#in volts
vad=vs-vr;#
vp=vtn-vad;#in volts

#Results
print "part (a)"
print "voltage available to trolley when it is at the far end without using boosters in volts is %.f"%va
print "part (b)"
print "positive booster should provide boost of ",(vp)," volts"

part (a)
voltage available to trolley when it is at the far end without using boosters in volts is 528
part (b)
positive booster should provide boost of  40  volts