Chapter 4, Electrical Features of lines - 1

Exa 4.1 : page 104

In [105]:
from __future__ import division
from numpy import log, pi
import math
#Given Data :
r=1.213/2 #in cm
f=60 #in Hz
ds=0.77888*r #in cm
spacing=1.25 #in meter
L=4*10**-7*log(spacing*100/ds) #in H/m
L*=1000  # in H/km
print "Inductance = %0.2e H/km" %L
XL=2*pi*f*L #in ohm/km
XL*=60  # ohm per 60 km
print "Inductive reactance for 60 km line = %0.1f ohm" %XL

Inductance = 2.23e-03 H/km
Inductive reactance for 60 km line = 50.5 ohm

Exa 4.2 : page 105

In [106]:
from __future__ import division
from math import log
#Given Data : 
l=20  # km (length of line)
d=2.8*100 #in cm(spacing)
r=0.5*1.5 #in cm
ds=0.77888*r #in cm
L=0.2*log(d/ds) #in H/m/phase
L*=l # mH for 20km line
print "Inductance per phase for a 20 km line = %0.2f mH" %L

Inductance per phase for a 20 km line = 24.69 mH

Exa 4.3 : page 105

In [107]:
from __future__ import division
from math import log
#Given Data :
a=1.5 #in cm**2
d=8 #in meter(spacing)
r=39.8/2 #in mm
l=1*10**5 #in cm
rho=1.73*10**-6 #in ohm-cm
R=rho*l/a #in ohm/km
print "Resistance of line = %0.4f ohms/km" %R
ds=0.77888*r #in cm
L=0.2*log(d/(ds*10**-3)) #in mH/km/phase
print "Inductance per phase for a 1 km line = %0.3f mH/km" %L

Resistance of line = 0.1153 ohms/km
Inductance per phase for a 1 km line = 1.249 mH/km

Exa 4.4 : page 106

In [108]:
from __future__ import division
#Given Data :
Cs=1/3 #in uF
Cc=(0.6-Cs)/2 #in uF
#Part (a) :
C1=(3/2)*Cc+(1/2)*Cs #in uF(between any two conductor)
print "Capacitance between any two conductor = %0.3f uF" %C1
#Part (b) :
C2=2*Cc+2*Cs/3
print "Capacitance between any shorted conductors = %0.3f uF" %C2

Capacitance between any two conductor = 0.367 uF
Capacitance between any shorted conductors = 0.489 uF

Exa 4.5 : page 107

In [109]:
from __future__ import division
#Given Data :
d1=3 #in meter
d2=3 #in meter
d3=d1+d2 #in meter
d=378 #in cm
dia=2.5 #in cm
r=dia/2 #in cm
epsilon_o=8.854*10**-12 #constnt
L=(0.5+2*log10(d/r))*10**-7 #in H/m
L*=60*1000*1000  # mH per 60 km
print "Inductance for 60 km line %0.2f mH" %L
C=2*pi*epsilon_o/log(d/r) #in F/m
C*=60*10**3*10**6 # uF per 60 km line
print "Capacitnce for 60 km line = %0.4f uF" %C
# Answer is not accurate in the textbook.

Inductance for 60 km line 32.77 mH
Capacitnce for 60 km line = 0.5844 uF

Exa 4.6 : page 107

In [110]:
from __future__ import division
from numpy import log10
#Given Data : 
dinner=6 #in meter
douter=12 #in meter
d=(dinner**2*douter)**(1/3) #in meter
r=2.8 #in meter
ds=0.7788*r #in cm
L=2*log10(d*100/ds) #in mH/phase/km
L*=100   # mH per 100 km
print "Inductance for 100 km line %0.f mH" %L

Inductance for 100 km line 508 mH

Exa 4.7 : page 108

In [111]:
from __future__ import division
from math import log
#Given Data :
dia=5 #in mm
d=1.5 #in meter(spacing)
r=dia/2 #in mm
r=r*10**-3 #in meter
epsilon_o=8.854*10**-12 #constnt
C=pi*epsilon_o/log(d/r) #in Farad per meter
C*=50*1000 # F per 50km line
print "Capacitance for 50 km line = %0.2e F" %C
#Note : answer is not accurate in the book. 

Capacitance for 50 km line = 2.17e-07 F

Exa 4.8 : page 108

In [112]:
from __future__ import division
from math import log
#Given Data :
d=300 #in cm(spacing)
r=1 #in cm
#Formula : L=10**-7*[mu_r+4*log10(d/r)] #in H/m
#Part (i) : mu_r=1
mu_r=1 #constant
L=10**-4*(mu_r+4*log(d/r)) #in H/m
L*=1000 # mh per km
print "Loop inductance per km for copper %0.2f mH" %L
#Part (ii) : mu_r=100
mu_r=100 #constant
L=10**-4*(mu_r+4*log(d/r)) #in H/m
L*=1000 # mH per km
print "Loop inductance per km for steel = %0.2f mH" %L
# Answer not calculated completely and calculation mistake.

Loop inductance per km for copper 2.38 mH
Loop inductance per km for steel = 12.28 mH

Exa 4.9 : page 109

In [113]:
from __future__ import division
from math import log
#Given Data :
d1=100 #in cm(spacing)
d2=100 #in cm(spacing)
d3=100 #in cm
r=1 #in cm
L=10**-7*(0.5+2*log((d1*d2*d3)**(1/3)/r)) #in H/m
L=L*1000*1000 #in mH/km
print "Inductance per km = %0.2f mH" %L
#Note : Answer in the book is wrong due to calculation mistake.
#Note : In the last line it should be multiply by 10**6 to convert from H/m to mH/km instead of 10**8.

Inductance per km = 0.97 mH

Exa 4.10 : page 109

In [114]:
from __future__ import division
from numpy import log
#Given Data : 
d1=2 #in cm
d2=2.5 #in cm
d3=4.5 #in cm
r=1.24/2 #in cm
L=10**-7*(0.5+2*log((d1*d2*d3)**(1/3)/r)) #in H/m
L=L*1000*1000 #in mH/km
print "Inductance per km per phase = %0.2f mH" %L
#Note : Answer in the book is wrong(calculation mistake).

Inductance per km per phase = 0.35 mH

Exa 4.11 : page 110

In [115]:
from __future__ import division
from numpy import log
#Given Data :
r=0.75*10 #in mm
d=1.5*10**3 #in mm
ds=0.7788*r #in mm
L=4*10**-7*log(d/ds) #in H/m
L=L*10**6 #in mH/km
print "Inductance of line = %0.2f mH/km" %L

Inductance of line = 2.22 mH/km

Exa 4.12 : page 110

In [116]:
from __future__ import division
from numpy import log
#Given Data :
d1=4*100 #in cm
d2=5*100 #in cm
d3=6*100 #in cm
r=1 #in cm
ds=0.7788*r #in cm
L=(0.2*log((d1*d2*d3)**(1/3)/ds)) #in mH
L*=10**3  # uH
print "Inductance per km = %0.2f uH" %L
#Note : answer in the book is wrong.

Inductance per km = 1290.20 uH

Exa 4.13 : page 110

In [117]:
from __future__ import division
from numpy import log
#Given Data :
d=300 #in cm(spacing)
r=1 #in cm
epsilon_o=8.854*10**-12 #constnt
C=pi*epsilon_o/log(d/r) #in Farad per meter
C*=30*1000*10**6  # uF per 30k km line
print "Capacitance for 30 km line = %0.2f uF" %C

Capacitance for 30 km line = 0.15 uF

Exa 4.14 : page 111

In [119]:
from __future__ import division
from numpy import log
#Given Data :
d=2.5*100 #in cm(spacing)
r=2/2 #in cm
epsilon_o=8.854*10**-12 #constnt
C=2*pi*epsilon_o/log(d/r) #in Farad per meter
C*=10*1000*10**6  # uF per 10 km line
print "Capacitance for 10 km line = %0.2f uF" %C
#Note : answer given in the book is wrong but calculated is right.

Capacitance for 10 km line = 0.10 uF

Exa 4.15 : page 111

In [120]:
#Given Data :
VL=33 #in KV
f=50 #in hz
d1=4 #in meter
d2=4 #in meter
d3=8 #in meter
d=(d1*d2*d3)**(1/3) #in meter
epsilon_o=8.854*10**-12 #constnt
d=d*100 #in cm
r=0.62 #in cm
C=2*pi*epsilon_o/log(d/r) #in Farad per meter
C*=50*1000*10**6  # uF per  m line
print "Capacitance for 50 km line %0.3f uF" %C
Vp=VL/sqrt(3) #in KV
Vp=Vp*10**3 #in volt
Ic=2*pi*f*(C*10**-6)*Vp #in Ampere
print "The charging current = %0.2f Ampere" %Ic

Capacitance for 50 km line 0.415 uF
The charging current = 2.48 Ampere