Chapter8 - Underground cables and faults

Exa 8.1 - page 222

In [4]:
from numpy import exp, pi
#given data
R=500 #in Mohm/Km
R=R*10**6 #in ohm
r1=2.5/2 #in cm
r1=r1*10**-2 #in meter
rho=4.5*10**16 #in ohm/cm
rho=rho*10**-2 #in ohm/m
l=1 #in Km
l=l*1000 #in meter
#Formula : R=(rho/(2*pi*l))*log(r2/r1)
r2=(exp(R/(rho/(2*pi*l))))*r1 #in meter
thickness=r2-r1 #in meter
thickness=thickness*100 #in cm
print "Thickness of Insulation = %0.3f cm" %thickness
# Answer in the textbook is wrong due to accuracy.

Thickness of Insulation = 0.009 cm

Exa 8.2 - page 223

In [6]:
from numpy import log10
#given data
d=1 #in cm
d=d*10**-2 #in meter
D=1.8 #in cm
D=D*10**-2 #in meter
epsilon_r=4 #permittivity of insulation
C=0.024*epsilon_r/log10(D/d) #in uF/Km
print "Capacitance/km of the fibre = %0.3f uF" %C

Capacitance/km of the fibre = 0.376 uF

Exa 8.3 - page 223

In [10]:
from numpy import log
#given data
V=33 #in KV
d=1 #in cm
D=4 #in cm
#Part (a) :
gmax=2*V/(d*log(D/d)) #in KV/cm
print "Maximum Stress = %0.1f KV/cm" %gmax
#Part (b) :
gmin=2*V/(D*log(D/d)) #in KV/cm
print "Minimum Stress = %0.0f KV/cm" %round(gmin)

Maximum Stress = 47.6 KV/cm
Minimum Stress = 12 KV/cm

Exa 8.4 - page 224

In [13]:
from math import sqrt
#given data
Vrms=66 #in KV
gmax=40  #in KV/cm
V=sqrt(2)*Vrms #in Volt
#Part (a) : 
d=2*V/gmax #in cm
print "The most economical diameter = %0.3f cm" %d
#Part (b) : 
PeakVoltage=sqrt(2)*Vrms/sqrt(3) #in Volt
V=PeakVoltage #in Volt
d=2*V/gmax #in cm
print "The most economical diameter for 3 phase system  = %0.1f m" %d

The most economical diameter = 4.667 cm
The most economical diameter for 3 phase system  = 2.7 m

Exa 8.5 - page 224

In [17]:
from __future__ import division
from math import sqrt
#given data
d=2 #in cm
D=2.5*2 #in cm
d1=(5/4)*d #in cm
d2=(5/3)*d #in cm
gmax=40 #in KV/cm
PeakVoltage=(gmax/2)*(d*log(d1/d)+d1*log(d2/d1)+d2*log(D/d2)) #in KV
print "The safe Working Potential = %0.1f KV" %(PeakVoltage/sqrt(2))

The safe Working Potential = 35.6 KV

Exa 8.6 - page 25

In [18]:
from __future__ import division
from numpy import pi, sqrt
#given data
CN=0.4 #in uF
V=33 #in KV
VP=V/sqrt(3) #in KV
f=25 #in Hz
#Capacitance between 2 cores for 15 Km length
CN_1=15*CN #in uF
#Capacitance of each core to neutral
CN=2*CN_1 #in uF
#Charging current per phase
I=2*pi*f*VP*1000*CN*10**-6 #in Ampere
print "Charging current per phase = %0.2f A" %round(I)

Charging current per phase = 36.00 A

Exa 8.7 - page 225

In [21]:
from __future__ import division
from numpy import pi, sqrt
#given data
l=10 #in Km
C=0.3 #in uF
V=22 #in KV
VP=V/sqrt(3) #in KV
VP=VP*1000 #in Volt
f=50 #in Hz
Capacitance=C*l #in uF
CN=2*Capacitance #in uF
KVA_Taken=3*VP*2*pi*f*VP*CN*10/1000 #in KVA
print "KVA taken by the 10 Km cable = %0.3e KVA" %KVA_Taken

KVA taken by the 10 Km cable = 9.123e+09 KVA

Exa 8.8 - page 225

In [23]:
#given data
P=10 #in Ohm
Q=80 #in Ohm
S2=3400 #in Ohm
S1=2400 #in Ohm
X=P*(S2-S1)/(P+Q) #in Ohm
LoopResistance=P*S2/Q #in Ohm
ResistancePerKm=LoopResistance/10 #in Ohm
Distance=X/ResistancePerKm #in Km
print "Distance of fault from testing end = %0.3f Km" %Distance

Distance of fault from testing end = 2.614 Km

Exa 8.9 - page 226

In [24]:
#given data
Resistance=1.6 #in ohm/Km
l=1000 #in meter
PbyQ=3 #unitless
PplusQbyQ=4 #unitless
LoopResistance=(Resistance/1000)*2*l #in Ohm
X=(1/PplusQbyQ)*LoopResistance #in Ohm
Distance=X/(Resistance/1000) #in meter
print "Distance of Fault from testing end = %0.2f meters" %Distance

Distance of Fault from testing end = 500.00 meters