Chapter 11: Underground Cables

Example 11.1, Page Number: 273

In [1]:
from __future__ import division
import math

#Variable declaration:
r1 = 0.5                       #conductor radius(cm)
l = 2000                       #conductor length(m)
rho = 5*10**12                 #Resistivity of insulation(ohm-m)
t = 0.4                        #insulation thickness(cm)


#Calculation:
r2 = r1+t                      #Internal sheath radius(cm)
R = rho*math.log(r2/r1)/(2*math.pi*l)      #Insulation resistance of cable(ohm)


#Result:
print "Insulation resistance of cable is",round(R/10**6),"Mohm"

Insulation resistance of cable is 234.0 Mohm

Example 11.2, Page Number: 274

In [2]:
from __future__ import division
import math


#Variable declaration:
r1 = 1.25                      #conductor radius(cm)
l = 1000                       #conductor length(m)
rho = 4.5*10**12                 #Resistivity of insulation(ohm-m)
R = 495*10**6                 #Cable insulation resistance(ohm)


#Calculation:
#Let r2 cm be the internal sheath radius,

r2 = r1*math.exp(R*2*math.pi*l/rho)

#Result:
print "Insulation thickness is",round(r2-r1,2),"cm"

Insulation thickness is 1.25 cm

Example 11.3, Page Number: 274

In [3]:
from __future__ import division
import math


#Variable declaration:
r1 = .10                      #conductor radius(cm)
l = 5000                       #conductor length(m)
r2 = 0.25                      #Internal sheath radius(cm)
R = 0.4*10**6                 #Cable insulation resistance(ohm)



#Calculation:
rho = R*2*3.14*l/(math.log(r2/r1)*10**9)    #resistivity(ohm-m)



#Result:
print "Resistivity of the insulating material is",round(rho,2),"* 10**9  ohm-m"

Resistivity of the insulating material is 13.71 * 10**9  ohm-m

Example 11.4, Page Number: 275

In [4]:
from __future__ import division
import math


#Calculation:
er = 4                       #relative permittivity
D = 1.8                       #internal sheath diameter(cm)
l = 1000                      #cable length(m)
d = 1                         #conductor diameter(cm)


#Calculation:
C = er*l/(41.4*math.log10(D/d))*10**-9     #Capacitance(F)



#Result:
print "The capacitance of the cable is",round(C*10**6,3),"uF"

The capacitance of the cable is 0.378 uF

Example 11.5, Page Number: 276

In [5]:
from __future__ import division
import math

#Variable declaration:
er = 4                       #relative permittivity
d = 10                       #core diameter(cm)
l = 1000                      #cable length(m)
t = 7                         #insulation thickness(cm)
Vl = 66000                     #line voltage(V)
f = 50                       #frequency(Hz)


#Calculation:
D = d+2*t                         #conductor diameter(cm)
C = 4*1000/(41.4*math.log10(D/d))*10**-3     #Capacitance(uF)

Vp = Vl/3**0.5
I = 2*3.14*f*C*Vp*10**-6                   #Carging current(A)


#Result:
print "The capacitance is",round(C,3),"uF"
print "Charging current of a single core cable is",round(I,2),"A"

The capacitance is 0.254 uF
Charging current of a single core cable is 3.04 A

Example 11.6, Page Number: 276

In [6]:
from __future__ import division
import math

#Calculation:
er = 3                       #relative permittivity
d = 2.5                       #core diameter(cm)
l = 4000                      #cable length(m)
t = 0.5                         #insulation thickness(cm)
Vl = 33000                     #line voltage(V)
f = 50                       #frequency(Hz)


#Calculation:
D = d+2*t                         #conductor diameter(cm)
C = er*l/(41.4*math.log10(D/d))*10**-3    #Capacitance(uF)
Vp = Vl/3**0.5
I = 2*3.14*f*C*Vp*10**-6                   #Carging current(A)
kVAR = 3*Vp*I                    #Total charging kVAR


#Result:
print "(i)  The capacitance is",round(C*10**3),"* 10**-9 F"
print "(ii) Charging current of a single core cable is",round(I,2),"A"
print "(iii)Total charging kVAR is",round(kVAR/1000,1),"* 10**3 kVAR"

(i)  The capacitance is 1984.0 * 10**-9 F
(ii) Charging current of a single core cable is 11.87 A
(iii)Total charging kVAR is 678.3 * 10**3 kVAR

Example 11.7, Page Number: 278

In [7]:
from __future__ import division
import math


#Variable Declaration:
V = 33                      #voltage of cable(V)
d = 1                          #conductor diameter(cm)
D = 4                          #sheath diameter(cm)


#Calculation:
gmax = 2*V/(d*math.log(D/d))         #maximum stress,rms(kV/cm)
gmin = gmax*d/D                      #minimum stress,rms(kV/cm)

#Result:
print "The maximum and minimum stress in the insulation are"
print "gmax =",round(gmax,2),"kV/cm rms & gmin =",round(gmin,2),"kV/cm rms"

The maximum and minimum stress in the insulation are
gmax = 47.61 kV/cm rms & gmin = 11.9 kV/cm rms

Example 11.8, Page Number: 278

In [8]:
from __future__ import division
import math


#Variable declaration:
gmax = 40                    #kV/cm
gmin = 10                    #kV/cm
d = 2                        #conductor diameter(cm)


#Calculation:
D = gmax/gmin*d              #cm
t = (D-d)/2                  #thickness of insulation(cm)
V = gmax*d*math.log(D/d)/2

#Result:
print "(i) Thickness of insulation is",t,"cm"
print "(ii)Operating voltage is",round(V,2),"kV rms"

(i) Thickness of insulation is 3.0 cm
(ii)Operating voltage is 55.45 kV rms

Example 11.9, Page Number: 279

In [9]:
from __future__ import division
import math

#Variable declaration:
V = 11                         #voltage of cable(V)
a = 0.645                         #conductor area(cm**2)
D = 2.18                          #internal diameter of sheath(cm)
er = 3.5                           #relative permitivity
l = 1000                          #conductor length(m)



#Calculation:
d = (4*a/3.14)**0.5                      #Diameter of the conductor(cm)
gmax = 2*V/(d*math.log(D/d))         #Maximum electrostatic stress(kV/cm rms)
gmin = 2*V/(D*math.log(D/d))         #Minimum electrostatic stress(kV/cm rms)
C = er*l/(41.4*math.log10(D/d))*10**-9         #Capacitance of cable(F)
I = 2*3.14*f*C*V*1000                   #Carging current(A)


#Result:
print "(i)  Maximum electrostatic stress in the cable is",round(gmax,2),"kV/cm rms"
print "(ii) Minimum electrostatic stress in the cable is",round(gmin,2),"kV/cm rms"
print "(iii)Capacitance of the cable per km length is",round(C*10**6,2),"* 10**-6 F"
print "(iv) Charging current is",round(I,3),"A"

(i)  Maximum electrostatic stress in the cable is 27.66 kV/cm rms
(ii) Minimum electrostatic stress in the cable is 11.5 kV/cm rms
(iii)Capacitance of the cable per km length is 0.22 * 10**-6 F
(iv) Charging current is 0.766 A

Example 11.10, Page Number: 280

In [11]:
from __future__ import division


#Variable declaration:
V = 50                       #Cable voltage(kV)
gmax = 40                    #Maximum permissible stress(kV/cm)


#Calculation:
Vp = V*2**0.5                 #Peak value of cable voltage(kV)
d = 2*Vp/gmax                 #Most economical conductor diameter(cm)


#Result:
print "The most economical value of diameter is",round(d,2),"cm"

The most economical value of diameter is 3.54 cm

Example 11.11, Page Number: 280

In [12]:
from __future__ import division


#Variable declaration:
Vl = 132                       #Cable voltage(kV)
gmax = 60                    #Maximum permissible stress(kV/cm)


#Calculation:
Vph = Vl/3**0.5                #phase voltage(kV)
Vp = Vph*2**0.5                 #Peak value of cable voltage(kV)
d = 2*Vp/gmax                 #Most economical conductor diameter(cm)
D = 2.718*d                      #Internal diameter of sheath(cm)

#Result:
print "Most economical conductor diameter is",round(d,1),"cm"
print "Internal diameter of sheath, D is",round(D,2),"cm"

Most economical conductor diameter is 3.6 cm
Internal diameter of sheath, D is 9.76 cm

Example 11.12, Page Number: 282

In [13]:
from __future__ import division
import math


#Variable declaration:
d = 2                            #conductor diameter(cm)
e3 = 3                             #relative permittivity
e2 = 4
e1 = 5
D = 8                              #overall diameter(cm)
gmax = 40                           #kV/cm

#Calculation:
#Graded cable: As the maximum stress in the three dielectrics is the same,
d1 = e1*d/e2                      #diameter of 1st layer(cm)
d2 = e1*d/e3                      #diameter of 2nd layer(cm)
#Permissible peak voltage for the cable:
Vp1 = gmax/2*(d*math.log(d1/d)+d1*math.log(d2/d1)+d2*math.log(D/d2))  #kV
Vs1 = Vp1/2**0.5                   #Safe working voltage (r.m.s.) for cable(kV)

#Ungraded cable:
Vp2 = gmax/2*d*log(D/d)             #kV
Vs2 = Vp2/(2**0.5)                 #kV


#Result:
print "For Graded cable, safe working voltage is",round(Vs1,2),"kV"
print "For Ungraded cable, safe working voltage is",round(Vs2,1),"kV"

For Graded cable, safe working voltage is 57.75 kV
For Ungraded cable, safe working voltage is 39.2 kV

Example 11.13, Page Number: 283

In [14]:
from __future__ import division
import math


#Variable declaration:
d = 3                            #conductor diameter(cm)
e2 = 4                       #relative permittivity
e1 = 5
D = 9                              #overall diameter(cm)
g1max = 30                           #kV/cm
g2max = 20                         #kV/cm


#Calculation:
d1 = g1max/g2max*e1*d/e2              #cm
t1 = (d1-d)/2                   #Radial thickness of inner dielectric(cm)
t2 = (D-d1)/2                   #Radial thickness of outer dielectric(cm)
Vp = g1max/2*d*math.log(d1/d)+g2max/2*d1*math.log(D/d1)
Vsf = Vp/2**0.5                   #Safe working voltage(r.m.s.)for the cable(kV)


#Result:
print "Radial thickness of inner dielectric is",round(t1,3),"cm"
print "Radial thickness of outer dielectric is",round(t2,2),"cm"
print "Safe working voltage (r.m.s.) for the cable is",round(Vsf,2),"kV"

Radial thickness of inner dielectric is 1.313 cm
Radial thickness of outer dielectric is 1.69 cm
Safe working voltage (r.m.s.) for the cable is 38.7 kV

Example 11.14, Page Number: 284

In [16]:
from __future__ import division
import math

#Variable Declaration:
e1 = 5                          #relative permittivity
e2 = 3
t = 1                          #thickness(cm)
d = 2                          #core diameter(cm)
V = 66                         #cable voltage(kV)

#Calculation:
d1 = d+2*t
D = d+4*t                     #total diameter of cable(cm)
Vpk = V/3**0.5*2**0.5            #Peak voltage per phase(kV)
g1max = 2*Vpk/(d*(math.log(d1/d)+e1/e2*math.log(D/d1)))      #kV/cm
g2max = 2*Vpk/(d1*(e2/e1*math.log(d1/d)+math.log(D/d1)))      #kV/cm

#Result:
print "Maximum stresses in two dielectrics are:"
print "g1max =",round(g1max,2),"kV/cm     g2max =",round(g2max,2),"kV/cm"

Maximum stresses in two dielectrics are:
g1max = 39.37 kV/cm     g2max = 32.8 kV/cm

Example 11.15, Page Number: 285

In [18]:
from __future__ import division
import math
from sympy import *

#Variable declaration:
d = 2                           #cm
d1 = 3.1                        #cm
d2 = 4.2                        #cm
D = 5.3                         #cm
V = 66                          #cable voltage(kV)


#Calculation:
Vpk = V/3**0.5*2**0.5           #peak voltage/phase(kV)
#let V1,V2 & V3 are the voltages at different grades.
V1,V2,V3 = symbols('V1 V2 V3')
g1max = V1/(d/2*math.log(d1,d))
g2max = V2/(d1/2*math.log(d2,d1))
g3max = V3/(d2/2*math.log(D,d2))

#As the maximum stress in the layers is the same,
#∴     g1max = g2max = g3max
#or    2·28 V1 = 2·12 V2 = 2·04 V3
#∴     V2 = (2·28/2·12) V1 = 1·075 V1
#and   V3 = (2·28/2·04) V1 = 1·117 V1
#Now   V1 + V2 + V3 = Vpk
#or    V1 + 1·075 V1 + 1·117 V1 = 53·9
V1 = 53.9/3.192
V2 = 1.075*V1
V1s = Vpk-V1                   #Voltage on first intersheath(near to core)(kV)
V2s = Vpk-V1-V2                #Voltage on second intersheath(kV)


#Result:
print "Voltage on first intersheath is",round(V1s,2),"kV"
print "Voltage on second intersheath is",round(V2s,2),"kV"

Voltage on first intersheath is 37.0 kV
Voltage on second intersheath is 18.85 kV

Example 11.16, Page Number: 286

In [1]:
from __future__ import division
import math


#Variable declaration:
d = 2                       #core diameter(cm)
D = 5.3                       #cm
V = 66                      #cable voltage(kV)



#Calculation:
Vpk = V/3**0.5*2**0.5           #peak voltage/phase(kV)
#(i) Positions of intersheaths.

#Let the diameters of intersheaths are d1 and d2 cm respectively.
#Let V1 = voltage b/w conductor & intersheath 1
#    V2 = voltage b/w intersheaths 1 and 2
#    V3 = voltage b/w intersheath 2 & outer lead sheath


#Given the maximum stress in the three layers is the same,
#we get the relation as given below:
#            d1**2 = d * d2 = 2*d2      [∵ d = 2 cm]
#or          d2 = d1**2/2
#and         d1*d2 = D * d = 5·3 × 2 = 10.6 cm
#or          d1 * d1**2/2 = 10·6
d1 = 21.2**(1/3)                       #cm
d2 = d1**2/2                           #cm


#(ii) Voltage on intersheaths,
#        V = V1 + V2 + V3
#or      53·9 = V1+d1/d*V1+d2/d*V1
#             = 4.28*V1
V1 = 53.9/4.28                        #kV
V2 = d1/d*V1                          #kV

#(iii) Stresses in dielectrics,
gmax = V1/(d/2*math.log(d1/d))            #max stress(kV/cm)
gmin = V1/(d1/2*math.log(d1/d))            #min stress(kV/cm

#Result:
print "(i) Positions of intersheaths are:"
print "\td1 =",round(d1,2),"cm d2 =",round(d2,1),"cm"
print "(ii) Voltage on the intersheaths are:"
print "\tVoltage on first intersheath is",round(Vpk-V1,2),"kV"
print "\tVoltage on second intersheath is",round(Vpk-V1-V2,1),"kV"
print "(iii) Maximum and minimum stress are:"
print "\tgmax =",round(gmax),"kV/cm\tgmin =",round(gmin,2),"kV/cm"

(i) Positions of intersheaths are:
	d1 = 2.77 cm d2 = 3.8 cm
(ii) Voltage on the intersheaths are:
	Voltage on first intersheath is 41.3 kV
	Voltage on second intersheath is 23.9 kV
(iii) Maximum and minimum stress are:
	gmax = 39.0 kV/cm	gmin = 28.01 kV/cm

Example 11.17, Page Number: 289

In [3]:
from __future__ import division
import math

#Variable Declaration:
c = 0.3                          #capacitance per kilometre(uF/km)
V = 11                           #line voltage(kV)
l = 5                            #length of the cable(km)
f = 50                           #Hz

#Calculation:
C3 = c*l                         #capacitance between a pair of cores with third core
                                 #earthed for a length of 5 km (uF)
Vph = V*1000/3**0.5              #phase voltage(V)
#core to neutral capacitance Cn of this cable is given by :
Cn = 2*C3                        #uF
Ic = 2*math.pi*f*Vph*Cn*10**-6


#Result:
print " The charging current is",round(Ic,2),"A"

 The charging current is 5.99 A

Example 11.18, Page Number: 290

In [4]:
from __future__ import division
import math

#Variable Declaration:
V = 66                           #line voltage(kV)
C1 = 12.6                         #uF
C2 = 7.4                          #uF
f = 50                           #Hz

#Calculation:
Vph = V*1000/3**0.5               #phase voltage(V)
Ce = C1/3                         #core-earth capacitances(uF)
Cc = (C2-Ce)/2                    #core-core capacitances(uF)
Cn = Ce+3*Cc                      #Core to neutral capacitance(uF)
Ic = 2*math.pi*f*Vph*Cn*10**-6


#Result:
print "The charging current is",round(Ic,2),"A"

The charging current is 107.74 A

Example 11.19, Page Number: 290

In [5]:
from __future__ import division
import math

#Variable Declaration:
c = 0.18                          #capacitance per kilometre(uF/km)
V = 3300                           #line voltage(V)
l = 20                            #length of the cable(km)
f = 50                           #Hz



#Calculation:
C3 = c*l                         #capacitance between a pair of cores with third core
                                 #earthed for a length of 20 km (uF)
Vph = V/3**0.5                  #phase voltage(V)
Cn = 2*C3                       #Core to neutral capacitance(uF)
Ic = 2*math.pi*f*Cn*Vph*10**-6                   #charging current(A)
kVA = 3*Vph*Ic/1000                    #kVA taken by the cable


#Result:
print "The kVA taken by 20 km long cable is",round(kVA,2),"kVA"

The kVA taken by 20 km long cable is 24.63 kVA

Example 11.20, Page Number: 292

In [6]:
from __future__ import division
import math

#Variable declaration:
k = 5                #thermal resistivity of the dielectric(ohm-m)
S2 = 0.45        #thermal resistance b/w the sheath and the ground surface
R = 110           #electrical resistance of the cable(u-ohm)
r = 15                #core radius(mm)
t = 40                #dielectric thickness(mm)
T = 55                  #temperature(deg. C)
n = 1                   #no. of conductors

#Calculation:
r1 = r+t                  #mm
S1 = k/(2*math.pi)*math.log(r1/r)         #ohm/m
S = S1+S2                          #ohm/m
I = (T/(n*R*10**-6*S))**0.5               #current loading(A)



#Result:
print "Maximum permissible current loading is",round(I,1),"A"

Maximum permissible current loading is 580.5 A

Example 11.21, Page Number: 296

In [23]:
from __future__ import division

#Variable declaration:
Q = 15                            #ohm
P = 45                             #ohm
l = 300                           #length of faulty cable(m)


#Calculation:
L = 2*l                         #loop length(m)
d = Q/(P+Q)*L             #Distance of the fault point from test end(m)

#Result:
print " The distance of the fault point from the test end is",d,"m"

 The distance of the fault point from the test end is 150.0 m

Example 11.22, Page Number: 297

In [24]:
from __future__ import division


#Variable declaration:
l = 500                           #length of faulty cable(m)
#P:Q = 3

#Calculation:
#Let:
P = 3;                  Q = 1            #ohm
#then,
d = Q/(P+Q)*2*l             #Distance of the fault point from test end(m)

#Result:
print "The distance of the fault from the testing end of cables",d,"m"

The distance of the fault from the testing end of cables 250.0 m

Example 11.23, Page Number: 297

In [32]:
from __future__ import division

#Variable declaration:
l = 500                           #length of faulty cable(m)
rp = 0.001                        #resistance of cable(ohm/m)
rq = 0.00225                      #resistance of sound cable(ohm/m)
#P:Q = 2.75:1

#Calculation:
#Let:
P = 2.75;                  Q = 1            #ohm
#then,
R = rp*l+rq*l                    #Resistance of loop(ohm)
X = Q/(P+Q)*R       #Resistance of faulty cable from test end upto fault point(ohm)
d = X/rp            #Distance of the fault point from test end(m)

#Result:
print "The distance of the fault from the testing end of cables",round(d),"m"

The distance of the fault from the testing end of cables 433.0 m

Example 11.24, Page Number: 297

In [33]:
from __future__ import division

#Variable declaration:
S = 200                        #ohm
r = 20                         #resistance per km(ohm)
l = 20                         #length of cable(km)



#Calculation:
#       R+X = 20*(20+20)        #ohm
#         P = Q
X = (800-200)/2                 #ohm
d = X/r                         #m


#Result:
print "The distance of the fault from the test end is",d,"km"

The distance of the fault from the test end is 15.0 km