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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```