# CHAPTER 2.7: UNDERGROUND CABLES¶

## Example 2.7.1, Page number 211¶

In :
import math

#Variable declaration
d = 2.5              #Core diameter(cm)
t = 1.25             #Insulation thickness(cm)
rho = 4.5*10**14     #Resistivity of insulation(ohm-cm)
l = 10.0**5          #Length(cm)

#Calculation
D = d+2*t                                #Overall diameter(cm)
R_i = rho/(2*math.pi*l)*math.log(D/d)    #Insulation resistance(ohm)

#Result
print('Insulation resistance per km, R_i = %.2e ohm' %R_i)
print('\nNOTE: ERROR: Mistake in final answer in textbook')

Insulation resistance per km, R_i = 4.96e+08 ohm

NOTE: ERROR: Mistake in final answer in textbook


## Example 2.7.2, Page number 211¶

In :
import math
import cmath

#Variable declaration
R = 495.0*10**6      #Insulation resistance(ohm/km)
d = 3.0              #Core diameter(cm)
rho = 4.5*10**14     #Resistivity of insulation(ohm-cm)

#Calculation
l = 1000.0                               #Length of cable(m)
Rho = rho/100.0                          #Resistivity of insulation(ohm-m)
r1_r2 = cmath.exp((2*math.pi*l*R)/Rho)   #r1/r2
thick = r_1-r_2                          #Insulation thickness(cm)

#Result
print('Insulation thickness = %.1f cm' %thick)

Insulation thickness = 1.5 cm


## Example 2.7.3, Page number 212¶

In :
import math

#Variable declaration
V = 66.0*10**3        #Line Voltage(V)
l = 1.0               #Length of cable(km)
d = 15.0              #Core diameter(cm)
D = 60.0              #Sheath diameter(cm)
e_r = 3.6             #Relative permittivity
f = 50.0              #Frequency(Hz)

#Calculation
C = e_r/(18.0*math.log(D/d))*l          #Capacitance(µF)
I_ch = V/3**0.5*2*math.pi*f*C*10**-6    #Charging current(A)

#Result
print('Capacitance, C = %.3f µF' %C)
print('Charging current = %.2f A' %I_ch)

Capacitance, C = 0.144 µF
Charging current = 1.73 A


## Example 2.7.4, Page number 212¶

In :
#Variable declaration
V_l = 132.0         #Line Voltage(kV)
g_max = 60.0        #Maximum Line Voltage(kV)

#Calculation
V = V_l/3**0.5*2**0.5     #Phase Voltage(kV)
d = 2*V/g_max             #Core diameter(cm)
D = 2.718*d               #Overall diameter(cm)

#Result
print('Overall diameter of the insulation, D = %.3f cm' %D)

Overall diameter of the insulation, D = 9.765 cm


## Example 2.7.6, Page number 212-213¶

In :
import math

#Variable declaration
V = 11.0*10**3        #Line Voltage(V)
dia_out = 8.0         #Outside diameter(cm)

#Calculation
D = dia_out/2.0                      #Overall diameter(cm)
d = (D)/2.718                        #Conductor diameter(cm)
g_m = 2*V/(d*math.log(D/d)*10)       #Maximum value of electric field strength(kV/m)

#Result
print('Conductor radius, r = %.3f cm' %r)
print('Electric field strength that must be withstood, g_m = %.f kV/m' %g_m)

Conductor radius, r = 0.736 cm
Electric field strength that must be withstood, g_m = 1495 kV/m


## Example 2.7.7, Page number 214¶

In :
#Variable declaration

#Calculation
R_2 = (R_1*R_3)**0.5      #Location of intersheath(cm)
alpha = R_1/R_2           #α
ratio = 2.0/(1+alpha)     #Ratio of maximum electric field strength with & without intersheath

#Result
print('Location of intersheath, R_2 = %.2f cm' %R_2)
print('Ratio of maximum electric field strength with & without intersheath = %.3f ' %ratio)

Location of intersheath, R_2 = 1.58 cm
Ratio of maximum electric field strength with & without intersheath = 0.775


## Example 2.7.8, Page number 215¶

In :
import math

#Variable declaration
V = 33.0              #Line Voltage(kV)
D_2 = 2.0             #Conductor diameter(cm)
D_1 = 3.0             #Sheath diameter(cm)

#Calculation
g_max = V/(R_2*math.log(R_1/R_2))   #RMS value of maximum stress in the insulation(kV/cm)
g_min = V/(R_1*math.log(R_1/R_2))   #RMS value of minimum stress in the insulation(kV/cm)

#Result
print('Maximum stress in the insulation, g_max = %.2f kV/cm (rms)' %g_max)
print('Minimum stress in the insulation, g_min = %.2f kV/cm (rms)' %g_min)

Maximum stress in the insulation, g_max = 81.39 kV/cm (rms)
Minimum stress in the insulation, g_min = 54.26 kV/cm (rms)


## Example 2.7.9, Page number 215¶

In :
import math

#Variable declaration
d = 2.5             #Conductor diameter(cm)
D = 6.0             #Sheath diameter(cm)
V_l = 66.0          #Line Voltage(kV)

#Calculation
alpha = (D/d)**(1.0/3)               #α
d_1 = d*alpha                        #Best position of first intersheath(cm)
d_2 = d_1*alpha                      #Best position of second intersheath(cm)
V = V_l/3**0.5*2**0.5                #Peak voltage on core(kV)
V_2 = V/(1+(1/alpha)+(1/alpha**2))   #Peak voltage on second intersheath(kV)
V_1 = (1+(1/alpha))*V_2              #Voltage on first intersheath(kV)
stress_max = 2*V/(d*math.log(D/d))   #Maximum stress without intersheath(kV/cm)
stress_min = stress_max*d/D          #Minimum stress without intersheath(kV/cm)
g_max = V*3/(1+alpha+alpha**2)       #Maximum stress with intersheath(kV/cm)

#Result
print('Maximum stress without intersheath = %.2f kV/cm' %stress_max)
print('Best position of first intersheath, d_1 = %.2f cm' %d_1)
print('Best position of second intersheath, d_2 = %.3f cm' %d_2)
print('Maximum stress with intersheath = %.2f kV/cm' %g_max)
print('Voltage on the first intersheath, V_1 = %.2f kV' %V_1)
print('Voltage on the second intersheath, V_2 = %.2f kV' %V_2)
print('\nNOTE: Changes in the obtained answer is due to more precision here')

Maximum stress without intersheath = 49.24 kV/cm
Best position of first intersheath, d_1 = 3.35 cm
Best position of second intersheath, d_2 = 4.481 cm
Maximum stress with intersheath = 39.13 kV/cm
Voltage on the first intersheath, V_1 = 40.85 kV
Voltage on the second intersheath, V_2 = 23.38 kV

NOTE: Changes in the obtained answer is due to more precision here


## Example 2.7.10, Page number 215-216¶

In :
import math

#Variable declaration
e_1 = 3.6           #Inner relative permittivity
e_2 = 2.5           #Outer relative permittivity
d = 1.0             #Conductor diameter(cm)
d_1 = 3.0           #Sheath diameter(cm)
D = 5.0             #Overall diameter(cm)
V_l = 66.0          #Line Voltage(kV)

#Calculation
V = V_l/3**0.5*2**0.5                                          #Peak voltage on core(kV)
g1_max = 2*V/(d*(math.log(d_1/d)+e_1/e_2*math.log(D/d_1)))     #Maximum stress in first dielectric(kV/km)
g2_max = 2*V/(d_1*(e_2/e_1*math.log(d_1/d)+math.log(D/d_1)))   #Maximum stress in second dielectric(kV/km)

#Result
print('Maximum stress in first dielectric, g_1_max = %.2f kV/km' %g1_max)
print('Maximum stress in second dielectric, g_2_max = %.2f kV/km' %g2_max)

Maximum stress in first dielectric, g_1_max = 58.76 kV/km
Maximum stress in second dielectric, g_2_max = 28.20 kV/km


## Example 2.7.11, Page number 216-217¶

In :
#Variable declaration
V = 85.0          #Line Voltage(kV)
g_max = 55.0      #Maximum stress(kV/cm)

#Calculation
V_1 = 0.632*V         #Intersheath potential(kV)
d = 0.736*V/g_max     #Core diameter(cm)
d_1 = 2*V/g_max       #Intersheath diameter(cm)
D = 3.76*V/g_max      #Overall diameter(cm)
d_un = 2*V/g_max      #Core diameter of ungraded cable(cm)
D_un = 2.718*d_1      #Overall diameter of ungraded cable(cm)

#Result
print('Diameter of intersheath, d_1 = %.2f cm' %d_1)
print('Voltage of intersheath, V_1 = %.2f kV, to neutral' %V_1)
print('Conductor diameter of graded cable, d = %.2f cm' %d)
print('Outside diameter of graded cable, D = %.2f cm' %D)
print('Conductor diameter of ungraded cable, d = %.2f cm' %d_un)
print('Outside diameter of ungraded cable, D = %.2f cm' %D_un)

Diameter of intersheath, d_1 = 3.09 cm
Voltage of intersheath, V_1 = 53.72 kV, to neutral
Conductor diameter of graded cable, d = 1.14 cm
Outside diameter of graded cable, D = 5.81 cm
Conductor diameter of ungraded cable, d = 3.09 cm
Outside diameter of ungraded cable, D = 8.40 cm


## Example 2.7.12, Page number 219¶

In :
import math

#Variable declaration
c = 0.3         #Capacitance b/w any 2 conductor & sheath earthed(µF/km)
l = 10.0        #Length(km)
V = 33.0        #Line Voltage(kV)
f = 50.0        #Frequency(Hz)

#Calculation
C_eq = l*c                            #Capacitance b/w any 2 conductor & sheath earthed(µF)
C_p = 2.0*C_eq                        #Capacitance per phase(µF)
kVA = V**2*2*math.pi*f*C_p/1000.0     #Three-phase kVA required(kVA)

#Result
print('Equivalent star connected capacity, C_eq = %.f µF' %C_eq)
print('kVA required = %.1f kVA' %kVA)

Equivalent star connected capacity, C_eq = 3 µF
kVA required = 2052.7 kVA


## Example 2.7.13, Page number 219¶

In :
import math

#Variable declaration
V = 11.0*10**3    #Line Voltage(V)
f = 50.0          #Frequency(Hz)
C_c = 3.7         #Measured capacitance(µF)

#Calculation
C_0 = 2*C_c                                   #Capacitance(µF)
I_ch = 2*math.pi*f*C_0*V/3**0.5*10**-6        #Charging current per phase(A)

#Result
print('Charging current drawn by a cable = %.2f A' %I_ch)

Charging current drawn by a cable = 14.76 A


## Example 2.7.14, Page number 219-220¶

In :
import math

#Variable declaration
c_s = 0.90        #Capacitance b/w all conductors(µF)
C_0 = 0.4         #Capacitance b/w two conductor(µF)
V = 11.0*10**3    #Line Voltage(V)
f = 50.0          #Frequency(Hz)

#Calculation
C_s = c_s/3.0                              #Capacitance measured(µF)
C_c = (C_0-C_s)/2.0                        #Capacitance(µF)
C_a = 3.0/2*(C_c+(1/3.0)*C_s)              #Capacitance b/w any two conductors(µF)
C_b = 2.0*C_c+(2.0/3)*C_s                  #Capacitance b/w any two bounded conductors and the third conductor(µF)
C_o = 3.0*C_c+C_s                          #Capacitance to neutral(µF)
I_c = 2.0*math.pi*f*C_o*V/3**0.5*10**-6    #Charging current(A)

#Result
print('Case(a): Capacitance between any two conductors = %.3f µF' %C_a)
print('Case(b): Capacitance between any two bounded conductors and the third conductor = %.1f µF' %C_b)
print('Case(c): Capacitance to neutral, C_0 = %.2f µF' %C_o)
print('         Charging current taken by cable, I_c = %.3f A' %I_c)
print('\nNOTE: ERROR: Calculation mistakes in textbook answer')

Case(a): Capacitance between any two conductors = 0.225 µF
Case(b): Capacitance between any two bounded conductors and the third conductor = 0.3 µF
Case(c): Capacitance to neutral, C_0 = 0.45 µF
Charging current taken by cable, I_c = 0.898 A

NOTE: ERROR: Calculation mistakes in textbook answer


## Example 2.7.15, Page number 220-221¶

In :
import math

#Variable declaration
V = 13.2*10**3    #Line Voltage(V)
f = 50.0          #Frequency(Hz)
C_BC = 4.2        #Capacitance b/w two cores(µF)

#Calculation
C_n = 2.0*C_BC                              #Capacitance to neutral(µF)
V_ph = V/3**0.5                             #Operating phase voltage(V)
I_c = 2.0*math.pi*f*C_n*V/3**0.5*10**-6     #Charging current(A)

#Result
print('Charging current drawn by cable, I_c = %.2f A' %I_c)

Charging current drawn by cable, I_c = 20.11 A


## Example 2.7.16, Page number 222-223¶

In :
import math

#Variable declaration
V = 33.0*10**3    #Line Voltage(V)
f = 50.0          #Frequency(Hz)
l = 4.0           #Length(km)
d = 2.5           #Diameter of conductor(cm)
t = 0.5           #Radial thickness of insulation(cm)
e_r = 3.0         #Relative permittivity of the dielectric
PF = 0.02         #Power factor of unloaded cable

#Calculation
#Case(a)
r = d/2.0                                         #Radius of conductor(cm)
e_0 = 8.85*10**-12                                #Permittivity
C = 2.0*math.pi*e_0*e_r/math.log(R/r)*l*1000      #Capacitance of cable/phase(F)
#Case(b)
V_ph = V/3**0.5                                   #Phase voltage(V)
I_c = V_ph*2.0*math.pi*f*C                        #Charging current/phase(A)
#Case(c)
kVAR = 3.0*V_ph*I_c                               #Total charging kVAR
#Case(d)
phi = math.acos(PF)*180/math.pi                   #Φ(°)
delta = 90.0-phi                                  #δ(°)
P_c = V_ph*I_c*math.sin(delta*math.pi/180)/1000   #Dielectric loss/phase(kW)
#Case(e)
E_max = V_ph/(r*math.log(R/r)*1000)               #RMS value of Maximum stress in cable(kV/cm)

#Result
print('Case(a): Capacitance of the cable, C = %.3e F/phase' %C)
print('Case(b): Charging current = %.2f A/phase' %I_c)
print('Case(c): Total charging kVAR = %.4e kVAR' %kVAR)
print('Case(d): Dielectric loss/phase, P_c = %.2f kW' %P_c)
print('Case(e): Maximum stress in the cable, E_max = %.1f kV/cm (rms)' %E_max)

Case(a): Capacitance of the cable, C = 1.983e-06 F/phase
Case(b): Charging current = 11.87 A/phase
Case(c): Total charging kVAR = 6.7847e+05 kVAR
Case(d): Dielectric loss/phase, P_c = 4.52 kW
Case(e): Maximum stress in the cable, E_max = 45.3 kV/cm (rms)