15 files changed, 439 insertions, 0 deletions
diff --git a/3472/CH14/EX14.1/Example14_1.sce b/3472/CH14/EX14.1/Example14_1.sce
new file mode 100644
index 000000000..8f06c32ef
--- /dev/null
+++ b/3472/CH14/EX14.1/Example14_1.sce
@@ -0,0 +1,26 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.1 :
+// Page number 211
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+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)
+
+// Calculations
+D = d+2*t                        // Overall diameter(cm)
+R_i = rho/(2*%pi*l)*log(D/d)     // Insulation resistance(ohm)
+
+// Results
+disp("PART II - EXAMPLE : 7.1 : SOLUTION :-")
+printf("\nInsulation resistance per km, R_i = %.2e ohm\n", R_i)
+printf("\nNOTE: ERROR: Mistake in final answer in textbook")
diff --git a/3472/CH14/EX14.10/Example14_10.sce b/3472/CH14/EX14.10/Example14_10.sce
new file mode 100644
index 000000000..c16e707c4
--- /dev/null
+++ b/3472/CH14/EX14.10/Example14_10.sce
@@ -0,0 +1,29 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.10 :
+// Page number 215-216
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+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)
+
+// Calculations
+V = V_l/3**0.5*2**0.5                                // Peak voltage on core(kV)
+g1_max = 2*V/(d*(log(d_1/d)+e_1/e_2*log(D/d_1)))     // Maximum stress in first dielectric(kV/km)
+g_max = 2*V/(d_1*(e_2/e_1*log(d_1/d)+log(D/d_1)))    // Maximum stress in second dielectric(kV/km)
+
+// Results
+disp("PART II - EXAMPLE : 7.10 : SOLUTION :-")
+printf("\nMaximum stress in first dielectric, g_1_max = %.2f kV/cm", g1_max)
+printf("\nMaximum stress in second dielectric, g_max = %.2f kV/cm", g_max)
diff --git a/3472/CH14/EX14.11/Example14_11.sce b/3472/CH14/EX14.11/Example14_11.sce
new file mode 100644
index 000000000..22ebce036
--- /dev/null
+++ b/3472/CH14/EX14.11/Example14_11.sce
@@ -0,0 +1,32 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.11 :
+// Page number 216-217
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 85.0          // Line Voltage(kV)
+g_max = 55.0      // Maximum stress(kV/cm)
+
+// Calculations
+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)
+
+// Results
+disp("PART II - EXAMPLE : 7.11 : SOLUTION :-")
+printf("\nDiameter of intersheath, d_1 = %.2f cm", d_1)
+printf("\nVoltage of intersheath, V_1 = %.2f kV, to neutral", V_1)
+printf("\nConductor diameter of graded cable, d = %.2f cm", d)
+printf("\nOutside diameter of graded cable, D = %.2f cm", D)
+printf("\nConductor diameter of ungraded cable, d = %.2f cm", d_un)
+printf("\nOutside diameter of ungraded cable, D = %.2f cm", D_un)
diff --git a/3472/CH14/EX14.12/Example14_12.sce b/3472/CH14/EX14.12/Example14_12.sce
new file mode 100644
index 000000000..eae925cdd
--- /dev/null
+++ b/3472/CH14/EX14.12/Example14_12.sce
@@ -0,0 +1,27 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.12 :
+// Page number 219
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+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)
+
+// Calculations
+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*%pi*f*C_p/1000.0         // Three-phase kVA required(kVA)
+
+// Results
+disp("PART II - EXAMPLE : 7.12 : SOLUTION :-")
+printf("\nEquivalent star connected capacity, C_eq = %.f µF", C_eq)
+printf("\nkVA required = %.1f kVA", kVA)
diff --git a/3472/CH14/EX14.13/Example14_13.sce b/3472/CH14/EX14.13/Example14_13.sce
new file mode 100644
index 000000000..b216b785d
--- /dev/null
+++ b/3472/CH14/EX14.13/Example14_13.sce
@@ -0,0 +1,24 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.13 :
+// Page number 219
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 11.0*10**3    // Line Voltage(V)
+f = 50.0          // Frequency(Hz)
+C_c = 3.7         // Measured capacitance(µF)
+
+// Calculations
+C_0 = 2*C_c                               // Capacitance(µF)
+I_ch = 2*%pi*f*C_0*V/3**0.5*10**-6        // Charging current per phase(A)
+
+// Results
+disp("PART II - EXAMPLE : 7.13 : SOLUTION :-")
+printf("\nCharging current drawn by a cable = %.2f A", I_ch)
diff --git a/3472/CH14/EX14.14/Example14_14.sce b/3472/CH14/EX14.14/Example14_14.sce
new file mode 100644
index 000000000..62d9e231e
--- /dev/null
+++ b/3472/CH14/EX14.14/Example14_14.sce
@@ -0,0 +1,33 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.14 :
+// Page number 219-220
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+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)
+
+// Calculations
+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*%pi*f*C_o*V/3**0.5*10**-6        // Charging current(A)
+
+// Results
+disp("PART II - EXAMPLE : 7.14 : SOLUTION :-")
+printf("\nCase(a): Capacitance between any two conductors = %.3f µF", C_a)
+printf("\nCase(b): Capacitance between any two bounded conductors and the third conductor = %.1f µF", C_b)
+printf("\nCase(c): Capacitance to neutral, C_0 = %.2f µF", C_o)
+printf("\n         Charging current taken by cable, I_c = %.3f A \n", I_c)
+printf("\nNOTE: ERROR: Calculation mistakes in textbook answer")
diff --git a/3472/CH14/EX14.15/Example14_15.sce b/3472/CH14/EX14.15/Example14_15.sce
new file mode 100644
index 000000000..487b6a5a4
--- /dev/null
+++ b/3472/CH14/EX14.15/Example14_15.sce
@@ -0,0 +1,25 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.15 :
+// Page number 220-221
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 13.2*10**3    // Line Voltage(V)
+f = 50.0          // Frequency(Hz)
+C_BC = 4.2        // Capacitance b/w two cores(µF)
+
+// Calculations
+C_n = 2.0*C_BC                              // Capacitance to neutral(µF)
+V_ph = V/3**0.5                             // Operating phase voltage(V)
+I_c = 2.0*%pi*f*C_n*V/3**0.5*10**-6         // Charging current(A)
+
+// Results
+disp("PART II - EXAMPLE : 7.15 : SOLUTION :-")
+printf("\nCharging current drawn by cable, I_c = %.2f A", I_c)
diff --git a/3472/CH14/EX14.16/Example14_16.sce b/3472/CH14/EX14.16/Example14_16.sce
new file mode 100644
index 000000000..e8ce1dea8
--- /dev/null
+++ b/3472/CH14/EX14.16/Example14_16.sce
@@ -0,0 +1,46 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.16 :
+// Page number 222-223
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+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
+
+// Calculations
+// Case(a)
+r = d/2.0                                         // Radius of conductor(cm)
+R = r+t                                           // External radius(cm)
+e_0 = 8.85*10**-12                                // Permittivity
+C = 2.0*%pi*e_0*e_r/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*%pi*f*C                            // Charging current/phase(A)
+// Case(c)
+kVAR = 3.0*V_ph*I_c                               // Total charging kVAR
+// Case(d)
+phi = acosd(PF)                                   // Φ(°)
+delta = 90.0-phi                                  // δ(°)
+P_c = V_ph*I_c*sind(delta)/1000                   // Dielectric loss/phase(kW)
+// Case(e)
+E_max = V_ph/(r*log(R/r)*1000)                    // RMS value of Maximum stress in cable(kV/cm)
+
+// Results
+disp("PART II - EXAMPLE : 7.16 : SOLUTION :-")
+printf("\nCase(a): Capacitance of the cable, C = %.3e F/phase", C)
+printf("\nCase(b): Charging current = %.2f A/phase", I_c)
+printf("\nCase(c): Total charging kVAR = %.4e kVAR", kVAR)
+printf("\nCase(d): Dielectric loss/phase, P_c = %.2f kW", P_c)
+printf("\nCase(e): Maximum stress in the cable, E_max = %.1f kV/cm (rms)", E_max)
diff --git a/3472/CH14/EX14.2/Example14_2.sce b/3472/CH14/EX14.2/Example14_2.sce
new file mode 100644
index 000000000..c3553c459
--- /dev/null
+++ b/3472/CH14/EX14.2/Example14_2.sce
@@ -0,0 +1,28 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.2 :
+// Page number 211
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+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)
+
+// Calculations
+l = 1000.0                               // Length of cable(m)
+r_2 = d/2.0                              // Core radius(cm)
+Rho = rho/100.0                          // Resistivity of insulation(ohm-m)
+r1_r2 = exp((2*%pi*l*R)/Rho)             // r1/r2
+r_1 = 2*r_2                              // Cable radius(cm)
+thick = r_1-r_2                          // Insulation thickness(cm)
+
+// Results
+disp("PART II - EXAMPLE : 7.2 : SOLUTION :-")
+printf("\nInsulation thickness = %.1f cm", thick)
diff --git a/3472/CH14/EX14.3/Example14_3.sce b/3472/CH14/EX14.3/Example14_3.sce
new file mode 100644
index 000000000..a2975e342
--- /dev/null
+++ b/3472/CH14/EX14.3/Example14_3.sce
@@ -0,0 +1,28 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.3 :
+// Page number 212
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+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)
+
+// Calculations
+C = e_r/(18.0*log(D/d))*l           // Capacitance(µF)
+I_ch = V/3**0.5*2*%pi*f*C*10**-6    // Charging current(A)
+
+// Results
+disp("PART II - EXAMPLE : 7.3 : SOLUTION :-")
+printf("\nCapacitance of single-core cable, C = %.3f µF", C)
+printf("\nCharging current of single-core cable = %.2f A", I_ch)
diff --git a/3472/CH14/EX14.4/Example14_4.sce b/3472/CH14/EX14.4/Example14_4.sce
new file mode 100644
index 000000000..60b14d907
--- /dev/null
+++ b/3472/CH14/EX14.4/Example14_4.sce
@@ -0,0 +1,26 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.4 :
+// Page number 212
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V_l = 132.0         // Line Voltage(kV)
+g_max = 60.0        // Maximum Line Voltage(kV)
+
+// Calculations
+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)
+
+// Results
+disp("PART II - EXAMPLE : 7.4 : SOLUTION :-")
+printf("\nMost economical diameter of a single-core cable, d = %.1f cm", d)
+printf("\nOverall diameter of the insulation, D = %.3f cm\n", D)
+printf("\nNOTE: Slight change in obtained answer due to precision")
diff --git a/3472/CH14/EX14.6/Example14_6.sce b/3472/CH14/EX14.6/Example14_6.sce
new file mode 100644
index 000000000..9e573631b
--- /dev/null
+++ b/3472/CH14/EX14.6/Example14_6.sce
@@ -0,0 +1,26 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.6 :
+// Page number 212-213
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 11.0*10**3        // Line Voltage(V)
+dia_out = 8.0         // Outside diameter(cm)
+
+// Calculations
+D = dia_out/2.0                      // Overall diameter(cm)
+d = (D)/2.718                        // Conductor diameter(cm)
+r = d/2                              // Conductor radius(cm)
+g_m = 2*V/(d*log(D/d)*10)            // Maximum value of electric field strength(kV/m)
+
+// Results
+disp("PART II - EXAMPLE : 7.6 : SOLUTION :-")
+printf("\nConductor radius, r = %.3f cm", r)
+printf("\nElectric field strength that must be withstood, g_m = %.f kV/m", g_m)
diff --git a/3472/CH14/EX14.7/Example14_7.sce b/3472/CH14/EX14.7/Example14_7.sce
new file mode 100644
index 000000000..3d0fb350e
--- /dev/null
+++ b/3472/CH14/EX14.7/Example14_7.sce
@@ -0,0 +1,25 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.7 :
+// Page number 214
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+R_3 = 1.00            // Cable radius(cm)
+R_1 = 2.5             // Cable radius(cm)
+
+// Calculations
+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
+
+// Results
+disp("PART II - EXAMPLE : 7.7 : SOLUTION :-")
+printf("\nLocation of intersheath, R_2 = %.2f cm", R_2)
+printf("\nRatio of maximum electric field strength with & without intersheath = %.3f ", ratio)
diff --git a/3472/CH14/EX14.8/Example14_8.sce b/3472/CH14/EX14.8/Example14_8.sce
new file mode 100644
index 000000000..6082725f5
--- /dev/null
+++ b/3472/CH14/EX14.8/Example14_8.sce
@@ -0,0 +1,27 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.8 :
+// Page number 215
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 33.0              // Line Voltage(kV)
+D_2 = 2.0             // Conductor diameter(cm)
+D_1 = 3.0             // Sheath diameter(cm)
+
+// Calculations
+R_2 = D_2/2                         // Conductor radius(cm)
+R_1 = D_1/2                         // Sheath radius(cm)
+g_max = V/(R_2*log(R_1/R_2))        // RMS value of maximum stress in the insulation(kV/cm)
+g_min = V/(R_1*log(R_1/R_2))        // RMS value of minimum stress in the insulation(kV/cm)
+
+// Results
+disp("PART II - EXAMPLE : 7.8 : SOLUTION :-")
+printf("\nMaximum stress in the insulation, g_max = %.2f kV/cm (rms)", g_max)
+printf("\nMinimum stress in the insulation, g_min = %.2f kV/cm (rms)", g_min)
diff --git a/3472/CH14/EX14.9/Example14_9.sce b/3472/CH14/EX14.9/Example14_9.sce
new file mode 100644
index 000000000..155fe19f0
--- /dev/null
+++ b/3472/CH14/EX14.9/Example14_9.sce
@@ -0,0 +1,37 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 7: UNDERGROUND CABLES
+
+// EXAMPLE : 7.9 :
+// Page number 215
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+d = 2.5             // Conductor diameter(cm)
+D = 6.0             // Sheath diameter(cm)
+V_l = 66.0          // Line Voltage(kV)
+
+// Calculations
+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*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)
+
+// Results
+disp("PART II - EXAMPLE : 7.9 : SOLUTION :-")
+printf("\nMaximum stress without intersheath = %.2f kV/cm", stress_max)
+printf("\nBest position of first intersheath, d_1 = %.2f cm", d_1)
+printf("\nBest position of second intersheath, d_2 = %.3f cm", d_2)
+printf("\nMaximum stress with intersheath = %.2f kV/cm", g_max)
+printf("\nVoltage on the first intersheath, V_1 = %.2f kV", V_1)
+printf("\nVoltage on the second intersheath, V_2 = %.2f kV \n", V_2)
+printf("\nNOTE: Changes in the obtained answer is due to more precision here")