14 files changed, 646 insertions, 0 deletions
diff --git a/1820/CH5/EX5.1/Example5_1.sce b/1820/CH5/EX5.1/Example5_1.sce
new file mode 100755
index 000000000..d628dec4b
--- /dev/null
+++ b/1820/CH5/EX5.1/Example5_1.sce
@@ -0,0 +1,32 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES
+
+// EXAMPLE : 5.1 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+d = 2 ; // Diameter  of conductor in cm
+D = 5 ; // Inside diameter of lead sheath in cm
+V = 24.9 ; // Line-to-neutral voltage in kV
+
+// CALCULATIONS
+// For case (a)
+r = d/2 ;
+R = D/2 ;
+E_max = V/( r * log(R/r) ) ; // Maximum electric stress in kV/cm
+E_min = V/( R * log(R/r) ) ; // Minimum electric stress in kV/cm
+
+// For case (b)
+r_1 = R/2.718 ; // Optimum conductor radius in cm . From equ 5.15
+E_max1 = V/( r_1 * log(R/r_1) ) ; // Min value of max stress in kV/cm
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 5.1 : SOLUTION :-") ;
+printf("\n (a) Maximum value of electric stress , E_max = %.2f kV/cm \n",E_max) ;
+printf("\n     Minimum value of electric stress , E_min = %.2f kV/cm \n",E_min) ;
+printf("\n (b) Optimum value of conductor radius , r = %.2f cm \n",r_1) ;
+printf("\n     Minimum value of maximum stress , E_max = %.2f kV/cm \n",E_max1) ;
diff --git a/1820/CH5/EX5.10/Example5_10.sce b/1820/CH5/EX5.10/Example5_10.sce
new file mode 100755
index 000000000..9cc60eb38
--- /dev/null
+++ b/1820/CH5/EX5.10/Example5_10.sce
@@ -0,0 +1,53 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES
+
+// EXAMPLE : 5.10 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+f= 60 ; // frequency in hertz
+t = 245 ; // insulation thickness in mils
+t_s = 95 ; // Lead/metal sheath thickness in mils
+d = 0.575 ; // diameter of conductor in inches
+r_s = 1.72 ; // sheath resistance in Ω/mi
+r_a = 0.263 ; // Conductor resistance in Ω/mi
+r = 100 ; // earth resistivity in Ω-mi
+D_s = 0.221 ; // GMR of one conductor in inches
+D_ab = 24 ; // distance b/w conductor a & b in inch . refer fig 5.30
+D_bc = 24 ; // distance b/w conductor b & c in inch . refer fig 5.30
+D_ca = 48 ; // distance b/w conductor c & a in inch . refer fig 5.30
+
+// CALCULATIONS
+T = t/1000 ; // insulation thickness in inch . [1 mils = 0.001 inch]
+T_s = t_s/1000 ; // Lead/metal sheath thickness in mils
+r_i = (d/2) + T ; // Inner radius of metal sheath in inches
+r_0 = r_i + T_s ; // Outer radius of metal sheath in inches
+r_e = 0.00476 * f ; //  AC resistance of earth return in Ω/mi
+D_e = 25920 * sqrt(r/f) ; // Equivalent depth of earth return path in inches
+D_eq = (D_ab*D_bc*D_ca)^(1/3) ; // Mean distance among conductor centers in inches
+Z_0a = (r_a + r_e) + (%i) * (0.36396) * log(D_e/((D_s*D_eq^2)^(1/3))) ;
+D_s_3s = (D_eq^2 * (r_0+r_i)/2)^(1/3) ; // GMR of conducting path composed of 3 sheaths in parallel in inches
+Z_0s = (r_s + r_e) + (%i) * 0.36396 * log (D_e/D_s_3s) ; // Zero sequence impedance of sheath in inches
+D_m_3c_3s = D_s_3s ; // Zero sequence mutual impedance b/w conductors & sheaths in inches
+Z_0m = r_e + (%i)*(0.36396)*log(D_e/D_m_3c_3s) ;
+
+// For case (a)
+Z_00 = Z_0a - (Z_0m^2/Z_0s) ; // Total zero sequence impedance when ground and return paths are present in Ω/mi/phase
+
+// For case (b)
+Z_0 = Z_0a + Z_0s - 2*Z_0m ; // Total zero sequence impedance when there is only sheath return path in Ω/mi/phase
+
+// For case (c)
+Z_01 = Z_0a ; // Total zero sequence impedance when there is only ground return path in Ω/mi/phase
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 5.10 : SOLUTION :-") ;
+printf("\n (a) Total zero sequence impedance when both ground & return paths are present , Z_00 = %.3f<%.1f Ω/mi/phase \n",abs(Z_00),atand(imag(Z_00),real(Z_00))) ;
+printf("\n (b) Total zero sequence impedance when there is only sheath return path , Z_0 = %.3f<%.1f Ω/mi/phase \n",abs(Z_0),atand(imag(Z_0),real(Z_0))) ;
+printf("\n (c) Total zero sequence impedance when there is only ground return path , Z_0a = %.4f<%.1f Ω/mi/phase \n",abs(Z_01),atand(imag(Z_01),real(Z_01))) ; 
+
+printf("\n NOTE : ERROR : There are mistakes in units in the Textbook \n") ;
diff --git a/1820/CH5/EX5.11/Example5_11.sce b/1820/CH5/EX5.11/Example5_11.sce
new file mode 100755
index 000000000..33abedbcf
--- /dev/null
+++ b/1820/CH5/EX5.11/Example5_11.sce
@@ -0,0 +1,40 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES
+
+// EXAMPLE : 5.11 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+f= 60 ; // frequency in hertz
+T = 0.175 ; // insulation thickness in inches
+d = 0.539 ; // diameter of conductor in inches
+G = 0.5 ; // Geometric factor from fig 5.3
+K = 3.7 ; // Dielectric constant
+V_LL = 13.8 ; // Line-to-line voltage in kV
+
+// CALCULATIONS
+D = d + 2 * T ; // Inside diameter of sheath in inches
+G = 2.303 * log10 (D/d) ; // Geometric factor for a single conductor
+sf = 0.710 ; // sector factor From Table 5.3 . For (T+t/d) obtained
+V_LN = V_LL/sqrt(3) ; // Line-to-neutral voltage in kV
+
+// For case (a)
+C_0 = 0.0892 * K/(G * sf) ; // shunt capacitances in μF/mi/phase . C_0 = C_1 = C_2 . From equ 5.161
+
+// For case (b)
+X_0 = 1.79 * G * sf/( f * K ) ; // shunt capacitive reactance in MΩ/mi/phase .X_0 = X_1 = X_2. From equ 5.162
+
+// For case (c)
+I_0 = 0.323 * f * K * V_LN/( 1000 * G * sf ) ; // Charging current in A/mi/phase .I_0 = I_1 = I_2. From equ 5.163
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 5.11 : SOLUTION :-") ;
+printf("\n (a) Shunt capacitances for zero , positive & negative sequences , C_0 = C_1 = C_2 = %.2f μF/mi/phase \n",C_0) ;
+printf("\n (b) Shunt capacitive reactance for zero , positive & negative sequences , X_0 = X_1 = X_2 = %.2e MΩ/mi/phase \n",X_0) ;
+printf("\n (c) Charging current for zero , positive & negative sequences , I_0 = I_1 = I_2 = %.3f A/mi/phase \n",I_0) ;
+
+printf("\n NOTE : 2.87e-03 MΩ/mi/phase can also be written as 2.87 kΩ/mi/phase as in textbook case (b) \n") ;
diff --git a/1820/CH5/EX5.12/Example5_12.sce b/1820/CH5/EX5.12/Example5_12.sce
new file mode 100755
index 000000000..d8103cae7
--- /dev/null
+++ b/1820/CH5/EX5.12/Example5_12.sce
@@ -0,0 +1,63 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES
+
+// EXAMPLE : 5.12 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+f= 60 ; // frequency in hertz
+r_a = 0.19 ; // Conductor resistance in Ω/mi
+l = 10 ; // length in mi
+D_s = 0.262 ; // GMR of one conductor in inches
+d = 18 ; // conductors spacing in inches
+
+// CALCULATIONS
+// For case (a)
+X_a = %i * 0.1213 *log (12/D_s) ; // reactance of individual phase conductor at 12 inch spacing in Ω/mi
+Z_aa = l * ( r_a + X_a ) ; // Z_aa = Z_bb = .... = Z_zz
+Z_bb = Z_aa ;
+Z_zz = Z_aa ;
+Z_cc = Z_aa ;
+D_eq1 = d * 2 ;
+Z_ab = (l) * ( %i * 0.1213 * log(12/D_eq1) ) ;
+Z_bc = Z_ab ;
+Z_xy = Z_ab ; // Z_xy = Z_yx
+Z_yz = Z_ab ;
+Z_ba = Z_ab ;
+Z_cb = Z_ab ;
+D_eq2 = d * 3 ;
+Z_bz = (l) * ( %i * 0.1213 * log(12/D_eq2) ) ;
+Z_ay = Z_bz ; // Z_ya = Z_ay
+Z_cx = Z_bz ; // Z_cx = Z_xc
+Z_yz = Z_bz ; // Z_zy = Z_yz
+D_eq3 = d * 4 ;
+Z_ac = (l) * ( %i * 0.1213 * log(12/D_eq3) ) ;
+Z_ca = Z_ac ; // Z_ac = Z_xz = Z_zx
+D_eq4 = d * 1 ;
+Z_ax = (l) * ( %i * 0.1213 * log(12/D_eq4) ) ;
+Z_bx = Z_ax ; // Z_ax = Z_xa ; Z_bx = Z_xb
+Z_by = Z_ax ; // Z_by = Z_yb
+Z_cy = Z_ax ; // Z_cy = Z_yc
+Z_cz = Z_ax ;
+D_eq5 = d * 5 ;
+Z_az = (l) * (%i*0.1213*log(12/D_eq5)) ; // Z_za= Z_az
+
+Z_s = [Z_aa Z_ab Z_ac ; Z_ba Z_bb Z_bc ; Z_ca Z_cb Z_cc] ;
+Z_tm = [Z_ax Z_bx Z_cx ; Z_ay Z_by Z_cy ; Z_az Z_bz Z_cz] ;
+Z_M = [Z_ax Z_ay Z_az ; Z_bx Z_by Z_bz ; Z_cx Z_cy Z_cz] ;
+Z_N = [Z_aa Z_xy Z_ac ; Z_xy Z_aa Z_ab ; Z_ac Z_ab Z_aa] ;
+Z_new = (Z_s)-(Z_M)*(Z_N)^(-1)*(Z_tm) ;
+
+// For case (b)
+a = 1*exp(%i*120*%pi/180) ; // By symmetrical components theory to 3-Φ system
+A = [1 1 1; 1 a^2 a ;1 a a^2] ;
+Z_012 = inv(A) * Z_new * A ; // Sequence-impedance matrix
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 5.12 : SOLUTION :-") ;
+printf("\n (a) Phase Impedance Matrix , [Z_abc] = \n") ; disp(Z_new) ;
+printf("\n (b) Sequence-Impedance Matrix , [Z_012] = \n") ; disp(Z_012) ;
diff --git a/1820/CH5/EX5.15/Example5_15.sce b/1820/CH5/EX5.15/Example5_15.sce
new file mode 100755
index 000000000..95fe702ba
--- /dev/null
+++ b/1820/CH5/EX5.15/Example5_15.sce
@@ -0,0 +1,63 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES
+
+// EXAMPLE : 5.15 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+L = 50 ; // length of transmission line in km
+P_l_oh = 820 ; // Power loss at peak load for overhead transmission line in kW/km
+P_l_g = 254 ; // Power loss at peak load for gas insulated transmission line in kW/km
+cost_kwh = 0.10 // cost of electric energy in $ per kWh
+lf_ann = 0.7 ; // Annual load factor
+plf_ann = 0.7 ; // Annual Power loss factor
+h_yr = 365*24 ; // Time in Hours for a year
+total_invest = 200000000 ; // Investment cost of GIL in $ ( for case (j) )
+
+// CALCULATIONS
+// For case (a)
+Power_loss_OHline = P_l_oh * L ; // Power loss of overhead line at peak load in kW
+
+// For case (b)
+Power_loss_GILline = P_l_g * L ; // Power loss of gas-insulated transmission line at peak load in kW
+
+// For case (c)
+energy_loss_OH = Power_loss_OHline * h_yr ; // Total annual energy loss of OH line at peak load in kWh/yr
+
+// For case (d)
+energy_loss_GIL = Power_loss_GILline * h_yr ; // Total annual energy loss of GIL at peak load in kWh/yr
+
+// For case (e)
+energy_ann_OH = lf_ann * energy_loss_OH ; // Average energy loss of OH line at peak load in kWh/yr
+
+// For case (f)
+energy_ann_GIL = lf_ann * energy_loss_GIL ; // Average energy loss of GIL line at peak load in kWh/yr
+
+// For case (g)
+cost_ann_OH = cost_kwh * energy_ann_OH ; // Average annual cost of losses of OH line in $ per year
+
+// For case (h)
+cost_ann_GIL = cost_kwh * energy_ann_GIL ; // Average annual cost of losses of GIL line in $ per year
+
+// For case (i)
+P_loss_ann = cost_ann_OH - cost_ann_GIL ; // Annual resultant savings of losses per yr
+
+// For case (j)
+break_period = total_invest/P_loss_ann ; // Payback period if GIL alternative period is selected
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 5.15 : SOLUTION :-") ;
+printf("\n (a) Power loss of Overhead line at peak load , (Power loss)_OH_line = %d kW \n",Power_loss_OHline) ;
+printf("\n (b) Power loss of Gas-insulated transmission line , (Power loss)_GIL_line = %d kW \n",Power_loss_GILline) ;
+printf("\n (c) Total annual energy loss of Overhead transmission line at peak load = %.4e kWh/yr \n",energy_loss_OH) ;
+printf("\n (d) Total annual energy loss of Gas-insulated transmission line at peak load = %.5e kWh/yr \n",energy_loss_GIL);
+printf("\n (e) Average energy loss of Overhead transmission line = %.5e kWh/yr \n",energy_ann_OH);
+printf("\n (f) Average energy loss of Gas-insulated transmission line at peak load = %.5e kWh/yr \n",energy_ann_GIL);
+printf("\n (g) Average annual cost of losses of Overhead transmission line  = $ %.5e/yr \n",cost_ann_OH);
+printf("\n (h) Average annual cost of losses of Gas-insulated transmission line = $ %.5e/yr \n",cost_ann_GIL);
+printf("\n (i) Annual resultant savings in losses using Gas-insulated transmission line = $ %.6e/yr \n",P_loss_ann);
+printf("\n (j) Breakeven period when GIL alternative is selected = %.1f years \n",break_period);
diff --git a/1820/CH5/EX5.16/Example5_16.sce b/1820/CH5/EX5.16/Example5_16.sce
new file mode 100755
index 000000000..1101e8681
--- /dev/null
+++ b/1820/CH5/EX5.16/Example5_16.sce
@@ -0,0 +1,80 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES
+
+// EXAMPLE : 5.16 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+n = 40 ; // useful life in years
+i = 10/100 ; // carrying charge rate
+A_P = (i*(1+i)^n)/((1 + i)^n - 1) ; // Refer page 642
+A_F = 0.00226 ; // A_F = A/F
+pr_tax = 3/100 ; // Annual ad property taxes is 3% of 1st costs of each alternative
+
+// FOR OVERHEAD TRANSMISSION
+L_OH = 50 ; // length of route A in mi
+cost_b_A = 1 * 10^6 ; // cost per mile to bulid in $
+salvage_A = 2000 ; // salvage value per mile at end of 40 years
+cost_mait_OH = 500 ; // cost in $ per mile to maintain
+
+// SUBMARINE TRANSMISSION LINE
+L_S = 30 ; // length of route B in mi
+cost_b_B = 4*10^6 ; // cost per mile to bulid in $
+salvage_B = 6000 ; // salvage value per mile at end of 40 years
+cost_mait_S = 1500 ; // cost in $ per mile to maintain
+
+// GIL TRANSMISSION
+L_GIL = 20 ; // length of route C in mi
+cost_b_C = 7.6*10^6 ; // cost per mile to bulid in $
+salvage_C = 1000 ; // salvage value per mile at end of 40 years
+cost_mait_GIL = 200 ; // cost in $ per mile to maintain
+savings = 17.5*10^6 ; // relative savings in power loss per year in $
+
+
+// CALCULATIONS
+n = 25 ; // useful life in years
+i = 20/100 ; // carrying charge rate
+p = ((1 + i)^n - 1)/(i*(1+i)^n) ; // p = P/A
+// FOR OVERHEAD TRANSMISSION
+P_OH = cost_b_A * L_OH ; // first cost of 500 kV OH line in $
+F_OH = salvage_A * L_OH ; // Estimated salvage value in $
+A_1 = P_OH * A_P - F_OH * A_F ; // Annual equivalent cost of capital in $
+A_2 = P_OH * pr_tax + cost_mait_OH * L_OH ; // annual equivalent cost of tax and maintainance in $
+A = A_1 + A_2 ; // total annual equi cost of OH line in $
+
+// SUBMARINE TRANSMISSION LINE
+P_S = cost_b_B * L_S ; // first cost of 500 kV OH line in $
+F_S = salvage_B * L_S ; // Estimated salvage value in $
+B_1 = P_S * A_P - F_S * A_F ; // Annual equivalent cost of capital in $
+B_2 = P_S * pr_tax + cost_mait_S * L_S ; // annual equivalent cost of tax and maintainance in $
+B = B_1 + B_2 ; // total annual equi cost of OH line in $
+
+// GIL TRANSMISSION
+P_GIL = cost_b_C * L_GIL ; // first cost of 500 kV OH line in $
+F_GIL = salvage_C * L_GIL ; // Estimated salvage value in $
+C_1 = P_GIL * A_P - F_GIL * A_F ; // Annual equivalent cost of capital in $
+C_2 = P_GIL * pr_tax + cost_mait_GIL * L_GIL ; // annual equivalent cost of tax and maintainance in $
+C = C_1 + C_2 ; // total annual equi cost of OH line in $
+A_net = C - savings ; // Total net annual equi cost of GIL
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 5.16 : SOLUTION :-") ;
+printf("\n OVERHEAD TRANSMISSION LINE : \n") ;
+printf("\n  Annual equivalent cost of capital invested in line , A_1 = $ %d \n",A_1) ;
+printf("\n  Annual equivalent cost of Tax and maintainance , A_2 = $ %d \n",A_2) ;
+printf("\n  Total annual equivalent cost of OH transmission , A = $ %d \n",A) ;
+printf("\n \n SUBMARINE TRANSMISSION LINE : \n") ;
+printf("\n  Annual equivalent cost of capital invested in line , A_1 = $ %d \n",B_1) ;
+printf("\n  Annual equivalent cost of Tax and maintainance , A_2 = $ %d \n",B_2) ;
+printf("\n  Total annual equivalent cost of Submarine power transmission , A = $ %d \n",B) ;
+printf("\n \n GIL TRANSMISSION LINE : \n") ;
+printf("\n  Annual equivalent cost of capital invested in line , A_1 = $ %d \n",C_1) ;
+printf("\n  Annual equivalent cost of Tax and maintainance , A_2 = $ %d \n",C_2) ;
+printf("\n  Total annual equivalent cost of Submarine power transmission , A = $ %d \n",C) ;
+printf("\n  Total net equivalent cost of GIL transmission = $ %d \n",A_net) ;
+printf("\n \n The result shows use of GIL is the best choice \n") ;
+printf("\n The next best alternative is Overhead transmission line \n") ;
diff --git a/1820/CH5/EX5.2/Example5_2.sce b/1820/CH5/EX5.2/Example5_2.sce
new file mode 100755
index 000000000..6a8cab487
--- /dev/null
+++ b/1820/CH5/EX5.2/Example5_2.sce
@@ -0,0 +1,36 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES
+
+// EXAMPLE : 5.2 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+r = 1 ; // Radius of conductor in cm
+t_1 = 2 ; // Thickness of insulation layer in cm
+r_1 = r + t_1 ;
+r_2 = 2 ; // Thickness of insulation layer in cm . r_2 = t_1 = t_2 
+R = r_1 + r_2 ;
+K_1 = 4 ; // Inner layer Dielectric constant 
+K_2 = 3 ; // Outer layer Dielectric constant 
+kv = 19.94 ; // potential difference b/w inner & outer lead sheath in kV
+
+// CALCULATIONS
+// E_1 = 2q/(r*K_1) & E_2 = 2q/(r_1*K_2) . Let E = E_1/E_2
+E = ( r_1 * K_2 )/( r * K_1 ) ; // E = E_1/E_2
+V_1 = poly(0,'V_1') ; // defining unknown V_1
+E_1 = V_1/( r * log(r_1/r) ) ;
+V_2 = poly(0,'V_2') ; // defining unknown V_2
+V_2 = kv - (V_1) ;
+E_2 = V_2/( r_1 * log(R/r_1) ) ;
+E_3 = E_1/E_2 ;
+// Equating E = E_3 . we get the value of V_1
+V_1 = 12.30891068 ; // Voltage in kV
+E_1s = V_1/( r * log(r_1/r) ) ; // Potential gradient at surface of conductor in kV/cm . E_1 = E_1s
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 5.2 : SOLUTION :-") ;
+printf("\n Potential gradient at the surface of conductor , E_1 = %.2f kV/cm \n",E_1s) ;
diff --git a/1820/CH5/EX5.3/Example5_3.sce b/1820/CH5/EX5.3/Example5_3.sce
new file mode 100755
index 000000000..4925188cc
--- /dev/null
+++ b/1820/CH5/EX5.3/Example5_3.sce
@@ -0,0 +1,31 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES
+
+// EXAMPLE : 5.3 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+D = 1.235 ; // Inside diameter of sheath in inch
+d = 0.575 ; // Conductor diameter in inch
+kv = 115 ; // Voltage in kV
+l = 6000 ; // Length of cable in feet
+r_si = 2000 ; // specific insulation resistance is 2000 MΩ/1000ft . From Table 5.2
+
+// CALCULATIONS
+// For case (a)
+r_si0 = r_si * l/1000 ;
+R_i = r_si0 * log10 (D/d) ; // Total Insulation resistance in MΩ
+
+// For case (b)
+P = kv^2/R_i ; // Power loss due to leakage current in W
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 5.3 : SOLUTION :-") ;
+printf("\n (a) Total insulation resistance at 60 degree F , R_i= %.2f MΩ \n",R_i) ;
+printf("\n (b) Power loss due to leakage current , V^2/R_i = %.4f W \n",P) ;
+
+printf("\n NOTE : ERROR : Mistake in textbook case (a) \n") ;
diff --git a/1820/CH5/EX5.4/Example5_4.sce b/1820/CH5/EX5.4/Example5_4.sce
new file mode 100755
index 000000000..db5012266
--- /dev/null
+++ b/1820/CH5/EX5.4/Example5_4.sce
@@ -0,0 +1,25 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES
+
+// EXAMPLE : 5.4 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+C_a = 2 * 10^-6 ; // Capacitance b/w two conductors in F/mi
+l = 2 ; // length in mi
+f = 60 ; // Frequency in Hz
+V_L_L = 34.5 * 10^3 ; // Line-to-line voltage in V
+
+// CALCULATIONS
+C_a1 = C_a * l ; // Capacitance for total cable length in F
+C_N = 2 * C_a1 ; // capacitance of each conductor to neutral in F . From equ 5.56
+V_L_N = V_L_L/sqrt(3) ; // Line-to-neutral voltage in V
+I_c = 2 * %pi * f * C_N * (V_L_N) ; // Charging current of cable in A
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 5.4 : SOLUTION :-") ;
+printf("\n Charging current of the cable , I_c = %.2f A \n",I_c) ;
diff --git a/1820/CH5/EX5.5/Example5_5.sce b/1820/CH5/EX5.5/Example5_5.sce
new file mode 100755
index 000000000..eb55c260b
--- /dev/null
+++ b/1820/CH5/EX5.5/Example5_5.sce
@@ -0,0 +1,40 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES
+
+// EXAMPLE : 5.5 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+C_a = 0.45 * 10^-6 ; // Capacitance b/w two conductors in F/mi
+l = 4 ; // length of cable in mi
+f = 60 ; // Freq in Hz
+V_L_L = 13.8 * 10^3 ; // Line-to-line voltage in V
+pf = 0.85 ; // lagging power factor
+I = 30 ; // Current drawn by load at receiving end in A
+
+// CALCULATIONS
+// For case (a)
+C_a1 = C_a * l ; // Capacitance for total cable length in F
+C_N = 2 * C_a1 ; // capacitance of each conductor to neutral in F
+V_L_N = V_L_L/sqrt(3) ; // Line-to-neutral voltage in V
+I_c = 2 * %pi * f * C_N * (V_L_N) ; // Charging current in A
+I_c1 = %i * I_c ; // polar form of Charging current in A
+
+// For case (b)
+phi_r = acosd(pf) ; // pf angle
+I_r = I * ( cosd(phi_r) - sind(phi_r) * %i ) ; // Receiving end current in A
+I_s = I_r + I_c1 ; // sending end current in A
+
+// For case (c)
+pf_s = cosd( atand( imag(I_s),real(I_s) ) ) ; // Lagging pf of sending-end
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 5.5 : SOLUTION :-") ;
+printf("\n (a) Charging current of feeder , I_c = %.2f A \n",I_c) ;
+printf("\n     Charging current of feeder in complex form  , I_c = i*%.2f A \n",imag(I_c1)) ;
+printf("\n (b) Sending-end current , I_s = %.2f<%.2f A\n",abs(I_s),atand( imag(I_s),real(I_s) )) ;
+printf("\n (c) Sending-end power factor ,cos Φ_s = %.2f  Lagging power factor \n",pf_s) ;
diff --git a/1820/CH5/EX5.6/Example5_6.sce b/1820/CH5/EX5.6/Example5_6.sce
new file mode 100755
index 000000000..87b052a06
--- /dev/null
+++ b/1820/CH5/EX5.6/Example5_6.sce
@@ -0,0 +1,33 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES
+
+// EXAMPLE : 5.6 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+f = 60 ; // Freq in Hz
+V_L_L = 138 ; // Line-to-line voltage in kV
+T = 11/64 ; // Thickness of conductor insulation in inches
+t = 5/64 ; // Thickness of belt insulation in inches
+d =  0.575 ; // Outside diameter of conductor in inches
+
+// CALCULATIONS
+// For case (a)
+T_1 = (T + t)/d ; // To find the value of geometric factor G for a single-conductor cable
+G_1 = 2.09 ; // From table 5.3 , by interpolation
+sf = 0.7858 ; // sector factor obtained for T_1 from table 5.3
+G = G_1 * sf ; // real geometric factor
+
+// For case (b)
+V_L_N = V_L_L/sqrt(3) ; // Line-to-neutral voltage in V
+K = 3.3 ; // Dielectric constant of insulation for impregnated paper cable
+I_c = 3 * 0.106 * f * K * V_L_N/(1000 * G) ; // Charging current in A/1000ft
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 5.6 : SOLUTION :-") ;
+printf("\n (a) Geometric factor of cable using table 5.3 , G_1 = %.3f \n",G) ;
+printf("\n (b) Charging current , I_c = %.3f A/1000ft \n",I_c) ;
diff --git a/1820/CH5/EX5.7/Example5_7.sce b/1820/CH5/EX5.7/Example5_7.sce
new file mode 100755
index 000000000..42fd374bd
--- /dev/null
+++ b/1820/CH5/EX5.7/Example5_7.sce
@@ -0,0 +1,57 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES
+
+// EXAMPLE : 5.7 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+V_L_N = 7.2 ; // Line-to-neutral voltage in kV
+d = 0.814 ; // Conductor diameter in inches
+D = 2.442 ; // inside diameter of sheath in inches
+K = 3.5 ; // Dielectric constant
+pf = 0.03 ; // power factor of dielectric
+l = 3.5 ; // length in mi
+f = 60 ; // Freq in Hz
+u = 1.3 * 10^7 ; // dielectric resistivity of insulation in MΩ-cm
+
+// CALCULATIONS
+// For case (a)
+r = d * 2.54/2 ; // conductor radius in cm . [1 inch = 2.54 cm]
+R = D * 2.54/2 ; // Inside radius of sheath in cm
+E_max = V_L_N/( r * log(R/r) ) ; // max electric stress in kV/cm
+
+// For case (b)
+C = 0.0388 * K/( log10 (R/r) ) ; // capacitance of cable in μF/mi . From equ 5.29 
+C_1 = C * l ; // capacitance of cable for total length in μF
+
+// For case (c)
+V_L_N1 = 7.2 * 10^3 ; // Line-to-neutral voltage in V
+C_2 = C_1 * 10^-6 ; // capacitance of cable for total length in F
+I_c = 2 * %pi * f * C_2 * (V_L_N1) ; // Charging current in A
+
+// For case (d)
+l_1 = l * 5280 * 12 * 2.54 ; // length in cm . [1 mi = 5280 feet] ; [1 feet = 12 inch] 
+R_i = u * log(R/r)/( 2 * %pi * l_1) ; // Insulation resistance in MΩ
+
+// For case (e)
+P_lc = V_L_N^2/R_i ; // power loss in W
+
+// For case (f)
+P_dl = 2 * %pi * f * C_1 * V_L_N^2 * pf ; // Total dielectric loss in W
+
+// For case (g)
+P_dh = P_dl - P_lc ; // dielectric hysteresis loss in W
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 5.7 : SOLUTION :-") ;
+printf("\n (a) Maximum electric stress occuring in cable dielectric , E_max = %.2f kV/cm \n",E_max) ;
+printf("\n (b) Capacitance of cable , C = %.4f μF \n",C_1) ;
+printf("\n (c) Charging current of cable , I_c = %.3f A \n",I_c) ;
+printf("\n (d) Insulation resistance , R_i = %.2f MΩ \n",R_i) ;
+printf("\n (e) Power loss due to leakage current , P_lc = %.2f W \n",P_lc) ;
+printf("\n (f) Total dielectric loss , P_dl = %.2f W \n",P_dl) ;
+printf("\n (g) Dielectric hysteresis loss , P_dh = %.2f W \n",P_dh) ;
diff --git a/1820/CH5/EX5.8/Example5_8.sce b/1820/CH5/EX5.8/Example5_8.sce
new file mode 100755
index 000000000..4629ff7ba
--- /dev/null
+++ b/1820/CH5/EX5.8/Example5_8.sce
@@ -0,0 +1,30 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES
+
+// EXAMPLE : 5.8 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+l = 3 ; // underground cable length in mi
+f = 60 ; // frequency in hertz
+
+// CALCULATIONS
+// For case (a)
+R_dc = 0.00539 ; // dc resistance of cable in Ω/1000ft , From table 5.5 
+R_dc1 = (R_dc/1000) * 5280 * 3 ; // Total dc resistance in Ω . [1 mi = 5280 feet]
+
+// For case (b)
+s_e = 1.233 ; // skin effect coefficient
+R_eff = s_e * R_dc1 ; // Effective resistance in Ω
+percentage  = ( (R_eff - R_dc1)/(R_dc1) ) * 100 ; // skin effect on effective resistance in %
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 5.8 : SOLUTION :-") ; 
+printf("\n (a) Total dc resistance of the conductor , R_dc = %.4f Ω \n",R_dc1) ;
+printf("\n (b) Effective resistance at 60 hz , R_eff = %.4f Ω \n",R_eff) ;
+printf("\n     Skin effect on the Effective resistance in percent at 60 hz , R_eff = %.1f percent greater than for direct current\n",percentage) ;
+printf("\n (c) Percentage of reduction in cable ampacity in part (b) = %.1f percent \n",percentage) ;
diff --git a/1820/CH5/EX5.9/Example5_9.sce b/1820/CH5/EX5.9/Example5_9.sce
new file mode 100755
index 000000000..733e774f8
--- /dev/null
+++ b/1820/CH5/EX5.9/Example5_9.sce
@@ -0,0 +1,63 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 5 : UNDERGROUND POWER TRANSMISSION AND GAS-INSULATED TRANSMISSION LINES
+
+// EXAMPLE : 5.9 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+kV = 35 ; // voltage in kV
+f = 60 ; // operating frequency of cable in hertz
+d = 0.681 ; // diameter of conductor in inches
+t_i = 345 ; // Insulation thickness in cmil
+t_s = 105 ; // Metal sheet thickness in cmil
+r_c = 0.190 ; // Conductor ac resistance in Ω/mi
+l = 10 ; // Length of cable in mi
+
+// CALCULATIONS
+// For case (a)
+T_i = t_i/1000 ; // insulation thickness in inch
+T_s = t_s/1000 ; // Metal sheet thickness in inch
+r_i = (d/2) + T_i ; // Inner radius of metal sheath in inches
+r_0 = r_i + T_s ; // Outer radius of metal sheath in inches
+S = r_i + r_0 + T_s ; // Spacing b/w conductor centers in inches
+X_m = 0.2794 * (f/60) * log10 ( 2*S/(r_0 + r_i) ) ; // Mutual reactance b/w conductor & sheath per phase in Ω/mi . From Equ 5.78
+X_m1 = X_m * l ; // Mutual reactance b/w conductor & sheath in Ω/phase
+
+// For case (b)
+r_s = 0.2/((r_0+r_i)*(r_0-r_i)) ; // sheet resistance per phase in Ω/mi/phase . From equ 5.79
+r_s1 = r_s * l ; // sheet resistance per phase in Ω/phase
+
+// For case (c)
+d_r = r_s * (X_m^2)/( (r_s)^2 + (X_m)^2 ) ; // increase in conductor resistance due to sheath current in Ω/mi/phase . From equ 5.77
+d_r1 = d_r * l ; //  // increase in conductor resistance due to sheath current in Ω/phase
+
+// For case (d)
+r_a = r_c + ( r_s * X_m^2 )/( (r_s)^2 + (X_m)^2 ) ; // Total positive or negative sequence resistance including sheath current effects in Ω/mi/phase . From equ 5.84
+r_a1 = r_a * l ; // Total positive or negative sequence resistance including sheath current effects in Ω/phase
+
+// For case (e)
+ratio = d_r/r_c ; // ratio = sheath loss/conductor loss
+
+// For case (f)
+I = 400 ; // conductor current in A ( given for case (f) )
+P_s = 3 * (I^2) * ( r_s * X_m^2)/( r_s^2 + X_m^2 ) ; // For three phase loss in W/mi
+P_s1 = P_s * l ; // Total sheath loss of feeder in Watts
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 5.9 : SOLUTION :-") ;
+printf("\n (a) Mutual reactance b/w conductors & sheath , X_m = %.5f Ω/mi/phase \n",X_m) ;
+printf("\n    or Mutual reactance b/w conductors & sheath , X_m = %.4f Ω/phase \n",X_m1) ;
+printf("\n (b) Sheath resistance of cable , r_s = %.4f Ω/mi/phase \n",r_s) ;
+printf("\n    or Sheath resistance of cable , r_s = %.3f Ω/phase \n",r_s1) ;
+printf("\n (c) Increase in conductor resistance due to sheath currents , Δr = %.5f Ω/mi/phase \n",d_r) ;
+printf("\n    or Increase in conductor resistance due to sheath currents , Δr = %.4f Ω/phase \n",d_r1) ;
+printf("\n (d) Total resistance of conductor including sheath loss , r_a = %.5f Ω/mi/phase \n ",r_a) ;
+printf("\n    or Total resistance of conductor including sheath loss , r_a = %.4f Ω/phase \n ",r_a1) ;
+printf("\n (e) Ratio of sheath loss to conductor loss , Ratio = %.4f \n",ratio) ;
+printf("\n (f) Total sheath loss of feeder if current in conductor is 400A , P_s = %.2f W \n",P_s1) ;
+
+printf("\n NOTE : ERROR : There are mistakes in some units in the Textbook \n") ;