80 files changed, 3865 insertions, 0 deletions
diff --git a/1820/CH10/EX10.1/Example10_1.sce b/1820/CH10/EX10.1/Example10_1.sce
new file mode 100755
index 000000000..52fc514ff
--- /dev/null
+++ b/1820/CH10/EX10.1/Example10_1.sce
@@ -0,0 +1,25 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 10 : PROTECTIVE EQUIPMENT AND TRANSMISSION SYSTEM PROTECTION
+
+// EXAMPLE : 10.1 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+X_d = 0.14*%i ; // Reactance of generator in pu
+E_g = 1*exp(%i*0*%pi/180) ;
+S_B = 25*10^3 ; // voltage in kVA
+V_BL_V = 13.8 ; // low voltage in kV
+
+// CALCULATIONS
+I_f = E_g/X_d ; // Subtransient fault current in pu
+I_BL_V = S_B/( sqrt(3)*V_BL_V) ; // Current base for low-voltage side
+I_f1 = abs(I_f)*I_BL_V ; // magnitude of fault current in A
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 10.1 : SOLUTION :-") ;
+printf("\n Subtransient fault current for 3-Φ fault in per units = pu \n") ; disp(I_f) ;
+printf("\n Subtransient fault current for 3-Φ fault in ampere = %.f A \n",I_f1) ;
diff --git a/1820/CH10/EX10.2/Example10_2.sce b/1820/CH10/EX10.2/Example10_2.sce
new file mode 100755
index 000000000..3438dc066
--- /dev/null
+++ b/1820/CH10/EX10.2/Example10_2.sce
@@ -0,0 +1,42 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 10 : PROTECTIVE EQUIPMENT AND TRANSMISSION SYSTEM PROTECTION
+
+// EXAMPLE : 10.2 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+// For case (a)
+I_f = 7.1428571 ; // Subtransient fault current in pu . Result of exa 10.1
+
+// For case (d)
+V_pf = 13800 ; // voltage in V
+zeta = 1.4 ;
+I_f1 = 7471 ; // magnitude of fault current in A
+
+// CALCULATIONS
+// For case (a)
+I_fdc_max = sqrt(2)*I_f ; // Max dc current in pu
+
+// For case (b)
+I_f_max = 2*I_fdc_max ; // Total max instantaneous current in pu
+
+// For case (c)
+I_momt = 1.6*I_f ; // Total rms momentary current
+
+// For case (d)
+S_int = sqrt(3)*(V_pf)*I_f1*zeta*10^-6 ; // Interrupting rating in MVA
+
+// For case (e)
+S_momt = sqrt(3)*(V_pf)*I_f1*1.6*10^-6 ; // Momentary duty of CB in MVA
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 10.2 : SOLUTION :-") ;
+printf("\n (a) Maximum possible dc current component , I_fdc_max = %.1f pu \n",I_fdc_max) ;
+printf("\n (b) Total maximum instantaneous current , I_max = %.1f pu \n",I_f_max) ;
+printf("\n (c) Momentary current , I_momentary = %.2f pu \n",I_momt) ;
+printf("\n (d) Interrupting rating of a 2-cycle CB , S_interrupting = %.f MVA \n",S_int) ;
+printf("\n (e) Momentary duty of a 2-cycle CB , S_momentary = %.2f MVA \n",S_momt) ;
diff --git a/1820/CH10/EX10.4/Example10_4.sce b/1820/CH10/EX10.4/Example10_4.sce
new file mode 100755
index 000000000..2e166e257
--- /dev/null
+++ b/1820/CH10/EX10.4/Example10_4.sce
@@ -0,0 +1,41 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 10 : PROTECTIVE EQUIPMENT AND TRANSMISSION SYSTEM PROTECTION
+
+// EXAMPLE : 10.4 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+z_l = 0.2 + %i * 0.7 ; // Line impedance in pu
+f_l = 0.7 ; // Fault point at a distance from A in pu
+f_m = 1.2 ; // magnitude of fault current in pu
+l = 10.3 ; // Line spacing in ft
+p = 100 ; // Power in MVA
+v = 138 ; // voltage in kV
+i = 418.4 ; // current in A
+z = 190.4 ; // Impedance in Ω
+
+// CALCULATIONS
+// For case (a)
+I = f_m * i ; // Current in arc in A
+R_arc = 8750 * l/(I^1.4) ; // Arc resistance in Ω
+R_arc1 = R_arc/z ; // Arc resistance in pu
+
+// For case (b)
+Z_L = z_l * f_l ;
+Z_r = Z_L + R_arc1 ; // Impedance seen by the relay in pu
+
+// For case (c)
+phi_1 = atand( imag(Z_L),real(Z_L) ) ; // Line impedance angle without arc resistance in degree
+phi_2 = atand( imag(Z_r),real(Z_r) ) ; // Line impedance angle with arc resistance in degree
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 10.4 : SOLUTION :-") ;
+printf("\n (a) Value of arc resistance at fault point in Ω , R_arc = %.2f Ω \n",R_arc) ;
+printf("\n     Value of arc resistance at fault point in pu , R_arc = %.2f pu \n",R_arc1) ;
+printf("\n (b) Value of line impedance including the arc resistance , Z_L + R_arc = pu \n") ; disp(Z_r) ;
+printf("\n (c) Line impedance angle without arc resistance , Φ = %.2f degree \n",phi_1) ;
+printf("\n     Line impedance angle with arc resistance , Φ = %.2f degree \n",phi_2) ;
diff --git a/1820/CH10/EX10.5/Example10_5.sce b/1820/CH10/EX10.5/Example10_5.sce
new file mode 100755
index 000000000..a9361d6d5
--- /dev/null
+++ b/1820/CH10/EX10.5/Example10_5.sce
@@ -0,0 +1,104 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 10 : PROTECTIVE EQUIPMENT AND TRANSMISSION SYSTEM PROTECTION
+
+// EXAMPLE : 10.5 :
+clear ; clc ; close ; // Clear the work space and console
+
+// CALCULATIONS
+// For case (a)
+// Coordinate Values taken here are only for reference . Refer exa 10.5
+
+T = 0:0.01:300 ;
+
+for i = 1:int(length(T)/1.1) ; 
+    po(i) = 4 ;
+end
+for i = int(length(T)/1.1):length(T)
+    po(i) = 5 ;
+end
+for i = 1:int(length(T)/1.1)
+    io(i) = 4 ;
+    end
+for i = int(length(T)/1.1):length(T)
+    io(i) = 3 ;
+end
+
+a= gca() ;
+subplot(2,1,1) ; // To plot 2 graph in same graphic window
+a.thickness = 2 ; // sets thickness of plot of points
+plot2d(T,po,3,'012','',[0 0 310 7]) ;
+plot2d(T,io,3,'012','',[0 0 310 7]) ;
+xtitle("Fig 10.5 (a)   Zones of protection for relay R_12") ;
+xset('thickness',2); // sets thickness of axes
+xstring(25,3.8,'[]') ;
+xstring(45,4.2,'(1)') ;
+plot(45,4,'+') ;
+xstring(60,3.8,'[]') ;
+xstring(60,4.2,'B_12') ;
+xstring(120,3.8,'[]') ;
+xstring(120,4.2,'B_21') ;
+xstring(140,4.2,'(2)') ;
+plot(140,4,'+') ;
+xstring(155,3.8,'[]') ;
+xstring(155,4.2,'B_23') ;
+xstring(220,3.8,'[]') ;
+xstring(220,4.2,'B_32') ;
+xstring(270,5.0,'(3)') ;
+xstring(285,2.8,'[]') ;
+xstring(285,3.2,'B_35') ;
+xstring(285,4.8,'[]') ;
+xstring(285,5.2,'B_34') ;
+xstring(85,3.4,'TL_12') ;
+xstring(180,3.4,'TL_23') ;
+xstring(60,3,'ZONE 1') ;
+xstring(100,2,'ZONE 2') ;
+xstring(190,1,'ZONE 3') ;
+
+// For case (b)
+
+for i = 1:int(length(T)/4) ;
+    vo(i) = 0.5;
+end
+for i = int(length(T)/4):length(T/1.7)
+    vo(i) = 2;
+end
+for i = int(length(T)/1.7):length(T)
+    vo(i) = 4
+end
+
+for i = int(length(T)/2.14):length(T/1.35) ; // plotting Voltage values
+    uo(i) = 0.5;
+end
+for i = int(length(T)/1.35):length(T)
+    uo(i) = 2;
+end
+
+a = gca() ;
+a.thickness = 2 ;
+subplot(2,1,2)
+plot2d(T,vo,2,'012','',[0 0 310 7]) ;
+plot2d(T,uo,2,'012','',[0 0 310 7]) ;
+ylabel("OPERATING TIME") ;
+xlabel("IMPEDANCE") ;
+xtitle("Fig 10.5 (b) Coordination of distance relays , Operating time v/s Impedance") ;
+xset('thickness',2); // sets thickness of axes
+xstring(0.1,0.3,'T_1') ;
+xstring(30,0.6,'R_12') ;
+xstring(58,1.3,'T_2') ;
+xstring(100,2.0,'R_12') ;
+xstring(160,3.0,'T_3') ;
+xstring(230,4.0,'R_12') ;
+xstring(160,0.6,'R_23') ;
+xstring(260,2.1,'R_23') ;
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 10.5 : SOLUTION :-") ;
+printf("\n (a) The zone of protection for relay R_12 is shown in Fig 10.5 (a) \n") ;
+printf("\n  ZONE 1 lies b/w (1) & B_21 \n") ;
+printf("\n  ZONE 2 lies b/w (1) & TL_23 \n") ;
+printf("\n  ZONE 3 lies after (1) \n") ;
+printf("\n (b) The coordination of the distance relays R_12 & R_21 in terms of Operating time v/s Impedance is shown in Fig 10.5 (b)") ;
diff --git a/1820/CH10/EX10.6/Example10_6.sce b/1820/CH10/EX10.6/Example10_6.sce
new file mode 100755
index 000000000..07e13e031
--- /dev/null
+++ b/1820/CH10/EX10.6/Example10_6.sce
@@ -0,0 +1,55 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 10 : PROTECTIVE EQUIPMENT AND TRANSMISSION SYSTEM PROTECTION
+
+// EXAMPLE : 10.6 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+kv = 230 * 10^3 ; // transmission system voltage in V
+VA = 100 * 10^6 ; // Maximum peak load supplied by TL_12 in VA
+ZTL_12 = 2 + %i * 20 ; // Positive-sequence impedances of line TL_12
+ZTL_23 = 2.5 + %i * 25 ; // Positive-sequence impedances of line TL_23
+pf = 0.9 ; // Lagging pf
+
+// CALCULATIONS 
+// For case (a)
+I_max = VA/(sqrt(3)*kv) ; // Maximum load current in A
+
+// For case (b)
+CT = 250/5 ; // CT ratio which gives about 5A in secondary winding under the maximum loading
+
+// For case (c)
+vr = 69 ; // selecting Secondary voltage of 69 V line to neutral
+VT = (kv/sqrt(3))/vr ; // Voltage ratio
+
+// For case (d)
+Z_r = CT/VT ; // impedance measured by relay . Z_r = (V/VT)/(I/CT)
+Z_TL_12 = Z_r * ZTL_12 ; // Impedance of lines TL_12 as seen by relay
+Z_TL_23 = Z_r * ZTL_23 ; // Impedance of lines TL_23 as seen by relay
+
+// For case (e)
+Z_load = vr * CT * (pf + %i*sind(acosd(pf)))/(I_max) ; // Load impedance based on secondary ohms
+
+// For case (f)
+Z_r1 = 0.80 * Z_TL_12 ; // Zone 1 setting of relay R_12
+
+// For case (g)
+Z_r2 = 1.20 * Z_TL_12 ; // Zone 2 setting of relay R_12
+
+// For case (h)
+Z_r3 = Z_TL_12 + 1.20*(Z_TL_23) ; // Zone 3 setting of relay R_12
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 10.6 : SOLUTION :-") ;
+printf("\n (a) Maximum load current , I_max = %.2f A \n",I_max) ;
+printf("\n (b) CT ratio , CT = %.1f \n",CT) ;
+printf("\n (c) VT ratio , VT = %.1f \n",VT) ;
+printf("\n (d) Impedance measured by relay  = %.3f Z_line \n",Z_r) ;
+printf("\n (e) Load impedance based on secondary ohms , Z_load = Ω(secondary) \n") ; disp(Z_load) ;
+printf("\n (f) Zone 1 setting of relay R_12 , Z_r = Ω(secondary) \n") ; disp(Z_r1) ;
+printf("\n (g) Zone 2 setting of relay R_12 , Z_r = Ω(secondary) \n") ; disp(Z_r2) ;
+printf("\n (h) Zone 3 setting of relay R_12 , Z_r = Ω(secondary) \n") ; disp(Z_r3) ;
diff --git a/1820/CH10/EX10.7/Example10_7.sce b/1820/CH10/EX10.7/Example10_7.sce
new file mode 100755
index 000000000..13610f8ae
--- /dev/null
+++ b/1820/CH10/EX10.7/Example10_7.sce
@@ -0,0 +1,33 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 10 : PROTECTIVE EQUIPMENT AND TRANSMISSION SYSTEM PROTECTION
+
+// EXAMPLE : 10.7 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+Z_r1 = 0.0415692 + %i*0.4156922 ; // Required zone 1 setting . From result of exa 10.6
+Z_r2 = 0.0623538 + %i*0.6235383 ; // Required zone 2 setting . From result of exa 10.6
+Z_r3 = 0.1299038 + %i*1.2990381 ; // Required zone 3 setting . From result of exa 10.6
+
+// CALCULATIONS 
+// For case (a)
+theta1 = atand(imag(Z_r1),real(Z_r1)) ;
+Z_1 = abs(Z_r1)/cosd(theta1 - 30) ; // Zone 1 setting of mho relay R_12
+
+// For case (b)
+theta2 = atand(imag(Z_r2),real(Z_r2)) ;
+Z_2 = abs(Z_r2)/cosd(theta2 - 30) ; // Zone 2 setting of mho relay R_12
+
+// For case (b)
+theta3 = atand(imag(Z_r3),real(Z_r3)) ;
+Z_3 = abs(Z_r3)/cosd(theta3 - 30) ; // Zone 3 setting of mho relay R_12
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 10.7 : SOLUTION :-") ;
+printf("\n (a) Zone 1 setting of mho relay R_12 = %.4f Ω(secondary) \n",Z_1) ;
+printf("\n (b) Zone 2 setting of mho relay R_12 = %.4f Ω(secondary) \n",Z_2) ;
+printf("\n (c) Zone 3 setting of mho relay R_12 = %.4f Ω(secondary) \n",Z_3) ;
diff --git a/1820/CH12/EX12.1/Example12_1.sce b/1820/CH12/EX12.1/Example12_1.sce
new file mode 100755
index 000000000..d1e7e0bd1
--- /dev/null
+++ b/1820/CH12/EX12.1/Example12_1.sce
@@ -0,0 +1,36 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 12 : CONSTRUCTION OF OVERHEAD LINES
+
+// EXAMPLE : 12.1 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+cost_avg = 1500 ; // Average cost on each repair in $
+r_0 = 0 ; // No. of times repair required for damage to line
+r_1 = 1 ; // No. of times repair required
+r_2 = 2 ; // No. of times repair required
+r_3 = 3 ; // No. of times repair required
+P_r_0 = 0.4 ; // Probability of exactly no. of repairs for r_0
+P_r_1 = 0.3 ; // Probability of exactly no. of repairs for r_1
+P_r_2 = 0.2 ; // Probability of exactly no. of repairs for r_2
+P_r_3 = 0.1 ; // Probability of exactly no. of repairs for r_3
+R_0 = 0 ; // No. of times repair required for relocating & rebuilding
+R_1 = 1 ; // No. of times repair required
+P_R_0 = 0.9 ; // Probability of exactly no. of repairs for R_0
+P_R_1 = 0.1 ; // Probability of exactly no. of repairs for R_1
+n = 25 ; // useful life in years
+i = 20/100 ; // carrying charge rate
+p = ((1 + i)^n - 1)/(i*(1+i)^n) ; // p = P/A . Refer page 642
+
+// CALCULATIONS
+B = cost_avg*(r_0*P_r_0 + r_1*P_r_1 + r_2*P_r_2 + r_3*P_r_3 - R_0*P_R_0 - R_1*P_R_1)*p ; // Affordable cost of relocating line
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 12.1 : SOLUTION :-") ;
+printf("\n Affordable cost of relocating line , B = $ %.1f \n",B) ;
+printf("\n  Since actual relocating & rebuilding of line would cost much more than amount found \n") ;
+printf("\n  The distribution engineer decides to keep the status quo \n") ;
diff --git a/1820/CH12/EX12.2/Example12_2.sce b/1820/CH12/EX12.2/Example12_2.sce
new file mode 100755
index 000000000..98533881e
--- /dev/null
+++ b/1820/CH12/EX12.2/Example12_2.sce
@@ -0,0 +1,39 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 12 : CONSTRUCTION OF OVERHEAD LINES
+
+// EXAMPLE : 12.2 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+V = 40 ; // Actual wind velocity in mi/hr
+c_pg = 40 ; // Circumference at ground level in inches
+c_pt = 28 ; // Circumference at pole top in inches
+l = 35 ; // height of pole in feet
+l_g = 6 ; // Height of pole set in ground in feet
+d_c = 0.81 ; // dia. of copper conductor in inches
+span_avg = 120 ; // Average span in ft
+no_c = 8 ; // NO. of conductors
+
+// CALCULATIONS
+// For case (a)
+p = 0.00256 * (V^2) ; // Buck's Formula to find wind pressure on cylindrical surface in lb/ft^2 
+d_pg = c_pg/(%pi) ; // dia. of pole at ground line in inches
+d_pt = c_pt/(%pi) ; // dia. of pole at pole top in inches
+h_ag = ( l - l_g ) * 12 ; // Height of pole above ground in inch
+S_pni = (1/2) * (d_pg + d_pt) * h_ag ; // projected area of pole in square inch
+S_pni_ft = S_pni * 0.0069444 ; // projected area of pole in square ft
+P = S_pni_ft * p ; // Total pressure of wind on pole in lb
+
+// For case (b)
+S_ni = d_c * span_avg * 12 ; // Projected area of conductor in square inch . [1 feet = 12 inch]
+S_ni_ft = S_ni * 0.0069444 ; // Projected area of conductor in square ft . [1 sq inch = (0.0833333)^2 sq feet ≅ 0.069444 sq feet]
+P_C = S_ni_ft * p * no_c ; // Total pressure of wind on conductor in lb
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 12.2 : SOLUTION :-");
+printf("\n (a) Total pressure of wind on pole , P = %.2f lb \n",P);
+printf("\n (b) Total pressure of wind on conductors , P = %.2f lb \n",P_C);
diff --git a/1820/CH12/EX12.3/Example12_3.sce b/1820/CH12/EX12.3/Example12_3.sce
new file mode 100755
index 000000000..0689f6586
--- /dev/null
+++ b/1820/CH12/EX12.3/Example12_3.sce
@@ -0,0 +1,44 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 12 : CONSTRUCTION OF OVERHEAD LINES
+
+// EXAMPLE : 12.3 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+a = 45 ; // OH line to be bulit on wood poles in ft
+b = 6.5 ; // Ground depth in ft
+c = 1 ; // Top cross-arm below pole top in ft
+d = 3 ; // Lower cross-arm below pole top in ft
+m_t = 0.6861 ; // Transverse wind load on top cross-arm in lb/ft
+m_l = 0.4769 ; // Transverse wind load on lower cross-arm in lb/ft
+u_s = 8000 ; // Ultimate strength of wood pole in lb/sq.in
+sf = 2 ; // Safety factor
+span_avg = 250 ; // Average span in ft
+p = 9 ; // Transverse wind load on wood poles in clb/sq.ft
+
+// CALCULATIONS
+h_1j = a - b - c ; // Moment arms for top arm in ft
+h_2j = a - b - d ; // Moment arms for top arm in ft
+M_tc1 = 1 * 4* m_t * span_avg * h_1j ; // Total bending moment for top arm in lb-ft
+M_tc2 = 1 * 4* m_l * span_avg * h_2j ; // Total bending moment for lower arm in lb-ft
+M_tc = M_tc1 + M_tc2 ; // Total bending moment for both cross-arms together in lb-ft
+S = u_s/sf ; // Allowable max fiber stress in pounds per sq.inch
+c_pg = ( M_tc/( 2.6385*10^-4*S ) )^(1/3) ; // circumference of pole at ground line in inch
+
+c_pt = 22 ; // From proper tables , for 8000 psi , 
+h_ag = a - b ; // Height of pole above ground in ft
+d_pg = c_pg/(%pi) ; // circumference of pole at ground line in inches
+d_pt = c_pt/(%pi) ; // circumference of pole at pole top in inches
+M_gp = (1/72)*p *(h_ag^2)*(d_pg + 2*d_pt) ; // Bending moment due to wind on pole in pound ft . using equ 12.9
+M_T = M_tc + M_gp ; // Total bending moment due to wind on conductor & pole
+c_pg1 = (M_T/( 2.6385 * 10^-4 * S ) )^(1/3) ; // using equ 12.11
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 12.3 : SOLUTION :-") ;
+printf("\n Minimum required pole circumference at the ground line , c = %.1f in \n",c_pg1) ;
+printf("\n Therefore , the nearest standard size pole,which has a ground-line circumference larger than c = %.1f in , has to be used \n",c_pg1) ;
+printf("\n Therefore required pole circumference at the ground line to be used is , c = %.f inch \n",c_pg1) ;
diff --git a/1820/CH12/EX12.4/Example12_4.sce b/1820/CH12/EX12.4/Example12_4.sce
new file mode 100755
index 000000000..19011b15f
--- /dev/null
+++ b/1820/CH12/EX12.4/Example12_4.sce
@@ -0,0 +1,40 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 12 : CONSTRUCTION OF OVERHEAD LINES
+
+// EXAMPLE : 12.4 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+T1 = 3000 ; // Bending moments in lb
+T2 = 2500 ; // Bending moments in lb
+h1 = 37.5 ; // Bending moments at heights in ft
+h2 = 35.5 ; // Bending moments at heights in ft
+h_g = 36.5 ; // Height at which Guy is attached to pole in ft
+L = 15 ; // Lead of guy in ft
+
+// CALCULATIONS
+// For case (a)
+T_h = ( T1*h1 + T2*h2 )/h_g ; // Horizontal component of tension in guy wire in lb . From equ 12.26
+
+// For case (b)
+bet = atand(h_g/L) ; // beta angle in degree . From equ 12.28
+
+// For case (c)
+T_v = T_h * tand(bet) ; // Vertical component of tension in guy wire in lb . From equ 12.34
+
+// For case (d)
+T_g = T_h/( cosd(bet )) ; // Tension in guy wire in lb . From equ 12.29
+T_g1 = sqrt( T_h^2 + T_v^2 ) ; // Tension in guy wire in lb
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 12.4 : SOLUTION :-") ;
+printf("\n (a) Horizontal component of tension in guy wire , T_h = %.1f lb \n",T_h) ;
+printf("\n (b) Angle β , β = %.2f degree \n",bet) ;
+printf("\n (c) Vertical component of tension in guy wire , T_v = %.2f lb \n",T_v) ;
+printf("\n (d) Tension in guy wire , T_g = %.1f lb \n",T_g) ;
+printf("\n     (or)  From another equation , \n") ;
+printf("\n     Tension in guy wire , T_g = %.1f lb \n",T_g1) ;
diff --git a/1820/CH13/EX13.1/Example13_1.sce b/1820/CH13/EX13.1/Example13_1.sce
new file mode 100755
index 000000000..d5d0d97eb
--- /dev/null
+++ b/1820/CH13/EX13.1/Example13_1.sce
@@ -0,0 +1,39 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 13 : SAG AND TENSION ANALYSIS
+
+// EXAMPLE : 13.1 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+c = 1600 ; // Length of conductor in feet
+L = 500 ; // span b/w conductors in ft
+w1 = 4122 ; // Weight of conductor in lb/mi
+
+// CALCULATIONS
+// For case (a)
+l = 2 * c *( sinh(L/(2*c)) ) ; // Length of conductor in ft using eq 13.6
+l_1 = L * (1 + (L^2)/(24*c^2) ) ; // Length of conductor in ft using eq 13.8
+
+// For case (b)
+d = c*( cosh( L/(2*c) ) - 1 ) ; // sag in ft
+
+// For case (c)
+w = w1/5280 ; // Weight of conductor in lb/ft . [1 mile = 5280 feet]
+T_max = w * (c + d) ; // Max conductor tension in lb
+T_min = w * c ; // Min conductor tension in lb
+
+// For case (d)
+T = w * (L^2)/(8*d) ; // Appr value of tension in lb using parabolic method
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 13.1 : SOLUTION :-") ;
+printf("\n (a) Length of conductor using eq 13.6 , l = %.3f ft \n",l) ;
+printf("\n  &  Length of conductor using eq 13.8 , l = %.4f ft \n",l_1) ;
+printf("\n (b) Sag , d = %.1f ft \n",d) ;
+printf("\n (c) Maximum value of conductor tension using catenary method , T_max = %.1f lb \n",T_max) ;
+printf("\n     Minimum value of conductor tension using catenary method , T_min = %.1f lb \n",T_min) ;
+printf("\n (d) Approximate value of tension using parabolic method , T = %.2f lb \n",T) ;
diff --git a/1820/CH13/EX13.2/Example13_2.sce b/1820/CH13/EX13.2/Example13_2.sce
new file mode 100755
index 000000000..4ff3d6e49
--- /dev/null
+++ b/1820/CH13/EX13.2/Example13_2.sce
@@ -0,0 +1,47 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 13 : SAG AND TENSION ANALYSIS
+
+// EXAMPLE : 13.2 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+L = 500 ; // span b/w conductors in ft
+p = 4 ; // Horizontal wind pressure in lb/sq ft
+t_i = 0.50 ; // Radial thickness of ice in inches
+d_c = 1.093 ; // outside diameter of ACSR conductor in inches
+w1 = 5399 ; // weight of conductor in lb/mi
+s = 28500 ; // ultimate strength in lb
+
+// CALCULATIONS
+// For case (a)
+w_i = 1.25 * t_i * (d_c + t_i) ; // Weight of ice in pounds per feet
+
+// For case (b)
+w = w1/5280 ; // weight of conductor in lb/ft . [1 mile = 5280 feet]
+W_T = w + w_i ; // Total vertical load on conductor in pounds per feet
+
+// For case (c)
+P = ( (d_c + 2*t_i)/(12) )*p ; // Horizontal wind force in lb/ft
+
+// For case (d)
+w_e = sqrt( P^2 + (w + w_i)^2 ) ; // Effective load on conductor in lb/ft
+
+// For case (e)
+T = s/2 ;
+d = w_e * L^2/(8*T) ; // sag in feet
+
+// For case (f)
+d_v = d * W_T/w_e ; // vertical sag in feet
+
+// DISPLAY RESULTS
+disp("EXAMPLE :13.2 : SOLUTION :-") ;
+printf("\n (a) Weight of ice in pounds per feet , w_i = %.4f lb/ft \n",w_i) ;
+printf("\n (b) Total vertical load on conductor in pounds per feet , W_T = %.4f lb/ft \n",W_T) ;
+printf("\n (c) Horizontal wind force in pounds per feet , P = %.4f lb/ft \n",P) ;
+printf("\n (d) Effective load acting in pounds per feet , w_e = %.4f lb/ft \n",w_e) ;
+printf("\n (e) Sag in feet , d = %.2f ft \n",d) ;
+printf("\n (f) Vertical Sag in feet  = %.2f ft \n",d_v) ;
diff --git a/1820/CH14/EX1.C/ExampleC_1.sce b/1820/CH14/EX1.C/ExampleC_1.sce
new file mode 100755
index 000000000..c9d43c0cb
--- /dev/null
+++ b/1820/CH14/EX1.C/ExampleC_1.sce
@@ -0,0 +1,31 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// APPENDIX C : REVIEW OF BASICS
+
+// EXAMPLE : C.1 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+z = 100 * exp(60*%i*%pi/180) ; // Impedance of transmission line in Ω
+v1 = 73034.8 * exp(30*%i*%pi/180) ; // Bus voltages in V
+v2 = 66395.3 * exp(20*%i*%pi/180) ; // Bus voltages in V
+
+// CALCULATIONS
+// For case (a)
+S_12 = v1 * ( conj(v1) - conj(v2) )/( conj(z) ) ; // Complex power per phase in VA
+
+
+// For case (b)
+P_12 = real(S_12) ; // Active power per phase in W
+
+// For case (c)
+Q_12 = imag(S_12) ; // Reactive power per phase in vars
+
+// DISPLAY RESULTS
+disp("EXAMPLE : C.1 : SOLUTION :-") ;
+printf("\n (a) Complex power per phase that is being transmitted from bus 1 to bus 2 , S12 = %.2f<%.2f VA \n",abs(S_12), atan(imag(S_12),real(S_12))*(180/%pi)) ;
+printf("\n (b) Active power per phase that is being transmitted , P12 = %.2f W \n",P_12) ;
+printf("\n (b) Reactive power per phase that is being transmitted , Q12 = %.2f vars \n",Q_12) ;
diff --git a/1820/CH14/EX2.C/ExampleC_2.sce b/1820/CH14/EX2.C/ExampleC_2.sce
new file mode 100755
index 000000000..44d7b5300
--- /dev/null
+++ b/1820/CH14/EX2.C/ExampleC_2.sce
@@ -0,0 +1,43 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// APPENDIX C : REVIEW OF BASICS
+
+// EXAMPLE : C.2 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+X_pu = 12/100 ; // Leakage reactance in pu
+kV_B_HV = 345 ; // HV side ratings in Y kV
+kV_B_LV = 34.5 ; // LV side ratings in Y kV
+MVA_B = 20 ; // selected Base on HV side in MVA
+
+// CALCULATIONS
+// For case (a)
+X_pu = 12/100 ; // Reactance of transformer in pu
+
+// For case (b)
+Z_B_HV = (kV_B_HV)^2/MVA_B ; // HV side base impedance in Ω
+
+// For case (c)
+Z_B_LV = (kV_B_LV)^2/MVA_B ; // LV side base impedance in Ω
+
+// For case (d)
+X_HV = X_pu * Z_B_HV ; // Reactance referred to HV side in Ω
+
+// For case (e)
+X_LV = X_pu * Z_B_LV ; // Reactance referred to LV side in Ω
+n = (kV_B_HV/sqrt(3))/(kV_B_LV/sqrt(3)) ; // Turns ratio of winding
+X_LV1 = X_HV/n^2 ; // From equ C.89
+
+// DISPLAY RESULTS
+disp("EXAMPLE : C.2 : SOLUTION :-") ;
+printf("\n (a) Reactance of transformer in pu , X_pu = %.2f pu \n",X_pu) ;
+printf("\n (b) High-voltage side base impedance , Z_B_HV = %.2f Ω \n",Z_B_HV) ;
+printf("\n (c) Low-voltage side base impedance , Z_B_LV = %.4f Ω \n",Z_B_LV) ;
+printf("\n (d) Transformer reactance referred to High-voltage side , X_HV = %.2f Ω \n",X_HV) ;
+printf("\n (e) Transformer reactance referred to Low-voltage side , X_LV = %.4f Ω \n",X_LV) ;
+printf("     (or) From another equation C.89 ,") ;
+printf("\n     Transformer reactance referred to Low-voltage side , X_LV = %.4f Ω \n",X_LV1) ;
diff --git a/1820/CH14/EX3.C/ExampleC_3.sce b/1820/CH14/EX3.C/ExampleC_3.sce
new file mode 100755
index 000000000..cd350fd5a
--- /dev/null
+++ b/1820/CH14/EX3.C/ExampleC_3.sce
@@ -0,0 +1,43 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// APPENDIX C : REVIEW OF BASICS
+
+// EXAMPLE : C.3 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+X_pu = 12/100 ; // Leakage reactance in pu
+kV_B_HV = 345 ; // HV side ratings in Y kV
+kV_B_LV = 34.5 ; // LV side ratings in Δ kV
+MVA_B = 20 ; // Base on HV side in MVA
+
+// CALCULATIONS
+// For case (a)
+n = ( kV_B_HV/sqrt(3) )/kV_B_LV ; // Turns ratio of windings
+
+// For case (b)
+Z_B_HV = (kV_B_HV)^2/MVA_B ; // HV side base impedance in Ω
+X_HV = X_pu * Z_B_HV ; // Reactance referred to HV side in Ω
+X_LV = X_HV/(n^2) ; // transformer reactance referred to delta LV side in Ω
+
+// For case (c)
+Z_dt = X_LV ;
+Z_Y = Z_dt/3 ; // Reactance of equi wye connection
+Z_B_LV = kV_B_LV^2/MVA_B ; // LV side base impedance in Ω
+X_pu1 = Z_Y/Z_B_LV ; // reactance in pu referred to LV side
+
+// Alternative method For case (c)
+n1 = kV_B_HV/kV_B_LV ; // Turns ratio if line-to-line voltages are used
+X_LV1 = X_HV/(n1^2) ; // Reactance referred to LV side in Ω
+X_pu2 = X_LV1/Z_B_LV ; // reactance in pu referred to LV side
+
+// DISPLAY RESULTS
+disp("EXAMPLE : C.3 : SOLUTION :-") ;
+printf("\n (a) Turns ratio of windings , n = %.4f \n",n) ;
+printf("\n (b) Transformer reactance referred to LV side in ohms ,X_LV = %.4f Ω \n",X_LV) ;
+printf("\n (c) Transformer reactance referred to LV side in per units ,X_pu = %.2f pu \n",X_pu1) ;
+printf("\n    (or) From another equation if line-to-line voltages are used ,") ;
+printf("\n     Transformer reactance referred to LV side in per units ,X_pu = %.2f pu \n",X_pu2) ;
diff --git a/1820/CH14/EX4.C/ExampleC_4.sce b/1820/CH14/EX4.C/ExampleC_4.sce
new file mode 100755
index 000000000..f346c474f
--- /dev/null
+++ b/1820/CH14/EX4.C/ExampleC_4.sce
@@ -0,0 +1,103 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// APPENDIX C : REVIEW OF BASICS
+
+// EXAMPLE : C.4 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+I_1 = 1000 ; // Physical current in A for 2.4 kV circuit
+Z_pu = 0.04 ; // Leakage reactance in pu
+I_pu = 2.08*exp(%i*(-90)*%pi/180) ; // Generator supply for pure inductive load 
+kVA_Bg1 = 6000 ; // Rated kVA values for T1
+kVA_Bg2 = 4000 ; // Rated kVA values for T2
+N2 = 2.4 ; // N2 = V2 in Y kV ,refer fig C.4
+N1 = 24 ; // N1 = V1 in Y kV ,refer fig C.4
+N3 = 24 ; // N3 = V3 = N1 in Y kV ,refer fig C.4
+N4 = 12 ; // N4 = V4 in Y kV ,refer fig C.4
+
+// CALCULATIONS
+// For case (a)
+kVA_B = 2080 ; // arbitrarily selected kVA values for all 3 ckt
+
+// For case (b)
+n1 = N2/N1 ; // Turns ratio of transformer T1 & T2 i.e N2/N1
+n2 = N3/N4 ; // Turns ratio N1'/N2'
+kV_BL_L1 = 2.5 ; // arbitrarily selected Base voltage for 2.4 kV ckt in kV
+kV_BL_L2 = kV_BL_L1/n1 ; // arbitrarily selected Base voltage for 24 kV ckt in kV
+kV_BL_L3 = kV_BL_L2/n2 ; // arbitrarily selected Base voltage for 12 kV ckt in kV
+
+// For case (c)
+Z_B1 = (kV_BL_L1)^(2) * 1000/(kVA_B) ; // Base impedance in Ω for 2.4 kV ckt
+Z_B2 = (kV_BL_L2)^(2) * 1000/(kVA_B) ; // Base impedance in Ω for 24 kV ckt
+Z_B3 = (kV_BL_L3)^(2) * 1000/(kVA_B) ; // Base impedance in Ω for 12 kV ckt
+
+// For case (d)
+I_B1 = kVA_B/(sqrt(3)*kV_BL_L1) ; // Base current in A for 2.4 kV ckt
+I_B2 = kVA_B/(sqrt(3)*kV_BL_L2) ; // Base current in A for 24 kV ckt
+I_B3 = kVA_B/(sqrt(3)*kV_BL_L3) ; // Base current in A for 12 kV ckt
+
+// For case (e)
+I_2 = (n1) * I_1 ; // Physical current in A for 24 kV circuit
+I_4 = (n2) * I_2 ; // Physical current in A for 12 kV circuit
+
+// For case (f)
+I_pu_3ckt = abs(I_pu) ; // per-unit current values for all 3-ckt
+
+// For case (g)
+kV_B1 = N2 ; // Given voltage in kV
+kV_B2 = N4 ; // Given voltage in kV
+Z_pu_T1 = (%i)*Z_pu*(kVA_B/kVA_Bg1)*(kV_B1/kV_BL_L1)^(2) ; // New reactance of T1
+Z_pu_T2 = (%i)*Z_pu*(kVA_B/kVA_Bg2)*(kV_B2/kV_BL_L3)^(2) ; // New reactance of T2
+
+// For case (h)
+V1 = kV_B1/kV_BL_L1 ; // voltage in pu at bus 1
+V2 = V1 - I_pu * (Z_pu_T1) ; // voltage in pu at bus 2
+V4 = V2 - I_pu * (Z_pu_T2) ; // voltage in pu at bus 3
+
+// For case (i)
+S1 = V1 * abs(I_pu) ; // Apparent power value at bus 1 in pu
+S2 = V2 * abs(I_pu) ; // Apparent power value at bus 2 in pu
+S4 = V4 * abs(I_pu) ; // Apparent power value at bus 4 in pu
+
+// DISPLAY RESULTS
+disp("EXAMPLE : C.3 : SOLUTION :-") ;
+printf("\n (a) Base kilovoltampere value for all 3-circuits is , kVA_B = %.1f kVA \n",kVA_B) ;
+printf("\n (b) Base line-to-line kilovolt value for 2.4 kV circuit , kV_BL_L = %.1f kV \n",kV_BL_L1) ;
+printf("\n     Base line-to-line kilovolt value for 24 kV circuit , kV_BL_L = %.1f kV \n",kV_BL_L2) ;
+printf("\n     Base line-to-line kilovolt value for 24 kV circuit , kV_BL_L = %.1f kV \n",kV_BL_L3) ;
+printf("\n (c) Base impedance value of 2.4 kV circuit , Z_B = %.3f Ω \n",Z_B1) ;
+printf("\n     Base impedance value of 24 kV circuit , Z_B = %.1f Ω \n",Z_B2) ;
+printf("\n     Base impedance value of 12.5 kV circuit , Z_B = %.1f Ω \n",Z_B3) ;
+printf("\n (d) Base current value of 2.4 kV circuit , I_B = %d A \n",I_B1) ;
+printf("\n     Base current value of 24 kV circuit , I_B = %d A \n",I_B2) ;
+printf("\n     Base current value of 2.4 kV circuit , I_B = %d A \n",I_B3) ;
+printf("\n (e) Physical current of 2.4 kV circuit , I = %.f A \n",I_1) ;
+printf("\n     Physical current of 24 kV circuit , I = %.f A \n",I_2) ;
+printf("\n     Physical current of 12 kV circuit , I = %.f A \n",I_4) ;
+printf("\n (f) Per unit current values for all 3 circuits , I_pu = %.2f pu \n",I_pu_3ckt) ;
+printf("\n (g) New transformer reactance of T1 , Z_pu_T1 =  j%.4f pu \n",abs(Z_pu_T1)) ;
+printf("\n     New transformer reactance of T2 , Z_pu_T2 =  j%.4f pu \n",abs(Z_pu_T2)) ;
+printf("\n (h) Per unit voltage value at bus 1 ,V1 = %.2f<%.1f pu \n",abs(V1),atand(imag(V1),real(V1))) ;
+printf("\n     Per unit voltage value at bus 2 ,V2 = %.4f<%.1f pu \n",abs(V2),atand(imag(V2),real(V2))) ;
+printf("\n     Per unit voltage value at bus 4 ,V4 = %.4f<%.1f pu \n",abs(V4),atand(imag(V4),real(V4))) ;
+printf("\n (i) Per-unit apparent power value at bus 1 , S1 = %.2f pu \n",S1) ;
+printf("\n     Per-unit apparent power value at bus 2 , S2 = %.4f pu \n",S2) ;
+printf("\n     Per-unit apparent power value at bus 4 , S4 = %.4f pu \n",S4) ;
+printf("\n (j) TABLE C.2 \n") ;
+printf("\n     Results Of Example C.4 \n") ;
+printf("\n     ___________________________________________________________________________________") ;
+printf("\n     QUANTITY      \t  2.4-kV circuit   \t  24-kV circuit   \t  12-kV circuit   ");
+printf("\n     ___________________________________________________________________________________") ;
+printf("\n     kVA_B(3-Φ)    \t  %d kVA           \t  %d kVA          \t  %d kVA \n",kVA_B,kVA_B,kVA_B) ;
+printf("\n     kV_B(L-L)     \t  %.1f kV          \t  %d kV           \t  %.1f kV \n",kV_BL_L1,kV_BL_L2,kV_BL_L3) ;
+printf("\n     Z_B           \t  %.3f Ω           \t  %.1f Ω          \t  %.1f Ω \n",Z_B1,Z_B2,Z_B3) ;
+printf("\n     I_B           \t  %d A             \t  %d A            \t  %d A \n",I_B1,I_B2,I_B3) ;
+printf("\n     I_physical    \t  %d A             \t  %.f A           \t  %.f A \n",I_1,I_2,I_4) ;
+printf("\n     I_pu          \t  %.2f pu          \t  %.2f pu         \t  %.2f pu \n",I_pu_3ckt,I_pu_3ckt,I_pu_3ckt) ;
+printf("\n     V_pu          \t  %.2f pu          \t  %.4f pu         \t  %.4f pu \n",abs(V1),abs(V2),abs(V4)) ;
+printf("\n     S_pu          \t  %.2f pu          \t  %.4f pu         \t  %.4f pu \n",S1,S2,S4) ;
+printf("     ___________________________________________________________________________________") ;
diff --git a/1820/CH14/EX5.C/ExampleC_5.sce b/1820/CH14/EX5.C/ExampleC_5.sce
new file mode 100755
index 000000000..351e319d6
--- /dev/null
+++ b/1820/CH14/EX5.C/ExampleC_5.sce
@@ -0,0 +1,30 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// APPENDIX C : REVIEW OF BASICS
+
+// EXAMPLE : C.5 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+D_ab = 6.8 ; // distance b/w conductors center-to-center in ft
+D_bc = 5.5 ; // distance b/w conductors center-to-center in ft
+D_ca = 4 ; // distance b/w conductors center-to-center in ft
+
+// CALCULATIONS
+// For case (a)
+D_eq = (D_ab * D_bc * D_ca)^(1/3) ; // Equi spacing for pole top in ft
+D_s = 0.01579 ; // GMR in ft From Table A.1
+X_L = 0.1213 * log(D_eq/D_s) ; // Inductive reactance in Ω/mi . From equ C.135
+
+// For case (b)
+X_a = 0.503 ; // Inductive reactance in Ω/mi From Table A.1 
+X_d = 0.2026 ; // From Table A.8 for D_eq,by linear interpolation in Ω/mi
+X_L1 = X_a + X_d ; // Inductive reactance in Ω/mi
+
+// DISPLAY RESULTS
+disp("EXAMPLE : C.5 : SOLUTION :-") ;
+printf("\n (a) Inductive reactance using equation C.135 , X_L = %.4f Ω/mi \n",X_L );
+printf("\n (b) Inductive reactance using tables , X_L = %.4f Ω/mi \n",X_L1) ;
diff --git a/1820/CH14/EX6.C/ExampleC_6.sce b/1820/CH14/EX6.C/ExampleC_6.sce
new file mode 100755
index 000000000..f19dfad32
--- /dev/null
+++ b/1820/CH14/EX6.C/ExampleC_6.sce
@@ -0,0 +1,33 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// APPENDIX C : REVIEW OF BASICS
+
+// EXAMPLE : C.6 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+D_ab = 6.8 ; // distance b/w conductors center-to-center in ft
+D_bc = 5.5 ; // distance b/w conductors center-to-center in ft
+D_ca = 4 ; // distance b/w conductors center-to-center in ft
+l = 100 ; // Line length in miles
+
+// CALCULATIONS
+// For case (a)
+D_m = (D_ab * D_bc * D_ca)^(1/3) ; // Equi spacing for pole top in ft
+r = 0.522/(2 * 12) ; // feet
+X_C = 0.06836 * log10 (D_m/r) ; // Shunt capacitive reactance in MΩ*mi
+
+// For case (b)
+X_a = 0.1136 ; // Shunt capacitive reactance in MΩ*mi , From table A.1
+X_d = 0.049543 ; // Shunt capacitive reactance spacing factor in MΩ*mi , From table A.9
+X_C1 = X_a + X_d ; // Shunt capacitive reactance in MΩ*mi
+X_C2 = X_C1/l ; // Capacitive reactance of 100 mi line in MΩ
+
+// DISPLAY RESULTS
+disp("EXAMPLE : C.6 : SOLUTION :-") ;
+printf("\n (a) Shunt capacitive reactance using equation C.156 , X_C = %.6f MΩ*mi \n",X_C) ;
+printf("\n (b) Shunt capacitive reactance using tables , X_C = %.6f MΩ*mi \n",X_C1) ;
+printf("\n (c) Capacitive reactance of total line , X_C = %.5e MΩ \n",X_C2) ;
diff --git a/1820/CH2/EX2.1/Example2_1.sce b/1820/CH2/EX2.1/Example2_1.sce
new file mode 100755
index 000000000..ece1104e6
--- /dev/null
+++ b/1820/CH2/EX2.1/Example2_1.sce
@@ -0,0 +1,37 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION 
+
+// CHAPTER : 2 : TRANSMISSION LINE STRUCTURES AND EQUIPMENT
+
+// EXAMPLE : 2.1 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+t_s = 0.49 ; // Human body is in contact with 60 Hz power for 0.49 sec
+r = 100 ; // Resistivity of soil based on IEEE std 80-2000 
+
+// CALCULATIONS
+// For case (a)
+v_touch50 = 0.116*(1000+1.5*r)/sqrt(t_s) ; // Maximum allowable touch voltage for 50 kg body weight in volts
+
+// For case (b)
+v_step50  = 0.116*(1000+6*r)/sqrt(t_s) ; // Maximum allowable step voltage for 50 kg body weight in volts
+// Above Equations of case (a) & (b) applicable if no protective surface layer is used
+
+// For metal to metal contact below equation holds good . Hence resistivity is zero
+r_1 = 0 ; // Resistivity is zero
+
+// For case (c)
+v_mm_touch50 = 0.116*(1000)/sqrt(t_s) ; // Maximum allowable touch voltage for 50 kg body weight in volts for metal to metal contact
+
+// For case (d)
+v_mm_touch70 = 0.157*(1000)/sqrt(t_s) ; // Maximum allowable touch voltage for 70 kg body weight in volts for metal to metal contact
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 2.1 : SOLUTION :-") ;
+printf("\n (a) Tolerable Touch potential , V_touch50 = %.f V , for 50 kg body weight \n",v_touch50) ;
+printf("\n (b) Tolerable Step potential , V_step50 = %.f V , for 50 kg body weight \n",v_step50) ;
+printf("\n (c) Tolerable Touch Voltage for metal-to-metal contact , V_mm_touch50 = %.1f V , for 50 kg body weight \n",v_mm_touch50) ;
+printf("\n (d) Tolerable Touch Voltage for metal-to-metal contact , V_mm_touch70 = %.1f V , for 70 kg body weight \n",v_mm_touch70) ;
diff --git a/1820/CH3/EX3.1/Example3_1.sce b/1820/CH3/EX3.1/Example3_1.sce
new file mode 100755
index 000000000..50032258e
--- /dev/null
+++ b/1820/CH3/EX3.1/Example3_1.sce
@@ -0,0 +1,25 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 3 : FUNDAMENTAL CONCEPTS
+
+// EXAMPLE : 3.1 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+kV = 345 ; // Three phase transmission line voltage in kV
+Z_s = 366 ; // Surge impedance of line in Ω
+a = 24.6 ; // Spacing between adjacent conductors in feet
+d = 1.76 ; // Diameter of conductor in inches
+
+// CALCULATIONS
+SIL = (kV)^2/Z_s ; // Surge Impedance loading of line in MW
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 3.1 : SOLUTION :-") ;
+printf("\n Surge Impedance Loading of line , SIL = %.f MW \n",SIL) ;
+
+printf("\n NOTE: Unit of SIL is MW and surge impedance is Ω") ;
+printf("\n ERROR: Mistake in unit of SIL in textbook \n") ;
diff --git a/1820/CH3/EX3.2/Example3_2.sce b/1820/CH3/EX3.2/Example3_2.sce
new file mode 100755
index 000000000..5c0b950bf
--- /dev/null
+++ b/1820/CH3/EX3.2/Example3_2.sce
@@ -0,0 +1,52 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 3 : FUNDAMENTAL CONCEPTS
+
+// EXAMPLE : 3.2 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+SIL = 325 ; // Surge impedance Loading in MW . From exa 3.1
+kV = 345 ; // Transmission line voltage in kV . From exa 3.1
+
+// For case (a)
+t_shunt1 = 0.5 ; // shunt capacitive compensation is 50%
+t_series1 = 0 ; // no series compensation
+
+// For case (b)
+t_shunt2 = 0.5 ; // shunt compensation using shunt reactors is 50%
+t_series2 = 0 ; // no series capacitive compensation
+
+// For case (c)
+t_shunt3 = 0 ; // no shunt compensation
+t_series3 = 0.5 ; // series capacitive compensation is 50%
+
+// For case (d)
+t_shunt4 = 0.2 ; // shunt capacitive compensation is 20%
+t_series4 = 0.5; // series capacitive compensation is 50%
+
+// CALCULATIONS
+// For case (a)
+SIL1 = SIL*(sqrt( (1-t_shunt1)/(1-t_series1) )) ; // Effective SIL in MW
+
+// For case (b)
+SIL2 = SIL*(sqrt( (1+t_shunt2)/(1-t_series2) )) ; // Effective SIL in MW
+
+// For case (c)
+SIL3 = SIL*(sqrt( (1-t_shunt3)/(1-t_series3) )) ; // Effective SIL in MW
+
+// For case (d)
+SIL4 = SIL*(sqrt( (1-t_shunt4)/(1-t_series4) )) ; // Effective SIL in MW
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 3.2 : SOLUTION :-") ;
+printf("\n (a) Effective SIL , SIL_comp = %.f MW \n",SIL1) ;
+printf("\n (b) Effective SIL , SIL_comp = %.f MW \n",SIL2) ;
+printf("\n (c) Effective SIL , SIL_comp = %.f MW \n",SIL3) ;
+printf("\n (d) Effective SIL , SIL_comp = %.f MW \n",SIL4) ;
+
+printf("\n NOTE: Unit of SIL is MW and surge impedance is Ω ") ;
+printf("\n ERROR: Mistake in unit of SIL in textbook \n") ;
diff --git a/1820/CH3/EX3.3/Example3_3.sce b/1820/CH3/EX3.3/Example3_3.sce
new file mode 100755
index 000000000..1d5daec8b
--- /dev/null
+++ b/1820/CH3/EX3.3/Example3_3.sce
@@ -0,0 +1,30 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 3 : FUNDAMENTAL CONCEPTS
+
+// EXAMPLE : 3.3 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+// For case (c)
+I_normal = 1000 ; // Normal full load current in Ampere
+
+// CALCULATIONS
+// For case (a) equation is (1.5pu)*I_rated = (2 pu)*I_normal
+// THEREFORE
+// I_rated = (1.333pu)*I_normal ; // Rated current in terms of per unit value of the normal load current
+
+// For case (b) 
+Mvar = (1.333)^2 ; // Increase in Mvar rating in per units
+
+// For case (c)
+I_rated = (1.333)*I_normal ; // Rated current value
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 3.3 : SOLUTION :-") ;
+printf("\n (a) Rated current , I_rated = (1.333 pu)*I_normal \n") ;
+printf("\n (b) Mvar rating increase = %.2f pu \n",Mvar) ;
+printf("\n (c) Rated current value , I_rated = %.f A \n",I_rated) ;
diff --git a/1820/CH4/EX4.1/Example4_1.sce b/1820/CH4/EX4.1/Example4_1.sce
new file mode 100755
index 000000000..fc905cafb
--- /dev/null
+++ b/1820/CH4/EX4.1/Example4_1.sce
@@ -0,0 +1,114 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 4 : OVERHEAD POWER TRANSMISSION
+
+// EXAMPLE : 4.1 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+V_RL_L = 23*10^3 ; // line to line voltage in volts
+z_t = 2.48+%i*6.57 ; // Total impedance in ohm/phase
+p = 9*10^6 ; // load in watts
+pf = 0.85 ; // lagging power factor
+
+// CALCULATIONS
+// METHOD I : USING COMPLEX ALGEBRA
+
+V_RL_N = (V_RL_L)/sqrt(3) ; // line-to-neutral reference voltage in V
+I = (p/(sqrt(3)*V_RL_L*pf))*( pf - %i*sind(acosd(pf))) ; // Line current in amperes
+IZ = I*z_t ;
+V_SL_N = V_RL_N + IZ // Line to neutral voltage at sending end in volts
+V_SL_L = sqrt(3)*V_SL_N ; // Line to line voltage at sending end in volts
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 4.1 : SOLUTION :-") ;
+disp("METHOD I : USING COMPLEX ALGEBRA") ; 
+printf("\n (a) Line-to-neutral voltage at sending end , V_SL_N = %.f<%.1f V \n",abs(V_SL_N),atand( imag(V_SL_N),real(V_SL_N) )) ;
+printf("\n i.e Line-to-neutral voltage at sending end , V_SL_N = %.f V \n",abs(V_SL_N)) ;
+printf("\n     Line-to-line voltage at sending end , V_SL_L = %.f<%.1f V \n",abs(V_SL_L),atand( imag(V_SL_L),real(V_SL_L) )) ;
+printf("\n i.e Line-to-line voltage at sending end , V_SL_L = %.f V \n",abs(V_SL_L)) ;
+printf("\n (b) load angle , δ = %.1f degree \n",atand( imag(V_SL_L),real(V_SL_L) )) ;
+printf("\n") ;
+
+
+// CALCULATIONS
+// METHOD II : USING THE CURRENT AS REFERENCE PHASOR
+theta_R = acosd(pf) ; 
+V1 = V_RL_N*cosd(theta_R) + abs(I)*real(z_t) ; // unit is volts
+V2 = V_RL_N*sind(theta_R) + abs(I)*imag(z_t) ; // unit is volts
+V_SL_N2 = sqrt( (V1^2) + (V2^2) ) ; // Line to neutral voltage at sending end in volts/phase
+V_SL_L2 = sqrt(3) * V_SL_N2 ; // Line to line voltage at sending end in volts
+theta_s = atand(V2/V1) ; 
+delta = theta_s - theta_R ; 
+
+// DISPLAY RESULTS
+disp("METHOD II : USING THE CURRENT AS REFERENCE PHASOR");
+printf("\n (a) Line-to-neutral voltage at sending end , V_SL_N = %.f V \n",V_SL_N2) ;
+printf("\n     Line-to-line voltage at sending end , V_SL_L = %.f V \n",V_SL_L2) ;
+printf("\n (b) load angle , δ = %.1f degree \n",delta) ;
+printf("\n") ;
+
+// CALCULATIONS
+// METHOD III : USING THE RECEIVING-END VOLTAGE  AS REFERENCE PHASOR
+// for case (a)
+V_SL_N3 = sqrt( (V_RL_N + abs(I) * real(z_t) * cosd(theta_R) + abs(I) * imag(z_t) * sind(theta_R))^2 + (abs(I)*imag(z_t) * cosd(theta_R) - abs(I) * real(z_t) * sind(theta_R))^2) ;
+V_SL_L3 = sqrt(3)*V_SL_N3 ;
+
+// for case (b)
+delta_3 = atand( (abs(I)*imag(z_t) * cosd(theta_R) - abs(I) * real(z_t) * sind(theta_R))/(V_RL_N + abs(I) * real(z_t) * cosd(theta_R) + abs(I) * imag(z_t) * sind(theta_R)) ) ;
+
+// DISPLAY RESULTS
+disp("METHOD III : USING THE RECEIVING END VOLTAGE  AS REFERENCE PHASOR") ;
+printf("\n (a) Line-to-neutral voltage at sending end , V_SL_N = %.f V \n",V_SL_N3) ;
+printf("\n     Line-to-line voltage at sending end , V_SL_L = %.f V \n",V_SL_L3) ;
+printf("\n (b) load angle , δ = %.1f degree \n",delta_3) ;
+printf("\n") ;
+
+// CALCULATIONS
+// METHOD IV : USING POWER RELATIONSHIPS
+P_4 = 9 ; // load in MW (Given)
+P_loss = 3 * (abs(I))^2 * real(z_t) * 10^-6 ; // Power loss in line in MW
+P_T = P_4 + P_loss ; // Total input power to line in MW
+Q_loss = 3 * (abs(I))^2 * imag(z_t) * 10^-6 ; // Var loss of line in Mvar lagging
+Q_T = ( (P_4*sind(theta_R))/cosd(theta_R) ) + Q_loss ; // Total megavar input to line in Mvar lagging
+S_T = sqrt( (P_T^2)+(Q_T^2) ) ; // Total megavoltampere input to line
+// for case (a)
+V_SL_L4 = S_T*10^6/(sqrt(3) * abs(I)) ; // line to line voltage in volts
+V_SL_N4 = V_SL_L4/sqrt(3) ; // Line to line neutral in volts
+
+// for case (b)
+theta_S4 = acosd(P_T/S_T) ; // Lagging
+delta_4 = theta_s - theta_R ; 
+
+// DISPLAY RESULTS
+disp("METHOD IV : USING POWER RELATIONSHIPS");
+printf("\n (a) Line-to-neutral voltage at sending end , V_SL_N = %.f V \n",V_SL_N4) ;
+printf("\n (a) Line-to-line voltage at sending end , V_SL_L = %.f V \n",V_SL_L4) ;
+printf("\n (b) load angle , δ = %.1f degree \n",delta_4) ;
+printf("\n");
+
+// CALCULATIONS
+// METHOD V : Treating 3-Φ line as 1-Φ line having having V_S and V_R represent line-to-line voltages not line-to-neutral voltages
+// for case (a)
+I_line = (p/2)/(V_RL_L * pf) ; // Power delivered is 4.5 MW
+R_loop = 2*real(z_t) ;
+X_loop = 2*imag(z_t) ;
+V_SL_L5 = sqrt( (V_RL_L * cosd(theta_R) + I_line*R_loop)^2 + (V_RL_L * sind(theta_R) + I_line * X_loop)^2) ; // line to line voltage in V
+V_SL_N5 = V_SL_L5/sqrt(3) ; // line to neutral voltage in V
+
+// for case (b)
+theta_S5 = atand((V_RL_L * sind(theta_R) + I_line * X_loop)/(V_RL_L * cosd(theta_R) + I_line*R_loop)) ;
+delta_5 = theta_S5 - theta_R ;
+
+// DISPLAY RESULTS
+disp("METHOD V : TREATING 3-Φ LINE AS 1-Φ LINE") ;
+printf("\n (a) Line to neutral voltage at sending end , V_SL_N = %.f V \n",V_SL_N5) ;
+printf("\n (a) Line to line voltage at sending end , V_SL_L = %.f V \n",V_SL_L5) ;
+printf("\n (b) load angle , δ = %.1f degree \n",delta_5) ;
+printf("\n") ;
+
+printf("\n NOTE : ERROR : Change in answer because root(3) = 1.73 is considered in Textbook ") ;
+printf("\n But here sqrt(3) = 1.7320508 is considered \n") ;
diff --git a/1820/CH4/EX4.10/Example4_10.sce b/1820/CH4/EX4.10/Example4_10.sce
new file mode 100755
index 000000000..153a1adca
--- /dev/null
+++ b/1820/CH4/EX4.10/Example4_10.sce
@@ -0,0 +1,27 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 4 : OVERHEAD POWER TRANSMISSION
+
+// EXAMPLE : 4.10 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+Z_1 = 10 * exp(%i*(30)*%pi/180) ; // Impedance in Ω
+Z_2 = 40 * exp(%i*(-45)*%pi/180) ; // Impedance in Ω
+
+// CALCULATIONS
+P = [1 ,Z_1 ; 0 , 1]; // For network 1
+Y_2 = 1/Z_2 ; // unit is S
+Q = [1 0 ; Y_2 1]; // For network 2
+EQ = P * Q ;
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 4.10 : SOLUTION :-") ;
+printf("\n Equivalent A , B , C , D constants are \n") ;
+printf("\n A_eq = %.3f<%.1f \n",abs( EQ(1,1) ),atand( imag(EQ(1,1)),real(EQ(1,1)) )) ;
+printf("\n B_eq = %.3f<%.1f \n",abs( EQ(1,2) ),atand( imag(EQ(1,2)),real(EQ(1,2)) )) ;
+printf("\n C_eq = %.3f<%.1f \n",abs( EQ(2,1) ),atand( imag(EQ(2,1)),real(EQ(2,1)) )) ;
+printf("\n D_eq = %.3f<%.1f \n",abs( EQ(2,2 )),atand( imag(EQ(2,2)),real(EQ(2,2)) )) ;
diff --git a/1820/CH4/EX4.11/Example4_11.sce b/1820/CH4/EX4.11/Example4_11.sce
new file mode 100755
index 000000000..3c4882fb6
--- /dev/null
+++ b/1820/CH4/EX4.11/Example4_11.sce
@@ -0,0 +1,38 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 4 : OVERHEAD POWER TRANSMISSION
+
+// EXAMPLE : 4.11 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+Z_1 = 10*exp(%i*(30)*%pi/180) ; // Impedance in Ω
+Z_2 = 40*exp(%i*(-45)*%pi/180) ; // Impedance in Ω
+Y_2 = 1/Z_2 ;
+A_1 = 1 ;
+B_1 = Z_1 ;
+C_1 = 0 ;
+D_1 = 1 ;
+A_2 = 1 ;
+B_2 = 0 ;
+C_2 = Y_2 ;
+D_2 = 1 ;
+
+// CALCULATIONS
+P = [A_1 B_1 ; C_1 D_1]; // For network 1
+Q = [A_2 B_2 ; C_2 D_2]; // For network 2
+A_eq = ( A_1*B_2 + A_2*B_1 )/( B_1 + B_2 ) ; // Constant A
+B_eq = ( B_1*B_2 )/(B_1 + B_2) ; // Constant B
+C_eq = C_1 + C_2 + ( (A_1 - A_2) * (D_2 -D_1)/(B_1 + B_2) ) ; // Constant C
+D_eq = ( D_1*B_2 + D_2*B_1 )/(B_1+B_2) ; // Constant D
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 4.11 : SOLUTION :-") ;
+printf("\n Equivalent A , B , C , D constants are \n") ;
+printf("\n A_eq = %.2f<%.f \n",abs(A_eq),atand( imag(A_eq),real(A_eq) )) ;
+printf("\n B_eq = %.2f<%.f \n",abs(B_eq),atand( imag(B_eq),real(B_eq) )) ;
+printf("\n C_eq = %.3f<%.f \n",abs(C_eq),atand( imag(C_eq),real(C_eq) )) ;
+printf("\n D_eq = %.2f<%.f \n",abs(D_eq),atand( imag(D_eq),real(D_eq) )) ;
diff --git a/1820/CH4/EX4.12/Example4_12.sce b/1820/CH4/EX4.12/Example4_12.sce
new file mode 100755
index 000000000..e97d48289
--- /dev/null
+++ b/1820/CH4/EX4.12/Example4_12.sce
@@ -0,0 +1,51 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 4 : OVERHEAD POWER TRANSMISSION
+
+// EXAMPLE : 4.12 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+Z = 2.07 + 0.661 * %i ; // Line impedance in Ω
+V_L = 2.4 * 10^3 ; // Line voltage in V
+p = 200 * 10^3; // Load in VA
+pf = 0.866 ; // Lagging power factor
+
+// CALCULATIONS
+// for case (a)
+A = 1 ;
+B = Z ;
+C = 0 ;
+D = A ;
+theta = acosd(pf) ;
+S_R = p * ( cosd(theta) + %i * sind(theta) ) ; // Receiving end power in VA
+I_L1 = S_R/V_L ;
+I_L = conj(I_L1) ;
+I_S = I_L ; // sending end current in A
+I_R = I_S ; // Receiving end current in A
+
+// for case (b)
+Z_L = V_L/I_L ; // Impedance in Ω
+V_R = Z_L * I_R ;
+V_S = A * V_R + B * I_R ; // sending end voltage in V
+P = [A B ;C D] * [V_R ; I_R] ;
+
+// for case (c)
+V_S = P(1,1) ;
+I_S = P(2,1) ;
+Z_in = V_S/I_S ; // Input impedance in Ω
+
+// for case (d)
+S_S = V_S * conj(I_S) ;
+S_L = S_S - S_R ; // Power loss of line in VA
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 4.12 : SOLUTION :-") ;
+printf("\n (a) Sending-end current , I_S = %.2f<%.2f A \n",abs(I_S),atand( imag(I_S),real(I_S) )) ;
+printf("\n (b) Sending-end voltage , V_S = %.2f<%.2f V \n",abs(V_S),atand( imag(V_S),real(V_S) )) ;
+printf("\n (c) Input impedance , Z_in = %.2f<%.2f Ω \n",abs(Z_in),atand( imag(Z_in),real(Z_in) )) ;
+printf("\n (d) Real power loss in line , S_L = %.2f W \n",real(S_L)) ;
+printf("\n     Reactive power loss in line , S_L = %.2f var \n",imag(S_L)) ;
diff --git a/1820/CH4/EX4.13/Example4_13.sce b/1820/CH4/EX4.13/Example4_13.sce
new file mode 100755
index 000000000..4d710fa45
--- /dev/null
+++ b/1820/CH4/EX4.13/Example4_13.sce
@@ -0,0 +1,49 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 4 : OVERHEAD POWER TRANSMISSION
+
+// EXAMPLE : 4.13 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+KV = 345 ; // Transmission line voltage in kV
+V_R = KV ;
+V_S = KV ; 
+x_L = 0.588 ;// Inductive reactance in Ω/mi/phase 
+b_c = 7.20*10^-6 ;// susceptance S phase to neutral per phase
+l = 200 ;// Total line length in mi
+
+// CALCULATIONS
+// for case (a)
+x_C = 1/b_c ;// Ω/mi/phase
+Z_C = sqrt(x_C * x_L) ;
+SIL = KV^2/Z_C ; // Surge impedance loading in MVA/mi . [1MVA = 1MW]
+SIL1 = (KV^2/Z_C) * l ; // Surge impedance loading of line in MVA . [1MVA = 1MW]
+
+// for case (b)
+delta = 90 ; // Max 3-Φ theoretical steady-state power flow limit occurs for δ = 90 degree
+X_L = x_L * l ; // Inductive reactance Ω/phase 
+P_max = V_S * V_R * sind(delta)/(X_L) ;
+
+// for case (c)
+Q_C = V_S^2 * (b_c * l/2) + V_R^2 *( b_c * l/2) ; // Total 3-Φ magnetizing var in Mvar
+
+// for case (d)
+g = %i * sqrt(x_L/x_C) ; // rad/mi
+g_l = g * l ; // rad
+V_R_oc = V_S / cosh(g_l) ; // Open-circuit receiving-end voltage in kV
+X_C = x_C * 2 / l ;
+V_R_oc1 = V_S * ( - %i * X_C/( - %i * X_C + %i * X_L) ) ; // Alernative method to find Open-circuit receiving-end voltage in kV
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 4.13 : SOLUTION :-") ;
+printf("\n (a) Total 3-Φ SIL of line , SIL = %.2f MVA/mi \n",SIL) ;
+printf("\n     Total 3-Φ SIL of line for total line length , SIL = %.2f MVA \n",SIL1) ;
+printf("\n (b) Maximum 3-Φ theortical steady-state power flow limit , P_max = %.2f MW \n",P_max) ;
+printf("\n (c) Total 3-Φ magnetizing var generation by line capacitance , Q_C = %.2f Mvar \n",Q_C) ;
+printf("\n (d) Open-circuit receiving-end voltage if line is open at receiving end , V_R_oc = %.2f kV \n",V_R_oc) ;
+printf("\n    From alternative method ,") ;
+printf("\n     Open-circuit receiving-end voltage if line is open at receiving end , V_R_oc = %.2f kV \n",V_R_oc1) ;
diff --git a/1820/CH4/EX4.14/Example4_14.sce b/1820/CH4/EX4.14/Example4_14.sce
new file mode 100755
index 000000000..3a10b0ae2
--- /dev/null
+++ b/1820/CH4/EX4.14/Example4_14.sce
@@ -0,0 +1,58 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 4 : OVERHEAD POWER TRANSMISSION
+
+// EXAMPLE : 4.14 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+KV = 345 ; // Transmission line voltage in kV
+V_R = KV ; // Sending end voltage in kV
+x_L = 0.588 ;// Inductive reactance in Ω/mi/phase 
+b_c = 7.20*10^-6 ;// susceptance S phase to neutral per phase
+l = 200 ;// Total line length in mi
+per = 60/100 ; // 2 shunt reactors absorb 60% of total 3-Φ magnetizing var
+cost = 10 ; // cost of each reactor is $10/kVA
+
+// CALCULATIONS
+// For case (a)
+x_C = 1/b_c ;// Ω/mi/phase
+Z_C = sqrt(x_C * x_L) ;
+SIL = KV^2/Z_C ; // Surge impedance loading in MVA/mi
+SIL1 = (KV^2/Z_C) * l ; // Surge impedance loading of line in MVA . [1MVA = 1MW]
+
+// For case (b)
+delta = 90 ; // Max 3-Φ theoretical steady-state power flow limit occurs for δ = 90 degree
+V_S = V_R ; // sending end voltage in kV
+X_L = x_L * l ; // Inductive reactance Ω/phase 
+P_max = V_S * V_R * sind(delta)/(X_L) ;
+
+// For case (c)
+Q_C = V_S^2 * (b_c * l/2) + V_R^2 *( b_c * l/2) ; // Total 3-Φ magnetizing var in Mvar
+Q = (1/2) * per * Q_C ; // 3-Φ megavoltampere rating of each reactor . Q = (1/2)*Q_L
+
+// For case (d)
+Q_L1 = Q * 10^3 ; // Total 3-Φ magnetizing var in Kvar
+T_cost = Q_L1 * cost ; // Cost of each reactor in $
+
+// For case (e)
+g = %i * sqrt(x_L * (1-per)/x_C) ; // rad/mi
+g_l = g * l ; // rad
+V_R_oc = V_S/cosh(g_l) ; // Open circuit receiving-end voltage in kV
+X_L = x_L *l ; 
+X_C = (x_C * 2) / (l * (1 - per)) ;
+V_R_oc1 = V_S * ( -%i*X_C/(-%i*X_C + %i*X_L) ) ; // Alernative method to find Open-circuit receiving-end voltage in kV
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 4.14 : SOLUTION :-") ;
+printf("\n (a) Total 3-phase SIL of line , SIL = %.2f MVA/mi \n",SIL) ;
+printf("\n     Total 3-Φ SIL of line for total line length , SIL = %.2f MVA \n",SIL1) ;
+printf("\n (b) Maximum 3-phase theortical power flow , P_max = %.2f MW \n",P_max) ;
+printf("\n (c) 3-phase MVA rating of each reactor , (1/2)Q_L = %.2f MVA \n",Q) ;
+printf("\n (d) Cost of each reactor at $10/kVA = $ %.2f \n",T_cost) ;
+printf("\n (e) Open circuit receiving voltage , V_Roc= %.2f kV \n",V_R_oc) ;
+printf("\n    From alternative method ,") ;
+printf("\n     Open-circuit receiving-end voltage if line is open at receiving end , V_R_oc = %.2f kV \n",V_R_oc1) ;
diff --git a/1820/CH4/EX4.15/Example4_15.sce b/1820/CH4/EX4.15/Example4_15.sce
new file mode 100755
index 000000000..93ac6eccf
--- /dev/null
+++ b/1820/CH4/EX4.15/Example4_15.sce
@@ -0,0 +1,55 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 4 : OVERHEAD POWER TRANSMISSION
+
+// EXAMPLE : 4.15 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+D_12 = 26 ; // distances in feet
+D_23 = 26 ; // distances in feet
+D_31 = 52 ; // distances in feet
+d = 12 ; // Distance b/w 2 subconductors in inches
+f = 60 ; // frequency in Hz
+kv = 345 ; // voltage base in kv
+p = 100 ; // Power base in MVA
+l = 200 ; // length of line in km
+
+// CALCULATIONS
+// For case (a)
+D_S = 0.0435 ; // from A.3 Appendix A . Geometric mean radius in feet
+D_bS = sqrt(D_S * 0.3048 * d * 0.0254) ; // GMR of bundled conductor in m .[1 ft = 0.3048 m ; 1 inch = 0.0254 m]
+D_eq = (D_12 * D_23 * D_31 * 0.3048^3)^(1/3) ; // Equ GMR in meter
+L_a = 2 * 10^-7 * log(D_eq/D_bS); // Inductance in H/meter
+
+// For case (b)
+X_L = 2 * %pi * f * L_a ; // inductive reactance/phase in ohms/m 
+X_L0 = X_L * 10^3 ; // inductive reactance/phase in ohms/km 
+X_L1 = X_L0 * 1.609 ;// inductive reactance/phase in ohms/mi [1 mi = 1.609 km]
+
+// For case (c)
+Z_B = kv^2 / p ; // Base impedance in Ω
+X_L2 = X_L0 * l/Z_B ; // Series reactance of line in pu
+
+// For case (d)
+r = 1.293*0.3048/(2*12) ; // radius in m . outside diameter is 1.293 inch given in A.3
+D_bsC = sqrt(r * d * 0.0254) ;
+C_n = 55.63 * 10^-12/log(D_eq/D_bsC) ; // capacitance of line in F/m
+
+// For case (e)
+X_C = 1/( 2 * %pi * f * C_n ) ; // capacitive reactance in ohm-m
+X_C0 = X_C * 10^-3 ; // capacitive reactance in ohm-km
+X_C1 = X_C0/1.609 ; // capacitive reactance in ohm-mi
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 4.15 : SOLUTION :-") ;
+printf("\n (a) Average inductance per phase , L_a = %.4e H/m \n",L_a) ;
+printf("\n (b) Inductive reactance per phase , X_L = %.4f Ω/km \n",X_L0) ;
+printf("\n     Inductive reactance per phase , X_L = %.4f Ω/mi \n",X_L1) ;
+printf("\n (c) Series reactance of line , X_L = %.4f pu \n",X_L2) ;
+printf("\n (d) Line-to-neutral capacitance of line , C_n = %.4e F/m \n",C_n);
+printf("\n (e) Capacitive reactance to neutral of line , X_C = %.3e Ω-km \n",X_C0) ;
+printf("\n     Capacitive reactance to neutral of line , X_C = %.3e Ω-mi \n",X_C1) ;
diff --git a/1820/CH4/EX4.2/Example4_2.sce b/1820/CH4/EX4.2/Example4_2.sce
new file mode 100755
index 000000000..e6b80c945
--- /dev/null
+++ b/1820/CH4/EX4.2/Example4_2.sce
@@ -0,0 +1,33 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 4 : OVERHEAD POWER TRANSMISSION
+// EXAMPLE : 4.2 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+// for case (a)
+V_S = 14803 ; // sending end phase voltage at no load in volts . From exa 4.1
+V_R = 13279.056 ; // receiving end phase voltage at full load in volts . From exa 4.1
+
+// for case (b)
+I_R = 265.78785 ; // Line current in amperes . From exa 4.1
+z_t = 2.48+%i*6.57 ; // Total impedance in ohm/phase
+pf = 0.85 ; // power factor
+theta_R = acosd(pf) ;
+
+// CALCULATIONS
+// for case (a)
+V_reg1 = ( (V_S - V_R)/V_R )*100 ; // percentage voltage regulation using equ 4.29
+
+// for case (b)
+V_reg2 = (I_R * ( real(z_t) * cosd(theta_R) + imag(z_t) * sind(theta_R) )/ V_R)*100 ; // percentage voltage regulation using equ 4.31
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 4.2 : SOLUTION :-") ;
+printf("\n (a) Percentage of voltage regulation using equ 4.29 = %.1f \n",V_reg1) ;
+printf("\n (b) Percentage of voltage regulation using equ 4.31 = %.1f \n",V_reg2) ;
+
+printf("\n NOTE : ERROR : The question is with respect to values given in Exa 4.1 not 4.5 \n") ;
diff --git a/1820/CH4/EX4.3/Example4_3.sce b/1820/CH4/EX4.3/Example4_3.sce
new file mode 100755
index 000000000..28ce14e24
--- /dev/null
+++ b/1820/CH4/EX4.3/Example4_3.sce
@@ -0,0 +1,23 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 4 : OVERHEAD POWER TRANSMISSION
+
+// EXAMPLE : 4.3 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+Z_xy = 0.09 + %i*0.3 ; // Mutual impedance between two parallel feeders in Ω/mi per phase
+Z_xx = 0.604*exp(%i*50.4*%pi/180) ; // Self impedance of feeders in Ω/mi per phase
+Z_yy = 0.567*exp(%i*52.9*%pi/180) ; // Self impedance of feeders in Ω/mi per phase
+
+// SOLUTION
+Z_2 = Z_xx - Z_xy ; // mutual impedance between feeders
+Z_4 = Z_yy - Z_xy ; // mutual impedance between feeders
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 4.3 : SOLUTION :-") ;
+printf("\n  Mutual impedance at node 2 , Z_2 = %.3f + j%.3f Ω\n",real(Z_2),imag(Z_2)) ;
+printf("\n  Mutual impedance at node 4 , Z_4 = %.3f + j%.3f Ω\n",real(Z_4),imag(Z_4)) ;
diff --git a/1820/CH4/EX4.4/Example4_4.sce b/1820/CH4/EX4.4/Example4_4.sce
new file mode 100755
index 000000000..87957ca66
--- /dev/null
+++ b/1820/CH4/EX4.4/Example4_4.sce
@@ -0,0 +1,60 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 4 : OVERHEAD POWER TRANSMISSION
+
+// EXAMPLE : 4.4 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+V = 138*10^3 ; // transmission line voltage in V
+P = 49*10^6 ; // load power in Watts
+pf = 0.85 ; // lagging power factor
+Z = 95 * exp(%i*78*%pi/180) ; // line constants in Ω
+Y = 0.001 * exp(%i*90*%pi/180) ; // line constants in siemens
+
+// CALCULATIONS
+V_RL_N = V/sqrt(3) ;
+theta_R = acosd(pf) ; 
+I_R = P/(sqrt(3)*V*pf)*( cosd(theta_R) - %i*sind(theta_R) ) ; // receiving end current in ampere
+
+// for case (a)
+// A,B,C,D constants for nominal-T circuit representation
+A = 1 + (1/2)*Y*Z ; 
+B = Z + (1/4)*Y*Z^2 ;
+C = Y ;
+D = A ;
+
+// for case (b)
+P = [A B ; C D] * [V_RL_N ; I_R] ;
+V_SL_N = P(1,1) ; // Line-to-neutral Sending end voltage in V
+V_SL_L = sqrt(3) * abs(V_SL_N) * exp(%i* ( atand( imag(V_SL_N),real(V_SL_N) ) + 30 )* %pi/180) ; // Line-to-line voltage in V
+// NOTE that an additional 30 degree is added to the angle since line to line voltage is 30 degree ahead of its line to neutral voltage
+
+
+// for case (c)
+I_S = P(2,1); // Sending end current in A
+
+// for case (d)
+theta_s = atand( imag(V_SL_N),real(V_SL_N) ) - atand( imag(I_S),real(I_S) ) ;
+
+// for case (e)
+n = (sqrt(3) * V * abs(I_R) * cosd(theta_R)/(sqrt(3) * abs(I_S) * abs(V_SL_L) * cosd(theta_s) ))*100 ; // Efficiency
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 4.4 : SOLUTION :-") ;
+printf("\n (a) A constant of line , A = %.4f<%.1f \n",abs(A),atand( imag(A),real(A) )) ;
+printf("\n     B constant of line , B = %.2f<%.1f Ω \n",abs(B),atand( imag(B),real(B) )) ;
+printf("\n     C constant of line , C = %.3f<%.1f S \n",abs(C),atand( imag(C),real(C) )) ;
+printf("\n     D constant of line , D = %.4f<%.1f \n",abs(D),atand( imag(D),real(D) )) ;
+printf("\n (b) Sending end line-to-neutral voltage , V_SL_N = %.1f<%.1f V \n",abs(V_SL_N),atand( imag(V_SL_N),real(V_SL_N) )) ;
+printf("\n     Sending end line-to-line voltage , V_SL_L = %.1f<%.1f V \n",abs(V_SL_L),atand( imag(V_SL_L),real(V_SL_L) )) ;
+printf("\n (c) sending end current , I_S = %.2f<%.1f A \n",abs(I_S),atand( imag(I_S),real(I_S) )) ;
+printf("\n (d) sending end power factor , cosθ_s = %.3f \n",cosd(theta_s)) ;
+printf("\n (e) Efficiency of transmission , η = %.2f Percentage \n",n) ;
+
+printf("\n NOTE : From A = 0.9536<0.6 , magnitude is 0.9536 & angle is 0.6 degree") ;
+printf("\n ERROR : Change in answer because root(3) = 1.73 is considered in Textbook ") ;
+printf("\n But here sqrt(3) = 1.7320508 is considered \n") ;
diff --git a/1820/CH4/EX4.5/Example4_5.sce b/1820/CH4/EX4.5/Example4_5.sce
new file mode 100755
index 000000000..14c4be030
--- /dev/null
+++ b/1820/CH4/EX4.5/Example4_5.sce
@@ -0,0 +1,59 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 4 : OVERHEAD POWER TRANSMISSION
+
+// EXAMPLE : 4.5 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+V = 138*10^3 ; // Transmission line voltage in V
+P = 49*10^6 ; // load power in Watts
+pf = 0.85 ; // lagging power factor
+Z = 95 * exp(%i*78*%pi/180) ; // line constants in Ω
+Y = 0.001 * exp(%i*90*%pi/180) ; // line constants in siemens
+
+// CALCULATIONS
+V_RL_N = V/sqrt(3) ;
+theta_R = acosd(pf) ;
+I_R = P/(sqrt(3)*V*pf) * ( cosd(theta_R) - %i*sind(theta_R) ) ; // Receiving end current in A
+
+// for case (a)
+// A,B,C,D constants for nominal-π circuit representation
+A = 1 + (1/2)*Y*Z ;
+B = Z ;
+C = Y + (1/4)*(Y^2)*Z ;
+D = 1 + (1/2)*Y*Z ;
+
+// for case (b)
+P = [A B ; C D] * [V_RL_N ; I_R] ;
+V_SL_N = P(1,1) ; // Line-to-neutral Sending end voltage in V
+V_SL_L = sqrt(3) * abs(V_SL_N) * exp(%i* ( atand( imag(V_SL_N),real(V_SL_N) ) + 30 )* %pi/180) ; // Line-to-line voltage in V
+// NOTE that an additional 30 degree is added to the angle since line-to-line voltage is 30 degree ahead of its line-to-neutral voltage
+
+
+// for case (c)
+I_S = P(2,1); // Sending end current in A
+
+// for case (d)
+theta_s = atand( imag(V_SL_N),real(V_SL_N) ) - atand( imag(I_S),real(I_S) ) ;
+
+// for case (e)
+n = (sqrt(3) * V * abs(I_R) * cosd(theta_R)/(sqrt(3) * abs(I_S) * abs(V_SL_L) * cosd(theta_s) ))*100 ; // Efficiency
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 4.5 : SOLUTION :-") ;
+printf("\n (a) A constant of line , A = %.4f<%.1f \n",abs(A),atand( imag(A),real(A) )) ;
+printf("\n     B constant of line , B = %.2f<%.1f Ω \n",abs(B),atand( imag(B),real(B) )) ;
+printf("\n     C constant of line , C = %.3f<%.1f S \n",abs(C),atand( imag(C),real(C) )) ;
+printf("\n     D constant of line , D = %.4f<%.1f \n",abs(D),atand( imag(D),real(D) )) ;
+printf("\n (b) Sending end line-to-neutral voltage , V_SL_N = %.1f<%.1f V \n",abs(V_SL_N),atand( imag(V_SL_N),real(V_SL_N) )) ;
+printf("\n     Sending end line-to-line voltage , V_SL_L = %.1f<%.1f V \n",abs(V_SL_L),atand( imag(V_SL_L),real(V_SL_L) )) ;
+printf("\n (c) sending end current , I_S = %.2f<%.1f A \n",abs(I_S),atand( imag(I_S),real(I_S) )) ;
+printf("\n (d) sending end power factor , cosθ_s = %.3f \n",cosd(theta_s)) ;
+printf("\n (e) Efficiency of transmission , η = %.2f Percentage \n",n) ;
+
+printf("\n NOTE : ERROR : Change in answer because root(3) = 1.73 is considered in Textbook ") ;
+printf("\n But here sqrt(3) = 1.7320508 is considered \n") ;
diff --git a/1820/CH4/EX4.6/Example4_6.sce b/1820/CH4/EX4.6/Example4_6.sce
new file mode 100755
index 000000000..9cb1d61fd
--- /dev/null
+++ b/1820/CH4/EX4.6/Example4_6.sce
@@ -0,0 +1,91 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 4 : OVERHEAD POWER TRANSMISSION
+
+// EXAMPLE : 4.6 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+V_RL_L = 138*10^3 ; // transmission line voltage in V
+R = 0.1858 // Line constant in Ω/mi
+f = 60 // frequency in Hertz
+L = 2.60*10^-3 // Line constant in H/mi
+C = 0.012*10^-6 // Line constant in F/mi
+pf = 0.85 // Lagging power factor
+P = 50*10^6 // load in VA
+l = 150 // length of 3-Φ transmission line in mi
+
+// CALCULATIONS
+z = R + %i*2*%pi*f*L ; // Impedance per unit length in Ω/mi
+y = %i*2*%pi*C*f ; // Admittance per unit length in S/mi
+g = sqrt(y*z) ; // Propagation constant of line per unit length
+g_l = real(g) * l + %i * imag(g) * l ; // Propagation constant of line
+Z_c = sqrt(z/y) ; // Characteristic impedance of line
+V_RL_N = V_RL_L/sqrt(3) ;
+theta_R = acosd(pf) ; 
+I_R = P/(sqrt(3)*V_RL_L)*( cosd(theta_R) - %i*sind(theta_R) ) ; // Receiving end current in A
+
+// for case (a)
+// A,B,C,D constants of line
+A = cosh(g_l) ;
+B = Z_c * sinh(g_l) ;
+C = (1/Z_c) * sinh(g_l) ;
+D = A ;
+
+// for case (b)
+P = [A B ; C D] * [V_RL_N ; I_R] ;
+V_SL_N = P(1,1) ; // Line-to-neutral Sending end voltage in V
+V_SL_L = sqrt(3) * abs(V_SL_N) * exp(%i* ( atand( imag(V_SL_N),real(V_SL_N) ) + 30 )* %pi/180) ; // Line-to-line voltage in V
+// NOTE that an additional 30 degree is added to the angle since line-to-line voltage is 30 degree ahead of its line-to-neutral voltage
+
+// for case (c)
+I_S = P(2,1); // Sending end current in A
+
+// for case (d)
+theta_s = atand( imag(V_SL_N),real(V_SL_N) ) - atand( imag(I_S),real(I_S) ) ; // Sending-end pf
+
+// For case (e)
+P_S = sqrt(3) * abs(V_SL_L) * abs(I_S) * cosd(theta_s) ; // Sending end power
+
+// For case (f)
+P_R = sqrt(3)*abs(V_RL_L)*abs(I_R)*cosd(theta_R) ; // Receiving end power
+P_L = P_S - P_R ; // Power loss in line
+
+// For case (g)
+n = (P_R/P_S)*100 ; // Transmission line efficiency
+
+// For case (h)
+reg = (( abs(V_SL_N) - V_RL_N )/V_RL_N )*100 ; // Percentage of voltage regulation
+
+// For case (i)
+Y = y * l ; // unit is S
+I_C = (1/2) * Y * V_SL_N ; // Sending end charging current in A
+
+// For case (j)
+Z = z * l ;
+V_RL_N0 = V_SL_N - I_C*Z ;
+V_RL_L0 = sqrt(3) * abs(V_RL_N0) * exp(%i* ( atand( imag(V_RL_N0),real(V_RL_N0) ) + 30 )* %pi/180) ; // Line-to-line voltage at receiving end in V
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 4.6 :SOLUTION :-") ;
+printf("\n (a) A constant of line , A = %.4f<%.2f \n",abs(A),atand( imag(A),real(A) )) ;
+printf("\n     B constant of line , B = %.2f<%.2f Ω \n",abs(B),atand( imag(B),real(B) )) ;
+printf("\n     C constant of line , C = %.5f<%.2f S \n",abs(C),atand( imag(C),real(C) )) ;
+printf("\n     D constant of line , D = %.4f<%.2f \n",abs(D),atand( imag(D),real(D) )) ;
+printf("\n (b) Sending end line-to-neutral voltage , V_SL_N = %.2f<%.2f V \n",abs(V_SL_N),atand( imag(V_SL_N),real(V_SL_N) )) ;
+printf("\n     Sending end line-to-line voltage , V_SL_L = %.2f<%.2f V \n",abs(V_SL_L),atand( imag(V_SL_L),real(V_SL_L) )) ;
+printf("\n (c) sending-end current , I_S = %.2f<%.2f A \n",abs(I_S),atand( imag(I_S),real(I_S) )) ;
+printf("\n (d) sending-end power factor , cosθ_s = %.4f \n",cosd(theta_s)) ;
+printf("\n (e) sending-end power , P_S = %.5e W \n",P_S) ;
+printf("\n (f) Power loss in line , P_L = %.5e W \n",P_L) ;
+printf("\n (g) Transmission line Efficiency , η = %.1f Percentage\n",n) ;
+printf("\n (h) Percentage of voltage regulation = %.1f Percentage \n",reg) ;
+printf("\n (i) Sending-end charging current at no load , I_C = %.2f A \n",abs(I_C)) ;
+printf("\n (j) Receiving-end voltage rise at no load ,V_RL_N = %.2f<%.2f V \n",abs(V_RL_N0),atand(imag(V_RL_N0),real(V_RL_N0)));
+printf("\n     Line-to-line voltage at receiving end at no load ,V_RL_L = %.2f<%.2f V \n",abs(V_RL_L0),atand(imag(V_RL_L0),real(V_RL_L0)));
+
+printf("\n NOTE : ERROR : Change in answer because root(3) = 1.73 is considered in Textbook & change in α & β values ") ;
+printf("\n But here sqrt(3) = 1.7320508 is considered \n") ;
diff --git a/1820/CH4/EX4.7/Example4_7.sce b/1820/CH4/EX4.7/Example4_7.sce
new file mode 100755
index 000000000..c4ab75628
--- /dev/null
+++ b/1820/CH4/EX4.7/Example4_7.sce
@@ -0,0 +1,50 @@
+
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 4 : OVERHEAD POWER TRANSMISSION
+
+// EXAMPLE : 4.7 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+R = 0.1858 // Line constant in Ω/mi
+f = 60 // frequency in Hertz
+L = 2.60*10^-3 // Line constant in H/mi
+C = 0.012*10^-6 // Line constant in F/mi
+l = 150 // length of 3-Φ transmission line in mi
+
+// CALCULATIONS
+z = R + %i*2*%pi*f*L ; // Impedance per unit length in Ω/mi
+y = %i*2*%pi*C*f ; // Admittance per unit length in S/mi
+g = sqrt(y*z) ; // Propagation constant of line per unit length
+g_l = real(g) * l + %i * imag(g) * l ; // Propagation constant of line
+Z_c = sqrt(z/y) ; // Characteristic impedance of line
+
+A = cosh(g_l) ;
+B = Z_c * sinh(g_l) ;
+C = (1/Z_c) * sinh(g_l) ;
+D = A ;
+Z_pi = B ;
+Y_pi_by2 = (A-1)/B ; // Unit in Siemens
+Z = l * z ; // unit in ohms
+Y = y * l ;
+Y_T = C ;
+Z_T_by2 = (A-1)/C ; // Unit in Ω
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 4.7 : SOLUTION :-") ;
+printf("\n FOR EQUIVALENT-π CIRCUIT ") ;
+printf("\n   Z_π = B = %.2f<%.2f Ω \n",abs(Z_pi),atand( imag(Z_pi),real(Z_pi) )) ;
+printf("\n   Y_π/2 = %.6f<%.2f S \n",abs(Y_pi_by2),atand( imag(Y_pi_by2),real(Y_pi_by2) )) ;
+printf("\n FOR NOMINAL-π CIRCUIT ") ;
+printf("\n   Z = %.3f<%.2f Ω \n",abs(Z),atand( imag(Z),real(Z) )) ;
+printf("\n   Y/2 = %.6f<%.1f S \n",abs(Y/2),atand( imag(Y/2),real(Y/2) )) ;
+printf("\n FOR EQUIVALENT-T CIRCUIT ") ;
+printf("\n   Z_T/2 = %.2f<%.2f Ω \n",abs(Z_T_by2),atand( imag(Z_T_by2),real(Z_T_by2) )) ;
+printf("\n   Y_T = C = %.5f<%.2f S \n",abs(Y_T),atand( imag(Y_T),real(Y_T) )) ;
+printf("\n FOR NOMINAL-T CIRCUIT ") ;
+printf("\n   Z/2 = %.2f<%.2f Ω \n",abs(Z/2),atand( imag(Z/2),real(Z/2) )) ;
+printf("\n   Y = %.6f<%.1f S \n",abs(Y),atand( imag(Y),real(Y) )) ;
diff --git a/1820/CH4/EX4.8/Example4_8.sce b/1820/CH4/EX4.8/Example4_8.sce
new file mode 100755
index 000000000..1e2071f42
--- /dev/null
+++ b/1820/CH4/EX4.8/Example4_8.sce
@@ -0,0 +1,65 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 4 : OVERHEAD POWER TRANSMISSION
+
+// EXAMPLE : 4.8 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+V_RL_L = 138*10^3 ; // transmission line voltage in V
+R = 0.1858 // Line constant in Ω/mi
+f = 60 // frequency in Hertz
+L = 2.60*10^-3 // Line constant in H/mi
+C = 0.012*10^-6 // Line constant in F/mi
+pf = 0.85 // Lagging power factor
+P = 50*10^6 // load in VA
+l = 150 // length of 3-Φ transmission line in mi
+
+// CALCULATIONS
+// For case (a)
+z = R + %i*2*%pi*f*L ; // Impedance per unit length in Ω/mi
+y = %i*2*%pi*C*f ; // Admittance per unit length in S/mi
+g = sqrt(y*z) ; // Propagation constant of line per unit length
+
+// For case (b)
+lamda = (2 * %pi)/imag(g) ; // Wavelength of propagation in mi
+V = lamda * f ; // Velocity of propagation in mi/sec
+
+ // For case (c)
+Z_C = sqrt(z/y) ;
+V_R = V_RL_L/sqrt(3) ;
+theta_R = acosd(pf) ; 
+I_R = P/(sqrt(3)*V_RL_L) * ( cosd(theta_R) - %i*sind(theta_R) ) ; // Receiving end current in A
+V_R_incident = (1/2)*(V_R + I_R*Z_C) ; // Incident voltage at receiving end in V
+V_R_reflected = (1/2)*(V_R - I_R*Z_C) ; // Reflected voltage at receiving end in V
+
+// For case (d)
+V_RL_N = V_R_incident + V_R_reflected ; // Line-to-neutral voltage at receiving end in V
+V_RL_L = sqrt(3)*V_RL_N // Receiving end Line voltage in V
+
+// For case (e)
+g_l = real(g) * l + %i * imag(g) * l ; // Propagation constant of line
+a = real(g) ; // a = α is the attenuation constant
+b = imag(g) ; // b = β is the phase constant
+V_S_incident = (1/2) * (V_R+I_R*Z_C) * exp(a*l) * exp(%i*b*l) ; // Incident voltage at sending end in V
+V_S_reflected = (1/2) * (V_R-I_R*Z_C) * exp(-a*l) * exp(%i*(-b)*l) ; // Reflected voltage at sending end in V
+
+// For case (f)
+V_SL_N = V_S_incident + V_S_reflected ; // Line-to-neutral voltage at sending end in V
+V_SL_L = sqrt(3)*V_SL_N ; // sending end Line voltage in V
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 4.8 : SOLUTION :-") ;
+printf("\n (a) Attenuation constant , α = %.4f Np/mi \n",real(g)) ;
+printf("\n     Phase change constant, β = %.4f rad/mi \n",imag(g)) ;
+printf("\n (b) Wavelength of propagation = %.2f mi \n",lamda) ;
+printf("\n     velocity of propagation = %.2f mi/s \n",V) ;
+printf("\n (c) Incident voltage receiving end , V_R(incident) = %.2f<%.2f V \n",abs(V_R_incident),atan(imag(V_R_incident),real(V_R_incident))*(180/%pi));
+printf("\n     Receiving end reflected voltage , V_R(reflected) = %.2f<%.2f V \n",abs(V_R_reflected),atan(imag(V_R_reflected),real(V_R_reflected))*(180/%pi)) ;
+printf("\n (d) Line voltage at receiving end , V_RL_L = %d V \n",V_RL_L) ;
+printf("\n (e) Incident voltage at sending end , V_S(incident) = %.2f<%.2f V \n",abs(V_S_incident),atan(imag(V_S_incident),real(V_S_incident))*(180/%pi)) ;
+printf("\n     Reflected voltage at sending end , V_S(reflected) = %.2f<%.2f V \n",abs(V_S_reflected),atan(imag(V_S_reflected),real(V_S_reflected))*(180/%pi)) ;
+printf("\n (f) Line voltage at sending end , V_SL_L = %.2f V \n",abs(V_SL_L)) ;
diff --git a/1820/CH4/EX4.9/Example4_9.sce b/1820/CH4/EX4.9/Example4_9.sce
new file mode 100755
index 000000000..c30e58801
--- /dev/null
+++ b/1820/CH4/EX4.9/Example4_9.sce
@@ -0,0 +1,26 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 4 : OVERHEAD POWER TRANSMISSION
+
+// EXAMPLE : 4.9 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+L = 2.60 * 10^-3 ; // Inductance of line in H/mi
+R = 0.1858 ; // Resistance of line in Ω/mi
+C = 0.012 * 10^-6 ; // Capacitance in F/mi
+kV = 138 ; // Transmission line voltage in kV
+Z_c1 = 469.60085 // Characteristic impedance of line in Ω . Obtained from example 4.6
+
+// CALCULATIONS
+Z_c = sqrt(L/C) ; // Approximate value of surge Impedance of line in ohm
+SIL =  kV^2/Z_c ; // Approximate Surge impedance loading in MW
+SIL1 = kV^2/Z_c1 ; // Exact value of SIL in MW
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 4.9 : SOLUTION :-") ;
+printf("\n Approximate value of SIL of transmission line , SIL_app = %.3f MW\n",SIL) ;
+printf("\n Exact value of SIL of transmission line , SIL_exact = %.3f MW\n",SIL1) ;
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") ;
diff --git a/1820/CH6/EX6.1/Example6_1.sce b/1820/CH6/EX6.1/Example6_1.sce
new file mode 100755
index 000000000..5d557a2a8
--- /dev/null
+++ b/1820/CH6/EX6.1/Example6_1.sce
@@ -0,0 +1,39 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.1 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+K_1 = 2.5 ; // Factor
+K_2 = 1.7 ; // Factor
+
+// CALCULATIONS
+// For case (b)
+I_d = poly(0,'I_d') ; // since P_loss(dc) = P_loss (ac) 
+I_L = poly(0,'I_L') ; // i.e 2*I_d^2*R_dc = 3*I_L^2*R_ac 
+I_d = sqrt(3/2)*I_L ; // Ignoring skin effects R_dc = R_ac
+I_d1 = 1.225*I_L ; // Refer Equ 6.23
+
+// For case (a)
+V_d = poly(0,'V_d') ; // Defining a ploynomial V_d
+E_p = poly(0,'E_p') ; // since P_dc = P_ac (or) V_d*I_d = 3*E_p*I_L
+V_d = 2.45*E_p ; // Refer Equ 6.25
+
+// For case (c)
+ins_lvl = (K_2*(V_d/2))/(K_1*E_p) ; // Ratio of dc insulation level to ac insulation level
+ins_lvl_1 = (K_2*2.45/2)/K_1 ; // simplifying above equ
+dc_i = poly(0,'dc_i') ; // dc_i = dc insulation level
+ac_i = poly(0,'ac_i') ; // ac_i = ac insulation level
+dc_i = ins_lvl_1 * ac_i ;
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.1 : SOLUTION :-") ;
+printf("\n (a) Line-to-line dc voltage of V_d in terms of line-to-neutral voltage E_p , V_d = \n") ; disp(V_d) ;
+printf("\n (b) The dc line current I_d in terms of ac line current I_L , I_d = \n"); disp(I_d1) ;
+printf("\n (c) Ratio of dc insulation level to ac insulation level =  \n") ; disp(dc_i/ac_i) ;
+printf("\n  (or) dc insulation level = \n") ; disp(dc_i) ;
diff --git a/1820/CH6/EX6.10/Example6_10.sce b/1820/CH6/EX6.10/Example6_10.sce
new file mode 100755
index 000000000..c53593ff8
--- /dev/null
+++ b/1820/CH6/EX6.10/Example6_10.sce
@@ -0,0 +1,64 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT-CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.10 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+X_C = 6.2292 ; // commutating reactance when all 3 breakers are closed
+I_db = 1600 ; // dc current base in A
+V_db = 125 * 10^3 ; // dc voltage base in V
+I_d = I_db ; // Max continuous current in A
+V_d = 100 * 10^3 ; // dc voltage in V
+alpha = 0 ; // Firing angle α = 0 degree
+
+// CALCULATIONS
+// For case (a)
+R_c = (3/%pi) * X_C ;
+R_cb = V_db/I_db ; // Resistance base in Ω
+V_d_pu = V_d/V_db ; // per unit voltage
+I_d_pu = I_d/I_db ; // per unit current
+R_c_pu = R_c/R_cb ; // per unit Ω
+E_pu = (V_d_pu + R_c_pu * I_d_pu)/cosd(alpha) ; // Open ckt dc voltage in pu
+V_d0 = E_pu * V_db ; // Open ckt dc voltage in V
+
+// For case (b)
+E = V_d0/2.34; // Open ckt ac voltage on wye side of transformer in V               
+
+// For case (c)
+E_1LN = 92.95 * 10^3 ; // voltage in V
+E_1B = E_1LN ;
+E_LN = 53.44 * 10^3 ; // voltage in V
+a = E_1LN/E_LN ;
+n = a ; // when LTC on neutral
+X_c_pu = 2 * R_c_pu ;
+E_1_pu = E_1LN / E_1B ; // per unit voltage
+cos_delta = cosd(alpha) - ( (X_c_pu * I_d_pu)/( (a/n) *E_1_pu) ) ;
+delta = acosd(cos_delta) ;
+u = delta - alpha ; 
+
+// For case (d)
+cos_theta = V_d/V_d0 ; // pf of rectifier
+theta = acosd(cos_theta) ;
+
+// For case (e)
+Q_r = V_d*I_d*tand(theta) ; // magnetizing var I/P
+
+// For case (f)
+d_V = E_LN - E ; // necessary change in voltage in V
+p_E_LN = 0.00625 * E_LN ; // one buck step can change in V/step
+no_buck = d_V / p_E_LN ; // No. of steps of buck
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.10 : SOLUTION :-") ;
+printf("\n (a) Open circuit dc Voltage , V_d0 = %.2f V \n",V_d0);
+printf("\n (b) Open circuit ac voltage on wye side of transformer , E = %.2f V \n",E);
+printf("\n (c) Overlap angle , u = %.2f degree \n",u)
+printf("\n (d) Power factor , cosθ = %.3f  \n",cos_theta);
+printf("\n     and   θ = %.2f degree \n ",theta);
+printf("\n (e) Magnetizing var input to rectifier , Q_r = %.4e var \n",Q_r);
+printf("\n (f) Number of 0.625 percent steps of buck required , No. of buck = %.f steps \n",no_buck);
diff --git a/1820/CH6/EX6.2/Example6_2.sce b/1820/CH6/EX6.2/Example6_2.sce
new file mode 100755
index 000000000..a6f4a7fbe
--- /dev/null
+++ b/1820/CH6/EX6.2/Example6_2.sce
@@ -0,0 +1,36 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.2 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+K = 3 ; // factor
+
+// CALCULATIONS
+// For case (a)
+V_d = poly(0,'V_d') ; // defining a polynomial
+E_p = poly(0,'E_p') ;
+V_d = K*2*E_p ; // From equ 6.18
+
+// For case (b)
+P_dc = poly(0,'P_dc') ;
+P_ac = poly(0,'P_ac') ;
+P_dc = 2*P_ac ;
+
+// For case (c)
+P_ld = poly(0,'P_ld') ; // P_loss(dc)
+P_la = poly(0,'P_la') ; // P_loss(ac)
+P_ld = (2/3)*P_la ; 
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.2 : SOLUTION :-") ;
+printf("\n (a) Maximum operating V_d in terms of voltage E_p , V_d = \n") ; disp(V_d) ;
+printf("\n (b) Maximum power transmission capability ratio,i.e,ratio of P_dc to P_ac , P_dc/P_ac = \n") ; disp(P_dc/P_ac) ;
+printf("\n (or) P_dc = \n") ; disp(P_dc) ;
+printf("\n (c) Ratio of total I^2*R losses , i.e ,Ratio of P_loss(dc) to P_loss(ac),which accompany maximum power flow = \n") ; disp(P_ld/P_la) ;
+printf("\n (or) P_loss(dc) = \n") ; disp(P_ld) ;
diff --git a/1820/CH6/EX6.3/Example6_3.sce b/1820/CH6/EX6.3/Example6_3.sce
new file mode 100755
index 000000000..1e22f6b1c
--- /dev/null
+++ b/1820/CH6/EX6.3/Example6_3.sce
@@ -0,0 +1,28 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT-CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.3 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+V_d0 = 125 ; // voltage rating of bridge rectifier in kV
+V_dr0 = V_d0 ; // Max continuos no-load direct voltage in kV
+I = 1600 ; // current rating of bridge rectifier in A
+I_d = I ; // Max continuous current in A
+
+// CALCULATIONS
+// For case (a)
+S_B = 1.047 * V_d0 * I_d ; // 3-phase kVA rating of rectifier transformer
+
+// For case (b)
+// SINCE   V_d0 = 2.34*E_LN
+E_LN = V_d0/2.34 ; // Wye side kV rating
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.3 : SOLUTION :-") ;
+printf("\n (a) Three-phase kilovolt-ampere rating , S_B = %d kVA \n",S_B) ;
+printf("\n (b) Wye-side kilovolt rating , E_L-N = %.4f kV \n",E_LN) ;
diff --git a/1820/CH6/EX6.4/Example6_4.sce b/1820/CH6/EX6.4/Example6_4.sce
new file mode 100755
index 000000000..72b982f6c
--- /dev/null
+++ b/1820/CH6/EX6.4/Example6_4.sce
@@ -0,0 +1,49 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT-CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.4 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+E_LN = 53.418803 ; // Wye-side kV rating . From exa 6.3
+I = 1600 ; // current rating of bridge rectifier in A
+I_d = I ; // Max continuous current in A
+X_tr = 0.10 ; // impedance of rectifier transformer in pu Ω
+
+// For case (a)
+sc_MVA1 = 4000 ; // short-ckt MVA
+
+// For case (b)
+sc_MVA2 = 2500 ; // short-ckt MVA
+
+// For case (c)
+sc_MVA3 = 1000 ; // short-ckt MVA
+
+// CALCULATIONS
+nom_kV = sqrt(3) * E_LN ; // Nominal kV_L-L
+I_1ph = sqrt(2/3) * I_d ; // rms value of wye-side phase current
+E_LN1 = E_LN * 10^3 ; // Wye-side rating in kV
+X_B = (E_LN1/I_1ph) ; // Associated reactance base in Ω
+
+// For case (a)
+X_sys1 = nom_kV^2/sc_MVA1 ; // system reactance in Ω
+X_tra = X_tr * X_B ; // Reactance of rectifier transformer
+X_C = X_sys1 + X_tra ; // Commutating reactance in Ω
+
+// For case (b)
+X_sys2 = nom_kV^2/sc_MVA2 ; // system reactance in Ω
+X_C2 = X_sys2 + X_tra ; // Commutating reactance in Ω
+
+// For case (b) When breaker 1 & 2 are open
+X_sys3 = nom_kV^2/sc_MVA3 ; // system reactance in Ω
+X_C3 = X_sys3 + X_tra ; // Commutating reactance in Ω
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.4 : SOLUTION :-") ;
+printf("\n (a) Commutating reactance When all three breakers are closed, X_C = %.4f Ω \n",X_C) ; 
+printf("\n (b) Commutating reactance When breaker 1 is open, X_C = %.4f Ω \n",X_C2) ;
+printf("\n (c) Commutating reactance When breakers 1 and 2 are open, X_C = %.4f Ω \n",X_C3) ;
diff --git a/1820/CH6/EX6.5/Example6_5.sce b/1820/CH6/EX6.5/Example6_5.sce
new file mode 100755
index 000000000..919af5d8a
--- /dev/null
+++ b/1820/CH6/EX6.5/Example6_5.sce
@@ -0,0 +1,45 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT-CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.5 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+X_C = 6.2292017 ; // commutating reactance when all 3 breakers are closed
+E_LN = 53.418803 * 10^3 ; // Wye-side volt rating
+V_d0 = 125 * 10^3 ; // voltage rating of bridge rectifier in V
+V_dr0 = V_d0 ; // Max continuos no-load direct voltage in V
+I = 1600 ; // current rating of bridge rectifier in A
+I_d = I ; // Max continuous current
+nom_kV = sqrt(3) * E_LN ; // Nominal kV_L-L
+X_tr = 0.10 ; //impedance of rectifier transformer in pu Ω
+alpha = 0 ; // delay angle α = 0 degree
+
+// CALCULATIONS
+// For case (a)
+E_m = sqrt(2) * E_LN ;
+u = acosd(1 - (2*X_C*I_d)/(sqrt(3)*E_m)); // overlap angle when delay angle α = 0 degree
+
+// For case (b)
+R_C = (3/%pi) * X_C ; // Equ commutation resistance per phase
+V_d = V_d0 * cosd(alpha) - R_C * I_d ; // dc voltage of rectifier in V
+
+// For case (c)
+cos_theta = V_d/V_d0 ; // Displacement or power factor of rectifier
+
+// For case (d)
+Q_r = V_d * I_d * tand( acosd(cos_theta) ) ; // magnetizing var I/P
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.5 : SOLUTION :-") ;
+printf("\n (a) Overlap angle u of rectifier, u = %.2f degree\n",u) ;
+printf("\n (b) The dc voltage V_dr of rectifier, V_dr = %.2f V \n",V_d) ;
+printf("\n (c) Displacement factor of rectifier, cosθ = %.3f  \n",cos_theta) ;
+printf("\n     and   θ = %.1f degree \n ",acosd(cos_theta)) ;
+printf("\n (d) Magnetizing var input to rectifier, Q_r = %.4e var \n",Q_r) ;
+
+printf("\n NOTE : In case(d) 7.6546e+07 var is same as 7.6546*10^7 var = 76.546 Mvar \n") ;
diff --git a/1820/CH6/EX6.6/Example6_6.sce b/1820/CH6/EX6.6/Example6_6.sce
new file mode 100755
index 000000000..179317862
--- /dev/null
+++ b/1820/CH6/EX6.6/Example6_6.sce
@@ -0,0 +1,42 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT-CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.6 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+I_d = 1600 ; // Max continuous dc current in A
+V_d0 = 125 * 10^3 ; // voltage rating of bridge rectifier in V
+V_d = 100 * 10^3 ; // dc voltage of rectifier in V
+X_C = 6.2292017 ; // commutating reactance when all 3 breakers are closed
+
+// CALCULATIONS
+// For case (a)
+R_C = (3/%pi) * X_C ;
+cos_alpha = (V_d + R_C*I_d)/V_d0 ; // Firing angle α
+alpha = acosd(cos_alpha) ;
+
+// For case (b)
+// V_d = (1/2)*V_d0*(cos_alpha + cos_delta)
+cos_delta = (2 * V_d/V_d0) - cos_alpha ;
+delta = acosd(cos_delta) ;
+u = delta - alpha ; // Overlap angle u in degree
+
+// For case (c)
+cos_theta = V_d/V_d0 ; // power factor
+theta = acosd(cos_theta) ;
+
+// For case (d)
+Q_r = V_d * I_d * tand(theta) ; // magnetizing var I/P
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.6 : SOLUTION :-") ;
+printf("\n (a) Firing angle α of rectifier, α = %.2f degree\n",alpha) ;
+printf("\n (b) Overlap angle u of rectifier, u = %.2f degree\n",u) ;
+printf("\n (c) Power factor , cosθ = %.2f  \n",cos_theta) ;
+printf("\n     and   θ = %.2f degree \n ",theta) ;
+printf("\n (d) Magnetizing var input , Q_r = %.2e var \n",Q_r) ;
diff --git a/1820/CH6/EX6.7/Example6_7.sce b/1820/CH6/EX6.7/Example6_7.sce
new file mode 100755
index 000000000..6200c30ca
--- /dev/null
+++ b/1820/CH6/EX6.7/Example6_7.sce
@@ -0,0 +1,32 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT-CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.7 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+X_C = 12.649731 ; // commutating reactance when 2 breakers are open
+alpha = 0 ;
+I_d = 1600 ; // DC current in A
+E_LN = 53.4188 * 10^3 ; // Wye-side rating in V
+V_d0 = 125 * 10^3 ; // voltage rating of bridge rectifier in V
+
+// CALCULATIONS
+// For case (a)
+E_m = sqrt(2) * E_LN ;
+u = acosd(1 - (2 * X_C * I_d)/(sqrt(3) * E_m)) ; // overlap angle u = δ
+
+// For case (b)
+// since rectifier operates in first mode i.e doesn't operate in second mode
+R_C = (3/%pi) * X_C ;
+V_dr = ( V_d0 * cosd(alpha) ) - (R_C*I_d) ; // dc voltage of rectifier in V
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.7 : SOLUTION :-") ;
+printf("\n (a) u = %.1f degree \n",u) ;
+printf("\n  since u < 60 degree . The rectifier operates at FIRST mode , the normal operating mode \n") ;
+printf("\n (b) When dc current is 1600 A , V_dr = %.2f V \n",V_dr) ;
diff --git a/1820/CH7/EX7.1/Example7_1.sce b/1820/CH7/EX7.1/Example7_1.sce
new file mode 100755
index 000000000..18680cf68
--- /dev/null
+++ b/1820/CH7/EX7.1/Example7_1.sce
@@ -0,0 +1,26 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 7 : TRANSIENT OVERVOLTAGES AND INSULATION COORDINATION
+
+// EXAMPLE : 7.1 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+V = 1000 ; // surge voltage in kV
+Z_c = 500 ; // surge impedance in Ω
+
+// CALCULATIONS
+// For case (a)
+P = V^2/Z_c ; // Total surge power in MW
+
+// For case (b)
+V1 = V*10^3 ; // surge voltage in V
+i = V1/Z_c ;// surge current in A
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 7.1 : SOLUTION :-") ;
+printf("\n (a) Total surge power in line , P = %d MW \n",P) ;
+printf("\n (b) Surge current in line , i = %d A \n",i) ;
diff --git a/1820/CH7/EX7.2/Example7_2.sce b/1820/CH7/EX7.2/Example7_2.sce
new file mode 100755
index 000000000..79436652e
--- /dev/null
+++ b/1820/CH7/EX7.2/Example7_2.sce
@@ -0,0 +1,26 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 7 : TRANSIENT OVERVOLTAGES AND INSULATION COORDINATION
+
+// EXAMPLE : 7.2 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+V = 1000 ; // surge voltage in kV
+Z_c = 50 ; // surge impedance in Ω
+
+// CALCULATIONS
+// For case (a)
+P = V^2/Z_c ; // Total surge power in MW
+
+// For case (b)
+V1 = V*10^3 ; // surge voltage in V
+i = V1/Z_c ;// surge current in A
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 7.1 : SOLUTION :-") ;
+printf("\n (a) Total surge power in line , P = %d MW \n",P) ;
+printf("\n (b) Surge current in line , i = %d A \n",i) ;
diff --git a/1820/CH7/EX7.4/Example7_4.sce b/1820/CH7/EX7.4/Example7_4.sce
new file mode 100755
index 000000000..3e973bc83
--- /dev/null
+++ b/1820/CH7/EX7.4/Example7_4.sce
@@ -0,0 +1,56 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 7 : TRANSIENT OVERVOLTAGES AND INSULATION COORDINATION
+
+// EXAMPLE : 7.4 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+R = 500 ; // Resistance in Ω
+Z_c = 400 ; // characteristic impedance in Ω
+v_f = 5000 ; // Forward travelling voltage wave in V
+i_f = 12.5 ; // Forward travelling current wave in A
+
+// CALCULATIONS
+// For case (a)
+r_v = (R - Z_c)/(R + Z_c) ; // Reflection coefficient of voltage wave
+
+// For case (b)
+r_i = -(R - Z_c)/(R + Z_c) ; // Reflection coefficient of current wave
+
+// For case (c)
+v_b = r_v * v_f ; // Backward-travelling voltage wave in V
+
+// For case (d)
+v = v_f + v_b ; // Voltage at end of line in V
+v1 = (2 * R/(R + Z_c)) * v_f ; // (or) Voltage at end of line in V
+
+// For case (e)
+t1 = (2 * R/(R + Z_c)) ; // Refraction coefficient of voltage wave
+
+// For case (f)
+i_b = -( v_b/Z_c ) ; // backward-travelling current wave in A
+i_b1 = -r_v * i_f ; // (or) backward-travelling current wave in A
+
+
+// For case (g)
+i = v/R ; // Current flowing through resistor in A
+
+// For case (h)
+t2 = (2 * Z_c/(R + Z_c)) ; // Refraction coefficient of current wave
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 7.4 : SOLUTION :-") ;
+printf("\n (a) Reflection coefficient of voltage wave , ρ = %.4f \n",r_v) ;
+printf("\n (b) Reflection coefficient of current wave , ρ = %.4f \n",r_i) ;
+printf("\n (c) Backward-travelling voltage wave , v_b = %.3f V \n",v_b) ;
+printf("\n (d) Voltage at end of line , v = %.3f V \n",v) ;
+printf("\n     From alternative method ")
+printf("\n     Voltage at end of line , v = %.3f V \n",v) ;
+printf("\n (e) Refraction coefficient of voltage wave , Γ = %.4f \n",t1) ;
+printf("\n (f) Backward-travelling current wave , i_b = %.4f A \n",i_b) ;
+printf("\n (g) Current flowing through resistor, i = %.4f A \n",i) ;
+printf("\n (h) Refraction coefficient of current wave , Γ = %.4f \n",t2) ;
diff --git a/1820/CH7/EX7.5/Example7_5.sce b/1820/CH7/EX7.5/Example7_5.sce
new file mode 100755
index 000000000..e4b703268
--- /dev/null
+++ b/1820/CH7/EX7.5/Example7_5.sce
@@ -0,0 +1,96 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 7 : TRANSIENT OVERVOLTAGES AND INSULATION COORDINATION
+
+// EXAMPLE : 7.5 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+Z_c1 = 400 ; // Surge impedance of line in Ω
+Z_c2 = 40 ; // Surge impedance of cable in Ω
+v_f = 200 ; // Forward travelling surge voltage in kV
+
+// CALCULATIONS
+// For case (a)
+v_f1 = v_f * 10^3 ; // surge voltage in V
+i_f = v_f1/Z_c1 ; // Magnitude of forward current wave in A
+
+// For case (b)
+r = (Z_c2 - Z_c1)/(Z_c2 + Z_c1) ; // Reflection coefficient
+
+// For case (c)
+t = 2 * Z_c2/(Z_c2 + Z_c1) ; // Refraction coefficient
+
+// For case (d)
+v = t * v_f ; // Surge voltage transmitted forward into cable in kV
+
+// For case (e)
+v1 = v * 10^3 ; // Surge voltage transmitted forward into cable in V
+I = v1/Z_c2 ; // Surge current transmitted forward into cable in A
+
+// For case (f)
+v_b = r * v_f ; // surge voltage reflected back along overhead line in kV
+
+// For case (g)
+i_b = -r * i_f ; // surge current reflected back along overhead line in A
+
+// For case (h)
+// Arbitrary values are taken in graph.Only for reference not for scale 
+T = 0:0.1:300 ;
+
+for i = 1:int(length(T)/3) ; // plotting Voltage values
+    vo(i) = 3;
+end
+for i = int(length(T)/3):length(T)
+    vo(i) = 1 ;
+end
+for i = int(length(T))
+    vo(i) = 0 ;
+end
+
+
+a=gca() ;
+ylabel("CURRENT             SENDING END               VOLTAGE           ") ;
+b = newaxes() ; // creates new axis
+b.y_location = "right" ; // Position of axis
+ylabel ("RECEIVING END") ; // Labelling y-axis
+b.axes_visible = ["off","off","off"] ; 
+e = newaxes() ; 
+e.y_location = "middle" ; 
+e.y_label.text = "JUNCTION" ;
+subplot(2,1,1) ;
+plot2d(T,vo,2,'012','',[0,0,310,6]) ;
+
+for i = 1:int(length(T)/3) ; // Plotting current surges value
+    io(i) = 1 ;
+end
+for i = int(length(T)/3):length(T)
+    io(i) = 3 ;
+end
+for i = int(length(T))
+    io(i) = 0 ;
+end
+
+
+c=gca() ;
+d = newaxes() ;
+d.y_location = "right" ;
+d.filled = "off" ;
+f.y_location = "middle" ; 
+f.y_label.text = "JUNCTION" ;
+subplot(2,1,2) ;
+plot2d(T,io,5,'012','',[0,0,310,6]) ;
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 7.5 : SOLUTION :-") ;
+printf("\n (a) Magnitude of forward current wave , i_f = %d A \n",i_f) ;
+printf("\n (b) Reflection coefficient , ρ = %.4f \n",r) ;
+printf("\n (c) Refraction coefficient , Γ = %.4f \n",t) ;
+printf("\n (d) Surge voltage transmitted forward into cable , v = %.2f kV \n",v) ;
+printf("\n (e) Surge current transmitted forward into cable , i = %.f A \n",I) ;
+printf("\n (f) Surge voltage reflected back along the OH line , v_b = %.2f kV \n",v_b) ;
+printf("\n (g) Surge current reflected back along the OH line , i_b = %.f A \n",i_b) ;
+printf("\n (h) Graph shows plot of voltage & current surges after arrival at the junction \n") ;
diff --git a/1820/CH7/EX7.6/Example7_6.sce b/1820/CH7/EX7.6/Example7_6.sce
new file mode 100755
index 000000000..50799022d
--- /dev/null
+++ b/1820/CH7/EX7.6/Example7_6.sce
@@ -0,0 +1,140 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 7 : TRANSIENT OVERVOLTAGES AND INSULATION COORDINATION
+
+// EXAMPLE : 7.6 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+v = 1000 ; // ideal dc voltage source in V
+Z_s = 0 ; // internal impedance in Ω
+Z_c = 40 ; // characteristic impedance in Ω
+Z_r = 60 ; // Cable is terminated in 60Ω resistor
+
+// CALCULATIONS
+// For case (a)
+r_s = (Z_s - Z_c)/(Z_s + Z_c) ; // Reflection coefficient at sending end
+
+// For case (b)
+r_r = (Z_r - Z_c)/(Z_r + Z_c) ; // Reflection coefficient at receiving end
+
+// For case (c)
+T = 0:0.001:10.6 ; // // plotting values
+for i = 1:length(T) ;
+    if(T(i)<=1)
+        x(i) = (1.2)*T(i) - 1 ;
+    elseif(T(i)>=1 & T(i)<=2)
+        x(i) = (-1.2)*T(i) + 1.4 ;
+    elseif(T(i)>=2 & T(i)<=3)
+        x(i) = (1.2)*T(i)- 3.4 ;
+    elseif(T(i)>=3 & T(i)<=4)
+        x(i) = (-1.2)*T(i) + 3.8 ;
+    elseif(T(i)>=4 & T(i)<=5)
+        x(i) = (1.2)*T(i)- 5.8 ;
+    elseif(T(i)>=5 & T(i)<=6)
+        x(i) = (-1.2)*T(i) + 6.2 ;
+    elseif(T(i)>=6 & T(i)<=7)
+        x(i) = (1.2)*T(i)- 8.2 ;
+    elseif(T(i)>=7 & T(i)<=8)
+        x(i) = (-1.2)*T(i) + 8.6 ;
+    elseif(T(i)>=8 & T(i)<=9)
+        x(i) = (1.2)*T(i)- 10.6 ;
+    elseif(T(i)>=9 & T(i)<=10)
+        x(i) = (-1.2)*T(i) + 11 ;
+    elseif(T(i)>=10 & T(i)<=10.6)
+        x(i) = (1.2)*T(i) - 13 ;
+        end 
+end
+
+subplot(2,1,1) ; // Plotting two graph in same window
+plot2d(T,x,5,'012','',[0,-1,11,0.2]) ;
+
+a = gca() ;
+xlabel("TIME") ;
+ylabel("ρ_s = -1             DISTANCE             ρ_r = 0.2") ;
+xtitle("Fig 7.6 (c) Lattice diagram") ;
+a.thickness = 2 ; // sets thickness of plot
+xset('thickness',2) ; // sets thickness of axes
+xstring(1,-1,'T') ;
+xstring(2,-1,'2T') ;
+xstring(3,-1,'3T') ;
+xstring(4,-1,'4T') ;
+xstring(5,-1,'5T') ;
+xstring(6,-1,'6T') ;
+xstring(7,-1,'7T') ;
+xstring(8,-1,'8T') ;
+xstring(9,-1,'9T') ;
+xstring(10,-1,'10T') ;
+xstring(0.1,0.1,'0V') ;
+xstring(2,0.1,'1200V') ;
+xstring(4,0.1,'960V') ;
+xstring(6,0.1,'1008V') ;
+xstring(8,0.1,'998.4V') ;
+xstring(1,-0.88,'1000V') ;
+xstring(3,-0.88,'1000V') ;
+xstring(5,-0.88,'1000V') ;
+xstring(7,-0.88,'1000V') ;
+xstring(9,-0.88,'1000V') ;
+
+// For case (d)
+q1 = v ; // Refer Fig 7.11 in textbook
+q2 = r_r * v ;
+q3 = r_s * r_r * v ;
+q4 = r_s * r_r^2 * v ;
+q5 = r_s^2 * r_r^2 * v ;
+q6 = r_s^2 * r_r^3 * v ;
+q7 = r_s^3 * r_r^3 * v ;
+q8 = r_s^3 * r_r^4 * v ;
+q9 = r_s^4 * r_r^4 * v ;
+q10 = r_s^4 * r_r^5 * v ;
+q11 = r_s^5 * r_r^5 * v ;
+V_1 = v - q1 ;
+V_2 = v - q3 ;
+V_3 = v - q5 ;
+V_4 = v - q7 ; // voltage at t = 6.5T & x = 0.25l in Volts
+V_5 = v - q9 ;
+
+// For case (e)
+t = 0:0.001:9 ;
+
+for i= 1:length(t)
+    if(t(i)>=0 & t(i)<=1)
+        y(i) = V_1 ;
+    elseif(t(i)>=1 & t(i)<=3)
+        y(i) = V_2 ;
+    elseif(t(i)>=3 & t(i)<=5)
+        y(i)= V_3 ;
+    elseif(t(i)>=5 & t(i)<=7)
+        y(i)= V_4 ;
+    elseif(t(i)>=7 & t(i)<=9)
+        y(i)= V_5 ;
+    end
+end
+subplot(2,1,2) ;
+a = gca() ;
+a.thickness = 2 ; // sets thickness of plot
+plot2d(t,y,2,'012','',[0,0,10,1300]) ;
+a.x_label.text = 'TIME (T)' ; // labels x-axis
+a.y_label.text = 'RECEIVING-END   VOLTAGE (V)' ; // labels y-axis
+xtitle("Fig 7.6 (e) . Plot of Receiving end Voltage v/s Time") ;
+xset('thickness',2); // sets thickness of axes
+xstring(1,0,'1T') ; // naming points
+xstring(3,0,'3T') ;
+xstring(5,0,'5T') ;
+xstring(7,0,'7T') ;
+xstring(1,1200,'1200 V') ;
+xstring(4,960,'960 V') ;
+xstring(6,1008,'1008 V') ;
+xstring(8,998.4,'998.4 V') ;
+
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 7.6 : SOLUTION :-") ;
+printf("\n (a) Reflection coefficient at sending end , ρ_s = %.f \n",r_s) ;
+printf("\n (b) Reflection coefficient at sending end , ρ_r = %.1f \n",r_r)
+printf("\n (c) The lattice diagram is shown in Fig 7.6 (c) \n") ;
+printf("\n (d) From Fig 7.6 (c) , the voltage value is at t = 6.5T & x = 0.25 l is = %.d Volts \n",V_4) ;
+printf("\n (e) The plot of the receiving-end voltage v/s time is shown in Fig 7.6 (e) \n") ;
diff --git a/1820/CH8/EX8.1/Example8_1.sce b/1820/CH8/EX8.1/Example8_1.sce
new file mode 100755
index 000000000..ed8980a05
--- /dev/null
+++ b/1820/CH8/EX8.1/Example8_1.sce
@@ -0,0 +1,33 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 8 : LIMITING FACTORS FOR EXTRA-HIGH AND ULTRAHIGH VOLTAGE TRANSMISSION
+
+// EXAMPLE : 8.1 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+m_0 = 0.90 ; // Irregularity factor
+p = 74 ; // Atmospheric pressure in Hg
+t = 10 ; // temperature in degree celsius
+D = 550 ; // Equilateral spacing b/w conductors in cm
+d = 3 ; // overall diameter in cm
+
+// CALCULATIONS
+// For case (a)
+r = d/2 ;
+delta = 3.9211 * p/( 273 + t ) ; // air density factor
+V_0_ph = 21.1 * delta * m_0 * r * log(D/r) ; // disruptive critical rms line voltage in kV/phase
+V_0 = sqrt(3) * V_0_ph ; // disruptive critical rms line voltage in kV 
+
+// For case (b)
+m_v = m_0 ;
+V_v_ph = 21.1*delta*m_v*r*(1 + (0.3/sqrt(delta*r) )) * log(D/r) ; // visual critical rms line voltage in kV/phase
+V_v = sqrt(3)*V_v_ph ; // visual critical rms line voltage in kV 
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 8.1 : SOLUTION :-") ;
+printf("\n (a) Disruptive critical rms line voltage , V_0 = %.1f kV \n",V_0) ;
+printf("\n (b) Visual critical rms line voltage , V_v = %.1f kV \n",V_v) ;
diff --git a/1820/CH8/EX8.2/Example8_2.sce b/1820/CH8/EX8.2/Example8_2.sce
new file mode 100755
index 000000000..e188f128b
--- /dev/null
+++ b/1820/CH8/EX8.2/Example8_2.sce
@@ -0,0 +1,33 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 8 : LIMITING FACTORS FOR EXTRA-HIGH AND ULTRAHIGH VOLTAGE TRANSMISSION
+
+// EXAMPLE : 8.2 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+f = 60 ; // freq in Hz
+d = 3 ; // overall diameter in cm
+D = 550 ; // Equilateral spacing b/w conductors in cm
+V1 = 345 ; // operating line voltage in kV
+V_0 = 172.4 ; // disruptive critical voltage in kV
+L = 50 ; // line length in mi
+p = 74 ; // Atmospheric pressure in Hg
+t = 10 ; // temperature in degree celsius
+m_0 = 0.90 ; // Irregularity factor
+
+// CALCULATIONS
+r = d/2 ;
+delta = 3.9211 * p/( 273 + t ) ; // air density factor
+V_0 = 21.1 * delta * m_0 * r * log(D/r) ; // disruptive critical rms line voltage in kV/phase
+V =V1/sqrt(3) ; // Line to neutral operating voltage in kV
+P_c = (390/delta)*(f+25)*sqrt(r/D)*(V - V_0)^2 * 10^-5 ; // Fair weather corona loss per phase in kW/mi/phase
+P_cT = P_c * L ; // For total line length corona loss in kW/phase
+T_P_c = 3 * P_cT ; // Total corona loss of line in kW
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 8.2 : SOLUTION :-") ;
+printf("\n (a) Total fair weather corona loss of the line , P_c = %.1f kW \n",T_P_c) ;
diff --git a/1820/CH9/EX9.1/Example9_1.sce b/1820/CH9/EX9.1/Example9_1.sce
new file mode 100755
index 000000000..8c0e92a2b
--- /dev/null
+++ b/1820/CH9/EX9.1/Example9_1.sce
@@ -0,0 +1,42 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 9 : SYMMETRICAL COMPONENTS AND FAULT ANALYSIS
+
+// EXAMPLE : 9.1 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+V_a = 7.3 * exp(%i*12.5*%pi/180) ; // Phase voltage in V
+V_b = 0.4 * exp(%i*(-100)*%pi/180) ; // Phase voltage in V
+V_c = 4.4 * exp(%i*154*%pi/180) ; // Phase voltage in V
+a = 1 * exp(%i*120*%pi/180) ; // operator 'a' by application of symmetrical components theory to 3-Φ system . Refer section 9.3 for details
+
+// CALCULATIONS
+V_a0 = (1/3) * (V_a + V_b + V_c) ; // Analysis equ in V
+V_a1 = (1/3) * (V_a + a*V_b + a^2*V_c) ;
+V_a2 = (1/3) * (V_a + a^2*V_b + a*V_c) ;
+V_b0 = V_a0 ;
+V_b1 = a^2 * V_a1 ;
+V_b2 = a * V_a2 ;
+V_c0 = V_a0 ;
+V_c1 = a * V_a1 ;
+V_c2 = a^2 * V_a2 ;
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 9.1 : SOLUTION :-") ;
+printf("\n The symmetrical components for the phase voltages V_a , V_b & V_c are\n") ;
+printf("\n V_a0 = %.2f<%.1f V \n",abs(V_a0),atand( imag(V_a0),real(V_a0) )) ;
+printf("\n V_a1 = %.2f<%.1f V \n",abs(V_a1),atand( imag(V_a1),real(V_a1) )) ;
+printf("\n V_a2 = %.2f<%.1f V \n",abs(V_a2),atand( imag(V_a2),real(V_a2) )) ;
+printf("\n V_b0 = %.2f<%.1f V \n",abs(V_b0),atand( imag(V_b0),real(V_b0) )) ;
+printf("\n V_b1 = %.2f<%.1f V \n",abs(V_b1),atand( imag(V_b1),real(V_b1) )) ;
+printf("\n V_b2 = %.2f<%.1f V \n",abs(V_b2),atand( imag(V_b2),real(V_b2) )) ;
+printf("\n V_c0 = %.2f<%.1f V \n",abs(V_c0),atand( imag(V_c0),real(V_c0) )) ;
+printf("\n V_c1 = %.2f<%.1f V \n",abs(V_c1),atand( imag(V_c1),real(V_c1) )) ;
+printf("\n V_c2 = %.2f<%.1f V \n",abs(V_c2),atand( imag(V_c2),real(V_c2) )) ;
+
+printf("\n NOTE : V_b1 = 3.97<-99.5 V & V_c2 = 2.52<-139.7 V result obtained is same as textbook answer V_b1 = 3.97<260.5 V & V_c2 = 2.52<220.3 V \n") ;
+printf("\n Changes is due to a^2 = 1<240 = 1<-120  where 1 is the magnitude & <240 is the angle in degree \n") ;
diff --git a/1820/CH9/EX9.10/Example9_10.sce b/1820/CH9/EX9.10/Example9_10.sce
new file mode 100755
index 000000000..5782c9c3e
--- /dev/null
+++ b/1820/CH9/EX9.10/Example9_10.sce
@@ -0,0 +1,84 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 9 : SYMMETRICAL COMPONENTS AND FAULT ANALYSIS
+
+// EXAMPLE : 9.10 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+Z_0 = 0.2619 * %i ;
+Z_1 = 0.25 * %i ;
+Z_2 = 0.25 * %i ;
+v = 1 * exp(%i*0*%pi/180) ;
+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] ;
+
+// CALCULATIONS
+// For case (b)
+I_a0 = v/(Z_0 + Z_1 + Z_2) ; // Sequence currents at fault point F in pu A
+I_a1 = I_a0 ;
+I_a2 = I_a0 ;
+
+// For case (c)
+I_a1g1 = (1/2) * I_a1 ; // Sequence current at terminals of generator G1 in pu A
+I_a2g1 = (1/2) * I_a2 ;
+I_a0g1 = 0.5/(0.55 + 0.5)*I_a0 ; // By current division in pu A
+
+// For case (d)
+I_f = [A] * [I_a0g1 ; I_a1g1 ; I_a2g1] ; // Phase current at terminal of generator G1 in pu A
+
+// For case (e)
+V_a = [0 ; v ; 0] - [Z_0 0 0 ; 0 Z_1 0 ; 0 0 Z_2]*[I_a0g1 ; I_a1g1 ; I_a2g1] ; // Sequence voltage in pu V
+
+// For case (f)
+V_f = [A]*[V_a] ; // Phase voltage at terminal of generator G1 in pu V
+
+// For case (g)
+I_a1g2 = (1/2) * I_a1 ; // By symmetry for Generator G2
+I_a2g2 = (1/2) * I_a2 ;
+I_a0g2 = 0 ; // By inspection
+// V_a1(HV) leads V_a1(LV) by 30 degree & V_a2(HV) lags V_a2(LV) by 30 degree
+I_a0G2 = I_a0g2 ;
+I_a1G2 = abs(I_a1g2)*exp(%i * (atand( imag(I_a1g2),real(I_a1g2) ) - 30) * %pi/180) ; // (-90-30) = (-120)
+I_a2G2 = abs(I_a2g2)*exp(%i *(atand( imag(I_a2g2),real(I_a2g2) ) + 30) * %pi/180)  ; // (-90+30) = (-60)
+
+I_f2 = [A] * [I_a0G2 ; I_a1G2 ; I_a2G2] ; // Phase current at terminal of generator G2 in pu A
+
+ // Sequence voltage at terminal of generator G2 in pu V
+V_a0G2 = 0 ;
+V_a1G2 = abs(V_a(2,1))*exp(%i * (atand( imag(V_a(2,1)),real(V_a(2,1)) ) - 30) * %pi/180) ; // (0-30) = (-30)
+V_a2G2 = abs(V_a(3,1))*exp(%i * (atand( imag(V_a(3,1)),real(V_a(3,1)) ) + 30) * %pi/180) ; // (180+30)=(210)=(-150)
+
+V_f2 = A * [V_a0G2 ; V_a1G2 ; V_a2G2] ; // Phase voltage at terminal of generator G2 in pu V
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 9.10 : SOLUTION :-") ;
+printf("\n (b) The sequence current at fault point F , I_a0 = I_a1 = I_a2 = %.4f<%.f pu A \n",abs(I_a0),atand(imag(I_a0),real(I_a0) )) ;
+printf("\n (c) Sequence currents at the terminals of generator G1 , \n") ;
+printf("\n     I_a0,G_1 = %.4f<%.f pu A ",abs(I_a0g1),atand( imag(I_a0g1),real(I_a0g1) )) ;
+printf("\n     I_a1,G_1 = %.4f<%.f pu A ",abs(I_a1g1),atand( imag(I_a1g1),real(I_a1g1) )) ;
+printf("\n     I_a2,G_1 = %.4f<%.f pu A ",abs(I_a2g1),atand( imag(I_a2g1),real(I_a2g1) )) ;
+printf("\n \n (d) Phase currents at terminal of generator G1 are , [I_af ; I_bf ; I_cf] = pu A \n") ;
+printf("\n       %.4f<%.f ",abs(I_f),atand(imag(I_f),real(I_f) )) ;
+printf("\n \n (e) Sequence voltages at the terminals of generator G1 , [V_a0 ; V_a1 ; V_a2 ] = pu V \n") ;
+printf("\n       %.4f<%.1f ",abs(V_a),atand(imag(V_a),real(V_a) )) ; 
+printf("\n \n (f) Phase voltages at terminal of generator G1 are , [V_af ; V_bf ; V_cf] = pu V \n") ;
+printf("\n       %.4f<%.1f ",abs(V_f),atand(imag(V_f),real(V_f) )) ; 
+printf("\n \n (g) Sequence currents at the terminals of generator G2 , \n") ;
+printf("\n     I_a0,G_2 = %.f<%.f pu A ",abs(I_a0G2),atand( imag(I_a0G2),real(I_a0G2) )) ;
+printf("\n     I_a1,G_2 = %.4f<%.f pu A",abs(I_a1G2),atand( imag(I_a1G2),real(I_a1G2) )) ;
+printf("\n     I_a2,G_2 = %.4f<%.f pu A",abs(I_a2G2),atand( imag(I_a2G2),real(I_a2G2) )) ;
+printf("\n \n     Phase currents at terminal of generator G2 are , [I_af ; I_bf ; I_cf] = pu A \n") ;
+printf("\n       %.4f<%.f ",abs(I_f2),atand(imag(I_f2),real(I_f2) )) ;
+printf("\n \n     Sequence voltages at the terminals of generator G2 , [V_a0 ; V_a1 ; V_a2 ] = pu V\n") ;
+printf("\n        %.f<%.f ",abs(V_a0G2),atand( imag(V_a0G2),real(V_a0G2) )) ;
+printf("\n        %.4f<%.f ",abs(V_a1G2),atand( imag(V_a1G2),real(V_a1G2) )) ;
+printf("\n        %.4f<%.f ",abs(V_a2G2),atand( imag(V_a2G2),real(V_a2G2) )) ;
+printf("\n \n     Phase voltages at terminal of generator G2 are , [V_af ; V_bf ; V_cf] = pu V \n") ;
+printf("\n        %.4f<%.1f ",abs(V_f2),atand(imag(V_f2),real(V_f2) )) ; 
+
+printf("\n \n NOTE : ERROR : Calclation mistake in textbook case(f) ") ;
+printf("\n In case (g) V_a2 = 0.1641<-150 is same as textbook answer V_a2 = 0.1641<210 , i.e (360-150)=210 \n") ;
diff --git a/1820/CH9/EX9.11/Example9_11.sce b/1820/CH9/EX9.11/Example9_11.sce
new file mode 100755
index 000000000..31f34cbbb
--- /dev/null
+++ b/1820/CH9/EX9.11/Example9_11.sce
@@ -0,0 +1,58 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 9 : SYMMETRICAL COMPONENTS AND FAULT ANALYSIS
+
+// EXAMPLE : 9.11 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+kv = 230 ; // Line voltage in kV from Exa 9.9
+Z_0 = 0.56*%i ; // Zero-sequence impedance in pu
+Z_1 = 0.2618*%i ; // Zero-sequence impedance in pu
+Z_2 = 0.3619*%i ; // Zero-sequence impedance in pu
+z_f = 5 ; // Fault impedance in Ω
+v = 1*exp(%i*0*%pi/180) ; // 
+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] ;
+
+// CALCULATIONS
+// For case (b)
+I_a0 = 0 ; // Sequence current in A
+Z_B = kv^2/200 ; // Base impedance of 230 kV line
+Z_f = z_f/Z_B ; // fault impedance in pu
+I_a1 = v/(Z_1 + Z_2 + Z_f) ; // Sequence current in pu A
+I_a2 = - I_a1 ; // Sequence current in pu A
+I_f = [A] * [I_a0 ; I_a1 ; I_a2] ; // Phase current in pu A
+
+// For case (c)
+V_a = [0 ; v ; 0]-[Z_0 0 0 ; 0 Z_1 0 ; 0 0 Z_2]*[I_a0 ; I_a1 ; I_a2] ; // Sequence voltages in pu V
+V_f = A*V_a ; // Phase voltages in pu V
+
+// For case (d)
+V_abf = V_f(1,1) - V_f(2,1) ; // Line-to-line voltages at fault points in pu V
+V_bcf = V_f(2,1) - V_f(3,1) ; // Line-to-line voltages at fault points in pu V
+V_caf = V_f(3,1) - V_f(1,1) ; // Line-to-line voltages at fault points in pu V
+
+
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 9.11 :SOLUTION :-") ;
+printf("\n (b) Sequence currents are , \n") ;
+printf("\n  I_a0 = %.f pu A ",I_a0) ;
+printf("\n  I_a1 = %.4f<%.2f pu A ",abs(I_a1),atand( imag(I_a1),real(I_a1) )) ;
+printf("\n  I_a2 = %.4f<%.2f pu A ",abs(I_a2),atand( imag(I_a2),real(I_a2) )) ;
+printf("\n \n Phase currents are , [I_af ; I_bf ; I_cf] =  pu A \n") ;
+printf("\n       %.4f<%.1f ",abs(I_f),atand(imag(I_f),real(I_f) )) ;
+printf("\n \n (c) Sequence voltages are , [V_a0 ; V_a1 ; V_a2] =  pu V \n") ;
+printf("\n       %.4f<%.1f ",abs(V_a),atand(imag(V_a),real(V_a) )) ;
+printf("\n \n  Phase voltages are , [V_af ; V_bf ; V_cf] =  pu V \n") ;
+printf("\n       %.4f<%.1f ",abs(V_f),atand(imag(V_f),real(V_f) )) ;
+printf("\n \n (d) Line-to-line voltages at the fault points are \n") ;
+printf("\n     V_abf = %.4f<%.1f pu V \n",abs(V_abf),atand( imag(V_abf),real(V_abf) )) ;
+printf("\n     V_bcf = %.4f<%.1f pu V \n",abs(V_bcf),atand( imag(V_bcf),real(V_bcf) )) ;
+printf("\n     V_caf = %.4f<%.1f pu V \n",abs(V_caf),atand( imag(V_caf),real(V_caf) )) ;
+
+printf("\n \n NOTE : ERROR : Minor calclation mistake in textbook ") ;
diff --git a/1820/CH9/EX9.12/Example9_12.sce b/1820/CH9/EX9.12/Example9_12.sce
new file mode 100755
index 000000000..b887a9f22
--- /dev/null
+++ b/1820/CH9/EX9.12/Example9_12.sce
@@ -0,0 +1,56 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 9 : SYMMETRICAL COMPONENTS AND FAULT ANALYSIS
+
+// EXAMPLE : 9.12 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+z_f = 5 ; // Fault-impedance in Ω
+z_g = 10 ; // Ground-impedance in Ω
+kv = 230 ; // Line voltage in kV from Exa 9.9
+Z_0 = 0.56*%i ; // Zero impedance in pu Ω
+Z_1 = 0.2618*%i ; // Positive sequence Impedance in pu Ω
+Z_2 = 0.3619*%i ; // Negative sequence Impedance in pu Ω
+v = 1*exp(%i*0*180/%pi) ;
+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] ;
+
+// CALCULATIONS
+// For case (b)
+Z_B = kv^2/200 ; // Base impedance of 230 kV line
+Z_f = z_f/Z_B ; // fault impedance in pu Ω
+Z_g = z_g/Z_B ;
+I_a1 = v/( (Z_1 + Z_f) + ( (Z_2 + Z_f)*(Z_0 + Z_f + 3*Z_g)/((Z_2 + Z_f)+(Z_0 + Z_f + 3*Z_g)) )) ; // Sequence current in pu A
+I_a2 = -[(Z_0 + Z_f + 3*Z_g)/( (Z_2 + Z_f )+(Z_0 + Z_f + 3*Z_g) )]*I_a1 ; // Sequence current in pu A
+I_a0 = -[(Z_2 + Z_f)/( (Z_2 + Z_f)+(Z_0 + Z_f + 3*Z_g) )]*I_a1 ; // Sequence current in pu A
+I_f = A*[I_a0 ; I_a1 ; I_a2] ; // Phase currents in pu A
+
+// For case (c)
+V = [0 ; v ; 0] - [Z_0 0 0 ; 0 Z_1 0 ; 0 0 Z_2]*[I_a0 ; I_a1 ; I_a2] ; // Sequence Voltages in pu V
+V_f = A*[V] ; // Phase voltages in pu V
+
+// For case (d)
+V_abf = V_f(1,1) - V_f(2,1) ; // Line-to-line voltages at fault points a & b
+V_bcf = V_f(2,1) - V_f(3,1) ; // Line-to-line voltages at fault points b & c
+V_caf = V_f(3,1) - V_f(1,1) ; // Line-to-line voltages at fault points c & a
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 9.12 : SOLUTION :-") ;
+printf("\n (b) Sequence currents are , \n") ;
+printf("\n   I_a0 = %.4f<%.2f pu A ",abs(I_a0),atand( imag(I_a0),real(I_a0) )) ;
+printf("\n   I_a1 = %.4f<%.2f pu A ",abs(I_a1),atand( imag(I_a1),real(I_a1) )) ;
+printf("\n   I_a2 = %.4f<%.2f pu A ",abs(I_a2),atand( imag(I_a2),real(I_a2) )) ;
+printf("\n \n   Phase currents are , [I_af ; I_bf ; I_cf] = pu A \n ") ;
+printf("\n       %.4f<%.1f ",abs(I_f),atand(imag(I_f),real(I_f) )) ;
+printf("\n \n (c) Sequence voltages , [V_a0 ; V_a1 ; V_a2] = pu V \n ") ;
+printf("\n       %.4f<%.1f ",abs(V),atand(imag(V),real(V) )) ;
+printf("\n \n  Phase voltages , [V_af ; V_bf ; V_cf] = pu V \n ") ;
+printf("\n       %.4f<%.1f ",abs(V_f),atand(imag(V_f),real(V_f) )) ;
+printf("\n \n (d) Line-to-line voltages at the fault points are , \n") ;
+printf("\n   V_abf = %.4f<%.1f pu V \n",abs(V_abf),atand( imag(V_abf),real(V_abf) )) ;
+printf("\n   V_bcf = %.4f<%.1f pu V \n",abs(V_bcf),atand( imag(V_bcf),real(V_bcf) )) ;
+printf("\n   V_caf = %.4f<%.1f pu V \n",abs(V_caf),atand( imag(V_caf),real(V_caf) )) ;
diff --git a/1820/CH9/EX9.13/Example9_13.sce b/1820/CH9/EX9.13/Example9_13.sce
new file mode 100755
index 000000000..2d314373a
--- /dev/null
+++ b/1820/CH9/EX9.13/Example9_13.sce
@@ -0,0 +1,56 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 9 : SYMMETRICAL COMPONENTS AND FAULT ANALYSIS
+
+// EXAMPLE : 9.13 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+z_f = 5 ; // Fault-impedance in Ω
+Z_0 = 0.56*%i ; // Zero impedance in pu Ω
+Z_1 = 0.2618*%i ; // Positive sequence Impedance in pu Ω
+Z_2 = 0.3619*%i ; // Negative sequence Impedance in pu Ω
+kv = 230 ; // Line voltage in kV from Exa 9.9
+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] ;
+
+// CALCULATIONS
+// For case (b)
+Z_B = kv^2/200 ; // Base impedance of 230 kV line
+Z_f = z_f/Z_B ; // fault impedance in pu Ω
+v = 1*exp(%i*0*%pi/180) ;
+I_a0 = 0 ; // Sequence current in pu A
+I_a1 = v/(Z_1 + Z_f) ; // Sequence current in pu A
+I_a2 = 0 ; // Sequence current in pu A
+I_f = A*[I_a0 ; I_a1 ; I_a2] ; // Phase-current in pu A
+
+// For case (c)
+V = [0 ; v ; 0] - [Z_0 0 0 ; 0 Z_1 0 ; 0 0 Z_2]*[I_a0 ; I_a1 ; I_a2] ; // Sequence Voltages in pu V
+V_f = A*[V] ; // Phase voltages in pu V
+
+// For case (d)
+V_abf = V_f(1,1) - V_f(2,1) ; // Line-to-line voltages at fault points a & b
+V_bcf = V_f(2,1) - V_f(3,1) ; // Line-to-line voltages at fault points b & c
+V_caf = V_f(3,1) - V_f(1,1) ; // Line-to-line voltages at fault points c & a
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 9.13 : SOLUTION :-") ;
+printf("\n (b) Sequence currents are , \n") ;
+printf("\n     I_a0 = %.1f pu A ",I_a0) ;
+printf("\n     I_a1 = %.4f<%.1f pu A ",abs(I_a1),atand( imag(I_a1),real(I_a1) )) ;
+printf("\n     I_a2 = %.1f pu A ",I_a2) ;
+printf("\n \n   Phase currents are , [I_af ; I_bf ; I_cf] = pu A \n ") ;
+printf("\n       %.4f<%.1f ",abs(I_f),atand(imag(I_f),real(I_f) )) ;
+printf("\n \n (c) Sequence voltages , [V_a0 ; V_a1 ; V_a2] = pu V \n ") ;
+printf("\n       %.4f<%.1f ",abs(V),atand(imag(V),real(V) )) ;
+printf("\n \n    Phase voltages , [V_af ; V_bf ; V_cf] = pu V \n ") ;
+printf("\n       %.4f<%.1f ",abs(V_f),atand(imag(V_f),real(V_f) )) ;
+printf("\n \n (d) Line-to-line voltages at the fault points are , \n") ;
+printf("\n    V_abf = %.4f<%.1f pu V \n",abs(V_abf),atand( imag(V_abf),real(V_abf) )) ;
+printf("\n    V_bcf = %.4f<%.1f pu V \n",abs(V_bcf),atand( imag(V_bcf),real(V_bcf) )) ;
+printf("\n    V_caf = %.4f<%.1f pu V \n",abs(V_caf),atand( imag(V_caf),real(V_caf) )) ;
+
+printf("\n \n NOTE : ERROR : Calclation mistake in textbook case(d) ") ;
diff --git a/1820/CH9/EX9.14/Example9_14.sce b/1820/CH9/EX9.14/Example9_14.sce
new file mode 100755
index 000000000..6fe7597fc
--- /dev/null
+++ b/1820/CH9/EX9.14/Example9_14.sce
@@ -0,0 +1,40 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 9 : SYMMETRICAL COMPONENTS AND FAULT ANALYSIS
+
+// EXAMPLE : 9.14 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+VG_1 = 1*exp(%i*0*%pi/180) ;
+VG_2 = 1*exp(%i*0*%pi/180) ;
+
+// CALCULATIONS
+// For case (a)
+I_1 = 1*exp(%i*0*%pi/180) ;
+I_2 = 1*exp(%i*0*%pi/180) ;
+V_1 = 0.4522*exp(%i*90*%pi/180) ;
+V_2 = 0.4782*exp(%i*90*%pi/180) ;
+Y_11 = I_1/V_1 ; // When V_2 = 0 
+Y_21 = (-0.1087)*Y_11 ; // When V_2 = 0 
+Y_22 = I_2/V_2 ; // When V_1 = 0 
+Y_12 = Y_21 ;
+Y = [Y_11 Y_12 ; Y_21 Y_22] ; // Admittance matrix associated with positive-sequence n/w
+
+// For case (b)
+I_S1_12 = 2.0193*exp(%i*90*%pi/180) ; // Short-ckt F & F' to neutral & by superposition theorem
+I_S1_10 = 0.2884*exp(%i*90*%pi/180) ; // Short-ckt F & F' to neutral & by superposition theorem
+I_S2_12 = 0.4326*exp(%i*90*%pi/180) ;
+I_S2_10 = 1.4904*exp(%i*90*%pi/180) ;
+I_S1 = I_S1_12 + I_S1_10 ;
+I_S2 = I_S2_12 + I_S2_10 ;
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 9.14 :SOLUTION :-") ;
+printf("\n (a) Admittance matrix associated with positive-sequence network , Y = \n") ; disp(Y) ;
+printf("\n (b) Source currents Two-port Thevenin equivalent positive sequence network are , \n") ;
+printf("\n     I_S1 = %.4f<%.f pu ",abs(I_S1),atand( imag(I_S1),real(I_S1) )) ;
+printf("\n     I_S2 = %.4f<%.f pu \n",abs(I_S2),atand( imag(I_S2),real(I_S2) )) ;
diff --git a/1820/CH9/EX9.15/Example9_15.sce b/1820/CH9/EX9.15/Example9_15.sce
new file mode 100755
index 000000000..e4ccd682a
--- /dev/null
+++ b/1820/CH9/EX9.15/Example9_15.sce
@@ -0,0 +1,37 @@
+
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 9 : SYMMETRICAL COMPONENTS AND FAULT ANALYSIS
+
+// EXAMPLE : 9.15 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+Y_11 = -2.2115*%i ;
+Y_12 = 0.2404*%i ;
+Y_21 = 0.2404*%i ;
+Y_22 = -2.0912*%i ;
+Y = [Y_11 Y_12 ; Y_21 Y_22] ;
+I_S1 = 2.3077*%i ;
+I_S2 = 1.9230*%i ;
+
+I_a1 = poly(0,'I_a1') ;
+I_a2 = poly(0,'I_a2') ;
+a = Y_12*I_S2 - Y_22*I_S1 ;
+b = (Y_12+Y_22)*I_a1 ;
+c = Y_12*I_S1 - Y_11*I_S2 ;
+d = (Y_12 + Y_11)*I_a1 ;
+V1 = (1/det(Y))*[(a-b) ; (c+d)] ; // Gives the uncoupled positive sequence N/W
+A = (Y_12+Y_22)*I_a2 ;
+B = (Y_12 + Y_11)*I_a2 ;
+V2 = (1/det(Y))*[A ; B] ; // Gives the uncoupled negative sequence N/W
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 9.15 : SOLUTION :-") ;
+printf("\n (a) [V_a1 ; V_a11] = ") ; disp(V1) ;
+printf("\n     Values of Uncoupled positive-sequence network \n") ;
+printf("\n (b) [V_a2 ; V_a22] = ") ; disp(V2) ;
+printf("\n     Values of Uncoupled negative-sequence network \n") ;
diff --git a/1820/CH9/EX9.16/Example9_16.sce b/1820/CH9/EX9.16/Example9_16.sce
new file mode 100755
index 000000000..a30c5b2d6
--- /dev/null
+++ b/1820/CH9/EX9.16/Example9_16.sce
@@ -0,0 +1,58 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 9 : SYMMETRICAL COMPONENTS AND FAULT ANALYSIS
+
+// EXAMPLE : 9.16 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+H_aa = 81.5 ;
+D_aa = 1.658 ;
+f = 60 ; // Freq in Hz
+I = 20 ;
+kV = 69 ; // Line voltage in kV
+MVA = 25 ; // Transformer T1 rating in MVA
+
+// CALCULATIONS
+// For case (a)
+C_0 = 29.842*10^-9/(log(H_aa/D_aa)) ; // Capacitance in F/mi
+b_0 = 2*%pi*f*C_0 ; // Susceptance in S/mi
+B_0 = b_0*I ; // For total system
+X_C0 = (1/B_0) ; // Total zero-sequence reactance in Ω
+TC_0 = B_0/(2*%pi*f) ; // Total zero-sequence capacitance in F
+
+// For case (c)
+X_1 = 0.05 ; // Leakage reactance of transformer T1 in pu
+X_0 = X_1 ;
+X_2 = X_1 ;
+Z_B = kV^2/MVA ;
+X_01 = X_0*Z_B ; // Leakage reactance in Ω
+V_F = 69*10^3/sqrt(3) ;
+I_a0PC = V_F/(17310.8915*%i) ; // Zero-sequence current flowing through PC in A
+I_PC = 3*abs(I_a0PC) ; // Continuous-current rating of the PC in A
+
+// For case (d)
+X_PC = (17310.8915 - X_01)/3 ; // Required reactance value for PC in Ω
+
+// For case (e)
+L_PC = X_PC/(2*%pi*f) ; // Inductance in H
+
+// For case (f)
+S_PC = (I_PC^2)*X_PC ; // Rating in VA
+S_PC1 = S_PC*10^-3 ; // Continuous kVA rating in kVA
+
+// For case (g)
+V_PC = I_PC * X_PC ; // continuous-voltage rating for PC in V
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 9.16 :SOLUTION :-") ;
+printf("\n (a) Total zero-sequence susceptance per phase of system at 60 Hz , ΣX_C0 = %.4f Ω \n",X_C0) ;
+printf("\n     Total zero-sequence capacitance per phase of system at 60 Hz , ΣC_0 = %.4e F \n",TC_0) ;
+printf("\n (c) Continuous-current rating of the PC , I_PC = 3I_a0PC = %.4f A \n",abs(I_PC)) ;
+printf("\n (d) Required reactance value for the PC , X_PC = %.4f Ω \n",X_PC) ;
+printf("\n (e) Inductance value of the PC , L_PC = %.4f H \n",L_PC) ;
+printf("\n (f) Continuous kVA rating for the PC , S_PC = %.2f kVA \n",S_PC1) ;
+printf("\n (g) Continuous-voltage rating for PC , V_PC = %.2f V \n",V_PC) ;
diff --git a/1820/CH9/EX9.2/Example9_2.sce b/1820/CH9/EX9.2/Example9_2.sce
new file mode 100755
index 000000000..148df0c1d
--- /dev/null
+++ b/1820/CH9/EX9.2/Example9_2.sce
@@ -0,0 +1,39 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 9 : SYMMETRICAL COMPONENTS AND FAULT ANALYSIS
+
+// EXAMPLE : 9.2 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+V_abc = [0 ; 50 ; -50] ; // Phase voltages of a 3-Φ system in V
+I_abc = [-5 ; 5*%i ; -5] ; // Phase current of a 3-Φ system in A
+
+// CALCULATIONS
+// For case (a)
+S_3ph = (V_abc)' * conj(I_abc) ; // 3-Φ complex power in VA
+
+// 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] ;
+V_012 = inv(A) * (V_abc) ; // Sequence voltage matrices in V
+I_012 = inv(A) * (I_abc) ; // Sequence current matrices in A
+
+// For case (c)
+S_3ph1 = 3 * ([V_012(1,1) V_012(2,1) V_012(3,1)]) * (conj(I_012)) ; // Three-phase complex power in VA . Refer equ 9.34(a)
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 9.2 : SOLUTION :-") ;
+printf("\n (a) Three-phase complex power using equ 9.30 , S_3-Φ = %.4f<%.f VA \n",abs(S_3ph) , atand(imag(S_3ph),real(S_3ph) )) ;
+printf("\n (b) Sequence Voltage matrices , [V_012] =  V \n") ;
+printf("\n     %.f<%.f ",abs(V_012(1,1)),atand( imag(V_012(1,1)),real(V_012(1,1)) )) ;
+printf("\n     %.4f<%.f ",abs(V_012(2,1)),atand( imag(V_012(2,1)),real(V_012(2,1)) )) ;
+printf("\n     %.4f<%.f ",abs(V_012(3,1)),atand( imag(V_012(3,1)),real(V_012(3,1)) )) ;
+printf("\n \n    Sequence current matrices , [I_012] = A \n") ;
+printf("\n     %.4f<%.1f ",abs(I_012(1,1)),atand( imag(I_012(1,1)),real(I_012(1,1)) )) ;
+printf("\n     %.4f<%.f ",abs(I_012(2,1)),atand( imag(I_012(2,1)),real(I_012(2,1)) )) ;
+printf("\n     %.4f<%.f ",abs(I_012(3,1)),atand( imag(I_012(3,1)),real(I_012(3,1)) )) ;
+printf("\n \n (c) Three-phase complex power using equ 9.34 , S_3-Φ = %.4f<%.f VA \n",abs(S_3ph1) , atand(imag(S_3ph1),real(S_3ph1) )) ;
diff --git a/1820/CH9/EX9.3/Example9_3.sce b/1820/CH9/EX9.3/Example9_3.sce
new file mode 100755
index 000000000..bd12b9d7e
--- /dev/null
+++ b/1820/CH9/EX9.3/Example9_3.sce
@@ -0,0 +1,44 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 9 : SYMMETRICAL COMPONENTS AND FAULT ANALYSIS
+
+// EXAMPLE : 9.3 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+l = 40 ; // line length in miles
+// Conductor parameter from Table A.3
+r_a = 0.206 ; // Ohms per conductor per mile in Ω/mi
+r_b = r_a ; // r_a = r_b = r_c in Ω/mi
+D_s = 0.0311 ; // GMR in ft where D_s = D_sa = D_sb = D_sc
+D_ab = sqrt(2^2 + 8^2) ; // GMR in ft
+D_bc = sqrt(3^2 + 13^2) ; // GMR in ft
+D_ac = sqrt(5^2 + 11^2) ; // GMR in ft
+D_e = 2788.5 ; // GMR in ft since earth resistivity is zero
+r_e = 0.09528 ; // At 60 Hz in Ω/mi
+
+// CALCULATIONS
+// For case (a)
+Z_aa =[(r_a + r_e) + %i * 0.1213*log(D_e/D_s)]*l ; // Self impedance of line conductor in Ω
+Z_bb = Z_aa ;
+Z_cc = Z_bb ;
+Z_ab = [r_e + %i * 0.1213*log(D_e/D_ab)]*l ; // Mutual impedance in Ω
+Z_ba = Z_ab ;
+Z_bc = [r_e + %i * 0.1213*log(D_e/D_bc)]*l ;
+Z_cb = Z_bc ;
+Z_ac = [r_e + %i * 0.1213*log(D_e/D_ac)]*l ;
+Z_ca = Z_ac ;
+Z_abc = [Z_aa Z_ab Z_ac ; Z_ba Z_bb Z_bc ; Z_ca Z_cb Z_cc] ; // Line impedance matrix
+
+// 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_abc*A ; // Sequence impedance matrix
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 9.3 : SOLUTION :-") ;
+printf("\n (a) Line impedance matrix , [Z_abc] = \n") ; disp(Z_abc) ;
+printf("\n (b) Sequence impedance matrix of line , [Z_012] = \n") ; disp(Z_012) ;
diff --git a/1820/CH9/EX9.4/Example9_4.sce b/1820/CH9/EX9.4/Example9_4.sce
new file mode 100755
index 000000000..8565dbd34
--- /dev/null
+++ b/1820/CH9/EX9.4/Example9_4.sce
@@ -0,0 +1,36 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 9 : SYMMETRICAL COMPONENTS AND FAULT ANALYSIS
+
+// EXAMPLE : 9.4 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+l = 40 ; // line length in miles
+// Conductor parameter from Table A.3
+r_a = 0.206 ; // Ohms per conductor per mile in Ω/mi
+r_b = r_a ; // r_a = r_b = r_c in Ω/mi
+D_s = 0.0311 ; // GMR in ft where D_s = D_sa = D_sb = D_sc
+D_ab = sqrt(2^2 + 8^2) ; // GMR in ft
+D_bc = sqrt(3^2 + 13^2) ; // GMR in ft
+D_ac = sqrt(5^2 + 11^2) ; // GMR in ft
+D_e = 2788.5 ; // GMR in ft since earth resistivity is zero
+r_e = 0.09528 ; // At 60 Hz in Ω/mi
+
+// CALCULATIONS
+// For case (a)
+Z_s =[(r_a + r_e) + %i*0.1213*log(D_e/D_s)]*l ; // Self impedance of line conductor in Ω . From equ 9.49
+D_eq = (D_ab * D_bc * D_ac)^(1/3) ; // Equ GMR
+Z_m = [r_e + %i*0.1213*log(D_e/D_eq)]*l ; // From equ 9.50
+Z_abc = [Z_s Z_m Z_m ; Z_m Z_s Z_m ; Z_m Z_m Z_s] ; // Line impedance matrix
+
+// For case (b)
+Z_012 = [(Z_s+2*Z_m) 0 0 ; 0 (Z_s-Z_m) 0 ; 0 0 (Z_s-Z_m)] ; // Sequence impedance matrix . From equ 9.54
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 9.4 : SOLUTION :-") ;
+printf("\n (a) Line impedance matrix when line is completely transposed , [Z_abc] = \n") ; disp(Z_abc) ;
+printf("\n (b) Sequence impedance matrix when line is completely transposed , [Z_012] = \n") ; disp(Z_012) ;
diff --git a/1820/CH9/EX9.5/Example9_5.sce b/1820/CH9/EX9.5/Example9_5.sce
new file mode 100755
index 000000000..15e6dc2c2
--- /dev/null
+++ b/1820/CH9/EX9.5/Example9_5.sce
@@ -0,0 +1,45 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 9 : SYMMETRICAL COMPONENTS AND FAULT ANALYSIS
+
+// EXAMPLE : 9.5 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+Z_012 = [(19.6736 + 109.05044*%i) (0.5351182 + 0.4692097*%i) (- 0.5351182 + 0.4692097*%i) ; (- 0.5351182 + 0.4692097*%i) (8.24 + 28.471684*%i) (- 1.0702365 - 0.9384195*%i) ; (0.5351182 + 0.4692097*%i) (1.0702365 - 0.9384195*%i) (8.24 + 28.471684*%i)] ; // Line impedance matrix . result of exa 9.3
+Y_012 = inv(Z_012) ; // Sequence admittance of line
+
+// CALCULATIONS
+// For case (a)
+Y_01 = Y_012(1,2) ;
+Y_11 = Y_012(2,2) ;
+m_0 = Y_01/Y_11 ; // Per-unit unbalance for zero-sequence in pu from equ 9.67b
+m_0_per = m_0 * 100 ; // Per-unit unbalance for zero-sequence in percentage
+
+// For case (b)
+Z_01 = Z_012(1,2) ;
+Z_00 = Z_012(1,1) ;
+m_01 = -(Z_01/Z_00) ; // Per-unit unbalance for zero-sequence in pu from equ 9.67b
+m_01_per = m_01 * 100 ; // Per-unit unbalance for zero-sequence in percentage
+
+// For case (c)
+Y_21 = Y_012(3,2) ;
+Y_11 = Y_012(2,2) ;
+m_2 = (Y_21/Y_11) ; // Per-unit unbalance for zero-sequence in pu from equ 9.67b
+m_2_per = m_2 * 100 ; // Per-unit unbalance for zero-sequence in percentage
+
+// For case (d)
+Z_21 = Z_012(3,2) ;
+Z_22 = Z_012(3,3) ;
+m_21 = -(Z_21/Z_22) ; // Per-unit unbalance for zero-sequence in pu from equ 9.67b
+m_21_per = m_21 * 100 ; // Per-unit unbalance for zero-sequence in percentage
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 9.5 : SOLUTION :-") ;
+printf("\n (a) Per-unit electromagnetic unbalance for zero-sequence , m_0 = %.2f<%.1f percent pu \n",abs(m_0_per),atand( imag(m_0_per),real(m_0_per) )) ;
+printf("\n (b) Approximate value of Per-unit electromagnetic unbalance for negative-sequence , m_0 = %.2f<%.1f percent pu \n",abs(m_01_per),atand( imag(m_01_per),real(m_01_per) )) ;
+printf("\n (c) Per-unit electromagnetic unbalance for negative-sequence , m_2 = %.2f<%.1f percent pu \n",abs(m_2_per),atand( imag(m_2_per),real(m_2_per) )) ;
+printf("\n (d) Approximate value of Per-unit electromagnetic unbalance for negative-sequence , m_2 = %.2f<%.1f percent pu \n",abs(m_21_per),atand( imag(m_21_per),real(m_21_per) )) ;
diff --git a/1820/CH9/EX9.6/Example9_6.sce b/1820/CH9/EX9.6/Example9_6.sce
new file mode 100755
index 000000000..846a349c3
--- /dev/null
+++ b/1820/CH9/EX9.6/Example9_6.sce
@@ -0,0 +1,56 @@
+
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 9 : SYMMETRICAL COMPONENTS AND FAULT ANALYSIS
+
+// EXAMPLE : 9.6 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+kv = 115 ; // Line voltage in kV
+
+// For case (a)
+h_11 = 90 ; // GMD b/w ground wires & their images
+r_a = 0.037667 ; // Radius in metre
+p_aa = 11.185 * log(h_11/r_a) ; // unit is F^(-1)m
+p_bb = p_aa ;
+p_cc = p_aa ;
+l_12 = sqrt(22 + (45 + 37)^2) ;
+D_12 = sqrt(2^2 + 8^2) ; // GMR in ft
+p_ab = 11.185*log(l_12/D_12) ; // unit is F^(-1)m
+p_ba = p_ab ;
+D_13 = sqrt(3^2 + 13^2) ; // GMR in ft
+l_13 = 94.08721051 ;
+p_ac = 11.185 * log(l_13/D_13) ; // unit is F^(-1)m
+p_ca = p_ac ;
+l_23 = 70.72279912 ;
+D_23 = sqrt(5^2 + 11^2) ; // GMR in ft
+p_bc = 11.185 * log(l_23/D_23) ; // unit is F^(-1)m
+p_cb = p_bc ;
+P_abc = [p_aa p_ab p_ac ; p_ba p_bb p_bc ; p_ca p_cb p_cc] ; // Matrix of potential coefficients
+
+// For case (b)
+C_abc = inv(P_abc) ; // Matrix of maxwells coefficients
+
+// For case (c)
+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] ;
+C_012 = inv(A) * C_abc * A ; // Matrix of sequence capacitances
+
+// For case (d)
+C_01 = C_012(1,2) ;
+C_11 = C_012(2,2) ;
+C_21 = C_012(3,2) ;
+d_0 = C_01/C_11 ; // Zero-sequence electrostatic unbalances . Refer equ 9.115
+d_2 = -C_21/C_11 ; // Negative-sequence electrostatic unbalances . Refer equ 9.116
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 9.6 : SOLUTION :-") ;
+printf("\n (a) Matrix of potential coefficients , [P_abc] = \n") ; disp(P_abc) ;
+printf("\n (b) Matrix of maxwells coefficients , [C_abc] = \n") ; disp(C_abc) ;
+printf("\n (c) Matrix of sequence capacitances , [C_012] = \n") ; disp(C_012) ;
+printf("\n (d) Zero-sequence electrostatic unbalances , d_0 = %.4f<%.1f \n",abs(d_0),atand( imag(d_0),real(d_0) )) ;
+printf("\n     Negative-sequence electrostatic unbalances , d_2 = %.4f<%.1f \n",abs(d_2),atand( imag(d_2),real(d_2) )) ;
diff --git a/1820/CH9/EX9.9/Example9_9.sce b/1820/CH9/EX9.9/Example9_9.sce
new file mode 100755
index 000000000..b712f6b58
--- /dev/null
+++ b/1820/CH9/EX9.9/Example9_9.sce
@@ -0,0 +1,52 @@
+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 9 : SYMMETRICAL COMPONENTS AND FAULT ANALYSIS
+
+// EXAMPLE : 9.9 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+kv = 230 ; // Line voltage in kV
+Z_0 = 0.56 * %i ; // impedance in Ω
+Z_1 = 0.2618 * %i ; // Impedance in Ω
+Z_2 = 0.3619 * %i ; // Impedance in Ω
+z_f = 5 + 0*%i ; // fault impedance in Ω
+v = 1 * exp(%i*0*%pi/180) ;
+
+// CALCULATIONS
+// For case (a)
+Z_B = kv^2/200 ; // Imedance base on 230 kV line
+Z_f = z_f/Z_B ; // fault impedance in pu Ω
+I_a0 = v/(Z_0 + Z_1 + Z_2 + 3*Z_f) ; // Sequence currents in pu A
+I_a1 = I_a0 ;
+I_a2 = I_a0 ;
+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] ;
+I_f = A * [I_a0 ; I_a1 ; I_a2] ; // Phase currents in pu A
+
+// For case (b)
+V_a = [0 ; v ; 0] - [Z_0 0 0 ; 0 Z_1 0 ; 0 0 Z_2]*[I_a0 ; I_a1 ; I_a2] ; // Sequence voltage in pu V
+V_f = A*V_a ; // Phase voltage in pu V
+
+// For case (c)
+V_abf = V_f(1,1) - V_f(2,1) ; // Line-to-line voltages at fault points in pu V
+V_bcf = V_f(2,1) - V_f(3,1) ; // Line-to-line voltages at fault points in pu V
+V_caf = V_f(3,1) - V_f(1,1) ; // Line-to-line voltages at fault points in pu V
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 9.9 : SOLUTION :-") ;
+printf("\n (b) Sequence currents , I_a0 = I_a1 = I_a2 = %.4f<%.1f pu A \n",abs(I_a0),atand(imag(I_a0),real(I_a0) )) ;
+printf("\n  Phase currents in pu A , [I_af ; I_bf ; I_cf] = pu A \n") ;
+printf("\n    %.4f<%.1f ",abs(I_f),atand(imag(I_f),real(I_f) )) ;
+printf("\n \n (c) Sequence voltages are , [V_a0 ; V_a1 ; V_a2 ] = pu V \n") ;
+printf("\n    %.4f<%.1f ",abs(V_a),atand(imag(V_a),real(V_a) )) ;
+printf("\n \n    Phase voltages are , [V_af ; V_bf ; V_cf ] = pu V \n") ;
+printf("\n    %.4f<%.1f ",abs(V_f),atand(imag(V_f),real(V_f) )) ;
+printf("\n \n (d) Line-to-line voltages at fault points are , V_abf = %.4f<%.1f pu V \n",abs(V_abf),atand(imag(V_abf),real(V_abf) )) ;
+printf("\n     Line-to-line voltages at fault points are , V_abf = %.4f<%.1f pu V \n",abs(V_bcf),atand(imag(V_bcf),real(V_bcf) )) ;
+printf("\n     Line-to-line voltages at fault points are , V_caf = %.4f<%.1f pu V \n",abs(V_caf),atand(imag(V_caf),real(V_caf) )) ;
+
+printf("\n NOTE : ERROR : Calclation mistake in textbook from case(c) onwards \n") ;