11 files changed, 391 insertions, 0 deletions
diff --git a/3472/CH12/EX12.1/Example12_1.sce b/3472/CH12/EX12.1/Example12_1.sce
new file mode 100644
index 000000000..6e781e9b6
--- /dev/null
+++ b/3472/CH12/EX12.1/Example12_1.sce
@@ -0,0 +1,29 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES
+
+// EXAMPLE : 5.1 :
+// Page number 198
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+u = 5758.0       // Ultimate strength(kg)
+S = 2.0          // Sag(m)
+s = 2.0          // Factor of safety
+L = 250.0        // Span length(m)
+
+// Calculations
+T = u/s                                // Allowable max tension(kg)
+w = S*8.0*T/L**2                       // weight(kg/m)
+l = L/2                                // Half span length(m)
+half_span = l+(w**2*l**3/(6*T**2))     // Half span length(m)
+total_length = 2*half_span             // Total length(m)
+weight = w*total_length                // Weight of conductor(kg)
+
+// Results
+disp("PART II - EXAMPLE : 5.1 : SOLUTION :-")
+printf("\nWeight of conductor = %.2f kg", weight)
diff --git a/3472/CH12/EX12.10/Example12_10.sce b/3472/CH12/EX12.10/Example12_10.sce
new file mode 100644
index 000000000..bcb91a03e
--- /dev/null
+++ b/3472/CH12/EX12.10/Example12_10.sce
@@ -0,0 +1,33 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES
+
+// EXAMPLE : 5.10 :
+// Page number 201-202
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+L = 250.0          // Span(m)
+d = 1.42           // Diameter(cm)
+w = 1.09           // Dead weight(kg/m)
+wind = 37.8        // Wind pressure(kg/m^2)
+r = 1.25           // Ice thickness(cm)
+f_m = 1050.0       // Maximum working stress(kg/sq.cm)
+
+// Calculations
+w_i = 913.5*%pi*r*(d+r)*10**-4          // Weight of ice on conductor(kg/m)
+w_w = wind*(d+2*r)*10**-2               // Wind load of conductor(kg/m)
+w_r  = ((w+w_i)**2+w_w**2)**0.5         // Resultant pressure(kg/m)
+a = %pi*d**2/4.0                        // Area(cm^2)
+T_0 = f_m*a                             // Tension(kg)
+S = w_r*L**2/(8*T_0)                    // Total sag(m)
+vertical_sag = S*(w+w_i)/w_r            // Vertical component of sag(m)
+
+// Results
+disp("PART II - EXAMPLE : 5.10 : SOLUTION :-")
+printf("\nCase(i) : Sag in inclined direction = %.f m", S)
+printf("\nCase(ii): Sag in vertical direction = %.2f m", vertical_sag)
diff --git a/3472/CH12/EX12.11/Example12_11.sce b/3472/CH12/EX12.11/Example12_11.sce
new file mode 100644
index 000000000..cbb405ed7
--- /dev/null
+++ b/3472/CH12/EX12.11/Example12_11.sce
@@ -0,0 +1,47 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES
+
+// EXAMPLE : 5.11 :
+// Page number 202-203
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+a = 120.0                 // Area(mm^2)
+ds = 2.11                 // Diameter of each strand(mm)
+W = 1118.0/1000           // Weight of conductor(kg/m)
+L = 200.0                 // Span(m)
+stress = 42.2             // Ultimate tensile stress(kg/mm^2)
+wind = 60.0               // Wind pressure(kg/m^2)
+t = 10.0                  // Ice thickness(mm)
+
+// Calculations
+n = 3.0                                          // Number of layers
+d = (2*n+1)*ds                                   // Overall diameter of conductor(mm)
+u = stress*a                                     // Ultimate strength(kg)
+T = u/4.0                                        // Working stregth(kg)
+// Case(a)
+S_a = W*L**2/(8*T)                               // Sag in still air(m)
+// Case(b)
+area = d*100*10.0*10**-6                         // Projected area to wind pressure(m^2)
+w_w = wind*area                                  // Wind load/m(kg)
+w_r = (W**2+w_w**2)**0.5                         // Resultant weight/m(kg)
+S_b = w_r*L**2/(8*T)                             // Total sag with wind pressure(m)
+w_i = 0.915*%pi/4*((d+2*t)**2-(d**2))/1000.0     // Weight of ice on conductor(kg/m)
+area_i = (d+2*t)*1000.0*10**-6                   // Projected area to wind pressure(m^2)
+w_n = wind*area_i                                // Wind load/m(kg)
+w_r_c = ((W+w_i)**2+w_n**2)**0.5                 // Resultant weight/m(kg)
+S_c = w_r_c*L**2/(8*T)                           // Total sag with wind pressure and ice coating(m)
+S_v = S_c*(W+w_i)/w_r_c                          // Vertical component of sag(m)
+
+// Results
+disp("PART II - EXAMPLE : 5.11 : SOLUTION :-")
+printf("\nCase(a) : Sag in still air, S = %.2f m", S_a)
+printf("\nCase(b) : Sag with wind pressure, S = %.2f m", S_b)
+printf("\n          Sag with wind pressure and ice coating, S = %.2f m", S_c)
+printf("\n          Vertical sag, S_v = %.2f m \n", S_v)
+printf("\nNOTE: ERROR: calculation mistake in the textbook")
diff --git a/3472/CH12/EX12.2/Example12_2.sce b/3472/CH12/EX12.2/Example12_2.sce
new file mode 100644
index 000000000..14d967e14
--- /dev/null
+++ b/3472/CH12/EX12.2/Example12_2.sce
@@ -0,0 +1,31 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES
+
+// EXAMPLE : 5.2 :
+// Page number 198
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+L = 250.0             // Span length(m)
+h = 10.0              // Difference in height(m)
+r = 1.0               // Radius of conductor(cm)
+w = 2.5               // Weight of conductor(kg/m)
+wind = 1.2            // Wind load(kg/m)
+s = 3.0               // Factor of safety
+tensile = 4300.0      // Maximum tensile strength(kg/sq.cm)
+
+// Calculations
+W = (w**2+wind**2)**0.5       // Total pressure on conductor(kg/m)
+f = tensile/s                 // Permissible stress in conductor(kg/sq.cm)   
+a = %pi*r**2                  // Area of the conductor(sq.cm)
+T = f*a                       // Allowable max tension(kg)
+x = (L/2)-(T*h/(L*W))         // Point of maximum sag at the lower support(m)
+
+// Results
+disp("PART II - EXAMPLE : 5.2 : SOLUTION :-")
+printf("\nPoint of maximum sag at the lower support, x = %.2f metres", x)
diff --git a/3472/CH12/EX12.3/Example12_3.sce b/3472/CH12/EX12.3/Example12_3.sce
new file mode 100644
index 000000000..04d7dd31a
--- /dev/null
+++ b/3472/CH12/EX12.3/Example12_3.sce
@@ -0,0 +1,31 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES
+
+// EXAMPLE : 5.3 :
+// Page number 198-199
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+a = 2.5             // Cross-sectional area(sq.cm)
+L = 250.0           // Span(m)
+w_c = 1.8           // Weight of conductor(kg/m)
+u = 8000.0          // Ultimate strength(kg/cm^2)
+wind = 40.0         // Wind load(kg/cm^2)
+s = 3.0             // Factor of safety
+
+// Calculations
+d = (4.0*a/%pi)**0.5           // Diameter(cm)
+T = u*a/s                      // Allowable max tension(kg)
+w_w = wind*d/100.0             // Horizontal wind force(kg)
+w_r = (w_c**2+w_w**2)**0.5     // Resultant force(kg/m)
+S = w_r*L**2/(8*T)             // Slant sag(m)
+vertical_sag = S*(w_c/w_r)     // Vertical sag(m)
+
+// Results
+disp("PART II - EXAMPLE : 5.3 : SOLUTION :-")
+printf("\nVertical sag = %.3f metres", vertical_sag)
diff --git a/3472/CH12/EX12.4/Example12_4.sce b/3472/CH12/EX12.4/Example12_4.sce
new file mode 100644
index 000000000..1da68593b
--- /dev/null
+++ b/3472/CH12/EX12.4/Example12_4.sce
@@ -0,0 +1,35 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES
+
+// EXAMPLE : 5.4 :
+// Page number 199
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+a = 110.0               // Cross-sectional area(sq.mm)
+w_c = 844.0/1000        // Weight of conductor(kg/m)
+U = 7950.0              // Ultimate strength(kg)
+L = 300.0               // Span(m)
+s = 2.0                 // Factor of safety
+wind = 75.0             // Wind pressure(kg/m^2)
+h = 7.0                 // Ground clearance(m)
+d = 2.79                // Diameter of copper(mm)
+n = 7.0                 // Number of strands
+
+// Calculations
+dia = n*d                        // Diameter of conductor(mm)
+w_w = wind*dia/1000.0            // Horizontal wind force(kg)
+w = (w_c**2+w_w**2)**0.5         // Resultant force(kg)
+T = U/2.0                        // Allowable tension(m)
+l = L/2.0                        // Half-span(m)
+D = w*l**2/(2*T)                 // Distance(m)
+height = h+D                     // Height above ground at which the conductors should be supported(m)
+
+// Results
+disp("PART II - EXAMPLE : 5.4 : SOLUTION :-")
+printf("\nHeight above ground at which the conductors should be supported = %.2f metres", height)
diff --git a/3472/CH12/EX12.5/Example12_5.sce b/3472/CH12/EX12.5/Example12_5.sce
new file mode 100644
index 000000000..71b843d2b
--- /dev/null
+++ b/3472/CH12/EX12.5/Example12_5.sce
@@ -0,0 +1,30 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES
+
+// EXAMPLE : 5.5 :
+// Page number 199
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+w_w = 1.781          // Wind pressure on conductor(kg/m)
+w_i = 1.08           // Weight of ice on conductor(kg/m)
+D = 6.0              // Maximum permissible sag(m)
+s = 2.0              // Factor of safety
+w_c = 0.844          // Weight of conductor(kg/m)
+u = 7950.0           // Ultimate strength(kg)
+
+// Calculations
+w = ((w_c+w_i)**2+w_w**2)**0.5       // Total force on conductor(kg/m)
+T = u/s                              // Allowable maximum tension(kg)
+l = ((D*2*T)/w)**0.5                 // Half span(m)
+L = 2.0*l                            // Permissible span between two supports(m)
+
+// Results
+disp("PART II - EXAMPLE : 5.5 : SOLUTION :-")
+printf("\nPermissible span between two supports = %.f metres \n", L)
+printf("\nNOTE: ERROR: Horizontal wind load, w_w = 1.781 kg/m, not 1.78 kg/m as mentioned in problem statement")
diff --git a/3472/CH12/EX12.6/Example12_6.sce b/3472/CH12/EX12.6/Example12_6.sce
new file mode 100644
index 000000000..db90f76f1
--- /dev/null
+++ b/3472/CH12/EX12.6/Example12_6.sce
@@ -0,0 +1,45 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES
+
+// EXAMPLE : 5.6 :
+// Page number 199-200
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+a = 0.484           // Area of conductor(sq.cm)
+d = 0.889           // Overall diameter(cm)
+w_c = 428/1000.0    // Weight(kg/m)
+u = 1973.0          // Breaking strength(kg)
+s = 2.0             // Factor of safety
+L = 200.0           // Span(m)
+t = 1.0             // Ice thickness(cm)
+wind = 39.0         // Wind pressure(kg/m^2)
+
+// Calculations
+// Case(i)
+l = L/2.0                                         // Half span(m)
+T = u/s                                           // Allowable maximum tension(kg)
+D_1 = w_c*l**2/(2*T)                              // Maximum sag due to weight of conductor(m)
+// Case(ii)
+w_i = 913.5*%pi*t*(d+t)*10**-4                    // Weight of ice on conductor(kg/m)
+w = w_c+w_i                                       // Total weight of conductor & ice(kg/m)
+D_2 = w*l**2/(2*T)                                // Maximum sag due to additional weight of ice(m)
+// Case(iii)
+D = d+2.0*t                                       // Diameter due to ice(cm)
+w_w = wind*D*10**-2                               // Wind pressure on conductor(kg/m)
+w_3 = ((w_c+w_i)**2+w_w**2)**0.5                  // Total force on conductor(kg/m)
+D_3 = w_3*l**2/(2*T)                              // Maximum sag due to (i), (ii) & wind(m)
+theta = atand(w_w/(w_c+w_i))                      // θ(°)
+vertical_sag = D_3*cosd(theta)                    // Vertical sag(m)
+
+// Results
+disp("PART II - EXAMPLE : 5.6 : SOLUTION :-")
+printf("\nCase(i)  : Maximum sag of line due to weight of conductor, D = %.2f metres", D_1)
+printf("\nCase(ii) : Maximum sag of line due to additional weight of ice, D = %.2f metres", D_2)
+printf("\nCase(iii): Maximum sag of line due to (i),(ii) plus wind, D = %.2f metres", D_3)
+printf("\n           Vertical sag = %.2f metres", vertical_sag)
diff --git a/3472/CH12/EX12.7/Example12_7.sce b/3472/CH12/EX12.7/Example12_7.sce
new file mode 100644
index 000000000..ea011306d
--- /dev/null
+++ b/3472/CH12/EX12.7/Example12_7.sce
@@ -0,0 +1,28 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES
+
+// EXAMPLE : 5.7 :
+// Page number 200
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+W = 428/1000.0      // Weight(kg/m)
+u = 1973.0          // Breaking strength(kg)
+s = 2.0             // Factor of safety
+l = 200.0           // Span(m)
+h = 3.0             // Difference in tower height(m)
+
+// Calculations
+T = u/s                               // Allowable maximum tension(kg)
+x_2 = (l/2.0)+(T*h/(W*l))             // Point of minimum sag from tower at higher level(m)
+x_1 = l-x_2                           // Point of minimum sag from tower at lower level(m)
+
+// Results
+disp("PART II - EXAMPLE : 5.7 : SOLUTION :-")
+printf("\nPoint of minimum sag, x_1 = %.1f metres", x_1)
+printf("\nPoint of minimum sag, x_2 = %.1f metres", x_2)
diff --git a/3472/CH12/EX12.8/Example12_8.sce b/3472/CH12/EX12.8/Example12_8.sce
new file mode 100644
index 000000000..ce1b55025
--- /dev/null
+++ b/3472/CH12/EX12.8/Example12_8.sce
@@ -0,0 +1,37 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES
+
+// EXAMPLE : 5.8 :
+// Page number 200-201
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+h_1 = 50.0          // Height of tower P1(m)
+h_2 = 80.0          // Height of tower P2(m)
+L = 300.0           // Horizontal distance b/w towers(m)
+T = 2000.0          // Tension in conductor(kg)
+w = 0.844           // Weight of conductor(kg/m)
+
+// Calculations
+h = h_2-h_1                           // Difference in height of tower(m)
+x_2 = (L/2.0)+(T*h/(w*L))             // Point of minimum sag from tower P2(m)
+x_1 = (L/2.0)-(T*h/(w*L))             // Point of minimum sag from tower at lower level(m)
+P = (L/2.0)-x_1                       // Distance of point P(m)
+D = w*P**2/(2*T)                      // Height of P above O(m)
+D_2 = w*x_2**2/(2*T)                  // Height of P2 above O(m)
+mid_point_P2 = D_2-D                  // Mid-point below P2(m)
+clearance = h_2-mid_point_P2          // Clearance b/w conductor & water(m)
+D_1 = w*x_1**2/(2*T)                  // Height of P1 above O(m)
+mid_point_P1 = D-D_1                  // Mid-point above P1(m)
+clearance_alt = h_1+mid_point_P1      // Clearance b/w conductor & water(m)
+
+// Results
+disp("PART II - EXAMPLE : 5.8 : SOLUTION :-")
+printf("\nClearance between conductor & water at a point midway b/w towers = %.2f m above water\n", clearance)
+printf("\nALTERNATIVE METHOD:")
+printf("\nClearance between conductor & water at a point midway b/w towers = %.2f m above water", clearance_alt)
diff --git a/3472/CH12/EX12.9/Example12_9.sce b/3472/CH12/EX12.9/Example12_9.sce
new file mode 100644
index 000000000..dbe03be1a
--- /dev/null
+++ b/3472/CH12/EX12.9/Example12_9.sce
@@ -0,0 +1,45 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 5: MECHANICAL DESIGN OF OVERHEAD LINES
+
+// EXAMPLE : 5.9 :
+// Page number 201
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+L = 300.0            // Span(m)
+T_still = 45.0       // Temperature in still air(°C)
+a = 226.0            // Area(mm^2)
+d = 19.53/10         // Overall diameter(cm)
+w_2 = 0.844          // Weight of conductor(kg/m)
+u = 7950.0           // Ultimate strength(kg)
+alpha = 18.44*10**-6 // Co-efficient of linear expression(/°C)
+E = 9.32*10**3       // Modulus of elasticity(kg/mm^2)
+t = 0.95             // Ice thickness(cm)
+wind = 39.0          // Wind pressure(kg/m^2)
+T_worst = -5.0       // Temperature in worst condition(°C)
+
+// Calculations
+w_i = 915.0*%pi*t*(d+t)*10**-4          // Weight of ice on conductor(kg/m)
+w_w = wind*(d+2*t)*10**-2               // Wind load of conductor(kg/m)
+w_1  = ((w_2+w_i)**2+w_w**2)**0.5       // Total force on conductor(kg/m)
+t = T_still-T_worst                     // Temperature(°C)
+l = L/2.0                               // Half span(m)
+T = u/2.0                               // Allowable tension(kg)
+A = 1.0                                 // Co-efficient of x^3
+B = a*E*(alpha*t+((w_1*l/T)**2/6))-T    // Co-efficient of x^2
+C = 0                                   // Co-efficient of x
+D = -(w_2**2*l**2*a*E/6)                // Co-efficient of constant
+T_2_sol = roots([A,B,C,D])              // Roots of tension of a line
+T_2_s = T_2_sol(3)                      // Feasible solution of tension of 
+T_2 = 1710.0                            // Tension in conductor(kg). Obtianed directly from textbook
+sag = w_2*l**2/(2*T_2)                  // Sag at erection(m)
+
+// Results
+disp("PART II - EXAMPLE : 5.9 : SOLUTION :-")
+printf("\nSag at erection = %.2f metres", sag)
+printf("\nTension of the line, T_2 = %.f kg (An app. solution as per calculation) = %.f kg (More correctly as standard value)", T_2_s,T_2)