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diff --git a/3472/CH24/EX24.1/Example24_1.sce b/3472/CH24/EX24.1/Example24_1.sce
new file mode 100644
index 000000000..6ca53dff1
--- /dev/null
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@@ -0,0 +1,41 @@
+// 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 17: ELECTRIC POWER SUPPLY SYSTEMS

+

+// EXAMPLE : 17.1 :

+// Page number 422-423

+clear ; clc ; close ; // Clear the work space and console

+

+// Given data

+no_phase = 3.0          // Number of phases in ac transmission system

+V = 380.0*10**3         // Voltage b/w lines(V)

+load = 100.0            // Load(MW)

+PF = 0.9                // Power factor

+l = 150.0               // Line length(km)

+n = 0.92                // Efficiency

+r = 0.045               // Resistance(ohm/km/sq.cm)

+w_cu_1 = 0.01           // Weight of 1 cm^3 copper(kg)

+

+// Calculations

+// Case(i)

+P_loss = (1-n)*load                  // Power loss in the line(MW)

+I_L = load*10**6/(3**0.5*V*PF)       // Line current(A)

+loss_cu = P_loss/no_phase*10**6      // I^2*R loss per conductor(W)

+R = loss_cu/I_L**2                   // Resistance per conductor(ohm)

+R_km = R/l                           // Resistance per conductor per km(ohm)

+area = r/R_km                        // Conductor area(Sq.cm)

+volume = area*100.0                  // Volume of copper per km run(cm^3)

+W_cu_km = volume*w_cu_1              // Weight of copper per km run(kg)

+W_cu = no_phase*l*1000*W_cu_km       // Weight of copper for 3 conductors of 150 km(kg)

+// Case(ii)

+W_cu_dc = 1.0/2*PF**2*W_cu           // Weight of copper conductor in dc(kg)

+

+// Results

+disp("PART II - EXAMPLE : 17.1 : SOLUTION :-")

+printf("\nWeight of copper required for a three-phase transmission system = %.f kg", W_cu)

+printf("\nWeight of copper required for the d-c transmission system = %.f kg \n", W_cu_dc)

+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision")

diff --git a/3472/CH24/EX24.2/Example24_2.sce b/3472/CH24/EX24.2/Example24_2.sce
new file mode 100644
index 000000000..5a018c867
--- /dev/null
+++ b/3472/CH24/EX24.2/Example24_2.sce
@@ -0,0 +1,22 @@
+// 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 17: ELECTRIC POWER SUPPLY SYSTEMS

+

+// EXAMPLE : 17.2 :

+// Page number 423

+clear ; clc ; close ; // Clear the work space and console

+

+// Given data

+P_1 = 1.0    // Assume P1 to be 1

+

+// Calculations

+P_2 = (3.0*2)**0.5            // 3-phase power transmitted in terms of P_1

+inc_per = (P_2-P_1)/P_1*100   // Increase in power transmitted(%)

+

+// Results

+disp("PART II - EXAMPLE : 17.2 : SOLUTION :-")

+printf("\nPercentage increase in power transmitted = %.f percent", inc_per)

diff --git a/3472/CH24/EX24.3/Example24_3.sce b/3472/CH24/EX24.3/Example24_3.sce
new file mode 100644
index 000000000..96520138e
--- /dev/null
+++ b/3472/CH24/EX24.3/Example24_3.sce
@@ -0,0 +1,23 @@
+// 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 17: ELECTRIC POWER SUPPLY SYSTEMS

+

+// EXAMPLE : 17.3 :

+// Page number 424

+clear ; clc ; close ; // Clear the work space and console

+

+// Given data

+PF = 0.95    // Lagging power factor

+

+// Calculations

+P_1 = 1.0                                     // Power in terms of V*I_1

+P_2 = 2.0*PF**2                               // Power in terms of V*I_1

+P_additional_percentage = (P_2-P_1)/P_1*100   // Percentage additional power transmitted in a 3-phase 3-wire system

+

+// Results

+disp("PART II - EXAMPLE : 17.3 : SOLUTION :-")

+printf("\nPercentage additional power transmitted in a 3-phase 3-wire system = %.f percent", P_additional_percentage)

diff --git a/3472/CH24/EX24.4/Example24_4.sce b/3472/CH24/EX24.4/Example24_4.sce
new file mode 100644
index 000000000..adc332977
--- /dev/null
+++ b/3472/CH24/EX24.4/Example24_4.sce
@@ -0,0 +1,22 @@
+// 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 17: ELECTRIC POWER SUPPLY SYSTEMS

+

+// EXAMPLE : 17.4 :

+// Page number 424-425

+clear ; clc ; close ; // Clear the work space and console

+

+// Given data

+n = 3.0       // 3-phase 4 wire ac system

+

+// Calculations

+a2_a1 = 1.0/6            // Ratio of cross-sectional area of 2 wire dc to 3-phase 4-wire system

+ratio_cu = 3.5/2*a2_a1   // Copper for 3 phase 4 wire system to copper for 2 wire dc system

+

+// Results

+disp("PART II - EXAMPLE : 17.4 : SOLUTION :-")

+printf("\nCopper for 3-phase 4-wire system/Copper for 2-wire dc system = %.3f : 1", ratio_cu)

diff --git a/3472/CH24/EX24.5/Example24_5.sce b/3472/CH24/EX24.5/Example24_5.sce
new file mode 100644
index 000000000..6975ac3c5
--- /dev/null
+++ b/3472/CH24/EX24.5/Example24_5.sce
@@ -0,0 +1,41 @@
+// 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 17: ELECTRIC POWER SUPPLY SYSTEMS

+

+// EXAMPLE : 17.5 :

+// Page number 425

+clear ; clc ; close ; // Clear the work space and console

+

+// Given data

+L = 60.0            // Line length(km)

+P = 5.0             // Load(MW)

+PF = 0.8            // Lagging power factor

+V = 33.0*10**3      // Voltage(V)

+n = 0.85            // Transmission efficiency

+rho = 1.73*10**-8   // Specific resistance of copper(ohm-mt)

+density = 8900.0    // Density(kg/mt^3)

+

+// Calculations

+I = P*10**6/(3**0.5*V*PF)               // Line current(A)

+line_loss = (1-n)*P*1000/n              // Line loss(kW)

+line_loss_phase = line_loss/3.0         // Line loss/phase(kW)

+R = line_loss_phase*1000/I**2           // Resistance/phase(ohm)

+a = rho*L*1000/R                        // Area of cross section of conductor(m^2)

+volume = 3.0*a*L*1000                   // Volume of copper(m^3)

+W_cu = volume*density                   // Weight of copper in 3-phase system(kg)

+I_1 = P*10**6/V                         // Current in single phase system(A)

+R_1 = line_loss*1000/(2*I_1**2)         // Resistance in single phase system(ohm)

+a_1 = rho*L*1000/R_1                    // Area of cross section of conductor in single phase system(m^2)

+volume_1 = 2.0*a_1*L*1000               // Volume of copper(m^3)

+W_cu_1 = volume_1*density               // Weight of copper in 1-phase system(kg)

+reduction_cu = (W_cu-W_cu_1)/W_cu*100   // Reduction in copper(%)

+

+// Results

+disp("PART II - EXAMPLE : 17.5 : SOLUTION :-")

+printf("\nWeight of copper required for 3-phase 2-wire system = %.2e kg", W_cu)

+printf("\nReduction of weight of copper possible = %.1f percent \n", reduction_cu)

+printf("\nNOTE: ERROR: Calculation mistakes in the textbook solution")

diff --git a/3472/CH24/EX24.6/Example24_6.sce b/3472/CH24/EX24.6/Example24_6.sce
new file mode 100644
index 000000000..c7a194cbc
--- /dev/null
+++ b/3472/CH24/EX24.6/Example24_6.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 17: ELECTRIC POWER SUPPLY SYSTEMS

+

+// EXAMPLE : 17.6 :

+// Page number 427-428

+clear ; clc ; close ; // Clear the work space and console

+

+// Given data

+L = 250.0                 // Cable length(m)

+P = 80.0*10**3            // Load(W)

+V = 400.0                 // Voltage(V)

+PF = 0.8                  // Lagging power factor

+time = 4000.0             // Time of operation(hours/annum)

+a = poly(0,'a')           // Area of each conductor(Sq.cm)

+cost_instal = 15.0*a+25   // Cost of cable including installation(Rs/m)

+interest_per = 0.1        // Interest & depreciation

+cost_waste_per = 0.1      // Cost of energy wasted(Rs/unit)

+r = 0.173                 // Resistance per km of 1 cm^2(ohm)

+

+// Calculations

+I = P/(3**0.5*V*PF)                                     // Line current(A)

+energy_waste = 3.0*I**2*r/a*L*10**-3*time*10**-3        // Energy wasted per annum(kWh)

+cost_energy_waste = cost_waste_per*energy_waste         // Annual cost of energy wasted as losses(Rs)

+capitaL_cost_cable = cost_instal*L                      // Capital cost of cable(Rs)

+annual_cost_cable = capitaL_cost_cable*cost_waste_per   // Annual cost on cable(Rs)

+area = (1081.25/375)**0.5                               // Area = a(Sq.cm). Simplified and taken final answer

+

+// Results

+disp("PART II - EXAMPLE : 17.6 : SOLUTION :-")

+printf("\nEconomical cross-section of a 3-core distributor cable, a = %.1f cm^2", area)

diff --git a/3472/CH24/EX24.7/Example24_7.sce b/3472/CH24/EX24.7/Example24_7.sce
new file mode 100644
index 000000000..73082f6ce
--- /dev/null
+++ b/3472/CH24/EX24.7/Example24_7.sce
@@ -0,0 +1,40 @@
+// 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 17: ELECTRIC POWER SUPPLY SYSTEMS

+

+// EXAMPLE : 17.7 :

+// Page number 428

+clear ; clc ; close ; // Clear the work space and console

+

+// Given data

+V = 110.0*10**3               // Voltage(V)

+l_1 = 24.0*10**6              // Load(MW)

+t_1 = 6.0                     // Time(hours)

+l_2 = 8.0*10**6               // Load(MW)

+t_2 = 6.0                     // Time(hours)

+l_3 = 4.0*10**6               // Load(MW)

+t_3 = 12.0                    // Time(hours)

+PF = 0.8                      // Lagging power factor

+a = poly(0,'a')               // Cross-section of each conductor(Sq.cm)

+cost_line = 12000.0+8000*a    // Cost of line including erection(Rs/km)

+R = 0.19/a                    // Resistance per km of each conductor(ohm)

+cost_energy = 8.0/100         // Energy cost(Rs/unit)

+interest_per = 0.1            // Interest & depreciation. Assumption

+

+// Calculations

+annual_charge = interest_per*cost_line     // Total annual charge(Rs)

+I_1 = l_1/(3**0.5*V*PF)                    // Line current for load 1(A)

+I_2 = l_2/(3**0.5*V*PF)                    // Line current for load 2(A)

+I_3 = l_3/(3**0.5*V*PF)                    // Line current for load 3(A)

+I_2_t = I_1**2*t_1+I_2**2*t_2+I_3**2*t_3   // I^2*t

+annual_energy = 3.0*R*365/1000*I_2_t       // Annual energy consumption on account of losses(kWh)

+cost_waste = annual_energy*cost_energy     // Cost of energy wasted per annum(Rs)

+area = (2888.62809917355/800.0)**0.5       // Economical cross-section = a(Sq.cm). Simplified and taken final answer

+

+// Results

+disp("PART II - EXAMPLE : 17.7 : SOLUTION :-")

+printf("\nMost economical cross-section, a = %.2f cm^2", area)

diff --git a/3472/CH24/EX24.8/Example24_8.sce b/3472/CH24/EX24.8/Example24_8.sce
new file mode 100644
index 000000000..5522c76d1
--- /dev/null
+++ b/3472/CH24/EX24.8/Example24_8.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 17: ELECTRIC POWER SUPPLY SYSTEMS

+

+// EXAMPLE : 17.8 :

+// Page number 428-429

+clear ; clc ; close ; // Clear the work space and console

+

+// Given data

+cost_km_cu = 2800.0         // Cost per km for each copper conductor of sq.cm(Rs)

+LF_I = 80.0/100             // Load factor of load current

+LF_loss = 65.0/100          // Load factor of losses

+interest_per = 10.0/100     // Rate of interest and depreciation

+cost_energy = 5.0/100       // Cost of energy(Rs/kWh)

+rho = 1.78*10**-8           // Resistivity(ohm-m)

+

+// Calculations

+P_2 = cost_km_cu*interest_per                   // Cost in terms of L(Rs)

+time_year = 365.0*24                            // Total hours in a year

+P_3 = cost_energy*rho*10**4*time_year*LF_loss   // Cost in terms of I^2 & L(Rs)

+delta = (P_2/P_3)**0.5                          // Economical current density for the transmission line(A/sq.cm)

+    

+// Results

+disp("PART II - EXAMPLE : 17.8 : SOLUTION :-")

+printf("\nMost economical current density for the transmission line, δ = %.f A/sq.cm", delta)

diff --git a/3472/CH24/EX24.9/Example24_9.sce b/3472/CH24/EX24.9/Example24_9.sce
new file mode 100644
index 000000000..47de007ac
--- /dev/null
+++ b/3472/CH24/EX24.9/Example24_9.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 17: ELECTRIC POWER SUPPLY SYSTEMS

+

+// EXAMPLE : 17.9 :

+// Page number 429

+clear ; clc ; close ; // Clear the work space and console

+

+// Given data

+MD = 1000.0                 // Maximum demand(kW)

+energy_cons = 5.0*10**6     // Annual energy consumption(kWh)

+PF = 0.85                   // Power factor

+capital_cost = 80000.0      // Capital cost of cable(Rs/km)

+cost_energy = 5.0/100       // Energy cost(Rs/kWh)

+interest_per = 10.0/100     // Rate of interest and depreciation

+r_specific = 1.72*10**-6    // Specific resistance of copper(ohm/cubic.cm)

+V = 11.0                    // Voltage(kV)

+

+// Calculations

+I = MD/(3**0.5*V*PF)                                             // Line current corresponding to maximum demand(A)

+hours_year = 365.0*24                                            // Total hours in a year

+LF = energy_cons/(MD*hours_year)                                 // Load factor

+loss_LF = 0.25*LF+0.75*LF**2                                     // Loss load factor

+P_2 = capital_cost*interest_per                                  // Cost in terms of L(Rs)

+P_3 = 3.0*I**2*r_specific*10**4*hours_year*loss_LF*cost_energy   // Cost in terms of I^2 & L(Rs)

+a = (P_3/P_2)**0.5                                               // Most economical cross-section of conductor(sq.cm)

+

+// Results

+disp("PART II - EXAMPLE : 17.9 : SOLUTION :-")

+printf("\nMost economical cross-section of the conductor, a = %.2f cm^2 \n", a)

+printf("\nNOTE: ERROR: Calculation mistake in the textbook solution")