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diff --git a/3472/CH43/EX43.1/Example43_1.sce b/3472/CH43/EX43.1/Example43_1.sce
new file mode 100644
index 000000000..2f257d607
--- /dev/null
+++ b/3472/CH43/EX43.1/Example43_1.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 IV : UTILIZATION AND TRACTION

+// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT

+

+// EXAMPLE : 5.1 :

+// Page number 778

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

+

+// Given data

+speed = 45.0      // Scheduled speed(kmph)

+D = 1.5           // Distance between 2 stops(km)

+t = 20.0          // Time of stop(sec)

+alpha = 2.4       // Acceleration(km phps)

+beta = 3.2        // Retardation(km phps)

+

+// Calculations

+t_total = D*3600/speed                   // Total time(sec)

+T = t_total-t                            // Actual time for run(sec)

+k = (alpha+beta)/(alpha*beta)            // Constant

+V_m = (T/k)-((T/k)**2-(7200*D/k))**0.5   // Maximum speed over the run(kmph)

+

+// Results

+disp("PART IV - EXAMPLE : 5.1 : SOLUTION :-")

+printf("\nMaximum speed over the run, V_m = %.f kmph", V_m)

diff --git a/3472/CH43/EX43.10/Example43_10.sce b/3472/CH43/EX43.10/Example43_10.sce
new file mode 100644
index 000000000..b44ae441e
--- /dev/null
+++ b/3472/CH43/EX43.10/Example43_10.sce
@@ -0,0 +1,29 @@
+// A Texbook on POWER SYSTEM ENGINEERING

+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar

+// DHANPAT RAI & Co.

+// SECOND EDITION 

+

+// PART IV : UTILIZATION AND TRACTION

+// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT

+

+// EXAMPLE : 5.10 :

+// Page number 784

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

+

+// Given data

+W = 350.0      // Weight of train(tonne)

+G = 1.0        // Gradient

+alpha = 0.8    // Acceleration(km phps)

+u = 0.25       // Co-efficient of adhesion

+r = 44.5       // Train resistance(N/tonne)

+I = 10.0       // Rotational inertia(%)

+

+// Calculations

+W_e = W*(100+I)/100                  // Accelerating weight of train(tonne)

+F_t = 277.8*W_e*alpha+W*r+98.1*W*G   // Tractive effort(N)

+adhesive_weight = F_t/(u*9.81*1000)  // Adhesive weight(tonnes)

+

+// Results

+disp("PART IV - EXAMPLE : 5.10 : SOLUTION :-")

+printf("\nMinimum adhesive weight of the locomotive = %.1f tonnes\n", adhesive_weight)

+printf("\nNOTE: ERROR: Train resistance is 44.5 N per tonne & not 45 N per tonne as mentioned in textbook problem statement")

diff --git a/3472/CH43/EX43.11/Example43_11.sce b/3472/CH43/EX43.11/Example43_11.sce
new file mode 100644
index 000000000..04866946a
--- /dev/null
+++ b/3472/CH43/EX43.11/Example43_11.sce
@@ -0,0 +1,34 @@
+// A Texbook on POWER SYSTEM ENGINEERING

+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar

+// DHANPAT RAI & Co.

+// SECOND EDITION 

+

+// PART IV : UTILIZATION AND TRACTION

+// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT

+

+// EXAMPLE : 5.11 :

+// Page number 784

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

+

+// Given data

+W = 400.0       // Weight of train(tonne)

+G = 100.0/75    // Gradient

+alpha = 1.6     // Acceleration(km phps)

+r = 66.75       // Train resistance(N/tonne)

+I = 10.0        // Rotational inertia(%)

+V = 48.0        // Speed(kmph)

+n = 0.7         // Overall efficiency of equipment

+

+// Calculations

+W_e = W*(100+I)/100                  // Accelerating weight of train(tonne)

+F_t = 277.8*W_e*alpha+W*r+98.1*W*G   // Tractive effort(N)

+t = V/alpha                          // Time(sec)

+energy_a = F_t*V*t/(2*3600**2)       // Energy usefully employed(kWh)

+G_r = 98.1*G+r                       // Force(N)

+work_tonne_km = G_r*1000             // Work done per tonne per km(Nw-m)

+energy_b = work_tonne_km/(n*3600)    // Energy consumption(Wh per tonne-km)

+

+// Results

+disp("PART IV - EXAMPLE : 5.11 : SOLUTION :-")

+printf("\nCase(a): Energy usefully employed in attaining speed = %.2f kWh", energy_a)

+printf("\nCase(b): Specific energy consumption at steady state speed = %.1f Wh per tonne-km", energy_b)

diff --git a/3472/CH43/EX43.12/Example43_12.sce b/3472/CH43/EX43.12/Example43_12.sce
new file mode 100644
index 000000000..6daa0b040
--- /dev/null
+++ b/3472/CH43/EX43.12/Example43_12.sce
@@ -0,0 +1,27 @@
+// A Texbook on POWER SYSTEM ENGINEERING

+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar

+// DHANPAT RAI & Co.

+// SECOND EDITION 

+

+// PART IV : UTILIZATION AND TRACTION

+// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT

+

+// EXAMPLE : 5.12 :

+// Page number 784-785

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

+

+// Given data

+W = 200.0       // Trailing weight(tonne)

+G = 1.0         // Gradient(%)

+alpha = 1.0     // Acceleration(km phps)

+u = 0.2         // Co-efficient of adhesion

+r = 50.0        // Train resistance(N/tonne)

+I = 10.0        // Rotational inertia(%)

+

+// Calculations

+W_L = ((277.8*(100+I)/100*alpha)+98.1*G+r)*W/(u*9.81*1000-((277.8*(100+I)/100*alpha)+98.1*G+r))  // Weight of locomotive(tonnes)

+

+// Results

+disp("PART IV - EXAMPLE : 5.12 : SOLUTION :-")

+printf("\nMinimum adhesive weight of a locomotive, W_L = %.1f tonnes\n", W_L)

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

diff --git a/3472/CH43/EX43.2/Example43_2.sce b/3472/CH43/EX43.2/Example43_2.sce
new file mode 100644
index 000000000..f7b0f6502
--- /dev/null
+++ b/3472/CH43/EX43.2/Example43_2.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 IV : UTILIZATION AND TRACTION

+// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT

+

+// EXAMPLE : 5.2 :

+// Page number 778

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

+

+// Given data

+V_m = 65.0        // Maximum speed(kmph)

+t = 30.0          // Time of stop(sec)

+speed = 43.5      // Scheduled speed(kmph)

+alpha = 1.3       // Acceleration(km phps)

+D = 3.0           // Distance between 2 stops(km)

+

+// Calculations

+t_total = D*3600/speed                                 // Total time of run including stop(sec)

+T = t_total-t                                          // Actual time for run(sec)

+V_a = D/T*3600                                         // Average speed(kmph)

+beta = 1/((7200.0*D/V_m**2*((V_m/V_a)-1))-(1/alpha))   // Value of retardation(km phps)

+

+// Results

+disp("PART IV - EXAMPLE : 5.2 : SOLUTION :-")

+printf("\nValue of retardation, β = %.3f km phps\n", beta)

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

+printf("\n      ERROR: β unit is km phps & not km phps as mentioned in textbook solution")

diff --git a/3472/CH43/EX43.3/Example43_3.sce b/3472/CH43/EX43.3/Example43_3.sce
new file mode 100644
index 000000000..b34a37915
--- /dev/null
+++ b/3472/CH43/EX43.3/Example43_3.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 IV : UTILIZATION AND TRACTION

+// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT

+

+// EXAMPLE : 5.3 :

+// Page number 778-779

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

+

+// Given data

+speed = 25.0       // Scheduled speed(kmph)

+D = 800.0/1000     // Distance between 2 stations(km)

+t = 20.0           // Time of stop(sec)

+V_m_per = 20.0     // Maximum speed higher than(%)

+beta = 3.0         // Retardation(km phps)

+

+// Calculations

+t_total = D*3600/speed                                 // Total time of run including stop(sec)

+T = t_total-t                                          // Actual time for run(sec)

+V_a = D/T*3600                                         // Average speed(kmph)

+V_m = (100+V_m_per)*V_a/100                            // Maximum speed(kmph)

+alpha = 1/((7200.0*D/V_m**2*((V_m/V_a)-1))-(1/beta))   // Value of acceleration(km phps)

+

+// Results

+disp("PART IV - EXAMPLE : 5.3 : SOLUTION :-")

+printf("\nRate of acceleration required to operate this service, α = %.2f km phps", alpha)

diff --git a/3472/CH43/EX43.4/Example43_4.sce b/3472/CH43/EX43.4/Example43_4.sce
new file mode 100644
index 000000000..99e20524e
--- /dev/null
+++ b/3472/CH43/EX43.4/Example43_4.sce
@@ -0,0 +1,32 @@
+// A Texbook on POWER SYSTEM ENGINEERING

+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar

+// DHANPAT RAI & Co.

+// SECOND EDITION 

+

+// PART IV : UTILIZATION AND TRACTION

+// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT

+

+// EXAMPLE : 5.4 :

+// Page number 779

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

+

+// Given data

+D = 2.0          // Distance between 2 stations(km)

+V_a = 40.0       // Average speed(kmph)

+V_1 = 60.0       // Maximum speed limitation(kph)

+alpha = 2.0      // Acceleration(km phps)

+beta_c = 0.15    // Coasting retardation(km phps)

+beta = 3.0       // Braking retardation(km phps)

+

+// Calculations

+t_1 = V_1/alpha                                        // Time for acceleration(sec)

+T = 3600*D/V_a                                         // Actual time of run(sec)

+V_2 = (T-t_1-(V_1/beta_c))*beta*beta_c/(beta_c-beta)   // Speed at the end of coasting period(kmph)

+t_2 = (V_1-V_2)/beta_c                                 // Coasting period(sec)

+t_3 = V_2/beta                                         // Braking period(sec)

+

+// Results

+disp("PART IV - EXAMPLE : 5.4 : SOLUTION :-")

+printf("\nDuration of acceleration, t_1 = %.f sec", t_1)

+printf("\nDuration of coasting, t_2 = %.f sec", t_2)

+printf("\nDuration of braking, t_3 = %.f sec", t_3)

diff --git a/3472/CH43/EX43.5/Example43_5.sce b/3472/CH43/EX43.5/Example43_5.sce
new file mode 100644
index 000000000..c20ad42d1
--- /dev/null
+++ b/3472/CH43/EX43.5/Example43_5.sce
@@ -0,0 +1,26 @@
+// A Texbook on POWER SYSTEM ENGINEERING

+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar

+// DHANPAT RAI & Co.

+// SECOND EDITION 

+

+// PART IV : UTILIZATION AND TRACTION

+// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT

+

+// EXAMPLE : 5.5 :

+// Page number 781-782

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

+

+// Given data

+r = 1.0    // Tractive resistance(N/tonne)

+

+// Calculations

+tractive_res_i = 0.278*r     // Tractive resistance(N/tonne) = Energy consumption(Wh/tonne-km)

+beta = 1/277.8               // Tractive resistance(N/tonne) = Retardation(km kmps/tonne)

+energy = 98.1*1000/3600      // 1% gradient = energy(Wh per tonne km)

+

+// Results

+disp("PART IV - EXAMPLE : 5.5 : SOLUTION :-")

+printf("\nCase(i)  : Tractive resistance of 1 N per tonne = %.3f Wh per tonne-km", tractive_res_i)

+printf("\nCase(ii) : Tractive resistance of 1 N per tonne = %.5f km phps per tonne", beta)

+printf("\nCase(iii): 1 percent gradient = %.2f Wh per tonne km\n", energy)

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

diff --git a/3472/CH43/EX43.6/Example43_6.sce b/3472/CH43/EX43.6/Example43_6.sce
new file mode 100644
index 000000000..7f09be3e1
--- /dev/null
+++ b/3472/CH43/EX43.6/Example43_6.sce
@@ -0,0 +1,36 @@
+// A Texbook on POWER SYSTEM ENGINEERING

+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar

+// DHANPAT RAI & Co.

+// SECOND EDITION 

+

+// PART IV : UTILIZATION AND TRACTION

+// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT

+

+// EXAMPLE : 5.6 :

+// Page number 782

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

+

+// Given data

+W = 254.0      // Weight of motor-coach train(tonne)

+no = 4.0       // Number of motor

+t_1 = 20.0     // Time(sec)

+V_m = 40.25    // Maximum speed(kmph)

+G = 1.0        // Gradient(%)

+gamma = 3.5    // Gear ratio

+n = 0.95       // Gear efficiency

+D = 91.5/100   // Wheel diameter(m)

+r = 44.0       // Train resistance(N/tonne)

+I = 10.0       // Rotational inertia(%)

+

+// Calculations

+W_e = W*(100+I)/100                  // Accelerating weight of train(tonne)

+alpha = V_m/t_1                      // Acceleration(km phps)

+F_t = 277.8*W_e*alpha+W*r+98.1*W*G   // Tractive effort(N)

+T = F_t*D/(2*n*gamma)                // Torque developed(N-m)

+T_each = T/no                        // Torque developed by each motor(N-m)

+

+// Results

+disp("PART IV - EXAMPLE : 5.6 : SOLUTION :-")

+printf("\nTorque developed by each motor = %.f N-m\n", T_each)

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

+printf("\n      ERROR: W = 254 tonne, not 256 tonne as mentioned in textbook problem statement")

diff --git a/3472/CH43/EX43.7/Example43_7.sce b/3472/CH43/EX43.7/Example43_7.sce
new file mode 100644
index 000000000..2afb0db70
--- /dev/null
+++ b/3472/CH43/EX43.7/Example43_7.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 IV : UTILIZATION AND TRACTION

+// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT

+

+// EXAMPLE : 5.7 :

+// Page number 782

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

+

+// Given data

+W = 203.0      // Weight of motor-coach train(tonne)

+no = 4.0       // Number of motors

+T = 5130.0     // Shaft torque(N-m)

+V_m = 42.0     // Maximum speed(kmph)

+G = 100.0/250  // Gradient

+gamma = 3.5    // Gear ratio

+n = 0.93       // Gear efficiency

+D = 91.5/100   // Wheel diameter(m)

+r = 45.0       // Train resistance(N/tonne)

+I = 10.0       // Rotational inertia(%)

+

+// Calculations

+W_e = W*(100+I)/100                     // Accelerating weight of train(tonne)

+F_t = n*4*T*2*gamma/D                   // Tractive effort(N)

+alpha = (F_t-W*r-98.1*W*G)/(277.8*W_e)  // Acceleration(km phps)

+t_1 = V_m/alpha                         // Time taken by train to attain speed(sec)

+

+// Results

+disp("PART IV - EXAMPLE : 5.7 : SOLUTION :-")

+printf("\nTime taken by train to attain speed, t_1 = %.1f sec", t_1)

diff --git a/3472/CH43/EX43.8/Ex43_8.png b/3472/CH43/EX43.8/Ex43_8.png
new file mode 100644
index 000000000..be2c075fb
--- /dev/null
+++ b/3472/CH43/EX43.8/Ex43_8.png
diff --git a/3472/CH43/EX43.8/Example43_8.sce b/3472/CH43/EX43.8/Example43_8.sce
new file mode 100644
index 000000000..4e609fd39
--- /dev/null
+++ b/3472/CH43/EX43.8/Example43_8.sce
@@ -0,0 +1,44 @@
+// A Texbook on POWER SYSTEM ENGINEERING

+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar

+// DHANPAT RAI & Co.

+// SECOND EDITION 

+

+// PART IV : UTILIZATION AND TRACTION

+// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT

+

+// EXAMPLE : 5.8 :

+// Page number 782-783

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

+

+// Given data

+V_a = 42.0          // Average speed of train(kmph)

+D = 1400.0/1000     // Distance(km)

+alpha = 1.7         // Acceleration(km phps)

+beta = 3.3          // Retardation(km phps)

+r = 50.0            // Tractive resistance(N/tonne)

+I = 10.0            // Rotational inertia(%)

+

+// Calculations

+T = D*3600/V_a                                       // Time for run(sec)

+k = (alpha+beta)/(alpha*beta)                        // Constant

+V_m = (T/k)-((T/k)**2-(7200*D/k))**0.5               // Maximum speed over the run(kmph)

+t_1 = V_m/alpha                                      // Time of acceleration(sec)

+t_3 = V_m/beta                                       // Time(sec)

+t_2 = T-(t_1+t_3)                                    // Time(sec)

+D_1 = D-(V_a*t_1/(2*3600))                           // Distance(km)

+We_W = (100+I)/100                                   // W_e/W

+energy = (0.0107*V_m**2*We_W/D)+(0.278*r*D_1/D)      // Energy consumption(Wh per tonne-km)

+a = gca() ;

+a.thickness = 2                                      // sets thickness of plot

+plot([0,t_1,t_1,(t_1+t_2),(t_1+t_2),(t_1+t_2+t_3)],[0,V_m,V_m,V_m,V_m,0])       // Plotting speed-time curve

+plot([t_1,t_1],[0,V_m],'r--')

+plot([t_1+t_2,t_1+t_2],[0,V_m],'r--')

+a.x_label.text = 'Time(seconds)'                     // labels x-axis

+a.y_label.text = 'Speed (km/h)'                      // labels y-axis

+xtitle("Fig E5.1 . Speed-time curve for the run") 

+xset('thickness',2)                                  // sets thickness of axes

+

+// Results

+disp("PART IV - EXAMPLE : 5.8 : SOLUTION :-")

+printf("\nSpeed-time curve for the run is shown in Figure E5.1")

+printf("\nEnergy consumption at the axles of train = %.1f Wh per tonne-km", energy)

diff --git a/3472/CH43/EX43.9/Example43_9.sce b/3472/CH43/EX43.9/Example43_9.sce
new file mode 100644
index 000000000..fa0ce06e7
--- /dev/null
+++ b/3472/CH43/EX43.9/Example43_9.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 IV : UTILIZATION AND TRACTION

+// CHAPTER 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT

+

+// EXAMPLE : 5.9 :

+// Page number 783

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

+

+// Given data

+V_A = 48.0       // Speed(kmph)

+t_1 = 24.0       // Time taken to accelerate from rest to speed(sec)

+t_2 = 69.0       // Coasting time(sec)

+r = 58.0         // Constant resistance(N/tonne)

+beta = 3.3       // Retardation(km phps)

+t_3 = 11.0       // Retardation time(sec)

+t_iii_a = 20.0   // Station stop time(sec)

+t_iii_b = 15.0   // Station stop time(sec)

+I = 10.0         // Rotational inertia(%)

+

+// Calculations

+alpha = V_A/t_1                                                   // Acceleration(km phps)

+V_B = beta*t_3                                                    // Speed at B(km phps)

+beta_c = (V_A-V_B)/t_2                                            // Retardation during coasting(km phps)

+distance_acc = 1.0/2*t_1*V_A/3600                                 // Distance covered during acceleration(km)

+distance_coasting = (V_A**2-V_B**2)/(2*beta_c*3600)               // Distance covered during coasting(km)

+distance_braking = t_3*V_B/(3600*2)                               // Distance covered during braking(km)

+distance_total = distance_acc+distance_coasting+distance_braking  // Total distance(km)

+speed_iii_a = distance_total*3600/(t_1+t_2+t_3+t_iii_a)           // Scheduled speed with a stop of 20 sec(kmph)

+speed_iii_b = distance_total*3600/(t_1+t_2+t_3+t_iii_b)           // Scheduled speed with a stop of 15 sec(kmph)

+

+// Results

+disp("PART IV - EXAMPLE : 5.9 : SOLUTION :-")

+printf("\nCase(i)  : Acceleration, α = %.f km phps", alpha)

+printf("\nCase(ii) : Coasting retardation, β_c = %.2f km phps", beta_c)

+printf("\nCase(iii): Scheduled speed with a stop of 20 seconds = %.2f kmph", speed_iii_a)

+printf("\n           Scheduled speed with a stop of 15 seconds = %.2f kmph\n", speed_iii_b)

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