diff options
Diffstat (limited to '3472/CH46')
| -rw-r--r-- | 3472/CH46/EX46.1/Example46_1.sce | 36 | ||||
| -rw-r--r-- | 3472/CH46/EX46.2/Example46_2.sce | 35 | ||||
| -rw-r--r-- | 3472/CH46/EX46.3/Example46_3.sce | 35 | ||||
| -rw-r--r-- | 3472/CH46/EX46.4/Example46_4.sce | 32 | ||||
| -rw-r--r-- | 3472/CH46/EX46.5/Example46_5.sce | 29 |
5 files changed, 167 insertions, 0 deletions
diff --git a/3472/CH46/EX46.1/Example46_1.sce b/3472/CH46/EX46.1/Example46_1.sce new file mode 100644 index 000000000..0f9cd203d --- /dev/null +++ b/3472/CH46/EX46.1/Example46_1.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 8: BRAKING
+
+// EXAMPLE : 8.1 :
+// Page number 806
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 525.0 // Voltage of motor(V)
+I_1 = 50.0 // Current(A)
+T_1 = 216.0 // Torque(N-m)
+I_2 = 70.0 // Current(A)
+T_2 = 344.0 // Torque(N-m)
+I_3 = 80.0 // Current(A)
+T_3 = 422.0 // Torque(N-m)
+I_4 = 90.0 // Current(A)
+T_4 = 500.0 // Torque(N-m)
+V_m = 26.0 // Speed(kmph)
+R_b = 5.5 // Resistance of braking rheostat(ohm)
+R_m = 0.5 // Resistance of motor(ohm)
+
+// Calculations
+I = 75.0 // Current drawn at 26 kmph(A)
+back_emf = V-I*R_m // Back emf of the motor(V)
+R_t = R_b+R_m // Total resistance(ohm)
+I_del = back_emf/R_t // Current delivered(A)
+T_b = T_3*I_del/I_3 // Braking torque(N-m)
+
+// Results
+disp("PART IV - EXAMPLE : 8.1 : SOLUTION :-")
+printf("\nBraking torque = %.f N-m", T_b)
diff --git a/3472/CH46/EX46.2/Example46_2.sce b/3472/CH46/EX46.2/Example46_2.sce new file mode 100644 index 000000000..e7b2523b3 --- /dev/null +++ b/3472/CH46/EX46.2/Example46_2.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 IV : UTILIZATION AND TRACTION
+// CHAPTER 8: BRAKING
+
+// EXAMPLE : 8.2 :
+// Page number 806
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 525.0 // Voltage of motor(V)
+I_1 = 50.0 // Current(A)
+N_1 = 1200.0 // Speed(rpm)
+I_2 = 100.0 // Current(A)
+N_2 = 950.0 // Speed(rpm)
+I_3 = 150.0 // Current(A)
+N_3 = 840.0 // Speed(rpm)
+I_4 = 200.0 // Current(A)
+N_4 = 745.0 // Speed(rpm)
+N = 1000.0 // Speed opearting(rpm)
+R = 3.0 // Resistance(ohm)
+R_m = 0.5 // Resistance of motor(ohm)
+
+// Calculations
+I = 85.0 // Current drawn at 1000 rpm(A)
+back_emf = V-I*R_m // Back emf of the motor(V)
+R_t = R+R_m // Total resistance(ohm)
+I_del = back_emf/R_t // Current delivered(A)
+
+// Results
+disp("PART IV - EXAMPLE : 8.2 : SOLUTION :-")
+printf("\nCurrent delivered when motor works as generator = %.f A", I_del)
diff --git a/3472/CH46/EX46.3/Example46_3.sce b/3472/CH46/EX46.3/Example46_3.sce new file mode 100644 index 000000000..9e8f5b048 --- /dev/null +++ b/3472/CH46/EX46.3/Example46_3.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 IV : UTILIZATION AND TRACTION
+// CHAPTER 8: BRAKING
+
+// EXAMPLE : 8.3 :
+// Page number 810
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+W = 400.0 // Weight of train(tonne)
+G = 100.0/70 // Gradient(%)
+t = 120.0 // Time(sec)
+V_1 = 80.0 // Speed(km/hr)
+V_2 = 50.0 // Speed(km/hr)
+r_kg = 5.0 // Tractive resistance(kg/tonne)
+I = 7.5 // Rotational inertia(%)
+n = 0.75 // Overall efficiency
+
+// Calculations
+W_e = W*(100+I)/100 // Accelerating weight of train(tonne)
+r = r_kg*9.81 // Tractive resistance(N-m/tonne)
+energy_recuperation = 0.01072*W_e*(V_1**2-V_2**2)/1000 // Energy available for recuperation(kWh)
+F_t = W*(r-98.1*G) // Tractive effort during retardation(N)
+distance = (V_1+V_2)*1000*t/(2*3600) // Distance travelled by train during retardation period(m)
+energy_train = abs(F_t)*distance/(3600*1000) // Energy available during train movement(kWh)
+net_energy = n*(energy_recuperation+energy_train) // Net energy returned to supply system(kWh)
+
+// Results
+disp("PART IV - EXAMPLE : 8.3 : SOLUTION :-")
+printf("\nEnergy returned to lines = %.2f kWh\n", net_energy)
+printf("\nNOTE: ERROR: Calculation mistakes & more approximation in textbook solution")
diff --git a/3472/CH46/EX46.4/Example46_4.sce b/3472/CH46/EX46.4/Example46_4.sce new file mode 100644 index 000000000..550c2a55c --- /dev/null +++ b/3472/CH46/EX46.4/Example46_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 8: BRAKING
+
+// EXAMPLE : 8.4 :
+// Page number 810
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+W = 355.0 // Weight of train(tonne)
+V_1 = 80.5 // Speed(km/hr)
+V_2 = 48.3 // Speed(km/hr)
+D = 1.525 // Distance(km)
+G = 100.0/90 // Gradient(%)
+I = 10.0 // Rotational inertia(%)
+r = 53.0 // Tractive resistance(N/tonne)
+n = 0.8 // Overall efficiency
+
+// Calculations
+beta = (V_1**2-V_2**2)/(2*D*3600) // Braking retardation(km phps)
+W_e = W*(100+I)/100 // Accelerating weight of train(tonne)
+F_t = 277.8*W_e*beta+98.1*W*G-W*r // Tractive effort(N)
+work_done = F_t*D*1000 // Work done by this effort(N-m)
+energy = work_done*n/(1000*3600) // Energy returned to line(kWh)
+
+// Results
+disp("PART IV - EXAMPLE : 8.4 : SOLUTION :-")
+printf("\nEnergy returned to the line = %.1f kWh", energy)
diff --git a/3472/CH46/EX46.5/Example46_5.sce b/3472/CH46/EX46.5/Example46_5.sce new file mode 100644 index 000000000..fecc3fbf5 --- /dev/null +++ b/3472/CH46/EX46.5/Example46_5.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 8: BRAKING
+
+// EXAMPLE : 8.5 :
+// Page number 811-812
+clear ; clc ; close ; // Clear the work space and console
+funcprot(0)
+
+// Given data
+area = 16.13 // Area of brakes(sq.cm/pole face)
+phi = 2.5*10**-3 // Flux(Wb)
+u = 0.2 // Co-efficient of friction
+W = 10.0 // Weight of car(tonnes)
+
+// Calculations
+a = area*10**-4 // Area of brakes(sq.m/pole face)
+F = phi**2/(2*%pi*10**-7*a) // Force(N)
+force = F*u // Braking effect considering flux and coefficient of friction(N)
+beta = u*F/(W*1000)*100 // Rate of retardation produced by braking effect(cm/sec^2)
+
+// Results
+disp("PART IV - EXAMPLE : 8.5 : SOLUTION :-")
+printf("\nBraking effect, F = %.f N", force)
+printf("\nRate of retardation produced by this braking effect, β = %.2f cm/sec^2", beta)
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