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-rw-r--r--3472/CH30/EX30.1/Example30_1.sce77
-rw-r--r--3472/CH30/EX30.10/Example30_10.sce49
-rw-r--r--3472/CH30/EX30.11/Example30_11.sce40
-rw-r--r--3472/CH30/EX30.12/Example30_12.sce54
-rw-r--r--3472/CH30/EX30.13/Example30_13.sce37
-rw-r--r--3472/CH30/EX30.14/Example30_14.sce48
-rw-r--r--3472/CH30/EX30.2/Example30_2.sce48
-rw-r--r--3472/CH30/EX30.3/Example30_3.sce59
-rw-r--r--3472/CH30/EX30.4/Example30_4.sce55
-rw-r--r--3472/CH30/EX30.5/Example30_5.sce46
-rw-r--r--3472/CH30/EX30.6/Example30_6.sce63
-rw-r--r--3472/CH30/EX30.7/Example30_7.sce85
-rw-r--r--3472/CH30/EX30.8/Example30_8.sce46
-rw-r--r--3472/CH30/EX30.9/Example30_9.sce33
14 files changed, 740 insertions, 0 deletions
diff --git a/3472/CH30/EX30.1/Example30_1.sce b/3472/CH30/EX30.1/Example30_1.sce
new file mode 100644
index 000000000..9a5b769b8
--- /dev/null
+++ b/3472/CH30/EX30.1/Example30_1.sce
@@ -0,0 +1,77 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART III : SWITCHGEAR AND PROTECTION
+// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.1 :
+// Page number 510-512
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+MVA = 15.0 // Generator rating(MVA)
+kV = 6.9 // Generator voltage(kV)
+X_1 = 25.0 // Positive sequence reactance(%)
+X_2 = 25.0 // Negative sequence reactance(%)
+X_0 = 8.0 // Zero sequence reactance(%)
+X = 6.0 // Reactor placed in line(%)
+
+// Calculations
+a = exp(%i*120.0*%pi/180) // Operator
+Z_1 = %i*X_1/100 // Positive sequence impedance(p.u)
+Z_2 = %i*X_2/100 // Negative sequence impedance(p.u)
+Z_g0 = %i*X_0/100 // Impedance(p.u)
+Z = %i*X/100 // Impedance(p.u)
+Z_0 = Z_g0+3*Z // Zero sequence impedance(p.u)
+E_a = 1.0 // Voltage(p.u)
+E_b = a**2*E_a // Voltage(p.u)
+// Case(a)
+I_a0_a = 0 // Current(A)
+I_a1_a_pu = 1.0/(Z_1+Z_2) // Current(p.u)
+I_a1_a = I_a1_a_pu*MVA*1000/(3**0.5*kV) // Current(A)
+I_a2_a = -I_a1_a // Current(A)
+I_b0_a = 0 // Current(A)
+I_b1_a = a**2*I_a1_a // Current(A)
+I_b2_a = a*I_a2_a // Current(A)
+I_a_a = I_a1_a+I_a2_a // Line current(A)
+I_b_a = I_b1_a+I_b2_a // Line current(A)
+I_c_a = -I_b_a // Line current(A)
+I_g_a = 0 // Ground wire current(A)
+V_a_a = (E_a-I_a1_a*Z_1-I_a2_a*Z_2-I_a0_a*Z_0)*kV*1000/3**0.5 // Voltage(V)
+V_b_a = (a**2*E_a+%i*3**0.5*I_a1_a_pu*Z_1)*kV*1000/3**0.5 // Voltage(V)
+V_c_a = V_b_a // Voltage(V)
+// Case(b)
+I_a1_b_pu = E_a/(Z_1+(Z_2*Z_0/(Z_2+Z_0))) // Current(p.u)
+I_a1_b = I_a1_b_pu*MVA*1000/(3**0.5*kV) // Current(A)
+I_a2_b_pu = -Z_0*Z_2/(Z_2*(Z_0+Z_2))*I_a1_b_pu // Current(p.u)
+I_a2_b = -Z_0*Z_2/(Z_2*(Z_0+Z_2))*I_a1_b // Current(A)
+I_a0_b_pu = -Z_0*Z_2/(Z_0*(Z_0+Z_2))*I_a1_b_pu // Current(p.u)
+I_a0_b = -Z_0*Z_2/(Z_0*(Z_0+Z_2))*I_a1_b // Current(A)
+I_a_b = I_a0_b+I_a1_b+I_a2_b // Line current(A)
+I_b_b = I_a0_b+a**2*I_a1_b+a*I_a2_b // Line current(A)
+I_c_b = I_a0_b+a*I_a1_b+a**2*I_a2_b // Line current(A)
+I_0_b = 3*I_a0_b // Current in the ground resistor(A)
+V_a_b_pu = E_a-I_a1_b_pu*Z_1-I_a2_b_pu*Z_2-I_a0_b_pu*Z_0 // Voltage(p.u)
+V_a_b = abs(V_a_b_pu)*kV*1000/(3**0.5) // Voltage(V)
+V_b_b = 0 // Voltage(V)
+V_c_b = 0 // Voltage(V)
+
+// Results
+disp("PART III - EXAMPLE : 4.1 : SOLUTION :-")
+printf("\nCase(a): Initial symmetrical rms line current when ground is not involved in fault, I_a = %.f A", abs(I_a_a))
+printf("\n Initial symmetrical rms line current when ground is not involved in fault, I_b = %.f A", real(I_b_a))
+printf("\n Initial symmetrical rms line current when ground is not involved in fault, I_c = %.f A", real(I_c_a))
+printf("\n Ground wire current = %.f A", I_g_a)
+printf("\n Line to neutral voltage, V_a = %.f V", real(V_a_a))
+printf("\n Line to neutral voltage, V_b = %.f V", real(V_b_a))
+printf("\n Line to neutral voltage, V_c = %.f V", real(V_c_a))
+printf("\nCase(b): Initial symmetrical rms line current when fault is solidly grounded, I_a = %.f A", abs(I_a_b))
+printf("\n Initial symmetrical rms line current when fault is solidly grounded, I_b = (%.f+%.fj) A", real(I_b_b),imag(I_b_b))
+printf("\n Initial symmetrical rms line current when fault is solidly grounded, I_c = (%.f+%.fj) A", real(I_c_b),imag(I_c_b))
+printf("\n Ground wire current = %.fj A", imag(I_0_b))
+printf("\n Line to neutral voltage, V_a = %.f V", V_a_b)
+printf("\n Line to neutral voltage, V_b = %.f V", V_b_b)
+printf("\n Line to neutral voltage, V_c = %.f V\n", V_c_b)
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here and approximation in textbook")
diff --git a/3472/CH30/EX30.10/Example30_10.sce b/3472/CH30/EX30.10/Example30_10.sce
new file mode 100644
index 000000000..819d7598b
--- /dev/null
+++ b/3472/CH30/EX30.10/Example30_10.sce
@@ -0,0 +1,49 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART III : SWITCHGEAR AND PROTECTION
+// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.10 :
+// Page number 519-520
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+MVA_A = 30.0 // Alternator rating(MVA)
+kV_A = 11.0 // Alternator rating(kV)
+X_1 = 2.5 // Reactance to positive sequence current(ohm)
+X_2 = 0.8*X_1 // Reactance to negative sequence current(ohm)
+X_0 = 0.3*X_1 // Reactance to zero sequence current(ohm)
+
+// Calculations
+// Case(a)
+a = exp(%i*120.0*%pi/180) // Operator
+Z_1 = %i*X_1 // Positive sequence impedance(ohm)
+Z_2 = %i*X_2 // Negative sequence impedance(ohm)
+Z_0 = %i*X_0 // Zero sequence impedance(ohm)
+Z_02 = Z_0*Z_2/(Z_0+Z_2) // Impedance(ohm)
+E_a = kV_A*1000/3**0.5 // Phase voltage(V)
+I_a1 = E_a/(Z_1+Z_02) // Positive sequence current(A)
+I_a2 = -Z_0/(Z_0+Z_2)*I_a1 // Negative sequence current(A)
+I_a0 = -Z_2/(Z_0+Z_2)*I_a1 // Zero sequence current(A)
+I_0 = I_a0 // Zero sequence current(A)
+I_a = I_a0+I_a1+I_a2 // Line current(A)
+I_b = I_0+a**2*I_a1+a*I_a2 // Line current(A)
+I_c = I_0+a*I_a1+a**2*I_a2 // Line current(A)
+// Case(b)
+I_n = 3*abs(I_0) // Current through ground(A)
+// Case(c)
+V_a2 = Z_02*I_a1 // Negative sequence voltage(V)
+V_a = 3*abs(V_a2) // Voltage of healthy phase to neutral(V)
+
+// Results
+disp("PART III - EXAMPLE : 4.10 : SOLUTION :-")
+printf("\nCase(a): Currents in the faulted phase are")
+printf("\n I_a = %.f A", abs(I_a))
+printf("\n I_b = %.f∠%.1f° A", abs(I_b),phasemag(I_b))
+printf("\n I_c = %.f∠%.1f° A", abs(I_c),phasemag(I_c))
+printf("\nCase(b): Current through ground, I_n = %.f A", I_n)
+printf("\nCase(c): Voltage of healthy phase to neutral, V_a = %.f V\n", V_a)
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
diff --git a/3472/CH30/EX30.11/Example30_11.sce b/3472/CH30/EX30.11/Example30_11.sce
new file mode 100644
index 000000000..bb7909999
--- /dev/null
+++ b/3472/CH30/EX30.11/Example30_11.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 III : SWITCHGEAR AND PROTECTION
+// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.11 :
+// Page number 520-521
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+n = 6.0 // Number of alternator
+kV_A = 6.6 // Alternator rating(kV)
+X_1 = 0.9 // Positive sequence reactance(ohm)
+X_2 = 0.72 // Negative sequence reactance(ohm)
+X_0 = 0.3 // Zero sequence reactance(ohm)
+Z_n = 0.2 // Resistance of grounding resistor(ohm)
+
+// Calculations
+E_a = kV_A*1000/3**0.5 // Phase voltage(V)
+// Case(a)
+Z_1_a = %i*X_1/n // Positive sequence impedance when alternators are in parallel(ohm)
+Z_2_a = %i*X_2/n // Negative sequence impedance when alternators are in parallel(ohm)
+Z_0_a = %i*X_0/n // Zero sequence impedance when alternators are in parallel(ohm)
+I_a_a = 3*E_a/(Z_1_a+Z_2_a+Z_0_a) // Fault current assuming 'a' phase to be fault(A)
+// Case(b)
+Z_0_b = 3*Z_n+%i*X_0 // Zero sequence impedance(ohm)
+I_a_b = 3*E_a/(Z_1_a+Z_2_a+Z_0_b) // Fault current(A)
+// Case(c)
+Z_0_c = %i*X_0 // Zero sequence impedance(ohm)
+I_a_c = 3*E_a/(Z_1_a+Z_2_a+Z_0_c) // Fault current(A)
+
+// Results
+disp("PART III - EXAMPLE : 4.11 : SOLUTION :-")
+printf("\nCase(a): Fault current if all alternator neutrals are solidly grounded, I_a = %.f A", imag(I_a_a))
+printf("\nCase(b): Fault current if one alternator neutral is grounded & others isolated, I_a = %.1f∠%.1f° A", abs(I_a_b),phasemag(I_a_b))
+printf("\nCase(c): Fault current if one alternator neutral is solidly grounded & others isolated, I_a = %.2fj A\n", imag(I_a_c))
+printf("\nNOTE: ERROR: Calculation mistakes in the textbook solution")
diff --git a/3472/CH30/EX30.12/Example30_12.sce b/3472/CH30/EX30.12/Example30_12.sce
new file mode 100644
index 000000000..46d32187e
--- /dev/null
+++ b/3472/CH30/EX30.12/Example30_12.sce
@@ -0,0 +1,54 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART III : SWITCHGEAR AND PROTECTION
+// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.12 :
+// Page number 521-522
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+MVA_A = 30.0 // Alternator rating(MVA)
+kV_A = 6.6 // Alternator rating(kV)
+X_G = 10.0 // Reactance of alternator(%)
+kV_lv_T = 6.6 // Transformer lv side rating(kV)
+kV_hv_T = 33.0 // Transformer hv side rating(kV)
+X_T = 6.0 // Reactance of transformer(%)
+kV_line = 33.0 // Transmission line voltage(kV)
+X_line = 4.0 // Transmission line reactance(ohm)
+X_g2 = 70.0 // Negative sequence reactance is 70% of +ve sequence reactance of generator(%)
+
+// Calculations
+MVA_base = 30.0 // Base MVA
+kV_base = 6.6 // Base kV
+Z_base = kV_base**2/MVA_base // Base impedance(ohm)
+Z_g1 = %i*Z_base*X_G/100 // Positive sequence impedance of alternator(ohm)
+Z_T1 = %i*Z_base*X_T/100 // Positive sequence impedance of transformer(ohm)
+Z_L1 = %i*(kV_base/kV_line)**2*X_line // Positive sequence impedance of transmission line(ohm)
+Z_g2 = X_g2/100*Z_g1 // Negative sequence impedance of alternator(ohm)
+Z_T2 = %i*Z_base*X_T/100 // Negative sequence impedance of transformer(ohm)
+Z_T0 = %i*Z_base*X_T/100 // Zero sequence impedance of transformer(ohm)
+Z_L2 = Z_L1 // Negative sequence impedance of transmission line(ohm)
+Z_1 = Z_g1+Z_T1+Z_L1+Z_T1 // Positive sequence impedance(ohm)
+Z_2 = Z_g2+Z_T2+Z_L2+Z_T2 // Negative sequence impedance(ohm)
+Z_0 = Z_T0 // Zero sequence impedance(ohm)
+E_a = kV_base*1000/3**0.5 // Base voltage(V)
+// Case(a)
+I_sc = E_a/Z_1 // Fault current if all 3 phases short circuited(A)
+// Case(b)
+I_a = 3*E_a/(Z_1+Z_2+Z_0) // Fault current if single line is grounded assuming 'a' to be grounded(A)
+// Case(c)
+I_b = %i*3**0.5*E_a/(Z_1+Z_2) // Fault current for a short circuit between two lines(A)
+I_c = -%i*3**0.5*E_a/(Z_1+Z_2) // Fault current for a short circuit between two lines(A)
+
+// Results
+disp("PART III - EXAMPLE : 4.12 : SOLUTION :-")
+printf("\nCase(a): Fault current if all 3 phases short circuited, I_sc = %.f∠%.f° A", abs(I_sc),phasemag(I_sc))
+printf("\nCase(b): Fault current if single line is grounded, I_a = %.fj A", imag(I_a))
+printf("\nCase(c): Fault current for a short circuit between two lines, I_b = %.f A", real(I_b))
+printf("\n Fault current for a short circuit between two lines, I_c = %.f A\n", real(I_c))
+printf("\nNOTE: ERROR: (1).Calculation mistake in Z_2 in the textbook solution")
+printf("\n (2).Transformer reactance is 6 percent, not 5 percent as in problem statement")
diff --git a/3472/CH30/EX30.13/Example30_13.sce b/3472/CH30/EX30.13/Example30_13.sce
new file mode 100644
index 000000000..f9de20838
--- /dev/null
+++ b/3472/CH30/EX30.13/Example30_13.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 III : SWITCHGEAR AND PROTECTION
+// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.13 :
+// Page number 522
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+kV = 6.9 // Alternator rating(kV)
+MVA = 10.0 // Alternator rating(MVA)
+X_st = 0.15 // Sub-transient reactance(p.u)
+X_2 = 0.15 // Negative sequence reactance(p.u)
+X_0 = 0.05 // Zero sequence reactance(p.u)
+X = 0.397 // Grounding reactor(ohm)
+
+// Calculations
+MVA_base = 10.0 // Base MVA
+kV_base = 6.9 // Base kV
+Z_base = kV_base**2/MVA_base // Base impedance(ohm)
+Z_n = X/Z_base // Grounding reactor(p.u)
+Z_1 = %i*X_st // Positive sequence impedance(p.u)
+Z_2 = %i*X_2 // Negative sequence impedance(p.u)
+Z_0 = %i*(X_0+3*Z_n) // Zero sequence impedance(p.u)
+E_a = 1.0 // Phase voltage(p.u)
+I_a_pu = 3*E_a/(Z_1+Z_2+Z_0) // Sub-transient current in the faulty phase(p.u)
+I_base = kV_base*1000/(3**0.5*Z_base) // Base current(A)
+I_a = abs(I_a_pu)*I_base // Sub-transient current in the faulty phase(A)
+
+// Results
+disp("PART III - EXAMPLE : 4.13 : SOLUTION :-")
+printf("\nSub-transient current in the faulty phase, I_a = %.f A\n", I_a)
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
diff --git a/3472/CH30/EX30.14/Example30_14.sce b/3472/CH30/EX30.14/Example30_14.sce
new file mode 100644
index 000000000..e49b7bba0
--- /dev/null
+++ b/3472/CH30/EX30.14/Example30_14.sce
@@ -0,0 +1,48 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART III : SWITCHGEAR AND PROTECTION
+// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.14 :
+// Page number 522-523
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+kVA = 10000.0 // Generator rating(kVA)
+kV = 13.8 // Generator rating(kV)
+X_st = 10.0 // Sub-transient reactance(%)
+X_2 = 10.0 // Negative sequence reactance(%)
+X_0 = 5.0 // Zero sequence reactance(%)
+X = 8.0 // Grounding reactor(%)
+X_con = 6.0 // Reactance of reactor connecting generator & transformer(%)
+
+// Calculations
+a = exp(%i*120.0*%pi/180) // Operator
+Z_1 = %i*(X_st+X_con)/100 // Positive sequence impedance(p.u)
+Z_2 = %i*(X_2+X_con)/100 // Negative sequence impedance(p.u)
+Z_0 = %i*X_con/100 // Zero sequence impedance(p.u)
+E_a = 1.0 // Phase voltage(p.u)
+I_a1 = E_a/(Z_1+Z_2+Z_0) // Sub-transient current in the faulty phase(p.u)
+I_A1 = %i*I_a1 // Positive sequence current(p.u)
+I_A2 = -%i*I_a1 // Negative sequence current(p.u)
+I_A = I_A1+I_A2 // Initial symmetrical r.m.s current in phase a(p.u)
+I_B1 = a**2*I_A1 // Positive sequence current(p.u)
+I_B2 = a*I_A2 // Negative sequence current(p.u)
+I_B = I_B1+I_B2 // Initial symmetrical r.m.s current in phase b(p.u)
+I_C1 = a*I_A1 // Positive sequence current(p.u)
+I_C2 = a**2*I_A2 // Negative sequence current(p.u)
+I_C = I_C1+I_C2 // Initial symmetrical r.m.s current in phase c(p.u)
+I_base = kVA/(3**0.5*kV) // Base current(A)
+I_A_amp = I_A*I_base // Initial symmetrical r.m.s current in phase a(p.u)
+I_B_amp = I_B*I_base // Initial symmetrical r.m.s current in phase b(p.u)
+I_C_amp = I_C*I_base // Initial symmetrical r.m.s current in phase c(p.u)
+
+// Results
+disp("PART III - EXAMPLE : 4.14 : SOLUTION :-")
+printf("\nInitial symmetrical r.m.s current in all phases of generator are,")
+printf("\n I_A = %.f A", abs(I_A_amp))
+printf("\n I_B = %.f∠%.f° A", abs(I_B_amp),phasemag(I_B_amp))
+printf("\n I_C = %.f∠%.f° A", abs(I_C_amp),phasemag(I_C_amp))
diff --git a/3472/CH30/EX30.2/Example30_2.sce b/3472/CH30/EX30.2/Example30_2.sce
new file mode 100644
index 000000000..e9d1dfc99
--- /dev/null
+++ b/3472/CH30/EX30.2/Example30_2.sce
@@ -0,0 +1,48 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART III : SWITCHGEAR AND PROTECTION
+// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.2 :
+// Page number 512
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+kVA = 10000.0 // Generator rating(kVA)
+f = 50.0 // Frequency(Hz)
+I_1 = 30.0 // Positive sequence current(%)
+I_2 = 10.0 // Negative sequence current(%)
+I_0 = 5.0 // Zero sequence current(%)
+d = 1.0/100 // Diameter of conductor(m)
+D = 5.0 // Triangular spacing(m)
+kV = 30.0 // Generator voltage on open-circuit(kV)
+l = 20.0 // Distance of line at short circuit occurance(km)
+
+// Calculations
+a = exp(%i*120.0*%pi/180) // Operator
+Z_g1 = kV**2*I_1*I_2/kVA // Positive phase sequence reactance of generator(ohm)
+Z_g2 = Z_g1*I_2/I_1 // Negative phase sequence reactance of generator(ohm)
+Z_g0 = Z_g1*I_0/I_1 // Zero phase sequence reactance of generator(ohm)
+r = d/2 // Radius of conductor(m)
+Z_l1 = 2.0*%pi*f*(0.5+4.606*log10(D/r))*10**-7*l*1000 // Positive phase sequence reactance of line(ohm)
+Z_l2 = 2.0*%pi*f*(0.5+4.606*log10(D/r))*10**-7*l*1000 // Negative phase sequence reactance of line(ohm)
+Z_1 = %i*(Z_g1+Z_l1) // Z1 upto the point of fault(ohm)
+Z_2 = %i*(Z_g2+Z_l2) // Z2 upto the point of fault(ohm)
+E_a = kV*1000/3**0.5 // Phase voltage(V)
+I_a1 = E_a/(Z_1+Z_2) // Positive sequence current in line a(A)
+I_a2 = -I_a1 // Negative sequence current in line a(A)
+I_a0 = 0 // Zero sequence current in line a(A)
+I_b0 = 0 // Zero sequence current in line b(A)
+I_c0 = 0 // Zero sequence current in line c(A)
+I_a = I_a0+I_a1+I_a2 // Current in line a(A)
+I_b = I_b0+a**2*I_a1+a*I_a2 // Current in line b(A)
+I_c = I_c0+a*I_a1+a**2*I_a2 // Current in line c(A)
+
+// Results
+disp("PART III - EXAMPLE : 4.2 : SOLUTION :-")
+printf("\nCurrent in line a, I_a = %.f A", abs(I_a))
+printf("\nCurrent in line b, I_b = %.f A", real(I_b))
+printf("\nCurrent in line c, I_c = %.f A", real(I_c))
diff --git a/3472/CH30/EX30.3/Example30_3.sce b/3472/CH30/EX30.3/Example30_3.sce
new file mode 100644
index 000000000..ac8ab68d8
--- /dev/null
+++ b/3472/CH30/EX30.3/Example30_3.sce
@@ -0,0 +1,59 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART III : SWITCHGEAR AND PROTECTION
+// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.3 :
+// Page number 512-513
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+kVA = 10000.0 // Alternator rating(kVA)
+Z_g1 = complex(0.5,4.7) // Positive sequence impedance(ohm/phase)
+Z_g2 = complex(0.2,0.6) // Negative sequence impedance(ohm/phase)
+Z_g0 = complex(0,0.43) // Zero sequence impedance(ohm/phase)
+Z_l1 = complex(0.36,0.25) // Impedance(ohm)
+Z_l2 = complex(0.36,0.25) // Impedance(ohm)
+Z_l0 = complex(2.9,0.95) // Impedance(ohm)
+V = 6600.0 // Voltage(V)
+
+// Calculations
+a = exp(%i*120.0*%pi/180) // Operator
+// Case(a)
+E_a = V/3**0.5 // Phase voltage(V)
+Z_1 = Z_g1+Z_l1 // Z1 upto the point of fault(ohm)
+Z_2 = Z_g2+Z_l2 // Z2 upto the point of fault(ohm)
+Z_0 = Z_g0+Z_l0 // Z0 upto the point of fault(ohm)
+I_a = 3*E_a/(Z_1+Z_2+Z_0) // Fault current(A)
+// Case(b)
+I_a0 = abs(I_a)/3 // Zero sequence current of line a(A)
+I_a1 = abs(I_a)/3 // Positive sequence current of line a(A)
+I_a2 = abs(I_a)/3 // Negative sequence current of line a(A)
+I_b0 = I_a0 // Zero sequence current of line b(A)
+I_b1 = a**2*I_a1 // Positive sequence current of line b(A)
+I_b2 = a*I_a2 // Negative sequence current of line b(A)
+I_c0 = I_a0 // Zero sequence current of line c(A)
+I_c1 = a*I_a1 // Positive sequence current of line c(A)
+I_c2 = a**2*I_a2 // Negative sequence current of line c(A)
+// Case(c)
+V_b = E_a/(Z_1+Z_2+Z_0)*((a**2-a)*Z_2+(a**2-1)*Z_0) // Voltage of the line b(V)
+V_c = E_a/(Z_1+Z_2+Z_0)*((a-a**2)*Z_2+(a-1)*Z_0) // Voltage of the line c(V)
+
+// Results
+disp("PART III - EXAMPLE : 4.3 : SOLUTION :-")
+printf("\nCase(a): Fault current, |I_a| = %.f A", abs(I_a))
+printf("\nCase(b): Zero sequence current of line a, I_a0 = %.f A", I_a0)
+printf("\n Positive sequence current of line a, I_a1 = %.f A", I_a1)
+printf("\n Negative sequence current of line a, I_a2 = %.f A", I_a2)
+printf("\n Zero sequence current of line b, I_b0 = %.f A", I_b0)
+printf("\n Positive sequence current of line b, I_b1 = (%.1f%.1fj) A", real(I_b1),imag(I_b1))
+printf("\n Negative sequence current of line b, I_b2 = (%.1f+%.1fj) A", real(I_b2),imag(I_b2))
+printf("\n Zero sequence current of line c, I_c0 = %.f A", I_c0)
+printf("\n Positive sequence current of line c, I_c1 = (%.1f+%.1fj) A", real(I_c1),imag(I_c1))
+printf("\n Negative sequence current of line c, I_c2 = (%.1f%.1fj) A", real(I_c2),imag(I_c2))
+printf("\nCase(c): Voltage of the sound line to earth at fault, |V_b| = %.f V", abs(V_b))
+printf("\n Voltage of the sound line to earth at fault, |V_c| = %.f V\n", abs(V_c))
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
diff --git a/3472/CH30/EX30.4/Example30_4.sce b/3472/CH30/EX30.4/Example30_4.sce
new file mode 100644
index 000000000..5d21f554a
--- /dev/null
+++ b/3472/CH30/EX30.4/Example30_4.sce
@@ -0,0 +1,55 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART III : SWITCHGEAR AND PROTECTION
+// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.4 :
+// Page number 513-514
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 11000.0 // Alternator voltage(V)
+kVA = 50000.0 // Alternator rating(kVA)
+Z_l1 = complex(0.4,0.7) // Positive sequence impedance of feeder(ohm)
+Z_l2 = complex(0.4,0.7) // Negative sequence impedance of feeder(ohm)
+Z_l0 = complex(0.7,3.0) // Zero sequence impedance of feeder(ohm)
+Z_g1_A = complex(0,0.6) // Positive sequence reactance(ohm)
+Z_g1_B = complex(0,0.6) // Positive sequence reactance(ohm)
+Z_g2_A = complex(0,0.4) // Negative sequence reactance(ohm)
+Z_g2_B = complex(0,0.4) // Negative sequence reactance(ohm)
+Z_g0_A = complex(0,0.2) // Zero sequence reactance(ohm)
+Z_g0_B = complex(0,0.2) // Zero sequence reactance(ohm)
+Z_n_A = complex(0,0.2) // Neutral reactance(ohm)
+Z_n_B = complex(0,0.2) // Neutra reactance(ohm)
+
+// Calculations
+a = exp(%i*120.0*%pi/180) // Operator
+Z_g1 = 1.0/((1/Z_g1_A)+(1/Z_g1_B)) // Equivalent positive sequence impedance(ohm)
+Z_g2 = 1.0/((1/Z_g2_A)+(1/Z_g2_B)) // Equivalent negative sequence impedance(ohm)
+Z_g0 = 1.0/((1/Z_g0_A)+(1/Z_g0_B)) // Equivalent zero sequence impedance(ohm)
+Z_n = 1.0/((1/Z_n_A)+(1/Z_n_B)) // Equivalent neutral impedance(ohm)
+Z_1 = Z_l1+Z_g1 // Positive sequence impedance(ohm)
+Z_2 = Z_l2+Z_g2 // Negative sequence impedance(ohm)
+Z_0 = Z_l0+Z_g0+3*Z_n // Zero sequence impedance(ohm)
+Z = Z_0*Z_2/(Z_0+Z_2) // Impedance(ohm)
+E_R = V/3**0.5 // Phase voltage(V)
+I_R1 = E_R/(Z_1+Z) // Postive sequence current(A)
+I_R2 = -Z*I_R1/Z_2 // Negative sequence current(A)
+I_R0 = -Z*I_R1/Z_0 // Zero sequence current(A)
+I_R = I_R0+I_R1+I_R2 // Fault current in line(A)
+I_Y = I_R0+a**2*I_R1+a*I_R2 // Fault current in line(A)
+I_B = I_R0+a*I_R1+a**2*I_R2 // Fault current in line(A)
+I_earth = 3.0*I_R0 // Current through earth reactance(A)
+V_neutral = abs(I_earth*Z_n) // Magnitude of potential above earth attained by generator neutral(V)
+
+// Results
+disp("PART III - EXAMPLE : 4.4 : SOLUTION :-")
+printf("\nFault current in the line R, I_R = %.f A", abs(I_R))
+printf("\nFault current in the line Y, I_Y = (%.f%.fj) A", real(I_Y),imag(I_Y))
+printf("\nFault current in the line B, I_B = (%.f+%.fj) A", real(I_B),imag(I_B))
+printf("\nPotential above earth attained by the alternator neutrals = %.f V\n", V_neutral)
+printf("\nNOTE: ERROR: Voltage is 11000 not 11000 kV as given in textbook statement")
+printf("\n Changes in the obtained answer from that of textbook is due to more precision here")
diff --git a/3472/CH30/EX30.5/Example30_5.sce b/3472/CH30/EX30.5/Example30_5.sce
new file mode 100644
index 000000000..cb7d7aaf1
--- /dev/null
+++ b/3472/CH30/EX30.5/Example30_5.sce
@@ -0,0 +1,46 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART III : SWITCHGEAR AND PROTECTION
+// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.5 :
+// Page number 514-515
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 6600.0 // Alternator voltage(V)
+kVA = 10000.0 // Alternator rating(kVA)
+x_1 = 15.0 // Reactance to positive sequence current(%)
+x_2 = 75.0 // Reactance to negative sequence current(%)
+x_0 = 30.0 // Reactance to zero sequence current(%)
+R_earth = 0.3 // Earth resistance(ohm)
+
+// Calculations
+a = exp(%i*120.0*%pi/180) // Operator
+E_g = V/3**0.5 // Phase voltage(V)
+// Case(a)
+I = kVA*1000/(3**0.5*V) // Full load current of each alternator(A)
+X = x_1*V/(100*3**0.5*I) // Positive sequence reactance(ohm)
+Z_g1 = %i*X // Equivalent positive sequence impedance(ohm)
+Z_g2 = Z_g1*x_2/100 // Equivalent negative sequence impedance(ohm)
+Z_g0 = Z_g1*x_0/100 // Equivalent zero sequence impedance(ohm)
+Z_1 = Z_g1/3 // Positive sequence impedance(ohm)
+Z_2 = Z_g2/3 // Negative sequence impedance(ohm)
+Z_0 = Z_g0/3 // Zero sequence impedance(ohm)
+I_a_a = 3*E_g/(Z_1+Z_2+Z_0) // Fault current(A)
+// Case(b)
+Z_0_b = Z_g0 // Impedance(ohm)
+I_a_b = 3*E_g/(Z_1+Z_2+Z_0_b) // Fault current(A)
+// Case(c)
+Z_0_c = R_earth*3+Z_g0 // Impedance(ohm)
+I_a_c = 3*E_g/(Z_1+Z_2+Z_0_c) // Fault current(A)
+
+// Results
+disp("PART III - EXAMPLE : 4.5 : SOLUTION :-")
+printf("\nCase(a): Fault current if all the alternator neutrals are solidly earthed, I_a = %.fj A", imag(I_a_a))
+printf("\nCase(b): Fault current if only one of the alternator neutrals is solidly earthed & others isolated = %.fj A", imag(I_a_b))
+printf("\nCase(c): Fault current if one of alternator neutrals is earthed through resistance & others isolated = %.f A\n", abs(I_a_c))
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
diff --git a/3472/CH30/EX30.6/Example30_6.sce b/3472/CH30/EX30.6/Example30_6.sce
new file mode 100644
index 000000000..cc919ad90
--- /dev/null
+++ b/3472/CH30/EX30.6/Example30_6.sce
@@ -0,0 +1,63 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART III : SWITCHGEAR AND PROTECTION
+// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.6 :
+// Page number 515-516
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+kVA_G = 2000.0 // Generator rating(kVA)
+X_G = 10.0 // Generator reactance(%)
+kVA_T1 = 2000.0 // Transformer rating(kVA)
+lv_T1 = 6.6 // LV side voltage(kV)
+hv_T1 = 11.0 // HV side voltage(kV)
+X_T1 = 5.0 // Transformer reactance(%)
+X_cable = 0.5 // Cable reactance(ohm)
+V_cable = 11.0 // Cable voltage(V)
+kVA_T2 = 2000.0 // Transformer rating(kVA)
+lv_T2 = 6.6 // LV side voltage(kV)
+hv_T2 = 11.0 // HV side voltage(kV)
+X_T2 = 5.0 // Transformer reactance(%)
+
+// Calculations
+a = exp(%i*120.0*%pi/180) // Operator
+kVA_base = 2000.0 // Base kVA
+kV = 6.6 // Base voltage(kV)
+X_1 = X_G*kV**2*10/kVA_base // 10% reactance at 6.6 kV(ohm)
+X_2 = X_T1*kV**2*10/kVA_base // 5% reactance at 6.6 kV(ohm)
+X_3 = (kV/hv_T1)**2*X_cable // 0.5 ohm at 11kV when referred to 6.6kV(ohm)
+Z_g1 = %i*X_1 // Positive sequence impedance of generator(ohm)
+Z_g2 = Z_g1*0.7 // Negative sequence impedance of generator equal to 70% of +ve sequence impedance(ohm)
+T1_Z_T1_1 = %i*X_2 // Positive sequence impedance of transformer(ohm)
+T1_Z_T1_2 = %i*X_2 // Negative sequence impedance of transformer(ohm)
+Z_C1 = %i*X_3 // Positive sequence impedance of cable(ohm)
+Z_C2 = %i*X_3 // Negative sequence impedance of cable(ohm)
+T2_Z_T2_1 = %i*X_2 // Positive sequence impedance of transformer(ohm)
+T2_Z_T2_2 = %i*X_2 // Negative sequence impedance of transformer(ohm)
+Z_1 = Z_g1+T1_Z_T1_1+Z_C1+T2_Z_T2_1 // Positive sequence impedance(ohm)
+Z_2 = Z_g2+T1_Z_T1_2+Z_C2+T2_Z_T2_2 // Negative sequence impedance(ohm)
+Z_0 = %i*X_2 // Zero sequence impedance(ohm)
+E_a = kV*1000/3**0.5 // Phase voltage(V)
+// Case(a)
+I_a1 = E_a/(Z_1+Z_2) // Positive sequence current(A)
+I_a2 = -I_a1 // Negative sequence current(A)
+I_a0 = 0 // Zero sequence current(A)
+I_a = I_a1+I_a2+I_a0 // Fault current in line a(A)
+I_b = (a**2-a)*I_a1 // Fault current in line b(A)
+I_c = -I_b // Fault current in line c(A)
+// Case(b)
+I_a_b = 3*E_a/(Z_1+Z_2+Z_0) // Fault current for line to ground fault(A)
+
+// Results
+disp("PART III - EXAMPLE : 4.6 : SOLUTION :-")
+printf("\nCase(a): Fault current for line fault are")
+printf("\n I_a = %.f A", abs(I_a))
+printf("\n I_b = %.f A", abs(I_b))
+printf("\n I_c = %.f A", abs(I_c))
+printf("\nCase(b): Fault current for line to ground fault, |I_a| = %.f A\n", abs(I_a_b))
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
diff --git a/3472/CH30/EX30.7/Example30_7.sce b/3472/CH30/EX30.7/Example30_7.sce
new file mode 100644
index 000000000..383ba9512
--- /dev/null
+++ b/3472/CH30/EX30.7/Example30_7.sce
@@ -0,0 +1,85 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART III : SWITCHGEAR AND PROTECTION
+// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.7 :
+// Page number 516-518
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+MVA_G1 = 40.0 // Generator rating(MVA)
+kV_G1 = 13.2 // Generator voltage(kV)
+X_st_G1 = 0.15 // Sub-transient reactance(p.u)
+X_2_G1 = 0.15 // Negative sequence reactance(p.u)
+X_0_G1 = 0.08 // Zero sequence reactance(p.u)
+MVA_G3 = 60.0 // Generator rating(MVA)
+kV_G3 = 13.8 // Generator voltage(kV)
+X_st_G3 = 0.20 // Sub-transient reactance(p.u)
+X_2_G3 = 0.20 // Negative sequence reactance(p.u)
+X_0_G3 = 0.08 // Zero sequence reactance(p.u)
+MVA_T1 = 40.0 // Transformer rating(MVA)
+kV_lv_T1 = 13.8 // Transformer low voltage(kV)
+kV_hv_T1 = 138 // Transformer high voltage(kV)
+X_1_T1 = 0.10 // Positive sequence reactance(p.u)
+X_2_T1 = 0.10 // Negative sequence reactance(p.u)
+X_0_T1 = 0.08 // Zero sequence reactance(p.u)
+MVA_T5 = 30.0 // Transformer rating(MVA)
+kV_lv_T5 = 13.8 // Transformer low voltage(kV)
+kV_hv_T5 = 138 // Transformer high voltage(kV)
+X_1_T5 = 0.10 // Positive sequence reactance(p.u)
+X_2_T5 = 0.10 // Negative sequence reactance(p.u)
+X_0_T5 = 0.08 // Zero sequence reactance(p.u)
+X_neutral = 0.05 // Reactance of reactor connected to generator neutral(p.u)
+
+// Calculations
+MVA_base = 100.0 // Base MVA
+kV_line = 138.0 // Base voltage for line(kV)
+kV_G = 13.8 // Base voltage for generator(kV)
+X_st_G1_pu = %i*X_st_G1*(kV_G1/kV_G)**2*MVA_base/MVA_G1 // Impedance of G1 & G2(p.u)
+X_2_G1_pu = %i*X_2_G1*(kV_G1/kV_G)**2*MVA_base/MVA_G1 // Impedance of G1 & G2(p.u)
+X_g0_G1_pu = %i*X_0_G1*(kV_G1/kV_G)**2*MVA_base/MVA_G1 // Impedance of G1 & G2(p.u)
+X_gn_G1_pu = %i*X_neutral*(kV_G1/kV_G)**2*MVA_base/MVA_G1 // Impedance of G1 & G2(p.u)
+X_st_G3_pu = %i*X_st_G3*(kV_G3/kV_G)**2*MVA_base/MVA_G3 // Impedance of G3(p.u)
+X_2_G3_pu = %i*X_2_G3*(kV_G3/kV_G)**2*MVA_base/MVA_G3 // Impedance of G3(p.u)
+X_g0_G3_pu = %i*X_0_G3*(kV_G3/kV_G)**2*MVA_base/MVA_G3 // Impedance of G3(p.u)
+X_gn_G3_pu = %i*X_neutral*(kV_G3/kV_G)**2*MVA_base/MVA_G3 // Impedance of G3(p.u)
+X_1_T1_pu = %i*X_1_T1*MVA_base/MVA_T1 // Impedance of T1,T2,T3 & T4(p.u)
+X_2_T1_pu = %i*X_2_T1*MVA_base/MVA_T1 // Impedance of T1,T2,T3 & T4(p.u)
+X_0_T1_pu = %i*X_0_T1*MVA_base/MVA_T1 // Impedance of T1,T2,T3 & T4(p.u)
+X_1_T5_pu = %i*X_1_T5*MVA_base/MVA_T5 // Impedance of T5 & T6(p.u)
+X_2_T5_pu = %i*X_2_T5*MVA_base/MVA_T5 // Impedance of T5 & T6(p.u)
+X_0_T5_pu = %i*X_0_T5*MVA_base/MVA_T5 // Impedance of T5 & T6(p.u)
+X_1_line_20 = %i*20.0*100/kV_line**2 // Impedance of 20 ohm line(p.u)
+X_2_line_20 = %i*20.0*100/kV_line**2 // Impedance of 20 ohm line(p.u)
+X_0_line_20 = 3.0*X_1_line_20 // Impedance of 20 ohm line(p.u)
+X_1_line_10 = %i*10.0*100/kV_line**2 // Impedance of 10 ohm line(p.u)
+X_2_line_10 = %i*10.0*100/kV_line**2 // Impedance of 10 ohm line(p.u)
+X_0_line_10 = 3.0*X_1_line_10 // Impedance of 10 ohm line(p.u)
+// Positive,negative and zero sequence network
+Z_1_1 = X_1_T1_pu+X_1_T1_pu+X_1_line_20 // Impedance(p.u)
+Z_2_1 = X_1_T1_pu+X_1_T5_pu+X_1_line_10 // Impedance(p.u)
+Z_3_1 = X_1_T1_pu+X_1_T5_pu+X_1_line_10 // Impedance(p.u)
+Z_4_1 = Z_1_1*Z_2_1/(Z_1_1+Z_2_1+Z_3_1) // Impedance after star-delta transformation(p.u)
+Z_5_1 = Z_3_1*Z_1_1/(Z_1_1+Z_2_1+Z_3_1) // Impedance after star-delta transformation(p.u)
+Z_6_1 = Z_3_1*Z_2_1/(Z_1_1+Z_2_1+Z_3_1) // Impedance after star-delta transformation(p.u)
+Z_7_1 = X_st_G1_pu+Z_4_1 // Impedance(p.u)
+Z_8_1 = X_st_G1_pu+Z_5_1 // Impedance(p.u)
+Z_9_1 = Z_7_1*Z_8_1/(Z_7_1+Z_8_1) // Impedance in parallel(p.u). Refer Fig E4.14(e) & E4.14(f)
+Z_10_1 = Z_9_1+Z_6_1 // Impedance(p.u). Refer Fig E4.14(f) & E4.14(g)
+Z_11_1 = Z_10_1*X_st_G3_pu/(Z_10_1+X_st_G3_pu) // Impedance in parallel(p.u). Refer Fig E4.14(g) & E4.14(h)
+Z_1 = Z_11_1 // Positive sequence impedance(p.u)
+Z_2 = Z_1 // Negative sequence impedance(p.u)
+Z_0 = X_g0_G3_pu+3.0*X_gn_G3_pu // Zero sequence impedance(p.u)
+E_g = 1.0 // Voltage(p.u)
+I_f_pu = 3*E_g/(Z_1+Z_2+Z_0) // L-G fault current(p.u)
+I_f = abs(I_f_pu)*MVA_base*1000/(3**0.5*kV_G) // Actual fault current(A)
+MVA_fault = abs(I_f_pu)*MVA_base // Fault MVA
+
+// Results
+disp("PART III - EXAMPLE : 4.7 : SOLUTION :-")
+printf("\nFault current for a L-G fault at C = %.f A\n", I_f)
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
diff --git a/3472/CH30/EX30.8/Example30_8.sce b/3472/CH30/EX30.8/Example30_8.sce
new file mode 100644
index 000000000..d4967851a
--- /dev/null
+++ b/3472/CH30/EX30.8/Example30_8.sce
@@ -0,0 +1,46 @@
+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART III : SWITCHGEAR AND PROTECTION
+// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.8 :
+// Page number 518-519
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+kV_G = 11.0 // Generator rating(kV)
+X_1_G = %i*0.1 // Positive sequence reactance of generator(p.u)
+X_2_G = %i*0.1 // Negative sequence reactance of generator(p.u)
+X_0_G = %i*0.02 // Zero sequence reactance of generator(p.u)
+Z = 1.0 // Earthing resistor(ohm)
+X_1_T1 = %i*0.1 // Positive sequence reactance of 2-winding transformer(p.u)
+X_2_T1 = %i*0.1 // Negative sequence reactance of 2-winding transformer(p.u)
+X_0_T1 = %i*0.1 // Zero sequence reactanc of 2-winding transformere(p.u)
+X_1_T2_hv = %i*0.05 // Positive sequence reactance of hv 3-winding transformer(p.u)
+X_2_T2_hv = %i*0.05 // Negative sequence reactance of hv 3-winding transformer(p.u)
+X_0_T2_hv = %i*0.05 // Zero sequence reactanc of hv 3-winding transformere(p.u)
+X_1_T2_lv_1 = %i*0.02 // Positive sequence reactance of lv 3-winding transformer(p.u)
+X_2_T2_lv_1 = %i*0.02 // Negative sequence reactance of lv 3-winding transformer(p.u)
+X_0_T2_lv_1 = %i*0.02 // Zero sequence reactanc of lv 3-winding transformere(p.u)
+X_1_T2_lv_2 = %i*0.05 // Positive sequence reactance of lv 3-winding transformer(p.u)
+X_2_T2_lv_2 = %i*0.05 // Negative sequence reactance of lv 3-winding transformer(p.u)
+X_0_T2_lv_2 = %i*0.05 // Zero sequence reactanc of lv 3-winding transformere(p.u)
+
+// Calculations
+MVA_b = 10.0 // Base MVA
+kV_b = 11.0 // Base voltage(kV)
+Z_n = Z*MVA_b/kV_b**2 // Impedance(p.u)
+Z_1 = X_1_G+X_1_T1+X_1_T2_hv+((X_1_T2_lv_1*X_1_T2_lv_2)/(X_1_T2_lv_1+X_1_T2_lv_2)) // Positive sequence impedance(p.u)
+Z_2 = X_2_G+X_2_T1+X_2_T2_hv+((X_2_T2_lv_1*X_2_T2_lv_2)/(X_2_T2_lv_1+X_2_T2_lv_2)) // Negative sequence impedance(p.u)
+Z_0 = ((X_0_T1+X_0_T2_hv)*X_0_T2_lv_2/(X_0_T1+X_0_T2_hv+X_0_T2_lv_2))+X_0_T2_lv_1+3*Z_n // Zero sequence impedance(p.u)
+E = 1.0 // Voltage(p.u)
+I_f_pu = 3*E/(Z_1+Z_2+Z_0) // Fault current(p.u)
+I_f = MVA_b*1000*abs(I_f_pu)/(3**0.5*kV_b) // Fault current(A)
+
+// Results
+disp("PART III - EXAMPLE : 4.8 : SOLUTION :-")
+printf("\nFault current, I_f = %.f A\n", I_f)
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
diff --git a/3472/CH30/EX30.9/Example30_9.sce b/3472/CH30/EX30.9/Example30_9.sce
new file mode 100644
index 000000000..5a6a0ae4c
--- /dev/null
+++ b/3472/CH30/EX30.9/Example30_9.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 III : SWITCHGEAR AND PROTECTION
+// CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
+
+// EXAMPLE : 4.9 :
+// Page number 519
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+MVA_G = 10.0 // Generator rating(MVA)
+kV_G = 11.0 // Generator rating(kV)
+X_1_G = 27.0 // Positive sequence reactance of generator(p.u)
+X_2_G = 9.0 // Negative sequence reactance of generator(p.u)
+X_0_G = 4.5 // Zero sequence reactance of generator(p.u)
+X_1_L = 9.0 // Positive sequence reactance of line upto fault(p.u)
+X_2_L = 9.0 // Negative sequence reactance of line upto fault(p.u)
+X_0_L = 0 // Zero sequence reactance of line upto fault(p.u)
+
+// Calculations
+E_a = kV_G*1000/3**0.5 // Phase voltage(V)
+Z_1 = %i*(X_1_G+X_1_L) // Positive sequence reactance(p.u)
+Z_2 = %i*(X_2_G+X_2_L) // Negative sequence reactance(p.u)
+I_b = %i*3**0.5*E_a/(Z_1+Z_2) // Fault current in line b(p.u)
+I_c = -I_b // Fault current in line c(p.u)
+
+// Results
+disp("PART III - EXAMPLE : 4.9 : SOLUTION :-")
+printf("\nFault current in line b, I_b = %.f A", abs(I_b))
+printf("\nFault current in line c, I_c = %.f A", real(I_c))
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