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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3472/CH30/EX30.3/Example30_3.sce | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
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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")
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