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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))
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