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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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