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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")
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