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diff --git a/3472/CH28/EX28.4/Example28_4.sce b/3472/CH28/EX28.4/Example28_4.sce new file mode 100644 index 000000000..0a72b383e --- /dev/null +++ b/3472/CH28/EX28.4/Example28_4.sce @@ -0,0 +1,35 @@ +// 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 2: FAULT LIMITING REACTORS
+
+// EXAMPLE : 2.4 :
+// Page number 481
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+kVA = 20000.0 // Rating of generator(kVA)
+f = 50.0 // Frequency(Hz)
+V = 11.0*10**3 // Voltage of generator(V)
+X_G = 20.0 // Generator short-circuit reactance(%)
+x = 60.0 // Reactance falls to 60% normal value
+
+// Calculations
+kVA_base = 20000.0 // Base kVA
+X = poly(0,"X") // Reactance of each reactors E,F,G & H(%)
+X_AE = X+X_G // Reactances of A & E in series(%)
+X_BF = X+X_G // Reactances of B & F in series(%)
+X_CD = X+X_G // Reactances of C & D in series(%)
+X_eq = X_AE/3 // X_eq = X_AE*X_BF*X_CD/(X_BF*X_CD+X_AE*X_CD+X_AE*X_BF). Combined reactances of 3 groups in parallel(%)
+X_f = X_eq+X // Reactances of these groups to fault via tie-bar(%)
+X_sol = roots(6.66666666666667-(100-x)/100*(X_f)) // Value of reactance of each reactors E,F,G & H(%)
+I_fl = kVA_base*1000/(3**0.5*V) // Full load current corresponding to 20000 kVA & 11 kV(A)
+X_ohm = X_sol*V/(3**0.5*100*I_fl) // Ohmic value of reactance X(ohm)
+
+// Results
+disp("PART III - EXAMPLE : 2.4 : SOLUTION :-")
+printf("\nReactance of each reactor = %.4f ohm \n", X_ohm)
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
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