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+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.10 :
+// Page number 336
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+load_1 = 10000.0      // Total balanced load(kW)
+V = 33000.0           // Voltage(V)
+PF_1 = 0.8            // Lagging power factor
+R = 1.6               // Resistance of feeder(ohm/phase)
+X = 2.5               // Reactance of feeder(ohm/phase)
+load_2 = 4460.0       // Load delivered by feeder(kW)
+PF_2 = 0.72           // Lagging power factor
+
+// Calculations
+I = load_1*1000/(3**0.5*V*PF_1)*exp(%i*-acos(PF_1))       // Total line current(A)
+I_1 = load_2*1000/(3**0.5*V*PF_2)*exp(%i*-acos(PF_2))     // Line current of first feeder(A)
+I_2 = I-I_1                                               // Line current of first feeder(A)
+Z_1 = complex(R,X)                                        // Impedance of first feeder(ohm)
+Z_2 = I_1*Z_1/I_2                                         // Impedance of second feeder(ohm)
+
+// Results
+disp("PART II - EXAMPLE : 11.10 : SOLUTION :-")
+printf("\nImpedance of second feeder, Z_2 = %.2f∠%.1f° ohm \n", abs(Z_2),phasemag(Z_2))
+printf("\nNOTE: ERROR: Changes in the obtained answer from that of textbook is due to wrong values of substitution")