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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+clc,clear
+printf('Example 5.7\n\n')
+
+//case(i)
+V_L=440
+V_ph=V_L/sqrt(3)
+phi=acos(0.8)
+
+//armature resistance drop from the graph
+//RS=1.1 cm and scale =50 V/cm
+arm_leak_resis= 1.1*50 //armature leakage resistance
+
+OB=V_ph*cos(phi)
+AB=V_ph*sin(phi) + arm_leak_resis
+E_1ph= sqrt( OB^2+AB^2 )
+
+F_f1=6.1 //corresponding value from OCC
+F_AR=3.1*1
+
+F_R= sqrt( F_f1^2 + F_AR^2 -2*F_f1*F_AR*cosd(90+acosd(0.8)) )
+E_ph=328 //voltage corresponding to F_R=8.33 A from OCC graph
+regulation1= 100*(E_ph - V_ph)/V_ph
+printf('(i)Regulation for 0.8 pf lagging is %.2f percent \n',regulation1)
+
+//case(ii)
+
+OC=V_ph*cos(phi)
+BC=V_ph*sin(phi) - arm_leak_resis
+E_1ph= sqrt( OC^2+BC^2 )
+
+F_f1=6.1 //corresponding value from OCC
+F_R= sqrt( F_f1^2 + F_AR^2 -2*F_f1*F_AR*cosd(90-acosd(0.8)) )
+E_ph=90 //volatge corresponding to F_R=3.34 A from OCC graph
+regulation2= 100*(E_ph - V_ph)/V_ph
+printf('(ii)Regulation for 0.8 pf leading is %.2f percent \n',regulation2)
+printf('\nThe answer in part (ii) doesnt match with textbook because of calculation mistake done in last step in the textbook')