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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+// Example 8.2
+// Determine (a) The minimum value of excitation that will maintain
+// synchronism (b) Repeat (a) using eq.(8.16) (c) Repeat (a) using eq.(8.21)
+// (d) Power angle if the field excitation voltage is increased to 175% of the
+// stability limit determined in (c)
+// Page No. 322
+
+clc;
+clear;
+close;
+
+// Given data
+Pin=40; // Input power
+Pin1phase=40/3; // Single phase power
+Xs=1.27; // Synchronous reactnace
+VT=220/sqrt(3); // Voltage
+delta=-90; // Power angle
+
+f=60; // Operating frequency
+P=4; // Number of poles
+Pmech=100; // Mechanical power
+eta=0.96; // Efficiency
+FP=0.80; // Power factor leading
+V=460; // Motor voltage
+Xs_Mag=2.72; // Synchronous reactnace magnitude
+Xs_Ang=90; // Synchronous reactnace magnitude
+deltaPull=-90; // Pullout power angle
+
+// (a) The minimum value of excitation that will maintain synchronism
+Ef=98; // From the graph (Figure 8.13)
+
+// (b) The minimum value of excitation using eq.(8.16)
+Ef816=-Pin*Xs*746/(3*VT*sind(delta));
+
+
+// (c) The minimum value of excitation using eq.(8.21)
+Ef821=Xs*Pin1phase*746/(VT);
+
+// (d) Power angle if the field excitation voltage is increased to 175%
+delta2=Ef816*sind(delta)/(1.75*Ef816);
+delta2=asind(delta2);
+
+// Display result on command window
+printf("\n The minimum value of excitation = %0.0f V ",Ef);
+printf("\n The minimum value of excitation using eq.(8.16) = %0.0f V ",Ef816);
+printf("\n The minimum value of excitation using eq.(8.21) = %0.0f V ",Ef821);
+printf("\n Power angle = %0.0f deg ",delta2);