// 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);