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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 6.43\n\n')
+
+V_l=6.6*10^3
+V_t=V_l/sqrt(3)
+X_d=23.2,X_q=14.5,R_a=0 //armature resistance and synchronous reactance of direct,quadrature axis
+VA=1800*10^3
+phi=acos(0.8) //lag
+
+I_a=VA/(V_l*sqrt(3))
+
+psi=atan( (V_t*sin(phi)-I_a*X_q)/(V_t*cos(phi)-I_a*R_a) ) //minus sign in numerator and denomenator for motors
+delta=psi+phi
+I_d=I_a*sin(psi)
+I_q=I_a*cos(psi)
+E_f=V_t*cos(delta)-I_d*X_d-I_q*R_a
+printf('Excitation emf = %.4f V\n',E_f)
+//P_m= ( V_t*E_f*sin(delta)/X_d ) + ((1/X_q)-(1/X_d))*0.5*sin(2*delta)*V_t^2
+//P_m=0.4996*cos(delta)+0.1877*sin(2*delta)
+//for maximum power output, differenciate and equate to zero
+
+delta_max=63.4 //degree
+
+P_m_max=((1/X_q)-(1/X_d))*0.5*sind(2*delta_max)*V_t^2 //Maximuum load supplied with E_f=0
+printf('Maximum load the motor can supply is %.4f MW per phase ',P_m_max*10^-6 )