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+// chapter 7
+// example 7.2
+// Design dual convertor
+// page-432
+clear;
+clc;
+// given
+Ea=220; // in V
+Ia=30; // in A
+N=1500; // in rpm (speed of motor)
+Eac=400; // in V (3-phase supply)
+f=50; // in Hz (supply frequency)
+drop=15; // in percent (drop in the circuit)
+// calculate
+E_drop=(drop/100)*Ea;// calculation of total drop in the system
+Edc_alpha=Ea+E_drop;// calculation of total dc voltage
+// since Edc_alpha=1.35*Eac*cosd(alpha), therefore we get
+alpha1=acosd(Edc_alpha/(1.35*Eac));// calculation of firing angle
+Iac=0.817*Ia;// calculation of AC line current
+Pac=sqrt(3)*Eac*Iac;// calculation of AC terminal power
+// since Pac=1.35*Pdc, therefore we get
+Pdc=Pac/1.35;// calculation of DC average power
+I_ripple=Ia/5;// calculation of ripple current
+w=2*%pi*f;// calculation of angular velocity
+Lc=(2*1.35*Eac/(6*w*I_ripple))*(1/7+1/5);// calculation of current limiting inductance
+alpha2=180-alpha1;// calculation of firing angle
+PIV=2*sqrt(2)*Eac;// calculation of peak inverse voltage of SCR
+I_T=2*sqrt(2)*Iac;// calculation of current rating of SCR
+printf("\nThe total drop in the system is \t E_drop=%.f V",E_drop);
+printf("\nThe total dc voltage is \t\t Edc_alpha=%.f V",Edc_alpha);
+printf("\nThe firing angle is \t\t\t alpha1=%.f degree",alpha1);
+printf("\nThe AC line current is \t\t\t Iac=%.2f A",Iac);
+printf("\nThe AC terminal power is \t\t Pac=%.2f kW",Pac*1E-3);
+printf("\nThe DC average power is \t\t Pdc=%.2f kW",Pdc*1E-3);
+printf("\nThe current limiting inductance is \t Lc=%.f mH",Lc*1E3);
+printf("\nThe firing angle is \t\t\t alpha2=%.f degree",alpha2);
+printf("\nThe peak inverse voltage of SCR is \t PIV=%.f V",PIV);
+printf("\nThe current rating of SCR is \t\t I_T=%.f A",I_T);
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