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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+//Book name: Fundamentals of electrical drives by Mohamad A. El- Sharkawi
+//chapter 5
+//example 5.5
+//edition 1
+//publisher and place:Nelson Engineering
+clc;
+clear;
+P=40; //load of an industrial plant in Mw
+pf=.85; //power factor lagging
+pfnew=.95 //To improve new power factor
+V=5000; //motor rated voltage in volts
+Xs=5; //synchronous reactance in ohm
+c=200; //constant value given
+Vt=V/3^(1/2);
+a=acosd(pf); //power factor angle of the load in degree
+Ql=P*tand(a); //load reactive power in KVAR
+Qtot=P*tand(acosd(pfnew)); //total reactive power for .95 power factor lagging
+disp(Qtot,'The total reactive power for .95 power factor lagging in KVAR is')
+Qm=Qtot-Ql;
+Vt=(V/sqrt(3));
+Ef=((Qm*Xs)/(3*Vt))+Vt;
+If=Ef/c;
+disp(If,'The excitation current required to improve overall power factor of the plant in A is')