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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')
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