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//CHAPTER 2- STEADY-STATE ANALYSIS OF SINGLE-PHASE A.C. CIRCUIT
//Example 7
disp("CHAPTER 2");
disp("EXAMPLE 7");
//VARIABLE INITIALIZATION
s1=300; //apparent power absorbed by the plant in kVA
pf1=0.65; //power factor(lagging)
pf2=0.85; //power factor(lagging)
//SOLUTION
//solution (a)
p=s1*pf1; //active power P=S.cos(Φ)
q1=sqrt((s1^2)-(p^2)); //Q=sqrt(S^2-P^2) in kVAr
disp(sprintf("(a) To bring the power factor to unity, the capacitor bank should have a capacity of %3.0f kVAR",q1));
//solution (b)
s2=p/pf2; //since P=S.cos(Φ)
q2=sqrt((s2^2)-(p^2)); //Q=sqrt(S^2-P^2) in kVAr
disp(sprintf("(b) To bring the power factor to 85%% lagging, the capacitor bank should have a capacity of %3.0f kVAR",q2));
//END
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