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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+clc;
+v=3300; // balanced supply voltage
+p=500000; // rated power of induction motor
+f=50; // frequency of supply
+m=3; // number of phases
+pf=0.7; // lagging power factor
+pfn=0.9; // improved power factor
+vc=420; // rated voltage of capacitor
+n=0.86; // motor efficiency
+i=p/(sqrt(3)*v*pf*n); // line current
+i=i*(pf-%i*sqrt(1-pf^2)); // complex form of line current
+il=real(i)/pfn; // line current at improved power factor
+il=il*(pfn-%i*sqrt(1-pfn^2)); // complex form of new line current
+//from fig. 6.39
+ic=-(imag(i)-imag(il)); // reactive component of current to be neutralised
+// capacitor bank is delta connected
+// capacitor bank is delta connected, converting into equivalent star Xstar=Xdelta/3
+xcd=v/(ic*sqrt(3)); // capacitance reactance
+Cd=1/(2*%pi*f*xcd*m); // capacitance
+// now each capacitor is rated at 420 V, number of capacitor connected in series is
+n=ceil(v/vc);
+C=Cd*n;
+printf('Per phase value of each capacitance for delta connected capacitor bank is %f microfarad\n',C*10^6);
+// let R be resistance of distribution circuit
+// power lost without capacitor bank is m*abs(i)^2*R
+// power lost with capacitor bank is m*abs(il)^2*R therefore
+ps=(abs(i)^2-abs(il)^2)/abs(i)^2
+printf('Percentage saving in losses is %f percent',ps*100);