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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+clc;
+p=500*10^3; // rated power of alternator
+v=11000; // rated voltage of alternator
+m=3; // number of phases
+l1=1500; // friction and windage losses
+l2=2500; // open circuit core losses
+ra=4; // armature resistance per phase
+l3=1000; // field copper loss
+pf=0.8; // power factor
+disp('case a: Half load');
+ia=p/(sqrt(3)*v); // armature current
+l4=(m*ia^2*ra)/4; // short circuit load loss at half load
+lt=l1+l2+l3+l4; // net loss
+n=(1-(lt/((p/2)*pf+lt)))*100;
+printf('Efficiency is %f percent\n',n);
+disp('case b');
+// for maximum efficiency variable losses=constant losses
+iam=sqrt((l1+l2+l3)/(m*ra)); // armature current at maximum efficiency
+pout=m*(v/sqrt(3))*iam*pf; // output power ta maximum efficiency
+lt=2*(l1+l2+l3); // net losses
+nm=(1-(lt/(pout+lt)))*100;
+printf('Maximum efficiency is %f percent\n',nm);