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//Swinburne test on a dc shunt motor
clc;
clear;
V=500;
I=5;
Rf=250;
Ra=0.5;
P=V*I;
If=V/Rf;
Ia=I-If;
Pfc=(If^2)*Rf;// Field Copper Loss
Pac=(Ia^2)*Ra; // Armature Copper Loss
Pil=P-Pfc-Pac;// Iron loss
// Generator
Vg=500;
Ig=100;
Pog=Vg*Ig; // Power Output
Iag=Ig+If; //Armature current
Pgac=(Iag^2)*Ra; // Armature Copper loss
slg=0.01*Pog;//stray loss
Pgtl=Pgac+Pfc+slg+Pil; // Total losses
effg=Pog*100/(Pog+Pgtl);
// Motor
Vm=500;
Im=100;
Pim=Vm*Im; // Power input to the motor
Iam=Ig-If; // Armature current
Pmac=(Iam^2)*Ra; // Armature Copper Loss
Pom=Pim-Pmac-Pil-Pfc;// Ouput of the motor
slm=0.01*Pom;// Stray loss
Pmtl=Pmac+Pil+Pfc+slm; // Total loss of the motor
effm=(Pom-slm)*100/(Pim);
printf('i) The Efficiency of the machine as a generator delivering 100A at 500V = %g percent \n',effg)
printf('ii) The Efficiency of the machine as a motor having a line current 100A at 500V = %g percent \n',effm)
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