Unsymmetrical Fault

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Information about Unsymmetrical Fault
Education

Published on March 14, 2014

Author: anuragyadav94

Source: slideshare.net

EXPERIMENT – 6 OBJECT: -To perform unsymmetrical fault analysis. CODE clc clearall z1=.2i z2=.2i z0=.05i Ea=1 a=-.5+.866i Ib=17.32 zf=input('Please Enter the value of zf'); k=input('Enter 1 for 1LG, Enter 2 for LL, Enter 3 for 2LG Enter 4 for 3phase fault'); switch(k) case 1 Ia=(3*Ea/(z1+z2+z0+3*zf)); If=Ia*Ib case 2 Ia1=(Ea/z1+z2+zf); Ia2=((a*a-a)*Ia1); If=Ia2*Ib case 3 Ia3=(Ea/z1+(z2*(z0+3*zf)/(z2+z0+3*zf))); Ia4=(-(Ea-Ia3*z1)/(z0+3*zf)); Ia5=3*Ia4; If=Ia5*Ib case 4 Ia6=Ea/z1; If=Ia6*Ib end [Theta,rho]=cart2pol(real(If),imag(If)) a=Theta*180/pi i=[rho a]

RESULT:- z1 =0 + 0.2000i z2 =0 + 0.2000i z0 = 0 + 0.0500i Ea =1 a = -0.5000 + 0.8660i Ib =17.3200 Please Enter the value of zf0 Enter 1 for 1LG, Enter 2 for LL, Enter 3 for 2LG Enter 4 for 3phase fault1 If =0 -1.1547e+002i Theta =-1.5708 rho = 115.4667 a = -90 i = 115.4667 -90.0000

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