Question

I need help in matlab codes please

Single line to ground fault analysis Example 9.4 Problem 1 A three-phase, il kV, 30 MVA alternator with grounded neutral has a direct axis subtransien pu respectively based on system base. A single line to ground fault has occurred in phase a. Determine the fault current, line to ground voltages in pu. t, negative sequence and zero sequence reactances of 0.2 pu, 0.3 pu and 0.1 Solution: Phase voltage E.0 pu From equation (9.131) 3E pu Z, +2 0.2+0.3+0.1 0.6 9.16.2 MATLAB Program of Single Line to Ground Fault Analysis a--0.5+0.866i: p-input('Enter the faulted phase number:,); Vs=zeros (3,1); Va (p, 1)-input 'Enter the pre-fault voltage of the faulted phase(in pu):' vil-zeros (3,1) Iph-zeros3,1) for self and H-input( 'Enter 1 if data 1sin sequen mutual impedances: nce impedance else enter 2o-input('Enter the zero sequence impedancetin pu: 21-input 'Enter the positive sequence impedance fin pu) 2 input( 'Enter the negative sequence impedance (in pu):' Z faj put ('Enter the fault İmpedance(in Pu):リ; Power System Analysis: Operation and Control 242 From equation (9.122) From equation (9.124) From equao (9.125) From equation 9.123) Line to ground voltage as given by equation (9.126) 1Z167) 0.3)0.5 pu 0 pu as given by equation (P.13) is zero becasue the fault has occurred on phase a Now phase b and c voltages are r, = (-0.1 67) + (-05-j 0.866) (0.67) + (-0.5 +j 0.866) (-0.5) ,--0.252-j l '0 l 3 pu and r s(-0 1 67) (-0.5ザ0.866) (0.67)十(-05-j 0866) (-0.5) r=-0.252 +/1.013 pu

Answer 1

kindly read the comments as well for better understanding of code

MATLAB CODE:

a=-0.5+0.866i; % operator 'a' in fault analysis
p=input('enter the faulted phase number:');
Vs=zeros(3,1); %array to store prefualt voltages
Vs(p,1)=input('enter the pre-fault voltage of faulted phase(in pu):');
Vsy=zeros(3,1); %symmetrical component of phase(line to ground) voltages
Iph=zeros(3,1); %array of phase currents
H=input('enter 1 if data is in sequence impedance else enter 0 for self and mutual impedances:');
if(H==1)
   Zo=input('enter zero sequence impedance (in pu):'); %enter zero sequence impedance value= 0.1i at command prompt for given problem as it is reactance only
   Z1=input('enter positive sequence impedance (in pu):'); % positive sequence impedance value= 0.2i at command prompt for given problem as it is reactance only
   Z2=input('enter negative sequence impedance (in pu):'); % negative sequence impedance value= 0.3i at command prompt for given problem as it is reactance only
   Zf=input('enter fault impedance (in pu):'); %fault impedance value=0 at command prompt since no value is given in problem
else
   Zf=input('enter fault impedance (in pu):');
   Zs=input('enter self impedance of transmission line(in pu):'); %enter self impedance value in terms of i at command prompt if it is reactance only
   Zm=input('enter mutual impedance between transmission line(in pu):'); %enter mutual impedance in terms of i at command prompt if it is reactance only
   Zo=Zs+(2*Zm); %zero sequence impedance
   Z1=Zs-Zm;   % positive sequence impedance
   Z2=Z1;       % negative sequence impedance
end
Iph(p,1)=(3*Vs(p,1))/(Zo+Z1+Z2+Zf);
fprintf('fault current is (%f)+(%fi) \n',real(Iph(p,1)),imag(Iph(p,1)));
T=(1/3)*[1 1 1;1 a a*a;1 a*a a]; %T is inverse of symmetrical component transformation matrix
Isy=T*Iph;
Vsy(1,1)=-Isy(1,1)*Zo;
Vsy(2,1)=Vs(p,1)-Isy(2,1)*Z1;
Vsy(3,1)=-Isy(3,1)*Z2;
Vph=inv(T)*Vsy;
fprintf('line to ground voltage of phase a is \n');
disp(Vph(1,1));
fprintf('line to ground voltage of phase b is \n');
disp(Vph(2,1));
fprintf('line to ground voltage of phase c is \n');
disp(Vph(3,1));

RESULT:

fault current is (0.000000)+-5.000000i) line to ground voltage of phase a is 7.3334e-06 9.2519e-17i line to ground voltage of phase b is 0.25003 1.01039i line to ground voltage of phase c is -0.249981.01039i F-4

pic of code in editor:

a--0,5+0. 8 6 6 1 ; operator . a. 1n fault analyฮา 2 P-input('enter the taulted phase number:') Va-zeros (3,1); %array to 8tore pretualt voltages Vs (p,1)-input ('enter the pre-fault voltage of faulted phase (in pu):") Vsy-zeros (3,1); 香symmetrical component of phase (line to ground) voltages Iph-zeros (3,1): Sarray of phase currents H-input ('encer 1 if data 1ธ in jequence impedance e1 7 encer 0 for ฮืelf and mutual impedancc:'); Zo-input ('enter zero equence impedance (în pu) :'); %enter zero sequence impedance value= o.11 at command prompt for given problem as it 18 reactance only 21=input('enter positive sequence impedance (in pu) :'); positive Bequence impedance value« o.21 at eammand prompt for given problem as it is reactance onl Z2=input ('enter negative sequence impedance (in pu):"); % negative sequence impedance value-0.3i at command prompt for given problem as it is reactance onl Zf-input ('enter fault impedance (in pu):"); %fault impedance value-0 at command prompt since no value is given in problem 10 12 13 14 clsc 2f-input ('enter fault impedance (in pu): Z8-input ('enter self Impedance of transmission line {in pu) :'); %enter self 1mpedance value 1n terms of 1 at command prompt if it is reactance only 16 17 18 19 20 21 22 23 input(' enter mutual i pedance between transni88ion line (in pu):'); enter mutual impedance in terms of i at eammand prompt if it is reactance only Zo-Zs+(2#2m); 21-2s-2m: 22-21 impedance zero sequence % positive sequence impedance negative sequence impedance end lph (p,L)-(3*Va (p,L))/(20+21+22+2f); rprintf ('fault current i8 ( 2)+/%f1) Ts(:/3) *[1 1 1;1 a a*a; 1 ata a];钉 \n ''real (Iph (p ,L)), inag (Iph (p,L))); inverse of symmetrical component transformation matrix 25 Vsy (,1)--Isy(,1)2o: 26 Vay (2,1)-Va (p,1)-Isy (2,1) 21; 27 Vay (3,1)--Isy (3,1)"22 ; fprintf('line to ground voltage of phase a is n' 30 disp (Vph(,1)) 31 fprintf('line to ground voltage of phase b is In') 32 disp (Vph (2,1)) 33 fprintt('line to ground voltage of phase c 1s n 3 diap (Vp(3,1