Matlab Program

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1. To find ABCD constants of equivalent PI transmission line and hence1. To find ABCD constants of equivalent PI transmission line and hence efficiency and regulation

efficiency and regulation ALGORITHM:

ALGORITHM: 1.

1. Read Read length, length, impedance/km(zkm), impedance/km(zkm), admittance/km admittance/km of of transmission transmission line(ykm)line(ykm) 2.

2. Enter Enter choice choice 1-for 1-for short short lineline 2-for medium line 2-for medium line 3-for long line 3-for long line 3. 3. If If choice=1choice=1 Read length Read length Set B = Z Set B = Z A = D = 1 A = D = 1 C = 0 C = 0 If choice=2 If choice=2 Read length Read length A=D=1 + yz/2 A=D=1 + yz/2 B = Z B = Z C = y (1 + yz/4) C = y (1 + yz/4) If choice=3 If choice=3 Read length Read length Zeq Zeq zc = zc = z/yz/y   == z*y*lengthz*y*length A = D = Cosh ( A = D = Cosh ()) Yeq Yeq Yeq Yeq B = Zc Sinh ( B = Zc Sinh ()) C = 1/Zc Sinh ( C = 1/Zc Sinh ())

Find generalized equivalent circuit elements of

Find generalized equivalent circuit elements of  networknetwork zeq = z*length*sinh (

zeq = z*length*sinh ()/)/  yeq = y*length/2*tanh (

yeq = y*length/2*tanh (/2)/ (/2)/ (/2)/2) Display Zeq & Yeq

Display Zeq & Yeq 4.

4. Display Display A, A, B, B, C, C, & & D D constantsconstants 5.

5. Read Read choice choice 1 1 – – To To read read V V r, r, I I r r and and compute compute V V s s , , I I ss 2 – To read V s, I s and compute V r, I r

2 – To read V s, I s and compute V r, I r 6.

6. If If choice choice = = 11

Find V s = A V r + B I r Find V s = A V r + B I r

I

I s s = = C C V V r r + + D D I I r r where where V V r r & & I I r r are are in in per per phph ase ase Display V s & I s. Display V s & I s. if choice = 2 if choice = 2 Find V r = D Vs – B Is Find V r = D Vs – B Is I

I r r = = -C -C Vs+ Vs+ A A Is Is where where Vs Vs & & Is Is are are in in per per phasephase Display V r & I r.

Display V r & I r. 7.

7. Find Find (i) (i) VSVS SV & IsSV & Is SISI

(ii) Sending end Power factor Pf-send = cos (

(ii) Sending end Power factor Pf-send = cos (SV -SV -SI)SI) (iii) Sending end Power, PS = 3 VS x  I s x Pf-send (iii) Sending end Power, PS = 3 VS x  I s x Pf-send 8.

8. Find Find (i) (i) V V rr rv & I rrv & I r rIrI

(ii) Receiving end power factor Pf-rec = cos (

(ii) Receiving end power factor Pf-rec = cos (rV -rV -rI)rI) (iii)

(iii) Receiving Receiving end end power, power, Pr=3 Pr=3 V V r r x x I I r r x x Pf-recPf-rec 9.

9. Find Find efficiency,efficiency,  = Pr = Pr x x 100100 Ps

Ps 10.

10. Find Find voltage voltage regulation regulation VR VR = = Vs Vs - - V V rr X 100

X 100

V r V r 11.

11. Display Display (i) (i) efficiency,efficiency, 

(ii) Voltage regulation, VR (ii) Voltage regulation, VR 12. Stop

PROGRAM: PROGRAM:

% ABCD of eqt. PI Network % ABCD of eqt. PI Network zkm

zkm = = 0.2+0.408i 0.2+0.408i ; ; ykm= ykm= 0+3.14e-6i;0+3.14e-6i; k1=input

k1=input ('\n ('\n Enter Enter 1-for 1-for short short line line 2-for 2-for medium medium line line 3-for 3-for long long line line ');'); switch k1

switch k1 case 1, case 1,

length

length = = input input ('\n ('\n Enter Enter length length of of short short line line ');'); Z=zkm*length; Y=ykm*length; Z=zkm*length; Y=ykm*length; A=1; B=Z; C=0; D=1; A=1; B=Z; C=0; D=1; case 2, case 2, length

length = = input input ('\n ('\n Enter Enter length length of of medium medium line line ');'); Z=zkm*length; Y=ykm*length; Z=zkm*length; Y=ykm*length; A=1+Y*Z/2; A=1+Y*Z/2; B=Z; B=Z; C=Y*(1+Y*Z/4); C=Y*(1+Y*Z/4); D=A; D=A; case 3, case 3, length

length = = input input ('\n ('\n Enter Enter length length of of long long line line ');'); zc=sqrt(zkm/ykm); zc=sqrt(zkm/ykm); gam=sqrt(zkm*ykm)*length; gam=sqrt(zkm*ykm)*length; A=cosh (gam); A=cosh (gam); D=A; D=A; B=zc*sinh(gam); B=zc*sinh(gam); C=1/zc*sinh(gam); C=1/zc*sinh(gam);

fprintf ('\n The equivalent PI circuit constants :'); fprintf ('\n The equivalent PI circuit constants :'); zeq = zkm*length*sinh (gam)/gam;

zeq = zkm*length*sinh (gam)/gam;

yeq = ykm*length/2*tanh (gam/2) / (gam/2); yeq = ykm*length/2*tanh (gam/2) / (gam/2); fprintf

fprintf ('\n ('\n Zeq = Zeq = %15.4f %15.4f %+15.4fi',real(zeq),imag(zeq));%+15.4fi',real(zeq),imag(zeq)); fprintf

fprintf ('\n ('\n Yeq/2= Yeq/2= %15.4f %15.4f %+15.4fi',real(yeq),imag(yeq));%+15.4fi',real(yeq),imag(yeq)); otherwise

otherwise

disp('wrong choice of tr.line'); disp('wrong choice of tr.line'); end

end

fprintf('\nA,B,C and D constants : \n'); fprintf('\nA,B,C and D constants : \n');

fprintf('---'); fprintf('---'); fprintf('\nA = %15.4f %+15.4fi',real(A),imag(A)); fprintf('\nA = %15.4f %+15.4fi',real(A),imag(A)); fprintf('\nB = %15.4f %+15.4fi',real(B),imag(B)); fprintf('\nB = %15.4f %+15.4fi',real(B),imag(B)); fprintf('\nC = %15.4f %+15.4fi',real(C),imag(C)); fprintf('\nC = %15.4f %+15.4fi',real(C),imag(C)); fprintf('\nD = %15.4f %+15.4fi',real(D),imag(D)); fprintf('\nD = %15.4f %+15.4fi',real(D),imag(D)); fprintf('\n The product AD-BC=%f',A*D-B*C);

fprintf('\n The product AD-BC=%f',A*D-B*C); k2=input('\n

k2=input('\n Enter Enter 1 1 - - To To read read Vr, Vr, Ir Ir and and compute compute Vs Vs , , Is Is \n \n 2 2 - -To

To read read Vs, Vs, Is Is and and compute compute Vr, Vr, Ir Ir ');'); switch k2,

switch k2, case 1, case 1,

vr=input('enter

vr=input('enter Vr/line Vr/line ');'); ir=input('enter

ir=input('enter Ir/phase Ir/phase ');'); vr=vr*1e3/sqrt(3); vr=vr*1e3/sqrt(3); vs = (A*vr+B*ir)/1e3; vs = (A*vr+B*ir)/1e3; is = C*vr+D*ir; is = C*vr+D*ir;

fprintf('\nSending end Voltage/ph=%f %+fi KV',real(vs),imag(vs)); fprintf('\nSending end Voltage/ph=%f %+fi KV',real(vs),imag(vs)); fprintf('\nSending end Current/ph=%f %+fi AMP',real(is),imag(is)); fprintf('\nSending end Current/ph=%f %+fi AMP',real(is),imag(is)); vs=vs*1e3;

vs=vs*1e3; case 2,

case 2,

vs=input('enter

vs=input('enter Vs/line Vs/line ');'); is=input('enter

is=input('enter Is/phase Is/phase ');'); vs=vs*1e3/sqrt(3.0); vs=vs*1e3/sqrt(3.0); vr=(D*vs-B*is)/1e3; vr=(D*vs-B*is)/1e3; ir= -C*vs+A*is; ir= -C*vs+A*is;

fprintf('\nReceiving end Voltage/ph=%f %+fi KV',real(vr),imag(vr)); fprintf('\nReceiving end Voltage/ph=%f %+fi KV',real(vr),imag(vr));

fprintf('\nReceiving end Current/ph=%f %+fi AMP',real(ir),imag(ir)); vr=vr*1e3; otherwise disp('wrong choice'); end rec_pow=3*real(vr*conj(ir))/1e6; send_pow=3*real(vs*conj(is))/1e6; eff=rec_pow/send_pow*100; reg=(abs(vs)/abs(A)-abs(vr))/abs(vr)*100;

fprintf('\n Receiving end power=%.2f KVA',rec_pow); fprintf('\n Sending end power=%.2f KVA',send_pow); fprintf('\n Efficiency=%.2f %%',eff);

fprintf('\n Voltage Regulation=%.2f%%',reg);

2. To find ABCD constants of equivalent T transmission line and hence efficiency and regulation

ALGORITHM:

1. Read length, impedance/km, admittance/km of transmission line 2. Enter choice 1-for short line

2-for medium line

3-for long line 3. If choice=1

Read length

Set B = Z Zeq Zeq

A = D = 1 C = 0

If choice=2 Yeq

Read length

Set A=D=1 + YZ/2

B = Z* (1 + YZ/4) C =YZ If choice=3 Read length zc = z/y  = (z*y)*length A = D = Cosh () B = zc Sinh () C = 1/zc Sinh () zeq=z*length/2*tanh(/2)/( /2); yeq=y*length*sinh()/; Display Zeq & Yeq

4. Display A, B, C, & D constants

5. Read choice 1 – To read V r, I r and compute V s , I s 2 – To read V s, I s and compute V r, I r

6. If choice = 1

Find V s = A V r + B I r

I s = C V r + D I r where V r & I r are in per p hase

Display V s & I s. if choice = 2

Find V r = D Vs – B Is

I r = -C Vs+ A Is where Vs & Is are in per phase Display V r & I r.

7. Find (i) VS SV & Is SI

(ii) Sending end Power factor Pf-send = Cos (SV -SI) (iii) Sending end Power, PS = 3 VS x  I s x Pf-send 8. Find (i) V r rv & I r rI

(iii) Receiving end power, Pr=3 V r x I r x Pf-rec 9. Find efficiency,  = Pr x 100

Ps

10. Find voltage regulation VR = Vs - V r X 100

V r 11. Display (i) efficiency, 

(ii) Voltage regulation, VR 12. STOP

PROGRAM:

%ABCD of eqt. T Network

zkm =0.2+0.408i; ykm=0+3.14e-6i;

k1=input('\n Enter 1-for short line 2-for medium line 3-for long line '); switch k1

case 1,

length=input('\n Enter length of short line '); Z=zkm*length; Y=ykm*length;

A=1;B=Z;C=0;D=1; case 2,

length=input('\n Enter length of medium line '); Z=zkm*length; Y=ykm*length; A=1+Y*Z/2; B=Z; C=Y*(1+Y*Z/4); D=A; case 3,

length=input('\n Enter length of long line '); zc=sqrt(zkm/ykm); gam=sqrt(zkm*ykm)*length; A=cosh(gam); D=A; B=zc*sinh(gam); C=1/zc*sinh(gam); otherwise

disp('wrong choice of tr.line'); end

fprintf('\nA,B,C and D constants : \n');

fprintf('---'); fprintf('\nA = %15.4f %+15.4fi',real(A),imag(A));

fprintf('\nB = %15.4f %+15.4fi',real(B),imag(B)); fprintf('\nC = %15.4f %+15.4fi',real(C),imag(C)); fprintf('\nD = %15.4f %+15.4fi',real(D),imag(D));

k2=input('\n Enter 1 - To read Vr, Ir and compute Vs , Is \n 2 -To read Vs, Is and compute Vr, Ir ');

switch k2, case 1, vr=input('enter Vr/phase '); ir=input('enter Ir/phase '); vr=vr*1e3/sqrt(3); vs=(A*vr+B*ir)/1e3; is=C*vr+D*ir;

fprintf('\nSending end Voltage/ph=%f %+fi KV',real(vs),imag(vs)); fprintf('\nSending end Current/ph=%f %+fi AMP',real(is),imag(is)); vs=vs*1e3;

case 2,

vs=input('enter Vs/phase '); is=input('enter Is/phase '); vs=vs*1e3/sqrt(3.0);

vr=(D*vs-B*is)/1e3; ir=-C*vs+D*is;

fprintf('\nReceiving end Voltage/ph=%f %+fi KV',real(vr),imag(vr)); fprintf('\nReceiving end Current/ph=%f %+fi AMP',real(ir),imag(ir)); vr=vr*1e3; end rec_pow=3*real(vr*conj(ir))/1e6; send_pow=3*real(vs*conj(is))/1e6; eff=rec_pow/send_pow*100; reg=(abs(vs)/abs(A)-abs(vr))/abs(vr)*100;

fprintf('\n Receiving end power=%.2f KVA',rec_pow); fprintf('\n Sending end power=%.2f KVA',send_pow); fprintf('\n Efficiency=%.2f %%',eff);

fprintf('\n Voltage Regulation=%.2f%%\n',reg);

3. Power Angle Characteristics of Cylindrical Rotor Generator ALGORITHM:

1. Read x, V and Ia 2. E = V + j x Ia

Pm =  E   V / Xd

3. Plot power angle characteristics from delta 0 to pi Indicate Pmax and corresponding delta

E - V 4. Regulation R = --- x 100 V Display R 4. Stop PROGRAM:

%power angle characteristics cylindrical rotor syn.m/c % enter the values of x,v,Ia;

x=0.3; v=1.0; ia=0.5000-0.375i; e=v+x*ia*i;

pm=abs(e)*abs(v)/x; del=0:0.1:pi;

plot(del,pm*sin(del),'r');

title('power angle characteristics cylindrical rotor syn.m/c'); xlabel('Delta'); ylabel('power ,p'); legend('pm*sin(delta)'); del=0:0.01:pi/2; line(del,pm); y=0:0.01:pm; line(pi/2,y); reg=(abs(e)-abs(v))/abs(v)*100;

fprintf('\n the excitation voltage=%f%%\n',e); fprintf('\n the peak power=%f%%\n',pm);

fprintf('\n the voltage regulation=%f%%\n',reg);

4. Power Angle characteristics of salient rotor generator ALGORITHM:

1. Read Xd, Xq, Xd1, V, P & Pf 2. Ө = Cos-1(pf)

 I  = P / (V x pf)

Ia =  I  cosӨ - j I  SinӨ

delta = tan((XE)  Ia  CosӨ ) / (IV) + Xq  Ia  SinӨ))  E  = V Cos(delta) + Xd /  Ia  Sin (delta + Ө) P = E V / Xd

Pr =  V 2 (Xd – Xq) / (2 x d x q) Pe = P Sm(delta) + Pr Sin ( 2 delta) 3. Plot P, Pr, Pe from delta = 0 to Pi

Find max Pe and the corresponding delta and indicate the same E - V

4. Regulation R = --- x 100 V

Display  E  , maxPe, delta ,Reluctance power (Pr) & R 5. Stop

PROGRAM:

%power angle characteristics salient pole rotor syn.m/c % enter the values of xd,xq,xdd,v,Ia;

xd=1.0 ; xq=0.6; v=1.0; ia=0.5000-0.375i; theta=abs(angle(ia)); del=atan((xq*abs(ia)*cos(theta))/(abs(v)+xq*abs(ia)*sin(theta))); e=abs(v)*cos(del)+xd*abs(ia)*sin(del+theta); p=e*abs(v)/xd; pr=abs(v)^2*(xd-xq)/(2*xd*xq); del=0:0.01:pi; line(del,0);hold on; pe=p*sin(del)+pr*sin(2*del); plot(del,p*sin(del), 'r',del,pr*sin(2*del),'m',del,p*sin(del)+pr*sin(2*del),'b'); title('power angle characteristics salient pole syn.m/c'); xlabel('Delta');

ylabel('power ,p');

legend('p*sin(delta)','pr*sin(2delta)','p*sin(delta)+pr*sin(2delta)'); [pmax,k]=max(pe);

fprintf('\n maximum power=%f and the corresponding angle=%f',pmax,del(k)*180/pi) del=0:0.01:del(k); line(del,pmax); y=0:0.01:pmax; line(del(k),y); reg=(abs(e)-abs(v))/abs(v)*100; text(-0.4,pmax,'pmax'); text(del(k),-0.1,'deltamax');

fprintf('\n the excitation voltage =%f pu',abs(e)); fprintf('\n the voltage regulation=%f%%',reg);

fprintf('\n Reluctance power= %f p.u\n',pr); 5. Swing curve under sustained fault

ALGORITHM:

1. Read Input power (P s), voltages (e & v), M, External reactance (xe), line reactance (xl)

2. Pre fault reactance,x1=xe+xL/2

3. Read choice 1-for fault at the beginning 2-for fault at the middle

if choice=1 x2=infinity

if choice=2 x2= (xe*xL+xe*xL/2+xl*xL/2)/(xL/2); 4. Set dt=0.001

r r=180/ (required to convert deg into radius etc) f=(dt)2 / M

Iteration Count, it=1 Time, t(1)=0 =0 5. Find Pm1 = e1*e2/X1 0 = sin-1(Ps/Pm1) Pm2 = Pm=e1*e2/X2 Pe = Pm Sin(0)

Pa = (Ps-Pe)/2

6. Repeat step no. i to step no. vi from time=0sec to 1 sec in steps of 0.0 5 sec

I. Find ft = f*Pa/rr II.  = +ft

III. Display time (t), Pmax (Pm),  in degrees. IV. Increment iteration count it=it + 1

V. Find t(it)=t(it-1) + dt VI. (it) =  (it-1)+  VII. Set Pm=Pm2

VIII. Find Pe=Pm Sin IX. Pa=Ps-Pe

7. Plot swing curve

8. Find critical clearing angle

c=cos-1(ps(m-o)+pm3*cos(m)-pm2*cos(o)) / (pm3-pm2) Where pm3=e*v/x3, x3=xe+x1, m=pi-o.

9. Find the critical clearing time(Tc), corresponding to c and dislay Tc 10. Stop

PROGRAM:

%swing curve for sustained fault and critical clearing angle & time. %ps=mech.power input xe=xg+xt, x1=reactance before fault x2=reactance after %fault

clc;

clear all;

% ps = 0.9; e1 = 1.1; v = 1.0; m = 0.00028; x1 = 0.45; x2 = 1.25; x3 = 0.55;

data = input( ‘ps = , e1= , v= , m= , x1= , x2= , x3= in matrix form’);

ps = data(1); e1 = data(2); v = data(3);

m = data(4); x1 = data(5); x2 = data(6); x3= data(7); rr = 180/pi; dt = 0.05; rr=180/pi; f=dt^2/m; it=1; t(it)=0; deld=0; pm1=e1*v/x1; del(it)=asin(ps/pm1); pm2=e1*v/x2; pm=pm2; pe=pm*sin(del(it)); pa=(ps-pe)/2;

fprintf( \n SUSTAINED FAULT );

fprintf( \n--- ); fprintf( \n TIME PMAX DELTA ); fprintf( \n--- ); t1=0; while(t(it)<=1.0) ft=f*pa/rr; deld=deld+ft; if(t1-t(it)<0.0001) fprintf( \n %5.3f %5.2f %5.2f ,t(it),pm,del(it)* rr); t1=t1+0.05; end it=it+1; t(it)=t(it-1)+dt; del(it)=del(it-1)+deld; pm=pm2; pe=pm*sin(del(it));                    

pa=ps-pe; end

plot(t,del, r );

title( swing curve ); xlabel( Delta ); ylabel( power ,p ); pm3=e1*v/x3; delm=pi-del(1); cdc=(ps*(delm-del(1))+pm3*cos(delm)-pm2*cos(del(1)))/(pm3-pm2); delc=acos(cdc);

fprintf( \n critical clearing angle-%f ,delc*rr); it=1; while(t(it)<1.0) if (del(it)>=delc) break; end it=it+1; end

fprintf( \n critical clearing time-%f ,t(it));

6. Swing curve with fault being cleared ALGORITHM:

1. Read Input power (P s), voltages (e & v), M, External reactance(xe),line reactance (xl)

pre fault reactance,x1=xe+xL/2

2. Read choice 1-for fault at the beginning 2-for fault at the middle

if choice=1 x2=infinity

if choice=2 x2= (xe*xL+xe*xL/2+xL*xL/2)/(xL/2); 3. Set x3=xe+xL

dt=0.05

4. Read clearing time ct in sec’s K=ct/dt (integer division) r=ct-k*dt

If r=0 then

Print ‘Fault is cleared at the beginning of an interval’ else

Print ‘Fault is cleared in the middle of an interval’ 5. Set rr=180/

f=dt2/M

Iteration Count, it=0 Time, t=0 6. Find Pm1=e1*e2/x1 0=Sin-1 Ps/Pm1 Pm2=e1*e2/x2 Pm3=e1*e2/x3 Pm=Pm2 Pe=Pm * Sin (0) Pa=Ps-Pe

7. Repeat step no. i to step no.vii from time t=0 to 1 sec in steps of 0.05 sec

i. Find ft=f * Pa/rr  = +ft

ii. Display time (t), Pmax (Pm),  in degrees. iii. Increment iteration count it=it+1

iv. Find t(it)=t(it-1) + dt i. (it)=  (it-1)+  v. If r=0 then                        

i. If(time < clearing time) then 1. Pm=Pm2

ii. else if(time=clearing time) then 1. Pm=(Pm2+Pm3)/2

b. If(r! =0) then

i. If (time < clearing time) then 1. Pm=Pm2

ii. else 1. Pm=Pm3

vi. Find Pe=Pm Sin vii. Pa=Ps-Pe

8. Plot swing curve

9. If  decreases at any point then the system is stable otherwise unstable. Find whether the system is stable or not and display the

same.

10. Stop PROGRAM:

clc; clear all;

%program for swing curve when the fault is cleared % ps=0.9; e=1.1; v=1.0; m=0.00028;

% x1=0.45; x2=1.25; x3=0.55;

data = input( ‘ps = , e1= , v= , m= , x1= , x2= , x3= in matrix form’);

ps = data(1); e1 = data(2); v = data(3);

m = data(4); x1 = data(5); x2 = data(6); x3= data(7); dt=0.05;

ct=input( \n enter clearing time in secs ); k=ct/dt;

r=ct-floor(k)*dt; if(r==0)

fprintf( FAULT IS CLEARED AT THE BEGINNING OF AN INTERVEL ); else

fprintf( FAULT IS CLEARED AT THE MIDDLE OF AN INTERVEL ); end ct = ct/50; rr=180/pi; f=dt*dt/m; deld=0; it=1; t(it)=0; pm1=e*v/x1; del(it)=asin(ps/pm1); pm2=e*v/x2; pm3=e*v/x3; pm=pm2; pe=pm*sin(del(it)); pa=(ps-pe)/2;

fprintf( \n TIME PMAX DELTA ); while(t(it)<=1.05) ft=f*pa/rr; deld=deld+ft; fprintf( \n %5.2f %5.2f %5.2f ,t(it),pm,del(it)* rr); it=it+1; t(it)=t(it-1)+dt; del(it)=del(it-1)+deld; if(r==0) if(t(it)<ct) pm=pm2; elseif(t(it)==ct) pm=(pm2+pm3)/2;                    

else pm=pm3; end end if(r~=0) if(t(it)<ct) pm=pm2 ; else pm=pm3; end end pe=pm*sin(del(it)); pa=ps-pe; end plot(t,del, r );

title( swing curve ); xlabel( Delta ); ylabel( power ,p ); pr=0; for k=2:it if (del(k)<del(k-1)) pr=1; break; end end if(pr)

text(0.5,1.0, system is stable ); else

text(0.5,1.0, system is unstable ); end

7. Formation of Y bus by direct method ALGORITHM:

1. Read the no. of buses(nbus) and no. of lines (nline)

2. Read the following line data at each line from bus no, to bus no, line impedance

3. Convert line impedance to admittance(y) of each line 4. Repeat step i to iii for K=1 to nline

i. P = from bus no. q=to bus no. Set y = y/a

ii. Diagonal elements Ypp = Ypp + y Yqq = Yqq + y Ypq = Ypq – y Yqp = Ypq 5. Display Y bus 6. Stop PROGRAM

% Program to obtain Ybus formulation % by inspection method z=[ 0 1 0 0.08 0 2 0 0.6 1 2 0.02 0.4 1 3 0.01 0.5 2 3 0.02 0.3 3 4 0.01 0.5 ]; fbus = z(:,1); tbus = z(:,2);                        

R = z(:,3); X = z(:,4); nbr = length(fbus); nbus = max(max(fbus),max(tbus)); Z = R+j*X; y = ones(nbr,1)./Z; Ybus = zeros(nbus,nbus); for k = 1: nbr

if fbus(k)>0 & tbus(k)>0

Ybus(fbus(k),tbus(k)) = -y(k); Ybus(tbus(k),fbus(k)) = Ybus(fbus(k),tbus(k)); else, end end Ybus; for n = 1:nbus for k = 1:nbr if fbus(k)==n  tbus(k)==n; Ybus(n,n) = Ybus(n,n)+y(k); else, end end end Ybus

8. Formation of Ybus using singular transformation method with half line charging (or shunt admittance)

ALGORITHM:

1. Read no. of buses( nbus) and no. of lines( nline). 2. Initialize shunt admittance matrix to zeros.

3. Repeat step no i. to step no iv for I=1 to nline.

i. Read line data i.e., frombusno, tobus no, Zseries, half line charging. ii. Shunt admittance [from bus]=shunt admittance [from bus]+half line chargi ng.

iii. Shunt admittance [to bus]=shunt admittance [to bus]+half line charging. iv. Series admittance [I]=1/series impedance [I].

4. Form the incidence matrix A

(i) Initialize all the elements of matrix A of size [(nbus + nline) x nbus] to zeros

(ii) Form A = Where AI is identity matrix of size (nbus x nbus) AA is a matrix of size (nline x nbus) and elements of AA = [aij]

Where aij = 0 if ith branch is not incident on jth bus.

aij = 1 if ith branch is incident to and oriented away from jth bus. aij = -1 if ith branch is incident to and oriented away

from jth bus.

5. Form Yprimitive matrix

Initialize all elements of matrix Ypr of size (nbus+nline x nbus+nline) to zeros Form Ypr =

Where Y is a matrix of size (nbus x nbus) with diagonal elements Yii = total shunt admittance at bus i and all other elements being

zeros.YY is a matrix of size (nbus x nline) with all elements being zeros.Y1 is matrix of size (nline x nline) with diagonal elements of Y1ii =Yseries of line i=1/Zseries of line i.

6. Display A matrix & Ypr matrix.

7. Form Transpose of matrix A i.e., AT. 8. Form ATY = AT x Ypr.

9. Form Ybus = ATY x A and display. 10. Form Zbus = [Ybus]-1 and display. 11. Stop

PROGRAM:

%FORMATION OF ybus USING SINGULAR TRANSFORMATION METHOD %WITHOUT MUTUAL COUPLING:

% p q Z hlcY(ADM) z=[ 1 2 0.02+0.06i 0.03i 1 3 0.08+0.24i 0.025i 2 3 0.06+0.18i 0.02i 2 4 0.06+0.18i 0.02i 2 5 0.04+0.12i 0.015i 3 4 0.01+0.03i 0.01i 4 5 0.08+0.24i 0.025i ]; fb=z(:,1); tb=z(:,2); Z=z(:,3); hlcy=z(:,4); y=1./Z; nbus=max(max(fb),max(tb)); Y=zeros(nbus); nline=length(fb); nlb=nline+nbus; A=zeros(nlb,nbus); for k=1:nbus A(k,k)=1; end for k=1:nline A(nbus+k,fb(k))=1; A(nbus+k,tb(k))=-1; end sh=zeros(nbus); for k=1:nline sh(fb(k))=sh(fb(k))+hlcy(k); sh(tb(k))=sh(tb(k))+hlcy(k); end ypr=zeros(nlb,nlb); for k=1:nbus ypr(k,k)=sh(k); end for k=1:nline ypr(nbus+k,nbus+k)=y(k); end format short; Ybus=A'*ypr*A

ALGORITHM:

1. Read no. of buses( nbus) and no. of lines (nline).

2. Road line data i.e., Frombus no, tobus no, Zseries, mutual line no, Zmutual at all lines.

3. Form bus incidence matrix A

(i) Initialize all the elements of A of size (nline x nline) to zeros (ii) Form A = [aij]

Where aij = 0 if ith branch is not incident on jth bus.

aij = 1 if ith branch is incident to and oriented away from jth bus. aij = -1 if ith branch is incident to and oriented away from jth bus. 4. Form Zprimitive matrix, Zpr

(iii) Initialize all the elements of Zpr of size (nline x nline) to zeros (iv) The diagonal elements of Zpr are assigned with respective Zseries i.e., Zpr (i, i) = Zseries of ith line

(v) The off diagonal element is zero if the line i has no mutual coupling with line j

(vi) The off diagonal element Zpr (i, j) = Zmutual if the line i has mutual coupling with line j

5. Form Y primitive matrix Ypr = [Zpr]-1 6. Display A matrix & Ypr matrix.

7. Form Transpose of matrix A i.e., AT. 8. Form ATY = AT x Ypr.

9. Form Ybus = ATY x A and display. 10. Form Zbus = [Ybus]-1 and display. 11. Stop

PROGRAM:

%FORMATION OF ybus USING SINGULAR TRANSFORMATON METHOD WITH MUTUAL COUPLING & WITHOUT LINE CHARGING:

% p q Z mno mutual(imp) z=[ 0 1 0.6i 0 0 0 2 0.5i 1 0.1i 2 3 0.5i 0 0 0 1 0.4i 1 0.2i 1 3 0.2i 0 0 ]; p=z(:,1); q=z(:,2); Z=z(:,3); mno=z(:,4); zmc=z(:,5); nbus=max(max(p),max(q)); Y=zeros(nbus); nline=length(p); A=zeros(nline,nbus); for k=1:nline, if(q(k) = = 0) A(k,p(k))=1; elseif(p(k) = = 0) A(k,q(k)) = -1; end if(p(k)~=0 & q(k)~=0) A(k,p(k))=1; A(k,q(k))=-1; end end zpr=zeros(nline,nline); for k=1:nline zpr(k,k)=Z(k); if(mno(k))

zpr(k,mno(k))=zmc(k); zpr(mno(k),k)=zmc(k); end end ypr=inv(zpr); format short; Ybus=A'*ypr*A

10. Bus cuurents, Bus power and Line Flows ALGORITHM:

1. Read the following line data at each line from bus no(p), to bus no(q), line impedance(Zpq) and half line charging admittance(Ychpq) Read voltages at all buses

2. Convert line impedance to admittance of each line Ypq 3. Repeat step i to ix for K=1 to no. of lines

i. P = from bus no q = tobusno ii, Line current, Ipq = (Vp-Vq)Ypq + Vq x Ychpq iii. Line loss = SLpq =VpIpq*

iv. Generation at bus p = SGp = SGp + S2P2 v. Iqp = (Vq-Vp)Ypq + Vq x Ychqp

vi. Iq = Iq + Iqp vii. SLqp + Vq Iqp*

viii. Loss (k) = Loss + SLpq + SLqp ix. Total loss = Total loss + loss(k) 5. Display current at each bus

Display line flows and line loss at each line Display generations at each bus

Display the total losses 6. Stop PROGRAM: % p q z hlc(ADM) z= [ 5 4 0.02+0.06i 0.03i 5 1 0.08+0.24i 0.025i 4 1 0.06+0.18i 0.02i 4 2 0.06+0.18i 0.02i 4 3 0.04+0.12i 0.015i 1 2 0.01+0.03i 0.01i 2 3 0.08+0.24i 0.025i ];

v(1)=1+0.2i; v(2)=1.01-0.5i; v(3)=1.05-0.3i; v(4)=1.01-0.5i; v(5)=1.0-0i; fb=z(:,1); tb=z(:,2); Z=z(:,3); ysh=z(:,4); y=1./Z; nbus=max(max(fb),max(tb)); nline=length(fb); II=zeros(nbus,1);SG=zeros(nbus,1); totloss=0; for k=1:nline p=fb(k);q=tb(k);

I(p,q)=(v(p)-v(q))*y(k)+v(p)*ysh(k); II(p)=II(p)+I(p,q); SL(p,q)=v(p)*conj(I(p,q)); SG(p)=SG(p)+SL(p,q); I(q,p)=(v(q)-v(p))*y(k)+v(q)*ysh(k); II(q)=II(q)+I(q,p); SL(q,p)=v(q)*conj(I(q,p)); SG(q)=SG(q)+SL(q,p); loss(k)=SL(p,q)+SL(q,p); totloss=totloss+loss(k); end f=fopen('lf.out','w'); fprintf(f,'bus currents'); fprintf(f,'\nBus no Current'); for k=1:nbus fprintf(f,'\n %d %10.4f %+10.4fi',k,real(II(k)),imag(II(k))); end fprintf(f,'\nLine flows');

fprintf(f,'\nFrom bus To bus Lineflows Line Loss'); for k=1:nline p=fb(k);q=tb(k); fprintf(f,'\n %d %d %10.4f%+10.4fi %10.4f%+10.4fi',p,q,real(SL(p,q)),imag(SL(p,q)),real(loss(k)),imag(loss(k))); fprintf(f,'\n %d %d %10.4f%+10.4fi %10.4f%+10.4fi',q,p,real(SL(q,p)),imag(SL(q,p)),real(loss(k)),imag(loss(k))); end fprintf(f,'\nBus Generations'); fprintf(f,'\nBus no Generation'); for k=1:nbus fprintf(f,'\n%d %10.4f%+10.4fi',k,real(SG(k)),imag(SG(k))); end fprintf(f,'\n\nTotal Losses=%10.4f%+10.4fi',real(totloss),imag(totloss)); fclose(f);

11. Load flow analysis using Gauss-Seidal method. ALGORITHM:

1. Read Ybus, no.of buses (nbus) (Assume bus1 as slack bus)

No.of pq buses (Npq), No.of pv buses (Npv) Voltage at slack bus

Real power (P) and reactive power (Q) at all pq buses

Real power (P) and specified voltage (Vsp) and Qlimits (Qmin and Qmax) at PV buses.

Convergance critirian ( ). 2. Set iteration count k=1.

3. Calculate voltages at all the pq buses( p=2 to Npq+1), using

4. To calculate voltages at all the pv buses( p=Npq+2 to nbus) Find QCp= -Im( ) whare

If QCp is within Qmin and Qmax, treat the bus p as pv bus and calculate where at an angle p corresponding to previous iteration

Else if QCp violates Qlimits, treat bus p as pq bus and if QCP< Qmin , set Qp=Qmin else set Qp=Qmax and find

5. Display voltages at all buses and QCp at pv buses 6. Determine the largest absolute value of .

7. If goto next step otherwise set k=k+1and goto step no.3. 8. Stop

PROGRAM:

% enter Y-bus of the system

Y=[ 20-50i -10+20i -10+30i -10+20i 26-52i -16+32i -10+30i -16+32i 26-62i ];

%enter no.of buses,slack bus no,no.of pq buses,no of pv buses nbus=3;

sbno=1; npq=1; npv=1;

% assume bus no 1 as slack bus ,next npq buses as pq buses and the remaining as pv buses

%enter slack bus voltage v(1)=1.05+0i;

%assume voltages at all other buses as 1+j0 v(2)=1; v(3)=1;

%enter p & q at all pq buses P(2)= -4.0; Q(2)= -2.5;

%enter p ,vsp and q limits at pv buses P(3)=2.0;

vsp(3)=1.04; Qmin(3)=0.1; Qmax(3)=1.0; %enter accuracy of convergence

acc=0.001; for it=1:5

%to find voltages at pq buses for p=2:npq+1, v1(p)=v(p); pq=(P(p)-Q(p)*i)/conj(v(p)); ypq=0; for q=1:nbus if(p= =q) continue; end ypq=ypq+Y(p,q)*v(q); end v(p)=(pq-ypq)/Y(p,p); end

%to find voltages at pv buses for p=npq+2:nbus, v1(p)=v(p); s=0; for q=1:nbus, if(p~=q) s=s+Y(p,q)*v(q); else vp=v(q); ang=angle(v(q)); v(q)=complex(vsp(q)*cos(ang),vsp(q)*sin(ang)); s=s+Y(p,q)*v(q); end end Qc(p)=-1*imag(conj(v(p))*s); if(Qc(p)>=Qmin(p)& Qc(p)<=Qmax(p)) pq=(P(p)-Qc(p)*i)/conj(v(p)); ypq=0;

for q=1:nbus if(p= =q) continue; end ypq=ypq+Y(p,q)*v(q); end v(p)=(pq-ypq)/Y(p,p); ang=angle(v(p)); v(p)=complex(vsp(p)*cos(ang),vsp(p)*sin(ang)); else if(Qc(p)<Qmin(p)) Q(p)=Qmin(p); else Q(p)=Qmax(p); end pq=(P(p)-Q(p)*i)/conj(vp); ypq=0; for q=1:nbus if(p= =q) continue; end ypq=ypq+Y(p,q)*v(q); end v(p)=(pq-ypq)/Y(p,p); end end

%to find the votages at all buses and Q at pv busses

fprintf('\nThe votages at all buses and Q at pv busses after iteration no %d',it); for p=1:npq+1 fprintf('\nV(%d)=%.4f at%.2fdeg',p,abs(v(p)),angle(v(p))*180/pi); end for p=npq+2:nbus fprintf('\nV(%d)=%.4f at%.2fdeg Q(%d)=%+.3f\n',p,abs(v(p)),angle(v(p))*180/pi,p,Qc(p)); end

%to check for convergence for p=2:nbus diff(p)=abs(v(p)-v1(p)); end err=max(diff); if(err<=acc) break; end end

12. Formation of Jacobin Matrix in polar form without PV Bus ALGORITHM:

1. Read (i) Y bus

(ii) No. of buses,n

(iii) Voltages at all buses 2. . Repeat for p =1 to (n-1)

Өpq = Өp - Өq where Өp = ∟Vp & Өq = ∟Vq Gpq + real (Ypq), Bpq = image (Ypa)

Qcal(p) = Vp (GpqSm Өpq - BpqCos Өpq)Vq

3. Off diagonal of J1 = VpVq(Gpq SinӨpq-BpqCosӨpq) = off diagonal elements of J4 Off diagonal of J2 = VpVq(Gpq CosӨpq + Bpq Sin Өpq)

= - Off diagonal elements of J3 Diagonal of J1 = -Qcal (p) – Bpp Vp2

J2 = Pcal (p) + Gpp  Vp2 J3 = Pcal (p) – Gpp  Vp  2 J4 = Qcal (p) – Bpp  Vp2 4. Combine J1, J2, J3 & J4 into a single matrix J 5. Display J

6. Stop

PROGRAM:

% enter Y-bus of the system

Y=[ 20-50i -10+20i -10+30i -10+20i 26-52i -16+32i

-10+30i -16+32i 26-62i ]; %enter no.of buses

nbus=3;

% assume last bus as slack bus ,and the remaining as pq buses %assume voltages at all other buses as 1+j0

v(1)=1+0i; v(2)=1+0i; %enter slack bus voltage v(3)=1.05+0i;

%to find pcal and qcal for p=1:nbus-1 vp=abs(v(p)); pc(p)=0;qc(p)=0; for q=1:nbus vq=abs(v(q)); thpq=angle(v(p))-angle(v(q)); Gpq=real(Y(p,q));Bpq=imag(Y(p,q)); pc(p)=pc(p)+(Gpq*cos(thpq)+Bpq*sin(thpq))*vq; qc(p)=qc(p)+(Gpq*sin(thpq)-Bpq*cos(thpq))*vq; end pcal(p)=pc(p)*vp;qcal(p)=qc(p)*vp; end %formation of J1 J2 J3 & J4 for p=1:nbus-1 vp=abs(v(p));Gpp=real(Y(p,p));Bpp=imag(Y(p,p)); for q=1:nbus-1 vq=abs(v(q)); thpq=angle(v(p))-angle(v(q)); Gpq=real(Y(p,q));Bpq=imag(Y(p,q)); if(p~=q) J1(p,q)=vp*vq*(Gpq*sin(thpq)-Bpq*cos(thpq)); J2(p,q)=vp*vq*(Gpq*cos(thpq)+Bpq*sin(thpq)); J3(p,q)=-J2(p,q); J4(p,q)=J1(p,q); end end J1(p,p)=-qcal(p)-Bpp*vp^2; J2(p,p)=pcal(p)+Gpp*vp^2; J3(p,p)=pcal(p)-Gpp*vp^2; J4(p,p)=qcal(p)-Bpp*vp^2; end

%to combine J1,J2,J3&J4 into single matrix J J = [J1 J2; J3 J4];

fprintf('The Jacobian matrix is \n'); disp(J);

13. Z bus by Z- BUS Building Algorithm ALGORITHM:

Type 1 : Line added between New bus to Reference bus Type 2 : Line added between New bus to Old bus

Type 3 : Line added between Old bus to reference bus Type 4 : Line added between two Old buses

Note :

[ Treat BUS – 1 as Reference bus Form the input data from single line diagram. Start with type 1 only.

Select starting bus (SB) as either Reference bus or existing bus ( for type 1 or 2 respectively ) and Select ending bus (EB) as new bus .]

Initialize zbus with the size 1 X 1 size =nsize=rows=column=1;

Read line data , depending on the type of the line specified, form the Z-bus as follows

Type - 1 : Increase the size of zbus by 1

zbus( size,size) = zline; % diagonal element all the other off diagonal elements (added row & column) being zero

Type – 2: Increase the size of zbus by 1

zbus( size,size) = zbus ( sb , sb ) + zline; --- diagonal element for off diagonal elements

Copy contents of the row corresponding to sb to new row and to new column

Type - 3 : size remains the same

zsum = zbus( sb , sb ) + zline(i);

repeat the steps i to ii for k= 2 to nsize step i : repeat the step ii for j = 2 to nsize

step ii : zmod(k,j) = -1 * zbus(k,m) * zbus( m, j) / zsum; then the modified z bus is

zbus=zbus+zmod; Type - 4 : size remains the same

Calculate

zsum = + zline (i) + zbus ( sb , sb )+zbus ( eb , eb )-2 * zbus( sb , eb )

repeat the steps i to ii for k= 2 to nsize step i : repeat the step ii for j = 2 to nsize

step ii :

zmod (k , j )=-1* ( zbus( k , sb ) - zbus(k , eb ) ) * ( zbus (sb , j)-zbus( eb , j )) / zsum

the modified z bus is zbus = zbus + zmod PROGRAM:

% Z-BUS --- using Z-Bus Building Algorithm

% data for line = line no , type, SB, EB, line_impedance zdata=[ 1 1 1 2 1.0

3 2 2 3 1.0 4 3 3 1 1.0 5 3 4 1 1.0]; line=zdata(:,1); type=zdata(:,2); sb=zdata(:,3); eb=zdata(:,4); zline=zdata(:,5); nb=max(max(sb),max(eb)); nl= length(line); %zbus=zeros(nb); % formation of Z-BUS nsize=1; for i= 1:nl m=sb(i); p=eb(i); % type 1 if (type(i)==1) nsize=nsize+1; zbus(nsize,nsize)=zline(i); end % type 2 if (type(i)==2) nsize=nsize+1; for j=2:(nsize-1) zbus(nsize,j)=zbus(m,j); zbus(j,nsize)=zbus(j,m); end zbus(nsize,nsize)= zbus(m,m)+zline(i); end % type 3 if (type(i)==3) zsum=zbus(m,m)+zline(i); for k=2:nsize for j=2:nsize zmod(k,j)=-1*zbus(k,m)*zbus(m,j)/zsum; end end zbus=zbus+zmod; end % type 4 if (type(i)==4) zsum=zbus(m,m)+zbus(p,p)-2*zbus(m,p)+zline(i); for k=2:nsize for j=2:nsize zmod(k,j)=-1*(zbus(k,m)-zbus(k,p))*(zbus(m,j)-zbus(p,j))/zsum; end end zbus=zbus+zmod; end display(zbus); end display(zbus);