PSS Lab Manual_ssit

POWER SYSTEM SIMULATION LAB

Subject Code : 06EEL78 IA Marks : 25

No. of Practical Hrs/Week : 03 Exam Hrs. : 03Total No. of Practical Hrs : 42 Exam Marks : 50

Power system simulation using MA TLAB, Software Packages and C++

1. a) Y bus formation for power systems, with & without mutual coupling, by singular transformation & inspection method.

b) Determination of Bus currents, bus power & line flows for a specified system voltage (Bus) profile.

2. Formation of Z-bus, using Z-bus build Algorithm without mutual coupling.

3. ABCD parameters: i) Formation for symmetric /T configuration. ii) Verification of AD-BC = 1. iii) Determination of co-efficient & regulation.

4. Determination of power angle diagrams for salient and non-salient pole synchronous m/cs, reluctance power, excitation emf & regulation.

5. To determine i) swing curve ii) critical clearing time for a single m/c connected to infinite bus through a pair of identical transmission lines, for a 3 fault on one of the lines for variation of inertia constant/line parameters/fault location/clearing time/prefault electrical output.

6. Formation of Jacobian for a system not exceeding 4 buses * (no PV buses) in polar co-ordinates.

7. Write a program to perform load flow analysis using Gauss-Siedel method(only PQ bus).

8. To determine fault currents and voltages in a single transmission line systems with star- delta transformers at a specified location for SLGF, DLGF.

9. Load flow analysis using Gauss Siedel method, NR method, Fast Decoupled flow method for both PQ and PV buses.

10. Optimal generator scheduling for thermal power plants.

Note: 1, 2, 3, 4, 5, 6, 7, 8 – Simulation experiments using MATLAB.

4, 6, 9, 10 – Using Mi-Power package.

VII Semester Power System Simulation Lab

1. a) YBUS FORMATION FOR A GIVEN POWER SYSTEM

i) Formation of YBus without mutual coupling by InspectionMethod

Single Line Diagram of Transmission Network:

Half Line Charging Admittance Data: Shunt Admittance Element Data:

Program for Ybus formation without mutual coupling by Inspection method:

clc;clear; linedata = [1 2 0.2j 0.03i; 1 4 0.25j 0.025i; 2 3 0.4j 0.02i; 2 4 0.5j 0.015i; 3 4 0.2j 0.01i];Sh=[4 0.02j];fb=linedata(:,1); %From bustb=linedata(:,2); %To bus z=linedata(:,3); %Series line impedanceBc=linedata(:,4); %Half line charging admittance nbus=max(max(fb),max(tb));nbr=length(z);y=ones(nbr,1)./z; %branch admittanceYbus=zeros(nbus,nbus); % initialize Ybus to zero% formation of the off diagonal elementsfor k=1:nbr; Ybus(fb(k),tb(k))=Ybus(fb(k),tb(k))-y(k); Ybus(tb(k),fb(k))=Ybus(fb(k),tb(k));

Department of E & EE, SSIT, Tumkur

Bus Shunt Admittance4 j0.02

Bus Code HLC Admittance1-2 j 0.031-4 j0.0252-3 j0.022-4 j0.0153-4 J0.01

1

(2)

(3)

(1)

(4)

j 0.2

j 0.2

j 0.5j 0.25 j 0.4


end% formation of the diagonal elementsfor n=1:nbus for k=1:nbr if fb(k)==n Ybus(n,n) = Ybus(n,n)+y(k) + Bc(k); elseif tb(k)==n Ybus(n,n) = Ybus(n,n)+y(k) + Bc(k); end endend%Addition of shunt admittance data[r,c]=size(Sh);for x=1:r for y=1:nbr p=fb(y); q=tb(y); if Sh(x,1)==p Ybus(p,p)=Ybus(p,p)+Sh(x,2); elseif Sh(x,1)==q Ybus(q,q)=Ybus(q,q)+Sh(x,2); end endend Ybus

OUTPUT:

Ybus = 0 - 8.9450i 0 + 5.0000i 0 0 + 4.0000i 0 + 5.0000i 0 - 9.4350i 0 + 2.5000i 0 + 2.0000i 0 0 + 2.5000i 0 - 7.4700i 0 + 5.0000i 0 + 4.0000i 0 + 2.0000i 0 + 5.0000i 0 -10.8900i

Signature of the Staff Incharge

Department of E & EE, SSIT, Tumkur 2


ii) Formation of Ybus without mutual coupling using Singular Transformation

Single Line Diagram of Transmission Network:

Impedances and line charging admittances for the power system:

Bus Code Impedance HLC Admittance1-2 0.02 + j 0.06 0.0 + j 0.0301-3 0.08 + j 0.24 0.0 + j 0.0252-3 0.06 + j 0.18 0.0 + j 0.0202-4 0.06 + j 0.18 0.0 + j 0.0202-5 0.04 + j 0.12 0.0 + j 0.0153-4 0.01 + j 0.03 0.0 + j 0.0104-5 0.08 + j 0.24 0.0 + j 0.025

Program for Ybus formation without mutual coupling using Singular Transformation:

clc;clear;Dm=input('Enter the Line data : ');fb=Dm(:,1);tb=Dm(:,2);z=Dm(:,3);hlc=Dm(:,4);nb=max(max(fb),max(tb));e=length(z);%Formation of Bus Incidence matrixa=zeros(e,nb);for i=1:1:e p=fb(i); a(i,p)=1; q=tb(i); a(i,q)=-1;end a

%Formation of the Primitive Admittance matrix


(5)(2)

(3)(1) (4)


Zpri=zeros(e,e);for j=1:1:e Zpri(j,j)=z(j);endYpri=inv(Zpri); %Ybus formation Ybus=a'*Ypri*a; %Addition of Half Line Chaging Admittancefor r=1:1:nb-1 for s=1:1:e if fb(s)==r|tb(s)==r Ybus(r,r)=Ybus(r,r)+hlc(s); end endendYbus

INPUT Data:

Enter the Line data : [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]

OUTPUT:a =

1 -1 0 0 0 1 0 -1 0 0 0 1 -1 0 0 0 1 0 -1 0 0 1 0 0 -1 0 0 1 -1 0 0 0 0 1 -1



Ybus =

6.2500 -18.6950i -5.0000 +15.0000i -1.2500 + 3.7500i 0 0

-5.0000 +15.0000i 10.8333 -32.4150i -1.6667 + 5.0000i -1.6667 + 5.0000i -2.5000 + 7.5000i

-1.2500 + 3.7500i -1.6667 + 5.0000i 12.9167 -38.6950i -10.0000 +30.0000i 0

0 -1.6667 + 5.0000i -10.0000 +30.0000i 12.9167 -38.6950i -1.2500 + 3.7500i

0 -2.5000 + 7.5000i 0 -1.2500 + 3.7500i 3.7500 -11.2500i


iii) Formation of Ybus with mutual coupling using Singular Transformation

Program for Ybus formation with mutual coupling using Singular Transformation:

clc;clear;Dm=input('Enter the Line data : ');Mc=input('Enter the Mutual Coupling data : ');fb=Dm(:,1);tb=Dm(:,2);z=Dm(:,3);hlc=Dm(:,4);nb=max(max(fb),max(tb));e=length(z);%Formation of Bus Incidence Matrixa=zeros(e,nb);for i=1:1:e p=fb(i); a(i,p)=1; q=tb(i); a(i,q)=-1;enda%Formation of the Primitive Admittance matrixZpri=zeros(e,e);for j=1:1:e Zpri(j,j)=z(j);end%Addition of Mutual Coupling elements[x,y]=size(Mc);for i=1:1:x m=Mc(i,1); n=Mc(i,2); Zpri(m,n)=Mc(i,3); Zpri(n,m)=Zpri(m,n);endYpri=inv(Zpri);%Ybus formation Ybus=a'*Ypri*a;%Addition of Half Line Charging Admittancefor r=1:1:nb-1 for s=1:1:e if fb(s)==r|tb(s)==r Ybus(r,r)=Ybus(r,r)+hlc(s); end endendYbus

INPUT Data:



Enter the Line data : [1 2 0.6j 0;1 3 0.5j 0;3 4 0.5j 0;1 2 0.4j 0;2 4 0.2j 0]

Enter the Mutual Coupling data : [1 2 0.1j;1 4 0.2j]

OUTPUT: a = 1 -1 0 0 1 0 -1 0 0 0 1 -1 1 -1 0 0 0 1 0 -1

Ybus = 0 - 4.6875i 0 + 2.8125i 0 + 1.8750i 0 0 + 2.8125i 0 - 8.0208i 0 + 0.2083i 0 + 5.0000i 0 + 1.8750i 0 + 0.2083i 0 - 4.0833i 0 + 2.0000i 0 0 + 5.0000i 0 + 2.0000i 0 - 7.0000i




1. b) DETERMINATION OF BUS CURRENTS, BUS POWER & LINE FLOWS FOR A SPECIFIED SYSTEM VOLTAGE (BUS) PROFILE

Case study:

Program:

clc;clear;linedata = [1 2 0.02 0.04 0 1 3 0.01 0.03 0 2 3 0.0125 0.025 0];fr=linedata(:,1);to=linedata(:,2);R=linedata(:,3);X=linedata(:,4);B_half=linedata(:,5);nb=max(max(fr),max(to));nbr=length(R); bus_data = [1 1.05 0 2 0.98183 -3.5 3 1.00125 -2.8624];bus_no=bus_data(:,1);volt_mag=bus_data(:,2);angle_d=bus_data(:,3); Z = R + j*X; %branch impedancey= ones(nbr,1)./Z; %branch admittanceVbus=(volt_mag.*cos(angle_d*pi/180))+ (j*volt_mag.*sin(angle_d*pi/180));%line currentfor k=1:nbrIfr(k)=(Vbus(fr(k))-Vbus(to(k)))*y(k) + (Vbus(fr(k))*j*B_half(k));Ito(k)=(Vbus(to(k))-Vbus(fr(k)))*y(k) + (Vbus(to(k))*j*B_half(k));end%line flows% S=P+jQ


(2)0.98183<-3.5

(3)1.00125<-2.8624

(1)1.05<0

G1

0.02 + j 0.04

0.0125 + j 0.0250.01 + j 0.03

256.6 + j 110.2

138.6 + j 45.2

7


for k=1:nbrSfr(k)=Vbus(fr(k))*Ifr(k)';Sto(k)=Vbus(to(k))*Ito(k)';end%line lossesfor k=1:nbrSL(k)=Sfr(k)+Sto(k);end%bus powerSbus=zeros(nb,1);for n=1:nb for k=1:nbr if fr(k)==n Sbus(n)=Sbus(n)+Sfr(k); elseif to(k)==n Sbus(n)=Sbus(n)+Sto(k); end endend %Bus current%P+jQ=V*I'Ibus=(Sbus./Vbus)';fprintf(' OUTPUT DATA \n'); fprintf('------------------------------\n'); fprintf(' Line Currents\n'); fprintf('------------------------------\n'); for k=1:nbr fprintf(' I%d%d= %4.3f (+) %4.3f \n',fr(k),to(k),real(Ifr(k)),imag(Ifr(k))); fprintf(' I%d%d= %4.3f (+) %4.3f \n', to(k),fr(k),real(Ito(k)),imag(Ito(k))); end fprintf('------------------------------\n'); fprintf(' Line Flows\n') fprintf('------------------------------\n'); for k=1:nbr fprintf(' S%d%d= %4.3f (+) %4.3f \n',fr(k),to(k),real(Sfr(k)),imag(Sfr(k))); fprintf(' S%d%d= %4.3f (+) %4.3f \n', to(k),fr(k),real(Sto(k)),imag(Sto(k))); end fprintf('------------------------------\n'); fprintf(' Line Losses\n') fprintf('------------------------------\n'); for k=1:nbr fprintf(' SL%d%d= %4.3f (+) %4.3f \n',fr(k),to(k),real(SL(k)),imag(SL(k))); end fprintf('------------------------------\n'); fprintf(' Bus Power\n') fprintf('------------------------------\n'); for k=1:nb fprintf(' P%d= %4.3f (+) %4.3f \n',k,real(Sbus(k)),imag(Sbus(k))); end fprintf(' Bus Current\n') fprintf('------------------------------\n'); for k=1:nb



fprintf(' I%d= %4.3f (+) %4.3f \n',k,real(Ibus(k)),imag(Ibus(k))); end fprintf('------------------------------\n');

OUTPUT: OUTPUT DATA ------------------------------ Line Currents------------------------------ I12= 1.899 (+) -0.801 I21= -1.899 (+) 0.801 I13= 2.000 (+) -1.000 I31= -2.000 (+) 1.000 I23= -0.638 (+) 0.481 I32= 0.638 (+) -0.481 ------------------------------ Line Flows------------------------------ S12= 1.994 (+) 0.841 S21= -1.909 (+) -0.671 S13= 2.100 (+) 1.050 S31= -2.050 (+) -0.900 S23= -0.654 (+) -0.433 S32= 0.662 (+) 0.449 ------------------------------ Line Losses------------------------------ SL12= 0.085 (+) 0.170 SL13= 0.050 (+) 0.150 SL23= 0.008 (+) 0.016 ------------------------------ Bus Power------------------------------ P1= 4.094 (+) 1.891 P2= -2.563 (+) -1.104 P3= -1.388 (+) -0.451 Bus Current------------------------------ I1= 3.899 (+) -1.801 I2= -2.537 (+) 1.282 I3= -1.362 (+) 0.519 ------------------------------




2. FORMATION OF Z-BUS, USING Z-BUS BUILDING ALGORITHM WITHOUT MUTUAL COUPLING

Program:

clc;clear;disp('Zbus Building Algorithm');zprimary = [1 1 0 0.25 2 2 1 0.1 3 3 1 0.1 4 2 0 0.25 5 2 3 0.1];[elements,columns]=size(zprimary);zbus=[];currentbusno=0;for count=1:elements [rows cols]=size(zbus); from=zprimary(count,2); to=zprimary(count,3); value=zprimary(count,4); newbus=max(from,to); ref=min(from,to); %Type-1 Modification%A new element is added from new bus to reference busif newbus>currentbusno & ref==0 disp('Adding Z ='); disp(value); disp('between buses:'); disp(from); disp(to); disp('This impedance is added between a new bus and reference(Type1)'); zbus=[zbus zeros(rows,1) zeros(1,cols) value] currentbusno=newbus; continueend %Type-2 Modification%A element is added from new bus to old bus other than reference busif newbus>currentbusno & ref~=0 disp('Adding Z ='); disp(value); disp('between buses:'); disp(from); disp(to); disp('This impedance is added between a new bus and an existing bus(Type2)'); zbus=[zbus zbus(:,ref) zbus(ref,:) value+zbus(ref,ref)] currentbusno=newbus;



continueend %Type-3 Modification%A new element is added between an old bus and reference busif newbus<=currentbusno & ref==0 disp('Adding Z ='); disp(value); disp('between buses:'); disp(from); disp(to); disp('This impedance is added between an existing bus and reference(Type3)'); zbus=zbus-1/(zbus(newbus,newbus)+value)*zbus(:,newbus)*zbus(newbus,:) continueend %Type-4 Modification%A new element is added between two old buses(bus-2 to 3) if newbus<=currentbusno & ref~=0 disp('Adding Z ='); disp(value); disp('between buses:'); disp(from); disp(to); disp('This impedance is added between two existing buses(Type4)'); zbus=zbus-1/(value+zbus(from,from)+zbus(to,to)-2*zbus(from,to))*((zbus(:,from)-zbus(:,to))*((zbus(from,:)-zbus(to,:)))) continue endend

OUTPUT:

Zbus Building AlgorithmAdding Z = 0.2500

between buses: 1

0

This impedance is added between a new bus and reference(Type1)

zbus =

0.2500

Adding Z = 0.1000



between buses: 2

1

This impedance is added between a new bus and an existing bus(Type2)

zbus =

0.2500 0.2500 0.2500 0.3500

Adding Z = 0.1000

between buses: 3

1

This impedance is added between a new bus and an existing bus(Type2)

zbus =

0.2500 0.2500 0.2500 0.2500 0.3500 0.2500 0.2500 0.2500 0.3500

Adding Z = 0.2500

between buses: 2

0

This impedance is added between an existing bus and reference(Type3)

zbus =

0.1458 0.1042 0.1458 0.1042 0.1458 0.1042 0.1458 0.1042 0.2458

Adding Z = 0.1000

between buses: 2

3

This impedance is added between two existing buses(Type4)



zbus =

0.1397 0.1103 0.1250 0.1103 0.1397 0.1250 0.1250 0.1250 0.1750




3. ABCD PARAMETERS

i) Formation of Symmetric T/PI Configurationii) Verification of AD-BC =1iii) Determination of Efficiency & Regulation

Program :

z=0.2+0.408i; y=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=40; Z=z*length;Y=y*length; A=1;B=Z;C=0;D=1; case 2, length=140; Z=z*length;Y=y*length; A=1+Y*Z/2; B=Z; C=Y*(1+Y*Z/4); D=A; case 3, length=300; zc=sqrt(z/y); gam=sqrt(z*y)*length; A=cosh(gam); D=A; B=zc*sinh(gam); C=1/zc*sinh(gam); fprintf('\n the equivalent pi circuit constants:'); zeq=z*length*sinh(gam)/gam; yeq=y*length/2*tanh(gam/2)/(gam/2); fprintf('\n Zeq = %15.4f%+15.4fi',real(zeq),imag(zeq)); fprintf('\n yeq/2 = %15.4f%+15.4fi',real(yeq),imag(yeq)); otherwise disp('wrong choice of transmission line');endfprintf('\n A,B,C and D constants:\n');fprintf('---------------------------------------');fprintf('\n A = %15.4f%+15.4fi',real(A),imag(A));fprintf('\n B = %15.4f%+15.4fi',real(B),imag(B));fprintf('\n C = %15.4f%+15.4fi',real(C),imag(C));fprintf('\n D = %15.4f%+15.4fi',real(D),imag(D));fprintf('\n The product AD-BC = %f',A*D-B*C);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=132+0.0i; ir=174.96-131.22i;



vr=vr*1e3/sqrt(3); vs=(A*vr+B*ir)/1e3; is=C*vr+D*ir; fprintf('\n Sending end voltage/ph=%f%+fi KV', real(vs),imag(vs)); fprintf('\n Sending end current/ph=%f%+fi Amp', real(is),imag(is)); vs=vs*1e3; case 2, vs=132+0.0i; is=174.96-131.22i; vs=vs*1e3/sqrt(3.0); vr=(D*vs-B*is)/1e3; ir=-C*vs+D*is; fprintf('\n Receiving end voltage/ph=%f%+fi KV', real(vr),imag(vr)); fprintf('\n Receiving end current/ph=%f%+fi Amp', real(ir),imag(ir)); vr=vr*1e3; otherwise disp('wrong choice');endrec_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 Recieving end power=%.2fKVA',rec_pow);fprintf('\n sending end power=%.2fKVA',send_pow);fprintf('\n efficiency=%.2f%%',eff);fprintf('\n voltage regulation=%.2f%%',reg);

Note: For Equivalent T circuit constants

fprintf('\n the equivalent T circuit constants:'); yeq=y*length*sinh(gam)/gam; zeq=z*length/2*tanh(gam/2)/(gam/2); fprintf('\n Zeq/2 = %10.4f%+10.4fi',real(zeq),imag(zeq)); fprintf('\n yeq = %10.4f%+10.4fi',real(yeq),imag(yeq));

OUTPUT :Equivalent PIenter 1-for short line 2- for medium line 3- for long line3

the equivalent pi circuit constants: Zeq = 57.7123 +120.6169i yeq/2 = 0.0000 +0.0005i A,B,C and D constants:--------------------------------------- A = 0.9428 +0.0277i B = 57.7123 +120.6169i C = -0.0000 +0.0009i D = 0.9428 +0.0277i The product AD-BC = 1.000000 enter 1- to read Vr, Ir and compute Vs, Is 2- to read Vs, Is and compute Vr, Ir1



Sending end voltage/ph=97.773395+15.642655i KV Sending end current/ph=167.915908-48.443896i Amp Recieving end power=40.00KVA sending end power=46.98KVA efficiency=85.15% voltage regulation=37.75%>>

Equivalent Tenter 1-for short line 2- for medium line 3- for long line3

the equivalent T circuit constants: Zeq/2 = 30.5858 +61.6486i yeq = -0.0000 +0.0009i A,B,C and D constants:--------------------------------------- A = 0.9428 +0.0277i B = 57.7123 +120.6169i C = -0.0000 +0.0009i D = 0.9428 +0.0277i The product AD-BC = 1.000000 enter 1- to read Vr, Ir and compute Vs, Is 2- to read Vs, Is and compute Vr, Ir2

Receiving end voltage/ph=45.924027-11.417589i KV Receiving end current/ph=169.252896-189.276902i Amp Recieving end power=29.80KVA sending end power=40.00KVA efficiency=74.50% voltage regulation=70.75%>>




4. DETERMINATION OF POWER ANGLE DIAGRAMS

A) Determination of power angle diagrams for salient and non-salient pole synchronous m/cs, reluctance power, excitation emf & regulation using MATLAB.

Example:

A 34.64 kV,60 MVA synchronous generator has a direct axis reactance of 13.5 ohms and quadrature axis reactance of 9.333 ohms is operating at 0.8 p.f determine the excitation e.m.f, regulation, reluctance power and also plot the power angle diagram.

Program:

clc;clear;P=input('Power in MW = ');pf=input('Power Factor = ');Vt=input('Line to Line Voltage in kV = ');Xd=input('Xd in Ohms = ');Xq=input('Xq in Ohms = ');Vtph=Vt*1000/sqrt(3); % Per phase Voltagepf_a=acos(pf);Q=P*tan(pf_a);I=(P-j*Q)*1000000/(3*Vtph); % Current in Ampsdelta=0:1:180;delta_rad=delta*(pi/180);if Xd~=Xq %Salient Pole Synchronous Motor Eq=Vtph+(j*I*Xq); Id_mag=abs(I)*sin(angle(Eq)-angle(I)); Ef_mag=abs(Eq)+((Xd-Xq)*Id_mag); Exitation_emf=Ef_mag Reg=(Ef_mag-abs(Vtph))*100/abs(Vtph) PP=Ef_mag*Vtph*sin(delta_rad)/Xd; Reluct_Power=Vtph^2*(Xd-Xq)*sin(2*delta_rad)/(2*Xd*Xq); Net_Reluct_Power=3*Reluct_Power/1000000; Power_sal=PP+Reluct_Power; Net_Power_sal=3*Power_sal/1000000; plot(delta,Net_Reluct_Power,'K'); hold on plot(delta,Net_Power_sal,'r'); xlabel('\Delta(deg)-------->'); ylabel('Three Phase Power(pu)-------->'); title('Plot:Power Angle Curve for Salient Synchronous M/c'); legend('Reluct Power','Salient Power');end if Xd==Xq

%Non-Salient Pole Synchronous Motor Ef=Vtph+(j*I*Xd); Exitation_emf=abs(Ef)



Reg=(abs(Ef)-abs(Vtph))*100/abs(Vtph) Power_non=abs(Ef)*Vtph*sin(delta_rad)/Xd; Net_Power=3*Power_non/1000000; plot(delta,Net_Power); xlabel('\Delta(deg)-------->'); ylabel('Three Phase Power(MW)-------->'); title('Plot:Power Angle Curve for Non-Salient Synchronous M/c'); legend('Non-Salient Power'); end grid;

Input Data: from the keyboard: (Salient Pole)

P = 48 Power in MW

Pf = 0.8 Power Factor

V = 34.64 Line to Line Voltage in kV

Xd = 13.5 Xd in ohm

Xq = 9.33 Xq in ohm

Output:

Excitation EMF = 30000 Volts/ phase

Regulation = 50.0024 %

Input Data: from the keyboard: (Non-Salient Pole)

P = 48 Power in MW

Pf = 0.8 Power Factor

V = 34.64 Line to Line Voltage in kV

Xd = 10Xd in ohm

Xq = 10Xq in ohm

Output:

Excitation EMF = 27203 Volts/ phase

Regulation = 36.01 %




5. SOLUTION OF SWING EQUATION

Example:A 50 Hz, synchronous generator is having inertia constant H = 2.52 MJ/MVA and Xd’= 0.32 p.u. is connected to an infinite bus delivering 16 MW over a double circuit line. The reactance of the connecting HT transformer is 0.2 p.u. and reactance of each line is 0.4 p.u. on a 20 MVA base. |Eg| = 1.2 p.u. and |V| = 1.0 p.u. A three phase circuit occurs at the mid point of one of the transmission line. The fault is cleared by simultaneous opening of breakers at both ends of line at 2.5 cycles and 6.25 cycles after the occurrence of the fault. Find the critical clearing angle and critical clearing time by plotting swing curve using Point by Point method.

Program:

clcclear allH = input('Generator Inertia constant H in MJ/MVA = ');Xpre = input('Pre fault reactance Xpre in p.u. = ');Xfault = input('Reactance during fault Xfault in p.u. = ');Xpost = input('Post fault reactance Xpost in p.u. = ');E = input('Sending end voltage E in p.u. = ');V = input('Receiving end voltage V in p.u. = ');MVA = input('Base MVA = ');MW = input('Generator output power in MW = ');delta = input('Initial displacement angle in degree = ');tc = input('Fault clearing time in sec = ');tfinal = input('Final time for swing equation in sec = ');M = H/(180*50);Pm = MW/MVA;Ppre = (E*V)/Xpre;Pfault = (E*V)/Xfault;Ppost = (E*V)/Xpost;ddelta = 0;tstep = 0.05;t = 0;tf = 0;i = 2;time(1) = 0;ang(1) = delta;while t<tfinal,if (t = = tf),Paa = Pm-Ppre*sin(delta*pi/180);Pab = Pm-Pfault*sin(delta*pi/180);Pa = (Paa+Pab)/2;endif (t = = tc),Paa = Pm-Pfault*sin(delta*pi/180);Pab = Pm-Ppost*sin(delta*pi/180);Pa = (Paa+Pab)/2;endif(t>tf & t<tc),Pa = Pm-Pfault*sin(delta*pi/180);



endif(t>tc),Pa = Pm-Ppost*sin(delta*pi/180);endddelta = ddelta+(tstep*tstep*Pa/M);delta = (delta+ddelta);t = t+tstep;time(i) = t;ang(i) = delta;i = i+1;endplot(time,ang,'bo-');grid;title(['One-machine system swing curve by Point by Point method. Fault cleared at ', num2str(tc),'s']);xlabel('t, sec'), ylabel('Delta, degree');deltamax = pi-asin(Pm/Ppost);dmax = deltamax; d0 = asin(Pm/Ppre);dc = acos((Pm*(dmax-d0)+Ppost*cos(dmax)-Pfault*cos(d0))/(Ppost-Pfault));fprintf('\nCritical clearing angle = %6.2f degrees ',dc*180/pi);it = 1;while(time(it)<1.0) if (ang(it)> = (dc*180/pi)) break; end it = it+1;endfprintf('\n Critical clearing time = %4.2f seconds \n', time(it));

Input:Sustained fault:1. Generator Inertia constant H in MJ/MVA = 5.22. Pre fault reactance Xpre in p.u. = 0.73. Reactance during fault Xfault in p.u. = 1.94. Post fault reactance Xpost in p.u. = 0.95. Sending end voltage E in p.u. = 1.26. Receiving end voltage V in p.u. = 1.07. Base MVA = 208. Generator output power in MW = 169. Initial displacement angle in degree = 27.8210. Fault clearing time in sec = 1.511. Final time for swing equation in sec = 1.5

** For Fault cleared in 2.5 cycles & Fault cleared in 6.5 cycles, Fault clearing time is 0.05 seconds & 0.125 seconds respectively.

Output:



Sustained fault:

Critical clearing angle = 91.24 degrees

Critical clearing time = 0.45 seconds

Fault cleared in 2.5 cycles (0.05secs):



Fault cleared in 6.5 cycles (0.125secs):






6. FORMATION OF JACOBIAN FOR A SYSTEM NOT EXCEEDING



4 BUSES (NO PV BUSES) IN POLAR CO-ORDINATES USING SOFTWARE PACKAGE

A Three-Bus system is given below. The system parameters are given in the Table A and the load and generation data in Table B. Line impedances are in per unit on a 100 MVA Base. Line charging susceptances are neglected. Taking bus 1 as slack bus, obtain the Jacobian

matrix.

Table A

Bus Code (i-k) Impedance (p.u) ZikLine charging

Admittance (p.u) Yi

1-2 0.08+j0.24 01-3 0.02+j0.06 02-3 0.06+j0.18 0

Table B

Bus No Bus Voltage Generation LoadMW Mvar MW Mvar

1 1.05 + j0.0 ---- --- 0 02 ------ 0 0 50 203 ----- 0 0 60 25

Procedure to enter the data for performing studies using Mi Power Mi Power - Database Configuration Open Power System Network Editor. Select menu option Database→Configure. Configure Database dialog is popped up. Click Browse button.


Bus 3 (PQ)

Slack bus (1)G

Bus 2 (PQ)

23


Open dialog box is popped up as shown below, where you are going to browse the desired directory and specify the name of the database to be associated with the single line diagram. Click Open button after entering the desired database name. Configure Database dialog will appear with path chosen.

Click on OK button in the Configure database dialog, the following dialog appears.


Enter here to specify the name

Select the folder and give database name in File name window with .Mdb extension. And now click on Open.

Click OK

24


Uncheck the Power System Libraries and Standard Relay Libraries. For this example these standard libraries are not needed, because all the data is given on p u for power system libraries (like transformer, line\cable, generator), and relay libraries are required only for relay co-ordination studies. If Libraries are selected, standard libraries will be loaded along with the database. Click Electrical Information tab.

Procedure to Draw First Element - Bus Click on Bus icon provided on power system tool bar. Draw a bus and a dialog appears prompting to give the Bus ID and Bus Name. Click OK. Database manager with corresponding Bus Data form will appear. Modify the Area number, Zone number and Contingency Weightage data if it is other than the default values. If this data is not furnished, keep the default values. Usually the minimum and maximum voltage ratings are ± 5% of the rated voltage. If these ratings are other than this, modify these fields. Otherwise keep the default values. Bus description field can be effectively used if the bus name is more than 8 characters. If bus name is more than 8 characters, then a short name is given in the bus name field and the bus description field can be used to abbreviate the bus name. For example let us say the bus name is Northeast, then bus name can be given as NE and the bus description field can be North East.


Since the impedances are given on 100 MVA base, check the pu status. Enter the Base MVA and Base frequency as shown below. Click Breaker Ratings tab. If the data is furnished, modify the breaker ratings for required voltage levels. Otherwise accept the default values. Click OK button to create the database to return to Network Editor.

Bus Base Voltage Configuration In the network editor, configure the base voltages for the single line diagram. Select menu option Configure Base voltage. Dialog shown below appears. If necessary change the Base-voltages, color, Bus width and click OK.


After entering data click save, which invokes Network Editor. Follow the same procedure for remaining buses. Following table gives the data for other buses.

Bus Number Bus Name Nominal Voltage (KV)1 Bus 1 112 Bus 2 113 Bus 3 11

Note: Since the voltages are in pu, any kV can be assumed. So the base voltage is chosen as 11kV.

Enter Element ID number and click OK. Database manager with corresponding Line\Cable Data form will be open. Enter the details of that line as shown below.

Enter Structure Ref No. as 1 and click on Transmission Line Library >> button. Line & Cable Library form will appear. Enter transmission line library data in the form as shown.

Transmission Line Element Data

Line No From Bus To Bus Structure Ref No1 1 2 12 1 3 23 2 3 3


Procedure to Draw Transmission Line Click on Transmission Line icon provided on power system tool bar. To draw the line click in between two buses and to connect to the from bus, double click LMB (Left Mouse Button) on the From Bus and join it to another bus by double clicking the mouse button on the To Bus Element ID dialog will appear.


Transmission Line Library Data

Structure Ref No

Structure Ref Name

Resistance ReactanceLine

Charging B/2Thermal

rating1 Line 1-2 0.08 0.24 0 1002 Line 1-3 0.02 0.06 0 1003 Line 2-3 0.18 0.18 0 100

Procedure to Draw Generator Click on Generator icon provided on power system tool bar. Draw the generator by clicking LMB (Left Mouse Button) on the Bus1. Element ID dialog will appear. Enter ID number and Click OK. Database with corresponding Generator Data form will appear. Enter details as given below.





Procedure to Enter Load DataClick on Load icon provided on power system tool bar. Connect load 1 at Bus 2 by clicking the LMB (Left Mouse Button) on Bus 2. Element ID dialog will appear. Give ID no. as 1 and say OK. Load Data form will appear. Enter load details as shown below. Then click Save button, which invokes Network Editor.

Load Data


BusNo

LoadMW Mvar

1 0 02 50 203 60 25

29


When Study info button is clicked, following dialog will open. Select Newton Raphson method and enter P-Tolerance and Q-tolerance as 0.001. Select print option as Detailed results with Data. Click OK.

Click on Execute to run the power flow. When Report button is clicked, following dialog box will open. Select Standard and then click OK.


1. Select standard

30

1. Select this option

2. Click OK


Report:














7. LOAD FLOW USING GAUSS-SEIDEL METHOD

The power system used for load flow studies:

Line impedance and line charging data:

Line (bus to bus) Impedance Line charging (Y /2)

1-2 0.02 + j 0.10 j 0.030

1-5 0.05 + j 0.25 j 0.020

2-3 0.04 + j 0.20 j 0.025

2-5 0.05 + j 0.25 j 0.020

3-4 0.05 + j 0.25 j 0.020

3-5 0.08 + j 0.40 j 0.010

4-5 0.10 + j 0.50 j 0.075

Ybus matrix of the system:

1 2 3 4 5

1 2.6923 - j 13.4115

- 1.9231 + j 9.6154

0 0 - 0.7692 + j 3.8462

2 - 1.9231 + j 9.6154

3.6538 - j 18.1942

- 0.9615 + j 4.8077

0 - 0.7692 + j 3.8462

3 0 - 0.9615 + j 4.8077

2.2115 - j 11.0027

- 0.7692 + j 3.8462

- 0.4808 + j 2.4038

4 0 0 - 0.7692 + j 3.8462

1.1538 - j 5.6742

- 0.3846 + j 1.9231



5 - 0.7692 + j 3.8462

- 0.7692 + j 3.8462

- 0.4808 + j 2.4038

- 0.3846 + j 1.9231

2.4038 - j 11.8942

Program:%Power flow solution by Gauss-Seidel methodclcclear alld2r=pi/180;w=100*pi;% The Y_bus matrix isybus=[2.6923-13.4115j -1.9231+9.6154j 0 0 -0.7692+3.8462j; -1.9231+9.6154j 3.6538-18.1942j -0.9615+4.8077j 0 -0.7692+3.8462j; 0 -0.9615+4.8077j 2.2115-11.0027j -0.7692+3.8462j -0.4808+2.4038j; 0 0 -0.7692+3.8462j 1.1538-5.6742j -0.3846+1.9231j; -0.7692+3.8462j -0.7692+3.8462j -0.4808+2.4038j -0.3846+1.9231j 2.4038-11.8942j]; g = real(ybus);b = imag(ybus); % The given parameters and initial conditions arep = [0;-0.96;-0.35;-0.16;0.24];q = [0;-0.62;-0.14;-0.08;-0.35];mv = [1.05;1;1;1;1.02];th = [0;0;0;0;0];v = [mv(1);mv(2);mv(3);mv(4);mv(5)];acc = input('Enter the acceleration constant: ');del = 1; indx =0;nb = length(p);% The Gauss-Seidel iterations starts herewhile del>1e-6% P-Q buses for i = 2:4 tmp1= (p(i)-j*q(i))/conj(v(i)); tmp2 = 0; for k =1:5 if (i = = k) tmp2 = tmp2+0; else tmp2 = tmp2+ybus(i,k)*v(k); end end vt = (tmp1-tmp2)/ybus(i,i); v(i) = v(i)+acc*(vt-v(i)); end % P-V bus q5 = 0; for i =1:5 q5 = q5+ybus(5,i)*v(i); end q5 = -imag(conj(v(5))*q5); tmp1 = (p(5)-j*q5)/conj(v(5)); tmp2 = 0;



for k =1:4 tmp2 = tmp2+ybus(5,k)*v(k); end vt = (tmp1-tmp2)/ybus(5,5); v(5) = abs(v(5))*vt/abs(vt); % Calculate P and Q for i =1:5 sm = 0; for k =1:5 sm = sm+ybus(i,k)*v(k); end s(i) = conj(v(i))*sm; end % The mismatch delp = p-real(s)'; delq = q+imag(s)'; delpq = [delp(2:5);delq(2:4)]; del = max(abs(delpq)); indx = indx+1; V = abs(v)'; V_angle = angle(v)'/d2r; Real_pow = (real(s)+[0 0 0 0 0.24])*100; Reactive_pow = (-imag(s)+[0 0 0 0 0.11])*100; fprintf('\nGS LOAD FLOW ITERATION NO. %d',indx); fprintf('\nVOLTAGES ARE:'); for k =1:nb fprintf('\nV(%d) = %.4f<%.2f',k,V(k),V_angle(k)); end fprintf('\n\nTHE REAL POWERS IN EACH BUS IN MW ARE:'); for k =1:nb fprintf('\nP(%d) = %.4f',k,Real_pow(k)); end fprintf('\n\nTHE REACTIVE POWERS IN EACH BUS IN MVar ARE:'); for k =1:nb fprintf('\nQ(%d) = %.4f',k,Reactive_pow(k)); end fprintf('\n'); if indx = = 4 break endend

OUTPUT:Enter the acceleration constant: 1.02

GS LOAD FLOW ITERATION NO. 1VOLTAGES ARE:V(1) = 1.0500<0.00V(2) = 0.9926<-2.65



V(3) = 0.9881<-2.91V(4) = 0.9967<-3.61V(5) = 1.0200<-0.94

THE REAL POWERS IN EACH BUS IN MW ARE:P(1) = 67.3133P(2) = -68.1565P(3) = -8.3389P(4) = -13.9469P(5) = 48.6426

THE REACTIVE POWERS IN EACH BUS IN MVar ARE:Q(1) = 55.0891Q(2) = -58.1404Q(3) = -17.0080Q(4) = -7.6435Q(5) = 5.5171

GS LOAD FLOW ITERATION NO. 2VOLTAGES ARE:V(1) = 1.0500<0.00V(2) = 0.9875<-3.50V(3) = 0.9842<-4.72V(4) = 0.9928<-5.05V(5) = 1.0200<-1.82














8. FAULT STUDIES FOR A GIVEN POWER SYSTEM USING SOFTWARE PACKAGE

Figure shows a single line diagram of a 6-bus system with two identical generating units, five lines and two transformers. Per-unit transmission line series impedances and shunt susceptances are given on 100 MVA base, generator's transient impedance and transformer leakage reactances are given in the accompanying table.

If a 3-phase to ground fault occurs at bus 5, find the fault MVA. The data is given below.

Bus-code Impedance Line Chargingp-q Zpq Y’pq/23-4 0.00+j0.15 03-5 0.00+j0.10 03-6 0.00+j0.20 05-6 0.00+j0.15 04-6 0.00+j0.10 0



Generator details:G1= G2 = 100MVA, 11KV with X’

d =10%

Transformer details:T1= T2 = 11/110KV , 100MVA, leakage reactance = X = 5%

**All impedances are on 100MVA base

Mi Power Data Interpretation: SOLUTION: In transmission line data, elements 3 – 4 & 5 – 6 have common parameters. Elements 3 - 5 & 4 – 6 have common parameters. Therefore 3 libraries are required for transmission line. As generators G1 and G2 have same parameters, only one generator library is required. The same applies for transformers also. Procedure to enter the data for performing studies using Mi Power Mi Power - Database Configuration Open Power System Network Editor. Select menu option Database→Configure. Configure Database dialog is popped up. Click Browse button.

Open dialog box is popped up as shown below, where you are going to browse the desired directory and specify the name of the database to be associated with the single line diagram. Click Open button after entering the desired database name. Configure Database dialog will


Enter here to specify the name

43


appear with path chosen.

Click on OK button in the Configure database dialog, the following dialog appears.

Uncheck the Power System Libraries and Standard Relay Libraries. For this example these standard libraries are not needed, because all the data is given on p u for power system libraries (like transformer, line\cable, generator), and relay libraries are required only for relay co-ordination studies. If Libraries are selected, standard libraries will be loaded along with the database. Click Electrical Information tab.


Select the folder and give database name in File name window with .Mdb extension. And now click on Open.

Click OK

44


Procedure to Draw First Element - Bus Click on Bus icon provided on power system tool bar. Draw a bus and a dialog appears prompting to give the Bus ID and Bus Name. Click OK. Database manager with corresponding Bus Data form will appear. Modify the Area number, Zone number and Contingency Weightage data if it is other than the default values. If this data is not furnished, keep the default values. Usually the minimum and maximum voltage ratings are ± 5% of the rated voltage. If these ratings are other than this, modify these fields. Otherwise keep the default values. Bus description field can be effectively used if the bus name is more than 8 characters. If bus name is more than 8 characters, then a short name is given in the bus name field and the bus description field can be used to abbreviate the bus name. For example let us say the bus name is Northeast, then bus name can be given as NE and the bus description field can be North East.

After entering data click save, which invokes Network Editor. Follow the same procedure for remaining buses. Following table gives the data for other buses.

Bus dataBus Number 1 2 3 4 5 6Bus Name Bus1 Bus2 Bus3 Bus4 Bus5 Bus6


Since the impedances are given on 100 MVA base, check the pu status. Enter the Base MVA and Base frequency as shown below. Click Breaker Ratings tab. If the data is furnished, modify the breaker ratings for required voltage levels. Otherwise accept the default values. Click OK button to create the database to return to Network Editor.

Bus Base Voltage Configuration In the network editor, configure the base voltages for the single line diagram. Select menu option Configure Base voltage. Dialog shown below appears. If necessary change the Base-voltages, color, Bus width and click OK.


Nominal voltage 11 11 110 110 110 110Area number 1 1 1 1 1 1Zone number 1 1 1 1 1 1Contingency Weightage 1 1 1 1 1 1

Enter Element ID number and click OK. Database manager with corresponding Line\Cable Data form will be open. Enter the details of that line as shown below. Enter Structure Ref No. as 1 and click on Transmission Line Library >> button. Line & Cable Library form will appear. Enter transmission line library data in the form as shown for Line3-4.

After entering data, Save and Close. Line\Cable Data form will appear. Click Save, which invokes network editor. Data for remaining elements given in the following table. Follow the same procedure for rest of the elements.

Transmission Line Element DataLine Number 1 2 3 4 5Line Name Line3-4 Line3-5 Line3-6 Line4-6 Line5-6 De-Rated MVA 100 100 100 100 100No. Of Circuits 1 1 1 1 1From Bus No. 3 3 3 4 5To Bus No. 4 5 6 6 6Line Length 1 1 1 1 1From Breaker Rating 5000 5000 5000 5000 5000To Breaker Rating 5000 5000 5000 5000 5000


Procedure to Draw Transmission Line Click on Transmission Line icon provided on power system tool bar. To draw the line click in between two buses and to connect to the from bus, double click LMB (Left Mouse Button) on the From Bus and join it to another bus by double clicking the mouse button on the To Bus .Element ID dialog will appear.


Structure Ref No. 1 2 3 2 1

Transmission Line Library DataStructure Ref. No. 1 2 3Structure Ref. Name Line3-4 & 5-6 Line3-5 & 4-6 Line3-6 Positive Sequence Resistance 0 0 0Positive Sequence Reactance 0.15 0.1 0.2Positive Sequence Susceptance 0 0 0Thermal Rating 100 100 100

Procedure to Draw Transformer Click on Two Winding Transformer icon provided on power system tool bar. To draw the transformer click in between two buses and to connect to the from bus, double click LMB (Left Mouse Button) on the From Bus and join it to another bus by double clicking the mouse button on the To Bus Element ID dialog will appear. Click OK.

Transformer Element Data form will be open. Enter the Manufacturer Ref. Number as 30. Enter transformer data in the form as shown below. Click on Transformer Library >>button. Transformer library form will be open. Enter the data as shown below. Save and close library screen.

Transformer element data form will appear. Click Save button, which invokes network editor. In the similar way enter other transformer details.



2nd Transformer detailsTransformer Number 2Transformer Name 2T2From Bus Number 6To Bus Number 2Control Bus Number 2Number of Units in Parallel 1Manufacturer ref. Number 30De Rated MVA 100From Breaker Rating 5000To Breaker Rating 350Nominal Tap Position 5

Procedure to Draw Generator Click on Generator icon provided on power system tool bar. Draw the generator by clicking LMB (Left Mouse Button) on the Bus1. Element ID dialog will appear. Click OK.

Generator Data form will be opened. Enter the Manufacturer Ref. Number as 20. Enter Generator data in the form as shown below.



Click on Generator Library >> button. Enter generator library details as shown below. Save and Close the library screen. Generator data screen will be reopened. Click Save button, which invokes Network Editor. Connect another generator to Bus 2. Enter its details as given in the following table.

2nd Generator detailsName GEN-2Bus Number 2Manufacturer Ref. Number 20Number of Generators in Parallel 1Capability Curve Number 0De-Rated MVA 100Specified Voltage 11Scheduled Power 20Reactive Power Minimum 0Reactive Power Maximum 60Breaker Rating 350Type of Modeling Infinite

Note: To neglect the transformer resistance, in the multiplication factor table give the X to R



Ratio as 9999. TO solve short circuit studies choose menu option Solve →Short Circuit Analysis or click on SCS button on the toolbar on the right side of the screen. Short circuit analysis screen appears.

Study Information.

In Short Circuit Output Options select the following.


1 Click here to select case no

2 Click here to open short circuit studies screen

3. Click here . Click here



50


Afterwards click Execute. Short circuit study will be executed. Click on Report to view the report file.

Exercise: Simulate single line to ground fault, line to line fault for the above case.

Exercise problems:

1. Using the available software package determine (a) fault current (b) post fault voltages at all buses (c) fault MVA , when a line to ground fault with / with out Zf = ___ occurring at _____ bus for a given test system.( Assume pre fault voltages at all buses to be 1 pu) compare the results with the symmetrical fault.


Click OK

1 Click here to execute

2 Click here for report

51


2. Using the available software package determine (a) fault current (b) post fault voltages at all buses (c) fault MVA , when a line to line & ground fault with / with out Zf = ___ occurring at _____ bus for a given test system.( Assume pre fault voltages at all buses to be 1 pu) compare the results with the symmetrical fault.

Refer below question for Q.(1) & Q(2)

Reactance of all transformers and transmission lines = 0.1 puZero sequence reactance of transmission line = 0.25 puNegative sequence reactance of generator = Subtransient reactaceZero sequence reactance of generator = 0.1 pu

3. For the given circuit, find the fault currents, voltages for the following type of faults at Bus3.1. Single Line to Ground Fault2. Line to Line Fault3. Double line to Ground FaultFor the transmission line assume X 1=X 1, X 0=2.5 XL

4. For the given circuit, find the fault currents, voltages for the following type of faults at Specified location.

1. Single Line to Ground Fault2. Double line to Ground Fault3. Line to Line Fault


G1

X’’= 0.3

G3

X’’= 0.1

G2

X’’= 0.2

X’t = 0.1 XL= 0.4

G

X’d= 0.3

Bus 1 Bus 2 Bus 3

∆ Y

52



9. LOAD FLOW STUDIES FOR A GIVEN POWER SYSTEM USING

SOFTWARE PACKAGE

Figure below shows a single line diagram of a 5bus system with 2 generating units, 7 lines. Per unit transmission line series impedances and shunt susceptances are given on 100MVA Base, real power generation, real & reactive power loads in MW and MVAR are given in the accompanying table with bus1 as slack, obtain a load flow solution with Y-bus using Gauss-Siedel method and Newton Raphson method. Take acceleration factors as 1.4 and tolerances of 0.0001 and 0.0001 per unit for the real and imaginary components of voltage and 0.01 per unit tolerance for the change in the real and reactive bus powers.


100MVA13.8kV/138kVX = 0.1pu

X1= X2= 20ΩX0= 60Ω

G1

100MVA13.8 kVX’’= 0.15X2 = 0.17X0= 0.05

Bus 1 Bus 2 Bus 3

∆ Y Y ∆Bus 4

G2

100MVA13.8 kVX’’= 0.2X2 = 0.21X0= 0.1Xn = 0.05pu

53


IMPEDANCES AND LINE CHARGING ADMITTANCES FOR THE SYSTEM

Table: 1.1Bus coneFrom-To

ImpedanceR+jX

Line ChargingB/2

1-2 0.02+j0.06 j 0.0301-3 0.08+j0.24 j 0.0252-3 0.06+j0.18 j 0.0202-4 0.06+j0.18 j 0.0202-5 0.04+j0.12 j 0.0153-4 0.01+j0.03 j 0.0104-5 0.08+j0.24 j 0.025

GENERATION, LOADS AND BUS VOLTAGES FOR THE SYSTEM

Table: 1.2Bus No

Bus Voltage

Generation MW

GenerationMVAR

LoadMW

LoadMVAR

1 1.06+j0.0 0 0 0 02 1.00+j0.0 40 30 20 103 1.00+j0.0 0 0 45 154 1.00+j0.0 0 0 40 55 1.00+j0.0 0 0 60 10

Procedure to enter the data for performing studies using MiPower


(5)(2)

(3)(1)

(4)

G

G


MiPower - Database ConfigurationOpen Power System Network Editor. Select menu option Database Configure. Configure Database dialog is popped up as shown below. Click Browse button.

Open dialog box is popped up as shown below, where you are going to browse the desired directory and specify the name of the database to be associated with the single line diagram. Click Open button after entering the hired database name. Configure Database dialog will appear with path chosen.

Note: Do not work in the MiPower directory.Click OK button on the Configure database dialog. The dialog shown below appears.



Uncheck the Power System Libraries and Standard Relay Libraries. For this example these standard libraries are not needed, because all the data is given on pu for power system libraries (like transformer, line\cable, generator), and relay libraries are required only for relay co-ordinate studies. If Libraries are selected, standard libraries will be loaded along with the database. Click Electrical Information tab. Since the impedances are given on 100 MVA base, check the pu status. Enter the Base and Base frequency as shown below. Click on Breaker Ratings button to give breaker ratings. Click OK to create the database to return to Network Editor.

Bus Base Voltage Configuration

In the network editor, configure the base voltages for the single line diagram. Select menu option Configure→Base voltage. The dialog shown below appears. If necessary change the Base-voltages, color, Bus width and click OK.



Procedure to Draw First Element - Bus

Click on Bus icon provided on power system tool bar. Draw a bus and a dialog appears prompting to give the Bus ID and Bus Name. Click OK. Database manager with corresponding Bus Data form will appear. Modify the Area number, Zone number and Contingency Weightage data if it is other than the default values. If this data is not furnished, keep the default values. Usually the minimum and maximum voltage ratings are ± 5% of the rated voltage. If these ratings are other than this, modify these fields. Otherwise keep the default values.Bus description field can be effectively used if the bus name is more than 8 characters. If bus name is more than 8 characters, then a short name is given in the bus name field and the bus description field can be used to abbreviate the bus name. For example let us say the bus name is Northeast, then bus name can be given as NE and the bus description field can be North East.


















Exercise Problems:1. Using the available software package conduct load flow analysis for the given power

system using Gauss-Siedel / Newton-Raphson / Fast decoupled load flow method. Determine

(a) Voltage at all buses(b) Line flows(c) Line losses and(d) Slack bus power.

Also draw the necessary flow chart ( general flow chart)

2. Using the available software package conduct load flow analysis for the given power system using Gauss-Siedel / Newton-Raphson / Fast decoupled load flow method. Determine


Also draw the necessary flow chart ( general flow chart), Compare the results with the results obtained when (i) a line is removed (ii) a load is removed.

Refer below question for Q.(1) & Q(2)Consider the 3 bus system of figure each of the line has a series impedence of (0.02 + j0.08) p.u. & a total shunt admittance of j 0.02 p.u. The specified quantities at the buses are tabulated below on 100 MVA base.



Bus No.

Real LoadDemand PD

Reactive LoadDemand QD

Real PowerGeneration PG

Reactive PowerGeneration QG

VoltageSpecification

1 2.0 1.0 Unspesified Unspesified V1 = (1.04 + j 0)(Slack Bus)

2 0.0 0.0 0.5 1.0 Unspesified(PQ Bus)

3 0.5 0.6 0.0 QG3 = ? V3 = 1.04(PV Bus)

Controllable reactive power source is available at bus 3 with the constraint0 ≤ QG3 ≤ 1.5 p.u.

Find the load flow solution using N R method. Use a tolerance of 0.01 for power mismatch.

3. Figure shows the one line diagram of a simple three bus system with generation at bus 1, the voltage at bus 1 is V1 = 1.0<00 pu. The scheduled loads on buses 2 and 3 are marked on diagram. Line impedances are marked in pu on a 100MVA base. For the purpose of hand calculations line resistances and line charging susceptances are neglected. Using Gauss-Seidel method and initial estimates of V2 = 1.0<00 pu & V3 = 1.0<00 pu. Conduct load flow analysis.


(2)

(3)

(1)

G1 G2

G3

Bus 3(300+j270) MW

Slack busV1=1.0+j0.0

GBus 2(400+j320) MWj0.03333

j0.0125 j0.05


10. OPTIMAL GENERATOR SCHEDULING FOR THERMAL POWER PLANTS USING SOFTWARE PACKAGE












Exercise Problems:

1. Data given:Total No. of Generators : 2Total Demand : 100MW

GENERATOR RATINGS: GENERATOR 1: Pscheduled = 70 MW ; Pmin= 0.0MW; Pmax= 100MW

C0 = 50 ; C1 = 16 ; C2 = 0.015

Cg1 = 50 + 16 P1 + 0.015 P12 Rs / hr

GENERATOR 2: Pscheduled = 70 MW; Pmin= 0.0MW; Pmax= 100MW

C0 = 30 ; C1 = 12 ; C2 = 0.025

Cg1 = 30 + 12 P1 + 0.025 P12 Rs / hr

Loss ( or B) Co- efficients :B11 = 0.005; B12 = B21 = 0.0012; B22 = 0.002



2. Given the cost equation and loss Co-efficient of different units in a plant determine economic generation using the available software package for a total load demand of _____ MW ( typically 150 MW for 2 units and 250 MW for 3 Units) neglecting transmission losses.

3. Given the cost equation and loss Co-efficient of different units in a plant determine penalty factor, economic generation and total transmission loss using the available software package for a total load demand of _____ MW ( typically 150 MW for 2 units and 250 MW for 3 Units) .

Refer below questions for Q(2) & Q(3)A)

Loss ( or B) Co- efficients :

B11 = 0.0005 B12 = 0.00005 B13 = 0.0002

B22 = 0.0004 B23 = 0.00018 B33 = 0.0005

B21 = B12 B32 = B23 B13 = B31

B)

Loss ( or B) Co- efficients : B11 = 0.005 B12 = B21 = -0.0012 B22 = 0.002


Unit No. Cost of Fuel input in Rs./ hr.1 C1 = 800 + 20 P1 + 0.05 P1

2 0 ≤ P1 ≤ 100

2 C2 = 1000 + 15 P2 + 0.06 P22 0 ≤ P2 ≤ 100

3 C3 = 900 + 18 P3 + 0.07 P32 0 ≤ P3 ≤ 100


2 0 ≤ P1 ≤ 100

2 C2 = 30 + 12 P2 + 0.025 P22 0 ≤ P2 ≤ 100

73


VIVA QUESTIONS

FORMATION OF Z BUS & Y BUS:

1. What is meant by graph and sub graph?2. Define tree and co-tree of a network?3. What is meant by cut set and basic cut set?4. What is ment by bus incidence matrix (A)?5. Write the performance equation of an element both in impedance form and in

admittance form.

DETERMINATION OF ABCD CONSTANTS:

1. Define Regulation and efficiency of transmission line.2. Write the expression for Vs and Is in terms of Vr, Ir and ABCD constants.

SWING EQUATION:

1. Write the swing equation for a synchronous m/c connected to an infinite bus and explain the terms.

2. What is swing curve and what are its uses?3. What are the assumptions made in the solution of swing equation?

LOAD FLOW STUDIES:

1. What is load flow analysis?2. How the bases are classified in load flow studies?3. What is the data required to carry out load flow analysis?4. What is the significance of slack bus?5. What are the methods of load flow analysis?6. Write down the equation for the calculation of voltage at any bus in Gauss & Gauss

sidle iterative method.7. What are acceleration factors? What for they are used?8. What are the advantages of NR method over Gauss seidel method of load flow

analysis?9. What is meant by Fast decoupled load flow analysis?

FAULT STUDIES:

1. What are the various types of faults that occur normally on power system?2. What are the parameters considered for a fault study?3. How the sequence networks are connected for (i) L-L-G fault (ii) LL fault (iii) L-G

fault.4. Write down the fault current equation for (i) single line to ground fault (ii) double line

to ground fault (iii) LL fault5. What are the limitations of ungrounded system?6. Write zero sequence equivalent for (i) Y-Y (ii) Y-Δ (iii) Δ –Δ (iv)Y-Y



ECONOMIC OPERATION OF POWER SYSTEM:

1. What do you mean by Economic operation of a power system? 2. Name the performance characteristic curves of a turbo generator boiler unit.3. What are heat rate, incremental fuel rate and incremental fuel cost of a unit?4. Write down the co-ordination equation for economic operation of generators in a

thermal plant when (i) transmission losses are neglected (ii) transmission losses are considered.

5. What is penalty factor? Write down the total transmission loss formula in terms of B Co-efficient.

QUESTION BANK

3. Using singular transformation analysis, determine the bus admittance matrix of given test system.

4. Using singular transformation analysis, determine Ybus for a given test system with mutual coupling of j0.01 pu between the line numbers _____ & ______ Neglect line charging.

5. Using singular transformation analysis, determine the bus admittance matrix of given test system & hence find Zbus.

6. Using singular transformation analysis, determine Ybus for a given test system with mutual coupling of j0.01 pu between the line numbers _____ & ______ & hence find Zbus _______ . Neglect line charging.

Refer below figure for Q.(1) – Q.(4)

7. Given any two of the I/O parameters given either (Vs, Is) or (Vr ,Ir) of a two port short transmission line network ( 0-80km or 0-50 miles), determine the following

. Regulation . Efficiency

( The line impedance and admittance values are to be specified)

8. Given any two of the I/O parameters given either (Vs, Is) or (Vr ,Ir) of a two port short transmission line network ( >80km – 24 km or > 50 miles to 150 miles), determine the following




1

4

2

3

(0.02 + j0.04)

(0.01 + j0.03)

(0.0125 + j0.025)(0.0125 + j0.025)


9. Given any two of the I/O parameters given either (Vs, Is) or (Vr ,Ir) of a two port short transmission line network ( >240km or >150 miles), determine the following



10. Given any two of the I/O parameters of a two port short transmission line system and for a given set of line parameters ( Z & Y) and length _______km( >240km or >150 miles),determine the following

(i) ABCD constants and hence the generalized equivalent circuit (ii) Regulation, (iii) Efficiency

11. Determine the generalized circuit constants for a transmission line with the following Data:

(a) Line length: _________________ km(b) Line impedance : _____________ Ω/km(c) Total shunt admittance : ___________________ mho / km

12. The system having a single machine connected to an infinite bus has the following data.Pi =0.9 M= 0.016 ( 2.8 × 10 -4 S / electrical Degree)E1 = 1.1 Transfer reactance before fault – Xo = 0.45 puE2 = 1.0 Transfer reactance during fault – X1 = 1.25 puPlot the swing curve ( either using graph sheet or any grapher software) for sustained fault step by step method up to 1 second.

13. The system having a single machine connected to an infinite bus has the following data.Write and execute a program to plot the swing curve when the fault is cleared in 0.125 seconds.

Pi =0.9 M= 0.016 ( 2.8 × 10 -4 S / electrical Degree)E1 = 1.1 Transfer reactance before fault – Xo = 0.45 puE2 = 1.0 Transfer reactance during fault – X1 = 1.25 pu Transfer reactance after clearing the fault – X2 = 0.55 pu

Plot the swing curve ( either using graph sheet or any grapher software) for a fault cleared in _______ ( Typically 0.3 & 0.42 second) or ______ cycle using step by step method .



Also draw the necessary flow chart ( general flow chart)


(a) Voltage at all buses



(b) Line flows(c) Line losses and(d) Slack bus power.

Also draw the necessary flow chart ( general flow chart), Compare the results with the results obtained when (i) a line is removed (ii) a load is removed.

Refer below question for Q.(12) & Q(13)Consider the 3 bus system of figure each of the line has a series impedence of (0.02 + j0.08) p.u. & a total shunt admittance of j 0.02 p.u. The specified quantities at the buses are tabulated below on 100 MVA base.

Bus No.

Real LoadDemand PD

Reactive LoadDemand QD

Real PowerGeneration PG

Reactive PowerGeneration QG

VoltageSpecification

1 2.0 1.0 Unspesified Unspesified V1 = (1.04 + j 0)(Slack Bus)

2 0.0 0.0 0.5 1.0 Unspesified(PQ Bus)

3 0.5 0.6 0.0 QG3 = ? V3 = 1.04(PV Bus)

Controllable reactive power source is available at bus 3 with the constraint0 ≤ QG3 ≤ 1.5 p.u.

Find the load flow solution using N R method. Use a tolerance of 0.01 for power mismatch.

14. Given the cost equation and loss Co-efficient of different units in a plant determine economic generation using the available software package for a total load demand of _____ MW ( typically 150 MW for 2 units and 250 MW for 3 Units) neglecting transmission losses.

15. Given the cost equation and loss Co-efficient of different units in a plant determine penalty factor, economic generation and total transmission loss using the available software package for a total load demand of _____ MW ( typically 150 MW for 2 units and 250 MW for 3 Units) .

Refer below questions for Q.(14) & Q(15)A)



2 0 ≤ P1 ≤ 100

2 C2 = 1000 + 15 P2 + 0.06 P22 0 ≤ P2 ≤ 100

3 C3 = 900 + 18 P3 + 0.07 P32 0 ≤ P3 ≤ 100

77

(2)

(3)

(1)

G1 G2

G3


Loss ( or B) Co- efficients : B11 = 0.0005 B12 = 0.00005 B13 = 0.0002

B22 = 0.0004 B23 = 0.00018 B33 = 0.0005

B21 = B12 B32 = B23 B13 = B31

B)

Loss ( or B) Co- efficients : B11 = 0.005 B12 = B21 = -0.0012 B22 = 0.002

16. Using the available software package determine (a) fault current (b) post fault voltages at all buses (c) fault MVA , when a line to ground fault with / with out Zf = ___ occurring at _____ bus for a given test system.( Assume pre fault voltages at all buses to be 1 pu) compare the results with the symmetrical fault.

17. Using the available software package determine (a) fault current (b) post fault voltages at all buses (c) fault MVA , when a line to line & ground fault with / with out Zf = ___ occurring at _____ bus for a given test system.( Assume pre fault voltages at all buses to be 1 pu) compare the results with the symmetrical fault.

Refer below question for Q.(16) & Q(17)

Reactance of all transformers and transmission lines = 0.1 puZero sequence reactance of transmission line = 0.25 puNegative sequence reactance of generator = Subtransient reactaceZero sequence reactance of generator = 0.1 pu



2 0 ≤ P1 ≤ 100

2 C2 = 30 + 12 P2 + 0.025 P22 0 ≤ P2 ≤ 100

G1

X’’= 0.3

G3

X’’= 0.1

G2

X’’= 0.2

78

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