CYCLE I

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PSS Laboratory Manual

Department of E & EE, NIEIT, Mysore-18 Page 1

CYCLE I

Exercise 01

Aim: A) To familiarize with General Purpose MATLAB Commands.

1. Write a program to determine vector X for the matrix equation AX=B.

2. Find the Eigen values of matrix A

3. Evaluate the following function V= ZC cosh( g) + sinh (g/ZC), where ZC = 200+ j300 and

g= 0.02+ j1.5.

4. In the circuit shown below, determine the node voltages V1 and V2 and the powerdelivered by the each source.

5. Find the roots of the following polynomial.

6. The roots of a polynomial are -1,-2, -3j4. Determine the polynomial equation.

7. Determine the roots of the characteristic equation of the following matrix.

8. Create a linear X-Y plot for the following variables.

x 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5y 10 10 16 24 30 38 52 68 82 96 123

9. Given A=

and B=

Find C= A+B and D= A-B by writing a program . And Comment on Matrix Type of C and D.

10. Given A=

and B=

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Determine C=AB, D=BA and E= ATBT. What is the relationship between E and D.

11. Given A=

and B=

. Determine

a) C= BT. A. B b) D=C-1 c) E= C. C-1

What type of matrix C is? What type of matrix E is?12. Given A=

Determine B=0.1A + 0.5 U. What type of matrix A is? Show that B is orthogonal matrix.

13. Given A=

and B=

and C=

. Determine D=A- CT. B. C

14. Given A=

=

Determine B= A1- A2. A4-1. A3

15. Given A=

. Show that A is a singular matrix. And Determine its Rank.

16. Given A=

where A1=

and A4=

and A2 and A3 are null matrix. Show

that inverse of C is

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17. Given

A =1 2 0 0 0 0

1 1 0 0 0 0

0 0 11 3 2 0

0 0 2 3 4 0

0 0 1 5 2 0

0 0 0 0 0 5

and B= 1 3 2 5 6 1

4 7 1 3 2 4

3 4 2 6 5 1

4 6 3 1 2 52 6 7 3 8 1

7 2 3 1 4 5

Find C= A.B by Method of matrix partitioning.

18. A =

Determine a) B= (A*)T. What type of matrix A is?

19. Given A= , B= and y= .

Determine X from (A+jB) X* =y. What type of matrices are A and B.

20. Given the partitioned matrix A=

Determine B= A-1 using the formula for the inverse of the partitioned matrix.

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21. Given the matrix.

Determine A2.

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B) To Familiarize Plotting Curves in MATLAB

1. Plot y= Sinx, 0 x 2. Taking 200 linearly spaced points in the given interval.

2. Plot y=e-0.4xSinx, 0 x 4, taking 10, 50 and 100 points in the interval.

3. Plot y= Cos(x) and z=1-(x2/2) +-(x4/24) for 0 x on the same plot.

Function File and Script File

Solve the following Linear Equation.

Soln:Script File: File name solvex.m

A= [5 2*r r; 3 6 2*r-1;2 r-1 3*r];

B= [2;3;5];

det_A=det(A);

inv_A=inv(A);

[E, V]=eig(A);

x=A\B;

>> r=1;

>>solvex

Function File: File name solvexf.m

function(det_A, x)=solvexf(r)

A= [5 2*r r; 3 6 2*r-1;2 r-1 3*r];

B= [2;3;5];

det_A=det(A);

x=A\B;

>>[det_A, y]=solvexf(1);

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Exercise 2:

Aim: To determine ABCD parameters for a given problem to:

i) Form symmetric /T configurationii) Verify AD-BC = 1,iii) Determine co-efficient & regulation.

Software Used: MATLAB ver. R2010a

Theory: The Transmission System can also be assumed to be a four terminal network withtwo input terminals where power enters the network and two output terminals where powerleaves the network.

Let Vs= Sending End Voltage; Is= Sending end current; Vr= Receiving End Voltage; Ir=Receiving end current;

The sending end parameters can be expressed in terms of receiving end parametersthrough the set of parameters known as transmission line parameters or ABCD parameters.

Thus, Vs= AVr+ BIr;

Is= C Vr+ DIr

The transmission network should be linear, passive and bilateral. The parameters A, B, Cand D are complex numbers and are called as generalized circuit constants. The methodwhich is used for analysis of transmission line has influence on these constants.Performance calculation of the line can be done using these constants.

Terms Related to Performance of Transmission Line:

i) Voltage Regulation:% VR= 100* (VNL VFL)/ VFL

But VNL= Vs (as there is no drop) and VFL = VR(on load) , hencepercentage voltage regulation equation becomes

% VR= 100* (Vs VR)/ VR.

ii) Transmission Efficiency:

% Transmission efficiency, =

* 100

receiving end power= VrIrcos(r)sending end power= VsIscos(s)

cos(r)= receiving end power factor

cos(s)=sending end power factor

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Case A: To find constants in medium transmission line represented by Nominal circuit.

In nominal T method of analysis of medium transmission line the total line capacitance isassumed to be lumped or concentrated at the center point of the line whereas the half ofthe line resistance and reactance are lumped on either side of the line. The constants are:

A=1+(Y*Z)/2B=Z*(1+Y*Z/4)C=YD= (1+Y*Z/2)

Case B: To find constants in medium transmission line represented by Nominal Tcircuit.

in nominal method, the total capacitance is divided into two halves with one half at thereceiving end and the other half at the sending end. The constants are:

A=1+(Y*Z)/2B=ZC=Y*(1+Y*Z/4)D= (1+Y*Z/2)

Problems:

1.A 200km long 3-phase overhead line has a resistance of 48.7 ohms per phase , inductivereactance of 80.20 ohms per phase and capacitance (line to neutral) 8.42pF per km. Itsupplies a load of 13.5 MW at a voltage of 88kV and power factor 0.9 lagging. Using

nominal and nominal T circuit, find the sending end voltage, current, regulation.2. A 3 phase 50Hz 100km long overhead line has the following line constants,R=0.153/phase/km, L=1.21mH/phase/km and C=0.00958microfarad/phase/km. The linesupplies a load of 20MW at 0.9pf lagging at a line voltage of 110kV at the receiving end.Using nominal and nominal T representation calculate sending end voltage, current,power factor, regulation and efficiency.

3. A 345kV, three phase transmission line is 130km long has the line constants,R=0.036/phase/km, L=0.8mH/phase/km and C=0.0112microfarad/phase/km. The receivingend load is 270 MVA with 0.8pf lagging at 325kV. Use the medium line model to find thevoltage and power at the sending end and the voltage regulation.

4. A 345kV, three phase transmission line is 130km long. The series impedance is z=0.036+j0.3 ohms per phase per km, and shunt admittance is y=j4.22x10 -6 seimens per phase perkm. The sending end voltage is 345kV, and the sending end current is 400A at 0.95 pflagging. Use the medium line model to find the voltage and power at the recieving end andthe voltage regulation.

Input required:

Line Type, Line length, line impedance, shunt admittance, (Vs, Is) or ( Vr, Ir).

Output Expected:

% Voltage Regulation, % efficiency, A, B, C and D,

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MATLAB CODE:

% abcd parameterclczreal=input('enter impedance(real)/phase/km:')zimag=input('enter impedance(imag)/phase/km:')z=zreal+i*zimag

yreal=input('enter shunt(real)admittance/phase/km:')yimag=input('enter shunt(imag)admittance/phase/km:')y=yreal+i*yimag

k1=input('enter 1-for short line 2-for medium line 3-for long line:')

switch(k1)

case 1,length=input('enter length of short transmission line:')

Z=z*length;Y=y*length;A=1B=ZC=0D=1

case 2,length=input('enter length of medium transmission line:')Z=z*length;Y=y*length;

A=1+(Y*Z)/2B=Z*(1+Y*Z/4)C=YD=(1+Y*Z/2)

case 3,length=input('enter length of long transmission line:')zc=sqrt(z/y);gam=sqrt((z*y)*length);A=cosh(gam)D=AC=1/zc*sinh(gam)B=zc*sinh(gam)

otherwisedisp('wrong choice of transmission line')

endfprintf('\n the product AD-BC=%f',A*D-B*C')

k2=input('\n enter 1-to read Vr Ir and compute Vs Is 2-to read Vs Is and compute Vr Ir:');

switch(k2)case 1,vr=input('enterreceiving end voltage in KV:');Vr=vr*1e3/sqrt(3);

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ir=input('receiving end current in KA:');Ir=ir*1e3;Vs=(A*Vr+B*Ir)Is=(C*Vr+D*Ir)

case 2,vs=input('enter sending end voltage in KV:');

Vs=vs*1e3/sqrt(3);is=input('sending end current in KA:');Is=is*1e3;Vr=(D*Vs-B*Is)Ir=(-C*Vs)+(D*Is)

otherwisedisp('wrong choice')

endrp=3*real(Vr*conj(Ir))/1e6;sp=3*real(Vs*conj(Is))/1e6;eff=(rp/sp)*100;

reg=((abs(Vs)-abs(Vr))/abs(Vr))*100;

%fprintf('\n receiving end power=%0.2fMW',rp)%fprintf('\n sending end power=%0.2fMW',sp)%fprintf('\n efficiency=%0.2f',eff)%fprintf('\n regulation=%0.2f',reg)regeffrpsp

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Exercise 3:

Aim: To draw the power angle diagrams for salient and non-salient pole synchronousmachines and to determine reluctance power, excitation emf & regulation.

Software: MATLAB ver. R2010a or MiPower

Theory:The equation relating the electrical power generated to the angular displacement of therotor is called as the power angle equation. A graphical Plot showing the variation pfelectrical power E against the load angle for fixed values of E, V and reactance is calledpower-angle curve. The power- angle relationship plays a vital role in the solution of theswing equation.

Non-Salient Pole Machines:

This type of machines has approximately equal magnetic reluctance, regardless of theangular position of the rotor, with respect to the armature mmf.

P= (3*E*V*sin)/Xd

P=Electrical power;

E=Generator EMF;

V=Terminal Voltage;

= Machine Angle;

Xd=Machine Direct axis reactance;

Salient Pole Machines:

P= 3*{(E*V*sin)/Xd + ( V2 * (Xd-Xq) * sin2)/(2*Xd*Xq)}

The First term in the above equation constitutes major power which is similar to non- salientpole machines with Xs = Xd. The second term is known as reluctance power, which does notdepend on the excitation and makes the maximum power greater than in the classicalmodel. Here, because of the salient poles, the reluctance of the magnetic circuit in whichflows the flux produced by an armature mmf in line with the quadrature axis is higher thanthat of the magnetic circuit in which flows the flux produced by the armature mmf in line withthe direct axis.

Reluctance Power=3* ( V2 * (Xd-Xq) * sin2)/(2*Xd*Xq)}

Xd=Machine Direct axis reactance;

Xq=Machine quadrature axis reactance;

Excitation EMF: |E|=|Eq|+(Xd-Xq)*|Id|

Id=|I| * sin(-)

= Power Factor Angle

Eq=V + j* Xq* I

I=MVA/(1.731 * V)

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% Regulation=100 * (|E|-|V|)/|V|

Problems:

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

Input required:

1. Power in MW

2. Power factor

3. Line to Line Voltage level in kV

4. Xd in ohms

5. Xq in ohms

Output Expected:

1. Excitation e.m.f in kV

2. Regulation in %

3. Plot of Delta v/s Reluctance Power for Salient pole & Non salient pole m/cs.

A) Determination of power angle diagrams using MATLAB CODE:

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 MotorEq=Vtph+(j*I*Xq);Id_mag=abs(I)*sin(angle(Eq)-angle(I));

Ef_mag=abs(Eq)+((Xd-Xq)*Id_mag);Exitation_emf=Ef_magReg=(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 onplot(delta,Net_Power_sal,'r');xlabel('\Delta(deg)-------->');ylabel('Three Phase Power(pu)-------->');title('Plot:Power Angle Curve for Salient Synchronous M/c');

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legend('Reluct Power','Salient Power');end

if Xd==Xq%Non-Salient Pole Synchronous MotorEf=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');

endgrid;

B) Determination of power angle diagrams using MiPower package:

Procedure to enter the data for performing power angle diagram studies using MiPower:

1. To solve power angle curve by using MiPowerPackage, invoke Tools in the MiPower

main screen and select Power Angle Curve.

2. Select the Save option.

3. Select the location to save the file and give the file name.

4. Enter the values of Terminal Voltage, Terminal Angle, Transient Reactance, Resistance,

Reactance, Susceptance & Infinite bus voltage magnitude, angle.

5. Click on Execute. Output file will appear.

6. To plot the graph, select Graph.

7. In the graph tool bar go to Window, click on Show 2nd Pane.

8. Select angle as X-axis parameter & Power as Y-axis parameter & click on Plot to plot the

curve.

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Exercise 4:

Aim: To Plot Swing Curve and To determine Critical Clearing time for SMIB through apair of identical transmission lines for a 3-phase fault on one of the lines for variationof inertia constant/Line Parameters/Fault Location/Clearing Time/Pre-Fault ElectricalOutput.

Software: MATLAB ver. R2010aTheory:

Swing Equation describes the relative motion of the rotor (load angle or torque angleor power angle ) with respect to the stator field as a function of time. It is the fundamentalequation governing the rotor dynamics of the synchronous machine. The solution of swingequation gives the relation between rotor angle as a function of time t. Normally, it issolved in digital computers using step-by- step method or employing numerical solutiontechniques like Eulers method or Runge-Kuttas method. The Plot of versus t is called asthe swing curve. For simple systems like single machine connected to infinite bus or a twomachine system, it is not necessary to solve the swing equation for finding the transient

stability. It can be conveniently determined using the method known as Equal AreaCriterion. Swing curves are useful in designing the protective devices for the system. Evenin an SMIB system, we have to resort to a numerical technique to evaluate the variation of with time and to determine CCT. All numerical methods use the concept of discretizationof the variables, over suitable time intervals.

Procedure:

1. Excitation EMF E and transfer reactance X0 between line generators and infinite bus aredetermined for the specified output of the generator taking infinite bus voltage V asreference.

2. Pre-fault power characteristics is determined as

Pa = ( |E|* |V| * sin) / X0

Where E and V are magnitude of excitation EMF and voltage of infinite bus.

3. Pre-fault power angle is obtained as

0 = sin-1 (Pmech / Pm)

where Pmech is electrical output of generator before fault.

4. For the specified fault location the new transfer reactance X1 is determined assumingconstant excitation EMF and infinite bus voltage. The P- characteristics during the fault isobtained as

P1 = ( |E|* |V| * sin) / X1 = P1m sin

5. For the system configuration after the isolation of faulty line, the transfer reactance X 2and the corresponding post fault. The P- characteristics during the fault is obtained as

P2 = ( |E|* |V| * sin) / X2 = P2m sin

6. The total time of transient stability study T, time at the instant of fault clearance tc, inertiaconstant H of the generator and normal system frequency f are all identified from thesystem data.

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7. Critical clearing time is determined from the equation

8. For determining critical clearing time, solution for swing equation is obtained forsustained fault using point by point method for above equations. Critical clearing time istaken for the time corresponding to c.

9. For calculation of swing curve for sustained fault, it is enough to assume tc >T

for ex: tc =T+0.01s

10. For obtaining swing curve when the fault is cleared the procedure is similar, i.e.,modified Euler method is applied to equations. to obtain incremental value of rotor swing

during successive time steps and the rotor swing at the end of respective slips. It isimportant to use Pc= P1 for intervals before fault clearance and Pc= P2. for intervals afterfault clearance.

11. A plot of vs t gives the swing curve in both cases (sustained fault /fault cleared).

12. Procedure is repeated for different values of inertia constant, fault location, fault clearingtime, line reactance and pre-fault electrical output to study their effect on swing curve bychanging the value of one of them at a time keeping other constant.

Problems:

Input required:

1. Inertia constant H

2. Pre fault reactance Xpre in p.u

3. Reactance during fault Xfault in p.u

4. Post fault reactance Xpost in p.u

5. Sending end voltage E in p.u

6. Receiving end voltage V in p.u

7. MVA

8. MW

9. Initial displacement angle

10. Fault clearing time in sec

Output Expected:

Plot of rotor angle versus time

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1. A generator operating at 60Hz delivers 1pu power to an infinite bus through a pair oftransmission lines in which resistance is ignored. A fault takes place in the system, reducingthe maximum power transfer to 0.4 pu where as before the fault this power is 2.0pu. andafter the clearance of the fault it is 1.5pu. By the use of equal area criterion, determine thecritical clearing angle. If the machine constant is 1 and the inertia constant is 7. Obtain theswing curve step by step with the change in time of 0.05s. Repeat the same problem, whenthe power after the clearance of the fault is 0.5pu.

2. Find the critical clearing angle for the system shown below for a three phase fault at apoint P. The generator is delivering 1.0pu power under prefault condition.

3. The figure below represents the single line diagram of a generator connected to aninfinite bus through a pair of transmission lines. The transient reactance of the generator isincluded in the diagram . Assume a 3 phase fault when generator is delivering 1pu power.Find the critical clearing angle assuming the voltage behind transient reactance to be 1.2 pu

and that of the infinite bus to be 1pu.

4. The single line diagram below shows a generator connected through paralleltransmission lines to a large metropolitan system considered as an infinite bus. Themachine is delivering1.0pu power and both the terminal voltage and the infinite bus voltageare 1.0 p.u. calculate the critical clearing angle and clearing time when the system issubjected to a 3 phase fault at point p on the short transmission line . Take H=5MJ/MVA.

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MATLAB CODE for solution of swing equation by point by point method:clcswdat=[2 20 50 18 2.52 0.35 0 0.02 1.25 1.1 1 2.5 .025 .5];ckt=swdat(1); mva=swdat(2);f=swdat(3);mw=swdat(4);h=swdat(5);xdd=swdat(6);r=swdat(7);x=swdat(8);xdur=swdat(9);ed=swdat(10);v=swdat(11);fclear=swdat(12);delta_t=swdat(13);tmax=swdat(14)m=h/(180*f);xpre=xdd+x/ckt;xpost=xdd+x;

tclr=fclear*1/f%tclr=tmax+0.01pm=mw/mvappre=ed*v/xprepfault=ed*v/xdurppost=ed*v/xpostdelta(1)=asin(pm/ppre)*180/pi;i=1;del_delta=0;for t=0:delta_t:(tmax-delta_t)

if t==0,paa=0;pab=pm-pfault*sin(delta(i)*pi/180);pa=(paa+pab)/2;

elseif t==tclr,paa=pm-pfault*sin(delta(i)*pi/180);pab=pm-ppost*sin(delta(i)*pi/180);pa=(paa+pab)/2;

elseif t>tclr,pa=pm-ppost*sin(delta(i)*pi/180);

else pa=pm-pfault*sin(delta(i)*pi/180);end

del_delta=del_delta+pa*delta_t^2/m;delta(i+1)=delta(i)+del_delta; i=i+1;

endt=[0:delta_t:tmax];plot(t,delta)xlabel('time in seconds'); ylabel('delta in degrees');[t,'delta']delta_max=pi-asin(pm/ppost);delta_0=delta(1)*pi/180;pcr1=pm*(delta_max-delta_0);pcr2=pfault*cos(delta_0);pcr3=ppost*cos(delta_max);

delta_crt=(acos(pcr1-pcr2+pcr3)/(ppost-pfault))*180/pi;

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Exercise 5:

Aim: To determine fault currents and voltages in a single transmission line systemswith star-delta transformers at a specified location for SLGF, DLGF.

Software: MATLAB ver. R2010a

Theory:A fault in a circuit is any failure, which interferes with the normal flow of control. the faultsoccurs in power system due to insulation failure of equipments, flashover of lines initiatedby a lightning stroke, due to permanent damage to conductors and towers or due toaccidental faulty operations. The faults can be broadly classified into shunt faults and seriesfault. The shunt type of faults involves short circuit between conductor and ground or shortcircuit between two or more conductors. The shunt faults are characterized by increase incurrent and fall in voltage and frequency. The shunt faults can be classified as shownbelow.

1. Line to ground fault

2. Line to line fault

3. Double line to ground fault

4. Three phase fault

The Series fault may occur with one or two broken conductors, which create open circuits. Italso happens in circuits controlled by fuses or breakers, which do not open all the threephases, i.e., one or two phases of the circuit may open and the other phases may beclosed. The series faults are characterized by increase in voltage and frequency and fall incurrent. The series faults can be classified as one conductor fault and two open conductorfault.

Problems:

1. Three phase generator with an open circuit voltage of 400V is subjected to an LG faultthrough a fault impedance of j2 ohms. Determine the fault current if Z 1= j4 ohms, Z2=j*2ohms and Z0=j1 ohms. Repeat the problem for LL and LLG fault.

2. A double line to ground fault occurs at point p on the transmission line. Determine thefault current such that both machines are rated 1250KVA, 600V at UPF with X1= X2=10%and X0=4% . Each 3 phase transformer is rated 1250KVA, 600/4160V -Y with leakage

reactance of 5%. The reactances of transmission line are X1= X2=15% and X0 =50% on abase of 1250kVA and 4.16kV. Neglect the pre-fault current.

3. A synchronous motor is receiving 10MW of power at 0.8pf lag at 6kV. An LG fault takesplace at the middle point of the transmission line as shown in fig. Find the fault current. Theratings of the generator, motor and transformer are as under:

Generator: 20MVA, 11kV, X1= 0.2pu, X2=0.1 pu , X0=0.1 pu.

Transformer T1: 18MVA, 11.5/34.5kV Y-Y, X=0.1pu

Transformer T2: 15MVA, 6.9/34.5kV Y-Y, X=0.1pu

Transmission line: X1=X2=5 ohms, X0=10ohms

Motor: 15MVA, 6.9kV, X1= 0.2pu, X2=0.1 pu , X0=0.1 pu.

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Input required: Generator, transformers and motor rating.

Output Expected: Fault current.

MATLAB CODE for Ex 3

data=[20 6 10 .8 0 20 18 15 20 15 11.5 34.5 6.9 34.5 6.9 .2i .1i .1i .1i .1i .1i .1i .1i .1i 5i 5i 10i .2i .1i .1i 11 6.633];

mva_base=data(1);vm_actual=data(2);pm=data(3);pf=data(4);zf=data(5);mva_base_old_g=data(6);mva_base_old_t1=data(7);mva_base_old_t2=data(8);mva_base_old_tl=data(9);mva_base_old_m=data(10);kv_base_lv_t1=data(11);kv_base_hv_t1=data(12);kv_base_lv_t2=data(13);

kv_base_hv_t2=data(14);kv_base_old_m=data(15);xg1=data(16);xg2=data(17);xg0=data(18);xt11=data(19);xt12=data(20);xt10=data(21);xt21=data(22);xt22=data(23);xt20=data(24);xtl1=data(25);xtl2=data(26);

xtl0=data(27);xm1=data(28);xm2=data(29);xm0=data(30);kv_base_new_g=data(31);kv_base_new_m=data(32);kv_base_new_tl=data(33);

% seq reactaces of TLxtl1=data(25)*mva_base/(kv_base_new_tl*kv_base_new_tl);xtl2=xtl1;xtl0=data(27)*mva_base/(kv_base_new_tl*kv_base_new_tl);% seq reactaces of tr1xt11=xt11*(mva_base/mva_base_old_t1)*((kv_base_lv_t1/kv_base_new_g)^2);xt12=xt11;xt10=xt12;% seq reactaces of tr2xt21=xt21*(mva_base/mva_base_old_t2)*((kv_base_lv_t2/kv_base_new_m)^2);xt22=xt21;xt20=xt22;% seq reactaces of motorxm1=xm1*(mva_base/mva_base_old_m)*((kv_base_old_m/kv_base_new_m)^2);xm2=xm2*(mva_base/mva_base_old_m)*((kv_base_old_m/kv_base_new_m)^2);xm0=xm0*(mva_base/mva_base_old_m)*((kv_base_old_m/kv_base_new_m)^2);%psr, nsr and zsr of systemz1=((xg1+xt11+(xtl1)/2)*((xtl1)/2+xt21+xm1))/(xg1+xt11+(xtl1)/2+(xtl1)/2+xt21+xm1); z2=((xg2+xt12+(xtl2)/2)*((xtl2)/2+xt22+xm2))/(xg2+xt12+(xtl2)/2+(xtl2)/2+xt22+xm2); z0=((xg0+xt10+(xtl0)/2)*((xtl0)/2+xt20+xm0))/(xg0+xt10+(xtl0)/2+(xtl0)/2+xt20+xm0); im=(pm*1e6/(1.73*vm_actual*1e3*pf))*exp(i*-acos(pf)); im_pu=im/(mva_base*1e6/(1.73*6.6*1e3));

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vm_pu=vm_actual/kv_base_new_m;va1=vm_pu+im_pu*((xtl1)/2+xt21);fault_type=input(' 1-SLGF and 2-DLGF');switch(fault_type)

case 1ia0=va1/(z1+z2+z0+3*zf);ia1=ia0;ia2=ia1;

If=3*abs(ia0);case 2ia1=(va1/(sqrt(3)))/(z1+(z2(3*zf+z0)/(z2+3*zf+z0)));ia2=-ia1*(z0+3*zf)/(z2+3*zf+z0);ia0=-ia1*z2/(z2+3*zf+z0);If=-3*abs(ia0);

end

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CYCLE II

Exercise 6(a):

Aim: Formation of Y Bus for a given power system without having line chargingadmittances by inspection method.

Theory:

The formulation of a suitable mathematical model is the first step in the analysis ofan electrical network. A network matrix equation provides a convenient mathematical modelfor a numerical solution. The elements of a network matrix will be either impedance oradmittance, depending on the selection of the independent variables, which can be eithercurrents or voltages.

Inspection Method:

In this method the admittance matrix may be assembled as follows:

1. The diagonal element of each node is the sum of admittances connected to it.

2. The off-diagonal element is the negated admittances between the nodes.

This method is applicable only if there is no mutual coupling between elements.

Input Required:

System Specification data:

no. of buses

no. of series elements

no. of shunt admittances

Element Data:

For Each Series Element

From Bus no.

To Bus no.

resistance in pu

reactance in pu

shunt admittance B/2 in pu

For each shunt element

Bus number

resistance in pu

reactance in pu

Output Expected: YBUS Matrix

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Power system network:

Power System data:

Bus Code p-q Impedance (z ) in pu

1-2 0.25+j1.0

1-3 0.2+j0.8

1-4 0.3+j1.2

2-3 0.2+j0.8

4-3 0.15+j0.6

MATLAB program for formation of Ybus for a given power systemwithout having line charging admittances by inspection method:

clcclear alldata=[4 5 0 0;

1 2 0.25 1;1 3 0.2 0.8;1 4 0.3 1.2;

2 3 0.2 0.8;4 3 0.15 0.6];

nb=data(1,1);nl=data(1,2);for i=1:nlsb(i)=data(i+1,1);eb(i)=data(i+1,2);serz(i)=complex(data(i+1,3),data(i+1,4));end

for j=1:nbfor k=1:nb

ybus(j,k)=0;

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endendfor i=1:nl

j=sb(i);k=eb(i);ybus(j,j)=ybus(j,j)+1/serz(i);

ybus(k,k)=ybus(k,k)+1/serz(i);ybus(j,k)=ybus(j,k)-1/serz(i);ybus(k,j)=ybus(j,k);

endybus

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Exercise 6(b):

Aim: Formation of Y Bus for a given power system with having line chargingadmittances by inspection method.


Power System data:

Bus code Impedance Line charging

admittances (ypr/2)

1-2 0.06+j0.18 j0.05

1-3 0.02+j0.06 j0.06

2-3 0.04+j0.12 j0.05

MATLAB program for formation of Ybus for a given power systemwithhaving line charging admittances by inspection method:

clcclear alldata=[1 2 0.06 0.18 0.05;

1 3 0.02 0.06 0.06;2 3 0.04 0.12 0.05];

sb=data(:,1);eb=data(:,2);nl=max(size(sb));nb=max(max(sb,eb));serz=complex(data(:,3),data(:,4));shty=complex(0,data(:,5));ybus=zeros(nb,nb);for i=1:nl

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j=sb(i);k=eb(i);ybus(j,j)=ybus(j,j)+1/serz(i)+shty(i);ybus(k,k)=ybus(k,k)+1/serz(i)+shty(i);ybus(j,k)=-1/serz(i);ybus(k,j)=ybus(j,k);

endybus

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Exercise 7

Aim: Formation of Ybus for a given power system with having mutual couplingimpedances by singular transformation method.

Theory

The electrical characteristics of the individual network components can be

represented conveniently in the form of a primitive network matrix. It is necessary thereforeto transform primitive network matrix that describes the performance of the interconnectednetwork. The form of the network matrix used in the performance equation depends on theframe of reference namely bus or loop. Normally the bus frame of reference is employed.

If mutual coupling is present in any system, the bus admittance matrix has to be obtainedby suitable transformation of the primitive network matrices.

YBUS=A.[y].A

where A is the Bus incidence matrix and [y] is the primitive admittance matrix.

Input Required:

System Specification data:

no. of buses

no. of series elements

no. of shunt admittances

Element Data:

For Each Series Element

From Bus no. To Bus no.

resistance in pu

reactance in pu

shunt admittance B/2 in pu

For each shunt element

Bus number

resistance in pu

reactance in pu

Output Expected: YBUS Matrix

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Power System data:

Element

number

Self Mutual

Bus code

p-q

Impedance

zpq,pq

Bus code

r-s

Impedance

zpq,rs

1 1-2(1) 0.6 --- ---

2 1-3 0.5 1-2(1) 0.1

3 3-4 0.5 --- ---

4 1-2(2) 0.4 1-2(1) 0.2

5 2-4 0.2 --- ---

Consider bus 1 as reference bus.

MATLAB program for formation of Ybus for a given power systemwith

having mutual coupling impedances by singular transformation method:

clcclear alldata=[1 2 0.6 0 0;

1 3 0.5 1 0.1;3 4 0.5 0 0;1 2 0.4 1 0.2;2 4 0.2 0 0];

fb=data(:,1);tb=data(:,2);z=data(:,3);mc=data(:,4);

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mz=data(:,5);nnode=max(max(fb),max(tb));nbus=max(max(fb),max(tb))-1;nline=length(fb);zpr=zeros(nline,nline);for k=1:nline

zpr(k,k)=z(k);if mc(k)~=0

zpr(k,mc(k))=mz(k);zpr(mc(k),k)=mz(k);

endendypr=inv(zpr);Acap=zeros(nline,nnode);for k=1:nline

Acap(k,fb(k))=1;Acap(k,tb(k))=-1;

endA=Acap(:,2:nnode);ybus=A'*ypr*A;ybus

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Exercise 8

Aim: Determination of bus currents, bus power, line flows and line losses for a

specified power system network.

Theory:

IBUS=YBUS. EBUS ; Bus Currents

Iik=Yik.(Vi-Vk) ; Line Currents

Sik=Vi.(Iik)* ; Line Flows

SLik= Sik+ Ski ; Line Losses

SBUS= VBus. (IBUS)* ; Bus Powers


Power system data:

Line Impedance Bus Voltage

Bus code Impedance Bus No Voltage

1-2 0.02+j0.04 1 1.05+j0.00

1-3 0.01+j0.03 2 0.98-j0.06

2-3 0.0125+j0.025 3 1.00-j0.05

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MATLAB program for determination of bus currents, bus power, line flows and

line losses for a specified power system network:

clcclear all

data=[1 2 0.02+0.04i;1 3 0.01+0.03i;2 3 0.0125+0.025i];

vb=[1.05+0.00i;0.98-0.06i;1-0.05i];

fb=data(:,1);tb=data(:,2);z=data(:,3);

nl=max(size(fb));y=1./z;for k=1:nl

il(fb(k),tb(k))=y(k)*(vb(fb(k))-vb(tb(k)));il(tb(k),fb(k))=-il(fb(k),tb(k));

endfprintf('The line currents\n')ilfor k=1:nl

lf(fb(k),tb(k))=vb(fb(k))*conj(il(fb(k),tb(k)));lf(tb(k),fb(k))=vb(tb(k))*conj(il(tb(k),fb(k)));ll(k)=lf(fb(k),tb(k))+lf(tb(k),fb(k));

endfprintf('The line flows\n')lffprintf('The line losses\n')llfor k=1:nl

bp(k)=sum(lf(k,:));ibus(k)=conj(bp(k)/vb(k));

endfprintf('The bus power\n')bpfprintf('The bus currents\n')ibus

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Exercise 9

Aim: To determine i) swing curve ii) critical clearing time for a single machineconnected to infinite bus through a pair of identical transmission lines, for a 3sustained fault on one of the lines.


System data:

Mechanical power input, Ps 0.9 pu

Induced emf of the machine, E 1.1 pu

Terminal voltage of the infinite bus, V 1 pu

Angular momentum of the machine, M 0.00028 pu

Transient reactance of the machine, Xt 0.35 pu

Series reactance of transmission lines, Xe 0.2 pu

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MATLAB program to determine i) swing curve ii) critical clearing time for a singlemachine connected to infinite bus through a pair of identical transmission lines, fora 3 sustained fault on one of the lines.

clc

clear all

ps=0.9;e=1.1;v=1;m=0.00028;xt=0.35;xe=0.2;x1=xt+xe/2;

ch=input('enter 1 for fault at the beginning & 2-for fault at the middle: ');

switch(ch)

case 1,

x2=inf;

case 2,

x2=(xt*xe+xt*xe/2+xe*xe/2)/(xe/2);

otherwise

disp('wrong input');end

dt=0.05;

it=1;

t(it)=0;

del_d=0;

pm1=e*v/x1;

del(it)=asin(ps/pm1);

pm2=e*v/x2;

pe=pm2*sin(del(it));

pa=(ps-pe)/2;

while(t(it)

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delm=pi-del(1);

cdc=(ps*(delm-del(1))+pm3*cos(delm)-pm2*cos(del(1)))/(pm3-pm2);

delc=acos(cdc);

D=delc*(180/pi);

it=1;

while(t(it)=delc)

break;

end

it=it+1;

end

T=t(it);

fprintf('critical clearing angle=%f degree\n',D)

fprintf('critical clearing time=%f sec\n',T)

CYCLE I

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