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ANNASAHEB DANGE COLLEGE OF ENGINEERING AND TECHNOLOGY ASHTA

Power System Stability & Control Laboratory Manual FOR TE – II, Semester – VI, ELECTRICAL ENGINEERING

Editor Iranna Korachagaon Assistant Professor

Department of Electrical Engineering

January 2011, Second Edition February 2010, First Edition

As per Shivaji University Syllabus effective from the year 2009-10

Power System Stability and Control Laboratory Manual

Syllabus Power System Stability and Control T.E. -II (Electrical Engineering) (Semester- VI) Teaching Scheme: Lectures: 04 Hours/Week Practical: 02 Hours/Week

Examination Scheme: Paper: 100 Marks T.W.: 25 Marks

SECTION - I 1. Symmetrical fault analysis: (04 Hrs.): Short circuit transients on transmission line, short circuit currents and reactance of a Synchronous Machine, Internal voltages of loaded Synchronous machine under transient conditions, Numericals expected. 2. Symmetrical Components: (07 Hrs.): Fundamentals of Symmetrical Components, sequence impedances and sequence networks of Synchronous machine, star connected loads, transmission lines and transformer. 3. Unsymmetrical fault analysis: (06 Hrs.): Analysis of Single Line to Ground (LG) fault, Line-ToLine (LL) fault, Double-Line-To-Ground (LLG) fault, One conductor open fault, Two conductors open fault, Numericals expected. 4. Power System Control: (07 Hrs.): Load frequency control (Single and two area), modeling of Generator, Governor, prime mover, Load, Load frequency control and economic dispatch, Automatic generation control, Steady state analysis and dynamic response of an isolated power system, Automatic voltage control, reactive power control. SECTION - II 5. Optimal Power System Operation: (08 Hrs.): System constraints, Generator operating cost, Inputoutput and incremental fuel characteristics of a generating unit, optimal operation of generators on a bus bar, algorithm and flow chart for optimal power flow study, optimal unit commitment, spinning reserve, thermal and hydro constraints, Numericals expected. 6. Power System Stability: (12 Hrs.): Dynamics of Synchronous machine, Swing equation for single machine connected to infinite bus, Steady state stability and transient state stability, Equal area criterion, Numerical solution of swing equation, factors affecting transient stability, methods for improving stability of system. Voltage stability analysis, mathematical formulation, voltage collapse, Numericals expected. 7. Power system Security: (04 Hrs.): Brief Introduction to- System state classification, Security analysis, Contingency analysis, Sensitivity actors.


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Term Work: The laboratory exercise consists of (minimum 08 exercises): • Modeling and Simulation of Problems based on theoretical data. • This simulation is to be carried out using software like MiPower, PSIM, ETAP, EMTP, MATLAB.

Reference Books: 1. Modern Power System Analysis by I. J. Nagrath, D. P. Kothari, 3rd Edition, Tata McGraw Hill Publishing Co. Ltd., 2003 2. Electrical power System by Ashfaq Husain, CBS Publishers and Distributors, Fifth Edition 2007 3. Power System Analysis by Grainger John J and W D Stevenson Jr. McGraw,Hill, 1994. 4. Power System Analysis by Hadi Sadat, McGraw Hill International, 1999. 5. Power System Analysis and Design, Third Edition by J. Duncan Glover and Mulukuta S. Sarma, Prentice Hall, 2002 6. Power System Analysis by A.R. Bergen and Vijay Vittal, 2nd edition, Pearson Education 7. Computer Methods in Power System Analysis - M.A. Pai


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Power System Stability and Control Laboratory T.E.-II, (Electrical Engineering) (Semester- VI)

List of Experiments

Date Conducted

Remarks

1. MATLAB Primer

………………..

………………..

2. Transients on a Transmission Line

………………..

………………..

3. Symmetrical Short Circuit on a Synchronous Machine

………………..

………………..

4. Symmetrical Components from Unbalance Currents

………………..

………………..

5. Unbalanced Voltages from Symmetrical Components

………………..

………………..

6. Optimal Power Dispatch

………………..

………………..

7. Transient Power-angle Curve for Synchronous Machine

………………..

………………..

8. Equal –area Criterion

………………..

………………..

9. Simulating RLC series Circuit (Out of Syllabus)

………………..

………………..

10. SIMULINK Basics

………………..

………………..

11. LFC (Load Frequency Control) with Simulink

………………..

………………..

This is to certify that, Mr./Ms. ______________________________________________________________ Roll No. ___________ of _______________________ class has completed satisfactorily the ____________ Experiments in ___________________________________ during the year __________________________ Date:

/

/20

____________ Subject in-charge


_____________ H.O.D.

______________ Principal

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Experiment 1: MATLAB Primer Practice the following examples. >> a = 2 a = 2 >> x = [1;2;3] x = 1 2 3 » x = [1 2 3 4 5] x = 1 2 3 4 5 >> A = [1 2 3;4 5 6;7 8 0] A = 1 2 3 4 5 6 7 8 0 » y = [6;7;8;9;10] y = 6 7 8 9 10 » y = [6,7,8,9,10] y = 6 7 8 9 10 » y' ans = 6 7 8 9 10 » u = [0:8] u = 0 1 2 3 4 5 6 7 8 » v = [0:2:8] Department of Electrical Engineering

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v = 0 2 4 6 8 » v(1:3) ans = 0 2 4 » v(1:2:4) ans = 0 4 » sin(pi/2) ans = 1 » cos(pi/2) ans = 6.1230e-017

» x = [0:0.1:1] x = Columns 1 through 7 0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 Columns 8 through 11 0.7000 0.8000 0.9000 1.0000 » y = exp(x) y = Columns 1 through 7 1.0000 1.1052 1.2214 1.3499 1.4918 1.6487 1.8221 Columns 8 through 11 2.0138 2.2255 2.4596 2.7183 » A 9 0 7

A = 7 8 1

= [9,7,0;0,8,6;7,1,-6] 0 6 -6

» size(A) ans = 3 3 » det(A) ans = -192 » inv(A) Department of Electrical Engineering

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ans = 0.2812 -0.2187 -0.2187 -0.2187 0.2812 0.2812 0.2917 -0.2083 -0.3750 » A*inv(A) ans = 1.0000 0.0000 0.0000 0.0000 1.0000 0.0000 0.0000 0.0000 1.0000 » inv(A)*A ans = 1.0000 0.0000 0 0.0000 1.0000 0 0.0000 0 1.0000 » eig(A) ans = 12.6462 3.1594 -4.8055 » x=-pi:0.01:pi; » y=cos(x); » plot(x,y)


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» xlabel('x') » ylabel('y=cos(x)') » title('Graph of cosine from - \pi to \pi') » title('Graph of cosine from -pi to pi')

Various line types, plot symbols and colors can be used. If these are not specified (as in the case above) MATLAB will assign (and cycle through) the default ones as given in the table below. y yellow . point m magenta o circle c cyan x x-mark r red + plus g green - solid b blue * star w white : dotted k black -. dashdot -- dashed So, to obtain the same graph but in green, we type » plot(x,y,’g’) where the third argument indicating the color, appears within single quotes. We could get a dashed line instead of a solid one by typing » plot(x,y,’--’) or even a combination of line type and color, say a blue dotted line by typing » plot(x,y,’b:’) Multiple curves can appear on the same graph. If for example we define another vector » z = sin(x); we can get both graphs on the same axis, distinguished by their line type, using Department of Electrical Engineering

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» plot(x,y,'r--',x,z,'b:') The resulting graph can be seen below, with the red dashed line representing y = cos(x) and the blue dotted line representing z = sin(x).

When multiple curves appear on the same axis, it is a good idea to create a legend to label and distinguish them. The command legend does exactly this. » legend('cos(x)','sin(x)') The text that appears within single quotes as input to this command, represents the legend labels. We must be consistent with the ordering of the two curves, so since in the plot command we asked for cosine to be plotted before sine, we must do the same here.


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At any point during a MATLAB session, you can obtain a hard copy of the current plot by either issuing the command print at the MATLAB prompt, or by using the command menus on the plot window. In addition, MATLAB plots can by copied and pasted (as pictures) in your favorite word processor (such as Microsoft Word). This can be achieved using the Edit menu on the figure window.


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Experiment 2: Transients on a transmission line Consider a transmission line model with a simple series RL circuit. Let R=0.125 ohms, L=10mH, and the source voltage is given by v(t) = 151 sin (377t+α). Determine the current response after closing the switch for the following cases. (a) No dc offset (b) For maximum dc offset [Ref: Example 8.1, Power System Analysis, Haadi Sadat]


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Now create a file name ‘Experiment2.m’ and save the following. Execute it. gamma=88.1*pi/180; alpha1=88.1*pi/180; alpha2=1.9*pi/180; t=0:0.001:0.3; i1=40*sin(377*t+alpha1 -gamma)-40*exp(-t/.08).*sin(alpha1-gamma); i2=40*sin(377*t+alpha2 -gamma)-40*exp(-t/.08).*sin(alpha2-gamma); subplot(2,1,1), plot(t, i1), grid xlabel('t, sec'), ylabel('i(t)') subplot(2,1,2), plot(t, i2), grid xlabel('t, sec'), ylabel('i(t)') subplot(111);

The output is


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Experiment 3: Symmetrical short circuit on a synchronous machine A 500 MVA, 30-kV, 60 Hz synchronous generator is operating at no-load with a constant excitation voltage of 400V. a three-phase short circuit occurs at the armature terminals. Obtain the transient waveforms for the current in each phase and the field current. Assume the short circuit is applied at the instant when the rotor direct axis is along the magnetic axis of phase a, i.e., δ=0. Also assume that the rotor speed remains constant at the synchronous value. The machine parameters are – Ld = 0.0072 H LD = 0.0068 H MD= 0.0054 H r =0.0020 Ohms rQ = 0.0150 Ohms


Generator parameters for Experiment 3 Lq = 0.0070 H LF = 2.500 H LQ = 0.0016 H MF = 0.100 H MQ= 0.0026 H MR= 0.1250 H rF =0.400 Ohms rD =0.0150 Ohms LO = 0.0010 H The dc field voltage is VF = 400V [Ref: Example 8.2, Power System Analysis, Haadi Sadat]

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The function iprime = symshort(t, i) defines the differential equations of the synchronous machine during a balanced three-phase fault. The function returns the state derivatives of the current. Save the following function in the file name as ‘symshort.m’ function iprime =symfault(t,i) f=60.;

w=2.*pi*f;

d=0.;

d=d*pi/180.;

%

theta=w*t+d +pi/2;

Parameters of a 160 MVA, 15 kV Synchronous Machine

LF = 2.500; 0.0070;

LD = 0.0068;

LQ = 0.0016;

Ld = 0.0072;

MF = 0.100;

MD = 0.0054;

MQ = 0.0026;

MR = 0.1250;

ra = 0.002;

rF = 0.4000;

rD

rQ = 0.0150;

= 0.015;

VF = 400;

% DC field voltage

V = [0; -VF; 0; 0; 0];

% Voltage column vector

Lq =

k=sqrt(1.5);

R= [ra

0

0

w*Lq

w*k*MQ

0

rF

0

0

0

0

0

rD

0

0

-w*Ld -w*k*MF -w*k*MD

ra

0

0

0

L =[Ld

0

0

rQ];

k*MF

k*MD

0

0

k*MF

LF

MR

0

0

k*MD

MR

LD

0

0

0

0

0

Lq

0

0

0

k*MQ

k*MQ LQ];

Li=inv(L); iprime=- Li*V - Li*R*i; Department of Electrical Engineering

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Now create a file name ‘Experiment3.m’ and save the following. Execute it. fprintf('\n Wait simulation is in process. \n') VF = 400; rF = 0.4; iF0 = VF/rF; f = 60;

w=2.*pi*f;

d = 0;

d=d*pi/180;

t0 = 0 ; tfinal = 0.80; i0 = [0; iF0; 0; 0; 0 ];

% Initial currents

tspan = [t0, tfinal]; [t, i] = ode45('symshort', tspan, i0); theta = w*t+d +pi/2; ia = sqrt(2/3)*(cos(theta).*i(:,1)+sin(theta).*i(:,4)); ib = sqrt(2/3)*(cos(theta-2*pi/3).*i(:,1)+sin(theta-2*pi/3).*i(:,4)); ic = sqrt(2/3)*(cos(theta+2*pi/3).*i(:,1)+sin(theta+2*pi/3).*i(:,4)); iF = i(:,2);

% ia = Armature current

iF=Field current

figure(1), plot(t,ia), xlabel('t, sec'), ylabel('ia, A'), grid title(['Three-phase short circuit

ia,

','delta = ', num2str(d)])

figure(2), plot(t,ib), xlabel('t, sec'), ylabel('ib, A'), grid title(['Three-phase short circuit

ib,


figure(3), plot(t,ic), xlabel('t, sec'), ylabel('i c, A'), grid title(['Three-phase short circuit

ic,


figure(4), plot(t,iF), xlabel('t, sec'), ylabel('iF, A'), grid title(['Three-phase short circuit iF, ','delta = ',num2str(d)])

Results of the simulation are shown below – Armature currents in the various phases vary with time in a rather complicated way. Analysis of the waveforms show that they consists of , • • •

A fundamental-frequency component A dc component A double-frequency component.


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16



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Experiment 4: Symmetrical Components from Unbalance Currents Obtain the symmetrical components of a set of unbalanced currents – Ia = 1.6 @ 25º, Ib = 1.0 @ 180º and Ic = 0.9 @ 132º. [Ref: Example 10.1, Power System Analysis, Haadi Sadat]


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This function transforms a three-phase unbalanced phasor into its symmetrical components. The Three phasors can be expressed in a one column array in rectangular complex form or in a two column array in polar form, with 1st column magnitude and 2nd column phase angle in degree. The function returns the symmetrical components in rectangular form. The function plots the original unbalanced phasors and the symmetrical components. function [symcomp] = abc2sc(fabc) rankfabc=length(fabc(1,:)); if rankfabc == 2 mag= fabc(:,1); ang=pi/180*fabc(:,2); fabcr=mag.*(cos(ang)+j*sin(ang)); elseif rankfabc ==1 fabcr=fabc; else fprintf('\n Three phasors must be expressed in a one column array in rectangular complex form \n') fprintf(' or in a two column array in polar form, with 1st column magnitude & 2nd column \n') fprintf('

phase angle in degree. \n')

return, end a=cos(2*pi/3)+j*sin(2*pi/3); A = [1

1

1; 1 a^2

a; 1 a

a^2];

fa012=inv(A)*fabcr; symcomp= fa012; %scpolar = [abs(fa012)

180/pi*angle(fa012)];

%fprintf(' \n Symmetrical components \n') %fprintf(' Magnitude

Angle Deg.\n')

%disp(scpolar) fabc0=fa012(1)*[1; 1; 1]; fabc1=fa012(2)*[1; Department of Electrical Engineering

a^2;

a]; 19


fabc2=fa012(3)*[1;

a;

a^2];

figure subplot(221); [Px, Py, Vscale]= phasor3(fabcr); [Px0, Py0, Vscale0]= phasor3(fabc0); [Px1, Py1, Vscale1]= phasor3(fabc1); [Px2, Py2, Vscale2]= phasor3(fabc2); Vscle=max([Vscale, Vscale0, Vscale1, Vscale2]); plot(Px', Py','r') title('a-b-c set') axis([-Vscle Vscle -Vscle Vscle]); axis('square') subplot(222); plot(Px0', Py0','g') title('Zero-sequence set') axis([-Vscle Vscle -Vscle Vscle]); axis('square') subplot(223); plot(Px1', Py1','m') title('Positive-sequence set') axis([-Vscle Vscle -Vscle Vscle]); axis('square') subplot(224); plot(Px2', Py2','b') title('Negative-sequence set') axis([-Vscle Vscle -Vscle Vscle]); axis('square') subplot(111) Department of Electrical Engineering

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% The function returns the symmetrical components in polar form function Fpol = rec2pol(Frec) Fpol = [abs(Frec)

angle(Frec)*180/pi];

Create the main file by name ‘Experiment4.m’and save the following. Execute it. Iabc = [1.6

25

1.0

180

0.9 132]; I012 = abc2sc(Iabc);

% Symmetrical components of phase a

I012p= rec2pol(I012) % Converts rectangular phasors into polar form

Result in I012p = 0.4512 0.9435 0.6024

96.4529 -0.0550 22.3157


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Experiment 5: Unbalanced Voltages from Symmetrical Components The symmetrical components of a set of unbalanced three-phase voltages are V0 = 0.6 @ 90º, V1 = 1.0 @ 30º, and V2 = 0.8 @ 30º. [Ref: Example 10.2, Power System Analysis, Haadi Sadat]


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This function transforms the symmetrical components into its original abc phasors. The symmetrical components can be expressed in a one column array in rectangular complex form or in a two column array in polar form, with 1st column magnitude and 2nd column phase angle in degree. First, second and third rows are Zero-, positives- and negative-sequence components respectively. The function returns the abc phsors in rectangular form. The function plots the original symmetrical components and the unbalanced phasors.

function [fabc] = sc2abc(fa012)

rankf012=length(fa012(1,:)); if rankf012 == 2 mag= fa012(:,1); ang=pi/180*fa012(:,2); fa012r=mag.*(cos(ang)+j*sin(ang)); elseif rankf012 ==1 fa012r=fa012; else fprintf('\n Symmetrical components must be expressed in a one column array in rectangular complex form \n') fprintf(' or in a two column array in polar form, with 1st column magnitude & 2nd column \n') fprintf('

phase angle in degree. \n')

return, end a=cos(2*pi/3)+j*sin(2*pi/3); A = [1

1

1; 1 a^2

a; 1 a

a^2];

fabc= A*fa012r; %fabcp= [abs(fabc)

180/pi*angle(fabc)];

%fprintf(' \n Unbalanced phasors \n') %fprintf(' Magnitude

Angle Deg.\n')

%disp(fabcp) fabc0=fa012r(1)*[1; 1; fabc1=fa012r(2)*[1; Department of Electrical Engineering

1];

a^2;

a]; 23


fabc2=fa012r(3)*[1;

a;

a^2];

figure subplot(221); [Px, Py, Vscale]= phasor3(fabc); [Px0, Py0, Vscale0]= phasor3(fabc0); [Px1, Py1, Vscale1]= phasor3(fabc1); [Px2, Py2, Vscale2]= phasor3(fabc2); Vscle=max([Vscale, Vscale0, Vscale1, Vscale2]); plot(Px', Py','r') title('a-b-c set') axis([-Vscle Vscle -Vscle Vscle]); axis('square') subplot(222); plot(Px0', Py0','g') title('Zero-sequence set') axis([-Vscle Vscle -Vscle Vscle]); axis('square') subplot(223); plot(Px1', Py1','m') title('Positive-sequence set') axis([-Vscle Vscle -Vscle Vscle]); axis('square') subplot(224); plot(Px2', Py2','b') title('Negative-sequence set') axis([-Vscle Vscle -Vscle Vscle]); axis('square') subplot(111) Department of Electrical Engineering

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% The function returns the symmetrical components in polar form function Fpol = rec2pol(Frec) Fpol = [abs(Frec)

angle(Frec)*180/pi];

Create the main file by name ‘Experiment5.m’and save the following. Execute it. V012 = [.6

90

1

30

0.8

-30];

Vabc = sc2abc(V012); % Unbalanced phasors from symmetrical components Vabcp= rec2pol(Vabc)

% Converts rectangular phasors into polar form

Results in Vabcp = 1.7088 0.4000 1.7088

24.1825 90.0000 155.8175


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Experiment 6: Optimal Power Dispatch The fuel-cost function for three thermal plants in Rs/h are given by – C1 = 500 + 5.3 P1 + 0.004 P12;

C2 = 400 + 5.5 P2 + 0.006 P22;

C3 = 200 + 5.8 P3 + 0.009 P32

where P1, P2 and P3 are in MW. The total load, PD is 800 MW. Neglecting line losses and generator limits, find the optimal dispatch and the total cost in Rs/h. [Ref: Example 7.4, Power System Analysis, Haadi Sadat]


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Create a file name ‘Experiment6.m’and save the following. Execute it. % Iterative solution Using Newton method alpha =[500; 400; 200]; beta = [5.3; 5.5; 5.8]; gama=[.004; .006; .009]; PD=800; DelP = 10;

% Error in DelP is set to a high value

lambda = input('Enter estimated value of Lambda = '); fprintf('\n ') disp(['

Lambda

'

grad

P1

P2

P3

DP'...

Delambda'])

iter = 0;

% Iteration counter

while abs(DelP)

>= 0.001

% Test for convergence

iter = iter + 1;

% No. of iterations

P = (lambda - beta)./(2*gama); DelP =PD - sum(P);

% Residual

J = sum( ones(length(gama), 1)./(2*gama)); Delambda = DelP/J;

% Gradient sum % Change in variable

disp([lambda, P(1), P(2), P(3), DelP, J, Delambda]) lambda = lambda + Delambda;

% Successive solution

end totalcost = sum(alpha + beta.*P + gama.*P.^2) %Graphical Demonstration of Example 7.4 axis([0 450

6.5 10.5]);

P1=250:10:450; P2 = 150:10:350; P3=100:10:250; IC1= 5.3 + 0.008*P1; IC2= 5.5 + 0.012*P2; IC3= 5.8 + 0.018*P3; Px = 0:100:400; Department of Electrical Engineering

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plot(P1, IC1, P2, IC2, P3, IC3, Px, lambda*ones(1, length(Px)),'-m'), xlabel('P, MW'), ylabel(' Rs/MWh'), grid

When the program is run, the result is Enter estimated value of Lambda = 6

Lambda

P1

P2

P3

DP

grad

Delambda

6.0000

87.5000

41.6667

11.1111

659.7222

263.8889

2.5000

8.5000

400.0000

250.0000

150.0000

0

263.8889

0

totalcost = 6.6825e+003


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Experiment 7: Transient Power-angle Curve for Synchronous Machine Consider a synchronous machine characterized by the following parameters: Xd = 1.0, Xq = 0.6, Xd’ = 0.3 per unit and negligible armature resistance. The machine is connected directly to an infinite bus of voltage 1.0 per unit. The generator is delivering a real power of 0.5 per unit at 0.8 power factor lagging. Determine the voltage behind transient reactance and the transient power-angle equation for the following cases – (a) Neglecting the saliency effect (b) Including the effect of saliency. [Ref: Example 11.1, Power System Analysis, Haadi Sadat]


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Create a file name ‘Experiment7.m’and save the following. Execute it. V=1.0; Xd = 1; Xdd = 0.3; Xq=0.6; th = acos(0.8); S=0.5/0.8*(0.8 +j*0.6) Ia = conj(S)/V disp('Neglecting Saliency') Ed = V + j*Xdd*Ia Ed = abs(Ed) Pmax = Ed*V/Xdd delta = 0:.01:pi; P = Pmax*sin(delta); % disp('Considering the saliency effect') d = atan(Xq*abs(Ia)*.8/(V + Xq* abs(Ia)*0.6)); dd=d*180/pi E = V*cos(d)+Xd*abs(Ia)*sin(d+th) Edq = (Xdd*E-(Xdd - Xd)*V*cos(d))/Xd Pmax1 = Edq*V/Xdd, Pmax2=V^2*(Xdd - Xq)/(2*Xdd*Xq) Ps =Pmax1*sin(delta)+ Pmax2*sin(2*delta); %plot(delta*180/pi, P, delta*180/pi, Ps), grid subplot(1,2,1), plot(delta*180/pi, P), grid ylabel('Pe') subplot(1,2,2), plot(delta*180/pi, Ps), grid ylabel('Pe') [Pmax, k]=max(Ps) dmax=delta(k)*180/pi subplot(111)


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When executed the above program, which results in -S = 0.5000 + 0.3750i

Ia = 0.5000 - 0.3750i Neglecting Saliency Ed = 1.1125 + 0.1500i

Ed = 1.1226

Pmax = 3.7419 Considering the saliency effect dd = 13.7608

E = 1.4545

Edq = 1.1162

Pmax1 = 3.7208 Department of Electrical Engineering

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Pmax2 = -0.8333

Pmax = 4.0321

k = 193

dmax = 110.0079


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Experiment 8: Equal –area Criterion Consider a synchronous machine characterized by the following parameters: Xd = 1.0, Xq = 0.6, Xd’ = 0.3 per unit and negligible armature resistance. The machine is connected directly to an infinite bus of voltage 1.0 per unit. The generator is delivering a real power of 0.5 per unit at 0.8 power factor lagging. Determine the voltage behind transient reactance and the transient power-angle equation for the following cases – (a) Neglecting the saliency effect (b) Including the effect of saliency. [Ref: Example 11.1, Power System Analysis, Haadi Sadat]


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This program obtains the power angle curve for a one-machine system during normal operation. Using equal area criterion the maximum input power that can be suddenly applied for the machine to remain critically stable is obtained. Create a file in the name ‘eacpower.m’ and save the following. function eacpower(P0, E, V, X) if exist('P0')~=1 P0 = input('Generator initial power in p.u. P0 = '); else, end if exist('E')~=1 E = input('Generator e.m.f. in p.u. E = '); else, end if exist('V')~=1 V = input('Infinite bus-bar voltage in p.u. V = '); else, end if exist('X')~=1 X = input('Reactance between internal emf and infinite bus in p.u. X = '); else, end Pemax= E*V/X; if P0 >= Pemax fprintf('\nP0 must be less than the peak electrical power Pemax = %5.3f p.u. \n', Pemax) fprintf('Try again. \n\n') return, end d0=asin(P0/Pemax); delta = 0:.01:pi; Pe = Pemax*sin(delta); dmax=pi; Ddmax=1; while abs(Ddmax) > 0.00001 Df = cos(d0) - (sin(dmax)*(dmax-d0)+cos(dmax)); J=cos(dmax)*(dmax-d0); Ddmax=Df/J; dmax=dmax+Ddmax; end Department of Electrical Engineering

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dc=pi-dmax; Pm2=Pemax*sin(dc); Pmx =[0

pi-d0]*180/pi; Pmy=[P0 P0];

Pm2x=[0

dmax]*180/pi; Pm2y=[Pm2 Pm2];

x0=[d0 d0]*180/pi; y0=[0 Pm2]; xc=[dc dc]*180/pi; yc=[0 Pemax*sin(dc)]; xm=[dmax dmax]*180/pi; ym=[0 Pemax*sin(dmax)]; d0=d0*180/pi; dmax=dmax*180/pi; dc=dc*180/pi; x=(d0:.1:dc); y=Pemax*sin(x*pi/180); %y1=Pe2max*sin(d0*pi/180); %y2=Pe2max*sin(dc*pi/180); x=[d0 x dc]; y=[Pm2 y Pm2]; xx=dc:.1:dmax; h=Pemax*sin(xx*pi/180); xx=[dc xx

dmax];

hh=[Pm2 h

Pm2];

delta=delta*180/pi; %clc fprintf('\nInitial power

=%7.3f p.u.\n', P0)

fprintf('Initial power angle

=%7.3f degrees \n', d0)

fprintf('Sudden additional power

=%7.3f p.u.\n', Pm2-P0)

fprintf('Total power for critical stability =%7.3f p.u.\n', Pm2) fprintf('Maximum angle swing

=%7.3f degrees \n', dmax)

fprintf('New operating angle

=%7.3f degrees \n\n\n', dc)

fill(x,y,'m') hold; fill(xx,hh,'c') Department of Electrical Engineering

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plot(delta, Pe,'-', Pmx, Pmy,'g', Pm2x,Pm2y,'g', x0,y0,'c', xc,yc, xm,ym,'r'), grid Title('Equal-area criterion applied to the sudden change in power') xlabel('Power angle, degree'), ylabel(' Power, per unit') axis([0 180

0 1.1*Pemax])

hold off;

Create a file name ‘Experiment8.m’and save the following. Execute it. % (a) Initial real power P0 = 0.60 P0 = 0.6; E = 1.35; V = 1.0; X = 0.65; eacpower(P0, E, V, X) h=figure; % (b) Zero initial power P0 = 0; eacpower(P0, E, V, X)

When executed the above program, which results in --

Initial power = 0.600 p.u. Initial power angle = 16.791 degrees Sudden additional power = 1.084 p.u. Total power for critical stability = 1.684 p.u. Maximum angle swing =125.840 degrees New operating angle = 54.160 degrees

Current plot held Initial power = 0.000 p.u. Initial power angle = 0.000 degrees Sudden additional power = 1.505 p.u. Total power for critical stability = 1.505 p.u. Maximum angle swing =133.563 degrees New operating angle = 46.437 degrees Current plot held Department of Electrical Engineering

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Experiment 9: Simulating RLC series Circuit (Out of Syllabus) The circuit elements in the shown figure are R = 1.4 Ohms, L = 2 H, and C = 0.32 F. the initial inductor current is zero and the initial capacitor voltage is 0.5 volts. A step voltage of 1.0 volt is applied at time t = 0. Determine i(t) and v(t) over the range 0 < t < 15 seconds. Also obtain a plot of current versus capacitor voltage. [Ref: Example A.19, Power System Analysis, Haadi Sadat]


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This program returns the state variables. Save the program in a file name ‘electsys.m’ function xdot = electsys(t,x); V = 1;

% returns the state derivatives % Step input

R =1.4; L = 2; C = 0.32; xdot = [x(2)/C ; 1/L*( V - x(1) - R*x(2) ) ];

Create a file name ‘Experiment9.m’and save the following. Execute it. t0= 0;

tfinal =15;

x0 = [0.5, 0];

% time interval % initial conditions

tspan = [t0, tfinal];

% use with MATLAB 5

[t,x] = ode23('electsys', tspan, x0);

% use with MATLAB 5

subplot(2,1,1),plot(t,x) title('Time response of an RLC series circuit') xlabel('t, sec') text(8,1.15,'Capacitor voltage') text(8, 0.1,'Current') vc= x(:,1);

i = x(:,2);

subplot(2,1,2),plot(vc, i) title('Current versus capacitor voltage ') xlabel('Capacitor voltage') subplot(111)


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Experiment 10: SIMULINK Basics Model the equation that converts Celsius temperature to Fahrenheit. Obtain a display of Fahrenheit- Celsius temperature graph over a range of 0 to 100ºC. ி

9 32 5 ஼

[Ref: Example A.21, Power System Analysis, Haadi Sadat]

Write down the steps to create the following SIMULINK model. Save the model as ‘Experiment10.mdl’

Steps:


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The output appears in the scope as under –


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Experiment 11: LFC (Load Frequency Control) with Simulink An isolated power station has the following parameters – Turbine time constant τT = 0.5 seconds Governor time constant τg = 0.2 seconds Governer inertia constant H = 5 seconds Governor speed regulation = R per unit. The load varies by 0.8 percent for a 1 percent change in frequency, i.e., D = 0.8. (a) Use MATLAB rlocus function to obtain the root locus plot. (b) The governor speed regulation is set to R = 0.5 per unit. The turbine rated output is 250 MW at nominal frequency of 60 Hz. A sudden load change of 50 MW (∆PL = 0.2 per unit) occurs. (i) Find the steady-state frequency deviation in Hz. (ii) Use MATLAB to obtain the time-domain performance specifications and the frequency deviation step response. (c) Construct the SIMULINK block diagram and obtain the frequency deviation response for the condition in part (b). [Ref: Example 12.1, Power System Analysis, Haadi Sadat]


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Create a file name ‘Experiment11.m’and save the following. Execute it.

disp('Experiment 11(a)

Root-locus')

num = 1; den = [1

7.08

10.56 .8];

figure (1), rlocus(num, den) disp('Experiment 11(b)

Frequency deviation step response')

PL = 0.2; numc = [0.1 denc = [1

0.7 7.08

1]; 10.56

20.8];

t = 0:.02:10; c = -PL*step(numc, denc, t); figure(2), plot(t, c), grid xlabel('t, sec'), ylabel('pu') title('Frequency deviation step response') timespec(numc, denc)

When executed the output appears as –

Experiment 11(a)

Root-locus

Experiment 11(b)

Frequency deviation step response

Peak time = 1.22311

Percent overshoot = 54.8019

Rise time = 0.418873 Settling time = 6.80249


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Create a file name ‘Experiment11simulink.mdl’and create the following SIMULINK model and save it. Execute it.

The output in the scope appears as under –


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