Eee598 P3

TRANSIENT STABILITY ANALYSIS OF A SAMPLE
FOUR MACHINE SYSTEM USING MATLAB

A PROJECT REPORT

Submitted by

Members of Group 5

Fei Gao
Supriya Chathadi
Changxu Chen
Habibou Maiga

Submitted to

Dr. Vijay Vittal

In partial fulfillment for completion of the course

EEE598: POWER SYSTEM STABILITY

at

ARIZONA STATE UNIVERSITY, TEMPE

EEE598-Group 5

Table of Contents

1. Introduction ............................................................................................................................. 3
2. Machine Modelling................................................................................................................... 3
2.1. Two-Axis Model ...................................................................................................................3
2.1.1. Initial Conditions ..................................................................................................... 3
2.2. E’’ Model ..............................................................................................................................4
2.3. Classical Model .....................................................................................................................5
3. Network Parameters ................................................................................................................ 6
4. Simulation using MATLAB ...................................................................................................... 7
5. Results ..................................................................................................................................... 7
5.1. Without Damping..................................................................................................................7
5.2. With Damping.......................................................................................................................9
6. Conclusion ............................................................................................................................. 10
7. References ............................................................................................................................. 10

APPENDIX: MATLAB CODES .................................................................................................... i

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1. Introduction

The sample four-machine two-area system [1] was analyzed in project 1, and the critical
clearing time was determined, for a three-phase fault at bus 5. The same system is now taken
into account for transient stability analysis using Matlab.

Many simplified models such as E’’, two-axis and classical models are used for the purpose
of study. In this specific case, the machines are modelled as shown below.

Machine 1 – Two-axis model
Machines 2 and 3 – E’’ model
Machine 4 – Classical model

The machine parameters are taken from the dynamic file of project 1. And the missing
parameters are assumed from example 12.6 of reference [1].

2. Machine Modelling

Different machines are modelled differently as specified in the previous section. The
differential equations and the initial conditions which define every model are explained
below.

2.1. Two-Axis Model

The two-axis model is a simplified way to model a synchronous machine, in which only the
transient effects are taken into account and the sub-transient effects are neglected. The
machine is represented by an equivalent circuit which has transient emf and reactance.

2.1.1. Initial Conditions

Before doing the initial condition calculations, all the generator parameters are converted into
S G
a common system base, i.e., X sys  X gen base , H sys  H gen base .
Gbase Sbase

The initial conditions of the two-axis model are calculated from the equations shown below.

Currents, voltages along with their angles from the q-axis are calculated.

,


EEE598-Group 5

Ir  Ia cos , Ix  Ia sin 
xq I r  rI x
    tan 1
Va  rI r  xq I x
     
I q  I a cos 
I d   I a sin 
Vq  Vt cos(   )
Vd  Vt sin(   )

The field current is calculated, from which all the flux linkages are obtained.

E  Vq  rIQ  xd I d  EFD
3E
iF 
LAD
d  Ld id  LADiF
q  Lqiq
D  LADid  LADiF
Q  LAQiq
 D  D / 3
Q  Q / 3

The two-axis model is represented by the following equations.

d '  d  Ld 'id
q '  q  Lq 'iq
eq '  d '
ed '  q '
wR  1
Ed '  ed ' / 3
Eq '  eq ' / 3

2.2. E’’ Model

Both transient and sub-transient effects are accounted for in this case. The machine is
modeled by neglecting the transformer voltage terms compared to the speed voltage terms in
the stator voltage equations. The E” model is relatively a detailed model among the various
simplified models, where a machine is represented by the sub-transient emf and reactance.


EEE598-Group 5

Where:


The initial condition calculations are quite similar to the procedure followed for two-axis
model. Currents, voltages and flux linkages are obtained from the same equations, as
mentioned earlier and E” model is represented by the following.

d ''  d  Ld ''id
q ''  q  Lq ''iq
eq ''  d ''
ed ''  q ''
Ed ''  ed '' / 3
Eq ''  eq '' / 3

2.3. Classical Model


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Classical model is the simplest way of modeling a machine, which assumes that the
mechanical power input remains constant during the period of transient and the damping
power is neglected. Thus, the machine is modeled as constant voltage behind transient
reactance, with network parameters in steady state.


For classical model, the initial value for power angle is obtained from internal voltage
Pgen  jQgen
E  Vt  ,   angle( E ) , the direction of E is set as q axis. The values for the Vt,
Vt
Pgen and Qgen are obtained from the power flow data. The angle between current and q axis
is     angle(V t )   , and then
I q  I a cos  ,
I d   I a sin  ,

3. Network Parameters

Step 1: Calculate Zeq for all the generators, based on the model used.

 r  xd  1
 Z eq   T  '
r 
  T
 xq 
 cos   sin  
T  
 sin  cos  

The sub-transient reactances are used in the above equation if it is an E” model.

Step 2: As the values of Eq and Ed are known, the current equations can be written as
follows.
1
r  xd  1
Yeq  T  '
r 
T
 xq 
1
 IQ  r  xd   Eq '  VQ 
I   T x '   '   Yeq  
 D  q r   Ed  VD 

Step 3: From the values of passive network parameters, the equations can be further modified
as shown.
 I Q  G  B  VQ 
 I    B G  V 
 D   D


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IQ  GVQ  BVD
I D  BVQ  GVD

Step 4: Thus the voltages can be calculated from the below equation.

VQ  1  I eq 
V   Y augument  0 
 D  

4. Simulation using MATLAB

Step 1: Read the data from the dynamic data file and the power-flow data file.

Step 2: Generate Y-bus for pre-fault, faulted and post-fault conditions.

Step 3: Model the generators using the appropriate differential equations and initial values,
corresponding to the model used.

Step 4: Integrate the differential equations to calculate the values of angle δ and angular
speeds, and use Euler’s method to find their values at various time steps.

Step 5: Plot the relative rotor angles, which would give a measure of the stability of the
system.

Step 6: By trial and error method, find out the critical clearing time for a 3-phase fault at bus
5 of the system.

The Mat lab codes are shown in Appendix

5. Results

From the simulation, it is seen that the clearing time for a 3-phase fault is dependent on the
damping co-efficient used. When there is no damping (as specified in [1]), the critical
clearing time for a 3-phase fault at bus 5 is approximately 9 cycles. If damping is included
(say, D=2 p.u. from Appendix D of reference [2]), the critical clearing time appears to be
10.6 cycles. The relative power angle plots for stable, critically stable and unstable cases;
with and without damping are shown below.

5.1. Without Damping


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Relative Power Angles
80
Gen1
Gen2
60
Gen3
Gen4
40

20
Delta, degrees

0

-20

-40

-60

-80
0 0.5 1 1.5 2 2.5 3 3.5 4
Time, seconds

Fig. 1: Stable System (4 cycles)

150
Gen1
Gen2
Gen3
100 Gen4

50
Delta, degrees

0

-50

-100
0 0.5 1 1.5 2 2.5 3 3.5 4
Time, seconds

Fig. 2: Marginally Stable System (9 cycles)


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200
Gen1
Gen2
Gen3
150
Gen4

100
Delta, degrees

50

0

-50

-100
0 0.5 1 1.5 2 2.5 3 3.5 4
Time, seconds

Fig. 3: Unstable System (9.1 cycles)

5.2. With Damping

120
Gen1
100 Gen2
Gen3
80 Gen4

60

40
Delta, degrees

20

0

-20

-40

-60

-80
0 0.5 1 1.5 2 2.5 3 3.5 4
Time, seconds

Fig. 4: Stable System (8 cycles)


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150
Gen1
Gen2
Gen3
100 Gen4

50
Delta, degrees

0

-50

-100
0 0.5 1 1.5 2 2.5 3 3.5 4
Time, seconds

Fig. 5: Marginally Stable System (10.6 cycles)

400
Gen1
350 Gen2
Gen3
300 Gen4

250

200
Delta, degrees

150

100

50

0

-50

-100
0 0.5 1 1.5 2 2.5 3 3.5 4
Time, seconds

Fig. 6: Unstable System (10.7 cycles)

6. Conclusion

To conclude, it is seen that a three phase fault applied at bus 5 of the sample four-machine,
two area system has a critical clearing time of about 9 cycles without damping and 10.6
cycles, when damping is included. This is also dependant on the type of model used for
modeling each machine.

7. References

[1] P. Kundur, “Power System Stability and Control”, McGraw-Hill, Inc., Publications, 1994


EEE598-Group 5

[2] P.M. Anderson and A.A. Fouad, “Power System Control and Stability”, A John Wiley
and Sons, Inc., Publications, 2003.


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APPENDIX: MATLAB CODES

MAIN.M
%********* Transient simulation Controls and options*****************
%********************************************************************
%System base
Sbn=100;
w=1;
%initial angluar speed and frequency
omega = [1,1,1,1];
f=60;
%faulted bus
bfault=5;
%simualation duration in seconds
tsend=5;
% Time when fault is applied in seconds
tfault=0.1;
%Duration of the fault in cycles
tclear=9;
%Time when fault is cleared in Cyles
tfclear = tfault + tclear/60;
%simulation steps
dtime = 0.001;

%************************Initial condition calculations*****************
%***********************************************************************

% Machine dynamic data
[r,xd,xdtr,xdstr,xq,xqstr,xqtr,xl,H,D,td0tr,td0str,tq0tr,tq0str,x,Ldstr,Ldt
r,Ld,Lqstr,Lqtr,Lq,ld,lq,LAD,LF,LD,LAQ,LG,LQ,lF,lD,lG,lQ,K1,K2,K3,K4,Kd,Kq,
xxd,xxq,tj] = inputdyn(Sbn);

% Y matrice and power flow input data
[nbus,ngen,Y,B,G,v,vang,Pg,Qg] = inputpf(Sbn);

% Initial conditon values of Machines
[Vt,It,Ia,E,PF,Ir,Ix,deltabeta,gamma,delta,Id,Iq,Vq,Vd,Efd,iq,id,vq,vd,iF,L
amd,LamF,LamD,LamAD,Lamq,LamG,LamQ,LamAQ,Tm,Eq,Ed,Edtr,Eqtr,Edstr,Eqstr] =
inicond(w,ngen,v,vang,Pg,Qg,xqtr,xdtr,Lqstr,Ldstr,xq,xd,r,LAD,Ld,LAQ,Lq,LF)
;

%Ieq calculations and injected current vector. initial injected currents.
Ii=zeros(2*nbus,1);
[Ii] =
ieqcalc(Ii,ngen,delta,xqtr,xdtr,xqstr,xdstr,Ed,E,Edtr,Eqtr,Edstr,Eqstr);

%Y augmented matrix
[Yeq,Y1,Y2] = yaugmented(nbus,ngen,B,G,bfault,x);

%********************Fault Simulation*************************************
%*************************************************************************

Vexy=eye(12)/(Y1)*(Ii);

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for k=1:nbus
Ve(k)=Vexy(2*k-1)+1i*Vexy(2*k);
end

%*********** start itegrations******************************

%pointer recording data
cur = 1;

for t = 0:dtime:tsend

for k = 4; % gen4, Classical model

Te(k) =abs(E(k))*Iq(k);
Ve(k) = abs(E(k));

end
for k=2 % gen2 and 3, E" model

dLamD = (-LamD(k) + Eqtr(k) + (xdtr - xl)*Id(k))/td0str;
dEqtr = (-Eqtr(k) + Efd(k) -Kd*Eqtr(k) + Kd*LamD(k) +
xxd*Id(k))/td0tr;
LamD(k) = LamD(k) + dLamD*dtime;
Eqtr(k) = Eqtr(k) + dEqtr*dtime;
Eqstr(k) = K1*Eqtr(k) + K2*LamD(k);
dLamQ = (-LamQ(k) - Edtr(k) + (xqtr - xl)*Iq(k))/tq0str;
dEdtr = (-Edtr(k) - Kq*Edtr(k) - Kq*LamQ(k) - xxq*Iq(k))/tq0tr;
LamQ(k) = LamQ(k) + dLamQ*dtime;
Edtr(k) = Edtr(k) + dEdtr*dtime;
Edstr(k) = K3*Edtr(k) - K4*LamQ(k);
Te(k) = (Eqstr(k)*Iq(k) + Edstr(k)*Id(k));
t;
dEdtr;
end
for k=3
xxd*Id(k))/td0tr;
end
for k = 1;% gen1, two axis model
dEdtr(k) = (-Edtr(k) - (xq-xqtr)*Iq(k))/tq0tr;
dEqtr(k) = (Efd(k) - Eqtr(k) + (xd-xdtr)*Id(k))/td0tr;
Edtr(k) = Edtr(k) + dEdtr(k)*dtime;
Eqtr(k) = Eqtr(k) + dEqtr(k)*dtime;

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Te(k) = Edtr(k)*Id(k) + Eqtr(k)*Iq(k);
end

xxd*Id(k))/td0tr;
Ve(k) = Eqii(k) + j*Edii(k);
%*****************************************************************

% torque equations and delta is updated
domega = (Tm - D*(omega-1) - Te)/tj;
omega = omega + domega*dtime;
ddelta = omega - 1;
delta = delta + ddelta*dtime*2*pi*f;

IDQ = Ii(1:2*ngen) - Yeq*Vexy(1:2*ngen);
if(t >= tfault && t <= tfclear)
Vexy = eye(2*nbus)/(Y2)*Ii;
else
Vexy = eye(2*nbus)/(Y1)*Ii;
end

%update values with new delta for next iteration

for k=1:ngen
if k == 4 %gen4, Classical model
gx(k) = cos(delta(k))/xqtr;
bx(k) = sin(delta(k))/xdtr;
by(k) = sin(delta(k))/xqtr;
gy(k) = -cos(delta(k))/xdtr;
Ixi(k) = gx(k)*Ed(k) + bx(k)*abs(E(k));
Iyi(k) = by(k)*Ed(k) + gy(k)*abs(E(k));
Ii(2*k-1) = Ixi(k);
Ii(2*k) = Iyi(k);
elseif k == 1 %gen1, Two axis model
gx(k) = cos(delta(k))/xqtr;
bx(k) = sin(delta(k))/xdtr;
by(k) = sin(delta(k))/xqtr;
gy(k) = -cos(delta(k))/xdtr;
Ixi(k) = gx(k)*Edtr(k)+bx(k)*Eqtr(k);
Iyi(k) = by(k)*Edtr(k)+gy(k)*Eqtr(k);
Ii(2*k-1) = Ixi(k);
Ii(2*k) = Iyi(k);
else % gen2 and 3, E" model
gx(k) = cos(delta(k))/xqstr;
bx(k) = sin(delta(k))/xdstr;
by(k) = sin(delta(k))/xqstr;
gy(k) = -cos(delta(k))/xdstr;

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Ixi(k) = gx(k)*Edstr(k)+bx(k)*Eqstr(k);
Iyi(k) = by(k)*Edstr(k)+gy(k)*Eqstr(k);
Ii(2*k-1) = Ixi(k);
Ii(2*k) = Iyi(k);
end
end

for k=1:ngen
Iq(k) = cos(delta(k))*IDQ(2*k-1) + sin(delta(k))*IDQ(2*k);
Id(k) = -sin(delta(k))*IDQ(2*k-1) + cos(delta(k))*IDQ(2*k);
end
% record resultomegas
resultomega(1,cur) = t;
resultomega(2:5,cur) = omega;
resultdelta(1,cur) = t;
resultdelta(2:5,cur) = delta*180/pi; % conversion to degree
cur = cur + 1;
end

%******************* plot the curves**************************
%*************************************************************

close all

figure('name', 'relative power angles');
for k = 1:4
subplot(2,2,k);
plot(resultomega(1,:),resultdelta(k+1,:)-resultdelta(4,:),'LineWidth',1.5);
title(strcat('gen #',int2str(k)));
xlabel('time (second)');
ylabel('delta (degree)');
grid
end
figure('name', 'relative power angles');
plot(resultomega(1,:),resultdelta(1+1,:)-
resultdelta(4,:),':',resultomega(1,:),resultdelta(2+1,:)-
resultdelta(4,:),'--',resultomega(1,:),resultdelta(3+1,:)-
resultdelta(4,:),resultomega(1,:),resultdelta(4+1,:)-resultdelta(4,:),'-
.');
legend('Gen1','Gen2','Gen3','Gen4')
grid

INPUTDYN.M
function
[r,xd,xdtr,xdstr,xq,xqstr,xqtr,xl,H,D,td0tr,td0str,tq0tr,tq0str,x,Ldstr,Ldt
r,Ld,Lqstr,Lqtr,Lq,ld,lq,LAD,LF,LD,LAQ,LG,LQ,lF,lD,lG,lQ,K1,K2,K3,K4,Kd,Kq,
xxd,xxq,tj] = inputdyn(Sbn)

Sbold= 900;

% Machine dynamic data from kundur book example 12.6
r=0;
xd=1.8;
xdtr=0.3;
xdstr=0.25;
xq=1.7;
xqtr=0.55;

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xqstr=0.25;
xl=0.2;
H=6.5;
D=0;
td0tr=8;
td0str=0.03;
tq0tr = 0.4;
tq0str = 0.05;

% conversion to new base

r=r*Sbn/Sbold;
xd=xd*Sbn/Sbold;
xdtr=xdtr*Sbn/Sbold;
xdstr=xdstr*Sbn/Sbold;
xq=xq*Sbn/Sbold;
xqtr=xqtr*Sbn/Sbold;
xqstr=xqstr*Sbn/Sbold;
xl=xl*(Sbn/Sbold);
H=H/(Sbn/Sbold);
D=D/(Sbn/Sbold);

% vector that stores all the reactances

x=[xdtr;xdstr;xdstr;xdtr];%Two-axis,E",E",Classical

% Machine inductances calculations in per unit

Ldstr = xdstr;
Ldtr = xdtr;
Ld = xd;
Lqstr = xqstr;
Lqtr = xqtr;
Lq = xq;
ld = xl;
lq = ld;
LAD = Ld - ld;
LF = LAD^2/(Ld - Ldtr);
LD = LAD^2/LF + (Ldtr - ld)^2/(Ldtr - Ldstr);
LAQ = Lq - lq;
LG = LAQ^2/(Lq - Lqtr);
LQ = LAQ^2/LG + (Lqtr - lq)^2/(Lqtr - Lqstr);
lF = LF - LAD;
lD = LD - LAD;
lG = LG - LAQ;
lQ = LQ - LAQ;

%Constantes described in AA found book on page 137

K1 = (xdstr - xl)/(xdtr - xl);
K2 = 1 - K1;
K3 = (xqstr - xl)/(xqtr - xl);
K4 = 1 - K3;
Kd = (xd - xdtr)*(xdtr - xdstr)/(xdtr - xl)^2;
Kq = (xq - xqtr)*(xqtr - xqstr)/(xqtr - xl)^2;
xxd = (xd - xdtr)*(xdstr - xl)/(xdtr - xl);
xxq = (xq - xqtr)*(xqstr - xl)/(xqtr - xl);
tj = 2*H;

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end

INPUTPF.M
function [nbus,ngen,Y,B,G,v,vang,Pg,Qg] = inputpf(Sbn)

% *********************Power data**********************

% data are taken from the 4mflow file

% number of bus and generators
nbus=6;
ngen=4;

v=[1.03,1.01,1.03,1.01,0.9773,0.9898];
vang=[-44.25,-55.09,0,-9.85,-64.04,-18.08];
Pga=[790,790,719.19,740,0,0];
Qga=[77.59,363.59,71.85,212.38,0,0];
Pload=[0,0,0,0,1241,1699];
Qload=[0,0,0,0,100,100];

Pg=Pga/Sbn;
Qg=Qga/Sbn;
Pl=Pload/Sbn;
Ql=Qload/Sbn;

vang=vang*pi/180; % convert angles from degrees to radians

% shunt impedances

yc=[0,0,0,0,1i*2.235,1i*2.58];

% line impedances

z12=2.5E-3+1i*2.5E-2;
z25=1E-3+1i*1E-2;
z34=2.5E-3+1i*2.5E-2;
z46=1E-3+1i*1E-2;
z56=2.2E-3+1i*2.2E-2;

%***** original Y matrix*************
for k = 1:nbus

Yl(k) = (Pl(k) - 1i*Ql(k))/v(k)^2 - yc(k);

end

Y=zeros(nbus,nbus);

% off diagonal elements
Y(1,2)=-1/z12;
Y(2,1)=Y(1,2);
Y(2,5)=-1/z25;
Y(5,2)=Y(2,5);

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Y(5,6)=-1/z56;
Y(6,5)=Y(5,6);
Y(6,4)=-1/z46;
Y(4,6)=Y(6,4);
Y(4,3)=-1/z34;
Y(3,4)=Y(4,3);

%**********diagonal elements

for m=1:nbus

for n=1:nbus
Y(m,m) = (Y(m,m)-Y(m,n));
end

Y(m,m) = Y(m,m) + Yl(m);
end

for m=1:nbus
for n=1:nbus
G(m,n)=real(Y(m,n));
B(m,n)=imag(Y(m,n));
end
end
end

INICOND.M
function
[Vt,It,Ia,E,PF,Ir,Ix,deltabeta,gamma,delta,Id,Iq,Vq,Vd,Efd,iq,id,vq,vd,iF,L
amd,LamF,LamD,LamAD,Lamq,LamG,LamQ,LamAQ,Tm,Eq,Ed,Edtr,Eqtr,Edstr,Eqstr] =
inicond(w,ngen,v,vang,Pg,Qg,xqtr,xdtr,Lqstr,Ldstr,xq,xd,r,LAD,Ld,LAQ,Lq,LF)

% Initial conditon values of Machines

for j=1:ngen
Vt(j) = v(j)*(cos(vang(j)) + 1i*sin(vang(j)));
It(j) = (Pg(j) - 1i*Qg(j))/conj(Vt(j));
Ia(j)=abs(It(j));
E(j) = Vt(j) + It(j)*1i*xdtr; %internal voltage
PF(j) = atan(Qg(j)/Pg(j));
Ir(j) = Ia(j)*cos(PF(j));
Ix(j) = -Ia(j)*sin(PF(j));
deltabeta(j) = atan((xq*Ir(j) + r*Ix(j))/(v(j) + r*Ir(j) - xq*Ix(j)));
gamma(j) = deltabeta(j) + PF(j);
delta(j) = deltabeta(j) + vang(j);
Id(j)=-Ia(j)*sin(gamma(j)); Iq(j)=Ia(j)*cos(gamma(j));
Vq(j) = v(j)*cos(deltabeta(j)); Vd(j) = -v(j)*sin(deltabeta(j));
Efd(j)= Vq(j) - xd*Id(j);
end

iq = sqrt(3)*Iq;
id = sqrt(3)*Id;
vq = sqrt(3)*Vq;
vd = sqrt(3)*Vd;

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iF = sqrt(3)*Efd/LAD;

%****** FLux linkages*********************

Lamd = (Ld*id + LAD*iF)/sqrt(3);
LamF = (LAD*id + LF*iF)/sqrt(3);
LamD = (LAD*id + LAD*iF)/sqrt(3);
LamAD = (LAD*(id + iF))/sqrt(3);
Lamq = (Lq*iq)/sqrt(3);
LamG = (LAQ*iq)/sqrt(3);
LamQ = (LAQ*iq)/sqrt(3);
LamAQ = (LAQ*iq)/sqrt(3);

Tm=zeros(ngen,1);
Tm = (iq.*Lamd - id.*Lamq)/sqrt(3); % mechanical torque is equal to the
steady state electrical torque

% initial conditions
% CASE : gen2 classical, gen 1 and 3 are E", gen 4 is two axis

% gen 2, classical model
E1 = abs(E(2));
delta(2) = angle(E(2));
gamma(2) = delta(2) - vang(2) + PF(2);
Iq(2) = Ia(2)*cos(gamma(2));
Id(2) = -Ia(2)*sin(gamma(2));
delta(1) = deltabeta(1) + vang(1);
% ***************************************************************

% Ed and Eq initial conditions chap 4 page 132, 138,

for k=1:ngen
if k == 4 % gen 4, classical model % we need t0 check this statement
Eq(k) = E(k);
Ed(k) = 0;
elseif k == 1 % gen 1, two axis model
Edtr(k)= Vd(k) + xqtr*Iq(k);
Eqtr(k)= Vq(k) - xdtr*Id(k);
else % gen 2 and 3, E” model
Edtr(k)= Vd(k) + xqtr*Iq(k);
Eqtr(k)= Vq(k) - xdtr*Id(k);
Edstr(k) = -w*(Lamq(k) - Lqstr*Iq(k));
Eqstr(k) = w*(Lamd(k) - Ldstr*Id(k));
end
end
end

IEQCALC.M
function[Ii] =
ieqcalc(Ii,ngen,delta,xqtr,xdtr,xqstr,xdstr,Ed,E,Edtr,Eqtr,Edstr,Eqstr)

%Ieq calculations and injected current vector

for k=1:ngen

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EEE598-Group 5

% used to calculate Ieq
%***** used to also build yeq***
if k == 4 % gen 4, classical model
gx(k) = cos(delta(k))/xqtr; bx(k) = sin(delta(k))/xdtr;
by(k) = sin(delta(k))/xqtr; gy(k) = -cos(delta(k))/xdtr;
Ixi(k) = gx(k)*Ed(k) + bx(k)*abs(E(k));
Iyi(k) = by(k)*Ed(k) + gy(k)*abs(E(k));
Ii(2*k-1) = Ixi(k); Ii(2*k) = Iyi(k); %storing all the currents to
one vector Ii
elseif k == 1 % gen 1, two axis model
gx(k) = cos(delta(k))/xqtr; bx(k) = sin(delta(k))/xdtr;
by(k) = sin(delta(k))/xqtr; gy(k) = -cos(delta(k))/xdtr;
Ixi(k) = gx(k)*Edtr(k)+bx(k)*Eqtr(k);
Iyi(k) = by(k)*Edtr(k)+gy(k)*Eqtr(k);
Ii(2*k-1) = Ixi(k);
Ii(2*k) = Iyi(k);
else % gen 2 and 3, E” model
gx(k) = cos(delta(k))/xqstr; bx(k) = sin(delta(k))/xdstr;
by(k) = sin(delta(k))/xqstr; gy(k) = -cos(delta(k))/xdstr;
Ixi(k) = gx(k)*Edstr(k)+bx(k)*Eqstr(k);
Iyi(k) = by(k)*Edstr(k)+gy(k)*Eqstr(k);
Ii(2*k-1) = Ixi(k);
Ii(2*k) = Iyi(k);
end
end
end

YAUGMENTED.M
function [Yeq,Y1,Y2] = yaugmented(nbus,ngen,B,G,bfault,x)

%Y augmented matrix

Y1=zeros(2*nbus,2*nbus);
Y2=zeros(2*nbus,2*nbus);

for m=1:nbus

for n=1:nbus
if(m==n && n<=ngen)
Y1(2*m-1,2*n)=-B(m,n)+1/x(m);
Y1(2*m,2*n-1)=B(m,n)-1/x(m);

Yeq(2*m-1,2*n)=+1/x(m);
Yeq(2*m,2*n-1)=-1/x(m);
else
Y1(2*m-1,2*n)=-B(m,n);
Y1(2*m,2*n-1)=B(m,n);
end

Y1(2*m-1,2*n-1)=G(m,n);
Y1(2*m,2*n)=G(m,n);
end

%Y augmented matrix during the three phase fault at bus bfault

Arizona State University ix

EEE598-Group 5

for m=1:2*nbus
for n=1:2*nbus
if( m~=n && (m==(2*bfault)|| n==(2*bfault)|| m==(2*bfault-1) ||
n==(2*bfault-1)))
Y2(m,n)=0;
elseif ((m==n && m==(2*bfault))|| (m==n && m==(2*bfault-1)))
Y2(m,n)=99999999999;
else
Y2(m,n)=Y1(m,n);
end
end
end
end

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