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

Arizona State University 2

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 2.2.1. Initial Conditions ..................................................................................................... 5

2.3. Classical Model .....................................................................................................................5 2.3.1. Initial Conditions ..................................................................................................... 6

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

a common system base, i.e., basesys gen

base

SX X

G , base

sys gen

base

GH H

S .

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.

,

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sin,cos axar IIII

1tanq r x

a r q x

x I rI

V rI x I

cosq aI I

sind aI I

cos( )q tV V

sin( )d tV V

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

q Q d d FDE V rI x I E

3F

AD

d d d AD F

Ei

L

L i L i

q q qL i

D AD d AD FL i L i

Q AQ qL i

/ 3D D

/ 3Q Q

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

' '

d d d dL i

' '

q q q qL i

' '

q de

' '

d qe

1Rw

' ' / 3d dE e

' ' / 3q qE e

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.

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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 d dL i '' ''

q q q qL i

'' ''

q de

'' ''

d qe

'' '' / 3d dE e

'' '' / 3q qE e

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

gen gen

t

t

P jQE V

V

, ( )angle E , the direction of E is set as q axis. The values for the Vt,

Pgen and Qgen are obtained from the power flow data. The angle between current and q axis

is ( )tangle V , and then

cos ,q aI I

sin ,d aI I

3. Network Parameters

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

1

'

d

eq

q

r xZ T T

x r

cos sin

sin cosT

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

1

'

d

eq

q

r xY T T

x r

1'

' '

dQ Qq

eq

qD Dd

r xI VET Y

x rI VE

Step 3: From the values of passive network parameters, the equations can be further modified

as shown.

Q Q

D D

I VG B

I VB G

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Q Q DI GV BV

D Q DI BV GV

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

1

0

Q eq

augument

D

V IY

V

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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Fig. 1: Stable System (4 cycles)

Fig. 2: Marginally Stable System (9 cycles)

0 0.5 1 1.5 2 2.5 3 3.5 4-80

-60

-40

-20

0

20

40

60

80Relative Power Angles

Time, seconds

Delta,

degre

es

Gen1

Gen2

Gen3

Gen4

0 0.5 1 1.5 2 2.5 3 3.5 4-100

-50

0

50

100


Time, seconds

Delta,

degre

es

Gen1

Gen2

Gen3

Gen4

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Fig. 3: Unstable System (9.1 cycles)

5.2. With Damping

Fig. 4: Stable System (8 cycles)

0 0.5 1 1.5 2 2.5 3 3.5 4-100

-50

0

50

100

150


Time, seconds

Delta,

degre

es

Gen1

Gen2

Gen3

Gen4

0 0.5 1 1.5 2 2.5 3 3.5 4-80

-60

-40

-20

0

20

40

60

80

100


Time, seconds

Delta,

degre

es

Gen1

Gen2

Gen3

Gen4

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Fig. 5: Marginally Stable System (10.6 cycles)

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

0 0.5 1 1.5 2 2.5 3 3.5 4-100

-50

0

50

100


Time, seconds

Delta,

degre

es

Gen1

Gen2

Gen3

Gen4

0 0.5 1 1.5 2 2.5 3 3.5 4-100

-50

0

50

100

150

200

250

300

350


Time, seconds

Delta,

degre

es

Gen1

Gen2

Gen3

Gen4

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[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 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)); 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

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)); 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); delta(3) = deltabeta(3) + vang(3); delta(4) = deltabeta(4) + vang(4); % ***************************************************************

% 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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Arizona State University ix

% 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

EEE598-Group 5

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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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Eee598 P3