Lab Record_power System Dynamics

8/18/2019 Lab Record_power System Dynamics

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EXP. NO: 4

DATE: 11.04.2016

SMALL-SIGNAL STABILITY ANALYSIS OF SINGLE MACHINE-

INFINITE BUS (SMIB) SYSTEM USING TYPE 1B MACHINE MODEL

EFFECT OF EXCITATION SYSTEM

AIM:

To develop a MATLAB program to study Small-signal stability analysis of single machine-infinite

bus (SMIB) system using Type 1B machine model effect of excitation system.

SOFTWARE REQUIRED:

MATLAB

THEORY:

Small signal (or small disturbance) stability is the ability of the power system to maintain synchronismunder small disturbances. The disturbances are considered sufficiently small for linearization ofsystem equations to be permissible for purpose of analysis. Instability that may result can be of twoforms.

I. Steady increase in rotor angle due to lack of sufficient synchronizing torque.II. II. Rotor oscillations of increasing amplitude due to lack of sufficient damping torque.

Generator Represented by Variable Voltage behind Transient Reactance Effect of Field CircuitDynamics

We consider variation of field flux linkage in this model. We will consider only one winding in therotor, f winding.

The equations for this model are:

Vq=-R aIq + X d Id + Eq

Vd=-R aId + X qIq

pE q =-(1/T d0 )[Eq -(Xd-Xd )Id-EFD]

The system equations in state space form:



EFFECT OF EXCITATION SYSTEM:



The feedback of terminal voltage for purpose of voltage regulation results in negative damping. Thenegative damping is caused by delays in excitation system and synchronous machine field circuit (T d0 )with the latter exerting a predominant influence.



LARGE

SYSTEM

Time Response: Assuming

11 12 12 22 + 13 32

3t

21 12 22 22 + 23 32

3t

PROBLEM:

Single-line-diagram :

System data:Power Pl ant: 3x210 MW. The rating of each unit given below:

Rated kVA = 247,000 kVA; rated power factor = 0.85; rated voltage = 15.75 kV; rated frequency = 50Hz. Machine electrical parameters in per unit on the alternator rating ( i.e., on a base of 247,000 kVAand 15.75 kV) are: X d = 2.225; X d = 0.266; X d = 0.214; X leakage = 0.179; X q = 2.11; X q = 0.2454 pu;

X q (estimated) = 0.25; M af (estimated) = 0.004341= M aD ; R f (estimated) = 0.00106; L ff (estimated) =0.004534; M ag (estimated) = 0.004098= M aQ .

H -constant = 3.0 sec. on machine base.Initial loading conditions: P = 0.85 pu; Q = 0.5268 pu on rated MVA ;V = 1.0 pu.Step-up tr ansformer: 3x 15.75/400 kV, leakage: 9% for each transformer on generator MVA.Tr ansmission li ne: Each line X = 0.3 Ohm/km; length: 300 km.The K-constant values are: K 1 = 0.6821; K 2 = 0.9411; K 3 = 0.3010; K 4 = 1.9702

Neglect the damping due to the other sources [ K D in the acceleration equation equals zero].[p.u. reactance of transmission line=0.1389, base impedance=(kV) 2/kVA*1000]

The K-constant values are: K 1 = 0.6821; K 2 = 0.9411; K 3 = 0.3010; K 4 = 1.9702; K 5 = -0.1102; K 6 =0.5467.The excitation system considered is a simple one consisting of a single transfer function representingthe amplifier block (IEEE Type 1 1968) with typical value of gain and time constant.

G

(a)



PROGRAM:

%power factor=0.85 so base MVA=241 MVA, base voltage=15.75kV %on LT & 400kV on HT clc;k=180/pi; %for converting rad to deg p=0.85;q=0.5268;xd=2.225/3;xdd=0.266/3;xl=0.179/3;xe=0.0994;xq=2.11/3;f=50; v=1.0;kd=0;h=9; %xdd=xd'

k1=0.6821;k2=0.9411;k3=0.3010;k4=1.9702;k5 = -0.1102;k6 = 0.5467;Rf= 0.00106; Lff = 0.004534;ka=200;tr=0.02;tds0=Lff/Rft3=k3*tds0fprintf( 'STATE SPACE MATRIX' );A=[-kd/(2*h) -(k1)/(2*h) -k2/(2*h) 0; %type 1Bmachine with effect of excitation

(2*pi*f) 0 0 0;0 -k4/tds0 -1/(k3*tds0) -ka/tds0;0 k5/tr k6/tr -1/tr ]

B=[1/(2*h); 0; ka/tds0; 0]

fprintf( 'EIGEN VALUES OF A' );lamda=eig(A) %eigen values of matrix A fprintf( 'V--> RIGHT MODAL MATRIX D--> DIAGONAL MATRIX' );[v d]=eig(A) % v is right modal matrix fprintf( 'LEFT MODAL MATRIX' );l=inv(v) % left modal matrix m=l';fprintf( 'PARTICIPATION MATRIX' );p=(v.*m) %participation matrix abs(p)fprintf( 'ANGLE OF P MATRIX (IN RADIAN)' );

angle(p)*kwn=abs(imag(lamda(1,1))) %natural frequencyzita=abs(real(lamda(1,1)))/abs(lamda(1,1))fprintf( 'VALUE OF KA JUST TO COMPENSATE THE DEMAGNETISING EFFECT OFARMATURE REACTION' );-k4/k5s=i*wn;m=(-k3*(k4*(1+(i*wn*tr))+(k5*ka)))/((k3*tds0*tr*i*i*wn*wn)+(((k3*tds0)+tr)*(i*wn))+(1+(k3*k6*ka)));fprintf( 'SYNCHRONIZING & DAMPING TORQUE COEFFICIENT AT ROTOR

OSCILLATION FREQUENCY' );kseq=real(m)ksnet=kseq+k1kdeq=imag(m)fprintf( 'DAMPING TORQUE COEFFICIENT IN P.U.TORQUE/ELEC.RADIAN' );kdeq*((2*pi*f)/wn)fprintf( 'wd=damped frequency zita damping ratio' );zita %damping ratio wd=wn*(sqrt(1-(zita*zita))) %damped frequency wdh=wd/(2*pi)%for plots using zero input response

t=0:0.01:20;dd=(v(2,1)*l(1,2)*5*exp(lamda(1,1)*t))+(v(2,2)*l(2,2)*5*exp(lamda(2,1)*t))+(v(2,3)*l(3,2)*5*exp(lamda(3,1)*t))+(v(2,4)*l(4,2)*5*exp(lamda(4,1)*t)); %assuming initial condition



dw=(v(1,1)*l(1,2)*5*exp(lamda(1,1)*t))+(v(1,2)*l(2,2)*5*exp(lamda(2,1)*t))+(v(1,3)*l(3,2)*5*exp(lamda(3,1)*t))+(v(1,4)*l(4,2)*5*exp(lamda(4,1)*t)); %del=5 deg or 0.0873 rad subplot(3,1,1),plot(t,dd),gridxlabel( 't sec' ),ylabel( 'delta degre' )subplot(3,1,2),plot(t,dw),gridxlabel( 't sec' ),ylabel( 'rotor speed ' )

OUTPUT:

STATOR CURRENT OF ONE GEN.

tds0 =

4.2774

t3 =

1.2875

STATE SPACE MATRIXA =

0 -0.0379 -0.0523 0

314.1593 0 0 0

0 -0.4606 -0.7767 -46.7578

0 -5.5100 27.3350 -50.0000

B =

0.0556

0

46.7578

0

EIGEN VALUES OF A

lamda =

0.0591 + 3.8461i

0.0591 - 3.8461i

-25.4475 +25.9329i

-25.4475 -25.9329i

V--> RIGHT MODAL MATRIX D--> DIAGONAL MATRIX

v =

0.0002 + 0.0121i 0.0002 - 0.0121i 0.0008 + 0.0008i 0.0008 - 0.0008i

0.9843 0.9843 0.0001 - 0.0099i 0.0001 + 0.0099i

0.1728 - 0.0273i 0.1728 + 0.0273i 0.7940 0.7940

-0.0150 - 0.0137i -0.0150 + 0.0137i 0.4190 - 0.4403i 0.4190 + 0.4403i



d =

0.0591 + 3.8461i 0 0 0

0 0.0591 - 3.8461i 0 0

0 0 -25.4475 +25.9329i 0

0 0 0 -25.4475 -25.9329i

LEFT MODAL MATRIX

l =

0.0005 -41.3940i 0.5068 - 0.0078i 0.0059 + 0.0821i -0.0114 - 0.0758i

0.0005 +41.3940i 0.5068 + 0.0078i 0.0059 - 0.0821i -0.0114 + 0.0758i

1.4207 - 0.0612i -0.1100 + 0.1222i 0.6256 - 0.5976i 0.0051 + 1.1327i

1.4207 + 0.0612i -0.1100 - 0.1222i 0.6256 + 0.5976i 0.0051 - 1.1327i

PARTICIPATION MATRIX

p =

-0.4988 + 0.0077i -0.4988 - 0.0077i 0.0011 + 0.0012i 0.0011 - 0.0012i

0.4988 + 0.0077i 0.4988 - 0.0077i -0.0012 + 0.0011i -0.0012 - 0.0011i

-0.0012 - 0.0144i -0.0012 + 0.0144i 0.4967 + 0.4745i 0.4967 - 0.4745i

0.0012 - 0.0010i 0.0012 + 0.0010i -0.4966 - 0.4768i -0.4966 + 0.4768i

ans =

0.4989 0.4989 0.0016 0.00160.4989 0.4989 0.0016 0.0016

0.0144 0.0144 0.6870 0.6870

0.0016 0.0016 0.6885 0.6885

ANGLE OF P MATRIX (IN RADIAN)

ans =

179.1186 -179.1186 47.4918 -47.4918

0.8801 -0.8801 138.5743 -138.5743-94.8268 94.8268 43.6907 -43.6907

-39.0585 39.0585 -136.1656 136.1656

wn =

3.8461

zita =

0.0154

VALUE OF KA JUST TO COMPENSATE THE DEMAGNETISING EFFECT OF ARMATUREREACTION

ans = 17.8784



SYNCHRONIZING & DAMPING TORQUE COEFFICIENT AT ROTOR OSCILLATIONFREQUENCY

kseq =

0.1760

ksnet =

0.8581kdeq =

-0.0278

DAMPING TORQUE COEFFICIENT IN P.U.TORQUE/ELEC.RADIAN

ans =

-2.2672

wd=damped frequency zita damping ratio

zita =

0.0154

wd =

3.8456

wdh =

0.6120

RESULT: E 1 and 2 corresponds to rotor mode. In this mode state variables E q X1 participate very little. E q X1 are associated with modes that decay very rapidly(real part=-25.4475). There is hardly any participation from in this mode. The effect ofAVR is to increase synchronizing torque and decrease damping torque component at the rotoroscillation frequency. The feedback of terminal voltage for purpose of voltage regulation results innegative damping. The negative damping is caused by delays in excitation system and synchronous

machine field circuit (T d0 ) with the latter exerting a predominant influence. A MATLAB programwas written to analyze the small-signal stability of a single-machine-infinite bus (SMIB) system(type1B).



EXP. NO: 5

DATE: 15.04.2016

SMALL-SIGNAL STABILITY ANALYSIS OF SINGLE MACHINE-

INFINITE BUS (SMIB) SYSTEM USING TYPE 1B MACHINE MODEL

WITH A SIMPLE EXCITATION SYSTEM EFFECT OF PSS

AIM:

To develop a MATLAB program to study Small-signal stability analysis of single machine-infinite

bus (SMIB) system using Type 1B machine model with a simple excitation system effect of PSS.

SOFTWARE REQUIRED:

MATLAB

THEORY:

Small signal (or small disturbance) stability is the ability of the power system to maintain synchronismunder small disturbances. The disturbances are considered sufficiently small for linearization ofsystem equations to be permissible for purpose of analysis. Instability that may result can be of twoforms.

I. Steady increase in rotor angle due to lack of sufficient synchronizing torque.II. II. Rotor oscillations of increasing amplitude due to lack of sufficient damping torque.

Generator Represented by Variable Voltage behind Transient Reactance Effect of Field CircuitDynamics

We consider variation of field flux linkage in this model. We will consider only one winding in therotor, f winding.

The equations for this model are:

Vq=-R aIq + X d Id + Eq

Vd=-R aId + X qIq

pE q =-(1/T d0 )[Eq -(Xd-Xd )Id-EFD]

The system equations in state space form:



EFFECT OF EXCITATION SYSTEM AND INCLUSION OF POWER SYSTEM STABILIZER (PSS) :

The feedback of terminal voltage for purpose of voltage regulation results in negative damping. Thenegative damping is caused by delays in excitation system and synchronous machine field circuit (T d0 )with the latter exerting a predominant influence. The negative damping torque component can beneutralized if we can superimpose a sufficient positive damping torque component. The positivedamping torque can be produced by feeding back the speed deviation (from synchronous speed) signalat the excitation system input after providing to it appropriate gain and phase advance for the range ofexpected rotor oscillation frequencies. The basic function of the PSS is to introduce positive dampingtorque to rotor oscillations by controlling the excitation using auxillary stabilizing signal such as thespeed deviation. The PSS consists of three major blocks: (i) Gain (ii) Phase compensation (iii) Signalwashout.



.



In state variable form,

Time Response: Assuming deg.

11 12 12 22 + 13 32

3t

21 12 22 22 + 23 32

3t +



LARGE

SYSTEM

PROBLEM:

Single-line-diagram :

System data:Power Pl ant: 3x210 MW. The rating of each unit given below:

Rated kVA = 247,000 kVA; rated power factor = 0.85; rated voltage = 15.75 kV; rated frequency = 50Hz. Machine electrical parameters in per unit on the alternator rating ( i.e., on a base of 247,000 kVAand 15.75 kV) are: X d = 2.225; X d = 0.266; X d = 0.214; X leakage = 0.179; X q = 2.11; X q = 0.2454 pu;

X q (estimated) = 0.25; M af (estimated) = 0.004341= M aD ; R f (estimated) = 0.00106; L ff (estimated) =0.004534; M ag (estimated) = 0.004098= M aQ .

H -constant = 3.0 sec. on machine base.Initial loading conditions: P = 0.85 pu; Q = 0.5268 pu on rated MVA ;V = 1.0 pu.Step-up tr ansformer: 3x 15.75/400 kV, leakage: 9% for each transformer on generator MVA.Tr ansmission li ne: Each line X = 0.3 Ohm/km; length: 300 km.The K-constant values are: K 1 = 0.6821; K 2 = 0.9411; K 3 = 0.3010; K 4 = 1.9702

Neglect the damping due to the other sources [ K D in the acceleration equation equals zero].[p.u. reactance of transmission line=0.1389, base impedance=(kV) 2/kVA*1000]

The K-constant values are: K 1 = 0.6821; K 2 = 0.9411; K 3 = 0.3010; K 4 = 1.9702; K 5 = -0.1102; K 6 =0.5467.The excitation system considered is a simple one consisting of a single transfer function representing

the amplifier block (IEEE Type 1 1968) with typical value of gain and time constant.The PSS data is as follows: K PSS = 10; T W = 1.4 s; T 1 = 0.1; T 2 = 0.03

PROGRAM:

%power factor=0.85 so base MVA=241 MVA, base voltage=15.75kV %on LT & 400kV on HT clc;k=180/pi; %for converting rad to deg

p=0.85;q=0.5268;xd=2.225/3;xdd=0.266/3;xl=0.179/3;xe=0.0994;xq=2.11/3;f=50;v=1.0;kd=0;h=9; %xdd=xd' k1=0.6821;k2=0.9411;k3=0.3010;k4=1.9702;k5 = -0.1102;k6 = 0.5467;Rf = 0.00106; Lff= 0.004534;ka=200;tr=0.02; KPSS = 10; TW = 1.4 ; T1 = 0.1; T2 = 0.03;a=(-KPSS*kd)/(2*h);b=(-KPSS*k1)/(2*h);c=(-KPSS*k2)/(2*h);d=(-(T1/T2)*(1/TW))+(1/T2); tds0=Lff/Rf t3=k3*tds0 fprintf( 'STATE SPACE MATRIX' ); A=[-kd/(2*h) -(k1)/(2*h) -k2/(2*h) 0 0 0; %type 1Bmachine with effect of excitation

(2*pi*f) 0 0 0 0 0; 0 -k4/tds0 -1/(k3*tds0) -ka/tds0 0 ka/tds0; 0 k5/tr k6/tr -1/tr 0 0; a b c 0 -1/TW 0;

(a*T1)/T2 (b*T1)/T2 (c*T1)/T2 0 d -1/T2] B=[1/(2*h); 0; ka/tds0; 0; KPSS/(2*h); (KPSS/(2*h))*(T1/T2)] fprintf( 'EIGEN VALUES OF A' ); lamda=eig(A) %eigen values of matrix A

G

(a)



fprintf( 'V--> RIGHT MODAL MATRIX D--> DIAGONAL MATRIX' ); [v d]=eig(A) % v is right modal matrix

fprintf( 'LEFT MODAL MATRIX' ); l=inv(v) % left modal matrix m=l'; fprintf( 'PARTICIPATION MATRIX' ); p=(v.*m) %participation matrix abs(p) fprintf( 'ANGLE OF P MATRIX (IN RADIAN)' ); angle(p)*k wn=abs(imag(lamda(1,1))) %natural frequencyzita=abs(real(lamda(1,1)))/abs(lamda(1,1)) s=i*wn; delt=(k2*k3*ka)/(1+(k3*k6*ka)+(k3*tds0*s))abs(delt) fprintf( 'MINIMUM PHASE LEAD THAT SHOULD BE PROVIDED BY THE PSS AT ROTOROSCILLATION FREQUENCY' ); angle(delt)*(-k) m=(-

k3*(k4*(1+(i*wn*tr))+(k5*ka)))/((k3*tds0*tr*i*i*wn*wn)+(((k3*tds0)+tr)*(i*wn))+(1+(k3*k6*ka))); kseq=real(m); kdeq=imag(m)*((2*pi*f)/wn); fprintf( 'GAIN REQUIRED TO NEUTRALISE THE NEGATIVE DAMPING OF AVR AT ROTOROSCILLATION FREQUENCY' ); -kdeq/abs(delt) fprintf( 'wd=damped frequency zita damping ratio' ); zita %damping ratio wd=wn*(sqrt(1-(zita*zita))) %damped frequency wdh=wd/(2*pi) n=delt*KPSS*((s*TW)/(1+(s*TW)))*((1+(s*T1))/(1+(s*T2))); fprintf( 'EFFECT OF PSS AT ROTOR OSCILLATION' );

abs(n) angle(n)*k fprintf( 'VALUES OF KSPSS & KDPSS' ); kdpss=real(n) kspss=imag(n)*(wn/314.159) fprintf( 'NET SYNCHRONIZING AND DAMPING TORQUE COEFFICIENT' ); ksnet=k1+kseq+kspss kdnet=kdeq+kdpss %for plots using zero input response t=0:0.01:20; dd=(v(2,1)*l(1,2)*5*exp(lamda(1,1)*t))+(v(2,2)*l(2,2)*5*exp(lamda(2,1)*t))+(v(2,3)*l(3,2)*5*exp(lamda(3,1)*t))+(v(2,4)*l(4,2)*5*exp(lamda(4,1)*t))+(v(2,5)*l(5,2)*5*exp(lamda(5,1)*t))+(v(2,6)*l(6,2)*5*exp(lamda(6,1)*t)); %assuming initial condition

dw=(v(1,1)*l(1,2)*5*exp(lamda(1,1)*t))+(v(1,2)*l(2,2)*5*exp(lamda(2,1)*t))+(v(1,3)*l(3,2)*5*exp(lamda(3,1)*t))+(v(1,4)*l(4,2)*5*exp(lamda(4,1)*t))+(v(1,5)*l(5,2)*5*exp(lamda(5,1)*t))+(v(1,6)*l(6,2)*5*exp(lamda(6,1)*t)); %del=5 deg or 0.0873 rad subplot(3,1,1),plot(t,dd),grid xlabel( 't sec' ),ylabel( 'delta degre' ) subplot(3,1,2),plot(t,dw),grid xlabel( 't sec' ),ylabel( 'rotor speed ' )



OUTPUT:

tds0 =

4.2774

t3 =

1.2875

STATE SPACE MATRIX

A =

0 -0.0379 -0.0523 0 0 0

314.1593 0 0 0 0 0

0 -0.4606 -0.7767 -46.7578 0 46.7578

0 -5.5100 27.3350 -50.0000 0 0

0 -0.3789 -0.5228 0 -0.7143 0

0 -1.2631 -1.7428 0 30.9524 -33.3333

B =

0.0556

0

46.7578

0

0.55561.8519

EIGEN VALUES OF A

lamda =

-24.4035 +27.2772i

-24.4035 -27.2772i

-0.3725 + 3.6515i

-0.3725 - 3.6515i-0.7463

-34.5260

V--> RIGHT MODAL MATRIX D--> DIAGONAL MATRIX

v = 0.0008 + 0.0009i 0.0008 - 0.0009i -0.0011 + 0.0112i -0.0011 - 0.0112i 0.0015 0.0007

0.0010 - 0.0098i 0.0010 + 0.0098i 0.9645 0.9645 -0.6402 -0.0060

0.8067 0.8067 0.0757 + 0.1597i 0.0757 - 0.1597i 0.4857 0.4369

0.4043 - 0.4288i 0.4043 + 0.4288i -0.0586 + 0.0923i -0.0586 - 0.0923i 0.3412 0.77390.0077 + 0.0088i 0.0077 - 0.0088i -0.0330 + 0.1079i -0.0330 - 0.1079i 0.3541 0.0067

-0.0033 + 0.0417i -0.0033 - 0.0417i -0.0609 + 0.0996i -0.0609 - 0.0996i 0.3352 0.4585



d = -24.4035 +27.2772i 0 0 0 0 0

0 -24.4035 -27.2772i 0 0 0 0

0 0 -0.3725 + 3.6515i 0 0 0

0 0 0 -0.3725 - 3.6515i 0 0

0 0 0 0 -0.7463 0

0 0 0 0 0 -34.5260

LEFT MODAL MATRIX

l =

1.0008 + 0.3091i -0.1046 + 0.0629i 0.6507 - 0.5480i -0.0571 + 1.0618i -0.5400 + 1.0574i -0.5186 - 1.2852i

1.0008 - 0.3091i -0.1046 - 0.0629i 0.6507 + 0.5480i -0.0571 - 1.0618i -0.5400 - 1.0574i -0.5186 + 1.2852i

-9.8334 -45.5794i 0.5414 - 0.0603i 0.0040 + 0.0847i -0.0096 - 0.0791i 1.0070 - 0.0649i 0.0188 + 0.1181i

-9.8334 +45.5794i 0.5414 + 0.0603i 0.0040 - 0.0847i -0.0096 + 0.0791i 1.0070 + 0.0649i 0.0188 - 0.1181i-29.6103 - 0.0000i 0.0703 + 0.0000i -0.0022 + 0.0000i 0.0021 - 0.0000i 3.0822 + 0.0000i -0.0032 + 0.0000i

-0.6931 + 0.0000i 0.0762 + 0.0000i -0.0510 + 0.0000i 0.1540 -1.8294 + 0.0000i 1.9984 - 0.0000i


p =

0.0010 + 0.0006i 0.0010 - 0.0006i -0.4997 - 0.1624i -0.4997 + 0.1624i -0.0450 + 0.0000i -0.0005 - 0.0000i

-0.0007 + 0.0010i -0.0007 - 0.0010i 0.5222 + 0.0581i 0.5222 - 0.0581i -0.0450 + 0.0000i -0.0005 + 0.0000i

0.5249 + 0.4420i 0.5249 - 0.4420i 0.0138 - 0.0058i 0.0138 + 0.0058i -0.0011 - 0.0000i -0.0223 - 0.0000i

-0.4784 - 0.4048i -0.4784 + 0.4048i -0.0067 - 0.0055i -0.0067 + 0.0055i 0.0007 + 0.0000i 0.1192

0.0051 - 0.0129i 0.0051 + 0.0129i -0.0402 + 0.1065i -0.0402 - 0.1065i 1.0915 - 0.0000i -0.0122 - 0.0000i

-0.0519 - 0.0258i -0.0519 + 0.0258i 0.0106 + 0.0091i 0.0106 - 0.0091i -0.0011 - 0.0000i 0.9162 + 0.0000i


ans = 0.0012 0.0012 0.5255 0.5255 0.0450 0.0005

0.0012 0.0012 0.5255 0.5255 0.0450 0.0005

0.6862 0.6862 0.0150 0.0150 0.0011 0.0223

0.6267 0.6267 0.0087 0.0087 0.0007 0.1192

0.0139 0.0139 0.1138 0.1138 1.0915 0.0122

0.0580 0.0580 0.0140 0.0140 0.0011 0.9162

ANGLE OF P MATRIX (IN RADIAN)

ans = 30.5141 -30.5141 -162.0015 162.0015 180.0000 -180.0000

126.8793 -126.8793 6.3505 -6.3505 180.0000 180.0000

40.1008 -40.1008 -22.6562 22.6562 -180.0000 -180.0000

-139.7581 139.7581 -140.6724 140.6724 0.0000 0

-68.5327 68.5327 110.6851 -110.6851 -0.0000 -180.0000

-153.5428 153.5428 40.4821 -40.4821 -180.0000 0.0000



wn =

27.2772

zita =

0.6668

delt =

0.8061 - 0.8348i

ans =

1.1605

MINIMUM PHASE LEAD THAT SHOULD BE PROVIDED BY THE PSS AT ROTOROSCILLATION FREQUENCY

ans =

46.0022

GAIN REQUIRED TO NEUTRALISE THE NEGATIVE DAMPING OF AVR AT ROTOROSCILLATION FREQUENCY

ans =

1.4672

wd=damped frequency zita damping ratio

zita =

0.6668

wd =20.3290

wdh =

3.2355

EFFECT OF PSS AT ROTOR OSCILLATION

ans =

26.0834

ans =

-13.9295

VALUES OF KSPSS & KDPSS

kdpss =

25.3163

kspss =

-0.5452

NET SYNCHRONIZING AND DAMPING TORQUE COEFFICIENT

ksnet = 0.1890

kdnet = 23.6136



RESULT: E 3 and 4 corresponds to rotor mode. In this mode state variables E q X1

participate very little. E q X1 are associated with eigen values 1 and 2. There is hardly any

participation from in this mode. X 2 is associated with 5. X s is associated with 6. The

effect of PSS is evident from increase in damping torque component at the rotor oscillation frequency.

A MATLAB program was written to analyze the small-signal stability of a single-machine-infinite bus

(SMIB) system(type 1B) with PSS.

Lab Record_power System Dynamics

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Transcript of Lab Record_power System Dynamics