M.E record 2Sem.docx

7/28/2019 M.E record 2Sem.docx

1/63

AIM

(i) To understand modeling and analysis of small signal stability of a SMIB system.(ii)To obtain linearized swing equation and to determine the roots of characteristic equation,

damped frequency and undamped natural frequency of oscillation.

SOFTWARE REQUIRED

MATLAB 7.0

THEORY

Small signal stability is the ability of the power system to maintain synchronism when subjected to

small disturbances. A disturbance is considered to be small if the equations that describe the resulting

response of the system can be linearized for the purpose of analysis. Instability is of two forms

1) Steady increase in generator rotor angle due to synchronizing torque.

2) Rotor oscillations of increasing amplitude due to insufficient damping torque.

For small disturbance, the nonlinear equations characterizing the dynamics of the system may be

linearized around an operating point for the purpose of analysis. Small signal stability using linear

techniques provides valuable information about the dynamics of the system and assists in its design. The

location of the roots of the characteristic equation of the linearized differential equation will reveal the

stability status of the system. We can also obtain the response of rotor angle and frequency of the rotor of

the synchronous machines.

Modelling For Small Signal Stability

Let us consider a system as shown in the Fig.

Fig 1.1

Ex. No: SMALL SIGNAL STABILITY ANALYSIS OF A SINGLE

MACHINE INFINITE BUS SYSTEM USING CLASSICAL

MACHINE MODELDate :


2/63


3/63

Where ris the per unit speed deviation, is the rotor angle in electrical radians o is the base rotor

electrical speed in radians per second and p is the differential Operator d/dt with time t in seconds

Linearizing (1.4) and substituting for Te we get

(1.5)

Where Ks is the synchronizing torque coefficient given by

(1.6)

Writing the equations (1.5) and (1.6) in vector matrix form we get

(1.7)This is of the form X=AX +bu. The elements of the state matrix are seen to be dependent on t he

system parameters KD ,H, XTand the initial operating condition represented by the values of E and o .

The block diagram representation shown in the Fig 1.3 can be used to describe small signal performance.

From the block diagram (1.3) we have

On rearranging, we get,

(1.8)

Therefore the characteristic equation is given by

(1.9)

This is of the general form

Therefore the undamped natural frequency is


4/63

and the damping ratio is

As the synchronizing torque coefficient Ks increases, the natural frequency increases and the damping

ratio decreases. An increase in damping torque coefficient KD increases the damping ratio, whereas an

increase in inertia constant decreases both nand .

Fig 1.3 Block diagram of SMIB System with classical generator model

Where,

PROCEDURE

1. Enter the command window of the MATLAB.

2. Create a new Mfile by selecting File - NewMFile.

3. Type and save the program in the editor window.

4. Execute the program by pressing ToolsRun.

5. View the results.


5/63

EXERCISE

Fig. shows a system representation applicable to a thermal generating station consisting of four

555MVA, 24 KV, 60 Hz units.

The network reactances are in per unit on 2220 MVA, 24 KV base (referred to the LT side of thestep up transformer). Resistances are assumed to be negligible. The objective of this example is to

analyze the small signal stability characteristics of the system about the steady state operating condition

following the loss of circuit 2. The post fault condition in per unit on the 2220 MVA, 24 KV base is as

follows:

The generators are to be modeled as a single equivalent generator by the classical model with the

following parameters expressed in per unit on 2220 MVA, 24 KV base:

Xd'=.65 H=9.94MWs/MVA


6/63

PROGRAM

disp('SMALL SIGNAL STABILITY ANALYSIS OF SMIB');

% ks=synchronizing torque coefficient in pu torque/rad

% kd=damping torque coefficient in pu torque/pu speed deviation

% H=inertia constant in MW-s/MVA

% wr =speed deviation in pu

% del=rotor angle deviation in elec. rad = Laplace operator =

rated speed in elec. rad/s =

% 2nf0 = 377 for a 60 Hz system

% S=Laplace operator

% wo=rated speed in elec. rad/s = 2nf0 = 377 for a 50 Hz system

f=input('enter the no of kd');

for k=1:f

kd(k)=input(':');

end

ET=1;

ETang=36;

%Eb=0.995;

Eb=1;xd=.3i;x1=.15i;x2=.5i;p=0.9;q=0.3i;sdel=.0873;swr=0;m=p+q;

IT=conj(m)/ET;xt=xd+x1+x2;

E=ET+(xd*(IT));

Eang=angle(E)*(180/pi);

delnot=ETang+Eang;

ks=((abs(E)*Eb)/abs(xt))*cos(delnot*(pi/180));

wo=2*pi*60;

for k=1:f

ss=[-.143*kd(k) -.108

wo 0];

lh=poly(ss);

wn=sqrt(lh(3))/(2*pi) ;del=(lh(2))/(2*sqrt(lh(3))) ;

lamda=roots(lh);

wd=wn*sqrt(1-(del*del));

res(:,k)=[lamda' wd' del' wn'];

%disp([lamda' wd' del' wn']);

end

kd=input('enter the kd value:');

ss=[-.143*kd -.108

wo 0];

[vb,DB] = eig(ss);

shi=inv(vb);% ppp=VB*transpose(shi)

vb

shi

par=[vb(1,1)*shi(1,1) vb(1,2)*shi(2,1)

vb(2,1)*shi(1,2) vb(2,2)*shi(2,2) ]

c=(shi)*[swr;sdel]


7/63

OUTPUT:

SMALL SIGNAL STABILITY ANALYSIS OF SMIBenter the no of kd3

:0

:10

:-10enter the kd value:10

vb =

-0.0019 + 0.0168i -0.0019 - 0.0168i

0.9999 0.9999

shi =

-0.0000 -29.7324i 0.5001 - 0.0564i0.0000 +29.7324i 0.5001 + 0.0564i

par =

0.5000 + 0.0564i 0.5000 - 0.0564i0.5000 - 0.0564i 0.5000 + 0.0564i

c =

0.0437 - 0.0049i

0.0437 + 0.0049i

RESULT


8/63

AIM

(i)

To understand the modeling and analysis of small signal stability of a multi machinepower system.

(ii) To obtain linearized swing equation and to determine the roots of characteristic equation

damped frequency of oscillation and undamped natural frequency.

SOFTWARE REQUIRED

MATLAB 7.8

THEORY

Small signal stability is the ability of the power system to maintain synchronism when

subjected to small disturbances. A disturbance is considered to be small if the equations that describe the

resulting response of the system can be linearized for the purpose of analysis. Instability is of two forms

1) Steady increase in generator rotor angle due to synchronizing torque.

2) Rotor oscillations of increasing amplitude due to insufficient damping torque.

For small disturbance, the nonlinear equations characterizing the dynamics of the system

may be linearized around an operating point for the purpose of analysis. Small signal stability using linear

techniques provides valuable information about the dynamics of the system and assists in its design. The

location of the roots of the characteristic equation of the linearized differential equation will reveal the

stability status of the system. We can also obtain the response of rotor angle and frequency of the rotor of

the synchronous machines.

Modelling For Small Signal Stability

For system stability studies it is appropriate to neglect the transmission network and machine stator

transients. The dynamics of machine stator transients, machine rotor circuits, excitation systems, prime

mover and other devices are represented by ODE. The result is that the complete system model consists of

a large number of ordinary differential and algebraic equations.

Ex. No: SMALL SIGNAL STABILITY ANALYSIS OF MULTI-

MACHINE CONFIGURATION WITH CLASSICAL

MACHINE MODEL

Date :


9/63

Fig.2.1 Structure of a complete power system

Formulation State Equation

The linearized model of each dynamic device is expressed in the following form

Modeling of synchronous machines

Classical model of synchronous machines is considered with the q axis leading the d axis. The

dynamic equations are

sii

i

D

i

ei

i

mi

dt

d

H

K

H

T

H

T

dt

dii

222

(2.1)

where,

i - 1 m number of machines

- Rotor angle in electrical radians


10/63

s - Synchronous speed in electrical rad/s

- Rotor angular velocity in rad/s

Tm - Mechanical torque of synchronous generator

H - Inertia constant

D - Damping constant

The Equations are linearized around a steady-state operating point in order to see the effect

of small disturbances on the system. The steady-state operating point is obtained from the load flowresults. The linearized state Equations are:

isi

i

iD

i

e

i

mi

dt

d

H

K

H

T

H

T

dt

dii

222

1

(2.2)

The Equations (2.2) are functions of network variables. To express the network variables in

terms of the state variables coordinated transformation is carried out.

Coordinate Transformation

Network equations for the 14 bus system with reduced Y-bus are given as:

mmmmm

m

m

m

i

E

E

E

YYY

YYY

YYY

I

I

I

'

2'

1'

21

22221

11211

2

Let T be the transformation matrix that transforms the d and q quantities of all machines to system

frame, which a common reference frame is moving at synchronous speed. T is given as:

mj

j

e

e

T

00

00

001

Network equations in individual machine d-q coordinates are:

1 I M E (2.3)

where, M is given as

Linearizing Equation (2.3) we get

(2.4)

where,

mm

j

m

j

m

YeY

eYY

M

m

m

10

01

1

111

0

1 M T Y T

1 1

0 0 0 0 I M E j M M E


11/63

40''

40

30''

30

20

''

20

10''

10

'0

mr

dq

dq

dq

jEE

jEEjEE

jEE

E

44

33

22

11

mr

dq

dq

dq

EjI

IjIIjI

IjI

I

From Equation (2.4) the real and imaginary parts can be separated out and the q and d components of I

can be obtained. They are then substituted in the state equations to eliminate the algebraic variables.

Expressing the state equations in the standard canonical form ,

BuAxdt

dx

where,

tiix ][ and A is the state space matrix.

PROCEDURE

1) Write down the algebraic equations in network coordinates in the nodal form. While doing

this the machine internal nodes are explicitly included. The loads are represented as constant

admittance matrices.

2) Eliminate all passive nodes. This results in a reduced set of algebraic equations.

3) Transform the reduced set of equations obtained in step 2 to individual machine qd

coordinates.

4) Linearize the transformed equations obtained at the end of step 3.

5) Linearize the differential equations for the machine. The resulting equations will contain

incremental changes in algebraic variables.

6) Eliminate the incremental changes in algebraic variables from the linearized the differential

equations obtained in step 5 using the expressions obtained for incremental changes in step

4. The resulting equation will be in the state variable canonical form.

7) Apply eigenvalue technique for stability assessment.

1

2

3

I

I I

I


12/63

EXERCISE

To determine the small signal stability of the power system shown in Fig

Fig 2.2 Single line diagram of sample bus system

TABLE 2.1 ANDERSON AND FOUAD 3 MACHINE 9 BUS SYSTEM DATA

Generator 1 2 3

Rated MVA 247.5 192 128

kV 16.5 18 13.8

Power factor 1.0 0.85 0.85

Type hydro steam steam

Speed 180 r/min 3600 r/min 3600 r/min

Xd (p.u) 0.1460 0.8958 1.3125

Xd (p.u) 0.0608 0.1198 0.1813

Xq (p.u) 0.0969 0.8645 1.2578

Xq(p.u) 0.0969 0.1969 0.25

Xl (p.u) 0.0336 0.0521 0.0742

Td0 (s) 8.96 6 5.89

Tq0 (s) 0 0.535 0.6

Stored energy at rated speed (MW-s) 2364 640 301


13/63

TABLE 2.2 BUS DATA

Bus No Pg (p.u) PD (p.u) QD (p.u) V (p.u) Qmax

(p.u)

Qmin

(p.u)

1 - - - 1.04 1 -1

2 1.63 - - 1.025 1 -1

3 0.85 - - 1.025 1 -1

4 - 0 0 - - -

5 - 1.25 0.5 - - -

6 - 0.9 0.3 - - -

7 - 0 0 - - -

8 - 1.0 0 - - -

9 - 0 0 - - -

Type 1-Slack Bus; Type 2-PV Bus; Type 3-PQ Bus


14/63

PROGRAM

clear all;

clc;

%'multimachine system'

Mva=160;

Xdt=0.49;

Xt=0.8;

Ke=379.2;W0=(2*60*3.14);

Pf1=0.85;

Pf2=0.85;

P1=0.5;

P2=0.5;

Vt=1.0+j*0.0;

q1=0.5*(tan(acos(Pf1)));

q2=0.5*(tan(acos(Pf2)));

i1=(P1-j*q1)/(conj(Vt));

i2=(P2-j*q2)/(conj(Vt));

eq1=Vt+j*Xdt*i1;

eq2=Vt+j*Xdt*i2;d10=angle(eq1);

d20=angle(eq2);

il=i1+i2;

Vl=Vt-j*Xt*i1;

yl=il/Vl;

Z=[j*1.29 inf -j*1.29;

inf j*1.29 -j*1.29;

-j*1.29 -j*1.29 (1/(yl+(-1.5504*j)))];

for i=1:3

for j=1:3

y(i,j)=1/Z(i,j);

end

end

y

eq=[eq1+j*0 eq2+j*0];

ygg=y(1:2,1:2);

ygng=y(1:2,3:3);

yngg=y(3:3,1:2);

yngng=y(3:3,3:3);

yggr=ygg-(ygng*(inv(yngng))*yngg);

d12=d10-d20;

m=[yggr(1,1) yggr(1,2)*exp(-j*d12);

yggr(2,1)*exp(j*d12) yggr(2,2)]

b=imag(m)

t13=((b(2,1))*abs(eq1)*abs(eq2));t23=-((b(2,1))*abs(eq1)*abs(eq2));

h=Ke/(2*Mva);

dw12=-(t13-t23)/(2*h)

a=[0 dw12;

W0 0]

lamda=eig(a)

wn=imag(lamda(1))

fn=(wn/(2*3.14))


15/63

OUTPUT:

y =

0 - 0.7752i 0 0 + 0.7752i

0 0 - 0.7752i 0 + 0.7752i0 + 0.7752i 0 + 0.7752i 1.3781 - 1.6415i

m =

0.1803 - 0.5605i 0.1803 + 0.2147i0.1803 + 0.2147i 0.1803 - 0.5605i

b =

-0.5605 0.2147

0.2147 -0.5605

dw12 =

-0.2513

a =

0 -0.2513

376.8000 0

lamda =

0 + 9.7309i0 - 9.7309i

wn =

9.7309

fn =

1.5495

RESULT


16/63

AIM

(i) To understand the significance of relay coordination in power system protection.

(ii) To manually perform relay coordination of a radial distribution system, to determine the

T.M.S, PSM settings and validate the results using MATLAB.

SOFTWARE USED

MATLAB 7.8

THEORY

Relay

A protective relay is a complex electromechanical apparatus, often with more than one coil,

designed to calculate operating conditions on an electrical circuit and trip circuit breakers when a fault is

detected. Unlike switching type relays with fixed and usually ill-defined operating voltage thresholds andoperating times, protective relays have well-established, selectable, time/current (or other operating

parameter) curves.

Such relays may be elaborate, using arrays of induction disks, shaded-pole magnets, operating and

restraint coils, solenoid-type operators, telephone-relay style contacts, and phase-shifting networks.

Protection relays respond to such conditions as over-current, over-voltage, reverse power flow, over- and

under- frequency, and even distance relays that would trip for faults up to a certain distance away from a

substation but not beyond that point. An important transmission line or generator unit will have cubicles

dedicated to protection, with a score of individual electromechanical devices. The various protective

functions available on a given relay are denoted by standard ANSI device numbers.

Table- List of commonly used relay device numbers

Relay device function number Protective function

21 Distance

25 Synchronizing

27 Undervoltage

32 Directional power

40 Loss of excitation(field)

46 Phase balance (current balance , negative sequence balance)

47 Phase sequence voltage

49 Instantaneous overcurrent

50 Time overcurrent51 Time - Overcurrent

59 Over voltage

60 Voltage balance

67 Dirctional Overcurrent

81 Frequency (Under and Over frequency)

86 Lockout

87 Differential

Ex. No: CO-ORDINATION OF OVER-CURRENT AND

DISTANCE RELAYS FOR RADIAL LINE PROTECTIONDate :


17/63

The different types of overcurrent relays used are

Instantaneous over current relay -Operation is instantaneous in the sense that the relay

has no intentional time delay. Time setting is absent.

Definite time over current relay - A definite time delay can be set after which the relay

operates. It has both time setting and plug setting.

Inverse definite time overcurrent relay- The operating time of the relay is inverselyproportional to the magnitude of fault current.

Need For Relay Coordination

Relay coordination is the process of determining the settings for the relays that will provide

an orderly shutdown and determine the sequence of relay operation with sufficient margin. It is by means

of relay coordination that we ensure that if a fault occurs in a network only the breaker nearest to the fault

operates leaving the healthy sections in service. This can be achieved by time grading or discrimination by

time in which the relay farthest from the source is set to operate first and others follows it. If relay

coordination is not performed it may result in

a) Loss of selectivity resulting in nuisance tripping.b) Loss of reliability of power supply to the healthy parts of the power system.

To Perform Relay Coordination

The mathematical expression to arrive at the required characteristics of a relay has been

specified by IEEE ad IEC

where,

I = Fault current level in CT secondaryIp = Pick up current selected

t =Tripping time

Tp= TMS (Time Multiplier Setting).

The values of , , L for the different types of over current relays have been specified by IEC as follows

Table-Relay types and the constant values

Curve description Standard alpha beta L

Moderately Inverse IEEE .02 .0515 .114

Very Inverse IEEE 2.0 19.61 .491

Extremely Inverse IEEE 2 28.2 .1217

Standard Inverse IEC .02 .14 0

Very Inverse IEC 1 13.5 0

Extremely Inverse IEC 2 80 0

Long time inverse IEC 1 120 0

[sec]( / ) 1

p

p

t T LI I


18/63

PROCEDURE

1. Draw the single line diagram of the radial distribution system with the ratings of all the

concerned equipments noted down.

2. Compute the nominal current that would flow through the system.

3. Compute the short circuit current (IEC standard or IEEE standard red book or buff book )

4. CT ratio is selected depending on the magnitude of the load current.

5. The relay to be used and its placement is then determined

6. Compute the pickup setting and time setting.

7. Plot the time Vs current characteristics of all relay and verify their coordination. (The plot

should include the thermal characteristics of the relays also).8. If characteristics intersect the procedure must be repeated for another suitable value of

TMS.

EXERCISE

To perform relay coordination of the radial distribution systems given below i.e., to suitably

places the relays and determine the T.M.S (Time Multiplier Setting) and P.S.M (Plug Setting Multiplier)

Fig 4.1 Single line diagram of radial distribution system


19/63

PROGRAM

'FOR RELAY R4'

TMS4=0.044;

psm=1;

for i=1:10

top=(0.14*TMS4)/((psm^0.02)-1);

top4(i)=top;

psm4(i)=psm;

psm=psm+10;end

'FOR RELAY R3'

TMS3=0.0129;

psm=1;

for i=1:10

top=(0.14*TMS3)/((psm^0.02)-1);

top3(i)=top;

psm3(i)=psm;

psm=psm+10;

end

'FOR RELAY R2'

TMS2=0.178;psm=1;

for i=1:10

top=(0.14*TMS2)/((psm^0.02)-1);

top2(i)=top;

psm2(i)=psm;

psm=psm+10;

end

'FOR RELAY R1'

TMS1=0.221;

psm=1;

for i=1:10

top=(0.14*TMS1)/((psm^0.02)-1);

top1(i)=top;

psm1(i)=psm;

psm=psm+10;

end

plot(psm4,top4,'-',psm4,top3,'-',psm4,top2,'-',psm4,top1,'-')


20/63

OUTPUT:

RESULT


21/63

AIM

To understand the starting characteristics of slip ring induction motor.

SOFTWARE REQUIRED

MATLAB 7.8

THEORY

1. Induction Motor DataRequired data for motor starting calculations for induction motors includes:

Induction Motor ID.

Rated KW/HP and KV.

Power factors and efficiencies at 100%, 75%, and 50% loading.

Loading Category ID and % loading.

Equipment cable data

2. Static Load DataRequired data for motor starting calculation for static loads includes:

2.1 Static Load ID

Rated KVA/MVA and KV.

Power factors at 100%, 75%, and 50% loading

Loading Category ID and % Loading

Equipment cable data.

2.2 MOV Data

Required data for motor starting calculation for MOV includes:

2.2.1MOV ID

Rated kW/hp and kV

Current, PF, and time for each operation stage

Equipment cable data

3. Capacitor DataRequired data for motor starting calculation for capacitor includes:

Capacitor ID

Rated kV, KVAR/bank and number of banks

Loading category ID and % loading

Equipment cable data.

Ex. No:INDUCTION MOTOR STARTING ANALYSIS

Date :


22/63

4. Lumped Load DataRequired data for motor starting calculation for lumped load includes:

Load ID

Rated KV, MVA, power factor, and % for motor load

Loading category ID and % loading

Additional Data for Starting MotorsFor static motor starting studies, the additional data includes:

Motor locked-rotor impedance and power factor

Motor acceleration time at no load and full load

Start and final percent loading and begin and end of load change time

Starting device data when needed

No load and full load accelerated time (for static motor starting)

For Dynamic Motor Acceleration studies, the additional data includes:

Dynamic motor model for induction motors

LR model for synchronous motors

Load torque model

Motor inertia

STUDY CASE DATA

There are some study case related data, which must also be provided. This data includes:

Study Case ID

Maximum number of iteration

Precision of solution

Total simulation time, simulation time step, and plot time step

Prestart loading (loading category)

Initial condition

Transformer LTC data

Equipment adjustment options

Alert option


23/63

PROGRAM

clc;

clear all;

v1=1.026+j*0.0;

BaseMVA=1e7;

MVA=125e6;

zs=BaseMVA/MVA;

zc=0.0;

zst=0.0;

ztr=0.0001+j*0.0001;

J=63.87;

R1=0.029;

R2=0.022;

X1=0.226;

X2=0.226;

Ns=1800;

ws=(2*pi*Ns/60);% speed PF ang Iratio

data=[ 0.0 6.0 9.0 0.0;

10.0 5.9 18.0 0.0;

20.0 5.8 20.0 0.0;

30.0 5.7 36.0 0.0;

40.0 5.6 50.0 0.0;

50.0 5.5 60.0 0.0;

60.0 5.5 71.0 0.0;

70.0 5.4 81.0 0.0;

80.0 5.2 86.0 0.0;

90.0 4.0 88.0 0.0;

96.5 3.2 89.0 0.0;

98.5 1.7 91.0 0.0;];

n=length(data);

%PF=data(:,2)

%ang=data(:,3)

%Iratio=data(:,4)

t=0;

y=0;

for i=1:n

%for k=1:4

%Zm=(1/PF(i))*((ang/100)+sin(acos(ang/100))*j)

Zm=(1/data(i,2))*((data(i,3)/100)+sin(acos(data(i,3)/100))*j);

Zl=Zm*10/(2.098*3);

I=v1/(zs+ztr+Zl);

v3=abs(I*Zl*2300/1.732);

w=data(i,1)*ws/100;

s=(ws-w)/ws;


24/63

R=(R1+R2/s)*(R1+R2/s);

X=(X1+X2)*(X1+X2);

Tm=((3*v3*v3*R2)/s)/(ws*(R+X));

Tacc=Tm-(data(i,4)/100);

del=w-y;

delt=(J*del)/Tacc;

t=t+delt;

t1(i)=t;

w1(i)=w;

I1(i)=I;

Tm1(i)=Tm;

s1(i)=s;

y=w;

%end

end

t1;

I1;

Tm1;

subplot(2,2,1);plot(t1,Tm1);title('Torque Vs time');

subplot(2,2,2);plot(s1,Tm1);

title('Torque Vs slip');

subplot(2,2,3);plot(t1,I1);

title('current vs time');

subplot(2,2,4);


25/63

OUTPUT:

RESULT


26/63

AIM

(i) To understand the modeling of STATCOM (static synchronous compensator) for power flow

studies using Newton Raphson method.

(ii)To investigate the operating principle and power injection modeling of STATCOM.

(iii)To investigate the control and operating limits of STATCOM.

SOFTWARE REQUIRED

MATLAB 7.8

THEORY

Power Injection Model of STATCOM

Accurate power flow calculation should consider the steady state losses of STATCOM. The

converter losses include mainly three parts: the power losses in the dc capacitor, the switching losses, and

conduction losses. The percentage of each loss component relates to the control mode of STATCOM and

the steady-state operating point.

The equivalent circuit of STATCOM is shown in Fig 5.1. In this circuit, the inductance, xs

represents the leakage inductance of the transformer, the series resistance, rs represents the sum of the

switching and conduction losses, and the shunt resistance, rp represents the power losses in the dc capacitor.

From Fig.5.1, it can be seen that the VSC acts as an ac voltage source behind equivalent

impedance, where both magnitude and Phase angle of the source are controllable. The voltage vector is

given as,

(5.1)

where,

Vdc = dc voltage across the capacitor

kc = converter dc to ac gain

kt = Ratio of the coupling transformer.

m = Modulation ratio of PWM control

= Control angle of VSC

Fig.5.1 Equivalent circuit of STATCOM

c c t dcV k k mV

Ex. No: LOAD FLOW ANALYSIS OF TWO-BUS SYSTEM WITH

STATCOMDate :


27/63

The current injected by STATCOM into the ac system is given as,

(5.2)

Where, is the voltage of ac system buss, and

(5.3)The active and reactive power injections of STATCOM into the ac system are given as,

(5.4)

(5.5)

Then the power balance equations at the ac system bus, are expressed as,

(5.6)

(5.7)

where, = PSPD (5.8)

= QS - QD (5.9)

and are calculated real and reactive powers respectively at bus s. and are fixed for

load buses. is fixed for generation buses. As the STATCOM being connected to the ac system at bus

s, and are not fixed. Since real and reactive power exchange between the ac system andSTATCOM is varying.

The active power output at the ac side of VSC is given as,

(5.10)

The real power of dc side of VSC is given as,

(5.11)

The power balance between the ac and dc side of the STATCOM is given as,

(5.12)

The active power exchange equation is given as,

(5.13)

c ss

s

V VI

Z

s s sV V

s s s sZ Z r jx

2* c t dc s s s

s s s

s

k k mV V cos( ) V cosP Re(V I )

Z

2* c t dc s s s

s s ss

k k mV V sin( ) V sinQ Im(V I )

Z

sch cal

s s sP P P sch cals s sQ Q Q

schsP

schsQ

calsP

calsQ

schsP

schsQ

schsP

sch

sPsch

sQ

2* c c s s

ac c ss

V cos V V cos( )P Re(V I )

Z

dc dcdc dc dc dc dc

p

dV VP V I V (C )

dt r

ac dcP P

ex ac dcP P P


28/63

By substituting Equations (5.10) and (5.11) in Eq.(5.13) and setting , the active power

exchange equation is reduced to,

(5.14)

Control Strategies of STATCOM

With the developments on power electronic switches, PWM control is becoming a more practicaltechnology for power system applications. In PWM control, switching losses associated with the relatively

fast switching of the electronic devices have an important effect on the power flow calculation, as these

have a direct effect on the charging and discharging of the dc capacitor, and hence should be considered in

the modeling. The proposed model can readily be adapted to represent PWM and phase control of

STATCOM. In Fig.5.1, eliminating the dc voltage control loop will yield the basic diagram of a controller

with a typical phase control strategy.

The output voltage magnitude of the VSC relates to the dc side voltage and the control strategy of

itself. If the VSC applies the phase control, the magnitude is a function of the dc side voltage and the phase

angle of the VSC. If the PWM mode is applied, the output voltage magnitude is a function of the control

phase angle and the modulation ratio of the PWM.

Operating limits of STATCOM

It is very important to adequately address the system limits when studying voltage stability. Voltage

collapse often occurs as a consequence of some devices hitting their limits in a heavily stressed power

system. For a STATCOM, the limits on the ac current, Is modulation ratio, m phase angle, and dc voltage,

Vdc. can be directly introduced into this model. Because the current limit is the main limiting factor in VSC-

based devices, this thesis will implement it in NR power flow. Either the maximum limit, or minimum

limit, is reached, depending on whether the controller is operating in the inductive or capacitive

domain.

Newton Raphson Equations with Control and Operating Limits of STATCOM

In order to implement the STATCOM model in NR power flow, two augmented bus types are

introduced, which are used to representing the specific control strategy and the operating limits of

STATCOM.

One is an augmented PV bus, namely APV. This is a controlled bus where the active and reactive

power injections and the voltage magnitude are specified, while the voltage phase angle of ac system bus,

control angle and/or modulation ratio and/or dc voltage of STATCOM are treated as the state variables.

dcV 0

2ex p c t dc s dc p c t s sP r (k k m) V cos Z V r k k mV cos( ) 0

maxsI

minsI


29/63

Fig 5.2 Bus types conversion and enforcement of the operating limit

The other is an augmented PQ bus, namely APQ. This is an uncontrolled bus where the active and

reactive power injections are specified, while the voltage magnitude and angle of ac system bus and the dc

voltage of STATCOM are treated as the state variables in its place. When the ac current Is lies between

and the bus is treated as APV type, otherwise it is APQ type. Fig.5.2 shows the conversion of bus

types and the enforcement of the current limit.

In order to implement the model in NR power flow, the STATCOM control strategy must be properly

expressed. As mentioned above, there are two control strategies for STATCOM and different variables must

be selected depending on whether a PWM or phase control is used in the controller. For PWM control, the

control variables are modulation ratio, m and control phase angle , and the control objective is to maintain

the dc voltage Vdc and ac voltage Vs at their reference value. For phase control, the control variable is phase

angle , and control objective is to maintain the ac voltage Vs at the reference value.

According to the operating principles, control strategies and operating limit of STATCOM, the

augmented NR power flow equations are described as follows:

(I) Phase Control Strategy

In unconstraint operation and phase control strategy of STATCOM, the modulation ratio m is given a

certain value, and ac buss is set as APV type. The augmented NR power flow equations are given by,

(5.15)

minsI

maxsI

s s s

s dcs s

s s ss dc

s dcex

ex ex ex

s dc

P P P

VPQ Q Q

Q VV

PP P P

V


30/63

(II) PWM Control Strategy

In unconstraint operation and PWM control strategy of STATCOM, the dc voltage V dc is assigned a

certain value, and buss is set as APV type, Jacobian elements in detail are illustrated in Appendix B and the

augmented NR power flow equations are given by,

(5.16)

(III) Enforcement of the current limit

Whether STATCOM is operating with PWM control or phase control, if the ac current limit ( or

) is violated, it will act as an unregulated voltage compensator whose production or absorption reactivepower capabilities will be a function of the nodal voltage at the STATCOM point of connection. In this

situation, the phase angle and/or modulation ratio m of the VSC are fixed at the values corresponding with

or , and the dc voltage, Vdc is uncontrollable. Then the ac system buss is converted to APQ type,

and the augmented NR power flow equations are given as,

(5.17)

s s s

ss s

s s ss

sex

ex ex ex

s

P P P

mPQ Q Q

Qm

P mP P P

m

minsI

max

sI

minsI

maxsI

s s s

s s dcs s

s s ss ss s dc

ex dc

ex ex ex

s s dc

P P P

V VP

Q Q QQ VV V

P VP P P

V V


31/63

256.6+110.2i.0125+.025i.01+.03i

.02+.04i 21

3

138.6+45.2i

EXERCISE

Fig 5.3 Single line diagram of three bus system with STATCOM

To conduct the load flow analysis for three bus system shown in figure 5.3 with and without

STATCOM (placed at bus three) and demonstrate the voltage control by STATCOM for various loading

conditions:

STATCOM parameters

rs = 0.01

Xs = 0.1

rp =200

K=Kc*Kt=0.9

Data of Three Bus System

Table A.1: Line Data of three bus system

Bus Resistance (R) Reactance (X) Susceptance(B)From To

1 2 0.02 0.04 0.05

1 3 0.01 0.03 0.03

2 3 0.0125 0.025 0.06

Table A.2 Bus data of three bus system

Bus

no

Bus

code

V

(p.u)

AngleLoad Generator

MW MVAR MW MVAR1 1 1 0 0 0 0 0

2 0 1 0 256.6 110.2 0 0

3 0 1 0 138.6 45.2 0 0


32/63

PROGRAM

% Program to form Admittance And Impedance Bus Formation....

% with Transformer Tap setting..

function ybus = ybusnr1() % Returns ybusclear all;

linedata = linedata3(); % Calling "linedata3.m" for Line Data...

fb = linedata(:,1); % From bus number...

tb = linedata(:,2); % To bus number...

r = linedata(:,3); % Resistance, R...

x = linedata(:,4); % Reactance, X...

b = linedata(:,5); % Ground Admittance, B/2...

a = linedata(:,6); % Tap setting value..

z = r + 1i*x; % Z matrix...

y = 1./z; % To get inverse of each element...

b = 1i*b; % Make B imaginary...

nbus = max(max(fb),max(tb)); % no. of buses...

nbranch = length(fb); % no. of branches...

ybus = zeros(nbus,nbus); % Initialise YBus...

% Formation of the Off Diagonal Elements...

for k = 1:nbranch

ybus(fb(k),tb(k)) = ybus(fb(k),tb(k))-y(k)/a(k);

ybus(tb(k),fb(k)) = ybus(fb(k),tb(k));

end

% Formation of Diagonal Elements....

for m = 1:nbus

for n = 1:nbranch

if fb(n) == m

ybus(m,m) = ybus(m,m) + y(n)/(a(n)^2) + b(n);

elseif tb(n) == m

ybus(m,m) = ybus(m,m) + y(n) + b(n);

end

end

end

line data

function linedata = linedata3()

linedata= [1 2 0.02 0.04 0.0 1;

1 3 0.01 0.03 0.0 1;

2 3 0.0125 0.025 0.0 1;];

end


33/63

STATCOM PROGRAM

function busdata = busdata3()% Bus| Type | Vsp | theta | PGi |QGi | PLi | QLi | Qmin | Qmax

busdata= [1 1 1.05 0 0.0 0 0 0 0 0;2 3 1.0 0 0.0 0 400 250 0 0;3 2 1.04 0 200 0 0 0 -50 100;];

End

function statdata = statdata3()statdata = [ 2 1 0 0.5 -0.5; ];end

% Program for Newton-Raphson Load Flow with STATCOM

% Program for Newton-Raphson Load Flow Analysis% Bus number 1 is assumed to be slack bus..

clc;Y = ybusnr1(); % Get Y-Bus..busdata = busdata3(); % Get Bus Data..statdata = statdata3();

baseMVA = 100; % Base MVA..bus = busdata(:,1); % Bus Number..type = busdata(:,2); % Type of Bus 1-Slack, 2-PV, 3-PQ..V = busdata(:,3); % Specified Voltage..del = busdata(:,4); % Voltage Angle..Pg = busdata(:,5); % PGi..Qg = busdata(:,6); % QGi..Pl = busdata(:,7); % PLi..Ql = busdata(:,8); % QLi..Qmin = busdata(:,9); % Minimum Reactive Power Limit..Qmax = busdata(:,10); % Maximum Reactive Power Limit..nbus = max(bus); % To get no. of buses..P = Pg - Pl; % Pi = PGi - PLi..Q = Qg - Ql; % Qi = QGi - QLi..P = P/baseMVA; % Converting to p.u..Q = Q/baseMVA;Qmin = Qmin/baseMVA;Qmax = Qmax/baseMVA;Tol = 10; % Tolerence kept at high value.Iter = 1; % iteration startingPsp = P;Qsp = Q;G = real(Y); % Conductance..

B = imag(Y); % Susceptance..

pv = find(type == 2 | type == 1); % Index of PV Buses..pq = find(type == 3); % Index of PQ Buses..

npv = length(pv); % Number of PV buses..npq = length(pq); % Number of PQ buses..

while (Tol > 1e-5) % Iteration starting..P = zeros(nbus,1);


34/63

Q = zeros(nbus,1);

% Calculate P and Qfor i = 1:nbusfor k = 1:nbus

P(i) = P(i) + V(i)* V(k)*(G(i,k)*cos(del(i)-del(k)) +

B(i,k)*sin(del(i)-del(k)));Q(i) = Q(i) + V(i)* V(k)*(G(i,k)*sin(del(i)-del(k)) -

B(i,k)*cos(del(i)-del(k)));

end

end

% Checking Q-limit violations..if Iter = 2for n = 2:nbusif type(n) == 2if Q(n) < Qmin(n)

V(n) = V(n) + 0.01;elseif Q(n) > Qmax(n)

V(n) = V(n) - 0.01;end

endendend

% Calculate change from specified valuedPa = Psp-P;dQa = Qsp-Q;dQ = zeros(npq,1);k = 1;

for i = 1:nbusif type(i) == 3

dQ(k,1) = dQa(i);k = k+1;

endenddP = dPa(2:nbus);M = [dP; dQ]; % Mismatch Vector

% Jacobian% J1 - Derivative of Real Power Injections with Angles..

J1 = zeros(nbus-1,nbus-1);for i = 1:(nbus-1)

m = i+1;for k = 1:(nbus-1)

n = k+1;if n == mfor q = 1:nbus

J1(i,k) = J1(i,k) + V(m)* V(q)*(-G(m,q)*sin(del(m)-del(q)) +

B(m,q)*cos(del(m)-del(q)));end

J1(i,k) = J1(i,k) - V(m)^2*B(m,m);else

J1(i,k) = V(m)* V(n)*(G(m,n)*sin(del(m)-del(n)) -

B(m,n)*cos(del(m)-del(n)));endendend


35/63

% J2 - Derivative of Real Power Injections with V..J2 = zeros(nbus-1,npq);for i = 1:(nbus-1)

m = i+1;for k = 1:npq

n = pq(k);

if n == mfor q = 1:nbus

J2(i,k) = J2(i,k) + V(q)*(G(m,q)*cos(del(m)-del(q)) +

B(m,q)*sin(del(m)-del(q)));end

J2(i,k) = J2(i,k) + V(m)*G(m,m);else

J2(i,k) = V(m)*(G(m,n)*cos(del(m)-del(n)) + B(m,n)*sin(del(m)-

del(n)));endendend

% J3 - Derivative of Reactive Power Injections with Angles..J3 = zeros(npq,nbus-1);for i = 1:npq

m = pq(i);for k = 1:(nbus-1)

n = k+1;if n == mfor q = 1:nbus

J3(i,k) = J3(i,k) + V(m)* V(q)*(G(m,q)*cos(del(m)-del(q)) +

B(m,q)*sin(del(m)-del(q)));end

J3(i,k) = J3(i,k) - V(m)^2*G(m,m);else

J3(i,k) = V(m)* V(n)*(-G(m,n)*cos(del(m)-del(n)) -

B(m,n)*sin(del(m)-del(n)));endendend

% J4 - Derivative of Reactive Power Injections with V..J4 = zeros(npq,npq);

for i = 1:npqm = pq(i);

for k = 1:npqn = pq(k);

if n == mfor q = 1:nbus

J4(i,k) = J4(i,k) + V(q)*(G(m,q)*sin(del(m)-del(q)) -B(m,q)*cos(del(m)-del(q)));end

J4(i,k) = J4(i,k) - V(m)*B(m,m);else

J4(i,k) = V(m)*(G(m,n)*sin(del(m)-del(n)) - B(m,n)*cos(del(m)-

del(n)));endendend


36/63

J = [J1 J2; J3 J4]; % JacobianX = inv(J)*M; % Correction VectordTh = X(1:nbus-1);dV = X(nbus:end);del(2:nbus) = dTh + del(2:nbus);k = 1;

for i = 2:nbusif type(i) == 3

V(i) = dV(k) + V(i);k = k+1;

endend

Iter = Iter + 1;Tol = max(abs(M));

end

Iter = Iter - 1; % Number of Iterations took..V;Del = 180/pi*del;

E1 = [V Del]; % Bus Voltages and angles..disp('------------------------------');disp('| Bus | V | Angle | ');disp('| No | pu | Degree | ');disp('------------------------------');for m = 1:nbus

fprintf('%4g', m), fprintf(' %8.4f', V(m)), fprintf(' %8.4f', Del(m));

fprintf('\n');enddisp('-----------------------------');disp('-----------------------------');disp('----------------------------------------');disp('| STATCOM | Vsh | Thst | Qsh |');disp('| Bus | pu | Degree | pu |');disp('----------------------------------------');for m = 1:np

fprintf(' %4g',statb(m)), fprintf(' %8.4f', Vsh(m)), fprintf(' %8.4f',

Thst(m)),fprintf(' %8.4f', Qsh(m)), fprintf('\n');end

disp('----------------------------------------');


37/63

RESULT


38/63

+

AIM

To understand modeling of STATCOM for transient stability analysis using explicit numerical

integration method.

SOFTWARE REQUIRED

MATLAB 7.8

THEORY

Voltage collapse of the transmission system has caused more than 90% of the blackouts. Voltage

magnitude regulation in the network can be achieved by controlling the production, absorption, and flow of

reactive power throughout the system. The industry needs fast-acting dynamic sources of reactive power to

counter fast voltage-collapse events. Sources or sinks are the Flexible AC Transmission System (FACTS)

sometimes permanently connected and others are automatically adjusted to maintain fixed voltagemagnitude at connection points.

The static synchronous compensator is a main member of the FACTS family of voltage source

converter (VSC) based devices. It is being to improve voltage regulation as well as to increase loadability

margin. The STATCOM is a shunt connected reactive power compensation device that is capable of

generating and/or absorbing reactive power and its output can be varied to control the specific parameters of

an electric power system. It is a solid state switching converter capable of generating or absorbing

independently controllable real and reactive power at its output terminals when it is fed from an energy

source at its input terminals.

Modelling of Power System ComponentsThe modelling of various power system components are presented below.

Synchronous Machine model

The synchronous generators are modelled as classical machines with rotor angle ( ) and

speed ( ) as state variables.

(6.1)

Fig 6.1 Synchronous Machine model

Ex. No: TRANSIENT ANALYSIS OF TWO-BUS SYSTEM WITH

STATCOMDate :


39/63

The equivalent circuit for classical model of the generator is given in Fig.6.2.a.This equivalent

circuit is the Thevenin form. The Norton form of the equivalent circuit is shown in Fig.6.2.b.

E'= V + (Ra+jXd

')I INorton= YNortonV + I

Fig.6.2. Modelling of Generators and Representation for Network Solution

INorton is the Norton current source. The expressions for the current source and the admittance is given by

INorton = E'/(Ra+jXd

') (6.2)

YNorton = 1 / (Ra+jXd') (6.3)

Transmission line model

Transmission line is modeled as lumped - model. The transmission lines are modeled as -

circuits using positive sequence parameters: series resistance, series reactance and half line charging.

Load Model

The loads are modeled as admittances. The admittances are computed from the initial load flow

solution as shown below.

YL = ( PLj QL) / |VL|2 (6.4)

where, PL and QL are the active and reactive powers of the load.

|VL| is the magnitude of the voltage of the load bus computed by load flow analysis.YL is the load admittance and it gets added to the diagonal of bus admittance matrix Y

corresponding to the node where the load is connected.

Modeling of STATCOM

The STATCOM consists of a voltage source converter which produces a set of 3 phase ac output

voltages each in phase with and coupled to the corresponding ac system voltage through a leakage

inductance of coupling transformer. The dc input voltage is provided by an energy storage capacitor. The

functional model of STATCOM is shown in Fig.6.3 where a VSI is connected to a utility bus through

magnetic coupling. The exchange of reactive power between the converter and the ac system can becontrolled by varying the amplitude of 3-phase voltage of the VSI. If the amplitude of the output voltage

of VSI is greater than the amplitude of utility bus voltage, then a current flows from VSI to ac system. If

the amplitude of the output voltage of VSI is lesser than the amplitude of utility bus voltage, then a

current flows from ac system to converter. Adjusting the phase shift between converter output voltage

and ac system voltage can control real power exchange between converter and ac system voltage.

(a) Thevenin equivalent

E'

Ra + jXd' I

V

(b) Norton equivalent

INorton

V

YNorton

I


40/63

Fig 6.3 Functional Model of STATCOM

Transient stability model

Assuming balanced, fundamental frequency voltages, STATCOM with PWM voltage control can

be accurately represented in transient stability studies using the basic model shown in Fig.6.4.

Pac = P dc + P loss (6.6)

Equation (6.6) basically represents the balance between the controllers ac power Pac and dc power Pdc

under balanced operation at fundamental frequency.

Magnitude

Magnitude

Controller

PWM

K (PWM)

Filters

+

+

-

-

+

-

i iV

s sI

P + j Q

Rt + j Xt

dcV - C CR

a : 1

refV

Vdcref

Fig 6.4 Transient stability model of STATCOM

AC System

Coupling

Transformer

KVdc

VdcIdc

C

Voltage

Source

Inverter

+-


41/63

1

mdc

mdc

K

sT

KidcKpdc

S

+

_

+

+refdcV

dcV

m axdcV

m indcV

Fig 6.6 DC voltage controller

1

Vdc

Vdcd

1

mac

mac

K

sT

1

2

(1 )p

D

K s T

K s T

0m

m

refV

m axI

minI

+

Vi

_

+

+

Fig.6.5 Voltage magnitude controller

m1Vd ac

Vpac

+

Voltage magnitude controller

This controller controls the AC output voltage magnitude of the voltage source inverter by

controlling the modulation index, m of the PWM controller. The controller has a bias corresponding to

the steady state value of the modulation index, 0m .

The differential equations are derived from the controller block shown in Fig.6.5

(6.7)

(6.8)

DC Voltage Controller

The DC voltage controller controls the phase angle of the output voltage of the voltage source

inverter. The phase angle determines the exchange of active power between the controllers with the ac

system and is used to directly control the DC voltage magnitude. The controller has a bias corresponding

to the phase angle, .The differential equations are derived from the controller blocks given in Fig 6.6 as,

(6.9);(6.10)

1 1 1

2

1dpac p p dpac D

m V K K T V K mT

1pac i mac pac

mac

V V K V T

1

1

dcp dc mdc dcp

mdc

dcdp idc dcdp pdc

V V K V T

V K V K


42/63

From power balance condition,

Pac = P dc + P loss (6.11)

From Fig 6.4, Pac= (6.12)

Pdc= Vdc I dc (6.13)

The capacitor dynamic equation is

(6.14)

Discretized form of these equations and interfacing STATCOM as a current with network are given in

appendix.

ALGORITHM TO ADVANCE SIMULATION BY ONE TIME STEP

The stepwise computations to be performed to advance the simulation by one step from t-t to t are

as follows. The time step width (t) used in the algorithm is 0.0001 sec.

Assumptions

The machines are considered to be classical (no controllers).

Damping ignored.

Loads are assumed as constant admittances.

Preparation

The initial conditions for , and voltages are obtained from load

Note:

i. The loads are converted into the constant admittances and these are pushed in to diagonal

elements of the corresponding load buses.

Y TRANBUSii = Y ii+ Y Li Here i is for all load buses.

ii. The diagonal elements of Y bus corresponding to all synchronous generators are modified

as:

Y TRANBUSii = Y ii+ 1/ (R1i+ j X d)

iii. The diagonal elements of Ybus corresponding to STATCOM are modified as:

Y TRANBUSii = Y ii+ 1/(Rt+ j X t).

i s i sV I cos( )


43/63

Explicit Method - Runge-Kutta Fourth Order Method

The general formula giving the value of x for the (n+1)thstep is

Xn+1 = Xn + 1/6*(k1 + 2 k2 +2 k3 + k4)

where,

k1 = f(xn , tn)t,

k2 = f(xn+ k1/2 , tn+t/2)t

k3 = f(xn+ k2/2 , tn+t/2)t

k4 = f(xn+k3 , tn+t)t

Explicit algorithm

The stepwise computations to be performed to advance the simulation by one step from t-t to t are

as follows. The time step width (t) used in the algorithm is 0.0001 sec.

Step 1: Find the Norton current injections at all the generator buses;

For synchronous machine :

For STATCOM connected bus :

Step2: Solve the network equations, which yields approximate bus voltage vector Vapprox

.

[Y] TRANBUS [V] = [I] NOR

Step3: Application of Fourth Order Runge-Kutta (RK) method involves the following sequence of steps.For synchronous generator:

First estimate. k1 )*t, l1 )*t

Second estimate. k2 +l1/2)*t , l2 +k1/2)*t

Third estimate. k3 +l2/2)*t , l3 +k2/2)*t

Fourth estimate. k4 +l3)*t, l4 +k3)*t

= 1/6*(k1+2k2 2k3 k4), = 1/6*(l1+2l2 2l3 l4)

= 0+ ; = 0+ .

Similarly include the STATCOM state variables also.

Step4: Repeat steps with updated values of state variables.

Step5: Introduce the fault and repeat step 6 and later the fault will be cleared and repeat the procedure.

1 '

q

NOR

d

EI

R jX

dcsh sh

t t

KVI

R jX


44/63

Handling network discontinuities

For the normal time advance and the first solution at the discontinuity, the induction generator is

represented by the Norton equivalent. The second solution is iterative as far as terminal voltage is

concerned since slip, being a state variable, cannot change across a discontinuity. The second solution is

computed using the model described by Norton equivalent. The pre-disturbance solution at a discontinuity

is called the first solution and the post disturbance solution is called the second solution.

Fig 6.7 Handling network discontinuities; dots denote the points computed by the algorithm

Solution (td-)

V

t

Second

Solution (td+)

td td+t

First

td - t


45/63

EXERCISE

To conduct the load flow analysis for three bus system shown in figure 7.1 with and without

STATCOM (placed at bus three) and demonstrate the voltage control by STATCOM for various loading

conditions

STATCOM parameters

rs =.01, Xs=0.1, rp=200,K=Kc*Kt=.9

1 .02+.04i 2

256.6+110.2i

.01+.03i .0125+.025i

3

138.6+45.2i

Fig 7.1 Single line diagram of three bus system with STATCOM

Data Of Three Bus System

Table A.1: Line Data of three bus system

Bus Resistance(r)

Reactance (x) Susceptance(b)From To

1 2 .02 .04 .05

1 3 .01 .03 .03

2 3 .0125 .025 .06

Table A.2 Bus data of three bus system

Bus

no

Bus code V

(pu)

Angle load Generator

MW MVAR MW MVAR

1 1 1 0 0 0 0 0

2 0 1 0 256.6 110.2 0 0

3 0 1 0 138.6 45.2 0 0


46/63

PROGRAM

clc;Pd=[0 2.562 1.386];Qd=[0 1.102 0.452];Ybus=[20-50j -10+20j -10+30j;

-10+20j 26-52j -16+32j;-10+30j -16+32j 26-62j];

Zs=0.01+0.1j;Kc=0.9;

Kt=0.9;V=[1.05 0.9810 1.000];theta=[0 -0.0640 -0.0547];Vdc=1;Vdco=1;Mo=0.2189;M=0.2189;alpha=0.1150;Msh=M-Mo;Vc=(Kc*Kt*M*Vdc);alpha1=alpha-theta(3);

% Ybus Modification

Ybusnew=Ybus;Xd=0.3j;Ybusnew(1,1)=Ybus(1,1)+1/Xd;Ybusnew(2,2)=Ybus(2,2)+(Pd(2)-Qd(2)*j)/(V(2)^2);Ybusnew(3,3)=Ybus(3,3)+1/Zs+(Pd(3)-Qd(3)*j)/(V(3)^2);for i=1:3

Vrect(i)=V(i)*(cos(theta(i))+j*sin(theta(i)));endI=Ybusnew*Vrect;% Transient stability%faultdisp('During fault')count =1;c=0.043;r=0;Kp=10;Ki=10;G=0;Pm=0.85;f=50;wo=2*pi*f;dt=0.0001;Vref=V(3);Y=V(3);w=wo;H=4;Inorton=(1.0500-j*2.1155)/V(1);

Eg=V(1)+Inorton*Xd;

del=angle(Eg);Pe=real(Eg*conj(Inorton));Ybusnew(3,3)=1000000000;VV=inv(Ybusnew)*I;

V(3)=abs(VV(3));theta(3)=angle(VV(3));c3(count)=abs(I(3));


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Vcc(count)=Kc*Kt*M*Vdc;V3(3)=V(3);t(1)=0;while(count


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V

V3(count)=V(3);t(count)=t(count-1)+dt;c3(count)=abs(I(3));Vcc(count)=Kc*Kt*M*Vdc;end

%After clear the faultdisp('After the fault')

Ybusnew(3,3)=Ybus(3,3)+1/Zs;Vrect=inv(Ybusnew)*I;for i=1:3

V(i)=abs(i)theta(i)=angle(i);

end

while(count


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alpha=alpha1+theta(3);Y=V(3);Msh=Msh+(x1+2*(x2+x3)+x4)/6;M=Msh+Mo;

for i=1:3Vrect(i)=V(i)*(cos(theta(i))+j*sin(theta(i)));

end

Eg=(abs(Eg))*(cos(del)+j*sin(del));

Inorton=(Eg-V(1))/Xd;

Pe=real(Eg*conj(Inorton));VV=inv(Ybusnew)*I;V(3)=abs(VV(3));theta(3)=angle(VV(3));V

V3(count)=V(3);Vcc(count)=Kc*Kt*M*Vdc;t(count)=t(count-1)+dt;c3(count)=abs(I(3));end

plot(t,V3)xlabel('time in msec')ylabel('Bus3 voltage in p.u')title('Voltage at Bus3 Vs Time')


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OUTPUT:

During faultcount = 2

V =

1.0500 0.9810 0.0000

count = 3

V =

1.0500 0.9810 0.0000

count = 4

V =

1.0500 0.9810 0.0000

count = 5

V =

1.0500 0.9810 0.0000

After the fault

V =

1.0000 0.9810 0.0000

V =

1.0000 2.0000 0.0000

V =

1 2 3

count = 6

V =

1.0000 2.0000 1.0564

count = 7

V =

1.0000 2.0000 1.0564


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count = 8

V =

1.0000 2.0000 1.0564

count = 9

V =

1.0000 2.0000 1.0564

count = 10

V =

1.0000 2.0000 1.0564

RESULT


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AIM

(i) To estimate the available transfer capability between the sending area and receiving area

through the tie-lines connecting the two areas.

(ii) To study the methods of determining the ATC and the importance of it.

SOFTWARE USED

MATLAB 7.8

THEORY

Definitions

Available Transfer Capability (ATC) is a measure of the transfer capability remaining in the

physical transmission network for further commercial activity over and above the already committed uses.

ATC can be expressed as,

ATC=TTC-TRM-{ETC+CBM}

where,

TTC (Total Transfer Capability) is defined as the amount of electric power that can be

transferred over the interconnected transmission network or particular path or interface in a reliable

manner while meeting all of a specific set of defined pre and post contingency conditions.

TRM (Transfer Reliability Margin) is defined as the amount of transmission transfer

capability necessary to ensure that the interconnected network is secure under a reasonable range of

uncertainties in the system conditions.

CBM (Capacity Benefit Margin) is the amount of transmission transfer capability reserved

by the load serving entities to ensure access to generation from interconnected systems to meet generation

reliability requirements. It also helps to reduce the installed capacity of the plant.

ETC (Existing Transfer Capability) refers to the power transfer capability that must be

reserved for already committed transactions.

Existing transaction is the power flow over the transmission paths at the desired time at

which ATC should be calculated. This is the already committed used power on the transmission path.

Utilities would have to determine adequately their ATCs to ensure that system reliability is maintained

while serving a wide range of transmission transactions. ATC between and within areas of the

interconnected power system and ATC for critical transmission paths between areas would be

continuously updated and posted changes in scheduled power transfers between areas.

Ex. No: AVAILABLE TRANSFER CAPABILITY CALCULATION

USING AN EXISTING LOAD FLOW PROGRAMDate :


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Governing Principles

The following are the ATC principles that govern the development of the definition and

determination method of ATC:

The ATC value produced by the calculation must give a reasonable and dependable

indication of transfer capabilities available to the electric power market.

ATC calculation must recognize the dependency of ATC on the points of electric power

injection, the direction of transfer across the interconnected transmission network and the

points of power extraction.

Regional or wide-area co-ordination is necessary to develop the post information that

reasonably reflects the ATC of the interconnected transmission network.

ATC calculations must confirm to regional, sub-regional, power pool and individual system

reliability planning and operating policies, criteria or guides.

The determination of ATC can accommodate reasonable uncertainties in system conditions

and provide operating flexibility to ensure the secure operation of the interconnected

network.

Significance of ATC

The information of ATC as an important indicator of the system performance, is useful in

restructured energy market in the following ways:

It provides the knowledge of power system capability about the present system condition.

Running the system under ATC limits also ensures system security and reliability to some

extent, since the calculation of ATC is based on the security constraints with the

consideration of critical contingencies that can lead the system normal state to alert state.

The ATC is required in making decisions for the transactions between market participants.

The market participants check for the power contract themselves.

The ATC is also useful in enhancing the system stability. With the knowledge of the

limiting condition for the ATC, the system operator can take some operating or planningdecision to avoid this limiting condition and thus enhance system capability.

The ATC can also serve as an indicator of power congestion through transmission lines.

The ATC is useful in transmission costing function. The ISO can put more transmission

cost for the transaction through transmission path having low value of ATC.


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Methods of Static ATC Determination

Method based on linear sensitivity factors

Method based on Continuation power flow

Method based on Optimization power flow

Method based on linear sensitivity factors

It offers a great potential for real time calculation of ATC. Use of these factors offers an

approximate but extremely fast model for static ATC determination. In this we are having DC power

transfer distribution factor, AC power transfer distribution factors, and line outage distribution factors for

ATC determination.

PTDF-Definition

For ATC determination the MW flows must be allocated to each line or group of line in proportion

to the MWs being transmitted by each transaction. This is accomplished through the use of the linear

Power Transfer Distribution Factors (PTDF). From the power flow point of view, a transaction is a

specific amount of power that is injected into the system at one bus by a generator and removed at another

bus by a load.

The coefficient of linear relationship between amount of transaction and flow on a line is called

PTDF. It is also called as sensitivity because it relates the amount of one change transaction amount to

another change line power flow. The PTDF is the fraction of the amount of transaction from one bus to

another that flows over a given transmission line. PTDFij, mn is the fraction of transmission from bus m to n

that flows over a transmission line connecting bus i to j.

Linear sensitivity factors method use DC power flow. The DC power flow model assumes thatonly the angles of the complex bus voltage vary, and the variation in small voltage magnitudes is assumed

to be constant. Transmission lines are assumed to have no resistance and hence no losses.

The equation for PTDF is

PTDFij,mn= (Xim- XjmXin + Xjn) / xij

where,

xij is the reactance of transmission line connecting bus i and bus j.

Xim is entry in the ith row and mth column of the bus reactance matrix X.

The maximum amount of power transfer is given by,

Pmn,ijmax = (Pijmax - Pijo )/ PTDFij,mn

where,

Pijmax

is the lines flow limit.

Pijo

is the base case flow on the line.

ATC is the minimum of the maximum allowable transactions over all the lines and is expressed as,

ATCmn = minij { Pmn,ijmax

}


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Method Based on Continuation of Power Flow

This method runs AC load flow for each increment of transaction between an interface and checks

whether any of the operating conditions such as line flow limit or bus voltage limit is violated. For the

increment of transactions, continuation power flow is also used to find out maximum load ability or

voltage instability point.

Method Based on Optimization of Power Flow

ATC determination can be formulated as an optimization problem which finds out maximum value

of the transaction between given interface while satisfying the network power balance and security

constraints such as line flow, voltage limits and voltage instability points.

ALGORITHM

Step 1: Start

Step 2: Read the number of buses and their type.

Step 3: Read the power generation (Pg) and power demand (Pd) at all the buses and calculate the

net power injection (P).

Step 4: Read the voltage at each buses or if not available initialize to flat start.

Step 5: Read the voltage limits Vmin and Vmax.

Step 6: Read the bus load angles or if not available initialize to flat start.

Step7: Read the Transmission Reliability Margin (TRM) & Capacity Benefit Margin (CBM)

values for all lines.

Step 8: Read or compute the Y-bus matrix.

Step 9: Set a suitable convergence tolerance and run Fast Decoupled Load Flow (FDLF).

Step 10: From FDLF find the new power flows at all the buses.

Step 11: Calculate the power flows through all lines P ij=Pi-Pj.

Step 12: Existing Transfer Commitments (ETC) = Flow through the lines.

Step 13: Increase the power demand at all the load buses by a small set value.

Step 14: Run steps 8,9,10.

Step 15: Check for voltage limit violation, if not violated go to 12 else 16.

Step 16: Calculate the power flows through all lines P ij=Pi-Pj.

Step 17:Total Transfer Commitments (TTC) = Flow through the lines.

Step 18: Calculate Available Transfer Capability ATC=TTC-TRM-{ETC+CBM}


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EXERCISE

To determine the ATC of the given sample bus system.

Data of Three Bus System

Table A.1 Line Data of Three Bus System

Bus Resistance

(R)

Reactance

(X)

Susceptance

(B)From To

1 2 .02 .04 .05

1 3 .01 .03 .03

2 3 .0125 .025 .06

Table A.2 Bus Data of Three Bus System

Bus

No

Bus

Code

V

(p.u)Angle

Load Generator

MW MVAR MW MVAR

1 1 1 0 0 0 0 0

2 0 1 0 256.6 110.2 0 0

3 0 1 0 138.6 45.2 0 0

256.6+110.2i

.0125+.025i.01+.03i

.02+.04i 21

3

138.6+45.2i

Fig 7.1 Single line diagram of three bus system


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PROGRAM

clear all

clc;

n=2;

Pd=[0 1.7];

Qd=[0 -1.0535];

Pg=[0 0];

v=[1 1];

vmin=[0.95 0.85];

vmax=[1.05 1.00];

TRM=0;

CBM=0;

theta=[0 0];

ybus=[3.815629-19.078144*i -3.815629+19.078144*i; -3.815629+19.078144*i

3.81569-19.078144*i];

b=-imag(ybus);

g=real(ybus);an=angle(ybus);

magybus=abs(ybus);

Bmat(1:1,1:1)=b(2:2,2:2);

Cmat(1:1,1:1)=b(2:2,2:2);

increment=1;

exit=0;

while(increment0.01 && iter


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for k=1:n

qq(k)=0;

for l=1:n

qe(k)=-v(k)*magybus(k,l)*v(l)*sin(an(k,l)-

theta(k)+theta(l))+qq(k);

qq(k)=qe(k);

end

end

qq;delq(1:1)=(Qd(2:2)-qq(2:2));

delv=inv(Cmat)*delq';

dv(2:2)=delv(1:1);

v=v+dv;

con=max(abs(dv));

iter=iter+1;

end

flow=((theta(1)-theta(2))/0.6);

atc(increment)=flow;

for i= 1:n

if(v(i)>vmax(i))

exit=1;

end

if (v(i)


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OUTPUT:

increment =

573

ans =

4.5600

ETC =

0.1377

ATC =

0.2965

RESULT


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AIM

(i) To compute the harmonic analysis of the output voltage and output current.

(ii)To find the harmonic order by conducting FFT analysis.

SOFTWARE REQUIRED

MATLAB SIMULINK

THEORY

Harmonics

Harmonics are sinusoidal components of a periodic wave or quantity having a frequencythat is an integral multiple of the fundamental frequency. The fundamental together with its frequency

multiples is referred as harmonic series. The frequency multiples of 3f, 5f, 7f etc are odd harmonics while,

the frequency multiples of 2f, 4f, 6f etc are even harmonics. The FFT analysis is done to find the harmonic

order. The sinusoidal voltage or current which is dependent on time is represented by the following

expression

v(t) = Vsin(wt)

i(t) = Isin(wt+)

Non sinusoidal voltage wave is represented in Fourier series as follows:

v(t) =V0+ V1 sin(wt) + V2 sin(2wt) + V3sin(3wt) ..

Sources of Harmonics

Any electrical load that presents a constant impedance to the power source throughout the

cycle is a linear load. Linear loads draw sinusoidal current from the source and do not cause harmonics.

A non-linear load can be defined as one which draws current discontinuously or its

impedance varies with each cycle of the ac waveform. It is the source of harmonics. Some of the non-

linear loads which cause harmonics are given below:

Diode and thyristor rectifiers.

Switched mode power supplies.

Variable speed drives.

Electronic power supplies.

DC motor drives.

Battery charges.

Electronic ballasts.

Arc furnace.

Ex. No: COMPUTATION OF HARMONIC INDICES

GENERATED BY A RECTIFIER FEEDING A R-L LOADDate :


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Effects of Harmonics

Increase in system line losses.

Interaction with power system equipment and causes additional line losses, over heating,

over loading.

Leads to voltage distortion.

Leads to low power factor.

Interference with telecommunication lines.

Errors in metering device.

Low efficiency.

Low power factor.

Harmful disturbances to neighboring appliances.

SYSTEM MODEL

Fig 8.1 Single-phase bridge rectifier with RL load

System Parameters

Frequency=60Hz.

Load resistance=1000.

Load inductance=1H.


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SIMULINK DIAGRAM


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OUTPUT

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