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 :
7/28/2019 M.E record 2Sem.docx
2/63
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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.
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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]
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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 :
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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))
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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 :
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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,'-')
7/28/2019 M.E record 2Sem.docx
20/63
OUTPUT:
RESULT
7/28/2019 M.E record 2Sem.docx
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 :
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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;
7/28/2019 M.E record 2Sem.docx
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);
7/28/2019 M.E record 2Sem.docx
25/63
OUTPUT:
RESULT
7/28/2019 M.E record 2Sem.docx
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 :
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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);
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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('----------------------------------------');
7/28/2019 M.E record 2Sem.docx
37/63
RESULT
7/28/2019 M.E record 2Sem.docx
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 :
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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
+-
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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( )
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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
7/28/2019 M.E record 2Sem.docx
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));
7/28/2019 M.E record 2Sem.docx
47/63
Vcc(count)=Kc*Kt*M*Vdc;V3(3)=V(3);t(1)=0;while(count
7/28/2019 M.E record 2Sem.docx
48/63
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
7/28/2019 M.E record 2Sem.docx
49/63
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')
7/28/2019 M.E record 2Sem.docx
50/63
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
7/28/2019 M.E record 2Sem.docx
51/63
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
7/28/2019 M.E record 2Sem.docx
52/63
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 :
7/28/2019 M.E record 2Sem.docx
53/63
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.
7/28/2019 M.E record 2Sem.docx
54/63
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
}
7/28/2019 M.E record 2Sem.docx
55/63
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}
7/28/2019 M.E record 2Sem.docx
56/63
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
7/28/2019 M.E record 2Sem.docx
57/63
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
7/28/2019 M.E record 2Sem.docx
58/63
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)
7/28/2019 M.E record 2Sem.docx
59/63
OUTPUT:
increment =
573
ans =
4.5600
ETC =
0.1377
ATC =
0.2965
RESULT
7/28/2019 M.E record 2Sem.docx
60/63
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 :
7/28/2019 M.E record 2Sem.docx
61/63
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.
7/28/2019 M.E record 2Sem.docx
62/63
SIMULINK DIAGRAM
7/28/2019 M.E record 2Sem.docx
63/63
OUTPUT