Power System Modeling and Simulation lab manual

SSCE, Udaipur Lab: PSMS Lab (7EE8A)
1
Dheeraj Kr. Dhaked B.Tech [EE, VII Semester]
Experiment No. 1
OBJECTIVE
Simulate Swing Equation in Simulink (MATLAB).
APPARATUS REQUIRED
MATLAB 7.8.0 (R2009-a).
THEORY
Swing equation being a non-linear equation, numerical methods are used to solve it. Point by
Point method is one of the classical solutions to solve swing equation.
Below is a solution of swing equation for a machine connected to infinite bus through two
parallel lines. Swing equation is drawn for a persisting fault in one of the parallel line and also
after fault is cleared. Stability of system is concluded after analyzing the swing curve and
clearing angle is calculated for system stability.
f = 50 Hz generator 50 MVA supplying 50 MW with inertia constant 'H' = 2.7 MJ/MVA at rated
speed. E = 1.05 pu ,V = 1 pu, X1 = X2 = 0.4 pu. three phase fault at line 2.
(a) Plot swing curve for a sustained fault up to a time of 5 secs.
(b) Plot swing curve if fault is cleared by isolating line in 0.1 seconds.
(c) Find the critical clearing angle
Fig1: One line diagram

2
Programming
%% MVA base = 50
% given
E = 50; V =1; Xd = 0.2; X1 =0.4; X2 = 0.4;H = 2.7;
% prefault condition
del = 0:pi/10:pi;
del1 =del;
del2 = del;
M = 2.7/(180*50); % angular momentum = H/180*f
Peo = (1.05/0.4)*sin(del); % Initial power curve
Po = 1 ; % power output in pu = 50 MW/50 MVA
delo = asind(0.4/1.05); % initial load angle in degrees //Pe =
(E*V/X) sin(delo)
% During fault
Pe2 = 1.05*sin(del1); % Power curve during fault
%Post fault condition
Pe3 = (1.05/0.6)*sin(del2); % Power curve after clearing fault
%% Primary Power curve plot Figure-1
plot(del,Peo);
set(gca,'XTick',0:pi/10:pi);
set(gca,'XTickLabel',{'0','','','','','pi/2','','','','','pi'});
title('Power Curve');
xlabel('Load angle');
ylabel('Genpower');
text((2/3)*pi,(1.05/0.4)*sin((2/3)*pi),'leftarrow intial
curve','HorizontalAlignment','left');
text(pi/2,2.75,'2.625*sindelta','HorizontalAlignment','center');
hold all
plot(del1,Pe2);
text((2/3)*pi,1.05*sin((2/3)*pi),'leftarrow during
fault','HorizontalAlignment','left');
plot(del2,Pe3);
text((2/3)*pi,(1.05/0.6)*sin((2/3)*pi),'leftarrow fault
cleared','HorizontalAlignment','left');
hold off
%% ------------
t = 0.05; % time step preferably 0.05 seconds
t1 = 0:t:0.5;

3
%% (a) sustained fault at t = 0
% for discontinuity at t = 0 , we take the average of accelerating power
% before and after the fault
% at t = 0-, Pa1 = 0
% at t = 0+. Pa2 = Pi - Pe2
% at t = 0 ,Pa =Pa1+Pa2/2
Pao = (1 - (1.05*sind(delo)))/2; % at the instant of fault del1 = delo
Pa(1) = Pao;
cdel(1) = 0;
d1 = t^2/M;
for i = 1:11
if i == 1
d2(i) = d1*Pa(i);
del(i) = delo;
else
cdel(i) = cdel(i-1)+d2(i-1);
del(i) = del(i-1)+cdel(i);
Pe(i) = 1.05*sind(del(i));
Pa(i) = 1 - Pe(i);
d2(i) = d1*Pa(i);
end
end
%% swing curve 1 plot
figure (2);
plot(t1,del);
set(gca,'Xtick',0:0.05:0.5);
set(gca,'XtickLabel',{'0','0.05','0.10','0.15','0.20','0.25','0.30','0.35','0
.40','0.45','0.50'});
title('Swing Curve');
xlabel('seconds');
ylabel('degrees');
text(0.30,150,' Sustained fault','HorizontalAlignment','right');
text(0.001,130,' load angle increases with time -- Unstable
state','HorizontalAlignment','left');
%% (b) Fault cleared in 0.10 seconds ,2nd step ---- 3rd element [1]0
[2]0.05,[3]0.10
Pafo = (1 - (1.05*sind(delo)))/2; % at the instant of fault del1 = delo
Paf(1) = Pao;
cdelf(1) = 0;

4
d1f = t^2/M;
for i = 1:2
if i == 1
d2f(i) = d1*Pa(i);
delf(i) = delo;
else
cdelf(i) = cdelf(i-1)+d2f(i-1);
delf(i) = delf(i-1)+cdelf(i);
Pef(i) = 1.05*sind(delf(i));
Paf(i) = 1 - Pef(i);
d2f(i) = d1*Paf(i);
end
end
% after clearing fault, power curve shift to Pe3
for i = 3:11
if i == 3
a1 = Paf(i);
d2f(i) = d1*Paf(i);
a2 = d2f(i);
d2f(i) = d1*Paf(i);
Paf(i) = (Paf(i)+ a1)/2;
d2f(i) = (d2f(i) + a2)/2;
else

5
d2f(i) = d1*Paf(i);
end
end
%% ------
figure (3);
plot(t1,delf);
set(gca,'Xtick',0:0.05:0.5);
set(gca,'XtickLabel',{'0','0.05','0.10','0.15','0.20','0.25','0.30','0.35','0
.40','0.45','0.50'});
title('Swing Curve');
xlabel('seconds');
ylabel('degrees');
text(0.25,57,' Fault Cleared in 0.10 sec','HorizontalAlignment','right');
text(0.15,30,' load angle decreases with time -- Stable
state','HorizontalAlignment','left');
%% (c) critical clearing angle
delo = degtorad(delo); % initial load angle in rad
delm = pi - asin(1/1.75); % angle of max swing
c1 = ((delm-delo)-(1.05*cos(delo))+(1.75*cos(delm)))/(1.75-1.05);
cclang = acos(c1); % critical clearing angle
in rad
cclang = radtodeg(cclang); % critical clearing angle
in degree
cclang = int16(cclang); % converting to integer
fprintf('nntt Critical Clearing angle is %d degree ',cclang);
Results

6
Fig 2: Power Curve
Fig 3: Swing Curve
Fig 4: Swing curve for fault clearing in 0.10 sec
Viva Questions
1. What is the unit of Inertia constant?
Ans: Mega-Joule /MVA.

7
2. What is swing equation?
Ans: Sin
X
VV
P
sr
r
.

3. For enhancing the power transmission in along EHV transmission line, the most preferred
method is to connect a-
(A) Series inductive compensator in the line
(B) Shunt inductive compensator at the receiving end
(C) Series capacitive compensator in the line
(D) Shunt capacitive compensator at the sending end
4. What is a disturbance and what are the two types of disturbances?
Ans: If a sudden change or sequence of changes occurs in one or more of the system
parameters or one or more of its operating quantities, the system is said to have undergone a
disturbance from its steady state operating condition. The two types of disturbances in a
power system are, i) Large disturbance ii) Small disturbance
5. What is a small disturbance? Give example.
Ans: If the power system is operating in a steady state condition and it undergoes change,
which can be properly analyzed by linearized versions of its dynamic and algebraic
equations, a small disturbance is said to have occurred. Example of small disturbance is a
change in the gain of the automatic voltage regulator in the excitation system of a large
generating unit.
6. What is a large disturbance? Give some examples.
Ans: A large disturbance is one for which the nonlinear equations describing the dynamics of
the power system cannot be validly linearized for the purpose of analysis. Examples of large
disturbances are transmission system faults, sudden load changes, loss of generating units
and line switching. 6. When is a power system said to be steady-state stable? The power
system is steady state stable for a particular steady-state operating condition if, following a
small disturbance, it returns to essentially the same steady state condition of operation.
7. When is a power system said to be transiently stable?

8
Ans: If the machines of the system are found to remain essentially in synchronism within the
first second following a system fault or other large disturbance, the system is considered to
be transiently stable.
8. What is transient state of the power system?
Ans: The state of the system in the first second following a system fault or large disturbance
is called the transient state of the power system.

9
Experiment No. 2
Objective
Modeling of synchronous machine in MATLAB software.
Apparatus Required
MATLAB 7.8.0 (R2009-a)
Theory:
Circuit Description
A three-phase generator rated 200 MVA, 13.8 kV, 112.5 rpm is connected to a 230 kV, 10,000
MVA network through a Delta-Wye 210 MVA transformer. At t = 0.1 s, a three-phase to ground
fault occurs on the 230 kV bus. The fault is cleared after 6 cycles (t = 0.2 s). During this demo,
you will initialize the system in order to start in steady-state with the generator supplying 150
MW of active power and observe the dynamic response of the machine and of its voltage and
speed regulators.
Demonstration
1. Start Simulation and observe the three machine currents on the Iabc scope. If the 9 parameters
defining initial conditions for the Synchronous Machine are set at zero or not set correctly, the
simulation will not start in steady state.
2. In order to start the simulation in steady-state you must initialize the synchronous machine for
the desired load flow. Open the Powergui and select 'Load Flow and Machine Initialization' . A
new window appears. The machine 'Bus type' should be already initialized as 'PV generator',
indicating that the load flow will be performed with the machine controlling the active power
and its terminal voltage. Specify the desired values by entering the following parameters :
Load flow: Terminal voltage (Vrms) = 13800; Active Power = 150e6. Then press the 'Execute
Load Flow' button.

10
Fig 1: Simulink model of synchronous machine
Once the load flow has been solved the phasors of AB and BC machine voltages as well as the
currents flowing out of phases A and B are updated. The machine reactive power, mechanical
power and field voltage requested to supply the electrical power should also be displayed: Q =
3.4 Mvar; Pmec = 150.32 MW (0.7516 pu); field voltage Ef = 1.291 pu.
3. In order to start the simulation in steady state with the HTG and excitation system connected,
these two Simulink blocks must also be initialized according to the values calculated by the load
flow. This initialization is automatically performed when you execute the Load Flow, as long as
you connect at the Pm and Vf inputs of the machine either Constant blocks or regulation blocks
from the machine library (HTG, STG, or Excitation System). Open the HTG block menu and
notice that the initial mechanical power has been automatically set to 0.5007 pu (100.14 MW) by
the Load Flow. Then, open the Excitation System block menu and note that the initial terminal
voltage and field voltage have been set respectively to 1.0 and 1.1291 pu.
4. Open the 4 scopes and restart the simulation. The simulation now starts in steady state.
Observe that the terminal voltage Va is 1.0 p.u. at the beginning of the simulation. It falls to
about 0.4 pu during the fault and returns to nominal quickly after the fault is cleared. This quick
response in terminal voltage is due to the fact that the Excitation System output Vf can go as
high as 11.5 pu which it does during the fault. The speed of the machine increases to 1.01 pu
during the fault then it oscillates around 1 p.u. as the governor system regulates it. The speed
Synchronous Machine
1
Continuous
1.0
-K-
Volts > pu
Vf (pu)
Va (pu)
v
+
-
Va
A
B
C
a
b
c
Three-phase
Transformer
210 MVA 13.8 kV / 230 kV
Pm
Vf _
m
A
B
C
Synchronous Machine
200 MVA 13.8 kV
0.7516
wref
Pref
we
Pe0
dw
Pm
gate
HTG
v ref
v d
v q
v stab
Vf
Excitation
System
A
B
C
5 MW
A
B
C
A
B
C
3-Phase Fault
A
B
C
10,000 MVA, 230 kV
source
A
B
C
10 MW
Speed (pu)
Iabc (pu)
<Stator current>
<Stator v oltage v q (pu)>
<Stator v oltage v d (pu)>
<Rotor speed wm (pu)>
<Rotor speed dev iation dw (pu)>
<Output activ e power Peo (pu)>

11
takes much longer than the terminal voltage to stabilize mainly because the rate of valve
opening/closing in the governor system is limited to 0.1 pu/s.
Fig 2: Stator current of synchronous machine

12
Fig 3: Stator voltage in pu at Phase a

13
Fig 4: Synchronous machine field voltage
Fig 5: Rotor Speed of synchronous machine
Viva-Questions
1. If supply frequency increases, the skin effect is?
Ans: Increased
2. Arc lamp operates at?
Ans: Low lagging power factor
3. The transformer used for AC welding set is?
Ans: step down type
4. The No-Load speed of DC motor is?
Ans: Very high
5. A Dynamometer type wattmeter can be used on?
Ans: A.C. as well as D.C.
6. The field winding of an alternator requires?
Ans: D. C. Supply
7. Torque angle of synchronous motor is?

14
Ans: angle between stator field axis and rotor field axis.
8. Zero power factor method for an alternator is generally used to determine?
Ans: Voltage regulation of alternator.
9. Core losses are determined in transformer by which test?
Ans: Open circuit Test.
10. Iron losses in transformer are determined by?
Ans: Short circuit Test.

15
Experiment No. 3
Objective:
Modeling of Induction Machine in MATLAB.
Apparatus Required:
MATLAB 7.8.0 (R2009a)
Programming:
Rr =0.39; %Rotor resistance
Rs =0.19; %Stator resistance
Lls =0.21e-3; %Stator inductance
Llr =0.6e-3; %Rotor inductance
Lm =4e-3; %Magnetizing Inductance
fb =100; %Base frequency
p =4; %Number of poles
J =0.0226; %Moment of inertia
Lr = Llr + Lm;
Tr = Lr / Rr;
% Impedance and angular speed calculations
wb =2*pi*fb; %Base speed
Xls =wb*Lls; %Stator impedance
Xlr =wb*Llr; %Rotor impedance
Xm =wb*Lm; %Magnetizing impedance
Xmstar =1/(1/Xls+1/Xm+1/Xlr);

16
Fig 1: Simulink model of Induction machine
Fig 2: Simulink model of Induction machine abc to dq axis
we & wr
o/p Current
Te & Tl
100
Stator angular frequency
Rotor angular speed
Phase3
Phase2
Phase1
5
Load Torque
Input Voltage
v ao
v bo
v co
Tl
we
Te
wr
ia
ib
ic
Induction Machine
Electrical o/p Torque
5
ic
4
ib
3
ia
2
wr
1
Te
we theta-e
theta-e
iqs
ids
cos(theta-e)
sin(theta-e)1
ia
ib
ic
syn-abc
vao
vbo
vco
van
vbn
vcn
o-to-n
van
vbn
vcn
cos(theta-e)
sin(theta-e)
vqs
vds
abc-syn
sin(u)
cos(u)
SinCos
Vqs
Vds
we
Tl
Te
wr
iqs
ids
iqr
idr
Induction Motor d-q model
5
we
4
Tl
3
vco
2
vbo
1
vao
6
idr
5
iqr
4
ids
3
iqs
2
wr
1
Te
Te
Tl
wr
WR
Fqs
iqs
Fds
ids
Te
TE
Fqs
Fmq
iqs
Iqs
Fmq
Fqr
iqr
Iqr
Fds
Fmd
ids
Ids
Fmd
Fdr
idr
Idr
Vqs
we
Fds
Fqr
Fqs
Fqs
we
wr
Fdr
Fqs
Fqr
Fqr
Fqs
Fqr
Fmq
Fmq
Fds
Fdr
Fmd
Fmd
Vds
we
Fqs
Fdr
Fds
Fds
we
wr
Fqr
Fds
Fdr
Fdr
4
Tl
3
we
2
Vds
1
Vqs

17
Fig 3: Simulink model of Induction machine dq axis model
Results:
Fig 4: Output Current of Induction Machine
Fig 5: Input Voltage of Induction Machine

18
Fig 5: We and Wr of induction motor
Fig 6: Te and Ti induction motor

19
Viva-Questions
1. A Dynamometer type wattmeter can be used on?
Ans: A.C. as well as D.C.
2. The field winding of an alternator requires?
Ans: D. C. Supply
3. Torque angle of synchronous motor is?
Ans: angle between stator field axis and rotor field axis.
4. Zero power factor method for an alternator is generally used to determine?
Ans: Voltage regulation of alternator.
5. Core losses are determined in transformer by which test?
Ans: Open circuit Test.
6. Iron losses in transformer are determined by?
Ans: Short circuit Test.
7. If supply frequency increases, the skin effect is?
Ans: Increased
8. Arc lamp operates at?
Ans: Low lagging power factor
9. The transformer used for AC welding set is?
Ans: step down type

20
Experiment No. 4
Objective:
Simulate simple circuits using Circuit Maker.
Apparatus Required:
MATLAB 7.8.0 (R2009-a)
Theory:
Circuit Description
This circuit is a simplified model of a 230 kV three-phase power system. Only one phase of the
transmission system is represented. The equivalent source is modeled by a voltage source (230
kV rms/sqrt(3) or 187.8 kV peak, 60 Hz) in series with its internal impedance (Rs Ls)
corresponding to a 3-phase 2000 MVA short circuit level and X/R = 10. (X = 230e3^2/2000e6 =
26.45 ohms or L = 0.0702 H, R = X/10 = 2.645 ohms). The source feeds a RL load through a 150
km transmission line. The line distributed parameters (R = 0.035ohm/km, L = 0.92 mH/km, C =
12.9 nF/km) are modeled by a single pi section (RL1 branch 5.2 ohm; 138 mH and two shunt
capacitances C1 and C2 of 0.967 uF). The load (75 MW - 20 Mvar per phase) is modeled by a
parallel RLC load block.
A circuit breaker is used to switch the load at the receiving end of the transmission line. The
breaker which is initially closed is opened at t = 2 cycles, then it is reclosed at t = 7 cycles.
Current and Voltage Measurement blocks provide signals for visualization purpose.
Demonstration
1. Simulation using a continuous solver (ode23tb)
Start the simulation and observe line voltage and load current transients during load switching
and note that the simulation starts in steady-state. Use the zoom buttons of the oscilloscope to
observe the transient voltage at breaker reclosing.
2. Using the Powergui to obtain steady-state phasors and set initial states

21
Open the Powergui block and select "Steady State Voltage and Currents" to measure the steady-
state voltage and current phasors. Using the Powergui select now "Initial States Setting" to
obtain the initial state values (voltage across capacitors and current in inductances). Now, reset
all the initial states to zero by clicking the "to zero" button and then "Apply" to confirm changes.
Restart the simulation and observe transients at simulation starting. Using the same Powergui
window, you can also set selected states to specific values.
3. Discretizing your circuit and simulating at fixed steps
The Powergui block can also be used to discretize your circuit and simulate it at fixed steps.
Open the Powergui. Select "Discretize electrical model" and specify a sample time of 50e-6 s.
The state-space model will now be discretized using trapezoidal fixed step integration. The
precision of results is now imposed by the sample time. Restart the simulation and compare
simultion results with the continuous integration method. Vary the sample time of the discrete
system and note the impact on precision of fast transients.
4. Using the phasor simulation method
You will now use a third simulation technique. The "phasor simulation" method consists to
replace the circuit state-space model by a set of algebraic equations evaluated at a fixed
frequency and to replace sinusoidal voltage and current sources by phasors (complex numbers).
This method allows a fast computation of voltage and current phasors at a selected frequency,
disregarding fast transients. It is particularly efficient to study electromechanical transients of
generators and motors involving low frequency oscillation modes. Open the Powergui block and
select "Phasor simulation". Restart the simulation. Observe that the magnitude of 60 Hz voltage
and current is now displayed on the scope. If you double click on the voltage or current
measurement block you can choose to output phasor signals in four different formats: Complex,
Real/Imag, Magnitude/Angle (in degres), or just Magnitude (default value). Notice that you
cannot send a complex signal to an oscilloscope.
Fig 1: Simple circuits using circuit maker (closed)
150 km
transmission line
Simple circuits using Circuit Maker
Continuous
Scope
Rs Ls RL1
RL Load
75 MW 20 Mvar
i
+
-
Load current
v
+
-
Line voltage
C2C1
Breaker
132.8 kV rms
60 Hz

22
Fig 2: Waveform of load current and voltage when circuit is closed
Fig 3: Simple circuits using circuit maker (open)
Fig 4: Waveform of load current and voltage when circuit is open
150 km
transmission line
Simple circuits using Circuit Maker
Continuous
Scope
Rs Ls RL1
RL Load
75 MW 20 Mvar
i
+
-
Load current
v
+
-
Line voltage
C2C1
Breaker
132.8 kV rms
60 Hz

23
Experiment No. 5
Objective:
(a) Modeling of Synchronous Machine with FACTS device
(b) Simulation of Synchronous Machine with FACTS devices.
Apparatus Required:
MATLAB 7.8.0 (R2009-a)
Theory:
The detailed model includes detailed representation of power electronic IGBT converters. In
order to achieve an acceptable accuracy with the 1680 Hz switching frequency used in this demo,
the model must be discretized at a relatively small time step (5 microseconds). This model is
well suited for observing harmonics and control system dynamic performance over relatively
short periods of times (typically hundreds of milliseconds to one second).
The average model (discrete) such as the one presented in the “power_dstatcom_avg.mdl” model
in the FACTS library of demos. In this type of model the IGBT Voltage-Sourced Converters
(VSC) are represented by equivalent voltage sources generating the AC voltage averaged over
one cycle of the switching frequency. This model does not represent harmonics, but the
dynamics resulting from control system and power system interaction is preserved. This model
allows using much larger time steps (typically 50 microseconds), thus allowing simulations of
several seconds.
Alternatively, a third type of model can be used for simulating on larger time frames: the phasor
model. This type of model is not available for the D-STATCOM , but it is available for the
STATCOM, a similar device, in the "power_statcom.mdl" demo.
Model Description
A Distribution Static Synchronous Compensator (D-STATCOM) is used to regulate voltage on
a 25-kV distribution network. Two feeders (21 km and 2 km) transmit power to loads connected
at buses B2 and B3. A shunt capacitor is used for power factor correction at bus B2. The 600-V

24
load connected to bus B3 through a 25kV/600V transformer represents a plant absorbing
continuously changing currents, similar to an arc furnace, thus producing voltage flicker. The
variable load current magnitude is modulated at a frequency of 5 Hz so that its apparent power
varies approximately between 1 MVA and 5.2 MVA, while keeping a 0.9 lagging power factor.
This load variation will allow you to observe the ability of the D-STATCOM to mitigate voltage
flicker
The D-STATCOM regulates bus B3 voltage by absorbing or generating reactive power. This
reactive power transfer is done through the leakage reactance of the coupling transformer by
generating a secondary voltage in phase with the primary voltage (network side). This voltage is
provided by a voltage-sourced PWM inverter. When the secondary voltage is lower than the bus
voltage, the D-STATCOM acts like an inductance absorbing reactive power. When the
secondary voltage is higher than the bus voltage, the D-STATCOM acts like a capacitor
generating reactive power.
Fig 1: Synchronous machine with FACTS device
The 'InitFcn' automatically sets
the sample time Ts= 5e-6 s
Synchronous machine with FACTS device
Discrete,
Ts = 5e-006 s.
A
B
C
Variable
Load
Scope3
Scope2
Scope1
N
A
B
C
Programmable
Voltage Source
VaIa
Va_Inv
Iq_Iqref
PQ
Vdc
m
PQ_B3
VmagB1B3
Data Acquisition
A
B
C
D-STATCOM
+/- 3Mvar
A
B
C
a
b
cBstatcom
A
B
C
a
b
c
B3
A
B
C
a
b
c
B2
A
B
C
a
b
c
B1
A
B
C
3 MW
0.2 Mvar
A
B
C
a
b
c
n2
25kV / 600V
A
B
C
A
B
C
25 kV, 100MVA
System
A
B
C
A
B
C
21-km
Feeder
2-km
Feeder
A
B
C
1 MW
Va,Ia (pu)
Va Inv (V)
Vdc (V)
P,Q (MVA)
Iq,Iqref (pu)
modulation index
Ia STAT (pu)
PQ_B3 (MW Mv ar)
V_B1 V_B3 (pu)
+
-
D-STATCOM
25kV, +/- 3Mvar
3
C
2
B
1
A
1 2
T r C
1 2
T r B
1 2
T r A
3
Multimeter
Vaa_Inv
Iabc_Bstatcom
Vabc_Bstatcom
Demux
DC Link
Vdc
Vabc
Iabc
P1
P2
Controller
g
A
B
C
+
-
Bridge 2
g
A
B
C
+
-
Bridge 1

25
Fig 2: Circuit diagram of D-STATCOM
The D-STATCOM consists of the following components:
a 25kV/1.25kV coupling transformer which ensures coupling between the PWM inverter and the
network. a voltage-sourced PWM inverter consisting of two IGBT bridges. This twin inverter
configuration produces fewer harmonic than a single bridge, resulting in smaller filters and
improved dynamic response. In this case, the inverter modulation frequency is 28*60=1.68 kHz
so that the first harmonics will be around 3.36 kHz. LC damped filters connected at the inverter
output. Resistances connected in series with capacitors provide a quality factor of 40 at 60 Hz. a
10000-microfarad capacitor acting as a DC voltage source for the inverter a voltage regulator
that controls voltage at bus B3 a PWM pulse generator using a modulation frequency of 1.68
kHz anti-aliasing filters used for voltage and current acquisition.
Fig 4: Circuit diagram of IGBT
The D-STATCOM controller consists of several functional blocks:
a Phase Locked Loop (PLL). The PLL is synchronized to the fundamental of the transformer
primary voltages. two measurement systems. Vmeas and Imeas blocks compute the d-axis and q-
axis components of the voltages and currents by executing an abc-dq transformation in the
synchronous reference determined by sin(wt) and cos(wt) provided by the PLL. an inner current
regulation loop. This loop consists of two proportional-integral (PI) controllers that control the d-
5
-
4
+
3
C
2
B
1
A
tp0d97dd5f_4411_4c67_8366_fec5ab2b7d06
VF
tpa41254f8_165e_40ef_b37b_66c9d329546c
powersysdomain
UniversalBridge
tp05f31448_e0da_4bd2_b1a8_260ccec973c0
Status
Template
Model
discrete IGBT
tp6b937b89_0510_4fbe_b635_e536ee42e76d
ITAIL
ISWITCH
tp8bc8aa42_d4e7_4eb7_85a6_d2c695fbfbc8
Goto
1
g

26
axis and q-axis currents. The controllers outputs are the Vd and Vq voltages that the PWM
inverter has to generate. The Vd and Vq voltages are converted into phase voltages Va, Vb, Vc
which are used to synthesize the PWM voltages. The Iq reference comes from the outer voltage
regulation loop (in automatic mode) or from a reference imposed by Qref (in manual mode). The
Id reference comes from the DC-link voltage regulator. an outer voltage regulation loop. In
automatic mode (regulated voltage), a PI controller maintains the primary voltage equal to the
reference value defined in the control system dialog box. a DC voltage controller which keeps
the DC link voltage constant to its nominal value (Vdc=2.4 kV).
The electrical circuit is discretized using a sample time Ts=5 microseconds. The controller uses a
larger sample time (32*Ts= 160 microseconds).
Fig 3: Controller diagram for D-STATCOM
Demonstration
1. D-STATCOM dynamic response
During this test, the variable load will be kept constant and you will observe the dynamic
response of a D-STATCOM to step changes in source voltage. Check that the modulation of the
Variable Load is not in service (Modulation Timing [Ton Toff]= [0.15 1]*100 > Simulation Stop
time). The Programmable Voltage Source block is used to modulate the internal voltage of the
25-kV equivalent. The voltage is first programmed at 1.077 pu in order to keep the D-
STATCOM initially floating (B3 voltage=1 pu and reference voltage Vref=1 pu). Three steps are
programmed at 0.2 s, 0.3 s, and 0.4 s to successively increase the source voltage by 6%, decrease
it by 6% and bring it back to its initial value (1.077 pu).
Start the simulation. Observe on Scope1 the phase A voltage and current waveforms of the D-
STATCOM as well as controller signals on Scope2. After a transient lasting approximately 0.15
sec., the steady state is reached. Initially, the source voltage is such that the D-STATCOM is
inactive. It does not absorb nor provide reactive power to the network. At t = 0.2 s, the source
voltage is increased by 6%. The D-STATCOM compensates for this voltage increase by
absorbing reactive power from the network (Q=+2.7 Mvar on trace 2 of Scope2). At t = 0.3 s, the
2
P2
1
P1
w theta
w->theta
m_Phi
theta
Vabc (t)
m_Phi->Vabc(t)
1/z
1/z
Ust
theta
P1
P2
PWM Modulator
Vabc
Iabc
Vdc
m_Phi
w
Controller
(sample time=Ts*32)
Vabc
Iabc
Vdc
Vabc
Iabc
Vdc
Anti-aliasing
Filters (7)
3
Iabc
2
Vabc
1
Vdc

27
source voltage is decreased by 6% from the value corresponding to Q = 0. The D-STATCOM
must generate reactive power to maintain a 1 pu voltage (Q changes from +2.7 MVAR to -2.8
MVAR). Note that when the D-STATCOM changes from inductive to capacitive operation, the
modulation index of the PWM inverter is increased from 0.56 to 0.9 (trace 4 of Scope2) which
corresponds to a proportional increase in inverter voltage. Reversing of reactive power is very
fast, about one cycle, as observed on D-STATCOM current (magenta signal on trace 1 of
Scope1).
2. Mitigation of voltage flicker
During this test, voltage of the Programmable Voltage Source will be kept constant and you will
enable modulation of the Variable Load so that you can observe how the D-STATCOM can
mitigate voltage flicker. In the Programmable Voltage Source block menu, change the "Time
Variation of" parameter to "None". In the Variable Load block menu, set the Modulation Timing
parameter to [Ton Toff]= [0.15 1] (remove the 100 multiplication factor). Finally, in the D-
STATCOM Controller, change the "Mode of operation" parameter to "Q regulation” and make
sure that the reactive power reference value Qref (2nd line of parameters) is set to zero. In this
mode, the D-STATCOM is floating and performs no voltage correction.
Run the simulation and observe on Scope3 variations of P and Q at bus B3 (1st trace) as well as
voltages at buses B1 and B3 (trace 2). Without D-STATCOM, B3 voltage varies between 0.96
pu and 1.04 pu (+/- 4% variation). Now, in the D-STATCOM Controller, change the "Mode of
operation" parameter back to "Voltage regulation" and restart simulation. Observe on Scope 3
that voltage fluctuation at bus B3 is now reduced to +/- 0.7 %. The D-STATCOM compensates
voltage by injecting a reactive current modulated at 5 Hz (trace 3 of Scope3) and varying
between 0.6 pu capacitive when voltage is low and 0.6 pu inductive when voltage is high.
Results:

28

29
Viva Questions
1. What is meant by reactive power control in electrical power transmission lines?
Ans: To make transmission networks operate within desired voltage limits, methods of
making up or taking away reactive-power is called reactive-power control.
2. States uses of series compensation. Series compensation used in the improvement of the
maximum power-transmission capacity of the line.
Ans: The net effect is a lower load angle for a given power-transmission level and, therefore,
a higher-stability margin. The series compensation effectively reduces the overall line
reactance, it is expected that the net line-voltage drop would become less susceptible to the
loading conditions.
3. What is the necessity of compensation?
Ans: The objectives of line compensation are invariably 1. to increase the power-
transmission capacity of the line, and/or 2. to keep the voltage profile of the line along its
length within acceptable bounds to ensure the quality of supply to the connected customers as
well as to minimize the line-insulation costs.

30
4. How is reactive power controlled in the electrical networks?
Ans: To keep the voltages in the network at nearly the rated value, two control actions seem
possible: 1. load compensation, and 2. system compensation.
5. What is meant by Power System Stabilizer (PSS)?
Ans: A power-system stabilizer (PSS) is implemented by adding auxiliary damping signals
derived from the shaft speed, or the terminal frequency, or the power—an effective and
frequently used technique for enhancing small-signal stability of the connected system.
6. What is meant by STATCOM?
Ans: A control on the output voltage of this converter—lower or higher than the connecting
bus voltage— controls the reactive power drawn from or supplied to the connected bus. This
FACTS controller is known as a static compensator (STATCOM).
7. What is meant by load compensation?
Ans: It is possible to compensate for the reactive current Ix of the load by adding a parallel
capacitive load so that Ic c `Ix. Doing so causes the effective power factor of the combination
to become unity. The absence of Ix eliminates the voltage drop DV1, bringing Vr closer in
magnitude to Vs; this condition is called load compensation. 8. What is meant by system
compensation?

31
Experiment No. 6
Objective:
FACTS Controller designs with FACTS devices for SMIB system.
(300 Mvar/-100 Mvar Static Var Compensator (SVC) ; 1 TCR - 3 TSCs)
Apparatus Required:
MATLAB 7.8.0 (R2009-a)
Theory:
Circuit Description: A 300-Mvar Static Var Compensator (SVC) regulates voltage on a 6000-
MVA 735-kV system. The SVC consists of a 735kV/16-kV 333-MVA coupling transformer, one
109-Mvar thyristor-controlled reactor bank (TCR) and three 94-Mvar thyristor-switched
capacitor banks (TSC1 TSC2 TSC3) connected on the secondary side of the transformer.
Switching the TSCs in and out allows a discrete variation of the secondary reactive power from
zero to 282 Mvar capacitive (at 16 kV) by steps of 94 Mvar, whereas phase control of the TCR
allows a continuous variation from zero to 109 Mvar inductive. Taking into account the leakage
reactance of the transformer (15%), the SVC equivalent susceptance seen from the primary side
can be varied continuously from -1.04 pu/100 MVA (fully inductive) to +3.23 pu/100 Mvar
(fully capacitive). The SVC controller monitors the primary voltage and sends appropriate pulses
to the 24 thyristors (6 thyristors per three-phase bank) in order to obtain the susceptance required
by the voltage regulator.

32
Fig 1: SVC (Static Var Compensator) with SMIB (Single Machine Infinite Bus) System
Use Look under Mask to see how the TCR and TSC subsystems are built. Each three-phase bank
is connected in delta so that, during normal balanced operation, the zero-sequence tripplen
harmonics (3rd, 9th... ) remain trapped inside the delta, thus reducing harmonic injection into the
power system. The power system is represented by an inductive equivalent (6000 MVA short
circuit level) and a 200-MW load. The internal voltage of the equivalent can be varied by means
of programmable source in order to observe the SVC dynamic response to changes in system
voltage. Open the voltage source menu and look at the sequence of voltage steps which are
programmed.
SVC(Static Var Compensator) with SMIB(Single Machine Infinite Bus) system
+300 Mvar/-100 Mvar Static Var Compensator (SVC) ;1 TCR- 3 TSCs
TSC1
94 Mvar
TCR
109 Mvar
TSC2
94 Mvar
TSC3
94 MVar
Q
<------
Discrete,
Ts = 5e-005 s.
P
A
B
C
P
A
B
C
P
A
B
C
P
A
B
C
Va_Ia
Q(Mvar)
Vmeas Vref
alpha TCR (deg)
nTSC
Signals&
Scopes
A
B
C
a
b
c
Secondary
(16 kV)
Vabc_prim
Vabc_sec
TCR
TSC1
TSC2
TSC3
SVC Controller
SVC
N
A
B
C
Programmable
Voltage Source
A
B
C
a
b
c
Primary
(735 kV)
[Vabc_Prim]
[Vabc_Sec]
A
B
C
A
B
C
735kV 6000 MVA
A
B
C
a
b
c
735/16 kV
333 MVA
A
B
C
200 MW
Va (pu) Ia (pu/100MVA)
Vmeas Vref (pu)
number of TSCs
Q (Mvar)
alpha TCR (deg)

33
Fig 2: Circuit diagram of TCR (Thyristor Controlled Rectifier)
Fig 3: Circuit diagram of TSC (Thyristor Static Compensator)
Demonstration:
Firing pulses TCR
3 C2 B
1 A
g
a
k
ThCA-
g
a
k
ThCA+
g
a
k
ThBC-
g
a
k
ThBC+
g
a
k
ThAB-
g
a
k
ThAB+
[Cm]
[Cp]
[Bm]
[Bp]
[Am]
[Ap]
[Cm]
[Cp]
[Bm]
[Bp]
[Am]
[Ap]
Demux1
P
Firing pulses TSC1
3 C2 B
1 A
g
a
k
ThCA-
g
a
k
ThCA+
g
a
k
ThBC-
g
a
k
ThBC+
g
a
k
ThAB-
gm
ak
ThAB+
LcaLbc
Lab
Vth_TSC1ab
[Cm]
[Cp]
[Bm]
[Bp]
[Am]
[Ap]
[Cm]
[Cp]
[Bm]
[Bp]
[Am]
[Ap]
Demux
CcaCbcCab
1
P

34
Dynamic response of the SVC :
Run the simulation and observe waveforms on the SVC scope block. The SVC is in voltage
control mode and its reference voltage is set to Vref=1.0 pu. The voltage droop of the regulator is

35
0.01 pu/100 VA (0.03 pu/300MVA). Therefore when the SVC operating point changes from
fully capacitive (+300 Mvar) to fully inductive (-100 Mvar) the SVC voltage varies between 1-
0.03=0.97 pu and 1+0.01=1.01 pu.
Initially the source voltage is set at 1.004 pu, resulting in a 1.0 pu voltage at SVC terminals when
the SVC is out of service . As the reference voltage Vref is set to 1.0 pu, the SVC is initially
floating (zero current). This operating point is obtained with TSC1 in servive and TCR almost at
full conduction (alpha=96 degrees). At t=0.1s voltage is suddenly increased to 1.025 pu. The
SVC reacts by absorbing reactive power (Q=-95 Mvar) in order to bring the volatge back to 1.01
pu. The 95% settling time is approximately 135 ms. At this point all TSCs are out of service and
the TCR is almost at full conduction (alpha = 94 degrees) . At t=0.4 s the source voltage is
suddenly lowered to 0.93 pu. The SVC reacts by generating 256 Mvar of reactive power, thus
increasing the voltage to 0.974 pu. At this point the three TSCs are in service and the TCR
absorbs approximately 40% of its nominal reactive power (alpha =120 degrees). Observe on the
last trace of the scope how the TSCs are sequentially switched on and off. Each time a TSC is
switched on the TCR alpha angle changes suddenly from 180 degrees (no conduction) to 90
degrees (full conduction). Finally, at t=0.7 s the voltage is increased to 1.0 pu and the SVC
reactive power is reduced to zero.
Misfiring of TSC-1 :
Each time a TSC is switched off a voltage remains trapped across the TSC capacitors. If you
look at the 'TSC1 Misfiring' scope inside the "Signals and Scope" subsystem you can observe the
TSC1 voltage (first trace) and the TSC1 current (second trace) for branch AB. The voltage across
the positive thyristor (thyristor conducting the positive current) is shown on the 3rd trace and the
pulses sent to this thyristor are shown on the 4th trace. Notice that the positive thyristor is fired at
maximum negative TSC voltage, when the valve voltage is minimum. If by mistake the firing
pulse is not sent at the right time, very large overcurrent’s can be observed in the TSC valves.
Look inside the SVC Controller block how a misfiring can be simulated on TSC1. A Timer
block and a OR block are used to add pulses to the normal pulses coming from the Firing Unit.
Open the Timer block menu and remove the 100 multiplication factor. The timer is now
programmed to send a misfiring pulse lasting one sample time at time t= 0.121 s. Restart
simulation. Observe that the misfiring pulse is sent when the valve voltage is maximum positive
immediately after the TSC has blocked. This thyristor misfiring produces a large thyristor
overcurrent (18 kA or 6.5 times the nominal peak current). Also, immediately after the thyristor
has blocked, the thyristor voltage reaches 85 kV (3.8 times the nominal peak voltage). In order to
prevent such overcurrent’s and overvoltage’s, thyristor valves are normally protected by metal
oxide arresters

36
Viva Questions
1. Define passive and active VAR control.
Ans: When fixed inductors and / or capacitors are employed to absorb or generate reactive
power they constitute passive control. An active var control, is produced when its reactive
power is changed irrespective of terminal voltage to which the var controller is connected.
2. What is meant by shunt compensation? Shunt devices may be connected permanently or
through a switch.
Ans: Shunt reactors compensate for the line capacitance, and because they control
overvoltage’s at no loads and light loads, they are often connected permanently to the line, not
to the bus.
3. Define series compensation. Series capacitors are used to partially offset the effects of the
series inductances of lines.
Ans: Series compensation results in the improvement of the maximum power-transmission
capacity of the line. The net effect is a lower load angle for a given power-transmission level
and, therefore, a higher-stability margin. The reactive-power absorption of a line depends on
the transmission current, so when series capacitors are employed, automatically the resulting
reactive-power compensation is adjusted proportionately.
4. What are the factors are affecting the application of series compensation?
Ans: The following factors need careful evaluation:-
a. The voltage magnitude across the capacitor banks (insulation);
b. The fault currents at the terminals of a capacitor bank;
c. The placement of shunt reactors in relation to the series capacitors (resonant
overvoltages); and d. The number of capacitor banks and their location on a long line
(voltage profile).
6. What is SVC?
Ans: Static var compensators (SVCs) are used primarily in power systems for voltage control
as either an end in itself or a means of achieving other objectives, such as system
stabilization.

37
7. What is TCSC?
Ans: Thyristor switches may be used for shorting capacitors; hence they find application in
step changes of series compensation of transmission lines. A blocked thyristor switch
connected across a series capacitor introduces the capacitor in line, whereas a fully
conducting thyristor switch removes it. In reality, this step control can be smoothed by
connecting an appropriately dimensioned reactor in series with the thyristor switch as shown
in to yield vernier control. This application of thyristor switches creates the thyristor
controlled series capacitor (TCSC) FACTS controller.

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