PSS Lab Manual PDF
PSS Lab Manual PDF
PSS Lab Manual PDF
IV YEAR I SEM
EEE
By
Dr. J. Sridevi
G.Sandhya Rani
Bachupally
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Power Systems Simulation Lab GRIET/EEE
(Autonomous)
CERTIFICATE
This is to certify that it is a record of practical work done in
the Power Systems Simulation Laboratory in I sem of IV
year during the year ________________________
Name:
Roll No:
Branch: EEE
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INDEX
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18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
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Date: Experiment-1
Apparatus: MATLAB
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Circuit diagram:
Procedure:
1. Open Matlab-->Simulink--> File ---> New---> Model
2. Open Simulink Library and browse the components
3. Connect the components as per circuit diagram
4. Set the desired voltage and required frequency
5. Simulate the circuit using MATLAB
6. Plot the waveforms.
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Graph:
Calculations:
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Result:
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Date: Experiment-2
Apparatus: MATLAB
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primary current has two components or branches, so there must be a parallel path with
primary winding of transformer. This parallel path of electric current is known as excitation
branch of equivalent circuit of transformer. The resistive and reactive branches of the
excitation circuit can be represented as
Now if we see the voltage drop in secondary from primary side, then it would be K times
greater and would be written as K.Z2.I2.
Again I2.N1 = I2.N2
Therefore,
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So, the complete equivalent circuit of transformer referred to primary is shown in the figure
below,
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Circuit Diagram:
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Procedure:
1. Open Matlab-->Simulink--> File ---> New---> Model
2. Open Simulink Library and browse the components
3. Connect the components as per circuit diagram
4. Set the desired voltage and current
5. Simulate the circuit using MATLAB
6. Plot the waveforms
Graph:
Calculations:
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Result:
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Date: Experiment-3
Aim: To determine voltage and power at the sending end and to regulate the voltage using
medium line model.
Apparatus: MATLAB
Theory: The transmission line having its effective length more than 80 km but less than
250 km, is generally referred to as a medium transmission line. Due to the line length being
considerably high, admittance Y of the network does play a role in calculating the effective
circuit parameters, unlike in the case of short transmission lines. For this reason the modelling
of a medium length transmission line is done using lumped shunt admittance along with the
lumped impedance in series to the circuit.
These lumped parameters of a medium length transmission line can be represented using two
different models, namely-
1)Nominal representation.
2)Nominal T representation.
Lets now go into the detailed discussion of these above mentioned models.
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As we can see here, VS and VR is the supply and receiving end voltages respectively, and
I1 and I3 are the values of currents flowing through the admittances. And
..
Comparing equation (4) and (5) with the standard ABCD parameter equations
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Voltage regulation of transmission line is measure of change of receiving end voltage from
no-load to full load condition.
Circuit Diagram:
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PROCEDURE:
1. Open Matlab-->Simulink--> File ---> New---> Model
2. Open Simulink Library and browse the components
3. Connect the components as per circuit diagram
4. Set the desired voltage and required frequency
5. Simulate the circuit using MATLAB
6. Calculate the voltage regulation of medium line model.
Calculations:
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Result:
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Date: Experiment-4
Apparatus: MATLAB
As we can see here, VS and VR is the supply and receiving end voltages respectively, and
I1 and I3 are the values of currents flowing through the admittances. And
..........(1)
..........(2)
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Comparing equation (4) and (5) with the standard ABCD parameter equations
Voltage regulation of transmission line is measure of change of receiving end voltage from
no-load to full load condition.
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Circuit Diagram:
PROCEDURE:
1. Open Matlab-->Simulink--> File ---> New---> Model
2. Open Simulink Library and browse the components
3. Connect the components as per circuit diagram
4. Set the desired voltage and required frequency
5. Simulate the circuit using MATLAB
6. Obtain the line performance of a line.
Graph:
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Result
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Date: Experiment-5
Aim: To determine the bus admittance matrix for the given power system Network
Theory:
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For i = 1,2,.n, n = no. of buses of the given system, yij is the admittance of element
connected between buses i and j and yii is the admittance of element connected
between bus i and ground (reference bus).
START
Consider line l = 1
i = sb(1); I= eb(1)
NO YES
Is l =NL?
Stop
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MATLAB PROGRAM
function[Ybus] = ybus(zdata)
nl=zdata(:,1); nr=zdata(:,2); R=zdata(:,3); X=zdata(:,4);
nbr=length(zdata(:,1)); nbus = max(max(nl), max(nr));
Z = R + j*X; %branch impedance
y= ones(nbr,1)./Z; %branch admittance
Ybus=zeros(nbus,nbus); % initialize Ybus to zero
for k = 1:nbr; % formation of the off diagonal elements
if nl(k) > 0 & nr(k) > 0
Ybus(nl(k),nr(k)) = Ybus(nl(k),nr(k)) - y(k);
Ybus(nr(k),nl(k)) = Ybus(nl(k),nr(k));
end
end
for n = 1:nbus % formation of the diagonal elements
for k = 1:nbr
if nl(k) == n | nr(k) == n
Ybus(n,n) = Ybus(n,n) + y(k);
else, end
end
end
Calculations:
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Result:
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Date: Experiment-6
Aim:
To carry out load flow analysis of the given power system network by Gauss Seidel
method
Apparatus: MATLAB
Theory:
Load flow analysis is the study conducted to determine the steady state operating
condition of the given system under given conditions. A large number of numerical algorithms
have been developed and Gauss Seidel method is one of such algorithm.
Problem Formulation
The performance equation of the power system may be written of
[I bus] = [Y bus][V bus] (1)
Selecting one of the buses as the reference bus, we get (n-1) simultaneous equations. The bus
loading equations can be written as
Ii = Pi-jQi / Vi* (i=1,2,3,..n) (2)
Where,
n
Pi=Re [ Vi*Yik Vk] . (3)
k=1
n
Qi= -Im [ Vi*Yik Vk]. (4)
k=1
The bus voltage can be written in form of
n
Vi=(1.0/Yii)[Ii- Yij Vj] (5)
j=1
ji(i=1,2,n)& islack bus
Substituting Ii in the expression for Vi, we get
n
Vi new=(1.0/Yii)[Pi-JQi / Vio* - Yij Vio] (6)
J=1
The latest available voltages are used in the above expression, we get
n n
Vi new=(1.0/Yii)[Pi-JQi / Voi* - YijVj - Yij Vio]
n
(7)
J=1 j=i+1
The above equation is the required formula .this equation can be solved for voltages in
interactive manner. During each iteration, we compute all the bus voltage and check for
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convergence is carried out by comparison with the voltages obtained at the end of previous
iteration. After the solutions is obtained. The stack bus real and reactive powers, the reactive
power generation at other generator buses and line flows can be calculated.
Algorithm
Step1: Read the data such as line data, specified power, specified voltages, Q limits at the
generator buses and tolerance for convergences
Step2: Compute Y-bus matrix.
Step3: Initialize all the bus voltages.
Step4: Iter=1
Step5: Consider i=2, where i is the bus number.
Step6: check whether this is PV bus or PQ bus. If it is PQ bus goto step 8 otherwise go to next
step.
Step7: Compute Qi check for q limit violation. QGi=Qi+QLi.
7).a).If QGi>Qi max ,equate QGi = Qimax. Then convert it into PQ bus.
7).b).If QGi<Qi min, equate QGi = Qi min. Then convert it into PQ bus.
Step8: Calculate the new value of the bus voltage using gauss seidal formula.
i=1 n
Vi=(1.0/Yii) [(Pi-j Qi)/vi0*- Yij Vj- YijVj0]
J=1 J=i+1
Adjust voltage magnitude of the bus to specify magnitude if Q limits are not violated.
Step9: If all buses are considered go to step 10 otherwise increments the bus no. i=i+1 and Go to
step6.
Step10: Check for convergence. If there is no convergence goes to step 11 otherwise go to
step12.
Step11: Update the bus voltage using the formula.
Vinew=Vi old+ (vinew-Viold) (i=1,2,..n) i slackbus , is the acceleration factor=1.4
Step12: Calculate the slack bus power, Q at P-V buses real and reactive give flows real and
reactance line losses and print all the results including all the bus voltages and all the
bus angles.
Step13: Stop.
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START
Read
1. Primitive Y matrix
2. Bus incidence matrix A
3. Slack bus voltages
4. Real and reactive bus powers Pi& Qi
5. Voltage magnitudes and their limits
Form Ybus
Test for
type of bus
Compute Qi(r+1)
Compute Ai
Compute i(r+1) and
Vi(r+1) =|Vis|/i(r+1)
Compute Vi(r+1)
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r (r+1)
Replace Vi by Vi and
advance bus count i = i+1
B Is i<=n
Advance iteration
A Is
count, r = r+1
Vmax<=
Procedure
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Calculations:
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Result
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Date: Experiment-7
Aim
To carry out load flow analysis of the given power system by Newton Raphson method.
Apparatus: MATLAB 7.7
Theory:
The Newton Raphson method of load flow analysis is an iterative method which approximates
the set of non-linear simultaneous equations to a set of linear simultaneous equations using
Taylors series expansion and the terms are limited to first order approximation. The load flow
equations for Newton Raphson method are non-linear equations in terms of real and imaginary
part of bus voltages.
Step5: If the bus is generator (PV) bus, check the value of Qp*is within the limits.If it Violates
the limits, then equate the violated limit as reactive power and treat it as PQ bus. If limit is not
isolated then calculate,
|vp|^r=|vgp|^rspe-|vp|r ; Qp*=qsp-qp*
Step6: Advance bus count by 1 and check if all the buses have been accounted if not go to step5.
Step7: Calculate the elements of Jacobean matrix.
Step8: Calculate new bus voltage increment pk and fpk
Step9: Calculate new bus voltage ep*h+ ep*
Fp^k+1=fpK+fpK
Step10: Advance iteration count by 1 and go to step3.
Step11: Evaluate bus voltage and power flows through the line .
Procedure
Enter the command window of the MATLAB.
Create a new M file by selecting File - New M File.
Type and save the program in the editor Window.
Execute the program by pressing Tools Run.
View the results.
MATLAB program:
clear
basemva=100;accuracy=0.001;maxiter=100;
% no code mag degree MW Mvar MW Mvar Qmin Qmax Mvar
busdata=[1 1 1.05 0.0 0.0 0.0 0.0 0.0 0 0 0
2 0 1.0 0.0 256.66 110.2 0.0 0.0 0 0 0
30 1.0 0.0 138.6 45.2 0.0 0.0 0 0 0];
% bus bus R X 1/2B =1 for lines
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];
lfybus
lfnewton
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busout
lineflow
Calculations:
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Result
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Date: Experiment-8(a)
Apparatus: MATLAB
Theory:
As the losses are neglected, the system model can be understood as shown in Fig, here n number
of generating units are connected to a common bus bar, collectively meeting the total power
demand PD. It should be understood that share of power demand by the units does not involve
losses.
Since transmission losses are neglected, total demand PD is the sum of all generations of n-
number of units. For each unit, a cost functions Ci is assumed and the sum of all costs computed
from these cost functions gives the total cost of production CT.
where the cost function of the ith unit, from Eq. (1.1) is:
Ci = i + iPi + iPi2
Now, the ED problem is to minimize CT, subject to the satisfaction of the following equality and
inequality constraints.
Equality constraint
The total power generation by all the generating units must be equal to the power demand.
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Inequality constraint
Pimin Pi Pimax i = 1, 2, , n
Economic dispatch problem can be carried out by excluding or including generator power
limits, i.e., the inequality constraint.
The constrained total cost function can be converted into an unconstrained function by using
the Lagrange multiplier as:
The conditions for minimization of objective function can be found by equating partial
differentials of the unconstrained function to zero as
Since Ci = C1 + C2++Cn
From the above equation the coordinate equations can be written as:
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For a known value of , the power generated by the ith unit from can be written as:
The value of can be calculated and compute the values of Pi for i = 1, 2,, n for optimal
scheduling of generation.
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Procedure :
Create a new file in edit mode by selecting File - New File.
Browse the components and build the bus sytem
Execute the program in run mode by selecting tools-opf areas-select opf
Run the primal lp
View the results in case information-Generator fuel costs.
Tabulate the results.
Results:
Generator Gen Min Max Cost
Name MW IOA IOB IOC MW MW $/Hr Lambda
1 79.52 150 5 0.11 10 250 1243.12 22.49
2 126 600 1.2 0.085 10 300 2100.66 22.62
3 114 335 1 0.1225 10 270 2041.01 28.93
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Calculations
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Result
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Date: Experiment-8(b)
Apparatus: MATLAB
Theory : When the transmission losses are included in the economic dispatch problem, we can
modify (5.4) as
PT P1 P2 PN PLOSS
where PLOSS is the total line loss. Since PT is assumed to be constant, we have
In the above equation dPLOSS includes the power loss due to every generator, i.e.,
PLOSS P P
dPLOSS dP1 LOSS dP2 LOSS dPN
P1 P2 PN
Also minimum generation cost implies dfT = 0 as given Multiplying by and combing we get
P P P
0 LOSS dP1 LOSS dP2 LOSS dPN
P1 P2 PN
N
f P
0 T LOSS dPi
i 1 Pi Pi
fT P
LOSS 0, i 1,, N
Pi Pi
Again since
fT dfT
, i 1,, N
Pi Pi
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we get
df1 df df
L1 2 L2 N LN
dPi dP2 dPN
1
Li , i 1,, N
1 PLOSS Pi
Consider an area with N number of units. The power generated are defined by the vector
P P1 P2 PN
T
PLOSS PT BP
The elements Bij of the matrix B are called the loss coefficients. These coefficients are not
constant but vary with plant loading. However for the simplified calculation of the penalty factor
Li these coefficients are often assumed to be constant.
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Circuit Diagram:
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Procedure :
Create a new file in edit mode by selecting File - New File.
Browse the components and build the bus sytem
Execute the program in run mode by selecting tools-opf areas-select opf
Run the primal lp
View the results in case information-Generator fuel costs.
Tabulate the results.
Results:
Total Cost =
5359.15$/hr
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Result
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Date: Experiment-9
Apparatus: MATLAB
Theory: RLC circuits are widely used in a variety of applications such as filters in
communications systems, ignition systems in automobiles, defibrillator circuits in biomedical
applications, etc. The analysis of RLC circuits is more complex than of the RC circuits we have
seen in the previous lab. RLC circuits have a much richer and interesting response than the
previously studied RC or RL circuits. A summary of the response is given below.
Lets assume a series RLC circuit. The discussion is also applicable to other RLC circuits such as
the parallel circuit.
By writing KVL one gets a second order differential equation. The solution consists of two parts:
in which is the damping ratio and is the undamped resonant frequency. The roots of the
quadratic equation are equal to,
For the example of the series RLC circuit one has the following characteristic equation for the
current iL(t) or vC(t),
s2 + R/L.s + 1/LC =0.
Depending on the value of the damping ratio one has three possible cases:
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Procedure:
Enter the command window of the MATLAB.
Create a new M file by selecting File - New M File.
Type and save the program in the editor Window.
Execute the program by pressing Tools Run.
View the results.
MATLAB Program:
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wd = wn*p;
h = [p/z];
k = atan(h)
m = pi-k;
tr = [m/wd]
tp = [pi/wd]
q = z*wn
ts = [4/q]
r = z*pi
f = [r/p]
mp = exp(-f)
num = [0 0 n]
den = [1 z*z*wn n]
s = tf(num,den)
hold on
step(s)
impulse(s)
hold off
Result:
Enter value of undamped natural frequency 16
wn = 16
z= 0.5000
n = 256
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p= 0.8660
k= 1.0472
tr = 0.1511
tp = 0.2267
q= 8
ts = 0.5000
r = 1.5708
f= 1.8138
mp = 0.1630
num = 0 0 256
den = 1 4 256
Transfer function:
256
---------------
s^2 + 4 s + 256
Graph:
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Calculations:
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Result
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Date: Experiment-10
Apparatus: MATLAB
Theory: The response of the armature current when a three-phase symmetrical short circuit
occurs at the terminals of an unloaded synchronous generator.
It is assumed that there is no dc offset in the armature current. The magnitude of the current
decreases exponentially from a high initial value. The instantaneous expression for the fault
current is given by
where Vt is the magnitude of the terminal voltage, is its phase angle and
with .
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we have neglected the effect of the armature resistance hence = /2. Let us assume that the
fault occurs at time t = 0. From (6.9) we get the rms value of the
current as
which is called the subtransient fault current. The duration of the subtransient current is
dictated by the time constant Td . As the time progresses and Td < t < Td , the first
exponential term will start decaying and will eventually vanish. However since t is still nearly
equal to zero, we have the following rms value of the current
This is called the transient fault current. Now as the time progress further and the second
exponential term also decays, we get the following rms value of the current for the sinusoidal
steady state
In addition to the ac, the fault currents will also contain the dc offset. Note that a symmetrical
fault occurs when three different phases are in three different locations in the ac cycle.
Therefore the dc offsets in the three phases are different. The maximum value of the dc offset
is given by
Procedure:
1. Open Matlab-->Simulink--> File ---> New---> Model
2. Open Simulink Library and browse the components
3. Connect the components as per circuit diagram
4. Set the desired voltage and required frequency
5. Simulate the circuit using MATLAB
6. Plot the waveforms
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Circuit Diagram:
Graph:
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Calculations:
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Result
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Date: Experiment-11
Apparatus: MATLAB
Theory:
Single Line-to-Ground Fault
The single line-to-ground fault is usually referred as short circuit fault and occurs when one
conductor falls to ground or makes contact with the neutral wire. The general representation
of a single line-to-ground fault is shown in Figure 3.10 where F is the fault point with
impedances Zf. Figure 3.11 shows the sequences network diagram. Phase a is usually
assumed to be the faulted phase, this is for simplicity in the fault analysis calculations. [1]
Ia0
F0
+
Va0 Z0
Iaf - N0
Ia1
F1
F
a + Z1
+
3Zf Va1
b - N1 1.0
-
c Ia2
Iaf Ibf = 0 Icf = 0
F2
+
Vaf Zf +
Va2 Z2
- N2
-
n
Since the zero-, positive-, and negative-sequence currents are equals as it can be observed
in Figure 3.11. Therefore,
1.00
I a 0 I a1 I a 2
Z 0 Z1 Z 2 3Z f
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With the results obtained for sequence currents, the sequence voltages can be obtained
from
Va 0 0 1 1 1 Ia0
V 1.00 1 a 2 a I a1
b1
Vc 2 0 1 a a 2 I a 2
By solving Equation
Va 0 Z 0 I a 0
Va1 1.0 Z1 I a1
Va 2 Z 2 I a 2
If the single line-to-ground fault occurs on phase b or c, the voltages can be found by the
relation that exists to the known phase voltage components,
Vaf 1 1 1 Va 0
Vbf 1 a
2
a Va1
Vcf 1 a a 2 Va 2
as
Line-to-Line Fault
A line-to-line fault may take place either on an overhead and/or underground
transmission system and occurs when two conductors are short-circuited. One of the
characteristic of this type of fault is that its fault impedance magnitude could vary over a wide
range making very hard to predict its upper and lower limits. It is when the fault impedance is
zero that the highest asymmetry at the line-to-line fault occurs
The general representation of a line-to-line fault is shown in Figure 3.12 where F is the
fault point with impedances Zf. Figure 3.13 shows the sequences network diagram. Phase b and
c are usually assumed to be the faulted phases; this is for simplicity in the fault analysis
calculations [1],
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Zf
Zf
I af 0
Ibf I cf
Vbc Z f Ibf
Ia0 0
1.00
I a1 I a 2
Z1 Z 2 Z f
If Zf = 0,
1.00
I a1 I a 2
Z1 Z 2
Ibf I cf 3 I a1 90
Va 0 0
Va1 1.0 - Z1 I a1
Va 2 Z 2 I a 2 Z 2 I a1
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F
a
b Ia0 Zf +3Zg Ia1 Zf Ia2 Zf
c
Iaf = 0 Ibf Icf F0 F1 F2
+ + +
Zf Zf Va0 Va1 Z1 Va2
Z0 + Z2
- -
N0 N1 1.0 0o - N2
Zg Ibf +Icf -
N
n
I af 0
Vbf ( Zf Zg ) Ibf ZgIcf
Vcf ( Zf Zg ) Icf ZgIbf
1.00
Ia1
( Z 2 Zf )( Z 0 Zf 3Zg )
( Z 1 Zf )
( Z 2 Zf ) ( Z 0 Zf 3Zg )
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( Z 0 Zf 3Zg )
Ia 2 [ ]Ia1
( Z 2 Zf ) ( Z 0 Zf 3Zg )
( Z 2 Zf )
Ia 0 [ ]Ia1
( Z 2 Zf ) ( Z 0 Zf 3Zg )
If Zf and Zg are both equal to zero, then the positive-, negative-, and zero-sequences can
be obtained from
1.00
Ia1
( Z 2)( Z 0)
( Z 1)
( Z 2 Z 0)
( Z 0)
Ia 2 [ ]Ia1
( Z 2 Z 0)
( Z 2)
Ia 0 [ ]Ia1
( Z 2 Z 0)
I af 0
I bf I a 0 a 2 I a1 aI a 2
I cf I a 0 aI a1 a 2 I a 2
I n 3I a 0 Ibf I cf
The resultant phase voltages from the relationship given in Equation 3.78 can be
expressed as
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Circuit Diagram:
Procedure:
1. Open Matlab-->Simulink--> File ---> New---> Model
2. Open Simulink Library and browse the components
3. Connect the components as per circuit diagram
4. Set the desired voltage and required frequency
5. Simulate the circuit using MATLAB
6. Plot the waveforms
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Graph:
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Result
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Date: Experiment-12
Aim: To determine the bus impedance matrix for the given power system Network
Theory:
Formation of Z BUS matrix
Z-bus matrix is an important matrix used in different kinds of power system study such as
short circuit study, load flow study etc. In short circuit analysis the generator uses transformer
impedance must be taken into account. In quality analysis the two-short element are neglected by
forming the z-bus matrix which is used to compute the voltage distribution factor. This can be
largely obtained by reversing the y-bus formed by inspection method or by analytical method.
Taking inverse of the y-bus for large system in time consuming; Moreover modification in the
system requires whole process to be repeated to reflect the changes in the system. In such cases
is computed by z-bus building algorithm.
Algorithm
Step 1: Read the values such as number of lines, number of buses and line data, generator data
and transformer data.
Step 2: Initialize y-bus matrix y-bus[i] [j] =complex.(0.0,0.0)
Step 3: Compute y-bus matrix by considering only line data.
Step 4: Modifies the y-bus matrix by adding the transformer and the generator admittance to the
respective diagonal elements of y-bus matrix.
Step 5: Compute the z-bus matrix by inverting the modified y-bus matrix.
Step 6: Check the inversion by multiplying modified y-bus and z-bus matrices to check whether
the resulting matrix is unit matrix or not.
Step 7: Print the z-bus matrix.
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START
STOP
Procedure:
Enter the command window of the MATLAB.
Create a new M file by selecting File - New M File.
Type and save the program in the editor Window.
Execute the program by pressing Tools Run.
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MATLAB Program:
function[Ybus] = ybus(zdata)
zdata(:,1)); nbus = max(max(nl), max(nr));
Z = R + j*X; %branch impedance
y= ones(nbr,1)./Z; %branch admittance
Ybus=zeros(nbus,nbus); % initialize Ybus to zero
for k = 1:nbr; % formation of the off diagonal elements
if nl(k) > 0 & nr(k) > 0
Ybus(nl(k),nr(k)) = Ybus(nl(k),nr(k)) - y(k);
Ybus(nr(k),nl(k)) = Ybus(nl(k),nr(k));
end
end
for n = 1:nbus % formation of the diagonal elements
for k = 1:nbr
if nl(k) == n | nr(k) == n
Ybus(n,n) = Ybus(n,n) + y(k);
else, end
end
end
Zbus= inv(Y)
Ibus=[-j*1.1;-j*1.25;0;0]
Vbus=Zbus*Ibus
OUTPUT:
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Result:
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Date: Experiment-13(a)
SYMMETRICAL COMPONENTS
Apparatus: MATLAB
Theory:
Before we discuss the symmetrical component transformation, let us first define the a-
operator.
1 3
a e j 120 j
0
2 2
Note that for the above operator the following relations hold
1 3
a 2 e j 240 j a
0
2 2
a 3 e j 360 1
0
1 3 1 3
1 a a2 1 j j 0
2 2 2 2
Similarly
Finally
Va 0 Vb 0 Vc 0
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Va 0 1 1 1 Va
V 1 1 a a 2 Vb
a1 3
Va 2 1 a 2 a Vc
Va 0 Va
Va 012 Va1 , Vabc Vb
Va 2 Vc
Program:
V012 = [0.6 90
1.0 30
0.8 -30];
rankV012=length(V012(1,:));
if rankV012 == 2
mag= V012(:,1); ang=pi/180*V012(:,2);
V012r=mag.*(cos(ang)+j*sin(ang));
elseif rankV012 ==1
V012r=V012;
else
fprintf('\n Symmetrical components must be expressed in a one column array in rectangular
complex form \n')
fprintf(' or in a two column array in polar form, with 1st column magnitude & 2nd column
\n')
fprintf(' phase angle in degree. \n')
return, end
a=cos(2*pi/3)+j*sin(2*pi/3);
A = [1 1 1; 1 a^2 a; 1 a a^2];
Vabc= A*V012r
Vabcp= [abs(Vabc) 180/pi*angle(Vabc)];
fprintf(' \n Unbalanced phasors \n')
fprintf(' Magnitude Angle Deg.\n')
disp(Vabcp)
Vabc0=V012r(1)*[1; 1; 1];
Vabc1=V012r(2)*[1; a^2; a];
Vabc2=V012r(3)*[1; a; a^2];
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Procedure:
1. Open Matlab--> File ---> New---> Script
2. Write the program
3. Enter F5 to run the program
4. Observe the results in MATLAB command window.
Result:
Vabc =
1.5588 + 0.7000i
-0.0000 + 0.4000i
-1.5588 + 0.7000i
Unbalanced phasors
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Result
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Date: Experiment-13(b)
Aim: To obtain the original unbalanced phase voltages from symmetrical components
Apparatus: MATLAB
Theory:
Va 012 CVabc
1 1 1
1
C 1 a a 2
3
1 a 2 a
The original phasor components can also be obtained from the inverse symmetrical
component transformation, i.e.,
Va 1 1 1 Va 0 Va 0
V 1 a 2 a Va1 C Va1
1
b
Vc 1 a a 2 Va 2 Va 2
Va Va 0 Va1 Va 2
Finally, if we define a set of unbalanced current phasors as Iabc and their symmetrical
components as Ia012, we can then define
I a 012 CI abc
I abc C 1I a 012
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Program:
Iabc = [1.6 25
1.0 180
0.9 132];
rankIabc=length(Iabc(1,:));
if rankIabc == 2
mag= Iabc(:,1); ang=pi/180*Iabc(:,2);
Iabcr=mag.*(cos(ang)+j*sin(ang));
elseif rankIabc ==1
Iabcr=Iabc;
else
fprintf('\n Three phasors must be expressed in a one column array in rectangular complex
form \n')
fprintf(' or in a two column array in polar form, with 1st column magnitude & 2nd column
\n')
fprintf(' phase angle in degree. \n')
return, end
a=cos(2*pi/3)+j*sin(2*pi/3);
A = [1 1 1; 1 a^2 a; 1 a a^2];
I012=inv(A)*Iabcr;
symcomp= I012
Result:
symcomp =
-0.0507 + 0.4483i
0.9435 - 0.0009i
0.5573 + 0.2288i
Symmetrical components
Magnitude Angle Deg.
0.4512 96.4529
0.9435 -0.0550
0.6024 22.3157
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Result
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Date: Experiment-14
Apparatus: POWERWORLD
Theory:
V f Z 44I f
Zi 4
Vi Zi 4 I f V f , i 1,2,3
Z 44
We further assume that the system is unloaded before the fault occurs and that the magnitude and
phase angles of all the generator internal emfs are the same. Then there will be no current
circulating anywhere in the network and the bus voltages of all the nodes before the fault will be
same and equal to Vf. Then the new altered bus voltages due to the fault will be given from by
Z
Vi V f Vi 1 i 4 V f , i 1,,4
Z 44
14 bus system :
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Procedure :
Create a new file in edit mode by selecting File - New File.
Browse the components and build the bus sytem
Execute the program in run mode by selecting tools-fault analysis
Select the fault on which bus and calculate
Tabulate the results.
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= 1861.67 A
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Result
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Date: Experiment-15
Aim: To obtain the frequency response of single and two area power system using MATLAB
Formula used:
1)
2)
3)
4)
Where
D = damping coefficient
GG gain of generator
GT - gain of turbine
GP - gain of power
KP power system constant
KT- turbine constant
KG- generator constant
TP power system time constant
TG- generator time constant
Procedure:
1. Open Matlab-->Simulink--> File ---> New---> Model
2. Open Simulink Library and browse the components
3. Connect the components as per circuit diagram
4. Set the desired voltage and required frequency
5. Simulate the circuit using MATLAB
6. Plot the waveforms
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SIMULINK RESULTS:
Single Area Power System Block Diagram
Graph:
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RESULT
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Date: Experiment-16
Aim: To obtain the frequency response of two area power system using MATLAB
Apparatus: MATLAB
Formula used:
1)
2)
3)
4)
Where
D = damping coefficient
GG gain of generator
GT - gain of turbine
GP - gain of power
KP power system constant
KT- turbine constant
KG- generator constant
TP power system time constant
TG- generator time constant
Procedure:
1. Open Matlab-->Simulink--> File ---> New---> Model
2. Open Simulink Library and browse the components
3. Connect the components as per circuit diagram
4. Set the desired voltage and required frequency
5. Simulate the circuit using MATLAB
6. Plot the waveforms
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SIMULINK RESULTS:
Single Area Power System Block Diagram
Graphs:
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Result
Date: Experiment-17
Aim: To obtain step response of rotor angle and generator frequency of a synchronous
machine
Apparatus: MATLAB
Circuit Diagram:
Procedure:
1. Open File ---> New---> Model
2. Open Simulink Library and browse the components
3. Connect the components as per circuit diagram
4. Set the desired voltage and required frequency
5. Simulate the circuit using MATLAB
6. Plot the waveforms
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Graph:
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Result
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