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LIST OF EXPERIMENTS
1) FORMATION OF BUS ADMITTANCE MATRIX
2) LOAD FLOW ANALYSIS BY GAUSS SEIDEL ETHOD
3) MODELLING OF AGC FOR A SINGLE AREA NETWORK USING
SIMULINK
4)MODELLING OF AGC FOR TWO AREA NETWORK USING
SIMULINK
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Exp No: 1
FORMATION OF BUS ADMITTANCE MATRIX
AIM : To determine the bus admittance matrix for the given power system Network
APPARATUS REQUIRED: MATLAB 7.7
THEORY :
Formation of Y BUS matrix
Bus admittance matrix is often used in power system studies.In most of power system
studies it is necessary to form Y-bus matrix of the system by considering certain power system
parameters depending upon the type of analysis. For example in load flow analysis it is
necessary to form Y-bus matrix without taking into account the generator impedance and load
impedance. In short circuit analysis the generator transient reactance and transformer
impedance taken in account, in addition to line data. Y-bus may be computed by inspection
method only if there is no natural coupling between the lines. Shunt admittance are added to
the diagonal elements corresponding to the buses at which these are connected. The off
diagonal elements are unaffected. The equivalent circuit of tap changing transformer may be
considered in forming[y-bus] matrix.
There are b independent equations (b = no. of buses) relating the bus vectors of currents
and voltages through the bus impedance matrix and bus admittance matrix:
EBUS = ZBUS IBUS
IBUS = YBUS EBUS
The relation of equation can be represented in the form
IBUS = YBUS EBUS
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Where, YBUS is the bus admittance matrix, IBUS & EBUS are the bus current and bus
voltage vectors respectively.
Diagonal elements: A diagonal element (Yii) of the bus admittance matrix, YBUS, is equal to
the sum total of the admittance values of all the elements incident at the bus/node i,Off
Diagonal elements: An off-diagonal element (Yij) of the bus admittance matrix, YBUS, is
equal to the negative of the admittance value of the connecting element present between the
buses I and j, if any.
This is the principle of the rule of inspection. Thus the algorithmic equations for the rule of
inspection are obtained as:
Yii = Σ yij (j = 1,2,…….n)
Yij = - yij (j = 1,2,…….n)
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).
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START
Read the no. Of buses , no of
lines and line data
Initialize the Y- BUS Matrix
Consider line l = 1
i = sb(1); I= eb(1)
Y(i,i) =Y(i,i)+Yseries(l) +0.5Yseries(l)
Y(j,j) =Y(j,j)+Yseries(l) +0.5Yseries(l)
Y(i,j) = -Yseries(l),Y(j,i) =Y(i,j)
NO YES
Is l =NL?
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l = l+1
STOP
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
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if nl(k) == n | nr(k) == n
Ybus(n,n) = Ybus(n,n) + y(k);
else, end
end
RESULT:
Exp No: 2
LOAD FLOW ANALYSIS BY GAUSS SEIDEL METHOD
AIM: To find load flow solution of the given power system using Gauss-Seidel method
theoretically for one iteration and obtain full solution using MATLAB.
Apparatus: MATLAB
THEORY
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 [ Σ Vi*Yik Vk] . (3)
k=1
N
Qi= -Im [ Σ Vi*Yik Vk]. (4)
k=1
The bus voltage can be written in form of
n
Vi=(1.0/Yii)[Ii- Σ Yij Vj] (5)
j=1
j≠i(i=1,2,…………n)& i≠slack bus
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Substituting Ii in the expression for Vi, we get
N
Vi new=(1.0/Yii)[Pi-JQi / Vio* - Σ Yij Vio] (6)
J=1
The latest available voltages are used in the above expression, we get
n N
o
Vi new=(1.0/Yii)[Pi-JQi / V i* - Σ YijVjn- Σ Yij Vio] (7)
J=1 j=i+1
The above equation is required formula .This equation can be solved foe voltage in
interactive manner. During each iteration, we compute all the bus voltage and
check forDuring each iteration, we compute all the bus voltage and check for
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.
PROBLEM:
For the sample power system shown below, the generators are connected at all the four buses,
while loads are at buses 2 and 3. Values of real and reactive powers are listed in the table. All
buses other than the slack are PQ type. Assuming a flat voltage start, find the voltages and bus
angles at the three buses at the end of first GS iteration.
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MATLAB PROGRAM:
% Load flow using gauss siedel method
clc clear n=4;
V=[1.04 1 1 1];
Y= [3-j*9 -2+j*6 -1+j*3 0
-2+j*6 3.666-j*11 -0.666+j*2 -1+j*3 -
1+j*3 -0.666+j*2 3.666-j*11 -2+j*6
0 -1+j*3 -2+j*6 3-j*9];
type=ones(n,1);
typechanged=zeros(n,1);
Qlimitmax=zeros(n,1);
Qlimitmin=zeros(n,1);
Vmagfixed=zeros(n,1); type(2)=2;
Qlimitmax(2)=1.0;
Qlimitmin(2)=-0.2;
Vmagfixed(2)=1.04;
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diff=10; noofiter=1;
Vprev=V;
while (diff>0.00001 | noofiter==1),
abs(V);
abs(Vprev);
Vprev=V;
P=[inf 0.5 -1 0.3];
Q=[inf -0.3 0.5 -0.1];
S=[inf 0.5-j*0.2 -1.0+j*0.5 0.3-j*0.1];
for i=2:n,
if type(i)==2 |typechanged(i)==1,
if (Q(i)>Qlimitmax(i)| Q(i)<Qlimitmin(i))
if (Q(i)<Qlimitmin(i))
Q(i)=Qlimitmin(i);
else
Q(i)=Qlimitmax(i);
end
type(i)=1;
typechanged(i)=1;
else
type(i)=2;
typechanged(i)=0;
end
end
sumyv=0;
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for k=1:n,
if(i~=k)
sumyv=sumyv+Y(i,k)*V(k);
end
end
V(i)=(1/Y(i,i))*((P(i)-j*Q(i))/conj(V(i))-sumyv)
if type(i)==2 & typechanged(i)~=1,
V(i)=PolarTorect(Vmagfixed(i),angle(V(i)*180/pi))
end
end
diff=max(abs(abs(V(2:n))-abs(Vprev(2:n))));
noofiter=noofiter+1
end
EXPECTED OUTPUT:
no of iterations V1 V2 V3 V4
1.0250 -
1 1.04 1.0191 + 0.0464i 1.0280 - 0.0870i 0.0092i
1.0334 -
2 1.04 1.0290 + 0.0269i 1.0352 - 0.0934i 0.0208i
1.0385 -
3 1.04 1.0335 + 0.0223i 1.0401 - 0.0999i 0.0269i
1.0413 -
4 1.04 1.0360 + 0.0193i 1.0427 - 0.1034i 0.0304i
1.0429 -
5 1.04 1.0374 + 0.0176i 1.0442 - 0.1053i 0.0324i
1.0437 -
6 1.04 1.0382 + 0.0167i 1.0450 - 0.1064i 0.0335i
1.0442 -
7 1.04 1.0386 + 0.0161i 1.0455 - 0.1070i 0.0341i
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1.0445 -
8 1.04 1.0388 + 0.0158i 1.0457 - 0.1074i 0.0344i
1.0446 -
9 1.04 1.0390 + 0.0157i 1.0459 - 0.1076i 0.0346i
1.0447 -
10 1.04 1.0390 + 0.0156i 1.0460 - 0.1077i 0.0347i
1.0448 -
11 1.04 1.0391 + 0.0155i 1.0460 - 0.1078i 0.0348i
1.0448 -
12 1.04 1.0391 + 0.0155i 1.0460 - 0.1078i 0.0348i
1.0448 -
13 1.04 1.0391 + 0.0155i 1.0460 - 0.1078i 0.0348i
1.0448 -
14 1.04 1.0391 + 0.0155i 1.0461 - 0.1078i 0.0349i
Pranita chavan
POWER SYSTEM OPERATION AND CONTROL
Electrical/PHCET
Exp No: 3
MODELLING OF AGC FOR A SINGLE AREA NETWORK
USING SIMULINK
AIM
To construct a SIMULINK block diagram of a system network and obtain the frequency
deviation response and power deviation response
APPARATUS REQUIRED: Matlab Software
THEORY
As the system load changes continuously the generation is adjusted automatically to
restore the frequency to nominal value. This scheme is known as automatic generation control.
The operation objectives of the LFC are to main reasonably uniform frequency to divide the
load between generators, and to control the tie-line interchange schedules. The change in
frequency and tie line real power are sensed, which is the measure of the change in rotor angle,
i.e. the rotor error to be corrected. The error signal i.e. f and Ptie ,are amplified, mixed,
and transformed into a real power command signal Pv , which is sent to the prime mover to
call for an increment in the torque.
Fig 3.1: Block diagram of a single area power system
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PROBLEM
The isolated power system in the given figure has the following parameters, turbine
time constant 𝛕𝐓 =0.8sec, governor time constant τg =0.1sec, inertia time constant
H=5sec,governor speed regulation R=0.05pu. The frequency changes by 1%,load changes by
0.8%.Plot the variation of frequency for a sudden load change of 0.02pu.Hence find the steady
state frequency deviation in hertz. The frequency dependency of load D=0.8
SOLUTION
Build the model as shown in figure. Two points have been noted.
1) To give a step input of 0.02 as ∆P double click on the block and give final value as 0.02.
2) Click on simulation-configuration parameter and set stop time as 20sec.
SIMULINK MODEL FOR SINGLE AREA WITHOUT AGC
Fig 3.2: Simulink model for single area power system without AGC.
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SIMULINK MODEL FOR SINGLE AREA WITH AGC
Fig 3.4: Simulink model for single area power system with AGC.
CALCULATION
The steady state frequency deviation without AGC in pu
−∆PL
∆wss = 1
[R + D]
RESULT
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POWER SYSTEM OPERATION AND CONTROL
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Exp No: 4
MODELLING OF AGC FOR TWO AREA NETWORK USING
SIMULINK
AIM
To construct a SIMULINK block diagram of a two area system network and obtain the power
deviation response and frequency deviation response .
APPARATUS REQUIRED: Matlab Software
THEORY In a two area power system when an additional load is added, in area
1, the frequency of entire system decreases. Hence, generation in both area increases because of
governor action. However, without AGC, the steady state frequency deviation will not be zero.
Moreover ,since generation in area two has increased there will be additional power flow(tie
line flow) from area two to area one to share the additional load in area1.This additional tie line
flow is denoted as ∆𝑃𝑡𝑖𝑒.
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Fig 4.1: Block diagram representation of a two area power system.
PROBLEM
Obtain the plot of frequency deviation response and power deviation step response for a
load change of 0.05pu in area 1 for the two area system given. The constants for the system are.
Constants Area 1 Area 2
Speed regulation R1=0.05 R2=0.0625
Frequency-sensitive load D1=0.6 D2=0.9
coefficient
Inertia constant H1=5 H2=4
Base power 1000MVA 1000 MVA
Governor time constant τg1 =0.2 sec τg2 =0.3 sec
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Turbine time constant τT1 =0.5 sec τT2 =0.6 sec
PS 2 2
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SIMULINK MODEL WITHOUT AGC
Fig 4.2: Simulink model of a two area power system without AGC.
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SIMULINK MODEL WITH AGC
Fig 4.5: Simulink model of a two area power system with AGC.
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CALCULATION
Steady state frequency deviation in pu
−∆PL
∆Ws = 1 1 =
+
R1 R2
RESULT
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FORMATION OF YBUS BY SINGULAR TRANSFORMATION
AIM
The matrix pair YBUS and ZBUS form the network models for load flow studies. The YBUS can
be alternatively assembled by use of singular transformations given by a graph theoretical
approach. This alternative approach is of great theoretical and practical significance and is
therefore discussed here.
To start with, the graph theory is briefly reviewed.
Graph
To describe the geometrical features of a network, it is replaced by single line segments
called elements whose terminals are called nodes. A linear graph depicts .the geometrical
interconnection of the elements of a network. A graph is connected if, and only if, there is a
path between every pair of nodes. If each element of a connected graph is assigned a
direction,*** it is called an oriented graph.
Power networks are so structured that out of the m total nodes, one node (normally
described by 0) is always at ground potential and the remaining n = m — 1 nodes are the
buses at which the source power is injected. Figure 6.2 shows the graph of the power
network of Fig. 6.1c. It may be noted
here that each source and the shunt admittance connected across it are represented by a
single element. In fact, this combination represents the most general network element and
is described under the subheading “Primitive Network”.
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A connected subgraph containing all the nodes of a graph but having no closed paths is
called a tree. The elements of a tree are called branches or tree branches. The number of
branches b that form a tree are given by
b = m — 1 = n (number of buses) (6.9)
Those elements of the graph that are not included in the tree are called links (or link
branches) and they form a subgraph, not necessarily connected, called
cotree. The number of links 1 of a connected graph with e elements is
Note that a tree (and
therefore, cotree) of a graph is not unique.
A tree and the corresponding cotree of the graph of Fig. 6.2 are shown in Fig. 6.3. The
reader should try and find some other tree and cotree pairs.
If a link is added to the tree, the corresponding graph contains one closed path called
a loop. Thus a graph has as many loops as the number of links
Primitive Network
A network element may in general contain active and passive components. Figures 6.4a
and b show, respectively, the alternative impedance and admittance form representation of
a general network element. The impedance form is a voltage source ers in series with an
impedance zrs while the admittance form is a current source jrs in parallel with an admittance
yrs. The element current is irs and the element voltage is
νrs = Er — Es
where Er and Es are the voltages of the element nodes r and s, respectively.
It may be remembered here that for steady state AC performance, all element
variables (νrs, Er, Es, irs, jrs) are phasors and element parameters (zrs, yrs) are complex
numbers.
The voltage relation for Fig. 6.4a can be written as
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Similarly, the current relation for Fig. 6.4b is
The forms of Figs. 6.4a and b are equivalent wherein the parallel source current in
admittance form is related to the series voltage in impedance form by
A set of unconnected elements is defined as a primitive network. The performance
equations of a primitive network are given below:
In impedance form
Here V and I are the element voltage and current vectors respectively, and J and E are the
source vectors. Z and Y are referred to as the primitive impedance and admittance
matrices, respectively. These are related as Z = Y -1. If there is no mutual coupling between
elements, Z and Y are diagonal where the diagonal entries are the impedances/admittances
of the network elements and are reciprocal.
Network Variables in Bus Frame of Reference
Linear network graph helps in the systematic assembly of a network model. The main
problem in deriving mathematical models for large and complex power networks is to select
a minimum or zero redundancy (linearly independent) set of current or voltage variables
which is sufficient to give the information about all element voltages and currents. One set
of such variables is the b tree voltages.* It may easily be seen by using topological
reasoning that these variables constitute a non-redundant set. The knowledge of b tree
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voltages allows us to compute all element voltages and therefore, all bus currents assuming
all element admittances being known.
Consider a tree graph shown in Fig. 6.3a where the ground node is chosen as the reference
node. This is the most appropriate tree choice for a power network. With this choice,
the b tree branch voltages become identical with the bus voltages as the tree branches are
incident to the ground node.
Bus Incidence Matrix
For the specific system of Fig. 6.3a, we obtain the following relations between
the nine element voltages and the four bus (i.e. tree branch) voltages V 1, V2, V3 and V4.
or, in matrix form
where the bus incidence matrix A is
This matrix is rectangular and therefore singular. Its elements aik are found as per the
following rules:
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aik =1 if ith element is incident to and oriented away from the kth node (bus)
= — 1 if ith element is incident to but oriented towards the kth node
= 0 if the ith element is not incident to the kth node
Substituting Eq. (6.16) into Eq. (6.14), we get
Premultiplying by AT,
Each component of the n-dimensional vector ATI is the algebraic sum of the element
currents leaving the nodes 1, 2, …, n.
Therefore, the application* of the KCL must result in
Similarly, each component of the vector ATJ can be recognized as the algebraic sum of all
source currents injected into nodes 1, 2, …, n. These components are therefore the bus
currents . Hence we can write
Equation (6.19) then is simplified to
Thus, following an alternative systematic approach, we have in fact, obtained the same
nodal current equation as (6.6). The bus admittance matrix can then be obtained from the
singular transformation of the primitive Y, i.e.
A computer programme can be developed to write the bus incidence matrix A from the
interconnection data of the directed elements of the power system. Standard matrix
transpose and multiplication subroutines can then be used to compute YBUS from Eq. (6.23
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POWER SYSTEM OPERATION AND CONTROL
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Pranita chavan
POWER SYSTEM OPERATION AND CONTROL
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