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61 views70 pagesCMPSA Pratical 1-8
Uploaded by
Vivek BendeCopyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
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a
s For
SSGMCE
ANAN MAN
‘SHRI SANT GAJANAN MAHARAJ COLLEGE OF ENGG,
PRACTICAL EXPERIMENT INSTRUCTION SHEET
EXPERIMENT TITLE: WOT APPLICABLE
LABORATORY MANUAL
Subject: CMPSA Lab
BORATORT MANUAL NO. . : SSGMCE/WI/ELE/O7/8E06 ISSUE NOL 00
JORATORY : PLC & FA Lab
ISSUE DATE : 13/01/2025
‘SEMESTER: vit | pace: tor,
MASTER LIST OF EXPERIMENT
No eeerimenr 5
No. EXPERIMENT DESCRIPTION
B Write a program to find bus incidence matrix
2 Write a program to find “ews lwtngntce Se basic Cut set incidence matrix & basic
loop incidence matrix
3 Write a program to find bus admittance (Ybus) matrix. by building algorithm
FP Write a Program to find bus Admittance (bus) matrix for a given Power System
‘Network using Singular ‘Transformation,
5 Perform short circuit analysis ofa Power system network by using simulation software.
| iE ‘Write a program for load flow Analysis of a given power system network using bus
| Admittance (Ybus) Matrix :
7 7o plot Power-Angle curve of Synchronous Machine for stability analysis
8 To plot Swing curve of Synchronous Machine for aa ty analysis,
\3 .
APPROVED BY‘ 6r. SR, Paraskar ae
FORM No. ssomce-rRM.32 B
SHRI SANT GADANAN MAHARAY COLLEGE OF —
G - ENGG, SHEGAON
crcnyeame_ PRACTICAL EXPERIMENT INST
EXPERIMENT TITLE: letermir
To determine bus incidi © Matrix ;
nee Incidence Matrix of a given power
GM/W1/ ELE/07/8EP06/01
ISSUE NO. : 00 | ISSUE DATE : 12.01.2023
tev. DATE: 00 REV. NO.: 00 | DEPTT. : ELECTRICAL ENGINEERING
QBORATORY: Computer Methods In Power S St
\nalysis Lab vystem
‘SEMESTER: VIII [Pe lof 5
Date:
01. Aim:
To determine bus incidence ma
02. Scope: By using this
mat
1A] of a given power system
periment students will be able to determi:
trix of a given power system using MATLAB software
03. Apparatus: PC with MATLAB Software
\¢ bus incidence
04. Theory:
The formulation ofa suitable mathematical model is the first step in the analysis of an
clectrical network. The model must describe the characteristics of individual network
SomPonents as well asthe relations that govem the interconnection of these elements.
A network matrix equation provides a convenient. mathematical model for a digital
computer solution. The elements of a network matrix depend on the selection of the
independent variables, which ean be either currents or voltages. Correspondingly, the
clements of the network matrix will be impedances or admittances, The electrical
characteristics of the individual network components can be presented conveniently in the
form ofa primitive network matrix, This matrix, while adequately describing the
characteristics of each component, does not provide any information pertaining to the
network connections. It is necessary, therefore, to transform the primitive network matrix
into a network matrix that describes the performance of the interconnected network.
The form of the network matrix used in the performance equation depends on the frame of
reference, namely, bus or loop. In the bus frame of reference, the variables are the nodal
voltages and nodal currents, In the loop frame of reference, the variables are loop voltages
A ix is an i tofa
and loop currents. The formation of the appropriate network matrix is an integral pa
digital computer program for the solution of power system problems. PAGE NO. 2/5]
(Leerno-or [
Im order to describe the geometrical structure of a network itis sufficient to replace the
network components by single line segments irrespective of the characteristics of the
components. These line segments are called elements and their terminals are called nodes.
A node and an element are incident if the node is a terminal of the element. Nodes can be
incident to one or more elements. A graph shows the geometrical interconnection of the
clements of a network, A subgraph is any subset of elements of the graph.
A path is a subgraph of connected elements with no more than two elements connected to
‘any one node. A graph is connected if and only if there is a path between every pair of
nodes. If each element of the connected graph is assigned a direction itis then oriented.
‘A representation of a power system and the corresponding oriented graph are shown in
Fig,
ig. Power system
representations,
a) Single line diagram
b) Positive sequence
network
Diagram
©) oriented conneeted
graph,
Fig: Tree and Co-tree of
connected oriented
connected graph
where
¢= number of elements“?
n= number of nodes =
Cincluding reference
node) a
b= number of branche
1 number of links =3
a
: : ne ye v
[EXPTNO. 01 PAGENO. 3/5
‘Acconnected subgraph containing all nodes of a graph hut no closed path is called
atree,
‘The elements of a tree are called branches and form a subset of the elements of the
connected graph. The number of branches b required to form a tree is
ben-l-
(1), where nis the number of nodes in the graph.
‘Those elements of the connected graph that are not included in the tree are called links and
form asubgraph, not necessarily connected, called the cotree. The cotree is the
complement ofthe tree. The number of links / of a connected graph with ¢ elemen
I=e-b
From equation (1) it follows that
J=e-n+] 2)
‘A tree and the corresponding cotree of the graph are shown in Fig
Talink is added to the tree, the resulting graph contains one closed path, called a loop.
‘The addition of each subsequent link forms one or more additional loops. Loops which
contain only one link are independent and are called basic loops. Consequently, the
number of basic loops is equal to the number of links given by equation (2). Orientation of
abasic loop is chosen to be the same as that of its link. The basic loops of the graph given
g. are shown in Fig. A cut-set is a set of elements that, if removed, divides a connected
gxaph into two connected subgraphs.
‘A unique independent group of cut-sets may be chosen jfeach eut-set contains only one
cut, sets, The number of basic cursets is
branch. Independent cut-sets are called basic
Je the same a5
equal to the number of branches. Orientation ofa basic cut-set is chosen to b
jven in Fig, and are showin in Fig
that ofits branch, The basic cut-sets of the graph
sie Cut set of
Fig: Ba
Graph.
oriented connected
Number of basic ut set=
Number of branches
tion of
jon of oriental
rection be ee
(EXPCNO Oa —————
Incidence Matrices:
Element-node incidence matrix [A* I:
The incidence of elements to nodes in a connected ‘graph is shown by the element-node
incidence mats. The element of the matrix are as follows:
81-1 ifthe ith element is incident to and oriented away from the jth node
li =-1 ifthe ith element. is incident to and oriented toward the jth node
$3 Oif the ith element isnot incident othe th node,
Flement-node incidence matrix [A™]
Bus incidence matrix [A];
tix ise X(n- 1)
and the rank ism =1 = b, where bis the number of branch
sin the graph, Selecting node 0
as reference for the graph shown in Fig, a PAGEN
515
“This matrix is rectangular andl therefore singular, Ifthe rows of [A] are arranged according
toa particular tree, the matrix can be partitioned into submatrices [Ab] of dimension
pX(a- Hand [Al] of dimension 1 X (n 1), where the rows of [Ab] correspond to
janches and the rows of [AJ ] to links. The partitioned matrix for the graph shown in
Fie
Bune
7 7 =
Asis «nonsingular equare matrix with rank (n= by
Partitioned matrix of bus ineidence Matrix [A]
05 Circuit Diagram: NA
06 Procedure:
’) open MATLAB
ii) open new M-File
iii) Type the program
iv) Save in current directory
) Compile and run the program
& ¥i) For output see output window / command window
07 Observations: NA
8 Sample calculations: Not Applicable
09 Graph if any: Not Applicable
10 Resu
ven power
in this way we have determined bus incidence matrix [Al ofagi
system using MATLAB software
Prepared By: R.K.Mankar ‘Approved By: H.0.D.
ines | EXPT NO. Ol PAGENO. 5/5
‘is matrix is rectangular and therefore singular. If the rows of [Al are arranged according
toa particular tee, the matrix ean be partitioned into submatices [Ab] of dimension
5X (n= Land [Al] of dimension 1 X (n 1), where the rows of [Ab] correspond to
tonches and the rows of [AJ] to links, The partitioned matrix forthe graph shown in
ie.
Asis a nonsingular equare matrix with rank (2 — 1
Pariitioned matrix of bus incidence Matrix [A]
(5 Cireuit Diagram: NA
06 Procedure:
’) open MATLAB
ii) open new M-File
i) Type the program
iv) Save in current directory
¥) Compile and run the program
vi) For output see output window / command window
07 Observations: NA
08 Sample calculations: Not Applicable
09 Graph if any: Not Applicable ven power
10 Result: In this way we have determined bus incidence matrix [A] of a given Po
system using MATLAB software
0.0.
Prepared By: 1K Mankar Approved By: H: % Expt 1:
% Program to find Bus incidence Matrix [_ a
lear all
cle
neinput(‘enter total no. of nodes(including ref. node) :
cinput(‘enter total no. of elements : ');
Feb;
A=zeros(e,b);
fprintf(Enter the nodes where elements are connected start with branches, \n');
{print{('Note:~ as per the direction enter from "p" toward "q" \n');
for
seq(
prinput('Enter the starting node : ');
input(Enter clement no, to be added
i-input('Enter the End node : ');
itp=0
AGp=1;
end
ifq=0
AGig=-13
end
end
Bus_Ined_Mat=[0:b ;seq' »A] I no. of nodes(including ref. node) : 5
enter total no, of elements : 6
Enter the nodes where elements are connected start with branches,
Note:-as per the direction enter from "p" toward "q"
Enter element no, to be added :3
Enter the starting node : 1
Enter the End node :3
Enter element no. to be added : 4
Enter the starting node : 3
Enter the End node : 4
Enter element no. to be added : 5
Enter the starting node :
Enter the End node : 4
Enter element no. to be added 16
Enter the Starting node : 0
Enter the End node: 1
Enter element no. to be added : 1
Enter the starting node : 1
Enter the End node ; 2
Enter element no. to be added : 2
Enter the starting node : 1
Enter the End node: 4 [
pxpt J: OUTPUT
Bus_Incd_Mat=[A]
RESULT:
6X magpeix. =p
FORM NO. SSGMCE-FRM-32 B
SHRI SANT GAJANAN MAHARAJ COLLEGE OF LABORATORY MANUAL
ENGG. SHEGAON
PRACTICAL EXPERIMENT INSTRUCTION SHEET
EXPERIMENT TITLE: To determine basic Cut set incidence Matrix and Basic
00p Incidence Matrix of a given power system Network
EXPERIMENT NO, : SGM/WI/ ELE/07/SEP06/02 | ISSUENO,: 00 | ISSUE DATE: 12.01.2023
REV. DATE: 00 REV. NO? 00 DEPTT. : ELECTRICAL ENGINEERING
LABORATORY : Computer Methods in Power System | SEMESTER: VIII
Analysis Lab PAGE: 1 0f8
Date:
01, Aim: =
To determine basic Cut set incidence matrix [B] and Basie loop incidence matrix [C]
of a given power system
02, Scope: By using this experiment students will be able to determine basic Cut set,
incidence matrix and basic loop incidence matrix of a given power system using MATLAB
software
03. Apparatus: PC with MATLAB Software
04, Theor
In order to describe the geometrical structure of a network it is sufficient to replace the
network components by single line segments irrespective of the characteristics of the
‘components. These line segments are called elements and their terminals are called nodes.
‘Anode and an element are incident if the node is a terminal of the element. A graph shows
the geometrical interconnection of the elements of a network. A subgraph is any subset of
elements of the graph.
‘A path is a subgraph of connected elements with no more than two elements connected to
any one node. A graph is connected if and only if there is a path between every pair of
nodes, If cach element of the connected graph is assigned a direction itis then oriented.
‘A representation of a power system and the corresponding oriented graph are shown in
Fig. ee
ig. Power system
Fi :
representations:
a) Single line diagran
by Positive sequence
network
Diagram
c) oriente¢
graph.
d connected
Tig: Tree and Co-tree of
connected oriented
connected graph
where
c= number of elements~7
n= number of nodes =
(including reference
node) Z
umber of branch
umber of links = PAGENO. 38
Fig: Basie Cut set of
oriented connected
Graph
Number of basic cut set
‘Number of branches
Direction of orientation
of basic cut sets =
Direction of orientation
of Branches
Incidence Matrices:
Element-node incidence matrix [A* ]
“The incidence of elements to nodes in a connested graph is shown by the clement-node
incidence matrix. The element of the matrix are as follows
aij=1 ifthe ith element is incident to and oriented away from the jth node
aij
sf the ith element. is incident to and oriented toward the jth node
aij =0 ifthe ith element is not incident to the jth node, ‘The dimension of the matrix [ A*]
nis the mumber of nodes in the graph.
jseXn,wheree is the number of elements and
‘The element-node incidence matrix for the graph shown in Fig. is
Flement-node incidence matrix [A*]
TA]: Any node of
selected as the
aeonnected graph can be
Bus incidence ma
reference node, Then, the variables of the
o the assigned reference. 7
ther nodes, referred to as buses, can be
ol
measured with respect ( ne matrix obtained from [A] by deleting incidence matrix [A] ,
: nt-bus incidence
the column Corresponding to the reference node is the elem
eae mn of this matrix is eX (n- 1)
led the bus fo
ich incidence matrix. The dimension of thi Pa
and the rank is,
-1=5,
© a8 reference forthe grap
7 h. Selecting node
i: f branches in the grapl
, where bis the number o
h shown i
Partitioned matrix of
where a path is oriented from a bus
“31 ifthe ith branch is in the pa
oriented in the same direc
Kij = -1 ifthe ith brana
'S oriented in the opposite
4+ 18 « nonalngslne wauare macnn
with rank og yy
us Ineldence Mata Tay
Branch Path Incidence Matrix I:
‘The incidence of branches to
Nis in the path
Paths ina tree is shown,
to the reference
th from the i
ion
by the branch.
Node. The
th bus to rete
Path incidence matrix,
Clements of this matrix are:
tence and is
th from
rection
the jth bus to Fefetence but
— O 4s
4, Fi
NO.
» pet : PAGENO, 5/6
ie wacoutine jth branch is not in the path from the jth bus to reference
ayant a reerenes the branch-path incidence matrix associated with the tree shown
in Fig.
Branch Path Incidence Matrix [K)
“The branch:path incidence matrix and the submatrix [AB] relate the branches to paths and
tc-one correspondence between paths
ranches to buses, respectively. Since there is a one-
and buses,
fab] (K] =]
‘Therefore
[K’] =[Ab-1]
Basic Cut-set incidence Matrix [B] =
“The incidence of elements to basic cu-sets oF connected graph s shown bythe Basie our
set incidence matrix B. The elements of this matrix are:
bij =1 ifthe ith clement is incident to and oriented in the same direction as the jth basic
cut-set
bij =-1 ifthe ith element is incident to and oriented in the opposite direction a the jth
basic cut-set
bij = 0 if the ith element is not
incident tothe jth asic cut-set
ee 6/8
ee _ PAGEN
The
basic cut-sot incidence Matrix, of dimension e X b, for the graph shown in Fig. is
[B] : Basic Cut Set Incidence Matrix
Number of basic cut set = Number
of branches ) |) (emo | PAGENO, 78
“The incidence of tnksto buses shown bythe submis [AT] andthe incidence or
N ranches to buses is shown by the submatrix [Ab ] . Since there is a one-to-one
correspondence of the branches and basic cut-sets,
[BN [Ab] , shows the incidence of links to buses, that is,
| [BN [Ab]= [A]
Therefore,
[BIJ = [AT] [AB]
In addition, as shown in equation, [Ab]-1 = [Kt]
Therefore,
BS pry = [AN Ke) (where [K] = Branch Path Incidence Matrix )
Basic Loop Incidence Matrix [C
——
The incidence of elements to basic loops of a connected graph is shown by basic loop
incidence matrix. The elements of the matrix are :
cij=1, if the ith element is incident to and oriented in the same direction as the jth basic
loop
cij =-1 , if the ith element is incident to and oriented in the opposite direction as the jth.
basic loop
0, if the ith element is not incident to the jth basic loop
‘The basic loop incidence matrix [C] of dimension e x I for the graph shown in figure
Fig: Basic Loop Incidence Matrix [C] PAGE NO. 88
EXPTNO. 07
. the rows of [Cb]
‘The matrix (C] can be partitioned into sub matrices [Cb] & [UI] _ “une
. iti mi
SorFesponding to branches and rows of {UL to links. The partition
1) Basie loops N Basic loops
‘NE oF @ eN\
SNF Ue
1 1
g
2 3 Cc
3 a
C= 4
a
U
05 Cireuit Diagram: NA
06 Procedure:
i) open MATLAB
ii) open new M-File
iii) Type the program
'N) save in current directory &
¥) Compile and run the program
vi) For output see output window / command window
07 Observations: NA,
08 Sample calculations: Not Applicable
09 Graph if any: Not Applicable
10 Result: In this way we have determine basic Cut
loop incidence matrix [C] ofa given Power system
St incidence matrix [B) and basic
using MATL AR Software
Prepared By: R.K.Mankar > 2
quexpt no 2% program to find B,C matrix from [A] matrix
age Branch path incidence matrix
9% B=Basic Cut set incidence matrix
1% C= Basic looP incidence matrix
clear all
input’enter total no, of nodes(including ref. node)
einptfonter total no. of elements:
beni
[ze-b;
Azzeros(e,b);
{printf(Enter the nodes where elements are connected start with branches, \n');
fprint{(‘Notes- as per the direction enter from "p" toward "q" \n');
for
seal
put(‘Enter element no. to be added
peinput('Enter the starting node :
qzinput(‘Enter the End node =");
end
tar
Alial=-35
end
end
Bus_Incd_Mat=[0:b ;seq’ A]
Abe A(L: bi
Alcalb#t: € 335
Kteinv((ab)*eve())i
Keke;
WK Cb:
Basic_Cut_Set=[eye(b);Bl]
Basic_Close_Loop=[Cb;eye(|)]} ~b
expt no 2!
center total no. of nodes(including ref. node}: 5
enter total no. of elements: 7
Enter the nodes where elements are connected start with branches,
Note: as per the direction enter from "p" towards
Enter element to be added: 2
Enter starting node: 1
Enter the end node: 0
Enter element to be added: 3
Enter starting node: 1
Enter the end node: 2
Enter element to be added: 5
Enter starting node: 3
Enter the end node: 0
Enter element to be added: 7
Enter starting node: 4
Enter the end node: 3
Enter element to be added: 1
Enter starting node: 4
Enter the end node: 1
Enter element to be added: 4
Enter starting node: 2
Enter the end node: 0
Enter element to be added: 6
Enter starting node: 3
Enter the end node: 2 ‘OUTPUT ( expt 2)
Bus_Incd_Mat =
Basic_Cut_Set=[B]
Basic_Loop = [C]
Oot
21000
31200
50010
yoort
a 0 02
40100
Ores
1000
o100
oo10
ooot
Pip Ota!
1100
a110
ena
oat
a) Ot
100
100
o10
oot »\\
FORM NO. SSGMCE-FRM-32 B
‘SHRI SANT GAJANAN MAHARAJ COLLEGE OF TABORATORY MANUAL
ENGG. SHEGAON
PRACTICAL EXPERIMENT INSTRUCTION SHEET
EXPERIMENT TITLE: To determine bus Admittance [Ypus] matrix for 2
given Power System Network by building algorithm
EXPERIMENT NO.7 SGM/ WIT ELE707/8EP06/03_[ TSSUENO.: 00 | ISSUE DATE ; 12.01.2025
REV. DATE : 00 REV. NO. + 00 DEPTT. # ELECTRICAL ENGINEERING
TABORATORY: Computer Methods in Power System SEMESTER: VIII] PAGE : Lf 2
Analysis Lab
01. Aim:
To determine bus Admittance [Ybus] matrix for a given Power System Network by
building algorithm
02. Scope: By using this experiment students will be able to determine bus Admittance
matrix [Ybus] of a given power system using MATLAB software
03. Apparatus: PC with MATLAB Software
04. Theory:
‘The bus impedance matrix is given by Zous = [Yous]
‘This transformation can be derived using the concept of power invariance . Ypys matrix is a
matrix that contains admittances of the power system. Ybus matrix can be inverted and
multiplied with currents to find voltages of buses. Although taking inverse of a large
system is computat
ally not efficient and there are various methods to find Zbus directly
instead of taking inverse of Ybus. Ybus is sparse matrix with a large number of zeroes in it.
ad So its a large size matrix with a large number of zeroes.
Off-diagonal elements are symmetric, the diagonal elements of each node are the sum of
the admittances connected to it. ¥ bus is a square matrix. For n bus power system, size of
Y bus matrix is nxn. Value of diagonal element corresponding to bus i, then. Yii = Sum of
the admittances connected to bus i.
Step-by-step algorithm for determining the bus admittance matrix [Ybus] for a power
system network: Initialize the [Ybus] matrix with zeros of size n x n, where n is the number
of buses in the network. : S number isi and the to bus number ij
1) locations in the [Ybus) matrie,
‘ork, apply following steps:
shunt admittance at each bus,
b. Add [Ysh] to [Ybus] to obtain the final [¥bus] matrix,
‘The [Ybus] matrix can now be used to solve various power system problems,
flow analysis and fault analysis,
05 Cireuit Diagram: NA
06 Procedure:
i) open MATLAB
ii) open new M-File
iii) Type the program
iv) Save in current directory
such as load
¥) Compile and run the program
Vi) For output see output window / command window
07 Observations: NA
08 Sample calculatior
lot Applicable
09 Graph if any: Not Applicable
10 Result: In this way we have determined bus Admittance matrix [You] of a given power
‘system by building algorithm using MATLAB software
[Prepared ny: RK Mankar wy ud | Approved ByEH.OD, Wr Expt 3: % Program to find Y Bus matrix by building algorithm
clear all
dle
nzinput(‘enter total no. of nodes (including ref. nodes):');
ecinput(' eneter total no. of elements:);
Ybus=zeros(b,b);
forielie
\einput(‘Enter the starting node & end node & Impedance in matrix form);
pei °
ov)
Ye)
if (p&a)
Yous(p,p)=Ybus(P,P)+¥s
Yous(q.q)=Ybus(a.a)+¥;
Yous(p,q)=Ybus(p.a)-¥s
Ybus(q,p)=Ybus(P,a)i
else
Yous(p+a,ptq)=Ybus(p+a,p+a)+¥s
end
end
Yous
Zouszinv{Vbus) Expt no 3:
Enter total no, of nodes (Including ref. nodes):4
Enter total no. of elements:5
Enter the starting node 8 end node & Impedance in matrix form: (0 1.0.4]
Enter the starting node & end node & Impedance in matrix form: [0 2.0.4]
Enter the starting node & end node & Impedance In matrix form: [2 3 1.0]
Enter the starting node & end node & Impedance in matrix form: [0 3 0.5]
Enter the starting node & end node & Impedance in matrix form: [1 20.5] FORM NO. SSGMCE-FRM-32 B
SHAY SANT GAJANAN MAHARAJ COLLEGE OF
ENGG. SHEGAON
PRACTICAL EXPERIMENT INSTRUCTION HEE
EXPERIMENT TITLE: To determine bus Admittance (You:
kcusing Singular Transformation.
DATE: 12.01.2023
given Power System Networt
EXPERIMENT NO.1 SGM/WH/ ELE/O7/SEPOG/04 [TSSUENO.;00 | 1SSUE
REV. DATE: 00 REV. NO. 00 DEPTT. 1 ELECTRICAL ENGINEERING
LABORATORY: Computer Methods in Power System SEMESTER VIKT | pAGE : Lof 3
J
Analysis Lab
Date:
01. Aim:
To determine bus Admittance (Ybus) matrix for a given Power System Network using
Singular Transformation.
(02. Scope: By using this experiment students will be able to determine bus Admittance
matrix [Ybus] of a given power system using MATLAB software
03. Apparatus: PC with MATLAB Software
04. Theory:
Determination of [Ybus ] Matrix in the bus frame of reference (Mathematically)
Performance equation of primitive network in admittance form
(1)
i4j- 17
where, j= External current source.
Ly ]§ primitive admittance matrix (element element matrix)
Select a matrix which is related to Bus
ie select [A] Matrix (where [A] is bus incidence matrix)
Pre multiplying eq (1) by [At]
fat] i #fAC0 i) = (Ad)
Q) PAGER
(EXPTNO.a
but [At](i)=0 (KCL is applied at various nodes )
(algebraic sum of current at node = 0)
(AUG) = TAY) eee @)
As [A (j)=[1 bus} (4)
(bus ]= [At] [y] 7
Convert % into [E]
Network Performance equation in bus fram
{bus} = [Ybus] [E bus] (6)
Z power consumed by individual el
1¢ of reference in admittance matrix,
lement = j*t V7
; Power in bus frame of reference = O* bus}t [E bus] - (8)
Equating equation (7) & (8) we get ,
Uttv = qe bus}t [E bus ]
Take conjugate trans
—
Orbus}t = j+
0)
Y
{Y bus) te bus)
Thetefoge FAN tye by
Is
LY bus) =A ly] 1A) ([EXPTNO. 04 I PAGE NO. 3/3
where , [Y bus] =Bus admittance matrix. (msn matrix)
[A] = Bus incidence matrix.
Ly] = Primitive admittance matrix.
05 Circuit Diagram: NA.
06 Procedut
i) open MATLAB
ii) open new M-File
iti) Type the program
iv) Save in current directory
-¥y) Compile and run the program
vi) For output see output window / command window
07 Observations: NA
08 Sample calculations: Not Applicable
09 Graph if any: Not Applicable
10 Results In this way we have determined bus Admittance matrix [Ybus] of a given
power system by singular Transformation using MATLAB software
[rover 203.
Prepared By: RK Marker ry
Exptno :4/Code/Ybus matrix/CMPSA-Lab
This program forms YBUS by singular transformation
athe data for this program is a pr.
imitive admittance matrix
y
&which is to be given in the following format and stored in
ydata.
Sground is given as bus no 0.
gif the element is not mutually coupled with any other
element, then the entry corresponding to 4th and 5th
% of ydatahas to be zero
olumn
%element no | connected | y(selg) | _ mutually |
y (mutual)
% | From | to | |coupled tol
% |Busno|Busno]
ao- —
1 2 1/(0.05+3*0.15) ° °
2 1 3 1/(0.1 +5*0.3 ) 0 °
3 2 3 1/(0.15+5*0.45) ° °
4 2 4 1/(0.10+3*0.30) 0 0
5 3 4 1/(0.05+3*0.15) °
Ol;
%form primitive y matrix from this data and intialize it to
zero.
&%to start with
elements=max (ydata(:,1))
yprimitive=zeros (elements ,elements)
gprocess ydata matrix rowwise to form yprimitive
for i=:
:elements,
yprimitive (4,4)=ydata (i, 4) yprimitive (i ,i)=ydata(i,4)
eae the element is mutually coupled with any other
oe clement is not indicated in 5th column of ydata
F teats eePOnding column no in ith row is made equal to
if (ydata(i,5)
o)
& ;
J is the A ; ji
mutually coupled element no with which its element is
J=ydata (i,5)
ymutual=ydata (i, 6)
yprimitive (i,j) = ymutual
end
end
% form bus incidence matrix A from ydata
%this gives a matrix of size element x buses
buses=max (max (ydata(2,:)) ,max(ydata(3,:)))
A=zeros (elements , 4)
for i:
elements;
sydata(i,2) gives the 'from' bus no.
&The entry corresponding to column correponding to this
bus
%in A matrix is made 1 if this is not ground bus
if ydata(i,2) ~= 0
A(i,ydata(i,2))=1;
%ydata(i,3) gives the 'to' bus no.
% the entry corresponding to column corresponding to this
bus
gin A matrix is made 1 if this is not ground bus
elseif ydata(i,3)~
A(i,ydata (i,3)
end
°
end
YBUS=A' *typrimitive*A i
FORM NO. SSGMCE-FRM-32 5
SHRY SANT GAJANAN MAHARAT COLLEGE OF LABORATORY MANUAL
ENGG, SHEGAON
Pl
RACTICAL EXPERIMENT INSTRUCTION SHEET |
EXPERIMENT Tr
TLE: Perform short circuit analysis of a Power system
etvork by using simulation software
SSM /WT/ ELE/07/sEP06/05 | ISSUENO-TO0 ISSUE DATE : 12.03.2025
REV. DATE : 00 REV N i
[Wiborarory
[ DEPTT. + ELECTRICAL ENGINEERING
Computer Meth SEMESTER: VITT
Analyse ‘ods in Power System
PAGE: 1 of
Date:
02. Scope: Perform shor circuit analysis of a Power system network by using using
PSCAD software
03. Apparatus: Personal Computer with PSCAD Software
04. Theory: Fault : Fault in an electrical equipment is defined as a defect in its electrical
Circuit due to which a current is diverted from the intended path. The fault impedance
being low, the fault currents are relatively high. During the faults, the voltage of the three
Phases becomes unbalanced. The fault current being excessive, th
ey can damage the faulty
auipment and the supply installation, During the faults, the ower flow is diverted
‘owards the fault and supply to the neighboring zone is affected,
depends on voltage at the faulty point and the total impedance u
voltage atthe fault location changes from its normal value. Di
The value of fault current
iP to the fault location. The
ing the fault, the current
and voltage undergo a continuous change and phenomena observed are called transient
phenomena,
Faulu inovetsead
tnnsainion yatem
Series Passes :
‘Shunt Fautts
(Open creuit fats) (Shon creat fatts)
One open Twolopen Apmmevical —— Symateaieal
contctorfudts —eonnetteaty ttt tutte
=
10 tat 1 faa,
Loft — LLL fue LLLe faut
(o.n0.co) —(AD,BECA) (anornce'taay «awe tance)
Figure. Various types of faults that can occur within thrce-phase power tranamlsion ayer > (EXPINO.05 PAGENO. 2/8
Causes of Faults : aon
i) Symmetrical Fault : Symmetrical Faults are usually occur due to wrong operation /
coordination between circuit Breaker and earthling switch. It occurs when line is energize
‘and earthling switch is inadvertently kept ON.
ii) Asymmetrical Fault :
L-G fault : It occur due to flashover / failure of insulator. One conductor comes in contact
with the ground or neutral conductor.
LL and LL-G fault : It occurs due to swinging of two conductors / shorting of wires due
to birds/ kite string / tree limb. LL-G fault occurs when two conductors fall on the ground
or come in contact with neutral conductor,
During monsoon season dielectric strength reduces , there is a fair chances of flashover or
Power are between the two conductors,
Consequences of Faults :
Faults cause two types of damage :
') Thermal damage: It occurs slowly as iti related to temp. ie class of insulation .
Instantaneous tripping is not required in case of thermal damage,
Probability of occurrence of Faults on different Element:
fs.
Equipment
Overhead transmission line
Underground Cable
Switchgear ineludi
Power Transformer
Miscellaneous
rs
Probability of Occurrence of Faults 0)
Probability of fue oF occurenge of
" Overhead tines:
abnormal cond
to their greater lengy ion i
"BIE Exposure tothe atmosph “smo on Oe ines Ds
ete, (EXPT .NO. 05 [PAGENO. 3/8
Most of the Power system. faults occur on transmission system networks , especially on
overhead lines. Due to their inherent characteristics of being exposed to atmospheric
conditions , transmission lines have the highest fault rate in the system.
‘Type of Fault ‘% of occurrence of fault
‘Asymmetrical Fault L-G 80 to 90 %
LL 6 to 10%
LLG 306%
Symmetrical Fault LLLLLL-G 1% or less
Occurrence of fault ean cause:
i) Interruption in the power supply due to consumers.
ii) Substantial loss of revenue due to interruption of service .
iii) Loss of Synchronism : wide spread blackout due to tripping of multiple lines /
generators.
iv) Extensive damage to equipment ( transformers, generators, Induction Motors etc)
v) Serious hazards to personnel,
Objectives of Short Circuit Analysis :
1) Design of power system and configuration of the network with taking into account the
expected level of fault current.
ii) Selection of interrupting capacity of the devices for operation on occurrence of fault
and isolation of faulty equipment /subsystem from healthy part of the power system .
) Selection of fault making rating of the interrupting, devices when required to close
against sustained fault.
iv) Analysis of Performance of protective relays and their setting for proper their
coordi
\n. Selection of short time rating of all the equipment which are required to carry
short circuit current from the instant of occurrence of fault tll the interrupting device
operates and isolates the faulty equipment /subsystem from the healthy part of the power
system.
v) Determination of the cross section area of conductors and cables based on fault clearing
time, Selecting the methods for limiting fault currents and design of respective current
limiting devices.
vi) Design of earthling and grounding systems.
i) Operation of the systems with particular references to security of supply and economic
consideration, =
(EXPTNO.05 PAGENO. 4/8
‘Symmetrical or Balanced three phase fault analysis :
Most preferred method of analysis of symmetrical three phase fault is per unit / percentage
reactance method by modelling the reactance of the equipment in the reactance diagram of
the system, All reactance of the equipment in the power system referred to a common base.
To simplify analysis of fault currents in the network, following simplifications are
made.
i) Transmission lines are represented by their series reactance calculated based on size of
Conductor used and spacing of the conductor inthe transmission line configuration,
ii) Transformers are represented by their leakage reactance.
Synchronous machines are modelled as constant voltage behind direct axis sub-
transient reactance
iv) Induction motors are ignored
In this type of faults all three phases are simultaneously short citcuited, Since the network
remains balanced, itis analysed on per phase basic. The remaining two phases carry
identical currents but with a phase shit of 120 degree with each other, A fault inthe
network is simulated by connecting impedance in the network at the fault location, The
faulted network i then solved using the Thevenin’s equivalent as seen from the fault point,
Prior tothe occurrence of fault ,the system is assumed to be in a balanced steady state,
Types of Bus : P-Q bus (Load bus),
Reference bus
POJECT SETTINGS IN PSCAD SOFTWARE :
Duration ofrun 2.0 see
P-V bus (voltage control bus), Slack / Swing/
Solution time step: 50.0 micro see
Channel plot step: 100.0 micro sec
Sampling Frequency ;20.0 kHz
Time Fault Logic :
Time of Apply Fault : 1.0 sec
Duration of Fault: 0.5 sec
Fault Resistance :
Fault ON Resistance : 0.01 ohm
Fault Type Control : Internal
Fault Type : L-G, L-L,LL-G,LLL,LLL-G EXETINO!05 PAC 5
05 Cireuit Diagram
vz
e- Po
2
1"
‘ T,
a
Fig I: Single Line Model of IBEE 9 Bus Test System
PSCAD Model of IEEE 9 bus Test System:
ewe -PSACD modelof EES BUNATNED
Fig 2: PSCAD Model of! TEEE 9 bus System Fig 3 : PSCAD Model of IEEE 9 bus system durins ‘No fault ‘condition.
source impedance i:
Each machine (Generator) is represented asa voltage source Where its
set arbitrarily as 1 ohm. Table 1 summarizes the per unit (P¥) terminal conditions of each
source, with 100 [MVA] base.
06 Procedure:
i) Select the required components from the master library
ii) Draw single line diagram of IEEE 9 bus system by using PSCAD Software.
iii) Enter the specifications of each component.
iv) Simulate the system for NO fault condition and create different types of Faults ( like L-
G, L-L,LL-G,LLL,LLL-G)) on Various buses of IEEE 9 bus system and (measure
voltage by using three phase RMS meter) observe the voltage waveform during-- without
fault and with fault condition.
v) Right click on output channel, select graph/meter/control > add overlay graph with
signal.
h properties) and insert horizontal and vertical
vi) Left lick on overlay graph (overlay grap
axis properties of graph for output channels of
ee
all quantities (or parameters). a ae
[EXPE NO-93 TP O78
Right click on workspace and select the project settings:
) Duration of run
) Solution time step
©) Channel plot step
@ Make save channel to disk 10 ‘yes’ and put same name (i. name of simulation model
file) for output file with (out) extension,
©) This creates (.emt) folder, which is useful for: analysis in MATLAB. ( if required )
vili) Click on run button from the toolbar, with required run time, and let the circuit to
simulate,
ix) Click on arrow at the top right comer on the graph window and click zoom in and zoom
ut as required to set the graph within the visible window.
07 Observati
voltage with respect to time at each bus of the power system during No fault and with
After the successful simulation observe the output waveforms of bus
fault condition. [Fe ‘4; PSCAD Model & bus voltage waveforms of IEEE 9 bus Test system during L-G fault condition. |
08 Sample calculations: Not Applicable é
09 Graph if any: Bus voltage wrt time at each bus of the power system during No fault
and fault condition.
10 Result: In this way we have Simulat
ted IEEE 9 Bus system by using PSCAD Software
and output waveforms are observed.
[ Prepared By: R.K.Mankar Bi l Approved By: H.0.D. i APPENDIX
1+ Terminal conditions of IEEE 9-bus system
[Bus
V (kV) [8 fae;
1 [7.1600 fo.b000— [ote | en Tapa Poa |
$ Heo 9.3507 Tee 713 MW 0.2791 2791 MVAr_|
0 [staf us| San —| ee Lew]
Table 1.2: Term iti
‘Terminal conditions of IEEE 9-bus syste
Voltage source Name
Base MVA( Three Pi Soure
hase Source 2 | Source 3
Base Frequency (Hz) ey ae 100 00 |
| Voltage magnitude (L-L, RM 0 60 |
“s IS.
Phase (degree) ) Ha kV 18.4500 kV | 14.1450 kV
ial Real Power (pu) ta 9.3507 5.1420 |
nitial Reactive Power (pu) tom nae oS 4
Transmission lines are modeled using the Bergeron model.
Table 2 : Transmission line characteristics of IEEE 9-bus system.
__lesasunission i a Ripwm] | X{puim] | B [pum]
a 5 0.0100 0.0680 0.1760
2 6 0.0170, 0.0920 0.1580 4
5 iu 0.0320 0.1610 0.3060
6 9 0.0390 0.1738 0.3580 |
7 8 0.0085 0.0576 0.1490
8 9 0.0119 0.1008 0.2090 |
Loads are modeled as a constant PQ load with parameters as shown in Table 3.
‘Table 3 - Load characteristics of IEEE 9-bus system
SS LmLmrmUmUmre Uva |
5 1.25 0.50 125 30
6 0.90 0:30 90 30
8 1.00 035 100 35 Table 4: Transformer Data
Transformer Name TRI TRE
Throe Phase Transformer (MVA) 100) 7
ase operating Frequency (Hz) 0
winding | Type star grounded | stat
[winding 2 Type star grounded —_| star
16.5 18
230
[winding [Line to Line voltage RMS (kV)
230
[winding I Line to Line voltage RMS (KV)
inger on Transformer
number)
Table 5: Bus Data
[Bus No Bus Voltage (kV) _| Bus Voltage (pu)
E 16.50 1.04000
2 18.00, 1.02500,
B 13.80. 1.02500
Ez 230 1.02531
5 230 0.99972
[6 230 1.01225
[7 230 1.02683
[s 230 1.01727
By 230 1.03269 FORM NO. SSGMCE-FRM.32 5
‘SHRI SANT GAJANAN MAHARAJ COLLEGE OF TABORATORY MANUAL
— _ENGG. SHEGAON
PRACTICAL EXPERIMENT INSTRUCTION SHEET
‘XPERIMENT TITLE: Analysis of load flow study of a given power System
etwork using MATLAB software.
EXPERIMENT NO,
PIPERIMENT NO. : SGH/WH/ ELE/07/8EP06/06 | TEUEWO.T00-[ ISSUE ORTE TOTO
REY. DATE: 09 [REV.NO. 00 [[DEPTT. + ELECTRICAL ENGINEERING |
{ABORATORY: Computer Methods in Power System [SeeeeTER VAT | race: TOFS \
01. Aim: a
Write a program for N-R load flow
bus Admittance (Yiu)
02. Scope:
Method of a given power system network using
ris
By using this experiment students will be able to analyze load flow study of a
given power system network using MATLAB sofware
03. Apparatus: PC with MATLAB Software
04. Theory:
‘The load centers are usually located away from generating stations. Therefore, the power is
‘transmitted to the load centers and is stepped down to distribution level, The load is,
supplied at various voltage levels. The load may be residential, industrial or commercial,
Depending on the requirement the loads are switched on and off. Therefore, there are peak
oad hours and off peak load hours. When there is a need, power is transmitted from one
area to the other area through the tie lines. The control of generation, transmission,
distribution and area exchange are performed from a centralized location. In order
to perform the control functions satisfactorily, the steady state power flow must be
known. Therefore, the entire system is modeled as electric networks and a solution
is simulated using a digital program. Such a problem solution practice is called
power flow analysis.
The power flow solution is used to evaluate the bus voltage, branch current, real power
flow, reactive power flow for the specified generation and load conditions. The results are
used to evaluate the line or transformer loading and the acceptability of bus voltages.
In general, the power flow solutions are needed for the system under the following,
conditions:
#) Various systems loading conditions (peak and off peak).
ii) With certain equipment out aged. AO
oe NO. 06
ime ‘on of new generators.
les.
fon tines of cab
iV) Addition of new transmission lines
; it =r systems,
”) Interconnection with other syst
vi) Load ‘Browth studies,
Vil) Loss of line. ‘evaluation,
"Sed (0 construct the Jacobian matrix,
For the three-bus pro latior iven by:
Wer relations are gi
* the three-bus py bem, the bus power relat
V1
an, or ep} aPl Tay
Siar ova
ep2 ep2 2 Av2
ap2 | vw ov3
ap3 SP3 Ps
O31. lev av2 v3 Lavs
The above “dations can be Written as,
l= 1 fay
Where Ulis the. Jacobian trix, Fop the three bus. flow Prot ™, the Voltage
the swin, erator jg SP&Cified as VI .OPU, and is Constant, crefore, AVI = 0
and therefore, the Clements of Jacobian Matrix cay ces to:
42) [apy P2 v2
=| vi av2
&P3 OP3
ae avi 2 | avs
hanges in Vand y3 can be Caleutate iterative Methog ir
arting val for V2 a Lop, i
as; [EXPTNO. 06 I PAGENO. 3/6]
AV2 v2 —1 | AP2
=|. [7a
AV3 V3 AP3
This is the basic equation for the calculation of the Newton-Raphson method.
For the three-bus system, the derivatives for the Jacobian matrix are calculated ast
2
re Y21V1+2¥22 V2+Y23 V3
aP2
= = Y23V2
6v3
P3
== = Y32V3
ev2
GP3
ee Yy31V1+ ¥32 V2+2¥33 V3
v3
Using the V2= V3 hres, the Jacobian matrix and
= 1.0 and the admittance values for the brane
the inverse are obtained as:
10 10 0.2 02
| 10 15 | ot DT! = Re “
ap2=1.0(-5-15 +10) “1 = -hd
Ap3=1.0(-10 - 10 + 20) +13 = 13
(" a e | i ie or . (|
V3. 1.0 0.2 0.3 JL 13 0.87
ove the iteration procedure will give @
in the same way as outlined ab
this much faster than the other approaches:
ution approacl
Proceeding
‘on. The Newton-Raphson sol
solutic 416
Seg
Algorithm: eck ue
except the slac
Step I: Assume a flat voltage profile | +j0bfor all buses excep!
ence criterion
Step 2: Assume a suitable value of ¢ called convergenc
ted as
1e buses are denot
Step 3: Set iteration coun, k=0 and assumed voltage profile ofthe
TEE ad?
Step 4: Set the bus count P
wise go to next step
Step S: Check for stack bus, itis aslack bus goto step 13 otherwise go
ng the equation
S96: Calculate the real and reactive power of bus P using the eq
Filsc,,+A%2,)+Re(A8G, eB.)
= Slr io,+ #8, )eet (Roy et,
Srp; Caealat the change in eal power
‘snerator bus goto next step otherwise go to step 12,
Si? 9Check for reactive power tint
Violation of generator bus for this compare the calculated
ep ere Oe ith specie tints eee Timitis violated go to step 11 otherwise go to next
Sep ese go t0 step 13,
Step 10: Cheek, 41é
Exes
Algorithm:
except the slack bus
SEDI: Assume 9 fat voage profte |+4J0bfor all buses except
nce criterion,
SteP2: Assume a suitable value of elle ‘convergence
Step 3: Set iteration count, kao.
snoted as,
wl assumed voltage profile of the buses are denot
ve
Step Setthebus count P= 1
ise go to next step
SIPS: Check for sack bus, it jen Slack bus go to step 13 otherwise go.
(on F oy) RURt6, eho,
a Elica beet, By)
SP 8\Cheek the generator
SITitis a ponerator bus go to ext
Step 9:Check for reactive Power fi
Taetve power QA span
Hep else gig
EP 10: Check or eactive Power limit Violation o
Pele goto net st
Se otherwise go to step 12 Expt: 6 5/6
‘Step16: Compare AE & ¢ if AE E go to next step
Step17: Determine the elements of jacobian matrix by partially differentiating the load flow equation
and evaluates using K"* iteration values
Step18: Calculate the increment
matrix B=JC
real and reactive prt of voltages Ae! & Ad by solving the
‘Step19: Calculate new bus voltages
VET CPS « oe -ww (Zea)
siege
Therefore
ye
‘Step20: Advance iteration count K=K+1 & goto step $
Step21: Calculate the line flows and stop the program EXPT NO. 06
05 Cirenit Diagram: NA
06 Procedure:
i) open MATLAB,
‘open new M-File
Type the program
iv) Save in current directory
¥) Compile and run the program.
vi) For output see output window / command window
07 Observations: NA
08 Sample calculations: Not Applicable
09 Graph if any: Not Applicable
10 Result: In this way we have analyse load flow study of a given power system network
using MATLAB software
Pa By: R.K.Mankar Peeves] Approved By: H.0.D. WW
\ [EXPTNO.06 PAGENO, 6/6
95 Circuit Dingram: NA
06 Procedure:
i) open MATLAB
ii) open new M-File
iii) Type the program
i) Save in current directory
V) Compile and run the program
Vi) For output see output window / command window
07 Observations: NA
0S Sample calculations: Not Applicable
09 Graph if any: Not Applicable
10 Result: In this way we have analyse load flow study of a given power system network
using MATLAB software
Prepared By: R.K.Mankar Wes —— [Areroved By: HOD, WW. Expr 6
% NEWTON RAPHSON LOAD FLOW METHOD.
Se
DATA INPU
format short g
disp ("TABLE 9.2 PAGE # 337 LINE DATA FOR EXAMPLE 9.2 ')
linedata=
[2 2 0.01008, o.05040, 3.815629, -19.078144, 10.25, 0.05125;
1 3 0.00744, 0.03720, 5.169561, -25.847809, 7.75, 0.03875;
2 4 0.00744, 0.03720, 5.169561, -25.847809, 7.75, 0.03875;
3 4 0.01272, 0.06360, 3.023705, -15.118528, 12.75, 0.06375]
disp (' TABLE 9.3 PAGE # 338 BUS DATA FOR EXAMPLE 9.2‘)
busdata=[1 0, 0, 50, 30.99, 1.00, 0 1;
20, 0, 170, 105.35,1.00, 0 2;
30, 0, 200, 123.94, 1.00, 0 2;
4 318, 0,80, 49.58, 1.02, 0 3)
% Last column shows Bus Type: 1.Slack Bus 2.PQ.Bus 3.PV Bus
yelinedata(:,5)+i*linedata(,6);
totalbuses = max(max(linedata(:,1)),max(linedata(;,2))); % total buses
totalbranches = length(linedata(:,1))
%
ROGRAM STARTS HERE
sszi*linedata(;8); %Y/2; 9%no. of branches
ybus = zeros(totalbuses,totalbuses);
w=0;
u=0;
forn=1:totalbuses _% total no of PV busses.
if busdata(n,2)>0
end
wewt;
end
forn=2:totalbuses —_% total no of PQ busses if busdata(n,2)
end
6 —~ INITIALLIZING YBUS ~~
for b=1:totalbranches
ybusl(linedata(b,1)),(linedata(b,2)))=-¥(b)
end
‘yus((inedata(b,2),{linedata(b,a) =ybusl (lin
. atab,aptinedatat®.2)):
end
for c=1:totalbuses
for d=t:totalbranches
if inedata(d,1)
|| linedata(d,2) == ¢
ybus(c,c) = ybus(c,c) + y(d) + ss(d);
end
end
end
disp(/TABLE 9.3 PAGE # 338 BUS ADMITTANCE MATRIX FOR EXAMPLE 9.2')
yous
yousR=zeros(totalbuses,totalbuses);
YousA=zeros(totalbuses,totalbuses);
2=zeros{totalbuses,8);
busnumber=busdata(,1);
PG=busdatal;,2);
QG=busdatat,3);
Pl=busdata(:4);
Al=busdatat5);
Vebusdatal:.6);
wey;
ANG=busdata(:7};
type = busdata(,8);
5=(2*totalbuses)-w-2; J=zeros(s,s); _ % initializing empty jacobian matrix
MMezeros(s,1);
AN=ANG;
P=(PG-PL)./100; % per unit active power at buses
Q=(QG-QL),/100; % per unit reactive power at buses
tole1;
iter=0;
kksinput(‘Enter the tolerance for iteration ')
%
fectangular to Polar ~~
for b=1:totalbranches
% Real part of ybus
ybusR((linedata(b,1)),(linedata(b,2)))=
ybusRi(linedatalb,2)),linedata(b,1))) =ybusR((linedata(b,1)},(linedatalb,2)));
‘ybusf(b,b}=abs(-ybus(b,b));
96 Angle of ybus
ybusA((linedata(b,1)),(linedata(b,2))}=angle(-y(b));
‘ybusA\(linedata(b,2)),linedata(b,1))) =ybusA((linedata(b, 1), (linedata(b,2)));
ybusA(b,b}=angle(-ybus(b,b));
end
ybusR;
ybush;
while tol > kk
YW = YW + (ybusR(in)* V(n)*V())*cos{(ybusA(in)}+busdata(n,7)-busdata(i,7));
% multiplying admittance & voltage
end
w; bu
PG.
as
Pl
Vebe
wev
ANG=
type =
se(2"
sierra
Forest?
neh
sorrows
fren
sane yt
sprite SOE
ene
nit
ane
weet ra as CoE SSIES
serena end
Peli) = Mabs(viyaays
“eallyous(uiy)syvy, % Compute Calculated P,
SekaPt)-PLy.Peqp,
end
Pe;
Sp=deltapr,
for i=2:totalbuses
MMU-1a} J214(i-1,}-1)= -abs(({ybusr(i,j)* ‘VU)*VG))*sin (ybusA(,j))+busdata(),7)-busdata(i)));
end
ifissj
for n
stotalbuses.
ifn
J146-4,1-1)=121(14-1}4{ abs{((ybusR(i,n)* V(i)*V(n)))*sin{(ybusA(n))+busdata(n, 7}
busdatali,7))));
end
end
end
end
= end
for j=2:totalbuses-w
if busdata(j,8)== 2 Ee
i705
,7)-busdatal
YuI2ey2 + abs{(ybusR(n)* Vn) *cos{(ybusa(in}}ebusdataln
ities]
for n=1:totalbuses
itn
end
end
722(-4,5-1) = abs(V(i)*(2*abs(V())*real(ybus(i,)}+yv22);
end
end
end
end
312;
for i=2:totalbuses.w
Wousdatali.g)== 2
for j=2:totalbuses
its
PAL} absl(ybusr(j)+ PMO Ob st. wsdaaty,
end
ies
forn=ttotalbuses
tarsi
2am
Lita “Abst ((yousRG nye
end
ttt Sey
end ‘
end
end
end
end
2; 322
for
:totalbuses-w
for j-2:totalbuses-w
if busdata(i,8)
&& busdatalj,8)
222(+-4)-1}= -abs(V(G))*abs{(yousR(j)* V(i)})*sin{(ybusA(j)}+busdatal,7}-busdata(7));
end
ywi2=0;
if
322{i-1,j-1)=-(11.1(i-1,i-4)) - (2*(abs(V(i))*2)*imaglybus{i,i)));
end
end
end
end
for k=1:totalbuses-1
for:
stotalbuses-1
SMH 1;
end
end
for k=!
2
for le1:totalbuses-4
ALkstotalbuses-t}=I12(1k);
end
end
for
for I=t:totalbuses-1
end
end forketiu
fortetu
Alstotalbuses-
‘Lkttotalbuses-1)q
end
122 (Lk);
end
format short g
inv) na;
for n=t:totalbuses-1
StP)=(ANG(n+2)sk(ny),
sa=radtodeg(s);
%—
busdatalnea,7)es(),
Yo
end
ANG=busdata(.7);
sa; DISPLAVING OUTPUTS.
‘%real(VACC})
2(d:totalbuses,1)=busdata(,1);
x(a
talbuses,2)=busdatal:,8);
2(1:totalbuses;
)=abs(busdata(:,6));
2(1:totalbuses,4)=radtodeg(busdata(:,7));
2a:totalbuses 5)
Gi
2(1:totalbuses,7)=busdatal
2(1:totalbuses,8)=busdata(:,5);
disp(’ [Bus No.|_ |8usType| [Voltage
Lanelet
IMvar(t)1;
format short g
z
mwa)
\Mvar(6)|
Imi currently being calculated and (k = 1) i
iteration. Thus, we see that the valu
this equation are the most rec
buses (or the estimated volta
that particular bus).
are specified, an ad
voltage magnitude~is to remain c
is_john_grainger_Istpdr file://1:/19-April~-2023/power_system_analysis_john_prainger_tst.
E: No.
Kp 6 92 THe causs-seioeL METHOD 337
ates the numberof the preceding
6r the voltages op-the right-hand side of
ly calculated valuéS for the corresponding
ration has yet been made at
if k is 1 and no
Since Eq. (9.19) applies only at load-buses where real and reactive power
onal step is negeSsary at voltage-controlied buses where
Stant, Before investigating this additional
step, let us loof~at an example of the calculations at a load bus.
Example 9.2. Figure 9.2 shows the one-line diagram of a simple power system.
Generators aire connected at buses (1) and (4) while loads are indicated at all
four buses. Base values for the transmission system are 100 MVA, 230 kV. The line
data of Table 9.2 give per-unit series impedances and line-charging susceptances
for the nominal-m equivalents of the four lines identified by the buses at which
they terminate. The bus data in Table 9.3 list values for P,Q, and V at each bus.
The Q values of load are calculated from the corresponding P values assuming a
power factor of 0.85. The net scheduled valucs, P, en and Q,, sn, are negative at
the load buses 2) and @). Generated Q,; is not specified where voltage
_ Ein 6
Rina: Maple @® FIGURE 92
One-line diagram for Example
9.2 showing the bus names and
numbers.
TABLE 9.2
Line data for Example 9.2¢
Series Z Series ¥ = Shunt ¥
bus t R x G a charging’ -¥/2
Sperunit —perunit per unit perunit — Mvare per unit
o.o10us = 00S040 «3.815629 -19.078144 10.28 oosi2s
0.00744 0.03720 © 5.169561 — 25.847809 15 0.03875
0.00744 0.03720 5.169561 — 25.847809 1 0.03875 i;
0.06360 3.023705 5.128528 * 12.75 0.06375
0.01272
Base JOOMVA, 230 kV,
fAt 230 kV. &
=a
&...
tem_analysis_john_srainger_Istpdt
““filei1/:/19-Aprit-2023/power_system_analysis_john_grainger_Istpd
338 CHAPTERS POWER-FLOW SOLUTIONS
TABLE 9.3 :
Bus data for Example 9.2
Generation Load
Bus P,MW Q,MvarP,MW Q,Mvart —-,perunit Remarks
1 - - 30 3099 1.00 /0° bus | Suing | Ae FRBeu¢ 9
2 0 0 170 10535 1.00/07 Load bus
. (inductive)
3 0 0 200 12394 1.00/07 Load bus
(inductive)
4 318 - 80 4958 1.02/0" Voltage controlled
The Q values of loud are calculated fom the corresponding P values assuming 2 power factor of 0.85.
magnitude is constant. In the voltage column the values for the load buses are
flat-start estimates. The slack bus voltage magnitude |V;| and angle 5,, and also
magnitude |V;| at bus G), are to be kept constant at the values listed. A
power-flow study is to be made by the Gauss-Seidel method. Assuming that the
iterative calculations start at bus @), find the value of Vz for the first iteration.
Solution. In order to approach the accuracy of a digital computer, we perform the
computations indicated below to six decimal places. From the line data given in
Table 9.2 we construct the system Y,,, of Table 9.4. For example, associated with
bus @ of Fig. 9.2 are the nonzero off-diagonal elements Yz, and Y,4, which are
equal to the negative of the respective line admittances
Yay = — (3.815629 — j19.078144);
Yq = - (5.169561 — j25.847809)
Since Yzy is the sum of all the admittances connected to bus @), including the
‘TABLE 94 .
Bus admittance matrix for Example 9.24
. Bus
no. ® ® ® ®
8.985190 3.815629 169561 0
® —j44.835953 +j19.078144
3.815629 8.985190 5.169561
® + /19.078144 4.835953 2 +425.847809
- 5.169561 8.193267, ~ 3.023705
® — spssireno a 40863838 + 15.118528
5.169561 = 3.023705 8.193267
@ o 425847809 «——_++/15.118528 j40.863838
‘tPer-unit values rounded to six decimal places %
FORM NO. SSGMCE-FRM-32 B
e
‘SHRI SANT GAANAN MAHARAJ COLLEGE OF
LABORATORY MANUAL
ENGG. SHEGAON
PRACTICAL EXPERIMENT INSTRUCTION SHEET
EXPERIMENT TITLE: To Plot Power-Angle eurve of Synchronous
Machine for stability analysis
EXPERIMENT NO. : SGM,
/WI/ ELE/07/8EPO6/ ~ 07 | ISSUENO, 100 | ISSUE DATE: 12.01.2035
REV. DATE: 00
REV. NO. 00 DEPT. : ELECTRICAL ENGINEERING
AABORATORY : Computer Methods in Power Systm SEMESTER: VIKI | PAGE: 1 of.” 3
Analysis Lab
Date:
01. Aim:
To Plot Power-Angle curve of Synchronous Machine for stability analysis
02. Scope:
By using this exper
iment students will be able to Plot Power-Angle curve of
Synchronous Machine for stabi
lity analysis using MATLAB software
03. Apparatus: PC with MATLAB Software
04. Theory:
Transient stability studies are related to the effects of transmission line faults on generator
synchronism. During the fault the electrical power from the nearby generators is reduced
and the power from remote generators remains relatively unchanged. The resultant
Aifferences in acceleration produce speed differences over the time interval of the fault and
itis important to clear the fault as quickly as possible, The fault clearing removes one or
‘ore transmission elements and weakens the system. The change in the transmission
System produces change in the generator rotor angles. Ifthe changes are such that the
accelerated machines pick up additional load, they slow down and a new equilibrium
Position is reached. The loss of synchronism will be evident within one second of the initial
disturbance,
.
Faults on heavily loaded lines are more likely to cause instability than the faults on lightly
loaded lines because they tend to produce more acceleration during the fault. Three phase
faults produce greater accelerations than those involving one or two phase conductors,
Faults not cl
generators.
‘eared by primary faults produce more angle deviations in the nearby
Also, the backup fault clearing is performed after a time delay and hence
produces severe oscillations. The loss of a major load or a major generating station
produces significant disturbance in the system, Steady State Stability
ine impedance Z. as shown in
Consider a generator connected to a remote souree Paraee de aa
Figure . The phasor diagram of the one machine syst
— 1
2
Zz
et
1
1
Figure : one machine system
P= Reale1i ener ELE,
z
oe Foe sis —
Wnete2=(R +390. Fore pur reactance, 23 and a0,
pa Bsa |
‘Thisis the generator electrical power, ‘here 8is the rotor angle. This isasine
function as shown in Figure.The Perating point occurs where the electrical Power output
of the generator Pe is balanced withthe mechanical power (Pm), A change inthe angle
Say from the operating point will es, ina power imbalance, which acts to accelerate ot
decelerate the rotor. The Prax point ig ‘he maximum power Possible from the generator
‘The (0 through 90) degree is the Steady state Perating range in a stable mode. The (90
through 180) degree is the unstable Operating region o oe .
ye |Z Pe Figure : Power-Angle curve
The steady state operating limit is given by relatioi
(4)
os
‘Prax is the steady state stability imi,
05 Circuit Diagram: NA
06 Procedure:
i) open MATLAB
open new M-File
iii) Type the program
iv) save in current directory
v) compile and run the program
vi) For output see output window / command window / plot curve
07 Observations: NA
08 Sample calculations: Not Applicable
09 Graph if any: Not Applicable
10 Result: In this way we have Plot Power-Angle curve of Synchronous Machine for
stability analysis using MATLAB software
Prepared By: R.K.Mankar ee __ Approved By: H.0.D. Nn Power = Angie Cusve
Exptno ‘Bicode/eutesn- tas
% To Plot power angie o, ;
ue) thine for
stability analysis ‘ve of synchronous Mac!
cle
clear all
delta =0:1:180;
Pm=ones (size (delta) ) *0.8;
=0.65;
(V1*V2/X*sind (delta) ) ;
plot(delta,Pm,'r') ;
hold on;
plot (delta,Pe,'b');
axis([0 180 0 21);
syms del;
xdel=solve (V1*V2/X*sin (del) -Pm==0 del)
legend('P Me chanical!,"PElectrical','Loc
ation', 'best') 7 7 ~~
xlabel('Angle ,degree') ;
ylabel('Power , pu');
% Reference: Computer aided power System analysis by
Ramasamy Natarajan,MARCEL DEKKER, INC. Page no 89 to 92) \
FORM NO. SSGMCE-FRM-32 B
‘SHRI SANT GAJANAN MAHARAJ COLLEGE OF ENGG.
SHEGAON
LABORATORY
MANUAL
PRACTICAL EXPERIMENT INSTRUCTION SHEET
EXPERIMENT TITLE : To Plot Swing curve of Synchronous Machine for
=a Stability analysis
EXPERIMENT NO, : SGM/WI/ ELE/07/SEP0G/08 ISSUE NO. : 00 | ISSUE DATE: 12.01.2023
REV. DATE : 00 REV.NO.: 00 DEPTT. : ELECTRICAL ENGINEERING
LABORATORY : Computer Methods in Power ‘SEMESTER: VIIT PAGE: 1 of 2
System Analysis
Date:
01. Aim:
To Plot Swing-curve of Synchronous Machine for stability analysis
02. Scope: By using this experiment students will be able to Plot Swing curve of
Synchronous Machine for stability analysis using MATLAB software
03. Apparatus: PC with MATLAB Software
* 04, Theory: Following a system disturbance or load change on a power system, a
generating unit tends to oscillate around its operating point until it reaches a steady state,
For a synchronous machine with constant field excitation, an equation for the dynamic
motion is obtained by relating the angular acceleration of the rotor to the rotor torque.
‘This relation is: (Inertia) (Angular acceleration) + Damping torque + (Te - Tm) = 0
For small deviations the characteristic equation can be written as :
2
288 6, DE cigs oo
aad
This is called the swing equation.
Figure : Swing Curve TimeGey
where: AB = Rotor angle deviation from the steady state, radians
Inertia constant of the generator unit, kW-sec/kVA
Damping coefficient representing friction EXPT NO. 08:
PAGE NO. 2/2
‘© = Synchronous frequency, 377 radians/s for 60 HZ system
KI = Synchronizing coefficient, P.U. power/Radian
Rpm = Synchronous machine speed in revolutions/minute
J= Moment of inertia, Ib-fT
H = 2231) (RPM 207%
Base MVA
The term Kl is called the synchronizing power that acts to accelerate or decelerate the
inertia towards the synchronous operating point. The synchronizing coefficient Kl is the
slope of the transient power angle curve,
fe2|
Coss
Where,El = Internal voltage behind transient reactance, per unit. E2 - Bus voltage, per unit
X= (Xd! + Xe) =Series reactance between the terminal voltage and the infinite bus,per unit
Xd = Generator transient reactance, per unit
8 = Angle between El and E2
The swing equation governs the power system dynamic response with a frequency given
by:
05 Cireuit Diagram: NA
06 Procedure:
i) openMATLAB
ii) open new M-File
iii) Type the program
iv) save in current directory
-¥) Compile and run the program
vi) For output see output window / command window
07 Observations: NA
08 Sample calculations: Not Applicable
09 Graph if any: Not Applicable
10 Result: In this way we have Plot Swing-curve of Synchronous Machine for stability
analysis using MATLAB software
‘Approved By: H.0.D. 7
Prepared By: R.K.Mankar Wee — 7d
\
Swing Cur cMpsa Lab
fo Expt: To Plot Swing Curve by using Euler Method
clear
ele
Pm-input(Prefault Power transfer Pm =
dettad=input(initial power angle deltad in degrees =;
pmax2input(’Max.Power transfer during fault Pmax2=");
Pmax3=input(’ Max.Power transfer after clearing fault Pmax
fecinput(’Time to clear the fault te=')s
tmax=0.33 dt=0,055 £505 H=2.525 w0= 2*pi"6
a=(pi* fH; £0.05 #1 t€)=ts WOO d=deltad;
while t