Power System NetworkMotrites

1.T INTRODUCTION i/Graphs are very useful in the slveral fields like engineering, physical and social, etc. Manyapplications of several electrical components such as machines and power system componentcharacteristics are representing in a simple way in graph form and for analysis of electrical circuitsalso it plays very important role. For small circuit analysis based on nodal and meshed equationmethods by using KirchhofT's law and Ohm's law are sufficient. But for complex networks thesemethods are difficult and take more time for solving the equations. In this chapter brief discussions ongraph theory and the applications of graph theory in power system networks are going to be presentedin detail.

t.2 DEFINITIOI{SGraph TheoryWhen the e-l-e$g-4lllike resistqrs, inductors,-capacitors and voltage sources [shown in Fig. 1.1(a)] ina nerwork are ieffied by lines. This type of network is called the 'graph' is shown in Fig. 1.1(D) andthe line segments are joined by means of nodes. The rank of the graph is 'n- 1 ' , where r is number ofnodes of the graph. The graph thus obta\ned is also known as undirected graph. While drawing graphseries elements can be replaced by a single edge in the graph.

QFig. f.f(a) Netwwh Fig. f .f (b) Equivnlcnt grqh

Elffiical Pouer Silstem Analgsls(hiented GiaphI srapn whose branches are oriented is called oriented graph and is shown in Fig. 1.2. It is also calledas directed graph.

Fig. '1.2 Grapb

NodeTlre meeting point ol'the two or more elements in the graph is called as .node, and denoted by ,n, .Itis also called as vertex.Element (Edge)An element is a line segment representing one hetwork element or a combination of

'elements

connected between two nodes. The number of elements is represente dby ,e, .Ex. : From Fig. 1.2,Number of nodes n = 2,Nunrber of elements e : 3.Degree: The number of elements connected to a node is called degree.

Sub-graphIt is a subset of branches and nodes of a graph. The sub-graph is said to be proper if it consists ofstrictly less than all the branches and nodes oi the graph.TreeA tree is a connected sub-graph of a connected-graph containing all the nodes of the graph butcontaining no closed paths or loops. If the graptf consistint 'n' number of node, the number ofbranches ofa tree is (z

- l). It can also be referred in this con-text as spanning tree sinci it spansoverall the verrices of the graph.

i A tree has the following properties :\ -' ') I Tree is a sub-graph containing ail the vertices of the originar graph.i 2. Tree is a connecred graph.

1 3. Tree does nor conraining anyloop. ,\ 4. Every connected graph has at least one tree.\ l. !*h rree has (n - 1) branches and rank is also (z _ 1).\9: ,n-. elemenrs in the spanning tree are called branches or hyigs.\grefore the number of tree branches ,6, : 1n _ l).

0Odentetl.

Potuer Sgstem Nefitork Mafrices 3

It is a sub-graph formed with all the links of the graph or, in other words co-tree is a complement ofa tree. The number orunk@ b : e

- (n

- l),.

Cotree

where e : number of elementsD : number oftree branchesn : number ofnodes

TwigsThese are the branches of a tree.

y = -rrr b -A

LinksThe elements that are not on a tree or elements of the co-tree are called links or chords. The numberoflinksisrepresentedbyLNumberoflinks,tmflwhere e : numbei of elements

n : number ofnodes

Gut SetIt is a rninimal set of elements of a connected graph, such that the removal of these elements fromgraph isolated at least one vertex.

Tle.setA fundamental tie.set of a graph with respect to a tree is arith other twigs.- since for eaqh link of the tree their coriesponaing ro i , thentmber of fundamental loops is equal to the number of links in the given tree. Number of fundamentalloqsorlinks, l:e-n+l/rfiere e : number of elements of graph,

n = number of nodes of a graph.

L:Il.2-3.4-5.6-7_

t3.1 Element llode lncldence MatrixE ircidence of elements to nodes in a connected graph is given-i is derioted Ly A'.

TYPES OF MATRI.CESincidence matrix [A']

Bus incidence matrix [A]Branch path incidence matrix [K]Basic cut-set incidence matrix [B]Augmented cut-set matrix IB') ,.Basic loop incidence matrix [C] /Augmented loop incidence matrix [C'],

forrned one link associated

by element node incidence matrix

I 4 Electrical Power System AnalyslsTo determine the elements of A' matrix for the Fig. 1.3 as follows

ai : l,If the ift element is incidence and oriented l$X$om theT'fr node.a,j :

- l, If the ith element is incidence and oriented towards the je node.

aij : 0,If the /h element is not incident to theTm node.The size of the matrix is (e x n)

z = number of node in the graph

era'@ O @oA':

00-1 0_0_ _1-1 0I -tI

-tIt is observed from the elements of the matrix that

3

Lo, :0, for P = 1, 2, ...,6.l=o

It can be inferred that the column of A' is linearly independent and the rank of A' is-Iess than 'n'.

,..3.2 Bus lncidence MatrlxAny node ofa connected graph can be selected as the reference node. Then the uariable ofthe othernodes, referred to as buses, can be measured with respect to the assigned reference(The matrix obtainedfrom the element-incidence matrix (A) by eliminating the column corresponding tb the reference no@is the bus incidence matrix 'A'r7The order of the matrix is [e x (r - 1)] and fte rank is (n - 1).

Selecting node (0) as reference node in the graph shown in Fig. 1.3.

l [l -l

zl r o:lr o-t---+ lo rr lo o610 0

,[-r 0 ol2l o

-l ol;i i -?-+l =(*)5l o r

-rl6L o r

-r.l

00-l-1 0l0 -rl

elbus @ O O

A_

where 4.6 : matrix formed by twigs.

l-- rIAr:10Io

Al : matrix formed by links.

I r -1

o,: lo I[or1.3.3 Branch Path lncldence MatrixThe branch path incidence matrix shows the incidence of branches to paths

jl

oriented from a bus to the refere tode.We can determine the elements of this matrix as follows:

K, : l, if the re branch is in the path from theTth bus to reference and is oriented in thesame direction.

Ku = - 1, if the ifr branch is in the path from theT'fr bus to reference and is oriented in' the opposite direction.

K, : 0, if the ifr branch is not in the path from the jth bus to reference with node '0' asreference.

The branch path incidence matrix associated with the tree shown in Fig. 1.4 is

branch/path C @ Ol[-1 o olr=21 o -r ol

sL o o -r.l

The non-singular square matrix with rank (n - 1)I : 3 f4,5,61b:3 Il,2,3l

Ite branch path incidence matrix and the sub TatrixArrelatethanches to pattrp and branch6b to bu$es, respectively. Sincek is one to one correspondence between paths.and buses.

A6I{ : UKT : Aa-l .

ft U-is an ldentities or Unit matrix.-

Baslc Cutet lncidenceis a minimal set of branches of the removal of which cuts the graph into two parts. It

being in one of the two parts is shown in Fig. 1.5-

in a tree, where a Path is

Fig. 1.4

groups,

6 Electrical Power Sustem AnnltrctcWe can determine the elements of this matrix

". rorrornF

,

B,J : 1, when the ifr branch is in the cut_setf, and orientation coincide.- -

l, when the i6 branch is in the cut-set,/, and orieirtation do not coincide.: 0, when the ifr branch is not/the cut_setl.tt-Set inciden.e rnqtriv nf rli---^:^-^ ,- .- ,-. . 6The basic cut-ser incidence matrix of dimensions 1a x ai forthe graph shown in Fig. I.5 is

U6 and B1, where the rows of U_6_correspond to

elb a b crf 10ol

'::rtl+;ls'lo

-1 tlu L o -1 1l

The matrix 'B' can be partitioned into sub matrixbranches and the-rows ofb, to links.

The partitioned matrix is

t;:]

cut-set

bc001I 0l0 1l

II 0l-r 1l-1 rl

Basic cut-sets

B, : matrix formed by links

IJl:l( 0 -1 1)

The identity matrix u6 shows the one to one correspondence of the branches and basic cut-sets.The sub matrix B, can be obtained from the bus incidence matrix 'A,. The incidence of links r

Basic

a

1[ I2lo3l o

':r[-I

oL Pwhere U, : unit matrix formed by twigs

(r o o)Io r ol[o ot)

,

lli

il'

Fig. 1.5

buses is given by the sub matrix Al and the incidence of elements to buses is given by the sub ma'46" since there is a one to one correspondence of the elements and basic cup-sets, B1 46 givelincidence of links to buses,

Pouer S,stem Netuork Matrices ToE., , BtAtr:At

Bl = Al Ar-lBr = Al Kr lsince Ar-1 - 11r,

135 Augmented Cutet lncidence MatrixryF:,",:.H[:f [.*:::,:j,T,,:T]::g,:: llayrres in the basic cur-ser incidence matrix. rhe*:j,iJi;:Xll;t: ,,a. .q,.ii" ir,eri#i.' ;ittHJ,il'ff l*:::::ffi';:;',1;f,*:l;H*:",li::::::..r:::l::*-,u,.o,,sinini'onrv.od.rink:,fi;:,.ffiJ$Htrrr

imaginary

[::]I:i,H,,*ff:ff9:S],*,,::::,I1" ""_;iil":t,:,,1",ff i::ji[:li::fi,il,1:#i:;1X,,1:':::.:':'i:::_1'_1;"ilr;ffi;**.Hd}l;l:tT-:"1H:i:ffJ'J:i?-,ffJffxi.T;frb '8" is a square marix of size (e x e). The B, ,ut.i* ,".ir*. -r.6 is given berow.-,,Basic cut-set fie sets

eleI2

a bc dr 0 0t0lo r o oi0 o t,oio o o. rl

-------------t0 -l 1 I

-l I 0 00

-l I 0

e/l

le f00000000100010001

o6

0'0olol

34r-567

n re Basic Loop lncidence MatrixFig. 1.6

HHfili:g.""t rnatrix gives the incidence of eremenrs ro basic loops of a connecred graph.!b elements of this matrix can be formed as follows

',,

: '.

-,_n.,

the erement 'j', is in basic loop J, and their directions coincide.: -

l, when the erement 'i' is in basic loop ,j, andtheir directions do not coincide.: 0, when the element is not in theTth basic loop.h rize of matrix .C, is (e x Ly. The C matrix for Fig. 1.7 qi given below()^(I2

3

4

5

67

-1-1-T

00

e

0I

fI

-l00

I0I

-t0 =f

I

cbU,0

1

0

00I 1.7

Basic cut-set I tie sets

= tJb i o

----B;----l--u,-

Fig.

8 Electrical Pouer Sgstem Analysrswhere C, : the matrix formed by twigs

U, : unit matrix formed by links. (t 0 0\rlU7:10 I 0l

[o o 1)The matrix 'C' can partition into sub matrix .cr, and .u,,, where the rows of ,c6, correspond tobranches and the rows of ,U/, to links.

L.9.7 Augmented Loop lncidence Matrix*:"1|:l:'-:.rj::f^Li::T:",,:d g.rph is equat to rh number of links. rhe total number of loops;;;"h*,;;rffiffiffi;*ffi:An nnpn lnnn then rtpfi-on ^d +L^ -^.L r^i.---^,- , '-*l:f".: j:T::T:*Iro,u:rylllbgfy.eenadj;;t;ile;;;;ffiy;ffi ;':i,ilffi;ffi;SilT s:^"p-h F,r"l ir.Fig:.I.8, thd orienration or an open ffi;&;;;;;;;; i;. il.;ffi.'#i;branch. It is denoted by C'.

This matrix C'for the graph shown in Fig. 1.g given belowOpen loops basic loops

eleI2

C': 34

5

67

a b c d,, fg

----------L0 0 0 0i

Fig. 1.8

(0 1 0\",:l-r i ll[-, o oJ

1000010000100001

00000000

0101

-l I-1 0 -1-1 0 0100010001

It is a square matrix of size (e x e) and is non-singular. The use of these augmented incmatrix will be seen when non-singular method of obtaining network matrices are discussed.7..4 CONSTRUCTTON OF NETWORK MATRICESPower system network consists of number of components, (generators, transformeter, transmislines, etc.)which are interconnected normally. Either ,r*iru.t ..r or design office suppliesdetails of individual components.

The information giving the characteristics of individual components are represented by .prinetwork'.[' Primitive network is a set of unconnected elements represented either impedance form[admittance form. 'v'vu Lrwr'E,]r rtrPrclI

D_lu, i .,- l' i....u;

Tbre [I] ,r, : Vector of impressed bus currents matrix =(n-l)xl

Pouser Srfstem Netusotk Matices 9To devglop the net{ork matrices the effect of interconnection between the elements are included

by suitable transformation.

1.5 NETWORK MATRICESThe mathematical model of the netwqrks can be developed in three different forms and these are

1. Bus frame of reference -rt'

2. Branch frame of reference /3. Loop frame of reference

1.5.1 &s Frame of ReferenceThe bus current and voltage vector equations of a network can be written as /

[I] r* : [Y] ru, [V] ",.

(Admittance form) / /r...(l.L)(n-l)xl (r-l)x(n-l) (z-l)xl ,/[V] ,,, : [Z] ,u, [I]

"* (Impedance form)

,r/ ...(t.Z)(n-l)xl (,r-l)x(,r-1) (n-1)xlIll2

ln -t

[V] ,* : Vector of bus voltagesmeasured with respect to reference bus(n-l)xl

v,

Y2

ur -,

[Y] ,,, : Bus admittance matrix z/tn-l)x(n-l)

[ZJ ,* : Bus impedance matrixln-t)x0,-l)

./Grnri&r 3-bus system as shown in Fig. !.9, thelZrus and YBus matrices are given below

Fig. 1.9 3-bas poror system networh

.-

,,/1O Etectricol Potaer Sgstem Analgsis

(Y,, Y,, Yrr)Yru.: IY, Y, Y, I["r, Y, vu)('" z"

'"\zsu": lZl 22 zr, Il.Z'' z" z")The impedance or admittance element on principal diagonal elements are called driving point

impedance or admittance of bus and the off diagonal elements are called transfer impedance oradmittance of the bus.

L.5.2 Branch Frame of ReferenceThe equations are written in branch currents and branch voltages

[I] n, : [Y] ".

Fl ,.(n-r)xr (,-t)x(,-t) (n-r)xr[v] ,. = lzl B, lrl ,,

(Admittance form)

(Impedance form)

...(1.3)

...(1.4)(,r-r)xr (,-r)*(,,-r) (n-r)xrwhere [t] ,, : Column vector of branch current{n-l)xl

[V] ,, : Column vector of branch voltage(r-l)xllYl

',(n-l)x(rr-l)Lz) u,(n-l)x(n-1)

: Branch admittance {$"}i.= Branch impedance matrix

1.5.3 Loop Frame of ReferenceThe following equations are written for loop currents and loop voltages

[I] '*,c-(n-l)xl

[v] -",e-(n-t)xt

where [I] *",e-(n-l)xl

[v] -",e-(n-l)xl

[Y] *",c-(n-l)xe-ln-l)

lz) ^"ne-(n-l)re-(n-l)

: [Y] r_ooo [V] *,0 (Admittance form) ...(1.5)e-(n- l)x e-(n-t\ e- (a

-l)x I: lZ) ,,,oo tI]** (Impedance form) ...(1.6)

" -

(, -

r) x e-(r -

r) e-(a-1)xl: Vector of loop current,

: Vector of loop voltage

: Loop admittance matrix

: Loop impedance matrix

Potaer System Network Mafiires 11

1.6 CONSTRUCTION OF PRIMITIYE NETWORK ETEMENTThere are two types of representation for primitive nefworks and they are

1. Impedance form2. Admittance form

1.6.1 lmpedance FormConsider the network having two nodes 'a' and'D' is shown in Fig. 1.10, the equations of primitivenetwork in impedance form can be written as

Fig. I.Iovn6:

o -

6

en* eny-Za;ioy: 6o- b * enu: Zo6in6

vab+e(tb:Zabinb,qs:lz)i

rrhre vou = voltage across element 'a - b'

eou: voltage source in series with element'a -

b'la6 : current through the element 'a

- b'

Zolr: impedance matrix of element 'a - b' .

L6.2 Admittance FormCmsider tire networt shown in Fig. 1.11, the equations of primitive network in admittance form canbe written as

...(1.7)

Fig. I.lliot * joo: labvab

t+j:[Y]vrrbre jn,

- current source between nodes a - D

yr6 : admittance matrix of branch a - D.

...>

...(1.t)

lF

12 Electrical Power Sgstem Analysis7..7 NETWORK MATRIX FORT|ATIONThe admittance matrix Y and impedance matrix Z canbe determined by using the following methods.

1. Based on the incidence matrices(a) Singular transformation method(D) Non-singular transformation method

2. Based on the network analysis equationsBy direct inspection method.

1.8 SINGULAR TRANSFORMATION METHODBy using singular transformation, we can derive bus, branch and loop impedance matrix as well asadmittance matrix for the primitive netwok.1.8.1 Determination of Bus lmpedance and Bus Admlttance MatrlcesThe bus impedance matrix Zsu, and bus admittance matrix Yso. coo be determined by using the busincidence matrix 'A' to related variable parameters of the primitive network quantities of the inter-connected network.

From the primitive network equation (1.8)i+i:$lv

Pre-multiplying the both sides with ArAri + At,/: e'[Y] v " '(1'10)

'*- According to Kirchhoff's current law the algebraic sum of currents meets at any node is equal tozeto.

i.e., sum of currents meeting at a node : Ar i : 0 ...(1.11)Similarly A'./ : sum of current sources of element incidence at a node . It is a column vector

AT,/ : Is*Substituting the equations (1.11) and (1.12) in equation (1.10), we get

IBu, : At [Y] '

Power into the network : [I*sur] t V"u,= the surn of powers in the primitive network, i.e.,1i"1r v

Therefore, [. srr]T VBu, : U-]t , ]Taking conjugate transpose of equation (L.12), it is modified as

[A']*r U*lt : [I*rr.]'But A is a real matrix so A* : A,From matrix property [Ar]r : AApplying these two conditions in equation (1.15), we get

[I.'u']' : A[i.]: $Substituting the eqgation (1.16) in equation (1.14) and simplify

;{ vBus : vSubstituting the'V'from equation (1.17) in equation (1.13)

...(1.e)

.. .(1.12)

...(1.13)

...(1.14)

...(1.ls)

...(1.16)

...(1.17)

Power System NetuorkMafrfces 13

IBu, : e' Dl A vso.Equating the equations (1.1) and (1.18), we get

[Ysur] : At [y] eAnd Zsu, can be determined by

[Zsu,] : [YsuJ-l : {Ar D] A}-1

1.8.2 Determlnation of Branch lmpedance andMatrlces

The branch impedance matrix 'Zrr' and branch admittance matrix 'Ysr' are to be determined byusing the branch incidence matrix 'B' to related variable parameters of the primitive networkquantities of the inter-connected network.

From the primitive network equation (1.8)/rl admittance form is

.. .(1.18)

.. .(1.1e)

" ,...(1.20)

Branch Admittance

i+i:[y]VPre-multiplying the both sides with 'Br'

Bri+B'"/:BtDlvAccording to Kirchhoff's current law (the algebraic sum of currents at any node is equal to zero)

Br I : sum of currents passing through the elements that are connected to basic

BTiBTJ

cut-set:Q: sum ofcurrent sources ofelement incidence to the basic cut-set and

represents the total source current in parallel with a branch. It is acolumn vector

BTJ : IS,Substituting these two conditions in equation (1.22), then

Isr:BrDlvPower into the network is given by (U-srlT Vsr) and this is equal to the sum of powers in the

primitive network, i.r., U"l' v. Since power is invariant.Power in the primitive network : powr in the interconnected network

.. .(r.21)

...(r.23)

...(1.24)

.. .(r.2s)

...(r.26)

(... BrDlv : e}) ...(1.27)

...(1.2E)

...(1.29)

[I-sJr VB. : [l-]''Taking conjugate transpose of equation (1.25) it is modified as

[B'].r U*]t = [I*r,]'But 'B' is a real matrix so, B* : B,From matrix property [Br]r : BApplying these two conditions in equation (1.27), then

[I*nr]r : B U*lrFrom the equations (1.26) and (1.28)

BVsr:vFinally substituting 'v : B Vsr' from equation (I.29), in eqn. (1.25)

Is, : Br [y] B Vs.The branch frame reference of admittance form can be written as

Ig, : [Yr.] VrtComparing the equations (1.30) and (1.31)

lYe.l : g'Dl nAfter calculating l[Yrr]' next from this expression we can calculate ,lzsrl,

...(1.30)

...(1.31)

.. .(1.32)

...(1.33)i.e., tZs,l : [ysJ-t : {Br D] B}-t1.8.3 Determinatioq of Loop lmpedance and Loop Admlttance MatrleesThe loop impedance matrix^'-Zruor' and loop admittance matrix 'Yloop' are to be determined by usingthe bus incidence matrix 'C' to relate variable parameters of *re p-iirlitive network quantities of thcinter-connected network.

From the primitive network equation (1.7) in impedance form isv+e:lzli

Pre-multiplying the both sides with 'Cr'Crv+ Cre: cr[z]iw e: \- tzl r ...(1.

According to Kirchhoff's voltage law (the algebraic sum of voltage in a closed loop is equaequalzero)

Cr v : sum of voltages in a closed loop : 6cr e : sum of voltage sources around each basic loop. It is a column vector

v,-oop : [Cr] eSubstituting these two values in equation (1.35) then

Vluop : cr lzl tFrom power invariance condition for the loop and the primitive networks is given by ,[Ii*p]r

VLo,,p' and this is equal to the power in the primitive network, i.e.,,i*r e, .Power in the primitive network = powr in the inter-connected network

ttl*lt vl-p : [i"lr eIn equation (1.38), substituting Vloop : lClr e

lll,,orl' [Clr e: [i*]r , . 'Eliminating 'e' and taking transpose of above equation, it is modified as

i : [C*]r [Ir_oop]But 'C' is a real matrix, C* : C,

i: 1C1a1l*or.;Fromrthe equations (1.35), (t.37) and (1.41)

vl-p : tCrl tel [C] Ir.*pThe loop frame reference of impedance form is

Vto,p : [Zrrop) Ir-oopComparing the equarions (1.42) and (1.43)

Power Sustem Netusork Matices .15

1.9 NON-SINGUUTR TRAT{SFORMATIOil METHODBy using non-singular transformation we can determine branch and loop impedance matrix andadmittance matrix for the primitive network. But in this section, we discuss about the branchimpedance and admittance matrices only.1,9.1 Determination of Branch Admlttance and lmpedance MatricesConsider the augmented incidence matrix, the augmented network is obtained by connecting afictitious branch in series with each link of the original network.

Fictitious node

Fig. l.12The admittance of each fictitious branch cannot allow altering inter-connected network, its

admittance is set to zero. The current source of fictitious branch carries the same current as thecurrent source ofthe link so that the voltage across the fictitious branch is zero. A tie cut-set, can betreated as a fictitious branch in series with cut-set containing a link.

From the branch frame reference form of performance equation

IZuopl = Cr [z] CAfter calculating 'fZrooo]' next from this expression we can calculate '[YLoop]'.This is given by

[Yr_oop] : [Zr_oop]-l : {Cr tzl C}-1

lln,l : [YnJ [VsJThe performance of primitive network is given by the equation (1.7) as

r+i;tJlVPre-Multiply the above equation on both sides with [B']r, we get

lBlr, + tBlri : [B']r Dl vWe know, n,:fU' olLB, urlSubstituting this in equation (1.47), we get

[UD] iD : i6,U1it: it

...(1.44)

...(1.4s)

...(1.46)

...(r.47')

...(1.4r)

...(1.4e)IrL

But

['; ]1lt?1.[? l:][f] : [? fl] ,, ,,u, + t_B.l ,,.l + [u,i, + BI i,-l [u, sl.l D] . tv,"o+u,il,'l'L o+u,7j I :Lor, .l

16 Electrical Pouser Systern AnalyslsIUalit = ju,Urit: jr

Substituting these values in equation (1.49)

[i, + nf i,1 ,l j,+ e,'r,l _ [u, Bfl ..., .,,,I lrlL ,, I L i,' "' ): Lo u,_] DI tvtwlrere ib + Bf i, : [Br] i = 0

ir, + Bf ,it: [Br1i = Ir,Substituting equations [1.51 (a), (D)l in equation (1.50)

[;].[?] : [? ";] ,,,,[,T;] : [? fl] ,,

'",I ln, I =

1,,*r;l:'u'where [,,,] : [? 'ul] ,,ru,

i, + j, is the total source currents of a fictitious branch and its associated link.

...[1.s1 (a)I...t1.s1 (D)l

...(1.52)

Is, : [Br] DI tv1and voltage across fictitious branch is zero. So the voltage vector of the augmented network is Vs.

And the voltage across original network is

V: [B] [Vsr] : IB'1 [VB,]Substituting equation (1.58) in equation (1.56) then

Ie. : [8,]r DI tBl [VB,]The pertbrmance equation of the augmented network is

comparing equarions,, .:;, ;!",T) (v* )

[Yr,J : tB]r t vl tBrl'fhe equation (1.61) can also write as follows :

["o Y,l : lu, BII [y. yal [u, o lLv. yol I o u,l [r" y,.] Ls, u,j

D1 : fy' lu') :Primitive admittance matrix- lfu, lu )

Itu,l : Primitive admittance matrix formed by branchesDtl : Primitive admittance matrix formed by links

Pou:er System Nefiuork Matrices 17

Iyrl : D,r]' : mutual admittance matrix formed by branches and links[Yo Yrl I yuo *Bl y,u ]r, + BI y,,l [U, 0 I1""

",.1 : I ,* trt .llt, u,l

[Yo v"l -

[[rr, * BT y,u + lnt Bt + Bf n, B,] [r, * gl y,,]lLv. Yrl L yn+ruB lu l

Y e,: !ur, + Bf yu * lutBt + B/ (y,,) tB,IYs: lut + BrrY,Yc=lt*)z'BYo=)a

But from singular transformation method,

, [Ys,1 : [Brl tyl tB]: [u, ,r][;; i;][::]= ltoo+ BI vra )ar + BI ,rt []]= !u, *Bl y* + lat Br + Bf y,, B,

From equations (1.63) and (1.67)

rhe branch impedance L]i";,I.1,;Lined bylZsJ: [YrJ-' : [Ye]-l

...(1.63)'.

..(1.64)...(1.6s)...(1.66)

...(1.67)

. ..(1.68)

...(1.6e)

LlO DTRECT INSPECTION METHoD FOR DETERMIilATION OF 'YBUs',By direct inspection of the network, Yru. can be obtained for any network, if there is no mutualimpedances between elements.

Apply the Kirchhoff's current law at every bus for developing the bus admittance matrix. In this.ay, iy.t"matic nodal equations are developed for every node except for the reference bus, which isoormally taken as ground bus. Let us write the nodal equations using Kirchhoff s current law for theoetwork shown in Fig. 1.13.

Fig. l.13

18 Electrica| power Analysi.s

.

Here 'I,' and 'Ir' are the extemal current sources at the bus '1 and 3'. In the nodal formulation,the voltage sources with the series impedance, which is usuaily the case in the power system netwoare replaced by the equivalent current sources with shunt irnpeoance by the following method.The two sources are equivalent if(l) Ee : l,Z,(ii) Zo:7r

Fig. l.l4 ld.cnl rtobagc sowrcc Fig. l.l5 ldeal cuwent sourceNow consider Fig. 1-13 and applying Kirchhoff's current law at the buses 1,2 and3. Weobtain the fbllowing nodal equations

Ir : Vr )r + (Vr -Y) yn0 : V, lz + (Yz- V3))zr + (Vz

-yt)ynIs : Vr): + (Vl

-yz)yzsylttt J.,' vr and v, are the voltages of bus 1, 2 and 3, respectively with respect to the reference'0' and these are also known as bus voltages.

The admittances)ii : lii. i'e.,!tz: )rr since network elements are linear and bilateral.Rearranging equation (1.70) and separating co-efficient of bus vortage variabres (vr, v, and

Let

Ir : ()r -ln)yr_ joyz

O : -lzr Vr * ()rz * lz * lzt)yz_lztyt

lt: -lzzyz * (jzz + ):) Vs

!r * !n: Ynln * lz * lzt: Yzz

:"zl * ll: Y3:From the above equations it is clear thatDiagonal elements : sum of the admittances of the elements that are creating the node.off-diagonal elements : Negative sign of admittance between the adjacent nodes.

t.e., yr2 = _ Jtz,yzt: _y23andyr, : _)13

Diagonal elements are called self admittances of the node or driving point admittances. Thediago,al elements are calred mutual admittances or transfer admittances.Substiruting equarion (t.72) in equation (1.71)

Ir = Yrr Vl + yt2 V20:Yz,Vt+yr2V2+y23V3lr : Y:tz V2 + y33 V3

zL

...11.71

zs

VL

Power Sgstem Nehoork Matrier':s 19

From the,above discussions we can write the 'Ysu.' general form for any power system networkwith'n' buses (n + 1 nodes) as follows

rhese equarions can _;:T.,;l::" T:,_T?;:1" ".*

*

IoI:l"r, Y, r, llv, I1,,] [ o Y,, v,,,lLu,][],I [Y,, Y,, Y,,,-l [v,lI ,, I 1",, Y,, v,,, I lr, I

The elements of bus admittance matrix can be form as follows :

...(1.74)

...(1.7s)

Y1; : Sum of the admittances of the elements that are creating the node at bus 'i'Y, : Negative sign of admittance between the adjacent busses 'i' and J'

...(t.76)

Itbte : (l) The 'Ysu,' matrix can be found by direct inspection method provided mtitual couplingbetween the elements of the given power system network is neglected.

(i0 The 'Yru.' matrix can be developed by direct inspection method, but not in the case of 'Zsor' matrix.

,I 'I'I DIRECT INSPECTIOT{ METHOD FOR DETERIYI]NATION OF 'Z,OON'In this method the elements of loop incidence matrix (Z1oor) can be calculated directly by inspectionmd hence 'ZLuop' has direct correspondence with the given primitive network.

I-et us take the one example for solving 'ZLoop'by using direct inspection method.The elements of 'Z.oor' matrix are developed by applying Kirchhoff's voltage law and writing

bop equations of the given power system network. However if there is any current sources in theretwork that is replaced by equivalent voltage source with series impedance.

nll

!l2

Fig. l.16

-

-

Writing the loop equation by using *O.nEt-Ez:IrZr + Gr _ I)22 + (It_\)24E2-\: (Iz

- t)22 + tzz3 + (12_\)Zs

0: (Ir -I)Zt + (Ir

- 12)Zs + \26Separating the coefficients of loop currents I,, I, and 13, *" g"r"

Er _ Ez : (Zr t 22 + Z) \ _ Zzt2_ Z3I3Et _ Et _ _ zt\ + (22 + 23 + z) 12 _ zst3

O : -

Z+\ -

Zsl2 + (23 + Zs + 26) 13Let the impedances

Zrt: Zt + Zz + Z',4Zzz:h+23+ZsZts: Zq + Zs + 26Zo:Zzr--22Zn:Ztr--24Zzs:Zzz--Zs

The resultant loop voltage sources in loop .l' isV1:E1-E2

The resultant loop voltage sources in loop .2' isyz:Ft_Et

By using above assumprions, equation (r.7g) can be modified as foilows:Yt: Zrt\ + ZnI2 + ZBI3Yz: Zzt\ + Zzz12 + 7.23130:Zzr\+23212+Znl3

Write equation (1.82) in matrix form

l-v,l lz,, 2,, zu1(t,\lr,l: lr,, 2,, ,"llr,lL o I Lz, z* z")ltr)

From above discussions we can write the elements of loop impedance matrix as follows :2,, : Sum of the impedances of the elements forming the loop .i, : Diagonal elements2,, : Negative sign of impedance that are conrmon to the loops 'i' and r' : off-diagonal

...U.77(all

...tL.77(bfl

...tt.77(c)l

l',, 2,, z,,f. '7

-

lzrt z* zr,l...(1.lloon- 1... I

Lr,, znz 2,,)Diagonal elements are called self impedances of the loop or driving point impedances. Theliagonal elements are called mufual impedances or transfer i*p"d"r..r.

sowED PROBLEMShoblem l.l. For network shown in Fig. I.l7 draw the graph, fromthatfind Ai A, B: B, C', C, K

@o

o+J)

Fig. r.r7

Sol. For Fig. 1.17(a) /Numberofelements, e:9 / ,/Numberof nodes, n:5 + l:6/Numberofbranches,b:5Number of links, I : 4

@Fig. I.l7 (a) Graph

Element node incidence matrix (A') elements can be obtained from Fig. l.l7 (a) as follows :ai : l, if the id' element is incidence and oriented away from the j6-node.ou : - 1, if the ith element is incidence and oriented towards tne;m-i"Oqe.oi : 0, if the ith element is not incidence to the je node.

Node

fi,,,r ry29 //*fr'*c'd)

a/1L Lu- ntol,r/ /tfiuou,n "( ,, @v ,_7

1

2J456789

00001

001

1

0000100

-10111100010

'1 .-1 00 1 -1

100100000000001001000

A':

For obtaining bus incidence matrix (A), eliminate the reference node column in A'.Bus incidence matrix (A)

Element @o1

2

3

4

-11

0

0000-1 0 0 00

-1 0 00 0 :[ 0

[: ,5@-

Fig. l.I7 (b) Tree

Branch path incidenle-qatlix (K) elements are obtiined from Fig. 1.17 (D).Puh t

Branch

K_

I2

J

4

5

-1 -1 0 0 00

-1 0 0 00 0

-1 0 00 0 0

-1 -10 0 0 0

-lBasic cut-setThe elements

(B)in this matrix can be found as follows from Fig. l.Ll (c)

11 ,r') \,'

ff,

@Fig. f.f7 (c) Cut-set

B, -

1, if the ith element is incidence to and oriented in the same direction as the/ basic cut-setB, :

- l. if the lth element is incidence to and oriented in the opposite direction as theT'fr basic cut-set

B, : 0, if the ift element is not incidence with theTs basic cut-set

5 I 0 0 0 I -t6 I 0 0 -1 1 07 | 0 1 -1 0 08 I 0 1 0 0 -l9 | 0 0 1 0 -t

Cut

rll 0 0 02l0 1 0 0310 0 I 04l0 0 0 Islo 0 0 0610 0 1-17 l-t -1 I 08 l-1 -1 0 I910 0

-1 I

0

0

0

0

1

0

01

1

Elr coiO 0 0010 0 0010 0 0010 0 0liO 0 0

--+-

tj*

Augmented cut-set matrix (B)Cut-set

Element A B C, Dlll 0 0 02l0 1 0 0310 0 1 04l0 0 0 1

B': 5 | o o o o610 0 1-17 l-t -1 1 08l-1 -1 0 r9 I 0 0 -t I

-4ta'00

0

0

00

0

1

1

100010001000

00

01

Basic loop incidence matrix (C)The elements can be found as follows from Fig. \.fi (A9,,

: 1, if the element is incidence to and oriehted in the same direction as theTe basic loop!, : ; 1;'lln:.:,_:."-ry is incidence to and oriented in the opposite direction_as trreT; uasic toopC,j : 0, if the element is not incidence to theTs basic loop

GI

HF

1,2,7,39, 5, 4,3

-+ 7 as link-+ 9 as link

L,2,8,5, 4 -+ 8 as link

Fig. r.l7 (d)

3, 4,6 + 6 as link

FGHI01100110

-l -l 0 II 0

-1 _l0 0

-l -11000010000100001

24 Electrical power System Analusis

't2

3

4

c- 56

7

8

9

Augmented loop incidence matrix (C') eremenrs are is obtained from Fig. r.r7 (aI-oop

c D El. c H rllt { 0 0 0i0 I 1 0210 I 0 0 0i0 1 I 0fl r -1\t',;'. \

u' \ Cb \( t2Yuo)\ -\-O \(,' \-/

3 J0 0 I 0 o l_r _l 0 I

tlo ,o o o olo i o o8,1 0 o 0 0 0lo 0 I 0elo 0 0 0 0i0 0 0 I

,

Probleml"2' ForthenetworkshowninFig. l.Isdrawthegraphandtree.Alsodeterminethe' matrix by direct inspection method. All the meniloned valuei aie i*pra"irri;;;.;:.

+ jl0+ j10

+j5

+j5

+ j20

+ jZO20t60"A 20 230" A

Power S-r,rstem Network Malrices 25

Fig. l.I9 (a) Graph Fig. t.I9 (b) TreeFrom the graph shown in Fig. L.I9 (a)Number of nodes, n:5Number of elements, e : 8Numberoftreebranches, b: n- 1:5

-l:4Number of links, l=e-n*l:8-5+l:4Therefore 'Yrur' has the dimension of '4 x 4'Given values are impedances and are given in p.u. quantities. We can determine admittance values

and then form the 'YBu.'.The elements of 'Ysu"' can calculate as follows:

Y,, : Sum of the admittances of the elements that are creating the node at bus 'i'Y, : Negative sign of admittance between the adjacent busses 'i' arrd'j'For i,7 = 1. 2,3 and 4

Diagonal elements'.Yrr :./0:1 + j0.1 + 7O.05 : j0.25 p.u.Yzz: jU.L + j0.2 + j0.2: j0.5 p.u.Y3r: j0.1 + j0.2 + JO.05 :70.35 p.u.

off-diagonar erements ," * :7o'05 + 70'05 + Jo'05' : i0"J5 ptu'

t.

Y tz : Y2r : - 70.1 p.u.Yr3 : Ygr : 0.0 p.u.Yr4 : Y4r :

-7O.05 p.u.Y23:Y1,2:-j0.2p.u.Yzq:Y+z:0p.u.Yl+:Yar:-j0.05p.u.

26 ElectricalPouerSystemAnalysls

I j0.25 - j0.1| -;o I jo.srBu':l o -jo.zl-;o.os o

0 -

j0.05-

j0.2 0j0.3s

-

j0.05-

j0.05 i0.1sProblem 1.3. For problem 1.2, determine the elements of 'Zurp' by direct inspection methd-Sol. Here we have to replace the current sources into equivalent voltage sources when applying

direct inspection method.The elements of loop impedance matrix as follows :

2,, : Sum of the impedances of the elements forming the loop .f,2,7 : Negative sign of impedance that are common to the loops ,i, ad,

Diagonal elements areZ,: 1725 p.u.Zzz: * 720 p.u.Ztt: * 750 p.u.Z++: * 755 p.u.

and off-diagonal elements are

jl0l_rslj20ljss_lC, C'

Zrz: Zzr: -

j5 p.u.Zn: Ztt: 0 p.u.Zu: Zq:

-

jlO p.u.Zzt : Ztz

- -710 p.u.Zz+ = Z+z: - j5 p.u.Zy = Zq: - j20 p.u.

l+1zs -js o _.7

_l -js +j2o -jro\ aloop - |I 0 -ilO +i50L-j10 -js -j20 +

and K. From that verifiProblem 1.4. Determine the incidence matrices A, B, 8,,,following relations shown in Fig. 1.20. Take I as ground bus.

(i) ct -_-

BI i r,fi c,B,T: t(iii) At) { : U and (iv) B, : A,I{

Fig. l.2o!

(ii) C, B,T = U(iv) B,: Ar {

Power SgstemNehaork Matrices 27Sol, Number of nodes, n : 4Number of tree branches, b : n

- |

Number of links,:4-I=3

l:e,-n*l:4-4 * 1:1.

Assume node (D as referenceFrom Fig. 1.20 (b), Bus incidence matrix [A]

A-

Fig. f.20 (a) Grapb

Fig. I.2O (b) Tree

I,)

3

4

0-1

1

-l00

00

-1----1

l^,)A,

dFrom Fig.

|--r o oleu:l 0 -t 0lI o I -r.l

A,:[l 0 -1]

1.20 (b), Branch path matrix (K)

K- 23

0to fs,rJI )8,

tlrI

I-r,r:lo

loBr: [-

o@0

-t -l0 0

-1I

From Fig. 1.20 (c), basic cut-set incidence matrix (B)

ABCI2

3

4

1

0

0

-l01

0

ll

0

1

0

IFig. r.2o (c)

d

28 ElectrimlPower SgstentAnafusisFrom Fig. l:20 (Q, Augmented cut-set incidence matrix @')

A B C INI o oloo 1 or6:[*]+]o o 1lo

B,=[-111]U,: [lJ

B':1

2

3

4

and rig. r.20 (d) i

c-t>

--A-\

Fig. r.20 (e)I-'11

c, =l-tlL-il

and C, = [1]From Fig. 1.20 (e), Augmented loop incidence matrix (C)

From Fig. 1.20 (e), basic loop incidence matrix (C)

C:

A B C IO

Verification :(i) Ci,--B1From the matrices C, and B, r,alues

Ca:-BlT

l:)c,

t

-1-1

1

1

2

3

4

I2

3

4

.,:i-il

-iua I cal-LT-TI[]

100110 1 0 l-ro o r l-r

i__000;1I

il

:l:l

00100100

1000100011il

C':

B':

(ii) c'B'T: u

C'(B)T _ U(iii) AbKr: u

l-1 0 0 -ll

,":l: ;: llLooo,ll-1 0 0 lll-l 0

.,,rr':ll ;i _lll: ;lo o o lllo o[r o o ollo 1 o ol:ttlo o r olLr o o ,l

[-r o olAr:l o -l olI o r -rl

":fi ; -j] =*

:fi -i l]

A' c:fi I :]fi -i l]

0 -r.l

o tlr tl0rl

_U

001I 0l0 r.l

:[iElectricolPower

,4,6KT: UBr = ArKr

A_

temAnalgsis

(iv)

:Bl

incidence matrices and3

A,:[t 0 _t][-r o ofl

x':lo -t I| 0l

L 0 _l _l_l[-t o ol l--rl

A,Kr:[l o -r]l o -l ol:l tl

I o -r -r_l L ,jProblem 1.5. For the graph shown in Fig . L 2l , form the necessary

verify the following relations :@ A6I{:U_ Gi) Bt:At{

riiit ir: - BI (iv) c,(B)r : s a(Sol. Numberof branches,b: n-l \

b:4-l-3

Numberoflinks =e-n*I:5-4*l:2

From Fig. 1.21, Bus incidence matrix (A) @Eig. l.2l Grapb

e6

I^,)o,[-r o otAr:lo o _ll

L r -r ,]l-r 0

-11o':1, -r il

l.2l (a), branch path matrix (K)and

From Fig.

Fig. I.2l (a) Tree

Power Sustem Nefioork Matires 31

(: 0010

-1 0From Fig. t.Zl (b), basic cut-set incidence matrix (B)

ABC

A B C IO

t--1 I ol @Br:l : '.:l Fig.r.2l(b)1 1 -r r.lFrom Fig. l.2l (b).Augmented cut-set incidence matrix (B')

2

4

iccCut

I 0 0'r0Llr,0 0 lJ

I2

B- 435

I2

B': 4aJ

5

I 0 0100 1 0100 0 I l0

- I -- I --0-T-l-I

-1 I I O

1",

lurlol_ l-J-l- Lnr lurj

l--l I 0ls,:Ll _l tl

From Fig. l.2l (c). Basic loop incidence matrix (C)

},,.": [-l l-l-'-

L-; _ilFrom Fig. l.2l (c), Augmented loop incidence matrix (C,)

@Fig. r.2I (c)

-r I 0 ir,I -l I )'

:[u, lcrlI o lur.]

001o

-rl-t olo

-rl [-r o ol-l ol=xt:l o -r olo-rl L-r o-,1o ol[-r o olo

-,ll o -r ol-1 0lL-l 0 -tl

l-- rer:10Ir

[-rK:10L0[- r

ArxKT: | 0lr[r 0: l0 rloo

l:[-l

[1 1 -11A,=l I

' L0 -1 0l[-r o ort:l o

-t of-r o -r-l--r o ol

-'ll o -r ololL_, o _rlR.H.S.

Verification :(i) AaKT: U

(ii) Br: Ar KrR.H.S.

[to[o -r

: L.H.S.

l0l l:B,-1 lj t

I: Cb.

l--l IB,: I' L r -l

I r-1.l_r,

.

L_; ll

(iii) cb--n/

Power SgstemNefitnrk Matrices 33

(iv) C'(B')r : U

C':

1 0 0 1 -t

0 1 0 -1 I

001 0 I000 1 0000 0 1I 0 0

-t I0 1 0 I

-1(B')T : 0 0 I 0\ r

000 1 0000 0 1100 1-1010-l I

C'(B1T : 0 0 1'0 I000 10000 0 I1000001000

:10 0 I 0 0l:000r000001

Problem 1.6. The incidence matrix is given below.Branchesr

nodes OAT: @@@

Sol.

2345

I 0 0 -l 1

0 l 0 I -1

001 0 1000 1 0000 0 l

U : Unit matrix.

From that draw the oriented graph.

6781000Jq001000

010010 *t I 0 00 0

-1 I -l1 0 0 -1 0

Eig. 1.22

Problem l.T.ForFig- 1.23' theimpedancedatais giveninTable l.l. Alttheimpedancevaluesare in p.u. values. Determine yr^ matrices by singulaiTransformation method.

Fig, 1.23Table 1.1

Sol. From Fig. 1.23, the Bus incidence matrix ^A, is

A_

1000010000100001l

-l 0 00 1

-l 00 0 I

-lI 0

-l 0

l

)

iil

23456T8

Primitive impedarce marrix isElementsr-+ I 2

J000000000.2000000

0 00.25 0 0 0 0 00000.50000o 0 0 0 0,1 0 0 0000000.400000' 0000.300 r0 0 0..0- 0 O 0.6/ .j

Element Bus Code Self impedanceinp.u.

I2345678

0- I0*20-30-4I -22-3

3-41-3

0.10.2

0.250.50.10.40.30.6

Foroer Sysffi"iYettuorft IftfB SPrimitive admittance is given by

Y: [zl-r

v:

y=

YB,.: tArl DI

0J0 000.2 00 0 a.25000000000000000

0.5

0 0.10 00.40 0 00.30 0 0 00.6

r0050040 0020 000100 000 0 2.s0 000 0 0 3.3300000001.67

tAI00 0 0 0 000 0 0 0 040 0 0 0 002 0 0 0 00010 0 0 000 0 2.5 0 000 0 0 3.33 000 0 0 0 t.67

0 0 I -1

1 0 -l 0

I

i

IJ] [A] :

10005000000000000

1000100010001

-1 0

000I0

I -i 0

100000s000040000210

-10 0 00 +25

- 2.5 0

0 0 3.33 -3.33

+1.67 0 -1.67 0

36 Elrcticalfuwer

tArl Dl [A] =

rBus -

00 I 0 0 1l0 0

-l I 0 0lI 0 0

-l I -110 I 0 0 -l ,l

- 10

- 1.67

t7 .5 -

2.5

-2.5 r2.50

- 3.33

Fig. 1.24Table I.2

- 10

- 1.67(s+ro+2.s)

-z.s- z.s (+ + z.s +3.33 +1.67)o

-3.33

l-1 0lo Ilo oLoo

[(,0*lo+1.67)| -ro:l | -r.otLo

00

- 3.33

5.33

I zt.et|

-,0| -r.u',

Lo

l0000

10

00

1.67

000,

0

a

-3.330

yo500400

-10 02.5

-2.50 3.330

-1.67

,r{,;,]

koblem l.E. For the system shown in Fig. 1.24, construct yru, by singutar Tmnsfornmethod. The parameters of various erements ari given;rir;,;;;.;. f;i;'*;'?;,} uyirrn"r,

Element Reactances in p.u.r-2t-62-4z-J3-44-55-6

0.u0.64.030.020.80.060.05

Sol.

Fig. 1.24 (t\ Oriented' grnPbElement node incidence matrix (A')

ELg,. L.24 (P) Tree

o@o@ @A,=

I -1 0 0 0 01 0 0 0 0 -10 1 0 -1 0 00 I -1 0 0 00 0 1 -l 0 00 0- 0 l--_1 00 0 0 0 1 -1

Fromtheelementnodeincidencematrixbusincidencematrixcanbedetermined, i.e-,by deletingthe row corresponding to reference node (6).

Bus incidence matrix (A) is

1

23

45

67

e@o@o1 I 1 -1 0 0 02l1 0 0 0 0{: 3 I O 1 0 -1 04 I o I -1 0 0s I 0 0 1 -1 06 I 0 0 0 I -1710 0 0 0 I

The bus incidence matrix (A) is rearranged by separating branches and links as follows :

,-(-l o@o@@I I I -1 0 0 0211 0 0 0 04 I 0 1 -1 0 0 A, (Branch)

--3-T o Is I 0 0 1' -1 0

e \t \2/ v v v v \

I I I -1 0 0 0 Izl1 o o o o l4 I o 1 -1 o o I6 I o o o 1 -1 I

o -r o i1' -1 o J Ar ( Link)

gg Elecffirr:rtfuu:erPrimitive impedance matrix is

j0.040 j0.060 0 j0.03000

000

Primitive admittance matrix isy : lzl-l

j0.020 j0.080 0 i0.06000

000

000 i0.05

y:

-

jzs0

-

j16.670000000000

25 16.67-25 0000000

- j33.33

0000

- iso0 -

jtz.s0 0

- jr6.67

0 0 0 _j2oYBu,: tAT] DI tA]

[A]T DJ : -I

11 0 0 0 00250 16.670 0 33.33000000000000

-l 0 1 I 0 0 00 0 0

-1 I 0 00 0

-l 0 _1 I 00 0 0 0 0 _t I

500 12.50 0 1667000 20

33.33 50 00

- s0 t2.5

- 33.33 0 _ 125 16.67 00 0 0 _16.67 20

000

000- -J

25 t667-25 0000000

000033.33 s0 0 00

-50 125 0-33.33 0

-12.5 16.670 0 0 _t6.67

I -l 0 0

10000 I 0

-10 1

-1 00 0 I

-tr0 0 0 1_0000

0000

20

tArl DI [A]: -,r

fuwer Sgstem,Netunrk kfiatrires 3g

YBr, : -,1

(zs + to.ot)_25

000

'' -zs

(zs +::.r + so)-50

- 33.330

050

(so + rz.s)-

12.50

0

-33.33-12.5

(33.33 +12.5 + 16.67)-

16.67

000

- 16.67

(a.at + zo)4167

-25- 25 108.330

-500

- 33.33

00

00-50 -333362.5

- 12.5

-125 62.50

- 16.67

000

-16.6736.67

koblem 1.9. Determine the Ys^ matrix by SingatnrTransformntion methodfor the network shownit Fig. 1 .25 . The parameters are given in Table t .3.

Table 1.3

Ele'ment Self impedance in p.u. Muttml impedance in p.u.Bus Code

p-q ImpedanceZrr-*Bus Code

r-sImpedance

z*_r,I2345

t-2(1)t-33-4

1-2(2)2-4

0.20.40.5

0.250.2

1-2 (1)

1-2 (1)

0.05

0.1

Fig. I.25

Sol. Number of elements, e : JNumberofbranches, b:n- l:4-l =3

l:e-nll:5-4*l:2i

Number of links,

Anafusls

Element-node incidence matrix (A')

I2

A'- 345

Bus-incidence matrix (A)

00-1 0tr -l000

-1

-l 0 00

-1 00 1

-l-l 0 0

I 0 -l

ele L 2 3 4 51 102 0.05 0 0.1 02 IO.OS 0.4 0 0 0310 0 05 0 04 l0.l 0 0 02s 0

I -1

l0001

-101

('.' @ is reference

[:

Primitive impedance matrix [z)

Primitive admittance matrixy j

o, \ o ol oI 0 0'2J

; o: foos /o o.r o,loor@o o oy:[z]-,:l 0 0 0.5 0 0

-l

To calculate al ttrere arc two methodsMethod I :

0.1 0 0 025.0000002

0.2 0.05 0 0r 00.050.4 0 0 0000.5000.1 0 00250000002

-10

0

Y : [2]-r :

-l

fuwa SgstemNetworkMaffies 41

0 l-t0l

J]y'=

oe 0.0s 0.1 i 0 0I0.0s 0.4 0 i0 00.1 0 02si0 0

0.05 0 0.10.4000 002500.50000

Stcp 1. Interchange rows 3 and 4

020.050.1

00

Step 2. Interchange columns 3 and 4

y"=

-l

[4, ierlLAr iArl--:-----------l-----0 ---0----0--l 05-- 0

I0 0 0lo 02

0l,]

I oz o.os o.l Ilo.os 0.4 o ILo.r o o2slfool roolo ol A,=lo oLOOI L[05 o-lLo oz)l- o.s

- osr

-zs1| - o.r, 2.6 033 |L -z.o 033 5.ol12 0lL, rl

Step 3. Above matrix can divide into 4 submatrices

Al:

A2=

A,:

A,-' :

Ant:Step 4.

65 -08r -2.6 0 0l

-o8l 26 0.33 o ol-26 033 5 o 0l

o o o2 olo o oos_l

,-: h-l-11-,-l' La, ia;'l

t''=

42 E:leetir:"\ryunr SgsternAnalysisStep 5. Interchange cofunrns 3 and 4

v':

-260.33

5

00

00005

Step 6. Interchange rows 3 and 4

y : [2-1] :

6.5 *081 0 -2.6 0

0.81 2.6 0 0.33 00 02 0 0

-2.6 0.33 0 5 00 00 05

Limitations : This method is applicable only the matrices A, and A, are having thezero's only. Otherwise it is not applicable.

Method 2 :

-l

6.5

- 0.81

-2.600

y : [2-1] =

A,Ao-,": ltt ;] t;

02 0.05 00.05 0.4 0000.5dl ---o--- o000

-081 02.6 00.33 00200

0.1 0000b

0.25 0002

[A, lArl-'-

t ---F--- |

LA: i Aol

Let

[o, o,1-' : [r, s,lLo, eol : Lr, B*l

The values of matrices 81, 82, 83 and 84 areBr : [Ar _ (Az A+-r Ar)]-1

o*-,: [o;" ;] : [; I[o.r ol

e,:lo ol- lo ,l

l-0.1 0 0tA^:l I' L0 0 0l

:lt'J : 3l

Power Systetn Netwo* Matrices 43

I-o.a ol l-0.1 o oll, olIo o olLo oJLo o ol[o.o+ o olI o o oll-o o olll oz o.o5 o I [o.o+ o oll-'tlt' eJ ,:l L: : :lll-o.ro o.o5 o l-'

L'f'T J,jl- os

- o.8r oll-orr 2.6 olLo o z)

- B, A, A11

[,;r :;: :] tT :lr; :ll-- o.s 0.81 0l I- 04 olL'f' :JL : :]l-2.6 ollo:z+ o ILo oJ

^ l-2.6 0.324 0l&':Lo , olAo-'

- (e;' e, nr)

lAr - (Az Rt' a;-'

Bt

82:

:

B2:

B3:B+:

.14Ao-'At

",

: Lo :t tT : :t[-i1 l]

-l - z.e olo ollorro olo oll o ol

l-0.4_l-lo

4 Elrcticalkru:q

A3 82)1.04 000

-r [Bt!:z':Lr,

Next after.calculating primitive admittance matrix,YBu,: tArl Dl tAl

l-- 1.04 0l:l Ilo o,l

Ba:A+1-(Atl:[; I i I-s.04 0l:lo s]

6.5 -

081- 081 2.60 ,-0

-2.6 A32400

0 -2.6

0 0324200 5.0400

00005

YBr. :

B"l.t

_

Bnl

Tl2 0.48 0I

= lo.8l -26 zL0 o -2

- 0.116

- 02081

- 0.1e3

- 0.028 I

- 0.028

- o.ol4l

l00

-l0t1010

h'Io

0

-l1

00

o o -l rl

I

-r I o 0l0

-l 0 -u

0.81 -

0.33

-2.60

06 -

0.81 0-

0.81 2.6 0002

- 2.6 0.33 0000

2.6

- 0.330

-2.6 0033 0000005

0-l

0-l

l- o.e + zs= | o.8tLo il

02

-2

-t00

-l1

00

-l0

-l

2.6

- 0.330

-t00

-lI

00-l 0I

-l000

-1

lBus -

a,zBus -

- 0.48

- sl

4.6 -rl

-2 '): {[Ar] tJI IAI]-'

- 0.48

- sl

4.6 -Zl

-2 ')

I o.+| -

o.+a

L _,IYuur]-'l- o.+| -

o.ot

L -'| -

o.zs

I -

0.,,ul- o.rot

.,lBus -

Problem 1.10. Determine Ys^ matrix for problem 1.7 by direct inspection method.Sol. Diagonal elements

y,, :l*l+ 1 : 10+ 10 + t.67 :21.67" 0l 0.1 06

y..:l*!+ I :5+10 t2.5:L7.5z1 0.2 0.1 o.4y..= I *!*l :4,-2.s +3.33=9.83-

'r 025 0.4 0.3Ytr= #.* :2 + 3.33 = 5.33.

Off-diagonal elementsIyr2:Yzt:-ut:-to

Yr3:Y3r:--#: -1.67

Yla:Y+t:0yzz:y32=-#:. r.tY2+: Yaz = 0

L : -3'33Y3+:Yur=-0.3

12r.67 -10 -t.67 0 Il-10 r7.5 -2.5 0 IV- :l I

'tsus | -t.ot -2.5 9.83 - 3.331L o o -333 s33l

koblem l.ll, \'he network shown in Fig. 1.26. Draw graph and ffee. Determine the Yau, W*c6 inspection method and verify Ysurby nodat equation analysis. The admittance values are givenip.u- quantities.

-

jl0

-

j10

-j5

-js

-i20

-

j2020 z-60"

Fig. r.26

20 z-30" A

6 ebmeatfuruerSgrstemArulystssol. l

oFig. 1.26 (a,\ Grapb

By lXr'ect Inspertion MethodDiagonal elements

Y,, = -710 + (-ilO) + (- jZO): _ j40Yzz :

-j5 + (-i5) + (-jlo) = _jZOYrr = -j5 + (-j20) + (-j2O\ :

-j4sofr-diagonal etemen*

Y*: -i20 + (-i20) + (-ilO) = -750

Ylr = Ylz: -(-jl0) = +jlOYr3=Y31 =0Yr+=Yar =-(j20):j20Y::: Y3z = -(-j5): j5Y2a=Ya-0Yrc:Y+r:-(j20):j20

Verification by Nodal Analysis Method

[-ioo iro o j2o I"*.: I 'lo

-t;' _'io, ,\l

I izo o j2o -

jso.l

l

1

-

j10

-

j10

-j5

-j5

@

-

j20

-

j20 20/-30.A

fuuer SgsterrNefiuork Matriees 47

Let voltages at node@ : V,node@: V,node @ : Y,node@: Yn

Writing nodal equations for Fig. 1.26 (c)At node@(Vr

- 0) (- j10) + (Vr

- V2) (- j10) + (Vr

- vJ C.120) :20 Z- 60"

v; t- j10 - jto - jzo) + v2 010) + v4 020) :20 t- 60"-

j40 vl + j10 v2 + jzov4: 20 l- 60"At node @

(vz -

vr) (- j10) + (vz -

vi_Ljs) + (vz -

0) (-js) : 0j10 vl + (- j10

-j5 -js) v' + 7i v3j10vr +Gjzo)v2+j5v::0At node @(Vr

- Vz) (- js) + (vr

- 0) (= j20) + (Vr

- V, (- j20) = 0

j5 v2 + G j5 - j20 -j20)v3 *720v0: g

At node @ 75 v' * (- i45\ v3 + i2o v+ : o

V2 (- j10) + (V+ -

Vl) (-j20) + (V+ -

V) (-720) : 20 Z-30'+ j20 vr + jz}v3 + (-jl0

- jz}

- j}O) Y t : 20 Z- 30"

+ j20 vr + j20 Y3 + (-j50) Y + : 20 z- 30"Writing the equations (1.85), (1;86), (1.87) and (1.8S) in matrix form.

jrO 0-

jzo jsjs -

j450 j20

...(1.8s)

...(1.86)

...(1.87)

...(1.88)

l^'l-- ;L I lil-TsoJ [v. ]l- iqoI .iro

:l loL ;,0

[In .] : [Ysu*] [VsuJ

l- i+o ilo o izoljro -jzo js ov _l "rBus-l o js -j4s j2ot"

L j20 0 j20 - isoProblem 1.12. For the graph given inFig. 1.27, draw the tree and corresponding co-tree. Choose

a tree and hence write the basic cut-set schedule. (April/may 2005,2ffi7)

Sol. Tree of Fig. 1.27 isEig. 1.27

shown in Fig. 1.28(a)7

Fig. 1.28(a)Corresponding co-tree is shown in Fig. l.2$(b\

Cut-set schedule:

Fig. r.28(b)

---'-l-.'-o.----.

Fig. r.28(c)

fuwer SgstemNetunrk Maffices 49

ABCD

h,oblem 1.13. For the givm network shown in Fig. I . 29,dtcdule.

draw the graph and tree. Write the cut-set(April2005)

10001-1 000-11-100100001010000-t0

, fJ Graph of Fig.i

1.29 is shown in Fig. 1.30(a)

Fig. l.3o(a)dFig. 1.30(a) is shown in Fig. 1.30(D)

@Fig. r.3o(b)

----?--- ^v-----r----

I etaregrtruwerslstenrCut-set schedule

A B CiO E F

hoblem 1.14. For the network shown in Fig. 1.31. DetermineB', B, C and C' and verify the followings:.(i) Ar, Kr : U(ii) 4=At{(iii)cb=-EITake node@ as reference and I, 2 and S as *ee branches

Sol. Oriented graph of Fig. l.3l

Fig. l.3l

is shown in Fig. 1.32(a)

I2

3

45

6

the incidence mntrices Ai

l 0 0i0 0 00 I 0t0 0 0o 0 lio 0 0

- 1- - -- -n- --;- *- -.- - -:- -- =-1 I oil 0 0

-l 0 rio I 00

-l li0 0 I

Fig. 1.32(a)

't

Tree of Fig. 1.32(a) is shown in Fig. 1.32(b)

Element node incidence matrix (A')

1

2

IAI= 34

5

1

2

A:53

4

Fig. r32@)

-1 00\0-1 0

'ol'1-1 00-l 0l

II0

I0

Bus incidence matrix [A]Take O as reference node, delete column corresponding'to reference node.

@o@

x1

2[A',] :

3

4

5

@0

0

1

@@-1 00-101

-1 ,0

-1 0

o

0

I

lonl@@

-l0-1

-l 0

@ol0i1l =[ it]

0lo1 0

_I

0

52 turri.",fuug-esrananatysrsH@iri&ncmatrix [K]km Fig. 1.32(b)

@

1l,l

From Fig. 1.32(c), the basic cut-setFig. r.32(c)

matrix [B]ABC100010001l--- r ---:l100

g

I

j;

IIiIi{I

1

I'l{I

III

itlt{l=l:i.

i J,l IllII

I

!

1

{tllI

I

ll

DE000000T---o0l

i-IIIII

ABC1000100011-l-1100

@ol [-l o

r:zl o -r

rlo o

I2

B- 53

4

=[*]

Augmented cut-set matrix [B']

I2

B,: 53

4

-----____l>3

\D)J1

Fis. r32(d)

Power SgstemNefitork Matr'rces 53 \From Fig. 132(4, the basic loop incident matrix [C]Basic loops: D : 2, 3, 5, 1, 3 as link

E : 1, 4, 4 as linkDE

I I -l I

2t I 0C:51 I 0

3l I 04t 0 1

Augmented loop incidence matrix [C']

=[i:]

I2

C'= 53

4

The rearranged matrices for verification

[-rAa: I 0[-rl-o

,,: L,[-r

Cr: I IIr

ABCDE1 0 ol-l

-10 l 0i1 o

I0 0 li I 0

0 0 0!0

Iu, I cr-lLo iu'I

ool[-rool; :l' *: L-: ; :j

-r rl I-o t tlo o.l ' At: [-, o o.J1lol

(i) AaKr: u

tl i :lti ;(u) Bl = Al Kr

Ir:[,-[-r o' -'ll o -r0 0ll'L-l 0i:

:]:[iol

:l

0I0

olol:u'l

-r -tll:B,00_J '

54 ElecticolfuioerSystemAnalysls

(iii1 go - -Bl

[-r -r I-sI:l r I, I ol:",[r o_]Hence all relations are verified.

Problem 1.15- The graph shown in Fig. t .33, select@ as reference node and tree T {2, 4, S,Determine the incidence and augmented rnatrices. From these verify the followings:(i) B, : 1,6r(ii) Cb:

- B!'

(iii) Br : -

Ar Ail

sol. Tree of Fig. 1.33 shown in Fig. 1.34(a) rree branches are given T {2,4,5,6}

Fi$ l.3a(a)Element node incidence matrix [A']

@-1-t-l

0

0

0

0

I)3

A'= 45

67

o@o0 0 0l0 0 rl0 I 0l0 0

-rl0 -r 1lI I ol

-r o o_]

OI00

I0

0

-l

Pouer Sgstem Network Matric.es 55

Bus incidence matrix [A]: node @ as referen.., .o delete the column corresponding @ in [A Ithen

o@I

-l0-l0-1100000-r0

,|

4

5

A- 61

3

7

Branch path incidence matrix [K]

(oR)o @ @0-1 01000000011-1 00

,-1 0-r0-l

I2

3

A:45

6

7

@o0 0l0 rl0 0l0

-1 I0 1lI 0l

-r 0]

o1

-11

0

@ooI 0 0l

I0 0 0l-1 0 1lo I o]

=[*]0

0

0

o210+l-r

IK: s I -t

6[-oFrom Fig. 1.34(b), basic cut-set matrix

2

4

5

B=61

3

7

00-1 0

I

-l -1

B

0

I

0

0

AI0

0

0

tB1

C0

0

I0

l;ibol0l0lrl

11100-l Fig. l.3a@)

t7I

56 Hffiimltuu:qAugmented cut-set matrix [B']

2

4

5B'=6

1

J7

2

45C-6I3

7

2

4

5

C':6IJ7

From Fig. 1.34(c), basic loop incidence marrEFG-l-10-r0t011001

-i--- o

--o

010001

Augmented loop incidence matrix [C,]

l_._\DtintlGI

6

Fig. I.3a(c)

ABCD1000010000100001

ix (c)

t;I

EFG000000000000 =fX: i-,ltl

l0-10-1-t

0

0

-l

100010001

The matrices reananged for verification

Itor: I 0

l-r

ABCDEFGI 0 o 0 | _r _1 00 1 0 0l-1 0 10' 0 I 0 0110 0 0 ll0 0 I0 0 0 0l r 0 00000 0100 0 0 0i0 0 I

-10

0

0

=;*' ; 3:1

-r 0l-1 0l0 rlr0l

-roor[o-r o ol, r,-lr; _; ;J [:L:

aaII

IIIII

NetuorkMatrices 57

t-r -r o otlo -r -r olI lll 0 -1 0llo -l o ollo o o ,lL-l o -1 oilo o 1 ol

loolIo

-r o l= [n,1-l -l -u-r 0l0 rl, ,l= t'0rl

m &=A/Kr =

-1 0000001

-l 0-t00-l

fr=lt

Lo

[-rc,-BI -l-'

L:tt:

- A, Aa-'

IoIror: loLo

oro.rl-1 o o rl0 0 o ,looro.]

-l-o l o rl,ll_; ; ; ;l [-r -, o ololl o o o 1l:l-' o I ol=-''oll o o r o-l Lo I I rl

J] A:Ir

A/x e;r = | 0[-r

1.16. For the power system shown in Fig. 1.35. Determine bus incidence matrix 'A',lpth incidence matrix 'K', basic cut-set maffix 'B' and basic loop incidence matrix 'C' anddrrlw that (i) lu Xr = U (ii) Bt : At t{. fate 1, 5, 4 as tee

(Aug.2fiI7)Fig. r.35

58 ElecticalPower S4stemAnalysrisSol. Selected tree of Fig.1.35 is shown in Fig. 1.36(a)From Fig. 1.36(a). bus incidence matrix (A),

o @ o1

4

A:52

3

I4B-5

2

3

Io 0[-r o

0-1 0-1 0 I01-100-1

-1 00

-l0

0

-l0

0

IIIII

zlt,l2=[it]| .,'

@Fig. r.36(a)

,/;

Branch path incidence matrix [K]o @ o

r[ r -, ,lr:+l-l o ol

,L r o ,lFrom Fig. 1.36(b), basic cut-set matrix [B]

ABC100010001

-i--- o ---:l-l 1 -1

=[:l

III2l,,

,,

l,)ci6Fig. 1.36(b)

Y'@

Fig. I.36(c)

From Fig. 1.36(c), basic loop incidence matrix (c)DE

-1 -10-1

-1 -1

0lI rilIl

,r[ 1,lL-l

l00l

rlolC:5

I

,l:lA,KT: U

I o -ri-l 0[u rBl=AlKr

a5^Ic, I

=Lr,]

[r o oltt=lo r ol:u

Lo o 1l

r'Izl l,q l .....""

(i)rlol,]

:]:[-r-rl-,1

: ''

0

I

(,,)

Power System Network Matrices 59

Problem 1.17. A 4 bus system is shown in Fig. 1.37.(i) Find the bus incidence matrix 'A'for 4 bus system. Take ground as reference. The reactance of

generators are 2 p-u.(ii) Find also primitive admittance matrix for thip iystem, it is given that all the lines are characterized

by a seriis impednnce of (0.! + j\.i) ohm/fun and shuni admittance of j0.35 x t0'5 mhos/ton.

(November 2004)Sol. The oriented graph of Fig. 1.37 is shown in Fig. 1.38(a).Frorn Fig. t.38(a), bus incidence matrix [A]

a

b

c

A- de

fo6

Given series impedance,

SimilarlyAnd shunt admittance.

0

I0

I

-l0

-1

1000l00-l001

-110

0

0

-l0

I0

0 @ ground/refFig. I.38(a)

z : (0.1 + j0.7) Q/kmZp: (0.1 + j0.7) x 100 : (10 + j70)0Zs: (0.1 + j0.7)x 110: (ll + j77)AZ1a: 15 + j105, Z"a: l0 + j70 and Ztq: 12 + j84

y :7O.35 x 10-s U/km)12..: j0.35 x l0-5 x 100 : j0.35 x 10-3O, )13s :i0.3S5 x 10-3 O)r4. : j0.525x 10-3U, yz*. :j0.35 x 10-3 O andyro, : iO.42x l0-3(I.

1 - ---!n: lzr: i

:0'002 -i0'014

Fig. 1.37

Mutual admittances

50 Electicaltuurer Analysis

Similarly

)r:)r: + :0.00182 -jl.OlzlLtt)ia : )+r : 0.0013 -j0.0093!z+: J+z: 0.002 - j0.014!as : !za.: 0.00166 -j0.01167

Diagonal admittances

Yrr : )rz * )r: * -v,o + * ** *+ : 0-00512 -j0.0354"222r-----

y:z : ):r * lzt * $*'?" : O.OO4 -j0.0276,22

Yr::)sr *):+ * {l*+ :0.00348-j0.023915,2 2Y+r:0.00496- j0.03432

I o.oos 12 - i0.0354 - 0.002 + .10.014 - 0.00182 + j0.0127v _ I -0.002 + j0.014 0.004 * j0.0276 0reus-l-0.00182+io.ot2l 0 0.00348-jo.0z3gr5

[ -o.oorr + j0.0093 -0.002 + j0.014 -0.00166+ j0.01167

-0.0013 + j0.0093.-0.002 + j0.014

-0.00166 + j0.011670.00496

- j0.o3432,

slroRT QUESTTONS AND ANSWERS1. Whcn does the need for using a curved line segment arise when drawing a graph for a

netrvork'JAns: For representing parallel connected network elements in a graph curved line segments

required.2. Wltat are the directions assigned to the elements of an oriented graph of a power network?

Ans: The direction assigned to an element of the graph of a power network is taken to the knownassumed direction of current and voltage in the corresponding network element.

3. What is primitive impedance matrix?Ans: The elements of the primitive impedance matrix are the self and mutual impedance of indivi

network elements with their interconnection not considered4. What is the dift'erence between an ordinary loop and a basic loop?

Ans: An ordinary loop is a closed path containing one or more links. A basic loop is a closedhaving only one link

5. Wlnt is a basic cut-set?Ans: A basic cut-set is a set of minimum number of elements, removal of which divides the

' graph into two connected sub graphs and this set contains only one tree branch.6. What does an incidence matrix'J

Ans: An incidence matrix represents, in general, interconnection of tlre elements, with respect tonodes.

7. What is a bus'lAns: When one of the nodes of a power network is taken as the reference, then the other nodes

called buses.

8. What are the rows of all the incidence matrices?Ans: The rows of all the incidence matrices correspond to the elements.

g. why is tlie sub matrix U6 of basic cur-set matrix (B) is a unity matrix and what is its dimension?Ans: Thi sub matrix U6 describes the incidence of branches to basic cut-sets. Since each basic cut- sets

.ontuiri, only one branch, this sub matrix U6 is a unity matrix' The number of basic cut-setsbeing'equal to the number of branches. Uu is a square matrix of dimension (b x b)

r0. what does the sub matrix c, of the incidence matrix c represent?Ans: The sub matrix C, describes the incidence of branches to basic loops.ll. What are the elements of Y5u, represent?

Ans: The diagonal elements of Ysu. represent the short-circuit driving point admittances and the offdiagonal elements, the short-circuit transfer admittances'How is a uetwork element represented in impedance form?

By a voltage source in series with its self impedance'What is a primitive admiftance matrix?A primitive admittance matrix is a matrix which is an inverse of primitive impedance matrix' thediagonal elements of the primitive impedance matrix being the self impedances of the networkelements and the off-diagonal elements, the mutual impedances between the network elements'

If there is no mutual coupling between network elements, what will be typeinature of the

t2.Are:

13.Are:

14.primitive imPedance matrix?

.t\ It will be a diagonal matrix.15. Why the determination of You. using an incidence matrix is called singular transformation?

,lns: The incidence matrix which is used in the determination of an interconnected network matrixtiom a primitive network matrix is called a transformation matrix antl is singular. Hence, thistransformation is called a singular transformation'

16. How Y5u. is determined if some of the network elements are coupled?Are: Y5,,, is determined by singular transtbrmation using Yhr. : At Dl e17. How Zt.,,,p is determined if all the network elements are uncoupled?

Am: All tliagonal elements are determined as the sum of the impedances of elements in the basic loop'corresponding to that diagonal element. An off-diagonal element is determined as the negative ofthe sum of the impedances of the network elements common to the basic loops pertaining to thisoff-diagonal element.

ft. If Znu. is required, how can it be determined using singular transformation?Ans: Znu" is determined by singular transtbrmation using Zbr, = {Arb]A}-'19. Detine directed graPtr?

Am: A graph is said to be directed or oriented graph when all the nodes and the branches arenumbered and directions are assigned to the branches by ariows.

20. Define a Notle?Ans: Node is the meeting point of the two or more elements in a graph.21. Define an element or edge ?

Ans: A edge is a line segment representing one network element or a combination of elementscomected between two nodes, in a graph.

22. What does a bus incidence matrix describe?Ans: A bus incidence matrix describes the incidence i.e., the connection of arbitrary oriented elements

to the buses of the graph of a network.

62 ElectrbalPouser Anatgsis23. The entries in the bus admittance matrix are either +l of

-l or 0. Is it a fact?Ans: Yes

24. what cloes an entry of -l in fh row and /th column of matrix A denote ?

.

Ans: It shows that the fh element is incident to the ts node and is directed towards it.25. Write down the pertbrmance equation of an interconnected network in the bus frame of

Ans: Ior. =Yr,rrVr,r.26. ln the transformation. used to obtain Ynu*, power variant

Ans: No27. What is a Tree?

Ans: A tree is a sub graph containing all the nodes of the original graph but no closed paths or loops.28. Define Twigs?

Ans: These are the braches of the trees.29. What are the type for Construction of primitive network element ?

Ans: There are two types of representation:1l) Impedance form (ii) Admittance form.

30. What is the Iimitation for direct inspection method?Ans: The impedance and admittances can be found by direct inspection method provided mutual

between the elements of the given power system network is neglected.31. what are the tbrmulae for determined you, from direct inspection method?

Ans: Self admiffance, Yii = the sum of the admittances of the elements that are creating the node at bus .i'Mutual admittance, Y,7 : Negative slgn of admittance berween the adjacent busses 'i' and 'jr.

32.Am:

what are the lormulae for determined znq, from direct inspection method?Self impedance

. Zii : the sum of the impedances of the elements that are creating the node at busand Mutual impedance, Z, : Negative sign of impedance between the adjacent basic loops 'i' and f

OBJECTIVE TYPE QUESTIONSA tree has(a) a closed path (D) no closeil paths

The number of branches in a tree is(c) botr (d) none

, the number ofbranches in a graph.(a) less than (D) more than (c) equal (d) none

The tie set schedule gives the relation between(a) branch currents and link currents (D) branch voltage and link currents(c) branch currenrs and link voltages (@ none

4. TliJ cut set schedule gives the relation between(a) branch currents and link currents (D) branch voltages aod tree branch voltages(c) branch voltages and link voltaggs (d) brauch curent and EE crurents

5. If a network contains 'e' elements 'n' nodes, then the number of mesl crrrcil equations would be(a) e-(n-l) (b') n{te-l\ (c) e-n-l (i, e+*l6. A network has seven nodes and five independent loop, then the rrumbcr of rlanr: in tb netrrork is(a\ 13 (b) 12 (c) 1l (4 x)

3. In branch path incidence matrix, if the ie branch is in the path from the jth bus to reference butis oriented in opposite direction then K,) =(a) | (b\

-1 (c) 0 (d) none(d) none

t. Among the following relations, which is correct(a) K,

- A, -t (D) It : Ar,

(b) 'f independent loop equations(d) none

7. In element nsde incidence matrix. the rank is(a) less than nodes (b) greater than nodes (c) equal to

(c) A't -

K, -'

17. A set of unconnected elements is defined as a network.(a) primitive (D) complete (c) symmetrical

It. The-branch admittance matrix can be obtained by, using _(a) a closed path (b) no closed paths (c) both

ANSWERS

ll. The dimension of basic loop incidence matrix is(a) links by nodes (b) elements by links (c) links by branches (d) none

ll. The performance equation in impedance form is(a) v * i = z\e (b), t, s - 1xi (c) both (d) none

Lll. In the loop frame of reference, the perforrnance of an interconnected network is describedby________(a) n-l independent nodal equations(c) 'b' independent branch equations

13. Zbu* expression in terms of branch path matrix(a) K'YBRK (D * zur. (c) K zBRlK (4 K' ZBRK

Itl. The meeting of various components in a power system is called(a) branches (b) bus (c) links (d) none

15. If You, is symmetrical then the corresponding Zru" is(a) unsymrnetrical (b) symmetrical (c) neither (a) nor (b) (d) none

15. The off-diagonal elements in You" are called as(c) self admittance (b) mutual admittance (c) symmetrical

(d) none

(d) att

(d) att(d) none

r. (b)7. (a\

13. (d)

2. (a)8. (D)

14. (b\

3.. (a)9. (a\

rs. (b)

4. (b)10. (r)16. (b)

5. (a)11. (r)17. (a)

6. (c)12. (c)8.(A

EXERCISE1. Define the following matrices with one example for each matrix.(a) Elements node incidence matrix

(c) Basic cut:St incidence matrix(e) Branch path incidence matrix.

2. What are the elements of the following:(a) Bus incidence matrix(c) Basic cut-set incidence matrix

(b) Bus incidence matrix(d) Basic loop incidence matrix

(b) Basic loop incidence matrix(d) Branch path incidence matrix.

il Etecfficalfuwer fustemAnalgsls3. Bus incidence matrix is given below.

Draw the graph and tree.

ooor[-l 0 0l

^ zl r o -llo:rl r -l ol

+L o -r 'j4. Prove the following relations with one example.

@) A,,kr : U (D) C'[B']r : U (c) Co - - B,t i5. Prove thar Yu,. : tArl tYl IAI6. Prove that Z,nu, : Cr tzl [C].7. Determine the Zsu. and Yuu. for the following network shown in Fig. I .39 and the admittance

in p.u. value.

15 -

j7.5

Fig. 1.39

Determine Ysu. flnd Zrn., by direct inspection method for the following network shown in Fig.

Fig. l.4O

Given values urlin p.u. of impedances.

is

i20 i15

j2s i10 j5

Power SgstemNehuorkMatnces 6t

9. Determine all the incidence matrices for following power system network shown in Fig. 1.41.

Fig. r.4l

For the problem 9 the impedance values are given below. Determine Y6u, and Zsu, by singulartransformation method.

r0.

Element No. Self impedance in p.u.1

2

3

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