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CHAPTER 2
GENERALISED BUS IMPEDANCE ALGORITHM
2*1 INTRODUCTION
Bub impedance matrix of a network is amathematical equivalent which relates the busvoltages explicitly in terms of the bus currents.The primitive impedance matrix and the networktopology are required to construct the bus impedancematrix Zg^. The elements of the bus impedancematrix can be interpreted as the open circuitimpedance parameters relating the bus voltages andcurrents. Alternatively using linear graph theorythe bus impedance matrix can be viewed as the multiterminal representation in impedance form correspon-
oc 07ding to the Lagrangian tree as the terminal graph * ' The topological formula for the computation of Zgug however requires, the matrix inversion of the order equal to the number of independent loops in the network and hence cannot be used to compute Zgug of large networks. The alternate method of constructing the bus impedance matrix by adding one element at a time has been very popular with power engineers and is known as Zgus building algorithm.
The bus impedance algorithm was originallydeveloped in I960 by Brown, Person, Kirchmayer and
28 29Stagg ' . This algorithm was developed interpreting
15
the elements of ZgUjg as the open circuit parametersof the network. While constructing the Zg^, theelements are added to the partial network one at atime. The added element will be either a branch ora link. When a branch is added the elements of thebus impedance matrix corresponding to the partialnetwork are not altered. This result was used whileobtaining the modified hue impedance matrix due tothe addition of a link also, by first considering theadded element as a branch creating a fictitious bus.This has resulted in a two step procedure for theaddition of a link algorithm. Furthermore, thealgorithm for the modification of the bus impedancematrix dup to the removal of elements from thecoupled group is much involved. This originalalgorithm has since been expanded upon by several
30-3'5authors for new applications ^ ,
Bhat has developed the bus impedance algorithm27based on linear graph theory . Using the theorem on
multi terminal representation, the modified bus impedance matrix due to changes in the network Was obtained by deriving the impedance representation of the modified network, taking the Lagrangian tree with ground as reference, as the terminal graph. When certain elements in the network are mutually coupled, it is necessary in this algorithm that the coupled group should be added as
16
a Whole and not one element at a time. Por removal of elements from coupled group, a two step procedure will be required. In the first step the coupled group should be removed as a whole and in the second step . the elements that are not to be removed should be added.
The existing algorithms do not effectively utilize the topology of the network and cannot deal with the effects of mutual coupling satisfactorily. Besides, the presence of phase shifting component is ignored except in reference 23. It is therefore desirable to have a comprehensive and generalised algorithm which does not have the drawbacks discussed above and yet is based on firm topological considerations. This chapter is devoted to the development of such an algorithm.
2.2 SOME USEFUL RESULTS PROM GRAPH THEORY
The bus impedance matrix of a network depends on the parameters of the elements that constitute the network and the network topology. The parameters of the elements are given by the primitive impedance matrix and the bus incidence matrix gives the complete information as to how the elements are interconnected. Some pertinent results of linear graph theory are briefly reviewed below.
17
!Dhe elements of the bus incidence matrix A
are defined as follows?
alJ = +1 If the element corresponding to the Jth
column is incident to the feus corresponding to.
the ith row and is directed away from it.
ai3 “ el®®®n,fc corresponding to the Jth
column is incident to the bus corresponding
to the ith row and is directed towards it.
aii = 0 if element corresponding to the Jth
column is not incident to the bus corresponding
to the ith row.
For a network with N buses and e elements the bus
incidence matrix A is of dimension N-l x *, the trow esasresponding<t© the reference bus being deleted.
29Consider an oriented connected graph for
which A is the bus incidence matrix. Let v#i be
the element voltage and current vectors and Vgus,
Igus be the bus voltage and current vectors,29W v, i and A are then related aa
Tv = A Y.Bias
Bus A i(2.1)(2.2)
T .where A is the transpose of A. If y denotes the
primitive admittance matrix, i.e. i = y v, then from (2.1) and (2.2), the well known expression for bus
admittance matrix is obtained as
Bus = A y A (2.3)
18
2.3 PRELIMINARY RESULTS USEFUL FOR THE DEVELOPMENTLet zoi Aq and Zg^o) be the primitive impedance
matrix, bus incidence matrix and bus impedance matrix respectively of the partial network shown in Figure 2.1.
FIGURE 2.1: Partial network(Interconnections are not shown).
When certain modifications such as switching in andout of certain line(s) in the network are made, we
would like to know the modified bus impedance matrixZ_ of the new network in terms of L , nus Bus (o)
For the partial network, using (2.3) we have
ZBus(o) “ YBus(o) = £Ao zoo Ao 3 (2.4)
Suppose element a having self impedance zaa and mutual impedance zQa with the elements of the partial network,
19
is added to the partial network. For the new network, the element voltages and currents are . deaofceddbjr
where vq and iQ are the element voltage and current vectors of elements in the partial network, v and
v«
ia are the eleaent voltage and current corresponding to the element a.
The primitive equations in the impedance form are
V zoo zQCC
1—otl!_
V zao Zaa vu wThe inverse form of (2.5) is given by
io oo y0a~ V'*« fa**- M
yao 7<UC vau. —1
In (2.5) and (2.6) zQQf yQQ are matrices, zQa, zao> yQa, yaQ are vectors and zaa, yaa are scalars. Using (2.5) and (2.6) the primitive equations in mixed form^
can be expressed as
(2.7)
20
Prom the discussion in the previous section, for the new network
A1 V,Bus (2.8)
where A is the bus incidence matrix of the With these preliminaries we now proceed to
(2.9)
new network, the specific
cases.
2.4 ADDITION OP LUffKfS)
Let element a he added between buses p and q with the orientation from p to q as shown in figure 2.2.
FIGURE 2.2: Near network due to addition of a link
21
In this case no new bus is created. The bus incidence matrix A for the new network can be partitioned as
* [a0 : ^3 (2.10)
where ® submatrix of the bus incidence matrix corresponding to the added element. Subscripts p and q in (2,10) refer to the pth row and the qth row
respectively. essentially shows how the new element is incident to the partial network. Making use of (2.10) in (2.8) and (2.9), for the new network,
the bus variables and primitive variables are related
as
A =
f 0
0A.o i
»
♦ •I 0
aAo *0 + % *«
A ^ v o Bus
jr ® w*1 VBus •*3 us
(2.11)(2,12)(2.13)
For the new network, the primitive equations in mixed form are as shown in (2.7). Now to get an explicit
relation between ?gug and IBug the other variables vQ, iQ, va and ia are to be eliminated from the set of equations (2.7), (2.11), (2.12) and (2.13).
22
Elimination of vQ, iQ and va from the above set of
equation gives** -IT
A_ z** A*0 00 0 ^l^o yoa ^aa VBus*b«7
*(% +^aa ^ao ^o ^ y “1Jaa hi _
ss0
(2.14)
Symbolically (2.14) can be written as%
B11 B12 M— —.vBus Bus1 sc .
_B21 B22_ ia 0.(2.15)
Finally eliminating the variable ±a from the above equation
[B11 " B12 ®22-1 B2l^ VBus * l̂US
(2.16)
From the above equation, the bus impedance matrix of the new network can be written as
ZBus B12 B22 1 (2.17)
To evaluate the inverse on the right hand side of(2.17), Householder's special matrix inversion formula^
given by[F+G H K]-1 = F^1 - F”1 GCH**1 + K F"*1 G]**1 K F”1
can be used. This gives(2.18)
ZBus ' B11 1+B11 1 B12^"B21 B11 1 B +B-]% Bn-112^22(2.19)
23
Making use of (2.4) it is observed
Bn i *"1 = [A zJ1 A^]"1 = .11 o oo o ** Bus(o)
Hence the expression for Zg^ reduces to
ZBus * ZBus(o) “ hu&(o) C12 D 1 °21 ^usCo)
whereAo »oa
C,„ * K« +12 1 yactv A
p _ v- T a. o°21 ~ *1 + yJ aa
^us^'21 “Bus"'7 12 yaa
(2.20)
(2.21)
(2.22)
(2.23)
(2.24)
When there is no mutual coupling between the
added element and the elements in the partial network
both y and y become null vectors and hence the bus oa ao
impedance matrix of the new network is given by
JBus Bus(o)
Wo) *1 glT Wo)
K1 . ZBus(o) % + zaa(2.25)
where zaa is the primitive impedance of the added element.
It is to be noted that C^g and Cg^ in (2.22) and
(2.23) need not be the transpose of each other. When the
element considered is a phase shifting component,T Tyoa 4 yao and hence C12 j C21 . Phase shifting components
such as delta-wye transformer, tap changing transformer with
24
complex turns ratio etc. can fee represented fey V-equivalents^ . The primitive admittance matrix of
the phase shifting component is a 2 x 2, unsymmetrie
and singular matrix. The presence of such unsymmetry
in the primitive admittance matrix will make the feus
impedance matrix unsymmetrie. So far in our develop
ment no assumption has feeen made regarding the
symmetry of the primitive impedance and the feus
impedance matrices. Hence (2.21) and (2.25) are
applicafele even when the added element is a phase
shifting component. However, the singular nature ef
the primitive admittance matrix calls for a special computational procedure^ while adding a phase shift
ing component. This is illustrated in the example 4
in Section 2.9.
In the subsequent discussion of this section it is assumed that the added element is not a phase
shifting component.
Addition of a link with mutual coupling
Since the added element is not a phase shiftingm
component, yQa = yao and hence (2.21) is simplified as
7-7 _ ^UBto) °1 °1 ZBus(o)ZBus ZBus(o) -r----------- ~ ~ } <2-26>
°i zbus(o)ci+'1/wwhere
A yn - v j. 0 0<x°1 ~ *1 + yJaa
(2.27)
25
Addition of a link with no~mutual coupling
When there is no mutual coupling between the added link and the elements in the partial network (2,26) reduces to
7 = 7^ . v ZBus(o) \ ZSus(o) (n .% ^BusCo) + Zaa
where zaa is the primitive impedance of the added element.
As a special case, if the added link is between the reference bus and the bus q, then vector reduces to
K1 (2.29)
This vector has to be used in the appropriate equations.
Addition of links with mutual coupling
When a number of links are added at a time, following a similar procedure, we get
ZBuS - *80.(0) - ZBUS(0) °1 C1T *80.(0) (2-30)
where the matricesS = °1T ZBue(o) °1 + yaa*1 <2-^
26
C1 - % + Ao yoain which is the .bus incidence matrix correspondingto the added links. Further, it is to be noticed thaty_ and y_ are now matrices obtained from (2,6).
oce aceAddition of links with no mutual coupling
When the added links are not mutually coupled with any element in the partial network, (2.30) reduces to
ZBus * ZBus(o) "'^BubCo) K1 25 K1 ZBus(o)
wherez = K, Z , . JL, + z1 j3us (o) 1 T aa
(2.33)
(2.34)
in which zaa is the primitive impedance matrix of the added elements.
2.5 ADDITION OF BRANCH(ES)
Let element a be added at the bus p creating a new bus q with the orientation from p to q as shown in Figure 2.3. The bus. incidence matrix for the new network is given by
A *
1 0 1 :
A.
0
i 01 1I 1 0
-410 ( -1
(2.35)
27
FIGURE 2.3: New network due to addition of a' branch.
Defining
(2.36)
(2.37)
let Vs(o)’ Wo) be the bus varlables for the P®1-*181network and V _ , I _ be the bus variables for theq Bus q Busnewly created bus q. Then the bus variables VL andBusI for the new network can be written as Bus
28
VBus(o) ■^Bus(o)
V =vBus\BUS
; ^US =
Xq Busi
Subst itut ing (2 .37) and (2.^6) in (2,8) and (2.9)# we
get
voTAo ^Bus(o) (2.39)
va = E2 ¥Bus(o) Vq Bus
(2.40)
Ao xa + K2 ia “ IBus(c>) (2.41)
•^a "*Xq Bus (2.42)
Again the primitive equation for the new network is
given by (2.7). To obtain the relation between Vx5USand Vs’ other tables T0, i0> and i„ are to be
eliminated from the set of equations (2.7)* (2.39) to
(2.42). After eliminating these variables, we get
-1 TAz A_0 oo 0~(^2+Ao^oa^aa ^ VBus(o)
mm"XBus(o)
tv T “1 a T\_2(K2 +*aa yaoAo > -1”^aa _ -Jq Bus__ -V_ q Bus_
(2.43)
Rearranging (2.43) to express vBus(0) and
explicitly in terms of IBus(0) and I^.^g and making
use of (2.20), we get
29
V , xBus(o) Z , VBus(o) ZBus(o)^12 "'‘Bus (o)
V „__ q Bus_ ^21^8(0) D21ZBus(o)D12+ y^ *q Bus
(2.44)where
A« yrtiVTV * 7 j, Q -Off-Dia K2 + -y^“TV - K T 4- ygQ A°D21 " K2 + y,
(2.45)
(2.46)oca
Hence the bus impedancejof the new network is given by
JBus^Bus (o) ZBus(o) D12
D21ZBus(o) D21ZBus(o)D12 + yaa
(2.47)
where D^2 and are as given by (2.45) and (2.46).
When there is no mutual coupling between the added element and the elements in the partial network, the bus impedance matrix of the new network is computed from
JBusJBus(o) ZBus(o) K2
*2* ZBus(o) K2TzBus(o)E2 + zaa (2.48)
where zaa is the primitive impedance of the addedelement.
It should be noted that D-^ and in (2.45) and (2.46) need not be the transpose of each other.
30
Expressions (2.47) and (2.48) are very general
expressions which can he used even if the added element is a phase shifting component. In the subsequent discussion it is assumed that the added
element is not a phase shifting component.
Addition of a branch with mutual coupling
Since the added element is not a phase shifting component, yoa = yaQT and hence (2.47) reduces to
husJBus(o) ZBus(o) C2
°2 ZBus(o) C2TzBus(o)C2 + yact
wheren = K + -^oa °2 2 + y
Jaa
(2.49)
(2.50)Addition of a branch with no mutual coupling
. When there is no mutual coupling between theadded element and the elements in the partial network
f(2.49) simplifies as
BusJBus (o) ^US ( 0 ) K,K2TzBus(o) K2 ZBus(o)K2+zaa
(2.51)
where zaa is the primitive impedance of the added element.
As a special case, the added branch may be from the reference bus. In such a case the vector
31
reduces to a null vector. This null vector has to be
used in the appropriate equations.
Addition of branches with mutual coupling
Similar procedure can be followed when a
number of branches are added at a time. In such a
case.
JBus
JBus(o) ZBus(o) C2
°2IzBus(o) 02TiWo)02+yaa'1(2.52)
where the matrix
C2 " E2 + Ao yoa y -1 ■aa 4-524 1
(2.53)
in which is the submatrix of the bus incidence
matrix corresponding to the added branches. Further
it is to be noticed ■
obtained from (2.6).
it is to be noticed that y and y„ are now matricesoa aa
Addition of branches with no mutual coupling
♦When the added branches are not mutually
coupled with any element in the partial network,
(2.52) reduces toZBus(o) ZBus(o) K2
JBus.K2 ZBus(o) K2*ZBus(o)K2 + zaa
(2.54)
in which zaa is the primitive impedance matrix of the
added elements.
*(>
32
2.6 REMOVAL OP LINK(S)
Removal of a link with mutual couplingRemoval of a link from a network makes the
•self and mutual admittances corresponding to the element "being removed as zero and hence can be simulated by adding a fictitious link in the network. This fictitious link has self and mutual admittance as the negative of those of the link to be removed and the mutual admittance between the fictitious link and the link to be removed as zero. To add this fictitious link, the algorithm for addition of a link can be used.
Because of the zero mutual coupling betweenthe fictitious link and the link to be removed, itis to be noticed that the product of the bus incidence
*matrix of the network and the mutual admittance vector corresponding to the fictitious element is equal to
network with element a removed and y is the vectoroaShowing the mutual admittance between element a andthe other elements in the network. Furthermore, theself admittance of the fictitious element is -y •aaTherefore the vector can still be computed from (2.27). Thus when a link a is*removed from the network whose bus impedance matrix is Z£Ug(0)r the modified bus impedance matrix ZgUj5 is given by
*-Aoyoa where Aq is the bus incidence trix of the
bUS7 - ^Bus(o) ^1 ^Bus(o)SusCo) “ 2 C -fl /v )
C1 ZBus(o) C1(2.55)
33
where%
Ao yoa yaa
(2.56)
Removal of a link with no mutual couplingIf the link to he removed is not mutually
coupled with any other element in the network, (2.55) reduces to
5?7,-7 - ZBus(o) K1 K1 husjo)i3us “ ^Bus(o) m
K1 ZBus(o) K1 ” zaa(2.57)
where 2(xa is the primitive impedance of the element to he removed.
Removal of links with mutual couplingThe algorithm for removal of links can he
derived along the similar line and is given hy
ZBue " ZBus(o) - *Bua(o) °1 ^ «L* <2-58>where
2 = °1T ZBus(o) °1 - ^aa'1 <2-59)
°1 * h + Ao yoa yaa'1 <2-60>
in which y_ is the matrix obtained from (2.6). Further,aa
it is to he noticed that K-. and y in (2.60) are no• OCXlonger vectors hut are matrices.
34
Removal of links with, no mutual coupling
When the links to he removed are not mutually
coupled with any element in the partial network,
(2.58) reduces to
ZBus * ZBus(o) * Wo) *1 ^ *1 ZBus(o) <2-61> where
fi = V Wo) *1 - z«« 1 (2-62)in which zaa is the primitive impedance matrix of the
links to he removed.
2.7 HINTS FOR IMPLEMENTATION OF THE ALGORITHM
The following remarks Serve to clarify the
implementation of the algorithm for numerical problems.
Consider the case of addition of a link. If the
*dded element has no mutual coupling, the vector yoa
becomes a null vector. In such a case, while computing
A0 yoa the hue incidence matrix A„ need not he consi-
dered at all. If the added element is mutually coupled,
then it is enough to consider the elements in the
coupled group alone since the components of yQa corres
ponding to the elements that are not in the coupled
group are zero. While computing AQ 7oar reduced yQa
vector denoted by yQa and reduced AQ matrix denoted
by Aq can be used. The rows of yQa and the columns«na»
of Aq correspond to the elements having mutual coupling
with the added element a. The order of the matrix to
55¥
■be inverted to find yQa and yaa is now limited to the number of elements in the coupled group. All these remarks are applicable not only for the case of addition of a link but also for all other cases discussed.
A word of caution is necessary regarding the orientations of the elements and the sign of the mutual impedances in the primitive impedance matrix of the coupled group. In circuit theory the proper dot convention is normally followed to specify the sign of mutual impedances. But in power systems mutual coupling arises w|ien two transmission lines
* 'V»run in parallel for a cirisiderable distance. The sign of mutual impedance depends on the orientations assigned to them in the oriented graph. If the elements have alike orientations, then the mutual impedance will have positive value and if the elements have opposite orientations then the corresponding mutual impedance will have negative value.
All these points are brought out in the numerical examples given in the Section 2.9.
2.8 SPECIAL CASES Removal of radial line
When an element corresponding to a radial line is removed one bus gets isolated and the number of buses in the network is reduced by one. From (2.49)
36
14t can be inferred that the modified bus impedance matrix due to removal of radial line can be obtained from the original bus impedance matrix by simply deleting the row and column corresponding to the isolated bus. If the isolated bus is the reference bus itself, then the bus impedance matrix of the new network is indefinite.
Isolated bus
When a set of elements are removed, some buses may become- isolated. For example let us assume that when r numbers of elements are removed, one bus becomes isolated. Iheh; from these r elements, it is possible to pick out ^elements which on removal will maintain the network graph as connected one. Removing these r-1 elements first, ty using the removal of links algorithm, the bus impedance matrix is modified. In this resulting bus impedance matrix, the row and column corresponding to the bus which would become isolated is now deleted to get the final modified bus impedance matrix due to the removal of all the r elements*
Parameter changes
When the parameter of an element is changed, the bus impedance matrix can be modified by simultaneously removing the element with the old parameter and adding an element with the revised parameter. Ihis
particular case has been illustrated through the
numerical example 2 in the following section.
2.9 NUMERICAL EXAMPLES
The networks that are considered in the first
three numerical examples are taken from reference 32
Example 1-
Consider the network shown in Figure 2.4,
whose line impedances are marked in the figure. The
bus impedance matrix of this network, is
1 2 3.4
1 0.0862 0.0372 0.0319 0.0660
2 0.0372 0.2074 0.1064 0.0532
3 0.0319 0.1064 0.2340 0.1170
4 0.0660 0.0532 0.1170 0.2085
Calculate the modified bus impedance matrix
when the line f is removed.
Solution:
It is to be noticed that the element f ‘lias
mutual coupling with the element d which?has mutual■m
coupling with the element e. Thus the coupled group
to be considered consists of elements d, e and f.
Let us assume the orientations of the elements d, e
and f as shown in the oriented graph in Figure 24>.
Then the primitive impedance matrix of the coupled
group is
38
14 3
FIGURE 2.4: Network of example 1.
e
*
* k#
*
39
d e f
d 0.4 0.1 -0.2
e 0.1 0.2 0.0
f -0.2 0.0 0.3
Taking the inverse of this, the primitive admittance
matrix of the coupled group is obtained as
d
e
f
Therefore
oa
d
4-6153
■2.3077
3.0769
3.0769
,^1.5385,
e f
-2.3077 3.0769
6.1538 -1.5385
-1.5385 5.3846
’’ yaa " 5-3846
Further, from the oriented graph shown in Figure 2.5
1 1" ~0~
-1 0 0—
0 0; % «
1
0 -1 f» H
Now substituting yQa, yaa, AQ and in (2.56)1
T** —90 1
*
*
r— -a» 1 1
■■ mr.c
*n;•*+w -*■
0.286.
0 1 -1 0 “3.0769 -0.576
1+ 5.3846 «
0 0 ■ -1.5385 ,1.*0♦
-1*-* -mA ' 0 -1mm*
-0.714
40*
Substituting the given Zgu8(0) and the calculated yaa and C1 in (2.55)> the modified bus impedance matrixis computed as
1 2 3 41 0.09 0.05 0.0 0.082 0.05 0.25 0.0 0.15
z =Bus ^ 0.0 0.0 0.5 0.04 0.08 0.15 0.0 0.26
Example 2For the network given in example 1, calculate
the modified bus impedance matrix when the self impedance of the line f is changed from 0.3 to 0.24 and also the mutual impedance between the lines d and f is changed from 0.2 to 0.0.
SolutionsThis case can be treated as simultaneous
removal of the line f with old impedance value (which can be viewed as the addition of fictitious link) and the addition of a line f' with new impedance yalue. ,Thus this belongs to the case of addition of Itfiks ,
*with mutual coupling. The orientations of the elements * in the coupled group are now assumed. Figure 2*6 shows the oriented graph. Then the primitive impedance matrix of the coupled group is
41f •
d e f t*'
'd ^0.4 0.1 -0.2 0,0e 0.1 0.2 0.0 0.0f -0.2 0.0 0.3 0.0f' 0.0 0.0 0.0 0.24
Inverting this, the primitive admittance matrixthe coupled group is obtained as
d e f f •d *-4.6153 -2.3077 3.0769 0.0
e -2.3077 6.1538 -1.5585 0.0f 3.0769 -1.5385 5.3846 0.0f * 0.0 0.0 0.0 4.1667
Remembering that f is the element to be removed, now v. c have- ♦ * "
“^3.0769 0.0 "-5.3846 0.0yoa 1.5385 0.0
•9 0.0 4.1667
Also fr^m’ the Oriented graph shown in Figure 2.6,
42
Ao =
• [HP**- **•
r 1 0 0-l 0
ti
M 0 0
0 0 1 1
0 -1 -1 -1mm*. --- mmm
For this case
-1aa
-0.186 0.0
0.0 0.240
Substituting the values of yoa, y„e' , A0 end ^ in (2.32)
Ci -
0.2855 0.0>0.5715 0.01.0 1.0■0.7140 -1.0
-1Knowing zgue(0)» 0^ and yaa 9 irom (2*3l) z is obtainedas
Z as
Hence
a-iz
•0.0369 0.1428
0.1428 0.4485
-12.230 3.890
3.890 1.007
43
Finally substituting the values of given andcalculated and z“^ in (2,30) the modified bus
impedance matrix is obtained as
1 2 3 41 0.0836 0.0420 0.0400 0.05922 0.0420 0.2400 0.5000 0.07403 0.0400 0,5000 0.2500 0.13004 0.0592 0.0740 0.1300 0.1920
mm
Example 5
Hie bus impedance matrix of the network shown in Figure 2.7 is
1 2 3 41 0.0831 0.0414 0.0430 0.0570*
2 0.0414 0.2392 0.0538 0.07213 0.0430 0.0538 0.2312 0.13974 0.0570 0.0721 0.1397 0.1810
Calculate the modified bus impedance matrix when the lines d and g are removed. Values that are marked in the figure are impedances.
*Solution:This case belongs to the removal of links with
mutual coupling discussed in Section 2.6, It is to be noticed that the element d is coupled with element e and the element g is coupled with element f. Hence
44
elements d and g as a group have mutual coupling with
elements e and f. let us assume the orientations of the coupled elements d, e, f and g as shown in the oriented graph in Figure 2.8. The primitive impedance
FIGURE 2.7: Network of example 3-
FIGURE 2.8: Oriented graph of example %
45
matrix of the coupled group is
d e f gd
MM*w
0.4 0.1 0.0 0.0 ~
e 0.1 0.2 OOO 0.0f 0.0 0.0 0.24 0.1g 0.0 0.0 0.1 0.24
Inverting this, the primitive admittance matrix of the coupled group is obtained as
d e f gd 2.85 71 -1.4286 0.0 0.0e -1.4286 5.7143 0.0 0.0f 0.0 0.0 5.0420 -2.1008g 0.0 0.0 -2.1008 5.0420
Therefore""-1.4286 0.0 "2.8571 0.0
J =0.0 -2.1008 0.0 5.0420
Prom the oriented graph shown in Figure 2.8
~i o~ ~1 0 ~0 0 -1 0; Kj =0 1 0 1
1 H 1 H 0 -1——- —ft
46
For this case
-1acc
0.3504 0,0
0.0 0*1987-1Substituting the values of AQ, yQa and yaa * in
(2.60)
0.5 0.01.0 0.00.0 0.5820.5 •-0.582
and y -1
obtained as
Hence
z
•0.12867 -0.00721
■0.00721 -0.15170
-7.800 0.37l"
0*371 -6.620
Finally substituting the values of the given 2gue(0) and the calculated and in (2.58), we get the modified bus_ impedance matrix as
Bus
0.0904 0.0 0.0481 0.07110.0 0.4998 0
•0 0.0
0.0481 0.0 0.2596 0.14420.0711 0.0 0.1442 0.2129
47
Example 4Construct the bus impedance matrix of the
*37 *38network shown in Figure 2.5r * , which contains aphase shifting component. The values marked in the figure are impedances.
Let us create bus 3 such that element having impedance of 0.07+30.1 be between buses 1 and 3 and the phase shifting component be between buses 3 and 2. Knowing the transformation ratio (t = 0.6) and the angle of phase shift (9 * 30°), the V-equivalent of the phase shifting component^'J shown in Figure 2.10 can be obtained. The primitive admittance matrix corresponding to the elements d and e is computed as follows:
d e d ey32 _hz~ d 7.650-314.36 -8.30+15.17t2 CD1
y32mm mmmmbmm
t/ei*.
y32 e 0.333+3 9.778 2.76-35.16
Now the bus impedance matrix of the network can be constructed as follows:
Step 1: Add the element between the buses 0 and 1.1
ZBus * 1 C i-0^0-0 J
48
FIGURE 2.9: Network of example 4.
3 2
FIGURE 2.10: V-equivalent of the phase shifting component.
49
Step 2: Add the element between the buses 1 and 2,1 2 l.O+JO.O l.O+jO.Ol.O+jO.O 1.06+jO.20
Step 3s Add the element between the buses 1 and 312 3
"l.O+jO.O l.O+jO.O l.O+jO.O
Bus
123
l.O+jO.O 1.06+j0.2 l.O+jO.O1.0+30.0 1.0+30.Q 1.07+30.1
Step 4: Add the element d (Figure 2.10).
This is a link between buses 0 and 3# and it has no mutual coupling with the partial network.Since the primitive admittance matrix corresponding to elements d and e is singular, the self impedance of the element d is infinity. Adding a link of infinite impedance will not cause any change in the bus impedance
matrix. Thus Zgu remains same as calculated in step 3.
Step 5; Add the element e (Figure 2.10).
This is a link which has unsymmetric mutual couplings with the element d in the partial network. For this case
1 “0“" > 1 0Kl" 2 1 f A0 = ^ 0
3 0 3 _JL_
50
yQa = -8.3+35.17
yao = 0.333+j9.778
yaa *= 2.76-35.16
Substituting these values in (2.22), (2.23) and (2.24)
C12
0.0+30.0 1.0+30.0 4..44-jO. 832
°21 ® CO.O+jO.O -1.0-30.0 1.44-30.832]
D *= 1.2361+30.6264
Now from (2.21) ZBug is computed as
Bus
0.4222+30.29530.5601+30.25700.2827+30.3048
0.3777+30.43090.5912+30.56520.2179+30.4634
0.3521+30.17480.4998+30.16570.2789+30.2545
Step 6: Add the element between the buses 0 and 2.
This is a link with no mutual coupling. For
this case
oca O.O+31.O
Substituting these values in (2,28)
511
zBus0.2117+JO.28020.2589+j0.2620
3 0.1057+JO.2398
2
0.1202+JO.3207 0.2112+JO.4409 0.0240+jO.3051
30.1719+jO.1818 0.2445+JO.1982 0.1225+JO.215 9_
Deleting the row and column corresponding to the hue 3>the additional bus that has been created, the busimpedance matrix of the given network is obtained as
1 2 0.2117+JO. 2802 0.1202+JO. 3207*"
0.2589+JO.2620 0.2112+jO.4409
2.10 CONCLUSIONSIn this chapter a generalised bus impedance
algorithm has been developed based on certain fundamental concepts of graph theory. The bus incidence matrix which shows how various elements are interconnected is essentially used at various stages. In other words this development exploits the topology of the network. The merits of this algorithm are summarised below.
i) In contrast to the algorithm given in reference 29^ the present algorithm for the addition of a link needs only one step instead of a two step procedure.
ii) Mutual couplings between the elements of the network do not pose any problem either at the addition or removal stage.
52
iii) The algorithm is amenable for the addition
of link and branch either one at a time or in groups.
iv) The algorithm is comprehensive so as to
include the effects of phase shifting components,
changes in network parameters etc*