2 Power System Network Matrices I

Transcription

1 Power System Analysis Power System Network Matrices I. INRODUCION he various terms used in Graph heory are presented in this chapter. Formulation of different network matrices are discussed. Primitive impedance and admittance matrices are explained. his chapter also deals with the formulation of Y BUS.. IMPORAN DEFINIIONS IN GRAPH HEORY A graph shows the geometrical interconnection of the elements of a network. A subgraph is any subset of elements of the graph. A path is a subgraph of connected elements with not more than two elements connected to any one node. A graph is connected if and only if there is a path between every pair of nodes. If each element of the connected graph is assigned a direction then it is an oriented graph. A representation of a power system and the corresponding graph and oriented graph are shown in Figs..,. and.. A tree is a connected subgraph of a connected graph having all the nodes of the graph but without any closed path (or) loop. he elements of a tree are called tree branches (or) twigs and are denoted by thick lines. G G Fig..: Single-line diagram of power system.

2 Power System Analysis (ref.) Fig..: ree of power system, with tree branches [,, ].. BASIC LOOPS (OR) FUNDAMENAL LOOPS Loops which contain only one link are independent and are called basic loops as shown in Fig... In other words, whenever a link element is added to the existing tree, basic loops or fundamental loops can be obtained. No. of fundamental loops = No. of links E D (ref.) Fig..: Basic loops, i th tree branches [,, ].. BASIC CU SES (OR) FUNDAMENAL CU SES A cut set is a set of elements that, if removed, divides a connected subgraphs. In other words, a basic or fundamental cut set of the graph is the set of elements consisting of only one branch (or) twig and minimal number of links (or) chords as shown in Fig..6. No. of basic cut sets = No. of twigs

3 Power System Network Matrices I C A (ref.) B Fig..6: Basic cut set, with tree branches, [,, ].. INCIDENCE MARICES Every element of a graph is incident between any two nodes. Incidence matrices give the information about incidence of elements may be incident to loops, cut sets etc. and this information is furnished in a matrix, known as incidence matrix as follows:.. Element-Node Incidence Matrix ˆ (A) he incidence matrix (A) describes whether an element is incident to a particular node (or) not. he elements of the matrix are as follows: a ij = If the i th element is incident to and oriented away from the j th node. a ij = If the i th element is incident to and oriented towards from the j th node. a ij = If the i th element is not incident to the j th node. he dimension of the matrix is e n where e is the number of elements and n is the number of nodes in the graph. he element-node incidence matrix (A) ˆ for the graph is shown in Fig..7. (ref.) Fig..7: ree of power system, with tree branches, [,, ].

4 8 Power System Analysis he transpose of this matrix is K = he branch-path incidence matrix and the submatrix A b relate the branches to paths and branches to buses respectively. Since, there is a one-to-one correspondence between paths and buses, we can prove that the relation A b K is an unity matrix. = Hence, where U is the unity matrix b A K = U... (.) we can write K = A b.. Basic Cut Set Incidence Matrix (B) he incidence of elements to basic cut sets of a connected graph is shown by the basic (or) fundamental cut set incidence matrix (B). he elements of this matrix are as follows: b ij = If the i th element is incident to and oriented in the same direction as the j th basic cut set. b ij = If the i th element is incident to and oriented in the opposite direction as the j th basic cut set. b ij = If the i th element is not incident to the j th basic cut set. C B A (ref.) Fig..9: Basic cut set.

5 Power System Network Matrices I 9 he basic cut est incidence matrix, of dimension e B, for the graph is shown in Fig..9. Matrix B can be partitioned into submatrices U b and B l where the rows of U b correspond to branches and the rows of B l to links. he partitioned matrix is shown above. e Basic A cut sets B C [B] = Branches (B b or U b ) Links ) (B = U b B... (.) Fig..9 he identity matrix U b shows the one-to-one correspondence of the branches and basic cut sets. he submatrix B l can be obtained from the bus incidence matrix A. he incidence of the links to buses is shown by the submatrix, A l and the incidence of branches to buses is shown by the sub-matrix A b. But B l Ab shows the incidence of links to buses, i.e., B A = A B = A A B A K A K l b l l l l = l b = b (.).. Augmented Cut Set Incidence Matrix (B) he basic cut set incidence matrix is of the size e b, therefore, a non-square matrix and hence, inverse does not exist. In other words, B is a singular matrix. In order to make the matrix B a nonsingular matrix, we augment the number of columns equal to the number of links by adding fictitious cut sets known as tie-cut sets, which contain only links. he tie-cut sets are added for the graph in Fig.. and are shown as, E D C A (ref.) B Fig..: Augmented basic cut sets.

6 Power System Analysis e e Basic cut sets A B C ie cut sets D E ^ [B] = Branches Links U b = B U... (.) his is a square matrix of dimension e e and is non-singular. Matrix (B) above. can be partitioned as..6 Basic Loop Incidence Matrix (C) he incidence of elements to basic loops of a connected graph is shown by the basic loop incidence matrix (C). he elements of this matrix are as follows: cij = If the i th element is incident to and oriented in the same direction as the j th basic loop. cij = If the i th element is incident to and oriented in the opposite direction as the j th basic loop. cij = If the i th element is not incident to the j th basic loop. he basic loop incidence matrix, of dimension e l, for the graph shown in Fig.. is, E D (ref.) Fig..: Basic loops.

7 Power System Network Matrices I where v, e and i are the column matrices of size e and e is the number of elements. z is a square matrix of size e e. he matrix z is known as primitive impedance matrix. A diagonal element of the matrix z of the primitive network is the self-impedance z ik ik. An off-diagonal element is the mutual impedance z ik ps between the elements ik and ps..6. Primitive Network in Admittance Form Let the element i k connected between the two nodes i and k. his is shown in Fig... j ik i V i y ik i ik + j ik = V V V k k ik i k Fig..: Primitive network. where V i, V k i th and k th node voltages respectively v ik, V i V ik Voltage across the element i k j ik Source current in parallel with element i k i ik Self-admittance of the element i k i k Current flowing through the element i k Hence, current flowing through the element, iik + jik = yik ν ik (.) he above Eq. (.), for all the elements in a condensed form can be written as, i+ j= y ν... (.) Where i, j and v are the column matrices of size e and matrix y is a square matrix of size e e. he matrix y is known as primitive admittance matrix. he diagonal elements of the matrix y ik ik represents self-admittances and off-diagonal elements of the matrix y ik ps represents the mutual admittances of the elements ik and ps. If there is no mutual coupling between the elements, the primitive admittance matrix y can be obtained by inverting the primitive impedance matrix z. he matrices z and y are diagonal matrices. In the case, the self-impedances are equal to the reciprocals of the corresponding self-admittances.

8 Power System Analysis e e e e Z Z Z [z] = (or) Z Z [y] = Z Z Z Z Z If the buses and have mutual element then the corresponding primitive impedance matrix is shown below: e e e e Z Z [z] = Z Z [y] = [z] = Z Z (or) Z Z Z Z Z Z Z Z Z Z Z Z.7 NEWORK EQUAIONS AND NEWORK MARICES A power system is a big complex network. herefore, we require transforming the primitive network matrices to be developed either in the bus frame, branch frame or loop frame of reference. In these frames of references network matrices can be written as, where VBus = ZBus I Bus (or) VBR = ZBR I BR (or) VLoop = ZLoop ILoop (.) V Bus Matrix contains bus voltages I Bus Matrix represents the injected currents into the buses Z Bus Bus impedance matrix For an n-bus power system, the dimensions of these matrices are n, n, and n, respectively. he network equations in admittance form can be written as, Bus Bus Bus Bus Bus I = Y V where Y = Z (.) he branch frame of reference performance equations can be written as, BR BR BR BR BR I = Y V where Y = Z (.) Here, V BR and I BR represent branch voltages and currents. Z BR and Y BR represent branch impedance and admittance matrices respectively. he dimensions of these matrices depend upon the number of branches in a graph of a given power system.

9 Power System Network Matrices I Finally, the loop frame of reference V Loop denotes the basic loop voltages, I Loop represents the basic loop currents and Z Loop is the loop impedance matrix. In the admittance form, the network equations can be written as, Loop Loop Loop Loop Loop I = Y V where Y = Z (.) he size of the matrices in the network equation based on loop frame depends on the number of basic links or loops in a graph of a given power system..8 BUS ADMIANCE MARIX Bus admittance matrix (Y Bus ) for an n-bus power system is square matrix of size n n. he diagonal elements represent the self or short circuit driving point admittances with respect to each bus. he off-diagonal elements are the short circuit transfer admittances (or) the admittances common between any two number of buses. In other words, the diagonal element y ii of the Y Bus is the total admittance value with respect to the i th bus and y ik is the value of the admittance that is present between i th and k th buses. Y Bus can be obtained by the following methods:. Direct inspection method. Step-by-step procedure. Singular transformation. Non-singular transformation.9 PROCEDURE OF FORMULAION OF Y Bus.9. By Direct Inspection Method Formulation of Y Bus by direct inspection method is suitable for the small size networks. In this method the Y Bus matrix is developed simply by inspecting structure of the network without developing any kind of equations. Let us consider a -bus power system shown in figure below: z z y z y y

10 8 Power System Analysis Since [B] * i is zero because, algebraic sum of all the currents meeting at a node is zero. he source current matrix [j] can be partitioned into, j b j = jl... (.8) Where j b is the source acting in parallel across the branches herefore, Eq. (.) is modified as, b B j = B j = IBR... (.9) IBR = B y B VBR... (.) I = Y V (or) I = Y V... (.) or BR BR BR BR BR BR where YBR = B y B, and BR { } BR Z = Y = B y B... (.).9. Loop Impedance Matrix he relation between element currents and the loop currents can be written as i = C I Loop... (.) he primitive equation in the impedance form is ν+ e= z i... (.) Multiplying both sides by [C], where C is the basic loop incidence matrix, C ν+ C e= C z i... (.) Since [C] * v is zero because, algebraic sum of all voltages in a closed loop is zero. Similarly, [C] * e gives the algebraic sum of series source voltages around each basic loop. herefore, Eq.(.) is modified as, ELoop = C e... (.6)

11 Power System Network Matrices I 9 ELoop = C z C ILoop... (.7) or E Loop = [Z Loop ] ILoop (or) ELoop = ZLoop ILoop... (.8) where ZLoop = C z C, and Loop Loop { } Y = Z = C z C... (.9). NUMERICAL PROBLEMS P... For the power system network shown in Fig..: (i) draw the oriented graph, (ii) formulate element node incidence matrix, bus incidence matrix, basic cut set incidence matrix, augmented cut set incidence matrix, basic loop incidence matrix, augmented loop incidence matrix and path incidence matrix, (iii) formulate primitive admittance matrix, and (iv) formulate Y Bus, Y Br and Z Loop. ake reactance of each element j. pu. Load G Load L L L G Fig..: Power system network. Solution: For the given power system network graph, oriented graph and tree graph are shown in Figs (.6), (.7) and (.8) respectively. No. of elements, e = No. of nodes, n = No. of buses = n = =

12 Power System Analysis L L L G G Fig..6: Connected graph of power system. L L L G G (ref.) Fig..7: Oriented graph of power system. (ref.) Fig..8: ree of power system with tree branches, [,, ]. No. of branches, b = n = = No. of links, l = e b= e (n ) = e n+ = =

13 Power System Network Matrices I Element node incidence matrix, (A) e n [A] = ^ Bus incidence matrix, (A) e ) n ( Branches (A ) b Links (A ) l A l A b [A] = = Bus cut set incidence matrix, (B) C B A e Basic cut sets Branches (B or U ) b b Links (B ) l [B] = B U b l = Fig..8a: ree of power system with tree branches [,, ]. (.) ref C A B

14 Power System Analysis Augmented cut set incidence matrix, (B) e e Basic cut sets A B C ie cut sets D E ^ [B] = Branches Links U b O = B l U l E A D C B (ref.) Fig..8b: ree of power system with tree branches, [,, ]. Basic loop incidence matrix, (C) [C] = e Basic loops D E Branches Links = C b U l Fig..8c: ree of power system with tree branches, [,, ]. D E

15 Power System Network Matrices I L G G L Load Solution: For the given power system network graph, oriented graph and tree graph are shown in Figs..,. and. respectively. No. of elements, e = No. of nodes, n = No. of buses = n = No. of branches, b = n = No. of links, l = e n + = Fig..9: Power system network. Load L L L L G G G G (ref.) (ref.) Fig..: Connected graph. Fig..: Oriented graph. L L G G (ref.) Fig..: ree graph.

16 8 Power System Analysis From given data, the self-impedances are he mutual impedances are Z = Z = j. Z = Z = j. Z = Z = Z = Z = Z = Z = ; Z = Z = j. Y = = = j = Y ( Z = Z ) Z j. Y Y Z Z j. j. j.97 j. Y Y = Z Z = = j. j. j. j.97 [ Z] j. j. = j. j. j. j. and [ Y] j j = j.97 j. j. j.97 P... Consider the power system as shown in Fig... Each generator and the line impedance of (. + j.) pu and (. + j.) pu respectively. Neglecting line charging admittances, form Y Bus by direct inspection method. Solution: he admittance of each generator is, yg = yg = yg = =. j. (. + j.) he admittance of each line is yl = yl = yl = = j (. + j.) Since, power system has three buses (n = ), the size of the Y Bus is. he elements of Y Bus by direct inspection is as follows:

17 Power System Network Matrices I 9 Load G G Load L L L G Load Fig..: Power system network. Diagonal elements: Off-diagonal elements: Similarly, Y = yg + yl + yl =. j. + j + j =. j. Y = yg + yl + yl =. j. + j + j =. j. Y = yg + yl + yl =. j. + j + j =. j. Y = Y = y L = ( j) = + j Y = Y = y L =( j) = + j Y = Y = y L = ( j) = + j Y Y Y. j. + j + j [ YBus ] = Y Y Y j. j. j = + + Y Y Y + j + j. j. P... Using data in the above problem, obtain Y Bus by singular transformation method.

18 Power System Analysis Solution: For the given power system network, the tree graph with tree branches [,,] is shown in Fig... 6 (ref.) G L G L L Fig..: ree of power system. e n 6 [A] = ^ = [ ] A [A] = = he primitive impedance matrix can be written as,

19 Power System Network Matrices I [ z] 6. + j.. j j. =. + j.. + j j. Primitive admittance matrix can be written as, 6. j.. j.. j. = = j j 6 j [ y] [ z] herefore, the Y Bus is. j. + j + j [ Y Bus ] = [A] [ y] [ A] = j. j. j j + j. j. P... For the power system shown in Fig.., build Y Bus matrix using: (i) by direct inspection, and (ii) singular transformation. he branch impedances of the lines are as follows: Line Impedances (Ω) L + j + j + j + j + j L L L L Fig..: Power system network.

20 Power System Network Matrices I Y = Y = Y = Y = Y = Y = Y = Y = Number of buses are, therefore, Y is matrix. [ Y ] Y Y Y Y Y Y Y Y Bus Bus = Y Y Y Y Y Y Y Y.7 j j j j.9. j j j.8 =.76 + j.9. j.. + j j j.8. + j.. j.9 (ii) By using singular transformation For the given power system network, the tree graph with tree branches [,,,] is shown in Fig..6. L L 6 L L L (ref.) L Fig..6: ree graph. ^ [A] = e 6 n =

21 Power System Analysis = = [ A ] [A] he primitive impedance matrix can be written as, [ z] 6 + j j + + j = + j + j 6 + j [] y [] z.9 j..88 j.7.86 j.8 = =. j..8 j j.9 [ ] Y = [A] [y] [A] = Bus.7 j j j j.9. j j j j.9. j.. + j j j.8. + j.. j.9 P..6. Form Y Bus for the network, using: (i) direct inspection method and (ii) singular transformation. JNU (Nov 8)

22 Power System Network Matrices I Element Positive sequence reactance Solution: (i) By direct inspection method: For the given data, E A. E B. A B. B C. A D. C F.7 D F. y EA = = = j z j. EA y EB = = = j z j. EB y AB = = = j z j. AB ybc = = = j. z j. BC y AD = = = j z j. AD ycf = = = j.87 z j.7 CF y DF = = = j z j. DF Assume that node E as reference,

23 6 Power System Analysis Diagonal elements: YAA = yae + yab + yad =j YBB = ybe + yba + ybc = j78. YCC = ycb + ycf = j7.69 YDD = yda + ydf =j6 Off-diagonal elements: YFF = yfc + yfd = j.87 YAB = YBA = yab = j Y = Y = AC CA YAD = YDA = yad = j Y = Y = AF FA YBC = YCB = ybc = j. Y = Y = BD BF DB Y = Y = CD FB Y = Y = DC YCF = YFC = ycf = j.87 YDF = YFD = ydf = j j j j j j78. j. [ YBus ] = j. j7.6 j.9 j j6 j j.9 j j.9

24 Power System Network Matrices I 9 P..7. From Y Bus for the network, using: (i) by direct inspection method and (ii) singular transformation. JNU (8) Element Positive sequence reactance j. j. j. j. j.8 j.8 Solution: (i) By direct inspection method: For the given data, y = = = j. = y z j. y = = = j. = y z j. y = = = j = y z j. y = = = j = y z j. y = = = j. = y z j.8 Diagonal elements: y = = = j. = y z j.8 Y = y + y = j. Y = y + y + y =j8. Y = y + y + y = j8.7 Y = y + y + y = j.

25 6 Power System Analysis Off-diagonal elements: Y = Y = y = j. Y = Y = y = j. Y = Y = Y = Y = y = j. Y = Y = y = j Y = Y = y = j [ Y ] j. j. j. j. j8. j. j. j. j. j8.7 j. j. j. j. Bus = (ii) Singular transformation: For the given data, power system network tree graph is shown in Fig..8. Assume node as reference, 6 z = j.8 z = j. z = j. z = j.8 z = j. ref. z = j.8 Fig..8: ree graph. A = 6

26 [ A ] [A] Power System Network Matrices I 6 = = [ z] 6 j.8 j. j. = j. j. 6 j.8 6 j. j j. = z = j. j 6 j. [ y] [ ] [ Y ] [A] [ y] [ A] j. j. j. j. j8. j. j. j. j. j8.7 j. j. j. j. Bus = = P..8. Form the Y Bus for the given network. JNU (8, 9) Bus code Z pq y pq. + j.8. + j..8 + j.. + j..6 + j.8. + j..6 + j.8. + j.. + j.. + j.. + j.. + j..8 + j.. + j.

27 6 Power System Analysis Solution: he admittance to ground is obtained by adding the admittances to ground at that bus Bus code, p Admittance to ground, y p j. + j. = j. = y j. + j. + j. + j. = j.8 = y j. + j. + j. = j. = y j. + j. + j. = j. = y j. + j. = j. = y he primitive admittances are: y = y = = =.9 j.767 z. + j.8 y = y = = =. j.7 = y = y z.8 + j. y = y = = =.6667 j = y = y z.6 + j.8 y = y = = =. j7. z. + j. y = y = = = j z.+ j. he diagonal elements are: Y = y + y + y =.9 j.697 Y = y + y + y + y + y = j9.97 Y = y + y + y + y =.967 j8.8 Y = y + y + y + y =.967 j8.8 Y = y + y + y =.7 j.9

28 Power System Network Matrices I 6 he off-diagonal elements are: Y = Y = y =.9 + j.767 Y = Y = y =. + j.7 = Y = Y Y = Y = y = j = Y = Y Y = Y = y =. + j7. Y = Y = y = + j Y = Y = Y = Y = Y = Y = herefore, the Y Bus is:.9 j j j j j j j. + j7. Y Bus =. + j j.967 j8.8 + j j + j.967 j j.7. + j7.. + j.7.7 j.9 P..9. For the power system network shown in Fig..9, formulate Y Bus by singular transformation method. he self and mutual reactances in pu are indicated. ake bus () as reference. Solution: he oriented network is shown in Fig.. and tree graph is shown in Fig... j. j. j. j. j. j. Fig..9: ree graph.

29 Power System Network Matrices I 6 A = e n A = [A] = [ ] j.67 j. Y Bus = [A] y [A] = j. j6.9 j.87 j.87 j6.87 [ ] [ ] P... Fig.. shows a bus power system network. he line impedances are given below: Line (bus to bus): () () Impedance in pu:. + j.. + j.. + j.. + j. () () Ref. Fig..: ree graph. Formulate Y Bus for the following cases:. Assume that the line shown dotted between bus and bus i.e., line () is not present.. A new line () is connected and no mutual coupling with other lines.. When line () which has mutual impedance of (. + j.) pu with line () is connected.

30 66 Power System Analysis Solution: Case (): By direct inspection: y = y = = =.9 j9.6 = y = y z. + j. y = y = = =.769 j.86 z. + j. Y = y + y =.69 j.66 Y = y + y =.69 j.66 Y = Y = y = j.86 Y.69 j j.86 = j j.66 Bus By singular transformation: A= = = [ A] [A] (. + j.) (. + j.) (. + j.) Ref Fig..(a): ree graph, [, ].. + j. z =. j j. [ ] [ ].9 j9.6 y =.9 j j j j.86 Y Bus = [A] [y] [A] = j j.66 [ ]

31 Power System Network Matrices I 67 Case (): By direct inspection: y = y = = =.9 j9.6 = y = y z. + j. y, = y, = = =.769 j.86 z. + j., y, = y, = = =. j. z. + j., Y = y + y + y =.8 j6.78,, Y = y + y + y =.8 j6.78,, Y = Y =y y =.98 + j7.89 Y,,.8 j j7.89 =.98 + j j6.78 Bus By singular transformation: A = (. + j.) (. + j.) (. + j.) (. + j.) Ref Fig..(b): ree graph, [, ].

32 Power System Network Matrices I 69 [ z]. + j.. + j. =. + j. j. j.. + j.. + j. j. z, = j.. j j. j..779 j j.7 y, = z, = = j.. + j..8 + j.7.7 j.69.9 j9.6.9 j9.6 y =.779 j j.7.8 j.7.7 j j j6.89 = =.9 + j j6.7 Y Bus [A] [y] [A] P... Form Y Bus, Y Br and Z Loop by singular transformation for the network connections given below: JNU(9) Element p-r () () SELF MUUAL Bus code Impedance Bus code Impedance ().6. ().. (). ()..

33 7 Power System Analysis Solution: he power network is shown in Fig... Element SELF MUUAL Bus code Impedance Bus code Impedance ().6. ().. (). () ().6 ().. Ref. Fig..: Power system network. Fig..(a): ree graph, [,, ]. [ z ] () =. (or). () y = [z] = [ ]

34 Power System Network Matrices I 7 A = A = [A] = [ ] Y Bus = [A] y [A] = [ ] [ ] ZBus = Y Bus = [ ] [B] = C A E Ref. B D Ref. Fig..(b): ree graph, [,, ]. Fig..(c): ree graph, [,, ].

35 7 Power System Analysis [Y Br ] = [B] [y] [B] = [C] =.6. [Z Loop ] = [C] [z] [C] =.. P... Find the bus admittance matrix of the following power system. Bus code Line data impedance - + j j j j - + j j. Solution: he primitive line admittances are obtained as y = y = = y = y =. j. ( + j) y = y = =. j. (.96 + j.8769) y = y = =. j. ( + j) y = y = =. j. (.89 + j.978) y = y = =.8 j. (.7 + j.)

36 Power System Network Matrices I 7 he diagonal elements are given by By direct inspection, the Y Bus is y = y + y =. j. y = y + y =.6 j.6 y = y + y + y =.9 j.6 y = y + y =. j. y = y + y =. j. y = y = y =. + j. y = y = y =. + j. y = y = y =.+ j. y = y = y =. + j. y = y = y =. + j. y = y = y =.8 + j.. j.. + j..8 + j.. + j..6 j.6. + j. YBus =. + j..9 j.6.+ j.. + j.. j.. j.. + j..8 + j.. + j.. + j.. j. P... Form the Y Bus for the given network GAE Element no. Positive sequence reactance - j. - j. - j. - j. - j.8 - j.8

37 7 Power System Analysis Solution: he primitive admittances are y = y = = j., y = y = = j., y = y = = j = y = y j. j. j. he diagonal elements are given by y = y = = j., y = y = =j. j.8 j.8 Y = (y + y ) =j. j. =j. Y = (y + y + y ) =j. j. j =j8. Y = (y + y + y ) =j. j. j =j8.7 Y = (y + y + y ) = j j j. =j. y = (y ) = j. he off-diagonal elements are given by y = y = y = j. y = y = y = j. y = y = y = j = y = y y = y = y = j. he Y BUS matrix is y = y = y = j. j. j j. j j8. j. j YBUS = j. j. j8.7 j j j j. j. j.. j. P... he single line-diagram of a network is shown below. he line reactance is. pu/km and shunt capacitance is.6 pu km. Assemble the bus admittance matrix neglecting the line resistance. GAE

38 Power System Network Matrices I 7 km km km Solution: he calculated line parameters are Line Series impedance in pu Shunt admittance In pu -.() = j..6() = j.6 -.() = j..6() = j. -.() = j..6() = j. he pu series admittance is Fig..: Single-line diagram of a -bus system. y = y = /(j.) = j,y = y = /(j.) = j and y = y = /(j.) =j Shunt admittances at the nodes are y = (j.6 + j.) = j.8 y = (j. + j.6) = j. y = (j. + j.) = j.6 he elements of bus admittance matrix are calculated as Y = (y + y + y ) = (j.8 j j) = j.7 Y = (y + y + y ) = (j. j j) = j.76 Y = (y + y + y ) = (j.6 j j) =j8.6

39 76 Power System Analysis Y = Y = y = j Y = Y = y = j Y = Y = y = j hen the bus admittance matrix is denoted by j.7 j j YBUS = j j.6 j j j j8.6 P... he line data of a -bus system is given below, obtain the Y BUS. GAE Bus code Line data impedance - j. - j. - j. - j. - j. Solution: By inspection the Y BUS is calculated. he values of the system are y = y = = j., y = y = y = y = =j6.667, j. j. y = y = y = y = = j. j. he value of Y = Y = ( j j6.667 j) = j.667 he Y BUS is shown as Y = Y = ( j6.667 j) = j6.667 Y = Y = y = j,y = Y = y = j6.667, Y = Y = y = j j.667 j. j6.667 j. j. j.667 j. j6.667 (Y BUS ) = j6.667 j. j6.667 j. j. j6.667 j. j6.667

40 Power System Network Matrices I 79 Solution: Admittance = /( + j) = ( j)/, the conductance is /. P..9. he z matrix of a two-port network is given by corresponding Y matrix is he element Y of the (a). (b). (c). (d).8 Solution: he determinant of z is = Y = Z = = hence Y =.8 P... he network shown in Fig..6 has impedance in pu as indicated, the shunt admittance is j at each bus. he diagonal element of Y GAE (a) j9.8 (b) j (c) j. (d) j9.9 j. j. j j j Solution: Fig..6: Power system network. y = = j, y = = j, y = = j. j. j. j Y = (y + y + y ) = (j. j j) =j9.9 P... he bus admittance matrix of the network shown in the Fig..7 is