Graphs Elements – Line Segments In A Graph That Represent

GRAPHS
Elements – line segments in a graph that represent network components.
Nodes – the terminals of the line segment.
Incident node and element – if the node is terminal of the element.
Graph – shows the geometrical interconnection of the elements of a network.
Subgraph – any subset of elements of a graph.
Path – a subgraph of connected elements with no more than two elements connected to any one node.
Connected graph – if and only if there is a path between every pair of nodes.
Oriented graph – if each element of a connected graph is assigned a direction. Example: Single-line diagram: 2

4

1 G G

3

G

Oriented connected graph:

1

7

2

6

3

4

5

4

2

3

1

0

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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Tree – a connected subgraph containing all nodes of a graph but no closed path.
Branches – the elements of a tree
Number of branches b required to form a tree: b = n −1

where n = no. of nodes in the graph


Links – elements of the connected graph that are not included in the tree.
Cotree – a subgraph formed by the links of a connected graph.
Number of links l of a connected graph: l =e−b

where e = no. of elements of a connected graph

It follows that l = e − n +1 Example: 7

1

2

6

3

4

5

4

2

3

1 Branch Link

0

e=7 n=5 b=4 l=3

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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INCIDENCE MATRICES ˆ Element-Node Incidence Matrix A The elements of the element-node incidence matrix of a connected graph are as follows: aij = 1

if the ith element is incident to and oriented away from the jth node. if the ith element is incident to and oriented towards from the jth node. if the ith element is not incident to the jth node.

aij = −1 aij = 0

The dimension of the matrix is e × n. Example:

For the previous example network, the element-node incidence matrix is (0) (1) (2) (3) (4)

Nodes

 1 − 1 1  −1   1 − 1   −1 1     1 −1   6  1 −1   7  1 − 1

1 2 3 ˆ = 4 A 5

Elements Since n−1

∑ aij = 0

i = 1,2, K , e

j =0

ˆ are linearly dependent ⇒ rank A ˆ < n. the columns of A

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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Bus Incidence Matrix A

ˆ by deleting the column corresponding to
Bus incidence matrix A − obtained from A the reference node. The dimension of this matrix is e × (n – 1) and the rank is n – 1 = b. Example:

For the previous example network, if node 0 is chosen as the reference node, the bus incidence matrix is,

1 2 3 A = 4

5 6 7

(1) (2) (3) (4) − 1    −1    − 1   −1 1     1 −1    1 −1   1 − 1 

Bus

Element The matrix is rectangular and therefore singular.

BINARY BUS CONNECTION MATRIX B


Binary valued matrix – a matrix whose entries are binary (Boolean) variables. The elements of the square, binary bus connection matrix B is given by

1  bij = 1 0 

bus i connected to bus j by a line i= j otherwise

i, j = 1,2, K , n “1” denotes Boolean TRUE “0” denotes Boolean FALSE


B s symmetric
Negation: A : All TRUE elements of A are replaced by FALSE, and all FALSE entries by TRUE Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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Boolean AND: A•B

(A • B )ij

= a i1 • b1 j + ai 2 • b2 j + L + aim • bmj

i = 1,2, K , r

j = 1,2, K , c

“•” denotes Boolean AND “+” denotes Boolean OR A•B is r × c A is r × m B is m × c


Boolean OR: A + B

(A + B )ij

= aij + bij

for all i, j

It is necessary that A and B have the same dimensions.




The operation B•B produces a square matrix of dimension n for an n-bus system where (B•B)ij is 1 when buses i and j are joined by a line or are joined through an intervening bus by a line. Otherwise (B•B)ij is 0.

Example:

(1) (3)

(5)

(2)

(4)

(1) (2) (3) (4) (5) (1) 1 1 0 1 0 (2) 1 1 1 1 0 B = (3) 0 1 1 0 1   (4) 1 1 0 1 0 (5) 0 0 1 0 1

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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(1) ( 2) B • B = (3)

( 4) (5)

(1) (2) (3) (4) (5) 1 1 1 1 0  1 1 1 1 1    1 1 1 1 1    1 1 1 1 0  0 1 1 0 1


Note that B•B•B is also square and contains “1” in positions corresponding to buses joined by three or fewer lines and two or less intervening buses. The diagonal entries of B, B•B, B•B•B, and so on, are all “1”. Generalization: Boolean AND Operations on the Binary Bus Connection Matrix If B is the binary bus connection matrix for an n-bus power system, and the notation (m) B , B ( m) = B • B • L • B

← m times → is used to denote repeated AND operations, and B

(1)

= B, then B

(m)

consists of all 0’s (m)

contains except in the diagonal position, where 1’s appear. Also, the position ij of B 1’s if and only if buses i and j are joined via m lines or less (hence giving m − 1 intervening buses). Furthermore, for an n-bus system B system buses are connected to the system.

(n−1)

consists of all 1’s when all

Applications if the Binary Bus Connection Matrix
To determine whether a given bus is close to another bus. The buses i and j may be (m) evaluated for “proximity” by examining B ; if this entry is a “1”, buses i and j are connected to each other through m or fewer lines. Potential application: If a fault study is to be done for a short circuit at bus k, only buses “nearby” bus k need to be examined. (n−1)


To determine whether a give system is connected. The matrix B must all be 1’s when an n-bus system is connected since the farthest possible configuration between bus 1 and bus n occurs when 1 and n are on opposite ends of a radial string of buses. If B

(n−1)

contains a “0”, the system is disconnected.

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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Bus Disconnection Matrix B
The disconnection matrix B and the matrices B • B , B • B • B , and so on, have (2) (3) properties similar to B , B , and so on. Allowing D to denote the bus disconnection matrix, it is easy to show that

[ ]

B ( k ) D ( l ) = B ( k −l )

LINE INCIDENCE MATRIX L

The line incidence matrix L is defined by

1  (L) ij = − 1 0 

line i starts at bus j line i ends at bus j otherwise

For a system of l lines and n buses, L is l × n. This matrix is used to calculate the voltage difference between the buses at the terminals of each system line. Let Vline be a l-vector of “line voltages” (i.e., voltage drops across each system line); then Vline = L Vbus

For the purpose of calculating line voltage drops, each line must be considered to be directed (i.e., having a start or higher voltage bus and an end or lower voltage bus). This convention is reflected in the definition of line start and end in matrix L. While it is unimportant which bus is selected as the start of a line and which as the end, once the convention is established, it must be consistent in the definition of elements of Vbus.

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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BUS ADMITTANCE MATRIX Ybus Ybus Formation Methods:

1. Ybus building rules 2. Building block approach 3. Network incidence matrix Node Equations




Node − a junction formed when two or more circuit elements (R, L, C, Vs, Is) are connected to each other at their terminals.

Consider the circuit diagram

Yb

2

Ye Yd

Yc

Yf

3

4 1

I3

Ya

0

Yg

I4

Reference

KCL at node 1: (V1 − V3 )Yc + (V1 − V2 )Yd + (V1 − V4 )Y f = 0 KCL at node 3: V3Ya + (V3 − V2 )Yb + (V3 − V1 )Yc = I 3 Rearranging,

(

)

V1 Yc + Yd + Y f − V2Yd − V3Yc − V4Y f = 0 − V1Yc − V2Yb + V3 (Ya + Yb + Yc ) = I 3 Similar equations can be formed for node 2 and 4.
NOTE: All branch currents can be found when the node voltages are known, and a node equation formed for the reference node would yield no further information. Hence, the required number of independent node equations is one less than the number of nodes.

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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General matrix format:

Y11 Y  21 Y31  Y41

Y12 Y22 Y32 Y42

Y13 Y14  V1   I1  Y23 Y24  V2   I 2  = Y33 Y34  V3   I 3      Y43 Y44  V4   I 4 

or Ybus V = I Usual rules when forming Ybus: 1. Diagonal element Yjj = sum of the admittances directly connected to node j. 2. Off-diagonal element Yij = the negative of the net admittance connected between nodes i and j. Yjj : self-admittance or driving-point admittance. Yij : mutual admittance or transfer admittance.

Using the rules,

(

 Yc  Ybus =    

+ Yd + Y f − Yd

)

− Yd

− Yc

(Yb + Yd + Ye )

− Yb

− Yc

− Yb

(Ya + Yb + Yc )

−Yf

− Ye

0

  − Ye   0  Ye + Y f + Yg  −Yf

(

)

Separating the entries for Yc,

(

 Yd + Y f  −Y d Ybus =   0   − Y f  Yc  0 + − Yc   0

)

− Yd (Yb + Yd + Ye ) − Yb

0 − Yb (Ya + Yb )

− Ye

0

0 − Yc 0 0 0 Yc 0

0

     Ye + Y f + Yg  −Yf − Ye 0

(

)

0 0 0  0

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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More compactly,  Yc  ⋅  − Yc   ⋅

⋅ − Yc ⋅ ⋅ ⋅

Yc

⋅

⋅

⋅ (1) (3)  ⋅  1 − 1 ⇔  Yc ⋅ − 1 1   ⋅

The smaller matrix on the right is a compact storage matrix for matrix contribution of Yc to Ybus. It is an important building block in forming Ybus for more general networks. Bus impedance matrix:

 Z11 Z12  −1  Z 21 Z 22 Z bus = Ybus =  Z 31 Z 32   Z 41 Z 42

Z13 Z 23 Z 33 Z 43

Z14  Z 24  Z 34   Z 44 

Branch and Node Admittances Consider a generator in steady-state:Voltage equation:

I +

Za Es

I

+

V

-

-

N e t w o r k

+ Is

Ya

V -

N e t w o r k

E s = IZ a + V Dividing voltage equation by Za, E 1 I s = s = I + VYa where Ya = Za Za

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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The voltage source Es and its series impedance Za can be interchanged with the current source Is and its shunt admittance Ya, provided that

Is =

Es Za

and Ya =

1 Za

Sources Es and Is may be considered externally applied at the nodes of the transmission network, which then consists of only passive branches. Suppose that only branch admittance Ya is connected between nodes m and n as part of a larger network of which only the reference node is shown, Za = branch impedance, primitive impedance

Im

Va Ia + + Ya = 1/Za Vm

m

-

n

In

+ Vn -

Reference node

Ya = branch admittance, primitive admittance Current and voltage equations: I m   1  Vm   I  = − 1 I a ; Va = [1 − 1] V   n    n Vm  YaVa = I a ; Ya [1 − 1]   = I a Vn  Vm   1  I m  1 − 1Ya [1 − 1] V  = − 1 I a =  I     n    n  Ya − Y  a

− Ya  Vm   I m  = Ya  Vn   I n 

nodal admittance equation for branch Ya

nodal admittance matrix

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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The nodal admittance matrix is singular because neither node m nor node n connects to the reference.

When n is the reference node, Vn = 0 and

[Ya ]Vm = I m This corresponds to removal of row n and column n from the coefficient matrix (nodal admittance matrix). Note that 1  1 − 1 − 1[1 − 1] = − 1 1     

building block








The nodal admittance matrices are simply storage matrices with row and column labels determined by the end nodes of the branch. To obtain the overall nodal admittance matrix of a network, we simply combine the individual branch matrices by adding together elements with identical row and column labels. Such addition causes the sum of the branch currents flowing from each node of the network to equal the total current injected into that node, as required by KCL. Provided at least one of the network branches is connected to the reference node, the net result is Ybus of the system.

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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Example: Single-line diagram: 2

3

4 1

Reactance diagram (per unit):

j0.25

2

j0.2 j0.125 j0.4

j0.25 3

4 1

j0.1 + -

j0.1

j1.15 1.25

j1.15 0o

0

0.85

-45 o

+ -

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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Admittance diagram:

Ib

Ie

2

-j4.0 Ic

Id

-j8.0

-j5.0

-j4.0

3

If -j2.5

Ia

1

-j0.8

Ig

-j0.8

0

1 -90

4

0.68 -135 (3)

(2)

(3) (1)

(4) (3)  1 − 1 (3)  1 - 1 (3) [1] Ya ; Y ; Y ; (4) [ 1 ]Yg b c (2) − 1 1  (1) - 1 1  (3)

( 2)

(1)

(4) (2)

(4) (1)

(2)  1 − 1 (4)  1 - 1 (4)  1 - 1 Yd ; Ye ; Yf     (1) − 1 1  (2) - 1 1  (1) - 1 1  Combining elements of the above matrices having identical row and column labels,

(1) ( 2) (3) ( 4)

(1)  Yc + Yd + Y f  − Yd   − Yc  −Yf 

(

)

(2) − Yd

(3) − Yc

(Yb + Yd + Ye )

− Yb

− Yb

(Ya + Yb + Yc )

− Ye

0

(4) −Yf

  − Ye  =Y bus  0  Ye + Y f + Yg 

(

)


The order in which the labels are assigned is not important here, provided the columns and rows follow the same order. Nodal admittance equations of the overall network:

j8.0 j 4.0 j 2.5  V1   0 − j14.5      j8.0   j 5.0  V2   0 − j17.0 j 4.0   =  j 4.0 j 4.0 − j8.8 0.0  V3   1.00∠ − 90o       j 5.0 0.0 − j8.3 V4  0.68∠ − 135o   j 2.5

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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Mutually Coupled Branches in Ybus

Two mutually coupled branches with impedance parameters or with admittance parameters:

Ia

Im

+ Va -

In

Za

Ia

Im

Zb

Ya

Ip

+ Vb -

Iq

Ib

YM

+ Va -

In

Ib

ZM

Yb

Ip

+ Vb -

Iq

Primitive impedance equations:

Va   Z a V  =  Z  b  M

Z M  Ia  Z b   I b 

where mutual impedance ZM is positive when Ia and Ib enter the dotted terminals. Inverse of the primitive impedance matrix:

 Za Z  M

ZM  Z b 

−1

=

 Zb  Z a Z b − Z M 2 − Z M 1

− Z M   Ya = Z a  YM

YM  Yb 

primitive admittance matrix

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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Voltage-drop equations: Vm  Vm    V  Va  Vm − Vn  1 − 1 0 0  Vn   n V  = V − V  = 0 0 1 − 1 V  = A V  q  p  p  b  p     Vq  Vq 

where A = coefficient matrix Current equations: I m   1 0 I     n  =  − 1 0   I a  = AT  I a  I   I p   0 1   I b   b     I  q   0 − 1 Substitute voltage drop equations into the primitive impedance equation, Vm     Ya YM  Vn   I a  =  Y A  M Yb  V p   I b    Vq  Premultiply by AT,

 Ya AT  YM

Vm  I m  V    YM   n  T I a   I n  A =A  = Yb  V p   Ib   I p      Vq   I q 

Nodal admittance equations of the two mutually coupled branches:

( m) (n) ( p) (q)

(m)  Ya  −Y  a  YM  − YM

( n) − Ya

(p) YM

Ya

− YM

− YM

Yb

YM

− Yb

(q) − YM  Vm   I m      YM  Vn   I n  = − Yb  V p   I p      Yb  Vq   I q 


The 4 × 4 submatrices above form part of the larger nodal admittance matrix of the overall system. The pointers m, n, p, q indicate the rows and columns of the system matrix to which the elements of the above nodal admittance matrix belong.

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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Nodal admittance matrix by inspection:

( m) (n) ( p) (q)

(m) 1   − 1   1   − 1

(n) − 1 Ya 1 

(p) (q)   1 − 1 − 1 1 YM       1 − 1  − 1 1 Yb    

− 1 YM 1 


If node n is the reference, we may eliminate the row and column of that node.
If n and q are one and the same node, columns n and q are combined (since Vn = Vq), and the corresponding rows are added because In and Iq are parts of the common injected current. Example:

j0.25 j0.25 j0.15

3

2

1

j0.25 Primitive admittances:  j 0.25  j 0.15 

j 0.15 j 0.25

−1

− j 6.25 j 3.75  =   j 3.75 − j 6.25

Nodal admittance matrix:

(3) (1) (3) ( 2)

(3) (1)   1 − 1   (− j 6.25) − 1 1       1 − 1   (+ j 3.75)  − 1 1 

(3) (2)   1 − 1 − 1 1  (+ j 3.75)       1 − 1 − 1 1  (− j 6.25)   

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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Adding the columns and rows of the common node 3, (1)

(2)

(3)

(1)  − j 6.25 j 3.75 j 6.25 − j 3.75    (2)  j 3.75 − j 6.25 − j 3.75 + j 6.25  (3)  j 6.25 − j 3.75 − j 3.75 + j 6.25 − j 6.25 − j 6.25 + j 3.75 + j 3.75 Nodal equations:

j 2.50  V1   I1  − j 6.25 j 3.75  j 3.75 − j 6.25 j 2.50  V  =  I     2  2  j 2.50 j 2.50 − j 5.00 V3   I 3 

Three branches with mutual coupling:

m

p

Ia Za

r

Ib Zb

Ic Zc

ZM1 ZM2 n

 Za Z  M1  Z M 2

ZM1 ZM 2  Zb 0  0 Z c 

q

−1

s

 Ya YM 1 YM 2  =  YM 1 Yb YM 3  YM 2 YM 3 Yc 

To form Ybus for a network with mutually coupled branches: 1. Invert the primitive impedance matrices of the network branches to obtain the corresponding primitive admittance matrices. A single branch has a 1 × 1 matrix. Two mutually coupled branches have a 2 × 2 matrix, three mutually coupled branches have a 3 × 3 matrix, and so on. 2. Multiply the elements of each primitive admittance matrix by the 2 × 2 building-block matrix.

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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3. Label the two rows and the two columns of each diagonal building-block matrix with the end-node numbers of the corresponding self-admittance. For mutually coupled branches it is important to label in the order of the marked (dotted) --- then --unmarked (undotted) node numbers. 4. Label the two rows of each off-diagonal building-block matrix with node numbers aligned and consistent with the row labels assigned in (3); then label the columns consistent with the column labels of (3). 5. Combine, by adding together, those elements with identical row and column labels to obtain the nodal admittance matrix of the overall network. If one of the nodes encountered is the reference node, omit its row and column to obtain the system Ybus.

Bus Admittance Matrix 1 -- A.C. Nerves, U.P. Dept. of Electrical & Electronics Engineering, Nov. 13 2003

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