2 Power System Matrices 4 - darshan.ac.in€¦ · 2 Power System Matrices Interconnected Power System (2170901) 4 • Now the term Aj represents a vector in which each element is

2 Power System Matrices

EE Department Interconnected Power System (2170901) 1

2.1 Basic explanation of graph theory

1

5

2

4

3

Figure 2.1 System

• Graph: - Set of n and e where n represents nodes (vertices / junctions) of a network and

e represents elements (edges) of a network.

1 2

3

5

4

0

Figure 2.2 Graph

• Oriented graph: - If directions are assigned to each element of graph it is known as an

oriented graph.

1 2

3

5

4

0

1

2

34

5

7

8

69

Figure 2.3 Oriented Graph



• Subgraph: - Subset of original graph is known as a subgraph.

• Connected graph: - A graph in which there is at least one path between each / every pair

of node.

• Incidence matrix ( A ): - Let “G” be a graph with “ n ” nodes and “ e ” elements. Then the

matrix A whose r rows correspond to “n ” nodes (i.e. nxeA nodes / junctions) and c

columns correspond to “ e ” elements, i.e. edges is known as incidence matrices.th

th

th th

1 if the j element is incident to but directed away from the node i.

1 if the j element is incident to but directed towards the node i.

0 if the j element is not incident to i.e. touch i

ija

node.

1 2 3 4 5 6 7 8 9

0 1 0 1 1 0 0 0 0 0

1 1 1 0 0 0 0 0 0 0

2 0 1 0 0 0 0 1 1 0

3 0 0 1 0 0 1 1 0 1

4 0 0 0 1 1 1 0 0 0

5 0 0 0 0 1 0 0 1 1

nxe

nxe

A

• Reduced \ Bus incidence matrix ( 1n xeA

): - Any node of the connected graph can be

selected as the reference node and then the variables of the remaining 1n nodes which

are termed as buses can be measured with respect to this assigned reference node. The

matrix “ A ” obtained from the incidence row is termed as reduced or bus incidence

matrix.

1

1 1 2 3 4 5 6 7 8 9

1 1 0 1 1 0 0 0 0 0

2 1 1 0 0 0 0 0 0 0

3 0 1 0 0 0 0 1 1 0

4 0 0 0 1 1 1 0 0 0

5 0 0 0 0 1 0 0 1 1

n xe

n xe

A

2.2 Primitive Network • The data obtained from electricity boards or the power companies is in the form of

primitive network.

• Primitive network is a set of uncoupled elements which gives information regarding the

characteristics of individual elements only.

• The primitive network can be represented in impedance form as well as in admittance

form.

• The performance equations of any element i in impedance form will be.



- +1 2iz

iv

ie

Figure 2.4 Impedance form

1 2 and i i i i iv e z i v v v

• Similarly, in admittance form, the performance equations will be.

1 2iy

iv

ij

ii

Figure 2. 5 Admittance form

( Multiplication of is not possible)i i i i i ii j y v v y

• The performance equations of power system network in the impedance and the

admittance form for a complete network will be as follows.

v e z i

i j y v

• Here the diagonal elements of z and y are the self impedances / admittances for

that element and the off-diagonal elements are mutual impedances / admittance between

elements.

• However, is there is no mutual coupling between the elements, the matrices z or y

will be diagonal.

2.3 BUSY formation methods

(a) Singular transformation method

• The performance equation of a primitive network in the admittance form will be

i j y v

• Multiplying both the sides by A

A i A j A y v

• As we know

0A i

Since the product of A i represents a vector in which each element gives the sum

of currents flowing through the elements terminating at a bus.


Interconnected Power System (2170901) 4

• Now the term A j represents a vector in which each element is the algebraic sum

of source currents injected into each bus and thus equals the vector of injected bus

currents. Therefore

BUS

BUS

A j I

I A y v

• Now, injected power in bus frame of reference is*T

BUS BUSS I V

• Injected power for primitive network is*

**

T

TT

BUS BUS

S j v

I V j v

• Now,

**

* * ** * *

.

. .

TT

T T T

BUS

BUS

BUS T T T

A j I

A j I

x y x yj A I

x y y x

• A is a real matrix*

A A

**

* *

TT

T T

T

BUS

T

BUS

T

BUS

j A I

j A V j v

A V v

• Now,T

BUS BUS

T

BUS

BUS BUS BUS

I A y A V

Y A y A

I Y V

• Here y is a diagonal matrix is no mutual coupling is assumed.



(b) Direct method

y12 y23

y2 y3y1

1 2 3

I1I3

Figure 2.6 Circuit representing power system

• Bus admittance matrix can be developed by applying Kirchhoff’s current law at

every bus / node. In this way, systematic nodal equations are written for every

node except for the reference node which is normally the ground node.

• In case of power system usually all the voltage sources with series impedances are

replaced by the equivalent current sources with shunt impedances.

• For node 1

1 1 1 12 1 2

1 1 12 1 12 2

I y V y V V

I y y V y V

• Now, for node 2

12 2 1 2 2 23 2 3 2

12 1 12 2 23 2 23 3

0 0

0

y V V y V y V V I

y V y y y V y V

• Now, for node 3

3 23 3 2 3 3

3 23 2 23 3 3

I y V V y V

I y V y y V

• Here 1 2 3, and V V V are the voltages of buses 1, 2 and 3 respectively with reference

to the reference bus “0” (which is ground in this case) and voltages are known as

bus voltages.

• Moreover, for network12 21 i.e. ij jiy y y y and so on, as the network elements are

linear bilateral.

1 1 12 1 12 2

12 1 12 2 23 2 23 3

3 23 2 23 3 3

0

I y y V y V

y V y y y V y V

I y V y y V

• Let,

11 1 12

22 12 2 23

33 23 3

Y y y

Y y y y

Y y y



• Similarly,

12 12 21

23 23 32

13 310

Y y Y

Y y Y

Y Y

• Now,

1 11 12 13 1

21 22 23 2

3 31 32 33 3

0

I Y Y Y V

Y Y Y V

I Y Y Y V

• Now, any diagonal element is the sum of the admittances (say 11 1 12Y y y ) of the

elements terminating at that node and the off-diagonal elements is always negative

of the admittance of that elements between the adjacent nodes (say between nodes

1 and 2, off-diagonal element is 12 12Y y ).

• The above equations can be written in general form for any power system network

with n buses i.e. 1n nodes.

1 11 12 1 1

2 21 22 2 2

1 2

n

n

n n n nn n

I Y Y Y V

I Y Y Y V

I Y Y Y V

BUS BUS BUSI Y V

• Its diagonal elements iiY is the sum of admittances of the elements terminating at

the node i . These elements are known as short circuit driving point admittances

and they correspond to self-admittances.

• Similarly, off-diagonal elements ijY is the negative of the admittances of elements

connected between nodes and i j . These are known as short circuit transfer

admittances and they correspond to mutual admittances.

2.4 Algorithm for BUSY formation(a) Assuming no mutual coupling between transmission lines: -

• Initially all the elements of BUSY are set to zero. Addition of an element of admittance

y between buses and i j affects four entries BUSY in viz, , , and ii ij ji jjY Y Y Y as

follows: -

iinew iiold

ijnew ijold

jinew jiold

jjnew jjold

Y Y y

Y Y y

Y Y y

Y Y y

• Addition of an element of admittance from bus i to ground will affect iiY i.e.,

iinew iioldY Y y

(b) Assuming mutual coupling between transmission lines



• The equivalent circuit of mutually coupled transmission lines is shown in fig. . Shunt

elements are omitted for simplicity.

• The mutual impedance between the transmission lines ismz , and the series

impedances are1 2 and s sz z .

1

2

i s i m k j

k s k m i l

V z I z I V

V z I z I V

1

2

j s mi i

m sk kl

V z z I

z zV IV

1

2

i js mi

m sk k l

V Vy yI

y yI V V

Similarly,

1

2

j j is m

m sl l k

I V Vy y

y yI V V

1

1 1

2 2

s m s m

m s m s

y y z z

y y z z

• The elements of BUSY become

1

1

2

2

1

2

iinew iiold s

jjnew jjold s

kknew kkold s

llnew llold s

ijnew jinew ijold s

klnew lknew klold s

iknew kinew ikold m

jlnew ljnew jlold m

ilnew linew ilold m

jknew kjnew

Y Y y

Y Y y

Y Y y

Y Y y

Y Y Y y

Y Y Y y

Y Y Y y

Y Y Y y

Y Y Y y

Y Y Y

jkold my

2.5 Algorithm for BUSZ formation

Notation: i, j – old buses; r – reference bus; k – new bus.

i) Type-1 Modification: - Branch bZ is added between new bus and reference bus

ii) Type-2 Modification: - Branch bZ is added between new bus and old bus

iii) Type-3 Modification: - Branch bZ is added between old bus to reference bus

iv) Type-4 Modification: - Branch bZ is added between two old buses



• Type-1 Modification: - Adding a branchbZ between new bus k and reference bus r as

shown in fig 2.11.

Passive Linear n-bus

network

1

ni

j

k

rkV bZ

kI

Figure 2.7 Type-1 Modification

0; 1,2,...,

k b k

ik ki

kk b

V Z I

Z Z i n

Z Z

0

(old)

(new)=

0 0

BUS

BUS

b

Z

Z

Z

• Type-2 Modification: - Adding a branch bZ between old bus j and new bus k as shown in

fig. 2.12.


network

1

ni

j

k

r kV

bZkI

jI j kI I


1 1 2 2 ... ...

k b k j

b k j j jj j k jn n

V Z I V

Z I Z I Z I Z I I Z I

Rearranging,

1 1 2 2 ... ...k j j jj j jn n jj b kV Z I Z I Z I Z I Z Z I

1

2

1 2

(old)

(new)=

j

BUS j

BUS

nj

j j jn jj b

Z

Z Z

Z

Z

Z Z Z Z Z



• Type-3 Modification: - Adding a branchbZ between old bus j and reference bus r as shown

in fig. 2.13. This case follows by connecting bus k to the reference bus r, i.e., by setting

0kV .


network

1

ni

j

r

bZ


11 1

22 2

1 2

(old)

=

0

j

BUS j

n nj n

kj j jn jj b

ZV IZ ZV I

V Z I

IZ Z Z Z Z

Eliminate kI in the set of equations contained in the matrix,

1 1 2 2

1 1 2 2

0 ...

1...

j j jn n jj b k

k j j jn n

jj b

Z I Z I Z I Z Z I

I Z I Z I Z IZ Z

Now,

1 1 2 2 ...i i i in n ij kV Z I Z I Z I Z I

1 1 1 2 2 2

1 1 1...i i ij j i ij j in ij jn n

jj b jj b jj b

V Z Z Z I Z Z Z I Z Z Z IZ Z Z Z Z Z

Similarly to what was done in type-2 modification, it follows that in matrix form,

1

2

1 2

1(new)= (old)

j

j

BUS BUS j j jn

jj b

nj

Z

ZZ Z Z Z Z

Z Z

Z

• Type-4 Modification: - Adding a branch bZ between old bus i and old bus j in fig. 2.14.


network

1

ni

j

r

bZkI

j kI Ii kI I




1 1 2 2 ... ...i i i ii i k ij j k in nV Z I Z I Z I I Z I I Z I

Similar equations follow for other buses. The voltages of the buses i and j are, however,

constrained by the equation

1 1 2 2

1 1 2 2

... ...

... ...

j b k i

j j ji i k jj j k jn n

b k i i ii i k ij j k in n

V Z I V

Z I Z I Z I I Z I I Z I

Z I Z I Z I Z I I Z I I Z I

Rearranging,

1 1 10 ... ...i j ii ji i ij jj j in jn n b ii jj ij ji kZ Z I Z Z I Z Z I Z Z I Z Z Z Z Z I

In matrix form,

1 11 1

2 22 2

1 1 2 2

(old)

=

02

i j

BUS i j

n nni nj

ji j i j in jn b ii jj ij

Z ZV I

Z Z ZV I

V IZ Z

IZ Z Z Z Z Z Z Z Z Z

Eliminate kI on lines similar to what was done in type-2 modification, it follows that

1 1

2 2

1 1 2 2

1(new)= (old)

2

i j

i j

BUS BUS i j i j in jn

ii jj b ij

ni nj

Z Z

Z ZZ Z Z Z Z Z Z Z

Z Z Z Z

Z Z

Documents

2 Power System Matrices 4 - darshan.ac.in€¦ · 2 Power System Matrices Interconnected Power System (2170901) 4 • Now the term Aj represents a vector in which each element is