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EXPERIMENT 2

FORMATION OF BUS ADMITTANCE AND IMPEDANCE MATRICES AND SOLUTION OF

NETWORKS

2.1 AIM

To understand the formation of network matrices, the bus admittance matrix Yand the bus impedance

matrix Zof a power network, to effect certain required changes on these matrices and to obtain network

solution using these matrices.

2.2 OBJECTIVES

i. To write a computer program to form bus admittance matrix Y,given the impedances of

the elements of a power network and their connectivity (mutual coupling between elements

neglected)

ii. To modify the matrix Yto effect specified changes in the configuration of the network.

iii. To obtain network solution, that is, to determine the bus voltages given bus current injections.

iv. To obtain certain specified columns of the bus impedance matrix Zor the full matrix Zusing the

factors of Yor the inverse of Y.

2.3 SOFTWARE REQUIRED

FORMATION OF NETWORK MATRICES module of AU Powerlab or equivalent

2.4 THEORETICAL BACKGROUND

2.4.1 Network Description of A Multinode Power System

The bus admittance matrix Yand bus impedance matrix Zare the two important network descriptions of

interconnected power system. The injected bus currents and bus voltages of a power system understeady state condition can be related through these matrices as

Y V = I (2.1)

Z I = V (2.2)

These matrices are important building blocks of power system modelling and analysis. The Zis mainly

used in fault analysis while Yis mainly used in power flow and stability analysis.

2.4.2 Formation of Bus Admittance MatrixTwo-Rule Method (Based on Node - Voltage Analysis)

Consider a three-bus power system shown in Fig.2.1 .The equivalent power network for the system is

shown in Fig.2.2 in which the generator is replaced by Norton equivalent, the loads by equivalent

- equivalent circuits.

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y12

y13 y23

y30 yd3

y 23

y 23

yg1 yd2I1GLoad

Generation

Short line

Short line Long line

Shunt

Compensation

Load

2

1 1 2

3

3

Fig.2.1 A Sample Power System Fig.2.2 Equivalent Power Network

In Fig.2.2, the admittances of the generator, loads and transmission lines are given in per unit tosystem MVA base. The ground is taken as reference node.

Applying Kirchhoffs current law (KCL) to nodes 1,2 and 3.

yg1V1+ y12(V1V2) + y13(V1-V3) = I1

yd2V2+ y12(V2V1) + y23(V2 V3) + y23V2= 0 (2.3)

yd3V3+ y30V3+ y23(V3 V2) + y23V3+ y13(V3 V1) = 0

Rearranging these equations

( yg1+ y12+y13) V1+ (-y12) V2+ (-y13) V3= I1

(-y12) V1+( yd2+ y12+ y23+ y23) V2+(-y23) V3=0 (2.4)

(-y13) V1+ (-y23) V2+ (yd3+ y30+ y23+ y

23+ y13) V3=0In matrix form

Y11 Y12 Y13 V1 I1

Y21 Y22 Y23 V2 = 0 (2.5)

Y31 Y32 Y33 V3 0

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Vk Vm

Ik a :1

.Vk/a

aIk y Imk m

o o

where

Y11= ( yg1+ y12+y13)

Y22= (yd2+ y12+ y23+ y23) (2.6)

Y33= (yd3+ y30+ y23+ y23+ y13)

Y12= Y21= -y12

Y13= Y31 = -y13

Y23= Y32= -y23

The matrix equation (2.5) can be extended to an n node system. The steps invol ved in assembling bus

admittance matrix may be extracted from equations (2.5) and (2.6) and are given below.

Two-Rule Method for Assembling Y matrix:

1. The diagonal element Yiiof the matrix is equal to the sum of the admittances of all elements

connected to the ithnode.

2. The off-diagonal element Yijof the matrix is equal to the negative of the sum of the admittances of

all elements connected between the nodes i and j.

2.4.3 Equivalent Circuit of A Transformer With Off-Nominal Tap for the Purpose of Formation

of Bus Admittance Matrix

A two winding transformer with off- nominal turns ratio, connected between nodes k and m is shown

in Fig 2.3. In this representation, the turns ratio is normalized as a:1 and the nonunity side is called thetap side which is taken as the sending end side. The series admittance of the transformer is connected to

the unity side.

k t m

Fig.2.3 Transformer with off-nominal tap

Imk Ik y/a m

y{1-(1/a)} VmVk y/a {(1/a)-1}

Fig.2.4 Equivalent Circuit for Transformer With Off Nominal tap .

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By applying the two-rule method to buses t and m in Fig.2.3 we get

y -y Vk/a aIk

-y y Vm = Im (2.7)

By elementary matrix operation equation (2.7) can be reduced to

y/a2

-y/a Vk Ik (2.8)

-y/a y Vm = Im

It can be checked by applying the Two-Rule method to buses k and m in Fig.2.4 that the bus admittance

- equivalent circuitfor transformer with offnominal tap is that in Fig.2.4 and the contribution of this transformer to busadmittance matrix Y of the power system to which it is connected is (refer equation (2.8))

Ykk= y/a2; Ymm= y ; Ykm= Ymk= -y/a (2.9)

2.4.4 Algorithm for Formation of Bus Admittance Matrix

The algorithm initializes the matrix Y with all the elements set to zero. Then read one element of the

network at a time and update the matrix Y by adding the contribution of that element. The contribution

of a transformer connected between nodes k and m is given in equation (2.9).

The contribution of transmission line connected between nodes k and m to Y is

Ykk= Ymm = ykm+ ykm (2.10)

Ykm= Ymk= -ykm

where ykmand ykmare respectively the series admittance in p.u and half line changing admittance in p.u

of the line.

The contribution of a shunt element connected to node k to Y is

Ykk= y (2.11)

where y is the admittance in p.u of the shunt element. If S and Soare respectively the MVA rating of the

shunt element (capacitor) and base MVA chosen for the system; then the shunt admittance is given by

y= 0+j(S/So) p.u (2.12)

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Algorithm

Step 1: Initialize Y with all elements set to zero

Step 2: Read the line list, one line at-a-time and update Y by adding the respective

contribution, equation (2.10)

Step 3: Read the transformer list, one transformer at-a-time and update Y by adding the

respective contribution, equation (2.9)

Step 4: Read the shunt element list, one element at-a-time and update Y by adding therespective contribution, equation (2.11)

2.4.5 Building Algorithm for Bus Impedance Matrix

A building algorithm for bus impedance matrix can be developed by first studying the rules required formodifying an existing Z matrix for addition of new elements. Let us start with a given partial power

network with r nodes whose bus impedance matrix Z is known. It is proposed to add new elements, one

at a time, to this network and get the modified matrix Zm. Any one of the following four rules can be

used depending upon the type of modification.

Modification 1:Add an element with impedance z, connected between the reference node of thepartial network and a new node (r+1).

Rule 1 :The modified matrix Zmof dimension (r+1) x (r+1) is given by

Zm= Z 0 (2.13)

0 z

where Zis the bus impedance matrix of the partial network.

Modification 2:Add an element with impedance z, connected between an existing node i and a

new node (r+1).

Rule 2 : The modified matrix Zmof dimension (r+1) x (r+1) is given by

Z Zi (2.14)

Z

m

= ZiT (Zii+ z)

where Ziis the ith column of Z

ZiTis the transpose of Zi

Ziiis the iith element of Z

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Modification 3 :Add an element with impedance z, connected between an existing node i and

the reference node of the partial network.

Rule 3 : The modified matrix Zmof dimension r x r is obtained through a two step process. In

the first step, assume that the added element is between the existing node i and a fictitious node (r+1)(instead of the reference node) and obtain the modified matrix Z of dimension (r+1) x (r+1) by

augmenting Zwith an extra row and column as in (2.14). The second step is to connect the fictitious

node (r+1) by zero impedance link to the reference node whose voltage is zero and to obtain the finalmatrix Zmof dimension r x r by applying Krons -reduction to the last row and column to obtain

Zm

jk= Zjk- Zj, (r+1)Z(r+1), k ; j,k = 1,2,.,r (2.15)

Zii+ z

Modification 4 :Add an element with impedance z, connected between existing nodes i and j.

Rule 4 :The modified matrix Zmof dimension r x r is given by

Zm= Z c b bT (2.16)

where b = Zi- Zj (2.17)

c= (z + Zii+Zjj 2Zij)-1

(2.18)

Zi, Zj: ith

and jth

columns of Z

Zii, Zjj, Zij: iith

, jjth

and ijth

elements of Z

Note:

b bT= b1 b1,b2..br b

21 b1b2 b1br

b2 = b2b1 b

2

2 b2br (2.19)

br brb1 brb2 b2

r

Building Algorithm for Z:

The above rules are built into the following step wise procedure to build Zmatrix:

Step 1:

Start with a partial network composed only of those elements connected directly to reference node. Let

the number of these elements be r. The corresponding bus impedance matrix Z(1)is of dimension r x r

and is diagonal with the impedance values of the elements appearing on the diagonal. This process is

equivalent to the repeated use of rule 1.

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Step 2:

Add a new element which brings a new node and modify Z(1)

using rule 2. Continue until all the nodes

of the complete network are brought in.

Step 3: Add a new element connected between existing nodes i and j using rule 4. Continue until all theelements are connected.

2.4.6. Network Solution Using Factorization And Repeat Solution

Network solution in a power network is concerned with the determination of the bus voltage vector V

from the network equation (2.20), given the bus admittance matrix Yand bus current source vector I

Y V = I (2.20)

In power system applications, during most of the studies, the network configuration and parametersremain the same implying that Yremains fixed. However the operating conditions change resulting in

changed bus current vector I. In such cases, while finding numerical solution for the bus voltage vector

V repeatedly for different bus current vectors I, the computations can be performed effectively resulting

in considerable savings in time if triangular factorization and repeat solution are adopted.

First the matrix Y in equation (2.20) is split into factor matrices L and U using the triangular

factorization process [1]. The equation (2.20) may be written as

Y V = L U V = I (2.21)

~

By defining an intermediate voltage vector Vas

~U V = V (2.22)

equation (2.18) may be written as

~L V = I (2.23)

The solution for voltage vector V in Eqn. (2.21) is obtained through the repeat solution process which

comprises the following two steps:

~

(i) Forward Elimination: This involves solving eqn. (2.23) for vector Vthrough elementary

transformations on vector Iusing the elements of L.

(ii) Back substitution : This involves solving eqn.(2.22) for vector Vthrough elementary

~

transformations on vector Vusing the elements of U.

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Factorization Algorithm:

Let us define the Land Ufactor matrices for a sample Ymatrix of dimension 3 x 3 as

Y11 Y12 Y13 1 0 0 U11 U12 U13

Y= Y21 Y22 Y23 = L U= L21 1 0 0 U22 U23Y31 Y32 Y33 L31 L32 1 0 0 U33

(2.24)

After multiplication of Land Umatrices, eqn. (2.24) becomes,

Y11 Y12 Y13 U11 U12 U13

Y= Y21 Y22 Y23 = L21U11 (L21U12+U22) (L21U13+U23)

Y31 Y32 Y33 L31U11 (L31U12+L32U22) (L31U13+L32U23+U33)

(2.25)It is seen from (2.24) and (2.25) that the first row of U is readily obtained from the first row of Y,

whereas the first column of L can be obtained by dividing the first column of Yby U11which is the

same as Y11. All the elements of L and U can be obtained from the Y matrix in Eqn. (2.25) by

performing two transformations. The first transformation on Ymatrix in eqn. (2.25) is as follows.

Y(1)22= Y22 (Y21Y12) / Y11 ; Y(1)

23= Y23- (Y21Y13) / Y11

Y(1)32= Y32 (Y31Y12) / Y11; Y(1)

33= Y33 (Y31Y13) / Y11

and Y(1)21= Y21/ Y11 ; Y(1)31= Y31/ Y11

At the end of this first transformation Ymatrix is given by

Y11 Y12 Y13 U11 U12 U13

Y(1)= Y(1)21 Y

(1)22 Y

(1)23 = L21 U22 U23

Y(1)31 Y(1)

32 Y(1)

33 L31 L32U22 (L32U23+U33) (2.26)

The second transformation on Y(1)in eqn. (2.26) is as follows:

Y(2)33= Y(1)

33 (Y(1)

32Y(1)

23) / Y(1)

22

and Y(2)

32= Y(1)

32/ Y(1)

22

At the end of the second transformation, the Ymatrix is given by

Y11 Y12 Y13 U11 U12 U13

Y(2)= Y(1)21 Y(1)

22 Y(1)

23 = L21 U22 U23

Y(1)

23 Y(1)

23 Y(1)

33 L31 L32 U33

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F11 F12 F13

= F21 F22 F23 = F

F31 F32 F33 (2.27)

Now the Y

(2)

matrix in (2.27) contains all the elements of Land U. This can be termed as factor matrix F

The above factorization process can be generalized for a Ymatrix of dimension nxn as

follows.

kth transformation : k = 1,2,(n -1)

Yij(k)

= Yij(k-1)

(Yik(k-1)

Ykj(k-1)

) / Ykk(k-1)

; i, j = (k+1) , .. , n (2.28)

and Yik(k)

= Yik(k-1)

/ Ykk(k-1)

; i = (k+1) , .. , n (2.29)

After the above (n-1) transformations, the Ymatrix is converted into the factor matrix F which containall the elements of Land U.

Repeat Solution Algorithm:

The forward elimination is performed on the given vector Iusing the elements of Lstored

~

in Fto obtain the intermediate vector Vas shown below.

~

1 0 0 V1 I1

~F21 1 0 V2 = I2 (2.30)

~

F31 F32 1 V3 I3~

V1= I1~

V2= I2(1)

; I2(1)

= I2 F21I1 (2.31)

~

V3= I3(2)

; I3(1)

= I3 F31I1; I3(2)

= I3(1)

F32I2(1)

The above steps can be generalized as :

~Vi= Ii

(i-1); i = 1,2,..n

Transformation on Ii; i = 2,3 ..n is performed using

Ii(j)

= Ii(j-1)

- FijIj(j-1)

; j= 1,2, .. (i -1) (2.32)

~The back substitution is performed on the intermediate vector V using the elements of Ustored in F to

obtain the solution vector Vas shown below:

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~

F11 F12 F13 V1 V1~

0 F22 F23 V2 = V2 (2.33)

~0 0 F33 V3 V3

~ ~ ~

V3= V3(1)

; V3(1)

= V3/ F33~ ~ ~ ~ ~ ~

V2= V2(2); V2

(1)= V2 F23V3(1); V2

(2)= V2(1)/ F22 (2.34)

~ ~ ~ ~ ~ ~ ~ ~ ~V1= V1

(3); V1(1)= V1 F12V2

(2); V1(2)= V1

(1) F13V3(1)

; V1(3)= V1

(2)/ F11

The above steps can be generalized as

Fig 2.5 Flow Chart for Repeat Solution

Z matrix through factorization of Y and repeat solution:

The relation between Yand Zis given by eqn. (2.35)

Y Z = I (2.35)

where Iis the identity matrix.

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i = n, .., 2, 1

j = n, .., (i+1), i

If (j = i)?

~ ~ ~

Vi= Vi FijVj

~ ~

Vi= Vi/ Fij

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Eqn. (2.35) can be written as

Y (Z1Z2. Zn) = ( I1I2.. In ) (2.36)

where Zi= ith column of Z matrix

and Ii= ith column of identity matrix.

Eqn. (2.36) can be split into n equations

Y Zi= Ii; i = 1, 2, . ,n (2.37)

From (2.37) it is clear that the ithcolumn Ziof Zmatrix can be obtained through factorization of Yand

repeat solution on Ii.

2.5 EXERCISES

2.5.1 Write a program in C language for formation of bus admittance matrix Yof a power network

using the Two Rule Method.

The program should have three sections: Input Section, Compute Section and Output Section.

I. Input Section:

Pre-requisite:

Before creating the input data file, draw a single-line diagram showing the buses, lines, transformers,shunt elements, bus generations and demands. (Refer Fig in Annexure 2.1) Bus Id numbers are serially

given from 1 to NB where NB is the total number of buses in the system.

The data to be read from an input file should contain the general data, transmission line data,

transformer data and shunt element (capacitor / reactor) data in the following sequence:

i. General Data: Number of buses, number of transmission lines, number of transformers, number ofshunt elements (capacitor / reactor) and the base MVA of the network.

ii. Transmission Line Data: The following data to be read for all lines, (one line for each transmissionline):

a. Identification number (serial number) of lineb. Identification number of sending end and receiving end buses of line

c. Series impedance (R,X)of the line in per unit.

d. Half-line-charging susceptance, Bc, in per unit.e. Maximum loadability limit (rating) of line in MVA

iii. Transformer Data:

The following data to be read for all transformers (one line for each transformer):

a. Identification number (serial number) of transformer

b. Identification number of sending end (tap side) bus and receiving end bus of transformer.c. Impedance (R,X) of transformer in per unit.

d. Off-nominal tap ratio

e. Maximum loadability limit (rating) of transformer in MVA

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iv. Shunt element (capacitor / reactor) Data:

The following data to be read for all shunt elements (one line for each element):

a. Identification number (serial number)of the shunt element

b. Identification number of the bus

c. Rated capacity in MVA (positive for capacitor and negative for inductor).

II. Compute Section:

To form the Ymatrix using the algorithm in section 2.4

III. Output Section:

To create an output file in a report form comprising the following:

(i) Student Information :

Title of the Experiment :

Case Title

Name of the Student :

Roll Number :

Semester :

Date of Experiment :

(ii) Input Data with proper headings :

(iii) Results with proper headings: Element value (in per unit) of Yprinted row wise.

) 2.5.2 The line series impedance (R,X) is given in ohm per km length of line and the linecharging susceptance is given in milli Siemens per km length. Further the length of line in kmas well as rated voltage of the line in kV are also given. Modify the program developed in

2.5.1 to read the line impedance / susceptance data in per unit as in 2.5.1 or in ohms / mhos

as in 2.5.2.

2.5.3 Using a text editor create an input file in the sequence given in 2.5.1 for formation of Ymatrix

for the 6 bus system given in Annexure 2.1, run the program developed in 2.5.1, compute Y

and print the output file. Check the results obtained using the available software.

2.5.4 Create necessary changes in the C program developed in 2.5.1 to modify the formed Ymatrix forthe outage (removal) of a transmission line, a transformer and a shunt capacitor / reactor.

2.5.5 Run the program in 2.5.4 and print the modified Ymatrix for the 6 bus system for the removal

of the following components, one at-a-time:

a. Line 4-6b. Transformer 4-3

c. Shunt capacitor at Bus 4

Check the results, using the available software.

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2.5.6 Create necessary changes in the C program in 2.5.1 to obtain bus impedance matrix Z either fullmatrix or only certain specified column from the formed Ymatrix either by factorizing Y and

going for one repeat solution for each column of Z or by inverting Y.

2.5.7 Run the program developed in 2.5.6 and obtain Zfor the six bus system in Annexure 2.1. Obtain

only the column of Z corresponding to bus 4. Check the results using the available software.Repeat this exercise again but after dropping the two shunt capacitors of the 6 bus system. What

is the inference?

2.5.8 Create necessary changes in the C program in 2.5.1 to obtain network solution V given the bus

current source vector I, either using the factors of Yor the inverse of Y.

2.5.9 In the 6 bus system in Annexure 2.1 replace the generators at buses 1 and 2 by equivalent ideal

current sources 1.0 op.u and the loads at buses 3,5 & 6 by equivalent ideal current sources

computed assuming a bus voltage of 1.0 op.u. For this network run the program in 2.5.8 and

obtain the solution V.

) 2.5.10 Write a program in C language for forming the bus impedance matrix Zstep by step, usingthe Building Algorithm, run the program for the 6 bus example in Annexure 2.1 and

compare the results with that obtained in 2.5.7.as well as with that obtained using theavailable software.

) These exercises are optional and not mandatory.

2.6 REFERENCE

[1] Arthur R.Bergen and Vijay Vittal, Power Systems Analysis, Pearson Education Asia, Delhi,Second Edition, 2000.

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ANNEXURE 2.1

6-BUS, 7-LINES / TRANSFORMER POWER SYSTEM

Single-Line Diagram

S1L2 T2

L3 a:1

L1 L5

T1 L4

a:1

Buses : 6, numbered serially from 1 to 6

Lines : 5, numbered serially from L1 to L5Transformers : 2, numbered serially as T1 and T2

Shunt Load : 2, numbered serially as s1 and s2

Base MVA : 100

Bus Data P-V Buses:

Generation, MW Demand Gen. LimitMVAR

Bus IDNo.

Schedule Max Min MW MVAR Max Min

ScheduledVolt (p.u)

1 ? 200 40 0.0 0.0 100.0 -50.0 1.02

2 50.0 100 20 0.0 0.0 50.0 -25.0 1.02

Bus Data P-Q Buses

DemandBus ID No

MW MVAR

Volt. Mag.

Assumed (p.u)

3 55.0 13.0 1.0

4 0.0 0.0 1.0

5 30.0 18.0 1.0

6 50.0 5.0 1.0

2-14

G

1

6

4 3

2

G

5

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Transmission Line Data:

Line ID.No

Send BusNo.

ReceiveBus No.

ResistP.U

ReactanceP.U.

Half Linecharging

Suscept.

P.U

RatingMVA

1 1 6 0.123 0.518 0.0 55

2 1 4 0.080 0.370 0.0 653 4 6 0.087 0.407 0.0 30

4 5 2 0.282 0.640 0.0 55

5 2 3 0.723 1.050 0.0 40

Transformer Data:

Transformer

ID.No

Send (*)

Bus No.

Receive

Bus No.

Resist.

P.U

Reactance

P.U.

Tap Ratio Rating

MVA

1 6 (*) 5 0.0 0.300 1.000 30

2 4 (*) 3 0.0 0.133 1.000 55

(*) Note: The sending end bus of a transformer should be the tap side.

Shunt Element Data:

Shunt ID No. Bus ID. No. Rated Capacity

MVAR (*)

1 4 2.0

2 6 2.5

(*) Note: Sign for capacitor : +ve

Sign for inductor : -ve

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