BENECO TCDP1 Vol 1 - Power System Modeling Part 1

PROF. ROWALDO R. DEL MUNDO

Benguet Electric Cooperative, Inc.Baguio City

Technical Competency Development Program

Power System Modeling

Training Course in

2Prof. Rowaldo R. del Mundo

BENECO Technical Competency Development Program

Course Outline

Numerical Methods for Power System Analysis

Electric Circuits and Power System Representation

Per Unit Quantities

Symmetrical Components

Utility Thevenin Equivalent Circuit

Generalized Machine Models

Line and Cable Models

Transformer and Voltage Regulator Models



Numerical Methods for Power System Analysis

1. Matrix Representation of System of Equations

2. Type of Matrices

3. Matrix Operations

4. Direct Solutions of System of Equations

5. Iterative Solutions of System of Equations



System of n Linear Equations

System of Equations in Matrix Form

Matrix Representation of System of Equations



⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

333231

232221

131211

aaaaaaaaa

A

which is called the Coefficient Matrix of the system of equations.

Matrix Representations of System of Equations

The coefficients form an array



Similarly, the variables and parameters can be written in matrix form as.

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

3

2

1

xxx

X and

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

3

2

1

yyy

Y




The system of equations in matrix notation is

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

3

2

1

3

2

1

333231

232221

131211

yyy

xxx

aaaaaaaaa

AX = Y

System of Equations in Matrix Form




Definition of a MATRIX

A matrix consists of a rectangular array of elements represented by a single symbol.

[A] is a shorthand notation for the matrix and aijdesignates an individual element of the matrix.

A horizontal set of elements is called a row and a vertical set is called a column.

The first subscript i always designates the number of the row in which the element lies.

The second subscript j designates the column.

For example, element a23 is in row 2 and column 3.



Definition of a MATRIX

[ ] [ ]

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

==

mnm3m2m1

3n333231

2n232221

1n131211

ij

aaaa

aaaaaaaaaaaa

aA

K

MMM

K

K

K

The matrix has m rows and n columns and is said to have a dimension of m by n (or m x n).

[aij]mxn



Definition of a Vector

A vector X is defined as an ordered set of elements. The components x1, X2…, Xn may be real or complex numbers or functions of some dependent variable.

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

n

2

1

x

xx

XM

“n” defines the dimensionality or size of the vector.



Matrices with only one row (n = 1) are called RowVectors while those with one column (m=1) are called Column Vectors. The elements of a vectors are denoted by single subscripts as the following:

[ ]n21 rrr R L=

Thus, U is a row vector of dimension n while W is a column vector of dimension m.

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

m

2

1

c

cc

CM



Square MatrixUpper Triangular MatrixLower Triangular MatrixDiagonal MatrixIdentity or Unit MatrixSymmetric MatrixSkew-symmetric MatrixNull Matrix

Type of Matrices



Type of Matrices

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

333231

232221

131211

aaaaaaaaa

A

A square matrix is a matrix in which m = n.

For a square, the main or principal diagonalconsists of the elements of the form aii; e.g., for the 3 x 3 matrix shown

the elements a11, a22, and a33 constitute the principal diagonal.



⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

33

2322

131211

u00uu0uuu

U

An upper triangular matrix is one where all the elements below the main diagonal are zero.

Type of Matrices



Type of Matrices

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

333231

2221

11

lll0ll00l

L

A lower triangular matrix is one where all elements above the main diagonal are zero.



⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

33

22

11

d000d000d

D

A diagonal matrix is a square matrix where all elements off the diagonal are equal to zero.Note that where large blocks of elements are zero, they are left blank.

Type of Matrices



⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

100010001

I

An identity or unit matrix is a diagonal matrix where all elements on the main diagonal are equal to one.

Type of Matrices



Type of Matrices

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

872731215

S

A symmetric matrix is one where aij = aji for all i’sand j’s.

7aa2aa1aa

3223

3113

2112

======



Type of Matrices

A skew-symmetric matrix is a matrix which has the property aij = -aji for all i and j; this implies aii = 0

063605350

K⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−=

5a5a

21

12

+=−=



Type of Matrices

The null matrix is matrix whose elements are

equal to zero.

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

000000000

N



Addition of MatricesProduct of a Matrix with a ScalarMultiplication of MatricesTranspose of a MatrixKron Reduction MethodDeterminant of a MatrixMinors and Cofactors of a MatrixInverse of a Matrix

Matrix Operations



Addition of Matrices

Two matrices A = [aij] and B = [bij] can be added together if they are of the same order (mxn). The sum C = A + B is obtained by adding the corresponding elements.

C = [cij] = [aij + bij]



⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

110625

B 372041

A

Example:

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡++++++

=+482666

1)(31)(70)(26)(02)(45)(1

B A

then,

⎥⎦

⎤⎢⎣

⎡ −−=⎥

⎦

⎤⎢⎣

⎡−−−−−−

=−262624

1)(31)(70)(26)(02)(45)(1

B A




⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−−++++−+

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−++++−−+

=5j64j55j74j56j41j25j71j22j3

B 9j81j13j61j13j51j43j61j42j1

A

Example:

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )⎥

⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++−+−−+++++++++++++−−+−+++

=+5j69j84j51j15j73j64j51j16j43j51j21j45j73j61j21j42j32j1

B A

then,




⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−−++++−+

=+14j145j62j135j69j92j62j132j64j4

B A


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++−+−−−−+−−++−

=−4j23j48j13j43j10j28j10j20j2

B A



Product of a Matrix with a Scalar

A matrix is multiplied by a scalar k by multiplying all elements mn by k , that is,

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

==

mn2m1m

n22221

n11211

kakaka

kakakakakaka

AkkA

L

MMMM

L

L



Example:

3 k and 162534

A =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

318615912

B


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

162534

3 Ak B



Example:

3 k and 4j1j365j2j2-56j3j14

A =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−++−+

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−++−−+

=12j39j1815j66j1518j93j12

B


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++

+==

j4-1j36j52j2-5j6-3j14

3 Ak B



Multiplication of Matrices

Two matrices A = [aij] and B = [bij] can be multiplied in the order AB if and only if the number of columns of A is equal to the number of rows of B .

That is, if A is of order of (m x l), then B should be of order (l x n).

If the product matrix is denoted by C = A B, then Cis of order (m x n). The elements cij are given by

∑=

=l

1kkjikij bac for all i and j.



[ ] [ ] [ ] n x mn x ll x m CBA =

An easy way to check whether two matrices can be multiplied.

Interior dimensions are equal multiplication is possible


Exterior dimensions definethe dimensions of the result




2x33231

2221

1211

aaaaaa

A⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡= and

2x22221

1211

bbbb

B ⎥⎦

⎤⎢⎣

⎡=

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2221

1211

3231

2221

1211

bbbb

aaaaaa

C = A B =

then

Example:

2112111111 babac +=∑= kjikij bac



⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++++++

==)baba()bab(a)baba()baba()bab(a)bab(a

AB C

2232123121321131

2222122121221121

2212121121121111


∑= kjikij bac




2x3635241

A⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

2x20987

B ⎥⎦

⎤⎢⎣

⎡=and

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

0987

635241

ABC

then

Example:




⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

24751659843

C

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++++++

==)0x68x3()9x67x3()0x58x2()9x57x2()0x48x1()9x47x1(

ABC




⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−−++++−+

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−++++−−+

=5j64j55j74j56j41j25j71j22j3

B 9j81j13j61j13j51j43j61j42j1

A

( )( ) ( )( ) ( )( ) 41j355j73j61j21j42j32j1c11 −=−−++−+++=

( )( ) ( )( ) ( )( ) 42j725j63j64j51j45j72j1c13 +=+−++−+++=

Example:

( )( ) ( )( ) ( )( ) 16j444j53j66j41j41j22j1c12 −=−−++−+−+=

( )( ) ( )( ) ( )( ) 24j295j71j11j23j52j31j4c21 +=−+++++++=




( )( ) ( )( ) ( )( ) 73j375j61j14j53j55j71j4c23 +=++++++++=( )( ) ( )( ) ( )( ) 41j204j51j16j43j51j21j4c22 +=−+++++−+=

Example:( )( ) ( )( ) ( )( ) 24j295j71j11j23j52j31j4c21 +=−+++++++=

( )( ) ( )( ) ( )( ) 144j395j69j84j51j15j73j6c33 +=++++−+++=( )( ) ( )( ) ( )( ) 15j1014j59j86j41j11j23j6c32 +=−+++−+−+=( )( ) ( )( ) ( )( ) 43j1165j79j81j21j12j33j6c31 +=−+++−+++=

[ ] [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++++++−−

=144j3915j10143j11673j3741j2024j2942j7216j4441j35

Bx A



Transpose of a Matrix

233231

2221

1211

aaaaaa

A

x⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

If the rows and columns of an m x n matrix are interchanged, the resultant n x m matrix is the transpose of the matrix and is designated by AT.

For the matrix

The transpose is

32322212

312111T

aaaaaa

Ax

⎥⎦

⎤⎢⎣

⎡=



Example:

2x3

T

3x2

654321

A

642531

A

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎦

⎤⎢⎣

⎡=

then,




Example:

then,


⎥⎦

⎤⎢⎣

⎡−+−+−+

=3j61j45j22j56j34j1

A

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−++−−+

=3j62j51j46j35j24j1

AT



Determinant of a Matrix

The solutions of two simultaneous equations:

can be obtained by eliminating the variables oneat a time. Solving for x2 in terms of x1 from the second equation and substituting this expression for x2 in the first equation, the following is obtained:

1

22

21

22

22 x

aa

ayx −=

Determinant of a 2 x 2 Matrix

)2(yxaxa)1(yxaxa

2222121

1212111

=+=+




21122211

2121221

212122121122211

1221211221212211

11

22

21

22

212111

aaaayaya x

yayax )aaaa( yaxaayaxaa

y)xaa

ay(a xa

−−

=

−=−=−+

=−+

substituting x2 and solving for x1




The expression (a11a22 – a12a21) is the value of the determinant of the coefficient matrix A, denoted by |A|.

2221

1211

aaaa

|A| =


Then, substituting x1 in either equation (1) or (2), x2 is obtained

21122211

1212112 aaaa

yayax−−

=




The determinant obtained by striking out the ith

row and jth column is called the minor element aij.

Example:

3332

1312

333231

232221

131211

aaaa

aaaaaaaaa

=


Minors and Cofactors of a Matrix

)aaaa(The 1332331221a ofminor −=



The cofactor of an element aij designated by Aij is

( )ijji

ij a of minor)1(A +−=

( )2121

213

2112

21

a of minor the 1- A )a of orminthe((-1)

)a of orminthe()1(A

==

−= +Example:



)aaaa(theSince 1332331221a ofminor −=

)aaaa(1A of cofactor the 1332331221 −−=∴



4)]6)(1()2)(1[(1246121211

)1(A 2112 −=−−−−=

−−−−= +

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

246-12-1-211

AExample:

8)]4)(1()2)(2[(1246121211

)1(A 1111 −=−−=

−−−−= +







14)]6)(2()2)(1[(1246121211

)1(A 2222 =−−=

−−−−= +

16)]6)(2()4)(1[(1246121211

)1(A 3113 −=−−−−=

−−−−= +

6)]2)(4()2)(1[(1246121211

)1(A 1221 =−−=

−−−−= +



10)]6)(1()4)(1[(1246121211

)1(A 3223 −=−−−=

−−−−= +

5)]2)(2()1)(1[(1246121211

)1(A 1331 =−−=

−−−−= +

3)]2)(1()1)(1[(1246121211

)1(A 2332 −=−−−=

−−−−= +





1A 10A 16A3A 14A 4A

5A 6A 8A

332313

322212

312111

−=−=−=−==−=

==−=

Therefore the cofactors of matrix A are:

1)]1)(1()2)(1[(1246121211

)1(A 3333 −=−−−=

−−−−= +







⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

333231

232221

131211

AAAAAAAAA

A

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−−

=135101461648

A

and in matrix form:



Inverse of a MatrixDivision does not exist in matrix algebra except in the case of division of a matrix by a scalar. However, for a given set of equations.

or in matrix form [AX] = [Y]. It is desirable to express x1, x2, and x3 a function of y1, y2, and y3, i.e.. [X] = [BY], where B is the inverse of Adesignated by A-1.

3333232131

2323222121

1313212111

yxaxaxayxaxa xay xaxaxa

=++

=++=++



If the determinant of A is not zero, the equations can be solved for x ’s as follows;

331

221

111

1 y|A|

Ay|A|

Ay|A|

Ax ++=

332

222

121

2 y|A|

Ay|A|

Ay|A|

Ax ++=

333

223

113

3 y|A|

Ay|A|

Ay|A|

Ax ++=

where A11, A12, …, A33 are cofactors of a11, a12,,a33and |A| is the determinant of A.

Inverse of a Matrix



Thus,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

==

|A|A

|A|A

|A|A

|A|A

|A|A

|A|A

|A|A

|A|A

|A|A

A B

332313

322212

312111

1-

A+ is called the adjoint of A. It should be noted that the elements of adjoint A+ are the cofactors of the elements of A, but are placed in transposed position.

Inverse of a Matrix

|A|AA 1-

+

=or



Inverse of a Matrix

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−

−=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=+

110163144

568

AAAAAAAAA

A

332313

322212

312111

Example: Get the inverse of A

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

246-12-1-211

A

the Adjoint of A is

|A|AA 1-

+

=



246121211

|A|−

−−=

The determinant of A is

4621

2)1(2611

1)1(2412

1)1(|A| 312111

−−−

−+−−

−+−

−= +++

Inverse of a Matrix

44|A|4622486244

)2)(6)(2()4)(1)(2()1)(6)(1( )2)(1)(1()1)(4)(1()2)(2(1|A|

−=−=−−−+−−=

−−−−+−+−−−−=



Inverse of a Matrix

44110163144

568

AAA 1

−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−

−

==+

−

Hence, the inverse of matrix A is

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

−−

−−

−−

−−

−−−

−−−−

441

4410

4416

443

4414

444

445

446

448

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−

−−=−

110163144

568

441A 1



Kron Reduction Method

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

4

3

2

1

44434241

34333231

24232221

14131211

3

2

1

xxxx

aaaaaaaaaaaaaaaa

0yyy

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

4

3

2

1

44434241

34333231

24232221

14131211

3

2

1

xxxx

aaaaaaaaaaaaaaaa

0yyy




⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥⎦

⎤⎢⎣

⎡

2

1

43

211

XX

AAAA

0Y

[ ] [ ] [ ][ ] [ ]( )[ ]13

1

4211 XAAAAY −−=

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

3

2

1

1

yyy

Y [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

3

2

1

1

xxx

X

[ ] [ ]42 xX =[ ] [ ]4342413 aaaA = [ ] [ ]444 aA =

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

333231

232221

131211

1

aaaaaaaaa

A [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

34

24

14

2

aaa

A




⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

4

3

2

1

3

2

1

xxxx

8742654356784321

0yyyExample:

[ ] [ ] [ ] [ ]1

1

1 X7428654

543678321

Y⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡= −




[ ] [ ] [ ]11 X74281

654

543678321

Y⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

[ ] [ ] [ ]11 X74275.0625.05.0

543678321

Y⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

Example:

[ ] [ ]11 X25.535.1375.45.225.15.321

543678321

Y⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=




[ ] [ ]11 X25.00.150.1

625.15.475.650.000

Y ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−=∴

The 4x4 matrix was reduced to a3x3matrix.



Cramer’s Rule

Matrix Inversion Method

Gaussian Elimination Method

Gauss-Jordan Method

Direct Solutions of System of Equations



Solutions of System of Equations by Cramer’ s Rule

3333232131

2323222121

1313212111

yxaxaxayxaxa xay xaxaxa

=++

=++=++

The system of three linear equations in three unknowns x1, x2, x3:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

3

2

1

3

2

1

333231

232221

131211

yyy

xxx

aaaaaaaaa

or AX = Y

written in matrix form as :



can be solved by Cramer’s Rule of determinants. The determinant of coefficient matrix A is

333231

232221

131211

aaaaaaaaa

|A| =

|A|aayaayaay

x 33323

23222

13121

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

x1 can be obtained by :

Note that values in the numerator are the values of the determinant of A with the first column were replaced by the Y vector elements.



Similarly, x2 and x3 can be obtained by:

|A|yaayaayaa

x 33231

22221

11211

3

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

and

|A|ayaayaaya

x 33331

23221

13111

2

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=



142x4x6x-7x2x x-3 2xx x

321

321

321

=++=+−=++Example:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

1473

xxx

246-12-1-211

3

2

1

Solutions of System of Equations by Cramer’s Rule



⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

246-12-1-211

A 44246121-211

|A| −=−

−=

244

88 44-

241412-7213

x1 −=−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=


|A|aayaayaay

x 33323

23222

13121

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=



144

44 44-

2146-171-231

x2 −=−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

344

132 44-

1446-72-1-311

x3 =−−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

3 x-1 x2- x 321 ===Therefore,


|A|ayaayaaya

x 33331

23221

13111

2

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

|A|yaayaayaa

x 33231

22221

11211

3

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=



Solutions of System ofEquations by Matrix Inversion

The system of equations in matrix form can be manipulated as follows:

YAXYA IXYA AXA

YAX

1-

1-

1-1-

=

=

=

=

Hence, the solution X can be obtained by multiplyingThe inverse of the coefficient matrix by the constant

matrix Y.



142x4x6x-7x2x x-3 2xx x

321

321

321

=++=+−=++Example:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

246-12-1-211

A




⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−

−−=

110163144

568

441A1-

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−

−−==

1473

110163144

568

441YA X 1-

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+−+−−++−

++−−==

)14(1 )7(10 )3(16(14)3 )7(14 )3(4

)14(5 )7(6 )3(8

441YA X 1-


From slide no.56



⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−==

312

1324488

441YA X 1-

3 x1- x2- x

3

2

1

===

Therefore:





The following matrix manipulation do not change the original sets of equations in matrix form:

(1) Interchange rows

(2) Multiply row by constant

(3) Add rows



Example:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

1424671213211

M

M

M

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

3214100103103211

M

M

M

Add row 1 to row 2 to get row 2.Add 6 times row 1 to row 3 to get row 3.


142x4x6x-7x2x x-3 2xx x

321

321

321

=++=+−=++



⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

3214100103103211

M

M

M

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

1324400103103211

M

M

M

Multiply row 2 by 10 then add to row 3 to obtained row 3.




By Back Substitution:

3)3(2)1(x10)3(3xx0132x44x00x

1

21

321

=+−+=+−=++

-2x 1- x 3 x 123 ===

Therefore:




Forwardelimination

1

Backsubstitution

The two phases ofGauss Elimination: forward elimination& back substitution.The primes indicate the number of timesthat the coefficients and constants havebeen modified.

131321211

'223

'23

'22

"33

"33

"3

"33

'2

'23

'22

1131211

3333231

2232221

1131211

a/)xaxac(xa/)xac(x

a/cx

cacaacaaa

caaacaaacaaa

−−=−=

=

⇓

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⇓

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

M

M

M

M

M

MGaussian Elimination Method



⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−1324400103103211

M

M

M

Multiply row 2 by -1.

Gauss-Jordan Method

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

1324400103103211

M

M

M

From Gauss Elimination Method slide no.75



⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−3100103103211

M

M

M

Divide row 3 by 44.

Gauss-Jordan Method

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−310010310

13501

M

M

M

Multiply row 2 by -1 then add to row 1 to get row 1.



Gauss-Jordan Method

Multiply row 3 by -5 then add to row 1 to get row 1.Multiply row 3 by 3 then add to row 2 to get row 2.

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−

310010102001

M

M

M



Gauss-Jordan Method

Therefore:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

312

xxx

100010001

3

2

1

3x 1x 2x

3

2

1

=−=−=Then,



Gauss-Jordan Method

The Gauss-Jordan method is a variation of Gauss Elimination. The major differences is that when an unknown is eliminated in the GJM, it is eliminated from all other equations rather than just the subsequent ones.

In addition, all rows are normalized by dividing them by their pivot elements. Thus, the elimination steps results in an identity matrix rather than a triangular matrix.



(n)3 3

(n)22

(n)11

(n)3

(n)2

(n)1

3333231

2232221

1131211

c x c x c x

c100c010c001

caaacaaacaaa

=

=

=

↓

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

↓

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

M

M

M

M

M

MGauss-Jordan Method

Graphical depiction of theGauss-Jordan Method.The superscript (n) meansthat the elements of the right-hand-side vector have been modified ntimes (for this case, n=3).



Gauss Iterative Method

Gauss-Seidel Method

Newton-Raphson Method

Iterative Solutions of System of Equations




An iterative method is a repetitive process for obtaining the solution of an equation or a system of equation. It is applicable to system of equations where the main-diagonal elements of the coefficient matrix are larger in magnitude in comparison to the off-diagonal elements.

The Gauss and Gauss-Seidel iterative techniques are for solving linear algebraic solutions and the Newton-Raphson method applied to the solution of non-linear equations.




The solutions starts from an arbitrarily chosen initial estimates of the unknown variables from which a new set of estimates is determined. Convergence is achieved when the absolute mismatch between the current and previous estimates is less than some pre-specified precision index for all the variables.



Example:

Assume a convergence index of ε = 0.001 and the following initial estimates:

5 3x x x6 x 4x x

4 x x 4x

3 21

32 1

321

=++=++=+−

0.5 x x x b)0.0 x x x a)

0 3

0 2

0 1

0 3

0 2

0 1

===

===




Solution:

a) The system of equation must be expressed in standard form.

)x- x 4 (41x k

3k2

1k1 +=+


) x -x - 5(31x k

2k1

1k3 =+

) x - x - 6 ( 41x k

3k1

1k2 =+



Iteration 1 (k = 0):

1.6667 x max6667.106667.1x

5.105.1x101x

1.6667 ) 0 -0 - 5(31x

1.5 ) 0 - 0 - 6 ( 41x

1.0 ) 0 - 0 4 (41x

03

03

02

01

1 3

1 2

1 1

=

=−=

=−=

=−=

==

==

=+=

Δ

ΔΔΔ

0 x x x witha) 0 3

0 2

0 1 ===




0.83334 x max 83334.06667.1833333.0x

66667.05.1833333.0x041667.01958325.0x

0.833333 ) 1.5 -1.0 - 5(31x

0.833333 ) 1.6667 - 1.0 - 6 ( 41x

0.958333 ) 1.6667 - 1.5 4 (41x

13

13

12

11

2 3

2 2

2 1

=

−=−=

=−=

=−=

==

==

=+=

Δ

ΔΔΔ






0.23617 x max 23617.08333.00695.1x

21877.0833325.00521.1x041667.0958325.01x

1.0695 ) 0.8333 -0.9583 - 5(31x

1.0521 ) 0.8333 - 0.9583 - 6 ( 41x

1.0 ) 0.8333 - 0.8333 4 (41x

23

23

22

21

3 3

3 2

3 1

=

=−=

=−=

=−=

==

==

=+=

Δ

ΔΔΔ




0.0869 x max 0869.00695.19826.0x0695.00521.19826.0x

0044.019956.0x

0.9826 ) 1.0521 -1.0 - 5(31x

0.9826 ) 1.0695 - 1.0 - 6 ( 41x

0.9956 ) 1.0695 - 1.0521 4 (41x

33

33

32

31

4 3

4 2

4 1

=

−=−=

−=−=

−=−=

==

==

=+=

Δ

ΔΔΔ






0.0247 x max

0247.09826.00073.1x0228.09826.00054.1x

0044.09956.01x

1.0073 0.9826) -0.9956 - 5(31x

1.0054 ) 0.9826 - 0.9956 - 6 ( 41x

1.0 ) 0.9826 - 0.9826 4 (41x

43

43

42

41

5 3

5 2

5 1

=

=−=

−=−=

=−=

==

==

=+=

Δ

ΔΔΔ




0.0091 x max 0091.00073.19982.ox0072.00054.19982.0x

0005.019995.0x

0.9982 ) 1.0054 -1.0 - 5(31x

0.9982 ) 1.0071 - 1.0 - 6 ( 41x

0.9995 ) 1.0073 - 1.0054 4 (41x

53

53

52

51

6 3

6 2

6 1

=

−=−=

−=−=

−=−=

==

==

=+=

Δ

ΔΔΔ






0.0026 xmax0026.09982.00008.1x0024.09982.00006.1x

0005.09995.01x

1.0008 0.9982) -0.9995 - 5(31x

1.0006 ) 0.9982 - 0.9995 - 6 ( 41x

1.0 ) 0.9982 - 0.9982 4 (41x

63

63

62

61

7 3

7 2

7 1

=

=−=

=−=

=−=

==

==

=+=

Δ

ΔΔΔ




0.0010 x max

0010.00008.19998.0x0008.00006.19998.0x

0005.019995.0x

0.9998 ) 1.0008 -1.0 - 5(31x

0.9998 ) 1.0008 - 1.0 - 6 ( 41x

0.9995 ) 1.0008 - 1.0006 4 (41x

73

73

72

71

8 3

8 2

8 1

=

−=−=

−=−=

−=−=

==

==

=+=

Δ

ΔΔΔ





The Gauss iterative method has converged at iteration 7. The method yields the following solution.

0.9998 x0.9998 x

0.9995 x

3

2

1

===





=

=

=

=

=

=

=

x max xxx

xxx

03

02

01

1 3

1 2

1 1

ΔΔΔΔ

0.5 x x x withb) 0 3

0 2

0 1 ===




x max xxx

xxx

1

13

12

11

2 3

2 2

2 1

=

=

=

=

=

=

=

Δ

ΔΔΔ







x max

xxx

xxx

2

23

22

21

3 3

3 2

3 1

=

=

=

=

=

=

=

Δ

ΔΔΔ





x max xxx

xxx

3

33

32

31

4 3

4 2

4 1

=

=

=

=

=

=

=

Δ

ΔΔΔ





x max xxx

xxx

4

43

42

41

5 3

5 2

5 1

=

=

=

=

=

=

=

Δ

ΔΔΔ





x max xxx

xxx

5

53

52

51

6 3

6 2

6 1

=

=

=

=

=

=

=

Δ

ΔΔΔ





x max xxx

xxx

6

63

62

61

7 3

7 2

7 1

=

=

=

=

=

=

=

Δ

ΔΔΔ





x max xxx

xxx

7

73

72

71

8 3

8 2

81

=

=

=

=

=

=

=

Δ

ΔΔΔ



x x x

3

2

1

===


Note: Number of iterations to achieve convergence is also dependent on initial estimates




Given the system of algebraic equations,

In the above equation, the x’s are unknown.

3nnn232131

2n2n222121

1n1n212111

yxaxaxa

yxaxa xay xaxaxa

=+++↓↓↓↓

=+++=+++

L

L

L



From the first equation,

)xaxay(a1x n1n2121

11

1 −−= K

n1n2121111 xaxayxa −−= K

Similarly, x2, x3…xn of the 2nd to the nth equations can be obtained.

↓↓↓↓

−−−=

)xaxaxa(ba1x n2n3231212

22

2 K

)xaxaxab(a1x 1-n1-nn,2n21n1n

nn

n −−−−= K




In general, the jth equation may be written

as

)xab(a1x i

n

1i jij

jj

jji

∑≠=−=

n2,1,j K=

equation “a”




In general, the Gauss iterative estimates are:

where k is the iteration count


k

n

11

1nk

3

11

13k

2

11

12

11

11k

1 xaa...x

aax

aa

ayx −−−−=+

xaa...x

aax

aa

ayx k

n22

2nk3

22

23k1

22

21

22

21k2 −−−−=+

k

1-n

nn

1-nn,k

2

nn

n2k

1

nn

n1

nn

n1k

n xa

a...x

aax

aa

ayx −−−−=+



From an initial estimate of the unknowns (x10,

x20,…xn

0), updated values of the unknown variables are computed using equation “a”. This completes one iteration. The new estimates replace the original estimates. Mathematically, at the kth iteration,

)xab(a1x kn

1i jij

jj

1kj i

ji∑

≠=

+ −=

n2,1,j K=

equation “b”




A convergence check is conducted after each iteration. The latest values are compared with their values respectively.

kj

1kj

k xxx −= +Δn2,1,j K=

equation “c”

The iteration process is terminated when

t)(convergen |x |max kj εΔ <

)convergent-(non itermax k =




Example: Solve the system of equations using the Gauss-Seidel method. Used a convergence index of ε = 0.001

5 3x x x6 x 4x x4 x x 4x

3 21

32 1

321

=++=++=+−

0.5 x x x 0 3

0 2

0 1 ===

Gauss-Seidel Method



Solution:

a) The system of equation must be expressed in standard form.

) x -x - 5(31x

) x - x - 6 ( 41x

)x- x 4 (41x

1k2

1k1

1k3

k3

1k1

1k2

k3

k2

1k1

+++

++

+

=

=

+=

Gauss-Seidel Method



Iteration 1 (k =0):

0.625 | x |max4583.050.09583.0x

625.050.0125.1x50.05.01x

0.9583 ) 1.125 -1.0 - 5(31x

1.125 ) 0.5 - 1.0 - 6 ( 41x

1.0 ) 0.5 - 0.5 4 (41x

02

03

02

01

1 3

1 2

1 1

=

=−=

=−=

=−=

==

==

=+=

ΔΔΔΔ

0.5 x x x with 0 3

0 2

0 1 ===

Gauss-Seidel Method



0.125 | x |max0323.09583.09861.0x

125.0125.11x0417.010417.1x

0.9861 ) 1.0 -1.0417 - 5(31x

1.0 ) 0.9583 - 1.0417 - 6 ( 41x

1.0417 ) 0.9583 - 1.125 4 (41x

12

13

12

11

2 3

2 2

2 1

=

=−=

−=−=

=−=

==

==

=+=

ΔΔΔΔ


Gauss-Seidel Method




0.0119 | x |max0119.09861.09980.0x

0026.010026.1x0382.00417.10035.1x

0.9980 ) 1.0026 -1.0035 - 5(31x

1.0026 ) 0.9891 - 1.0035 - 6 ( 41x

1.0035 ) 0.9861 - 1.0 4 (41x

23

23

22

21

3 3

3 2

3 1

=

=−=

=−=

−=−=

==

==

=+=

ΔΔΔΔ

Gauss-Seidel Method



0.0024 | x |max0015.09980.09995.0x

0024.00026.10002.1x0023.00035.10012.1x

0.9995 1.0002) -1.0012 -1.0 - 5(31x

1.0002 0.9980) - 1.0012 - 6 ( 41x

1.0012 )0.9980 -1.0026 4 (41x

32

33

32

31

4 3

4 2

4 1

=

=−=

−=−=

=−=

==

==

=+=

ΔΔΔΔ


Gauss-Seidel Method




εΔΔΔΔ

0.001 | x|max0004.09995.09999.0x

0001.00002.10001.1x001.00012.10002.1x

0.9999 1.0001) -1.0002 - 5(31x

1.0001 0.9995) - 1.0002 - 6 ( 41x

1.0002 )0.9995 - 1.0002 4 (41x

4

43

42

41

5 3

5 2

5 1

<=

=−=

−=−=

−=−=

==

==

=+=

Gauss-Seidel Method



The Gauss-Seidel Method has converged after 4 iterations only with the following solutions:

0.9999 x1.0001 x1.0002 x

3

2

1

===

Gauss-Seidel Method



The Gauss-Seidel method is an improvement over the Gauss iterative method. As presented in the previous section, the standard form of the jthequation may be written as follows.

)xab(a1x i

n

1i jij

jj

jji

∑≠=−= n2,1,j K=

Gauss-Seidel Method

From an initial estimates (x10, x2

0,…xn0), an updated

value is computed for x1 using the above equation with j set to 1.This new value replaces x1

0 and is then used together with the remaining initial estimates to compute a new value for x2. The process is repeated until a new estimate is obtained for xn. This completes one iteration.



Note that within an iteration, the latest computed values are used in computing for the remaining unknowns. In general, at iteration k,

)xab(a1x i

n

1i jij

jj

1k

jij

α∑≠=

+ −=

n2,1,j K=

Gauss-Seidel Method

j i if 1k j i if kwhere

<+=>=α

After each iteration, a convergence check is conducted. The convergence criterion applied is the same with Gauss Iterative Method.



An improvement to the Gauss Iterative Method

Gauss-Seidel Method

xaa...x

aa

ayx k

n11

1nk2

11

12

11

11k

1−−−=

+

xaa...x

aa

ayx k

n22

2n1k1

22

21

22

21k

2−−−= ++

1kn

ii

in1k1i

ii

1ii,1k1-i

ii

1-ii,1ki

ii

ij

ii

i xaax

aa

xaa

...xaa

ayx

1k

i

+++

+++ −−−−−=+

xa

a...x

aa

ayx 1k

1-nnn

1-nn,1k1

nn

n1

nn

n1k

n

++ −−−=+



The Newton-Raphson method is applied when the system of equations is non-linear.

Consider a set of n non-linear equations in n unknowns.


)x,,x,(xfy

)x,,x,x(fy)x,,x,x(fy

n21nn

n2122

n2111

K

M

K

K

=

==



0

n0

n

10

20

2

10

10

1

1011 x)x(

xfx)x(

xfx)x(

xf)x(fy ΔΔΔ

∂∂

++∂∂

+∂∂

+= K

0

n0

n

20

20

2

20

10

2

2022 x)x(

xfx)x(

xfx)x(

xf)x(fy ΔΔΔ

∂∂

++∂∂

+∂∂

+= K

0

n0

n

10

20

2

10

10

n

n0nn x)x(

xfx)x(

xfx)x(

xf)x(fy ΔΔΔ

∂∂

++∂∂

+∂∂

+= K

M MM M


The system of non-linear equations can be linearized using Taylor’s Series



Where:X0 = (x1

0,x20, …, xn

0)= set of initial estimates

fi(x0) = the function fi (x1,x2, …, xn)evaluated using the set of initial estimates.

= the partial derivatives of the function fi(x1,x2,…,xn) evaluated using the set of original estimates.

j

0i

x)x(f

∂∂




⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

−−

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

0n

02

01

x)x(f

x)x(f

x)x(f

x)x(f

x)x(f

x)x(f

x)(xf

x)(xf

x)x(f

0nn

022

011

x

xx

)x(fy

)x(fy)x(fy

n

0n

2

0n

1

0n

n

02

2

02

1

02

n

01

2

01

1

o1

Δ

ΔΔ

M

K

MMM

K

K

M

The equation may be written in matrix form as follows:

The matrix of partial derivatives is known as the Jacobian. The linearized system of equations may be solved for ∆x’s which are then used to update the initial estimates.




n..., 2, 1, j xxx k

j

k

j

1kj =+== Δ

At the kth iteration:

n..., 2, 1, j )x(fy 1k

jj =≤− ε

n ..., 2, 1, j x 2

k

j =≤ εΔor

Convergence is achieved when

Where ε1 and ε2 are pre-set precision indices.




Example: Solve the non-linear equation

1x ,1x :use

2xx24x4x

0

2

0

1

21

221

−==

=−=−

2211 x4xf −=

First, form the JacobianSolution:


11

1 x2xf

=∂∂ 4

xf

2

1 −=∂∂

212 xx2f −= 1xf

2

2 −=∂∂ 2

xf

1

2 =∂∂



⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=⎥⎦

⎤⎢⎣

⎡−−

2

1

2

02

1

02

2

01

1

01

022

011

xx

x)x(f

x)x(f

x)x(f

x)x(f

)x(fy)x(fy

ΔΔ

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

=⎥⎦

⎤⎢⎣

⎡−−−−

2

11

21

221

xx

124-x2

)xx2(2)x4x(4

ΔΔ


Jacobian Matrix

In Matrix form



4xf

2 (1)2xf

4y ,5)1(41)x(f

2

1

1

1

120

1

−=∂∂

==∂∂

==−−=

1xf

2xf

2y ,3)1()1(2)x(f

2

2

1

2

20

2

−=∂∂

=∂∂

==−−=

Newton-Raphson MethodIteration 0:



The equations are:

02

01

02

01

x)1(x)2(32

x)4(x)2(54

ΔΔ

ΔΔ

−+=−

−+=−⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−−

=⎥⎦

⎤⎢⎣

⎡−−

02

01

xx

1242

11

ΔΔ

In matrix form:

0x

5.0x0

2

0

1

=

−=

Δ

Δ

101x

5.0)5.0(1x1

2

11

−=+−=

=−+=

Solving,

Thus,




Repeating the process with the new estimates,Iteration 1:

4x

)x(f

0.1)5.0(2x

)x(f

4y ,25.4)1(4)5.0()x(f

2

11

1

11

121

1

−=∂

∂

==∂

∂

==−−=

1x

)x(f

2x

)x(f

2y ,2)1()5.0(2)x(f

2

12

1

12

21

2

−=∂

∂

=∂

∂

==−−=




The equations are: In matrix form:

Solving,

Thus,


1

2

1

1

1

2

1

1

xx222

x4x25.44

ΔΔ

ΔΔ

−=−

−=−⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−−

=⎥⎦

⎤⎢⎣

⎡−1

2

1

1

xx

1241

025.0

ΔΔ

07143.0x

03571.0x1

2

1

1

=

=

Δ

Δ

92857.007143.01x

53571.003571.05.0x2

2

2

1

−=+−=

=+=



Repeating the process with the new estimates,Iteration 2:

4x

)x(f

07142.1)53571.0(2x

)x(f4y ,001265.4)92857.0(4)53571.0()x(f

2

11

1

21

122

1

−=∂

∂

==∂

∂

==−−=

1x

)x(f

2y 2x

)x(f299999.1)92857.0()53571.0(2)x(f

2

22

2

1

22

22

−=∂

∂

==∂

∂

≅=−−=




The equations are:

In matrix form:

Solving,

Thus,


2

2

2

1

2

2

2

1

xx20.22

x4x07142.1001265.44

ΔΔ

ΔΔ

−=−

−=−

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−−

=⎥⎦

⎤⎢⎣

⎡−2

2

2

1

xx

12407142.1

0001265.0

ΔΔ

00036.0x

00018.0x2

2

2

1

=

−=

Δ

Δ

92893.000035.092857.0x

53553.000018.053571.0x3

2

3

1

−=−−=

=−=



4y ,2512400.4)92893.0(4)53553.0()x(f 123

11 ==−−=2y ,99928.1)92893.0()53553.0(2)x(f 2

3

12 ==−−=

00072.0fy0025.0fy

22

11

=−−=−

92893.0x53553.0x

2

1

−==

Substituting to the original equation:

Therefore,

Note the rapid convergence of the Newton-RaphsonMethod.




Electric Circuits and Power System Representation

1. Electric Circuits

2. Power System Representation

3. Network Model – Bus Admittance Matrix



Electric Circuits

LElectric Current

T D

G

VOLTAGE (V) is the electromotive force or the pressure which causes electric current to flow. The unit of measurement for Voltage is Volts

Electric CURRENT (I) is the rate at which electrons flow through a circuit (wires). The unit of measurement for Electric Current is Amperes.

Electric Energy Produced by Power Plants

Electric Energy Consumed by

Electrical Loads



V+

- Z

Electric Circuits

Water Pressure causes the

flow of water=

Voltage

Water Flow = Current

VI

Analogy

pedanceIm

VoltageCurrent =I

Z

OHM’s LAW



Electric Circuits

Power is the rate at which Energy is generated, transmitted, distributed or consumed (measured in watts, kW or MW)

)h(Time)kWh(Energy)kW(Power = kW

2

1 2

1

Hrs

5

3

4

10 k

WH

10 kWH

2

1 2

1

543

Energy is the Work done (e.g., light or heat produced) when electric current flows in electric circuits (measured in watt-hours or kWh)

Hrs

kW

10 kWH of Energy in 5 Hrs Vs. 2 Hrs



Voltage and Current Directions Polarity Marking of Voltage Source Terminals:

Plus sign (+) for the terminal where positive current comes out

Specification of Load Terminals: Plus sign (+) for the terminal where positive current enters

Specification of Current Direction:

Arrows for the positive current (i.e., from the source towards the load)

Vs ZB

+

-

Electric Circuits

++

-

-

VA

ZA

VB

a b

o n

I

II

I



The letter subscripts on a voltage indicate the nodes of the circuit between which the voltage exists.

The first subscript denotes the voltage of that node with respect to the node identified by the second subscript.

+

-

+ -VA

ZAa b

o n

VS = Vao Vb = VbnI = Iab Zb

+

-

Vao - IabZA - Vbn = 0

Iab =Vao - Vbn

ZA

Electric Circuits

VS - VA - Vb = 0

Double Subscript Notations



Electric Circuits

Voltage, Current and Phasor Notation

V

I

θ

V = |V| ∠ 0° volts

I = |I| ∠ - θ amperes



Complex Impedance and Phasor Notation

Electric Circuits

+

-

+

-

VLi(t)R (Resistance)

L (Inductance)tj

mS VV ωε=

The first order linear differential equation has a particular solution of the form . tjK)t(i ωε=

tjmV

dt)t(diL)t(Ri ωε=+Applying Kirchoff’s voltage law,

tjm

tjtj VLKjRK ωωω εεωε =+Hence, 1j −=

tsinjtcostj ωωε ω +=



Electric Circuits


Solving for the current

Dividing voltage by current to get the impedance,

tjm

LjRV

)t(i ωεω+

=

LjR

LjRVV

)t(i)t(vZ

tjm

tjm ω

εω

εω

ω

+=

+

==

Therefore, the impedance Z is a complex quantity with real part R and an imaginary (j) part ωL

ωL

Rθ

Z



+

-

+

-

VLi(t)R (Resistance)

C (Capacitance)tj

mS VV ωε=

For Capacitive Circuit, .

)C1(jRZω

−=


Electric Circuits

Z= |Z|ejφ or Z = |Z|(cosφ + jsinφ) or Z = |Z|∠φ

φR

Z

1ωC



v = 141.4 cos(ωt + 30°) voltsi = 7.07 cos(ωt) amperesVmax = 141.4 |V| = 100 V = 100∠30Imax = 7.07 |I| = 5 I = 5∠0


Electric Circuits

Z= |Z|ejφ or Z = |Z|(cosφ + jsinφ) or Z = |Z|∠φ

10

17.32

30°203020

0530100Z ∠=

∠∠

=

)10j32.17)30sinj30(cos20Z +=+=

I30o

5

10V



Complex Power and Power Triangle

Power = Voltage x Current*

• Delivering bulk power (large amount of energy in short time) will require large current.

• Delivering the same bulk power in higher voltage will result in lower current.

Electric Circuits



The Complex Power (S) can be obtained from the product VI*. It’s real part equals the average power Pand it’s imaginary part is equal to the reactive power Q.

Consider, V = |V|∠α and I = |I|∠β

S = VI* = |V| ejα |I| e-jβ

= |V| |I| ej(α -β)

= |V| |I| ∠(α -β)= |V| |I|cos(α -β) + j|V| |I|sin(α -β) = |V| |I|cosθ + j|V| |I|sin θ= P + jQ

Electric Circuits




The equations associated with the average, apparent andreactive power can be developed geometrically on a right triangle called the power triangle.

Apparent Power (S) = voltage x current = VI

Average Power (P) = voltage x in-phase component of current= VI cos θ

Reactive Power Q = voltage x quadrature comp. of current= VI sin θ

(note: Average Power is also called Real Power or Active Power)

Electric Circuits




V

I

S = VI

θ

θ

V

I

Q = VI sin θleading

S = VI

θ

θ


P = VI cos θ

Q = VI sin θlagging

P = VI cos θ

Inductive Circuit Capacitive Circuit

Electric Circuits

Power Factor

θθ cosVIcosVI

)S( Power pparentA)P( Power alRePF

==

=



Q = VI sin θ

Electric Circuits


P = VI cos θ

S = VI

θ

Power Factor

VIcosVI

)S( Power pparentA)P( Power alRePF θ

==

θcosPF =

Measure of Efficient Utilization of Power



EXAMPLE:Given a circuit with an impedance Z = 3 + j4 and anapplied phasor voltage V = 100∠ 30°, determine the power triangle.

Electric Circuits


I = V/Z = (100∠30°) / (5∠53.1°) = 20∠-23.1°S = VI = 100(20) = 2000 VAP = VI cos θ = 2000 cos 53.1° = 1200 WQ = VI sin θ = 2000 sin 53.1° = 1600 Vars laggingpf = cos θ = cos 53.1° = 0.6 lagging



Alternative Solution:S = VI* = (100∠30°) /

(20∠23.1°) = 2000∠53.1°= 1200 + j1600

We get,P = 1200 WQ = 1600 VArs laggingS = 2000 VA andpf = cos 53.1° = 0.6 lagging

P = 1200 W

Q = 1600 varslagging

S = 2000 VA

53.1°

Electric Circuits




Electric CircuitsBalanced Three-Phase System

Va = |V| ∠ 30° volts

Vb = |V| ∠ 270° volts

Vc = |V| ∠ 150° volts

Vc

Vb

Va



Balanced Three-Phase System

Electric Circuits

ZR

ZR

nZREao = |E|∠0° V

Ebo= |E| ∠240° V

Eco = |E| ∠120° V

abc

Eao

Eco

Ebo

Phase Sequence

Line currentsIa = Ian = Eao /ZR = Van / ZR = |I|∠(0-φ) Ib = Ibn = Ebo /ZR = Vbn / ZR = |I|∠(240-φ ) Ic = Icn = Eco /ZR = Vcn / ZR = |I|∠(120-φ)

a

b

c

Ia

Ic

Ib

φ



Ib Ic

Ia

Ib

Ic


Electric Circuits

Ia

Note: In is not equal to zero for unbalanced load

In = Ia + Ib = Ic = 0



Line-to-line voltage

Vab = Van + Vnb = Van – Vbn = |Van |∠0° – |Van |∠240° = |Van |[1 – a2 ]

Vab = √3 Van ∠30°

Van = | Van |∠0°

Vcn = aVanVca = √3 Vcn ∠30°

Vbc = √3 Vbn ∠30°

Vbn = a²Van

a = 1/120


Electric Circuits

o303∠



SOLUTION:With Vab as reference,Vab = 173.2 ∠ 0° Van = 100 ∠ -30°Vbc = 173.2 ∠ 240° Vbn = 100 ∠ 210°Vca = 173.2 ∠ 120° Vcn = 100 ∠ 90°

EXAMPLE:

Balanced 3φ circuit Vab = 173.2 ∠ 0°

Determine all the voltages and currents in a Y-connected load having ZL = 10 ∠ 20°Ω. Assume that the phase sequence is abc.


Electric Circuits

VanVbn

Vcn

a

c

b

nVca

Vab

Vbc



Each current lags the voltage across it’s load impedance by 20°and each current magnitude is 10 A.

Ian = 10∠ -50° Ibn = 10∠ 190° Icn = 10∠ 70°

Ian

Ibn

Icn

Van

70°120°

120°30°

20°


Electric Circuits



Choosing Iab as reference,Ic = √3 Ica ∠-30°

Ib = √3 Ibc ∠-30° Ia = √3 Iab ∠-30°

Ica = aIab

Iab

Ibc = a²Iab


Electric Circuits



Y - connected 3-phase System Δ - connected 3-phase System

VOLTAGE AND CURRENT RELATIONSHIPS


Electric Circuits

L

LNLL

IIV3V

==

θ LLLn

l

VV3/II

==θ



Single-Phase and Three-Phase Power

)tcos(I)t(itcosV)t(v

m

m

θωω

−==

t2sinsinIV21t2cosIV

21cosIV

21

t2sinsinIV21)t2cos1(cosIV

21

]sintsintcoscost[cosIV)sintsincost[costcosIV

)tcos(tcosIV)tcos(tIcosV)t(i)t(v

mmmmmm

mmmm

2mm

mm

mm

mm

ωθωθ

ωθωθ

θωωθωθωθωω

θωωθωω

++=

++=

+=

+=−=−=

t2sin21tsintcos ωωω =

Electric Circuits

)t2cos1(21tcos2 ωω +=




I2I

V2V

m

m

=

=

θ

θ

θ

sinVI

sin2

I2V2

sinIV21Q mm

=

=

=

Electric Circuits

θ

θ

θ

cosVI

cos2

I2V2

cosIV21P mmave

=

=

=



)120tcos(V)t(v)120tcos(V)t(v

tcosV)t(v

0mc

0mb

ma

+=

−=

=

ωωω

)120tcos(I)t(i)120tcos(I)t(i

)tcos(I)t(i

0mc

0mb

ma

θωθω

θω

−+=

−−=

−=

)t(i)t(v)t(i)t(v)t(i)t(v)VA(AmpereVolt ccbbaa3 ++=− φ

Electric Circuits

θ

θφφ

cosIV3

cosIV3P3P

LLL

pp

1,ave3,ave

=

=

=

θ

θφφ

sinIV3

sinIV3Q3Q

LLL

pp

13

=

=

=|I||I|

|V|3|V|

connectedY

pL

pLL

=

=

−

|I|3|I|

|V||V|connected

pL

pLL

=

=−Δ




Single Phase Equivalent of Balanced Three-Phase System

Electric Circuits

ZR

ZR

ZR

nEao = |E|∠0° V

Ebo= |E| ∠240° V

Eco = |E| ∠120° V

a

b

c

o

Ic

b

c

Iaa

Ia



Electric Circuits

ZR

nEao = |E|∠ 0° V

ao

ZRZR

n

Ebo= |E| ∠240° V b

o

ZR

ZR

n

Eco = |E| ∠120° V c

o

cb

a

Ic

Ib

Ia

θθ

−∠=∠∠

= IZ

0EIR

a

)240(IZ

240EIR

b θθ

−∠=∠

∠=

)120(IZ

120EIR

c θθ

−∠=∠

∠=

Note: Currents are Balanced



Single Phase Representation of a Balanced Three-Phase System

Electric Circuits

Eao = |E|∠ 0° V

n

a

o

a

Ia

ZR

ZR

ZR

nEao = |E|∠0° V

Ebo= |E| ∠240° V

Eco = |E| ∠120° V

a

b

c

o

Ic

b

c

Iaa

Ia

ZR



EXAMPLE:The terminal voltage of a Y-connected load consisting of three equal impedances of 20∠ 30° Ω is 4.4 kV line-to-line. The impedance of each of the three lines connecting the load to a bus at a substation is ZL = 1.4∠ 75° Ω. Find the line-to-line voltage at the substation bus.

1.4 ∠75° Ω

20 ∠30° Ω

+

-

+

-n

SubstationVLN

a


Electric Circuits

Load VLN



SOLUTION:The magnitude of the voltage to neutral at the load is 4400/√3 = 2540 V.

If Van (voltage across the load) is chosen asreference,Van = 2540 ∠ 0°

andIan = 2540 ∠ 0° / 20 ∠ 30° = 127 ∠ -30°

The line-to-neutral voltage at the substation is

Van + IanZL = 2540 ∠ 0° + 127 ∠ -30° (1.4 ∠ 75°)

= 2666 + j125.7 = 2670 ∠ 2.70° V


Electric Circuits



The magnitude of the voltage at the substation is √ 3 (2.67) = 4.62 kV

127 ∠-30° A

2670 ∠2.7°

1.4 ∠75° Ω

20 ∠30° Ω 2540 ∠0° V

+

-

+

-


Electric Circuits



Electrical Symbols

Generator

TransformerCircuit Breaker

Transmission or Distribution Line

Bus

or

G

Power System representation

Switch

Node

Fuse



Electrical Symbols

3-phase wye neutral grounded

3-phase delta connection

Ammeter

Voltmeter

3-phase wye neutral ungrounded

Protective Relay

V

A

R

Power System representation

Current Transformer

Potential Transformer



Three Line DiagramThe three-line diagram is used to represent each phase of a three-phase power system.

Relays

Cir

cuit

Bre

aker

Mai

n B

us

RR

R

R

CTs Distribution Lines

Power System Representation

Transformer

a b c



The three-line diagram becomes rather cluttered for large power systems. A shorthand version of the three-line diagram is referred to as the Single Line Diagram.

Single Line Diagram

R

CBTransformerBus

Distribution Line

CT and Relay


Node




Equivalent Circuit of Power System Components: Generator

Ec

Eb

b

a

c

Ea

Ic

Ib

Iasa jXR +

aI

-

+

aV+

-gE

Za

Zb Zc

Three-Phase Equivalent Single-Phase Equivalent



Power System RepresentationEquivalent Circuit of Power System Components: Transformer

A

BC

a

bc

Core Loss

Primary Secondary

6x6 Admittance Matrix

Y11 Y12 Y13 Y14 Y15 Y16

Y21 Y22 Y23 Y24 Y25 Y26

Y31 Y32 Y33 Y34 Y35 Y36

Y41 Y42 Y43 Y44 Y45 Y46

Y51 Y52 Y53 Y54 Y55 Y56

Y61 Y62 Y63 Y64 Y65 Y66

Three-Phase Equivalent



X2

HT RaRR +=

X2

HT XaXX +=HIr+

-HVr

+

-XVar

mjXcR

exIv

TT jXR +

Power System RepresentationEquivalent Circuit of Power System Components: Transformer

+ +

- -

TZ

Single-Phase Equivalent




Equivalent Circuit of Power System Components: Distribution Lines

ZccZcbZca

ZbcZbbZba

ZacZabZaa

Equivalent π-Network

ABC

abc

YccYcbYca

YbcYbbYba

YacYabYaa

1/2YccYcbYca

YbcYbbYba

YacYabYaa

1/2

Capture Unbalanced

Characteristics Three-Phase Equivalent




Equivalent Circuit of Power System Components: Transmission Lines

T&D Lines can be represented by an infinite series of resistance and inductance and shunt capacitance.

Equivalent circuit depends on the length of the Linesa) Short Lineb) Medium Linec) Long Line

and whether the system is balance or not.Δl




Equivalent Circuit of Power System Components: Long Transmission Lines

• •

••+

-

Vs

+

-

IS IR VR

Length = Longer than 240 km. (150 mi.)lsinhZ'Z c γ=

yzZ C =

Characteristic Impedance zy=γPropagation Constant

2ltanh

Z1

2'Y

c

γ=

2'Y





Equivalent Circuit of Power System Components: Medium-Length

Transmission LinesLength = 80 – 240 km. (50 - 150 mi.)

• •

••+ +

--

VsIS IR

( )ljxrZ L+=

2c/1

2Y ω=

2Y VR





Equivalent Circuit of Power System Components: Short Transmission Lines

Length = up to 80 km. (50 mi.)

• •

••+ +Is = IR

-

Vs VR

-Single-Phase Equivalent

( )ljxrZ L+=




Impedance and Admittance Diagrams

21 3

4

bcaBus

1234

Gen

abc

Line

1 - 32 - 31 - 42 - 43 - 4

Single Line Diagram





21 3

4

Impedance Diagram

0

0

0

1

31

Ea za

zd

za

ze

zf zg

zb

Ea Ec Eb

zhz13

zcGenerator

Line





VL

ILZpIs

Zg

+

-Eg

VL

IL

The two sources will be equivalent if VLand IL are the same for both circuits.

Eg = ISZPZg = Zp





21 3

4

Admittance Diagram

0

y13 y23

y14 y24

I1 I3 I2y03y01 y02

I1 = Ea/zay01 = 1/za

I2 = Eb/zby02 = 1/zb

I3 = Ec/zcy03 = 1/zc

y13 = 1/zd

y23 = 1/ze

y14 = 1/zf

y24 = 1/zg

y34 = 1/zh

y34



at node 1:( ) ( ) 144113310111 yVVyVVyVI −+−+=

at node 4:( ) ( ) ( ) 343424241414 yVVyVVyVV0 −+−+−=

at node 2:( ) ( ) 244223320222 yVVyVVyVI −+−+=

at node 3:( ) ( ) ( ) 1313344323230333 yVVyVVyVVYVI −+−+−+=


Applying Kirchoff’s Current Law



Rearranging the equations,( ) 14413314130111 yVyVyyyVI −−++=

( ) 3432421413424144 yVyVyVyyyV0 −−−++=( ) 1313442321334230333 yVyVyVyyyyVI −−−+++=

In matrix form,

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

++−−−+++

++++

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

4

3

2

1

342414342414

34133423032313

2423242302

1413141301

3

2

1

VVVV

yyyyyyy-yyyyy-y-y-y-yyy0y-y-0yyy

0III


( ) 24423324230222 yVyVyyyVI −−++=



The standard form of n independent equations:

Network ModelBus Admittance Matrix

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

n

3

2

1

I III

M

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

nn3n2n1n

n3333231

n2232221

n1131211

YYYY

YYYY

YYYY

YYYY

L

MMMM

L

L

L

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

n

3

2

1

V VVV

M

[ I ] = [Ybus][V]

Ybus is also called Bus Admittance Matrix



⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

nn3n2n1n

n3333231

n2232221

n1131211

YYYY

YYYY

YYYY

YYYY

L

MMMM

L

L

L

[YBUS] =

Ypp = self-admittance, the sum of all admittances terminating on the node (diagonal elements)

Ypq = mutual admittance, the negative of the admittances connected directly between the nodes identifed by the double subscripts

Network ModelBus Admittance Matrix



The Per Unit System

Per Unit Impedance

Changing Per Unit Values

Consistent Per Unit Quantities of Power System

Advantages of Per Unit Quantities

Per Unit Quantities



The Per Unit System

Base Value

Actual ValuePer Unit Value =

Per-unit Value is a dimensionless quantity

Per-unit value is expressed as decimal

100

Actual ValuePercent =

Per Unit Value



Base Power

Actual Value of PowerPer Unit Power =

Base Voltage

Actual Value of VoltagePer Unit Voltage =

Base Current

Actual Value of CurrentPer Unit Current =

Base Impedance

Actual Value of ImpedancePer Unit Impedance =

The Per Unit System

Per Unit Value



PU VoltagePU Current =

PU Impedance

PU Power = PU Voltage x PU Current

The Per Unit System

Per Unit Calculations



I

Zline = 1.4 ∠75° Ω

Zload = 20 ∠30° Ω 2540 ∠0° V

-

+

-

+

Example:

Vs = ?

Determine Vs

The Per Unit System

Per Unit Calculations



Choose: Base Impedance = 20 ohms (single phase)Base Voltage = 2540 volts (single phase)

PU Impedance of the load = 20∠30°/20 = ______ p.u.PU Impedance of the line = 1.4∠75°/20 = ______ p.u.PU Voltage at the load = 2540∠0° /2540 = ______ p.u.

Line Current in PU = PU voltage / PU impedance of the load= ______ / ______ = ______ p.u.

PU Voltage at the Substation = Vload(pu) + IpuZLine(pu)= ________ + _______ x _______ = _______ p.u.

The Per Unit SystemPer Unit Calculations



The magnitude of the voltage at the substation is 1.05 p.u. x 2540 Volts = _______ Volts

1.0 ∠-30° p.u.

1.05∠2.70°

0.07 ∠75° p.u.

1.0∠30° p.u. 1.0∠0° p.u.

+

-

+

-

The Per Unit SystemPer Unit Calculations



1. Base values must satisfy fundamental electrical laws (Ohm’s Law and Kirchoff’sLaws)

2. Choose any two electrical parameters

• Normally, Base Power and Base Voltage are chosen

3. Calculate the other parameters

• Base Impedance and Base Current

The Per Unit SystemEstablishing Base Values



Base PowerBase Current =

Base Voltage

Base Voltage (Base Voltage)2

Base Impedance = = Base Current Base Power

For a Given Base Power and Base Voltage,

Establishing Base Values

The Per Unit System



For Single Phase System,

Pbase(1φ)------------Vbase(1φ)

Ibase =

Vbase(1φ)------------Ibase(1φ)

Zbase =

[Vbase(1φ)]²= ------------Pbase(1φ)

For Three Phase System,

Pbase(3φ)------------√3Vbase(LL)

Ibase =

Vbase(LN)------------Ibase(L)

Zbase =

[Vbase(LL)]²= ------------Pbase(3φ)




3kV Base kV Base

MVABase31 MVA Base

LL1

31

=

=

φ

φφ

• Base MVA is the same base value for Apparent, Active and Reactive Power

• Base Z is the same base value for Impedance, Resistance and Reactance

• Base Values can be established from Single Phase or Three Phase Quantities




Base kVA1φ = 10,000 kVA= 10 MVA

Base kVLN = 69.282 kV

Base Z = (69.282)2/10= 480 ohms

Base kVA3φ = 30,000 kVA= 30 MVA

Base kVLL = 120 kV

Base Z = (120)2/30= 480 ohms

Amps 144.34 )120(3

1000x30 Current Base

=

=

Amps 144.34 282.691000x10 Current Base

=

=

Example:




Manufacturers provide the following impedance in per unit:1. Armature Resistance, Ra2. Direct-axis Reactances, Xd”, Xd’ and Xd3. Quadrature-axis Reactances, Xq”, Xq’ and Xq4. Negative Sequence Reactance, X25. Zero Sequence Reactance, X0

The Base Values used by manufacturers are:1. Rated Capacity (MVA, KVA or VA)2. Rated Voltage (kV or V)

{ }

}Positive Sequence Impedances

Per Unit Impedance Generators



base

)(L)pu(L Z

XX Ω=

base

)()pu( Z

RR Ω=

base

)(C)pu(C Z

XX Ω=

Per Unit ImpedanceTransmission and Distribution Lines



The ohmic values of resistance and leakage reactance of a transformer depends on whether they are measured on the high- or low-tension side of the transformer

The impedance of the transformer is in percent or per unit with the Rated Capacity and Rated Voltages taken as base Power and Base Voltages, respectively

The per unit impedance of the transformer is the same regardless of whether it is referred to the high-voltage or low-voltage side

The per unit impedance of the three-phase transformer is the same regardless of the connection

Per Unit ImpedanceTransformers



ExampleA single-phase transformer is rated 110/440 V, 2.5 kVA.The impedance of the transformer measured from the low-voltage side is 0.06 ohms. Determine the impedance in per unit (a) when referred to low-voltage side and (b) when referred to high-voltage side

Solution

Low-voltage Zbase = = ______ ohms 1000/5.2

110.0 2

PU Impedance, Zpu = = ______ p.u.

Per Unit Impedance



High Voltage, Zbase = = _______ ohms

PU Impedance, Zpu = = _______ p.u.

If impedance had been measured on the high-voltage side, the ohmic value would be

ohmsZ _______11044006.0

2

=⎟⎠⎞

⎜⎝⎛=

Note: PU value of impedance referred to any side of the transformer is the same

Per Unit Impedance



Example:Consider a three-phase transformer rated 20 MVA, 69 kV/13.2 kV voltage ratio and a reactance of 7%. The resistance is negligible.

a) What is the equivalent reactance in ohms referred to the high voltage side?

b) What is the equivalent reactance in ohms referred to the low voltage side?

c) Calculate the per unit values both in the high voltage and low voltage side at 100 MVA.

Changing Per-Unit Values



SOLUTION:

a) Pbase = 20 MVAVbase = 69 kV (high voltage)

( kV)² = ________ ohms

( MVA)Zbase =

Xhigh = Xp.u. x Zbase = _______ x _______= _______ ohms




b) Pbase = 20 MVAVbase = 13.2 kV (low voltage)

= ________ ohms( kV)²

( MVA)Zbase =

Xlow = Xp.u. x Zbase = _______ x _______= _______ ohms

SOLUTION:




c) Pbase = 100 MVAVbase,H = 69 kV

(69)² = ________ ohms

100Zbase,H =

ohms = ______ p.u.Xp.u.,H =

Vbase,L = 13.2 kV(13.2)²

= _______ ohms100

Zbase,L =

= ______ p.u.Xp.u.,L =Note that the per unit quantities are the same regardless of the voltage level.

ohms

ohms

ohms




Three parts of an electric system are designated A, B and C and are connected to each other through transformers,as shown in the figure. The transformer are rated as follows:

A-B 10 MVA, 3φ, 13.8/138 kV, leakage reactance 10%B-C 10 MVA, 3φ, 138/69 kV, leakage reactance 8%

Determine the voltage regulation if the voltage at the load is 66 kV.

SOURCE A B C

A-B B-C

300 Ω/ φPF=100 %LOAD




A-B B-C13.8/138 kV 138/69 kV

XBC=8%XAB=10%

VcVA

Solve using actual quantities

300 Ω/ φPF=100%




SOLUTION USING PER UNIT METHOD:

Pbase = 10 MVAVA,base = 13.8 kVVB,base = 138 kVVC,base = 69 kV

(69)² -------- = _____ ohms

10ZC,base =

---------- = ______ + j _____ p.u.ZLOAD,p.u. =

-------------- = _______ p.u.VC,p.u. =

---------------- = _______ p.u.Ip.u. =




VA = _______ + ( ________ ) x ( ________ + ________ )= ________ + j ________ p.u.= _________ p.u.

VNL - VL---------------- x 100%

VL

V.R. =


------------------------ x 100%V.R. =

=



Consider the previous example, What if transformer A-B is 20 MVA instead of 10 MVA. The transformer nameplate impedances are specified in percent or per-unit using a base values equal to the transformer nameplate rating.

The PU impedance of the 20 MVA transformer cannot be added to the PU impedance of the 10 MVA transformer because they have different base values

The per unit impedance of the 20 MVA can be referred to 10 MVA base power




Convert per unit value of 20 MVA transformer,

Pbase = 20 MVA (Power Rating)Vbase,H = 138 kV (Voltage Rating)

(138)² ---------- = _______ ohms

20Zbase,H =

0.10 p.u. x _______ ohms = _______ ohmsXactual,H =

At Pbase = 10 MVA (new base)(138)²

---------- = 1904.4 ohms10

Zbase,H =95.22

---------- = 0.05 p.u. 1904.4

Xp.u.(new) =

The per unit impedance of the 20MVA and 10 MVA transformer can now be added.




Zactual = Zpu1 • Zbase1 Zactual = Zpu2 • Zbase2

2base2pu1base1pu ZZZZ ⋅=⋅

2base

1base1pu2pu Z

ZZZ ⋅=

Note that the transformer can have different per unit impedance for different base values (i.e., the actual ohmicimpedances of the equipment is independent of the selected base values), then




Recall: ( )Powerbasevoltagebase

Z2

base =

( )

( )2,3

22,

1,3

21,

12

base

baseLL

base

baseLL

pupu

MVAkVMVAkV

ZZ

φ

φ=

Then,

or,⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

φ

φ

1base,3

2base,32

2base,LL

1base,LL1pu2pu MVA

MVAkVkV

ZZ




A three-phase transformer is rated 400 MVA, 220Y/22 ΔkV. The impedance measured on the low-voltage side of the transformer is 0.121 ohms (approx. equal to the leakage reactance). Determine the per-unit reactance of the transformer for 100 MVA, 230 kV base values at the high voltage side of the transformer.

Example


)given(base

)new(base

2)new(base

2)given(base

)given.(u.p)new.(u.p PP

x]V[]V[

xZZ =



Solution

On its own base the transformer reactance is

On the chosen base the reactance becomes

= ________ puX =( )

( )2

( )

X = ( ) x x = ________ pu( )2

( )2( )





Procedure:a) Establish Base Power and Base Voltages

• Declare Base Power for the whole Power System

• Declare Base Voltage for any one of the Power System components

• Compute the Base Voltages for the rest of the Power System Components using the voltage ratio of the transformers

Note: Define each subsystem with unique Base Voltage based on separation due to magnetic coupling



b) Compute Base Impedance and Base Current

• Using the Declared Base Power and Base Voltages, compute the Base Impedances and Base Currents for each Subsystem

c) Compute Per Unit Impedance

• Using the declared and computed Base Values, compute the Per Unit values of the impedance by:

Dividing Actual Values by Base ValuesChanging Per Unit Impedance with change in Base Values




Generator 1 (G1): 300 MVA; 20 kV; 3φ; Xd” = 20 %Transmission Line(L1): 64 km; XL = 0.5 Ω / kmTransformer 1 (T1): 3φ; 350 MVA; 230 / 20 kV; XT1 = 10 %Transformer 2 (T2): 3-1φ; 100 MVA; 127 / 13.2 kV; XT2 = 10 %Generator 2 (G2): 200 MVA; 13.8kV, Xd” = 20 %Generator 3 (G3): 100 MVA; 13.8kV, Xd” = 20 %

T1

L1

T2

G1G2

G3


Use Base Power = 100 MVA



E1

XT1

E2 E3

T1Transmission Line

T2

G1G2

G3XLINE XT2

XG1 XG2 XG3




a) Establish Base Power, Base Voltages, Base Impedance, and Base Current


Sub-System

Vbase (kV) Zbase (Ohm) Ibase (Amp)

Base Power:



b) Compute Per Unit Impedance


G2:

T1:

G1:

G3:

L1:

T2:



Advantages of Per-Unit Quantities

The computation for electric systems in per-unit simplifies the work greatly. The advantages of Per Unit Quantities are:

1. Manufacturers usually specify the impedances of equipments in percent or per-unit on the base of the nameplate rating.

2. The per-unit impedances of machines of the same type and widely different rating usually lie within a narrow range. When the impedance is not known definitely, it is generally possible to select from tabulated average values.



3. When working in the per-unit system, base voltages can be selected such that the per-unit turns ratio of most transformers in the system is equal to 1:1.

4. The way in which transformers are connected in three-phase circuits does not affect the per-unit impedances of the equivalent circuit, although the transformer connection does determine the relation between the voltage bases on the two sides of the transformer.

5. Per unit representation yields more meaningful and easily correlated data.




6. Network calculations are done in a much more handier fashion with less chance of mix-up • between phase and line voltages• between single-phase and three-phase powers,

and• between primary and secondary voltages.




Sequence Components of Unbalanced Phasor

Sequence Impedance of Power System Components

Practical Implications of Sequence Components of Electric Currents

Symmetrical Components




In a balanced Power System,Generator Voltages are three-phase balancedLine and transformer impedances are balancedLoads are three-phased balanced

Single-Phase Representation and Analysis can be used for the Balanced Three-Phase Power System



In a practical Power Systems,Lines are not transposed.Single-phase transformers used to form three-phase banks are not identical.Loads are not balanced.Presence of vee-phase and single phase lines.Faults

Single-phase Representation and Analysis cannot be use for an unbalanced three-phase power system.




Any unbalanced three-phase system of phasors may be resolved into three balanced systems of phasors which are referred to as the symmetrical components of the original unbalanced phasors, namely:

a) POSITIVE-SEQUENCE PHASOR

b) NEGATIVE-SEQUENCE PHASOR

c) ZERO-SEQUENCE PHASOR




Phase a

Phase b

Phase c

120°120°

120°

Positive Sequence Phasorsare three-phase, balanced and have the phase sequence as the original set of unbalanced phasors.

Negative Sequence Phasors are three-phase, balanced but with a phase sequence opposite to that original set of unbalanced phasors.

REFERENCE PHASE SEQUENCE: abc


Zero Sequence Phasors are single-phase, equal in magnitude and in the same direction.



Va = Va1 + Va2 + Va0

Vb = Vb1 + Vb2 + Vb0

Vc = Vc1 + Vc2 + Vc0

Each of the original unbalanced phasor is the sum of it’s sequence components. Thus,

Where,Va1 – Positive Sequence component of Voltage VaVa2 – Negative Sequence component of Voltage VaVa0 – Zero Sequence component of Voltage Va




a = 1 ∠ 120° a² = 1 ∠ 240° a³ = 1 ∠ 0°

OPERATOR “a”An operator “a” causes a rotation of 120° in the counter clockwise direction of any phasor.

Vb in terms of VaVb = a² VaVb1 = a² Va1Vb2 = a Va2Vb0 = Va0

Vc in terms of VaVc = a VaVc1 = a Va1Vc2 = a2 Va2Vc0 = Va0




Writing again the phasors in terms of phasor Va and operator “a”,

Va = Va0 + Va1 + Va2 Vb = Va0 + a²Va1 + aVa2 Vc = Va0 + aVa1 + a²Va2

Rearranging and writing in matrix form


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2a

1a

0a

2

2

c

b

a

VVV

aa1aa1111

VVV



Let

Using the numerical techniques, the inverse of A can be obtained as,

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

2

2

aa1aa1111

A


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=−

aa1aa1111

31A

2

21



The symmetrical sequence components can be obtained by pre-multiplying the original phasors (Va, Vb and Vc) by the inverse of A,

Thus,


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

c

b

a

2

2

2a

1a

0a

VVV

aa1aa1111

31

VVV

[ ] [ ]c2

ba1acba0a VaaVV31V VVV

31V ++=++=

[ ]cb2

a2a aVVaV31V ++=



EXAMPLE:

Determine the symmetrical components of the followingunbalanced voltages.

Vc = 8 ∠143.1

Vb = 3 ∠-90

Va = 4 ∠0




[ ]18.38 4.9

143.1) 240)(8 (1 90)- 120)(3 (1 0 431

)Va aV V(31 V c

2ba1a

∠=

∠∠+∠∠+∠=

++=

For Phasor VFor Phasor Vaa: :

143.05 1

143.1) 8 90- 3 0 4(31

)V V V(31 V cba0a

∠=

∠+∠+∠=

++=




For Phasor VFor Phasor Vaa: :

[ ]86.08 2.15

143.1) 120)(8 (1 90)- 240)(3 (1 0 431

)aV Va V(31 V cb

2a2a

−∠=

∠∠+∠∠+∠=

++=




Components of Vb can be obtained by operating the sequence components of phasor Va.

33.92 2.15 86.08)- 120)(2.15 (1

aV V101.62- 4.9

258.38 4.9 18.38) 240)(4.9 (1

Va V143.05 1 143.05 1

VV

a2b2

1a2

b1

a0b0

∠=∠∠=

=∠=∠=

∠∠==

∠=∠==




Similarly, components of phasor Vc can be obtained byoperating Va.

3.92512.15 86.08)- 0)(2.1542 (1

Va V8.3831 4.9

18.38) 0)(4.921 (1 Va V

143.05 1 143.05 1 VV

a22

c2

1ac1

a0c0

∠=∠∠=

=

∠=∠∠=

=∠=∠=

=




Positive Sequence ComponentsNegative Sequence Components

Zero Sequence ComponentsVa1

Vb1

Va2

Vb2Vc2

Va0 Vb0 Vc0Vc1




The results can be checked either mathematically or graphically.

143.1 8 153.92 2.15 138.38 4.9 05.143 1

V V V V90- 3

33.92 2.15 101.62- 4.9 143.05 1 V V V V

0 4 86.08- 2.15 18.38 4.9 05.143 1

V V V V

c2c1c0c

b2b1b0b

a2a1a0a

∠=∠+∠+∠=

++=∠=

∠+∠+∠=++=

∠=∠+∠+∠=


++=



Components of Vc

Components of Va

Components of Vb

Va0

Vb0

Vc0

Vc1

Va1

Vb1

Va2

Vb2

Vc2

Vc = 8 ∠143.1

Vb = 3 ∠-90

Va = 4 ∠0

Add Sequence Components Graphically





Positive Sequence Negative Sequence Zero Sequence

+

-

Z1

F

Ia1

Vf+

Va1

+

-

F

Ia2

Va2Z2

+

-

F

Ia0

Va0Z0

2a2a2ZI - V =1a1fa1

ZI – V V =oaoao

ZI - V =

Sequence Networks




In general,

Z1 ≠ Z2 ≠ Z0 for generators

Z1 = Z2 = Z0 for transformers

Z1 = Z2 ≠ Z0 for lines




ZERO-SEQUENCE CURRENTS:

Ia0

3Io

Ic0

Ib0

b

a

c

The neutral return carries the in-phase zero-sequence currents.

Ia0

Ic0

Ib0

ba

c

Zero-sequence currents circulates in the delta-connected transformers. There is “balancing ampere turns” for the zero-sequence currents.



Example: Consider a 100 MVA 230YG-20Δ kV transformer. A single line-to-ground fault at the 230-kV side results in the following line currents:

A 753IA =r

0II CB ==rr

Find the line currents in the 20-kV side. Use 100 MVA and 230 kV as bases in the HV side.

H1

H2

H3

CIrBIr

AIr

B

C

A

X1

X2

X3

cIr

bIr

aIrb

c

a




The Base Currents

L

3

kV Base 3

1000 x MVABaseI Base φ=

A 251(230) 3000,100

== at the 230 kV side

A 887,2(20) 3000,100

== at the 20 kV side

The Per-Unit Phase Fault Currents

p.u. 0.3251753I A =

r0II CB ==

rr




p.u. 0.1)III(I CBA31

0A =++=rrrr

p.u. 0.1)IaIaI(I C2

BA31

1A =++=rrrr

p.u. 0.1)IaIaI(I CB2

A31

2A =++=rrrr

The Per-Unit Sequence Fault Currents

The Sequence Fault Currents at the 20-kV side

0I 0a =r

p.u. 30- 0.130II oo1A1a ∠=−∠=

rr

p.u. 300.130II oo2A2a ∠=∠=

rr




p.u. 732.1IIII 2a1a0aa =++=rrrr

p.u. 732.1IaIaII 2a1a2

0ab −=++=rrrr

0IaIaII 2a2

1a0ac =++=rrrr

The Line Currents at the 20-kV side

The Line Currents in Amperes

A 000,5II ba =−=rr

0Ic =r

Check ampere turns: orXXHH ININrr

= XHX

H IINN rr

=

000,5)753(20

3/230= (Check)

Three-Phase Transformer




NEGATIVE-SEQUENCE CURRENTS:

A three-phase unbalanced load produces a reaction field which rotates synchronously with the rotor-field system of generators.

Any unbalanced condition will have negative sequence components. This negative sequence currents rotates counter to the synchronously revolving field of the generator.

The flux produced by sequence currents cuts the rotor field at twice the rotational velocity, thereby inducing double frequency currents in the field system and in the rotor body.

The resulting eddy-currents are very large and cause severe heating of the rotor.



Power System Modeling

Training Course in

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BENECO TCDP1 Vol 1 - Power System Modeling Part 1

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