BENECO TCDP1 Vol 1 - Power System Modeling Part 1

transcript

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

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.

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

The system of equations in matrix notation is

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

333231

232221

131211

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

aaaaaaaaaaaa

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” 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.

⎥⎥⎥⎥

⎢⎢⎢⎢

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

Type of Matrices

⎥⎥⎥

⎢⎢⎢

333231

232221

131211

aaaaaaaaa

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.

⎥⎥⎥

⎢⎢⎢

131211

u00uu0uuu

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

Type of Matrices

⎥⎥⎥

⎢⎢⎢

333231

lll0ll00l

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

⎥⎥⎥

⎢⎢⎢

d000d000d

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

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

Type of Matrices

⎥⎥⎥

⎢⎢⎢

872731215

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

7aa2aa1aa

======

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⎥⎥⎥

⎢⎢⎢

−−

+=−=

Type of Matrices

The null matrix is matrix whose elements are

equal to zero.

⎥⎥⎥

⎢⎢⎢

000000000

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

Example:

⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡++++++

=+482666

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

⎥⎦

⎤⎢⎣

⎡ −−=⎥

⎤⎢⎣

⎡−−−−−−

=−262624

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

⎥⎥⎥

⎢⎢⎢

+−−++++−+

=⎥⎥⎥

⎢⎢⎢

+−++++−−+

=5j64j55j74j56j41j25j71j22j3

B 9j81j13j61j13j51j43j61j42j1

Example:

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

⎥⎥

⎢⎢⎢

+++−+−−+++++++++++++−−+−+++

=+5j69j84j51j15j73j64j51j16j43j51j21j45j73j61j21j42j32j1

⎥⎥⎥

⎢⎢⎢

+−−++++−+

=+14j145j62j135j69j92j62j132j64j4

⎥⎥⎥

⎢⎢⎢

++−+−−−−+−−++−

=−4j23j48j13j43j10j28j10j20j2

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

Example:

3 k and 162534

A =⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

318615912

⎥⎥⎥

⎢⎢⎢

162534

3 Ak B

Example:

3 k and 4j1j365j2j2-56j3j14

A =⎥⎥⎥

⎢⎢⎢

−++−+

⎥⎥⎥

⎢⎢⎢

−++−−+

=12j39j1815j66j1518j93j12

⎥⎥⎥

⎢⎢⎢

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

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

aaaaaa

A⎥⎥⎥

⎢⎢⎢

⎡= and

2x22221

B ⎥⎦

⎤⎢⎣

⎥⎦

⎤⎢⎣

⎥⎥⎥

⎢⎢⎢

aaaaaa

C = A B =

Example:

2112111111 babac +=∑= kjikij bac

⎥⎥⎥

⎢⎢⎢

++++++

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

2232123121321131

2222122121221121

2212121121121111

∑= kjikij bac

2x3635241

A⎥⎥⎥

⎢⎢⎢

2x20987

B ⎥⎦

⎤⎢⎣

⎡=and

⎥⎦

⎤⎢⎣

⎥⎥⎥

⎢⎢⎢

635241

Example:

⎥⎥⎥

⎢⎢⎢

24751659843

⎥⎥⎥

⎢⎢⎢

++++++

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

⎥⎥⎥

⎢⎢⎢

+−−++++−+

=⎥⎥⎥

⎢⎢⎢

+−++++−−+

=5j64j55j74j56j41j25j71j22j3

B 9j81j13j61j13j51j43j61j42j1

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

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

Example:

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

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

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

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

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

[ ] [ ]⎥⎥⎥

⎢⎢⎢

+++++++−−

=144j3915j10143j11673j3741j2024j2942j7216j4441j35

Transpose of a Matrix

233231

aaaaaa

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

⎥⎦

⎤⎢⎣

Example:

654321

642531

⎥⎥⎥

⎢⎢⎢

⎥⎦

⎤⎢⎣

Example:

⎥⎦

⎤⎢⎣

⎡−+−+−+

=3j61j45j22j56j34j1

⎥⎥⎥

⎢⎢⎢

−++−−+

=3j62j51j46j35j24j1

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:

ayx −=

Determinant of a 2 x 2 Matrix

)2(yxaxa)1(yxaxa

2222121

1212111

21122211

2121221

212122121122211

1221211221212211

212111

aaaayaya x

yayax )aaaa( yaxaayaxaa

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|.

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:

333231

232221

131211

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

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

⎥⎥⎥

⎢⎢⎢

−−−−−−

=135101461648

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;

1 y|A|

Ax ++=

2 y|A|

Ax ++=

3 y|A|

Ax ++=

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

Inverse of a Matrix

⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

332313

322212

312111

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-

Inverse of a Matrix

⎥⎥⎥

⎢⎢⎢

−−−−−

−=⎥⎥⎥

⎢⎢⎢

110163144

AAAAAAAAA

332313

322212

312111

Example: Get the inverse of A

⎥⎥⎥

⎢⎢⎢

246-12-1-211

the Adjoint of A is

|A|AA 1-

246121211

|A|−

−−=

The determinant of A is

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

⎥⎥⎥

⎢⎢⎢

−−−−−

Hence, the inverse of matrix A is

⎥⎥⎥

⎢⎢⎢

−−

−−−

−−−−

⎥⎥⎥

⎢⎢⎢

−−−−−

−−=−

110163144

441A 1

Kron Reduction Method

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

44434241

34333231

24232221

14131211

aaaaaaaaaaaaaaaa

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

44434241

34333231

24232221

14131211

aaaaaaaaaaaaaaaa

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥⎦

⎤⎢⎣

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

4211 XAAAAY −−=

[ ]⎥⎥⎥

⎢⎢⎢

Y [ ]⎥⎥⎥

⎢⎢⎢

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

[ ]⎥⎥⎥

⎢⎢⎢

333231

232221

131211

aaaaaaaaa

A [ ]⎥⎥⎥

⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

8742654356784321

0yyyExample:

[ ] [ ] [ ] [ ]1

1 X7428654

543678321

Y⎟⎟⎟

⎜⎜⎜

⎥⎥⎥

⎢⎢⎢

⎡−⎥⎥⎥

⎢⎢⎢

⎡= −

[ ] [ ] [ ]11 X74281

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:

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

333231

232221

131211

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|aayaayaay

x 33323

⎥⎥⎥

⎢⎢⎢

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

⎥⎥⎥

⎢⎢⎢

|A|ayaayaaya

x 33331

⎥⎥⎥

⎢⎢⎢

142x4x6x-7x2x x-3 2xx x

=++=+−=++Example:

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

246-12-1-211

Solutions of System of Equations by Cramer’s Rule

⎥⎥⎥

⎢⎢⎢

246-12-1-211

A 44246121-211

|A| −=−

88 44-

241412-7213

x1 −=−

=⎥⎥⎥

⎢⎢⎢

|A|aayaayaay

x 33323

⎥⎥⎥

⎢⎢⎢

44 44-

2146-171-231

x2 −=−

=⎥⎥⎥

⎢⎢⎢

132 44-

1446-72-1-311

x3 =−−

=⎥⎥⎥

⎢⎢⎢

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

|A|ayaayaaya

x 33331

⎥⎥⎥

⎢⎢⎢

|A|yaayaayaa

x 33231

⎥⎥⎥

⎢⎢⎢

Solutions of System ofEquations by Matrix Inversion

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

YAXYA IXYA AXA

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

=++=+−=++Example:

⎥⎥⎥

⎢⎢⎢

246-12-1-211

⎥⎥⎥

⎢⎢⎢

−−−−−

−−=

110163144

441A1-

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

−−−−−

−−==

110163144

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

⎥⎥⎥

⎢⎢⎢

⎡−−

=⎥⎥⎥

⎢⎢⎢

−−==

1324488

441YA X 1-

3 x1- x2- x

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

⎥⎥⎥

⎢⎢⎢

⎡−

3214100103103211

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

=++=+−=++

⎥⎥⎥

⎢⎢⎢

⎡−

3214100103103211

⎥⎥⎥

⎢⎢⎢

⎡−

1324400103103211

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

By Back Substitution:

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

=+−+=+−=++

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

Therefore:

Forwardelimination

Backsubstitution

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

131321211

1131211

3333231

2232221

1131211

a/)xaxac(xa/)xac(x

cacaacaaa

caaacaaacaaa

−−=−=

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

MGaussian Elimination Method

⎥⎥⎥

⎢⎢⎢

⎡−−1324400103103211

Multiply row 2 by -1.

Gauss-Jordan Method

⎥⎥⎥

⎢⎢⎢

⎡−

1324400103103211

From Gauss Elimination Method slide no.75

⎥⎥⎥

⎢⎢⎢

⎡−−3100103103211

Divide row 3 by 44.

Gauss-Jordan Method

⎥⎥⎥

⎢⎢⎢

⎡−−310010310

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

Gauss-Jordan Method

Therefore:

⎥⎥⎥

⎢⎢⎢

⎡−−

=⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

100010001

3x 1x 2x

=−=−=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

3333231

2232221

1131211

c x c x c x

c100c010c001

caaacaaacaaa

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

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

=++=++=+−

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

Solution:

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

)x- x 4 (41x k

1k1 +=+

) x -x - 5(31x k

1k3 =+

) x - x - 6 ( 41x k

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

ΔΔΔ

0 x x x witha) 0 3

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

−=−=

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

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

−=−=

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

−=−=

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

−=−=

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

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

−=−=

ΔΔΔ

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

0.9998 x0.9998 x

0.9995 x

x max xxx

ΔΔΔΔ

0.5 x x x withb) 0 3

0 1 ===

x max xxx

ΔΔΔ

x max xxx

ΔΔΔ

x max xxx

ΔΔΔ

x max xxx

ΔΔΔ

x max xxx

ΔΔΔ

x max xxx

ΔΔΔ

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

=+++↓↓↓↓

=+++=+++

From the first equation,

)xaxay(a1x n1n2121

1 −−= K

n1n2121111 xaxayxa −−= K

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

↓↓↓↓

−−−=

)xaxaxa(ba1x n2n3231212

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

n −−−−= K

In general, the jth equation may be written

)xab(a1x i

1i jij

∑≠=−=

n2,1,j K=

equation “a”

In general, the Gauss iterative estimates are:

where k is the iteration count

1 xaa...x

ayx −−−−=+

xaa...x

21k2 −−−−=+

1-nn,k

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

+ −=

n2,1,j K=

equation “b”

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

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

=++=++=+−

0.5 x x x 0 3

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

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

ΔΔΔΔ

0.5 x x x with 0 3

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

−=−=

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

−=−=

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

−=−=

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

−=−=

Gauss-Seidel Method

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

0.9999 x1.0001 x1.0002 x

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

1i jij

∑≠=−= 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

1i 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

1−−−=

xaa...x

2−−−= ++

in1k1i

1ii,1k1-i

1-ii,1ki

i xaax

...xaa

+++ −−−−−=+

ayx 1k

1-nn,1k1

++ −−−=+

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

1011 x)x(

xfx)x(

xf)x(fy ΔΔΔ

∂∂

++∂∂

+∂∂

2022 x)x(

xfx)x(

xf)x(fy ΔΔΔ

∂∂

++∂∂

+∂∂

n0nn x)x(

xfx)x(

xf)x(fy ΔΔΔ

∂∂

++∂∂

+∂∂

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.

∂∂

⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢

−−

∂∂

)x(fy)x(fy

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

1kj =+== Δ

At the kth iteration:

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

jj =≤− ε

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

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

=−=−

2211 x4xf −=

First, form the JacobianSolution:

1 x2xf

=∂∂ 4

1 −=∂∂

212 xx2f −= 1xf

2 −=∂∂ 2

2 =∂∂

⎥⎦

⎤⎢⎣

⎥⎥⎥⎥

⎢⎢⎢⎢

∂∂

=⎥⎦

⎤⎢⎣

⎡−−

)x(fy)x(fy

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

=⎥⎦

⎤⎢⎣

⎡−−−−

124-x2

)xx2(2)x4x(4

Jacobian Matrix

In Matrix form

2 (1)2xf

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

−=∂∂

==∂∂

==−−=

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

−=∂∂

=∂∂

==−−=

Newton-Raphson MethodIteration 0:

The equations are:

x)1(x)2(32

x)4(x)2(54

−+=−

−+=−⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−−

=⎥⎦

⎤⎢⎣

⎡−−

In matrix form:

5.0)5.0(1x1

−=+−=

=−+=

Solving,

Repeating the process with the new estimates,Iteration 1:

0.1)5.0(2x

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

−=∂

==−−=

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

−=∂

==−−=

The equations are: In matrix form:

Solving,

x4x25.44

−=−

−=−⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−−

=⎥⎦

⎤⎢⎣

⎡−1

07143.0x

03571.0x1

92857.007143.01x

53571.003571.05.0x2

−=+−=

Repeating the process with the new estimates,Iteration 2:

07142.1)53571.0(2x

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

−=∂

==−−=

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

−=∂

≅=−−=

The equations are:

In matrix form:

Solving,

xx20.22

x4x07142.1001265.44

−=−

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−−

=⎥⎦

⎤⎢⎣

⎡−2

12407142.1

0001265.0

00036.0x

00018.0x2

92893.000035.092857.0x

53553.000018.053571.0x3

−=−−=

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

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

12 ==−−=

00072.0fy0025.0fy

=−−=−

92893.0x53553.0x

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

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

Electric Circuits

Water Pressure causes the

flow of water=

Voltage

Water Flow = Current

Analogy

pedanceIm

VoltageCurrent =I

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

10 kWH

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

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)

Electric Circuits

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.

VS = Vao Vb = VbnI = Iab Zb

Vao - IabZA - Vbn = 0

Iab =Vao - Vbn

Electric Circuits

VS - VA - Vb = 0

Double Subscript Notations

Electric Circuits

Voltage, Current and Phasor Notation

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=

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

tjtj VLKjRK ωωω εεωε =+Hence, 1j −=

tsinjtcostj ωωε ω +=

Electric Circuits

Solving for the current

Dividing voltage by current to get the impedance,

)t(i ωεω+

)t(i)t(vZ

tjm ω

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

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

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

30°203020

0530100Z ∠=

∠∠

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

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

S = VI

Q = VI sin θleading

S = VI

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

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

Balanced Three-Phase System

Electric Circuits

nZREao = |E|∠0° V

Ebo= |E| ∠240° V

Eco = |E| ∠120° V

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)

Electric Circuits

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

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°

70°120°

120°30°

Electric Circuits

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

Ib = √3 Ibc ∠-30° Ia = √3 Iab ∠-30°

Ica = aIab

Ibc = a²Iab

Electric Circuits

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

VOLTAGE AND CURRENT RELATIONSHIPS

Electric Circuits

θ LLLn

VV3/II

Single-Phase and Three-Phase Power

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

θωω

t2sinsinIV21t2cosIV

21cosIV

t2sinsinIV21)t2cos1(cosIV

]sintsintcoscost[cosIV)sintsincost[costcosIV

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

mmmmmm

ωθωθ

θωωθωθωθωω

θωωθωω

+=−=−=

t2sin21tsintcos ωωω =

Electric Circuits

)t2cos1(21tcos2 ωω +=

sinIV21Q mm

Electric Circuits

cosIV21P mmave

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

tcosV)t(v

ωωω

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

)tcos(I)t(i

θωθω

−−=

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

Electric Circuits

θφφ

cosIV3

cosIV3P3P

1,ave3,ave

θφφ

sinIV3

sinIV3Q3Q

=|I||I|

|V|3|V|

connectedY

|I|3|I|

|V||V|connected

=−Δ

Single Phase Equivalent of Balanced Three-Phase System

Electric Circuits

nEao = |E|∠0° V

Ebo= |E| ∠240° V

Eco = |E| ∠120° V

Electric Circuits

nEao = |E|∠ 0° V

Ebo= |E| ∠240° V b

Eco = |E| ∠120° V c

−∠=∠∠

)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

nEao = |E|∠0° V

Ebo= |E| ∠240° V

Eco = |E| ∠120° V

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

SubstationVLN

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

Power System representation

Switch

Electrical Symbols

3-phase wye neutral grounded

3-phase delta connection

Ammeter

Voltmeter

3-phase wye neutral ungrounded

Protective Relay

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

CTs Distribution Lines

Power System Representation

Transformer

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

CBTransformerBus

Distribution Line

CT and Relay

Equivalent Circuit of Power System Components: Generator

Iasa jXR +

Three-Phase Equivalent Single-Phase Equivalent

Power System RepresentationEquivalent Circuit of Power System Components: Transformer

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

HT RaRR +=

HT XaXX +=HIr+

TT jXR +

Power System RepresentationEquivalent Circuit of Power System Components: Transformer

Single-Phase Equivalent

Equivalent Circuit of Power System Components: Distribution Lines

ZccZcbZca

ZbcZbbZba

ZacZabZaa

Equivalent π-Network

YccYcbYca

YbcYbbYba

YacYabYaa

1/2YccYcbYca

YbcYbbYba

YacYabYaa

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

• •

••+

IS IR VR

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

yzZ C =

Characteristic Impedance zy=γPropagation Constant

2ltanh

Equivalent Circuit of Power System Components: Medium-Length

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

• •

••+ +

VsIS IR

( )ljxrZ L+=

2Y ω=

Equivalent Circuit of Power System Components: Short Transmission Lines

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

• •

••+ +Is = IR

-Single-Phase Equivalent

( )ljxrZ L+=

Impedance and Admittance Diagrams

bcaBus

1 - 32 - 31 - 42 - 43 - 4

Single Line Diagram

Impedance Diagram

Ea Ec Eb

zcGenerator

ILZpIs

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

Eg = ISZPZg = Zp

Admittance Diagram

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

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,

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

++−−−+++

⎥⎥⎥⎥

⎢⎢⎢⎢

342414342414

34133423032313

2423242302

1413141301

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

( ) 24423324230222 yVyVyyyVI −−++=

The standard form of n independent equations:

Network ModelBus Admittance Matrix

⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

nn3n2n1n

n3333231

n2232221

n1131211

⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

[ I ] = [Ybus][V]

Ybus is also called Bus Admittance Matrix

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

nn3n2n1n

n3333231

n2232221

n1131211

[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

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

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

• 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

)(L)pu(L Z

XX Ω=

)()pu( Z

RR Ω=

)(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

=⎟⎠⎞

⎜⎝⎛=

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.

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%

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%

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 ⋅=⋅

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

base =

( )2,3

baseLL

MVAkVMVAkV

or,⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

1base,3

2base,32

2base,LL

1base,LL1pu2pu MVA

MVAkVkV

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[

Solution

On its own base the transformer reactance is

On the chosen base the reactance becomes

= ________ puX =( )

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 %

Use Base Power = 100 MVA

T1Transmission Line

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

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°

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

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

aa1aa1111

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

⎥⎥⎥

⎢⎢⎢

aa1aa1111

⎥⎥⎥

⎢⎢⎢

⎡=−

aa1aa1111

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

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

aa1aa1111

[ ] [ ]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

∠∠+∠∠+∠=

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

−∠=

∠∠+∠∠+∠=

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

∠=∠∠=

=∠=∠=

∠∠==

∠=∠==

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

∠=∠∠=

=∠=∠=

Positive Sequence ComponentsNegative Sequence Components

Zero Sequence ComponentsVa1

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

Vc = 8 ∠143.1

Vb = 3 ∠-90

Va = 4 ∠0

Add Sequence Components Graphically

Positive Sequence Negative Sequence Zero Sequence

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:

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

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.

CIrBIr

The Base Currents

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 ==

p.u. 0.1)III(I CBA31

0A =++=rrrr

p.u. 0.1)IaIaI(I C2

1A =++=rrrr

p.u. 0.1)IaIaI(I CB2

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 ∠=−∠=

p.u. 300.130II oo2A2a ∠=∠=

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

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

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BENECO TCDP1 Vol 1 - Power System Modeling Part 1

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