Power System Analysis II

Instructor:Instructor: Yrd. Doc. Dr. Bunyamin TamyurekE-mail: btamyurek@ogu.edu.trOffice phone: 0(222) 239-3750 extension 3252

Text book: Text book: ““ Power System Analysis and Design,” by Glover and Sarma.

Midterm IMidterm I = 30%Midterm IIMidterm II = 30%FinalFinal = 40%

Electrical and Electronics Eng. Dept.Osmangazi University

February 28, 2005

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Power FlowsSuccessful power system operation under normal balanced three-phase steady-state conditions require the following:

Generation supplies the demand (load) plus losses.Bus voltage magnitudes remain close to rated values.Generators operate within specified real and reactive power limits.Transmission lines and transformers are not overloaded.

The power-flow computer program (commonly called load flow) is the basic tool for investigating these requirements.

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Power Flows (Cont.)

The power flow program computes the voltage magnitude and the angle at each bus in a power system under balanced three-phase steady-state conditions.

It also computes real and reactive power flows for all equipmentinterconnecting the busses, as well as equipment losses.

Conventional nodal or loop analysis is not suitable for power-flow studies. Therefore, the problem is formulated as a set of nonlinear algebraic equations suitable for computer solution.

Since balanced three-phase steady-state conditions are assumed, we use only positive-sequence networks in this chapter.

6.1 DIRECT SOLUTIONS TO LINEAR ALGERAIC EQUATIONS: GAUSS ELIMINATION

Consider the following set of linear algebraic equations in matrix format:

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

L

L

M M M M M M

L

11 12 1N 1 1

21 22 2N 2 2

N1 N2 NN N N

A A A x yA A A x y

A A A x y

(6.1.1)

or Ax=y (6.1.2)

Given A and y, we want to solve for x.

We assume the det(A) is nonzero, so a unique solution to (6.1.1) exits.

An easy solution is obtained when A is an upper triangular matrix with nonzero diagonal elements.

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⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

L

M M

L

L

11 12 1N 1 1

2N 2 2

N - 1, N - 1 N - 1, N N - 1 N - 1

NN N N

A A A x y0 A x y

0 0 A A x y0 0 A x y

M (6.1.3)

AN

NNN

yx =

Since the last equation involves only xN ,

(6.1.4)

After xN is computed, the next-to-last equation can be solved:

N 1 N 1,N NN 1

N 1,N 1

y A xx

A− −

−− −

−= (6.1.5)

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In general, with In general, with xN, , xN-1,,……., ., xk+1 already computed, already computed, the the k k th equation can solvedth equation can solved

k kn n

k

kk

N

n k 1

y A xx

A= +

−=

∑k=N, N-1,….,1 (6.1.6)

This procedure for solving equation (6.1.3) is called back substitution.

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11 12 1N 1

21 21 21122 12 2N 1N 2 1

11 11 112

31 31 31322 12 32 1N 3 1

11 11 11

NN1 N1 N1

N2 12 NN 1N N 1

11 11 11

A A A yA A Ax0 (A A ) (A A ) y yA A AxA A Ax0 (A A ) (A A ) y yA A A

xA A A0 (A A ) (A A ) y yA A A

−

−

−

⎡ ⎤ ⎡⎢ ⎥ ⎢

⎡ ⎤⎢ ⎥ ⎢− −⎢ ⎥⎢ ⎥ ⎢⎢ ⎥⎢ ⎥ ⎢⎢ ⎥ =− −⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦

⎢ ⎥− −⎢ ⎥⎣ ⎦ ⎣

K

L

L

MM M M

L

⎤⎥⎥⎥⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎦

(6.1.7)

If A is not upper triangular, (6.1.1) can be transformed to an equivalent equation with an upper triangular matrix. The transformation is called Gauss elimination and described by (N-1) steps. During step 1, we use the first equation in (6.1.1) to eliminate X1 from the remaining equations. That is, equation 1 is multiplied by An1/A11 and then subtracted from equation n, for n=2,3,….,N. After completing step 1,we have

Equation (6.1.7) has the following form:

8

1N

0

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢⎢ ⎥⎢ ⎥ ⎢

⎥ (6.18)⎥⎢ ⎥ =⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

L

MM M L M M

L

111 12

222 2N

332 3N

NN2 NN

(1) (1) (1) (1)1

(1) (1) (1)2

(1) (1) (1)3

(1) (1) (1)N

xA A A yx0 A A yxA A y

x0 A A y

Where the superscript (1) denotes step 1 of Gauss elimination.

1N⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ =⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

MM M M M M

111 12 13

222 23 2N

333 3N

NN3 NN

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

(2) (2) (2) (2)2

(2) (2) (2)3

(2) (2) (2)N

xA A A A yx0 A A A yx0 0 A A y

x0 0 A A y

(6.1.9)

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EAMPLE 6.1 Gauss elimination and back substitution: direct

solution to linear algebraic equations

1

2

10 5 x 62 9 x 3

⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Solve

Using Gauss elimination and back substitution.

SOLUTION: Since N=2 for this example, there is (N-1)=1 Gauss elimination step. Multiplying the first equation by A21/A11=2/10 and then subtracting from the second,

1

2

10 5 6x2 2x0 9 (5) 3 (6)

10 10

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦⎣ ⎦ ⎣ ⎦

10

or1

2

10 5 x 60 8 x 1.8

⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Which has the form A(1)x=y(1), where A(1) is upper triangular. Now, using back substitution (6.1.6) gives, for k=2:

2

(ı)2(ı)22

y 1.8x 0.225A 8

= = =

(ı) (ı)1 12 2

1 (ı)11

y A x 6 (5)(0.225)x 0.4875A 10− −

= = =

and, for k=1,

EAMPLE 6.2 Gauss elimination: triangularizing a matrix

Use Gauss elimination to triangularize

1

2

3

2 3 1 x 54 6 8 x 7

10 12 14 x 9

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

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SOLUTION: There are (N-1)=2 Gauss elimination steps.During step 1, we subtract A21/A11=-4/2=-2 times equation 1 from equation 2,

and we subtract A31/A11=10/2=5 times equation 1 from equation 3, to give

1

2

3

2 3 1 x 50 6 ( 2)(3) 8 ( 2)( 1) x 7 ( 2)(5)0 12 (5)(3) 14 (5)( 1) x 9 (5)(5)

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − − − = − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

or

1

2

3

2 3 1 x 50 12 6 x 70 3 19 x 16

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Which is A(1)x=y(1). During step 2, we subtract A(1)32/A(1)

22=-3/12=-0.25times equation 2 from equation 3, to give

12

1

2

3

2 3 1 x 50 12 6 x 170 0 19 ( 0.25)(6) x 16 ( 0.25)(17)

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

or

1

2

3

2 3 1 x 50 12 6 x 170 0 20.5 x 11.75

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Which is triangularized. The solution x can now be easily obtained via back substitution.

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6.2 ITERATIVE SOLUTIONS TO LINEAR ALGEBRAIC EQUATIONS: JACOBI AND GAUSS-SEIDEL

A general iterative solution to (6.1.1) proceeds as follows. First select an initial guess x(0). Then use

x(i+1) = g[x(i)] i = 0,1,2,3….

Where x(i) is the ith guess and g is an N vector of functions that specify the iteration method. Continue the procedure until the following stopping condition is satisfied:

k k

k

x (i 1) x (i)x (i)+ −

< ε For all k=1,2,….,N (6.2.2)

Where xk(i) is the kth component of x(i) and ε is a specified tolerance level.

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The following questions are important:

1. Will the iteration procedure converge to the unique solution?

2. What is the convergence rate (how many iterations are required)?

3. When using a digital computer, what are the computer storage and time requirements?

These questions are addressed for two specific iteration methods: Jacobi and Gauss-Seidel.

The Jacobi method is obtained by considering the kth equation of (6.1.1), as follows:

yk=Ak1x1+Ak2x2+….+Akkxk+….+AkNxN (6.2.3)

Solving for xk ,

xk=( 1/Akk )[ yk - (Ak1x1+….+Ak,k-1xk-1+Ak,k+1xk+1+….+AkNxN)]

11 A AA

k kn n kn n

kk

k - N

kn=1 n=k+1

x = y - x - x⎡ ⎤⎢ ⎥⎣ ⎦

∑ ∑ (6.2.4)

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k k kn n kn n

kk

k 1 N

n 1 n k 1

1x (i 1) y A x (i) A x (i)A

−

= = +

⎡ ⎤+ = − −⎢ ⎥⎣ ⎦

∑ ∑

11

22

NN

A 0 00 A 0

D

0 0 0 A

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

L

L

M M O M

The Jacobi method uses the “old” values of x(i) at iteration i on the right side of (6.2.4) to generate the “new ” value xk(i+1) on the left side of (6.2.4).That is,

K=1,2,….,N

(6.2.5)

The Jacobi method given by (6.2.5) can also be written in the following matrix format:

x(i+1)=Mx(i)+D-1y (6.2.6)

Where

M=D-1(D-A) and (6.2.7)

For Jacobi, D consist of the diagonal elements of the A matrix.

(6.2.8)

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EXAMPLE 6.3 Jacobi method: iterative solution to linearalgebraic equations

11

1 010 0 10D0 9 10

9

−

−

⎡ ⎤⎢ ⎥⎡ ⎤

= = ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥

⎢ ⎥⎣ ⎦

Solve Example 6.1 using the Jacobi method.

Start with x1(0)=x2(0)=0 and continue until

(6.2.2) is satisfied for ε=10-4.

SOLUTION: From (6.2.5) with N=2

k=1 x1(i+1)=(1/A11)[y1-A12x2(i)]=0.1[6-5x2(i)]

k=2 x2(i+1)=(1/A22)[y2-A21x1(i)]=(1/9)[3-2x1(i)]Alternatively, in matrix format using (6.2.6)-(6.2.8)

1

2

10 5 x 62 9 x 3

⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦

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1 50 00 510 10M1 2 0 20 09 9

⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥−⎡ ⎤= =⎢ ⎥ ⎢ ⎥⎢ ⎥−⎣ ⎦⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ 1 1

2 2

5 10 0x (i 1) x (i) 610 10x (i 1) x (i)2 1 30 0

9 9

⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥+⎡ ⎤ ⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ ⎣ ⎦⎣ ⎦ ⎣ ⎦⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

i 0 1 2 3 4 5 6 7 8 9 10

x1(i) 0 0.60000 0.43334 0.50000 0.48148 0.48889 0.48683 0.48766 0.48743 0.48752 0.48749

x2(i) 0 0.33333 0.20000 0.23704 0.22222 0.22634 0.22469 0.22515 0.22496 0.22502 0.22500

The above two formulations are identical. Starting with x1(0)=x2(0)=0, the iterative solutions is given in the following table:

JACOBI

As shown, the Jacobi method converges to the unique solution obtained in Example 6.1. The convergence criterion is satisfied at the 10th iteration, since

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51 1

1

x (10) x (9) 0.48749 0.48752 6.2 10x (9) 0.48749

−− −= = × < ε

and52 2

2

x (10) x (9) 0.22500 0.22502 8.9 10x (9) 0.22502

−− −= = × < ε

k 1 N

k k kn n knn 1 n k 1kk

1x (i 1) ny A x (i 1) A x (i)A

−

= = +

⎡ ⎤+ = − + −⎢ ⎥⎣ ⎦∑ ∑

The Gauss-Seidel method is given by

(6.2.9)

Gauss-Siedel is similar to Jacobi except that during each iteration, the “new”values, xn(i+1), for n<k are used on the right side of (6.2.9) to generate the “new” value xk(i+1) on the left side.

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The Gauss-Seidel method of (6.2.9) can also be written in the matrix format.

Where

11

21 22

N1 N2 NN

A 0 0 0A A 0 0

D

A A A

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

L

L

M M M

L

(6.2.10)

For Gauss-Seidel, D in (6.2.10) is the lower triangular portion of A, whereas for Jacobi, D in (6.2.8) is the diagonal portion of A.

EXAMPLE 6.4 Gauss-Seidel method: iterative solution to linearalgebraic equations

Rework Example 6.3 using the Gauss-Seidel method.

SOLUTION From (6.2.9)

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1 10 00 510 2M12 1 0 0 0990 9

⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥−⎡ ⎤= =⎢ ⎥ ⎢ ⎥⎢ ⎥

⎣ ⎦⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

11

1 010 0 10D2 9 2 1

90 9

−

−

⎡ ⎤⎢ ⎥⎡ ⎤

= = ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥−⎢ ⎥⎣ ⎦

k=1 x1(i+1) = 1/A11 [y1-A12x2(i)] = 0.1[6-5x2(i)]

k=2 x2(i+1) = 1/A22 [y2-A21x1(i+1)] = 1/9 [3-2x1(i+1)]

Using the equation for x1(i+1), x2(i+1) can also be written as

x2(i+1) = 1/9{3-0.2[6-5x2(i)]}

Alternatively, in matrix format:

1 1

2 2

11 00x (i 1) x (i) 6102x (i 1) x (i)1 2 1 30

9 90 9

⎡ ⎤⎡ ⎤− ⎢ ⎥⎢ ⎥+⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

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These two formations are identical. Starting with x1(0)=x2(0)=0, the solution is given in the following table:

i 0 1 2 3 4 5 6x1(i) 0 0.60000 0.50000 0.48889 0.48765 0.48752 0.48750x2(i) 0 0.20000 0.22222 0.22469 0.22497 0.22497 0.22500

For this example, Gauss-Seidel converges in 6 iterations, compared to 10 iterations withJacobi.

The convergence rate is faster with Gauss-Seidel for some A matrices, but faster with Jacobi for other A matrices. In some cases, one method diverges while the other converges.

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6.3 ITERATIVE SOLUTIONS TO NONLINEAR 6.3 ITERATIVE SOLUTIONS TO NONLINEAR

ALGEBRAIC EQUATIONS: NEWTONALGEBRAIC EQUATIONS: NEWTON--RAPHSONRAPHSON

A set of nonlinear algebraic equations in matrix format is given by

1

2

N

f (x)f (x)

f (x) y

f (x)

⎡ ⎤⎢ ⎥⎢ ⎥= =⎢ ⎥⎢ ⎥⎣ ⎦

M(6.3.1)

Where y and x are N vectors and f(x) is an N vector of functions. Given y and f(x), we want to solve for x. The iterative methods described in Section 6.2 can be extended to nonlinear equations as follows. Rewriting (6.3.1),

0 = y - f(x) (6.3.2)

Adding Dx to both sides of (6.3.2), where D is a square NxN invertible matrix,

Dx=Dx+y-f(x) (6.3.3)

Premultiplying by D-1,

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x=x+D-1[y-f(x)] (6.3.4)The old values x(i) are used on the right side of (6.3.4) to generate the new valuesx(i+1) on the left side. That is,

x(i+1)=x(i)+D-1{y-f[x(i)]} (6.3.5)

for linear equations, f(x)=Ax and (6.3.5) reduces to

x(i+1)=x(i)+D-1[y-Ax(i)]=D-1(D-A)x(i)+D-1y (6.3.6)Which is identical to the Jacobi an Gauss-Seidel methods of (6.2.6). For nonlinear equations, the matrix D in (6.3.5) must be specified.

One method for specifying D, called Newton-Rapson, is based on the following Taylor series expansion of f(x) about an operating point x0.

0

0 0x x

dfy f (x ) (x x )dx =

= + − L (6.3.7)

Neglecting the higher order terms in (6.3.7) and solving for x,

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[ ]0

1

0 0x x

dfx x y f (x )dx

−

=

⎡ ⎤= + −⎢ ⎥

⎢ ⎥⎣ ⎦(6.3.8)

1 1 1

1 2 N

2 2 2

1 2 Nx x(i)

N N N

1 2 N x x(i)

df df dfdx dx dxdf df df

df dx dx dxJ(i)dx

df df dfdx dx dx

=

=

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥= = ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

L

L

M M M

L

The Newton-Rapson method replaces x0 by the old value x(i) and x by the new value x(i+1) in (6.3.8). Thus,

x(i+1)=x(i)+J-1(i){y-f[x(i)]} (6.3.9)

where

(6.3.10)

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The NxN matrix J(i), whose elements are the partial derivatives shown in (6.3.10) is called the Jacobian matrix. The Newton-Rapson method is similar to extended Gauss-Seidel, except that D in (6.3.5) is replaced by J(i) in (6.3.9).

EXAMPLE 6.6 NewtonEXAMPLE 6.6 Newton--Rapson method: solution to polynomialRapson method: solution to polynomial

equationsequationsSolve the scalar equation f(x)=y, where y=9 and f(x)=x2. Starting with x(0)=1, use (a) Newton-Rapson and (b) extended Gauss-Seidel with D=3 until (6.2.2) is satisfied for ε=10-4. Compare the two methods.

SOLUTION

a. Using (6.3.10) with f(x)=x2 ,

2x x(i)

x x(i)

dJ(i) (x ) 2x 2x(i)dx =

=

= = =

Using J(i) in (6.3.9),

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21x(i 1) x(i) 9 x (i)2x(i)

⎡ ⎤+ = + −⎣ ⎦

Starting with x(0)=1, successive calculations of the Newton-Rapson equation are shown in the following table:

i 0 1 2 3 4 5

x1(i) 1 5,00000 3,40000 3,02353 3,00009 3,00000

Newton-Rapson

b. Using (6.3.5) with D=3, the Gauss-Seidel method is

x(i+1)=x(i)+(1/3)[9-x2(i)]

The corresponding Gauss-Seidel calculations are as follows:

Gauss-Seidel i 0 1 2 3 4 5 6

x1(i) 1 3,66667 2,18519 3,59351 2,28908 3,54245 2,35945

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EXAMPLE 6.7 Newton-Rapson method: solution to nonlinear

algebraic equations

Solve 4x(0)

9⎡ ⎤

= ⎢ ⎥⎣ ⎦

1 2

1 2

x x 15x x 50+⎡ ⎤ ⎡ ⎤

=⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

Use the Newton-Rapson method starting with the above x(0) and continue until ε=10-4.

SOLUTION

Using (6.3.10) to find the Jacobian matrix with f1=(x1+x2) and f2=x1x2

111 1

11 2 21

2 12 2 1 2

1 2 x x(i)

x (i) 1f fx x 1 1 x (i) 1

J(i)x (i) x (i)f f x (i) x (i)

x x

−

−

−

=

−∂ ∂ ⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥∂ ∂ −⎡ ⎤ ⎣ ⎦⎢ ⎥= = =⎢ ⎥∂ ∂⎢ ⎥ −⎣ ⎦

⎢ ⎥∂ ∂⎣ ⎦

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Using J(i)-1 in (6.3.9),x(i+1)=x(i)+J-1(i){y-f[x(i)]}

1

1 1 1 22

2 2 1 21 2

x (i) 1x (i 1) x (i) 15 x (i) x (i)x (i) 1x (i 1) x (i) 50 x (i)x (i)x (i) x (i)

−⎡ ⎤⎢ ⎥+ − −−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎣ ⎦= +⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ −−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Writing the preceding equation as two separate equations,

1 1 2 1 21 1

1 2

x (i)[15 x (i) x (i)] [50 x (i)x (i)]x (i 1) x (i)x (i) x (i)

− − − −+ = +

−

2 1 2 1 22 2

1 2

x (i)[15 x (i) x (i)] [50 x (i)x (i)]x (i 1) x (i)x (i) x (i)

− − − + −+ = +

−

Successive calculation of these equations are shown in the following table:

29

i 0 1 2 3 4

x1(i) 4 5,20000 4,99130 4,99998 5,00000

x2(i) 9 9,80000 10,00870 10,00002 10,00000

Newton-Rapson converges in 4 iterations for this example.

Equation (6.3.9) contains the matrix inverse J-1.

Instead of computing J-1. (6.3.9) can be rewritten as follows:

J(i) ∆x(i)=∆y(i) (6.3.11)

Where

∆x(i)=x(i+1)-x(i) (6.3.12)

And

∆y(i)=y-f[x(i)] (6.3.13)

Then, during each iteration, the following four steps are completed:

STEP 1 Compute ∆y(i) from (6.3.13).

STEP 2 Compute J(i) from (6.3.10).

30

STEP 3 Using Gauss elimination and back substitution, solve (6.3.11) for ∆x(i).

STEP 4 Compute x(i+1) from (6.3.12).

EXAMPLE 6.8 Newton-Rapson method in four steps

4x(0)

9⎡ ⎤

= ⎢ ⎥⎣ ⎦

1 2

1 2

x x 15x x 50+⎡ ⎤ ⎡ ⎤

=⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

Complete the above four steps for the first iteration of Example 6.7

SOLUTION:STEP 1 [ ] 15 4 9 2

y(0) y f x(0)50 (4)(9) 14

+⎡ ⎤ ⎡ ⎤ ⎡ ⎤∆ = − = − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦

2 1

1 1 1 1J(0)

x (0) x (0) 9 4⎡ ⎤ ⎡ ⎤

= =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

STEP 2

STEP 3 Using ∆y(0) and J(0), (6.3.11) becomes

31

1

2

x (0)1 1 2x (0)9 4 14

∆⎡ ⎤⎡ ⎤ ⎡ ⎤=⎢ ⎥⎢ ⎥ ⎢ ⎥∆⎣ ⎦ ⎣ ⎦⎣ ⎦

Using Gauss elimination, subtract J21/J11=9/1=9 times the first equation from the second equation, giving

1

2

x (0)1 1 2x (0)0 5 4∆⎡ ⎤⎡ ⎤ ⎡ ⎤

=⎢ ⎥⎢ ⎥ ⎢ ⎥∆− −⎣ ⎦ ⎣ ⎦⎣ ⎦

Solving by back substitution.

∆x2(0)=-4/-5=0.8

∆x1(0)=2-0.8=1.2

4 1.2 5.2x(1) x(0) x(0)

9 0.8 9.8⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= +∆ = + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

STEP 4

This is the same as computed in Example 6.7

32

6.4 THE POWER-FLOW PROBLEM

The power-flow problem is the computation of voltage magnitude and the angle at each bus in a power system under balanced three-phase steady-state conditions.

Start the problem by obtaining the single-line diagram of the power system

Single-line diagram is needed to extract the input data for computer solution.

Input data consist ofBus data,

Transmission line data, and

Transformer data

33

6.4 THE POWER-FLOW PROBLEM

There are 4 variables associated with each bus k:

Voltage magnitude Vk

Phase angle δk

Net real power Pk

Reactive power Qk supplied to the bus

At each bus, two of these variables are specified as input data, and the other two are unknowns to be computed by the power-flow program.

Gen Load

Pk Qk

Bus k

PGk QGk

PLk QLk

Vk k kV δ= ∠

34

Each bus k categorized into one of the following three bus types:Swing bus There is only one swing bus and numbered as bus 1. Swing bus is a reference bus for which , typically per unit, is input data.

The power-flow program computes P1 and Q1.

Load bus Pk and Qk are input data. Vk and δk are computed. Most buses in a typical power-flow program are load buses.

Voltage controlled bus Pk and Vk are input data. Qk and δk are computed.

Examples are buses to which generators, switched shunt capacitors, or static var systems, and tap-changing transformer are connected.

Maximum and minimum var limits QGkmax and QGkmin that this equipment can supply are also input data.

Power-flow program computes the tap setting of the tap-changing transformer.

1 1V δ∠ 1.0 0°∠

35

The power delivered to bus k is separated into generator and load terms.

Pk = PGk - PLkQk = QGk - QLk

(6.4.1)

When bus k is a load bus with no generation, Pk=-PLk. Real power supplied to bus k is negative.

When the load is inductive, Qk=-QLk is negative.

Transmission lines are represented by the equivalent π circuit.

Transformers are represented by equivalent circuits.

36

Input data for each transmission line include:Series impedance Z' and shunt admittance Y' of per-unit equivalent π circuitThe two buses to which line is connectedMaximum MVA rating

Input data for each transformer include:Per-unit winding impedances ZPer-unit exciting branch admittance YThe buses to which the windings are connectedMaximum MVA rating Maximum tap settings for tap-changing transformers.

37

The bus admittance matrix Ybus can be constructed from the line and transformer input data. The elements of Ybus are:

Diagonal elements: Ykk=sum of admittances connected to bus kOff-diagonal elements: Ykn=-(sum of admittances connected between buses k and n) k ≠ n

EXAMPLE 6.9 Power-flow input data and Ybus

Figure shows a single-line diagram of a five-bus power system. Input data are given in Tables 6.1, 6.2, and 6.3.Bus 1, to which a generator is connected, is the swing bus.Bus 3, to which a generator and a load are connected, is a voltage-controlled bus. Buses 2, 4, and 5 are load buses. The loads at buses 2 and 3 are inductive.

38

For each bus k, determine which of the variables Vk , δk , Pk , and Qk are input data and which are unknowns. Also, compute the elements of the second row of Ybus.

GB1

T1

400 MVA15 kV

400 MVA15/345 kV

B51

B52

GB3

T2

800 MVA345/15 kV

B41

B42

40 Mvar 80 MW

520 MW

800 MVA15 kV

Line 3345 kV

50 mi

280 Mvar 800 MW

B22B21

Line 2345 kV100 mi

Line 1345 kV200 mi

1 5

2

4 3

39

Sbase=100 MVA, Vbase=15 kV at buses 1, 3.

Sbase=100 MVA, Vbase=345 kV at buses 2, 4, 5.

Table 6.1

Bus input data

V δ PG QG PL QL QGmax QGmin

perunit degrees

perunit

perunit

perunit

perunit

perunit

perunit

1 Swing 1.0 0 - - 0 0 - -

2 Load - - 0 0 8.0 2.8 - -

3ConstantVoltage 1.05 - 5.2 - 0.8 0.4 4 -2.8

4 Load - - 0 0 0 0 - -

5 Load - - 0 0 0 0 - -

Bus Type

40

Table 6.2 Line input data

Bus-to-Bus R´(per-unit)

X´(per-unit)

G´(per-unit)

B´(per-unit)

Max. MVA (per-unit)

2-4 0.0090 0.100 0 1.72 12.0

2-5 0.0045 0.050 0 0.88 12.0

4-5 0.00225 0.025 0 0.44 12.0

Table 6.3 Transformer input data

Bus-to-BusR

(per-unit)X

(per-unit)Gc

(per-unit)Bm

(per-unit)Max. MVA(per-unit)

Max. TAP setting(per-unit)

1-5 0.00150 0.02 0 0 6.0 -

3-4 0.00075 0.01 0 0 10.0 -

SOLUTION: The input data and unknowns are listed in Table 6.4

41

Table 6.4 Input data and unknowns

BUS INPUT DATA UNKNOWNS

1 V1=1.0 δ1=0 P1 Q1

2P2=PG2-PL2= -8

Q2=QG2-QL2= -2.8V2 δ2

3V3=1.05

P3=PG3-PL3= 4.4Q3 δ3

4 P4=0 Q4=0 V4 δ4

5 P5=0 Q5=0 V5 δ5

The elements of Ybus : Y21 = Y23 = 0

2424 24

24

1 1 0.89276 9 per-un.91964R X 0.009 0.19.95

it

972 95.14 per-u i3 n t

Y jj j

Y

− −= = =− +

′ ′+ += °

42

( ) ( )

2525 25

25

252422

24 24 25 25

22

22

1 1 1.78552 19.83932R X 0.0045 0.0519.919595.143

BB1 1R X R X 2 2

1.72 0.880.89276 9.91964 1.78552 19.839322 2

2.67828 28.4590 2

per-unit

per-unit

8

Y jj j

Y

Y j jj j

Y j j j j

Y j

− −= = =− +

′ ′+ +

= °′′

= + + +′ ′ ′ ′+ +

= − + − + +

= − = .5847 84.62 per t4 -uni− °

Where half of the shunt admittance of each line connected to bus 2 is included in Y22

the other half is located at the other ends of these lines.

43

Using Ybus, The nodal equations for a power system network is

∑N

k kn nn=1

I = Y V (6.4.4)

busI = Y Vwhere I is N vector of source currents injected into each busand V is the N vector of bus voltages.

For bus k, the kth equation in (6.4.3) is

(6.4.3)

The complex power delivered to bus k is

Sk=Pk+jQk=VkI*k (6.4.5)

Power-flow solutions by Gauss-Seidel are based on nodal equations, (6.4.4), where each current source Ik is calculated from (6.4.5). Using (6.4.4) in (6.4.5),

k kP +jQ ⎡ ⎤⎢ ⎥⎣ ⎦∑

*N

k kn nn=1

= V Y V (6.4.6)k=1,2,….,N

44

With the following notation

knjknY e θ=knYnj

nV e δ=nV k,n=1,2,…,N

k n kn

Nj( )

k k k kn nn 1

P jQ V Y V e δ −δ −θ

=

+ = ∑(6.4.6) becomes

(6.4.9)

Taking the real and imaginary parts

N

k k kn n k n knn 1

P V Y V cos( )=

= δ − δ −θ∑ (6.4.10)

( )N

k k kn n k n knn 1

Q V Y V sin=

= δ − δ −θ∑ (6.4.11)k=1,2,….,N

Power-flow solutions by Newton-Rapson are based on the nonlinear equations given by (6.4.10) and (6.4.11).

45

6.5 POWER-FLOW SOLUTION BY GAUSS-SEIDEL

busI = Y V y = Axis analogous to

Note: The current source vector I is unknown

For each load bus, Ik can be calculated from (6.4.5), giving

k kk

P jQ−= *

k

IV (6.5.1)

Applying the Gauss-Seidel method:

k 1 Nk k

kn n kn nn 1 n k 1kk k

P jQ1(i 1) (i 1) (i)(i)

−

∗= = +

⎡ ⎤−+ = − + −⎢ ⎥

⎣ ⎦∑ ∑kV Y V Y V

Y V

(6.5.2)

46

For the load bus:(6.5.2) can be applied twice during each iterationFirst using V*

k(i), then replacing V*k(i) by V*

k(i+1) on the right side.

( )N

k k kn n k n knn 1

Q V (i) Y V (i)sin (i) (i)=

= δ − δ −θ∑

For the voltage-controlled bus:Qk is unknownFind Qk using

Then calculate QGk usingGk k LkQ Q Q= +

If calculated value of QGk is within its limits,

Use this Qk in (6.5.2) to calculate Vk(i+1)=Vk(i+1)∟δk(i+1).Change the magnitude Vk(i+1) to Vk , which is input data for the voltage-controlled bus. Thus we use (6.5.2) to compute only the angle δk(i+1) for voltage-controlled buses.

47

If calculated value exceeds its limit QGkmax or QGkmin during any iteration

Change the bus type from a voltage-controlled bus to a load bus,

with QGk set to its limit value.

Under this condition, the voltage controlling device is not capable

of maintaining Vk as specified by the input data. The power-flow

program then calculates a new value of Vk.

For the swing bus:

V1 and δ1 are input data. As such, no iterations are required for bus 1.

P1 and Q1 can be computed using 6.4.10 and 6.4.11.

48

EXAMPLE 6.10 Power-flow solutions by Gauss-Seidel

For the power system of Example 6.9, use Gauss-Seidel to calculate V2(1), the phasor voltage at bus 2 after the first iteration. Use zero initial phase angles and 1.0 per-unit initial voltage magnitudes (except at bus 3, where V3=1.05) to start the iteration procedure.

SOLUTION Bus 2 is a load bus. Using the input data and bus admittance values from Example 6.9 in (6.5.2),

[ ]2 22 21 1 23 3 24 4 25 5*

22 2

P jQ1(1) Y V (1) Y V (0) Y V (0) Y V (0)Y V (0)

⎧ ⎫−= − + + +⎨ ⎬

⎩ ⎭V

[ ]2

2

1 8 j( 2.8)(1) ( 1.78552 j19.83932)(1.0) ( 0.89276 j9.91964)(1.0)28.5847 84.624 1.0 0

( 8 j2.8) ( 2.67828 j29.7589)(1) 0.96132 16.54328.5847 84.624

per unit

° °

°°

⎧ ⎫− − −= − − + + − +⎨ ⎬− ⎩ ⎭

− + − −= −

−−

+=

V

V

49

Next, the above value is used in (6.5.2) to recalculate V2(1):

[ ]2

2

1 8 j2.8(1) 2.67828 j29.7582928.5847 84.624 0.9613216.543

4.4698 j24.5973(1) 0.87460 15.67528.5847 84.62

per unit4

° °

°°

⎧ ⎫− += − − +⎨ ⎬− ⎩

−

⎭

− −= = −

−

V

V

50

6.6 POWER-FLOW SOLUTION BY NEWTON-RAPHSON

( )N

k k kn n k n knn 1

Q V Y V sin=

= δ − δ −θ∑N

k k kn n k n knn 1

P V Y V cos( )=

= δ − δ −θ∑

Above equations are analogous to the nonlinear equation y=f(x), solved in Section 6.3 by Newton-Raphson.

Define x, y, and f vectors for the power-flow problem as

2

N

2

N

P (x)

P (x)P(x)(x)

Q (x)Q(x)

Q (x)

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎡ ⎤

= = ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

f

M

M

2

N

2

N

P

PPQQ

Q

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎡ ⎤

= = ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

y

M

M

2

N

2

N

VV

V

δ⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥δδ⎡ ⎤

= = ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

x

M

M

51

All V, P, and Q terms are in per-unit and δ terms are in radians. The swing bus variables δ1 and V1 are omitted from equations since they are already known. Power-flow equations for Newton-Raphson have the following form:

N

k k k k kn n k n knn 1

N

k N k k k kn n k n knn 1

y P P (x) V Y V cos( )

y Q Q (x) V Y V sin( )

=

+=

= = = δ −δ −θ

= = = δ −δ −θ

∑

∑

(6.6.2)

(6.6.3)

k=2,3,….,N

52

The Jacobian matrix of (6.3.10) has the form

2 2 2 2

2 N 2 N

N N N N

2 N 2 n

2 2 2 2

2 N 2 N

N N2 2

2 N 2 N

P P P PV V

P P P PV V

Q Q Q QV V

Q QQ QV V

δ δ δ δ⎡ ⎤⎢ ⎥∂δ ∂δ ∂ ∂⎢ ⎥⎢ ⎥⎢ ⎥δ δ δ δ⎢ ⎥⎢ ⎥∂δ ∂δ ∂ ∂

= ⎢ ⎥δ δ δ δ⎢ ⎥

⎢ ⎥∂δ ∂δ ∂ ∂⎢ ⎥⎢ ⎥⎢ ⎥δ δδ δ⎢ ⎥∂δ ∂δ ∂ ∂⎢ ⎥⎣ ⎦

J

L L

M M

L L

L L

M M

L L

J1 J2

J3 J4

(6.6.4)

J is partitioned into four blocks. The partial derivatives in each block are given in Table 6.5.

53

TABLE 6.5: Elements of the Jacobian matrix

kkn k kn n k n kn

n

kkn k kn k n kn

n

kkn k kn n k n kn

n

kkn k kn k n kn

n

Nk

kk k kn n k n knn 1nn k

kkk k kn kk kn n

k

PJ1 V Y V sin ( )

PJ 2 V Y cos( )VQJ3 V Y V cos( )

QJ 4 V Y sin( )V

PJ1 V Y V sin( )

PJ 2 V Y cos Y VV

=≠

∂= = δ − δ − θ∂δ∂

= = δ − δ − θ∂∂

= = − δ − δ − θ∂δ∂

= = δ − δ − θ∂

∂= = − δ − δ − θ∂δ

∂= = θ +∂

∑

N

k n knn 1

Nk

kk k kn n k n knn 1kn k

Nk

kk k kk kk kn n k n knn 1k

cos( )

QJ3 V Y V cos( )

QJ 4 V Y sin Y V sin( )V

=

=≠

=

δ − δ − θ

∂= = − δ − δ − θ

∂δ

∂= = − θ + δ − δ − θ

∂

∑

∑

∑

n ≠ k

n = k

k, n = 2,3,….,N

54

Apply the four Newton-Raphson steps to the power-flow problem

start with at the ith iteration.(i)x(i)

V(i)δ⎡ ⎤

= ⎢ ⎥⎣ ⎦

P(i) P P[x(i)]y(i)

Q(i) Q Q[x(i)]∆ −⎡ ⎤ ⎡ ⎤

∆ = =⎢ ⎥ ⎢ ⎥∆ −⎣ ⎦ ⎣ ⎦STEP 1 compute (6.6.5)

STEP 2 Use the equations in the Table 6.5 to calculate the Jacobian matrix.

STEP 3 Use Gauss elimination and back substitution to solve

J1(i) J2(i) (i) P(i)J3(i) J4(i) V(i) Q(i)

∆δ ∆⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥∆ ∆⎣ ⎦ ⎣ ⎦ ⎣ ⎦ (6.6.6)

(i 1) (i) (i)x(i 1)

V(i 1) V(i) V(i)δ + δ ∆δ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

+ = = +⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ ∆⎣ ⎦ ⎣ ⎦ ⎣ ⎦STEP 4 Compute (6.6.7)

55

Convergence criteria are often based on ∆y(i) (called power mismatches) rather than on ∆x(i) (phase angle and voltage magnitude mismatches).For each voltage-controlled bus

Vk is already known, andFunction Qk(x) is not needed

Omit Vk from x vectorQk from y vectorAlso omit the column and the row from Jacobian matrix that includes

Partial derivates with respect to Vk andPartial derivatives of Qk(x).

At the end of each iteration Qk(x) and QGk are calculated, and necessary checks are performed.

56

EXAMPLE 6.10 Jacobian matrix and power-flow solution by Newton-Raphson

Determine the dimension of the Jacobian matrix for the power system in Example 6.9. Also calculate ∆P2(0) in step 1 and J124(0) is Step 2 of the first Newton-Raphson iteration. Assume zero initial phase angles and 1.0 per-unit initial voltage magnitudes (except V3=1.05).

SOLUTION Since N=5, (6.6.2) and (6.6.3) constitute 2(N-1)=8 equations, for which J(i) has dimension 8x8. However, there is one voltage-controlled bus, bus 3. Therefore, V3 and the equation for Q3(x) could be eliminated, with J(i) reduced to a 7x7 matrix.

From Step 1 and (6.6.2),

∆P2(0)=P2-P2(x)=P2-V2(0){Y21V1cos[δ2(0)- δ1(0)-θ21]

+Y22V2cos[-θ22]+Y23V3cos[δ2(0)- δ3(0)-θ23]

+Y24V4cos [δ2(0)- δ4(0)-θ24]

+Y25V5cos[δ2(0)- δ5(0)-θ25]}

57

∆P2(0)=-8.0-1.0{28.5847(1.0) cos(84.624°) +9.95972(1.0) cos(-95.143°)+

+19.9159(1.0) cos(-95.143°)}

=-8.0-(-2.89x10-4)=-7.99972 per unit

From Step 2 and J1 given in the Table 6.5

J124(0)=V2(0)Y24V4(0) sin[δ2(0)- δ4(0)-θ24]

=(1.0)(9.95972)(1.0) sin(-95.143°)

=-9.91964 per unit

58

6.7 CONTROL OF POWER FLOW

The following means are used to control system power flows:

1. Prime mover and excitation control of generators.

2. Switching of shunt capacitor banks, shunt reactors, and static var systems

3. Control of tap-changing and regulating transformers

59

Generator Thevenin Equivalent

~E δg g= ∠E V 0t t= ∠V o

jXg

I

P

Q

60

Vt is the generator terminal voltage, Eg is the excitation voltage, δ is the power angle, and Xg is the positive-sequence synchronous reactance.

jg t

g

E e VjX

δ −=I (6.7.1)

and the complex power delivered by the generator is

jg t

t tg

E e VP jQ V ( )

jX

− δ∗ −

= + = =−

S V I

2t g t

g

V E ( jcos sin ) jVXδ + δ −

=S (6.7.2)

61

The real and reactive powers delivered are then

t g

g

V EP Re sin

X= = δS (6.7.3)

tg t

g

VQ Im (E cos V )X

= = δ−S (6.7.4)

δ ↑ P ↑ Q ↓

Large increase in P but very small change in Q when δ<15°

Eg ↑ Q ↑

62

Effect of adding a Shunt Capacitor Bank to a Power System Bus

~ThE tV

jXTh

C

IC

SWRTh

IC

Vt

Th CR IThE

Th CX Ij

63

6.8 SPARSITY TECHNIQUES

1.0 1.1 2.1 3.14.1 2.0 0 5.1

S6.1 0 3.0 07.1 0 0 4.0

− − −⎡ ⎤⎢ ⎥− −⎢ ⎥=⎢ ⎥−⎢ ⎥−⎣ ⎦

(6.8.1)

One method for compact storage of S consists of the following four vectors:

[ ]DIAG 1.0 2.0 3.0 4.0= (6.8.2)

[ ]OFFDIAG 1.1 2.1 3.1 4.1 5.1 6.1 7.1= − − − − − − − (6.8.3)

[ ]COL 2 3 4 1 4 1 1= (6.8.4)

[ ]ROW 3 2 1 1=(6.8.5)

64

40N instead of 8N2

For a 1000-bus power system this means40 kilobytes instead of 8000 kilobytes

Reordering the buses is another sparsity technique

The S matrix after one Gauss elimination step

(1)

1.0 1.1 2.1 3.10 2.51 8.61 7.61

S0 6.71 9.81 18.910 7.81 14.91 18.01

− − −⎡ ⎤⎢ ⎥− − −⎢ ⎥=⎢ ⎥− − −⎢ ⎥− − −⎣ ⎦

(6.8.6)

Original degree of sparsity is lost

65

Reorder the buses to start with those buses having the fewest connected branches and to end with those having the most connected branches.

Reorder buses 4, 3, 2, 1 instead of 1, 2, 3, 4.

reordered

4.0 0 0 7.10 3.0 0 6.1

S5.1 0 2.0 4.13.1 2.1 1.1 1.0

−⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥− −⎢ ⎥− − −⎣ ⎦

1.0 1.1 2.1 3.14.1 2.0 0 5.1

S6.1 0 3.0 07.1 0 0 4.0

− − −⎡ ⎤⎢ ⎥− −⎢ ⎥=⎢ ⎥−⎢ ⎥−⎣ ⎦

(6.8.7)

After one Gauss elimination step

(1)reordered

4.0 0 0 7.10 3.0 0 6.1

S0 0 2.0 13.150 2.1 1.1 4.5025

−⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥−⎢ ⎥− − −⎣ ⎦

(6.8.8)

66

6.9 FAST DECOUPLED POWER FLOW

The purpose is to reduce the time required to solve the power flow program

J1(i) J2(i) (i) P(i)J3(i) J4(i) V(i) Q(i)

∆δ ∆⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥∆ ∆⎣ ⎦ ⎣ ⎦ ⎣ ⎦

J2(i) 0J3(i) 0

≈≈

1

4

J (i) (i) P(i)

J (i) V(i) Q(i)

∆δ = ∆

∆ = ∆

(6.9.1)

(6.9.2)