power flow algorithms

7/27/2019 Power Flow Algorithms

1/9


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2

Keep on for (k)1 x , k=0,1,2,3..Until the algorithm converges, i.e. stopAt a k th iteration when:

tolerance)smallverysome(For

(k)

2

(k)

1

(k)2

(k)1

x x f K

x x f K

,(

,(

22

11

NR Algorithm normally converges in 3-5 iterations but, each iteration requires the evaluation of J -1, which is time-consuming for large matrix (Sparaty techniques).

Previous Example

22231212212

11132112211

)sin()sin()(),()sin()sin()(),(

P B B f

P B B f

II. Jacobian of functions of several variables:

0),.....,(...

0),.....,(

21

211

nn

n

x x x f

x x x f

Scalar Form

Vector form : f(x) =0f(x) =0 or f i(x) =0, I=1,2,.n

n

nn

n

nn

nnnn

nn

n

x

x f

x

x f

x x f

x x f

x x f

x x f

x J

where

t oh x X J x f x x f

So

terms

order Higher

For

x x

x f x

x

x f x f x x f

x x

x f x

x x f

x f x x f

)(....

)(.......

)(

)(...

)()(

)(

..)()(),(,

)(.......

)()(),(

...

...

...

)(.......

)()(),(

1

1

2

1

2

1

1

1

1

11

11

1

11

J abobian matrix of Partial derivatives


3/9

3

III. Applying N-R Method to power flow equations* J acobian of full load flow Eq:

Example:N-R Method:

(k)n

(k)2

(k)1

(k)n

(k)2

(k)1

.

.

K

K K

J

x

x x

k )1()(

.

.

Need to evaluate [J (k) ] and then [J (k) ]-1 V, given:

Jacobianbuses

QP

V

V

QQQV PPPV PPP

Q

PP

busbus

example

Busbus

slack

Bus

,

3,2

3

1

2

32

2

2

3

2

2

2

2

3

3

3

2

3

2

2

3

2

2

2

2

3

2

IV. Applying N-R Method to Power Flow Equations:(PF2)

)cos()sin(

,...,1)sin()cos(

1

1

k iik k iik k

n

k ii

k iik k iik k

n

k ii

BGV V Q

ni

BGV V P

P i,Q i can be regarded as functions:

),(),(vQQ

vPP

ii

ii

Take 1 as the ref. 1 =0, V 1 givenRight hand side are functions of 2, 3,.., n and V 2, V 3,..,V n

For case I. Al l P,Q buses except the 1st

bus

)(

)cos()sin(

)cos()sin(1

Summation Noik

BGV V P

BGV V P

k iik k iik k ik

i

k iik k iik k

n

ik

k i

i

i


4/9

4

ik

BGV V

P

BGV GV V

P

k iik k iik ik

i

k iik k iik

n

ik

k k iii

i

i

)sin()cos(

)sin()cos(21

)cos()cos(

)sin()cos(1

k iik k iik k ik

i

k iik k iik k

n

ik

k i

i

i

BGV V Q

BGV V Q

)cos()sin(

)cos()sin(21

k iik k iik ik

i

k iik k iik

n

ik

k k iii

i

i

BGV V

Q

BGV BV V

Q

For case 1, all load buses, except bus 1 (slack) the J acobian for Newton-Raphson:

equationsn

QV V Q

equationsn

PV V P

x x x

inni

inni

n

n2,.....,i

n2,.....,i

)1(2

,........,,,....,()1(2

,........,,,....,(

22

22

)1(21

or

nn

nn

Q xQ

Q xQ

P xP

P xP

k x f

)(..

)()(

.

.)(

)(22

22


5/9

5

x x

x x

xn

n

x

n

x x

xn

n

x

n

xn x

V QQ

V PP

V

QQ

V QQ

V

PP

V

PP

x x f

..........

)(

2

2

2

2

2

2

2

2

2

)()()()(

)(

)(

.....)(

)(

11

11

22

22

1

2

2

1

12

1

12

x f k x J x x x

x f k x J x x

xQQ

xQQ

xPP

xPP

V QQ

V PP

V

V

V

V

nn

nn

x

n

n

n

n

VII. Fast Decoupled Power Flow (FDLF method for the solution of PF 2)

smallElements

)sin( p 211221

Qand

V P

BV V

V QQ

V PP

x f


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6

hence,

V

Q

P

x f

0

0

So Newton-Raphson iteration:

Q

P

V V Q

P

byed approximat becan

Q

P

V V QQ

V PP

0

0

Of (PF2) is used, then the approximation applies to the iterations only. Since the mismatch will bechecked for convergence. The final solution should be same or close to the N-R solutionpreviously described.

Further Approximations

(small)esusceptancshuntof Sum

Then,

1)0(0

)cos()sin(

1

2)4(

1

2

1

22

)3(

2

1

1

)2()1(

1

n

k ik

iii

n

k iiik i

n

k iii

small

ik i

k i

iiiik k

n

k i

ik k

n

ik

k i

ik

k iik k iik k

n

ik k

ii

I

B

BV B BV

BV BV

V V

BV BV V

BV V

G

BGV V P


7/9

7

Similarly,

ik i

ik

k iik k iik i

k k

i

ik k i

ik

k iik k iijk i

k k

i

iii

n

k iii

small

ik k iii

n

ik

k ik k iii

ik

n

ik

k k iik k iik k iii

i

i

BV

G

BGV V

Q

BV V

G

BGV V P

BV

BV BV BV

BV BV

G

BGV BV V

Q

1)0(0

)cos()sin(

1)0(0

)cos()sin(

2

2

1)0(0

)cos()sin(2

)1(

)1(

1

1

1

Matrix Form

To check, nnn

n

n B B

B B B

B

V

V

V

V

V BV P

2

22322

3

2

,][

][ ][


8/9

8

00

0

0

000

i

iii

i

V

V V V

QB

)(][ )(][][

||

00

][

2

)(

xQV BV

xPV BV

BV BV BV

BV

I

V

V

Q

BV V Q

ik i

nk i

ik i

k ii

ik k

i

Further Approximation,

scaled diagonal

xQ xQV V B

xP xPV B

xQV BV

xP BV

)()(][

)()(][

)(][)(][

~1)(

~1)(

)(

)(

busesloadn,2,n

KassumesThis

B does not change over the iteration Therefore can be computed before the iteration.


9/9

9

Fast Decoupled Power Flow

Ex.

3

33

3

3

21

3

2

3

33

3

3

2

3

2

3

3333

3

3

2

3

2

3332

2322

332

23233333

0501.0

,....2,1,0

05.1

98.1910

1098.19

98.19

05.1

98.1910

1098.19

)(

05.1

,,]98.19)cos(5.10cos10[

V Q

V

V P

P

V Q

V

V P

P

V Q

V B

V P

P

B B

B B

equationsQPP Needs

V V V Q

01 00.1

1

V

S G

98.19

1005.1

666.0

2

2

ii

ij

G

B

B

V

P

2244.18653.2

2244.18653.2

3

3

3

Q

P

jV

power flow algorithms

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