EE 369 POWER SYSTEM ANALYSIS Lecture 13 Newton-Raphson Power Flow Tom Overbye and Ross Baldick 1.

EE 369POWER SYSTEM ANALYSIS

Lecture 13Newton-Raphson Power Flow

Tom Overbye and Ross Baldick

1

Announcements

• Read Chapter 12, concentrating on sections 12.4 and 12.5.

• Homework 12 is 6.43, 6.48, 6.59, 6.61, 12.19, 12.22, 12.20, 12.24, 12.26, 12.28, 12.29; due Tuesday Nov. 25.

2

Using the Power Flow: Example 1

slack

SLACK345

SLACK138

RAY345

RAY138

RAY69

FERNA69

A

MVA

DEMAR69

BLT69

BLT138

BOB138

BOB69

WOLEN69

SHI MKO69

ROGER69

UI UC69

PETE69

HI SKY69

TI M69

TI M138

TI M345

PAI 69

GROSS69

HANNAH69

AMANDA69

HOMER69

LAUF69

MORO138

LAUF138

HALE69

PATTEN69

WEBER69

BUCKY138

SAVOY69

SAVOY138

J O138 J O345

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

1.02 pu

1.01 pu

1.02 pu

1.03 pu

1.01 pu

1.00 pu1.00 pu

0.99 pu

1.02 pu

1.01 pu

1.00 pu

1.01 pu1.01 pu

1.01 pu

1.01 pu

1.02 pu

1.00 pu

1.00 pu

1.02 pu

0.997 pu

0.99 pu

1.00 pu

1.02 pu

1.00 pu1.01 pu

1.00 pu

1.00 pu 1.00 pu

1.01 pu

1.02 pu1.02 pu

1.02 pu1.03 pu

A

MVA

1.02 pu

A

MVA

A

MVA

LYNN138

A

MVA

1.02 pu

A

MVA

1.00 pu

A

MVA

218 MW 54 Mvar

21 MW 7 Mvar

45 MW 12 Mvar

140 MW 45 Mvar

37 MW

13 Mvar

12 MW 5 Mvar

150 MW 0 Mvar

56 MW

13 Mvar

15 MW 5 Mvar

14 MW

2 Mvar

42 MW 2 Mvar

45 MW 0 Mvar

58 MW 36 Mvar

36 MW 10 Mvar

0 MW 0 Mvar

22 MW 15 Mvar

60 MW 12 Mvar

20 MW 30 Mvar

23 MW 7 Mvar

33 MW 13 Mvar

16.0 Mvar 18 MW 5 Mvar

58 MW 40 Mvar 51 MW

15 Mvar

14.3 Mvar

33 MW 10 Mvar

15 MW 3 Mvar

23 MW 6 Mvar 14 MW

3 Mvar

4.8 Mvar

7.2 Mvar

12.8 Mvar

29.0 Mvar

7.4 Mvar

0.0 Mvar

106 MW 8 Mvar

20 MW 8 Mvar

150 MW 0 Mvar

17 MW 3 Mvar

0 MW 0 Mvar

14 MW 4 Mvar

Usingcasefrom Example6.13

3

Dishonest Newton-RaphsonSince most of the time in the Newton-Raphson

iteration is spent dealing with the Jacobian, one way to speed up the iterations is to only calculate (and factorize) the Jacobian occasionally:– known as the “Dishonest” Newton-Raphson or Shamanskii

method,– an extreme example is to only calculate the Jacobian for the

first iteration, which is called the chord method.

( 1) ( ) ( ) -1 ( )

( 1) ( ) (0) -1 ( )

( )

Honest: - ( ) ( )

Dishonest: - ( ) ( )

Stopping criterion ( ) used in both cases.

v v v v

v v v

v

x x J x f x

x x J x f x

f x 4

Dishonest Newton-Raphson Example

2

1( ) (0) ( )

( ) ( ) 2(0)

( 1) ( ) ( ) 2(0)

Use the Dishonest Newton-Raphson (chord method)

to solve ( ) 0, where:

( ) - 2

( ) ( )

1(( ) - 2)

21

(( ) - 2)2

v v

v v

v v v

f x

f x x

dfx x f x

dx

x xx

x x xx

5

Dishonest N-R Example, cont’d( 1) ( ) ( ) 2

(0)

(0)

( ) ( )

1(( ) - 2)

2

Guess 1. Iteratively solving we get

(honest) (dishonest)

0 1 1

1 1.5 1.5

2 1.41667 1.375

3 1.41422 1.429

4 1.41422 1.408

v v v

v v

x x xx

x

x x

We pay a pricein increased iterations, butwith decreased computationper iteration

6

Two Bus Dishonest ROCSlide shows the region of convergence for different initialguesses for the 2 bus case using the Dishonest N-RRed region

convergesto the highvoltage solution,while the yellow regionconvergesto the lowvoltage solution

7

Honest N-R Region of Convergence

Maximum of 15

iterations

8

Decoupled Power FlowThe “completely” Dishonest Newton-

Raphson (chord) is not usually used for power flow analysis. However several approximations of the Jacobian matrix are used that result in a similar approximation.

One common method is the decoupled power flow. In this approach approximations are used to decouple the real and reactive power equations.

9

Coupled Newton-Raphson Update

( ) ( )

( ) ( )( )

( )( ) ( ) ( )

( )2 2 2

( )

( )

Standard form of the Newton-Raphson update:

( )( )

( )

( )

where ( ) .

( )

Note

v v

v vv

vv v v

vD G

v

vn Dn Gn

P P P

P P P

P Pθθ V P x

f xQ xVQ Q

θ V

x

P x

x

that changes in angle and voltage magnitude

both affect (couple to) real and reactive power. 10

Decoupling Approximation( ) ( )

( )

( ) ( )( )

( ) ( ) ( )

Usually the off-diagonal matrices, and

are small. Therefore we approximate them as zero:

( )( )

( )

Then the update c

v v

v

v vv

v v v

P QV θ

P0

θ P xθf x

Q Q xV0V

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

an be decoupled into two separate updates:

( ), ( ).v v

vv v v

P Qθ P x V Q x

θ V11

Off-diagonal Jacobian TermsSo, angle and real power are coupled closely, and

voltage magnitude and reactive power are coupled cl

Justification for Jacobian approximations:

1. Usually , therefore

2. U

os

su

ely.

ally is s

ij ij

ij

r x G B

mall so sin 0

Therefore

cos sin 0

cos sin 0

ij

ii ij ij ij ij

j

ii j ij ij ij ij

j

V G B

V V G B

P

V

Qθ 12

Decoupled N-R Region of Convergence

13

Fast Decoupled Power FlowBy further approximating the Jacobian we obtain a

typically reasonable approximation that is independent of the voltage magnitudes/angles.

This means the Jacobian need only be built and factorized once.

This approach is known as the fast decoupled power flow (FDPF)

FDPF uses the same mismatch equations as standard power flow so it should have same solution if it converges

The FDPF is widely used, particularly when we only need an approximate solution. 14

FDPF Approximations

( ) 1 ( ) 1 ( )

( ) 1 ( ) 1 ( )

The FDPF makes the following approximations:

1. 0

2. 1 (for some occurrences),

3. sin 0 cos 1

Then: {| | } ( ),

{| | } ( )

Where is just the imaginary pa

ij

i

ij ij

v v v

v v v

G

V

diag

diag

θ B V P x

V B V Q x

B bus bus bus

bus

( )

rt of the ,

except the slack bus row/column are omitted. That is,

is , but with the slack bus row and column deleted.

Sometimes approximate {| | } by identity. v

j

diag

Y G B

B B

V 15

FDPF Three Bus Example

Line Z = j0.07

Line Z = j0.05 Line Z = j0.1

One Two

200 MW 100 MVR

Three 1.000 pu

200 MW 100 MVR

Use the FDPF to solve the following three bus system

34.3 14.3 20

14.3 24.3 10

20 10 30bus j

Y

16

FDPF Three Bus Example, cont’d

1

(0)(0)2 2

3 3

34.3 14.3 2024.3 10

14.3 24.3 1010 30

20 10 30

0.0477 0.0159

0.0159 0.0389

Iteratively solve, starting with an initial voltage guess

0 1

0 1

bus j

V

V

Y B

B

(1)2

3

0 0.0477 0.0159 2 0.1272

0 0.0159 0.0389 2 0.1091

17

FDPF Three Bus Example, cont’d(1)

2

3

1

(2)2

3

1 0.0477 0.0159 1 0.9364

1 0.0159 0.0389 1 0.9455

( )( cos sin )

0.1272 0.0477 0.0159

0.1091 0.0159 0.0389

ni Di Gi

k ik ik ik iki ik

V

V

P x P PV G B

V V

(2)2

3

0.151 0.1361

0.107 0.1156

0.924

0.936

0.1384 0.9224Actual solution:

0.1171 0.9338

V

V

θ V

18

FDPF Region of Convergence

19

“DC” Power Flow

The “DC” power flow makes the most severe approximations:– completely ignore reactive power, assume all the

voltages are always 1.0 per unit, ignore line conductance

This makes the power flow a linear set of equations, which can be solved directly:

1θ B P

20

DC Power Flow Example

21

DC Power Flow 5 Bus Example

slack

One

Two

ThreeFourFiveA

MVA

A

MVA

A

MVA

A

MVA

A

MVA

1.000 pu 1.000 pu

1.000 pu

1.000 pu

1.000 pu 0.000 Deg -4.125 Deg

-18.695 Deg

-1.997 Deg

0.524 Deg

360 MW

0 Mvar

520 MW

0 Mvar

800 MW 0 Mvar

80 MW 0 Mvar

Notice with the dc power flow all of the voltage magnitudes are 1 per unit.

22

Power System ControlA major problem with power system operation is

the limited capacity of the transmission system– lines/transformers have limits (usually thermal)– no direct way of controlling flow down a transmission

line (e.g., there are no low cost valves to close to limit flow, except “on” and “off”)

– open transmission system access associated with industry restructuring is stressing the system in new ways

We need to indirectly control transmission line flow by changing the generator outputs.

23

Indirect Transmission Line ControlWhat we would like to determine is how a change in generation at bus k affects the power flow on a line from bus i to bus j.

The assumption isthat the changein generation isabsorbed by theslack bus

24

Power Flow Simulation - Before•One way to determine the impact of a generator change is to compare a before/after power flow.•For example below is a three bus case with an overload.

Z for all lines = j0.1

One Two

200 MW 100 MVR

200.0 MW 71.0 MVR

Three 1.000 pu

0 MW 64 MVR

131.9 MW

68.1 MW 68.1 MW

124%

25

Power Flow Simulation - After

Z for all lines = j0.1Limit for all lines = 150 MVA

One Two

200 MW 100 MVR

105.0 MW 64.3 MVR

Three1.000 pu

95 MW 64 MVR

101.6 MW

3.4 MW 98.4 MW

92%

100%

•Increasing the generation at bus 3 by 95 MW (and hence decreasing generation at the slack bus 1 by a corresponding amount), results in a 31.3 MW drop in the MW flow on the line from bus 1 to 2.

26

Analytic Calculation of SensitivitiesCalculating control sensitivities by repeated power flow

solutions is tedious and would require many power flow solutions.

An alternative approach is to analytically calculate these values

The power flow from bus i to bus j is

sin( )

So We just need to get

i j i jij i j

ij ij

i j ijij

ij Gk

V VP

X X

PX P

27

Analytic Sensitivities1From the fast decoupled power flow we know: ( ).

Sign convention in definition of ( ) is that entry in ( )

is negative if change in net injection (generation) is positive.

So to get the chan

θ B P x

P x P x

ge in due to a change of generation at

bus , just set ( ) equal to all zeros except a minus one

at position :

0

For 1MW increase in generation at bus 1

0

k

k

k

θ

P x

P

28

Three Bus Sensitivity Exampleline

bus

12

3

For the previous three bus case with Z 0.1

20 10 1020 10

10 20 1010 20

10 10 20

Hence for a change of generation at bus 3

20 10 0 0.0333

10 20 1 0.0667

j

j

Y B

3 to 1

3 to 2 2 to 1

0.0667 0Changes in line flows are: 0.667 pu

0.10.333 pu 0.333 pu

P

P P

29