ELE B7 Power Systems Engineering ELE B7 Power Systems Engineering

NewtonNewton--RaphsonRaphson MethodMethod

Slide # 2

Newton-Raphson Algorithm

• The second major power flow solution method is the Newton-Raphson algorithm

• Key idea behind Newton-Raphson is to use sequential linearizationGeneral form of problem: Find an x such that

( ) 0ˆf x

Slide # 3

Newton-Raphson Method (scalar)

( )

( ) ( )

( )( ) ( )

2 ( ) 2( )2

1. For each guess of , , define ˆ

-ˆ2. Represent ( ) by a Taylor series about ( )ˆ

( )( ) ( )ˆ

( ) higher order terms

v

v v

vv v

vv

x x

x x xf x f x

df xf x f x xdx

d f x xdx

Slide # 4

Newton-Raphson Method, cont’d

( )( ) ( )

( )

1( )( ) ( )

3. Approximate ( ) by neglecting all termsˆexcept the first two

( )( ) 0 ( )ˆ

4. Use this linear approximation to solve for

( ) ( )

5. Solve for a new estim

vv v

v

vv v

f x

df xf x f x xdx

x

df xx f xdx

( 1) ( ) ( )

ate of x̂v v vx x x

Slide # 5

Newton-Raphson Example

2

1( )( ) ( )

( ) ( ) 2( )

( 1) ( ) ( )

( 1) ( ) ( ) 2( )

Use Newton-Raphson to solve ( ) - 2 0The equation we must iteratively solve is

( ) ( )

1 (( ) - 2)2

1 (( ) - 2)2

vv v

v vv

v v v

v v vv

f x x

df xx f xdx

x xx

x x x

x x xx

Slide # 6

Newton-Raphson Example, cont’d

( 1) ( ) ( ) 2( )

(0)

( ) ( ) ( )

3 3

6

1 (( ) - 2)2

Guess x 1. Iteratively solving we get

v ( )0 1 1 0.51 1.5 0.25 0.08333

2 1.41667 6.953 10 2.454 10

3 1.41422 6.024 10

v v vv

v v v

x x xx

x f x x

Slide # 7

Sequential Linear Approximations

Function is f(x) = x2 - 2 = 0.Solutions are points wheref(x) intersects f(x) = 0 axis

At each iteration theN-R methoduses a linearapproximationto determine the next valuefor x

Slide # 8

Newton-Raphson Comments

• When close to the solution the error decreases quite quickly -- method has quadratic convergence

• f(x(v)) is known as the mismatch, which we would like to drive to zero

• Stopping criteria is when f(x(v))

< • Results are dependent upon the initial guess.

What if we had guessed x(0) = 0, or x (0) = -1?• A solution’s region of attraction (ROA) is the set

of initial guesses that converge to the particular solution. The ROA is often hard to determine

Slide # 9

Multi-Variable Newton-Raphson

1 1

2 2

Next we generalize to the case where is an n-dimension vector, and ( ) is an n-dimension function

( )( )

( )

( )Again define the solution so ( ) 0 andˆ ˆ

n n

x fx f

x f

xf x

xx

x f x

xx f x

x

ˆ x x

Slide # 10

Multi-Variable Case, cont’d

i

1 11 1 1 2

1 2

1

n nn n 1 2

1 2

n

The Taylor series expansion is written for each f ( )f ( ) f ( )f ( ) f ( )ˆ

f ( ) higher order terms

f ( ) f ( )f ( ) f ( )ˆ

f ( ) higher order terms

nn

nn

x xx x

xx

x xx x

xx

xx xx x

x

x xx x

x

Slide # 11

Multi-Variable Case, cont’d

1 1 1

1 21 1

2 2 22 2

1 2

1 2

This can be written more compactly in matrix form( ) ( ) ( )

( )( ) ( ) ( )

( )( )ˆ

( )( ) ( ) ( )

n

n

nn n n

n

f f fx x x

f xf f f

f xx x x

ff f f

x x x

x x x

xx x x

xf x

xx x x

higher order terms

nx

Slide # 12

Jacobian Matrix

1 1 1

1 2

2 2 2

1 2

1 2

The n by n matrix of partial derivatives is knownas the Jacobian matrix, ( )

( ) ( ) ( )

( ) ( ) ( )( )

( ) ( ) ( )

n

n

n n n

n

f f fx x x

f f fx x x

f f fx x x

J xx x x

x x xJ x

x x x

Slide # 13

Multi-Variable N-R Procedure

1

( 1) ( ) ( )

( 1) ( ) ( ) 1 ( )

( )

Derivation of N-R method is similar to the scalar case( ) ( ) ( ) higher order termsˆ( ) 0 ( ) ( )ˆ

( ) ( )

( ) ( )

Iterate until ( )

v v v

v v v v

v

f x f x J x xf x f x J x x

x J x f x

x x x

x x J x f x

f x

Slide # 14

Multi-Variable Example

1

22 2

1 1 22 2

2 1 2 1 2

1 1

1 2

2 2

1 2

xSolve for = such that ( ) 0 where

x

f ( ) 2 8 0

f ( ) 4 0First symbolically determine the Jacobian

f ( ) f ( )

( ) =f ( ) f ( )

x x

x x x x

x x

x x

x f x

x

x

x x

J xx x

Slide # 15

Multi-variable Example, cont’d

1 2

1 2 1 2

11 1 2 1

2 1 2 1 2 2

(0)

1(1)

4 2( ) =

2 2Then

4 2 ( )2 2 ( )

1Arbitrarily guess

1

1 4 2 5 2.11 3 1 3 1.3

x xx x x x

x x x fx x x x x f

J x

xx

x

x

Slide # 16

Multi-variable Example, cont’d

1(2)

(2)

2.1 8.40 2.60 2.51 1.82841.3 5.50 0.50 1.45 1.2122

Each iteration we check ( ) to see if it is below our specified tolerance

0.1556( )

0.0900If = 0.2 then we wou

x

f x

f x

ld be done. Otherwise we'd continue iterating.

Slide # 17

NR Application to Power Flow

** * *

i1 1

We first need to rewrite complex power equationsas equations with real coefficients

S

These can be derived by defining

n n

i i i ik k i ik kk k

V I V Y V V Y V

ik ik ik

i

Y G jB

V

jRecall e cos sin

iji i i

ik i k

V e V

j

=

=

=

ik ik ik

i

Y G jB

V

jRecall e cos sin

iji i i

ik i k

V e V

j

=

=

=

Slide # 18

Real Power Balance Equations

* *i

1 1

1

i1

i1

S ( )

(cos sin )( )

Resolving into the real and imaginary parts

P ( cos sin )

Q ( sin cos

ikn n

ji i i ik k i k ik ik

k kn

i k ik ik ik ikk

n

i k ik ik ik ik Gi Dikn

i k ik ik ik ik

P jQ V Y V V V e G jB

V V j G jB

V V G B P P

V V G B

)k Gi DiQ Q

Slide # 19

Newton-Raphson Power Flow

i1

In the Newton-Raphson power flow we use Newton'smethod to determine the voltage magnitude and angleat each bus in the power system. We need to solve the power balance equations

P ( cosn

i k ik ikk

V V G

i1

sin )

Q ( sin cos )

ik ik Gi Di

n

i k ik ik ik ik Gi Dik

B P P

V V G B Q Q

Slide # 20

Power Flow Variables

Assume the slack bus is the first bus (with a fixedvoltage angle/magnitude). We then need to determine the voltage angle/magnitude at the other buses.

DnGnn

DG

DnGnn

DG

n

n

QQQ

QQQPPP

PPP

V

V

(x)

(x)(x)

(x)

ΔQ(x)ΔP(x)

f(x)X

222

222

2

2

,

Slide # 21

N-R Power Flow Solution

( )

( )

( 1) ( ) ( ) 1 ( )

The power flow is solved using the same procedurediscussed last time:

Set 0; make an initial guess of ,

While ( ) Do

( ) ( )1

End While

v

v

v v v v

v

v v

x x

f x

x x J x f x

Slide # 22

Power Flow Jacobian Matrix

1 1 1

1 2

2 2 2

1 2

1 2

The most difficult part of the algorithm is determiningand inverting the n by n Jacobian matrix, ( )

( ) ( ) ( )

( ) ( ) ( )( )

( ) ( ) ( )

n

n

n n n

n

f f fx x x

f f fx x x

f f fx x x

J xx x x

x x xJ x

x x x

Slide # 23

Power Flow Jacobian Matrix, cont’d

i

i

i1

Jacobian elements are calculated by differentiating each function, f ( ), with respect to each variable.For example, if f ( ) is the bus i real power equation

f ( ) ( cos sin )n

i k ik ik ik ik Gik

x V V G B P P

xx

i

1

i

f ( ) ( sin cos )

f ( ) ( sin cos ) ( )

Di

n

i k ik ik ik iki k

k i

i j ik ik ik ikj

x V V G B

x V V G B j i

Slide # 24

Line Flows and Losses

Slide # 25

Line Flows and Losses

Slide # 26

Two Bus Newton-Raphson Example

For the two bus power system shown below, use the Newton-Raphson power flow to determine the voltage magnitude and angle at bus two. Assumethat bus one is the slack and SBase = 100 MVA.

2

2

10 1010 10busj j

V j j

x Y

222 VV

1002002 jS

01 01V

Slide # 27

Two Bus Example, cont’d

i1

i1

2 2 1 22

2 2 1 2 2

General power balance equations

P ( cos sin )

Q ( sin cos )

Bus two power balance equationsP (10sin ) 2.0 0

( 10cos ) (10) 1.0 0

n

i k ik ik ik ik Gi Dikn

i k ik ik ik ik Gi Dik

V V G B P P

V V G B Q Q

V V

Q V V V

Slide # 28

Two Bus Example, cont’d

2 2 22

2 2 2 2

2 2

2 2

2 2

2 2

2 2 2

2 2 2 2

P ( ) (10sin ) 2.0 0

( ) ( 10cos ) (10) 1.0 0Now calculate the power flow Jacobian

P ( ) P ( )

( )Q ( ) Q ( )

10 cos 10sin10 sin 10cos 20

V

Q V V

VJ

V

VV V

x

x

x x

xx x

Slide # 29

Two Bus Example, First Iteration

(0)

2 2(0)2

2 2 2

2 2 2(0)

2 2 2 2

(1)

0Set 0, guess

1Calculate

(10sin ) 2.0 2.0f( )

1.0( 10cos ) (10) 1.0

10 cos 10sin 10 0( )

10 sin 10cos 20 0 10

0 10 0Solve

1 0 10

v

V

V V

VV V

x

x

J x

x1 2.0 0.2

1.0 0.9

Slide # 30

Two Bus Example, Next Iterations

(1)2

(1)

1(2)

0.9(10sin( 0.2)) 2.0 0.212f( )

0.2790.9( 10cos( 0.2)) 0.9 10 1.08.82 1.986

( )1.788 8.199

0.2 8.82 1.986 0.212 0.2330.9 1.788 8.199 0.279 0.8586

f(

x

J x

x

(2) (3)

(3)2

0.0145 0.236)

0.0190 0.85540.0000906

f( ) Done! V 0.8554 13.520.0001175

x x

x

Slide # 31

Two Bus Solved Values

Once the voltage angle and magnitude at bus 2 are known we can calculate all the other system values,such as the line flows and the generator reactive power output

02 52.13855.0 V

1002002 jS

3.16820012 jS

01 01V

10020021 jS

3.680211212 jSSSloss

Slide # 32

PV Buses

• Since the voltage magnitude at PV buses is fixed there is no need to explicitly include these voltages in x

or write the reactive power balance equations

– the reactive power output of the generator varies to maintain the fixed terminal voltage (within limits)

– optionally these variations/equations can be included by just writing the explicit voltage constraint for the generator bus

|Vi | – Vi setpoint = 0

Slide # 33

Three Bus PV Case Example

Line Z = 0.1j

Line Z = 0.1j Line Z = 0.1j

One Two 1.000 pu 0.941 pu

200 MW 100 MVR

170.0 MW 68.2 MVR

-7.469 Deg

Three 1.000 pu

30 MW 63 MVR

2 2 2 2

3 3 3 3

2 2 2

For this three bus case we have( )

( ) ( ) 0V ( )

G D

G D

D

P P PP P P

Q Q

xx f x x

x

Slide # 34

Solving Large Power Systems

• The most difficult computational task is inverting the Jacobian matrix– inverting a full matrix is an order n3 operation, meaning

the amount of computation increases with the cube of the size size

– this amount of computation can be decreased substantially by recognizing that since the Ybus is a sparse matrix, the Jacobian is also a sparse matrix

– using sparse matrix methods results in a computational order of about n1.5.

– this is a substantial savings when solving systems with tens of thousands of buses

Slide # 35

Newton-Raphson Power Flow

• Advantages– fast convergence as long as initial guess is close to

solution– large region of convergence

• Disadvantages– each iteration takes much longer than a Gauss-Seidel

iteration– more complicated to code, particularly when

implementing sparse matrix algorithms• Newton-Raphson algorithm is very common in power flow

analysis

ELE B7 Power Systems Engineering � � Newton-Raphson MethodNewton-Raphson AlgorithmNewton-Raphson Method (scalar)Newton-Raphson Method, cont’dNewton-Raphson ExampleNewton-Raphson Example, cont’dSequential Linear ApproximationsNewton-Raphson CommentsMulti-Variable Newton-RaphsonMulti-Variable Case, cont’dMulti-Variable Case, cont’dJacobian MatrixMulti-Variable N-R ProcedureMulti-Variable ExampleMulti-variable Example, cont’dMulti-variable Example, cont’dNR Application to Power FlowReal Power Balance EquationsNewton-Raphson Power FlowPower Flow VariablesN-R Power Flow Solution Power Flow Jacobian MatrixPower Flow Jacobian Matrix, cont’dLine Flows and LossesLine Flows and LossesTwo Bus Newton-Raphson ExampleTwo Bus Example, cont’dTwo Bus Example, cont’dTwo Bus Example, First IterationTwo Bus Example, Next IterationsTwo Bus Solved ValuesPV BusesThree Bus PV Case ExampleSolving Large Power SystemsNewton-Raphson Power Flow