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Divyanshu PrakashB.tech/3rd Yr./EEE

Mewar University, Chittorgarh

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Contentsa. Load-Flow analysisb. Introduction to Newton Raphson Method

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Gauss IterationThere are a number of different iterative methods

we can use. We'll consider two: Gauss and Newton.

With the Gauss method we need to rewrite our

equation in an implicit form: x = h(x)

To iterate we fir (0)

( +1) ( )

st make an initial guess of x, x ,

and then iteratively solve x ( ) until we

find a "fixed point", x, such that x (x).ˆ ˆ ˆ

v vh x

h

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Gauss Iteration Example

( 1) ( )

(0)

( ) ( )

Example: Solve - 1 0

1

Let k = 0 and arbitrarily guess x 1 and solve

0 1 5 2.61185

1 2 6 2.61612

2 2.41421 7 2.61744

3 2.55538 8 2.61785

4 2.59805 9 2.61798

v v

v v

x x

x x

k x k x

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Stopping Criteria

( ) ( ) ( 1) ( )

A key problem to address is when to stop the

iteration. With the Guass iteration we stop when

with

If x is a scalar this is clear, but if x is a vector we

need to generalize t

v v v vx x x x

( )

2i2

1

he absolute value by using a norm

Two common norms are the Euclidean & infinity

max x

v

j

n

i ii

x

x

x x

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Gauss Power Flow

** * *

i1 1

* * * *

1 1

*

*1 1,

*

*1,

We first need to put the equation in the correct form

S

S

S

S1

i i

i

i

n n

i i i ik k i ik kk k

n n

i i i ik k ik kk k

n ni

ik k ii i ik kk k k i

ni

i ik kii k k i

V I V Y V V Y V

V I V Y V V Y V

Y V Y V Y VV

V Y VY V

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Gauss Two Bus Power Flow Example

•A 100 MW, 50 Mvar load is connected to a generator •through a line with z = 0.02 + j0.06 p.u. and line charging of 5 Mvar on each end (100 MVA base). Also, there is a 25 Mvar capacitor at bus 2. If the generator voltage is 1.0 p.u., what is V2?

SLoad = 1.0 + j0.5 p.u.

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Gauss Two Bus Example, cont’d2

2 bus

bus

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The unknown is the complex load voltage, V .

To determine V we need to know the .

15 15

0.02 0.06

5 14.95 5 15Hence

5 15 5 14.70

( Note - 15 0.05 0.25)

jj

j j

j j

B j j j

Y

Y

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Gauss Two Bus Example, cont’d*2

2 *22 1,2

2 *2

(0)2

( ) ( )2 2

1 S

1 -1 0.5( 5 15)(1.0 0)

5 14.70

Guess 1.0 0 (this is known as a flat start)

0 1.000 0.000 3 0.9622 0.0556

1 0.9671 0.0568 4 0.9622 0.0556

2 0

n

ik kk k i

v v

V Y VY V

jV j

j V

V

v V v V

j j

j j

.9624 0.0553j

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Gauss Two Bus Example, cont’d

2

* *1 1 11 1 12 2

1

0.9622 0.0556 0.9638 3.3

Once the voltages are known all other values can

be determined, such as the generator powers and the

line flows

S ( ) 1.023 0.239

In actual units P 102.3 MW

V j

V Y V Y V j

1

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, Q 23.9 Mvar

The capacitor is supplying V 25 23.2 Mvar

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Slack Bus

In previous example we specified S2 and V1 and then solved for S1 and V2. We can not arbitrarily specify S at all buses

because total generation must equal total load + total lossesWe also need an angle reference bus.To solve these problems we define one bus as

the "slack" bus. This bus has a fixed voltage magnitude and angle, and a varying real/reactive power injection.

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Three Types of Power Flow Buses

There are three main types of power flow buses– Load (PQ) at which P/Q are fixed; iteration solves

for voltage magnitude and angle. – Slack at which the voltage magnitude and angle

are fixed; iteration solves for P/Q injections– Generator (PV) at which P and |V| are fixed;

iteration solves for voltage angle and Q injectionspecial coding is needed to include PV buses in the

Gauss-Seidel iteration

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Gauss-Seidel Advantages

Each iteration is relatively fast (computational order is proportional to number of branches + number of buses in the systemRelatively easy to program

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Gauss-Seidel Disadvantages

Tends to converge relatively slowly, although this can be improved with accelerationHas tendency to miss solutions, particularly

on large systemsTends to diverge on cases with negative

branch reactances (common with compensated lines)Need to program using complex numbers

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Newton-Raphson Algorithm Newton-Raphson method is a numerical technique for solving

non-linear equations. The second major power flow solution method is the Newton-

Raphson algorithm. Key idea behind Newton-Raphson is to use sequential

linearization

General form of problem: Find an x such that

( ) 0ˆf x

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Newton-Raphson Method (scalar)

( )

( ) ( )

( )( ) ( )

2 ( ) 2( )2

1. For each guess of , , define ˆ

-ˆ

2. Represent ( ) by a Taylor series about ( )ˆ

( )( ) ( )ˆ

1 ( )higher order terms

2

v

v v

vv v

vv

x x

x x x

f x f x

df xf x f x x

dx

d f xx

dx

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Newton-Raphson Method, cont’d

( )( ) ( )

( )

1( )( ) ( )

3. Approximate ( ) by neglecting all terms ˆ

except 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 x

dx

x

df xx f x

dx

( 1) ( ) ( )

ate of x̂v v vx x x

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Newton-Raphson Example2

1( )( ) ( )

( ) ( ) 2( )

( 1) ( ) ( )

( 1) ( ) ( ) 2( )

Use Newton-Raphson to solve ( ) - 2 0

The 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 x

dx

x xx

x x x

x x xx

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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 0.5

1 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

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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

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Newton-Raphson CommentsWhen close to the solution the error decreases

quite quickly -- method has quadratic convergencef(x(v)) is known as the mismatch, which we would

like to drive to zeroStopping 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?NR method is not sensitive to starting solution.

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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 ˆn n

x f

x f

x f

x

f x

x

xx f x

x

x f x

x

ˆ x x

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Multi-Variable Case, cont’di

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

x

x xx x

x

x xx x

x

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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 fx x x

x x x

xx x x

xf x

xx x x

higher order terms

nx

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Jacobian Matrix

1 1 1

1 2

2 2 2

1 2

1 2

The n by n matrix of partial derivatives is known

as 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 x

x x x

x x x

J x

x x x

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References

a. https://www.youtube.com/watch?v=WlQclObEAiAb. Book: power system engineering/Ngrath &

kothari/1st edition, 6.6

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