lecture 07 08 power flow analysis

8/8/2019 Lecture 07 08 Power Flow Analysis

1/68

PowerFlowSolution


2/68

Powerflow

equations

n

k

ikkikikii VVYP1

)cos(

n

kikkiiii VVYQ 1 33 )sin(

NetpowerenteringthenetworkatbusI

CalculatedfromV, ,Y


3/68

NonlinearSystemsMayHave

MultipleSolutions

or

No

Solution

Example1:x2 2=0hassolutionsx=1.414

Example2:

x

2

+2=0has

no

real

solution

f(x)=x2 2 f(x)=x2 +2

two

solutions

where

f(x)

=

0

nosolutionf(x)=0


4/68

MultipleSolution

Example

3

Thedcsystemshownbelowhastwosolutions:

wherethe18watt

loadisaresistive

load

2

2

Load

Load

Load

The equation we're solving is

9 voltsI 18 watts

1 +R

One solution is R 2

Other solution is R 0.5

Load Load

R R


5/68

GaussIteration

There 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).

v vh x

h


6/68

GaussIteration

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


7/68

StoppingCriteria

( ) ( ) ( 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 v x x x x

( )

2i

2 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


8/68

GaussPower

Flow

*

* * *i

1 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 k k k

n n

i i i ik k ik k k k

n ni

ik k ii i ik k

k k k i

ni

i ik k

ii k k i

V I V Y V V Y V

V I V Y V V Y V

Y V Y V Y V

V

V Y V

Y V


9/68

GaussTwoBusPowerFlowExample

A100MW,50Mvar loadisconnectedtoagenerator throughalinewithz=0.02+j0.06p.u.andlinechargingof5Mvar oneachend

(100MVA

base).

Also,

there

is

a25

Mvar capacitor

at

bus

2.

If

the

generatorvoltageis1.0p.u.,whatisV2?

SLoad= 1.0 + j0.5 p.u.


10/68


11/68

GaussTwo

Bus

Example,

contd

*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 k

k k i

v v

V Y VY V

jV j

j V

V

v V v V

j j

j j

.9624 0.0553j


12/68

GaussTwo

Bus

Example,

contd

2

* *1 1 11 1 12 2

1

0.9622 0.0556 0.9638 3.3

Once the voltages are known all other values canbe 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

22

, Q 23.9 Mvar

The capacitor is supplying V 25 23.2 Mvar


13/68

SlackBus

InpreviousexamplewespecifiedS2 andV1 and

thensolved

for

S1 and

V2.

WecannotarbitrarilyspecifySatallbusesbecausetotalgenerationmustequaltotalload+

totallosses

Wealsoneedananglereferencebus.

Tosolvetheseproblemswedefineonebusasthe

"slack"bus.

This

bus

has

afixed

voltage

magnitudeandangle,andavaryingreal/reactivepowerinjection.


14/68

Gausswith

Many

Bus

Systems

*( )( 1)

( )*1,

( ) ( ) ( )1 2

( 1)

With multiple bus systems we could calculate

new V ' as follows:

S1

( , ,..., )

But after we've determined we have a better

estimate of

i

i

nvv i

i ik kvii k k i

v v vi n

vi

s

V Y VY V

h V V V

V

its voltage , so it makes sense to use this

new value. This approach is known as the

Gauss-Seidel iteration.


15/68

GaussSeidel

Iteration

( 1) ( ) ( ) ( )

2 12 2 3( 1) ( 1) ( ) ( )

2 13 2 3

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

2 14 2 3 4

( 1) ( 1) ( 1)( 1) ( )2 1 2 3 4

Immediately use the new voltage estimates:

( , , , , )( , , , , )

( , , , , )

( , , , ,

v v v v

nv v v v

n

v v v v v

n

v v vv vn n

V h V V V V V h V V V V

V h V V V V V

V h V V V V V

)

The Gauss-Seidel works better than the Gauss, and

is actually easier to implement. It is used instead

of Gauss.


16/68

GaussTwo

Bus

Example,

contd

2

* *1 1 11 1 12 2

1

0.9622 0.0556 0.9638 3.3

Once the voltages are known all other values canbe 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

22

, Q 23.9 Mvar

The capacitor is supplying V 25 23.2 Mvar


17/68

SlackBus

InpreviousexamplewespecifiedS2 andV1 and

thensolved

for

S1 and

V2.

WecannotarbitrarilyspecifySatallbusesbecausetotalgenerationmustequaltotalload+

totallosses

Wealsoneedananglereferencebus.

Tosolvetheseproblemswedefineonebusasthe

"slack"bus.

This

bus

has

afixed

voltage

magnitudeandangle,andavaryingreal/reactivepowerinjection.


18/68

Gausswith

Many

Bus

Systems

*( )( 1)

( )*1,

( ) ( ) ( )1 2

( 1)

With multiple bus systems we could calculate

new V ' as follows:

S1

( , ,..., )

But after we've determined we have a better

estimate of

i

i

nvv i

i ik kvii k k i

v v vi n

vi

s

V Y VY V

h V V V

V

its voltage , so it makes sense to use this

new value. This approach is known as the

Gauss-Seidel iteration.


19/68

GaussSeidel

Iteration

( 1) ( ) ( ) ( )

2 12 2 3( 1) ( 1) ( ) ( )

2 13 2 3

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

4 2 3 4

( 1) ( 1) ( 1)( 1) ( )2 1 2 3 4

Immediately use the new voltage estimates:

( , , , , )( , , , , )

( , , , , )

( , , , ,

v v v v

nv v v v

n

v v v v vn

v v vv vn n

V h V V V V V h V V V V

V h V V V V V

V h V V V V V

)

The Gauss-Seidel works better than the Gauss, and

is actually easier to implement. It is used instead

of Gauss.


20/68

ThreeTypes

of

Power

Flow

Buses

Therearethreemaintypesofpowerflow

buses Load(PQ)atwhichP/Qarefixed;iterationsolves

forvoltagemagnitudeandangle.

Slackat

which

the

voltage

magnitude

and

angle

arefixed;iterationsolvesforP/Qinjections

Generator(PV)atwhichPand|V|arefixed;iterationsolvesforvoltageangleandQinjection

specialcodingisneededtoincludePVbusesintheGaussSeideliteration


21/68

Inclusionof

PV

Buses

in

G

S

i

i

* *

1

( )( ) ( )*

1

( ) ( )

To solve for V at a PV bus we must first make a

guess of Q :

Hence Im

In the iteration we use

k

n

i i ik k i ik

nvv v

i i ik k

v vi i i

S V Y V P jQ

Q V Y V

S P jQ


22/68

Inclusionof

PV

Buses,

cont'd

( 1)

( )*( )( 1)

( )*1,

( 1)i i

Tentatively solve for

S1

But since V is specified, replace by V

i

vi

v nvv i

i ik kvii k k i

vi

V

V Y VY V

V


23/68

TwoBus

PV

Example

Bus 1

(slack bus)

Bus 2V1 = 1.0 V2 = 1.05

P2 = 0 MW

z = 0 .02 + j 0.06

Considerthesametwobussystemfromtheprevious

example,excepttheloadisreplacedbyagenerator


24/68

TwoBus

PV

Example,

cont'd

*2

2 21 1*

22 2* *

2 21 1 2 22 2 2

2

( ) ( 1) ( 1)2 2 2

1

Im[ ]

Guess V 1.05 0

0 0 0.457 1.045 0.83 1.050 0.83

1 0 0.535 1.049 0.93 1.050 0.932 0 0.545 1.050 0.96 1.050 0.96

v v v

SV Y V

Y V

Q Y V V Y V V

v S V V

j

jj


25/68

GeneratorReactive

Power

Limits

Thereactivepoweroutputofgeneratorsvariestomaintaintheterminalvoltage;onarealgenerator

this

is

done

by

the

exciter

Tomaintainhighervoltagesrequiresmorereactivepower

Generatorshave

reactive

power

limits,

which

aredependentuponthegenerator'sMWoutput

These

limits

must

be

considered

during

the

powerflow

solution.


26/68

GeneratorReactive

Limits,

cont'd

Duringpowerflowonceasolutionisobtainedchecktomakegeneratorreactivepoweroutputiswithin

its

limits

Ifthereactivepowerisoutsideofthelimits,fixQatthemaxorminvalue,andresolvetreatingthe

generatoras

aPQ

bus

thisisknowas"typeswitching"

alsoneedtocheckifaPQgeneratorcanagainregulate

Ruleof

thumb:

to

raise

system

voltage

we

need

tosupplymorevars


27/68

AcceleratedGSConvergence

( 1) ( )

( 1) ( ) ( ) ( )

(

Previously in the Gauss-Seidel method we were

calculating each value x as

( )

To accelerate convergence we can rewrite this as

( )

Now introduce acceleration parameter

v v

v v v v

v

x h x

x x h x x

x

1) ( ) ( ) ( )

( ( ) )With = 1 this is identical to standard gauss-seidel.

Larger values of may result in faster convergence.

v v v

x h x x


28/68

AcceleratedConvergence,

contd

( 1) ( ) ( ) ( )

Consider the previous example: - 1 0

(1 )Comparison of results with different values of

1 1.2 1.5 2

0 1 1 1 1

1 2 2.20 2.5 3

2 2.4142 2.5399 2.6217 2.4643 2.5554 2.6045 2.6179 2.675

4 2.59

v v v v

x x

x x x x

k

81 2.6157 2.6180 2.596

5 2.6118 2.6176 2.6180 2.626


29/68

GaussSeidel

Advantages

Eachiterationisrelativelyfast(computational

orderis

proportional

to

number

of

branches

+

numberofbusesinthesystem

Relativelyeasytoprogram


30/68

GaussSeidel

Disadvantages

Tendstoconvergerelativelyslowly,although

thiscan

be

improved

with

acceleration

Hastendencytomisssolutions,particularlyon

largesystems

Tendstodivergeoncaseswithnegative

branchreactances(commonwith

compensatedlines)

Needtoprogramusingcomplexnumbers


31/68

NewtonRaphson

Algorithm

Thesecondmajorpowerflowsolution

methodis

the

Newton

Raphson

algorithm

KeyideabehindNewtonRaphsonistouse

sequentiallinearization

General form of problem: Find an x such that

( ) 0f x


32/68


33/68

NewtonRaphson

Method,

contd

( )( ) ( )

( )

1( )( ) ( )

3. Approximate ( ) by neglecting all terms

except the first two

( )( ) 0 ( )

4. Use this linear approximation to solve for

( )( )

5. Solve for a new estim

vv v

v

vv v

f x

df x f x f x x

dx

x

df x x f x

dx

( 1) ( ) ( )

ate of x

v v v x x x


34/68

NewtonRaphson

Example

2

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 v

v

v v v

v v v

v

f x x

df x x f x

dx

x xx

x x x

x x xx


35/68

NewtonRaphson

Example,

contd

( 1) ( ) ( ) 2

( )

(0)

( ) ( ) ( )

3 3

6

1(( ) - 2)

2Guess 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 v

v

v v v

x x x

x

x f x x


36/68

SequentialLinear

Approximations

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

At eachiteration theN-R methoduses a linearapproximationto determine

the next valuefor x


37/68

NewtonRaphson

Comments

Whenclosetothesolutiontheerror

decreasesquite

quickly

method

has

quadraticconvergence

f(x(v))isknownasthemismatch,whichwe

wouldlike

to

drive

to

zero

Stoppingcriteriaiswhenf(x(v))


38/68

MultiVariable

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 and

n n

x f

x f

x f

x

f xx

x

x f x

x

x f xx

x x


39/68

Multi

Variable

Case,

contdi

1 11 1 1 21 2

1

n nn n 1 2

1 2

n

The Taylor series expansion is written for each f ( )

f ( ) f ( )f ( ) f ( )

f ( )higher order terms

f ( ) f ( )

f ( ) f ( )

f ( )higher order terms

nn

n

n

x xx x

x

x

x xx x

x

x

x

x xx x

x

x x

x x

x


40/68

Multi

Variable

Case,

contd

1 1 1

1 21 1

2 2 2

2 21 2

1 2

This can be written more compactly in matrix form

( ) ( ) ( )

( )( ) ( ) ( )

( )( )

( )( ) ( ) ( )

n

n

n

n n n

n

f f f x x x

f x f f f

f x x x x

f f f f

x x x

x x x

xx x x

xf x

xx x x

higher order terms

nx


41/68

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 f

x x x

f f f

x x x

f f f

x x x

J xx x x

x x x

J x

x x x


42/68

Multi

Variable

N

R

Procedure

1

( 1) ( ) ( )

( 1) ( ) ( ) 1 ( )

( )

Derivation of N-R method is similar to the scalar case

( ) ( ) ( ) higher order terms( ) 0 ( ) ( )

( ) ( )

( ) ( )

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


43/68

Multi

Variable

Example1

2

2 21 1 2

2 22 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 0

First 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


44/68

Multi

variable

Example,

contd1 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 2.1

1 3 1 3 1.3

x x

x x x x

x x x f

x x x x x f

J x

x

x

x

x


45/68

Multi

variable

Example,

contd1

(2)

(2)

2.1 8.40 2.60 2.51 1.8284

1.3 5.50 0.50 1.45 1.2122

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

specified tolerance0.1556

( )0.0900

If = 0.2 then we wou

x

f x

f x

ld be done. Otherwise we'd

continue iterating.


46/68

NR

Application

to

Power

Flow

*

* * *i

1 1

We first need to rewrite complex power equations

as equations with real coefficients

S

These can be derived by defining

Recal

i

n n

i i i ik k i ik k k k

ik ik ik

j

i i i i

ik i k

V I V Y V V Y V

Y G jB

V V e V

jl e cos sinj


47/68

Real

Power

Balance

Equations* *

i1 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 k

n

i k ik ik ik ik k

n

i k ik ik ik ik Gi Dik

n

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


48/68

Newton

Raphson

Power

Flow

i1

In the Newton-Raphson power flow we use Newton's

method 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 ik k

V V G

i1

sin )

Q ( sin cos )

ik ik Gi Di

n


B P P

V V G B Q Q


49/68

Power

Flow

Variables

2 2 2

n

2

Assume the slack bus is the first bus (with a fixed

voltage angle/magnitude). We then need to determine

the voltage angle/magnitude at the other buses.

( )

( )

G

n

P P

V

V

x

x f x

2

2 2 2

( )

( )

( )

D

n Gn Dn

G D

n Gn Dn

P

P P P

Q Q Q

Q Q Q

x

x

x


50/68

N

R

Power

Flow

Solution

( )

( )

( 1) ( ) ( ) 1 ( )

The power flow is solved using the same procedure

discussed 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


51/68

Power

Flow

Jacobian

Matrix

1 1 1

1 2

2 2 2

1 2

1 2

The most difficult part of the algorithm is determining

and inverting the n by n Jacobian matrix, ( )( ) ( ) ( )

( ) ( ) ( )

( )

( ) ( ) ( )

n

n

n n n

n

f f f

x x x

f f f

x x x

f f f

x x x

J xx x x

x x x

J x

x x x


52/68

Power

Flow

Jacobian

Matrix,

contd

i

i

i 1

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

x

x

i

1

i

f ( )( sin cos )

f ( )( sin cos ) ( )

Di

n

i k ik ik ik ik

i kk i

i j ik ik ik ik

j

xV V G B

xV V G B j i


53/68

Two

Bus

Newton

Raphson

Example

Line Z = 0 .1j

One Two1.000 pu 1.000 pu

200 MW

100 MVR

0 MW

0 MVR

For the two bus power system shown below, use tNewton-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 10

10 10bus

j j

V j j

x Y


54/68

Two

Bus

Example,

contd

i1

i1

2 2 1 2

22 2 1 2 2

General power balance equations

P ( cos sin )

Q ( sin cos )

Bus two power balance equations

P (10sin ) 2.0 0

( 10cos ) (10) 1.0 0

n


n


V V G B P P

V V G B Q Q

V V

Q V V V


55/68

Two

Bus

Example,

contd2 2 2

2

2 2 2 2

2 2

2 2

2 2

22

2 2 2

2 2 2 2

P ( ) (10sin ) 2.0 0

( ) ( 10cos ) (10) 1.0 0

Now calculate the power flow Jacobian

P ( ) P ( )

( )Q ( ) Q ( )

10 cos 10sin

10 sin 10cos 20

V

Q V V

VJ

V

V

V V

x

x

x x

xx x


56/68

Two

Bus

Example,

First

Iteration(0)

2 2(0)

22 2 2

2 2 2(0)

2 2 2 2

(1)

0Set 0, guess

1

Calculate

(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

V

V V

x

x

J x

x

12.0 0.2

1.0 0.9


57/68

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 1.0

8.82 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 0.8554

0.0000906f( ) Done! V 0.8554 13.52

0.0001175

x x

x


58/68

Two

Bus

Solved

Values

Line Z = 0.1j

One Two1.000 pu 0.855 pu

200 MW100 MVR200 .0 MW168 .3 MVR

-13.522 De g

200.0 MW

168.3 MVR

-200.0 MW

-100.0 MVR

Once the voltage angle and magnitude at bus 2 areknown we can calculate all the other system values

such as the line flows and the generator reactivepower output

Two Bus Case Low Voltage Solution


59/68

TwoBusCaseLowVoltageSolution

(0)

2 2(0)

22 2 2

This case actually has two solutions! The second

"low voltage" is found by using a low initial guess.

0Set 0, guess

0.25

Calculate

(10sin ) 2.0f( )

( 10cos ) (10) 1.0

v

V

V V

x

x

2 2 2(0)

2 2 2 2

2

0.875

10 cos 10sin 2.5 0( )

10 sin 10cos 20 0 5

V

V V

J x


60/68

Low

Voltage

Solution,

cont'd1(1)

(2) (2) (3)

0 2.5 0 2 0.8Solve

0.25 0 5 0.875 0.075

1.462 1.42 0.921( )

0.534 0.2336 0.220

x

f x x x

Line Z = 0.1j

One Two1.000 pu 0 .261 pu

200 MW

100 MVR

200.0 MW

831.7 MVR

-49.914 Deg

200 .0 MW

831 .7 MVR

-200.0 MW

-100.0 MVR

Low voltage solution


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PV

BusesSincethevoltagemagnitudeatPVbusesis

fixed

there

is

no

need

to

explicitly

include

thesevoltagesinx orwritethereactive

powerbalanceequations

thereactive

power

output

of

the

generator

variestomaintainthefixedterminalvoltage

(withinlimits)

optionallythese

variations/equations

can

be

includedbyjustwritingtheexplicitvoltage

constraintforthegeneratorbus


62/68

Three

Bus

PV

Case

Example

Line Z = 0.1j

Line Z = 0.1j Line Z = 0.1j

One Two1.000 pu

0.941 pu

200 MW

100 MVR170.0 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

( )

( ) ( ) 0

V ( )

G D

G D

D

P P P

P P P

Q Q

x

x f x x

x


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Modeling

Voltage

Dependent

LoadSo far we've assumed that the load is independent of

the bus voltage (i.e., constant power). However, the

power flow can be easily extended to include voltage

depedence with both the real and reactive l

Di Di

1

1

oad. This

is done by making P and Q a function of :

( cos sin ) ( ) 0

( sin cos ) ( ) 0

i

n

i k ik ik ik ik Gi Di ik

n

i k ik ik ik ik Gi Di ik

V

V V G B P P V

V V G B Q Q V


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Voltage

Dependent

Load

Example

22 2 2 2

2 22 2 2 2 2

2 2 2 2

In previous two bus example now assume the load is

constant impedance, so

P ( ) (10sin ) 2.0 0

( ) ( 10cos ) (10) 1.0 0

Now calculate the power flow Jacobian

10 cos 10sin 4.0( )

10

V V

Q V V V

V VJ

x

x

x

2 2 2 2 2sin 10cos 20 2.0V V V


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Voltage

Dependent

Load,

cont'd(0)

22 2 2(0)

2 22 2 2 2

(0)

1(1)

0Again set 0, guess

1

Calculate

(10sin ) 2.0 2.0f( )

1.0( 10cos ) (10) 1.0

10 4( )

0 120 10 4 2.0 0.1667

Solve1 0 12 1.0 0.9167

v

V V

V V V

x

x

J x

x


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Voltage

Dependent

Load,

cont'd

Line Z = 0.1j

One Two1.000 pu

0.894 pu

160 MW

80 MVR

160 .0 MW

120.0 MVR

-10.304 Deg

160.0 MW120 .0 MVR

-160.0 MW-80.0 MVR

With constant impedance load the MW/Mvar load atbus 2 varies with the square of the bus 2 voltagemagnitude. This if the voltage level is less than 1.0,the load is lower than 200/100 MW/Mvar


67/68

Solving

Large

Power

SystemsThemostdifficultcomputationaltaskis

inverting

the

Jacobian

matrix invertingafullmatrixisanordern3 operation,

meaningtheamountofcomputationincreases

with

the

cube

of

the

size

size thisamountofcomputationcanbedecreased

substantiallybyrecognizingthatsincetheYbus isa

sparse

matrix,

the

Jacobian

is

also

a

sparse

matrix usingsparsematrixmethodsresultsina

computationalorderofaboutn1.5.

thisis

asubstantial

savings

when

solving

systems

Newton Raphson Power Flow


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NewtonRaphsonPowerFlow

Advantages

fastconvergence

as

long

as

initial

guess

is

close

to

solution

largeregionofconvergence

Disadvantages eachiterationtakesmuchlongerthanaGauss

Seideliteration

morecomplicated

to

code,

particularly

when

implementingsparsematrixalgorithms

Newton

Raphson

algorithm

is

very

common

in

lecture 07 08 power flow analysis

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