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# Aps_chap9 (Power Flow)

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Energy Conversion Lab

POWER FLOW ANALYSIS

balanced single-phase network

network may contain hundreds of nodes and

branches with impedance X specified in per unit onMVA base

Power flow equations

bus admittance matrix of node-voltage equation isformulated

currents can be expressed in terms of voltages

resulting equation can be in terms of power in MW

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Energy Conversion Lab

Nodal solution nodal solution is based on

the Kirchhoffs current law

impedance is converted to

the impedance diagram:see Fig.6.1

ijijij

ijjxrZ

y+

==11

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Energy Conversion Lab

bus: see Fig.6.2 if no connection

between bus-to-bus, leave as zero

node voltageequation is in theform

busbusbus VYI =

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Energy Conversion Lab

Node-voltage matrix Ibus=YbusVbus

Ibus is the vector of injected currents

Vbus is the vector of the bus voltage from referencenode

Ybus is the bus admittance matrix

=

n

i

n

i

V

V

V

V

I

I

I

I

2

1

2

1

nnnin2n1

iniii2i1

2n2i2221

1n1i1211

YYYY

YYYY

YYYY

YYYY

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Energy Conversion Lab

Node-voltage matrix diagonal element Yii: sum of admittance connected to bus i

off-diagonal matrix Yij: negative of admittance between nodes Iand j

when the bus currents are known, bus voltages are unknown,bus voltage can be solved as

inverse of bus admittance matrix is known as impedance matrixZbus

if matrix of Ybus is invertible, Ybus should be non-singular

ij0

==

n

j

ijii yY

ijjiij yYY ==

busbusbus IYV1

=

1= busbus YZ

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Energy Conversion Lab

which result in an upper diagonal nodal admittance matrix

a typical power system network, each bus is connected by a fewnearby bus, which cause many off-diagonal elements are zero

many zero off-diagonal matrix is called sparse matrix the bus admittance matrix in Fig.(6.2) by inspection is

=

5.125.1200

5.125.220.50.500.575.85.2

00.55.25.8

jj

j-jjjjjj

jjj

Ybus

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Energy Conversion Lab

SOLUTION OF NONLINEAR ALGEBRA EQUATIONS

Techniques for iterative solution of non-linearequations

Gauss-Seidal

Newton-Raphson

Quasi-Newton

Gause-Seidal method consider a nonlinear equation f(x)=0

rearrange f(x) so that x=g(x), f(x)=x-g(x) orf(x)=g(x)-x

guess an initial estimate of x = x(k)

use iteration, obtain next x value as x(k+1) = g(x(k))

criteria for stop iteration: |x(k+1)-x(k)|

is the desired accuracy

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Energy Conversion Lab

GAUSE-SEIDAL METHOD

Nature of Gause-Seidal method see Ex.(6.2) and Fig.(6.3) Gause-Seidal method needs many iterations to

achieve desired accuracy

no guarantee for the convergence, depend on the

location of initial x estimate

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Energy Conversion Lab

GAUSE-SEIDAL METHOD

Nature of Gause-Seidal method

solution: if initial estimate x is within convergentregion, solution will converge in zigzag fashion to oneof the roots

no solution: if initial estimate x is outside convergentregion, process will diverge, no solution found

in some case, an acceleration factor is added toimprove the rate of convergence:

x(k+1) = x(k) +[g(x(k))-x(k)], where >1

acceleration factor should not too large to produceovershoot

see Ex.(6.3) for the acceleration factor used

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Energy Conversion Lab

GAUSE-SEIDAL METHOD

Extend one variable to n variable equations using Gause-

Seidal method consider the system of n equations in n variables and solving for

one variable from each equation in one time of iteration

the updated variable x1(k+1) calculated in first equation in

Eq.(6.12) is used in the calculation of x2(k+1) in the secondequation

Ex: in the 2nd iteration x2(k+1) = c2+g2(x1

(k+1)+x2(k)+x3

(k)++xn(k))

at n iteration to complete n variables, the x1(k+1),,xn

(k+1) istested against x

1

(k),,xn

(k) for accuracy criterion

nnn

n

n

cxxxf

cxxxf

cxxxf

=

=

=

),,,(

),,,(

),,,(

21

2212

1211

),,,(

),,,(

),,,(

21

21222

21111

nnnn

n

n

xxxgcx

xxxgcx

xxxgcx

+=

+=

+=

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Energy Conversion Lab

POWER FLOW SOLUTION

operating condition: balanced, single phase model

quantities used in power flow equation are: voltagemagnitude |V|, phase angle , real power P, andreactive power Q

system bus classification:

slack bus (swing bus): taken as reference where |V| andV are specified. It makes up the loss between generatedpower and scheduled loads

load bus (PQ bus): P and Q are specified, |V| and V areunknown

regulated bus (PV bus): P and |V| are specified, V and Qare unknown

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Energy Conversion Lab

POWER FLOW EQUATION

Power flow formulation consider bus case in Fig.(6.7)

current flow into bus i:

express Ii in terms of P,Q:

the power flow equation becomes

the power flow problem results in algebraic nonlinear equationswhich must be solved by iteration methods

ij10

= ==

j

n

j

ij

n

j

ijii VyyVI

*

i

iii

V

jQPI =

ij10

* =

==j

n

j

ij

n

j

iji

i

ii VyyVVjQP

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Energy Conversion Lab

GAUSS-SEIDEL POWER FLOW EQUATION

Gauss-Seidel power flow solution solving Vi: for PQ bus, assume P,Q are known

solving Pi: for slack bus, assume V is known

solving Qi: for PV bus, assume |V| is known

ij

)(

)(*)1(

+

=

+

ij

k

kijk

i

sch

i

sch

i

k

iy

VyV

jQP

V

ijRe)(

10

)()(*)1(

= ==

+ kj

n

ijj

ij

n

j

ij

k

i

k

i

k

i VyyVVP

ijIm )(

10

)()(*)1(

=

==

+ kj

n

ijj

ij

n

j

ij

k

i

k

i

k

i VyyVVQ

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Energy Conversion Lab

GAUSS-SEIDEL POWER FLOW EQUATION

Instructions for Gauss-Seidel solution there are 2(n-1) equations to be solved for n bus

voltage magnitude of the buses are close to 1pu orclose to the magnitude of the slack bus

voltage magnitude at load buses is lowerthan the slackbus value

voltage magnitude at generator buses is higherthanthe slack bus value

phase angle of load buses are below the referenceangle

phase angle of generator buses are above thereference angle

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Energy Conversion Lab

INSTRUCTIONS FOR G-S SOLUTION

Instructions for PQ bus solution

real and reactive power Pisch, Qi

sch are known

starting with an initial estimate of voltage using Viequation

Instructions for PV bus solution

Pisch, |Vi| are specified

assume Vi = |Vi|0o, solve the Qi equation as below

ij

)(

)(*

)1(

+

=

+

ij

k

jijk

i

sch

i

sch

i

ki

y

Vy

V

jQP

V

ijIm )(

10

)()(*)1(

=

==

+ kj

n

ijj

ij

n

j

ij

k

i

k

i

k

i VyyVVQ

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Energy Conversion Lab

INSTRUCTIONS FOR G-S SOLUTION

Instructions for PV bus solution

when Qi(k+1) is available, solve Vi using equation below

since |Vi| is specified, keep imaginary part of Vi,calculate real part of Vi

solve Vi

stopping criteria

ij

)(

)(*

)(

)1( +

=

+

ij

k

kijk

i

k

i

sch

i

k

iy

VyV

jQP

V

{ } { }( )2)1(2)1(Re ++ = kiiki VimagVV

)1()1()1( ImRe +++ += kik

i

k

i VjVV

{ } { } { } { } ++ )()1()()1( ImIm,ReRe kikikiki VVVV

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Energy Conversion Lab

INSTRUCTIONS FOR G-S SOLUTION

Instructions for PV bus solution

to accelerate the convergence, using the followingapproximation after new Vi is obtained

is in the range between 1.3 to 1.7

voltage accuracy in |Vi| and is in the range between0.00001 to 0.00005

( ))()()()1( kikcalikiki VVVV +=+

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Energy Conversion Lab

INSTRUCTIONS FOR G-S SOLUTION

Instructions for V,

slack bus solution solve Pi

solve Qi

accuracy: the largest PQ is less than thespecified value, typically is about 0.001pu

ijRe )(

10

)()(*)1(

=

==

+ kj

n

ij

j

ij

n

j

ij

k

i

k

i

k

i VyyVVP

ijIm )(

10

)()(*)1(

=

==

+ kj

n

ij

j

ij

n

j

ij

k

i

k

i

k

i VyyVVQ

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G-S Power flow Homework

For the one-line diagram shown below, using the G-S methodto determine all bus voltages (magnitude and phase) andshow the power flow solution between the buses assume theregulated bus (#2) reactive power limits are between 0 and

600Mvar.

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Energy Conversion Lab

NEWTON RAPHSON METHOD

Newton Raphson method for solving one variable

consider the solution of one-dimensional equation f(x)=c

assume x = x(0)+x(0)

f(x)=f(x(0)+x(0))=c

use Taylors series expansion

assume x(0) is very small, higher order terms of expansion canbe neglected, Taylor series becomes

assume f(x(0))=c-c(0), the equation becomes c(0)(df/dx)(0)x(0)

the new approximation of x

( ) cxdxfdx

dxdfxfxxf =+

+

+=+

2)0(

)0(

2

2)0(

)0(

)0()0()0(

!21)()(

cxdx

df

xfxxf =

+=+

)0(

)0(

)0()0()0(

)()(

)0(

)0()0()1(

+=

dx

df

cxx

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Energy Conversion Lab

NEWTON RAPHSON METHOD

Newton Raphson method for solving one variable

the new approximation of x

Newton Raphson algorithm

Newtons method converges faster than Gauss-Seidal, the rootmay converge to a root different from the expected one ordiverge if the starting value is not close enough to the root

)0(

)0()0()1(

+=

dx

df

cxx

)()()1(

)(

)()(

)()( )(

kkk

k

kk

kk

xxx

dx

df

cx

xfcc

+=

=

=

+

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Energy Conversion Lab

NEWTON RAPHSON METHOD FOR n VARIABLES

Newton Raphson method for solving n variables

nn

n

nnnnn

n

n

n

n

cx

x

fx

x

fx

x

fxfxxf

cxxfx

xfx

xfxfxxf

cxx

fx

x

fx

x

fxfxxf

=

++

+

+=+

=

++

+

+=+

=

++

+

+=+

)0(

)0(

)0(

2

)0(

2

)0(

1

)0(

1

)0()0()0(

2)0(

)0(

2)0(2

)0(

2

2)0(1

)0(

1

2)0(2

)0()0(2

1

)0(

)0(

1)0(

2

)0(

2

1)0(

1

)0(

1

1)0(

1

)0()0(

1

)()(

)()(

)()(

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Energy Conversion Lab

NEWTON RAPHSON METHOD FOR n VARIABLES

Rearrange in matrix form

The matrix can be written as

C(k) = J(k) X(k)

=

)0(

)0(

2

)0(

1

)0()0(

2

)0(

1

)0(

2

)0(

2

2

)0(

1

2

)0(

1

)0(

2

1

)0(

1

1

)0(

)0(

22

)0(

11

n

n

nnn

n

n

nn x

x

x

x

f

x

f

x

f

x

f

x

f

x

f

x

f

x

f

x

f

fc

fc

fc

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Energy Conversion Lab

NEWTON RAPHSON METHOD FOR n VARIABLES

The Newton-Raphson algorithm for n-dimensional case is

X(k+1) = X(k) +X(k) = X(k) + [J(k)]-1C(k)

where

=

)(

)(

22

)(11

)(

k

nn

k

k

k

fc

fc

fc

C

=

)()(

2

)(

1

)(

2

)(

2

2

)(

1

2

)(

1

)(

2

1

)(

1

1

)(

k

n

n

k

n

k

n

k

n

kk

k

n

kk

k

x

f

x

f

x

f

x

f

x

f

x

f

x

f

x

f

x

f

J

=

)(

)(

2

)(

1

)(

k

n

k

k

k

x

x

x

X

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Energy Conversion Lab

NEWTON RAPHSON METHOD FOR n VARIABLES

The Newton-Raphson algorithm

J(k) is called the Jacobian matrix

solution to X(k+1) is inefficient because it involvesinverse of J(k) , a triangular factorization is used tofacilitate the computation

in MATLAB, the operator \ (i.e., X=J\C) is used toapply the triangular factorization

Newton-Raphson method converge to solutionquadratically when near a root

The limitation is that it does not generally converge toa solution from an arbitrary starting point

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Energy Conversion Lab

LINE FLOWS AND LOSSES

Complex power flow between bus i,j

for line model, see Fig. 6.8

current flow from bus i to bus j

current flow from bus j to bus i

complex power Sij from bus i to j and Sji from j to i

power loss in the line i-j

for more Gauss-Seidel method examples, see Ex. (6.7)and Ex. (6.8)

iijiijilij VyVVyIII 00 )( +=+=

jjijijjlji VyVVyIII 00 )( +=+=

** jijjiijiij IVSIVS ==

jiijjiL SSS += )(

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Energy Conversion Lab

NEWTON-RAPHSON POWER FLOW

Real power flow in terms of Vi , , and Yij

Reactive power flow

Newton-Raphson matrix form: C(k) = J(k) X(k)

diagonal and off-diagonal elements of J1

( )=

+=n

j

jiijijjii YVVP1

cos

( )=

+=n

j

jiijijjii YVVQ1

sin

=

VJJ

JJ

Q

P

43

21

( )

( ) ijsin

sin

+=

+=

jiijijji

j

i

ij

jiijijji

i

i

YVVP

YVVP

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Energy Conversion Lab

NEWTON-RAPHSON POWER FLOW

Newton-Raphson matrix form: C(k) = J(k) X(k)

diagonal and off-diagonal elements of J2

diagonal and off-diagonal elements of J3

=

VJJ

JJ

Q

P

43

21

( )( ) ijcos

coscos2

+=

++=

jiijiji

j

i

ij

jiijijjiiiii

i

i

YVV

P

YVYVV

P

( )

( ) ijcos

cos

+=

+=

jiijijji

j

i

ij

jiijijji

i

i

YVVQ

YVVQ

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Energy Conversion Lab

NEWTON-RAPHSON POWER FLOW

Newton-Raphson matrix form: C(k) = J(k) X(k)

diagonal and off-diagonal elements of J4

power residuals

Pi(k)

Qi(k)

new estimates for bus voltages

=

VJJ

JJ

Q

P

43

21

( )( ) ijsin

sinsin2

+=

+=

jiijiji

j

i

ij

jiijijjiiiii

i

i

YVV

Q

YVYVV

Q

)()()()( , kisch

i

k

i

k

i

sch

i

k

i QQQPPP ==

)()()1()()()1( , kik

i

k

i

k

i

k

i

k

i VVV +=+= ++

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Energy Conversion Lab

NEWTON-RAPHSON POWER FLOW

Procedure for Newton-Raphson method:

PQ bus: set |Vi(0)|=1.0, i

(0)=0.0

PV bus: set i(0)=0.0

set PQ bus equation for J matrix elements:

set PV bus equation for J matrix elements:

)()()()( , kisch

i

k

i

k

i

sch

i

k

i QQQPPP ==

( )=+=

n

jjiijijjii YVVP 1 cos

( )=

+=n

j

jiijijjii YVVQ1

sin

( )=

+=n

j

jiijijjii YVVP1

cos

)()( k

i

sch

i

k

i PPP =

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Energy Conversion Lab

NEWTON-RAPHSON POWER FLOW

Procedure for Newton-Raphson method:

use above equation to calculate Jacobian matrix (J1, J2,J3, J4)

solve |V| and using Newton-Raphson matrix

update |V| and by

repeat the calculation until

for example: see Ex.(6.10)

)()( , kik

i QP

=

VJJ

JJ

Q

P

43

21

)()()1()()()1( , kik

i

k

i

k

i

k

i

k

i VVV +=+= ++

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Energy Conversion Lab

FAST DECOUPLED POWER FLOW

Fast decoupled power flow solution:

the algorithm is based on Newton-Raphson method

when transmission lines has a high X/R ratio, theNewton-Raphson method could be further simplified

Consider the Newton-Raphson power flowequation

P are less sensitive to |V| and most sensitive to

Q is less sensitive to and most sensitive to |V|

we can reasonably eliminate J2 and J3 elements inJacobian matrix

=

VJJ

JJ

Q

P

43

21

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Energy Conversion Lab

FAST DECOUPLED POWER FLOW Consider the Newton-Raphson power flow equation

the power flow equation reduces to

P = J1 = [P/], Q = J4|V| = [Q/|V|]|V|

Pi/i = -Qi - |Vi|2Bii, Bii = |Yii|sinii is the imaginary part of the

diagonal elements

since Bii >> Qi, Pi/i (diagonal elements of J1) can be further

reduced to

Pi/

i = - |Vi|Bii (|Vi|

2

|Vi| )

off diagonal element of J1: Pi/i = - |Vi||Vj|Yijsin(ij-i+j), since j-iis quite small, ij-i+j = ij, J1 = Pi/j = - |Vi||Vj|Bij

since |Vj|

1, off diagonal elements of J1 =

Pi/

j = - |Vi|Bij

=

VJ

J

Q

P

4

1

0

0

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Energy Conversion Lab

FAST DECOUPLED POWER FLOW

Consider the Newton-Raphson power flow

equation similarly, diagonal elements of J4: Qi/|Vi| = - |Vi|Bii off diagonal elements of J4: Qi/|Vj| = - |Vi|Bij therefore, P and Q has the following forms

B and B are the imaginary part of Ybus the updated and |V| can be obtained from

to calculate PQ bus, use simplified J1 and J4 to obtainsolution

to calculate PV bus, J4 can be further eliminated, onlyJ

1is used to obtain solution

VBV

QB

V

P

ii=

=

''' ,

[ ] [ ] V

QBV

V

PB

=

= 11 ",'

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