module 6 introduction to power flow studies

8/3/2019 Module 6 Introduction to Power Flow Studies

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BEE 3243Electric Power Systems

BEE 3243BEE 3243Electric Power SystemsElectric Power Systems

Module 5

Introduction to Power Flow Studies

Module 5Module 5

Introduction to Power Flow StudiesIntroduction to Power Flow Studies


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BEE 3243 Electric Power Systems Module 5 22

OutlinesOutlines

IntroductionIntroduction

Basic Techniques for Power Flow StudiesBasic Techniques for Power Flow Studies

The Bus Admittance MatrixThe Bus Admittance Matrix Power Flow EquationsPower Flow Equations GaussGauss--Seidel MethodSeidel Method GaussGauss--SeidalSeidal Power Flow SolutionPower Flow Solution


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Power flow study is also known asPower flow study is also known as load flowload flow study.study.

It is an analysis duringIt is an analysis during steadysteady--statestate

conditions.conditions.

It is used forIt is used forplanningplanning andand controllingcontrolling a system.a system. Assumptions:Assumptions: balancedbalanced conditions andconditions and singlesinglephasephase

analysis.analysis.

Problems:Problems:

determine thedetermine the voltage magnitudevoltage magnitude

andand phase anglephase angle

atat

each bus.each bus.

determine the active and reactive (determine the active and reactive (P & QP & Q) power flow in) power flow ineach lineeach line

each bus haseach bus has fourfour

state variables:state variables: voltage magnitude,voltage magnitude,

voltage phase angle, real power injection, and reactivevoltage phase angle, real power injection, and reactivepower injectionpower injection


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Each bus hasEach bus has twotwo of the four state variablesof the four state variablesdefined or given.defined or given. Text book:Text book: HadiHadi SaadatSaadat Power System AnalysisPower System Analysis((Chapter 6Chapter 6).).


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Types of Buses in Power SystemsTypes of Buses in Power Systems

Types of network buses:Types of network buses: Load BusLoad Bus

ororPQ BusPQ Bus

known real (P) and reactive (Q) power injections.known real (P) and reactive (Q) power injections.

No generator attach to load bus.No generator attach to load bus.

Generator BusGenerator Bus

ororPV BusPV Bus

known real (P) power injection and the voltageknown real (P) power injection and the voltage

magnitude (V).magnitude (V).

Slack BusSlack Bus

ororSwing BusSwing Bus

known voltage magnitude (V) and voltage angle (known voltage magnitude (V) and voltage angle (),),

often it is taken to beoften it is taken to be 1100

p.up.u..

must have one generator as the slack bus.must have one generator as the slack bus. takes up the power slack due to losses in the network.takes up the power slack due to losses in the network.


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Power flow analysis is anPower flow analysis is an iterativeiterative problem.problem.StepsSteps

to be taken in power flow analysis:to be taken in power flow analysis:

1)1) One line diagram orOne line diagram orload flow dataload flow data of a power systemof a power system2)2)

Construct Bus Admittance Matrix (Construct Bus Admittance Matrix (YbusYbus))

3)3)

Calculate theCalculate the power flowpower flow

analysis equationsanalysis equations

Power flow is aPower flow is a nonlinearnonlinear problem and it isproblem and it iscommonly solved by the iterative solution ofcommonly solved by the iterative solution ofnonlinear algebraic equationsnonlinear algebraic equations::

GaussGauss--SeidalSeidal

NewtonNewton--RaphsonRaphson

Fast DecoupledFast Decoupled




Example ofExample ofload flow input dataload flow input data:: Bus dataBus data

Line dataLine data

Bus no Bus

code

Voltage Load Generator

Magnitude

(p.u.)

Angle

(degree)

P

(MW)

Q

(Mvar)

P

(MW)

Q

(Mvar)

1 1 1.06 0 0.0 0.0 0.0 0.0

2 2 1.043 0 21.70 12.7 40.0 0.03 0 1.0 0 2.4 1.2 0.0 0.0

Line bus no

(From)

Line bus no

(To)

Lineresistance

R (p.u.)

Linereactance

X (p.u.)

linesusceptance

B (p.u.)

Transfor-mer tapsetting

1 2 0.0192 0.0575 0.02640 0.978



The Bus Admittance MatrixThe Bus Admittance Matrix

The matrix equation for relating theThe matrix equation for relating the nodal voltagesnodal voltages to theto the currentscurrents

that flow into and out of a networkthat flow into and out of a network

using theusing the admittanceadmittance values of circuit branches.values of circuit branches.V1

Vi

Ii

yi1

yi2

yin

yi0

V2

Vn

Iinj = YbusVnodeIinj = YbusVnode

ijijij

ijjxrz

y

11

nnnnn

n

n

n V

V

V

YYY

YYY

YYY

I

I

I

2

1

21

22221

11211

2

1


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One line diagram of a power system




Impedance Diagram




Admittance Diagram




Kirchhoffs current law:




Rearranging the KCL Equations:Rearranging the KCL Equations:

Matrix Formation of the EquationsMatrix Formation of the Equations::


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Completed Matrix Equation:Completed Matrix Equation:

nnnnn

nn

n V

V

V

YYY

YYY

YYY

I

I

I

21

21

2222111211

21


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YY--Bus Matrix Building RulesBus Matrix Building Rules

Square matrixSquare matrix with dimensions equal to thewith dimensions equal to thenumber of buses.number of buses. Convert all network impedances intoConvert all network impedances into admittancesadmittances.. DiagonalDiagonal elements:elements:

OffOff--diagonaldiagonal

elements:elements:

Matrix isMatrix is symmetricalsymmetrical along the leading diagonal.along the leading diagonal.


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Power Flow EquationsPower Flow Equations

V1Vi

Ii

yi1

yi2

yin

yi0

V2

V3

KCL Equations:KCL Equations:


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Power Flow EquationsPower Flow Equations

Power flow equation:Power flow equation:


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GaussGauss--Seidel MethodSeidel Method

GaussGauss--SeidalSeidal is ais a nonlinearnonlinear algebraic equationalgebraic equationsolver. It is a method of successive displacements.solver. It is a method of successive displacements. ItsIts iterative stepsiterative steps::

take a function and rearrange it into the formtake a function and rearrange it into the form x =x = g(xg(x))

make an initial estimate of the variable x:make an initial estimate of the variable x: xx[0][0]

= initial value= initial value

find an iterative improvement offind an iterative improvement ofxx[k[k]]

, that is:, that is: xx[k+1][k+1]

== g(xg(x[k[k]]

))

a solution is reached when the difference between twoa solution is reached when the difference between two

iterations is less than a specified accuracy:iterations is less than a specified accuracy: |x|x[k+1][k+1]

xx[k[k]]||

AccelerationAcceleration factors (factors ())::

can improve the rate of convergence:can improve the rate of convergence:

> 1> 1

modified step: the improvement is found asmodified step: the improvement is found as

xx[k+1][k+1]

= x= x[k][k]

++

[[g( xg( x[k][k]

))

xx[k][k]]]


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Example of the GaussExample of the Gauss--Seidel method:Seidel method:Find x of the equation:Find x of the equation: f(xf(x) = x) = x33--6x6x22+9x+9x--4 = 0.4 = 0.

9x =9x = --xx33+6x+6x22+4+4

Start withStart with initialinitial

guess xguess x[0][0]

= 2,= 2,

)(9

4

9

6

9

123 xgxxx

5173.29

4)2222.2(

9

6)2222.2(

9

1)2222.2(

2222.29

4)2(

9

6)2(

9

1)2(

23]1[]2[

23]0[]1[

xgx

xgx


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0000.4

9988.3

9568.39

4)7398.3(

9

6)7398.3(

9

1)7398.3(

7398.39

4)3376.3(

9

6)3376.3(

9

1)3376.3(

3376.394)8966.2(

96)8966.2(

91)8966.2(

8966.29

4)5173.2(

9

6)5173.2(

9

1)5173.2(

]8[

]7[

23]5[]6[

23]4[]5[

23]3[]4[

23]2[]3[

x

x

xgx

xgx

xgx

xgx


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MatlabMatlab Results of all iterations:Results of all iterations:Iter g dx x1 2.2222 0.2222 2.2222

2 2.5173 0.2951 2.5173

3 2.8966 0.3793 2.8966

4 3.3376 0.4410 3.3376

5 3.7398 0.4022 3.7398

6 3.9568 0.2170 3.9568

7 3.9988 0.0420 3.9988

8 4.0000 0.0012 4.0000

9 4.0000 0.0000 4.0000


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Graphical illustration:Graphical illustration:

Iterations

x = g(x)

Initial value

Solution points


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GaussGauss--Seidel Method withSeidel Method with

Find the root of the equation:Find the root of the equation: f(xf(x) = x) = x33

-- 6x6x

22

+ 9x+ 9x -- 44= 0 with an= 0 with an acceleration factoracceleration factor

of 1.25.of 1.25.

Starting with anStarting with an initialinitial guess of xguess of x[0][0] = 2.= 2.


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MatlabMatlab results of all iterations:results of all iterations:Iter g dx x

1 2.2222 0.2222 2.2778

2 2.5902 0.3124 2.6683

3 3.0801 0.4118 3.1831

4 3.6157 0.4326 3.7238

5 3.9515 0.2277 4.0084

6 4.0000 -0.0085 3.9978

7 4.0000 0.0022 4.0005

8 4.0000 -0.0005 3.9999


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Do not use aDo not use a very largevery large number ofnumber of as the largeras the largerstep sizestep size

may result in an overshoot.may result in an overshoot.

If we take the previous example withIf we take the previous example with = 1.8= 1.8, we, wewill needwill need more iterationsmore iterations to obtain the answer:to obtain the answer:Iter g dx x1 2.2222 0.2222 2.4000

2 2.7484 0.3484 3.0272

3 3.4714 0.4442 3.8268

4 3.9806 0.1538 4.1036

5 3.9927 -0.1109 3.9040

Iter g dx x6 3.9940 0.0900 4.0659

7 3.9971 -0.0688 3.9420

8 3.9978 0.0558 4.0424

9 3.9988 -0.0436 3.9639

10 3.9991 0.0352 4.0273Overshoot


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Iter g dx x11 3.9995 -0.0278 3.9772

12 3.9997 0.0224 4.0176

13 3.9998 -0.0178 3.9856

14 3.9999 0.0143 4.0113

15 3.9999 -0.0114 3.9908

16 3.9999 0.0091 4.0073

17 4.0000 -0.0073 3.9941

18 4.0000 0.0058 4.0047

Iter g dx x19 4.0000 -0.0047 3.9963

20 4.0000 0.0037 4.0030

21 4.0000 -0.0030 3.9976

22 4.0000 0.0024 4.0019

23 4.0000 -0.0019 3.9985

24 4.0000 0.0015 4.0012

25 4.0000 -0.0012 3.9990

26 4.0000 0.0010 4.0008


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OVERSHOOT


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GaussGauss--SeidalSeidal for a System of n Equationsfor a System of n Equations Consider a system ofConsider a system ofn equationsn equations::

Rearrange each equation for each of the variables:Rearrange each equation for each of the variables:


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GaussGauss--SeidalSeidal for a System of n Equationsfor a System of n Equations Steps:Steps:Assume anAssume an approximate solutionapproximate solution

for the independentfor the independent

variables,variables,

Find the results in aFind the results in a new approximatenew approximate solutionsolution

In theIn the

GaussGauss--SeidelSeidel

method, the updated values of themethod, the updated values of the

variables calculated in the preceding equations are usedvariables calculated in the preceding equations are used

immediately in the solution of the subsequent equations.immediately in the solution of the subsequent equations.

The rate ofThe rate ofconvergenceconvergence can be increased by suitablecan be increased by suitable ..

G S id l P Fl S l i


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GaussGauss--SeidalSeidal Power Flow SolutionPower Flow Solution Previously derivedPreviously derived power flow equationpower flow equation,,

GaussGauss--SeidalSeidal form,form,

G S id l P Fl S l ti


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GaussGauss--SeidalSeidal Power Flow SolutionPower Flow Solution Rewriting the power equation to findRewriting the power equation to find P and QP and Q

thethe real and reactive powersreal and reactive powers

are scheduled for theare scheduled for the loadload

busesbuses

that is, they remain fixedthat is, they remain fixed

the currents and powers are expressed as going into thethe currents and powers are expressed as going into thebusbus

forforgenerationgeneration the powers arethe powers are positivepositive

forforloadsloads

the powers arethe powers are negativenegative

thethe scheduled powerscheduled power is theis the sumsum of the generation and loadof the generation and loadpowerspowers

GG S id lS id l P Fl S l iP Fl S l ti


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GaussGauss--SeidalSeidal Power Flow SolutionPower Flow Solution TheThe complete setcomplete set of equations become:of equations become:

GG S id lS id l P Fl S l tiP Fl S l ti


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GaussGauss--SeidalSeidal Power Flow SolutionPower Flow Solution Since theSince the offoff--diagonaldiagonal elements ofelements ofYYbusbus

,, YYijij

== --yyijij

and theand the diagonaldiagonal

elements,elements, YYiiii

== yyijij

,,

ii

n

ijj

kjijk

i

schi

schi

ki

Y

VYV

jQP

V

,1

][

][*]1[

n

ijj

kjijii

ki

ki

ki

n

ijj

kjijii

ki

ki

ki

VYYVVQ

VYYVVP

,1

][][][*]1[

,1

][][][*]1[



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GaussGauss--SeidalSeidal Power Flow SolutionPower Flow Solution System characteristics:System characteristics:Since both components (V &Since both components (V & ) are specified for the) are specified for the

slack busslack bus, there are 2(n, there are 2(n -- 1) equations which must be1) equations which must besolved iterativelysolved iterativelyFor theFor the load busesload buses, the real and reactive powers are, the real and reactive powers are

known/ scheduled:known/ scheduled:

the voltage magnitude and angle must be estimatedthe voltage magnitude and angle must be estimated

in per unit, the nominal voltage magnitude is 1in per unit, the nominal voltage magnitude is 1 pupu

the angles are generally close together, so an initial value ofthe angles are generally close together, so an initial value of00

degrees is appropriatedegrees is appropriate



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GaussGauss--SeidalSeidal Power Flow SolutionPower Flow SolutionFor theFor the generator busesgenerator buses, the real power and voltage, the real power and voltage

magnitude are known,magnitude are known,

the real power is scheduledthe real power is scheduled

the reactive power is computed based on the estimated voltagethe reactive power is computed based on the estimated voltage

valuesvalues

the voltage is computed by Gaussthe voltage is computed by Gauss--Seidel, only the imaginarySeidel, only the imaginary

part is keptpart is kept

the complex voltage is found from the magnitude and thethe complex voltage is found from the magnitude and the

iterative imaginary partiterative imaginary part

Li Fl d LLi Fl d L


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Line Flows and LossesLine Flows and Losses

After solving for bus voltages and angles,After solving for bus voltages and angles, powerpowerflowsflows

andand losseslosses

on the network branches areon the network branches are

calculated.calculated.Transmission lines and transformers are networkTransmission lines and transformers are network

branchesbranches

The direction of positive current flow are defined asThe direction of positive current flow are defined asfollows for a branch element (demonstrated on afollows for a branch element (demonstrated on a

medium length line)medium length line)

Power flow is defined for each end of the branchPower flow is defined for each end of the branch

Li Fl d LLi Fl d L


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ExampleExample: the power leaving: the power leaving bus ibus i and flowing toand flowing tobus jbus j::

Li Fl d LLi Fl d L


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Current and power flowsCurrent and power flows::

Power lossesPower losses

E lE ample


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ExampleExample

a)a) Using theUsing the GaussGauss--Seidel methodSeidel method, determine the, determine the phasorphasor values of thevalues of the voltagevoltage at the load buses 2 and 3, accurateat the load buses 2 and 3, accurate

to 4 decimal places.to 4 decimal places.

b)b) Find the slack busFind the slack bus PP andand QQ..c)c)

Determine theDetermine the line flowsline flows

andand line lossesline losses. Construct a. Construct a

power flow diagrampower flow diagram

showing the direction of the line flow.showing the direction of the line flow.

1

2

3

E lExample


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ExampleExample

Line admittances:Line admittances:

AtAt PP--Q busesQ buses, the complex loads in, the complex loads in p.up.u.:.:3216

025.00125.0

1

301003.001.0

1

2010

04.002.0

1

23

13

12

jj

y

jj

y

j

j

y

p.u.452.0386.1

p.u.102.1566.2

3

2

jS

jS

sch

sch

a)


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ExampleExample


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ExampleExample

For 2For 2ndnd

iteration,iteration,

0353.00011.1

)6226(

)031.09825.0)(3216()005.1)(3010(00.1

452.0386.1

2313

)1(223113)0(*

3

33

)1(3

j

j

jjjjjj

yy

VyVyV

jQP

V

schsch

052.09816.0

)5226(

)0353.00011.1)(3216()005.1)(2010(031.09825.0

102.1566.2

)2(2

j

j

jjjjj

j

V

ExampleExample


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ExampleExample

The remaining iterations until the solution isThe remaining iterations until the solution isconverged with an accuracy of 5 x 10converged with an accuracy of 5 x 10--55

p.up.u.:.:

0459.00008.1

)6226(

)052.09816.0)(3216()005.1)(3010(0353.00011.1452.0386.1

)1(3

j

j

jjjjjj

V

0500.00000.10600.09800.0

0500.00000.10599.09801.0

0499.00001.10598.09801.0

0497.00002.10594.09803.0

0488.00004.10578.09808.0

)7(3

)7(2

)6(3

)6(2

)5(3

)5(2

)4(3

)4(2

)3(3

)3(2

jVjV

jVjV

jVjV

jVjV

jVjV

Final solution

ExampleExample


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ExampleExample

The slack bus power:The slack bus power:b)

Mvar189MW5.409

p.u.890.1095.4

)]05.00.1)(3010()06.098.0)(2010()5020(05.1[05.1

)]()([ 31321213121*

111

j

j

jjjjj

VyVyyyVVjQP

ExampleExample


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ExampleExample

Line currents:Line currents:c)

48.064.0

48.064.0)]05.00.1()06.098.0)[(3216()(

0.10.2

0.10.2)]05.00.1()005.1)[(3010()(8.09.1

8.09.1)]06.098.0()005.1)[(2010()(

2332

322323

1331

311313

1221

211212

jII

jjjjVVyI

jII

jjjjVVyIjII

jjjjVVyI

ExampleExample


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ExampleExample

The line flows are:The line flows are:

Mvar8.44MW4.66

p.u.448.0664.0)48.064.0)(05.00.1(

Mvar243MW6.65

p.u.432.0656.0)48.0656.0)(06.098.0(

Mvar0.90MW0.205

p.u.90.005.2)0.10.2)(05.00.1(

Mvar0.105MW0.210

p.u.05.11.2)0.10.2)(0.005.1(

Mvar0.67MW0.191p.u.67.091.1)8.09.1)(06.098.0(

Mvar0.84MW5.199

p.u.84.0995.1)8.09.1)(0.005.1(

*

32332

*23223

*31331

*13113

*

21221

*12112

j

jjjIVS

.j

jjjIVS

j

jjjIVS

j

jjjIVS

jjjjIVS

j

jjjIVS

ExampleExample


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ExampleExample

Line losses are:Line losses are:

Mvar60.1MW8.0Mvar0.15MW0.5

Mvar0.17MW5.8

322323

311313

211212

jSSS

jSSS

jSSS

L

L

L

ExampleExample


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ExampleExample

409.5

191

65.6

66.4

44.8

67.0

43.2

205

90.0

199.5

84.0

210.0

105.0

189

256.6

138.6

110.2

45.2

(8.5)

(17.0)

(5)

(15)

(0.8)(1.6)

The power flow diagram:The power flow diagram:

module 6 introduction to power flow studies

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