Load flow PPT

Power flow calculations Dirk Van Hertem Hakan Ergun Priyanko Guha Thakurta Research group Electa Department of electrical engineering (ESAT) K.U.Leuven, Belgium

DVH, HE, PGT (KUL/ESAT/ELECTA)

Power flow calculations

September 19, 2011

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Introduction

Outline 1

Introduction Example

2

System representation

3

The load flow problem

4

Solving the problem Gauss-Seidel Newton-Raphson power flow Simplified Newton-Raphson

5

Software

6

State estimation



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Introduction

Introduction: load or power flow What are “power flow calculations” Calculating the power flow (active and reactive) through all the lines in the power system Calculating the voltages (amplitudes and angles) at every node (substation) Determination of the static state of a given system Knowing only: Grid configuration and parameters (R and X ) Power outputs of generator units Loads (active and reactive) Some voltages

“Load flow” and “power flow” are synonyms



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Introduction

Introduction: load or power flow

Why is load flow important? Assessing if the power system is: Within operational limits Safe (N-1)

Basis for other (e.g. dynamic) calculations Checking whether future situations are valid



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Introduction

Introduction: load or power flow

When is it used? System planning System operations State estimation Dynamic simulations (basis, first calculation) ...



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Introduction

Introduction: load or power flow 2˚30’ E . Greenwich

2˚40’

2˚50’

3˚00’

3˚20’

3˚10’

3˚30’

3˚40’

4˚00’

3˚50’

4˚10’

4˚30’

4˚20’

4˚40’

5˚20’

5˚10’

5˚00’

4˚50’

5˚30’

5˚50’

5˚40’

6˚20’

6˚10’

6˚00’

6˚30’

6˚40’

6˚50’

7˚00’ CBR

GEERTRUIDENBERG

51˚30’

51˚30’

BERNEAU

LIXHE

13

Meuse

Visé (SNCB)

BORSSELE KREEKRAK

Fooz 0)

Den aut

Esc

+70

MARCHE-LEZECAUSSINNES

50˚30’

70 150+

50) 70(1

70(15

Ciply

AUVELAIS

Namur

FLEURUS

as Ma 0)

Turon

Spa

LA PRAYE FOUR

Heid-deGoreux

Comblain

Bévercé

150 kV 70 kV

circuits multiples

2

meervoudige stroomketen

2

Butgenbach Am

Bronrome blèv

COO-TROIS-PONTS

Stéphanshof

BRUME Bomal

Florée

CENTRALES

CENTRALES

e

Wierde Sart-Bernard (SNCB)

Bois-deVillers

2de draadstel in aanbouw of in ontwerp

ONDERGRONDSE KABELS

70 kV

Warch e

Grands-Malades

FARCIENNES

existante

c. thermique

bestaande

en projet ferme

Trois-Ponts

70 (220)

Miécret

c. nucléaire

Amel

thermische c.

vast ontworpen bestaande

existante

kernenergie c.

d'Heure

MASTAING

380+15

Dorinne

150+3 6

Solre St-Géry

Hogne (SNCB)Marche-enFamenne

Dinant

bestaande in ontwerp

50˚10’

MONT-LEZ-HOUFFALIZE

36 (150)

Les

se

DAMPLEIN Belliard. Hovenier.

CHOOZ

2

Pondrôme

WOMMELGEM

Tabak.

2

VIREUX

ZURENBORG

3

Couvin

OELEGEM

Berchem(NMBS)

2

15(70)

Baileux

LesForges

50˚00’

Momignies FOURMIES

MORTSEL

Wilrijk

Herbaimont Meuse

Moons. ANTWERPEN ZUID

MHo

380+ 220(2 x380 )

Oever BURCHT

380+220

BEVEREN

St-Niklaas

Forrières (SNCB)

2

ST-PAUWELS 50˚00’

STATIONS existants en projet

MERKSEM ZWIJNDRECHT

Hatrival(SNCB)

Our

VILLEROUX Hoboken

MERCATOR

150+70

pomp-c.

vast ontworpen

Charneux

On

Romedenne

NEUVILLE SCHELDELAAN

BPCHM

existante

Hastière

PLATE-TAILLE KALLO

hydraulische c.

bestaande

POSTES

Cierreux

MARCOURT

220+150(220)

Renlies

9

HEIMOLEN

he

vast ontworpen

en projet ferme

St-Vith 380+220

Ourt

ACHENE

JAMIOLLE

0) 15(7

FINA

Soy

Ciney

Sommière

Clermont

12(70)

Beaumont

c. de pompage

(2x380

Hanzinelle 12(7 0)

EKEREN

bestaande

en projet ferme

Yvoir (SNCB)

SNCB

Eau

50˚20’

7eHAVENDOK

50˚10’

Warnant

Thy-le-Château

LILLO BAYER

KETENISSE

Gerpinnes

)

20) 150(2

150+36

THUILLIES

OORDEREN

existante

c. hydraulique

Lobbes

1:500000

SOLVAY

vast ontworpen

en projet ferme

SNCB BERENDRECHTSLUIS.

DOEL

50˚20’

0

MONCEAUMONTIGNIES

70 kV

in aanbouw of in ontwerp

CABLES SOUTERRAINS 150 kV

JEMEPPESOLVAY

TERGNEE

DAMPREMY

BINCHE

220 kV

70 kV

2

Sambre

380+150

AMERCŒUR

PERONNES

HARMIGNIES

GRAMME

SNCB

GOSSELIES

150 kV

150 kV

HteSARTE

Andenne

Marche-les-D.

St-Servais

GOUY

BASCOUP TRIVIERES

70

220 kV

en construction ou en projet 2e terne en construction ou en projet

SEILLES

Pâturages

BASF

Statte (SNCB)

COGNELEE

70(150)

0)

Elouges

2

+

380 kV

380 kV

Gileppe Gile ppe

ST-AMAND

LACROYERE V/HAINE

+ 70 150

20

150

QUAREGNON

Nominale spanning

Tension nominale Eupen

Stembert dre Ves

50˚30’

380+2

BAUDOUR 70(1 50)

+ 70

150

Harchies

BOVENGRONDSE LIJNEN

LIGNES AERIENNES

Garnstock

150+70

Les Plenesses PT-RECHAIN

50 220+1

RIMIERE

CHAMPION

GHLINOBOURG

TERTRE

CHEVALET

Battice

Soiron Pepinster

TIHANGE

Waret

COURCELLES

AIRLIQ.

HenriChapelle

Welkenraedt (SNCB)

150

Monsin

ROMSEE

LATROQUE LEVAL

CLERMONT

Leuze

Gembloux 12

Quevaucamps

ZANDVLIET

CroixChabot

2

PETROCHIM FELUY

Meuse

SERAING Ivoz AWIRS

70) 12(

Soignies Lens

4

15(70)

Aische-en-Refail

NIVELLES

AVELIN

3 BRESSOUX JUPILLE

Glatigny CORBAIS

LECHENOI

Dyle

Sauvenière

ANTOING

220+

150+36

e De ndr

CHERTAL

2

2

VIEUX GENAPPE

SNCB Baulers

SNCB

CHIEVRES

Fooz

50˚40’

50˚40’

0

Ronquières Braine-le-C.

GAURAIN Carr. Milieu

1

Court-St-Et.

BAISY-THY Ath (SNCB)

WATTINES SNCB

Saives

Hannut

OISQUERCQ Meslin

LIGNE

Tournai

0 +15 380

0)

150+7

+70

Montzen (SNCB)

BERNEAU

Ottignies(SNCB)

Ceroux THIEULAIN 70 (150) 70(150) 150

LIXHE

AVERNAS Jodoigne

WATERLOO

BRAINE-L'ALLEUD

+ 70

Sche

der

2 NMBS Landen

RHODEST-G. ST-G.RODE

10(7

150

150

HERDEREN Borgloon Tongeren

BUIZINGEN

Blandain

Lanaken

BRUSTEM

St-Truiden

LABORELEC

MEKINGEN

Enghien (SNCB) Hoves

MARQUAIN

70(15

70(150)

lde

TIENEN

Herfelingen

Deux-Acren +70 150

150

380+150

51˚20’ 51˚10’ 50˚50’

WOLUWE-ST-L. ST-L.WOLUWE

IXELLES ELSENE

FOREST VORST

19 Geraardsbergen Ronse

DOTTIGNIES

WARANDE

KNP

50˚50’

DROGENBOS

RUIEN

AVELGEM

DHANIS

Q.DEMETSk. ZUID/MIDI

EIZERINGEN Appelterre

MOEN

MOUSCRON

Rivage (SNCB)

ZUTENDAAL

70(150)

Bilzen

150

2

NINOVE

OUDENAARDE

70(150)

+70

1:500000

Poulseur

Anthisnes

2 15

150(380)

16

Abée-Scry

Maasmechelen

2

GODSHEIDE

(NMBS)

MARCHIN

STALEN GENKLANGERLO SIKEL

Hasselt

Alken

Heverlee

ZWEVEGEM

150

Halen

Kersbeek

Pellenberg

380+

150+70

HEULE Kortrijk -NMBS WEVELGEM

ZONHOVEN

Paalsteenstr. 70

150+70

PEKKE

MENEN WEST

Lummen

DIEST

Dorenberg

Kessel-lo

2 Leuven (NMBS)

e Dyl

DILBEEKHELIPORTWIERTZ

Bas Warneton

Ourthe

HteSARTE

EISDEN Demer

Aarschot

WIJGMAAL

50) 70(1

Gasthuisberg

Esneux

RIMIERE

GRAMME Bekaert

WILSELE

ZAVENTEM HARENHEIDE

Vesdre

70(150)

TIHANGE

HOUTHALEN

HOENDERVELD

3

MACHELEN MOLENBEEK

BRUEGEL

Sart-Tilman

Ehein Hermalle s/Huy

AmpsinNeuville

OBERZIER

Opglabbeek

11

WESPELAAR

2

GRIMBERGEN

RELEGEM

Essene Denderleeuw (NMBS)

ST-Denijs Boekel

Magotteaux

Maaseik

BERINGEN

2

TESSENDERLO

KOBBEGEM Welle Zottegem

WORTEGEM

14

CLERMONT

HERCULES Dowchemical TIP

Muizen

VERBRANDEBRUG

Gavere

DESSELGEM

ROMSEE

Chénée

Ramet Croix-Chabot 70(1 50)

ESSOCHEM

Nete

Grote

KRUISBAAN

NMBS

TERLINDEN

AALST 150+70

ST-BAAFS-VIJVE Oostrozebeke

HARELBEKE KUURNE Bekaert 2 K.Oost

70+150

BELLAIRE 16

Grivegnée

AWIRS

MAASBRACHT

Gerdingen

Hechtel

MEERHOUT AMOCO

Langveld

PUTTE MECHELEN

70(150)

MERCHTEM

AALST NOORD Amylum

MUIZELAAR

IZEGEM

Angleur

Ougr.Sclessin

51˚00’

70 150+

FLORA

DEINZE

150+70

RUMBEKE

IEPER

16

150 +70

JUPILLE

70(1 50)

LATROQUE

LEVAL Flémalle

REPPEL HEZE Geel/Oevel

HEIST/BERG SIDAL Duffel

1

LEEST Tisselt

Leie

TIELT

150+70

Westrozebeke

IEPERNOORD

VIEILLE MONTAGNE

Herenthout LIER

LINT 7 el

MALDEREN

PITTEM 150+70

BEVEREN

STADEN Noordschote

BALEN HerentalsOlen

Nijlen

MORTSEL

Rup

Willebroek

BUGGENHOUT ST-GILLISDENDERMONDE

SERAING 17

Profondval IvozRamet

(I.E.)

380 + 150

ANTWERPEN ZUID

SCHELLE -DORP

Bornem

ZELE

Baasrode

IJzer

70(150)

MOL

MASSENHOVEN

WOMMELGEM

BURCHT

Temse Hamme

Lokeren

NIEUWEVAART HAM

Jemeppe

St-Huibr.-Lille

OVERPELT

POEDERLEE 2

MERCATOR

WALGOED

HEIMOLEN

KENNEDYLAAN RINGVAART

9

Pouplin

Montegnée

MHO

LOMMEL

OELEGEM

51˚10’

LANGERBRUGGESADA.

ZOMERGEM

DRONGEN

CHERATTE 16

FN 4

BRESSOUX Glain

8 Tilleur 70(220)

MERKSEM

2

St-Niklaas

AALTER

3

Vottem

Ans

St-JOB

DODEWAARD

SCHELDELAAN

ZURENBORG

SIDMAR

RODENHUIZE ZEDELGEM

51˚00’

7eHAVENDOK

FINA

DAMPLEIN

ST-PAUWELS EEKLO 6

70(15

Monsin

Voroux (SNCB) Hollogne

BAYER

ZWIJNDRECHT

BEERST

Saives Turnhout

EKEREN

+36

KOKSIJDE

BEERSE

MALLE

LILLO 150

KALLO

EEKLO NOORD

51˚20’

Alleur Rijkevorsel

BERENDRECHTSLUIS.

SOLVAY

KETENISSE

BEVEREN MALDEGEM

70(150)

EINDHOVEN

Herstal

BASF

HERDERSBRUG

BRUGGE

150+70

380+ 150

Ravels

ZANDVLIET

DOEL

OOSTBURG

3 BLAUWE TOREN

SLIJKENS

CHERTAL

FN

Kalmthout

NIEDERSTEDEM

VIANDEN S.E.O.BAULER

WALGOED

Aartselaar

Kontich

SCHELLE

MAZURES

LINT 7

Recogne

REVIN

LIER -SCH.DORP

Temse

FLEBOUR

Monceau-en-Ard. S˚re

Respelt

WAARLOOS

PETROCHIM

Orgeo

Sombreffe

50) 70(1

FELUY

LUMES

Gembloux

2

Vierre

Chassart

12

NeufchâteauLonglier(SNCB)

49˚50’

BUISSERET

MARCHE-LEZ-ECAUSSINNES

70(220)

Fays-les-Veneurs ois Sem

49˚50’

LONNY

VESLE Marbais (SNCB)

Seneffe

ROOST

Liberchies

TRIER Marbehan (SNCB)

Vierre

COURCELLES

ST-AMAND

Maisières LACROYERE

70 150+

GOSSELIES Arlon 70(1

Heppignies (30) sud

50)

TRIVIERES

70 + 30(150 )

AUVELAIS

Jumet

70(150)

380+150

70( 150 )

BERTRANGE Samb

re

AUBANGE

LATOUR

LAPRAYE FOUR

1:1000000

Echelle

CAB

Schaal

St-MARD

PONT-de-LOUP LAPRAYE

4

0

10

20

MONTST.MARTIN

30km

HERSERANGE

MOULAINE

PosteEst MONTIGNIES

Fosses-la-Ville

MONCEAU

1:500000

Differd. Arbed OXYLUX Belv.Arbed

BELVAL ESCH-SUR-ALZETTE SCHIFFLANGE

Situation au stand op

Schif.

VIGY Institut Géographique National Nationaal Geografisch Instituut

LANDRES 2˚40’ E . Greenwich

1-1-2005 49˚30’

150 +70

) 70(150

Marchienne 150+70

49˚30’

CSTMM F.DEFER 2 BLANCHISSERIE Charleroi

50) 70(1

MALFALISE

JEMEPPESOLVAY

TERGNEE FARCIENNES

Gilly

Fontaine l'Evêque BINCHE

+70

Moustier

70(150)

AMERCŒUR

DAMPREMY CARAL FOC

Distrigaz

HARMIGNIES

HEISDORF

150

FLEURUS

PERONNES

30(150)

70(150)

Ciply

Pâturages

HEINSCH SNCB

GOUY BASCOUP

elle

BOELTCC BOELHF LaLouvière

Piéton (SNCB) 150+70

Boussu

Elouges

Bonnert

Villers-s/Semois

Chiny

Keumiée

BOEL FOUR

BOEL LL 150+ 70

50) 70(1

+ 70

VILLE/HAINE Mons

JEMAPPES

Mos

150 + 70

OBOURG

50)

150+70

150+70

70(1

49˚40’

70(150)

GHLIN

QUAREGNON

16

49˚40’

AIR LIQUIDE

Alze tte

150

150+30 30(150)

TERTRE

BAUDOUR

2˚50’

3˚00’

3˚10’

3˚20’

3˚30’

3˚40’


3˚50’

4˚00’

4˚10’

4˚20’

4˚30’

4˚40’

4˚50’

5˚00’

5˚10’

5˚20’

5˚30’


5˚40’

5˚50’

6˚00’

6˚10’

6˚20’

6˚30’

6˚40’

6˚50’

September 19, 2011

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Introduction

Example

Example File: case6 wh.m, from the book “computational methods for electric power systems”, M. Crow. Bus data bus 1 2 3 4 5 6

type 3 2 2 1 1 1

Pd 25 15 27.5 0 15 25

Qd 10 5 11 0 9 15

Vm 1.05 1.05 1 1 1 1

Va 0 0 0 0 0 0

Vmax 1.05 1.05 1.05 1.05 1.05 1.05

Generator data Gen Pg Pq 1 0 0 2 50 0

Vmin 1.05 1.05 0.95 0.95 0.95 0.95

Pmax 200 150

G 1

4

#1

#6

3

Branch data

#2

#3

#5 #7 6

#4 5

line 1 2 3 4 5 6 7

from 1 1 2 2 4 3 5

to 4 6 3 5 6 4 6

R 0.020 0.031 0.006 0.071 0.024 0.075 0.025

X 0.185 0.259 0.025 0.320 0.204 0.067 0.150

B 0.009 0.010 0.000 0.015 0.010 0.000 0.017

2 G



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Introduction

Example

Example File: case6 wh.m, from the book “computational methods for electric power systems”, M. Crow. Bus data bus 1 2 3 4 5 6

type 3 2 2 1 1 1

Pd 25 15 27.5 0 15 25

Qd 10 5 11 0 9 15

Vm 1.05 1.05 1 1 1 1

Va 0 0 0 0 0 0

Vmax 1.05 1.05 1.05 1.05 1.05 1.05

Generator data Gen Pg Pq 1 0 0 2 50 0

Vmin 1.05 1.05 0.95 0.95 0.95 0.95

Pmax 200 150

G 1

4

#1

#6

3

Branch data

#2

#3

#5 #7 6

#4 5

line 1 2 3 4 5 6 7

from 1 1 2 2 4 3 5

to 4 6 3 5 6 4 6

R 0.020 0.031 0.006 0.071 0.024 0.075 0.025

X 0.185 0.259 0.025 0.320 0.204 0.067 0.150

B 0.009 0.010 0.000 0.015 0.010 0.000 0.017

2 G

How would you solve this simple example by hand?



September 19, 2011

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Outline 1


2


3


4


5

Software

6

State estimation



September 19, 2011

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Most power systems are three phase AC Normal power flow uses one phase equivalents ⇒ We only focus on this one today One phase power flow only valid for balanced systems Systems are usually given in per unit values Lines can be represented by a π-equivalent



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Per-unit calculations Normalized representation of the four basic properties: voltage, current, impedance and complex power Of these, two can be chosen independently Normally rated phase voltage and one phase rated power are taken as basis Upu =

Ibase = Zbase =

U Ubase

and Spu =

S Sbase

Sbase Ubase Ubase Ibase

or Zbase =

2 Ubase Sbase

Logical values: for a 11.8 kV , 60 MVA machine, √ kV and Sbasis = 60 MVA Ubasis = 11.8 3 3 √ For a 400 kV line, with 100 MVA: Ubase = 400/ 3, 2 3 √ Sbase = 100/3 ⇒ Rbase = 400 · 100 = 1600Ω 3



September 19, 2011

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Per-unit calculations Normalized representation of the four basic properties: voltage, current, impedance and complex power Of these, two can be chosen independently Normally rated phase voltage and one phase rated power are taken as basis Upu =

Ibase = Zbase =

U Ubase

and Spu =

S Sbase

Sbase Ubase Ubase Ibase

or Zbase =

2 Ubase Sbase

Logical values: for a 11.8 kV , 60 MVA machine, √ kV and Sbasis = 60 MVA Ubasis = 11.8 3 3 √ For a 400 kV line, with 100 MVA: Ubase = 400/ 3, 2 3 √ Sbase = 100/3 ⇒ Rbase = 400 · 100 = 1600Ω 3 Why are voltage and complex power chosen as fixed values?



September 19, 2011

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Example per-unit

Generator example of before Ubasis =

11.8 √ 3

kV and Sbasis =

60 3

MVA

Sbase Ubase

=

√ 60· 3 11.8·3 = 2.9357 2

Basis for current: Ibase =

Basis for impedance: Zbase =

2 Ubase Sbase

=

11.8 √ 3 60 3

kA

= 2.3207 Ω

Line connecting load: 0, 5 + 1 Ω = 0.21546 + 0.43091 pu Afterwards, calculate using per-unit instead of original values



September 19, 2011

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Per-unit and transformers Z1

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111

U1

Z2

n1 : n2 11 00 00 11 11 00 11 00 00 11 00 11 00 11 00 11 11 00

11 00 00 11 11 00 11 00 00 11 00 11 00 11 00 11 11 00

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111

U2

Voltage at both sides of the transformer is different → different basis One of the major advantages of per-unit calculations because of simplification



September 19, 2011

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Per-unit and transformers Z1

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111

Z20

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111

n1 : n2

11 00 00 11 11 00 11 00 00 11 00 11 11 00 00 11 00 11

U1

11 00 00 11 11 00 11 00 00 11 00 11 11 00 00 11 00 11

U2

Voltage at both sides of the transformer is different → different basis One of the major advantages of per-unit calculations because of simplification 2 2 U n Z20 = Z2 · n12 = Z2 · U12 2


2


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Per-unit and transformers Zp

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111

U1

n1 : n2 11 00 00 11 11 00 11 00 00 11 00 11 00 11 00 11 11 00

11 00 00 11 11 00 11 00 00 11 00 11 00 11 00 11 11 00

U2

Voltage at both sides of the transformer is different → different basis One of the major advantages of per-unit calculations because of simplification 2 2 n U Z20 = Z2 · n12 = Z2 · U12 2

Zp = Z1 + Z20


2


September 19, 2011

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Per-unit and transformers Zs

n1 : n2

U1

11 00 00 11 11 00 11 00 00 11 00 11 00 11 00 11 11 00

11 00 00 11 11 00 11 00 00 11 00 11 00 11 00 11 11 00

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111

U2

Voltage at both sides of the transformer is different → different basis One of the major advantages of per-unit calculations because of simplification 2 2 n U Z20 = Z2 · n12 = Z2 · U12 2

Zp = Z1 +Z20

2

U2

Zs = Zp · U22 and Zs (pu) = ZbaseZs(sec) 1 The per-unit impedance is the same on both sides of the transformer ⇒ can be replaced by one series impedance! DVH, HE, PGT (KUL/ESAT/ELECTA)


September 19, 2011

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Representation of a transmission line X

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 B 2

G 2

R

B 2

G 2

π-equivalent Valid for lines up to 240 km All values are normally small Other equivalents exist and are sometimes used in practice Normally, G can be neglected With overhead lines, B can be neglected as well, for cables this is not the case (see chapters on lines and cables)



September 19, 2011

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Outline 1


2


3


4


5

Software

6

State estimation



September 19, 2011

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A bit of algebra: incidence matrix Directed Graph

Incidence matrix (A0 ) line # 1

-1

nodes 1 0 0

# lines (branches) × # nodes



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Incidence matrix (A0 ) line # 1 line # 2

-1 -1

nodes 1 0 0 0 1 0




September 19, 2011

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Incidence matrix (A0 ) line # 1 line # 2 line # 3

-1 -1 0

nodes 1 0 0 1 -1 1

0 0 0




September 19, 2011

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Incidence matrix (A0 ) line line line line

# # # #

1 2 3 4

-1 -1 0 0

nodes 1 0 0 1 -1 1 -1 0

0 0 0 1




September 19, 2011

12 / 33



Incidence matrix (A0 ) line line line line line

# # # # #

1 2 3 4 5

-1 -1 0 0 0

nodes 1 0 0 1 -1 1 -1 0 0 -1

0 0 0 1 1




September 19, 2011

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Incidence matrix (A0 ) line line line line line

# # # # #

1 2 3 4 5

-1 -1 0 0 0

nodes 1 0 0 1 -1 1 -1 0 0 -1

0 0 0 1 1

# lines (branches) × # nodes the columns are dependent



September 19, 2011

12 / 33


Incidence matrix Incidence matrix is written as A0

Some symbols Meaning of the incidence matrix Describes the directed graph

Iij : Current from node i to node j

Produces differences  −1 −1  0  −1 0

1 0 −1 0 −1

   0 0   U2 − U1 U1   1 0  U2  U3 − U1    = U3 − U2  (1) 1 0 ·  U3    U4 − U1  0 1 U4 0 1 U4 − U2


Ii : Current injected at node i


Ui : Potential of node i Eij : Potential difference (voltage) between nodes i and j Cij : Conductance of the line between nodes i and j

September 19, 2011

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Meaning of the incidence matrix


Describes the directed graph Produces differences  −1 −1  0  −1 0

1 0 −1 0 −1

   0 0   U2 − U1 U1   1 0  U2  U3 − U1    = U3 − U2  (1) 1 0 ·  U3    U4 − U1  0 1 U4 0 1 U4 − U2

Setting U4 = 0 Resulting matrix is the incidence matrix: A


Some symbols


Iij : Current from node i to node j Ui : Potential of node i Eij : Potential difference (voltage) between nodes i and j Cij : Conductance of the line between nodes i and j

September 19, 2011

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Meaning of the incidence matrix


Describes the directed graph Produces differences  −1 −1  0  −1 0



    1 0 U2 − U1 U1   0 1   U2  U3 − U1    = U3 − U2  (1) −1 1  ·  U3     − U1  0 0  −1 0 − U2

Setting U4 = 0 Resulting matrix is the incidence matrix: A


Some symbols


Iij : Current from node i to node j Ui : Potential of node i Eij : Potential difference (voltage) between nodes i and j Cij : Conductance of the line between nodes i and j

September 19, 2011

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Meaning of the incidence matrix 1

2

3

4

5

Incidence matrix is A0 with one node removed (grounded, reference) Ii is a nodal current injection, Iij is a branch flow  T   −1 1 0 I12   −1 0   I1 1   I13  I2  =  0 −1 1  · I23       0 −1 0  I24  I3 0 0 −1 I34 Ui is a nodal voltage/potential, Eij represents a potential drop over line ij The relation between the voltage difference (e) and line flows (f): Ohms law (take Cij the conductance of i to j) Link on youtube



September 19, 2011

14 / 33



2 3

4

5

Incidence matrix is A0 with one node removed (grounded, reference) Ii is a nodal current injection, Iij is a branch flow Ui is a nodal voltage/potential, Eij represents a potential over line ij   drop  E12 −1 1 0   E13  −1 0 1     U1 E23  =  0 −1 1  · U2      E24   0 −1 0  U3 E34 0 0 −1 The relation between the voltage difference (e) and line flows (f): Ohms law (take Cij the conductance of i to j) Link on youtube



September 19, 2011

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2 3

4

5

Incidence matrix is A0 with one node removed (grounded, reference) Ii is a nodal current injection, Iij is a branch flow Ui is a nodal voltage/potential, Eij represents a potential drop over line ij The relation between the voltage difference (e) and line flows (f): Ohms law (take Cij the conductance    of i to j)    I12 C12 0 0 0 0 E12 I13   0 C13  E13  0 0 0       I23  =  0   0 C23 0 0      · E23  I24   0  E24  0 0 C24 0 I34 0 0 0 0 C34 E34 Link on youtube



September 19, 2011

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Putting it together AT · C · A · U = I Ybus · U = I Ybus is the bus admittance matrix Representation of the entire network by an admittance matrix, a vector of nodal voltages and a vector of nodal current injections Yij = −yij (admittance between node i and j) Pn Yii = j yij (sum of the rest of the row + yii , the impedance to the reference)



September 19, 2011

15 / 33


The power system represented

The power system consists of: Generators: delivering P and Q Loads: consuming P and Q Lines or branches: connecting generation and load Nodes or busbars: connections points in the power system



September 19, 2011

16 / 33


The power system represented

The power system consists of: Generators: delivering P and Q Loads: consuming P and Q Lines or branches: connecting generation and load Wanted: Power flow of P and Q through these lines

Nodes or busbars: connections points in the power system Wanted: Voltage amplitude (U) and voltage angle (θ) at each node



September 19, 2011

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Mathematical statement of the problem Ia a

Iab

Iac c

yac

Ic

yab ybc

b

Ibc

Ib

Uc ∠θc

Ub ∠θb

Ua ∠θa

Neutral



September 19, 2011

17 / 33


Mathematical statement of the problem

Ia

Iac c

yac

Ic

Voltage of node i to neutral is Ui ∠θi

Iab

Admittance between i and j is yij

yab ybc

Ibc

a

Current from i to j is Iij The injected current at i is Ii

b

Ib

Uc ∠θc

Ub ∠θb Neutral


Ua ∠θa

Ia = Iab + Iac Ia = (Ua − Ub ) · yab + (Ua − Uc ) · yac Ia = Ua · (yab + yac ) − Ub · yab − Uc · yac


September 19, 2011

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Mathematical statement of the problem Last equation repeated: Ia = Ua · (yab + yac ) − Ub · yab − Uc · yac Pn We take Yaa = yab + yac = yaa + i6=a yai yaa = ya0 = the parallel branches to node a (in this example, yaa = 0) We take Yai = −yai ⇒ as with Ybus , the bus admittance matrix Which results in: Ia = Yaa · Ua + Yab · Ub + Yac · Uc

(2)

Or for the entire system:       Ia Yaa Yab Yac Ua  Ib  =  Yba Ybb Ybc  ·  Ub  Ic Yca Ycb Ycc Uc or and

(3)

I = Ybus · U

(4)

Yij = Yji in symmetrical systems (e.g. not with phase shifting transformers)



September 19, 2011

17 / 33


Mathematical statement of the problem Resulting equations for a general system with n nodes Ii =

n X

Yij · Uj

∀i ∈ N ≤ n

(5)

j=1

Si∗ = Ui∗ · Ii

(6)

Above equations form the basis of power flow There are 4 basic quantities for each node in power flow calculations: Voltage amplitude |U| Voltage angle θ between the voltage vector and the voltage reference Active power injection, withdrawal at a node Reactive power injection, withdrawal at a node



September 19, 2011

17 / 33


Mathematical statement of the problem Where do shunt elements fit? 

  Ia Yaa  Ib  =  Yba Ic Yca n X Yii = yij

=

j=1 n X

Yab Ybb Ycb

   Yac Ua Ybc  ·  Ub  Ycc Uc

−Yij + yii

(7)

(8)

(9)

j=1 j6=i

yii is the term to the node that has been “grounded” In practice: shunt elements



September 19, 2011

17 / 33


Types of nodes Three distinct types of nodes (important) PV bus: A generating source is connected to the bus; the nodal voltage is controlled at a certain magnitude U by injecting or absorbing reactive energy. The generated power PG is set at a specified value. θ and QG are computed. Constant voltage operation is only possible when the generator is within its reactive energy generation limits. PQ bus: P and Q are the control variables. This is the case when there is only a load connected to the bus or the generator is outside its reactive power limits. Slack (swing) bus: one of the generator busses is chosen to be the slack bus where the nodal voltage magnitude, Uslack , and phase angle θslack are specified. This bus is needed to provide a “compensation” for the electrical losses that are not known in advance. The bus forms a reference for the voltage angle.



September 19, 2011

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What is the mathematical meaning of the slack bus? DVH, HE, PGT (KUL/ESAT/ELECTA)


September 19, 2011

18 / 33



Ii =

Pn

j=1

Yij · Uj


∀i ∈ N ≤ n and i 6= nslack Power flow calculations

September 19, 2011

18 / 33

Solving the problem

Outline 1


2


3


4


5

Software

6

State estimation



September 19, 2011

19 / 33

Solving the problem

The problem Known data: Active power injections in my system at generator nodes Voltages at generator nodes Active and reactive withdrawals (load) at PQ nodes Slack node voltage and angle Impedances (Ybus )

Unknowns: Rest of P (slack), Q (slack and PV), voltage amplitude (PQ nodes) and voltage angle (all but slack)

Equations I =Y ·U

(10)

S∗ = U∗ · I

(11)

S∗ = U∗ · Y · U

(12)

∗

(13)

P − Q = U · Y · U n X Pi − Qi = Ui∗ · Yij · Uj

(14)

j=1 DVH, HE, PGT (KUL/ESAT/ELECTA)


September 19, 2011

20 / 33

Solving the problem

Gauss-Seidel

Gauss-Seidel Algorithm Ii =

Pn

j=1

Yij · Uj and Si∗ = Ui∗ · Ii give:       n  1  Si∗ X Ui = Yij · Uj · − ∗  Yii  U   j=1   i

(15)

j6=i

This is solved bus by bus, and solutions of previous calculations are filled in directly       i n  X X 1  Si∗ (i+1) (i+1) (i) Ui = · − Y · U − Y · U (16) ij ij j j ∗(i)  Yii  U   j=1 j=i+1  i  j6=i



j6=i

September 19, 2011

21 / 33

Solving the problem

Gauss-Seidel

Gauss-Seidel 1

2

For busbar 21 , calculate I2 = Pn Calculate j=1 Y2j · Uj

S2∗ U2∗

j6=2 3

Subtract solution 2 from solution 1 and divide the result by Y22 to obtain a new value for U2

4

For busbar 3, calculate I3 =

5

Using the new value of U2 of step 3, calculate

6

Subtract solution 5 from solution 4 and divide the result by Y33 to obtain a new value for U3

7

Repeat for all busses

8

Compare latest set of voltages with previous and check tolerance: U(i+1) − U(i) < ε? If not, go to step 1.

1 when

S3∗ U3∗

Pn

j=1 j6=3

Y3j · Uj

1 is the reference bus



September 19, 2011

21 / 33

Solving the problem

Gauss-Seidel

Gauss-Seidel

Convergence and acceleration The Gauss-Seidel method converges linearly (slow) with system size Each iteration itself requires limited processing power Often, the method is corrected with an acceleration factor (new ) (new ) (old) Ui(acc) = α · Ui − (α − 1) · Ui (new )

= α · Ui =

(old) Ui

(old)

− α · Ui

+α·

(new ) (Ui

−

(old)

(17)

+ Ui

(18)

(old) Ui )

(19)

1<α<2 For large systems, often a value of 1.6 is chosen



September 19, 2011

21 / 33

Solving the problem

Gauss-Seidel

Gauss-Seidel

Gauss-Seidel properties A starting vector must be chosen Often, the starting voltages are set to 1∠0 pu called “Flat start”

If the voltages are calculated in block (and not replaced after one has calculated the former one), we call the method the Jacobi method. The Jacobi method has a slower convergence

The Gauss-Seidel method is not often used anymore



September 19, 2011

21 / 33

Solving the problem

Gauss-Seidel

Gauss-Seidel

Gauss-Seidel properties A starting vector must be chosen Often, the starting voltages are set to 1∠0 pu called “Flat start”

If the voltages are calculated in block (and not replaced after one has calculated the former one), we call the method the Jacobi method. The Jacobi method has a slower convergence

The Gauss-Seidel method is not often used anymore



September 19, 2011

21 / 33

Solving the problem

Newton-Raphson power flow

Newton-Raphson method: theory

Solution of equation y = f (x) Taylor series expansion: df d2 f dx x=x(0) dx 2 x=x(0) y = f [x(0)]+ ·[x −x(0)]+ ·[x −x(0)]2 +. . . (20) 1! 2!



September 19, 2011

22 / 33

Solving the problem



Solution of equation y = f (x) Taylor series expansion: df d2 f dx x=x(0) dx 2 x=x(0) y = f [x(0)]+ ·[x −x(0)]+ ·[x −x(0)]2 +. . . (20) 1! 2!



September 19, 2011

22 / 33

Solving the problem



Solution of equation y = f (x) Taylor series expansion:

y = f [x(0)] +

df dx x=x(0) 1!

· [x − x(0)]

(20)

Solving this for x: x = x(0) +


1 · [y − f (x(0))] df dx x=x(0)


(21)

September 19, 2011

22 / 33

Solving the problem



f (U, θ)

(U, θ)∗ (U, θ)0



September 19, 2011

22 / 33

Solving the problem



f (U, θ)0 f (U, θ)

(U, θ)∗ (U, θ)0



September 19, 2011

22 / 33

Solving the problem


Newton-Raphson method: theory derivative f (U, θ)0 f (U, θ)

(U, θ)∗ (U, θ)0



September 19, 2011

22 / 33

Solving the problem



(U, θ)1

(U, θ)∗ (U, θ)0

f (U, θ)1



September 19, 2011

22 / 33

Solving the problem



f (U, θ)2 (U, θ)1

(U, θ)∗ (U, θ)2

(U, θ)0

f (U, θ)1



September 19, 2011

22 / 33

Solving the problem



Solution for multivariable nonlinear equations y1 = f1 (x1 , x2 , . . . , xn ) y2 = f2 (x1 , x2 , . . . , xn ) .. . y3 = fn (x1 , x2 , . . . , xn )



September 19, 2011

22 / 33

Solving the problem


Newton-Raphson method: theory Solution for multivariable nonlinear equations     

y1 y2 .. . yn





    =        +    


 f1 (x1 (0), x2 (0), . . . , xn (0)) f2 (x1 (0), x2 (0), . . . , xn (0))    ..  . fn (x1 (0), x2 (0), . . . , xn (0))  ∂f1 ∂f1 ∂f1 ···  ∂x1 ∂x2 ∂xn  x1 − x1 (0) ∂f2 ∂f2 ∂f2    x − x (0) ··· 2   2 ∂x1 ∂x2 ∂xn  ·  .. .. .. ..  ..   . . . . .   x − xn (0) n ∂fn ∂fn ∂fn ··· ∂x1 ∂x2 ∂xn


    

(22)

September 19, 2011

22 / 33

Solving the problem



Solution for multivariable nonlinear equations Summarized, we can write the following: y = f [x(0)] + J(0) · [x − x(0)]

(23)

or solving for x: x = x(0) + J(0)−1 · [y − f (x(0))]

(24)

or in its recursive form: xi+1 = xi + J−1 · [y − f (xi )] i



(25)

September 19, 2011

22 / 33

Solving the problem


Newton-Raphson and load flow Remember Ii = Si∗

Pn

=

j=1 Yij Ui∗ · Ii

· Uj

Equivalents Si = f (Ui ) is equivalent to y = f (x) Ui+1 = Ui + Ji · [S − f (Ui )]

(26)

S is here the specified complex power at any busbar f (Ui ) is here the specified complex power at any busbar ∆Si = Ji · Ui+1

(27)

You normally know the active and reactive power injections in each node (load and generation) You want to know the complex voltages at the nodes DVH, HE, PGT (KUL/ESAT/ELECTA)


September 19, 2011

23 / 33

Solving the problem


Newton-Raphson and load flow Si∗ = Ui∗

Pn

j=1

Yij · Uj

and

i = 1, . . . , n

Newton-Raphson in rectangular coordinates Pi = Ui

n X

Uj · (Gij · cos(θi − θj ) + Bij · sin(θi − θj ))

j=1 n X

Qi = −Ui

Uj · (Gij · sin(θi − θj ) − Bij · cos(θi − θj ))

(28)

(29)

j=1

Newton-Raphson in polar coordinates Pi = Ui

n X

Uj · Yij · cos(θi − θj − φij )

(30)

j=1

Qi = −Ui

n X

Uj · Yij · sin(θi − θj − φij )

(31)

j=1

note: Yij = Gij +  · Bij = |Yij |∠φij DVH, HE, PGT (KUL/ESAT/ELECTA)


September 19, 2011

23 / 33

Solving the problem


Newton-Raphson and load flow

Pi,

∆Pi = Pi,

scheduled

− Pi,

calc

∀PQ and PV

(32)

∆Qi = Qi,

scheduled

− Qi,

calc

∀PQ

(33)

scheduled

and Qi,

scheduled

are known from the input data

Pi, calc and Qi, calc are obtained from the calculation in rectangular or polar coordinates



September 19, 2011

23 / 33

Solving the problem


Newton-Raphson and load flow Putting it in one equation Writing the power flow equations (both rectangular and polar) in the form of equation (25): (i)  ∂P ∂P U (i) (i)  ∂θ ∂U  ∆θ ∆P   = − (34)  · ∆U ∆Q  ∂Q ∂Q  U U ∂U | ∂θ {z } J((U,θ)(i−1) )

Or written in a simplified form: (i) (i) (i) ∆θ ∆P H N · ∆U =− ∆Q M L U {z } |

(35)

J((U,θ)(i−1) )



September 19, 2011

23 / 33

Solving the problem


Newton-Raphson and load flow Equation explained (i+1)

∆θi = θi

(i)

− θi

(i+1)

∆Ui = Ui

(i)

− Ui

Voltages and angles (i + 1) are updated after each iteration and used for the following step J is the Jacobian, and forms the derivative (tangent, gradient) of the power flow equations ∂P ∂Q U and U simplify the equations and results in fewer ∂U ∂U computations There are n − 1 equations for ∆P There are n − #pv − 1 equations for ∆Q The Jacobian is a square matrix (2 · n − #pv − 2) × (2 · n − #pv − 2) The Jacobian is a sparse matrix (Special techniques can be used when numerical calculations are performed) DVH, HE, PGT (KUL/ESAT/ELECTA)


September 19, 2011

23 / 33

Solving the problem


Newton-Raphson and load flow Equation explained (i+1)

∆θi = θi

(i)

− θi

(i+1)

∆Ui = Ui

(i)

− Ui

Voltages and angles (i + 1) are updated after each iteration and used for the following step J is the Jacobian, and forms the derivative (tangent, gradient) of the power flow equations ∂P ∂Q U and U simplify the equations and results in fewer ∂U ∂U computations There are n − 1 equations for ∆P ⇒ Why? There are n − #pv − 1 equations for ∆Q ⇒ Why? The Jacobian is a square matrix (2 · n − #pv − 2) × (2 · n − #pv − 2) The Jacobian is a sparse matrix (Special techniques can be used when numerical calculations are performed) DVH, HE, PGT (KUL/ESAT/ELECTA)


September 19, 2011

23 / 33

Solving the problem



Full equations (polar form): off-diagonal ∂∆Pi = Hij ∂θj ∂∆Pi Uj = Nij ∂Uj ∂∆Qi = Mij ∂θj ∂∆Qi Uj = Lij ∂Uj


= −Uj · Ui · Yij · sin(θi − θj − φij ) = −Uj · Ui · Yij · cos(θi − θj − φij ) = Uj · Ui · Yij · cos(θi − θj − φij ) = −Uj · Ui · Yij · sin(θi − θj − φij )


September 19, 2011

23 / 33

Solving the problem


Newton-Raphson and load flow Full equations (polarn form): diagonal

X ∂∆Pi = Hii = Ui · Uj Yij sin(θi − θj − φij ) + Ui2 · Yii · sin(φij ) ∂θi j=1 n X

∂∆Pi Ui = Nii = −Ui · ∂Ui

∂∆Qi = Mii = −Ui · ∂θi ∂∆Qi Ui = Lii = −Ui · ∂Ui

Uj Yij cos(θi − θj − φij ) − Ui2 · Yii · cos(φij )

j=1 n X j=1 n X

Uj Yij cos(θi − θj − φij ) + Ui2 · Yii · cos(φij ) Uj Yij sin(θi − θj − φij ) + Ui2 · Yii · sin(φij )

j=1

Notice the symmetry Notice that the off-diagonal elements are also in the diagonal elements M = −N for off-diagonal elements DVH, HE, PGT (KUL/ESAT/ELECTA)


September 19, 2011

23 / 33

Solving the problem



Figure: Sparsity of the Jacobian matrix shown DVH, HE, PGT (KUL/ESAT/ELECTA)


September 19, 2011

23 / 33

Solving the problem


Newton-Raphson and load flow Numerical aspects Iterative process until mismatch is below threshold (max(∆P(i) ; ∆Q(i) ) = ε < εlimit ) Quadratic convergence Major computational effort is calculating the inverse of the Jacobian The Jacobian is sparse, so special techniques can be used (less storage) Ordering schemes can increase speed Convergence is not guaranteed A good starting point is needed Flat start? Previous outcome DC load flow as starting point

Simplifications exist DVH, HE, PGT (KUL/ESAT/ELECTA)


September 19, 2011

23 / 33

Solving the problem


Newton-Raphson and load flow derivative f (U, θ)0 f (U, θ)

(U, θ)∗ (U, θ)0



September 19, 2011

23 / 33

Solving the problem


Newton-Raphson and load flow derivative f (U, θ)0 f (U, θ)

(U, θ)∗ (U, θ)0



September 19, 2011

23 / 33

Solving the problem

Simplified Newton-Raphson

Decoupled load flow In a power system with mostly inductive lines, the power flow equations can be decoupled. (φij ≈ 90◦ ) Active power is related to the angle between nodes Reactive power is related to the voltage

∆P ∆Q

(i)

=− |

H M

N L {z

(i) (i) ∆θ · ∆U U }

(36)

J((U,θ)(i−1) )

Advantages and disadvantages + Two small inverses instead of one big + Faster as only 2 · n3 calculations are needed, and not (2 · n)3 = 8 · n3 - The two subsystems may converge differently - Convergence rate is slightly reduced - Not often used nowadays DVH, HE, PGT (KUL/ESAT/ELECTA)


September 19, 2011

24 / 33

Solving the problem


Decoupled load flow In a power system with mostly inductive lines, the power flow equations can be decoupled. (φij ≈ 90◦ ) Active power is related to the angle between nodes Reactive power is related to the voltage

∆P ∆Q

(i)

H 0 =− 0 L {z |

(i) (i) ∆θ · ∆U U }

(37)

J((U,θ)(i−1) )

Advantages and disadvantages + Two small inverses instead of one big + Faster as only 2 · n3 calculations are needed, and not (2 · n)3 = 8 · n3 - The two subsystems may converge differently - Convergence rate is slightly reduced - Not often used nowadays DVH, HE, PGT (KUL/ESAT/ELECTA)


September 19, 2011

24 / 33

Solving the problem


Decoupled load flow derivative f (U, θ)0 f (U, θ)

(U, θ)∗ (U, θ)0 Approximation



September 19, 2011

24 / 33

Solving the problem


Fast decoupled load flow In decoupled load flow, a new reduced Jacobian is determined during each iteration Of each new Jacobian, the inverse needs to be calculated Fast decoupled does not calculate a new Jacobian for each iteration h i h i ∆P(i) = [B0 ] · ∆θ(i+1) (38) (i) ∆Q = [B00 ] · ∆Ui+1 (39) U B0 and B00 are real, sparse and constant matrices Only series elements are involved (no shunts) If the system has high R/X -ratio, large voltage angle deviations or voltages which seriously differ from 1 pu, convergence problems can arise Slower convergence (more iterations) but each iteration is much faster DVH, HE, PGT (KUL/ESAT/ELECTA)


September 19, 2011

25 / 33

Solving the problem


Fast decoupled load flow derivative f (U, θ)0 f (U, θ)

(U, θ)∗ (U, θ)0 Approximation



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Solving the problem


DC load flow If we consider the system to be lossless (Y = B) And voltages to be around 1.0 pu (∆U = 0) Voltage angles between busses are small (sin(θi − θj ) ≈ (θi − θj ) and cos(θi − θj ) ≈ 1) One equation of Newton-Raphson: n X ∆P = Ui Uj · (Gij · cos(θi − θj ) + Bij · sin(θi − θj ))

(40)

j=1

[∆P] = [B0 ] · [∆δ]

(41)

B0 is real Linear system One calculation, no iterations Easy for optimizations Not correct (approximation) DVH, HE, PGT (KUL/ESAT/ELECTA)


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Software

Outline 1


2


3


4


5

Software

6

State estimation



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Software

Power flow software Demo’s + try at home Free available (open source) Matpower (matlab based): http://www.pserc.cornell.edu/matpower/ PSAT (matlab based): http://www.power.uwaterloo.ca/~fmilano/psat.htm InterPSS (Java based): http://www.interpss.org/

Professional software PSS/E Eurostag DigSilent Powerworld (demo at http://www.powerworld.com/downloads.asp)



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State estimation

Outline 1


2


3


4


5

Software

6

State estimation



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State estimation

State estimation Known and unknown variables in the real power system Lines, cables, transformers, location of generation and load ⇒ all known and constant in time Voltages, currents, actual generation and load (at that moment), position of power switches, tap-changer settings,. . . ⇒ mostly unknown or variable Measurements: P, Q: Generation and load, some lines Voltage: |U| every substation. θ only with PMU (phasor measurement unit) Tap-changer settings Incomplete Measurement errors



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State estimation

State estimation

State estimation: what? Monitoring or supplementing data for load flow Many measurements in the system Determining measurement errors, estimate and (statistically) analyze If needed, certain measurements should be rejected Least Squares approach Another Youtube video: least squares Has to be solved iteratively for power systems



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State estimation

State estimation Weighted least-square method (measurements of the state x)   z, with errors e, h(x) isthe true model  z1 h1 (x1 , x2 , . . . , xn ) e1  z2   h2 (x1 , x2 , . . . , xn )   e2       z=  z3  =  h3 (x1 , x2 , . . . , xn )  +  e3  = h(x) + e (42) z4 h4 (x1 , x2 , . . . , xn ) e4 With errors having a zero average, and each independent we get a covariance matrix R:  2  σ1 0 · · · 0  0 σ22 · · · 0    R= . (43) .. ..  ..  .. . . .  0 0 · · · σn2 R is the inverse of what we could call the weighting matrix R = inv (W)



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State estimation

State estimation Solving the state estimation The expected  values are: −1  xˆ1  xˆ2    T · W · H · HT · W · z = G−1 · HT · W · z ˆ x=  ...  = H | {z } G xˆn (44) ˆ x = G−1 · HT · W · (H · x + e) ˆ x=G

−1

T

· (H · W · H) ·x + G | {z }

−1

(45) T

·H ·W·e

(46)

G

ˆ z = H·ˆ x


(47)


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State estimation

State estimation: simple example

Figure: Example network



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State estimation

State estimation: simple example We want to know x1 and x2 , which are voltages U1 and U2 Two amp`eremeters measuring z1 = 9.01 A and z2 = 3.02 A U1 = 16.0233 V and U2 = 8.0367 V

Two voltmeters measuring z3 = 6.98 V and z4 = 5.01 V U1 = 15.93 V and U2 = 8.05 V

The system equations can be written as: z1 z2 z3 z4 |{z}

measurements


5 8

· x1 − 18 · x1 3 8 · x1 1 8 · x1

= = = = |

− + + + {z

1 8 5 8 1 8 3 8

· x2 · x2 · x2 · x2

true values from model


+ + + + }

e1 e2 e3 e4 |{z}

(48)

errors

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State estimation

State estimation: simple example We want to know x1 and x2 , which are voltages U1 and U2 Two amp`eremeters measuring z1 = 9.01 A and z2 = 3.02 A ⇒ Conflict

U1 = 16.0233 V and U2 = 8.0367 V

Two voltmeters measuring z3 = 6.98 V and z4 = 5.01 V ⇒ Conflict

U1 = 15.93 V and U2 = 8.05 V

The system equations can be written as: z1 z2 z3 z4 |{z}

measurements


5 8

· x1 − 18 · x1 3 8 · x1 1 8 · x1

= = = = |

− + + + {z

1 8 5 8 1 8 3 8

· x2 · x2 · x2 · x2

true values from model


+ + + + }

e1 e2 e3 e4 |{z}

(48)

errors

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State estimation

State estimation: simple example

Calculating the expected values ˆ x We take the following weighting matrix (1/sigma): W = diag([100, 100, 50, 50]) The most probable values for U1 and U2 are 16.00719 and 8.02614 resp.   0.00877  0.00456   The expected error will be: ˆ e=  −0.02596  −0.00070



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State estimation

State estimation: simple example Measurement 4 changes z4 = 4.4 instead of z4 = 5.01 The best estimate for the voltages: U1 = 15.86807 and U2 = 7.75860  0.06228  0.15438  In that case, the expected error will be: ˆ e= 0.05964 −0.49298

   

When the expected error is too high, measurements can/should be disregarded Statistical test are needed to determine when errors are “high”



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State estimation

State estimation: simple example Measurement 4 changes z4 = 4.4 instead of z4 = 5.01 The best estimate for the voltages: U1 = 15.86807 and U2 = 7.75860  0.06228  0.15438  In that case, the expected error will be: ˆ e= 0.05964 −0.49298

   

When the expected error is too high, measurements can/should be disregarded Statistical test are needed to determine when errors are “high” The weight matrix also has a serious influence on the results



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State estimation

State estimation for power flow calculations

State estimator calculates voltage magnitudes and relative phase angles of the system buses Redundancy in input data With errors on all measurement data Non-DC circuit ⇒ non-linear equations: h = h(x) Iterative solutions (as in the Newton-Raphson method) are needed The principle is the same



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State estimation

References

Power System Analysis; Grainger, John J. and Stevenson, William D., Jr. Computational Mehods for Electric Power Systems; Crow, Mariesa Power System Load Flow Analysis; Powell, Lynn



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Load flow PPT

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