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In simple language,the book providesa modernintroductionto power system operation,controland analysis. Key Features of the Third
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.PowerSystemSecurity State Estimation Powersystem compensationincludingsvs and FACTS Load Forecasting VoltageStability
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From
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lJ Nagrath
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The book is very comprehensive,well organised,uptodateand (above all) lucid and easyto follow for selfstudy.lt is ampiy illustratedw1h solved examplesfor everyconceptand technique.
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iltililttJffiillililil www.muslimengineer.info
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Modern PorroerSYstem Third Edition
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About the Authors D P Kothari is vice chancellor, vIT University, vellore. Earlier,he was Professor, Centre for Energy Studies, and Depufy Director (Administration) Indian Instituteof Technology,Delhi. He has uiro t."n the Head of the centre for Energy Studies(199597)and Principal(l gg7g8),Visvesvaraya Regional Engineeringcollege, Nagpur. Earlier lflazs: and 19g9),he was a visiting fellow at RMIT, Melbourne, Australia. He obtained his BE, ME and phD degreesfrom BITS, Pilani. A fellow of the Institution Engineers(India), prof. Kothari has published/presented450 papers in national and international journals/conferences.He has authored/coauthored more than 15 books, including Power system Engineering, Electric Machines, 2/e, power system Transients, Theory and problems of Electric Machines, 2/e., and. Basic Electrical Engineering. His researchinterestsinclude power system control, optimisation,reliability and energyconservation. I J Nagrath is Adjunct Professor,BITS Pilani and retired as professor of Electrical Engineeringand Deputy Director of Birla Institute of Technology and Science, Pilani. He obtained his BE in Electrical Engineering from the university of Rajasthanin 1951 and MS from the Unive.rity of Wi"sconsinin 1956' He has coauthoredseveral successfulbooks which include Electric Machines 2/e, Power system Engineering, signals and systems and.systems: Modelling and Analyns. He has also puulistred ,"rr.ui researchpapers in prestigiousnationaland internationaljournats.
Modern Power System Analysis Third Edition
D P Kothari Vice Chancellor VIT University Vellore Former DirectorIncharge, IIT Delhi Former Principal, VRCE, Nagpur
I J Nagrath Adjunct Professor, and Former Deputy Director, Birla Ins1i1y7"of Technologt and Science Pilani
Tata McGraw Hill Education private Limited NEWDELHI McGrawHill Offices
New Delhi Newyork St Louis San Francisco Auckland Bogot6 Caracas KualaLumpur Lisbon London Madrid Mexicocity Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto
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Preface to the Third Edition
Information contained in this work has been obtained by Tata McGrawHill, from sources believed to be reliable. However, neither Tata McGrawHill nor its authors guaranteethe accuracy or completenessof any information published hereiir, and neittier Tata McGrawHill nor its authors shall be responsiblefor any errors, omissions, or damages arising out of use of this information. This work is publi'shedwith the understandingthat Tata McGrawHill and its authorsare supplying information but are not attempting to render enginecring or other professionalservices. If such seryicesare required, the assistanceof an appropriate professional should be sought
TataMcGrawHill O 2003,1989,1980,TataMcGrtrwI{ill EducationPrivateIimited Sixteenthreprint2009 RCXCRRBFRARBQ No part of this publicationcan be reproducedin any form or by any "un, withoutthe prior writtenpermissionof the publishers This edition can be exportedfrom India only by the publishers, TataMcGraw Hill EducationPrivateLimited ISBN13: 9780070494893 ISBN 10: 0070494894 Publishedby TataMcGrawHill EducationPrivateLimited, 7 WestPatelNagat New Delhi I l0 008, typesetin TimesRomanby Script Makers, Vihar,New Delhi ll0 063 andprintedat Al8, ShopNo. 19, DDA Markct,Paschim Delhi ll0 053 GopaljeeEnterprises, Coverprinter:SDR Printcrs
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Sincethe appearanceof the secondedition in 1989,the overall energy situation has changed considerably and this has generatedgreat interest in nonconventionaland renewableenergy sources,energyconservationand management, power reforms and restructuringand distributedarrddispersedgeneration. Chapter t has been therefore,enlargedand completely rewritten. In addition, the influences of environmentalconstraintsare also discussed. The present edition, like the earlier two, is designed for a twosemester course at the undergraduatelevel or for firstsemesterpostgraduatestudy. Modern power systemshave grown larger and spreadover larger geographical areawith many interconnections betweenneighbouringsystems.Optimal planning,operationand control of such largescalesystemsrequire advanced computerbasedtechniquesmany of which are explainedin the studentoriented and readerfriendlymannerby meansof numericalexamplesthroughout this book. Electric utility engineerswill also be benefittedby the book as it will preparethem more adequatelyto face the new challenges.The style of writing is amenableto selfstudy.'Ihe wide rangeof topicsfacilitatesversarileselection of chaptersand sectionsfbr completion in the semestertime frame. Highlights of this edition are the five new chapters.Chapter 13 deals with power system security. Contingency analysis and sensitivity factors are described.An analytical framework is developedto control bulk power systems in sucha way that securityis enhanced.Everythingseemsto have a propensity to fail. Power systemsare no exception.Power systemsecuritypracticestry to control and operatepower systemsin a defensivepostureso that the effects of theseinevitable failures are minimized. Chapter 14 is an introduction to the use of stateestimationin electric power systems.We have selectedLeast SquaresEstimationto give basic solution. External system equivalencing and treatment of bad data are also discussed. The economics of power transmissionhas always lured the planners to transmit as much power as possible through existing transmission lines. Difficulty of acquiring the right of way for new lines (the corridor crisis) has always motivated the power engineersto develop compensatorysystems. Therefore, Chapter 15 addressescompensationin power systems.Both series and shunt compensationof linqs have been thoroughly discussed.Concepts of SVS, STATCOM and FACTS havcbeen briefly introduced. Chapter 16 covers the important topic of load forecasting technique. Knowing load is absolutelyessentialfor solving any power systemproblem. Chapter 17 dealswith the important problem of voltagestability.Mathematical formulation, analysis, stateofart, future trends and challenges are discussed.
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Wl
Preracero rne lhlrd Edrtion
for powersystemanalysisare MATLAB andSIMULINK, idealprograms illustrating includedin thisbookasan appendixalongwith 18solvedexamples tem problems. The help rendered tive theiruse in solvin by Shri Sunil Bhat of VNIT, Nagpur in writing this appendix is thankfully acknowledged. Tata McGrawHill and the authors would like to thank the following reviewers of this edition: Prof. J.D. Sharma,IIT Roorkee; Prof. S.N. Tiwari, MNNIT Allahabad; Dr. M.R. Mohan, Anna University, Chennai; Prof. M.K. Deshmukh,BITS, Pilani; Dr. H.R. Seedhar,PEC, Chandigarh;Prof.P.R. Bijwe and Dr. SanjayRoy, IIT Delhi. While revising the text, we have had the benefit of valuable advice and suggestionsfrom many professors,studentsand practising engineerswho used the earlier editions of this book. All these individuals have influenced this edition.We expressour thanks and appreciationto them. We hope this support/ responsewould continue in the future also. D P Kors[m I J Nlcn+rn
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Preface to the First Mathematicalmodellingand solutionon digital computersis the only practical approach to systems analysis and planning studies for a modern day power system with its large size, complex and integrated nature. The stage has, therefore,been reachedwhere an undergraduatemust be trained in the latest techniquesof analysisof largescalepower systems.A similar needalso exists in the industry wherea practisingpower systemengineeris constantlyfacedwith the challengeof the rapidly advancingfield. This book has bedndesignedto fulfil this need by integratingthe basic principlesof power systemanalysisillustrated through the simplestsystemstructurewith analysistechniquesfor practical size systems.In this book largescalesystemanalysisfollows as a naturalextension of the basicprinciples.The form and level of someof the wellknown techniques are presented in such a manner that undergraduatescan easily grasp and appreciatethem. The book is designedfor a twosemestercourse at the undergraduatelevel. With a judicious choice of advancedtopics, some institutionsmay also frnd it useful for a first course for postgraduates. The readeris expectedto have a prior grounding in circuit theory and electrical machines. He should also have been exposed to Laplace transform, linear differential equations, optimisation techniquesand a first course in control theory. Matrix analysisis applied throughoutthe book. However, a knowledge of simple matrix operations would suffice and these are summarisedin an appendixfbr quick reference. The digital computerbeing an indispensabletool for power systemanalysis, computationalalgorithms for various systemstudiessuch as load flow, fault level analysis,stability, etc. have been included at appropriateplacesin the book. It is suggestedthat where computerfacilities exist, studentsshouldbe encouraged to build computer programs for these studies using the algorithms provided. Further, the students can be asked to pool the various programs for more advancedand sophisticatedstudies,e.g. optimal scheduling.An importantnovel featureof the book is the inclusion of the latestand practicallyuseful topics like unit commitment, generation reliability, optimal thermal scheduling,optimal hydrothermalschedulingand decoupledload flow in a text which is primarily meantfor undergraduates. The introductory chapter contains a discussion on various methods of electricalenergygenerationand their technoeconomic comparison.A glimpse is given into the future of electricalenergy.The readeris alsoexposedto the Indian power scenariowith facts and figures. Chapters2 and3 give the transmissionline parametersand theseare included for the sakeof completnessof the text. Chapter4 on the representation of power systemcomponentsgives the steadystatemodelsof the synchronous machineand the circuit modelsof compositepower systemsalong with the per unit method.
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
Contents
preface ro rhe Frrst Edition W Chapter5 deals with the performanceof transmissionlines. The load flow problem is introducedright at this stagethroughthe simple twobus systemand basicconceptsof watt and var control are illustrated.A brief treatmentof circle concept of load flow and line compensation. ABCD constants are generally well covered in the circuit theory course and are, therefore, relegated to an appendix. Chapter 6 gives power network modelling and load flow analysis, while Chapter 7 gives optimal system operation with both approximate and rigorous treatment. Chapter 8 deals with load frequency control wherein both conventional and modern control approaches have been adopted for analysis and design. Voltage control is briefly discussed. Chapters 9l l discuss fault studies (abnormal system operation). The synchronous machine model for transient studies is heuristically introduced to the reader. Chapter l2 emphasisesthe concepts of various types
(kV)?x 1,000
BasecurrentIr
n^n
(MVA)r, ototo (MVA)a' n'*' andtV base when MVA baseis changecltrom nt* p"t unit impedancefrom Eq' s changedfrom (kv)r, orotJ (kV)n, new'the .4.9) is given bY
Base kilovoltamperes= (kVA)B
_
, i
=z(pu)o.,. Z(pu)n"* ffi"ffi
or
Base current 1,
12
transformer of singlephase (a) '' Representation neglected) impedance (mignetizing
l'ooox(kv)u J3rB
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(4.e)
transformer circuitof singlephase (b) Perunitequivalent Fig. 4.7
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W
todern
I
Power Srrstem Analrrsie Y
vr
v.vr.;
, rr rqt
ystr
Representationof Power System Uomponents
I
Therefore 'I 2 8
= a (as (VA)a is common) Vrn z 1,,
LaD
.
Vza


1,,
(4.1i b) (4.11c)
From Fig. 4.7a we can write Vz=(VtIflp)alrZ,
(4.12)
We shall convertEq. (4.12) into per unit form Vz(pu)Vzn= [Vr(pu) Vw  I r(pu)I 6Zo(pu)Zru]a Ir(pu)IruZ,(pu)Z* Dividing by vzn throughout and using base relations (4. rra, b, c), we get Vz(pu)= Vr(pu)  I,(pu)Zr(pu)_ Ir(pu)Z,(pu) (4.13)
Now or
+=+=, Ir
12
I'o
I'u
I r ( p u ) = 1 2 ( p u ) =1 ( p u ) Equation (4.13) can thereforebe written as
[email protected])= Vr(pu) (pu)Z(pu)
(4.r4)
where
Z(pu)=Z,,(pu)+ Z,(pu) Equation (4.I4) can be representedby the simple equivalent circuit of Fig. 4'7b which doesnot requirean idealtransfurmet.Coniia.rable simplification has thereforebeenachievedby the per unit method with a common voltampere base and voltagebaseson the two sidesin the ratio of transformation. Z(pu) can be determineddirectlyfrom the equivalentimpedanceon primary or secondaryside of a transformerby using the appropriaie impedancl base. On primary side: Zr=Zp+ Z,/a2
Z ( p' = u+ ) :!+L*I zru
But
Zro
z,o"a2
a2Ztn = Zzr
Zr(pu)= Zo(pu)+ Z,(pu) Z(pu) On secondary side: zz= Z, + o2zo
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(4.1s)
zz(pu)= ' +++o'1' zzu zB
[,.t'ff.
l
zru
Thus the per unit impedanceof a transformeris the samewhethercomputed from primary or secondaryside so long as the voltage baseson the two sides are in the ratio of transformation(equivalentper phaseratio of a threephase transformer which is the same as the ratio of linetoline voltage rating). The pu transformerimpedanceof a threephasetransformeris conveniently obtained by direct use of threephaseMVA base and linetoline kV base in relation (4.9). Any other impedanceon either side of a transformeris converted to pu value just like Zo or Zr. Per Unit Impedance
Diagram of a Power System
From a oneline diagramof a power systemwe can directly draw the impedance diagramby following the stepsgiven below: 1. Choosean appropriatecommon MVA (or kVA) basefor the system. 2. Consider the system to be divided into a number of sections by the transfbrmers.Choose an appropriatekV base in one of the sections. Calculate kV bases of other sectionsin the ratio of transformation. 3. Calculateper unit valuesof voltagesand impedancesin eqch sectionand connectthem up as per the topologyof the onelinediagram.The result is the singlephaseper unit impedancediagram. The above stepsare illustrated by the fallowing examples.
j Example4 1 Obtain the per unit impedance(reactance)diagram of the power system of Fig. 4.5. Solution The per phaseimpedancediagramof the power system of Fig. 4.5 has been drawn in Fig. 4.6. We shall make some further simplifying assumptions. 1. Line capacitanceand resistanceare neglectedso that it is representedas a series reactanceonly. 2. We shall assumethat the impedancediagram is meant for short circuit studies.Current drawn by static loadsunder short circuit conditions can be neglected.Loads A and B are thereforeignored. Let us convert all reactancesto per unit form. Choosea common threephase MVA base of 30 and a voltagebaseof 33 kV linetolineon the transmission line. Then the voltagebasein the circuit of generator1 is 11 kV linetoline and that in the circuits of generators2 and 3 is 6.2 kV. The per unit reactancesof various componentsare calculatedbelow:
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Flepresentation of power System eomponents
Transmission line:
2o.5x3o GT2
Transform er T,: TransformerTt:
Generator2:
Transformer Z :
16x30
= 0.44
0.209 x
0.22 x ff =0.++
Transformet Tr:
1 . 6x 3 0
u t7 = o'396
1 . 2x 3 0
tazr Generator3:
= 0.564
l!,?I l_0_= 0 . 4 1 8
aiT Generator1:
I iliffij IExample 4.I, we now calculatethe pu values of the reactancesof transfonners and generatorsas per relation (4.10):
Generator2:
The reactancediagramof the systemis shown in Fig. 4.g.
(ll)'
U000._J
0.44
I
0 . 4 1 3r *
I)
*(6'612 =0.936 6.2)'
(6'6)1 = 0.431 0.3214 * i9 x /.r 6.21' obviously these values are the same as obtained already in Exampre 4.r. Generator3:
{)frL_/XfX) Reactivepower(pu)lagging
Leading
Fig. 4.33 operatingchartfor rargesynchronous generator , Considera point h onthe theoreticallimit on the lETl= 1.0 pu excitations arc, the pcrwerMh is reducedby 0.1 pu to Mk; the op"ruiing point must, however, still be on rhe on the desiredlimiting curve. This is repeatedfor other excitationsgiving the curve afg. The completeworking area.shown shaded.is gfabcde. ,{ working point placed within this area at once defines the MVA, Mw, MVAR, current, power factor and excitation.The load angle 6 canbe measuredas shown in the figure. 4.7
REPRESENTATION
OF LOADS
Load drawnby consutnersis the toughestparameterto assess scientifically.The magniiudeof the ioad, in iact, changescontinuouslyso that the load forecasting problern is truly a statistical one. A typical daily load curve is shown in Fig' 1.1. The loads are generally composed of industrial and domestic components.An industrial load consistsmainly of largethreephase induction nlotors with sulficient load constancyand predictableduty .y.lr, whereas the domestic lclad mainly consists of lighting, heating and many single_phase devicesusedin a randomway by houscholders. The designun6 upJrotionof power systemsboth economically and electrically are greatly influenced by ttrp nature and magnitude of loads.
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
.sa I ._IaZ I
Modern Power SvstemAnatrrsic
Representation of p
In representationof loads for various systemstudies such as load flow and stability studies,it is essentialto know the variationof real and reactivepower with variation of voltage. Normatly in such studies the load is of composite nature with both industrialand domesticcomponents.A typical composition of Induction motors 55757o Synchronousmotors 5:75Vo Lighting and heating 2030Vo Though it is always better to consider the PV and QV characteristicsof each of these loads for simulation, the analytic treatment would be very cumbersome and complicated. In most of the analytical work one of the following three ways of load representationis used. (i) Constant
BasekV = 24, lineto_line Load volt
)L
ASa= i
Full load (MVA) = I pu, 0.9 pf lagging Load current = generatorcurrent
Power Representation
Current
Io= 7 pu, 0.9 pf lagging = 0.9  7 0.436 pu (a) Excitation emf (see Fig. 4.Ig)
Representation
Ef = V,+ j XJ"
Here the load current is given by Eq. (4.17), i.e.
= 1 1 0 " + j 1 . 3 4 4( 0 . 9  j 0 . 4 3 6 )
"'l(60) I' = P : i Q  t n V{< where V = lVl 16and 0= tanl QlP is the power factor angle. It is known as constantcurrent representationbecausethe magnitudeof current is regarded as constantin the .study. (iii) Constant
Impedance
= 1 . 5 8 6 j l . 2 l = 1 9 9 1 3 7 . 1 " E, (actual) = 1.99 x 24 = 47.76 kV (line) 6= 3j. 1" ( leading) (b) Reactive power drawn by load
Representation
Q = VJ,, sin r/
This is quite olten used in stability studies.The load specified in MW and MVAR at nominal voltageis used ro compurethe load impedance(Eq. (4.?2b)). Thus I "
z = !I:  P w =* J O 
= 1 pu
Synchronous reactance X," = +# = 1.344  ' r  pu (24)z
This is usedin load flow studies.Both the specifiedMW and MVAR are taken to be constant. (ii) Constant
JlfZJ.t
Sotution BaseMV A = 645,3_phase
= 1 x I x 0.436 = 0.436 pu or 0.436 x M5 = 281 MVAR
l v l 2 t PJQ:T
which then is regarded as constant throughout the study. f ll Fvattrt F^sr..lsrv I
li
.l a e zrv
I I
t


T
A synchronousgeneratoris rated 645 MVA , 24 kv,0.9 pf lagging. It has a syrrchronous reactancel.z o. The generatoris feeding full loadat 0.9 pf lagging at rated voltage.Calculate: (a) Excitation emf (E1) andpower angle 6 (b) Reactivepower drawn by the load Carry out calculationsin pu form and convert the result to actual values.
The generator of Example 4.3 is carrying full load at rated voltage but its excitationemf is (i) increasedby 20vo and (ii) reducedby 20vo. Calculatein each case (a) load pf (b) reactive power drawn by load (c) load angle 6 Solution Full load, P _ 1 x 0 . 9 = 0 . 9 p u Ef=
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r.99
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lM
tation of Power
ftrodernPower System Analysis
I
(b)
V,=7 (i)
Et is
x 0.9 x sin 1.5= 0.024
or
d by 20Voat samereal load.Now
Q = 0.024x 645= 15.2MVAR
As per Eq. (a.28) P=
l E', l l v', l
x,
0.9= (2'388xr) ,i' d \ 1.344) sin d= 0.5065
or
6 = 30.4"
or
2.388130.4"110" jr.344 = 0 . 8 9 j 0 . 7 9 = 1 . 1 8 3l 
PROB IEIvIS
(i)
sind
4'r Figure P4'l shows the schematic diagram of a radial transmission system' The ratings and reactancesof the various componentsare shown rherein.A load of 60 MW at 0.9power factor ragging is tappedfrom the 66 kv substation which is to be mainraineo alt oo kv. calcurate rhe terminal voltage of the synchronousmachine. Representthe transmission line and the transforrnersby series reactances ontu. . 11t220kv 220t66kV
kv
4L2" V1
(a) pf = cos 4I.2" = 0.75 lagging (b)
=1x1.183x0.659 = 0.78 pu or 502.8 MVAR E, decreasedby 20o/oor E f = 1 . 9 9x 0 . 8 = 1 . 5 9 Substitutingin Eq. (i)
o e =( L 2 \ L ) , i n 6 \ 1.344 )
4.2 Draw the pu inrpedancediagram fbr the power system shown in nig. e4.2. Neglectresistance,and usea baseof ioo vrre , 220kv in 50 () rine. The ratings of the generator,motor and transformers are Generator40 MVA, 25 kV, Xu = 20Vo Motor 50 MVA, I I kV, X,t = 30Vo YIltransformer, 40 MVA, 33 y_220 y kV, X = I5Vo Yl transformer,30 MVA, ll L_220 y kV, X = l5*o
,50o UvGir+1,__iFrF
.l
l   Y Y I
t.59149,5"rlo" .i1.344 =0.9j0.024
In=
= 0.9 l1.5" (a) pf = cos 1.5" = 1; unity pf
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.*fF., [y^
which gives 6  49.5"
60 MW 0.9pf tagging
Fig. p_4.1
Reactive power drawn bY load Q = lV,lllolsin /
(ii)
100MVA X = 10o/o
160 dh ___EF_l___> 5F  100MVA V2 X = Bo/o
1f 1 \M ) 'Y l_ 2
) /
I
Fig. p4.2
4 . 3 A synchronousgenerator is rated 60 MVA, 1r kv. It has a resistance Ro = 0l pu and xo 7 r.65 pu. It is feedinginto an infinite bus bar ar 11 kV delivering a cirrrent 3.15 kA at 0.g ff hgging. (a) Determine E, and angte d (b) Draw a phasor diagram for this operation. (c) Bus bar voltage fails to 10 kv while the mechanicalpower input to generator and its excitation remains unchanged. wtrat is the value and pf of the current delivered to the bus. In this case assumethe
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
I
126 
ModernPower SystemAnalysis
T
generatorresistanceto be negligible. 4.4 A 250 MVA, 16 kV ratedgeneratoris feedinginto an infinite bus bar at 15 kV. The generator hasa synchronous reactance of 1.62pu.lt is found that the machine excitation and mechanicalpower input are adiustedto give E, = 24 kY and power angle 6 = 30o. (a) Determine the line current and active and reactivepowers fed to the bus bars. (b) The mechanicalpower input to the generatoris increasedby 20Vo from that in part (a) but its excitation is not changed.Find the new line current and power factor. (c) With referenceto part (a) current is to be reduced by 20Voat the same power factor by adjusting mechanical power input to the generatorand its excitation. Determine Ey, 6 and mechanicalpower rnput. (d) With the reducedcurrent as in part (c), the power is to be delivered to bus bars at unity pf, what are ttre correspondingvalues of El and d and also the rnechanicalpower input to the generator. 4.5 The generatorof Problern4.4 is feeding 150 MVA at 0.85 pf lagging to infinite bus bar at 15 kV. (a) Determine Ey and d for the above operation.What are P and Q fed to the bus bars? (b) Now E, is reducedby l0o/okeeping mechanicalinput to generator same, find new dand Q delivered. (c) Et is now maintainedas in part (a) but mechanicalpower input to generatoris adjustedtill Q = 0. Find new d and P. (d) For the value of Eyin part (a) what is the maximum Qthat can be deliveredto bus bar. What is the corresponding6and {,? Sketchthe phasor diagram for each part.
Representationoflgygr_gyg!"_!L_qg[p!g$q

REFERE I.ICES Books l' Nagrath,I'J. and D.P Kothari, Electric Machines,2nd edn Tata McGrawHill, New Delhi, 7997. 2' van E' Mablekos,Electric MachineTheory PowerEngineers,Harperlno for Raw, New York, 1980. 3. Delroro, v., Electric Machines and power systems,prentice_Hall, Inc., New Jersey,1985. 4' Kothari, D.P. and I.J. Nagrath, Theory and.Problems of Electric Machines, 2nd Edn, Tata McGrawHill, New De\hi, 2002. 5. Kothari, D.p. and I.J. Nagrath, Basic Electicar Engineering, 2nd Edn., Tata McGrawHill, New Delhi. 2002. Paper 6' IEEE CornitteeReport,"The Effect of Frequencyand Voltage on power System Load", Presentedat IEEE winter po,ver Meeting,New york, 1966.
Answers 4.1 12kV 4.3 (a) 26.8 kV (line),42.3" leading (c ) i .i 3 l 2 8 .8 " k A; 0 .8 7 6l a g 4.4 (a) 0.5ll"l 25.6"kA; 108 MW, 51.15MVAR (b) 6.14 kA, 0.908lagging ( c ) 1 . 5 7 8 ,1 3 . 5 o5, 3 . 3 M W ( d ) 1 8 . 3 7k v , ' 3 5 . 5 " ,9 6 M W 4. 5 (a ) 2 5 .2 8 k V, 2 0 .2 ' ,1 2 7 .5MW , 79.05MV A R (b) 33.9", 54 14 MVAR (c ) 4 1 .1 " , 1 5 0 .4MW (d) 184.45,MVAR,53.6", 7 0.787 pu
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eharacteristics and Perforr"nance of povrerTransmissionLines The following nomenclaturehas been adoptedin this chapter: z = seriesimpedance/unitlength/phase y = shunt admittance/unitlength/phaseto neutral
;=;l_*#:,T.":Jff:: C = cepacitance/unitlength/phaseto neutral / = transmissionline length Z = zl = total series impedance/phase Y = ll = total shunt admittance/phaseto neutral Note: Subscript,Sstandsfor a sendingendquantityand subscriptR standsfor a receivingend quantity 5.1
INTRODUCTION 5.2
This chapter deals primarily with the characteristics and performance of transmissionlines.A problemof major importancein power systemsis the flow of load over transmissionlines such that the voltage at various nodes is maintained within specified limits. While this general interconnectedsystem problem will be dealt with in Chapter 6, attention is presently focussed on performance of a single transmissionline so as to give the reader a clear understandingof the principle involved. Transmissionlines are normally operatedwith a balancedthreephaseload; the analysiscan thereforeproceedon a per phasebasis.A transmissionline on a per phasebasis can be regardedas a twoport network, wherein the sendingend voltage Vr and current 15are related to the receivingendvoltage Vo and current 1o through ABCD constants"as
lyrl lA BlIy*l L1,J Lc plLroJ Also the following identity holds for ABCD constants: ADBC=l
(sr) /< t\
These constants can be determined easily for short and mediumlength lines by suitable approximations lumping the line impedance and shunt admittance. For long lines exact analysis has to be carried out by considering the distribution of resistance, inductance and capacitance parameters and the ABCD constants of the line are determined therefrom. Equations for power flow on a line and receiving and sendingend circle diagrams will also be developed in this chapter so that various types of end conditions can be handled. *R"f'..
SHORT TRANSMISSION
LINE
For short lines of length 100 km or less, the total 50 Hz shunt admittance* QwCl) is small enoughto be negligible resultingin the simple equivalentcircuit of Fig. 5.1.
Fig. 5.1 Equivalent circuitof a shortline This being a simple series circuit, the relationship between sendingend receivingend voltagesand currents can be immediately written as:
=I l z f l v o l Lr,J Lo rJLr.J lyrl
(5'3)
The phasor diagram for the short line is shown in Fig. 5.2 for the lagging current case.From this figure we can write lV5l = l(ly^l cos /o + lllR)z + (lV^l sin dn + lllxyzlr/2
l v 5l = [ t v R P +l l t 2 ( R z+ f )
* z t v R  i l ( R c o sQ ^ + X s i nw f{u ) r l 2 ( 5 . 4 ) It2
. rl/ 1il21n2+ x2 r l ?JlJr cos/o = tvnt r * + t / A Qp f 4#rin , I / i " "L lVRl tvRP J
; to Appenclix B.
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For overhead transmissionlines, shunt admittanceis mainly capacitive susceptance (iwcl) as the line conductance(also called leakance)is always negligible.
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Lines J"#, Characteristics and Performanceof PowerTransmission 1 l/l R cos/^+l1l X sind^ x 100 I yRl
The last term is usually of negligible order.
(s.7)
In the above derivation, Q*has been consideredpositive for a lagging load. Expanding binomially and retaining first order terms, we get
Ivsl=lv^rfr.
ff
cos /^ +
ff
per cenrregularion !!4 9"t!t:l n x tin 4r x 100 lvRl
sinf^)'''
l V 5l = l V ^ l + l X ( R c o s / o + X s i n / o ) The above equation is quite accurate for the normal load range.
(s.8)
(for leading load)
(s.s)
Voltage regulationbecomesnegative(i.e. load voltage is more than no load voltage), when in Eq. (5.8) X sin Qo> R cos /p, or tan + X It also follows from Eq. (5.8) that for zero voltage regulation
tan/n= i.e.,
X
/^(teading= [
=cot d e
(5.e)
where d is the angle of the transmission line impedance. This is, however, an approximate condition. The exact condition for zero regulation is determinedas follows:
Fig. 5.2 Phasordiagramof a shortlinefor laggingcurrent Voltage
Regulation
Voltage regulationof a transmissionline is defined as the rise in voltage at the receivingend,expressedas percentageof full load voltage, when full load at a specifiedpower factor is thrown off, i.e. Per cent regutation= where
''^?):'Yu'x lvRLl
100
lVool= magnitude of no load receivingendvoltage lVprl = magnitude of full load receivingendvoltage (at a specified power factor)
= lVsl, lVsl = lVpl For short line, lV^61 Per cent regutation=
'u1,'%'
(5.6)
condition Fig. 5.3 Phasordiagramunderzeroregulation Figure 5.3 shows the phasor diagram under conditions of zero voltage regulation, i.e. lV5I = lVpl or
O C= O A AD  AClz _ llllzl sn /AOD _ zlvRl lyR oA
lvRl
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and Performanceof PowerTransmissionLines Characteristics
', or
IAOD= sinrU!4 zlvRl lt follows from the geometryof anglesat A, that for zero voltaeeregulation, /p (leading) =
or
new value of lVr  = 11.09 kV
Figure 5.4 showsthe equivalentcircuit of the line with a capacitivereactance placed in parallel with the load. R+jx
(5.10)
From the above discussion it is seen that the voltage regulation 6f a line is heavily dependent upon load power factor. voltage regulation improves (decreases) as the power factor of a lagging load is increased and it becomes zero at a leading power factor given by Eq. (5.10).
A singlephase50 Hz generatorsuppliesan inductive load of 5,000 kw at a power factor of 0'707 lagging by means of an overheadtransmissionline 20 km long. The line resistanceand inductanceare 0.0195ohm and 0.63 mH per km' The voltageat the receivingendis required to be kept constantat 10
kv.
Find (a) the sendingendvoltage and voltage regulation of the line; (b) the value of the capacitorsto be placed in parailet viittr the load such that the regulation is reducedto 50vo of that obtained in part (a); and (c) compare the transmissionefficiency in parrs (a) and (b). Solution The line constantsare R = 0 .0 1 9 5x 2 0 = 0.39 f)
l
l V 5l =
l V o l + l 1 l ( R c o sQ * + X s i n / ^ ) = 1 0 ,0 0 0+ 7 0 7(0.39x 0.701 + 3.96 x 0.707,\y = 72.175kV
Voltage regulation= pfTL l9 x roo :Zt.j7vo 10 (b) Voltage regulariondesired= ?+t
!
,L

l
Fig. 5.4 Assuming cos y''^now to be the power factor of load and capacitive reactance taken together, we can write
(11.09 10) x 103= l1n ( R cos d^+ X sin dn) Since the capacitancedoes not draw any real power, we have l/ol=
5000
(ii)
and l l a l = 5 4 9A Now Ic= In I = 549( 0. 911 j0'412) 707( 0. 107  j0. 70'7) = 0.29 + j273'7 Note that the real part of 0.29 appearsdue the approximationin (i) Ignoring it, we have I, = j273.7 A
' or (c)
'Y ^L
:ltl loxlooo 3wxc
ll. I
273'7
C81 P'F Efficiency of transmission;
= l0.9Vo Case (a)
lyst 10 = 0.109 t0
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(i)
10 x cos/^ Solving Eqs. (i) and (ii), we get cos dn= 0'911 lagging
(a) This is the case of a short line with I = Ia= 1, given by
From Eq. (5.5),
l
l
X = 3I4 x 0.63 x 103x Z0 = 3.96 e
l 1 l=   5 0 0 0 = 7 0 7 A 1 0 x 0 .7 0 i
133
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characteristicsand Pedormanceof Power TransmissionLines LI{S Case (b)
f*
Per unit transformerimpedance,
r/= '
5000  g7.7%o ' 5000+ (549)2 x 0.39x 103
/u Lrr6f.LuJ prdvrtg
, uapacltor
ln parallel
wlth
Zr= the load,
the
receivingendpower factor improves (from 0.707 iug to 0.911 lag), the line current reduces(from 707 A to 549 A), the line voitage regulationdecreases (one half the previous value) and the transmission iproves (from "ffi"i"nJy 96'2 to 97'7vo)' Adding capacitorsin parallel with load is a powerful method of improving the performanceof a transmissionsystem and will be discussed further towards the end of this chapter.
A substationas shown in Fig. 5.5 receives5 MVA at 6 kv, 0.g5lagging power factor on the low voltage side of a transforner from a power stationthrough a cablehaving per phaseresistanceandreactanceof 8 and 2.5 ohms,respectively. Identical 6.6/33 kV transfoffnersare installed at each end of the line. The 6'6 kV side of the transfonnersis delta connectedwhile the 33 kV side is star connected.The resistanceand reactanceof the star connected windings are 0.5 and 3'75 ohms, respectivelyand for the delta connected windings arJ0.06 and 0.36 ohms. what is the voltage at the bus at the power station end?
6.6/33kV
33/6.6kV
Fig. 5.5 Solution lt is convenient here to employ the per unit method. Let us choose, BaseMVA = 5 Base kV = 6.6 on low voltageside = 33 on high voltage side Cabie impeciance= (8 + jZ.S) e/phase
_ ( 8 + r 2 . s ): x s= (0'037+ io'0r15)Pu (33)' Equivalent star impedance of 6.6 kv winding of the transformer
=
(O.OU = (0.02+ /0.12) O/phase + 7O.36) f
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( 0 . 0 2 +j 0 . 1 2 ) x 5 ( 0 . 5 + 1 3 . 7 5 ) x 5 _ 6.q2 Q'2
= (0.0046+ j0.030) pu Total seriesimpedance= (0.037 + j0.0115)+ 2(0.0046+ j0.030) = (0.046 + j0.072) pu Given: Load MVA = 1 pu = 0. 91 pu Loadvolt age= + 6. 6 Load current= .1 = 1.1 pu 0.91 Using Eq. (5.5),we get lVs = 0. 91 + 1. 1( 0. 046x 0. 85 + 0. 072x 0. 527) = 0.995 pu = 0.995 x 6.6  6.57 kV (linetoline)
Input to a singlephase short line shown in Fig. 5.6 is 2,000kw at 0.8 lagging power factor. The line has a seriesimpedanceof (0.4 + j0.a) ohms. If the load voltage is 3 kV, find the load and receivingendpower f'actor.Also tind the supply voltage. I
I
2,000 kw I at 0.8 pf *Vs
lassins
3kv
L
__1
I
Fig. 5.6 Solution It is a problem with mixedend conditionsload voltage and input power are specified.The exact solution is outlined below: Sendingendactive/reactivepower = receivingendactive/reactivepower + activekeactiveline losses For active power l ys I l1l cos ds= lVRllll cos / a + l1l2p For reactivepower
( i)
l y s I l / l s i n g 5 5 =l V p ll 1 l s i n Q o + l t l 2 X
(ii)
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Modern PowerSystem Anatysis
'F6, I
Squaring(i) and (ii), adding and simplifying, we get lvrl2 lll2 = lVnlzlll2 + zlvRl lll2 (l1lR cos
5.3 /o
+ tItX sh /n) + tlta @2 + f)
(iii)
For lines morethan 100 km long, chargingcurrentsdueto shuntadmittance 100 km to 250 km leneth. it is line admittance at the receivingend the all to lump accurate sufficiently resulting in the equivalent diagram shown in Fig. 5.7. Starting frorn fundamental circuit equations,it is fairly straightforward to write the transmissionline equationsin the ABCD constantform given below:
the numerical values given l Z l 2 =( R 2 +f ) = 0 . 3 2 lysl l1l
Lines l. l3?.,{ of PowerTransmission and Performance Characteristics IMEDIUM TRANSMISSION I,INE
2,ooox1o3  2 , 5 0 0x 1 0 3 0.8
[:]=l'*;'llVl
lVs I l1l cos /,  2,000 x 103
( s . 11 )
l y s l l l l s i n / 5 = 2 , 5 0 0x 1 0 3x 0 . 6 = 1 , 5 0 0x 1 0 3 From Eqs. (i) and (ii), we get 2 0 0 0 x 1 0 3 0 . 4 l t P ,n l 1 cl o s P o = F
(iv) Fig. 5.7 Mediumline,localizedloadendcapacitance
1 5 0 0 x 1 0 30 . 4 t I f 3000
l1l sin /o
Substitutingall the known valuesin Eq. (iii), we have
p'{lt (2,500x 103;2 = (3,000)'til2+ 2 x 3,000t210.4*29W"19! 3000 L 1s00xlq1r0.4112 +0.4x l+ o.zztt ta 3000 J
(v)
Nominalf
Representation
If all the shunt capacitanceis lumped at the middle of the line, it leads to the nominalZ circuit shown in Fig. 5.8.
Simplifying, we get 0 . 3 2 V f  1 1 . 8x 1 0 6l l l 2 + 6 . 2 5 x 1 0 1 2= 0 which upon solution yields t_725 A Substitutingfor l1l in Eq. (iv), we ger cos Qp= 0.82 Load Pn = lVRl lll cos /a = 3,000 x 725 x 0.82 = 1,790kW Now l V 5  = l 1 l c o s d s = 2 ,000 l V 5l =
2000 :3 .44 kY
Fig. 5.8 Mediumline,nominalTrepresentation For the nominalZ circuit, the following circuit equationscan be written: Vc= Vn+ I o( Zl2) Is = In + VrY = In r Wo+
IR(ZL)Y
Vs = Vc + it (ZiZ) Substitutingfor Vg and 1, in the last equation, we get
YvR) . vs= vn+ I^ (zt2)+ (zD) [r^(t +)+ = v n ( , . t { ) + rn z ( t . t f )
7 2 5 x 0 .8
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l.
r
Characteristicsand Performanceof Power Tralq4lqslon Lines [l$k t
Rearrangingthe results, we get the following equations
+
MVA at 0.8 lagging power factor to a balancedload at 132 kV. The line conductorsare spacedequilaterally3 m apart. The conductorresistanceis 0.11 ohmlkm and its effective diameteris 1.6 cm. Neglect leakance.
(s.12 2 = 0.0094 pFlkrrr
Nominal zr Representation In this methoc the total line capacitanceis divided into two equal parts which are lumped at the sending and receivingendsresulting in the nominal zr representation as shown in Fie. 5.9.
R = 0. 11 x 250 = 27. 5 C) X  Zr f L = 2r x
50 x I . 24 x 10 3x 250 = 97. 4 Q
Z= R + jX = 27.5 + j97^4 = I0I.2 174.T Q Y = jutl
= 314 x 0.0094 x 10{ x 250 lg0"
= 7.38 x 104 lW
li6.g"o = 109.3I 369" A r^ = #9 "13xl32 vo (per phase)= (I32/d, ) 10" = 76.2 l0 kv
Fig. S.9 Mediumline,nominal_7r representation From Fig. 5.9, we have
/
_ t Is= In + VoY + Vs=
tvoy
+
\
\ 2 ) " r 1 I i +: x 7.38x ioa 190"x10r.2134.2"\to.z \ 2 ) + 101.2174.2" x 109.3x 103l36.9"
t;trr(t+vz)
v o v( t + t ^ v z ) . + ( , +
1
v s =[ t + + Y Z l v R +z I *
l 2Vsy
vo+ eo*,)vov>z = vn(r.*
1s=r**
U
7 6 . 2+ 2 . 8 5l 1 & . 2 ' + 1 1 . 0 61 3 7 . 3 " 82.26+ j7.48  825 15.2" lV5 (line)= 82.6xJl  143 kv
)vz)
I + L Y Z = 1 + 0.0374ll&.2" = 0.964+ 70.01 2
Finally, we have
'r i4 't ', I ('*  f /
1
\
/
z l,u.r ., ll "  
= (rine noload) rv^or (5.13)
LrU+;rt) [t*r")]L1nr It slhou uldJ bt el noted d tha at nominalzand nominalrrwith the aboveconstantsare not equ vale3nt to e each t other. The readershould verify this fact by applying star_ lurva delt:a tr trans nsfcormati< ion to t either one.
Using the nominaiz method, find the sendingendvoltage and voltage regulation of a 250 km, threephase,50 Hz, transmission line delivering 25
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= 148.3 kv ffid:;# t z l
Voltage regulation = W#2
5.4
x 100  l2.3vo
THE LONG TRANSMISSION LINERIGOROUS
SOLUTION
For lines over 250 km, the fact that the parametersof a line are not lumped but distributeduniformally throughoutits length, must be considered.
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mooetnPo*", Sv
149 1
sis
characteristicsand Performanceof Power TransmissionLines
I
C17e)r'Czle)'21., 9t ,r, C, ,',', Z, Z,
(5.1e)
Fig. 5.10 Schematicdiagramof a longline
(5.20)
Figure 5.10 shows one phase and the neutral return (of zero impedance) of a transmission line. Let dx be an elemental section of the line at u dirtunr. " from the receivingend having a series impedance zdx and a shunt admittance ydr. The rise in voltage* to neutral over the elemental section in the direction of increasing "r is dV". We can write the following differential relationships acrossthe elemental section:
dVx = Irzdx o, Y
= ZI,
The constantsC, and C2 may be evaluatedby using the end conditions, i.e. when .r = 0, Vr= Vn and 1r= In. Substitutingthese values in Eqs. (5.17) and (5.19) gives Vn= Cr + Cz
(ct cz)
r^=!
Lc
(s.14)
which upon solving yield
d,lx= v*ldx o,
!1'
= yvx
(s.1s)
It may be noticed that the kind of connection (e.g. T or r) assumedfor the elemental section, does not affect these first order differential relations. Differentiating Eq. (5.14) with respecrro tr,we obtain drv, dT =
Ea
i''
= rZv'
(5.16)
This is a linear differentiai equation whose general solution can be written as follows: V*=CpI**Cre1x
(s.r7)
where
7= ,lW
Cr=
*
(vn+zJn)
1
2Un
(s.18)
and C, and C, are arbitraryconstantsto be evaluated. DifferentiatingEq. (5.17)with respectto x:
,,=(Yn+/o),,. *(h.? ),,. ,._(bITk),,._(w*),,.
Here V' is the complex expressionof the rms voltage, whose magnitude and phase vary with distance along the line.
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(s.2r)
Here Z, is called the charqcteristic impedanceof the line and 7is called the propagatton constant. Knowing vp, In and the parametersof the line, using Eq. (5.21) complex number rms values of I/, and I, at any distance x along the line can be easily found out. A more convenientform of expressionfbr voltageand currentis obtainedby introducing hyperbolic functions. Rearranging Eq. (5.21), we get ( o7* +
o7'\
v*=vnt+l+ \
*_
ZJ*)
with 9r ant c, as determinedabove,Eqs. (5.r7) and (5.19) yield the solutionfor V.and 1. as
dI.
Substituting the valueof + from Eq. (5.15),we ger dx d2v
,r=
I. = VoL( " " 2 " \ "*
2
.t" 2
/
/
+
r"^' 2( , ( e : J : : ) 2
)* ," ( "\ )
e1' + et' \ 2 )
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)
These can be rewritten after intloducing hyperbolic functions, as Vr= Vn cosh 1r + I^2, sinh 1r
Characteristics and Performance of PowerTransmission Lines ffi 
sinh 7/ .+*4* 3!
(5,22)
..=Jyz(H+)
)!
\^'
\(s.28a)
6.)
This seriesconvergesrapidly for values of 7t usually encountered for power r,Lqrrrrrnwio.l.r
I,= Io cosh rr + V l.i
lines
and
can
he convenienflw
^l^i^.^

expressionsfor ABCD constanis art when x = l, Vr= V, Ir= Is
':::i;:l;;t =l;::,, Hl
(s.23)
Bx,z (t.+)
Here A=D cosh 7/ B = Z, sinh 7/
(s.24)
\ 6 ) The above approximation is computationally convenient and quite accuratefor lines up to 400/500 km. Method
case [vs 1s] is known, fvn Inl can be easily found by inverting Eq. (5.23).
:l =li, Hl [:]
Evaluation
(s.2s)
of ABCD Constants
7 a+ jp
(s.26)
The hyperbolic function of.complex numbers involved in evaluating ABCD constantscan be computedby any one of the three methods given uJtow. Method
3
cosh (o/ + i7t) 
tat"iot +'?t'iot _=
sinh (o/ + jpl) 
,at"i0t _e:deipt
5.5
The ABCD constantsof a long line can be evaluatedfiorn the results given in F4. 6.24). It must be noted that 7 Jw is in generala complex number and can be expressedas ^
(s.28b)
c x Y(r*V\
c = Jsinh :r/ Z, *In
YZ 2
A=D=l*
Z
INTERPRETATION
1
;(e"
tpt + e"1 lBt1
(s.2e) :
){r",
tpt  td lBt1
OF THE LONG LINE EOUATIONS
As alreadysaid in Eq. (5.26), 7is a comprexnumber which can be expressed as 7= a+ jp The real part a is called the attenuation constantand the imaginary part ^ Bis called the phase constant.Now v, of Eq. (5.2r) can be writtin as Z,lRlr.,,ritgx+n V^, = I V R + *lVnZ,lnl

I
2


z
oc,x,tt/,xe2)
I'
where cosh (cr./+ j1l) = cosh ul cos gl+ j sinh a/ sn pt
(5.27)
sinh (a/ + jQl) = sinh al cos gl + j cosh a/ sn pt Note that sinh, cosh, sin and cos of real numbers as in Eq. (5.27) can be looked tp in standard tables. Method
Qr= I (Izn+ I&,)
v* (t) = xd Ol! + 3/tl L t 2
2
(5.30)
dz= I (Vn InZ,) The instantaneous voltagev*(t) canbe writtenfrom Eq. (5.30)as ,o' ,
[email protected]+r,,.+o,) l
cosh 7/=r ++* #*. .=(t.+) www.muslimengineer.info
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nrodern Power System Analysis
144 
 ZrI
characteristics and Performanceof PowerTransmissionLines
'1
*l ^* j(at/tx+t , nlV^ +Jlltrle*ett''tLtrh)
(5.31)
I
voltageconsistsof two termseachof which is a function The instantaneous of two variablestime and distance.Thus they representtwo travelling waves, Vx=
Vrl *
(s.32)
VxZ
 A t ( f+ 4 0 Sendingendx=/
Now
 JIlh?\,* "',,
gx+ h) "or{'l+
(5.33)
At any instant of time t, v.rl is sinusoidally distributed along the distance from the receivingendwith amplitude increasingexponentially with distance, as shown in Fig. 5.11 (a > 0 for a line having resistance). Envelop eox
Fig. 5.12 Reflectedwave If the load impedance Zr = +
IR
.
urAf
B x=0Receiving end ' ' Direction of travellingwave
Sendingendx=/
w ave F i g . 5 .1 1 l nci dent After timeAt, the distribution advancesin distancephaseby (u'Atlfl. Thus this wave is travellingtowarclsthe receivingendand is the incidentv'ave'Line lossescauseits arnplitucleto decreaseexponentiallyin gclingl}onr tht: sendirlg to the receivingend. Now
u,r=Elt+tlu^
 0x+dz) (az cos
(s.34)
After time At the voltage distribrrtionretardsin distancephaseby (uAtl4. This is the reflecterlwave travelling trom the receivingendto the sendingend with amplitudedecreasingexponentiallyin going from the receivingendto the as shownin Fig' 5.12. sendingend, point along the line, the voltage is the sum of incident and reflected At any voltage waves presentat the point tEq. (5.32)1.The same is true of current waves.Expressionsfor incident and reflected current waves can be similarly written down by proceedingfrom Eq. (5.21).If Z" is pure resistance,current wavescan be simply obtainedfrom voltage wavesby dividingby Zr.
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= 2,, i.e. the line is terminated in its
characteristic impedance,the reflected voltage wave is zero (vn zJn= 0). A line terminatedin its characteristicimpedanceis called the infinite line. The incident wave under this condition cannot distinguish betweena termination and an infinite continuation of the line. Power systemengineersnormally caII Zrthe surge impedance.It hasa value of about 400 ohms for an overhead line and its phase angle normally varies from 0" to  15o.For undergroundcables Z. is roughly onetenthof the value for overheadlines. The term surge impedanceis, however, used ih connection with surges(dueto lightningor switching)or transmission lines.wherethe lines loss can be neglectedsuch that
z, = z,=
(
;,.,1
l:i:r)
rl/2
i/1
(;)"'. ,, purcrcsist''cc
SLtrge Impetlance Loading (SIL) of a transmission line is.defined as the power delivered by a line to purely resistive load equal in value to the surge impedance of the line. Thus for a line having 400 ohms surge impedance,
SIL= JT y!tv, I x looo lvo t00okkw J3 x aoo = 2.5 lyRl2kw
(s.3s)
where lVol is the linetoline receivingendvoltagein kV. Sometirnes,it is found convenientto expressline loading in per unit of SIL, i.e. as the ratio of the power transmittedto surge impedanceloading. At any time the voltage and current vary harmonically along the linc with respectto x, the spacecoordinate.A completevoltageor currentcycle along the line correspondsto a change of 2r rad in the angular argument Bx. The correspondingline length is defined as the wavelength. It 0 i.s expressedin radlm,
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
iAd
147
ModernPo*g!Sy$e!l Anelysis
I
(s.36)
)Zn/gm line power transmission a typical for Now length)= 0 g (shuntconductancelunit
*fu=
oflight velocitv
(s.42)
of wave along the line would be The actual velocity of the propagation of light' somewhatless than the velocity
7= (yz)1/2= Qu,C(r+ juL))rt2
 iu (LQ'''(r t
' =
., = 1" lot = 6,000km "50(usuallya few lines are much shorterthan this Practicaltransmission tlrau'ttin
i)'''
pointed out heyethat the vyaves hundredkilometres). It needsto be Figs,5.11and5.]'2areforitlustrationonlyanddonotpertainnareal power transmissionline'
7= a+ jg = ju\Lc),,,(tt#) r ( C\ltz a   l  l
(s.37)
0 = a (Lq'''
(s.38)
2\L)
Now time for a phasechangeof 2n is 1f s, where/= cul2r is the frequency in cycles/s.During this time the wave travels a distanceequal to ). i.e. one wavelensth. \ ' (5.3e) Velocity of propagationof wave, , = 4=: f^ m/s "' ri f
at the line is 400 km long' The voltage A threephase50 Hz transmission = ohm/ 0.4 x are r = o.!25 ohnr/km, sending.end ts 220kV. The line parameters tm ani y = 2.8 x 106rnho/km' Find the following: (i )Thesending enclcur r ent andr ebeiving endvolt agewhent her eisno load
which is a well known result. For a losslesstransmissionline (R = 0, G = 0), ,= 7yz)'''  iu(LC)tlz such that e. = A, 0 = . (Lq'''
)2110 ,?n=,=:
1,, m
(s.40)
are: Solution The total line parameters R  0.125 x 400 = 50.0 f) X = O.4 x 400 = 160'0 fl
and
(s.41)
v = fA = ll(LC)rlz ms For a singlephasetransmissionline L=
C_
U  l'12 x IA3 lxf Y = 2.8 x 106 x 400 lg)" = 172'6" Q Z = R + i X = ( 5 0 ' 0+ j 1 6 0 ' 0 ) 1 6 8 ' 0 YZ = l.l2 x 1O3/90" x 168 172'6"
lto ,n D r' 2r
= 0.1881162'6"
2ffi0
(i)
l n D /r
v' = 4 (Pol^D
It;,.t
2*o
'lt"nG
)t/2
'
)
Since r and rt arequite close to each other, when log is taken, it is sufficiently q, = h D/r. accurateto assumethat ln
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At noload Vs= AVn' ls = CVa
A and C are comPutedas follows: /AL = 
l*
tl
Y Z= l +
l
*x
0 . 1 8 81 1 6 2 ' , 6 "
2
)
= 0.91+ j0.028
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
.r14Ef
Modern Power Svstem Analvsis Characteristics. and Performanceof Power TransmissionLines I 14g^ tsimplifying, we obtain the maximum permissiblefrequency as f = 57.9 Hz
l Al = 0 .9 1
C = Y(l + YZ/6) = = 1.09x 103I
= lvnhn' #:#
'
= 242kY
1 1 l5= l c l l V R l = 1.09 x 103xry * 103= 152A "J3 It is to be noted that under noload conditions, the receivingend voltage (242 kV) is more than the sendingendvoltage. This phenomenonis known as the Ferranti elfect and is discussedat length in Sec. 5.6. (ii) Maximum permissible noload receivingend voltage = 235 kv. " l l :2f2f0i = 0 . e 3 6 ^ ,= llV% ,r, Now
2
1 ^ = 1 + ; t', .i2.8 x 106 x (0.125+ 70.4) z
,l
th of the real part. lzll can be
,A,=
1 . 0 6 gx l 0  3
;
Vn I
Incident vcrltage
?cl n 2
V,,
vRzclR

)
VR
= 63.51 10" kV (to neurral) (b) At 200 km from the receivingend:
L( to+160,.I) s0) J0\
Neglecting the imaginary part, we can write
tAl= 1 +" r.r2x1o3 x 160 x &=
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a = 0 . 1 6 3x t O  3 ;f =
Reflected voltage 
220= 0.88 250 I Al*;xi1.l2xl03x
= ( 0. 163+ i1. 068) x t O   r 0+ j{J
 220/J3 = 63.5110" kV (to neurral) 2
r0.936 0 . 5 6x l 0 6
/ = 3 3 8k m ;;,,
From Example 5.5, we have following line parameters: r = 0.725 Qlkm; x = 0.4 Olkm; y = j2.g x 10{ Uncm z = (0.125 + j0.4) Olkm = 0.42 172.6 CI/km
For open circuit 1n= 0
l A l =   0 . 5 6x l r J 6 P= 0 . 9 3 6 p_
solution
(a) At the receivingend;
= (1  0._56x t06P1+ j0.t75 x r06P
approxirnatedas
(b) The incident and reflected voltages to neutral at 200 km from the receivingend. (c) The resultant voltage at 200 km from the receivingend. Note: Use the receivingendline to neutral voltage as reference.
y 1yz)t'2= (2.8 x 106x 0.42 /.(g0 + 72.6))t/2 = 1 . 0 8x 1 0  3l g I . 3
1 A = l + L Y Z
S in c eth c i rn a g i n a ry p a rt w i l l b c l e ssthan,
If in Example 5.5 the line is open circuited with a receivingendvoltag e of 220 kV, find the rms value and phase angle of the following: (a) The incident and reflected voltages to neutral at the receivingend.
0.88
l
/
l
Incident voltage = :!u+urltlxl 2 lr:2oorm
= 63. 51exp ( 0. 163x l0 3 x 200) x exp 01.068 x 103x 200)
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
ffi0'.I
characteristics andperformance of powerTransmission LinesI rsr / v ^ , \ u"'J andturnsrhrough a positiveanglepr (represenred by phasoroB); I i.
ModernPo*e, SystemAnatysis
f
= 65.62 112.2" kV (to neutral) Reflectedvoltage
Ys"'uritt'l
while the reflected voltage wave decreasesin magnitude exponentiaily
l' ,,,n,u
2
YEu, It is apparentfrom the geometry of this figure that the resultantphasor voltage Vs QF) is such that lVol > lysl. A simple explanationof the Ferranti effect on an approximate basis can be advancedby lumping the inductanceand capacitanceparameters of the line. As shown in Fig.5.14 the capacitanceis lumpld at the ieceivingend of the line.
= 6I.47 l12.2" kV (to neutral) (c)
Resultantvoltage at 200 km from the receivingend  65.62 112.2" + 61.47 I  12.2" = 124.2+ j0.877 = 124.2 10.4"
Resultantlinetoline voltage at 200 km = 124.2 x J3  215.1 kV 5.6
FERRANTI EFFECT
As has beenillustratedin Exarnple5.5, the eff'ectof the line capacitanceis to cause the noload receivingend voltage to be more than the sendingend voltage.The effect becomesmore pronouncedas the line length increases.This phenomenonis known as the Ferranti. effect. A general explanation of this effect is advancedbelow: Substitutingx = / and In = 0 (noload)in Eq. (5.21), we have t7 /5 
Vn
,at4gt * 
Vn
2
,at
(< A2\
ifl
\J.tJ
I
Fig.5. 14 Here
V,
(+.
Ir= ,
\ juCt ) Since c is small comparedto L, uLl can be neglected in comparisohto yc,tl. Thus 15  jVruCl Now
Vn= Vs  Is QwLl) = V, + V,tj
CLlz
= vs0+ Jctt2)
lncreasino/ \.
(s.M)
Magnitude of voltage rise _ lvrltJ CLf Locus of V5 wlth /
=olv.rt+ D Vpfor I = 0
\ En= Ero= Vpl2
Increasing/ Fig. 5.13
The above equationshows that at I = 0, the incident (E,o)and reflected (E o) voltage waves are both equal to V^/2. With reference to Fig. 5.13, as I increases,the incident voltage wave increasesexponentiallyin magnitude
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(s.4s)
where v = 7/J LC. i1 velocity of propagationof the electromagneticwave the along the line, which is nearly equal to trr" velocity of light. 5.7
TUNED POWER LINES
Equation (523) characterizesthe performanceof a long line. For an overhead line shunt conductanceG is always negligible and it is sufficiently accurateto neglect line resistanceR as well. with this approximation 7 Jyz = jalLC cosh 7/ = cosh jwlJTC
= cos Lt,lrc
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
;fiTi I
I
uooernpowersystemRnarysis sinh ?/ = sinh jalJLC
eharacteristicsand performanceof power
[email protected]
= j sin wtJLC
t Z = Z,
Hence Eq. (5.23) simplifies to
cosulJ LC . rT sina'tJ LC Z,
jZ, sinutJE
=!
_1
cosutE
i.e. the receivingend voltage and current are numerically equal to the colrespondingsendingendvalues,so that thereis no voltagedrop on load. Such a line is called a tuned line. For 50 Hz, the length of line for tuning is
1 !r_.
,,1_ (t. It,t,)
z,
.Jru. ,
fi:I=filH;':"! l'lt,^t
. cosh 7/ L1,_, I z, ___.. JLl^J
(s.47)
= 3,000km, 6,000km,... It is too long a distance of transmissionfrorn the point of view of cost and efficiency (note that line resistancewas neglectedin the above analysis).For a given line, length and freouencytuning can be aehievedby increasingL or C, i.e. by adding series inductancesor shunt capacitancesat severalplaces along the line length. The methodis impracticaland uneconomicalfor power frequencylines ancl is adoptedfor tclephonywhere higher frecluencies are employed. A methodof tuning power lines which is being presentlyexperimentedwith, uses seriescapacitorsto cancel the effect of the line inductanceand shunt inductors to neutralize line capacitance.A long line is divided into several sectionswhich are individuatly tuned. However, so far the practical method of improving line regulationand power transfercapacity is to add seriescapacitors to reduceline inductance;shunt capacitorsunder heavy load conditions;and shunt inductors under light or noload conditions.
( 5. 48)
According to exact solurion of a long line [refer to Eq. (5.23)].
= y, the velocity of light
THE EOUIVALENT
t(tU2) Y ganh1il2)
_ _ ___{l
Lr,J +!v,2,) (r*ir,r,)lL,_) Lr,(,
2nrfJ LC
5.8
.o,nh..,tt2
For a zrnetworkshown in Fig. 5.15 [refer ro Eq. (5.13)].
llsl = llal
,i^,^,?rx...
[email protected]))=
=
Fig. 5.15 Equivarentz networkof a transmission rine
N o wi f a l J t C = h r , n = I , 2 , 3 , . . . lV5l= lVpl
Since ll^frc

sinhl/ =
For exact equivalence,we must have Z/= Z, sin h 7/ f * *YtZ=cosh 2 From Eq. (5.50)
7/
Z' = ,l:. sinn1/ = Z, tlnh'' : ( z[*init rr 1 vY tJyz 7t )
(s.4e)
(s.s0) (s.51)
(5.52)
11"'' sln!24 is thc fitct.r by which thc scrics irupcdunceol'the no'ri'alz must be multiplied to obtain the z parameter of the equivalenta Substituting 7 from Eq. (5.50) in Eq. (5.51),we ger 1*
I
; YtZ, sitrh 7/ = cosh fl
CIRCUIT OF A LONG LINE
So far as the end conditions are concerned,the exact equivalentcircuit of a transmissionline can be establishedin the form of a T or zrnetwork. The parametersof the equivalentnetwork are easily obtainedby comparingthe perfbrmanceequationsof a znetworkand a transmissionline in terms of end quantities.
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Lv,
(s.53)
2
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
ModernPowerSystemAnqlysis
Wl I
(tanhfll2\. d rnus I is thc lactor by which thc shunt aclmittanccanr ol' the
\ il/z )
nominalnmustbe multiplied to obtain the shunt parameter (Ytl2) of the z. equivalent/
1
\
l
Note that Ytl  + +Y' Z' I : j\ 4 ) 2 , abovevaluesof Y/and Z/.
V R =+ 2 0 VJ
 l z 7 1 0k v
(a) Short line approximation:
Vs= t27 + 0.164/._36.9 x 131.2 172.3 = 14514.9
sinh 7/ is a consistentequationin termsof the
= 25L2 kV lYsltin"
sinh 7/ = tuthl!/2 1 so that the 1 and For a line of medium length
1t
fll2
equivalent n network reduces to that of nominaln.
Is= In= 0.764/_36.9kA Sendingend powerfactor= cos (4.9"+ 36.9"_ 41.g.) = 0.745laggrng Sendingend power_ JT x 251.2x 0.764x 0.745 = 53.2MW
[email protected])Nominaltrmethod:
Fig. 5.16' EquivalentTnetwork of a transmissionline EquivalentT network parameters of a transmission line are obtained on similar lines. The equivalentT network is shown in Fig. 5.16. As we shall see in Chapter 6 equivalentr (or nominalr) network is easily ^.1^^+^l
Sn=5l SDI= 15 +/5 (b)
G) {
zo.ro+ i16.12 V 2 =1 . 0l o "
Powerfactorat starionI = cos [run' t'f:t') = 0.963lagging \ 2 0 ) Total load on station2 (25 + jls) (5  j0.638) =20+i15.638
15+j5
25 + j1S (c)
Fig. 5.18 Twobussystem
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Reactivc powcr loss can also be cornputedas l/l2X=
:t +(9.!1q)1 o.os= t.z7 pu. " I
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
characteristicsand Performanceof power TransmissionLines
toctern power Svstem ,Analvsis
164 I
') Powerfactor ar station2 = coS (ron*r 15'638 = O.zgslagging \ 2 0 ) The stationloads,load demands,and line flows are shownin Fig.5.1g(b).
ir;t { iz i, r"rp rt v. "" "case "r Cableimpedance= 0.005 +70.05 = 0.0502 lB4.3 pu. In rhis
Case(b): the cable resistancecausesreal power loss which is not known a priori. The real load flow is thus not obvious as was in the case of R = 0. we specify the generation at station I as Pcr= 20 Pu The considerationfor fixing this generation is economic as we shall see in Chapter 7. The generationat station 2 will be 20 pu plus the cable loss. The unknown variables in the problem are
165
6r = 14.4" Substitutingdr in Eqs.(ii), (iii) and (iv), we ger Q c t = 5 . 1 3 ,Q c 2= 1 6 . 1 2P, G z= 2 0 . 1 0 ^'

^D'
"'^'
It may be noted that the real power loss of 0. i pu is supplied by Gz(Pcz  20.10). The above presentedproblem is a twobus load flow problem. Explicit solution is always possiblein a twobus case.The readershould try the case when Q c z = 7 1 0 a n d l V 2 l= ' The general load flow problem will be taken up in Chapter 6. It will be seen that explicit solution is not possiblein the generalcaseand iterative techniques have to be resortedto.
P62, 6p Qcp Qcz Let us now examine as to how many system equationscan be formed. From Eqs. (5.68)and (5.69)
= ', = Pct Por
#cos
d V#*
[email protected]+
5 = = cos 84.3"U.U)UZ
A 275 kV transmissionline has the following line constants: A = 0.85 15": B  200 175"
61)
cos (84.3"+ 4) ^= U.U)UZ
(i)
e c r  e o t =e s = f f r t ^ t  t r r , \ i , s i n ( d +4 ) sin 84.3" =+sin Qcr  S = ;j0.0502 0.0502 FromEqs.(5.66)and (5.67)
(84.3"* Ur)
(ii)
*;rcos
(84.3" A,l 
#rcos
84.3o
 4) sin(84.3"
#,
sin (75" d) 
ij;
x Q75)2sin(75"5")
0=378sin(75"A302
sin84.3"
whichgives 6 ))"
(v)
Thus we havefour equations,Eqs. (i) to (iv), in four unknownsp52, 51,e61, Q62.Eventhoughtheseare nonlinearalgebraicequations,solutionis possible in this case.Solving Eq. (i) for d,, we have
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o  !5x]75200
(iii)
 ,Yl!rr", eorecz=en=Uffiri, (o 6t> lzl 15 Qcz= #,
(c) With the load as in part (b), what r,vouldbc thc receivingend voltageif the compensationequipmentis not installed? Solution (a) Given lV5l= lVal = 215 kY; e, = 5o,C = J5". Since the power is receivedat unity power factor, Qn= o Substitutingthesevalues in Eq. (5.62), we can write
P o z  PG z =PP n =l v ftf l l v z cl o s( d  6 r )  l 1v :'Pc o s e 2s  Pc2=
(a) Determine the power at unity power factor that can be received if the \ voltage profile at each end is to be maintained at 275 ky . (b) What type andrating of compensation equipmentwould be requiredif the load is 150 MW at unity power factor with the samevoltage profile as in part (a).
From Eq. (5.61)
275^x275 cos(75o 22") 9 85 * (2712cos70o 200 200  227.6 109.9= 117.7D{W
Pn= "
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
todern power SvstemAnalvsis
166 
characteristicsand Performanceof Power TransmissionLines
I
(b) Now lV5l= lVpl = 275 kV
0 = !:y*t sin (75' .5) *tvot2 200 200 From Eq. (ii), we get sin(75" A=0.00291VR1
Power dernanded by load = 150 MW at UPF P n = P R = 1 5 0 M W ;Q o = 0
150=
215zq7cos ( 7 5 o 5 ) 200
0 . 8 5 . ^  , ) x (275)'cos70o ,*
1 5 0 = 3 7 8 c o s ( 7 5 " 4  1 1 0 5 = 28.46"
or
sin 70'
(ii)
_ 0.00l45tyRl2 150 = 1.375 lynt (1 _ (0.002912lv^121u2 Solving the quadratic and retaining the higher value of lv^, we obtain lVal = 244.9 kV
From Eq. (5.62)
sin (75.  28.46") 0'85x e75)2 sin 70o 200 "#  274.46 302=  2l.56 MVAR
en=
Thusin orderto maintain2T5kV at a receivingend,en= 27.56 MVAR must be drawn alongwith the real power of Po = 150 MW. The load being 150 MW at unity power factor, i.e. Qo = 0, compensationequipmentmust be installed at the receivingend.With referenceto Fig. 5.19, we have  2 7 . 5 6 +Q c = 0 or Qc = + 27.56 MVAR i.e. the compensationequipment nnustfeed positive VARs into the line. See subsection5.10 for a more detailedexplanation. 1 5 0 j 2 7 . 5 6
1 5 0+ / 0
Note: The second and lower value solution of lVol though feasible, is impractical as it correspondsto abnormally low voltage and efficiency. It is to be observedfrom the results of this problem that larger power can be transmitted over a line with a fixed voltage profile by installing compensationequipmentat the receivingendcapableof feedingpositive VARs into the line. Circle Diagrams It has been shown above thatthe flow of active and reactive power over a transmissionline can be handledcomputationally.It will now be shown that the Iocusof complex sendingand receivingendpower is a circle.Sincecircles are convenientto draw, the circle diagramsare a useful aid to visualize the load flow problem over a single transmission. The expressionsfor complex numberreceivingand sendingend powers are reproducedbelow from Eqs. (5.58) and (5.59).
sR=4/rr. +P 4r b) lfl rv^r' n=l+lvs(t(ro)+3 4/r+o
Since no compensationequipmentis provided P n = 1 5 0 M W ,Q n = 0
Now,
l V 5l = 2 7 5 k V , l V a l =? Substitutingthis data in Eqs. (5.61)and (5.62),we have
t':\!'"os 150= (75" A  ggtv^t2 cos70" 200 200
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(s.s8) (5.5e)
The units for Sp and S, are MVA (threepha,se)with voltages in KV line. As per the above equations, So and ,9, are each composedof two phasor componenfs6ne a constantphasorand the other a phasorof fixed magnitude but variable angle. The loci lor S^ and S, would, therefore,be circles drawn from the tip of constantphasorsas centres. It follows from Eq. (5.58) that the centre of receivingendcircle is located at the tip of the.phasor.
F i g .5 . 1 9 (c)
Ifil
(i)
l+l vRP ' \ t(r  a)
(s.16)
= llrvf cos (// a)MW
(s.77)
IB l in polar coordinatesor in terms of rectangularcoordinates, Horizontal coordinate of the centre
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
e.'1,,
,
.
I
.d6Eif
Modernpower SystemAnalysis
_
.,.,,
characteristicsand performanceof power Transmission Lines
t69
Vertical coordinate of the centre
Vertical coordinateof the centre
= l+ilvrt2sin(F a) MVAR
= i{ltv*t2sin(tJ a) MVAR
tB t The radiusof the sendingend circleis
lBl The radius of the receivingendcircle is tysllyRlMVA
(5.78)
tBl
(s.81) The sendingendcircle diagramis shownin Fig. 5.2I.Thecenrre is located by drawing OC, at angle rt? a) from the positive MWaxis. From the centre the sendingendcircle is drawn with a .uoi.rr${e(same as in the case of \ lBt receivingend).The operatingpoint N is located by measuring the torque angle d(as read from the recefvingendcircle diagram) in ttredirection indicated from thc re'fi'rcncc' Iinc.
MVAR
MVAR C5
Roforoncolino  for angled lAll t2 lallu"l
Radius lYsllVnl lBl Phasor55 = P5 +ie5
Referenceline f o r a n g l e6
Fig. 5.20
1
Receivingendcircle diagram
Qs
F'or constant lVol, the centre Co rernains fixed and concentric circles result for varying l7rl. However, for the case of constant ll{l and varying lvol the centrcso1'circlesmovc along the line OCoaru),have raclii in accordance to ltzrl lvRtABt. Similarly, it follows from Eq. (5.59) that the centre of the sendingend circle is located at the tip of the phasor
l4l,u,,'t(0. a)
IB l
cos(f  a)Mw Bltv;2
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Fig. 5.21 Sendingendcircle diagram
(s.7e)
=u':;=o i,i,=',!"i
in the polar coordinates or in terms of rectangular coordinales. Horizontal coordinate of the centre
=
+
(s.80)
The corresponding receiving and sendingend circle diagrams have been clrawn
in Figs5.22 and5.23.
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
characteristicsand Performanceof power TransmissionLines [giM t Resistance= 0.035 Olkm per phase Inductance= 1.1 mHlkm per phase Capacitance= 0.012 pFlkm per phase If the line is supplied at 275 kV, determine the MVA rating of a shunt
lvnl2
tzl
Pa=oK Qn= KM
Fig.522 Receivingend circrediagramfor a short rine MVAR
Ps=oL Qs=LN
thereceivingend whentheline is deliveringno load.Usenominalzrmethod. Solution R=0.035x400=14Q X = 314x 1.1x l03 x 400 = 138.2O Z = 14 + 7138 138.1184.2" Q Y= 314 x 0.012x 106x 400 lg0"  1.507x 103/_W U A = ( t + L v z \ = 1 +  l  x t . 5 0 7x 1 0  3x r 3 g . i l r i 4 . z " \ 2 ) 2 = (0.896+ 70.0106) = 0.896 lj.l' B=Z138.7 184.2" lV5 = 275 kV, lVpl= 275kV lysllyRl 275x275  545.2MVA Radiusof receivingend circletBl 138.7 Locationof the cenre of receivingend circle, ,, = 275x275x0.896== '488.5 r d d ' ) rMVA ll l4:l1 i l l IB I
\
138J
[email protected]  a) = 84.2" 0.7" = 83.5"
lvsl2
lzl i,\/A P
55 MVAR
_
> M W
L 488.5 MVA
Fig. 5.23 Sendingendcirclediagramfor a stror.tline
oJl:
.tuou.rrro
useof circlediagrams is i$ustratedby meansof therwo examplesgiven Cp
Fi1.5.24 Circlediagramfor Example5.10 A 50 Hz, threephasg,275 kY,400 km transmissionline has the following parameters:
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From the circle diagramof Fig. 5.24, + 55 MVAR must be drawn from the receivingendof the line in order to maintain a voltageof 275 kV. Thus rating of shunt reactor needed= 55 MVA.
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Modernpower SystemAnalysis
X12,:jl
and Performanceof PowerTransmissionLines  ,i?3;Characteristics
t
(a) Locate OP conespondingto the receivingendload of 250 MW at 0.85 laggingpf (+ 31.8). Then
Example5.11
, ^ A  0.93 11.5", B = Il5 ll7" If the receivingend voltage is 2'r5kV, determine: (a) The sendingendvoltagerequired if a load of 250 MW at 0.g5 lagging pf is being delivered at the receivingend. (b) The maximum power that can be delivered if the sendingendvoltage is held at 295 kV. (c) The additional MVA that has to be provided at the receivihgend when delivering 400 MVA at 0.8 lagging pt the suppty voltage being maintained at 295 kV. Solution
In Fig. 5.25 the centre of the receivingendcircle is located at
2t5x275x0'93611.6 MVA l4li Rt'' lBl
tt5
or^
lysllyRl
275lvsl
l V s l = 3 5 5 . 5k V (b) Given lV5  = 295 kV. t".l?"  705.4 MVA 115 Drawing the receivingend circle (see Fig. 5.25) and the line C^Q parallel to the MWaxis, we read PR o = RQ = 556 M W Radius of circle diagram
(c) Locate OPt conespondingto 400 MVA at 0.8 lagging pf (+ 36.8"). Draw P/S parallel to MVARaxis to cut the circle drawn in part (b) at S. For the specified voltage profile, the line load should be O^S.Therefore, additional MVA to be drawn from the line is P/S = 295 MVAR or 295 MVA leading
c o s l 0 .8 5= 3 1 .8 "
[email protected]
a ) = 7 7 o  1 . 5 "= 7 5 . 5 " MVAR
I
5.10
METHODS OF VOLTAGE CONTROL
Practically each equipmentusedin power systemare ratedfor a cprtain voltage with a permissible band of voltage variations.Voltage at various buses must, therefore,be controlled within a specified regulation figure. This article will discussthe two methodsby meansof whieh voltage at a bus can be controlled. lvslt6
a)
lvRltj
I p".io,
P^iio* 
Fig. 5.26 A twobussystem Considerthe twobus systemshown in Fig. 5.26 (akeady exemplifiedin Sec. 5.9). For the sake of simplicity let the line be characterizedby a series reactance(i.e. it has negligible resistance).Further, since the torque angle d is small under practicalconditions,real and reactivepowersdeliveredby the line for fixed sendingendvoltage lVrl and a specifiedreceivingenclvoltagel{  can bewritten as below from Eqs. (5.71) and (5.73).
(5.82) Fig. 5.25 Circle diagram for ExampleS.11
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nsmtssion Linesl"ffie
'li, ef, rl ' = X ,,u,r rv,i Equation (5.83) upon quadraticsolution*can also be written
rr{ r=
(5.83) as
. +tys  (1  4xesnAvrtzlt/z
}vrt
(5.84)
Since the real power demandedby the loacl must be delivered by the line, Pn= Po varying real power demandp, is met by consequent changesin the rorque angle d. It is, however' tobe noted that the receivedreactive power of the line must remainfixed at esnas given by Eq. (5.g3) for fixed rv, I and specifiedr4r. il" line would, therefore,operatewith specified receivingend voltage for only one value of Qo given by Qo = Qsn
Reactive Power Injection It follows from the above discussion that in order to keep the receivingend voltageat a specifiedvalue l{1, a fixed amountof VARs drawn tai I *;;; from the line. To accomplish this under conditions Qn, a local VAR generator (controlled reactive power source/compensating equipment)must be usedas shown in Fig. 5.27.fle vAR balanceequationat the receivingendis now
Oi * Qc= Qo Fluctuationsin Qo ue absorbedby the local vAR generator o6 such that the vARs drawn from the line remain fixed at esn.The receivingendvoltage would thus remain at l4l (this of a fixed sending_end {1ed voltage lVrl). L,ocal VAR compensation"ourr"lrrumes can, in fact, be made automatic by using the signal from the VAR meter installedat the receivingend of the line.
Practical loads are generally lagging in nature and are such that the vAR demandQn may exceedet*.rt easily follows from Eq. (5.g3) that for or; otthe receivingendvoltagemust changefrom the specified value 'n'i some value lTol to meet the demandedVARs. Thus 'Jo= ^ == l v * l ' (lYsl  lVol) for (QD> QsR) Q^= o Qn ;i The modified lVol is then given by
= rvlql
lvrt

+ty3
(1  4xeRltvrt )r,,
(s.85)
crrmparisonof Eqs. (5.84)ancl(5.85) rcvcalsthut r^r. n.  n  .,),s,!,., YD vR vp' tttt: receivingendvoltageis r{r, butior bo= Oo; A:,"'
tvo < t4l
Thus a VAR demandlarger than Qf is met by a consequent fall in receivingfrom the specifiedvalue. similarly, if the vAR demandis less than 11d ":t!q," Q " *, it fo l l o w s th a t
tyRt> tr4l Indeed, under light load conditions, the charging of the line may "upu.lance cause the VAR demand to become negative resuliing in the receivingend voltage exceeding the sendingendvoltage (this is the Ferranti effect already illustraredin Section5.6). In order to regulate the line voltage under varying demandsof VARs, the two methodsdiscussedbelow are employed.
Fig. szr use of rocarvAR generatorat the roadbus Trryotypes of vAR generators are employed in practicestatic type and rotating type. These are discussedbelow. Static
It is nothing but a bank of threephase static capacitors and/or inductors. With referenceto Fig. 5.28,if lV^l is in line kV, and Xg is the per phase capacitive reac_ tance of the capacitor bank on an equiva_ Ient star basis, the expression for the VARs fed into the line can be derived as under.
r,=iH
'Negative
sign in the quadratic solution is rejected because otherwise the solution would not match the specified receivingend voltage which is only slightly less than the sendingendvortage(the differenceis ress thai nqo).
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VAR grenerator
lIc
Fig. 5.28
Static capacitor bank
kA
'of
course, sincct{tis spccificcl withina buntl,Ql rury vrry withil a corresponding band.
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: iii,',1,1
ModernpowersystemAnarysis t Figure 5'29 shows a synchronous motor connectedto the receivingend bus bars and running at no load. since the motor o.u*, n.grigible real power from
3ry ( IF) iQcGPhase)J3
i3x #.HMVA t v P
QsQPhase)+
XC
MVAR
in pr,use. ; th" ,yn.r,ronous reachnce $",.t":::H:,.,:o,^fl !:T? T*ly wtr,icrr i, usium"a tohave ", ;rr,r*,n
;r;:r*::':il;
:j 3:^t1o1 (s.86)
.C__ I
(lvRl  IEGD/0. ,. ^ KA
If inductors are employed instead,vARs fed into the line are
Q{3phase)='F''tuo* XL
Rotating
VAR grenerator
It is nothing but a synchronousmotor running at noload and having excitation adjustableover a wide range. It feeds positive VARs into the line uncler overexcitedconditionsand f'eedsnegativeVARs when underexcited. A machine thus running is called a synchronouscondenser. lvnl
Fig. 5.29 RotatingVAR generation
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i e c_ = 3Jt v R3l 4 oG i l
(s.87)
Under heavy load conditions,when positive VARs are needed,capacitor banks are employed; while under light load conditions, when negative vARs are needed,inductor banks are switchedon. The following observationscan be rnade for.static vAR generators. (i) Capacitor and inductor banks can be switched on in steps.However, stepless(smooth) VAR control can now be achieved using SCR (Silicon Controlled Rectifier) circuitrv. (ii) Since Qg is proportionalto the squareof terminal voltage,for a given capacitor bank, their effectivenesstends to decreaseas the voltage sags under full load conditions. (iii) If the system voltage containsappreciableharmonics, rhe fifth being the most troublesome,the capacitorsmay be overloadedconsiderably. (iv) capacitors act as short circuit when switched on. (v) There is a possibility of series resonancewith the line incluctance particuia.riyat harmonic frequencies.
J5;E
= 3 W (  l Y R rrr c l ) J3 ( _jxsJl ) = jlVpt(tE6t _ IVRt)lXs MVA ec= tVRt(EGt _ tVRt)lXs MVAR (5.8g) It immediately follows from the above relationship that the machine feeds positive vARs into the line when rEGt> tv^r (werexcited case) and injects negarive VARs if lEGl continuously adjustableby adjusting machine which controls tE6l. rn contrastto statrcvAR generators, "Jtution the following observations are made in respect of rotating VAR generators. ., (i) These can provide both positive and negativevARs which are continu_ ously adjustable. (ii) vAR inje*ion ar a given .. excirarionis ressst tlt""r""
volrage.As Irzo r deJeases and(rE  rv^rr i,Lt
6r J":ff i::: {.::: smallerreductionin Qc comparedto the .0.r. or static capacitors. From rhe observarionsua" in ,.rp..t of ,tuti. vAR generators,it seemsthat rotating ibr_"" ""d;;;;;ing vAR g.n"ruro* would be preferred. However, economic considerations, install.tion and 'rai'r.rrun.. problerns limit their buses in the svstem *r,"i"' a taigeu.oun,.orvAR
il]ffi'ir'$ t:L::"::ch Control
by Transformers
The vAR injectionmethoddiscussed abovelacksthe flexibility and economy of voltage control by transformer tap changing. The transformer tap changing is obviously rimited to a narrow range of voltage control. If the vortage correctionneededexceedsthis range, tap changingIs usedin conjunction with the VAR injection method. Receivingendvoltage which tends to sagowing to vARs demanded by the load, can be raised by simultaneousry .t,uogir; th. taps of sending_and receivingend transformers.Such tap changesniust"bemade ,on_road, and can be done either manua'y or automaiically,the oo*io.,o"r being ca'ed a Tap Changing Under Load ifCUf_l transformer.
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Characteristicsand Performanceof Power TransmissionLines fi.l7!J,
ModernPowerSystemAnalysis
,tig;'J
I
I
Considerthe operationof a transmissionline with a tap changingtransformer at eachend as shown in Fig. 5.30. Let /5 and r^ be the fractions of the nominal transformationratios, i.e. the tap ratio/nominal ratio. For example, a transkV input has rr  I2lll
l.{Vl which is to be compensated.Thus merely tap setting as a method of voltage drop compensationrvould give rise to excessivelylarge tap setting if compensationexceedscertain limits. Thus, if the tap setting dictated by Eq. setting range (usually not more than + 20Vo), it would be necessary to simultaneously inject VARs at the receivingend in order to maintain the desiredvoltage level.
= z=R+jx
Compensation
1 : fsnl
at eachend linewithtap changingtransformer Fig.5.30 Transmission With referenceto Fig. 5.30 let the impedancesof the transformerbe lumped tn Z alongwith the line impedance.To compepsatefor voltage in the line and transformers,let the transformer taps be set at off nominal values, rr and ro. With referenceto the circuit shown. we have
(s.8e)
trn rV s = t^ n rVo + IZ
From Eq. (5.75) the voltage drop ref'erredto the high voltage side is given by
tAvl = !I,!jIQs
(s.e0)
t on,rlVol
Now
lAVl  tsn, lTsl 
trnrlvrl tonrlvol +
ton2lVol
RPR+xQR
(s.e1)
t* n r l V o l
In order that the voltage on the HV side of the two transformersbe of the sameorder and the tap setting of eachtransformerbe the minimum, we choose (5.92) tstn= 1 SubstitutinEtn= llttin Eq. (5.91) and reorganising,we obtain .r( ,
RPR+xgo ) _ n2 lvRl
"['
"r"rWW ) ", W
(s.93)
For complete voltage drop compensation,the right hand side of Eq. (5.93) shouldbe unity. It is obvious from Fig. 5.30 that rr > 1 and tn 1 I for voltage drop compensation.Equation (5.90) indicatesthat /^ tends to increase*the voltage This
is so becausefn < 1 increasesthe line current / and hence voltage drop.
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of Transmission
Lines
The perfonnance of long EHV AC transmissionsystemscan be improved by reactive compensationof series or shunt (parallel) type. Seriescapacitorsand shunt reactors are used to reduce artificially the series reactanceand shunt susceptanceof lines and thus they act as the line compensators.Compensation of lines results in improving the system stability (Ch. 12) and voltage conffol, in increasingthe efficiency of power transmission,facilitating line energization and reducing temporaryand transient overvoltages. Series compensationreduces.the seriesimpedanceof the line which causes voltage drop and is the most important factor in finding the maximum power transmission capability of a line (Eq. (5.70)). A, C and D constants are functions of Z and therefore the also affected by change in the value of.Z, but these changes are small in comparison to the change in B as B = Z for the ., for the equivalent zr. nominal rr and equalsZ (sinh 4ll) The voltage drop AV due to series compensationis given by
AV = 1Rcos S, + I(X,. X.) sin ,!,
(s.e4)
Here X, = capacitivereactanceof the seriescapacitor bank per phaseancl X, is thc total incluctivercactanceof the line/phasc.In practice,X. may be so selected that the factor (XL  X.) sin Q, becomes negative and equals (in magnitude) R cos /, so that AV becomes zero. The ratio X=IXL is called "compensation factor" and when expressedas a percentageis known as the "percentagecompensation". The extent of effect of compensationdependson the number, location and circuit arrangementsof series capacitor and shunt reactor stations.While planning longdistance lines, besides the average degree of compensation required, it is requiredto find out the most appropriatelocation of the reactors and capacitor banks, the optimum connection scheme and the number of intermediate stations.For finding the operating conditions along the line, the ABCD constantsof the portions of line on eachside of the capacitorbank, and ABCD constants of the bank may be first found out and then equivalent constantsof the seriescombination of linecapacitorlinecan then be arrived at by using the formulae given in Appendix B. In India, in stateslike UP, seriescompensationis quite importantsincesuper thermal plants are located (east) several hundred kilometers from load centres (west) and large chunks of power must be transmitted over long distances. Seriescapacitors also help in balancing the voltage drop of two parallel lines.
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rhOt" l
Characteristics and Performanceof PowerTransmissionLines
uodern Power SvstemAnalysis
When seriescompensationis used, there are chancesof sustainedovervoltage to the ground at the series capacitor terminals. This overvoltage can be the power limiting criterion at high degree of compensation.A spark gap with a high speed contactor is used to protect the capacitors under overvoltage trons. Under light load or noload conditions, charging current should be kept less than the rated fullload current of the line. The charging current is approximately given by BrltA where B. is the total capacitive susceptanceof the line and lVl is the rated voltageto neutral.If the total inductive susceptanceis Br due to several inductors connected(shunt compensation)from line to neutral at appropriateplaces along the line, then the charging current would be
(Bc Br) lvl= BclVf[r I,he,=
+)
(s.es)
Reduction of the charging current is by the factor of (1  Br lBc) and 81lBg is the shunt compensationfactor. Shunt compensationat noload also keeps the receiving end voltage within limits which would otherwise be quite high becauseof the Ferranti Effect. Thus reactors should be introduced as load is removed,f = u.JJ4 pu
(viii)
Now, let us find the line flows. Equation (5.6g) can be written in the form
( l z l =X , 0 = 9 0 " )
lvi]lvkl Pik =  Pki sin ({  6o)
x,o
Using the above Y"u, and bus powers as shown in Fig. 6.6, approximate load flow Eqs. (6.37) are expressedas (all voltage magnitudesare equal to 1.0 pu)
(6 6q) P z =3 = 5 ( 6  6 ) + 1 0( 6  4 ) + 6 . 6 6 7 P t =  2  6 . 6 6 (76  4 ) + l 0 ( 4  6 ) P + =  2  1 0 ( 4  4 ) + 6 . 6 6(76 q : 6 )
(ii) (iii )
(iv) = (iv), (ii), (iii) we 0, and solving and Taking bus 1 as a referencebus, i.". 4 get O.0ll rad = 4.4I" 4. 4=0.074rad=4.23'
(v)
6q=0.089rad=5.11' Substituting 6s in Eqs. (6.38), we have Qr =  5 cos 4.4I"  6.667 cos 4.23"  10 cos 5.11' + 21.667 Qz =  5 cos 4.41"  10 cos 8.64"  6.667 cos 9.52o+ 21.667 Qz =  6.667 cos 4.23o  10 cos 8.64' + 16.667
where P* is the real power flow from bus j to bus k.
sin1.23" pn =  pz,= r/ = 0.492pu \ r  q)+0 . 1 5 sin(d, 0.1: 4'41o Pt z =  Pzr =  0'385 Pu ( ix) L'L=  1 sin ( 4  6) =  $n 0. 2 02 Pqt= + s i n ( {  6 o )= 1 0 s i n 5 . 1 1 o = 0 . g 9 1p u 0.1 Real power flows on other lines can be similarly calculated. For reactive power flow, Eq. (5.69) can be written in the general form ( l Z l =X , 0 = 9 0 o ) Pru=
 lvillvkl cos(,{  6o) ei* =W Xik Xik where Q* ir the reactive power flow from bus i to bus ft.
I Q p = _1, Q,zr=+cos(d,_ hl = 0.015pu 0.2 i.,
Qq =  l 0 c o s 5 .1 1 "  6 .6 6 7cos 9.52o+ 16.667
@
OI'
Qr = 0'07Pu Qz = 0'22Pu Qz = 0'732Pu Q+ = 0.132Pu Reactivepower generationat the four busesare Qa = Qt + 0.5 = 0.57pu Qcz = Qz + 0.4 = 0.62 ptr Qa = Qs + 1.0= 1.L32pu Qc+= Q++ 1.0= 1.132pu
(vi)
2 + j0.ET 1 0.891+/O.04
* l i't ',t, , 3 8 5 7 0 .0 1 5
0.49270. 018
3 1 . s 0 2 i 0 . 1 1 3
0.385+,p. 01S 0.8917O.04
1.502+ 10.113
1.103 j0.0s2
4
1 . 1 0 3+ p . 0 9 2
n:n l
2 t, jD.4
(vii), Fig. 6.7 Loadflow solutionfor the four'bus system
Q r c= Q u=
#
#
c o s( d 1 ,O = 0 . 0 1p8u I
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"
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ModernPowerSystemAnalysis
.2Oftril I
I en*.
+ ' = +  1 .o, (6r  64)= 0.04 pu eA = *e+t 0.1 0.1 lines can be similarlycalculated. other on power flows Reactive and load demandsat all the busesand all the line flows are Generations
carried out at the end of a completeiteration, the processis known as the Gauss iterative method. It is much slower to convergeand may sometimesfail to do so. Algorithm
6.5 GAUSS.SEIDEL METHOD The GaussSeidel(GS) method is an iterative algorithm for solving a set of nonlinear algebraic equations.To start with, a solution vector is assumed, basedon guidancefrom practical experiencein a physical situation. One of the equationsis then used to obtain the revised value of a particular variable by substituting in it the present values of the remaining variables. The solution vectoris immecliatelyupdateclin respectof this variable.The processis then repeatedfor all the variablesthereby completing one iteration. The iterative processis then repeatedtill the solution vector convergeswithin prescribed accuracy.The convergenceis quite sensitive to the starting values assumed' Fortunately,in a load flow study a starting vector closeto the final solution can be easily identified with previousexperience. To explain how the GS method is applied to obtain the load flow solution, let it be assumedthat all busesother than the slack bus are PB buses.We shall seelater that the methodcanbe easily adoptedto include PV buses as well. The slackbus voltage being specified,there are(n  1) bus voltages starting values of whose magnitudesand angles are assumed.These values are then updated through an iterativeprocess.During the courseof any iteration, the revised voltage at the ith bus is obtained as follows: (6.3e) J, = (P,  jQ)lvi [from Eq. (6.25a)] From Eq. (6.5)
,[
I
v , =,,' *lL,_ fv,ovol
L T:i I
Substitutingfbr J, from Eq. (6.39) into (6.40)
t^ .
v,= *l '':jq
,
(6.40)
l
 t Yit' vr'l = 2'3""' n Y" I vr* ilt I" k*t I I
(6.4r)
The voltages substitutedin the right hand side of Eq. (6.41) are the most recently calcuiated (updated)values for the correspondingbuses. During each iterationvoltagesat buses i = 2,3, ..., n aresequentiallyupdated throughuse of Eq. (6.41). Vr, the slack bus voltage being fixed is not required to be updated.Iterations are repeatedtill no bus voltage magnitudechangesby more than a prescribed value during an iteration. The computation processis then said to converge to a solution.
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for Load Flow Solution
Presentlywe shall continueto considerthe casewhere all busesother than the slack are PQ buses.The stepsof a computationalalgorithm'aregiven below: 1. With the load profile known at each bus (i.e. P^ and 0p; known), allocate* Po, and.Q5; to all generatingstations. While active and reactive generationsare allocatedto the slack bus, these are permitted to vary during iterative computation.This is necessaryas voltagemagnitudeand angle are specifiedat this bus (only two variables can be specified at any bus). With this step,bus iniections(P, + jQ) are known at all busesother than the slack bus. 2. Assembly of bus' admittance matrix rsus: with the line and shunt admittancedata storedin the computer, Yru, is assembledby using the rule for self and mutual admittances(Sec. 6.2). Alternatively yru, is assembledusing Eq. (6.23) where input data are in the form of primitive matrix Y and singularconnectionmatrix A. 3. Iterative computationof bus voltages(V;; L = 2, 3,..., n): To start the iterationsa set of initial voltage values is assumed.Since, in a power systemthe voltagespreadis not too wide. it is normalpracticero use a flat voltage start,** i... initialiy ali voltagesare set equal to (r + 70) exceptthe voltageof the slack bus which is fixed. It shouldbe noted that (n  l) equations(6.41)in cornplexnumbersare to be solvediteratively for findin1 @  1) complex voltages V2,V3, ..., V,. If complex number operationszlreDot availablein a computer.Eqs (6.41)can be converted rnto 2(n  1) equationsin real unknowns(ei,fr or lV,l, 5) by writing Vr = €i + ifi = lV,l ei6' (6.42) A significant reduction in the computer time can be achieved by performing in advance all the arithmetic operationsthat do not change with the iterations. Define i = 2,3, ...,ft 'Active
(6.43)
and reactive generation allocations are made on econorfc considerations discussedin Chapter 7. .*A flat voltage start means that to start the iteration set the voltage magnitudesand anglesat all busesother than the PV busesequal to (i + l0). The slack bus angle is conveniently taken as zero. The voltage magnitudesat the PV buses and slack bus are set equal to the specified values.
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Modernpower SvstemAnatvsis
206 
I*rl.d (6.44)
flows on the lines terminatingat the slack bus. Acceleration
vi'
17(r+l) 
Ai
'S
n
ilt
k=irl
 t B , o v or t,( ,, * t , 
(Vl")
r , o v l , ,i = 2 , 3 , . . . , n
f
(6.4s) The iterative process is continued till the change in magnitude of bus voltage, lav.('*r), between two consecutiveiterations is less than a certain tolerancefor all bus voltages,i.e. IAV.G*r)l_ 1y.t+r)_ V,e)l < 6, ; i = 2, 3, ..., n (6.46) 4. computation of slack bus power: substitution of all bus voltages computedin step 3 along with V, in Eq. (6.25b) yields Sf = pr jey 5 . computtttionofline .flows:This is thc last step in thc loaclflow analysis whereinthe power flows on the variouslines of the network are computed. Considerthe line connectingbuses i and k. The line and transformersat each end can be representedby a circuit with series admittance y* and two shunt admittances1l;roand )no as shown in Fig. 6.8. B u si
up by the use of the acceleration factor. For the tth bus, the accelerated value of voltage at the (r + l)th iteration is given by y(r+r)(accelerated) = V,Q'* a(v.G+r) V,Ql, where a is a real number called the acceleration factor. A suitablevalue of a for any system can be obtained by trial load flow sfudies. A generally recommendedvalue is a = 1.6. A wrong choice of o. may indeed slow down convergenceor even cause the method to diverge. This concludesthe load flow analysisfor thJ case pe of busesonry. Algorithm
when pv Buses are arso present
Modification
At the PVbuses, p andrv]rarespecified and ancr dare the unknownsto be e detcnnincd.'l'hererirre, the valuesof e and d are to be updatedin every GS iteration through appropriate bus equations.This is accomplishedin the following stepsfor the ith pV bus. l. From Eq. (6.26b)
Busk l*io
of convergence
f
"
)
L
ft:l
)
ei =  Im j yr* D, y,ovof
sm
The revised value of ei is obtained from the above equation by substitutingmost updatedvalues of voltageson rhe right hand side. In fact, for the (.r + 1)th iteration one can write from the above equation I
g.(r+t) = Ln Fig. 6.8 7i'representation of a line and transformers connectedbetweentwo buses The current fed by bus i into the line can be expressedas Iit = Iitt + Iirc = (Vi  V) !it,+ V,y,oo The power fed into the line from bus i is. S* = Pir* jQiF Vi lfr= Vi(Vf  Vr\ yft+ V!,*y,f, Similarly, the power fed into the line from bus k is
Sri = Vk (V*k Vf) yfo+ VoVfyi,l,
(6.47)
2' The revisedvalueof {.is obtainedfrom Eq. (6.45) immediatelyfollowing step 1. Thus 6Q+r)_ ay!+r)
= Ansle fei]l  i Bovo(,+r)  D B,ovo,,,l (6.s1) "t L ( t 1 ' ' ' ) * , r :r k:,+r J
(6.48) where
(6.4e)
The power loss in the (t  t)th line is the sum of the power flows determined frorn Eqs. (6.48) and (6.49). Total transmissionloss can be computed by sununingall the line llows (i.e. 5';a+ Sri fbr all i, /)r, "h
+ + lr^ ^i{zu
Zti
2,,1)r, (s.4t)
Equation (9.37) can be written in matrix form as
Zzj :
(9'39)
(e.40)
V; = 2; 111+Z, r I , + "' + Z'nI n+ Z; / r SubstitutingEq. (9.40) in Eq. (9.39)
Consequently
.
(e.37)
zjj + zb Type
Modification
zu connectsan old bus (l) to the referencebus (r) as in Fig. 9.26. This case follows from Fig.9.25 by connectingbus ft to the referencebus r, i.e. by setting
t"f
: ltz't"Zv) (new)= znvs(old ZsLr5 ;+;l "jjT"ulz,, l
Znj
IYpe3
1
Now
Vo= Zdo + V,
V*=4lt+
'o= 12
or
(e.42)
Modification
two old busesasin Fig. 9.27. Equationscanbe writtenas follows zo connects for all the networkbuses.
v*=o'
(l;+ lp
J
Passive linear nbuS network
Fig. 9.27 TYPe4modification Fig. 9.26 Type3 modification
"' + Zr i ( I i + / ) + ZU Q i /o)+ ...+ ZiJ"(9.43) Similar equationsfollow for other buses. Vi= Z; 11,+ Z, lr +
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Modern power System Analysis
360 
SymmetricalFault Analysis
I he voltages of the buses i and j are, however, constrained by the equation (Fig. 9.27)
V ,= Z ,rl o + V i
(e.44)
or
2 1 1 1+1 \ 2 1 2 + . . . + Z i t ( 1 , + I ) * Z i i ( I i  I ) + . . . + zi,In = Z t l * + Z , r l , + Z , r l "+ . . . + Z l U , + I _I earTangtng 0 =(Zit 21) Ir+ ... + (Zti Z) Ii+ eu Z) Ii +...+ (Z*  21,) In + (Zu * Zii * Zii  Zi  Z) 11,
Step I: Add branch 2,,  0.25 (from bus I (new) to bus r) ZBUS= t}.25l
e.45)
Step 3: Add branch ,rt = 0.1 (from bus 3 (new) to bus I (old)); typeZ modification
lo.2s o.zs 0.2s1
Step 4: Eliminating 1o in Eq. (9.46) on lines similar to whar was done modification,it follows that
l'"
(new) = Zsuts (old) Z131,s
zb+zii*Ziizzij
lZ,,
lzit  z1t)... (zi,  zi))
with the useof fbur relarionshipsEqs (9.36), (9.37),(9.42) and (9.47) bus impedancematrix can be built by a stepbystepprocedure(bringing in one hci ng bra n c h a t a ti n l e ) a s i l l trs tra tc 'itln l i xrrrnpl e9.8. Thi s pror.cdrrrc a mechanicalone can be easily computerized. When the network undergoeschanges,the modiflcationproceclurescan be cttrploycclto rcviscthc bLrsirtrpcdance rnatrixul'thc nctwol'k.Thc opeling o1 a line (Ztt) is equivalentto adding a branch in parallel to it with impedance  2,, (see Example9.8).
(iii)
zsvs= o.zs o.3s o.2s   lo.2s 0.2s 0.3s..l
(9.46)
0
(ii)
zsvs=',lZ:11, 31i]
V vn
(i)
Add branch Zzt = 0.1 (from bus 2 (new) to bus I (old)); typeZ
Step
Collecting equationssimilar to Eq. (9.43) and Eq. (9.45) we can write vl
I fOt
Add branch zz, (from bus 2 (old) to bus r); type3 modification
10.2s o.zs 0.25'l = o.zs 0.3s 0.2s Zet)s  o'3s +o
Lo.2so.2s o3sl
[0.25l  o.tsI to.zs0.3s0.251
" Ln.trj
[0.14s8 0.1042 0.14s8l = I o.ro+z o.r4s8 o.lo42
fo.tott o.ro42o.24s8l Step5: Acldbranchizr= 0.1 (from bus 2 (olct)to brrs31old)): type4 modification [o.l4s8 0.1042 0.l4s8l
= o.'oo, 0.1458o.to42lzsvs  0 . 1+ 0 . 1 4 5+80 . 2 4 5 8  2 x O . l O 4 2 l ) 10.l4s8 O.tO42 0.24s8
1042.l [0 = 0.0417 [o.to+z 0.0411 0.0417] I 
[o.o+tzl
j Example9.8 For the 3busnetworkshownin Fig. 9.28build Zsus I I I
!,0.25l,'
(I l
r z 'irl9 l I l L,trili ,  6"11. I 3 0.1
o.t
I
Ref bus r
I i,J
or.
I
0.1103 0.12501 lo.r3e7 = 0.1ro3 o.I3e7 0.1250  I  0.1250 0.1250 0.17sol  0.I Oltenirtgo tinu (line 32): This is equivalentto connectingan intpeclance betweenbus 3 (old) and bus 2 (old) i.e. type4 modification.
Zsus=Zuur(olcl) (01)+ol?5#
Fig. 9.28
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ModernPower System Anatysis
filf.lid,
rr^r :)ymmelrlual
[t 0.0147l 0.0147 l 10.0147  0.01470.0s001  
L o.osoo_J
r^..tr raull
Ana*raio rurqryo's
ffir r**
Now
I
  0.1042 0.14s8 0.1042l; (sameas in step 4) 0.1458 0.1042 0.2458
and
i' li =2 ( F[ ^Z E  ) = 0 . 2 8 6 zu)
symmetry of the given power These two voltages are equal becauseof the network (c) From Eq. (9.29)
Iri = Y,j(vl  vl)
For the power system shown in Fig. 9.29 the pu reactancesare shown therein. For a solid 3phasefault on bus 3, calculate the following (a) Fault current
(b) V\ and vt (c) It,r, I'\, uxl Il, (d) 1fr, andIf, Assumeprefaultvoltageto be I pu. 0.2
0.09 1
ffr
JIc)  i,,
(d)
//
o '' \
\
/ , ,/
\.. . l 3 '
'l
Solution The Thevenin passivenetwork for this systemis drawn in Fig. 9.28 with its Zru, given in Eq. (iv) of Example 9.8. (a) As per Eq. (9.25)
V: 2,, + Zl
Y o  : j0.175 . _ 1    j s . j l
zn
(b) As per Eq. (9.26) rr.f v i 
But
,/'0.,
Fig. 9.29
1f 
  i2.86 As per Eq. (9.32) ,r r Gr
\,
or
I\t=tt= ro.186o) fr
and
0.1
F
Jt 
=g  0.286; Irtz=+(0.286 ' j0.1
E' orvrf ixtic + ixr
E'Gr= 1 Pu (Prefaultno load)
IL.= Gt
10'286= j2'86 io.2+io.os
SimilarlY I f cz= i2. 86
LEMS PROB anclresistance5 ohms is suddenly 9 . 1 A transmissionline of inductance0.1 H in Fig' P9'1' Write the as shortcircuitedat t =0 at the bar end shown the value of approximately Find i(r). expressionfor short circuit current current)' momentary (maximum the first current maximum currentntaximum occurs at the sametime as [Hint: Assumethat the first ttrcuit current') the first current maximum of the symmgtricl.:non
vl'  Ztyu
zrr+zl'r
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564,'l
power System Analvsis rr]logern
I
errrnrnorrinal Farrlt Analvsis "' " J
vyrrrrlvt"vE'
I
geS
I
I fl
'i
0.1H
SO
64A' v^/v\
i
On(100n f + 15.) u = 1 Osi
Fig. p9.1
Fig. P9.5
9 '2 (a) What should the instantof short circuit be in Fig. p9.1 so that the DC offset current is zero? (b) what should the instantof short circuit be in Fig. p9.1 so rhar the DC offset current is maximum? 9 . 3 For the system of Fig. 9.8 (Example 9.2) find the symmetrical currents to be interrupted by circuit breakersA and B for a fault ar (i) p and (ii) e. 9 . 4 For the system in Fig. p9.4 the ratings of the various componenrs are: Generator: 25 MVA, 12.4 kV, l\Vo subtransientreactance Motor: TransformerT,: Transfbrmer Tr: Line:
20 MVA, 3.8 kV, l5%o subtransientreactance 25 MVA, 11/33 ky, gVoreactance 20 MVA, 33/3.3 kV, I \Vo reactance 20 ohms reactance
The systemis loadedso that the motor is drawing 15 Mw at 0.9 loading power factor, the motor terminal voltage being 3.1 kv. Find the subtransientcurrent in generatorand motor for a fault at generator bus. lHint: Assumea suitablevoltagebasefbr the generator.The voltagebase for transformers,line and motor would then be given by the transforma_ ti 0 n ri l ti < l sF. rl r c x a n rp l ci,f' w c chooscgcrrcrator. vol tagcbascas I I kv, the line voltage baseis 33 kv and motor voltagebaseis 3.3 kv. per unit reactances are calculatedaccordingly.]
T
,1
T2
Fig. p9.4 9 . 5 Two synchronousmotors are connectedto the bus of a large system through a short transmissionline as shown in Fig. p9.5. The ratings of vanouscomponentsare: Motors (each): 1 MVA, 440 v,0.1 pu transientreactance Line: 0.05 ohm reacrance Large system: Shortcircuit MVA at its bus at 440 V is g. when the motors are operating at 440 v, calculate the short circuit cuffent (symmetrical) fed into a threephasefault at motor bus.
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ansient 9 . 6 A s y n c h r o n o u s g e n e r a t o f r a t e d 5 0 0 k V A , 4 4 0 v , 0 . l p u s u b t rpowel at 0.8 lagging ."u"'tun". is supptying a passiveload of 400 kW a threephasefault for factor. Calculate tfr" in'itiui symmetrical rms current at generator terminals. a line through a circuit 9 . 7 A generatortransformerunit is connected to breaker.The unit ratings are: pu, Xi= o,2o pu and Generator: 10 MVA, 6.6 kV; X,,d= 0.1 = 0'80 X,r Ptt
Transformer:10MVA,6.9133kV,reactance0.08pu of 30 kv, when a threeThe systemis operatingno load at a line voltage breaker'Find phasetault occurson ih" tin" just beyondthe circuit breaker' the in current rms symmetrical (a) the initial DC offset currenr in rhe breaker, iui tt. maximum possible rating of the breaker' current (c) the momentary kVA' (d) the current,o U. intemrpted by the breakerand the intemrpting and (e) the sustainedshort circuit current in the breaker' 50 MVA at 1 1 k v , 0 . 8 9 . 8 The systemshown in Fig' P9'8 is delivering ils infinite. laggirrgp0w0rl.itctttrillttlltbrrswhichrtriryberegirrdecl Particularsof various systemcomponentsare: 60 MVA, 12 kV' X,/ = 0'35 Pu Generator; = 0'08 pu Transtbrmers(each): 80 MVA, 12166kV' X Reactance12 ohms' resistanc:negligible' Line: breakersA and B will Calculatethe syrnmetricalcurrentthat the circuit fault occurring at threephase a of event the in be called upon to interrupt B' F near the circuit breaker Line
Fig. P9.8
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tgdern power Svstem Anatvsis t . 9'9 A two generatorstaticln suppliesa f'eederthrougha bus as shown in Fig. P9'9' Additional power is fed to the bus throJgh a transfoilner from a large system which may be regardedas infinite. A reactor X is included
*Sf,S
and,1"!T to limit theSCrupruring capacity of :,'jT:^"1^:t::1u1.',f"'Ter 4v 0 (Ya + Xr ll X).The systemis operatinginitially with a steadypower transfer Pr= P^ at a torque angle 4 on curve I. Immediately on switching off line 2, the electrical operating point shifts to curve II (point b). Acceleratingenergycorrespondingto areaA, is put into rotor followed by decelerating energy for 6 > q. Assuming that an area A2 correspondingfo deceleratingenergy (energy out of rotor) can be found such that At = Az, the system will be stable and will finally operate at c correspondingto a new,rotor angle 6, > 60"This is so becausea single line offers larger reactanceand larger rotor angle is neededto transfer the same steadypower. ,"
(both lines in)
/
Sudden Loss of One of parallel Lines considernow a singlemachinetied to infinite bus throughtwo parallellinesas in Fig. 12.25a.circuit model of the sysremis given in Fig. r2.25b. Let us study the transientstability of the ,yir"rn when one of the lines is suddenly switched off with the system operating at a steady road. Before switchingoff, poweranglecurve is given by P"r=
lE'llvl
sin d= Pm*l sin d xa i xt llx2 Immediately on switching off line 2, power angrecurve is given by
P"n= g:+ ,\d
T
rt7
sin d= pmaxrsin d
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Fig.12.26
Equalareacriterionappliedto the openingof one of the two linesin parallel
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Mod"rnpo*rr.surt*
ffiffi4l
nrryr', Power Syt St"blllry via the healthy line (through higher line reactance X2 in place of Xl ll Xz)7;with power angle curve sln
4=4o*_T_6, which is the samecondition as in the previous example. Sudden Short Circuit on One of parallel Case a: Short circuit
Lines
at one end of line
Let us now assumethe disturbanceto be a short circuit at the generatorend of line 2 of a double circuit line as shown in Fig. 12.27a.We shall assumethe fault to be a threephaseone.
sin d
obviously, Po[ ( P"*r. The rotor now starts to decelerate as shown in Fig. 12.28. The system will be stable if a decelerating areaA, can be found equal to accelerating area A, before d reachesthe maximum allowable value 4o*.At areaA, dependsupon clearing time /. (correspondingto clearing angle {), clearing time must be less than a certain value (critical clearing time) for the system to be stable. It is to be observedthat the equal area criterion helps to determine critical clearing angle and not critical clearing time. Critical clearing time can be obtained by numerical solution of the swing equation (discussedin Section 12.8).
P"y, prefault (2 lines)
P6n1,postfault (1 line)
X2 ,
?
(b)
F19.12.27 Shoftcircuitat one end of the line Before the occurrenceof a fault, the power angle curve is given by
'
p"t' = sind= p_*, sind ,)4,'rlr,,,a, xi + xltx2 which is plotted in Fig. 12.25. Upon occulrenceof a threephasefault at the generatorend of line 2 (see Fig. I2.24a), the generatorgetsisolatedfrom the power systemfor purposesof power flow as shown by Fig. 12.27b.Thus during the period the fauit lasts, The rotor therefore accelerate, .i;t:;i"s dincreases. synchronism will be lost unless the fault is cleared in time. The circuit breakers at the two ends of the faulted line open at time tc (correspondingto angle 4), the clearing time, disconnectingthe faulted line.
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t
6
Ffg. 12.28 Equalarea criterionappliedto the systemof Fig. 12.24a, I systemnormal,ll faultapplied,lll faultedline isolated. It also easily follows that larger initial loading (P.) increasesA, for a given clearing angle (and time) and thereforequicker fault clearing would be needed to maintain stable operation. Case b: Short circuit
away from line ends
When the fault occurs away from line ends(say in the middle of a line), there is some power flow during the fault thoughconsiderablyreduced,as different from case a where Pen= 0. Circuit model of the system during fault is now shown in Fig. 12.29a.This circuit reducesto that of Fig. 12.29cthrough one deltastar and one stardelta conversion. Instead,node elimination technique of Section 12.3 could be employed profitably. The power angle curve during fault is therefore given by P"t=
 Ellvl '1r'II
s i n d = P m a xsr irn d
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ffiil
ls#z#i
Modern Power svstem Analvsis '!'vev"
+t*F;4l
'
'

'
v'
'"
v,
'
'
"
'',
'
I
system operationis shown in Fig. 12.30,wherein it is possibleto find an area A, equalto A, for q. < 4nu*. At the clearing angle d. is increased, area ai increat"t und to nna Az = Ar, 4. increasestill it has a value 4n*' t6" ooi*,, ollnrvohlefnr stahilitv This caseof critical clearineangleis shown in Fig. 12.3L Pe
x,t
Pr1,prefault (2 lines)
X6
postfault (1 line) P6111,
xc
G
@
(b)
Xr
Fig. 12.31 Faulton middleof one lineof the systemof Fig. t2.l4a, case of criticalclearingangle (c)
Annlvins eoual areaeriterion to the caseof critical clearing angle of Fig. 12.31' we can wnte
Fig. 12.29
Pe
dntn'
4,
P"rand P,u as in Fig. 12.28 and Per as obtained above are all plotted in Fig. I2.3O. Accelerating area A, corresponding to a given clearing angle d is less
j (P^ 4n*u sinfldd= J {r^*r sind P^)d6 6,,
60
where P"'Prefault (2 lines)
 sinr (:t_) 4,,* =T
,.a
V
postfault(1 line) P"11;,
maxIII
(r2.66) ./
Integrating, we get l6* (P^a + Pmaxrrcos d)  * (P'*,,1 cos d +
P"11, duringfault
= Q
16o
or P^ (6",  6) * P.u*u (cos '[.  cos do) I P* (6**  6"r) * Pom (cos fi*  cos 4J = 0 Fig. 12.30 Faulton middleof one line of the systemof Fig. 12.24a with d"< {,
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t
cos {r = 4naxtn
 PmaxII
:otd",*
(12.67)
critical clearinganglecanbe calculated from Eq.(12.67)above.The anglesin are in radians.The equationmooiiiesas belowif the angJes are :lt::t1on ln oegrees. cos {. 
ft
r.(6 ^i*  do)  Pmaxrr cosdo * prnu*ru cosd,ou*
Give the system of Fig. 12,33 where a threephascfault is applied at rhe point P as shown. i0.s
Pmaxltr  Prnaxn
Infinite bus vFlloo
Case c: Reclosure If the circuit breakersof line 2 arereclosed successfully (i.e., the fault was a transient one and therefore vanishedon clearing the faurty line), the power transfer once again becomes P"N = P"r= p*u*I sin d Since reclosure restores power transfer, the chances of stable operati'n improve. A case of stable operationis indicated by Fig. 12.32. For critical clearing angle
4 = 4r* = 1T sinl 1p_/p*.*r; t ucr
6rc
sin 0 dd = pm) d,6 J @r, Pmaxrr J (p.*m sin d_
60
6r,
Flg. 12.33 Find the critical clearing angle for clearing the fault with simultaneousopening of the breakers I and 2. T\e reactance values of'various components are indicated on the diagram. The generatoris delivering 1.0 pu power at the instanr preceding the fault. Solution With referenceto Fig. 12.31, three separatepower angle curves are involved. f. Normal
dru, t
+ J (P,*r sin d_ p^) d6
operation
(prefault)
Xr=0.2s+ffi+0.05
6.
= 0.522pu
p,t=rysind:
ffirino
= 2.3 sin d
(il
Prefault operatingpower angle is given by 1.0 = 2.3 sin 6 or IL During
6o =25.8" = 0.45 radians fault
It is clear from Fig. 12.31 that no power is ffansferredduring fault, i.e., = o (Clearing angle)
\
(Angleof reclosure)
,0.:"o
Fig 1232 Faurtin middreof a rineof the systemof Fig. 12.27a where trrj tr, + r; T = time betw,een clearing anclreclosure.
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U
ModernpowerSystemAnatysis
iffi
E
=  1.5 (cos2.41 cos 6,)  (2.41 6")
rrr. Post fault operation (fault cleared by openingr the faulted Iiae)
= 1.5cos 6", + 6r, I.293 Setting A = Az and solving
6r,  0.45 = 1.5cos 6r,+ 6r, 1.293 l.2xl.0 n Perrr=
ff
sin d= 1.5sin 6
(iii)
 0.562 cos {,. = 0.84311.5
or
or 4, = 55.8" poweranglediagramsare shownin Fig. 12.35. The corresponding
Pe
Find the critical clearing angle for the system shown in Fig. 12.36 for a threephase fault at the point P. The generator is delivering 1.0 pu power under prefault conditions.
Pn=1 'O
i 0. 1s
i0.15
lnfinite bus
66=0.45 rad
Fig. 12.35 TTto rrrw
ooi*"* urour.rLurr
^*:^^:Ll^ psllluDDlulc
^l^ alilBrtr
C Omax
f^l()f
lvF1.otoo
lF,l=1.2Pu
6^rr=2.41rdd
jo.15
.10.15 afea
^ Al
=
. A2
given by
(Sge
flg.
Flg. 12.36
12.35)is Solution
I = 2.4Lradians 1.5 Applying equal area criterion for critical clearing angle { Ar = P^ (6",  6) = 1.0 (6",  0.45) = 6c, 0.45 4ou*=rsinl
dr*
Az= !{r,np^)d,6
f, Prefault bus is
2.41 I 6.,
Transfer reactance between generator and infinite
& = 0.25+ 0.17+
0 . 1 5 + 0 . 2 8 + 0 . 1=5 0.71 2
Pc'  , = r ' Z x l s i n d = 1 . 6 9s i n 6 0. 71 power angle is given by The operating
or
do = 0.633 rad
The positive sequencereactancediagram during fault is IL Durtng fault presentedin Fig. 12.37a.
r2.41
=  1 . 5 c o ds _ d l  6",
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(i)
1.0= 1.69sin ,fr
6,,
=  ( 1 . 5s i n 6  1 ) d d J
operation
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PowerSystemStabilitv J
/ 000 Lr \000
000 L
j0.25 0 '
Perrr=U! ' l
Mi#ffi
r*
sin d = r'2 sin 6
(iii)
jo.17 j0.15
+
j0.14
j0.14
With referenceto Fig. 12.30and Eq. (12.66),we have
j0.'15
) E1=t.z
+
V=1.0
To find the cqitical clearing angle, areas A1 and A, arc to be equated. 6",
 , At = l.o (6,, 0.633)
(a) Positivesequence reactancediagramduringfault j0.25
j0.145
j0.145
',
J
o.+e5 sin d dd
60
j0.17
lE'l=1.2
and
Az =
V=1.OlOo
dmax f J
 1 . 2s i n d d d  1 . 0( 2 . 1 5 5 4 )
6  cr
Now At =Az (b) Networkafterddltastar conversion
or
J
6r, = 0.633
0.495sin d dd
o.63J
l9l=1'z
V=1.0100
2.155
=
[ t.Z rin 6 d6  2.t55 + 6,, '
J 6cr
(c)Network afterstardelta conversion .
or  0.633+ 0.495cos olo' =  1.2cos ol"tt 2.155
Ftg. 12.32
lo.orr
Converting delta to star*, the reactancenetwork is changed to that of Fig. 12.37(b). Further, upon converting star to delta, we obtain the reactance network of Fig. .r2.37(c). The transfer reactanceis given 6y
(0.2s 0.072s + 0.145) + (0.145 + 0.17)0.0725 + (0.25 + 0.145) (0.14s 0.17) + Xu=
la.,
or  0.633 + 0.495 cos 6,,  0.399 = 0.661 + 1.2 cos 6",  2.155 or cos 6r, = 0.655 U 6r, = 49.I"
0.075
_ 2.424 rpeI =
fir. Postfault
lal! i.+Z+
operation
uur sin vd = 0.495 v.a/, sin
(faulty
6
line switched
(ii) off)
X r l = 0 . 2 5 + 0 . 1 5+ 0 . 2 8 + 0 . 1 5 + 0 . 1 7= 1 . 0 *Node elirnination techniquewould be'used for complex network.
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A generatoroperating at 50 Hz delivers 1 pu power to an infinite bus through a transmission circuit in which resistance is ignored. A fault takes place reducingthe maximum power transferableto 0.5 pu whereasbefore the fault, this power was 2.0 pu and after the clearanceof the fault, it is 1.5 pu. By the use of equal area criterion, determine the critical clearing angle. Solution All the three power angle curves are shown in Fig. 12.30.
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
,'ffi
Mod"rnPo*.. sEl!"nAn"lytit
Ilere
= 1.5 pu P"*r =2.0 pu, Pmaxl= 0.5 pu and Pmaxrrr
2.The angular rotor velocity u= d6ldt (over and above synchronous velocity
Initial loading P^ = 1.0 pu
( p 6r,ro= zrsinI tffiJ
\
1 :2.4!rad
E7sinl
1.5
Applying Eq. (r2.67)
Continuous solution
U
 0.523)  0.5cos0.523 1.0(2.41 + 1.5cos2.41= o ??? cos {, 1.5 0.5 6r, = 70'3"
t Af
n
n1
r>2
Discretesolution un/2
t
u13/2 unI/T+tsn4l2
T2.9
NUMERICAT
SOTUTION
OF SWING EOUATION n2
In most practical systems,after machine lumping has been done, there are still more than two machines to be considered from the point of view of system stability. Therefore, there is no choice but to solve thp swing equation of each machine by a numerical techniqueon the digital computer. Even in the case of a single machine tied to infinite bus bar, the critical clearing time cannot be obtained from equal area criterion and we have to make this calculation ! . .rr
numerlca[y
rr
l
mrougn
:

swulg
Ll^,
equauulr.
zFL^^
 ^Ll^+i^^+^l *^+L^lt rttrIc aIU ssvtrIilr JuPurDtruilL('(l lllELlluLlD
now available for the solution of the swing equation including the powerful RungeKutta method. He.rewe shall treat the pointbypoint method of solution which is a conventional, approximatemethod like all numerical methods but a well tried and proven one. We shall illustrate the pointbypoint method for one machine tied to infinite bus bar. The procedure is, however, general and can be appliedto every machine of a multimachine system. Consider the swing equation d26 1 = ;T ;e*P^*sind):
PolM;
(*  9H orinpu systemM = +) \
7t
iTf)
The solution c(r) is obtained at discrete intervals of time with interval spread of At uniform throughout. Accelerating power and changein speed which are continuous functions of time are discretrzedas below: 1. The acceleratingpower Po computed at the beginning of an interval is assumed to remain constant from the middle of the preceding interval to the middle of the interval being consideredas shown in Fig. t2.38.
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p3l2
n'l
r>112
n
t Af
$ni
6nz
Jn1
n2 Fig. 12.38
n
Af
Pointbypoint solution of swing equation
In Fig. L2.38, the numbering on tl\t axis pertains to the end of intervals. At the end of the (n l)th interval, the accelerationpower is Pa (n_r)Pm P* sh 4r Q2.68) where d_1 has been previously calculated.The changein velocit! (a= d6ldt), causedby the
[email protected]),assumedconstant over At from (n312) to (nll2) is w n  '2
wn 3t 2=( Lt / M )
[email protected] r )
The change in d during the (nl)th interval is L6rt= 6r1 6n2= A'tun4'2
(12.6e) (12.70a)
and during the nth interval L6r 
6n 6n t = / \ t un 112
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
(12.70b)
,ir Yirtvt r t Ar lr.rl b r )
power SystemStabilitv [i{8il;r
SubtractingEq. (12.70a\ from Eq. (12.70b) and using Eq. (12.69), we get
L'6,= A6,t +
(A r)2 D r M
a(.nI)
Q^t,,1;^^
\rvtu.,v,t
Using this, we can write 6n= 6nt + L,6n
7 , ,, 6=,
(Ar)t M * P a2o +
where Pos* is the acceleratingpower immediately after occurrence of fault. Immediately before the fault the system is in steady state, so that Poo= 0 and ds is a known value. If the fault is cleared at the beginning of the nth interval, in calculation for this interval one should use for
[email protected])the value llP"6r>+ Po6_9*), where
[email protected]_r)is the acceleratingpower immediately before clearing and Po6_r)+is that immediately after clearing the fault. If the discontinuity occurs at ihe miciciie of an intervai, no speciai proceciure is neecled.The increment of angle during such an interval is calculated, as usual, from the value of Po at the beginning of the interval. The procedure of calculating solution of swing equation is illustrated in the following example.
A 20 MVA, 50 Hz generator delivers 18 MW over a double circuit line to an infinite bus. The generatorhas kinetic energy of 2.52 MJA4VA at rated speed. The generator transient reactanceis X/o = 0.35 pu. Each transmission circuit has R = 0 and a reactance of 0.2 pu on a 20 MVA bgqe.lE/l = 1.1 pu and infinite bus voltage V = 7.0 10". A threephaseshort circuit occurs at the mid point of one of the transmissionlines. Plot swing curves with fault cleared by simrrltaneousopening of breakersat both ends of the line at2.5 cycles and 6.25 cycles after the occuffence of fault. Also plot the swing curve over the period of 0.5 s if the fault is sustained.
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^f^^
ws
^*1,,
+L^
Lall aPPt.y ultt
^+^
L..
^+^
stEPUyslttP
^rl.^l
lIIculUU,
 ^ t
^
 r
ra
Wtr lltrC(l t() Calculate
the inertia constant M and the power angle equations under prefault and postfault conditions. Base MVA = 20
(12.7r)
G2.72) The processof computation is now repeatedto obtain Pa61, L6r*tand d*t. The time solution in discrete form is thus carried out over the desiredlength of time, normally 0.5 s. Continuous form of solution is obtained by drawing a ;mooth curve through discrete values as shown in Fig. 12.38. Greater accuracy of solution can be achi.evedby reducing the time duration of intervals. The occurrence or removal of a fault or initiation of any switching event causesa discontinuity in acceleratingpower Po.lf such a discontinuityoccurs at the beginning of an interval, then the averageof the values of Po before and after the discontinuity must be used.Thus, in computing the incrementof angle occurring durirrg the first interval after a fault is applied at t = 0, Eq. (I2.7I) becomes
E
nsluls
IneRia coflstant, Mepu\ =
1.0x L52 1 8 0x 5 0
1 8 0/
= 2.8 x 10+ s2le\ectdegree I
I
Prefault &' = 0 . 3 5 +
0'2 =0.45 2
Pd= Pr.*r sin d !,.lxt . r = = '.;;sin 2.M sin 5 d
(i)
Prefault power transf'er = + = 0.9 pu 20 Initial power angle is given by 2.44sin4=0.9 or
\
6o= 21.64"
II During fault A positive sequence reactancediagram is shown in Fig. 12.39a.Converting star to delta, we obtain the network of Fig. 12.39b, in which 2 , , = 0. 35x 0. i + A. 2x0. i + 0. 35x0.= trtr 0l
1 AI..Z) pu
P.u = Pmaxtt sin d
 1'1x1 r;n d = 0.88sin 6 1.25
Fig. 12. 39
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
(ii)
lil. Postfault
With the faulted line switched off,
055 J;=:::,J;2
which it is obvious that the systernis unstable'
: : .: 1t ' x '1s.i n dI = 2 . 0 s i n d = 1
Let us choose Al = 0.05 s The recursive relationshi'ps for stepbystep swing curve calculation are reproducedbelow. Pa(n_r)=P^  P** sin 4_r (iv) L6n= L6nt*
t ;100
o E @
(Lt)z
' 'oa(nl)
M 6n= 6nt + A,6n
(v)
o 8 0 o)
(vi)
ioo P
c (5 o
Sincethereis a discontinuity in P, andhencein Po, the averagevalueof po mustbe usedfor the first interval. P"(0)= 0 pu and Po (0*) = 0.9  0.88 sin 2I.64" = 0.576pu = 9t#ZQ Po(ouu.,us"l
fault cleared at 2.5 cycles
= 0.288pu
L
I o
Sustained Fault Calculations are carried out in Table 12.2 in accordance with the recursive relationship (iv), (v) and (vi) above. The secondcolumn of the table showsP* the maximum power that can be transferredat time r given in the first column. Pn * in the case of a sustainedfault undergoesa sudden change at t = 0* and remains constant thereafter.The procedure of calculations is illustrated below by calculating the row correspondingto t = 0.15 s. ( 0 ' l s e c )= 3 1 . 5 9 " P."* = 0.88 sin d (0.1 s) = 0.524 P, (0.1 s) = P,,,u*sin 6 (0.1 s) = 0.88 x 0.524  0.46I P, (0.1 s) = 0.9  0.46I  0.439 ( At\2
YP, M
(0.1 s) = 8.929x 0.439 3.92 6 (0.1ss) = Ad (0.1s) + qL M
P, (0.1s) U
\
= J.38"+ 3.92"= 11.33" d ( 0 . 1 5s )= d ( 0 . 1s ) + A d ( 0 . 1 5s ) = 31.59"+ 11.30'= 42.89"
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t
l 0.1
l
t l l 0.3 0.2 f (s) 
l i 0.4
i 0.5
_ r l
0.6
fault and for Fig. 12.40 Swingcuryesfor Example12.10for a sustained clearingin 2.5 and 6.25 cYcles aaDle
!A A aZ.Z
n^i. l., ^aia* fUllll'UyPulllt
rratinnc t/vlllPutqrrvrro
n{ vr
ce rr r' rr ri n r Vn
n r r' n / a vu
fnr
qtrctainarl
fattlt
/ f = 0. 05s
t
P^u
sec
pu
0+ o^u, 0.05 0.10 0 .t 5 0.20 0.25 0.30 0.35 0.40 0.45 0.50
2.44 0. 88 0. 88 0.88 0.88 0. 88 0. 88 0. 88 0.88 0.88 0.88 0.88
sin 6
Pu
0. 368 0. 368 0. 368 0. 41 0.524 0.680 0. 837 0.953 0.999 0.968 0. 856 0.657
a
P"=Prrr,*sin6 P,,= 0'9 P,
0324 0.361 0.46r 0. 598 0. 736 0. 838 0. 879 0. 852 0.154 0.578
deg
Pu
0. 0 0. 576 0. 288 0. 539 0.439 0. 301 0.163 0. 06 0.021 0.048 0.145 0.32r
6
2.57 4.8r 3. 92 2.68 r . 45 0. 55 0.18 0.426 1.30 2. 87
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
6 deg
2r.64 2r . 64 74.21 31.59 42.89 56.87 72. 30 r 6 . r 6 88. 28 1 6 . 5 8 r04.44
2.57 7.38 11.30 13.98 15.43 15.98
r 7 . 8 8 r2r.02 20.75 138. 90 1s9.65
I
f ttlt F . a su.a
Modern
[email protected]
Flaae^'J vtcrete.t
:tlt
ltr
o E ntA,i, fryCIeS
Time to clear f'ault= 2.5 = 0.05 s 50 Pu^ suddenly
to 2.0 at t = 0.05. Since the
be assumedto remain constant from 0.025 s to 0.075 s. The rest of the prtrcedureis the sameanclcomplete calculations are shown in Table 12.3.The swing curve is plotted in Fig. 12.40 from which we find that the generator undergoesa maximum swing of 37.5" but is stable as c5finaily begins to decrease.
progressively greater clearing time till the torque angle d increases without bound. In this example, however, we can first find the critical clearing angle using Eq. (12.67) and then read the critical clearing time from the swing curve corresponding to the sustainedfault case. The values obtained are: Critical clearing angle = 118.62 Critical clearing time = 0.38 s Table 12.4 Computations of swing curve for fault clearedat , = 0. 05s 6. 25cycles( 0. 125s)Af
Table 12'3 Computations of swing curyesfor fauttcleared at2.s cvcles (0 .0 5s ), At = 0.05s
P,no
sin 6
P"=P^ *sin6
4
Po= 0.9 P"
pu P,,,,,^
.sin.5
pu
0 2.44 0. 0 .8 8 ouu, 0. 05 0 . 8 8 0. 05+ 2 .0 0 0.05uus 0 . 1 0 2.00 0 . i 5 Z .U U o. 20 2 .0 0 0.25 2 .0 0 0 . 3 0 2 .0 0 0 . 3 5 2 .0 0 o. 40 2 .0 0 0.45 2 .C 0 0. 50 T
Pr,=P,rr.,*,tin5 pr,= 0.9 pu pu
0.368 0 .3 6 8 0 .3 6 8 0.41 0 .4 1
0 .3 6 0 .8 2
0.493 0 .5 6 0.s91 0.597 0.561 0.494 0.41 0.337
0 .9 8 6 I.t2 T.I9 r.t9 I.t2 0 .9 8 9 0.82 0.615
0 .9 0.324
pu
0.0 0.576 0.288 0.54 0.08 0.31  0.086  0.22  0.29  0.29  0 .22  0.089 0 .08 0.225

Fault Cleared in 6.25 Cycles Time to clearfault= ua?t = 0.125 s s0
A6
6
deg
deg
2.57
2.57
2.767 0.767 1.96 2.s8 2.58 1.96 0.79 0.71 2.0
5.33 4.56 2.60 0.02 2.56 4.52 5.31 4.60 2.6
_ _ 
21.64 21.64 21.64 24.21 24.21 24.21 29.54 34.10 36.70 37.72 34.16 29.64 24.33 19.73 17.13
0 0+ ouu, 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 A
A ?
u.4J
0. 88 0.88 2.O0 2.00 2.00 2. 00 2. 00 2.00 ' \ N
Z.W
deg 0. 368 0. 368 0. 368 0.41 0.524 0.680 0.767 0. 78 0. 734 0.613 0.430
0.36r o.46t 1.36 1.53 1.56 1.46 r.22 0. 86
u. z11
u.4c)0
0.9 0.324

0.0 0.576 0. 288 0.539 0.439 4.46 0.63 0.66 0. 56 0.327 0.04 u.4J4

T2.TO MULTIMACHINE
6 deg
. 2.57 2.57 4.81 7. 38 3. 92 11.30 4.10 7.20 5.66 r . 54 5.89  4.35 5. 08  9.43 2.92  12.35 0. 35  12.00  u.t  J  r . 6/
0.50 2.oo
2r.& 21.64 zt.U 24. 2r 3t.59 42.89 50. 09 51.63 47.28 37.85 25.50 l_J. ) u 5.37
STABITITY
From what has been discussedso far, the following stepseasily follow for determiningmultimachinestability. 1. From the prefault load flow data determine E/ovoltage behind transient . reactancefor all generators.This establishesgeneratoremf magnitudes lEll which remain constant during the study and initial rotor angle 6f = lEt. Also record prime mover inputs to generators,P*o  PoGk 2. Augment the load flow network by the generator transient reactances. Shift network busesbehind the transient reactances. 3. End Inus for various network conditions*during fault, post fault (faulted line cleared), after line reclosure. 4.
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2.44 0.88
6
For faulted mode, find generator outputs from power angle equations (generalizedforms of Eq. (12.27)) and solve swing equationsstep,by step (pointbypoint method).
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
Modern power SystemAnalysis
4ffi 
5. Keep repeating the above step for post fault mode and after line reclosuremode. 6. Examine d(r) plots of all generators and establish the answer to the stability question. The above stgps are illustrated in the following example.
A 50 Hz, 220 kV transmissionline has two generatorsand an infinite bus as shown in Fig. 12.4I. The transformer and line data are given in Table I2.5. A threephasefault occurs as shown. The prefault load flow solution is presented in Table 12.6. Find the swing equation fbr each generatorcluring the fault period.
o
o
Vz=1.0328.2350
) v i F1.O217.16o
([5, ) v s = 1 . o l l z
lltI' 
22.490
i
Se ).5+j(
6) i6,u
l,,m 220kV, 100M VAbase
0.1r 0.0235 0.04 o.022 0.04
0.018 0.004 0.007
Line 45 Line 51 Line 4I Trans;24 Trans:35
S.No. and
Voltage Polar
Bus No.
Form
Bus Upe
Voltage Real e
Slack  1.010" PV 2 t.0318.35" 3 t.0217.16 PV 4 1.017414.32"PQ 5 1 . 0 1 1 2 1 2 . 6 9 "P Q
1.00 t.0194 1.0121 1.0146 1.0102
Data are given below for the two generatorson a 100 MVA base. Gen 1 500 MVA, 25 kV, XJ = 0.067 pu, H = 12 MJAyIVA Gen 2 300 MVA, 20 kV, X,j = 0.10 pu, H = 9 MJA4VA Plot the swing curves for the machines at buses 2 and 3 for the above fault which is cleared by simultaneousopening of the circuit breakersat the ends of the faulted line at (i) 0.275 s and (ii) 0.0g s.
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Generation
Load
Imaginary f
 3. 8083 0.2199 0 0 0. 0 0 0.6986 0 3. 25 0. 1475 0 0.3110 0 0.1271 2.10 1. 0 1. 0 0. 44 0 0.167 0 0.5 0.16 0 0.0439
Solution Before determining swing equations, we have to find transient internal voltages. The current into the network at bus 2 basecion the rn n u r.If n o t, r r+ r and repeatfromstep5 above. Otherwise print results and stop. The swing curves of all the machines are plotted. If the rotor angle of a machine (or a group of machines) with r"rp".t to other machines increases without bound, such a machine (or grouf of machines) is unstable and eventually falls out of step. The computational algorithm given above can be easily modified to include simulation of voltage regulator, field excitation response, saturation of flux paths and governor action Study of Large Systems
To limit the computer memory and the time requirements and for the sake of computationalefficiency, a large multimachine system is divided into a study subsystemand an external system.The study subsystem is modelled in detail whereas approximate modelling is carried out for the external subsystem.The total study is renderedby ihe modern techniqueof dynamic equivalencing.In the externalsubsysteln,nutnber o Ail J3
System e.:^ s€sstlt€rrt involves two func tions: (i) system monitoring and (ii) contingencyanalysis.Systemmonitoring provides the operator of the power system with pertinent uptodate information on the current condition:;clf the power system. In its simplest form, this just detects violations in the actual systemoperating state.Contingency analysisis much
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اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
,.sfe ;l
Modern power Svstem Analvsis
Only a small proportion of work on optimal power flow (OPF) has taken into account the security constraints.The most successfulapplications have beento the security constrained MW dispatch OPF subproblem. The contingencyconstrainedvoltageivar reschedulingproblem, as of the writing of this text, still remains to be solved to a satisf desree. The total number of contingency constraintsimposed on SCO is enormous. The SCO or contingency constrained OPF problem is solved with or without first optimizing with respectto the base case(precontingency)constraints.The general procedure adopted is as follows: (i) Contingency analysis is carried out and cases with violations or near violations are identified. (ii) The SCO problem is solved. (iii) The rescheduling in Step 1 might have created new violations, and therefore step 1 should be repeatedtill no violations exist. Hence, SCO representsa potentially massiveadditional computing effort. An excellent comprehensiveoverview of various available methods is presentedby Stott et. al [15]. There is still great potential for further improvement in power system security control. Better problem formulations, theory, computer solution methods and implementation techniquesare required. T3.4
I_ srai
Power System Security SLACKBUS
t +s*y ts
f+o*ys
1.024215.0" 3 1 . O 6 t O "1
.>40.7+ j1.2 _ 39.5_i3.0:_ ,l
,f88.ey8.6
_ + 1 8 . 9 y 5 . 2 _ 1 6 . 9+ 1 3 . 2  <
Ja.a/e.8 +
i6.3j2.3
zt.si5.9 I z+.2*js.a
'L?:'q'''/
t az.s*
t
* .i;n;l+s 53.7i7.2 '
*S4.9 +17.3
41.0236t5.3"
1.0474t2
'"'' l+u 5 i1.o17gt6.2" Y
*oo*ito
) , zo+1t o !G) t40 +i30 Fig. 13.2
Base Case AC Lineflow for sample5 bus system
SLACKBUS
{+s*yts
f+o+7s
1 . 0 1 0 7 1 5 . 9 " 3 1.0610" 1
>48.6 + j5.2 46.7  j5.3
u,
(r4.49b)
where a, is a preselectedconstant threshold level. Obviously, the performance indcx ./("r)may bc cxpressedas m
I(i) = Dy, c;) ; .  
It 549. The main advantage of the choice of the form (14.49) for the estimation index is that it is still a quadraticin the function g(.i) and so the LSE theory may be mimicked in order to derive the following iterative formuia for the state to State Estimationof Power Systems An tntroduction
ModernpowerSystemAnalysis
I
(14.s0)
I I
and each componenthas a quadratic nature for small values of the residual but has a constantmagnitudefor residual magnitudesin excessof the threshold. Figure 14.5 shows a typicai variation of J, (.r) for the quadratic and the nonquadratic choices.
(74.5ra) wherethe matrix C is diagonaland its elementsarecomputedas
Comparing this solution with that given in Eq. (14.22), it is seen that the main effect of the particularchoice of the estimationindex in Eq. (14.49) is to ensurethat the data producingresidualsin excessof the thresholdlevel will not changethe estimate. This is achieved by the production of a null value for the matrix C for large values of the residual. I4.8
NETWORK OBSERVABILITY MEASUREMENTS
I
Fig. 14.5
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AND PSEUDO
A minimum amount of data is necessaryfor State Estimation (SE) to be effective. A more analytical way of determining whether a given data is enough for SE is called observability analysis.It forms an integral part olany real time state estimator.The ability to perform state estimation depends on whether .sufficientmeasurementsare well distributedthroughoutthe system. When sufficient measurementsare available SE can obtain the state vector of the whole system. In this casethe network is observable.As explained earlier in Sec. 14.5 this is true when the rank of measurententJacobiantlratrix is equal to the number of unknown state variables. The rank of the measurement Jacobian matrix is dependent on the locations and types of available rneasurementsas well as on the network topology. An auxiliary problem in state estimation is where to add additional data or pseudomeasurementsto a power systemin orderto improvethe accuracyof the calculated state i.e. to improve observability. The additional measurements represent a cost for the physicat transducers,remote terminal or telemetry sy.stem,and software data processing in the central computer. Selection of pseudomeasurements,filling of missing data,providing appropriateweightage are the functions of the observability analysis algorithm. UbsefvaDlllty
\
( 1 4 . s1 b )
Ct = l, f ot l; lo, S ai = 0, f or lilo, > a'
I
 r
I
l:
^
Can De CIICCKCU UUtrrlB
C^^t^i^+l^
lilulullzittLltrll.
Tf
^.'
^i"^+
Il 4IrJ Prv\rr
L^^^^.
'^.' r.rvtrrr.rrrlr .YvrJ
small or zero during factoization, the gain matrix may be singular, and the systemmay not be observable. To finil the value ol an injection without nteasuringit, we tlrust know the power system beyond the measurementscurrently being made.For example, we pormally know the generated MWs and MV ARs at generators through telemetry channels (i.e. these measurementswould generally be known to the
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
An introduction to state Estimation of powqr systems
i,
t,
stateestimator). If thesechannelsare out, we can perhapscommunicate with the operatorsin the plant control room by telephoneand ask for these values and enter them manually. Similarly, if we require a loadbus MW and MVAR for a pseudo measurement'we could use past recordsthat show the relationship between an individual load and the total qyrtemload. We canestimate the tstal systetn load quite accuratelyby f)nding the total power being generated and estimating the line losses.Further, if we have just had a telernetry failure, we could use the most recentlyestimatedvaluesfiom the estimator(a.ssgming that i','is run periodic:ally)as pseudorlreasureftlents. Thus, if required,we can give . the state estimatorwith a reasonablevalue to use as a pseudomeasurement at any bus in the system. Pseudomeasurements increasethe data redundancyof SE. If this approach is adapted, care must be taken in assigning weights to various types of measurements. Techniquesthat can he useclto cleternrine the rnctcror.pseu4g measurementlocationsfbr obtaininga completeobservabilityof the sysrem are available in Ref. [251.A review of the principal observability analysis and m et erp l a c c rn c nat l g ' ri th rn si s a v u i ra hrci n I(cr' . [261. T4.9
APPLICATION ESTIMATION
U ' E
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changingand hcncel.hccurrerttr.clentetered breaker'/switch stertus must be used to restructurethe electricalsystem model. This is called the network tr,tpop.tgy program or system status proce,\.tor or netwc;rk configurator.
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c o o o a U)
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i l_
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z 8
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topology,
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In realtime environmentthe state estimatorconsistsof different modulessuch as network topologyprocessor,observabilityanalysis,stateestimation and bad data processing.The network topology processoris required for ail power systemanalysis.A conventionalnetwork topologyprogram usescircuit breaker statusinformation and network connectivity datato determinethe connectivity of the network. Figtrrc 1r4.6is it schcrnaticcliagtarnshowing the info.rnationflow between the variorrsfunctions to be performed in an operationscontrol centre compurer system. The system gets information from remote terminal unit (RTU) that cncodc lllcasLlrclllcnt tl'ullsduccr0rrtputs anclopcnccl/closccl staLusinlbmration into digital signals which are sent to the operation centre over communications circuits. Control centre can also transmit commands such as raiseflower to generators and open/closeto circuit breakers and switches. The analog measurementsof generatoroutput would be rJirectlyused by the AGC program (Chapter 8). However, rest of the data will be processedby the stateestimator befbre being used fbr other functions such as OLF (Optimal Loacl Flow) etc. Before running the SE, we must know how the transmission lines are
I
I FSI t 
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
An lntroductionto State Estimationof Power
Po Modern
ilr l I
The output of the stateestimator i.e. lvl, 6, P,j, Q,jtogether.with latestmodel form the basis for the economic dispatch (ED) or minimum emission dispatch (MED), contingency analysis program etc. Further
Readin
The weighted leastsquaresapproach to problems of static state estimation in power systems was introduced by Schweppe 11969741. It was earlier originated in the aerospaceindustry. Since 1970s, state estimators have been installed on a regular basis in nern'energy (power system or load dispatch) control centres and have proved quite helpful. Reviews of the state of the art in stateestimation algorithms basedon this modelling approach were published by Bose and Clements[27] and Wu [28]. Reviewsof externalsystemmodelling are available in 1291.A generalisedstate estimator with integrated state,stutus and parameterestimation capabilities has recently been proposed by Alsac et al [30]. The new role of stateestimationand other advancedanalytical functions in competitive energy markets was discussedin Ref. [31].A comprehensive bibliography on SE from 196889 is available in Ref. [32].
F i g .P . 1 4 . 2
14.3Given a single line as shown in Fig. p 14.3, two measurementsare available. using DC load flow, calculate the best estimate of the power flowing through the line.
4= 0 rad. /1

( ,r1r t t1
LEMS PROB
Mt2
[l
y0.1pu (1ooMVAbase)
Mzt
l 
2
Fig. P. 14. 3 14.1 For Ex. 6.6 if the power injected at buses are given as Sr = 1.031 j0.791, Sz = 0.5 + 71.0and 'S3=  1.5  j0.15 pu. Consider Wt= Wz= Wz = l. Bus I is a referencebus. Using flat start, find the estimatesof lV,l and {. Tolerance= 0.0001. lAns: Vl = l/0", V', = t.04223 10.4297",V\ = 0.9982q l2.1864";
Meter
Mrz
l1.356, Va  1.03831 Final values: Vt  1.04./.0", V2 = 1.080215 l 3.736"1. 14.2 For samplesystemshown in Fig. P. 14.2,assumethat the threemeters have the followine characteristics. Meter
Full scale (MW)
100 r00 100
Mrz Mtt Mzz
Accuracy (MW)
o (pu)
+ 8 +4 + 0.8
0.02 0.01 0.002
Calculate the best estimate for the phase angles 4 *d f,'11,...,:  1rl l\., W   l5
 I lU{lDLll
nmnnf t, t/l I lUl lL,)
Meter
Meusured value (MW)
Mtz
70.0 4.0 30.5
Mrz Mt,
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d2 given the
Mzr
Full scale (MW)
100 100
Meter Standard Deviation ( o ) in full scale
Reading
1 4
32 26
Meter
(Mw)
REFEREN CES Books l. Mahalanabis,A.K., D.P. Kothari and S.L Ahson, ComputerAidetl pr;wer System Analysis and Control, Tata McGrawHill, New Delhi, 19gg. 2. Nagrath, I.J. and D.P. Kothari, Power SystemEngineering, Tata McGrawHill. Ncw Dclhi, I q94. 3. Mtrnticclli' A., State li.rtinrution in Eler:tric:Power SysremsA Gcnerali.recl Apprctat:h, Kluwcr Academic Publishers,Boston, 1999. 4. Kusic, G'L., ComputerAidedPower SystemsAnalysis,PrenticeHall,N.J. 19g6. 5. Wood, A'J. and B.F. Wollenberg, Power Generation,Operation and Control. Znd Ed., John Wiley, NY, 1996.
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
'l I
I j
t

ModernPowerSystemAnalysls
;554.;l
An Introductionto State Estrmationof Power Systems
6. Grainger, J.J. and W.D. Stevenson, Power System Analysis,McGrawHill, NY', 1994. 7. Deautsch,R., Estimation Theory, PrenticeHall Inc' NJ, 1965 g. Lawson, C.L. and R.J. Hanson, Solving Least SquaresProblens, PrenticeHall.
26. Clements, K.A., "Observability Methods and optimal Meter Placement", Int. J. Elec. Power, Vol. 12, no. 2, April 1990,pp 8993. 27. BoSe,A. and Clements,K.A., "RealtimeModelling of Power Networks", IEEE Proc., Special Issue on Computersin Power Svstem Operations,Vol. 75, No. 12,
inc., NJ., i974. g. Sorenson,H.W., ParameterEstimation,Mercel Dekker, NY, 1980'
Dee 1987;aP 76ff7=1ffi2: 28. Wu, F.F., "Power System State Estimation: A Survey", Int: J. Elec. Power and Energy Syst.,Vol. 12, Jan 1990, pp 8087. 29. Wu, F.F. and A. Monticelli, "A Critical Review on External Network Medelling for Online Security Analysis", Int. J. Elec. Power and Energy Syst.,Vol. 5, Oct
Papers 10. Schweppe,F.C., J. Wildes, D. Rom, "Power SystemStatic StateEstimation,Parts l, ll and lll", IEEE Trans', Vol. PAS89, 1970,pp 120135' ll. Larson, R.E., et. al., "State Estimation in Power Systems", Parts I and II, ibid, pp 345359. lZ. Schweppe,F.C. and E.J. Handschin,"static State Estimation in Electric Power System", Proc. of the IEEE, 62, 1975, pp 972982' 13. Horisbcrgcr,H.P., J.C. Richarcland C. Rossicr,"A Fast DccouplcdStatic Statc Estimator for Electric Power Systems", IEEE Trans. Vol. PAS95, Jan/Feb1976,
pp 2 O 8 2 1 5 .
1983,pp 222235. 30. Alsac, O., et. aI., "Genetalized State Estimation", IEEE Trans. on Power Systems, Vol. 13, No. 3, Aug. 1998,pp 10691075. D. et. al., "TransmissionDispatch and CongestionManagement 31. Shirmohammadi, Market Structures",IEEE Trans. Power System.,Vol 13, Energy Emerging in the No. 4, Nov 1998,pp 14661474. 32. Coufto, M.B. et. aI, "Bibliography on Power System State Estimation (19681989)",IEEE Trans.Power Sysr.,Vol.7, No.3, Aug. 1990,pp 950961.
1 4 . Monticelli, A. and A. Garcia, "Fast DecoupledEstimators", IEEE Trans' Power Sys/, 5, May 1990,pP 556564. 1 5 . Dopazo,J.F. et. al., "State Calculation of Power Systems from Line Flow PartsI and ll", IEEE Trans.,89, pp. 16981708,91, 1972,pp Measurcments, 145151. A.S. anclR.E. Larson, "A Dynamic Estimatorfor Tracking the Stateof a Dcbs, 16. Power System", IEEE Trans' 89, 1970,pp 16701678' iZ. K1upholz, G.R. et. al.,"Power SystemObservability:A PracticalAlgorithm Using Nctwork Topology", IEEE Trans. 99' 1980' pp 15341542' A and V.H. Quintana,"A Robust Numerical Techniquefor Power lg. SirnoesCosta, SystemStateEstimation", IEEE Trans. 100, 1981, pp 691698' A and V.H. Quintana,"An OrthogonalRow ProcessingAlgorithm 19. SimgcsCosta, for Power System SequentialState Estimation", IEEE Trans., 100, 1981' pp
ii
I
_TI 555
3'79r3799. 20. Debs, A.S., "Estimation of External Network Equivalentsfrom Internal System Data", IEEE Trans., 94, 1974, pp 12601268' 21. Garcia.A., A. Monticelli and P. Abreu, "Fast DecoupledStateEstimationand Bad Dara Processing",IEEE Trans. PAS98, Sept/Oct 1979, pp 16451652. 22. Handschin,E. et. al., "Bad Data Analysis for Power System State Estimation", IEEE Trans., PAS94, 1975, pp 329337.t2 4J.
r.^r:6 r\V6lrrrt
rJ r Lt.r.
or @nle . ,
eL.
"Flqd T)cre Detecfion
and ldentification".
Int. J. EleC. POWer,
Vol. 12, No. 2, April 1990, PP 94103' 24. MeJjl, H.M. and F.C. Schweppe,"Bad Data Suppressionin Power SystemState Estimation",IEEE Trans. PAS90, 1971' pp 27182725' 25. Mafaakher, F., et. al, "Optimum Metering Design Using Fast Decoupted Estimator", IEEE Trans. PAS98, 7979, pp 6268.
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اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
in lglver Systems Compensation
I
55?
can be connecteciin the system in two ways, in series and irr shunt at ihe line ends (or even in the midPoint)' Apart from the wellknown technologies of compensation, the latest technology of Flexible AC TransmissionSystem (FACTS) will be introduced towards the end of the chaPter. 15.2
LOADING
CAPABILITY
There are three kinds of limitations for loading capability of transmission system: (i) Thermal (ii) Dielectric (iii) Stability Thermal capabitity of an overhead line is a function of the ambient temperature, wind conditions, conditions of the conductor, and ground
15.1
INTRODUCTION
For reduction of cost and improved reliability, most of the world,s electric power systemscontinueto be intercor of diversity of loads,availability of sor to loads at minimum cost and pollr deregulatedelectric service environme to the competitive environmentof reli Nowadays,greaterdemandshave been placedon the transmissionnetwork, and these demandswill continue to rise becauseof the increasingnumber of nonutility generatorsand greatercompetition among utilitics the'rselves.It is not easy to acquire new rights of way. Increased demands on transmission, absence of longterm planning, and the need to provide open access to generatingcompaniesand customers have resultedin less securityand reduced quality of supply. compensationin power systemsis, therefore, essentialto alleviate some of theseproblems'series/shuntcompensation has beenin usefor past many years to achieve this objective. In a power system, given the insignificant erectricarstorage, the power generationand load mttst balance at all times. To some extent,the electrical system is self'regulating. If generationis less than roacl,voltage anclfrequency d'op, and thereby reducing the road. However, there is only a few percent margin for such selfregulation.lf ,,^r+^ : :_,
support, rhenloadincrease withcon"se;:;iilTi: ,:t#'ji;l:ffL;
systemcollapse'Alternatively,if there isinadequatereactivepower,the system rnay havc voltagecollapse. This chapter is devoteclto the stucly of various methods.f compensating power systemsand varioustypesclf compensating crevices, cailccr.,,,rrj",rru,urr, to alleviatethe problemsof power system outlined above.Thesecompensators
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clearance. There is a possibility of converting a singlecircuit to a doublecircuit line to increasethe loading caPabilitY. Dieletric Limitations From insulationpoint of view, many lines are designed very conservatively. For a given nominal voltage rating it is often possible to increasenormal operatingvoltagesby l07o (i.e. 400 kV 440 kV). One should, however,ensurethat dynamic and transientovervoltagesarewithin limits. [See Chapter 13 of Ref. 71. Stability Issues.There are certain stability issues that limit the tlansmission capability. These include steadystatestability, transient stability, dynamic stability, frequency collapse, voltage collapse and subsynchronousresonance. Severalgooclbooks l, l, 2, 6,7 ,8) are availableon thesetopics.The FACTS technology can certainly be used to overcome any of the stability limits, in which casethe final limits would be thermal and dielectric' 15.3
LOAD COMPENSATION
Load compensation is the managementof reactive pcwer to improve power quality i.e. V profile and pf. Here the reactive power flow is controlled by at the load end installing shunt compensating devices (capacitors/reactors) reactive power. consurned and generated bringing about proper balancebetween of the system capability power transfer the This is most effective in improving to technically and economically both ancl its voltagestability. It is desirable impose utilities why some is This power factor. operatethe systemnear uriity a penalty on low pf loads. Yet another way of irnproving the system performaneeis to operate it under near balancedconditions so as to reduce the ho* of legative sequence currents thereby increasing the system's load capabilityand reducingpower loss' line hast hr cccr it icalloadings( i) nat ur alloading( ii) st eadyA transm ission line the starestability limit and (iii) thermallimit loading.For a compensated reached, is limit loading the thermal before and lowest the is natural loading steadystatestability limit is arrived.
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
esJ I
T5.4 LINE COMPENSATION Ideal voltage profile for a transmissionline is flat, which can only be achieved by loading the line with its surge impedance loading while this may not be achievable,the characteristicsof the line can be modi so that (i) Ferranti effect is minimized. (ii) Underexcitedoperationof synchronousgeneratorsis not required. (iii) The power transfer capability of the line is enhanced.Modifying the characteristicsof a line(s) is known as line compensation. Various compensatingdevicesare: o Capacitors . Capacitors and inductors
compensated,it will behave as a purely resistive element and would. cause series resonance even at fundamental frequency. The location of series capacitorsis decidedby economical factors and severity of fault currents.Series capacitor reducesline reactancethereby level of fault currents. on on vanous lssues ln in series and shunt compensatorsnow follows. 15.5
A capacitor in series rvith a line gives control over the effective reactance between line ends.This effective reactanceis eiven bv Xr=XX, where
. Active voltage source (synchronousgenerator)
I
ii tit i
When a number of capacitors are connected in parallel to get the desired capacitance,it is known as a bank of capacitors,similarly a Uant< of incluctors. A bank of capacitors and/or inductors can be adjusted in stepsby switching (mechanical). . Capacitors and inductors as such are passive line compensators,while synchronous generator is an active compensator.When solidstatedevices are used for switching off capacitorsand inductors, this is regardeclas active compensation. Before proceedingto give a detailedaccount of line compensator,we shall briefly discussboth shunt and seriescompensation. Shunt compensation is more or less like load compensation with all the advantagesassociatedwith it and discussedin Section 15.3. It needs to be pointed out here that shunt capacitors/inductors can not be distributed uniformally along the line. These are normally connected at the end of the line and/or at midpoint of the line. Shunt capacitors raise the load pf which greatly increases the power transmitted over the line as it is not required to carry the reactivepower. There is a limit to which transmitted power can be increased by shunt compensation as it would require very iarge size capacitorbank, which would be impractical. For increasing power transmitted over the line other and better means can be adopted.For example,seriescompensation, higher transmissionvoltage,HVDC etc. When switched capacitors are employed for compensation,these shculd be disconnected irnmediatelyunderlight loaclconclitionsto avoicler.cessive voltage rise and ferroresonancein presenceof transformers. The purpose of series compensationis to cancel part of the seriesinductive reactanceof the line using series capacitors.This helps in (i) increase of maximum power transfer (ii) reduction in power angle for a given amount of power transfer (iii) increasedloading. From practical point of view, it is
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SERIES COMPENSATION
Xl = line reactance Xc = capacrtor reactance It is easy to see that capacitorrcduccsthe cffectivc line rcactance*. This results in improvement in performanceof the system as below. (i) Voltage drop in the line reduces(getscompensated)i.e. minimization of endvoltagevariations. (ii) Preventsvoltage collapse. (iii) Steadystttte power transferincreases;it is inversclyproportionalto Xl. (iv) As a result of (ii) rransientstability limit increasbs. The benefitsof the seriescapacitorcompensatorare associatedwith a problem. The capacitive reactanceXg fbrms a seriesresonantcircuit with the total series reactance X = Xt * X*.n * Xoun, The natural frequency of oscillation of this circuit is given by. rfL^ 
 2rJrc
'2n where
l=
system frequency
xReactive voltage drops of a series reactance added in a line is I2x It is positive if X is inductive and negative if X is capacitive. So a series capacitive reactance reduces the reactance voltage drop of the line, which is an alternative wav of saying that X't= \
X,
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
I
, ,
ModernPower System AnalYsis rL X
I
=degree of compensation = 25 to J5Vo (recommended)
fc R i.e. 0 = 90".It then follows that
o = EY.o,a 11 X
X
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
(17.s)
a
or V"  EV cos 6+ QX = 0 Taking derivativewrt V gives
(r7.6)
d Q = E c o s6  2 V d V X
Using ihe decoupiing principle 1.". dP = 0, we set dV
+=."'r[#.+]
(17.7)
The QV characteristic on normalized basrs(etf**. VIE) for various values of P/P^ are plotted in Fig. 17.3.The system is voltage stable in the region where dQldv is positive, while the voltage stability limit is reached at d,eldV = 0 which may also be termed as the critical operating point.
o
or
I s c = c odsl o O + 1 1 LdV
Ers6= E cos5l !9 + 2Y]
or
LdV Voltagestability is achievedwhen
,D max
E cos , (#
Unstable. . operation
P I P ^ " , =9 . 5
0.2
0.6
0.8
1.0
Xsource a o2
(17.8)
The inferencesdrawn from the simple radial system qualitatively apply to a practicalsize system.Other factors that contribute to system voltage collapse are: strengthof transmissionsystem,power transfer levels, load chaiacteristics, generatorreactive power limits and characteristicsof reactive power compensating devices.
(i) 34
(17.10)
(r7.1r) dZ dV Application of this criterion gives value of z";. \ (iii) Ratio of source to load reactance is very important dnd for voltage stability
VIE
= t"o'26 Q,n^
of Voltage
(shortcircuitMVA of powersource)
v r  = uM d= o o o
The limiting value of the reactive power transfer at the limiting stageof voltage stability is given by
Criteria
> Ers,
cntenon dV voltage instability occurs when the system Z is such that
Fig 17.3 QV characteristics for the systemof Fig. 17.1 tor variousvaluesof plp^^r.
Other
+)
X J
,..\ dz (lU 
0.75
0
.
Pr", is the maxlmum powertransferat upf
1.0
X J
Stability
..it.rion: (E'=generatorvoltage; V=load voltage). Using this crite
rion, the voltage stability limit is reachedwhen
cos d{#.#}+sindffi}=o www.muslimengineer.info
(17.r2)
X load
a indicates the offnominal tap ratio of the OLTC transformer at the load end. T7.5
VOLTAGE
STABILITY
ANALYSIS
The voltage stability analysis for a given system state involves examining following two aspects. (i) Proximity to voltage instabitity: Distance to instability may be measured in terms of physical quantities,such as load level, rea power flow through a critical interface, and reactive power reserye.posiible contingencies such as a line outage,loss of a generating unit or.a reactiu. po*"rio*.. must be given due consideration. (ii) iuiechanismof voitage instabiiity: How and why does voltage instabitity take place? What are the main factors leading to instability?'ulhat are the voltageweak areas?What are the most effective ways to improv" stability? "of,ug,
(r7.e)
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
.ffiiffi
.
Analysis ModernPowerSystem
of system The static analysis techniques permit examination of a wide range give the main and problem the of nature the conditions and can descriUe specific of study detailed for useful is analysis contributing factors. Dynamic testing and controls, and protection of coordination voltage coliapse situations, the how and whether us tell further simulations of remedial rneasures.Dynamic wr point steadystateequilibrium Requirements
Modelling
of various
Power system
components
systemconditionsat variousdme The staticapproachcaptures.snapshotsof frames along the timedomain trajectory.
At each of these time frames' X in
to purely redPce frame.Thus,theoverausystemequations ii h?lrJdil. allowing'h.t..Ytt:f ,t'1tit Tlv:i: ::*:1,::t"t' vlp qnrtvo equations algebraic andvQ by computin s "'?'J"6;ffiffi;;;g."riuulirv is determrned
.vP
Loads Detailed Load modelling is very critical in voltage stability analysis. required' be may area voltageweak a in representation subiransmissionsystem and This may includeiransformer ULTC action, reactive power compensation' voltage regulators of loads' It is essential.to consider the voltage and frequency dependence Induction motors should also be modelled' and their
Generators
excitation
controls
the AVR, load It is necessary to consider the droop chatacteristics of controls should and protection AGC, compensation,SVSs (static var system), also be modelled appropriatelyl4l. Dynamic
Static AnalYsis
AnalYsis
analysis is generalstructurc of the systcm moclel for voltage stability may be equations system overall similar to that for transient stability analysis. cxprefihcdas 'lhe
*=f(X,n and a set of algebraicequations I (X, V) = YxV (Xo' Ve)' with a set of known initial conditions X = system state vector where Y
= but voltage vector
1= current injeition vector Ilv = rletwork node admittance matrix' be s Equations(17.13)and (17'14)can methods integration oi tt e numerical in Ch' 6' T clescribed analysisnrethocls repl models special the ,ninutes. As bee have collapse ieading to voltage considerably is differential equations models.Stiffnessis also called synchrc
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(r7.r3) (r7.r4)
using static analysis havb been curves at selected load buses.Special techniques VQ sensitivity such as eigenvalue (or reported in literature. Methods based on methods give stabilityrelated modal) analysis have been devised. These and also identify areasof potential information from a systemwide perspective problemst13151. Proximity
to InstabilitY
by increasing proximity to smalldisturbance voltage instability is determined flow fails load the or unstable in steps until,the system becomes il;^";ruiion the point determining for techniques to converge.Refs. t16181 discussspecial instability' of voltage collapse and proximity to volage The Continuation
Powerflow
Analysis
at the voltage stability limit. A! a result' The Jacobian matril becomes singular have convergenceproblems at operating conventional toadflow algorith*, rnuy continuation powerflow analysis The conditions near the stability limit. loadflow equationsso that they the overcomesthis problem by reformulating the possible loading conditions. This allows remain wellconditioned at all PV the of forioth upper *d lo*"t portions solution of loadflow problem *fi"t:lltinuationmethod and flexible and of powerflow analysis is ,obrrrt However' difficulties' with convergence is to approach better the suming. Hence continuaand (NR/FDLF) flow irethod :ase, LF is solved using a conventional levels ns for successivelyincreasingload is method Hereafter, the continuation is method ns. Normally, the continuation point' critical I exactly at and past the Voltage
StabilitY
with
HVDC Links
for extremely long distance (HVDC).litt tl:1r:t.:d High voltage direct current* ffansmissiclnanclftrrasynchronousinterconnections.AnHVDClinkcanbe maY refer to [3]' *For dctailcd accountof HVDC, the reader
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
'
ModernpowerSystemAnatysis
f00 
either a backtoback rectifier/inverter link or can include long distance dc transmission. Multiterfninal HVDC links are also feasible. The technology has come to such a level that HVDC terminals can be connected even at voltageweak points in power systems. HVDC links mav present unfavourable "load" characterisficsfo fhe nnrr/rr converter consumesreactive power equal to 5060vo of the dc power. HVDCrelated voltage control (voltage stability and fundamental frequency temporary over voltages)may be studied using a transient stability program. Transient stability is often interrelated with voltage stability. Ref. t2i .onJia.r, this problem in greaterdetail. 17.6
PREVENTION
OF VOLTAGE
COLLAPSE
(i) Application of reactive powercompensating devices. Adequate stability margins should be ensured by proper selection of compensationschemesin terms of their size, ratingrund locations. (ii) control of network voltage and generator reactive output Several utilities in rhe world such as EDF (France), ENEL (Italy) are developing specialschemesfor control of network voltages and reactive power. (iii) Coordination of protections/controls Adequatecoordinationshould be ensuredbetweenequipment protections/ controls basedon dynamic simulation studies. Tripping of equipment to avoid an overloadedcondition should be the last alternative. Controlled system s€parationand adaptive or irrtelligent control could also be used. (iv) Control of transfurmer tap chan.gers T'apchangerscan be controlled, either locally or centrally, so as to reduce the risk of voltage collapse. MicroprocessorbasedOLTC controls offer almost unlimitedflexibility for implementingULTC control strategies so as to take advantageof the load characteristics. (v) Under voltage load shedding For unplanned or extreme situations, it may be necessary to use undervoltageloadsheddingschemes.This is similar to under ir"q1r"rr.y load shedding,which is a common practiceto deal with extreme situationsresulting from generationdeficiency. Strategic load sheddingprovides cheapestway of preventing widespread VOltaSg
CO l l a n s_e ____r
Lnarl
sherldino
cnl,aoo uvrrvrrrvJ
olrn,,r,r L^ orr\_rrll\.t ug
r^^: ^^) ugsrBlrtru
differentiate berween faults, transient voltage dips unJ lo* conditions leading to voltage collapse. (vi) Operators' role
^^ su
as
t()
voltage
Operators must be able to recognise voltage stabiiityrelated symptoms and take requiredremedial actions to prevenr voltage collapse.Online
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VoltaseStability
I
__{tr*g
monitoringand analysisto identify potentialvoltage stabilityproblems are extremelyhelpful. and appropriateremedialmeasures T7.7 STATEOFTHE.ART, FUTURE TRENDS AND CHALLENGES The present day transmission networks are getting more and more stresseddue to economic and environmental constraints.The trend is to operatethe existing networks optimally close to their loadability limit. This consequentlymeansthat the system operation is also near voltage stability limit (nose point) and there is increasedpossibility of voltage instability and even collapse. Offline and online techniquesof determining state of voltage stability and when it entersthe unstable state,provide the tools for system planning and real time control. Energy managementsystem (EMS) provide a variety of measured and computer processed data. This is helpful to system operators in taking critical decisions inter alia reactive power management and control. In this regard autornationand specializedsoftware relieve the operator of good part of the burden of system management but it does add to the complexity of the systemoperation. Voltage stability analysis and techniques have been pushed forward by several researchers and several of these are in commercial use as outlined in this chapter.As it is still hot topic, considerableresearcheffort is being devoted , to it. Pw et al. l8l considered an exponential type voltage dependentload model and a new index called condition number for staticvoltage stability prediction. Eigenvalue analyseshas been used to find critical group of busesresponsible for voltage collapse. Some researchers[26] have also investigated aspects of bifurcations (local, Hopf, global) and chaos and their implications on power systemvoltage stability. FACTS devicescan be effectively used for controlling the occurrence of dynamic bifurcations and chaos by proper choice of error signal and controllergains. Tokyo Electric Power Co. has developed a pPbased controller for coordinated control of capacitor bank switching and network transformer tap ctranging.HVDC power control is used to improve stability. More systematic approach is still required for optimal siting and sizing of FACTS devices.The availability of FACTS controllers allow operationclose to the thermal limit of the lines without jeopardizing security. The reactivepower compensationclose to the load centresand at the critical busesis essentialfor overcoming voltage instability. Better and probabilistic load modelling [11] should be tried. It will be worthwhile developingtechniquesand models for study of nonlineardynamics of large size systems.This may requireexploring new methods to obtain network equivalents suitable for the voltage stability analysis. AI is another approach to centralized reactive power and voltage control. An expert system [9] could assistoperatorsin applying Cbanksso that
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
Modern Po
a00.l,*l
I
. .
t
generatorsoperate near upf. The design of suitable protective measuresin the event of voltage instability is necessary. So far, computed PV curves are the most widely used method of estimating voltage security, providing MW margin type indices. PostdisturbanceMW or MVAr margins should be translated to predisturbance operating limits that operators can monitor. Both control centre and power.plant operators should be trained in the basics of voltage stability. For operator training simulator [10] a realtime dynamic model of the power system that interfaceswith EMS controls such as AGC is of great help. Voltage stability is likely to challengeutility plannersand operatorsfor the foreseable future. As load grows and as new transmission and load area generationbecome increasingly difficult to build, rnore and more utilities will face the voltage stability'challenge.Fortunately, many creative researchersand plannersare working on new analysis methods and an innovative solutions to the voltage stability challenge.
AQ = reactive power variation (i.e. the size of the compensator) Srr.= system short circuit capacity
Then
AV =
trg d,.
AQ = AVSsk =1(0.05x5000) = + 250 MVAR The capacityof the staticvAR compensator is +250 MVAR.
REFERE NCES Books
A load bus is composed of induction motor where the nominal reactive power is I pu. The shunt compensationis K,n. Find the reactive power sensitivityat the bus wrt change in voltage. Solution Qrcot= Qno V2 [given] ' Qcomp=
Qn"t= Qnua t
.'. Here,
[ve sign denotesinductive reactivepowerinjection.l
Krt V2
Qn r= v' 
Qcomp
Krn vz lQoo^ =
1.0givenl
dQn"t = 2v2v K.., i',
dv Sensitivity increasesor decreaseswith Krn as well as the magnitude of the voltage.Say at V  1.0 pu, Krn = 0.8 dQn t
21.6=0.4pu.
dv Example:,17.2 Find the capacity of a static VAR compensator to be installed at a bus with x 5Vo voltage fluctuation. The short circuit capacity is 5000 MVA. Solution For the switching of static shunt compensator,
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1. Chakrabarti,A., D.P. Kothari and A.K. Mukhopadhyay,Perfurmance,Operation and Control of EHV Power Transmission Systerts,Wheeler Publishing, New Delhi, 1995. 2. Taylor, c.w., Power system voltage stability,McGrawHill, New york, 1994. 3. Nagrath, I.J. and D.P. Kothari, Power SystemEngineering, Tata McGrawHill, New Delhi, 1994. 4. Kunclur,P., Powcr Sy,stem Stubilityatul Control, McGrawHill, New york, 1994. 5. Padiyar, K.R., Power System Dynamics: Stability and Control, John Wilev. Singapore,1996. 6. Cutsem,T Van and CostasVournas, Voltage Stabitityof Electric power Systems, Kluwer Int. Series, 1998.
Papers 7 . Concordia, C (Ed.), "special Issue on Voltage Stability and Collapse,', Int. J. of Electrical Power and Energy systems, voi. i5, no. 4, August 1993. 8 . Pai, M.A. and M.G.O, Grady,"Voltage CollpaseAnalysis with ReactiveGeneration and voltage Dependentcons,traints", J. of Elect Machines and power systems, V o l . 1 7 , N o . 6 , 1 9 8 9 ,p p 3 7 9  3 9 0 . 9 . cIGRE/ Task Forcc 38060l,"Expcrt Systems Applied to voltage and var Control//,1991. 1 0 . "operator Training simulator", EpRI Final Report EL7244, May 1991, prepared by EMPROS SystemsInternational. l l . Xu, W and Y Mansour,"Voltage Stability using GenericDynamic Load Models", IEEE Trans. on Power Systems,Vol 9, No l, Feb 1994, pp 479493.
اﻹﺗﺠﺎه اﻹﺳﻼﻣﻲ ﻟﺠﻨﺔ اﻟﺒﻮر واﻹﺗﺼﺎﻻت
Modern Po t
'Yoltage JlaDrtrty DnnansEulenruy IZ. VerTna,tl.lt., L.IJ. Arya ano u.r. I\.oman, ReactivePower Loss Minimization", JIE (I), Vol. 76, May 1995, pp.4449. 13. IEEE, special Publication 90 TH 03582 PWR, "Voltage Stability of Power Systems:Concepts,Analytical Tools, and Industry Experience" 1990. 14. Flatabo,N., R. Ogncdaland T. Carlsen,"Voltagc StabilityCondition,in a Power Tiansmisiion System calculated by Sensilivity Methods", IEEE Tians. VoI. .
. . t ?
i 
d 
t
! l ! 
F
L 
   ,     ^
L  
AppnNDrx A
PWRS5,No. 5, Nov 1990,pp 128693.
1 5 . Gao, 8., G.K. Morison and P. Kundur, "Voltagc Stability EvaluationUsing Modal Analysis", IEEE Trans. Vol. PWRS7, No. 4, Nov. 1992, pp 15291542.
T6, Cutsem, T. Van, "A Method to Compute Rbactive Power Margins wrt Voltage Collapse",IEEE Trans.,Vol. PWRS6, No. 2, Feb 1991, pp 145156. 1 7 . Ajjarapu, V. and C. Christy, "The continuation Power Flow: A Tool for Steady StateVoltageStabilityAnalysis", IEEE PICA Conf.Proc., May 1991,pp 304311. 1 8 . Ldf, AP, T. Sined, G. Anderson and D.J. Hill, "Fast Calculationof a Voltage Stability Index", IEEE Trans., Vol. PWRS7, No. 1, Feb 1992, pp 5464. 1 9 . Arya, L.D., S.C. Chaube and D.P. Itothari, "Line Outage Ranking based on EstimatedLower Bound on Minimum Eigen Value of Load Flow Jacobian", JIE (1),Vol. 79, Dec 1998,pp 126129. '20. Bijwe, P.R,, S.M. Kelapure,D,P. Kothari and K.K. Saxena,"Oscillatory Stability Limit Enhancementby Adaptive Control Rescheduling",Int J. of Electrical Power and Energy Systems,Vol. 21, No. 7, 1999, pp 507514.
2t, .)
Arya, L.D., S,C, Chaube and D.P. Kothari, "Linc Switching for Alleviating Overloadsunder Line OutageCondition taking Bus Voltage Limits into Account", Int J. of Electric Power and Energy System,YoL 22, No. 3, 2000, pp 213221. Bijwc, P.R., D.P. Kothari and S. Kclapure, "An Efficient Approach to Voltage Sccurity Analysis and Enhancement",Int J. of EP and,ES., Vol. 22, No. 7, Oct. 2000, pp 483486.
2 3 . Arya, L.D., S.C., Chaubeand D.P. Kothari, "Reactive Power Optimization using Static Stability Index (VSD", Int J. of Electric Power Components and Systems, Vol. 29, No. 7, July 2001, pp 615628. 24. Arya, L.D., S.C. Chaube and D.P. Kothari, "Line Outage Ranking for Voitage Limit Violations witti Conective Rescheduling Avoiding Masking", Int. J. of EP and ES,Vol. 23, No. 8, Nov. 2001, pp 837846. 2 5 . CIGRE Task Foice 380210, "Cigre Technical Brochure: Modelling of Voltage CollapseIncludihgDynamicPhenomena",Electa, No. 147, April 1993,pp' 7l77. 2 6 . Mark J. Laufenbergand M.A. Pai, "Hopf bifurcation control in power system with
In this appendix, our aim is to present definitions and elementary operationsof vectors and matrices necessaryfor power system analysis. VECTORS
A vectorx is definedas an orderedset of numbers(realor complex),i.e.
,
xp ...t xn are known as the components of the vector r. Thus the vector x is a ndimensional column vector. Sometimestransposedform is found to be more convenient and is written as the row vector. rT A fx1, x2,..., xrf Some Special Vectors The null vector 0 is one whose each component is zero, i.e.
static var compensators",Electrical Power and Energy Systems,Vol. 19, No. 5, 1997, pp 339347.
equaito unity, i.e. The sum vector i has each of its components
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(A1)
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(A2)

.
flTY'
t I
A
0x.= 0 The unit 'tector e the.rest of the componentbare zero, i.e.
The multiplication of two vectors x and y of same dimensions results in a very important product known as inner or j9g!g! pfodqcij.e.
*tv AD",y,Aytx
0
€k:
A
0 1 kth component 0 0
Some Fundamental Vector Operations Two vectors x and y are known as equal if, and only if, .yk= !*for k = r,2, ..., n. Then we sav
(A_3)
i:l
Also, it is interesting to note that xTx = lx 12
(A4)
cos d' 4 "tY lxllyl ,
(A5)
wtere Q is angle between vectors, lxl and lyl are the geometric lengths of vectors x and y, respectively.Two nonzero vectors are said to be orthJgonral, if
*ty= o
(A6)
MATRICES
x = y The product of a vector by a scalar is carried out by multiplying each component of the vector by that scalar, i.e.
Definitions '
Matrix
'
'\n m x n (ot m' n) matrix is an orderedrectangulararray of elementswhich may be real numbers,complexnumbers,functionsor operators. The matrix
(A7) If a vector y is to be addedto or subtractedfrom anothervector x of the same dimension, then each component of the resulting vector will consist of addition or subtraction of the correspondingcomponents of the vectors x and vr,i.e.
The
f n l l n u r i nYcv r r r 6 y nr vr nPnvAr !r af ivoD o 4 r v 4 y^ I* ^, u U^>Gt=[1 3;45;63]' also stores the transposeof matrix given in square brackets in matrix Gl In case one does not want to take conjugate of elements while taking transposeof complex matrix, one should use . 'operator instead of ' operator.
eps This variabie is preinitiiized to 2s2. Identity
i
I
Matrix
To generate an identity matrix and store it in variabte K give the following command
't
t
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I
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[ait+
Modernpo*er SystemAnalysis
#ffif
So K after this becomes
K=[1 0 0 0
"
0
1
0
0
1l
Zeros Matrix generates a 3 x 2 matrix whoseall elementsare zero and,storesthem in L. Ones Matrix generates a 3 x 2 matrix whoseall elementsare one and storesthem in M. : (Colon) operator This is an important operator which can extract a submatrix from a given matrix. Matrix A is given as below
A= [1 ?
5 6
5
4
7 8 9 1 0 3 1
e
3
r
2l
(.. oPERATOR)
This operation unlike complete matrix multiplication, multiplies element of one eiement of other matrixlaving same index. However matdx WitlreOnrespond'rng; in latter case both the matrices must have equal dimensions. We have used this operator in calculating complex powers,at the buses. Say V = [0.845 + j*0.307 0.921 + j*0.248 0.966 + 7*0.410] And I = [0.0654 j*0.432 0.876  j*0.289 0.543 + j*0.210]' Then the complex power S is calculated as Here, conj is a builtin t'unctionwhich gives complex conjugateof its argument. So S is obtainedas [ 0.0774+ 0.385Li 0.7404 + 0.4852t 0.6106+ 0.0198i] Note here, that if the result is complex MATLAB automatically assignsi in the result without any multiplication sign(*). But while giving the input as complex number we have to use multiplication (*) along with i or 7. G.4
COMMON
BUILT.IN
FUNCTIONS
sum sum(A) gives the sum or total of all the elements of a given matrix A. min
This command extracts all columns of matrix A correspondingto row and Z stores them in B. SoBbecomes1269l,}lt Now try this command. The abovecofirmand extracts all the rows correspondingto column 3 of matrix A and stores them in matrix C. So C becomes
l7 9 3 1l
This function can be used in two forms (a) For comparing two scalar quantities
if either a or b is complex number, then its absolute value is taken for comparison (b) For finding minimum amongst a matrix or ilray e . g .i f A = [ 6  3 ; 2  5 ] >> min (A) results in  5
Now try this command
abs If applied to'a scalar, it gives the absolute positive value of that element (magnitude).For example,
This command extracts a submatrix such that it contains all the elements corresponding to row number 2 to 4 and column number 1 to 3.
>>x=3+j*4
SoD=[2
6
9
5
4
3
3
1l
9
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abs(x) gives 5 If applied to a matrix, it results in a matrix storing absolute values of all the elementSof a given matrix.
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t '__ I 041/
G,5
CONTROL STRUCTURES comes to an end when k reachesor exceedsthe final value c. For example, for i  1:1:10, a(i)  1 end This initializes every element of a to 1. If increment is of 1, as in this case,then the increment part may as well be omitted and the above loop could be written
IF Statement The generalfrom of the IF statementis IF expression statements E L S Ee x p r e s s i o n statements ELSEIF statements END
AS
f o ri  1 : 1 0 , a(i) = 1 end
E x pr es s io ni s a l o g i c a l e x p re s s i o nre sul ti ng i n an answ er,true,(l ) or 'false'(0). The logical expressioncan consist of (i) an expressioncontaining relational operators tabulatedalong with their m e a n i n g si n T a b l eG . l . Table G.1 Relational Operator
Meaning Greater than Greater than or Equal to Less than Less than or Equal to Equal to Not equal to
(ii) or many logical expressions combincd with Logical operators.Various logical operators and their meanings are given in Table G.2.
While Loop This loop repeatsa block of statementstill the condition given in the loop is true while expression statements end For example, j  1 while i 6 II J 6
E E i l $ n E
o o o Il f
U'
o o a o .tt
o
tt
sE 3
8 o
l
.= l ct
o c
.9 U'
c, 6 F o
.= .c, ct
G'
Ex. G. 17 (Example
E =
IZ,II)
%Examp7 l e2 . 1 1i E s o l v e du s i n g S I M U L I N K % T h ec o d eg i v e nb e l o ws h o u l db e r u n p r i o r t o s i m u ' l a t i osnh o w n in Fig.G.6. cl ear alI globa1 n ryyr yngg globalPmfHE global r t d d tr % c o n v e rs j ofa n c to r ra d /d egree 91obal Y bf Y d f y a f
q (J dt IL
f=50; ngg=2; r=5; nbus=r; rtd=180/pi; d t r = p i/ 1 8 0 i 9o
Gen. gendata=[ 1
2 3
Ra Xd' 0 0 0 0.067 0 0.100
H inf 12.00 9.00 l;
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Modern Power svstem Anatvsts
iltl ydf'I * 3 5 .6 3 0 1 5. 798 6 J 0  0 . 0 6 8 1 +* 5 i .1 6 6 1 Y af =[ 1.3932j*13.8731 0.22L4+j*7 .6289  0 . 0 9 0 1 +* 6 j .0975
0 j*11.236 0 0.2 2 I4 + j * 7 .6?89 0.5+j*7.7898 0

Appendix G I
0.0681+J*5.1661 0 0 . 1 3 6 2  j * 6 . 2 7 3i 7 1 0.0901+* j6 .0975 0 0.1591j*6.116 ; 81
f c t = i n p u t ( ' f a u l t c l e a r i n gt i m e f c t = ' ) ;
Complex Yaf (2,1) Product21
&Dampi ng factors d a m p 2 =0O; . dam p 3 =. 00;
E2
%Ini t i a l g e n e ra to rA n ge' ls d2 0, 3 3 7 7 * rtd l d3=0.31845*rtd; %Initial Powers Pn2=3.25; Pm3=2 . 10; %G en e ra toIn r te rn a l V o l ta g e s El.= 1. 0 ; E 2 =.10 3 ; E3=1.02; % M a c h i nI e nertia Constants; H 2 = g e n daa( 2 t , 4 )t 1 1 3 = g e n d(a3t,a4 ); & M a cn he l X d '; , 3 ); 1 6 6 1 = g e n d (a2t a , 3 ); 1 6 6 1 = g e n d (a3t a Note : For the simulation for multimachine stability the'two summation boxes sum 2 & sum 3 give the net acceleating powers Po2 and Por. T}iregains of the gain blocks Gl, G2 G3 and G4 are set equal to pi*fAlZ, dampZ,pi*f/Fl3 and damp3, respectively. The accelerating power P" is then integrated twice for each mahcine to give the rotor angles 4 ^d 4. Th" initial conditions for integrator biocks integrater i, integrator 2,integrator 3 anciintegrator 4 are set to 0, d2/r+d, 0 and d3lrtd, respectively.The gain blocks G5 and G6 convert the angles $ and d, into degrees and hence their gains are set to rtd. The electrical power P"2ts calculatedby using two subsystems1 and 2.The detailed diagram for subsystem 1 is shown in Fig. G.7. Subsystem I gives two outputs (i) complex voltage Etl6 and (ii) current of generator 12 which is equal to
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E
Flg. G.7 Multimachine transientStability: Subsystem1
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vodernpowersysterRnalysis
ffi
E/61 *Yar(2,1)+ Erlq*Yq{z,z) + ErlErYar(2,3).The switchesareusedto switch betweenprefault and post fault conditionsfor eachmachineand their thresholdvaluesare adjustedto fct, i.e. the fault clearingtime.
cl ear 'l % T h i s Program fi nds t h e r e d u c e d matri x for stabi i ty s t u d ie s % w h i c h el imnates the I oad I b u s e s a n d r e t a i n s o n l y t h e generator b u s e s % lIgll Y l l Y L ? . Yln Yln+l YLn+Z. Y 1 n + m l IV 1 I % lIszl YZn+l YZn+Z. Y l n + m l IV 2 . YZn iIY21. Y?r
% l . l
I
v o l . l I Y o l l g n l = l y n l YnZ % lrLll IY n + 1 2 yo lrL2l I I % l . l I v o l . l I % lILml I Yn+m1
t
.
Ynn
Y n n + 1 Y n n + 2Y n n + m Yn+Lm
Yfull= _j*5 0 7*5 j*7.5 j*2.5 0 j*12.51 j*5 j*?.5 (Yfull) [row columns]=sjze n=2 Y A A = Y f u l (l 1 : n , 1 : n ) YAB;Yfull (1:n, n+1;columns) Y B A = Y f u1l ( n + 1 :r o w s ,n + L :c o lu m n s ) Y B B = Y f u1l ( n + 1 :r o w s ,n + 1: c o 1u m s ) % T h i s g i v e s t h e r e d u c e dm a t r i x Y r e d u c e d = Y A A  Y AnBv*(iY B B*)Y B A
rs to Problems CHAPTER 2
l
Vn V n+ 1 V n+ 1
2.1 Lirt =
1x 2
ro7 x *Vlr,:
,lt+n,e,,,)+arlr?]
2.2 0.658ohmlkm F loR w^ 2r r 2.4 260.3Vlkm 2.3 L=
Yn+mn+m
Vn*m
 .,:I 2.5 He n = 3 n d, AT/mz(directedupwards) 2.6X=
(XtXn)(X2Xn)
\ * x2zxn
2.7 0.00067m}J/krlt,0.0314V/r