CH3 - Fundamental Theory

7/29/2019 CH3 - Fundamental Theory

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etwork Protection & Alltomation Gllide

. 3 . Fundamental TheoryN-:-k:JSUCTi::;tl

The Protection Engineer is concerned with limiting theeffects of disturbances in a power system. Thesedisturbances, if allowed to persist, may damage plantand interrupt the supply of electric energy. They aredescribed as faults (short and open circuits) or powerswings, and result from natural hazards (for instancelightning],plant failure or human error.Tofacilitate rapid removalof a disturbance froma powersystem, the system is divided into 'protection zones'.Relaysmonitor the system quantities (current, voltage)appearing in these zones; if a fault occurs insidea zone,the relaysoperate to isolate the zonefrom the remainderof the power system.The operating characteristic of a relay depends on theenergizingquantities fed to it such as current or voltage,or various combinationsof these two quantities, and onthe manner in whichthe relay is designed to respond tothis information. For example, a directional relaycharacteristic would be obtained by designing the relayto comparethe phase angle between voltage and currentat the relaying point. An impedance-measuringcharacteristic, on the other hand, would be obtained bydesigningthe relay to dividevoltage by current. Manyother more complex relay characteristics may beobtained by supplyingvarious combinations of currentand voltage to the relay. Relaysmayalso be designedtorespond to other system quantities such as frequency,power,etc.In order to applyprotection relays,it is usuaIlynecessaryto know the limitingvalues of current and voltage, andtheir relative phase displacement at the relay location,for various types of short circuit and their position inthesystem. Thisnormallyrequiressome systemanalysisforfaults occurringat various points in the system.The maincomponents that make up a power system aregenerating sources, transmission and distributionnetworks,and loads.Manytransmissionand distributioncircuits radiate from key points in the system and thesecircuits are controlled by circuit breakers. For thepurpose of analysis, the power system is treated as anetwork of circuit elements contained in branchesradiating from nodes to form closed loops or meshes.The system variables are current and voltage, and in

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.......'"-Q~

~;:~;:~;:;::

3.

steady state analysis. they are regarded as time varyingquantities at a single and constant frequency. Thenetwork parameters are impedance and admittance;these are assumed to be linear, bilateral (independent ofcurrent direction) and constant for a constant frequency.

3.2 JECT:'R ;. ...GEBP,.,A vector represents a quantity in both magnitude anddirection. In Figure 3.1 the vector OP has a magnitudeIZI at an angle 8 with the reference axis OX.

y

n_=,P_nnnnn_nnn'_n- i

nnnn n nn :_om IZ~ i,\ !8 :

~ : x0 xFigure 3.7: Vector OP "It may be resolvedinto two components at right anglesto each other, in this case x and y. The magnitude orscalarvalue ofvectorZ is knownas the modulus IZI.andthe ang~ 8 is the argument, or amplitude,and iswrittenas argoZ. Theconventionalmethodof expressinga vectorZ is to write simply IZIL8.This form completely specifies a vector for graphicalrepresentation or conversioninto other forms.For vectors to be useful, they m~t be expressedalgebraically.In Figure3.1, the vectorZ is the resultantof vectorially adding its components x and y;algebraicallythis vector may be written as:

Z=x+jy ...Equotion 3.7where the operatorj indicates that the component y isperpendicular to component x. In electricalnomenclature, the axisOCis the 'real' or 'in-phase' axis,and the vertical axis OY is called the 'imaginary' or'quadrature' axis. The operatorj rotates a vector anti-clockwisethrough 90. Ifa vector ismadeto rotate anti-clockwise through 180, then the operator j hasperformed its function twice, and since the vector hasreversed its sense, then:

j x j orp = -1whence j = ~-1

--

The representation of a vector quantity algebraically interms of its rectangular co-ordinates iscalleda 'complexquantity'. Therefore,x + jy is a complexquantity and isthe rectangular formof the vector IZIL8 where:

Izl=~(2 +y2)8=tan-7 ;:x=IZlcos8y=IZlsin8 ...Equation 3.2

From Equations 3.1 and 3.2:Z = IZI (cos8 + jsin8) ...Equation 3.3

and since cos8 and sin8 may be expressed inexponential form bythe identities:

sin8 ejS-e-jS2je jS .scos8 -e-J2

it follows that Z may also be written as:2 = 1ZI ejS ...Equation 3.4

Therefore, a vector quantity may also be representedtrigonometrically and exponentially.

,., u...M"1= ~~~ - ~ ~...E :~/-\ ~"'ES

Complexquantities may be represented in any of thefour co-ordinate systems given below:a. Polar ZL8b. Rectangular x + jyc. Trigonometric IZI(cos8 + jsin8)d. Exponential Izi ejS

ThemodulusIZIand the argument 8 are together knownas 'polar co-ordinates', and x and yare described as'cartesian co-ordinates'. Conversion between co-ordinate systems is easily achieved. As the operator jobeysthe ordinarylawsof algebra.complexquantities inrectangular form can be manipulated algebraically,ascan be seen bythe following:

21 + 22 = (X1+X2)+ j(Y1+Y2)21 - 22= (XrX2)+j(YrY2)

(see Figure3.2)

...Equation 3.5

...Equation 3.6

----. 18 . Network Protection & Automation Guide


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2122=lzd \z21L81+82

t1 =EJL8 -822 IZ21 1 2 ...Equatian 3.7y

oj.3.2: Addition of vectors

':!,"Y' j e va"~" "'~mecomplex quantities arevariable with, for example,; when manipulating such variables in differential

uations it is expedientto write the complexquantityexponential form.hen dealing with such functions it is important topreciatethat the quantity contains real and imaginary

omponents. If it is required to investigate only oneomponent of the complex variable, separation intomponentsmust becarried out after the mathematicaleration has taken place.xample: Determine the rate of change of the realmponent of a vector IZILwt with time.

IZILwt = Izl (coswt + jsinwt)= Izl ejwt

he real component of the vector is IZlcoswt.fferentiating Izi ejwtwith respect to time:

~Izi ejwt = jwlZI ejwtdt= jwlZI (coswt + jsinwt)

into real and imaginary components:~( IZI ejwt )=Izl (-wsinwt+ jwcoswt)dt

us, the rate of change of the real component of aIZILwt is:-IZI w sinwt

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l~ < L ~::;. "'P ex 1I...n"'f->p.sA complexnumbermaybe definedas a constant thatrepresents the real and imaginary components of aphysical quantity. The impedance parameter of anelectric circuit is a complex number having real andimaginarycomponents,which aredescribedas resistanceand reactancerespectively.Confusion often arises between vectors and complexnumbers. A vector, as previously defined, may be acomplex number. In this context, it is simply a physicalquantity of constant magnitude acting in a constantdirection. A complex number, which, being a physicalquantity relating stimulus and response in a givenoperation, is known as a 'complex operator'. In thiscontext, it is distinguishedfrom a vector by the fact thatit has no direction of its own.Becausecomplex numbersassumea passiverole in anycalculation, the form taken by the variables in theproblemdeterminesthe method of representingthem.

a e d.,~ ~ ~!-IF' at-{,t:..,Mathematical operators are complex numbers that areused to move a vector through a given angle withoutchanging the magnitude or character of the vector. Anoperator is not a physicalquantity; it is dimensionless.The symbol j, which has been compounded withquadrature components of complex quantities, is anoperator that rotates a quantity anti-clockwise through90. Another useful operator is one which moves avector anti-clockwise through 120, commonlyrepresentedby the symbola.Operatorsare distinguished by one further feature; theyare the roots of unity. UsingDeMoivre's theorem, thenth root of unity isgiven by solving the expression:

~;I..Q~..:::::N-~.............~~~~~;.:::~

Jl/n = (cos2nm + jsin2nm)l/nwherem is any integer.Hence: .3.

1/2nm. . 2nm

1n=COS-+Jsm-n nwhere m hasvalues 1,2,3,... (n-J)Fromthe aboveexpression is found to be the 4th rootand a the 3rd root of unity, as they havefour and threedistinctvaluesrespectively.Table3.1 givessomeusefulfunctions of the a operator.

. 19 . ---'

IZ2 Y2

rLJ 1Xl l. X2 X


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- -- ---1 .'"\13 #a=--+ )-=e 32 2

2 1 . -f] j41ta =--- J-=e 32 21=1+ jO = ejO1+ a + a2= 0l-a= j-J];2l-a2 =- ja-a2=j-f]

a-a2j= -f]

Table3.1:Propertiesof thea operator

Circuit analysis may be described as the study of theresponse of a circuit to an imposed condition, forexamplea short circuit. Thecircuit variablesare currentand voltage. Conventionally,current flow results fromthe application of a driving voltage, but there iscomplete duality between the variables and either maybe regardedas the causeof the other.Whena circuit exists,there is an interchangeof energy;a circuit maybe describedas being madeup of 'sources'and 'sinks'for energy.Thepartsof a circuit aredescribedas elements; a 'source' may be regardedas an 'active'element and a 'sink' asa 'passive'element. Somecircuitelements are dissipative, that is, they are continuoussinks for energy, for example resistance. Other circuitelements may be alternately sources and sinks, forexamplecapacitanceand inductance. Theelementsof acircuit are connectedtogether to form a network havingnodes (terminals or junctions) and branches (seriesgroupsof elements)that form closedloops(meshes).In steadystate a.c.circuit theory, the ability of a circuitto accept a current flow resulting from a given drivingvoltage is called the impedance of the circuit. Sincecurrent and voltage are duals the impedanceparametermust also havea dual, called admittance.

As current and voltage are sinusoidal functions of time,varying at a single and constant frequency, they areregardedas rotating vectors and can be drawn as planvectors (that is, vectors defined by two co-ordinates) ona vector diagram.

For example, the instantaneous value, e, of a voltagevarying sinusoidallywith time is:e=Emsin(wt+8) ...Equation3.8

where:Em is the maximumamplitude of the waveform;ffi=21tf, the angularvelocity,8 is the argument defining the amplitude of thevoltage at a time t=O

At t=O, the actual value of the voltage isEmsin8. So ifEm is regarded as the modulus of a vector, whoseargument is8, then Emsin8 is the imaginarycomponentof the vector IEmIL8. Figure3.3 illustrates this quantityasa vector and asa sinusoidalfunction of time.

y e

X'

,,,~ r-0 i,01 x

Y' t = 0

Figure 3.3: Representationof a sinusoidalfunction

Thecurrent resulting from applyinga voltage to a circuitdependsupon the circuit impedance. If the voltage is asinusoidal function at a given frequency and theimpedance is constant the current will also varyharmonically at the samefrequency,so it can beshownon the samevector diagramasthe voltage vector, and isgiven by the equation

i= IEmIsin( wt+8-


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component',X. isthe circuit reactance. Whencircuit reactance is inductive (that is, wL>l/wC), therrent 'lags' the voltage by an angle wL) it 'leads' the voltage by ang Ie


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Voltage drops are also positive when acting in thedirection of current flow. From Figure 3.4(a) it can beseenthat (Zl+Z2+ziYis the total voltage drop in theloop in the direction of current flow, andmust equatetothetotalvoltageriseErE2' In Figure3.4(b.hhevoltagedrop between nodes a and b designated Vab indicatesthat point b isat a lower potential than a, and is20sitivewhen current flows from a to b. ConverselyVba is anegativevoltage drop.Symbolically:

~ab = ~an - ~bn }ba = Vbn - Van ..Equation 3.14where n is a common reference point.

3".3 Pc,Me'The product of the potential difference acrossand the

.~current through a branchof a circuit isa measureof therate at which energyis exchangedbetween that branchand the remainder of the circuit. If the potentialdifference is a positive voltage drop, the branch ispassiveand absorbsenergy. Conversely,if the potentialdifference is a positivevoltage rise, the branch is activeandsuppliesenergy.The rate at which energy is exchanged is known aspower, and by convention, the power is positive whenenergy is being absorbed and negative when beingsupplied. With a.c.circuits the power alternates, so, toobtain a rate at which energyis suppliedor absorbed,itis necessaryto take the average power over one wholecycle.If e=Emsin(wt+o)and i=Imsin(wt+o-~),then the powerequation is:

p=ei=P[1-cos2 (wt+0)]+ Qsin2 (wt+o)where: ...Equation 3.15

P=IEIII/cos~Q=IEIIIlsin~

From Equation 3.15 it can be seen that the quantity Pvariesfrom 0 to 2P and quantity Qvariesfrom -Q to +Qin one cycle, and that the waveform is of twice theperiodic frequency of the current voltage waveform.The averagevalue of the power exchangedin one cycleisa constant, equal to quantity P, and asthis quantity isthe product of the voltage and the componentof currentwhich is 'in phase'with the voltage it is known as the'real' or 'active' power.Theaveragevalueof quantity Q iszerowhen taken overa cycle,suggestingthat energyis storedin onehalf-cycleand returned to the circuit in the remaining half-cycle.Q is the product of voltage and the quadrature

and

component of current, and is known as 'reactive power'.As P and Q are constants which specify the powerexchange in a given circuit, and are E!ducts of thecurrent ~'29 voltage vectors, ~en if S is the vectorproductE I it follows that wi~hE a~the referencevectorand ~ as the angle between E and I:

S = P + jQ ...Equation3.16The quantity S is described as the 'apparent power', andis the term used in establishing the rating of a circuit.S has units of VA.

ard PovphaseSystemsA systemis singleor polyphasedependinguponwhetherthe sourcesfeeding it are single or polyphase. A sourceis singleor polyphaseaccordingto whether there areoneor several driving voltages associated with it. Forexample, a three-phase source is a source containingthree alternating driving voltages that are assumedtoreach a maximum in phaseorder,A, B, C. Eachphasedriving voltage is associatedwith a phasebranch of thesystemnetwork as shown in Figure3.5(a).If a polyphase system has balanced voltages, that is,equal in magnitude and reachinga maximum at equallydisplaced time intervals, and the phase branchimpedancesare identical, it is calleda 'balanced' system.It will become 'unbalanced' if any of the aboveconditions are not satisfied. Calculations using abalanced polyphasesystem are simplified, as it is onlynecessaryto solvefor a single phase,the solution for theremaining phasesbeing obtained by symmetry.Thepowersystemis normally operatedasa three-phase,balanced,system. Forthis reasonthe phasevoltagesareequal in magnitude and can be represented by threevectors spaced1200 or 2n/3 radiansapart, as shown inFigure3.5(b).

[a) Three-phasesystemEa

/0Direction1200 1200 of rotation( )E~E "--120J Eb=a2EC 11 a

(b)Balancedsystemof vectors

Figure3.5: Three-phasesystems

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the voltages are symmetrical, they may bepressed in terms of one, that is:

Ea = Ea- -Eb = a2EaEc = aEa ...Equatian 3. 17

a isthe vector operator ej21C/3.urther, if the phaseanch impedances are identical in a balanced system, itllows that the resulting currents are also balanced.

PEDANCE\jO~AT'O!\1can be seen by inspection of any power systemam that:

a. several voltage levelsexist in a systemb. it is common practice to refer to plant MVAinterms of per unit or percentage values

c. transmission line and cable constants are given inohms/kmefore any system calculations can take place, thestem parameters must be referredto 'base quantities'nd represented as a unified system of impedances inther ohmic, percentage, or per unit values.he base quantities are power and voltage. Normally,ey are given in terms of the three-phase power inMVAd the linevoltage in kV.Thebase impedance resultingom the above base quantities is:

(kV)2 .Zb= ohms ...EquatlOn.18MVAnd, provided the system is balanced, the basepedance may be calculated using either single-phasethree-phase quantities.he per unit or percentage value of any impedance in thestem is the ratio of actual to base impedance values.

Z(p.u.)=Z( ohms)x MVAb

}(kVb)2

Z(%)=Z(p.u.)X1O0 ...Equatian3.19

MVAb = baseMVAkVb = base kV

imple transposition of the above formulae will refer thehmic value of impedance to the per unit or percentagealues and base quantities.aving chosen base quantities of suitable magnitude all

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system impedances may be converted to those basequantities by using the equations given below:

MVAb2Zb2 =ZbJ X MVAbJ

( )2

kVbJZb2=ZbJX -k Vb2 J ...Equation 3.20

where suffixbl denotes the value to the originalbaseand b2 denotes the value to new base

The choice of impedance notation depends upon thecomplexityof the system, plant impedancenotation andthe nature of the systemcaIculations envisaged.If the system is relatively simple and contains mainlytransmission line data, given in ohms, then the ohmicmethod can be adopted with advantage. However,theper unit method of impedance notation is the mostcommonfor general systemstudies since:

1. impedancesare the same referredto either side ofa transformer if the ratio of base voltages on thetwo sides of a transformer is equal to thetransformer turns ratio2. confusion caused by the introduction of powers of100 in percentage calculations is avoided

3. by a suitable choice of bases, the magnitudes ofthe data and results are kept within a predictablerange, and hence errors in data and computationsare easier to spot

;; ,to..c::.~~E-..-~"""~~~~~~;:::~

Most power system studies are carried out usingsoftware in per unit quantities. Irrespective of themethod of calculation, the choice of base voltage, andunifying system impedances to this base, should beapproached with caution, as shown in the followingexample.

.3.J1.8kV 11.8/141 kV 132/11kV

132kVOverhead line

llkVDistribution

Wrongselection of basevoltageJ1.8kV 132kV llkV

Right selection11.8kV 141kV ill x 11=J1.7kV132

Figure3.6: Selection of basevoltages

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FromFigure3.6 it can beseenthat the basevoltages inthe three circuits are related by the turns ratios of theintervening transformers. Care is required as thenominal transformation ratios of the transformersquoted may be different from the turns ratios- e.g. a110j33kV(nominal) transformer may havea turns ratioof 110j34.5kV.Therefore,the rule for hand calculationsis: 'to refer an impedancein ohms from one circuit toanother multiply the given impedanceby the squareofthe turns ratio (open circuit voltage ratio) of theintervening transformer'.Where power system simulation software is used,thesoftware normally has calculation routines built in toadjust transformer parameters to take account ofdifferences between the nominal primary and secondaryvoltagesand turns ratios. In this case,the choiceof basevoltagesmay bemoreconvenientlymadeasthe nominalvoltages of each section of the power system. Thisapproach avoids confusion when per unit or percentvalues are used in calculations in translating the finalresults into volts, amps,etc.Forexample,in Figure3.7, generatorsG} and G2haveasub-transient reactance of 26% on 66.6MVA rating at11kV, and transformers T} and T2 a voltage ratio of11145kV and an impedance of 12.5% on 75MVA.Choosing 100MVA as base MVA and 132kV as basevoltage, find the percentage impedancesto new basequantities.

a. Generatorreactancesto new basesare:

(11)226x 100 x -0.27%66.6 (132)2

b.Transformerreactancesto newbasesare:

12.5x 100 x (145)2 -20.1%75 (132)2

NOTE: The basevoltages of the generator and circuitsare 11kV and 145kV respectively, that is, the turnsratio of the transformer. The corresponding per unitvalues can befound by dividing by 100, and the ohmicvalue can be found by using Equation 3.19.

,--- -IIIIII

----

T}

132kVoverheadlines

T211-Figure3.7: Section of a power system

Most practical power system problems are solved byusing steadystate analytical methods. Theassumptionsmade are that the circuit parameters are linear andbilateral and constant for constant frequency circuitvariables. In some problems,describedas initial valueproblems, it is necessaryto study the behaviour of acircuit i'n the transient state. Such problems can besolved using operational methods. Again, in otherproblems, which fortunately are few in number, theassumption of linear, bilateral circuit parameters is nolonger valid. Theseproblemsare solved using advancedmathematical techniques that are beyond the scope ofthis book.

In linear, bilateral circuits, three basic network lawsapply, regardless of the state of the circuit, at anyparticular instant of time. Theselaws are the branch,junction and mesh laws, due to Ohm and Kirchhoff, andare stated below,usingsteadystate a.c. nomenclature.

The current I in a given branch of imp~danceZ isproportional to the pote~ial_c!lfference V appearingacrossthe branch,that is,V = I Z .

The algebraic sum of all currents entering any junction(or node) in a network is zero,that is:2)=0

The algebraic sum of all the driving voltages in anyclosed path (or mesh) in a network is equal to thealgebraicsumof all the passivevoltages (productsof theimpedances and the currents) in the componentsbranches,that is:

IE=IZIAlternatively, the total change in potential around aclosedloop is zero.

Fromthe abovenetwork laws,many theoremshavebeenderived for the rationalisation of networks, either toreach a quick, simple, solution to a problem or torepresenta complicated circuit by an equivalent. Thesetheorems are divided into two classes:those concernedwith the general properties of networks and those



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cernedwith network reduction.f the many theorems that exist, the th ree mostmportant are given. These are: the Superposition

em,Thevenin'sTheoremand Kennelly'sStar/Delta6;: '::;u..,e ..,vs c c~ .,- ee e~

;e' e N ~oh ~- /-ho~ Qhe resultant current that flows in any branch of aetwork due to the simultaneous action of severalriving voltages is equal to the algebraic sum of themponent currents due to each driving voltage actingnewith the remaindershort-circuited.2.L .,- 1'"e .'75 Thee 'e~

2 ~-, '0'" ~hG~ e J e~ey active network that may be viewed from two

erminals can be replaced by a single driving voltagecting in series with a single impedance. The drivingoltage is the open-circuit voltage between the tworminals and the impedance is the impedance of theetwork viewed from the terminals with all sources.6.2':; e e s 5cu ::e cu .,-. e e

;JOSS"e ~p/-' .~ -educt';]" t ev'ey three-terminal network can bereplacedbya delta ortar impedance equivalent without disturbing the

network. Theformulae relating the replacementf a delta network by the equivalent star network is aslows (Figure3.8):

Zco = Z13 Z23 / (Z12 + Z13 + Z23)d so on.

'~b r ;~2yzn3(b)Delta networka) Star network

Figure3.8: Star-Delta network transformation

he impedance of a delta network corresponding to andeplacing any star network is:

- - - Zao ZboZ12 = Zao+ Zbo+Zco

nd so on.

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e C' ' R~~ ~~"...The aim of network reduction is to reduce a system to asimple equivalent while retaining the identity of thatpart of the systemto bestudied.For example,consider the system shown in Figure3.9.The network has two sourcesE' and E", a line AOBshunted by an impedance,which maybe regardedasthereductionof a further network connectedbetweenA andB, anda load connectedbetween 0 and N. The object ofthe reduction is to study the effect of openinga breakerat A or B during normal system operations, or of a faultat A or B. Thus the identity of nodesA and B must beretained together with the sources, but the branch ONcan be eliminated, simplifying the study. Proceeding,.B, N, forms a star branchand can therefore be convertedto an equivalent delta.

2.55Q1.6Q O.4QA B

00.75Q 0.45Q

;:..,~c~~N

18.85Q

N

Figure3.9: Typicalpowersystem network -~.....;::ZAN =ZAO+ZNO+ ZAOZNOZBO

~~~"'!::::;::=0.75+18.85+ 0.75x18.850.45 ;::t~= 51 ohms. 3.

ZEN =ZBO+ZNO+ ZBOZNOZAO=0.45+18.85+ 0.45x18.850.75=30.6 ohms

ZAOZEOZAN =ZAO +ZBO+ ZNO= 1.2 ohms (sinceZNO> ZAOZBO)

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2.5QI.6Q OAQ.2Q

A B5IQ 30.6Q

N

Figure 3. 10: Reduction usingstar/delta transform

The network is now reducedas shown in Figure3.10.By applying Thevenin'stheorem to the active loops,thesecan be replacedby a single driving voltage in serieswithan impedanceasshown in Figure3.11.

B

30.6Q

N

I.6Q1.6 x 51-Q52.6A

N

A

5IQ - ~E'- 52.6

N(a) Reduction of left active mesh

OAQ OAx30.6 Q31B

N(b) Reduction of right active mesh

Figure 3. 11: Reduction of active meshes:Thevenin's Theorem

The network shown in Figure3.9 is now reducedto thatshown in Figure3.12 with the nodesA andB retainingtheir identity. Further, the load impedance has beencompletely eliminated.The network shown in Figure3.12 may now be usedtostudy system disturbances, for example power swingswith and without faults.

i?

2.5QI.55Q O.39QA B

I.2Q

NFigure 3.12: Reduction of tvpical

power system network

Most reductionproblemsollow the samepatternastheexample above.The rules to apply in practical networkreduction are:

a. decide on the nature of the disturbance ordisturbancesto be studiedb.decide on the information required, for examplethe branch currents in the network for a fault at a

fJarticular locationc. reduce all passive sections of the network notdirectly involved with the section underexaminationd. reduce all active meshesto a simple equivalent,

that is, to a simple source in serieswith a singleimpedanceWith the widespread availability of computer-basedpower systemsimulation software, it is now usualto usesuch software on a routine basisfor network calculationswithout significant network reduction taking place.However,the network reduction techniques given aboveare still valid, as there will be occasions where suchsoftware is not immediately available and a handcalculation must be carried out.In certain circuits, for exampleparallel lines on the sametowers, there is mutual coupling between branches.Correct circuit reduction must take account of thiscoupling.

p~ ~: i ~QZbb(a) Actual circuit

IP 0---.- c=JZ- ZaaZbb-Z2abZaa+Zbb-2Zab

oQ

(b) Equivalent when Zaa",ZbbIP0---.- c=J

z=1-(Zaa+Zbb)2(c) Equivalent when Zaa=Zbb

oQ

with mutual coupling

Threecasesare of interest. Theseare:a. two branchesconnectedtogether at their nodesb. two branchesconnectedtogetherat onenodeonlyc. two branchesthat remain unconnected



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sideringeach casein turn:a. consider the circuit shown in Figure 3.13(a).The

application of a voltage Vbetween the terminalsPand Qgives:v = 1aZaa + hZabV = 1aZab+ hZbb

where 1aand1bare the currents in branchesa andb, respectivelyand 1= 1a+ h , the total currententering at terminal P and leavingat terminal Q.Solving for 1aandh :

(Zbb-Zab)VI =a ZaaZbb - Z;b

from which(Zaa-Zab)VI -

b ZaaZbb-Z;band

1=1 +1 - V(Zaa+Zbb-2Z )a b abZaaZbb-Z2 abso that the equivalent impedance of the originalcircuit is:

Z=~- ZaaZbb-Z;bI Zaa+Zbb-2Zab ...Equation 3.21(Figure 3.13(b)), and, if the branch impedancesareequal, the usual case, then:

7Z=-(Zaa+Zab)2(Figure3.13(c)).

...Equation 3.22

b. considerthe circuit in Figure3.14(a).

cB

Zb=Zbb-Zab(b) Equivalent circuit

3.14:Reductionwith acommonterminal

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~I,,he assumption is made that an equivalent starnetwork can replace the network shown. Frominspectionwith one terminal isolated in turn and avoltage Vimpressedacrossthe remainingterminalsit can be seenthat:

Za+Zc=ZaaZb+Zc=ZbbZa+Zb=Zaa+ Zbb- 2Zab

ISolving these equations gives:

za=Zaa-Zab

jZb =Zbb-ZabZc =Zab

'"...Equotion 3.23

-see Figure3.14(b).

c. considerthe four-terminal network given in Figure3.15(a), in which the branches 11' and 22' areelectrically separateexcept for a mutual link. Theequationsdefining the network are:

VI =ZIl11+Z1212V2=Z2111+Zn1211=Yll V1+Y12 V2

~""'..1:::1~

-.1::1~2=Y21 V1+ Yn V2where Z12=Z21 and Y12=Y21 , if the network isassumedto be reciprocal. Further,by solving theaboveequations it can beshown that:

-~"'"'~~:;~~~;:=~

Y11=Z22 /!J.Y22 =Z11 /!J.Y12 =Z12 /!J.!J.=Z11Z22 -Z12 ...Equation 3.24

There are three independent coefficients, namelyZ12' Zll' Zn, so the original circuit may bereplaced by an equivalent mesh containing fourexternal terminals, each terminal being connectedto the other three by branch impedancesasshownin Figure3.15(b).

.3.

10 Zll Zll 1 '1 ',

2' ii,z,~20 02'Z22(a) Actual circuit

Z12

2 Z22(b) Equivalent circuit

/ Figure 3.J5: Equivalent circuitsorfour terminal network with mutual coupling

. 27 .

Zaai j ocZbb(a)ActualcircuitZa=Zaa-Zab-A


13/13

.

Inorder to evaluate the branches of the equivalentmesh let all points of entry of the actual circuit becommoned except node 1 of circuit 1, as shown inFigure 3.15(c). Then all impressed voltages exceptV1 willbe zero and:

11= Yl1V1h = Y12V1

Ifthe sameconditionsare appliedto the equivalentmesh, then:11= V1Z1112= -V1/Z12 = -V1/Z12

These relations follow from the fact that the branchconnecting nodes 1 and l' carries current 11 andthe branches connecting nodes 1 and 2' and 1 and2 carry current h This must be true since branchesbetween pairs of commoned nodes can carry nocurrent.By considering each node in turn with theremainder commoned, the following relationshipsare found:

Zl1' = 11Y11Z22' = 11Y22Z12' = -lIY12Z12 = Z1'2' = -Z21' = -Z12'

Hence:Zl1' = Zl1Z22-Z212

Z22Z22' = Zl1Z22-Z212

Zl1Z12 = Zl1Z22-Z212

Z 12 J ...Equatian.25A similar but equally rigorousequivalent circuit isshown in Figure3.15(d). This circuit [3.2] followsfrom the fact that the self-impedance of any circuitis independent of all other circuits. Therefore, itneed not appear in any of the mutual branches if itis lumped as a radial branch at the terminals. Soputting Zl1 and Z22 equal to zero in Equation 3.25,

2111J2d 1212c(e!Equivalentwith allnodescommonedexcept10 / Figure 3.15: Equivalent circuits forfour terminal network with mutual coupling

defining the equivalent mesh in Figure 3.15(b), andinserting radial branches having impedances equalto Zl1 and Z22 in terminals 1 and 2, results inFigure3.15(d).I-

3.1 Power System Analysis. J. R. Mortlock and- M.W.HumphreyDavies. Chapmana Hall.3.2 Equivalent Circuits I.Frank M.Starr, Proc. A.I.E.E.Vol. 51. 1932, pp. 287-298.

211 l'

-212 -212

2 2'212(d) Equivalent circuit

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CH3 - Fundamental Theory