I nt roduc t i on 5 .1 S ynchr onous machines 5.2 Armature re ac tion 5. 3 St e ady s tate theory 5.4 S al i ent pol e rot or 5 .5 Transient analy s is 5.6 As ymmet ry 5.7 Mac hine re ac tanc es 5.8 Ne gativ e s e qu enc e reac tanc e 5.9 Ze ro s e que nc e re ac tanc e 5.10 Direct and qua dr ature axis values 5.11 E ff e c t of s atura ti on on mac hine reac tanc es 5.1 2 Transformers 5.13 T rans former p os it ive s eq uen c e e quiv alen t c irc uit s 5.14 Transformer zero sequenc e e quiv alen t c ircuits 5.15 Auto- trans formers 5.16 Trans former imp edances 5.17 Ov er he ad line s and c ab les 5.18 C alc ulation of s er ies impe da nce 5.1 9 C alc ula tion o f s hunt impe da nce 5.20 Ov er he ad line c irc uits wit h o r without e ar th wires 5.21 OHL equivalent circuits 5.22 C able c irc ui ts 5.23 Overhead line and cable data 5.24 R eferenc es 5.25 • 5 • Equi v alent C irc uit s and Parameters o f Po wer Sys tem Pla nt
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e • 4 7 •
5.1 INTRODUCTION
Knowledge of the behaviour of the principal electricalsystem plant items under normal and fault conditions isa prerequisite for the proper application of protection.This chapter summarises basic synchronous machine,
transformer and transmission line theory and givesequivalent circuits and parameters so that a fault studycan be successfully completed before the selection andapplication of the protection systems described in laterchapters. Only what might be referred to as 't raditional'synchronous machine theory is covered, as that is all t hatcalculations for fault level studies generally require.Readers interested in more advanced models ofsynchronous machines are referred to the numerouspapers on the subject, of which reference [5.1] is a goodstarting point.
Power system plant may be divided into two broadgroups - static and rotating.
The modelling of static plant for fault level calculationsprovides few difficulties, as plant parameters generallydo not change during the period of interest followingfault inception. The problem in modell ing rotat ing plantis that the parameters change depending on theresponse to a change in power system conditions.
5 .2 SYNCHRONOUS M ACHIN ES
There are two main types of synchronous machine:cylindrical rot or and salient pole. In general, the formeris confined to 2 and 4 pole turbine generators, whilesalient pole types are built with 4 poles upwards andinclude most classes of dut y. Both classes of machineare similar in so far that each has a stator carrying athree-phase winding distributed over i ts inner periphery.Within the stator bore is carried the rotor which ismagnetised by a winding carrying d.c. current.
The essential difference between the two classes of
machine lies in the rotor construction. The cylindricalrotor type has a uniformly cylindrical rotor that carriesits excitation winding distributed over a number of slots
• 5 • Equiv alent Ci rcuit s and Parameters of Pow er System Plant
most common. Two-stroke diesel engines are oftenderivatives of marine designs with relatively large output s(circa 30MW is possible) and may have running speeds ofthe order of 125rpm. This requires a generator wit h alarge number of poles (48 for a 125rpm, 50Hz generator)and consequently is of large diameter and short axiallength. This is a contrast t o turbine-driven machines thatare of small diameter and long axial length.
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e
5 •
E q u i v a l e n t C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m P l a n t
around its periphery. This construction is unsuited tomulti-polar machines but it is very sound mechanically.Hence it is particularly well adapted for the highestspeed electrical machines and is universally employed for2 pole units, plus some 4 pole unit s.
The salient pole type has poles that are physicallyseparate, each carrying a concentrated excitation
winding. This type of construction is in many wayscomplementary to that of the cylindrical rotor and isemployed in machines having 4 poles or more. Except inspecial cases it s use is exclusive in machines having morethan 6 poles. Figure 5.1 illustrates a typical largecylindrical rotor generator installed in a power plant.
Two and four pole generators are most often used inapplications where steam or gas turbines are used as thedriver. This is because the steam turbine tends to besuit ed to high rotational speeds. Four pole steam turbinegenerators are most often found in nuclear power
stations as the relative wetness of the steam makes thehigh rotational speed of a two-pole design unsuitable.Most generators with gas turbine drivers are four polemachines to obtain enhanced mechanical strength in therotor- since a gearbox is often used to couple the powerturbine to the generator, the choice of synchronousspeed of the generator is not subject to the sameconstraints as with steam turbines.
Generators with diesel engine drivers are invariably offour or more pole design, to match the running speed ofthe driver without using a gearbox. Four- stroke diesel
engines usually have a higher running speed than two-stroke engines, so generators having four or six poles are
Strong
N S
Di rec t ion of rot a t ion
(a )
(b )
S N N
W ea k ea k Strong
Figure 5.2: Distort ion of flux due to armature reaction
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e • 4 9 •
• 5 •
5 .3 ARM ATURE REACTION
Armature reaction has the greatest effect on theoperation of a synchronous machine wit h respect both t othe load angle at which it operates and to the amount ofexcitation that it needs. The phenomenon is most easilyexplained by considering a simplified ideal generatorwith full pitch winding operating at unity p.f., zero lag
p.f. and zero lead p.f. When operating at uni ty p.f., thevoltage and current in the stator are in phase, the statorcurrent producing a cross magnetising magneto-motiveforce (m.m.f.) which interacts with that of the rotor,result ing in a distortion of f lux across the pole face. Ascan be seen f rom Figure 5.2(a) the tendency is to weakenthe flux at the leading edge or effectively to distort thefield in a manner equivalent to a shift against thedirection of rotation.
If the power factor were reduced to zero lagging, thecurrent in the stator would reach its maximum 90°after
the voltage and the rotor would therefore be in theposit ion shown in Figure 5.2(b). The stator m.m.f. is nowacting in direct opposition to the field.
Similarly, for operation at zero leading power factor, thestator m.m.f. would directly assist t he rotor m.m.f. Thism.m.f. arising from current flowing in the stator is knownas 'armature reaction'.
5.4. STEADY STATE THEORY
The vector diagram of a single cylindrical rotorsynchronous machine is shown in Figure 5.3, assumingthat the magnetic circuit is unsaturated, the air-gap isuniform and all variable quantities are sinusoidal.Further, since the reactance of machines is normally verymuch larger than the resistance, the latter has beenneglected.
The excitation ampere-turns, AT e, produces a flux Φacross the air-gap thereby inducing a voltage, E t , in thestator. This voltage drives a current I at a power factorcos-1φ and gives rise to an armature reaction m.m.f. AT ar in phase wi th it . The m.m.f. AT f resulting from thecombination of these two m.m.f. vectors (see Figure5.3(a)) is the excitation which must be provided on therotor to maintain fluxΦacross the air-gap. Rotat ing therotor m.m.f. diagram, Figure 5.3(a), clockwise untilcoincides with E t and changing the scale of the diagramso that AT e becomes the basic unit, where AT e = E t =1,result s in Figure 5.3(b). The m.m.f. vectors have thusbecome, in eff ect, volt age vectors. For example
AT ar /AT e is a unit of voltage that is directly proportionalto the stator load current. This vector can be ful lyrepresented by a reactance and in practice this is called
'armature reaction reactance' and is denoted by X ad .Similarly, the remaining side of the triangle becomes
AT f / AT e , which is the per unit voltage produced onopen circuit by ampere-turns AT f . It can be consideredas the internal generated voltage of the machine and isdesignated E o .
The true leakage reactance of the stator winding whichgives rise to a voltage drop or regulation has beenneglected. This reactance is designated X L (or X a insome texts) and the voltage drop occurring in it, IX L, isthe difference between the terminal voltage V and thevoltage behind the stator leakage reactance, E L.
IZ L is exactly in phase with the voltage drop due to X ad ,as shown on the vector diagram Figure 5.3(c). It shouldbe noted that X ad and X L can be combined to give asimple equivalent reactance; this is known as the'synchronous reactance', denoted by X d .
The power generated by the machine is given by theequation:
…Equat ion 5.1
where δ is the angle between the internal voltage andthe terminal voltage and is known as the load angle ofthe machine.
P VI VE X d
= =cos sinφ δ
E q u i v a l e n t
C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m
It follows from the above analysis that, for steady stateperformance, the machine may be represented by theequivalent circuit shown in Figure 5.4, where X L is a truereactance associated with flux leakage around the statorwinding and X ad is a fictitious reactance, being the ratioof armature reaction and open-circuit excitationmagneto-motive forces.
In practice, due to necessary constructional features of acylindrical rotor to accommodate the windings, thereactance X a is not constant irrespective of rotorposition, and modelling proceeds as for a generator witha salient pole rotor. However, the numerical dif ferencebetween the values of X ad and X aq is small, much lessthan for the salient pole machine.
5.5 SALIENT POLE ROTORThe preceding theory is limited to the cylindrical rotorgenerator. The basic assumption that the air- gap isuniform is very obviously not valid when a salient polerotor is being considered. The eff ect of t his is that t he fluxproduced by armature reaction m.m.f. depends on theposit ion of the rotor at any instant, as shown in Figure 5.5.
LagArmaturereaction M.M.F.
Lead
FluxFlux
r a t u r e a x i s
u a d
D i r
l e
e c t a x i s p
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e
When a pole is aligned with the assumed sine wavem.m.f. set up by the stator, a corresponding sine waveflux will be set up, but when an inter-polar gap is alignedvery severe distort ion is caused. The difference is treatedby considering these two axes, that is thosecorresponding to the pole and the inter-polar gap,separately. They are designated the 'di rect ' and'quadrature' axes respectively, and the general theory isknown as the 'two axis' theory.
The vector diagram for the salient pole machine is similarto that for the cylindrical rotor except that the reactanceand currents associated with them are split into twocomponents. The synchronous reactance for the directaxis is X d = X ad + X L, while that in the quadrature axisis X q = X aq + X L. The vector diagram is construct ed asbefore but the appropriate quantities in this case areresolved along two axes. The result ant internal volt age
is E o, as shown in Figure 5.6.In passing it should be noted that E
’
0 is the internalvoltage which would be given, in cylindrical rotor theory,by vectorially adding t he simple vectors IX d and V . Thereis very little difference in magnitude between E 0 and E
’
0but a substantial difference in internal angle; the simpletheory is perfectly adequate for calculation of excitationcurrents but not for stability considerations where loadangle is significant.
5 •
E q u i v a l e n t C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m P
l a n t
• 5 0 •
Figure 5.5: Variat ion of armature reaction m.m.f.wit h pole position
V
I d
I q
I d X dI q X q
E OIX d
E 'O
I
Pole axis
Figure 5.6: Vector diagram for salient pole machine
Figure 5.4: Equivalent circuit of elementary machine
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e • 5 1 •
5.6 TRANSIENT ANALYSIS
For normal changes in load conditions, steady statetheory is perfectly adequate. However, there areoccasions when almost instantaneous changes areinvolved, such as fault s or switching operations. Whenthis happens new factors are introduced within themachine and to represent these adequately a
corresponding new set of machine characteristics isrequired.
The generally accepted and most simple way toappreciate the meaning and derivation of thesecharacteristics is to consider a sudden three-phase shortcircuit applied to a machine initially running on opencircuit and excited to normal voltage E 0 .
This voltage will be generated by a flux crossing the air-gap. It is not possible to confine the flux to one pathexclusively in any machine, and as a result there will bea leakage flux Φ L that will leak from pole to pole andacross the inter-polar gaps without crossing the mainair- gap as shown in Figure 5.7. The flux in the pole willbe Φ+ Φ L.
If the stator winding is then short-circuited, the powerfactor in it will be zero. A heavy current wil l tend toflow, as the resulting armature reaction m.m.f. isdemagnetising. This will reduce the flux and condit ionswill settle until the armature reaction nearly balancesthe excitation m.m.f., the remainder maintaining a verymuch reduced flux across the air-gap which is justsufficient to generate the voltage necessary to overcomethe stator leakage reactance (resistance neglected). Thisis the simple steady state case of a machine operating onshort circuit and is fully represented by the equivalent of
Figure 5.8(a); see also Figure 5.4.
It might be expected that the fault current would begiven by E 0 /(X L+ X ad ) equal to E 0 /X d , but this is verymuch reduced, and the machine is operating with nosaturation. For this reason, the value of voltage used isthe value read from the air-gap line corresponding tonormal excitation and is rather higher than the normalvolt age. The steady state current is given by:
…Equat ion 5.2
where E g = voltage on air gap line
An important point to note now is that between the
initial and final conditions there has been a severereduction of f lux. The rotor carries a highly inductivewinding which links the flux so that the rotor fluxlinkages before the short circuit are produced by(Φ + Φ L). In practice the leakage flux is distributed overthe whole pole and all of it does not link all the winding.Φ L is an equivalent concentrated flux imagined to link allthe winding and of such a magnitude that the totallinkages are equal to those actually occurring. It is afundamental principle that any attempt to change theflux linked with such a circuit wil l cause current to flowin a direction that will oppose the change. In the presentcase the flux is being reduced and so the inducedcurrents wil l t end to sustain i t.
I E
X d
g
d =
• 5 •
E q u i v a l e n t
C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e
For the position immediately following the application ofthe short circuit, it is valid to assume that the flux linkedwith the rotor remains constant, this being broughtabout by an induced current in the rotor which balancesthe heavy demagnetising effect set up by the short-circuit ed armature. So ( Φ + Φ L) remains constant, butowing to the increased m.m.f. involved, the flux leakagewill increase considerably. With a constant total rotorflux, this can only increase at the expense of that fluxcrossing the air- gap. Consequently, this generates areduced voltage, which, acting on the leakage reactance,gives the short circuit current.
It is more convenient for machine analysis to use therated voltage E 0 and to invent a fictitious reactance thatwill give rise to the same current . This reactance iscalled the ' t ransient reactance' X ’d and is defined by theequation:
Transient current…Equat ion 5.3
It is greater than X L, and the equivalent circuit isrepresented by Figure 5.8(b) where:
and X f is the leakage reactance of the field winding
The above equation may also be written as:
X ’d = X L + X ’ f where X ’ f = effective leakage reactance of field winding
The flux will only be sustained at it s relat ively high valuewhile the induced current flows in the field winding. Asthis current decays, so conditions will approach thesteady state. Consequently, the duration of this phasewill be determined by the time constant of the excitationwinding. This is usually of the order of a second or less- hence the term 'transient' applied to characteristicsassociated with it.
A further point now arises. All synchronous machineshave what is usually called a ‘damper winding’ orwindings. In some cases, this may be a physical winding(like a field winding, but of fewer turns and locatedseparately), or an ‘effective’ one (for instance, the solidiron rotor of a cylindrical rotor machine). Sometimes,both physical and effective damper windings may exist(as in some designs of cylindrical rotor generators,having both a solid iron rotor and a physical damperwinding located in slots in the pole faces).
Under short circuit conditions, there is a transfer of fluxfrom the main air- gap to leakage paths. This diversion is,to a small extent, opposed by the excitation winding andthe main transfer will be experienced towards the pole tips.
X X X
X X X d
ad f
ad f L
' = + +
I E
X d o
d
'
' =
The damper winding(s) is subjected to the full effect offlux transfer to leakage paths and will carry an inducedcurrent t ending to oppose it . As long as this current canflow, the air-gap flux will be held at a value slightlyhigher than would be the case if only the excitationwinding were present, but still less than the originalopen circuit fluxΦ.
As before, it is convenient to use rated voltage and tocreate another fictitious reactance that is considered tobe effective over this period. This is known as the 'sub-transient reactance' X ’’d and is defined by the equation:
Sub-transient current I ’’d …Equat ion 5.4
where
or X ’’d
= X L
+ X ’ kd
and X kd = leakage reactance of damper winding(s)
X ’kd = effective leakage reactance of damper winding(s)
It is greater than X L but less than X ’d and thecorresponding equivalent circuit is shown in Figure5.8(c).
Again, the duration of this phase depends upon the timeconstant of the damper winding. In practice this isapproximately 0.05 seconds - very much less than thetransient - hence the term 'sub-transient'.
Figure 5.9 shows the envelope of the symmetricalcomponent of an armature short circuit currentindicating the values described in the preceding analysis.The analysis of the stator current waveform resultingfrom a sudden short circuit test is traditionally the
X X X X X
X X X X X X d L
ad f kd
ad f kd f ad kd
'' = + + +
= E
X
o
d ''
5 •
E q u i v a l e n t C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m
P l a n t
• 5 2 •
C u r r e n t
Time
E I '' d
X '' d=
E oI ' d X ' d
=E ir gapI d
X d=
Figure 5.9: Transient decay envelope of short- circuit current
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e
method by which these reactances are measured.However, the major limitation is that only direct axisparameters are measured. Detailed test methods forsynchronous machines are given in references [5.2] and[5.3], and include other tests that are capable ofproviding more detailed parameter information.
5 .7 ASYMM ETRYThe exact instant at which the short circuit is applied tothe stator winding is of signif icance. If resistance isnegligible compared with reactance, the current in a coilwill lag the voltage by 90°, that is, at t he instant whenthe voltage wave attains a maximum, any currentflowing through would be passing through zero. If ashort circuit were applied at this instant, the resultingcurrent would rise smoothly and would be a simple a.c.component . However, at the moment when the inducedvoltage is zero, any current flowing must pass through amaximum (owing to the 90°lag). If a fault occurs at thismoment, the resulting current will assume thecorresponding relationship; it will be at its peak and inthe ensuing 180° will go through zero to maximum inthe reverse direction and so on. In fact the current mustactually start from zero and so will follow a sine wavethat is completely asymmetrical. Intermediate posit ionswill give varying degrees of asymmetry.
This asymmetry can be considered to be due to a d.c.component of current which dies away becauseresistance is present.
The d.c. component of stator current sets up a d.c. fieldin the stator which causes a supply frequency ripple onthe field current, and this alternating rotor flux has afurt her effect on the stator. This is best shown byconsidering the supply frequency flux as beingrepresented by two half magnitude waves each rotating
in opposit e directions at supply frequency relative to therotor. So, as viewed from the stator, one is stationaryand the other rotating at twice supply frequency. Thelatter sets up second harmonic currents in the stator.Further development along these lines is possible but theresulting harmonics are usually negligible and normallyneglected.
5 .8 M ACHINE REACTANCES
Table 5.1 gives values of machine reactances for salientpole and cylindrical rotor machines typical of latestdesign practice. Also included are parameters forsynchronous compensators – such machines are nowrarely built, but significant numbers can still be found inoperation.
5.8.1 Synchronous Reactance X d = X L + X ad
The order of magnitude of X L is normally 0.1-0.25p.u.,while that of X ad is 1.0-2.5p.u. The leakage reactance X Lcan be reduced by increasing the machine size (derating),or increased by artificially increasing the slot leakage,but it will be noted that X L is only about 10% of thetotal value of X d and cannot exercise much influence.
The armature reaction reactance can be reduced bydecreasing the armature reaction of the machine, whichin design terms means reducing the ampere conductor orelectrical (as distinct from magnetic) loading - this willoft en mean a physically larger machine. Alternativelythe excitation needed to generate open-circuit voltagemay be increased; this is simply achieved by increasingthe machine air- gap, but is only possible if the excitationsystem is modified to meet the increased requirements.
In general, control of X d is obtained almost entirely byvarying X ad , and in most cases a reduction in X d willmean a larger and more costly machine. It is also worth
• 5 •
E q u i v a l e n t
C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m
Type of machineCylindrical rotor turbine generators Salient pole generators
4 Pole IAir Cooled Hydrogen Hydrogen/ 4 Pole Multi-pole
Multi-Pole Cooled Water Cooled
Short circuit ratio 0.5-0.7 1.0-1.2 0.4-0.6 0.4-0.6 0.4-0.6 0.4-0.6 0.6-0.8Direct axis synchronous reactance X d (p.u.) 1.6-2.0 0.8-1.0 2.0-2.8 2.1-2.4 2.1-2.6 1.75-3.0 1.4-1.9Quadrature axis synchronous reactance X q (p.u.) 1.0-1.23 0.5-0.65 1.8-2.7 1.9-2.4 2.0-2.5 0.9-1.5 0.8-1.0Direct axis transient reactance X ’d (p.u.) 0.3-0.5 0.2-0.35 0.2-0.3 0.27-0.33 0.3-0.36 0.26-0.35 0.24-0.4Direct axis sub-transient reactance X ’’ d (p.u.) 0.2-0.4 0.12-0.25 0.15-0.23 0.19-0.23 0.21-0.27 0.19-0.25 0.16-0.25Quadrature axis sub-transient reactance X ’’ q (p.u.) 0.25-0.6 0.15-0.25 0.16-0.25 0.19-0.23 0.21-0.28 0.19-0.35 0.18-0.24Negative sequence reactance X 2 (p.u.) 0.25-0.5 0.14-0.35 0.16-0.23 0.19-0.24 0.21-0.27 0.16-0.27 0.16-0.23Zero sequence reactance X 0 (p.u.) 0.12-0.16 0.06-0.10 0.06-0.1 0.1-0.15 0.1-0.15 0.01-0.1 0.045-0.23Direct axis short circuit transient time constant T ’d (s) 1.5-2.5 1.0-2.0 0.6-1.3 0.7-1.0 0.75-1.0 0.4-1.1 0.25-1Direct axis open circuit transient t ime constant T ’do (s) 5-10 3-7 6-12 6-10 6-9.5 3.0-9.0 1.7-4.0Direct axis short circuit sub-transient- time constant T ’’ d (s) 0.04-0.9 0.05-0.10 0.013-0.022 0.017-0.025 0.022-0.03 0.02-0.04 0.02-0.06Direct axis open circuit sub-transient t ime constant T ’’do(s) 0.07- 0.11 0.08- 0.25 0.018- 0.03 0.023- 0.032 0.025- 0.035 0.035- 0.06 0.03- 0.1
Quadrature axis short circuit sub- transient t ime constant T ’’q (s) 0.04- 0.6 0.05- 0.6 0.013- 0.022 0.018- 0.027 0.02- 0.03 0.025- 0.04 0.025- 0.08Quadrature axis open circuit sub-transient time constant T ’’qo (s) 0.1-0.2 0.2-0.9 0.026-0.045 0.03-0.05 0.04-0.065 0.13-0.2 0.1-0.35
noting that X L normally changes in sympathy with X ad ,but that it is completely overshadowed by it.
The value 1/ X d has a special significance as itapproximates to the short circuit ratio (S.C.R.), the onlydifference being that the S.C.R. takes saturation intoaccount whereas X d is derived from the air-gap line.
5.8.2 Transient Reactance X ’d = X L + X ’ f
The transient reactance covers the behaviour of amachine in the period 0.1-3.0 seconds after adisturbance. This generally corresponds to the speed ofchanges in a system and therefore X ’d has a majorinfluence in transient stability studies.
Generally, the leakage reactance X L is equal to theeffective field leakage reactance X ’ f , about 0.1-0.25p.u.The principal factor determining the value of X ’ f is thefield leakage. This is largely beyond the control of the
designer, in that other considerations are at present moresignificant than field leakage and hence take precedencein determining the field design.
X L can be varied as already outlined, and, in practice,control of transient reactance is usually achieved byvarying X L
5.8.3 Sub-transient Reactance X ’’ d = X L + X ’kd
The sub-transient reactance determines the initialcurrent peaks following a disturbance and in the case ofa sudden fault is of importance for selecting the breakingcapacit y of associated circuit breakers. The mechanicalstresses on the machine reach maximum values thatdepend on this constant. The effective damper windingleakage reactance X ’kd is largely determined by theleakage of the damper windings and control of this isonly possible to a limited extent. X ’kd normally has avalue between 0.05 and 0.15 p.u. The major f actor is X Lwhich, as indicated previously, is of the order of 0.1-0.25p.u., and control of the sub-transient reactance isnormally achieved by varying X L.
It should be noted that good transient stability isobtained by keeping the value of X ’d low, whichtherefore also implies a low value of X ’’d . The fault ratingof switchgear, etc. will therefore be relatively high. It i snot normally possible to improve transient stabilityperformance in a generator without adverse effects onfault levels, and vice versa.
5.9 NEGATIVE SEQUENCE REACTANCE
Negative sequence currents can arise whenever there isany unbalance present in t he system. Their effect is toset up a field rotating in the opposite direction to themain field generated by the rotor winding, so subjectingthe rotor to double frequency flux pulsations. This gives
rise to parasitic currents and heating; most machines arequite limited in the amount of such current which theyare able to carry, both in the steady – state andtransiently.
An accurate calculation of the negative sequence currentcapability of a generator involves consideration of thecurrent paths in the rotor body. In a turbine generator
rotor, for instance, they include the solid rotor body, slotwedges, excitation winding and end-winding retainingrings. There is a tendency for local over- heating to occurand, although possible for the stator, continuous localtemperature measurement is not practical in the rotor.Calculation requires complex mathematical techniquesto be applied, and involves specialist software.
In practice an empirical method is used, based on thefact that a given type of machine is capable of carrying,for short periods, an amount of heat determined by itsthermal capacity, and for a long period, a rate of heat
input which it can dissipate continuously. Synchronousmachines are designed to be capable of operatingcontinuously on an unbalanced system such that, withnone of the phase currents exceeding the rated current,the ratio of the negative sequence current I 2 to t he ratedcurrent I N does not exceed the values given in Table 5.2.Under fault conditions, the machine shall also be capable
of operat ion wi th the product of and time in
seconds (t) not exceeding the values given.
I I N
2
2
5 •
E q u i v a l e n t C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e • 5 5 •
5.10 ZERO SEQUENCE REACTANCE
If a machine is operating with an earthed neutral, asystem earth fault will give rise to zero sequencecurrents in the machine. This reactance represents themachine's contribution to the total impedance offered tothese currents. In practice it is generally low and oftenoutweighed by other impedances present in the circuit .
5.11 DIRECT AND QUADRATURE AXIS VALUES
The transient reactance is associated with the fieldwinding and since on salient pole machines this isconcentrated on the direct axis, there is nocorresponding quadrature axis value. The value ofreactance applicable in the quadrature axis is thesynchronous reactance, that is, X ’q = X q.
The damper winding (or its equivalent) is more widelyspread and hence the sub- transient reactance associated
with this has a definit e quadrature axis value X ”
q, whichdiffers significantly in many generators from X ”d .
5.12 EFFECT OF SATURATIONON M ACHIN E REACTANCES
In general, any electrical machine is designed to avoidsevere saturation of its magnetic circuit . However, it isnot economically possible to operate at such low fluxdensities as to reduce saturation to negligibleproportions, and in practice a moderate degree ofsaturation is accepted.
Since the armature reaction reactance X ad is a ratio AT ar / AT e it is evident that AT e will not vary in a linearmanner for different voltages, while AT ar will remainunchanged. The value of X ad will vary with the degree ofsaturation present in the machine, and for extremeaccuracy should be determined for the particularconditions involved in any calculation.
All the other reactances, namely X L , X ’d and X ’’ d aretrue reactances and actually arise from flux leakage.Much of this leakage occurs in the iron parts of themachines and hence must be affected by saturation. Fora given set of conditions, the leakage flux exists as aresult of the net m.m.f. which causes it . If t he ironcircuit is unsaturated its reactance is low and leakageflux is easily established. If t he circuit s are highlysaturated the reverse is true and the leakage flux isrelatively lower, so the reactance under saturatedconditions is lower than when unsaturated.
Most calculation methods assume infinite ironpermeability and for this reason lead to somewhatidealised unsaturated reactance values. The recognit ionof a finite and varying permeability makes a solutionextremely laborious and in practice a simple factor ofapproximately 0.9 is taken as representing the reductionin reactance arising from saturation.
It is necessary to distinguish which value of reactance isbeing measured when on test . The normal instantaneousshort circuit test carried out from rated open circuitvoltage gives a current that is usually several times fullload value, so that saturation is present and thereactance measured will be the saturated value. Thisvalue is also known as the 'rated voltage' value since it ismeasured by a short circuit applied with the machineexcited to rated voltage.
In some cases, if it is wished to avoid the severemechanical strain to which a machine is subjected bysuch a direct short circuit, the test may be made from asuitably reduced voltage so that the initial current isapproximately full load value. Saturat ion is very muchreduced and the reactance values measured are virtuallyunsaturat ed values. They are also known as 'ratedcurrent' values, for obvious reasons.
5 .1 3 TRANSFORM ERSA transformer may be replaced in a power system by anequivalent circuit representing the self-impedance of,and the mutual coupling between, the windings. A two-winding transformer can be simply represented as a 'T'network in which the cross member is the short-circuitimpedance, and the column the excitation impedance. Itis rarely necessary in fault studies to consider excitationimpedance as this is usually many times the magnitudeof the short-circuit impedance. With these simplifyingassumptions a three-winding transformer becomes a starof three impedances and a four-winding transformer amesh of six impedances.
The impedances of a transformer, in common with otherplant, can be given in ohms and qualified by a basevoltage, or in per unit or percentage terms and qualifiedby a base MVA. Care should be taken wit h mult i-winding transformers to refer all impedances to acommon base MVA or to state the base on which each isgiven. The impedances of stat ic apparatus areindependent of the phase sequence of the appliedvoltage; in consequence, transformer negative sequenceand posit ive sequence impedances are ident ical. Indetermining the impedance to zero phase sequencecurrents, account must be taken of the windingconnections, earthing, and, in some cases, theconstruct ion type. The existence of a path for zerosequence currents implies a fault to earth and a flow ofbalancing currents in the windings of the transformer.
Practical three-phase transformers may have a phaseshift between primary and secondary windingsdepending on the connections of the windings – delta or
star. The phase shift t hat occurs is generally of nosignificance in fault level calculations as all phases areshif ted equally. It is therefore ignored. It is normal tofind delta-star transformers at the transmitting end of a
• 5 •
E q u i v a l e n t
C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e
transmission system and in distribution systems for thefollowing reasons:
a. at the transmitting end, a higher step-up voltageratio is possible than with other windingarrangements, while t he insulation to ground of t hestar secondary winding does not increase by thesame ratio
b. in distribution systems, the star winding allows aneutral connection to be made, which may beimportant in considering system earthingarrangements
c. the delta winding allows circulation of zerosequence currents within the delta, thuspreventing transmission of these from thesecondary (star) winding into the primary circuit.This simplifies protection considerations
5 .1 4 TRAN SFORM ER POSITIV E SEQUENCEEQUIVALENT CIRCUITS
The transformer is a relat ively simple device. However,the equivalent circuits for fault calculations need notnecessarily be quite so simple, especially where earthfault s are concerned. The following two sections discussthe equivalent circuits of various types of transformers.
5.14.1 Two-winding Transformers
The two-winding transformer has four terminals, but inmost system problems, two-terminal or three-terminalequivalent circuits as shown in Figure 5.10 can representit . In Figure 5.10(a), terminals A ' and B' are assumed tobe at the same potential. Hence if the per unit self-impedances of the windings are Z 11 and Z 22 respectivelyand the mutual impedance between them Z 12 , the
transformer may be represented by Figure 5.10(b). Thecircuit in Figure 5.10(b) is similar t o that shown in Figure3.14(a), and can therefore be replaced by an equivalent'T ' as shown in Figure 5.10(c) where:
…Equat ion 5.5
Z 1 is described as the leakage impedance of winding AA '
and Z 2 the leakage impedance of winding BB ' .
Impedance Z 3 is the mutual impedance between thewindings, usually represented by X M , the magnetizingreactance paralleled wit h the hysteresis and eddy currentloops as shown in Figure 5.10(d).
If the secondary of the transformers is short-circuited,and Z 3 is assumed to be large with respect to Z 1 and Z 2 ,then the short-circuit impedance viewed from theterminals AA ’ is Z T = Z 1 + Z 2 and the transformer canbe replaced by a two-terminal equivalent circuit asshown in Figure 5.10(e).
The relative magnitudes of Z T and X M are of the order of10% and 2000% respectively. Z T and X M rarely have tobe considered together, so that the transformer may berepresented either as a series impedance or as anexcitation impedance, according to the problem beingstudied.
A typical power transformer is illustrated in Figure 5.11.
5.14.2 Three-winding Transformers
If excitation impedance is neglected the equivalentcircuit of a three-winding transformer may berepresented by a star of impedances, as shown in Figure5.12, where P , T and S are the primary, tertiary andsecondary windings respectively. The impedance of anyof these branches can be determined by considering theshort-circuit impedance between pairs of windings withthe third open.
Z Z Z
Z Z Z
Z Z
1 11 12
2 22 12
3 12
= −= −
=
5 •
E q u i v a l e n t C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m
P l a n t
• 5 6 •
Zero bus(d) 'π' equivalent circuit
Zero bus(b) Equivalent circuit of model
Zero bus(c) 'T' equivalent circuit
Zero bus(e) Equivalent circuit: secondary winding s/c
R j X M
B'
B' C '
B'B' A '
B'
A'
A '
A '
B C A
A '
B
B A
B A
B A
AZ T =Z 1+Z 2
Z 1 =Z 11 -Z 12 Z 2=Z 22 -Z 12
Z 3=Z 12
r 1+ jx 1 r 2+ jx 2
Z 12Z 11 Z 22LoadE
(a) Model of transformer
~
Figure 5.10: Equivalent circuit s for a two- winding transformer
Zero bus
S
P
T
Z t
Z s
Z p
Tertiary
Secondary
Primary
Figure 5.12: Equivalent circuit for a t hree-w inding transformer
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e • 5 7 •
The exceptions to the general rule of neglectingmagnetising impedance occur when the transformer isstar/star and either or both neutrals are earthed. Inthese circumstances the transformer is connected to thezero bus through the magnetising impedance. Where athree-phase transformer bank is arranged withoutinterlinking magnetic flux (that is a three-phase shelltype, or three single-phase unit s) and provided there is apath for zero sequence currents, the zero sequenceimpedance is equal to the positive sequence impedance.In the case of three-phase core type units, the zero
sequence fluxes produced by zero sequence currents canfind a high reluctance path, the effect being to reducethe zero sequence impedance to about 90% of thepositive sequence impedance.
However, in hand calculations, it is usual to ignore thisvariation and consider the positive and zero sequenceimpedances to be equal. It is common when usingsoftware to perform fault calculations to enter a value ofzero-sequence impedance in accordance with the aboveguidelines, if the manufacturer is unable to provide avalue.
• 5 •
E q u i v a l e n t C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m P l a n t
5.15 TRANSFORM ER ZERO SEQUENCEEQUIVALENT CIRCUITS
The flow of zero sequence currents in a transformer isonly possible when the transformer forms part of aclosed loop for uni-directional currents and ampere-turnbalance is maintained between windings.
The positive sequence equivalent circuit is stillmaintained to represent the transformer, but now thereare certain condit ions attached to it s connection into theexternal circuit . The order of excitation impedance is
very much lower than for the positive sequence circuit;it wil l be roughly between 1 and 4 per unit , but sti ll highenough to be neglected in most fault studies.
The mode of connection of a transformer to the externalcircuit is determined by taking account of each windingarrangement and its connection or otherwise to ground.If zero sequence currents can flow into and out of awinding, the winding terminal is connected to theexternal circuit (that is, link a is closed in Figure 5.13). Ifzero sequence currents can circulate in the windingwithout flowing in the external circuit, the winding
terminal is connected directly to the zero bus (that is,link b is closed in Figure 5.13). Table 5.3 gives the zerosequence connections of some common two- and three-winding t ransformer arrangements applying the above rules.
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e • 5 9 •
5 .1 6 AUTO- TRANSFORM ERS
The auto-transformer is characterised by a singlecontinuous winding, part of which is shared by both thehigh and low voltage circuits, as shown in Figure 5.14(a).The 'common' winding is the winding between the lowvoltage terminals whereas the remainder of the winding,belonging exclusively to the high voltage circuit, is
designated the 'series' winding, and, combined with the'common' winding, forms the 'series-common' windingbetween the high volt age terminals. The advantage ofusing an auto-transformer as opposed to a two-windingtransformer is that the auto-transformer is smaller andlighter for a given rating. The disadvantage is thatgalvanic isolation between the two windings does notexist, giving rise to the possibility of large overvoltageson the lower voltage system in the event of majorinsulation breakdown.
Three-phase auto- transformer banks generally have star
connected main windings, the neutral of which isnormally connected solidly to earth. In addition, it iscommon practice to include a third winding connected indelta called the tertiary winding, as shown in Figure5.14(b).
5.16.1 Positive Sequence Equivalent Circuit
The positive sequence equivalent circuit of a three-phaseauto-transformer bank is the same as that of a two- orthree-winding transformer. The star equivalent for athree-winding transformer, for example, is obtained inthe same manner, with the difference that theimpedances between windings are designated as follows:
…Equation 5.8
where: Z sc-t = impedance between 'series common' and tertiary
windings
Z sc-c = impedance between 'series common' and'common' windings
Z sc-t = impedance between 'common' and tert iarywindings
When no load is connected to the delt a tertiary, the pointT will be open-circuited and the short-circuit impedanceof the transformer becomes Z L + Z H = Z sc-c ’ , that is,similar to the equivalent circuit of a two-windingtransformer, with magnetising impedance neglected; seeFigure 5.14(c).
Z Z Z Z
Z Z Z Z
Z Z Z Z
L sc c c t sc t
H sc c sc t c t
T sc t c t sc c
= + −( )
= + −( )
= + −( )
− − −
− − −
− − −
1
2
1
2
1
2 • 5 •
E q u i v a l e n t
C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m
P l a n t
Zero potential bus
a
b
Z e
Z t
Z p
Z s
Z e
bb
a
a
(b) Three windings
(a) Two windings
Zero potential bus
b
aa
b
Z T 2
Z T 2
Figure 5.13: Zero sequence equivalent circuits
H
L
N
L
N
T
H
T
H L H L
T
Zero potential bus
Zero potential bus(e) Equivalent circuit with isolated neutral
L H
T
Z HT
Z LH
Z X
Z Z
Z Y Z H
I H I L
I L-I H
I L-I H
V LV H
I H
I H
I LI L I T
I N
Z N
Z L
Z T
Z LT
I L0 I H0
I L0
I T0I T1
I L1 I H0I H1
I T0
(c) Positive sequence impedance (d) Zero sequence equivalent circuit
(a) Circuit diagram (b) Circuit diagram with tertiary winding
Figure 5.14: Equivalent circuit of auto- transformer
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e
5.16.2 Zero Sequence Equivalent Circuit
The zero sequence equivalent circuit is derived in asimilar manner to the positive sequence circuit, exceptthat, as there is no identity for the neutral point, thecurrent in the neutral and the neutral voltage cannot begiven direct ly. Furt hermore, in deriving the branchimpedances, account must be taken of an impedance in
the neutral Z n, as shown in the following equations,where Z x, Z y and Z z are the impedances of the low, highand tertiary windings respectively and N is the ratiobetween the series and common windings.
…Equation 5.9
Figure 5.14(d) shows the equivalent circuit of thetransformer bank. Currents I LO and I HO are thosecirculating in the low and high voltage circuits respectively.The difference between these currents, expressed inamperes, is the current in the common winding.
The current in the neutral impedance is three times thecurrent in the common winding.
5.16.3 Special Conditions of Neutral EarthingWith a solidly grounded neutral, Z n = O, the branchimpedances Z x, Z y, Z z, become Z L, Z H , Z T , that is,identical to the corresponding positive sequenceequivalent circuit, except that the equivalent impedance
Z T of t he delta t ertiary is connected to the zero potentialbus in the zero sequence network.
When the neutral is ungrounded Z n = ∞ and theimpedances of the equivalent star also become infinitebecause there are apparently no paths for zero sequence
currents between the windings, although a physicalcircuit exists and ampere-turn balance can be obtained.A solution is to use an equivalent delta circuit (see Figure5.14(e)), and evaluate the elements of the delta directlyfrom the actual circuit . The method requires threeequations corresponding to three assumed operatingcondit ions. Solving these equations will relate the deltaimpedances to the impedance between the series andtertiary windings, as follows:
…Equation 5.10
Z Z N
N
Z Z N
Z Z N N
LH s t
LT s t
HT s t
=+( )=
= +( )
−
−
−
2
1
1
Z Z Z N
N
Z Z Z N
N
Z Z Z N
x L n
y H n
z T n
= + +( )
= −+( )
= + +( )
3 1
3 1
3
1
1
2
With the equivalent delta replacing the star impedancesin the auto- transformer zero sequence equivalent circuitthe transformer can be combined with the systemimpedances in the usual manner to obtain the systemzero sequence diagram.
5 .1 7 TRANSFORM ER IM PEDANCES
In the vast majority of fault calculations, the ProtectionEngineer is only concerned with the transformer leakageimpedance; the magnetising impedance is neglected, asit i s very much higher. Impedances for transformersrated 200MVA or less are given in IEC 60076 andrepeated in Table 5.4, together with an indication of X/Rvalues (not part of IEC 60076). These impedances arecommonly used for transformers installed in industrialplants. Some variat ion is possible to assist in controllingfault levels or motor starting, and typically up to ±10%variation on the impedance values given in the table is
possible wit hout incurring a signif icant cost penalt y. Forthese transformers, the tapping range is small, and thevariation of impedance with tap position is normallyneglected in fault level calculations.
For transformers used in electricity distributionnetworks, the situation is more complex, due to anincreasing trend to assign importance to the standing (orno-load) losses represented by the magnetisingimpedance. This can be adjusted at the design stage butthere is often an impact on the leakage reactance inconsequence. In addit ion, it may be more import ant tocontrol fault levels on the LV side than t o improve motorstart ing voltage drops. Therefore, departures from theIEC 60076 values are commonplace.
IEC 60076 does not make recommendations of nominalimpedance in respect of transformers rated over200MVA, while generator transformers and a.c. tractionsupply transformers have impedances that are usuallyspecified as a result of Power Systems Studies to ensuresatisfactory performance. Typical values of transformerimpedances covering a variety of transformer designs aregiven in Tables 5.5 – 5.9. Where appropriate, theyinclude an indication of the impedance variation at theextremes of the taps given. Transformers designed towork at 60Hz will have substantially the sameimpedance as their 50Hz counterparts.
5 •
E q u i v a l e n t C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e • 6 3 •
In the formula for Z 0 the expression 3√dcD 2 is thegeometric mean radius of the conductor group.Where the circuit is not symmetrical, the usual case,symmetry can be maintained by transposing theconductors so that each conductor is in each phaseposit ion for one third of the circuit length. If A , B and C are the spacings between conductors bc, ca and ab then
D in the above equations becomes the geometric meandistance between conductors, equal to 3√ ABC .
Writing Dc = 3√dcD 2 , the sequence impedances inohms/km at 50Hz become:
…Equation 5.14
5 .2 0 CALCULATION OF SHUNT IM PEDANCE
It can be shown that the potential of a conductor aabove ground due to its own charge qa and a charge - qaon its image is:
…Equation 5.15
where h is the height above ground of the conductor andr is the radius of the conductor, as shown in Figure 5.16.
Similarly, it can be shown that the potential of aconductor a due to a charge qb on a neighbouringconductor b and the charge -qb on its image is:
…Equation 5.16
where D is the spacing between conductors a and b and D’ is the spacing between conductor b and the image ofconductor a as shown in Figure 5.14.
Since the capacitance C =q/V and the capacitivereactance X c =1/ ωC , it follows that the self and mutualcapacitive reactance of the conductor system in Figure5.16 can be obtained directly from Equations 5.15 and5.16. Furt her, as leakage can usually be neglected, theself and mutual shunt impedances Z ’ p and Z ’m inmegohm-km at a system frequency of 50Hz are:
…Equation 5.17
Where the distances above ground are great in relation
Z jh
r
Z j D D
p
m
'
'
'
=−
=−
0 132 2
0 132
10
10
. log
. log
V qb D D
a e'
'
=2 log
V qa hr
a e=2 2
log
Z Z R jABC dc
Z R j D D
oe
c
1 2 10
3
10
0 145
0 148 0 434
= = +
= +( )+
. log
. . log
to the conductor spacing, which is the case wit h overheadlines, 2h=D ’. From Equat ion 5.12, the sequence
impedances of a symmetrical three-phase circuit are:
…Equation 5.18
It should be noted that the logarithmic terms above aresimilar to those in Equation 5.13 except that r is theactual radius of the conductors and D’ is the spacingbetween the conductors and their images.
Again, where the conductors are not symmetricallyspaced but transposed, Equation 5.18 can be re-writtenmaking use of the geometric mean distance betweenconductors, 3√ ABC , and giving the distance of eachconductor above ground, that is, h a , h2 , h c , as follows:
…Equation 5.19
Z Z j ABC r
Z jh h h
r A B C a b b
1 2 10
3
0 10 2 2 2 3
0 132
0 132 8
= =−
=−
. log
. log
Z Z j Dr
Z j DrD
o
1 2 10
10 2 3
0 132
0 396
= =−
=−
. log
. log'
• 5 •
E q u i v a l e n t
C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m
P l a n t
Ea rth
a '
h
h
a b
D '
D
ConductorRadiusr
Figure 5.16 Geometry of tw o parallel conductors a and b and the image of a (a')
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e
5 .21 OVERHEAD LINE CIRCUITSWITH OR WITHOUT EARTH WIRES
Typical configurations of overhead line circuits are givenin Figure 5.17. Tower heights are not given as they varyconsiderably according to the design span and nature ofthe ground. As indicated in some of t he tower outlines,some tower designs are designed with a number of base
extensions for this purpose. Figure 5.18 shows a typicaltower.
In some cases, the phase conductors are notsymmetrically disposed to each other and therefore, aspreviously indicated, electrostatic and electromagneticunbalance will result , which can be largely eliminated bytransposit ion. Modern practice is to build overhead lineswithout transposition towers to reduce costs; this mustbe taken into account in rigorous calculations of theunbalances. In other cases, lines are formed of bundledconductors, that is conductors formed of two, three orfour separate conductors. This arrangement minimiseslosses when voltages of 220kV and above are involved.
It should be noted that the line configuration andconductor spacings are influenced, not only by voltage,but also by many other factors including type ofinsulators, type of support, span length, conductor sagand the nature of terrain and external climatic loadings.Therefore, there can be large variations in spacingsbetween different line designs for the same voltage level,so those depicted in Figure 5.17 are only typicalexamples.
When calculating the phase self and mutual impedances,Equations 5.11 and 5.17 may be used, but it should beremembered that in this case Z p is calculated for eachconductor and Z m for each pair of conductors. Thissection is not, therefore, intended to give a detailedanalysis, but rather to show the general method offormulating the equations, taking the calculation ofseries impedance as an example and assuming a singlecircuit line with a single earth wire.
The phase voltage drops V a ,V b,V b of a single circuit linewith a single earth wire due to currents I a, I b, I b flowingin the phases and I e in the earth wire are:
…Equation 5.20
where:
and so on.
The equation required for the calculation of shuntvoltage drops is identical to Equation 5.20 in form,except that primes must be included, the impedancesbeing derived from Equation 5.17.
Z f j f D D
abe= +0 000988 0 0029 10 . . log
Z R f j f Ddc
aae= + +0 000988 0 0029 10 . . log
V Z I Z I Z I Z I
V Z I Z I Z I Z I
V Z I Z I Z I Z I
Z I Z I Z I Z I
a aa a ab b ac c ae e
b ba a bb b bc c be e
c ca a cb b cc c ce e
ea a eb b ec c ee e
= + + += + + += + + +
= + + +0 5 •
E q u i v a l e n t C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e • 6 7 •
From Equation 5.20 it can be seen that:
Making use of this relation, the self and mutualimpedances of the phase conductors can be modifiedusing the following formula:
…Equation 5.21
For example:
and so on.
So Equation 5.20 can be simplified while still taking accountof the effect of the earth wire by deleting the fourth row andfourth column and substituting J aa for Z aa , J ab for Z ab , andso on, calculated using Equation 5.21. The single circuit linewith a single earth wire can therefore be replaced by anequivalent single circuit line having phase self and mutualimpedances J aa , J ab and so on.
It can be shown from the symmetrical component theorygiven in Chapter 4 that the sequence voltage drops of ageneral three-phase circuit are:
…Equation 5.22
And, from Equation 5.20 modif ied as indicated above andEquation 5.22, the sequence impedances are:
V Z I Z I Z I
V Z I Z I Z I
V Z I Z I Z I
0 00 0 01 1 02 2
1 10 0 11 1 12 2
2 20 0 21 1 22 2
= + += + += + +
J Z Z Z Z
ab abae be
ee= −
J Z Z
Z aa aa
ae
ee= −
2
J Z Z Z Z
nm nmne me
ee= −
− = + + I Z Z
I Z Z
I Z Z
I e ea
eea eb
eeb ec
eec
The development of these equations for double circuitlines with two earth wires is similar except that moreterms are involved.
The sequence mutual impedances are very small and canusually be neglected; this also applies for double circuitlines except for the mutual impedance between the zerosequence circuits, namely (Z OO’ = Z O’O). Table 5.10 givestypical values of all sequence self and mutual impedancessome single and double circuit l ines with earth wires. Allconductors are 400mm 2 ACSR, except for the 132kVdouble circuit example where they are 200mm 2.
5.22 OHL EQUIVALENT CIRCUITS
Consider an earthed, infinite busbar source behind alength of transmission line as shown in Figure 5.19(a).An earth fault involving phase A is assumed to occur atF . If the driving volt age is E and the fault current is I a
Z J J J J J J
Z J J J J J J
Z J a J aJ aJ a J J
Z J aJ a J a J aJ
aa bb cc ab bc ac
aa bb cc ab bc ac
aa bb cc ab ac bc
aa bb cc ab ac
00
11
12 2 2
212 2
1
3
2
3
1
3
1
3
1
3
2
3
1
3
2
3
= + +( )+ + +( )
= + +( )− + +( )
= + +( )+ + +( )
= + +( )+ + ++( )
= + +( )− + +( )
= + +( )− + +( )===
J
Z J a J aJ aJ a J J
Z J aJ a J a J aJ Jbc Z Z
Z Z
Z Z
bc
aa bb cc ab ac bc
aa bb cc ab ac
20 2 2
10
2 2
22 11
01 20
02 10
1
3
1
3
1
3
1
3
• 5 •
E q u i v a l e n t
C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m
P l a n t
132kV 380kV 132kV 275kVSequence impedance Single circuit line Single circuit line Double circuit line Double circuit line
(400 mm2) (400 mm2) (200 mm2) (400 mm2) Z 00 = ( Z 0 ’
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e
then the earthfault impedance is Z e . From symmetrical componenttheory (see Chapter 4):
thus
since, as shown, Z 1 = Z 2 for a transmission circuit . FromEquations 5.12, Z 1= Z p- Z m and Z O= Z p+2Z m. Thus,substituting these values in the above equation gives
Z e= Z p. This relation is physically valid because Z p is theself- impedance of a single conductor with an earth return.Similarly, for a phase fault between phases B and C at F:
where √ _3E is the voltage between phases and 2Z is the
impedance of the fault loop.
Making use of the above relations a transmission circuitmay be represented, without any loss in generality, bythe equivalent of Figure 5.19(b), where Z 1 is the phaseimpedance to the fault and (Z 0-Z 1)/3 is the impedanceof the earth path, there being no mutual impedancebetween the phases or between phase and earth. The
equivalent is valid for single and double circuit linesexcept that for double circuit lines there is zero sequencemutual impedance, hence Z 0=(Z 00 - Z 0 ’0).
The equivalent circuit of Figure 5.19(b) is valuable in
I I E Z
b c=− = 3 2 1
Z Z Z
e = +2
3 1 0
I E
Z Z Z a =
+ +
3
1 2 0
distance relay applications because the phase and earthfault relays are set to measure Z 2 and are compensatedfor the earth return impedance (Z 0-Z 1)/3 .
It is customary to quote the impedances of atransmission circuit in terms of Z 1 and the ratio Z 0 /Z 1 ,since in this form they are most directly useful. Bydefinition, the positive sequence impedance Z 1 is a
function of the conductor spacing and radius, whereasthe Z 0 /Z 1 ratio is dependent primarily on the level ofearth resistivity ρ . Further detail s may be found inChapter 12.
5.23 CABLE CIRCUITS
The basic formulae for calculating the series and shuntimpedances of a transmission circuit, Equations 5.11 and5.17 may be applied for evaluating cable parameters;since the conductor configuration is normallysymmetrical GMD and GMR values can be used withoutrisk of appreciable errors. However, the formulae mustbe modified by the inclusion of empirical factors to takeaccount of sheath and screen effects. A useful generalreference on cable formulae is given in reference [5.4];more detailed information on particular types of cablesshould be obtained direct f rom the manufacturers. Theequivalent circuit for determining the positive andnegative sequence series impedances of a cable is shownin Figure 5.20. From this circuit it can be shown that:
…Equation 5.24
where Rc, Rs are the core and sheath (screen) resistancesper unit length, X c and X s core and sheath (screen)reactances per unit length and X cs the mutual reactancebetween core and sheath (screen) per unit length. X cs isin general equal to X s.
The zero sequence series impedances are obtained
directly using Equation 5.11 and account can be taken ofthe sheath in the same way as an earth wire in the caseof an overhead line.
The shunt capacitances of a sheathed cable can becalculated from the simple formula:
…Equation 5.25
where d is the overall diameter for a round conductor, T core insulation thickness and ε permittivity of dielectric.When the conductors are oval or shaped, an equivalent
C d T
d
F km= +0 0241
1
2 .
log / ε µ
Z Z R RX
R X
j X X X R X
c scs
s s
c scs
s s
1 2
2
2 2
2
2 2
= = ++
+ − +
5 •
E q u i v a l e n t C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m
P l a n t
• 6 8 •
(a) Actual circuit
B
A
E
ource LineF
B
C F S I c Z
Z
Z 1
(Z 0-Z )/3
I
I a A
E
(b) Equivalent circuit
3E
Figure 5.19: Three-phase equivalent of a transmission circuit
N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e • 6 9 •
• 5 •
E q u i v a l e n t
C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m
P l a n t
diameter d’ may be used where d’ =( 1/ π ) x periphery ofconductor. No simple formula exists for belted orunscreened cables, but an empirical formula that givesreasonable results is:
…Equation 5.26
where G is a geometric factor which is a function of core andbelt insulation thickness and overall conductor diameter.
5.24 OVERHEAD LINE AND CABLE DATA
The following tables contain typical data on overheadlines and cables that can be used in conjunction with thevarious equations quoted in this text. It is not intendedthat this data should replace that supplied bymanufacturers. Where the result s of calculations areimportant, reliance should not be placed on the data in
these Tables and data should be sourced directly from amanufacturer/supplier.
At the conceptual design stage, init ial selection of overheadline conductor size will be determined by four factors:
a. maximum load to be carried in MVAb. length of linec. conductor material and hence maximum
temperatured. cost of losses
Table 5.21 gives indicative details of the capability ofvarious sizes of overhead lines using the above factors,
for AAAC and ACSR conduct or materials. It is based oncommonly used standards for voltage drop and ambienttemperature. Since these factors may not be appropriat efor any particular project, the Table should only be usedas a guide for initial sizing, with appropriately detailedcalculations carried out to arrive at a final proposal.
66kV* Series Reactance X (Ω /km) - - - - - - - - - - - 0.117 0.113 0.109 0.102Susceptance ω C (mS/km) 0.079 0.082 0.088 0.11
Series Resistance R (Ω /km) - - - - - - - - - - - 0.0387 0.031 0.0254 0.0215145kV* Series Reactance X (Ω /km) - - - - - - - - - - - 0.13 0.125 0.12 0.115
Susceptance ω C (mS/km) 0.053 0.06 0.063 0.072Series Resistance R (Ω /km) 0.0487 0.0387 0.0310 0.0254 0.0215 0.0161 0.0126
245kV* Series Reactance X (Ω /km) 0.145 0.137 0.134 0.128 0.123 0.119 0.113Susceptance ω C (mS/km) 0.044 0.047 0.05 0.057 0.057 0.063 0.072
Series Resistance R (Ω /km) 0.0310 0.0254 0.0215 0.0161 0.0126420kV* Series Reactance X (Ω /km) 0.172 0.162 0.156 0.151 0.144
Susceptance ω C (mS/km) 0.04 0.047 0.05 0.057 0.063
For aluminium conductors of the same cross-section, the resistance increases by 60-65 percent, the series reactance and shunt capacitance is virtually unaltered.* - single core cabDifferent values apply if laid in spaced flat formation.Series Resistance - a.c. resistance @ 90°C. Series reactance - equivalent star reactance.Data for 245kV and 420kV cables may vary significantly from that given, dependent on manufacturer and construction.
Table 5.18: Characteristics of polyethylene insulated cables (XLPE)
C i r c u i t s a n d P a r a m e t e r s o f P o w e r S y s t e m
P l a n t
5.25 REFERENCES
5.1 Physical significance of sub-subtransient quantit ies in dynamic behaviour of synchronous machines. I.M. Canay. Proc. IEE, Vol. 135, Pt. B,November 1988.
5.2 IEC 60034-4. Methods for determiningsynchronous machine quantities from tests.
5.3 IEEE Standards 115/115A. IEEE Test Proceduresfor Synchronous Machines.
5.4 Power System Analysis. J.R.Mortlock andM.W.Humphrey Davies (Chapman & Hall,London).