电力系统保护与分析 ELEC4617 章节3、4

Instrument transformers, circuit breakers andSection1Section2Section3Section4Section 1 Instrument transformers: Current transformer Cut View ofTwo typesWound primaryaround a core.in switchgearBar primarya toroid whichsingle primarynecessary primarytransformer isBushing-type bar-primary CTHigh current bushing type CTOil-immersed bar primary CTs•There are two main types of oil-immersed bar primary CTs:1.Tankaprimary(hair-pin)Oil-immersed bar primary CTsABB Tank Type CT IMB 420-550 Bar primary CT types manufacturedby ABBDetails on oil-immersed•Bushingpower transformersno appreciable•Difference between dead tank and live-tankare grounded,while the latter ones are not.•Live-tankmounted•CT for live-tank breakers are constructed in form (consisting of insulators),isolating them from the ground.usually submerged in oil(e.g.ABB CT IMB420-550).Operating•Currentdevicevoltagesystem•The operation requirements–Measuringnormalcurrent);in order–Protectivecurrents;•The main–primarymeasuring–secondaryinstruments •Ideal CT–Themagnetizing–the primarysecondary;the primary•In an actual−the magnetizing−ampere-turnsmagnetizingCT Equivalent Circuit:Phasor diagram:13•The currentif the CTwhere rated transformation ratio:•The angulardisplacementCT selectionStandard CT and 1A in Standard CT ratios:The secondary current of a CT should be just below 5A (USA type CT)or 1A (Europe type CT)for the maximum load.CT Operating RMS value of exciting currentV o l t a g e i n r m s a t s e c o n d a r y s i d eThe exciting value of the the secondary17•The remanence (Ψr)of the closed iron core CTs (TPX)is very large as shown in the figure.Such design can cause severe saturation in the core if core size is not large enough.•The air-gap in the CT-core dramatically reduces the magnitude of the remanent flux >>time required for de-magnetisation is reduced.But it has lower magnetization reactance,thereby producing higher magnetizing current.This may result in large error .closed iron core CTair-gap CTTransient Performance (TP )CTs are included in IEC 60044-6as types TPX,TPY and TPZ.TPX is the closed iron-core CT while TPY and TPZ are CTs with air-gaps in the core.The rated output and the corresponding accuracy class:•Measuring CT specificationFor this protective CT rated at 5A of the secondary,15VA/5A=3V permissible burden is 15VA/52=0.6Ω.It will have no more than 10%to 20X5AX0.6Ω=60V or 20X3V=60V at the secondary across burden.CT accuracy •Protective CT specification•More on standard values for relaying CTs or protective CTs:Continuous VA: 2.5,5,10,15and30Accuracy classes:5%and10%Accuracy-limit factor:5,10,15,20,and30Rated secondaryFor a CT rated10%error30VA/52=1.2With the burdenside is30X5=150Aburden iswithout exceedingRating Plate•For protectionsecond(ie.the faultformat:•For protection it is necessarypresentin termsAn example23CT accuracy classification according to ANSI or IEEE standards The American National Standards Institute (ANSI)relaying accuracy classes are specified in ANSI Standard C57.13-1993.The letter designation codes are given as:Class C;Class T .Knee point definitionTypical excitation curves for a multiratio class C current transformer:Point A is the ANSI knee.Point B is the IEC knee.(Figure of ANSI=IEEE Standard C57.13-1978.)Class T CT The accuracy 4(4Ω)and shown below are the times of rated primary and secondary currents respectively.Example 3-1:Find the error%of a class T CT with 4-ohm burden when carrying 17times of the rated primary current.The corresponding I s =13from the 4ohm curve on the R.H.S.Class C CT accuracy evaluation Standard Classstandard burdensC indicatesburden atwithout exceedingFor example,4ΩburdenThe error isFor the sameV’=5Ω×5A×If CT burdennot saturated.is400V/3ΩPerformanceExample3-2The load andCT is selectedCT copperthe CT ratiorequired.*Note thatselected ratioSolution:For C100(maximum)V’maxFor the maximumV’=This causesChanging theV’maxFor the maximumV’=Performance evaluation using the excitation curveIn this method,CT performance is evaluated using the excitation curve and the equivalent circuit of CT as shown in Figs.3.1and3.2respectively.Fig.3.1Excitation curveof a multi ratio class C CTTo obtain curves:Before knee point,I e is small and negligible,especially for large turns ratio,and I’is almostan exact replica of primary current I p’.After knee point,I e increase significantly with V e,and I’is not exact replica of primary current I p’.Figs. 3.2 Equivalentcircuit of a CTEvaluationStep1:Voltagecircuit asequation:ignored and sStep2:Check saturation from the excitation curve to given CTCT is saturatedCT is not saturatedStep3:IfStep2.OtherwiseExample100:5CT ratioΩ.Solution:(a)I’=I/20=5A.V e=5×(0.082+0.5)=2.91VFrom Fig.CT is not saturated.(b)I’=I/20=8A,V e=8×(0.082+1.6)=13.46VFrom Fig.3-1,saturated.31Exercise :Consider a 13.2kV feeder that is carrying a load of 10MVA at 1.0power factor .Associated with this circuit is a 500/5CT feeding a measurement system whose total load is 10VA.The equivalent circuit of the CT referred to the secondary side is shown in the figure on the right-hand side.Calculate the voltage that would occur in the secondary circuitof the CT if the measurement system was accidentally opened.CT equivalent circuitreferred to secondary side Solution :Single-line diagramReferring the values to the secondary side of the CT givesComplete equivalent circuit with load consideredBurden:Z b =10/52=0.4Ω.When the secondary circuit is closed,the voltage across the measurement system can be calculated approximately,ignoring the shunt branch,asIf the secondary the shunt terminals of the CT isTherefore the voltage increases by almost 120times.CT Symbols:•The polarity marks–P1, P2, S1, S2 for IEC standard;–square–K,L,standard);•Accordingflowingwindingof the otherWe also use the following symbolswhere two dots or two crossesinto the primary winding via secondary winding via dot terminalare approximatelySTAR (WYE) connection provides the line currents at the secondary load.DELTA connection provides delta (difference) currents to the secondary load.Three-phase Connections:CT performance impedance voltage cross on CT connection For a three calculated Y connected Y-connection ∆-connection∆connected A’B’B C’B A’ B.Example 3-4:Select CT ratio of the multi-ratio class-C CT as shown in Fig.3.1for a protection system.The maximum load current,and the minimum and maximum fault currents are 90A,360A and 1200A,respectively.Assume that burden impedance of the CT is 0.5Ω.Check CT saturation and adjust CT ratio if necessary.Assume that three-phase CTs are in Y-connection.Solutions :Select 100:5CT ratio because maximum load current is 90A.Change the ratio to 200:5:For the maximum fault current 1200A,I’=1200/40=30A,V e =30(0.5+0.125)=18.75V .The CT is not saturated.For maximum load current 90A, I’=90/20=4.5A,V e =4.5(0.5+0.082)=2.619V . The CT is not saturated.For the maximum fault current 1200A, I’=1200/20=60A, V e =60(0.5+0.082)=34.92V . The CT is saturated.37Practice problems:1.Find the error%of a class T CT with 4-ohm burden and carrying 16times rated primary current.(21%)2.Determine the maximum voltage supplied by a multi ratio 400/5C400CT without saturation.What is the maximum current which can be supplied if selected CT ratio is 200/5and its burden is 1ohm and.(400V ,200A)3.Figure below shows the connections of the Y and ∆connected current transformers and relays.Z B and Z BG are the impedances of phase relay and ground relay respectively.Calculate the CT burdens for phase A under single phase to ground fault for Y and ∆connected CT respectively,(Z B +Z BG ,2Z B )Section 2VT SelectionSince the impedance current provided When standard volts,the of the protected example,the 1:140:1600:1Standard VT ratios39Types of VTs–Electromagnetic (inductive)voltage transformers (VTs),–Capacitor voltage transformers (CVTs),–Cascade type of inductive VT .Cut View of a Single Pole electromagnetic VT:Equivalent circuits and principle of Operation Electromagnetic VTs•Electromagnetic VT is connected across the points at which voltage measured,and are therefore much like power transformers with secondary winding operating close open-circuit.•For such equal to•The parameters error and •An inductive no-load error is and leakage thus negligible.•In practice rated flux saturation –>>consequently,impedance operating •It is extremely •VT contains Three-phase Star connectionDelta connectionThree-phaseOpen Delta connection Broken Delta connectionDesign Examples of Inductive VTs•VTs have only one iron core with attached secondary windings.•Theyinsulated44single pole double poleDesign Examples Principle of Operation of Capacitor VTs (CVT)•The cost of inductive (electromagnetic)VTs tends to increase at exponential rate with the rated voltage.•The CVT combines a capacitive potential divider with an auxiliary (intermediate)electromagnetic Voltage Transformer .•In the case of higher voltage levels (>145kV)either Capacitor VT (CVT)or cascade type of inductive VT are used.–The CVT was developed to reduce the high cost of conventional VT by compromising on the frequency and transient response.–Connecting two or more inductive VT units together forms the cascade VT .•The CVT equivalent circuit:L is chosen to cancel reactance of C or V L cancels V C.Then small and Vs’is almost in phase with V i or V p.Design Example (ABB CVT)Section 3•The circuit the circuit.•Once a •It takes trip command.•When two appears the medium •The contacts mechanism releases Timing Diagram •This voltage flow of •This current •Different cooled and •Therefore instant of Circuit breaker contact parting:The gap between insulating medium,•Oil;•Pressurized •Air .•Sulphur •Vacuum;Summary of Types of circuit 6.Direct-current The most important 1.Oil circuit 2.Air-blast 3.SF6circuit 4.Vacuum 5.Solid-state There are mainly the Miniature circuit Molded case Residual current-operated53Molded case circuit breaker(MCCB)Gas-insulated switchgear (GIS) 1.The space 5.In addition SF6has 2.It is totally saline air 3.It possesses personnel.4.It is easier in seismically IntroductionGas -insulated (SF6)gas HV switchgearGIS enclosureSectional view of a typical SF6gas-insulated,145kV circuit breaker one enclosure for the three phase.NameplateMain ratings Rated voltageIt designates should be this rating.Rated continuous The maximum continuously is determined Rated interrupting The time in to the instant breaker part cycles.Rated short-circuit asymmetrical It is the maximum rated voltage.generally calculated symmetrical sub-transient The asymmetrical interrupting determined and the circuit-breaker’s Example 3-5installed in 2)15,3)20,determining system frequency is 50Hz.1)X/R=102)X/R=153)X/R=204)X/R=305)X/R=40One can see from above results that when X/R ratio is less than multiplying ratio is less other words,Rated momentary It is defined the breaker position.It should be short-circuit three-phase Circuit breaker The breaking of a three-phase where V is line-to-line voltage current or rated Example 3-6The neutrals system data below.The system The load level is 100MVA with a lagging power factor of 1)Determine 2)Calculate 3)Calculate 4)Calculate Item Base MVA GENSolution:1)Since thevoltage ofkV.2)Since theSo the minimum circuit breaker is2.887 3)CalculateThevenin equivalent impedanceviewed towards source side atplace where circuit breakerinstalled:So the rms value of ac component of fault current which results bolted three-phase symmetrical fault at load side during sub-transient isFrom the conditionthe fault locationone.TheinterruptingTherefore,6.0768kA.4)CalculateThe ratedSection4•Simplest•If the currentrating,theconducted•For low•Easiest•Fuse worksfor opening.Notched fuse elementFuses–Specifications•VoltageA blown fusedeterminesafter the fuse•ContinuousThis is the•InterruptingThis is the largest current that the fuse can safely interrupt.•Time ResponseThe meltingovercurrentThe curve haslike an exponentialconductorcreated.Fuses–Benefits•Benefit–low cost and simplicity of operation •Limitations–current–process–additionalitself.•These limitationscircuits by•The individualExample of high current fuses:Expulsion Fuses•The expulsion fuse is a protection method for medium voltage level with lower costthan circuit breakers.•They are typically mounted at the top of the pole for11-33kV systems.•The fuse element is kept in tension with spring.•Arc extinguished by expanding gasses which blow out of end of tube.•Extinction aided by spring separating contacts.67Transient waveformsof fuses:Oil circuit Oil circuit breakers an overload the insulated separate,pressure of causes cool Air-blast circuit breakersThese circuit supersonic reservoirs at located in the SF6 circuit breakers: Vacuum circuit breakersThese circuit because there hermetically polluted.Their several circuit often used AppendixSolid-state circuit breakerSolid-state semiconductors,considerable respect to much faster that get eroded or replaced.material costs Circuit breakers to bring the be used as DC Direct-current (DC) circuit breakersGas-insulated switchgear (GIS) 1.The space 5.In addition SF6has 2.It is totally saline air 3.It possesses personnel.4.It is easier in seismically IntroductionGas -insulated (SF6)gas HV switchgearGas-insulated switchgear (cont’d)2.It is nonflammable.1.It is nontoxic.4.It has and low effectively 3.It is noncorrosive.At 3atm,the Chemical At atmospheric that of air ,In addition physical and Furthermore,and is therefore of oil or air Gas-insulated switchgear (cont’d)A (cont’d)1—busbar 2—busbar two; 10—density monitor3—disconnector;4—current transformer;5—circuit breaker;6—voltage transformer;7—maintenance grounding switch;8—high-speed grounding switch;9—gas connector;4.1Introduction Protection incoming signal Construction •electromechanical •solid state •microprocessor •non-electricIncoming signal •current•voltage•power •frequency •temperature •pressure•speed•others75Armature-type relaySolenoid relayAttraction relays effectively have no time delay widely used when instantaneous operation is required.77These relays motor .As such,relays are induction cup to the rotor motor ,the moving element is completed Induction-type sources of in which the turn.The phase difference otherwise,produced.pole faces one of the motor action.sources of sources of permanent to provide characteristics.Typical electromechanicalinduction cup or cylinder relayLet us assume in the coils sinusoidal:Typical electromechanicalinduction cup or cylinder relayUsing motor working mechanism,one can derive torque on where K is a constant.The above derived both induction relay to compute by currents The direction coil with the leading current to the one with theThe induction-type of the source characteristic relay or a Induction disc-produce under-a current source made to respond Table below common over-current the time to pickup is given for the same multiple M of pickup current.where M=I/I p multiple of pick-up current.For more details on the derivation of the formula above,see appendix 4.2Relays:All relays design.Two amplitude parator compared two signals by their phase difference than by their amplitudes.The following analysis shows that for two signals,S A and S B ,which compared in magnitude,there exist two other signals S 1and S 2compared by phase.The relationship between the signals is as follows:SoThe comparison of the amplitude Defining S 181Given that thenSois satisfied whenThe above relationships demonstrate that two signals,obtained for use with an amplitude comparator ,can be converted in order to be used by a phase-angle comparator .Symbols :Conversion of current to voltage by air-gap transformer (also called transactor or compensator):Choosing suitable values for design constantsenables a wide range of comparator characteristics to be obtained.or834.2.1Amplitude comparator characteristics Plain impedance relay Hence the input signals are The relay operates wheneverConventional mho relayThe characteristic of the conventional mho relay is shown in the figure below:HereHence the input signals areFrom one can haveor85Example k =1;Prove :Admittance locus of the conventional mho relay is a straight line.(For this reason it is referred to as a mho characteristic)LetThen one can obtain boundary equality equation:OrProof :Exercise 4-1The input signals are thusProve:the wheneverAn amplitude874.2.3Phase comparator characteristicsThe Characteristics of phase comparators are conveniently derived by taking the argument of the complex ratio S 1/S 2,i.e.whereTaking the argument of the complex ratio given byOne hasConventional mho relay89Conventional mho relay (Cont’d)For point P on KMO,SoFor point P on KNL,SoFor point P within semicircle KMO,For point P within semicircle KNO,Compared with conventional impedance relay,mho relay is directional.It can be used independently to carry out distance protection and there is no need of directional relay.But the conventional impedance relay needs.Exercise 4-2An phase comparator based reactance relay Prove:when ,relay operates and corresponding imaginary or reactance of the detectedSummary4.3 Directional relay Relays thatlocation and, practical example Directionalwith otheruse the outputfault sensors, overcurrentdesired operatingthe fault sensor’s operate.Ifdirection orcurrent is higherA directional-sensing quantity)that protected circuitfor phase-typeFor all practical positions significantly around180of the circuit934.3.1Relay structure and operating principleFig. 4-1 Input to directional relay •Relay signals:current and voltage•Two coils (C1,C2)with four poles,where coil C2is modeled by R V and L V .V r is the voltage applied to coil C2via some other circuit elements which are to form phase shifter .Itproduces •is the voltage across ideal winding or L V as shown in Fig.4-1.Flux ФVproduced by coil C2in the structure is in phase with current I V and 900lagging .•R V can be made negligible compared with reactance of L V at power frequency.•I r applied on coil C1producesFig. 4-2 Four-pole induction cup directional relay•An electromagnetic torque T e on the induction cup:’4.3.2 Phasor diagram and operating areaPhasor diagram of current and voltage and relay characteristic are shown in figure below•Фi is in phase with the current I r . •Фv’lags the voltage V’r by an angle close to 900.••T e is the maximum at α=90o or when I r is in phase with V’r .•Maximum torque line (MTL)is aligned with V’r .•T eis zero at α=0o .•Zero torque line (ZTL)is the line at α=0o and ZTL and MTL are always perpendicular to each other .•Maximum torque angle MTA or angle of maximum torque ATM (denoted by β)is the angle between polarizing or reference signal (normally voltage)Fig. 4-3•Restrain area:αis from 1800to 3600(sin αand T e are negative.)•Trip area:αis from 00to 1800(sin αand T e are positive.)4.3.3Directional Directional or βvalue are 30o ,45reference voltage When the constant while operating discussed 30o -MTA relay unit:The operating directional voltage V r and 30o .The relay unit will operate for the current from leading .Fig. 4-4 3030o -MTA relay:Feeding the Meaning of and reference MTL.9760o -MTA relay unit:The operating characteristic of 60o type relay is shown in Fig.4-5.For this directional relay,MTA=−60o which means that the angle between reference or polarizing voltage V r and MTL is 600,and maximum torque occurs when I r lags V r by 60o .The relay unit will operate for the current from 150o lagging V r to 30o leading .Figure 4-5 60o -type or 60o -MTA relay unitFeeding the relays:V r =-3V 0,I r =3I 00o -MTA relay unitThe operating characteristic 0o -MTA type relay shown in Fig.4-6.For this type directional relay,MTA=0o which means that the maximum torque occurs when actual I r and V r are in phase.Fig. 4-6 0o -MTA relay unit0o -MTA relay unit: 300connection Feeding the relays:990o -MTA relay connection)Feeding the relays:0o -MTA relay connection)Feeding the relays:This is equivalent connection 45o -MTA relay。