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 |  موضوع: Over-current and Earth Fault Protection *****part 3 الأربعاء أكتوبر 27, 2010 10:06 am | |
| Over-current and Earth Fault Protection *****part 3
Parallel feedersIf non-directional relays are applied to parallel feeders,any faults that might occur on any one line will, regardless of the relay settingsused, isolate both lines andcompletely disconnect the power supply. With this type of system configurationit is necessary to apply directional relays at the receiving end and to gradethem with the non-directional relays at the sending end, to ensure correctdiscriminative operation of the relays during line. faults. This isdone by setting the directional relays R'1 and R'2 as shown in (Fig.18) with their directional elementslooking into the protected line, and giving them lower time and currentsettings than relays R1 and R2. The usual practice is to set relays R'1 and R'2 to 50% of the normal full load of theprotected circuit and 0.1 TMS, but care must be taken to ensure that their continuous thermal rating of twice rated currentis not exceeded. v Ring mainsDirectional relays are more commonlyapplied to ring mains. In the case of a ring main fed at one point only, therelays at the supply end and at the mid-point substation, where the setting ofboth relays are identical, can be made non-directional, provided that in thelatter case the relays are located on the same feeder, that is, one at each endof the feeder.It is interesting to note that when thenumber of feeders round the ring is an even number, the two relays with thesame operating time are at the same substation and will have to be directional,whereas when the number of feeders is an odd number, the two relays with thesame operating time are at different substations and therefore do not need tobe directional. It may also be noted that, atinter-mediate substations, whenever the operating times of the relays at eachsubstation are different, the difference between their operating times is neverless than the grading margin, so the relay with the longer operating time canbe non-directional. v Grading of ring mainsThe usual procedure for grading relaysin an inter-connected system is to open the ring at the supply point and tograde the relays first clockwise and then anti-clockwise; that is, the relayslooking in a clock-wise direction round the ring are arranged to operate in the sequence 1—2—3—4—5—6 and therelays looking in theanti-clockwise direction are arranged to operate in the sequence 1'—2'—3'—4'—5'—6', as shown in (Fig.19)[ندعوك للتسجيل في المنتدى أو التعريف بنفسك لمعاينة هذه الصورة]The arrows associated with the relayingpoints indicate the direction of current flow that will cause the relays to operate. A double-headed arrow is used to indicate a non-directional relay,such as those at the supply point where the power can flow only in onedirection, and a single-headed arrow a directional relay, such as those at intermediate substations around the ring where the power canflow in either direction. Thedirectional relays are set in accordance with the invariable rule, applicable to all forms ofdirectional protection that the current in the system must flow from thesubstation bus-bars into the protected line in order that the relays mayoperate.Disconnection of the faulty line is carried out accordingto time and fault current direction. As in any parallel system, the faultcurrent has two parallel paths and divides itself in the inverse ratio of theirimpedances.
Thus, at each substation in the ring, one set of relays will be made inoperativebecause of the direction of current flow, and the other set operative. It willalso be found that the operating times of the relays that are inoperative are fasterthan those of the operativerelays, with the exception of the mid-point substation, where the operating times ofrelays 3 and 3' happen to bethe same.
The relays which are operative are graded downwards towards the fault and the last to beaffected by the fault operates first. This applies to both paths to the fault.Consequently, the faulty line is the only one to be disconnected from the ringand the power supply is maintained to all the substations.
When two or more power sources feed into a ring main, timegraded over current protection is difficult to apply and full discriminationmay not be possible. With two sources of supply, two solutions are possible.The first is to open the ring at one of the supply points, whichever is moreconvenient, by means of a suitable high set instantaneous over-current relayand then to proceed to grade the ring as in the case of a single infeed, thesecond to treat the section of the ring between the two supply points as acontinuous bus separate from the ring and to protect it with a unit system ofprotection, such as pilot wire relays, and then proceed to grade the ring as inthe case of a single infeed. Directional Earth-Fault ProtectionIn the directional over-currentprotection the current coil of relay is actuated from secondary current of lineCT. whereas the current coil of directional earth fault relay is actuated byresidual current.In directional over-current relay, thevoltage coil is actuated by secondary ofline VT. In directional earth fault relay, the voltage coil is actuated by the residualvoltage. Directional earth fault relays sense the direction in whichearth fault occurs with respect to the relaylocation and it operates for fault in a particular direction.The directional earth fault relay (single phase unit) has two coils. Thepolarizing quantity is obtained either from residual currentIRS = (Ia+ Ib + Ic)or residual voltage VRs = Va + Vb + Vc Where Va, Vb and Vc are phase voltages.Referring to (Fig. 11) thedirectional earth-fault relay has two coils. Oneto the coils is connected in residual current circuits (Ref. Fig. 5). Thiscoil gets current during earth-faults. The other coil gets residual voltage, VRS= Va + Vb + VcWhere Va, Vb and Vc are secondary voltages of the potential transformer('Three phase five limb potential transformer or threeseparate single phase potential transformersconnected as shown in Fig. 20). The coil connected inpotential-transformer secondary circuit gives a polarizing field.[ندعوك للتسجيل في المنتدى أو التعريف بنفسك لمعاينة هذه الصورة]The residual current IRS i.e. the out of balance current is given tothe current coil and the residualvoltage VRsisgiven to the voltage coil of the relay. The torque is proportional to T =IRS * VRS* cos (Φ - α) Φ = angle between IRS and VRs α = angle of maximum torque. v SummaryOver-current protection responds toincrease in current above the pick-upvalue over-currents are caused by overloads and short-circuits.The over-current relays are connected the secondary ofcurrent transformer. The characteristic of over-current relays include inversetime characteristic, definite time characteristic.Earth fault protection responds tosingle line to ground faults and doubleline to ground faults. The current coil ofearth-fault relay is connected either in neutral to ground circuit or inresidually connected secondary CT circuit.Core balance CTs are used for earth-fault protection.Frame leakage protection can be used for metal cladswitchgear.Directional over-current relay andDirectional Earth fault relay respondsto fault in which power flow is in the set direction from the CT and PTlocations. Such directional relays are used when power can flow from bothdirections to the fault point.v Co-ordination Correct current relay application requires knowledgeof the fault current that can flow in each part of the network. Since large scale tests are normally impracticable, systemanalysis must be used. It is generally sufficient to use machine transientreactance X'd and to work on the instantaneous symmetrical currents.The data required for a relay setting study are:1. A one-line diagram of the power system involved, showingthe type and rating of the protective devices and their associated currenttransformers. 2. The impedances in ohms, per cent or per unit, of all powertransformers, rotating machines and feeder circuits. 3. The maximum and minimum values of short circuit currentsthat are expected to flow through each protective device. 4. The starting current requirements of motors and thestarting and stalling times of induction motors. 5. The maximum peak load current through protective devices. 6. Decrement curves showing the rate of decay of the faultcurrent supplied by the generators.7. Performance curves of the current transformers. 8. The relay settings are first determined so as to give theshortest operating times at maximum fault levels and then checked to see ifoperation will also be satisfactory at the minimum fault current expected. It isalways advisable to plot the curves of relays and other protective devices,such as fuses, that are to operate in series, on a common scale. It is usuallymore convenient to use a scale corresponding to the current expected at thelowest voltage base or to use the predominant voltage base. The alternativesare a common MVA base or a separate current scale for each system voltage. 9. The basic rules for correct relay co-ordination cangenerally be stated as follows:10. Whenever possible, use relays with thesame operating characteristic in series with each other.11. Make sure that the relay farthest fromthe source has current settings equal to or less than the relays behind it,that is, that the primary current required operating the relay in front is always equal toor less than the primarycurrent required operating the relay behind it. v PRINCIPLES OF TIME/CURRENTGRADINGAmong the various possible methods usedto achieve correct relay co-ordination are thoseusing either time or over current or a combination of bothtime and over-current. Thecommon aim of all three methods is to give correct discrimination. That is to say, each onemust select and isolate only the faulty section of the power system network,leaving the rest of the system undisturbed.1. Discrimination by time In this method an appropriate timeinterval is given by each of the relays controlling the circuit breakers in apower system to ensure that the breaker nearest to the fault opens first. Asimple radial distribution system is shown in (Fig. 21) to illustrate the principle.[ندعوك للتسجيل في المنتدى أو التعريف بنفسك لمعاينة هذه الصورة]Circuit breaker protection is providedat B, C, D and E, that is, at the infeed end ofeach section of the power system. Each protection unit comprises a definitetime delay over current relay in which the operation of the current sensitiveelement simply initiates the time delay element. Provided the setting of the current element is below thefault current value this element plays no part in the achievement of discrimination. For this reason, therelay is sometimes described as an'independent definite time delay relay' since its operating time is for practical purposes independent of the level of overcurrent.It is the time delay element, therefore,which provides the means ofdiscrimination. The relay at B is set at the shortest time delay permissible to allow a fuse to blow for a fault on the secondary side oftrans-former A. Typically,a time delay of 0.25s is adequate.If a fault occurs at F, the relay at B will operate in 0.25s, and the subsequent operation of thecircuit breaker at B will clear the fault before the relaysat C, D and E have time to operate. The main disadvantage of this method of discrimination isthat the longest fault clearance time occurs for faults in the section closestto the power source, where the fault level (MVA) is highest. 1. Discrimination by currentDiscrimination by current relies on thefact that the fault current varies with the position of the fault, because ofthe difference in impedance values between the source and the fault. Hence,typically, the relays controlling the various circuit breakers are set tooperate at suitably tapered values such that only the relay nearest to thefault trips its breaker. (Fig. 22) illustrates the method.[ندعوك للتسجيل في المنتدى أو التعريف بنفسك لمعاينة هذه الصورة]For a fault at F1, the system short circuit current isgiven by:I = 6350 /(Zs + ZL1) A
Where Zs = source impedance = 112 / 250 = 0.485 ohms ZL1= cable impedance between C and B =0.24 ohms Hence I=6350/0.725 = 8800 ASo a relay controlling the circuitbreaker at C and set to operate at a fault current of 8800 A would in simple theory protect the whole of thecable section between C andB. However, there are two important practical points which affect this methodof co-ordination.1. It is not practical to distinguish between a fault at Fl and a fault at F2, since the distance between these points can be only a few meters,corresponding to a change infault current of approximately 0.1%.2. In practice, there would be variations in the source faultlevel, typically from 250 MVA to 130 MVA. At this lower fault level the faultcurrent would not exceed 6800 A even for a cable fault close to C, so a relay set at 8800 Awould not protect any of thecable section concerned.
Discrimination by current is thereforenot a practical proposition for correct grading between the circuit breakers at C and B. However, the problem changes appreciably when there is significantimpedance between the twocircuit breakers concerned. This can be seen by considering the grading required between the circuit breakers at B and A in (Fig. 22). Assuming a fault at F4, theshort-circuit current is given by:I =6350 /(Zs + ZL1 + ZL2 +ZT) A
Where
ZS= source impedance =112 / 250 = 0.485 ohms ZL1 = cable impedance between C and B 0.24 ohmsZL2 = cable impedance between B and 4 MVA
transformer 0.04 ohmsZT = transformer impedance =0.07(112/4) =2.12 ohmsHenceI = 6350/ 2.885 = 2200 A For this reason, a relay controlling thecircuit breaker at B and set to operate at a current of 2200A plus a safetymargin would not operate for a fault at F4 and would thusdiscriminate with the relay at A. Assuming a safety margin of 20% to allow forrelay errors and afurther 10% for variations in the system impedance values, it is reasonable to choose a relay setting of 1.3 x 2200, that is, 2860 Afor the relay at B. Now, assuming a fault at F3, that is, at the end of the 11 kV cablefeeding the 4 MVA transformers, the short-circuit current is given by:
I =6350 /(Zs + ZL1 + ZL2 +ZT)
I =6350 /(0.485 + 0.24 + 0.04)=8300 Amp.
Alternatively,assuming a source fault level of 130 MVA:
I =6350 /(0.93 + 0.24 + 0.004)=5250 Amp.In other words, for either value ofsource level, the relay at B would operate correctly for faultsanywhere on the 11 kVcable feeding the transformer. | |
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