Presentation on theme: "Announcements Be reading Chapters 9 and 10 HW 8 is due now."— Presentation transcript:
0 ECE 476 POWER SYSTEM ANALYSIS Lecture 21Symmetrical Components, Unbalanced Fault AnalysisProfessor Tom OverbyeDepartment of Electrical and Computer Engineering
1 Announcements Be reading Chapters 9 and 10 HW 8 is due now. HW 9 is 8.4, 8.12, 9.1,9.2 (bus 2), 9.14; due Nov 10 in classStart working on Design Project. Tentatively due Nov 17 in classSecond exam is on Nov 15 in class. Same format as first exam, except you can bring two note sheets (e.g., the one from the first exam and another)
2 Single Line-to-Ground (SLG) Faults Unbalanced faults unbalance the network, but only at the fault location. This causes a coupling of the sequence networks. How the sequence networks are coupled depends upon the fault type. We’ll derive these relationships for several common faults.With a SLG fault only one phase has non-zero fault current -- we’ll assume it is phase A.
7 Line-to-Line (LL) Faults The second most common fault is line-to-line, which occurs when two of the conductors come in contact with each other. With out loss of generality we'll assume phases b and c.
19 Unbalanced Fault Summary SLG: Sequence networks are connected in series, parallel to three times the fault impedanceLL: Positive and negative sequence networks are connected in parallel; zero sequence network is not included since there is no path to groundDLG: Positive, negative and zero sequence networks are connected in parallel, with the zero sequence network including three times the fault impedance
20 Generalized System Solution Assume we know the pre-fault voltagesThe general procedure is thenCalculate Zbus for each sequenceFor a fault at bus i, the Zii values are the thevenin equivalent impedances; the pre-fault voltage is the positive sequence thevenin voltageConnect and solve the thevenin equivalent sequence networks to determine the fault currentSequence voltages throughout the system are
21 Generalized System Solution, cont’d Sequence voltages throughout the system are given byThis is solvedfor eachsequencenetwork!5. Phase values are determined from the sequence values
22 Unbalanced System Example For the generators assume Z+ = Z = j0.2; Z0 = j0.05For the transformers assume Z+ = Z =Z0 = j0.05For the lines assume Z+ = Z = j0.1; Z0 = j0.3Assume unloaded pre-fault, with voltages =1.0 p.u.
23 Positive/Negative Sequence Network Negative sequence is identical to positive sequence
25 For a SLG Fault at Bus 3The sequence networks are created using the pre-faultvoltage for the positive sequence thevenin voltage,and the Zbus diagonals for the thevenin impedancesPositive Seq Negative Seq Zero Seq.The fault type then determines how the networks areinterconnected
28 Faults on LinesThe previous analysis has assumed that the fault is at a bus. Most faults occur on transmission lines, not at the busesFor analysis these faults are treated by including a dummy bus at the fault location. How the impedance of the transmission line is then split depends upon the fault location
29 Line Fault Example Assume a SLG fault occurs on the previous system on the line from bus 1 to bus 3, one third of the wayfrom bus 1 to bus 3. To solve the system we add adummy bus, bus 4, at the fault location
30 Line Fault Example, cont’d The Ybusnow has4 buses
31 Power System Protection Main idea is to remove faults as quickly as possible while leaving as much of the system intact as possibleFault sequence of eventsFault occurs somewhere on the system, changing the system currents and voltagesCurrent transformers (CTs) and potential transformers (PTs) sensors detect the change in currents/voltagesRelays use sensor input to determine whether a fault has occurredIf fault occurs relays open circuit breakers to isolate fault
32 Power System Protection Protection systems must be designed with both primary protection and backup protection in case primary protection devices failIn designing power system protection systems there are two main types of systems that need to be considered:Radial: there is a single source of power, so power always flows in a single direction; this is the easiest from a protection point of viewNetwork: power can flow in either direction: protection is much more involved
33 Radial Power System Protection Radial systems are primarily used in the lower voltage distribution systems. Protection actions usually result in loss of customer load, but the outages are usually quite local.The figure showspotential protectionschemes for aradial system. Thebottom scheme ispreferred since itresults in less lost load
34 Radial Power System Protection In radial power systems the amount of fault current is limited by the fault distance from the power source: faults further done the feeder have less fault current since the current is limited by feeder impedanceRadial power system protection systems usually use inverse-time overcurrent relays.Coordination of relay current settings is needed to open the correct breakers
35 Inverse Time Overcurrent Relays Inverse time overcurrent relays respond instan-taneously to a current above their maximum settingThey respond slower to currents below this value but above the pickup current value