ECE 476 POWER SYSTEM ANALYSIS Lecture 21 Symmetrical

ECE 476 POWER SYSTEM ANALYSIS Lecture 21 Symmetrical Components, Unbalanced Fault Analysis Professor Tom Overbye Department of Electrical and Computer Engineering

Announcements l l l 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 class Start working on Design Project. Tentatively due Nov 17 in class Second 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) 1

Single Line-to-Ground (SLG) Faults l l 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. 2

SLG Faults, cont’d 3

SLG Faults, cont’d 4

SLG Faults, cont’d With the sequence networks in series we can solve for the fault currents (assume Zf=0) 5

Example 9. 3 6

Line-to-Line (LL) Faults l 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. 7

LL Faults, cont'd 8

LL Faults, con'td 9

LL Faults, cont'd 10

LL Faults, cont'd 11

LL Faults, cont'd 12

Double Line-to-Ground Faults l With a double line-to-ground (DLG) fault two line conductors come in contact both with each other and ground. We'll assume these are phases b and c. 13

DLG Faults, cont'd 14

DLG Faults, cont'd 15

DLG Faults, cont'd 16

DLG Faults, cont'd l The three sequence networks are joined as follows Assuming Zf=0, then 17

DLG Faults, cont'd 18

Unbalanced Fault Summary l l l SLG: Sequence networks are connected in series, parallel to three times the fault impedance LL: Positive and negative sequence networks are connected in parallel; zero sequence network is not included since there is no path to ground DLG: Positive, negative and zero sequence networks are connected in parallel, with the zero sequence network including three times the fault impedance 19

Generalized System Solution l l 1. 2. 3. 4. Assume we know the pre-fault voltages The general procedure is then Calculate Zbus for each sequence For a fault at bus i, the Zii values are thevenin equivalent impedances; the pre-fault voltage is the positive sequence thevenin voltage Connect and solve thevenin equivalent sequence networks to determine the fault current Sequence voltages throughout the system are 20

Generalized System Solution, cont’d Sequence voltages throughout the system are given by 4. This is solved for each sequence network! 5. Phase values are determined from the sequence values 21

Unbalanced System Example For the generators assume Z+ = Z = j 0. 2; Z 0 = j 0. 05 For the transformers assume Z+ = Z =Z 0 = j 0. 05 For the lines assume Z+ = Z = j 0. 1; Z 0 = j 0. 3 Assume unloaded pre-fault, with voltages =1. 0 p. u. 22

Positive/Negative Sequence Network Negative sequence is identical to positive sequence 23

Zero Sequence Network 24

For a SLG Fault at Bus 3 The sequence networks are created using the pre-fault voltage for the positive sequence thevenin voltage, and the Zbus diagonals for thevenin impedances Positive Seq. Negative Seq. Zero Seq. The fault type then determines how the networks are interconnected 25

Bus 3 SLG Fault, cont’d 26

Bus 3 SLG Fault, cont’d 27

Faults on Lines l l The previous analysis has assumed that the fault is at a bus. Most faults occur on transmission lines, not at the buses For 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 28

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 way from bus 1 to bus 3. To solve the system we add a dummy bus, bus 4, at the fault location 29

Line Fault Example, cont’d The Ybus now has 4 buses 30

Power System Protection l l Main idea is to remove faults as quickly as possible while leaving as much of the system intact as possible Fault sequence of events 1. 2. 3. 4. Fault occurs somewhere on the system, changing the system currents and voltages Current transformers (CTs) and potential transformers (PTs) sensors detect the change in currents/voltages Relays use sensor input to determine whether a fault has occurred If fault occurs relays open circuit breakers to isolate fault 31

Power System Protection l l 1. 2. Protection systems must be designed with both primary protection and backup protection in case primary protection devices fail In 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 view Network: power can flow in either direction: protection is much more involved 32

Radial Power System Protection l 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 shows potential protection schemes for a radial system. The bottom scheme is preferred since it results in less lost load 33

Radial Power System Protection l l l 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 impedance Radial power system protection systems usually use inverse-time overcurrent relays. Coordination of relay current settings is needed to open the correct breakers 34

Inverse Time Overcurrent Relays l l Inverse time overcurrent relays respond instantaneously to a current above their maximum setting They respond slower to currents below this value but above the pickup current value 35