Symmetrical Components, Unbalanced Fault Analysis

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ECE 476POWER SYSTEM ANALYSIS

• Lecture 21
• Symmetrical Components, Unbalanced Fault Analysis
• Professor Tom Overbye
• Department of Electrical andComputer Engineering

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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 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)

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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. Well
  derive these relationships for several common
  faults.
• With a SLG fault only one phase has non-zero
  fault current -- well assume it is phase A.

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SLG Faults, contd
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SLG Faults, contd
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SLG Faults, contd
With the sequence networks in series we
can solve for the fault currents (assume Zf0)
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Example 9.3
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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.

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LL Faults, cont'd
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LL Faults, con'td
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LL Faults, cont'd
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LL Faults, cont'd
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LL Faults, cont'd
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Double Line-to-Ground Faults

• 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.

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DLG Faults, cont'd
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DLG Faults, cont'd
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DLG Faults, cont'd
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DLG Faults, cont'd

• The three sequence networks are joined as follows

Assuming Zf0, then
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DLG Faults, cont'd
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Unbalanced Fault Summary

• 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

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Generalized System Solution

• 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 the
  thevenin equivalent impedances the pre-fault
  voltage is the positive sequence thevenin voltage
• Connect and solve the thevenin equivalent
  sequence networks to determine the fault current
• Sequence voltages throughout the system are

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Generalized System Solution, contd

• Sequence voltages throughout the system are given
  by

This is solved for each sequence network!
5. Phase values are determined from the
sequence values
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Unbalanced System Example
For the generators assume Z Z? j0.2 Z0
j0.05 For the transformers assume Z Z? Z0
j0.05 For the lines assume Z Z? j0.1 Z0
j0.3 Assume unloaded pre-fault, with voltages
1.0 p.u.
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Positive/Negative Sequence Network
Negative sequence is identical to positive
sequence
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Zero Sequence Network
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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 the
thevenin impedances
Positive Seq. Negative Seq.
Zero Seq.
The fault type then determines how the networks
are interconnected
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Bus 3 SLG Fault, contd
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Bus 3 SLG Fault, contd
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Faults on Lines

• 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

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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
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Line Fault Example, contd
The Ybus now has 4 buses
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Power System Protection

• Main idea is to remove faults as quickly as
  possible while leaving as much of the system
  intact as possible
• Fault sequence of events
• 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

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Power System Protection

• 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

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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 shows potential protection schemes for
a radial system. The bottom scheme is preferred
since it results in less lost load
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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 impedance
• Radial power system protection systems usually
  use inverse-time overcurrent relays.
• Coordination of relay current settings is needed
  toopen the correct breakers

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Inverse Time Overcurrent Relays

• Inverse time overcurrent relays respond
  instan-taneously to a current above their maximum
  setting
• They respond slower to currents below this value
  but above the pickup current value