Fault Analysis ECE4334

8/11/2019 Fault Analysis ECE4334

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ECE4334

Dr. C.Y. Evrenosoglu

ECE4334

FAULT ANALYSIS

Dr. E


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ECE4334Content

System representation Single line-to-ground fault

Line-to-line fault

Double line-to-ground fault

Balanced (3) faults



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ECE4334System representation and assumptions

The power system operates under balanced steady-state conditions before

the fault occurs. Thus, the sequence networks are uncoupled before the

fault occurs. During unsymmetrical faults they are interconnected at the

fault location.

Prefault load current is neglected. This, prefault voltage, VF at the faultpoint is close to its nominal value (i.e. it can be take to be 10). The

prefault voltage at each bus (and machine internal voltages) in the

positive-sequence network equals VF. This assumption is called flat

profile. In defining the prefault flat voltage profile, we do take into account the

connection-induced phase shifts in Y xfmr.

Transformer winding resistances and shunt admittances are ignored.

Load impedances can be ignored

Series resistance and shunt elements in lines are ignored.

Armature resistance in generators is ignored.

The resulting circuit model consists only of sources and pure reactances.Dr. C.Y. Evrenosoglu


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ECE4334Using symmetrical components



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ECE4334Using symmetrical components



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ECE4334Example 9.1



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ECE4334Example 9.1



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ECE4334Example 9.2


At bus 2 all the voltages are 0!


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ECE4334Example 9.2



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ECE4334Single line-to-ground fault



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ECE4334Single line-to-ground fault on the example


Remember?


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Remember?


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ECE4334Example 9.3


No fault impedance


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ECE4334REMEMBER: Example 9.1



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ECE4334Back to Example 9.3



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ECE4334Example 9.3



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ECE4334Line-to-line fault


{


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ECE4334Line-to-line fault



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ECE4334Line-to-line fault on the example



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ECE4334Line-to-line fault on the example



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ECE4334Example 9.4



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ECE4334Double line-to-ground fault



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D bl li d f l


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ECE4334D bl li d f l h l


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ECE4334Double line-to-ground fault on the example


ECE4334D bl li t d f lt th l


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ECE4334Double line-to-ground fault on the example


ECE4334E l 9 5


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ECE4334Example 9.5


ECE4334Example 9 5 (a b)


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ECE4334Example 9.5 (a,b)


ECE4334Example 9 5 (a b)


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ECE4334Example 9.5 (a,b)


ECE4334Example 9 5 (c)


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ECE4334Example 9.5 (c)




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ECE4334Example 9.5 (c)




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C 33Example 9.5 (c)




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Example 9.5 (c)


ECE4334Example 9.6


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Example 9.6




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REMEMBER: Example 9.1


ECE4334Example 9.6


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p


ECE4334Example 9.6


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p


ECE4334Example 9.6


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p


ECE4334Example 9.6


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p


ECE4334Example 9.6


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Results by ignoringthe xfmr shifts

ECE4334Generalized solution for networks


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Assume that we know the pre-fault voltages (solution

of power flow)

The general procedure is given as follows:

1. Calculate Zbus for each sequence (first calculateYbus)

2. For a fault at bus i, the Zii values are the Thvenin

equivalent impedances; the pre-fault voltage is thepositive sequence Thvenin voltage.

3. Connect and solve the Thvenin equivalent

sequence networks to determine the fault currents

in each sequence.

Dr. C.Y. Evrenosoglu Thanks to Dr. Tom Overbye, University of Illinois for the content

ECE4334Generalized solution for networks


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4. Sequence voltages throughout the system can be calculated by

5. Phase values are determined from the sequence values


0

0

0

0

prefaultfIZ

= +

V V

M

M

This is solved

for each

sequencenetwork!

The entry

corresponds to the

faulted bus

ECE4334Unbalanced fault example for a network


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Pre-fault voltages are 1 pu at all buses. Pre-fault loads are ignored.

Transformer phase-shifts are ignored.

Calculate the bus voltages after a single-line-to-ground fault at bus

3.


ECE4334Unbalanced fault example in network


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First step: Sequence networks


ECE4334Ex.: Positive/negative sequence networks


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Form Ybus and subsequently calculate Zbus

j0.2 j0.05 j0.2j0.05j0.1

j0.1 j0.1

ECE4334Ex.: Positive/negative sequence networksF Y d b tl l l t Z


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j0.2 j0.05 j0.2j0.05j0.1

j0.1 j0.1

Thvenin equivalent

impedance at bus 1

of positive & neg.

sequence networks

Thvenin equivalent

impedance at bus 2 Thvenin equivalent

impedance at bus 3

ECE4334Ex.: Zero sequence network


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j0.05 j0.05 j0.05j0.05j0.3

j0.3 j0.3

ECE4334Ex.: Zero sequence network


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j0.05 j0.05 j0.05j0.05j0.3

j0.3 j0.3

Thvenin equivalent

impedance at bus 1

of zero sequence

network

Thvenin equivalent

impedance at bus 2Thvenin equivalent

impedance at bus 3

ECE4334EX: Sequence networks


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For a single-line-to-ground fault at bus 3 the Thvenin equivalent

impedances are used for bus 3 and since it is a bolted fault we do not have3Zf in the final circuit.


For SLG fault the

sequence networks

are connected in

series

ECE4334EX: Calculate the sequence currents & voltages


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Bus voltages at each bus in each sequence:

ECE4334EX: Calculate the sequence voltages


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ECE4334EX: Calculate the sequence voltages


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ECE4334EX: Calculate the phase voltages at bus 1


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ECE4334What about faults on lines?


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The previous analysis has assumed that thefault is at a bus. Most faults occur on

transmission lines, not at the buses.

These faults are treated by including a

dummy bus at the fault location. How the

impedance of the transmission line is thensplit depends upon the fault location.


ECE4334Power system protection

M i id i t f lt i kl


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Main idea is to remove faults as quickly as

possible while leaving as much of the system as

intact as possible

Fault sequence of events1. Fault occurs somewhere on the system, changing the

system currents and voltages

2. Current transformers (CTs) and potential transformers(PTs) sensors detect the change in currents/voltages

3. Relays use sensor input to determine whether a fault

has occurred

4. When a fault occurs the relays open the circuit

breakers to isolate fault


ECE4334Power system protection

P t ti t t b d i d ith b th


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Protection systems must be designed with both

primary protection and backup protection in

case primary protection devices fail.

In designing power system protection systemsthere are two main types of systems that need to be

considered:

1. Radial: there is a single source of power, so poweralways flows in a single direction; this is the easiest

from a protection point of view

2. Network: power can flow in either direction:protection is much more complicated


Documents

Fault Analysis ECE4334