Chapter 7: Symmetrical Faults

Chapter 7:
Symmetrical
Faults

Dr. Mohamad Arnaout

 Chapter 7: Symmetrical Faults

Course Content
⚫ Main purpose from this course is:
⚫ Power System Analysis and Faults Calculation
⚫ Transient Operations
⚫ Control in Power Systems

⚫ Content:
⚫ Ch. 7: Symmetrical Faults
⚫ Ch. 8: Symmetrical Components
⚫ Ch. 9: Unsymmetrical Faults
⚫ Ch. 11: Transient Stability
⚫ Ch. 12: Power Systems Control

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Chapter 7: Symmetrical Faults

Introduction
⚫ Fault Analysis
⚫ The cause of electric power system faults is insulation
breakdown
⚫ This breakdown can be due to a variety of different factors
⚫ Fault currents cause equipment damage due to both thermal
and mechanical processes
⚫ Goal of fault analysis is to determine the magnitudes of the
currents present during the fault

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 Chapter 7: Symmetrical Faults

Introduction
⚫ Fault Types
⚫ There are two main types of faults
– symmetric faults: system remains balanced; these
faults are relatively rare, but are the easiest to analyze
so we’ll consider them first.
– unsymmetrical faults: system is no longer balanced;
very common, but more difficult to analyze
⚫ The most common type of fault on a three phase system
by far is the single line-to-ground (SLG), followed by the
line-to-line faults (LL), double line-to-ground (DLG)
faults, and balanced three phase faults

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 Chapter 7: Symmetrical Faults

Introduction
⚫ Lightning Strike Event
⚫ Lighting hits line, setting up an ionized path to ground
⚫ Conduction path is maintained by ionized air after
lightning stroke energy has dissipated
⚫ Within one to two cycles (16 ms) relays at both ends of
line detect high currents
⚫ Circuit breakers open to de-energize line in an additional
one to two cycles
⚫ Circuit breakers may reclose after several seconds, trying
to restore faulted line to service

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Chapter 7: Symmetrical Faults

Series R–L Circuit Transients

⚫ To understand fault analysis we need to review the
behavior of an RL circuit

v(t ) = 2 V cos(t +  )

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Chapter 7: Symmetrical Faults

Series R–L Circuit Transients

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Chapter 7: Symmetrical Faults

Series R–L Circuit Transients

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Chapter 7: Symmetrical Faults

Series R–L Circuit Transients

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Chapter 7: Symmetrical Faults

Three Phase Short Circuits

⚫ During a fault the only devices that can contribute fault
current are those with energy storage
⚫ Thus the models of generators (and other rotating
machines) are very important since they contribute the
bulk of the fault current.
⚫ Generators can be approximated as a constant voltage
behind a time-varying reactance

'
Ea

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Chapter 7: Symmetrical Faults

Three Phase Short Circuits

The time varying reactance is typically approximated
using three different values, each valid for a different
time period:
X"d = direct-axis subtransient reactance
X 'd = direct-axis transient reactance
Xd = direct-axis synchronous reactance

T’’d T’d

X’’d < X’d < Xd 10

Chapter 7: Symmetrical Faults

Generator Modeling, cont’d

For a balanced three-phase fault on the generator
terminal the ac fault current is (see page 245)
 1  1 1  −
t
Td'

 + ' − e +
X
'  d  Xd Xd  
iac (t ) = 2Ea 
− t "
 sin( t +  )
 1 1  Td 
−
 X " X '  e 
 d d  
where
Td" = direct-axis subtransient time constant (  0.035sec)
Td' = direct-axis transient time constant (  1sec)
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Chapter 7: Symmetrical Faults

Generator Modeling, cont’d

The phasor current is then
 1  1 1  − t
Td'

 + ' − e +
X
'  d  Xd Xd  
I ac = Ea  
− t "
 1 1  Td 
 X " − X '  e 
 d d  
The maximum DC offset is
2 Ea' − t
TA
I DC (t ) = "
e
Xd
where TA is the armature time constant (  0.2 seconds)
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Chapter 7: Symmetrical Faults

Generator Short Circuit Currents

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Chapter 7: Symmetrical Faults

Generator Short Circuit Example

⚫ A 500 MVA, 20 kV, 3 is operated with an internal
voltage of 1.05 pu. Assume a solid 3 fault occurs on the
generator's terminal and that the circuit breaker operates
after three cycles. Determine the fault current. Assume

X d" = 0.15, X d' = 0.24, X d = 1.1 (all per unit)

Td" = 0.035 seconds, Td' = 2.0 seconds
TA = 0.2 seconds

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Chapter 7: Symmetrical Faults

Generator S.C. Example, cont’d

Substituting in the values
1  1 1  − t 2.0 
1.1 +  0.24 − 1.1  e +
I ac (t ) = 1.05  
 1 − 1  e − t 0.035 
 0.15 0.24  
I ac (0) = 1.05 = 7 p.u.
0.15
500  106
I base = 3
= 14,433 A I ac (0) = 101,000 A
3 20  10
t
I DC (0) = 101 kA  2 e 0.2 = 143 k A I RMS (0) = 175 kA
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Chapter 7: Symmetrical Faults

Generator S.C. Example, cont’d

Evaluating at t = 0.05 seconds for breaker opening
1  1 1  − 0.05 2.0 
+  −
1.1  0.24 1.1   e + 
I ac (0.05) = 1.05  
−
 1 − 1  e 0.05
0.035 
 0.15 0.24  
I ac (0.05) = 70.8 kA
−0.05
I DC (0.05) = 143  e 0.2 kA = 111 k A
I RMS (0.05) = 70.82 + 1112 = 132 kA

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 Chapter 7: Symmetrical Faults

Power System Three Phase Short

Circuits
⚫ Network Fault Analysis Simplifications
To simplify analysis of fault currents in networks we'll make several
simplifications:
1. Transmission lines are represented by their series reactance
2. Transformers are represented by their leakage reactances
3. Synchronous machines are modeled as a constant voltage
behind direct-axis subtransient reactance
4. Induction motors are ignored or treated as synchronous
machines
5. Other (non spinning) loads are ignored

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Chapter 7: Symmetrical Faults

Power System Three Phase Short

Circuits
Network Fault Example:
For the following network assume a fault on the terminal of
the generator; all data is per unit except for the transmission
line reactance. Generator has 1.05 terminal voltage &
supplies 100 MVA, with 0.95 lag PF

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Chapter 7: Symmetrical Faults

Power System Three Phase Short Circuits

Faulted network per unit diagram

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Chapter 7: Symmetrical Faults

Power System Three Phase Short Circuits

Fault Analysis Solution Techniques
⚫ Circuit models used during the fault allow the network to
be represented as a linear circuit
⚫ There are two main methods for solving for fault
currents:
1. Direct method: Use prefault conditions to solve for
the internal machine voltages; then apply fault and
solve directly
2. Superposition: Fault is represented by two opposing
voltage sources; solve system by superposition
– first voltage just represents the prefault operating
point
– second system only has a single voltage source
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Chapter 7: Symmetrical Faults

Power System Three Phase Short Circuits

Superposition Approach

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Chapter 7: Symmetrical Faults

Power System Three Phase Short Circuits

Superposition Approach

Fault is represented by two equal and opposite voltage

sources, each with a magnitude equal to the pre-fault voltage
Since this is now a linear network, the faulted voltages and
currents are just the sum of the pre-fault conditions [the (1)
component] and the conditions with just a single voltage source
at the fault location [the (2) component]. Pre-fault (1)
component is equal to the pre-fault power flow solution

Obvious the
pre-fault
“fault current”
is zero! 22
Chapter 7: Symmetrical Faults

Power System Three Phase Short Circuits

Superposition Approach
Fault (1) component due to a single voltage source
at the fault location, with a magnitude equal to the
negative of the pre-fault voltage at the fault location.

(1) (2) (1) (2)

Ig = I + Ig Im = Im + Im
g

I f = I (1)
f + I (2)
f = 0 + I (2)
f
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Chapter 7: Symmetrical Faults

Power System Three Phase Short Circuits

Example

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Chapter 7: Symmetrical Faults

Power System Three Phase Short Circuits

Example

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Chapter 7: Symmetrical Faults

Power System Three Phase Short Circuits

Example

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Chapter 7: Symmetrical Faults

Bus Impedance Matrix

The superposition approach can be easily extended
to larger systems. Using the Ybus we have
Ybus V = I
For the second (2) system there is only one voltage
source so I is all zeros except at the fault location
 
 0 
  However to use this
I = − I f 
  approach we need to
 0  first determine If
 

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Chapter 7: Symmetrical Faults

Determination of Fault Current

Define the bus impedance matrix Z bus as
−1
Z bus = Ybus V = Z busI

   V1
(2) 

   (2) 
 Z11 Z1n  0  V2 
 
Then   − I  =  
  f   
 Z n1 Z nn   0  V (2) 
n −1
   (2) 
Vn 
(1)
For a fault a bus i we get -If Zii = −V f = −Vi
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 Chapter 7: Symmetrical Faults

Determination of Fault Current, cont’d

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Chapter 7: Symmetrical Faults

Steps
• Use the line diagram of the original circuit showing the pu
admittances rather than the pu impedances
• Compute Ybus:
• Diagonal elements: Ykk= Sum of admittances connected to
bus k
• Off-diagonal elements Ykn= - (Sum of admittances
connected between buses k and n
• Deduce matrix Zbus which is the invert of Ybus matrix
• Apply previous equations

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Chapter 7: Symmetrical Faults

Bus Impedance Matrix

Example

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Chapter 7: Symmetrical Faults

Bus Impedance Matrix

Example

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Chapter 7: Symmetrical Faults

Bus Impedance Matrix

Example

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Chapter 7: Symmetrical Faults

Bus Impedance Matrix

Example

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Chapter 7: Symmetrical Faults

Bus Impedance Matrix

Example 2

For simplicity assume the system is unloaded

before the fault with
E g1 = Eg 2 = Eg 3 = 1.050
Hence all the prefault currents are zero.
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Chapter 7: Symmetrical Faults

Three Gen Example, cont’d

 −15 10 0
Ybus = j  10 −20 5 
 
 0 5 −9 
−1
 −15 10 0
Z bus = j  10 −20 5 
 
 0 5 −9 
 0.1088 0.0632 0.0351
= j  0.0632 0.0947 0.0526 
 
 0.0351 0.0526 0.1409 
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Chapter 7: Symmetrical Faults

Three Gen Example, cont’d

−1.05
For a fault at bus 1 we get I1 = = j 9.6 = − I f
− j 0.1088
 0.1088 0.0632 0.0351  j 9.6 
V (2) = j 0.0632 0.0947 0.0526   0 
  
 0.0351 0.0526 0.1409   0 
 −1.050 
=  −0.600 
 
 −0.3370 

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Chapter 7: Symmetrical Faults

Three Gen Example, cont’d

1.050  −1.050   00 
V = 1.050 +  −0.6060  = 0.4440 
     
1.050  −0.3370  0.7130 

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Chapter 7: Symmetrical Faults

Analysis of Unsymmetric Systems

⚫ Except for the balanced three-phase fault, faults result in
an unbalanced system.
⚫ The most common types of faults are single line-ground
(SLG) and line-line (LL). Other types are double line-
ground (DLG), open conductor, and balanced three phase.
⚫ System is only unbalanced at point of fault!
⚫ The easiest method to analyze unbalanced system
operation due to faults is through the use of symmetrical
components

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