Lecture 8
Short Circuit Analysis (Faults)
By Dr. Mostafa Elshahed
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Objectives
•
Learn a brief history of the electrical power systems construction
•
List and describe different components of electrical power systems
•
Learn the per-unit system calculations
•
Draw the single line and impedance diagram
•
Define the transformers and synchronous machines representations
•
Calculate the transmission lines parameters
•
Formulate the various transmission lines models
•
Establish the transmission lines performance
 Calculate the symmetrical faults variables
 Learn the principles of symmetrical components transformations
 Calculate the unsymmetrical faults variables
1
Preliminaries
A fault in a circuit is any failure that
interferes with the normal flow of current
to the load. In most faults, a current path
forms between two or more phases, or
between one or more phases and the
neutral (ground). Since the impedance of
a new path is usually low, an excessive
current may flow.
High-voltage transmission lines have
strings of insulators supporting each
phase. The insulators must be large
enough to prevent flashover – a condition
when the voltage difference between the
line and the ground is large enough to
ionize the air around insulators and thus
provide a current path between a phase
and a tower.
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Preliminaries
High currents due to a fault must be detected by protective circuitry and
the circuit breakers on the affected transmission line should automatically
open for a brief period (about 1/3 second).
Selecting an appropriate circuit breaker (type, size, etc.) is important…
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Preliminaries
 This breakdown can be due to a variety of different
factors
–
–
–
–
lightning
wires blowing together in the wind
animals or plants coming in contact with the wires
salt spray or pollution on insulators
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Fault Types
– Symmetrical 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
– On very high voltage lines faults are
practically always single line to ground
due to large conductor spacing
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RL Circuit Analysis
To understand fault analysis we need to
review the behavior of an RL circuit
v(t ) 
2 V cos( t   )
Before the switch is closed obviously i(t) = 0.
When the switch is closed at t=0 the current will
have two components: 1) a steady-state value
2) a transient value
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RL Circuit Analysis, cont’d
1. Steady-state current component (from standard
phasor analysis)
iac (t ) 
where Z 
I ac
2 V cos(t   )
Z
R 2  ( L)2  R 2  X 2
V

Z
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RL Circuit Analysis, cont’d
2. Exponentially decaying dc current component
i dc (t )  C1e
t
T
where T is the time constant, T  L R
The value of C1 is determined from the initial
conditions:
t
2V
i (0)  0  i ac (t )  idc (t ) 
cos(t     Z )  C1e T
Z
2V
C1  
cos(   Z ) which depends on 
Z
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RL Circuit Analysis, cont’d
Hence i(t) is a sinusoidal superimposed on a decaying
dc current. The magnitude of idc (0) depends on when
the switch is closed. For fault analysis we're just
2V
concerned with the worst case: C1 
Z
i (t )  i ac (t )  i dc (t )
i (t )


2V
2V t T
cos(t ) 
e
Z
Z
t
2V
(cos(t )  e T )
Z
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RMS for Fault Current
t
2V
The function i(t) 
(cos(t )  e T ) is not periodic,
Z
so we can't formally define an RMS value. However,
as an approximation define
I RMS (t ) 

2
2
iac
(t )  idc
(t )
2
I ac
2t
2  T
 2 I ac e
This function has a maximum value of 3 I ac
Therefore the dc component is included simply by
multiplying the ac fault currents by 3
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Generator Short Circuit Currents
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Generator Modeling During Faults
 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
Sub-transient
Transient
Steady
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Generator Modeling, cont’d
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
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Internal generated voltages of synchronous motors
When the motor is short-circuited, it does not receive power from the line but its
field circuit is still energized and still spinning (due to inertia in the machine and
in its load).
Therefore, the motor acts as a generator, supplying power to the fault. The
equivalent circuit of a synchronous motor is the same as the one of the synchronous
generator except that the direction of the current flow is reversed. Therefore, the
equations for the internal generated voltage, voltage behind the subtransient
reactance, and voltage behind transient reactance become
E A  V  jX s I A
E A"  V  jX " I A
E A'  V  jX ' I A
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Large System Modeling
• Given the short circuit MVA at the bus of the large system.
• The large system is simulated by a voltage source 1 pu in series with
a reactance XS
MVASC  3  V kV  I SC  10 3
MVABase  3  kVBase  I Base 10 3
10
MVASC , pu  I SC , pu
XS 
V
I SC , pu

1
I SC , pu
1

MVASC , pu
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Internal generated voltages of loaded machines
under transient conditions
Example 12-3: A 100 MVA, 13.8 kV, 0.9 PF lagging, Y-connected, 3 phase, 60 Hz
synchronous generator is operating at rated voltage and full load when a
symmetrical 3 phase fault occurs at its terminals. The reactances in per-unit to the
machine’s own base are
X s  1.00 X '  0.25 X "  0.12
a) If the generator operates at full load when the fault develops, what is the
subtransient fault current produced by this generator?
b) If the generator operates at no load and rated voltage when the fault develops,
what is the subtransient fault current produced by this generator? Observe that
this calculation is equivalent to ignoring the effects of prefault load on fault
currents.
c) How much difference does calculating the voltage behind subtransient
reactance make in the fault current calculations?
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Internal generated voltages of loaded machines
under transient conditions
The base current of the generator can be computed as
I L ,base
Sbase
100, 000, 000


 4,184 A
3VL ,base
3 13,800
a) Before the fault, the generator was working at rated conditions and the per-unit
current was:
I A  1.0  25.84 pu
The voltage behind subtransient reactance is
E A"  V  jX " I A  10   j 0.12  1  25.84  1.0585.86
Therefore, the per-unit fault current when terminals are shorted is
1.0585.86
IF 
 8.815  84.1 pu36,880 A
j 0.12
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Internal generated voltages of loaded machines
under transient conditions
b) Before the fault, the generator was assumed to be at no-load conditions and the
per-unit current was:
I A  0.00 pu
The voltage behind subtransient reactance is
E A"  V  jX " I A  10   j 0.12   00  1.00
Therefore, the per-unit fault current when the terminals are shorted is
1.00
IF 
 8.333  90 pu34,870 A
j 0.12
c) The difference in fault current between these two cases is
Difference
36,880  34,870
100%  5.76%
34,870
The difference in the fault current when the voltage behind subtransient reactance is
considered and when it is ignored is small and usually systems are assumed unloaded.
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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 (nonspinning) loads are ignored
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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
Convert to per unit: X line
19.5

 0.1 per unit
2
138
100
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Network Fault Example, cont'd
Faulted network per unit diagram
To determine the fault current we need to first estimate
the internal voltages for the generator and motor
For the generator VT  1.05, SG  1.018.2
*
I Gen
1.018.2 


  0.952  18.2
 1.05 
'
Ea
 1.1037.1
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Network Fault Example, cont'd
The motor's terminal voltage is then
1.050 - (0.9044 - j 0.2973)  j 0.3  1.00  15.8
The motor's internal voltage is
1.00  15.8  (0.9044 - j 0.2973)  j 0.2
 1.008  26.6
We can then solve as a linear circuit:
1.1037.1 1.008  26.6
If 

j 0.15
j 0.5
 7.353  82.9  2.016  116.6  j 9.09
1.050 1.050
If 

  j 9.1
j 0.15
j 0.5
At no-load conditions
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Fault current transients
Two generators are connected in
parallel to the low-voltage side of a
transformer. Generators G1 and G2 are
each rated at 50 MVA, 13.8 kV, with a
subtransient resistance of 0.2 pu.
Transformer T1 is rated at 100 MVA,
13.8/115 kV with a series reactance of
0.08 pu and negligible resistance.
Assume that initially the voltage on the high side of the transformer is 120 kV, that
the transformer is unloaded, and that there are no circulating currents between the
generators.
Calculate the subtransient fault current that will flow if a 3 phase fault occurs at the
high-voltage side of transformer.
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Fault current transients
Let choose the per-unit base values for this power system to be 100
MVA and 115 kV at the high-voltage side and 13.8 kV at the lowvoltage side of the transformer.
The subtransient reactance of the two generators to the system base is
2
Therefore:
 Vgiven   S new 
Z new  Z given 

 
 Vnew   S given 
2
 13,800   100, 000 
X 1"  X 2"  0.2  
 0.4 pu



 13,800   50, 000 
The reactance of the transformer is already given on the system base,
it will not change
X T  0.08 pu
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Fault current transients
The per-unit voltage on the high-voltage side of the transformer is
actualvalue 120, 000
V pu 

 1.044 pu
basevalue 115, 000
Since there is no load on the system,
the voltage at the terminals of each
generator, and the internal generated
voltage of each generator must also
be 1.044 pu. The per-phase per-unit
equivalent circuit of the system is
We observe that the phases of
internal generated voltages are
arbitrarily chosen as 00. The phase
angles of both voltages must be the
same since the generators were
working in parallel.
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Fault current transients
To find the subtransient fault current, we need to solve for the voltage at the bus 1
of the system. To find this voltage, we must convert first the per-unit impedances to
admittances, and the voltage sources to equivalent current sources. The Thevenin
impedance of each generator is ZTh = j0.4, so the short-circuit current of each
generator is
Voc 1.0440
I sc 

 2.61  90
Z th
j 04
The equivalent circuit
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Fault current transients
Then the node equation for voltage V1
V1   j 2.5   V1   j 2.5   V1   j12.5   2.61  90  2.61  90
V1 
5.2290
 0.2980
 j17.5
Therefore, the subtransient current in the fault is
I F  V1   j12.5   3.729  90 pu
Since the base current at the high-voltage side of the transformer is
I base 
S3 ,base
3VLL ,base
100, 000, 000

 502 A
3 115, 000
The subtransient fault current will be:
I F  I F , pu I base  3.729  502  1,872 A
Ig 
IF
100,000,000
 1.8645  90 pu 
 7800.5 A
2
3 13,800
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Three Generator Example
I2
I1
1.05
I1 
  j 5.25 pu
j 0. 2
1.05
I2 
  j 4.4 pu

j 0.1   j 0.2  j 0.45
 j 0.2  j 0.45

1.05
1.05
V
If 

  j 9.65 pu 
j 0.2 j 0.1   j 0.2  j 0.45
Z Thevenin

 j 0.2  j 0.45

V1  0
V2  I 2  j 0.1  0.44
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Element
Generator 1
Generator 2
All Transmission Lines
Z
j 0.1
j 0.05
j 0.1
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CB Selection
From the current view point two factors to be considered in selecting
circuit breakers are:
1. The maximum instantaneous current which the breaker must carry
(with stand)
2. The total current when the breaker contacts part to interrupt the
circuit
Up to this point we have devoted most of our attention to the subtransient current
called the initial symmetrical current, which does not include the DC component.
Inclusion of the DC component results in RMS value of current immediately after
the fault, which is higher than the subtransient current.
For oil circuit breakers above 5 kV the subtransient current multiplied by 1. 6 is
considered to be the RMS value of the current whose disruptive forces the breaker
must with stand during the first half cycle after the fault occurs.
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CB Selection
From the current view point two factors to be considered in selecting
circuit breakers are:
1. The maximum instantaneous current which the breaker must carry
(with stand)
2. The total current when the breaker contacts part to interrupt the
circuit
The interrupting rating of a circuit breaker was specified in kVA or MVA.
The interrupting kVA equal √3 x kV x I
I = current which the breaker must be capable of interrupting when its contacts
part)
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CB Selection
This interrupting current is, of course , lower than
the momentary current and depends on the
speed of the breaker, such as 8, 5, 3, or 2 cycles,
which is a measure of the time from the
occurrence of the fault to the extinction of the
arc. Breakers of different speeds are classified by
their rated interrupting times.
The rated interrupting time of a circuit breaker is
the period between the instant of energizing the
trip circuit and the arc extinction on an opening
operation. Preceding this period is the tripping
delay time, which is usually assumed to be 0.5
cycle for relays to pick up.
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