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EEE3121: ADVANCED POWER SYSTEM ANALYSIS
Power system faults By
K Kaberere
2
Causes of power system faults
Some causes of system faults are:
Equipment insulation fails
System over-voltages caused by lightning and switching surges
Insulation contamination
Falling trees and tree branches shorting overhead lines
Vehicles and aircraft hitting towers and poles
Vandalism
Birds shorting overhead lines
Winds and gales
Wild fires in Australia, 2009
3 4
5 6
2
7
Relay detects the fault and circuit breaker isolates or
disconnects the faulty section (Fuse detects and isolates)
For proper relay setting and CB interrupting capacity – the
values of expected fault currents and voltages should be
known – short circuit calculations
When a fault occurs in the power system, large currents
and/or abnormal voltages are developed.
Short circuit currents may cause thermal damage to
equipment
Windings and bus-bars may suffer mechanical damage due
to high magnetic forces during faults
Hence, faulty section should be disconnected immediately
so that normal operation of the rest of the system is not
affected.
8
Short circuit calculations are broadly classified as:
Three-phase / symmetrical / balanced faults – involve all
the 3 phases and sometimes even the ground – most severe
but not common
Unsymmetrical / unbalanced faults – involve one or two
phases – voltages and currents become unbalanced –
symmetrical components used for calculating fault
currents
Short circuit current is determined by:
The internal emfs of the machines in the network
The machines internal impedances
The impedance of the network between the machine and
the fault location
Unsymmetrical Faults
Short circuits
9 10
Introduction
Most power system faults are unsymmetrical.
They may be unsymmetrical short circuits or
open conductors.
They may be classified as:
1. Single line-to-ground (SLG or L-G)
2. Line-to-line (L-L or DL)
3. Double line-to-ground (DLG or L-L-G).
Unsymmetrical faults result in the flow of
unbalanced currents in the system
Symmetrical components are used to determine
post fault system currents and voltages
11
Use Thevenin's theorem to reduce the system
to a circuit with a single source and
impedances
Network impedances are reduced to a single
sequence impedance as seen at the fault
location
System is assumed to operate under steady
state conditions before the fault occurs
Prefault load current is neglected
12
1. Single line-to-ground fault on an unloaded generator (S-L-G)
The fault
occurs on
phase 'a'.
The fault
conditions
are:
1. Ib = Ic = 0
2. Va = 0
3
13
The symmetrical components of the currents are
02 2
12 2
2
1 1 1 1 1 11 1
1 1 03 3 01 1
a a a
a b
a c
I I II a a I a aI Ia a a a
0 1 2 3a
a a a
II I I …………………………….(1)
From the generator sequence network drawn earlier
0 0 0
1 1 1
22 2
0 000 00 00
a a
a a a
a a
V Z IV E Z I
ZV I
Substitute Ia0 = Ia1 = Ia2 in the [Is] vector
14
0 0 1
1 1 1
22 1
0 000 00 00
a a
a a a
a a
V Z IV E Z I
ZV I
……………….(2)
0 1 2But a a a aV V V V
From (2), and knowing Va = 0
1 0 1 1 1 2 0a a a a aV I Z E I Z I Z
10 1 2
aa
EI
Z Z Z
…………………………….(3)
15 15
From (1) and (3), it can be
seen that the sequence
networks are connected in
series after the occurrence of
the fault.
Current flowing is Ia1 and
the voltage across the series
circuit is Ea.
10 1 2
The fault current
33 a
f a a
EI I I
Z Z Z
Ia0 = Ia1
Ia1
Ia2 = Ia1
Va1
Va2
Va0
Ea
16
If the neutral of the generator is ungrounded, the
zero sequence network is open-circuited.
Then, Ia1 and hence Ia = 0.
After determining Ia1
Va1 = Ea – Ia1Z1
Va2 = – Ia1Z2
Va0 = – Ia1Z0
20 1 2
20 1 2
b a a a
c a a a
V V a V aV
V V aV a V
Phase voltage of the healthy phases
17
2. Line-to-line fault on an unloaded generator
(L-L or DL)
Let the fault be
on phase ‘b‘
and ‘c’.
The fault
conditions are:
1. Ib = -Ic
2. Ia = 0
3. Vb = Vc
18
The symmetrical components of the voltage are
02
12
2
1 1 11
13
1
a a
a b
a b
V VV a a VV Va a
1 2 a aV V ………………….…………….(1)
0
21
22
12
31
31
3
a a b
a a b
a a b
V V V
V V V a a
V V V a a
4
19
The symmetrical components of current are
02
12
2
1 1 1 011
31
a
a c
a c
II a a II Ia a
0
21
22
0
1
31
3
a
a c
a c
I
I I a a
I I a a
0 0 0
1 1 1
22 2
0 00But 0 0
0 00
a a
a a a
a a
V Z IV E Z I
ZV I
Ia1 = -Ia2 …………………………………….…….(2)
20
1 1 1 1 2a a a aV E I Z I Z
From (1) and (2), it can be concluded that the
positive and negative sequence networks must be
connected in parallel without the zero sequence
network since Z0 does not appear in (3).
11 2
aa
EI
Z Z
…...………………………….…….(3)
0 0
1 1 1
21 1
0 00 00 00 00
a
a a a
a a
V ZV E Z I
ZV I
21
The fault current
f b cI I I
The grounding of the
generator neutral does
not affect the fault
current.
2 20 1 2 1b a a a aI I a I aI I a a
Ia1 Ia2 = -Ia1
2
1 2
af
EI a a
Z Z
…...…………………….…….(4)
1 2 0 and 0a a aV V V 22
0 1 2 12a a a a aV V V V V Healthy phase
20 1 2
21
b c a a a
a
V V V a V aV
a a V
faulted phases
23
3. Double line-to-ground fault on an unloaded generator
(L-L-G or DLG)
Ib
Ic
Let the fault be
on phase ‘b‘
and ‘c’.
The fault
conditions are:
1. Ia = 0
2. Vb = Vc = 0
If = Ib + Ic
24
02
12
2
1 1 11 1
1 03 301
a a a
a a
a a
V V VV a a VV Va a
0 1 2 3a
a a a
VV V V ………….………………….(1)
0 1 2
0 1 2
0a a a a
a a a
I I I I
I I I
………….………………….(2)
20 1 2
20 1 2
0
0
b a a a
c a a a
V V a V aV
V V aV a V
………….………….(3)
But Va0 = – Ia0Z0
Substituting (2) for Ia0 gives
Va0 = (Ia1 + Ia2)Z0
Va1 = Ea – Ia1Z1
Va2 = – Ia2Z2
………….……………………….(4)
5
25
After making the necessary substitutions and simplifying,
we obtain
12 0
12 0
aa
EI
Z ZZ
Z Z
02 1
2 0a a
ZI I
Z Z
20 1
2 0a a
ZI I
Z Z
It can be seen from (1) that the sequence voltages are equal.
Therefore, their networks must be connected in parallel.
26
Ia1 Ia2 Ia0
Va1 = Ea – Ia1Z1
Va2 = – Ia2Z2
Va0 = – Ia0Z0
0 1 2a a a aV V V V Healthy phase
20 1 2
20 1 2
b a a a
c a a a
I I a I aI
I I aI a I
The fault current
f b cI I I
27
Unsymmetrical faults in power systems
Prefault voltage is designated Vf and only appears in the positive sequence network
The sequence networks for the system are first drawn
Then Thevenin equivalent for each network as seen from the fault location is derived
The value of equivalent impedance is applied in the equations derived earlier for each type of fault.
28
Example 1
Consider the power system whose one-line diagram is shown
below.
Bus 5 is at the mid-point of transmission line TL2.
a) Draw the sequence networks and clearly mark the buses.
b) Reduce the networks in (a) to their Thevenin equivalents as
seen at Bus 5.
System data are as given below.
29 29
Item MVA rating
Voltage rating (kV)
X1
(pu)
X2
(pu) X0
(pu)
G1 100 25 0.2 0.2 0.05
G2 100 13.8 0.2 0.2 0.05
T1 100 25/230 0.05 0.05 0.05
T2 100 13.8/230 0.05 0.05 0.05
TL1 100 230 0.1 0.1 0.3
TL2 100 230 0.2 0.2 0.6
30
Positive sequence network
30
ref
j0.175
6
31
Negative sequence network
31
The negative sequence impedances of the power system
components are equal to the positive sequence impedances.
Hence, the Thevenin equivalents for the two sequence
networks are similar with the exception of the emf source, as
shown below.
j0.175
32
Zero sequence network
32
33
Example 2
For the power system described in Example 1:
a) Determine the fault current and the phase voltages for the
following faults at bus 5; 3 phase, SLG, DL, and DLG
b) For each case, find the current In flowing through the
neutral.
j0.175
Positive sequence
j0.175
Negative sequence Zero sequence
Solution
34
Case 1: Balanced three phase to ground fault at bus 5
0 2 0a aI I 1
1 05.71
0.175aI jj
2
2
1 1 1 0 5.71 901 5.71 90 5.71 150 pu
0 5.71 301
a
b
c
II a aI a a
000
a
b
c
V
V
V
0
1
2
0 0.199 0 0 0 0 1 0 0.175 0 5.71 0
0 0 0 0.175 0 0
a
a
a
V jV j j
jV
0nI
35
Case 2: Single-line-to-ground fault at bus 5
0 2 1
1 01.821
0.175 0.175 0.199a a aI I I jj j j
2
2
1 1 1 1.821 5.461 1.821 0 pu
1.821 01
a
b
c
I j jI a a j
jI a a
0
1
2
0 0.199 0 0 1.82 0.362 1 0 0.175 0 1.82 0.681
0 0 0 0.175 1.82 0.319
a
a
a
V j jV j j
j jV
2
2
1 1 1 0.362 01 0.681 1.022 238 pu
0.319 1.022 1221
a
b
c
VV a aV a a
5.46 punI j
36
Case 3: Line-to-line fault at bus 5
1 2
1 02.857
0.175 0.175a aI I jj j
0 0aI
2
2
1 1 1 0 01 2.857 4.95 pu
2.857 4.951
a
b
c
II a a j
jI a a
1 2
0
2.857 0.175 0.5
0a a
a
V V j j
V
2
2
1 1 1 0 1.01 0.5 0.5 pu
0.5 0.51
a
b
c
VV a aV a a
0nI
7
37
Case 4: Double-line-to-ground fault at bus 5
1
1 03.73
0.175 0.1990.175
0.175 0.199
aI jj j
jj j
2
0.1993.73 1.99
0.175 0.199a
jI j j
j j
0
0.1753.73 1.75
0.175 0.199a
jI j j
j j
2
2
1 1 1 1.75 01 3.73 5.60 152.1 pu
1.99 5.60 27.91
a
b
c
I jI a a j
jI a a
5.25 punI j 38
0 1 2 1.99 0.175 0.348a a aV V V j j
2
2
1 1 1 0.348 1.0441 0.348 0 pu
0.348 01
a
b
c
VV a aV a a
Unsymmetrical Faults
Open circuits
39 40
1. One phase open-circuit fault
Let the open-circuit fault be on phase ‘a’ as shown in
Fig. 1. The fault conditions are:
1.Ia = 0
2.Vff’b = Vff’
c = 0
Bus i
a b c
Bus j
a b c
f f’
f f’
f f’
Fig. 1: Unbalanced one phase open circuit fault
41
The symmetrical components of the voltage are
02
12
2
1 1 11
1 03 01
ffa a
a
a
V VV a aV a a
0 1 2
1
1
3
And
ffa a a a
ffa L
V V V V
V I Z
…………….(6)
0 1 2
0 1 2
0a a a a
a a a
I I I I
I I I
………….………………….(7)
Where IL is the pre-fault load current flowing in the
balanced system
42
20 1 2
20 1 2
0
0
ffb a a a
ffc a a a
V V a V aV
V V aV a V
………….………….(8)
But Va0 = – Ia0Z0
Substituting (7) for Ia0 gives
Va0 = (Ia1 + Ia2)Z0
Va1 = ILZ1 – Ia1Z1
Va2 = – Ia2Z2
………….……………………….(9)
8
43
After making the necessary substitutions and simplifying,
we obtain
It can be seen from (7) and (10) that the sequence networks
must be connected in parallel as shown in Fig. 2.
11
2 01
2 0
La
I ZI
Z ZZ
Z Z
02 1
2 0a a
ZI I
Z Z
20 1
2 0a a
ZI I
Z Z
………….………………..(10)
44
2 00 1 2 1
1 2 2 0 1 0a a a L
Z ZV V V I Z
Z Z Z Z Z Z
The phase voltage across the open circuit is given by
2 01
1 2 2 0 1 0
3ff
a L
Z ZV I Z
Z Z Z Z Z Z
ILZ1
Z1 f’
f
Va1
f’
Va2
Z2
f
Va0
f
f’ Z0
Fig. 2: Thevenin equivalent for one phase open circuit fault
………….………(11)
45
Current flowing in the healthy phases is given by
2. Two phases open-circuit fault
Bus i
a b c
Bus j
a b c
f f’
f f’
f f’
Fig. 3: Unbalanced two phase open circuit fault
Let the open-circuit fault
be on phases ‘b’ and ‘c’
as shown in Fig. 3.
The fault conditions are:
1. Ib = Ic = 0
2. Vff’a = 0
20 1 2
20 1 2
b a a a
c a a a
I I a I aI
I I aI a I
……….………….……(12)
46
The symmetrical components of the currents are
02 2
12 2
2
1 1 1 1 1 11 1
1 1 03 3 01 1
a a a
a b
a c
I I II a a I a aI Ia a a a
0 1 2 3a
a a a
II I I …………………………….(13)
0 1 2But 0ff
a a a aV V V V ………..….(15)
0 0 1
1 1 1 1
22 1
0 000 0
0 0 0
a a
a L a
a a
V Z IV I Z Z I
ZV I
………….(14)
1 0 1 1 1 1 2 0a
ffa L a aV I Z I Z I Z I Z
47
11
0 1 2
La
I ZI
Z Z Z
…….(16)
From (13) and (16), it can
be seen that the sequence
networks are connected in
series after the occurrence
of the fault, as in Fig. 4
Current flowing is Ia1 and
the voltage across the
series circuit is IL Z1.
ILZ1
Z1 f’
f
Va1
f’
Va2
Z2
f
Va0
f
f’ Z0
Fig. 4: Thevenin equivalent for
two phase open circuit fault 48
11
0 1 2
33 L
a a
I ZI I
Z Z Z
Current flowing in the healthy phase is given by
After determining Ia1
Va1 = ILZ1 – Ia1Z1
Va2 = – Ia1Z2
Va0 = – Ia1Z0
Phase voltage of the open phases is given by
………………..….(17)
20 1 2
20 1 2
ffb a a a
ffc a a a
V V a V aV
V V aV a V
……….…………..….(18)
9
Simultaneous faults
49 50
1.Single-line-to-ground and line-to-line faults
Let the SLG fault be
on phase ‘a’ and DL
fault on phases ‘b’ and
‘c’ as shown in Fig. 5.
The fault conditions
are:
1. Vb = Vc
2. Ib = -Ic
3. Va = 0
Balanced 3-phase
power system
a
b
c
Fig. 5: Unbalanced 3-phase
short-circuit fault
51
The symmetrical components of current are
02
12
2
1 1 11
13
1
a a
a c
a c
I II a a II Ia a
0
21
22
1 2 0
1
31
31
32
23
a a
a a c c
a a c c
a a a a
I I
I I aI a I
I I a I aI
I I I I
…………………………………….…….(19)
……………………….(20)
52
0 03 2 2a b c a b cV V V I Z V V
0 0 0
1 1 1
22 2
0 00But 0 0
0 00
a a
a f a
a a
V Z IV V Z I
ZV I
…...………………….…….(22) 0 03
2a
b c
I ZV V
Knowing Va = 0 and Vb = Vc, then
0 0 02 2
1 1 12 2
2 2 2
1 1 1 1 1 1
1 1
1 1
a a a
b a f a
c a a
V V I ZV a a V a a V I ZV V I Za a a a
… (21a)
… (21b)
… (21c)
53
Multiplying (21c) by a and subtracting the result
from (21b) i.e.
1b c bV aV a V
2 2 0 0
2 2 0 0
1 a a
b a a
a I Z I Z
V I Z I Z
…...………………….(23)
Equating (22) and (23),
0 02 2 0 0
20 2
0
3
22
ab a a
a a
I ZV I Z I Z
ZI I
Z
…………….………………….(24)
54
Substituting (24) in (20) for Ia0 gives
1 22 0 1
2 0 2
0
2 and
41 4
aa a a
I ZI I I
Z Z Z
Z
………….(25)
Substituting (25) in (21a) for the sequence currents gives
0 21
1 2 2 0 1 0
02
1 2 2 0 1 0
20
1 2 2 0 1 0
4
4
4
2
4
a f
a f
a f
Z ZI V
Z Z Z Z Z Z
ZI V
Z Z Z Z Z Z
ZI V
Z Z Z Z Z Z
10
55
02
12
2
1 1 1
1
1
a a
b a
c a
I II a a II Ia a
The fault currents are given by
2
1 2 2 0 1 0
2 0
1 2 2 0 1 0
6
4
3 2
4
a f
b c f
ZI V
Z Z Z Z Z Z
j Z ZI I V
Z Z Z Z Z Z
………………. (26a)
………………. (26b)
56
2 01 2
1 2 2 0 1 0
2 00
1 2 2 0 1 0
4
2
4
a a f
a f
Z ZV V V
Z Z Z Z Z Z
Z ZV V
Z Z Z Z Z Z
0 0 0
1 1 1
22 2
0 000 00 00
a a
a f a
a a
V Z IV V Z I
ZV I
2 0
1 2 2 0 1 0
3
4b c f
Z ZV V V
Z Z Z Z Z Z
…………. (27)
57
2.Simultaneous faults at different locations
Consider a SLG fault on phase ‘a’ at location J and
another SLG fault be on either phase ‘a’ or ‘b’ or ‘c’ at
location K.
At location J
0 1 2 0 1 20 and J J J J J J J
a a a a a a aV V V V I I I ……. (28)
At location K and fault on phase ‘a’
2 20 1 2 0 1 20 and
K K K K K K Kb a a a a a aV V a V aV I a I aI …. (30)
At location K and fault on phase ‘b’
0 1 2 0 1 20 and K K K K K K K
a a a a a a aV V V V I I I ……. (29)
58
2 20 1 2 0 1 20 and
K K K K K K Kc a a a a a aV V aV a V I aI a I …. (31)
At location K and fault on phase ‘c’
Fig. 6: Simultaneous SLG
faults at different locations
The analysis is done
using Kirchoff’s voltage
and current laws.
Symmetrical Faults
Short circuits
59 60
Transients in RL series circuits
max sin( )e V t
max sin( ) for t 0di
V t L Ridt
/max
2 2 1
sin( ) sin .............(1)
where ( ) and tan /
Rt LVi t
Z
Z R L L R
11
61
61
The 1st term of (1) varies sinusoidally with time (ac
component) whereas the 2nd term is non-periodic and
decays exponentially with a time constant of L/R (dc
component)
Current i for - = 0 Current i for - = -/2
62
3- short circuit
The current that flows when a generator which is on no
load is shorted at the terminals is similar to that which
flows when the switch in the RL circuit is closed.
However, due to armature reaction, the current flowing
immediately after a fault, a few cycles later, and the steady
state value differ considerable
Since the short circuit occurs at different points of the
voltage wave for the 3 phases (120° separation,
different for each phase), the dc current is different in each
phase. Furthermore, if CB opens in reasonable time,dc
component will have decayed. Therefore neglect
Resulting plot of phase current Vs. time is as shown below.
1. Unloaded synchronous machine
63
The ac fault current in a synchronous machine can be
modeled by an RL circuit with time varying inductance L(t).
This is represented by
Xd" = direct-axis sub-transient reactance
Xd' = direct-axis transient reactance
Xd = direct-axis synchronous reactance
Xd" < Xd' < Xd 64
Initial fault current after fault occurrence is high due to
reduced armature reaction but reduces to a lower value as
armature reaction increases. If the pre-fault rms phase
voltage at the generator terminal (on no load) is |Eg|,
| || | rms steady state current
| || | rms transient current
| || | rms subtransient current
g
d
g
d
g
d
EI
X
EI
X
EI
X
The q-axis reactances do not significantly affect fault
current since armature resistance is small.
65
EXAMPLE 1
65
Two generators are connected in parallel to the L.V. side
of a - transformer. G1 is rated 50 MVA, 13.8 kV and
G2 25 MVA, 13.8 kV. Each generator has a subtransient
reactance of 25%. The transformer is rated 75 MVA, 13.8
/ 69 kV with a reactance of 10%. Before the fault
occurs, the H.V. side of the transformer is at 66 kV and
the transformer is unloaded with no circulating current
between the generators. Find the subtransient fault
current in each generator when a short circuit occurs on
the H.V. side of the transformer.
66
Solution
66
75 MVA
13.8 /69 kV
25 MVA
13.8 kV
50 MVA
13.8 kV
Let kVbase = 69 kV and
MVAbase = 50 MVA
2
( ) ( )
( ) ( )
base new base old
new oldbase old base new
MVA kVX X
MVA kV
1
2
0.25 pu
0.50 pu
0.067 puTx
X
X
X
Pre-fault voltage
= 66/69 = 0.957 pu
12
67
67
1 2//
0.233
eq TxX X X X
pu
4.101f
feq
VI pu
X
voltage across gen. 0.684 pu
1
2
I 2.734
I 1.367
pu
pu
3
base
1
2
10I
3
2091.85
I 5.72
I 2.86
base
base
MVA
kV
A
kA
kA
68
68
2. Loaded synchronous machine
Consider a generator that is loaded before a fault occurs at
point P on the system as shown in the circuit below.
The pre-fault current and voltage at point P are IL and Vf
respectively, and the generator terminal voltage is Vt.
Before the fault occurs and
immediately after, the internal
voltage of the generator is E"g.
E"g = Vt + j IL X"g
The generator feeds I"g to the
fault where
I"g = I"g1 + IL
69
69
If the load is a motor, after the occurrence of the
fault, the motor behaves like a generator and
supplies a current I“m = I“m1 - IL
E“m = Vt - j IL X“m
The total fault current I“f = I“g + I“m = I“g1 + I“m1
70
EXAMPLE 2 The synchronous motor in the power system shown below is
drawing 80 MW at 0.8 p.f. leading and a terminal voltage of
13.4 kV, when a symmetrical 3 phase fault occurs at Bus 2.
Find the sub-transient current in the generator, motor, and
fault.
100 MVA
13.8 kV
X“m = 0.2
100 MVA
13.8 /138 kV
XT1 = 0.10
100 MVA
13.8 kV
X"g = 0.15
G M
T1 T2
100 MVA
138 / 13.8 kV
XT2 = 0.10
1 2 XTL= j20
71
Solution
Let kVbase at bus 1 = 13.8 kV and MVAbase = 100 MVA
2
base(line)Z
190.44
base
base
kV
MVA
XTL = j0.105 pu
ZTh = j0.2//j0.455 pu
= j0.139 pu
Vf = 0.9710 pu 72
72
6.986f
fTh
VI j pu
Z
3
base
10I 4183.7
3
base
base
MVAA
kV
29.2fI j kA
1
1
8.92
20.28
g
m
I j kA
I j kA
13
73
73
4.309 36.873 cos
3.447 2.585
L
PI kA
V
j kA
1
1
7.212 61.45
23.14 98.57
g g L
m m L
I I I kA
I I I kA
NB: If the pre-fault voltage is equal to the base voltage,
then the fault current may be calculated as follows.
Fault level at fault location
3
3
3 10
10
3
basef base f
pu
f basef
pubase
MVAMVA kV I
Z
MVA II
ZkV
74
Limiting short circuit current
For a fault that occurs at the generator bus bar, the impedance presented to the fault current is low and hence the fault current is very high.
This may cause thermal and mechanical damage to equipment.
To limit fault current, high impedance reactors are connected between the source and the fault location.
Possible locations of the reactors include:
75
2. Bus-bar reactor – between sections of busbars
75
Not common in modern design –generators
have inherently high reactance windings
and they are protected by very fast relays
1. Generator reactor – between generator and busbar
Prevents total voltage collapse at the
station in case of a fault on one bus bar.
Limits circulating current in case of
imbalance e.g. fault on one section of the
bus bar. 3. Feeder
reactor Prevents voltage collapse in case of a fault
near the generating station.
APPENDIX
76
77
Fault analysis and choice of reference frame
Va1
Vb1
Vc1
Positive sequence Zero sequence Negative sequence
Va2
Vb2
Vc2
For a SLG fault on phase ‘a’ with phase ‘a’ chosen as
reference
0 1 2 0 1 20 and a a a a a a aV V V V I I I …..…. (1)
78
Fault analysis and choice of reference frame
For a SLG fault on phase ‘b’ with phase ‘b’
chosen as reference
0 1 2 0 1 20 and b b b b b b bV V V V I I I …..…. (2)
Expressing the sequence voltages and currents in
terms of phase ‘a’ gives 2
0 1 22
0 1 2
0
and
b a a a
a a a
V V a V aV
I a I aI
…..…. (3)
For a SLG fault on phase ‘c’ with phase ‘c’
chosen as reference
0 1 2 0 1 20 and c c c c c c cV V V V I I I …..…. (4)
14
79
20 1 2
20 1 2
0
and
c a a a
a a a
V V aV a V
I aI a I
………………..........…. (5)
Faulted
phase
Ref phase k1
k2
k0
k1
k2
k0
a-g a 1 1 1 1 1 1
b-g b a2 a 1 1 a2 a
c-g c a a2
1 1 a a2
For SLG fault on phases ‘b’ and ‘c’ the sequence
voltages and currents have to be multiplied by the
following k-factors to be expressed in terms of phase ‘a’
components – phase shifting.
80
Faulted
phases
Ref phase k1
k2
k0
k1
k2
k0
b-c-g a 1 1 1 1 1 1
a-c-g b a2 a 1 1 a2 a
a-b-g c a a2
1 1 a a2
Similar transformations can be derived for DLG faults to give
NB: For the same reference phase, the multipliers are the same
irrespective of the type of fault.
