2. INTRODUCTION OF UNSYMMETRICAL
FAULTS
Most Of the faults that occur on power systems are unsymmetrical faults, which
may consist of unsymmetrical short circuits, unsymmetrical faults through
impedances, or open conductors.
Unsymmetrical Faults occur as single line--‐to--‐ground faults, line--‐to--‐line
faults, or double line--‐to--‐ground faults. The Path of the fault current from line
to line or line to ground may or may not contain impedance. One Or two open
conductors result in unsymmetrical faults, through either the breaking of one or
two conductors or the achon of fuses and other devices that may not open the
three phases simultaneously
3. Connection of sequence network for
LG fault 𝑖 𝑡ℎ bus
Consider that there are n number of buses in a
System and a LG fault occurs 𝑖 𝑡ℎ bus of this system
The positive sequence network is replaced by its
Thevenin’s equivalent i.e. the prefault voltage. 𝑉1−𝑖
0
Of bus I in series with the passive positive sequence
Network.
As there is no prefault negative and zero sequence
Voltage ,both are passive network only
5. According to sequence network connection , current −𝐼1−𝑖
𝑓
is injected only at the
faulted 𝑖 𝑡ℎ bus of the positive sequence network. Hence,
I1−𝑏𝑢𝑠 =
0
0..
.
−𝐼1−𝑖
𝑓
.
.
0
Hence the positive sequence voltage at the 𝑖 𝑡ℎ bus of the passive positive
sequence network is
𝑉1−𝑖 = −Z1−𝑖𝑖 . I1−𝑖
𝑓
Thus, the passive positive sequence network present an impedance Z1−𝑖𝑖 to the
positive sequence current I1−𝑖
𝑓
6. For a negative sequence network.
𝑉2−𝑏𝑢𝑠 = Z2−𝑏𝑢𝑠 . I2−𝑏𝑢𝑠
The negative sequence network is injected with current I1−𝑖
𝑓
at the 𝑖 𝑡ℎ bus
only.hence,
I2−𝑏𝑢𝑠 =
0
0..
.
−𝐼2−𝑖
𝑓
.
.
0
∴𝑉2−𝑖 = −Z2−𝑖𝑖 . I2−𝑖
𝑓
Thus negative sequence network offers an impedance Z2−𝑖𝑖 to the negative
sequence current 𝐼2−𝑖
𝑓
7. For zero sequence network
𝑉0−𝑏𝑢𝑠 = Z0−𝑏𝑢𝑠 . I0−𝑏𝑢𝑠
I0−𝑏𝑢𝑠 =
0
0..
.
−𝐼0−𝑖
𝑓
.
.
0
𝑉0−𝑖 = −Z0−𝑖𝑖 . I0−𝑖
𝑓
Thus zero sequence network offers an impedance Z0−𝑖𝑖 to the zero sequence
current 𝐼0−𝑖
𝑓
8. From sequence network connection, we can write,
𝐼1−𝑖
𝑓
= 𝐼2−𝑖
𝑓
= 𝐼0−𝑖
𝑓
=
𝑉1−𝑖
0
𝑍1−𝑖𝑖 + 𝑍2−𝑖𝑖 + 𝑍0−𝑖𝑖 + 3𝑍 𝑓
Similarly we can compute sequence current for LL & LLG fault.
For passive positive sequence network, the voltage developed at bus k due to
injection of current −𝐼1−𝑖
𝑓
at bus I is ,
𝑉1−𝑘 = −Z1−𝑖𝑘 . I1−𝑖
𝑓
Hence the postfault positive sequence voltage at bus K is,
𝑉1−𝑘
𝑓
= 𝑉1−𝑘
0
−Z1−𝑖𝑘 . I1−𝑖
𝑓
; k=1,2,…n.
Where, 𝑉1−𝑘
0
=prefault positive sequence voltage at bus K
Z1−𝑖𝑘 = 𝑖𝑘 𝑡ℎ component of 𝑉1−𝑏𝑢𝑠
9. As the prefault negative sequence bus voltage is zero, the postfault negative
sequence bus voltage is,
𝑉2−𝑘
𝑓
= 0 + V2−𝑘
𝑉2−𝑘
𝑓
= −Z2−𝑖𝑘 . I2−𝑖
𝑓
Where Z2−𝑖𝑘=𝑖𝑘 𝑡ℎ component of 𝑍2−𝑏𝑢𝑠
The postfault zero sequence bus voltage is
𝑉0−𝑘
𝑓
= 𝑉0−𝑘
0
−Z0−𝑖𝑘 . I1−𝑖
𝑓
;k=1,2,…n.
Where Z0−𝑖𝑘=𝑖𝑘 𝑡ℎ component of 𝑍0−𝑏𝑢𝑠
10. After computing postfault sequence voltage , the sequence current in the line can
be computed as , for line pg. having sequence admittance 𝑌1−𝑝𝑔, 𝑌2−𝑝𝑔and 𝑌0−𝑝𝑔
𝐼1−𝑝𝑔
𝑓
= 𝑌1−𝑝𝑔(𝑉1−𝑝
𝑓
− 𝑉1−𝑔
𝑓
)
𝐼2−𝑝𝑔
𝑓
= 𝑌2−𝑝𝑔(𝑉2−𝑝
𝑓
− 𝑉2−𝑔
𝑓
)
𝐼0−𝑝𝑔
𝑓
= 𝑌0−𝑝𝑔(𝑉2−𝑝
𝑓
− 𝑉2−𝑔
𝑓
)
After computing sequence voltage and current, phase volage and current can be
easily computed as ,
𝑉𝑝 = 𝐴 𝑣𝑠
𝐼 𝑝 = 𝐴𝑖𝑠
11. As this method requires computation of bus impedance matrices of all the three
sequence network, it seems to be more tedious and time consuming . But once the
bus impedance matrices have been formed, fault analysis can be easily can be
easily carried out for all the buses which is the aim of fault analysis.
Also for any changes in power network bus impedance matrix can be easily
modified