Single line to ground (LG) fault analysis - Chapter Notes, Power system Analysis Notes

: Single line to ground (LG) fault analysis - Chapter Notes, Power system Analysis Notes

``` Page 1

4.9.5 Single line to ground (LG) fault analysis :
¯
Z
n
is shown in Fig. 4.54. Suppose a single line to ground fault (LG) occurs on phase ‘a’ though
an impedance
¯
Z
f
.
Figure 4.54: LG fault on phase ‘a’ of an unloaded generator
Since the generator is unloaded, the following terminal conditions exist at the fault point:
¯
V
a
=
¯
Z
f
¯
I
a
¯
I
b
= 0 (4.96)
¯
I
c
= 0
Substituting
¯
I
b
=
¯
I
c
= 0 in equation (4.86), the symmetrical components of currents can be
calculated as:
?
?
?
?
?
?
?
?
¯
I
a0
¯
I
a1
¯
I
a2
?
?
?
?
?
?
?
?
=
1
3
?
?
?
?
?
?
?
?
1 1 1
1 a a
2
1 a
2
a
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
I
a
0
0
?
?
?
?
?
?
?
?
(4.97)
162
Page 2

4.9.5 Single line to ground (LG) fault analysis :
¯
Z
n
is shown in Fig. 4.54. Suppose a single line to ground fault (LG) occurs on phase ‘a’ though
an impedance
¯
Z
f
.
Figure 4.54: LG fault on phase ‘a’ of an unloaded generator
Since the generator is unloaded, the following terminal conditions exist at the fault point:
¯
V
a
=
¯
Z
f
¯
I
a
¯
I
b
= 0 (4.96)
¯
I
c
= 0
Substituting
¯
I
b
=
¯
I
c
= 0 in equation (4.86), the symmetrical components of currents can be
calculated as:
?
?
?
?
?
?
?
?
¯
I
a0
¯
I
a1
¯
I
a2
?
?
?
?
?
?
?
?
=
1
3
?
?
?
?
?
?
?
?
1 1 1
1 a a
2
1 a
2
a
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
I
a
0
0
?
?
?
?
?
?
?
?
(4.97)
162
Solving the above equation, the values of the symmetrical components of fault current
¯
I
a
are:
¯
I
a0
=
¯
I
a1
=
¯
I
a2
=
1
3
¯
I
a
(4.98)
The voltage of phase a can be expressed in terms of symmetrical components from equation (4.83),
as
¯
V
a
=
¯
V
a0
+
¯
V
a1
+
¯
V
a2
(4.99)
Substituing in the equation the values of
¯
V
a0
,
¯
V
a1
and
¯
V
a2
from equation (4.94) into equation (4.99),
¯
V
a
can be written as (with
¯
I
a0
=
¯
I
a1
=
¯
I
a2
from equation (4.98)):
¯
V
a
=
¯
E
a
-(
¯
Z
0
+
¯
Z
1
+
¯
Z
2
)
¯
I
a0
(4.100)
From equations (4.96) and (4.98),
¯
V
a
=
¯
Z
f
¯
I
a
= 3
¯
Z
f
¯
I
a0
. Hence, equation (4.100) can be expressed as:
3
¯
Z
f
¯
I
a0
=
¯
E
a
-(
¯
Z
0
+
¯
Z
1
+
¯
Z
2
)
¯
I
a0
or,
¯
I
a0
=
¯
E
a
¯
Z
0
+
¯
Z
1
+
¯
Z
2
+ 3
¯
Z
f
(4.101)
The fault current, therefore, is:
¯
I
f
=
¯
I
a
= 3
¯
I
a0
=
3
¯
E
a
¯
Z
0
+
¯
Z
1
+
¯
Z
2
+ 3
¯
Z
f
(4.102)
From equations (4.98) and (4.101), it be easily interpreted that the three sequence networks are
connected in series as shown in Fig. 4.55.
Figure 4.55: Connection of sequence networks for LG fault
Note that, for solidly grounded generator,
¯
Z
n
= 0 and for bolted fault
¯
Z
f
= 0.
Extending the above concept to the analysis of LG fault in a power system, the Thevenin’s equiv-
163
Page 3

4.9.5 Single line to ground (LG) fault analysis :
¯
Z
n
is shown in Fig. 4.54. Suppose a single line to ground fault (LG) occurs on phase ‘a’ though
an impedance
¯
Z
f
.
Figure 4.54: LG fault on phase ‘a’ of an unloaded generator
Since the generator is unloaded, the following terminal conditions exist at the fault point:
¯
V
a
=
¯
Z
f
¯
I
a
¯
I
b
= 0 (4.96)
¯
I
c
= 0
Substituting
¯
I
b
=
¯
I
c
= 0 in equation (4.86), the symmetrical components of currents can be
calculated as:
?
?
?
?
?
?
?
?
¯
I
a0
¯
I
a1
¯
I
a2
?
?
?
?
?
?
?
?
=
1
3
?
?
?
?
?
?
?
?
1 1 1
1 a a
2
1 a
2
a
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
I
a
0
0
?
?
?
?
?
?
?
?
(4.97)
162
Solving the above equation, the values of the symmetrical components of fault current
¯
I
a
are:
¯
I
a0
=
¯
I
a1
=
¯
I
a2
=
1
3
¯
I
a
(4.98)
The voltage of phase a can be expressed in terms of symmetrical components from equation (4.83),
as
¯
V
a
=
¯
V
a0
+
¯
V
a1
+
¯
V
a2
(4.99)
Substituing in the equation the values of
¯
V
a0
,
¯
V
a1
and
¯
V
a2
from equation (4.94) into equation (4.99),
¯
V
a
can be written as (with
¯
I
a0
=
¯
I
a1
=
¯
I
a2
from equation (4.98)):
¯
V
a
=
¯
E
a
-(
¯
Z
0
+
¯
Z
1
+
¯
Z
2
)
¯
I
a0
(4.100)
From equations (4.96) and (4.98),
¯
V
a
=
¯
Z
f
¯
I
a
= 3
¯
Z
f
¯
I
a0
. Hence, equation (4.100) can be expressed as:
3
¯
Z
f
¯
I
a0
=
¯
E
a
-(
¯
Z
0
+
¯
Z
1
+
¯
Z
2
)
¯
I
a0
or,
¯
I
a0
=
¯
E
a
¯
Z
0
+
¯
Z
1
+
¯
Z
2
+ 3
¯
Z
f
(4.101)
The fault current, therefore, is:
¯
I
f
=
¯
I
a
= 3
¯
I
a0
=
3
¯
E
a
¯
Z
0
+
¯
Z
1
+
¯
Z
2
+ 3
¯
Z
f
(4.102)
From equations (4.98) and (4.101), it be easily interpreted that the three sequence networks are
connected in series as shown in Fig. 4.55.
Figure 4.55: Connection of sequence networks for LG fault
Note that, for solidly grounded generator,
¯
Z
n
= 0 and for bolted fault
¯
Z
f
= 0.
Extending the above concept to the analysis of LG fault in a power system, the Thevenin’s equiv-
163
alent circuit (as seen from the fault point) is obtained, individually for the three sequence networks.
For the positive sequence network
¯
V
th
, the open circuit pre-fault voltage at the fault point, and
¯
Z
1th
,
the positive sequence Thevenin’s equivalent impedance as seen from the fault point are determined.
For negative and zero sequence networks, only the Thevenin’s equivalent impedances
¯
Z
2th
and
¯
Z
0th
,
respectively are calculated. The three Thevenin’s equivalent networks are then connected in series.
4.9.6 Line to Line (LL) fault analysis :
Fig. 4.56 shows a line to line fault (LL) between phases ‘b’ and ‘c’ through an impedance
¯
Z
f
, on
an unloaded three phase generator. The terminal conditions at the fault point are:
Figure 4.56: LL fault between phases ‘b’ and ‘c’ of an unloaded generator
¯
V
b
-
¯
V
c
=
¯
Z
f
¯
I
b
¯
I
b
+
¯
I
c
= 0 (4.103)
164
Page 4

4.9.5 Single line to ground (LG) fault analysis :
¯
Z
n
is shown in Fig. 4.54. Suppose a single line to ground fault (LG) occurs on phase ‘a’ though
an impedance
¯
Z
f
.
Figure 4.54: LG fault on phase ‘a’ of an unloaded generator
Since the generator is unloaded, the following terminal conditions exist at the fault point:
¯
V
a
=
¯
Z
f
¯
I
a
¯
I
b
= 0 (4.96)
¯
I
c
= 0
Substituting
¯
I
b
=
¯
I
c
= 0 in equation (4.86), the symmetrical components of currents can be
calculated as:
?
?
?
?
?
?
?
?
¯
I
a0
¯
I
a1
¯
I
a2
?
?
?
?
?
?
?
?
=
1
3
?
?
?
?
?
?
?
?
1 1 1
1 a a
2
1 a
2
a
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
I
a
0
0
?
?
?
?
?
?
?
?
(4.97)
162
Solving the above equation, the values of the symmetrical components of fault current
¯
I
a
are:
¯
I
a0
=
¯
I
a1
=
¯
I
a2
=
1
3
¯
I
a
(4.98)
The voltage of phase a can be expressed in terms of symmetrical components from equation (4.83),
as
¯
V
a
=
¯
V
a0
+
¯
V
a1
+
¯
V
a2
(4.99)
Substituing in the equation the values of
¯
V
a0
,
¯
V
a1
and
¯
V
a2
from equation (4.94) into equation (4.99),
¯
V
a
can be written as (with
¯
I
a0
=
¯
I
a1
=
¯
I
a2
from equation (4.98)):
¯
V
a
=
¯
E
a
-(
¯
Z
0
+
¯
Z
1
+
¯
Z
2
)
¯
I
a0
(4.100)
From equations (4.96) and (4.98),
¯
V
a
=
¯
Z
f
¯
I
a
= 3
¯
Z
f
¯
I
a0
. Hence, equation (4.100) can be expressed as:
3
¯
Z
f
¯
I
a0
=
¯
E
a
-(
¯
Z
0
+
¯
Z
1
+
¯
Z
2
)
¯
I
a0
or,
¯
I
a0
=
¯
E
a
¯
Z
0
+
¯
Z
1
+
¯
Z
2
+ 3
¯
Z
f
(4.101)
The fault current, therefore, is:
¯
I
f
=
¯
I
a
= 3
¯
I
a0
=
3
¯
E
a
¯
Z
0
+
¯
Z
1
+
¯
Z
2
+ 3
¯
Z
f
(4.102)
From equations (4.98) and (4.101), it be easily interpreted that the three sequence networks are
connected in series as shown in Fig. 4.55.
Figure 4.55: Connection of sequence networks for LG fault
Note that, for solidly grounded generator,
¯
Z
n
= 0 and for bolted fault
¯
Z
f
= 0.
Extending the above concept to the analysis of LG fault in a power system, the Thevenin’s equiv-
163
alent circuit (as seen from the fault point) is obtained, individually for the three sequence networks.
For the positive sequence network
¯
V
th
, the open circuit pre-fault voltage at the fault point, and
¯
Z
1th
,
the positive sequence Thevenin’s equivalent impedance as seen from the fault point are determined.
For negative and zero sequence networks, only the Thevenin’s equivalent impedances
¯
Z
2th
and
¯
Z
0th
,
respectively are calculated. The three Thevenin’s equivalent networks are then connected in series.
4.9.6 Line to Line (LL) fault analysis :
Fig. 4.56 shows a line to line fault (LL) between phases ‘b’ and ‘c’ through an impedance
¯
Z
f
, on
an unloaded three phase generator. The terminal conditions at the fault point are:
Figure 4.56: LL fault between phases ‘b’ and ‘c’ of an unloaded generator
¯
V
b
-
¯
V
c
=
¯
Z
f
¯
I
b
¯
I
b
+
¯
I
c
= 0 (4.103)
164
¯
I
a
= 0
Substituting
¯
I
a
= 0 and
¯
I
b
=-
¯
I
c
in equation (4.86), the symmetrical components of cuurents can be
calculated as:
?
?
?
?
?
?
?
?
¯
I
a0
¯
I
a1
¯
I
a2
?
?
?
?
?
?
?
?
=
1
3
?
?
?
?
?
?
?
?
1 1 1
1 a a
2
1 a
2
a
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
¯
I
b
-
¯
I
b
?
?
?
?
?
?
?
?
(4.104)
Solving the above equation, the values of the symmetrical components of the current
¯
I
a
are:
¯
I
a0
= 0
¯
I
a1
=
1
3
(a- a
2
)
¯
I
b
(4.105)
¯
I
a2
=
1
3
(a
2
- a)
¯
I
b
=-
¯
I
a1
From equation (4.83), we have
¯
V
b
-
¯
V
c
= (a
2
- a)(
¯
V
a1
-
¯
V
a2
)=
¯
Z
f
¯
I
b
(4.106)
Substituting
¯
V
a1
and
¯
V
a2
from equation (4.94) and noting that
¯
I
a1
=-
¯
I
a2
, one can write:
(a
2
- a)
¯
E
a
-(
¯
Z
1
+
¯
Z
2
)
¯
I
a1
=
¯
Z
f
¯
I
b
(4.107)
Also from equation (4.105),
¯
I
b
=
3
¯
I
a1
(a- a
2
)
(4.108)
Substituting this value of
¯
I
b
in equation (4.107), we get:

¯
E
a
-(
¯
Z
1
+
¯
Z
2
)
¯
I
a1
=
3
¯
Z
f
¯
I
a1
(a- a
2
)(a
2
- a)
Since, (a- a
2
)(a
2
- a)= 3, the above expression can be simpli?ed and written as:
¯
I
a1
=
¯
E
a
(
¯
Z
1
+
¯
Z
2
+
¯
Z
f
)
(4.109)
The phase currents during fault can be calculated as:
?
?
?
?
?
?
?
?
¯
I
a
¯
I
b
¯
I
c
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
1 1 1
1 a
2
a
1 a a
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
¯
I
a1
-
¯
I
a1
?
?
?
?
?
?
?
?
(4.110)
Solving for the phase currents, the expressions for
¯
I
b
and
¯
I
c
can be written as:
165
Page 5

4.9.5 Single line to ground (LG) fault analysis :
¯
Z
n
is shown in Fig. 4.54. Suppose a single line to ground fault (LG) occurs on phase ‘a’ though
an impedance
¯
Z
f
.
Figure 4.54: LG fault on phase ‘a’ of an unloaded generator
Since the generator is unloaded, the following terminal conditions exist at the fault point:
¯
V
a
=
¯
Z
f
¯
I
a
¯
I
b
= 0 (4.96)
¯
I
c
= 0
Substituting
¯
I
b
=
¯
I
c
= 0 in equation (4.86), the symmetrical components of currents can be
calculated as:
?
?
?
?
?
?
?
?
¯
I
a0
¯
I
a1
¯
I
a2
?
?
?
?
?
?
?
?
=
1
3
?
?
?
?
?
?
?
?
1 1 1
1 a a
2
1 a
2
a
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
¯
I
a
0
0
?
?
?
?
?
?
?
?
(4.97)
162
Solving the above equation, the values of the symmetrical components of fault current
¯
I
a
are:
¯
I
a0
=
¯
I
a1
=
¯
I
a2
=
1
3
¯
I
a
(4.98)
The voltage of phase a can be expressed in terms of symmetrical components from equation (4.83),
as
¯
V
a
=
¯
V
a0
+
¯
V
a1
+
¯
V
a2
(4.99)
Substituing in the equation the values of
¯
V
a0
,
¯
V
a1
and
¯
V
a2
from equation (4.94) into equation (4.99),
¯
V
a
can be written as (with
¯
I
a0
=
¯
I
a1
=
¯
I
a2
from equation (4.98)):
¯
V
a
=
¯
E
a
-(
¯
Z
0
+
¯
Z
1
+
¯
Z
2
)
¯
I
a0
(4.100)
From equations (4.96) and (4.98),
¯
V
a
=
¯
Z
f
¯
I
a
= 3
¯
Z
f
¯
I
a0
. Hence, equation (4.100) can be expressed as:
3
¯
Z
f
¯
I
a0
=
¯
E
a
-(
¯
Z
0
+
¯
Z
1
+
¯
Z
2
)
¯
I
a0
or,
¯
I
a0
=
¯
E
a
¯
Z
0
+
¯
Z
1
+
¯
Z
2
+ 3
¯
Z
f
(4.101)
The fault current, therefore, is:
¯
I
f
=
¯
I
a
= 3
¯
I
a0
=
3
¯
E
a
¯
Z
0
+
¯
Z
1
+
¯
Z
2
+ 3
¯
Z
f
(4.102)
From equations (4.98) and (4.101), it be easily interpreted that the three sequence networks are
connected in series as shown in Fig. 4.55.
Figure 4.55: Connection of sequence networks for LG fault
Note that, for solidly grounded generator,
¯
Z
n
= 0 and for bolted fault
¯
Z
f
= 0.
Extending the above concept to the analysis of LG fault in a power system, the Thevenin’s equiv-
163
alent circuit (as seen from the fault point) is obtained, individually for the three sequence networks.
For the positive sequence network
¯
V
th
, the open circuit pre-fault voltage at the fault point, and
¯
Z
1th
,
the positive sequence Thevenin’s equivalent impedance as seen from the fault point are determined.
For negative and zero sequence networks, only the Thevenin’s equivalent impedances
¯
Z
2th
and
¯
Z
0th
,
respectively are calculated. The three Thevenin’s equivalent networks are then connected in series.
4.9.6 Line to Line (LL) fault analysis :
Fig. 4.56 shows a line to line fault (LL) between phases ‘b’ and ‘c’ through an impedance
¯
Z
f
, on
an unloaded three phase generator. The terminal conditions at the fault point are:
Figure 4.56: LL fault between phases ‘b’ and ‘c’ of an unloaded generator
¯
V
b
-
¯
V
c
=
¯
Z
f
¯
I
b
¯
I
b
+
¯
I
c
= 0 (4.103)
164
¯
I
a
= 0
Substituting
¯
I
a
= 0 and
¯
I
b
=-
¯
I
c
in equation (4.86), the symmetrical components of cuurents can be
calculated as:
?
?
?
?
?
?
?
?
¯
I
a0
¯
I
a1
¯
I
a2
?
?
?
?
?
?
?
?
=
1
3
?
?
?
?
?
?
?
?
1 1 1
1 a a
2
1 a
2
a
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
¯
I
b
-
¯
I
b
?
?
?
?
?
?
?
?
(4.104)
Solving the above equation, the values of the symmetrical components of the current
¯
I
a
are:
¯
I
a0
= 0
¯
I
a1
=
1
3
(a- a
2
)
¯
I
b
(4.105)
¯
I
a2
=
1
3
(a
2
- a)
¯
I
b
=-
¯
I
a1
From equation (4.83), we have
¯
V
b
-
¯
V
c
= (a
2
- a)(
¯
V
a1
-
¯
V
a2
)=
¯
Z
f
¯
I
b
(4.106)
Substituting
¯
V
a1
and
¯
V
a2
from equation (4.94) and noting that
¯
I
a1
=-
¯
I
a2
, one can write:
(a
2
- a)
¯
E
a
-(
¯
Z
1
+
¯
Z
2
)
¯
I
a1
=
¯
Z
f
¯
I
b
(4.107)
Also from equation (4.105),
¯
I
b
=
3
¯
I
a1
(a- a
2
)
(4.108)
Substituting this value of
¯
I
b
in equation (4.107), we get:

¯
E
a
-(
¯
Z
1
+
¯
Z
2
)
¯
I
a1
=
3
¯
Z
f
¯
I
a1
(a- a
2
)(a
2
- a)
Since, (a- a
2
)(a
2
- a)= 3, the above expression can be simpli?ed and written as:
¯
I
a1
=
¯
E
a
(
¯
Z
1
+
¯
Z
2
+
¯
Z
f
)
(4.109)
The phase currents during fault can be calculated as:
?
?
?
?
?
?
?
?
¯
I
a
¯
I
b
¯
I
c
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
1 1 1
1 a
2
a
1 a a
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
¯
I
a1
-
¯
I
a1
?
?
?
?
?
?
?
?
(4.110)
Solving for the phase currents, the expressions for
¯
I
b
and
¯
I
c
can be written as:
165
¯
I
b
=-
¯
I
c
= (a
2
- a)
¯
I
a1
(4.111)
Substituting
¯
I
b
from equation (4.111) in equation (4.106) one gets:
(
¯
V
a1
-
¯
V
a2
)=
¯
Z
f
¯
I
a1
The equivalent circuit of the fault in terms of the sequence networks is shown in Fig. 4.57. The
circuit has been drawn on the basis of equation (4.105) and the above equation. It shows that the
positive sequence and negative sequence networks are connected in phase opposition bridged by the
fault impedance
¯
Z
f
. Also, since
¯
I
a0
= 0, the zero sequence network is open circuited and hence is
not shown in the diagram.
Figure 4.57: Connection of sequence networks for an LL fault between phases ‘b’ and ‘c’ of an
Extending the above concept to LL fault calculations in a power system, it can be concluded that,
the Thevenin’s equivalent positive and negative sequence networks, as seen from the fault point, can
be connected in phase opposition through the fault impdedance for calculating fault current.
4.9.7 Double Line to ground (LLG) fault analysis :
Fig. 4.58 shows a double line to ground (LLG) fault on phases ‘b’ and ‘c’ through an impedance
¯
Z
f
on an unloaded three phase generator. The terminal conditions at the fault point are:
¯
V
b
=
¯
V
c
=
¯
Z
f
¯
I
f
=
¯
Z
f
(
¯
I
b
+
¯
I
c
)
¯
I
a
=
¯
I
a1
+
¯
I
a2
+
¯
I
a0
= 0 (4.112)
From equation (4.83),
¯
V
b
and
¯
V
c
can be written as:
¯
V
b
=
¯
V
a0
+ a
2
¯
V
a1
+ a
¯
V
a2
¯
V
c
=
¯
V
a0
+ a
¯
V
a1
+ a
2
¯
V
a2
(4.113)
166
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