4.9.5
Single line to ground (LG) fault analysis :
An unloaded balanced three-phase synchronous generator with neutral grounded through an impedance
Z̄n is shown in Fig. 4.54. Suppose a single line to ground fault (LG) occurs on phase ‘a’ though
an impedance Z̄f .
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:
V̄a = Z̄f 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:
⎡1 1 1 ⎤ ⎡I¯ ⎤
⎡I¯ ⎤
⎢
⎥ ⎢ a⎥
⎢ a0 ⎥
⎢ ⎥ 1⎢
⎥⎢ ⎥
⎢I¯a1 ⎥ = ⎢1 a a2 ⎥ ⎢ 0 ⎥
⎢ ⎥ 3⎢
⎥⎢ ⎥
⎢
⎥⎢ ⎥
⎢¯ ⎥
⎢Ia2 ⎥
⎢1 a 2 a ⎥ ⎢ 0 ⎥
⎣
⎦⎣ ⎦
⎣ ⎦
162
(4.97)
Solving the above equation, the values of the symmetrical components of fault current I¯a are:
1
I¯a0 = I¯a1 = I¯a2 = I¯a
3
(4.98)
The voltage of phase a can be expressed in terms of symmetrical components from equation (4.83),
as
V̄a = V̄a0 + V̄a1 + V̄a2
(4.99)
Substituing in the equation the values of V̄a0 , V̄a1 and V̄a2 from equation (4.94) into equation (4.99),
V̄a can be written as (with I¯a0 = I¯a1 = I¯a2 from equation (4.98)):
V̄a = Ēa − (Z̄0 + Z̄1 + Z̄2 )I¯a0
(4.100)
From equations (4.96) and (4.98), V̄a = Z̄f I¯a = 3Z̄f I¯a0 . Hence, equation (4.100) can be expressed as:
3Z̄f I¯a0 = Ēa − (Z̄0 + Z̄1 + Z̄2 )I¯a0
or,
I¯a0 =
Ēa
Z̄0 + Z̄1 + Z̄2 + 3Z̄f
(4.101)
The fault current, therefore, is:
I¯f = I¯a = 3I¯a0 =
3Ēa
Z̄0 + Z̄1 + Z̄2 + 3Z̄f
(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, Z̄n = 0 and for bolted fault Z̄f = 0.
Extending the above concept to the analysis of LG fault in a power system, the Thevenin’s equiv163
alent circuit (as seen from the fault point) is obtained, individually for the three sequence networks.
For the positive sequence network V̄th , the open circuit pre-fault voltage at the fault point, and Z̄1th ,
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 Z̄2th and Z̄0th ,
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 Z̄f , 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
V̄b − V̄c = Z̄f I¯b
I¯b + I¯c = 0
164
(4.103)
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¯ ⎤
⎡1 1 1 ⎤ ⎡ 0 ⎤
⎥⎢ ⎥
⎢ a0 ⎥
⎢
⎥⎢ ⎥
⎢ ⎥ 1⎢
⎢I¯a1 ⎥ = ⎢1 a a2 ⎥ ⎢ I¯b ⎥
⎥⎢ ⎥
⎢ ⎥ 3⎢
⎥⎢ ⎥
⎢¯ ⎥
⎢
2
⎢Ia2 ⎥
⎢1 a a ⎥ ⎢−I¯b ⎥
⎦⎣ ⎦
⎣ ⎦
⎣
(4.104)
Solving the above equation, the values of the symmetrical components of the current I¯a are:
I¯a0 = 0
1
I¯a1 = (a − a2 )I¯b
3
1
I¯a2 = (a2 − a)I¯b = −I¯a1
3
(4.105)
From equation (4.83), we have
V̄b − V̄c = (a2 − a)(V̄a1 − V̄a2 ) = Z̄f I¯b
(4.106)
Substituting V̄a1 and V̄a2 from equation (4.94) and noting that I¯a1 = −I¯a2 , one can write:
(a2 − a) [Ēa − (Z̄1 + Z̄2 )I¯a1 ] = Z̄f I¯b
(4.107)
Also from equation (4.105),
I¯b =
3I¯a1
(a − a2 )
(4.108)
Substituting this value of I¯b in equation (4.107), we get:
[Ēa − (Z̄1 + Z̄2 )I¯a1 ] =
3Z̄f I¯a1
(a − a2 )(a2 − a)
Since, (a − a2 )(a2 − a) = 3, the above expression can be simplified and written as:
I¯a1 =
Ēa
(Z̄1 + Z̄2 + Z̄f )
(4.109)
The phase currents during fault can be calculated as:
⎡I¯ ⎤ ⎡1 1 1 ⎤ ⎡ 0 ⎤
⎢ a⎥ ⎢
⎥⎢
⎥
⎢ ⎥ ⎢
⎥⎢
⎥
⎢ I¯b ⎥ = ⎢1 a2 a ⎥ ⎢ I¯a1 ⎥
⎢ ⎥ ⎢
⎥⎢
⎥
⎢¯⎥ ⎢
⎥⎢
⎥
⎢ Ic ⎥ ⎢1 a a2 ⎥ ⎢−I¯a1 ⎥
⎣ ⎦ ⎣
⎦⎣
⎦
Solving for the phase currents, the expressions for I¯b and I¯c can be written as:
165
(4.110)
I¯b = −I¯c = (a2 − a)I¯a1
(4.111)
Substituting I¯b from equation (4.111) in equation (4.106) one gets:
(V̄a1 − V̄a2 ) = Z̄f 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 Z̄f . 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
unloaded generator
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
Z̄f on an unloaded three phase generator. The terminal conditions at the fault point are:
V̄b = V̄c = Z̄f I¯f = Z̄f (I¯b + I¯c )
I¯a = I¯a1 + I¯a2 + I¯a0 = 0
(4.112)
From equation (4.83), V̄b and V̄c can be written as:
V̄b = V̄a0 + a2 V̄a1 + aV̄a2
V̄c = V̄a0 + aV̄a1 + a2 V̄a2
166
(4.113)
Figure 4.58: LLG fault between phases ‘b’ and ‘c’ of an unloaded generator
Since V̄b = V̄c , from equation (4.113), one can write
V̄a1 = V̄a2
(4.114)
Substituting I¯b and I¯c in terms of their sequence components from equation (4.85), voltage of phase
’b’ can be expressed as:
V̄b = Z̄f (I¯a0 + a2 I¯a1 + aI¯a2 + I¯a0 + aI¯a1 + a2 I¯a2 )
167
V̄b = Z̄f (I¯a0 + a2 I¯a1 + aI¯a2 + I¯a0 + aI¯a1 + a2 I¯a2 )
= Z̄f (2I¯a0 + (a2 + a)(I¯a1 + I¯a2 ))
= Z̄f (2I¯a0 − (I¯a1 + I¯a2 ))
Since 1 + a + a2 = 0 and I¯a = I¯a1 + I¯a2 + I¯a0 = 0, hence
V̄b = 3Z̄f I¯a0
(4.115)
Further substituting V̄b from equation (4.115) and the condition of equation (4.114) in equation
(4.113), we get:
3Z̄f I¯a0 = V̄a0 + (a2 + a)V̄a1
= V̄a0 − V̄a1
(4.116)
Substituting V̄a0 and V̄a1 from equation (4.94) in equation (4.116), the zero sequence component of
Figure 4.59: Connection of sequence networks for an LLG fault between phases ‘b’ and ‘c’ of an
unloaded generator
current I¯a0 is given by:
(Ēa − Z̄1 I¯a1 )
I¯a0 = −
(Z̄0 + 3Z̄f )
(4.117)
For calculating the negative sequence component of current, I¯a2 , substitute V̄a1 and V̄a2 from equation
(4.94) in equation (4.114). The expression for I¯a2 is:
(Ēa − Z̄1 I¯a1 )
I¯a2 = −
Z̄2
(4.118)
Finally, by substituting I¯a0 and I¯a2 from equations (4.117) and (4.118) in equation (4.112), the value
168
of the positive sequence component of current I¯a1 is found out as:
I¯a1 =
Ēa
Z̄2 (Z̄0 + 3Z̄f )
Z̄1 +
(Z̄0 + Z̄2 + 3Z̄f )
(4.119)
Since V̄b = Z̄f I¯f , from equation (4.115) one can conclude that
I¯f = 3I¯a0
(4.120)
The equivalent circuit for the fault in terms of the sequence networks is shown in Fig. 4.59. The
circuit shown in Fig. 4.59 is based on equations (4.114) and (4.116). For LLG fault calculations in
a power system, the Thevenin’s equivalent of the three sequence networks, as seen from the fault
point, are found out. The positive and negative sequence equivalents are connected in parallel and
the combination is then connected to the zero sequence network through 3Z̄f . In the next lecture,
we will study the procedure of unbalanced fault analysis using Z̄BUS matrix.
169