4.13
Open Conductor Faults:
When one or two phases of a balanced three-phase line opens it creates an unbalance in the
system and results in the flow of unbalanced currents. Such conditions occur in the system when
one or two conductors of a trnansmission line are broken due to storm or if fuses, isolators or
circuit breakers operate only on one or two phases leaving others connected. Such open conductor
faults can also be analysed with the help of [Z̄Bus ] matrices of sequence networks.
Figure 4.75: Open Conductor faults on a section of three phase system
In Fig. 4.75, a section of a three phase system between buses i and j is shown. Fig. 4.75 (a)
shows one conductor open while Fig. 4.75 (b) shows two conductors open between points k and
k′ . The positive direction of currents I¯a , I¯b and I¯c are shown in the figure. For the analysis of
such faults, the Thevenin’s impedance between two buses i and j is required and the relationship
between the elements of [Z̄Bus ] and Thevenin’s impedances at each bus of the network needs
to be established.
Let [V̄0 ] be the vector of open-circuit bus voltages corresponding to the initial (pre-fault) value
of bus current vector [Ī0 ] injected in a network with bus impedance matrix [Z̄Bus ]. We can
then write
208
[V̄0 ] = [Z̄Bus ][Ī0 ]
(4.134)
If the bus currents are changed to a new value, [Ī0 + ∆Ī], the new bus voltage [V̄] can be
expressed as:
[V̄] = [Z̄Bus ] [Ī0 + ∆Ī]
= [Z̄Bus ][Ī0 ] + [Z̄Bus ][∆Ī]
(4.135)
= [V̄0 ] + [∆V̄]
where,[∆V̄] represents the change in the values of the original bus voltage [V̄0 ].
Fig. 4.76 represents a power system with buses i and j taken out along with the reference node.
The circuit is not energised so that [V̄0 ] and [Ī0 ] are zero. Currents [∆I¯i ] and [∆I¯j ] are
injected into the ith and jth buses respectively, through current sources connected between the
node and the reference node.
Figure 4.76: Change in bus voltage [∆V̄] due to current [∆I¯i ] and [∆I¯j ]
209
The changes in bus voltage [∆V̄] can be calculated from equation (4.135) as
⎡ ∆V̄ ⎤ 1
⎢
1⎥
⎥
⎢
⎢ ⋮ ⎥
⎥
⎢
⎥
⎢
⎢ ∆V̄i ⎥ i
⎥
⎢
⎢ ∆V̄ ⎥ = j
⎢
j⎥
⎥
⎢
⎢ ⋮ ⎥
⎥
⎢
⎥
⎢
⎢∆V̄N ⎥ N
⎦
⎣
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1
Z̄11
⋮
Z̄i1
Z̄j1
⋮
Z̄N 1
i
⋯ Z̄1i
⋮
⋯ Z̄ii
⋯ Z̄ji
⋮
⋯ Z̄N i
j
Z̄1j
⋮
Z̄ij
Z̄jj
⋮
Z̄N j
N
⋯ Z̄1N
⋮
⋯ Z̄iN
⋯ Z̄jN
⋮
⋯ Z̄N N
⎤ ⎡ ∆I¯ ⎤
⎥⎢ 1⎥
⎥
⎥⎢
⎥⎢ ⋮ ⎥
⎥
⎥⎢
⎥⎢ ¯ ⎥
⎥ ⎢ ∆Ii ⎥
⎥
⎥⎢
⎥ ⎢ ∆I¯ ⎥
⎥⎢ j ⎥
⎥
⎥⎢
⎥⎢ ⋮ ⎥
⎥
⎥⎢
⎥⎢ ¯ ⎥
⎥ ⎢∆IN ⎥
⎦
⎦⎣
⎡ ∆V̄ ⎤ ⎡ Z̄ ∆I¯ + Z̄ ∆I¯ ⎤
⎢
⎢
1j
j ⎥
1⎥
⎥
⎥ ⎢ 1i i
⎢
⎥
⎢ ⋮ ⎥ ⎢
⋮
⎥
⎥ ⎢
⎢
⎥
⎥ ⎢
⎢
¯
¯
⎢ ∆V̄i ⎥ ⎢ Z̄ii ∆Ii + Z̄ij ∆Ij ⎥
⎥
⎥=⎢
⎢
⎢ ∆V̄ ⎥ ⎢ Z̄ ∆I¯ + Z̄ ∆I¯ ⎥
⎢
⎢
jj
j ⎥
j⎥
⎥
⎥ ⎢ ji i
⎢
⎥
⎢ ⋮ ⎥ ⎢
⋮
⎥
⎥ ⎢
⎢
⎥
⎥ ⎢
⎢
⎢∆V̄N ⎥ ⎢Z̄N i ∆I¯i + Z̄N j ∆I¯j ⎥
⎦
⎦ ⎣
⎣
(4.136)
The modified voltage at ith bus can be written as :
V̄i = V̄i0 + ∆V̄i = V̄i0 + Z̄ii ∆I¯i + Z̄ij ∆I¯j
(4.137)
adding and subtracting Z̄ij ∆I¯i in equation (4.137), one obtains
V̄i = V̄i0 + (Z̄ii − Z̄ij )∆I¯i + Z̄ij (∆I¯i + ∆I¯j )
(4.138)
Similarly the modified voltage at jth bus can be written as
¯ j = V̄j0 + Z̄ji ∆I¯i + Z̄jj ∆I¯j
V̄j = V̄j0 + ∆V
(4.139)
adding and subtracting Z̄ji ∆I¯j in equation (4.139) , one obtains
V̄j = V̄j0 + (Z̄jj − Z̄ji )∆I¯j + Z̄ji (∆I¯i + ∆I¯j )
(4.140)
Since the network is symmetrical
Z̄ji = Z̄ij
(4.141)
Thus the equations (4.138) and (4.140) can be represented by an equivalent circuit shown in
Fig. 4.77, which is also the Thevenin’s Equivalent circuit of the network as seen from the ith
and jth buses.
From the figure it can be observed that the Thevenin’s open circuit voltage between ith and jth
buses is (V̄i0 − V̄j0 ).
210
Figure 4.77: The Thevenin’s Equivalent of the original network
To calculate the open circuit impedance between ith and jth buses, the initial voltages V̄i0 and
V̄j0 are set equal to zero and an ideal current source I¯ is connected between the two busses.
Next, the resulting voltages V̄i and V̄j are calculated. Note that ∆I¯i = I¯ and ∆I¯j = −I¯.
V̄i = (Z̄ii − Z̄ij )I¯
(4.142)
¯
V̄j = (Z̄jj − Z̄ji )(−I)
(4.143)
Next, calculate the voltage difference ∆V̄ij between ith and jth buses as:
∆V̄ij = V̄i − V̄j = (Z̄ii + Z̄jj − 2Z̄ij )I¯
(4.144)
Hence,
∆Z̄T hevenin,ij =
∆V̄ij
= (Z̄ii + Z̄jj − 2Z̄ij )
I¯
(4.145)
Once the Thevenin’s equivalent is established, the analysis of open-conductor faults can proceed
further. The opening of all the three phases is equivalent to the removal of the line i → j totally
(0)
(1)
(2)
from the network. If z̄ij , z̄ij and z̄ij are the the three sequence impedance of the line i →
j, then the removal of this line from the network can be simulated by adding −z̄ij(0) , −z̄ij(1) and
−z̄ij(2) to the corresponding Thevenin’s equivalent network of the three sequence networks of the
original network as seen from ith and jth buses.
Let x represents the fractional length of the broken line i → j from ith bus to the break point
‘k’, where 0 ≤ x ≤ 1.
211
The positive sequence impedance of the conductor segment between the ith bus and the point of
(1)
break k is xz̄ij , and the positive sequence impedance of the remaining conductor from point k to
j th bus is (1 − x)z̄ij(1) . These two impedances are then added to represent the broken conductor.
This is illustrated in Fig. 4.78
Figure 4.78: Positive sequence equivalent network with line open between buses k and k’
(a)
(b)
(c)
If V̄kk′ , V̄kk′ and V̄kk′ represent the phase component of voltage drops between points k and
(0)
(1)
(2)
k’, then V̄kk′ , V̄kk′ and V̄kk′ represent the sequence components of the voltage drops between
points k and k’. These sequence voltages have different values depending on the type of open
conductor fault.
(1)
(1)
To further simplify the circuit, the voltage V̄kk′ and the total series impedance [xz̄ij + (1 −
x)z̄ij(1) ] = z̄ij(1) is replaced by a current source
V̄kk(1)′
(1)
ij
z̄
(1)
and a parallel impedance z̄ij as shown in
Fig. 4.79.
(1)
(1)
Further, the parallel combination of z̄ij and −z̄ij is ∞ and hence, is replaced by an open circuit.
The final simplified positive sequence impedance equivalent circuit is shown in Fig. 4.80
Similarly, the negative sequence and zero sequence equivalent networks are shown in Fig.4.81 (a)
and (b) repectively. These equivalent networks are identical to the positive sequence equivalent
network but do not contain any internal voltage sources.
The equivalent currents
V̄kk(1)′ V̄kk(2)′
,
(1)
z̄ij
z̄ij(2)
and
V̄kk(0)′
z̄ij(0)
are due to open conductor fault between k and
k’. If no conductor is open then the sequence voltages are all zero and the current sources are
not present in the equivalent circuit. Further, the current sources can be regarded as current
injections into the buses i and j of the original sequence networks. All through the calculation
212
Figure 4.79: Thevenin’s Equivalent with transformed current source
Figure 4.80: Final positive sequence Thevenin’s Equivalent circuit repesenting the opening of line i
→ j between buses k and k’
(0)
(1)
(2)
process, the bus impedance matrices [Z̄Bus ], [Z̄Bus ] and [Z̄Bus ] of the original network are
used.
The current injections at the buses i and j can be tabulated as:
213
Figure 4.81: Final (a) negative sequence (b) zero sequence Thevenin’s Equivalent circuit repesenting
the opening of line i → j between k and k’
Positive Sequence Negative Sequence Zero sequence
at ith bus
at j th bus
V̄kk(1)′
V̄kk(2)′
z̄ij(1)
V̄kk(1)′
− (1)
z̄ij
V̄kk(0)′
z̄ij(2)
V̄kk(2)′
− (2)
z̄ij
(0)
z̄ij(0)
V̄kk(0)′
− (0)
z̄ij
(1)
(2)
The sequence voltage drops ∆V̄n , ∆V̄n and ∆V̄n at any bus ‘n’ due to the current injections
at the buses ‘i’ and ‘j’ can be calculates from equation (4.136) as:
∆V̄
(0)
n
∆V̄
(1)
n
∆V̄
(2)
n
=
=
=
(0)
(0)
(Z̄ni
− Z̄nj
)V̄kk(0)′
z̄ij(0)
(1)
(1)
(Z̄ni
− Z̄nj
)V̄kk(1)′
z̄ij(1)
(4.146)
(2)
(2)
(Z̄ni
− Z̄nj
)V̄kk(2)′
z̄ij(2)
Next, the Thevenin’s equivalent impedances for each sequence network, as seen from the busesk
and k’, are calculated as follows:
(1)
From Fig.4.78, the positive sequence equivalent impdeance Z̄kk′ is found out as :
(1)
kk′
Z̄
(1)
ij
= xz̄
+
(1)
Z̄th,ij
(−z̄ij(1) )
(1)
th,ij
Z̄
(1)
ij
+ (−z̄ )
214
+ (1 − x)z̄ij(1)
(1)
kk′
=
Z̄
−(z̄ij(1) )2
(1)
Z̄th,ij
− z̄ij(1)
(4.147)
Similarly from Fig. 4.81 (a) and (b), the negative sequence and zero sequence Thevenin’s equivalent impedances can be expressed as:
(2)
kk′
=
Z̄
−(z̄ij(2) )2
(2)
Z̄th,ij
− z̄ij(2)
(4.148)
(0)
kk′
=
Z̄
−(z̄ij(0) )2
(0)
Z̄th,ij
− z̄ij(0)
The open-circuit voltage from point k to k’ can be calculated as:
V̄
substituting
(1)
Z̄kk
′
z̄ij(1)
=−
(1)
V̄th,kk
′ is obtained as:
z̄ij(1)
(1)
th,kk′
(1)
Z̄th,ij
− z̄ij(1)
=
(−z̄ij(1) )2
(1)
th,ij
(1)
ij
− z̄
Z̄
(V̄i(1) − V̄j(1) )
(4.149)
from equation (4.147) in equation (4.149), the final value of
(1)
V̄th,kk
′ =
(1)
Z̄kk
′
(1)
ij
z̄
(V̄i(1) − V̄j(1) )
(4.150)
(1)
Also prior to the occurance of open-conductor fault on any conductor, the current I¯ij flowing
in phase a is the positive sequence component and is given by the relation:
I¯ij(1) =
(V̄i(1) − V̄j(1) )
z̄ij(1)
(4.151)
Substituting equation (4.151) in equation (4.150), one gets
(1)
(1) ¯(1)
V̄th,kk
′ = Z̄kk ′ Iij
(4.152)
The Thevenin’s equivalent network as seen from points k and k’ for the three sequence networks
are shown in Fig. 4.82
We are now ready to discuss the two possible cases of open-circuit fault i.e. (a) open phase open
(b) two phases open. We will be discussing them in the next lecture.
215
Figure 4.82: Thevenin’s Equivalent networks as seen from k and k’
4.13.1
One Phase open:
Consider that phase a conductor is open as shown in Fig. 4.75(a), hence phase a current I¯a = 0.
As a result:
I¯a(1) + I¯a(2) + I¯a(0) = 0
(1)
(2)
(4.153)
(0)
where, I¯a , I¯a and I¯a are the symmetrical components of phase a current. Since phases b and
c are closed , the voltage drops .
V̄kk′ ,b = 0 V̄kk′ ,c = 0
216
(4.154)
The symmetrical components of voltage drops across the fault point can be calculated as :
⎡V̄ (0) ⎤
⎡1 1 1 ⎤ ⎡V̄ ′ ⎤
⎡V̄ ′ ⎤
⎥ ⎢ kk ,a ⎥
⎢ a ⎥
⎢
⎢ kk ,a ⎥
⎥
⎢
⎥
⎥
⎢ (1) ⎥ 1 ⎢
⎢
1
⎢V̄a ⎥ = ⎢1 a a2 ⎥ ⎢ 0 ⎥ = ⎢Vkk′ ,a ⎥
⎥ 3⎢
⎥
⎥ 3⎢
⎥⎢
⎢
⎥
⎥
⎥⎢
⎢ (2) ⎥
⎢
⎢
2
⎢V̄a ⎥
⎢1 a a ⎥ ⎢ 0 ⎥
⎢Vkk′ ,a ⎥
⎦
⎦
⎦
⎦⎣
⎣
⎣
⎣
(4.155)
1
V̄a(0) = V̄a(1) = V̄a(2) = V̄kk′ ,a
3
(4.156)
Hence,
It implies that open conductor in phase a causes equal voltages to appear across points k and
k’ of each sequence network. Hence, the three equivalent sequence networks can be connected in
parallel across points k and k’ as shown in Fig. 4.83.
Figure 4.83: Connection of Equivalent sequence networks to represent open phase a between k and
k’
(1)
The current I¯a is given as :
I¯a(1) = I¯ij
(1)
Z̄kk
′
(1)
kk′
Z̄
+
(2) (0)
Z̄kk
′ Z̄kk ′
(2)
(0)
Z̄kk
′ + Z̄kk ′
simplifying
I¯a(1) = I¯ij
(1)
(1)
(2)
(0)
Z̄kk
′ [Z̄kk ′ + Z̄kk ′ ]
(0) (1)
(1) (2)
(2) (0)
Z̄kk
′ Z̄kk ′ + Z̄kk ′ Z̄kk ′ + Z̄kk ′ Z̄kk ′
(2)
(0)
(4.157)
The sequence voltage drops V̄kk′ , V̄kk′ and V̄kk′ can be calculated with reference to Fig.4.83 as:
V̄
(1)
kk′
(2) (0)
(1) Z̄kk′ Z̄kk′
¯
= Ia
(2)
(0)
Z̄kk
′ + Z̄kk ′
(1)
Substituting I¯a and simplifying we get:
217
V̄
(1)
(2)
(1)
kk′
= V̄
(2)
kk′
= V̄
(0)
kk′
= I¯ij
(1) (2) (0)
Z̄kk
′ Z̄kk ′ Z̄kk ′
(0) (1)
(1) (2)
(2) (0)
Z̄kk
′ Z̄kk ′ + Z̄kk ′ Z̄kk ′ + Z̄kk ′ Z̄kk ′
(4.158)
(0)
• Z̄kk′ , Z̄kk′ , and Z̄kk′ are obtained from the impedance parameters of the sequence networks
[equation (4.147) and equation (4.148)].
• I¯ij is the pre-fault current or load current in phase a of the line i → j
Next, the equivalent injected currents
(0)
(1)
(2)
V̄kk(1)′ V̄kk(2)′
,
(1)
z̄ij
z̄ij(2)
and
V̄kk(0)′
z̄ij(0)
are calculated.
Further, ∆V̄i ,∆V̄i and ∆V̄i representing the changes in the symmetrical components of
bus voltage are calculated using equation (4.146).
Finally, the bus voltages after fault are calculated using superposition principle as :
V̄i(1) (F ) = V̄i(1) (0) + ∆V̄i(1)
V̄i(2) (F ) = ∆V̄i(1)
V̄
4.13.2
(0)
i
(F ) = ∆V̄
(4.159)
(2)
i
Two Phases open:
When two phases b and c are open then,
V̄kk(1)′ ,a = V̄a(0) + V̄a(1) + V̄a(2) = 0
I¯b = 0
I¯c = 0
(4.160)
The sequence components of line current are:
⎡I¯(0) ⎤
⎡1 1 1 ⎤ ⎡I¯ ⎤
⎢
⎥ ⎢ a⎥
⎢ a ⎥
⎥⎢ ⎥
⎢ (1) ⎥ 1 ⎢
⎢I¯a ⎥ = ⎢1 a a2 ⎥ ⎢ 0 ⎥
⎥⎢ ⎥
⎢ ⎥ 3⎢
⎢
⎥⎢ ⎥
⎢ ¯(2) ⎥
2
⎢1 a a ⎥ ⎢ 0 ⎥
⎢Ia ⎥
⎣
⎣ ⎦
⎦⎣ ⎦
Simplifying one gets the condition:
1
I¯a(0) = I¯a(1) = I¯a(2) = I¯a
3
(4.161)
equation (4.161) indicates that the three equivalent sequence networks are in series and to ensure
V̄ + V̄a(1) + V̄a(2) = 0 the circuit should be closed.The interconnection of the sequence networks is
shown in Fig.4.84.
From the equivalent circuit of Fig.4.84, the sequence currents can be calculated as :
(0)
a
218
Figure 4.84: Connection of Equivalent sequence networks to represent open phases b and c between
k and k’
I¯a(1) = I¯a(2) = I¯a(0) = I¯ij
(1)
Z̄kk
′
(0)
(1)
(2)
Z̄kk
′ + Z̄kk ′ + Z̄kk ′
(4.162)
I¯ij is the pre-fault current in phase a.
The sequence voltage can be calculated as:
V̄
(1)
kk′
(1)
(0)
(2)
Z̄kk
′ (Z̄kk ′ + Z̄kk ′ )
(0)
(2)
(1)
¯
¯
= Ia (Z̄kk′ + Z̄kk′ ) = Iij (0)
(1)
(2)
Z̄kk′ + Z̄kk
′ + Z̄kk ′
(2)
¯
V̄kk(2)′ = −I¯a(2) Z̄kk
′ = −Iij
(0)
¯
V̄kk(0)′ = −I¯a(0) Z̄kk
′ = −Iij
(1) (2)
Z̄kk
′ Z̄kk ′
(0)
(1)
(2)
Z̄kk
′ + Z̄kk ′ + Z̄kk ′
(4.163)
(1) (0)
Z̄kk
′ Z̄kk ′
(0)
(1)
(2)
Z̄kk
′ + Z̄kk ′ + Z̄kk ′
Remaining calculations are similar to those performed for single conductor open case. In the
next lecture we will be looking at an example of open conductor fault analysis.
219