Chapter 10: Symmetrical Components and
Unbalanced Faults, Part III
10.8 Unbalanced Fault Analysis Using Bus Impedance Matrix
We showed that the symmetrical components are independent, and for each component
0
1
2
there is an impedance matrix Z bus . Thus we can find Z bus
, Z bus
and Z bus
in order to deal
with each component separately. If the fault is at bus k , then we know that the k-th
diagonal element of the bus is the Thevenin impedance of the network viewed from that
bus. This then allows us to find the sequence networks, which we connect according to
the fault, and then go on to solve for the fault currents and voltages. The impedances
obtained from the Z bus matrices are called Z kk0 , Z 1kk and Z kk2 which are connected
according to figures 10.11, 10.13, and 10.15 in the book. In the material to follow, the
subscript "a" is implied, but it is left out. It is understood that the symmetrical
components refer to phase a.
10.8.1 Single Line-To-Ground Fault Using Zbus
Consider the single line to ground fault on line "a" through
impedance Z f to ground on bus k as shown in the figure. This
fault requires the zero- positive- and negative-sequence networks
for phase "a" be placed in series in addition to 3Z f in order to
compute the sequence currents for phase "a", thus:
Vk ( 0 )
I k0 = I k1 = Ik2 = 0
Z kk + Z kk1 + Z kk2 + 3Z f
a
b
c
Zf
1
Where Z kk0 , Z kk
,and Z kk2 are the diagonal elements in the k axis of the corresponding
impedance matrix, and Vk ( 0) is the pre-fault voltage at bus k. The fault phase current is:
I kabc = AI k012
10.8.2 Line-To-Line Fault Using Zbus
Consider a fault between phases b and c through an impedance Z f
at bus k as shown in the figure. Connecting the sequence networks
for phase "a" as found earlier (positive- and negative-sequence
networks in opposition,) the symmetrical components of the fault
current are:
a
b
Zf
c
I k0 = 0
I 1k = − I k2 =
Vk ( 0 )
Z + Z kk2 + Z f
1
kk
Where Z 1kk , and Z kk2 are on the diagonal elements in the k-axis of the corresponding bus
impedance matrix. The fault phase current is then obtained using the A matrix as usual.
10.8.3 Double Line-To-Line Fault Using Zbus
Here phases b and c are shorted and connected to ground through a
fault impedance Z f . From before, the equations for the sequence
currents of phase a at bus k are given by:
Vk ( 0 )
I 1k =
Z kk2 ( Z kk0 + 3Z f )
1
Z kk + 2
Z kk + Z kk0 + 3Z f
a
b
c
Zf
Vk ( 0) − Z kk1 I k1
I =−
Z kk2
2
k
I k0 = −
1 1
Vk ( 0 ) − Z kk
Ik
0
Z kk + 3Z f
1
,and Z kk2 are the diagonal elements in the k axis of the
As usual, the Z kk0 , Z kk
corresponding impedance matrix, and Vk ( 0) is the pre-fault voltage at bus k. The phase
currents are obtained using the A matrix, and the fault current is I k ( F ) = I kb + I kc .
10.8.4 Bus Voltages and Line Currents During Fault
Knowing the fault currents at bus k, the symmetrical components of the i-th bus voltage
during fault are obtained using the equation (++):
Vi 0 ( F ) = 0 − Zik0 I k0
Vi 1 ( F ) = Vi1 ( 0 ) − Z ik1 I 1k
Vi 2 ( F ) = 0 − Z ik2 Ik2
Where Vi 1 (0 ) = Vi (0 ) is the pre-fault phase voltage at bus i and Z ik is the ik-th element
of the proper Z matrix. The phase voltages during the fault are found using the A matrix
thus:
Vi abc = AVi 012
The symmetrical components of fault current in line i to j are given by:
2
I =
0
ij
Vi 0 ( F ) −V j0 (F )
zij0
Vi1 ( F ) − V j1 ( F )
I =
z1ij
1
ij
I =
2
ij
Vi 2 ( F ) − V j2 ( F )
zij2
Where zij0 , z1ij ,and zij2 are the zero-, positive-, and negative-sequence components of the
actual line impedance between buses i and j. Using the A matrix the phase fault currents
in line i to j are:
I ijabc = AI ij012
Example 10.6
Solve Example 10.5 using the bus impedance matrix. In addition, for each type of fault
determine the bus voltages and line currents during the fault.
Go over the example in detail, page 435.
10.9 Unbalanced Fault Programs
Go over the various programs for unbalanced fault analysis.
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