Chapter 9 Balanced Faults
9.1 Introduction
The most common types of fault are (in order) single-line-to-ground fault, line-to-line
fault, and double-line-to-ground fault. All of these are unbalanced faults. The balanced
(or three phase) fault is the one when all three lines are shorted to ground. It is usually
rare, but can happen. When a fault occurs it is important to isolate it by opening protective breakers. To properly set the breakers, the magnitude of the fault currents need to be
known. This is the major point of study in this chapter.
The life of a fault can be divided into the following times which occur in sequence:
1. The sub-transient period which lasts for only a few cycles
2. The transient period which lasts for a much longer period (tens of cycles)
3. The steady state period which lasts till a major change in the transmission network
takes place (like a circuit breaker opening or a line failing)
The magnitude of the current in each period needs to be known so as to program breakers
to operate at the right time. To do this in the balanced fault case we need to know how
to construct the bus impedance matrix using the building algorithm which is detailed later
on in the chapter. (It is possible to invert the Ybus matrix, but it is more expedient to
"build" the Zbus directly.
The author presents three programs that are useful in the solution of three phase faults.
These are: Zbus=zbuild(data) and Zbus=zbuildpi(linedata, gendata, yload). The latter one is compatible with the power-flow input/output files. The third program
symfault(zdata, Zbus,V) is developed for systematic computation of three phase balanced faults of a large interconnected power system.
9.2 Balanced Three-Phase Fault
During a three-phase fault, the impedance of a generator is a time varying quantity. It is
X d′′ in the sub-transient period (one to four cycles), X d′ in the transient period (about 30
cycles), and the synchronous reactance X d after that. Which one to use depends on the
purpose of the study. Keep in mind that the sub-transient currents can be very large due
to the small size of X d′′ . Due to symmetry, the three phase currents during a symmetrical
fault can be solved using ordinary circuit theory. In this case we use Thevenin's theorem
to analyze the fault. If the fault has zero impedance to ground, it is called a solid fault or
bolted fault (all three lines shorted to ground with zero impedance).
Example 9.1
The one-line diagram of a three bus system is shown below. Each generator is represented by an emf behind the transient reactance. All impedances are in per unit on a 100
MVA base and for simplicity, resistances are neglected. The following assumptions are
made:
1. Shunt capacitances are neglected and the system is considered on no-load.
2. All generators are running at their
rated voltage and rated frequency with
j0.1
j0.2
their emfs in phase.
Determine the fault current, bus voltages,
and the line currents during the fault when
a balanced three-phase fault with impedance Z f = 0.16 pu occurs on
(a) Bus 3.
(b) Bus 2.
(c) Bus 1.
Part (a): The fault on bus 3 is simulated
by connecting a fault impedance Zf =
j0.16 pu as shown in the figure below.
j0.1
j0.2
j0.8
2
1
j0.4
According to Thevenin's theorem, the change due to the short
j0.2
j0.4
circuit is equivalent to that caused
by the added source Vth = V3(0)
j0.8
with all other sources shorted as
shown on the right, where V3(0)
1
2
is the pre-fault voltage at bus 3
(where the fault will occur). It is
j0.4
j0.4
now easy to solve for the changes
3
due to the fault. These changes
are then added to the pre-fault
values to get the post-fault values.
The fault current at bus 3 is
V (0 )
Zf = j0.16
I3 ( F ) = 3
where Z 33 is
Z 33 + Z f
the Thevenin impedance seen at
the faulted bus. To solve for this,
the delta is first changed to an equivalent Y then the circuit
is simplified as shown next page. Note that all values are pu.
j0.4
3
j0.2
j0.4
j0.8
1
2
j0.4
3
j0.4
Vth=V 3(0)
I 3 (F)
Zf = j0.16
2
j0.2
j0.4
1
2
j0.2
j0.24
j0.2
j0.1
j0.34
j0.1
3
3
V th
I3(F)
3
Vth
I 3(F)
Zf = j0.16
V th
I3(F)
Z f = j0.16
Z f = j0.16
Note that the delta is not balanced, hence the Y is not either. Simplification above goes
from left to right with the result that Z33 is equal to j0.34. Now we can compute the fault
V ( 0)
1
=
= − j 2 (note: due to the assumptions made
current as I 3 ( F ) = 3
Z 33 + Z f
j0.34 + j 0.16
in the statement of the problem, V1 ( 0 ) = V2 ( 0 ) = V3 ( 0 ) = 1 pu ). Now we go backward
through the circuit to compute the currents in the original circuit. First the current divides
j 0.6
into two parts through generators 1 and 2 thus: IG1 =
I ( F ) = − j1.2 pu and
j 0.4 + j 0.6 3
j0.4
IG 2 =
I ( F ) = − j 0.8 pu . Thus the changes in bus voltages are given by:
j0.4 + j0.6 3
∆V1 = 0 − ( j0.2 )( − j1.2 ) = −0.24 pu
∆V2 = 0 − ( j 0.4 )( − j 0.8) = −0.32 pu
∆V3 = ( j 0.16 )( − j 2 ) − 1.0 = −0.68 pu
Adding these changes to the pre-fault values we have the post-fault voltages thus:
V1 ( F ) = V1 ( 0 ) + ∆V1 = 1.0 − 0.24 = 0.76 pu
V2 ( F ) = V2 ( 0 ) + ∆V2 = 1.0 − 0.32 = 0.68 pu
V3 ( F ) = V3 ( 0 ) + ∆V3 = 1.0 − 0.68 = 0.32 pu
The short circuit currents in the lines are found as follows:
3
I12 ( F ) =
I13 ( F ) =
I 23 ( F ) =
V1 ( F ) − V2 ( F )
z12
V1 ( F ) − V3 ( F )
z13
V2 ( F ) − V3 ( F )
z23
=
0.76 − 0.68
= − j 0.1 pu
j 0.8
=
0.76 − 0.32
= − j1.1pu
j 0.4
=
0.68 − 0.32
= − j 0.9 pu
j0.4
Part (b). The pre-fault circuit is shown below followed by the Thevenin equivalent then
the reduced circuits. These are shown from left to right:
j0.2
j0.24
j0.4
j0.4
j0.2
j0.2
j0.4
j0.4
2
1
j0.8
2
j0.8
1
1
2
2
j0.4
j0.4
3
V th
V th
j0.4
Z f = j0.16
3
I2(F)
j0.4
I2 (F)
V th
Zf = j0.16
I2 (F)
Zf = j0.16
Zf = j0.16
The Thevenin impedance Z22 at the fault point is given by j0.24 as shown in the figure on
the right above. The fault current is given by:
V ( 0)
1.0
I2 ( F ) = 2
=
= − j 2.5 pu
Z 22 + Z f
j0.24 + j0.16
This current divides between the two generators as follows:
j 0.4
I G1 =
I 2 ( F ) = − j1.0 pu
j 0.4 + j 0.6
j0.6
IG 2 =
I 2 ( F ) = − j1.5 pu
j0.4 + j0.6
Now we find the changes due to the fault as:
∆V1 = 0 − ( j0.2 )( − j1.0 ) = −0.2 pu
∆V2 = 0 − ( j 0.4 )( − j1.5 ) = −0.6 pu
− j1.0 
∆V3 = −0.2 − ( j 0.4 ) 
 = −0.4 pu
 2 
By superposition we find the post-fault voltages thus:
4
V1 ( F ) = V1 ( 0 ) + ∆V1 = 1.0 − 0.2 = 0.8 pu
V2 ( F ) = V2 ( 0 ) + ∆V2 = 1.0 − 0.6 = 0.4 pu
V3 ( F ) = V3 ( 0 ) + ∆V3 = 1.0 − 0.4 = 0.6 pu
And the short circuit current in the lines are found to be:
V ( F ) − V2 ( F ) 0.8 − 0.4
I12 ( F ) = 1
=
= − j 0.5 pu
z12
j 0.8
I13 ( F ) =
I 32 ( F ) =
V1 ( F ) − V3 ( F )
z13
V3 ( F ) − V2 ( F )
z23
=
0.8 − 0.6
= − j 0.5 pu
j 0.4
=
0.6 − 0.4
= − j 0.5 pu
j0.4
Part (c) is similar, and the three currents would be equal to one another but equal to
− j0.3125 instead of –j0.5. Please go over part (c) carefully…
In this example the voltages were assumed equal to 1 pu. For more accurate calculations,
a load flow study is made and the pre-fault voltages are accurately obtained. Also, the
example above did not have any loads at the buses. If there were any loads they would
be represented as constant impedances to ground. This is a good approximation (the
loads change after a fault, but that may be neglected). The general process is summarized
below:
•
The pre-fault bus voltages are obtained from the results of a load flow analysis.
•
Loads are converted to constant impedances to ground using the bus voltages.
•
The faulted network is reduced into a Thevenin equivalent circuit viewed from the
faulted bus. Applying Thevenin's theorem, changes in bus voltages are obtained.
•
Bus voltages during the fault are obtained by superposition of the prefault bus voltages and the changes in the bus voltages computed in the previous step.
•
The currents during the fault may now be obtained in all branches of the faulted system.
5
9.3 Short-Circuit Capacity (SCC)
The short-circuit capacity of a bus is a measure of the strength of a bus. The SCC is defined as the product of the magnitudes of the rated bus voltage and the fault current. It is
used to determine the size of the bus bar and to size the breaker used at the bus.
From the above definition we have for the short-circuit MVA at bus k:
SCC = 3VLk Ik ( F ) × 10−3 MVA
where the line-to-line voltage is VLk is in kV and Ik(F) in A. The symmetrical threephase current in per unit is given by
V ( 0)
I k ( F ) pu = k
X kk
where Vk ( 0 ) is the per unit prefault bus voltage and X kk is the per unit reactance at the
point of fault. N.B. System resistance is neglected, thus the current computed is on the
pessimistic (larger) side. The base current is given by
S B × 103
IB =
3VB
where S B is the base MVA and VB is the line-to-line base voltage in kV. Thus the fault
current in ampere is
I k ( F ) = I k ( F ) pu I B
=
Vk (0 ) S B × 103
X kk
3VB
Thus we have for the SCC:
Vk (0 ) S BVLk
X kk VB
If we assume the base voltage is equal to the line voltage (i.e. VLk = VB ) then
SCC =
Vk (0 ) S B
X kk
The prefault bus voltage is usually assumed 1.0 pu, thus we have the following approximate equation for the short-circuit capacity or the short-circuit MVA
S
SCC ≈ B
X kk
SCC =
6