EE 340 EE 340 Symmetrical Faults Symmetrical Faults

johnnepaleseElectronics - Devices

Oct 10, 2013 (4 years and 1 month ago)

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1
EE 340
EE 340
Symmetrical Faults
Symmetrical Faults
2
Introduction
Introduction


A
A
fault
fault
in a circuit is any failure that
in a circuit is any failure that
interferes with the normal system
interferes with the normal system
operation.
operation.


Lighting strokes cause most faults on
Lighting strokes cause most faults on
high
high
-
-
voltage transmission lines
voltage transmission lines
producing a very high transient that
producing a very high transient that
greatly exceeds the rated voltage of
greatly exceeds the rated voltage of
the line.
the line.


This voltage usually causes flashover
This voltage usually causes flashover
between the phases and/or the ground
between the phases and/or the ground
creating an arc.
creating an arc.


Since the impedance of this new path
Since the impedance of this new path
is usually low, an excessive current
is usually low, an excessive current
may flow.
may flow.


Faults involving ionized current paths
Faults involving ionized current paths
are also called transient faults. They
are also called transient faults. They
usually clear if power is removed from
usually clear if power is removed from
the line for a short time and then
the line for a short time and then
restored.
restored.
3
Introduction
Introduction
•If one, or two, or all three phases break or
if insulators break due to fatigue or
inclement weather This fault is called a
permanent fault since it will remain after a
quick power removing.
•Approximately 75% of all faults in power
systems are transient in nature.
•Selecting an appropriate circuit breaker
(type, size, etc.) is important…
4
3
3
-
-
Phase fault current transients in
Phase fault current transients in
synchronous generators
synchronous generators
When a symmetrical 3-phase fault
occurs at the terminals of a
synchronous generator, the resulting
current flow in the phases of the
generator can appear as shown.
The current can be represented as a
transient DC component added on top
of a symmetrical AC component.
Therefore, while before the fault, only
AC voltages and currents were present
within the generator, immediately after
the fault, both AC and DC currents are
present.
5
Fault current transients in machines
Fault current transients in machines


When the fault occurs, the AC component
When the fault occurs, the AC component
of current jumps to a very large value, but
of current jumps to a very large value, but
the total current cannot change instantly
the total current cannot change instantly
since the series inductance of the machine
since the series inductance of the machine
will prevent this from happening.
will prevent this from happening.


The transient DC component of current is
The transient DC component of current is
just large enough such that the sum of the
just large enough such that the sum of the
AC and DC components just
AC and DC components just
after
after
the fault
the fault
equals the AC current just
equals the AC current just
before
before
the fault.
the fault.


Since the instantaneous values of current
Since the instantaneous values of current
at the moment of the fault are different in
at the moment of the fault are different in
each phase, the magnitude of DC
each phase, the magnitude of DC
components will be different in different
components will be different in different
phases.
phases.
Th
DC
t
d
f
i
l
Th
DC
t
d
f
i
l
6
Symmetrical AC component of the fault current:
Symmetrical AC component of the fault current:


There are three periods of time:
There are three periods of time:


Sub
Sub
-
-
transient period: first cycle or so after the fault
transient period: first cycle or so after the fault


AC current is very
AC current is very
large and falls rapidly;
large and falls rapidly;


Transient period: current falls at a slower rate;
Transient period: current falls at a slower rate;


Steady
Steady
-
-
state period: current reaches its steady value.
state period: current reaches its steady value.


It is possible to determine the time constants for the sub
It is possible to determine the time constants f
o
r the sub
-
-
transient
transient
and transient periods .
and transient periods .
7
Fault current transients in machines
Fault current transients in machines


The AC current flowing in the generator during the sub
The AC current flowing in the generator during the sub
-
-
transient period is called the sub
transient period is called the sub
-
-
transient current and is
transient current and is
denoted by
denoted by
I
I


.
.
The time constant
The time constant
of the sub
of the sub
-
-
transient current is
transient current is
denoted by
denoted by
T
T


and it can be determined from the slope. This
and it can be determined from the slope. This
current may be 10 times the steady
current may be 10 times the steady
-
-
state fault current.
state fault current.


The AC current flowing in the generator during the transient
The AC current flowing in the generator during the transient
period is called the transient current and is denoted by
period is called the transient current and is denoted by
I
I


. The
. The
time constant
time constant
of the transient current is denoted by
of the transient current is denoted by
T
T


. This
. This
current is often as much as 5 times the steady
current is often as much as 5 times the steady
-
-
state fault
state fault
current.
current.


After the transient period, the fault current reaches a steady
After the transient period, the fault current reaches a steady
-
-
state condition
state condition
I
I
ss
ss
.
.
This current is obtained by dividing the
This current is obtained by dividing the
induced voltage by the synchronous reactance:
induced voltage by the synchronous reactance:
A
ss
s
E
I
X
=
8
Fault current transients in machines
Fault current transients in machines


The
The
rms
rms
magnitude of the AC fault current in a
magnitude of the AC fault current in a
synchronous generator varies over time as
synchronous generator varies over time as


The sub
The sub
-
-
transient and transient
transient and transient
reactances
reactances
are defined as
are defined as
the ratio of the internal generated voltage to the sub
the ratio of the internal generated voltage to the sub
-
-
transient and transient current components:
transient and transient current components:
"
"
A
E
X
I
=
(
)
(
)
(
)
"'
"''
tTtT
s
sss
I
tIIeIIeI
−−
=−+−+
'
'
A
E
X
I
=
s
A
s
I
E
X=
9
Fault current transients
Fault current transients
Example 12-1:
A 100 MVA, 13.8 kV, Y-connected, 3 phase 60 Hz synchronous
generator is operating at the rated voltage and no load when a 3phase fault occurs
at its terminals. Its reactancesper unit to the machine’s own base are
1.00'0.25"0.12
s
XXX= = =
and the time constants are
'1.10"0.04TsTs
=
=
The initial DC component in this machine averages 50 percent of the initial AC
component.
a)What is the AC component of current in this generator the instant after the fault?
b)What is the total current (AC + DC) in the generator right afterthe fault occurs?
c)What will the AC component of the current be after 2 cycles? After 5 s?
10
Fault current transients
Fault current transients
The base current of the generator can be computed as
,
,
100,000,000
4,184
3313,800
base
Lbase
Lbase
S
I
A
V
===

The subtransient, transient, and steady-state currents are (per-unit and Amps)
1.0
"8.33334,900
"0.12
1.0
'4,700
'0.25
1.0
14,184
1.0
A
A
A
ss
s
E
IpuA
X
E
IpuA
X
E
IpuA
X
=== =
=== = 16
=== =
11
Fault current transients
Fault current transients
a)The initial AC component of current is I”= 34,900 A.
b)The total current (AC and DC) at the beginning of the fault is
1.5"52,350
tot
I
IA
=
=
c) The AC component of current as a function of time is
()()
()
0.041.1
"'
"''18,20012,5164,184
tt
tt
TT
ssss
I
tIIeIIeIeeA
− −
− −
=−+−+=⋅+⋅+
After 2 cycles t= 1/30 s and the total AC current is
1
7,91012,1424,18424,236
30
IA
⎛⎞
=
++=
⎜⎟
⎝⎠
At 5 s, the current reduces to
(
)
501334,1844,317IA
=
++=
12
Fault current transients
Fault current transients
Example 12-2:
Two generators are
connected in parallel to the low-
voltage side of a transformer.
Generators G1
and G2
are each rated
at 50 MVA, 13.8 kV, with a
subtransientresistance of 0.2 pu.
Transformer T1
is rated at 100 MVA,
13.8/115 kV with a series reactance of
0.08 puand negligible resistance.
Assume that initially the voltage on the high side of the transformer is 120 kV, that
the transformer is unloaded, and that there are no circulating currents between the
generators.
Calculate the subtransientfault current that will flow if a 3 phase fault occurs at the
high-voltage side of transformer.
13
Fault current transients
Fault current transients
Let choose the per-unit base values for this power system to be 100 MVA and 115
kV at the high-voltage side and 13.8 kV at the low-voltage side of the transformer.
The subtransientreactance of the two generators to the system base is
2
given
new
newgiven
newgiven
V
S
ZZ
VS
⎛⎞
⎛⎞
=
⎜⎟
⎜⎟
⎜⎟
⎝⎠
⎝⎠
The reactance of the transformer is already given on the system base, it will not
change
2
""
12
13,800100,000
0.20.4
13,80050,000
XXpu
⎛⎞⎛⎞
==⋅=
⎜⎟⎜⎟
⎝⎠⎝⎠
0.08
T
X
pu
=

Therefore:
14
Fault current transients
Fault current transients
The per-unit voltage on the high-voltage side of the transformer is
120,000
1.044
115,000
pu
actualvalue
Vpu
basevalue


===

Since there is no load on the system,
the voltage at the terminals of each
generator, and the internal generated
voltage of each generator must also
be 1.044 pu. The per-phase per-unit
equivalent circuit of the system is
We observe that the phases of
internal generated voltages are
arbitrarily chosen as 00. The phase
angles of both voltages must be the
same since the generators were
working in parallel.
15
Fault current transients
Fault current transients
To find the subtransientfault current, we need to solve for the voltage at the
bus 1 of the system. To find this voltage, we must convert firstthe per-unit
impedances to admittances, and the voltage sources to equivalentcurrent
sources. The Theveninimpedance of each generator is ZTh
= j0.4, so the
short-circuit current of each generator is
1.0440
2.6190
04
oc
sc
th
V
I
Zj

°
=
==∠−°
The
equivalent
circuit
16
Fault current transients
Fault current transients
Then the node equation for voltage V1
(
)
(
)
(
)
111
1
2.52.512.52.61902.6190
5.2290
0.2980
17.5
VjVjVj
V
j
−+−+−=∠−°+∠−°
∠°
==∠°

Therefore, the subtransientcurrent in the fault is
(
)
1
12.53.72990
F
I
Vjpu
=
−=∠−°
Since the base current at the high-voltage side of the transformer is
3,
,
100,000,000
502
33115,000
base
base
LLbase
S
I
A
V
φ
===

,
3.7295021,872
FFpubase
I
IIA
=
=⋅=
the subtransientfault current will be
17
Fault current calculations using the bus
Fault current calculations using the bus
impedance matrix
impedance matrix


So far, we considered simple circuits. To
So far, we considered simple circuits. To
determine the fault current in a system:
determine the fault current in a system:


Create a per
Create a per
-
-
phase per
phase per
-
-
unit equivalent circuit of the power
unit equivalent circuit of the power
system using either sub
system using either sub
-
-
transient
transient
reactances
reactances
(if
(if
subtransient
subtransient
currents are needed) or transient
currents are needed) or transient
reactances
reactances
(if transient
(if transient
currents are needed).
currents are needed).


Add a short circuit between one node of the equivalent circuit
Add a short circuit between one node of the equivalent circuit
and the neutral and calculate the current flow through that shor
and the neutral and calculate the current flow through that shor
t
t
by standard analysis.
by standard analys
i
s.


This approach always works but can get
This approach always works but can get
complex while dealing with complex
complex while dealing with complex
systems. Therefore, a nodal analysis
systems. Therefore, a nodal analysis
technique will be used.
technique will be used.


We introduce a new voltage source
We introduce a new voltage source
in the
in the
system to represent the effects of a fault
system to represent the effects of a fault
at a bus. By solving for the currents
at a bus. By solving for the currents
i
t
d
d
b
t
hi
dditi
l
lt
i
t
d
d
b
t
hi
dditi
l
lt
18
Fault current calculations using the
Fault current calculations using the
impedance matrix
impedance matrix
Let us consider a power system
shown.
Assuming that we need to find the
sub-transient fault current at some
node in the system, we need to create
a per-phase, per-unit equivalent
circuit using sub-transient reactances
X”. Additionally, we assume that the
system is initially unloaded, making
the voltages behind sub-transient
reactances
1
2
"
"
10
10
A
A
Epu
Epu
=∠°
=∠°
19
Fault current calculations using the
Fault current calculations using the
impedance matrix
impedance matrix
The resulting equivalent circuit is shown.
Suppose that we need to determine the
sub-transient fault current at bus 2 when
a symmetrical 3 phase fault occurs on
that bus.
20
Fault current calculations using the
Fault current calculations using the
impedance matrix
impedance matrix
Before the fault, the voltage on bus 2
was Vf. If we introduce a voltage source
of value Vf
between bus 2 and the
neutral, nothing will change in the
system.
Since the system operates normally
before the fault, there will be no current
If”through that source.
21
Fault current calculations using the
Fault current calculations using the
impedance matrix
impedance matrix
Assume that we create a short circuit on
bus 2, which forces the voltage on bus 2 to
0. This is the same as inserting an
additional voltage source of value -V
f
in
series with the existing voltage source. This
make the total voltage at bus 2 equal to 0.
With this additional voltage source, there
will be a fault current If”, which is entirely
due to the insertion of the new voltage
source to the system. Therefore, we can
use superposition to analyze the effects of
the new voltage source on the system.
The resulting current If”will be the current
for the entire power system, since the other
sources in the system produced a net zero
current.
22
Fault current calculations using the
Fault current calculations using the
impedance matrix
impedance matrix
If all voltage sources except –Vf”are
set to zero and the impedances are
converted to admittances, the power
system appears as shown.
For this system, we can construct
the bus admittance matrix as
discussed previously:
16.2125.006.667
5.012.55.02.5
05.013.3335.0
6.6672.55.014.167
bus
jjj
jjjj
Y
jjj
jjjj

⎡⎤
⎢⎥

⎢⎥
=
⎢⎥

⎢⎥

⎣⎦
The nodal equation describing this power system is
bus
YV=I
23
Fault current calculations using the
Fault current calculations using the
impedance matrix
impedance matrix
With all other voltage sources set to zero, the voltage at bus 2is –V
f,
and the current entering the bus 2 is –I
f”. Therefore, the nodal
equation becomes
11121314
1
"
21222324 31323334
3
41424344
4
0
0
0
f
f
YYYY
V
YYYY
V
I
YYYY
V
YYYY
V
Δ
⎡⎤
⎡⎤


⎢⎥
⎢⎥




⎢⎥
⎢⎥


=
⎢⎥
⎢⎥


Δ
⎢⎥
⎢⎥


Δ


⎣⎦
⎣⎦
where ΔV1, ΔV3, and ΔV4
are the changes in the voltages at those
busses due to the current –If”injected at bus 2 by the fault.
The solution to is found as
-1
busbus
V=YI=ZI
24
Fault current calculations using the
Fault current calculations using the
impedance matrix
impedance matrix
11121314
1
"
21222324 31323334
3
41424344
4
0
0
0
f
f
ZZZZ
V
ZZZZ
V
I
ZZZZ
V
ZZZZ
V
Δ
⎡⎤
⎡⎤


⎢⎥
⎢⎥




⎢⎥
⎢⎥


=
⎢⎥
⎢⎥


Δ
⎢⎥
⎢⎥


Δ


⎣⎦
⎣⎦
where Zbus
= Ybus-1. Since only bus 2 has current injected at it, the
system reduces to
"
"
112
"
332
"
2
4
2
42
Which, in the case considered, is
f
f
f
f
f
VZI
VZI
V
I
ZI
VZ
=−
=−
=−
−=−



25
Fault current calculations using the
Fault current calculations using the
impedance matrix
impedance matrix
Therefore, the fault current at bus 2 is just the prefaultvoltage V
f
at
bus 2 divided by Z22, the driving point impedance at bus 2.
"
22
f
f
V
I
Z
=
The voltage differences at each of the nodes due to the fault current
can be calculated by
"
12
112
22
2
"
32
332
22
"
42
442
22
f
f
ff
f
f
f
f
Z
VZIV
Z
VVV
Z
VZIV
Z
Z
VZIV
Z
=−=−
=−=−
=−=−
=−=−




26
Fault current calculations using the Z
Fault current calculations using the Z
-
-
bus matrix
bus matrix
Assuming that the power system was running at no load conditions
before the fault, it is easy to calculate the voltages at every bus
during the fault. At no load, the voltage will be the same on every
bus in the power system, so the voltage on every bus in the system
is Vf. The change in voltage on every bus caused by the fault current
–If”is specified , so the total voltage during the fault is
12
22
12
22
1
2
32
3
22
4
1
2
32
3
22
4
442
22
2
22
1
0
1
1
f
ff
f
ff
f
ff
ff
f
f
Z
V
Z
VV
V
V
VV
V
Z
V
VV
V
Z
V
Z
Z
V
V
V
Z
V
Z
V
Z
Z
V
V
Z
V
Z
⎡⎤

⎢⎥
⎢⎥
Δ
⎡⎤⎡⎤
⎡⎤

⎢⎥
⎢⎥⎢⎥
⎢⎥
Δ
⎢⎥
⎢⎥⎢⎥
⎢⎥
=







⎡⎤


⎢⎥


⎢⎥
=



⎢⎥


⎢⎥


⎣⎦






+=+
⎢⎥
⎢⎥⎢⎥

⎢⎥
Δ
⎢⎥
⎢⎥⎢⎥
⎢⎥
Δ
⎢⎥
⎢⎥⎢⎥
⎣⎦
⎣⎦⎣⎦
⎢⎥

⎢⎥
⎣⎦⎦
Therefore, we can calculate the voltage at every bus in the
power system during the fault from a knowledge of the pre-fault
voltage at the faulted bus and the bus impedance matrix!
27
Fault current calculations using the Z
Fault current calculations using the Z
-
-
bus matrix
bus matrix


Once these bus voltages are known, we can calculate the fault
Once these bus voltages are known, we can calculate the fault
current flowing in the transmission line using bus voltages and
current flowing in the transmission line using bus voltages and
the bus admittance matrix.
the bus admittance matrix.


The general procedure for finding the bus voltages and line
The general procedure for finding the bus voltages and line
currents during a symmetrical 3 phase fault is as follows:
currents during a symmetrical 3 phase fault is as follows:


Create a per
Create a per
-
-
unit equivalent circuit of the power system. Include
unit equivalent circuit of the power system. Include
subtransient
subtransient
reactances
reactances
of each synchronous and induction
of each synchronous and induction
machine when looking for
machine when looking for
subtransient
subtransient
fault currents; include
fault currents; include
transient
transient
reactances
reactances
of each synchronous machine when looking
of each synchronous machine when looking
for transient fault currents.
for transient fault currents.


Calculate the bus admittance matrix. Include the admittances of
Calculate the bus admittance matrix. Include the admittances of
all transmission lines, transformers, etc. between busses includ
all transmission lines, transformers, etc. between busses includ
ing
ing
the admittances of the loads or generators themselves at each
the admittances of the loads or generators themselves at each
bus.
bus.


Calculate the bus impedance matrix
Calculate the bus impedance matrix Zbus
Zbus
as inverse of the bus
as inverse of the bus
admittance matrix.
admittance matrix.


Assume that the power system is at no load and determine the
Assume that the power system is at no load and determine the
voltage at every bus, which will be the same for every bus and t
voltage at every bus, which will be the same for every bus and t
he
he
same as the internal voltage of the generators in the system. Th
same as the internal voltage of the generators in the system. Th
is
is
is the pre
is the pre
-
-
fault voltage
fault voltage
Vf
Vf
.
.
28
Fault current calculations using the Z
Fault current calculations using the Z
-
-
bus matrix
bus matrix
5. Calculate the current at the faulted bus ias
"
,
f
fi
ii
V
I
Z
=
6. Calculate the voltages at each bus during the fault. The voltage at
bus jduring a symmetrical 3 phase fault at the bus iis found as
1
ji
jf
ii
Z
VV
Z
⎛⎞
=−
⎜⎟
⎝⎠
7. Calculate the currents in any desired transmission line during the
fault. The current through a line between bus iand bus jis found as
(
)
i
j
i
j
i
j
I
YVV=−−
29
Fault current calculations using the Z
Fault current calculations using the Z
-
-
bus matrix
bus matrix
•Example 12-4:
The previous power system is working at no load
when a symmetrical 3 phase fault is developed on bus 2. a)
Calculate the per-unit subtransientfault current I
f”at bus 2. b)
Calculate the per-unit voltage at every bus in the system during the
subtransientperiod. c) Calculate the per-unit current I1
flowing in line
1 during the subtransientperiod of the fault.
Following the 7 steps discussed before, we write:
1.The per-phase per-unit equivalent circuit was shown earlier.
2.The bus admittance matrix was previously calculated as
16.2125.006.667
5.012.55.02.5
05.013.3335.0
6.6672.55.014.167
bus
jjj
jjjj
Y
jjj
jjjj

⎡⎤
⎢⎥

⎢⎥
=
⎢⎥

⎢⎥

⎣⎦
30
Fault current calculations using the Z
Fault current calculations using the Z
-
-
bus matrix
bus matrix
3. The bus impedance matrix calculated using Matlabas the inverse
of Ybus
is
00.151500.123200.093400.1260 00.123200.210400.132100.1417
00.093400.132100.172600.1282
00.126000.141700.128200.2001
bus
jjjj jjjj
Z
jjjj
jjjj
++++
⎡⎤
⎢⎥
++++
⎢⎥
=
⎢⎥
++++
⎢⎥
++++
⎣⎦
4. For the given power system, the no-load voltage at every bus is
equal to the pre-fault voltage at the bus that is
1.000
f
Vpu
=
∠°
5. The current at the faulted bus is computed as
"
,2
22
1.000
4.75390
0.2104
f
f
V
Ipu
Zj
∠°
===∠−°
31
Fault current calculations using the Z
Fault current calculations using the Z
-
-
bus matrix
bus matrix
6. The voltage at bus jduring a symmetrical 3 phase fault at bus I
can be found as
12
1
22
2
32
3
22
42
4
22
0.1232
111.000.4140
0.2104
0.00
0.1321
111.000.3720
0.2104
0.1417
111.000.3270
0.2104
f
f
f
Z
j
VVpu
Zj
Vpu
Z
j
VVpu
Zj
Zj
VVpu
Zj
⎛⎞
⎛⎞
=−=−⋅∠°=∠°
⎜⎟
⎜⎟
⎝⎠
⎝⎠
=∠°
⎛⎞
⎛⎞
=−=−⋅∠°=∠°
⎜⎟
⎜⎟
⎝⎠
⎝⎠
⎛⎞
⎛⎞
=−=−⋅∠°=∠°
⎜⎟
⎜⎟
⎝⎠
⎝⎠
7. The current through the transmission line 1 is computed as
(
)
(
)
121212
0.41400.005.02.0790IYVVjpu=−−=∠°−∠°⋅=∠°
That’s it!