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7/27/2019 Lecture Sc
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7/27/2019 Lecture Sc
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5
Symmetrical short circuit transienton a transmission line
Resulting currents
6
New features in this course
9 DC components
Top values
Modeling of synchronous machines
Other types of faults
7
Maximum current (R is small)
Resulting currents
2 2peak x y ac RMSI I I I = + =
8
Maximum RMS current (R is small)
2 2
RMS ac RMS dc RMSI I I = +
2
dc RMS Y Y I I I = =
2
2 2 2 31 322 2
X XRMS Y X X ac RMSI II I I I I
= + = + = =
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Different fault angles in differentphases => Maximum not in allphases at the same time
10
New features in this course
9 DC components
9 Top values
Modeling of synchronous machines
Other types of faults
11
Short circuit ona synchronousmachine
12
Short circuit on a synchronous machine
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Short circuit on a synchronous machine
14
New features in this course
9 DC components
9 Top values
9 Modeling of synchronous machines
Other types of faults
15
Differentfaulttypes
16
Single line-to-ground fault, SLGF
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Double line-to-ground faults
18
Symmetrical components (basic course)
The transformation matrix for symmetrical
components
19
Example 12.4
Find the subtransient currents and the line-to-line voltages at the fault under subtransientconditions when a double line-to-ground fault
with Zf=0 occurs at the terminals of machine 2in the system below. Assume that the system isunloaded and operating at rated voltage whenthe fault occurs. Neglect resistance.
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1. Build an impedance diagram each for the positive-, negativeand zero-sequence systems, for the entire network except for
the un-symmetrical load.2. Calculate the Thvenin-equivalent for the positive-,negative-
and zero-sequences at the bus where the un-symmetricalload is located.
3. Calculate the positive-, negative- and zero- sequencecurrents through the load.
4. Calculate the positive-, negative- and zero- sequencevoltages at the locations that are of interest
5. Calculate the positive-, negative- and zero-sequence currentsthrough components that are of interest
6. Transform those symmetrical components to phase-quantities that are asked for.
Analysis of a network with one un-symmetrical load (basic course)
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Example12.4
Pos. seq.
Zero seq.
Neg. seq.= pos. -Vf
22
Example12.4
Pos. seq.
Zero seq.
Neg. seq.= pos. -Vf
23
Considerations regardingneutral grounding
Fault currents Voltage on healthy phases
Suppression of arcing faults
Interference with communication (EMC)
24
Assume a grid
Z=?
YZ
Loads
Z=?
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Neutral grounding
Isolated neutral (no grounding)
Grounding through resistor
Grounding through reactor (Petersen coil)
Solidly grounded
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Isolated NeutralResistor grounded
Very high Z
1
1 2
33
1shU
I j CU
Z Zj C
= + +
YZ
Long lines => large C => large capacitive Ish-1
C
Resistor grounded => identify faulty part
Healthy phases 3 phaseU U=
Ua=0
27
Isolated Neutral (Resistorgrounded)
Small systems Small fault current (SLGF)
High voltages on healthy phases at faults(SLGF)
28
Petersen coilZ = jX
0
1 13
1
3
Z j Xj C
j CjX
= = = +
&
YZ
Long lines => large C
C
if
Healthy phases 3 phaseU U=
1 1
3 3C X
X C
= =
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Grounded through reactor(Petersen coil)
Distribution networks
Arcing faults could be clearedautomatically
Small fault currents (SLGF)
High voltages on healthy phases (SLGF)
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Solidly groundedZ = 0
YZ
Long lines => large C
C
Healthy phasesphase
U U=
1
1 21 2
3 3
1(3 0 )
sh
TT
U UI
Z Z ZZ Z Z
j C
= =+ ++ + +&
= large
31
Solidly grounded systems
Subtransmission and transmission systems
Large fault currents (SLGF)
Voltages on healthy phases limited (SLGF)