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7/27/2019 Lecture Sc

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2/8

5

Symmetrical short circuit transienton a transmission line

Resulting currents

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New features in this course

9 DC components

Top values

Modeling of synchronous machines

Other types of faults

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Maximum current (R is small)

Resulting currents

2 2peak x y ac RMSI I I I = + =

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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

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9 DC components

9 Top values

Modeling of synchronous machines


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Short circuit ona synchronousmachine

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Short circuit on a synchronous machine


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Short circuit on a synchronous machine

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9 DC components

9 Top values

9 Modeling of synchronous machines


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Differentfaulttypes

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Single line-to-ground fault, SLGF


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Double line-to-ground faults

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Symmetrical components (basic course)

The transformation matrix for symmetrical

components

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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

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Example12.4

Pos. seq.

Zero seq.

Neg. seq.= pos. -Vf

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Considerations regardingneutral grounding

Fault currents Voltage on healthy phases

Suppression of arcing faults

Interference with communication (EMC)

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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

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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

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Isolated Neutral (Resistorgrounded)

Small systems Small fault current (SLGF)

High voltages on healthy phases at faults(SLGF)

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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

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Solidly grounded systems

Subtransmission and transmission systems

Large fault currents (SLGF)

Voltages on healthy phases limited (SLGF)

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