Exercise Earth Fault in Cable Network
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Transcript of Exercise Earth Fault in Cable Network
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7/30/2019 Exercise Earth Fault in Cable Network
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Technology for Decentralised Generation and Storage Part II Project work
Grid Assets
Brandenburg University of Technology
Department of Power Distribution and High Voltage Technology Pfeiffer 2
Solution
First step: Calculation of impedances
110-kV-grid( )
6MVA2000
kV110Z1Q 05
2
,==
1102 = 0,05110
6,05Z1Q ==
0,0498Z0,995X 1Q1Q ==
0,0049X0,1R 1Q1Q ==
TransformerCorrection factor for transformer impedance will be neglected
( )0,551
MVA20
kV10,50,1Z
2
1T ==
( )0,055
MVA20
kV10,50,01R
2
1T ==
0,548RZX2
1T
2
1T1T ==
0,521X0,95X 1T0T == (Z0m is there included)
Cable 10,48km10/km0,0967
2
1R1K ==
0,9km10/km0,1792
1X1K ==
F9,12km10F/km0,4562C1K ==
Cable 2 0,48km5/km0,0967R1K ==
0,9km5/km0,179X1K ==
F2km5F/km0,456C1K 28,==
Zero-sequence impedances of the cable:
Because data for the zero-sequence impedance of the cable are not available, we use the
following estimated values:
1K0K R4R =
1K0K X4X =
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Technology for Decentralised Generation and Storage Part II Project work
Grid Assets
Brandenburg University of Technology
Department of Power Distribution and High Voltage Technology Pfeiffer 3
Second step: Definition of the symmetrical component networks
Zero-sequence system
Positive-sequence system
Negative-sequence system
Calculation of the impedances (from capacitances)
698jF4,56314sj
1
Cj
1Z
1C=
==
2792jF1,14314sj
1
Cj
1Z
1C=
==
559jF5,7314sj
1
Cj
1
Z 1C ===
Third step: Interconnection of the symmetrical component networks
From lecture EDS II we know:
210 III ==
0F210 IZ3UUU =++
This means, the symmetrical component networks have to be connected in series at faultlocation.
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Technology for Decentralised Generation and Storage Part II Project work
Grid Assets
Brandenburg University of Technology
Department of Power Distribution and High Voltage Technology Pfeiffer 4
In the next step we can simplify the three equivalent circuit diagrams.
Zero-sequence system
Due to the very large values of the impedances of the capacitances we can neglect the
impedances of cable 1 and cable 4 and get the parallel connection of the four capacitances:
( )279j
F1,142F4,562314sj
1
Cj
1Z
1C=
+==
Positive-sequence system
Negative-sequence system
The simplified equivalent component networks we can now interconnect:
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Technology for Decentralised Generation and Storage Part II Project work
Grid Assets
Brandenburg University of Technology
Department of Power Distribution and High Voltage Technology Pfeiffer 5
It is to be seen, that
- the impedance in the zero-sequence system is equivalent to the zero capacitance
- the shunt impedance in the positive sequence system is equivalent to the effective
capacitance
- the positive current is not flowing over the effective capacitance, because the
impedance in the parallel branch is smaller
- the impedance in the negative sequence system is very small and can therefore be
neglected
Fourth step:Calculation of the currents in 012-system
Due to the fact, that the voltage in negative-sequence system is negligible we can assume,
that10 UU =
( )Aj21kA0,02069
j2793
kV10I0 =
= (completely capacitive current)
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Technology for Decentralised Generation and Storage Part II Project work
Grid Assets
Brandenburg University of Technology
Department of Power Distribution and High Voltage Technology Pfeiffer 6
Fifth step:Calculation of earth fault current
Aj63I3I 0CE ==
Verification: A623kV10F11,4s3143C3I 1ECE === YU
Fifth step:Calculation of the voltages in 012-system
3
kV10U0
3
kV10U1
0U2
Sixth step:Calculation of the voltages in abc-system
003
kV10
3
kV10UUUU 210a =++=++=
03
kV10a
3
kV10aUUaUU 221
20b ++=++=
( )kVja10
3
kV103ja
3
kV101aU 2
b
===
03
kV10a
3
kV10UaaUUU 2
210c ++=++=
( ) kV10ja3
kV103ja
3
kV101aU 22c ===
Seventh step:Calculation of earth fault factor
3
3
kV10
kV10==
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Technology for Decentralised Generation and Storage Part II Project work
Grid Assets
Brandenburg University of Technology
Department of Power Distribution and High Voltage Technology Pfeiffer 7
Eighth step:Phasor diagrams