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