ground fault protection isolated neutral

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    Power System Protection

    systems

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    2

    Professor at Norwegian Univ. Science and

    Technology Dept. Electrical Engineering Power system transients and protection

    g vo age engneerng, s ress cacua ons

    Recent focus on Power Transformers

    Sabbatical at MTU

    , - [email protected]

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    Solid vs. isolated neutral

    + No over-voltages in faultsituations

    - Undefined voltages toground. Sqrt(3) rise

    - g au curren

    + Easy fault detection

    + Ground faults ersist

    .MOV dimensioning.

    + Low fault current

    Fast trip and reclosing

    - Poor power quality

    - Difficult faultdetection/location

    + Can continue to o erate- Extra stress

    during ground fault.Increased power quality

    down/broken conductors

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    Why is the fault current low insoate neutra systems

    capacitances. High return impedance.

    The most common fault

    Often temporary rees ranc es,

    snow/ice/wind,

    Birds

    Simultaneous faults.

    3 V

    1 2 03

    f

    fZ Z Z R

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    In Norway DyDy

    LV system 230 V IT MV system 12-24 kV

    -

    The 230 V IT system is gradually replaced by400 V TN solidl rounded due to risk ofundetected ground faults, double ground faults,and lower loss in a 400 V TN system.

    n ; requremen s o e ec a groun au Power quality is very important to the industry

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    CurrentI0: SumIa,Ib,Ic

    Summing the current numerically

    I0Loweraccuracy

    Residual connection

    Summation transformer

    3I0

    differences

    3I0

    (Toroidal/Rogowski coil)

    a, ,

    c

    Open delta 3U0

    Neutral point (isolated or resonance earth)U0

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    Normal to only consider ground capacitance

    symmetrical system,

    ignore conductive line charging no voltage drop along line

    imbalance in loads canceled b D-cou led transformers

    Fault current mainly returns in the two healthy phases

    Fault resistance important

    hVpa h = ej120VN

    Vpa

    h2Vpa

    gff

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    occurs For calculation of feeder current and volta es

    capacitances can be concentrated

    The volta e to round will rise for the healthphases

    Easy to detect that there is a ground fault (by

    looking atV0), difficult to detect where (bylooking at I0 in various feeders)

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    N A

    V

    +If

    3pa gV j CI

    fg 1 3 g fj C R

    V I C0

    1 3 3N

    g f g

    V Vj C R j C

    f

    A

    UN

    Rf=Rf=0

    BBefore

    Line-voltages

    unchanged

    B After

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    Same zero-sequence voltage in the entire

    system; selectivity is challenging The fault is traditionally located by

    sectionalizing (often manual)

    Z r n rr ntv ri l n th lin :I0

    If

    3I0U0 constant

    Rf

    pos.

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    From fault to zero sequence current

    VpaVN

    x l-x

    h2Vpa

    hVpa

    CgxCg(l-x)

    2

    ( ) ( ) ' ( )a N pa gI x V V j C l x

    Fault locationx=dx

    ( ) ( ) ' ( ) ( )

    3 '

    b N pa g

    c N pa g f

    x xI x V h V j C l x I H d x

    V C x x d

    03 ( ) ( ) 3 ' ( ) ( )

    3 ' ( ),a b c N g f

    N g

    I x I I I V j C l x I H d xV j C l x x d

    ignore voltage drop, no load-effect

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    Three radials fed by the same transformer

    Fault current in the faulty feeder is the negative sumof the current in the two healthy feeders

    01 1N gI V j C

    02 2N gI V j C

    03 01 02( )I I I

    With two radials the currents are equal in size sodirectional protection is required

    -

    current protection becomes possible

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    - Often the preferred solution

    Measure zero sequence voltage and zerose uence current in each radial

    Then for each radial:

    01

    I02

    Healthyfeeder

    V0 03 V0

    I0nForward

    rppng zoneFaulty feeder

    direction

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    V0>Vlim (type 59G over-voltage relay)

    and in correct quadrant (3rd)? > m ype rec ona ove -curren re ay

    +

    67N (I0>Ilim)

    52 trip-

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    -22 kV

    y28 mi

    50 miles line in totalTotal capacitance is50 miCg [F/mi]

    14 mi

    8 mi

    5 nF/mi assumed in

    this case.

    Fault current inde endent on location

    cable systems.

    Depends on fault resistance

    2.44 54.7 A 3 kIU ,'

    ,max 3.0 90 A @ 042441/ ( 3 50 )f

    fff g

    IIR jR j mi C

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    ar on o e ear au curren measure oreach radial

    28 mi

    22/5050/50

    14 mi

    8 mi8/50

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    Partition of the earth fault current measured for

    each radial Health lines 1&3 feed 28 + 8 arts into line 2

    28/5028 mi

    8 mi

    36/50

    8/50

    50/50

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    Partition of the earth fault current measured for

    each radial Health lines 1&2 feed 28 + 14 arts into line 3

    28/5028 mi

    14 mi

    14/50

    8 mi

    42/50