Siemens - Machine Protection Setting Exercises

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Transcript of Siemens - Machine Protection Setting Exercises

  • Machine Protection -Setting Exercises

  • Exercise 1: Single line diagram

    7UM62.

    -T150 MVA, YNd11

    110 52.5% / 11 kVuT(1) = 8 %

    3 110 kV, 50 Hz

    side 1iL1,2,3

    uL1,2,3

    G3~

    -G146.6 MVA

    11kV 7.5%50Hz

    300/1A, 20VA5P20, Rct=1.2

    3000/1A, 20VA5P20, Rct=12

    300/1A, 20VA5P20, Rct=1.2

    side 2iL1,2,3

    IEE2

    UE

    IEE1

    Excit.7XT71

    TD1 (REF)TD2 (REF)TD3 (

  • Exercise 1: Neutral transformer circuits

    400/5A1FS515VA

    P1

    RL

    P2

    1B1

    1A4

    1A3

    1A1

    7XT34

    1B4 4A1

    4A3

    7XT33

    20 HzBandpass

    20 HzGenerator

    S1S2

    7UM62.

    R13 UER14

    J8 IEE1J7

    max. 3A (20Hz)

    Protection cubicle

    .. mA

    20Hz500V 3kV/ 11

    1A2

    330

    660

    660

    Burden < 0.5

  • Exercise 1: Required protection elements

    -- Threshold supervision (for Decoupling)64G-1 90% Stator Earth Fault U0> (calculated)64G-2 100% Stator Earth Fault (20Hz principle)64R Rotor Earth Fault (1-3Hz principle)87 Differential Protection46 Unbalanced Load (negative sequence)40 Under excitation49 Thermal Overload (Stator)24 Overexcitation (V/Hz)21 Impedance Protection78 1) Out of Step (loss of synchronism)

    1) Option

  • Exercise 1: Device configuration (partly)

    for Decoupling

    for Decoupling

  • Exercise 1: Power System Data 1 (1/4)

    1)

    1) Neutral transformer is high resistive

    (CT) ratio)(U ratioTransf.) Neutral (ratioSEF RFactor 0275 divider2 =

    5.04400/55/2

    500V311000V/SEF RFactor 0275

    2

    =

    =

  • Exercise 1: Power System Data 1 (2/4)

  • Exercise 1: Power System Data 1 (3/4)

    )/ratio(UUU

    UU

    Factor UE 0224dividerNTsec

    NTprim

    Esec

    VTprim==

    31.7V/(5/2) 500

    3V/ 11000Factor UE 0224 ==

    Refer to Setting Options for the UE Input and their Impacton the Protection Functions (refer to slide No. 10)

  • Exercise 1: Power System Data 1 (4/4)

  • Setting Options for the UE Input and their Impacton the Protection Functions

  • Exercise 2: Generator Electrical Data (1/2)

  • Exercise 2: Generator Electrical Data (2/2)

  • Exercise 2.1: Calculation of load resistor and neutral transf. (1/3)

    CK = 10 nFUHV = 110 kVULV = 11 kVprotected zone (stator) = 90% K = (100%-90%)/100% = 0.1

    500V 3kV/ 11

    90% (K=0.1)

    CG

    1L1

    1L2

    1L3

    CL CTr

    CK

    RL

    UHV

    7UM62.

    UE

    ULV

    equivalent circuit

    ~

    CK

    UE0RL primCE

    CE = CG + CL + CTr neglectedRL prim

  • Exercise 2.1: Calculation of load resistor and neutral transf. (2/3)

    Formula symbols and definitions used:UE0 Displacement voltage on HV side of unit transformerFe Earthing factor, here: solid earthed Fe = 0.8ICprim Interference current on neutral transformer primary sideICsec Interference current on neutral transformer secondary sideCK Total capacitance (3x phase capacitance) between HV and LV side of

    unit transformer (coupling capacitance)f Rated frequencyTRNT Transformation ratio of the neutral transformerUNTPrim Primary rated voltage of neutral transformerUNTSec Secondary rated voltage of neutral transformerRL Load resistorK Protected zone factorFS Safety factor FS = 2SNT(20s)Required output of neutral transformer when burdened by RL for 20 sIRLmax Current of load resistor R at 100 % UE

  • Exercise 2.1: Calculation of load resistor and neutral transf. (3/3)

    [ ]

    L

    NTsecRLmax

    L

    2NTsec

    NT(20s)

    sec C

    NTsec

    s

    L

    NTprim Csec C

    NTsec

    NTprimNT

    KE0prim C

    HVeE0

    RUI

    VAR

    U =S

    IU

    FK

    = R

    TR I =IUU

    TR

    Cfpi2 U= I3/UFU

    =

    =

    =

    [ ]A 40.65

    12.3V 500I

    kVA 20.3VA 12.3V500

    =S

    12.3A 2.03V 500

    20.1

    = R

    A 2.037.12A 0.16 =I

    12.7V 500

    3V/ 11000TR

    A 0.16V

    sA 101050s2pi 50800V= I

    kV 50.83kV/ 1100.8U

    RLmax

    22

    NT(20s)

    L

    sec C

    NT

    91-prim C

    E0

    ==

    =

    =

    =

    ==

    =

    ==

    TRRUI

    :terminalsgenerator at fault Earth

    NTL

    NTsecprim E

    = ok 10A A 3.212.7 12.3V 500I prim E

  • Exercise 2.2: Decoupling - Example with Threshold supervision (1/3)

    -dP (-50%)

    (3IN, p.u.)

    &

    CFCMV2

  • Exercise 2.2: Decoupling - Example with Threshold supervision (2/3)

  • Exercise 2.2: Decoupling - Example with Threshold supervision (3/3)

    Settings: primary values

  • Exercise 2.3: Settings for 90% Stator Earthfault - U0>calculated (1/2)

    UL1

    UL3

    UL2

    90%

    L1

    L2

    L3

    Settings (primary value):5002: U0prim> = 100% - 90% = 10.0 %5003: T = 0.30 sec

    Fuse Failure Monitor (FFM) to be enabledto block 90% SEF Element via CFC

  • Exercise 2.3: Settings for 90% Stator Earthfault - U0>calculated (2/2)

    Settings: primary values

  • Exercise 2.4: Settings for unbalanced load (1/3)

    (from Manufacturer)

  • Exercise 2.4: Settings for unbalanced load (2/3)

    Settings in primary values:1702: I2prim> = I2perm prim / IN Machine = 8.0 %1704: Kprim = (I2/IN)2t = 20 s1705: tCooldown = Kprim /(I2perm prim /IN Machine)2 = 20s/0.082 = 3125 s1706: I2prim>> = 60.0%1707: T I2>> = 3.00 sec

    Conversion to secondary values:1702: I2sec> = I2prim> IN Machine/IN CT = 8.0% 2446A/3000A = 6.5 %1704: Ksec = Kprim (IN Machine/IN CT)2 = 20s(2446A/3000A)2 = 20s0.664 = 13.30 s1706: I2sec>> = I2prim>> IN Machine/IN CT = 60.0% 2446A/3000A 49 %

  • Exercise 2.4: Settings for unbalanced load (3/3)

    Settings: primary values

  • Exercise 2.5: Settings for under excitation protection (1/4)Generator capability diagram

  • Exercise 2.5: Settings for under excitation protection (2/4)

    Settings in primary values (from Capability Diagram) 3002: 1/xd1prim = 0.58tan(1) = 0.7/0.2 = 3.5 , arctan(1) = 1.2923003: 1 743005: 1/xd2prim = 0.443006: 2 = 90Conversion to secondary values:

    UNMACH = 11kV , UN VTprim = 11 kVINMACH = 46600kVA/(311kV) = 2446A IN CTprim = 3000A

    CTprim N

    VTprim N

    NMACH

    NMACH

    dMachdsec IU

    UI

    x

    1x

    1=

    0.815A 3000V11k

    kV 11A 2446IU

    UI

    CTprim N

    VTprim N

    NMACH

    NMACH=

    = 3002: 1/xd1sec = 0.580.815 0.473005: 1/xd2sec = 0.440.815 0.36

  • Exercise 2.5: Settings for under excitation protection (3/4)

    generatorshaft

    7XR6004 7UM62.

    K13 +

    TD1

    K14Excit.

    K15 +TD2

    K16

    20 k

    7XT71

    27

    25

    1921

    1517

    A3

    A11

    A6

    G3~

    20 k

    20 kB11

    B18

    B1420 k

    control

    meas.

    Example:LiYCY 4x1.5

    Protection cubicle

    3PP1326

    5001

    4 9 k

    2

    500

    3

    7UM62.

    K17 +

    TD3

    K18

    10F250V

    1)

    u

    0 EXC

    kU0.5Exc U:3013