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Transient Stability

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The Swing EquationApplication to Synchronous Machines Step-by-Step Solution Method

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  • TRANSMISSION & DISTRIBUTIONA Division of Global Power

    POWER SYSTEM STABILITY CALCULATION TRAININGD 2 T i t St bilitDay 2 - Transient Stability

    July5,2013Prepared by: Peter Anderson

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  • OUTLINE

    2

    OUTLINE

    The Swing Equation

    Application to Synchronous Machines

    Step-by-Step Solution Method eBoo

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  • THE SWING EQUATION

    3

    THE SWING EQUATIONPower-Angle Relationship

    d2 = d = 0.(Pm Pe)dt2 dt 2Hdt2 dt 2HH = Inertia Constant = J.2/2S in MW-s/MVA or seconds

    Metric: GD2 in kg-m2 622

    10.kVA

    )RPM()GD(48163.5=H

    Imperial: WR2 in lb.ft2 622

    10.kVA

    )RPM()WR(231.0=H

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  • ANALYSIS OF THE SWING EQUATION

    4

    ANALYSIS OF THE SWING EQUATIONd = 0.(Pm Pe)dt 2Hdt 2H

    In terms of short-term transient stability studiesIn terms of short-term transient stability studies,

    Pm is constantm

    Rotor acceleration is a function of the Electrical Power

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  • APPLICATION TO SYNCHRONOUS

    5

    MACHINESIncrease in Mechanical Power InputPm increased Rotor accelerates from 25 deg. to new operating point at 40 deg. 1.2

    1.4

    gRotor overshoots to 60 deg, where area above Pm equals the area below PmNow Pe >Pm and rotor

    0.8

    1

    Now Pe Pm and rotor decelerates towards =40 deg.Rotor will oscillate around =40 deg. 0.2

    0.4

    0.6

    40 deg.If the overshoot reaches =140 deg. The rotor will not be able to return to the new operating point and will slip

    0

    0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180

    operating point and will slip to the next pole position

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  • STEP BY STEP SOLUTION METHOD

    6

    STEP-BY-STEP SOLUTION METHODPm Increased

    (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12)T Pm Pe Pa Acceleration t1 t2

    (3)(4) (5)*k (6)*(7) deg/s (9)*(10) (2)+(11)0.0 0 Pm0 Pe0 Pm0Pe0 0 0 0=0 +0.0 0 Pm1 Pe0 Pm1Pe0 0 t/2 1 0+1=1 t 1 10.5 1 Pm1 Pe1 Pm1Pe1 1 t 2 1+2=2 t 2 21 2 Pm1 Pe2 Pm1Pe2 2 t 3 2+3=3 t 3 31.5 3 Pm1 Pe3 Pm1Pe3 3 t 4 3+4=4 t 4 42 4 Pm1 Pe4 Pm1Pe4 4 t 5 4+5=5 t 5 5

    0 8

    1

    1.2

    1.4

    0.2

    0.4

    0.6

    0.8

    0

    0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180

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  • STEP BY STEP SOLUTION METHOD

    7

    STEP-BY-STEP SOLUTION METHODPm Increased from 0.42 to 0.8 pu

    1.00

    1.20

    0.60

    0.80

    0.20

    0.40

    0.00

    0 30 60 90 120 150 180

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  • STEP BY STEP SOLUTION METHOD

    8

    STEP-BY-STEP SOLUTION METHODT Pm Pe Pa t1 t2 0 0 25 0 0 42 0 42 0 0 0 0 0 00.0 25.0 0.42 0.42 0 0 0 0 0.00 25.0 0.80 0.42 0.38 10.87 0.25 2.72 2.7 0.5 1.4 26.40.5 26.4 0.80 0.44 0.36 10.25 0.5 5.13 7.8 0.5 3.9 30.31 30.3 0.80 0.50 0.30 8.52 0.5 4.26 12.1 0.5 6.1 36.31.5 36.3 0.80 0.59 0.21 5.98 0.5 2.99 15.1 0.5 7.5 43.92 43.9 0.80 0.69 0.11 3.08 0.5 1.54 16.6 0.5 8.3 52.22.5 52.2 0.80 0.79 0.01 0.29 0.5 0.14 16.8 0.5 8.4 60.63 60.6 0.80 0.87 0.07 2.05 0.5 1.02 15.8 0.5 7.9 68.53.5 68.5 0.80 0.93 0.13 3.75 0.5 1.87 13.9 0.5 6.9 75.44 75.4 0.80 0.97 0.17 4.83 0.5 2.41 11.5 0.5 5.7 81.14.5 81.1 0.80 0.99 0.19 5.42 0.5 2.71 8.8 0.5 4.4 85.55 85.5 0.80 1.00 0.20 5.67 0.5 2.84 5.9 0.5 3.0 88.55.5 88.5 0.80 1.00 0.20 5.75 0.5 2.87 3.0 0.5 1.5 90.06 90.0 0.80 1.00 0.20 5.76 0.5 2.88 0.2 0.5 0.1 90.16.5 90.1 0.80 1.00 0.20 5.76 0.5 2.88 2.7 0.5 1.4 88.77 88.7 0.80 1.00 0.20 5.75 0.5 2.88 5.6 0.5 2.8 85.97 5 85 9 0 80 1 00 0 20 5 69 0 5 2 84 8 4 0 5 4 2 81 77.5 85.9 0.80 1.00 0.20 5.69 0.5 2.84 8.4 0.5 4.2 81.78 81.7 0.80 0.99 0.19 5.46 0.5 2.73 11.2 0.5 5.6 76.18.5 76.1 0.80 0.97 0.17 4.92 0.5 2.46 13.6 0.5 6.8 69.39 69.3 0.80 0.94 0.14 3.90 0.5 1.95 15.6 0.5 7.8 61.59.5 61.5 0.80 0.88 0.08 2.27 0.5 1.14 16.7 0.5 8.4 53.210 53.2 0.80 0.80 0.00 0.01 0.5 0.00 16.7 0.5 8.4 44.8

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  • STEP BY STEP SOLUTION METHOD

    9

    STEP-BY-STEP SOLUTION METHODPm Increased from 0.42 to 0.8 pu

    80 0

    90.0

    100.0

    50.0

    60.0

    70.0

    80.0

    A

    n

    g

    l

    e

    (

    d

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    )

    10 0

    20.0

    30.0

    40.0

    M

    a

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    h

    i

    n

    e

    0.0

    10.0

    0 2.5 5 7.5 10 12.5 15

    Time(s)

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  • TRANSMISSION & DISTRIBUTIONA Division of Global Power

    POWER SYSTEM STABILITY CALCULATION TRAININGD 3 T i t St bilitDay 3 - Transient Stability

    July8,2013Prepared by: Peter Anderson

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  • CASE STUDY: ANALYSIS OF A FAULT CASE

    2

    CASE STUDY: ANALYSIS OF A FAULT CASEL1

    L2

    3 h f lt li L2 l t t b3-phase fault on line L2, close to generator busFrequency = 60 HzInertia Constant = 3.0 sP f lt P A l C i /0 8Pre-fault Power Angle Curve = sin/0.8Post-fault Power Angle Curve = sin/1.1Fault duration = 0.1 sTime Step = 0.02 s

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  • CASE STUDY: ANALYSIS OF A FAULT CASE

    3

    CASE STUDY: ANALYSIS OF A FAULT CASE

    Step 1: Construct Power-Angle Curvesp g

    1.40

    1.00

    1.20

    0 40

    0.60

    0.80

    0.00

    0.20

    0.40

    0 30 60 90 120 150 180

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  • CASE STUDY: ANALYSIS OF A FAULT CASE

    4

    CASE STUDY: ANALYSIS OF A FAULT CASEStep 1: Construct Step-by-Step Table

    K 3600 =180*f/(H*S)/( )1 2 3 4 5 6 7 8 9 10 11 12

    Pa t1 t2 (3)(4) (5)*K (6)*(7) (9)N1+(8)N (9)*(10) (2)+(11)

    s deg pu pu pu deg/s/s s deg/s deg/s s deg deg0.0 39.8 0.80 0.80

    T Pm Pe

    0.0 39.8 0.80 0.800 39.8 0.80 0.00 0.80 2880 0.01 28.80 28.8 0.02 0.6 40.4

    0.02 40.4 0.80 0.00 0.80 2880 0.02 57.60 86.4 0.02 1.7 42.10.04 42.1 0.80 0.00 0.80 2880 0.02 57.60 144.0 0.02 2.9 45.00.06 45.0 0.80 0.00 0.80 2880 0.02 57.60 201.6 0.02 4.0 49.00.08 49.0 0.80 0.00 0.80 2880 0.02 57.60 259.2 0.02 5.2 54.20.08 49.0 0.80 0.00 0.80 2880 0.02 57.60 259.2 0.02 5.2 54.20.1 54.2 0.80 0.00 0.80 2880 0.01 28.80 288.00.1 54.2 0.80 0.76 0.04 152 0.01 1.52 289.5 0.02 5.79 60.00.12 60.0 0.80 0.81 0.01 52 0.02 1.04 200.6 0.02 4.0 64.00.14 64.0 0.80 0.85 0.05 179 0.02 3.58 197.0 0.02 3.9 67.90.16 67.9 0.80 0.88 0.08 291 0.02 5.83 191.2 0.02 3.8 71.80.16 67.9 0.80 0.88 0.08 291 0.02 5.83 191.2 0.02 3.8 71.80.18 71.8 0.80 0.91 0.11 389 0.02 7.78 183.4 0.02 3.7 75.40.2 75.4 0.80 0.93 0.13 471 0.02 9.42 174.0 0.02 3.5 78.90.22 78.9 0.80 0.95 0.15 539 0.02 10.77 163.2 0.02 3.26 82.20.24 82.2 0.80 0.96 0.16 592 0.02 11.85 151.3 0.02 3.03 85.20.26 85.2 0.80 0.98 0.18 634 0.02 12.68 138.7 0.02 2.77 88.00.26 85.2 0.80 0.98 0.18 634 0.02 12.68 138.7 0.02 2.77 88.00.28 88.0 0.80 0.98 0.18 665 0.02 13.30 125.4 0.02 2.51 90.50.3 90.5 0.80 0.99 0.19 687 0.02 13.74 111.6 0.02 2.23 92.7

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  • CASE STUDY: ANALYSIS OF A FAULT CASE

    5

    CASE STUDY: ANALYSIS OF A FAULT CASEStep 1: Plot Results

    100.0

    120.0

    60.0

    80.0

    h

    i

    n

    e

    A

    n

    g

    l

    e

    (

    d

    e

    g

    )

    20.0

    40.0

    M

    a

    c

    h

    0.0

    0 0.1 0.2 0.3 0.4 0.5 0.6

    Time(s)

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  • THREE PHASE SHORT CIRCUIT

    6

    THREE PHASE SHORT-CIRCUITSteady-state:Leakage Reactance of the machine (X )Leakage Reactance of the machine (Xl) Armature Reaction to the Fault Current (Xa)Synchronous Reactance Xd = Xl + Xay d l a

    Xl Xa

    E

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  • THREE PHASE SHORT CIRCUIT

    7

    THREE PHASE SHORT-CIRCUITAt the instant of the Fault:

    Leakage Reactance of the machine (Xl)Armature Reaction to the Fault Current (Xa)Since the Air Gap flux cannot change instantaneously, currents are induced in the field and damper windings (Xdw and Xf)f)Sub-transient Reactance Xd = Xl + 1/(1/Xa +1/Xdw + 1/Xf)

    Xl Xa

    XdwE Xf

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  • THREE PHASE SHORT CIRCUIT

    8

    THREE PHASE SHORT-CIRCUITShortly after the instant of the Fault:Leakage Reactance of the machine (Xl)Leakage Reactance of the machine (Xl)Armature Reaction to the Fault Current (Xa)Damper winding currents (low X/R ratio) die out quicklyq yTransient Reactance Xd = Xl + 1/(1/Xa + 1/Xf)

    E

    Xl Xa

    E Xf

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  • THREE PHASE SHORT CIRCUIT

    9

    THREE PHASE SHORT-CIRCUIT

    6

    2

    4

    (

    p

    u

    )

    Subtransient

    4

    2

    0

    C

    u

    r

    r

    e

    n

    t

    Transient

    Synchronous

    6

    4

    0 0.05 0.1 0.15 0.2 0.25 0.3Time(s)

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  • SYNCHRONOUS MACHINE MODELS

    10

    SYNCHRONOUS MACHINE MODELSSingle Phase Equivalent of a 3-phase Generator

    jXd I

    E U~

    Im

    EjXd IjXd.I

    Re U

    II

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  • SYNCHRONOUS MACHINE MODELS

    11

    SYNCHRONOUS MACHINE MODELSThree Possible Generator Models:Synchronous Model (Constant voltage behind Xd)Synchronous Model (Constant voltage behind Xd)Transient Model (Constant voltage behind Xd)Sub-transient Model (Constant/Variable voltage behind Xd))

    Sub-transient model allows exciter effects to be explicitly represented

    For each model, the prime mover can be represented as a fixed power model or a variable power model under the control of governor p gaction

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  • MODEL APPLICATION

    12

    MODEL APPLICATIONUse of Mixed Generator Models:Complex models used for machines of interestSimpler models used for remote machines

    Requires less dataRequires less dataSignificantly reduces the computing burden

    With the removal of computer limitations, it is recommended to use the sub-transient model for all generators

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  • TRANSMISSION & DISTRIBUTIONA Division of Global Power

    POWER SYSTEM STABILITY CALCULATION TRAININGD 4 T i t St bilitDay 4 - Transient Stability

    July9,2013Prepared by: Peter Anderson

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  • OUTLINE

    2

    OUTLINE Machine Differential Equations

    Exciter Differential Equations

    G Diff ti l E ti Governor Differential Equations

    Solution of Differential Equations

    Network Solution

    Sample Cases

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  • BASIC MODELS IN STABILITY STUDIES

    3

    BASIC MODELS IN STABILITY STUDIESSynchronous MachinesSwing Equations:Swing Equations:

    ( )aem0 p.DPP.H

    =p

    0=p

    H

    Flux Linkage Differential Equations:

    ( )[ ] ( )[ ]'dq'qq'd"dq"qq"d EIXX1=pEEIXX1=pE ( )[ ] ( )[ ]

    ( )[ ] ( )[ ]'qd'ddfd''q"qd"d'd'q""q

    dqqq'0q

    ddqqq"0q

    d

    EIXXE1

    =pEEIXXE1

    =pE

    EIXXT

    pEEIXXT

    pE

    ( )[ ] ( )[ ]qdddfd'0d

    qqdddq"0d

    qT

    pT

    p

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  • BASIC MODELS IN STABILITY STUDIES

    4

    BASIC MODELS IN STABILITY STUDIESSynchronous MachinesAlgebraic Model:Algebraic Model:

    VE.Y=II

    .XR

    =VE dqadd

    ( )FrameferenceReSystemtoFrameferenceReMachinefromtionTransforma

    VE.YII

    .RXVE qadqq

    ( ) ( )( )2/jexpI+I=jI+I qdMRI

    YY.E

    Y.V V

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  • BASIC MODELS IN STABILITY STUDIES

    5

    BASIC MODELS IN STABILITY STUDIESExciter

    ( )[ ]EVVVK1=pE fdfbtsEfd ( )[ ]

    ( )VpEKT1

    =pV

    EVVVKT

    pE

    fbfdff

    fb

    fdfbtsEE

    fd

    maxDpE;maxEVminE;V+KE

    =V fdE0tE

    0fds

    f

    Vs Exciter Limiter RateLimiter+

    Vt KE EfdEmax, Dmax1+TEp

    ,Emin Dmax

    Vfb pKf1+Tfp

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  • BASIC MODELS IN STABILITY STUDIES

    6

    BASIC MODELS IN STABILITY STUDIESGovernor

    ( ) ( )[ ]1 ( ) ( )[ ]

    [ ]mons

    m

    on00sc

    on

    PCT1

    =pP

    CR/PT1

    =pC

    limM

    s

    GP

    T

    0 Regulator Limiter Control Valve Turbine

    + 1 1 Con 1 Pm0.R 1+Tcp 1+Tsp

    Ps

    Glim

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  • SOLUTION OF DIFFERENTIAL EQUATIONS

    7

    SOLUTION OF DIFFERENTIAL EQUATIONSMethod of Integration:Implicit Trapezoidal Rule 70

    80

    90

    Implicit Trapezoidal Rule( ) ( ) ( )( )

    VariableIntegrable=Y

    pY+pY.2h

    +Y=Y th+tt)h+t(

    20

    30

    40

    50

    60

    Y

    LengthStepnIntegratio=h

    VariableIntegrableNon=X

    g

    I i T f i f Diff i l E i

    0

    10

    20

    2 4 6 8 10

    X

    Iterative Transformation of Differential Equations to Algebraic Equations:

    ( )h+t)h+t( X.YX+YC=Y ( )

    ( ) ( )( ) ( )( )[ ]

    ( )[ ]

    ttt

    h+t)h+t(

    h

    XF,YF.2h

    +Y=ttanCons=YC

    X.YX+YCY

    ( ) ( )( )[ ]h+th+t XF.2h

    =X.YX

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  • SOLUTION OF DIFFERENTIAL EQUATIONS

    8

    SOLUTION OF DIFFERENTIAL EQUATIONSSolution Algorithm:

    1 Establish Initial Conditions by Solving for Y and X at t=01. Establish Initial Conditions by Solving for Y and X at t=02. Solve for Y(t+h) using values of X(t) by means of Trapezoidal

    Rule3 Solve for X using values of Y from above3. Solve for X(t+h) using values of Y(t+h) from above4. Resolve for Y(t+h) using values of X(t+h)5. Repeat Steps 2, 3 & 4 until convergence is achieved6 Proceed to next time step6. Proceed to next time step

    Numerically StableAllows Larger Time Steps to be used than other methodsgIn Step 2 in practice, an extrapolated value of X(t) is used except immediately after a discontinuity usingX = 2X XX(t+h) = 2X(t) X(t-h)

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  • NETWORK SOLUTION

    9

    NETWORK SOLUTION1. Calculate Integrable Variables2 T f f M hi R f F t S t2. Transform from Machine Reference Frame to System

    Reference Frame3. Calculate Norton equivalent Nodal Injected Current4. Solve for V using I = Y.V where Y = Nodal Admittance4. Solve for V using I Y.V where Y Nodal Admittance

    Matrix5. Use New values for V to move to next time step6. Network Solution carried out simultaneously with

    Integration ProcessIntegration Process eBook

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    ou

  • SAMPLE CASES

    10

    SAMPLE CASES4x500MWST 250MWH

    GBUS1 GBUS2

    600MVA 0 975pf

    4x590MVA,21/400kV 1x295MVA,21/400kV

    600MVA,0.975pf

    100km

    HVBUS1

    400km

    100km 50km

    GBUS3

    HVBUS31x590MVA,21/400kV

    600MVA,0.975pf1200MVA,0.975pf

    500MWST

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  • FAULT POSITION 1

    11

    FAULT POSITION-1Line Fault close to Generator HV Bus

    BaseFaultatGBus1 BaseFaultatGBus2

    0

    30

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    90

    e

    (

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    1

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    G

    1

    )

    90

    60

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    0 50 100 150 200

    M

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    0 50 100 150 200

    M

    a

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    l

    e

    Time(cycles) G2 G3 Time(cycles) G2 G3

    60

    90

    w

    r

    t

    G

    1

    )

    BaseFaultatGBus3

    60

    30

    0

    30

    c

    h

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    n

    e

    A

    n

    g

    l

    e

    (

    d

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    r

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    e

    s

    w

    90

    60

    0 50 100 150 200

    M

    a

    c

    Time(cycles) G2 G3

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  • FAULT POSITION 2

    12

    FAULT POSITION-2Line Fault close to System HV Bus

    90BaseFaultatHVBus1

    90BaseFaultatHVBus3

    0

    30

    60

    g

    l

    e

    (

    d

    e

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    s

    w

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    G

    1

    )

    0

    30

    60

    n

    g

    l

    e

    (

    d

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    G

    1

    )

    90

    60

    30

    0 50 100 150 200

    M

    a

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    g

    90

    60

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    0 50 100 150 200

    M

    a

    c

    h

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    A

    n

    Time (cycles) G2 G3Time(cycles) G2 G3 Time(cycles)G2 G3

    10

    20

    s

    w

    r

    t

    G

    1

    )

    BaseAllFaults

    40

    30

    20

    10

    0

    a

    c

    h

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    A

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    g

    l

    e

    (

    d

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    r

    e

    e

    s

    50

    0 10 20 30 40 50

    M

    a

    Time(cycles)

    HVBus1 HVBus3 GBus1 GBus2 GBus3

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  • GENRATOR INERTIA

    13

    GENRATOR INERTIALine Fault close to Generator HV Bus (GBUS-3_

    10

    20

    w

    r

    t

    G

    1

    )

    G3Hincreased

    20

    10

    0

    g

    l

    e

    (

    d

    e

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    r

    e

    e

    s

    w

    40

    30

    20

    0 50 100 150 200

    M

    a

    c

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    e

    A

    n

    g

    0 50 100 150 200

    Time(cycles)

    G3 G30

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  • GENERATOR EXCITATION CONTROL

    14

    GENERATOR EXCITATION CONTROLLine Fault close to Generator HV Bus (GBUS-3_

    10

    20

    r

    t

    G

    1

    )

    Excitersadded

    20

    10

    0

    l

    e

    (

    d

    e

    g

    r

    e

    e

    s

    w

    r

    60

    50

    40

    30

    M

    a

    c

    h

    i

    n

    e

    A

    n

    g

    l

    60

    0 50 100 150 200

    Time(cycles) G30 G3

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  • GENERATOR REACTANCE

    15

    GENERATOR REACTANCELine Fault close to Generator HV Bus (GBUS-3_

    60

    90

    t

    G

    1

    )

    G3Xdreduced

    0

    30

    60

    e

    (

    d

    e

    g

    r

    e

    e

    s

    w

    r

    t

    60

    30

    M

    a

    c

    h

    i

    n

    e

    A

    n

    g

    l

    e

    90

    0 50 100 150 200

    Time(cycles) G3 G30

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  • GENERATOR MODEL TYPE

    16

    GENERATOR MODEL TYPELine Fault close to Generator HV Bus (GBUS-3_

    60

    90

    r

    t

    G

    1

    )

    G2FixedV/Xd'

    0

    30

    g

    l

    e

    (

    d

    e

    g

    r

    e

    e

    s

    w

    r

    90

    60

    30

    M

    a

    c

    h

    i

    n

    e

    A

    n

    g

    0 50 100 150 200Time(cycles)G2 G3 G20 G30

    eBoo

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  • FAULT CLEARING TIME

    17

    FAULT CLEARING TIMELine Fault close to Generator HV Bus (GBUS-3)

    120150180

    w

    r

    t

    G

    1

    )

    BaseAllFaults

    306090

    120

    l

    e

    (

    d

    e

    g

    r

    e

    e

    s

    w

    9060300

    M

    a

    c

    h

    i

    n

    e

    A

    n

    g

    90

    0 10 20 30 40 50Time(cycles)

    5cycles 7cycles 10cycles 12cycles

    eBoo

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  • LOAD REPRESENTATION

    18

    LOAD REPRESENTATIONLine Fault close to Generator HV Bus (GBUS-3_

    60

    90

    r

    t

    G

    1

    )

    ConstantPowerAllFaults

    0

    30

    60

    e

    (

    d

    e

    g

    r

    e

    e

    s

    w

    r

    30

    0

    M

    a

    c

    h

    i

    n

    e

    A

    n

    g

    l

    e

    60

    0 10 20 30 40 50

    M

    Time(cycles)

    3cycles 4cycles 5cycles3 cycles 4 cycles 5 cycles

    eBoo

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  • SUMMARY OF CASES

    19

    SUMMARY OF CASESParameter Stability

    Increase DecreaseIncrease DecreaseFaultPosition SystemDependentIncreasedInertia Y GeneratorExcitationControl Y DecreasedSynchronousReactance Y Si lifi d G t M d l YSimplifiedGeneratorModel Y

    IncreasedFaultClearingTime Y

    Decreased VoltageDependencyofLoad Y

    Fault Clearing Time & Voltage Dependency of Load have the most significant impacts on stabilitythe most significant impacts on stability

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