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5.3 the Optimal Power Flow Problem

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    OPF IntroductionThe idea of minimizing the total generation cost under full consideration of the

    power flow equations goes back to the decade between 1960 and 1970. The

    mathematical formulation and first solutions of this optimization problem have been

    given by Carpentier (1962) and Tinney, Dommel (1967).

    5.3.15.3 The optimal power flow problem

    Looking back to the Lagrangian function which has been used in economic dispatch;

    N1i)PPP()P(FLi

    GilossLGii

    i K=++=

    We can realize, that the power flow in the network has been reduced to one simple

    equality constraint: =+

    iGilossL 0PPP

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    The idea of Carpentier was to replace the simplifying constraint

    load plus losses equals generation by the power flow equations for each node in

    the network. This formulation of the minimization problem including the power flow

    equations is called Optimal Power Flow (OPF).

    Important applications of the OPF today:

    Calculation of the optimum generation pattern to achieve the minimum total cost

    of generation. Calculation of the optimum generation pattern to minimize air pollution by thermal

    generating units.

    Using the network losses as objective, the OPF algorithm can find the optimum

    reactive power injections of generators, optimum settings of transformer taps and

    switched capacitors.

    5.3.25.3 The optimal power flow problem

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    5.3.35.3 The optimal power flow problem

    PG2PL2 =100 MW

    V2 = 224 kV

    PG1PL1 =100 MW

    V1 = 224 kV; 1=0

    PG3PL3 = 600 MW

    V3 = 224 kV

    1 2

    3

    ~

    ~ ~L12

    L23L13

    OPF Example:Find the operating pattern with minimum total fuel costs if the total load of 800 MW is

    distributed on the nodes PL1 = 100 MW, PL2 = 100 MW, PL3 = 600 MW.

    ( ) [ ]( ) [ ]

    ( ) [ ]h$2

    3G3G3G3

    h$2

    2G2G2G2

    h

    $2

    1G1G1G1

    P012.0P162000PF

    P01.0P141500PFP008.0P121000PF

    ++=

    ++=

    ++=

    Line impedances:( )+=== 6164841412

    2 31 31 2

    .j.ZZZ

    Line shunt admittances:

    0=+ ikik jBSGS

    Network voltage:

    kV224VVV 321 ===

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    5.3.45.3 The optimal power flow problem

    Solution:

    2

    33

    2

    22

    2

    11

    01201620000101415000080121000 GGGGGG P.PP.PP.PF ++++++++=

    Objective function:

    Equality constraints (power flow equations, active power):

    ( ) ( )[ ] 32102 ,,iPPsinBcosGVVGV LiGiiiiiiiii ==+++

    ( ) ( ) ( ) ( ) ( ) 01001311 3311 3211 2211 21 31 2

    =++++ GPsinbcosasinbcosaaa

    ( ) ( ) ( ) ( ) ( ) 01002322 3322 3122 1122 12 32 1

    =++++ GPsinbcosasinbcosaaa

    ( ) ( ) ( ) ( ) ( ) 06003233 2233 2133 1133 13 23 1

    =++++ GPsinbcosasinbcosaaa

    MW.aaaaaa 42473 22 33 11 32 11 2

    ======

    MW.bbbbbb 99683 22 33 11 32 11 2

    ======

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    5.3.55.3 The optimal power flow problem

    Variables:

    Lagrange function:

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

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

    ( ) ( ) ( ) ( ) ( )[ ]600xxxsinbxxcosaxsinbxcosaaa

    100xxxsinbxxcosaxsinbxcosaaa

    100xxsinbxcosaxsinbxcosaaa

    x012.0x162000x01.0x141500x008.0x121000L

    34532453253153132313

    25423542342142123212

    151351341241213121

    233

    212

    211

    +++++

    +++++

    +++++

    ++++++++=

    3

    2

    1

    3

    2

    3

    2

    1

    3

    2

    1

    5

    4

    3

    2

    1

    =

    G

    G

    G

    P

    PP

    x

    x

    x

    xx

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    5.3.65.3 The optimal power flow problem

    Necessary conditions for extremum:

    ( ) ( )[ ]

    ( ) ( ) ( ) ( )[ ]

    ( ) ( )[ ] 0453 2453 23

    542 3542 342 142 12

    41 241 21

    =+

    ++++

    xxcosbxxsina

    xxcosbxxsinaxcosbxsina

    xcosbxsina:x

    L0

    4

    =

    :x

    L0

    5

    =

    :x

    L0

    1

    =

    0016012

    11

    =+ x. (1)

    :xL 0

    2

    = 002014

    22

    =+ x. (2)

    :x

    L0

    3

    =

    0024016

    33

    =+ x. (2)

    (5)

    (4)

    ( ) ( )[ ]

    ( ) ( )[ ]

    ( ) ( ) ( ) ( )[ ] 0xxcosbxxsinaxcosbxsina

    xxcosbxxsina

    xcosbxsina

    453245325315313

    542354232

    5135131

    =++++

    +

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    5.3.75.3 The optimal power flow problem

    :L

    0

    1

    =

    ( ) 0

    5411

    =x,x,xh

    (6)( ) ( ) ( ) ( ) ( ) 0100151 351 341 241 21 31 2 =++++ xxsinbxcosaxsinbxcosaaa

    :L

    0

    2

    =

    (7)

    ( ) 05422

    =x,x,xh

    ( ) ( ) ( ) ( ) ( ) 01002542 3542 342 142 12 32 1

    =++++ xxxsinbxxcosaxsinbxcosaaa

    :

    L0

    3

    =

    (8)

    ( )0

    5433

    =x,x,xh

    ( ) ( ) ( ) ( ) ( ) 06003453 2453 253 153 13 23 1

    =++++ xxxsinbxxcosaxsinbxcosaaa

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    5.3.85.3 The optimal power flow problem

    000

    X

    h

    X

    h100

    000X

    h

    X

    h010

    000X

    h

    X

    h001

    X

    h

    X

    h

    X

    h

    XX

    L

    XX

    L

    000

    X

    h

    X

    h

    X

    h

    XX

    L

    XX

    L000

    10000024.000

    010000020.00

    0010000016.0

    0G

    GL

    5

    3

    4

    3

    5

    2

    4

    2

    5

    1

    4

    1

    5

    3

    5

    2

    5

    1

    55

    2

    54

    24

    3

    4

    2

    4

    1

    45

    2

    44

    2

    T

    2

    =

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    5.3.95.3 The optimal power flow problem

    [ ] [ ] [ ]

    [ ] [ ] [ ]( )

    ( )

    ( )5433

    5422

    5411

    321

    321

    33

    22

    11

    024016

    020014

    016012

    x,x,xh

    x,x,xh

    x,x,xh

    x.

    x.

    x.

    h

    L

    ++

    ++

    +

    +

    +

    =

    Vector of right

    hand side:

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    5.3.105.3 The optimal power flow problem

    63901

    .x =

    Result of

    iterative

    procedure:

    62372

    .x =

    91953

    .x =

    052104

    .x =

    242405 .x =

    250181

    .=

    752182 .=

    701203

    .=

    MW.PG 63901 =

    MW.PG 62372 =

    MW.PG 91953 =

    = 98722

    .

    = 889133 .

    MWh$.25018

    1=

    MWh

    $

    .75218

    2

    =

    MWh$.70120

    3=

    h

    $

    .F9617892=

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    5.3.115.3 The optimal power flow problem

    PG1=390.6 MW

    V1 = 224 kV ; 1=0

    PL1 =100 MW

    1=18.250 $/MWh 1 2

    3

    ~

    ~ ~

    Ploss=24.06 MW

    F = 17892.96 $/h

    PG2=237.6 MWV2 = 224 kV ; 2=-2.987

    PL2 =100 MW

    2=18.752 $/MWh

    PG3=195.9 MW

    V3 = 224 kV ; 3=-13.889

    PL3 =600 MW3=20.701 $/MWh

    50.82 50.15

    239.81

    225.35

    187.72

    178

    .79

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    5.3.125.3 The optimal power flow problem

    Locational Marginal Price (LMP)

    Cost to supply the next unit of energy in the most economic way at a particular location in

    the network

    PG1 = 0.351; PG2 = 0.248; PG3 = 0.420PL3=1

    PG1 = 0.320; PG2 = 0.414; PG3 = 0.249PL2=1

    PG1 = 0.471; PG2 = 0.256; PG3 = 0.234PL1=1

    LMPPGi (result of OPF calculation)PLiNode

    1

    1

    25018 ==

    MWh

    $

    L

    .P

    F

    2

    2

    75218 ==

    MWh

    $

    L

    .P

    F

    3

    3

    70120 ==

    MWh

    $

    L

    .P

    F

    1

    2

    3


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