An Overviewof HVDC Transmission Systems by fiw10869

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									U       Department of Electrical
    I    Engineering                             ECE529: Session 44



           An Overview of HVDC Transmission Systems


                                    ECE 529
                                   Spring 2009




HVDC Transmission Systems                                         1/24
U       Department of Electrical
    I    Engineering                                                      ECE529: Session 44



                 Steady-State HVDC Converter Representation
  • Steady state equivalent circuit
                                        I dc
                                                R
                                                 dc        Inverter


                                         V                         V
                                          dcr                       dci



                            Rectifier                 f (α ,Idc , |V |)
                                                                   ac

  • Have fast, direct control over α (firing delay angle)
  • Vdc = Vdo ∗ cosα (firing delay angle) where Vdo = const ∗ |VLL|
  • Some control of |Vac| with tap changing transformer
  • DC current indirectly controlled by changing α


HVDC Transmission Systems                                                                  2/24
U       Department of Electrical
    I    Engineering                                                                          ECE529: Session 44



                                   Basic Six-Pulse Converter
  • Based on line commutated, current source converter
  • Thyristors used as devices
  • Converter with stiff current source on dc side
  • Stiff voltage source on ac side (turns off thyristors)
  • Basic 6-pulse bridge:
                                                                            Smoothing Reactor
                                                                                                    Ls            I
                                                      A                                                               dc
                                                          +
                                                                                     1    3     5
                                                          ean (t)           Xc
                                                          -
                                                              -                                                            +
                                                  -               ebn (t)                                V                 -
                                                      -                                                      dc
                                              ecn (t)                 +
                                          +                       B                 4     6     2
                                      C


                                                          Transformer Inductance


HVDC Transmission Systems                                                                                                  3/24
U       Department of Electrical
    I    Engineering                                           ECE529: Session 44



                                   Basic Six-Pulse Converter
  • Initially assume: 1) Ideal ac sources, 2) ideal switches,
    3) Xc = 0, and 4) Ls → ∞ source)
  • AC side of converter has an ideal voltage source, dc side of converter
    has an ideal current source
  • Apply Kirchhoff’s Current Law:
    i1 + i3 + i5 = Idc (one switch always closed)
    i2 + i4 + i6 = Idc
  • Apply Kirchhoff’s Voltage Law:
    ean + ebn + ecn = 0 (balanced 3 phase set)
  • Since Xc = 0, only one switch in (1,3,5) can be closed with a switch in
    (2,4,6)



HVDC Transmission Systems                                                       4/24
U       Department of Electrical
    I    Engineering                                            ECE529: Session 44



                            Basic Six-Pulse Converter (cont.)
  • Allowable combinations:
        1 with (2 or 6) (4 shorts dc bus)
        3 with (2 or 4)
        5 with (4 or 6)
        2 with (1 or 5)
        4 with (1 or 3)
        6 with (3 or 5)
  • Need to determine a switching sequence
  • Start from assumption of positive phase sequence
  • Typical current waveforms:
    ia       |     |
    ib       |     |
    ic      |      |

HVDC Transmission Systems                                                        5/24
U       Department of Electrical
    I    Engineering                                            ECE529: Session 44



                            Basic Six-Pulse Converter (cont.)
  • Possible sequences:
    Top three switches: 1-3-5-1 or 1-5-3-1
    Bottom three switches: 4-6-2-4 or 4-2-6-4
                    +     −
  • Assume: Vdc = Vdc - Vdc
                 +                   −
    Switch # Vdc         Switch # Vdc
        1    ean(t)         4     ean(t)
        3    ebn(t)         6     ebn(t)
        5    ecn(t)         2     ecn(t)




HVDC Transmission Systems                                                        6/24
U       Department of Electrical
    I    Engineering                                                                              ECE529: Session 44



                                               Basic Six-Pulse Converter (cont.)
  • Positive sequence (α = 0, 1-3-5-1 and 4-6-2-4)


                                    10.00
                                    6.00
                                    2.00
                      Voltage (V)

                                    -2.00
                                    -6.00
                                    -10.00




                                             0.0    6.67   13.33               20.00   26.66   33.33

                                                                   Time (mS)




  • Negative sequence (α = 0, 1-5-3-1 and 4-2-6-4)
                                    10.00
                                    6.00
                                    2.00
                      Voltage (V)

                                    -2.00
                                    -6.00
                                    -10.00




                                             0.0    6.67   13.33               20.00   26.66   33.33

                                                                   Time (mS)




HVDC Transmission Systems                                                                                          7/24
U       Department of Electrical
    I    Engineering                                                        ECE529: Session 44



                            Basic Six-Pulse Converter (cont.)
                                                                                    I dc
                                                 5        1        3           5

                                                                                    I dc
                                                     6         2           4
  • Phase currents:
                                                                               +     −
                                           Switch Combination             Vdc=Vdc - Vdc

                                                         1-6           eab= ean - ebn = Vdc
                                                         1-2              eac= ean - ecn
  • Look at the line voltages:
                                                         3-2              ebc= ebn - ecn
                                                         3-4              eba = ebn - ean
                                                         5-4              eca= ecn - ean
                                                         5-6              ecb= ecn - ebn
                                    √
  • If α = 0, then Vdc =           3 2
                                    π
                                       |VLL |   = 1.35|VLL| We define this as Vdo

HVDC Transmission Systems                                                                     8/24
U       Department of Electrical
    I    Engineering                                                                                 ECE529: Session 44



                                                   Controlled Firing of Thyristors
  • Now add a firing delay (α) for the thyristors. Same delay for all 6
    switches

                                    10.00
                                    6.00
                                    2.00
                      Voltage (V)

                                    -2.00
                                    -6.00
                                    -10.00




                                             0.0       6.67   13.33               20.00   26.66   33.33

                                                                      Time (mS)




HVDC Transmission Systems                                                                                             9/24
U       Department of Electrical
    I    Engineering                                                  ECE529: Session 44



                             Controlled Firing of Thyristors
                  Ê π +α √                      √                π +α
  • Vdc =     3
              − π +α
              π
                    6
                       2|VLL |cos(θ)dθ     =   3 2
                                                π  |VLL |sin(θ) |− π +α
                                                                 6
                6                                                  6
                   √
  •   Then Vdc = π |VLL |cosα
                  3 2
                     √
  •   Define Vdo = π |VLL|
                    3 2

  • Therefore Vdc = Vdo cosα
  • α = 0 → diode bridge Vdc = Vdo
    0 ≤ α < 90 → rectifier Vdc > 0
    α = 90 → P =0 Vdc = 0
    90 < α ≤ 180 → inverter Vdc < 0
  • Current does not reverse




HVDC Transmission Systems                                                             10/24
U       Department of Electrical
    I    Engineering                                                 ECE529: Session 44



                                   Commutation Overlap
  • Now add source inductance (Lc = 0)
                                               Ls           I
                                                                dc

                                   1   3   5            +
                    Lc

                                                                                    +
                                                    V                     V         -
                                                    dc r                    dc i
                                   4   6   2
                                                    -




HVDC Transmission Systems                                                            11/24
U       Department of Electrical
    I    Engineering                                                    ECE529: Session 44



                            Current Transfer Between Switches
  • Current does not fall to zero immediately in ac side inductance
  • Temporarily create line to line short
                                               Ls


                1                   3


             Xc                     Xc

          ean (t)                    ebn (t)


                                ecn (t)                     α   µ

                            Xc                                                I dc
                                                    1               3
                            2                  Ls
                                                        0


HVDC Transmission Systems                                                               12/24
U       Department of Electrical
    I    Engineering                                          ECE529: Session 44



                    Current Transfer Between Switches (cont.)
  • What happens if α gets to big (i.e. α ⇒ 180◦)?
                          I dc
        1                   1
            0                        This is called a commutation failure
  • Thyristor 3 fails to turn on and thyristor 1 fails to turn off
  • This is more common if Lc is large, which is the case looking into a
    “weaker” ac system
  • Normally corrects during next interval, although often have a second
    failure when thyristor 5 turns on, “double commutation failure”




HVDC Transmission Systems                                                     13/24
U       Department of Electrical
    I    Engineering                                                            ECE529: Session 44



                        Output Voltage During Commutation
                            +
  • Switch 1 contribution: Vdc1 = ean - Lc di1
                                           dt
                            +
  • Switch 3 contribution: Vdc3 = ebn - Lc di3
                                           dt
                                               +     +
  • During overlap we see the average between Vdc1 &Vdc3
                    +     +
                   Vdc1 +Vdc3
         +
     So Vdc    =        2       = ean+ebn -
                                     2
                                              Lc
                                              2
                                                    di1
                                                    dt    + di3
                                                            dt
                            di1 +di3
  • i1 + i3 = Idc, so          dt      =0
                                                   di1 +di3
     But since its a linear network:                  dt      =   di1
                                                                  dt    + di3 = 0
                                                                          dt
                                                           eac+ebc
  • So: Vdc = Vdc − ecn = ean+ebn − ecn =
               +
                             2                                2




HVDC Transmission Systems                                                                       14/24
U       Department of Electrical
    I    Engineering                                                    ECE529: Session 44



                             Average DC Voltage with Overlap
                             √            √           √
  • Recall: Vdo =          = π |VLL |
                            3 2
                                     =    π |Vφ |
                                         3 6
                                                      π |Em |
                                                     3 3

    where Em is peak line to neutral voltage
  • Then we find:
                    3           α+µ 3                  α+ π √
                                                          3                 π
              Vdc =                     Emcosθdθ +              3|Em|cos(θ − )dθ
                    π         α     2                 α+µ                   6
                                   √
  • Leading to: Vdc =             3 3
                                      E [cosα + cos(α + µ)]
                                   2π m




  • Or Vdc = V2 [cosα + cos(α + µ)]
              do




HVDC Transmission Systems                                                               15/24
U       Department of Electrical
    I    Engineering                                                                             ECE529: Session 44



                           Average DC Current
  • Start out with Lc = 0 and α = 0 for now
                                                              Fundamental Current Component
                                            120


                                                                   210               330 360

                                   0   30                150 180




  • Firing delay simply adds a phase shift to the current (always lagging),
    and cosα = cosφ
                                                         E                              E
                                                         an                                 an
                                            30
                                                                     90
                                                 I
                                                     a
                                                                     Ia


  • Fundamental Component
                     π                                                          π            √
                 2 2               2                                            3           2 3
          i1pk =       iacos(θ)dθ = Idc                                          cos(θ)dθ =     Idc
                 π −π
                    2
                                   π                                          −π
                                                                               3
                                                                                             π
                   √
  • |I1RMS| =        6
                    π Idc

HVDC Transmission Systems                                                                                        16/24
U       Department of Electrical
    I    Engineering                                          ECE529: Session 44



                                     Average DC Current
                         √
  • Then i1(t) =        2 π3 Idccos(ωt        − α)
  • Also: P = 3I1RMSVPcosφ = VdcIdc
                        √
    So: 3I1RMSVPcosφ = π VPcosαIdc
                       3 6
                            √
  • So: |Ia1RMS| =           6
                               I
                            π dc
                                    as expected
                                               √
  • During overlap: Idc = Ic =                  3Em
                                               2ωLc
                                                      = 2Xc
                                                        eLL


  • i3(t) = Ic(cosα − cosωt) with α ≤ ωt ≤ α + µ
    where ωt = α + µ at the end of the commutation interval
  • So average current is: Idc = Ic(cosα − cos(α + µ))
                     √
                                    3 |Vp |
  • Also: Ic =        3Em
                     2ωLc
                                =   2 Xc
                                                |VLL
                                              = √2X|
                                                      c




HVDC Transmission Systems                                                     17/24
U       Department of Electrical
    I    Engineering                                                 ECE529: Session 44



                                  Average DC Circuit Equations
  • We have the following equations:
                              Vdo
                       Vdc =       [cosα + cos(α + µ)]
                                2
                        Idc = Ic(cosα − cos(α + µ))
                                √
                              3 2
                       Vdo =         |VLL|
                                 π
                               |VLL|     πVdo
                         Ic = √        =
                                 2Xc       6Xc
  • Substitute for the cos(α + µ) in the Vdc equation
  • Then Vdc = Vdo cosα − Vdo Idc
                          2Ic

  • Where Vdo =
          2Ic
                            Vdo
                         πVdo     = π Xc = Rc (called the commutating “resistance”)
                                    3
                        2 6X
                            c

  • So Vdc = Vdo cosα − IdcRc

HVDC Transmission Systems                                                            18/24
U       Department of Electrical
    I    Engineering                                                            ECE529: Session 44



                                        Average DC circuit
  • Rc represents a current dependent voltage drop due to overlap
  • Rc does not represent any energy dissipation!
  • So using Vdc = Vdo cosα − IdcRc we get:

                                                  R line

                                                                 Rc
                                   Rc




                                        V cos α            V cos α
                                                            do
                                         do




                            RECTIFIER                                INVERTER




HVDC Transmission Systems                                                                       19/24
U       Department of Electrical
    I    Engineering                                         ECE529: Session 44



                                   Inverter Operation
  • α + µ + γ = π Covers positive half cycle of voltage
  • γ is defined as the extinction angle
  • γo is minimum extinction angle for proper turn-off
    Typical values: 15◦ → 20◦
  • So α + µ ≤ 180◦ − γo gives limits for control settings
  • Replace α with 180◦ − µ − γ in averaged equations
    * note: cos(180◦ − θ) = −cos(θ)




HVDC Transmission Systems                                                    20/24
U       Department of Electrical
    I    Engineering                                                 ECE529: Session 44



                                   Inverter Operation (cont.)
  • Generate equations in terms on γ instead of α

                                   Vdo
                      Vdc =            [cosα + cos(α + µ)]
                                    2
                                   Vdo
                      Vdc =            [cos(180◦ − µ − γ) + cos(180◦ − γ)]
                                    2
                                     Vdo
                      Vdc =        − [cos(γ + µ) + cos(γ)]
                                      2
                       Idc =       Ic(cosα − cos(α + µ))
                       Idc =       Ic(cosγ − cos(γ + µ))

  • Sign reversal in voltage equation expected for inverter




HVDC Transmission Systems                                                            21/24
U       Department of Electrical
    I    Engineering                                                    ECE529: Session 44



                            Effect of Overlap on Power Transfer
  • Pac = 3I1RMSVp cosφ
                                                 √
                                                3 6Vp
  •   Pdc = Idc ∗Vdo cosα+cos(α+µ)
                           2             =   Idc π      cosα+cos(α+µ)
                                                              2
                 √
  •   I1RMS = Idc π6
  • Then cosφ = cosα+cos(α+µ) = Vdo
                      2
                                Vdc


  • Note: overlap equations change if µ > 60◦, covered in
    Kimbark, Direct Current Transmission: Volume I. Wiley, 1971.




HVDC Transmission Systems                                                               22/24
U       Department of Electrical
    I    Engineering                                                              ECE529: Session 44



                                       Transformer Loading
                                                                      DC System
                            AC system



                                             x (p.u.)

  • Xc = ωLc where X → 12 − 20%
  • ZB = VB
         I
           B

                                          2
                                        3VBφ                     2
                                                                VLL
                        VBφ
  • Xc = ZBX =           IB
                            X      =   3VBφ IB
                                               X   , this is      B
                                                               MVA3φB
               2
              VBL
  • Xc =     MVAB3φ
                    X
                        √
  • IB = I1RMS =         6
                           I
                        π dcB
                                                    √
  • So MVAB3φ = 3VBφIB = 3VBφ                         6
                                                     π IdcB     = Vdo IdcB
                                                                   R



HVDC Transmission Systems                                                                         23/24
U       Department of Electrical
    I    Engineering                                                      ECE529: Session 44



                               Transformer Loading (cont.)
  • Need to use true RMS (transformers sees all harmonic components)
                     Ê 2π
  • IRMS =         1 3 2
                       I dθ
                   π o dc
                                   = Idc   2
                                           3

  • So MVAB3φ = 3IRMSVφ = πVdo IdcB
                          3
                            R

                       2                         √
                      VBL
  • Then Xc =        MVAB3φ X      with Vdo =   3 2
                                                 π VBL
              πVdo
  • Then ZB = 6IdcB
  • Then from Vdc = Vdo cosα − Idc π Xc we get
                                   3

    Vdc = Vdo cosα − Idc π (ZBX )
                         3

    Vdc = Vdo cosα −Vdo X IIdcB
                           2
                             dc



  • Leading to a “per unit” expression:            Vdc
                                                   Vdo   = cosα − X Idc
                                                                  2




HVDC Transmission Systems                                                                 24/24

								
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