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Increase of transmission line losses caused by higher

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					    Increase of transmission line losses caused by higher harmonic components
evaluated by orthogonal decomposition of three-phase currents in the time domain
                            1
                             Miran Rošer, 2Gorazd Štumberger, 2Matej Toman, 2Drago Dolinar
                                                        1
                                                          Elektro Celje d.d.
                                                 Vrunčeva 2a, 3000 Celje, Slovenia
                        2
                       University of Maribor, Faculty of Electrical Engineering and Computer Science
                                          Smetanova 17, 2000 Maribor, Slovenia
                   phone:+386 2 220 7075, fax:+386 2 220 7272, e-mail: Gorazd.Stumberger.@uni-mb.si



Abstract. This work analyses transmission line losses in a 35        additional losses can be evaluated by the norm or RMS
kV distribution network where currents and voltages contain          value of the current vector component orthogonal to the
higher harmonic components. They are caused by a nonlinear           voltage vector.
load – a rectifier with rated power of 4.7 MVA. In order to
evaluate increase of transmission line losses due to the current
higher harmonic components, the three-phase current vector is
decomposed into two orthogonal components. The first                 2.                    Network with nonlinear load
orthogonal component of the three-phase current vector is
collinear with the three-phase line voltage vector, while the        Figure 1 shows a part of discussed distribution network
second one is orthogonal to it. The first one contributes to the     with a nonlinear load. Our observation focuses on the
active power while the second one does not. It causes additional     losses in the 35 kV line, connecting 35 kV bus and
transmission line losses which could be avoided if the second        nonlinear load.
orthogonal component of the three-phase current vector is
minimized. The RMS values of both orthogonal components of
the three-phase current vector are used to evaluate the                                             SUBSTATION
                                                                                                        TR
additional transmission line losses.                                                                  110/35 kV
                                                                                                                       BUS
                                                                                                                       35kV       M2
                                                                                                                    L1 L2 L3
                                                                                      D
Key words: transmission line losses, higher harmonic                         ~                                                        M3
components, orthogonal decomposition, three-phase                                   Y yn
                                                                               SUPPLY
                                                                                                                                                                NONLINEAR
                                                                                                                                                35 kV LINE       LOAD
                                                                                          110kV
currents
                                                                                                    M1       DV1

1.   Introduction
                                                                     Figure 1: Part of the distribution network with a nonlinear load
One of the very effective tools used for analysis of three-
phase currents and voltages [1]-[8] are orthogonal                   The line voltages and currents measured in 35 kV
decompositions of three-phase currents in time domain                network in points M2 and M3, respectively, are shown in
[1]-[5], [8]. In this work, this tool is applied to analyze          Figure 2, while their amplitude spectra are shown in
increase of transmission line losses in a 35 kV                      Figures 3 and 4.
distribution network where currents and voltages contain
higher harmonics. These higher harmonics are caused by
                                                                      uL1 [kV],iL1 [A]




                                                                                          50                                                                           uL1
a nonlinear load – a rectifier with rated power of 4.7                                     0
                                                                                                                                                                       iL1

MVA.
                                                                                          -50


The aforementioned three-phase line currents and                                                0     0.01   0.02    0.03      0.04   0.05    0.06   0.07    0.08   0.09     0.1
                                                                      uL2 [kV],iL2 [A]




voltages are given in the form of current vector and                                      50                                                                           uL2
voltage vector [8]. The current vector is decomposed into                                  0
                                                                                                                                                                       iL2

two orthogonal components. The first one is collinear                                     -50
with the voltage vector while the second one is
                                                                                                0     0.01   0.02    0.03      0.04   0.05    0.06   0.07    0.08   0.09     0.1
orthogonal to it. Only the current vector component
                                                                      uL3 [kV],i L3 [A]




collinear with the voltage vector contributes to the energy                               50                                                                           uL3

transmission and to active power while the current vector                                  0
                                                                                                                                                                       iL3

component orthogonal to the voltage vector causes                                         -50
reciprocal energy exchange between source and load                                              0     0.01   0.02    0.03      0.04   0.05    0.06   0.07    0.08   0.09     0.1
whose average value equals zero. Thus, the current vector                                                                             t [s]
component orthogonal to the voltage causes additional
transmission line losses which can be avoided. These                 Figure 2: Line currents and voltages measured in a 35 kV
                                                                     network
                                                                                                          calculated by (4), while (5) introduces the equivalent
           30
                                                                                             UL1
                                                                                                          conductivity of the three-phase system Ge(t):
           25                                                                                UL2
                                                                                             UL3
                                                                                                                                       t
                                                                                                                                   1
                                                                                                                                          u T (τ ) i (τ ) dτ
                                                                                                                                   T t ∫T
           20
                                                                                                                   P(t ) =                                                                  (4)
   [kV]




           15
      35kV




                                                                                                                                       -
   u




           10
                                                                                                                                     P(t )
                                                                                                                   Ge (t ) =                                                                (5)
                                                                                                                                    U 2 (t )
             5

             0
              0                5                 10           15   harmonik        20               25
Figure 3: Amplitude spectra of line voltages on 35 kV level
                                                                                                          Two orthogonal components of the current vector i(t),
           60                                                                                             marked with iu(t) and iuo(t), are introduced in (6) and (7).
                                                                                             I L1
                                                                                                                                               P(t )
                                                                                                                  i u ( t ) = Ge (t )u ( t ) = 2 u ( t )
                                                                                             I L2
           50
                                                                                             I L3                                                                   (6)
           40                                                                                                                                 U (t )
                                                                                                                   i uo ( t ) = i ( t ) − i u ( t )
i35kV[A]




           30                                                                                                                                                                               (7)
           20

           10                                                                                             The first one is collinear with the voltage vector and is
             0
              0               5                  10           15   harmonik        20               25
                                                                                                          indispensable for energy transmission and active power
Figure 4: Amplitude spectra of line currents on 35 kV level                                               generation. The second one is orthogonal to the current
                                                                                                          vector. It is responsible for reciprocal energy exchange
The RMS values of individual line currents and voltages                                                   between source and load. The average value of this
are shown in Table I together with active power                                                           reciprocal energy exchange equals zero.
measured in individual lines.
                                                                                                          If the voltage vector u(t) is given, the current vector i(t)
             TABLE I: The RMS values of measured line currents and                                        and its orthogonal components iu(t) and iuo(t) produce,
                          voltages and active power                                                       together with the voltage vector, instantaneous powers
                                                                                                          p(t), ps(t) and pq(t) (8).
                                                 Line L1        Line L2                 Line L3
     Current [A]                                 36.64          36.48                   37.23
     Voltage [kV]                                20.69          20.80                   20.75                       ps ( t ) = u T ( t ) i ( t )
     Active Power [kW]                           742            749                     751                         p ( t ) = uT ( t ) iu ( t )                                             (8)
                                                                                                                    pq ( t ) = u      T
                                                                                                                                           ( t ) i uo ( t )
3.                Orthogonal decomposition
                                                                                                          The instantaneous power losses in the discussed 35 kV
Orthogonal decomposition in the time domain [8] is                                                        line caused by the current vectors i(t), iu(t) and iuo(t) are
performed in order to decompose currents measured at 35                                                   denoted as psi(t), pi(t) and pqi(t), respectively. They are
kV level into two orthogonal components. To allow for                                                     introduced in (9) to (11):
orthogonal decomposition, current and voltage vectors,
i(t) and u(t), are introduced by (1):                                                                               psi ( t ) = R i T ( t ) i ( t )                                         (9)
                                                                                                                  pi ( t ) = R i (t ) iu (t )
                                                                                                                                      T
                                                                                                                                                                                           (10)
                            ⎡iL1 ( t ) ⎤              ⎡uL1 ( t ) ⎤
                                                                                                                                      u

                            ⎢           ⎥             ⎢           ⎥                                               pqi ( t ) = R i ( t ) i uo ( t )
                                                                                                                                           T
                                                                                                                                                                                           (11)
                  i ( t ) = ⎢iL 2 ( t ) ⎥ , u ( t ) = ⎢uL 2 ( t ) ⎥ ,
                                                                                                                                           uo
                                                                                                    (1)
                            ⎢i t ⎥                    ⎢u t ⎥
                            ⎣ L3 ( ) ⎦                ⎣ L3 ( ) ⎦                                          where R is the resistance of the 35 kV line. Integrals of
                                                                                                          instantaneous powers (9) to (11) represent corresponding
where iL1(t), iL2(t), iL3(t) and uL1(t), uL2(t), uL3(t) denote                                            energy losses in time interval [t-T, t]. These losses are
measured line currents and voltages. The norms or RMS                                                     calculated by (12) to (14).
values of both vectors, I(t) and U(t), are given by (2) and
(3):                                                                                                                           t                              t
                                                                                                                  Wsi =    ∫       psi (τ ) dτ =              ∫     R i T (τ ) i (τ ) dτ   (12)
                                                                                                                          t −T                            t −T
                                         t
                                    1                                    L3
                       I (t ) =         ∫ i (τ ) i (τ ) dτ =
                                    T t −T
                                           T
                                                                        ∑I         2
                                                                                   k                (2)                    t                             t

                                                                                                                          ∫        pi (τ ) dτ =          ∫ R i (τ ) i (τ ) dτ
                                                                        k = L1
                                                                                                                  Wi =                                                T
                                                                                                                                                                      u        u           (13)
                                                                                                                         t −T                           t −T
                                             t
                                     1                                        L3
                       U (t ) =          ∫ u (τ ) u (τ ) dτ =
                                            T
                                                                           ∑U           2
                                                                                                    (3)                        t                               t
                                                                                                                                    pqi (τ ) dτ =              ∫ R i (τ ) i (τ ) dτ
                                                                                        k
                                     T t −T                                                                       Wqi =        ∫
                                                                                                                                                                          T
                                                                           k = L1
                                                                                                                                                                          uo       uo      (14)
                                                                                                                          t −T                               t −T

where the time interval of interest is given as [t-T, t]
while Uk and Ik denote RMS values of individual line                                                      The average power losses in the given time interval, are
currents and voltages. The average active power P(t) is                                                   calculated by (15) to (17).
                                                             t
                                                     1
                                        Psi = R          ∫ i (τ ) i (τ ) dτ                            (15)
                                                            T
                                                                                                              The transmission line instantaneous power losses, given
                                                     T t −T
                                                                                                              by expressions (9) to (11), are shown in Figure 7 together
                                                                                                              with their integrals (12), (13) and (14).
                                                     t
                                                  1
                                        Pi = R        ∫ iu (τ ) iu (τ ) dτ                             (16)
                                                         T

                                                  T t −T
                                                                                                                                   2000
                                                                                                                                                                                                                  psi
                                                         t
                                               1                                                                                   1500                                                                           pi




                                                                                                              psi, pi, pqi [W]
                                        Pqi = R ∫ iT (τ ) i uo (τ ) dτ
                                                      uo                                               (17)                                                                                                       pqi
                                               T t −T                                                                              1000

                                                                                                                                           500

where the power losses Psi are caused by the current                                                                                            0
vector i(t), the power losses Pi are caused by the current                                                                                          0     0.002 0.004 0.006 0.008    0.01   0.012 0.014 0.016 0.018        0.02

vector iu(t), while the power losses Pqi are caused by the
                                                                                                                                           10
current vector iuo(t). Only the power losses Pi are                                                                                                                                                               wsi
                                                                                                                                                8
unavoidable. They are caused by the current vector iu(t)




                                                                                                                       wsi , wi, wqi [J]
                                                                                                                                                                                                                  wi

which is indispensable for energy transmission. The                                                                                             6                                                                 w
                                                                                                                                                                                                                      qi

current vector iuo(t) does not contribute to the transmitted                                                                                    4

energy, therefore, losses caused by this current vector                                                                                         2

represent additional losses which can be avoided.                                                                                               0
                                                                                                                                                    0     0.002 0.004 0.006 0.008    0.01   0.012 0.014 0.016 0.018        0.02
                                                                                                                                                                                    t [s]
                                                                                                              Figure 6: Power and energy losses in 35 kV transmission line
4.                             Results
                                                                                                              The energy losses (12) to (14) were used to calculate
The results presented in this section are given for currents                                                  average power losses Psi, Pi and Pqi, which can be
and voltages presented in Figure 2, section 2.                                                                calculated also by (15) to (17). The obtained results are
Components of vectors u(t), i(t), iu(t) and iuo(t) are shown                                                  shown in Table II.
in Figure 5. They were determined by (1) to (7).
Components of the voltage vector u(t) are line voltages                                                                                                             TABLE II: Power losses
uL1, uL2, uL3, while the line currents iL1, iL2, iL3 are
components of the current vector i(t). Components of the                                                                                                Psi [W]              Pi [W]                   Pqi [W]
current vectors iu(t) and iuo(t) in individual lines are iL1u,                                                                                          1371.4               1250.7                    120.7
iL2u, iL3u and iL1uo, iL2uo, iL3uo.
                                                                                                              A similar ratio between individual losses, as it is show in
                                                                                                              Table II for one period of fundamental, was confirmed by
                                    x 10
                                         4                                                                    measurement performed for one day. In order to evaluate
                                                                                                              monthly loss of energy due to the current vectors i(t), iu(t)
          uL1, uL2, uL3 [V]




                               2
                                                                                                              and iuo(t), an average daily energy transmission was
                               0
                                                                                                              analyzed for a period of one month. Considering ratio
                               -2
                                                                                                              between individual power losses given in Table II and
                                    0      0.002 0.004 0.006 0.008   0.01    0.012 0.014 0.016 0.018
                                                                                                              measured average daily transmission of energy in the
                                                                                                              discussed 35 kV line, transmission line losses were
 iL1, iL2, iL3 [A]




                              50

                               0
                                                                                                              evaluated.
                              -50
                                                                                                              Figure 7 shows the average daily transmission line losses,
                                    0      0.002 0.004 0.006 0.008   0.01    0.012 0.014 0.016 0.018
                                                                                                              due to the current vectors i(t), iu(t) and iuo(t). The daily
 iL1u, iL2u, iL3u [A]




                              50                                                                              losses caused by i(t) are marked as Wsid, the daily losses
                               0                                                                              caused by iu(t) are denoted as Wid while Wqid represents
                              -50
                                                                                                              losses due to the current vector iuo(t).
                                    0      0.002 0.004 0.006 0.008   0.01    0.012 0.014 0.016 0.018
 iL1uo, iL2uo, iL3uo [A]




                              50                                                                                                   1.6

                                                                                                                                   1.4
                               0
                                                                                                                     Wloss [MWh]




                                                                                                                                   1.2
                              -50
                                                                                                                                           1
                                    0      0.002 0.004 0.006 0.008    0.01   0.012 0.014 0.016 0.018
                                                                                                                                   0.8
                                                                     t [s]                                                                                                                                            W sid

Figure 5: Components of vectors u(t), i(t), iu(t) in iuo(t)                                                                        0.6                                                                                W
                                                                                                                                                                                                                        id
                                                                                                                                                                                                                      W qid
                                                                                                                                   0.4

                                                                                                                                   0.2


The results presented in Figure 5 show that components                                                                                     0
                                                                                                                                            0                 5        10           15         20         25               30
                                                                                                                                                                               Day in month
of the current vector iu(t) in individual lines (iL1u, iL2u,
iL3u) are collinear with the corresponding voltage vector
                                                                                                              Figure 7: Daily energy losses in 35 kV transmission line
components uL1, uL2, and uL3 .
                                                                  References
For the given case, the total monthly transmission line
losses caused by the current vectors i(t), iu(t) and iuo(t) are   [1] L. S. Czarnecki, “On some misinterpretations of the
given in Table III.                                                   instantaneous reactive power p-q theory,” IEEE Trans. on
                                                                      Power Electronics., vol. 19, no. 3, pp. 828-836, 2004.
                                                                  [2] H. Akagi, Y. Kanazawa, A. Nabae, “Instantaneous reactive
TABLE III: Total monthly transmission line losses due to
                                                                      power compensators comprising switching devices without
                     i(t), iu(t) and iuo(t)                           energy storage components,” IEEE Trans. on Industry
                                                                      Applications, vol. 20, no. 3, pp. 625–631, 1984.
       Total monthly transmission line losses [MWh]               [3] A. Nabae, T. Tanaka, “A new definition of instantaneous
     Due to i(t)        Due to iu(t)         Due to iuo(t)            active- reactive current and power based on instantaneous
      35.59               32.45                 3.14                  space vectors on polar coordinates in three-phase circuits,”
                                                                      IEEE Trans. on Power Delivery, vol. 11, no. 3, pp. 1238–
                                                                      1243, 1996.
The transmission losses due to the current iu(t) cannot be        [4] J. L. Willems, “A new interpretation of the Akagi-Nabae
avoided, because iu(t) is the minimal current vector                  power components for nonsinusoidal three-phase
                                                                      situations,” IEEE Trans. on Instr. and Meas., vol. 41, no.
needed for energy transmission. On the contrary, the
                                                                      4, pp. 523–527, 1992.
current vector iuo(t) represents only a reciprocal energy         [5] Gorazd Štumberger, Treatment of three-phase systems
exchange between the source and load, which must be                   using vector spaces. Ph.D. Thesis, Faculty of Electrical
avoided in order to minimize transmission line losses.                Engineering and Computer Science, Maribor, 1996.
                                                                  [6] D. L. Milanez, A. E. Emmanuel, “The instantaneous-space
                                                                      phasor a powerful diagnosis tool,” IEEE Trans. on Instr.
5.   Conclusion                                                       and Meas., vol. 52, no. 1, pp. 143–148, 2003.
                                                                  [7] A. E. Emmanuel, “Summary of IEEE Standard 1459:
                                                                      definitions for measurement of electric power quantities
This paper discusses transmission line losses in the case             under sinusoidal, nonsinusoidal, balanced or unbalanced
of 35 kV network, where currents and voltages contain                 conditions,” IEEE Trans. on Industry Applications, vol.
higher harmonic components caused by a nonlinear load.                40, no. 3, pp. 869–876, 2004.
Measured three phase currents are decomposed into two             [8] G. Štumberger, B. Polajžer, M. Toman, D. Dolinar,
orthogonal components. Only one of them contributes to                Orthogonal decomposition of currents, power definitions
the energy transmission, while the other one increases                and energy transmission in three-phase systems treated in
                                                                      the time domain, Proceedings of the International
transmission losses. In the given case transmission losses
                                                                      conference on renewable energies and power quality
could be decreased for 10 percents by rejection of current            (ICREPQ'06), Palma de Mallorca, Spain, April 206.
component that does not contribute to the energy
transmission.

				
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