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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.