Fizyczne przyczyny redukcji współczynnika mocy

Transcription

1 Fizyczne przyczyny redukcji współczynnika mocy Prof. dr hab. inż. Leszek S. Czarnecki, IEEE Life Fellow Distinguished Professor at School of Electrical Engineering and Computer Science Louisiana State University, USA Internet Page:

2 Cost of the electric energy production and delivery is related to a main degree to the apparent power S At the supply source terminals. it is a product of RMS values of the supply voltage U and the load current I S=U I For customers the importance has only the energy delivered τ W = 0 Pdt The inequality S thus the power factor P P = S λ are of the major importance for electrical systems economy

3 By the end of XIX century it was concluded that S P = Q where Q denotes the reactive power of the load

4 Steinmetz Experiment: 892 P 2 + Q 2 < S 2, Q = 0 Why the apparent power S is higher than the active power P? How the difference between S and P can be reduced?

5 Charles Proteus Steinmetz In Einstain s company

6 Present day Steinmetz Experiment with line currents up to 625 ka S = 750 MVA P = 35 MW Current not only distorted, but also asymmetrical and random Power factor: λ ~ 0.42 Annual bill for energy ~ 500 Million $

7 Why the apparent power S is higher than the active power P? How the difference between S and P can be reduced? These apparently simle questions have occurred to be ones of the most difficult questions of electrical engineering for the whole XX century. Hundreds of scientists were involved in the quest for the answer. Hundreds of papers have been published.

8 Major disscussion forums: International Workshop on Reactive Power Definition and Measurements in Nonsinusoidal Systems, Bi-annual meetings in Italy, Chaired by A. Ferrero International School on Nonsinusoidal Currents and Compensation (ISNCC) Bi-annual meetings in Poland Chaired by L.S. Czarnecki

9 927: Budeanu: A number of different answers to the Steinmetz s question have been suggested: = B + S P Q D QB = UnInsinϕn Endorsed by the IEEE Standard Dictionery of Electrical and Electronics Terms in 992 and German Standards DIN in 972 n= 93: Fryze: S P = Q F QF = u irf Endorsed by German Standards DIN in R S 97: Shepherd: S S = Q QS = u irs K r 975: Kusters: S P = Q + Q QK = u irc Endorsed by the International Electrotechnical Commission in : Depenbrock: = + + S P Q V N T T Q = U I sinϕ 2003: Tenti: S P = Q + D QT = u irt

10 Since the Steinmetz s experiment in 892, after 9 years of Power Theory development for single-phase, LTI loads, such as u = U0+ 2 Re U e n= n jnω t jnω t i = G0 U0+ 2 Re Yn Un e n= we have had five different power equations and five different reactive powers Compensation problem was not solved The problem was eventually solved in frame of the Currents Physical Components (CPC) based power theory by L.S. Czarnecki in 984

11 This, eventually a positive result was a conclusion of a specific approach to the power theory development. Its core is the concept of Currents Physical Components (CPC) According to the CPC concept, the load current can be decomposed into - components associated with distinctive physical phenomena which are - mutually orthogonal

12 CPC of three-phase Harmonics Generating Loads (HGL) supplied by a four-wire line with nonsinusoidal and asymmetrical voltages: n p z Ca Cr Cs Cu Cu Cu G i() t = i () t + i () t + i ()+ t i () t + i () t + i () t + i () t These currents are associated with distinctive physical phenomena in the load All of them are mutually orthogonal so that their three-phase RMS value satisfy the relationship n 2 p 2 z 2 2 Ca Cr Cs Cu Cu Cu G i = i + i + i + i + i + i + i

13 Teoria mocy oparta na koncepcji Składowych Fizycznych Prądu (CPC) jest obecnie jedyną teorią mocy, która wyjaśnia wszystkie zjawiska fizyczne w systemach o odkształconych i asymetrycznych przebiegach prądów i napięć niezależnie od złożoności systemu Teoria ta tworzy podstawy kompensacji reaktancyjnej oraz podstawy sterowania kompensatorow kluczujących

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16 93 S. Fryze, Professor of Lwow University, Poland, defined the reactive power in a time-domain, based on the load current orthogonal decomposition into active and reactive currents i = i a + i rf ia() t = P ut () 2 u = Ge ut (), irf () t = it () ia() t T T z ia ( t) irf( t) dt = ( i a,irf ) = i = i + i a 2 rf 2 Fryze's power equation: Fryze s definition of reactive power: S = P + Q F Q = u i F rf

17 97 Shepherd & Zakikhani, England, developed power theory in the frequency-domain: u = U + 2 U cos( nω t+ α ) 0 n n= i = I + 2 I cosϕ cos( nω t+ α ) + 2 I sinϕ sin( nω t+ α ) 0 n n n n n n n= n= i = i + i R, r n 2 2 ir = In cos ϕ 2 2 n ir = In sin ϕn n= 0 n= = R + r i i i

18 984 Leszek S. Czarnecki: Scattered Current (prąd rozrzutu) u = U0+ 2 Re U e n= n jnω t jnω t i = G0 U0+ 2 Re Yn Un e n= i = i a + i r + i s Fryze: (93) i = G u, G = P u a e e 2 Active current Shepherd: (97) Czarnecki: (984) ir = 2 Re jbnune n= jnω t i = ( G G ) U + 2 Re ( G G ) U e s 0 e 0 n e n= n jnω t Reactive current Scattered current This decomposition reveals a new power phenomenon: the existance of the scattered current, i s, that occurs when the load conductance, G n, changes with harmonic order, n.

19 u = U0+ 2 Re U e n= n jnω t jnω t i = G0 U0+ 2 Re Yn Un e n= Y I = = + = n n Gn jbn Un R + jnω L G n R = Re{ } = R + jnω L R + n L 2 2 ( ω ) ω L = Ω, R = Ω G0 G. G 2. G 3. G = S, = 0 5 S, = 0 2 S, = 0 S,... > 0

20 990 Leszek S. Czarnecki: Load Generated Current (prąd generowany) e = 00 2 sinωt V j = 50 2 sin3ω t A i = 2(20 sinω t + 40 sin 3 ω t) A u = 2(80 sinω t 40 sin 3 ω t) V i = 44.7 A u = 89.4 V S = 4.0 kw P = = 0, Q = 0 How to write power equation for such a circuit? This circuit reveals that at some harmonic frequencies the energy flows from the load back to the supply source thus P n < 0 20

21 P 0, n NC, energy flows to the load = U I cos{ ϕ } < 0, n NG, energy flows to the supply n n n n Permanent flow of energy at some harmonic frequencies from the load back to the supply source can be regarded as a physical phenomenon P 0, n N, energy flows to the load n C P < 0, n N, energy flows to the supply n G

22 CPC in single-phase circuits with Harmonics Generating Loads (HGL) i i i i i () t = () t + () t + () t + ( t) Ca Cr Cs G C Ca = e C Ce = 2 uc i Gu, G i = 2 Re ( G G ) U e Cs n NC n Ce P n jnω t Active current Scattered current i = 2 Re jb U e i Cr G n N G n N n C n n jnω t Reactive current = i Load generated current

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24 Current Physical Components (CPC) in three-phase systems with linear time-invariant (LTI) loads with sinusoidal voltage L.S. Czarnecki, "Orthogonal decomposition of the current in a three-phase non-linear asymmetrical circuit with nonsinusoidal voltage," IEEE Trans. Instr. Measur., Vol. IM-37, No., March 988. L.S. Czarnecki, "Reactive and unbalanced currents compensation in three-phase circuits under nonsinusoidal conditions," IEEE Trans. Instr. Measur., Vol. IM-38, No. 3, June 989. L.S. Czarnecki, "Minimization of unbalanced and reactive currents in three-phase asymmetrical circuits with nonsinusoidal voltage," Proc. IEE, Vol. 39, Pt. B, No. 4, July 992. L.S. Czarnecki, Power factor improvement of three-phase unbalanced loads with nonsinusoidal voltage," European Trans. on Electrical Power Systems, ETEP, Vol. 3, No., Jan./Febr L.S. Czarnecki, Equivalent circuits of unbalanced loads supplied with symmetrical and asymmetrical voltage and their identification, Archiv fur Elektrotechnik, 78 pp , 995. L.S. Czarnecki, "Energy flow and power phenomena in electrical circuits: illusions and reality," Archiv fur Elektrotechnik, (82), No. 4, pp. 0-5, 999.

25 Apparent power definitions: SA = URIR+ USS I + UTT I G 2 2 S = P + Q B S = U + U + U I + I + I R S T R S T Which of these three definitions is right? 25

26 P = 72.3 kw S = S = U I + U I + U I = kva A G R R SS TT 2 2 S = S = P + Q = kva B S = S = U + U + U I + I + I = = kva R S T R S T P. λ A = = 0 86 S G A λ = P = λ B = P = 0. 7 S S G Which of these three is the right value of the power factor? B 26

27 Apparent power definition selection: S A = S G = S B =00 kva λ A = λ G = λ B = S A = 9 kva, λ A = 0.84 S G = 00 kva, λ G = S B = 49 kva, λ B = 0.67 Geometrical definition of the apparent power SG results in unity power factor in spite of increase of energy loss at delivery The apparent power should not be calculated according to geometrical definition G 2 2 S = P + Q It is wrong

28 S A = 9 kva, λ A = 0.84 S G = 00 kva, λ G = S B = 49 kva, λ B = 0.67 S A = S G = S B =49 kva λ A = λ G = λ B = 0.67 Power loss in the supply of unbalanced load is the same as power loss of a balanced load when the apparent power is calculated according to formula S = U + U + U I + I + I R S T R S T L.S. Czarnecki, "Energy flow and power phenomena in electrical circuits: illusions and reality," Archiv fur Elektrotechnik, (82), No. 4, pp. 0-5,

29 S A = 9 kva, λ A = 0.84 S G = 00 kva, λ G = S B = 49 kva, λ B = 0.67 S A = S G = S B =49 kva λ A = λ G = λ B = 0.67 Also arithmetical definition of the apparent power SA results in a wrong value of the power factor. The apparent power should not be calculated according to arithmetical definition SA = URIR+ USS I + UTT I It is wrong 29

30 Numerical illustration: S = U + U + U I + I + I = kva R S T R S T P = 72.6 kw Q = 0 The relationship: S 2 = P 2 + Q 2 is not fulfilled by these values This power equation is erroneous 30

31 Three phase vectors and their scalar product Three-phase structure, combined with Fourier decomposition makes equations of three-phase nonsinusoidal systems complex and illegible More compact mathematical symbols are needed to simplify such equations u () t i () t u() t = u () t = u, i( t) = i () t = i R R S S ut() t it() t T ( ) T P = u i + u i + u i dt = ( t ) ( t ) dt = (, ) T u i ui R R SS TT T 0 0 T Scalar product of three-phase vectors, x (t) and y (t) T (, ) T x y = ( t ) ( t ) dt T x y 0 Vectors, x(t) and y(t) are orthogonal when their scalar product (x, y) = 0 3

32 RMS value of a Three-Phase quantity Heat released in such equipment is proportional to the active power T T P = R ( i i i ) dt R ( t ) ( t ) dt R (, ) R T + + = = = T i i ii i T 2 R S T 0 0 The quantity: T i = ( i + i + i ) dt = i + i + i T R S T R S T is the RMS value of the three phase current.. Similarly T u = ( u + u + u ) dt = u + u + u T R S T R S T is the RMS value of the three phase voltage 32

33 Apparent power Apparent power is a conventional quantity S = u i Product of RMS values of the voltage and current needed for the load supply S = u i At sinusoidal voltages and currents S = U + U + U I + I + I R S T R S T 33

34 CPC at sinusoidal supply voltage ir IR jωt i = i = 2 Re I e = 2 ReI e S S i T I T jωt I = ( Y + Y + Y ) U ( Y + αy + α * Y ) U R RS ST TR R ST TR RS R I = ( Y + Y + Y ) U ( Y + αy + α * Y ) U S RS ST TR S ST TR RS T I = ( Y + Y + Y ) U ( Y + αy + α * Y ) U T RS ST TR T ST TR RS S j ω t p n e e UR YU u R i = 2 Re{ Ie } = 2 Re{[ ( G + jb ) + ] e } jωt Y = G + jb = Y + Y + Y e e e RS ST TR Equivalent admittance Y Y Y Y u = ( ST + α TR + α RS) * Unbalanced admittance 34

35 j ω t p n e e UR YU u R i = 2 Re{ Ie } = 2 Re{[ ( G + jb ) + ] e } jωt CPC of three-phase LTI load at sinusoidal supply voltage i = i + i + i a r u p a = GeU R i 2 Re{ e } p r = jbeu R jω t j ω t i 2 Re{ e } n u = YU u R jω t i 2 Re{ e } Active current Reactive current Unbalanced current The unbalanced current is associated with the phenomenon of the load current asymmetry 35

36 Powers in Three-Phase systems under Sinusoidal Conditions i = ia + ir + iu CPC are mutually orthogonal, so that their three-phase RMS value satisfy relationship a r u i = i + i + i, 2 u We obtain right power equation of three-phase systems: S = P + Q + D u D u = u i u Unbalanced power 36

37 Current Physical Components(CPC) in linear, time-invariant (LTI) systems with symmetrical nonsinusoidal voltage L.S. Czarnecki, "Orthogonal decomposition of the current in a three-phase non-linear asymmetrical circuit with nonsinusoidal voltage," IEEE Trans. Instr. Measur., Vol. IM-37, No., pp , March 988. L.S. Czarnecki, "Reactive and unbalanced currents compensation in three-phase circuits under nonsinusoidal conditions," IEEE Trans. Instr. Measur., Vol. IM-38, No. 3, pp , June 989.

38 u = u = 2 Re U n N n n N n jnω t e, i = i + i + i + i a s r u i = i = 2 Re I e n N Active Current: a e e 2 i n = G u, G = P u n N CPC of three-phase LTI loads currents at nonsinusoidal supply voltage n jn ω t Scattered current: Reactive current: Unbalanced current: df ( t) = 2Re ( G G ) U e i s en e n n n N df ( t) 2 Re jb U e ir = df n N en n n jnω t * jnωt ( t) 2 Re Y U e iu = n N un n n jnω t df Yu n = ( Y Y Y ) = p jn 2π jn 2π Y 3 3 u n ST n + e TR n + e RSn n Yu n, for n=3 k+, for n=3k 38

39 Current Physical Components (CPC) of three-phase LTI loads supplied by four-wire line with sinusoidal symmetrical voltage L.S. Czarnecki, P.M. Haley, Power properties of four-wire systems at nonsinusoidal supply voltage, IEEE Trans. on Power Delivery, Vol. 3, No. 2, 206, pp L.S. Czarnecki, P.M. Haley, Unbalanced power in four-wire systems and its reactive compensation, IEEE Trans. on Power Delivery, Vol. 30, No., Feb. 205, pp L.S. Czarnecki, P.H. Haley, Currents Physical Components (CPC) in Four-Wire Systems with Nonsinusoidal Symmetrical Voltage, Przegląd Elektrotechniczny R.9, No. 6/205, pp

40 Currents Physical Components in four-wire unbalanced system P jq Ye = G e+ jbe = = 2 ( YR + YS + YT ) u 3 df n u R + α S+ α* T Y = ( Y Y Y ) 3 df z u R + α* S+ α T Y = ( Y Y Y ) 3 i = i a r n u + i + i + i z u i () t = G u() t a e ir() t = B d e u (). t d( ωt) df n n n u Yu UR jωt i = 2Re{ e } df z z z u Yu UR jωt i = 2Re{ e }. Active Current: Reactive current: Unbalanced current of negative sequence: Unbalanced current of zero sequence

41 CPC of three-phase Harmonics Generating Loads (HGL) supplied by a four-wire line with nonsinusoidal and asymmetrical voltages: n p z Ca Cr Cs Cu Cu Cu G i() t = i () t + i () t + i ()+ t i () t + i () t + i () t + i () t These currents are associated with distinctive physical phenomena in the load All of them are mutually orthogonal so that their three-phase RMS value satisfy the relationship n 2 p 2 z 2 2 Ca Cr Cs Cu Cu Cu G i = i + i + i + i + i + i + i

42 Teoria mocy oparta na koncepcji Składowych Fizycznych Prądu (CPC) jest obecnie jedyną teorią mocy, która wyjaśnia wszystkie zjawiska fizyczne w systemach o odkształconych i asymetrycznych przebiegach prądów i napięć niezależnie od złożoności systemu Teoria ta tworzy podstawy kompensacji reaktancyjnej oraz podstawy sterowania kompensatorow kluczujących

43 Working energy-based economic incentives for the supply and loading qualities improvement in islanded micro-grids

44 The active power, P T df df P = u() t i () t dt = ( u,i) T 0 df T df P = ( urir+ uss i + utt i ) dt (, ) T = ui 0 is the most fundamental power quantity, introduced to electrical engineering in the last decades of XIX century

45 The terms: The active power and the useful power are regarded as synonyms The active energy: month 0 P dt = W a is regarded as a useful energy and it is the major component of bills for energy provided to customers

46 T 0 ut= () un = u+ u n N h P = u () t i () t dt = P+ P2+ P3+ P T it= () in = i+ i n N P h = > UI 0 2 n n n s n n s n P = U I = ( RI ) I = R I < 0 To operate the load, the power P has to be provided. It is A WORKING ACTIVE POWER, P w ( P + P + P +...) = P r Is A REFLECTED ACTIVE POWER P= P P w r

47 Harmonics generating load with active power P has to be supplied with working power P w P w > P

48 It might be more fair, if the customer is billed for energy dissipated by harmonics he produces month month P dt = ( P + P ) dt = W w r w 0 0 To deliver not only active power P, but also the reflected active power P r, the supply current fundamental harmonic I has to increase. This causes additional loss of active power in the supply 2 s r s w P= P+ RI

49 α = 0 deg, Rs = 5% of R Active power P = 000 W Working active power P w = 536 W Reflected active power Pr = 536 W Active power loss in the supply: P s = 900 W

50 Working and reflected powers in three-phase systems. u R df p u S = = + u T u u u n i R df p i S = = + i T i i i n T P = dt = (, ) = (, ) + (, ) = P + P T ui ui u i u i 0 T p p n n p n df n n n n n n 2 u i si i s i r P = (, ) = ( R, ) = R = P < 0 p n df P = P + P = P P w r

51 Unbalanced load that has active power P has to be supplied with the working active power P w

52 i = = A Active power loss s s 2 2 P = R i = = 5. 0 kw

53 0 p UR j 0 U, α, α* e U = S V 0 n 3 =, *, j20 U α α U. e T 0 p IR j 0 I, α, α* 68. 5e I = S A 0 n 3 =, *, j 60 I α α I 68. 5e T p p p p p Pw = P = ( u, i ) = 3U I = = kw n n n n n* Pr = P = ( u, i ) = 3Re{ U I } = = 5. 6 kw P 3 w uw iw = = = 29. 8A Active power loss 2 2 s s iw r P = R + P = =. 2 kw

54 P P w r > 0 < 0 P w > P Energy account based on the working active power P w INCREASES the bill for energy

55 The supply voltage is asymmetrical P P w r p = P > 0 n = P > 0 P w < P Energy account based on the working active power P w REDUCES bill for energy The same is when the supply voltage contains harmonics

56 LOSS OF REVENUE by utilities for the electric energy sold to customers with harmonics and asymmetry producing loads is proportional to the difference between the working active power P w and the active power P P= P P w

57 OVERPAYMENT for the electric energy of customers with high quality loads, supplied with asymmetrical and distorted voltage is proportional to the difference between the active power P and the working active power P w P= P P w

58 Smart grids should have built in a mechanism of economic incentives for improving loading quality and supply quality Energy accounts based on the working active power creates such a mechanism

59 To keep a healthy distance to what I told, a power system engineer after such a presentation told: It does not matter that the apparent power S is higher than the active power P. Eventually customers pay for everything

60 Dziękuję za uwagę!

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