Power analysis of static VAr compensators

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1 Available online at Electrical Power an Energy ystems 0 (008) Power analysis of static VAr comensators F.. Quintela *, J.M.G. Arévalo,.. eono Escuela écnica uerior e Ingeniería Inustrial, Universia e alamanca, 7700 Béjar, ain eceive 1 March 007; receive in revise form 17 December 007; accete 18 December 007 Abstract Analysis of three-hase s usually assume them to be three imeances in a star or triangle connection. his is the reason why obtaine results can only be consiere vali for assive s, strictly seaking. Analysis leaing to the roosal of some static comensators is usually erforme in this way, which inuces to believe that this comensators are only vali for assive s. An analysis roceure, which uses only owers to escribe s, is exoune in this aer. If alie to the analysis of static comensators, it reveals unequivocally their usefulness with active an assive three-hase s. herefore, this metho is more general an, as it will be seen, easier. Ó 007 Elsevier t. All rights reserve. Keywors: Power analysis; VAr comensator; oa balance 1. Introuction It is usual to suose three-hase s exclusively forme by imeances in a star or triangle connection while stuying them. For instance, a lot of textbooks o this while eucing the active, reactive, an aarent ower s formulas of balance three-hase systems [1,]. his euction metho assures the valiity of the results for assive s, but it oes not justify their valiity for active s, strictly seaking. Every three-hase assive forme only by imeances connecte in any manner is equivalent to three imeances in a star or triangle connection. evertheless, if the contains motors or generators it is necessary to escribe it by means of voltage or current sources along with imeances. hat is to say, the result is an active, which is not equivalent to a forme just by three imeances. * orresoning author. el.: x7; fax: aresses: felixrq@usal.es (F.. Quintela), jumagar@usal.es (J.M.G. Arévalo), roberrm@usal.es (.. eono). ome solutions, which are obtaine only for assive s, are also vali for active s even though euction roceures o not justify them. heir aarent limitation is ue exclusively to the euction metho. For this reason it is imortant to use analysis methos that are caable of clearly showing the scoe of the solutions they lea to. One of those roceures consists in using relations between the owers of three-hase systems instea of escribing the s by means of imeances. hree avantages of this metho are: (a) all variables involve are external to the, (b) there is no nee for an assumtion concerning the internal connections of the, its constituting recetors, or whether it is a assive or active, thereby conferring great generality to the results, an (c) oerations ten to be easier than those involving imeances []. ome roceures have been roose to balance threewire three-hase s, an to comensate the reactive ower they absorb, using static VAr comensators. everal algorithms have been create to obtain the suscetances of the static VAr comensator. hose algorithms erive from methos euce for assive s consisting only of certain imeance connections [ ]. hey are euce in this aer from ower relations, thus showing that the solution 01-01/$ - see front matter Ó 007 Elsevier t. All rights reserve. oi:10.101/j.ijees

2 F.. Quintela et al. / Electrical Power an Energy ystems 0 (008) is unequivocally vali not only for assive but also for active s, whichever their internal connection might be.. elation between active an reactive owers elivere by each hase to a three-wire three-hase he relation between the active an reactive owers elivere by each hase to a three-wire three-hase is the basis of the metho shown here. Fig. 1 reresents a three-hase connecte to a three-hase line with balance voltages. If voltage V is chosen as reference hasor (Fig. 1), then V ¼ V =0, V ¼ V = 10, V ¼ V =10, an V V V P +jq P +jq P +jq hree-hase Fig. 1. Power elivere to a three-wire three-hase by each hase. 0 ¼ I þ I þ I ¼ P þ jq þ P þ jq V =0 V = 10 þ P þ jq ð1þ V =10 where I is the conjugate of, an P þ jq ¼ V I is the comlex ower elivere to the through its -hase. he same alies ffiffi to the rest of the hases. ffiffias 1=10 ¼ 1= þ j = an 1= 10 ¼ 1= j =, from (1), two new equations of real numbers are obtaine P P ffiffiffi Q P þ ffiffiffi Q ¼ 0 Q Q þ ffiffi P Q ffiffi ðþ P ¼ 0 hese are the relations between the owers elivere by each hase looke for. Whichever the connecte to a three-wire three-hase line with balance voltages might be, its owers must satisfy the relations given in (). If Q ¼ Q ¼ Q ðþ one erives from () that P ¼ P ¼ P ðþ an the system is balance. hat is to say, it suffices to make the reactive owers elivere by the hases equal to each other to attain a balance three-wire three-hase system. his is the basis of the metho roose in this aer, an use to obtain the algorithms of the comensator which balances any three-hase : the metho consists on a comensator, constitute entirely by reactances, which makes the reactive ower elivere by the hases equal to each other. hus, if Q is the reactive ower elivere to the set forme by the comensator an the, then each hase elivers Q = Q /. he active ower elivere now by each hase is P, an the active ower elivere to the comensator grou is P, which is also the active ower elivere to the, since no active ower is elivere to the comensator as it consists of reactances only. hus, the new ower factor is P P cos u ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðþ ðpþ þðqþ P þ Q If Q = 0 is achieve, the ower factor woul be unity.. eactive ower elivere by each hase to a three-wire urely-reactive connecte in triangle et the owers jq, jq, an jq, which are absorbe by the mono-hase reactive s that constitute the triangle in Fig., be known. It is very useful for our uroses to fin the relation between those owers an the comlex owers elivere by each hase, which will be esignate by ¼ P þ jq, ¼ P þ jq, an ¼ P þ jq. eeing that I ¼ I 1 I, we have jq I ¼ P þ jq ¼ jq ffi V =0 ffi V =0 ¼ Q Q V =10 ffiffiffi V ffi ðq þ Q þ j Þ ffiffiffi ðþ V omaring the secon an fourth members of this equation it is clear that Q ¼ Q þ Q imilarly Q ¼ Q þ Q ð8þ Q ¼ Q þ Q ð9þ he reactive owers elivere by each hase can be fixe at any esire value using three aroriate reactances, as shown in these last three equations. For instance, it is ossible to achieve equal values for the three reactive owers elivere by the hases to a, thus balancing it. Furthermore, all the reactive energy is comensate if those three owers are set to zero. his result will be use in a moment. V V V P +jq P +jq P +jq I 1 jq Fig.. By choosing the aroriate value for Q, Q, an Q, the reactive owers elivere by each hase, Q, Q, an Q, might be fixe at any esire value. I jq jq I ð7þ

3 78 F.. Quintela et al. / Electrical Power an Energy ystems 0 (008) 7 8. eactance comensator Fig. shows an unbalance three-wire three-hase. Q 1, Q 1 an Q 1 are the reactive owers elivere by each hase to the ; thus, the reactive ower absorbe by the is Q 1 = Q 1 + Q 1 + Q 1 [7]. A reactance comensator is connecte in a triangle configuration to balance hase currents. In orer to o it, it is enough to make the reactive owers elivere by each hase to the comensator grou equal, as seen before. hat is to say, if Q is the reactive ower elivere to the comensator grou, each hase must eliver Q / to it. Q can be any arbitrary value. he reactive owers of the comensator, Q, Q, an Q, which make the reactive ower elivere by each hase to the comensator grou equal to Q /, will be etermine. he ower elivere to the comensator grou by each hase is the sum of the ower elivere to the by each hase an the ower elivere to the comensator. aking into account (7) (9), this ower is Q ¼ Q 1 þ Q þ Q Q ¼ Q 1 þ Q þ Q Q ¼ Q 1 þ Q þ Q ð10þ he values looke for are obtaine from this set of equations Q ¼ Q þ Q 1 Q 1 Q 1 Q ¼ Q þ Q 1 Q 1 Q 1 Q ¼ Q þ Q 1 Q 1 Q 1 ð11þ hese are the reactive ower values of the comensator s reactances that balance the three-hase system. he reactive ower absorbe now is Q. If the only urose is to balance the system without changing the value of the reactive ower elivere to the comensator grou, then we use Q = Q 1 = Q 1 + Q 1 + Q 1. But if this reactive ower, Q, is intene to be zero, then the owers of the reactances shoul be Q / Q / Q / jq Unbalance three- hase Fig.. his comensator, forme by reactances, balances the system. hree varmeters measure Q 1, Q 1 an Q 1, an allow to obtain which ower values the comensator s reactances must absorb. Q 1 Q 1 Q 1 Q ¼ Q 1 Q 1 Q 1 Q ¼ Q 1 Q 1 Q 1 Q ¼ Q 1 Q 1 Q 1 ð1þ hese are the values which balance the system an make the ower factor unity. Uner this conitions the ower lost in the line is minimum. he revious formulas constitute a very easy algorithm to euce the values of the reactances of the comensator. he reactive owers elivere to the initial by each hase can be measure using three VAr meters connecte as shown in Fig. [7]. ffiffi Knowing that V is the effective value of the voltage between the hases of the, the reactance values of the comensator are X ¼ V ; X ¼ V ; X ¼ V ð1þ Q an suscetances are B ¼ Q V ; Q B ¼ Q V ; Q B ¼ Q V ð1þ he sign of each reactance is the same as the sign of its corresoning reactive ower. For that reason, if the reactive ower is ositive, the corresoning reactance is inuctive. If the ower is negative, the corresoning reactance is caacitive. he sign of each suscetance is the oosite to that of its corresoning reactive ower an reactance.. Other algorithms Other algorithms can be euce to obtain the reactive ower values of the comensator s reactances. For instance, reactive owers of each hase can be obtaine from the active an reactive owers measure as shown in Fig. [7]. In this figure, 1 ¼ P 1 þ jq 1 an 1 ¼ P 1 þ jq 1 are the owers acquire by the meters reresente. As V ¼ V =0, V ¼ V = 10 ¼ ffiffi ffiffi V 1= j =, V ¼ V =10 ¼ V 1= þ j =, I ¼ I =u,ani ¼ I = 10 þ u, then P 1 þ jq 1 ¼ U I ¼ðV V ÞI ¼ðV =0 V = 10 ÞI =u ¼ ffi!! P 1 Q 1 þ j P 1 þ Q 1 ð1þ If the first an last members are comare P 1 ¼ ffi P 1 Q 1 ð1þ Q 1 ¼ P 1 þ Q 1 ð17þ

4 By subtracting ffiffi the secon equation, reviously multilie by, from the first equation, P1 is eliminate an the result is Q 1 ¼ 1 ffiffi P 1 þ 1 Q 1 ð18þ imilarly Q 1 ¼ 1 ffiffi P 1 þ 1 Q 1 ð19þ An, seeing that Q 1 + Q 1 + Q 1 = Q 1 + Q 1 [7], the value for Q 1 is Q 1 ¼ 1 ffiffi ðp 1 P 1 Þþ 1 ðq 1 þ Q 1 Þ ð0þ By substituting (18) (0) in (1), these values are obtaine Q ¼ P 1 ffi Q 1 Q ¼ P 1 ffi Q 1 Q ¼ P 1 P 1 ð1þ hese are the owers of the reactances of the comensator which are able to balance the currents an fully comensate the reactive ower. If (18) (0) are substitute in (11), the comensator reactances, which balance the system, are obtaine; nevertheless, the reactive energy absorbe in this case by the comensator grou is Q, which can be arbitrarily fixe. If Q is the reactive ower elivere to the, the static comensator oes not absorb reactive ower (nor active ower, of course).. Examle he meters in Fig. show the following owers in kw an kvar: 1 ¼ P 1 þ jq 1 ¼ þ j an 1 ¼ P 1 þ jq 1 ¼ 8 þ j. V ¼ U= ffiffiffi ¼ 0 V. U is the line-to-line voltage. hus, currents are jq P 1,Q 1 P 1,Q 1 P 1 +jq 1 +jq 1 P 1 P 1 +jq 1 Unbalance three- hase Fig.. In orer to use the measures from two wattmeters an two varmeters, other algorithms coul be obtaine. F.. Quintela et al. / Electrical Power an Energy ystems 0 (008) I ¼ I ¼ 1 U 1 U ¼ 1:1= :87 ¼ :7=7:99 I ¼ I I ¼ :7= 1:9 he reactive ower elivere to the is Q 1 = Q 1 + Q 1 = 8 kvar. If the objective is to balance the system without altering the reactive ower it elivers to the, (1) shoul be use by aing Q / = Q 1 / = 8 10 / to them Q ¼ Q þ P 1 Q 1 ¼ 8:7 VAr Q ¼ Q P 1 Q 1 ¼ :7 VAr Q ¼ Q þ P 1 P 1 ¼ 7:7 VAr (1) gives the values of the suscetances B ¼ Q V ¼ 9:1 10 B ¼ Q V ¼ 1:97 10 B ¼ Q V ¼ : 10 he hase currents of the comensator are I ½I 7 Š¼ I I jðb þ B Þ jb jb ¼ jb jðb þ B Þ 7 jb jb jb jðb þ B Þ 11:7= :9 ¼ 11:7=:0 7 11:7=17:0 V V V 7 which is a negative-sequence symmetrical current set (more comments on this result will come later). he hase currents before the comensator are ½IŠ ¼ 7 ¼ I I I 7 þ I I I 7 ¼ 1:8= :9 1:8= 1:9 7 1:8=8:1 which is a ositive-sequence symmetrical current set, as intene. If ½V Š t ¼½V ; V ; V Š, then the comlex ower elivere to the comensator grou is [7] ½V Š t ½IŠ ¼ 1000 þ j8000 which is equal to that of the alone. If the objective is to comensate the reactive ower, then Q = 0. In that case

5 80 F.. Quintela et al. / Electrical Power an Energy ystems 0 (008) 7 8 B ¼ 11:1 10 ; B ¼ 0: 10 ; an B ¼ 1:91 10 ; 18:18=0 7 ½IŠ ¼ ¼ 18:18= :18=10 an the comlex ower elivere to the comensator grou is ½V Š t ½IŠ ¼ 1000 þ j0 7. urrents of the comensator he revious examle showe that the currents of the comensator form a negative-sequence symmetrical current set, when only the line currents are balance without comensating the reactive ower. his is not acciental, for it haens whenever the value of the comensator s reactive energy is zero. Inee, as the hase-neutral voltages are balance, they only have ositive-sequence symmetrical comonent V, therefore only the currents ositive-sequence symmetrical comonent of the comensator I intervenes in the formula of the ower it absorbs. hat is to say, the value of the comlex ower sulie to the comensator at all times is ¼ V I where I ¼ P þ is the conjugate of I. he value of the active ower P is always zero, ue to the comensator being forme only by reactances. If the value of the reactive ower Q is also zero, then ¼ V I ¼ 0 eeing that V ¼ 0, it follows that I ¼ 0 an I ¼ 0. herefore, if Q = 0, the comensator s currents lack of ositive-sequence symmetrical comonent. hose currents are also lacking zero-sequence symmetrical comonent, seeing that they a u to zero, consequently only the negative-sequence symmetrical comonent cannot be zero. It is conclue that the comensator s currents are a negativesequence symmetrical current set when the value of its reactive ower is zero. 8. omensator for four-wire three-hase s Fig. shows a roceure to cancel out the current through the neutral wire of any four-wire three-hase. wo reactances, X 1 an X, are connecte between the an hases, an the neutral wire, resectively, so that V =0 jx 1 V = 10 jx ¼ I ðþ Voltages of the initial are not affecte by the connection of X 1 an X. he active ower is not increase either. evertheless, the current through the neutral wire, I 0,is cancele out, which, in fact, transforms the into a three-wire three-hase. he reviously seen comensator can now be connecte to the three hases, as shown in Fig.. Phasor I can be etermine by using a watt varmeter connecte as inicate in Fig.. his meter reas ð V Þð I Þ ¼ V I ¼ P þ jq where P an Q are the wattmeter an varmeter reaings, resectively. From this formula it can be seen that I ¼ P þ jq ¼ P V V j Q V If this result is substitute in (), then V X 1 ¼ P V X ¼ P Q ffiffi which are the reactances that cancel out the current through the neutral wire. 9. oa voltage J 1 J I' jx 1 jx o far, the voltages of the have been suose to be symmetrical. But, if the currents are not balance, then the voltage ros of the hases, Z, Z, an Z, are ifferent, which leas to unbalance voltages near the. However, the iterative oeration of the comensator quickly rouces symmetrical voltages, when connecte. Fig. 7 summarizes one of the simulations mae. It shows a three-wire three-hase system forme by three symmetrical voltage sources V ¼ 100=0, V ¼ 100= 10, an I P,Q Fig.. If J 1 þ J ¼ I, then I 0 is zero, an the four-wire behaves as a three-wire. jq jx 1 jx I his behaves as a three-wire three-hase Fig.. tatic VAr comensator for a four-wire three-hase.

6 F.. Quintela et al. / Electrical Power an Energy ystems 0 (008) V V V Z Z I V V V Z Z Z ignal conitioning network (A) tatic comensator ignal conitioning network (B) Unbalance three- hase Z jq I/O ars Fig. 8. Block iagram of the exerimental evice. omuter (abview) Fig. 7. imulation network. able 1 Phase voltages an currents of the network simulation (Fig. 7) for each iteration of the comensator Iteration V (V) V (V) V (V) (A) (A) (A) V ¼ 100=10. he imeance of each hase wire is Z ¼ 1 þ j1. he is a triangle of imeances Z ¼ 10 þ j1, Z ¼ 0 þ j, an Z ¼ 80 þ j0. he effective values of the hase currents without the comensator are ¼ 9:7 A; ¼ 10:8 A; an ¼ : A: he voltages are V ¼ 8:70 V; V ¼ 8: V; an V ¼ 9:7 V: able 1 shows the values of the hase currents an voltages when the comensator is connecte, which have been obtaine from the iterate use of (1) an (1). V has been substitute each time by U = ffiffiffi in these algorithms. After the fifth iteration of the comensator the currents an voltages are balance. he tenency to balance is fast, because right after the secon iteration of the algorithm the effective value of the currents are very close to each other, as occurs with the effective values of the voltages. 10. Exerimental test An exerimental automatic evice was esigne an built at our laboratory to test the results of this aer. Fig. 8 shows its block iagram. It consists of a assive forme by imeances connecte between every two hases, which is reresente in Fig. 8 by the unbalance three-hase block. he connection of imeances can be moifie manually, thus allowing to obtain balance or unbalance s easily. Using a ata acquisition (DAQ) car an the alication abview running on a eskto comuter, the instantaneous currents i an i, Fig. 9. creenshots showing the unbalance hase currents near the (to) an the balance currents resulting from using the comensator (bottom). an the instantaneous voltages v an v, are measure before an after the comensator. o, by rogramming the aroriate algorithms in abview, the values of the reactances of the comensator are obtaine an the instruction to connecte them to the original is given through a igital outut car. Fig. 9 is a screenshot from the tests. he uer art of the icture shows the unbalance currents near the, an the lower art shows the balance currents resulting from using the comensator. 11. onclusion Avantages of the analysis of three-hase s escribe by relations between owers, instea of imeances, are shown in this aer. When its use is ossible, this analysis metho exlicitly inicates if results can, or cannot, be alie to active s. More secifically, the ossibility to use it to obtain the static VAr comensators algorithms is shown. hen, it is mae clear that the comensators function is vali whatever kin of three-hase it is connecte to, articularly, whether they are active or assive s. It is also shown that, if a comensator oes not eliver reactive ower, its three currents are always a negative-sequence symmetrical current set.

7 8 F.. Quintela et al. / Electrical Power an Energy ystems 0 (008) 7 8 eferences [1] ilsson James W, ieel usan. Electric ircuits. 7th e. Prentice Hall; 00. [] Howatson AM. Electric ircuits an ystems. Oxfor: Oxfor University Press; 199. [] Quintela F, Melchor, Melchor MM. Power of balance olyhase systems: an alication of the multi-terminal network ower theorem. Int J Elect Eng Euc 00;(): 7. [] ee, an-yi, hang, Wei-an, Wu, hi-jui. A comact algorithm for three-hase three-wire system reactive ower comensation an balancing. In: International conference on energy management an ower elivery, 199. Proceeings of EMPD 9, vol. 1, ovember, [] ingh, Bhim, axena, Anuraha, Kothari, DP. Power factor correction an balancing in three-hase istribution systems. In: EO IEEE region 10 international conference on global connectivity in energy, comuter, communication an control, vol., [] ee, Wu J, hang W. A comact control algorithm for reactive ower comensation an balancing with static Var comensator. Elect Power yst es 001;8: 70. [7] eono Quintela F, eono Melchor. Multi-terminal network ower measurement. Int J Elect Eng Euc 00:18 1.