Three Phase Theory - Professor J R Lucas

Transcription

1 Three Phae Theory - Profeor J Luca A you are aware, to tranit power with ingle phae alternating current, we need two wire live wire and neutral. However you would have een that ditribution line uually have only 4 wire. Thi i becaue ditribution i done uing three phae and the 4th wire i the neutral. How doe thi help? Since the three phae are uually o out of phae, their phaor addition will be ero if the upply i balanced. V N V V Y Figure a Three phae wavefor Figure b Phaor Diagra It i een fro figure a that in the balanced yte hown, the three phae, uually deignated, Y, correponding to ed, Yellow and lue, are equal in agnitude and differ in phae angle by. The correponding phaor diagra i hown in figure b. The voltage between any of the phae and the neutral i called the phae-to-neutral voltage or phae voltage V p. It i uual to call the voltage between any two line a the line-to-line voltage or line voltage V L. If the -phae voltage i V V p, then the reaining phae voltage would be V Y V p -π/ and V V p -4π/. alanced Supply A balanced three phae upply can be connected either in tar a in figure a or in delta a in figure b. V I P I L V Y I L I d V P V L V L V V V Y V Y Figure a Star-connected upply Figure b Delta-connected upply Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber

2 In a balanced three phae yte, knowledge of one of the phae give the other two phae directly. However thi i not the cae for an unbalanced upply. In a tar connected upply, it can be een that the line current current in the line i equal to the phae current current in a phae. However, the line voltage i not equal to the phae voltage. The line voltage are defined a V Y V V Y, V Y V Y V, and V V V. V -V Y V V V V Y -V V V Y V V Y V Y -V V Y V Y Figure a Parallelogra addition Figure b Triangular addition Figure a how how the line voltage ay be obtained uing the noral parallelogra addition. It can alo be een that triangular addition alo give the ae reult fater. For a balanced yte, the angle between the phae i o and the agnitude are all equal. Thu the line voltage would be leading the nearet phae voltage. Calculation will eaily how that the agnitude of the line voltage i tie the phae voltage. I L I P, V L V P, I L I d Siilarly in the cae of a delta connected upply, the current in the line i tie the current in the delta. It i iportant to note that the three line voltage in a balanced three phae upply i alo out of phae, and for thi purpoe, the line voltage ut be pecified in a equential anner. i.e. V Y, V Y and V. [Note: V Y i 8 out of phae with V Y o that the correponding angle if thi i choen ay appear to be 6 rather than ]. Thu if the direction of V i elected a reference, then V V p, V Y V p -π/ and V V p -4π/ and V Y V p π/6, V Y V p -π/ and V V p -7π/6 Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber

3 alanced Load A balanced load would have the ipedance of the three phae equal in agnitude and in phae. Although the three phae would have the phae angle differing by in a balanced upply, the current in each phae would alo have phae angle differing by with balanced current. Thu if the current i lagging or leading the correponding voltage by a particular angle in one phae, then it would lag or lead by the ae angle in the other two phae a well Figure 4a. V I V Z Z Z D Z D I Y I Z Z D V Y Figure 4a Phaor Diagra Figure 4b Star connection Figure 4c Delta connection The balanced load can take one of two configuration tar connection, or delta connection. For the ae load, tar connected ipedance and the delta connected ipedance will not have the ae value. However in both cae, each of the three phae will have the ae ipedance a hown in figure 4b and 4 c. It can be hown, for a balanced load uing the tar delta tranforation or otherwie, that the equivalent delta connected ipedance i tie that of the tar connected ipedance. The phae angle of the ipedance i the ae in both cae. Z D Z. Note: Thi can alo be reebered in thi anner. In the delta, the voltage i tie larger and the current tie aller, giving the ipedance tie larger. It i alo een that the equivalent power i unaffected by thi tranforation. Three Phae Power In the cae of ingle phae, we learnt that the active power i given by P V I co φ In the cae of three phae, obviouly thi ut apply for each of the three phae. Thu P V p I p co φ However, in the cae of three phae, the neutral ay not alway be available for u to eaure the phae voltage. Alo in the cae of a delta, the phae current would actually be the current inide the delta which ay alo not be directly available. Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber

4 It i uual practice to expre the power aociated with three phae in ter of the line quantitie. Thu we will firt conider the tar connected load and the delta connected load independently. For a balanced tar connected load with line voltage V L and line current I L, V tar V L /, I tar I L Z tar V tar / I tar V L / I L S tar V tar I tar * V L I L * Thu P tar V L I L co φ, Q tar V L I L in φ For a balanced delta connected load with line voltage V line and line current I line V delta V L, I delta I L / Z delta V delta / I delta V L / I L S delta V delta I delta V L I L Thu P tar V L I L co φ, Q tar V L I L in φ It i worth noting here, that although the current and voltage inide the tar connected load and the delta connected load are different, the expreion for apparent power, active power and reactive power are the ae for both type of load when expreed in ter of the line quantitie. Thu for a three phae yte in fact we do not even have to know whether it i a load or not, or whether it i tar-connected or delta-connected Apparent Power Active Power S V L I L P V L I L co φ eactive Power Q V L I L in φ Analyi of three phae balanced yte Since we know that the three phae are balanced and that the voltage and current are related to each other by, we do not have to do calculation for each of the three phae unnecearily. We could calculate for jut one phae uually the A phae in a yte with phae equence A--C. There are two coon ethod of doing thi. line Z L E N S Z L E E Y line line Z L Figure 5 Three phae yte Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber

5 a Single circuit of a three phae yte Conider the phae wire yte hown in figure 5 with the neutral wire abent. For a balanced yte, the upply voltage E, E Y and E will be out of phae. Uing Millann theore or otherwie, it can be eaily een that the potential of the tar point S of the load i equal to the potential of the neutral N of the upply. Thu whether a neutral wire i preent or not in the yte, the analyi of the yte can be identical. Thu we will draw a neutral wire between S and N of ero ipedance and do our analyi in that anner. Once the neutral wire i in place, and there i no potential difference between S and N, we could analye only one ingle phae of the yte, naely the A phae. Thi ay be redrawn a in figure 6. I p line V p Z L P p E p Figure 6 One phae of Three phae yte In thi cae, the upply voltage E p i a phae voltage, the upply current i the phae current I p, the load voltage V p i a phae voltage and the power P p i the power i one phae. If we copare with the line quantitie, we have E L E p, I L Ι p and P p P/ Uually, we are ore intereted in knowing the voltage and the power fro practical conideration, rather than the current. [For exaple, if I ak you the voltage and power rating of a bulb in your hoe, you would know it. However if I aked you for the current taken by the bulb, you would norally not be aware of the value but would probably obtain it fro the wattage and the voltage. The ae i true in a large power yte]. Thu we would like to reforulate the proble o a to give the voltage and the power at the deired value, even at the expene of a wrong current. b Equivalent circuit for three phae balanced yte I L I p P P p line V L V p Z L E L E p Figure 7 Equivalent ingle phae of Three phae yte Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber

6 Conider ultiplying the ource voltage E p in figure 6 by. Thi would increae both the line current I p and the load voltage V p by a factor of. Since both the load voltage and the load current ha increaed by tie, the load power would increae or tie. Thu we ee that uch a circuit figure 7 would have all voltage correponding to the line voltage, and all power correponding to the total three phae power a deired. The only quantity that would be in error i the line current which would appear a a current tie too high. Thi circuit i known a the equivalent ingle phae diagra and give the voltage and power a for the three phae yte but with the current being in error by tie. Let u conider an exaple to illutrate the ue of the circuit. Exaple A three phae 4V, 5 H, balanced upply feed a balanced load coniting of a three equal ingle phae load of 4 j Ω connected in tar, and b a three phae heating load purely reitive of.8 kw. Deterine the upply current, upply power factor, active and reactive power upplied and the value of the capacitance that ut be connected in delta to iprove the overall power factor to.95 lag. Obtain the reult uing i one phae of the three phae yte, and ii the equivalent ingle phae circuit. Solution I i Uing one-phae diagra figure 8 p I p I p Z L 4 j Ω E p Z L Z L V p E p 4/.9 P p.8/.6 kw 6 W.9.9 Figure 8 Single phae diagra I p j [Note: Quite often, we take the phae voltage of the three phae 4 V yte to be V rather than the calculated value of.9 V. You would then of coure get a lightly different anwer.] In order to calculate I p, we need not calculate Z L, but can ue P V I co φ. 6 I p.598 [Note: angle i ero becaue it i purely reitive].9 Thu I p I p I p j A upply current A upply power factor co lag active power upplied V L I L co φ W reactive power upplied V L I L in φ in var Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber

7 The capacitance are connected to iprove the power factor. Why do we want to iprove the power factor? Thi i becaue of the power factor i low and to tranfer the ae aount of active power we need a greater aount of current which in ter ean a uch greater aount of power loe in the yte, ince power lo i proportional to the quare of the current. Why do we ue capacitor to correct the power factor? Thi i becaue ot noral load have an inductive coponent for exaple, a fan ha a winding and a fluorecent lap ha a choke, both of which are baically inductor. We can copenate for the reactive power of an inductive reactance by the reactive power of a capacitive reactance. In thi particular exaple, we are required to iprove the power factor to.95. Why to.95? Why not to., which would give the lowet power factor? Thi i becaue iproveent of power factor ean additional capacitance. We try to ue only o uch capacitance a would give u a financial benefit. When we iprove the power factor angle fro 6.87 o correponding to power factor.8 lag to 8.9 o correponding to a power factor of.95 we get an iproveent of or.5/ %. Wherea when we iprove a iilar aount of angle fro 8.9 o to o, we get an iproveent of only or.5/.95 5.%. In fact if we iproved a iilar aount fro 5. o correponding to a power factor of.6 to 6.87 o we get an even larger iproveent.8.6. or./.6.%. Thu we can ee that a we coe cloer and cloer to unity power factor, the benefit rapidly decreae. Thu in indutry it i uual to iprove the power factor to a value lightly le than unity power factor, and thi value can be theoretically calculated uing uch inforation a the cot of capacitor, the electricity tariff etc. Let u get back to doing the calculation. When the power factor i iproved to.95 lag, uing a pure capacitance, then the aount of active power doe not change but reain the ae a before. i.e. P 4.6 kw. However, the reactive power will decreae uch that the overall power factor i.95. P [.95 i obviouly power factor lag, a to correct to.95 power factor lead would be even ore cotly Q than to iprove to unity power factor and hence new would abolutely have no advantage] Q cap Figure 9 how the diagra howing the active and Q reactive power during power factor correction. Figure 9 Power factor correction The capacitance ut add the difference between original aount of reactive power upplied and the new aount of reactive power upplied. Q.9 k var with the new power factor, Q new i calculated fro Q P tanφ 4.6 tan8.9.4 k var new new Q cap k var each of the capacitor would provide one-third thi reactive power. Capacitance required.487/.6 k var V C ω Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber

8 If the capacitor are connected in delta, then the line voltage would appear acro each. 6 4 C π C. µf each connected in delta If the capacitor are connected in tar, then the phae voltage would appear acro each. 6.9 C π C µf each connected in tar ii Uing equivalent ingle-phae diagra figure 9 Z L 4 j Ω E L 4 P.8 kw 8 W 4 4 j 4 I I L I I L ZL I I E L Z L V L Figure 9 Equivalent diagra In order to calculate I, we can ue P V L I L co φ V L I co φ. 8 I 4.5 [Note: angle i ero becaue it i purely reitive] 4 Thu I L I I I j A upply current / A which i the ae a in the earlier ethod. eaining calculation will be iilar. Unbalanced three phae yte An unbalanced three phae yte i one which i not perfectly balanced. It ay be caued by the upply being unbalanced, or ore uually the load being unbalanced or both. In uch a cae, knowledge of the current or voltage in one phae doe not tell u the current or voltage in the other phae. Thu all phae quantitie ut be independently deterined. Let u conider oe of the coon unbalanced ituation to ee how thi ay be done. a Star connected upply feeding a tar connected load line Z L E N Z neutral S Z LY E E Y line line Z L Figure Unbalanced Three phae yte Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber

9 i If Z neutral i conidered ero, each individual phae current can be independently deterined fro the upply voltage in that phae and the ipedance of that phae. I L E line Z L, I LY Then the load voltage etc can be deterined. E Y line Z LY, I L E Z line L ii If there i a neutral ipedance, then uing Millann theore, we will firt have to deterine the voltage of the tar point of the load with repect to the upply neutral.. E. E Y. V line Z L line Z LY VSN Y line Z L line Z LY fro which V SN i known. Thu the load current can be deterined fro E VSN EY VSN I L, I LY line Z L line Z Hence the reaining quantitie can be deterined. LY Y, I L line Z Z E line V line SN Z L L L. E neutral neutral. iii If the yte i a -wire yte, rather than a 4-wire yte, the analyi i the ae a if neutral were i.e. / neutral. Thu again Millann theore i ued to deterine V SN and the load current are then deterined. b Delta connected upply feeding a tar connected load If the upply wa connected, not in tar but in delta figure, which i not the cae in practice, then we would have to write the Kirchoff current law for the loop and olve a a noral circuit proble. E Y line E S Z L Z LY E Y line Z L line Figure Delta upply feeding tar load Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber

10 c Delta connected upply feeding a delta connected load E Y line E E Y line line Figure - Delta connected upply feeding a delta connected load When a delta connected upply feed a delta connected load figure, which i not uual, then the line voltage are known o that the current inide the delta can be obtained directly fro Oh Law. The line current can then be obtained by phaor uing of the current inide the delta. The reaining variable are then obtained directly. d Star connected upply feeding a delta connected load line E E Y E line line Figure - Star connected upply feeding a delta connected load When a tar connected upply feed a delta connected load figure, then fro the phae voltage the the line voltage are known o that the current inide the delta can be obtained directly fro Oh Law. The line current can then be obtained by phaor uing of the current inide the delta. The reaining variable are then obtained directly. Thu baically, any unbalanced yte can be calculated uing the baic network theore. Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber

11 Syetrical Coponent or Sequence Coponent Phae Sequence A three phae yte of voltage or current ha a equence or order in which the phae reach a particular poition for exaple peak value. Thi i the natural equence of the upply. According to uual notation, we would call the equence -Y- or A--C. If we conider a balanced yte of voltage or current they will have only the natural equence, and there will no other coponent preent. However, Fortecue ha forulated that any unbalanced yte can be plit up into a erie of balanced yte. [Thi i like aying that any force can be broken up into it coponent along the x-axi, y-axi and -axi. The advantage of uch a decopoition i in the analyi of ore than one quantity] In the cae of unbalanced three phae yte, uch a hown in figure 4, the unbalanced yte can be plit up into coponent: i a balanced yte having the ae phae equence a the unbalanced yte, ii a balanced yte having the oppoite phae equence to the unbalanced yte rotation of phaor i alway anticlockwie whether they are in the ae equence or oppoite, o that it i the order of the phae that change, and not the direction of rotation, and iii a balanced yte of inphae quantitie. Poitive Sequence Negative Sequence Zero Sequence Y Y Y,Y, Figure 4 Decopoition of unbalanced three phae In any three phae yte, the phae quantitie, Y and or A, and C ay be expreed a the phaor u of: - a et of balanced poitive phae equence quantitie A, and C phae equence a-b-c : ae phae equence a original unbalanced quantitie, - a et of balanced negative phae equence current A, and C phae equence a-c-b: oppoite phae equence to original unbalanced quantitie, - a et of identical ero phae equence current A, and C inphae, no phae equence. Figure 5 - egrouping Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber

12 It i to be noted that the original unbalanced yte effectively ha coplex unknown quantitie A, and C agnitude and phae angle of each i independent. It i alo to be noted that each of the balanced coponent have only one independent coplex unknown each, a the other can be written by yetry. Thu the three et of yetrical coponent alo have effectively coplex unknown quantitie. Thee are uually elected a the coponent of the firt phae A i.e. A, A and A. One of the other phae could have been elected a well, but all coponent hould be elected for the ae phae. A can be obtained by the phaor addition of A, A and A. Siilarly and C. Thu A A A A C C C C If the balanced coponent are conidered, we ee that the ot frequently occurring angle i. In coplex nuber theory, we defined j a the coplex operator which i equal to - and a agnitude of unity, and ore iportantly, when operated on any coplex nuber rotate it anti-clockwie by an angle of 9. i.e. j - 9 In like anner, we can define a new coplex operator which ha a agnitude of unity and when operated on any coplex nuber rotate it anti-clockwie by an angle of. i.e. -.5 j.866 Let u again exaine the equence coponent of the unbalanced quantity. A A Y A C C A A C A C A Figure 6 Decopoition of unbalanced three phae We can expre all the equence coponent in ter of the quantitie for A phae uing the propertie of rotation of, or 4. Thu A A A A A A A C A A A Thi can be written in atrix for. Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber

13 Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber [ ] Sy Ph A A A C A Λ Thi give the baic yetrical coponent atrix equation, which how the relationhip between the phae coponent vector Ph and the yetrical coponent vector Sy uing the yetrical coponent atrix [Λ]. oth the phae coponent vector Ph and the yetrical coponent vector Sy can be either voltage or current, but in a particular equation, they ut of coure all be of the ae type. Since the atrix i a [ ] atrix, it i poible to invert it and expre Sy in ter of Ph. ut to do thi, it would be convenient to firt expre oe propertie of. Soe Propertie of π/ or 4π/ or 4 or - π or 6 or i.e. - - Since i coplex, it cannot be equal to, o that - cannot be ero. Thi alo ha the phyical eaning that the three ide of an equilateral triangle ut cloe. Alo and Now let u look at inverting the yetrical coponent atrix. Invere of Syetrical coponent atrix [ ] Λ, o that [ ] Λ 4 Figure 7 Phaor Addition

14 Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber [Λ] - and the dicriinent - - Subtituting, the atrix equation iplifie to give [Λ] - Since, and, the atrix equation further iplifie to [Λ] - It i een that i the coplex conjugate of, and i the coplex conjugate of. Thu the above atrix [ ] - i one-third of the coplex conjugate of [ ]. i.e. [Λ] - [Λ] * Thi can now be written in the expanded for a [ ] Ph Sy C A A A A Λ Exaple Find the yetrical coponent of the unbalanced yte of the following voltage V, - V and 9 V. Solution Writing the atrix equation 9 V V V Expanding the equation give V [ - 9 ] [ - / j/ j] [.4 j.5].45 j V V V V V V Y V 6 o Figure 8 Unbalanced Phaor

15 V [ ] [ j] [4.464 j].488 j V V [ 4-9 ] [ - / j/ - j] [-.598 j.5] -.5 j V Graphical Method of Solution The ethod of deterining the equence coponent fro the phae coponent and the phae coponent fro the equence coponent are iilar except that the correct equation ut be ued. The olution of the equation ay alo be done graphically. The advantage of a graphical olution i that it give an inight to the coponent very quickly without the need of a rigorou analyi. In the graphical analyi, in addition to phaor addition, ultiplication by or would correpond to an anticlockwie rotation of or 4 repectively. Thi can be bet undertood by an exaple. V V Exaple For the unbalanced et of phaor hown in figure 9, verify graphically the equence coponent obtained in exaple. Solution V V V V V V V V Y V [ V VY V ] or V [ V V V ] Y It i ore coon to plot V rather than V and to obtain one-third the reult. Thi i hown in figure. It can be een that the reultant ha a agnitude of about half that of V ay.5 and an angle of lightly greater than that of the V Y ay 75. Thi correpond to V. If we copare the reult with that of the analytical ethod in exaple, we ee that the value for V hould be V agreeing with the obervation. Siilarly, V [ V V V ] Y 6 o V Y V Figure 9 Unbalanced Phaor Figure Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber

16 or V [ V V V ] Y The phaor addition i hown in figure. It can be een that the reultant ha a agnitude of lightly greater than one and half tie that of V & V y ay 4. together and an angle cloe to 5 below the horiontal axi. Thi correpond to V. If we copare the reult with that of the analytical ethod in exaple, we ee that the value for V hould be V roughly agreeing with the obervation. 4 Figure [An accurate reult could have been obtained if actual eaureent had been done ]. Siilarly the negative equence can be obtained a follow. V [ V VY V ] or V [ V V V ] Thi phaor addition i hown in figure. It can be een that the reultant ha a agnitude of lightly le than that of V y ay.7 together and an angle cloe to 5 above the negative horiontal axi. Thi correpond to V. If we copare the reult with that of the analytical ethod in exaple, we ee that the value for V hould be V roughly agreeing with the obervation. Y Figure 4 Suary V p [Λ] V, V >Λ] * V p, I p [Λ] I I >Λ] * I p [ Λ], [Λ] - [Λ] * Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber

17 Sequence Ipedance Let u now conider how the ipedance appear in equence coponent. To do thi we ut firt look at the ipedance atrix in phae coponent which we know. V p [Z p ].I p Subtituting for V p and I p in ter of the yetrical coponent we have [Λ] V [Z p ]. [Λ] I pre-ultiplying equation by [Λ] - we have V [Λ] -.[Z p ]. [Λ] I Thi give the relationhip between the yetrical coponent voltage V and the yetrical coponent current I, and hence define the yetrical coponent ipedance atrix or Sequence Ipedance atrix. Thu [Z ] [Λ] -.[Z p ]. [Λ] [Λ] *.[Z p ]. [Λ] In a iilar anner, we could expre the phae coponent ipedance atrix in ter of the yetrical coponent ipedance atrix a follow. [Z p ] [Λ].[Z ]. [Λ] - [Λ].[Z ]. [Λ] * The for of the equence ipedance atrix for practical proble give one of the ain reaon for ue of yetrical coponent in practical power yte analyi. If we conider the iple arrangeent of a phae traniion line figure, we would have the equivalent circuit a a L a b M ab L b M ca c M bc L c Figure phae traniion line If we think of an actual line uch a fro Victoria to Kotale, we would realie that all phae wire would have approxiately the ae length other than due to difference in agging and hence we can aue the elf ipedance coponent to be equal for each phae. i.e. a b c and L a L b L c When a current pae in one phae conductor, there would be induced voltage in the other two phae conductor. In practice all three phae conductor behave iilarly, o that we could conider the utual coupling between phae alo to be equal. Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber

18 Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber i.e. M ab M bc M ca In uch a practical ituation a above, the phae coponent ipedance atrix would be fully yetrical, and we could expre the uing a elf ipedance ter and a utual ipedance ter. Thu we ay write the phae coponent ipedance atrix a [ ] p Z We ay now write the yetrical coponent ipedance atrix a [ ] [ ] [ ][ ] Λ Λ.. * p Z Z Thi can be iplified uing the property a follow [ ] Z i.e. [ ] Z Z Z Z We ee an iportant reult here. While the phae coponent ipedance atrix wa a full atrix, although it had copletely yetry, the equence coponent ipedance atrix i diagonal. The advantage of a diagonal atrix i that it allow decoupling for eae of analyi. To undertand the iportance of decoupling or a diagonal atrix, let u look at the following iple algebraic proble. 5 x y 6 x 5 y 4 x y 5 -

19 You can ee that there i a lot of yetry in the proble, if forulated a a atrix equation. 5 x 6 5 y 4 5 However, I a alot ure you would not be able to give the olution x, y, -5 of thi equation entally. However if I give you the following et of equation 4 x y x 5 y x y -5 which correpond to a diagonal atrix, I a ure all of you would have been able to get the correct olution entally and in a flah. Thi i becaue the olution of x require only the firt equation, that of y require only the econd equation and that of only the third equation. Power aociated with Sequence Coponent With phae coponent, power in a ingle phae i expreed a P phae V I co φ Thu in three phae, we ay either write P V L I L co φ V p I p co φ for a balanced three phae yte. However, with an unbalanced yte thi i not poible and we would have to write the power a the addition of the power in the three phae. Thu Apparent Coplex Power S V a I * a V b I * * b V c I c The active power P i obtained a the eal part of the coplex variable S. Thi equation ay be re-written in atrix for a follow. * I a * T * S [ V a Vb Vc ] I b V p. I p * I c Let u now convert it to yetrical coponent, a follow. S V T p. I * T p [ Λ ] V ].[ Λ]. I ] *. which ay be expanded a follow. T T * S [ ] [ ] * T * V Λ Λ. I V [ ] [ ] Λ. I.. Λ V T. I * i.e. S V a I * a V a I * a V a I * a Thi reult can alo be expected, a there are phae in each of the equence coponent taking the ae power. Thu P V a I a co φ V a I a co φ V a co φ Three Phae Theory & Syetrical Coponent Profeor J Luca Noveber