Chapter 10: Symmetrical Components and Unbalanced Faults, Part II

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1 Chpter : Symmetricl Components nd Unblnced Fults, Prt.4 Sequence Networks o Loded Genertor n the igure to the right is genertor supplying threephse lod with neutrl connected through impednce n to ground. The genertor genertes threephse blnced voltges deined s: bc E = E The terminl voltge o the genertor is ound using Kircho's voltge lw: V = E s n n V = E b b s b n n V = E c c s c n n And since n = b c, we hve: V E s n n n Vb = Eb n s n n b V E c c n n s n c bc bc bc bc Or in compct orm: V = E. This eqution cn be trnsormed to the "" or symmetricl component orm thus: bc AV = AE A Multiplying by Where A we hve: bc = A A V = E A A bc = E s n n n = n s n n n n s n Perorming the bove mtrix multipliction we hve: s n = s = s n n E c c E E b b V V V b c

2 t is importnt to note tht the mtrices bove re digonl, nd the equtions re completely decoupled. Expnding into three voltge equtions we hve: V = These equtions represent three seprte circuits s shown to the right, one or the positivesequence, one or the negtivesequence, nd one or the zerosequence networks. V = E ( ) V E = V V Positivesequence Negtivesequence erosequence V The ollowing importnt observtions re mde: The three sequences re independent. The positivesequence network is the sme one used in the oneline digrm or studying blnced threephse currents nd voltges. Only the positive sequencenetwork hs voltge source. Thus the positivesequence voltge will generte only positivesequence currents. There is no voltge source in the negtive nd zerosequence networks. Negtive nd zerosequence currents cuse only negtive nd zerosequence voltges. The neutrl o the system is the reerence point or the positive nd negtivesequence networks. The ground is the reerence point or the zerosequence; hence the zerosequence current cnnot low unless there is connection to ground. The grounding impednce is relected in the zero sequence network s n. Ech o the three sequence networks cn be solved seprtely on per phse bsis. The phse currents nd voltges cn then be ound by superposition, by dding their symmetricl components o current nd voltge respectively..5 Single LineToGround Fult This is the most common ult on threephse system. t is illustrted using the ollowing simple network:

3 Here the ult occurs between line "" nd ground through ult impednce. Assuming the genertor ws initilly t no lod, the boundry conditions t the ult re: V = b = c = Using the second eqution in the symmetricl components o currents we hve: = Thereore = = =. Thus the three symmetricl components re equl nd ech is equl to onethird phse "" current. Phse "" voltge in terms o symmetricl components is: V = V V V Using eqution lst pge, nd the ct the three symmetricl components o current re Where = = = we hve: V = E s n = nd bove eqution becomes: = =. Recll tht V = nd tht = E =, the Or: E = ( ) And the ult current is: E = = Using the symmetricl components o current in eqution the symmetricl components o voltge re ound, hence by trnsormtion, the voltges during the ult. n n E c E c = E b b = V V V b c Observing the equtions nd we notice tht the sequence networks derived erlier my be connected s shown here which stisies these two equtions. E V V V

4 the genertor neutrl is grounded, then n = nd = s, nd or bolted ult =..6 LineToLine Fult This ult is shown in the igure to the right. Note the neutrl is not grounded. The ult is shown between phses b nd c. t is ssumed the genertor is initilly on nolod. The boundry conditions t the ult point re: V V = b c b b = c = Using = nd c = b, the symmetricl components o currents become: = b b Expnding the mtrix eqution we hve: = = x = ( ) b b E c E c E b b = V b V c V Note tht = rom (x) bove. From the equtions or phse voltges (reproduced below) we hve: V = V V V Hence Using the vlues o V nd V = V V V b V = V V V c V V = V V b c = V rom, nd note tht b =, we hve: E = And substituting or b rom (x) bove we hve: b 4

5 Since ( )( ) E = =, solving or The phse currents re: The ult current is: we hve: E = ( )( ) b = c = = b c = j Using the symmetricl components o currents in eqution, the symmetricl components o voltge (hence phse voltges by trnsormtion,) t the ult point re obtined. Note tht = nd =, hence rom xx bove we see tht the sequence network or the linetoline ult is s shown on the right. E V V.7 Double LineToGround Fult This ult is not s common s the previous two, but more likely thn the symmetricl ult (ll lines bolted, or connected through ult to the neutrl.) The double linetoground ult is illustrted on the right. The ult is short between lines b nd c, which re grounded through ult impednce. Agin, the genertor is ssumed initilly t nolod. The boundry conditions t the ult point re: Vb = Vc = b c (.) = = n E c E c E b b = V c V b V We know tht the voltges V b nd V c re given by: 5

6 (.) V = V V V b V = V V V c And since Vb = Vc, we hve: (.) V = V Substituting "" or the "bc" currents in (.) we hve: (.4) ( ) V = b = = Using (.4) nd (.) in the top eqution o (.) we hve: = V V (.5) = V V Using () or the symmetricl components o voltge in the eqution bove, nd solving or we hve: E (.6) = Using the eqution () in (.) we hve: E (.7) = Now using the bove two equtions in the second eqution o (.) we hve (ter some lgebr:) E (.8) = ( ) Note tht E drives the current ( ). Clerly,, hence the impednce seen by E is: this is the impendence in series with the prllel combintion o nd E. Thus the sequence V V networks cn be connected s shown in the circuit on the right. Once the sequence currents re ound rom (.8), (.7) nd (.6) [in tht order,] the phse currents re ound using the trnsormtion mtrix A. Finlly the ult current is ound rom: (.9) = = b c V 6

7 [This is seen rom the eqution ( ) Exmple.5 = nd knowing in this cse =.] b c The oneline digrm o simple power system is shown to the right. The neutrl o ech genertor is grounded through currentlimiting rector o.5/ per unit n MVA bse. The system dt expressed in per unit on common MVA bse is tbulted below. The genertors re running on nolod t their rted voltge nd rted requency with their ems in phse. Determine the ult current or the ollowing ults: G G T T () A blnced threephse ult t bus through ult impednce = j. per unit. (b) A single linetoground ult t bus through ult impednce = j. per unit. (c) A linetoline ult t bus through ult impednce = j. per unit. (d) A double linetoground ult t bus through ult impednce = j. per unit. tem Bse MVA Voltge rting X X X G kv G kv T / kv... T / kv... L kv.5.5. L kv L kv Preliminry clcultions: First we ind the Thevenin impednce viewed rom bus, the ulted bus. First the delt is chnged to Y s cn be seen in the igure below on the right. Now the Yimpednces re clculted thus: 7

8 S S S ( j.5)( j.5) = = j.55 ( j.5)( j.5) = = j.55 ( j.5)( j.5) = = j.55 j.574 j.5958 j.7486 Combining prllel brnches, nd dding the series brnch, the positivesequence Thevenin impednce is: j.5 j.5 j.5 j.5 j.5 j.5 j.5 j.574 S j.5954 j.748 ( j.8574)( j.958) = j.5958 j.7486 = j. The positive nd negtivesequence networks re shown to the right. The only dierence between them in this cse is tht the source is missing in the negtivesequence network. Hence we hve: = = j. Now, the zerosequence network is constructed bsed on the trnsormer j.5 connections nd is shown j.5 in the igures to the right. j.5 Note tht the delt is j.5 reduced to its equivlent Y irst, nd then the impednces re combined to ind the equivlent zerosequence circuit. The j. j. result is the simple circuit shown below. j. j.5 j.5 j.75 E j. j.5 j.5 j.5 j.5 j. Positivesequence Negtivesequence j. j. j.7764 S j.5688 erosequence j.86 8

9 Now we re redy to solve the problem! () Blnced threephse ult t bus. We ssume the genertors t nolod hve voltge o. per unit, hence the ult current is: V. = = = j.5 pu j. j. =8. 9 V (b) Single linetogroud ult t bus. We know the sequence currents re the sme nd given by: V = = = =. j. j. j.5 j. = j.974 pu And using the mtrix A we hve:.75 j b = = = pu c (c) Linetoline ult t bus. The zerosequence component o current is zero, i.e. =. Also rom the nlysis o the linetoline done previously we hve: V = = = = j.859 pu j. j. j. The ult current is: b j = = c j (d) Double linetoground ult t bus pu From the equtions developed erlier or the sequence currents o this ult we hve: 9

10 V = = = j.67 pu ( ) j.( j.5 j.) j. j. j.5 j. (.)(.67) V j j = = = j. (.)(.67) V j j = = = j.5 j. And using the A mtrix we hve the phse currents: The ult current is:.6579 j b j = = c j b c F = = j.948 j pu

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