Power System Engineering Unblnced Fult Anlysis
Anlysis of Unblnced Systems Except for the blnced three-phse fult, fults result in n unblnced system. The most common types of fults re single lineground (SLG) nd line-line (LL). Other types re double line-ground (DLG), open conductor, nd blnced three phse. The esiest method to nlyze unblnced system opertion due to fults is through the use of symmetricl components Slide # 1
Symmetricl Components The key ide of symmetricl component nlysis is to decompose the unblnced system into three sequence of blnced networks. The networks re then coupled only t the point of the unblnce (i.e., the fult) The three sequence networks re known s the positive sequence (this is the one we ve been using) negtive sequence zero sequence Slide # 2
Symmetricl Components Unblnce Currents Blnce Systems Sequence Currents zero sequence Unsymmetricl Fult Unblnce System C A Symmetricl components Zero Sequence Positive Sequence Three blnced Systems positive sequence B Negtive Sequence negtive sequence Slide # 3
Slide # 4 Assuming three unblnce voltge phsors, A, B nd C hving positive sequence (bc). Using symmetricl components it is possible to represent ech phsor voltge s: C C C C B B B B A A A A Where the symmetricl components re: Positive Sequence Component Negtive Sequence Component Zero Sequence Component Symmetricl Components
Symmetricl Components, The Positive Sequence Components ( A B C ) Three phsors Equl in mgnitude Displced by 12 o in phse Hving the sme sequence s the originl phsors (bc), C o 12 o o 12 12 B A,, The Negtive Sequence Components ( ) A B C Three phsors Equl in mgnitude Displced by 12 o in phse Hving the opposite sequence s the originl phsors (cb) B o 12 o o 12 12 C A The zero Sequence Components ( A B ) Three phsors Equl in mgnitude Hving the sme phse shift ( in phse),, C B C A Slide # 5
Exmple A Zero Sequence A B C A B C A B C A B C A B C A A C Positive Sequence C o 12 o 12 A A A B C A B Unblnce oltge Negtive Sequence B o 12 o 12 C B Synthesis Unsymmetricl phsors using symmetricl components Slide # 6
Sequence Set Representtion Any rbitrry set of three phsors, sy, b, c cn be represented s sum of the three sequence sets b b b b c c c c where b c b c b c,, is the zero sequence set,, is the positive sequence set,, is the negtive sequence set Slide # 7
Conversion Sequence to Phse Only three of the sequence vlues re unique,,, ; the others re determined s follows: b c 2 3 3 α 1 12 α α α α 1 (since by definition they re ll equl) 2 b α c α b α c α 2 1 1 1 1 1 1 2 2 b 1 α α 1 α α 1 2 2 c α α 1 α α Slide # 8
Conversion Sequence to Phse Define the symmetricl components trnsformtion mtrix 1 1 1 2 A 1 α α 2 1 α α Then A A A b s c Slide # 9
Conversion Phse to Sequence By tking the inverse we cn convert from the phse vlues to the sequence vlues s 1 A 1 1 1 1 1 2 with A 1 3 α α 1 2 α α Sequence sets cn be used with voltges s well s with currents Slide # 1
Exmple f the vlues of the fult currents in three phse system re: 15 45 25 15 1 3 A B Find the symmetricl components? Solution: C O Slide # 11
Exmple f the vlues of the sequence voltges in three phse system re: o 1 2 6 1 12 Find the three phse voltges Solution: A 2 6 1 12 A 3 6 1 B 1 24( 2 6 ) 1 12( 1 12 ) 1 B 3 6 C 1 12( 2 6 ) 1 24( 1 12 ) 1 C Slide # 12
Use of Symmetricl Components Consider the following wye-connected lod: n b c Z Z g y n n ( Z Z ) Z Z g Y n n b n c Z ( Z Z ) Z bg n Y n b n c Z Z ( Z Z ) cg n n b Y n c g Zy Zn Zn Zn Z Z Z Z bg n y n n b cg Zn Zn Zy Z n c Slide # 13
Use of Symmetricl Components g Zy Zn Zn Zn Z Z Z Z bg n y n n b cg Zn Zn Zy Z n c Z A A 1 1 s s s s A ZA A ZA A Zy 3Zn Z A Z y Z y s s Slide # 14
Networks re Now Decoupled Zy 3Zn Zy Z y Systems re decoupled ( y 3 n) y Z Z Z Z y Slide # 15
Grounding When studying unblnced system opertion how system is grounded cn hve mjor impct on the fult flows Ground current only impcts zero sequence system n previous exmple if lod ws ungrounded the zero sequence network is (with Z n equl infinity): Slide # 16
Sequence digrms for lines Similr to wht we did for lods, we cn develop sequence models for other power system devices, such s lines, trnsformers nd genertors For trnsmission lines, ssume we hve the following, with mutul impednces Slide # 17
Sequence digrms for lines, cont d Assume the phse reltionships re Δ Zs Zm Zm Δ b Zm Zs Z m b Δ c Zm Zm Zs c where Z Z s m self impednce of the phse mutul impednce between the phses Writing in mtrix form we hve Δ Z Slide # 18
Sequence digrms for lines, cont d Similr to wht we did for the lods, we cn convert these reltionships to sequence representtion Δ Z Δ A Δ A 1 s s s s A Δ Z A Δ A Z A Zs 2Zm 1 A Z A Zs Zm Zs Zm Thus the system is gin decoupled. A rule of thumb is tht Z Z nd Z is pproximte 3 times Z. s s Slide # 19
Sequence digrms for genertors Key point: genertors only produce positive sequence voltges; therefore only the positive sequence hs voltge source During fult Z Z X d. The zero sequence impednce is usully substntilly smller. The vlue of Z n depends on whether the genertor is grounded Slide # 2
Sequence digrms for Trnsformers The positive nd negtive sequence digrms for trnsformers re similr to those for trnsmission lines. The zero sequence network depends upon both how the trnsformer is grounded nd its type of connection. The esiest to understnd is double grounded wye-wye Slide # 21
Trnsformer Sequence Digrms Slide # 22
Unblnced Fult Anlysis The first step in the nlysis of unblnced fults is to ssemble the three sequence networks. Consider the following exmple T 1.5 G1 T1 Trnsmission Line T2 T 1.5 fult G2 Δ Δ MA oltge X X - X o G1 1 11 k.15.17.5 G2 1 11 k.2.21.1 T1 1 11/22k. 1.1.1 T2 1 11/22k.1.1.1 Line 1 22k.15.15.315 J. 5 Slide # 23
Sequence Digrms for Exmple Positive Sequence Network J.15 J.1 J.15 J.1 J.2 o o 1.5 1.5 Negtive Sequence Network J.17 J.1 J.15 J.1 J.21 Slide # 24
Sequence Digrms for Exmple Zero Sequence Network J.5 J.1 J.315 J.1 J.1 J.15 Slide # 25
Crete Thevenin Equivlents Second is to clculte the Thevenin equivlents s seen from the fult loction. n this exmple the fult is t the terminl of the right mchine so the Thevenin equivlents re: 1.5 E j.1389 - j.1456 - j.25 Z Z th th j.2 in prllel with j.455 j.21 in prllel with j.475 Slide # 26
Single Line-to-Ground (SLG) Fults Unblnced fults unblnce the network, but only t the fult loction. This cuses coupling of the sequence networks. How the sequence networks re coupled depends upon the fult type. We ll derive these reltionships for severl common fults. With SLG fult only one phse hs non-zero fult current -- we ll ssume it is phse A. Slide # 27
SLG Fults, cont d gnoring prefult currents, the SLG fult cn be described by the following voltge nd current reltionships: b c & Z f The terminl unblnce currents t the fult point cn be trnsferred into their sequence components s follows: 1 1 1 1 2 1 α α 3 2 1 α α Z f b b c 3 c Slide # 28
SLG Fults, cont d During fult, Z f nd o 3Z f The terminl voltge t phse cn be trnsferred into its sequence components s: 3Z f 3Z f Slide # 29
SLG Fults, cont d The only wy tht these two constrint cn be stisfied is by coupling the sequence networks in series Z f b b c c Zero Sequence Circuit Positive Sequence Circuit o o Zero Sequence Circuit Positive Sequence Circuit o o 3Z f 1 E Z o Z o o 3Z f 3Z f 3Z f Negtive Sequence Circuit Negtive Sequence Circuit Z Slide # 3
Exmple: Consider the following system T 1.5 G1 T1 Trnsmission Line T2 T 1.5 fult G2 Δ Δ J.5 ts Thevenin equivlents s seen from the fult loction re: 1.5 E j.1389 f - j.1456 f - j.25 f Slide # 31
Exmple, cont d j. 1389 With the sequence networks in series, we cn solve for the fult currents E 1. 5 j. 1456 - - A s 1.5 j(.1389.1456.25) j5.8 (of course, b c j1.964 NOTE 1: These re the currents t the SLG fult point. The currents in the system during the SLG fult should be computed by nlyzing the sequence circuits. ) j. 25 Slide # 32
Exmple, cont d From the sequence currents we cn find the sequence voltges s follows: 1.5 A s Z,, b Z, o 1.166 j.178, c Z o 1.166 j.178 NOTE 2: These re the voltges t the SLG fult point. The voltges t other loctions in the system (during the SLG fult) should be computed by nlyzing the sequence circuits. Slide # 33
Line-to-Line (LL) Fults The second most common fult is line-to-line, which occurs when two of the conductors come in contct with ech other. With out loss of generlity we'll ssume phses b nd c. b b Z f c c Current reltionships: oltge reltionships: b c & Z b f b c Slide # 34
LL Fults, cont'd Using the current reltionships, we get Therefore, Hence 1 3 1 1 1 α α 1 2 1 ( α α 2 b ) 3 1 ( α 2 α b ) 3 1 2 α b α b 2 α NOTE α 1 12 α.5 j.866 2 α 1 24.5 2 α α j 2 α α j.866 j 3 3 Slide # 35
LL Fults, con'td Therefore, it is obvious tht, during LL Fults there is no zero sequence components in the sequence circuit tht represents this fult. During LL fult, we hve: b c bz f Using the symmetricl components, then: α 2 α b Z f c b 2 α α 2 Z ( f α α ) _ Z o Z b b _ o Z f c _ Slide # 36
LL Fults, con'td Therefore, α 2 α 2 2 α α Z ( ) f α α Z 2 2 2 ( α α) ( α α) ( α α) Z f Substitute for Then, b b Z f c c Z f To stisfy, Z f nd, the positive nd negtive sequence networks must be connected in prllel Zero Sequence Circuit Positive Sequence Circuit Negtive Sequence Circuit o o Z f Zero Sequence Circuit Positive Sequence Circuit Negtive Sequence Circuit o o Z f 1 E Z Z Slide # 37 Z f
LL Fults-Exmple n the previous exmple, ssume phse-b-to-phse-c fult occurs t the busbr of genertor 2 (G 2 ) j.1389 j.1389 Z f 1.5 E Z f 1.5 E j.1456 j.1456 j.25 Note: Z f Solving the network for the currents, we get b c 1 3 1.5 3.691 9 j(.1389.1456) 1 1 1 1 α 2 α 1 2 α α 3.691 9 3.691 9 6.39 6.39 Slide # 38
LL Fults-Exmple, cont'd Solving the network for the voltges we get f f 1.5 j.1389 3.691 9.537 j.1452 3.691 9.537 f 1 1 1 1.74 f 2 b 1 α α.537.537 f 2 1.537.537 c α α Slide # 39
Double Line-to-Ground Fults With double line-to-ground (DLG) fult two line conductors come in contct both with ech other nd ground. We'll ssume these re phses b nd c. The voltge nd the current reltionships re: b c ( ) Z b c b c f b b Z f c c Note, becuse of the pth to ground the zero sequence current is no longer zero. Slide # 4
Slide # 41 DLG Fults, cont'd Using the symmetricl components, the terminl voltges re: b b b b b α α 2 2 2 α α α α b c ) ( ) ( 2 2 α α α α c 2 α α
Slide # 42 DLG Fults, cont'd f Z ) ( 2 2 2 α α α α α α f c b c b Z ) ( Using the symmetricl components, the terminl currents re: b α α 2 c 2 α α The voltge between fult terminl nd ground is: Express the bove eqution in terms of its symmetricl components: 1 2 α α, Using & f Z 3 Then
DLG Fults, cont'd To stisfy, the three symmetricl circuits, during double line to ground fult, re connected s follows: & o o 3 Z b c b c Z f 3 Z f f Zero Sequence Circuit Positive Sequence Circuit Negtive Sequence Circuit o o Zero Sequence Circuit Positive Sequence Circuit Negtive Sequence Circuit o 3Z f 1 E Z o Z Z o 3Z f Slide # 43
DLG Fults-Exmple n previous exmple, ssume DLG fult occurred t G2 bus. 1.5 E j.25 j.1389 3Z f 1.5 j.1389 E j.1456 j.25 j.1456 Assuming Z f, then 3Z f Z Z //( Z 3Z f ) 1.5 j(.1389 j.92) 4.547 9 Slide # 44
DLG Fults, cont'd j.1389 j.1456 j.25 1.5 E f f f f f 3Z f 1.5 4.547 9 j.1389.4184.4184/ j.1456 j2.874 j4.547 j2.874 j1.673 Converting to phse: 1.4 j6.82 f b f c 1.4 j6.82 Slide # 45
Unblnced Fult Summry SLG: Sequence networks re connected in series, prllel to three times the fult impednce LL: Positive nd negtive sequence networks re connected in prllel; zero sequence network is not included since there is no pth to ground DLG: Positive, negtive nd zero sequence networks re connected in prllel, with the zero sequence network including three times the fult impednce Slide # 46
Generlized System Solution Assume we know the pre-fult voltges The generl procedure is then 1. Clculte Z bus for ech sequence 2. For fult t bus i, the Z ii vlues re the Thevenin equivlent impednces; the pre-fult voltge is the positive sequence Thevenin voltge 3. Connect nd solve the Thevenin equivlent sequence networks to determine the fult current; how the sequence networks re interconnected depends upon the fult type Slide # 47
Generlized System Solution, cont d 4. Sequence voltges throughout the system re given by M This is solved for ech prefult Z sequence f network! M 5. Phse vlues re determined from the sequence vlues Slide # 48
Unblnced System Exmple Bus 1 Bus 2 G1 G2 fult Bus 3 Δ For the genertors ssume Z Z j.2; Z j.5 For the trnsformers ssume Z Z Z j.5 For the lines ssume Z Z j.1; Z j.3 Assume unloded pre-fult, with voltges 1. p.u. Slide # 49
Positive/Negtive Sequence Network Bus 1 j.2 j.5 j.3 Bus 2 j.5 j.2 o o 1. j.3 fult j.3 Bus 3 1. Y 24 1 1.1397.113.125 j1 24 1 Zbus.113.1397.125 1 1 2.125.125.175 bus Negtive sequence is identicl to positive sequence Slide # 5
Zero Sequence Network Bus 1 j.5 j.5 j.3 Bus 2 j.5 j.5 j.3 fult j.3 Bus 3 Y bus 16.66 3.33 3.33 j 3.33 26.66 3.33 3.33 3.33 6.66 Z bus.732.148.44.148.435..292.44.292.1866 Slide # 51
For SLG Fult t Bus 3 The sequence networks re creted using the pre-fult voltge for the positive sequence thevenin voltge, nd the Z bus digonls for the thevenin impednces j.175 E 1. j.175 j. 1866 Positive Seq. Negtive Seq. Zero Seq. The fult type then determines how the networks re interconnected Slide # 52
Bus 3 SLG Fult, cont d f 1. j(.175.175.1866) f f f j1.863 j1.863 1..7671 1. Z bus.7671 1. j 1.863.674.2329 Z bus.2329 j1.863.326 Slide # 53
Bus 3 SLG Fult, cont d.82 Z bus.544 j1.863.3479 We cn then clculte the phse voltges t ny bus 3 1.3479 A.674.522 j.866.326.522 j.866.82.4522 A.7671.3491 j.866.2329.3491 j.866 Slide # 54
Fults on Lines The previous nlysis hs ssumed tht the fult is t bus. Most fults occur on trnsmission lines, not t the buses For nlysis these fults re treted by including dummy bus t the fult loction. How the impednce of the trnsmission line is then split depends upon the fult loction Slide # 55
Line Fult Exmple Assume SLG fult occurs on the previous system on the line from bus 1 to bus 3, one third of the wy from bus 1 to bus 3. To solve the system we dd dummy bus, bus 4, t the fult loction Bus 1 Bus 2 j.25 j.1 j.25 o j.333 o 1. Dummy fult bus Bus 4 j.677 j.1 Bus 3 1. Slide # 56
Line Fult Exmple, cont d 44 1 3 The Y bus 1 24 1 now hs Ybus j 4 buses 1 25 15 3 15 45 Adding the dummy bus only chnges the new row/column entries ssocited with the dummy bus Z bus.1397.113.125.1348.113.1397.125.1152 j.125.125.175.1417.1348.1152.1417.1593 Slide # 57