A method for studying sequential faults on a three phase distribution transformer

Transcription

1 Scholrs' Mine Msters Theses Student Reserch & Cretive Works 1971 A method for studying sequentil fults on three phse distribution trnsformer Shshi Knt Pndey Follow this nd dditionl works t: Prt of the Electricl nd Computer Engineering Commons Deprtment: Recommended Cittion Pndey, Shshi Knt, "A method for studying sequentil fults on three phse distribution trnsformer" (1971). Msters Theses This Thesis - Open Access is brought to you for free nd open ccess by Scholrs' Mine. It hs been ccepted for inclusion in Msters Theses by n uthorized dministrtor of Scholrs' Mine. This work is protected by U. S. Copyright Lw. Unuthorized use including reproduction for redistribution requires the permission of the copyright holder. For more informtion, plese contct scholrsmine@mst.edu.

2 A METHOD FOR STUDYING SEQUENTIAL FAULTS ON A THREE PHASE DISTRIBUTION TRANSFORMER BY SHASHI KANT PANDEY, A THESIS Presented to the Fculty of the Grdute School of the UNIVERSITY OF MISSOURI-ROLLA In Prtil Fulfillment of the Requirements for the Degree MASTER OF SCIENCE IN ELECTRICAL ENGINEERING 1971 Approved by

3 ii ABSTRACT A method hs been presented for studying sequentil fults on opposite sides of three-phse distribution trnsformer using the theory of symmetricl components nd Alph, Bet, nd Zero components of three-phse systems. The effect of the fults on the distribution system hs lso been nlyzed nd improvements in the protection scheme suggested. The sequentil fults occurred, one fter the other, on the distribution system of the Centrl Electric Power Coopertive of Jefferson City in the stte of Missouri resulting in double unblnced fult condition. The dt used in the exmple problem hs been supplied through the courtesy of the bove Coopertive. The problem hs been solved in three stges, nmely, for two line to ground fult, n open conductor nd the resulting double fult conditions with grounded trnsformer s well s with the ungrounded trnsformer when the grounding wire of the trnsformer burns out.

4 iii ACKNOWLEDGEMENT The professionl dvice nd ledership of my dvisor, Dr. J.D. Morgn, is grtefully cknowledged nd sincerely pprecited. The vluble dvice nd constructive criticism of Dr. C.A. Gross is sincerely pprecited. The morl encourgement nd vluble suggestions of Professor George McPherson, Jr., re highly pprecited. The help rendered by Mr. Crl Herren of Centrl Electric Power Coopertive, Jefferson City in supplying the system dt is lso pprecited. Thnks go to Mrs. Mrgot Lewis for typing the mnuscript.

5 iv TABLE OF CONTENTS Pge ABSTRACT ACKNOWLEDGEMENT ii iii LIST OF FIGURES v LIST OF TABLES vi I. INTRODUCTION I I. REVIEW OF LITERATURE III. THEORETICAL DEVELOPHENT OF THE l'1athe:t-1atical MODEL A. Two Line to Ground Fult B. Simultneous Fults on Grounded Trnsformer C. Simultneous Fults with Ungrounded Trnsformer IV. APPLICATION OF THE MATHEMATICAL MODEL TO THE CENTRAL ELECTRIC SYSTEM V. CONCLUSION BIBLIOGRAPHY VITA APPENDICES A. Symmetricl Components of Three Phse System B. Alph, Bet nd Zero Components of Three Phse System... 64

6 v LIST OF FIGURES Figure Pge 1. Configurtion for Three-Phse Trnsformer Single Line Digrn Sequence Networks Thevenin Equivlent of Sequence Networks Hypotheticl Stubs Digrm for Double Line to Ground Fult Connection of Sequence Networks for Double Line to Ground Fult t D Phse Shift with Delt-Wye Trnsformer Direct Connections of the Alph, Bet nd Zero Networks for Two Line to Ground Fult One Open Conductor (Phse ) Connection of Alph, Bet nd Zero Equivlent Circuits for Open Conductor in Phse Equivlent Circuit for n Open Conductor t C nd Two Line to Ground Fult t D with Grounded Trnsformer Equivlent Circuit for n Open Conductor t C nd Two Line to Ground Fult t D with Ungrounded Trnsformer Grounding Wire Chrcteristics Fuse PE-4 Chrcteristics... 39

7 vi LIST OF TABLES Tble I II. SYSTEM DATA. o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o RES~ TS o o o o o o o o o o o o o o o o o o o o Pge 33 56

8 1 I. INTRODUCTION The configurtion of the three-phse distribution trnsformer, with its primry winding delt connected nd the secondry winding wye connected with solid neutrl grounding, is shown in Figure 1. The system ws protected by fuses in ech phse of the primry side of the trnsformer. At time t =, two line to ground fult on phses B nd C occurred on the secondry side of the system trnsformer. As result of this fult, the fuse of phse ws blown on the primry side of the trnsformer t time t = t nd thus the system developed double unsymmetricl fult condition on the opposite sides of the distribution 1 trnsformer. With the bove two fults existing, the grounding wire (neutrl wire) of the secondry wye connected winding burned through t time t = t becuse of excessive current which further unblnced the system, 2 resulting in the destruction of customers' pplinces on phse A. The theory of symmetricl components hs been pplied to solve the two line to ground fult by using Thevenin equivlent circuit ssuming the circuit to be liner nd bilterl nd neglecting the sturtion effects s well s the mgnetizing current of the trnsformer. The system currents under fult conditions were clculted in the primry winding, tking the pproprite phse-shift into considertion, nd the time (t ) for melting the phse fuse ws then obtined from the vilble current ginst elpsed time chrcteristics of the fuse. 1 Now, with n open conductor in the primry circuit nd two line to ground fult in the secondry circuit, the system ws reduced to

9 Primry Secondry c Fuse c A A c Fuse B Fuse b c Grounded Two Line to Ground Fult Figure 1. Configurtion for Three-Phse Trnsformer N

10 3 double unsymmetricl fult, occurring on opposite sides of the distribution trnsformer. The solution of the double fult condition ws obtined by using the Alph, Bet, nd Zero components of the threephse system. An equivlent circuit connecting the Alph, Bet, nd Zero networks together hs been developed to stisfy both the fult conditions. The vrious system currents nd voltges hve been clculted using the equivlent circuit. Once the fult current flowing through the grounding wire of the trnsformer is known, the melting time (t ) for this fult current is 2 computed from the chrcteristics of the prticulr ground wire used for the trnsformer. With the grounding wire opened, the zero sequence network is ltered nd new equivlent circuit is formed. Thus, the finl vlues of vrious system currents nd voltges re gin found. Voltges resulting from the bove double fult condition re found to be much bove the norml voltge of the system nd, consequently, the customers' pplinces re dmged. Thus, n nlyticl method for studying the double unsymmetricl fult conditions hs been developed to determine the system conditions, nd improvements in the protection scheme of such distribution systems hve lso been suggested to void repeting such fult conditions. An exmple problem hs lso been solved with the dt supplied by the Centrl Electric Power Coopertive.

11 4 II. REVIEW OF LITERATURE 1 In 1918, Dr. C. L. Fortescue developed method of symmetricl coordintes pplied to the solution of polyphse networks. Unsyrnmetricl fults on trnsmission systems, which my consist of short circuits, impednce between lines, impednce from one line or two lines to ground, or open conductors, re studied by the method of symmetricl components. In the method of symmetricl components, n unsymmetricl set of vector currents or voltges is resolved into three blnced sets of components. The theory of symmetricl components hs been used extensively for solving power system problems with single unsymmetricl fults. For cses of double fults, ech fult ws solved seprtely, nd then by using the superposition theorem, the finl vlues of the system currents nd voltges were obtined. The results thus obtined were only pplicble for simple system networks nd the sme technique, when pplied to complicted networks, proved to be incorrect nd pprecible error ws generted. In the yer 1931, pper presented t n A.I.E.E. meeting indicted tht pproximtely 2 percent of the fults on doublecircuit overhed trnsmission lines on the sme tower involved both circuits. Occsionlly, two fults occurred simultneously t points which were seprted geogrphiclly s well s electriclly, prticulrly on systems which were grounded through high impednces. In the sme yer Miss Edith Clrke 2 developed method deling with simultneous fults. This ws published s "Simultneous Fults on Three-Phse Systems" in the A.I.E.E. Trnsctions in September, In her pper, the method of symmetricl components ws extended to

12 5 pply to three-phse systems during simultneous fults t two or more points of the system. A generl equivlent circuit ws developed to replce, in the positive sequence digrm, two simultneous fults involving ny combintion of the six conductors. A double unblnce cnnot be correctly represented by connecting the sequence networks t ech of the two points of fult s they would be connected for single unblnces t those points. Miss Clrke's pper proved to be very useful in solving problen1s of double unblnces by using the method of equivlent circuits of vrious sequence networks. Subsequently, the problems of double unblnces were lso solved 3 4 by Wgner nd Evns in 1933 nd by Lyon in 1937 in their books deling with symmetricl components nd its ppliction to the nlysis of unblnced electricl circuits. In 1935, Kimbrk 5 developed prcticl wy of solving the problem of double unblnces by setting up miniture experiments to represent the fulted networks. Among ll these studies, Miss Clrke's method presents unique wy of solving ny combintion of double unblnces in three-phse power system. This sme method ws pursued for the problem initilly, but filed becuse of the complicted system developments which mke it difficult to obtin solution using the symmetricl components theory, 6 Clrke In the yer 1938, nother method ws developed by Miss Edith to solve simultneous fults titled "Modified Symmetricl Components" using Alph, Bet, nd Zero components. One yer lter Dr. Kimbrk 7 developed nother technique, quite sin1ilr to Alph, Bet, nd Zero components, nd clled them x, y, nd z components, respectively, to study unsymmetricl system conditions. The use of Alph, Bet,

13 6 nd Zero components gve simpler solutions thn the use of symmetricl components in power systems when equl positive nd negtive sequence impednces re used for representing rotting mchines. Thus, in n unsymmetricl circuit where the impednces of the two phses re equl, or two phses re symmetricl with respect to the third phse, Alph, Bet nd Zero components give n lmost immedite solution to mny systems. The theory of Alph, Bet, nd Zero components re lso de- 8 tiled in the book by Miss Clrke. This theory of Alph, Bet nd Zero components is used to solve problems when it becomes difficult to use the symmetricl components theory nd the sme hs been used to solve the double unblnced fult condition in the present problem. In 1941, Clrke, Peterson, nd Light 9 determined the condition under which bnorml voltges of sufficient mgnitude developed to dmge equipment when one or two conductors opened in circuits supplying ungrounded trnsformers. Investigtions were mde by mens of clcultions nd lbortory tests. But the results obtined by these tests cnnot be used in the present problem becuse of sustined simultneous fults nd, s such, the theory of Alph, Bet, nd Zero components hs been used to determine the vlue of overvoltges which dmged the customers' equipment during the conditions under study.

14 7 III. THEORETICAL DEVELOPMENT OF THE }MTHEMATICAL MODEL The distribution system of the Centrl Electric Power Coopertive, Jefferson City, cn be represented by the one line digrm given in Figure 2. The positive sequence source impednce nd the positive sequence trnsformer lekge rectnce re equl to their negtive sequence vlues respectively s per the system specifictions. The problem will be solved in three stges s the vrious fults occur. Sturtion nd the mgnetizing current hve been neglected. A. Two Line to Ground Fult At time t =, two line to ground fult occurred on the secondry side of the distribution trnsformer on phses B nd C t the point mrked D in Figure 2. This portion of the problem is solved using 1 symmetricl component theory s presented by Stevenson in his book nd detiled in Appendix A. We will pply Thevenin's theorem, which llows us to find the current in the fult by replcing the entire system by single genertor nd series impednce, ssuming the circuit to be lumped, liner nd bilterl. Now, let IA, IB, nd IC be the currents flowing out of the originl blnced system t the fult (D) fro~ phses A, B, nd C, respectively. The line-to-ground voltges t the fult will be designted VA' VB, nd VC. Before the fult occurs, the line-to-neutrl voltge of phse A t the fult will be clled vf, which is positive-sequence voltge since the system is ssumed to be blnced before the fult. Figure 3 shows the three sequence networks with the system impednces represented nd the fult mrked t point D. Since linerity hs been ssumed in drwing the sequence networks, ech of the networks cn be

15 System Single phse fuses Three-phse trnsformer,6~ "::" D 2-L-G Fult Z =Z =Z s s s 1 2 z so Source Sequence Impednce Z =Z =Z t tl t2 zt Trnsformer Sequence Impednce z z sl s2 z so Z tl Z t2 Z to = Positive Sequence Source Impednce = Negtive Sequence Source Impednce = Zero Sequence Source Impednce = Positive Sequence Trnsformer Lekge Impednce = Negtive Sequence Trnsformer Lekge Impednce = Zero Sequence Trnsformer Lekge Impednce Figure 2. Single Line Digrm CX>

16 Reference Bus Reference Bus Reference Bus z sl zt l z s2 z t2 z so zt D D Positive-Sequence Network Negtive-Sequence Network Zero-Sequence Network Figure 3. Sequence Networks "'

17 1 replced by its Thevenin equivlent between the two terminls composed of its reference bus nd the point of ppliction of the fult. The Thevenin equivlent circuit of ech sequence network hs lso been shown in Figure 4. The internl voltge of the single genertor of the equivlent circuit for the positive-sequence network is vf, the prefult voltge to neutrl t the point of ppliction of the fult. The impednce z 1 of the equivlent circuit is the impednce mesured between point D nd the reference bus of the positive-sequence network with ll the internl emf's short circuited. Since no negtive- or zero-sequence currents re flowing before the fult occurs, the prefult voltge between point D nd the reference bus is zero in the negtive- nd zero-sequence networks. Therefore, no emf's hve been shown in the equivlent circuits of the negtivend zero-sequence networks. The impednces z 2 nd z re mesured between point D nd the reference bus in their respective networks. Since IA is the current flowing from line A of the system into the fult, its components IA, IA, nd IA flow out of their respective 1 2 sequence networks nd out of the equivlent circuits of the networks t D, s shown in Figure 3 nd 4. Now the mtrix equtions for the symmetricl components of voltges t the fult cn be written from the equivlent circuits s follows: VA zo IA VA = vf zl IA (3.1) 1 1 VA z2 IA 2 2

18 Reference Bus Reference Bus Reference Bus z ==z +Z 1 sl tl ---+-I Al v J D z ::::z +Z 2 s2 t2 _..I A2 J D Z =Z o t J _..I Ao D Positive-Sequence Network Negtive-Sequence Network Zero-Sequence Network Figure 4. Thevenin Equivlent of Sequence Networks 1-' 1-'

19 12 or Now, for double line to ground fult (the fulted phses being Bnd C), the following reltions exist t the fult: v = v = B C The contribution of current from phse A to the fult is IA = nd the fult condition is represented s shown by the connection digrm of Figure 5. Now, with VB = nd VC, the symmetricl components of the voltge re given by VA VA VA VA VA 1 2 VA VA (3.2) therefore VA VA VA 1 2 ( 3. 3) hence, Now, s IA I + IA + IA Ao 1 2 (3.4) therefore,

20 A B c IA IB lf Ill 1 c lltj J In ~ Figure 5. Hypotheticl Stubs Digrm for Double Line to Ground Fult f-' w

21 14 IA - - (IA + IA ) 1 2 Now, from Equtions ( 3.1) nd (3.3), IA IA 2 VA VA 1 zo zo VA VA 2 1 z2 z2 Substituting these vlues in Eqution (3.4) gives Therefore, Substituting this vlue of VA into Eqution (3.1) 1 There fore, nd ( 3. 5) Equtions (3.3) nd (3.5) indicte tht the three equivlent sequence networks should be connected in prllel t the fult point in order to simulte double line to ground fult, nd the sme hs been shown in Figure 6. The prefult current (lod current), being very smll, is not being tken into ccount here nd hs been neglected

22 IAl t vf Positive-Sequence Network zl v Al D l IAl I V )I( D A2 ~ IA 2 ~ IA D VA z2 zo Negtive-Sequence Network Zero-Sequence Network -+-- IA 2 ~IA Figure 6. Connection of Sequence Networks for Double Line to Ground Fult t D t-' VI

23 16 ltogether s it will hve negligible effect on the fult current. Now, knowing Vf, z, z 1, nd z 2 in per unit vlues, the positive sequence current IA cn be clculted from Eqution (3.5) nd then 1 IA, IA, VA, VA, nd VA cn ll be found from vrious equtions lredy derived. From the sequence vlues, the currents nd voltges in ll three phses cn be found on the secondry side of the trnsformer by the symmetricl components reltions given in Appendix A. Once the trnsformer secondry vlues of current nd voltges re known, the primry currents nd voltges cn be clculted by tking the phse shift of the trnsformer into ccount. With the primry phse current known, the time (t 1 ) for phse fuse to blow nd cler cn be determined. The phse fuse blows erlier thn the other two fuses of phses b nd c becuse the current, I' is greter thn Ib nd Ic. Once the fuse of phse blows on the primry side of the trnsformer, there exists n open conductor on phse. The system with n open conductor on phse nd double line to ground fult on phses B nd C presents double unblnced fult condition. B. Simultneous Fults on Grounded Trnsformer Now the system, hving two fults, nmely two line-to-ground fult on the secondry side of the trnsformer nd n open conductor on the primry side of the trnsformer, presents condition with simultneous fults. This simultneous fult is combintion of short circuit involving ground nd n open conductor. Ech exists on the opposite side of distribution trnsformer, creting phse shift problem becuse of the delt-wye connection of the trnsformer. Since

24 17 ech fult ffects the voltges nd currents resulting from the other, the simultneous fults cnnot be treted seprtely s presented erlier. In order to solve this problem, the phse shift of the deltwye trnsformer must be tken into ccount. A delt-wye trnsformer bnk usully divides the zero sequence system into two prts which hve no connection with ech other in the zero sequence network. However, the positive nd negtive sequence equivlent circuits for three-phse power systems re bsed on equivlent wye-wye trnsformer connections. As such, the zero sequence quntities re unffected on either side of the trnsformer but the positive nd negtive sequence quntities thus obtined re bsed on wye-wye connection. Hence, phse correction is needed for both positive nd negtive sequence quntities in order to get correct vlues bsed on the ctul delt-wye connection of the intervening trnsformer. To find the phse correction, let the trnsformer connection be given s in Figure 7. Let the secondry circuit of the trnsformer with phses A, B, nd C be chosen s the reference circuit, nd phse A s the reference phse. In the primry circuit, let the reference phse be designted, nd so chosen tht the line-to-neutrl voltge V t no lod without ny fult is 9 degrees out of phse with the line to neutrl voltge VA. From the Figure, it is seen tht V lgs VA by 9 degrees for the connection shown. Hence, the positive nd negtive sequence quntities of current nd voltge bsed on equivlent wye-wye trnsformer connections with the secondry circuit s the reference, re multiplied by -j nd j, respectively, in order to get the ctul vlues of the positive nd negtive sequence quntities. The

25 c Fuse,. _,c A b Fuse c B b Open Conductor (Phse ) Grounded Two Line to Ground Fult (Phses B nd C)--- Figure 7. Phse Shift with Delt-Wye Trnsformer 1-'

26 19 phse correction is thereby obtined for the positive nd negtive sequence quntities bsed on the delt-wye connection. Now, in order to solve the double unblnced fult condition the theory of Alph, Bet, nd Zero components of three phse system is pplied nd the sme hs been presented in Appendix B. These components cn be esily pplied s long s the positive- nd negtivesequence impednces of the rotting mchines re equl, which is true in the system under study. Before pplying the theory of Alph, Bet, nd Zero components to the simultneous fults, ech fult condition will be discussed seprtely using the Alph, Bet, nd Zero theory. For double line to ground on phses B nd C, conditions t the fult s delt erlier, re Now, from Appendix B, certin reltions between line currents nd phse voltges nd their Alph, Bet, nd Zero components exist. All the equtions hve been derived in detil in Appendix B. The sme numbers will be used for ech eqution s in Appendix B. VA v + vo VB + 2 vc v J (VA 2 ex ) (B.1) (B. 4) vs 1 /3 (VB - VC) (B. 5) nd IA I + ro ex (B. 7) Applying the bove fult conditions to these equtions, the equtions stisfying two line to ground fult cn be obtined. These re, with

27 2 VB vc from Eqution (B.4) v 2 J VA or VA l.v 2 Now, from Eqution (B. 1), we hve or or l.v 2 l:.v 2 vo v + vo v 2v (3. 6) Also, from Eqution (B.S), with VB vc, VB = (3. 7) nd, from Eqution (B. 7), with IA, I -I (3. 8) In the circuit under study, the positive- nd negtive-sequence impednces re equl nd, s such, the Alph, Bet& nd Zero impednces re vilble from the sequence impednces,

28 21 The connection of the Alph, Bet, nd Zero networks tht stisfy the fult condition equtions, is shown in Figure 8. Here, Vf, is the prefult voltge of phse A to ground. In order to hve zero-sequence current equl to the Alph current, ll the zero-sequence impednces must be multiplied by two in the zero-sequence network. The theory of Alph, Bet, nd Zero components is lso pplied to the open conductor on phse on the primry side of the trnsformer. Figure 9 shows the configurtion for one open conductor. Here let v be the series voltge drop nd I' be the line current through the series impednce. Now the reltions between v, v 6, v, the components of the series voltge drop, nd between I', Is', nd I 1, the components of line current, will be developed considering the fult conditions. The conditions t the fult re v = v b c nd I I Now, from Appendix B, the vrious reltions mong the components nd the phse vlues re given by the following equtions: v v + vo (B.1) vb + v 2 v - (v c) (B. 4) v = (vb - v ) (B.S) c B /3 I I I nd I I + Io (B. 7) Now s vb v c Therefore, from Eqution (B. 4),

29 Alph-network vf Z =Z sl s Z =Z tl t Bet-network 2Z =22 so so 22 =2Z 2V O to to v ~ I = -I D + -jvf =Z ss z t D v = s Zero-network Figure 8. Direct Connections of the Alph, Bet nd Zero Networks for Two Line to Ground Fult N N

30 23 ~.w.. C1l u t t t H H H ::l " C1l QJ (f) C1l...c: P-< c u c QJ p.. QJ c QJ ~ ::l bd.,., u ~ '\

31 v v 3 or v 3 2 v Now, from Eqution (B.1) 3 2 v v + v or v 2v (3.9) Also from Eqution (B.S), (3.1) nd from Eqution (B.7), with I 1 : I ' -I I ( 3.11) Direct connection of Alph nd Zero networks is mde possible by multiplying the zero impednces by two s shown in Figure 1. In this figure, the currents in the zero network re zero but the voltges re twice the zero voltges. The Bet network is unffected by one open conductor on phse. The reltions between the Alph, Bet nd Zero components of voltge nd current in terms of syit~etricl components of the voltge nd current re lso given in Appendix B. Now the reltions derived re v = -j(v l

32 Alph-Network vf 2z z =Z z "'z sl s t t --+tv~l~ I I I= -I I + so "'2z so + Bet-Network Z =Z Z =Z s 2 s 6 ~v 6 =o~ t 2 t 6 ---t--is Zero-Network Figure 1. Connection of Alph, Bet nd Zero Equivlent Circuits for Open Conductor in Phse N Vl

33 26 v I I As erlier discussed, since delt-wye trnsformer is involved in the system, phse correction must be pplied by turning the positive-sequence components of line current nd voltge to neutrl bckwrd 9 degrees nd the negtive sequence components of current nd voltge forwrd 9 degrees in pssing through the trnsformer bnk, v (V + V ) becomes -j(v l 2 l v " B =- j (V V ) becomes - (V l 2 l v " Similrly, the corresponding currents re I =(I +I ) becomes -j(i I " B I = -j (J B l becomes - (I + I ) l 2 I II In order to pply phse correction to the present problem of simultneous fults on opposite sides of delt-wye trnsformer, let the loction of the double line to ground fult on the secondry side be represented by D nd tht of n open conductor on the primry side by C s shown in Figure 11 nd let the circuit t D be ssumed to be the reference circuit.

34 Zero Potentil for -Network vf Alph-Network s c -jv f Zero Potentil for -Network Zero Network Open (Delt Side) c c Bet-Network 2Z =2Z so so Z =Z D t2 ts Zero Potentil for (3-Network ~ v "= 2Zt Zero-Network (Wye Side) Zero Potentil for -Network Figure 11. Equivlent Circuit for n Open Conductor t C nd Two Line to Ground Fult t D with Grounded Trnsformer N...

35 28 For n open conductor t C, the existing reltions s derived erlier re Equtions (3.9), (3.1), nd (3.11). For two line to ground fult t D, the derived reltions re Equtions (3.6), (3.7), nd (3. 8). Let the double primed symbols indicte the vlues of the components of the line current nd series voltges t C (for n open conductor) referred to the circuit t D (secondry side of delt-wye trnsformer), then pplying the phse correction, s given erlier, to these vlues t C, v v " s v = -v " (3 I I = I " S I I = -I II (3 Substituting these vlues into Equtions (3.9), (3.1), nd (3.11), 2v 11 ( 3.12) v " (3. 13) I " [3 -I II (3.14) Now, theequtions (3.6), (3.7), (3.8), (3.12), (3.13), nd (3.14) describe the simultneous fults bsed on n equivlent wye-wye trnsformer bnk nd Figure 11 is the equivlent circuit used for nlytic clcultions which stisfies ll these equtions. Becuse of infinite series impednces in the zero sequence network t C, the Zero nd Bet networks re both open t C. The Alph network is closed t C, since v 11 =.

36 29 As there is only one fult connection between the point D in the Alph network nd the zero-potentil bus for the network, the generted voltge in the network cn be replced by the prefult voltge Vf in phse t the fult point before the dissernmetry occurred nd the vrious components of the current nd the voltges re clculted. The phse vlues re then obtined by using the reltions of Appendix B. The component impednces re equl to their sequence counterprts nd hve been used ccordingly in Figure 11. Now, with the method developed for clculting the current, the current through the neutrl wire of the trnsformer cn be clculted nd the time (t ) in seconds 2 for the ground wire to melt cn finlly be computed using vilble wire chrcteristics. The system is now n ungrounded delt-wye trnsformer which must now be treted differently. C. Simultneous Fults with Ungrounded Trnsformer The system is still confronted with double fults, n open conductor in the primry circuit nd two line-to-ground fult in the secondry circuit of the delt-wye trnsformer. But, with the secondry neutrl wire burned through, the trnsformer is ungrounded. The problem is gin solved using Alph, Bet nd Zero components nd the six equtions (3.6), (3.7), (3.8), (3.12), (3.13), nd (3.14). The equivlent circuit of Figure 11 chnges to tht of Figure 12 which shows the infinite impednce in the zero network, becuse of the melted neutrl wire of the trnsformer. As no current cn flow through the neutrl wire, I =

37 Zero Potentil for -Network Zero Network Open (Delt Side) vf Zero Potentil for -Network Alph-Network v " =-+i c z =z tl t -jv f c c Bet-Network 2Z =2Z so Z =Z t2 t8 Zero Potentil for 8-Network Zero Network (Wye Side) Zero Potentil for -Network Figure 12. Equivlent Circuit for n Open Conductor t C nd Two Line to Ground Fult t D with Ungrounded Trnsformer w

38 31 Hence, from Eqution (3.8), I -I Therefore, the voltge of the lph network V V f the prefult voltge. Also from Eqution (3.6), V 2V o nd from Appendix B, Eqution (B.l) v + v o l.v 2 It is evident from the bove reltion tht the voltge of phse A increses bnormlly with the burning of the neutrl wire nd this bnorml voltge of phse A cn cuse dmge to the customers' electricl equipment.

39 32 IV. APPLICATION OF THE 11ATHE.HATICAL MODEL TO THE CENTRAL ELECTRIC SYSTEM The system dt ws supplied by the Centrl Electric Power Coopertive, Jefferson City, Missouri for their distribution system nd is shown in Tble I. The problem is solved using the method developed in Chpter III. The mgnetizing current nd the sturtion effects hve been neglected. The lod dt, s given in the Tble I, is estimted nd it does not involve ny rotting mchinery. As such, the lod current will hve little effect on the totl line current during the fult nd hs been neglected in our fult clcultions. The lod is ssumed to be t the bse voltge, hence, the voltge of phse A before the fult is 1. p.u. Therefore, the bse current for the secondry circuit is I 375 s. Bse /3 X mps. nd the bse current for the Primry circuit is I 375 P. Bse /3 X mps. Now, for two line to ground fult (on phses B nd C) in the secondry circuit, the positive sequence current IA 1 given by Eqution (3.5) of phse A is Now, Vf 1. p.u. (Pre-fult Voltge)

40 TABLE I. System Dt S. No. System Components Dt Remrks (i) Voltge Rting of the Trnsformer KV Delt-Wye with solidly grounded secondry. Voltge rting selected s the bse (ii) KVA Rting o.f the Trnsformer 375 KVA Selected s bse (iii) System Bse KVA 375 KVA (iv) Lekge Rectnce of the trnsformer Zt 1 =zt 2 =Zt=.722p.u. zt =.7p.u. Bsed on System Bse KVA ( v) System Impednce & Voltge z ~z =Z =.269p.u. s s s 1 2 Bsed on System Bse KVA Z so is not fctor V=lp.u. (vi) Trnsformer lod t fult inception, lod voltge nd power fctor 15 KW, V=l p.u..93 power fctor Estimted vlues, neglected in fult clcultions. (vii) Type & chrcteristic of the fuses used on the primry side P.E.-4E fuse kits Totl clering time in seconds versus current in mperes chrcteristic shown in Figure 14. (viii) Size nd type of ground wire used with the trnsformer secondry winding 7.7 MCM Cu-wire Totl melting time in seconds ginst the current in mperes chrcteristic given in Figure 11 w

41 34 zl z + z = j sl tl z2 = z + z = j 52 t2 zo = z = j.7 to p.u. p.u. p.u. IA 1 = -j 7.14 p.u. nd or IA = j p.u. 2 Similrly X z.,::2_ Zo + z2 or I = j p.u. Ao Now the current in line A, IA = p.u. The current mgnitude through line A,!rAJ = x 164 = mps.

42 35 The current in Line B, or IB ; j Therefore /144.3 p. u. The current mgnitude through line B, I rbi = 1.75 X 164 = 176 mps. Similrly the current in Line C, or rc = j Therefore IC = 1.75;!f5.7 p.u. The current mgnitude through line C, [rc[ = 1.75 x 164 = 176 mps. The current through the neutrl of the trnsformer secondry winding, I = 3IA = j p.u. nl

43 36 The current mgnitude through the grounding wire is, II I= x 164 = 26 mps. nl Ftom the chrcteristics of the grounding wire in Figure 13, the melting time t 2 ' = 24 seconds for this vlue of current. The sequence voltges t the fult re The phse voltges to ground t the fult re: or = In order to ccount for phse shift, the positive nd the negtive sequence components of the line current nd the phse voltges on the system side of the Delt-Wye trnsformer re shifted -9 nd +9 respectively from the corresponding components on the lod side. As such, on the system side, I = -j (IA ) = 7.14 p.u. l 1 nd I = j (IA ) = p.u. 2 2

44 37 1 \ \ 1 ' \ \ ' ' (}) '"d p (.) Q) U) 1 \ \ \ \ \ \ \ \ \ \ \ 1. IDO 1 Current in Amperes 1, Figure 13. Grounding Wire Chrcteristics

45 38 I o = (since there is no zero sequence current) The current through line on the primry side is I or I = k p.u. The current mgnitude through line on the primry side, \I l = 1.95 x 32.3 = 325 mps. Ftom the chrcteristics of the fuse in Figure 14, the time t 1 = 1.7 seconds. This time is for totl clering nd includes the melting period of the fuse in line. Now the currents in phses b nd c re clculted s follows: Ib = 2 I 1 l = j 6.18 p.u. Ib = i = j p.u. 2 2 Ib = p.u. The current through line b on the primry side is: or Ib = j 3.625

46 39 1 Ul '" ~ u ClJ C/) ~ r-1 ClJ s r-1 E-4 ClO ~ r-1 H C1l ClJ """' u """' C1l +.J E J 1,J Current in Amperes Figure 14. Fuse PE - 4 Chrcteristics

47 4 Therefore 6.22 /215.7 p.u. Now the current mgnitude through line b on the primry side is 6.22 X mps. Now for line c: I cl I c2 I l 2 I j 6.18 p.u j p.u. I p. u. co The current through line con the primry side is: I c I j p.u. c Therefore I = 6.22 c / p. u. The current mgnitude through line c on the primry side is, I I I= X c 21 mps. A check on the primry side of the trnsformer is tht (I + Ib + Ic) must be equl to zero. It ws found to check properly.

48 41 The voltge components on the primry side cn be clculted from the currents on the primry side nd the system rectnces of the sequence networks. Now V = V - I Z 1 f 1 s 1 or v = 1. l - j.2715 p.u. Similrly v = - I z 2 2 s2 v = 2 j.795 p.u. nd V = (since I = ) o o The voltge of line to ground v = v + v = 1. - j.192 p.u. l 2 Therefore v = 1.2 /-1o.9 p.u.

49 42 The voltge of line b to ground: or vb = j.77 Therefore And the voltge of line c to ground: v c or v c = j.962 Therefore v c = o.982 ;!ol.5 p.u. Thus ll the voltges nd currents re known for two line to ground fult (phses Bnd C) on the secondry side of the trnsformer. Now fter t = 1.7 seconds, the fuse of phse is blown nd 1 thus cretes n open conductor sitution in phse. With double line to ground fult still existing nd n open conductor, simultneous fult developes which will be solved using the theory of Alph, Bet nd Zero components (given in Appendix B) nd the equivlent circuit

50 43 of Figure 11 developed in the lst chpter. As the time t 2 ' = 24 seconds hs still to elpse, the grounding wire of the trnsformer still exists nd the trnsformer is still grounded. However the figure t 2 ' = 24 is meningless since the current through the grounding wire of the trnsformer will now chnge. From Figure 11 on pge 27, I or I z + s z vf + 2Z t to z z + s z t z sl + z tl zl nd Substituting these vlues in the bove expression, I -j 4.18 p.u. For two line to ground fult, I j 4.18 p.u. Since the Bet network in Figure 11 is open, there will be no Bet current.

51 44 I = s From Appendix B, the current in line A on the secondry side: Therefore IA = p.u. The current mgnitude in line A, IrA! x 164 = mps. Similrly the current in line B on the secondry side: Therefore IB !..J.st_ p. u. The current mgnitude in line B, 6.27 X mps. And the current in line C on the secondry side: - 1/2 I Therefore 6.27 b:t_ p.u.

52 45 The current mgnitude in line C, 6.27 X mps. Now the current through the grounding wire of the trnsformer, Therefore & p.u. The current mgnitude through the grounding wire of the trnsformer, X mps. From the chrcteristics of the grounding wire in Figure 13, the time t = seconds for the wire to melt completely fter the occurence 2 of the fuse opening in phse. Now from Eqution (3.6) we hve From Appendix B, we hve from Eqution (B.l) VA v + vo Therefore VA 3/2 v Now v v - I X z f

53 46 Therefore V.585 k p.u. The voltge of line A to ground, VA= 3/2 V =.878 k p.u. The voltge of line B to ground, Since v 2v nd VB = p. u. Similrly the voltge of line C to ground, Agin since v p. u.

54 47 In order to get the line current nd voltge to ground on the primry side, use is mde of the equivlent circuit of Figure 11 on pge 27. Here the double primed symbols indicte the vlues of the components of the line current nd the voltge to ground on the primry side referred to the circuit t the secondry side of the delt-wye trnsformer. Wheres the single primed symbols of the components of the line current nd the voltge to ground indicte the ctul vlues on the primry side of the delt-wye trnsformer. I II I --j 4.18 p.u. Now from equtions derived for phse shift corrections I ' I " s j 4.18 p.u. I ' I " i3 Now from Equtions (3.11) nd (3.14) we hve I ' - I ' (j, I " I " i3 Since the Bet nd Zero networks re open circuited, I [3 II Io II Also I ' = Is II

55 48 Therefore from Eqution (3.11) I I - I I nd Is' = 4.18 ~ p.u. The current in line on the primry side becomes I I ' + I ' = mps. Since line is open, I checks s it should. The current in line b on the primry side is: or j ~ Therefore the current mgnitude in line b, 3.62 X mps. The current in line c on the primry side is: or I = -j 3.62 c

56 49 Therefore Ic = 3.62!9 p.u. The current mgnitude in line c, Ire! = 3.62 x 32.3 = 117 mps. From the chrcteristics in Figure 14, the fuses of phse b nd c will tke t 1 = 95 seconds to blow. Since time t 1 is greter 1 thn t 2 = seconds, the grounding wire will burn out erlier thn the fuses getting blown out nd thus the trnsformer is still energized through phses b nd c supplying the power. In order to get the voltge to ground on the primry side, we hve from the equivlent circuit of Figure 11 nd the equtions derived for phse shift corrections, v II Therefore V" =.8875/_.!!!._ p.u. Since no current flows through the Bet network nd Zero network, therefore Also V II v " s - j 1.

57 5 Now from the equtions for phse shift correction, V 1 = V "= -j 1. p.u. S V 1 S V "= p.u. v I v 11 p.u. Now the voltge of line to ground is: V = V I + V I or v -j 1. Therefore v 1. ~ p.u. Now the voltge of line b to ground is: vb - 1/2 V I + 13/2 V I + V I S or vb =- o j o.so Therefore vb =.917 & p.u. Now the voltge of line c to ground is:

58 51 or v j.5 c Therefore v c.917 ~ p.u. The sum of V, Vb nd V must be equl to zero which checks s it should. c All the voltges nd currents on both sides of the trnsformer re now known for the simultneous fult with grounded trnsformer. After t 2 = seconds hs elpsed, the grounding wire of the trnsformer melts nd the distribution trnsformer is now ungrounded nd the equivlent circuit of Figure 12 on pge 3 is now pplicble. Since the Zero network is now open, I - I therefore, Also the Bet network is open, there will be no Bet current, Now, s the Alph, Bet nd Zero components of the current re ll equl to zero, the line currents on ll the phses will lso be zero on the secondry side s well s the primry side of the trnsformer. Now the voltge to ground on the secondry side is clculted. From Appendix B, we hve from Eqution (B.l)

59 52 Since v = 2v Therefore v = 3/2 v A Now since the Zero network is open, I = - r Therefore v = V f = 1 p u Now s VA= 1.5 p.u. The voltge of line B to ground is, v B = - 1/2 V + /3;2 V S + V Since nd Therefore VB= p.u.

60 53 The voltge of line C to ground is, Since v = 2V nd Therefore VC = p.u. Also the voltges to ground on the primry side is given by the equivlent circuit in Figure 12, v " Since the zero network is open I " = ' I " Therefore V" = 1. p.u. Since no current flows through the Bet nd Zero network, therefore, vo II = nd lso vs II = -j vf = -j 1. p.u.

61 54 Now from the equtions for phse shift correction, v I vs II = -j 1. p.u. VB I v II - 1. p.u. vo I vo II The voltge of line to ground is: V =V 1 +V 1 or V = -j 1. p.u. 1. /-9 p.u. The voltge of line b to ground is: or vb = j p.u. Therefore vb = 1. M p.u. Now the voltge of line c to ground is: V - 1/2 V I - 13/2 V I + V I c S or V ==j p.u. c

62 55 Therefore Vc 1. ~ p.u. The sum of V, Vb nd V is equl to zero. All the voltges nd currents c on both sides of the trnsformer re now known for the fult condition with n ungrounded trnsformer. The vrious results hve been tbulted in Tble II. With the two line to ground fult on phses B nd C on the secondry side nd open conductor on phse on the primry side with the trnsformer being ungrounded, overvoltge occurs on phse A to ground to 15 percent of norml, nd indictes the posibility of dmge to the custmers supplied by phse A due to this overvoltge under the bove fult conditions.

63 TABLE II. Results Primry Currents & Voltges Fuse time Secondry Currents & Voltges s. d ystem I I I V V V 1n secon s I I I I Condition b c b c Totl clering A B C n VA VB vc Amp. Amp. Amp. p.u. p.u. p.u. Time Amp. Amp. Amp. Amp. p.u. _ p.u. p.u. (i) Double Lineto-Ground Fult (Phses B & C) t 1 =1.7 sees. for phse fuse (ii) Fults with grounded trnsformer (Open Conductor phse nd double line-to-ground fult phses B & C) t 1 '=95 sees. for phses b & c fuses (iii) Fults with ungrounded trnsformer (Open Conductor phse nd double line to ground fult phses B & C) Tble II. Cont. V1 (j\

64 TABLE II. Results (cont.) Grounding wire-time Remrks in seconds (Effective time t to melt in seconds ) t 2 ' = 24 sees. t 2 =23.95 sees. O<t<l.7 for phse fuse to blow out. 1. 7<t< for-grounding wire to melt. U1 ""-1

65 58 V. CONCLUSION This study hs provided better understnding of the voltges nd currents in double unsymmetricl fult conditions. The opening of phse fuse on the primry side, due to excessive current in phse, cn be esily voided by using circuit breker on the primry side which cn open ll the three phses simultneously in cse of fult nd thus protect the customers from overvoltge situtions. The melting of the trnsformer neutrl wire cn be voided if grounding resistor is used with the grounding wire insted of solid grounding in order to limit the fult current. This would lso help void the overvoltge sitution tht developed s well. In summry, two seprte modifictions re suggested to effect better protection scheme for the distribution system of the Centrl Power Electric Coopertive. These re given below: 1) A set of three fuses of suitble rtings should be instlled on phses of the secondry circuit of the trnsformer so tht, in cse of double fult like this one, the secondry circuit fuses open erlier thn those of the primry circuit of the trnsformer. This protection scheme will help void dmge to the customer nd the trnsformer. 2) The fuses on the primry circuit of trnsformer should be replced by circuit breker or combintion of circuit breker nd fuses instlled in order to void such single conductor opening in the system nd dmge to the customers' pplinces due to the overvoltges.

66 59 BIBLIOGRAPHY 1. Fortescue, C. L., ''Method of Symmetricl Coordintes Applied to the Solution of Polyphse Networks," A. I.E. E. Trns., Vol. 3 7, Prt II, 1918, pp Clrke, E., "Simultneous Fults on Three-Phse Systems," A.I.E.E Trns., Vol. 5, 1931, pp Wgner, C.F. nd Evns, R.D., Symmetricl Components s Applied to the Anlysis of Unblnced Electricl Circuits, New York: McGrw-Hill, Lyon, W.V., Applictions of the Hethod of Symmetricl Components, New York, HcGrw-Hill, Kimbrk, E.W., "Experimentl Anlysis of Double Unblnces," Elect. Eng., Vol. 54, Februry 1935, pp Clrke, E., "Problems Solved by Modified Symmetricl Components, 11 Gen. Elec. Rev., November nd December, 1938, Vol. 41, pp , Kimbrk, E. W., "Two-Phse Co-ordintes of Three-Phse Circuit," A.I.E.E. Trns., Vol. 58, 1939, pp Clrke, E., Circuit Anlysis of A-C Power Systems, Vol. I, Symmetricl nd Relted Components, New York, John Wiley nd Sons, 1943, Chpter X. 9. Clrke, E., Peterson, H.A., nd Light, P.L.,"Abnorml Voltge Conditions in Three-Phse Systems Produced by Single-Phse Switching," AlEE Trns., Vol. 6, 1941, pp Stevenson, Jr., W.D., Elements of Power System Anlysis, New York, McGrw-Hill, 1962, Chpter 14.

67 6 VITA Shshi Knt Pndey ws born on July 6, 1944 in Lucknow, Indi. He completed his high school eduction from Boys' Anglo Bengli Intermedite College, Lucknow, Indi. He received his Bchelor's degree in Electricl Enginering from Bnrs Hindu University, Vrnsi, Indi, in June He ws employed s n Assistnt Engineer in the construction nd design circles, with U.P. Stte Electricity Bord, Lucknow, Indi until August, He ws mrried to Vndn Updhyy on My 28, The uthor hs been enrolled in the Grdute School of the University of Missouri-Roll since September 1969 nd hs been cndidte for the Mster of Science degree in Electricl Engineering.

68 APPENDICES 61

69 62 APPENDIX A Symmetricl Components of Three Phse System Fundmentl symmetricl component equtions: IA + IA + IA 1 2 IA 2 + IA + la 1 2 IA + la 2 + IA 1 2 VA +VA +VA VA + VA + VA l 2 VA + VA 2 + VA 1 2 (A.l) (A. 2) (A. 3) (A.4) (A.S) (A. 6) (A. 7) (A. 8) IA = -(I + IB + le) 3 A 1 = J(VA +VB+ VC) (A. 9) (A. 1) (A. 11) (A.l2) where, = - 2 1:_+./3 2 J 2 1. ~ =:: 1.~ = 1. ;<12 (A.13) (A.14) Also, re the three line currents t ny point of the system.

70 63 VA' VB' nd VC re the three voltges to ground t ny point of the system. re the positive sequence currents in the three conductors. By definition, these currents re equl in mgnitude nd IA leds IB by nd Ic by re the negtive sequence currents in the three conductors. By definition, these currents re equl in mgnitude nd IA leds IC by nd IB by re the zero sequence currents in the three conductors. By definitions, these currents re equl in mgnitude nd in phse. Nottion for components of voltges corresponds to tht for the components of currents s described bove.

71 64 APPENDIX B Alph, Bet nd Zero Components of Three Phse System With phse A s the reference phse in three phse system, the Alph, Bet nd Zero components of current nd voltge re defined s follows: Alph components in phses B nd C re equl. Also, they re opposite in sign nd of hlf the mgnitude of the Alph components of phse A. Bet components in phses B nd C re equl in mgnitude nd opposite in sign; lso, they re zero in phse A. Zero components re equl in the three phses. Thus, the mtrix eqution for set of three voltge vectors VA, VB, nd VC of three phse system cn be written in terms of their Alph, Bet, nd Zero components s given below: 1 1 v -1/2 13!2-1/2-13/2 1 1 Also the three vectors VA, VB, nd VC cn be expressed in terms of their Alph, Bet, nd Zero components by equtions: VA v + vo (B.l) VB.lv 2 + /3v 2 B + vo (B.2) vc.lv 2 Ct /3v 2 B + vo (B. 3) The three voltge vectors V, v 6, nd v re lso expressed in terms

72 65 of the system vectors VA' VB' nd VC by solving the bove three equtions. Subtrcting one-hlf the sum of Equtions (B.2) nd (B.3) from Eqution (B.l) nd solving for V. gives: v 2 3 (VA (B.4) Subtrcting Eqution (B.3) from Eqution (B.2) nd solving for v 8 gives: (B.5) Adding Equtions (B.l), (B.2), nd (B.3) nd solving for v gives: (B.6) Also the mtrix eqution for the three voltge vectors V' v 6 nd v cn be written in terms of the system vectors VA, VB nd VC s given below: v 2/3-1/3-1/3 VA vs 1 /3-1//3 VB vo 1/3 1/3 1/3 vc The corresponding current equtions re: IA I l3 IB I s + 2 ro (B. 7) (B.S) Ic I ro (B.9)

73 66 Also, 2 IB + Ic I ) (IA ) 2 (B. 1) Is 1 13 (IB I c) (B. 11) I = 1 + IB + IC) ) (IA (B. 12) Substituting the bove equtions in the symmetricl components equtions given in Appendix A, the reltions between the symmetricl components nd the Alph, Bet, nd Zero components of voltge nd current re s given below: v VA + VA 1 2 vs -j (VA VA ) 1 2 vo VA Also, I IA + IA 1 2 nd Is -j (IA IA ) 1 2 Io IA