llll ‘ l 1 HI I! 1 HI 1 l s THS AN ANALYTICAL SOLUTION OF SUBSEQUENT FAULTS IN THREE-PHASE POWER SYSTEMS These: for Tim Degas of M. S. MFCHEGAN STATE COLLEGE WtLLARD DALE FREEDLE 1953 I ‘ 'FY—fi-‘F -' I This is to certify that the thesis entitled An Analytical Solution of Subsequent Faults in Three-pm.” Power Systems L presented by V1 110.111 Dsle Friedle has been accepted towards fulfillment ‘ of the requirements for . i1 ’ .71" . j} . Major profogor Due_.1n1;_21_.1953___ 0-169 } - l ' ’ . . '> -a W . . _ ~ . . l . “1‘5 . fijddb \ ‘ ‘8_t1;’ ‘/ ‘v- ..\ 3(“L4vlf‘. .'~, -m- 0.1]-‘f‘fi ‘ ‘le I" 'hk_‘.. \_: - "-fi_‘ Z£‘-;}_ .‘6 {‘lv‘. '.-?"l’;"‘~w SJ ‘2 e..'-‘. -;l .01 - '. '4' "t. :. '5‘. ' ‘% C - .~ . .13.;3 ' V's. . '2}; 4‘- -A."‘i" " . "\‘1, .‘tl .F‘ - ~ ' t?" ’: timid; ‘ s Al "fl-i, ~ , "76‘? w ' I i . I. I 4.16% é’.‘ Ll“ ~ . .a. ‘ 9" "‘2‘ ,D". ' ' alt/1‘31." trellis. 3: . ‘ 1' -- g ' 3’; 2 I 1.»), . r I" ‘."»-- ‘x ,".‘. r 1'.- I\ v‘ ‘1 v .V‘ .. l f", I‘ a :' 4~1éff§ { A: n. 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General Page IntrOduCtion e e e e e e e e e e e e e e e e e e e e e e e e e l Symmetrical Component Equations. . . . . . . . . . . . . . . . 4 Fault COHditions and Equation. 0 e e e e e e e e e e e e e e e 6 Determination of Current and Voltage Expressions . . . . . . . 11 Part ;;L,Solution to g_Problem ThoPr‘oblcmooooo..................... 16 Phase Diagrams . . . . . . . . . . . . . . . . . . . . . . . . 18 Normal Operating Currents. . . . . . . . . . . . . . . . . . . 21 Fault Voltages and Currents. . . . . . . . . . . . . . . . . . 25 Currents throughout the System . . . . . . . . . . . . . . . . 27 Three Phase Diagram. . . . . . . . . . . . . . . . . . . . . . 31 a) .\ . -.. t_. ~ \ ’1‘ - . . L” k, u)l_) l4) PnEFACE It is the purpose of this thesis to present in an organized manner the conditions, data, and computations for an analytical solution of the voltages and currents in a power system involving a simultaneous fault. The problem.will be solved by using the method of symmetrical components. This paper can be used as a guide for solving fault problems of a similar nature. The problem is worked for a line-to-line fault at One point in the system with the other line grounded at another point in the system. The author wishes to express his sincere thanls to Dr. J. A. Strel- zoff for suggesting the tOpic of this thesis and for his assistance in its preparation. The analysis of a three-phase circuit in which phase voltages and currents are balanced, and in which all circuit elements in each phase are balanced and symmetrical, is relatively simple since the treatment of a single-phase leads directly to the three-phase solution. The analysis by Kirchoff’s laws is much more difficult, however, when the circuit is not symmetrical, as the result of unbalanced loads or faults that are not symmetrical in the three phases. Symmetrical components is the method now generally adapted for calculating such circuits. This method of analysis makes possible the prediction, readily and accurately, of the behavior of a power system.during unbalanced short- circuit or unbalanced load conditions. The engineer's knowledge of such phenomena has been greatly increased and rapidly develOped since its introduction. Modern ideas of protective relaying and fault pro- tection grew from.an understanding of the symmetrical component methods. The extensive use of the network calculator for the determination of short-circuit currents and voltages has been furthered by the deveIOpment of symmetrical components, since each sequence network may be set up independently as a single-phase system, It is in this connection that the calculator has become indispensable in the analysis of power system performance and design. Not only has the method of symmetrical components been a valuable tool in system.analysis, but also, by providing new and simpler concepts, the understanding of power system.behavior has been clarified. The method is responsible for an entirely different procedure in the approach to predicting and analyzing power-system performance. 2. The material to be presented here involves the solution, by the method of symmetrical components, of a problem.involving a simultaneous fault condition. The solution Of such problems by network analyzers has been by now fully develOped. Occasionally it is desired to solve such a problem by analytical methods only. The object of this thesis is to develOp an analytical procedure for solving problems involving subsequent or simultaneous faults. The term fault will be defined as any condition occurring which is a departure from normal Operation. 4 short circuit or an Open circuit are two Of the more common conditions which constitute a fault. A simultaneous fault may consist of two or more short circuits on the same or different circuits, Open conductors, or any combination of diort .circuits and Open conductors. Occasionally two or'more faults will occur simultaneously at points which are widely separated in a three-phase power system” a combination of an unsymmetrical short circuit with an Open circuit may occur, for example, when the short circuit is partially cleared by the Opening of a fuse or a single pole circuit breaker. This condition may be described as a subsequent fault, one condition being the result of the other. These faults, although not occurring too frequently, are important from.the standpoint of protection. Relay systems which give satisfactory Operation for a fault at a single IOOation may fail to isolate simul- taneous faults. In order to simplify later derivations, a brief discussion of symr metrical components and the equations expressing the voltages and cur- rents in terms of these components will be given at this time. 3. The fundamental principle of symmetrical components, as applied to three-phase circuits, is that an unbalanced group of three related vectors, for example, voltage or current, can be resolved into three sets of vectors. The three vectors of each set are of equal magnitude and spaced either zero or 120 degrees apart. Each set is a "symmetrical component“ Of the original unbalanced vectors. Another way to express this fundamental principle is, any three co-planer vectors V3, Vb, and V6 can be expressed in terms Of three new vectors V1, V2, and V3 by three simultaneous linear equations with con- stant coefficients, where the choice of coefficients is arbitrary except for the restriction that the determinant made up of the coefficients must not be zero. However, there is only one choice for these coefficients so that the system Of three vectors can be replaced by three systems, each consisting of three symmetrical vectors. A system.of three symmetrical vectors is one in which the three vectors are equal in magnitude and displaced from each other by equal angles. For convenience in notation and manipulation, a vector Operator is introduced. Through usage it has come to be known as vector a. It is a vector of unit length and is oriented 120 degrees in a positive (counter- clockwise) direction from the reference axis or a ‘élaéy . A vector multiplied by it is not changed in magnitude but is simply rotated in position 120 degrees in a forward direction. In two Of the systems of revolving vectors, there is a sequence of phases, while in the third system.there is none, the three vectors being in phase. 4. The voltages and currents over an entire system.are then expressed 4 in terms of their components, all referred to the components of the reference phase. The choice of which phase to use as reference is en- tirely arbitrary, but once it is selected, this phase must be kept as the reference for voltages and currents throughout the system.and its analysis. It has become customary is symmetrical component notation to denote the reference phase as "phase a." The double subscript notation will be used with the small letters a, b, and c to indicate the phases and the nwmerals O, l, and 2 to designate the zero, positive, and negative sequence respectively. Now using “phase a” as the reference phase and making use of the Operator a, the following relations exist: For positive sequence vectors, vbl ' ‘2Va13 Vcl ' ava1 For negative sequence vectors, vbz = avgz; ch = azvaz For zero sequence vectors, Vbo ' Vac: Vco ‘ Vao Using these values for the positive, negative, and zero sequence, in writing the equations for the phase voltages in terms of the reference phase components, the following equations result: Va ' va1 f VaZ / V30 (1) Vb ' “zval f aVa2 f va0 (2) v6 = .val r azvaz f v80 (3) many times it is necessary to express the positive, negative, and zero components in terms of the unbalanced voltages. This can be done 5. by adding the above equations together, and equation (4) below is ob- tained. multiplying equation (1), (2), and (3) by l, a, and a2, res- pectively, and adding, expression (5) is obtained. Then multiply by 1, 2 a , and a, respectively, and add to obtain equation (6). v30 ! 1/3 (v, f vb f v,) (4) Val = 1/3 (v. f aVb i save) (5) v82 = 1/3 (v, f azvb f aVc) (s) The same expressions can be shown to exist for the vectors repre- senting the currents in a three-phase system. These will now be set down for later reference. Is s 131 f Ia2 f Ia0 (7) lb = «21.1 f alga f 1,0 (a) 1c : “Isl f ‘zlaz f Inc (9) 18° 3 1/3(Ia } Ib f Io) (10) I31 = 1/3(Ia f aIb f azxc) (11) IaZ = 1/3(I. I .sz f 81c) (12) These voltage and current vectors with the apprOpriate subscripts will be used to indicate various voltage and current vectors depending on the relationships to be found. In the following expressions, they will represent the conditions at the fault point. In the consideration of simultaneous faults, it is known that each fault affects the voltages and currents resulting from the other; there- fore, they cannot be treated independently. However, the conditions at each fault point can be set down expressing the fault voltages and cur- rents in terms of their symmetrical components and reference phase. 'Hfli 6. Using phase a as the reference phase, the currents as those flowing into the fault, and the voltages as those appearing as voltage to ground at the point of fault, the fault equations eXpressing the relations be- tween the symmetrical components of Ia and V8‘ will be develOped for the conditions to be used in this problem. Consider Fig. l for the cases given. X I ‘ :\IA ! —_"‘\‘<’a :- Vk' i ‘\\‘fih\;[¢ X 'x s: g: <1 51 I TIALJ I; ?‘\\fih\\#1c a; Fig..l Fault current and line-to-ground voltage representation at some point X in a system. If line a is grounded, Va 3 0 and lb = Tc :0. Substitution of these fault current conditions into equations (10), (11), and (12) will give, 1&0 - 1/3 I8 131 1/3 Ia 1&2 3 1/3 Ia From.which Iao ‘ Isl : IaZ (13) Substitution of Va 3 0 into equation (1) gives, Val = (Vao f V32) ' (14) Case B. Line-tg-Line Fault getweeg b and g. The equations for this fault are Vsz» » 31. \g. H < 0" H H H p I Substitution of the voltage (3) will give, fault conditions into equations (2) and (Vac f .2 / avaz) = (vao / aVal / azvaz Val (a2 - a) 3 V32 (a2 r a) v81 : VaZ (15) If the current conditions are substituted into equations (10), (11), and (12), we have IaO 3 O (16) 1,1 = 1/3 1b (a - a2) 1,2 = 1/3 1b (a2 - a) From.which Ib : 3 Ia1 (8 '62) 1,2 = 131(’a2 -a‘) . I 1 (a ~19 a -32 a (l -a I = I 3 150 - '1so “2 1gb: 1“ L“ 132 3 'Ial (17) In a similar manner, the expressions for the symmetrical component relations can be obtained for a line-to-line fault between phase a and b. They are H O Ia0 I! I O N Iaz Vaz (18) (19) (20) l'. .Na anilrlthflss. t . l: . Paul? 35/ Hedi a n a a ,. 8. Case C. Double Line-to-Ground Fault on Phases a and c. We again refer to Figure l and write immediately that Va ‘ V 0. c Ib 8 0 From equations (1), (3), and (8), we can write Va 'Vc ' val f V82 f Va0 "(aval * “eves f va0 8 (1 -.) val / (1 ~a2) v82 I o v‘ / vc = (1 f a) v81 f (1 f a2) v82 f avao = o Ib : ‘zlal f aIaZ f Ia0 8 0 From which we can write V32 : aval (21) = z ' e‘ Vao a V81 ($2) 2 . 131 = -a 132 -aIaO (as) The equations (13)--(23) give the relationship for the symmetrical components of Ia and Va at any point in the circuit for an unsymmetrical fault. For any one fault, three equations can be obtained, so that for a simultaneous fault, six equations can be written which are independent of the system impedances. The other six equations necessary for an analytical solution give the relations between fault currents and volt- ages for each sequence and therefore depend upon the impedances of the three sequence networks of the system. The.equiva1ent circuit for replacing a single fault in the positive- sequence network is a single impedance connected in shunt (or series) with the positive-sequence network at the point of fault. The equiva- lent circuit for a double fault is a three-terminal network. The three terminals are two points of fault and the zero potential bus. The simplest forms of three-terminal networks are the delta and the eye. 9. after the reduction of the sequence circuits into their equivalent wye form, the equations relating the sequence voltages and currents to the respective sequence impedances can be written. Fig. 2. hero Sequence Wye The notation used in the above diagram represents the two faults at point X and Y in the systemu The impedances are represented by 80, Co, and Do and the currents by 110 and Iyo’ From this equivalent wye, we can write the following equations: V10 8 ~1x0 (30 f co) -Iyo (so) (24) vyo . -1,o (so) -Iyo (50 f Do) (25) Solving these for the currents I10 and Iyo IIO : .VXO m1 1‘ vyO—én— 00 D00 D (24a) Doo Doo (25a) Where D00 3 (30 f Co)(Do f 30) ’(30)2 ’- SODO f 0090 f (3'08o This gives us two more equations necessary to make the analytical solution for a simultaneous fault condition. The other equations neces- sary are the relations between the positive and negative sequence fault currents and voltages and their impedances. Similarly for the negative sequence diagram two of the remaining equations can be written. . ESAWIHh a. Is I . Fig. 3. Negative Sequence Wye. sz ‘ “112 (Cs f 32) ’Iyz (52) Vyz ' -Ixz (52) -Iyz (52 f Dz) Solving for the currents 112 and Iyz 112 8 -sz ‘32 f D2) f Vyz §2 D22 022 Iyz 3 sz $2 ~vy2 32 t ca D22 D22 Where D22 3 0282 f D232 f CZDZ 10. (26) (27) (26a) (27a) Also, the positive sequence equivalent wye circuit can be drawn and the last two equations written. The equivalent circuit drawn here for the positive sequence is for s system.under load with the voltages V1: and ny being the voltages in the system at the points x and y before the faults occur. Vi. VVI Fig. 4. Positive Sequence Wye. vx1 = vfx ~111 (01 f 51) -Iy1 (31) ny 'Iy1 (91 f 51) “111 (51) yl (28) (29) 11. Now for the analytical solution to a problem.there are twelve equa- tions and twelve unknowns. Since the solution of these equations is wanted for a line-to-line fault at point x in the system between phase b and c anda line-to-ground fault at point y in the system.on phase a, equations (13) and (14) will apply for point y, and equations (15), (16), and (17) will apply for point I. (See Fig. 5.) These equations will be written here now for convenience and with the subscript a emitted since they all refer to phase a. The x or y subscript is added depending on to which fault point they apply. X \QA: \6n I _ I :\ 1,, :Nvg I I VQsI \fls (\WIXO I\ IV! we: ______ Vw. I I I\“*~115e I\\“\~.Ivz Fig. 5. Fault Conditions on System.(Line-to-Line between Phases b and c at one point in system, and Phase a to Ground at another Point). Iyo 3 1,1 = Iyz (13.) Vyl : '(Vyo f. VyZ) (1%) v11 3 vxz (15a) 110 3 0;,; (16:) 112 ‘ 'le (17a) From these six equations and equations (24) to (27), the positive and negative components of the voltage and currents are elhminated. If equa- tions (13a), (16a), and (17s) are substituted into equations (24), (25), and (27) for the zero and negative sequence currents, the result is shown below. A». [:14 .9.I II I! I. g 111.... Vi 12. V10 = 0 -Iy1 (so) (30) vyo = o ~1y1 (DO f so) (31) v12 2 111 (52 f CZ) -Iyl I32) (32) vyz = 1,1 s, -1y1 (s2 / D2) (33) These expressions are then substituted into equations (14a) and (15a) and there results an eXpression for the positive sequence voltages in terms of positive sequence currents and negative and zero sequence impedances. this vyl : -(Vyo f vyZ) : Iyl (Do f SO) -1X182 f Iyl (SZDZ) Vxl : v12 ' 1x1 (52 f Cg) -Iyl (32) Factoring and rearranging gives V11 ' 1,1 (33 f Oz) -Iy1 (52) (34) Vyl : '11; (32) f Iyl (Do f so I Dz f 32) (35) Equations (34) and (35) may now be simplified by being written in form: V11 : K 111. f m I yl (36) Vyl : 11 I11 f 1 Iyl Where k = 52 f 02' m.= -Sz, n 3 ~82, and l 3 (Do f So f D2 f 52) If the values of V11 and vyl from (36) are now substituted into equations (28) and (29), the following expression results: er 3 I11 (01 f 31) f Iyl (31) f x 1,, / miyl vy, = 1,1 (31) f Iyl (D1 f 31) f n Ix1 f 1 I yl Solving these for 111 and Iyl' h1=wrmuamial-V,(s/m) 1 A. y (37) 14. 1,1 = -V;g (s) f n) / vfig (g, f g, f k]) rig £50 13. Where Ac: (Cl f 31 f k“191 f 31 i‘ 1) “(51 f ““31 f m) The expression (37) can now be used for calculating the fault currents in the positive sequence at the two fault points. ane these are deter- mined, the voltages Vxl and vyl are determined from equation (36). The values of V10, Vyo' V12, and Vy2 are then found from equations (30)-- (33). There remains only Ixo' Iyo' 112' Iyz, which can be calculated from equations (24a)--(27a). This then gives all of the six fault volt- ages and the six fault currents. knowing these will enable the calcu- lations to be made of the voltages and current at any point in the cir- cuit due to the faults. The values of k, m, n, and 1 will now be computed for another simul- taneous fault condition(see Fig. 6). Y Vhd I : In“ ______ E IYR \a‘: VMM _# ______ r\‘*~\~IvB \AJ I\ Inc _ ————— I\ IYC Fig. 6. Simultaneous Fault Condition on a Power System (Line-to-Line Fault Between Line a and b at one point in system, with Phase a to Ground at another Point in System). The equations for the relation between the sequence voltages and currents at point x and y are (13), (14), (18), (19), and (20). I,o = 0 (18a) 1x2 = "2 Ix1 (193) v = azv (20a) x2 xl 14. Again the subscript a is omitted to avoid writing a triple sub- script. The subscript x or y has been added to determine which equations express the conditions at which fault point. Using equations (24)--(27) and substituting equations (13a), (18a), and (19a) into them for the zero and negative components of current, the result is on : 0 “Iyl (So) vyO : o ~1y1 (D0 / so) (38) sz ‘ 821:1 (02 f 32) 'Iyl (32) vyz = a2 1,1 (32) ~1y1 (Dz / 32) Substituting these values of (38) into equations (14a) and (15a), the following expressions result: an = .21 (c i s ) -I (s ) 11 11 2 2 y1 2 vyl = ~a2111 (33) f Iyl (D0 / 50 f D2 f 32) Rearranging and factoring gives, v11 = 1,1 (c2 / 32) ~1y1 (aSZ) V : 'le (‘zSZ) f Iyl (DO f 30 f D2 f 52) yl hewriting the above, they become, (39) vyl - n 1x1 / 1 Iyl Where k 3 oz f 52, m.‘ -a32, n 3 '323L' and The equations from.(39) can then be substituted into the equations for the positive sequence equivalent wye and the expressions for the positive sequence fault currents determined like (37). The equations 15. in (37) are valid for any one system for any variety of faults which are located at the same points. Just the values of k, m, n, and 1 need be determined for each condition. If the faults occur at different points every time, then new values of the equiValent wye circuits must be computed also. A problem.will now be solved illustrating the use of these equations for a short circuit on phases b and c with phase a to ground. :h. ale“! bsaaahau .h . 16. Part I; The problem which is to be solved by the method of symmetriCal com- ponents was suggested to Dr. J. a. Strelzoff by Consumers Power Company. It is a typical problem involving a portion of a large distribution system with small generating units connected into the system through the distri- bution. It has been separated from the rest of the system with the equi- valent voltage and impedance values of the system.determined at the point of separation. In these types of systems, there must be adequate protection pro- vided these small generating units for the different types of faults which may occur. Many of these conditions are now being neglected due to the cost of obtaining the use of networa analyzers and of setting the problems up. Therefore, the analytical solutions to simultaneous fault conditions on portions of large systems are very much desired. It is not necessarily a case of obtaining the exact solution for each current in each branch due to the faults but more to find the trend of the currents for many fault conditions and then setting the protective equipment from this over- all data. The problem worked here is for phase B and C short circuited together at some point with phase A shorted directly to ground at some point in the system isolated from the former fault point. The one-line diagram is shown in Fig. 7. Below the figure are the impedance values of the loads, transmission lines, and generators figured on one base KUa. Figure eight then shows the one-line diagram.with the values of the impedances included at each load. "’1: '- ‘ '_ .- . l7. 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