An Introduction to Symmetrical Components, System Modeling and Fault Calculation

Transcription

1 An ntrductin t Symmetrical Cmpnents, System Mdeling and Fault Calculatin Presented at the 3 th Annual HANDS-ON Relay Schl March - 5, 3 Washingtn State University Pullman, Washingtn By Stephen Marx, and Dean Bender Bnneville Pwer Administratin Symmetrical Cmpnents March, 3

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3 C C ntrductin The electrical pwer system nrmally perates in a balanced three-phase sinusidal steady-state mde. Hwever, there are certain situatins that can cause unbalanced peratins. The mst severe f these wuld be a fault r shrt circuit. Examples may include a tree in cntact with a cnductr, a lightning strike, r dwned pwer line. n 98, Dr. C. L. Frtescue wrte a paper entitled Methd f Symmetrical Crdinates Applied t the Slutin f Plyphase Netwrks. n the paper Dr. Frtescue described hw arbitrary unbalanced 3-phase vltages (r currents) culd be transfrmed int 3 sets f balanced 3-phase cmpnents, Fig.. He called these cmpnents symmetrical cmpnents. n the paper it is shwn that unbalanced prblems can be slved by the reslutin f the currents and vltages int certain symmetrical relatins. B B Fig. By the methd f symmetrical crdinates, a set f unbalanced vltages (r currents) may be reslved int systems f balanced vltages (r currents) equal in number t the number f phases invlved. The symmetrical cmpnent methd reduces the cmplexity in slving fr electrical quantities during pwer system disturbances. These sequence cmpnents are knwn as psitive, negative and zer-sequence cmpnents, Fig. Fig. Symmetrical Cmpnents Page

4 The purpse f this paper is t explain symmetrical cmpnents and review cmplex algebra in rder t manipulate the cmpnents. Knwledge f symmetrical cmpnents is imprtant in perfrming mathematical calculatins and understanding system faults. t is als valuable in analyzing faults and hw they apply t relay peratins.. Cmplex Numbers The methd f symmetrical cmpnents uses the cmmnly used mathematical slutins applied in rdinary alternating current prblems. A wrking knwledge f the fundamentals f algebra f cmplex numbers is essential. Cnsequently this subject will be reviewed first. Any cmplex number, such as a jb, may be represented by a single pint p, pltted n a Cartesian crdinates, in which a is the abscissa n the x axis f real quantities and b the rdinate n the y axis f imaginary quantities. This is illustrated in Fig.. θ Fig.. Referring t Fig.., let r represent the length f the line cnnecting the pint p t the rigin and θ the angle measured frm the x-axis t the line r. t can be bserved that a r csθ (.) b r sinθ (.). Prperties f Phasrs A vectr is a mathematical quantity that has bth a magnitude and directin. Many quantities in the pwer industry are vectr quantities. The term phasr is used within the steady state alternating linear system. t is used t avid cnfusin with spatial vectrs: the angular psitin f the phasr represents psitin in time, nt space. n this dcument, phasrs will be used t dcument varius ac vltages, currents and impedances. A phasr quantity r phasr, prvides infrmatin abut nt nly the magnitude but als the directin r angle f the quantity. When using a cmpass and giving directins t a huse, frm a given lcatin, a distance and directin must be prvided. Fr example ne culd say that a huse is miles at an angle f 75 degrees (rtated in a clckwise directin frm Nrth) frm where am standing. Just as we dn t say the ther huse is - miles away, the magnitude f Symmetrical Cmpnents Page

5 the phasr is always a psitive, r rather the abslute value f the length f the phasr. Therefre giving directins in the ppsite directin, ne culd say that a secnd huse is miles at an angle f 55 degrees. The quantity culd be a ptential, current, watts, etc. Phasrs are written in plar frm as Y Y θ (.) Y csθ jy sinθ (.) where Y is the phasr, Y is the amplitude, magnitude r abslute value and θ is the phase angle r argument. Plar numbers are written with the magnitude fllwed by the symbl t indicate angle, fllwed by the phase angle expressed in degrees. Fr example Z 9. This wuld be read as at an angle f 9 degrees. The rectangular frm is easily prduced by applying Eq. (.) The phasr can be represented graphically as we have demnstrated in Fig.., with the real cmpnents cinciding with the x axis. When multiplying tw phasrs it is best t have the phasr written in the plar frm. The magnitudes are multiplied tgether and the phase angles are added tgether. Divisin, which is the inverse f multiplicatin, can be accmplished in a similar manner. n divisin the magnitudes are divided and the phase angle in the denminatr is subtracted frm the phase angle in the numeratr. Example. Multiply A B where A 5 35 and B Slutin A B ( ) ( ) Example. Slve D C where C 35 5 and D 5 3. Slutin C 5 35 D ( 35 5 ) Symmetrical Cmpnents Page 3

6 3. The j and a peratr Recall the peratr j. n plar frm, j 9. Multiplying by j has the effect f rtating a phasr 9 withut affecting the magnitude. Table 3. - Prperties f the vectr j. j. j 9 j 8 j 3 7 j 9 j j Example 3. Cmpute jr where Slutin jr 9 ( 6 ) 5 R 6. Ntice that multiplicatin by the j peratr rtated the Phasr R by magnitude. Refer t Fig. 3. R 9, but did nt change the (a) R jr R (b) R j Fig. 3.. j effects Symmetrical Cmpnents Page 4

7 n a similar manner the a peratr is defined as unit vectr at an angle f, written as a. The peratr a, is als a unit vectr at an angle f 4, written a 4. Example 3. Cmpute ar where Slutin ar ( 6 ) 8 R 6. R (a) A R ar (b) j R Fig. 3.. a effects Table 3. - Prperties f the vectr a. j. a a 4 3 a 36 a a a a a 6 a 6 a a j 3 a a j 3 a 3 3 a 3 3 Symmetrical Cmpnents Page 5

8 4. The three-phase System and the relatinship f the 3 n a Wye cnnected system the vltage measured frm line t line equals the square rt f three, 3, times the vltage frm line t neutral. See Fig. 4. and Eq. (4.). The line current equals the phase current, see Eq. (4.) Fig. 4. VLL 3V LN (4.) L Φ (4.) n a Delta cnnected system the vltage measured frm line t line equals the phase vltage. See Fig. 4. and Eq. (4.3). The line current will equal the square rt f three, 3, times the phase current, see Eq. (4.4) L Φ Φ VLL Fig. 4. V LL V Φ (4.3) L 3 Φ (4.4) Symmetrical Cmpnents Page 6

9 The pwer equatin, fr a three phase system, is S 3V LL L (4.5a) P 3 LL L csψ (4.5b) Q 3 LL L sinψ (4.5c) where S is the apparent pwer r cmplex pwer in vlt-amperes (VA). P is the real pwer in Watts (W, kw, MW). Q is the reactive pwer in VARS (Vars, kvars, MVars). 5. The per-unit System 5. ntrductin n many engineering situatins it is useful t scale, r nrmalize, dimensined quantities. This is cmmnly dne in pwer system analysis. The standard methd used is referred t as the per-unit system. Histrically, this was dne t simplify numerical calculatins that were made by hand. Althugh this advantage is eliminated by the calculatr, ther advantages remain. Device parameters tend t fall int a relatively narrw range, making errneus values cnspicuus. Using this methd all quantities are expressed as ratis f sme value r values. The per-unit equivalent impedance f any transfrmer is the same when referred t either the primary r the secndary side. The per-unit impedance f a transfrmer in a three-phase system is the same regardless f the type f winding cnnectins (wye-delta, delta-wye, wye-wye, r delta-delta). The per-unit methd is independent f vltage changes and phase shifts thrugh transfrmers where the vltages in the winding are prprtinal t the number f turns in the windings. Manufactures usually specify the impedance f equipment in per-unit r percent n the f its nameplate rating f pwer (usually kva) and vltage (V r kv). The per-unit system is simply a scaling methd. The basic per-unit scaling equatin is actual _ value per unit (5.) _ value The value always has the same units as the actual value, frcing the per-unit value t be dimensinless. The value is always a real number, whereas the actual value may be cmplex. The subscript pu will indicate a per-unit value. The subscript will Symmetrical Cmpnents Page 7

10 indicate a value, and n subscript will indicate an actual value such as Amperes, Ohms, r Vlts. Per-unit quantities are similar t percent quantities. The rati in percent is times the rati in per-unit. Fr example, a vltage f 7kV n a f kv wuld be 7% f the vltage. This is equal t times the per unit value f.7 derived abve. The first step in using per-unit is t select the (s) fr the system. S pwer, in VA. Althugh in principle S may be selected arbitrarily, in practice it is typically chsen t be MVA. V vltage in V. Althugh in principle V is als arbitrary, in practice V is equal t the nminal line-t-line vltage. The term nminal means the value at which the system was designed t perate under nrmal balanced cnditins. Frm Eq. (4.5a) it fllws that the pwer equatin fr a three-phase system is: Slving fr current: S 3V 3 Φ (5.) S 3Φ 3V Because S 3Φ can be written as kva r MVA and vltage is usually expressed in kilvlts, r kv, current can be written as: kva amperes (5.3) 3kV Slving fr impedance: Z V V S r Z Z kv x hms (5.4a) kva kv hms (5.4b) MVA Symmetrical Cmpnents Page 8

11 Given the values, and the actual values: V Z, then dividing by the we are able t calculate the pu values V Z Vpu puz pu V Z After the values have been selected r calculated, then the per-unit impedance values fr system cmpnents can be calculated using Eq. (5.4b) r Z( Ω) MVA Z ( Ω) pu Z Z kv (5.5a) kva Z ( Ω) pu Z kv (5.5b) t is als a cmmn practice t express per-unit values as percentages (i.e. pu %). (Transfrmer impedances are typically given in % at the transfrmer MVA rating.) The cnversin is simple per unit Then Eq. (5.5a) can be written as percent _ value % Z ( Ω) kva Z( Ω) MVA Z (5.6) kv kv t is frequently necessary, particularly fr impedance values, t cnvert frm ne (ld) t anther (new). The cnversin is accmplished by tw successive applicatin f Eq. (5.), prducing: Substituting fr Z Z Z ld new ld pu Z pu new ld Z and new Z and re-arranging the new impedance in per-unit equals: new ld new ld kva kv Z pu Z pu (5.7) ld new kva kv n mst cases the turns rati f the transfrmer is equivalent t the system vltages, and the equipment rated vltages are the same as the system vltages. This means that the vltage-squared rati is unity. Then Eq. (5.7) reduces t Symmetrical Cmpnents Page 9

12 new new ld MVA Z pu Z pu (5.8) ld MVA We can quickly change frm ne impedance value in hms, t anther impedance value in hms by dividing by the ld vltage and multiplying by the new vltage in hms. This is shwn in Eq. (5.9) new new ld kv Z hm Zhm (5.9) ld kv Example 5. A system has S MVA, calculate the current fr a) V 3 kv b) V 55 kv Then using this value, calculate the actual line current and phase vltage where pu, and V. 5 pu at bth 3 kv and 55 kv. Slutin Using Eq. (5.3) kva amperes 3kV a) amperes 5A 3 3 Frm Eq. (5.) b) amperes. A 3 55 V actual actual (5.9) pu pu V V (5.) At 3 kv actual A 4 c) ( ) ( ) A d) V actual (.5) ( 3kV ) 5kV At 55 kv actual 4.95.A 544 e) ( ) ( ) A f) V actual (.5) ( 55kV ) 63kV Symmetrical Cmpnents Page

13 Example 5. A 9 MVA 55/4.5 auttransfrmer has a nameplate impedance f.4% a) Determine the impedance in hms, referenced t the 55 kv side. b) Determine the impedance in hms, referenced t the 4.5 kv side Slutin First cnvert frm % t pu. Z% Zpu.4 Arranging Eq. (5.5a) and slving fr Z actual gives kv Z( Ω ) Z pu MVA a) Z 55 kv Ω ; therefre 55 9 b) Z 4.5kV Ω A check can be made t see if the high-side impedance t the lw-side impedance equals the turns rati squared Applicatin f per-unit Appling this t relay settings, a practical example can be shwn in calculatin f the settings fr a relay n a transmissin line. Fr distance relays a cmmn setting fr zne is 85% f the line impedance. Zne shuld be set nt less than 5% f the line, with care t nt ver reach the zne f the next line sectin. f this des then zne will need t be crdinated with the next line sectin zne. Referring t Fig. 5. the line impedance fr the 6 kv line is Z hms. Using the abve criteria f 85% fr zne and 5 % fr zne the relays wuld be set at Fr zne Z Ω ) 85%( ) ( Z ( Ω ) Symmetrical Cmpnents Page

14 Fr zne Z Ω ) 5%( ) ( Z ( Ω ) kv x x a 6 kv a x x Z 8 hms Z 8 hms Z% MVA 6/5kV Fig 5. Fr the relays n the 5 kv side f the transfrmer, the impedance f the transfrmer needs t be calculated. Frm example 5. we see that 5 Z 5 kv Ω Next the line impedance needs referenced t the 5 kv side f the transfrmer. Using equatin 5.9 new new ld kv Z hm Zhm (5.9) ld kv Substituting, the line impedance equals 5 kv 5 hm Z hms 6 Adding this t the transfrmer, the impedance setting fr the relays n the 5 kv side f the transfrmer is Z Using the same criteria fr zne and zne reach. Fr zne Z Ω ) 85%( ) ( Z ( Ω ) Fr zne Z Ω ) 5%( ) ( Z ( Ω ) Given these values, ne can easily see that by ignring the values f the vltages the relay settings wuld nt be adequate. Fr example if the 6 kv settings were applied t the 5 kv relays, zne wuld ver reach the remte terminal. Cnversely, if the 5 Symmetrical Cmpnents Page

15 kv settings were applied t the 6 kv relays zne wuld nt reach past the remte terminal and wuld thus nt prtect the full line. 5 kv x x a 6 kv a x x Z 8 hms Z 8 hms Z 8 hms Z 8 hms Z 8 hms Z 8 hms Z% MVA 6/5kV Fig Calculating actual values frm per-unit n the fllwing sectins we will discuss symmetrical faults. The analysis f the faults uses the per-unit. A impedance and vltage f the system is express in per-unit. Then the fault current and fault vltage is slved and that value will be given in per unit. Next we need t cnvert frm per-unit t actual amps and vlts by using the values. Using the abve equatins it is easy t prve the fllwing equatins. The MVA fr a three phase fault is given as MVABase MVAFault (5.) ZFault PU Or MVAFault fr a MVA Base (5. a) Z PU Or Fault Base _ Current (5.) Z PU Fault Fault _ Current 5.3 Cnverting per-unit Fault, (5. a) ( Z PU ) 3( kv ) Fault Befre using the per-unit impedance f a transfrmer frm a manufacture nameplate yu must first cnvert it t a per-unit value f yur system. Typically the three-phase pwer f MVA is used. This is dne by first cnverting the per unit impedance t an actual impedance (in hms) at 55kV and then cnverting the actual impedance t a perunit impedance n the new. Repeat, this time cnverting the per unit impedance t Base Symmetrical Cmpnents Page 3

16 an actual impedance (in hms) at 4.5kV and then cnverting the actual impedance t a per-unit impedance n the new. n the prblem 3 at the end f this dcument, the transfrmer nameplate data is fr a rati f 55/4.5kV r.74, whereas BPA s ASPEN mdel uses nminal vltages f 55kV and 3kV fr a rati f.83. Because BPA used a transfrmer rati in ASPEN mdel that was different than the transfrmer nameplate values, we have a discrepancy in the per-unit impedance values that we btained. The prblem arises because when a transfrmer is applied t the BPA system the transfrmer tap used will ften be different than the ne used in the nameplate calculatins. What is the crrect way t cnvert the per-unit impedance t the BPA? Because the actual impedance f the transfrmer will vary when different taps are used, the mst accurate way t mdel the impedance wuld be t actually measure the impedance with the transfrmer n the tap that will nrmally be used n the BPA system. This impedance wuld then be cnverted t a per-unit value n the BPA mdel. Since this isn t nrmally pssible, a clse apprximatin can be made by assuming that the per-unit impedance given n the nameplate will remain the same fr the different tap psitins f the transfrmer. Find the transfrmer tap psitin that mst clsely matches the rati f the ASPEN mdel (.83 fr a 55/3kV transfrmer), then cnvert the nameplate per-unit impedance t an actual value d n either the high- r lw-side vltage given fr that tap psitin. This actual impedance is then cnverted t a per-unit value n the BPA mdel, using the high-side BPA vltage if the high-side vltage was used fr the cnversin t actual impedance, r using the lw-side BPA vltage if the lw-side vltage was used fr the cnversin t actual impedance. See prblem Sequence Netwrks Refer t the basic three-phase system as shwn in Fig. 6.. There are fur cnductrs t be cnsidered: a, b, c and neutral n. a b c Van Vbn Vcn n Fig. 6. Symmetrical Cmpnents Page 4

17 The phase vltages, V p, fr the balanced 3Φ case with a phase sequence abc are V an Va Vp (6.a) V bn Vb Vp (6.b) V 4 cn Vc Vp Vp (6.c) The phase-phase vltages, V LL, are written as V 3 ab Va Vb VLL (6.a) bc Vb Vc VLL (6.b) ca Vc Va VLL (6.c) V 9 V 5 Equatin (6.) and (6.) can be shwn in phasr frm in Fig. 6.. Ψ Ψ Ψ Fig. 6. There are tw balanced cnfiguratins f impedance cnnectins within a pwer system. Fr the wye case, as shwn in Fig. 4., and with an impedance cnnectin f Z Ψ, the current can be calculated as V VP a ψ (6.3) Z Z Y Y Where Ψ is between 9 and 9. Fr Ψ greater than zer degrees the lad wuld be inductive ( a lags V a ). Fr ψ less than zer degrees the lad wuld be capacitive ( a leads V a ). Symmetrical Cmpnents Page 5

18 The phase currents in the balanced three-phase case are ψ (6.4a) a p ψ (6.4b) b p 4 ψ (6.4c) c p See Fig. 6.. fr the phasr representatin f the currents. 7. Symmetrical Cmpnents Systems The electrical pwer system perates in a balanced three-phase sinusidal peratin. When a tree cntacts a line, a lightning blt strikes a cnductr r tw cnductrs swing int each ther we call this a fault, r a fault n the line. When this ccurs the system ges frm a balanced cnditin t an unbalanced cnditin. n rder t prperly set the prtective relays, it is necessary t calculate currents and vltages in the system under such unbalanced perating cnditins. n Dr. C. L. Frtescue s paper he described hw symmetrical cmpnents can transfrm an unbalanced cnditin int symmetrical cmpnents, cmpute the system respnse by straight frward circuit analysis n simple circuit mdels, and transfrm the results back int riginal phase variables. When a shrt circuit fault ccurs the result can be a set f unbalanced vltages and currents. The thery f symmetrical cmpnents reslves any set f unbalanced vltages r currents int three sets f symmetrical balanced phasrs. These are knwn as psitive, negative and zer-sequence cmpnents. Fig. 7. shws balanced and unbalanced systems. B Fig. 7. Cnsider the symmetrical system f phasrs in Fig. 7.. Being balanced, the phasrs have equal amplitudes and are displaced relative t each ther. By the definitin f symmetrical cmpnents, V b always lags V a by a fixed angle f and always has the same magnitude as V a. Similarly V c leads V a by. t fllws then that V V (7.a) a a b ( 4 ) Va a Va c ( ) Va ava V (7.b) V (7.c) Symmetrical Cmpnents Page 6

19 Where the subscript () designates the psitive-sequence cmpnent. The system f phasrs is called psitive-sequence because the rder f the sequence f their maxima ccur abc. Similarly, in the negative and zer-sequence cmpnents, we deduce V V (7.a) a a b ( ) Va ava c ( 4 ) Va a Va V (7.b) V (7.c) V V (7.3a) a a V V (7.3b) b a V V (7.3c) c a Where the subscript () designates the negative-sequence cmpnent and subscript () designates zer-sequence cmpnents. Fr the negative-sequence phasrs the rder f sequence f the maxima ccur cba, which is ppsite t that f the psitive-sequence. The maxima f the instantaneus values fr zer-sequence ccur simultaneusly. Fig.7. n all three systems f the symmetrical cmpnents, the subscripts dente the cmpnents in the different phases. The ttal vltage f any phase is then equal t the sum f the crrespnding cmpnents f the different sequences in that phase. t is nw pssible t write ur symmetrical cmpnents in terms f three, namely, thse referred t the a phase (refer t sectin 3 fr a refresher n the a peratr). V V V V (7.4a) a b a a a V V V V (7.4b) c b b b V V V V (7.4c) c c c Symmetrical Cmpnents Page 7

20 We may further simplify the ntatin as fllws; define V Va (7.5a) V V (7.5b) a V V (7.5c) a Substituting their equivalent values V a V (7.6a) V V V b V a V av (7.6b) V c V (7.6c) av a V These equatins may be manipulated t slve fr V, V, and V in terms f V a, V b, and V c. ( V V V ) V 3 a b c (7.7a) ( V av a V ) V a b c (7.7b) 3 V ( Va a Vb avc ) (7.7c) 3 t fllws then that the phase currents are a (7.8a) b a a (7.8b) c (7.8c) a a The sequence currents are given by ( ) 3 a b c (7.9a) ( a a ) a b c (7.9b) 3 ( a a b ac ) (7.9c) 3 The unbalanced system is therefre defined in terms f three balanced systems. Eq. (7.6) may be used t cnvert phase vltages (r currents) t symmetrical cmpnent vltages (r currents) and vice versa [Eq. (7.7)]. Symmetrical Cmpnents Page 8

21 Example 7. Given Va 5 53, Vb 7 64, Vc 7 5, find the symmetrical cmpnents. The phase cmpnents are shwn in the phasr frm in Fig. 7.3 Vc Va Vb Unbalanced cnditin Fig. 7.3 Slutin Using Eq. (7.7a) Slve fr the zer-sequence cmpnent: V V V V 3 a ( ) a b c ( ) Frm Eq. (7.3b) and (7.3c) V 3.5 b c 3.5 V Slve fr the psitive-sequence cmpnent: Va ( Va avb a Vc ) ( ( ) ( ) Frm Eq. (7.b) and (7.c) V 5. 3 b c 5. V Slve fr the negative-sequence cmpnent: V a ( Va a Vb avc ) 3 Symmetrical Cmpnents Page 9

22 ( 5 53 ( ) ( 7 5 ) Frm Eq. (7.b) and (7.c) V.9 48 b c.9 V 8 The sequence cmpnents can be shwn in phasr frm in Fig Fig. 7.4 Using Eq. (7.6) the phase vltages can be recnstructed frm the sequence cmpnents. Example 7. Given V 3.5, V 5., V.9 9, find the phase sequence cmpnents. Shwn in the phasr frm in Fig. 7.4 Slutin Using Eq. (7.6) Slve fr the A-phase sequence cmpnent: V a V V V Slve fr the B-phase sequence cmpnent: 9 V b V a V av Symmetrical Cmpnents Page

23 Slve fr the C-phase sequence cmpnent: V c V av a V This returns the riginal values given in Example 5.. This can be shwn in phasr frm in Fig Vc Vc Vc Vc Va Va Va Vb Va Vb Vb Vb Fig. 7.5 Ntice in Fig. 7.5 that by adding up the phasrs frm Fig. 7.4, that the riginal phase, Fig. 7.3 quantities are recnstructed. 8. Balanced and Unbalanced Fault analysis Let s tie it tgether. Symmetrical cmpnents are used extensively fr fault study calculatins. n these calculatins the psitive, negative and zer-sequence impedance netwrks are either given by the manufacturer r are calculated by the user using vltages and pwer fr their system. Each f the sequence netwrks are then cnnected tgether in varius ways t calculate fault currents and vltages depending upn the type f fault. Symmetrical Cmpnents Page

24 Given a system, represented in Fig. 8., we can cnstruct general sequence equivalent circuits fr the system. Such circuits are indicated in Fig. 8.. Fig. 8. The psitive-sequence impedance system data fr this example in per-unit is shwn in Fig. 8.. Fig. 8. Assuming the negative-sequence equals the psitive-sequence, then the negativesequence is shwn in Fig 8.3 Fig. 8.3 The zer-sequence impedance is greater then the psitive and fr ur purpse is assumed t be three times greater. Als because f the wye-delta transfrmer, zer-sequence frm the generatr will nt pass thrugh the transfrmer. This will be shwn in sectin.. Zer-sequence is shwn in Fig 8.4 Symmetrical Cmpnents Page

25 Fig. 8.4 The Thevenin equivalents fr each circuit is reduced and shwn in Fig. 8.5 V V V Fig. 8.5 Each f the individual sequence may be cnsidered independently. Since each f the sequence netwrks invlves symmetrical currents, vltages and impedances in the three phases, each f the sequence netwrks may be slved by the single-phase methd. After cnverting the pwer system t the sequence netwrks, the next step is t determine the type f fault desired and the cnnectin f the impedance sequence netwrk fr that fault. The netwrk cnnectins are listed in Table 8. Table 8. - Netwrk Cnnectin Three-phase fault - The psitive-sequence impedance netwrk is nly used in three-phase faults. Fig. 8.3 Single Line-t-Grund fault - The psitive, negative and zer-sequence impedance netwrks are cnnected in series. Fig. 8.5 Line-t-line fault - The psitive and negative-sequence impedance netwrks are cnnected in parallel. Fig. 8.7 Duble Line-t-Grund fault - All three impedance netwrks are cnnected in parallel. Fig. 8.9 Symmetrical Cmpnents Page 3

26 The system shwn in Fig. 8. and simplified t the sequence netwrk in Fig. 8.5 and will be used thrughut this sectin. Example 8. Given Z.99 9 pu, Z.75 9 pu, Z.75 9 pu, cmpute the fault current and vltages fr a Three-phase fault. Nte that the sequence impedances are in per-unit. This means that the slutin fr current and vltage will be in per-unit. Slutin The sequence netwrks are intercnnected, and shwn Z V - Nte that fr a three phase fault, there are n negative r zer-sequence vltages. V V The current is the vltage drp acrss Z V Z j.75 j5.7 Z Z V - V The phase current is cnverted frm the sequence value using Eq. (7.8). - a j pu b a ( j5.7) a() pu c a( j5.7) a () pu Calculating the vltage drp, the sequence vltages are V V V Z V j.75 j5.7. pu ( ). Symmetrical Cmpnents Page 4

27 The phase vltages are cnverted frm the sequence value using Eq. (7.6). V a.... pu V b. a (.) a(.). pu b c V c. a(.) a (.). pu The per-unit value fr the current and vltage wuld nw be cnverted t actual values using Eq. (5.9) and Eq. (5.) and knwing the pwer and vltage fr the given system. See example 5. fr a reference. Vc Vb Va The currents and vltages can be shwn in phasr frm. a Example 8. Given Z.99 9 pu, Z.75 9 pu, Z.75 9 pu, cmpute the fault current and vltages fr a Single line-t-grund fault. Nte that the sequence impedances are in per-unit. This means that the results fr current and vltage will be in per-unit. Slutin The sequence netwrks are intercnnected in series, as shwn. Because the sequence currents are in series, and using hms law. V ( Z Z Z ) ( j.99 j.75 j.75) Z Z Z V - V - j. 8 pu V The phase currents are cnverted frm the sequence value using Eq. (7.8). Substituting int - Symmetrical Cmpnents Page 5

28 Eq. (7.8) gives a 3 a b c a a a Refer t Table 3.: ( a a ) Nte that a 3. This is the quantity that the relay see s fr a Single Line-t- Grund fault. Substituting j. 8 pu a 3 3( j.8) j5. 46 pu Calculating the vltage drp, the sequence vltages are V Z V Z V V Z Substituting in the impedance and current frm abve Vc V j.99( j.8).36 V j.75( j.8). 68 V j.75 j.8. ( ) 39 Va The phase vltages are cnverted frm the sequence value using Eq. (7.6). Vb V a Vb.36 a (.68) a(.39). 38 pu Vc.36 a(.68) a (.39). pu a The per-unit value fr the current and vltage wuld nw be cnverted t actual values using Eq. (5.9) and Eq. (5.) and knwing the pwer and vltage fr the given system. See example 5. fr a reference. The currents and vltages can be shwn in phasr frm. Symmetrical Cmpnents Page 6

29 Example 8.3 Given Z.99 9 pu, Z.75 9 pu, Z.75 9 pu, cmpute the fault current and vltages fr a Line-t-Line fault. Nte that the sequence impedances are in per-unit. This means that the slutin fr current and vltage will be in perunit. Slutin The sequence netwrks are intercnnected, as shwn. Z Z V - Because the sequence currents sum t ne nde, it fllws that The current is the vltage drp acrss Z in series with Z V Z Z j.75 j.75 j. 86 pu Z V - V - j. 86 pu The phase current is cnverted frm the sequence value using Eq. (7.8). a j.86 j.86 pu b a ( j.86) a( j.86) 4. 95pu c a( j.86) a ( j.86) pu Calculating the vltage drp, and referring t Fig. 8.7, the sequence vltages are V V V Z ( j.75)( j.86).5pu V Symmetrical Cmpnents Page 7

30 The phase vltages are cnverted frm the sequence value using Eq. (7.6). V a pu V b. a (.5) a(.5). 5pu V c. a(.5) a (.5). 5 pu The per-unit value fr the current and vltage wuld nw be cnverted t actual values using Eq. (5.9) and Eq. (5.) and knwing the pwer and vltage fr the given system. See example 5. fr a reference. b Vc Vb Va c The currents and vltages can be shwn in phasr frm. Example 8.4 Given Z.99 9 pu, Z.75 9 pu, Z.75 9 pu, cmpute the fault current and vltages fr a Duble Line-t-Grund fault. Nte that the sequence impedances are in per-unit. This means that the slutin fr current and vltage will be in per-unit. Slutin The sequence netwrks are intercnnected, as Z shwn in Fig. 8.9 Because the sequence currents sum t ne nde, it fllws that V ( ) - The current is the vltage drp acrss Z in series with the parallel cmbinatin f Z and Z Z V V ZZ Z Z Z Z - Substituting in V, and Z, Z, and Z, then slving fr V - Symmetrical Cmpnents Page 8

31 j3. 73pu Z ( Z Z) j.75 Z ( Z Z) j.99 The phase current is cnverted frm the sequence value using Eq. (7.8). a j.75 j3.73 j.99 pu j.75 a ( j3.73) a( j.99) b j.75 a( j3.73) a ( j.99) c 9 Calculating the vltage drp, and referring t Fig. 8.9, the sequence vltages are pu pu V V V V Z ( j.75)( j.99).348 pu The phase vltages are cnverted frm the sequence value using Eq. (7.6). V a pu V b.348 a (.348) a(.348) pu V c.348 a(.348) a (.348) pu Refer t Table 3.: ( a a ) R The per-unit value fr the current and vltage wuld nw be cnverted t actual values using Eq. (5.9) and Eq. (5.) and knwing the pwer and vltage fr the given system. See example 5. fr a reference. b Va c The currents and vltages can be shwn in phasr frm. Symmetrical Cmpnents Page 9

32 9. Oscillgrams and Phasrs Attached are fur faults that were inputted int a relay and then captured using the relay sftware. Three-phase fault. Cmpare t example (8.) Fig 9.a Fig 9.b Fig 9.c Symmetrical Cmpnents Page 3

33 Single Line-t-Grund fault. Cmpare t example (8.) Fig 9.a Fig 9.b Fig 9.c Symmetrical Cmpnents Page 3

34 Line-t-Line fault. Cmpare t example (8.3) Fig 9.3a Fig 9.3b Fig 9.3c Symmetrical Cmpnents Page 3

35 Duble Line-t-Grund fault. Cmpare t example (8.4) Fig 9.4a Fig 9.4b Fig 9.4c Symmetrical Cmpnents Page 33

36 . Additin Symmetrical Cmpnents cnsideratins. Symmetrical Cmpnents int a Relay Using a directinal grund distance relay it will be demnstrated hw sequential cmpnents are used in the line prtectin. T determine the directin f a fault, a directinal relay requires a reference against which the line current can be cmpared. This reference is knwn as the plarizing quantity. Zer-sequence line current can be referenced t either zer-sequence current r zer-sequence vltage, r bth may be used. The zer-sequence line current is btained by summing the three-phase currents. See Fig.. Frm Eq. (7.9) ( a b c ) r This is knwn as the residual current r simply 3. 3 (.) The zer-sequence vltage at r near the bus can be used fr directinal plarizatin. The plarizing zer-sequence vltage is btained by adding an auxiliary ptential transfrmer t the secndary vltage. The auxiliary transfrmer is wired as a brkendelta and the secndary inputted t the relay. See Fig. Symmetrical Cmpnents Page 34

37 AΦ BΦ CΦ V A V a 3V V B V b V C V c Frm Eq. (7.7a) the zer-sequence vltage equals ( V V V ) V 3 a b c (.a) ( V V V ) 3 V (.a) a b c Example. Using the values btained frm example 8., calculate 3V. Slutin V a Vb. 38 pu Vc. pu V pu The zer-sequence vltage is.8 8 pu. By cnnecting the value in the reverse gives 3V which equals.8 pu. Pltting this, we can shw in phasr frm what the relay see s, a lagging 3V by the line angle. n this case resistance is neglected, therefre a lags by 9. (see Fig.3). Symmetrical Cmpnents Page 35

38 Vc 3V -3V Va Vb a Fig.3. Symmetrical Cmpnents thrugh a Transfrmer This sectin will lk at current flw thrugh a wye-delta transfrmer bank. t will be shwn in the next chapter that fr faults that include grund that zer-sequence quantities will be generated. t can be shwn using symmetrical cmpnents that zer-sequence cmpnents cannt pass thrugh delta-wye transfrmer banks. f zer-sequence is flwing n the wye side, the currents will be reflected t the ther side, but circulate within the delta. Fig.4 The current n the left side is a n ( ) A B Fig.4 Symmetrical Cmpnents Page 36

39 Frm equatin 7. we have (.3 a) A B A A A (.3 b) B B B Substituting n the right side f the equatin 8. gives ( B ) ) ( ) ( ) (.4) A ( A B A B A B The zer-sequence currents are in-phase, therefre equatin.3 simplifies t ( B ) ) ( ) (.5) A ( A B A B Where ( A B) 3 A 3 and ( A B) 3 B 3 a ( 3 A 3 ) ( 3 B 3 ) n 3 a ( A 3 B 3 ) (.6) n n a balanced system where there is n negative r zer-sequence current then equatin.6 reduces t 3 a ( A 3 ) (.7) n As can be seen the current will shift by 3 when transferring thrugh a transfrmer cnnected delta-wye. The same can be prve when lking at the vltages. Nw cnsider the cnnectin in Fig.5. n a n b n c A n ( ) a c Fig.5 Symmetrical Cmpnents Page 37

40 Substituting equatin 7. and reducing gives ( ) C ) ( ) ( ) (.8) a a A ( A C A C A C n( 3 A 3 ) ( 3C 3 ) n ( 3 3 ) (.9) 3 A C As seen frm the prir example equatin.9 will reduce t n 3( a A 3 ) if there is n negative r zer-sequence current, which is the case fr a balanced system. By inspectin f the equatins abve fr ANS standard cnnected delta-wye transfrmer banks if the psitive-sequence current n ne side leads the psitive current n the ther side by 3, the negative-sequence current crrespndingly will lag by 3. Similarly if the psitive-sequence current lags in passing thrugh the bank, the negative-sequence quantities will lead 3. The directin f the phase shifts between the delta-cnnected winding and the wyecnnected winding depends n the winding cnnectins f the transfrmer. The winding cnfiguratins f a transfrmer will determine whether r nt zer-sequence currents can be transfrmed between windings. Because zer-sequence currents d nt add up t zer at a neutral pint, they cannt flw in a neutral withut a neutral cnductr r a grund cnnectin. f the neutral has a neutral cnductr r if it is grunded, the zer-sequence currents frm the phases will add tgether t equal 3 at the neutral pint and then flw thrugh the neutral cnductr r grund t make a cmplete path. Fllwing are sme different transfrmer winding cnfiguratins and their effect n zersequence currents. Transfrmers with at least tw grunded wye windings When a transfrmer has at least tw grunded-wye windings, zer-sequence current can be transfrmed between the grunded-wye windings. The currents will add up t 3 in the neutral and return thrugh grund r the neutral cnductr. The currents will be transfrmed int the secndary windings and flw in the secndary circuit. Any impedance between the transfrmer neutral pints and grund must be represented in the zer-sequence netwrk as three times its value t crrectly accunt fr the zer-sequence vltage drp acrss it. Belw n the left is a three-phase diagram f a grunded-wye, grunded-wye transfrmer cnnectin with its zer-sequence netwrk mdel n the right. Ntice the resistance in the neutral f the secndary winding is mdeled by 3R in the zer-sequence netwrk mdel. Symmetrical Cmpnents Page 38

41 P Z 3R S P 3 3 S R Reference Bus. Transfrmers with a grunded-wye winding and a delta winding When a transfrmer has a grunded-wye winding and a delta winding, zersequence currents will be able t flw thrugh the grunded-wye winding f the transfrmer. The zer-sequence currents will be transfrmed int the delta winding where they will circulate in the delta withut leaving the terminals f the transfrmer. Because the zer-sequence current in each phase f the delta winding is equal and in phase, current des nt need t enter r exit the delta winding. Belw n the left is a three-phase diagram f a grunded-wye-delta transfrmer cnnectin with its zer-sequence netwrk mdel n the right. Symmetrical Cmpnents Page 39

42 3. Auttransfrmers with a grunded neutral Auttransfrmers can transfrm zer-sequence currents between the primary and secndary windings if the neutral is grunded. Zer-sequence current will flw thrugh bth windings and the neutral grund cnnectin. Belw n the left is a three-phase diagram f a grunded neutral auttransfrmer with its zer-sequence netwrk mdel n the right. 4. Auttransfrmers with a delta tertiary f an auttransfrmer has a delta tertiary, zer-sequence current can flw thrugh either the primary r secndary winding even if the ther winding is pen circuited in the same manner that zer-sequence current can flw in a grundedwye-delta transfrmer. f the grund is remved frm the neutral, zer-sequence current can still flw between the primary and secndary windings, althugh there will nt be any transfrmatin f currents between the primary and secndary windings nly between the partial winding between the primary and secndary terminals and the delta tertiary. This is nt a nrmal cnditin thugh, s it will nt be analyzed here. Nte that when mdeling three-winding transfrmers the impedance needs t be brken int the impedance f the individual windings. Symmetrical Cmpnents Page 4

43 5. Other transfrmers Other transfrmer cnfiguratins, such as ungrunded wye-ungrunded wye, grunded wye-ungrunded wye, ungrunded wye-delta, and delta-delta will nt allw zer-sequence currents t flw and will have an pen path in the zersequence netwrk mdel. Sme f these cnfiguratins are shwn belw with their zer-sequence netwrk mdels. n the preceding transfrmer cnnectin diagrams the values f at the terminals f the primary and secndary windings will be equal n a per-unit basis. They will als have the same per-unit values within the wye and delta windings; hwever, the per-unit values f current within the windings f an auttransfrmer are smewhat mre difficult t determine because part f the winding carries bth primary and secndary currents. f the magnitude f current within the winding f an auttransfrmer needs t be knwn, it can be determined by equating the ampere turns f the primary winding t thse f the secndary winding and slving. f a tertiary is invlved, it will need t be included in the equatin als. Symmetrical Cmpnents Page 4

44 Magnitude f transfrmer zer-sequence impedance The zer-sequence impedance f a single-phase transfrmer is equal t the psitivesequence impedance. When three single-phase units are cnnected as a three-phase unit in a cnfiguratin that will transfrm zer-sequence currents (grunded wye-grunded wye, grunded wye-delta, etc.), the zer-sequence impedance f the three-phase unit will nrmally be equal t the psitive-sequence impedance. n transfrmers built as three-phase units, i.e. with a three-phase cre, in a cnfiguratin capable f transfrming zer-sequence currents, the zer-sequence impedance will be the same as the psitive-sequence impedance if the transfrmer cre is f the shell type. f the cre is f the cre type, the zer-sequence impedance will be different than the psitive-sequence impedance. This is because the zer-sequence excitatin flux des nt sum t zer where the three legs f the cre cme tgether and is frced t travel utside f the irn cre, thrugh the il r the transfrmer tank where the magnetic permeability is much less than the irn cre. This results in a lw impedance (high cnductance) in the magnetizing branch f the transfrmer mdel. The larger zer-sequence magnetizing current results in a lwer apparent zer-sequence impedance. Using a lwer value f zer-sequence impedance in the transfrmer zer-sequence mdel is sufficient fr mst fault studies, but t btain a highly accurate zer-sequence mdel f a three-phase crefrm transfrmer, the magnetizing branch can nt be neglected.. System Mdeling. System Mdeling: Transmissin Lines Transmissin lines are represented n a ne-line diagram as a simple line cnnecting busses r ther circuit elements such as generatrs, transfrmers etc. Transmissin lines are als represented by a simple line n impedance diagrams, but the diagram will include the impedance f the line, in either hm r per-unit values. Smetimes the resistive element f the impedance is mitted because it is small cmpared t the reactive element. Here is an example f hw a transmissin line wuld be represented n an impedance diagram with impedances shwn in hms: n a balanced three-phase system the impedance f the lines and lads are the same, and the surce vltages are equal in magnitude. We can calculate the single-phase current, but must take int accunt the vltage drp acrss the mutual impedance caused by the ther phase currents. Frm Fig., the vltage drp in A-phase is Symmetrical Cmpnents Page 4

45 AΦ BΦ Zs Zs Zm Zm CΦ Zs Zm Fig. V Z Z Z (.a) a S Fr the case f a balanced three-phase current A m B m C ( ). Thefre: B C A V a ( Z S Z m ) A (.b) Dividing by A shws the psitive-sequence impedance f the line equals the self impedance minus the mutual impedance. Z V ( Z Z ) A a S m (.) A The negative-sequence current encunters a negative-sequence impedance which is equal t the psitive-sequence impedance Z V ( Z Z ) A a S m (.3) A Fr the zer-sequence impedance, because a, b and c are in phase with each ther, A B C then zer-sequence vltage drp is given in equatin.4 ( Zm Zm ) V Z Z Z Z (.4a) a S A m B m C S A A ( ZS Zm ) V (.4b) a A Dividing each side by A give the zer-sequence impedance: Z V ( Z Z ) A a S m (.5) A The result gives the zer-sequence impedance as functin f the self and mutual impedance f the line. The zer-sequence impedance is always larger than the psitivesequence because we are adding tw times the mutual impedance t the self impedance, instead f subtracting the mutual impedance frm the self impedance. Symmetrical Cmpnents Page 43

46 . System Mdeling: Subtransient, Transient, and Synchrnus Reactance f Synchrnus Generatrs A synchrnus generatr is mdeled by an internal vltage surce in series with an internal impedance. Belw is a typical ne-line diagram symbl fr a generatr. The circle represents the internal vltage surce. The symbl t the left f the circle indicates that the three phases f the generatr are wye-cnnected and grunded thrugh a reactance. The symbl fr a synchrnus mtr is the same as a synchrnus generatr. A typical impedance diagram representatin f a synchrnus generatr is shwn in Fig... Xg Rg Eg Vt - Fig.. When mdeling the impedance f a synchrnus generatr (r mtr), the resistive cmpnent is usually mitted because it is small cmpared t the reactive cmpnent. When a fault is applied t a pwer system supplied by a synchrnus generatr, the initial current supplied by the generatr will start at a larger value, and ver a perid f several cycles it will decrease frm its initial value t a steady state value. The initial value f current is called the subtransient current r the initial symmetrical rms current. Subtransient current decreases rapidly during the first few cycles after a fault is initiated, but its value is defined as the maximum value that ccurs at fault inceptin. After the first few cycles f subtransient current, the current will cntinue t decrease fr several cycles, but at a slwer rate. This current is called the transient current. Althugh, like the subtransient current, it is cntinually changing, the transient current is defined as its maximum value, which ccurs after the first few cycles f subtransient current. After several cycles f transient current, the current will reach a final steady state value. This is called the steady state current r the synchrnus current. Symmetrical Cmpnents Page 44

47 The reasn why the current supplied by the synchrnus generatr is changing after a fault is because the increased current thrugh the armature f the generatr creates a flux that cunteracts the flux prduced by the rtr. This results in a reduced flux thrugh the armature and therefre a reduced generated vltage. Hwever, because the decrease in flux takes time, the generatr vltage will be initially higher and decrease ver time. We accunt fr the changing generatr vltage in ur mdel by using different values f reactance in series with the internal generatr vltage. We use three values f reactance t mdel the generatr during the perid after fault inceptin: the subtransient reactance (Xd ) is used during the initial few cycles; the transient reactance (Xd ) is used fr the perid fllwing the initial few cycles until a steady state value is reached; the synchrnus reactance (Xd) is used fr the steady state perid. The impedance diagrams fr a synchrnus generatr (r mtr) during the subtransient, transient, and synchrnus perids are shwn in Fig..3. Fig..3 The reactance f synchrnus mtrs are the same as fr synchrnus generatrs. f the line t a synchrnus mtr develps a three-phase fault, the mtr will n lnger receive electrical energy frm the system, but its field remains energized and the inertia f its rtr and cnnected lad will keep the rtr turning fr sme time. The mtr is then acting like a generatr and cntributes current t the fault Symmetrical Cmpnents Page 45

48 .3 System Mdeling: Transfrmers Transfrmers are represented in ne-line diagrams by several symbls. Belw are sme typical nes. The first is a tw-winding transfrmer cnnected delta- grunded wye, and the secnd is a three-winding transfrmer cnnected grunded wye-delta-grunded wye. An impedance mdel f a practical tw-winding transfrmer is shwn in Fig..4. Fig..4 n the mdel, a: represents the winding rati f the ideal transfrmer shwn by the tw cupled cils, BL in parallel with G represents the magnetizing susceptance and cnductance which make up the magnetizing branch, E represents the excitatin current, r and x represent the leakage impedance f winding,r and x represent the leakage impedance f winding, V and represents the primary vltage and current respectively, and V and represent the secndary vltage and current respectively. Because nrmal fault and lad currents are very much larger than the magnetizing current, E, we can mit the magnetizing branch frm ur mdel. We can als mit the ideal transfrmer if we refer the leakage impedances t either the primary- r secndary-side f the transfrmer. The leakage impedance f ne side f the transfrmer can be referred t the ther side f the transfrmer by multiplying it by the square f the turns rati. Belw is the simplified impedance diagram with the magnetizing branch remved and the leakage impedance f the secndary winding referred t the primary side f the transfrmer. Symmetrical Cmpnents Page 46

49 Our impedance mdel can be further simplified by letting R r a r X x a x When using this simplified mdel, any impedances and vltages cnnected t the secndary side f the circuit must nw be referred t the primary side. As an example, the fllwing transfrmer mdel will be cnverted t the simplified impedance mdel. The magnetizing branch and the leakage resistances have been mitted t simplify the prblem. The secndary-side impedance is multiplied by the square f the turns rati befre being transferred t the primary side. j6. * 8.33 j46.3ω Symmetrical Cmpnents Page 47

50 This is added t the high side t get an impedance f j5ω j46.3ω j466.3ω The simplified mdel is shwn in Fig..5.4 Sme additinal pints DC Offset Fig..5 n a transmissin netwrk, the sudden ccurrence f a shrt circuit will result in a sinusidal current that is initially larger and decreases due t the changing air gap flux in the synchrnus generatrs. We ve seen that this is mdeled by subtransient, transient, and synchrnus reactances in ur generatr mdel. n a circuit cntaining resistance and inductance (RL circuit), such as in a transmissin netwrk, the sudden ccurrence f a shrt circuit will als result in DC ffset in the current that ccurs after a fault is applied. Cnsider the RL circuit belw: f the switch is clsed at time t, the vltage arund the circuit is Vmaxsin(ωtφ) Ri Ldi/dt Slving this differential equatin fr the instantaneus current, i, gives i Vmax [sin(ωtφ-θ) e-rt/lsin(φ-θ)] / Z Where Z (R (ωl) and θ tan-(ωl/r) Symmetrical Cmpnents Page 48

51 The imprtant thing t nte frm the slutin is that there is a sinusidal cmpnent that represents the steady-state slutin fr the current (Vmax sin(ωtφ-θ) / Z ) and a expnentially decaying cmpnent (-Vmax e-rt/lsin(φ-θ) / Z ). Sme pints t nte abut the expnentially decaying r DC ffset cmpnent: The initial value f the DC ffset is determined by what pint in the cycle the vltage wavefrm is at when the fault ccurs (the value f φ) and will range frm up t the value f the steady state cmpnent. The dc cmpnent will decrease with a time cnstant f L/R. The larger the rati f inductance t resistance in the circuit, the larger the time cnstant, and the slwer the dc cmpnent will decay. Three time cnstants after the switch is clsed, the dc ffset will have decayed t 5% f its initial value. DC ffset is an imprtant cnsideratin in sizing breakers. Mst mdern micrprcessr-d relays are immune t DC ffset because after the analg signals are cnverted t digital signals, they can be mathematically filtered t remve the DC cmpnent. Therefre the DC cmpnent desn t need t be cnsidered in the relay settings. Sme electrmechanical relays are immune t DC ffset, and sme aren t. Clapper and plunger type units are generally nt immune, and DC ffset will have t be allwed fr in the relay settings (ne guideline is t set pickup at 6% f the desired ac pickup current). Cylinder type units, used in distance relays, are immune t DC ffset. The different values f the AC fault current shuld be cnsidered in the relay settings. The subtransient fault current shuld be used in setting instantaneus current elements, whereas the synchrnus fault current shuld be used in current elements with lng time delays. Symmetrical Cmpnents Page 49

52 Prblem Prblems BPA s system mdel uses a three-phase pwer f MVA. The line-t-line vltage is 55kV fr the 5 system, 3kV fr the 3 system, and 5kV fr the 5 system. a) An undervltage relay n the 5 system is set t pick up at.85 pu (per unit) f the phase-t-grund vltage. What is the phase-t-grund vltage that the undervltage relay will pick up at? b) A three-phase fault n the 5 system results in a fault current f 75A. What is the per unit value f this current? c) What is the impedance fr the 5 system? d) What is the impedance fr the 3 system? e) What is the impedance fr the 5 system? Prblem Frm ur example 5., the percent impedance f a 55/4.5kV auttransfrmer is.4% d n its nameplate value f 9MVA. Suppse we need t mdel this transfrmer in BPA s ASPEN mdel which uses a MVA pwer. What wuld the per-unit impedance be? Prblem 3 Frm ur example in 5., cnvert the per-unit impedance t a per-unit value in a threephase pwer f MVA. a) First cnvert the per unit impedance t an actual impedance (in hms) at 55kV and then cnvert the actual impedance t a per-unit impedance n the new. b) Repeat, this time cnverting the per unit impedance t an actual impedance (in hms) at 4.5kV and then cnverting the actual impedance t a per-unit impedance n the new Prblem 4 Cnvert the per-unit impedance f the transfrmer in the example t a per-unit value in the BPA mdel with a three-phase pwer f MVA by first cnverting the per unit impedance t an actual impedance (in hms) at 3 kv and then cnverting the actual impedance t a per-unit impedance n the new. Symmetrical Cmpnents Page 5