SESSION 3. Short Circuit Calculations, Unsymmetrical Faults. Leonard Bohmann, Michigan State University Elham Makram, Clemson University

Transcription

1 SESSON Short Circuit Calculations, Unsymmetrical Faults Leonard Bohmann, Michigan State University Elham Makram, Clemson University

2 Short Circuit Calculations Leonard Bohmann Michigan State University

3 Line to Ground Short EEE T & D CONFERENCE

4 Short Circuit Fault Currents Amps (ka) dc offset t high frequency transient decaying sinusoid EEE T & D CONFERENCE 4

5 Short Circuit Fault Currents (continued) X(t) R v(t) v(t) (t) θ then m i(t) [cos( ωt + α θ) (t) cos( ωt m if cos( ωt + α) R + jx(t) + α θ)e Rωt X(t) ] - - EEE T & D CONFERENCE 5

6 dc offset i(t) m cos( ωt + α θ) cos( ωt EEE T & D CONFERENCE + α θ) e Rωt X 6

7 ariable magnitude i(t) m cos( ωt + α θ) (t) t EEE T & D CONFERENCE 7

8 Reactance of Synchronous Machine reactance (pu) X d X' d X" d t EEE T & D CONFERENCE 8

9 Use sub-transient reactance to calculate fault currents worst case for when CB is opening Do not calculate dc offset directly use X/R to approximate offset (asymmetry ratio) at opening time X/R is the time constant of the decay in cycles EEE T & D CONFERENCE 9

10 G L 5 T T5 T5 L5 T T4 f 4 How do you calculate the fault current ( f ) at bus? G4 EEE T & D CONFERENCE

11 Use Thevenin equivalent Th Th f Th : open circuit voltage also called the prefault voltage, f oftenuse EEE T & D CONFERENCE

12 Use for: symmetrical faults single phase networks three phase faults in three phase networks (per phase equivalent) concept for solving all faults EEE T & D CONFERENCE

13 This method calculates the current going into the fault Often want the current elsewhere in the network (through a circuit breaker, in bus bars, in other lines, etc.) EEE T & D CONFERENCE

14 A current source, with the magnitude of the fault current, placed at the fault, has the same effect on the rest of the circuit as the short circuit. f Th f f Th f EEE T & D CONFERENCE 4

15 Adding a current source to the original circuit has the same electrical effect as adding it to the equivalent circuit G L 5 T T5 T T5 f Th T4 f f L5 4 G4 EEE T & D CONFERENCE 5

16 G L All the voltages and all the currents can be found using the new circuit. 5 T T5 T L5 T5 T4 4 G4 f Use superposition: Use the original sources (or assume currents are and voltages are ) Use just the current source, gives changes due to the fault EEE T & D CONFERENCE 6

17 Calculation details Bus Admittance Matrix: Y bus G L Y bus T T5 T vector of current injections vector of bus voltages 5 L5 T5 T4 4 f G4 EEE T & D CONFERENCE 7

18 G T5 5 L5 T T5 4 L Diagonals: Σ of Y connected at node Y / T +/ L +/ T Y Y Y5 T Y Y Y Y bus Y Y Y4 Y5 f Y4 Y44 T4 Y5 Y5 Y55 G4 Off diagonals: - of Y connected between nodes Y 5 -/ T5 EEE T & D CONFERENCE 8

19 Y bus is sparse (lots of zero) smallsize easy to build quick algorithms to manipulate EEE T & D CONFERENCE 9

20 Bus mpedance Matrix: bus bus G vector of bus voltages vector of current injections L T T5 T 5 T5 f bus Y bus - L5 4 T4 G4 EEE T & D CONFERENCE

21 bus bus is a full matrix slowtoinvert slow to manipulate avoid using if possible Diagonal element is the Thevenin mpedance at that bus is the Thevenin mpedance at bus EEE T & D CONFERENCE

22 To find the change in voltages due to the fault current at bus : f EEE T & D CONFERENCE

23 To calculate current in line -: f T f T 5 f 4 EEE T & D CONFERENCE

24 Asymmetry ratio: S Method calculates the RMS worst case of the 6 Hz component this is needed to set protective relays For circuit breaker selection, need to know the total RMS current (6 Hz and dc offset) Circuit breakers are rated on symmetrical fault current, assuming an X/R ratio of 5 EEE T & D CONFERENCE 4

25 Asymmetry ratio RMS ac + dc RMS ac + ac e ω t X R RMS ac RMS S ac + e ωt X R ωt () X R t where S(t) is + e EEE T & D CONFERENCE 5

26 Calculating X/R G L 5 T T5 T5 L5 T f T4 4 G4 Equations presented are for a single order system For higher order systems, ignore R and find equivalent X, ignore X, and find equivalent R Actual decay is typically quicker than this approximation EEE T & D CONFERENCE 6

27 Algorithm Calculate Thevenin mpedance Calculate current into fault Find the needed elements of bus (not all of them) Find the needed changes in bus voltages Find the line currents of of interest Find the dc offset EEE T & D CONFERENCE 7

28 Example The oneline drawing of a system, along with bus are shown below. i) Calculate the fault current for a three phase solid fault at bus. j pu ii) Calculate the voltages at bus,, and during this fault. iii) Calculate the current flowing j.8pu j.5 pu j.5 pu from bus to bus. j.pu 5 j5 j. pu 4 f j.5 pu j. pu bus j EEE T & D CONFERENCE pu

29 Part (i): Calculate the fault current for a three phase solid fault at bus. Assume all initial bus voltages are therefore all initial currents are Use the, element of bus for the Thevenin impedance. f f j.7 j. 68 j. 7 f pu EEE T & D CONFERENCE 9

30 Multiply the negative of the fault current by elements in the rd column to get the change in voltages EEE T & D CONFERENCE Part (ii): Calculate the voltages at bus,, and 4 during this fault j j

31 Part (ii): Calculate the voltages at bus,, and 4 during this fault. (continued) The voltages during the fault are the initial voltages minus the changes in the voltages f f f i i i pu pu pu EEE T & D CONFERENCE

32 Part (iii): Calculate the current flowing from bus to bus. The current is the difference between the voltages divided by the impedance f f f T f. 766 j. 5 j. 86pu 5 -j.86 pu -j.68 pu 4 EEE T & D CONFERENCE

33 Problem A small power system is sketched below along with bus. j. pu j. pu j.5 pu j.8 j.5 pu 4 j. pu j.5 pu bus j pu a) What is the fault current for a three-phase fault to ground at bus? b) What are the voltages at busses,, and 4? c) What is the current due to the fault from bus 4 to and from to? EEE T & D CONFERENCE

34 Unsymmetrical Faults Elham Makram Clemson University EEE T & D CONFERENCE

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42 SNGLE LNE TO GROUND FAULT a b c af bf cf f n Fig.. Single line to ground fault on phase A Boundary Conditions at fault location bf f (a Or, cf + a Thus f Now a) a * + af From the current equations, + a + ( a f (a * + a) + a) af + a f + a * EEE T & D CONFERENCE

43 SNGLE LNE TO GROUND FAULT (CONTD.) To satisfy the voltage and current boundary conditions, the sequence networks should b connected in series as shown in Fig.. ero Sequence Network + - ero Sequence Network + - f ero Sequence Network + - Fig.. Sequence connection for A-G fault EEE T & D CONFERENCE

44 a b c n + af af bf cf bf f cf Fig.. Line to line fault LNE TO LNE FAULT Current boundary Conditions a af bf (a (a bf + - Thus, EEE T & D CONFERENCE cf + a + a) - + a cf - a) + a + a + This leads to + f f (a - (a oltage boundary conditions + a + a + a bf - a) + a) f ( f + a + (a + a - a) )

45 LNE TO LNE FAULT (CONTD.) To satisfy the voltage and current boundary conditions, the sequence networks should be connected as shown in Fig. 4. Positive Sequence Network + - Negative Sequence Network + - f Fig.4. Connection of sequence network for BC fault EEE T & D CONFERENCE 4

46 a b c n + af DOUBLE LNE TO GROUND FAULT bf f cf af bf cf Fig.5. Double line to ground fault EEE T & D CONFERENCE CurrentboundaryConditions at thefaultlocation, af bf and, bf a Thus, + ( Thus, Also, + cf bf oltageboundaryconditions + Thus cf + a + a Thus - ) f + a + a + a + a + a + a ( + a + ) f f 5

47 DOUBLE LNE TO GROUND FAULT (CONTD.) Considering the voltage and current boundary conditions, the sequence networks are connected as shown in Fig. 6. ero Sequence Network + - f Positive Sequence Network + - Negative Sequence Network + - Fig.6. Sequence network connection for double line to ground fault EEE T & D CONFERENCE 6

48 Consider the power system shown in Fig. 7. G T Example P T. L. T G Fig. 7. Example Ratings G and G: MA, k X X.pu X.4pu T and T : MA, 45/ k X T X T.8 pu T. L. X X.5pu X.5pu EEE T & D CONFERENCE Find the voltages at Bus and the currents from T to P due to (i) (ii) Single line to ground fault at P Line to line fault at P (iii) Double line to ground fault at P 7

49 Positive sequence network j. j. j.8 j.5 j. j.8 j.5 j.8 P P P f th E E (. + j.8) (j.+ j.8 + j.5) j. Negative sequence network ero sequence network j. P j.8 j.5 th j. P EEE T & D CONFERENCE th j.8 8

50 (i) Single line to ground fault E - j.94 (th + th + th ) (j. + j. + j.8) The following are the sequence current components from T- - > P The sequence voltage components at the fault location are : - j.4 - j.94 E * j.4 (.5+.8) -. (-j.4)(j.8) (j.4) (j.8) j.9 (j.8) -.54 The current phase components from T- - > P a b c a a a * a a a a EEE T & D CONFERENCE 9

51 a b c a (-j.4) + a(-j.4) j.9 - a(-j.4) + a (-j.4) j.9 - j.799 j.799 c c b -j.799 -j.799 -j.7 -j.7 -j5.79 The phase voltages at the fault location are : a b -j.4- j.4 - j.9 - j j.866 c + + a + a + + a + a j ( 4 )(.57668) + ( )(-.4) ( )(.57668) + ( 4 )(-.4) EEE T & D CONFERENCE

52 (ii) Line to line fault Currents: E j.6 ( + ) (j. + j.).5 (T- - > P) j.6 ( ) (T- - > P) j.6 a b c a a (-j.6) + a oltages : a a a b c E - (-j.6) + a(j.6) a a th. - (-j.6)(j.8).5.5 th (j.6) j.6 EEE T & D CONFERENCE

53 SEQUENCE PHASE SHFT THROUGH WYE/DELTA TRANSFORMER BANK Positive and negative sequence impedance is independent of connection These networks ignore phase shift When current and voltages are transformed from one side to the other phase shift must be considered. Standard ANS connection has been shown in figs. 8 (a) and 8 (b). High side Low side High side Low side a A n: n: A a n a A A A A a /n b a B B b B n B B b B b /n c b C C c C n C c C C c /n c A A a a C B c b C B c b Fig. 8 (a) Wye/Delta EEE T & D CONFERENCE Fig. 8 (b) Delta/Wye

54 EEE T & D CONFERENCE unit per n or n a) ( - n ) a - n( ) - n( : voltage sequence Negative unit per n or n ) a ( - n ) a - n( ) - n( : current sequence Negative unit, per in or n ) a ( - n ) a - n( : volatge sequence Positive unit, per n - n a) ( - n ) a - n( ) - n( : current sequence Positive 8(a) fig. Refer to a A a a a a b a A a A A A A A C A a a A a a a a A a A A A A A C A a

55 4 EEE T & D CONFERENCE voltage sequence Negative unit per n or n ) a - ( n : current sequence Negative unit, per in or, n ) - a ( n : voltage sequence Postitive unit, per in or, n ) a - ( n ) - ( n : current sequence Postitive 8 (b) fig. Refer to a A a A a a a A a A A A A a a A a a a b a A

56 n general,in Y - or - Y connections : (i) Positive sequence High voltage Positive sequence Low voltage + (ii) Negative sequence High voltage Negative sequence Low voltage - EEE T & D CONFERENCE 5

57 Example Find the currents and voltages at G of Example, due to single line to ground fault at P. (i) Currents : ag ag ag ag bg cg - j j j.4 + a ag ag + + ag a j j ag 4 (.4 - ) + (.4-6 ) (.4 ) + (.4 6 ) j.995 j j.995 EEE T & D CONFERENCE 6

58 Example ( contd.) (ii) oltages : ag ag ag ag ag bg cg ag bg cg E (from sequence Network neglecting phase shift) a ag a a ag + ag a + + a g a a g. - (-j.4)(j.) (from sequence Network neglecting phase shift) ag a ag ag a * a ag ag ag EEE T & D CONFERENCE 7

59 PROBLEMS PROBLEM # For the system shown below, find the currents in the transformer and generator windings in per unit and amperes for a single-line-to-ground fault at (i) P (ii) R G 5/5 k R P G: MA, 5 k X X., X.5 Transformer: MA, 5/5 k X. EEE T & D CONFERENCE 8

60 PROBLEMS PROBLEM # Solve examples and for a double-line-to-ground fault at P. EEE T & D CONFERENCE 9

61 PROBLEMS PROBLEM # Solve examples and for a line-line fault at P. EEE T & D CONFERENCE