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# ECNG 3015 Industrial and Commercial Electrical Systems

Symmetrical Components and Sequence Networks

Symmetrical Components and Sequence Networks

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### ECNG 3015 Industrial and Commercial Electrical Systems

1. 1. 1 Prof Chandrabhan sharma
2. 2. SYMMETRICAL COMPONENTS End of this topic, you should be able to: - Explain the importance of symmetrical component. - Construct positive sequence networks for a power system. - Construct negative sequence networks for a power system. - Construct zero sequence networks for a power system. - Use the networks to calculate fault current and voltages under SLG, DL, DLG and OC conditions 2
3. 3. SYMMETRICAL COMPONENTS Symmetrical Components: - Theory originally expounded by Stokvis L.G. (1915) but was given practical application to Electrical Engineering by C.L. Fortescue in 1918. - A mathematical model for representing an unbalanced system of ‘n’ phasors as ‘n’ systems of balanced phasors known as symmetrical components. N.B. : this is only a model for analysis since symmetrical components do not exist in real life 3
4. 4. Used for analyzing unbalanced fault conditions in a power system, since fault on a balanced 3 system is usually NOT symmetrical. They are usually: - SLG (Single Line – Ground) - DL (Double Line or Line - Line) - DLG (Double Line – Ground) - Open conductor 4
5. 5. If either SLG, DLG, LL, OC occur then - they will cause currents and voltages to be of unequal magnitude, and - the usual 3 phasor diagram is no longer displaced by 120apart from each other. It is imperative that values of short circuit currents be obtained for these types of faults since the information is necessary for: (a) Ordering protective equipment (b) the determination of protective relay settings (b) transient stability studies for interconnected systems 5
6. 6. THEORY The theory postulates that a 3 unbalanced system of voltage and current phasors can be resolved into three balanced system of phasors. These are : A positive phase sequence system - comprising three phasors of equal magnitude - displaced by 120 from each other - and with the same phase sequence as the original phasors. 6
7. 7. A negative phase sequence system - comprising three phasors of equal magnitude - displaced by 120 from each other - but with a phase sequence opposite to the original phasors. A zero phase sequence system - comprising three phasors equal in magnitude and phase, - and revolving in the positive phase rotation. 7
8. 8. Equivalent unbalanced system B1    = + + B Y R R1 Y1 R2 Y2 B2 Y0 B0 R0 120 120 120 120 120 120       Faulted System +ve sequence -ve sequence zero sequence From the theory: VR = VR1 + VR2 + VR0………………(1) VY = VY1 + VY2 + VY0………………(2) VB = VB1 + VB2 + VB0……………….(3) 8
9. 9. For a 3 system, the theory states that: R1 Y1 B1 V  V  V R2 Y2 B2 R0 Y0 B0 V V V and V V V Also,     The ‘a’ operator (or ‘h’) Since we will be dealing with three phasors displaced by 120, it would be helpful/convenient to define a new operator ‘a’ such that: a  1120 = ej2/3 = - 0.5 + j0.866 9
10. 10. And: a2 = 1240 = ej4/3 = - 0.5 – j0.866, and a3 = 1360 = 1 + j0 Note, unlike the ‘j’ operator, -a ≠ 1-120 Recall -j = 1-90  = (j2)(j) = j3 = 1-90  but -a = (j2)(a) = 1(120 + 180) = 1300  = 160  Note also a2 + a + 1 = 0 = a3 + a2 + a 10
11. 11. From the positive sequence diagram we have: VY1 = a2 VR1……………………(4) VB1 = a VR1 …………………...(5) From the negative sequence, VY2 = a VR2…………………….(6) VB2 = a2 VR2 ……………………(7) And from the zero sequence, VR0 = VY0 = VB0 ………………..(8) 11
12. 12. Substitute eqs. 4-8 into eqs. 1-3 we obtain: VR = VR0 +VR1 + VR2 VY = VR0 + a2 VR1 + aVR2 VB = VR0 + aVR1 + a2 VR2 Written in matrix :   ...........................(9) V R0 V R1 V 1 1 1     1 a a  1 a a V R V Y V R2 2 2 B                         12
13. 13. 1 1 1    1 a a  1 a a T Let, 2 2        'T' is known as the symmetrical component transformation matrix 1 1 1   -1 2 1 a a  1 a a 1 3 T 2         13 From matrix inversion
14. 14. Applying T-1 to eq. (9):   .............................1(0) V R V Y V 1 1 1   1 a a  1 a a 1 3 V R0 V R1 V B 2 2 R2                           1     V a V aV ............................1(3) 3    R0 R Y B 1    1 V V aV a V ............................1(2) 3 V V V V .................................(11) 3 V OR Y B 2 R2 R B 2 R1 R Y    14
15. 15. Similarly:   .............................1(4) I I I 1 1 1   1 a a  1 a a 1 3 I I I R Y B 2 2 R0 R1 R2                           1     I a I aI ............................1(7) 3    R0 R Y B 1    1 I I aI a I ............................1(6) 3 I I I I .................................(15) 3 I OR Y B 2 R2 R B 2 R1 R Y    15
16. 16. Analyzing equations (11) and (15) ….. (11) → for balanced 3 system VR +VY + VB = 0 Balanced  voltage magnitude and angle normal (current unbalanced) (15) → IR + IY + IB = IN  IN = 3IR0  if the system is ungrounded  IN = 0 = IR0 → no zero sequence currents Hence  connected loads have no zero sequence currents. 16
17. 17. Example: (Higher Electrical Engineering – Sheppard, Morton and Spence) The currents flowing in an unbalanced 4-wire system are IR = 1000 A; IY = 200-90 A ; IB = 100120 A Find the +ve, -ve and zero sequence components of these currents. 17
18. 18. Solution:      R0 R1 1 I I      I I I R0 R Y B          1 1 1   1 a a  R Y I I   1 3 1          100 0 200 90 100 120 3 From which     I 41.3 66.2 A I R0               B 2 2 R2 I 1 a a 3 I Re call : 18
19. 19.   1    I aI a I 3                100 0 (1.0 120 ).(200 90 ) (1.0 240 ).(100 120 ) 3 I 128.79 15 A 1 Also, I R1 B 2 R1 R Y     2      R2 R Y B                     R2 And, 1 I I a I aI 3 1 100 0 (1.0 240 ).(200 90 ) (1.0 120 ).(100 120 ) 3 I 41.3 174 A 19
20. 20. SEQUENCE NETWORKS Since - component currents of one phase sequence cause - voltage drops of like sequence only and - independent of the other sequences (in a balanced system) - currents of any one sequence may be considered to flow in an independent network composed of impedances to the current of that sequence. 20
21. 21. SEQUENCE NETWORKS The single phase equivalent circuit composed of the impedances to current of any one sequence only is called the sequence network for that particular sequence. In static symmetrical systems, conversion of a +ve sequence to a –ve sequence network is accomplished by changing only the impedances representing rotating machinery and by omitting any generated e.m.fs 21
22. 22. Synchronous Machines: - +ve sequence ≠ -ve sequence - Xd”, Xd’ and Xd are +ve sequence - the generated e.m.fs are +ve sequence only - the ref. bus for +ve and –ve sequence n/w is the generator neutral - the ref. Bus for the zero sequence n/w is the ground of the generator or earth - there are no generated e.m.fs in the –ve and zero sequence n/w 22
23. 23. Positive Sequence Networks: - For a given power system the positive sequence network shows all the paths for the flow of positive sequence currents in the system. - The one-line diagram of the system is converted to an impedance diagram that shows the equivalent circuit of each component under balanced operating conditions - Each generator in the system is represented by a source voltage in series with the appropriate reactance and resistance. - To simplify the calculations, all resistance and the magnetizing current for each transformer are neglected. 23
24. 24. Positive Sequencenetwork of Synchronous generator: 24
25. 25. Negative Sequence Networks: - Three-phase generators and motors have only positive sequence generated voltages. Thus, the negative sequence network model will not contain voltage sources associated with rotating machinery. - The negative sequence impedance will in general be different from the positive sequence values. - For static devices such as transmission lines and transformers, the negative sequence impedances have the same values as the corresponding positive sequence impedances. 25
26. 26. Negative Sequence network of Synchronous generator: 26
27. 27. Zero Sequence Networks: - The zero sequence network of a system depends on the nature of the connections of the three-phase windings for each of the system’s components. 27
28. 28. Zero Sequence Networks: 28
29. 29. Zero Sequence Networks: 29
30. 30. Zero Sequence Networks: 30
31. 31. Lines and cables: - +ve and –ve sequence impedances are the normal balanced values. - zero sequence impedances depends on the nature of the return path through earth. The following factors can be used in the absence of manufacturer data: Cable Zero Seq. Impedance 3 – core Z0 =3Z1 to 5Z1 1 – core Z0 =1.5Z1 31
32. 32. Single Circuit Double Circuit No ground wire  Z0 =3.5Z1 Z0 = 5.5Z1 Steel ground wire  Z0 =3.5Z1 Z0 =5Z1 Non-metallic ground wire  Z0 =3.5Z1 Z0 =3Z1 Lines: 32
33. 33. Transformers: - +ve and –ve sequence impedances are the same i.e. (Z1 = Z2) - zero sequence cct depends on whether a return path exists through which a completed circuit can be provided Recall the equivalent cct for a transformer is: 33
34. 34. The equivalent circuit becomes: For +ve and –ve sequence Z1 = Z2 For zero sequence the equivalent circuit changes depending on the configuration of the transformer. 34
35. 35. Zero Sequence Network (Transformer) - Close ‘a’ contact whenever zero sequence currents can flow, i.e. earth connection exists. - Close ‘b’ contact whenever zero sequence currents can circulate, i.e.  connection. - circuit is left open when neither condition exists, for example unearthed star connection. 35
36. 36. Transformer zero sequence models 36
37. 37. Transformer zero sequence models 37
38. 38. Static Loads 38
39. 39. Static Loads 39
40. 40. The A-phase of the unloaded generator shown below is subjected to a SLG fault. The pre-faulted L-N voltage of the generator is EA Volts. The generator has positive, negative and zero sequence impedance of Z1, Z2 and Zo respectively. Find the fault current and phase voltages under the fault condition. 40 Line to Ground Fault (SLG)
41. 41. Conditions under SLG fault: VA = 0 ; and IB = IC = 0 (i.e. neglecting load currents)           I I  0       1 1 1   1 a a                B I 0 1 a a 1 3 I I I C A 2 2 A0 A1 A2 1 A0 A1 A2 A I 3 I  I  I  Since IA1 = IA2 = IA0 then the seq. n/ws must be connected in series 41
42. 42. From the interconnected networks 42 A E 1 A0 A1 A2 A Z Z Z 1 2 0 I 3 I I I       Also VA = VA0 +VA1 + VA2 VB= VA0 + a2 VA1 + aVA2 VC= VA0 + aVA1 + a2 VA2 Where: VA1= EA - IA1 Z1 VA2= 0 - IA2 Z2 VA0= 0 - IA0 Z0
43. 43. A 3-phase star connected solidly earthed, unloaded generator has it’s A-phase subjected to a solidly grounded earth fault. Given that the pre-faulted L-N voltage of the generator is 100 V and Z1 = j1.0Ω, Z2 = j0.5Ω and Zo = j0.1Ω, calculate the fault currents developed and the L-L voltages attained due to a fault. 43 Example
44. 44. 100 j1.6 From analysis of the ckt we obtain E A A0 A1 A2 A  Z Z Z 1 I 3 I I I   1 2 0          I I I j62.5 A A0 A1 A2       I I 3( j62.5) A j187.5 A F A                we also know that  1 1 1     1 a a            V A0 V A1 A2 2 2 V A V B C V 1 a a V 44
45. 45.  VA = VA0 +VA1 + VA2………………(1) VB= VA0 + a2 VA1 + aVA2…………..(2) VC= VA0 + aVA1 + a2 VA2…………...(3) From the ckt: VA1= EA - IA1 Z1= 100 – [(-j62.5)(j1.0)] = 37.5 V/gn Similarly, VA2= 0 - IA2 Z2 = - (-j62.5) (j0.5) = -31.25 V/gn And, VA0= 0 – IA0 Z0 = - (-j62.5) (j0.1) = -6.25 V/gn 45
46. 46. Substitute V1, V2 and V0 into equations (1), (2) and (3) VA = VA0 +VA1 + VA2 = 0 And VB= VA0 + a2 VA1 + aVA2 = - 9.37 – j59.5 = 60.23-98.9 V VC= VA0 + aVA1 + a2 VA2 = - 9.37 + j59.5 = 60.23 98.9 V Also, VAB= VA - VB = 9.37 + j59.5 = 60.23 81.1 V VBC= VB - VC = - j119 = 119 -90 V VCA= VC - VA = - 9.37 + j59.5 = 60.23 98.9 V 46
47. 47. DOUBLE LINE FAULT (DL) Condition when fault occur IA = 0 ; IB + IC = 0 ; and phase voltage VB = VC 47
48. 48. 1 1 1 1   1 a a 1          B B  A0 A1 I I          A0 B B 2 A2 B 2 1   A1 B I  0 B A I C B 2 2 A2 1 0 a I - aI 3 I    0 aI - a I and 3 I 0 I - I 0; 3 I I I 1 a a 3 I                      1 a - a 2  I 3 A0 A1 A2     I 0 a nd I - I B 48 j But a - a j1.732 j 3 I - I A1 A2 B I 3 2      Since IA0 = 0 then the zero sequence ckt is open
49. 49. Equivalent sequence ckt becomes From which we obtain E A A1  Z Z I 1 2  IA1 = - j66.67 IA2 = j66.67 - 114.94 A  A1 B - j 66.67     j0.58 I - I j I 3 But I B C 49
50. 50. Also, VA1 = EA - IA1Z1 = 100 - (-j66.67)(j1.0) = 100 - 66.67 = 33.33 V/gn Therefore, VA1 = VA2 = 33.33 V/gn Now                1 1 1     1 a a             V A0 V A1 A2 2 2 V A V B C V 1 a a V 50
51. 51.  VA =VA1 + VA2 = 66.67 V/gn VB =VA1(a2 + a) = VC But a2 + a = -1  VB = -33.33 V/gn = VC 51
52. 52. DOUBLE LINE TO GROUND FAULT (DLG) Condition when fault occur IA = 0 ; VB = VC = 0 ; IB + IC = IF 52
53. 53. From inspection IA = 0 and VB = VC           V V  0       1 1 1   1 a a               B V 0 1 a a 1 3 V 0 V 1 V C A 2 2 2 1 0 1 2 A V 3 V  V  V  Hence, the sequence networks are in parallel 53
54. 54. Since Z2 //Z0 then the circuit reduces to: Z Z 2 0 P Z Z 2 0 where Z   A E 1 Z Z 1 P I    54 2 Z   where I  0 I Z Z Z   I and  2 I 1 0 Z Z 0 2      1 0 2     
55. 55. Example: The diagram shows two generators connected to a transmission system comprising two parallel 66kV transmission lines via transformers. The data of this system is given below GA and GB – 13.8kV, 100 MVA, Z1 = j0.15, Z2 = j0.1 and Z0 = j0.03 (all p.u) TFA – 13.8/66kV, 100 MVA, Z1 = Z2 = Z0 = j0.15 p.u TFB – 13.8/66kV, 100 MVA, Z1 = Z2 = Z0 = j0.1 p.u Each Transmission Line - Z1 = j0.4, Z2 = j0.4 and Z0 = j0.7 (all p.u) All star points are solidly earthed, except that of generator B which is unearthed. All p.u. values are on the 100 MVA base. Calculate the fault current for a DLG fault at the feeder terminals of transformer A. 55
56. 56. Soln + ve - ve zero 56
57. 57.   + ve seq n/w reduces to - ve seq n/w reduces to zero seq n/w reduces to 57
58. 58. For DLG fault, the three sequence networks are connected in parallel (as seen earlier).  j3.8 p.u. 1 I pu 1   j(0.18 0.083) V Z    j2.05 p.u. j0.18   2 1    j0.334 (j3.8) Z 0 Z Z I I 0 2        58
59. 59. j1.75 p.u. j0.154   0 1    j0.334 (j3.8) Z 2 Z Z I I 0 2         I I A B         I I I I j1.75 ( j3.8 ) j2.05 0 A 0 1 2          I 0 I I 1 1 1  1 a a  1 a a I Substitute the seq current values in the matrix 1 2 2 2 C                      2 I  I  a I  aIB 0 1 2            j1.75 (1 240 )(3.8 90 ) (1 120 )(2.05 90 )    5.72 152.7 p.u. 59
60. 60. 874.8A 100 x 10 pu    3 x 66 x 10 MVA 3V But I 3 6 base base I 874.8 (5.72 152.7 ) 5003.9 152.7 A B        Similarly, I 5003.9 27.37 A And I  I  I  j5.26 p.u. F B C I j4601.5 A or F C     60
61. 61. One line Open Circuit 61 Under this condition consider the elemental values of V. From inspection IA = 0 and VB = VC
62. 62. Applying these conditions:            V   V 0 B        1 1 1   1 a a               V 0 1 a a 1 3 V 0 V 1 V C A 2 2 2 1 0 1 2 A V 3 V  V  V  Hence, the sequence networks are in parallel as shown below: 62
63. 63. From the example before: EA = 1000 and Z1 = j1.0 ; Z2 = j0.5 ; Z0 = j0.1 Since Z2 //Z0 then the circuit reduces to: j0.083 (j0.5)(j0. 1) P   j0.6 Z Z 2 0 Z Z where Z  2 0  j92.31 100 1   1 j0.083 E A Z Z I 1 P           V E - I Z 100 [( j92.31)(j1.0)]    2 0 1 A 1 1 7.69 V/gn V V 63
64. 64. j15.38 A 7.69 2      j0.5 V Z Also, I 2 2 j76.9 A 7.69 0      j0.1 V Z And, I 0 0 But, 3V1 = VA  VA = 23.07 V/gn                1 1 1     1 a a              I  I  0 I I 1 2 2 2 A B C B C I 1 a a I To determine I and I we use: I I I I 0 A 0 1 2      Also I I a (-j92.31) a(j15.38) - 93.3 j115.4 2 B 0       And I I a(-j92.31) a (j15.38) 93.3 j115.4 2 C 0       64