1 / 36

Lesson 8 Symmetrical Components

Lesson 8 Symmetrical Components. Symmetrical Components. Due to C. L. Fortescue (1918): a set of n unbalanced phasors in an n-phase system can be resolved into n balanced phasors by a linear transformation The n sets are called symmetrical components

taline
Télécharger la présentation

Lesson 8 Symmetrical Components

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lesson 8Symmetrical Components Notes on Power System Analysis

  2. Symmetrical Components • Due to C. L. Fortescue (1918): a set of n unbalanced phasors in an n-phase system can be resolved into n balanced phasors by a linear transformation • The n sets are called symmetrical components • One of the n sets is a single-phase set and the others are n-phase balanced sets • Here n = 3 which gives the following case: Notes on Power System Analysis

  3. Symmetrical component definition • Three-phase voltages Va, Vb, and Vc (not necessarily balanced, with phase sequence a-b-c) can be resolved into three sets of sequence components: Zero sequence Va0=Vb0=Vc0 Positive sequence Va1, Vb1, Vc1 balancedwith phase sequence a-b-c Negative sequence Va2, Vb2, Vc2 balanced with phase sequence c-b-a Notes on Power System Analysis

  4. Notes on Power System Analysis

  5. where a = 1/120°= (-1 + j 3)/2 a2 = 1/240°= 1/-120° a3 = 1/360°= 1/0 ° Notes on Power System Analysis

  6. Vp = A VsVs= A-1Vp Notes on Power System Analysis

  7. We used voltages for example, but the result applies to current or any other phasor quantity Vp = A VsVs= A-1Vp Ip = A Is Is= A-1Ip Notes on Power System Analysis

  8. Va = V0 + V1 + V2 Vb = V0 + a2V1 + aV2 Vc = V0 + aV1 + a2V2 V0 = (Va + Vb + Vc)/3 V1 = (Va + aVb + a2Vc)/3 V2 = (Va + a2Vb + aVc)/3 These are the phase a symmetrical (or sequence) components. The other phases follow since the sequences are balanced. Notes on Power System Analysis

  9. a Zy b Zy c Zn g Sequence networks • A balanced Y-connected load has three impedances Zy connected line to neutral and one impedance Zn connected neutral to ground Notes on Power System Analysis

  10. Sequence networks or in more compact notation Vp= ZpIp Notes on Power System Analysis

  11. a Zy b Zy c Zn g Zy Vp = ZpIp Vp = AVs = ZpIp= ZpAIs AVs = ZpAIs Vs= (A-1ZpA) Is Vs = Zs Is where Zs = A-1ZpA n Notes on Power System Analysis

  12. V0 = (Zy + 3Zn) I0 = Z0 I0 V1 = ZyI1 = Z1 I1 V2 = ZyI2 = Z2 I2 Notes on Power System Analysis

  13. I2 I0 I1 Zy Zy Zy n a a a V2 V0 V1 3 Zn n g n Zero- sequence network Positive- sequence network Negative- sequence network Sequence networks for Y-connected load impedances Notes on Power System Analysis

  14. I2 I0 I1 ZD/3 ZD/3 ZD/3 n a a a V2 V0 V1 n g n Positive- sequence network Negative- sequence network Zero- sequence network Sequence networks for D-connected load impedances. Note that these are equivalent Y circuits. Notes on Power System Analysis

  15. Remarks • Positive-sequence impedance is equal to negative-sequence impedance for symmetrical impedance loads and lines • Rotating machines can have different positive and negative sequence impedances • Zero-sequence impedance is usually different than the other two sequence impedances • Zero-sequence current can circulate in a delta but the line current (at the terminals of the delta) is zero in that sequence Notes on Power System Analysis

  16. General case unsymmetrical impedances Notes on Power System Analysis

  17. Z0 = (Zaa+Zbb+Zcc+2Zab+2Zbc+2Zca)/3 Z1 = Z2 = (Zaa+Zbb+Zcc–Zab–Zbc–Zca)/3 Z01 = Z20 = (Zaa+a2Zbb+aZcc–aZab–Zbc–a2Zca)/3 Z02 = Z10 = (Zaa+aZbb+a2Zcc–a2Zab–Zbc–aZca)/3 Z12 = (Zaa+a2Zbb+aZcc+2aZab+2Zbc+2a2Zca)/3 Z21 = (Zaa+aZbb+a2Zcc+2a2Zab+2Zbc+2aZca)/3 Notes on Power System Analysis

  18. Special case symmetrical impedances Notes on Power System Analysis

  19. Z0 = Zaa+ 2Zab Z1 = Z2 = Zaa– Zab Z01=Z20=Z02=Z10=Z12=Z21= 0 Vp = ZpIpVs = ZsIs • This applies to impedance loads and to series impedances (the voltage is the drop across the series impedances) Notes on Power System Analysis

  20. Power in sequence networks Sp = VagIa* + VbgIb* + VcgIc* Sp = [VagVbgVcg] [Ia*Ib*Ic*]T Sp = VpTIp* = (AVs)T(AIs)* = VsTATA* Is* Notes on Power System Analysis

  21. Power in sequence networks Sp = VpTIp* = VsT ATA* Is* Sp = 3 VsT Is* Notes on Power System Analysis

  22. Sp= 3 (V0 I0* + V1 I1* +V2 I2*) = 3 Ss In words, the sum of the power calculated in the three sequence networks must be multiplied by 3 to obtain the total power. This is an artifact of the constants in the transformation. Some authors divide A by 3 to produce a power-invariant transformation. Most of the industry uses the form that we do. Notes on Power System Analysis

  23. Sequence networks for power apparatus • Slides that follow show sequence networks for generators, loads, and transformers • Pay attention to zero-sequence networks, as all three phase currents are equal in magnitude and phase angle Notes on Power System Analysis

  24. N E V1 Positive I1 Z1 N V2 I2 Negative Zn Z2 G Y generator V0 3Zn Zero I0 Z0 N Notes on Power System Analysis

  25. N V1 Positive I1 Z N V2 I2 Negative Ungrounded Y load Z G V0 Zero I0 Z N Notes on Power System Analysis

  26. Zero-sequence networks for loads G Y-connected load grounded through Zn V0 3Zn I0 Z N G D-connected load ungrounded V0 Z Notes on Power System Analysis

  27. Y-Y transformer H1 X1 A a Zeq+3(ZN+Zn) A a B I0 Va0 VA0 b g c C Zero-sequence network (per unit) N n ZN Zn Notes on Power System Analysis

  28. Y-Y transformer Zeq H1 X1 A a A a I1 Va1 VA1 B b n c C Positive-sequence network (per unit) Negative sequence is same network N n ZN Zn Notes on Power System Analysis

  29. D-Y transformer H1 X1 A a Zeq+3Zn A a B VA0 I0 Va0 b g c C Zero-sequence network (per unit) n Zn Notes on Power System Analysis

  30. D-Y transformer H1 X1 A a Zeq A a B I1 Va1 VA1 b n c C Positive-sequence network (per unit) Delta side leads wye side by 30 degrees n Zn Notes on Power System Analysis

  31. D-Y transformer H1 X1 A a Zeq A a B I2 Va2 VA2 b n c C Negative-sequence network (per unit) Delta side lags wye side by 30 degrees n Zn Notes on Power System Analysis

  32. Three-winding (three-phase) transformers Y-Y-D H and X in grounded Y and T in delta Zero sequence Positive and negative Neutral Ground ZT X ZX ZH ZH H H X ZX ZT T T Notes on Power System Analysis

  33. Three-winding transformer data: Windings Z Base MVA H-X 5.39% 150 H-T 6.44% 56.6 X-T 4.00% 56.6 Convert all Z's to the system base of 100 MVA: Zhx = 5.39% (100/150) = 3.59% ZhT= 6.44% (100/56.6) = 11.38% ZxT= 4.00% (100/56.6) = 7.07% Notes on Power System Analysis

  34. Calculate the equivalent circuit parameters: Solving: ZHX= ZH+ ZX ZHT= ZH+ ZT ZXT= ZX+ZT Gives: ZH= (ZHX+ ZHT- ZXT)/2 = 3.95% ZX= (ZHX+ ZXT- ZHT)/2 = -0.359% ZT= (ZHT+ ZXT- ZHX)/2 = 7.43% Notes on Power System Analysis

  35. Typical relative sizes of sequence impedance values • Balanced three-phase lines: Z0 > Z1 = Z2 • Balanced three-phase transformers (usually): Z1 = Z2 = Z0 • Rotating machines: Z1 Z2 >Z0 Notes on Power System Analysis

  36. Unbalanced Short Circuits • Procedure: • Set up all three sequence networks • Interconnect networks at point of the fault to simulate a short circuit • Calculate the sequence I and V • Transform to ABC currents and voltages Notes on Power System Analysis

More Related