Download presentation

1
**Lesson 8 Symmetrical 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: 8 Symmetrical Components 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 balanced with phase sequence a-b-c Negative sequence Va2, Vb2, Vc2 balanced with phase sequence c-b-a 8 Symmetrical Components Notes on Power System Analysis

4
**Notes on Power System Analysis**

8 Symmetrical Components Notes on Power System Analysis

5
**Notes on Power System Analysis**

Va = 1 V0 Vb a2 a V1 Vc V2 where a = 1/120° = (-1 + j 3)/2 a2 = 1/240° = 1/-120° a3 = 1/360° = 1/0 ° 8 Symmetrical Components Notes on Power System Analysis

6
**Notes on Power System Analysis**

1 a2 a Vs = V0 V1 V2 Vp = Va Vb Vc Vp = A Vs Vs = A-1 Vp 8 Symmetrical Components Notes on Power System Analysis

7
**Notes on Power System Analysis**

(1/3) 1 a a2 We used voltages for example, but the result applies to current or any other phasor quantity Vp = A Vs Vs = A-1 Vp Ip = A Is Is = A-1 Ip 8 Symmetrical Components Notes on Power System Analysis

8
**Notes on Power System Analysis**

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. 8 Symmetrical Components Notes on Power System Analysis

9
**Notes on Power System Analysis**

Sequence networks A balanced Y-connected load has three impedances Zy connected line to neutral and one impedance Zn connected neutral to ground Zy g c b a Zn 8 Symmetrical Components Notes on Power System Analysis

10
**Notes on Power System Analysis**

Sequence networks Vag = Zy+Zn Zn Ia Vbg Ib Vcg Ic or in more compact notation Vp = Zp Ip 8 Symmetrical Components Notes on Power System Analysis

11
**Notes on Power System Analysis**

Zy Vp = Zp Ip Vp = AVs = Zp Ip = ZpAIs AVs = ZpAIs Vs = (A-1ZpA) Is Vs = Zs Is where Zs = A-1ZpA Zy g c b a Zn n 8 Symmetrical Components Notes on Power System Analysis

12
**Notes on Power System Analysis**

Zs = Zy+3Zn Zy V0 = (Zy + 3Zn) I0 = Z0 I0 V1 = Zy I1 = Z1 I1 V2 = Zy I2 = Z2 I2 8 Symmetrical Components Notes on Power System Analysis

13
**Notes on Power System Analysis**

Zy n a V2 I2 Negative- sequence network Zy n g a 3 Zn V0 I0 Zero- sequence network Zy n a V1 I1 Positive- sequence network Sequence networks for Y-connected load impedances 8 Symmetrical Components Notes on Power System Analysis

14
**Notes on Power System Analysis**

ZD/3 n a V2 I2 Negative- sequence network ZD/3 n g a V0 I0 Zero- sequence network ZD/3 n a V1 I1 Positive- sequence network Sequence networks for D-connected load impedances. Note that these are equivalent Y circuits. 8 Symmetrical Components Notes on Power System Analysis

15
**Notes on Power System Analysis**

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 8 Symmetrical Components Notes on Power System Analysis

16
**Notes on Power System Analysis**

General case unsymmetrical impedances Zp = Zaa Zab Zca Zbb Zbc Zcc Zs=A-1ZpA = Z0 Z01 Z02 Z10 Z1 Z12 Z20 Z21 Z2 8 Symmetrical Components Notes on Power System Analysis

17
**Notes on Power System Analysis**

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 8 Symmetrical Components Notes on Power System Analysis

18
**Notes on Power System Analysis**

Special case symmetrical impedances Zp = Zaa Zab Zs = Z0 Z1 Z2 8 Symmetrical Components Notes on Power System Analysis

19
**Notes on Power System Analysis**

Z0 = Zaa + 2Zab Z1 = Z2 = Zaa – Zab Z01=Z20=Z02=Z10=Z12=Z21= 0 Vp = Zp Ip Vs = Zs Is This applies to impedance loads and to series impedances (the voltage is the drop across the series impedances) 8 Symmetrical Components Notes on Power System Analysis

20
**Power in sequence networks**

Sp = Vag Ia* + Vbg Ib* + Vcg Ic* Sp = [Vag Vbg Vcg] [Ia* Ib* Ic*]T Sp = VpT Ip* = (AVs)T (AIs)* = VsT ATA* Is* 8 Symmetrical Components Notes on Power System Analysis

21
**Power in sequence networks**

Sp = VpT Ip* = VsT ATA* Is* ATA* = 1 = 3 a2 a Sp = 3 VsT Is* 8 Symmetrical Components Notes on Power System Analysis

22
**Notes on Power System Analysis**

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. 8 Symmetrical Components 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 8 Symmetrical Components Notes on Power System Analysis

24
**Notes on Power System Analysis**

V1 I1 Positive Z1 N V2 I2 Negative Zn Z2 G Y generator V0 3Zn Zero I0 Z0 N 8 Symmetrical Components Notes on Power System Analysis

25
**Notes on Power System Analysis**

V1 I1 Positive Z N V2 I2 Negative Ungrounded Y load Z G V0 Zero I0 Z N 8 Symmetrical Components Notes on Power System Analysis

26
**Notes on Power System Analysis**

Zero-sequence networks for loads G V0 3Zn Y-connected load grounded through Zn I0 Z N G D-connected load ungrounded V0 Z 8 Symmetrical Components Notes on Power System Analysis

27
**Notes on Power System Analysis**

Y-Y transformer H1 X1 A a Zeq+3(ZN+Zn) A a B VA0 I0 Va0 b g c C Zero-sequence network (per unit) N n ZN Zn 8 Symmetrical Components Notes on Power System Analysis

28
**Notes on Power System Analysis**

Y-Y transformer Zeq H1 X1 A a A a VA1 I1 Va1 B b n c C Positive-sequence network (per unit) Negative sequence is same network N n ZN Zn 8 Symmetrical Components Notes on Power System Analysis

29
**Notes on Power System Analysis**

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 8 Symmetrical Components Notes on Power System Analysis

30
**Notes on Power System Analysis**

D-Y transformer H1 X1 A a Zeq A a B VA1 I1 Va1 b n c C Positive-sequence network (per unit) Delta side leads wye side by 30 degrees n Zn 8 Symmetrical Components Notes on Power System Analysis

31
**Notes on Power System Analysis**

D-Y transformer H1 X1 A a Zeq A a B VA2 I2 Va2 b n c C Negative-sequence network (per unit) Delta side lags wye side by 30 degrees n Zn 8 Symmetrical Components 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 Ground Neutral ZT X H ZH H ZH ZX ZX X ZT T T 8 Symmetrical Components Notes on Power System Analysis

33
**Notes on Power System Analysis**

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% 8 Symmetrical Components Notes on Power System Analysis

34
**Notes on Power System Analysis**

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 = % ZT = (ZHT + ZXT - ZHX)/2 = 7.43% 8 Symmetrical Components 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 8 Symmetrical Components 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 8 Symmetrical Components Notes on Power System Analysis

Similar presentations

Presentation is loading. Please wait....

OK

EET 103 Chapter 3 (Lecture 1) Three Phase System.

EET 103 Chapter 3 (Lecture 1) Three Phase System.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on computer networking for class 9 Ppt on media literacy Ppt on switching networking Ppt on centroid and centre of gravity Ppt on c language fundamentals inc Ppt on conceptual art definition Ppt on iso 9000 Ppt on applied operational research definition Ppt on services in android Ppt on infrared remote control system