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8 Symmetrical ComponentsNotes on Power System Analysis1 Lesson 8 Symmetrical Components.

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Presentation on theme: "8 Symmetrical ComponentsNotes on Power System Analysis1 Lesson 8 Symmetrical Components."— Presentation transcript:

1 8 Symmetrical ComponentsNotes on Power System Analysis1 Lesson 8 Symmetrical Components

2 8 Symmetrical ComponentsNotes on Power System Analysis2 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:

3 8 Symmetrical ComponentsNotes on Power System Analysis3 Symmetrical component definition Three-phase voltages V a, V b, and V c (not necessarily balanced, with phase sequence a-b-c) can be resolved into three sets of sequence components: Zero sequence V a0 =V b0 =V c0 Positive sequence V a1, V b1, V c1 balanced with phase sequence a-b-c Negative sequence V a2, V b2, V c2 balanced with phase sequence c-b-a

4 8 Symmetrical ComponentsNotes on Power System Analysis4

5 8 Symmetrical ComponentsNotes on Power System Analysis5 where a = 1/120° = (-1 + j  3)/2 a 2 = 1/240° = 1/-120° a 3 = 1/360° = 1/0 ° VaVa = 111V0V0 VbVb 1a2a2 aV1V1 VcVc 1aa2a2 V2V2

6 8 Symmetrical ComponentsNotes on Power System Analysis6 V p = A V s V s = A -1 V p A = 111 1a2a2 a 1aa2a2 V p = VaVa VbVb VcVc V s = V0V0 V1V1 V2V2

7 8 Symmetrical ComponentsNotes on Power System Analysis7 A -1 =(1/3) 111 1aa2a2 1a2a2 a I p = A I s I s = A -1 I p We used voltages for example, but the result applies to current or any other phasor quantity V p = A V s V s = A -1 V p

8 8 Symmetrical ComponentsNotes on Power System Analysis8 V a = V 0 + V 1 + V 2 V b = V 0 + a 2 V 1 + aV 2 V c = V 0 + aV 1 + a 2 V 2 V 0 = (V a + V b + V c )/3 V 1 = (V a + aV b + a 2 V c )/3 V 2 = (V a + a 2 V b + aV c )/3 These are the phase a symmetrical (or sequence) components. The other phases follow since the sequences are balanced.

9 8 Symmetrical ComponentsNotes on Power System Analysis9 Sequence networks –A balanced Y-connected load has three impedances Z y connected line to neutral and one impedance Z n connected neutral to ground ZyZy ZyZy g c b a ZnZn

10 8 Symmetrical ComponentsNotes on Power System Analysis10 Sequence networks V ag = Z y +Z n ZnZn ZnZn IaIa V bg ZnZn Z y +Z n ZnZn IbIb V cg ZnZn ZnZn Z y +Z n IcIc or in more compact notation V p = Z p I p

11 8 Symmetrical ComponentsNotes on Power System Analysis11 ZyZy n V p = Z p I p V p = AV s = Z p I p = Z p AI s AV s = Z p AI s V s = (A -1 Z p A) I s V s = Z s I s where Z s = A -1 Z p A ZyZy ZyZy g c b a ZnZn

12 8 Symmetrical ComponentsNotes on Power System Analysis12 ZsZs = Z y +3Z n 00 0ZyZy 0 00ZyZy V 0 = (Z y + 3Z n ) I 0 = Z 0 I 0 V 1 = Z y I 1 = Z 1 I 1 V 2 = Z y I 2 = Z 2 I 2

13 8 Symmetrical ComponentsNotes on Power System Analysis13 ZyZy n g a 3 Z n V0V0 I0I0 Zero- sequence network ZyZy n a V1V1 I1I1 Positive- sequence network ZyZy n a V2V2 I2I2 Negative- sequence network Sequence networks for Y-connected load impedances

14 8 Symmetrical ComponentsNotes on Power System Analysis14 Z  /3 n a V1V1 I1I1 Positive- sequence network Z  /3 n a V2V2 I2I2 Negative- sequence network Sequence networks for  -connected load impedances. Note that these are equivalent Y circuits. Z  /3 n g a V0V0 I0I0 Zero- sequence network

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 8 Symmetrical ComponentsNotes on Power System Analysis15

16 8 Symmetrical ComponentsNotes on Power System Analysis16 General case unsymmetrical impedances Z s =A -1 Z p A= Z0Z0 Z 01 Z 02 Z 10 Z1Z1 Z 12 Z 20 Z 21 Z2Z2 ZpZp = Z aa Z ab Z ca Z ab Z bb Z bc Z ca Z bc Z cc

17 8 Symmetrical ComponentsNotes on Power System Analysis17 Z 0 = (Z aa +Z bb +Z cc +2Z ab +2Z bc +2Z ca )/3 Z 1 = Z 2 = (Z aa +Z bb +Z cc –Z ab –Z bc –Z ca )/3 Z 01 = Z 20 = (Z aa +a 2 Z bb +aZ cc –aZ ab –Z bc –a 2 Z ca )/3 Z 02 = Z 10 = (Z aa +aZ bb +a 2 Z cc –a 2 Z ab –Z bc –aZ ca )/3 Z 12 = (Z aa +a 2 Z bb +aZ cc +2aZ ab +2Z bc +2a 2 Z ca )/3 Z 21 = (Z aa +aZ bb +a 2 Z cc +2a 2 Z ab +2Z bc +2aZ ca )/3

18 8 Symmetrical ComponentsNotes on Power System Analysis18 Special case symmetrical impedances ZsZs = Z0Z0 00 0Z1Z1 0 00Z2Z2 ZpZp = Z aa Z ab Z aa Z ab Z aa

19 8 Symmetrical ComponentsNotes on Power System Analysis19 Z 0 = Z aa + 2Z ab Z 1 = Z 2 = Z aa – Z ab Z 01 =Z 20 =Z 02 =Z 10 =Z 12 =Z 21 = 0 V p = Z p I p V s = Z s I s This applies to impedance loads and to series impedances (the voltage is the drop across the series impedances)

20 8 Symmetrical ComponentsNotes on Power System Analysis20 Power in sequence networks S p = V ag I a * + V bg I b * + V cg I c * S p = [V ag V bg V cg ] [I a * I b * I c * ] T S p = V p T I p * = (AV s ) T (AI s ) * = V s T A T A * I s *

21 8 Symmetrical ComponentsNotes on Power System Analysis21 Power in sequence networks A T A * = = 300 1a2a2 a1aa2a aa2a2 1a2a2 a003 S p = 3 V s T I s * S p = V p T I p * = V s T A T A* I s *

22 8 Symmetrical ComponentsNotes on Power System Analysis22 S p = 3 (V 0 I 0 * + V 1 I 1 * +V 2 I 2 * ) = 3 S s 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.

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 ComponentsNotes on Power System Analysis23

24 8 Symmetrical ComponentsNotes on Power System Analysis24 Y generator Zero I1I1 V1V1 Z1Z1 Z2Z2 I2I2 Z0Z0 I0I0 V0V0 N G N Negative N Positive ZnZn V2V2 3Z n E

25 8 Symmetrical ComponentsNotes on Power System Analysis25 Ungrounded Y load Zero I1I1 V1V1 Z Z I2I2 V2V2 Z I0I0 V0V0 N G N Negative N Positive

26 8 Symmetrical ComponentsNotes on Power System Analysis26 Zero-sequence networks for loads Z I0I0 V0V0 N G 3Z n Z V0V0 G Y-connected load grounded through Z n  -connected load ungrounded

27 8 Symmetrical ComponentsNotes on Power System Analysis27 Y-Y transformer A B C N H1X1 a b c n ZnZn ZNZN Z eq +3(Z N +Z n ) g A V A0 I0I0 Zero-sequence network (per unit) V a0 a

28 8 Symmetrical ComponentsNotes on Power System Analysis28 Y-Y transformer A B C N H1X1 a b c n ZnZn ZNZN Z eq n A V A1 I1I1 Positive-sequence network (per unit) Negative sequence is same network V a1 a

29 8 Symmetrical ComponentsNotes on Power System Analysis29  -Y transformer A B C H1X1 a b c n ZnZn Z eq +3Z n g A V A0 I0I0 Zero-sequence network (per unit) V a0 a

30 8 Symmetrical ComponentsNotes on Power System Analysis30  -Y transformer A B C H1X1 a b c n ZnZn Z eq n A V A1 I1I1 Positive-sequence network (per unit) Delta side leads wye side by 30 degrees V a1 a

31 8 Symmetrical ComponentsNotes on Power System Analysis31  -Y transformer A B C H1X1 a b c n ZnZn Z eq n A V A2 I2I2 Negative-sequence network (per unit) Delta side lags wye side by 30 degrees V a2 a

32 8 Symmetrical ComponentsNotes on Power System Analysis 32 Three-winding (three-phase) transformers Y-Y-  ZXZX ZTZT ZHZH H X Ground Zero sequence ZXZX ZHZH H X Neutral ZTZT T Positive and negative T H and X in grounded Y and T in delta

33 8 Symmetrical ComponentsNotes on Power System Analysis33 Three-winding transformer data: WindingsZBase MVA H-X5.39%150 H-T6.44%56.6 X-T4.00%56.6 Convert all Z's to the system base of 100 MVA: Z hx = 5.39% (100/150) = 3.59% Z hT = 6.44% (100/56.6) = 11.38% Z xT = 4.00% (100/56.6) = 7.07%

34 8 Symmetrical ComponentsNotes on Power System Analysis34 Calculate the equivalent circuit parameters: Solving: Z HX = Z H + Z X Z HT = Z H + Z T Z XT = Z X +Z T Gives: Z H = (Z HX + Z HT - Z XT )/2 = 3.95% Z X = (Z HX + Z XT - Z HT )/2 = % Z T = (Z HT + Z XT - Z HX )/2 = 7.43%

35 8 Symmetrical ComponentsNotes on Power System Analysis35 Typical relative sizes of sequence impedance values Balanced three-phase lines: Z 0 > Z 1 = Z 2 Balanced three-phase transformers (usually): Z 1 = Z 2 = Z 0 Rotating machines: Z 1  Z 2 > Z 0

36 8 Symmetrical ComponentsNotes on Power System Analysis36 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


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