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8 Symmetrical ComponentsNotes on Power System Analysis1 Lesson 8 Symmetrical Components

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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:

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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

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8 Symmetrical ComponentsNotes on Power System Analysis4

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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

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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

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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

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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.

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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 *

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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 *

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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.

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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%

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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%

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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

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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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