lecture16

lecture16 - three-phase single-line to ground...

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Unformatted text preview: three-phase ￿ ￿ ￿ single-line to ground double-line to ground line-to-line faults unbalanced faults ￿ balanced faults 60-75% 15-25% 5-15% <5% Percentage of total faults ￿ ￿ symmetrical components augmented component models Unbalance fault analysis requires new tools ￿ ￿ F a u l t ty p e s : Power Systems I ￿ ￿ Fault Analysis Ib1 120° 120° abc sequence positive sequence 120° Ic1 Ia1 120° 120° Ia2 acb sequence negative sequence Ic2 120° Ib2 zero sequence 0° Ic0 Ib0 Ia0 applicable to current and voltages permits modeling of unbalanced systems and networks Representative symmetrical components ￿ ￿ Allow unbalanced three-phase phasor quantities to be replaced by the sum of three separate but balanced symmetrical components Power Systems I ￿ ￿ Symmetrical Components = I a1 1+ a + a2 = 0 a 3 = 1∠0° = 1 + j 0 a 2 = 1∠240° = −0.5 − j 0.866 a = 1∠120° = −0.5 + j 0.866 Operator a identities I c1 = I a1 ∠(δ + 120°) = a I a1 I b1 = I a1 ∠(δ + 240°) = a 2 I a1 I a1 = I a1 ∠(δ + 0°) Positive sequence phasors Power Systems I ￿ ￿ Symmetrical Components I c 0 = I a 0 ∠(δ + 0°) = I a 0 I b 0 = I a 0 ∠(δ + 0°) = I a 0 I a 0 = I a 0 ∠(δ + 0°) = I a 0 Zero sequence phasors I c 2 = I a 2 ∠(δ + 240°) = a 2 I a 2 I b 2 = I a 2 ∠(δ + 120°) = a I a 2 I a 2 = I a 2 ∠(δ + 0°) = I a 2 Negative sequence phasors Power Systems I ￿ ￿ Symmetrical Components = I a 0 + a I a1 + a 2 I a 2 I c = I c 0 + I c1 + I c 2 Ⱥ I a Ⱥ Ⱥ1 1 Ⱥ I Ⱥ = Ⱥ1 a 2 Ⱥ b Ⱥ Ⱥ Ⱥ I c Ⱥ Ⱥ1 a Ⱥ Ⱥ Ⱥ 1 Ⱥ Ⱥ I a 0 Ⱥ a Ⱥ Ⱥ I a1 Ⱥ Ⱥ Ⱥ Ⱥ 2 a Ⱥ Ⱥ I a 2 Ⱥ Ⱥ Ⱥ Ⱥ = I a 0 + a 2 I a1 + a I a 2 I b = I b 0 + I b1 + I b 2 In matrix notation = I a 0 + I a1 + I a 2 I a = I a 0 + I a1 + I a 2 Relating unbalanced phasors to symmetrical components Power Systems I ￿ ￿ Symmetrical Components 1 Ⱥ Ⱥ a Ⱥ a 2 Ⱥ Ⱥ Ⱥ1 1 1 Ⱥ −1 A = Ⱥ1 a 3 Ⱥ1 a 2 Ⱥ I 012 = A −1 I abc 1 Ⱥ 1* 2 Ⱥ a Ⱥ = A 3 a Ⱥ Ⱥ Solving for the symmetrical components leads to I abc = A I 012 Ⱥ1 1 Ⱥ1 a 2 A = Ⱥ Ⱥ1 a Ⱥ [A] is known as the symmetrical components transformation matrix Power Systems I ￿ ￿ Symmetrical Components ) + aI ) Ia2 = 1 3 + a 2 Ib a c I a1 = 1 I a + aI b + a 2 I c 3 ( (I I a 0 = 1 (I a + I b + I c ) 3 In component form, the calculation for symmetrical components are Power Systems I ￿ Symmetrical Components * A T A* = 3 T * * * S 3φ = 3V012 I * = 3 Va 0 I a 0 + 3 Va1 I a1 + 3 Va 2 I a 2 012 T S 3φ = V012 A T A *I * 012 S 3φ = (AV012 ) (AI 012 ) T T S 3φ = Vabc I * abc The apparent power may also be expressed in terms of symmetrical components V012 = A −1 Vabc Vabc = A V012 Similar expressions exist for voltages Power Systems I ￿ ￿ Symmetrical Components I a0 = Solution (1.6∠25°) + (1.0∠180°) + (0.9∠132°) = 0.45∠96.5° I c = 0.9∠132° I b = 1.0∠180° I a = 1.6∠25° Obtain the symmetrical components of a set of unbalanced currents 3 (1.6∠25°) + a(1.0∠180°) + a 2 (0.9∠132°) = 0.94∠ − 0.1° I a1 = 3 (1.6∠25°) + a 2 (1.0∠180°) + a(0.9∠132°) = 0.60∠22.3° I a2 = 3 Power Systems I ￿ ￿ Example Ib1 Ic1 abc set Ia Ia1 positive sequence set Power Systems I Ib Ic Example Ib2 Ic2 negative sequence set Ia2 zero sequence set Ia0, Ib0, Ic0 Obtain the original unbalanced voltages: Va 2 = 0.8∠ − 30° Va1 = 1.0∠30° Va 0 = 0.6∠90° The symmetrical components of a set of unbalanced voltages are Power Systems I Vc = (0.6∠90°) + a (1.0∠30°) + a 2 (0.8∠ − 30°) = 1.7088∠155.8° Vb = (0.6∠90°) + a 2 (1.0∠30°) + a (0.8∠ − 30°) = 0.4∠90° Va = (0.6∠90°) + (1.0∠30°) + (0.8∠ − 30°) = 1.7088∠24.2° ￿ Example Vc1 Va2 negative sequence set Power Systems I Vc2 Vb2 zero sequence set Va0, Vb0, Vc0 Example Vc Vb1 abc set Vb positive sequence set Va1 Va positive, negative, and zero sequence impedances ￿ ￿ ￿ ￿ wye-connected balanced loads transmission line 3-phase transformers generators Augmented network models ￿ The impedance offered to the flow of a sequence current creating sequence voltages Power Systems I ￿ ￿ Sequence Impedances Va Vc Ib Ic Power Systems I Vb Ia Zs Zs In Zs Zn ZM ZM ZM Balanced Loads Vabc = Z abc I abc ȺVa Ⱥ Ⱥ Z S + Z n ȺV Ⱥ = Ⱥ Z + Z n Ⱥ b Ⱥ Ⱥ M ȺVc Ⱥ Ⱥ Z M + Z n Ⱥ Ⱥ Ⱥ I n = I a + Ib + Ic ZS + Zn ZM + Zn ZM + Zn Z M + Z n Ⱥ Ⱥ I a Ⱥ Z M + Z n Ⱥ Ⱥ I b Ⱥ Ⱥ Ⱥ Ⱥ Z S + Z n Ⱥ Ⱥ I c Ⱥ Ⱥ Ⱥ Ⱥ Vc = Z M I a + Z M I b + Z S I c + Z n I n Vb = Z M I a + Z S I b + Z M I c + Z n I n Va = Z S I a + Z M I b + Z M I c + Z n I n M o de l a n d governing equations A] Z abc Power Systems I 1 Ⱥ Ⱥ Z S + Z n Z M + Z n Z M + Z n Ⱥ Ⱥ1 1 a 2 Ⱥ Ⱥ Z M + Z n Z S + Z n Z M + Z n Ⱥ Ⱥ1 a 2 Ⱥ Ⱥ Ⱥ Ⱥ a Ⱥ Ⱥ Z M + Z n Z M + Z n Z S + Z n Ⱥ Ⱥ1 a Ⱥ Ⱥ Ⱥ Ⱥ 0 0 Ⱥ Ⱥ Z S + 3Z n + 2 Z M = Ⱥ 0 ZS − ZM 0 Ⱥ Ⱥ Ⱥ Ⱥ 0 0 Z S − Z M Ⱥ Ⱥ Ⱥ −1 Ⱥ1 1 1 Ⱥ = Ⱥ1 a 3 Ⱥ1 a 2 Ⱥ Z 012 V012 = A −1Z abc A I 012 → V012 = Z 012 I 012 [ = [A Vabc = Z abc I abc → (A V012 ) = Z abc (A I 012 ) Balanced Loads 1 Ⱥ a Ⱥ Ⱥ 2 a Ⱥ Ⱥ Ib Ic Power Systems I Vc1 In Vb1 Va1 Ia Zn Zs Zs Vn Vabc1 = Z abc I abc + Vabc 2 Zn ZS + Zn Zn Vc2 Vb2 Va2 ȺVa1 Ⱥ Ⱥ Z S + Z n ȺV Ⱥ = Ⱥ Z n Ⱥ b1 Ⱥ Ⱥ ȺVc1 Ⱥ Ⱥ Z n Ⱥ Ⱥ Ⱥ Zs Transmission Line Ⱥ Ⱥ I a Ⱥ ȺVa 2 Ⱥ Z n Ⱥ Ⱥ I b Ⱥ + ȺVb 2 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Z S + Z n Ⱥ Ⱥ I c Ⱥ ȺVc 2 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Zn I n + I a + Ib + Ic = 0 Vn = 0 + Z n I n Vc1 = Z S I c − Z n I n + Vc 2 Vb1 = Z S I b − Z n I n + Vb 2 Va1 = Z S I a − Z n I n + Va 2 Power Systems I 1 Ⱥ Ⱥ Z S + Z n a 2 Ⱥ Ⱥ Z n Ⱥ Ⱥ a Ⱥ Ⱥ Z n Ⱥ Ⱥ 0 Ⱥ Ⱥ Z S + 3Z n 0 0 Z S 0 Ⱥ = Ⱥ Ⱥ Ⱥ Ⱥ 0 0 Z S Ⱥ Ⱥ Ⱥ Ⱥ1 1 1 Ⱥ = Ⱥ1 a 3 Ⱥ1 a 2 Ⱥ Z 012 = A −1Z abc A ZS + Zn Zn Zn Ⱥ Ⱥ1 1 Z n Ⱥ Ⱥ1 a 2 Ⱥ Ⱥ Z S + Z n Ⱥ Ⱥ1 a Ⱥ Ⱥ Zn V012−1 = A −1Z abc A I 012 + V012− 2 = Z 012 I 012 + V012− 2 1 Ⱥ a Ⱥ Ⱥ 2 a Ⱥ Ⱥ Vabc1 = Z abc I abc + Vabc 2 → A V012−1 = Z abc A I 012 + A V012− 2 Transmission Line ￿ ￿ ￿ ￿ X2 ~ Xd” ￿ ￿ the zero sequence reactance is approximated to the leakage reactance X0 ~ XL zero sequence values are isolated from the airgap of the machine ￿ remember that the transient fault impedance is a function of time positive sequence values are the same as Xd, Xd’, and Xd” negative sequence values are affected by the rotation of the rotor Similar modeling of impedances to sequence impedances Typical values for common generators Power Systems I ￿ ￿ Generators Power Systems I Negative Sequence Positive Sequence Zero Sequence E1 Generator Model X2 X1 X0 VT2 VT1 VT0 Ec Power Systems I Zn Ea Eb Ⱥ Z S + 3Z n 0 = Ⱥ Ⱥ Ⱥ 0 Ⱥ ZS ZS Z 012 ZS Impedance Grounded Generators 0 0 ZS 0 Ⱥ 0 Ⱥ Ⱥ Z S Ⱥ Ⱥ the magnetization current and core losses represented by the shunt branch are neglected (they represent only 1% of the total load current) the transformer is modeled with the equivalent series leakage impedance the series leakage impedance is the same for all the sequences ￿ the series leakage impedance is the same for the positive and negative sequence only Z1 = Z 2 = Z ￿ Z 0 = Z1 = Z 2 = Z ￿ Three-legged core three-phase units ￿ Three single-phase units & five-legged core three-phase u n it s ￿ ￿ Series Leakage Impedance Power Systems I ￿ ￿ ￿ Transformers the positive sequence quantities rotate by +30 degrees the negative sequence quantities rotate by -30 degrees the zero sequence quantities can not pass through the transformer ￿ ￿ ￿ independent of the winding order (Δ-Y or Y- Δ) the positive sequence line voltage on the HV side leads the corresponding line voltage on the LV side by 30° consequently, for the negative sequence voltages the corresponding phase shift is -30° USA standard ￿ ￿ ￿ Wye-delta transformers create a phase shifting pattern for the various sequences Power Systems I ￿ ￿ Transformers ￿ ￿ primary winding - wye / wye-grounded / delta secondary winding - wye / wye-grounded / delta Zero-sequence network connections of the transformer d e p e n d s o n t h e w i n d i n g c o n n e c ti o n Power Systems I ￿ Transformers delta wye-grounded Power Systems I wye wye-grounded wye-grounded wye-grounded Transformers delta delta Power Systems I delta wye Transformers ...
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This note was uploaded on 02/01/2012 for the course EEL 4213 taught by Professor Thomasbaldwin during the Fall '11 term at FSU.

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