Unformatted text preview: threephase singleline to ground
doubleline to ground
linetoline faults unbalanced faults balanced faults 6075%
1525%
515% <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 threephase 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 wyeconnected balanced loads
transmission line
3phase 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 Threelegged core threephase units Three singlephase units & fivelegged core threephase
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 (ΔY 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 Wyedelta transformers create a phase shifting pattern for
the various sequences Power Systems I Transformers primary winding  wye / wyegrounded / delta
secondary winding  wye / wyegrounded / delta Zerosequence 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 wyegrounded Power Systems I wye wyegrounded wyegrounded wyegrounded 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.
 Fall '11
 THOMASBALDWIN

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