Short Circuit Lecture Using EDSA-ETAP-PSSE-Matlab
description
Transcript of Short Circuit Lecture Using EDSA-ETAP-PSSE-Matlab
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Analysis of Faulted Power System (Shunt Fault)
Sequence Model for Transformer
Roel B. Calano
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Sequence Model for AC Generator
Generator Connection Diagram
Zero Sequence Network Diagram
Z0
Z0
Z0
a
c
b
Z0
N0
Z0
Z0
Zn
Z0
a
c
b
Z0
3Zn
N0
Z0
Z0
Z0
Z0
a
c
N0
b
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Sequence Networks
One of the most useful concepts about the symmetrical components is the sequence network. A sequence network is an equivalent network for power system under the assumption that only one sequence component of voltages and currents is presented in the system. There will be no interaction between each sequence network and each of them is independent of each other. The positive sequence network is the only one containing voltage source since generators produce only voltages of positive sequence. Negative and zero sequence networks contain only their corresponding impedances and these impedances are obtained based on the location of the fault under investigation. These sequence networks are shown in Figure. The types of fault conditions will determine the connections between the sequence networks. The positive sequence impedance, Z1, is the impedance looking into the positive sequence network from the fault point. Similarly, the negative sequence impedance, Z2, is the impedance looking into the negative sequence network from the fault point, and the zero sequence impedance, Z0, is the impedance looking into the zero sequence networks from the fault point.
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Example: Draw the Zero sequence network of the power system shown
GG
Y grounded - Y grounded
Y grounded - Y grounded
∆ - ∆
∆ - ∆
Y grounded Reactance grounded
Solution:
j0.2 j0.3 j0.3
j0.25 j0.5 j0.35
2 3
5 6
1 4
j0.06 j0.06
N0 (Neutral or Zero-potential bus)
j0.09
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Example: Draw the Zero sequence network of the power system shown
GG
Y ungrounded - Y grounded
Y ungrounded - Y ungrounded ∆ - ∆
Y ungrounded Reactance grounded
Y ungrounded - Y ungrounded
Solution: j0.2 j0.3 j0.3
j0.25 j0.5 j0.35
j0.06 j0.06
N0 (Neutral or Zero-potential bus)
j0.09
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Y grounded- Y grounded
Y grounded - Y grounded
∆ - ∆
∆ - Y ungrounded
GG
G
∆ - ∆ Y ungrounded - Y grounded
Y ungrounded Reactance grounded
Y grounded
j0.2 j0.3 j0.3
j0.25 j0.35
2 3
5 6
1 4
j0.06 j0.06
NB0B (Neutral or Zero-potential bus)
j0.09
j0.09
j0.5
j0.09
j0.09
j0.09
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Y grounded - Y grounded
∆ - Y grounded
∆ - Y ungrounded
Y ungrounded - Y grounded
GG
G
∆ - ∆ ∆ - Y grounded
Solidly grounded
Y ungrounded
2 3
1 4
5
6
NB0B (Neutral or Zero-potential bus)
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Problem: Draw the Positive, Negative and zero sequence model for the power system shown.
GG
Y grounded - Y grounded
Y grounded - Y grounded
∆ - ∆
∆ - ∆
Y grounded Reactance grounded
2 3
5 6
1 4
GG
∆ - Y grounded
∆ - Y grounded
∆ - Y grounded
∆ - Y grounded
G
∆ - Y grounded ∆ - Y grounded
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Solution:
Reactance Diagram Simplification
j0.2 j0.15 j0.3
j0.25 j0.22 j0.35
2 3
5 6
1 4
j0.14 j0.14
N2 (Neutral or Zero-potential bus)
j0.2 j0.15 j0.3
j0.25 j0.22 j0.35
2 3
5 6
1 4
j0.2 j0.2
N1 (Neutral or Zero-potential bus)
1.0 angle 0° G 1.0 angle 0° G
j0.2 j0.3 j0.3
j0.25 j0.5 j0.35
2 3
5 6
1 4
j0.06 j0.06
NB0B (Neutral or Zero-potential bus)
j0.09
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Positive Sequence Network
j0.35 j0.3
j0.82
3
1 4
j0.2 j0.2
N1
1.0 angle 0° G 1.0 angle 0° G
j0.1952
j0.0.0714
3
1 4
N1
j0.2
1.0 angle 0° G
j0.2
1.0 angle 0° G
j0.1673
3
j0.0.0714
j0.3673 j0.3952
N1
1.0 angle 0° G G 1.0 angle 0°
NB1B
j0.3952 j0.3673
j0.0.0714
3
j0.3952
1.0 angle 0° G
NB1B
j0.1904
3
j0.0.0714
1.0 angle 0° G
N1
3
j0.0.2618
G 1.0 angle 0°
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Negative Sequence Network
j0.35 j0.3
j0.82
3
1 4
j0.14 j0.14
N2
j0.1952
j0.0.0714
3
1 4
N2
j0.14 j0.14
j0.1673
j0.0.0714
3
N2
j0.3352 j0.3073
j0.3073
N B2B
j0.3952
j0.0.0714
3
j0.3352
N B2B
j0.1603
j0.0.0714
3
N2
j0.0.2317
3
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Zero Sequence Network
N0
j0.0.56
3
j0.2
N0
j0.06
j0.3
2
3
Problem: Draw the connection diagram of Positive, Negative and Zero sequence network of the 4 bus shown.
GG∆ - Y grounded
∆ - Y grounded
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Short Circuit Current under Transient Condition
α
e= VS sin (ωt + α)
G
G
G Three Phase Symmetrical Fault
Consider a Three phase Fault occurs near the terminal of the generator
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Derivation of Short Circuit Current under Transient Condition
VR + V = VL S R + = VS sin (ωt + α)
Ldi
dt R + = V
Ldi S [sin ωt cosα + cosωt sinα]
dt Taking the Laplace Transform:
RI(s) + L [sI(s) – I(0) ] = VS [cosα ( ) + sinα ( ) ]
s
ω
s2 + ω2
ω 2 + ω2
I (0) = 0
Is [R + Ls] = VS [ + ] s sinα
2 + ω2
ω cosα
2 + ω2s s
Is [ R + Ls] = VS [ ]
s
ω cosα + s sinα
2 + ω2
Is = VS [ ]
ω cosα + s sinα
(s2 + ω2 ) (R + Ls) Simplification using partial fraction method:
Is = VS/ L [ + ] C
(s + R/L)
As + B
(s2 + ω2 ) Solving for A, B and C: A = Ls/Z [sin (α – θ)] B = ωl/Z [cos (α – θ)] C = L/Z [sin (α – θ)] Z = √ (R2 2 + (ωl) θ = arctan ( ωL/R)
Is = VS/L [ + ] L/Z [sin (α – θ)]
(s + R/L)
Ls/Z [sin (α – θ)]+ ωl/Z [cos (α – θ)]
(s2 + ω2 )
Is = VS/L [sin (α –θ){ } + cos (α – θ) { } + ]
s
s Using the Inverse Laplace Transform:
s 2 + ω2
s 2 + ω2
sin (α – θ)
(s + R/L)
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It = VS/Z [{sin (α – θ)} (cos ωt) + [{cos (α – θ)} (sin ωt) - {sin (α – θ)}e –Rt/L]
It = VS/Z [{sin (ωt + α – θ - {sin (α – θ)}e –Rt/L] It = VS/Z {sin (ωt + α – θ - VS/Z {sin (α – θ)}e –Rt/L
From the given equation it can be seen that the short circuit current has two component, the transient current which decays with respect to time ( at time = infinity) and the steady current.
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Transient Analysis of Three Phase Short Circuit Three Phase Short Circuit at Phase a
Three Phase Short Circuit at Phase b
Three Phase Short Circuit at Phase c
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Three Phase Short Circuit at the Field
Line to Line Short Circuit at Phase b
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Line to Line Short Circuit at the field
Line to Ground Short Circuit at Phase a
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Line to Ground Short Circuit at the Field
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Different Level of Short Circuit Current with respect to time
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Difference between Symmetrical and Unsymmetrical Fault
Short Circuit Current near the Generator
Symmetrical
Unsymmetrical
Example: Draw the wave form of short circuit current of different ωL/R and power factor angle Solution: ωL/R program
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Important Notes when Conducting Short Circuit Analysis
• The Three Sequence are independent • The positive-sequence network is the same as the one line diagram used in
studying balanced three-phase current and voltages • The Positive-sequence network has a voltage source. Therefore, the positive-
sequence current causes only positive-sequence voltage drops • There is no voltage source in the negative or zero sequence networks • Negative and Zero sequence currents cause negative and zero sequence
voltage drops only • The Neutral of the System is the Reference for positive and negative sequence
networks, but ground is the reference for the zero sequence networks. Therefore, the zero sequence current can flow only if the circuit from the system neutrals to ground is complete
• The grounding impedance is reflected in the zero sequence network as 3 Z0 • The threes sequence systems can be solved separately on a per phase basis.
The phase currents and voltages can be determined by superposing the symmetrical components of current and the voltages respectively.
Simplifications When computing short circuits in a power system further simplifications can be made. The following simplifications are also used for the analysis
• All line capacitances are ignored. • All non-motor shunt impedances are ignored; motor loads are treated the same
way as generators.
• The voltage magnitude and phase angle of generators and in feeds are all set to the same value
• All tap changing transformers are in middle position.
These simplifications are indicated for studies regarding medium- and long-term network planning. In the planning stage, the calculations are based on estimated and hence inaccurate data. Therefore, the demands on the short circuit computation algorithm are lower than for real-time applications in the network operation, where accurate results are desired. Studies have shown that the shunt elements and loads have little influence on the short circuit currents (0.5%. . . 4%) and may compensate each other. However, disregarding the actual generator pole voltages and the actual positions of tap changing transformers may sometimes lead to errors of up to 30%.
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Single Line to Ground Fault
Zf
a b c
Fault point
Iaf Ibf = 0 Icf = 0
Model Interconnection of sequence network
F2
N2
Z2Va2 Ia2
Negative sequence network
F0
N0
Z0Va0 Ia0
Zero - sequence network
Positive sequence network
F1
N1
Z1Va1 Ia1
G
3Zf
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Analysis of Single Line – to - Ground Fault Ia0 = Ia1 = I = 1.0 ∠ 0 ° a2 Z + Z0 1 + Z + 3Z2 f Iaf 1 1 1 Ia0
2 Ibf = 1 a a Ia12 Icf 1 a a I a2
Iaf = I + I + I a0 a1 a2 Iaf = 3Ia0 = 3I = 3Ia1 a2
Vaf = Z If af
Vaf = 3Z If a1
Vaf = V + V + Va0 a1 a2
V + V + V = 3Z Ia0 a1 a2 f a1
V 0 Z 0 0 Ia0 0 a0
V = 1.0 ∠ 0 ° - 0 Z 0 Ia1 1 a1 V 0 0 0 Z Ia2 2 a2
V = - Z Ia0 0 a0
V = 1.0 – Z Ia1 1 a1
V = -Z Ia2 2 a2
V 1 1 1 Vaf a02 Vbf = 1 a a Va1
2 V 1 a a Vcf a2
Vbf = V + a2V + aVa0 a1 a2 2
Vcf = V + aV + a Va0 a1 a2
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Line-Line Fault Model
a b c
Fault point
Iaf = 0 Ibf Icf Zf
Interconnection of sequence network
F2
N2
Z2Va2 Ia2
Negative sequence network
F0
N0
Z0Va0 Ia0 = 0Zero - sequence network
Positive sequence network
F1
N1
Z1Va1 Ia1
G
Zf
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Analysis of Line – to – line Fault I = 0 af I = -Ibf cf
Vbc = Vb – Vc = Z If bf
I = 0 a0
I = -I = 1.0 ∠ 0 ° a1 a2 Z1 + Z + Z2 f
with fault impedance
Ia1 = -I = 1.0 ∠ 0 ° a2 Z1 + Z 2
I = -I = √3I ∠ -90 ° bf cf a1 V = 0 a0 V = 1.0 – Z Ia1 1 a1
V = -Z Ia2 2 a2 = Z I2 a1
Vaf = Va1 + Va2
Vaf = 1.0 + Ia1 (Z – Z2 1 ) Vbf = a2V + aVa1 a2 Vbf = a2 2+ I (aZa1 2– a Z1 )
2Vcf = aV + a Va1 a2 Vcf = a + Ia1(a2Z – aZ2 1 )
V ab = Vaf – Vbf
V = √3 (Vab a1 ∠ 30 ° + Va2 ∠ -30 °) Vbc = Vbf – Vcf
Vbc = √3 (Va1 ∠ -90 ° + Va2 ∠ 90 °) Vca = Vcf – Vaf Vca = √3 (Va1 ∠ 150 ° + Va2 ∠ -150 °)
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Double Line to Ground Fault Model
a b c
Fault point
Iaf = 0 Ibf Icf
ZfZf
Zg Ibf + Icf
Interconnection of sequence network
F0
N0
Z0Va0 Ia0
F1
N1
Z1Va1 Ia1
G
F2
N2
Z2Va2 Ia2
Zf + 3ZGZf
Zf
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Analysis of Double Line – to – ground Fault I = 0 af
Vbf = (Z +Zf g ) Ibf + Z Ig bf
Ia1 = 1.0 ∠ 0 ° (Z1 +Zf ) + (Z +Z2 f )(Z +Z0 f + 3Z ) g
Z0 + Z + 2Z2 f + 3Zg I = Ia2 a1 Z0 + Z + 3Zf g
( Z0 + Z + 3Zf g ) + (Z2 +Zf )
Ia0 = I (Za1 2 + Zf ) ( Z + Z2 f ) + (Z0 + Z +3Z ) f g
I = 0 = Iaf a0 + I + Ia1 a2
Ia0 = - (Ia1 + Ia2)
I = 0 af
2Ibf = I + a I + aIa0 a1 a2
Icf = Ia0 + aI + a2Ia1 a2
Ia = Ibf + I = 3Icf a0
Va0 = - Z I0 a0 Va1 = 1.0 – Z I1 a1
Va2 = - Z I2 a2
Vaf = V + V + Va0 a1 a2
Vbf = V + a2V + aVa0 a1 a2 2
Vcf = V + aV + a Va0 a1 a2
Vab = V – Vaf bf
Vbc = V – Vbf cf
Vca = V – Vcf af
Va0 = Va1 = Va2 = 1.0 – Z I1 a1
I = - Va2 a2
Z1
I = -Va0 a0
Z0
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Vaf = Va0 + Va1 + Va3 = 3Va1
Vbf = Vcf = 0
Vab = Vaf – Vbf = Vaf
Vbc = Vbf – Vcf = 0
Vca = Vcf – Vaf = -Vaf
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Three Phase Fault
Model
abc
Fault point
Iaf Ibf Icf ZfZf
Zg Iaf + Ibf + Icf = 3Z0
Zf
Interconnection of sequence network
F0
N0
Z0Va0 Ia0 = 0
F1
N1
Z1Va1 Ia1
G
F2
N2
Z2Va2 Ia2 = 0
Zf
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Analysis of Three phase fault (symmetrical) I = 0 ao I = 0 a2
I = 1.0 ∠ 0 ° a1 Z1 + Zf
Iaf 1 1 1 0 2 Ibf = 1 a a Ia1
2 Icf 1 a a 0
I = I = 1.0 ∠ 0 ° a1 af Z1 + Z f
I = a2 I = 1.0 ∠240 ° bf a1 Z1 + Zf
I = a I = 1.0 ∠ 120 ° cf a1
Z1 + Zf V = 0 a0 V = Z Ia1 f a1
V = 0 a2
V 1 1 1 0 af2 Vbf = 1 a a Va1
2 V 1 a a 0 cf
Vaf = Va1 = Z If a1
Vbf = a2V = Z I ∠240 a1 f a1 Vbf = aV = Z I ∠120 a1 f a1
Vab = Vaf – Vbf = V ( 1- a2 ) = √3 Z I ∠30 a1 f a12Vbc = V – V = V ( a -a ) = √3 Zbf cf a1 f I ∠-90 a1
V ca = V – V = cf af Va1 ( a -1 ) = √3 Zf I ∠150 a1
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Three phase fault (unsymmetrical)
a b c
Fault point
Zf3Zf2
Zg
Zf1
a b c
Fault point
Zf2Zf1
Zg
Zf1
a b c
Fault point
Zf3Zf2Zf1
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a b c
Fault point
Zf1 Zf2 Zf3
a b c
Fault point
Zf1 Zf2 Zf3
a b c
Fault point
Zf1 Zf2 Zf3
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a b c
Fault point
Zf
Zg
Zf
a b c
Fault point
ZfZf
a b c
Fault point
Zf
Zg
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a b c
Fault point
ZfZf
Zg
a b c
Fault point
Zf
a b c
Fault point
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a b c
Fault point
Zg
Fault point
a b c
Fault point
ZfZfZf
Fault point
a b c
Fault point Fault point
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