Analysis of Faulted Power System (Shunt Fault)

Sequence Model for Transformer

Roel B. Calano

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

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.

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

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

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

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)

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

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

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°

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

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

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

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)

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.

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

Three Phase Short Circuit at the Field

Line to Line Short Circuit at Phase b

Line to Line Short Circuit at the field

Line to Ground Short Circuit at Phase a

Line to Ground Short Circuit at the Field

Different Level of Short Circuit Current with respect to time

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

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%.

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

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

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

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 °)

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

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

Vaf = Va0 + Va1 + Va3 = 3Va1

Vbf = Vcf = 0

Vab = Vaf – Vbf = Vaf

Vbc = Vbf – Vcf = 0

Vca = Vcf – Vaf = -Vaf

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

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

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

a b c

Fault point

Zf1 Zf2 Zf3

a b c

Fault point

Zf1 Zf2 Zf3

a b c

Fault point

Zf1 Zf2 Zf3

a b c

Fault point

Zf

Zg

Zf

a b c

Fault point

ZfZf

a b c

Fault point

Zf

Zg

a b c

Fault point

ZfZf

Zg

a b c

Fault point

Zf

a b c

Fault point

a b c

Fault point

Zg

Fault point

a b c

Fault point

ZfZfZf

Fault point

a b c

Fault point Fault point