Short Circuit Lecture Using EDSA ETAP PSSE Matlab

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

a Z0 c Z0 Z0

Z0  N0  b

a Z0 c Z0 Z0

3Zn Zn Z0

 N0  b

a

Z0

c Z0 Z0

 N0

Z0  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, Z 0, is the impedance looking into the zero sequence networks from the fault point.

 Example: Draw the Zero sequence network of the power system shown ∆

Y grounded - Y grounded

-∆

G

G

Y grounded

Reactance grounded

Y grounded - Y grounded

∆

-∆

Solution:

 j0.2

2

 

3

j0.3

1

 j0.06

j0.3

4

 

 j0.25

5

j0.5

 j0.06j0.35

6  j0.09

N0 (Neutral or Zero-potential bus)

 Example: Draw the Zero sequence network of the power system shown Y ungrounded - Y grounded

Y ungrounded - Y ungrounded

G

G

Reactance grounded

Y ungrounded

Y ungrounded - Y ungrounded

∆

-∆

Solution:

 j0.2

 j0.06

 j0.25

 j0.3

 j0.3

 j0.5

 j0.35

 j0.06

 j0.09

N0 (Neutral or Zero-potential bus)

∆

Y grounded- Y grounded

-∆

G

G

Y ungrounded

Reactance grounded

∆

Y grounded - Y grounded

∆

Y ungrounded - Y grounded

- Y ungrounded

-∆

G Y grounded 2

 j0.2

 

3

j0.3

1

 j0.06

j0.3

4

 j0.35

 j0.25

5

6

 j0.5  j0.09

 j0.09

 j0.09

 j0.09

N0 (Neutral or Zero-potential bus) B

B

 j0.06

 j0.09

∆

Y grounded - Y grounded

- Y ungrounded

Y ungrounded

G

G

∆

- Y grounded

∆

Y ungrounded - Y grounded

∆

- Y grounded

G

2

-∆

Solidly grounded

3

1

4

5

6

N0 (Neutral or Zero-potential bus) B

B

 Problem: Draw the Positive, Negative and zero sequence model for the power system shown.

∆

Y grounded - Y grounded

-∆

1

4 3

2

G

G 5

6 Reactance grounded

Y grounded

Y grounded - Y grounded

∆

∆

∆

- Y grounded

-∆

- Y grounded

G

G

∆

- Y grounded

∆

∆

∆

- Y grounded

G

- Y grounded

- Y grounded

Solution:

Reactance Diagram Simplification

2

 j0.2

3

 

j0.15

1

4

 j0.2

 

 j0.25

j0.22

5 G

j0.3

 j0.2j0.35

6

1.0 angle 0°

G

N1 (Neutral or Zero-potential bus)

2

 j0.2

3

 

j0.15

1

 j0.14

j0.3

4

 

 j0.25

j0.22

5

 j0.14 j0.35

6

N2 (Neutral or Zero-potential bus)

2

 j0.2

 

3

j0.3

1

 j0.06

j0.3

4

 

 j0.25

5

j0.5

 j0.06j0.35

6  j0.09

N0 (Neutral or Zero-potential bus) B

B

1.0 angle 0°

Positive Sequence Network   j0.35

 j0.3

3

1

4

 j0.2

G

 j0.82

 j0.2

1.0 angle 0°

G

1.0 angle 0°

N1

3

3

 j0.0.0714

 j0.0.0714 4

1  j0.1952

 j0.1673  j0.2

 j0.2

G

1.0 angle 0°

1.0 angle 0°

 j0.3673

 j0.3952

G

G

N1

1.0 angle 0°

1.0 angle 0°

G

N1

3

3

3  j0.0.0714

 j0.0.0714

 j0.0.2618

 j0.3952

 j0.3952

 j0.3673

 j0.1904

1.0 angle 0° 1.0 angle 0°

G

1.0 angle 0°

G

G N1

N1 B

N1 B

B

B

 Negative Sequence Network   j0.35

 j0.3

3

1

4

 j0.14

 j0.82

 j0.14

N2

3

3

 j0.0.0714

 j0.0.0714 4

1  j0.1952

 j0.1673

 j0.14

 j0.3073

 j0.3352

 j0.14

N2

N2

3

3

3  j0.0.0714

 j0.0.0714

 j0.0.2317

 j0.3352

 j0.3952

 j0.3073

 j0.1603

N2 N2 B

N2 B

B

B

Zero Sequence Network 

2  j0.3

 j0.2

3 3

 j0.06  j0.0.56

N0

N0

 Problem:

Draw the connection diagram of Positive, Negative and Zero sequence network of the 4 bus shown.

G

G ∆

- Y grounded

∆

- Y grounded

Short Circuit Current under Transient Condition

G

G

Three Phase Symmetrical Symmetrical Fault

G

Consider a Three phase Fault occurs near the terminal of the generator 

e= VS sin (ωt + α) α

Derivation of Short Circuit Current under Transient Condition VR  + VL = VS Ldi

R+

= VS sin (ωt + α)

dt Ldi

R+

= VS [sin ωt cosα + cosωt sinα]

dt Taking the Laplace Transform: Transform: ω

ω

RI(s) + L [sI(s) – I(0) ] = VS [cosα (

2

2

s +ω

) + sinα (

2

2

s +ω

)]

I (0) = 0

Is [R + Ls] = VS

[

Is [ R + Ls] = VS

[

Is = VS

ω

[

ω

cosα

2

2

s sinα +

s +ω ω

2

cosα + s sinα s2 + ω2

cosα + s sinα

2

]

2

s +ω

]

]

2

(s + ω ) (R + Ls)

Simplification using partial fraction method:

Is = VS/ L

[

As + B

]

C +

(s2 + ω2 )

(s + R/L) R/L)

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 = √ (R 2 + (ωl)2

θ

= arctan ( ωL/R)

Is = VS/L

Is = VS/L

[ [

Ls/Z [sin (α – θ)]+ 2

ωl/Z

(s +

{

sin (α – θ)

s s2 + ω2

2

ω

+ (s + R/L) R/L)

)

} + cos (

Using the Inverse Laplace Transform:

L/Z [sin (α – θ)]

[cos (α – θ)]

α

– θ)

{

s s2 + ω2

}+

sin (α – θ) (s + R/L) 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

Symmetrical

Unsymmetrical

Short Circuit Current near the Generator 

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 positivesequence 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 Z 0 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 Fault point

Model

a b c Zf 

Iaf 

I bf  = 0

Icf  = 0

Interconnection of sequence network 

F0 Va0

Z0

Ia0

Zero - sequence network 

N0

F1 3Zf 

Va1

Z1

Ia1

Positive sequence network 

G N1

F2 Va2

Ia2

Z2

N2

 Negative sequence network 

Analysis of Single Line – to - Ground Fault

1.0 ∠ 0 ° Z0 + Z1 + Z2 + 3Zf 

Ia0 = Ia1 = Ia2 =

Iaf  I bf  Icf 

=

1 1 1

1 a2 a

1 a a2

I a0 Ia1 I a2

Iaf = Ia0 + Ia1 + Ia2 Iaf = 3Ia0 = 3Ia1 = 3Ia2 Vaf = Zf Iaf  Vaf = 3Zf Ia1 Vaf = Va0 + Va1 + Va2 Va0 + Va1 + Va2 = 3Zf Ia1 Va0 Va1 Va2

=

0 1.0 ∠ 0 ° 0

Z0 - 0 0

0 Z1 0

Va0 = - Z0 Ia0 Va1 = 1.0 – Z1 Ia1 Va2 = -Z2 Ia2 Vaf V bf  Vcf

=

1 1 1

1 a2 a

V bf = Va0 + a2Va1 + aVa2 Vcf = Va0 + aVa1 + a2Va2

1 a a2

Va0 Va1 Va2

0 0 Z2

I a0 Ia1 I a2

Line-Line Fault Model Fault point

a b c

Iaf  = 0

Zf 

I bf 

Icf 

Interconnection Interconnection of sequence network 

F0 Va0

Ia0 = 0

Z0

Zero - sequence network 

N0

Zf 

F1 Va1

Z1

Ia1

F2

Positive sequence network 

Va2

Ia2

Z2

G N1

N2

 Negative sequence network 

Analysis of Line – to – line Fault

Iaf  = 0 I  = bf -I cf  V bc = V b – Vc = Zf Ibf  Ia0 = 0 Ia1 = -Ia2 =

1.0 ∠ 0 ° Z1 + Z2 + Zf 

with fault impedance

Ia1 = -Ia2 =

I 

1.0 ∠ 0 ° Z 1 + Z2

= -bf I = √cf3I ∠ -90 a1 °

Va0 = 0 Va1 = 1.0 – Z1 Ia1 Va2 = -Z2 Ia2 = Z2 Ia1

Vaf = Va1 + Va2 Vaf = 1.0 + Ia1 (Z2– Z1 ) V bf = a2Va1 + aVa2 V bf = a2 + Ia1(aZ2– a 2Z 1 ) Vcf = aVa1 + a2Va2 Vcf = a + Ia1(a2Z2– aZ1 )

Vab = Vaf  – V bf  Vab = √3 (Va1 ∠ 30 ° + Va2 ∠ -30 °) V bc = V bf  – Vcf  V bc = √3 (Va1 ∠ -90 ° + Va2 ∠ 90 °) Vca = Vcf  – Vaf  Vca = √3 (Va1 ∠ 150 ° + Va2 ∠ -150 °)

Double Line to Ground Fault Model Fault point

a b c Zf 

Zf 

I bf 

Iaf  = 0

Icf  Zg

I bf  + Icf 

Interconnection of sequence network 

Zf  + 3ZG

Zf 

F0 Va0

Ia0

Z0

N0

Zf 

F1 Va1

Z1

Ia1 G

F2 Va2

Ia2

Z2

N2 N1

Analysis of Double Line – to – ground Fault

Iaf  = 0 V bf = ((Z Zf +Zg ) I bf  + Zg Ibf 

Ia1 =

1.0 ∠ 0 ° (Z1 +Zf ) + (Z2+Zf )(Z0 +Zf + 3Zg) Z0 + Z2 + 2Zf + 3Zg

Ia2 = Ia1

Z0 + Zf + 3Z 3Zg ( Z0 + Zf + 3Z 3Zg ) + (Z2 +Zf )

Ia0 = Ia1

(Z2 + Zf ) ( Z2 + Zf ) + (Z0 + Zf +3Zg)

Iaf = 0 = Ia0 + Ia1 + Ia2 Ia0 = - (Ia1 + Ia2) Iaf  = 0 I bf = Ia0 + a2Ia1 + aIa2 Icf = Ia0 + aIa1 + a2Ia2 Ia = I bf  + Icf = 3Ia0

Va0 = - Z0 Ia0 Va1 = 1.0 1.0 – Z1 Ia1 Va2 = - Z2 Ia2

Vaf = Va0 + Va1 + Va2 V bf = Va0 + a2Va1 + aVa2 Vcf = Va0 + aVa1 + a2Va2

Vab = Vaf – Vbf V bc = Vbf – Vcf Vca = Vcf – Vaf

     

Va0 = Va1 = Va2 = 1.0 1.0 – Z1 Ia1 Ia2 = - Va2 Z1 Ia0 =

-Va0 Z0

Vaf = Va0 + Va1 + Va3 = 3Va1 V bf = Vcf = 0

Vab = Vaf  – V bf  = Vaf  V bc = V bf  – Vcf  = 0 Vca = Vcf  – Vaf  = -Vaf 

Three Phase Fault Model

Fault point

a b c Iaf 

Zf 

I bf 

Zf 

Zf 

Zg

Icf 

Iaf  + I bf  + Icf  = 3Z0

Interconnection of sequence network 

F2

F0 Va0

Ia0 = 0

Zf 

Va2

Z0

F1

N0 Va1

Z1

Ia1 G

N1

Ia2 = 0

Z2

N2

Analysis of Three phase fault (symmetrical)

Iao = 0 Ia2 = 0 Ia1 =

Iaf  I bf  Icf 

1.0 ∠ 0 ° Z1 + Zf 

=

Ia1 = Iaf  =

I 

= a2bfI =

1 1 1

1 a2 a

1 a a2

0 Ia1 0

1 a a2

0 Va1 0

1.0 ∠ 0 ° Z1 + Zf  a1

1.0 ∠240 ° Z1 + Zf  1.0 ∠ 120 ° Z1 + Zf 

Icf = a Ia1 =

Va0 = 0 Va1 = Zf Ia1 Va2 = 0 Vaf  V bf  Vcf 

=

1 1 1

1 a2 a

Vaf = Va1 = Zf Ia1 V bf = a2Va1= Zf Ia1∠240 V bf = aVa1 = Zf I a1 a1∠120 Vab = Vaf  – V bf  = Va1 ( 1- a2 ) = √3 Zf Ia1∠30 V bc = V – V =bV (caf 2 -a )a1= √3 Z Ia1∠-90 f   f Vca = Vcf – Vaf =  Va1 ( a -1 ) = √3 Zf  Ia1∠150

Three phase fault (unsymmetrical) (unsymmetrical) Fault point

a b c Zf1

Zf2

Zf3

Zg

Fault point

a b c Zf1

Zf1

Zf2

Zg

Fault point

a b c Zf1

Zf2

Zf3

Fault point

a b c Zf1

Zf2

Zf3

Fault point

a b c Zf1

Zf2

Zf3

Fault point

a b c Zf1

Zf2

Zf3

Fault point

a b c Zf 

Zf 

Zg

Fault point

a b c Zf 

Zf 

Fault point

a b c Zf 

Zg

Fault point

a b c Zf 

Zf 

Zg

Fault point

a b c Zf 

Fault point

a b c

Fault point

Fault point

a b c

Zg

Fault point

Fault point

a b c Zf 

Zf 

Fault point

a b c

Zf 

Fault point