fault currents

8/14/2019 Fault Currents

1/17

1J.Partanenwww.ee.lut.fi/fi/lab/sahkomarkkina/

Electric Power Transmission

Fault currents

Fault current issuesThe fault current withstand capacity of components has to be

sufficient. Fault currents can be affected by network planning.

Benefits of low short-circuit current: Low temperature rise

Lower force effect Better circuit-breaking capacity

Lower hazardous voltages Lower disturbance voltages

electric arc

Benefits of high short-circuit current low impedance

stability voltage rigidity

reactive power

operation of protection

2J.Partanen

www.ee.lut.fi/fi/lab/sahkomarkkina/

Electric Power Transmission, fault currents

Nature of short-circuit current

Maximum asymmetric short-circuit current

At high voltage 1.8

R L

U( ) += tUU sin2

= switching instant after the zero of the voltage, time t

calculated from the switching instant of the short circuit

( ) +=+ tUdt

diLRi sin2 ( ) ( )

+=

sinsin

2 tL

R

etZ

Ui

( )22 LRZ +=R

L =tan

KS II = 2

factordampingX

Rf =

= currentcircuitshortinitialIK =

07.0

X

RKS II 5.2


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Direct component of the short-circuit current and damping

- = - 900 - = 00

Ik> Ik

> Ik (X < X < X) especially if there aregenerators near to the fault point

4J.Partanen



Stages of a short-circuit state:Initial state

Ik , initial short-circuit current Xd, direct initial reactance of a synchronous machine T

d , time constant (ca. 0.1 s) protection of the main grid operates in 0.10.5 s

Transient state Ik , transient short-circuit current Xd , transient reactance of a synchronous machine Td , time constant (ca. 36 s) Ik and Ik determines the temperature rise of conductors

and componentsSteady state

Ik

, steady-state short-circuit current Xd , direct synchronous reactance of a synchronous

machine


3/17


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210210UUUUUUU RRRR ++=++=

21

2

0UaUaUU

S++=

2

2

10UaUaUU

T ++=

( )TSR UUUU ++=3

10

( )TSR UaUaUU

2

13

1++=

( )TSR UaUaUU ++= 223

1

( )TSR IIII ++=3

10

( )TSR IaIaII 213

1++=

( )TSR IaIaII ++= 223

1

Symmetric system

00 0 ==++ IIII TSR

RS IIa =

021 == III RRT IIa =

2

8J.Partanen



a. b.

UR0 US0 UT0

UT1

US1 UR1

c.

UR2 US2

UT2

d.

UT1

UT2UT0

UT

US1

US2

US0US

UR1

UR0UR2UR


5/17



ER Z IR

ES Z IS

ET Z IT

UR

US

UT

ZM

( )TSRMRRR IIIZIZEU ++=

( )TSRMSSS IIIZIZEU ++=( )TSRMTTT IIIZIZEU ++=

( )

( )( )

+

+

+

=

MMM

MMM

MMM

ZZZZ

ZZZZ

ZZZZ

E

a

a

U

U

U

aa

aa 12

2

1

0

2

2

1

1

1

111

2

1

0

2

2

1

1

111

I

I

I

aa

aa

+

=

2

1

0

1

2

1

0

00

00

003

0

1

0

I

I

I

Z

Z

ZZ

E

U

U

U M

10J.Partanen



0000 )3( IZIZZU M =+=

111111IZEIZEU ==

222 IZU =

I0

U0

Z

3ZME1

I1 I2

U2U1

Z2Z1

Zero-

sequencenetwork

Positive-sequencenetwork

Negative-sequence

network


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Example: 1-phase earth fault

ER Z UR

ES Z US

ET Z UT

ZM Zf

IR

RfR IZU =

0=SI

0=TI

210 UUUIZ Rf ++=

0

0

2

2

10

21

2

0

=++

=++

IaIaI

IaIaI

00201100 3

210

IZIZIZEIZ fUUU

=+32143421321

fMf ZZZ

E

ZZZZ

E

I 3333

1

210

1

0 ++=+++=

fM

fRZZZ

EIII

++=== 103

210 III ==

if Zjo

= Zj1

= Zj2

= Z

12J.Partanen



Fault current calculation

1. Determine the equation for fault current (search inliterature) e.g. single-phase earth fault in neutral earthed system

2. Construct the equivalent circuits for the requiredcomponent networks positive-, negative- and zero-sequence network

(transformer models)

3. Simplify the equivalent circuits; Z1, Z2, Z0 in a mesh network, this is not possible in manual calculation

4. Calculate the fault current


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Equations for fault current in parallel faults, node k1. Three-phase short circuit

2. Two-phase short circuit

3. Single-phase short circuit

4. Two-phase earth fault

fkk

kf

ZZ

UI

+

=1

(If= IR= IS = IT)

fkkkk

kf

ZZZ

UjI

++=

21

3

(= IS = -IT)

IR= 0

fkkkkkk

kf

ZZZZ

UI

3

3

021 +++=

(= IR)

IS = IT = 0

( )( )( )2102102

3

33

kkkkfkkkkkk

kfkkkkR

ZZZZZZ

UZZZajI

+++

=

( )( )( )21021

02

2

3

33

kkkkfkkkkkk

kfkkkk

SZZZZZZ

UZZZajI

+++

= 0=TI

14J.Partanen



GZ

ZY ZM

Zero-sequencenetwork

ZG0 ZM0 Z0

3ZY 3ZM

Formation of the equivalent circuit of a zero-sequence network:

Zero-sequence network is connected to earthonly at the points where the neutral point isearthed either directly or through impedance

the impedance between earth and the neutralpoint and the fault impedance are multiplied bythree in the equivalent circuit

the zero-sequence network of the transformer isconnected to

earth if the neutral point is earthed, orthe transformer is delta-star connected

via earth through the transformer if thetransformer is star-star connected and ifthe neutral points are earthed

Example of the equivalent circuit of the zero-sequence network


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Example

Construct the equivalent circuit of the zero-sequence networkwith the network illustration below, when the fault occursat the point g.

G

M

M

Zn Zm Zn

X0=300

g

g

Xm0 Xj0 Xm0ZG0 Zm0 Zm0

3Zn3Zn

3Zm 3Zn

g

Zm0

Z0=Zm0+3Zn

16J.Partanen



Example

A two-phase short circuit on a line without any rotating machinesconnected to the network

The three-phase short-circuit current becomes

If there are rotating machines on the grid, the above is not valid,since for generators

Z1 Z2.

ER Z UR

ES Z US

ET Z UT

VTSR UUUU ===

UURS =

Z

U

Z

UI Vf

2

3

2==

3232

3kk

Vk II

Z

UI ==


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1. Synchronous machines

source voltage only to positive-sequence network

Z1 and Z2 inequal (Z2 low) Z0 depends on the neutral point connection

Z0 Z0g + 3Z (Z0g itself low)2. Conductors

usually Z2 Z1 Z0 is clearly higher than these

depending on conductor type

3. Transformers

usually Z2 Z1 Z0 depends on the vector group of transformer and the

impedances to earth

e.g. an non-earthed star connection breaks the equivalentcircuit of the zero-sequence network

Representations of network components inpositive-, negative- and zero-sequence networks

5.5...11

0 =Z

Z

18J.Partanen



Earthing method in the Finnish power transmission network

400 and 220 kV are effectively earthed k 1.4A part of the neutral points of 400 kV and 220 kV

transformers earthed through a reactor, a part notearthed

sufficient, reasonable fault currents e.g. fordistance relay protection

110 kV network is earthed at the neutral points

of only a few transformers. k = 1.7

fault currents sufficient for earth fault protection


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Short-circuit currents in 400 kV and 220 kV networks

The highest short-circuit currents at the supply stations oflarge power plants are of the magnitude of 15 kA

Elsewhere in the 400 kV network ca. 10 kA

When new large power plants are connected to the network,local short-circuit currents may reach a value of 20 kA

220 kV network is not extended

20J.Partanen



Short-circuit currents in 110 kV networks

110 kV maximum short-circuit currents at the centralstations in South Finland ca. 2030 kA, in NorthFinland below 20 kA

In the future, when the 400/110 kV stations arelocated closer to each other and the 110 kV ismeshed, at some stations, a short-circuit current levelof 40 kA is reached

At central stations, the rated short-circuit currents

(I1S

/S) of 110 kV buses are normally 40/100 kA


11/17



Limiting the short-circuit current

Short-circuit currents increase as the demand forelectricity increases, because more power plants are built

transformer sizes increase more lines are built

the existing line impedances decrease

Means to limit short-circuit currents: application of inductors selection of the uk values of the transformers

current-limiting fuses isolation of neutral points of transformers from earth

choosing higher voltage level

splitting up the network

22J.Partanen



Fault current calculation in a mesh netw ork

1. Description of network components lines: equivalent circuit (positive-, zero-sequence) transformers: short-circuit impedance (positive-, zero-

sequence)

vector groupimpedances to earth

generators: initial and transient reactances (positive-,negative-, zero-sequence)

neutral point connection

impedance to earth

loads: motors presented similarly as generators

2. Description of the network

Bus impedance matrix (positive-, negative-, zero-) gathered from the component data


12/17



Bus impedance matrix

ZI

I

I

I

ZZZ

ZZZ

ZZZ

U

U

U

U

NNNNN

N

N

N

=

=

=...

...

...............

...

...

2

1

21

22221

11211

2

1

Network

UN Ui U1

I1

Ii

IN

R N

1

i

iiii IZU =

ii

ii

Z

UI =

iiNN IZU =

Zii is the self-impedance = internalimpedance at a node i= Thevninimpedance, by which for instance athree-phase fault current can be

calculated

ZiN is the transfer impedance, bywhich for instance a voltage changein a node N in the case of a fault in anode i can be calculated

24J.Partanen



Example

Determine

a) fault current

b) node voltages during the fault when at the node 4 there occurs asymmetrical three-phase short-circuit, the fault resistance of which isRf = 0,05 pu. Before the fault, there is a voltage of 1.0 pu in all nodes.

c) calculate the fault current and the voltage at the fault point during thetime period when the fault is a two-phase short circuit ( [Z] 1 = [Z] 2 )

~

~

1

5

2

3 4

line reactance (pu)

1-2 0.40

1-4 0.60

1-5 0.20

2-3 0.20

2-4 0.40

3-5 0.20

[ ]

=

1770.00794.00689.00756.00852.0

0794.03668.00617.01468.00972.0

0689.00617.00933.00732.00445.0

0756.01468.00732.01928.00780.0

0852.00972.00445.00780.01258.0

jZ


13/17



Symmetrical three-phase short circuit

Fault occurs at the node 4.The self-impedance Zkk1 represents the total impedance betweenthe fault point and the reference node. It is obtained from thediagonal of the Z-matrix: Zkk1=j0.3668

Fault impedance Zf = Rf = 0.05 and UV = 1.00

a.

b. The fault causes voltage changes at the nodes of the network.Also these are obtained with the elements of the Z-matrix

Ui = -Zik1If1 ; when i = k, we obtain the voltage change at thefault point

fkk

Vff

ZZ

UII

+==

1

1

positive-sequence component of the faultcurrent also represents the total faultcurrent (symmetry)

=+

== 24.82701.2

05.03668.0

00.11

jII ff

continued...

26J.Partanen



Symmetrical three-phase short circuit

This change is superposed on the node voltage that prevailed before thefault.

Voltages during the fault

Elements Zik1, when i k, are transfer impedances. They describe theeffect of the source (fault) current at the node in question on other nodes.The higher the impedance, the more the voltage changes at the othernode.

c.

=== 2.82135.01338.00183.024.82701.23668.00.14 jjU f

=== 74.2741.00354.07399.024.82701.20972.00.11 jjU f

=== 04.5609.00535.06071.024.82701.2146.00.12 jjU f

00 =I

{

0

442441

421

fZZZUII

++==

continued...


14/17



Voltage components at the node 4 during the fault:

1

2

1

22

23

0

1

1

1110

IjI

I

I

aa

aa

I

I

I

I

I

f

f

f

T

S

R

=

=

=

puj

jIf 36.23668.02

13 =

=

}

00

0

044040 == IZU 144441 IZUU = 24442 0 IZU =(before fault U40=U42=0)

5.03668.02

13668.0141 =

=j

jU 5.03668.02

13668.042 =

=j

jU

=

=

5,0

5,0

1

5,0

5,0

0

1

1

111

2

2

4

4

4

aa

aa

U

U

U

T

S

R

28J.Partanen



Construction of the Z-matrix

By inverting the Y-matrix; laborious calculation, requires

plenty of storage capacity By constructing the Z-matrix with an algorithm; requires

plenty of storage capacity

By constructing the required part of the Z-matrix applyingthe sparse-matrix technique; effective, small amount of

storage capacity required


15/17



Construction of the Z-matrix with an algorithmA branch from the reference node to a new node k:

Zkk = ZbranchZik = Zki = 0 , i k (if added to an already existing matrix)

A branch from the node k to a new node j:

Zij = Zik , i jZjj = Zkk + Zbranch

A branch from the node k to an existing node j:

an extra s-th vertical and horizontal row added

Zis = Zik - Zij , i jZss = Zkk + Zjj - 2Zkj + Zbranch the extra vertical and horizontal row eliminated

A branch from the existing node k to the reference node: special case of item c

(Zij = Zjj = Zkj = 0)

ss

jsis

oldijnewijZ

ZZZZ = )()(

30J.Partanen



Problem

Construct an impedance matrix for the network illustrated

below

How does the matrix change, if a branch, for which

Z= 100 , is added between the nodes 2 and 3?

~Z=300

Z=100

Z=50

1

3

2


16/17


Electric Power Transmission, fault currentsApplication of sparse-matrix technique to fault currentcalculation

In theory, the nodal impedance matrix is obtained by invertingthe nodal admittance matrix.

Z= Y-1In fault current programs, the Z-matrix row corresponding to thefault is solved, i.e., the matrix is not inverted (storage capacity,speed).The Z-matrix row corresponding to the fault can be obtainede.g. by the bi-factor method.The idea of the method: 2n factor matrices are constructed so thatcondition (1) is met.

1. LnLn-1Ln-2...L2L1 Y R1 R2...Rn-2Rn-1Rn = I (I = unit matrix) we obtain the Z-matrix as followsZ = Y-1 = R1R2...Rn-2Rn-1RnLnLn-1Ln-2...L2L1

32J.Partanen



Application of sparse-matrix technique to fault currentcalculation

Factor matrices Ri and Li are constructed step by step so that at each step,

one row and one column is eliminated from the admittance matrix.

Elimination takes place as follows:

Y0 = Y Y1 = L1Y0R1 Y2 = L2Y1R2 ....... Yn = LnYn-1Rn = I

The structure of the matrices Ri and Li is the following:

Although the matrices Li and Ri are very rare, the Z-matrix resulting astheir product is a full matrix. Thus, when applying this method,a whole Zmatrix is not constructed in one go, but only the elements

required at the fault situation in question are formed. If the fault is at thenode i, to calculate the effects of the fault, the self-impedance and the

transfer impedances of the corresponding column are required.They are obtained by multiplying the impedance matrix by the i-th columnof the unit matrix according to the following equation.

1

11

11

1

XX

X

1

32

n

.

.

.

.

.

21 . . . i . n

Li=

1

11

11

11

X X0

1

32

n

.

.

.

.

.

21 . . . i . n

Ri=


17/17



Application of sparse-matrix technique to fault current calculation

In fault calculation, the factor matrices Ri and Li of the Z-matrices

of the component networks are stored in the memory, and the self-

and transfer impedances are calculated with the above equation.

When the fault point transfers to another node, new self- and

transfer impedances are determined.

=

=

=

0

...

1

...

0

0

......

0

...

1

...

0

0

...

...1221

2

1

LLLRRRZ

Z

Z

Z

Z

Z nn

ni

ii

i

i

i

fault currents

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