Bahir Dar University

Institute of Technology

School of Computing and Electrical Engineering

Eeng-5161: Power System II

Nov 2011

1. FAULT ANALYSIS1. 1 Introduction

● Fault (short circuit ) current calculation is an important part of power system (PS) design, which enable the system engineer to set relay protection and selection of various equipments such us circuit breaker, bus bar etc.

● Short circuit (SC) occur in PS when equipment insulation fails, due to:- system over voltages (lighting or switching); insulation contamination; other mechanical causes.● SC current may be (Isc)>>normal current Inom and if it is allowed to persists, may cause thermal and mechanical damage to equipments. It is therefore necessary to remove faulted sections of a system from service as soon as possible.

● commonly occurring types of fault are (Fig.1.1):- three phase, phase- to-phase, two-phase-to ground, and phase-to ground fault. From these phase-to ground is the most frequent and three-phase is the most sever.

a) Three-phase S.C (k (3))b) L-L-L, E (k (3E))

C) L-L. (k (2))

d) L-L, E (k (2E))

e) L-E (k (1))

f) L-E fault with fault impedance

Fig .1.1. Common types of fault or short circuit (S.C). L-Line, E-earth

or

Rf

● Current flowing to faulted place is termed fault current (If). The fault MVA is frequently used as a rating (the fault MVA = VLIf *10-6, where VL is the nominal line voltage of the faulted part). The fault MVA is often referred to as the fault level. ● fault current calculations can be divided into the

following two main types:

All three lines are short-circuited, when the network remains balanced electrically. For these calculations, normal single-phase equivalent circuits may be used as in ordinary network calculation;

Faults other than three-phase short circuits when the network is electrically unbalanced. To facilitate these calculations a special method for dealing with unbalance network is used, known as the method of symmetrical components.

● The main objects of fault analysis :

1. To determine maximum and minimum three-phase short circuit current;2. To determine the unsymmetrical fault current for single and double line to ground faults, and sometimes for open circuit faults;3. Investigations of operation of protective relays; 4. Determination of rated rupturing capacity of breakers;5. To determine fault-current distribution and bus bar-voltage levels during fault condition.

1.2 Symmetrical Fault Analysis

1.2.1 Series R-L circuit transient

● Closing switch SW(Fig.1.2) at t = 0 represent bolted three- phase short circuit.

Fig.1.2. Current-voltage wave forms in R-L circuit transient

Using KVL for the circuit,

α)tVsin(2Ri(t)dt

di(t) L t>0 (1.1)

Solution for (1.1) is

T

t

θ)esin(αθ)-αtsin(Z

V2(t)dc

i(t)ac

ii(t) A (1.2)

Whereθ)-αtsin(

ZV2(t)

aci T

t

θ)esin(αZ

V2(t)dc

i

Ω2(X)2R)2RZ 2 L(ω RωL1tanθ =

RX1tan

fL2X

ωLX

RLT Eq.(1.3)…(1.6)

The total fault current in (1.2), called the asymmetrical fault current (fig.1.2). iac(t) – called symmetrical or steady state fault current and sinusoid The dc offset current idc(t) decay exponentially with time constant T = L/R.

● The rms ac fault current is Iac = V/Z. The magnitude of the dc offset which depend on α, varies from 0 when α = θ to when α = (θ/2). For α = (θ - /2) maximum fault current is

T

t

e)2π-tsin(

ZV2i(t) A (1.7)

acI2

The RMS of i(t) is

Tt2

2e1I

2

Tt-

eac

I22ac

I2(t)dc

I2ac

I(t)rms

I ac

(1.8)

Eq.(1.8) can be written as

Irms() = K()Iac (1.9)

Where K() =

RX

4π

2e1

per unit (1.10)

t= t*f - is time in cycle. K() is called asymmetrical factor and t varies from 1 to 3 as it can be seen from (1.10).

1.2.2 Balanced fault calculation in power system

● The balanced three-phase fault in unloaded synchronous machine can be modelled by the series R-L circuit of fig.1.2, if time varying inductance L(t) or XL(t) is employed. Oscillogram of current in one of phase for 3-ph fault is in Fig.1.3.

Fig.1.3

● As we know from machine modelling there are three reactances Xd, Xd’’ and Xd’. There for different currents flow from time of occurrence of faults. These currents are: Sub-transient current which flow immediately after the occurrence of fault (1)

''d

X

gE

''I(0)Iac 'd

X

gE

'I d

X

gE

I I’’>I’>I(1) (2) (3)

Transient current which flow few cycle later the occurrence of fault (2)The sustained (steady state) fault current (3)

● The sub-transient current I’’ is initial symmetrical current which does not include dc components. To consider dc components some coefficient (value of 1.4 to 1.8)is used.

● For practical system a simplified fault current calculation procedure called E/X method is used. In this method the Thevenin theorem is applied. Then the fault current If is

If = Vf/Xth or If = I/Xth p.u.

Where Vf is pre-fault voltage at point of fault and its value is taken 1p.u.; I-normal load current. The three phase short- circuit volt-amperes

P.UXB

MVA

x.p.u

IL

V3

fI

LV3

MVAF

● The following assumption will be made in practical short circuit current calculations:

All generated voltages of synchronous machines are equal & in phase and represented by sub-transient reactances, hence no load is considered for fault calculation. Generally, load has negligible impact on the faulted phase.

Neglect resistance except at the lower voltage where resistance is larger.

All shunt reactance’s neglected (loads, charging & magnetizing reactances).

All mutual reactance’s neglected.

Induction motors are either neglected (<38kW) or represented as synchronous machines.

1.2.3 The bus impedance matrix for fault calculation

● For general network fault calculation Zbus is used. By definition,

1bus

Ybus

Z (1.11)

Zbus is symmetrical

Using Zbus, the sub-transient fault current can be calculated as:

kkZ

FV''

FkI (1.11)

Where, VF is pre-fault voltage at bus k, Zkk is self impedance of bus k.

The voltage at any bus n during fault can be calculated:

FV

kkZ

nkZ

1n

V

n = 1, 2 …N (1.12)

Where, Znk is transfer impedance between bus k and n.

The bus impedance equivalent circuit is shown in fig. 1.4.

Fig.1.4. Bus impedance equivalent circuit (rake equivalent)

1.3 Unsymmetrical Fault (short circuit)

1.3.1 Method of symmetrical component

● To solve problems in unsymmetrical conditions of power system, method of symmetrical component is used. Any unbalanced phasors of a three-phase system can be resolved in to three-balanced system of phasors. The balanced sets of components are:

+ve sequence ;- equal in magnitude and 120o phase displacement; a, b, c phase sequence as original phasers. This sequence is always representing by subscript 1 (Va1, Ia1 etc.)

-ve sequence ;– equal in magnitude and 120o phase displacement; a, c, b phase sequence and have subscript 2 (Va2,Ia2, etc)

Zero sequence components consisting of three phasors equal in magnitude and with zero phase displacement from each other and have subscript 0 (Va0,Ia0 etc) .● Each of the original unbalanced phasors is the sum of its components (Fig.1.5) and the original phasors expressed in terms of their components are:

c0V

c2V

c1V

cV

b0

Vb2

V b1

Vb

V a0

Va2

V a1

Va

V

(1.13)

Using operator a, which is rotating the phasor 120o anticlockwise.

a=1120o= –0.5+j0.866;a2=1240o= -0.5-j0.866 ;

a3= 1 3600= 100=1+jo.

Fig.1.5

Vb1= a2Va1; Vb2 = aVa2; Vb0 = Vao

Vc1 = aVa1; Vc2 = a2Va2; Vco = Vao (1.14)

Substituting in eq (1.13) yield;

a0V

a2Va

a1V

cV

a0

Va2

aV a1

Vb

V

a0V

a2V

a1V

aV

2

2

a

a(1.15)

By solving the eq (1.15) for sequence component:

cV

bVa

aV

31

a2V

c

Vb

aV a

V31

a1V

cV

bV

aV

31

a0V

2

2

a

a

(1.16)

Sequence components for other phases can be found byconsidering eq.(1.14.)●The above equation similarly works for current too.

In three-phase system the sum of line current equal to current through neutral In.

In the absence of neutral In = 0. For connected system I0=0.

In=Ia+Ib+IC ; Ia0=(1/3)In In=3Ia0

Zero- sequence network is always different depending on way of system grounding. Z0=Z1 for transformer; Z0=2-3.5Z1 for line and Z0 of generator is small. See fig.1.7, for zero sequence current flow through transformer.

1.3.3. Unsymmetrical fault on power system

● Sequence component voltage drop equation can be written as:

0Z

a0I - 0

a0V

2Z

a2I - 0

a2V

1Z

a1I -

fV

a1V

(1.17)

Ia1Vf

3Zn Zg0

z0

F

Va0

Va2

+F

-

Va1

-

+-

Where F is point of fault

I0

Fig.1.6. equivalent circuit for sequence impedance network

Fig.1.7 Zero sequence equivalent circuits of 3-phase transformer

Single Line- to- ground fault on a power system

For 1-E (K(1)), Ia1 = Ia2 = Ia0 = Ia/3

Then

0Z

2Z

1Z

fV

a1I

(1.18) c

a

b

With Zf

F3Z

0Z

2Z

1Z

fV

a1I

(1.19)

Finally fault current isa1

3I(1)a

I (1.20)

●It can be seen from eq.(1.19),that I = 0, when ZfYh. For ungrounded system no path exist for current flow. Sequence network connection is shown in fig.1.8.

Vf Z1

Z2 Z0

Fig.1.8. Sequence connection for single-line to ground fault

Ia1

- +

Ia1 Ia2Ia0

Line-to- Line fault on a power system

2Z

1Z

fV

a1I

(1.21)

Ia=0

Ib

Ic

With ZF ,

FZ

2Z

1Z

fV

a1I

(1.22)

The fault currents: Ia(2) =0 ;Ib(2) = -3 Ia1 ;Ic(2) = 3 Ia1

In most case Z1=Z2 ; (3)f

I2

3

12Z

fV3(2)

fI

Finally line-to-line fault current

(3)f

I 0.866 (2)f

I (1.23)

Fig.1.9. Sequence network connection for line-to-line fault

Z1Ia1 Ia2

Va1 Va2 Z2-

+Vf

- +

+ -

Double line to ground fault on a power system

Vb=Vc= 0; Ia=0

Va0 =1/3(Va+0+0) =Va1=Va2

2Z

0Z

2Z

0Z

1Z

fV

a1I

(1.24) IF

Ib

Ic

From eq.(1.24), the network sequence Z connection is in fig.1.10

Fig.1.10. Sequence network connection for double line-to-ground fault

+

Z1Ia1

Va1

Va2 Z2-

+Vf

-+

-

Z0 Va0

+

-

Ia0Ia2

1.3.4. Sequence bus impedance matrices

Single line-to-ground fault:

(1.25) F3Z

2kkZ

1kkZ

0-kkZ

fV

2kI

1kI

0kI

Line-to-line fault:

FZ

2kkZ

1kkZ

fV

2kI

1kI

(1.26)

00k

I (1.27)

Double line-to-ground fault:

F3Z

0kkZ

2kkZ

)F

3Z0kk

(Z2kk

Z

1kkZ

fV

1kI

(1.28)