Balanced Faults

ELE B7 Power Systems ELE B7 Power Systems Engineering Engineering

Balanced (Symmetrical) FaultsBalanced (Symmetrical) Faults

Fault Analysis• Fault?

When the insulation of the system breaks down or a conducting object comes in touch with a live point, a short circuit or a fault occurs.

• Sources?• This breakdown can be due to a variety of different factors

• lightning• wires blowing together in the wind• animals or plants coming in contact with the wires• salt spray or pollution on insulators

Fault Types• There are two classes of faults

– Balanced faults (symmetric faults): A Fault involving all the three phases

• system remains balanced; • these faults are relatively rare, but are the easiest to analyze so

we’ll consider them first.– Unbalanced faults (unsymmetric faults): A fault involving only one

or two phases• The majority of the faults are unsymmetrical. • system is no longer balanced; • very common, but more difficult to analyze

• The most common type of fault on a three phase system is the single line-to-ground (SLG), followed by the line-to- line faults (LL), double line-to-ground (DLG) faults, and balanced three phase faults

Fault Types

IFb

Transformer

IFa

IFc

IF

Transformer

IF

Single Line to Ground Fault

IF

IF

IF

Transformer IF

Line to Line Fault

IF

IFb

Transformer

IFIF

IF

IFc

Double line to ground Fault

Three Phases Fault (Balanced Fault)

Fault Analysis• Fault Analysis involve finding the voltage and current

distribution throughout the system during fault condition.• Fault currents cause equipment damage due to both

thermal and mechanical processes• Why do we need that?

– To adjust and set the protective devices so we can detect any fault and isolate the faulty portion of the system.

– To protect the human being and the equipment during the abnormal operating conditions.

– need to determine the maximum current to insure devices can survive the fault

Symmetrical Faults (Balanced Faults)

• It is the most severe fault but can be easily calculated

• It is an important type of fault

• The circuit breaker rated MVA breaking capacity is selected based on the three phase short circuit MVA.

A Fault involving all the three phases.

Transformer

Three Phase

Short Circuit

1. Synchronous machines are represented by a constant voltage sources behind subtransient reactances. The EMF of all generators are 1 per unit making zero angle.

oV 01

The following assumptions are made in three phase fault calculation :

This means that the system voltage is at its nominal value and the system is operating at no-load conditions at the time of fault.

Therefore, All the generators can be replaced by a single generator since all EMFs are equal and are in phase.

2. Transformers are represented by their leakage reactances

Winding resistances, shunt admittances and Δ-Y phase shifts are neglected

3. The shunt capacitances of the transmission

Lines

are neglected.

4. The system resistances are neglected

and only

the inductive reactance of different elements are

taken into account.

Symmetrical

Fault Calculation

Transformer

Three Phase

Short Circuit

Simple CircuitsLoad Ignored

Use Thevenin's Equivalent

Simple CircuitsLoad not Ignored

Use Thevenin's EquivalentFind Pre-fault voltage

Large CircuitsConstruct

The Bus Impedance Matrix

Three Phase FaultCalculation

2

3

1

1. Simple Circuits and Load is ignored

1. Draw a single line diagram for the system.

The Calculations for the three phase fault are easy because the

circuit is completely symmetrical and calculations can be done

for only one phase.

T1

T2

G1

G2

1

2

3 phasefault e

Steps For Calculating Symmetrical Faults:

2. Select a common base and find out the per unit reactances of all generators, transformers, transmission lines, etc.

4. Reduce the single line reactance diagram by using series, parallel and Delta-Why transformations keeping the identity of the fault point intact. Find the total reactance of the system as seen from the fault point (Using Thevenin’s

Theorem)

3. From the single line diagram of the system draw a single line reactance diagram showing one phase and neutral. Indicate all the reactances, etc. on the single line reactance diagram.

GG15.0j

20.0j 16.0j

10.0j23.0j

GG

01

01 23.0j

5. Find the fault current and the fault MVA in per unit.

Convert the per unit values to actual values.

6. Retrace the steps to calculate the voltages and the currents throughout different parts of the power system

T1

T2

G1

G2

X=0.2 ohm/phase/km

1

2

50 MVA, 11 kV10% Reactance


100 MVA, 11/132 kV10% Reactance


200 km

3 phasefault e

Example:

A three phase fault occurs in the system as shown in the Figure.Find the total fault current, the fault level and fault current supplied by each generator.

Step 1: Draw a single line diagram for the system.

The single line diagram for the system is given in the example as shown.

Solution:

T1

T2

G1

G2

X=0.2 ohm/phase/km

1

2





200 km

3 phasefault e

Step 2: Select a common base and find the per unit reactances of all generators, transformers, etc.

Select the common base as:

100 MVA (100,000 kVA)

11 kV for Transformer low voltage side (LV)

132 kV for Transformer high voltage side (HV)

T1

T2

G1

G2

X=0.2 ohm/phase/km

1

2





200 km

3 phasefault e

kVVBase 11

kVVBase 132

T1

T2

G1

G2

X=0.2 ohm/phase/km

1

2





200 km

3 phasefault e

kVVBase 11

kVVBase 132

G1

G2

15.01 jXG The per unit reactance

2.050

100*1.02 jjXG The per unit reactance

1.01 jXT The per unit reactance

16.050

100*08.02 jjXT The per unit reactance

23.0)132(

100)200*2.0( 2 jkVMVAj

ZX

XBase

LINELINE The per unit reactance

T1

T2

TL

From the single line diagram of the system draw a single line reactance diagram showing one phase and neutral. Indicate all the reactances, etc. on the single line reactance diagram.

T1

T2

G1

G2

1

2

3 phasefault e

G15.0j

20.0j 16.0j

10.0j23.0j

G

01

01 23.0j

G25.0j

36.0jG

01

01115.0j

Find the total impedance

(reactance) of the system as seen from the fault side.

25.0j

36.0j115.0j

pujX

jjjjjX

Total

Total

2625.0

115.036.025.036.0*25.0

pujj

IF 8095.32625.0

01

AI sidekVBase 4.437132*31000*100)( 132

AjI oActualF 9027.16664.437*8095.3)(

01

2625.0j

LLBase

BaseLbase

V

SI

||3

|||| 3

BasepuFActualF III *)()(

The Fault Level is:

100)( BaseMVA

95.380100*8095.3)( LevelActualMVA

puMVA Level 8095.3)(

T1

T2

G1

G2

X=0.2 ohm/phase/km

1

2





200 km

3 phasefault e

1)( GFI

2)( GFI

FI

kVVBase 132kVVBase 11

The Fault Current at 11 kV side supplied by the two generators is:

AI sidekVBase 8.524811*3

1000*100)( 11

AjI oActualF 90199958.5248*8095.3)(

At 11 kV Side:

G15.0j

20.0j 16.0j

10.0j23.0j

G

01

01 23.0j

TotalFI )(

1)( GFI

2)( GFI

kVVBase 132kVVBase 11

111,2 )()()( GFkVTFGF III

Ajj

jI GF 903.1180025.036.0

36.0)9019995()( 1

AI GF 907.8194)( 2

G25.0j

36.0jG

01

01115.0j

T1

T2

G1

G2

1

2

1)( GFI

2)( GFI

1)( GFI

2)( GFI

TotalFI )(

Ajj

jII kVTFGF 903.1180025.036.0

36.0)()( 11,1

Example:Consider the single-line diagram of a power system shown below. The transient reactance of each part of the system is as shown and expressed in pu on a common 100 MVA base.

G1 G22.0j

2.0j

4.0j

8.0j

4.0j

1.0j

1.0j

Bus 1

Bus 3 Bus 4

Bus 2

Bus 5

Assuming that all generators are working on the rated voltages, when a three-phase fault with impedance of j0.16 pu occurs at bus 5. Find: The fault currents and buses voltages.

G1 G22.0j

2.0j

4.0j

8.0j

4.0j

1.0j

1.0j

Bus 1

Bus 3 Bus 4

Bus 2

Bus 5

2.0j

2.0j

4.0j

8.0j

4.0j

1.0j

1.0j

Bus 1

Bus 3 Bus 4

Bus 2

Bus 516.0j

2.0j4.0j

4.0j

8.0j

4.0j

Bus 3 Bus 4

Bus 516.0j

5.016.034.0 jjjZTH

pu0.2j5.0j

1)I(0

5BusF

pujjj

jII FGF 2.14.06.0

6.0)( 1

pujjj

jII FGF 8.04.06.0

4.0)( 2

16.0j

Bus 5

34.0j

FI

16.0jBus 5

34.0j

FI

2.0j 4.0j

16.0j

Bus 3 Bus 4

Bus 5

2.0j 2.0j

1.0j

FI

1FGI 2FGI

16.0jBus 5

6.0j

1.0j

4.0j

FI

2FGI1FGI

0V&01V:VoltagesPrefault

Groundo

BUS

pu

VprefaultVVoltageFault

o 76.024.001

:

33

pujj

jIVisFaultDuringVariationVoltage

G

24.0)2.1()2.0(0

)2.0(0:

13

68.0)0.2)(16.0(01

:

5

jjV

FaultduringVariationVoltageo

pu

VprefaultVVoltageFault

o 32.068.001

:

55

NOTE: If the fault impedance is zero.

0.1)I)(0.0j(01V:FaultduringVariationVoltage

Fo

5 pu0.00.101VprefaultV o55 and

1GI 2GI

FI

o01

2.0j 4.0j

16.0j

Bus 3 Bus 4

Bus 5

2.0j 2.0j

1.0j+_

5V

3V

G1 G22.0j

2.0j

4.0j

8.0j

4.0j

1.0j

1.0j

Bus 1

Bus 3 Bus 4

Bus 2

Bus 5

Problem:For the network shown in the figure, find the current flowing between buses 3 and 4 during symmetrical three phase fault at bus 5 as in the previous example.

16.0j

2.0j

2.0j

4.0j

8.0j

4.0j

1.0j

1.0j

Bus 1

Bus 3 Bus 4

Bus 2

Bus 5

pu

jjprefaultVprefaultV

pu

VprefaultVVoltageFaultDuring

o

o

68.032.001

)8.0)(4.0(0

76.024.001

:

44

33

pujjj

jII FGF 2.14.06.0

6.0)( 1

pujjj

jII FGF 8.04.06.0

4.0)( 2

pujj

VVIF 1.0

8.0)( 4334

4.0)( 5335 j

VVIF

4.0)( 5445 j

VVIF

and

2.0j

2.0j

4.0j

8.0j

4.0j

1.0j

1.0j

Bus 1

Bus 3 Bus 4

Bus 2

Bus 516.0j

1GI2GI

34F )I(

45F )I(35F )I(

4V

5V

Symmetrical

Fault Calculation

Transformer

Three Phase

Short Circuit

Three Phase FaultCalculation

Simple CircuitsLoad Ignored

Use Thevenin's Equivalent

Simple CircuitsLoad not Ignored

Use Thevenin's EquivalentFind Pre-fault voltage

Large CircuitsConstruct

The Bus Impedance Matrix

2

2. Symmetrical Faults Considering the load current

Generally, the fault currents are much larger than the load currents. Therefore, the load current can be neglected during fault calculations.

The pre-fault voltage and the ZTH are used in calculating the fault current.

The fault current represents two components:1. The Load Current2. The Short Circuit Current

There are some cases where considering the load current is an essential factor in fault calculations. Superposition technique is proposed for such cases to compute the fault current.

For such case, it is necessary to compute the terminal voltage at fault location before fault takes place. This terminal voltage is known as the pre-fault voltage.

Connecting the load to the system

causes the current to flow in the network. Voltage drop due to system impedance cause the voltage magnitude at different buses to be deviated from 1.0 pu.

Consider a load (synchronous Motor) connected to a synchronous generator through a transmission line. The circuit model of the network could be represented as shown in the Figure

MIGI

jXG jXM

G M

jXL

FI

G M

FMIFGI jXL

jXG jXM

Normal Operation Condition:

LMG III

Fault condition at Load terminal:

LGFG III )(

LMFM III )(

MFGFF III )()(

During fault condition in the system, the generator as well as the load (synchronous motor) will supply the faulted terminals with power from the energy stored in their windings.

The Motor is drawing 40 MW at 0.8 pf. Leading with terminal

voltage of 10.95 kV.

Calculate the total current in the generator and motor during 3

phase s.c.

Example

100 MVA, 11 kV Generator0.25 pusub-transient reactance

50 MVA, Motor0.2 pusub-transient reactance

0.05 pu. Reactance

On 100 MVA Base

3 phase fault

G M

AkV

MVACurretBase

kVkVBaseMVAMVABase

88.5248)11(3

10011

100

Solution:The network could be represented as shown in the Fig.

A4.2636)8.0)(kV95.10(3

MW40CurretLoadThe

pujIjI

I

L

L

L

)301.0402.0()6.08.0(502.0

)8.0(cos502.0 1

puI puL 502.088.52484.2636)(

Under Normal Operation Condition (Prefault condition)

leading0.8isfactorpowerThe

25.0j 4.0j

G M

05.0j MIGI

Consider the motor terminal voltage as a reference.

puvoltage)terminal(Motor pu 995.011

95.10

pu)0201.0j9799.0(V995.0)050.0j)(301.0j402.0(V

V)050.0j(IV

G

G

MLG

pu)(V opuM 0995.0

Using KVL, calculate the pre-fault voltage at the generator terminal.

Find the equivalent reactance as seen from the fault terminals.

pujXjj

jjX

TH

TH

1607.0)45.0()25.0(

)45.0)(25.0(

25.0j 4.0j

05.0j

THX

25.0j 4.0j

G M

05.0j

MV

Calculate the fault current

puj)(Ij

j)(I

XVoltaegefaultpre)(I

puF

puF

THpuF

)09.6125.0(1607.0

0201.09799.0

pu(I)

puj(I)jj(I)

I)(I(I)

oG

G

G

LGFG

4.8264.3

)609.3482.0()301.0402.0()91.308.0(

Fault Current from Generator Side

puj)(Ipujj)(I

)(I)(I)(I

MF

MF

GFpuFMF

)18.2045.0()91.308.0()09.6125.0(

Fault Current from Motor Side

Current from Generator Side:

puj)(Ij0.25j0.45

j0.45)(I)(I

GF

puFGF

)91.308.0(

pu(I)

puj(I)jj(I)

I)(I(I)

oM

M

M

LMFM

8.261505.2

)48.2357.0()301.0402.0()18.2045.0(

Current from Motor Side:

25.0j 4.0j

G M

05.0j

FI

G(I) M(I)

Consider the system shown in the Figure.

The two generators are supplying the buses 1, 2, 3

and 4.

Let the pre-fault voltage between bus 2 and ref. Bus

be (Vf ).

3 - Symmetrical Fault Calculation Using Bus Impedance Matrix

3

T1

T2

G1

G2

1

2

4fV

If a 3 phase short circuit occurs a bus 2.

A huge current will run in the system

during the fault condition.

"fI

3

2

1

4"fI

3

T1

T2

G1

G2

1 2

4

fV

3 phasefault e

If the pre-fault voltage between bus 2 and ref. bus is Vf , then the voltage during

fault is zero.

Then, the system can be analyzed using the reactance

circuit.

During the three-phase fault, the voltage between bus 2 and the reference becomes zero. Therefore, the 3 phase fault at bus 2 can be simulated by making the voltage between bus 2 and Ref. is equal to zero during the fault condition.

3

2

1

4

"fI

fVfV

This is simulated by inserting a source with a magnitude

equal to

Therefore, the voltage between bus 2 and reference

is zero and fault current runs in the system.

)( fV

This means that, the Fault current during the three-phase short circuit results from the voltage source

"fI

)( fV

Therefore, to compute the short circuit current, we have to find the current entering

bus 2 due to only )( fV

4

3

2

1

VVVV

4

3

1

VVVfV

BUSZ

44434241

34333231

24232221

14131211

ZZZZZZZZZZZZZZZZ

=

00I0

''f=

The Nodal Impedance Equation during the fault is:

Note:is the voltage difference between node i and the reference node

due to the current .iV

"fI

Ignoring all other currents entering other buses (ignore sources).

3

2

1

4

"fI

fV

Or

4

3

2

1

VVVV

4

3

1

VVVfV

''f42

''f32

''f22

''f12

IZIZIZIZ

= =

And from the second row, the fault current is:

22

''ZV

I ff

Substituting in the previous equation:

4

3

2

1

VVVV

4

3

1

VVVfV

f22

42

f22

32

f

f22

12

VZZ

VZZ

V

VZZ

= =

Assuming no load is connected in the system, no current will flow before fault and all bus voltages are the same and equal to the pre- fault voltage )V( f

Using Superposition, then during the fault:

4

3

2

1

VVVV

fV

22

4222

32

22

12

ZZ1

ZZ10ZZ1

= +

i)voltagefaultPre(f)voltageiBus(i VVV

iV is the voltage change in bus i due to fault current in bus 2.

4

3

2

1

VVVV

f

f

f

f

V

V

V

V

=

Example:

A three-phase fault occurs at bus 2 of the network shown in the Figure.

Determine:1. the sub-transient current during the fault, 2.the voltages at all the buses3.The current flow during the fault from buses 3-2, 1-2 and 4-2.

Consider the pre-fault voltage at bus 2 equal to 1 pu. and neglect the pre-fault currents.

3

T1

T2

G1

G2

1 2

4

fV

3 phasefault e

Solution:

Construct the reactance circuit under normal operation condition.

pu2.0j pu1.0j

pu125.0j

pu25.0j

pu4.0j

pu25.0j

up2.0j

3

2

1

4pu2.0j pu1.0j

Construct the bus impedance Matrix (ZBUS )

1954.0j1046.0j1506.0j1456.0j1046.0j1954.0j1494.0j1544.0j1506.0j1494.0j2295.0j1938.0j1456.0j1544.0j1938.0j2436.0j

ZBUS

NOTE:Constructing ZBus will be discussed later.

Since there are no-load currents, the pre-fault voltage at all the buses is equal to 1.0 pu. When the fault occurs at bus 2,

pu357.4j2295.0j0.1

ZV

I22

f''f

fV

22

4222

32

22

12

ZZ

1

ZZ

10ZZ

1

=

4

3

2

1

VVVV

fV

22

4222

32

22

12

ZZ

1

ZZ

10ZZ

1

=

4

3

2

1

VVVV

=

2295.0j1506.0j1

2295.0j1494.0j1

02295.0j1938.0j1

343.0349.00

155.0

=

1954.0j1046.0j1506.0j1456.0j1046.0j1954.0j1494.0j1544.0j1506.0j1494.0j2295.0j1938.0j1456.0j1544.0j1938.0j2436.0j

ZBUS

3

T1

T2

G1

G2

1 2

4

fV

3 phasefault e

The current flows from bus 1 to bus 2 is:

125.0jVVI 21

12

pu2448.1jI12

125.0j1556.0I12

pu2.0j pu1.0j

pu125.0j

pu25.0j

pu4.0j

pu25.0j

pu2.0j

3

2

1

4"fI

pu2.0j pu1.0j


25.0jV3VI 2

32

pu396.1jI32

pu719.1jI42



25.0jVV

I 1331

pu7736.0j

Balanced Faults

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