Lecture 8Short Circuit Analysis (Faults)

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By Dr. Mostafa Elshahed

Objectives• Learn a brief history of the electrical power systems construction

• List and describe different components of electrical power systems

• Learn the per-unit system calculations

• Draw the single line and impedance diagram

• Define the transformers and synchronous machines representations

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• Calculate the transmission lines parameters

• Formulate the various transmission lines models

• Establish the transmission lines performance

Calculate the symmetrical faults variables

Learn the principles of symmetrical components transformations

Calculate the unsymmetrical faults variables

PreliminariesA fault in a circuit is any failure thatinterferes with the normal flow of currentto the load. In most faults, a current pathforms between two or more phases, orbetween one or more phases and theneutral (ground). Since the impedance ofa new path is usually low, an excessivecurrent may flow.

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current may flow.

High-voltage transmission lines havestrings of insulators supporting eachphase. The insulators must be largeenough to prevent flashover – a conditionwhen the voltage difference between theline and the ground is large enough toionize the air around insulators and thusprovide a current path between a phaseand a tower.

PreliminariesHigh currents due to a fault must be detected by protective circuitry and the circuit breakers on the affected transmission line should automatically open for a brief period (about 1/3 second).

3Selecting an appropriate circuit breaker (type, size, etc.) is important…

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

Preliminaries

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– salt spray or pollution on insulators

Fault Types

– Symmetrical faults: system remainsbalanced; these faults are relativelyrare, but are the easiest to analyze sowe’ll consider them first.

– Unsymmetrical faults: system is nolonger balanced; very common, butmore difficult to analyze

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– On very high voltage lines faults arepractically always single line to grounddue to large conductor spacing

RL Circuit Analysis

To understand fault analysis we need to review the behavior of an RL circuit

( )

2 cos( )

v t

V t

6

2 cos( )V t

Before the switch is closed obviously i(t) = 0.When the switch is closed at t=0 the current willhave two components: 1) a steady-state value2) a transient value

RL Circuit Analysis, cont’d

ac

1. Steady-state current component (from standard

phasor analysis)

2 cos( )i ( )

V tt

Z

7

2 2 2 2where ( )

ac

Z R L R X

VI

Z

RL Circuit Analysis, cont’d

dc 1

2. Exponentially decaying dc current component

i ( )

where T is the time constant,

The value of is determined from the initial

tTt C e

LT R

C

8

1

ac dc 1

1

The value of is determined from the initial

conditions:

2(0) 0 i ( ) i ( ) cos( )

2

tT

Z

C

Vi t t t C e

Z

VC

Z

cos( ) which depends on Z

RL Circuit Analysis, cont’d

dc

1

Hence i(t) is a sinusoidal superimposed on a decaying

dc current. The magnitude of i (0) depends on when

the switch is closed. For fault analysis we're just

2concerned with the worst case:

VC

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1concerned with the worst case:

( )

CZ

i t

ac dci ( ) i ( )

2 2( ) cos( )

2(cos( ) )

tT

tT

t t

V Vi t t e

Z Z

Vt e

Z

RMS for Fault Current2

The function i(t) (cos( ) ) is not periodic,

so we can't formally define an RMS value. However,

as an approximation define

tTV

t eZ

10

2 2RMS

22 2

I ( ) ( ) ( )

2

This function has a maximum va

ac dc

tT

ac ac

t i t i t

I I e

lue of 3

Therefore the dc component is included simply by

multiplying the ac fault currents by 3

acI

Generator Short Circuit Currents

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Generator Modeling During Faults

Thus the models of generators (and other rotating machines) are very important since they contribute the bulk of the fault current.

Generators can be approximated as a constant voltage behind a time-varying reactance

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SteadyTransientSub-transient

Generator Modeling, cont’d

"d

The time varying reactance is typically approximated

using three different values, each valid for a different

time period:

X direct-axis subtransient reactance

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d

'd

d

X direct-axis subtransient reactance

X direct-axis transient reactance

X dire

ct-axis synchronous reactance

Internal generated voltages of synchronous motors

When the motor is short-circuited, it does not receive power from the line but itsfield circuit is still energized and still spinning (due to inertia in the machine andin its load).Therefore, the motor acts as a generator, supplying power to the fault. Theequivalent circuit of a synchronous motor is the same as the one of the synchronousgenerator except that the direction of the current flow is reversed. Therefore, theequations for the internal generated voltage, voltage behind the subtransientreactance, and voltage behind transient reactance become

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reactance, and voltage behind transient reactance become

"

'

"

'

A s A

A A

A A

E V jX I

E V jX I

E V jX I

Large System Modeling• Given the short circuit MVA at the bus of the large system.

• The large system is simulated by a voltage source 1 pu in series with a reactance XS

01

3103 SCSC IkVVMVA

3103 BaseBaseBase IkVMVA

15

puSCpuSCpuSC

SMVAII

VX

,,,

11

01103 BaseBaseBase IkVMVA

puSCpuSC IMVA ,,

Internal generated voltages of loaded machines under transient conditions

Example 12-3: A 100 MVA, 13.8 kV, 0.9 PF lagging, Y-connected, 3 phase, 60 Hz synchronous generator is operating at rated voltage and full load when a symmetrical 3 phase fault occurs at its terminals. The reactances in per-unit to the machine’s own base are

1.00 ' 0.25 " 0.12sX X X

a) If the generator operates at full load when the fault develops, what is the

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a) If the generator operates at full load when the fault develops, what is the subtransient fault current produced by this generator?

b) If the generator operates at no load and rated voltage when the fault develops, what is the subtransient fault current produced by this generator? Observe that this calculation is equivalent to ignoring the effects of prefault load on fault currents.

c) How much difference does calculating the voltage behind subtransient reactance make in the fault current calculations?

Internal generated voltages of loaded machines under transient conditions

1.0 25.84I pu

The base current of the generator can be computed as

,

,

100,000,0004,184

3 3 13,800base

L base

L base

SI A

V

a) Before the fault, the generator was working at rated conditions and the per-unit current was:

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1.0 25.84AI pu

" " 1 0 0.12 1 25.84 1.058 5.86A AE V jX I j

The voltage behind subtransient reactance is

Therefore, the per-unit fault current when terminals are shorted is

1.058 5.868.815 84.1 36,880

0.12FI pu A

j

Internal generated voltages of loaded machines under transient conditions

b) Before the fault, the generator was assumed to be at no-load conditions and the per-unit current was:

0.0 0AI pu

The voltage behind subtransient reactance is

" " 1 0 0.12 0 0 1.0 0A AE V jX I j

Therefore, the per-unit fault current when the terminals are shorted is

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Therefore, the per-unit fault current when the terminals are shorted is

1.0 08.333 90 34,870

0.12FI pu A

j

c) The difference in fault current between these two cases is

36,880 34,870100% 5.76%

34,870Difference

The difference in the fault current when the voltage behind subtransient reactance is considered and when it is ignored is small and usually systems are assumed unloaded.

Network Fault Analysis Simplifications

To simplify analysis of fault currents in networks we'll make several simplifications:

1. Transmission lines are represented by their series reactance

2. Transformers are represented by their leakage reactances

3. Synchronous machines are modeled as a constant voltage

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3. Synchronous machines are modeled as a constant voltage behind direct-axis subtransient reactance

4. Induction motors are ignored or treated as synchronous machines

5. Other (nonspinning) loads are ignored

Network Fault Example

For the following network assume a fault on the terminalof the generator; all data is per unit except for thetransmission line reactance

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2

19.5Convert to per unit: 0.1 per unit

138100

lineX

Generator has 1.05terminal voltage &supplies 100 MVAwith 0.95 lag pf

Network Fault Example, cont'dFaulted network per unit diagram

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*'a

To determine the fault current we need to first estimate

the internal voltages for the generator and motor

For the generator 1.05, 1.0 18.2

1.0 18.20.952 18.2 E 1.103 7.1

1.05

T G

Gen

V S

I

Network Fault Example, cont'dThe motor's terminal voltage is then

1.05 0 - (0.9044 - 0.2973) 0.3 1.00 15.8

The motor's internal voltage is

1.00 15.8 (0.9044 - 0.2973) 0.2

1.008 26.6

j j

j j

22

f

1.008 26.6

We can then solve as a linear circuit:

1I

.103 7.1 1.008 26.6

0.15 0.5

7.353 82.9 2.016 116.6 9.09

j j

j

1.95.0

005.1

15.0

005.1j

jjI f

At no-load conditions

Fault current transientsTwo generators are connected in parallel to the low-voltage side of a transformer. Generators G1 and G2 are each rated at 50 MVA, 13.8 kV, with a subtransient resistance of 0.2 pu. Transformer T1 is rated at 100 MVA, 13.8/115 kV with a series reactance of 0.08 pu and negligible resistance.

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0.08 pu and negligible resistance.

Assume that initially the voltage on the high side of the transformer is 120 kV, that the transformer is unloaded, and that there are no circulating currents between the generators.

Calculate the subtransient fault current that will flow if a 3 phase fault occurs at the high-voltage side of transformer.

Fault current transientsLet choose the per-unit base values for this power system to be 100 MVA and 115 kV at the high-voltage side and 13.8 kV at the low-voltage side of the transformer.

The subtransient reactance of the two generators to the system base is2

given newV S

Z Z

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given newnew given

new given

Z ZV S

The reactance of the transformer is already given on the system base, it will not change

2

" "1 2

13,800 100,0000.2 0.4

13,800 50,000X X pu

0.08TX pu

Therefore:

Fault current transientsThe per-unit voltage on the high-voltage side of the transformer is

120,0001.044

115,000pu

actual valueV pu

base value

Since there is no load on the system, the voltage at the terminals of each generator, and the internal generated voltage of each generator must also

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voltage of each generator must also be 1.044 pu. The per-phase per-unit equivalent circuit of the system is

We observe that the phases of internal generated voltages are arbitrarily chosen as 00. The phase angles of both voltages must be the same since the generators were working in parallel.

Fault current transientsTo find the subtransient fault current, we need to solve for the voltage at the bus 1 of the system. To find this voltage, we must convert first the per-unit impedances to admittances, and the voltage sources to equivalent current sources. The Theveninimpedance of each generator is ZTh = j0.4, so the short-circuit current of each generator is

1.044 02.61 90

04oc

sc

VI

Z j

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04thZ j

The equivalent circuit

Fault current transientsThen the node equation for voltage V1

1 1 1

1

2.5 2.5 12.5 2.61 90 2.61 90

5.22 900.298 0

17.5

V j V j V j

Vj

Therefore, the subtransient current in the fault is

1 12.5 3.729 90FI V j pu

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1 12.5 3.729 90FI V j pu

Since the base current at the high-voltage side of the transformer is

3 ,

,

100,000,000502

3 3 115,000

base

base

LL base

SI A

V

, 3.729 502 1,872F F pu baseI I I A

The subtransient fault current will be:

ApuI

I Fg 5.7800

800,133

000,000,100908645.1

2

Three Generator Example

05.1 pujI 4.405.1

I1

I2

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pujj

I 25.52.0

05.11

01 V 44.01.022 jIV

puj

jjjjj

I 4.4

45.02.045.02.01.0

05.12

Thevenin

fZ

Vpuj

jjjjjj

I

65.9

45.02.045.02.01.0

05.1

2.0

05.1

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Element Z

Generator 1 j 0.1

Generator 2 j 0.05

All Transmission Lines j 0.1

CB SelectionFrom the current view point two factors to be considered in selecting circuit breakers are:

1. The maximum instantaneous current which the breaker must carry (with stand)

2. The total current when the breaker contacts part to interrupt the circuit

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Up to this point we have devoted most of our attention to the subtransient current called the initial symmetrical current, which does not include the DC component. Inclusion of the DC component results in RMS value of current immediately after the fault, which is higher than the subtransient current.

For oil circuit breakers above 5 kV the subtransient current multiplied by 1. 6 is considered to be the RMS value of the current whose disruptive forces the breaker must with stand during the first half cycle after the fault occurs.

CB SelectionFrom the current view point two factors to be considered in selecting circuit breakers are:

1. The maximum instantaneous current which the breaker must carry (with stand)

2. The total current when the breaker contacts part to interrupt the circuit

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The interrupting rating of a circuit breaker was specified in kVA or MVA.

The interrupting kVA equal √3 x kV x I

I = current which the breaker must be capable of interrupting when its contacts part)

CB Selection

This interrupting current is, of course , lower thanthe momentary current and depends on thespeed of the breaker, such as 8, 5, 3, or 2 cycles,which is a measure of the time from theoccurrence of the fault to the extinction of thearc. Breakers of different speeds are classified bytheir rated interrupting times.

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The rated interrupting time of a circuit breaker isthe period between the instant of energizing thetrip circuit and the arc extinction on an openingoperation. Preceding this period is the trippingdelay time, which is usually assumed to be 0.5cycle for relays to pick up.