Short Circuit 2011 2

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POWER SYSTEMFAULTS


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POWER SYSTEM FAULTS

Power substations as a target of study consists of someelements like Transmission Lines, Bus Bars, PowerTransformers, Outgoing Feeders, and Bus Couplers.

Regardless of the design and the systematic preventivemaintenance procedures instituted, failures due toabnormal or fault conditions do occur

Fault are intolerable power conditions (other thansteady-state or rated ones) to which the power system orrequirement are subjected.


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General causes of power system

faults

1- Fault Current

Healthy insulation in the equipment is subjected to either transientover voltages of small time duration due to switching and lightningstrokes, direct or indirect. Failure of insulation may be happened,resulting in very high fault current. This current may be more than10 times the rated or nominal current of the equipment.

2- Insulation Aging

Aging of power equipments may cause breakdown of its insulationeven at normal power frequency voltage.


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General cause of power system

faults (contd)

3- External Causes

External object such as bird, kite, or tree branch areconsidered as external cause of fault. These objectsmay span one conductor and ground causing singleline to ground fault (phase-earth) or span twoconductors causing phase-phase fault


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Fault Effects

The fault must be cleared as fast as possible. Manyequipments may be destroyed if the fault is not cleared

rapidly. The dangerous of the faults depends on thetype of the fault, as example the three phase shortcircuit is the most dangerous fault because the shortcircuit current is maximum. Some of the effects of short

circuit current are listed here under.


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Fault Effects

1- Due to overheating and the mechanical forcesdeveloped by faults, electrical equipments such as bus

bars, generators, transformers will be damaged

2- Negative sequence current arises fromunsymmetrical faults will lead to overheating.

3- Voltage profiles may be reduced to unacceptable

limits as a result of faults. A frequency drop may leadto instability


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Fault Types

Fault can be classified due t as:

1- Permanent

2- Transient

Or due to PARTICIPATED PHASES as

1- Phase-Earth

2- Phase-Phase3- Phase-Phase-Earth

4- Three-Phase or Three-Phase-Earth


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Broad categorization of Fault Types

Fault may be categorized broadly into 2 types:

Symmetrical or balanced faults

Asymmetrical or unbalanced faults


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3 FAULTS MAY BE REPRESENTED BY 1 CIRCUIT

This is valid because system is maintained in a BALANCED state

during the faultVoltages equal and 120 apart

Currents equal and 120 apart

Power System Plant Symmetrical

Phase Impedances EqualMutual Impedances Equal

Shunt Admittances Equal


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Unbalanced FaultsUnbalanced Faults may be classified into

SHUNT FAULTS and SERIES FAULTS.

SHUNT FAULTS:

Line to Ground

Line to Line Line to Line to Ground


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Causes :

1) Insulation Breakdown

2) Lightning Discharges and other Overvoltages

3) Mechanical Damage

During unbalanced faults, symmetry of system Is

lost therefore single phase representation is nolonger Valid


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SERIES FAULTS OR OPEN CIRCUIT:

Single Phase Open Circuit

Double Phase Open Circuit

Causes:

1) Broken Conductor

2) Operation of Fuses3) Maloperation of Single Phase Circuit Breakers


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Characteristics of Faults

A fault is characterized by:

Magnitude of the fault current

Power factor or phase angle of the fault current

The magnitude of the fault current depends upon:

The capacity and magnitude of the generating sources

feeding into the fault


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The system impedance up to the point of fault orsource impedance behind the fault

Type of fault

System grounding, number and size of overheadground wires

Fault resistance or resistance of the earth in the caseof ground faults and arc

resistance in the case of both phase and ground faults


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The phase angle of the fault current is dependent upon:

For phase faults: - the nature of the source andconnected circuits up to the fault location and

For ground faults: - the type of system grounding inaddition to above.


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Necessity for fault calculations

Fault calculations are done primarily for the following:

To determine the maximum fault current at the point ofinstallation of a circuit breaker and to choose a

standard rating for the circuit breaker (rupturing)

To select the type of circuit breaker depending uponthe nature and type of fault.


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The calculation is not only limited to present system

requirements but also meet:

The future expansion schemes of the system such asaddition of new generating units

Construction of new transmission lines to evacuatepower.

Construction of new lines to meet the load growth andor Construction of interconnecting tie lines


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Basically, there are two approaches to faultcalculations. These are:

(a) Actual reactance or impedance method

(b) Percentage reactance or impedance method or perunit (p.u) reactance or impedance method.

Machine and Transformer impedance or reactance arealways noted in percentage values on the nameplate.Hence the latter method is considered for ourcalculation.


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Per Unit System

Power system quantities such as voltage,current and impedance are often expressed inper unit or percent of specified values.

Per unit quantities are calculated as:

ValueBaseValueActualValuePer Unit


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Per Unit Values

base

puSS

S

basepu

II

I

base

puVV

V

base

puZZ

Z

ZZ

2

base

base

base

puV

S

ZZ

pu

base

2

base

pubase ZS

VZZ Z

Conversion of Per Unit Values

Per Unit System


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Per Unit System

Usually, the nominal apparent power (S) andnominal voltage (V) are taken as the base values forpower (Sbase) and voltage (Vbase).

The base values for the current (Ibase) andimpedance (Zbase) can be calculated based on thefirst two base values.


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100%Z

base

actual

%

Z

Z

The percent impedance e.g. in a synchronous generator with 13.8

kV as its nominal voltage, instead ofsaying the voltage is 12.42 kV, we say thevoltage is 0.9 p.u.


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Simplified: Concerns about using phase or line voltages are

removed in the per-unit system

Actual values of R, XC and XL for lines, cables, and

other electrical equipment typically phase values. It is convenient to work in terms of base VA (base

volt-amperes)


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,3

3

B

BB

BBB

VSI

IVS

B

B

B

BB

BBB

SV

IVZ

ZIV

2

3/

3

Usually, the 3-phase SB or MVAB and line-to-line VB or kVB

are selected

IB and ZB dependent on SB and VB


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The impedance of individual generators &transformer, are generally in terms ofpercent/per unit based on their own ratings.

Impedance of transmission line in ohmicvalue

When pieces of equipment with variousdifferent ratings are connected to a system, itis necessary to convert their impedances to aper unit value expressed on the same base.


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ZV

S

Z

ZZ old

B

old

Bold

B

oldpu 2

Z

V

S

Z

ZZ

new

B

new

B

new

B

new

pu 2

new

B

new

B

Vbasevoltagenew&

Sbasepowernewon theimpedanceunitpernewthebenew

puZ

old

B

old

B

Vbasevoltage&

Sbasepoweron theimpedanceunitperthebeold

puZ

1

2

Change of Base


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2

new

B

old

B

old

B

new

Bold

pu

new

puVV

SSZZ

old

B

new

Bold

pu

new

puS

SZZ

From (1) and (2), the relationship between the old

and the new per unit value

If the voltage base are the same,

Change of Base


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2

)(

2

)(

)(

)(

)()(**

newbase

oldbase

oldbase

newbase

oldpunewpu

KV

KV

MVA

MVAZZ


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Other base quantites :-

kAinkV.3

MVACurrentBase

OhmsinMVA

)(kVZImpedanceBase

b

bb

b

2b

b


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(1) To calculate the p.u impedance and % impedanceof a transmission line at 100 MVA

base Line voltage 330 KV

Line length 200 Kms

Line resistance /Km = 0.06 ohms/KmLine reactance /Km = 0.4 ohms/Km


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Z = R + jXFor the 200kms line length

Z = 200 (0.06 + j 0.4)

= 12 + j 80

|Z| =[(12)2+ (80)2] = 80.895 ohms


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Zp.u = Z x MVA base

(KV)2base

= 80.895 x 100

(330)2

= 0.0743 p.u

%Z = 0.074 x 100

= 7.43

(2)T l l h i d 100 MVA b


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(2)To calculate the p.u impedance to a 100 MVA base

Given four generators; 90MVA, 11KV of 15%impedance each connected to step up transformers

of 90MVA 11KV/330KV of 14% impedance. Calculatethe fault current at F.

A d MVA 100


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Assumed MVA = 100

%Z generators = 15 on 90 MVA base

or Zg p.u = 0.15 on 90 MVA base

Zg p.u on 100 MVA base will be:

(Zg p.u) base2 = (Zg p.u) base1 x MVA base2

MVA base1

(Zg p.u) 100 = 0.15 x 100

90

= 0.167

%Z transformers = 14 on 90 MVA base


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% t a s o e s o 90 base

or Zt p.u = 0.14 on 90 MVA base

Zt p.u on 100 MVA base will be:(Zt p.u) base2 = (Zt p.u) base1 x MVA base2

MVA base1

(Zt p.u) 100 = 0.14 x 100

90= 0.156


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The system reduces as follows


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Ztotal = 0.323

4

= 0.08075

Total p.u impedance at F = 0.08075 = Ztotal

Fault MVA at F = Base MVA

Ztotal

= 100 MVA0.08075

= 1238.4 MVA


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MORE EXAMPLES IN YOUR LECTURE BOOK,PLEASE.


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Short Circuit 2011 2

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