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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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Broad categorization of Fault Types
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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Broad categorization of Fault Types
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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Broad categorization of Fault Types
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Broad categorization of Fault Types
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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Broad categorization of Fault Types
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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Characteristics of Faults
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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Characteristics of Faults
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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Necessity for fault calculations
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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Necessity for fault calculations
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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