Symmetrical Fault Analysis
Symmetrical Fault
Analysis
Chapter 8
Dr Senthil Krishnamurthy
Senior Lecturer
Programme Coordinator BET Electrical Engineering
Deputy Leader CSAEMS
Department of Electrical, Electronics and Computer Engineering
Cape Peninsula University of Technology
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SHORT CIRCUIT ANALYSIS
Symmetrical Fault Analysis
Lesson Summary
1. Introduction
2. Cause and Consequences
3. Types of Faults
4. Symmetrical Short Circuit Analysis
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Instructional Objective
On completion of this lesson a student
should be able to:
Symmetrical Fault Analysis
1. Explain the significance of Short Circuit
2. Explain the causes and consequences of
Short Circuit
3. Classify different types of faults
4. Compute currents for symmetrical faults
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Cause of Short Circuit Insulation Failure
Over-voltages caused by Lightning or Switching
Symmetrical Fault Analysis
Surges
Insulation contamination salt spray, pollution
Mechanical Causes Over-heating, abrasion
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Faults on Transmission Lines
Most Common Lines are exposed to elements of
nature (60-70%).
Symmetrical Fault Analysis
Lightning strokes Over voltages cause
insulators to flash over line to ground short
circuit or line to line short circuit.
High winds --Topple tower, tree falls on line.
Winds and ice loading Mechanical failure of
insulator.
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Fog, salt spray, dirty insulator Conduction
Symmetrical Fault Analysis
path insulation failure
Short circuit in other elements
Cables (10-15%), circuit breakers (10-12%),
generators, motors, transformers etc (10-15%).
much less common Over loading for
extended periods deterioration of insulation
Mechanical failure.
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Consequences of Short Circuit
• Currents several magnitude larger than normal
operating current.
Symmetrical Fault Analysis
• Thermal damage to equipment.
• Windings and busbars Mechanical damage due
to high magnetic forces caused by high current.
Faulted section must be removed from service as soon
as possible (3-5 cycles).
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Types of Short Circuit
Symmetrical Fault Analysis
L–G L–L L–L–G 3φ – G
75 - 80% 5 – 7% 10 – 12% 8 – 10%
Asymmetrical Faults
Symmetrical Faults
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What is symmetrical fault?
That fault on the power system which gives rise to symmetrical
fault currents (i.e. equal fault currents in the lines with 120o
displacement) is called a symmetrical fault.
Symmetrical Fault Analysis
(i) The symmetrical fault rarely occurs in practice as majority
of the faults are of unsymmetrical nature.
(ii) The symmetrical fault is the most severe and imposes
more heavy duty on the circuit breaker.
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What is Percentage reactance?
It is the percentage of the total phase-voltage dropped in the
circuit when full-load current is flowing
Symmetrical Fault Analysis
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Steps for symmetrical fault calculations
(i) Draw a single line diagram of the complete network indicating the
rating, voltage and percentage reactance of each element of the
network.
Symmetrical Fault Analysis
(ii) Choose a numerically convenient value of base kVA and convert all
percentage reactances to this base value.
(iii) Corresponding to the single line diagram of the network, draw the
reactance diagram showing one phase of the system and the neutral.
Indicate the % reactances on the base kVA in the reactance diagram.
The transformer in the system should be represented by a reactance
in series.
(iv) Find the total % reactance of the network upto the point of fault. Let
it be X%.
(v) Find the full-load current corresponding to the selected base kVA and
the normal system voltage at the fault point. Let it be I.
(vi) Then various short-circuit calculations are :
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Symmetrical fault analysis on the four bus
power system network
• To calculate the fault current, post fault voltage and fault current
through the branches for a three phase to ground fault in a small
Symmetrical Fault Analysis
power system and also study the effect of neighboring system.
Equipment ratings for the four-bus power system shown in Figure 2.1
are as follows:
Generator Gl: 500 MVA, 13.8 kV, X”= 0.20 per unit
Generator G2: 750 MVA, 18 kV, X"= 0.18 per unit
Generator G3: 1000 MVA, 20 kV, X" = 0.17 per unit
Transformer Tl: 500 MVA, 13.8 /5OOY kV, X= 0.12 per unit
Transformer T2: 750 MV A, 18 /500Y kV, X=0.10perunit
Transformer T3: 1000 MVA, 20 /500 Y kV, X= 0.10 per unit
Each 500-kV line: X1 =50Ω
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Formulae
Formulae:
Base kV on HT side Voltagerating on HT side
=
Symmetrical Fault Analysis
Base kV on LT side Voltagerating on LT side
2
NewbaseKVA old baseKV
Per-unit reactance to new base, Xpu new= Xpu old x X
Old base KVA newbase KV
X old .KVAB
Per-unit impendence of transmission line =
( KV )
2
B
1000
BaseKVA
Base current I B =
3 base KV
fault MVA1000
Fault current I sc = I sc pu (.I B ) or
3 baseKV
BaseMVA
Fault MVA =
X pu
Per-unit fault current I sc Pu =
pu volatge 13
pu reac tan ce
Fault power = 3 I sc sourcevolatge 10−6 MVA
Symmetrical fault
A three-phase short circuit occurs at bus 1, where the prefault
voltage is 525 kV. Prefault load current is neglected.
Symmetrical Fault Analysis
Draw the positive-sequence reactance diagram in per-unit on a
1000-MVA, 20-kV base in the zone of generator 3.
Determine
(a) The Thevenin reactance in per-unit at the fault,
(b.) The sub transient fault current in per unit and in kA RMS, and
(c) Contributions to the fault current from generator G1 and from
line 1-2.
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Symmetrical fault
Symmetrical Fault Analysis
Calculating the new pu values for the symmetrical faults(on busbar1):
Zone KV
Zone 1 =13.8 kV Zone 3 =20 kV
Zone 2 =500 kV
13.8/500 kV 500/20 kV
500/18 kV
Zone 4 =18 kV
a) The Thevenin reactance in per-unit at the fault. 15
13.8 1000 𝑋𝑋 1000 𝑘𝑘𝑘𝑘𝑘𝑘
𝑥𝑥𝐺𝐺1 = 0.2 × (
13.8
)²×
500 𝑋𝑋 1000 𝑘𝑘𝑘𝑘𝑘𝑘
= j0.4 pu
•
Symmetrical fault
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•
Symmetrical fault
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•
Symmetrical fault
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Symmetrical Fault Analysis
Reactor Control of Short-Circuit Currents
Advantages
(i) Reactors limit the flow of short-circuit current and thus protect
the equipment from overheating as well as from failure due to
Symmetrical Fault Analysis
destructive mechanical forces.
(ii) Troubles are localised or isolated at the point where they
originate without communicating their disturbing effects to
other parts of the power system. This increases the chances of
continuity of supply.
(iii) They permit the installation of circuit breakers of lower rating.
Location of Reactors
Short circuit current limiting reactors may be connected
(i) in series with each generator
(ii) in series with each feeder and
iii) in bus-bars. 19
Reactor Control of Short-Circuit Currents
Generator reactors
Disadvantages
(i) There is a constant voltage drop and power loss in the
Symmetrical Fault Analysis
reactors even during normal operation.
(ii) If a bus-bar or feeder fault occurs close to the bus-bar, the
voltage at the bus-bar will be reduced to a low value,
thereby causing the generators to fall out of step.
(iii) If a fault occurs on any feeder, the continuity of supply to
other is likely to be affected.
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Reactor Control of Short-Circuit Currents
Feeder reactors
Advantages
(i) If a fault occurs on any feeder, the voltage drop in its reactor will
Symmetrical Fault Analysis
not affect the bus-bars voltage so that there is a little tendency
for the generator to lose synchronism.
(ii) The fault on a feeder will not affect other feeders and
consequently the effects of fault are localised.
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Reactor Control of Short-Circuit Currents
Feeder reactors
Disadvantages
(i) There is a constant power loss and voltage drop in the reactors
even during normal operation.
Symmetrical Fault Analysis
(ii) If a short-circuit occurs at the bus-bars, no protection is provided
to the generators.
(iii) If the number of generators is increased, the size of feeder
reactors will have to be increased to keep the short-circuit
currents within the ratings of the feeder circuit breakers.
Bus-bar reactors
In the previous case there is a considerable voltage drop and
power loss in the reactors even during normal operation.
This disadvantage can be overcome by locating the reactors in the
bus-bars.
There are two methods for this purpose, namely ;
(i) Ring system and (ii) Tie-Bar system. 22
Reactor Control of Short-Circuit Currents
Bus bar reactors
(i) Ring system
Symmetrical Fault Analysis
(ii) Tie-Bar system
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%X =(IX/V)100
Isc=I(100/%X)
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