Types of Faults:
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Sn
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Types of Faults
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Symbols
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Causes
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1
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Single
line to ground fault
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On line
any how touches to ground or support.
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2
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Line to
line fault.
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Two
lines connected together due to insufficient distance or wind pressure.
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3
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Double
line to ground fault.
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Two
conductors touches together and touches to ground or pole/tower.
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4
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Three
phase short circuitl
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All the
three lines make contract in abnormal situation such as storm, falling of
trees.
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5
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Open
circuit fault.
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Conductor
may be broken.
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6
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Simultaneous
faults.
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Two
conductors touches and 3rd conductor touches to support/tower.
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Fault Statistic: Due to various causes as explained
above following is the collected statistical data for faults on various
elements.
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Sn
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Element
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%Total
Faults.
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1
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O.H. Lines
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50%
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2
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Under Ground Lines
(cables)
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9%
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3
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Generators
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7%
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4
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Transformers
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10%
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5
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Switchgears
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12%
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6
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Protective devices like
CT, PT, relay etc.
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12%
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Sn
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Types of Fault
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%Total faults.
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1
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Line to ground
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85%
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2
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Line to line
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8%
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3
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Line to line to ground
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5%
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4
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3-phase
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2% very rare
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SHORT CIRCUIT CALCULATIONS (SYMMETRICAL FAULTS ONLY)
Most of
the faults in power systems are due to short circuits which lead to a very
heavy short circuiting current which may damage the equipments. Hence, it
becomes necessary to determine the SC currents which enable us to determine the
capacity of CB relays, CTs, etc.
The faults
may be (1) Symmetrical, (2) Unsymmetrical.
1. Symmetrical Fault: In the system, if
the short circuit currents are equal in different conductors, if the short
circuit occurs in bothe the conductors of a single phase line or across three
wires sin a 3-phase system, then the fault is a symmetrical fault. Current in
all conductors are equal in magneitude and displaced by 1200.
2. Unsymmetrical fault: The fault in a
3-phase system which occurs across the two of the three-phases or between one
or two-phases and earth causing unequal currnts in the conductors is called as
an unsymmetrial fault. Our study is limited to calculations of SC currents in
symmetrical faults only. The calculations become less laborious because of the
balanced nature of fault. Only one phase need to be considered in calculations.
The results of the other two will be same.
Note that
most of the faults are unsymmetrical whereas the symmetrical faults are rare.
Symmetrical faults are most severe and hence CB shall be of bulk capacity.
Take the
case of a following fault.
Say short
circuit occurs at point ‘P’ in the power system. Current is inversely
proportional to the impedance. So SC current is limited by impedance of the
system up to the point of fault (at P). In this , the current limiting
impedance is the impedance of generator plus impedance of transformer and the
impedance of the line from G to P of the line. For calculating the SC current,
we should know these impedances. The equipment like transformer, generator,
reactors have the impedance mostly reactive whereas the lines, cables are
resistive. When the total reactance in calculations exceeds 3-times the
resistance, the resistance is usually negleced (as error in neglecting
resistance is hardly 5%, if resistance is not considered.
Symbols Used in Representing a System.
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Sn
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Components
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Symbols
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1
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Star connected generator with neutral
solidly grounded.
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2
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Star connected generator with neutral
grounded through resistance (R)
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3
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Star connected generator with neutral
grounded through reactance coil (R)
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4
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Transformer
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5
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Current Transformer (CT)
Potential Transformer (PT)
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6
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Circuit breaker
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7
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Air-circuit breaker
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8
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Bus Bar
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9
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Isolator
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10
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Earthing switch
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11
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Lightning arrestor
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12
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Fuse.
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Single Line Diagram
A balanced
3-phase system is always solved as a single phase circuit consissting of one of
the three lines and neutral return. Such simplified diagram of the system drawn
to represent different components of a power system by standard symbols is
called as a single line diagram. On the
diagram the information about loads, rating of the generator transformer
and reactance of different components of the circuit is written. Braking
capacity of CB is also written on the diagram.
See the
following single line figure and its electric circuit of reactance figure below
it.
n the reactance
diagram the compontne and their reactances are shown
Percentage Reactance:
For saving of time in
calculations the reactance is expressed in percentage. This is defined as the
percentage of the total phse dropped in the circuit when full load current is
flowing.
%X = IX / V x 100
I = F.L. current in
amphere
X = Reactance in Ω per phase
V = Phase voltage
& reactance can
also be expressed in t4erms of kVA and kV as given below from the above
(%X)V
X = -------------
100
– 1
Simplifying
i.e. multiplying numerator and denominator by V.
% X. V. V %X.(V/100) (V/100) x 100 %X.(Kv)2 X 10
X =
---------------- = ------------------------------------- = ---------------------
100 I.V.
100 I (V/1000) X I Kva
Kva.x
%X = -------------
Where, X is reactance in Ω.
10(kV)2
If only
reactance is in the circuit (omitting resistance, capacitance etc). Then shor
circuiting current,
V
%X.V x 100
Isc = -----
= I(100/%X) by putting x = ---------------- from eqn. (i), where I is full load current.
X I
i.e. short circuiting
is obtained by multiplying full load current with 100/%X.
It is seen that for calculations, use of %X rather than ohmic
value as reactance makes calculation more quick and simple.
Base kVA: kVA
ratings of different elements in the system are different. Percentage reactance
depends on kVA rating of the equipment. It becomes necessary to find percentage
reactance of all the elements on a common kVA rating. This kVA rating is known
as Base kVa.
1.
Generally, base kVa is selected by sum of all kVAs of the
system components .OR
2.
Highest kVa rating amongst the elements is taken as base kVa
ratings. OR
3.
Any arbitrary value can be taken as base kVa. Generally, (ii)
is taken as base kVa rating.
Therefore, % reactance at base k”Va = % reactance at rated
kVA x Base kVA/Rated kVA.
Illustration 500 kVA transformer has %X = 5%, highest kVa
equipment in the system is 1000 kVA. The
100
% X at
base kVa = 5 x ------- = 10%
500
1.
Taking 2000 kVa as base kVA, Therefore, % reactance of
Generateor same as 8%.
2000
So %
reactance of transformer = 4 x -------- = 8% , Total %X on the common base = 8
+ 8 = 16%.
1000
Full load
current corresponding to 2000 kVA base.
KVa = √3. V.I./1000
kVA X 1000 2000 x 1000
I =
--------------- = ------------------- = 17.5 Amp
√3.V √3 x 66 x 1000
100 17.5 x 100
Isc
= I x ------- = --------------- = 109.36
Amp
%X 16
2. Taking
base kVa as 4000. Then,
4000 4000
% X of
generator = 8 x ------- = 16%, so % X = 4 x --------- = 16%
2000 1000
Total of X
= 16 + 16 = 32%
4000 x 1000
F.L.
current corresponding to 4000 kVa. = ----------------- = 35 Amp
√3 x 66 x 1000
100 100
Isc
= I x ----- = 35 x ------ = 109.36 same as in case (i)
%X 32
Short Circuit kVA
The product of normal system voltage and the short circuiting
current at the point of fault expressed in kVA is known as sshort circuit kVA.
Let V = System
normal voltage
I = F.L.
current in Amp. at base kVA
%X = %
reactance of the system on the base kVA up to the point of fault.
100
Isc
= I x -----
% X
Now short
circuit kVA for 3-phase circuit,
3.V.Isc
3V 100 3 VI
100 100
=
---------- = ------- x I x ------ = ------- x ------ = Base kVA x -------
1000
1000 % X 1000
% X % X
So S.C. kVa = Base kVa x 100/%X
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