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Overcurrent Protection

Types of overcurrent system
 Where a source of electrical energy feeds directly to a single load, little
complication in the circuit protection is required beyond the provision of an
overcurrent device which is suitable in operating characteristics for the
load in question, i
...
appropriate current setting possibly with a time-lag to
permit harmless short time overloads to be supplied
...
It is usually not good enough to shut down the
whole system for every fault along the line
...
Circuit
breakers are installed at the feeding end of each line section and a graded
scheme of protection is applied
...
Where relays are used, they will usually be energised via currenttransformers, which latter are included in the above statement
...


 The magnitude of an interphase fault current will normally be governed by
the known impedances of the power plant and transmission lines; such
currents are usually large
...
g
...

In consequence, earth-fault current may be of low or moderate value and
also often rather uncertain in magnitude particularly on account of item (c)
above
...


 Faults, however, rarely occur in this way; a short circuit on the
system will immediately establish a large current of many times the
trip settings likely to be adopted, and would cause all the tripping
devices to operate simultaneously
...


 One might attempt to set the circuit breaker trips to just operate with
the expected fault current at the end of the associated feeder
section, but this would not be successful since:

a) It is not practicable to distinguish in magnitude between faults at
F1 and F2, since these two points may, in the limit, be separated
by no more than the path through the circuit breaker
...
g
...

b) In the diagram, the fault current at F1 is given as 8800 A when
the source busbar fault rating is 250MVA
...
A minimum source power of
130MVA may be assumed for illustration in this example, for
which the corresponding fault current at F2 would be only 5400
A and for a fault close up to Station A the current would be 6850
A
...
It is,
therefore, clear that discrimination by current setting is not in
general feasible
...

 In this arrangement, instantaneous overcurrent relays are used in
the role of starters or fault detectors
...
All settings are subject to a
tolerance; if all relays were given the same nominal setting, some
would in fact operate at a lower current value than others
...
This relay
might be located at the supply end of the feeder and would shut
down the entire system
...
A relay at D might have
instantaneous operation with a high current setting which will not
permit operation with a fault at E
...


 Usually a time-lag of 0
...

 The relay at C may be set to operate in 0
...
e
...
7 s, and those at B and A will be progressively slower by
the same amount, giving an operating time for relay A of 1
...

 Since the timing is not related to fault current, but is based only on
position, the difficulty that was discussed for current grading does
not exist here
...

 The minimum amount of power system is thereby isolated, although
for a fault on any but the last section, some disconnection unfaulted
sections and loss of load is inevitable
...

 At substations B, C and D, loads are connected through
transformers
...

Grading by both time and current: inverse-time overcurrent systems

 The disadvantage of the definite time system mentioned above, is
reduced by the use of protective devices with an inverse timecurrent characteristic in a graded system
...
An
exact inverse ratio has been used in this diagram in order that it
may be seen that the effects obtained in the grading are general
and not related to any specific device other than that the
device has an inverse type of characteristic
...
The current scale is in multiples
of setting current
...
protected at A and B by
the corresponding devices, is shown with a fault in alternative
positions immediately after stations A and B
...
5 and 1
...
There
is therefore a 'grading margin' of 0
...

 Alternatively, if fault F2 occurred, the fault current through A is now
8
...
59 s which is little
slower than device B with close-up fault F1
...

 This gain can be made at every stage of grading for a multisection
feeder so that the tripping time, for a fault close up to the power
source, may be very much shorter than would have been possible
with a definite-time system
...

Curve A is the most commonly used characteristic and is
standardised by BS 142: Parts 1-4, 1990 (1993)
...

 Steeper characteristics curves B and C are known as 'very inverse'
and 'extremely inverse' characteristics, respectively, and have
special applications
...
This cumbersome state of affairs is
simplified by realising that a relay of a given type always has the
same ampere-turn loading of its winding at setting and for each

operating time point, the coil turns being in inverse ratio to the
setting
...

 During study of electrical protective relays, some special terms are
frequently used
...
Such terms are,





Pick up current
Current setting
Plug setting multiplier (PSM)
Time setting multiplier (TSM)

 In all electrical relays, the moving contacts are not free to move
...
This force is called controlling
force of the relay
...
The force applied on
the relay’s moving parts for changing the normal position of the
contacts, is called deflecting force
...

 Although the deflecting force always presents in the relay directly
connected to live line, but as the magnitude of this force is less than
controlling force in normal condition, the relay does not operate
...
Once,
the deflecting force crosses the controlling force, the moving parts
of the relay initiate to move to change the position of the contacts in
the relay
...

Current Setting of Relay
 The minimum pick up value of the deflecting force of an electrical
relay is constant
...

 The current setting of relay is expressed in percentage ratio of relay
pick up current to rated secondary current of CT
...
If the relay is
rated with 1 A, the normal pick up current of the relay is 1 A and it
should be equal to secondary rated current of current transformer
connected to the relay
...
25 A
...

The current setting of over current relay is generally ranged from
50% to 200%, in steps of 25%
...

 Plug setting multiplier of relay is referred as ratio of fault current in
the relay to its pick up current
...
Hence, pick up current of the relay is, 1
× 150 % = 1
...
Hence,
fault current in the CT secondary i
...
in the relay coil is, 1000
× 1/200 = 5 A
 Therefore PSM of the relay is, 5 / 1
...
33
 Primary setting currents should be graded so that the relay furthest
from the power source has the lowest setting and each preceding
relay back towards the source has a higher setting than that
following
...
t
...


Time Setting Multiplier of Relay
 The operating time of an electrical relay mainly depends upon two factors :
a) How long distance to be travelled by the moving parts of the relay for
closing relay contacts and
b) How fast the moving parts of the relay cover this distance
...
The
adjustment of travelling distance of an electromechanical relay is commonly
known as time setting
...
The time setting dial is calibrated from 0 to 1 in steps 0
...

 But by adjusting only time setting multiplier, we cannot set the actual time of
operation of an electrical relay
...
The speed of moving parts of relay
depends upon the force due to current in the relay coil
...

 In other words, time of operation of relay depends upon plug setting multiplier
...
05
...
1, the moving parts of the relay has to travel only
0
...
So, if
we get total operating time of the relay for a particular PSM from time / PSM
graph and if we multiply that time with the time setting multiplier, we will get,
actual time of operation of relay for said PSM and TSM
...
1 and you have to calculate actual
time of operation for PSM 10
...
That means, the moving parts of the
relay take total 3 seconds to travel 100% travelling distance
...
1 here, actually the moving parts of the relay have to
travel only 0
...

 Hence, actual operating time of the relay is 3 × 0
...
3 sec
...
e
...

Calculation of Relay Operation Time
 For calculating actual relay operating time, we need to know these following
operation
...

Fault current level
...

Time / PSM curve
...


 Step – 1
From CT ratio, we first see the rated secondary current of CT
...
e
...

 Step – 2
From current setting we calculate the pick up current of the relay
...
5 A
...
For that,
we have to first divide primary faulty current by CT ratio to get relay faulty
current
...
From the curve, say we found the time of
operation of relay is 3 second for PSM = 10
...
Hence say time
setting of the relay is 0
...

Therefore, actual time of operation of the relay for PSM 10, is 3 × 0
...
3
sec or 300 ms

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