Steps in Solving
Numerical Examples on Symmentrical Faults
1.
Drae single line diagram representing all elements with
symbols.
2.
Put all details such as kVA capacity, voltage, % reactance
near the elements.
3.
Choose base kVA/MVA and convert all % X to this base kVA/MVA.
4.
Draw reactance diagram indicating %Xs.
5.
Find % X of the network up to the point of fault by network
reduction.
6.
Find full load current (I) correspondig to selected base kVa
and the normal system voltage at the fault point.
100
7.
Find Isc = I x ------
%X
S.C. KVA =
Base kVA x 100/%X
USE OF CURRENT
LIMINTING REACTORS AND THEIR ARRANGEMENT
In the previous articles, we noted theat the short circuiting
currents under fault conditions arae of very high values. These heavy currents
suggest to use the C.Bs. of higher capacity which will not be economical more
over there is a possibility of damaging the equipment in the power system.
This suggests that these fault currents (short circuiting
currents) as they are not controllable by the system and equipment reactances
some additional reactances shall be introduced in the systsem to limit the
short circuiting current to the safe
value such that C.Bs. are not overburdoned and find capable of
sustaining these currents.
The solution is to introduce inductive coils having
sufficient number of turns so that their reactace is higher. Such inductive
coil located (connected) at proper places in the system are called as
“reactors”. So reactors are high reactance coils with negligibly small
resistane.
The short circuit currents induce vey high stresses due to
the fault forces and it is likely that the coil turns of the reactors jmay be
affected. Due to this reason the reactor coils are mechanically protected and
solidly braced.
Advantages of
Reactors
1.
By intorducing reactors in the power system the magnitude of
Isc is controlled, hence suitable small capacity of CBs. can work
safely and hence economical.
2.
Reactors limit the Isc and hence provide
protection to the equipment from overheating and avoid failure due to
mechanical stresses.
3.
Introduction of reactors in the system ensures continuity of
supply. Troubles are isolated.
Location of Reactors
The reactors are therefore known as:
1)
Generator Reactors, 2) Feeder Reactors, 3) Bus bar Reactors
(Ring and Tie bar system)
1. Generator Reactor: The coil
of the reactor is connected electrically in series with the generator. This
reactor is considered as part of leakage reactance of the generator. Thus,
reactor protects the generator in case of short circuit beyond the reactor.
This type of reactor has the following drawbacks.
i)
Even in the normal operation (no fault) there is a constant
voltage drop (IZ) and power loss (I2R) in the system, hence
efficiency is affected.
ii)
If the fault occurs on any feeder continuity of supply to the
remaining feeders may be affected.
iii)
Generator may be pulled out to step if the fault occurs close
to bus bar which reduces the voltage at bus bar.
So separate reactors are rarely used as modern generators are
design with ‘considerable reactance’.
2. Feeder Reactors:
Many of the faults occur in the feeder lines and hence many reactors arae
connected in feeders to control the short circuit current to the safe value to
protect the system.
Advantages:
i)
Fault on one feeder does not affect the remaining feeders.
ii)
Due to fault the voltage drop in the reactror does not affect
the bus bar voltage. So synchronism of generators is not affected and all will
remain in the system.
Disadvantages:
I)
In normal operation also there is a voltage drop due to
reactance of reactor and also a power loss (I2R).
II)
If short circuit occurs at bus bars no protection is provided
to generators but it is not that serious as modern generators are designed for
high reactance.
III)
For increasing the units of generators in the existing
system, more reactors have to be provided to keep I within limits.
3. Bus-bar Reactors : As discussed in the
generator reactors and feeder reactors the scheme suffers from voltage drop and
power loss. This drawback is removed in the bus-bar reactors locating them in
the bus-bar itself.
According
to the location the bus-bar reactors are:
i)
Ring system bus-bar reactor.
ii)
Tie-bar system bus-bar reactor.
(i) Ring system: As shown in figure the reactors are
connected between the sections fo the bus-bar. One feeder if generally fed from
one generator only. In the normal working of the system each generator supplies
its own section and a very little power is supplkied to that section by
remaining generators. Hence, low power loss and less voltage drop in he
reactors.
If the fault occurss on any feeder only one generator to
which that feeder is connected mainly feeds that fault. Very little curreent is
supplied by the remaining generators to the fault due to preesence of reactors
in the feeder sections. Only that section of bus-bar is affected to which he
feeder is connected. Other sections work normal.
(ii) Tie-bar System: As seen in the
figure there are two reactors in series act between the sections. So reactance
of reactors are ½ value as that of in ring main system. One more advantage in
the system is that on adding the generator units no change is necessary in the
existing reactors.
The only
drawback is that additional bus bars (Tie bus-bars) are required.
EXERCISE
1.
State
the procedure to calculate short circuit kVA and current.
2.
What
are the possible faults on power system?
3.
Compare
reactors for merits and demerits.
4.
State
the functions of protective system.
5.
What
are the different types of faults occur in power system?
6.
What
are the reasons for fault occurrence?
7.
Derive
the relations for calculating:
a)
Short
circuit current.
b)
Short
circuit kVA.
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