CURRENT INTERRUPTING DEVICES                             CHAPTER - 2

INTRODUCTION

Electrical supply system has different stages like Generation, Transmission, Distribution and Utilization. The system is very complex. The systems are often to be made ON  or OFF  for different purposes. These are also to be protected from faults by automatically making off and resuming after clearing the faults. For these purposes circuit interrupting devices are needed to be introduced in the system at proper places.

Circuit interrupting Devices

1. Fuses                                              2. Drop-out-fuse link

3. Expulsion fuse                                4. Isolators

5. Air break switches                          6. Circuit breakers.

FUSES

A fuse is a short piece of metal inserted in the circuit, which melts when excessive current flows through it and thus breaks the circuit and protects the equipment.

Materials of Fuse: Most commonly used materials for fuse elements are lead, tin, copper, zinc, silver. Silver is used no-a-days. It is best fuse material.

Important Terms Related to Fuses

Current rating of fuse element: It is the (r.m.s. value of) current which the fuse element can carry without over heating or melting.

Fusing current: It is the minimum current at which the fuse element melts and disconnects the circuit protected by it. This value is obviously more than current rating.

Fusing current    I – kd^3/2 ….. d = diameter, k= constant

Fusing current depends upon material of fuse element, length, diameter and type of enclosure.

Fusing factor: It is the ratio of minimum fusing current to current rating of fuse.

        Minimum fusing current

=   ---------------------------------- 

        Current rating of fuse

The value is always greater than one. The smaller the fusing factor the greater the difficulty in avoiding deterioration due to overheating and oxidation at rated current.

Prospective current; it is the r.m.s. value of the first loop of the fault current obtained if the fuse is replaced by an ordinary conductor of negligible resistance.

                                   

Cut-off current: It is the maximum value of fault current actually reached before the fuse melts.

The cut-off value depends upon:

1.    Normal current rating of fuse

2.    Prospective current.

3.    The asymmetry of circuit waveform

Thus, the fuse breaks the circuit before the short circuit current reaches to its peak. This is an advantage of fuse over CB because peak fault current produces most severe electro-dynamic stresses and thermal effects. And because the fuse breaks the fault current (S.C. current) before its peak value, the circuit to be protected can be designed according to cut-off value. And cut-off value is less than peak value of S.C. current. The system is relatively cheaper and thus saving in cost is achieved.

Pre-arcing time: It is the time between commencement of fault and instant when cut-off occurs. Typical value 0.001 sec.

Arcing time: It is the time between the end of pre-arcing time and instant when arc is extinguished.

Total operating time; It is the sum of pre-arcing time and arcing time.

The operating time of fuse is quite low (0.002 sec) as compared to C.B. (0.2 sec). This is an added advantage.

Breaking capacity: It is the r.m.s. value of AC component of maximum prospective current that a fuse can deal with a rated service voltage.

Desirable Characteristic of a Fuse Element:

1.    Low melting point

2.    High conductivity

3.    Free from oxidation

4.    Low cost.

Advantages and Disadvantages of Fuse

Advantages:

1.    Very cheap in cost

2.    Maintenance is minimum

3.    Automatic operation

4.    Breaks the circuit without smoke and noise

5.    Inverse time current characteristic therefore suitable in over current protection

6.    Minimum time of operation.

Disadvantages:

1.    Refusing requires considerable time

2.    Discrimination between fuses in series cannot be obtained in short circuits.

3.    Current-time characteristic of a fuse cannot always be co-related with that of the protected apparatus.

Time Current Characteristic of Fuse

The time required to blow out the fuse depends on the strength of excessive current. The greater the current, smaller is the time taken by the fuse to blow off i.,e. fuse has time-current inverse characteristic. Study of this characteristic facilitates the selection of fuse for over current protection. The characteristic of different fuses are shown below in figure.

                                     

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 I_sc = 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 I_sc is controlled, hence suitable small capacity of CBs. can work safely and hence economical.

2.    Reactors limit the I_sc 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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