Armature reaction and commutation:Commutation

Commutation

It was shown in Art 26.2 that currents induced in armature conductors of a d.c. generator are alternating. To make their flow unidirectional in the external circuit, we need a commutator. More- over, these currents flow in one direction when armature conductors are under N-pole and in the opposite direction when they are under S-pole. As conductors pass out of the influence of a N-pole and enter that of S-pole, the current in them is reversed. This reversal of current takes place along magnetic neutral axis or brush axis i.e. when the brush spans and hence short- circuits that particular coil undergoing reversal of current through it. This process by which current in the short-circuited coil is reversed while it crosses the M.N.A. is called commutation. The brief period during which coil remains short-circuited is known as commutation period Tc.

If the current reversal i.e. the change from + I to zero and then to – I is completed by the end of short circuit or commutation period, then the commutation is ideal. If cur- rent reversal is not complete by that time, then sparking is produced between the brush and the commutator which re- sults in progressive damage to both.

Let us discuss the process of commutation or current reversal in more detail with the help of Fig. 27.10 where ring winding has been used for simplicity. The brush width is equal to the width of one commutator segment and one mica insulation. In Fig. 27.10 (a) coil B is about to be short circuited because brush is about to come in touch with commutator segment ‘a’. It is assumed that each coil carries Commutation 20 A, so that brush current is 40 A. It is so because every coil meeting at the brush supplies half the brush current lap wound or wave wound. Prior to the beginning of short circuit, coil B belongs to the group of coils lying to the left of the brush and carries 20 A from left to right. In Fig. 27.10 (b) coil B has entered its period of short-circuit and is approximately at one-third of this period. The current through coil B has reduced down from 20 A to 10 A because the other 10 A flows via segment ‘a’. As area of contact of the brush is more with segment ‘b’ than with segment ‘a’, it receives 30 A from the former, the total again being 40 A.

Fig. 27.10 (c) shows the coil B in the middle of its short-circuit period. The current through it has decreased to zero. The two currents of value 20 A each, pass to the brush directly from coil A and C as shown. The brush contact areas with the two segments ‘b’ and ‘a’ are equal.

In Fig. 27.10 (d), coil B has become part of the group of coils lying to the right of the brush. It is seen that brush contact area with segment ‘b’ is decreasing rapidly whereas that with segment ‘a’ is increasing. Coil B now carries 10 A in the reverse direction which combines with 20 A supplied by coil A to make up 30 A that passes from segment ‘a’ to the brush. The other 10 A is supplied by coil C and passes from segment ‘b’ to the brush, again giving a total of 40 A at the brush.

Fig. 27.10 (e) depicts the moment when coil B is almost at the end of commutation or short- circuit period. For ideal commutation, current through it should have reversed by now but, as shown, it is carrying 15 A only (instead of 20 A). The difference of current between coils C and B i.e. 20.15 = 5 A in this case, jumps directly from segment b to the brush through air thus producing spark.

If the changes of current through coil B are plotted on a time base (as in Fig. 27.11) it will be represented by a horizontal line AB i.e. a constant current of 20 A up to the time of beginning of commutation. From the finish of commutation, the current will be represented by another horizontal line CD. Now, again the current value is FC = 20 A, although in the reversed direction. The way in which current changes from its positive value of 20 A (= BE) to zero and then to its negative value of 20 A (= CF) depends on the conditions under which the coil B undergoes commutation. If the current varies at a uniform rate i.e. if BC is a straight line, then it is referred to as linear commutation. However, due to the production of self-induced e.m.f. in the coil (discussed below) the variations follow the dotted curve. It is seen that, in that case, current in coil B has reached only a value of KF = 15 A in the reversed direction, hence the difference of 5 A (20 A – 15 A) passes as a spark.

So, we conclude that sparking at the brushes, which results in poor commutation is due to the inability of the current in the short-circuited coil to reverse completely by the end of short-circuit period (which is usually of the order of 1/500 second).

At this stage, the reader might ask for the reasons which make this current reversal impossibly in the specified period i.e. what factors stand in the way of our achieving ideal commutation. The main cause which retards or delays this quick reversal is the production of self-induced e.m.f. in the coil undergoing commutation. It may be pointed out that the coil possesses appreciable amount of self inductance because it lies embedded in the armature which is built up of a material of high magnetic permeability. This self-induced e.m.f. is known as reactance voltage whose value is found as given below. This voltage, even though of a small magnitude, produces a large current through the coil whose resistance is very low due to short circuit. It should be noted that if the brushes are set so that the coils undergoing short-circuit are in the magnetic neutral plane, where they are cutting no flux and hence have no e.m.f. induced in them due to armature rotation, there will still be the e.m.f. of self- induction which causes severe sparking at the brushes.

Value of Reactance Voltage

Reactance voltage = coefficient of self-inductance ´ rate of change of current.

It should be remembered that the time of short-circuit or commutation is the time required by the commutator to move a distance equal to the circumferential thickness of the brush minus the thick- ness of one insulating plate of strip of mica.

As said earlier, the reactance e.m.f. hinders the reversal of current. This means that there would be sparking at the brushes due to the failure of the current in short-circuited coil to reach its full value in the reversed direction by the end of short-circuit. This sparking will not only damage the brush and the commutator but this being a cumulative process, it may worsen and eventually lead to the short- circuit of the whole machine by the formation of an arc round the commutator from brush to brush.

Example 27.9. The armature of a certain dynamo runs at 800 r.p.m. The commutator consists of 123 segments and the thickness of each brush is such that the brush spans three segments. Find the time during which the coil of an armature remains short-circuited.

Solution. As Wm is not given, it is considered negligible.

Wb = 3 segments and n = (800/60) ´ 123 segments/second

Example 27.10. A 4-pole, wave-wound, d.c. machine running at 1500 r.p.m. has a commutator of 30 cm diameter. If armature current is 150 A, thickness of brush 1.25 cm and the self-inductance of each armature coil is 0.07 mH, calculate the average e.m.f. induced in each coil during commutation. Assume linear commutation.

Methods of Improving Commutation

There are two practical ways of improving commutation i.e. of making current reversal in the short-circuited coil as sparkless as possible. These methods are known as (i) resistance commutation and (ii) e.m.f. commutation (which is done with the help of either brush lead or interpoles, usually the later).

Resistance Commutation

This method of improving commutation consists of replacing low-resistance Cu brushes by comparatively high-resistance carbon brushes.

From Fig. 27.12, it is seen that when current I from coil’ C reaches the commutator segment b, it has two parallel paths open to it. The first part is straight from bar ‘b’ to the brush and the other parallel path is via the short-circuited coil B to bar ‘a’ and then to the brush. If the Cu brushes (which have low contact resistance) are used, then there is no inducement for the current to follow the second longer path, it would preferably follow the first path. But when carbon brushes having high resistance are used, then current I coming from C will prefer to pass through the second path because

(i) the resistance r1 of the first path will increase due to the diminishing area of contact of bar ‘b’ with the brush and because (ii) resistance r2 of second path will decrease due to rapidly increasing contact area of bar ‘a’ with the brush.

Hence, carbon brushes have, usually, replaced Cu brushes. However, it should be clearly under- stood that the main cause of sparking commutation is the self-induced e.m.f. (i.e. reactance voltage), so brushes alone do not give a sparkless commutation; though they do help in obtaining it.

The additional advantages of carbon brushes are that (i) they are to some degree self-lubricating and polish the commutator and (ii) should sparking occur, they would damage the commutator less than when Cu brushes are used.

But some of their minor disadvantages are : (i) Due to their high contact resistance (which is beneficial to sparkless commutation) a loss of approximately 2 volt is caused. Hence, they are not much suitable for small machines where this voltage forms an appreciable percentage loss. (ii) Owing to this large loss, the commutator has to be made some what larger than with Cu brushes in order to dissipate heat efficiently without greater rise of temperature. (iii) because of their lower current density (about 7-8 A/cm2 as compared to 25-30 A/cm2 for Cu brushes) they need larger brush holders.

E.M.F. Commutation

In this method, arrangement is made to neutralize the reactance voltage by producing a reversing e.m.f. in the short-circuited coil under commutation. This reversing e.m.f., as the name shows, is an e.m.f. in opposition to the reactance voltage and if its value is made equal to the latter, it will completely wipe it off, thereby producing quick reversal of current in the short-circuited coil which will result in sparkless commutation. The reversing e.m.f. may be produced in two ways : (i) either by giving the brushes a forward lead sufficient enough to bring the short-circuited coil under the influence of next pole of opposite polarity or (ii) by using interpoles.

The first method was used in the early machines but has now been abandoned due to many other difficulties it brings along with.

Interpoles of Compoles

These are small poles fixed to the yoke and spaced in between the main poles. They are wound with comparatively few heavy gauge Cu wire turns and are connected in series with the armature so that they carry full armature current. Their polarity, in the case of a generator, is the same as that of the main pole ahead in the direction of rotation (Fig. 25.13).

The function of interpoles is two-fold :

(i) As their polarity is the same as that of the main pole ahead, they induce an e.m.f. in the coil (under commutation) which helps the reversal of current. The e.m.f. induced by the compoles is known as commutating or reversing e.m.f. The commutating e.m.f. neutralizes the reactance e.m.f. thereby making commutation sparkless. With interpoles, sparkless commutation can be obtained up to 20 to 30% overload with fixed brush position. In fact, interpoles raise sparking limit of a machine to almost the same value as heating limit. Hence, for a given output, an interpole machine can be made smaller and, therefore, cheaper than a non-interpolar machine.

As interpoles carry armature current, their commutating is proportional to the armature current. This ensures automatic neutralization of reactance voltage which is also due to armature current. Connections for a shunt generator with interpoles are shown in Fig. 27.14.

(ii) Another function of the interpoles is to neutralize the cross-magnetising effect of armature reaction. Hence, brushes are not to be shifted from the original position. In Fig 27.15, OF as before, represents the m.m.f. due to main poles. OA represents the cross-magnetising m.m.f. due to armature. BC which repre- sents m.m.f. due to interpoles, is obviously in opposition to OA, hence they cancel each other out. This cancellation of cross magnetisation is automatic and for all loads because both are produced by the same armature current.

The distinction between the interpoles and compensating windings should be clearly understood. Both are connected in series and thier m.m.fs. are such as to neutralize armature reaction. But compoles additionally supply m.m.f. for counteracting the reactance voltage induced in the coil undergoing commutation. Moreover, the action of the compoles is localized, they have negligible effect on the armature reaction occurring on the remainder of the armature periphery.

Example 27.13. Determine the number of turns on each commutating pole of a 6-pole machine, if the flux density in the air-gap of the commutating pole = 0.5 Wb/m2 at full load and the effective length of the air-gap is 4 mm. The full-load current is 500 A and the armature is lap-wound with 540 conductors. Assume the ampere turns required for the remainder of the magnetic circuit to be one- tenth of that the air gap. (Advanced Elect. Machines AMIE Sec.B, 1991)

Solution. It should be kept in mind that compole winding must be sufficient to oppose the armature m.m.f. (which is directed along compole axis) and to provide the m.m.f. for compole air-gap and its magnetic circuit.

Equalizing Connections

It is characteristic of lap -winding that all conductors in any parallel path lie under one pair of poles. If fluxes from all poles are exactly the same, then e.m.f. induced in each parallel path is the same and each path carries the same current. But despite best efforts, some inequalities in flux inevitably occur due either to slight variations in air-gap length or in the magnetic properties of steel. Hence, there is always a slight imbalance of e.m.f. in the various parallel paths. The result is that conductors under stronger poles generate greater e.m.f. and hence carry larger current. The current distribution at the brushes becomes unequal. Some brushes are overloaded i.e. carry more than their normal current whereas others carry less. Overloaded brushes spark badly whatever their position may be. This results in poor commutation and may even limit the output of the machine.

By connecting together a number of symmetrical points on armature winding which would be at equal potential if the pole fluxes were equal, the difference in brush currents is diminished. This requires that there should be a whole number of slots per pair of poles so that, for example, if there is a slot under the centre of a N–pole, at some instant, then there would be one slot under the centre of every other N–pole. The equalizing conductors, which are in the form of Cu rings at the armature back and which connect such points are called Equalizer Rings. The circulating current due to the slight difference in the e.m.fs. of various parallel paths, passes through these equalizer rings instead of passing through the brushes.

Hence, the function of equalizer rings is to avoid unequal distribution of current at the brushes thereby helping to get sparkless commutation.

One equalizer ring is connected to all conductors in the armature which are two poles apart (Fig. 27.17). For example, if the number of poles is 6, then the number of connections for each equalizer ring is 3 i.e. equal to the number of pair of poles. Maximum number of equalizer rings is equal to the number of conductors under one pair of poles. Hence, number of rings is

In practice, however, the number of rings is limited to 20 on the largest machines and less on smaller machines. In Fig.

27.16 is shown a developed armature winding. Here, only 4 equalizing bars have been used. It will be seen that the number of equalizing connections to each bar is two i.e. half the number of poles. Each alternate coil has been connected to the bar. In this case, the winding is said to be 50% equalized. If all conductors were connected to the equalizer rings, then the winding would have been 100% equalized.

Equalizer rings are not used in wave-wound armatures, because there is no imbalance in the e.m.fs. of the two parallel paths. This is due to the fact that armature conductors in either parallel path are not confined under one pair of poles as in lap-winding but are distributed under all poles. Hence, even if there are inequalities in the pole flux, they will affect each path equally.

Parallel Operation of Shunt Generators

Power plants, whether in d.c. or a.c. stations, will be generally found to have several smaller generators running in parallel rather than large single units capable of supplying the maximum peak load. These smaller units can be run single or in various parallel combinations to suit the actual load demand. Such practice is considered extremely desirable for the following reasons :

(i) Continuity of Service

Continuity of service is one of the most important requirements of any electrical apparatus. This would be impossible if the power plant consisted only of a single unit, because in the event of break- down of the prime mover or the generator itself, the entire station will be shut down. In recent years, the requirement of uninterrupted service has become so important especially in factories etc. that it is now recognized as an economic necessity.

(ii) Efficiency

Usually, the load on the electrical power plant fluctuates between its peak value sometimes during the day and its minimum value during the late night hours. Since generators operate most efficiently when delivering full load, it is economical to use a single small unit when the load is light. Then, as the load demand increases, a larger generator can be substituted for the smaller one or another smaller unit can be connected to run in parallel with the one already in operation.

(iii) Maintenance and Repair

It is considered a good practice to inspect generators carefully and periodically to forestall any possibility of failure or breakdown. This is possible only when the generator is at rest which means that there must be other generators to take care of the load. Moreover, when the generator does actually breakdown, it can be repaired with more care and not in a rush, provided there are other generators available to maintain service.

(iv) Additions to Plant

Additions to power plants are frequently made in order to deliver increasingly greater loads. Provision for future extension is, in fact, made by the design engineers fight from the beginning. It becomes easy to add other generators for parallel operation as the load demand increases.

Paralleling DC Generator

Whenever generators are in parallel, their +ve and -ve terminals are respectively connected to the +ve and -ve sides of the bus-bars. These bus-bars are heavy thick copper bars and they act as +ve and -ve terminals for the whole power station. If polarity of the incoming generator is not the same as the line polar- ity, as serious short-circuit will occur when S1, is closed.

Moreover, paralleling a generator with reverse polarity effectively short-circuits it and results in damaged brushes, a damaged commutator and a blacked-out plant. Generators that have been tripped off the bus-because of a heavy fault current should always be checked for reversed polarity before paralleling.

In Fig. 27.18 is shown a shunt generator No. 1 connected across the bus-bars BB and supplying some of the load. For putting generator No. 2 in parallel with it the following procedure is adopted.

The armature of generator No. 2 is speeded by the prime-mover up to its rated value and then switch S2 is closed and circuit is completed by putting a voltmeter V across the open switch S1. The excitation of the incoming generator No. 2 is changed till V reads zero. Then it means that its terminal voltage is the same as that of generator No. 1 or bus-bar voltage. After this, switch S1 is closed and so the incoming machine is paralleled to the system. Under these conditions, however, generator No. 2 is not taking any load, because its induced e.m.f. is the same as bus-bar voltage and there can be no flow of current between two points at the same potential. The generator is said to be ‘floating’ on the bus- bar. If generator No. 2 is to deliver any current, then its induced e.m.f. E should be greater than the bus-bar voltage V. In that case, current supplied by it is I = (E – V)/Ra where Ra is the resistance of the armature circuit. The induced e.m.f. of the incoming generator is increased by strengthening its field till it takes its proper share of load. At the same time, it may be found necessary to weaken the field of generator No. 1 to maintain the bus-bar voltage V constant.