Commutation in dc machines

COMMUTATION

The process of reversal of current in the short circuited armature coil is called ‘Commutation’. This process of reversal takes place when coil is passing through the interpolar axis (q-axis), the coil is short circuited through commutator segments. Commutation takes place simultaneously for ‘P’ coils in a lap-wound machine and two coil sets of P/2 coils each in a wave-wound machine.

The process of commutation of coil ‘B’ is shown below. In figure ‘1.29’ coil ‘B’ carries current from left to right and is about to be short circuited in figure ‘1.30’ brush has moved by 1/3 rd of its width and the brush current supplied by the coil are as shown. In figure ‘1.31’ coil ‘B’ carries no current as the brush is at the middle of the short circuit period and the brush current in supplied by coil C and coil A. In figure ‘1.32’ the coil B which was carrying current from left to right carries current from right to left. In fig ‘1.33’ spark is shown which is due to the reactance voltage. As the coil is embedded in the armature slots, which has high permeability, the coil posses appreciable amount of self inductance. The current is changed from +I to –I. So due to self inductance and variation in the current from +I to –I, a voltage is induced in the coil which is given by L dI/dt. Fig ‘1.34’ shows the variation of current plotted on the time axis. Sparking can be avoided by the use of interpoles or commutating-poles.

VALUE OF REACTANCE VOLTAGE:

Reactance voltage = co-efficient of self inductance X rate of change of current

Time of short circuit = Tc = (time required by commutator to move a distance equal to the circumferential thickness of brush)–(one mica insulating strip).

If brush width is given in terms of commutator segments, then commutator velocity should be converted in terms of commutator segments/seconds.

METHODS OF IMPROVING COMMUTATION:

There are two methods of improving commutation. They are (i) resistance commutation(ii) E.M.F commutation.

(i) Resistance commutation: In this method low resistance copper brushes are replaced by high resistance carbon brushes. From the figure 1.35. It is seen that when current ‘I’ from coil ‘C’ reaches the commutator segment ‘b’, it has two parallel paths opened to it. The first path is straight from bar ‘b’ to the brush and the other is via short circuited coil B to bar ‘a’ and then to brush. If copper brushes are used the current will follow the first path because of its low contact resistance. But when carbon brushes having high resistance are used, then current ‘I’ will prefer the second path because the resistance r1 of first path will increase due to reducing area of contact with bar ‘b’ and the resistance r2 of second path decreases due to increasing area of contact with bar ‘c’. Hence carbon brushes help in obtaining sparkles commutation. Also, carbon brushes lubricate and polish commutator. But, because of high resistance the brush contact drop increases and the commutator has to be made larger to dissipate the heat due to loss. Carbon brushes require larger brush holders because of lower current density.

(ii) E.M.F commutation: in this method, reactance voltage which is the cause for sparking is neutralized by producing an emf which is in opposite direction to that of reactance voltage, so that the reactance voltage is completely eliminated. The reversing emf may be produced in two ways (i) by giving a forward lead sufficient enough to bring the short circuited coil under the influence of next pole of opposite polarity or (ii) by using inter poles or compoles. The second method is commonly employed.

INTERPOLES OR COMPOLES

These are small poles fixed to the yoke and placed in between the main poles as shown in figure 1.36. They are wound with few turns of heavy gauge copper wire and are connected in series with the armature so that they carry full armature current. Their polarity in case of generator is that of the main pole ahead in the direction of rotation.

The function of interpoles is (i) to induce an emf which is equal and Figure 1.36 opposite to that of reactance emf thereby making commutation sparkles. (ii) Interpoles neutralize the cross magnetizing effect of armature reaction in fig 1.36. ‘OF’ represents mmf due to main poles and ‘OA’ represents the cross magnetizing mmf due to armature. ‘BC’ represents mmf due to Interpoles and is in opposite to that of ‘OA’ resulting in the cancellation of cross magnetization.

EQUILISER RINGS

Equalizer rings are used in connection with the lap winding. It is the characteristics of lap winding that all conductors in any parallel path will be under one pair of poles. If the fluxes from all poles are exactly the same, then emf induced in each parallel path is same and carries the same current. If there is any inequality in the flux/pole due to slight variations in the air gap or in the magnetic properties of steel, there will be imbalance of emf in various parallel paths resulting in unequal distribution of current at the brushes. This leads to poor commutation. By connecting together a number of symmetrical points on armature winding which would be at equal potential, the difference in brush current will be minimized. The equalizer conductors which are in the form of copper rings are connected to equi-potential points on the backside of the armature. Such rings are called as ‘Equalizer rings’.

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

Equalizer rings are not used in wave wound armatures because there is no imbalance in the emf of the two parallel paths. This is due to the fact that armature conductors are distributed under all poles. Hence even if there are inequalities in the flux/pole they will affect all the paths equally.

USES OF DC GENERATOR

(i) Shunt Generators with field regulators are used for lighting and power supply purposes. They are also used for charging batteries because their terminal voltage is almost constant.

(ii) Series Generators are not used for power supply because of their rising characteristics but their rising characteristic suits to be used as boosters.

(iii) Compound Generators maintain almost constant terminal voltage over a large range of load.

Hence they are used where large load is suddenly thrown on and off. Also they are used for power supplies.

PROBLEMS

1. A wave wound 4 pole D.C Generator with 480 armature conductors supplies a current of 144A, the brushes are given an actual lead of 100. Calculate the demagnetizing and cross magnetizing Amp turns/pole.

2. An 8 pole generator has an output of 200A at 500V; the lap connected arm has 1280 conductors, 160 commutator segments. If the brushes are advanced by 4 segments from the no load neutral axis, estimate the armature demagnetizing and cross-magnetizing ampere turns/pole.

3.A 100kW 500V 6 pole D.C shunt generator has a lap armature of 600conductors. If the brushes are given an actual lead of 100, determine the demagnetizing & distorting ampere turns/pole. The shunt field winding has a resistance of 50.0.

4. A pole wave wound motor armature has 880 conductors & delivers 120A.The brushes have been displaced through 3 angular degrees from the geometrical axis. calculate (a) demagnetizing ampere-turns/pole, (b) cross-magnetizing ampere-turns/pole, (c) the additional field current for neutralizing the demagnetization if the field winding has 1100turns/pole.

5. The armature of a dynamo runs at 800rpm. The commutator consists of 123 segments & the thickness of each brush is such that the brush spans three segments. Find the time during which the coil of armature remains short circuited.

6. A 4 pole wave wound D.C machine running at 1500rpm has a commutator of 30cm diameter. If armature current is 150A, thickness of brush 1.25cm & the self inductance of each armature coil is 0.07mH. Calculate the average value of emf induced in each coil during commutation. Assume linear commutation.

7. A 4 pole lap wound arm running at 1500rpm delivers a current of 150A & has 64 commutator segments. The brush spans 1.2 segments & inductance of each arm coil is 0.05mH. Calculate the value of reactance voltage assuming (i) linear commutation & (ii) sinusoidal commutation.

8. Calculate the No of conductors on each pole piece required in a compensating winding for a 6pole lap wound D.C armature containing 286 conductors. The compensating winding carries full armature current. Assume ratio of pole arc to pole pitch = 0.7.

SOLUTION:ATcw/pole = compensating winding ampere turns per pole

9. The open circuit characteristics of a D.C shunt Generator driven at rated speed is as follows.

If Rsh is adjusted to 53 Ω, calculate the O.C voltage and load current when the terminal voltage is 100 V. Neglect armature reaction and assume an armature resistance of 0.1 Ω.

10. A 240 V compound (cumulative) DC motor has the following O.C.C at normal full load speed of 850 rev/min.

Excitation AT/Pole	1200	2400	3600	4800	6000
Generated emf	76	135	180	215	240

The resistance voltage drop in the armature circuit at full load is 25 V at fullload. Theshunt

and the series winding provide equal AT excitation. Calculate the mmfper pole on No - load. Estimate the value to which the speed will rise when full load is removed the resistance voltage drop in the armature circuit under that condition being 3V. ignore armature reaction and brush contact effects. Assume long shunt cumulative compounding.

SOLUTION: