DC Machines & Synchronous Machines 10EE54
Question Bank Solutions
UNIT-1: DC Generator:
1.Derive the expression for emf induced in a DC generator with usual notations.
Dec 2013, Dec 2014
Let, = Flux / pole in webers
Change in flux d = P webers
Z = Total number armature conductors
= Number of slots x Number of conductors per slot
P = Number of poles ; A = Number of parallel paths in the armature.
N = Rotational speed of armature in revolutions per minute (r.p.m)
Time taken to complete one revolution = 60/N sec. E = e.m.f induced / parallel path in armature.
Generated e.m.f Eg = e.m.f generated /parallel path
By Faraday‟s Laws of Electromagnetic induction W K T ,
E.M.F generatedper conductor d PN
dt60 volts
Z
Number of armature conductorsper parallel path
A
Eg e.m.f generatedper conductorNumber of conductors ineachparallel path
Eg 60PN AZ volts...... (i)
For a Simplex Wave-Wound Generator, Number of parallel paths A=2;
E PN. Z2 ZPNvolts g60120
For Simplex Lap-Wound Generator:
Number of parallel paths, A = P.
2.Draw the no load characteristic of a DC shunt generator and comment on its shape.
Explain how critical resistance can be found out from the no load characteristics.
Dec 2014, June 2015,
Sol: When the armature is rotating with armature open circuited, an emf is induced in the armature because of the residual flux. When the field winding is connected with the armature, a =rent flows through the field winding ( in case of shunt field winding, field current flows even at No-load and in case of series field winding only with load) and produces additional flux. This additional flux along with the residual flux generates higher voltage. This higher voltage circulates more current to generate further higher voltage. This is a cumulative process till the saturation is attained. The voltage to which it builds is decided by the resistance of the field winding as shown in the figure 1.11. If field circuit resistance is increased such that the resistance line does not cut OCC like om in the figure 1.11, then the machine will fail to build up voltage to the rated value. The slope of the as gap lire dawn as a tangent to the initial linear portion of the curve represents the maximum resistance that the field circuit can have beyond which the machine fails to build up voltage. This value of field circuit resistance is called critical feld resistance. The field circuit is generally designed to have a resistance value less than this so that the machine builds up the voltage to the aced value.
Fig1. Critical Field Resistance of a Dc Shunt Generator
3.A 6 pole, 148 shunt shunt generator has 480 conductors and is wave wound. Its field current is 2A. Find the magnetizing and cross magnetizing amp turns /pole at full load if i) the brushes are at the geometrical neutral axis (GNA) ii) the brushes are shifted from GNA by 5° elec. iii) The brushes are shifted from GNA by 5° mech.
Dec 2013, Dec2014,June 2013
4.What is meant by commutation? Briefly explain the steps involved in commutation process. June 2013 , Dec 2013, June 2014
Sol; Fig.below shows the schematic diagram of 2-pole lap-wound generator. There are two parallel paths between the brushes. Therefore, each coil of the winding carries one half (Ia/2 in this case) of the total current (Ia) entering or leaving the armature.
Note that the currents in the coils connected to a brush are either all towards the brush (positive brush) or all directed away from the brush (negative brush). Therefore, current in a coil will reverse as the coil passes a brush. This reversal of current as the coil passes & brush is called commutation.
The reversal of current in a coil as the coil passes the brush axis is called commutation.
When commutation takes place, the coil undergoing commutation is short circuited by the brush. The brief period during which the coil remains short circuited is known as commutation period Tc. If the current reversal is completed by the end of commutation period, it is called ideal commutation. If the current reversal is not completed by that time, then sparking occurs between the brush and the commutator which results in progressive damage to both.
Ideal commutation
Let us discuss the phenomenon of ideal commutation (i.e., coil has no inductance) in one coil in the armature winding shown in Fig. (2.6) above. For this purpose, we consider the coil A. The brush width is equal to the width of one commutator segment and one mica insulation. Suppose the total armature current is 40 A. Since there are two parallel paths, each coil carries a current of 20 A.
(i)In Fig. (i), the brush is in contact with segment 1 of the commutator. The commutator segment 1 conducts a current of 40 A to the brush; 20 A from coil A and 20 A from the adjacent coil as shown. The coil A has yet to undergo commutation
(ii)As the armature rotates, the brush will make contact with segment 2 and thus short-circuits the coil A as shown in Fig. (2.7) (ii). There are now two parallel paths into the brush as long as the short-circuit of coil A exists. Fig. (2.7) (ii) shows the instant when the brush is one-fourth on segment 2 and three-fourth on segment 1. For this condition, the resistance of the path through segment 2 is three times the resistance of the path through segment 1 (Q contact resistance varies inversely as the area of contact of brush with the segment). The brush again conducts a current of 40 A; 30 A through segment 1 and 10 A through segment 2. Note that current in coil A (the coil undergoing commutation) is reduced from 20 A to 10 A.
(iii)Fig. (2.7) (iii) shows the instant when the brush is one-half on segment 2 and one-half on segment 1. The brush again conducts 40 A; 20 A through segment 1 and 20 A through segment 2 (Q now the resistances of the two parallel paths are equal). Note that now. current in coil A is zero. (iv) Fig. (2.7) (iv) shows the instant when the brush is three-fourth on segment 2 and one-fourth on segment 1. The brush conducts a current of 40 A; 30 A through segment 2 and 10 A through segment 1. Note that current in coil A is 10 A but in the reverse direction to that before the start of commutation. The reader may see the action of the commutator in reversing the current in a coil as the coil passes the brush axis.
Fig. (v) shows the instant when the brush is in contact only with segment 2. The brush again conducts 40 A; 20 A from coil A and 20 A from the adjacent coil to coil A. Note that now current in coil A is 20 A but in the reverse direction. Thus the coil A has undergone commutation. Each coil undergoes commutation in this way as it passes the brush axis. Note that during commutation, the coil under consideration remains short circuited by the brush. Fig. (2.8) shows the current-time graph for the coil A undergoing commutation. The horizontal line AB represents a constant current of 20 A upto the beginning of commutation. From the finish of commutation, it is represented by another horizontal line CD on the opposite side of the zero line and the same distance from it as AB
i.e., the current has exactly reversed (- 20 A). The way in which current changes from B to C depends upon the conditions under which the coil undergoes commutation. If the current changes at a uniform rate (i.e., BC is a straight line), then it is called ideal commutation as shown in Fig. (2.8). under such conditions, no sparking will take place between the brush and the commutator.
5.How would you classifies the types of generators (D.C) and explain them in brief, along
with their diagrams. Dec 2013, June 2015
Sol: Depending on the type of excitation of field winding, there are two basic types of DC machine. 1. Separately excited machine: In this type of machines the field flux is produced by connecting the field winding to an external source.
2. Self excited machine: The field flux is produced by connecting the field winding with the armature in this type. A self excited machine requires residual magnetism for operation. Depending on the type of field winding connection DC machines can be further classified as:
1.Shunt machine: The field winding consisting of large number of turns of thin wire is usually excited in parallel with armature circuit and hence the name shunt field winding. This winding will be having more resistance and hence carries less current.
2.Series machine: The field winding has a few turns of thick wire and is connected in series with armature.
3.Compound machine: Compound wound machine comprises of both series and shunt windings and
In separate excited DC machine, the field winding is connected to a separate DC source. This type of machine is most flexible as full and independent control of both armature and field circuit is possible. Figure 1.1 shows separately excited DC generator. Permanent magnet machines also fall in this category.
A self excited DC generator could be excited by its armature voltage as shown in figure 1.2 (i.e., shunt excitation) or by its armature current as shown in figure 1.3 (series excitation). Compound wound generator comprises of both series and shunt windings and can be either short shunt (figure 1.4) or long shunt (figure 1.5), cumulative or differential or flat compounded
6. Explain what lap and wave windings arc and where they are preferred in DC machines.
June 2014
1.Armature winding is a closed winding. Depending on the type of winding, the closed path gets divided into number of parallel paths and is available between the positive and negative brushes.
2.Wave winding is used for high voltage low current machines.
3.Equalizing rings are not required in wave winding where as there are used in lap winding.
4.Lap winding is suitable for low voltage high current machines because of more number of parallel paths.
In case of lap winding, the number of parallel path (A) = number of poles (P) In case of wave winding, the number of parallel path (A) = 2 irrespective of number of poles. Each path will have -AZ conductors connected in series.
7. Explain armature reaction with neat figures and derive equations for, i) DC magnetizing ampere tams/pole and ii) Gas magnetizing ampere turns/pole. Dec 2014
Sol: The current flowing through armature conductors also creates a magnetic flux (called armature flux) that distorts and weakens the flux coming from the poles. This distortion and field weakening takes place in both generators and motors. The action of armature flux on the main flux is known as armature reaction.The phenomenon of armature reaction in a d.c. generator is shown in Fig. Only one
distribution produces the following two effects:
(i)The main flux is distorted.
(ii)Due to higher flux density at pole tip B, saturation sets in.
Consequently, the increase in flux at pole tip B is less than the decrease in flux under pole tip A. Flux at full load is, therefore, less than flux at no load. As we shall see, the weakening of flux due to armature reaction depends upon the position of brushes.
Demagnetizing and Cross-Magnetizing Conductors
With the brushes in the G.N.A. position, there is only cross-magnetizing effect of armature reaction. However, when the brushes are shifted from the G.N.A. position, the armature reaction will have both demagnetizing and cross magnetizing effects. Consider a 2-pole generator with brushes shifted (lead) qm mechanical degrees from G.N.A. We shall identify the armature conductors that produce demagnetizing effect and those that produce cross-magnetizing effect.
(i)The armature conductor‟s oqm on either side of G.N.A. produce flux in direct opposition to main flux as shown in Fig. (2.4) (i). Thus the conductors lying within angles AOC = BOD = 2qm at the top and bottom ofthe armature produce demagnetizing effect. These are called demagnetizing armature conductors and constitute the demagnetizing ampere-turns of armature reaction (Remember two conductors constitute a turn).
(ii)The axis of magnetization of the remaining armature conductors lying between angles AOD and COB is at right angles to the main flux as shown in Fig. (2.4) (ii). These conductors produce the crossmagnetizing (or distorting) effect i.e., they produce uneven flux distribution on each pole. Therefore, they are called cross-magnetizing conductors and constitute the cross-magnetizing ampere-turns of armature reaction.
8. A long shunt compound generator delivers a load current of 50A at 500V, and has armature, series field resistance of 0.05 ohm and 0.03 ohm, 250 ohm respectively. Calculate the generated voltage and the armature current, allow 1 volt per brush for contact drop. Dec2013, June 2015
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