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Chapter (1), , D.C. Generators, Introduction, Although a far greater percentage of the electrical machines in service are a.c., machines, the d.c. machines are of considerable industrial importance. The, principal advantage of the d.c. machine, particularly the d.c. motor, is that it, provides a fine control of speed. Such an advantage is not claimed by any a.c., motor. However, d.c. generators are not as common as they used to be, because, direct current, when required, is mainly obtained from an a.c. supply by the use, of rectifiers. Nevertheless, an understanding of d.c. generator is important, because it represents a logical introduction to the behaviour of d.c. motors., Indeed many d.c. motors in industry actually operate as d.c. generators for a, brief period. In this chapter, we shall deal with various aspects of d.c., generators., , 1.1 Generator Principle, An electric generator is a machine that converts mechanical energy into, electrical energy. An electric generator is based on the principle that whenever, flux is cut by a conductor, an e.m.f. is induced which will cause a current to flow, if the conductor circuit is closed. The direction of induced e.m.f. (and hence, current) is given by Fleming’s right hand rule. Therefore, the essential, components of a generator are:, (a) a magnetic field, (b) conductor or a group of conductors, (c) motion of conductor w.r.t. magnetic field., , 1.2 Simple Loop Generator, Consider a single turn loop ABCD rotating clockwise in a uniform magnetic, field with a constant speed as shown in Fig.(1.1). As the loop rotates, the flux, linking the coil sides AB and CD changes continuously. Hence the e.m.f., induced in these coil sides also changes but the e.m.f. induced in one coil side, adds to that induced in the other., (i) When the loop is in position no. 1 [See Fig. 1.1], the generated e.m.f. is, zero because the coil sides (AB and CD) are cutting no flux but are, moving parallel to it
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(ii) When the loop is in position no. 2, the coil sides are moving at an angle, to the flux and, therefore, a low e.m.f. is generated as indicated by point, 2 in Fig. (1.2)., (iii) When the loop is in position no. 3, the coil sides (AB and CD) are at, right angle to the flux and are, therefore, cutting the flux at a maximum, rate. Hence at this instant, the generated e.m.f. is maximum as indicated, by point 3 in Fig. (1.2)., (iv) At position 4, the generated e.m.f. is less because the coil sides are, cutting the flux at an angle., (v) At position 5, no magnetic lines are cut and hence induced e.m.f. is zero, as indicated by point 5 in Fig. (1.2)., (vi) At position 6, the coil sides move under a pole of opposite polarity and, hence the direction of generated e.m.f. is reversed. The maximum e.m.f., in this direction (i.e., reverse direction, See Fig. 1.2) will be when the, loop is at position 7 and zero when at position 1. This cycle repeats with, each revolution of the coil., , Fig. (1.1), , Fig. (1.2), , Note that e.m.f. generated in the loop is alternating one. It is because any coil, side, say AB has e.m.f. in one direction when under the influence of N-pole and, in the other direction when under the influence of S-pole. If a load is connected, across the ends of the loop, then alternating current will flow through the load., The alternating voltage generated in the loop can be converted into direct, voltage by a device called commutator. We then have the d.c. generator. In fact,, a commutator is a mechanical rectifier., , 1.3 Action Of Commutator, If, somehow, connection of the coil side to the external load is reversed at the, same instant the current in the coil side reverses, the current through the load
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will be direct current. This is what a commutator does. Fig. (1.3) shows a, commutator having two segments C1 and C2. It consists of a cylindrical metal, ring cut into two halves or segments C1 and C2 respectively separated by a thin, sheet of mica. The commutator is mounted on but insulated from the rotor shaft., The ends of coil sides AB and CD are connected to the segments C1 and C2, respectively as shown in Fig. (1.4). Two stationary carbon brushes rest on the, commutator and lead current to the external load. With this arrangement, the, commutator at all times connects the coil side under S-pole to the +ve brush and, that under N-pole to the −ve brush., (i) In Fig. (1.4), the coil sides AB and CD are under N-pole and S-pole, respectively. Note that segment C1 connects the coil side AB to point P, of the load resistance R and the segment C2 connects the coil side CD to, point Q of the load. Also note the direction of current through load. It is, from Q to P., (ii) After half a revolution of the loop (i.e., 180° rotation), the coil side AB is, under S-pole and the coil side CD under N-pole as shown in Fig. (1.5)., The currents in the coil sides now flow in the reverse direction but the, segments C1 and C2 have also moved through 180° i.e., segment C1 is, now in contact with +ve brush and segment C2 in contact with −ve brush., Note that commutator has reversed the coil connections to the load i.e.,, coil side AB is now connected to point Q of the load and coil side CD to, the point P of the load. Also note the direction of current through the, load. It is again from Q to P., , Fig.(1.3), , Fig.(1.4), , Fig.(1.5), , Thus the alternating voltage generated in the loop will appear as direct voltage, across the brushes. The reader may note that e.m.f. generated in the armature, winding of a d.c. generator is alternating one. It is by the use of commutator that, we convert the generated alternating e.m.f. into direct voltage. The purpose of, brushes is simply to lead current from the rotating loop or winding to the, external stationary load.
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The variation of voltage across the brushes, with the angular displacement of the loop, will be as shown in Fig. (1.6). This is not a, steady direct voltage but has a pulsating, character. It is because the voltage, appearing across the brushes varies from, zero to maximum value and back to zero, twice for each revolution of the loop. A, Fig. (1.6), pulsating direct voltage such as is produced, by a single loop is not suitable for many, commercial uses. What we require is the steady direct voltage. This can be, achieved by using a large number of coils connected in series. The resulting, arrangement is known as armature winding., , 1.4 Construction of d.c. Generator, The d.c. generators and d.c. motors have the same general construction. In fact,, when the machine is being assembled, the workmen usually do not know, whether it is a d.c. generator or motor. Any d.c. generator can be run as a d.c., motor and vice-versa. All d.c. machines have five principal components viz., (i), field system (ii) armature core (iii) armature winding (iv) commutator (v), brushes [See Fig. 1.7]., , Fig. (1.7), , (i), , Fig. (1.8), , Field system, , The function of the field system is to produce uniform magnetic field within, which the armature rotates. It consists of a number of salient poles (of course,, even number) bolted to the inside of circular frame (generally called yoke). The
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yoke is usually made of solid cast steel whereas the pole pieces are composed of, stacked laminations. Field coils are mounted on the poles and carry the d.c., exciting current. The field coils are connected in such a way that adjacent poles, have opposite polarity., The m.m.f. developed by the field coils produces a magnetic flux that passes, through the pole pieces, the air gap, the armature and the frame (See Fig. 1.8)., Practical d.c. machines have air gaps ranging from 0.5 mm to 1.5 mm. Since, armature and field systems are composed of materials that have high, permeability, most of the m.m.f. of field coils is required to set up flux in the air, gap. By reducing the length of air gap, we can reduce the size of field coils (i.e., number of turns)., , (ii), , Armature core, , The armature core is keyed to the machine shaft and rotates between the field, poles. It consists of slotted soft-iron laminations (about 0.4 to 0.6 mm thick) that, are stacked to form a cylindrical core as shown in Fig (1.9). The laminations, (See Fig. 1.10) are individually coated with a thin insulating film so that they do, not come in electrical contact with each other. The purpose of laminating the, core is to reduce the eddy current loss. The laminations are slotted to, accommodate and provide mechanical security to the armature winding and to, give shorter air gap for the flux to cross between the pole face and the armature, “teeth”., , Fig. (1.9), , Fig. (1.10), , (iii) Armature winding, The slots of the armature core hold insulated conductors that are connected in a, suitable manner. This is known as armature winding. This is the winding in, which “working” e.m.f. is induced. The armature conductors are connected in, series-parallel; the conductors being connected in series so as to increase the
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voltage and in parallel paths so as to increase the current. The armature winding, of a d.c. machine is a closed-circuit winding; the conductors being connected in, a symmetrical manner forming a closed loop or series of closed loops., , (iv) Commutator, A commutator is a mechanical rectifier which converts the alternating voltage, generated in the armature winding into direct voltage across the brushes. The, commutator is made of copper segments insulated from each other by mica, sheets and mounted on the shaft of the machine (See Fig 1.11). The armature, conductors are soldered to the commutator segments in a suitable manner to give, rise to the armature winding. Depending upon the manner in which the armature, conductors are connected to the commutator segments, there are two types of, armature winding in a d.c. machine viz., (a) lap winding (b) wave winding., Great care is taken in building the commutator because any eccentricity will, cause the brushes to bounce, producing unacceptable sparking. The sparks may, bum the brushes and overheat and carbonise the commutator., , (v), , Brushes, , The purpose of brushes is to ensure electrical connections between the rotating, commutator and stationary external load circuit. The brushes are made of carbon, and rest on the commutator. The brush pressure is adjusted by means of, adjustable springs (See Fig. 1.12). If the brush pressure is very large, the friction, produces heating of the commutator and the brushes. On the other hand, if it is, too weak, the imperfect contact with the commutator may produce sparking., , Fig. (1.11), , Fig. (1.12), , Multipole machines have as many brushes as they have poles. For example, a 4pole machine has 4 brushes. As we go round the commutator, the successive, brushes have positive and negative polarities. Brushes having the same polarity
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are connected together so that we have two terminals viz., the +ve terminal and, the −ve terminal., , 1.5 General Features OF D.C. Armature Windings, (i), , A d.c. machine (generator or motor) generally employs windings, distributed in slots over the circumference of the armature core. Each, conductor lies at right angles to the magnetic flux and to the direction of its, movement Therefore, the induced e.m.f. in the conductor is given by;, , e = Bl v, where, , volts, , B = magnetic flux density in Wb/m2, l = length of the conductor in metres, v = velocity (in m/s) of the conductor, , (ii) The armature conductors are connected to form coils. The basic component, of all types of armature windings is the armature coil. Fig. (1.13) (i) shows, a single-turn coil. It has two conductors or coil sides connected at the back, of the armature. Fig. 1.13 (ii) shows a 4-turn coil which has 8 conductors or, coil sides., , Fig. (1.13), The coil sides of a coil are placed a pole span apart i.e., one coil side of the coil, is under N-pole and the other coil side is under the next S-pole at the, corresponding position as shown in Fig. 1.13 (i). Consequently the e.m.f.s of the, coil sides add together. If the e.m.f. induced in one conductor is 2.5 volts, then, the e.m.f. of a single-turn coil will be = 2 × 2.5 = 5 volts. For the same flux and, speed, the e.m.f. of a 4-turn coil will be = 8 × 2.5 = 20 V.
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(iii) Most of d.c. armature windings are, double layer windings i.e., there are, two coil sides per slot as shown in, Fig. (1.14). One coil side of a coil, lies at the top of a slot and the other, coil side lies at the bottom of some, Fig. (1.14), other slot. The coil ends will then lie, side by side. In two-layer winding, it is desirable to number the coil sides, rather than the slots. The coil sides are numbered as indicated in Fig., (1.14). The coil sides at the top of slots are given odd numbers and those at, the bottom are given even numbers. The coil sides are numbered in order, round the armature., As discussed above, each coil has one side at the top of a slot and the other, side at the bottom of another slot; the coil sides are nearly a pole pitch, apart. In connecting the coils, it is ensured that top coil side is joined to the, bottom coil side and vice-versa. This is illustrated in Fig. (1.15). The coil, side 1 at the top of a slot is joined to coil side 10 at the bottom of another, slot about a pole pitch apart. The coil side 12 at the bottom of a slot is, joined to coil side 3 at the top of another slot. How coils are connected at, the back of the armature and at the front (commutator end) will be, discussed in later sections. It may be noted that as far as connecting the, coils is concerned, the number of turns per coil is immaterial. For, simplicity, then, the coils in winding diagrams will be represented as, having only one turn (i.e., two conductors)., , Fig. (1.15), , Fig. (1.16)
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(iv) The coil sides are connected through commutator segments in such a, manner as to form a series-parallel system; a number of conductors are, connected in series so as to increase the voltage and two or more such, series-connected paths in parallel to share the current. Fig. (1.16) shows, how the two coils connected through commutator segments (A, R, C etc), have their e.m.f.s added together. If voltage induced in each conductor is 25 V, then voltage between segments A and C = 4 × 2.5 = 10 V. It may be, noted here that in the conventional way of representing a developed, armature winding, full lines represent top coil sides (i.e., coil sides lying at, the top of a slot) and dotted lines represent the bottom coil sides (i.e., coil, sides lying at the bottom of a slot)., (v) The d.c. armature winding is a closed circuit winding. In such a winding, if, one starts at some point in the winding and traces through the winding, one, will come back to the starting point without passing through any external, connection. D.C. armature windings must be of the closed type in order to, provide for the commutation of the coils., , 1.6 Commutator Pitch (YC), The commutator pitch is the number of commutator segments spanned by each, coil of the winding. It is denoted by YC., In Fig. (1.17), one side of the coil is connected to commutator segment 1 and the, other side connected to commutator segment 2. Therefore, the number of, commutator segments spanned by the coil is 1 i.e., YC = 1. In Fig. (1.18), one, side of the coil is connected to commutator segment 1 and the other side to, commutator segment 8. Therefore, the number of commutator segments spanned, by the coil = 8 − 1 = 7 segments i.e., YC = 7. The commutator pitch of a winding, is always a whole number. Since each coil has two ends and as two coil, connections are joined at each commutator segment,, , Fig. (1.17), , Fig. (1.18)
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∴, , Number of coils = Number of commutator segments, , For example, if an armature has 30 conductors, the number of coils will be 30/2, = 15. Therefore, number of commutator segments is also 15. Note that, commutator pitch is the most important factor in determining the type of d.c., armature winding., , 1.7 Pole-Pitch, It is the distance measured in terms of number of armature slots (or armature, conductors) per pole. Thus if a 4-pole generator has 16 coils, then number of, slots = 16., , ∴, Also, , Pole pitch =, , 16, = 4 slots, 4, , Pole pitch =, , No. of conductors 16 × 2, =, = 8 conductors, No. of poles, 4, , 1.8 Coil Span or Coil Pitch (YS), It is the distance measured in terms of the number of armature slots (or armature, conductors) spanned by a coil. Thus if the coil span is 9 slots, it means one side, of the coil is in slot 1 and the other side in slot 10., , 1.9 Full-Pitched Coil, If the coil-span or coil pitch is equal to pole pitch, it is called full-pitched coil, (See Fig. 1.19). In this case, the e.m.f.s in the coil sides are additive and have a, phase difference of 0°. Therefore, e.m.f. induced in the coil is maximum. If, e.m.f. induced in one coil side is 2-5 V, then e.m.f. across the coil terminals = 2, × 2.5 = 5 V. Therefore, coil span should always be one pole pitch unless there is, a good reason for making it shorter., Fractional pitched coil. If the coil span or coil pitch is less than the pole pitch,, then it is called fractional pitched coil (See Fig. 1.20). In this case, the phase, difference between the e.m.f.s in the two coil sides will not be zero so that the, e.m.f. of the coil will be less compared to full-pitched coil. Fractional pitch, winding requires less copper but if the pitch is too small, an appreciable, reduction in the generated e.m.f. results.
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Fig. (1.19), , Fig. (1.20), , 1.10 Types of D.C. Armature Windings, The different armature coils in a d.c. armature Winding must be connected in, series with each other by means of end connections (back connection and front, connection) in a manner so that the generated voltages of the respective coils, will aid each other in the production of the terminal e.m.f. of the winding. Two, basic methods of making these end connections are:, 1. Simplex lap winding, 2. Simplex wave winding, , 1., , Simplex lap winding., , For a simplex lap winding, the commutator pitch YC = 1 and coil span YS ~, pole pitch. Thus the ends of any coil are brought out to adjacent commutator, segments and the result of this method of connection is that all the coils of the, armature .ire in sequence with the last coil connected to the first coil., Consequently, closed circuit winding results. This is illustrated in Fig. (1.21), where a part of the lap winding is shown. Only two coils are shown for, simplicity. The name lap comes from the way in which successive coils overlap, the preceding one., , 2., , Simplex wave winding, , For a simplex wave winding, the commutator pitch YC ~ 2 pole pitches and coil, span = pole pitch. The result is that the coils under consecutive pole pairs will be, joined together in series thereby adding together their e.m.f.s [See Fig. 1.22]., After passing once around the armature, the winding falls in a slot to the left or
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right of the starting point and thus connecting up another circuit. Continuing in, this way, all the conductors will be connected in a single closed winding. This, winding is called wave winding from the appearance (wavy) of the end, connections., , Fig. (1.21), , Fig. (1.22), , 1.11 Further Armature Winding Terminology, Apart from the terms discussed earlier, the following terminology requires, discussion:, , (i) Back Pitch (YB), It is the distance measured in terms of armature conductors between the two, sides of a coil at the back of the armature (See Fig. 1.23). It is denoted by YB For, example, if a coil is formed by connecting conductor 1 (upper conductor in a, slot) to conductor 12 (bottom conductor in another slot) at the back of the, armature, then back pitch is YB = 12 − 1 = 11 conductors., , Fig. (1.23)
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(ii) Front Pitch (YF), It is the distance measured in terms of armature conductors between the coil, sides attached to any one commutator segment [See Fig. 1.23]. It is denoted by, YF For example, if coil side 12 and coil side 3 are connected to the same, commutator segment, then front pitch is YF = 12 − 3 = 9 conductors., , (iii) Resultant Pitch (YR), It is the distance (measured in terms of armature conductors) between the, beginning of one coil and the beginning of the next coil to which it is connected, (See Fig. 1.23). It is denoted by YR. Therefore, the resultant pitch is the, algebraic sum of the back and front pitches., , (iv) Commutator Pitch (YC), It is the number of commutator segments spanned by each coil of the armature, winding., For simplex lap winding, YC = 1, For simplex wave winding, YC ~ 2 pole pitches (segments), , (v) Progressive Winding, A progressive winding is one in which, as one traces through the winding, the, connections to the commutator will progress around the machine in the same, direction as is being traced along the path of each individual coil. Fig. (1.24) (i), shows progressive lap winding. Note that YB > YF and YC = + 1., , (vi) Retrogressive Winding, A retrogressive winding is one in which, as one traces through the winding, the, connections to the commutator will progress around the machine in the opposite, direction to that which is being traced along the path of each individual coil. Fig., (1.24) (ii) shows retrogressive lap winding. Note that YF > YB and YC = − 1. A, retrogressive winding is seldom used because it requires more copper., , Fig. (1.24)
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1.12 General Rules For D.C. Armature Windings, In the design of d.c. armature winding (lap or wave), the following rules may be, followed:, (i) The back pitch (YB) as well as front pitch (YF) should be nearly equal to, pole pitch. This will result in increased e.m.f. in the coils., (ii) Both pitches (YB and YF) should be odd. This will permit all end, connections (back as well as front connection) between a conductor at, the top of a slot and one at the bottom of a slot., (iii) The number of commutator segments is equal to the number of slots or, coils (or half the number of conductors)., No. of commutator segments = No. of slots = No. of coils, It is because each coil has two ends and two coil connections are joined, at each commutator segment, (iv) The winding must close upon itself i.e. it should be a closed circuit, winding., , 1.13 Relations between Pitches for Simplex Lap Winding, In a simplex lap winding, the various pitches should have the following relation:, (i) The back and front pitches are odd and are of opposite signs. They differ, numerically by 2,, ∴, YB = Y B = Y F ± 2, YB =YF + 2, for progressive winding, YB =YF − 2, for retrogressive winding, (ii) Both YB and YF should be nearly equal to pole pitch., (iii) Average pitch =(YB + YF)/2. It equals pole pitch (= Z/P)., (iv) Commutator pitch, YC = ± 1, YC = + 1 for progressive winding, YC = − 1 for retrogressive winding, (v) The resultant pitch (YB) is even, being the arithmetical difference of two, odd numbers viz., YB and YF., (vi) If Z = number of armature conductors and P = number of poles, then,, , Polr - pitch =, , Z, P, , Since YB and YF both must be. about one pole pitch and differ numerically by 2,, , Z, +1, P, Z, YF = − 1, P, , YB =, , , , , , , , For progressive winding
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Z, −1, P, Z, YF = + 1, P, YB =, , , , , , , , For retrogressive winding, , It is clear that Z/P must be an even number to make the winding possible., , Developed diagram, Developed diagram is obtained by imagining the cylindrical surface of the, armature to be cut by an axial plane and then flattened out. Fig. (1.25) (i) shows, the developed diagram of the winding. Note that full lines represent the top coil, sides (or conductors) and dotted lines represent the bottom coil sides (or, conductors)., The winding goes from commutator segment 1 by conductor 1 across the back, to conductor 12 and at the front to commutator segment 2, thus forming a coil., Then from commutator segment 2, through conductors 3 and 14 back to, commutator segment 3 and so on till the winding returns to commutator segment, 1 after using all the 40 conductors., , Position and number of brushes, We now turn to find the position and the number of brushes required. The, brushes, like field poles, remain fixed in space as the commutator and winding, revolve. It is very important that brushes are in correct position relative to the, field poles. The arrowhead marked “rotation” in Fig. (1.25) (i) shows the, direction of motion of the conductors. By right-hand rule, the direction of e.m.f., in each conductor will be as shown., In order to find the position of brushes, the ring diagram shown in Fig. (1.25) (ii), is quite helpful. A positive brush will be placed on that commutator segment, where the currents in the coils are meeting to flow out of the segment. A, negative brush will be placed on that commutator segment where the currents in, the coils are meeting to flow in. Referring to Fig. (1.25) (i), there are four, brushes—two positive and two negative. Therefore, we arrive at a very, important conclusion that in a simplex lap winding, the number of brushes is, equal to the number of poles. If the brushes of the same polarity are connected, together, then all the armature conductors are connected in four parallel paths;, each path containing an equal number of conductors in series. This is illustrated, in Fig. (1.26)., Since segments 6 and 16 are connected together through positive brushes and, segments 11 and 1 are connected together through negative brushes, there are, four parallel paths, each containing 10 conductors in series. Therefore, in a, simplex lap winding, the number of parallel paths is equal to the number of, poles.
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(i), , (ii), Fig. (1.25)
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Fig. (1.26), , Conclusions, From the above discussion, the following conclusions can be drawn:, (i) The total number of brushes is equal to the number of poles., (ii) The armature winding is divided into as many parallel paths as the, number of poles. If the total number of armature conductors is Z and P is, the number of poles, then,, Number of conductors/path = Z/P, In the present case, there are 40 armature conductors and 4 poles., Therefore, the armature winding has 4 parallel paths, each consisting of, 10 conductors in series., (iii), E.M.F. generated = E.M.F. per parallel path, Z, = average e.m.f. per conductor ×, P, (iv) Total armature current, Ia = P × current per parallel path, (v) The armature resistance can be found as under:, Let l = length of each conductor; a = cross-sectional area, A = number of parallel paths = P, for simplex lap winding, Resistance of whole winding, R =, Resistance per parallel path =, , ρl, ×Z, a, , R ρlZ, =, A a×A, , Since there are A (= P) parallel paths, armature resistance Ra is given by:, , Ra =, ∴, , Ra =, , Resistance per parallel path 1 ρlZ , = , , A, Aa×A, , ρlZ, aA 2
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1.14 Simplex Wave Winding, The essential difference between a lap winding and a wave winding is in the, commutator connections. In a simplex lap winding, the coils approximately pole, pitch apart are connected in series and the commutator pitch YC = ± 1 segment., As a result, the coil voltages add. This is illustrated in Fig. (1.27). In a simplex, wave winding, the coils approximately pole pitch apart are connected in series, and the commutator pitch YC ~ 2 pole pitches (segments). Thus in a wave, winding, successive coils “wave” forward under successive poles instead of, “lapping” back on themselves as in the lap winding. This is illustrated in Fig., (1.28)., The simplex wave winding must not close after it passes once around the, armature but it must connect to a commutator segment adjacent to the first and, the next coil must be adjacent to the first as indicated in Fig. (1.28). This is, repeated each time around until connections are made to all the commutator, segments and all the slots are occupied after which the winding automatically, returns to the starting point. If, after passing once around the armature, the, winding connects to a segment to the left of the starting point, the winding is, retrogressive [See Fig. 1.28 (i)]. If it connects to a segment to the right of the, starting point, it is progressive [See Fig. 1.28 (ii)]. This type of winding is called, wave winding because it passes around the armature in a wave-like form., , Fig. (1.27), , Fig. (1.28), , Various pitches, The various pitches in a wave winding are defined in a manner similar to lap, winding., (i) The distance measured in terms of armature conductors between the two, sides of a coil at the back of the armature is called back pitch YB (See, Fig. 1.29). The YB must be an odd integer so that a top conductor and a, bottom conductor will be joined.
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(ii) The distance measured in, terms, of, armature, conductors between the coil, sides attached to any one, commutator segment is, called front pitch YB (See, Fig. 1.29). The YB must be, an odd integer so that a top, Fig. (1.29)), conductor and a bottom, conductor will be joined., (iii) Resultant pitch, YR = YB + YF, (See Fig. 1.29), The resultant pitch must be an even integer since YB and YF are odd., Further YR is approximately two pole pitches because YB as well as YF is, approximately one pole pitch., Y + YF, (iv) Average pitch, YA = B, . When one tour of armature has been, 2, completed, the winding should connect to the next top conductor, (progressive) or to the preceding top conductor (retrogressive). In either, case, the difference will be of 2 conductors or one slot. If P is the number, of poles and Z is the total number of armature conductors, then,, , P × YA = Z ± 2, , YA =, , or, , Z+2, P, , (i), , Since P is always even and Z = PYA ± 2, Z must be even. It means that Z, ± 2/P must be an integer. In Eq.(i), plus sign will give progressive, winding and the negative sign retrogressive winding., (v) The number of commutator segments spanned by a coil is called, commutator pitch (YC) (See Fig. 1.29). Suppose in a simplex wave, winding,, P = Number of poles; NC = Number of commutator segments;, YC = Commutator pitch., ∴, , Number of pair of poles = P/2, , If YC × P/2 = NC, then the winding will close on itself in passing once around, the armature. In order to connect to the adjacent conductor and permit the, winding to proceed,, , or, , YC ×, , P, = NC ± 1, 2, , YC =, , 2 N C ± 2 N C ± 1 No. of commutator seg. ± 1, =, =, P, P/2, Number of pair of poles
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YC =, , Now, , ∴, , 2N C ± 2 Z ± 2, =, = YA, P, P, , (Q, , Commutator pitch, YC = YA =, , 2NC = Z), , YB + YF, 2, , In a simplex wave winding YB, YF and YC may be equal. Note that YB, YF and, YB are in terms of armature conductors whereas YC is in terms of commutator, segments., , 1.15 Design of Simplex Wave Winding, In the design of simplex wave winding, the following points may be kept in, mind:, (i) Both pitches YB and YF are odd and are of the same sign., (ii), , Average pitch, YA =, , Z±2, P, , (i), , (iii) Both YB and YF are nearly equal to pole pitch and may be equal or differ, by 2. If they differ by 2, they are one more and one less than YA., (iv) Commutator pitch is given by;, , YC = YA =, , Number of commutator segments ± 1, Number of pair of poles, , The plus sign for progressive winding and negative for retrogressive, winding., (v), , YA =, , Z±2, P, , Since YA must be a whole number, there is a restriction on the value of, Z. With Z = 180, this winding is impossible for a 4-pole machine, because YA is not a whole number., (vi), , Z = P YA ± 2, , ∴, , Numnber of coils =, , Z PYA ± 2, =, 2, 2, , Developed diagram, Fig. (1.30) (i) shows the developed diagram for the winding. Note that full lines, represent the top coil sides (or conductors) and dotted lines represent the bottom, coil sides (or conductors). The two conductors which lie in the same slot are, drawn nearer to each other than to those in the other slots.
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(i), , (ii), Fig. (1.30)
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Referring to Fig. (1.30) (i), conductor 1 connects at the back to conductor 12(1 +, 11) which in turn connects at the front to conductor 23 (12 + 11) and so on, round the armature until the winding is complete. Note that the commutator, pitch YC = 11 segments. This means that the number of commutator segments, spanned between the start end and finish end of any coil is 11 segments., , Position and number of brushes, We now turn to find the position and the number of brushes. The arrowhead, marked “rotation” in Fig. (1.30) (i) shows the direction of motion of the, conductors. By right hand rule, the direction of e.m.f. in each conductor will be, as shown., In order to find the position of brushes, the ring diagram shown in Fig. (1.30) (ii), is quite helpful. It is clear that only two brushes—one positive and one, negative—are required (though two positive and two negative brushes can also, be used). We find that there are two parallel paths between the positive brush, and the negative brush. Thus is illustrated in Fig. (1.31)., Therefore, we arrive at a very important conclusion that in a simplex wave, winding, the number of parallel paths is two irrespective of the number of poles., Note that the first parallel path has 11 coils (or 22 conductors) while the second, parallel path has 10 coils (or 20 conductors). This fact is not important as it may, appear at first glance. The coils m the smaller group should supply less current, to the external circuit. But the identity of the coils in either parallel path is, rapidly changing from moment to moment. Therefore, the average value of, current through any particular coil is the same., , Fig. (1.31), , Conclusions, From the above discussion, the following conclusions can be drawn:, (i) Only two brushes are necessary but as many brushes as there are poles, may be used., (ii) The armature winding is divided into two parallel paths irrespective of, the number of poles. If the total number of armature conductors is Z and, P is the number of poles, then,
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Number of conductors/path =, , Z, 2, , (iii) E.M.F. generated = E.M.F. per parallel path, = Average e.m.f. per conductor x —, (iv) Total armature current, Ia = 2 × current per parallel path, (v) The armature can be wave-wound if YA or YC is a whole number., , 1.16 Dummy Coils, In a simplex wave winding, the average pitch YA (or commutator pitch YC), should be a whole number. Sometimes the standard armature punchings, available in the market have slots that do not satisfy the above requirement so, that more coils (usually only one more) are provided than can be utilized. These, extra coils are called dummy or dead coils. The dummy coil is inserted into the, slots in the same way as the others to make the armature dynamically balanced, but it is not a part of the armature winding., Let us illustrate the use of dummy coils with a numerical example. Suppose the, number of slots is 22 and each slot contains 2 conductors. The number of poles, is 4. For simplex wave wound armature,, , YA =, , Z ± 2 2 × 22 ± 2 44 ± 2, 1, 1, =, =, = 11 or 10, P, 4, 4, 2, 2, , Since the results are not whole numbers, the number of coils (and hence, segments) must be reduced. If we make one coil dummy, we have 42 conductors, and, , YA =, , 42 ± 2, = 11 or 10, 4, , This means that armature can be wound only if we use 21 coils and 21 segments., The extra coil or dummy coil is put in the slot. One end of this coil is taped and, the other end connected to the unused commutator segment (segment 22) for the, sake of appearance. Since only 21 segments are required, the two (21 and 22, segments) are connected together and considered as one., , 1.17 Applications of Lap and Wave Windings, In multipolar machines, for a given number of poles (P) and armature, conductors (Z), a wave winding has a higher terminal voltage than a lap winding, because it has more conductors in series. On the other hand, the lap winding, carries more current than a wave winding because it has more parallel paths.
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In small machines, the current-carrying capacity of the armature conductors is, not critical and in order to achieve suitable voltages, wave windings are used., On the other hand, in large machines suitable voltages are easily obtained, because of the availability of large number of armature conductors and the, current carrying capacity is more critical. Hence in large machines, lap windings, are used., Note: In general, a high-current armature is lap-wound to provide a large, number of parallel paths and a low-current armature is wave-wound to, provide a small number of parallel paths., , 1.18 Multiplex Windings, A simplex lap-wound armature has as many parallel paths as the number of, poles. A simplex wave-wound armature has two parallel paths irrespective of the, number of poles. In case of a 10-pole machine, using simplex windings, the, designer is restricted to either two parallel circuits (wave) or ten parallel circuits, (lap). Sometimes it is desirable to increase the number of parallel paths. For this, purpose, multiplex windings are used. The sole purpose of multiplex windings is, to increase the number of parallel paths enabling the armature to carry a large, total current. The degree of multiplicity or plex determines the number of, parallel paths in the following manner:, (i), , A lap winding has pole times the degree of plex parallel paths., Number of parallel paths, A = P × plex, Thus a duplex lap winding has 2P parallel paths, triplex lap winding has 3P, parallel paths and so on. If an armature is changed from simplex lap to, duplex lap without making any other change, the number of parallel paths, is doubled and each path has half as many coils. The armature will then, supply twice as much current at half the voltage., , (ii) A wave winding has two times the degree of plex parallel paths., Number of parallel paths, A = 2 × plex, Note that the number of parallel paths in a multiplex wave winding, depends upon the degree of plex and not on the number of poles. Thus a, duplex wave winding has 4 parallel paths, triplex wave winding has 6, parallel paths and so on., , 1.19 Function of Commutator and Brushes, The e.m.f. generated in the armature winding of a d.c. generator is alternating, one. The commutator and brushes cause the alternating e.m.f. of the armature, conductors to produce a p.d. always in the same direction between the terminals, of the generator. In lap as well as wave winding, it will be observed that currents
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in the coils to a brush are either all directed towards the brush (positive brush) or, all directed away from the brush (negative brush). Further, the direction of, current in coil reverses as it passes the brush. Thus when the coil approaches the, contact with the brush, the current through the coil is in one direction; when the, coil leaves the contact with the brush, the current has been reversed. This, reversal of current in the coil as the coil passes a brush is called commutation, and fakes place while the coil is short-circuited by the brush. These changes, occur in every coil in turn. If, at the instant when the brush breaks contact with, the commutator segment connected to the coil undergoing commutation, the, current in the coil has not been reversed, the result will be sparking between the, commutator segments and the brush., The criterion of good commutation is that it should be sparkless. In order to, have sparkless commutation, the brushes on the commutator should be placed at, points known as neutral point where no voltage exists between adjacent, segments. The conductors connected to these segments lie between the poles in, position of zero magnetic flux which is termed as magnetic neutral axis (M.N.A), , 1.20 E.M.F. Equation of a D.C. Generator, We shall now derive an expression for the e.m.f. generated in a d.c. generator., Let, φ = flux/pole in Wb, Z = total number of armature conductors, P = number of poles, A = number of parallel paths = 2 ... for wave winding, = P ... for lap winding, N = speed of armature in r.p.m., Eg = e.m.f. of the generator = e.m.f./parallel path, Flux cut by one conductor in one revolution of the armature,, dφ = Pφ webers, Time taken to complete one revolution,, dt = 60/N second, Pφ N, dφ, Pφ, e.m.f generated/conductor =, =, =, volts, dt 60 / N, 60, e.m.f. of generator,, Eg = e.m.f. per parallel path, = (e.m.f/conductor) × No. of conductors in series per parallel path, Pφ N Z, ×, =, 60, A, Pφ ZN, ∴, Eg =, 60 A, where, A=2, for-wave winding
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=P, , for lap winding, , 1.21 Armature Resistance (Ra), The resistance offered by the armature circuit is known as armature resistance, (Ra) and includes:, (i) resistance of armature winding, (ii) resistance of brushes, The armature resistance depends upon the construction of machine. Except for, small machines, its value is generally less than 1Ω., , 1.22 Types of D.C. Generators, The magnetic field in a d.c. generator is normally produced by electromagnets, rather than permanent magnets. Generators are generally classified according to, their methods of field excitation. On this basis, d.c. generators are divided into, the following two classes:, (i) Separately excited d.c. generators, (ii) Self-excited d.c. generators, The behaviour of a d.c. generator on load depends upon the method of field, excitation adopted., , 1.23 Separately Excited D.C. Generators, A d.c. generator whose field magnet winding is supplied from an independent, external d.c. source (e.g., a battery etc.) is called a separately excited generator., Fig. (1.32) shows the connections of a separately excited generator. The voltage, output depends upon the speed of rotation of armature and the field current (Eg =, Pφ ZN/60 A). The greater the speed and field current, greater is the generated, e.m.f. It may be noted that separately excited d.c. generators are rarely used in, practice. The d.c. generators are normally of self-excited type., , Fig. (1.32)
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Armature current, Ia = IL, Terminal voltage, V = Eg − IaRa, Electric power developed = EgIa, Power delivered to load = E g I a − I 2a R a = I a E g − I a R a = VI a, , (, , ), , 1.24 Self-Excited D.C. Generators, A d.c. generator whose field magnet winding is supplied current from the output, of the generator itself is called a self-excited generator. There are three types of, self-excited generators depending upon the manner in which the field winding is, connected to the armature, namely;, (i), Series generator;, (ii) Shunt generator;, (iii) Compound generator, , (i), , Series generator, , In a series wound generator, the field winding is connected in series with, armature winding so that whole armature current flows through the field, winding as well as the load. Fig. (1.33) shows the connections of a series wound, generator. Since the field winding carries the whole of load current, it has a few, turns of thick wire having low resistance. Series generators are rarely used, except for special purposes e.g., as boosters., Armature current, Ia = Ise = IL = I(say), Terminal voltage, V = EG − I(Ra + Rse), Power developed in armature = EgIa, Power delivered to load, = E g I a − I 2a (R a + R se ) = I a E g − I a (R a − R se ) = VI a or VI L, , [, , Fig. (1.33), , ], , Fig. (1.34)
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(ii), , Shunt generator, , In a shunt generator, the field winding is connected in parallel with the armature, winding so that terminal voltage of the generator is applied across it. The shunt, field winding has many turns of fine wire having high resistance. Therefore,, only a part of armature current flows through shunt field winding and the rest, flows through the load. Fig. (1.34) shows the connections of a shunt-wound, generator., Shunt field current, Ish = V/Rsh, Armature current, Ia = IL + Ish, Terminal voltage, V = Eg − IaRa, Power developed in armature = EgIa, Power delivered to load = VIL, , (iii) Compound generator, In a compound-wound generator, there are two sets of field windings on each, pole—one is in series and the other in parallel with the armature. A compound, wound generator may be:, (a) Short Shunt in which only shunt field winding is in parallel with the, armature winding [See Fig. 1.35 (i)]., (b) Long Shunt in which shunt field winding is in parallel with both series, field and armature winding [See Fig. 1.35 (ii)]., , Fig. (1.35), Short shunt, Series field current, Ise = IL, V + Ise R se, Shunt field current, I sh =, R sh, Terminal voltage, V = Eg − IaRa − IseRse, Power developed in armature = EgIa, Power delivered to load = VIL
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Long shunt, Series field current, Ise = Ia = IL + Ish, Shunt field current, Ish = V/Rsh, Terminal voltage, V = Eg − Ia(Ra + Rse), Power developed in armature = EgIa, Power delivered to load = VIL, , 1.25 Brush Contact Drop, It is the voltage drop over the brush contact resistance when current flows., Obviously, its value will depend upon the amount of current flowing and the, value of contact resistance. This drop is generally small., , 1.26 Losses in a D.C. Machine, The losses in a d.c. machine (generator or motor) may be divided into three, classes viz (i) copper losses (ii) iron or core losses and (iii) mechanical losses., All these losses appear as heat and thus raise the temperature of the machine., They also lower the efficiency of the machine., , 1., , Copper losses, These losses occur due to currents in the various windings of the machine., (i) Armature copper loss = I 2a R a, 2, (ii) Shunt field copper loss = I sh, R sh, 2, (iii) Series field copper loss = I se, R se, , Note. There is also brush contact loss due to brush contact resistance (i.e.,, resistance between the surface of brush and surface of commutator). This, loss is generally included in armature copper loss.
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2., , Iron or Core losses, These losses occur in the armature of a d.c. machine and are due to the rotation, of armature in the magnetic field of the poles. They are of two types viz., (i), hysteresis loss (ii) eddy current loss., (i) Hysteresis loss, Hysteresis loss occurs in the, armature of the d.c. machine since, any given part of the armature is, subjected to magnetic field reversals, as it passes under successive poles., Fig. (1.36), Fig. (1.36) shows an armature, rotating in two-pole machine. Consider a small piece ab of the armature. When, the piece ab is under N-pole, the magnetic lines pass from a to b. Half a, revolution later, the same piece of iron is under S-pole and magnetic lines pass, from b to a so that magnetism in the iron is reversed. In order to reverse, continuously the molecular magnets in the armature core, some amount of, power has to be spent which is called hysteresis loss. It is given by Steinmetz, formula. This formula is, Hysteresis loss, Ph = η B16, max f V, where, , watts, , Bmax = Maximum flux density in armature, f = Frequency of magnetic reversals, = NP/120 where N is in r.p.m., V = Volume of armature in m3, η = Steinmetz hysteresis co-efficient, , In order to reduce this loss in a d.c. machine, armature core is made of such, materials which have a low value of Steinmetz hysteresis co-efficient e.g.,, silicon steel., (ii) Eddy current loss, In addition to the voltages induced in the armature conductors, there are also, voltages induced in the armature core. These voltages produce circulating, currents in the armature core as shown in Fig. (1.37). These are called eddy, currents and power loss due to their flow is called eddy current loss. The eddy, current loss appears as heat which raises the temperature of the machine and, lowers its efficiency., If a continuous solid iron core is used, the resistance to eddy current path will be, small due to large cross-sectional area of the core. Consequently, the magnitude, of eddy current and hence eddy current loss will be large. The magnitude of, eddy current can be reduced by making core resistance as high as practical. The
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core resistance can be greatly increased by constructing the core of thin, round, iron sheets called laminations [See Fig. 1.38]. The laminations are insulated, from each other with a coating of varnish. The insulating coating has a high, resistance, so very little current flows from one lamination to the other. Also,, because each lamination is very thin, the resistance to current flowing through, the width of a lamination is also quite large. Thus laminating a core increases, the core resistance which decreases the eddy current and hence the eddy current, loss., , Fig. (1.37), Eddy current loss, Pe = K e B 2max f 2 t 2 V, where, , Fig. (1.38), , watts, , Ke = Constant depending upon the electrical resistance of core and, system of units used, Bmax = Maximum flux density in Wb/m2, f = Frequency of magnetic reversals in Hz, t = Thickness of lamination in m, V = Volume of core in m3, , It may be noted that eddy current loss depends upon the square of lamination, thickness. For this reason, lamination thickness should be kept as small as, possible., , 3., , Mechanical losses, These losses are due to friction and windage., (i) friction loss e.g., bearing friction, brush friction etc., (ii) windage loss i.e., air friction of rotating armature., These losses depend upon the speed of the machine. But for a given speed, they, are practically constant., Note. Iron losses and mechanical losses together are called stray losses.
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1.27 Constant and Variable Losses, The losses in a d.c. generator (or d.c. motor) may be sub-divided into (i), constant losses (ii) variable losses., (i) Constant losses, Those losses in a d.c. generator which remain constant at all loads are known as, constant losses. The constant losses in a d.c. generator are:, (a) iron losses, (b) mechanical losses, (c) shunt field losses, (ii) Variable losses, Those losses in a d.c. generator which vary with load are called variable losses., The variable losses in a d.c. generator are:, (a) Copper loss in armature winding ( I 2a R a ), (b), , 2, Copper loss in series field winding ( I se, R se ), , Total losses = Constant losses + Variable losses, Note. Field Cu loss is constant for shunt and compound generators., , 1.28 Power Stages, The various power stages in a d.c. generator are represented diagrammatically in, Fig. (1.39)., A − B = Iron and friction losses, B − C = Copper losses, , Fig. (1.39), (i), , Mechanical efficiency, , ηm =, , EgIa, B, =, A Mechanical power input, , (ii) Electrical efficiency
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ηe =, (iii), , C V IL, =, B Eg Ia, , Commercial or overall efficiency, , ηc =, , V IL, C, =, A Mechanical power input, , ηc = η m × ηe, , Clearly, , Unless otherwise stated, commercial efficiency is always understood., C output input − losses, Now,, commercial efficiency, ηc = =, =, A input, input, , 1.29 Condition for Maximum Efficiency, The efficiency of a d.c. generator is not constant but varies with load. Consider a, shunt generator delivering a load current IL at a terminal voltage V., Generator output = V IL, Generator input = Output + Losses, = V IL + Variable losses + Constant losses, = VI L + I a2 R a + WC, , = VI L + (I L + Ish )2 R a + WC, , [Q, , I a + I L + I sh ], , The shunt field current Ish is generally small as compared to IL and, therefore,, can be neglected., ∴ Generator input = VI L + I 2L R a + WC, Now, , η=, =, , VI L, output, =, input VI L + I 2LR a + WC, 1, , (i), , W , I R, 1+ L a + C , VI L , V, , The efficiency will be maximum when the denominator of Eq.(i) is minimum, i.e.,, , d I L R a WC , +, , =0, dI L V, VI L , or, , R a WC, −, =0, V VI 2L, , Fig. (1.40)
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or, , R a WC, =, V VI 2L, , or, , I 2L R a = WC, , i.e., , Variable loss = Constant loss, , (Q, , I L ~ Ia ), , The load current corresponding to maximum efficiency is given by;, , IL =, , WC, Ra, , Hence, the efficiency of a d.c. generator will be maximum when the load current, is such that variable loss is equal to the constant loss. Fig (1.40) shows the, variation of η with load current.
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Chapter (2), , Armature Reaction and Commutation, Introduction, In a d.c. generator, the purpose of field winding is to produce magnetic field, (called main flux) whereas the purpose of armature winding is to carry armature, current. Although the armature winding is not provided for the purpose of, producing a magnetic field, nevertheless the current in the armature winding will, also produce magnetic flux (called armature flux). The armature flux distorts, and weakens the main flux posing problems for the proper operation of the d.c., generator. The action of armature flux on the main flux is called armature, reaction., In the previous chapter (Sec 1.19), it was hinted that current in the coil is, reversed as the coil passes a brush. This phenomenon is termed as commutation., The criterion for good commutation is that it should be sparkless. In order to, have sparkless commutation, the brushes should lie along magnetic neutral axis., In this chapter, we shall discuss the various aspects of armature reaction and, commutation in a d.c. generator., , 2.1 Armature Reaction, So far we have assumed that the only flux acting in a d.c. machine is that due to, the main poles called main flux. However, 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. (2.1)., Only one pole is shown for clarity. When the generator is on no-load, a smal1, current flowing in the armature does not appreciably affect the main flux φ1, coming from the pole [See Fig 2.1 (i)]. When the generator is loaded, the current, flowing through armature conductors sets up flux φ1. Fig. (2.1) (ii) shows flux, due to armature current alone. By superimposing φ1 and φ2, we obtain the, resulting flux φ3 as shown in Fig. (2.1) (iii). Referring to Fig (2.1) (iii), it is clear, that flux density at; the trailing pole tip (point B) is increased while at the
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leading pole tip (point A) it is decreased. This unequal field 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 φ3 at full load is, therefore, less than flux φ1 at no load., As we shall see, the weakening of flux due to armature reaction depends, upon the position of brushes., , Fig. (2.1), , 2.2 Geometrical and Magnetic Neutral Axes, (i), , The geometrical neutral axis (G.N.A.) is the axis that bisects the angle, between the centre line of adjacent poles [See Fig. 2.2 (i)]. Clearly, it is, the axis of symmetry between two adjacent poles., , Fig. (2.1), (ii) The magnetic neutral axis (M. N. A.) is the axis drawn perpendicular to, the mean direction of the flux passing through the centre of the armature., Clearly, no e.m.f. is produced in the armature conductors along this axis, because then they cut no flux. With no current in the armature, conductors, the M.N.A. coincides with G, N. A. as shown in Fig. (2.2).
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(ii). In order to achieve sparkless commutation, the brushes must lie, along M.N.A., , 2.3 Explanation of Armature Reaction, With no current in armature conductors, the M.N.A. coincides with G.N.A., However, when current flows in armature conductors, the combined action of, main flux and armature flux shifts the M.N.A. from G.N.A. In case of a, generator, the M.N.A. is shifted in the direction of rotation of the machine. In, order to achieve sparkless commutation, the brushes have to be moved along the, new M.N.A. Under such a condition, the armature reaction produces the, following two effects:, 1. It demagnetizes or weakens the main flux., 2. It cross-magnetizes or distorts the main flux., Let us discuss these effects of armature reaction by considering a 2-pole, generator (though the following remarks also hold good for a multipolar, generator)., (i) Fig. (2.3) (i) shows the flux due to main poles (main flux) when the, armature conductors carry no current. The flux across the air gap is, uniform. The m.m.f. producing the main flux is represented in magnitude, and direction by the vector OFm in Fig. (2.3) (i). Note that OFm is, perpendicular to G.N.A., (ii) Fig. (2.3) (ii) shows the flux due to current flowing in armature, conductors alone (main poles unexcited). The armature conductors to the, left of G.N.A. carry current “in” (×) and those to the right carry current, “out” (•). The direction of magnetic lines of force can be found by cork, screw rule. It is clear that armature flux is directed downward parallel to, the brush axis. The m.m.f. producing the armature flux is represented in, magnitude and direction by the vector OFA in Fig. (2.3) (ii)., (iii) Fig. (2.3) (iii) shows the flux due to the main poles and that due to, current in armature conductors acting together. The resultant m.m.f. OF, is the vector sum of OFm and OFA as shown in Fig. (2.3) (iii). Since, M.N.A. is always perpendicular to the resultant m.m.f., the M.N.A. is, shifted through an angle θ. Note that M.N.A. is shifted in the direction of, rotation of the generator., (iv) In order to achieve sparkless commutation, the brushes must lie along, the M.N.A. Consequently, the brushes are shifted through an angle θ so, as to lie along the new M.N.A. as shown in Fig. (2.3) (iv). Due to brush, shift, the m.m.f. FA of the armature is also rotated through the same, angle θ. It is because some of the conductors which were earlier under, N-pole now come under S-pole and vice-versa. The result is that, armature m.m.f. FA will no longer be vertically downward but will be
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rotated in the direction of rotation through an angle θ as shown in Fig., (2.3) (iv). Now FA can be resolved into rectangular components Fc and, Fd., , Fig. (2.3), (a) The component Fd is in direct opposition to the m.m.f. OFm due to main, poles. It has a demagnetizing effect on the flux due to main poles. For, this reason, it is called the demagnetizing or weakening component of, armature reaction., (b) The component Fc is at right angles to the m.m.f. OFm due to main poles., It distorts the main field. For this reason, it is called the crossmagnetizing or distorting component of armature reaction., It may be noted that with the increase of armature current, both demagnetizing, and distorting effects will increase.
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Conclusions, With brushes located along G.N.A. (i.e., θ = 0°), there is no demagnetizing, component of armature reaction (Fd = 0). There is only distorting or crossmagnetizing effect of armature reaction., (ii) With the brushes shifted from G.N.A., armature reaction will have both, demagnetizing and distorting effects. Their relative magnitudes depend on, the amount of shift. This shift is directly proportional to the armature, current., (iii) The demagnetizing component of armature reaction weakens the main flux., On the other hand, the distorting component of armature reaction distorts, the main flux., (iv) The demagnetizing effect leads to reduced generated voltage while crossmagnetizing effect leads to sparking at the brushes., (i), , 2.4 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 crossmagnetizing effects. Consider a 2-pole generator with brushes shifted (lead) θm, 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 conductors θom 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 = 2θm at the top and bottom of, the 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)., , Fig.(2.4)
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(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 cross-magnetizing (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., , 2.5 Calculation of Demagnetizing Ampere-Turns Per Pole, (ATd/Pole), It is sometimes desirable to neutralize the demagnetizing ampere-turns of, armature reaction. This is achieved by adding extra ampere-turns to the main, field winding. We shall now calculate the demagnetizing ampere-turns per pole, (ATd/pole)., Let, Z = total number of armature conductors, I = current in each armature conductor, = Ia/2 ... for simplex wave winding, = Ia/P ... for simplex lap winding, θm = forward lead in mechanical degrees, Referring to Fig. (2.4) (i) above, we have,, Total demagnetizing armature conductors, = Conductors in angles AOC and BOD =, , 4θ m, ×Z, 360, , Since two conductors constitute one turn,, , 2θ, , 1 4θ, Total demagnetizing ampere-turns = m × Z × I = m × ZI, 2 360, 360, , These demagnetizing ampere-turns are due to a pair of poles., θ, ∴, Demagnetizing ampere-turns/pole = m × ZI, 360, θ, i.e.,, ATd / pole = m × ZI, 360, ∴, , As mentioned above, the demagnetizing ampere-turns of armature reaction can, be neutralized by putting extra turns on each pole of the generator., ATd, ∴ No. of extra turns/pole =, for a shunt generator, I sh, ATd, =, for a series generator, Ia, Note. When a conductor passes a pair of poles, one cycle of voltage is, generated. We say one cycle contains 360 electrical degrees. Suppose there are P
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poles in a generator. In one revolution, there are 360 mechanical degrees and, 360 × P/2 electrical degrees., P, ∴ 360° mechanical = 360 × electrical degrees, 2, P, or, 1° Mechanical =, electrical degrees, 2, θ(electrical), ∴, θ (mechanical) =, Pair of pols, θ, 2θ, or, θm = e, ∴, θm = e, P/2, P, , 2.6 Cross-Magnetizing Ampere-Turns Per Pole, (ATc/Pole), We now calculate the cross-magnetizing ampere-turns per pole (ATc/pole)., Total armature reaction ampere-turns per pole, , =, , Z/ 2, Z, ×I=, ×I, P, 2P, , (Q, , two conductors make one turn)), , Demagnetizing ampere-turns per pole is given by;, , ATd / pole =, , θm, × ZI, 360, (, f, o, u, n, d, a, b, o, v, e, ), , ∴, , Cross-magnetizing ampere-turns/pole are
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ATd / pole =, ∴, , θ, θ , 1, Z, × I − m × ZI = ZI, − m, 2P, 360, 2P 360 , , θ , 1, − m, ATd / pole = ZI, 2P 360 , , 2.7 Compensating Windings, The cross-magnetizing effect of armature reaction may cause trouble in d.c., machines subjected to large fluctuations in load. In order to neutralize the crossmagnetizing, effect, of, armature, reaction,, a, compensating winding is, used., A compensating winding is, an, auxiliary, winding, embedded in slots in the, pole faces as shown in Fig., (2.5). It is connected in, series with armature in a, Fig. (2.5), manner, so, that, the, direction of current through the compensating conductors in any one pole face, will be opposite to the direction of the current through the adjacent armature, conductors [See Fig. 2.5]. Let us now calculate the number of compensating, conductors/ pole face. In calculating the conductors per pole face required for, the compensating winding, it should be remembered that the current in the, compensating conductors is the armature current Ia whereas the current in, armature conductors is Ia/A where A is the number of parallel paths., Let, Zc = No. of compensating conductors/pole face, Za = No. of active armature conductors, Ia = Total armature current, Ia/A = Current in each armature conductor, , ∴, or, , Zc Ia = Za ×, Zc =, , Ia, A, , Za, A, , The use of a compensating winding considerably increases the cost of a machine, and is justified only for machines intended for severe service e.g., for high speed, and high voltage machines.
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2.8 AT/Pole for Compensating Winding, Only the cross-magnetizing ampere-turns produced by conductors under the, pole face are effective in producing the distortion in the pole cores. If Z is the, total number of armature conductors and P is the number of poles, then,, Z, No. of armature conductors/pole =, P, Z, No. of armature turns/pole =, 2P, Z, Pole arc, No. of armature turns under pole face =, ×, 2P Pole pitch, If I is the current through each armature conductor, then,, ZI, Pole arc, AT/pole required for compensating winding =, ×, 2P Pole pitch, Pole arc, = Armature AT/pole ×, Pole pitch, , 2.9 Commutation, Fig. (2.6) 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., , Fig. (2.6)
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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 shortcircuited by the brush. The brief period during which the coil remains shortcircuited 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. (2.7) (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.
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Fig. (2.7), (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.
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(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., (v) Fig. (2.7) (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 shortcircuited 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, Fig. (2.8), 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., , Practical difficulties, The ideal commutation (i.e., straight line change of current) cannot be attained, in practice. This is mainly due to the fact that the armature coils have, appreciable inductance. When the current in the coil undergoing commutation, changes, self-induced e.m.f. is produced in the coil. This is generally called, reactance voltage. This reactance voltage opposes the change of current in the, coil undergoing commutation. The result is that the change of current in the coil, undergoing commutation occurs more slowly than it would be under ideal
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commutation. This is illustrated in Fig. (2.9). The straight line RC represents the, ideal commutation whereas the curve BE represents the change in current when, self-inductance of the coil is taken into account. Note that current CE (= 8A in, Fig. 2.9) is flowing from the commutator segment 1 to the brush at the instant, when they part company. This results in sparking just as when any other currentcarrying circuit is broken. The sparking results in overheating of commutatorbrush contact and causing damage to both., Fig. (2.10) illustrates how sparking takes place between the commutator, segment and the brush. At the end of commutation or short-circuit period, the, current in coil A is reversed to a value of 12 A (instead of 20 A) due to, inductance of the coil. When the brush breaks contact with segment 1, the, remaining 8 A current jumps from segment 1 to the brush through air causing, sparking between segment 1 and the brush., , Fig. (2.9), , Fig. (2.10), , 2.10 Calculation of Reactance Voltage, Reactance voltage = Coefficient of self-inductance × Rate of change of current, When a coil undergoes commutation, two commutator segments remain shortcircuited by the brush. Therefore, the time of short circuit (or commutation, period Tc) is equal to the time required by the commutator to move a distance, equal to the circumferential thickness of the brush minus the thickness of one, insultating strip of mica., Let, Wb = brush width in cm; Wm = mica thickness in cm, v = peripheral speed of commutator in cm/s, ∴, , Commutation period, Tc =, , Wb − Wm, v, , seconds, , The commutation period is very small, say of the order of 1/500 second.
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Let the current in the coil undergoing commutation change from + I to − I, (amperes) during the commutation. If L is the inductance of the coil, then, reactance voltage is given by;, Re, a, c, t, a, n, c, e, v, o, l, t, a, g, e, ,, , ER = L ×, f, o, r, l, i, n, e, a, r, c, o, m, m, u, t, a, , 2I, Tc
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t, i, o, n, , 2.11 Methods of Improving Commutation, Improving commutation means to make current reversal in the short-circuited, coil as sparkless as possible. The following are the two principal methods of, improving commutation:, (i) Resistance commutation, (ii) E.M.F. commutation, We shall discuss each method in turn., , 2.12 Resistance Commutation, The reversal of current in a coil (i.e., commutation) takes place while the coil is, short-circuited by the brush. Therefore, there are two parallel paths for the, current as long as the short circuit exists. If the contact resistance between the, brush and the commutator is made large, then current would divide in the, inverse ratio of contact resistances (as for any two resistances in parallel). This, is the key point in improving commutation. This is achieved by using carbon, brushes (instead of Cu brushes) which have high contact resistance. This method, of improving commutation is called resistance commutation., Figs. (2.11) and (2.12) illustrates how high contact resistance of carbon brush, improves commutation (i.e., reversal of current) in coil A. In Fig. (2.11) (i), the, brush is entirely on segment 1 and, therefore, the current in coil A is 20 A. The, coil A is yet to undergo commutation. As the armature rotates, the brush shortcircuits the coil A and there are two parallel paths for the current into the brush., Fig. (2.11) (ii) shows the instant when the brush is one-fourth on segment 2 and, three-fourth on segment 1. The equivalent electric circuit is shown in Fig. (2.11), (iii) where R1 and R2 represent the brush contact resistances on segments 1 and, 2. A resistor is not shown for coil A since it is assumed that the coil resistance is
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Fig. (2.11), , Fig.(2.12)
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negligible as compared to the brush contact resistance. The values of current in, the parallel paths of the equivalent circuit are determined by the respective, resistances of the paths. For the condition shown in Fig. (2.11) (ii), resistor R2, has three times the resistance of resistor R1. Therefore, the current distribution in, the paths will be as shown. Note that current in coil A is reduced from 20 A to, 10 A due to division of current in (he inverse ratio of contact resistances. If the, Cu brush is used (which has low contact resistance), R1 R2 and the current in, coil A would not have reduced to 10 A., As the carbon brush passes over the commutator, the contact area with segment, 2 increases and that with segment 1 decreases i.e., R2 decreases and R1, increases. Therefore, more and more current passes to the brush through, segment 2. This is illustrated in Figs. (2.12) (i) and (2.12) (ii), When the break, between the brush and the segment 1 finally occurs [See Fig. 2.12 (iii)], the, current in the coil is reversed and commutation is achieved., It may be noted that the main cause of sparking during commutation is the, production of reactance voltage and carbon brushes cannot prevent it., Nevertheless, the carbon brushes do help in improving commutation. The other, minor advantages of carbon brushes are:, (i) The carbon lubricates and polishes the commutator., (ii) If sparking occurs, it damages the commutator less than with copper, brushes and the damage to the brush itself is of little importance., , 2.13 E.M.F. Commutation, In this method, an arrangement is made to neutralize the reactance voltage by, producing a reversing voltage in the coil undergoing commutation. The, reversing voltage acts in opposition to the reactance voltage and neutralizes it to, some extent. If the reversing voltage is equal to the reactance voltage, the effect, of the latter is completely wiped out and we get sparkless commutation. The, reversing voltage may be produced in the following two ways:, (i) By brush shifting, (ii) By using interpoles or compoles, , (i), , By brush shifting, , In this method, the brushes are given sufficient forward lead (for a generator) to, bring the short-circuited coil (i.e., coil undergoing commutation) under the, influence of the next pole of opposite polarity. Since the short-circuited coil is, now in the reversing field, the reversing voltage produced cancels the reactance, voltage. This method suffers from the following drawbacks:
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(a) The reactance voltage depends upon armature current. Therefore, the, brush shift will depend on the magnitude of armature current which, keeps on changing. This necessitates frequent shifting of brushes., (b) The greater the armature current, the greater must be the forward lead for, a generator. This increases the demagnetizing effect of armature reaction, and further weakens the main field., , (ii), , By using interpoles or compotes, , The best method of neutralizing reactance voltage is by, using interpoles or, compoles. This method is discussed in Sec. (2.14)., , 2.14 Interpoles or Compoles, The best way to produce reversing voltage to neutralize the reactance voltage is, by using interpoles or compoles. These are small poles fixed to the yoke and, spaced mid-way between the main poles (See Fig. 2.13). They are wound with, comparatively few turns and connected in series with the armature so that they, carry armature current. Their polarity is the same as the next main pole ahead in, the direction of rotation for a generator (See Fig. 2.13). Connections for a d.c., generator with interpoles is shown in Fig. (2.14)., , Fig. (2.13), , Fig. (2.14), , Functions of Interpoles, The machines fitted with interpoles have their brushes set on geometrical neutral, axis (no lead). The interpoles perform the following two functions:, (i) As their polarity is the same as the main pole ahead (for a generator), they, induce an e.m.f. in the coil (undergoing commutation) which opposes
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reactance voltage. This leads to sparkless commutation. The e.m.f. induced, by compoles is known as commutating or reversing e.m.f. Since the, interpoles carry the armature current and the reactance voltage is also, proportional to armature current, the neutralization of reactance voltage is, automatic., (ii) The m.m.f. of the compoles neutralizes the cross-magnetizing effect of, armature reaction in small region in the space between the main poles. It is, because the two m.m.f.s oppose each other in this region., Fig. (2.15) shows the circuit, diagram of a shunt generator with, commutating, winding, and, compensating winding. Both, these windings are connected in, series with the armature and so, they carry the armature current., However, the functions they, perform must be understood, clearly. The main function of, commutating winding is to, produce, reversing, (or, commutating) e.m.f. in order to, Fig. (2.15), cancel the reactance voltage. In, addition to this, the m.m.f. of the commutating winding neutralizes the crossmagnetizing ampere-turns in the space between the main poles. The, compensating winding neutralizes the cross-magnetizing effect of armature, reaction under the pole faces., , 2.15 Equalizing Connections, We know that the armature circuit in lap winding of a multipolar machine has as, many parallel paths as the number of poles. Because of wear in the bearings, and, for other reasons, the air gaps in a generator become unequal and, therefore, the, flux in some poles becomes greater than in others. This causes the voltages of, the different paths to be unequal. With unequal voltages in these parallel paths,, circulating current will flow even if no current is supplied to an external load. If, these currents are large, some of the brushes will be required to carry a greater, current at full load than they were designed to carry and this will cause sparking., To relieve the brushes of these circulating currents, points on the armature that, are at the same potential are connected together by means of copper bars called, equalizer rings. This is achieved by connecting to the same equalizer ring the, coils that occupy the same positions relative to the poles (See Fig. 2.16). Thus, referring to Fig. (2.16), the coil consisting of conductor 1 and conductor 8
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occupies the same position relative to the poles as the coil consisting of, conductors 13 and 20. Therefore, the two coils are connected to the same, equalizer ring. The equalizers provide a low resistance path for the circulating, current. As a result, the circulating current due to the slight differences in the, voltages of the various parallel paths passes through the equalizer rings instead, of passing through the brushes. This reduces sparking., , Fig. (2.16)
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Equalizer rings should be used only on windings in which the number of coils is, a multiple of the number of poles. For best results, each coil should be, connected to an equalizer ring but this is seldom done. Satisfactory results are, obtained by connecting about every third coil to an equalizer ring. In order to, distribute the connections to the equalizer rings equally, the number of coils per, pole must be divisible by the connection pitch., Note. Equalizer rings are not used in wave winding because there is no, imbalance in the voltages of the two parallel paths. This is due to the fact that, conductors in each of the two paths pass under all N and S poles successively, (unlike a lap winding where all conductors in any parallel path lie under one pair, of poles). Therefore, even if there are inequalities in pole flux, they will affect, each path equally.
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Chapter (3), , D.C. Generator Characteristics, Introduction, The speed of a d.c. machine operated as a generator is fixed by the prime mover., For general-purpose operation, the prime mover is equipped with a speed, governor so that the speed of the generator is practically constant. Under such, condition, the generator performance deals primarily with the relation between, excitation, terminal voltage and load. These relations can be best exhibited, graphically by means of curves known as generator characteristics. These, characteristics show at a glance the behaviour of the generator under different, load conditions., , 3.1 D.C. Generator Characteristics, The following are the three most important characteristics of a d.c. generator:, , 1., , Open Circuit Characteristic (O.C.C.), This curve shows the relation between the generated e.m.f. at no-load (E0) and, the field current (If) at constant speed. It is also known as magnetic characteristic, or no-load saturation curve. Its shape is practically the same for all generators, whether separately or self-excited. The data for O.C.C. curve are obtained, experimentally by operating the generator at no load and constant speed and, recording the change in terminal voltage as the field current is varied., , 2., , Internal or Total characteristic (E/Ia), This curve shows the relation between the generated e.m.f. on load (E) and the, armature current (Ia). The e.m.f. E is less than E0 due to the demagnetizing effect, of armature reaction. Therefore, this curve will lie below the open circuit, characteristic (O.C.C.). The internal characteristic is of interest chiefly to the, designer. It cannot be obtained directly by experiment. It is because a voltmeter, cannot read the e.m.f. generated on load due to the voltage drop in armature, resistance. The internal characteristic can be obtained from external, characteristic if winding resistances are known because armature reaction effect, is included in both characteristics.
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3., , External characteristic (V/IL), This curve shows the relation between the terminal voltage (V) and load current, (IL). The terminal voltage V will be less than E due to voltage drop in the, armature circuit. Therefore, this curve will lie below the internal characteristic., This characteristic is very important in determining the suitability of a generator, for a given purpose. It can be obtained by making simultaneous measurements, of terminal voltage and load current (with voltmeter and ammeter) of a loaded, generator., , 3.2 Open Circuit Characteristic of a D.C. Generator, The O.C.C. for a d.c. generator is determined as follows. The field winding of, the d.c. generator (series or shunt) is disconnected from the machine and is, separately excited from an external d.c. source as shown in Fig. (3.1) (ii). The, generator is run at fixed speed (i.e., normal speed). The field current (If) is, increased from zero in steps and the corresponding values of generated e.m.f., (E0) read off on a voltmeter connected across the armature terminals. On plotting, the relation between E0 and If, we get the open circuit characteristic as shown in, Fig. (3.1) (i)., , Fig. (3.1), The following points may be noted from O.C.C.:, (i) When the field current is zero, there is some generated e.m.f. OA. This is, due to the residual magnetism in the field poles., (ii) Over a fairly wide range of field current (upto point B in the curve), the, curve is linear. It is because in this range, reluctance of iron is negligible as, compared with that of air gap. The air gap reluctance is constant and hence, linear relationship., (iii) After point B on the curve, the reluctance of iron also comes into picture. It, is because at higher flux densities, µr for iron decreases and reluctance of
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iron is no longer negligible. Consequently, the curve deviates from linear, relationship., (iv) After point C on the curve, the magnetic saturation of poles begins and E0, tends to level off., The reader may note that the O.C.C. of even self-excited generator is obtained, by running it as a separately excited generator., , 3.3 Characteristics of a Separately Excited D.C., Generator, The obvious disadvantage of a separately excited d.c. generator is that we, require an external d.c. source for excitation. But since the output voltage may, be controlled more easily and over a wide range (from zero to a maximum), this, type of excitation finds many applications., , (i) Open circuit characteristic., The O.C.C. of a separately excited generator is, determined in a manner described in Sec. (3.2). Fig., (3.2) shows the variation of generated e.m f. on no, load with field current for various fixed speeds. Note, that if the value of constant speed is increased, the, steepness of the curve also increases. When the field, current is zero, the residual magnetism in the poles, will give rise to the small initial e.m.f. as shown., , Fig. (3.2), , (ii) Internal and External Characteristics, The external characteristic of a separately excited generator is the curve between, the terminal voltage (V) and the load current IL (which is the same as armature, current in this case). In order to determine the external characteristic, the circuit, set up is as shown in Fig. (3.3) (i). As the load current increases, the terminal, voltage falls due to two reasons:, (a) The armature reaction weakens the main flux so that actual e.m.f., generated E on load is less than that generated (E0) on no load., (b) There is voltage drop across armature resistance (= ILRa = IaRa)., Due to these reasons, the external characteristic is a drooping curve [curve 3 in, Fig. 3.3 (ii)]. Note that in the absence of armature reaction and armature drop,, the generated e.m.f. would have been E0 (curve 1)., The internal characteristic can be determined from external characteristic by, adding ILRa drop to the external characteristic. It is because armature reaction, drop is included in the external characteristic. Curve 2 is the internal
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characteristic of the generator and should obviously lie above the external, characteristic., , Fig. (3.3), , 3.4 Voltage Build-Up in a Self-Excited Generator, Let us see how voltage builds up in a self-excited generator., , (i) Shunt generator, Consider a shunt generator. If the generator is run at a constant speed, some, e.m.f. will be generated due to residual magnetism in the main poles. This small, e.m.f. circulates a field current which in turn produces additional flux to, reinforce the original residual flux (provided field winding connections are, correct). This process continues and the generator builds up the normal, generated voltage following the O.C.C. shown in Fig. (3.4) (i)., The field resistance Rf can be represented by a straight line passing through the, origin as shown in Fig. (3.4) (ii). The two curves can be shown on the same, diagram as they have the same ordinate [See Fig. 3.4 (iii)]., Since the field circuit is inductive, there is a delay in the increase in current upon, closing the field circuit switch The rate at which the current increases depends
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upon the voltage available for increasing it. Suppose at any instant, the field, current is i (= OA) and is increasing at the rate di/dt. Then,, , E0 = i R f + L, where, , di, dt, , Rf = total field circuit resistance, L = inductance of field circuit, , At the considered instant, the total e.m.f. available is AC [See Fig. 3.4 (iii)]. An, amount AB of the c.m.f. AC is absorbed by the voltage drop iRf and the, remainder part BC is available to overcome L di/dt. Since this surplus voltage is, available, it is possible for the field current to increase above the value OA., However, at point D, the available voltage is OM and is all absorbed by i Rf, drop. Consequently, the field current cannot increase further and the generator, build up stops., , Fig. (3.4), We arrive at a very important conclusion that the voltage build up of the, generator is given by the point of intersection of O.C.C. and field resistance line., Thus in Fig. (3.4) (iii), D is point of intersection of the two curves. Hence the, generator will build up a voltage OM., , (ii) Series generator, During initial operation, with no current yet flowing, a residual voltage will be, generated exactly as in the case of a shunt generator. The residual voltage will, cause a current to flow through the whole series circuit when the circuit is, closed. There will then be voltage build up to an equilibrium point exactly, analogous to the build up of a shunt generator. The voltage build up graph will, be similar to that of shunt generator except that now load current (instead of, field current for shunt generator) will be taken along x-axis.
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(iii) Compound generator, When a compound generator has its series field flux aiding its shunt field flux,, the machine is said to be cumulative compound. When the series field is, connected in reverse so that its field flux opposes the shunt field flux, the, generator is then differential compound., The easiest way to build up voltage in a compound generator is to start under no, load conditions. At no load, only the shunt field is effective. When no-load, voltage build up is achieved, the generator is loaded. If under load, the voltage, rises, the series field connection is cumulative. If the voltage drops significantly,, the connection is differential compound., , 3.5 Critical Field Resistance for a Shunt Generator, We have seen above that voltage build up in a shunt, generator depends upon field circuit resistance. If, the field circuit resistance is R1 (line OA), then, generator will build up a voltage OM as shown in, Fig. (3.5). If the field circuit resistance is increased, to R2 (tine OB), the generator will build up a, voltage OL, slightly less than OM. As the field, circuit resistance is increased, the slope of, resistance line also increases. When the field, Fig. (3.5), resistance line becomes tangent (line OC) to, O.C.C., the generator would just excite. If the field circuit resistance is increased, beyond this point (say line OD), the generator will fail to excite. The field circuit, resistance represented by line OC (tangent to O.C.C.) is called critical field, resistance RC for the shunt generator. It may be defined as under:, The maximum field circuit resistance (for a given speed) with which the shunt, generator would just excite is known as its critical field resistance., It should be noted that shunt generator will build up voltage only if field circuit, resistance is less than critical field resistance., , 3.6 Critical Resistance for a Series, Generator, Fig. (3.6) shows the voltage build up in a series, generator. Here R1, R2 etc. represent the total, circuit resistance (load resistance and field winding, resistance). If the total circuit resistance is R1, then, series generator will build up a voltage OL. The, , Fig. (3.6)
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line OC is tangent to O.C.C. and represents the critical resistance RC for a series, generator. If the total resistance of the circuit is more than RC (say line OD), the, generator will fail to build up voltage. Note that Fig. (3.6) is similar to Fig. (3.5), with the following differences:, (i) In Fig. (3.5), R1, R2 etc. represent the total field circuit resistance., However, R1, R2 etc. in Fig. (3.6) represent the total circuit resistance, (load resistance and series field winding resistance etc.)., (ii) In Fig (3.5), field current alone is represented along X-axis. However, in, Fig. (3.6) load current IL is represented along Y-axis. Note that in a, series generator, field current = load current IL., , 3.7 Characteristics of Series Generator, Fig. (3.7) (i) shows the connections of a series wound generator. Since there is, only one current (that which flows through the whole machine), the load current, is the same as the exciting current., , Fig. (3.7), , (i) O.C.C., Curve 1 shows the open circuit characteristic (O.C.C.) of a series generator. It, can be obtained experimentally by disconnecting the field winding from the, machine and exciting it from a separate d.c. source as discussed in Sec. (3.2)., , (ii) Internal characteristic, Curve 2 shows the total or internal characteristic of a series generator. It gives, the relation between the generated e.m.f. E. on load and armature current. Due to, armature reaction, the flux in the machine will be less than the flux at no load., Hence, e.m.f. E generated under load conditions will be less than the e.m.f. E0, generated under no load conditions. Consequently, internal characteristic curve
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lies below the O.C.C. curve; the difference between them representing the effect, of armature reaction [See Fig. 3.7 (ii)]., , (iii) External characteristic, Curve 3 shows the external characteristic of a series generator. It gives the, relation between terminal voltage and load current IL:., , V = E − I a (R a + R se ), Therefore, external characteristic curve will lie below internal characteristic, curve by an amount equal to ohmic drop [i.e., Ia(Ra + Rse)] in the machine as, shown in Fig. (3.7) (ii)., The internal and external characteristics of, a d.c. series generator can be plotted from, one another as shown in Fig. (3.8). Suppose, we are given the internal characteristic of, the generator. Let the line OC represent the, resistance of the whole machine i.e. Ra +, Rse. If the load current is OB, drop in the, machine is AB i.e., , Fig. (3.8), , AB = Ohmic drop in the machine = OB(Ra + Rse), Now raise a perpendicular from point B and mark a point b on this line such that, ab = AB. Then point b will lie on the external characteristic of the generator., Following similar procedure, other points of external characteristic can be, located. It is easy to see that we can also plot internal characteristic from the, external characteristic., , 3.8 Characteristics of a Shunt Generator, Fig (3.9) (i) shows the connections of a shunt wound generator. The armature, current Ia splits up into two parts; a small fraction Ish flowing through shunt field, winding while the major part IL goes to the external load., , (i) O.C.C., The O.C.C. of a shunt generator is similar in shape to that of a series generator, as shown in Fig. (3.9) (ii). The line OA represents the shunt field circuit, resistance. When the generator is run at normal speed, it will build up a voltage, OM. At no-load, the terminal voltage of the generator will be constant (= OM), represented by the horizontal dotted line MC.
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Fig. (3.9), , (ii) Internal characteristic, When the generator is loaded, flux per pole is reduced due to armature reaction., Therefore, e.m.f. E generated on load is less than the e.m.f. generated at no load., As a result, the internal characteristic (E/Ia) drops down slightly as shown in Fig., (3.9) (ii)., , (iii) External characteristic, Curve 2 shows the external characteristic of a shunt generator. It gives the, relation between terminal voltage V and load current IL., , V = E − I a R a = E − (I L + Ish )R a, Therefore, external characteristic curve will lie below the internal characteristic, curve by an amount equal to drop in the armature circuit [i.e., (IL + Ish)Ra] as, shown in Fig. (3.9) (ii)., Note. It may be seen from the external characteristic that change in terminal, voltage from no-load to full load is small. The terminal voltage can always be, maintained constant by adjusting the field rheostat R automatically, , 3.9 Critical External Resistance for, Shunt Generator, If the load resistance across the terminals of a, shunt generator is decreased, then load current, increase? However, there is a limit to the, increase in load current with the decrease of, load resistance. Any decrease of load, resistance beyond this point, instead of, increasing the current, ultimately results in, , Fig. (3.10)
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reduced current. Consequently, the external characteristic turns back (dotted, curve) as shown in Fig. (3.10). The tangent OA to the curve represents the, minimum external resistance required to excite the shunt generator on load and, is called critical external resistance. If the resistance of the external circuit is less, than the critical external resistance (represented by tangent OA in Fig. 3.10), the, machine will refuse to excite or will de-excite if already running This means that, external resistance is so low as virtually to short circuit the machine and so, doing away with its excitation., Note. There are two critical resistances for a shunt generator viz., (i) critical, field resistance (ii) critical external resistance. For the shunt generator to build, up voltage, the former should not be exceeded and the latter must not be gone, below., , 3.10 How to Draw O.C.C. at Different Speeds?, If we are given O.C.C. of a generator at a constant speed N1, then we can easily, draw the O.C.C. at any other constant speed N2. Fig (3.11) illustrates the, procedure. Here we are given O.C.C. at a constant speed N1. It is desired to find, the O.C.C. at constant speed N2 (it is assumed that n1 < N2). For constant, excitation, E ∝ N., , ∴, or, , E 2 N2, =, E1 N1, E 2 = E1 ×, , N2, N1, , As shown in Fig. (3.11), for If = OH, E1 = HC., Therefore, the new value of e.m.f. (E2) for the, same If but at N2 i, , E 2 = HC ×, , Fig. (3.11), , N2, = HD, N1, , This locates the point D on the new O.C.C. at N2. Similarly, other points can be, located taking different values of If. The locus of these points will be the O.C.C., at N2., , 3.11 Critical Speed (NC), The critical speed of a shunt generator is the minimum speed below which it, fails to excite. Clearly, it is the speed for which the given shunt field resistance, represents the critical resistance. In Fig. (3.12), curve 2 corresponds to critical, speed because the shunt field resistance (Rsh) line is tangential to it. If the
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generator runs at full speed N, the, new O.C.C. moves upward and the, R'sh line represents critical resistance, for this speed., ∴, , Speed ∝ Critical resistance, , In order to find critical speed, take, any convenient point C on excitation, axis and erect a perpendicular so as to, cut Rsh and R'sh lines at points B and, A respectively. Then,, , BC N C, =, AC N, or, , NC = N ×, , Fig. (3.12), , BC, AC, , 3.12 Conditions for Voltage Build-Up of a Shunt, Generator, The necessary conditions for voltage build-up in a shunt generator are:, (i) There must be some residual magnetism in generator poles., (ii) The connections of the field winding should be such that the field current, strengthens the residual magnetism., (iii) The resistance of the field circuit should be less than the critical, resistance. In other words, the speed of the generator should be higher, than the critical speed., , 3.13 Compound Generator Characteristics, In a compound generator, both series and shunt excitation are combined as, shown in Fig. (3.13). The shunt winding can be connected either across the, armature only (short-shunt connection S) or across armature plus series field, (long-shunt connection G). The compound generator can be cumulatively, compounded or differentially compounded generator. The latter is rarely used in, practice. Therefore, we shall discuss the characteristics of cumulativelycompounded generator. It may be noted that external characteristics of long and, short shunt compound generators are almost identical., , External characteristic, Fig. (3.14) shows the external characteristics of a cumulatively compounded, generator. The series excitation aids the shunt excitation. The degree of
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compounding depends upon the increase in series excitation with the increase in, load current., , Fig. (3.13), , Fig. (3.14), , (i), , If series winding turns are so adjusted that with the increase in load, current the terminal voltage increases, it is called over-compounded, generator. In such a case, as the load current increases, the series field, m.m.f. increases and tends to increase the flux and hence the generated, voltage. The increase in generated voltage is greater than the IaRa drop so, that instead of decreasing, the terminal voltage increases as shown by, curve A in Fig. (3.14)., (ii) If series winding turns are so adjusted that with the increase in load, current, the terminal voltage substantially remains constant, it is called, flat-compounded generator. The series winding of such a machine has, lesser number of turns than the one in over-compounded machine and,, therefore, does not increase the flux as much for a given load current., Consequently, the full-load voltage is nearly equal to the no-load voltage, as indicated by curve B in Fig (3.14)., (iii) If series field winding has lesser number of turns than for a flatcompounded machine, the terminal voltage falls with increase in load, current as indicated by curve C m Fig. (3.14). Such a machine is called, under-compounded generator., , 3.14 Voltage Regulation, The change in terminal voltage of a generator between full and no load (at, constant speed) is called the voltage regulation, usually expressed as a, percentage of the voltage at full-load., , % Voltage regulation =, where, , VNL − VFL, × 100, VFL, , VNL = Terminal voltage of generator at no load
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VFL = Terminal voltage of generator at full load, Note that voltage regulation of a generator is determined with field circuit and, speed held constant. If the voltage regulation of a generator is 10%, it means that, terminal voltage increases 10% as the load is changed from full load to no load., , 3.15 Parallel Operation of D.C. Generators, In a d.c. power plant, power is usually supplied from several generators of small, ratings connected in parallel instead of from one large generator. This is due to, the following reasons:, , (i) Continuity of service, If a single large generator is used in the power plant, then in case of its, breakdown, the whole plant will be shut down. However, if power is supplied, from a number of small units operating in parallel, then in case of failure of one, unit, the continuity of supply can be maintained by other healthy units., , (ii) Efficiency, Generators run most efficiently when loaded to their rated capacity. Electric, power costs less per kWh when the generator producing it is efficiently loaded., Therefore, when load demand on power plant decreases, one or more generators, can be shut down and the remaining units can be efficiently loaded., , (iii) Maintenance and repair, Generators generally require routine-maintenance and repair. Therefore, if, generators are operated in parallel, the routine or emergency operations can be, performed by isolating the affected generator while load is being supplied by, other units. This leads to both safety and economy., , (iv) Increasing plant capacity, In the modern world of increasing population, the use of electricity is, continuously increasing. When added capacity is required, the new unit can be, simply paralleled with the old units., , (v) Non-availability of single large unit, In many situations, a single unit of desired large capacity may not be available., In that case a number of smaller units can be operated in parallel to meet the, load requirement. Generally a single large unit is more expensive.
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2.16 Connecting Shunt Generators in Parallel, The generators in a power plant are connected in parallel through bus-bars. The, bus-bars are heavy thick copper bars and they act as +ve and −ve terminals. The, positive terminals of the generators are .connected to the +ve side of bus-bars, and negative terminals to the negative side of bus-bars., Fig. (3.15) shows shunt generator 1 connected to the bus-bars and supplying, load. When the load on the power plant increases beyond the capacity of this, generator, the second shunt generator 2 is connected in parallel wish the first to, meet the increased load demand. The procedure for paralleling generator 2 with, generator 1 is as under:, (i) The prime mover of generator 2 is brought up to the rated speed. Now, switch S4 in the field circuit of the generator 2 is closed., , Fig. (3.15), (ii) Next circuit breaker CB-2 is closed and the excitation of generator 2 is, adjusted till it generates voltage equal to the bus-bars voltage. This is, indicated by voltmeter V2., (iii) Now the generator 2 is ready to be paralleled with generator 1. The main, switch S3, is closed, thus putting generator 2 in parallel with generator 1., Note that generator 2 is not supplying any load because its generated, e.m.f. is equal to bus-bars voltage. The generator is said to be “floating”, (i.e., not supplying any load) on the bus-bars.
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(iv) If generator 2 is to deliver any current, then its generated voltage E, should be greater than the bus-bars voltage V. In that case, current, supplied by it is I = (E − V)/Ra where Ra is the resistance of the armature, circuit. By increasing the field current (and hence induced e.m.f. E), the, generator 2 can be made to supply proper amount of load., (v) The load may be shifted from one shunt generator to another merely by, adjusting the field excitation. Thus if generator 1 is to be shut down, the, whole load can be shifted onto generator 2 provided it has the capacity to, supply that load. In that case, reduce the current supplied by generator 1, to zero (This will be indicated by ammeter A1) open C.B.-1 and then, open the main switch S1., , 3.17 Load Sharing, The load sharing between shunt generators in parallel can be easily regulated, because of their drooping characteristics. The load may be shifted from one, generator to another merely by adjusting the field excitation. Let us discuss the, load sharing of two generators which have unequal no-load voltages., Let, E1, E2 = no-load voltages of the two generators, R1, R2 = their armature resistances, V = common terminal voltage (Bus-bars voltage), Then, , I1 =, , E1 − V, and, R1, , I2 =, , E2 − V, R2, , Thus current output of the generators depends upon the values of E1 and E3., These values may be changed by field rheostats. The common terminal voltage, (or bus-bars voltage) will depend upon (i) the e.m.f.s of individual generators, and (ii) the total load current supplied. It is generally desired to keep the busbars voltage constant. This can be achieved by adjusting the field excitations of, the generators operating in parallel., , 3.18 Compound Generators in Parallel, Under-compounded generators also operate satisfactorily in parallel but overcompounded generators will not operate satisfactorily unless their series fields, are paralleled. This is achieved by connecting two negative brushes together as, shown in Fig. (3.16) (i). The conductor used to connect these brushes is, generally called equalizer bar. Suppose that an attempt is made to operate the, two generators in Fig. (3.16) (ii) in parallel without an equalizer bar. If, for any, reason, the current supplied by generator 1 increases slightly, the current in its, series field will increase and raise the generated voltage. This will cause, generator 1 to take more load. Since total load supplied to the system is constant,, the current in generator 2 must decrease and as a result its series field is
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weakened. Since this effect is cumulative, the generator 1 will take the entire, load and drive generator 2 as a motor. Under such conditions, the current in the, two machines will be in the direction shown in Fig. (3.16) (ii). After machine 2, changes from a generator to a motor, the current in the shunt field will remain in, the same direction, but the current in the armature and series field will reverse., Thus the magnetizing action, of the series field opposes that of the shunt field., As the current taken by the machine 2 increases, the demagnetizing action of, series field becomes greater and the resultant field becomes weaker. The, resultant field will finally become zero and at that time machine 2 will shortcircuit machine 1, opening the breaker of either or both machines., , Fig. (3.16), When the equalizer bar is used, a stabilizing action exist? and neither machine, tends to take all the load. To consider this, suppose that current delivered by, generator 1 increases [See Fig. 3.16 (i)]. The increased current will not only pass, through the series field of generator 1 but also through the equalizer bar and, series field of generator 2. Therefore, the voltage of both the machines increases, and the generator 2 will take a part of the load.
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Chapter (4), , D.C. Motors, Introduction, D. C. motors are seldom used in ordinary applications because all electric supply, companies furnish alternating current However, for special applications such as, in steel mills, mines and electric trains, it is advantageous to convert alternating, current into direct current in order to use d.c. motors. The reason is that, speed/torque characteristics of d.c. motors are much more superior to that of a.c., motors. Therefore, it is not surprising to note that for industrial drives, d.c., motors are as popular as 3-phase induction motors. Like d.c. generators, d.c., motors are also of three types viz., series-wound, shunt-wound and compoundwound. The use of a particular motor depends upon the mechanical load it has to, drive., , 4.1 D.C. Motor Principle, A machine that converts d.c. power into mechanical power is known as a d.c., motor. Its operation is based on the principle that when a current carrying, conductor is placed in a magnetic field, the conductor experiences a mechanical, force. The direction of this force is given by Fleming’s left hand rule and, magnitude is given by;, , F = BIl newtons, Basically, there is no constructional difference between a d.c. motor and a d.c., generator. The same d.c. machine can be run as a generator or motor., , 4.2 Working of D.C. Motor, Consider a part of a multipolar d.c. motor as shown in Fig. (4.1). When the, terminals of the motor are connected to an external source of d.c. supply:, (i) the field magnets are excited developing alternate N and S poles;, (ii) the armature conductors carry ^currents. All conductors under N-pole, carry currents in one direction while all the conductors under S-pole, carry currents in the opposite direction., Suppose the conductors under N-pole carry currents into the plane of the paper, and those under S-pole carry currents out of the plane of the paper as shown in, Fig.(4.1). Since each armature conductor is carrying current and is placed in the
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magnetic field, mechanical force acts on it., Referring to Fig. (4.1) and applying, Fleming’s left hand rule, it is clear that force, on each conductor is tending to rotate the, armature in anticlockwise direction. All these, forces add together to produce a driving, torque which sets the armature rotating., Fig. (4.1), When the conductor moves from one side of a, brush to the other, the current in that conductor is reversed and at the same time, it comes under the influence of next pole which is of opposite polarity., Consequently, the direction of force on the conductor remains the same., , 4.3 Back or Counter E.M.F., When the armature of a d.c. motor rotates under the influence of the driving, torque, the armature conductors move through the magnetic field and hence, e.m.f. is induced in them as in a generator The induced e.m.f. acts in opposite, direction to the applied voltage V(Lenz’s law) and in known as back or counter, e.m.f. Eb. The back e.m.f. Eb(= P φ ZN/60 A) is always less than the applied, voltage V, although this difference is small when the motor is running under, normal conditions., Consider a shunt wound motor shown in, Fig. (4.2). When d.c. voltage V is applied, across the motor terminals, the field magnets, are excited and armature conductors are, supplied with current. Therefore, driving, torque acts on the armature which begins to, rotate. As the armature rotates, back e.m.f., Eb is induced which opposes the applied, Fig. (4.2), voltage V. The applied voltage V has to, force current through the armature against, the back e.m.f. Eb. The electric work done in overcoming and causing the, current to flow against Eb is converted into mechanical energy developed in the, armature. It follows, therefore, that energy conversion in a d.c. motor is only, possible due to the production of back e.m.f. Eb., Net voltage across armature circuit = V − Eb, If Ra is the armature circuit resistance, then, I a =, , V − Eb, Ra, , Since V and Ra are usually fixed, the value of Eb will determine the current, drawn by the motor. If the speed of the motor is high, then back e.m.f. Eb (= P φ
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ZN/60 A) is large and hence the motor will draw less armature current and viceversa., , 4.4 Significance of Back E.M.F., The presence of back e.m.f. makes the d.c. motor a self-regulating machine i.e.,, it makes the motor to draw as much armature current as is just sufficient to, develop the torque required by the load., Armature current, I a =, , V − Eb, Ra, , (i), , When the motor is running on no load, small torque is required to, overcome the friction and windage losses. Therefore, the armature, current Ia is small and the back e.m.f. is nearly equal to the applied, voltage., (ii) If the motor is suddenly loaded, the first effect is to cause the armature to, slow down. Therefore, the speed at which the armature conductors move, through the field is reduced and hence the back e.m.f. Eb falls. The, decreased back e.m.f. allows a larger current to flow through the, armature and larger current means increased driving torque. Thus, the, driving torque increases as the motor slows down. The motor will stop, slowing down when the armature current is just sufficient to produce the, increased torque required by the load., (iii) If the load on the motor is decreased, the driving torque is momentarily, in excess of the requirement so that armature is accelerated. As the, armature speed increases, the back e.m.f. Eb also increases and causes, the armature current Ia to decrease. The motor will stop accelerating, when the armature current is just sufficient to produce the reduced torque, required by the load., It follows, therefore, that back e.m.f. in a d.c. motor regulates the flow of, armature current i.e., it automatically changes the armature current to meet the, load requirement., , 4.5 Voltage Equation of D.C. Motor, Let in a d.c. motor (See Fig. 4.3),, V = applied voltage, Eb = back e.m.f., Ra = armature resistance, Ia = armature current, Since back e.m.f. Eb acts in opposition to the, , Fig. (4.3)
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applied voltage V, the net voltage across the armature circuit is V− Eb. The, armature current Ia is given by;, , Ia =, , V − Eb, Ra, , V = Eb + Ia R a, , or, , (i), , This is known as voltage equation of the d.c. motor., , 4.6 Power Equation, If Eq.(i) above is multiplied by ly throughout, we get,, , VI a = E b I a + I a2 R a, This is known as power equation of the d.c. motor., VIa = electric power supplied to armature (armature input), EbIa = power developed by armature (armature output), 2, I a R a = electric power wasted in armature (armature Cu loss), Thus out of the armature input, a small portion (about 5%) is wasted as I 2a R a, and the remaining portion EbIa is converted into mechanical power within the, armature., , 4.7 Condition For Maximum Power, The mechanical power developed by the motor is Pm = EbIa, , Pm = VI a − I 2a R a, , Now, , Since, V and Ra are fixed, power developed by the motor depends upon, armature current. For maximum power, dPm/dIa should be zero., , ∴, , dPm, = V − 2I a R a = 0, dI a, , V, 2, , or, , Ia R a =, , Now,, , V = Eb + Ia R a = Eb +, , ∴, , Eb =, , V, 2, , ∴ I R = V , a a, , 2 , , V, 2, , Hence mechanical power developed by the motor is maximum when back e.m.f., is equal to half the applied voltage.
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Limitations, In practice, we never aim at achieving maximum power due to the following, reasons:, (i) The armature current under this condition is very large—much excess of, rated current of the machine., (ii) Half of the input power is wasted in the armature circuit. In fact, if we, take into account other losses (iron and mechanical), the efficiency will, be well below 50%., , 4.8 Types of D.C. Motors, Like generators, there are three types of d.c. motors characterized by the, connections of field winding in relation to the armature viz.:, (i) Shunt-wound motor in which the field winding is connected in parallel, with the armature [See Fig. 4.4]. The current through the shunt field, winding is not the same as the armature current. Shunt field windings are, designed to produce the necessary m.m.f. by means of a relatively large, number of turns of wire having high resistance. Therefore, shunt field, current is relatively small compared with the armature current., , Fig. (4.4), , Fig. (4.5), , (ii) Series-wound motor in which the field winding is connected in series with, the armature [See Fig. 4.5]. Therefore, series field winding carries the, armature current. Since the current passing through a series field winding is, the same as the armature current, series field windings must be designed, with much fewer turns than shunt field windings for the same m.m.f., Therefore, a series field winding has a relatively small number of turns of, thick wire and, therefore, will possess a low resistance., (iii) Compound-wound motor which has two field windings; one connected in, parallel with the armature and the other in series with it. There are two, types of compound motor connections (like generators). When the shunt, field winding is directly connected across the armature terminals [See Fig., 4.6], it is called short-shunt connection. When the shunt winding is so
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connected that it shunts the series combination of armature and series field, [See Fig. 4.7], it is called long-shunt connection., , Fig. (4.6), , Fig. (4.7), , The compound machines (generators or motors) are always designed so that the, flux produced by shunt field winding is considerably larger than the flux, produced by the series field winding. Therefore, shunt field in compound, machines is the basic dominant factor in the production of the magnetic field in, the machine., , 4.9 Armature Torque of D.C. Motor, Torque is the turning moment of a force about an axis and is measured by the, product of force (F) and radius (r) at right angle to which the force acts i.e., D.C. Motors 113, T=F×r, In a d.c. motor, each conductor is acted upon by a circumferential force F at a, distance r, the radius of the armature (Fig. 4.8). Therefore, each conductor exerts, a torque, tending to rotate the armature. The sum of the torques due to all, armature conductors is known as gross or armature torque (Ta)., Let in a d.c. motor, r = average radius of armature in m, l = effective length of each conductor in m, Z = total number of armature conductors, A = number of parallel paths, i = current in each conductor = Ia/A, B = average flux density in Wb/m2, φ = flux per pole in Wb, P = number of poles, Force on each conductor, F = B i l newtons, , Fig. (4.8)
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Torque due to one conductor = F × r newton- metre, Total armature torque,, , Ta = Z F r newton-metre, =ZBi l r, , Now i = Ia/A, B = φ/a where a is the x-sectional area of flux path per pole at, radius r. Clearly, a = 2πr l /P., , ∴, , φ I , Ta = Z × × a × l × r, 2 A , I, Zφ I a P, φ, = Z×, × a ×l×r =, N-m, 2π r l / P A, 2π A, P, Ta = 0.159 Z φ I a N - m, A, , or, , (i), , Since Z, P and A are fixed for a given machine,, , ∴, , Ta ∝ φI a, , Hence torque in a d.c. motor is directly proportional to flux per pole and, armature current., (i) For a shunt motor, flux φ is practically constant., , ∴, , Ta ∝ I a, , (ii) For a series motor, flux φ is directly proportional to armature current Ia, provided magnetic saturation does not take place., , ∴, , Ta ∝ I a2, u, p, t, o, m, a, g, n, e, t, i, c, s, a
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t, u, r, a, t, i, o, n, Alternative expression for Ta, , Eb =, , ∴, , PφZN, 60 A, , PφZ 60 × E b, =, A, N, , From Eq.(i), we get the expression of Ta as:, , 60 × E b , Ta = 0.159 × , × Ia, N , or, , Ta = 9.55 ×, , E b Ia, N-m, N, , Note that developed torque or gross torque means armature torque Ta., , 4.10 Shaft Torque (Tsh), The torque which is available at the motor shaft for, doing useful work is known as shaft torque. It is, represented by Tsh. Fig. (4.9) illustrates the concept, of shaft torque. The total or gross torque Ta, Fig. (4.9), developed in the armature of a motor is not available, at the shaft because a part of it is lost in overcoming, the iron and frictional losses in the motor. Therefore, shaft torque Tsh is, somewhat less than the armature torque Ta. The difference Ta − Tsh is called lost, torque., Clearly,, , Ta − Tsh = 9.55 ×, , Iron and frictional losses, N, , For example, if the iron and frictional losses in a motor are 1600 W and the, motor runs at 800 r.p.m., then,, , Ta − Tsh = 9.55 ×, , 1600, = 19.1 N - m, 800
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As stated above, it is the shaft torque Tsh that produces the useful output. If the, speed of the motor is N r.p.m., then,, , 2π N Tsh, 60, , Output in watts =, , Output in watts, N-m, 2π N / 60, , or, , Tsh =, , or, , Tsh = 9.55 ×, , Q 60 = 9.55 , , 2π, , , , Output in watts, N-m, N, , 4.11 Brake Horse Power (B.H.P.), The horse power developed by the shaft torque is known as brake horsepower, (B.H.P.). If the motor is running at N r.p.m. and the shaft torque is Tsh newtonmetres, then,, W.D./revolution = force x distance moved in 1 revolution, = F × 2π r = 2π × Tsh J, W.D./minute = 2π N Tsh J, , W.D. / sec . =, ∴, , Useful output power =, , B.H.P. =, , or, , 2π N Tsh −1, 2π N Tsh, Js or watts =, H.P., 60, 60 × 746, 2 π N Tsh, H.P., 60 × 746, , 2π N Tsh, 60 × 746, , 4.12 Speed of a D.C. Motor, E b = V − Ia R a, , Eb =, , But, , ∴, , Pφ Z N, 60 A, , Pφ Z N, = V − Ia R a, 60 A, , (V − I a R a ) 60 A, , or, , N=, , or, , N=K, , φ, , PZ, , (V − I a R a ), φ, , where, , K=, , 60 A, PZ
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V − Ia R a = E a, , But, , ∴, , N=K, N∝, , or, , Eb, φ, , Eb, φ, , Therefore, in a d.c. motor, speed is directly proportional to back e.m.f. Eb and, inversely proportional to flux per pole φ., , 4.13 Speed Relations, If a d.c. motor has initial values of speed, flux per pole and back e.m.f. as N1, φ1, and Eb1 respectively and the corresponding final values are N2, φ2 and Eb2, then,, , N1 ∝, ∴, (i), , E b1, φ1, , and, , N2 ∝, , E b2, φ2, , N 2 E b 2 φ1, =, ×, N1 E b1 φ 2, , For a shunt motor, flux practically remains constant so that φ1 = φ2., , ∴, , N 2 E b2, =, N1 E b1, , (ii) For a series motor, φ ∝ Ia prior to saturation., , ∴, where, , N 2 E b 2 I a1, =, ×, N1 E b1 I a 2, Ia1 = initial armature current, Ia2 = final armature current, , 4.14 Speed Regulation, The speed regulation of a motor is the change in speed from full-load to no-loud, and is expressed as a percentage of the speed at full-load i.e., % Speed regulation =, , =, where, , N.L. speed − F.L. speed, × 100, F.L. speed, , N0 − N, × 100, N, , N0 = No - load .speed, N = Full - load speed
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4.15 Torque and Speed of a D.C. Motor, For any motor, the torque and speed are very important factors. When the torque, increases, the speed of a motor increases and vice-versa. We have seen that for a, d.c. motor;, , N=K, , (V − I a R a ), φ, , =, , K Eb, φ, , Ta ∝ φ I a, , (i), (ii), , If the flux decreases, from Eq.(i), the motor speed increases but from Eq.(ii) the, motor torque decreases. This is not possible because the increase in motor speed, must be the result of increased torque. Indeed, it is so in this case. When the flux, decreases slightly, the armature current increases to a large value. As a result, in, spite of the weakened field, the torque is momentarily increased to a high value, and will exceed considerably the value corresponding to the load. The surplus, torque available causes the motor to accelerate and back e.m.f. (Ea = P φ Z N/60, A) to rise. Steady conditions of speed will ultimately be achieved when back, e.m.f. has risen to such a value that armature current [Ia = (V − Ea)/Ra] develops, torque just sufficient to drive the load., , Illustration, Let us illustrate the above point with a numerical example. Suppose a 400 V, shunt motor is running at 600 r.p.m., taking an armature current of 50 A. The, armature resistance is 0.28 Ω. Let us see the effect of sudden reduction of flux, by 5% on the motor., Initially (prior to weakening of field), we have,, Ea = V − IaRa = 400 − 50 × 0.28 = 386 volts, We know that Eb ∝ φN. If the flux is reduced suddenly, Eb ∝ φ because inertia, of heavy armature prevents any rapid change in speed. It follows that when the, flux is reduced by 5%, the generated e.m.f. must follow suit. Thus at the instant, of reduction of flux, E'b = 0.95 × 386 = 366.7 volts., Instantaneous armature current is, , I 'a =, , V − E'b 400 − 366.7, =, = 118.9 A, Ra, 0.28, , Note that a sudden reduction of 5% in the flux has caused the armature current, to increase about 2.5 times the initial value. This will result in the production of, high value of torque. However, soon the steady conditions will prevail. This will, depend on the system inertia; the more rapidly the motor can alter the speed, the, sooner the e.m.f. rises and the armature current falls.
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4.16 Armature Reaction in D.C. Motors, As in a d.c. generator, armature reaction also occurs in a d.c. motor. This is, expected because when current flows through the armature conductors of a d.c., motor, it produces flux (armature flux) which lets on the flux produced by the, main poles. For a motor with the same polarity and direction of rotation as is for, generator, the direction of armature reaction field is reversed., (i) In a generator, the armature current flows in the direction of the induced, e.m.f. (i.e. generated e.m.f. Eg) whereas in a motor, the armature current, flows against the induced e.m.f. (i.e. back e.m.f. Eb). Therefore, it should, be expected that for the same direction of rotation and field polarity, the, armature flux of the motor will be in the opposite direction to that of the, generator. Hence instead of the main flux being distorted in the direction, of rotation as in a generator, it is distorted opposite to the direction of, rotation. We can conclude that:, Armature reaction in a d.c. generator weakens the jinx at leading pole, tips and strengthens the flux at trailing pole tips while the armature, reaction in a d. c. motor produces the opposite effect., (ii) In case of a d.c. generator, with brushes along G.N.A. and no, commutating poles used, the brushes must be shifted in the direction of, rotation (forward lead) for satisfactory commutation. However, in case, of a d.c. motor, the brushes are given a negative lead i.e., they are shifted, against the direction of rotation., With no commutating poles used, the brushes are given a forward lead, in a d.c. generator and backward lead in a d.c. motor., (iii) By using commutating poles (compoles), a d.c. machine can be operated, with fixed brush positions for all conditions of load. Since commutating, poles windings carry the armature current, then, when a machine, changes from generator to motor (with consequent reversal of current),, the polarities of commutating poles must be of opposite sign., Therefore, in a d.c. motor, the commutating poles must have the same, polarity as the main poles directly back of them. This is the opposite of, the corresponding relation in a d.c. generator., , 4.17 Commutation in D.C. Motors, Since the armature of a motor is the same as that of a generator, the current from, the supply line must divide and pass through the paths of the armature windings.
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In order to produce unidirectional force (or torque) on the armature conductors, of a motor, the conductors under any pole must carry the current in the same, direction at all times. This is illustrated in Fig. (4.10). In this case, the current, flows away from the observer in the conductors under the N-pole and towards, the observer in the conductors under the S-pole. Therefore, when a conductor, moves from the influence of N-pole to that of S-pole, the direction of current in, the conductor must be reversed. This is termed as commutation. The function of, the commutator and the brush gear in a d.c. motor is to cause the reversal of, current in a conductor as it moves from one side of a brush to the other. For, good commutation, the following points may be noted:, (i) If a motor does not have commutating poles (compoles), the brushes, must be given a negative lead i.e., they must be shifted from G.N.A., against the direction of rotation of, the motor., (ii) By using interpoles, a d.c. motor can be operated with fixed brush, positions for all conditions of load. For a d.c. motor, the commutating, poles must have the same polarity as the main poles directly back of, them. This is the opposite of the corresponding relation in a d.c., generator., Note. A d.c. machine may be used as a motor or a generator without changing, the commutating poles connections. When the operation of a d.c. machine, changes from generator to motor, the direction of the armature current reverses., Since commutating poles winding carries armature current, the polarity of, commutating pole reverses automatically to the correct polarity., , Fig. (4.10), , 4.18 Losses in a D.C. Motor, , Fig. (4.11)
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The losses occurring in a d.c. motor are the same as in a d.c. generator [See Sec., 1.26]. These are [See Fig. 4.11]:, (i) copper losses (n) Iron losses or magnetic losses, (ii) mechanical losses, As in a generator, these losses cause (a) an increase of machine temperature and, (b) reduction in the efficiency of the d.c. motor., The following points may be noted:, (i) Apart from armature Cu loss, field Cu loss and brush contact loss, Cu, losses also occur in interpoles (commutating poles) and compensating, windings. Since these windings carry armature current (Ia),, Loss in interpole winding = I 2a × Resistance of interpole winding, Loss in compensating winding = I 2a × Resistance of compensating winding, (ii) Since d.c. machines (generators or motors) are generally operated at, constant flux density and constant speed, the iron losses are nearly, constant., (iii) The mechanical losses (i.e. friction and windage) vary as the cube of the, speed of rotation of the d.c. machine (generator or motor). Since d.c., machines are generally operated at constant speed, mechanical losses are, considered to be constant., , 4.19 Efficiency of a D.C. Motor, Like a d.c. generator, the efficiency of a d.c. motor is the ratio of output power, to the input power i.e., Efficiency, η =, , output, output, × 100 =, × 100, output + losses, input, , As for a generator (See Sec. 1.29), the efficiency of a d.c. motor will be, maximum when:, Variable losses = Constant losses, Therefore, the efficiency curve of a d.c. motor is similar in shape to that of a d.c., generator., , 4.20 Power Stages, The power stages in a d.c. motor are represented diagrammatically in Fig., (4.12)., A − B = Copper losses, B − C = Iron and friction losses
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Fig. (4.12), Overall efficiency, ηc = C/A, Electrical efficiency, ηe = B/A, Mechanical efficiency, ηm = C/B, , 4.21 D.C. Motor Characteristics, There are three principal types of d.c. motors viz., shunt motors, series motors, and compound motors. Both shunt and series types have only one field winding, wound on the core of each pole of the motor. The compound type has two, separate field windings wound on the core of each pole. The performance of a, d.c. motor can be judged from its characteristic curves known as motor, characteristics, following are the three important characteristics of a d.c. motor:, , (i) Torque and Armature current characteristic (Ta/Ia), It is the curve between armature torque Ta and armature current Ia of a d.c., motor. It is also known as electrical characteristic of the motor., , (ii) Speed and armature current characteristic (N/ia), It is the curve between speed N and armature current Ia of a d.c. motor. It is very, important characteristic as it is often the deciding factor in the selection of the, motor for a particular application., , (iii) Speed and torque characteristic (N/Ta), It is the curve between speed N and armature torque Ta of a d.c. motor. It is also, known as mechanical characteristic., , 4.22 Characteristics of Shunt Motors, Fig. (4.13) shows the connections of a d.c. shunt motor. The field current Ish is, constant since the field winding is directly connected to the supply voltage V, which is assumed to be constant. Hence, the flux in a shunt motor is, approximately constant.
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Fig. (4.13), (i), , Fig. (4.14), , Ta/Ia Characteristic. We know that in a d.c. motor,, , Ta ∝ φ I a, Since the motor is operating from a constant supply voltage, flux φ is, constant (neglecting armature reaction)., , ∴, , Ta ∝ I a, , Hence Ta/Ia characteristic is a straight line passing through the origin as, shown in Fig. (4.14). The shaft torque (Tsh) is less than Ta and is shown by, a dotted line. It is clear from the curve that a very large current is required, to start a heavy load. Therefore, a shunt motor should not be started on, heavy load., (ii) N/Ia Characteristic. The speed N of a. d.c. motor is given by;, , N∝, , Eb, φ, , The flux φ and back e.m.f. Eb in a shunt motor are almost constant under, normal conditions. Therefore, speed of a shunt motor will remain constant, as the armature current varies (dotted line AB in Fig. 4.15). Strictly, speaking, when load is increased, Eb (= V− IaRa) and φ decrease due to the, armature resistance drop and armature reaction respectively. However, Eb, decreases slightly more than φ so that the speed of the motor decreases, slightly with load (line AC)., (iii) N/Ta Characteristic. The curve is obtained by plotting the values of N and, Ta for various armature currents (See Fig. 4.16). It may be seen that speed, falls somewhat as the load torque increases.
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Fig. (4.15), , Fig. (4.16), , Conclusions, Following two important conclusions are drawn from the above characteristics:, (i) There is slight change in the speed of a shunt motor from no-load to fullload. Hence, it is essentially a constant-speed motor., (ii) The starting torque is not high because Ta ∝ Ia., , 4.23 Characteristics of Series Motors, Fig. (4.17) shows the connections of a series motor. Note that current passing, through the field winding is the same as that in the armature. If the mechanical, load on the motor increases, the armature current also increases. Hence, the flux, in a series motor increases with the increase in armature current and vice-versa., , Fig. (4.17), (i), , Fig. (4.18), , Ta/Ia Characteristic. We know that:, , Ta ∝ φ I a, Upto magnetic saturation, φ ∝ Ia so that Ta ∝ I 2a, After magnetic saturation, φ is constant so that Ta ∝ Ia, Thus upto magnetic saturation, the armature torque is directly proportional, to the square of armature current. If Ia is doubled, Ta is almost quadrupled.
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Therefore, Ta/Ia curve upto magnetic saturation is a parabola (portion OA, of the curve in Fig. 4.18). However, after magnetic saturation, torque is, directly proportional to the armature current. Therefore, Ta/Ia curve after, magnetic saturation is a straight line (portion AB of the curve)., It may be seen that in the initial portion of the curve (i.e. upto magnetic, saturation), Ta ∝ I 2a . This means that starting torque of a d.c. series motor, will be very high as compared to a shunt motor (where that Ta ∝ Ia)., (ii) N/Ia Characteristic. The speed N of a series motor is given by;, , N∝, , Eb, φ, , where, , E b = V − I a (R a + R se ), , When the armature current increases, the back e.m.f. Ed decreases due to, Ia(Ra + Rse) drop while the flux φ increases. However, Ia(Ra + Rse) drop is, quite small under normal conditions and may be neglected., , ∴, , N ∝, ∝, , 1, φ, 1, upto magnetic saturation, Ia, , Thus, upto magnetic saturation, the N/Ia curve follows the hyperbolic path, as shown in Fig. (4.19). After saturation, the flux becomes constant and so, does the speed., , Fig. (4.19), , Fig. (4.20), , (iii) N/Ta Characteristic. The N/Ta characteristic of a series motor is shown in, Fig. (4.20). It is clear that series motor develops high torque at low speed, and vice-versa. It is because an increase in torque requires an increase in, armature current, which is also the field current. The result is that flux is, strengthened and hence the speed drops (Q N ∝ 1/φ). Reverse happens, should the torque be low., , Conclusions
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Following three important conclusions are drawn from the above characteristics, of series motors:, (i) It has a high starting torque because initially Ta ∝ I 2a ., (ii) It is a variable speed motor (See N/Ia curve in Fig. 4.19) i.e., it, automatically adjusts the speed as the load changes. Thus if the load, decreases, its speed is automatically raised and vice-versa., (iii) At no-load, the armature current is very small and so is the flux. Hence,, the speed rises to an excessive high value (Q N ∝ 1/φ). This is, dangerous for the machine which may be destroyed due to centrifugal, forces set up in the rotating parts. Therefore, a series motor should never, be started on no-load. However, to start a series motor, mechanical load, is first put and then the motor is started., Note. The minimum load on a d.c. series motor should be great enough to keep, the speed within limits. If the speed becomes dangerously high, then motor must, be disconnected from the supply., , 4.24 Compound Motors, A compound motor has both series field and shunt field. The shunt field is, always stronger than the series field. Compound motors are of two types:, (i) Cumulative-compound motors in which series field aids the shunt field., (ii) Differential-compound motors in which series field opposes the shunt, field., Differential compound motors are rarely used due to their poor torque, characteristics at heavy loads., , 4.25 Characteristics of Cumulative Compound Motors, Fig. (4.21) shows the connections of a cumulative-compound motor. Each pole, carries a series as well as shunt field winding; the series field aiding the shunt, field., (i), , Ta/Ia Characteristic. As the load increases, the series field increases but, shunt field strength remains constant. Consequently, total flux is increased, and hence the armature torque (Q Ta ∝ φIa). It may be noted that torque of, a cumulative-compound motor is greater than that of shunt motor for a, given armature current due to series field [See Fig. 4.22].
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Fig. (4.21), , Fig. (4.22), , (ii) N/Ia Characteristic. As explained above, as the lead increases, the flux per, pole also increases. Consequently, the speed (N ∝ 1/φ) of the motor tails as, the load increases (See Fig. 4.23). It may be noted that as the load is added,, the increased amount of flux causes the speed to decrease more than does, the speed of a shunt motor. Thus the speed regulation of a cumulative, compound motor is poorer than that of a shunt motor., Note: Due to shunt field, the motor has a definite no load speed and can be, operated safely at no-load., , Fig. (4.23), , Fig. (4.24), , (iii) N/Ta Characteristic. Fig. (4.24) shows N/Ta characteristic of a cumulative, compound motor. For a given armature current, the torque of a cumulative, compound motor is more than that of a shunt motor but less than that of a, series motor., , Conclusions, A cumulative compound motor has characteristics intermediate between series, and shunt motors., (i) Due to the presence of shunt field, the motor is prevented from running, away at no-load., (ii) Due to the presence of series field, the starting torque is increased.
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4.26 Comparison of Three Types of Motors, (i), , The speed regulation of a shunt motor is better than that of a series motor., , Fig. (4.25), , Fig. (4.26), , However, speed regulation of a cumulative compound motor lies between, shunt and series motors (See Fig. 4.25)., (ii) For a given armature current, the starting torque of a series motor is more, than that of a shunt motor. However, the starting torque of a cumulative, compound motor lies between series and shunt motors (See Fig. 4.26)., (iii) Both shunt and cumulative compound motors have definite no-load speed., However, a series motor has dangerously high speed at no-load., , 4.27 Applications of D.C. Motors, 1., , Shunt motors, The characteristics of a shunt motor reveal that it is an approximately constant, speed motor. It is, therefore, used, (i) where the speed is required to remain almost constant from no-load to, full-load, (ii) where the load has 10 be driven at a number of speeds and any one of, which is required to remain nearly constant, Industrial use: Lathes, drills, boring mills, shapers, spinning and weaving, machines etc., , 2., , Series motors, It is a variable speed motor i.e., speed is low at high torque and vice-versa., However, at light or no-load, the motor tends to attain dangerously high speed., The motor has a high starting torque. It is, therefore, used, (i) where large starting torque is required e.g., in elevators and electric, traction
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(ii) where the load is subjected to heavy fluctuations and the speed is, automatically required to reduce at high torques and vice-versa, Industrial use: Electric traction, cranes, elevators, air compressors, vacuum, cleaners, hair drier, sewing machines etc., , 3., , Compound motors, Differential-compound motors are rarely used because of their poor torque, characteristics. However, cumulative-compound motors are used where a fairly, constant speed is required with irregular loads or suddenly applied heavy loads., Industrial use: Presses, shears, reciprocating machines etc., , 4.28 Troubles in D.C. Motors, Several troubles may arise in a d.c. motor and a few of them are discussed, below:
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1., , Failure to start, This may be due to (i) ground fault (ii) open or short-circuit fault (iii) wrong, connections (iv) too low supply voltage (v) frozen bearing or (vi) excessive, load., , 2., , Sparking at brushes, This may be due to (i) troubles in brushes (ii) troubles in commutator, (iii) troubles in armature or (iv) excessive load., (i) Brush troubles may arise due to insufficient contact surface, too short a, brush, too little spring tension or wrong brush setting., (ii) Commutator troubles may be due to dirt on the commutator, high mica,, rough surface or eccentricity., (iii) Armature troubles may be due to an open armature coil. An open, armature coil will cause sparking each time the open coil passes the, brush. The location of this open coil is noticeable by a burnt line, between segments connecting the coil., , 3., , Vibrations and pounding noises, These maybe due to (i) worn bearings (ii) loose parts (iii) rotating parts hitting, stationary parts (iv) armature unbalanced (v) misalignment of machine (vi) loose, coupling etc., , 4., , Overheating, The overheating of motor may be due to (i) overloads (ii) sparking at the brushes, (iii) short-circuited armature or field coils (iv) too frequent starts or reversals, (v) poor ventilation (vi) incorrect voltage.
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Chapter (5), , Speed Control of D.C. Motors, Introduction, Although a far greater percentage of electric motors in service are a.c. motors,, the d.c. motor is of considerable industrial importance. The principal advantage, of a d.c. motor is that its speed can be changed over a wide range by a variety of, simple methods. Such a fine speed control is generally not possible with a.c., motors. In fact, fine speed control is one of the reasons for the strong, competitive position of d.c. motors in the modem industrial applications. In this, chapter, we shall discuss the various methods of-speed control of d.c. motors., , 5.1 Speed Control of D.C. Motors, The speed of a d.c. motor is given by:, , N∝, , Eb, φ, , (V − I a R ), , or, , N=K, , where, , R = Ra, = Ra + Rse., , φ, , r.p.m., , (i), for shunt motor, for series motor, , From exp. (i), it is clear that there are three main methods of controlling the, speed of a d.c. motor, namely:, (i) By varying the flux per pole (φ). This is known as flux control method., (ii) By varying the resistance in the armature circuit. This is known as, armature control method., (iii) By varying the applied voltage V. This is known as voltage control, method., , 5.2 Speed Control of D.C. Shunt Motors, The speed of a shunt motor can be changed by (i) flux control method, (ii) armature control method (iii) voltage control method. The first method (i.e., flux control method) is frequently used because it is simple and inexpensive.
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1., , Flux control method, It is based on the fact that by varying the flux φ, the motor speed (N ∝ 1/φ) can, be changed and hence the name flux control method. In this method, a variable, resistance (known as shunt field rheostat) is placed in series with shunt field, winding as shown in Fig. (5.1)., , Fig. (5.1), , Fig. (5.2), , The shunt field rheostat reduces the shunt field current Ish and hence the flux φ., Therefore, we can only raise the speed of the motor above the normal speed (See, Fig. 5.2). Generally, this method permits to increase the speed in the ratio 3:1., Wider speed ranges tend to produce instability and poor commutation., Advantages, (i) This is an easy and convenient method., (ii) It is an inexpensive method since very little power is wasted in the shunt, field rheostat due to relatively small value of Ish., (iii) The speed control exercised by this method is independent of load on the, machine., Disadvantages, (i), , Only speeds higher than the normal speed can be obtained since the total, field circuit resistance cannot be reduced below Rsh—the shunt field, winding resistance., (ii) There is a limit to the maximum speed obtainable by this method. It is, because if the flux is too much weakened, commutation becomes poorer., Note. The field of a shunt motor in operation should never be opened because its, speed will increase to an extremely high value., , 2., , Armature control method, This method is based on the fact that by varying the voltage available across the, armature, the back e.m.f and hence the speed of the motor can be changed. This
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is done by inserting a variable resistance RC (known as controller resistance) in, series with the armature as shown in Fig. (5.3)., , Fig. (5.3), , Fig. (5.4), , N ∝ V − I a (R a + R C ), where, , RC = controller resistance, , Due to voltage drop in the controller resistance, the back e.m.f. (Eb) is, decreased. Since N ∝ Eb, the speed of the motor is reduced. The highest speed, obtainable is lhat corresponding to RC = 0 i.e., normal speed. Hence, this method, can only provide speeds below the normal speed (See Fig. 5.4)., Disadvantages, (i), , A large amount of power is wasted in the controller resistance since it, carries full armature current Ia., (ii) The speed varies widely with load since the speed depends upon the, voltage drop in the controller resistance and hence on the armature current, demanded by the load., (iii) The output and efficiency of the motor are reduced., (iv) This method results in poor speed regulation., Due to above disadvantages, this method is seldom used to control tie speed of, shunt motors., Note. The armature control method is a very common method for the speed, control of d.c. series motors. The disadvantage of poor speed regulation is not, important in a series motor which is used only where varying speed service is, required., , 3., , Voltage control method, In this method, the voltage source supplying the field current is different from, that which supplies the armature. This method avoids the disadvantages of poor, speed regulation and low efficiency as in armature control method. However, it
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is quite expensive. Therefore, this method of speed control is employed for large, size motors where efficiency is of great importance., (i) Multiple voltage control. In this method, the shunt field of the motor is, connected permanently across a-fixed voltage source. The armature can be, connected across several different voltages through a suitable switchgear., In this way, voltage applied across the armature can be changed. The speed, will be approximately proportional to the voltage applied across the, armature. Intermediate speeds can be obtained by means of a shunt field, regulator., (ii) Ward-Leonard system. In this method, the adjustable voltage for the, armature is obtained from an adjustable-voltage generator while the field, circuit is supplied from a separate source. This is illustrated in Fig. (5.5)., The armature of the shunt motor M (whose speed is to be controlled) is, connected directly to a d.c. generator G driven by a constant-speed a.c., motor A. The field of the shunt motor is supplied from a constant-voltage, exciter E. The field of the generator G is also supplied from the exciter E., The voltage of the generator G can be varied by means of its field, regulator. By reversing the field current of generator G by controller FC,, the voltage applied to the motor may be reversed. Sometimes, a field, regulator is included in the field circuit of shunt motor M for additional, speed adjustment. With this method, the motor may be operated at any, speed upto its maximum speed., , Fig. (5.5), Advantages, (a) The speed of the motor can be adjusted through a wide range without, resistance losses which results in high efficiency., (b) The motor can be brought to a standstill quickly, simply by rapidly, reducing the voltage of generator G. When the generator voltage is reduced, below the back e.m.f. of the motor, this back e.m.f. sends current through, the generator armature, establishing dynamic braking. While this takes
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place, the generator G operates as a motor driving motor A which returns, power to the line., (c) This method is used for the speed control of large motors when a d.c., supply is not available., The disadvantage of the method is that a special motor-generator set is required, for each motor and the losses in this set are high if the motor is operating under, light loads for long periods., , 5.3 Speed Control of D.C. Series Motors, The speed control of d.c. series motors can be obtained by (i) flux control, method (ii) armature-resistance control method. The latter method is mostly, used., , 1., , Flux control method, In this method, the flux produced by the series motor is varied and hence the, speed. The variation of flux can be achieved in the following ways:, (i) Field divertcrs. In this method, a, variable resistance (called field, diverter) is connected in parallel with, series field winding as shown in Fig., (5.6). Its effect is to shunt some, portion of the line current from the, series field winding, thus weakening, the field and increasing the speed, (Q N ∝ 1/φ). The lowest speed, obtainable is that corresponding to, Fig. (5.6), zero current in the diverter (i.e.,, diverter is open). Obviously, the lowest speed obtainable is the normal, speed of the motor. Consequently, this method can only provide speeds, above the normal speed. The series field diverter method is often employed, in traction work., (ii) Armature diverter. In order to obtain speeds below the normal speed, a, variable resistance (called armature diverter) is connected in parallel with, the armature as shown in Fig. (5.7). The diverter shunts some of the line, current, thus reducing the armature current. Now for a given load, if Ia is, decreased, the flux φ must increase (Q T ∝ φIa). Since N ∝ 1/φ, the motor, speed is decreased. By adjusting the armature diverter, any speed lower, than the normal speed can be obtained., (iii) Tapped field control. In this method, the flux is reduced (and hence speed, is increased) by decreasing the number of turns of the series field winding, as shown in Fig. (5.8). The switch S can short circuit any part of the field
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winding, thus decreasing the flux and raising the speed. With full turns of, the field winding, the motor runs at normal speed and as the field turns are, cut out, speeds higher than normal speed are achieved., , Fig. (5.7), , Fig. (5.8), , (iv) Paralleling field coils. This method is usually employed in the case of fan, motors. By regrouping the field coils as shown in Fig. (5.9), several fixed, speeds can be obtained., , Fig. (5.9), , 2., , Armature-resistance control, In this method, a variable resistance, is directly connected in series with, the supply to the complete motor as, shown in Fig. (5.10). This reduces, the voltage available across the, armature and hence the speed falls., By changing the value of variable, resistance, any speed below the, Fig. (5.10), normal speed can be obtained. This, is the most common method, employed to control the speed of d.c. series motors. Although this method has, poor speed regulation, this has no significance for series motors because they are, used in varying speed applications. The loss of power in the series resistance for, many applications of series motors is not too serious since in these applications,
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the control is utilized for a large portion of the time for reducing the speed under, light-load conditions and is only used intermittently when the motor is carrying, full-load., , 5.4 Series-Parallel Control, Another method used for the speed control of d.c. series motors is the seriesparallel method. In this system which is widely used in traction system, two (or, more) similar d.c. series motors are mechanically coupled to the same load., , Fig. (5.11), When the motors are connected in series [See Fig. 5.11 (i)], each motor armature, will receive one-half the normal voltage. Therefore, the speed will be low. When, the motors are connected in parallel, each motor armature receives the normal, voltage and the speed is high [See Fig. 5.11 (ii)]. Thus we can obtain two, speeds. Note that for the same load on the pair of motors, the system would run, approximately four times the speed when the machines are in parallel as when, they are in series., , Series-parallel and resistance control, In electric traction, series-parallel method is usually combined with resistance, method of control. In the simplest case, two d.c. series motors are coupled, mechanically and drive the same vehicle., (i) At standstill, the motors are connected is series via a starting rheostat. The, motors are started up in series with each other and starting resistance is cut, out step by step to increase the speed. When all the resistance is cut out, (See Fig. 5.12), the voltage applied to each motor is about one-half of the, line voltage. The speed is then about one-half of what it would be if the full, line voltage were applied to each motor.
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(ii) To increase the speed further, the two motors are connected in parallel and, at the same time the starting resistance is connected in series with the, combination (See Fig. 5.12). The starting resistance is again cut out step by, step until full speed is attained. Then field control is introduced., , Fig. (5.12), , 5.5 Electric Braking, Sometimes it is desirable to stop a d.c. motor quickly. This may be necessary in, case of emergency or to save time if the motor is being used for frequently, repeated operations. The motor and its load may be brought to rest by using, either (i) mechanical (friction) braking or (ii) electric braking. In mechanical, braking, the motor is stopped due to the friction between the moving parts of the, motor and the brake shoe i.e. kinetic energy of the motor is dissipated as heat., Mechanical braking has several disadvantages including non-smooth stop and, greater stopping time.
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In electric braking, the kinetic energy of the moving parts (i.e., motor) is, converted into electrical energy which is dissipated in a resistance as heat or, alternativley, it is returned to the supply source (Regenerative braking). For d.c., shunt as well as series motors, the following three methods of electric braking, are used:, (i) Rheostatic or Dynamic braking, (ii) Plugging, (iii) Regenerative braking, It may be noted that electric braking cannot hold the motor stationary and, mechanical braking is necessary. However, the main advantage of using electric, braking is that it reduces the wear and tear of mechanical brakes and cuts down, the stopping time considerably due to high braking retardation., , (i) Rheostatic or Dynamic braking, In this method, the armature of the running motor is disconnected from the, supply and is connected across a variable resistance R. However, the field, winding is left connected to the supply. The armature, while slowing down,, rotates in a strong magnetic field and, therefore, operates as a generator, sending, a large current through resistance R. This causes the energy possessed by the, rotating armature to be dissipated quickly as heat in the resistance. As a result,, the motor is brought to standstill quickly., Fig. (5.13) (i) shows dynamic braking of a shunt motor. The braking torque can, be controlled by varying the resistance R. If the value of R is decreased as the, motor speed decreases, the braking torque may be maintained at a high value. At, a low value of speed, the braking torque becomes small and the final stopping of, the motor is due to friction. This type of braking is used extensively in, connection with the control of elevators and hoists and in other applications in, which motors must be started, stopped and reversed frequently., , Fig. (5.13)
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We now investigate how braking torque depends upon the speed of the motor., Referring to Fig. (5.13) (ii),, Armature current, I a =, , Eb, k Nφ, = 1, R + Ra R + Ra, , k Nφ, Braking torque, TB = k 2 I a φ = k 2 φ 1, R + Ra, , (Q, , E b ∝ φN ), , , 2, = k 3 Nφ, , , where k2 and k3 are constants, For a shunt motor, φ is constant., ∴ Braking torque, TB ∝ N, Therefore, braking torque decreases as the motor speed decreases., , (ii), , Plugging, , In this method, connections to the armature are reversed so that motor tends to, rotate in the opposite direction, thus providing the necessary braking effect., When the motor comes to rest, the supply must be cut off otherwise the motor, will start rotating in the opposite direction., , Fig. (5.14), Fig. (5.14) (ii) shows plugging of a d.c. shunt motor. Note that armature, connections are reversed while the connections of the field winding are kept the, same. As a result the current in the armature reverses. During the normal, running of the motor [See Fig. 5.14 (i)], the back e.m.f. Eb opposes the applied, voltage V. However, when armature connections are reversed, back e.m.f. Eb, and V act in the same direction around the circuit. Therefore, a voltage equal to
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V + Eb is impressed across the armature circuit. Since Eb ~ V, the impressed, voltage is approximately 2V. In order 10 limit the current to safe value, a, variable resistance R is inserted in the circuit at the time of changing armature, connections., We now investigate how braking torque depends upon the speed of the motor., Referring to Fig. (5.14) (ii),, Armature current, I a =, , V + Eb, k Nφ, V, =, + 1, R + Ra R + Ra R + Ra, , (Q, , E b ∝ φN ), , k Nφ , V, 2, + 1, Braking torque, TB = k 2 I a φ = k 2 φ, = k 3φ + k 4 Nφ, R + Ra R + Ra , For a shunt motor, φ is constant., ∴, , Braking torque, TB = k5 + k6N, , Thus braking torque decreases as the motor slows down. Note that there is some, braking torque (TB = k5) even when the motor speed is zero., , (iii) Regenerative braking, In the regenerative braking, the motor is run as a generator. As a result, the, kinetic energy of the motor is converted into electrical energy and returned to, the supply. Fig. (5.15) shows two methods of regenerative braking for a shunt, motor., , Fig. (5.15), (a) In one method, field winding is disconnected from the supply and field, current is increased by exciting it from another source [See Fig. 5.15 (i)]., As a result, induced e.m.f. E exceeds the supply voltage V and the machine, feeds energy into the supply. Thus braking torque is provided upto the, speed at which induced e.m.f. and supply voltage are equal. As the machine, slows down, it is not possible to maintain induced e.m.f. at a higher value
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than the supply voltage. Therefore, this method is possible only for a, limited range of speed., (b) In a second method, the field excitation does not change but the load causes, the motor to run above the normal speed (e.g., descending load on a crane)., As a result, the induced e.m.f. E becomes greater than the supply voltage V, [See Fig. 5.15 (ii)]. The direction of armature current I, therefore, reverses, but the direction of shunt field current If remains unaltered. Hence the, torque is reversed and the speed falls until E becomes less than V., , 5.6 Speed Control of Compound Motors, Speed control of compound motors may be obtained by any one of the methods, described for shunt motors. Speed control cannot be obtained through, adjustment of the series field since such adjustment would radically change the, performance characteristics of the motor., , 5.7 Necessity of D.C. Motor Starter, At starting, when the motor is stationary, there is no back e.m.f. in the armature., Consequently, if the motor is directly switched on to the mains, the armature, will draw a heavy current (Ia = V/Ra) because of small armature resistance. As, an example, 5 H.P., 220 V shunt motor has a full-load current of 20 A and an, armature resistance of about 0.5 Ω. If this motor is directly switched on to, supply, it would take an armature current of 220/0.5 = 440 A which is 22 times, the full-load current. This high starting current may result in:, (i) burning of armature due to excessive heating effect,, (ii) damaging the commutator and brushes due to heavy sparking,, (iii) excessive voltage drop in the line to which the motor is connected. The, result is that the operation of other appliances connected to the line may, be impaired and in particular cases, they may refuse to work., In order to avoid excessive current at starting, a variable resistance (known as, starting resistance) is inserted in series with the armature circuit. This resistance, is gradually reduced as the motor gains speed (and hence Eb increases) and, eventually it is cut out completely when the motor has attained full speed. The, value of starting resistance is generally such that starting current is limited to, 1.25 to 2 times the full-load current., , 5.8 Types of D.C. Motor Starters, The stalling operation of a d.c. motor consists in the insertion of external, resistance into the armature circuit to limit the starting current taken by the, motor and the removal of this resistance in steps as the motor accelerates. When
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the motor attains the normal speed, this resistance is totally cut out of the, armature circuit. It is very important and desirable to provide the starter with, protective devices to enable the starter arm to return to OFF position, (i) when the supply fails, thus preventing the armature being directly across, the mains when this voltage is restored. For this purpose, we use no-volt, release coil., (ii) when the motor becomes overloaded or develops a fault causing the, motor to take an excessive current. For this purpose, we use overload, release coil., There are two principal types of d.c. motor starters viz., three-point starter and, four-point starter. As we shall see, the two types of starters differ only in the, manner in which the no-volt release coil is connected., , 5.9 Three-Point Starter, This type of starter is widely used for starting shunt and compound motors., , Schematic diagram, Fig. (5.16) shows the schematic diagram of a three-point starter for a shunt, motor with protective devices. It is so called because it has three terminals L, Z, and A. The starter consists of starting resistance divided into several sections, and connected in series with the armature. The tapping points of the starting, resistance are brought out to a number of studs. The three terminals L, Z and A, of the starter are connected respectively to the positive line terminal, shunt field, terminal and armature terminal. The other terminals of the armature and shunt, field windings are connected to the negative terminal of the supply. The no-volt, release coil is connected in the shunt field circuit. One end of the handle is, connected to the terminal L through the over-load release coil. The other end of, the handle moves against a spiral spring and makes contact with each stud, during starting operation, cutting out more and more starting resistance as it, passes over each stud in clockwise direction., , Operation, (i), , To start with, the d.c. supply is switched on with handle in the OFF, position., (ii) The handle is now moved clockwise to the first stud. As soon as it comes in, contact with the first stud, the shunt field winding is directly connected, across the supply, while the whole starting resistance is inserted in series, with the armature circuit., (iii) As the handle is gradually moved over to the final stud, the starting, resistance is cut out of the armature circuit in steps. The handle is now held
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magnetically by the no-volt release coil which is energized by shunt field, current., (iv) If the supply voltage is suddenly interrupted or if the field excitation is, accidentally cut, the no-volt release coil is demagnetized and the handle, goes back to the OFF position under the pull of the spring. If no-volt, release coil were not used, then in case of failure of supply, the handle, would remain on the final stud. If then supply is restored, the motor will be, directly connected across the supply, resulting in an excessive armature, current., (v) If the motor is over-loaded (or a fault occurs), it will draw excessive, current from the supply. This current will increase the ampere-turns of the, over-load release coil and pull the armature C, thus short-circuiting the novolt release coil. The no-volt coil is demagnetized and the handle is pulled, to the OFF position by the spring. Thus, the motor is automatically, disconnected from the supply., , Fig. (5.16)
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Drawback, In a three-point starter, the no-volt release coil is connected in series with the, shunt field circuit so that it carries the shunt field current. While exercising, speed control through field regulator, the field current may be weakened to such, an extent that the no-volt release coil may not be able to keep the starter arm in, the ON position. This may disconnect the motor from the supply when it is not, desired. This drawback is overcome in the four point starter., , 5.10 Four-Point Starter, In a four-point starter, the no-volt release coil is connected directly across the, supply line through a protective resistance R. Fig. (5.17) shows the schematic, diagram of a 4-point starter for a shunt motor (over-load release coil omitted for, clarity of the figure). Now the no-volt release coil circuit is independent of the, shunt field circuit. Therefore, proper speed control can be exercised without, affecting the operation of novolt release coil., Note that the only difference, between a three-point starter, and a four-point starter is the, manner in which no-volt, release coil is connected., However, the working of the, two starters is the same. It, may be noted that the threepoint starter also provides, protection against an openfield circuit. This protection is, Fig. (5.17), not provided by the four-point, starter., , 5.11 Grading of Starting Resistance—Shunt Motors, For starting the motor satisfactorily, the starting resistance is divided into a, number of sections in such a way that current fluctuates between maximum (Im), and minimum (I) values. The upper limit is that value established as the, maximum permissible for the motor; it is generally 1.5 times the full-load, current of the motor. The lower limit is the value set as a minimum for starting, operation; it may be equal to full-load current of the motor or some, predetermined value. Fig. (5.18) shows shunt-wound motor with starting, resistance divided into three sections between four studs. The resistances of
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these sections should be so selected that current during starting remains between, Im and I as shown in Fig. (5.19)., , Fig. (5.18), (i), , Fig. (5.19), , When arm A is moved from OFF position to stud 1, field and armature, circuits are energized and whole of the starting resistance is in series, with the armature. The armature current jumps to maximum value given, by;, , Im =, where, , V, R1, , R1 = Resistance of starter and armature, , (ii) As the armature accelerates, the generated e.m.f. increases and the, armature current decreases as indicated by curve ab. When the current, has fallen to I, arm A is moved over to stud 2, cutting out sufficient, resistance to allow the current to rise to Im again. This operation is, repeated until the arm A is on stud 4 and the whole of the starting, resistance is cut out of the armature circuit., (iii) Now the motor continues to accelerate and the current decreases until it, settles down at some value IL such that torque due to this current is just, sufficient to meet the load requirement., , 5.12 Starter Step Calculations for D.C. Shunt Motor, Fig. (5.20) shows a d.c. shunt motor starter with n resistance sections and (n + 1), studs., Let R1 = Total resistance in the armature circuit when the starter arm is on, stud no. 1 (See Fig. 5.20), R2 = Total resistance in the armature circuit when the starter arm is on, stud no. 2 and so on
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Im =, I=, n=, V=, Ra =, , Upper current limit, Lower current limit, Number of sections in the starter resistance, Applied voltage, Armature resistance, , On stud 1. When the starter arm-moves to stud 1, the total resistance in the, armature circuit is R1 and the circuit current jumps to maximum values Im given, by;, , Im =, , V, R1, , (i), , Since torque ∝ φ Ia, it follows that the maximum torque acts on the armature to, accelerate it. As the armature accelerates, the induced e.m.f. (back e.m.f.), increases and the armature current decreases When the current has fallen to the, predetermined value I, the starter arm is moved over to stud 2. Let the value of, back e.m.f. be Eb1 at the instant the starter arm leaves the stud 1. Then I is given, by;, , I=, , V − E b1, R1, , (ii), , Fig. (5.20), On stud 2. As the starter arm moves over to stud 2, sufficient resistance is cut, out (now total circuit resistance is R2) and current rises to maximum value Im, once again given by;
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Im =, , V − E b1, R2, , (iii), , The acceleration continues and the back e.m.f. increases and the armature, current decreases. When the current has fallen to the predetermined value I, the, starter arm is moved over to stud 3. Let Eb2 be the value of back e.m.f. at the, instant the starter arm leaves the stud 2. Then,, , I=, , V − E b2, R2, , (iv), , On stud 3., As the starter arm moves to stud 3, I m =, As the starter arm leaves stud 3, I =, , V − E b2, R3, , V − E b3, R3, , (v), (vi), , On nth stud., As the starter arm leaves nth stud, I =, , V − E bn, Rn, , On (n + 1)th stud. When the starter arm moves over to (n + 1)th stud, all the, external starting resistance is cut out, leaving only the armature resistance Ra., , ∴, , Im =, , V − E bn, Ra, , and, , I=, , V − Eb, Ra, , Dividing Eq.(ii) by Eq.(iii), we get,, , I, Im, , =, , R2, R1, , Dividing Eq.(iv) by Eq. (v), we get,, , I, Im, , =, , R3, R2, , Continuing these divisions, we Lave finally,, , I, Im, Let, , I, Im, , =, , Ra, Rn, , = k., , Then, , R, R2 R3, =, = ... = a = k, R1 R 2, Rn, , If we multiply these n equal ratios together, then,
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R 2 R3 R 4, R, ×, ×, × ... × a = k n, R1 R 2 R 3, Rn, ∴, , Ra, = kn, R1, , Thus we can calculate the values of R2, R3, R4 etc. if the values of R1, Ra and n, are known.
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Chapter (6), , Testing of D.C. Machines, Introduction, There are several tests that are conducted upon a d.c. machine (generator or, motor) to judge its performance. One important test is performed to measure the, efficiency of a d.c. machine. The efficiency of a d.c. machine depends upon its, losses. The smaller the losses, the greater is the efficiency of the machine and, vice-versa. The consideration of losses in a d.c. machine is important for two, principal reasons. First, losses determine the efficiency of the machine and, appreciably influence its operating cost. Secondly, losses determine the heating, of the machine and hence the power output that may be obtained without undue, deterioration of the insulation. In this chapter, we shall focus our attention on the, various methods for the determination of the efficiency of a d.c. machine., , 6.1 Efficiency of a D.C. Machine, The power that a d.c. machine receives is called the input and the power it gives, out is called the output. Therefore, the efficiency of a d.c. machine, like that of, any energy-transferring device, is given by;, , Efficiency =, , Output, Input, , Output = Input − Losses, , (i), and, , Input = Output + Losses, , Therefore, the efficiency of a d.c. machine can also be expressed in the, following forms:, , Efficiency =, , Input − Losses, Input, , (ii), , Efficiency =, , Output, Output + Losses, , (iii), , The most obvious method of determining the efficiency of a d.c. machine is to, directly load it and measure the input power and output power. Then we can use, Eq.(i) to determine the efficiency of the machine. This method suffers from, three main drawbacks. First, this method requires the application of load on the, machine. Secondly, for machines of large rating, the loads of the required sizes
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may not be available. Thirdly, even 'fit is possible to provide such loads, large, power will be dissipated, making it an expensive method., The most common method of measuring the efficiency of a d.c. machine is to, determine its losses (instead of measuring the input and output on load). We can, then use Eq.(ii) or Eq.(iii) to determine the efficiency of the machine. This, method has the obvious advantage of convenience and economy., , 6.2 Efficiency By Direct Loading, In this method, the d.c. machine is loaded and output, and input are measured to find the efficiency. For this, purpose, two simple methods can be used., , (i) Brake test, In this method, a brake is applied to a water-cooled, pulley mounted on the motor shaft as shown in Fig., (6.1). One end of the rope is fixed to the floor via a, spring balance S and a known mass is suspended at, the other end. If the spring balance reading is S kg-Wt, and the suspended mass has a weight of W kg-Wt,, then,, , Fig. (6.1), , Net pull on the rope = (W − S) kg-Wt = (W − S) × 9.81 newtons, If r is the radius of the pulley in metres, then the shaft torque Tsh developed by, the motor is, , Tsh = (W − S) × 9.81 × r N - m, If the speed of the pulley is N r.p.m., then,, Output power =, Let, , 2π N Tsh 2 π N × (W − S) × 9.81 × r, =, 60, 60, , watts, , V = Supply voltage in volts, I = Current taken by the motor in amperes, ∴, , ∴, , Input to motor = V I, , Efficiency =, , watts, , 2π N ( W − S) × r × 9.81, 60 × VI, , (ii) In another method, the motor drives a calibrated generator i.e. one whose, efficiency is known at all loads. The output of the generator is measured, with the help of an ammeter and voltmeter.
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∴, Let, , Output of motor =, , Generator output, Generator efficiency, , V = Supply voltage is volts, I = Current taken by the motor is amperes, Input to motor = VI, , Thus efficiency of the motor can be determined., Because of several disadvantages (See Sec. 6.1), direct loading method is used, only for determining the efficiency of small machines., , 6.3 Swinburne’s Method for Determining Efficiency, In this method, the d.c. machine (generator or motor) is run as a motor at noload and losses of the machine are determined. Once the losses of the machine, are known, its efficiency at any desired load can be determined in advance. It, may be noted that this method is applicable to those machines in which flux is, practically constant at all loads e.g., shunt and compound machines. Let us see, how the efficiency of a d.c. shunt machine (generator or motor) is determined by, this method. The test insists of two steps:, , (i) Determination of hot resistances of windings, The armature resistance and shunt field resistance are measured using a battery,, voltmeter and ammeter. Since these resistances are measured when the machine, is cold, they must be converted to values corresponding to the temperature at, which the machine would work on full-load. Generally, these values are, measured for a temperature rise of 40°C above the room temperature. Let the hot, resistances of armature and shunt field be Ra and Rsh respectively., , Fig. (6.2)
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(ii) Determination of constant losses, The machine is run as a motor on no-load with supply voltage adjusted to the, rated voltage i.e. voltage stamped on the nameplate. The speed of the motor is, adjusted to the rated speed with the help of field regulator R as shown is Fig., (6.2)., Let V = Supply voltage, I0 = No-load current read by ammeter A1, Ish = Shunt-field current read by ammeter A2, ∴, , No-load armature current, Ia0 = I0 − Ish, No-load input power to motor = V I0, No-load power input to armature = V Ia0 = V(I0 − Ish), , Since the output of the motor is zero, the no-load input power to the armature, supplies (a) iron losses in the core (b) friction loss (c) windage loss (d) armature, Cu loss I 2a 0 R a or (I 0 − Ish )2 R a ., , [, , ], , Constant losses, WC = Input to motor − Armature Cu loss, or, , WC = V I 0 − (I 0 − I sh )2 R a, , Since constant losses are known, the efficiency of the machine at any other load, can be determined. Suppose it is desired to determine the efficiency of the, machine at load current I. Then,, Armature current, Ia = I − Ish, = I + Ish, , ... if the machine is motoring, ... if the machine is generating, , Efficiency when running as a motor, Input power to motor = VI, Armature Cu loss = I 2a R a = (I − Ish )2 R a, Constant losses = WC, , found above, , Total losses = (I − Ish )2 R a + WC, 2, Input − Losses VI − (I − I sh ) R a − WC, =, ∴ Motor efficiency, ηm =, Input, VI, Efficiency when running as a generator, , Output of generator = VI, Armature Cu loss = (I + Ish )2 R a, Constant losses = WC, , found above
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Total losses = (I + I sh ) R a + WC, 2, , ∴, , Generator efficiency, ηg =, , Output, VI, =, Output + Losses VI + (I + I )2 R + W, sh, a, C, , Advantages of Swinburne’s test, (i), , The power required to carry out the test is small because it is a no-load test., Therefore, this method is quite economical., (ii) The efficiency can be determined at any load because constant losses are, known., (iii) This test is very convenient., , Disadvantages of Swinburne's test, (i), , It does not take into account the stray load losses that occur when the, machine is loaded., (ii) This test does not enable us to check the performance of the machine on, full-load. For example, it does not indicate whether commutation on fullload is satisfactory and whether the temperature rise is within the specified, limits., (iii) This test does not give quite accurate efficiency of the machine. It is, because iron losses under actual load are greater than those measured. This, is mainly due to armature reaction distorting the field., , 6.4 Regenerative or Hopkinson’s-Test, This method of determining the efficiency cf a d.c. machine saves power and, gives more accurate results. In order to carry out this test, we require two, identical d.c. machines and a source of electrical power., , Principle, Two identical d.c. shunt machines are mechanically coupled and connected in, parallel across the d.c. supply. By adjusting the field excitations of the machines,, one is run as a motor and the other as a generator. The electric power from the, generator and electrical power from the d.c. supply are fed to the motor. The, electric power given to the motor is mostly converted into mechanical power,, the rest going to the various motor losses. This mechanical power is given to the, generator. The electrical power of the generator is given to the motor except that, which is wasted as generator losses. Thus the electrical power taken from the, d.c. supply is the sum of motor and generator losses and this can be measured, directly by a voltmeter and an ammeter. Since the power input from the d.c., supply is equal to the power required to supply the losses of the two machines,, this test can be carried out with a small amount of power. By adjusting the field
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strengths of the machines, any load can be put on the machines. Therefore, we, can measure the total loss of the machines at any load. Since the machines can, be tested under full-load conditions (of course at the expense of power equal to, the losses in the two machines), the temperatures rise and commutation qualities, of the machines can be observed., , Circuit, Fig. (6.3) shows the essential connections for Hopkinson’s test. Two identical, d.c. shunt machines are mechanically coupled and are connected in parallel, across the d.c. supply. By adjusting the field strengths of the two machines, the, machine M is made to run as a motor and machine G as a generator. The motor, M draws current I1 from the generator G and current I2 from the d.c. supply so, that input current to motor M is (I1 + I2). Power taken from the d.c. supply is VI2, and is equal to the total motor and generator losses. The field current of motor M, is I4 and that of generator G is I3., , Fig. (6.3), , Calculations, If V be the supply voltage, then,, Motor input = V(I1 + I2), Generator output = VI1, We shall find the efficiencies of the machines considering two cases viz., (i) assuming that both machines have the same efficiency η (ii) assuming iron,, friction and windage losses are the same in both machines.
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(i) Assuming that both machines have the same efficiency η, Motor output = η × motor input = η V(I1 + I2) = Generator input, Generator output = η × generator input = η × η V(I1 + I2) = η2 V(I1 + I2), But generator output is VI1, , ∴, or, , η2 V(I1 + I 2 ) = VI1, η=, , I1, I1 + I 2, , This expression gives the value of efficiency sufficiently accurate for a rough, test. However, if, accuracy is required, the efficiencies of the two machines, should be calculated separately as, below., , (ii) Assuming that iron, friction and windage losses are same in, both machines., It is not, to assume that the two machines have the same efficiency. It, is because armature and field, in the two machines are not the same., However, iron, friction and windage losses in the two machines will be the same, because the machines are identical. On this assumption, we can find the, of, each machine as under:, Let Ra = armature resistance of each machine, I3 = field current of generator G, I4 = field current of motor M, Armature Cu loss in generator = (I1 + I 3 )2 R a, Armature Cu loss in motor = (I1 + I 2 − I 4 )2 R a, Shunt Cu loss in generator = V I3, Shunt Cu loss in motor = V I4, Power drawn from the d.c. supply is VI2 and is equal to the total losses of the, motor and generator, VI2 = Total losses of motor and generator, If we subtract armature and shunt Cu losses of the two machines from VI2, we, get iron, friction, windage losses of the two machines., Iron, friction and windage losses of two machines (M and G), = VI 2 − (I1 + I3 )2 R a + (I1 + I 2 − I 4 )2 R a + VI3 + VI 4 = W (say), , [, , ∴, , Iron, friction and windage losses of each machine = W/2, , ]
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For generator, Output of generator = VI1, Total losses =, ∴, , W, + (I1 + I3 )2 R a + VI3 = Wg (say), 2, , Generator efficiency, ηg =, , VI1, VI1 + Wg, , For motor, Input to motor = V(I1 + I2), Total losses = (I1 + I 2 − I 4 ) R a + VI 4 +, 2, , ∴, , Motor efficiency, ηm =, , W, = Wm (say), 2, , Input − Losses V (I1 + I 2 ) − Wm, =, Input, V (I1 + I 2 ), , 6.5 Alternate Connections for Hopkinson’s Test, Fig. (6.4) shows the alternative connections for Hopkinson’s test. The main, difference is that now the shunt field windings are directly connected across the, lines. Therefore, the input line current is I1, excluding the field currents. The, power VI1 drawn from the d.c. supply is equal to the total losses of the two, machines except the shunt field losses of the two machines i.e.,, V, I, I, , =, T, o, t, a, l, l, o, s, s, e, s, o
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f, t, h, e, t, w, o, m, a, c, h, i, n, e, s, e, x, c, e, p, t, s, h, u, n, t, f, i, e, l, d, l, o, s, s, e
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s, o, f, , t, h, e, t, w, o, m, a, c, h, i, n, e, s, , Fig. (6.4), Motor armature Cu loss = (I1 + I 2 ) R a, 2, , Generator armature Cu loss = I 22 R a, Iron, friction and windage losses of the two machines are VI1 minus armature, Cu losses of the two machines i.e..
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Iron, friction and windage losses of the two machines, = VI1 − (I1 + I 2 )2 R a + I 22 R a = W (say), , [, , ], , Iron, friction and windage losses of each machine = W/2, Motor efficiency, Motor input, Pi = V(I1 + I 2 + I3 ), Motor losses = (I1 + I 2 ) R a + VI 3 +, 2, , ∴, , Motor efficiency, ηm =, , W, = Wm (say), 2, , Motor input − Losses Pi − Wm, =, Motor input, Pi, , Generator efficiency, Generator output = VI2, Generator losses = I 22 R a + VI 4 +, ∴, , Generator efficiency, ηg =, , W, = Wg (say), 2, , VI 2, VI 2 + Wg, , 6.6 Advantages of Hopkinson’s Test, The advantages of Hopkinson’s test are :, (i) The total power required to test the two machines is small compared, with the full-load power of each machine., (ii) The machines can be tested under full-load conditions so that, commutation qualities and temperature rise can be checked., (iii) It is more accurate to measure the loss directly than to measure it as the, difference of the measured input and output., (iv) All the measurements are electrical which are simpler and more accurate, than mechanical measurements., The main disadvantage is that two similar d.c. machines are required., , 6.7 Retardation or Running down Test, This is the best and simplest method to find the efficiency of a constant-speed, d.c. machine (e.g., shunt generator and motor). In this method, we find the, mechanical (friction and windage) and iron losses of the machine. Then, knowing the armature and shunt Cu losses at any load, the efficiency of the, machine can be calculated at that load.
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Principle, Consider a d.c. shunt motor running at no-load., (i) If the supply to the armature is cut off but field remains normally, excited, the motor slows down gradually and finally stops. The kinetic, energy of the armature is used up to overcome friction, windage and iron, losses., (ii) If the supply to the armature as well as field excitation is cut off, the, motor again slows down and finally stops. Now the kinetic energy of the, armature is used up to overcome only the friction and windage losses., This is expected because in the absence of flux, there will be no iron, losses., By carrying out the first test, we can find out the friction, windage and iron, losses and hence the efficiency of the machine. However, if we perform the, second test also, we can separate friction and windage losses from the iron, losses., , Theory of retardation test, In the retardation test, the d.c. machine is run as a motor at a speed just above, the normal. Then the supply to the armature is cut off while the field is normally, excited. The speed is allowed to fall to some value just below normal. The time, taken for this fall of speed is noted. From these observations, the rotational, losses (i.e., friction, windage and iron losses) and hence the efficiency of the, machine can be determined., Let, , N = normal speed in r.p.m., ω = normal angular velocity in rad/s = 2π N/60, ∴, , Rotational losses, W = Rate of loss of K.E. of armature, , W=, , or, , d 1 2, dω, Iω = Iω, , dt 2, dt, , , Here I is the moment of inertia of the armature. As ω = 2π N/60,, , ∴, or, , 2, , 2πN d 2πN 2π , dN, W = I×, = IN, × , , 60 dt 60 60 , dt, , W = 0.011 IN, , dN, dt, , Let us illustrate the application of retardation test with a numerical example., Suppose the normal speed of a d.c. machine is 1000 r.p.m. When retardation test, is performed, the time taken for the speed to fall from 1030 r.p.m. to 970 r.p.m.
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is 15 seconds with field normally excited. If the moment of inertia of the, armature is 75 kg m , then,, Rotational losses, W = 0.011 IN, , I = 75 kg m 2 ;, , Here, , N = 1000 r.p.m, , dN = 1030 − 970 = 60 r.pm.;, , ∴, , dN, dt, , W = 0.011× 75 × 1000 ×, , dt = 15 sec, , 60, = 3300 watts, 15, , The main difficulty with this method is the accurate determination of the speed, which is continuously changing., , 6.8 Moment of Inertia (I) of the Armature, In retardation test, the rotational losses are given by;, , W = 0.011 IN, , dN, dt, , In order to find W, the value of I must be known. It is difficult to determine I, directly or by calculation. Therefore, we perform another experiment by which, either I is calculated or it is eliminated from the above expression., , (i) First method, It is a fly-wheel method in which the value of I is calculated. First, retardation, test is performed with armature alone and dN/dt1 is determined. Next, a flywheel of known moment of inertia I1 is keyed on to the shaft of the machine. For, the same change in speed, dN/dt2 is noted. Since the addition of fly-wheel will, not materially affect the rotational losses in the two cases,, ∴, , For the first case, W = 0.011 IN, , dN, dt 1, , For the second case, W = 0.011 (I + I1 )N, , ∴, , 0.011 IN, , dN, dN, = 0.011 (I + I1 )N, dt1, dt 2, , dN, dN, = (I + I1 ), dt 1, dt 2, , or, , I, , or, , I + I1 dN / dt1 dt 2, =, =, I, dN / dt 2 dt1, , dN, dt 2
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or, , I1 dt 2 − dt1 t 2 − t 1, =, =, I, dt 1, t1, , or, , I = I1 ×, , t 2 − t1, t1, , Since the values of I1, t1 and t2 are known, the moment of inertia I of the, armature can be determined., , (ii) Second method, In this method, I is eliminated from the expression by an experiment. First,, retardation test is performed with armature alone. The rotational losses are given, by;, , W = 0.011 IN, , dN, dt 1, , Next the motor is loaded with a known amount of power W' with a brake. For, the same change in speed, dN/dt2 is noted. Then,, , W + W ' = 0.011 IN, ∴, , dN, dt 2, , t, W + W ' dt, =, = 1, W, dt 2 t 2, W ' t1 − t 2, =, W, t2, , or, , ∴, , W = W '×, , t1 − t 2, t2, , Since the values of W', t1 and t2 are known, the value of W can be determined., , 6.9 Electric Loading in Retardation Test, In a retardation test, the rotational losses W are given by;, , W = 0.011 IN, , dN, dt, , As discussed in Sec. (6.8), we can eliminate I (moment of inertia of armature), from the above expression by applying either mechanical or electric loading to, the armature. The electric leading is preferred because of convenience and, reliability. Fig. (6.5) illustrates how electric loading is applied to slow down the, armature. The double throw switch S is thrown to the supply and the machine is, brought to full-load speed. Then the switch S is thrown to the other side, connecting a non-inductive resistance R across the armature. The supply now is
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cut off and the power dissipated in R acts as a retarding torque to slow down the, armature., Let V' = average voltage across R, I'a = average current through R, , Fig. (6.5), The electric loading W' (or extra power loss) is given by;, W' = average voltage x average current = V' I'a
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Chapter (7), , Transformer, Introduction, The transformer is probably one of the most useful electrical devices ever, invented. It can change the magnitude of alternating voltage or current from one, value to another. This useful property of transformer is mainly responsible for, the widespread use of alternating currents rather than direct currents i.e., electric, power is generated, transmitted and distributed in the form of alternating, current. Transformers have no moving parts, rugged and durable in construction,, thus requiring very little attention. They also have a very high efficiency—as, high as 99%. In this chapter, we shall study some of the basic properties of, transformers., , 7.1 Transformer, A transformer is a static piece of equipment used either for raising or lowering, the voltage of an a.c. supply with a corresponding decrease or increase in, current. It essentially consists of two windings, the primary and secondary,, wound on a common laminated magnetic core as shown in Fig. (7.1). The, winding connected to the a.c. source is called primary winding (or primary) and, the one connected to load is called secondary winding (or secondary). The, alternating voltage V1 whose magnitude is to be changed is applied to the, primary. Depending upon the number of turns of the primary (N1) and secondary, (N2), an alternating e.m.f. E2 is induced in the secondary. This induced e.m.f. E2, in the secondary causes a secondary current I2. Consequently, terminal voltage, V2 will appear across the load. If V2 > V1, it is called a step up-transformer. On, the other hand, if V2 < V1, it is called a step-down transformer., , Fig.(7.1), 124
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Working, When an alternating voltage V1 is applied to the primary, an alternating flux φ is, set up in the core. This alternating flux links both the windings and induces, e.m.f.s E1 and E2 in them according to Faraday’s laws of electromagnetic, induction. The e.m.f. E1 is termed as primary e.m.f. and e.m.f. E2 is termed as, secondary e.m.f., , dφ, dt, , Clearly,, , E1 = − N1, , and, , E 2 = −N2, ∴, , dφ, dt, , E 2 N2, =, E1 N1, , Note that magnitudes of E2 and E1 depend upon the number of turns on the, secondary and primary respectively. If N2 > N1, then E2 > E1 (or V2 > V1) and, we get a step-up transformer. On the other hand, if N2 < N1, then E2 < E1 (or V2, < V1) and we get a step-down transformer. If load is connected across the, secondary winding, the secondary e.m.f. E2 will cause a current I2 to flow, through the load. Thus, a transformer enables us to transfer a.c. power from one, circuit to another with a change in voltage level., The following points may be noted carefully:, (i) The transformer action is based on the laws of electromagnetic, induction., (ii) There is no electrical connection between the primary and secondary., The a.c. power is transferred from primary to secondary through, magnetic flux., (iii) There is no change in frequency i.e., output power has the same, frequency as the input power., (iv) The losses that occur in a transformer are:, (a) core losses—eddy current and hysteresis losses, (b) copper losses—in the resistance of the windings, In practice, these losses are very small so that output power is nearly equal to the, input primary power. In other words, a transformer has very high efficiency., , 7.2 Theory of an Ideal Transformer, An ideal transformer is one that has, (i) no winding resistance, (ii) no leakage flux i.e., the same flux links both the windings, (iii) no iron losses (i.e., eddy current and hysteresis losses) in the core, 125
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Although ideal transformer cannot be physically realized, yet its study provides, a very powerful tool in the analysis of a practical transformer. In fact, practical, transformers have properties that approach very close to an ideal transformer., , Fig.(7.2), Consider an ideal transformer on no load i.e., secondary is open-circuited as, shown in Fig. (7.2 (i)). Under such conditions, the primary is simply a coil of, pure inductance. When an alternating voltage V1 is applied to the primary, it, draws a small magnetizing current Im which lags behind the applied voltage by, 90°. This alternating current Im produces an alternating flux φ which is, proportional to and in phase with it. The alternating flux φ links both the, windings and induces e.m.f. E1 in the primary and e.m.f. E2 in the secondary., The primary e.m.f. E1 is, at every instant, equal to and in opposition to V1, (Lenz’s law). Both e.m.f.s E1 and E2 lag behind flux φ by 90° (See Sec. 7.3)., However, their magnitudes depend upon the number of primary and secondary, turns., Fig. (7.2 (ii)) shows the phasor diagram of an ideal transformer on no load., Since flux φ is common to both the windings, it has been taken as the reference, phasor. As shown in Sec. 7.3, the primary e.m.f. E1 and secondary e.m.f. E2 lag, behind the flux φ by 90°. Note that E1 and E2 are in phase. But E1 is equal to V1, and 180° out of phase with it., , 7.3 E.M.F. Equation of a Transformer, Consider that an alternating voltage V1 of frequency f is applied to the primary, as shown in Fig. (7.2 (i)). The sinusoidal flux φ produced by the primary can be, represented as:, φ = φm sinωt, The instantaneous e.m.f. e1 induced in the primary is, , 126
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or, , I 2 V1 1, =, =, I1 V2 K, , Hence, currents are in the inverse ratio of voltage transformation ratio. This, simply means that if we raise the voltage, there is a corresponding decrease of, current., , 7.5 Practical Transformer, A practical transformer differs from the ideal transformer in many respects. The, practical transformer has (i) iron losses (ii) winding resistances and (iii), magnetic leakage, giving rise to leakage reactances., (i), , Iron losses. Since the iron core is subjected to alternating flux, there occurs, eddy current and hysteresis loss in it. These two losses together are known, as iron losses or core losses. The iron losses depend upon the supply, frequency, maximum flux density in the core, volume of the core etc. It, may be noted that magnitude of iron losses is quite small in a practical, transformer., , (ii) Winding resistances. Since the windings consist of copper conductors, it, immediately follows that both primary and secondary will have winding, resistance. The primary resistance R1 and secondary resistance R2 act in, series with the respective windings as shown in Fig. (7.4). When current, flows through the windings, there will be power loss as well as a loss in, voltage due to IR drop. This will affect the power factor and E1 will be less, than V1 while V2 will be less than E2., , Fig.(7.4), (iii) Leakage reactances. Both primary and secondary currents produce flux., The flux φ which links both the windings is the useful flux and is called, mutual flux. However, primary current would produce some flux φ which, would not link the secondary winding (See Fig. 7.5). Similarly, secondary, current would produce some flux φ that would not link the primary, winding. The flux such as φ1 or φ2 which links only one winding is called, leakage flux. The leakage flux paths are mainly through the air. The effect, 128
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of these leakage fluxes would be the same as though inductive reactance, were connected in series with each winding of transformer that had no, leakage flux as shown in Fig. (7.4). In other words, the effect of primary, leakage flux φ1 is to introduce an inductive reactance X1 in series with the, primary winding as shown in Fig. (7.4). Similarly, the secondary leakage, flux φ2 introduces an inductive reactance X2 in series with the secondary, winding. There will be no power loss due to leakage reactance. However,, the presence of leakage reactance in the windings changes the power factor, as well as there is voltage loss due to IX drop., , Fig.(7.5), Note. Although leakage flux in a transformer is quite small (about 5% of φ), compared to the mutual flux φ, yet it cannot be ignored. It is because leakage, flux paths are through air of high reluctance and hence require considerable, e.m.f. It may be noted that energy is conveyed from the primary winding to the, secondary winding by mutual flux φ which links both the windings., , 7.6 Practical Transformer on No Load, Consider a practical transformer on no load i.e., secondary on open-circuit as, shown in Fig. (7.6 (i)). The primary will draw a small current I0 to supply (i) the, iron losses and (ii) a very small amount of copper loss in the primary. Hence the, primary no load current I0 is not 90° behind the applied voltage V1 but lags it by, an angle φ0 < 90° as shown in the phasor diagram in Fig. (7.6 (ii))., No load input power, W0 = V1 I0 cos φ0, , Fig.(7.6), 129
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As seen from the phasor diagram in Fig. (7.6 (ii)), the no-load primary current I0, can be resolved into two rectangular components viz., (i) The component IW in phase with the applied voltage V1. This is known, as active or working or iron loss component and supplies the iron loss, and a very small primary copper loss., IW = I0 cos φ0, (b), , The component Im lagging behind V1 by 90° and is known as, magnetizing component. It is this component which produces the, mutual flux φ in the core., Im = I0 sin φ0, Clearly, I0 is phasor sum of Im and IW,, , ∴, , I 0 = I 2m + I 2W, , cos φ0 =, , No load p.f.,, , IW, I0, , It is emphasized here that no load primary copper loss (i.e. I 20 R 1 ) is very small, and may be neglected. Therefore, the no load primary input power is practically, equal to the iron loss in the transformer i.e.,, No load input power, W0 = Iron loss, Note. At no load, there is no current in the secondary so that V2 = E2. On the, primary side, the drops in R1 and X1, due to I0 are also very small because of the, smallness of I0. Hence, we can say that at no load, V1 = E1., , 7.7 Ideal Transformer on Load, Let us connect a load ZL across the secondary of an ideal transformer as shown, in Fig. (7.7 (i)). The secondary e.m.f. E2 will cause a current I2 to flow through, the load., , I2 =, , E 2 V2, =, Z L ZL, , The angle at which I2 leads or lags V2 (or E2) depends upon the resistance and, reactance of the load. In the present case, we have considered inductive load so, that current I2 lags behind V2 (or E2) by φ2., The secondary current I2 sets up an m.m.f. N2I2 which produces a flux in the, opposite direction to the flux φ originally set up in the primary by the, magnetizing current. This will change the flux in the core from the original, value. However, the flux in the core should not change from the original value., 130
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In order to fulfill this condition, the primary must develop an m.m.f. which, exactly counterbalances the secondary m.m.f. N2I2. Hence a primary current I1, must flow such that:, , N1I1 = N 2 I 2, or, , I1 =, , N2, I = KI 2, N1 2, , Fig.(7.7), Thus when a transformer is loaded and carries a secondary current I2, then a, current I1, (= K I2) must flow in the primary to maintain the m.m.f. balance. In, other words, the primary must draw enough current to neutralize the, demagnetizing effect of secondary current so that mutual flux φ remains, constant. Thus as the secondary current increases, the primary current I1 (= KI2), increases in unison and keeps the mutual flux φ constant. The power input,, therefore, automatically increases with the output. For example if K = 2 and I2 =, 2A, then primary will draw a current I1 = KI2 = 2 × 2 = 4A. If secondary current, is increased to 4A, then primary current will become I1 = KI2 = 2 × 4 = 8A., Phaser diagram: Fig. (7.7 (ii)) shows the phasor diagram of an ideal, transformer on load. Note that in drawing the phasor diagram, the value of K has, been assumed unity so that primary phasors are equal to secondary phasors. The, secondary current I2 lags behind V2 (or E2) by φ2. It causes a primary current I2 =, KI2 = 1 × I2 which is in antiphase with it., (i), , φ1 = φ2, , or, , cos φ1 = cos φ2, , Thus, power factor on the primary side is equal to the power factor on the, secondary side., (ii) Since there are no losses in an ideal transformer, input primary power is, equal to the secondary output power i.e.,, , V1I1 cos φ1 = V2 I 2 cos φ 2, , 131
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7.8 Practical Transformer on Load, We shall consider two cases (i) when such a transformer is assumed to have no, winding resistance and leakage flux (ii) when the transformer has winding, resistance and leakage flux., , (i) No winding resistance and leakage flux, Fig. (7.8) shows a practical transformer with the assumption that resistances and, leakage reactances of the windings are negligible. With this assumption, V2 = E2, and V1 = E1. Let us take the usual case of inductive load which causes the, secondary current I2 to lag the secondary voltage V2 by φ2. The total primary, current I1 must meet two requirements viz., (a) It must supply the no-load current I0 to meet the iron losses in the, transformer and to provide flux in the core., (b) It must supply a current I'0 to counteract the demagnetizing effect of, secondary currently I2. The magnitude of I'2 will be such that:, , N1I'2 = N 2 I 2, or, , I' 2 =, , N2, I = KI 2, N1 2, , The total primary current I1 is the phasor sum of I'2 and I0 i.e.,, , I1 = I'2 + I 0, where, , I'2 = − KI 2, , Note that I'2 is 180° out of phase with I2., , Fig.(7.8), Phasor diagram. Fig. (7.9) shows the phasor diagram for the usual case of, inductive load. Both E1 and E2 lag behind the mutual flux φ by 90°. The current, I'2 represents the primary current to neutralize the demagnetizing effect of, secondary current I2. Now I'2 = K I2 and is antiphase with I2. I0 is the no-load, current of the transformer. The phasor sum of I'2 and I0 gives the total primary, current I1. Note that in drawing the phasor diagram, the value of K is assumed to, be unity so that primary phasors are equal to secondary phasors., Primary p.f. = cos φ1, , 132
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Secondary p.f. = cos φ2, Primary input power = V1 I1 cos φ1, Secondary output power = V1 I2 cos φ2, , (ii) Transformer with resistance and, leakage reactance, Fig. (7.10) shows a practical transformer, having winding resistances and leakage, reactances. These are the actual conditions that, exist in a transformer. There is voltage drop in, Fig.(7-9), R1 and X1 so that primary e.m.f. E1 is less than, the applied voltage V1. Similarly, there is voltage drop in R2 and X2 so that, secondary terminal voltage V2 is less than the secondary e.m.f. E2. Let us take, the usual case of inductive load which causes the secondary current I2 to lag, behind the secondary voltage V2 by φ2. The total primary current I1 must meet, two requirements viz., (a) It must supply the no-load current I0 to meet the iron losses in the, transformer and to provide flux in the core., (b) It must supply a current I'2 to counteract the demagnetizing effect of, secondary current I2. The magnitude of I'2 will be such that:, , N1I'2 = N 2 I 2, or, , I' 2 =, , N2, I = KI 2, N1 2, , Fig.(7.10), The total primary current I1 will be the phasor sum of I'2 and I0 i.e.,, I1 = I'2 + I0, , where, , V1 = −E1 + I1(R1 + jX1), , I'2 = −KI2, where, , = −E1 + I1Z1, 133, , I1 = I0 + (−KI2)
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i.e., impedance ratio (Z2/Z1) is equal to the square of voltage transformation, ratio. In other words, an impedance Z2 in secondary becomes Z2/K2 when, transferred to primary. Likewise, an impedance Z1 in the primary becomes K2 Z1, when transferred to the secondary., R2, X2, Similarly,, = K2, and, = K2, R1, X1, Note the importance of above relations. We can transfer the parameters from, one winding to the other. Thus:, (i) A resistance R1 in the primary becomes K2 R1 when transferred to the, secondary., (ii) A resistance R2 in the secondary becomes R2/K2 when transferred to the, primary., (iii) A reactance X1 in the primary becomes K2 X1 when transferred to the, secondary., (iv) A reactance X2 in the secondary becomes X2/K2 when transferred to the, primary., Note: It is important to remember that:, (i) When transferring resistance or reactance from primary to secondary,, multiply it by K2., (ii) When transferring resistance or reactance from secondary to primary,, divide it by K2., (iii) When transferring voltage or current from one winding to the other, only, K is used., , 7.10 Shifting Impedances in A Transformer, Fig. (7.13) shows a transformer where resistances and reactances are shown, external to the windings. The resistance and reactance of one winding can be, transferred to the other by appropriately using the factor K2. This makes the, analysis of the transformer a simple affair because then we have to work in one, winding only., , Fig.(7.13), , 135
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(i) Referred to primary, When secondary resistance or reactance is transferred to the primary, it is, divided by K2. It is then called equivalent secondary resistance or reactance, referred to primary and is denoted by R'2 or X'2., Equivalent resistance of transformer referred to primary, , R2, , R 01 = R 1 + R '2 = R 1 +, , K2, , Equivalent reactance of transformer referred to primary, , X 01 = X1 + X '2 = X1 +, , X2, K2, , Equivalent impedance of transformer referred to primary, 2, 2, Z 01 = R 01, + X 01, , Fig. (7.14), Fig. (7.14) shows the resistance and reactance of the secondary referred to the, primary. Note that secondary now has no resistance or reactance., , (ii), , Referred to secondary, , When primary resistance or reactance is transferred to the secondary, it is, multiplied by K2. It is then called equivalent primary resistance or reactance, referred to the secondary and is denoted by R'1 or X'1., Equivalent resistance of transformer referred to secondary, , R 02 = R 2 + R '1 = R 2 + K 2 R 1, Equivalent reactance of transformer referred to secondary, , X 02 = X 2 + X'1 = X 2 + K 2 X1, Equivalent impedance of transformer referred to secondary, 2, 2, Z 02 = R 02, + X 02, , 136
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Fig. (7.15), Fig. (7.15) shows the resistance and reactance of the primary referred to the, secondary. Note that primary now has no resistance or reactance., , 7.11 Importance of Shifting Impedances, If we shift all the impedances from one winding to the other, the transformer is, eliminated and we get an equivalent electrical circuit. Various voltages and, currents can be readily obtained by solving this electrical circuit., , Fig.(7.16), Consider an ideal transformer having an impedance Z2 in the secondary as, shown in Fig. (7.16)., , (i) Referred to primary, When impedance Z2 in the secondary is transferred to the primary, it becomes, Z2/K2 as shown in Fig. (7.17 (i)). Note that in Fig. (7.17 (i)), the secondary of, the ideal transformer is on open-circuit. Consequently, both primary and, secondary currents are zero. We can, therefore, remove the transformer, yielding, the equivalent circuit shown in Fig. (7.17 (ii)). The primary current can now be, readily found out., , I1 =, , (Z, , V1, 2, , / K2, , ), , The circuits of Fig. (7.16) and Fig. (7.17 (ii)) are electrically equivalent. Thus, referring to Fig. (7.16),, , I1 = KI 2, Also if we refer to Fig. (7.17 (ii)). we have,, 137
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I1 =, , (Z, , =K, , V1, 2, , / K2, , ), , K 2 V1 K (K V1 ), =, =, Z2, Z2, V , , Q I 2 = 2 , Z2 , , , V2, = KI 2, Z2, , Fig. (7.17), Thus the value of primary current I1 is the same whether we use Fig. (7.16) or, Fig. (7.17 (ii)). Obviously, it is easier to use Fig. (7.17 (ii)) as it contains no, transformer., , (ii) Referred to secondary, Refer back to Fig. (7.16). There is no impedance on the primary side. However,, voltage V1 in the primary when transferred to the secondary becomes K V1 as, shown in Fig. (7.18 (i)). Note that in Fig. (7.18 (ii)), the primary of the, transformer is on open circuit. Consequently, both primary and secondary, currents are zero. As before, we can remove the transformer yielding the, equivalent circuit shown in Fig. (7.18 (ii)). The secondary current I2 can be, readily found out as:, , I2 =, , K V1, Z2, , Fig. (7.18), The circuits of Fig. (7.16) and Fig. (7.18 (ii)) are electrically equivalent. Thus, referring back to Fig. (7.16), we have,, , I2 =, , V2, Z2, 138
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Also if we refer to Fig. (7.18 (ii)), we have,, , I1 =, , K V1 K (V2 / K ) V2, =, =, Z2, Z2, Z2, , Thus the value of secondary current I2 is the same whether we use Fig. (7.16) or, Fig. (7.18 (ii)). Obviously, it is easier to use Fig. (7.18 (ii) as it contains no, transformer., , 7.12 Exact Equivalent Circuit of a Loaded Transformer, Fig. (7.19) shows the exact equivalent circuit of a transformer on load. Here R1, is the primary winding resistance and R2 is the secondary winding resistance., Similarly, X1 is the leakage reactance of primary winding and X2 is the leakage, reactance of the secondary winding. The parallel circuit R0 − X0 is the no-load, equivalent circuit of the transformer. The resistance R0 represents the core losses, (hysteresis and eddy current losses) so that current IW which supplies the core, losses is shown passing through R0. The inductive reactance X0 represents a, loss-free coil which passes the magnetizing current Im. The phasor sum of IW and, Im is the no-load current I0 of the transformer., , Fig. (7.19), Note that in the equivalent circuit shown in Fig. (7.19), the imperfections of the, transformer have been taken into account by various circuit elements. Therefore,, the transformer is now the ideal one. Note that equivalent circuit has created two, normal electrical circuits separated only by an ideal transformer whose function, is to change values according to the equation:, , E 2 N 2 I' 2, =, =, E1 N1 I 2, The following points may be noted from the equivalent circuit:, (i) When the transformer is on no-load (i.e., secondary terminals are opencircuited), there is no current in the secondary winding. However, the, primary draws a small no-load current I0. The no-load primary current I0, , 139
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is composed of (a) magnetizing current (Im) to create magnetic flux in, the core and (b) the current IW required to supply the core losses., (ii) When the secondary circuit of a transformer is closed through some, external load ZL, the voltage E2 induced in the secondary by mutual flux, will produce a secondary current I2. There will be I2 R2 and I2 X2 drops, in the secondary winding so that load voltage V2 will be less than E2., , V2 = E 2 − I 2 (R 2 + j X 2 ) = E 2 − I 2 Z 2, (iii) When the transformer is loaded to carry the secondary current I2, the, primary current consists of two components:, (a) The no-load current I0 to provide magnetizing current and the, current required to supply the core losses., (b) The primary current I'2 (= K I2) required to supply the load, connected to the secondary., , ∴ Total primary current I1 = I 0 + (− KI 2 ), (iv) Since the transformer in Fig. (7.19) in now ideal, the primary induced, voltage E1 can be calculated from the relation:, , E1 N1, =, E2 N2, If we add I1R1 and I1X1 drops to E1, we get the primary input voltage V1, , V1 = −E1 + I1 (R 1 + j X1 ) = −E1 + I1Z1, or, , V1 = −E1 + I1Z1, , 7.13 Simplified Equivalent Circuit of a Loaded, Transformer, The no-load current I0 of a transformer is small as compared to the rated primary, current. Therefore, voltage drops in R1 and X1 due to I0 are negligible. The, equivalent circuit shown in Fig. (7.19) above can, therefore, be simplified by, transferring the shunt circuit R0 − X0 to the input terminals as shown in Fig., (7.20). This modification leads to only slight loss of accuracy., , 140
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Fig.(7.20), , (i) Equivalent circuit referred to primary, If all the secondary quantities are referred to the primary, we get the equivalent, circuit of the transformer referred to the primary as shown in Fig. (7.21 (i)). This, further reduces to Fig. (7.21 (ii)). Note that when secondary quantities are, referred to primary, resistances/reactances/impedances are divided by K2,, voltages are divided by K and currents are multiplied by K., , ∴, , K '2 =, , R2, K, , ;, 2, , X'2 =, , X2, K, , ;, 2, , Z'L =, , ZL, K, , ;, 2, , V'2 =, , V2, ;, K, , I' 2 = K I 2, , Z 01 = R 01 + j X 01, where, , R 01 = R 1 + R '2 ;, , X 01 = X1 + X'2, , Fig. (7.21), Phasor diagram. Fig. (7.22) shows the phasor, diagram corresponding to the equivalent circuit, shown in Fig. (7.21 (ii)). The referred value of, load voltage V'2 is chosen as the reference, phasor. The referred value of load current I'0 is, shown lagging V'2 by phase angle φ2. For a, given value of V'2 both I'2 and φ2 are, determined by the load. The voltage drop I'2 R01, , 141, , Fig.(7.22)
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is in phase with I'2 and the voltage drop I'2 X01, leads I'2 by 90°. When these, voltage drops are added to V'2, we get the input voltage V1., The current IW is in phase with V1 while the magnetization current Im lags, behind V1 by 90°. The phasor sum of IW and Im is the no-load current I0. The, phasor sum of I0 and I'2 is the input current I1., , (ii), , Equivalent circuit referred to secondary., , If all the primary quantities are referred to secondary, we get the equivalent, circuit of the transformer referred to secondary as shown in Fig. (7.23 (i)). This, further reduces to Fig. (7.23 (ii)). Note that when primary quantities are referred, to secondary resistances/reactances/impedances are multiplied by K2, voltages, are multiplied by K. and currents are divided by K., , Fig. (7.23), , ∴, , R '1 = K 2 R 1 ;, , X '1 = K 2 X1 ;, , V '2 = K V1; I'1 =, , I1, K, , Z 02 = R 02 + j X 02, where, , R 02 = R 2 + R '1 ;, , X 02 = X 2 + X'1, , Phasor diagram. Fig. (7.24) shows the phasor, diagram of the equivalent circuit shown in Fig., (7.23 (ii)). The load voltage V2 is chosen as the, 142, Fig.(7-24)
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reference phasor. The load current I2 is shown lagging the load voltage V2 by, phase angle φ2. The voltage drop I2 R02 is in phase with I2 and the voltage drop I2, X02 leads I2 by 90°. When these voltage drop are added to V2, we get the referred, primary voltage V'1 (= KV1)., The current I'W is in phase with V'1 while the magnetizing current I'm lags behind, V'1 by 90°. The phasor sum of I'W and I'm gives the referred value of no-load, current I'0. The phasor sum of I'0 and load current I2 gives the referred primary, current I'1 (= I1/K)., , 7.14 Approximate Equivalent Circuit of a Loaded, Transformer, The no-load current I0 in a transformer is only 1-3% of the rated primary current, and may be neglected without any serious error. The transformer can then be, shown as in Fig. (7.25). This is an approximate representation because no-load, current has been neglected. Note that all the circuit elements have been shown, external so that the transformer is an ideal one., , Fig. (7.25), As shown in Sec. 7.11, if we refer all the quantities to one side (primary or, secondary), the ideal transformer stands removed and we get the equivalent, circuit., , (i) Equivalent circuit of transformer referred to primary, If all the secondary quantities are referred to the primary, we get the equivalent, circuit of the transformer referred to primary as shown in Fig. (7.26). Note that, when secondary quantities are referred to primary, resistances/reactances are, divided by K2, voltages are divided by K and currents are multiplied by K., The equivalent circuit shown in Fig. (7.26) is an electrical circuit and can be, solved for various currents and voltages. Thus if we find V'2 and I'2, then actual, secondary values can be determined as under:, Actual secondary voltage, V2 = K V'2, 143
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Actual secondary current, I2 = I'2/K, , Fig.(7.26), , (ii), , Equivalent circuit of transformer referred to secondary, , If all the primary quantities are referred to secondary, we get the equivalent, circuit of the transformer referred to secondary as shown in Fig. (7.27). Note, that when primary quantities are referred to secondary, resistances/reactances, are multiplied by K2, voltages are multiplied by K and currents are divided by K., , Fig. (7.27), The equivalent circuit shown in Fig. (7.27) is an electrical circlet and can be, solved for various voltages and currents. Thus is we find V'1 and I'1, then actual, primary values can be determined as under:, Actual primary voltage, V1 = V'1/K, Actual primary current,, , I1 = KI'1, , Note: The same final answers will be obtained whether we use the equivalent, circuit referred to primary or secondary. The use of a particular equivalent, circuit would depend upon the conditions of the problem., , 7.15 Approximate Voltage Drop in a Transformer, The approximate equivalent circuit of transformer referred to secondary is, shown in Fig. (7.28). At no-load, the secondary voltage is K V1. When a load, having a lagging p.f. cos φ2 is applied, the secondary carries a current I2 and, voltage drops occur in (R2 + K2 R1) and (X2 + K2 X1). Consequently, the, secondary voltage falls from K V1 to V2. Referring to Fig. (7.28), we have,, , 144
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[(, , ) (, , V2 = KV1 − I 2 R 2 + K 2 R 1 + j X 2 + K 2 X1, = KV1 − I 2 (R 02 + j X 02 ), , )], , = KV1 − V2 = I 2 Z 02, Drop in secondary voltage = KV1 − V2 = I 2 Z 02, The phasor diagram is shown in Fig. (7.29). It is clear from the phasor diagram, that drop in secondary voltage is AC = I2 Z02. It can be found as follows. With O, as centre and OC as radius, draw an arc cutting OA produced at M. Then AC =, AM = AN. From B, draw BD perpendicular to OA produced. Draw CN, perpendicular to OM and draw BL || OM., , Fig.(7.28), , Fig.(7.29), , Approximate drop in secondary voltage, , = AN = AD + DN, = AD + BL, = I 2 R 02 cos φ 2 + I 2 X 02 sin φ 2, , (Q BL = DN ), , For a load having a leading p.f. cos φ2, we have,, Approximate voltage drop = I 2 R 02 cos φ 2 − I 2 X 02 sin φ 2, Note: If the circuit is referred to primary, then it can be easily established that:, Approximate voltage drop = I1R 01 cos φ 2 ± I1X 01 sin φ 2, , 7.16 Voltage Regulation, The voltage regulation of a transformer is the arithmetic difference (not phasor, difference) between the no-load secondary voltage (0V2) and the secondary, voltage V2 on load expressed as percentage of no-load voltage i.e., %age voltage regulation =, where, , − V2, × 100, 0 V2, , 0 V2, , 0V2, , = No-load secondary voltage = K V1, V2 = Secondary voltage on load, , 145
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As shown in Sec. 7.15, 0 V2, , − V2 = I 2 R 02 cos φ 2 ± I 2 X 02 sin φ 2, , The +ve sign is for lagging p.f. and −ve sign for leading p.f., It may be noted that %age voltage regulation of the transformer will be the same, whether primary or secondary side is considered., , 7.17 Transformer Tests, The circuit constants, efficiency and voltage regulation of a transformer can be, determined by two simple tests (i) open-circuit test and (ii) short-circuit lest., These tests are very convenient as they provide the required information without, actually loading the transformer. Further, the power required to carry out these, tests is very small as compared with full-load output of the transformer. These, tests consist of measuring the input voltage, current and power to the primary, first with secondary open-circuited (open-circuit test) and then with the, secondary short-circuited (short circuit test)., , 7.18 Open-Circuit or No-Load Test, This test is conducted to determine the iron losses (or core losses) and, parameters R0 and X0 of the transformer. In this test, the rated voltage is applied, to the primary (usually low-voltage winding) while the secondary is left opencircuited. The applied primary voltage V1 is measured by the voltmeter, the noload current I0 by ammeter and no-load input power W0 by wattmeter as shown, in Fig. (7.30 (i)). As the normal rated voltage is applied to the primary,, therefore, normal iron losses will occur in the transformer core. Hence, wattmeter will record the iron losses and small copper loss in the primary. Since, no-load current I0 is very small (usually 2-10 % of rated current). Cu losses in, the primary under no-load condition are negligible as compared with iron losses., Hence, wattmeter reading practically gives the iron losses in the transformer. It, is reminded that iron losses are the same at all loads. Fig. (7.30 (ii)) shows the, equivalent circuit of transformer on no-load., Iron losses, Pi = Wattmeter reading = W0, No load current = Ammeter reading = I0, Applied voltage = Voltmeter reading = V1, Input power, W0 = V1 I0 cos φ0, , 146
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∴, , No - load p.f., cos φ 0 =, I W = I 0 cos φ0 ;, R0 =, , V1, IW, , and, , W0, I, V1 0, I m = I 0 sin φ0, X0 =, , V1, Im, , Thus open-circuit test enables us to determine iron losses and parameters R0 and, X0 of the transformer., , Fig.(7.30), , 7.19 Short-Circuit or Impedance Test, This test is conducted to determine R01 (or R02), X01 (or X02) and full-load, copper losses of the transformer. In this test, the secondary (usually low-voltage, winding) is short-circuited by a thick conductor and variable low voltage is, applied to the primary as shown in Fig. (7.31 (i)). The low input voltage is, gradually raised till at voltage VSC, full-load current I1 flows in the primary., Then I2 in the secondary also has full-load value since I1/I2 = N2/N1. Under such, conditions, the copper loss in the windings is the same as that on full load., There is no output from the transformer under short-circuit conditions., Therefore, input power is all loss and this loss is almost entirely copper loss. It is, because iron loss in the core is negligibly small since the voltage VSC is very, small. Hence, the wattmeter will practically register the full-load copper losses, in the transformer windings. Fig. (7.31 (ii)) shows the equivalent circuit of a, transformer on short circuit as referred to primary; the no-load current being, neglected due to its smallness., , 147
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Fig. (7.31), Full load Cu loss, PC = Wattmeter reading = WS, Applied voltage = Voltmeter reading = VSC, F.L. primary current = Ammeter reading = I1, , PC = I12 R 1 + I12 R '2 = I12 R 01, ∴, , R 01 =, , PC, I12, , where R01 is the total resistance of transformer referred to primary., V, Total impedance referred to primary, Z 01 = SC, I1, 2, 2, Total leakage reactance referred to primary, X 01 = Z 01, − R 01, P, Short-circuit p.f, cos φ 2 = C, VSC I1, Thus short-circuit lest gives full-load Cu loss, R01 and X01., , Note: The short-circuit test will give full-load Cu loss only if the applied voltage, VSC is such so as to circulate full-load currents in the windings. If in a shortcircuit test, current value is other than full-load value, the Cu loss will be, corresponding to that current value., , 7.20 Advantages of Transformer Tests, The above two simple transformer tests offer the following advantages:, (i) The power required to carry out these tests is very small as compared to, the full-load output of the transformer. In case of open-circuit lest, power, required is equal to the iron loss whereas for a short-circuit test, power, required is equal to full-load copper loss., (ii) These tests enable us to determine the efficiency of the transformer, accurately at any load and p.f. without actually loading the transformer., (iii) The short-circuit test enables us to determine R01 and X01 (or R02 and, X02). We can thus find the total voltage drop in the transformer as, 148
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referred to primary or secondary. This permits us to calculate voltage, regulation of the transformer., , 7.21 Separation of Components of Core Losses, The core losses (or iron losses) consist of hysteresis loss and eddy current loss., Sometimes it is desirable to find the hysteresis loss component and eddy current, loss component in the total core losses., Hysteresis loss, Ph = k h f B1m.6, , watts / m 3, , Eddy current loss, Pe = k e f 2 B 2m t 2, , watts / m 3, , where Bm = maximum flux density; f = frequency; kh, ke = constants, For a given a.c. machine and maximum flux density (Bm),, , or, , Ph ∝ f, , and, , Pe ∝ f 2, , Ph = a f, , and, , Pe = b f 2, , where a and b are constants., Total core loss, Pi = af + bf2, Hence if the total core loss for given Bm is known at two frequencies, the, constants a and b can be calculated. Knowing the values of a and b, the, hysteresis loss component and eddy current loss component of the core loss can, be determined., Pt/f and f curve, , Pi = af + bf 2, or, , Pi, = a + bf, f, , The total core losses are measured at various, frequencies while the other factors upon which core, losses depend are maintained constant. If a graph is, plotted between Pi/f and f, it will be a straight line, with slope tan θ = b (See Fig. 7.32). Therefore,, Fig.(7.32), constants a and b can be evaluated. Hence the, hysteresis and eddy current losses at a given frequency (say f1) can be found out., , 7.22 Why Transformer Rating in kVA?, , 149
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An important factor in the design and operation of electrical machines is the, relation between the life of the insulation and operating temperature of the, machine. Therefore, temperature rise resulting from the losses is a determining, factor in the rating of a machine. We know that copper loss in a transformer, depends on current and iron loss depends on voltage. Therefore, the total loss in, a transformer depends on the volt-ampere product only and not on the phase, angle between voltage and current i.e., it is independent of load power factor., For this reason, the rating of a transformer is in kVA and not kW., , 7.23 Sumpner or Back-to-Back Test, This test is conducted simultaneously on two identical transformers and provides, data for finding the efficiency, regulation and temperature rise. The main, advantage of this test is that the transformers are tested under full-load, conditions without much expenditure of power. The power required to conduct, this test is equal to the losses of the two transformers. It may be noted that two, identical transformers are needed to carry out this test., , Circuit, Fig. (7.33) shows the connections for back-to-back test on two identical, transformers T1 and T2. The primaries of the two transformers are connected in, parallel across the rated voltage V1 while the two secondaries are connected in, phase opposition. Therefore, there will be no circulating current in the loop, formed by the secondaries because their induced e.m.f.s are equal and in, opposition. There is an auxiliary low-voltage transformer which can be adjusted, to give a variable voltage and hence current in the secondary loop circuit. A, wattmeter W1, an ammeter A1 and voltmeter V1 are connected to the input side., A wattmeter W2 and ammeter A2 are connected in the secondary circuit., , Operation, (i), , The secondaries of the transformers are in phase opposition. With switch S1, closed and switch S2 open (i.e., regulating transformer not in the circuit),, there will be no circulating current (I2 = 0) in the secondary loop circuit. It, is because the induced e.m.f.s in the secondaries are equal and in, opposition. This situation is just like an open-circuit test. Therefore, the, current drawn from the supply is 2 I0 where I0 is the no-load current of each, transformer. The reading of wattmeter W1 will be equal to the core losses, of the two transformers., W1 = Core losses of the two transformers, , 150
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Fig.(7.33), (ii) Now switch S2 is also closed and output voltage of the regulating, transformer is adjusted till full-load current I2 flows in the secondary loop, circuit. The full-load secondary current will cause full-load current I1 (= K, I2) in the primary circuit. The primary current I1 circulates in the primary, winding only and will not pass through W1. Note that full-load currents are, flowing through the primary and secondary windings. Therefore, reading of, wattmeter W2 will be equal to the full-load copper losses of the two, transformers., W2 = Full-load Cu losses of two transformers, ∴, , W1 + W2 = Total losses of two transforms at full load, , The following points may be noted:, (a) The wattmeter W1 gives the core losses of the two transformers while, wattmeter W2 gives the full-load copper losses (or at any other load, current I2) of the two transformers. Therefor, power required to conduct, this test is equal to the total losses of the two transformers., (b) Although transformers are not supplying any load, yet full iron loss and, full-load copper losses are occurring in them., (c) There are two voltage sources (supply voltage and regulating, transformer) and there is no Interference between them. The supply, voltage gives only 2I0 while regulating transformer supplies I2 and hence, I1 (= K I2)., , Advantages, (i) The power required to carry out the test is small., (ii) The transformers are tested under full-load conditions., 151
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(iii) The iron losses and full-load copper losses are measured simultaneously., (iv) The secondary current I2 can be adjusted to any current value. Therefore,, we can find the copper loss at full-load or at any other load., (v) The temperature rise of the transformers can be noted., , 7.24 Losses in a Transformer, The power losses in a transformer are of two types, namely;, 1. Core or Iron losses, 2. Copper losses, These losses appear in the form of heat and produce (i) an increase in, temperature and (ii) a drop in efficiency., , 1., , Core or Iron losses (Pi), These consist of hysteresis and eddy current losses and occur in the transformer, core due to the alternating flux. These can be determined by open-circuit test., Hysteresis loss, = k h f B1m.6 watts / m 3, Eddy current loss, = k e f 2 B 2m t 2, , watts / m 3, , Both hysteresis and eddy current losses depend upon (i) maximum flux density, Bm in the core and (ii) supply frequency f. Since transformers are connected to, constant-frequency, constant voltage supply, both f and Bm are constant. Hence,, core or iron losses are practically the same at all loads., Iron or Core losses, Pi = Hysteresis loss + Eddy current loss, = Constant losses, The hysteresis loss can be minimized by using steel of high silicon content, whereas eddy current loss can be reduced by using core of thin laminations., , 2., , Copper losses, These losses occur in both the primary and secondary windings due to their, ohmic resistance. These can be determined by short-circuit test., , 152
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Total Cu losses, PC = I12 R 1 + I 22 R 2, , = I12R 01 or I22 R 02, It is clear that copper losses vary as the square of load current Thus if copper, losses are 400 W at a load current of 10 A, then they will be (1/2)2 × 400 = 100, W at a load current of 5A., Total losses in a transformer = P1 + PC, = Constant losses + Variable losses, It may be noted that in a transformer, copper losses account for about 90% of the, total losses., , 7.25 Efficiency of a Transformer, Like any other electrical machine, the efficiency of a transformer is defined as, the ratio of output power (in watts or kW) to input power (watts or kW) i.e.,, Efficiency =, , Output power, Input power, , It may appear that efficiency can be determined by directly loading the, transformer and measuring the input power and output power. However, this, method has the following drawbacks:, (i) Since the efficiency of a transformer is very high, even 1% error in each, wattmeter (output and input) may give ridiculous results. This test, for, instance, may give efficiency higher than 100%., (ii) Since the test is performed with transformer on load, considerable, amount of power is wasted. For large transformers, the cost of power, alone would be considerable., (iii) It is generally difficult to have a device that is capable of absorbing all of, the output power., (iv) The test gives no information about the proportion of various losses., Due to these drawbacks, direct loading method is seldom used to determine the, efficiency of a transformer. In practice, open-circuit and short-circuit tests are, carried out to find the efficiency., Efficiency =, , Output, Output, =, Input, Output + Losses, , The losses can be determined by transformer tests., , 153
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7.26 Efficiency from Transformer Tests, F.L. Iron loss = P1, , ...from open-circuit test, , F.L. Cu loss = PC, , ...from short-circuit test, , Total F.L. losses = P1 + PC, We can now find the full-load efficiency of the transformer at any p.f. without, actually loading the transformer., F.L. efficiency, ηF.L. =, , Full - load VA × p.f ., (Full - load VA × p.f .) + Pi + PC, , Also for any load equal to x x full-load,, Corresponding total losses = Pi + x2 PC, Corresponding η x =, , (xx Full - load VA ) × p.f ., (xx Full - load VA × p.f .) + Pi + x 2 PC, , Note that iron loss remains the same at all loads., , 7.27 Condition for Maximum Efficiency, Output power = V2I2 cos φ2, If R02 is the total resistance of the transformer referred to secondary, then,, Total Cu loss, PC = I 22 R 02, Total losses = Pi + PC, ∴, , Transformer η =, , =, , V2 I 2 cos φ 2, V2 I 2 cos φ 2 + Pi + I 22 R 02, V2 cos φ2, V2 cos φ2 + Pi / I2 + I2 R 02, , (i), , For a normal transformer, V2 is approximately constant. Hence for a load of, given p.f., efficiency depends upon load current I2. It is clear from exp (i) above, that numerator is constant and for the efficiency to be maximum, the, denominator should be minimum i.e.,, , d, (denominat or) = 0, dI 2, or, , d, (V cos φ 2 + Pi / I 2 + I 2 R 02 ) = 0, dI 2 2, , 154
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Pi, , or, , 0−, , or, , Pi = I 22 R 02, , i.e.,, , Iron losses = Copper losses, , I 22, , + R 02 = 0, (ii), , Hence efficiency of a transformer will be maximum when copper losses are, equal to constant or iron losses., From eq. (ii) above, the load current I2 corresponding to maximum efficiency is, given by;, , I2 =, , Pi, R 02, , The relative value of these losses is in the control of the designer of the, transformer according to the relative amount of copper and iron he uses. A, transformer which is to operate continuously on full-load would, therefore, be, designed to have-maximum efficiency at full-load. However, distribution, transformers operate for long periods on light load. Therefore, their point of, maximum efficiency is usually arranged to be between three-quarter and half, lull-load., Note. In a transformer, iron losses are constant whereas copper losses are, variable. In order to obtain maximum efficiency, the load current should be such, that total Cu losses become equal to iron losses., , 7.28 Output kVA Corresponding to Maximum Efficiency, Let, , PC = Copper losses at full-load kVA, Pi = Iron losses, x = Fraction of full-load kVA at which efficiency is maximum, Total Cu losses = x2 PC, x2 PC = Pi, , or, , x=, , ...for maximum efficiency, , Pi, Iron loss, =, PC, F.L. Cu loss, , 155
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∴, , Output kVA corresponding to maximum efficiency, , = xx Full - load kVA = Full - load kVA ×, , Iron loss, F.L. Cu loss, , It may be noted that the value of kVA at which the efficiency is maximum is, independent of p.f. of the load., , 7.29 All-Day (or Energy) Efficiency, The ordinary or commercial efficiency of a transformer is defined as the ratio of, output power to the input power i.e.,, Commercial efficiency =, , Output power, Input power, , There are certain types of transformers whose performance cannot be judged by, this efficiency. For instance, distribution transformers used for supplying, lighting loads have their primaries energized all the 24 hours in a day but the, secondaries supply little or no load during the major portion of the day. It means, that a constant loss (i.e., iron loss) occurs during the whole day but copper loss, occurs only when the transformer is loaded and would depend upon the, magnitude of load. Consequently, the copper loss varies considerably during the, day and the commercial efficiency of such transformers will vary from a low, value (or even zero) to a high value when the load is high. The performance of, such transformers is judged on the basis of energy consumption during the, whole day (i.e., 24 hours). This is known as all-day or energy efficiency., The ratio of output in kWh to the input in kWh of a transformer over a 24-hour, period is known as all-day efficiency i.e.,, , ηall- day =, , kWh output in 24 hours, kWh input in 24 hours, , All-day efficiency is of special importance for those transformers whose, primaries are never open-circuited but the secondaries carry little or no load, much of the time during the day. In the design of such transformers, efforts, should be made to reduce the iron losses which continuously occur during the, whole day., Note. Efficiency of a transformer means commercial efficiency unless stated, otherwise., , 156
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7.30 Construction of a Transformer, We usually design a power transformer so that it approaches the characteristics, of an ideal transformer. To achieve this, following design features are, incorporated:, (i) The core is made of silicon steel which has low hysteresis loss and high, permeability. Further, core is laminated in order to reduce eddy current, loss. These features considerably reduce the iron losses and the no-load, current., (ii) Instead of placing primary on one limb and secondary on the other, it is a, usual practice to wind one-half of each winding on one limb. This, ensures tight coupling between the two windings. Consequently, leakage, flux is considerably reduced., (iii) The winding resistances R1 and R2 are minimized to reduce I2R loss and, resulting rise in temperature and to ensure high efficiency., , 7.31 Types of Transformers, Depending upon the manner in which the primary and secondary are wound on, the core, transformers are of two types viz., (i) core-type transformer and (ii), shell-type transformer., (i) Core-type transformer. In a core-type transformer, half of the primary, winding and half of the secondary winding are placed round each limb as, shown in Fig. (7.34). This reduces the leakage flux. It is a usual practice to, place the low-voltage winding below the high-voltage winding for, mechanical considerations., , Fig.(7.34), , Fig.(7.35), , (ii) Shell-type transformer. This method of construction involves the use of a, double magnetic circuit. Both the windings are placed round the central, limb (See Fig. 7.35), the other two limbs acting simply as a low-reluctance, flux path., , 157
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The choice of type (whether core or shell) will not greatly affect the efficiency, of the transformer. The core type is generally more suitable for high voltage and, small output while the shell-type is generally more suitable for low voltage and, high output., , 7.32 Cooling of Transformers, In all electrical machines, the losses produce heat and means must be provided, to keep the temperature low. In generators and motors, the rotating unit serves as, a fan causing air to circulate and carry away the heat. However, a transformer, has no rotating parts. Therefore, some other methods of cooling must be used., Heat is produced in a transformer by the iron losses in the core and I2R loss in, the windings. To prevent undue temperature rise, this heat is removed by, cooling., (i) In small transformers (below 50 kVA), natural air cooling is employed, i.e., the heat produced is carried away by the surrounding air., (ii) Medium size power or distribution transformers are generally cooled by, housing them in tanks filled with oil. The oil serves a double purpose,, carrying the heat from the windings to the surface of the tank and, insulating the primary from the secondary., (iii) For large transformers, external radiators are added to increase the, cooling surface of the oil filled tank. The oil circulates around the, transformer and moves through the radiators where the heat is released to, surrounding air. Sometimes cooling fans blow air over the radiators to, accelerate the cooling process., , 7.33 Autotransformer, An autotransformer has a single winding on an iron core and a part of winding is, common to both the primary and secondary circuits. Fig. (7.36 (i)) shows the, connections of a step-down autotransformer whereas Fig. (7.36 (ii)) shows the, connections of a step-up autotransformer. In either case, the winding ab having, N1 turns is the primary winding and winding be having N2 turns is the secondary, winding. Note that the primary and secondary windings are connected, electrically as well as magnetically. Therefore, power from the primary is, transferred to the secondary conductively as well as inductively (transformer, action). The voltage transformation ratio K of an ideal autotransformer is, , K=, , V2 N 2 I1, =, =, V1 N1 I 2, , Note that in an autotransformer, secondary and primary voltages are related in, the same way as in a 2-winding transformer., , 158
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Fig.(7.36), Fig. (7.37) shows the connections of a loaded step-down as well as step-up, autotransformer. In each case, I1 is the input current and I2 is the output or load, current. Regardless of autotransformer connection (step-up or step-down), the, current in the portion of the winding that is common to both the primary and the, secondary is the difference between these currents (I1 and I2). The relative, direction of the current through the common portion of the winding depends, upon the connections of the autotransformer. It is because the type of connection, determines whether input current I1 or output current I2 is larger. For step-down, autotransformer I2 > I1 (as for 2-winding transformer) so that I2 − I1 current, flows through the common portion of the winding. For step-up autotransformer,, I2 < I1. Therefore, I1 − I2 current flews in the common portion of the winding., , Fig.(7.37), In an ideal autotransformer, exciting current and losses are neglected. For such, an autotransformer, as K approaches 1, the value of current in the common, portion (I2 − I1 or I1 − I2) of the winding approaches zero. Therefore, for value, of K near unity, the common portion of the winding can be wound with wire of, smaller cross-sectional area. For this reason, an autotransformer requires less, copper., , 159
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i, ), From, e, q, s, ., (, i, ), a, n, d, (, i, i, ), ,, , V2 N 2 I1, =, = =K, V1 N1 I 2, Also, , V1I1 = V2I2, , (Input apparent power = Output apparent power), , Output, Since the primary and secondary windings of an autotransformer are connected, magnetically as well as electrically, the power from primary is transferred to the, secondary inductively (transformer action) as well as conductively (i.e.,, conducted directly from source to the load)., Output apparent power = V2I2, Apparent power transferred inductively = V2(I2 − I1) = V2(I2 − K I2), = V2I2(1 − K) = V1I1(1 − K), ∴, Power transferred inductively = Input × (1 − K), ∴, Power transferred conductively = Input − Input (1 − K), = Input [1 − (1 − K)], = K × Input, Suppose the input power to an ideal autotransformer is 1000 W and its voltage, transformation ratio K= 1/4. Then,, , 161
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1, Power transferred inductively = Input × (1 − K ) = 10001 − = 750 W, 4, 1, Power transferred conductively = K × Input = × 1000 = 250 W, 4, Note that input power to the autotransformer is 1000 W. Out of this, 750 W is, transferred to the secondary by transformer action (inductively) while 250 W is, conducted directly from the source to the load (i.e., it is transferred conductively, to the load)., , 7.35 Saving of Copper in Autotransformer, For the same output and voltage transformation ratio K(N2/N1), an, autotransformer requires less copper than an ordinary 2-winding transformer., Fig. (7.39 (i)) shows an ordinary 2-winding transformer whereas Fig. (7.39 (ii)), shows an autotransformer having the same output and voltage transformation, ratio K., The length of copper required in a winding is proportional to the number of, turns and the area of cross-section of the winding wire is proportional to the, current rating. Therefore, the volume and hence weight of copper required in a, winding is proportional to current x turns i.e.,, Weight of Cu required in a winding ∝ current × turns, , Fig.(7.39), Winding transformer, Weight of Cu required ∝ (I1 N1 + I 2 N 2 ), Autotransformer, Weight of Cu required in section 1-2 ∝ I1 ( N1 − N 2 ), Weight of Cu required in section 2-3 ∝ (I 2 − I1 ) N 2, ∴ Total weight of Cu required ∝ I1 ( N1 − N 2 ) + (I 2 − I1 ) N 2, , 162
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Weight of Cu in autotransf ormer, Weight of Cu in ordinary t ransformer, , =, , I1 ( N1 − N 2 ) + (I 2 − I1 ) N 2, I1 N1 + I 2 N 2, , =, , N1I1 − N 2 I1 + N 2 I 2 − N 2 I1, N1I1 + N 2 I 2, , =, , N1I1 + N 2 I 2 − 2 N 2 I1, N1I1 + N 2 I 2, , =1 −, , 2 N 2 I1, N1I1 + N 2 I 2, , =1 −, , 2 N 2 I1, 2 N1I1, , =1 −, , N2, =1− K, N1, , (Q N 2 I 2 = N1I1 ), , ∴ Wt. of Cu in autotransformer (Wa), = (1 − K) × Wt. in ordinary transformer (Wo), Wa = (1 − K) × Wo, , or, ∴, , Saving in Cu = Wo − Wa = Wa − (1 − K)Wo = K Wo, Saving in Cu = K × Wt. ofCu in ordinary transformer, , or, , Thus if K = 0.1, the saving of Cu is only 10% but if K = 0.9, saving of Cu is, 90%. Therefore, the nearer the value of K of autotransformer is to 1, the greater, is the saving of Cu., , 7.36 Advantages and Disadvantages of autotransformers, Advantages, (i), (ii), (iii), (iv), (v), , An autotransformer requires less Cu than a two-winding transformer of, similar rating., An autotransformer operates at a higher efficiency than a two-winding, transformer of similar rating., An autotransformer has better voltage regulation than a two-winding, transformer of the same rating., An autotransformer has smaller size than a two-winding transformer of, the same rating., An autotransformer requires smaller exciting current than a two-winding, transformer of the same rating., , It may be noted that these advantages of the autotransformer decrease as the, ratio of transformation increases. Therefore, an autotransformer has marked, , 163
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advantages only for relatively low values of transformation ratio (i.e. values, approaching 1)., , Disadvantages, (i), , There is a direct connection between, the primary and secondary. Therefore,, the output is no longer d.c. isolated, from the input., (ii) An autotransformer is not safe for, stepping down a high voltage to a low, voltage. As an illustration, Fig. (7.40), Fig.(7-40), shows, 11000/230, V, step-down, autotransformer. If an open circuit, develops in the common portion 2-3 of the winding, then full-primary, voltage (i.e., 11000 V in this case) will appear across the load. In such a, case, any one coming in contact with the secondary is subjected to high, voltage. This could be dangerous to both the persons and equipment. For, this reason, autotransformers are prohibited for general use., (iii) The short-circuit current is much larger than for the two-winding, transformer of the same rating. It can be seen from Fig. (7.40) that a, short-circuited secondary causes part of the primary also to be shortcircuited. This reduces the effective resistance and reactance., , 7.37 Applications of Autotransformers, (i), , Autotransformers are used to compensate for voltage drops in, transmission and distribution lines. When used for this purpose, they are, known as booster transformers., (ii) Autotransformers are used for reducing the voltage supplied to a.c., motors during the starting period., (iii) Autotransformers are used for continuously variable supply., , 7.38 Conversion of Two-Winding Transformer Into, Autotransformer, A two-winding transformer can be converted into an autotransformer, either, step-up or step-down. Consider a 10 kVA, 2300/230 V two-winding transformer, shown in Fig. (7.41 (i)). If we want to convert it into autotransformer, the two, windings of the transformer are connected in series. If we use the additive, polarity as shown in Fig. (7.41 (ii)), we get step-up autotransformer. The voltage, rating of the autotransformer is now 2300/2530 V. If we use subtractive polarity, as shown in Fig. (7.41 (iii)), we get a step-down autotransformer. The voltage, rating of the transformer is now 2300/2070 V., , 164
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Fig.(7.41), When a two-winding transformer is converted into autotransformer, the kVA, rating of the resulting autotransformer is greatly increased. This higher rating, results from the conduction connection., , 7.39 Parallel Operation of Single-Phase Transformers, Two transformers are said to be connected, in parallel if the primary windings are, connected to supply busbars and, secondary windings are connected to load, busbars., Fig., (7.42), shows, two, transformers A and B in parallel. While, connecting two or more than two, transformers in parallel, it is essential that, their terminals of similar polarities are, Fig.(7.42), joined to the same busbars as shown in, Fig. (7.42). The wrong connections may result in a dead short-circuit and, primary transformers may be damaged unless protected by fuses or circuit, breakers. There are three principal reasons for connecting transformers in, parallel. Firstly, if one transformer fails, the continuity of supply can be, maintained through other transformers. Secondly, when the load on the substation becomes more than the capacity of the existing transformers, another, transformer can be added in parallel. Thirdly, any transformer can be taken out, of the circuit for repair/routine maintenance without interrupting supply to the, consumers., , Conditions for satisfactory parallel operation, In order that the transformers work satisfactorily in parallel, the following, conditions should be satisfied:, (i) Transformers should be properly connected with regard to their, polarities., (ii) The voltage ratings and voltage ratios of the transformers should be the, same., (iii) The per unit or percentage impedances of the transformers should be, equal., 165
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(iv) The reactance/resistance ratios of the transformers should be the same., Condition (i), Condition (i) is absolutely essential because wrong connections may result in, dead short-circuit. Fig. (7.43 (i)) shows the correct method of connecting two, single-phase transformers in parallel. It will be seen that round the loop formed, by the secondaries, the two secondary e.m.f.s EA and EB oppose and there will, be no circulating current., , Fig.(7.43), Fig. (7.43 (ii)) shows the wrong method of connecting two single-phase, transformers is parallel. Here the two secondaries are so connected that their, e.m.f.s EA and EB are additive. This may lead to short-circuit conditions and a, very large circulating current will flow in the loop formed by the two, secondaries. Such a condition may damage the transformers unless they are, protected by fuses and circuit breakers., Condition (ii), This condition is desirable for the satisfactory parallel operation of transformers., If this condition is not met, the secondary e.m.f.s will not be equal and there will, be circulating current in the loop formed by the secondaries. This will result in, the unsatisfactory parallel operation of transformers. Let us illustrate this point., Consider two single-phase transformer A and B operating in parallel as shown in, Fig. (7.44). Let EA and EB be their no-load secondary voltages and ZA and ZB be, their impedances referred to the secondary. Then at no-load, the circulating, current in the loop formed by the secondaries is, Circulating current, I C =, , EA − EB, ZA + ZB, , assuming E A > E B, , Even a small difference in the, induced secondary voltages can, cause a large circulating current, in the secondary loop because, impedances of the transformers, 166, Fig.(7.44)
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are small. This secondary circulating current will cause current to be drawn from, the supply by the primary of each transformer. These currents will cause copper, losses in both primary and secondary. This creates heating with no useful output., When load is connected to the system, this circulating current will tend to, produce unequal loading conditions i.e., the transformers will not share the load, according to their kVA ratings. It is because the circulating current will tend to, make the terminal voltages of the same value for both transformers. Therefore,, transformer with smaller voltage ratio will tend to carry more than its proper, share of load. Thus, one transformer would tend to become overloaded than the, other and the system could not be loaded to the summation of transformer, ratings without overloading one transformer., Condition (iii), This condition is also desirable for proper parallel operation of transformers. If, this condition is not met, the transformers will not share the load according to, their kVA ratings. Sometimes this condition is not fulfilled by the design of the, transformers. In that case, it can be corrected by inserting proper amount of, resistance or reactance or both in series with either primary or secondary circuits, of the transformers where the impedance is below the value required to fulfil, condition (iii)., Condition (iv), If the reactance/resistance ratios of the two transformers are not equal, the power, factor of the load supplied by the transformers will not be equal. In other words,, one transformer will be operating with a higher and the other with a lower power, factor than that of the load. Condition (iii) is much more important than, condition (iv). Considerable deviation from condition (iv) will result in only a, small reduction in the satisfactory degree of operation. When desired, condition, (iv) also may be improved by inserting external impedance of proper value., , 7.40 Single-Phase Equal Voltage Ratio Transformers in, Parallel, Fig. (7.45) shows two single-phase equal voltage ratio transformers A and B in, parallel. The secondary e.m.f.s of the two transformers are equal (i.e., EA = EB =, E) because they have the same turns ratio and have their primaries connected to, the same supply., , 167
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Fig.(7.45), If the magnetizing current is ignored, the two transformers can be represented by, their equivalent circuits referred to secondary as shown in Fig. (7.46). It is clear, that the transformers will share total load in the same way as two impedances in, parallel., Let ZA, ZB = Impedances of transformers referred to secondary, IA,IB = Their respective currents, V2 = Common terminal voltage, I = Total load current, It is clear from Fig. (7.46) that:, IA + IB = I, , (i), , and, , or, , IAZA = IBZB, , ∴, , IA = IB, , ∴, , IB, , ZB, ZA, , ZB, + IB = I, ZA, , Z , , I B 1 + B = I, ZA , Fig.(7.46), , ∴, (, i, i, ), , 168, , IB = I, , ZA, ZA + Z
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Similarly, I A = I, , ZB, Z A + ZB, , (iii), , Thus the way in which the load current I is shared by the transformers is, independent of load impedance and depends only on the transformer, impedances., kVA carried by each transformer, Let, S = total load kVA = V2 I × 10-3, SA = kVA carried by transformer A, SB = kVA carried by transformer B, , ∴, , SA = V2 I A × 10 − 3 = V2 I × 10 − 3 ×, , ZB, ZB, =S, Z A + ZB, ZA + ZB, , ZB, ZA + Z B, , or, , SA = S, , Also, , SB = V2 I B × 10 − 3 = V2 I × 10 − 3 ×, , or, , SB = S, , ZA, ZA, =S, Z A + ZB, ZA + ZB, , ZA, ZA + ZB, , Therefore, SA and SB are obtained in magnitude as well as m phase from the, above expressions. It may be noted that in these expressions, ZA and ZB can be, expressed in ohms or in p.u. If p.u. values are to be used, they should be with, respect to common base kVA., , 169
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7.41 Single- Phase Unequal Voltage Ratio Transformers in, Parallel, Fig. (7.47) shows two single-phase unequal voltage ratio transformers A and B, in parallel. Since the voltage ratios of the transformers are unequal, their no-load, secondary voltages will also be unequal. We shall calculate how load current is, shared between the two transformers., , Fig.(7.47), Fig. (7.48) shows the equivalent circuit of the transformers referred to secondary, in a simplified way., , Fig.(7.48), Let, , EA, EB = no-load secondary voltages of the two transformers. It is, assumed that EA > EB., IA, IB = their respective currents, I = load current, ZA, ZB = impedances of the transformers referred to secondary, ZB = load impedance, V2 = load voltage, , At no-load, the circulating current IC is, , IC =, , EA − EB, Z A + ZB, , 170
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When the system is loaded, the load current I is shared by the two transformers., By kirchhof’s oitage law,, EA = V2 + IAZA, and, But, ∴, and, , EB = V2 + IBZB, V2 = I ZL = (IA + IB)ZL, , EA = (IA + IB)ZL + IAZA, , (i), , EB = (IA + IB)ZL + IBZB, , (ii), Now, or, , EA − EB = IAZA − IBZB, , IA =, , (E A − E B ) + I B Z B, ZA, , Putting this value of IA in eq. (ii), we get,, , (E − E B ) + I B Z B, , + I B ZL + I B ZB, EB = A, ZA, , , On solving, I B =, , E B Z A − (E A − E B ) Z L, Z A Z B + ZL (ZA + Z B ), , (iii), , From the symmetry of the expression, we get,, , IA =, Also, , E A Z B + (E A − E B ) Z L, Z A Z B + ZL (Z A + ZB ), , I = IA + IB =, , E A ZB + E B Z A, Z A Z B + ZL (Z A + ZB ), , E A Z B + E B ZA, , , V2 = I Z L = , ZL, , +, +, Z, Z, Z, (, Z, Z, ), A B, L, A, B , or, , V2 =, , E A Z B + E B ZA, ZA ZB, + Z A + ZB, ZL, , Since the transformers have a common primary voltage, EA and EB will be in, phase with each other., , 7.42 Three-Phase Transformer, A three-phase system in used to generate and transmit electric power. Threephase voltages are raised or lowered by means of three-phase transformers. A, three-phase transformer can be built in two ways viz., (i) by suitably connecting, a bank of three single-phase transformers or (ii) by constructing a three-phase, , 171
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transformer on a common magnetic structure. In either case, the windings may, be connected in Y − Y, ∆ − ∆, Y − ∆ or ∆ − Y., , (i) Bank of three single-phase transformers, Three similar single-phase transformers can be connected to form a three-phase, transformer. The primary and secondary windings may be connected in star (Y), or delta (∆) arrangement. Fig. (7.49 (i)) shows a Y - ∆ connection of a threephase transformer. The primary windings are connected in star and the, secondary windings are connected in delta. A more convenient way of showing, this connection is illustrated in Fig. (7-49 (ii)). The primary and secondary, windings shown parallel to each other belong to the same single-phase, transformer. The ratio of secondary phase voltage to primary phase voltage is, the phase transformation ratio K., Phase transformation ratio, K =, , Secondary phase voltage N 2, =, Primary phase voltage, N1, , Referring to Fig. (7.49 (ii)), primary line-to-line voltage is V and the primary, line current is I. The phase transformation ratio is K (= N2/N1). The secondary, line voltage and line current are also shown., , Fig.(7.49), , (ii) Three-phase transformer, A three-phase transformer can be constructed by having three primary and three, secondary windings on a common magnetic circuit. The basic principle of a 3phase transformer is illustrated in Fig. (7.50 (i)). The three single-phase coretype transformers, each with windings (primary and secondary) on only one leg, have their unwound legs combined to provide a path for the returning flux. The, primaries as well as secondaries may be connected in star or delta. If the primary, is energized from a 3-phase supply, the central limb (i.e., unwound limb) carries, the fluxes produced by the 3-phase primary windings. Since the phasor sum of, three primary currents at any instant is zero, the sum of three fluxes passing, through the central limb must be zero. Hence no flux exists in the central limb, , 172
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and it may, therefore, be eliminated. This modification gives a three leg coretype 3-phase transformer. In this case, any two legs will act as a return path for, the flux in the third leg. For example, if flux is φ in one leg at some instant, then, flux is φ/2 in the opposite direction through the other two legs at the same, instant. All the connections of a 3-phase transformer are made inside the case, and for delta-connected winding three leads are brought out while for starconnected winding four leads are brought out., , Fig.(7.50), For the same capacity, a 3-phase transformer weighs less, occupies less space, and costs about 20% less than a bank of three single-phase transformers., Because of these advantages, 3-phase transformers are in common use,, especially for large power transformations., A disadvantage of the three-phase transformer lies in the fact that when one, phase becomes defective, the entire three-phase unit must be removed from, service. When one transformer in a bank of three single-phase transformers, becomes defective, it may be removed from service and the other two, transformers may be reconnected to supply service on an emergency basis until, repairs can be made., , 7.43 Three-Phase Transformer Connections, A three-phase transformer can be built by suitably connecting a bank of three, single-phase transformers or by one three-phase transformer. The primary or, secondary windings may be connected in either star (Y) or delta (∆), arrangement. The four most common connections are (i) Y-Y (ii) ∆-∆ (iii) Y-∆, and (iv) ∆-Y. These four connections are shown in Fig. (7.51). In this figure, the, windings at the left are the primaries and those at the right are the secondaries., The primary and secondary voltages and currents are also shown. The primary, line voltage is V and the primary line current is I. The phase transformation ratio, K is given by;, , 173
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Fig.(7.51), K=, (i), , Secondary phase voltage N 2, =, Primary phase voltage, N1, , Y-Y Connection. In the Y-Y connection shown in Fig. (7.51 (i)), 57.7%, (or 1 / 3 ) of the line voltage is impressed upon each winding but full line, 174
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current flows in each winding. Power circuits supplied from a Y-Y bank, often create serious disturbances in communication circuits in their, immediate vicinity. Because of this and other disadvantages, the Y-Y, connection is seldom used., (ii) ∆-∆ Connection. The ∆-∆ connection shown in Fig. (7.51 (ii)) is often used, for moderate voltages. An advantage of this connection is that if one, transformer gets damaged or is removed from service, the remaining two, can be operated in what is known as the open-delta or V-V connection. By, being operated in this way, the bank still delivers three-phase currents and, voltages in their correct phase relationships but the capacity of the bank is, reduced to 57.7% of what it was with all three transformers in service., (iii) Y-∆ Connection. The Y-∆ connection shown in Fig. (7.51(iii)) is suitable, for stepping down a high voltage. In this case, the primaries are designed, for 57.7% of the high-tension line voltages., (iv) ∆-Y Connection. The ∆-Y connection shown in Fig. (7.51 (iv)) is, commonly used for stepping up to a high voltage., , 7.44 Three-Phase Transformation with Two Single-Phase, Transformers, It is possible to transform three-phase power by using two single-phase, transformers. Two methods of doing this are:, (i) the connection of two identical single-phase transformers in open delta, (or V-V connection)., (ii) the T-T connection (or Scott connection) of two nonidehtical singlephase transformers., Both of these methods of three-phase transformation result in slightly, unbalanced output voltages under load because of unsymmetrical relations. The, unbalance is negligible under usual conditions of operation., , 7.45 Open-Delta or V-V Connection, If one transformer breaks down in a star-star connected system of 3 single-phase, transformers, three-phase power cannot be supplied until the defective, transformer has been replaced or repaired. To eliminate this undesirable, condition, single-phase transformers are generally connected in ∆-∆. In this, case, if one transformer breaks down, it is possible to continue supplying threephase power with the other two transformers because this arrangement maintains, correct voltage and phase relations on the secondary. However, with two, , 175
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transformers, the capacity of the bank is reduced to 57.7% of what it was with, all three transformers in service (i.e., complete ∆-∆ circuit)., , Theory, If one transformer is removed in the ∆-∆ connection of three single-phase, transformers, the resulting connection becomes open delta or V-V connection. In, complete ∆-∆ connection, the voltage of any one phase is equal and opposite to, the sum of the voltages of the other two phases. Therefore, under no-load, conditions if one transformer is removed, the other two will maintain the same, three line voltages on the secondary side. Under load conditions, the secondary, line voltages will be slightly unbalanced because of the unsymmetrical relation, of the impedance drops in the transformers., Fig. (7.52 (i)) shows open delta (or V-V) connection; one transformer shown, dotted is removed. For simplicity, the load is considered to be star connected., Fig. (7.52 (ii)) shows the phasor diagram for voltages and currents. Here VAB,, VBC and VCA represent the line-to-line voltages of the primary; Vab, Vbc and Vca, represent line-to-line voltages of the secondary and Van, Vbn and Vcn represent, the phase voltages of the load. For inductive load, the load currents Ia, Ib and Ic, will lag the corresponding voltages Van, Vbn and Vcn by the load phase angle φ., , Fig.(7.52), , 176
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The transformer windings ab and bc will deliver power given by;, Pab = Vab Ia cos(30° + φ), Pbc = Vcb Ic cos(30° − φ), Let, , Vab = Vcb = V, the voltage rating of transformer secondary winding, Ia = Ic = I, current rating of the transformer secondary winding, p.f. = 1 i.e., , ∴, , φ = 0°, , ... For resistive load, , Power delivered to the resistive load by V-V connection is, , PV = Pab + Pbc = v IU cos 30° + V I cos 30° = 2 V I cos 30°, With all the three transformers connected in delta, the power delivered to the, resistive load is, , P∆ = 3 V I, , ∴, , PV 2 V I cos 30° 2 cos 30°, = 0.577, =, =, P∆, 3V I, 3, , Hence the power-handling capacity of a V-V circuit (without overheating the, transformers) is 57.7% of the capacity of a complete ∆-∆ circuit of the same, transformers., In a V-V circuit, only 86.6% of the rated capacity of the two transformers is, available. This can be readily proved., , Operating capacity 2 V I cos 30°, 3, =, = cos 30° =, = 0.866, Available capacity, 2V I, 2, Let us illustrate V-V connection with a numerical example. Suppose three, identical single-phase transformers, each of capacity 10 kVA, are connected in, ∆-∆. The total rating of the three transformers is 30 kVA. When one transformer, is removed, the system reverts to V-V circuit and can deliver 3-phase power to a, 3-phase load. However, the kVA capacity of the V-V circuit is reduced to 30 ×, 0.577 = 17.3 kVA and not 20 kVA as might be expected. This reduced capacity, can be determined in an alternate way. The available capacity of the two, transformers is 20 kVA. When operating in V-V circuit, only 86.6% of the rated, capacity is available i.e. 20 × 0.866 = 17.3 kVA., , 7.46 Power Factor of Transformers in V-V Circuit, When V-V circuit is delivering 3-phase power, the power factor of the two, transformers is not the same (except at unity p.f.). Therefore, the voltage, regulation of the two transformers will not be the same. If the load power factor, angle is φ, then,, 177
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p.f. of transformer 1 = cos (30° − φ), p.f. of transformer 2 = cos (30° + φ), (i), , When load p.f. = 1 i.e. φ = 0°, In this case, each transformer will have a power factor of 0.866., , (ii) When load p.f. = 0.866 i.e. φ = 30°, In this case, one transformer will have a p.f. of cos (30° − 30°) = 1 and the, other of cos (30° + 30°) = 0.5., (iii) When load p.f = 0.5 i.e. φ = 60°, In this case, one transformer will have a p.f. of cos (30° − 60°) = 0.866 and, the other of cos (30° + 60°) = 0. Thus at a load p.f. of 0.5, one transformer, delivers all the power at 0.866 p.f. and the other (although still necessary to, be in the circuit) delivers no power at all., , 7.47 Applications of Open Delta or V-V Connection, The V-V circuit has a number of features that arc advantageous. A few, applications are given below by way of illustration:, (i) The circuit can be employed in an emergency situation when one, transformer in a complete ∆-∆ circuit must be removed for repair and, continuity of service is required., (ii) Upon failure of the primary or secondary of one transformer of a, complete ∆-∆ circuit, the system can be operated as V-V circuit and can, deliver 3-phase power (with reduced capacity) to a 3-phase load., (iii) A circuit is sometimes installed as V-V circuit with the understanding, that its capacity may be increased by adding one more transformer to, form complete ∆-∆ circuit. As shown earlier,, , PV, 1, =, = 0.577, P∆, 3, ∴, , P∆ = 3 × PV, , Thus if a V-V circuit is changed to complete ∆-∆ circuit, the capacity is, increased by a factor of 3 = (= 1.732) ., , 7.48 Scott Connection or T-T Connection, Although there are now no 2-phase transmission and distribution systems, a 2phase supply is sometimes required. We can convert 3-phase supply into 2phase supply through scott or T-T connection of two single-phase transformers., One is called the main transformer M which has a centre-tapped primary; the, 178
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centre-tap being C [See Fig. (7.53 (i))]. The primary of this transformer has N1, turns and is connected between the terminals B and Y of the 3-phase supply., The other transformer is called teaser transformer T and its primary has 0.866 N1, turns. One end of this primary is connected to centre-tap C and the other end to, the terminal R of the 3-phase supply. The number of turns (N2) of the secondary, windings of the two transformers are equal. As we shall see, the voltages across, the secondaries are equal in magnitude having a phase difference of 90°. Thus, scott connection of two single-phase transformers enables us to convert 3-phase, supply to 2-phase supply., , Fig.(7.53), , Theory, Referring to Fig. (7.53 (i)), the centre-tapped primary of the main transformer, has line voltage VYB applied to its terminals. The secondary terminal voltage V2, of the main transformer is, , V2 =, , N2, N, VYB = 2 VL, N1, N1, , (VL = line voltage), , Fig. (7.53 (ii)) shows the relevant phasor diagram. The, line voltages of the 3-phase system VRY, VYB and VBR, are balanced and are shown on the phasor diagram as a, closed equilateral triangle. The voltages across the two, halves of the centre tapped primary of the main, transformer, VCB and VYC are equal and in phase with, VYB. Clearly, VRC leads VYB by 90°. This voltage (i.e.,, Fig.(7.54), VRC) is applied to the primary of the teaser transformer., Therefore, the secondary voltage V1 of the teaser transformer will lead the, 179
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secondary voltage V2 by 90° as shown in Fig. (7.54). We now show that, magnitudes of V2 and V1 are equal,, , VRC = VRY cos 30° =, V1 =, , 3, V = 0.866 VL, 2 L, , N2, N2, N, VRC =, × 0.866 VL = 2 VL = V2, 0.866 N1, 0.866 N1, N1, , Thus voltages V1 and V2 constitute balanced 2-phase system consisting of two, voltages of equal magnitude having a phase difference of 90°., , 7.49 Applications of Transformers, There are four principal applications of transformers viz., (i) power transformers, (ii) distribution transformers, (iii) autotransformers, (iv) instrument transformers, (i), , Power Transformers. They are designed to operate with an almost, constant load which is equal to their rating. The maximum efficiency is, designed to be at full-load. This means that full-load winding copper losses, must be equal to the core losses., , (ii) Distribution Transformers. These transformers have variable load which, is usually considerably less than the full-load rating. Therefore, these are, designed to have their maximum efficiency at between 1/2 and 3/4 of fullload., (iii) Autotransformers. An autotransformer has only one winding and is used, in cases where the ratio of transformation (K), either step-up or step down,, differs little from 1. For the same output and voltage ratio, an, autotransformer requires less copper than an ordinary 2-winding, transformer. Autotransformers are used for starting induction motors, (reducing applied voltage during starting) and in boosters for raising the, voltage of feeders., (iv) Instrument transformers. Current and voltage transformers are used to, extend the range of a.c. instruments., , (a) Current transformer, A current transformer is a device that is used to measure high alternating current, in a conductor. Fig. (7.55) illustrates the principle of a current transformer. The, conductor carrying large current passes through a circular laminated iron core., The conductor constitutes a one-turn primary winding. The secondary winding, , 180
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consists of a large number of turns of much fine wire wrapped around the core, as shown. Due to transformer action, the secondary current is transformed to a, low value which can be measured by ordinary meters., Secondary current, IS = I P ×, , NP, NS, , For example, suppose that IP = 100 A in Fig. (7.55) and the ammeter is capable, of measuring a maximum of 1 A. Then,, , NS = N P ×, , IP, 100, = 1×, = 100, IS, 1, , (b) Voltage transformer, It is a device that is used to measure high alternating voltage. It is essentially a, step-down transformer having small number of secondary turns as shown in Fig., (7.56). The high alternating voltage to be measured is connected directly across, the primary. The low voltage winding (secondary winding) is connected to the, voltmeter. The power rating of a potential transformer is small (seldom exceeds, 300 W) since voltmeter is the only load on the transformer., , VP = VS ×, , NP, NS, , Fig.(7.55), , Fig.(7.56), , 181
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Chapter (8), , Three Phase Induction Motors, Introduction, The three-phase induction motors are the most widely used electric motors in, industry. They run at essentially constant speed from no-load to full-load., However, the speed is frequency dependent and consequently these motors are, not easily adapted to speed control. We usually prefer d.c. motors when large, speed variations are required. Nevertheless, the 3-phase induction motors are, simple, rugged, low-priced, easy to maintain and can be manufactured with, characteristics to suit most industrial requirements. In this chapter, we shall, focus our attention on the general principles of 3-phase induction motors., , 8.1 Three-Phase Induction Motor, Like any electric motor, a 3-phase induction motor has a stator and a rotor. The, stator carries a 3-phase winding (called stator winding) while the rotor carries a, short-circuited winding (called rotor winding). Only the stator winding is fed, from 3-phase supply. The rotor winding derives its voltage and power from the, externally energized stator winding through electromagnetic induction and, hence the name. The induction motor may be considered to be a transformer, with a rotating secondary and it can, therefore, be described as a “transformertype” a.c. machine in which electrical energy is converted into mechanical, energy., , Advantages, (i), (ii), (iii), (iv), (v), , It has simple and rugged construction., It is relatively cheap., It requires little maintenance., It has high efficiency and reasonably good power factor., It has self starting torque., , Disadvantages, (i), , It is essentially a constant speed motor and its speed cannot be changed, easily., (ii) Its starting torque is inferior to d.c. shunt motor., , 181
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8.2 Construction, A 3-phase induction motor has two main parts (i) stator and (ii) rotor. The rotor, is separated from the stator by a small air-gap which ranges from 0.4 mm to 4, mm, depending on the power of the motor., , 1., , Stator, It consists of a steel frame which encloses a, hollow, cylindrical core made up of thin, laminations of silicon steel to reduce, hysteresis and eddy current losses. A number, of evenly spaced slots are provided on the, inner periphery of the laminations [See Fig., (8.1)]. The insulated connected to form a, Fig.(8.1), balanced 3-phase star or delta connected, circuit. The 3-phase stator winding is wound for a definite number of poles as, per requirement of speed. Greater the number of poles, lesser is the speed of the, motor and vice-versa. When 3-phase supply is given to the stator winding, a, rotating magnetic field (See Sec. 8.3) of constant magnitude is produced. This, rotating field induces currents in the rotor by electromagnetic induction., , 2., , Rotor, The rotor, mounted on a shaft, is a hollow laminated core having slots on its, outer periphery. The winding placed in these slots (called rotor winding) may be, one of the following two types:, (i) Squirrel cage type, (ii) Wound type, (i), , Squirrel cage rotor. It consists of a laminated cylindrical core having, parallel slots on its outer periphery. One copper or aluminum bar is placed, in each slot. All these bars are joined at each end by metal rings called end, rings [See Fig. (8.2)]. This forms a permanently short-circuited winding, which is indestructible. The entire construction (bars and end rings), resembles a squirrel cage and hence the name. The rotor is not connected, electrically to the supply but has current induced in it by transformer action, from the stator., Those induction motors which employ squirrel cage rotor are called, squirrel cage induction motors. Most of 3-phase induction motors use, squirrel cage rotor as it has a remarkably simple and robust construction, enabling it to operate in the most adverse circumstances. However, it, suffers from the disadvantage of a low starting torque. It is because the, rotor bars are permanently short-circuited and it is not possible to add any, external resistance to the rotor circuit to have a large starting torque., , 182
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Fig.(8.2), , Fig.(8.3), , (ii) Wound rotor. It consists of a laminated cylindrical core and carries a 3phase winding, similar to the one on the stator [See Fig. (8.3)]. The rotor, winding is uniformly distributed in the slots and is usually star-connected., The open ends of the rotor winding are brought out and joined to three, insulated slip rings mounted on the rotor shaft with one brush resting on, each slip ring. The three brushes are connected to a 3-phase star-connected, rheostat as shown in Fig. (8.4). At starting, the external resistances are, included in the rotor circuit to give a large starting torque. These, resistances are gradually reduced to zero as the motor runs up to speed., , Fig.(8.4), The external resistances are used during starting period only. When the motor, attains normal speed, the three brushes are short-circuited so that the wound, rotor runs like a squirrel cage rotor., , 8.3 Rotating Magnetic Field Due to 3-Phase Currents, When a 3-phase winding is energized from a 3-phase supply, a rotating, magnetic field is produced. This field is such that its poles do no remain in a, fixed position on the stator but go on shifting their positions around the stator., For this reason, it is called a rotating Held. It can be shown that magnitude of, this rotating field is constant and is equal to 1.5 φm where φm is the maximum, flux due to any phase., , 183
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To see how rotating field is produced, consider a 2-pole,, 3i-phase winding as shown in Fig. (8.6 (i)). The three, phases X, Y and Z are energized from a 3-phase source, and currents in these phases are indicated as Ix, Iy and Iz, [See Fig. (8.6 (ii))]. Referring to Fig. (8.6 (ii)), the fluxes, produced by these currents are given by:, , φ x = φ m sin ωt, , Fig.(8.5), , φ y = φ m sin (ωt − 120°), φ z = φ m sin (ωt − 240°), Here φm is the maximum flux due to any phase. Fig. (8.5) shows the phasor, diagram of the three fluxes. We shall now prove that this 3-phase supply, produces a rotating field of constant magnitude equal to 1.5 φm., , Fig.(8.6), 184
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(i), , At instant 1 [See Fig. (8.6 (ii)) and Fig. (8.6 (iii))], the current in phase X is, zero and currents in phases Y and Z are equal, and opposite. The currents are flowing outward, in the top conductors and inward in the bottom, conductors. This establishes a resultant flux, towards right. The magnitude of the resultant, flux is constant and is equal to 1.5 φm as, proved under:, At instant 1, ωt = 0°. Therefore, the three, fluxes are given by;, , φ x = 0;, , φ y = φ m sin (− 120°) = −, , φz = φ m sin (− 240°) =, , Fig.(8.7), , 3, φ ;, 2 m, , 3, φ, 2 m, , The phasor sum of − φy and φz is the resultant flux φr [See Fig. (8.7)]. It is, clear that:, Resultant flux, φr = 2 ×, , 3, 60°, 3, 3, φ m cos, = 2×, φm ×, = 1.5 φ m, 2, 2, 2, 2, , (ii) At instant 2, the current is maximum, (negative) in φy phase Y and 0.5 maximum, (positive) in phases X and Y. The, magnitude of resultant flux is 1.5 φm as, proved under:, At instant 2, ωt = 30°. Therefore, the three, fluxes are given by;, , φm, 2, φ y = φ m sin (−90°) = −φ m, , Fig.(8.8), , φ x = φ m sin 30° =, , φ z = φ m sin (−210°) =, , φm, 2, , The phasor sum of φx, − φy and φz is the resultant flux φr, φ, 120° φ m, Phasor sum of φx and φz, φ'r = 2 × m cos, =, 2, 2, 2, φ, Phasor sum of φ'r and − φy, φ r = m + φ m = 1.5 φ m, 2, Note that resultant flux is displaced 30° clockwise from position 1., , 185
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(iii) At instant 3, current in phase Z is zero and, the currents in phases X and Y are equal and, opposite (currents in phases X and Y arc, 0.866 × max. value). The magnitude of, resultant flux is 1.5 φm as proved under:, At instant 3, ωt = 60°. Therefore, the three, fluxes are given by;, , 3, φ x = φ m sin 60° =, φ ;, 2 m, 3, φ y = φ m sin (− 60°) = −, φ ;, 2 m, φz = φ m sin (− 180°) = 0, , Fig.(8.9), , The resultant flux φr is the phasor sum of φx and − φy (Q φ z = 0 ) ., , φr = 2 ×, , 3, 60°, φ m cos, = 1.5 φ m, 2, 2, , Note that resultant flux is displaced 60° clockwise from position 1., (iv) At instant 4, the current in phase X is, maximum (positive) and the currents in phases, V and Z are equal and negative (currents in, phases V and Z are 0.5 × max. value). This, establishes a resultant flux downward as, shown under:, At instant 4, ωt = 90°. Therefore, the three, fluxes are given by;, , Fig.(7.10), , φ x = φ m sin 90° = φ m, φm, 2, φ, φz = φ m sin (−150°) = − m, 2, φ y = φ m sin (−30°) = −, , The phasor sum of φx, − φy and − φz is the resultant flux φr, φ, 120° φ m, Phasor sum of − φz and − φy, φ'r = 2 × m cos, =, 2, 2, 2, φ, Phasor sum of φ'r and φx, φ r = m + φ m = 1.5 φ m, 2, Note that the resultant flux is downward i.e., it is displaced 90° clockwise, from position 1., 186
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It follows from the above discussion that a 3-phase supply produces a rotating, field of constant value (= 1.5 φm, where φm is the maximum flux due to any, phase)., , Speed of rotating magnetic field, The speed at which the rotating magnetic field revolves is called the, synchronous speed (Ns). Referring to Fig. (8.6 (ii)), the time instant 4 represents, the completion of one-quarter cycle of alternating current Ix from the time, instant 1. During this one quarter cycle, the field has rotated through 90°. At a, time instant represented by 13 or one complete cycle of current Ix from the, origin, the field has completed one revolution. Therefore, for a 2-pole stator, winding, the field makes one revolution in one cycle of current. In a 4-pole, stator winding, it can be shown that the rotating field makes one revolution in, two cycles of current. In general, fur P poles, the rotating field makes one, revolution in P/2 cycles of current., ∴, or, , Cycles of current =, , Cycles of current per second =, , P, × revolutions of field, 2, P, × revolutions of field per second, 2, , Since revolutions per second is equal to the revolutions per minute (Ns) divided, by 60 and the number of cycles per second is the frequency f,, , ∴, or, , f=, , P Ns Ns P, ×, =, 2 60 120, , Ns =, , 120 f, P, , The speed of the rotating magnetic field is the same as the speed of the, alternator that is supplying power to the motor if the two have the same number, of poles. Hence the magnetic flux is said to rotate at synchronous speed., , Direction of rotating magnetic field, The phase sequence of the three-phase voltage applied to the stator winding in, Fig. (8.6 (ii)) is X-Y-Z. If this sequence is changed to X-Z-Y, it is observed that, direction of rotation of the field is reversed i.e., the field rotates, counterclockwise rather than clockwise. However, the number of poles and the, speed at which the magnetic field rotates remain unchanged. Thus it is necessary, only to change the phase sequence in order to change the direction of rotation of, the magnetic field. For a three-phase supply, this can be done by interchanging, any two of the three lines. As we shall see, the rotor in a 3-phase induction, motor runs in the same direction as the rotating magnetic field. Therefore, the, , 187
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direction of rotation of a 3-phase induction motor can be reversed by, interchanging any two of the three motor supply lines., , 8.4 Alternate Mathematical Analysis for Rotating, Magnetic Field, We shall now use another useful method to find, the magnitude and speed of the resultant flux due, to three-phase currents. The three-phase sinusoidal, currents produce fluxes φ1, φ2 and φ3 which vary, sinusoidally. The resultant flux at any instant will, be the vector sum of all the three at that instant., The fluxes are represented by three variable, Fig.(8.11), magnitude vectors [See Fig. (8.11)]. In Fig. (8.11),, the individual flux directions arc fixed but their magnitudes vary sinusoidally as, does the current that produces them. To find the magnitude of the resultant flux,, resolve each flux into horizontal and vertical components and then find their, vector sum., , φh = φ m cos t ωt − φ m cos(ωt − 120°) cos 60° − φ m cos(ωt − 240°) cos 60°, 3, = φm cos ωt, 2, 3, φ v = 0 − φm cos(ωt − 120°) cos 60° + φ m cos(ωt − 240°) cos 60° = φ m sin ωt, 2, The resultant flux is given by;, , [, , ], , 1/ 2, 3, 3, φr = φ 2h + φ 2v = φ m cos 2 ωt + (− sin ωt ) 2, = φ m = 1.5 φ m = Constant, 2, 2, , Thus the resultant flux has constant magnitude (= 1.5 φm) and does not change, with time. The angular displacement of φr relative to, the OX axis is, , ∴, , 3, φ v 2 φ m sin ωt, = tan ωt, =, tan θ =, φh 3, φ cos ωt, 2 m, θ = ωt, , Fig.(8.12), , Thus the resultant magnetic field rotates at constant angular velocity ω( = 2 πf), rad/sec. For a P-pole machine, the rotation speed (ωm) is, , ωm =, , 2, ω rad / sec, P, , 188
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2 π Ns 2, = × 2πf, 60, P, , or, , ∴, , Ns =, , ... N s is in r.p.m., , 120 f, P, , Thus the resultant flux due to three-phase currents is of constant value (= 1.5 φm, where φm is the maximum flux in any phase) and this flux rotates around the, stator winding at a synchronous speed of 120 f/P r.p.m., For example, for a 6-pole, 50 Hz, 3-phase induction motor, N, = 120 × 50/6 =, 1000 r.p.m. It means that flux rotates around the stator at a speed of 1000 r.p.m., , 8.5 Principle of Operation, Consider a portion of 3-phase induction motor as shown in Fig. (8.13). The, operation of the motor can be explained as under:, (i) When, 3-phase, stator, winding is energized from, a 3-phase supply, a, rotating magnetic field is, set up which rotates round, the stator at synchronous, speed Ns (= 120 f/P)., (ii) The rotating field passes, through the air gap and, cuts the rotor conductors,, Fig.(1-), which, as, yet,, are, stationary. Due to the relative speed between the rotating flux and the, stationary rotor, e.m.f.s are induced in the rotor conductors. Since the, rotor circuit is short-circuited, currents start flowing in the rotor, conductors., (iii) The current-carrying rotor conductors are placed in the magnetic field, produced by the stator. Consequently, mechanical force acts on the rotor, conductors. The sum of the mechanical forces on all the rotor conductors, produces a torque which tends to move the rotor in the same direction as, the rotating field., (iv) The fact that rotor is urged to follow the stator field (i.e., rotor moves in, the direction of stator field) can be explained by Lenz’s law. According, to this law, the direction of rotor currents will be such that they tend to, oppose the cause producing them. Now, the cause producing the rotor, currents is the relative speed between the rotating field and the stationary, rotor conductors. Hence to reduce this relative speed, the rotor starts, running in the same direction as that of stator field and tries to catch it., 189
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8.6 Slip, We have seen above that rotor rapidly accelerates in the direction of rotating, field. In practice, the rotor can never reach the speed of stator flux. If it did,, there would be no relative speed between the stator field and rotor conductors,, no induced rotor currents and, therefore, no torque to drive the rotor. The, friction and windage would immediately cause the rotor to slow down. Hence,, the rotor speed (N) is always less than the suitor field speed (Ns). This difference, in speed depends upon load on the motor., The difference between the synchronous speed Ns of the rotating stator field and, the actual rotor speed N is called slip. It is usually expressed as a percentage of, synchronous speed i.e.,, % age slip, s =, , Ns − N, × 100, Ns, , (i) The quantity Ns − N is sometimes called slip speed., (ii) When the rotor is stationary (i.e., N = 0), slip, s = 1 or 100 %., (iii) In an induction motor, the change in slip from no-load to full-load is, hardly 0.1% to 3% so that it is essentially a constant-speed motor., , 8.7 Rotor Current Frequency, The frequency of a voltage or current induced due to the relative speed between, a vending and a magnetic field is given by the general formula;, Frequency =, , NP, 120, , where N = Relative speed between magnetic field and the winding, P = Number of poles, For a rotor speed N, the relative speed between the rotating flux and the rotor is, Ns − N. Consequently, the rotor current frequency f' is given by;, , ( N s − N)P, 120, s Ns P, =, 120, , f '=, , N −N, , Q s = s, , Ns , , N P, , Q f = s , 120 , , , = sf, , i.e., Rotor current frequency = Fractional slip x Supply frequency, (i) When the rotor is at standstill or stationary (i.e., s = 1), the frequency of, rotor current is the same as that of supply frequency (f' = sf = 1× f = f)., 190
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(ii) As the rotor picks up speed, the relative speed between the rotating flux, and the rotor decreases. Consequently, the slip s and hence rotor current, frequency decreases., Note. The relative speed between the rotating field and stator winding is Ns − 0, = Ns. Therefore, the frequency of induced current or voltage in the stator, winding is f = Ns P/120—the supply frequency., , 8.8 Effect of Slip on The Rotor Circuit, When the rotor is stationary, s = 1. Under these conditions, the per phase rotor, e.m.f. E2 has a frequency equal to that of supply frequency f. At any slip s, the, relative speed between stator field and the rotor is decreased. Consequently, the, rotor e.m.f. and frequency are reduced proportionally to sEs and sf respectively., At the same time, per phase rotor reactance X2, being frequency dependent, is, reduced to sX2., Consider a 6-pole, 3-phase, 50 Hz induction motor. It has synchronous speed Ns, = 120 f/P = 120 × 50/6 = 1000 r.p.m. At standsill, the relative speed between, stator flux and rotor is 1000 r.p.m. and rotor e.m.f./phase = E2(say). If the fullload speed of the motor is 960 r.p.m., then,, , s=, (i), , 1000 − 960, = 0.04, 1000, , The relative speed between stator flux and the rotor is now only 40 r.p.m., Consequently, rotor e.m.f./phase is reduced to:, , E2 ×, , 40, = 0.04 E 2, 1000, , or, , sE 2, , (ii) The frequency is also reduced in the same ratio to:, , 50 ×, , 40, = 50 × 0.04, 1000, , or, , sf, , (iii) The per phase rotor reactance X2 is likewise reduced to:, , X2 ×, , 40, = 0.04X 2, 1000, , or, , sX 2, , Thus at any slip s,, Rotor e.m.f./phase = sE2, Rotor reactance/phase = sX2, Rotor frequency = sf, where E2,X2 and f are the corresponding values at standstill., , 191
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8.9 Rotor Current, Fig. (8.14) shows the circuit of a 3-phase induction motor at any slip s. The rotor, is assumed to be of wound type and star connected. Note that rotor e.m.f./phase, and rotor reactance/phase are s E2 and sX2 respectively. The rotor, resistance/phase is R2 and is independent of frequency and, therefore, does not, depend upon slip. Likewise, stator winding values R1 and X1 do not depend, upon slip., , Fig.(8.14), Since the motor represents a balanced 3-phase load, we need consider one phase, only; the conditions in the other two phases being similar., At standstill. Fig. (8.15 (i)) shows one phase of the rotor circuit at standstill., Rotor current/phase, I 2 =, Rotor p.f., cos φ 2 =, , E2, E2, =, Z2, R 22 + X 22, , R2, R2, =, Z2, R 22 + X 22, , Fig.(8.15), When running at slip s. Fig. (8.15 (ii)) shows one phase of the rotor circuit, when the motor is running at slip s., Rotor current, I'2 =, , sE 2, sE 2, =, Z' 2, R 22 + (sX 2 )2, 192
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Rotor p.f., cos φ'2 =, , R2, R2, =, Z'2, R 22 + (sX 2 )2, , 8.10 Rotor Torque, The torque T developed by the rotor is directly proportional to:, (i) rotor current, (ii) rotor e.m.f., (iii) power factor of the rotor circuit, , ∴, , T ∝ E 2 I 2 cos φ 2, T = K E 2 I 2 cos φ 2, , or, where, , I2 = rotor current at standstill, E2 = rotor e.m.f. at standstill, cos φ2 = rotor p.f. at standstill, , Note. The values of rotor e.m.f., rotor current and rotor power factor are taken, for the given conditions., , 8.11 Starting Torque (Ts), Let, , E2 = rotor e.m.f. per phase at standstill, X2 = rotor reactance per phase at standstill, R2 = rotor resistance per phase, Rotor impedance/phase, Z 2 = R 22 + X 22, Rotor current/phase, I 2 =, Rotor p.f., cos φ 2 =, ∴, , ...at standstill, , E2, E2, =, Z2, R 22 + X 22, , ...at standstill, , R2, R2, =, Z2, R 22 + X 22, , ...at standstill, , Starting torque, Ts = K E 2 I 2 cos φ 2, , = K E2 ×, =, , K E 22 R 2, R 22 + X 22, , 193, , E2, R 22 + X 22, , ×, , R2, R 22 + X 22
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Generally, the stator supply voltage V is constant so that flux per pole φ set up, by the stator is also fixed. This in turn means that e.m.f. E2 induced in the rotor, will be constant., , ∴, , Ts =, , K1 R 2, R 22 + X 22, , =, , K1 R 2, Z 22, , where K1 is another constant., It is clear that the magnitude of starting torque would depend upon the relative, values of R2 and X2 i.e., rotor resistance/phase and standstill rotor, reactance/phase., It can be shown that K = 3/2 π Ns., , ∴, , E 22 R 2, 3, Ts =, ⋅, 2π N s R 22 + X 22, , Note that here Ns is in r.p.s., , 8.12 Condition for Maximum Starting Torque, It can be proved that starting torque will be maximum when rotor, resistance/phase is equal to standstill rotor reactance/phase., Now, , Ts =, , K1 R 2, , (i), , R 22 + X 22, , Differentiating eq. (i) w.r.t. R2 and equating the result to zero, we get,, , , dTs, R 2 (2 R 2 ) , 1, =0, = K1 2, −, dR 2, R 2 + X 22 R 2 + X 2 2 , , 2, 2 , , (, , or, , R 22 + X 22 = 2R 22, , or, , R 2 = X2, , ), , Hence starting torque will be maximum when:, Rotor resistance/phase = Standstill rotor reactance/phase, Under the condition of maximum starting torque, φ2 = 45° and rotor power, factor is 0.707 lagging [See Fig. (8.16 (ii))]., Fig. (8.16 (i)) shows the variation of starting torque with rotor resistance. As the, rotor resistance is increased from a relatively low value, the starting torque, increases until it becomes maximum when R2 = X2. If the rotor resistance is, increased beyond this optimum value, the starting torque will decrease., 194
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Fig.(8.16), , 8.13 Effect of Change of Supply Voltage, Ts =, , K E 22 R 2, R 22 + X 22, , Since E2 ∝ Supply voltage V, , ∴, , Ts =, , K2 V2 R 2, R 22 + X 22, , where K2 is another constant., , ∴, , Ts ∝ V 2, , Therefore, the starting torque is very sensitive to changes in the value of supply, voltage. For example, a drop of 10% in supply voltage will decrease the starting, torque by about 20%. This could mean the motor failing to start if it cannot, produce a torque greater than the load torque plus friction torque., , 8.14 Starting Torque of 3-Phase Induction Motors, The rotor circuit of an induction motor has low resistance and high inductance., At starting, the rotor frequency is equal to the stator frequency (i.e., 50 Hz) so, that rotor reactance is large compared with rotor resistance. Therefore, rotor, current lags the rotor e.m.f. by a large angle, the power factor is low and, consequently the starting torque is small. When resistance is added to the rotor, circuit, the rotor power factor is improved which results in improved starting, torque. This, of course, increases the rotor impedance and, therefore, decreases, the value of rotor current but the effect of improved power factor predominates, and the starting torque is increased., (i), , Squirrel-cage motors. Since the rotor bars are permanently shortcircuited, it is not possible to add any external resistance in the rotor circuit, at starting. Consequently, the stalling torque of such motors is low. Squirrel, 195
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cage motors have starting torque of 1.5 to 2 times the full-load value with, starting current of 5 to 9 times the full-load current., (ii) Wound rotor motors. The resistance of the rotor circuit of such motors, can be increased through the addition of external resistance. By inserting, the proper value of external resistance (so that R2 = X2), maximum starting, torque can be obtained. As the motor accelerates, the external resistance is, gradually cut out until the rotor circuit is short-circuited on itself for, running conditions., , 8.15 Motor Under Load, Let us now discuss the behaviour of 3-phase induction motor on load., (i) When we apply mechanical load to the shaft of the motor, it will begin to, slow down and the rotating flux will cut the rotor conductors at a higher, and higher rate. The induced voltage and resulting current in rotor, conductors will increase progressively, producing greater and greater, torque., (ii) The motor and mechanical load will soon reach a state of equilibrium, when the motor torque is exactly equal to the load torque. When this, state is reached, the speed will cease to drop any more and the motor will, run at the new speed at a constant rate., (iii) The drop in speed of the induction motor on increased load is small. It is, because the rotor impedance is low and a small decrease in speed, produces a large rotor current. The increased rotor current produces a, higher torque to meet the increased load on the motor. This is why, induction motors are considered to be constant-speed machines., However, because they never actually turn at synchronous speed, they, are sometimes called asynchronous machines., Note that change in load on the induction motor is met through the, adjustment of slip. When load on the motor increases, the slip increases, slightly (i.e., motor speed decreases slightly). This results in greater, relative speed between the rotating flux and rotor conductors., Consequently, rotor current is increased, producing a higher torque to, meet the increased load. Reverse happens should the load on the motor, decrease., (iv) With increasing load, the increased load currents I'2 are in such a, direction so as to decrease the stator flux (Lenz’s law), thereby, decreasing the counter e.m.f. in the stator windings. The decreased, counter e.m.f. allows motor stator current (I1) to increase, thereby, increasing the power input to the motor. It may be noted that action of, the induction motor in adjusting its stator or primary current with, , 196
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changes of current in the rotor or secondary is very much similar to the, changes occurring in transformer with changes in load., , Fig.(8.17), , 8.16 Torque Under Running Conditions, Let the rotor at standstill have per phase induced e.m.f. E2, reactance X2 and, resistance R2. Then under running conditions at slip s,, Rotor e.m.f./phase, E'2 = sE2, Rotor reactance/phase, X'2 = sX2, Rotor impedance/phase, Z'2 = R 22 + (sX 2 )2, Rotor current/phase, I'2 =, Rotor p.f., cos φ'm =, , E '2, sE 2, =, Z' 2, R 22 + (sX 2 )2, R2, R 22 + (sX 2 )2, , Fig.(8.18), Running Torque, Tr ∝ E'2 I'2 cos φ'2, , ∝ φ I'2 cos φ'2, , 197, , (Q E '2 ∝ φ)
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∝ φ×, , s E2, R 22 + (s X 2 )2, , ×, , R2, R 22 + (s X 2 )2, , φ s E2 R 2, , ∝, , R 22 + (s X 2 )2, K φ s E2 R 2, , =, , R 22 + (s X 2 )2, K1 s E 22 R 2, , =, , (Q E 2 ∝ φ), , R 22 + (s X 2 )2, , If the stator supply voltage V is constant, then stator flux and hence E2 will be, constant., , ∴, , Tr =, , K2 s R2, , R 22 + (s X 2 )2, , where K2 is another constant., It may be seen that running torque is:, (i) directly proportional to slip i.e., if slip increases (i.e., motor speed, decreases), the torque will increase and vice-versa., (ii) directly proportional to square of supply voltage (Q E 2 ∝ V ) ., It can be shown that value of K1 = 3/2 π Ns where Ns is in r.p.s., , ∴, , s E 22 R 2, s E 22 R 2, 3, 3, Tr =, ⋅, =, ⋅, 2π N s R 2 + (s X )2 2π N s (Z' )2, 2, 2, 2, , At starting, s = 1 so that starting torque is, , E 22 R 2, 3, Ts =, ⋅, 2π N s R 22 + X 22, , 8.17 Maximum Torque under Running Conditions, Tr =, , K2 s R 2, , (i), , R 22 + s 2 X 22, , In order to find the value of rotor resistance that gives maximum torque under, running conditions, differentiate exp. (i) w.r.t. s and equate the result to zero i.e.,, , [ (, , ), , ], , dTr K 2 R 2 R 22 + s 2 X 22 − 2s X 22 (s R 2 ), =, =0, ds, 2, 2, 2 2, R2 + s X2, , (, , 198, , )
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or, , (R, , or, , R 22 = s 2 X 22, , or, , R 2 = s X2, , 2, 2, 2+s, , ), , X 22 − 2s X 22 = 0, , Thus for maximum torque (Tm) under running conditions :, Rotor resistance/phase = Fractional slip × Standstill rotor reactance/phase, Now, , Tr ∝, , s R2, , … from exp. (i) above, , R 22 + s 2 X 22, , For maximum torque, R2 = s X2. Putting R2 = s X2 in the above expression, the, maximum torque Tm is given by;, , Tm ∝, , 1, 2 X2, , Slip corresponding to maximum torque, s = R2/X2., It can be shown that:, , E 22, 3, Tm =, ⋅, N-m, 2π N s 2 X 2, It is evident from the above equations that:, (i) The value of rotor resistance does not alter the value of the maximum, torque but only the value of the slip at which it occurs., (ii) The maximum torque varies inversely as the standstill reactance., Therefore, it should be kept as small as possible., (iii) The maximum torque varies directly with the square of the applied, voltage., (iv) To obtain maximum torque at starting (s = 1), the rotor resistance must, be made equal to rotor reactance at standstill., , 8.18 Torque-Slip Characteristics, As shown in Sec. 8.16, the motor torque under running conditions is given by;, , T=, , K2 s R 2, R 22 + s 2 X 22, , If a curve is drawn between the torque and slip for a particular value of rotor, resistance R2, the graph thus obtained is called torque-slip characteristic. Fig., (8.19) shows a family of torque-slip characteristics for a slip-range from s = 0 to, s = 1 for various values of rotor resistance., , 199
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Fig.(8.19), The following points may be noted carefully:, (i) At s = 0, T = 0 so that torque-slip curve starts from the origin., (ii) At normal speed, slip is small so that s X2 is negligible as compared to, R2., , ∴, , T ∝ s/R2, , ∝s, , ... as R2 is constant, , Hence torque slip curve is a straight line from zero slip to a slip that, corresponds to full-load., (iii) As slip increases beyond full-load slip, the torque increases and becomes, maximum at s = R2/X2. This maximum torque in an induction motor is, called pull-out torque or break-down torque. Its value is at least twice the, full-load value when the motor is operated at rated voltage and, frequency., (iv), to maximum torque, the term, 2 2, 2, s X 2 increases very rapidly so that R 2 may be neglected as compared, to s 2 X 22 ., , ∴, , T ∝ s / s 2 X 22, ∝ 1/ s, , ... as X2 is constant, , Thus the torque is now inversely proportional to slip. Hence torque-slip, curve is a rectangular hyperbola., (v) The maximum torque remains the same and is independent of the value, of rotor resistance. Therefore, the addition of resistance to the rotor, circuit does not change the value of maximum torque but it only changes, the value of slip at which maximum torque occurs., , 8.19 Full-Load, Starting and Maximum Torques, 200
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Tf ∝, Ts ∝, Tm ∝, , s R2, , R 22, , + (s X 2 )2, , R2, R 22 + X 22, 1, 2 X2, , Note that s corresponds to full-load slip., (i), , ∴, , Tm R 22 + (s X 2 )2, =, Tf, 2s R 2 X 2, , Dividing the numerator and denominator on R.H.S. by X 22 , we get,, , Tm (R 2 / X 2 )2 + s 2 a 2 + s 2, =, =, Tf, 2s(R 2 / X 2 ), 2a s, R2, Rotor resistance/phase, =, X 2 Standstill rotor reactance/phase, , where, , a=, , (ii), , Tm R 22 + X 22, =, Ts 2 R 2 X 2, , Dividing the numerator and denominator on R.H.S. by X 22 , we get,, , Tm (R 2 / X 2 )2 + 1 a 2 + 1, =, =, Tf, 2(R 2 / X 2 ), 2a, where, , a=, , R2, Rotor resistance/phase, =, X 2 Standstill rotor reactance/phase, , 8.20 Induction Motor and Transformer Compared, An induction motor may be considered to be a transformer with a rotating shortcircuited secondary. The stator winding corresponds to transformer primary and, rotor winding to transformer secondary. However, the following differences, between the two are worth noting:, (i) Unlike a transformer, the magnetic circuit of a 3-phase induction motor has, an air gap. Therefore, the magnetizing current in a 3-phase induction motor, is much larger than that of the transformer. For example, in an induction, motor, it may be as high as 30-50 % of rated current whereas it is only 15% of rated current in a transformer., (ii) In an induction motor, there is an air gap and the stator and rotor windings, are distributed along the periphery of the air gap rather than concentrated, 201
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on a core as in a transformer. Therefore, the leakage reactances of stator, and rotor windings are quite large compared to that of a transformer., (iii) In an induction motor, the inputs to the stator and rotor are electrical but the, output from the rotor is mechanical. However, in a transformer, input as, well as output is electrical., (iv) The main difference between the induction motor and transformer lies in, the fact that the rotor voltage and its frequency are both proportional to slip, s. If f is the stator frequency, E2 is the per phase rotor e.m.f. at standstill, and X2 is the standstill rotor reactance/phase, then at any slip s, these values, are:, Rotor e.m.f./phase, E'2 = s E2, Rotor reactance/phase, X'2 = sX2, Rotor frequency, f' = sf, , 8.21 Speed Regulation of Induction Motors, Like any other electrical motor, the speed regulation of an induction motor is, given by:, % age speed regulation =, where, , N 0 − N F.L., × 100, N F.L., , N0 = no-load speed of the motor, NF.L. = full-load speed of the motor, , If the no-load speed of the motor is 800 r.p.m. and its fall-load speed in 780, r.p.m., then change in speed is 800 − 780 = 20 r.p.m. and percentage speed, regulation = 20 × 100/780 = 2.56%., At no load, only a small torque is required to overcome the small mechanical, losses and hence motor slip is small i.e., about 1%. When the motor is fully, loaded, the slip increases slightly i.e., motor speed decreases slightly. It is, because rotor impedance is low and a small decrease in speed produces a large, rotor current. The increased rotor current produces a high torque to meet the full, load on the motor. For this reason, the change in speed of the motor from noload to full-load is small i.e., the speed regulation of an induction motor is low., The speed regulation of an induction motor is 3% to 5%. Although the motor, speed does decrease slightly with increased load, the speed regulation is low, enough that the induction motor is classed as a constant-speed motor., , 8.22 Speed Control of 3-Phase Induction Motors, , 202
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N = (1 − s) N s = (1 − s), , 120 f, P, , (i), , An inspection of eq. (i) reveals that the speed N of an induction motor can be, varied by changing (i) supply frequency f (ii) number of poles P on the stator, and (iii) slip s. The change of frequency is generally not possible because the, commercial supplies have constant frequency. Therefore, the practical methods, of speed control are either to change the number of stator poles or the motor slip., , 1., , Squirrel cage motors, The speed of a squirrel cage motor is changed by changing the number of stator, poles. Only two or four speeds are possible by this method. Two-speed motor, has one stator winding that may be switched through suitable control equipment, to provide two speeds, one of which is half of the other. For instance, the, winding may be connected for either 4 or 8 poles, giving synchronous speeds of, 1500 and 750 r.p.m. Four-speed motors are equipped with two separate stator, windings each of which provides two speeds. The disadvantages of this method, are:, (i) It is not possible to obtain gradual continuous speed control., (ii) Because of the complications in the design and switching of the, interconnections of the stator winding, this method can provide a, maximum of four different synchronous speeds for any one motor., , 2., , Wound rotor motors, The speed of wound rotor motors is changed by changing the motor slip. This, can be achieved by;, (i) varying the stator line voltage, (ii) varying the resistance of the rotor circuit, (iii) inserting and varying a foreign voltage in the rotor circuit, , 8.23 Power Factor of Induction Motor, Like any other a.c. machine, the power factor of an induction motor is given by;, Power factor, cos φ =, , Active component of current (I cos φ), Total current (I), , The presence of air-gap between the stator and rotor of an induction motor, greatly increases the reluctance of the magnetic circuit. Consequently, an, induction motor draws a large magnetizing current (Im) to produce the required, flux in the air-gap., (i) At no load, an induction motor draws a large magnetizing current and a, small active component to meet the no-load losses. Therefore, the, induction motor takes a high no-load current lagging the applied voltage, 203
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by a large angle. Hence the power factor of an induction motor on no, load is low i.e., about 0.1 lagging., (ii) When an induction motor is loaded, the active component of current, increases while the magnetizing component remains about the same., Consequently, the power factor of the motor is increased. However,, because of the large value of magnetizing current, which is present, regardless of load, the power factor of an induction motor even at fullload seldom exceeds 0.9 lagging., , 8.24 Power Stages in an Induction Motor, The input electric power fed to the stator of the motor is converted into, mechanical power at the shaft of the motor. The various losses during the energy, conversion are:, , 1., , Fixed losses, (i) Stator iron loss, (ii) Friction and windage loss, The rotor iron loss is negligible because the frequency of rotor currents under, normal running condition is small., , 2., , Variable losses, (i) Stator copper loss, (ii) Rotor copper loss, Fig. (8.20) shows how electric power fed to the stator of an induction motor, suffers losses and finally converted into mechanical power., The following points may be noted from the above diagram:, (i) Stator input, Pi = Stator output + Stator losses, = Stator output + Stator Iron loss + Stator Cu loss, (ii) Rotor input, Pr = Stator output, It is because stator output is entirely transferred to the rotor through airgap by electromagnetic induction., (iii) Mechanical power available, Pm = Pr − Rotor Cu loss, This mechanical power available is the gross rotor output and will, produce a gross torque Tg., (iv) Mechanical power at shaft, Pout = Pm − Friction and windage loss, Mechanical power available at the shaft produces a shaft torque Tsh., Clearly, Pm − Pout = Friction and windage loss, , 204
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Fig.(8.20), , 8.25 Induction Motor Torque, The mechanical power P available from any electric motor can be expressed as:, , P=, where, , 2π N T, 60, , watts, , N = speed of the motor in r.p.m., T = torque developed in N-m, , ∴, , T=, , 60 P, P, = 9.55, N-m, 2π N, N, , If the gross output of the rotor of an induction motor is Pm and its speed is N, r.p.m., then gross torque T developed is given by:, , Tg = 9.55, Similarly,, , Pm, N-m, N, , Tsh = 9.55, , Pout, N-m, N, , Note. Since windage and friction loss is small, Tg = Tsh,. This assumption hardly, leads to any significant error., , 8.26 Rotor Output, If Tg newton-metre is the gross torque developed and N r.p.m. is the speed of the, rotor, then,, Gross rotor output =, , 2π N Tg, 60, , watts, , If there were no copper losses in the rotor, the output would equal rotor input, and the rotor would run at synchronous speed Ns., , 205
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2π N s Tg, 60, , ∴, , Rotor input =, , ∴, , Rotor Cu loss = Rotor input − Rotor output, , =, (i), ∴, , watts, , 2π Tg, ( N s − N), 60, , Rotor Cu loss N s − N, =s, =, Rotor input, Ns, Rotor Cu loss = s × Rotor input, , (ii) Gross rotor output, Pm = Rotor input − Rotor Cu loss, = Rotor input − s × Rotor input, ∴, Pm = Rotor input (1 − s), , Gross rotor output, N, =1 − s =, Rotor input, Ns, Rotor Cu loss, s, =, Gross rotor output 1 − s, , (iii), (iv), , It is clear that if the input power to rotor is Pr then s Pr is lost as rotor Cu loss, and the remaining (1 − s)Pr is converted into mechanical power. Consequently,, induction motor operating at high slip has poor efficiency., Note., , Gross rotor output, =1 − s, Rotor input, If the stator losses as well as friction and windage losses arc neglected, then,, Gross rotor output = Useful output, Rotor input = Stator input, , ∴, , Useful output, = 1 − s = Efficiency, Stator output, , Hence the approximate efficiency of an induction motor is 1 − s. Thus if the slip, of an induction motor is 0.125, then its approximate efficiency is = 1 − 0.125 =, 0.875 or 87.5%., , 8.27 Induction Motor Torque Equation, The gross torque Tg developed by an induction motor is given by;, , 206
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Tg =, =, , Rotor input, 2π N s, , ... N s is r.p.s., , 60 × Rotor input, 2π N s, , (See Sec. 8.26), , ... N s is r.p.s., , 2, Rotor Cu loss 3(I'2 ) R 2, =, Rotor input =, s, s, , Now, , (i), , As shown in Sec. 8.16, under running conditions,, , I'2 =, where, , s E2, R 22 + (s X 2 ) 2, , =, , K = Transformation ratio =, ∴, , Rotor input = 3 ×, , s K E1, R 22 + (s X 2 ) 2, , Rotor turn s/phase, Stator tur ns/phase, , s 2 E 22 R 2, , 3 s E 22 R 2, 1, ×, =, R 22 + (s X 2 ) 2 s R 22 + (s X 2 ) 2, (Putting me value of I'2 in eq.(i)), , s 2 K 2 E12 R 2, , 2 2, 1 3 s K E1 R 2, Rotor input = 3 × 2, × =, R 2 + (s X 2 ) 2 s R 22 + (s X 2 ) 2, , Also, , (Putting me value of I'2 in eq.(i)), , ∴, , s E 22 R 2, Rotor input, 3, Tg =, =, ×, 2π N s, 2π N s R 22 + (s X 2 ) 2, s K 2 E12 R 2, 3, =, ×, 2π N s R 22 + (s X 2 ) 2, , ...in terms of E2, , ...in terms of E1, , Note that in the above expressions of Tg, the values E1, E2, R2 and X2 represent, the phase values., , 8.28 Performance Curves of Squirrel-Cage Motor, The performance curves of a 3-phase induction motor indicate the variations of, speed, power factor, efficiency, stator current and torque for different values of, load. However, before giving the performance curves in one graph, it is, desirable to discuss the variation of torque, and stator current with slip., , (i) Variation of torque and stator current with slip, Fig. (8.21) shows the variation of torque and stator current with slip for a, standard squirrel-cage motor. Generally, the rotor resistance is low so that full207
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load current occurs at low slip. Then even at full-load f' (= sf) and. therefore, X'2, (= 2π f' L2) are low. Between zero and full-load, rotor power factor (= cos φ'2), and rotor impedance (= Z'2) remain practically constant. Therefore, rotor current, I'2(E'2/Z'2) and, therefore, torque (Tr) increase directly with the slip. Now stator, current I1 increases in proportion to I'2. This is shown in Fig. (8.21) where Tr and, I1 are indicated as straight lines from no-load to full-load. As load and slip are, increased beyond full-load, the increase in rotor reactance becomes appreciable., The increasing value of rotor impedance not only decreases the rotor power, factor cos φ'2 (= R2/Z'2) but also lowers the rate of increase of rotor current. As a, result, the torque Tr and stator current I1 do not increase directly with slip as, indicated in Fig. (8.21). With the decreasing power factor and the lowered rate, of increase in rotor current, the stator current I1 and torque Tr increase at a lower, rate. Finally, torque Tr reaches the maximum value at about 25% slip in the, standard squirrel cage motor. This maximum value of torque is called the, pullout torque or breakdown torque. If the load is increased beyond the, breakdown point, the decrease in rotor power factor is greater than the increase, in rotor current, resulting in a decreasing torque. The result is that motor slows, down quickly and comes to a stop., , Fig.(8.21), In Fig. (8.21), the value of torque at starting (i.e., s = 100%) is 1.5 times the fullload torque. The starting current is about five times the full-load current. The, motor is essentially a constant-speed machine having speed characteristics about, the same as a d.c. shunt motor., , (ii) Performance curves, Fig. (8.22) shows the performance curves of 3-phase squirrel cage induction, motor., , 208
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Fig.(8.22), The following points may be noted:, (a) At no-load, the rotor lags behind the stator flux by only a small amount,, since the only torque required is that needed to overcome the no-load, losses. As mechanical load is added, the rotor speed decreases. A decrease, in rotor speed allows the constant-speed rotating field to sweep across the, rotor conductors at a faster rate, thereby inducing large rotor currents. This, results in a larger torque output at a slightly reduced speed. This explains, for speed-load curve in Fig. (8.22)., (b) At no-load, the current drawn by an induction motor is largely a, magnetizing current; the no-load current lagging the applied voltage by a, large angle. Thus the power factor of a lightly loaded induction motor is, very low. Because of the air gap, the reluctance of the magnetic circuit is, high, resulting in a large value of no-load current as compared with a, transformer. As load is added, the active or power component of current, increases, resulting in a higher power factor. However, because of the large, value of magnetizing current, which is present regardless of load, the power, factor of an induction motor even at full-load seldom exceeds 90%. Fig., (8.22) shows the variation of power factor with load of a typical squirrelcage induction motor., Output, (c) Efficiency =, Output + Losses, The losses occurring in a 3-phase induction motor are Cu losses in stator, and rotor windings, iron losses in stator and rotor core and friction and, windage losses. The iron losses and friction and windage losses are almost, independent of load. Had I2R been constant, the efficiency of the motor, would have increased with load. But I2R loss depends upon load., 209
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Therefore, the efficiency of the motor increases with load but the curve is, dropping at high loads., (d) At no-load, the only torque required is that needed to overcome no-load, losses. Therefore, stator draws a small current from the supply. As, mechanical load is added, the rotor speed decreases. A decrease in rotor, speed allows the constant-speed rotating field to sweep across the rotor, conductors at a faster rate, thereby inducing larger rotor currents. With, increasing loads, the increased rotor currents are in such a direction so as to, decrease the staler flux, thereby temporarily decreasing the counter e.m.f., in the stator winding. The decreased counter e.m.f. allows more stator, current to flow., (e) Output = Torque × Speed, Since the speed of the motor does not change appreciably with load, the, torque increases with increase in load., , 8.29 Equivalent Circuit of 3-Phase Induction Motor at, Any Slip, In a 3-phase induction motor, the stator winding is connected to 3-phase supply, and the rotor winding is short-circuited. The energy is transferred magnetically, from the stator winding to the short-circuited, rotor winding. Therefore, an, induction motor may be considered to be a transformer with a rotating secondary, (short-circuited). The stator winding corresponds to transformer primary and the, rotor finding corresponds to transformer secondary. In view of the similarity of, the flux and voltage conditions to those in a transformer, one can expect that the, equivalent circuit of an induction motor will be similar to that of a transformer., Fig. (8.23) shows the equivalent circuit (though not the only one) per phase for, an induction motor. Let us discuss the stator and rotor circuits separately., , Fig.(8.23), Stator circuit. In the stator, the events are very similar to those in the, transformer primal y. The applied voltage per phase to the stator is V1 and R1, and X1 are the stator resistance and leakage reactance per phase respectively., The applied voltage V1 produces a magnetic flux which links the stator winding, (i.e., primary) as well as the rotor winding (i.e., secondary). As a result, self-, , 210
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induced e.m.f. E1 is induced in the stator winding and mutually induced e.m.f., E'2(= s E2 = s K E1 where K is transformation ratio) is induced in the rotor, winding. The flow of stator current I1 causes voltage drops in R1 and X1., , ∴, , V1 = − E1 + I1 (R 1 + j X1 ), , ...phasor sum, , When the motor is at no-load, the stator winding draws a current I0. It has two, components viz., (i) which supplies the no-load motor losses and (ii), magnetizing component Im which sets up magnetic flux in the core and the airgap. The parallel combination of Rc and Xm, therefore, represents the no-load, motor losses and the production of magnetic flux respectively., , I0 = Iw + Im, Rotor circuit. Here R2 and X2 represent the rotor resistance and standstill rotor, reactance per phase respectively. At any slip s, the rotor reactance will be s X2, The induced voltage/phase in the rotor is E'2 = s E2 = s K E1. Since the rotor, winding is short-circuited, the whole of e.m.f. E'2 is used up in circulating the, rotor current I'2., , ∴, , E ' 2 = I' 2 ( R 2 + j s X 2 ), , The rotor current I'2 is reflected as I"2 (= K I'2) in the stator. The phasor sum of, I"2 and I0 gives the stator current I1., It is important to note that input to the primary and output from the secondary of, a transformer are electrical. However, in an induction motor, the inputs to the, stator and rotor are electrical but the output from the rotor is mechanical. To, facilitate calculations, it is desirable and, necessary to replace the mechanical load by an, equivalent electrical load. We then have the, transformer equivalent circuit of the induction, motor., It may be noted that even though the frequencies, of stator and rotor currents are different, yet the, magnetic fields due to them rotate at, synchronous speed Ns. The stator currents, produce a magnetic flux which rotates at a speed, Ns. At slip s, the speed of rotation of the rotor, field relative to the rotor surface in the direction, of rotation of the rotor is, , =, , 120 f ' 120 s f, =, = s Ns, P, P, , But the rotor is revolving at a speed of N relative, to the stator core. Therefore, the speed of rotor, 211, , Fig.(8.24)
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field relative to stator core, , = sN s + N = ( N s − N ) + N = N s, Thus no matter what the value of slip s, the stator and rotor magnetic fields are, synchronous with each other when seen by an observer stationed in space., Consequently, the 3-phase induction motor can be regarded as being equivalent, to a transformer having an air-gap separating the iron portions of the magnetic, circuit carrying the primary and secondary windings., Fig. (8.24) shows the phasor diagram of induction motor., , 8.30 Equivalent Circuit of the Rotor, We shall now see how mechanical load of the motor is replaced by the, equivalent electrical load. Fig. (8.25 (i)) shows the equivalent circuit per phase, of the rotor at slip s. The rotor phase current is given by;, , I'2 =, , s E2, R 22 + (s X 2 ) 2, , Mathematically, this value is unaltered by writing it as:, , I' 2 =, , E2, ( R 2 / s) 2 + ( X 2 ) 2, , As shown in Fig. (8.25 (ii)), we now have a rotor circuit that has a fixed, reactance X2 connected in series with a variable resistance R2/s and supplied, with constant voltage E2. Note that Fig. (8.25 (ii)) transfers the variable to the, resistance without altering power or power factor conditions., , Fig.(8.25), The quantity R2/s is greater than R2 since s is a fraction. Therefore, R2/s can be, divided into a fixed part R2 and a variable part (R2/s − R2) i.e.,, , R2, 1, = R 2 + R 2 − 1, s, s , , 212
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(i), , The first part R2 is the rotor resistance/phase, and represents the rotor Cu, loss., 1, (ii) The second part R 2 −1 is a variable-resistance load. The power, s , delivered to this load represents the total mechanical power developed in, the rotor. Thus mechanical load on the induction motor can be replaced by, 1, a variable-resistance load of value R 2 −1 . This is, s , , ∴, , 1, R L = R 2 − 1, s , , Fig. (8.25 (iii)) shows the equivalent rotor circuit along with load resistance RL., , 8.31 Transformer Equivalent Circuit of Induction Motor, Fig. (8.26) shows the equivalent circuit per phase of a 3-phase induction motor., Note that mechanical load on the motor has been replaced by an equivalent, electrical resistance RL given by;, , 1, R L = R 2 − 1, s , , (i), , Fig.(8.26), Note that circuit shown in Fig. (8.26) is similar to the equivalent circuit of a, transformer with secondary load equal to R2 given by eq. (i). The rotor e.m.f. in, the equivalent circuit now depends only on the transformation ratio K (= E2/E1)., Therefore; induction motor can be represented as an equivalent transformer, connected to a variable-resistance load RL given by eq. (i). The power delivered, to RL represents the total mechanical power developed in the rotor. Since the, equivalent circuit of Fig. (8.26) is that of a transformer, the secondary (i.e.,, rotor) values can be transferred to primary (i.e., stator) through the appropriate, use of transformation ratio K. Recall that when shifting resistance/reactance, from secondary to primary, it should be divided by K2 whereas current should be, multiplied by K. The equivalent circuit of an induction motor referred to, primary is shown in Fig. (8.27)., 213
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Fig.(8.27), Note that the element (i.e., R'L) enclosed in the dotted box is the equivalent, electrical resistance related to the mechanical load on the motor. The following, points may be noted from the equivalent circuit of the induction motor:, (i) At no-load, the slip is practically zero and the load R'L is infinite. This, condition resembles that in a transformer whose secondary winding is, open-circuited., (ii) At standstill, the slip is unity and the load R'L is zero. This condition, resembles that in a transformer whose secondary winding is short-circuited., (iii) When the motor is running under load, the value of R'L will depend upon, the value of the slip s. This condition resembles that in a transformer whose, secondary is supplying variable and purely resistive load., (iv) The equivalent electrical resistance R'L related to mechanical load is slip or, speed dependent. If the slip s increases, the load R'L decreases and the rotor, current increases and motor will develop more mechanical power. This is, expected because the slip of the motor increases with the increase of load, on the motor shaft., , 8.32 Power Relations, The transformer equivalent circuit of an induction motor is quite helpful in, analyzing the various power relations in the motor. Fig. (8.28) shows the, equivalent circuit per phase of an induction motor where all values have been, referred to primary (i.e., stator)., , Fig.(8.28), 214
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(i), , R', 1, Total electrical load = R '2 − 1 + R '2 = 2, s, s , Power input to stator = 3V1 I1 cos φ1, , There will be stator core loss and stator Cu loss. The remaining power will, be the power transferred across the air-gap i.e., input to the rotor., 3(I"2 )2 R '2, (ii) Rotor input =, s, 2, Rotor Cu loss = 3(I"2 ) R '2, Total mechanical power developed by the rotor is, Pm = Rotor input − Rotor Cu loss, , 3(I"2 )2 R '2, 1, =, 3(I"2 )2 R '2 = 3(I"2 )2 R '2 − 1, s, s , This is quite apparent from the equivalent circuit shown in Fig. (8.28)., (iii) If Tg is the gross torque developed by the rotor, then,, , Pm =, , 2π N Tg, 60, , or, , 2π N Tg, 1, 3(I"2 )2 R '2 − 1 =, 60, s , , or, , 1 − s 2π N Tg, 3(I"2 )2 R '2 , = 60, s , , or, , 1 − s 2π N s (1 − s) Tg, 3(I"2 )2 R '2 , =, 60, s , , ∴, or, , 3(I"2 )2 R '2 s, Tg =, 2π N s 60, , [Q N = N s (1 − s)], , N-m, , 3(I"2 )2 R '2 s, Tg = 9.55, Ns, , N-m, , Note that shaft torque Tsh will be less than Tg by the torque required to meet, windage and frictional losses., , 215
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8.33 Approximate Equivalent Circuit of Induction Motor, As in case of a transformer, the approximate equivalent circuit of an induction, motor is obtained by shifting the shunt branch (Rc − Xm) to the input terminals, as shown in Fig. (8.29). This step has been taken on the assumption that voltage, drop in R1 and X1 is small and the terminal voltage V1 does not appreciably, differ from the induced voltage E1. Fig. (8.29) shows the approximate equivalent, circuit per phase of an induction motor where all values have been referred to, primary (i.e., stator)., , Fig.(8.29), The above approximate circuit of induction motor is not so readily justified as, with the transformer. This is due to the following reasons:, (i) Unlike that of a power transformer, the magnetic circuit of the induction, motor has an air-gap. Therefore, the exciting current of induction motor (30, to 40% of full-load current) is much higher than that of the power, transformer. Consequently, the exact equivalent circuit must be used for, accurate results., (ii) The relative values of X1 and X2 in an induction motor are larger than the, corresponding ones to be found in the transformer. This fact does not, justify the use of approximate equivalent circuit, (iii) In a transformer, the windings are concentrated whereas in an induction, motor, the windings are distributed. This affects the transformation ratio., In spite of the above drawbacks of approximate equivalent circuit, it yields, results that are satisfactory for large motors. However, approximate equivalent, circuit is not justified for small motors., , 8.34 Starting of 3-Phase Induction Motors, The induction motor is fundamentally a transformer in which the stator is the, primary and the rotor is short-circuited secondary. At starting, the voltage, induced in the induction motor rotor is maximum (Q s = 1). Since the rotor, impedance is low, the rotor current is excessively large. This large rotor current, is reflected in the stator because of transformer action. This results in high, starting current (4 to 10 times the full-load current) in the stator at low power, 216
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factor and consequently the value of starting torque is low. Because of the short, duration, this value of large current does not harm the motor if the motor, accelerates normally. However, this large starting current will produce large, line-voltage drop. This will adversely affect the operation of other electrical, equipment connected to the same lines. Therefore, it is desirable and necessary, to reduce the magnitude of stator current at starting and several methods are, available for this purpose., , 8.35 Methods of Starting 3-Phase Induction Motors, The method to be employed in starting a given induction motor depends upon, the size of the motor and the type of the motor. The common methods used to, start induction motors are:, (i) Direct-on-line starting, (ii) Stator resistance starting, (iii) Autotransformer starting, (iv) Star-delta starting, (v) Rotor resistance starting, Methods (i) to (iv) are applicable to both squirrel-cage and slip ring motors., However, method (v) is applicable only to slip ring motors. In practice, any one, of the first four methods is used for starting squirrel cage motors, depending, upon ,the size of the motor. But slip ring motors are invariably started by rotor, resistance starting., , 8.36 Methods of Starting Squirrel-Cage Motors, Except direct-on-line starting, all other methods of starting squirrel-cage motors, employ reduced voltage across motor terminals at starting., , (i) Direct-on-line starting, This method of starting in just what the name implies—the motor is started by, connecting it directly to 3-phase supply. The impedance of the motor at, standstill is relatively low and when it is directly connected to the supply, system, the starting current will be high (4 to 10 times the full-load current) and, at a low power factor. Consequently, this method of starting is suitable for, relatively small (up to 7.5 kW) machines., Relation between starling and F.L. torques. We know that:, Rotor input = 2π Ns T = kT, Rotor Cu loss = s × Rotor input, , But, , ∴, or, , 3(I'2 )2 R 2 = s × kT, T ∝ (I'2 )2 s, 217
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(Q I'2 ∝ I1 ), , T ∝ I12 s, , or, , If Ist is the starting current, then starting torque (Tst) is, T ∝ I st2, , (Q at starting s = 1), , If If is the full-load current and sf is the full-load slip, then,, Tf ∝ I f2 s f, 2, , ∴, , Tst I st , = × sf, Tf I f , , When the motor is started direct-on-line, the starting current is the short-circuit, (blocked-rotor) current Isc., 2, , ∴, , Tst I sc , = × sf, Tf I f , , Let us illustrate the above relation with a numerical example. Suppose Isc = 5 If, and full-load slip sf =0.04. Then,, 2, , ∴, , 2, , Tst I sc , 5 I , = × s f = f × 0.04 = (5) 2 × 0.04 = 1, Tf I f , If , Tst = Tf, , Note that starting current is as large as five times the full-load current but, starting torque is just equal to the full-load torque. Therefore, starting current is, very high and the starting torque is comparatively low. If this large starting, current flows for a long time, it may overheat the motor and damage the, insulation., , (ii) Stator resistance starting, In this method, external resistances are connected in series with each phase of, stator winding during starting. This causes voltage drop across the resistances so, that voltage available across motor terminals is reduced and hence the starting, current. The starting resistances are gradually cut out in steps (two or more, steps) from the stator circuit as the motor picks up speed. When the motor, attains rated speed, the resistances are completely cut out and full line voltage is, applied to the rotor., This method suffers from two drawbacks. First, the reduced voltage applied to, the motor during the starting period lowers the starting torque and hence, increases the accelerating time. Secondly, a lot of power is wasted in the starting, resistances., , 218
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Fig.(8.30), Relation between starting and F.L. torques. Let V be the rated voltage/phase., If the voltage is reduced by a fraction x by the insertion of resistors in the line,, then voltage applied to the motor per phase will be xV., , I st = x I sc, 2, , Now, , Tst Ist , = × sf, Tf I f , , or, , Tst, I , = x 2 sc × s f, Tf, If , , 2, , Thus while the starting current reduces by a fraction x of the rated-voltage, starting current (Isc), the starting torque is reduced by a fraction x2 of that, obtained by direct switching. The reduced voltage applied to the motor during, the starting period lowers the starting current but at the same time increases the, accelerating time because of the reduced value of the starting torque. Therefore,, this method is used for starting small motors only., , (iii) Autotransformer starting, This method also aims at connecting the induction motor to a reduced supply at, starting and then connecting it to the full voltage as the motor picks up sufficient, speed. Fig. (8.31) shows the circuit arrangement for autotransformer starting., The tapping on the autotransformer is so set that when it is in the circuit, 65% to, 80% of line voltage is applied to the motor., At the instant of starting, the change-over switch is thrown to “start” position., This puts the autotransformer in the circuit and thus reduced voltage is applied, to the circuit. Consequently, starting current is limited to safe value. When the, motor attains about 80% of normal speed, the changeover switch is thrown to, , 219
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“run” position. This takes out the autotransformer from the circuit and puts the, motor to full line voltage. Autotransformer starting has several advantages viz, low power loss, low starting current and less radiated heat. For large machines, (over 25 H.P.), this method of starting is often used. This method can be used, for both star and delta connected motors., , Fig.(8.31), Relation between starting And F.L. torques. Consider a star-connected, squirrel-cage induction motor. If V is the line voltage, then voltage across motor, phase on direct switching is V 3 and starting current is Ist = Isc. In case of, autotransformer, if a tapping of transformation ratio K (a fraction) is used, then, phase voltage across motor is KV 3 and Ist = K Isc,, 2, , 2, , 2, , Tst Ist , KI , I , = × s f = sc × s f = K 2 sc × s f, Tf I f , If , If , , Now, , 2, , ∴, , Tst, I , = K 2 sc × s f, Tf, If , , Fig.(8.32), , 220
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The current taken from the supply or by autotransformer is I1 = KI2 = K2Isc. Note, that motor current is K times, the supply line current is K2 times and the starting, torque is K2 times the value it would have been on direct-on-line starting., , (iv) Star-delta starting, The stator winding of the motor is designed for delta operation and is connected, in star during the starting period. When the machine is up to speed, the, connections are changed to delta. The circuit arrangement for star-delta starting, is shown in Fig. (8.33)., The six leads of the stator windings are connected to the changeover switch as, shown. At the instant of starting, the changeover switch is thrown to “Start”, position which connects the stator windings in star. Therefore, each stator phase, gets V 3 volts where V is the line voltage. This reduces the starting current., When the motor picks up speed, the changeover switch is thrown to “Run”, position which connects the stator windings in delta. Now each stator phase gets, full line voltage V. The disadvantages of this method are:, (a) With star-connection during starting, stator phase voltage is 1 3 times the, , (, , )2, , line voltage. Consequently, starting torque is 1 3 or 1/3 times the value, it would have with ∆-connection. This is rather a large reduction in starting, torque., (b) The reduction in voltage is fixed., This method of starting is used for medium-size machines (upto about 25 H.P.)., Relation between starting and F.L. torques. In direct delta starting,, Starting current/phase, Isc = V/Zsc where V = line voltage, Starting line current =, In star starting, we have,, , 3 Isc, , Starting current/phase, I st =, , V 3 1, =, I sc, Z sc, 3, , Now, , 2, I sc, Tst Ist , = × s f = , Tf I f , 3 × If, , or, , Tst 1 Isc , = , × sf, Tf 3 I f , , 2, , , × sf, , , , 2, , where, , Isc = starting phase current (delta), If = F.L. phase current (delta), , 221
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Fig.(8.33), Note that in star-delta starting, the starting line current is reduced to one-third as, compared to starting with the winding delta connected. Further, starting torque, is reduced to one-third of that obtainable by direct delta starting. This method is, cheap but limited to applications where high starting torque is not necessary e.g.,, machine tools, pumps etc., , 8.37 Starting of Slip-Ring Motors, Slip-ring motors are invariably started by rotor resistance starting. In this, method, a variable star-connected rheostat is connected in the rotor circuit, through slip rings and full voltage is applied to the stator winding as shown in, Fig. (8.34)., , Fig.(8.34), (i), , At starting, the handle of rheostat is set in the OFF position so that, maximum resistance is placed in each phase of the rotor circuit. This, reduces the starting current and at the same time starting torque is, increased., (ii) As the motor picks up speed, the handle of rheostat is gradually moved in, clockwise direction and cuts out the external resistance in each phase of the, rotor circuit. When the motor attains normal speed, the change-over switch, is in the ON position and the whole external resistance is cut out from the, rotor circuit., , 222
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8.38 Slip-Ring Motors Versus Squirrel Cage Motors, The slip-ring induction motors have the following advantages over the squirrel, cage motors:, (i) High starting torque with low starting current., (ii) Smooth acceleration under heavy loads., (iii) No abnormal heating during starting., (iv) Good running characteristics after external rotor resistances are cut out., (v) Adjustable speed., The disadvantages of slip-ring motors are:, (i) The initial and maintenance costs are greater than those of squirrel cage, motors., (ii) The speed regulation is poor when run with resistance in the rotor circuit, , 8.39 Induction Motor Rating, The nameplate of a 3-phase induction motor provides the following information:, (i) Horsepower, (ii) Line voltage, (iii) Line current, (iv) Speed, (v) Frequency, (vi) Temperature rise, The horsepower rating is the mechanical output of the motor when it is operated, at rated line voltage, rated frequency and rated speed. Under these conditions,, the line current is that specified on the nameplate and the temperature rise does, not exceed that specified., The speed given on the nameplate is the actual speed of the motor at rated fullload; it is not the synchronous speed. Thus, the nameplate speed of the induction, motor might be 1710 r.p.m. It is the rated full-load speed., , 8.40 Double Squirrel-Cage Motors, One of the advantages of the slip-ring motor is that resistance may be inserted in, the rotor circuit to obtain high starting torque (at low starting current) and then, cut out to obtain optimum running conditions. However, such a procedure, cannot be adopted for a squirrel cage motor because its cage is permanently, short-circuited. In order to provide high starting torque at low starting current,, double-cage construction is used., , Construction, As the name suggests, the rotor of this motor has two squirrel-cage windings, located one above the other as shown in Fig. (8.35 (i))., (i) The outer winding consists of bars of smaller cross-section short-circuited, by end rings. Therefore, the resistance of this winding is high. Since the, 223
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outer winding has relatively open slots and a poorer flux path around its, bars [See Fig. (8.35 (ii))], it has a low inductance. Thus the resistance of, the outer squirrel-cage winding is high and its inductance is low., (ii) The inner winding consists of bars of greater cross-section short-circuited, by end rings. Therefore, the resistance of this winding is low. Since the, bars of the inner winding are thoroughly buried in iron, it has a high, inductance [See Fig. (8.35 (ii))]. Thus the resistance of the inner squirrelcage winding is low and its inductance is high., , Fig.(8.35), , Working, When a rotating magnetic field sweeps across the two windings, equal e.m.f.s, are induced in each., (i) At starting, the rotor frequency is the same as that of the line (i.e., 50 Hz),, making the reactance of the lower winding much higher than that of the, upper winding. Because of the high reactance of the lower winding, nearly, all the rotor current flows in the high-resistance outer cage winding. This, provides the good starting characteristics of a high-resistance cage winding., Thus the outer winding gives high starting torque at low starting current., (ii) As the motor accelerates, the rotor frequency decreases, thereby lowering, the reactance of the inner winding, allowing it to carry a larger proportion, of the total rotor current At the normal operating speed of the motor, the, rotor frequency is so low (2 to 3 Hz) that nearly all the rotor current flows, in the low-resistance inner cage winding. This results in good operating, efficiency and speed regulation., Fig. (8.36) shows the operating characteristics of double squirrel-cage motor., The starting torque of this motor ranges from 200 to 250 percent of full-load, torque with a starting current of 4 to 6 times the full-load value. It is classed as a, high-torque, low starting current motor., , 224
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Fig.(8.36), , 8.41 Equivalent Circuit of Double Squirrel-Cage Motor, Fig. (8.37) shows a section of the double squirrel cage motor., Here Ro and Ri are the per phase resistances of the outer cage, winding and inner cage winding whereas Xo and Xi are the, corresponding per phase standstill reactances. For the outer, cage, the resistance is made intentionally high, giving a high, Fig.(8.37), starting torque. For the inner cage winding, the resistance is low, and the leakage reactance is high, giving a low starting torque, but high efficiency on load. Note that in a double squirrel cage motor, the outer, winding produces the high starting and accelerating torque while the inner, winding provides the running torque at good efficiency., Fig. (8.38 (i)) shows the equivalent circuit for one phase of double cage motor, referred to stator. The two cage impedances are effectively in parallel. The, resistances and reactances of the outer and inner rotors are referred to the stator., The exciting circuit is accounted for as in a single cage motor. If the, magnetizing current (I0) is neglected, then the circuit is simplified to that shown, in Fig. (8.38 (ii))., , Fig.(8.38), 225
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From the equivalent circuit, the performance of the motor can be predicted., Total impedance as referred to stator is, , Z o1 = R 1 + j X1 +, , Z' Z', 1, = R 1 + j X1 + i o, 1 Z'i + 1 Z'o, Z'i + Z'o, , 226
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Chapter (9), , Single-Phase Motors, Introduction, As the name suggests, these motors are used on single-phase supply. Singlephase motors are the most familiar of all electric motors because they are, extensively used in home appliances, shops, offices etc. It is true that singlephase motors are less efficient substitute for 3-phase motors but 3-phase power, is normally not available except in large commercial and industrial, establishments. Since electric power was originally generated and distributed for, lighting only, millions of homes were given single-phase supply. This led to the, development of single-phase motors. Even where 3-phase mains are present, the, single-phase supply may be obtained by using one of the three lines and the, neutral. In this chapter, we shall focus our attention on the construction, working, and characteristics of commonly used single-phase motors., , 9.1 Types of Single-Phase Motors, Single-phase motors are generally built in the fractional-horsepower range and, may be classified into the following four basic types:, 1., , Single-phase induction motors, (i) split-phase type, (iii) shaded-pole type, , (ii) capacitor type, , 2., , A.C. series motor or universal motor, , 3., , Repulsion motors, (i) Repulsion-start induction-run motor, (ii) Repulsion-induction motor, , 4., , Synchronous motors, (i) Reluctance motor, , (ii) Hysteresis motor, , 9.2 Single-Phase Induction Motors, A single phase induction motor is very similar to a 3-phase squirrel cage, induction motor. It has (i) a squirrel-cage rotor identical to a 3-phase motor and, (ii) a single-phase winding on the stator., 227
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Unlike a 3-phase induction motor, a single-phase induction motor is not selfstarting but requires some starting means. The single-phase stator winding, produces a magnetic field that pulsates in strength in a sinusoidal manner. The, field polarity reverses after each half cycle but the field does not rotate., Consequently, the alternating flux cannot produce rotation in a stationary, squirrel-cage rotor. However, if the rotor of a single-phase motor is rotated in, one direction by some mechanical means, it will continue to run in the direction, of rotation. As a matter of fact, the rotor quickly accelerates until it reaches a, speed slightly below the synchronous speed. Once the motor is running at this, speed, it will continue to rotate even though single-phase current is flowing, through the stator winding. This method of starting is generally not convenient, for large motors. Nor can it be employed fur a motor located at some, inaccessible spot., Fig. (9.1) shows single-phase induction motor, having a squirrel cage rotor and a singlephase distributed stator winding. Such a, motor inherently docs not develop any, starting torque and, therefore, will not start to, rotate if the stator winding is connected to, single-phase a.c. supply. However, if the, Fig.(9.1), rotor is started by auxiliary means, the motor, will quickly attain me final speed. This strange behaviour of single-phase, induction motor can be explained on the basis of double-field revolving theory., , 9.3 Double-Field Revolving Theory, The double-field revolving theory is proposed to explain this dilemma of no, torque at start and yet torque once rotated. This theory is based on the fact that, an alternating sinusoidal flux (φ = φm cos ωt) can be represented by two, revolving fluxes, each equal to one-half of the maximum value of alternating, flux (i.e., φm/2) and each rotating at synchronous speed (Ns = 120 f/P, ω = 2πf), in opposite directions., The above statement will now be proved. The, instantaneous value of flux due to the stator current, of a single-phase induction motor is given by;, , φ = φ m cos ωt, Consider two rotating magnetic fluxes φ1 and φ2, each of magnitude φm/2 and rotating in opposite, directions with angular velocity ω [See Fig. (9.2)]., Let the two fluxes start rotating from OX axis at, 228, , Fig.(9.2)
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t = 0. After time t seconds, the angle through which the flux vectors have rotated, is at. Resolving the flux vectors along-X-axis and Y-axis, we have,, Total X-component =, , φm, φ, cos ωt + m cos ωt = φ m cos ωt, 2, 2, , Total Y-component =, , φm, φ, sin ωt − m sin ωt = 0, 2, 2, , Resultant flux, φ =, , (φ m, , cos ωt )2 + 0 2 = φ m cos ωt, , Thus the resultant flux vector is φ = φm cos, ωt along X-axis. Therefore, an alternating, field can be replaced by two relating fields, of half its amplitude rotating in opposite, directions at synchronous speed. Note that, the resultant vector of two revolving flux, vectors is a stationary vector that oscillates, in length with time along X-axis. When the, Fig.(9.3), rotating flux vectors are in phase [See Fig., (9.3 (i))], the resultant vector is φ = φm; when out of phase by 180° [See Fig. (9.3, (ii))], the resultant vector φ = 0., Let us explain the operation of single-phase induction motor by double-field, revolving theory., , (i) Rotor at standstill, Consider the case that the rotor is stationary and the stator winding is connected, to a single-phase supply. The alternating flux produced by the stator winding, can be presented as the sum of two rotating fluxes φ1 and φ2, each equal to one, half of the maximum value of alternating flux and each rotating at synchronous, speed (Ns = 120 f/P) in opposite directions as shown in Fig. (9.4 (i)). Let the flux, φ1 rotate in anti clockwise direction and flux φ2 in clockwise direction. The flux, φ1 will result in the production of torque T1 in the anti clockwise direction and, flux φ2 will result in the production of torque T2 In the clockwise direction. At, standstill, these two torques are equal and opposite and the net torque developed, is zero. Therefore, single-phase induction motor is not self-starting. This fact is, illustrated in Fig. (9.4 (ii))., Note that each rotating field tends to drive the rotor in the direction in which the, field rotates. Thus the point of zero slip for one field corresponds to 200% slip, for the other as explained later. The value of 100% slip (standstill condition) is, the same for both the fields., , 229
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Fig.(9.4), , (ii) Rotor running, Now assume that the rotor is started by spinning the rotor or by using auxiliary, circuit, in say clockwise direction. The flux rotating in the clockwise direction is, the forward rotating flux (φf) and that in the other direction is the backward, rotating flux (φb). The slip w.r.t. the forward flux will be, , sf =, where, , Ns − N, =s, Ns, , Ns = synchronous speed, N = speed of rotor in the direction of forward flux, , The rotor rotates opposite to the rotation of the backward flux. Therefore, the, slip w.r.t. the backward flux will be, , sb =, =, ∴, , N s − ( − N) N s + N 2 N s − N s + N, =, =, Ns, Ns, Ns, 2 N s ( N s − N), −, =2−s, Ns, Ns, , sb = 2 − s, , Thus fur forward rotating flux, slip is s (less than unity) and for backward, rotating flux, the slip is 2 − s (greater than unity). Since for usual rotor, resistance/reactance ratios, the torques at slips of less than unity arc greater than, those at slips of more than unity, the resultant torque will be in the direction of, the rotation of the forward flux. Thus if the motor is once started, it will develop, net torque in the direction in which it has been started and will function as a, motor., , 230
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Fig. (9.5) shows the rotor circuits for the forward and backward rotating fluxes., Note that r2 = R2/2, where R2 is the standstill rotor resistance i.e., r2 is equal to, half the standstill rotor resistance. Similarly, x2 = X2/2 where X2 is the standstill, rotor reactance. At standstill, s = 1 so that impedances of the two circuits are, equal. Therefore, rotor currents are equal i.e., I2f = I2b. However, when the rotor, rotates, the impedances of the two rotor circuits are unequal and the rotor current, I2b is higher (and also at a lower power factor) than the rotor current I2f. Their, m.m.f.s, which oppose the stator m.m.f.s, will result in a reduction of the, backward rotating flux. Consequently, as speed increases, the forward flux, increases, increasing the driving torque while the backward flux decreases,, reducing the opposing torque. The motor-quickly accelerates to the final speed., , Fig.(9.5), , 9.4 Making Single-Phase Induction Motor Self-Starting, The single-phase induction motor is not selfstarting and it is undesirable to resort to, mechanical spinning of the shaft or pulling a, belt to start it. To make a single-phase, induction motor self-starting, we should, somehow produce a revolving stator magnetic, field. This may be achieved by converting a, single-phase supply into two-phase supply, through the use of an additional winding., When the motor attains sufficient speed, the, starting means (i.e., additional winding) may, be removed depending upon the type of the, motor. As a matter of fact, single-phase, Fig.(9.6), induction motors are classified and named, according to the method employed to make them self-starting., (i) Split-phase motors-started by two phase motor action through the use of, an auxiliary or starting winding., , 231
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(ii) Capacitor motors-started by two-phase motor action through the use of an, auxiliary winding and a capacitor., (iii) Shaded-pole motors-started by the motion of the magnetic field produced, by means of a shading coil around a portion of the pole structure., , 9.5 Rotating Magnetic Field From 2-Phase Supply, As with a 3-phase supply, a 2-phase balanced supply also produces a rotating, magnetic field of constant magnitude. With the exception of the shaded-pole, motor, all single-phase induction motors are started as 2-phase machine. Once, so started, the motor will continue to run on single-phase supply., Let us see how 2-phase supply produces a rotating magnetic field of constant, magnitude. Fig. (9.10 (i)) shows 2-pole, 2-phase winding. The phases X and Y, are energized from a two-phase source and currents in these phases arc indicated, as Ix and Iy [See Fig. (9.10 (ii))]. Referring to Fig. (9.10 (ii)), the fluxes produced, by these currents arc given by;, , φY = φ m sin ωt, , and, , φ X = φ m sin( ωt + 90°) = φ m cos ωt, , Here φm is the maximum flux due to either phase. We shall now prove that this, 2-phase supply produces a rotating magnetic field of constant magnitude equal, to φm., (i) At instant 1 [See (Fig. 9.10 (ii)) and Fig. (9.10 (iii))], the current is zero in, phase Y and maximum in phase X. With the current in the direction shown,, a resultant flux is established toward the right. The magnitude of the, resultant flux is constant and is equal to φm as proved, under:, Fig.(9.7), At instant 1, ωt = 0°, ∴ φ Y = 0 and φ X = φ m, ∴ Resultant flux, φr = φ2X + φ 2Y = (φ m ) 2 + (0) 2 = φ m, (ii) At instant 2 [See Fig. (9.10 (ii)) and Fig. (9.10 (iii))], the current is still in, the same direction in phase X and an equal current flowing in phase Y., This establishes a resultant flux of the same value (i.e., φr =, φm) as proved under:, At instant 2, ωt = 45°, , ∴ φY =, , φm, and, 2, , φX =, , Fig.(9.8), , ∴ Resultant flux, φr = (φ X ) 2 + (φ Y ) 2, 2, , 2, , φ φ , = m + m = φm, 2 2, 232, , φm, 2
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Note that resultant flux has the same value (i.e. φm) but, turned 45° clockwise from position 1., (iii) At instant 3 [See Fig. (9.10 (ii)) and Fig. (9.10 (iii))], the, current in phase. X has decreased to zero and current in, phase Y has increased to maximum. This establishes a, resultant flux downward as proved under:, , Fig.(9.10), 233, , Fig.(9.9)
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At instant 3, ωt = 90°, ∴, , ∴ φ Y = φ m and, , φX = 0, , φ r = φ2X + (φY ) 2 = (0) 2 + (φ m ) 2 = φ m, , Note that resultant flux has now turned 90° clockwise from position 1., The reader may note that in the three instants considered above, the, resultant flux is constant and is equal to φm. However, this constant, resultant flux is shining its position (clockwise in this case). In other words,, the rotating flux is produced. We shall continue to consider other instants to, prove this fact., (iv) At instant 4 [See Fig. (9.10 (ii)) and Fig. (9.10 (iii))], the current in phase X, has reversed and has the same value as that of phase Y. This establishes a, resultant flux equal to φm turned 45° clockwise from position 3., At instant 4, ωt = 135° ∴ φ Y =, , φm, φ, and φX = m, 2, 2, , 2, , ∴, , φr =, , φ 2X, , + φ 2Y, , 2, , φ φ , = − m + m = φm, 2 2, , , Fig.(9.11), , (v) At instant 5 [See Fig. (9.10 (ii)) and Fig. (9.10 (iii))], the current in phase X, is maximum and in phase Y is zero. This establishes a resultant flux equal, to φm toward left (or 90° clockwise from position 3)., At instant 5, ωt = 180°, ∴, , ∴ φY = 0, , and, , φ X = −φ m, , φr = φ 2X + φ 2Y = (−φ m ) 2 + (0) 2 = φ m, Fig.(9.12), , (vi) Diagrams 6, 7, and 8 [See Fig. (9.10 (iii))] indicate the, direction of the resultant flux during the remaining successive instants., It follows from the above discussion that a 2-phase supply produces a rotating, magnetic field of constant value (= φm the maximum value of one of the fields)., Note: If the two windings arc displaced 90° electrical but produce fields that are, not equal and that are not 90° apart in time, the resultant field is still rotating but, is not constant in magnitude. One effect of this nonuniform rotating field is the, production of a torque that is non-uniform and that, therefore, causes noisy, operation of the motor. Since 2-phase operation ceases once the motor is started,, the operation of the motor then becomes smooth., , 234
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9.6 Split-Phase Induction Motor, The stator of a split-phase induction motor is provided with an auxiliary or, starting winding S in addition to the main or running winding M. The starting, winding is located 90° electrical from the main winding [See Fig. (9.13 (i))] and, operates only during the brief period when the motor starts up. The two, windings are so resigned that the starting winding S has a high resistance and, relatively small reactance while the main winding M has relatively low, resistance and large reactance as shown in the schematic connections in Fig., (9.13 (ii)). Consequently, the currents flowing in the two windings have, reasonable phase difference c (25° to 30°) as shown in the phasor diagram in, Fig. (9.13 (iii))., , Fig.(9.13), , Operation, (i), , When the two stator windings are energized from a single-phase supply,, the main winding carries current Im while the starting winding carries, current Is., (ii) Since main winding is made highly inductive while the starting winding, highly resistive, the currents Im and Is have a reasonable phase angle a (25°, to 30°) between them as shown in Fig. (9.13 (iii)). Consequently, a weak, revolving field approximating to that of a 2-phase machine is produced, which starts the motor. The starting torque is given by;, , Ts = kI m I s sin α, where k is a constant whose magnitude depends upon the design of the, motor., (iii) When the motor reaches about 75% of synchronous speed, the centrifugal, switch opens the circuit of the starting winding. The motor then operates as, a single-phase induction motor and continues to accelerate till it reaches the, 235
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normal speed. The normal speed of the motor is below the synchronous, speed and depends upon the load on the motor., , Characteristics, (i), (ii), (iii), , (iv), , (v), , The sinning torque is 15 to 2 times the full-loud torque mid (lie starting, current is 6 to 8 times the full-load current., Due to their low cost, split-phase induction motors are most popular singlephase motors in the market., Since the starting winding is made of fine wire, the current density is high, and the winding heats up quickly. If the starting period exceeds 5 seconds,, the winding may burn out unless the motor is protected by built-in-thermal, relay. This motor is, therefore, suitable where starting periods are not, frequent., An important characteristic of these motors is that they are essentially, constant-speed motors. The speed variation is 2-5% from no-load to fullload., These motors are suitable where a moderate starting torque is required and, where starting periods are infrequent e.g., to drive:, (a) fans (b) washing machines (c) oil burners (d) small machine tools etc., , The power rating of such motors generally lies between 60 W and 250 W., , 9.7 Capacitor-Start Motor, The capacitor-start motor is identical to a split-phase motor except that the, starting winding has as many turns as the main winding. Moreover, a capacitor, C is connected in series with the starting winding as shown in Fig. (9.14 (i))., The value of capacitor is so chosen that Is leads Im by about 80° (i.e., α ~ 80°), which is considerably greater than 25° found in split-phase motor [See Fig. (9.14, (ii))]. Consequently, starting torque (Ts = k Im Is sin α) is much more than that of, a split-phase motor Again, the starting winding is opened by the centrifugal, switch when the motor attains about 75% of synchronous speed. The motor then, operates as a single-phase induction motor and continues to accelerate till it, reaches the normal speed., , Characteristics, (i), , Although starting characteristics of a capacitor-start motor are better than, those of a split-phase motor, both machines possess the same running, characteristics because the main windings are identical., (ii) The phase angle between the two currents is about 80° compared to about, 25° in a split-phase motor. Consequently, for the same starting torque, the, current in the starting winding is only about half that in a split-phase motor., Therefore, the starting winding of a capacitor start motor heats up less, 236
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quickly and is well suited to applications involving either frequent or, prolonged starting periods., , Fig.(9.14), (iii) Capacitor-start motors are used where high starting torque is required and, where the starting period may be long e.g., to drive:, (a) compressors (b) large fans (c) pumps (d) high inertia loads, The power rating of such motors lies between 120 W and 7-5 kW., , 9.8 Capacitor-Start Capacitor-Run Motor, This motor is identical to a capacitor-start motor except that starting winding is, not opened after starting so that both the windings remain connected to the, supply when running as well as at starting. Two designs are generally used., (i) In one design, a single capacitor C is used for both starting and running as, shown in Fig.(9.15 (i)). This design eliminates the need of a centrifugal, switch and at the same time improves the power factor and efficiency of the, motor., , Fig.(9.15), (ii) In the other design, two capacitors C1 and C2 are used in the starting, winding as shown in Fig. (9.15 (ii)). The smaller capacitor C1 required for, optimum running conditions is permanently connected in series with the, 237
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starting winding. The much larger capacitor C2 is connected in parallel with, C1 for optimum starting and remains in the circuit during starting. The, starting capacitor C1 is disconnected when the motor approaches about 75%, of synchronous speed. The motor then runs as a single-phase induction, motor., , Characteristics, (i), , The starting winding and the capacitor can be designed for perfect 2-phase, operation at any load. The motor then produces a constant torque and not a, pulsating torque as in other single-phase motors., (ii) Because of constant torque, the motor is vibration free and can be used in:, (a) hospitals (6) studios and (c) other places where silence is important., , 9.9 Shaded-Pole Motor, The shaded-pole motor is very, popular for ratings below 0.05 H.P., (~ 40 W) because of its extremely, simple construction. It has salient, poles on the stator excited by, single-phase supply and a squirrelcage rotor as shown in Fig. (9.16)., A portion of each pole is, surrounded by a short-circuited, turn of copper strip called shading, coil., , Fig.(9.16), , Operation, The operation of the motor can be understood by referring to Fig. (9.17) which, shows one pole of the motor with a shading coil., (i) During the portion OA of the alternating-current cycle [See Fig. (9.17)], the, flux begins to increase and an e.m.f. is induced in the shading coil. The, resulting current in the shading coil will be in such a direction (Lenz’s law), so as to oppose the change in flux. Thus the flux in the shaded portion of, the pole is weakened while that in the unshaded portion is strengthened as, shown in Fig. (9.17 (ii))., (ii) During the portion AB of the alternating-current cycle, the flux has reached, almost maximum value and is not changing. Consequently, the flux, distribution across the pole is uniform [See Fig. (9.17 (iii))] since no, current is flowing in the shading coil. As the flux decreases (portion BC of, the alternating current cycle), current is induced in the shading coil so as to, oppose the decrease in current. Thus the flux in the shaded portion of the, 238
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pole is strengthened while that in the unshaded portion is weakened as, shown in Fig. (9.17 (iv))., , Fig.(9.17), (iii) The effect of the shading coil is to cause the field flux to shift across the, pole face from the unshaded to the shaded portion. This shifting flux is like, a rotating weak field moving in the direction from unshaded portion to the, shaded portion of the pole., (iv) The rotor is of the squirrel-cage type and is under the influence of this, moving field. Consequently, a small starting torque is developed. As soon, as this torque starts to revolve the rotor, additional torque is produced by, single-phase induction-motor action. The motor accelerates to a speed, slightly below the synchronous speed and runs as a single-phase induction, motor., , Characteristics, (i), , The salient features of this motor are extremely simple construction and, absence of centrifugal switch., (ii) Since starting torque, efficiency and power factor are very low, these, motors are only suitable for low power applications e.g., to drive:, (a) small fans (6) toys (c) hair driers (d) desk fans etc., The power rating of such motors is upto about 30 W., , 239
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9.10 Equivalent Circuit of Single-Phase Induction Motor, It was stated earlier that when the stator of a single-phase induction motor is con, heeled to single-phase supply, the stator current produces a pulsating flux that is, equivalent to two-constant-amplitude fluxes revolving in opposite directions at, the synchronous speed (double-field revolving theory). Each of these fluxes, induces currents in the rotor circuit and produces induction motor action similar, to that in a 3-phase induction motor Therefore, a single-phase induction motor, can to imagined to be consisting of two motors, having a common stator, winding but with their respective rotors revolving in opposite directions. Each, rotor has resistance and reactance half the actual rotor values., Let, R1 = resistance of stator winding, X1 = leakage reactance of stator winding, Xm = total magnetizing reactance, R'2 = resistance of the rotor referred to the stator, X'2 = leakage reactance of the rotor referred to the stator, (i), , revolving theory., At standstill. At standstill, the motor is simply a transformer with its, secondary short-circuited. Therefore, the equivalent circuit of single-phase, motor at standstill will be as shown in Fig. (9.18). The double-field, revolving theory suggests that characteristics associated with each, revolving field will be just one-half of the characteristics associated with, the actual total flux. Therefore, each rotor has resistance and reactance, equal to R'2/2 and X'2/2 respectively. Each rotor is associated with half the, , Fig.(9.18), , 240
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total magnetizing reactance. Note that in the equivalent circuit, the core, loss has been neglected. However, core loss can be represented by an, equivalent resistance in parallel with the magnetizing reactance., Ef = 4.44 f N φf;, , Now, At standstill,, , Eb = 4.44 f N φb, , φf = φb. Therefore, Ef = Eb., , V1 ~ Ef + Eb = I1Zf + I1Zb, where, , Zf = impedance of forward parallel branch, Zb = impedance of backward parallel branch, , (ii) Rotor running. Now consider that the motor is pinning at some speed in, the direction of the forward revolving field, the slip being s. The rotor, current produced by the forward field will have a frequency sf where f is, the stator frequency. Also, the rotor current produced by the backward field, will have a frequency of (2 − s)f. Fig. (9.19) shows the equivalent circuit of, a single-phase induction motor when the rotor is rotating at slip s. It is, clear, from the equivalent circuit that under running conditions, Ef becomes, much greater than Eb because the term R'2/2s increases very much as s, tends towards zero. Conversely, E^ falls because the term R'2/2(2 − s), decreases since (2 − s) tends toward 2. Consequently, the forward field, increases, increasing the driving torque while the backward field decreases, reducing the opposing torque., , Fig.(9.19), Total impedance of the circuit .is given by;, , 241
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Zr = Z1 + Zf + Zb, where, , Z1 = R1 + j X1, , X m R '2, X' , +j 2, , 2 2s, 2 , Zf =, R '2 X m X'2 , + j, +, , 2s, 2 , 2, j, , X m R '2, X' , +j 2, , 2 2(2 − s), 2 , Zb =, R '2, X' , X, + j m + 2 , 2( 2 − s ) 2, 2 , j, , ∴, , I1 = V1 / Z r, , 9.11 A.C. Series Motor or Universal Motor, A d.c. series motor will rotate in the same direction regardless of the polarity of, the supply. One can expect that a d.c. series motor would also operate on a, single-phase supply. It is then called an a.c. series motor. However, some, changes must be made in a d.c. motor that is to operate satisfactorily on a.c., supply. The changes effected are:, (i) The entire magnetic circuit is laminated in order to reduce the eddy current, loss. Hence an a.c. series motor requires a more expensive construction, than a d.c. series motor., (ii) The series field winding uses as few turns as possible to reduce the, reactance of the field winding to a minimum. This reduces the voltage drop, across the field winding., (iii) A high field flux is obtained by using a low-reluctance magnetic circuit., (iv) There is considerable sparking between the brushes and the commutator, when the motor is used on a.c. supply. It is because the alternating flux, establishes high currents in the coils short-circuited by the brushes. When, the short-circuited coils break contact from the commutator, excessive, sparking is produced. This can be eliminated by using high-resistance leads, to connect the coils to the commutator segments., , Construction, , 242
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The construction of en a.c. series motor is very similar to a d.c. series motor, except that above modifications are, incorporated [See Fig. (9.20)]. Such a, motor can be operated either on a.c. or d.c., supply and the resulting torque-speed curve, is about the same in each case. For this, reason, it is sometimes called a universal, motor., , Operation, When the motor is connected to an a.c., supply, the same alternating current flows, through the field and armature windings., Fig.(9.20), The field winding produces an alternating, flux φ that reacts with the current flowing in, the armature to produce a torque. Since both armature current and flux reverse, simultaneously, the torque always acts in the same direction. It may be noted, that no rotating flux is produced in this type of machines; the principle of, operation is the same as that of a d.c. series motor., , Characteristics, The operating characteristics of an a.c. series motor are similar to those of a d.c., series motor., (i) The speed increases to a high value with a decrease in load. In very small, series motors, the losses are usually large enough at no load that limit the, speed to a definite value (1500 - 15,000 r.p.m.)., (ii) The motor torque is high for large armature currents, thus giving a high, starting torque., (iii) At full-load, the power factor is about 90%. However, at starting or when, carrying an overload, the power factor is lower., , Applications, The fractional horsepower a.c. series motors have high-speed (and, corresponding small size) and large starting torque. They can, therefore, be used, to drive:, (a) high-speed vacuum cleaners, (b) sewing machines, (c) electric shavers, (d) drills, (e) machine tools etc., , 9.12 Single-Phase Repulsion Motor, A repulsion motor is similar to an a.c. series motor except that:, 243
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(i), , brushes are not connected to supply but are short-circuited [See Fig., (9.21)]. Consequently, currents are induced in the armature conductors by, transformer action., (ii) the field structure has non-salient pole construction., By adjusting the position of short-circuited brushes on the commutator, the, starting torque can be developed in the motor., , Construction, The field of stator winding is wound like the main winding of a split-phase, motor and is connected directly to a single-phase source. The armature or rotor, is similar to a d.c. motor armature with drum type winding connected to a, commutator (not shown in the figure). However, the brushes are not connected, to supply but are connected to each other or short-circuited. Short-circuiting the, brushes effectively makes the rotor into a type of squirrel cage. The major, difficulty with an ordinary single-phase induction motor is the low starting, torque. By using a commutator motor with brushes short-circuited, it is possible, to vary the starting torque by changing the brush axis. It has also better power, factor than the conventional single-phase motor., , Fig.(9.21), , Principle of operation, The principle of operation is illustrated in Fig. (9.21) which shows a two-pole, repulsion motor with its two short-circuited brushes. The two drawings of Fig., (9.21) represent a time at which the field current is increasing in the direction, shown so that the left-hand pole is N-pole and the right-hand pole is S-pole at, the instant shown., (i) In Fig. (9.21 (i)), the brush axis is parallel to the stator field. When the, stator winding is energized from single-phase supply, e.m.f. is induced in, the armature conductors (rotor) by induction. By Lenz’s law, the direction, of the e.m.f. is such that the magnetic effect of the resulting armature, currents will oppose the increase in flux. The direction of current in, armature conductors will be as shown in Fig. (9.21 (i)). With the brush axis, in the position shown in Fig. (9.21 (i)), current will flow from brush B to, 244
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brush A where it enters the armature and flows back to brush B through the, two paths ACB and ADB. With brushes set in this position, half of the, armature conductors under the N-pole carry current inward and half carry, current outward. The same is true under S-pole. Therefore, as much torque, is developed in one direction as in the other and the armature remains, stationary. The armature will also remain stationary if the brush axis is, perpendicular to the stator field axis. It is because even then net torque is, zero., (ii) If the brush axis is at some angle other than 0° or 90° to the axis of the, stator field, a net torque is developed on the rotor and the rotor accelerates, to its final speed. Fig. (9.21 (ii)) represents the motor at the same instant as, that in Fig. (9.21 (i)) but the brushes have been shifted clockwise through, some angle from the stator field axis. Now e.m.f. is still induced in the, direction indicated in Fig. (9.21 (i)) and current flows through the two, paths of the armature winding from brush A to brush B. However, because, of the new brush positions, the greater part of the conductors under the Npole carry current in one direction while the greater part of conductors, under S-pole carry current in the opposite direction. With brushes in the, position shown in Fig. (9.21 (ii), torque is developed in the clockwise, direction and the rotor quickly attains the final speed., (iii) The direction of rotation of the rotor, depends upon the direction in which, the brushes are shifted. If the brushes, are shifted in clockwise direction, from the stator field axis, the net, torque acts in the clockwise direction, and the rotor accelerates in the, Fig.(9.22), clockwise direction. If the brushes, are, shifted, in, anti-clockwise, direction as in Fig. (9.22). the armature current under the pole faces is, reversed and the net torque is developed in the anti-clockwise direction., Thus a repulsion motor may be made to rotate in either direction depending, upon the direction in which the brushes are shifted., (iv) The total armature torque in a repulsion motor can be shown to be, , Ta ∝ sin 2α, where, α = angle between brush axis and stator field axis, For maximum torque, 2α = 90° or α = 45°, , 245
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Thus adjusting α to 45° at starting, maximum torque can be obtained, during the starting period. However, α has to be adjusted to give a suitable, running speed., , Characteristics, (i), , The repulsion motor has characteristics very similar to those of an a.c., series motor i.e., it has a high starting torque and a high speed at no load., (ii) The speed which the repulsion motor develops for any given load will, depend upon the position of the brushes., (iii) In comparison with other single-phase motors, the repulsion motor has a, high starring torque and relatively low starting current., , 9.13 Repulsion-Start Induction-Run Motor, Sometimes the action of a repulsion motor is combined with that of a singlephase induction motor to produce repulsion-start induction-run motor (also, called repulsion-start motor). The machine is started as a repulsion motor with a, corresponding high starting torque. At some predetermined speed, a centrifugal, device short-circuits the commutator so that the machine then operates as a, single-phase induction motor., The repulsion-start induction-run motor has the same general construction of a, repulsion motor. The only difference is that in addition to the basic repulsionmotor construction, it is equipped with a centrifugal device fitted on the, armature shaft. When the motor reaches 75% of its full pinning speed, the, centrifugal device forces a short-circuiting ring to come in contact with the inner, surface of the commutator. This snort-circuits all the commutator bars. The rotor, then resembles squirrel-cage type and the motor runs as a single-phase induction, motor. At the same time, the centrifugal device raises the brushes from the, commutator which reduces the wear of the brushes and commutator as well as, makes the operation quiet., , Characteristics, (i), , The starting torque is 2.5 to 4.5 times the full-load torque and the starting, current is 3.75 times the full-load value., (ii) Due to their high starting torque, repulsion-motors were used to operate, devices such as refrigerators, pumps, compressors etc., However, they posed a serious problem of maintenance of brushes, commutator, arid the centrifugal device. Consequently, manufacturers have stopped making, them in view of the development of capacitor motors which are small in size,, reliable and low-priced., , 246
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9.14 Repulsion-Induction Motor, The repulsion-induction motor produces a high starting torque entirely due to, repulsion motor action. When running, it functions through a combination of, induction-motor and repulsion motor action., , 247
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Construction, Fig. (9.23) shows the connections of a 4-pole repulsion-induction motor for 230, V operation. It consists of a stator and a rotor (or armature)., (i) The stator carries a single distributed winding fed from single-phase, supply., (ii) The rotor is provided with two independent windings placed one inside the, other. The inner winding is a squirrel-cage winding with rotor bars, permanently short-circuited. Placed over the squirrel cage winding is a, repulsion commutator armature winding. The repulsion winding is, connected to a commutator on which ride short-circuited brushes. There is, no centrifugal device and the repulsion winding functions at all times., , Fig.(9.23), , Operation, (i), , When single-phase supply is given to the stator winding, the repulsion, winding (i.e., outer winding) is active. Consequently, the motor starts as a, repulsion motor with a corresponding high starting torque., (ii) As the motor speed increases, the current shifts from the outer to inner, winding due to the decreasing impedance of the inner winding with, increasing speed. Consequently, at running speed, the squirrel cage, winding carries the greater part of rotor current. This shifting of repulsionmotor action to induction-motor action is thus achieved without any, switching arrangement., (iii) It may be seen that the motor starts as a repulsion motor. When running, it, functions through a combination of principle of induction and repulsion;, the former being predominant., , Characteristics, (i), , The no-load speed of a repulsion-induction motor is somewhat above the, synchronous speed because of the effect of repulsion winding. However,, , 248
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the speed at full-load is slightly less than the synchronous speed as in an, induction motor., (ii) The speed regulation of the motor is about 6%., (iii) The starting torque is 2.25 to 3 times the full-load torque; the lower value, being for large motors. The starting current is 3 to 4 times the full-load, current., This type of motor is used for applications requiring a high starting torque with, essentially a constant running speed. The common sizes are 0.25 to 5 H.P., , 9.15 Single-Phase Synchronous Motors, Very small single-phase motors have been developed which run at true, synchronous speed. They do not require d.c. excitation for the rotor. Because of, these characteristics, they are called unexcited single-phase synchronous motors., The most commonly used types are:, (i) Reluctance motors, (ii) Hysteresis motors, The efficiency and torque-developing ability of these motors is low; The output, of most of the commercial motors is only a few watts., , 9.16 Reluctance Motor, It is a single-phase synchronous motor which does not require d.c. excitation to, the rotor. Its operation is based upon the following principle:, Whenever a piece of ferromagnetic material is located in a magnetic field; a, force is exerted on the material, tending to align the material so that reluctance, of the magnetic path that passes through the material is minimum., , Fig.(9.24), , Construction, A reluctance motor (also called synchronous reluctance motor) consists of:, , 249
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(i), , a stator carrying a single-phase winding along with an auxiliary winding to, produce a synchronous-revolving magnetic field., (ii) a squirrel-cage rotor having unsymmetrical magnetic construction. This is, achieved by symmetrically removing some of the teeth from the squirrelcage rotor to produce salient poles on the rotor. As shown in Fig. (9.24 (i)),, 4 sailent poles have been produced on me rotor. The salient poles created, on the rotor must be equal to the poles on the stator., Note that rotor salient poles offer low reductance to the stator flux and,, therefore, become strongly magnetized., , Operation, (i), , When single-phase stator having an auxiliary winding is energized, a, synchronously-revolving field is produced. The motor starts as a standard, squirrel-cage induction motor and will accelerate to near its synchronous, speed., (ii) As the rotor approaches synchronous speed, the rotating stator flux will, exert reluctance torque on the rotor poles tending to align the salient-pole, axis with the axis of the rotating field. The rotor assumes a position where, its salient poles lock with the poles of the revolving field [See Fig. (9.24, (ii))]. Consequently, the motor will continue to run at the speed of, revolving flux i.e., at the synchronous speed., (iii) When we apply a mechanical load, the rotor poles fall slightly behind the, stator poles, while continuing to turn at synchronous speed. As the load on, the motor is increased, the mechanical angle between the poles increases, progressively. Nevertheless, magnetic attraction keeps the rotor locked to, the rotating flux. If the load is increased beyond the amount under which, the reluctance torque can maintain synchronous speed, the rotor drops out, of step with the revolving field. The speed, then, drops to some value at, which the slip is sufficient to develop the necessary torque to drive the load, by induction-motor action., , Characteristics, (i) These motors have poor torque, power factor and efficiency., (ii) These motors cannot accelerate high-inertia loads to synchronous speed., (iii) The pull-in and pull-out torques of such motors are weak., Despite the above drawbacks, the reluctance motor is cheaper than any other, type of synchronous motor. They are widely used for constant-speed, applications such as timing devices, signalling devices etc., , 250
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9.17 Hysteresis Motor, It is a single-phase motor whose operation depends upon the hysteresis effect, i.e., magnetization produced in a ferromagnetic material lags behind the, magnetizing force., , Construction It consists of:, (i), , a stator designed to produce a synchronously-revolving field from a, single-phase supply. This is accomplished by using permanent-split, capacitor type construction. Consequently, both the windings (i.e., starting, as well as main winding) remain connected in the circuit during running, operation as well as at starting. The value of capacitance is so adjusted as to, result in a flux revolving at synchronous speed., (ii) a rotor consisting of a smooth cylinder of magnetically hard steel, without, winding or teeth., , Operation, (i), , When the stator is energized from a single-phase supply, a synchronouslyrevolving field (assumed in anti-clockwise direction) is produced due to, split-phase operation., (ii) The revolving stator flux magnetizes the rotor. Due to hysteresis effect, the, axis of magnetization of rotor will lag behind the axis of stator field by, hysteresis lag angle a as shown in Fig. (9.25). Thus the rotor and stator, poles are locked. If the rotor is stationary, the starting torque produced is, given by:, , Ts ∝ φs φ r sin α, where φs = stator flux., φr = rotor flux., From now onwards, the rotor accelerates to synchronous speed with a uniform, torque., (iii) After reaching synchronism, the motor continues to run at synchronous, speed and adjusts its torque angle so as to develop the torque required by, the load., , Characteristics, (i), , A hysteresis motor can synchronize any load which it can accelerate, no, matter how great the inertia. It is because the torque is uniform from, standstill to synchronous speed., (ii) Since the rotor has no teeth or salient poles or winding, a hysteresis motor, is inherently quiet and produces smooth rotation of the load., , 251
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Fig.(9.25), (iii) The rotor takes on the same number of poles as the stator field. Thus by, changing the number of stator poles through pole-changing connections,, we can get a set of synchronous speeds for the motor., , Applications, Due to their quiet operation and ability to drive high-inertia toads, hysteresis, motors are particularly well suited for driving (i) electric clocks (ii) timing, devices (iii) tape-decks (iv)from-tables and other precision audio-equipment., , 252
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Chapter (10), , Alternators, Introduction, A.C. system has a number of advantages over d.c. system. These days 3-phase, a.c. system is being exclusively used for generation, transmission and, distribution of power. The machine which produces 3-phase power from, mechanical power is called an alternator or synchronous generator. Alternators, are the primary source of all the electrical energy we consume. These machines, are the largest energy converters found in the world. They convert mechanical, energy into a.c. energy. In this chapter, we shall discuss the construction and, characteristics of alternators., , 10.1 Alternator, An alternator operates on the same, fundamental principle of electromagnetic, induction as a d.c. generator i.e., when the, flux linking a conductor changes, an e.m.f. is, induced in the conductor. Like a d.c., generator, an alternator also has an armature, winding and a field winding. But there is one, important difference between the two. In a, d.c. generator, the armature winding is placed, on the rotor in order to provide a way of, converting alternating voltage generated in the, winding to a direct voltage at the terminals, through the use of a rotating commutator. The, Fig.(10.1), field poles are placed on the stationary part of, the machine. Since no commutator is required in an alternator, it is usually more, convenient and advantageous to place the field winding on the rotating part (i.e.,, rotor) and armature winding on the stationary part (i.e., stator) as shown in Fig., (10.1)., , Advantages of stationary armature, The field winding of an alternator is placed on the rotor and is connected to d.c., supply through two slip rings. The 3-phase armature winding is placed on the, stator. This arrangement has the following advantages:, 252
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(i), , It is easier to insulate stationary winding for high voltages for which the, alternators are usually designed. Ii is because they are not subjected to, centrifugal forces and also extra space is available due to the stationary, arrangement of the armature., (ii) The stationary 3-phase armature can be directly connected to load without, going through large, unreliable slip rings and brushes., (iii) Only two slip rings are required for d.c. supply to the field winding on the, rotor. Since the exciting current is small, the slip rings and brush gear, required are of light construction., (iv) Due to simple and robust construction of the rotor, higher speed of rotating, d.c. field is possible. This increases the output obtainable from a machine, of given dimensions., Note: All alternators above 5 kVA employ a stationary armature (or stator) and a, revolving d.c. field., , 10.2 Construction of Alternator, An alternator has 3,-phase winding on the stator and a d.c. field winding on the, rotor., , 1., , Stator, , It is the stationary part of the machine and is built up of sheet-steel laminations, having slots on its inner periphery. A 3-phase winding is placed in these slots, and serves as the armature winding of the alternator. The armature winding is, always connected in star and the neutral is connected to ground., , 2., , Rotor, , The rotor carries a field winding which is supplied with direct current through, two slip rings by a separate d.c. source. This d.c. source (called exciter) is, generally a small d.c. shunt or compound generator mounted on the shaft of the, alternator. Rotor construction is of two types, namely;, (i) Salient (or projecting) pole type, (ii) Non-salient (or cylindrical) pole type, (i) Salient pole type, In this type, salient or projecting poles are mounted on a large circular steel, frame which is fixed to the shaft of the alternator as shown in Fig. (10.2). The, individual field pole windings are connected in series in such a way that when, the field winding is energized by the d.c. exciter, adjacent poles have opposite, polarities., , 253
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Low and medium-speed alternators (120-400 r.p.m.) such as those driven by, diesel engines or water turbines have salient pole type rotors due to the, following reasons:, (a) The salient field poles would cause .an excessive windage loss if driven at, high speed and would tend to produce noise., (b) Salient-pole construction cannot be made strong enough to withstand the, mechanical stresses to which they may be subjected at higher speeds., Since a frequency of 50 Hz is required, we must use a large number of poles on, the rotor of slow-speed alternators. Low-speed rotors always possess a large, diameter to provide the necessary spate for the poles. Consequently, salient-pole, type rotors have large diameters and short axial lengths., (ii) Non-salient pole type, In this type, the rotor is made of smooth solid forged-steel radial cylinder having, a number of slots along the outer periphery. The field windings are embedded in, these slots and are connected in series to the slip rings through which they are, energized by the d.c. exciter. The regions forming the poles are usually left, unslotted as shown in Fig. (10.3). It is clear that the poles formed are non-salient, i.e., they do not project out from the rotor surface., , Fig.(10.2), , Fig.(10.3), , High-speed alternators (1500 or 3000 r.p.m.) are driven by steam turbines and, use non-salient type rotors due to the following reasons:, (a) This type of construction has mechanical robustness and gives noiseless, operation at high speeds., (b) The flux distribution around the periphery is nearly a sine wave and hence, a better e.m.f. waveform is obtained than in the case of salient-pole type., 254
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Since steam turbines run at high speed and a frequency of 50 Hz is required, we, need a small number of poles on the rotor of high-speed alternators (also called, turboalternators). We can use not less than 2 poles and this fixes the highest, possible speed. For a frequency of 50 Hz, it is 3000 r.p.m. The next lower speed, is 1500 r.p.m. for a 4-pole machine. Consequently, turboalternators possess 2 or, 4 poles and have small diameters and very long axial lengths., , 10.3 Alternator Operation, The rotor winding is energized from the d.c. exciter and alternate N and S poles, are developed on the rotor. When the rotor is rotated in anti-clockwise direction, by a prime mover, the stator or armature conductors are cut by the magnetic flux, of rotor poles. Consequently, e.m.f. is induced in the armature conductors due to, electromagnetic induction. The induced e.m.f. is alternating since N and S poles, of rotor alternately pass the armature conductors. The direction of induced e.m.f., can be found by Fleming’s right hand rule and frequency is given by;, , f=, where, , NP, 120, , N = speed of rotor in r.p.m., P = number of rotor poles, , The magnitude of the voltage induced in each phase depends upon the rotor flux,, the number and position of the conductors in the phase and the speed of the, rotor., , Fig.(10.4), Fig. (10.4 (i)) shows star-connected armature winding and d.c. field winding., When the rotor is rotated, a 3-phase voltage is induced in the armature winding., The magnitude of induced e.m.f. depends upon the speed of rotation and the d.c., exciting current. The magnitude of e.m.f. in each phase of the armature winding, is the same. However, they differ in phase by 120° electrical as shown in the, phasor diagram in Fig. (10.4 (ii))., , 255
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10.4, , Frequency, , The frequency of induced e.m.f. in the armature conductors depends upon speed, and the number of poles., Let, N = rotor speed in r.p.m., P = number of rotor poles, f = frequency of e.m.f. in Hz, Consider a stator conductor that is successively swept by the N and S poles of, the rotor. If a positive voltage is induced when a N-pole sweeps across the, conductor, a similar negative voltage is induced when a S-pole sweeps by. This, means that one complete cycle of e.m.f. is generated in the conductor as a pair of, poles passes it i.e., one N-pole and the adjacent following S-pole. The same is, true for every other armature conductor., ∴ No. of cycles/revolution = No. of pairs of poles = P/2, No. of revolutions/second = N/60, ∴ No. of cycles/second = (P/2)(N/60) = N P/120, But number of cycles of e.m.f. per second is its frequency., , ∴, , f=, , NP, 120, , It may be noted that N is the synchronous speed and is generally represented by, Ns. For a given alternator, the number of rotor poles is fixed and, therefore, the, alternator must be run at synchronous speed to give an output of desired, frequency. For this reason, an alternator is sometimes called synchronous, generator., , 10.5 A.C. Armature Windings, A.C. armature windings are always of the nonsalient-pole type and are usually, symmetrically distributed in slots around the complete circumference of the, armature. A.C. armature windings are generally open-circuit type i.e., both ends, are brought out. An open-circuit winding is one that does not close on itself i.e.,, a closed circuit will not be formed until some external connection is made to a, source or load. The following are the general features of a.c. armature windings:, (i) A.C. armature windings are generally distributed windings i.e., they are, symmetrically distributed in slots around the complete circumference of the, armature. A distributed winding has two principal advantages. First, a, distributed winding generates a voltage wave that is nearly a sine curve., Secondly, copper is evenly distributed on the armature surface. Therefore,, heating is more uniform and this type of winding is more easily cooled., (ii) A.C. armature windings may use full-pitch coils or fractional-pitch coils. A, coil with a span of 180° electrical is called a full-pitch coil. In this case, the, two sides of the coil occupy identical positions under adjacent opposite, 256
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poles and the e.m.f. generated in the coil is maximum. A coil with a span of, less than 180° electrical is called a fractional-pitch coil. For example, a coil, with a span of 150° electrical would be called a 5/6 pitch coil. Although, e.m.f. induced in a fractional-pitch coil is less than that of a full-pitch coil,, fractional-pitch coils are frequently used in a.c. machines for two main, reasons. First, less copper is required per coil and secondly the waveform, of the generated voltage is improved., (iii) Most of a.c. machines use double layer armature windings. In a double, layer winding, one coil side lies in the upper half of one slot while the other, coil side lies in the lower half of another slot spaced about one-pole pitch, from the first one. This arrangement permits simpler end connections and it, is economical to manufacture., (iv) Since most of a.c. machines are of 3-phase type, the three windings of the, three phases are identical but spaced 120 electrical degrees apart., (v) A group of adjacent slots belonging to one phase under one pole pair is, known as phase belt. The angle subtended by a phase belt is known as, phase spread. The 3-phase windings are always designed for 60° phase, spread., , 10.6 Armature Winding of Alternator, With very few exceptions, alternators are 3-phase machines because of the, advantages of 3-phase service for generation, transmission and distribution. The, windings for an alternator are much simpler than that of a d c. machine because, no commutator is used. Fig. (10.5) shows a 2-pole, 3-phase double-layer, fullpitch, distributed winding for the stator of an alternator. There are 12 slots and, each slot contains two coil sides. The coil sides that are placed in adjacent slots, belong to the same phase such as a1, a3 or a2, a4 constitute a phase belt. Note that, in a 3-phase machine, phase belt is always 60° electrical. Since the winding has, double-layer arrangement, one side of a coil, such as a1, is placed at the bottom, of a slot and the other side − a1 is placed at the top of another slot spaced one, pole pitch apart. Note that each coil has a span of a full pole pitch or 180, electrical degrees. Therefore. the winding is a full-pitch winding., Note that there are 12 total coils and each phase has four coils. The four coils in, each phase are connected in series so that their voltages aid. The three phases, then may be connected to form Y or ∆-connection. Fig. (10.6) shows how the, coils ire connected to form a Y-connection., , 10.7 Winding Factors, The armature winding of an alternator is distributed over the entire armature., The distributed winding produces nearly a sine waveform and the heating is, 257
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Fig.(10.5), , Fig.(10.6), , more uniform. Likewise, the coils of armature winding are not full-pitched i.e.,, the two sides of a coil are not at corresponding points under adjacent poles. The, fractional pitched armature winding requires less copper per coil and at the same, time waveform of output voltage is unproved. The distribution and pitching of, the coils affect the voltages induced in the coils. We shall discuss two winding, factors:, (i) Distribution factor (Kd), also called breadth factor, (ii) Pitch factor (Kp), also known as chord factor, , (i) Distribution factor (Kd), A winding with only one slot per pole per phase is called a concentrated, winding. In this type of winding, the e.m.f. generated/phase is equal to the, arithmetic sum of the individual coil e.m.f.s in that phase. However, if the, coils/phase are distributed over several slots in space (distributed winding), the, e.m.f.s in the coils are not in phase (i.e., phase difference is not zero) but are, displaced from each by the slot angle α (The angular displacement in electrical, agrees between the adjacent slots is called slot angle). The e.m.f./phase will be, the phasor sum of coil e.m.f.s. The distribution factor Kd is defined as:, , e.m.f. with distribute d winding, e.m.f. with concentrat ed winding, phasor sum of coil e.m.f.s/phase, =, arithmetic sum of coil e.m.f.s/phase, , Kd =, , Note that numerator is less than denominator so that Kd < 1., Expression for Kd, 180° electrical, Let, α = slot angle =, No. of slots/pole, n = slots per pole per phase, 258
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The distribution factor can be determined by constructing a phasor diagram for, the coil e.m.f.s. Let n = 3. The three coil e.m.f.s are shown as phasors AB, BC, and CD [See Fig. (10.7 (i))] each of which is a chord of circle with centre at O, and subtends an angle α at O. The phasor sum of the coil e.m.f.s subtends an, angle n α (Here n = 3) at O. Draw perpendicular bisectors of each chord such as, Ox, Oy etc [See Fig. (10.7 (ii))]., , Kd =, , AD, 2 × Ax, Ax, =, =, n × AB n × (2 Ay) n × Ay, , OA × sin( n α / 2), n × OA × sin( α / 2), sin( n α / 2), Kd =, n sin( α / 2), =, , ∴, , Note that n α is the phase spread., , Fig.(10.7), , (ii) Pitch factor (Kp), A coil whose sides are separated by one pole pitch (i.e., coil span is 180°, electrical) is called a full-pitch coil. With a full-pitch coil, the e.m.f.s induced in, the two coil sides a in phase with each other and the resultant e.m.f. is the, arithmetic sum of individual e.m.fs. However the waveform of the resultant, e.m.f. can be improved by making the coil pitch less than a pole pitch. Such a, coil is called short-pitch coil. This practice is only possible with double-layer, type of winding The e.m.f. induced in a short-pitch coil is less than that of a fullpitch coil. The factor by which e.m.f. per coil is reduced is called pitch factor, Kp. It is defined as:, , Kp =, , e.m.f. induced in short - pitch coil, e.m.f. induced in full - pitch coil, , Expression for Kp. Consider a coil AB which is short-pitch by an angle β, electrical degrees as shown in Fig. (10.8). The e.m.f.s generated in the coil sides, 259
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A and B differ in phase by an angle β and can be represented by phasors EA and, EB respectively as shown in Fig. (10.9). The diagonal of the parallelogram, represents the resultant e.m.f. ER of the coil., , Fig.(10.8), , Fig.(10.9), , ER = 2Ea cos β/2, e.m.f. in short - pitch coil 2E A cos β / 2, Pitch factor, K p =, =, = cos β / 2, e.m.f. in full - pitch coil, 2E A, K p = cos β / 2, ∴, For a full-pitch winding, Kp = 1. However, for a short-pitch winding, Kp < 1., Note that β is always an integer multiple of the slot angle α., Since EA = EB,, , 10.8 E.M.F. Equation of an Alternator, Let, , Z = No. of conductors or coil sides in series per phase, φ = Flux per pole in webers, P = Number of rotor poles, N = Rotor speed in r.p.m., In one revolution (i.e., 60/N second), each stator conductor is cut by Pφ webers, i.e.,, dφ = Pφ;, ∴, , dt = 60/N, , Average e.m.f. induced in one stator conductor, , =, , Pφ N, dφ, Pφ, =, =, volts, dt 60 / N, 60, , Since there are Z conductors in series per phase,, Pφ N, ∴ Average e.m.f./phase =, ×Z, 60, Pφ Z 120 f, =, ×, 60, P, , 120 f , , , Q N =, P , , , = 2fφ Z volts, R.M.S. value of e.m.f./phase = Average value/phase x form factor, = 2fφ Z × 1.11 = 2.22 fφ Z volts, ∴, , E r.m.s. / phase = 2.22 fφ Z volts, 260, , (i)
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If Kp and Kd are the pitch factor and distribution factor of the armature winding,, then,, , E r.m.s. / phase = 2.22 K p K d fφ Z volts, , (ii), , Sometimes the turns (T) per phase rather than conductors per phase are, specified, in that case, eq. (ii) becomes:, , E r.m.s. / phase = 4.44 K p K d fφ T volts, , (iii), , The line voltage will depend upon whether the winding is star or delta, connected., , 10.9 Armature Reaction in Alternator, When an alternator is running at no-load, there will be no current flowing, through the armature winding. The flux produced in the air-gap will be only due, to the rotor ampere-turns. When the alternator is loaded, the three-phase currents, will produce a totaling magnetic field in the air-gap. Consequently, the air-gap, flux is changed from the no-load condition., The effect of armature flux on the flux produced by field ampere-turns (i. e.,, rotor ampere-turns) is called armature reaction., Two things are worth noting about the armature reaction in an alternator. First,, the armature flux and the flux produced by rotor ampere-turns rotate at the same, speed (synchronous speed) in the same direction and, therefore, the two fluxes, are fixed in space relative to each other. Secondly, the modification of flux in, the air-gap due to armature flux depends on the magnitude of stator current and, on the power factor of the load. It is the load power factor which determines, whether the armature flux distorts, opposes or helps the flux produced by rotor, ampere-turns. To illustrate this important point, we shall consider the following, three cases:, (i) When load p.f. is unity, (ii) When load p.f. is zero lagging, (iii) When load p.f. is zero leading, , (i) When load p.f. is unity, Fig. (10.10 (i)) shows an elementary alternator on no-load. Since the armature is, on open-circuit, there is no stator current and the flux due to rotor current is, distributed symmetrically in the air-gap as shown in Fig. (10.10 (i)). Since the, direction of the rotor is assumed clockwise, the generated e.m.f. in phase R1R2 is, at its maximum and is towards the paper in the conductor R1 and outwards in, conductor R2. No armature flux is produced since no current flows in the, armature winding., 261
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Fig.(10.10), Fig. (10.10 (ii)) shows the effect when a resistive load (unity p.f.) is connected, across the terminals of the alternator. According to right-hand rule, the current is, “in” in the conductors under N-pole and “out” in the conductors under S-pole., Therefore, the armature flux is clockwise due to currents in the top conductors, and anti-clockwise due to currents in the bottom conductors. Note that armature, flux is at 90° to the main flux (due to rotor current) and is behind the main flux., In this case, the flux in the air-gap is distorted but not weakened. Therefore, at, unity p.f., the effect of armature reaction is merely to distort the main field; there, is no weakening of the main field and the average flux practically remains the, same. Since the magnetic flux due to stator currents (i.e., armature flux) rotate;, synchronously with the rotor, the flux distortion remains the same for all, positions of the rotor., , (ii) When load p.f. is zero lagging, When a pure inductive load (zero p.f. lagging) is connected across the terminals, of the alternator, current lags behind the voltage by 90°. This means that current, will be maximum at zero e.m.f. and vice-versa., , Fig.(10.11), Fig. (10.11 (i)) shows the condition when the alternator is supplying resistive, load. Note that e.m.f. as well as current in phase R1R2 is maximum in the, position shown. When the alternator is supplying a pure inductive load, the, current in phase R1R2 will not reach its maximum value until N-pole advanced, 90° electrical as shown in Fig. (10.11 (ii)). Now the armature flux is from right, 262
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to left and field flux is from left to right All the flux produced by armature, current (i.e., armature flux) opposes be field flux and, therefore, weakens it. In, other words, armature reaction is directly demagnetizing. Hence at zero p.f., lagging, the armature reaction weakens the main flux. This causes a reduction in, the generated e.m.f., , (iii) When load p.f. is zero leading, When a pure capacitive load (zero p.f. leading) is connected across the terminals, of the alternator, the current in armature windings will lead the induced e.m.f. by, 90°. Obviously, the effect of armature reaction will be the reverse that for pure, inductive load. Thus armature flux now aids the main flux and the generated, e.m.f. is increased., , Fig.(10.12), Fig. (10.12 (i)) shows the condition when alternator is supplying resistive load., Note that e.m.f. as well as current in phase R1R2 is maximum in the position, shown. When the alternator is supplying a pure capacitive load, the maximum, current in R1R2 will occur 90° electrical before the occurrence of maximum, induced e.m.f. Therefore, maximum current in phase R1R2 will occur if the, position of the rotor remains 90° behind as compared to its position under, resistive load. This is illustrated in Fig. (10.12 (ii)). It is clear that armature flux, is now in the same direction as the field flux and, therefore, strengthens it. This, causes an increase in the generated voltage. Hence at zero p.f. leading, the, armature reaction strengthens the main flux., For intermediate values of p.f, the effect of armature reaction is partly distorting, and partly weakening for inductive loads. For capacitive loads, the effect of, armature reaction is partly distorting and partly strengthening. Note that in, practice, loads are generally inductive., , 263
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Summary, When the alternator is loaded, the, armature flux modifies the air-gap flux., Its angle (electrical) w.r.t. main flux, depends on the load p.t. This is, illustrated in Fig. (10.13)., (a) When the load p.f. is unity, the, effect of armature reaction is, wholly distorting. In other words,, the flux in the air-gap is distorted, but not weakened. As shown in, Fig. (10.13 (i)), the armature flux, is 90° electrical behind Ac main, flux. The result is that flux is, strengthened at the trailing pole, tips and weakened at the leading, Fig.(10.13), pole tips. However, the average, flux in the air-gap practically remains unaltered., (b) When the load p.f. is zero lagging, the effect of armature reaction is wholly, demagnetizing. In other words, the flux in the air-gap is weakened. As, shown in Fig. (10.13 (ii)), the wave representing the main flux is moved, backwards through 90° (elect) so that it is in direct opposition to the, armature flux. This considerably, reduces the air-gap flux and hence the, generated e.m.f. To keep the value of the generated e.m.f. the same, the, field excitation will have to be increased to compensate for the weakening, of the air-gap flux., (c) When the load p.f. is zero leading, the effect of armature reaction is wholly, magnetizing. In other words, the flux in the air-gap is increased. As shown, in Fig. (10.13 (iii)), the wave representing the main flux is now moved, forward through 90° (elect.) so that it aids the armature flux. This, considerably increases the air-gap flux and hence the generated e.m.f. To, keep the value of the generated e.m.f. the same, the field excitation will, have to be reduced., (d) For intermediate values of load p.f. the effect of armature reaction is partly, distorting and partly weakening for inductive loads. For capacitive loads,, the effect is partly distorting and partly strengthening. Fig. (10.13 (iv)), shows the effect of armature reaction for an inductive load. In practice,, load on the alternator is generally inductive., , 264
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10.10 Alternator on Load, Fig. (10.14) shows Y-connected alternator supplying inductive load (lagging, p.f.). When the load on the alternator is increased (i.e., armature current Ia is, increased), the field excitation and speed being kept constant, the terminal, voltage V (phase value) of the alternator decreases. This is due to, (i) Voltage drop IaRa where Ra is the armature resistance per phase., (ii) Voltage drop IaXL where XL is the armature leakage reactance per phase., (iii) Voltage drop because of armature reaction., , (i) Armature Resistance (Ra), Since the armature or stator winding has some resistance, there will be an IaRa, drop when current (Ia) flows through it. The armature resistance per phase is, generally small so that IaRa drop is negligible for all practical purposes., , (ii) Armature Leakage Reactance (XL), When current flows through the armature winding, flux is set up and a part of it, does not cross the air-gap and links the coil sides as shown in Fig. (10.15). This, leakage flux alternates with current and gives the winding self-inductance. This, is called armature leakage reactance. Therefore, there will be IaXL drop which is, also effective in reducing the terminal voltage., , Fig.(10.14), , Fig.(10.15), , (iii) Armature reaction, The load is generally inductive and the effect of armature reaction is to reduce, the generated voltage. Since armature reaction results in a voltage effect in a, circuit caused by the change in flux produced by current in the same circuit, its, effect is of the nature of an inductive reactance. Therefore, armature reaction, effect is accounted for by assuming the presence of a fictitious reactance XAR in, the armature winding. The quantity XAR is called reactance of armature reaction., The value of XAR is such that IaXAR represents the voltage drop due to armature, reaction., , 265
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Equivalent Circuit, Fig. (10.16) shows the equivalent circuit, of the loaded alternator for one phase., All the quantities are per phase. Here, E0 = No-load e.m.f., E = Load induced e.m.f. It is the, Fig.(10.16), induced e.m.f. after allowing, for armature reaction. It is, equal to phasor difference of E0 and IaXAR., V = Terminal voltage. It is less than E by voltage drops in XL and Ra., , ∴, and, , E = V + I a (R a + j X L ), E 0 = E + I a ( j X AR ), , 10.11 Synchronous Reactance (Xs), The sum of armature leakage reactance (XL) and reactance of armature reaction, (XAR) is called synchronous reactance Xs [See Fig. (10.17 (i))]. Note that all, quantities are per phase., , X s = X L + X AR, , Fig.(10.17), The synchronous reactance is a fictitious reactance employed to account for the, voltage effects in the armature circuit produced by the actual armature leakage, reactance and the change in the air-gap flux caused by armature reaction. The, circuit then reduces to the one shown in Fig. (10.17 (ii))., Synchronous impedance, Zs = Ra + j Xs, The synchronous impedance is the fictitious impedance employed to account for, the voltage effects in the armature circuit produced by the actual armature, resistance, the actual armature leakage reactance and the change in the air-gap, flux produced by armature reaction., , 266
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E 0 = V + Ia Zs = V + Ia (R + j Xs ), , 10.12 Phasor Diagram of a Loaded Alternator, Consider a Y-connected alternator supplying inductive load, the load p.f. angle, being φ. Fig. (10.18 (i)) shows the equivalent circuit of the alternator per phase., All quantities are per phase., , Fig.(10.18), Fig. (10.18 (ii)) shows the phasor diagram of an alternator for the usual case of, inductive load. The armature current Ia lags the terminal voltage V by p.f. angle, φ. The phasor sum of V and drops IaRa and IaXL gives the load induced voltage, E. It is the induced e.m.f. after allowing for armature reaction. The phasor sum, of E and IaXAR gives the no-load e.m.f. E0. The phasor diagram for unity and, leading p.f. is left as an exercise for the reader. Note that in drawing the phasor, diagram either the terminal voltage (V) or armature current (Ia) may be taken as, the reference phasor., , 10.13 Voltage Regulation, The voltage regulation of an alternator is defined as the change in terminal, voltage from no-load to full-load (the speed and field excitation being constant), divided by full-load voltage., % Voltage regulation =, , =, , No - load voltage − Full - load voltage, × 100, Full - load voltage, , E0 − V, × 100, V, , Note that E0 − V is the arithmetic difference and not the phasor difference. The, factors affecting the voltage regulation of an alternator are:, (i) IaRa drop in armature winding, (ii) IaXL drop in armature winding, (iii) Voltage change due to armature reaction, 267
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We have seen that change in terminal, voltage due to armature reaction depends, upon the armature current as well as, power-factor of the load. For leading load, p.f., the no-load voltage is less than the, full-load voltage. Hence voltage regulation, is negative in this case. The effects of, Fig.(10.19), different load power factors on the change, in the terminal voltage with changes of, load on the alternator are shown in Fig. (10.19). Since the regulation of an, alternator depends on the load and the load power factor, it is necessary to, mention power factor while expressing regulation., , 10.14 Determination of Voltage Regulation, The kVA ratings of commercial alternators are very high (e.g. 500 MVA). It is, neither convenient nor practicable to determine the voltage regulation by direct, loading. There are several indirect methods of determining the voltage, regulation of an alternator. These methods require only a small amount of power, as compared to the power required for direct loading method. Two such methods, are:, 1. Synchronous impedance or E.M.F. method, 2. Ampere-turn or M.M.F. method, For either method, the following data are required:, (i) Armature resistance, (ii) Open-circuit characteristic (O.C.C.), (iii) Short-Circuit characteristic (S.C.C.), , (i) Armature resistance, The armature resistance Ra per phase is determined by using direct current and, the voltmeter-ammeter method. This is the d.c. value. The effective armature, resistance (a.c. resistance) is greater than this value due to skin effect. It is a, usual practice to take the effective resistance 1.5 times the d.c. value (Ra = 1.5, Rdc)., , (ii) Open-circuit characteristic (O.C.C), Like the magnetization curve for a d.c. machine, the (Open-circuit characteristic, of an alternator is the curve between armature terminal voltage (phase value) on, open circuit and the field current when the alternator is running at rated speed., Fig. (10.20) shows the circuit for determining the O.C.C. of an alternator. The, alternator is run on no-load at the rated speed. The field current If is gradually, 268
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increased from zero (by adjusting field rheostat) until open-circuit voltage E0, (phase value) is about 50% greater than the rated phase voltage. The graph is, drawn between open-circuit voltage values and the corresponding values of If as, shown in Fig. (10.21)., , Fig.(10.20), , Fig.(10.21), , (iii) Short-circuit characteristic (S.C.C.), In a short-circuit test, the alternator is run at rated speed and the armature, terminals are short-circuited through identical ammeters [See Fig. (10.22)]. Only, one ammeter need be read; but three are used for balance. The field current If is, gradually increased from zero until the short-circuit armature current ISC is about, twice the rated current. The graph between short-circuit armature current and, field current gives the short-circuit characteristic (S.C.C.) as shown in Fig., (10.23)., , Fig.(10.22), , Fig.(10.23), , There is no need to take more than one reading because S.C.C. is a straight line, passing through the origin. The reason is simple. Since armature resistance is, much smaller than the synchronous reactance, the short-circuit armature current, lags the induced voltage by very nearly 90°. Consequently, the armature flux, and field flux are in direct opposition and the resultant flux is small. Since the, 269
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resultant flux is small, the saturation effects will be negligible and the shortcircuit armature current, therefore, is directly proportional to the field current, over the range from zero to well above the rated armature current., , 10.15 Synchronous Impedance Method, In this method of finding the voltage regulation of an alternator, we find the, synchronous impedance Zs (and hence synchronous reactance Xs) of the, alternator from the O.C.C. and S.S.C. For this reason, it is called synchronous, impedance method. The method involves the following steps:, (i) Plot the O.C.C. and S.S.C. on the same field current base as shown in Fig., (10.24)., (ii) Consider a field current If. The open-circuit voltage corresponding to this, field current is E1. The short-circuit armature current corresponding to field, current If is I1. On short-circuit p.d. = 0 and voltage E1 is being used to, circulate the snort-circuit armature current I1 against the synchronous, impedance Zs. This is illustrated in Fig. (10.25)., , ∴, , E1 = I1Zs, , or, , Zs =, , E1 (Open - circuit), I1 (Short - circuit), , Note that E1 is the phase value and so is I1., , Fig.(10.24), , Fig.(10.25), , (ii) The armature resistance can be found as explained earlier., ∴, , Synchronous reactance, X s = Zs2 − R 2a, , (iv) Once we know Ra and Xs, the phasor diagram can be drawn for any load, and any p.f. Fig. (10.26) shows the phasor diagram for the usual case of, inductive load; the load p.f. being cos φ lagging. Note that in drawing the, phasor diagram, current Ia has been taken as the reference phasor. The IaRa, , 270
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drop is in phase with Ia while IaXs, drop leads Ia by 90°. The phasor, sum of V, IaRa and IaXs gives the, no-load e.m.f. E0., , E 0 = (OB) 2 + (BC) 2, Now, , OB = V cos φ + I a R a, , and, , BC = V sin φ + I a X s, Fig.(10.26), , ∴, ∴, , % voltage regulation =, , E 0 = (V cos, , E0 − V, × 100, V, , Drawback, This method if easy but it gives approximate results. The reason is simple. The, combined effect of XL (armature leakage reactance) and XAR (reactance of, armature reaction) is measured on short-circuit. Since the current in this, condition is almost lagging 90°, the armature reaction will provide its worst, demagnetizing effect. It follows that under any normal operation at, say 0.8 or, 0.9 lagging power factors will produce error in calculations. This method gives a, value higher than the value obtained from an actual load test. For this reason, it, is called pessimistic method., , 10.16 Ampere-Turn Method, This method of finding voltage regulation considers the opposite view to the, synchronous impedance method. It assumes the armature leakage reactance to, be additional armature reaction. Neglecting armature resistance (always small),, this method assumes that change in terminal p.d. on load is due entirely to, armature reaction. The same two tests (viz open-circuit and short-circuit test) are, required as for synchronous reactance determination; the interpretation of the, results only is different. Under short-circuit, the current lags by 90° (Ra, considered zero) and the power factor is zero. Hence the armature reaction is, entirely demagnetizing. Since the terminal p.d. is zero, all the field AT (ampereturns) are neutralized by armature AT produced by the short circuit armature, current., (i) Suppose the alternator is supplying full-load current at normal voltage V, (i.e., operating load voltage) and zero p.f. lagging. Then d.c. field AT, required will be those needed to produce, 271
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normal voltage V (or if Ra is to be taken into account, then V + IaRa cos φ) on, no-load plus those to overcome the armature reaction,, Let, AO = field AT required to produce the normal voltage V (or V + IaRa, cos φ) at no-load, OB1 = fielder required to neutralize the armature reaction, Then total field AT required are the phasor sum of AO and OB1 [See Fig. (10.27, (i))] i.e.,, Total field AT, AB1 = AO + OB1, The AO can be found from O.C.C. and OB1 can be determined from S.C.C., Note that the use of a d.c. quantity (field AT) as a phasor is perfectly valid in, this case because the d.c. field is rotating at the same speed as the a.c. phasors, i.e., ω = 2 πf., , Fig.(10.27), (ii) For a full-load current of zero p.f. leading, the armature AT are unchanged., Since they aid the main field, less field AT are required to produce the, given e.m.f., ∴, , Total field AT, AB2 =AO − B2O, , where, , B2O = fielder required to neutralize armature reaction, , This is illustrated in Fig. (10.27 (ii)). Note that again AO are determined, from O.C.C. and B2O from S.C.C., (iii) Between zero lagging and zero leading power factors, the armature m.m.f., rotates through 180°. At unity p.f., armature reaction is cross-magnetizing, only. Therefore, OB3 is drawn perpendicular to AO [See Fig. (10.27 (iii))]., Now AB3 shows the required AT in magnitude and direction., , General case, We now discuss the case when the p.f. has any value between zero (lagging or, leading) and unity. If the power-factor is cos φ lagging, then φ is laid off to the, right of the vertical line OB3 as shown in Fig. (10.28 (i)). The total field required, are AB4 i.e., phasor sum of AO and OB4. If the power factor is cos φ leading,, then φ is laid off to the left of the vertical line OB3 as shown in Fig. (10.28 (ii))., The total field AT required are AB5 i.e., phase sum of AO and OB5., 272
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Fig.(10.28), Since current ∝ AT, it is more convenient to work, in terms of field current. Fig. (10.29) shows the, current diagram for the usual case of lagging, power factor. Here AO represents the field current, required to produce normal voltage V(or V + IaRa, cos φ) on no-load. The phasor OB represents the, field current required for producing full-load, Fig.(10.29), current on short-circuit. The resultant field current, is AB and is the phasor sum of AO and OB. Note that phasor AB represents the, field current required for demagnetizing and to produce voltage V and IaRa cos φ, drop (if Ra is taken into account)., , 10.17 Procedure for at Method, Suppose the alternator is supplying full-load current Ia at operating voltage V, and p.f. cos φ lagging. The procedure for finding voltage regulation for AT, method is as under:, (i) From the O.C.C., field current OA required to produce the operating load, voltage V (or V + IaRa cos φ) is determined [See Fig. (10.30)]. The field, current OA is laid off horizontally as shown in Fig. (10.31)., , Fig.(10.30), , Fig.(10.31), , 273
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(ii) From S.C.C., the field current OC required for producing full-load current, Ia on short-circuit is determined. The phasor AB (= OC) is drawn at an, angle of (90° + φ) i.e., ∠ OAB = (90° + φ) as shown in Fig. (10.31)., (iii) The phasor sum of OA and AB gives the total field current OB required., The O.C. voltage E0 corresponding to field current OB on O.C.C. is the noload e,m.f., E −V, ∴, % voltage regulation = 0, × 100, V, This method gives a regulation lower than the actual performance of the, machine. For this reason, it is known as Optimistic Method., , 10.18 Effect of Salient Poles, The treatment developed so far is applicable only to cylindrical rotor machines., In these machines, the air-gap is uniform so that the reluctance of the magnetic, path is the same in all directions. Therefore, the effect of armature reaction can, be accounted for by one reactance—the synchronous reactance Xs. It is because, the value of Xs is constant for all directions of armature flux relative to the rotor., However, in a salient-pole machine, the radial length of the air-gap varies [See, Fig. (10.32)] so that reluctance of the magnetic circuit along the polar axis, (called direct axis or d-axis) is much less than the reluctance along the interpolar, axis (called quadrature axis or q-axis). This is illustrated in Fig. (10.33)., , Fig.(10.32), , Fig.(10.33), , Because of the lower reluctance along the polar axis (i.e., d-axis), more flux is, produced along d-axis than along the q-axis. Therefore, reactance due to, armature reaction will be different along d-axis and q-axis. These are:, Xad = direct axis reactance due to armature reaction, Xaq = quadrature axis reactance due to armature reaction, , 274
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10.19 Two-Reactance Concept for Salient-Pole Machines, The effects of salient poles can be taken into, account by resolving the armature current into, two components; Id perpendicular to, excitation voltage E0 and Ia along E0 as shown, in phasor diagram in Fig. (10.34). Note that, this diagram is drawn for an unsaturated, salient-pole generator operating at a lagging, power factor cos φ. With each of the, component currents Id and Iq, there is, associated, a, component, synchronous, reactance Xd and Xq respectively., Xd = direct axis synchronous reactance, Xq = quadrature axis synchronous reactance, , Fig.(10.34), , If X1 is the armature leakage reactance and is assumed to be constant for direct, and quadrature-axis currents, then,, , X d = X ad + X l ;, , X q = X aq + X l, , Note that in drawing the phasor diagram, the armature resistance Ra is neglected, since it is quite small. Further, all values arc phase values. Here V is the terminal, voltage phase and E0 is the e.m.f. per phase to which the generator is excited., Referring to Fig. (10.34)., , I q = I a cos(δ + φ) and, , I d = I a sin(δ + φ), , The angle δ between E0 and V is called the power angle., , E 0 = V cos δ + I d X d, = V cos δ + I a X d sin( δ + φ), ∴, , E 0 = V cos δ + I a X d (sin δ cos φ + cos δ sin φ), , V sin δ = I q X q, , Also, , (i), , [Q I q = I a cos(δ + φ)], , = I a X q cos(δ + φ), ∴, , [Q I d = I a sin (δ + φ)], , V sin δ = I a X q (cos δ cos φ − sin δ sin φ), , 275, , (ii)
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∴, , V 2 (X d − X q ), E0V, Pd =, sin δ +, sin 2δ per phase, Xd, 2X d X q, , (ii), , The total power developed would be three times the above power. The following, points may be noted:, (i) If there is no saliency, Xd = Xq, E V, Pd = 0 sin δ per phase, ∴, Xd, This is the power developed by cylindrical rotor machine., (ii) The second term in eq. (ii) above introduces the effect of salient poles. It, represents the reluctance power i.e., power due to saliency., , V 2 (X d − X q ), (iii) If E0 = 0, then,, Pd =, sin 2δ per phase, 2X d X q, The power obtained with zero excitation is called reluctance power. It, should be noted that reluctance power is independent of the excitation., , Power angle characteristic, Fig (10.36) shows the power-angle, characteristic of a salient-pole machine. It, is clear that reluctance power varies with δ, at twice the rate of the excitation power., The peak power is seen to be displaced, towards δ = 0, the amount of displacement, depends upon the excitation. In Fig., (10.36), the excitation is such that the, excitation term has a peak value about 2.5, times that of the reluctance term. Under, steady-state conditions, the reluctance term, is positive because Xd > Xq., , Fig.(10.36), , 10.21 Parallel Operation of Alternators, It is rare to find a 3-phase alternator supplying its own load independently, except under test conditions. In practice, a very large number of 3-phase, alternators operate in parallel because the various power stations are, interconnected through the national grid. Therefore, the output of any single, alternator is small compared with the total interconnected capacity. For example,, the total capacity of the interconnected system may be over 40,000 MW while, the capacity of the biggest single alternator may be 500 MW. For this reason, the, performance of a single alternator is unlikely to affect appreciably the voltage, 277
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and frequency of the whole system. An alternator connected to such a system is, said to be connected to infinite busbars. The outstanding electrical, characteristics of such busbars are that they are constant-voltage, constantfrequency busbars., Fig. (10.37) shows a typical infinite bus system. Loads are tapped from the, infinite bus at various load centres. The alternators may be connected to or, disconnected from the infinite bus, depending on the power demand on the, system. If an alternator is connected to infinite busbars, no matter what power is, delivered by the incoming alternator, the voltage and frequency of the system, remain the same. The operation of connecting an alternator to the infinite, busbars is known as paralleling with the infinite busbars. It may be noted that, before an alternator is connected to an infinite busbars, certain conditions must, be satisfied., , Fig.(10.37), , 10.22 Advantages of Parallel Operation of Alternators, The following are the advantages of operating alternators in parallel:, (i) Continuity of service. The continuity of service is one of the important, requirements of any electrical apparatus. If one alternator fails, the, continuity of supply can be maintained through the other healthy units., This will ensure uninterrupted supply to the consumers., (ii) Efficiency. The load on the power system varies during the whole day;, being minimum during die late night hours. Since alternators operate most, efficiently when delivering full-load, units can be added or put off, depending upon the load requirement. This permits the efficient operation, of the power system., (iii) Maintenance and repair. It is often desirable to carry out routine, maintenance and repair of one or more units. For this purpose, the desired, unit/units can be shut down and the continuity of supply is maintained, through the other units., , 278
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(iv) Load growth. The load demand is increasing due to the increasing use of, electrical energy. The load growth can be met by adding more units without, disturbing the original installation., , 10.23 Conditions for Paralleling Alternator with Infinite, Busbars, The proper method of connecting an alternator to the infinite busbars is called, synchronizing. A stationary alternator must not be connected to live busbars. It, is because the induced e.m.f. is zero at standstill and a short-circuit will result. In, order to connect an alternator safely to the infinite busbars, the following, conditions are met:, (i) The terminal voltage (r.m.s. value) of the incoming alternator must be the, same as busbars voltage., (ii) The frequency of the generated voltage of the incoming alternator must be, equal to the busbars frequency., (iii) The phase of the incoming alternator voltage must be identical with the, phase of the busbars voltage. In other words, the two voltages must be in, phase with each other., (iv) The phase sequence of the voltage of the incoming alternator should be the, same as that of the busbars., The magnitude of the voltage of the incoming alternator can be adjusted by, changing its field excitation. The frequency of the incoming alternator can be, changed by adjusting the speed of the prime mover driving the alternator., Condition (i) is indicated by a voltmeter, conditions (ii) and (iii) are indicated by, synchronizing lamps or a synchroscope. The condition (iv) is indicated by a, phase sequence indicator., , 10.24 Methods of Synchronization, The method of connecting an incoming alternator safely to the live busbars is, called synchronizing. The equality of voltage between the incoming alternator, and the busbars can be easily checked by a voltmeter. The phase sequence of the, alternator and the busbars can be checked by a phase sequence indicator., Differences in frequency and phase of the voltages of the incoming alternator, and busbars can be checked by one of the following two methods:, (i) By Three Lamp (one dark, two bright) method, (ii) By synchroscope, , (i) Three lamp method, , 279
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In this method of synchronizing, three lamps L1, L2 and L3 are connected as, shown in Fig. (10.38). The lamp L1 is straight connected between the, corresponding phases (R1 and R2) and the other two are cross-connected, between the other two phases. Thus lamp L2 is connected between Y1 and B2, and lamp L3 between B1 and Y2. When the frequency and phase of the voltage of, the incoming alternator is the same as that of the busbars, the straight connected, lamps L1 will be dark while cross-connected lamps L2 and L3 will be equally, bright. At this instant, the synchronization is perfect and the switch of the, incoming alternator can be closed to connect it to the busbars., , Fig.(10.38), , Fig.(10.39), , In Fig. (10.39), phasors R1, Y1 and B1 represent the busbars voltages and phasors, R2, Y2 and B2 represent the voltages of the incoming alternator. At the instant, when R1 is in phase with R2, voltage across lamp L1 is zero and voltages across, lamps L2 and L3 are equal. Therefore, the lamp L1 is dark while lamps L2 and L3, will be equally bright. At this instant, the switch of the incoming alternator can, be closed. Thus incoming alternator gets connected in parallel with the busbars., , (ii) Synchroscope, A synchroscope is an, instrument that indicates by, means of a revolving pointer, the phase difference and, frequency difference between, the voltages of the incoming, alternator and the busbars, [See Fig. (10.40)]. It is, essentially-a small motor, the, , Fig.(10.40), , 280
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field being supplied from the busbars through a potential transformer and the, rotor from the incoming alternator. A pointer is attached to the rotor. When the, incoming alternator is running fast (i.e., frequency of the incoming alternator is, higher than that of the busbars), the rotor and hence the pointer moves in the, clockwise direction. When the incoming alternator is running slow (i.e.,, frequency of the incoming alternator is lower than that of the busbars), the, pointer moves in anti-clockwise direction. When the frequency of the incoming, alternator is equal to that of the busbars, no torque acts on the rotor and the, pointer points vertically upwards (“12 O’ clock”). It indicates the correct instant, for connecting the incoming alternator to the busbars. The synchroscope method, is superior to the lamp method because it not only gives a positive indication of, the time to close the switch but also indicates the adjustment to be made should, there be a difference between the frequencies of the incoming alternator and the, busbars., , 10.25 Synchronising Action, When two or more alternators have been connected in parallel, they will remain, in stable operation under all normal conditions i.e., voltage, frequency, speed, and phase equality will continue. In other words, once synchronized properly,, the alternators will continue to run in synchronism under all normal conditions., If one alternator tries to fall out of synchronism, it is immediately counteracted, by the production of a synchronizing torque which brings it back to, synchronism. This automatic action is called the synchronizing action of the, alternators., Consider two similar single-phase alternators 1 and 2 operating in parallel as, shown in Fig. (10.41 (i)). For simplicity, let us assume that the alternators are at, no-load. When in exact synchronism, the magnitudes of the e.m.f.s E1 (machine, 1) and E2 (machine 2) are equal. These e.m.f.s are acting in the same direction, with respect to the external circuit [See Fig. (10.41 (ii))]. But in relation to each, other, these e.m.f.s are in phase opposition i.e., if we trace the closed circuit, formed by the two alternators we find that the e.m.f.s oppose each other [See, Fig. (10.41 (iii))]. When the alternators are in exact synchronism, E1 and E2 are, in exact phase opposition. Since E1 = E2 in magnitude, no current flows in the, closed circuit formed by the two alternators., , 281
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Fig.(10.41), If one alternator drops out of synchronism, there is an automatic action to reestablish synchronism. Let us discuss this point., , 282
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(i) Effect of speed change, Suppose, due to any reason, the speed of machine 2 falls. Then e.m.f. E2 will fall, back by a phase angle of a electrical degrees as shown in Fig. (10.42) (though, still E1 = E2). There will be resultant e.m.f. Er in the closed circuit formed by the, two alternators. This e.m.f. Er will circulate current (known as synchronizing, current Isyt) in this closed circuit., Synchronizing current, I sy =, , Er, 2 Zs, , The current Isy lags behind Er by an angle θ given by;, , tan θ =, where, , 2X s X s, =, 2R a R a, , Ra = armature resistance of each alternator, Xs = synchronous reactance of each alternator, Zs = synchronous impedance of each alternator, , Fig.(10.42), , Fig.(10.43), , Since Ra is very small as compared to Xs, θ is nearly 90° so that the current Isy is, almost in phase with E1 and in phase opposition to E2. This means that machine, 1 is generating and machine 2 is motoring. Consequently, machine 1 tends to, slow down and machine 2, by accepting power, tends to accelerate. This restores, the status quo i.e., synchronism is re-established., Conversely, if E2 tends to advance in phase [See Fig. (10.43)], the directions of, Er and Isy are changed such that now machine 2 is generating and machine 1 is, motoring. Once again the synchronism is restored., , (ii) Effect of inequality of e.m.f.s., The automatic re-establishment of synchronism of two alternators operating in, parallel also extends to any changes tending to alter the individual e.m.f.s. When, in exact synchronism, then E1 = E2 (magnitude) and they are in exact phase, opposition as shown in Fig. (10.44 (i)). Suppose due to any reason, e.m.f. E1, increases. Then resultant e.m.f. Er exists in the closed circuit formed by the two, 283
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alternators. Then Er = E1 − E2 and is in phase with E1. The resultant e.m.f. Er, sends synchronizing current Isy in the closed circuit. Here again the current Isy, almost lags behind Er. by 90° (Q Zs ~ Xs) as shown in Fig. (10.44 (ii)). Also Isy, lags almost 90° behind E1 and leads E2 almost by 90°. The power produced is, practically zero; just enough to overcome copper losses. The current Isy lags, behind E1 and produces a demagnetizing armature reaction effect on machine 1., At the same time, Isy leads E2 and produces magnetizing armature reaction effect, on machine 2. Thus E1 tends to fall and E2 tends to rise. The result is that, synchronism is re-established. The converse is true for E2 > E1 as shown in Fig., (10.44 (iii))., , Fig.(10.44), , 10.26 Synchronizing Power, When two alternators are operating in parallel, each machine has an inherent, tendency to remain synchronized. Consider two similar single-phase alternators, 1 and 2 operating in parallel at no-load [See Fig. (10.45)]. Suppose, due to any, reason, the speed of machine 2 decreases. This will cause E2 to fall back by a, phase angle of a electrical degrees as shown in Fig. (10.46) (though still E1 =, E2). Within the local circuit formed by two alternators, the resultant e.m.f. Er is, the phasor difference E1 − E2. This resultant e.m.f. results in the production of, synchronizing current Isy which sets up synchronizing torque. The synchronizing, torque retards machine 1 and accelerates machine 2 so that synchronism is reestablished. The power associated with synchronizing torque is called, synchronizing power., In Fig. (10.45), machine 1 is generating and machine 2 is motoring. The power, supplied by machine 1 is called synchronizing power. Referring to Fig. (10.46),, we have,, Synchronizing power, Psy =E1Isy cos φ1 = E1Isy cos(90° − θ) = E1Isy sin θ, (Q θ ~ 90° so that sin θ = 1), , = E1 Isy, , 284
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10.29 Sharing of Load Currents by Two Alternators in, Parallel, Consider two alternators with identical, speed/load characteristics connected in, parallel as shown in Fig. (10.47)., Let E1,E2 = induced e.m.f.s per phase, Z1,Z2 = synchronous impedances per, phase, Z = load impedance per phase, I1,I2 = currents supplied by two machines, V = common terminal voltage per phase, V = E1 − I1Z1 = E2 − I2Z2, , ∴, , I1 =, , E1 − V, ;, Z1, , I = I1 + I 2 =, , I2 =, , Fig.(10.47), , E2 − V, Z2, , E1 − V E1 − V, +, Z1, Z1, , V = (I1 + I 2 ) Z = IZ, Circulating current on no-load is, , IC =, , E1 − E 2, Z1 + Z 2, , 10.30 Alternator on Infinite Busbars, Upto this point, we have considered only a single alternator working on an, isolated load or two such machines operating in parallel. In practice, generating, stations do not operate as isolated units but are interconnected by the national, grid. The result is that a very large number of alternators operate in parallel. An, alternator connected to such a network is said to be operating on infinite, busbars. The behaviour of alternators connected to an infinite busbars is as, under:, (i) Any change made in the operating conditions of one alternator will not, change the terminal voltage or frequency of the system. In other words,, terminal voltage (busbars voltage) and frequency are not affected by, changing the operating conditions of one alternator. It is because of large, size and inertia of the system., (ii) The kW output supplied by an alternator depends solely on the mechanical, power supplied to the prime mover of the alternator. An increase in, mechanical power to the prime mover increases the kW output of the, , 287
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alternator and not the kVAR. A decrease in the mechanical power to the, prime mover decreases the kW output of the alternator and not the kVAR., (iii) If the mechanical power to the prime mover of an alternator is kept, constant, then change in excitation will change the power factor at which, the machine supplies changed current. In other words, change of excitation, controls the kVAR and not kW., The change of driving torque controls the kW output and not kVAR of an, alternator. The change of excitation controls the kVAR and not the kW output of, an alternator., Note. An infinite busbars system has constant terminal voltage and constant, busbars frequency because of its large size and inertia. However, the busbars, voltage can be raised or lowered by increasing or decreasing simultaneously the, field excitation of a large number of alternators. Likewise, system frequency can, be raised or lowered by increasing or decreasing the speed of prime movers of a, large number of alternators., , 10.31 Effect of Change of Excitation and Mechanical, Input, Consider a star-connected alternator connected to an infinite busbars as shown, in Fig. (10.48). Note that infinite busbars means that busbars voltage will remain, constant and no frequency change will occur regardless of changes made in, power input or field excitation of the alternator connected to it., , Fig.(10.48), Let, , Fig.(10.49), , V = busbars voltage/phase, E = e.m.f. of alternator/phase, Xs = synchronous reactance of alternator/phase, Armature current/phase, I s =, , E−V, Xs, , Fig. (10.49) shows the equivalent circuit of alternator for one phase., 288
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(i) Effect of change of field excitation, Suppose the alternator connected to infinite busbars is operating at unity p.f. It is, then said to be normally excited. Suppose that excitation of the alternator is, increased (overexcited) while the power input to the prime mover is unchanged., The active power output (W or kW) of the alternator will thus remain unchanged, i.e., active component of current is unaltered. The overexcited alternator will, supply lagging current (and hence lagging reactive power) to the infinite, busbars. This action can be explained by the m.m.f. of armature reaction. When, the alternator is overexcited, it must deliver lagging current since lagging current, produces an opposing m.m.f. to reduce the over-excitation. Thus an overexcited, alternator supplies lagging current in addition to the constant active component, of current. Therefore, an overexcited alternator will operate at lagging power, factor. Note that excitation does not control the active power but it controls, power factor of the current supplied by the alternator to the infinite busbars. Fig., (10.50) shows the phasor diagram of an overexcited alternator connected to, infinite busbars. The angle δ between E and V is called power angle., , Fig.(10.50), , Fig.(10.51), , Now suppose that excitation of the alternator is decreased below normal, excitation (under-excitation) while the power input to the prime mover is, unchanged. Therefore, the active power output (W or kW) of the alternator will, remain unchanged L e., active component of current is unaltered. The, underexcited alternator supplies leading current (and hence leading reactive, power) to the infinite busbars. It is because when an alternator is underexcited, it, must deliver leading current since leading current produces an aiding m.m.f. to, increase the underexcitation. Thus an underexcited alternator supplies leading, current in addition to the constant active component of current. Therefore, an, underexcited alternator will operate at leading power factor. Fig. (10.51) shows, the phasor diagram of an underexcited alternator connected to infinite busbars., Conclusion. An overexcited alternator operates at lagging power factor and, supplies lagging reactive power to infinite busbars. On the other hand, an, underexcited alternator operates at leading power factor and supplies leading, reactive power to the infinite busbars., , 289
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(ii) Effect of change in mechanical input, Suppose the alternator is delivering power to infinite busbars under stable, conditions so that a certain power angle δ exists between V and E and E leads V., The phasor diagram for this situation is depicted in Fig. (10.52). Now, suppose, that excitation of the alternator is kept constant and power input to its prime, mover is increased. The increase in power input would tend to accelerate the, rotor and £ would move further ahead of V i.e., angle δ increases. Increasing δ, results in larger Ia (= E − V/Xs) and lower φ as shown in Fig. (10.53). Therefore,, the alternator will deliver more active power to the infinite busbars. The angle δ, assumes such a value that current Ia has an active power component, corresponding to the input: Equilibrium will be reestablished at the speed, corresponding to the frequency of the infinite busbars with a larger δ. Fig., (10.53) is drawn for the same d.c. field excitation and, therefore, the same E as, Fig. (10.52) but the active power output (= VIc cos φ) is greater than for the, condition of Fig. (10.52) and increase in 6 has caused the alternator to deliver, additional active power to the busbars. Note that mechanical input to the prime, mover cannot change the speed of the alternator because it is fixed by system, frequency. Increasing mechanical input increases the speed of the alternator, temporarily till such time the power angle δ increases to a value required for, stable operation. Once this condition is reached, the alternator continues to run, at synchronous speed., , Fig.(10.52), , Fig.(10.53), , Conclusion. Increasing the mechanical input power to the prime mover will not, change the speed ultimately but will increase the power angle δ. As a result, the, change of driving torque controls the output kW and not the kVAR. When this, change takes place, the power factor of the machine is practically not affected., , 10.32 Power Output Equation, Consider a star-connected cylindrical rotor alternator operating on infinite, busbars., Let, V = busbars voltage/phase, E = generated e.m.f./phase, 290
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Ia = armature current/phase delivered by the alternator, Zs = synchronous impedance/phase = Ra + j Xs, cosφ = lagging p.f. of the alternator, θ = internal angle = tan-1 Xs/Ra, , Fig.(10.54), , Fig.(10.55), , Fig. (10.54) shows the equivalent circuit for one phase of the alternator whereas, Fig. (10.55) shows the phasor diagram. Note that in drawing the phasor diagram,, V has been taken as the reference phasor. We shall now derive an expression for, the power delivered by the alternator. Referring to Fig. (10.55),, , E ∠ δ − V ∠0 E, V, Ia =, =, ∠ (δ − θ) −, ∠−θ, Zs ∠ θ, Zs, Zs, Power output/phase, P = V × Real part of Ia, E, , V, = V × cos (δ − θ) −, cos( − θ), Zs, Zs, , V, = [E cos (δ − θ) − V cos( − θ)], Zs, , =, , ∴, , V2 E, cos (δ − θ) − cos θ, , , Zs V, , EV, V2, P=, cos (δ − θ) −, cos θ, Zs, Zs, , (i), , Eq. (i) gives the electrical output of the alternator in terms of E, V, Z, θ and load, angle δ., , Maximum power output, For given E, V and frequency, the conditions for maximum power output can be, obtained by differentiating eq. (i) w.r.t. δ and equating the result to zero i.e.,, , dP, =0, dδ, , 291
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machine will lose synchronism. The dotted portion of the curve refers to, unstable operation, i.e., machine loses synchronism., Note that stability of the alternator is determined by the power/power angle, characteristic. Suppose the operating position of the alternator is represented by, point P on the curve. If unsteadiness occurs due to a transient spike of, mechanical input, then load angle δ increases by a small amount. The additional, electrical output caused by an increase in δ produces a torque which is not, balanced by the driving torque once the spike has passed. This torque causes, retardation of the rotor and the alternator returns to the operating point P. The, torque causing the return of the alternator to the steady-state position is called, the synchronizing torque and the power associated with it is known as, synchronizing power., , 10.34 Hunting, Sometimes an alternator will not operate satisfactorily with others due to, hunting. If the driving torque applied to an alternator is pulsating such as that, produced by a diesel engine, the alternator rotor may be pulled periodically, ahead of or behind its normal position as it rotates. This oscillating action is, called hunting. Hunting causes the alternators to shift load from one to another., In some cases, this oscillation of power becomes cumulative and violent enough, to cause the alternator to pull out of synchronism., In salient-pole machines, hunting is reduced by, providing damper winding. It consists of shortcircuited copper bars embedded in the pole faces, as shown in Fig. (10.57). When hunting occurs,, there is shifting of armature flux across the pole, faces, thereby inducing currents in the damper, winding. Since any induced current opposes the, action that produces it, the hunting action is, opposed by the flow of induced currents. The, following points may be noted:, (i) Hunting generally occurs in alternators driven, Fig.(10.57), by engines because the driving torque of, engines is not uniform., (ii) Alternators driven by steam turbines generally do not have a tendency to, hunt since the torque applied does not pulsate., (iii) In cylindrical rotor machines, the damper windings are generally not used., It is because the solid rotor provides considerable damping., Note. Under normal running condition damper winding does not carry any, current because rotor runs at synchronous speed., 293
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Chapter (11), , Synchronous Motors, Introduction, It may be recalled that a d.c. generator can be run as a d.c. motor. In like, manner, an alternator may operate as a motor by connecting its armature, winding to a 3-phase supply. It is then called a synchronous motor. As the name, implies, a synchronous motor runs at synchronous speed (Ns = 120f/P) i.e., in, synchronism with the revolving field produced by the 3-phase supply. The speed, of rotation is, therefore, tied to the frequency of the source. Since the frequency, is fixed, the motor speed stays constant irrespective of the load or voltage of 3phase supply. However, synchronous motors are not used so much because they, run at constant speed (i.e., synchronous speed) but because they possess other, unique electrical properties. In this chapter, we shall discuss the working and, characteristics of synchronous motors., , 11.1 Construction, A synchronous motor is a machine that operates at synchronous speed and, converts electrical energy into mechanical energy. It is fundamentally an, alternator operated as a motor. Like an alternator, a synchronous motor has the, following two parts:, (i) a stator which houses 3-phase armature, winding in the slots of the stator core and, receives power from a 3-phase supply [See, (Fig. (11.1)]., (ii) a rotor that has a set of salient poles excited, by direct current to form alternate N and S, poles. The exciting coils are connected in, series to two slip rings and direct current is, fed into the winding from an external exciter, mounted on the rotor shaft., The stator is wound for the same number of poles, as the rotor poles. As in the case of an induction, motor, the number of poles determines the, synchronous speed of the motor:, , 293, , Fig.(11.1)
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Synchronous speed, N s =, where, , 120f, P, , f = frequency of supply in Hz, P = number of poles, , An important drawback of a synchronous motor is that it is not self-starting and, auxiliary means have to be used for starting it., , 11.2 Some Facts about Synchronous Motor, Some salient features of a synchronous motor are:, (i) A synchronous motor runs at synchronous speed or not at all. Its speed is, constant (synchronous speed) at all loads. The only way to change its speed, is to alter the supply frequency (Ns = 120 f/P)., (ii) The outstanding characteristic of a synchronous motor is that it can be, made to operate over a wide range of power factors (lagging, unity or, leading) by adjustment of its field excitation. Therefore, a synchronous, motor can be made to carry the mechanical load at constant speed and at, the same time improve the power factor of the system., (iii) Synchronous motors are generally of the salient pole type., (iv) A synchronous motor is not self-starting and an auxiliary means has to be, used for starting it. We use either induction motor principle or a separate, starting motor for this purpose. If the latter method is used, the machine, must be run up to synchronous speed and synchronized as an alternator., , 11.3 Operating Principle, The fact that a synchronous motor has no starting torque can be easily explained., (i) Consider a 3-phase synchronous motor having two rotor poles NR and SR., Then the stator will also be wound for two poles NS and SS. The motor has, direct voltage applied to the rotor winding and a 3-phase voltage applied to, the stator winding. The stator winding produces a rotating field which, revolves round the stator at synchronous speed Ns(= 120 f/P). The direct (or, zero frequency) current sets up a two-pole field which is stationary so long, as the rotor is not turning. Thus, we have a situation in which there exists a, pair of revolving armature poles (i.e., NS − SS) and a pair of stationary rotor, poles (i.e., NR − SR)., (ii) Suppose at any instant, the stator poles are at positions A and B as shown, in Fig. (11.2 (i)). It is clear that poles NS and NR repel each other and so do, the poles SS and SR. Therefore, the rotor tends to move in the anticlockwise direction. After a period of half-cycle (or ½ f = 1/100 second),, the polarities of the stator poles are reversed but the polarities of the rotor, poles remain the same as shown in Fig. (11.2 (ii)). Now SS and NR attract, 294
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each other and so do NS and SR. Therefore, the rotor tends to move in the, clockwise direction. Since the stator poles change their polarities rapidly,, they tend to pull the rotor first in one direction and then after a period of, half-cycle in the other. Due to high inertia of the rotor, the motor fails to, start., , Fig.(10.2), Hence, a synchronous motor has no self-starting torque i.e., a synchronous, motor cannot start by itself., How to get continuous unidirectional torque? If the rotor poles are rotated by, some external means at such a speed that they interchange their positions along, with the stator poles, then the rotor will experience a continuous unidirectional, torque. This can be understood from the following discussion:, (i) Suppose the stator field is rotating in the clockwise direction and the rotor, is also rotated clockwise by some external means at such a speed that the, rotor poles interchange their positions along with the stator poles., (ii) Suppose at any instant the stator and rotor poles are in the position shown, in Fig. (11.3 (i)). It is clear that torque on the rotor will be clockwise. After, a period of half-cycle, the stator poles reverse their polarities and at the, same time rotor poles also interchange their positions as shown in Fig., (11.3 (ii)). The result is that again the torque on the rotor is clockwise., Hence a continuous unidirectional torque acts on the rotor and moves it in, the clockwise direction. Under this condition, poles on the rotor always, face poles of opposite polarity on the stator and a strong magnetic, attraction is set up between them. This mutual attraction locks the rotor and, stator together and the rotor is virtually pulled into step with the speed of, revolving flux (i.e., synchronous speed)., (iii) If now the external prime mover driving the rotor is removed, the rotor will, continue to rotate at synchronous speed in the clockwise direction because, the rotor poles are magnetically locked up with the stator poles. It is due to, 295
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this magnetic interlocking between stator and rotor poles that a, synchronous motor runs at the speed of revolving flux i.e., synchronous, speed., , Fig.(11.3), , 11.4 Making Synchronous Motor Self-Starting, A synchronous motor cannot start by itself. In order, to make the motor self-starting, a squirrel cage, winding (also called damper winding) is provided, on the rotor. The damper winding consists of copper, bars embedded in the pole faces of the salient poles, of the rotor as shown in Fig. (11.4). The bars are, short-circuited at the ends to form in effect a partial, Fig.(11.4), squirrel cage winding. The damper winding serves, to start the motor., (i) To start with, 3-phase supply is given to the stator winding while the rotor, field winding is left unenergized. The rotating stator field induces currents, in the damper or squirrel cage winding and the motor starts as an induction, motor., (ii) As the motor approaches the synchronous speed, the rotor is excited with, direct current. Now the resulting poles on the rotor face poles of opposite, polarity on the stator and a strong magnetic attraction is set up between, them. The rotor poles lock in with the poles of rotating flux. Consequently,, the rotor revolves at the same speed as the stator field i.e., at synchronous, speed., (iii) Because the bars of squirrel cage portion of the rotor now rotate at the same, speed as the rotating stator field, these bars do not cut any flux and,, therefore, have no induced currents in them. Hence squirrel cage portion of, the rotor is, in effect, removed from the operation of the motor., , 296
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It may be emphasized here that due to magnetic interlocking between the stator, and rotor poles, a synchronous motor can only run at synchronous speed. At any, other speed, this magnetic interlocking (i.e., rotor poles facing opposite polarity, stator poles) ceases and the average torque becomes zero. Consequently, the, motor comes to a halt with a severe disturbance on the line., Note: It is important to excite the rotor with direct current at the right moment., For example, if the d.c. excitation is applied when N-pole of the stator faces Npole of the rotor, the resulting magnetic repulsion will produce a violent, mechanical shock. The motor will immediately slow down and the circuit, breakers will trip. In practice, starters for synchronous motors arc designed to, detect the precise moment when excitation should be applied., , 11.5 Equivalent Circuit, Unlike the induction motor, the synchronous motor is connected to two, electrical systems; a d.c. source at the rotor terminals and an a.c. system at the, stator terminals., 1. Under normal conditions of synchronous motor operation, no voltage is, induced in the rotor by the stator field because the rotor winding is rotating, at the same speed as the stator field. Only the impressed direct current is, present in the rotor winding and ohmic resistance of this winding is the, only opposition to it as shown in Fig. (11.5 (i))., 2. In the stator winding, two effects are to be considered, the effect of stator, field on the stator winding and the effect of the rotor field cutting the stator, conductors at synchronous speed., , Fig.(11.5), (i), , The effect of stator field on the stator (or armature) conductors is, accounted for by including an inductive reactance in the armature, winding. This is called synchronous reactance Xs. A resistance Ra must, be considered to be in series with this reactance to account for the copper, losses in the stator or armature winding as shown in Fig. (11.5 (i)). This, , 297
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resistance combines with synchronous reactance and gives the, synchronous impedance of the machine., (ii) The second effect is that a voltage is generated in the stator winding by, the synchronously-revolving field of the rotor as shown in Fig. (11.5 (i))., This generated e.m.f. EB is known as back e.m.f. and opposes the stator, voltage V. The magnitude of Eb depends upon rotor speed and rotor flux, φ per pole. Since rotor speed is constant; the value of Eb depends upon, the rotor flux per pole i.e. exciting rotor current If., Fig. (11.5 (i)) shows the schematic diagram for one phase of a star-connected, synchronous motor while Fig. (11.5 (ii)) shows its equivalent circuit. Referring, to the equivalent circuit in Fig. (11.5 (ii))., Net voltage/phase in stator winding is, Er = V − Eb, Armature current/phase, I a =, where, , phasor difference, , Er, Zs, , Zs = R 2a + X s2, , This equivalent circuit helps considerably in understanding the operation of a, synchronous motor., A synchronous motor is said to be normally excited if the field excitation is such, that Eb = V. If the field excitation is such that Eb < V, the motor is said to be, under-excited. The motor is said to be over-excited if the field excitation is such, that Eb > V. As we shall see, for both normal and under excitation, the motor has, lagging power factor. However, for over-excitation, the motor has leading power, factor., Note: In a synchronous motor, the value of Xs is 10 to 100 times greater than Ra., Consequently, we can neglect Ra unless we are interested in efficiency or, heating effects., , 11.6 Motor on Load, In d.c. motors and induction motors, an addition of load causes the motor speed, to decrease. The decrease in speed reduces the counter e.m.f. enough so that, additional current is drawn from the source to carry the increased load at a, reduced speed. This action cannot take place in a synchronous motor because it, runs at a constant speed (i.e., synchronous speed) at all loads., What happens when we apply mechanical load to a synchronous motor? The, rotor poles fall slightly behind the stator poles while continuing to run at, 298
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synchronous speed. The angular displacement between stator and rotor poles, (called torque angle α) causes the phase of back e.m.f. Eb to change w.r.t. supply, voltage V. This increases the net e.m.f. Er in the stator winding. Consequently,, stator current Ia ( = Er/Zs) increases to carry the load., , Fig.(11.6), The following points may be noted in synchronous motor operation:, (i) A synchronous motor runs at synchronous speed at all loads. It meets the, increased load not by a decrease in speed but by the relative shift between, stator and rotor poles i.e., by the adjustment of torque angle α., (ii) If the load on the motor increases, the torque angle a also increases (i.e.,, rotor poles lag behind the stator poles by a greater angle) but the motor, continues to run at synchronous speed. The increase in torque angle α, causes a greater phase shift of back e.m.f. Eb w.r.t. supply voltage V. This, increases the net voltage Er in the stator winding. Consequently, armature, current Ia (= Er/Zs) increases to meet the load demand., (iii) If the load on the motor decreases, the torque angle α also decreases. This, causes a smaller phase shift of Eb w.r.t. V. Consequently, the net voltage Er, in the stator winding decreases and so does the armature current Ia (=, Er/Zs)., , 11.7 Pull-Out Torque, There is a limit to the mechanical load that can be applied to a synchronous, motor. As the load increases, the torque angle α also increases so that a stage is, reached when the rotor is pulled out of synchronism and the motor comes to a, standstill. This load torque at which the motor pulls out of synchronism is called, pull—out or breakdown torque. Its value varies from 1.5 to 3.5 times the full—, load torque., When a synchronous motor pulls out of synchronism, there is a major, disturbance on the line and the circuit breakers immediately trip. This protects, the motor because both squirrel cage and stator winding heat up rapidly when, the machine ceases to run at synchronous speed., 299
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11.8 Motor Phasor Diagram, Consider an under-excited ^tar-connected synchronous motor (Eb < V) supplied, with fixed excitation i.e., back e.m.f. Eb is constantLet, V = supply voltage/phase, Eb = back e.m.f./phase, Zs = synchronous impedance/phase, , (i) Motor on no load, When the motor is on no load, the torque angle α is small as shown in Fig. (11.7, (i)). Consequently, back e.m.f. Eb lags behind the supply voltage V by a small, angle δ as shown in the phasor diagram in Fig. (11.7 (iii)). The net voltage/phase, in the stator winding, is Er., Armature current/phase, Ia = Er/Zs, The armature current Ia lags behind Er by θ = tan-1 Xs/Ra. Since Xs >> Ra, Ia lags, Er by nearly 90°. The phase angle between V and Ia is φ so that motor power, factor is cos φ., Input power/phase = V Ia cos φ, , Fig.(11.7), Thus at no load, the motor takes a small power VIa cos φ/phase from the supply, to meet the no-load losses while it continues to run at synchronous speed., , (ii) Motor on load, When load is applied to the motor, the torque angle a increases as shown in Fig., (11.8 (i)). This causes Eb (its magnitude is constant as excitation is fixed) to lag, behind V by a greater angle as shown in the phasor diagram in Fig. (11.8 (ii))., The net voltage/phase Er in the stator winding increases. Consequently, the, motor draws more armature current Ia (=Er/Zs) to meet the applied load., Again Ia lags Er by about 90° since Xs >> Ra. The power factor of the motor is, cos φ., , 300
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Input power/phase, Pi = V Ia cos φ, Mechanical power developed by motor/phase, Pm = Eb × Ia × cosine of angle between Eb and Ia, = Eb Ia cos(δ − φ), , Fig.(11.8), , 11.9 Effect of Changing Field Excitation at Constant Load, In a d.c. motor, the armature current Ia is determined by dividing the difference, between V and Eb by the armature resistance Ra. Similarly, in a synchronous, motor, the stator current (Ia) is determined by dividing voltage-phasor resultant, (Er) between V and Eb by the synchronous impedance Zs., One of the most important features of a synchronous motor is that by changing, the field excitation, it can be made to operate from lagging to leading power, factor. Consider a synchronous motor having a fixed supply voltage and driving, a constant mechanical load. Since the mechanical load as well as the speed is, constant, the power input to the motor (=3 VIa cos φ) is also constant. This, means that the in-phase component Ia cos φ drawn from the supply will remain, constant. If the field excitation is changed, back e.m.f Eb also changes. This, results in the change of phase position of Ia w.r.t. V and hence the power factor, cos φ of the motor changes. Fig. (11.9) shows the phasor diagram of the, synchronous motor for different values of field excitation. Note that extremities, of current phasor Ia lie on the straight line AB., , (i) Under excitation, The motor is said to be under-excited if the field excitation is such that Eb < V., Under such conditions, the current Ia lags behind V so that motor power factor is, lagging as shown in Fig. (11.9 (i)). This can be easily explained. Since Eb < V,, the net voltage Er is decreased and turns clockwise. As angle θ (= 90°) between, Er and Ia is constant, therefore, phasor Ia also turns clockwise i.e., current Ia lags, behind the supply voltage. Consequently, the motor has a lagging power factor., 301
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(ii) Normal excitation, The motor is said to be normally excited if the field excitation is such that Eb =, V. This is shown in Fig. (11.9 (ii)). Note that the effect of increasing excitation, (i.e., increasing Eb) is to turn the phasor Er and hence Ia in the anti-clockwise, direction i.e., Ia phasor has come closer to phasor V. Therefore, p.f. increases, though still lagging. Since input power (=3 V Ia cos φ) is unchanged, the stator, current Ia must decrease with increase in p.f., , Fig.(11.9), Suppose the field excitation is increased until the current Ia is in phase with the, applied voltage V, making the p.f. of the synchronous motor unity [See Fig., (11.9 (iii))]. For a given load, at unity p.f. the resultant Er and, therefore, Ia are, minimum., , (iii) Over excitation, The motor is said to be overexcited if the field excitation is such that Eb > V., Under-such conditions, current Ia leads V and the motor power factor is leading, as shown in Fig. (11.9 (iv)). Note that Er and hence Ia further turn anti-clockwise, from the normal excitation position. Consequently, Ia leads V., From the above discussion, it is concluded that if the synchronous motor is, under-excited, it has a lagging power factor. As the excitation is increased, the, power factor improves till it becomes unity at normal excitation. Under such, conditions, the current drawn from the supply is minimum. If the excitation is, further increased (i.e., over excitation), the motor power factor becomes leading., Note. The armature current (Ia) is minimum at unity p.f and increases as the, power factor becomes poor, either leading or lagging., , 302
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11.10 Phasor Diagrams With Different Excitations, Fig. (11.10) shows the phasor diagrams for different field excitations at constant, load. Fig. (11.10 (i)) shows the phasor diagram for normal excitation (Eb = V),, whereas Fig. (11.10 (ii)) shows the phasor diagram for under-excitation. In both, cases, the motor has lagging power factor., Fig. (11.10 (iii)) shows the phasor diagram when field excitation is adjusted for, unity p.f. operation. Under this condition, the resultant voltage Er and, therefore,, the stator current Ia are minimum. When the motor is overexcited, it has leading, power factor as shown in Fig. (11.10 (iv)). The following points may be, remembered:, (i) For a given load, the power factor is governed by the field excitation; a, weak field produces the lagging armature current and a strong field, produces a leading armature current., (ii) The armature current (Ia) is minimum at unity p.f and increases as the p.f., becomes less either leading or lagging., , Fig.(11.10), , 11.11 Power Relations, Consider an under-excited star-connected synchronous motor driving a, mechanical load. Fig. (11.11 (i)) shows the equivalent circuit for one phase,, while Fig. (11.11 (ii)) shows the phasor diagram., , Fig.(11.11), 303
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Input power/phase, Pi = V Ia cos φ, Mechanical power developed by the motor/phase,, , (i), (ii), , Pm = Eb × Ia × cosine of angle between Eb and Ia, = Eb Ia cos(δ − φ), Armature Cu loss/phase = I 2a R a = Pi − Pm, Output power/phasor, Pout = Pm − Iron, friction and excitation loss., , (iii), (iv), , Fig. (11.12) shows the power flow diagram of the synchronous motor., , Fig.(11.12), , 11.12 Motor Torque, Gross torque, Tg = 9.55, where, , Pm, N-m, Ns, , Pm = Gross motor output in watts = Eb Ia cos(δ − φ), Ns = Synchronous speed in r.p.m., Shaft torque, Tsh = 9.55, , Pout, N-m, Ns, , It may be seen that torque is directly proportional to the mechanical power, because rotor speed (i.e., Ns) is fixed., , 11.13 Mechanical Power Developed By Motor, (Armature resistance neglected), Fig. (11.13) shows the phasor diagram of an, under-excited synchronous motor driving a, mechanical load. Since armature resistance Ra is, assumed zero. tanθ = Xs/Ra = ∞ and hence θ =, 90°., Input power/phase = V Ia cos φ, Fig.(11.13), 304
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Since Ra is assumed zero, stator Cu loss (I 2a R a ) will be zero. Hence input power, is equal to the mechanical power Pm developed by the motor., Mech. power developed/ phase, Pm = V Ia cos φ, , (i), , Referring to the phasor diagram in Fig. (11.13),, , AB = E r cos φ = I a X s cos φ, AB = E b sin δ, , Also, , ∴, or, , E b sin δ = I a X s cos φ, , I a cos φ =, , E b sin δ, Xs, , Substituting the value of Ia cos φ in exp. (i) above,, , Pm =, , V Eb, Xs, , per phase, , =, , VEb, Xs, , for 3-phase, , It is clear from the above relation that mechanical power increases with torque, angle (in electrical degrees) and its maximum value is reached when δ = 90°, (electrical)., , Pmax =, , V Eb, Xs, , per phase, , Under this condition, the poles of the rotor will be mid-way between N and S, poles of the stator., , 11.14 Power Factor of Synchronous Motors, In an induction motor, only one winding (i.e., stator winding) produces the, necessary flux in the machine. The stator winding must draw reactive power, from the supply to set up the flux. Consequently, induction motor must operate, at lagging power factor., But in a synchronous motor, there are two possible sources of excitation;, alternating current in the stator or direct current in the rotor. The required flux, may be produced either by stator or rotor or both., (i) If the rotor exciting current is of such magnitude that it produces all the, required flux, then no magnetizing current or reactive power is needed in, the stator. As a result, the motor will operate at unity power factor., , 305
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(ii) If the rotor exciting current is less (i.e., motor is under-excited), the deficit, in flux is made up by the stator. Consequently, the motor draws reactive, power to provide for the remaining flux. Hence motor will operate at a, lagging power factor., (iii) If the rotor exciting current is greater (i.e., motor is over-excited), the, excess flux must be counterbalanced in the stator. Now the stator, instead, of absorbing reactive power, actually delivers reactive power to the 3-phase, line. The motor then behaves like a source of reactive power, as if it were a, capacitor. In other words, the motor operates at a leading power factor., To sum up, a synchronous motor absorbs reactive power when it is underexcited and delivers reactive power to source when it is over-excited., , 11.15 Synchronous Condenser, A synchronous motor takes a leading current when over-excited and, therefore,, behaves as a capacitor., An over-excited synchronous motor running on no-load in known as, synchronous condenser., When such a machine is connected in parallel with induction motors or other, devices that operate at low lagging power factor, the leading kVAR supplied by, the synchronous condenser partly neutralizes the lagging reactive kVAR of the, loads. Consequently, the power factor of the system is improved., Fig. (11.14) shows the power factor improvement by synchronous condenser, method. The 3 − φ load takes current IL at low lagging power factor cos φL. The, synchronous condenser takes a current Im which leads the voltage by an angle, φm. The resultant current I is the vector sum of Im and IL and lags behind the, voltage by an angle φ. It is clear that φ is less than φL so that cos φ is greater than, cos φL. Thus the power factor is increased from cos φL to cos φ. Synchronous, condensers are generally used at major bulk supply substations for power factor, improvement., , Advantages, (i), , By varying the field excitation, the magnitude of current drawn by the, motor can be changed by any amount. This helps in achieving stepless, control of power factor., (ii) The motor windings have high thermal stability to short circuit currents., (iii) The faults can be removed easily., , 306
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Fig.(11.14), , Disadvantages, (i), (ii), (iii), (iv), , There are considerable losses in the motor., The maintenance cost is high., It produces noise., Except in sizes above 500 RVA, the cost is greater than that of static, capacitors of the same rating., (v) As a synchronous motor has no self-starting torque, then-fore, an auxiliary, equipment has to be provided for this purpose., , 11.16 Applications of Synchronous Motors, (i), , Synchronous motors are particularly attractive for low speeds (< 300, r.p.m.) because the power factor can always be adjusted to unity and, efficiency is high., (ii) Overexcited synchronous motors can be used to improve the power factor, of a plant while carrying their rated loads., (iii) They are used to improve the voltage regulation of transmission lines., (iv) High-power electronic converters generating very low frequencies enable, us to run synchronous motors at ultra-low speeds. Thus huge motors in the, 10 MW range drive crushers, rotary kilns and variable-speed ball mills., , 307
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11.17 Comparison of Synchronous and Induction Motors, S., Particular, No., 1. Speed, 2., , Power factor, , 3., , Excitation, , 4., , Economy, , 5., , Self-starting, , 6., 7., , Construction, Starting torque, , 3-phase Induction, Motor, Remains constant (i.e., Ns) from Decreases with load., no-load to full-load., Can be made to operate from, Operates at lagging, lagging to leading power factor. power factor., Requires d.c. excitation at the, No excitation for the, rotor., rotor., Economical fcr speeds below, Economical for, 300 r.p.m., speeds above 600, r.p.m., Self-starting, No self-starting torque., Auxiliary means have to be, provided for starting., Complicated, Simple, More, less, Synchronous Motor, , 308
