Induction machine characteristics and operation. Induction Machines

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1 Induction Machines 1.1 Introduction: An essential feature of the operation of the synchronous machine is that the rotor runs at the same speed as the rotating magnetic field produced by the stator winding. The magnetic field when observed from a point on the rotor does not vary with time. A very different type of machine results if the rotor is allowed to run more slowly than the rotating field. The rotor will experience a rotating field moving past it at the difference of the two speeds, and this will cause induced currents to flow in suitably arranged conductors on the rotor. These currents will interact with the rotating field to produce a torque, and this is the basis of the induction motor. In common with other rotating machines, induction machines can operate as motors or generators, but have as generators very limited use. Nearly all electric power is generated by synchronous machines. Induction motors, on the other hand, are used in far greater numbers than any other type of machine; they range in power rating from a few watts to tens of megawatts. The simplicity of the induction principle is reflected in the robust, reliable and relatively inexpensive construction of the machine itself, and die induction machine is the natural choice in the majority of motor applications. The induction machine consists essentially of a stator, which has a 3-phase winding, and a rotor. The stator winding is a 3'phase winding, often being a narrow-spread mesh-connected closed winding. The winding is laid in open or half-closed slots in a laminated silicon- steel core. The rotor winding is placed in half-closed or closed slots, the air- gap between stator and rotor being reduced to a minimum. There are two main types of rotor, the wound rotor and the squirrel-cage rotor. In the squirrel-cage rotor, solid conducting rods are inserted into closed slots, and at each end the rods are connected to a heavy short-circuiting ring. This forms a permanently short-circuited winding which is practically indestructible. In some smaller machines the conductors, rings and fan are cast in one piece in aluminum. The cage rotor is cheap and robust, but suffers from the disadvantage of a low starting torque. The wound rotor has a 3-phase winding with the same number of poles as the stator; the ends of the rotor winding may be brought out to three slip rings. The advantage of the wound-rotor machine is that an external starting resistance can be connected to the slip rings to give a large starting torque. This resistance is reduced to zero as the machine runs up to speed. In this report the characteristics of induction machines, as well as their effects on power systems operation are discussed. 1

2 Operational concepts: The axes of the stator and rotor distributions have an angular displacement, and as a result a torque acts on the rotor and causes it to accelerate in the same direction as the stator field. Should the speed of the rotor attain the synchronous value, the rotor conductors would be stationary in relation to the rotating flux. There would therefore be no e.m.f. and no current in the rotor conductors and consequently no torque on the rotor. Hence the latter could not continue rotating at synchronous speed. As the rotor speed falls more and more below the synchronous speed, the values of the rotor e.m.f. and current and therefore of the torque continue to increase until the latter is equal to that required by the rotor losses and by any load there may be on the motor. The speed of the rotor relative to that of the rotating flux is called slip. The steady-state rotor speed is normally slightly less than synchronous so that the motor runs with a slip, s, defined as n1 nr s = (1.1) n1 where nr is the rotor speed. At standstill, nr = 0 and s = 1. For the rotor to reach synchronous speed (nr = no and s = 0), an external drive is necessary. If the rotor is driven so that nr > no, the slip becomes negative, the rotor torque opposes the external driving torque and the machine acts as an induction generator. In all cases the slip speed is ns = no nr (1.) It was shown in the synchronous machine report that for a three-phase winding with p pairs of poles supplied at a frequency of f hertz, the speed of the rotating flux is given by n1 revolutions per second, where f = n1 p If nr is the rotor speed in revolutions per second, the speed at which the rotor conductors are being cut by the rotating flux is (n1-nr) revolutions per second, Frequency of rotor e.m.f: fr = ( n1 nr) p = sn1p = sf (1.3) Currents in the stator winding produce a resultant magnetic field, the axis of which rotates at synchronous speed, n1 revolutions per second, relative to the stator. Similarly the currents in the rotor winding produce a resultant magnetic field, the axis of which rotates (by expression(1.3)) at a speed sn1 revolutions per second relative to the rotor surface, in the direction of rotation of the rotor. But the rotor is revolving at a speed nr revolutions per second relative to the stator core. Hence the speed of the resultant rotor magnetic field relative to the stator core is sn1 + nr = ( n1 nr) + nr = n1rps The axis of the resultant rotor field e.m.f. is traveling at the same speed as that of the resultant stator field m.m.f., so that they are stationary relative to each other. Consequently the induction motor can be regarded as being equivalent to a transformer having an air-gap separating the steel portions of the magnetic circuit. Because of this gap, the magnetizing current and the

3 magnetic leakage for an induction motor are large compared with the corresponding values for a transformer of the same apparent power rating. Friction and windage losses contribute towards making the efficiency of the induction motor less than that of the corresponding transformer. Also, an increase of slip due to increase of load is accompanied by an increase of the rotor currents and therefore by a corresponding increase of the stator currents. The value of the slip at full load varies from about 6 per cent for small motors to about per cent for large machines. The induction motor can therefore he regarded as practically a constant-speed machine. The difficulty of varying its speed economically constitutes one of its main disadvantages. 3 Equivalent circuit. The similarity of the induction motor to a transformer with a closed secondary circuit should be noted. The relative motion between the primary and secondary is represented in the equivalent circuit by the factor 1 /s multiplying the secondary resistance R. By analogy with the transformer, the effects of stator resistance, stator leakage reactance and core loss may be included in the equivalent circuit. Fig. 1 shows the complete equivalent circuit of the machine. In this circuit the effect of a turns ratio 1: n between the stator and rotor windings is also included, by using referred values of the secondary R X parameters ( R` =, X ` ). n = n It is instructive to compare the equivalent circuit of the induction machine with thee complete equivalent circuit of the synchronous machine. In the synchronous machine equivalent circuit, the rotor is represented by an active element - a current generator, which is capable of supplying some or all of the magnetizing current. In the induction machine, on the other hand, there is no external rotor excitation. The rotor is represented by an impedance element, in the equivalent circuit, and all the magnetizing current must be drawn from the stator supply. This means that an induction motor necessarily behaves as an inductive load, taking current at a lagging power factor. I1 I` R1 jx1 Io jx V1 jxm Rl R`/s Figure 1: Equivalent circuit of induction machine 3

4 4 Basic electrical calculations. Let Vp be the voltage per phase applied to stator winding and Zs be the number of stator conductors in series per phase. If Eo = rotor e.m.f. generated per phase at standstill, and if Er is the rotor e.m.f. generated per phase when the per-unit slip is s and the rotor frequency is fr=sf Er = seo (1.4) If R = resistance per phase of the rotor winding and Xo = leakage reactance per phase of rotor winding at standstill =nfx leakage inductance per phase of rotor winding then for per-unit slip, corresponding reactance per phase is Xr = sxo and corresponding impedance per phase is ( ) Zr = R + sxo (1.5) If Io = rotor current per phase at standstill and Ir = rotor current per phase at slip s Eo Io = R + ( sxo) Er seo and Ir = = (1.6) R + ( sxo) R + ( sxo) If φ is the phase difference between Er and Ir Xr sxo tan φ r = R = R (1.7) R and cosφ r = (1.8) R + ( Xr) 4

5 5 Loses The following diagram, in Figure, indicates concisely what becomes of the power supplied to the stator of the induction motor: Input Power to stator winding I R loss in stator winding Core loss in stator core Power transferred to rotor via magnetic field of air gap I R loss in rotor winding Core loss in rotor core(very small at small slip) Total mechanical power developed by rotor Friction loss at bearings and slip-rings (if any) Windage loss Usefull mechanical poer obtained from rotor shaft Figure : Induction motor power loses The fundamental core loss, due to the fundamental component of the rotating magnetic field, is mainly confined to the stator core. This follows from the fact that the magnetic field at a point in the stator, alternates with the supply angular frequency ω, whereas the corresponding field at a point fixed in the rotor alternates with the slip angular frequency sω. Fundamental eddy-current and hysterisis losses are therefore insignificant in the rotor at the normal fullload slip. There is another component of core loss caused by (a) harmonic components of the rotating field which arise from the non-sinusoidal distribution of practical windings, and (b) pulsations in the field which arise from the relative motion of rotor and stator slots. This high-frequency component of core loss occurs in both the rotor and the stator, and the energy is supplied in a very complex way. It is customary to assume that it can be represented by a rotational loss term added to the mechanical (windage and friction) losses. If T = torque, in Newton-meters, exerted on the rotor by the rotating dux and n1= synchronous speed in revolutions per second, power transferred from stator to rotor is π Tn1 (watts) If nr = rotor speed in revolutions per second, total mechanical power developed by rotor is π Tnr (watts) But from the diagram above it is seen that total I R loss in rotor power transferred from stator to rotor - total mechanical power developed by rotor = πt ( n1 nr) (watts) Thus: πt ( n1 nr) = s (1.9) π Tn1 or total rotor I R loss (in watts) is: s x input power to rotor (in watts) 5

6 6 Torque influential factors If m is the number of rotor phases, then, using the symbols given in section, we have : Electrical power generated in rotor = mirer cosφr watts ms Eo R = R + ( sxo) All this power is dissipated as I R loss in the rotor circuits. Since input power to rotor= π Tn1 watts, hence, from equation (1.9), ms Eo R s π Tn1 = R + ( sxo) Consequently, for given synchronous speed and number of rotor phases, seo R sφ R T (1.10) R + ( sxo) R + ( sxo) Since Eo Φ If the impedance of the stator winding is assumed to be negligible, then for a given supply voltage, Φ and Eo remain constant. sr Torque (1.11) R + ( sxo) The value of Xo is usually far greater than the resistance of the rotor winding. If the slip/torque characteristic needs to be graphically displayed, an easy approach is to assume R = 1 Ω and Xo = 8 Ω, and calculate the value of sr for various values of the slip between 1 and It will be seen R + ( sxo) that, for small values of the slip, the torque is almost directly proportional to the slip, whereas for slips between about 0. and 1, the torque is almost inversely proportional to the slip. These relationships can be easily deduced from expression (1.11). Thus, in the case of the cage rotor, R is small compared to Xo, but for values of the slip less than about 0.1 per unit, ( sxo ) Is very small compared to R, so that sr s Torque R R (1.1) Thus the torque is directly proportional to the slip when the latter is very small. For large values of the slip, R is very small compared with ( sxo ) for the cage rotor and for the slip-ring rotor with no external resistance. sr R Torque ( sxo) s (1.13) since Xo is constant for a given motor, i.e. the torque is inversely proportional to the slip when the latter is large. 6

7 7 Effect of rotor resistance on the slip/torque relationship From expression (1.13), it is seen that when R is small compared with sxo. The torque for a given slip is directly proportional to the value of R. However, from expression (1.1) it follows that when R is large compared with sxo the torque for a given slip is inversely proportional to the value of R. The simplest sr method of demonstrating this effect is to repeat the calculation of [ R + ( sxo) ] with R = Ω, R = 4 Ω,and R = 8Ω,. The results, shown in Figure 3, demonstrate that for a slip of, say, 0.03 per unit, the effect of doubling the rotor resistance, is to reduce the torque by about 0.45 per unit, whereas for a slip of 1, the torque is nearly doubled when the resistance is increased from 1 Ω, to Ω,. Hence, if a large starting torque is required, the rotor must have a relatively high resistance. Also the maximum value of the torque is the same for the four values of R and that the larger the resistance the greater is the slip at maximum torque. The condition for maximum torque can be derived by differentiating (1.11) with respect to R, assuming s to remain constant, or with respect to s, assuming R to remain constant. Both methods give the same result; thus, with the second method, the torque is maximum when sr ( R + s Xo ) R sr sxo = s R + s Xo R + s Xo ( ) i.e R + s Xo = 0 so that sxo = R (1.14) Hence the torque is at maximum when the reactance is equal to the resistance. For instance, with R = 1 Ω and Xo = 8 Ω, maximum torque occurs when s = 0.I5 per unit, whereas with R=8 Ω and Xo = 8 Ω, maximum torque occurs when s=1, namely when the rotor is at standstill. Substituting R for sx0 In expression (1.11), sr 1 Maximum torque R Xo But Xo is the leakage reactance at standstill and is a constant for a given rotor; hence the maximum torque is the same whatever the value of the rotor resistance. T A B C D R=8, Xo=8 R=4, Xo=8 R=, Xo=8 R=1, Xo= Per unit Slip S Figure 3: Torque/Slip curves for an induction motor 7

8 At the instant of starting, s = 1, and with a motor having a low-resistance rotor, such as the usual type of cage rotor, the starting torque is small compared with the maximum torque available. On the other hand, if the bars of the cage rotor were made with sufficiently high resistance to give the maximum torque at standstill, the slip for full-load torque (usually about one-third to one half of the maximum torque) would be relatively large and the I R loss in the rotor winding would be high, with the result that the efficiency would be low. If this load was maintained for a period of time, the temperature rise would be excessive. Also, the variation of speed with load would be large. Hence, when a motor is required to exert its maximum torque at starting, the usual practice is to insert extra resistance into the rotor circuit and to reduce the resistance as the motor accelerates. Thus slip-ring machines are used, where the rotor resistance per phase is such as to give a satisfactory operating characteristic. The machine is started with all the resistances in, driving a high starting torque. As the machine runs up to speed the external resistance is reduced until the machine attains full speed with no external resistance. Large motors are often fitted with a short- circuiting and brush-lifting device which first short-circuits the three slip- rings and then lifts the brushes off the rings, thereby eliminating losses due to the brush-contact resistance and the brush friction and reducing the wear of the brushes and of the slip-rings. 8 Cage motor starting The cage-rotor machine suffers from the disadvantage that the starting torque is low if the resistance is low, while the efficiency is reduced if the rotor resistance is high. If this type of motor is started up by being switched directly across the supply, the starting current is about four to seven times the full-load current, the actual value depending upon the size and design of the machine. Such a large current can cause a relatively large voltage drop in the cables and thereby produce a voltage sag, observed by the momentary dimming of the lamps in the areas with small electrical distances from the motor. Consequently it is usual to start cage motors - except small machines, using one of the methods given under the headings below. - Star-Delta starter The two ends of each phase of the stator winding are brought out to the starter which, when moved to the 'starting' position, connects the winding in star. After the motor has accelerated, the starter is quickly moved to the 'running` position, thereby changing the connections to delta. Hence the 1 voltage per phase at starting is of the supply voltage, and the starting 3 torque is one-third of that obtained if the motor were switched directly across he supply with its stator winding delta-connected. Also, the starting current in 1 star is of its value in delta. So the line current for star connection is 1/3 of 3 the value for delta, restricting the voltage sag and its effects at the motor terminal. 8

9 - Auto transformer starting In autotransformer starting the transformer has at least three tappings giving open circuit voltages of not less than 40, 60 and 75 per cent of line voltage for starting. The stator is switched directly to the mains when the motor has run up to speed. If the fractional tapping is x, then the applied voltage per phase on starting is xvi (where V1 is the mains voltage), and the starting torque is reduced by x. The starting current from the mains will also be reduced by approximately x. The autotransformer is usually arranged with two or three tappings per phase so that the most suitable ratio can he selected for a given motor, but an autotransformer starter is more expensive than a star-delta starter. Where a high starting torque is required from a squirrel cage motor, it may be achieved by a double-cage arrangement of the rotor conductors, as shown in Fig. 4(a). The equivalent electrical rotor circuit is shown at (b), where X1 and X are leakage reactances. This equivalent circuit neglects mutual inductance between the cages. For the upper cage the resistance is made intentionally high, giving a high staring torque, while for the lower cage the resistance is low, and the leakage reactance is high, giving a low starting torque but high efficiency on load. The resultant characteristic will be approximately the sum of the two as shown in Fig. 4(c). If a 3-phase-induction motor starts in the wrong direction, this can be remedied by interchanging any two of the three supply leads to the stator. R R1 X X1 X1 R1/s X R/s Torque Combined Lower cage Upper cage Load torque Speed (a) (b) (c) Figure 4:Starting torque improvement in cage rotors. a) Double cage rotor b) Equivalent circuit c) Combined torque/speed characteristic The cage rotor possesses the following advantages: 1. Cheaper and more robust.. Slightly higher efficiency and power factor. 3. Explosion-proof since the absence of slip rings and brushes eliminates the risk of sparking. The advantages of the slip-ring rotor are: 1. The starting torque is much higher and the starting current much lower.. The speed can be varied especially by means of solid-state switching. 9

10 9 Speed control of induction machines In the majority of applications the speed of a driving motor is required to be almost constant, and hence the plain induction motor is very suitable. Sometimes, however, it is desirable to be able to control the speed of the motor, and this may be achieved in three main ways. The speed, may be varied by varying the number of stator poles, the resistance, or varying the supply frequency. It is rarely that variation of the supply frequency is used. In wound-rotor machines, the speed for a given load torque may be changed by varying the rotor resistance. This gives a range of speeds near full speed, four speeds being possible for the same load torque, with a four-position resistance. The disadvantages of this method are the heat lost in the regulating resistor, and the dependence of the sped variation on the load torque. If two or three different operating speeds are required near the synchronous, speeds may be achieved by having two or more stator windings each having a different number of pole-pairs. The required speed is obtained by switching on the appropriate winding. This is only possible with cage-rotor machines, since wound-rotor machines must have a fixed number of rotor poles. To avoid the added cost of separate windings with different numbers of polepairs, pole-changing windings were introduced. In these, by a series or parallel grouping of the coils (achieved by switching), the number of stator poles may be altered. Speed ratios of : 1 and 1.5: 1 may be achieved in this manner. The third method of speed control is achieved by variation of the supply angular frequency. This permits continuous variation of the speed. The slip can be kept low to maintain the efficiency, and the method can be applied to cage induction motors. 10

EEE3441 Electrical Machines Department of Electrical Engineering. Lecture. Introduction to Electrical Machines

EEE3441 Electrical Machines Department of Electrical Engineering. Lecture. Introduction to Electrical Machines Department of Electrical Engineering Lecture Introduction to Electrical Machines 1 In this Lecture Induction motors and synchronous machines are introduced Production of rotating magnetic field Three-phase