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Chapter 2 Induction, Or Asynchronous Machines Three-phase induction motors Unbalanced operation of three-phase induction motors 1 History & Application Of Induction Machines • The invention of induction machines in the l880s completed the ac system of electrical power production, transmission, and utilization, which at the time was in competition with the dc system for general acceptance. The whole concept of polyphase ac, including the induction motor, was the idea of the great Yugoslavian engineer Nikola Tesla. (Tesla became a U.S. citizen in 1889.) The system was patented in 1888. The first large-scale application of the Tesla polyphase ac system was the Niagara Falls hydroplant, completed in 1895. • Today, most industrial motors of one horsepower or larger are three-phase induction machines. Induction machines require no electrical connection to the rotor windings. Instead, the rotor windings are short-circuited. Magnetic flux flowing across the air gap links these closed rotor circuits. As the rotor moves relative to the air gap flux, Faraday's-law voltages are induced in the shorted rotor wind-ings, causing currents to flow in them. The fact that the rotor current arises from induction, rather than conduction, is the basis for the name of this class of machines. They are also called asynchronous (i.e., "not synchronous") machines because their operating speed is slightly less than synchronous speed in the motor mode and slightly greater than synchronous speed in the generator mode. 2 • Induction machines are usually operated in the motor mode, so they are usually called induction motors. As generators, their peculiar characteristics make them suitable for only a few special applications. As motors, they have many advantages. They are rugged, relatively inexpensive, and require very little maintenance. They range in size from a few watts to about 10,000 hp. The speed of an induction motor is nearly, but not quite, constant, dropping only a few percent in going from no load to full load. The chief disadvantages of induction motors are: 1. The speed is not easily controlled. 2. The starting current may be five to eight times full-load current. 3. The power factor is low and lagging when the machine is lightly loaded. For most applications, their advantages far outweigh their disadvantages. 3 4 5 6 Construction 1- STATOR A three-phase windings is put in slots cut on the inner surface of the stationary part. The ends of these windings can be connected in star or delta to form a three phase connection. These windings are fed from a three-phase ac supply. 7 8 9 10 Construction (Cont.) 2- Rotor it can be either: a- Squirrel-cage (brushless) The squirrel-cage winding consists of bars embedded in the rotor slots and shorted at both ends by end rings. The squirrel-cage rotor is the most common type because it is more rugged, more economical, and simpler. 11 Construction (Cont.) b- Slip ring (wound-rotor) The wound-rotor winding has the same form as the stator winding. The windings are connected in star. The terminals of the rotor windings are connected to three slip rings. Using stationary brushes pressing against the slip rings, the rotor terminals can be connected to an external circuit. 12 13 14 15 16 17 How The Induction Motor Works When the stator windings are connected to a three-phase supply; a rotating field will be produced in the air-gap. This field rotates at synchronous speed. This field is exactly like that of the stator of a synchronous machine. Its speed of rotation is given by: ωs = 4π f1/p rad/s ns = 120 f1/p rpm where ωs or ns is called the synchronous speed, f1 is the frequency of the stator voltages and currents, and p is the number of poles of the winding. 18 19 The strength of this field is proportional to the rms stator current, and is expressed by where Where: NI1 is the effective stator turns per pole, I1 is the rms stator phase current, and Nφ1 and kw1 are the series turns per phase and winding factor of the stator winding. 20 Suppose for the moment that the rotor is being driven by an adjustable-speed motor so that it is rotating at the same speed as the stator field. The relative velocity between the stator field and rotor conductors is zero. There is a mutual flux between the rotor and stator, due to the stator MMF F1.This flux links the rotor; but the flux linking any one conductor is constant, hence dλ /dt = 0, and no voltages are induced in the rotor circuits. There are no rotor currents, and no rotor MMF field. The situation is illustrated in Slide 22. 21 22 If the drive motor is now turned off, the friction and windage of the two machines will cause the rotor of the induction machine to slow down. The rotor conductors begin to slip backward through the stator flux field. The conductors now experience changing flux linkages, and voltages are produced in them according to Faraday's law. As a result, current flows in the closed rotor circuits. These rotor currents produce a MMF field having magnetic poles on the rotor surface. The new situation, with the drive motor turned off, is shown in Slide 24. Magnetic forces between the magnetic poles on the rotor surface and the flux field produce torque in the direction of field rotation. The rotor is wound to produce the same number of poles as the stator winding. As the machine slows down, the rotor conductors slip faster through the flux. Greater voltages are induced in the rotor circuits, heavier currents flow in the rotor circuits, stronger magnetic poles are produced on the rotor surface, and more torque is developed. When this torque balances friction and windage load on the machine, the rotor speed stabilizes at a value somewhat less than the field speed. 23 24 When the motor is lightly loaded as in Slide 24, the speed is only slightly less than synchronous speed (say, ω = O.999ωs). The conductors are slipping very slowly backward through the flux field, and the rotor voltages and frequency are both very low. Because the frequency is so low, rotor reactance is negligible, and the currents in the rotor conductors are practically in phase with the induced voltages. Consequently, the rotor MMF vector F2 is at right angles to the flux vector ɸ. With current flowing in the rotor, the flux is no longer due to the stator MMF F1 but rather to R, which is the resultant of F1 and F2. Note in Slide 24 that the position of the pattern of rotor currents is determined by the positions of the flux poles of the air gap field. The reason for this is that the rotor voltages are being induced by air gap fluxes. Since the air gap field is rotating at synchronous speed, the rotor current pattern is also revolving at synchronous speed. The rotor itself is rotating at less than synchronous speed, however. 25 The pattern of rotor conductor currents is thus moving forward, relative to the rotor, at a speed of (ωs – ω) rad/s. The position of the rotor MMF vector F2 is determined by the rotor current pattern and is directed by the right-hand rule along the axis of that pattern. The following statements thus may be made about the rotation of the rotor current pattern and its MMF vector F2: 1. They rotate at synchronous speed with respect to the stator. 2. They are stationary with respect to the air gap field vectors ɸ and R. 3. They rotate forward at (ωs – ω) rad/s, relative to the rotor. 26 The Concept Of Slip The magnitudes and frequency of the rotor voltages depend on the speed of the relative motion between the rotor and the flux crossing the air gap, ɸ. Let the angular speed of the rotor, in radians per second (rad/s) be assigned the symbol ω, and let n be its speed in revolutions per minute (rev/min). The slip speed expresses the speed of the rotor relative to the field, and is given by Slip speed = (ωs – ω) rad/s = ns – n rev/min (rpm) The per-unit slip, usually called simply the slip, is a very useful quantity in studying induction machines. It is given the symbol s and is defined as follows: 27 n = the rotor speed (the motor speed) w.r.t. stator ns = the speed of stator field w.r.t. stator or the synch. speed s = the slip = ns - n/ns nr = the speed of rotor field w.r.t rotor f1 = the frequency of the induced voltage in the stator (stator or supply frequency) f2 = the rotor circuit frequency or the slip frequency nr nS n snS f2 Slip rpm p p p (nr ) (nS n ) ( s nS ) s f1 120 120 120 28 The Frequency Of Rotor Voltages & Currents Consider a typical pair of rotor bars, labeled x and y in Slide 24. As the rotor slips backward through the flux field, the flux linking these bars will vary cyclically. In this Slide, these conductors are shown at the instant the rate of change of flux linkages is a maximum, and hence the voltage induced in the rotor circuit composed of these two bars and the end rings is at its peak. When y has slipped to the position now occupied by x, the polarity will have reversed. Conductor y will have moved relative to the field a distance of one pole pitch, and one half cycle of rotor voltage will have been generated. Thus one cycle of rotor voltage is generated as a given rotor conductor slips past two poles of the air gap flux field. In other words, one cycle of rotor voltage corresponds to 360 electrical degrees of slip. Then the frequency of the rotor voltages and currents is given by 29 The Induction Motor Under Load When a heavy mechanical load is placed on the induction motor, the rotor slows still more; that is, the slip increases somewhat. For most motors, the full-load slip will be about 0.03. The frequency of the rotor voltages and currents will still be quite small (say, about 2Hz), but the effect of leakage reactance can be neglected no longer. There will be a time phase lag between the rotor currents and voltages. If Ll2 is the effective leakage inductance of a typical rotor circuit composed of a pair of rotor bars and their end ring connection, the reactance of this circuit will be XRs = 2πf2Ll2 = 2πsf1 Ll2 The phase angle of lag between maximum voltage and maximum current in a typical rotor circuit will be the power-factor angle: θ2 = tan-1(XRs /RR) 30 Where RR is the ac resistance of the same circuit. Peak current in a pair of bars occurs θ2o in the cycle after peak induced voltage. During this interval, the rotor will have slipped θ2 electrical degrees relative to the rotating air gap field. If maximum voltage is being induced in conductors x and y, maximum current will not flow in these conductors until they have slipped an additional θ2 electrical degrees. Thus in Slide 32, maximum current occurs in conductors x' and y', which lag x and y in rotation by θ2 electrical degrees. Under load, then, the entire rotor current pattern is shifted from the lightly loaded position, with the result that the rotor MMF vector F2 is now at an angle of 90° + θ2 behind the flux, and therefore behind R, the MMF responsible for the flux. 31 32 Circuit Model Of The Induction Machine The induction machine, in certain aspects, is like a synchronous machine. In others, it is like a transformer. Charles Steinmetz developed the most widely used circuit model of the polyphase induction machine. In developing his model, Steinmetz took a transformer approach. This model has proved very useful in understanding the characteristics of the induction machines. Similarities between the induction motor and the transformer are readily apparent. Power from a sinusoidal source is supplied to a stator winding or primary. This winding establishes a flux that mutually couples a rotor winding or secondary. The cyclic mutual flux traverses a ferromagnetic material that gives rise to eddy current and hysteresis losses. Not all of the flux established by the primary winding necessarily links the secondary winding, suggesting that leakage reactance exists. Any current that flows in the secondary winding acts to oppose changes in the primary generated mutual flux, requiring the existence of an mmf balance. 33 As a result of these similarities, the practical transformer equivalent circuit seems like suitable candidate for modeling the induction motor. However, there are three notable dissimilarities between the induction motor and the transformer that would be expected to impact direct application of the practical transformer model. • The mutual flux path must cross the high-reluctance air gap, resulting in reduced magnetizing reactance and increased leakage reactance for the induction motor as compared to a transformer. • Also, there is relative motion between the induction motor and secondary coils. • Additionally, the secondary winding has its terminals shorted. The magnetic conditions in the motor at the instant the current in phase a is a maximum are illustrated in Slide 35. 34 35 The relative motion of the stator flux and the rotor conductors induces voltages of frequency f2 = s f1 called the slip frequency, in the rotor. Thus, the electrical behavior of an induction machine is similar to that of a transformer but with the additional feature of frequency transformation produced by the relative motion of the stator and rotor windings. In fact, a wound-rotor induction machine can be used as a frequency changer. The rotor terminals of an induction motor are short circuited; by construction in the case of a squirrel-cage motor and externally in the case of a wound-rotor motor. The rotating air-gap flux induces slip-frequency voltages in the rotor windings. The rotor currents are then determined by the magnitudes of the induced voltages and the rotor impedance at slip frequency. At starting, the rotor is stationary (n = 0), the slip is unity (s = 1), and the rotor frequency equals the stator frequency f1. The field produced by the rotor currents therefore revolves at the same speed as the stator field, and a starting torque results, tending to turn the rotor in the direction of rotation of the stator-inducing field. If this torque is sufficient to overcome the opposition of rotation created by the shaft load, the motor will come up to its operating speed. The operating speed can never equal the synchronous speed however, since the rotor conductors would then be stationary with respect to the stator field; no current would be induced in them, and hence no torque would be produced. 36 With the rotor revolving in the same direction of rotation as the stator field, the frequency of the rotor currents is sf1 and they will produce a rotating flux wave which will rotate at sns rev/min with respect to the rotor in the forward direction. But superimposed on this rotation is the mechanical rotation of the rotor at n rev/min. Thus, with respect to the stator, the speed of the flux wave produced by the rotor currents is the sum of these two speeds and equals sns + n = sns + ns(1-s) = ns Therefore, we see that the rotor currents produce an air-gap flux wave which rotates at synchronous speed and hence in synchronism with that produced by the stator currents. Because the stator and rotor fields each rotate synchronously, they are stationary with respect to each other and produce a steady torque, thus maintaining rotation of the rotor. Such torque, which exists for any mechanical rotor speed n other than synchronous speed, is called an asynchronous torque. 37 Induced EMF Where 38 Equivalent Circuit Per Phase At standstill (n = 0 , s = 1) The equivalent circuit of an induction motor at standstill is the same as that of a transformer with secondary short circuited. x1 r1 I1 V1 x2 I2 r2 Iex Ih+e Rc Iφ Xm E1 = E 2 All values are per phase Where E2 = per-phase induced voltage in the rotor at standstill referred to stator side. x2 = per-phase rotor leakage reactance at standstill referred to stator side. r2 = per-phase rotor resistance referred to stator side. 39 Equivalent Circuit Per Phase (cont.) At any slip s When the rotor rotates with speed n the rotor circuit frequency will be: f 2 s f1 Therefore induced voltage in the rotor at any slip s will be ERs = sERB , similarly XRs = s XRB and the rotor equivalent circuit per-phase will be: XRB XRs = sXRB ERs = sERB IR IR IR RR ERB XRB RR RR/s ERB RR (1-s)/s Where IR sERB ERB RR jsX RB ( RR / s) jX RB 40 Equivalent Circuit Per Phase (cont.) 1-The Complete Equivalent Circuit per phase Airgap & Magnetic Circuit Stator Circuit I1 V1 r1 Iex x1 E1 Rc X m Load + Rotation losses Rotor Circuit x2 I2 r2 r2(1-s)/s 41 Equivalent Circuit Per Phase (cont.) 3-Thevenin Equivalent Circuit In order to simplify the computations, IEEE recommended an equivalent circuit that can be replaced by the following Thevenin’s equivalent values: Vth Rth Xm r ( x1 X m ) 2 1 2 x2 Xth V1 I2 r2 /s Vth Zth jX m (r1 jx1 ) Rth jX th r1 j ( x1 X m ) gap Air 42 POWER FLOW IN INDUCTION MOTORS dev = 3V1I1cos 43 POWER FLOW IN INDUCTION MOTORS dev = 3V1I1cos r2 s PRCL 3( I 2 ) 2 r2 PAG 3( I 2 ) 2 Pdev 3( I 2 ) 2 r2 (1 s) s PAG : PRCL : Pdev 1 : s : (1 s) x1 r1 x2 I2 I I1 V1 Ic Rc Im r2 I2’ r2 (1-s) s Xm PAG 44 PERFORMANCE CHARACTERSTICS Rth x2 Xth Torque / speed Curve Consider Thevenin’s equivalent circuit The rotor current referred to stator side is: I2 I2 r 2 /s Vth Vth 2 r 2 Rth 2 ( X th x2 ) s gap Air And the developed torque T is : PAG (1 s) PAG 3( I 2 ) 2 r2 T m s (1 s) s s Pdev r T 3 2 ss r Rth 2 s Vth2 2 ( X th x2 ) 2 Speed p.u. 45 Torque/Speed Curve (cont.) Starting Torque: Maximum torque at starting s =1, so the starting torque is: Vth2 TSt 3 s Rth r2 2 ( X th x2 ) 2 r2 Unstable Stable For maximum torque dT 0 , ds sT max Tmax from which r2 [ Rth2 ( X th x2 ) 2 ]1/ 2 Vth2 2S Rth [( Rth ) 2 ( X th x2 ) 2 ]1/ 2 Starting torque STmax 3 46 Torque/Speed Curve (cont.) If the stator resistance is small, so Rth may be neglected sT max r2 ( X th x2 ) and Tmax Vth2 2S ( X th x2 ) 3 and the ratio between the maximum torque and the torque developed at any speed is given by: Tmax (r2 / s) 2 ( X th x2 ) 2 (r2 / s) 2 (r2 / sT max ) 2 s s ( ) ( ) 2 2 2 2 T (r2 / sT max ) ( X th x2 ) sT max (r2 / sT max ) (r2 / sT max ) sT max Tmax s 2T max s 2 T 2sT max s 47 Torque/Speed Curve for varying R2 Tmax Vth2 2S ( X th x2 ) sT max 3 ST max r2 ( X th x2 ) R2' ( X th X 2' ) The maximum torque is independent of the rotor resistance. However, the value of the rotor resistance determines the slip at which the maximum torque will occur. The torque-slip characteristics for various values of are shown. To get maximum torque at starting:: sT max 1 i.e. r2 ( X th x2 ) 48 Stator current and input power factor From the IEEE recommended equivalent circuit, the input impedance Zin jx 2 jx1 r1 is: r jX m 2 jx 2 s Z in r1 jx1 r2 j X m x2 s Z in Z in I1 V1 r2/ s jXm Maximum or pull-out torque T I1 The stator current I1 is : V1 I1 Z in And the input power factor is : p.f. = cos () p.f. h Tst art 1 s Starting Typical Induction Motor Characteristics 0 No-load 49 Example (1) 50 Solution 51 52 53 54 55 Control Of Performance By Rotor Design A simple way of obtaining a rotor resistance which is automatically vary with speed make use of the fact that at standstill the rotor frequency equals the stator frequency; as the motor accelerates, the rotor frequency decreases to a very low value, perhaps 2 or 3 Hz at full load in a 60 Hz motor. With suitable shapes and arrangements for rotor bars (deep-bar rotors [B] or double-cage rotors [C]), squirrelcage rotor can be designed so that their effective resistance at 60 Hz (at starting) is several times their resistance at 2 or 3 Hz (running condition). 56 57 Deep-Bar Rotors • The leakage flux paths for a deep-bar design are illustrated in Slide 59(a). Since inductance is the number of flux linkages produced per ampere (L = λ/I), it is obvious that the parts of each bar extending deeply into the iron have higher leakage inductances than those parts of the bar cross section near the air gap, because more of the leakage flux links the deeper parts of the bar. One may think of each bar as being composed of several layers of equal cross section, and thus of equal resistance, but with inductances increasing with depth, as in Slide 59(b). • Now under running conditions the slip is quite small, and therefore the frequency of the rotor currents is only 1 or 2 Hz. As a result, all of the leakage reactances are negligible, and the current distribution is essentially uniform throughout each rotor bar. The effective resistance of the rotor is that of all layers in parallel. This low resistance makes for low slip and high efficiency at full load. As the motor is overloaded, however, slip and rotor frequency increase. The reactances of the deeper levels of the bar come into play with two effects: (1) The value of x2 in the circuit model is larger than it would be for shallow bars like those of Design A, and the breakdown torque is reduced. (2) The rotor current begins to be crowded into those layers of the bars near the air gap, increasing the effective resistance of rotor circuits. Thus breakdown torque occurs at a somewhat greater slip than for Design A. 58 • When the motor is first connected to the line for starting, the slip is unity a frequency of the rotor currents equals the line frequency. Rotor-bar currents are concentrated near the air gap. Effective bar resistance and reactance are both high, with the result that starting current is only about 75% of that for Design A. • Thus a deep-bar rotor achieves good full-load efficiency at low slip, together with reduced starting current, at the expense of somewhat lower pull-out torque. A typical speed/torque curve for a deep-bar motor is shown in Slide 60 as the Design Class B curve, although this same characteristic may also be obtained with a double-cage rotor. 59 60 Double-Cage Rotors • A double-cage rotor has two squirrel cages. The two cages may connect to the same end rings, or there may be separate end rings for each cage. The usual arrangement is that of NEMA Design C, as illustrated in Slide 57. The outer cage for C is composed of small, high-resistance bars, while the inner cage is made up of larger, low-resistance bars. Since the inner bars are almost totally surrounded by the iron of the rotor core, their leakage inductance is very great, indeed. For this reason they have little effect on motor performance except at low slips; that is, they come into play when the motor has reached running speed. At running speed the two cages are essentially in parallel, and r2 in the circuit model is quite low. Full-load slip is low, almost as low as for Design A, and full-load efficiency is almost as high as for Design A, see Slide 60. 61 • At starting, the inner cage may nearly be disregarded, and the outer cage designed for optimum starting characteristics. The iron between the outer cage and the inner cage, together with the proximity of the outer cage with the stator windings, causes the leakage reactance to be low and permits a large maximum torque. If desired, the resistance of the outer cage may be chosen to realize this high torque at starting. The high outer-cage resistance keeps the starting current low. • Double-cage construction is the most costly of the squirrel cage designs, but it allows many of the advantages of the still more costly and less rugged wound rotor to be included in a squirrel cage machine. 62 Classes of squirrel-cage motors According to the National Electrical Manufacturing Association (NEMA) criteria, squirrel-cage motors are classified into class A, B, C or D. The torque-speed curves and the design characteristics for these classes are : Class Starting Current Starting Torque Rated Load Slip A Normal Normal < 5% (less than that of design B) B Low Normal < 5% (usually 0.02 to 0.03) C Low High < 5% (slightly larger than design B) D Low Very High 8-17 % 63 Tests To Determine Circuit Model Parameters Since Steinmetz's induction machine model is similar to his transformer model, the tests usually employed to evaluate its impedances are analogous to the open-circuit and short-circuit tests on transformers. In addition, a dc measurement of r1 is required. Many factors must be taken into account in making these tests. Winding resistances vary with temperature, and rotor resistance is a function of frequency. Test procedures are presented in great detail in IEEE Standard 112-1984. That standard should be consulted when accurate measurements are necessary. The following paragraphs, however, provide an understanding of the fundamental theory underlying these tests. 64 The No-Load Test The no-load test corresponds to the open-circuit test on a transformer. However the test measures friction and windage losses in addition to core losses. Slide 66 shows the test circuit and the effect of zero mechanical load on the model. The mechanical load per phase, represented by r2(1-s)/s, is simply the internal windage and friction of the machine. Since this amounts to only a few percent of the rated mechanical load, the slip is very small, say, 0.001, and r2(1 - s)/s is very large compared to x2 and r2 . It is thus essentially in parallel with Rc and jXm, where Rc represents the core loss per phase. In Slide 60, Rfwc represents the parallel combination of Rc and r2(1 - s)/s, and absorbs a power equal to one third of the total friction, windage, and core loss. Since under no-load conditions r2 is so much less than r2(1 - s)/s, the no-load rotor copper loss is negligible. The no-load input power Pnl , then, may be taken to be the sum of the stator copper loss and Pfwc, the friction, windage, and core loss. Pnl = P1 + P2 = SCLnl + Pfwc = 3I2nl r1 + Pfwc This permits the sum Pfwc to be evaluated: Pfwc = Pnl - 3I2nl r1 65 66 Where Pnl is the total three-phase power input to the machine at rated voltage and frequency, and Inl is the average of the three line currents, with no mechanical load connected to the shaft. Under no-load conditions, the power factor is quite low, 0.2 or less, so the circuit is essentially reactive. The current will be in the neighborhood of one third the rated current. These facts imply that r1 is small compared to Xm and that Rfwc is much larger than Xm. The input Impedance at no load is thus approximately jx1 and jXm in series: The value of r1 and x1 have yet to be determined 67 Direct-Current Test In the case of a transformer, it was only necessary to obtain Req1 or Req2 to solve practical problems. Knowledge of the individual winding resistances was not essential. In the induction machine, however, r2 plays a crucial role, and r1 must be known to evaluate the stator copper loss. With a steady direct current flowing in the stator windings, the stator magnetic field is fixed in space. If no external torque turns the shaft, the flux linkages with the circuits are constant in time. As a result, no voltages are induced in the rotor, no currents flow, and no torque is developed. There is no rotor MMF (F2 = 0), and the rotor circuits have no electrical effect on those of the stator. Thus the dc stator winding resistance may be measured independently of the rotor impedance. 68 The circuit is shown in Slide 70. The rheostat must be capable of handling rated current of the motor. The current is adjusted to approximately rated value, so that the temperature of the winding approximates that of running conditions. (The winding temperature is often measured so that the test resistance may be corrected to actual operating temperature.) Applying Ohm's law to the circuit of Slide 70, This value of r1 may now be substituted in Pfwc = Pnl - 3I2nl r1 to obtain Pfwc. 69 70 The Blocked-Rotor Test This test, which corresponds to the short-circuit test of a transformer, is also called the locked-rotor test. The shaft is clamped so that it cannot turn; thus ω = 0 and s = 1 during this test. If full voltage at rated frequency were applied, the current would be five to eight or more times rated value. For this reason, locked-rotor tests are not done at full voltage except for small motors. Instead, the applied voltage is usually adjusted upward from zero until the stator current is at rated value. Blocked-rotor test measurements are, then, taken at rated current rather than at rated voltage. Under running conditions the frequency of the stator currents is line frequency f1, while that of the rotor currents is sf1 only a few hertz. In the blocked-rotor test, however, the slip is unity and the rotor and stator currents are at the same frequency. Since both rotor and stator quantities are to be evaluated, the frequency of the blocked-rotor test voltage ft, is usually a compromise between the normal stator and rotor frequencies. 71 The test circuit and the effect of blocking the rotor on the model are shown in Slide 73. The voltage appearing across one phase of the stator winding VBR/√ 3, and the magnitude of the impedance looking into the terminals of one ph of the machine is Where IBR is the average of the three line currents and should be roughly equal to rated current. With unity slip, (r2 /s) = r2. The applied voltage is quite low and the iron core the machine is unsaturated. As a result, Pp is greater than at rated voltage and Xm is larger than normal. Even at rated voltage, Xm is typically 25 times the magnitude r2 + jx2. The effect of Xm may thus be neglected, and the assumption made that 72 73 Where X’BR = x’1 + x’2 is the blocked-rotor reactance at the test frequency ft .The real part of ZBR is given by the power per phase divided by the phase current squared: Now r1 was found by the dc test, then The imaginary part of ZBR is given by: If rated frequency is fB and the frequency of the blocked-rotor test is ft, then 74 The designer of the machine knows, on the basis of his or her calculations, how much of XBR is x1 and how much is x2. There is no way to determine this ratio from tests on a squirrel cage motor, however. Experience with these machines has been condensed into the table given which shows approximately how XBR divides between x1 and x2 for rotors of different designs. 75 Example (2) The following test data were obtained on a 5-hp, four-pole, 220-V, three-phase, 60-Hz Design B induction motor having a rated current of 12.9-A. Calculate the performance of this motor at a slip of 0.03. 76 Solution 77 78 79 80 81 Starting Induction Motors 82 83 84 Starting Induction Motors (cont.) 1. Direct-On-Line Starting Induction motors when started by connecting them directly across the supply line take 5 to 8 times their full-load current. This initial excessive current may cause large line voltage drop that affects the operation of other electrical equipment connected to the same lines. This method is suitable for small motors up to 10-hp rating. 2. Auto-transformer A three-phase step-down autotransformer may be employed as a reduced voltage starter. As the motor approaches full speed, the autotransformer is switched out of the circuit. 85 Starting Induction Motors (cont.) 3. Star-delta method The normal connection of the stator windings is delta while running. If these windings are connected in star at starting, the phase voltage is reduced, resulting in less current at starting. As the motor approaches the full speed, the windings will be connected in delta. 4. Solid-State Controller The controller can provide smooth starting. 86 Starting Induction Motors (cont.) 5. Using External Rotor Resistance For wound rotor induction motor only. The external resistance can be chosen to get high torque and low current at starting and can be decreased as motor starts up. 87 Speed control of induction motors 1. Line Voltage Control Recall that the torque developed in an induction motor is proportional to the square of the terminal voltage. A set of T – n characteristics with various terminal voltages is shown. If the motor drives a fan load the speed can be varied over the range n1 to n2 by changing the line voltage. Note that the class D motor will allow speed variation over a wider speed range. T Vs2 88 The terminal voltage V1 can be varied by using a three-phase autotrans-former or a solid-state voltage controller as shown. The auto-transformer provides a sinusoidal voltage for the induction motor, whereas the motor terminal voltage with a solid-state controller is nonsinusoidal. Speed control with a solid-state controller is commonly used with small squirrel-cage motors driving fan loads. In large power applications an input filter is required; otherwise, large harmonic currents will flow in the supply line. T1 ia iL A T4 T3 vAN vBN vCN Rotor C ib B N T6 T5 Rotor ic T2 89 The thyristor voltage controller shown in Slide 91 is simple. During the operation of the voltage controller the command signal for a particular set speed fires the thyristors at a particular firing angle (α) to provide a particular terminal voltage for the motor. If the speed command signal is changed, the firing angle (α) of the thyristors changes, which. results in a new terminal voltage and thus a new operating speed. Open-loop operation is not satisfactory if precise speed control is desired for a particular application. In such a case, closed-loop operation is needed. If the motor speed falls because of any disturbance, such as supply voltage fluctuation, the difference between the set speed and the motor speed increases. This changes the firing angle of the thyristors to increase the terminal voltage, which in turn develops more torque. The increased torque tends to restore the speed to the value prior to the disturbance. Note that for this method of speed control the slip increases at lower speeds, making the operation inefficient. However, for fans, or similar centrifugal loads in which torque varies approximately as the square of the speed, the power decreases significantly with decrease in speed. There-fore, although the power lost in the rotor circuit (=sPg) may be a significant portion of the air gap power, the air gap power itself is small and therefore the rotor will not overheat. The voltage controller circuits are simple and, although inefficient, are suitable for fan, pump, and similar centrifugal drives. 90 91 Speed control of induction motors 2. Line Frequency Control n=(1-s)ns , ns= 120 f1/p The synchronous speed and hence the motor speed can be varied by changing the frequency of the supply. Application of this speed control method requires a frequency changer. The Figure shows a block diagram of an open-loop speed control system in which the supply frequency of the induction motor can be varied. 92 From the emf equation the motor flux is If the voltage drop across r1 and x1 is small compared to the terminal voltage V1, that is, , then To avoid high saturation in the magnetic system, the terminal voltage of the motor must be varied in proportion to the frequency. This type of control is known as constant volts per hertz. At low frequencies, the voltage drop across r1 and x1 (Slide 41) is comparable to the terminal voltage V1 , and therefore is no longer valid. To maintain the same air gap flux density, the ratio V/f is increased for lower frequencies. The required variation of the supply voltage with frequency is shown in Slide 94. In Slide 92 the machine voltage will change if the input voltage to the inverter Vi is changed; Vi can be changed by changing the firing angle of the controlled rectifier. If the output voltage of the inverter can be changed in the inverter itself (as in pulse-widthmodulated inverters PWM), the controlled rectifier can be replaced by a simple diode rectifier circuit, which will make Vi constant. 93 The Torque-Speed characteristics for variable-frequency operation are shown in Slide 95. At the base frequency fbase the machine terminal voltage is the maximum that can be obtained from the inverter. Below this frequency, the air gap flux is maintained constant by changing V1 with f1 ; hence the same maximum torques are available. Beyond fbase, since V1 cannot be further increased with frequency, the air gap flux decreases, and so does the maximum available torque. This corresponds to the field-weakening control scheme used with dc motors. Constant-horsepower operation is possible in the fieldweakening region. 94 95 In Slide 95 the Torque-Speed characteristic of a load is superimposed on the motor Torque-Speed characteristic. Note that the operating speeds n1 …. n8 are close to the corresponding synchronous speeds. In this method of speed control, therefore, the operating slip is low and efficiency is high. The inverter in Slide 92 is known as a voltage source inverter. The motor line-to-line terminal voltage is a quasi-square wave of 120° width. However, because of motor inductance the motor current is essentially sinusoidal. 96 Speed control of induction motors 3. Rotor Resistance Control For wound-rotor induction motors only. a- Classical technique: It was pointed out that the speed of a wound-rotor induction machine can be controlled by connecting external resistance in the rotor circuit through slip rings, as shown in Slide 98. The Torque-Speed charac-teristics for four external resistances are shown. The load T - n characteristic is also shown by the dashed line. By varying the external resistance 0 < R-ex < Rex4, the speed of the load can be controlled in the range n1 < n < n5. Note that by proper adjustment of the external resistance (Rex = Rex2), maximum starting torque can be obtained for the load. The scheme requires a three-phase resistance bank, and for balanced operation all three resistances should be equal for any setting of the resistances. Manual adjustment of the resistances may not be satisfactory for some applications, particularly for a closed-loop feedback control system. 97 98 b- An alternative methods of varying rotor-circuit resistance Using three-phase diode bridge and a single variable resistor. Solid-state control of the external resistance may provide smoother operation. A block diagram of a solid-state control scheme with open-loop operation is shown in Slide 100. The three-phase rotor power is rectified by a diode bridge. The effective value Rex* of the external resistance Rex can be changed by varying the on-time (also called the duty ratio λ). When λ = 0.0, that is, when the chopper is off all the time, Rex* = Rex. When λ = 1.0, that is, the chopper is on all the time, Rex is short-circuited by the chopper and so Rex* = 0. In this case, the rotor circuit resistance consists of the rotor winding resistance only. Therefore, by varying λ in the range 1 > λ > 0, the effective resistance is varied in the range 0 < Rex* < Rex, and Torque-Speed characteristics similar to those shown Slide 98 are obtained. 99 100 Speed control of induction motors 4. Rotor Slip-Energy Recovery System Control For wound-rotor induction motors only. If the no-load slip is so, then 101 Speed control of induction motors 5. Pole-Changing Control For squirrel-cage induction motors only. Because the operating speed is close to the synchronous speed, the speed of an induction motor can be changed by changing the number of poles of the machine. This can be done by changing the coil connections of the stator winding. Normally, poles are changed in the ratio 2 to 1. This method provides two synchronous speeds. This method is not practical for wound-rotor motors, because the rotor windings have to be reconnected to have the same number of poles as the stator. A squirrel-cage rotor automatically develops a number of magnetic poles equal to those of the air gap field. It is obvious, however, that the speed can be changed only in discrete steps and that the elaborate stator winding makes the motor expensive. The basic scheme is illustrated in Slide 103, which shows what happens in a typical phase of a two-speed polyphase winding. 102 103 MODES OF OPERATION The induction motor can be operated in three modes: motoring, generating, and plugging as shown in Slide 105. To illustrate these three modes of operation, consider an induction machine mechanically coupled to a dc machine, as shown in Slide 105. 1. Motoring 0 < S < 1 In normal motor operation, the rotor revolves in the direction of rotation of the magnetic field produced by the stator currents. This is the natural (motoring) mode of operation of the induction machine. The steady-state speed n is less than the synchronous speed ns . 2. Generating S < 0 (Negative slip) The induction machine will operate as a generator if its stator terminals are connected to a three-phase supply and its rotor is driven above the synchronous speed by a prime mover (resulting in negative slip). The generating mode of operation is utilized in some drive applications to provide regenerative braking. For example, suppose an induction machine is fed from a variable-frequency supply to control the speed of a drive system. To stop the drive system, the frequency of the supply is gradually reduced. 104 105 In the process, the instantaneous speed of the drive system is higher than the instantaneous synchronous speed because of the inertia of the drive system. As a result, the generating action of the induction machine will cause the power flow to reverse and the kinetic energy of the drive system will be fed back to the supply. The process is known as regenerative braking. 3. Plugging (Braking) S > 1 Suppose an induction motor is running at a steady-state speed. If its terminal phase sequence is changed suddenly, the stator rotating field will rotate opposite to the rotation of the rotor, producing the plugging operation. The motor will come to zero speed rapidly and will accelerate in the opposite direction. This mode of operation is sometimes utilized in drive applications where the drive system is required to stop very quickly. Suppose an induction motor is running at a steady-state speed. If its terminal phase sequence is changed suddenly, the stator rotating field will rotate opposite to the rotation of the rotor, producing the plugging operation. The motor will come to zero speed rapidly and will accelerate in the opposite direction, unless the supply is disconnected at zero speed. 106 107 Example (4) 108 Solution 109 110 111 112 The Circle Diagram Of The Induction Machine The circle diagram is based on the approximate equivalent circuit of the induction motor shown in Slide 114. The shunt branch has been moved from points a, b to the input terminals and includes a resistor Rfwc to represent rotational losses. Since all rotational losses are taken care of by Rfwc, the approximate circuit assumes zero slip at no load on the motor shaft. Then at no load, r2/s is an open circuit, and the no-load line current Inl is V1/Zs. By Kirchhoff's current law, the line current under load is I1 = Inl + I2, where The current I2 will lag V1 by the impedance angle Therefore I2 may be rewritten as follows: 113 114 This is the polar equation of a circle of the form ρ = D sin θ, where D is the circle diameter. Thus the locus of I2 is a circle of diameter Where, (x1 + x2) is the blocked-rotor reactance when the blocked-rotor test is made at rated frequency. Therefore, the equation for the circle diagram may be rewritten as follows: 115 The resulting diagram is shown in Slides 117 or 118. Since I1 is the sum of Inl and I2, the point of the I1- phasor will coincide with that of I2. Thus the circle serves as the locus of both I1 and I2, though they radiate from different origins. At starting, I1 and I2 will be at some point H. As the motor accelerates, the tips of the current phasors travel around the circle in a CCW direction until the output torque matches the load torque. If there is no shaft load, the motor will continue to accelerate to very near synchronous speed (according to the approximate model, synchronous speed actually will be reached). At this point, I2 = 0 and I1 = Inl. 116 117 118 For example (Slide 123), the input power to the motor is given by Then, for constant applied phase voltage the input power is proportional to I1 cos θφm i.e. proportional to length AE. Then for an appropriate chosen scale, the length AE is calibrated in terms of power. Similarly the length AB represents the input no-load power (Prot = core loss + mechanical loss). Consequently, the length BE represents: 1) Stator copper loss + 2) Rotor copper loss + 3) Output mechanical power. The phasor OH represents blocked-rotor stator current I1BR, therefore FH represents (rotor + stator) copper losses (mechanical power output = zero in this case), i.e. Triangles O’FH & O’BD are similar, therefore 119 Therefore, BD represents the (stator + rotor) copper loss. Then DE represents output mechanical power. Therefore, line O’H is then called power line since the ordinate bounded by this line and the circumference O’EH represents output power by the motor. If added to these ordinates the rotor copper loss, then the new ordinates will be proportional to air gap power Pag which equals motor torque in synchronous watts. This is achieved by dividing FH in the ratio of r2 : r1, then line O’G is termed torque line. Therefore for the operating point E shown in Slide 123: 120 121 Measurements of slip in this way give poor accuracy. The method described in Slide 123 gives results with better accuracy. To evaluate the slip accurately point Q is joined by point H by the straight line QHS. Then from point S a line SK is drawn perpendicular to the torque line. This line is divided in equal divisions from 0 (point K) to 100 (point S). To determine the slip for the point E, for example, the line EQ is drawn which intersects SK at a point W graduated to the value of slip. 122 123 124 125 Example (5) A 22.38-kW, 500-V, 4-pole, 50-Hz, star connected three-phase induction motor has the following test data: No-load : 500-V 8.2-A 1.43-kW Blocked-rotor: 140-V 45-A 3.2-kW The equivalent rotor resistance is equal to the stator resistance. To a proper scale draw the circle diagram and find: (a) Stator current and power factor at full load, (b) Maximum torque and, (c) Starting torque in terms of full-load torque. Answers: (a) Stator full-load current =32-A, Full-load power factor = 0.91 lagging (b) Maximum torque = 374.9-N.m. (c) Starting Torque = 0.868 x Full-Load Torque 126 Unbalanced operation of three-phase Induction Motors Three-phase induction motors may operate under unbalanced conditions such as: Unequal magnitude of the three-phase supply voltages Unequal phase-shift between the three-phase supply voltages Disconnected one of the three phases. Evaluation of the three-phase induction motor performance under such conditions can be found using Symmetrical Components Technique. In such technique the unbalanced voltage supply can be analyzed into its equivalent three-sets of symmetrical components [ positive, negative, and zero sequence]. 127 Unbalanced Supply For unbalanced supply voltages Vas V11 Vbs V2 2 Vcs V3 3 Then the equivalent sequence voltages are given by: 1 ( Vas a.Vbs a 2 .Vcs ) 3 1 V ( Vas a 2 .Vbs a.Vcs ) 3 1 V 0 ( Vas Vbs Vcs ) 3 where a 1 120 , a 2 1 240 V 128 Symmetrical Components Vas Vas Vas Vcs Vas0 0 Vbs Vcs0 + Vbs Vcs cs V Vbs + Vbs 129 Unbalanced operation of threephase Induction Motors (cont.) The motor will then operates under the influence of the three balanced sets of voltages. The zero sequence voltage will produce no resultant torque. Regarding the resultant torque due to positive and negative sequence sets of voltages, it will depend on the relative magnitude of these sets. If V+ > V- then the motor will rotate with the positive sequence set of rotation, while if V+ < V- then the motor will rotate with the negative sequence set of rotation. Denoting the set of voltages of higher magnitude as the forward set the other will be denoted as the backward one. 130 Motor equivalent circuits The equivalent circuit per phase as seen by both forward and backward components will be as shown R1 X’2 X1 X’2 X1 ’ Ib If Xm Vf Forward eq. circuit where R1 ’ Sf n s nr S ns R2'/Sf Xm Vb R2'/Sb Backward eq. circuit Sb n s (nr ) 2S ns 131 Motor equivalent circuits The equivalent circuit per phase as seen by both forward and backward components may be redrawn as follows: R1 X1 R1 X1 If Vf Rf Ib Vb Xf R2 where Z f R f jX f ( jX 2 ) //( jX m ) S R Z b Rb jX b ( 2 jX 2 ) //( jX m ) 2S Vf Vb If and I b Z1 Z f Z1 Z b Rb Xb 132 Power and Torque From the equivalent circuits the air gap powers due to the forward and backward fields are: Pgf 3 I .R f 2 f and Pgb 3 I .Rb 2 b And the resultant developed torque is given by: T T f Tb PgF Pgb s 3 ( I 2f .R f I b2 .Rb ) s 133 Example (4) A 220 V, 3-phase, Y- connected, 50 Hz, 4-pole, squirrel cage induction motor develops its full load torque at slip of 0.04 when operated at rated voltage and frequency. The equivalent rotor resistance and reactance referred to stator side are 0.35 and 1.4 Ω respectively and the magnetizing reactance Xm is 30 Ω . The stator phases are aa', bb', and cc', star point is obtained by connecting a', b', and c'. Neglecting stator impedance, i. Find the full load torque ii. If the phase bb' is reversed (the star point is obtained by connecting a', b, and c') calculate the developed torque at 0.04 slip. 134 Solution i- With balanced supply, the equivalent circuit/phase is: Vcc' I1 Vaa' V Vbb' I'2 0.35/S J30 j1.4 220 0 1270 3 120 f1 120 * 50 ns 1500 rpm p 4 S 0.04 V 1270 I 2 14.34 9.1 0.35 ( j1.4) 0.04 2 Pg 3I 2 R2 3(14.34) 2 (0.35) T 34.36 Nm 2 .1500 s S s (0.04)( ) 60 135 ii- When reversing phase bb' (unbalanced supply) Vcs Vas 1270 ,Vbs 12760 ,Vcs 127120 1 V (Vas a.Vbs a 2 .Vcs ) 3 127 (1 160 . 1120 1120 . 1240) 3 127 (1 (1) 1) 42.340 3 1 (Vas a 2 .Vbs a.Vcs ) 3 127 (1 160 . 1240 1120 . 1120) 3 127 (1 1300 1240) 3 127 (1 j1.7321) 84.67 60 3 Since V V , V f 84.67 60 and Vb 42.340 Vbs Vas V 136 If I'2f Vf 0.35/S Ib Vb J30 I'2b J30 j1.4 j1.4 Forward 0.35/(2-S) Backward 0.35 j1.4)( j 30) Z f 0.04 8.1524.67 7.41 j 3.4 0.35 ( j 31.4) 0.04 ( 84.67 60 If 10.39 84.67 Z f 8.1524.67 Vf Pgf 3 . I 2f . R f 3 (10.39) 2 7.41 2399.7 W 137 Ib Vb I'2b 0.35/(2-S) J30 j1.4 Backward 0.35 j1.4)( j 30) Z b 2 0.04 1.3583.41 0.155 j1.34 0.35 ( j 31.4) 2 0.04 V 43.340 Ib b 31.36 83.41 Z b 1.3583.41 ( Pgb 3 . I b2 . Rb 3 (31.36) 2 0.155 457.3 W T Pgf Pgb s 2399.7 457.3 12.36 Nm 157.1 138