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UNIT 5 INDUCTION MOTOR Induction Motor Structure 5.1 Introduction Objectives 5.2 5.3 5.4 Construction of Induction Motor 5.2.1 Stator Parts 5.2.2 Rotor Parts 5.2.3 Other Parts Working of 3-Phase Induction Motor 5.3.1 Slip and Effect of Slip on Rotor Parameters 5.3.2 Torque and Torque-slip Characteristic 5.3.3 Power Flow 5.3.4 Equivalent Circuit of Inductor Motor Starting and Speed Control 5.4.1 Starters for Poly Phase Induction Motors 5.4.2 Speed Control of 3-phase Induction Motor 5.5 Measurement of Slip 5.6 Summary 5.7 Answers to SAQs 5.1 INTRODUCTION The most commonly encountered electric motors in industry are induction motors. In this unit, we are introducing the construction features of induction motors to explain slip ring induction motors and squirrel cage induction motors. Next, you will study various relations between rotor emf, rotor current and torque developed in rotor. The choice of selection of induction motor based on torque slip characteristics has been discussed. After that, we will discuss equivalent circuit, no-load and blocked rotor tests of induction motor. At the end of the unit, we will study speed control of induction motor which is very useful in industrial drives. Objectives After studying this unit, you should be able to • give elementary description of various parts and types of induction motor, • explain the principle of working of induction motor and various stator and rotor current relations, • describe torque-speed relations and condition for maximum torque, • describe power stages and efficiency of motor, • explain starting and speed control of induction motor, and • know effect of crawling and cogging of induction motor. 119 Electrical Technology 5.2 CONSTRUCTION OF INDUCTION MOTOR The three-phase induction motor differs constructionally from the DC machine on certain factors: its short air-gap, absence of a commutator, simple windings (particularly the cage) and laminated stator. It is not possible, as in the DC machine, to make the frame as a part of the magnetic circuit. The stator core must be carried in a shell or housing which provides a means for protecting the stator and carrying the end-covers, bearings and terminal box. The following parts normally constitute an induction motor (the typical constructional variation of slip-ring type is dealt under rotor construction). 5.2.1 Stator Parts Let us discuss, in detail, each of them separately. Frame The frame of an induction motor may be cast or fabricated, depending upon the size of the motor. For miniature and small motors, the frame has a single piece of short, hollow cylinder cast-iron casting. It is cheaper to use cast iron where losses and efficiency is of a lesser consideration than economy and where new designs and modifications are not to be done on the machine. However, for medium-sized and (in particular) large induction motors, fabricated frame structure is exclusively used. Four or more arc-shaped rolled steel pieces are joined together to form a cylindrical shape. The outer surface may be provided with cooling fins so as to increase the heat dissipating area without increasing the overall diameter. The chief advantage of fabricated construction is in its application to new designs and modifications, which can be made without reference to existing patterns. The principal constructional difficulty is the avoidance of distortion when the parts are welded together. The frame gives complete support and protection to the other parts (ruggedness) and an eye-bolt on its top is useful for transit purposes. Stator Core The stator core provides the space for housing for the three-phase stator windings and also forms the path for the rotating magnetic field. They are built up of thin sheets of thickness (called stampings or laminations) of 0.35 mm to 0.65 mm with of a special core of steel, insulated one from the other by means of paper. The gap facing inner circumference of the plates have suitable slots punched out, either open, semi-closed or completely closed. Normally, the stator core has semi-closed slots where the number of slots, S, is an integral multiple of 3 times the number of poles, p for which the induction motor is designed and constructed [S = n. (3 p)], where n = 1, 2, 3 etc. Every part of the stator core is subject to alternate changes in polarity of the magnetic field (due to its rotating nature) so hysteresis losses and eddy-current losses take place. To reduce the former, about 3-5% of silicon is added to high grade steel and to reduce the latter, a large number of thin laminations are stacked together. 5.2.2 Rotor Parts Rotor Core The construction of the rotor core should be separately discussed for (a) squirrel cage motors, and (b) slip-ring motors. Certain common features are as follows : (a) 120 Both types have rotors constructed to thin sheets of special core steel, but here the thickness is larger than that of stator stampings because no appreciable iron loss is incurred in the rotor. (b) In small motors, the rotor core is mounted directly on the shaft, to which it is keyed, and clamped between end-plates on the shaft and a shrink-ring. (c) For large motors, the rotor is built up on a fabricated spider, the crosssection of which is shown below : Induction Motor Terminal Box Eye Bolt Rib Stator core Stator slot Stator coil Air-gap Semi-closed slot for rotor Longitudinal vent-slot Shaft bore with key-way Frame Stand Figure 5.1(a) : Induction Motor Construction (d) The rotor core surface have to be ground to obtain an accurate air-gap. (e) For dynamic balancing of rotor, after the winding of cage has been done, there should be a provision in the rotor core, all along the periphery, for small amount of weights to be attached, which finally forms an integral part of the rotor. Rotor Core for Squirrel Cage Induction Motor Closed slots of either circular or rectangular shapes are provided in the stamping. After the rotor core is stacked by the required number of laminations, the rotor shaft is inserted through the space available at the centre and firmly fitted by the key-way. For machines of small and medium capacity, this assembly is then placed in a moulding machine, which forces molten aluminium under pressure through the slots to form rotor bars, end rings and cooling fan arrangement as an extension of end rings. This type of die-cast rotor is simple and rugged in construction as shown in Figure 5.1(b). Figure 5.1(b) : Fabricated Rotor Spider for Induction Motor 121 Electrical Technology For machines of larger capacity, copper or brass bars are driven through the slots manually and the ends on both sides are then welded or silver-soldered together to form end rings. The rotor bars are slightly inclined to the shaft axis due to the ‘skew’ provided while stacking the rotor stampings. Skewing is done because: (a) It helps the motor to run quietly by reducing magnetic hum. (b) It reduces the locking tendency of the rotor. Rotor Core for Slip Ring Induction Motor Semi-closed slots for three-phase winding are provided in the stamping. In cage-rotors, closed slots were used. This is the primary difference in the stamping. Secondly, the molten aluminium or copper bars are not used. Windings will be dealt separately. Stator Winding Three-phase stator winding is done on the stator core. One starting end and one finishing end for each of the phase windings is brought out for inter-connection in Y or ∆ fashion. Single layer mesh windings are used for machines of smaller capacity and mediumsized machines have double-layer lap windings. Large capacity motors employ single layer concentric winding. All the winding coils are placed in the slots, the slots are closed with wooden wedges or paper insulator, which keeps the coil sides intact. The whole stator winding may be impregnated with a thermosetting varnish and baked. Rotor Cage or Rotor Winding There is no rotor winding in squirrel cage motors where the short circuited aluminium, copper and brass bars form the rotor bars. In slip ring motors, there is a rotor winding on the rotor cage and hence are called the wound rotor motors. The rotor is wound for the same number of poles as that of stator. The winding is normally connected in star and the resultant three terminals are connected to three slip-rings provided on one end of the shaft as shown below : Wound Rotor Slip Rings RY B Figure 5.2(a) : Slip Ring Induction Motor When running normally, the slip-rings are short-circuited, but for starting purposes, they are connected to a three-phase star-connected starting resistance, as shown in Figure 5.2(b) alongside. Rheostat 122 Figure 5.2(b) : Starting Resistance In slip-ring rotors of large size, a bar winding can be used because the choice of voltage is usually free. For smaller machines, wire-wound rotors can be used, with coil arrangements similar to those of the stator. Induction Motor 5.2.3 Other Parts Shaft and Bearing The shaft of an induction motor is short and stiff, in order to keep as small an airgap as is mechanically possible. This helps in removing any significant deflection in the rotor. Even a small deflection in the shaft, and hence rotor, would create large irregularities in the air-gap length which would lead to production of unbalanced magnetic pull. For small and medium capacity motors, a roller bearing may be used at the driving end and a ball bearing at the non-driving end. Terminal Box and End Covers The terminal boxes have been shown in the diagram Figure 5.3 with the connection diagram for Y or ∆ connected stator. In squirrel cage motors, the rotor bars are short-circuited in the construction itself, so rotor terminals are not available. But in slip ring motors, it is essential to bring out the rotor terminals as it gives possibility of inserting an additional resistance in the rotor circuit. In the terminal box shown, only D, E, F constitute the rotor terminals, assuming that the other ends of it have already been connected inside the machine. Sometimes, it is desirable to bring out ends of each of the phases of the rotor. Then, together we have (6+6), i.e. 12 terminals on the terminal box. A1 A1 B1 C1 B2 Stator terminals C2 A2 A2 A1 B1 C1 C2 A2 B2 C2 B1 C1 Star connected stator B2 A1 C2 D E F Rotor terminals (in case of slip-ring motors) A2 A1 B1 C1 C2 A2 B2 C1 B1 B2 Delta connected stator Figure 5.3 : Parts of Induction Motor Two end covers with suitable bearings provide support for the rotor assembly. Brush Lifting and Short-Circuiting Gear Brushes sliding on the slip rings are used for making connection of rotor windings with the external circuit. Brush lifting and short-circuiting gear it used for operating the brushes. This is fitted with one of the end-covers which is nearer to the slip-ring side. Slip-rings are normally of brass or phosphor-bronze shrunk onto a cast-iron sleeve with moulded mica insulation. The assembly is pressed onto the rotor shaft and located either between the rotor core and the bearing, or onto the shaft extension. This part is exclusively in slip-ring motors and not in the common type squirrel cage motors. 5.3 WORKING OF 3-PHASE INDUCTION MOTOR When three phase supply is connected across stator windings then a rotating flux is 120 f produced in stator which rotates clockwise at synchronous speed giving N s = . P 123 Electrical Technology When the rotor of 3-phase motor is placed in air gap of stator, a emf is induced in rotor conductors. For squirrel cage rotor, conductors are shorted together and the four terminal wound rotor conductors are connected with additional resistance between slip rings. The induced emf causes a current to flow in these shorted conductors which produces a flux. Due to reaction of this flux with stator flux a torque is produced and rotor rotates in same direction of rotating flux with a speed less than synchronous speed. 5.3.1 Slip and Effect of Slip on Rotor Parameters The rotating field revolves with the speed of synchronism, and if the rotor conductors were to revolve at the same speed there would not be any torque. Hence, there is a difference between rotor and rotating field speeds. The rotor speed is less than the rotating field speed and the difference in speed is known as the slip of motor. Generally, slip lies between zero to five percent. Slip (s) = N s − Nr Ns where Ns is synchronous speed of rotating field in stator and Nr is speed of rotor (motor speed) Percent slip = or Ns − Nr × 100 Ns Rotor speed N = (1 − s) N s Three phase induction motor is a particular form of transformer which has secondary winding rotating. In transformer, the frequencies of primary and secondary windings are same but in induction motor the frequency of emf induced in rotor (secondary winding) depends on slip. Rotor emf frequency = s × frequency of applied voltage to stator = sf The induced emf in rotor also depends on slip. When rotor is stationary the emf induced in rotor is Rotor emf (E2 ) = n2 E E1 = 1 , n1 n where n = turns ration = n1/n2. where n2 and n1 are number of turns in rotor and stator windings and E1 is applied voltage in stator. Now for moving rotor, rotor emf becomes sE2. Important thing to remember here is that both E1 and E2 remain fixed by above relation even at different slip or load values. The rotor resistance (R2) is constant with change in slip unless external resistance is added or there is a change in temperature. Rotor reactance at slip one (standstill) is X2 or X2 = 2π f L2 where f is frequency of rotor emf at standfill and L2 is inductance of rotor winding. Now moving rotor frequency becomes sf, so Rotor reactance = 2π sf L2 = s X2 So, rotor impedance at motion Z 2 = R22 + s 2 X 22 The rotor current, I 2 = 124 sE2 = Z2 sE2 R22 + s 2 X 22 and rotor power factor cos φ2 = R2 = Z2 Induction Motor R2 R22 + s 2 X 22 5.3.2 Torque and Torque-slip Characteristic Torque The torque developed by the rotor is proportional to the product of rotor current and fundamental magnetic flux cutting the rotor. The total torque T = K φ I2 cos φ2 where φ is the useful fundamental flux per pole and is proportional to rotor emf (max.) So T = K E2 I2 cos φ2 We know I2 = T= so sE2 R22 + s 2 X 22 and cos φ2 = R2 R22 + s 2 X 22 K s E z2 R2 R22 + s 2 X 22 For small value of slip (s X2)2 is negligible compared to R22 . So torque is proportional to the slip when slip approaches unity or large values of slip, sX2 is large compared to R2 so (R2)2 is negligible compared to (sX2)2. Now torque is approximately inversely proportional to the slip as shown by rectangular hyperbola. K R E2 For slip, s = 1, i.e. standsfill the torque T = 2 2 22 is constant and is known as R2 + X 2 starting torque. The approximate torque slip curve is shown in dT Figure 5.4. The region where > 0 , is called stable region of operation where the ds motor normally operates. Torque Max. Torque Torque Starting Torque Slip Speed s=0 s=1 Figure 5.4 : Torque Slip Characteristics Maximum Torque By inspection of torque slip curve, we see that the maximum torque developed at particular slip, which is derived by differentiating T with respect to slip and by dT putting =0. ds Here After putting T = K s E22 R2 R22 + s 2 X 22 dT = 0 , we get ds R2 = s X 2 or smat = R2 X2 125 Electrical Technology The maximum torque is obtained when slip is equal to ratio of rotor resistance to rotor reactance : K E22 2X 2 Tmax = 1 and rotor power factor is 2 . The rotor reactance should be kept as low as possible otherwise torque developed is reduced. The maximum torque is independent of rotor resistance, but the value of slip at which maximum torque occurs is directly proportional to rotor resistance. If we increase the rotor resistance the value of slip, at which maximum torque occurs, approaches to one as shown in Figure 5.5. Small R2 Torque Max. Torque Medium R2 R2 is very large Large R2 Slip s=0 s=1 Speed Figure 5.5 : T-S Characteristic for Different Rotor Resistances Starting Torque We know that Torque, T = K s E22 R2 R22 + ( s X 2 )2 At the time of starting, s = 1 and E2 is constant then Tst = K1 R2 R22 + X 22 Maximum torque occurs when rotor resistance is equal to rotor reactance and the relation between starting torque and rotor resistance becomes Tst = K1 2 R2 Tst R2 Figure 5.6 126 Induction Motor Starting Torque in Slip Ring Motor Three resistors are used to start a slip-ring induction motor as shown in Figure 5.8. These resistors are variable and are varied by three armed handle which makes a star point. At the time of starting, all resistances are in the circuit and give high starting torque. When starting resistance is all cut out, except that in the leads, and the switch contacts remain closed, then this is comparable with the resistance of the rotor itself. It is necessary to short circuit the rotor with the slip rings. Stator Variable resistance 3-supply Slip rings Starting Rotor Figure 5.7 : Illustration for Starting Torque in Slip Ring Induction Motor 5.3.3 Power Flow Normally, out of the power supplied to stator, a portion is wasted in stator copper loss and another portion in stator iron loss. The remainder is supplied to rotor. Due to copper loss, there is some voltage drop and just voltage transfer from stator to rotor. Let I1, E1 and cos φ are phase values of current, voltage and power factor on stator side. Then power supplied to rotor is 3 I1 E1 cosφ. If n is turn ratio of stator to rotor turns then emf induced E s E1 in rotor is 1 at standstill. When rotor rotates at slip s then induced voltage is and n n rotor current becomes n I1. so rotor power losses = 3E2 I2 cos φ = 3s E1 I1 cos φ = s. Rotor input (i/p = input and o/p = output) Rotor copper loss =s Rotor i / p ⇒ then rotor output = 3 E1 I1 cos φ − 3 E1 I1 cos φ . s = 3 E1 I1 cos φ (1 − s) = Rotor input (1 – s) so Rotor o/p =1− s Rotor i/p and since Rotor copper loss =s Rotor i/p so Rotor copper loss s = Rotor o/p 1− s Thus, Rotor i/p : rotor copper loss : rotor o/p = 1 : s : (1 − s). We know so N Rotor o/p actual speed =1− s = r = Rotor i/p Ns synchronous speed Rotor o/p = rotor i/p × actual speed synchronou s speed 127 Electrical Technology But Rotor o/p = T × w = T × Torque = so = 2π N r 2π × Torque × Actual speed 60 60 × Rotor o/p 2π × actual speed 60 × Rotor i/p 2 π × synchronous speed Its unit is Newton-meter. The synchronous watt is another unit of torque which is the torque that will develop one watt at synchronous speed. Example 5.1 An induction motor is rated as 5 KW at 1440 r.p.m., 50 Hz, 400 V line to line. The rotational losses due to friction and windage are 279 watts. If the maximum torque is found at 900 rpm, determine the maximum torque developed. Solution Synchronous speed for 50 Hz supply nearest to 1440 r.p.m is 1500 r.p.m. ∴ Slip at full load s= 1500 − 1440 60 = = 0.04 1500 1500 Rotor output at full load = 5000 W ∴ Gross rotor output = 5000 + 279 = 5279 watts Rotor copper loss s 0.04 0.04 = = = Rotor gross output 1 - s 1 − 0.04 0.96 ∴ Rotor copper loss = 0.04 × 5279 0.96 ∴ Rotor input = 5279 + Torque at full load = 0.04 5279 × 5279 = 0.96 0.96 Rotor input at full load × 60 2π N s = 5279 × 60 N-m 0.96 × 2π × 1500 = 0.5834 × 60 = 35.004 N-m K S E22 R2 T fl R 2 + S 2 X 22 Torque at full load = = 2 Maximum Torque Tmax K E22 2X2 = = 128 R2 X 2 S R22 + S 2 X 22 R 2 2 S X2 2 R2 2 + S X 2 = 2as 2 a + S2 where a = Induction Motor R2 X2 and S is the slip at full load Slip at maximum torque = 1500 − 900 = 0.4 = smax T 1500 At maximum torque smax T = ∴ Maximum torque R2 = 0.4 X2 or a = 0.4 = a2 + S 2 full load torque 2aS = (0.4)2 + (0.04) 2 × 0.5834 × 60 2 × 0.4 × 0.04 = 2.94617 × 60 N-m = 17.666 N-m 5.3.4 Equivalent Circuit of Induction Motor The rotor current is given by I2 = Rotor emf Rotor Impedance s E2 I2 = R22 + s 2 X 22 If n is stator to rotor turns ratio then stator current I I1 = 2 n or I1 = s E2 n R22 + s 2 X 22 to produce I1 current in stator there is a requirement of voltage i.e. E1 = n E2 so that the rotor impedance referred in stator winding = = n R22 + s 2 X 22 E1 E1 = = nE2 × I2 I1 s E2 n n2 s R22 + s 2 X 22 2 n2R 2 + n2 X 2 = 2 s The rotor resistance n 2 R2 s ( ) can be divided as series combination of two resistance n2 R2 1 and n 2 R2 − 1. The n2 R2 part remains constant and represents the physical rotor s 1 resistance referred to stator side, i.e. R′2 . But n 2 R2 − 1 varies from zero to infinite as s s changes from unity to zero and represents the rotor output in the form of power in this resistance. The equivalent circuit referred to stator winding is shown in Figure 5.8. 129 Electrical Technology X1 n2 = X′2 R1 N2 R2 = R′2 Ι0 Ιm X0 1 – 1 = R′2 s – 1 n2 R2 1 s R0 Ιe Ι1 Ι0 Ι2 n Figure 5.8 The three phase induction motor may be compared with a transformer because it involves changing flux linkages with respect to the stator and rotor winding. The equivalent circuit parameters can be determined by no-load test and block rotor test which are similar to OC and SC test on transformers. This topic is not covered here. Example 5.2 A 3-phase, 500 volt, 6 pole, 50 Hz induction motor develops 30 b.h.p. at 990 r.p.m. at a power factor of 0.86 lagging. Calculate (i) the rotor copper loss, and (ii) the total power and the number of cycles per minute of the rotor e.m.f. The stator losses are equal at 2000 watts. Neglect mechanical and iron losses of the rotor. Solution Given Nr = 990 r.p.m., P = 6 and Synchronous speed N s = Slip s= f = 50 Hz 120 f 120 × 50 = = 1000 r.p.m. P 6 N s − N r 1000 − 990 = = 0.01 Ns 1000 Rotor output = 30 h.p. = 30 × 735.5 = 22065 watts Rotor copper loss s = Rotor output 1− s (i) ∴ Rotor copper loss = 0.01 × 22065 = 222.88 watts 1 − 0.01 Stator output = Rotor input = Rotor output + Rotor copper loss = 22065 + 222.88 = 22287.88 watts Stator input = Stator output + stator losses = 22287.88 + 2000 = 24287.88 watts (ii) ∴ Input power = 24.28788 kW Rotor frequency = s × f = 0.01 × 50 = 0.5 Hz = 0.5 cycles/sec 130 = 0.5 × 60 cycles/minutes = 30 cycles/min Induction Motor SAQ 1 A 3-phase, 400 volt, 4 pole, 50 Hz induction motor has a star connected stator and rotor. The rotor resistance and standstill reactance are 0.2 ohm and 1.0 ohm, respectively. The ratio of stator to rotor turns is 1.29. The full load slip is 4%. Calculate the torque and power developed at full load. Find also the maximum torque and speed at which it occurs. 5.4 STARTING AND SPEED CONTROL 5.4.1 Starters for Poly Phase Induction Motors If motor is started with full voltage, the starting torque is good but very large currents, of the order of 5-7 times the full-load current flow which causes objectionable voltage drop in the power supply lines and hence undesirable dip in the supply line voltage. Consequently, the operation of other equipment connected to the same supply line is affected considerably. If the motor is started with reduced voltage, there is no problem of high currents but it produces an objectionable reduction in the starting torque, on account of the fact that motor torque is proportional to the square of the applied voltage. Methods of Starting Squirrel Cage I.M There are basic four methods of starting the squirrel cage induction motor using (a) Direct online starters (b) Stator Resistor (or reactor) Starters (c) Auto-transformer Starters (d) Star-Delta Starters R Y B e3 b0 Stop Start S1 b1 A e1 M1 M2 M3 O1 O2 O3 UVRC e2 A1 C2 M A2 B1 C1 ∆-connected stator B2 Figure 5.9 : Direct Online Starter Rotor 131 Electrical Technology Methods of Starting Slip-Ring (Wound Rotor) I.M. Though all the above methods, except D.O.L. where the high currents may damage the rotor windings, can also be employed for starting slip-ring motors, but it is usually not done because the advantages of such motors can’t be fully realized. So, the method of adding resistance to the rotor circuit is the most common method for rotor wound I.M. starting. D.O.L. Starters The above Figure 5.9 shows a contactor type D.O.L. starter connected to a motor. As soon as the push-button S1 is pressed, the contactor coil is energized closing its contacts M1, M2 and M3 . Then, the motor windings get full supply through back-up fuses e1, e′ 2, e″1and bimetallic relays O1, O2 and O3 and the motor starts running. An auxiliary contact A in C1 retains the contactor in closed position after the release of start switch S1. An overload tripping device e1, working in conjunction with bimetallic relays, is placed in series with the contactor coil, so that during sustained overload, this opens and the motor stops automatically. For stopping the motor any time, a stop button is provided in series with the contactor coil. Primary Resistor Starter and Reactor Starter This method consists of connecting the motor to the line voltage through a series resistance in each phase. The resistors are short-circuited when the motor accelerates to the desired speed. Sequence of operation of switches shown in Figure 5.10 is : (a) Initially all switches are open. (b) Switches 1, 2 and 3 are closed simultaneously and motor starts running with full resistances in series. (c) Switches 4, 5 and 6 are closed when motor speed picks up and current becomes constant. Finally switches 7, 7, and 9 are closed to cut all resistances and motors attains final steady state speed. Advantages : (i) (ii) R It provides closed transition starting, resulting in smooth starting without any transition high current. A higher p.f. than auto-transformer starters. 4 7 5 8 1 2 M Y 6 9 3 B Figure 5.10 : Series Resistance Starting Sometimes as an alternative to resistor starting, reactor starting is used. This method is mainly used for large motors. Auto-Transformer in the First Step Starters In this type of starter, (A. T.) it attains the reduced voltage by means of an auto transformer at the start. After a definite time interval (about 15 sec.), and after the motor accelerates, it is transferred from the reduced voltage to 132 the full voltage in the second step. A. T. are generally provided with voltage drops to give 40%, 60%, 75% and 100% line voltage. The starting current and starting torque depends on the tapping selected. In the third step, the change-over switch may be hand operated or automatic through time relay which connects the motor finally to the line by changing over from position A to B. Induction Motor R 3ph Bus Bar Y B Auto-Transformer V1 A B Rotor of sq. cage I.M. Stator Figure 5.11 : Simple Diagram of Auto-transformer Starter Advantages of A.T. Starters (i) Greater efficiency. (ii) Taps on the transformer allow adjustment of starting torque to meet the particular requirement. Disadvantages of A. T. Starters (i) It opens the circuit before the motor is connected directly to the line, thus producing transient current and stresses. (ii) It reduces the p.f. of the circuit. (iii) The torque remains constant for the second step, resulting in acceleration which is not smooth. These disadvantages of open transition in A. T. may be overcome by the use of Korndorfer connection, which introduces another step in starting. On the second step, part of the A. T. remains in series with the stator windings. The third step involves the transfer of the full-voltage without open transition. Star-Delta Starter It is cheaper as compared to A. T. starter. This method of starting is used for motors designed to operate normally in delta. The six terminals from the three phases of the stator must be available : a, A : Terminals of phase A b, B : Terminals of phase B 133 Electrical Technology c, C : Terminals of phase C Commercially, the terminals are marked A1, A2; B1, B2 and C1, C2 respectively. The motor is started with TPDT switch in position 1 and subsequently switched to position 2. Position 1 : Starting-windings connected in Y Position 2 : Running-windings get connected in ∆ R Y B A C B B b a c A C a b c B C A 1 Start Y Run 2 Triple Pole Double Throw (TPDT) Switch Figure 5.12 : Star-Delta Starter Let VL = Line Voltage ISP., Y = Per phase motor current during start (in Y-connection) ∴ IL, y = ISP, Y = VL ... 3 Z1 (5.1) [Since rotor circuit behaves almost as short circuit on starting as 1 R2′ − 1 = 0 ]. s If stator winding is ∆-connected, then, with direct switching the per-phase motor current at start would be given by VL ISP., ∆ = ... Z1 (5.2) Starting line current I L ,∆ = 3 I SP.,∆ = (5.3) It is shown from Eq. (5.1) to (5.2) that 134 3 VL Z1 ... I SL.,Y Starting line current with star-delta starter = Starting line current with direct switching in delta I L ,∆ 1 VL 3 Z1 1 = = V 3 3 L Z1 Induction Motor 2 VL Starting torque with star-delta starter 1 3 = 2 = . Starting torque with direct switching in delta 3 VL and Example 5.3 Determine approximately the ratio of starting torque to full load torque of a three phase induction motor with (a) star-delta starter, and (b) an auto-transformer starter with 50% tapping used. The short circuit current of the motor at normal voltage is 5 times the full load current and the full load slip is 5%. Solution (a) Star-delta starter: In star-delta starter the voltage applied to each phase at the time of starting is 1 3 times the normal phase voltage. ∴ Starting current per phase = I st = 1 3 I sc where Isc is the short circuit current per phase, i.e. starting current per phase without Y-∆ starter. Since T α I 22 R2 I2 α 2 , s s Starting Torque Tst × I st2 S st where Sst = slip at starting = 1 Tst α I st2 ∴ or Tst ∝ T fl ∝ I st2 S fl 2 I sc 3 ... (5.4) Also, full load torque ... (5.5) where Ifl and Sfl are full load current and full load slip, respectively. Dividing Eq. (5.1) by Eq. (5.2) we get 2 1 I = sc S fl T fl 3 I fl Tst I It is given that sc = 5 and Sfl = 0.05 I fl ∴ Tst T fl = 1 2 (5) × 0.05 = 1.25 = 0.417 3 3 135 Electrical Technology (b) Auto-transformer starter: Let the auto-transformer tapping ratio be K. Then the phase voltage across the motor is K times the normal voltage. ∴ Starting current = I st = KI sc . At starting slip being unity, the starting torque Tst ∝ I st2 or (5.6) Tst ∝ ( KI sc ) 2 or Tst = K 2 ( I sc )2 ... Dividing Eq. (5.3) by Eq. (5.2), we get I Tst = K 2 sc I fl T fl 2 S fl I It is given that sc = 5, I fl S fl = 0.05 and K = 0.5 Tst 2 2 = ( 0.5) ( 5) × 0.05 = 0.3125 T fl ∴ Example 5.4 A 6 pole, 50 Hz, three phase induction motor running on full load develops a useful torque of 163 Newton-metres and the rotor electromotive force makes 90 complete cycles per minute. If BHP if the mechanical torque lost in friction be 14 Newton-metres, find the copper loss in the rotor windings, the input to the motor and the efficiency. Stator losses total 750 watts. Solution Frequency of rotor emf = 90 rps = 1.5 rps 60 But rotor emf frequency = slip × stator emf frequency ∴ Slip = 1.5 = 0.03 50 Synchronous speed = N s = But S= Ns − Nr Ns 120 f 120 × 50 = = 1000 rpm P 6 0.03 = or 1000 − N r 1000 Actual speed of motor N = 970 rpm. 2π N r T BHP = = 746 × 60 2π × 970 × 163 60 = 22.195 BHP 746 Gross torque produced = Useful torque + torque lost in friction or Gross torque = 163 + 14 = 177 Newton metres ∴ Mechanical power developed = 2π N r Tg 60 = 2π × 136 970 × 177 60 = 17979.33 watts Induction Motor Rotor copper loss S = Rotor gross output 1 − S or Rotor copper loss = 0.03 × 17979.33 1 − 0.03 = 556.06 watts Rotor input = Rotor output + Rotor copper loss = 17979.33 + 556.06 = 18535.39 watts Motor input = Rotor input + Stator losses = 18535.39 + 750 = 19285.39 watts Output = Efficiency of the motor = Input 970 × 163 60 × 100 19285.39 2π × = 85.85% 5.4.2 Speed Control of 3-Phase Induction Motor The stable operation of 3-phase induction motor is restricted within a small range of slips and because of this restriction its application is limited to substantially constant speed drives. We know that Ns = 120 f P Thus, we have the following methods of speed control: (a) Frequency Variation (b) Variation of Number of Poles (c) (d) (i) using multiple windings (ii) using consequent pole technique (iii) pole-amplitude modulation Supply Voltage Variation (i) Variation in Motor Parameters (ii) Variation of rotor resistance or stator reactance (iii) Variation of rotor reactance or both stator and rotor reactances Control of Rotor Slip Power Frequency Variation This method can be used for both squirrel cage and wound rotor type motors. This method provides smooth control of speed. In frequency variation method the variable frequency supply voltage is used. This voltage can be obtained from (a) variable frequency alternator, (b) rotating frequency converter, (c) static converters and cyclo-converters. Variation of Number of Poles This method is generally recommended for squirrel cage induction motors and for steady torque production where the stator and rotor poles should be equal. The 137 Electrical Technology change in the number of stator poles must be accompanied with a change in the number of rotor poles so the pole changing can be effected by the following methods : (a) Using multiple windings in stator for two different speeds, one for higher speed and another for lower speed. Change over from one speed to another is made by a switch. (b) Pole changing with a single stator winding also can be achieved by using consequent pole technique. In this technique the total stator winding coils are connected in different manner according to constant torque, constant power and variable torque and power applications. Supply Voltage Variation This method is used for both squirrel cage and wound rotor motors. In this method, there are some disadvantages like torque is reduced with the reduction in supply voltage. This method gives smooth stepless control, but the efficiency is comparatively lower. Variation in Motor Parameters The speed control can be achieved by changing motor parameters. For wound motor we can change rotor resistance, rotor reactance and stator reactance and for squirrel cage motors only stator reactance can be changed. By rotor resistance variation smooth speed control can be obtained. The drawback is that large copper losses occurs at lower speeds. Stator reactance method is not effective at lower slips with normal rotor resistance. Better performance can be achieved by using high resistance rotors. Control of Rotor Slip Power The wound rotor induction motor can be operated at lower speeds by inserting suitable resistance in the rotor circuit. In this method slip power is dissipated in the rotor circuit. Control or recovery of slip powers as a means of speed control of wound motor is based on the economic use of slip power and uses a slip power converter for this purpose. By this method, a motor can also be run at super synchronous speeds by feeding slip power into rotor. SAQ 2 (a) The power input to a 500 volts, 50 Hz, 6 pole, three phase induction motor running at 975 r.p.m is 40 kW. The stator losses are 1 kW. Calculate (i) slip (ii) rotor copper loss (iii) B.H.P., and (iv) efficiency. (b) A 4 pole, 200 V, 50 Hz, star-connected, 3-phase induction motor has primary leakage impedance of 0.08 + j 2 Ω, and an equivalent secondary leakage impedance at standstill of 1.2 + j 1.0 Ω per phase. The magnetizing current may be assumed to be negligible. Find the maximum torque in synchronous watts the motor can develop and the slip at which it occurs. 5.5 MEASUREMENT OF SLIP There are three main methods of measuring the slip viz: 138 (a) by measuring the synchronous and actual speeds, (b) by measuring the rotor frequency, and (c) by the stroboscopic method. The first method, depends upon the difference of two relatively large and nearly equal quantities. Induction Motor The measurement of the rotor frequency can be effected by inserting a moving-coil ammeter, preferably when rotor frequency is very low often not exceeding one or two cycles per second. The frequency can be determined by counting the oscillations of the pointer. If a central zero instrument is used, the number is a complete to and fro swings which must be counted, but in the ordinary type the pointer is pressing against the zero stop. The percentage slip is given by Rotor frequency × 100 Stator frequency This method must be modified for cage − type rotors, since there are no slip rings. There is usually a small portion of the magnetic flux going right through the centre of the rotor and cutting the shaft, which has a small e.m.f. induced in it. This can be detected by pressing a lead against each end of the shaft. The other ends being connected to a sensitive movingcoil millivoltmeter. The measurement is then made in the same way as with the wound rotor. In the stroboscopic method a disk, with alternate black and white sectors painted on it, is attached to the end of the motor shaft. This disk is illuminated by means of a neon lamp operated from the same supply. If the disk is normally in a poor light, an incandescent lamp is sufficient. Alternatively, the stroboscopic disk may be viewed through a slot cut in another disk driven by a small synchronous motor from the same supply. Assuming the number of black sectors to be equal to the number of poles on the motor, one sector moves forward a pole pitch in half a cycle if there is no slip at all. Then next black sector occupies the original position of the first. But the disk is illuminated once every half-cycle, so that the disk appears to be stationary. Since a certain amount of slip occurs, however, the second black sector does not quite reach the initial position of the first in half a cycle, with the result that the disk appears to rotate slowly in a direction opposite to that of the motor. A complete apparent revolution of the disk thus corresponds to a slip of as many cycles as there are pairs of poles on the motor. If the apparent revolutions of the disk per minute are counted, the frequency of slip is obtained from Rotor frequency = Apparent rpm × pairs of poles 60 The percentage slip is then calculated as before. For testing motors with different number of poles, a series of disks is required having different number of sectors painted on them. Example 5.5 The power input to a 415 V, 3-phase, 6-pole, 50 Hz induction motor is 50 kW when running at 970 r.p.m. The total stator losses are 2.0 kW and the mechanical losses are 1.5 kW. Determine (i) the rotor copper loss (ii) the gross torque in Newtonmeters, and (iii) the rotor resistance per phase if the rotor phase current is 110 A. Solution Rotor input = Stator output = Stator input − Stator loss = 50 − 2 = 48 kW Motor slip = N s − N r N s − 970 = Ns Ns The synchronous speed N s = 120 f 120 × 50 = = 1000 r.p.m. P 6 139 Electrical Technology Motor slip s = 1000 − 970 = 0.03 1000 Rotor copper loss = Slip Rotor input or Rotor copper loss = Slip × Rotor input = 0.03 × 48 kW = 1.44 kW Gross rotor output = Rotor input − Rotor copper loss = 48 − 1.44 = 46.56 kW Gross rotor output 2π × N r 60 Gross rotor torque = = 46.56 × 103 = 458.366 N-m 2π × 970 60 Rotor copper loss = 3I 22 R2 = 1.44 × 103 watts R2 = 1440 3 × (110) 2 = 0.0397 Ω per phase. SAQ 3 (a) A efficiency of a 6 pole, 3-phase, 45 kW, 50 Hz induction motor is 92 percent. The stator losses equal 102 kW. Determine (i) the slip, (ii) the rotor copper loss, (iii) the input, and (iv) the friction and windage losses. The motor runs at 975 r.p.m. (b) A squirrel-cage induction motor has a ratio of rotor stand still reactance to resistance of 4 to 1 and maximum torque is 2.25 times the normal full load torque. Calculate (i) the full load slip, and (ii) the ratio of starting torque to normal torque with direct-on line starting. 5.6 SUMMARY In Section 5.2 after a brief consideration of the constructional features of induction motors, you learnt how a revolving magnetic field is produced. Next the rotor current and torque-speed relations were derived and we also found the condition for maximum torque. You learnt how to calculate performance characteristics of induction motor, power stages and efficiency. Finally, you were introduced to various starters and speed control methods. 5.7 ANSWERS TO SAQs SAQ 1 140 Phase voltage = K= 400 3 Induction Motor Volts. rotor turn s 1 = stator turns 1.29 Standstill induced emf in the rotor per phase E2 = 400 × 3 1 = 179 Volts 1.29 Rotor impedance at full load = Z 2 = Z2 = or I2 = (R2 )2 + (SX 2 )2 (0.2)2 + (0.04 × 1)2 = 0.0416 = 0.204Ω SE2 179 × 0.04 = = 35.1 amp. Z2 0.204 Total copper losses = 3 I 22 R2 = 3 (35.1)2 × 0.2 = 739.2 watts We know, Rotor copper loss S = Rotor gross output 1 − S ∴ Rotor gross output = 739.2 (1 − 0.04 ) 0.04 = 17740 watts = 17.74 KW Rotor speed = (1 − 0.04) 1500 = 1440 r.p.m. ∴ Gross torque developed T = rotor output 2π N r 60 = 17740 × 60 = 117.6 N-m 2π × 1440 For maximum torque, R2 = S′ X2 S′ = ∴ R2 0.2 = = 0.2 X2 1 Rotor e.m.f. per phase Er′ = S ′ E2 = 179 × 0.2 = 35.8 Volts Z 2 = R22 + ( s X 2 ) 2 = (0.2) 2 + (0.2 × 1) 2 = 0.283 Ω I 2′ = Er′ 35.8 = = 126.5 amp Z 2 0.283 141 Electrical Technology Copper losses (Total) = 3 (126.5)2 × 0.2 = 9601 W Rotor gross output = 1− S′ × Rotor copper loss S′ = 1 − 0.2 × 9601 = 38404 Watts 0.2 Speed N′r at maximum torque = (1 − S′) Ns = 0.8 × 1500 = 1200 r.p.m. Maximum torque Tmax = Rotor output 2π N r′ 60 = 38402 2π × 1200 60 = 305.61 N-m SAQ 2 (a) Input to the stator = 40 kW Stator losses = 1 kW Stator output = Rotor input = 40 − 1 = 39 kW Synchronous speed corresponding to 6 poles Ns = 120 f 120 × 50 = = 1000 r.p.m. P 6 Ns − N Slip S = (ii) Rotor copper loss = slip × rotor input Ns = 1000 − 975 25 = = 0.025 = 2.5% 1000 1000 (i) = 0.025 × 39 kW = 0.975 kW Rotor output = Rotor input − Rotor copper loss = 39 − 0.975 = 38.025 kW (b) 38.025 = 50.972 BHP 0.746 (iii) B.H.P. = (iv) Efficiency = Output 38.025 × 100 = = 95.06% Input 40 The equivalent per phase circuit of motor is given below : 0.8 Ω j2 Ω j1 Ω E1 1.2 Ω s 142 200 E1 = or 3 Induction Motor volts E1 I1 = 0.8 + j 2 + j1 + 1.2 / s I1 = 200 3 (0.8 + 1.2 / s) + j 3 (200 3 )2 2 I1 = (0.8 + 1.2 / s) 2 + (3) 2 The input power to the rotor circuit is given by I12 × 1.2 / s watts which is also a measure of the torque developed in synchronous watts. Hence, torque developed per phase 2 T = 200 1.2 × s 3 2 0.8 + 1.2 2 + (3) s The maximum value of the toque is obtained at a slip which makes : dT =0 ds dT = ds = 200 3 2 − 1.2 s 2 2 2 1.2 d 1.2 1.2 2 2 0.8 + (0.8 + +3 +3 − s s ds s 2 1.2 2 1.8 + + (3) s 200 − 3 2 1.2 s 2 2 2 1.2 1.2 1.2 2 + × 2 0.8 + 0.8 +3 + s s s 2 1.2 2 1.8 + + (3) s − 1.2 2 s 2 2 = 200 1.2 1.2 1.2 1.2 2 − 2 0.8 + + 3 − 2 0.8 + s s s 3 s 2 1.2 2 1.8 + + (3) s 2 This is zero when 2 1.2 1.2 1.2 0.8 + + 9 = 20.8 + s s s i.e., 0.64 + 1.92 1.44 1.92 2.88 + +9= + 2 s s s s2 143 Electrical Technology 1.44 or 9.64 = or s2 = 1.44 9.64 or s= 1.44 = smax T 9.64 giving smax T = 0.387. s2 Therefore maximum torque will occur at a slip of 0.387. Substituting smax T = 0.387 in the torque expression, we get Maximum torque per phase = ∴ (200 3 )2 × 1.2 / 0.387 = 1707 (0.8 + 1.2 0.387)2 + 9 The maximum torque the motor can develop is 1707 × 3 = 5121 synchronous watts. SAQ 3 (a) (i) Motor speed N = 975 r.p.m. Synchronous speed N s = Slip = ∴ (ii) 120 f 120 × 50 = = 1000 r.p.m. P 6 Ns − N Ns = 1000 − 975 = 0.025 1000 s = 2.5 percent Motor output = 45 kW Efficiency = 92% Motor input = (iii) Output 45 = kW = 48.913 kW Efficiency 0.92 Rotor input = Stator output = Motor input − Stator losses = 48.913 − 1.2 = 47.713 kW Rotor copper loss = slip × rotor input = 0.025 × 47.713 kW = 1.1928 kW (iv) Rotor output = Rotor input − Rotor copper loss = 47.713 − 1.1928 = 46.5202 kW Friction and windage losses = Rotor output − Net output = 46.5202 − 45 = 1.5202 kW (b) 144 Rotor reactance at standstill X 2 4 = = Rotor resistance R2 1 ∴ a= R2 X2 = Induction Motor 1 = 0.25 4 Full load torque T f 2 aS = = Maximum torque Tm a 2 + S 2 But ∴ Tm Tf = 2.25 (given) 1 2 × 0.25 S = 2.25 (0.25) 2 + S 2 ... (1) or S2 − 1.125 S + 0.0625 = 0 S= S= 1.125 ± (1.125) 2 − 4 (0.0625) 2 1.125 ± 1.008 2 = 0.0585, 1.0665 Here value greater than 1 is inadmissible ∴ Full load slip = 0.0585 or 5.85 percent T Starting torque 2a 2 × 0.25 = s = = 2 Maximum torque Tm 1 + a 1 + (0.25) 2 or Ts Tm = 0.5 1.0625 ... (2) Multiplying Eq. (1) by Eq. (2), we have Tm T 0 .5 × s = 2.25 × T f Tm 1.0625 or ∴ Ts Tf = 1.059 Ratio of starting torque to normal torque = 1.059. 145