Electric Drive Braking Methods PDF

POPULAR PUBLICATIONS BRAKING OF ELECTRIC DRIVES Multiple Choice Type Questions 1. Regenerative braking squirrel cage induction motor takes place when 2014] in a...

POPULAR PUBLICATIONS BRAKING OF ELECTRIC DRIVES Multiple Choice Type Questions 1. Regenerative braking squirrel cage induction motor takes place when 2014] in a [WBUT 2007, a) the overhauling load drives the rotor at a speed greater than synchronous speed below the b) the stator frequency is reduced so that synchronous speed is rotor speed (a) and (b) cboth d)none of these Answer: (a) 2. In case of power failure, while a crane is in operation, the preferred electrical WBUT 2007, 2014] braking technique is b) dynamic c) counter current d) none of these a) regenerative Answer: (b) WBUT 2008] 3.Regenerative braking is a a) first quadrant (T-o) operation b) second quadrant operation c) muituquadrant operation d) third quadrant operation Answer: (b) 4. The slip of an induction.motor during d.c. rheostatic braking is WBUT 2008, 2017 a) s b) 2-s c) 1-s d) none of these Answer: (a) WBUT 2009] 5. Field control of a DC shunt motor gives a) constant torque drive b) constant kW drive c) constant speed drive d) variable load drive Answer: (b) WBUT 2009, 2013] 6. Theregenerative braking is not possible in b) induction motor a) DC series motor c) DC shunt motor d) DC separately excited motor Answer: (a) 7. The loss in energy during starting with m equal steps of voltage can be WBUT 2009] expressed as d) Lm Answer: (a) ED-52  ELECTRIC DRIVES 8. Most efficient braking is a) dynamic braking WBUT 2010] c) both (a) & (b) b) regenerating braking d) none of these Answer: (a) 9.In plugging of an electric motor effectively we apply WBUT 2011] a) a reverse voltage on the armature b) double voltage on the armature c) zero voltage on the armature d) zero magnetisation current Answer: (a) 10. The slip s for plugging is a) s-1 WBUT 2012] b) 25-1 c) 2-s d) 2+s Answer: (c) 11. To get the speed higher than the base speed of a dc shunt motor WBUT 2012, 2014] a) armature voltage control is used b) field control is used c) armature resistance control is used d) frequency control is used Answer: (6) 12. During dynamic braking employed for DC series motors, WBUT 2016] a) armature current is reversed b) field winding is reversed c) field current direction is unchanged. d) both (a) and (c) Answer: (d) 13. The magnetizing reactance of an induction motor during dc dynamic braking, increases with [WBUT 2017] a) increasing rotor speed b) decreasing rotor speed c) independent of rotor steed d) depends on the way the stator windings are connected during braking Answer: (a) 14. For an electric locomotive in the downward direction in a hilly region, the economical braking system will be WBUT 2017] a) counter-current braking b) dynamic braking ) regenerative braking d) mechanical braking Answer: (c) 15. Which braking is not possible in series motor? WBUT 20171 a) Regenerative braking b) Dynamic braking c) Counter-current braking d) Rheostatic braking Answer: (a) 'ED-53 POPULAR PUBLICATIONS Short Answer Type Questions 1. Explain, how regenerative braking is done in a 3-phase induction motor. Show graphically the four-quadrant operation of the motor. WBUT 2006, 2012] Answer: In regenerative braking of a 3- phase induction motor, the rotor speed of the motor must exceeds the synchronous speed of the motor. So far the procedural part is concermed it may be mentioned that when the number of poles of a pole changing induction motor is changed in the ratio 1: 2, for example, four pole to eight pole, regenerative braking takes place immediately after the change over, till the lower steady state speed is reached. Under such a situation, the machine acts as induction generator returning energy to the supply and taking only the reactive power for excitation. When the rotor speed exceeds the synchronous speed, the slip become negative. The four-quadrant operation of the motor shown graphically below: Regenerative Rho Forward motoring braking Rhi Forward braking Rh2 Natural characteristic Rh Rh Rhn T Reverse braking A. T IV Reverse braking The regenerative braking characteristic as shown above is the continuation of the motoring characteristic into the upper part of quadrants 11/V. The maximum regenerative braking torque is higher than the maximum motoring torque. 2. With appropriate characteristic curves explain the dynamic braking operation of a dc series motor WBUT 2011] Answer: In a dynamic braking of d.c. series motor with self-excitation, the supply to the motor i switched off and then the armature circuit includihg the series field winding is connected across a resistor ensuring that the excitation is not reversed during the change over. The dynamic braking torque is TR+R, Rn ED-54  ELECTRICDRIVES Here the flux o is dependent on the armature current I.. When braking is initiated, the current is high, thus, resulting in increased value of flux, and being approximaely proportional to the torque is also high, square of the current. The speed torque characteristics for dynamic braking are in the second quadrant as showm below: Forward brakine Forward motorine Dynamic braking R Reverse motoring Reverse braking At this instant, the driven unit may experience objectionable shocks due to a large value of braking torque. The machine now runs as a self-excited generator. 3. Describe the regenerative breaking operation of an 3-phase induction motor. [WBUT 2013] Answer: Regenerative braking An induction motor is subjected to regenerative braking. if the rotor speed exceeds the synchronous speed of the motor. Under regenerative braking, the machine acts as an induction generator returning energy to the supply and taking only the reactive power 1or excitation. When the rotor speed exceeds the symchronous speed, the slip becomes negative. The regenerative braking characteristic is the continuation of the motoring characteristic into the upper part of quadrants IlV as shown in Fig. below. The maximum regenerative braking torque is higher than the maximum motoring torque. When the number of poles of a pole-changing induction motor is changed in the ratio 1: 2, for example, four-pole to eight-pole regenerative braking takes place immediately after tne changeover, till the lower steady state speed is reached. ED-55  POPULARPUBLICATIONS Regenerative 11 braking Forward braking Forward braking Rho Natural characteristics Rha Rhi Rho NRhi cb Rh Dynamic Tu braking Tz Ta T (DM Reverse motoring - m IV Reverse braking Fig: Typical speed-torque characteristics of induction motor under different operating conditions. 4. How does the braking resistance control the dynamic braking torque in dc separately excited motor? How to employ dynamic braking in de series motors? [WBUT 2014] Answer: 1st Part: Rheostatic or dynamie braking Shunt Shunt field field winding Amature Armatur winding Es (a) (b) Fig: 1 Rheostatic or dynamic braking In this method of electric braking of shunt motors, the armature of the shunt is disconnected from the supply and is connected across a variable resistance Ras shown in ED-56  ELECTRIC DRIVES Fig. 1(b). The field winding is, however, left connected across the supply unalstul bed The braking effect is controlled by varying the series resistanceR. 2nd Part: Fig. shows dynamic braking scheme for a d.c. motor. During braking the motor is useu 1 as a scparately excited d.c. generator and the series field winding is connected to a low voltage high current converter. Propulsion Brake Brake resistor Series ficld Fig: 1 Dynamic braking for a d.c. motor 5.Describe with suitable diagram plugging operation of DC machine. WBUT 2015] Answer: Plugging or counter current braking for Shunt motor Shunt Armature field Fig: 1 Plugging or counter-current braking In this method, connection to the armature terminals are reversed so that motor tends to run in the opposite direction. (Fig. I). Due to reversal of armature connections, applied voltage & E, start acting in the same direction around the circuit. In order to limit the armature current to reasonable value, it is necessary to insert a resistor in the circuit while reversing armature connection. This method is commonly used in controlling: (i) Printing presses i) Rolling mills (ii) Machine tools (iv) Elevators etc ED-57 POPULAR PUBLICATIONS torque. As compare to rheostatic braking. plugging gives better braking (i) Plugging for Series Motor Scrics ficld winding Armature Fig: 2 Plugging In this method (as in the case of shunt motors) the connections of the armature are reversed and a variable resistance R is put in series with the armatur as shown in Fig (2) above. 6. Describe with suitable diagram dynamic braking operation of Induction Machine. [WBUT 2015] Answer The speed of an induction motor can be controlled by injecting D.C voltage in its stator winding. A variable resistance may be used in the rotor (in case of a slup ring inductio motor) for dissipating the required amount of power. Now-a-days thyristor bridges are used for supplying D.C which is controllable in nature. With the help of controlled D.C from a thyristors bridge the dynamic braking can be achieved in a more effective manner. The connection.diagram for scheme is shown in Fig. 5.10. 3-phase A.C is stepped down to lower voltage and fed to a 3-phase thyristor bridge which serves as the rectifier. This D.C is filtered by an L.C filter for minimizing the ripples. Ripple free D.C is then fed to the stator winding of the induction motor as shown in Fig ED-58  ELECTRIC DRIVES 3-phase.C. Rotor supply shaft Stator Rotor 3-phase slip ring indication motor S Slip Rotor rings variable resistance Step-down transforrme c Braking thyristor conroller Fig: Dynamic braking ofa 3-phase slip ring induction motor It is to be noted that while feeding D.C to the stator the 3-phase A.C input must be disconnected. A.C is disconnected with the help of S, and D.C is disconnected with the help of S,. Since, the input A.C oltage is stepped down to a lower value, the thyristor Converter may be of lower voltage rating. Long Answer Type Questions 1.a) With the help of relevant torque speed characteristics, discuss regenerative braking and reverse current braking of an induction motor. WBUT 2007 b)A 400 V, 3-phase, 50 Hz, 4 pole cage type induction motor has the following parameters: ==0.152, M =r=0.42, x=142 That motor was operating on full load slip 0.05 when the two stator terminals were suddenly interchanged. Calculate the primaty current and the braking torque immediately after application of plugging. ASSUme approximate equivalent circuit WBUT 2007, 20101 Answer: a)) Plugging (or counter curfent braking) an It isknown that plugging can be achieved in Induction motor merely by reversing two of the three phases which cause reversal of the direction of rotating magnetic field. At the ED-59 POPULAR PUBLICATIONS position the motor runs in the opposite instant of switching the motor to the plugging timnes] of is approximately twice [(2-s) direction to that of the field and the relative speed to two, being equal 1o (2-s). So synchronous sped i.e. the slip (s) is very nearly equal nornmally induced voltage at stand still and Voltage induced in the rotor will be twice of winding must be provided with the additional insulation to withstand this much the and it absorbs kinctic cnergy from the voltage. During plugging. the motor acts as a brake fall. The heat developed in the rotor during still revolving load causing its speed to during starting period. braking period are all out three times the heat developed In case of squirrel cage motor, energy is dissipated only within the machine where as in in the external resistance added case of wound rotor motor. This energy is dissipated also in the rotor circuit for this purpose. D Without external resistancC Ti B With extemal R Q resistancec 2 CT TL 0 +T -T 5 MT lLG H - (®r (i) Fig: (a) Speed-torque characteristics during plugging of (i) squirrel-cage motor and (ii) wound-rotor motor i) Regenerative braking When the load, forces the motor to run above synchronous speed, Regenerative braking takes place. When rotor speed, as when lowering of load in a crane or a hoist becomes more than rotating field speed, the slip becomes negative. The negative slip means that induction machine will operate as an induction generator in the 2" quadrant as shown in Fig. (b) below and returns power to the supply lines. The power delivered to the supply is given by the product of regenerative braking torque and the corresponding rotor speed ED-60  ELECTRIC DRIVES Regcnerative braking Spced 3 - 4 poles poles -Te TL T Fig: (b) regenerative braking of polyphase induction motor The Fig. (b) shows that the amount of power returned to the supply line depends upon how far is the rotor speed above synchronous speed when rotor speed falls synchronous speed. When rotor speed falis to synchronous speed, the regenerative braking comes to an end. In case of squirrel cage induction motor, stable speed is obtained at a speed considerable in excess of synchronous speed and the regenerative braking cannot be applied unless the motor is specially designed to withstand the excessive speed. b) Data given: =0.12, x =x=0.4, x =14.02, f=50Hz, p=4 400 230V, = 120 120x501500rpm ph N, N P 4 27T = 2Tx1500 N, =157.1 rad/s 60 60 Magnetizing Current I 230+JU=-jl6.43=16.432-90'A 0-0+j14 S=0.05, 2-s=2-0.05 =1.95, 2-s ==0.05132 1.95 2- +i(x +x) =(0.1+0.0513) +(0.4 +0.4) 0.1513+ j0.8 0.8142479.3°N Rotor curent referred to the stator ED-61 POPULAR PUBLICATIONS 23020° 282.5-79.3A Z 0.814279.3°. 1.+1. = 16.432-90°+282.52-79.3°= 299.02-79.88A = Primary current 1, = Braking torque T, 282.5) x0.0513 = 78.19N m. T= 2- 157.1 2. a) With the help of relevant torque-speed characteristics, discuss different methods of braking of D.C. shunt motor. WBUT 2008, 2010, 2012] Answer: The different types/methods of braking of DC shunt motor is discussed below with their torque-speed characteristics. 1. Regenerative Braking: In a regenerative braking. generated energy is supplied to the source, for which followingg condition should be satisfied. E>and negative.(1) Field flux cannot be increased substantially beyond rated value because of saturation. For a source of fixed voltage of rated value regenerative braking is possible only for speeds higher than rated and with a variable voltage source it is also possible below rated speeds. The speed-torque characteristic is show below in fig. 'a' for a DC shunt motor. In actual supply system when the machine regenerates its terminail voltage rises Consequently the regenerated power flows into the loads connected to the supply and the source is relieved from supplying this much amount of power. Natural This braking is therefore possible only when there are loads connected to the line and they are in need of power more or equal to the regenerated power. Braking Motoring When the capacity of the loads is less than the regenerated power, all the regenerated power will not be absorbed by the loads and remaining power is IncreasingV supplied to capacitors in line and the line voltage will rise to dangerous values leading to insulation breakdown. Hence regenerative braking should be Fig: (a) used where there are enough loads to absorb the regenerated power. The speed torque characteristics can be calculated from E Kpo Ra K K, (K T. ED-62  ELECTRICDRIVES 2. Dynamic Braking: In dynamic braking. when motor armature across a resistance is disconnected from the source say R,dynamic and connected dissipatcd in Ra & R,. In braking takes place. The generated energy 1s Fig. (b,) the specd-torque motoring to braking. As curves are shown & transition rom speed falls, sections torque, as shown in Fig.(b,) are cut out to maintain a high average for a shunt motor. converted as selt excited generator. During braking, shunt motor can be This permits braking even when suppiy ais. m Motoring Motoring RnRg,>R, >0 -Ru with fcw sections of Ra Braking - with few sections of Rs Fig: (bi) Fig: (b-) Shunt motor with variable amature resistance Dynamic braking speed-torque curves The characteristics are obtained from KK T and , for TK,K V=0 3. Plugging: In plugging, the supply voltage of a shunt motor is reversed so that it as ists the back emf in forcing armature current in reverse direction. A resistance Rp is also connected in series with armature to limit the current. In fig.(A) a particular case of plugging for motor rotation in reverse direction arises, when a motor connected for forward motoring, is driven by an active load in the reverse direction. Here again back emf & applied voltage act in the same direction. However, the direction of torque remains positive in fig. (B): This type of situation arises in crane & hoist applications & the braking is then called counter- torque braking Motorin Motoring Plugging T Plugging curves Fig: B Counter torque braking Fig: A Plugging speed torque ED-63 POPULAR PUBLICATIONS RRa a >0. If ar/3, it will be discontinuous. Three-phase full controlled bridge rectifier This type of thyristor driv circuits are used for armature voltage control with a view to change the speed of D.C. shunt motors and the field is fed from a diode bridgc. The ED-86  ELECTRIC DRIVES bridge circuits arc used for high power drives less than ratca motor current because and the averagc current in the thyriStorss cycle. The power circuits having the thyristors conduct for about one third of tne six thyristors with and the wavetormi in Fig. 2(b). d.c. motor load is shown in Fig 2{a) Ao- T T2 Bo- Vo O00 Co T S Fig: 2(a) power circuit with dc motor load =0 To T Ta Ts To Vcb Vha z/3 2/3 47/3 Sn/3 27 () 60° Ta Ts V V D, (ii) =t Fig: 2(b) voltage (i) a =0° (ii) a = 60° and current (a = 60° waveforms for continuous load current In this t system, the thyristors are fircd in sequence with delay angle a, with vith each htorristo conducting for angle 120°. Two thyristors conduct at a time. If thyristor T is Ts going off. Prior to LSEered, the thyristors T and T start conducting, with thyristor this, the thyristo when thyristor Ta is triggered after ne thyristors, Ts and Ts are conducting. Similarly ED-87 POPULAR PUBLICATIONS a delay of 60° from the instant Ti is triggered, the thyristors, Ti and T2 start conducting with thyristor, To going off. This sequence is repeated at an interval of 60°. Now. if a r/2, the output voltage is negative. The output voltage equation with continuous curTent can be expressed as: V=1.35V, cos a for T>a >0. Special feature: Except for large delay in motoring mode, the current is mostly continuous. ED-88  ELECTRIC DRIVES Electrical Drive 2 Dynamics of Electrical Drives 5 Motor Power Rating 20 Starting of Electric Drives 44 Braking of Electric Drives 52 DC Motor Drives (Rectifier and Chopper Fed) 69 Induction Motor Drives 89 Synchronous Motor Drives 135 Special Type Drives 148 Industrial Applications 152 Speed Torque Characteristics of D.C. Motors & Induction Motors 155 Speed Control of D.C. Shunt Motor by Ward Leonard Method & Buck Boost Method 158 Soft-Start 161.Electric Traction 162 Microprocessor-Based Controller for DC Motor Drives 170 Miscellaneous 171 POPULAR PUBLICATIONS ELECTRICAL DRIVE Multiple Choice Type Questions 1. A singlemotor which actuates several mechanisms or machines is called WBUT 2009, 2013, 2017] a) group drive b) individual drive c) multi-motor drive d) active drive Answer: (a) 2. Fourth quadrant operation of electric drive gives WBUT 2015] a) forward motoring b) forward braking reverse braking c) d) reverse motoring Answer: (C) 3. For multi-motor drives WBUT 2015] a) current source inverters are used b) voltage source inverters are used c) both inverters are used d) none of these Answer: (b) 4. The zone below base speed of an electric drive is known as [WBUT 2016] a) constant power zone b) constant torque zone c) constant voltage zone d) constant current zone Answer: (b) 5. A crane is used to more material horizontally and vertically. The type of drive used is WBUT 2016] a) multimotor b) groupP c) individual d) both (a) and (c) Answer: (d) 6. A four quadrant operation requires WBUT 2016] two ful converters connected in series a) b) two full converters connected in parallel c) two full converters connected in back to back d) two semi-converters connected in back to back Answer: (c) 7. Short time rating of an electric machine WBUT 20171 a) is equal to the name plate rating b) is less than the name plate rating c)is greaterthan the name plate rating d) has no bearing to its name plate rating Answer: (6) ED-2  ELECTRIC DRIVES Short Answer Type Questions 1. a) What do you mean by electric drives? WBUT 2007, 2013] b) What is group drive? Give examples. State advantages and disadvantages of such drive. WBUT 2007, 2013] Answer: a) The electromechanical device which converts electrical energy into mechanical energy to impart motion to different machines and mechanism for various kinds of process control the device is termed as electric drive. Various functions performed by electric drives include i) driving fans, ventilators, compressors and pumps ii) lifting goods by hoists and cranes ii) imparting motion to conveyors in factories, mines and ware houses, and iv) running excavators and escalators, electric locomotives trains, cars, trolley, lifts and drum winders. b) Group Drive If a group of machines or mechanisms are required to be operated by a single motor which will first impart motion to one or more line shaft supported on bearings and then the motions are transmitted to the machines or mechanisms to drive the same with the help of pulleys and belts or gears which are fitted on the line shafts, then the system may be called as group electric drive. Advantages: I. Generally an induction type large single motor is used instead of a number of small motor for which cost is reduced. 2. Taking into account the diversity factor of the loads, the rating of the motor is reduced to some extent 3. As the drive is an induction type, the motor can work at about full load thereby increasing the efficiency and power factor. Disadvantages: 1. The major disadvantage is that in case of any fault in the motor, the entire process will be at stand still. If some of the machines are equired to be kept in operative, the losses will increase thereby decreasing the efficiency and power factor. 2. It is difficult to add an extra machine to the main shaft. 2. What are the various factors that influence the choice of electric drives? WBUT 2016] Answer: i) Steady state operation requirements: Nature of speed torque characteristics, speed regulation, speed range, eficiency, duty cycle, quadrants of operation, speed fluctuations if any, ratings. ED-3 POPULAR PUBLICATIONS ii) Requirements related to source: Type of sourcc, and its capacity, magnitude of voltage, voltage fluctuations. power factor, harmonics and their cffect on other loads, ability to accept regenerative power. ii) Transient operation requirements: Values of acccleration and deceleration, starting, braking and reversing performance. iv) Capital and running cost. maintenance needs, life. v) Space and weight restrictions if any. vi) Environment and location vii) Reliability. Long Answer Type Questions 1. Write short note on Electrical drives and its components. WBUT 2013, 2015] Answer: So far the parts of electrical drives are concerned it may be noted that the major parts are load, motor, power modulator, control unit and source. Electrical motors commonly used in electrical drives are: a) D.C. motors shunt, series, compound and permanent magnet motors. - b) Induction motors- Squirel cage, wound rotor and linear. c)Synchronous motors- Wound field and permanent magnet. d) Other types-Brushless dc motors; stepper motors, switched reluctance motors. Previously, induction and synchronous motors were employed mainly in constant speed drives. Variable speed drives consisting these machines were either too expensive or had very poor efficiency. Consequently variable speed drive applications were dominated by dc motors. A.C motors are now employed in variable speed drives alšo due to development of semiconductor converters employing thyristors, power transistors, 1GBTs and GTOs. Brushless dc. motor is somewhat similar to a permanent magnet synchronous motor, but nas lower cost and requires simpler and cheaper converter. It is being considered for lo power high-speed drives and for servo applications, as an alternative to d. c servomotors, which has been very popular so far. At low power levels, the coulomb friction between the brushes and commutator is objectionable, as it adversely affects the steady Sta accuracy of the drive. Stepper motor is also becoming popular for position control and switched reluctance motor drive for speed control. ED-4  ELECTRIC DRIVES DYNAMIcs OF ELECTRICAL DRIVES Multiple Choice Type Questions 1. In a fan motor the load torque is proportional to WBUT 2006, 2007, 20101 a) speed b) (speed c) d) speed (speed) Answer: (6) 2. In constant power drive WBUT 2006, 2007, 2012] a) torque is proportional to the speed b) torque is proportional to the square of speed c) torque is inversely proportional to the speed d) torque is independent of speed Answer: (c) 3. A typical active load is WBUT 2007, 2016] a) hoist b) blower c) pump d) lathe Answer: (c) 4. A machine driving pulse torque load is equipped with a flywheel in order to [WBUT 2007 a) equalize the current demand during the operation b) equalize the torque requirement c) reduce the mechanical overload d) make the motor thermally suitable to drive the load Answer: (b) 5. A motor driving a passive load is said to be steady state stableif WBUT 2007, 2013] a d_dy=0 d7 dW dW b) dW dc0 dW c d7_d7i >0 dW dW dall of these Answer: (a) 6. A typical passive load is WBUT 2008, 2017] a) Hoist b) Friction c) Bloweer d) PumpP Answer: (b) 7. In constant torque drive WBUT 2008, 2012, 2013, 2017n a) power is proportional to the speed b) power is proportional to the square of speed c) power is inversely proportional to the speed d) power is independent of speed Answer: (a) ED-5 POPULAR PUBLICATIONS 8. A drive has following parameters: WBUT 2008] J=10kg-m, 7 =100-0.1 N, N-m T (passive) =0.05 N., N-m, where N is speed in rpm. Then the steady state speed is a) 700 rpm b) 800 rpm c) 667 rpm d) 680 rpm Answer: (c) 9. The speed-torque curve of a fan type load is given by WBUT 2009] +W W W a)T T b) +T -w - W Answer: (C) d 10. 7M +7 +J dt if T 3/W, consequently 0 of the drive increases. dw and therefore the speed ii) T-TL i.e., (1+2) B> 0°, both the speed and power factor are controlled. c) A y-connected cage rotor induction motor has the following parameters as given below (per phase): WBUT 20111 22,=32,x = r =6.52 and x=552, where symbols have their usual significance. The ratings are 400 v, 50 Hz, 4 pole and its rated speed is 1430 rpm. If VIf control is effected, calculate i) the motor torque, speed and current when the inverter frequency is 30 Hz and slip is rated ) inverter frequency and stator current at rated torque and motor speed of 1300 rpm. ED-112 Answer: yconnected cage rotor Induction Motor r22, r=32, x =x =6.52, x = 400 V, SO Hz, 4 pole 1430 r.p.m. 559 Synchronous speed n, 120f 120x 50 = 1500 r.p.m. , = P 9 50t rad/sec. Full load slip 1500-1430 Full load slip speed 1500 1500 = 1500- 1430 - 70 r.p.m. 0.0466 The equivalent circuit of induction motor = Stator resistance =Rotor resistance x = Stator inductance x=Rotor inductance =Magnetizing impedance Z=Total equivalentimpedance 3 j55|0.046J665 +x+ jXs+)-2+ j6.5+ +jx+) S 0.046(S5+6.5) 55(65.217+ j6.5) 2+ 2+ j6.5*65.217+ j61.5 J35(65.217+j6.5)(65.217- j61.s) = 2+ 6.4 (65.217+ j61.5)(65.217-j61.5) j5s(65.217 +j6.5x65.217-j61.5x65.217 +6.5x61.5) 2+j6.5+ 65.217+61.5 j55(4253.257089+ j423.9105 j4010.8455+399.75) 2+j6.5+ 4253.257089 +3782.25 4653.007089-j3586.935) 2+ 6.5+ 8035.507089 4653.007089 3586.935 2+ 2+ j6.5+ j5s j6.5+ j55| 8035.507089 8035.507089 2+ j6.5+ j31.848069-j*24.5521 ED-113 POPULARPUBLICATIONS 2+24.5521+ j38.348069 = 26.5521+j38.348069 Zo=26.5521 +38.348069 = 46.64322 Full load stator current 400/3 =4.9516 A L0 46.64 Full load torque T, 4.9516 x 0.046 4.9516 SOr x65.217|=30.5543N-m. 507L slip speed which remains constant so when i) As motor torque depends on the rated N-m. operating at 50 Hz also torque = T, = 30.5543 Stator curent 1, = 4.9516 A r.p.m. At 30 Hz synchronous speed=x1500 900 load slip speed= 90 r.p.m. constant Full Motor speed=900-90=810r.p.m. ii) At rated motor torque, slip speed andI, will be same as at 50 Hz operation. Therefore, , =4.9516 A Slip speed= 90 r.p.m. Synchronous speed=1500 +90 = 1390 r.p.m. When motor speed is 1300 r.p.m. 1390 x5046.333 HzZ Inverter frequency 1500 9. With the help of torque-speed characteristics discuss the different methods of braking an induction motor in appropriate details. WBUT 2011] Answer: i) Plugging (or counter current braking) Plugging can be achieved in an induction motor merely by reversing two of the three phases which cause reversal of the direction of rotating magnetic field. At the instant of switching the motor to the plugging position the motor runs in the opposite direction to that of the field and the relaivc speed is approximately 1wice [(2-s) times] of synchronous sped i.c. the slir (s) is very nearly equal to two, being equal to (2-s). So voltage induced in the rotor, will be twice of normally induced voltage at stand still and the winding must be provided with the additional insulation to withstand this mnuch voltage. During plugging, the motor acts as a brake and it absorbs kinetic energy from the ED-114  ELECTRIC DRIVES still revolving load causing its speed to fall. The heat developed in the rotor during braking period are all out three times the heat developed during starting period. In case ol squirrel cage motor. cnergy is dissipated only within the machine case ot ound rotor motor. This where as tn energy is dissipated also in the external resistance added in the rotor circuit for this purpose. Without external TB TCSIstance With extemal resistancee -T T T -T. T. G D TL M T L - (0) (ri) Fig: (a) Speed-torque characteristics during plugeing of (i) squirrel-cage motor and (ii) wound-rotor motor ii)Dynanic (or Rheostatic braking) The Rheostatic braking with a poly phase induction motor can be obtained by disconnecting the stator winding from the a.c supply and exciting it from a, d. c source to produce a stationary d.c. tield. For details refer chapter on "Starting & Braking of Electric Motors". ii) Regenerative braking Regenerative braking takes place when the load forces the motor to run above synchronous speed. When rotor speed, as when lowering of load in a crane or a hoist becomes more than rotating field speed, the slip becomes negative. The negative slip means that the induction machine will operate as an induction generator in the 2n quadrant as shown in Fig. (6) below and retiums pover to the supply lines. The power delivered to the supply is given by the product of regenerative braking torque and the corresponding rotor speed. The Fig. (b) shows that the amount of power returned to the supply line depends upon how far is the rotor specd above synchronous speed when rotor speed falls synchronous speed. When rotor speed talls to synchronous speed. the regenerative braking comes to an end. ED-115 POPULARPUBLICATIONS Regencrntive braking Speed A B 4 poles C 8 poles +T TL Fig: (b) regenerative braking of polyphase induction motor In case of squirrel cage induction motor, stable speed considerable is obtained at a speed in excess of synchronous speed and the regenerative braking cannot be applied unless the motor is specially designed to withstand the excessive speed. 10. Draw and explain the scheme for closed-loop speed control of a three phase induction motor by VIF control drive. WBUT 2012, 2014, 2016] Answer: In a closed loop system. speed control of a three-phase induction motor is made easy by monitoring the actual air-gap flux and control it by adjusting and ensuring a proper ratio. If the rotor frequency and the air-gap flux are maintained constant, the rotor current also remains constant and consequently, so does the stator current since it is the vector sun of the magnetising and the rotor currents. Therefore if S and o are kept constant, in a feedback control system, the stator current consequently remains constant developing a constant torque at all the speeds of the motor. On the other hand, if the stator curent is controlled in the feedback loop and also the air-gap flux, the rotor frequency automatically adjust itself. Again, if the stator current and rotor frequency are directly controlled, the air-gap flux is controlled automatically. The system in which the rotor frequency is controlled directly is called controlled slip system. A schematic six- step variably frequency dc link inverter with feedback controlled circuit is shown below in Figure. ED-116  ELECTRICDRIVES Fixed DC Supply DC Ink SCR Converter ollective turn-olt I Circuit In verter LLLLL4 Firing Circuit Amp: Speed ref Specd TN TCurTcnt limit Anpi Regulation Collective turn-off control 6f SCRA SCRa Ramp 6f Frequencyy Imverter Conrol Circut Ring Transnitter Counter SCR converter DClnk, Collective tum-off, Inverter monitor Schematic diagram of closed loop control of variable dc link(V/f=constant) inverter for speed control of IM a In the above scheme the converter is de link using a 3-phase controlled rectifier. The variable voltage is smoothed by an LC filter. The inverter is fed from the variable voltage of the converter and thus V/f requirement is achieved. ED-117 POPULAR PUBLICATIONS induction motor 11. Compare VSI and CSI drives. Show that a variable frequency a given slip-speed when drive develops at all frequencies the same torque for motor has the operating at constant flux. A Y-connected squirrel cage induction R, = 22, following ratings and parameters: 400V, 50Hz, 4pole, 1370rpm, =32. X, = 3.52. X, =552, where symbols have their usual meanings. It R X, = is controlled by CSI at a constant flux. Calculate i) Motor torque, speed and stator current when operating at 30 Hz and rated slip speed ii) Inverter frequency and stator current for rated motor torque and motor WBUT 2012] speed for 1200 rpm. Answer 1" Part: ) Current source inverter (CSI) is more reliable than voltage source inverter (VSI) because a) conduction of two devices in the same lag due to commutation failure does not lead to sharp risc of current through them and b) it has inherent protection against a short-circuit across motor terminals. i) Because of large inductance in the dc link and large inverter capacitors, CSI drives has higher cost, weight and volume, lower speed range and slower dynamic response. iii) The CSI drive is not suitable for multi-motor drives. Hence, each motor is fed from its own inverter and rectifier. A single converter can be used to feed a number of VSI motor systems connected in parallel. A single VSI can similarly feed a number of motors connected in parallel. 2 Part: The torque-slip relationship of a 3-phase induction motor can be represented as under: (1) 2 T(R,+R+s(%, +*:}] where p=number of pole pairs J stator frequency E, = stator voltage S=rotorfrequency = s In order to maintain a high torque through out the control range, the air-gap flux should be constant 1.., E,/J, should be constant. The speed torque characteristics for various supply frequencies. ED-118  ELECTRIC DRIVES Speed Speed In variable frequency control, the specd torque characteristics remain high hardness and at higher frequencies the maximum torque remains almost the same. However. at lower frequencies the stator resistance becomes comparable to or even larger than the stator Icakage reactance reducing the air-gap flux considerably. The stator voltage drop becomes an important factor and the maximum torque decreases. If we now write the equation for maximum or break down torque as under: Tmax 20 RR +(X, +¥) Then at higher frequencies. (K, + X. >>R and Ta 3E 2) 30,(X, +X;) Since o and (X, +A,) are proportional toS () The above cquation suggests that if E/S is maintained consant Tm also remains constant, while at lower frequencies becomes comparable to ,+; and 7ma mA decreases. Last Part: R, = 552 =22 R = 32 X , =3.5 Synchronous speed 120f 120x 50 =1500 r.p.m. , = 50r rad/sec. 4 ED-119 POPULAR PUBLICATIONS Full load slip 1500-1370130=0.0866 ° I500 1500 Full load s!ip speed 1500-1370=130 r.p.m. The equivalent circuit of induction motor R X X2 R Z =Total equivalent impedance =R+jX + +X;) j5s 3 0.0863.s 2+j3.5+ 3 +3.5) 0.086 j55(34.8837+j3.5) =2j3.5+ 34.8837+ j58.5 3.8857+J3.5)(34.8837-j58.5) =2+ 3.5+ (34.8837+j58.5)(34.8837-j58.5 j55 34.8837+j(35x34837)-(34.8837xj58.5)+-(3.5x58.5) 2+3.5 (34.8837+(58.5)* j122.09295- j2040.69645 + 204.751 =2+ 3.5+|1216.8/232+ 1216.8725+3422.25 j55| (1216.87252 +204.75)+j(122.09295-2040.09648) 2+j3.5+ 4639.1225 421.6225-j1918.60351 2+j3.5+ 4639.1225 55x1918.6035 = 2+/3.5+j2x1421.6225 =2+J3.* 4639.1225 4639.1225 2+j3.5+ jl6.8543+22.7463 ED-120  ELECTRIC DRIVES 24.7463+ j20.3543 Z=24.7463 +20.3543 = 32.047234.07982. Full load stator current 400/3 = 7.21 A 32.047 Full load rotor current j55 R/s,) +{X +X.) =7.21 3 5.865 A LO.0867 7(58.5) Full load torque, 7, = 35.865x 1813.3988-22.73 N-m s507L 0.086 S07 i) As motor torque depends on the rated slip speed which remains constant so when operating at 30 Hz also torque T, =22.73 Stator current = /,, =7.21 A 30 At 30 Hz synchronous speed =x1500=900 r.p.m. S0 Full load slip speed= 130 r.p.m. = constant. = Motor specd 900-130= 770 t.p.m. ii) At rated motor torque, slip speed and i, will be same as at 50 Hz operation. Therefore, ,1.21 A Slip speed=130 r.p.m. Synchronous speed=1200+ 130 =1330 r.p.m. When motor speed is 1290 r.p.m. 1330 133 Inverter frequency= 1500 x 50 x 50= 44.33 Hz. 150 12. a) When plugging is employed for stopping an induction motor, why is it necessary to disconnect it from supply when speed reaches close to zero? b) Explain the principle of slip power recovery scheme of controlling the speed of nduction motor, using static Scherbius Drive. WBUT 2013] Answer: a) When plugging is employed tor stopping tne induction motor, the braking torque is produced by interchanging any two supply terminals, so that the direction of rotation of the rotating magnetic field is reversed with respect to the rotation of the motor. The electromagnetic torgue developed provides the braking action and brings the rotor to a quick stop. If the rotor is of slip ring type. external resistance may be inserted into the rotor circuit to limit the plugging current. The motor is disconnected from the supply at ED-121 POPULARPUBLICATIONS the time when the speed drops to zero. Othervise the motor would continue to run in the opposite direction in reverse motoring mode. b) Refer to Question No. 3 of Short Answer Type Questions. 13. a) A 2.8 kW, 440 Volt, 50 Hz, 4 pole 1390 rpm, delta connected squirrel cage induction motor has following parameters referred to the stator R 32. R =602. X, =4.52. X =5.5S2, X = 752. Motor speed is controlled by stator voltage control. When driving a fan load it runs at rated voltage. Calculate motor terminal voltage, current and torque at 1230 rpm. WBUT 2015, 2017] Answer: T= rR/s Synchronous speed=E0_120x 50 = 1500 rpm=507 rad/sec P 4 Al full load 1500-1390 S=- 110 = 0.073 1500 1500 At full load 440 x 6 T 0.073 440 x82.19 0507 307 (3+82.191)+10* 3 0.073 +(4.5+5.5) 3 440 x82.19 3 440 x82.19 X 507 85.19110 507 7257.639+100 3 440x440x82.1941.32 N-m. 507 7357.639 For a fan load torque is proportional to (speed). Thus T=K(1-s At full load T T K(1-0.073) =41.32 41.32 41.32 K= = 48.08 0.927 0.927x0.927 Hence T =48.08(1-s. (1) At 1230 rpm ED-122  ELECTRIC DRIVES 1500-1230 270 S= = 0.18 1500 1500 At this speed from Eqn. (\) we get T 48.08(1-0.18) = 48.08x 0.822 =32.32N-m Since T=T. T 32.32 N-m 3 0.18 Now 07 507 -=32.32 318+(65+45 0.18 507 3 33.33 (3+33.33)+100 32.32- Vx33.33-32.32x 32.32x307 36.33+100 3 x33.33 507 320.11+10032.32xUT 3 Vx33.33 32.32x07 1420.11 1691.41x1420.1 72,066.996 33.33 V 268.452 V 268.452 268.452 268.452(25.33-j10) T-7 3 36.33+j10 (36.33+j10)(36.33-j10) 3+1+10 268.452(36.33-j10) 268.452 (36.33-j10) 268.452(36.33- j10) 36.33-100 36.33+100 1420.11 268.452 1420.11 1420.110.867-j1.89 A I= 268.452 268.452x j75268.452xj75 =-j3.579A jX, 75 7s-75 4, =,+ =6.867-jl.89-j3.579=6.867-j5.469 v6.867 + 5.469* tan = =v6.867 6.867 +5.469 tan (0.796) = 47.155+29.9092-38.534° = vW77.06492-38.534° =8.7782-38.534° A Line current= v3 x8.778 15.2039 Amp (Ans.) ED-123 POPULAR PUBLICATIONS D) Explain with suitable diagram variable voltage and variable frequency control or [WBUT 2015] three phase induction motor. Answer: t needless to mention that for induction motor, variable frequency input requires tat is As the Voitage to frequency ratio remains constant in order to maintain the flux conslant. to Irecquency inverter frequency is varied. the voltage must be varied to maintain voltage I) ratio constant. But this type of control requires two to thrce stages which results in higher losses and lower efficiency. Further, for low value of frequency, the clement of the filter circuit increase in size and weight wlhich results in their higlh cost and low efficiency. More-over the low values of d.c. input voltage, the voltage across the commutatüng capacitor decreases, thus reducing a circuit turm-off time for thyristors for constant load curent. The output voltagc can also be varied by control within the inverter. Generally there are two schemes of speed control of induction motor and synchronous motor by variable-voltage variable frequency method. 1" scheme Variable-voltage Transformer HReetifier Pulse-width Filter To ac motor variable- Modulated frequency Inverter 2nd scheme Variable-voltage Transformer Cy Cyclo-converter To ac motor variable- irequency Here the 1s scheme is discussed. To begin with, it may be told that in this method a three- phase bridge converter ac to variable d.c voltage which is impressed at the input ofa force commutated bridge inverter. The inverter generates a variable-voltage variable- frequency power supply to control the speed of the motor, the capacitor C as in the Fig. (i) supplies stiff voltage supply to the inverter and the inverter output voltage waves are not affected by nature of load. The inductance L is of large value to prevent discontinuity in output voltage and smooth the ripple. 360 IM) accO- Suppl Rectifier Filler Inverter Fig: () ED-124  ELECTRIC DRIVES The induced stator phase emf is given by V = 4.446fT where is the applied voltage, fthe applied frequency, the air gap flux and T the number of turns in the stator winding. In order to achieve constant torque operation below base speed, flux o has to be kept constant and for keeping ¢ constant, the ratiohas to be kept constant. Thus to control f the speed of a.c induction motor, below the rated value. not only frequency has to be decreased but also voltage has to be decreased in the same proportion. Theinverter oT Six thyristors connects each phase to positive and negative DC bus. Each thyristor is on for 180° and the switching sequence produces a 3h output voltage. The waveformat the output is a six-step waveform as shown below in Fig. (ii) OV Fig: (i) The output frequency is controlled by inverter and amplitude V is controlled by the phase controlled bridge. The shape of waveform is same at all frequencies. The inverter can also be used as a Pulse Width Modulator (PWM) but this requires a control logic. In the cycloconverter method, the DC link is not necessary. But the frequeney variation attainable is limited to l/3 of supply frequency. 14. a) With the help of relevant circuit diagram explain different methods of dynamic braking for a polyphase induction motor. b) A 3-phase, 440V, 50 Hz, 6 pole, Y-connected IM has the following parameters to referred to the stator: Rs 0.5 Q, Rr = 0.6 2, Xs = Xr'=10, Stator rotor ratio is 2. The motor is running on no load. The plugging is used to stop the motor. i) Determine the maximum braking current and initial and final braking torque when no external braking resistance is used. i) Calculate the additional braking resistor to be inserted into the rotor circuit so as to limit the maximum braking current to twice the rated value. The motor has a rated speed of 940 rpm. WBUT 2016] Answer a) The spced of an induction niotor can be controlled by injecting D.C voltage in its stator winding. A variable resistance may be used in the rotor (in case of a slip ring induction motor) for dissipating the required amount of power. Now-a-days thyristor bridges are used for supplying D.C which is controllable in nature. With the help of ED-125 POPULAR PUBLICATIONS controlled D.C from a thyristors bridge the dynamic braking can be achuevca effective manner. The connection diagram for scheme is shown in Fig. 1. thyristor 3pnase A.C IS stepped down to lower voltage and fed to a 3-phase bridge which serves as the rectifier. This D.C is filtered by an L.C filter for minimizing the ripples. Ripple fnree D.C is then fed to the stator winding of the induction motor as shown in Fig. 1. 3-phase Rotor AC. shaft supply Stator Rotor 3-phase slp ring indication motor S S Slip Rotor rings variable resistance Siep-down transformer L C Braking tdhyrstor controller Dynamic braking of a 3-phase slip ring induction motor Fig: 1 It is to be noted that while feeding D.C 1o the stator the 3-phase A.C input must be disconnected. A.C is disconnected with the help of S, and D.C is disconnected with the helpof S,. Since, the input A.C vollage is stepped down to a lower value, the thyristor converter may be of lower voltage rating. b) R =0.52 R, = 0.62 X, = N, =192 R, +R, = 0.5 +0.6 =1.12 Pole pairs P==3 w = 104.72 rad/sec N, = 1000 r.p.m. N=925 r.p.m. ED-126  ELECTRICDRIVES N, = 231V 1000-925=0.075 s, 1000 R0.6 s0.075=8 231 0.5+ 0.6 (2 0.075 231 231 26.46 Amp. V8.5)+(2) 8.731 3-(26.46) 8 T- 104.72 = 160.166 N-m Plugging Spt2-s =2-0.075=1.925 R0.6 1.925 0.3116 Sp Initial braking (plugging) current 231 231 (0.5+0.3116) +(2) Sp 231 231 23=107.44 A V(0.8116)+4 V4.8116 2.19 Therefore 107.44.06 26.46 Initial braking torque 3 3x(107.44x(0.310)=103.04 N-m Tp 104.72 ED-127 POPULARPUBLICATIONS Therefore 103.04 0.6433 160.166 In the next case N=-940r.p.m. N1000-(-940) 1940 =1.94 s. N, 1000 1000 As the maximum braking torque is at speed N,, we have 2.0.6-(0.5+(2} 1.925 2 0.6+R= v0.25+4x1.925 2.061x1.925 0.6+R= =1.984 2 or, R =1.984-0.6 1.384Q R. = 384 4 = 0.346 Sp=1.925 R+R0.6-1.384 1.984 =1.03 Sp 1.925 1.925 Maximum braking current with external resistance R = (0.346)2 inserted in rotor circuit 231 0.6+0.364 0.5+ 1.925 (2 Sp 231 231 V5.00155 23 103.309 A 0.5+ 0.964 2 , +4 2.236 1.925 ED-128  ELECTRIC DRIVES 15. a) What are the advantages of static rotor methods of rotor resistance control. resistance control over conventiona WBUT 20171 Answer: The major advantages of static rotor resistance control over conventíonal methods or rotor resistance control are 1. Speed control of induction motor of bigger sizes say 100-200 kW is easy in this process and by this system speed adjustment in the range from 15 to 30% below or above is possible. 2. variaon ot delay angle permits power flow in both directions and smooth speced control is achieved. 3. The slip power from the rotor circuit can be recovered and fecd back to the ac source so as to utilise it outside the motor. 4. There will be negligible power loss particularly at low speed which automatically improve the motor efficiency. b) Discuss the methods of ac dynamic braking operation of an induction motor. [WBUT 20171 Answer: AC dynamic braking of the induction motor is achieved by disconnecting the stator windings from ac supply and connecting it to the dc supply. When the machine is motoring, the stator magnetic field is stationary as it is fed from dc supply and the rotor continues to rotate in this stationary field. Alternating current is thus induced in the rotor windings. This current produces a rotating magnetic field, which rotates at the same speed as the rotor, but in the direction opposite to that of the rotor such that it becomes stationary with respect to the stator. The rotor current therefore, flows in the direction opposite to that corresponding to the motoring action. Hence, a braking torque is produced. Initially, the frequency of the rotor current corresponds to nearly the synchronous speed. The relative speed between the stationary magnetic field and the rotating rotor at any speed is given by (1-S)o, = So,. Therefore, as the speed drops, the frequency will also reduce and become zero at standstill. When the motor accelerates, the rotor frequency decreases and comes to zero at synchronous speed. The conditions in the rotor during dynamic braking are very similar to those when the motor accelerates from the standstill. The magnitude of the braking torque depends upon de excitation, the rotor speed and the rotor resistance. 16. Write short notes on the following: a) V/f control of Induction Motor WBUT 2010, 2012] DIsadvantages and advantages of inverter fed AC drives WBUT 2011] c) Armature voltage control vs rotor resistance control methods of speed control in W.R.I.M. WBUT 2011] d Vector control of Induction motor rotor WBUT 2013, 2016) e) Induction motor speed control by resistance control WBUT 2014] Source Inverter fed Induction Motor drive WBUT 2015] Current ED-129 POPULARPUBLICATIONS Answer a) Vif control of Induction Motor: Refer fo Question No. 5 of LongAnswer Type Questions. b) Disadvantages and advantages of inverter fed A.C. drive: motor,.In case of voltage source inverter fed drivc for example 3-phase induction modulated wave, it gives low when fed from this type of inverter with pulse width as for stepped harmonic content and therefore, low pulsation of the torque, where torque in the induction wavefom. harmonic gives rise to increased pulsation of the motor. by input voltage, which is 2 In voltage source inverters, the output voltage is controlled it can be adjusted. kept constant, while the output current is maintained constant, but source to For this purpose a high value of inductance is connected in series with the by load keep the current constant. The output voltage waveform is governed impedance. 3. CuTent source inverter fed drive is generally considered to be more reliable as the is minimal. chances of commutation failure compared to voltage source inverter Regeneration is also possible. The presence of a large inductance makes the current constant. The torque is controlled by link current , in the d.c. link which is obtained by varying the input voltage Vg using a full controlled bridge rectifier or chopper. c) Arm2ture voltage control vs rotor resistance control methods of speed control in W.RLML Armature voltage control vs. rotor resistance control methods of speed control in wound rotor induction motor (WRIM) Stator Voltage Variation: Although the torque developed in an induction motor at any slip is approximately proportional to the applied voltage, for an ordinary induction motor operating with constant torque. the range of speed control obtained is small, which may be seen from Fig. 1(a). ED-130  ELECTRIC DRIVES N lav? 08y 0 8V.. TL=Constant 6V T= Constant 05V 025V Speed N N, *: N,* Speed Slip Slip Fig: 1(a) Low rotor resistance, Fig: 1(b) High rotor resistance constant load torque constant load torque V N /Tu N IV 08V Torque 05V 0.5V Speed Slip Fig: 1(c) High rotor resistance fan type load (TL N) However, a continuous control of speed of an induction motor may be obtained by step 1ess adjustment of voltage. if its rotor resistance is high in which cause high resistance constant load torque eurve as shown in Fig. 1(b) may be referred to. The major draw back In this case is that the torque per ampere is small at low speed. It may however be noted that in a pump drive or in a pump fan, the load torque varies as the square of the speed and the power requirement for these drives decreases rapidly with the decrease in specd thereby removing the dificulties is as stated/mentioned. The situation may be understood by referring Fig. 1(¢). Rotor Resistance Variation: Where the motors drive loads with internittent type duty, such as cranes, ore or coal tunloaders, skip hoists, mine hoists lifis etc. stp ring induction motors with speed control Dy variation of resistance in the rotor circuit are frequently used. The speed is controlled in steps. The same resistance can be used for starting the motor. Figh starting torquc, low starting current and large pullout torque are obtained. Excessive ED-131 POPULAR PUBLICATIONS its major shortcomings. The torquc-specd POWer ioss at low specds is one of characteristics [Fig. 1(d)] loose hardness considerably at low speeds. T Ns N N N Speed Slip Fig: 1(d) Speed control of slip-ring indication motors by variation of rotor resistance The range of speed control is limited to 2:1 to 3:1. This method of speed control is employed for a motor-generator set with a flywheel (llgner set) used as an automatic skip regulator under shock loading conditions. d) Vector Control: This method is used for speed control of ac, both induction and synchronous motors. In closed loop control, either stator voltage or stator current is regulated by the error in speed but in magnitude only. The phase angle of the current is regulated by the error in speed, but in magnitude only. The phase angle of the current with respect to the flux is not considered. This is termed scalar control. But the stator current can be decomposed into two components in respect of rotor flux, the magnetizing and torque producing components which are separately controlled. The transient response improves to a great extent using this method of vector or field oriented control as this scheme resembles that of a normal de motor in which field excitation and armature voltage are separately controlled. The stator currents (phase) of the induction motor are decomposed into d- and q-axes components using a three-phase to two-phase transformation, with the rotor flux being available only along the d-axis which is taken as the reference. The quadrature component of flux is zero. The system is nowW decoupled. Two controllers are used, one of them computes the magnetizing components from the error in flux between the reference value (as set or may be raled) and the measured/observed or computed one, while the other controller is used i0 determine the lorque producing component from the error between the set value of lorque and the measured value. The phase angle 6 ED-132  ELECTRIC DRIVES between the a-phase stator current and the rotor flux on the d-axis must be copucu properly So as to determinc the phase currents in the stator to be fed to the motor, from the two components using a two-phase to three-phase transformation. The performaice of the closed loop system is superior to that of a system withaynd scaldr controller. Ihe main problem is to either mcasure the flux directly using search coils. or cstimate the flux indirectly using the motor model from terminal voltages and current. The first one is known as direct method, the second type is called the indirect method. In the direct method tlux observer has been used recently to estimate the rotor flux from the system (motor) modecl and use it to control the motor speed. This eliminates the neea search coils or Hall effect sensors. In the indirect method. the variation of rotor resistance affects the machine model. though various mcthods have been proposed to compensate this effect. With the advent of different types of processors PCs. it is now possible to design and then, implement a vector control scheme for speed control of ac (induction/synchronous) motors. e) Induction motor speed control by rotor resistance control: Refer to Question No. 14(¢) of Long Answer Type Questions. f Current Source Inverter: Speed control of AC motor L Ti CC2 T: Ve D: Load V, D Vc D T CC2 T Fig: 1 Single-phase current source inverter (Auto-sequential commutation) A single-phase current source inverter is shown above and in this case the input current is may be followed. maintained constant, but it can be adjusted. Iwo modes ot operations Mode I: It is initially assumed that the thyristor pair 7, & conducting and current T, is passes through the load and the diodes D, & Dg. The capacitors are charged to the same voltage, with the polarity such that the right hand plates are positive,and the left hand plates negative. ED-133 POPULAR PUBLICATIONS conducting thyristor pair 7, &, 8 ne other thyristor pair 7, & 7, is now gated. The the capacitors. Tie diodes tumed oft by the application of reverse voltages across charged through the load and D&D continue to conduct and both the capacitors are direction of currcnt shows, the the diodes D. & D, and thyristors 7; & 7,. As the be constant. The vol lages capacitors are charged lincarly by the load current. assumed to biased. But as the voltages decreases. The other diode D, &D, are initially reverse diodes D, & D, are across capacitors decrease with the polarity as shown earlier, the two then the polarity reverses forward biased.