MODULE - | CONCEPT & DYNAMICS OF ELECTRIC DRIVEELECTRIC DRIVES Several industrial and domestic applications like transportation systems, rolling mills, textile mills, fans, robots, pumps, machine tools etc involve motion control Systems employed for motion control are called Drives Drive systems use prime movers like diesel engines/petrol engines/hydraulic motors/electric motors to provide the necessary power The drives that use electric motor as the prime mover are known as Electric Drives An electric drive is an electro-mechanical energy conversion system with electrical control for this conversion Advantages of Electric es Flexible control characteristics Available in wide range of speed, torque and powerEnergy saving (power « speed?) Starting and braking are easy and simple Adaptable to any type of environmental conditions Do not cause any pollution Can be operated in all four quadrants High efficiency, low noise and low maintenance requirements PLC’s and computers can be employed for controlling the drive Block diagram of an electric drive Power modulator Control unit Input commandThe main components of an electric drive are, 1. Power source - It can be an AC or DC source - In India 5OHz, 1 phase and 3 phase AC supply are readily available in most locations at different voltage levels — 230V, AOOV, 11kV, 33kV etc. - So most of the drives are powered from AC source either directly or through a converter - With 50Hz AC supply, maximum speed of induction and synchronous motors are 3000 rpm. For higher speeds conversion of supply frequency is mandatory - Some drives are powered from DC source. Depending on the capacity of drive, the battery voltage may vary (6V, 12V, 24V, 48V and 110V) - The choice of motor depends on the type of supply, but there are many other factors which are more important2. Power modulator (Converter) * It act as an interface between source and motor * The main functions of power modulator are v¥ Modulate the flow of power from source to the motor as per load requirement Vv It will keep the motor current and source current within the safe limit during starting & braking Y Convert the supply as required by the motor * For the control of DC motors one require variable DC supply and for AC motors one require variable frequency variable voltage AC supply. Different type of power modulators are, a. Rectifiers or AC to DC converters It can be a diode rectifier to get a fixed DC or fully/half controlled rectifier to get a variable DC supply. Other configurations are also possible. Fixed voltage Fixed Fixed voltage —>| Diode rectifier | 1or3 phase AC voltageDC 1 or 3 phase ACb. Choppers — used to get variable voltage DC from fixed voltage DC supply. It uses semiconductor devices and output voltage can be varied by varying the duty ratio. Fixed voltage | ese Variable Fixed voltage ——>| Chopper ——>| pc voltage DC 1or3 phase AC c. AC voltage requlator — are employed to get variable voltage fixed frequency AC from a fixed voltage fixed frequency AC supply. It uses semiconductor devices as switches and output voltage can be varied by varying the firing angle. It perform the same function of an autotransformer, but here the output voltage and source current have harmonics and power factor is poor at low output voltages. Variable voltage Lor 3 phase AC d. Inverters — are employed to get a variable voltage variable frequency AC from a fixed voltage DC supply. Normally PWM control is used in inverters to reduce harmonics. Here semiconductor devices like MOSFET, IGBT are used as switches Fixed voltage Variable voltage & frequency & frequency ———+} 1or 3 phase AC lor 3 phase ACe. Cycloconverters — convert fixed voltage fixed frequency AC supply into variable voltage variable frequency AC. Uses semiconductor devices like SCR, MOSFET, IGBT as switches. Variable voltage Fixed voltage & frequency & frequency 1or3 phase AC 1or3 phase AC 3. Motor — Convert electrical energy to mechanical energy Commonly used motors are, Y DC motors (shunt, series, compound, PMDC) ¥ Induction motors (squirrel cage IM, slip ring IM, linear) v Synchronous motors (wound field, permanent magnet) v Special machines (Stepper motors, Brushless DC motors and Switched reluctance motors) - In the past DC motors were used for variable speed drives because speed control of AC motors were expensive and has poor efficiency. Now with the development of semiconductor converters using controlled devices like SCR, MOSFET, IGBT etc, AC motors are used in variable speed drives. AC motors have high efficiency, low cost, low maintenance and small size compared to DC motors. Selection of motor depends on application, cost, environmental factors, power requirement etc.4. Control unit - It will monitor & control the performance of entire drive system. - The type of control depends upon the desired drive performance, type of converter and motor - If semiconductor converters are used, then control unit consist of firing circuits - Acontrol unit can have a few digital ICs or several DSPs depends upon the type of control implemented - The basic function is to monitor system variables, compare them with desire values, and then adjust the converter output until the system achieves a desired performance. This feature is used in speed and position control. 5. Loads - Drives are commonly employed for loads like fans, pumps, conveyors, machine tools, trains, robots etc coupled to the motor shaft - Loads may employ either rotating or translation motion6. Sensing unit - Provide necessary feedback signals for the control unit - Different sensors employed in drive system are for speed sensing (from motor), torque sensing (from motor & load), position sensing, current sensing and voltage sensing (from source & from motor terminal), temperature sensing etc. Choice of an electric drive Choice of an electric drive is based on several factors. They are i, Steady state operation requirements (speed-torque characteristics, quadrant of operation, speed regulation, speed range, efficiency, ratings etc.) ii. Transient operation requirements (starting, braking, acceleration, deceleration, speed reversal etc.) iii.Requirements related to source (type, capacity and voltage level of source, regeneration, harmonics and their effects on other devices etc.)Choice of electric drive ctnd.. iv. Reliability v. Operating environment vi. Size & weight considerations vii.Capital, running & maintenance cost Classification of Electric Drives - Generally classified into 3 categories 1. Group drive - Several loads are connected to one shaft and driven by one motor - Loads may be coupled to motor shaft through belt or gear drive - This drive is more economical since a single motor with power rating smaller than the sum total of all connected loads - If fault occur in driving motor, all the driven equipments remain idle - Normally a single motor drives a number of machines through belts from a common shaft - E.g, floor mill drive system2. Individual drive - Here a single motor is used to drive a given mechanism and it does all the jobs connected with this - The power is transmitted to different parts of mechanism by means of gears, pulleys etc. - e.g, Ina lathe, all the operations are is performed by a single motor - The main drawback is power loss during transmission 3. Multi motor drive - Here each operation in a mechanism is taken care by separate drive motor - The system contain several individual drives - Separate motors are provided for actuating different parts of machine - e.g, paper mill, rolling mill - Having high reliability and flexibility - Initial cost is highTorque equation - Inan electric drive there will be a motor driving a load (machine). The motor will be coupled to load either directly or through some arrangement (belt, chain etc.). - Motor will rotate and the load coupled will undergo rotational or translational motion - For convenience consider a motor-load system by an equivalent rotational system as shown below —+) m) = nf’) } Motor Load Let, J = Moment of inertia of the motor-load systemW,, = Angular velocity of the motor shaft Tn = Torque developed by the motor T, = Load torque (opposing/resisting in nature) - Load torque include frictional torque, windage torque & torque required to do useful mechanical work (Frictional torque — the opposing torque produced by friction present in motor shaft and load arrangement, Windage torque — the opposing force produced by surrounding wind when motor and load rotates) - Under balanced condition, the torque produced by motor (T,,) is counter balanced by load torque (T,) - During unbalanced condition, another torque present in drive system d(Ja, given by rate of change of angular momentum = Ay) dt - So the fundamental torque equation can be written as d(J@,,) 5 (1) m T, =T,+7, 7, = Ged = 4m 4 oy HE dt dt dt Eqn (2) is applicable to variable inertia drives like mine winders and industrial robots For drives with constant inertia, dJ/dt = 0 da, 7-7, =m @) i.e, torque produced by motor is counter balanced by load torque T, & a dynamic torque. The term J i is called dynamic torque because It it is present only during transient operations. Drive accelerates/decelerate depending on whether motor torque is greater than or less than load torque If T,, > TL, drive accelerates. i.e, the motor is producing a dynamic torque in addition to load torque which help to overcome the drive inertia & accelerates the load- If T,, >T, - In drives requiring fast transient response, T,, >> T, & inertia of drive system should have a low value Multi quadrant operation (4 Quadrant operation) - A motor can operate in two modes, motoring & braking - In motoring it converts electrical energy to mechanical energy & supports motion - In braking it converts mechanical energy to electrical energy & opposes motion - A motor can provide motoring & braking operation in both forward & reverse direction - For considering multi quadrant operation, it is useful to define suitable conventions about sign of torque & speed- Motor speed is considered as positive, when it causes a forward / upward motion, otherwise it is considered as negative - Motor torque is considered as positive if it produces an acceleration, otherwise it is considered as negative - Load torque is always in opposition to Speed motor torque Forward Forward - Figure shows torque & speed aaeee aoe co-ordinates for forward & reverse @) motion Torque Reverse @) ©) are se Motoring Braking - For understanding the above notations let us consider the operation of a hoist in 4 quadrants as shown in next slide - The direction of motor & load torques & direction of speed are marked as arrowsMotion Counter weight Empty Cage Motion , Empty Cage Counter Load Tongue with weight empty cage Load Torque with loaded cage- Ahoist consist of a rope wound on a drum coupled to a motor shaft - One end of rope is tied to a cage which is used to transport man/material & other end is connected to a counter weight. - The weight of counter weight is chosen to be higher than weight of empty cage, but lower than fully loaded cage. - Load torque line T,, in quadrants | & IV represents speed-torque characteristics for loaded hoist - Load torque line T,, in quadrants II & Ill represents speed-torque characteristics for empty hoist Quadrant_| operation (forward motoring) - Loaded cage moves upward - Positive motor speed & motor torque - This will happen if motor produces positive torque in anticlockwise direction - Magnitude of motor torque = magnitude of load torque, T),Quadrant II operation (Forward braking) It is obtained when an empty cage is moved up Counter weight is heavier than empty cage. Therefore, in order to limit speed within safe value, motor must produce a braking torque with magnitude T,, in clockwise direction Quadrant Ill operation (Reverse motoring) It is obtained when an empty cage is lowered Since empty cage is lighter than counter weight, motor should produce a torque (T,,) in clockwise direction Quadrant IV operation (Reverse braking) It is obtained when a loaded cage is lowered Since the weight of the loaded cage is higher than counter weight, motor must produce a positive torque in anticlockwise direction to limit speed within safe valueEquivalent values of drive parameters Different parts of the loads may be coupled to motor through different mechanism such as gear, belt etc. These parts may have different speeds & different type of motion like rotational & translational motion We can find equivalent values of moment of inertia of drive system & torque components referred to motor shaft Loads with rotational motion Consider a motor driving 2 loads as shown in figure. ee es ‘ at a, wo, int Sica On Mot \ : nes ae en a onl a f Load «Noa pales ( Gear One load is directly coupled to motor through shaft & other connected to motor through a gear with n & n, as teeth number- Let moment of inertia of motor & load directly coupled = J, Motor speed = w,, Torque of directly coupled load = T,, - Let moment of inertia of load coupled through a gear = J, Speed = Wn; Torque =T,, NoW, Wmi/Wm = N/N, = az, gear tooth ratio - If transmission losses are neglected, then kinetic energy of drive system = kinetic energy of various moving parts Ep by = Ione FIO o) But Wy = Wy, Ay 1 2_1 2,1 202 (1)=> IQ. =F Ion +I On & 2 2 2Total Power in the system is equal to power in loads. If transmission efficiency is n,, then T,Q,, =Tp@y, +Ty, Pm we (3) Ui} 1 But, Wnt = Wm ay (3)=> TQ, =To@n+Ti oe i Ta, i.e, T,=T, ++ (4) 1 - In addition to load directly coupled to motor with moment of inertia Jo, if there are ‘m’ other loads with moment of inertia J,, J>,..Jm, gear ratio a,, a),...,a,,, load torques T),, T,2,...,.T}, and transmission efficiencies 14, Ny, Im Then, J = Jo + a42Jq + ay2Jy + ee + Amn TA, Bg Lin Mn And 7,Loads with translational motion - Consider a motor driving 2 loads, one coupled directly to shaft & other through a transmission system converting rotational motion to linear motion as shown in figure. Rotational to linear | ] em Lead = Moon) =e kb | Motoei gta i f motion transmission Ll “4 RIES atl Mass My \}y force Fy - Let moment of inertia of motor & load directly coupled = Jy Motor speed = w,, Torque of directly coupled load = Ti, - Let mass, velocity & force of load with translational motion be My, V, & Fy- If transmission losses are neglected, then kinetic energy due to equivalent inertia of system = kinetic energy of various moving parts. Then, 1 2_1 21 2 7m = ZI + yi ; sass {%) we 6) - Total power in the system is equal to power of loads. If transmission efficiency is n,, FY, T,Qy, = Tq + n 7-7, +n ©} Th Om Note — Moment of inertia can be calculated if dimension & weight of various parts of load & motor are known. It can be measured experimentally by retardation test‘Bxameig 2.1 A niotor drives two loads. One has rotational motion. It 1s coupled to the rotor through a reduction geat with a= 0.1 and efficiency of 90%. The load hus a moment of inertia of 10 kg- se? and a torque of 10 Nom. Other loud has anstotional motion and consists of 1000 kg weight to be lifted up ut an uniform speed of 1.3 m/s. Coupling between this losd and the motor has an ‘g-m’ und runs at & constant speed of 1420 1pm. efficiency of 859%, Motor has an inertie of 0.2 k Dovermine equivatent inertia referred o the motor shaft und power developed by the moior Sotution From Eqs. (2.8) and (2.12), the total moment of inertia referred to the motor shaft sensei em (Ze) © Here Jy = 0.2 kgem?, a, © 0.1, J, = mn ve hori r kgem!, a, Ot, Jy = 10 kee’, v 1S mts and ayy ~ (1420 % 9/30 = (48.7 subeusig 1m Ba, (1) sie - . 402+ (19 x 10+ 1000( say) = 04 beer ‘From Eqs. (2.9) and (2.13) 2 ne eed (35) ” Here 1 = 0.9, ay = O.1, Ty = 10 Nem, 1712 0.85, Fy = 1000 x 9.81 N, 0; = 1.5 mis and aceite ies ; "Substtuting in Bq. (2) gives = 21x10 , 1000x941 - = Ups oo 2st (ES 117.53 NomExamrce 2.2 A drive has following parameters: J = 10 kg-m?, T= 100 ~0,1N, N-m, Passive load torque 7; = 0.05N, N-m, where W is the speed in rpm, Initially the drive is operating in steady-state. Now it is to be reversed. For this motor characreristi¢ 1s chenged to T = — 100 -0.1N, N-m. Calculate the time of reversal. Sotution For steady-state speed T-%|=0 or 100-0.1N-0.05N=0 or Q.ISN = 100 or N= 666.7 rpm ‘After reversal, for steady-state speed, noting that the load ix passive - 100-01 -0.05N 20 or N= ~ 6667 ipm When reversing, from Eq. (2.2) dan J 2-100 - 0.1N-0.05N 0 x FE 100 - 0.15) = ~95.49 - 0.143 t= fanf” sao aa where N, = 666.7 rpm and Nz 0.95 x = 666.7 = = 633.4 (pm Integrating Eq. (1) yields ¢ = 25.58 5,Components of load torques - The load torque (T,) can be divided into following components i) Frictional torque (T,) — the opposing torque produced by friction present in motor shaft and load arrangement ii) Windage torque (Ty) — the opposing force produced by surrounding wind when motor and load rotates iii) Torque required to do useful mechanical work (T,) — Nature of this torque depends on type of application. It may be a constant or independent of speed or depend on speed Variation of frictional torque with speed is shown below - It’s value at stand still is much higher o, t than it’s value slight above zero speed. / - Frictional torque at zero speed is ¢ called striction or static friction >Frictional torque can be resolved into 3 components i) Viscous friction (Ty) — varies linearly with speed ii) Coulombs friction (T,) — independent of speed iii) T,, additional torque present at stand still (neglected in dynamic analysis) Windage torque, Ty, (speed)? ie, Ty = Cw,,? Therefore, load torque T, = T,+Ty+T+Ty, =T 4B Wat TctC Wp? [The term (T,+C w,,?) is very small & neglected] i.e, For a drive T,, = T,+ J(d w,,/dt) =T,+Bw,,+J(d w,,/dt)Nature & classification of load torques Nature of load torques — Nature of load torque depends on particular application. To explain nature, consider different applications i) Torque independent of speed — Example of such drive is paper mill drive. In a paper mill, drive is running at low speed, so windage torque is negligible & the net torque is mainly due to a gravity which is constant & independent of speed | ii) Torque is a function of speed — Fans, compressor, | pumps etc are example of cases where lV load torque is a function of w,,. mer saci - 7 - In fan, compressor, centrifugal pump etc, windage torque dominates & T, « w,,? Wj - Ina high speed hoist, viscous friction & windage has i appreciable magnitude along with gravity. Then the | [ speed — torque characteristics become as shown. fig eax T- intraction drives because of heavy mass, striction is large. This will disappear at a finite speed & windage & viscous friction dominates. Classification of load torques e vs ti Broadly classified into two i) Active load torque — load torque which has the potential to drive the load under equilibrium condition are called active load torque. - Such load torques retain their sign even if drive if the drive rotation is reversed. E.g, gravitational force ii) Passive load torque — is the load torque which always oppose the motion. - They change their sign on reversal of drive rotation. E.g — friction & windage torqueSteady state stability of electric drives It is important to investigate condition of stable operation of an electric drive Normally the drive is said to be in equilibrium if torque developed by motor is equal to load torque This equilibrium condition can be either stable or unstable Under stable equilibrium, the drive again returns to equilibrium state but in unstable equilibrium, the speed of drive either increases tremendously or decrease and comes to rest The condition for stable operation of a drive is that fora decrease in speed of system, motor torque must exceed load torque & for an increase in speed, the motor torque must be less than load torque This can be better understood by considering following torque- speed characteristics of different drive systems- First consider the figure shown, in ®, which point ‘A’ is the equilibrium point - Let some disturbances causes a reduction in speed by Aw,,. At new Ae, | speed, motor torque is greater than T,, consequently motor will accelerate Torque and operating point will be restored to ‘A’. - Similarly an increase in speed by Aw,, due to some disturbance will make T,>T,,, resulting into deceleration ©, and operating point is restored to point ‘A’. - Now consider the second figure in which point ‘B’ is the equilibrium point TorqueIn this case a decrease in speed causes the load torque to become greater than motor torque, drive decelerates & operating point moves away from ‘B’ An increase in speed causes motor torque greater than load torque & as a result motor accelerates & operating point moves away from ‘B’ Load equalization In some drive applications, load torque fluctuates widely within short interval of time For eg. In pressing machine large torque is required for short duration during pressing operation, otherwise torque is nearly zero Other eg. are electric hammer, steel rolling mills etc. In such drives, to produce a large torque for small period firstly the motor rating has to be high & secondly motor will draw a pulsed current from supply, which may gives rise to line voltage fluctuations ( it will affect other loads connected to the system)The above mentioned problem of fluctuating loads can be overcome by mounting a flywheel on motor shaft in the case of non-reversible drives During light load period, motor torque exceeds load torque & motor & flywheel accelerates and energy is stored in flywheel in the form of kinetic energy During high load period, load torque will be larger compared to motor torque Now deceleration occurs, producing a large dynamic torque This dynamic torque & load torque together produce the Tin} aT torque required by load Because of deceleration motor speed falls & motor again Tino accelerates during next light — Tixin k . 7 load period : pene 0 t